Applied Mathematical Sciences Volume 78

Editors

Applied Mathematical SciencesVolume 78

S.S. Antman J.E. Marsden L. Sirovich

J.K. Hale P. Holmes J. Keener

C.S. Peskin K.R. Sreenivasan

J. Keller B.J. Matkowsky A. Mielke

Advisors

1. John: Partial Differential Equations, 4th ed.

2. Sirovich: Techniques of Asymptotic Analysis.

3. Hale: Theory of Functional Differential

Equations, 2nd ed.

4. Percus: Combinatorial Methods.

5. von Mises/Friedrichs: Fluid Dynamics.

6. Freiberger/Grenander: A Short Course in

Computational Probability and Statistics.

7. Pipkin: Lectures on Viscoelasticity Theory.

8. Giacaglia: Perturbation Methods in

Non-linear Systems.

9. Friedrichs: Spectral Theory of Operators in Hilbert

Space.

10. Stroud: Numerical Quadrature and Solution

of Ordinary Differential Equations.

11. Wolovich: Linear Multivariable Systems.

12. Berkovitz: Optimal Control Theory.

13. Bluman/Cole: Similarity Methods for

Differential Equations.

14. Yoshizawa: Stability Theory and the

Existence of Periodic Solution and

Almost Periodic Solutions.

15. Braun: Differential Equations and Their

Applications, 3rd ed.

16. Lefschetz: Applications of Algebraic Topology.

17. Collatz/Wetterling: Optimization Problems

4th ed.

18. Grenander: Pattern Synthesis: Lectures

in Pattern Theory, Vol. I.

19. Marsden/McCracken: Hopf Bifurcation and Its

Applications.

20. Driver: Ordinary and Delay Differential

Equations.

21. Courant/Friedrichs: Supersonic Flow and

Shock Waves.

22. Rouche/Habets/Laloy: Stability Theory by

Liapunov’s Direct Method.

23. Lamperti: Stochastic Processes: A Survey

of the Mathematical Theory.

24. Grenander: Pattern Analysis: Lectures

in Pattern Theory, Vol. II.

25. Davies: Integral Transforms and Their

Applications, 2nd ed.

26. Kushner/Clark: Stochastic Approximation Methods for

Constrained and Unconstrained Systems.

27. de Boor: A Practical Guide to Splines: Revised

Edition.

28. Keilson: Markov Chain Models–Rarity and

Exponentiality.

29. de Veubeke: A Course in Elasticity.

30. Sniatycki: Geometric Quantization and

Quantum Mechanics.

31. Reid: Sturmian Theory for Ordinary


32. Meis/Markowitz: Numerical Solution

of Partial Differential Equations.

33. Grenander: Regular Structures: Lectures

in Pattern Theory, Vol. III.

34. Kevorkian/Cole: Perturbation Methods

in Applied Mathematics.

35. Carr: Applications of Centre Manifold Theory

36. Bengtsson/Ghil/Källén: Dynamic Meteorology:

Data Assimilation Methods.

37. Saperstone: Semidynamical Systems in Infi nite

Dimensional Spaces.

38. Lichtenberg/Lieberman: Regular and Chaotic

Dynamics, 2nd ed.

39. Piccini/Stampacchia/Vidossich: Ordinary

Differential Equations in Rn.

40. Naylor/Sell: Linear Operator Theory in

Engineering and Science.

41. Sparrow: The Lorenz Equations: Bifurcations,

Chaos, and Strange Attractors.

42. Guckenheimer/Holmes: Nonlinear

Oscillations, Dynamical Systems, and

Bifurcations of Vector Fields.

43. Ockendon/Taylor: Inviscid Fluid Flows.

44. Pazy: Semigroups of Linear Operators and

Applications to Partial Differential Equations.

45. Glashoff/Gustafson: Linear Operations and

Approximation: An Introduction to the

Theoretical Analysis and Numerical Treatment of

Semi-Infi nite Programs.

46. Wilcox: Scattering Theory for Diffraction

Gratings.

47. Hale: Dynamics in Infi nite Dimensions/Magalhāes/

Oliva, 2nd ed.

48. Murray: Asymptotic Analysis.

49. Ladyzhenskaya: The Boundary-Value Problems of

Mathematical Physics.

50. Wilcox: Sound Propagation in Stratifi ed Fluids.

51. Golubitsky/Schaeffer: Bifurcation and Groups

in Bifurcation Theory, Vol. I.

52. Chipot: Variational Inequalities and Flow

in Porous Media.

53. Majda: Compressible Fluid Flow and

Systems of Conservation Laws in

Several Space Variables.

54. Wasow: Linear Turning Point Theory.

55. Yosida: Operational Calculus: A Theory

of Hyperfunctions.

56. Chang/Howes: Nonlinear Singular Perturbation

Phenomena: Theory and Applications.

57. Reinhardt: Analysis of Approximation Methods for

Differential and Integral Equations.

58. Dwoyer/Hussaini/Voigt (eds): Theoretical

Approaches to Turbulence.

59. Sanders/Verhulst: Averaging Methods in

Nonlinear Dynamical Systems.

60. Ghil/Childress: Topics in Geophysical Dynamics:

Atmospheric Dynamics, Dynamo Theory and

Climate Dynamics.

Applied Mathematical Sciences

(continued after index)

Bernard Dacorogna

Calculus of Variations

Direct Methods in the

Second Edition

ABC

Editors:

J.E. Marsden L. Sirovich

USA

9 8 7 6 5 4 3 2 1

springer.com

Bernard DacorognaDepartement de MathematiquesEcole Polytechnique Federale de Lausanne

CH-1015 Lausanne, Switzerland

Brown UniversityProvidence, RI 02912

ISBN: 978-0-387-35779-9 e-ISBN: 978-0-387-55249-1

Mathematics Subject Classifi cation (2000): 74S05

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Control and Dynamical

California Institute of Systems, 107-81

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Contents

Preface xi

1 Introduction 1

1.1 The direct methods of the calculus of variations . . . . . . . . . . 1

1.2 Convex analysis and the scalar case . . . . . . . . . . . . . . . . . 3

1.2.1 Convex analysis . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Lower semicontinuity and existence results . . . . . . . . 5

1.2.3 The one dimensional case . . . . . . . . . . . . . . . . . . 7

1.3 Quasiconvex analysis and the vectorial case . . . . . . . . . . . . 9

1.3.1 Quasiconvex functions . . . . . . . . . . . . . . . . . . . . 9

1.3.2 Quasiconvex envelopes . . . . . . . . . . . . . . . . . . . . 12

1.3.3 Quasiconvex sets . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.4 Lower semicontinuity and existence theorems . . . . . . . 15

1.4 Relaxation and non-convex problems . . . . . . . . . . . . . . . . 17

1.4.1 Relaxation theorems . . . . . . . . . . . . . . . . . . . . . 18

1.4.2 Some existence theorems for differential inclusions . . . . 19

1.4.3 Some existence results for non-quasiconvexintegrands . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5.1 Holder and Sobolev spaces . . . . . . . . . . . . . . . . . . 23

1.5.2 Singular values . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5.3 Some underdetermined partial differentialequations . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.5.4 Extension of Lipschitz maps . . . . . . . . . . . . . . . . . 25

I Convex analysis and the scalar case 29

2 Convex sets and convex functions 31

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2 Convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2.1 Basic definitions and properties . . . . . . . . . . . . . . . 32

2.2.2 Separation theorems . . . . . . . . . . . . . . . . . . . . . 34

vi CONTENTS

2.2.3 Convex hull and Caratheodory theorem . . . . . . . . . . 38

2.2.4 Extreme points and Minkowski theorem . . . . . . . . . . 42

2.3 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.3.1 Basic definitions and properties . . . . . . . . . . . . . . . 44

2.3.2 Continuity of convex functions . . . . . . . . . . . . . . . 46

2.3.3 Convex envelope . . . . . . . . . . . . . . . . . . . . . . . 52

2.3.4 Lower semicontinuous envelope . . . . . . . . . . . . . . . 56

2.3.5 Legendre transform and duality . . . . . . . . . . . . . . . 57

2.3.6 Subgradients and differentiabilityof convex functions . . . . . . . . . . . . . . . . . . . . . . 61

2.3.7 Gauges and their polars . . . . . . . . . . . . . . . . . . . 68

2.3.8 Choquet function . . . . . . . . . . . . . . . . . . . . . . . 70

3 Lower semicontinuity and existence theorems 73

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.2 Weak lower semicontinuity . . . . . . . . . . . . . . . . . . . . . . 74

3.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.2.2 Some approximation lemmas . . . . . . . . . . . . . . . . 77

3.2.3 Necessary condition: the case without lowerorder terms . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3.2.4 Necessary condition: the general case . . . . . . . . . . . 84

3.2.5 Sufficient condition: a particular case . . . . . . . . . . . 94

3.2.6 Sufficient condition: the general case . . . . . . . . . . . . 96

3.3 Weak continuity and invariant integrals . . . . . . . . . . . . . . 101

3.3.1 Weak continuity . . . . . . . . . . . . . . . . . . . . . . . 101

3.3.2 Invariant integrals . . . . . . . . . . . . . . . . . . . . . . 103

3.4 Existence theorems and Euler-Lagrange equations . . . . . . . . 105

3.4.1 Existence theorems . . . . . . . . . . . . . . . . . . . . . . 105

3.4.2 Euler-Lagrange equations . . . . . . . . . . . . . . . . . . 108

3.4.3 Some regularity results . . . . . . . . . . . . . . . . . . . . 116

4 The one dimensional case 119

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4.2 An existence theorem . . . . . . . . . . . . . . . . . . . . . . . . 120

4.3 The Euler-Lagrange equation . . . . . . . . . . . . . . . . . . . . 125

4.3.1 The classical and the weak forms . . . . . . . . . . . . . . 125

4.3.2 Second form of the Euler-Lagrange equation . . . . . . . . 129

4.4 Some inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4.4.1 Poincare-Wirtinger inequality . . . . . . . . . . . . . . . . 132

4.4.2 Wirtinger inequality . . . . . . . . . . . . . . . . . . . . . 132

4.5 Hamiltonian formulation . . . . . . . . . . . . . . . . . . . . . . . 137

4.6 Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

4.7 Lavrentiev phenomenon . . . . . . . . . . . . . . . . . . . . . . . 148

CONTENTS vii

II Quasiconvex analysis and the vectorial case 153

5 Polyconvex, quasiconvex and rank one convex functions 1555.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.2 Definitions and main properties . . . . . . . . . . . . . . . . . . . 156

5.2.1 Definitions and notations . . . . . . . . . . . . . . . . . . 1565.2.2 Main properties . . . . . . . . . . . . . . . . . . . . . . . . 1585.2.3 Further properties of polyconvex functions . . . . . . . . . 1635.2.4 Further properties of quasiconvex functions . . . . . . . . 1715.2.5 Further properties of rank one convex functions . . . . . . 174

5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1785.3.1 Quasiaffine functions . . . . . . . . . . . . . . . . . . . . . 1795.3.2 Quadratic case . . . . . . . . . . . . . . . . . . . . . . . . 1915.3.3 Convexity of SO (n)× SO (n) and O (N)×O (n)

invariant functions . . . . . . . . . . . . . . . . . . . . . . 1975.3.4 Polyconvexity and rank one convexity of SO (n)× SO (n)

and O (N)×O (n) invariant functions . . . . . . . . . . . 2025.3.5 Functions depending on a quasiaffine function . . . . . . . 2125.3.6 The area type case . . . . . . . . . . . . . . . . . . . . . . 2155.3.7 The example of Sverak . . . . . . . . . . . . . . . . . . . . 2195.3.8 The example of Alibert-Dacorogna-Marcellini . . . . . . . 2215.3.9 Quasiconvex functions with subquadratic growth. . . . . . 2375.3.10 The case of homogeneous functions of degree one . . . . . 2395.3.11 Some more examples . . . . . . . . . . . . . . . . . . . . . 245

5.4 Appendix: some basic properties of determinants . . . . . . . . . 249

6 Polyconvex, quasiconvex and rank one convex envelopes 2656.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2656.2 The polyconvex envelope . . . . . . . . . . . . . . . . . . . . . . . 266

6.2.1 Duality for polyconvex functions . . . . . . . . . . . . . . 2666.2.2 Another representation formula . . . . . . . . . . . . . . . 269

6.3 The quasiconvex envelope . . . . . . . . . . . . . . . . . . . . . . 2716.4 The rank one convex envelope . . . . . . . . . . . . . . . . . . . . 2776.5 Some more properties of the envelopes . . . . . . . . . . . . . . . 280

6.5.1 Envelopes and sums of functions . . . . . . . . . . . . . . 2806.5.2 Envelopes and invariances . . . . . . . . . . . . . . . . . . 282

6.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2856.6.1 Duality for SO (n)× SO (n) and O (N)×O (n)

invariant functions . . . . . . . . . . . . . . . . . . . . . . 2856.6.2 The case of singular values . . . . . . . . . . . . . . . . . 2916.6.3 Functions depending on a quasiaffine function . . . . . . . 2966.6.4 The area type case . . . . . . . . . . . . . . . . . . . . . . 2986.6.5 The Kohn-Strang example . . . . . . . . . . . . . . . . . . 3006.6.6 The Saint Venant-Kirchhoff energy function . . . . . . . . 3056.6.7 The case of a norm . . . . . . . . . . . . . . . . . . . . . . 309

viii CONTENTS

7 Polyconvex, quasiconvex and rank one convex sets 3137.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3137.2 Polyconvex, quasiconvex and rank one convex sets . . . . . . . . 315

7.2.1 Definitions and main properties . . . . . . . . . . . . . . . 3157.2.2 Separation theorems for polyconvex sets . . . . . . . . . . 3217.2.3 Appendix: functions with finitely many gradients . . . . . 322

7.3 The different types of convex hulls . . . . . . . . . . . . . . . . . 3237.3.1 The different convex hulls . . . . . . . . . . . . . . . . . . 3237.3.2 The different convex finite hulls . . . . . . . . . . . . . . . 3317.3.3 Extreme points and Minkowski type theorem

for polyconvex, quasiconvex and rank one convex sets . . 3357.3.4 Gauges for polyconvex sets . . . . . . . . . . . . . . . . . 3427.3.5 Choquet functions for polyconvex and rank one

convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . 3447.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347

7.4.1 The case of singular values . . . . . . . . . . . . . . . . . 3487.4.2 The case of potential wells . . . . . . . . . . . . . . . . . . 3557.4.3 The case of a quasiaffine function . . . . . . . . . . . . . . 3627.4.4 A problem of optimal design . . . . . . . . . . . . . . . . 364

8 Lower semi continuity and existence theorems in thevectorial case 3678.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3678.2 Weak lower semicontinuity . . . . . . . . . . . . . . . . . . . . . . 368

8.2.1 Necessary condition . . . . . . . . . . . . . . . . . . . . . 3688.2.2 Lower semicontinuity for quasiconvex functions

without lower order terms . . . . . . . . . . . . . . . . . . 3698.2.3 Lower semicontinuity for general quasiconvex

functions for p =∞ . . . . . . . . . . . . . . . . . . . . . 3778.2.4 Lower semicontinuity for general quasiconvex

functions for 1 ≤ p < ∞ . . . . . . . . . . . . . . . . . . . 3818.2.5 Lower semicontinuity for polyconvex functions . . . . . . 391

8.3 Weak Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 3938.3.1 Necessary condition . . . . . . . . . . . . . . . . . . . . . 3938.3.2 Sufficient condition . . . . . . . . . . . . . . . . . . . . . . 394

8.4 Existence theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 4038.4.1 Existence theorem for quasiconvex functions . . . . . . . . 4038.4.2 Existence theorem for polyconvex functions . . . . . . . . 404

8.5 Appendix: some properties of Jacobians . . . . . . . . . . . . . . 407

III Relaxation and non-convex problems 413

9 Relaxation theorems 4159.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4159.2 Relaxation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 416

CONTENTS ix

9.2.1 The case without lower order terms . . . . . . . . . . . . 416

9.2.2 The general case . . . . . . . . . . . . . . . . . . . . . . . 424

10 Implicit partial differential equations 439

10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

10.2 Existence theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 440

10.2.1 An abstract theorem . . . . . . . . . . . . . . . . . . . . . 440

10.2.2 A sufficient condition for the relaxation property . . . . . 444

10.2.3 Appendix: Baire one functions . . . . . . . . . . . . . . . 449

10.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451

10.3.1 The scalar case . . . . . . . . . . . . . . . . . . . . . . . . 451

10.3.2 The case of singular values . . . . . . . . . . . . . . . . . 459

10.3.3 The case of potential wells . . . . . . . . . . . . . . . . . . 461

10.3.4 The case of a quasiaffine function . . . . . . . . . . . . . . 462

10.3.5 A problem of optimal design . . . . . . . . . . . . . . . . 463

11 Existence of minima for non-quasiconvex integrands 465

11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465

11.2 Sufficient conditions . . . . . . . . . . . . . . . . . . . . . . . . . 467

11.3 Necessary conditions . . . . . . . . . . . . . . . . . . . . . . . . . 472

11.4 The scalar case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

11.4.1 The case of single integrals . . . . . . . . . . . . . . . . . 483

11.4.2 The case of multiple integrals . . . . . . . . . . . . . . . . 485

11.5 The vectorial case . . . . . . . . . . . . . . . . . . . . . . . . . . 487

11.5.1 The case of singular values . . . . . . . . . . . . . . . . . 488

11.5.2 The case of quasiaffine functions . . . . . . . . . . . . . . 490

11.5.3 The Saint Venant-Kirchhoff energy . . . . . . . . . . . . . 492

11.5.4 A problem of optimal design . . . . . . . . . . . . . . . . 493

11.5.5 The area type case . . . . . . . . . . . . . . . . . . . . . . 494

11.5.6 The case of potential wells . . . . . . . . . . . . . . . . . . 498

IV Miscellaneous 501

12 Function spaces 503

12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503

12.2 Main notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503

12.3 Some properties of Holder spaces . . . . . . . . . . . . . . . . . . 506

12.4 Some properties of Sobolev spaces . . . . . . . . . . . . . . . . . 509

12.4.1 Definitions and notations . . . . . . . . . . . . . . . . . . 510

12.4.2 Imbeddings and compact imbeddings . . . . . . . . . . . . 510

12.4.3 Approximation by smooth and piecewise affine functions . 512

x CONTENTS

13 Singular values 51513.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51513.2 Definition and basic properties . . . . . . . . . . . . . . . . . . . 51513.3 Signed singular values and von Neumann type inequalities . . . . 519

14 Some underdetermined partial differential equations 52914.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52914.2 The equations div u = f and curlu = f . . . . . . . . . . . . . . . 529

14.2.1 A preliminary lemma . . . . . . . . . . . . . . . . . . . . 52914.2.2 The case div u = f . . . . . . . . . . . . . . . . . . . . . . 53114.2.3 The case curl u = f . . . . . . . . . . . . . . . . . . . . . . 533

14.3 The equation det∇u = f . . . . . . . . . . . . . . . . . . . . . . 53514.3.1 The main theorem and some corollaries . . . . . . . . . . 53514.3.2 A deformation argument . . . . . . . . . . . . . . . . . . . 53914.3.3 A proof under a smallness assumption . . . . . . . . . . . 54114.3.4 Two proofs of the main theorem . . . . . . . . . . . . . . 543

15 Extension of Lipschitz functions on Banach spaces 54915.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54915.2 Preliminaries and notation . . . . . . . . . . . . . . . . . . . . . . 54915.3 Norms induced by an inner product . . . . . . . . . . . . . . . . 55115.4 Extension from a general subset of E to E . . . . . . . . . . . . . 55815.5 Extension from a convex subset of E to E . . . . . . . . . . . . . 565

Bibliography 569

Notation 611

Index 615

Preface

The present monograph is a revised and augmented edition to Direct Methodsin the Calculus of Variations [179] which is now out of print. The core and thestructure of the present book are essentially the one of [179], although it has nowalmost doubled its size. While writing the present volume, it clearly appeared tome that a new subject has emerged and that it deserves to be called “quasiconvexanalysis”. This name, of course, refers to “convex analysis”, although the newsubject is still in its infancy when compared with the classical one.

The calculus of variations is an immense and very active field. It is therefore,when writing a book, necessary to make a severe selection. This was alreadythe case for [179] and is even more so for this new edition. Rather than super-ficially covering a lot of materials, I preferred to privilege only some aspectsof the field. Here are some main features of the book. I strongly emphasizedthe resemblances between convex and quasiconvex analysis as well as the “alge-braic” aspect of the field, notably through the determinants and singular values.Besides the classical results on lower semicontinuity and relaxation, an impor-tant feature of the monograph is the emphasis on the existence of minimizersfor non convex problems.

In doing so I missed several important aspects of the calculus of vari-ations such as regularity theory, study of stationary points, existence andrelaxation in BV spaces, minimal surfaces, Young measures and the mathe-matical study of microstructures, Γ convergence and homogenization. How-ever there are already several excellent books on these subjects, some ofthem very classical, such as: Almgren [18], Ambrosio-Fusco-Pallara [25],Braides-Defranceschi [101], Buttazzo [112], Buttazzo-Giaquinta-Hildebrandt[117], Dal Maso [217], Dierkes-Hildebrandt-Kuster-Wohlrab [248], Dolzmann[249], Ekeland [263], Ekeland-Temam [264], Evans [271], Fonseca-Leoni [284],Giaquinta [307], Giaquinta-Hildebrandt [309], Giaquinta-Modica-Soucek [312],Gilbarg-Trudinger [313], Giusti [315], [316], Ladyzhenskaya-Uraltseva [388],Mawhin-Willem [440], Morrey [455], Muller [462], Nitsche [476], Pedregal [492],Roubicek [517] or Struwe [546], [547]. I have also added in the bibliographyseveral articles which present important developments that I did not discuss inthe present monograph, but are still closely related.

For a reader not very familiar with the calculus of variations, it might beadvisable to start with an introductory book such as [180], which could be con-sidered as a companion to the present one. Nevertheless, the present monograph,

xii Preface

which is essentially a reference book on the subject of quasiconvex analysis, canbe used, as was [179], for an advanced course on the calculus of variations.

I would next like to reiterate my thanks to the people who helped me whilewriting the earlier version [179], namely J.M. Ball, L. Boccardo, P. Ciarlet,I. Ekeland, J.C. Evard, B. Kawohl, P. Marcellini, J. Moser, C.A. Stuart,E. Zehnder and B. Zwahlen.

However, since then I have benefited of many other important discussions.Surely the most influential ones were with P. Marcellini, with whom I have a longstanding collaboration. We have written together several articles and a book[202], which helped me in writing Part III of the present monograph. I want alsoto recall fruitful discussions with E. Acerbi, J.J. Alibert, N. Ansini, G. Aubert,S. Bandyopadhyay, A.C. Barroso, H. Brezis, G. Buttazzo, P. Cardaliaguet,A. Cellina, G. Croce, G. Dal Maso, F. De Blasi, E. De Giorgi, O. Dosly,J. Douchet, A. Ferriero, I. Fonseca, N. Fusco, W. Gangbo, N. Georgy,F. Gianetti, J.P. Haeberly, H. Hartwig, S. Hildebrandt, T. Iwaniec, O. Kneuss,H. Koshigoe, P.L. Lions, J. Maly, P. Marechal, A. Martinaglia, E. Mascolo,J. Matias, P. Metzener, G. Mingione, G. Modica, S. Muller, F. Murat,G. Pianigiani, G. Pisante, L. Poggiolini, A.M. Ribeiro, N. Rochat, C. Sbordone,K.D. Semmler, V. Sverak, M. Sychev, R. Tahraoui, C. Tanteri, L. Tartar,M. Troyanov and K. Zhang.

My thanks also go to Mme. G. Rime, who typed the manuscript of [179],and to Mme. M.F. De Carmine, who typed an earlier version of the presentmonograph. Finally, M. Hagler and C. Hebeisen prepared for me all the figuresincluded in the book.

During the past several years, I have benefited from grants from the FondsNational Suisse and the Troisieme Cycle Romand. Of course, particular thanksgo to the Section de Mathematiques of the Ecole Polytechnique Federale deLausanne.

Chapter 1

Introduction

1.1 The direct methods of the calculus

of variations

The main problem that we will be investigating throughout the present mono-graph is the following. Consider the functional

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx

where

- Ω ⊂ Rn, n ≥ 1, is a bounded open set and a point in Ω is denoted byx = (x1, ..., xn) ;

- u : Ω → RN , N ≥ 1, u =(u1, · · · , uN

), and hence

∇u =

(∂uj

∂xi

)1≤j≤N

1≤i≤n

∈ RN×n ;

- f : Ω× RN × RN×n → R, f = f (x, u, ξ) , is a given function.

We say that the problem under consideration is scalar if either N = 1 orn = 1; otherwise we speak of the vectorial case.

Associated to the functional I is the minimization problem

(P ) m := inf I (u) : u ∈ X ,

meaning that we wish to find u ∈ X such that

m = I (u) ≤ I (u) for every u ∈ X.

Here X is the space of admissible functions (in most parts, it is the Sobolevspace u0 + W 1,p

0

(Ω; RN

), where u0 is a given function).

2 Introduction

We now give several examples.

(1) The classical calculus of variations dealt essentially with the case n =N = 1, where the most celebrated examples are the Fermat principle in geo-metrical optics, where

f (x, u, ξ) := g (x, u)√

1 + ξ2,

the Newton problem, where

f (x, u, ξ) = f (u, ξ) := 2πuξ3

1 + ξ2,

or the brachistochrone problem, where

f (x, u, ξ) = f (u, ξ) :=

√1 + ξ2

√2gu

.

(2) When turning our attention to the case n > N = 1 (in our terminology,it is still part of the scalar case), the Dirichlet integral surely plays a centralrole; we have there

f (x, u, ξ) = f (ξ) :=1

2|ξ|2 .

A natural generalization is when 1 < p <∞ and

f (x, u, ξ) = f (ξ) :=1

p|ξ|p .

The minimal surface in non-parametric form enters also in this framework; wehave in this case

f (x, u, ξ) = f (ξ) :=

√1 + |ξ|2.

In geometrical terms, the integral represents the area of the surface given by(x, u (x)) ∈ Rn+1 when x ∈ Ω ⊂ Rn.

(3) In the vectorial case n, N ≥ 2, the first example is the case of minimalsurfaces in parametric form, a geometrical framework more general than thepreceding one. In this case, we have N = n + 1 and therefore the matrixξ ∈ R(n+1)×n. We denote by adjn ξ ∈ Rn+1 the vector formed by all the n × nminors of the matrix ξ. Finally, we let

f (x, u, ξ) = f (ξ) := |adjn ξ| ,

where |.| stands for the Euclidean norm. In geometrical terms, the integralrepresents the area of the surface given by u (x) ∈ Rn+1 when x ∈ Ω ⊂ Rn;moreover, adjn∇u represents the normal to the surface.

Other important examples in the vectorial case are motivated by non-linearelasticity. A particularly simple one is when N = n and

f (x, u, ξ) = f (ξ) := g (det ξ) ,

Convex analysis and the scalar case 3

where g : R → R is a given function.

We do not discuss the history of the calculus of variations and we refer for thismatter to the books of Dierkes-Hildebrandt-Kuster-Wohlrab [248], Giaquinta-Hildebrandt [309], Goldstine [319] and Monna [449].

The first question that arises in conjunction with problem (P ) is, of course,the existence of minimizers. This strongly depends on the choice of admissiblefunctions, which we denoted by X. A natural choice would be a subspace ofC1(Ω; RN

), or even C2

(Ω; RN

), if we want to be able to write the differential

equation naturally associated to the minimization problem and known as theEuler-Lagrange equation. This turns out to be a strategy too hard to implementin most problems, particularly those dealing with partial derivatives (i.e. n > 1).The essence of the direct methods of the calculus of variations is to split theproblem into two parts. First to enlarge the space of admissible functions, forexample by considering spaces such as the Sobolev spaces W 1,p so as to get ageneral existence theorem and then to prove some regularity results that shouldsatisfy any minimizer of (P ). In the present book, we are essentially concernedonly with the first problem. In most cases, the space of admissible functions is

X = u0 + W 1,p0

(Ω; RN

),

where u0 is a given function and the notation u ∈ X is a shortcut meaning thatu = u0 on ∂Ω and u ∈ W 1,p

(Ω; RN

).

The existence of minimizers in the above space relies on the fundamentalproperty of (sequential) weak lower semicontinuity, meaning that

uν u in W 1,p ⇒ lim infν→∞

I (uν) ≥ I (u) , (1.1)

where stands for weak convergence. This property is thoroughly investigated,notably in Chapters 3 and 8.

It turns out that the property (1.1) is intimately related to the convexity ofthe function ξ → f (x, u, ξ) in the scalar case where N = 1 or n = 1 and to thequasiconvexity (in the sense of Morrey) of the same function in the vectorialcase.

This leads us to the study of convex analysis in Chapter 2 and quasiconvexanalysis in Chapters 5, 6 and 7.

We now discuss in more details the content of the monograph and outlinesome of the main results in every chapter. We state them, most of the time,under slightly stronger hypotheses than needed, but we refer to the precisetheorems at each step.

1.2 Convex analysis and the scalar case

We start with the scalar case where n = 1 or N = 1. The first one correspondsto the case of one single independent variable and is much easier to deal with,in particular from the point of view of regularity. It is discussed in the general

4 Introduction

framework of the scalar case in Chapter 3 but also has a special treatment inChapter 4. The second case, n > N = 1, involves partial derivatives and isconsiderably harder; it is discussed in Chapter 3. However, since both cases usein a significant way many results of convex analysis, we start with the study ofthis classical subject.

1.2.1 Convex analysis

In Chapter 2, we present the most important results of convex analysis. Eventhough many excellent books exist on the subject, we have decided, for theconvenience of the reader, to state and to prove all the results that we need.Another motivation in the presentation of this chapter has been to stress boththe similarities and the differences with quasiconvex analysis, which is discussedin Part II.

Traditionally, convex analysis starts with the notion of a convex set and thencontinues with that of convex functions. This is also the path we have followed,in contrast with the quasiconvex case.

We start by recalling the notion of a convex set. A set E ⊂ RN is said to beconvex if for every x, y ∈ E and every t ∈ [0, 1]

tx + (1− t) y ∈ E.

We then give several elementary properties concerning the interior, closure andboundary of convex sets. We next turn to two of the most useful results for con-vex sets, namely the separation theorems (see Corollary 2.11) and Caratheodorytheorem (see Theorem 2.13). A typical separation theorem is, for example, thefollowing.

Theorem 1.1 Let E ⊂ RN be convex and x ∈ ∂E. Then there exists a ∈ RN ,a = 0, so that

〈x; a〉 ≤ 〈x; a〉 for every x ∈ E,

where 〈.; .〉 denotes the scalar product in RN .

We also recall that the convex hull of a set E ⊂ RN is the smallest convexset containing E and is denoted by coE. Caratheodory theorem then states thefollowing.

Theorem 1.2 Let E ⊂ RN . Then

coE =x ∈ RN : x =

∑N+1i=1 λixi , xi ∈ E, λi ≥ 0 with

∑N+1i=1 λi = 1

.

We then conclude this brief account on convex sets by recalling the notionof extreme points of a convex set and Minkowski theorem, ensuring that if E iscompact and Eext denotes the set of extreme points of co E, then

coE = coEext .


We next discuss the concept of a convex function. We recall that a functionf : RN → R ∪ +∞ is said to be convex if

f (tx + (1− t) y) ≤ tf (x) + (1− t) f (y)

for every x, y ∈ RN and every t ∈ [0, 1] . An important property of convex func-tions that take only finite values (i.e. f : RN → R) is that they are everywherecontinuous (see Theorem 2.31).

The notions of convex set and function are related through the indicatorfunction of a set E defined by

χE (x) =

0 if x ∈ E

+∞ if x /∈ E.

Indeed the function χE is convex if and only if the set E is convex.

As we defined the notion of a convex hull for a set, a natural concept isthe convex envelope of a given function f, which is, by definition, the largestconvex function below f and is denoted by Cf. We can therefore write, for everyx ∈ RN ,

Cf (x) := sup g (x) : g ≤ f and g convex .

Of central importance in convex analysis is the concept of a conjugate function(or Legendre transform). The conjugate of a function f is a function f∗ : RN →R ∪ +∞ defined by

f∗ (x∗) := supx∈RN

〈x; x∗〉 − f (x) ,

which is a convex function, independently of the convexity of f. Iterating theprocess, we define the biconjugate of f as f∗∗ : RN → R∪ ±∞ , it is given by

f∗∗ (x) = supx∗∈RN

〈x; x∗〉 − f∗ (x∗) .

It turns out that if f takes only finite values then (see Theorem 2.43)

Cf = f∗∗.

Finally, we also investigate the differentiability of convex functions, dis-cussing, in particular, the notion of a subgradient.

1.2.2 Lower semicontinuity and existence results

The main result of Chapter 3 is the following (more general ones are found inTheorem 3.15 and Corollary 3.24).

Theorem 1.3 Let n, N ∈ N, p ≥ 1, Ω ⊂ Rn be a bounded open set with aLipschitz boundary, f : Ω × RN × RN×n → R be a non-negative continuousfunction and

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx.

6 Introduction

Part 1. If the function ξ → f (x, u, ξ) is convex, then I is (sequentially)weakly lower semicontinuous in W 1,p (meaning that (1.1) is satisfied).

Part 2. Conversely, if either N = 1 or n = 1 and I is (sequentially) weaklylower semicontinuous in W 1,p, then the function ξ → f (x, u, ξ) is convex.

We should emphasize that in the vectorial case, n, N ≥ 2, Part 1 of thetheorem is valid but the conclusion of Part 2 does not hold.

This theorem, in the scalar case, has as a first direct consequence that thefunctional is (sequentially) weakly continuous in W 1,p, meaning that

uν u in W 1,p ⇒ limν→∞

I (uν) = I (u)

if and only if ξ → f (x, u, ξ) is affine. This result again strongly contrasts withthe vectorial case.

The main implication of the lower semicontinuity theorem is on the existenceof minimizers for the problem

(P ) infI (u) : u ∈ u0 + W 1,p

0

(Ω; RN

)= m.

Indeed we have, as a special case of our general theorem (see Theorem 3.30),the following result.

Theorem 1.4 Let Ω be a bounded open set of Rn with a Lipschitz boundary.Let f : Ω×RN×RN×n → R be continuous and satisfying the coercivity condition

f (x, u, ξ) ≥ α1 |ξ|p − α2 , ∀ (x, u, ξ) ∈ Ω× RN × RN×n,

for some α1 > 0, α2 ∈ R and p > 1. Assume that ξ → f (x, u, ξ) is convex andthat I (u0) <∞. Then (P ) has at least one minimizer.

This theorem is also valid in the vectorial case, but can then be improved agreat deal.

As is well known, associated with any variational problem is the differentialequation known as the Euler-Lagrange equation. Under appropriate regularityhypotheses on the function f and on a minimizer u of (P ) , we find that u shouldsatisfy, for every x ∈ Ω,

(E)

n∑

α=1

∂

∂xα[

∂f

∂ξiα

(x, u,∇u) ] =∂f

∂ui(x, u,∇u) , i = 1, · · · , N.

The differential equation is a second order ordinary differential equation if n =N = 1, a system of such equations if N > n = 1, a single second order partialdifferential equation if n > N = 1 and a system of such equations when n, N ≥ 2.In any case, the convexity of the function ξ → f (x, u, ξ) ensures the ellipticity ofthe Euler-Lagrange equations. The prototype example is the Dirichlet integralwhere n > N = 1,

f (x, u, ξ) = f (ξ) :=1

2|ξ|2 ,


and the associated equation is nothing other than the Laplace equation

∆u = 0.

1.2.3 The one dimensional case

In Chapter 4, we specialize to the case where N = n = 1, although most of theresults are also valid if N > n = 1. We are therefore considering the problem

(P ) inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

,

where f : [a, b]× R× R → R, p ≥ 1 and

X =u ∈W 1,p (a, b) , u (a) = α, u (b) = β

.

The Euler-Lagrange equation that should satisfy any minimizer u of (P ) isthen given by

(E)d

dx[fξ (x, u (x) , u′ (x))] = fu (x, u (x) , u′ (x)) , x ∈ [a, b] ,

where fξ = ∂f/∂ξ and fu = ∂f/∂u. When the function f does not dependexplicitly on the variable x, one can find a first integral of (E) that is known asthe second form of the Euler-Lagrange equation and can be written as

f (u (x) , u′ (x))− u′ (x) fξ (u (x) , u′ (x)) = constant, x ∈ [a, b] .

At this stage it might be enlightening to see some examples that showthat, even when n = N = 1, the hypotheses of the existence theorem (seeTheorem 1.4) are essentially optimal. Indeed non-existence of minimizers inSobolev spaces occurs in all the following cases.

(1) Let (see Example 4.4) f (ξ) = e−ξ2

and

(P ) inf

I (u) =

∫ 1

0

f (u′ (x)) dx : u ∈ X

,

where X = W 1,10 (0, 1) =

u ∈W 1,1 (0, 1) : u (0) = u (1) = 0

. Here both the

convexity and coercivity hypotheses of the theorem are violated.

(2) Consider (see Example 4.5) the case f (x, u, ξ) = f (u, ξ) =√

u2 + ξ2

and

(P ) inf

I (u) =

∫ 1

0

f (u (x) , u′ (x)) dx : u ∈ X

,

where X =u ∈ W 1,1 (0, 1) : u (0) = 0, u (1) = 1

. In this case, the coercivity

condition holds with p = 1 (and not, as it should, with p > 1).

8 Introduction

(3) The present example (see Example 4.6) is known as the Weierstrassexample. Let f (x, u, ξ) = f (x, ξ) = xξ2 and

(P ) inf

I (u) =

∫ 1

0

f (x, u′ (x)) dx : u ∈ X

,

where X =u ∈W 1,2 (0, 1) : u (0) = 1, u (1) = 0

. The coercivity hypothesis

is violated at just one point (namely at x = 0).

(4) Let (the example is known as the Bolza example, see Example 4.8)

f (x, u, ξ) = f (u, ξ) =(ξ2 − 1

)2+ u4

(P ) inf

I (u) =

∫ 1

0

f (u (x) , u′ (x)) dx : u ∈W 1,40 (0, 1)

.

Here it is the convexity assumption on the function ξ → f (x, u, ξ) that is notsatisfied.

Another advantage of the case N = n = 1 is that, under appropriate con-ditions on f, notably the convexity of ξ → f (x, u, ξ) , the solutions of (E) arealso solutions and conversely (see Theorem 4.29) of the Hamiltonian system

(H)

u′ (x) = Hv (x, u (x) , v (x))

v′ (x) = −Hu (x, u (x) , v (x)) ,

where v (x) = fξ (x, u (x) , u′ (x)) and H is the Legendre transform of ξ →f (x, u, ξ) , namely

H (x, u, v) = supξ∈R

v ξ − f (x, u, ξ) .

In classical mechanics, f is called the Lagrangian and H the Hamiltonian.

We conclude the study of Chapter 4 with a brief discussion on Lavrentievphenomenon. We just study the following example (see Theorem 4.41) exhibitedby Mania. We let

f (x, u, ξ) =(x− u3

)2ξ6,

I (u) =

∫ 1

0

f (x, u (x) , u′ (x)) dx.

Consider the two different Sobolev spaces

W∞ =u ∈ W 1,∞ (0, 1) : u (0) = 0, u (1) = 1

,

W1 =u ∈ W 1,1 (0, 1) : u (0) = 0, u (1) = 1

,

and the corresponding minimization problems

inf I (u) : u ∈ W∞ = m∞ and inf I (u) : u ∈ W1 = m1 .

Quasiconvex analysis and the vectorial case 9

We prove that

m∞ > m1 = 0

and that u (x) = x1/3 is a minimizer of I over W1 .

1.3 Quasiconvex analysis and the vectorial case

We next turn to the vectorial case n, N ≥ 2, which is the heart of our book anddeals with what we call quasiconvex analysis. The structure is similar to thatof Part I; namely, we develop the quasiconvex analysis in Chapters 5, 6 and 7and then discuss lower semicontinuity and existence results in Chapter 8.

A first striking difference between our presentations of convex and quasicon-vex analyses is the order in which we deal with sets and functions. In convexanalysis we first defined, as do essentially all other authors, the concept of con-vex sets and then that of convex functions. In the present context, we do exactlythe reverse. This has some historical reasons. The notion of a quasiconvex func-tion was introduced by Morrey in 1952, while the corresponding notion for setsappeared almost fifty years later and is, in some sense, in its infancy.

The main motivation for introducing the notion of quasiconvexity is to gen-eralize Theorem 1.3 to the vectorial case.

1.3.1 Quasiconvex functions

Unfortunately, when generalizing the notion of a convex function to the vectorialcase, several different concepts arise naturally. The notion of a quasiconvexfunction arises, as already said, in conjunction with (sequential) weak lowersemicontinuity of the corresponding integral. When dealing with the Euler-Lagrange equation, the right concept is the ellipticity and this leads to thedefinition of a rank one convex function. Finally, when one wants to generalizethe separation theorems, Caratheodory theorem, or the notion of duality, oneis driven to the concept of polyconvexity.

We now describe the content of Chapter 5 and we start with the followingdefinitions.

Definition 1.5 Let f : RN×n → R.

(i) The function f is said to be rank one convex if

f (λξ + (1− λ) η) ≤ λf (ξ) + (1− λ) f (η)

for every λ ∈ [0, 1] , ξ, η ∈ RN×n with rank ξ − η ≤ 1.

(ii) If f is Borel measurable and locally bounded, then it is said to be quasi-convex if

f (ξ) ≤ 1

measD

∫

D

f (ξ +∇ϕ (x)) dx

10 Introduction

for every bounded open set D ⊂ Rn, for every ξ ∈ RN×n and for every ϕ ∈W 1,∞

0

(D; RN

).

(iii) The function f is said to be polyconvex if there exists F : Rτ(n,N) → Rconvex, such that

f (ξ) = F (T (ξ)) ,

where T : RN×n → Rτ(n,N) is such that

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) .

In the previous definition, adjs ξ stands for the matrix of all s× s minors of thematrix ξ ∈ RN×n, 2 ≤ s ≤ n ∧N = min n, N , and

τ (n, N) =

n∧N∑

s=1

σ (s)

where

σ (s) =(Ns

)(ns

)=

N !n!

(s!)2 (N − s)! (n− s)!.

(iv) A function f is said to be rank one affine, quasiaffine or polyaffine if fand −f are rank one convex, quasiconvex or polyconvex respectively.

Remark 1.6 (i) Note that in the case N = n = 2, the notion of polyconvexitycan be read as follows:

τ (n, N) = τ (2, 2) = 5 (since σ (1) = 4, σ (2) = 1)

T (ξ) = (ξ,det ξ) , f (ξ) = F (ξ,det ξ) .

(ii) The first and third definitions extend in a straightforward manner tofunctions f : RN×n → R ∪ +∞ . However this is not the case for quasiconvexfunctions. At the moment, no good definition of quasiconvexity for such func-tions is available. This fact is a strong source of difficulty when dealing withthe definition of quasiconvex sets. ♦

The main properties of these functions are now given (see Theorems 5.3and 5.20).

Theorem 1.7 Let f : RN×n → R.

(i) The following implications hold

f convex ⇒ f polyconvex ⇒ f quasiconvex ⇒ f rank one convex.

(ii) If N = 1 or n = 1, then all these notions are equivalent.

(iii) If f ∈ C2(RN×n

), then rank one convexity is equivalent to the Legendre-

Hadamard condition (or ellipticity condition)

N∑

i,j=1

n∑

α,β=1

∂2f (ξ)

∂ξiα∂ξj

β

λiλjμαμβ ≥ 0


for every λ ∈ RN , μ ∈ Rn, ξ =(ξiα

)1≤i≤N

1≤α≤n∈ RN×n.

(iv) The notions of rank one affine, quasiaffine and polyaffine are equivalent.Moreover, any quasiaffine function is of the form

f (ξ) = α + 〈β; T (ξ)〉 ,

where α ∈ R and β ∈ Rτ(n,N) and 〈.; .〉 denotes the scalar product in Rτ(n,N).

We now give some significant examples. The first one (see Theorem 5.25)concerns quadratic forms and is one of the most important, since then theassociated Euler-Lagrange equations are linear.

Theorem 1.8 Let M be a symmetric matrix in R(N×n)×(N×n). Let

f (ξ) = 〈Mξ; ξ〉 ,

where ξ ∈ RN×n and 〈·; ·〉 denotes the scalar product in RN×n. The followingstatements then hold.

(i) f is rank one convex if and only if f is quasiconvex.

(ii) If N = 2 or n = 2, then

f polyconvex ⇔ f quasiconvex ⇔ f rank one convex.

(iii) If N, n ≥ 3, then in general

f rank one convex f polyconvex.

We next turn to some more examples.

1) Let N = n = 2. The function

f (ξ) = det ξ

is quasiaffine and thus polyconvex, quasiconvex or rank one convex, but notconvex.

2) When n ≥ 2 and N ≥ 3, Sverak (see Theorem 5.50) produced an exampleof a function that is rank one convex but not quasiconvex, answering a longstanding conjecture of Morrey. It is still not known if there are rank one convexbut not quasiconvex functions in the case N = n = 2, or more generally n ≥N = 2.

3) Let N = n = 2. The function studied by Alibert-Dacorogna-Marcellini(see Theorem 5.51) and given by fγ : R2×2 → R, for γ ∈ R, where

fγ (ξ) = |ξ|2 ( |ξ|2 − 2γ det ξ ),

12 Introduction

is such that

fγ is convex ⇔ |γ| ≤ γc = 2√

2/3,

fγ is polyconvex ⇔ |γ| ≤ γp = 1,

fγ is quasiconvex ⇔ |γ| ≤ γq and γq > 1,

fγ is rank one convex ⇔ |γ| ≤ γr = 2/√

3 .

It is not presently known if γq = 2/√

3 .

1.3.2 Quasiconvex envelopes

In Chapter 6, we define the convex Cf (already defined in Section 1.2.1) polycon-vex Pf, quasiconvex Qf and rank one convex envelope Rf, which are, respec-tively, defined as the largest convex, polyconvex, quasiconvex and rank oneconvex functions below f. We therefore have, for every ξ ∈ RN×n,

Cf (ξ) = sup g (ξ) : g ≤ f and g convex ,

Pf (ξ) = sup g (ξ) : g ≤ f and g polyconvex ,

Qf (ξ) = sup g (ξ) : g ≤ f and g quasiconvex ,

Rf (ξ) = sup g (ξ) : g ≤ f and g rank one convex .

Observe that Theorem 1.7 immediately implies

Cf ≤ Pf ≤ Qf ≤ Rf ≤ f.

Several representation formulas exist for computing these envelopes, we justgive a formula for the quasiconvex envelope (see Theorem 6.9).

Theorem 1.9 Let f : RN×n → R be locally bounded, non-negative and Borelmeasurable. Then, for every ξ ∈ RN×n,

Qf (ξ) = inf

1

meas D

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈W 1,∞0

(D; RN

),

where D ⊂ Rn is a bounded open set. In particular, the infimum in the formulais independent of the choice of D.

We now give some examples.

(1) Let f : RN×n → R, Φ : RN×n → R be quasiaffine not identically constantand g : R → R such that

f (ξ) = g (Φ (ξ)) .

Then (see Theorem 6.24)

Pf = Qf = Rf = Cg Φ

and in general Qf > Cf.


(2) Recall the area type case, where N = n + 1. Let f : R(n+1)×n → R andg : Rn+1 → R be such that

f (ξ) = g (adjn ξ) .

Then (see Theorem 6.26)

Pf = Qf = Rf = Cg adjn

and in general Qf > Cf.

(3) An interesting problem in optimal design is the following. Let N = n = 2and, for ξ ∈ R2×2,

f (ξ) =

1 + |ξ|2 if ξ = 0

0 if ξ = 0.

Then (see Theorem 6.28) Pf = Qf = Rf and

Qf (ξ) =

1 + |ξ|2 if |ξ|2 + 2 |det ξ| ≥ 1

2(|ξ|2 + 2 |det ξ|)1/2 − 2 |det ξ| if |ξ|2 + 2 |det ξ| < 1.

We also have

Cf (ξ) =

1 + |ξ|2 if |ξ| ≥ 1

2 |ξ| if |ξ| < 1.

1.3.3 Quasiconvex sets

We have seen in Section 1.2.1 that the connection between convex functions andsets is made via the indicator function. We recall that, for a set E, the indicatorfunction is defined by

χE (x) =

0 if x ∈ E

+∞ if x /∈ E.

Moreover, the function χE is convex if and only if the set E is convex.

The aim of Chapter 7 is to extend the definition of convexity for sets topolyconvexity, quasiconvexity and rank one convexity. A natural way to definepolyconvex, quasiconvex or rank one convex set E would be by requiring thatχE be polyconvex, quasiconvex or rank one convex. This is indeed so (seeProposition 7.5) for the first and third cases but not for quasiconvex sets, since,as we already said, we lack a good definition of quasiconvexity for functions thatare allowed to take the value +∞.

Before giving the definitions, let us introduce some notation. In this sectionwe let O(n) be the set of n× n orthogonal matrices,

D := (0, 1)n ⊂ Rn

14 Introduction

and W 1,∞per (D; RN ) be the space of periodic functions in W 1,∞(D; RN ), meaning

that

u(x) = u(x + ei), for every x ∈ D and i = 1, · · · , n,

where e1, · · · , en is the standard orthonormal basis of Rn. Finally, Wper

denotes the subspace of functions in W 1,∞per (D; RN ), whose gradients take only

a finite number of values.

We are now in a position to give the following definitions (see Definition 7.2).

Definition 1.10 (i) We say that E ⊂ RN×n is polyconvex if there exists aconvex set K ⊂ Rτ(N,n) such that

ξ ∈ RN×n : T (ξ) ∈ K

= E.

(ii) We say that E ⊂ RN×n is quasiconvex if we have

ξ +∇ϕ(x)R ∈ E, a.e. x ∈ D,

for some R ∈ O(n) and some ϕ ∈ Wper

⇒ ξ ∈ E.

(iii) We say that E ⊂ RN×n is rank one convex if for every λ ∈ [0, 1] andξ, η ∈ E such that rank ξ − η = 1, then

λξ + (1− λ)η ∈ E.

The best definition for quasiconvex sets is unclear. Several definitions havealready been considered by other authors. The one we propose here is consistentwith known properties for functions and has most properties that are desirableas witnessed by the following theorem (see Theorem 7.7).

Theorem 1.11 Let E ⊂ RN×n. The following implications then hold:

E convex ⇒ E polyconvex ⇒ E quasiconvex ⇒ E rank one convex.

All counter implications are false as soon as N, n ≥ 2.

We should draw attention to the last statement of the theorem. Surprisinglyit is better than the corresponding one for functions, where the example ofSverak provides a rank one convex function that is not quasiconvex only whenn ≥ 2 and N ≥ 3.

Before continuing, one main difference between convex sets and generalizedones should be emphasized. A set can be polyconvex, and thus quasiconvex andrank one convex, and be disconnected. Indeed, if ξ, η ∈ RN×n are such thatrank ξ − η ≥ 2, then E = ξ, η is polyconvex.

We next point out a fact (the second one in the next proposition) strikinglydifferent from the equivalent one for convex sets (see Proposition 7.24).


Proposition 1.12 (i) Let E ⊂ RN×n be, respectively, a polyconvex, quasi-convex or rank one convex set. Then intE is also, respectively, polyconvex,quasiconvex or rank one convex.

(ii) There exists a polyconvex and bounded set E ⊂ R2×2 such that E is notrank one convex (and hence neither quasiconvex nor polyconvex).

We next define the polyconvex, quasiconvex and rank one convex hulls of aset E ⊂ RN×n as the smallest polyconvex, quasiconvex and rank one convexsets containing E; they are respectively denoted by PcoE, QcoE and Rco E.

We clearly have

E ⊂ RcoE ⊂ QcoE ⊂ PcoE ⊂ co E.

Other hulls are also defined in Chapter 7.

We finally conclude this section by giving an example. We first recall thatthe singular values of a given matrix ξ ∈ Rn×n, denoted by

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) ,

are the eigenvalues of (ξξt)1/2

. Let 0 < γ1 ≤ · · · ≤ γn and consider the set

E =ξ ∈ Rn×n : λi (ξ) = γi, i = 1, · · · , n

.

We prove (see Theorem 7.43) that

coE =ξ ∈ Rn×n :

∑ni=ν λi (ξ) ≤∑n

i=ν γi , ν = 1, · · · , n

PcoE = QcoE = RcoE =ξ ∈ Rn×n :

∏ni=ν λi (ξ) ≤∏n

i=ν γi , ν = 1, · · · , n

.

1.3.4 Lower semicontinuity and existence theorems

In Chapter 8, we extend the lower semicontinuity results (see Theorem 1.3) tothe vectorial context. This is a delicate matter and, in Chapter 8, we deal withit in several steps. We now gather Theorems 8.1 and 8.11 to obtain the followingresult.

Theorem 1.13 Let 1 ≤ p <∞, Ω ⊂ Rn be a bounded open set with a Lipschitzboundary and let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a continuous function satisfying

0 ≤ f (x, u, ξ) ≤ g (x, u) (1 + |ξ|p) ,

16 Introduction

whereg : Ω× RN → R, g = g (x, u) ,

is a non-negative continuous function. Let

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx.

Then I is (sequentially) weakly lower semicontinuous in W 1,p(Ω; RN

)if and

only if ξ → f (x, u, ξ) is quasiconvex, i.e.

1

measD

∫

D

f (x0, u0, ξ0 +∇ϕ (x)) dx ≥ f (x0, u0, ξ0)

for every (x0, u0, ξ0) ∈ Ω × RN × RN×n, for every bounded open set D ⊂ Rn

and for every ϕ ∈W 1,∞0 (D; Rn) .

This result has as an immediate corollary that I is (sequentially) weaklycontinuous in W 1,p if and only if ξ → f (x, u, ξ) is quasiaffine, i.e. all minorsof the matrix ξ ∈ RN×n are weakly continuous. We now restate this result,in a more convenient and more general way, in the case where N = n = 2(see Theorem 8.20, Lemma 8.24 and Corollary 8.26). Let us start with thesimple but fundamental observation that Jacobian determinants can be writtenin divergence form. More precisely if u ∈ C2

(Ω; R2

), then letting

Det∇u :=∂

∂x1(u1 ∂u2

∂x2)− ∂

∂x2(u1 ∂u2

∂x1),

we find thatDet∇u (x) = det∇u (x) , for every x ∈ Ω,

since we trivially have

det∇u =∂u1

∂x1

∂u2

∂x2− ∂u2

∂x1

∂u1

∂x2

=∂

∂x1(u1 ∂u2

∂x2)− ∂

∂x2(u1 ∂u2

∂x1) = Det∇u.

The quantity Det∇u is called the distributional Jacobian of u. We can now statethe theorem (see Theorem 8.20, Lemma 8.24, Corollary 8.26 and Example 8.28).

Theorem 1.14 Let Ω ⊂ R2 be a bounded open set, 1 2, then

det∇uν det∇u in Lp/2 (Ω) .

Relaxation and non-convex problems 17

If p = 2, the result is false, but the following convergence holds

det∇uν det∇u in D′ (Ω) .

Part 2. If p ≥ 4/3, then Det∇u ∈ D′ (Ω) and if p ≥ 2, then

Det∇u = det∇u in D′ (Ω) .

Part 3. If p > 4/3, then

Det∇uν Det∇u in D′ (Ω) .

If p ≤ 4/3, the result is false.

Theorem 1.13 also has as a direct consequence the following existencetheorem (see Theorem 8.29).

Theorem 1.15 Let p > 1, Ω ⊂ Rn be a bounded open set with a Lipschitzboundary. Let f : Ω × RN × RN×n → R, f = f (x, u, ξ) , be a continuousfunction satisfying

ξ → f (x, u, ξ) is quasiconvex,

α1 |ξ|p + β1 ≤ f (x, u, ξ) ≤ α2 (|ξ|p + 1) ,

for every (x, u, ξ) ∈ Ω× RN × RN×n, where α2 ≥ α1 > 0, β1 ∈ R. Let

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

).

Then (P ) admits at least one minimizer.

Using Theorem 1.14, we can also prove some existence theorems for poly-convex functions (see Theorem 8.31).

1.4 Relaxation and non-convex problems

In Part III, we go back to the study of

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where

- Ω ⊂ Rn, n ≥ 1, is a bounded open set;

- u : Ω → RN , N ≥ 1 and u0 ∈ W 1,p(Ω; RN

)is a given function;

- f : Ω × RN × RN×n → R, f = f (x, u, ξ) , is a given non-convex (non-quasiconvex in the vectorial case) function.

The direct methods (see Theorems 1.4 and 1.15) do not apply and the generalrule is that (P ) has no minimizers, as already pointed out in Section 1.2.3.

18 Introduction

However, there is a way of defining generalized solutions of (P ) via the so calledrelaxed problem

(QP ) inf

I (u) =

∫

Ω

Qf (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where Qf is the quasiconvex envelope of f (with respect to the last variable∇u), defined in Section 1.3.2.

The relaxed problem is useful not only to define generalized solutions of (P ) ,but also to show that in many cases, although the direct methods do not apply,the problem (P ) does have minimizers.

1.4.1 Relaxation theorems

In Chapter 9, we prove the relaxation theorem (see Theorems 9.1 and 9.8) andwe state it here, as usual under stronger hypotheses, in the case where f doesnot depend on lower order terms.

Theorem 1.16 Let Ω ⊂ Rn be a bounded open set. Let f : RN×n → R be aBorel measurable function satisfying, for 1 ≤ p <∞,

0 ≤ f (ξ) ≤ α (1 + |ξ|p) , for every ξ ∈ RN×n,

where α > 0 is a constant while for p = ∞ it is assumed that f is locallybounded. For every ξ ∈ RN×n, let

Qf (ξ) = sup g (ξ) : g ≤ f and g quasiconvex

be the quasiconvex envelope of f.

Part 1. Theninf (P ) = inf(QP ).

More precisely, for every u ∈W 1,p(Ω; RN

), there exists a sequence uν∞ν=1 ⊂

u + W 1,p0

(Ω; RN

)such that

uν → u in Lp(Ω; RN

)as ν →∞,

∫

Ω

f (∇uν (x)) dx →∫

Ω

Qf (∇u (x)) dx as ν →∞.

Part 2. Assume, in addition to the hypotheses of Part 1, that, if 1 0, γ ∈ R such that

γ + β |ξ|p ≤ f (ξ) ≤ α (1 + |ξ|p) for every ξ ∈ RN×n.

Then, in addition to the conclusions of Part 1, the following holds:

uν u in W 1,p (Ω) as ν →∞.


1.4.2 Some existence theorems for differential inclusions

When we apply, in Section 1.4.3, the relaxation theorems to get existence ofminimizers for the problem (P ) , we need to find solutions of some differentialinclusions. This is achieved in Chapter 10, where we deal with the problem offinding a map u ∈W 1,∞ (Ω; RN

)that solves

∇u (x) ∈ E a.e. x ∈ Ω

u (x) = u0 (x) x ∈ ∂Ω,

where u0 ∈W 1,∞ (Ω; RN)

is a given map and E ⊂ RN×n is a given set.

In this introductory chapter, we do not give any general result but discussonly some significant examples. The first one concerns the scalar case, wherethe result takes an almost optimal form (see Theorem 10.18).

Theorem 1.17 Let Ω ⊂ Rn be a bounded open set and E ⊂ Rn. Let u0 ∈W 1,∞ (Ω) satisfy

∇u0 (x) ∈ E ∪ int coE a.e. x ∈ Ω (1.2)

(where int coE stands for the interior of the convex hull of E); then there existsu ∈ u0 + W 1,∞

0 (Ω) such that

∇u (x) ∈ E a.e. x ∈ Ω.

The theorem has as an immediate consequence the following result (seeCorollary 10.20). If F : Rn → R is continuous and such that

lim|ξ|→∞

F (ξ) = +∞

and u0 ∈ W 1,∞ (Ω) verifies

F (∇u0 (x)) ≤ 0 a.e. x ∈ Ω,

then there exists u ∈ u0 + W 1,∞0 (Ω) such that

F (∇u (x)) = 0 a.e. x ∈ Ω.

The condition (1.2) is also necessary when the boundary datum is affine (seeTheorem 10.24).

Theorem 1.18 Let Ω ⊂ Rn be a bounded open set, E ⊂ Rn and u0 be suchthat

∇u0 = ξ0

for some ξ0 ∈ Rn. If u ∈ u0 + W 1,∞0 (Ω) solves

∇u (x) ∈ E a.e. x ∈ Ω,

20 Introduction

thenξ0 ∈ E ∪ int co E.

The next result (see Theorem 10.25), which is now a vectorial one, shouldbe related to the example given in Section 1.3.3.

Theorem 1.19 Let Ω ⊂ Rn be a bounded open set, 0 < a1 ≤ · · · ≤ an andξ0 ∈ Rn×n be such that

n∏

i=ν

λi(ξ0) <

n∏

i=ν

ai , ν = 1, · · · , n.

If u0 is an affine map such that ∇u0 = ξ0 , then there exists u ∈ u0 +W 1,∞

0 (Ω; Rn) so that, for almost every x ∈ Ω,

λν(∇u (x)) = aν , ν = 1, · · · , n.

1.4.3 Some existence results for non-quasiconvexintegrands

We now apply (see Chapter 11) the results of the two previous sections to provethe existence of minimizers for

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ u0 + W 1,∞0

(Ω; RN

),

where u0 is an affine map such that ∇u0 = ξ0 ∈ RN×n, but without assumingany convexity or quasiconvexity hypothesis on the integrand f. We could alsotreat integrands depending on lower order terms as well as boundary data thatare not affine, but then only very few general results can be given and, moreover,they are often restricted to the scalar case.

Clearly, if the integrand f were quasiconvex, because of the special form ofthe boundary datum, we would trivially have that u0 is a minimizer of (P ) .

From the relaxation theorem, the following theorem easily follows (see The-orem 11.1).

Theorem 1.20 Let Ω ⊂ Rn be a bounded open set, f a non-negative, locallybounded and lower semicontinuous function and u0 be as above, in particular∇u0 = ξ0 . The problem (P ) has a solution if and only if there exists u ∈u0 + W 1,∞

0 (Ω; RN ) such that

f (∇u (x)) = Qf (∇u (x)) a.e. x ∈ Ω,

∫

Ω

Qf (∇u (x)) dx = Qf (ξ0) measΩ.

We do not continue here with general necessary and sufficient conditions forthe existence of minimizers for (P ) , but we rather give several examples.

We start with the very elementary case where n = N = 1 (see Theorem11.24). The result adapts in a straightforward manner to the case N > n = 1.


Theorem 1.21 Let f : R → R be non-negative, locally bounded and lowersemicontinuous. Let a < b, α, β ∈ R and

(P ) inf

I (u) =

∫ b

a

f (u′ (x)) dx : u ∈ X

,

whereX =

u ∈ W 1,∞ (a, b) : u (a) = α, u (b) = β

.

The following two statements are then equivalent.

(i) Problem (P ) has a minimizer.

(ii) There exist 0 ≤ λ ≤ 1 and γ, δ ∈ R such that

Cf(β − α

b− a) = λf (γ) + (1− λ) f (δ) and

β − α

b− a= λγ + (1− λ) δ, (1.3)

where Cf = sup g ≤ f : g convex .

Furthermore, if (1.3) is satisfied, then

u (x) =

γ (x− a) + α if x ∈ [a, a + λ (b− a)]

δ (x− a) + λ (γ − δ) (b− a) + α if x ∈ (a + λ (b− a) , b]

is a minimizer of (P ) .

Note that by Caratheodory theorem we always have

Cf(β − α

b− a) = infλf (γ) + (1− λ) f (δ) :

β − α

b− a= λγ + (1− λ) δ . (1.4)

Therefore (1.3) states that a necessary and sufficient condition for the existenceof solutions is that the infimum in (1.4) be attained. Note also that if f isconvex or coercive (in the sense that f (ξ) ≥ a |ξ|p + b with p > 1, a > 0),then the infimum in (1.4) is always attained. Hence, if f (x, u, ξ) = f (ξ) ,counterexamples to existence must be non-convex and non-coercive, as in theexample already considered in Section 1.2.3, where f (ξ) = e−ξ2

.Of course, if f depends explicitly on u, the example of Bolza (given in Section

1.2.3) shows that the theorem is then false.

We now give three examples in the vectorial case.

(1) The first one (see Theorem 11.32) deals with the minimization problem

(P ) inf

∫

Ω

g(Φ(∇u(x))) dx : u ∈ u0 + W 1,∞0 (Ω; RN )

,

where:

- g : R → R is a lower semicontinuous, locally bounded and non-negativefunction,

- Φ : RN×n → R is quasiaffine and non-constant (in particular, we can have,when N = n, Φ(ξ) = det ξ).

22 Introduction

The relaxed problem is then (see Section 1.3.2)

(QP ) inf

∫

Ω

Cg(Φ(∇u(x))) dx : u ∈ u0 + W 1,∞0 (Ω; RN )

,

where Cg is the convex envelope of g. The existence result is the following.

Theorem 1.22 Let Ω ⊂ Rn be a bounded open set, g : R → R as above andsatisfying

lim|t|→+∞

g(t)

|t| = +∞

and u0 (x) = ξ0x with ξ0 ∈ RN×n. Then there exists u ∈ u0 + W 1,∞0 (Ω; RN ) a

minimizer of (P ) .

(2) The second example deals with integrands of area type (see Section 1.3.2),where N = n + 1 and

f(ξ) = g(adjn ξ).

The minimization problem is then

(P ) inf

∫

Ω

g(adjn(∇u(x))) dx : u ∈ u0 + W 1,∞0 (Ω; Rn+1)

,

where Ω is a bounded open set of Rn, ∇u0 = ξ0 and g : Rn+1 → R is a non-negative, lower semicontinuous and locally bounded non-convex function.

From Section 1.3.2, we have

Qf(ξ) = Cg(adjn ξ).

We next setS = y ∈ Rn+1 : Cg(y) < g(y)

and assume, in order to avoid the trivial situation, that adjn ξ0 ∈ S.

The existence result (see Theorem 11.36) is then given by the following.

Theorem 1.23 If S is bounded, Cg is affine in S and rank ξ0 ≥ n − 1, then(P ) has a solution.

(3) The third problem is that of optimal design, already discussed in Section1.3.2, where

f (ξ) =

1 + |ξ|2 if ξ = 0

0 if ξ = 0.

Consider the problem

(P ) inf

∫

Ω

f (∇u (x)) dx : u ∈ u0 + W 1,∞0 (Ω; R2)

,

where Ω is a bounded open set of R2 and ∇u0 = ξ0 .

We then have the following (see Theorem 11.35).

Miscellaneous 23

Theorem 1.24 Let Ω ⊂ R2 be a bounded open set, f : R2×2 → R be as aboveand ξ0 ∈ R2×2. Then a necessary and sufficient condition for (P ) to have asolution is that one of the following conditions hold:

(i) ξ0 = 0 or |ξ0|2 + 2 |det ξ0| ≥ 1 (i.e. f (ξ0) = Qf (ξ0))

(ii) det ξ0 = 0.

1.5 Miscellaneous

In Part IV, we gather some notations and standard results on function spacesand on singular values. We also devote the last two chapters to results that playonly a marginal role in our analysis, but have some interest on their own.

1.5.1 Holder and Sobolev spaces

In Chapter 12, we only fix the notation concerning the main function spacesthat we use, namely the Holder spaces Cm,α

(Ω; RN

)and the Sobolev spaces

Wm,p(Ω; RN

), where m is an integer, 0 < α ≤ 1 and 1 ≤ p ≤ ∞. We recall

without proofs the most important results, for example the Sobolev imbeddingtheorem, that we use throughout the book.

1.5.2 Singular values

We recall in Chapter 13 the definition and some elementary properties of thesingular values of a matrix ξ ∈ Rn×n (in the present introduction, we discussonly the case N = n, but in Chapter 13 we consider general matrices in RN×n).We denote by

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ)

the eigenvalues of (ξξt)1/2

. The main result (see Theorem 13.3) is the following.

Theorem 1.25 Let ξ ∈ Rn×n and 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) be its singularvalues. Then there exist R, Q ∈ O (n) (the set of orthogonal matrices R ∈ Rn×n

so that RtR = I) such that

RξQ = diag (λ1 (ξ) , · · · , λn (ξ)) :=

⎛⎜⎝

λ1 · · · 0...

. . ....

0 · · · λn

⎞⎟⎠ .

In some applications, it might be better to replace the singular values λ1 (ξ) ≤· · · ≤ λn (ξ) of a given matrix ξ ∈ Rn×n by its signed singular values

0 ≤ |μ1 (ξ)| ≤ · · · ≤ μn (ξ)

24 Introduction

defined by

μ1 (ξ) = λ1 (ξ) sign (det ξ) and μj (ξ) = λj (ξ) , j = 2, · · · , n.

We then have the following inequality (see Theorem 13.10).

Theorem 1.26 Let ξ, η ∈ Rn×n. Then

maxQ,R∈SO(n)

trace(QξRtηt) =n∑

j=1

μj (ξ)μj (η)

and consequently

trace(ξηt) ≤n∑

j=1

μj (ξ) μj (η) .

1.5.3 Some underdetermined partial differentialequations

In Chapter 14, we prove the existence of solutions for three types of underdeter-mined partial differential equations that are encountered in mechanics. All threeproblems bear in common that, in general, there are infinitely many solutions.The first one concerns the divergence operator (see Theorem 14.2) as stated inthe next theorem.

Theorem 1.27 Let m ≥ 0 be an integer and 0 < α < 1. Let Ω ⊂ Rn be a smoothbounded connected open set. The following conditions are then equivalent.

(i) f ∈ Cm,α(Ω)

satisfies

∫

Ω

f (x) dx = 0.

(ii) There exists u ∈ Cm+1,α(Ω; Rn

)verifying

div u = f in Ω

u = 0 on ∂Ω

where div u =∑n

i=1∂ui

∂xi.

The second problem is related to the curl operator (see Theorem 14.4).

Theorem 1.28 Let m ≥ 1 be an integer and 0 < α < 1. Let Ω ⊂ R3 be asmooth bounded convex set and ν denote the outward unit normal. The followingconditions are then equivalent.

(i) f ∈ Cm,α(Ω; R3

)verifies

div f = 0 in Ω and 〈f ; ν〉 = 0 on ∂Ω.

Miscellaneous 25

(ii) There exists u ∈ Cm+1,α(Ω; R3

)satisfying

curlu = f in Ω

u = 0 on ∂Ω

where if u =(u1, u2, u3

), then

curlu =

(∂u3

∂x2− ∂u2

∂x3,

∂u1

∂x3− ∂u3

∂x1,

∂u2

∂x1− ∂u1

∂x2

).

Both cases are examples of a study of Dirichlet problems associated to du =f, where u is a k form and d is the exterior derivative.

The next theorem (see Theorem 14.6) is the nonlinear version of Theorem1.27. In terms of fluid mechanics, the first one is in Eulerian coordinates, whilethe second one is in Lagrangian coordinates.

For m ≥ 1 an integer, 0 < α < 1 and Ω ⊂ Rn a bounded open set with a suf-ficiently regular boundary, we denote by Diffm,α

(Ω)

the set of diffeomorphisms

u : Ω → Ω such that u, u−1 ∈ Cm,α(Ω; Rn

).

Theorem 1.29 Let m ≥ 0 be an integer and 0 < α < 1. Let Ω ⊂ Rn bea bounded connected open set with a sufficiently smooth boundary. Let f ∈Cm,α

(Ω), f > 0 in Ω and

∫

Ω

f (x) dx = meas Ω.

Then there exists u ∈ Diffm+1,α(Ω)

satisfying

det∇u (x) = f (x) x ∈ Ω

u (x) = x x ∈ ∂Ω.

The theorem can be applied in a straightforward manner to the minimizationproblem (see Corollary 14.9)

(P ) inf

I (u) =

∫

Ω

g (det∇u (x)) dx : u ∈ u0 + W 1,∞0 (Ω; Rn)

,

where g : R → R is convex and Ω ⊂ Rn and u0 satisfy appropriate smoothnessconditions.

1.5.4 Extension of Lipschitz maps

In Chapter 15, we consider the problem of extending a Lipschitz map defined ona set to the full space so as to preserve the Lipschitz constant. This is a classicalproblem and is of particular importance in the calculus of variations, where itis known as MacShane lemma in the scalar case and Kirszbraun theorem in thevectorial case.

26 Introduction

Let us describe the problem more precisely. We consider two Banach spaces(E, ‖.‖E) and (F, ‖.‖F ). We ask when a map u : D ⊂ E → F satisfying

‖u (x)− u (y)‖F ≤ ‖x− y‖E , x, y ∈ D,

can be extended to the whole of E so as to preserve the inequality.

We need the following three definitions.

Definition 1.30 (i) We say that u : E → F is a contraction on D or u is1–Lipschitz on D if

‖u(x)− u(y)‖F ≤ ‖x− y‖E for all x, y ∈ D.

In this case, we write that u ∈ Lip1(D, F ).

(ii) When u ∈ Lip1(E, F ), we simply say that u is a contraction.

Definition 1.31 (i) We say that [E; F ] has the extension property for contrac-tions on D if every u ∈ Lip1(D, F ) has an extension u ∈ Lip1(E, F ).

(ii) If [E; F ] has the extension property for contractions for every D ⊂ E,we simply say that [E; F ] has the extension property for contractions.

Definition 1.32 The unit sphere SF (i.e. the set of x ∈ F such that ‖x‖F = 1)is said to be strictly convex if it has no flat part, meaning that

‖(1− t)x + ty‖F < (1 − t) ‖x‖F + t ‖y‖F = 1

for all t ∈ (0, 1) and all x, y ∈ SF such that x = y.

A particularly interesting example is the Holder norms |x|p over Rn, 1 ≤p ≤ ∞, and they are defined as

|x|p :=

[∑n

i=1 |xi|p ]1/p

if 1 ≤ p < ∞max1≤i≤n |xi| if p = ∞.

When n ≥ 2, the unit sphere for | · |p is strictly convex if and only if 1 < p < ∞.

We can now state our main theorems (see Theorems 15.11 and 15.12). Part(i) of Theorem 1.33 is known as MacShane lemma and the implication (i)⇒ (ii)in Theorem 1.34 is known as Kirszbraun theorem.

Theorem 1.33 (i) Let (E, ‖.‖E) be a normed space. Then [E; R] has the exten-sion property for contractions.

(ii) Let (F, ‖.‖F ) be a Banach space. Then [R; F ] has the extension propertyfor contractions.

We now turn our attention to the case where both E and F have dimensionat least 2 and we give a theorem that characterizes the Banach spaces for which[E, F ] has the extension property for contractions.

Miscellaneous 27

Theorem 1.34 Assume that (E, ‖.‖E) and (F, ‖.‖F ) are Banach spaces suchthat dimE, dim F ≥ 2 and that the unit sphere in F is strictly convex. Assumealso that every closed set D ⊂ E contains a countable set Dc ⊂ D whose closureis D. Then, the following three properties are equivalent:

(i) ‖.‖E and ‖.‖F are induced by an inner product;

(ii) [E; F ] has the extension property for contractions;

(iii) for every x ∈ E and every S := x1, x2, x3 ⊂ E, every u ∈ Lip1(S, F )has an extension u ∈ Lip1(S ∪ x, F ).

Several comments are in order.

(i) If S consists of only two points x, y ∈ E, x = y, then the extension toany third point is always possible.

(ii) If one drops the assumption that SF is strictly convex, [E; F ] may havethe extension property for contractions even if none of the norms is induced byan inner product; for example, if F = RN , N ≥ 2, and ‖.‖F = | · |∞ , then [E; F ]has the extension property for any E.

(iii) In the case of Holder norms with (E = Rn, |.|p) and(F = RN , |.|q

)with

n, N ≥ 2, 1 < q < ∞ and 1 ≤ p ≤ ∞, then [E; F ] has the extension property ifand only if p = q = 2.

Chapter 2

Convex sets and convexfunctions

2.1 Introduction

We now give a brief introduction to convex analysis. The chapter is divided intotwo sections.

In Section 2.2, we give some of the most important theorems, namely theseparation theorems (sometimes also called Hahn-Banach theorem which is theirinfinite dimensional version), Caratheodory theorem and Minkowski theorem,also usually better known as Krein-Milman theorem, which is its infinite dimen-sional version.

In Section 2.3, we list some properties of convex functions such as Jenseninequality, the continuity of such functions, the notion of duality and of subdif-ferential.

The reference book on convex analysis is Rockafellar [514]. However one canalso consult Brezis [105], Ekeland-Temam [264], Fenchel [277], Hiriart Urruty-Lemarechal [342], Hormander [344], Ioffe-Tihomirov [351], Moreau [452] orWebster [597] for further references.

We adopt throughout this chapter the following notations.

- For a given set E ⊂ RN , E, ∂E, intE and Ec respectively stand for theclosure, the boundary, the interior and the complement of E respectively.

- 〈.; .〉 stands for the scalar product in RN and, unless explicitly specified, |.|denotes the Euclidean norm in RN .

- The ball centered at x ∈ RN and of radius r > 0 is denoted by

Br (x) :=y ∈ RN : |y − x| < r

.

32 Convex sets and convex functions

2.2 Convex sets

2.2.1 Basic definitions and properties

We recall the following definition.

Definition 2.1 (i) A set E ⊂ RN is said to be convex if for every x, y ∈ Eand every t ∈ [0, 1]

tx + (1− t) y ∈ E.

(ii) A set E ⊂ RN is said to be affine if for every x, y ∈ E and every t ∈ R

tx + (1− t) y ∈ E.

(iii) The affine hull of a set E ⊂ RN is the smallest affine set containing E.It is denoted by aff E.

(iv) A hyperplane H ⊂ RN is a set of the form

H =x ∈ RN : 〈x; a〉 = α

a ∈ RN , a = 0, and α ∈ R.

The next proposition is elementary.

Proposition 2.2 (i) The intersection of an arbitrary collection of convex setsis convex.

(ii) The intersection of an arbitrary collection of affine sets is affine.

Important concepts in convex analysis are the notions of relative interior andrelative boundary.

Definition 2.3 Let E ⊂ RN be convex.

(i) The relative interior of E, denoted by ri E, is the interior of the setrelative to its affine hull aff E.

(ii) The relative boundary of E, denoted by rbdE, is the set of points in Ebut not in ri E.

The following proposition is easily proved.

Proposition 2.4 Let E ⊂ RN be convex. Then E, riE and intE are convex.Moreover intE is empty if and only if E is contained in a hyperplane.

In a straightforward manner, we also deduce the next result (see Corollary1.4.1 in Rockafellar [514]).

Proposition 2.5 Every affine subset of RN is the intersection of a finite collec-tion of hyperplanes, where, by convention, the intersection of the empty familyis equal to RN .

Convex sets 33

Finally we have the following relations between the interior and closures ofconvex sets.

Theorem 2.6 Let E ⊂ RN be convex.

(i) intE = intE.

(ii) If intE = ∅, then intE = E.

(iii) ∂E = ∂E.

Remark 2.7 The results in the theorem remain valid if we replace the interiorby the relative interior and the boundary by the relative boundary (see Theorem6.3 in Rockafellar [514]). ♦

Proof. We divide the proof into five steps.

Step 1. We first show that if x ∈ intE and y ∈ E, then

z := λx + (1− λ) y ∈ intE, for every λ ∈ (0, 1] . (2.1)

Since x ∈ intE, we can find ǫ > 0 such that

Bǫ (x) :=b ∈ RN : |b− x| < ǫ

⊂ E. (2.2)

To prove (2.1) we show that

Bλǫ (z) ⊂ E, for every λ ∈ (0, 1] . (2.3)

So we choose a ∈ Bλǫ (z) and we let

b := x +1

λ(a− z) =

1

λa + (1− 1

λ)y. (2.4)

Since a ∈ Bλǫ (z) , we find that b ∈ Bǫ (x) and from (2.2) we deduce that b ∈ E.Consequently from (2.4) we obtain that

a = λb + (1− λ) y.

Since b, y ∈ E, we have that a ∈ E and hence (2.3) is satisfied. This achievesthe proof of Step1.

Step 2. Let us now show that, in fact, the result of Step 1 holds even ify ∈ E. If λ = 1, nothing is to be proved; so assume that λ ∈ (0, 1) and ǫ > 0 isso that Bǫ (x) ⊂ intE. We set

z := λx + (1− λ) y.

Let us prove that z ∈ intE. Since y ∈ E, we can find y ∈ E so that

|y − y| < λǫ

1− λ.


Then set

x :=1

λ[z − (1− λ) y] = x +

1− λ

λ(y − y) . (2.5)

We therefore have x ∈ Bǫ (x) ⊂ E and hence x ∈ intE. From (2.5) we deducethat

z = λx + (1− λ) y

and hence we apply Step 1 to get the claim, namely z ∈ intE.

Step 3. We now prove the first claim: intE = intE. We have to considertwo cases.

Case 1. If intE = ∅, then from Proposition 2.4 we find that E is containedin a hyperplane and thus E is contained in a hyperplane and hence intE = ∅.

Case 2. Consider now the case where intE = ∅. We then show that intE ⊂intE, the reverse inclusion being obvious. Let x ∈ intE; we want to prove thatx ∈ intE.

We choose z ∈ intE with z = x (if z = x, then the claim is established).Since x ∈ intE and z ∈ intE ⊂ intE, we can find μ > 1 with μ− 1 sufficientlysmall so that

y = μx + (1− μ) z = x + (1− μ) (z − x) ∈ intE ⊂ E.

We therefore get that

x =1

μy + (1− 1

μ)z

with y ∈ E and z ∈ intE. Applying Step 2, we have the claim, namely x ∈ intE.

Step 4. Let us prove that E ⊂ intE, the reverse inclusion being trivial. Solet x ∈ E and y ∈ intE (intE is assumed to be non empty). From Step 2 wehave that

xλ := λy + (1− λ) x ∈ intE, for every λ ∈ (0, 1] .

Since xλ → x as λ→ 0, we have indeed proved that x ∈ intE.

Step 5. We now show the last claim of the theorem. This follows at oncefrom Step 3, since

∂E = E − intE = E − intE = ∂E.

2.2.2 Separation theorems

In this section, we present different separation theorems, that in the infinitedimensional case are called Hahn-Banach theorem. We recall that given twosets E, F ⊂ RN , we let

E + F :=x ∈ RN : x = y + z, y ∈ E and z ∈ F

.

Convex sets 35

We now have some definitions.

Definition 2.8 (i) A hyperplane H, defined by 〈x; a〉 = α with a ∈ RN , a = 0,and α ∈ R, is said to separate the sets E, F ⊂ RN if either

〈x; a〉 ≤ α ≤ 〈y; a〉 for every x ∈ E and y ∈ F

or the same inequalities hold for every x ∈ F and y ∈ E.

(ii) The hyperplane is said to separate properly the sets E, F ⊂ RN if theyare separated by H and at least one of them is not contained in H itself.

(iii) A hyperplane H is said to separate strictly the sets E, F ⊂ RN if thereexists ǫ > 0 such that H separates E + ǫB1 and F + ǫB1 , where B1 is the unitball of RN .

Before proving the main theorem of this section we need to define the pro-jection onto a convex set.

Theorem 2.9 Let E ⊂ RN be a closed convex set, x ∈ RN and denote by |.|the Euclidean norm.

(i) There exists a unique x∞ ∈ E minimizing z → |x− z| over E. Moreover,if x /∈ intE, then x∞ ∈ ∂E. The map

x→ pE(x) := x∞

is well defined and is referred to as the projection map onto E.

(ii) Furthermore, the following inequalities hold for every x, y ∈ RN andz ∈ E

〈x− pE(x); z − pE(x)〉 ≤ 0; (2.6)

|pE(x)− pE(y)|2 ≤ 〈pE(x)− pE(y); x− y〉 ; (2.7)

|pE(x)− pE(y)| ≤ |x− y| . (2.8)

Proof. (i) Let x ∈ RN and let zν∞ν=1 ⊂ E be such

limν→+∞

|x− zν | = infz∈E

|x− z| .

The set zν∞ν=1 being bounded, we can extract a subsequence, which we stilllabel zν∞ν=1 , converging to some x∞ ∈ E and hence

|x− x∞| = limν→+∞

|x− zν | = infz∈E

|x− z| .

Let us now show that if x /∈ intE, then x∞ ∈ ∂E. By contradiction if x∞ ∈intE, we would have for t ∈ (0, 1) small enough that

xt := (1− t)x∞ + tx ∈ E


and thus

|x− xt| = (1− t) |x− x∞| < |x− x∞|

contradicting the definition of x∞ .

We now prove that the minimizer is unique. Assume for the sake of contra-diction that for a certain x /∈ E there exist two distinct minimizers x∞, x∞ ∈ Eof |x− z| over E. Since x∞, x∞ ∈ ∂E, we find that

x0 :=x∞ + x∞

2∈ E

is another minimizer of |x− z| . It is easily seen that since |x− x∞| = |x− x∞| >0, then (using the fact that ξ → |ξ|2 is strictly convex) |x− x0| < |x− x∞| ,which yields the desired contradiction. This proves that the minimizer of |x− z|over E is unique.

(ii) Since for every t ∈ [0, 1] and z ∈ E, we have

|x− pE(x)|2 ≤ g (t) := |x− [(1− t) pE(x) + tz]|2 ,

we find, from the fact that g′ (0) ≥ 0, that pE(x) should satisfy (2.6).

If x, y ∈ RN , we use (2.6), once with z = pE(y) and once with z = pE(x), toobtain that

〈x− pE(x); pE(y)− pE(x)〉 ≤ 0 and 〈y − pE(y); pE(x)− pE(y)〉 ≤ 0.

Adding up these two inequalities yields (2.7), namely

|pE(x)− pE(y)|2 ≤ 〈pE(x)− pE(y); x− y〉 .

This, together with Cauchy-Schwarz inequality, leads to (2.8).

We are now in a position to state the theorem.

Theorem 2.10 (Separation theorems) (i) Let E ⊂ RN be closed and con-vex and x /∈ E. Then there exists a ∈ RN , a = 0, such that

〈x; a〉 < inf 〈x; a〉 : x ∈ E .

(ii) Let E ⊂ RN be closed and convex and x ∈ ∂E. Then there exists a ∈ RN ,a = 0, so that

〈x; a〉 ≤ 〈x; a〉 for every x ∈ E.

(iii) Let E, F ⊂ RN be non-empty, disjoint and convex. Let E be closed andF compact. Then there exists a hyperplane that separates E and F strictly.

(iv) Every closed convex set is the intersection of the closed half spaces thatcontain it.

Convex sets 37

Proof. (i) We choose a := pE(x) − x and α := 〈x; a〉 . Observe that sincex /∈ E, then a = 0. We therefore get, for any x ∈ E, that, using (2.6),

〈x− x; a〉 = −〈x− pE(x); x− pE(x)〉+ |pE(x)− x|2 ≥ |a|2

and the claim follows.

(ii) Since x ∈ ∂E, we can find a sequence xν /∈ E with xν → x. Applying theabove result we can find aν ∈ RN , aν = 0 (and hence we can assume withoutloss of generality that |aν | = 1), such that

〈xν ; aν〉 < inf 〈x; aν〉 : x ∈ E .

Extracting a subsequence from aν we have the result by passing to the limit.

(iii) Define the set

G = E − F :=z ∈ RN : z = x− y with x ∈ E and y ∈ F

.

It is clearly convex and closed, since E is closed and F is compact. Moreoversince E ∩ F = ∅, we have that 0 /∈ G. We may then apply (i) of the theorem tofind a ∈ RN , a = 0, so that

0 < inf 〈z; a〉 : z ∈ G = inf 〈x; a〉 : x ∈ E − sup 〈y; a〉 : y ∈ F

which is the desired statement.

(iv) Let E ⊂ RN (E = ∅ and E = RN ) be closed and convex. For any x /∈ Ewe can find, from (i), a ∈ RN , a = 0, and α ∈ R, so that

〈x; a〉 < α < inf 〈x; a〉 : x ∈ E .

Therefore the closed half space

H =x ∈ RN : 〈x; a〉 ≥ α

contains E but does not contain x. Therefore the intersection of the closed halfspaces containing E does not contain any other point.

We next show that in the statement of (ii) of Theorem 2.10 we can removethe assumption on the closedness of E.

Corollary 2.11 Let E ⊂ RN be convex and x ∈ ∂E. Then there exists a ∈ RN ,a = 0, such that

〈x; a〉 ≤ 〈x; a〉 , for every x ∈ E.

Proof. From Theorem 2.6, we have that x ∈ ∂E. Therefore applying Theorem2.10 (ii) to E, we have the claim.


2.2.3 Convex hull and Caratheodory theorem

We start with the following definition.

Definition 2.12 The convex hull of a set E ⊂ RN , denoted by coE, is thesmallest convex set containing E.

According to Proposition 2.2, it is equivalent to say that coE is the inter-section of all the convex sets that contain E. In the sequel we denote for anyinteger s

Λs := λ = (λ1, · · · , λs) : λi ≥ 0 and∑s

i=1 λi = 1 .

One of the most important characterizations of the convex hull isCaratheodory theorem.

Theorem 2.13 (Caratheodory theorem) Let E ⊂ RN . Then

coE =x ∈ RN : x =

∑N+1i=1 λixi, xi ∈ E, λ ∈ ΛN+1

.

Proof. We decompose the proof into two steps.

Step 1. Observe first that if I is an integer,

FI :=x ∈ RN : x =

∑Ii=1 λixi, xi ∈ E, λ ∈ ΛI

and

F :=⋃

I∈N

FI

then obviously F is convex and E ⊂ F and therefore coE ⊂ F. Conversely letE ⊂ A, A convex, then F ⊂ A and therefore F ⊂ coE and thus F = coE.

Step 2. We now show that in fact we have

F =

N+1⋃

I=1

FI .

Let m ∈ F = coE, then trivially (1, m) ∈ 1 × coE = co (1 ×E) . ApplyingStep 1 to co (1 ×E) we have that there exist I, an integer, λ ∈ ΛI , mi ∈ Esuch that

I∑

i=1

λi (1, mi) = (1, m) . (2.9)

We wish to show that we can take I ≤ N + 1 in (2.9). Assume that I > N + 1,then there exist γi ∈ R not all zero such that

I∑

i=1

γi (1, mi) = 0, (2.10)

Convex sets 39

since (1, mi) ∈ RN+1 and I > N + 1.Let T := i ∈ 1, · · · , I : γi > 0 . We may assume without loss of generality

that T = ∅, otherwise replace γi by −γi . Let

β := mini∈T

λi/γi (2.11)

μi := λi − βγi, i = 1, · · · , I. (2.12)

We then deduce that

μi ≥ 0, i = 1, · · · , I. (2.13)

I∑

i=1

μi = 1 (2.14)

at least one of the μi = 0; (2.15)

where (2.13) has been obtained trivially if γi ≤ 0 and by (2.11) and (2.12) ifγi > 0; similarly (2.14) follows from (2.9) and (2.10). Finally (2.15) holds ifone takes the index i ∈ T which corresponds to the minimum in (2.11). Wemoreover have

I∑

i=1

λi (1, mi) =I∑

i=1

μi (1, mi) = (1, m) . (2.16)

In view of (2.13), (2.14), (2.15) and (2.16), we have therefore reduced the numberI to (I − 1) . Continuing this process up to I = N +1 we have indeed establishedthe theorem.

An important consequence of Caratheodory theorem is the following.

Theorem 2.14 Let E ⊂ RN .

(i) If E is compact, then coE is compact.

(ii) If E is open, then coE is open.

Before proceeding with the proof, let us mention that the analogous state-ment for closed sets is false.

Example 2.15 Let N = 2,

E = E1 ∪ E2 ,

where

E1 = (x1, 0) : x1 ∈ [0, 1] and E2 = (0, x2) : x2 ≥ 0.

Clearly E is closed, while

coE = (x1, x2) : x1 ∈ [0, 1) , x2 ≥ 0 ∪ (1, 0)

is not closed. ♦


Proof. (i) We need only show that coE is closed. So let xν ⊂ coE bea sequence converging to an element x ∈ RN and let us show that x ∈ coE.Appealing to Theorem 2.13 we can find λν ∈ ΛN+1 and xν

i ∈ E, 1 ≤ i ≤ N + 1,such that

xν =

N+1∑

i=1

λνi xν

i .

Since ΛN+1 and E are compact, we can find subsequences, still denoted λνand xν

i , such that

λν → λ ∈ ΛN+1 and xνi → xi ∈ E, i = 1, · · · , N + 1.

We therefore have

x =

N+1∑

i=1

λixi

and thus, by Theorem 2.13, x ∈ co E, as wished.

(ii) The proof is very similar in spirit to the preceding one. Let x ∈ coEand let us show that we can find ǫ > 0 such that

Bǫ (x) :=y ∈ RN : |y − x| < ǫ

⊂ co E.

From Theorem 2.13, we can find λ ∈ ΛN+1 and xi ∈ E, i = 1, · · · , N + 1, suchthat

x =N+1∑

i=1

λixi .

Since E is open, we can find ǫ > 0 so that Bǫ (xi) ⊂ E, i = 1, · · · , N + 1. Lety ∈ Bǫ (x) and let us show that y ∈ coE. Letting

yi := xi + y − x, i = 1, · · · , N + 1,

we find that yi ∈ Bǫ (xi) ⊂ E, i = 1, · · · , N + 1, and

y =

N+1∑

i=1

λiyi .

Using again Theorem 2.13, we find that y ∈ coE, as claimed.

Another direct application is the following corollary (see Lemma 2.11 in [202]and Theorem 20.4 in [514]).

Corollary 2.16 Let E ⊂ RN and x ∈ int coE. Then there exist

x1, x2, · · · , xm ∈ E, m ≥ N + 1,

such that x1 − x, x2 − x, · · · , xm − x spans the whole of RN ,

x ∈ int co x1, x2, · · · , xm , (2.17)

Convex sets 41

and there exist si > 0, i = 1, 2, · · · , m, with∑m

i=1 si = 1 such that

x =

m∑

i=1

sixi . (2.18)

Proof. Replacing E by −x + E, we can assume that x = 0. Since 0 ∈ int coEwe can find a cube Cǫ of side 2ǫ, with ǫ > 0 sufficiently small, so that

Cǫ =x = (a1, a2, · · · , aN ) ∈ RN : |ai| < ǫ, i = 1, 2, · · · , N

⊂ int co E.

We denote by y1, · · · , y2N the vertices of the cube Cǫ . Then

0 ∈ int co y1, · · · , y2N = Cǫ ⊂ int co E. (2.19)

Note that, since∑2N

i=1 yi = 0, then the 0-vector can be expressed by the followingconvex combination

0 =

2N∑

i=1

1

2Nyi . (2.20)

We next appeal to Caratheodory theorem (restricting attention only to strictlypositive coefficients) to find, for every i = 1, 2, · · · , 2N , integers Ni ≤ N so that

yi =

Ni+1∑

k=1

ti,k xi,k , xi,k ∈ E, ti,k > 0 and

Ni+1∑

k=1

ti,k = 1.

Since the corresponding set of vectors of RN

xi,k , k = 1, · · · , Ni + 1, i = 1, · · · , 2N

(2.21)

generate by convex combinations at least the whole cube Cǫ , then the span ofthe set in (2.21) is RN . By (2.20) we obtain

0 =

2N∑

i=1

1

2Nyi =

2N∑

i=1

Ni+1∑

k=1

1

2Nti,k xi,k =

∑

i,k

si,k xi,k ,

where

si,k =ti,k2N

> 0, ∀ i, k, and∑

i,k

si,k = 1,

which proves (2.18). By combining (2.19) with

co y1, y2, · · · , y2N ⊂ coxi,k , k = 1, · · · , Ni + 1, i = 1, · · · , 2N

we also have (2.17).It remains to prove that m ≥ N + 1. From (2.18) namely

0 =

m∑

i=1

sixi

it follows that x1, x2, · · · , xm are linearly dependent and since it spans thewhole of RN , we deduce that m ≥ N + 1. This achieves the proof of thecorollary.


2.2.4 Extreme points and Minkowski theorem

Definition 2.17 Let E ⊂ RN be convex. We say that x ∈ E is an extremepoint of E if

x = ta + (1− t) b

0 < t < 1, a, b ∈ E

⇒ a = b = x.

We denote the set of extreme points of E by Eext.

Note that the set of extreme points may be empty. This is indeed the case,for example, when E is an open convex set.

In dimension N = 2, the set of extreme points of a closed convex set is closed(see Exercise 2.6 in Webster [597]); however in higher dimensions this is not so,as the following classical example shows.

Example 2.18 Let E, E1, E2 ⊂ R3 be defined by

E1 =x = (x1, x2, 0) ∈ R3 : x2

1 + x22 ≤ 1

,

E2 =x = (1, 0, x3) ∈ R3 : |x3| ≤ 1

,

E = co (E1 ∪ E2) .

It is easy to see that

Eext = (∂E1 \ (1, 0, 0)) ∪ (1, 0, 1) ∪ (1, 0,−1)

and hence it is not closed since (1, 0, 0) is not an extreme point. ♦

The important fact about extreme points is given in the theorem below. Thisresult is often known as Krein-Milman theorem, which is the infinite dimensionalversion of the result due to Minkowski. Before proving this theorem, we give aproposition, whose proof is straightforward.

Proposition 2.19 Let E ⊂ RN .

(i) Let E be convex. Then

e ∈ Eext ⇔ E − e is convex.

(ii) Let K = co E. Then Kext ⊂ E.

We now have the main result.

Theorem 2.20 (Minkowski theorem) Let E ⊂ RN be compact and let Eext

denote the set of extreme points of coE. Then

coE = coEext .

Convex sets 43

Proof. The inclusion coEext ⊂ co E, follows at once from the above proposi-tion. To prove the reverse inclusion we proceed by induction on the dimensionN. If N = 1, the result is trivial since coE is a closed interval and its extremepoints are the end points of the interval.

So we now assume that the result has been established for any (N − 1)dimensional space. Moreover we can assume that int co E is non-empty other-wise the hypothesis of induction applies.

In view of Theorem 2.13, it is enough to show that any e ∈ coE can bewritten as

e =N+1∑

i=1

λixi , xi ∈ Eext , λ ∈ ΛN+1 .

We consider two cases.

Case 1. We first assume that e ∈ ∂ (co E) . Since coE is closed by Theorem2.14, we can apply Theorem 2.10 (ii) to get a ∈ RN , a = 0, and α ∈ R so that

〈e; a〉 = α ≤ 〈x; a〉 , for every x ∈ coE.

Next defineK := x ∈ co E : 〈x; a〉 = α

and observe that it is a non empty (since e ∈ K) convex and compact set thatlies in a subspace of dimension (N − 1) . We may therefore apply the inductionhypothesis and write

e =

N∑

i=1

λixi , xi ∈ Kext , λ ∈ ΛN .

Since, obviously, Kext ⊂ Eext , we have the claim.

Case 2. We then consider the case where e ∈ int coE. Note first that byCase 1 we have that the set of extreme points is non empty. So choose anyxN+1 ∈ Eext and consider the line containing the segment [e, xN+1] . Since coEis compact, we can find y ∈ ∂ (coE) , y = e, so that e ∈ (xN+1, y) and hence wecan find μ ∈ (0, 1) so that

e = μxN+1 + (1− μ) y.

Applying Case 1 to y we find that

y =

N∑

i=1

νixi , xi ∈ Eext , ν ∈ ΛN .

Writing λi = (1− μ) νi , i = 1, · · · , N and λN+1 = μ we have indeed obtainedthe claim, namely

e =

N+1∑

i=1

λixi , xi ∈ Eext , λ ∈ ΛN+1 .

This concludes the proof of the theorem.


2.3 Convex functions

2.3.1 Basic definitions and properties

While dealing with convex functions, it is convenient to allow the functions totake infinite values, so we always consider functions

f : RN → R ∪ ±∞ .

However, since convex functions taking the value −∞ are rather special andlead to difficulties of notation, such as ∞−∞, we mostly consider functions ofthe form

f : RN → R ∪ +∞ .

But functions taking the value −∞ arise naturally in the next sections.

We start with the definition of convexity.

Definition 2.21 (i) A function f : RN → R ∪ +∞ is said to be convex if

f (tx + (1− t) y) ≤ tf (x) + (1− t) f (y)

for every x, y ∈ RN and every t ∈ [0, 1] .

(ii) A function f : E ⊂ RN → R ∪ +∞ is said to be strictly convex on aconvex set E if

f (tx + (1− t) y) < tf (x) + (1− t) f (y)

for every x, y ∈ E, x = y, and every t ∈ (0, 1) .

Remark 2.22 (i) We also adopt the convention that

0. (±∞) = 0.

(ii) For functions f : RN → R ∪ ±∞ , we adopt the convention that f isconvex if and only if

f (tx + (1− t) y) < ta + (1− t) b

for every x, y ∈ RN , with f (x) < a and f (y) < b, and every t ∈ (0, 1) . But weconstantly avoid being in the situation where we have to refer to this definition.♦

We now give some important examples of convex functions.

Example 2.23 (i) Indicator function. The indicator function of a set E ⊂ RN

is defined by

χE (x) :=

0 if x ∈ E

+∞ if x /∈ E.

Convex functions 45

The function χE is convex if and only if the set E is convex.

(ii) Support function. For a given convex set E ⊂ RN , we define the supportfunction of E as

χ∗E (x∗) := sup

x∈E〈x; x∗〉 .

(iii) Gauge. For a given convex set E ⊂ RN , we define the gauge of E(see Section 2.3.7) as

ρE (x) := inf λ ≥ 0 : x ∈ λE .

(iv) Distance function. Given E ⊂ RN , we define

dE (x) := inf |x− e| : e ∈ E .

It is easily seen that if E is convex, then dE is convex . If E is closed, then

dE (x) = 0 ⇔ x ∈ E. ♦

We now recall some definitions and notations.

Definition 2.24 Let f : RN → R ∪ +∞ .

(i) f is said to be lower semicontinuous if

lim infxν→x

f (xν) ≥ f (x) .

(ii) The domain of f is defined as

dom f :=x ∈ RN : f (x) < +∞

.

(iii) The epigraph of f is defined as

epi f :=(x, α) ∈ RN × R : f (x) ≤ α

.

(iv) The level set of height α, α ∈ R, of f is defined as

levelα f :=x ∈ RN : f (x) ≤ α

.

Example 2.25 For the indicator function of a set E ⊂ RN , we have

domχE = E,

epiχE = E × [0, +∞) = E × R+ ,

levelα χE =

∅ if α < 0

E if α ≥ 0.♦

The proof of the following theorem is easy and we do not discuss the details.


Theorem 2.26 Let f : RN → R ∪ +∞ .

Part 1. The following three conditions are equivalent:

(i) f is lower semicontinuous;

(ii) epi f is closed;

(iii) levelα f is closed for every α ∈ R.

Part 2. The following two conditions are equivalent:

(i) f is convex;

(ii) epi f is convex.

Furthermore, if f is convex, then levelα is convex for every α ∈ R.

Part 3. Let fν : RN → R ∪ +∞ , ν ∈ I, be an arbitrary family of convex(respectively lower semicontinuous) functions. Then

f = supν∈I

fν

is a convex (respectively lower semicontinuous) function.

Remark 2.27 Note that in general the convexity of levelα f for every α ∈ Rdoes not imply the convexity of f, as the following example indicates. Let

f(x) =

0 if x ≤ 0

1 if x > 0

then

levelα f =

⎧⎪⎪⎨⎪⎪⎩

∅ if α < 0

(−∞, 0] if 0 ≤ α < 1

R if α ≥ 1

is convex for every α ∈ R, while f is not convex. In the context of optimization,functions whose level sets are convex are sometimes called quasiconvex ; wewill not be concerned with such functions and when we will use the notionof quasiconvexity in Part II it will have a different meaning. ♦

We close this very brief introduction by recalling Jensen inequality.

Theorem 2.28 (Jensen inequality) Let Ω ⊂ RN be a bounded open set, u ∈L1 (Ω) and f : R → R be convex. Then

f(1

measΩ

∫

Ω

u (x) dx) ≤ 1

measΩ

∫

Ω

f (u (x)) dx.

2.3.2 Continuity of convex functions

We now turn our attention to continuity of convex functions. Since the resultsof this section are available for separately convex functions, we start with thefollowing definition.

Convex functions 47

Definition 2.29 A function f : RN → R ∪ +∞ is said to be separatelyconvex, or convex in each variable, if, letting x = (x1, · · · , xN ) , the function

xi → f (x1, · · · , xi−1, xi, xi+1, · · · , xN ) is convex for every i = 1, · · · , N,

for every fixed (x1, · · · , xi−1, xi+1, · · · , xN ) ∈ RN−1.

Obviously any convex function is separately convex, but the converse is falseas seen in the next example.

Example 2.30 The function f : R2 → R defined by

f (x1, x2) = x1x2

is separately convex but not convex. ♦

The next theorem is usually stated for convex functions and not for sepa-rately convex functions, but in Part II and Part III we will need this strongerversion. The proof is however a straightforward adaptation of the classicalresults.

Theorem 2.31 Let f : RN → R ∪ +∞ be separately convex and f ≡ +∞.Then f is locally Lipschitz, and hence continuous, on int (dom f) .

Proof. We divide the proof into three steps. In the sequel, we define

|x|∞ := max |xi| : i = 1, · · · , N .

Step 1. We first prove that if x ∈ int (dom f) , then f is bounded above in aneighborhood of x.

There is no loss of generality, if we suppose that x = 0. Therefore, since0 ∈ int (dom f) , there exists ǫ > 0 such that

x = (x1, · · · , xN ) ∈ RN : |x|∞ ≤ ǫ

⊂ dom f. (2.22)

Letting

a := max f (ǫ1, ǫ2, .., ǫN) : ǫi = −ǫ, 0, ǫ, for every i = 1, · · · , N

we deduce from (2.22) that a < +∞. We now claim that

|x|∞ ≤ ǫ ⇒ f (x) ≤ a. (2.23)

In order to prove (2.23), observe that if 0 ≤ xN ≤ ǫ and ǫi = −ǫ, 0, ǫ, then theseparate convexity of f with respect to the last variable implies that

f (ǫ1, ǫ2, · · · , ǫN−1, xN )

≤ xN

ǫf (ǫ1, · · · , ǫN−1, ǫ) + (1− xN

ǫ)f (ǫ1, · · · , ǫN−1, 0)

≤ xN

ǫa + (1− xN

ǫ)a = a.


Using the above inequality and the separate convexity of f with respect to xN−1

we have, if 0 ≤ xN−1 ≤ ǫ, that

f (ǫ1, · · · , ǫN−2, xN−1, xN )

≤ xN−1

ǫf (ǫ1, · · · , ǫN−2, ǫ, xN ) + (1− xN−1

ǫ)f (ǫ1, · · · , ǫN−2, 0, xN ) ≤ a.

Iterating the process with respect to all the variables we have immediately (2.23)for 0 ≤ xi ≤ ǫ. A similar argument applies if some of the xi are negative.

The inequality (2.23) implies that if x ∈ int (dom f) , then f is boundedabove in a neighborhood of x, as claimed.

Step 2. We next show that if x ∈ int (dom f) , then f is continuous at x.There is no loss of generality if we assume that x = 0 and f (0) = 0. Since f isbounded above in a neighborhood of x = 0, there exist λ > 0 and a > 0 suchthat

|x|∞ ≤ λ ⇒ f (x) ≤ a. (2.24)

Fix ǫ > 0 and without loss of generality assume that ǫ ≤ aN2N (otherwisechoose a even larger). We now show that

|x|∞ ≤ ǫ

aN2Nλ ⇒ |f (x)| ≤ ǫ. (2.25)

We let

δ :=ǫ

aN2N≤ 1.

Using the separate convexity of f, we have

f (x) = f (x1, · · · , xN ) = f(δ(x1

δ, x2, · · · , xN ) + (1− δ) (0, x2, · · · , xN ))

≤ δf(x1

δ, x2, · · · , xN ) + (1− δ) f (0, x2, · · · , xN ) .

Repeating the process with the second variable we have

f (x) ≤ δf(x1

δ, x2, · · · , xN ) + (1− δ) δf(0,

x2

δ, · · · , xN )

+ (1− δ)2 f (0, 0, x3, · · · , xN ) .

Iterating the process, we obtain

f (x) ≤ δ

N∑

i=1

(1− δ)i−1

f(0, · · · , 0,xi

δ, xi+1 · · · , xN ) + (1− δ)

Nf (0, · · · , 0) .

If we now assume that

|x|∞ ≤ δλ =ǫλ

aN2N≤ λ,

Convex functions 49

we deduce immediately from the fact that f (0) = 0 and from (2.24) that

f (x) ≤ δa

N∑

i=1

(1− δ)i−1 ≤ δaN ≤ ǫ

which is one of the inequalities in (2.25).

In order to obtain (2.25), we still need to show that f (x) ≥ −ǫ and this isdone similarly. We have

0 = f (0, · · · , 0)

= f(1

1 + δ(0, · · · , 0, xN) +

δ

1 + δ(0, · · · , 0,

−xN

δ))

≤ 1

1 + δ[f (0, · · · , 0, xN) + δf(0, · · · , 0,

−xN

δ)].

Proceeding similarly with the xN−1 variable, we get

f (0, · · · , 0, xN ) = f(1

1 + δ(0, · · · , 0, xN−1, xN ) +

δ

1 + δ(0, · · · , 0,−xN−1

δ, xN ))

≤ 1

1 + δf(0, · · · , 0, xN−1, xN ) +

δ

1 + δf(0, · · · , 0,−xN−1

δ, xN )

and thus combining the two estimates, we obtain

0 ≤ 1

(1 + δ)2 f(0, · · · , 0, xN−1, xN ) +

δ

(1 + δ)2 f(0, · · · , 0,−xN−1

δ, xN )

+δ

1 + δf(0, · · · , 0,

−xN

δ).

Iterating the process as above we deduce that

0 ≤ 1

(1 + δ)N

f (x1, · · · , xN ) +N∑

i=1

δ

(1 + δ)N−i+1

f(0, , · · · , 0,−xi

δ, xi+1, · · · , xN )

and hence, if

|x|∞ ≤ δλ =ǫλ

aN2N≤ λ,

we find, from (2.24), that

f (x1, · · · , xN ) ≥ −δ

N∑

i=1

(1 + δ)i−1

f(0, , · · · , 0,−xi

δ, xi+1, · · · , xN )

≥ −δa

N∑

i=1

(1 + δ)i−1 ≥ −δaN2N = −ǫ.

From the above inequality, we thus infer that

|x|∞ ≤ ǫ

aN2Nλ ⇒ f (x) ≥ −ǫ.


Thus (2.25) holds and the continuity of f at x = 0 follows.

Step 3. It therefore remains to show that f is locally Lipschitz in the interiorof the domain of f. Let x ∈ int (dom f) . By continuity of f at x, there existα, β > 0 such that

|y − x|∞ ≤ 2β ⇒ |f (y)| ≤ α < +∞. (2.26)

Let z and z1 be such that

|z1 − z|∞ , |z1 − x|∞ ≤ β. (2.27)

Observe that (2.27) implies that |z − x|∞ ≤ 2β. Therefore (2.26) and (2.27)lead to

|z1 − z|∞ , |z1 − x|∞ ≤ β ⇒ f (z)− f (z1) ≤ 2α. (2.28)

Let ǫ > 0 be chosen later. Combining (2.25) and (2.28), we have immediately

|z1 − z|∞ , |z1 − x|∞ ≤ βǫ

2αN2N⇒ |f (z)− f (z1)| ≤ ǫ. (2.29)

Choosing

ǫ :=2αN2N

β|z1 − z|∞

we have from (2.27) and (2.29) that

|z1 − z|∞ , |z1 − x|∞ ≤ β ⇒ |f (z)− f (z1)| ≤2αN2N

β|z1 − z|∞ . (2.30)

Now let z2 be such that |z2 − x|∞ ≤ β. Let

u1, u2, · · · , uM ∈ [z1, z2]

(the segment in RN with endpoints z1 and z2) be such that

u1 = z1 , u2 , · · · , uM = z2 and |um − um+1|∞ ≤ β, m = 1, · · · , M − 1.

Note that, since |z1 − x|∞ , |z2 − x|∞ ≤ β, then

|um − x|∞ ≤ β, m = 1, · · · , M.

Using (2.30), we immediately get

|um − um+1|∞ ≤ β ⇒ |f (um)− f (um+1)| ≤2αN2N

β|um − um+1|∞ .

Summing the above inequalities, we obtain

|z1 − x|∞ , |z2 − x|∞ ≤ β ⇒ |f (z1)− f (z2)| ≤2αN2N

β|z1 − z2|∞

Convex functions 51

and whence the result.

We terminate this subsection with an elementary proposition concerningconvex functions (see Fusco [292], Marcellini [423], Morrey [455]) that shouldbe related to Theorem 2.31.

Proposition 2.32 Let f : RN → R be separately convex such that

|f (x)| ≤ α (1 + |x|p)

for every x ∈ RN , where α ≥ 0, p ≥ 1. Then there exists β ≥ 0 such that

|f (x)− f (y)| ≤ β(1 + |x|p−1+ |y|p−1

) |x− y|

for every x, y ∈ RN .

Proof. We divide the proof into three steps.

Step 1. We first prove that if g : R → R is convex, then, for every λ > μ > 0and every t ∈ R,

g (t± μ)− g (t)

μ≤ g (t± λ)− g (t)

λ. (2.31)

This follows at once from the convexity of g. Indeed write

g (t± μ) = g(μ

λ(t± λ) + (1− μ

λ)t)

≤ μ

λg (t± λ) + (1 − μ

λ)g (t)

and (2.31) follows.

Step 2. Fix x1 = (x2, · · · , xN ) ∈ RN−1 and define for t ∈ R

g (t) := f (t, x1)

and let us prove that there exists β1 ≥ 0 such that

|g (x1)− g (y1)| ≤ β1(1 + |x|p−1+ |y|p−1

) |x1 − y1| , (2.32)

Assume, without loss of generality that x1 < y1. Choose then in (2.31)

λ := 1 + |x|+ |y| and μ := y1 − x1

to get

g (y1)− g (x1) = g (x1 + (y1 − x1))− g (x1)

≤ (y1 − x1)g (x1 + 1 + |x|+ |y|)− g (x1)

1 + |x|+ |y|and

g (x1)− g (y1) = g (y1 − (y1 − x1))− g (y1)

≤ (y1 − x1)g (y1 − (1 + |x|+ |y|))− g (y1)

1 + |x|+ |y| .


Using the hypothesis on f, we have indeed obtained (2.32).

Step 3. Writing

f (x)− f (y)

= [f (x1, x2, · · · , xN )− f (y1, x2, · · · , xN )]

+N−2∑

i=1

[f (y1, · · · , yi, xi+1, xi+2, · · · , xN )− f (y1, · · · , yi, yi+1, xi+2, · · · , xN )]

+ [f (y1, · · · , yN−1, xN )− f (y1, · · · , yN−1, yN)]

and applying the proper adaptation of (2.32) to each term of the above identity,we have indeed obtained the proposition.

2.3.3 Convex envelope

We start with a definition to which we already alluded earlier.

Definition 2.33 Let f : RN → R ∪ +∞ . Then the convex envelope of f,denoted by Cf, is the largest convex function below f.

Remark 2.34 (i) The definition can be equivalently written, for every x ∈ RN ,as

Cf (x) = sup g (x) : g ≤ f and g convex .

(ii) It might be (see Example 2.45 below) that Cf takes the value −∞, eventhough f > −∞. An easy way to avoid this situation is to assume that thereexist a ∈ RN and α ∈ R such that

f (x) ≥ 〈a; x〉+ α for every x ∈ RN . ♦

An immediate consequence of Caratheodory theorem is the following char-acterization of the envelope.

Theorem 2.35 Let f : RN → R ∪ +∞ and, for every x ∈ RN ,

Cf (x) = sup g (x) : g ≤ f and g convex .

Assume that Cf > −∞. Then

Cf (x) = inf∑N+1i=1 αif (xi) :

∑N+1i=1 αixi = x, αi ≥ 0 with

∑N+1i=1 αi = 1.

Proof. We first define

C′f (x) := inf∑Ii=1 αif (xi) : I ∈ N,

∑Ii=1 αixi = x, α ∈ ΛI (2.33)

whereΛs = λ = (λ1, · · · , λs) : λi ≥ 0 and

∑si=1 λi = 1.

Convex functions 53

Note immediately that since f ≥ Cf and Cf is convex, then C′f ≥ Cf > −∞.

Step 1. Let us show that C′f is convex. So let y, z ∈ RN and t ∈ [0, 1] , wehave to prove that

C ′f (tz + (1− t) y) ≤ tC′f (z) + (1− t)C′f (y) . (2.34)

Fix ǫ > 0 and find I, J ≥ N + 1, λ ∈ ΛI , μ ∈ ΛJ and yi, zi ∈ RN such that

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ǫ + C′f (z) ≥I∑

i=1

λif (zi) , withI∑

i=1

λizi = z

ǫ + C′f (y) ≥J∑

i=1

μif (yi) , withI∑

i=1

μiyi = y.

Writing

αi := tλi, xi := zi, i = 1, · · · , I

αI+i := (1− t)μi, xI+i := yi, i = 1, · · · , J,

we find that α ∈ ΛI+J and we get that

ǫ + tC′f (z) + (1− t)C′f (y) ≥I+J∑

i=1

αif (xi) with

I+J∑

i=1

αixi = tz + (1− t) y.

Using (2.33) in the right hand side of the above inequality and the arbitrarinessof ǫ, we have indeed obtained (2.34). We have therefore shown that C′f isconvex.

Step 2. We next prove that C′f = Cf. We first observe that (see thebeginning of the proof)

Cf ≤ C′f ≤ f.

Since C′f is convex, we thus deduce that C′f = Cf.

Step 3. Finally we show that we can take I = N + 1. We first prove that wecan restrict attention to I ≤ N + 2. Consider the set

F := (xi , f (xi))Ii=1 ⊂ RN+1

and note that for any α ∈ ΛI we have

(∑I

i=1 αixi ,∑I

i=1 αif (xi)) ∈ coF.

Appealing to Theorem 2.13 we can find β ∈ ΛN+2 and yi ∈ x1, · · · , xI ,i = 1, · · · , N + 2, so that

(∑I

i=1 αixi ,∑I

i=1 αif (xi)) = (∑N+2

i=1 βiyi ,∑N+2

i=1 βif (yi)).

We can therefore choose I ≤ N + 2.


We finally further reduce I to N + 1. We show that we can find γ ∈ ΛN+2 ,with at least one of the γi = 0 (this implies in particular that, removing thisγi , we can assume, in fact, γ ∈ ΛN+1), and yi , i = 1, · · · , N + 2, so that

N+2∑

i=1

γif (yi) ≤N+2∑

i=1

βif (yi) . (2.35)

Since we have the convention that 0 . (+∞) = 0, the claim, I ≤ N + 1, willfollow from (2.35). Assume that β ∈ ΛN+2 is so that βi > 0, i = 1, · · · , N + 2,otherwise nothing is to be proved. Let us first denote by

x :=

N+2∑

i=1

βiyi

which implies that x ∈ co y1, · · · , yN+2 ⊂ RN . Applying Caratheodorytheorem, once more, we find

βi ≥ 0, for every 1 ≤ i ≤ N + 2 and

N+2∑

i=1

βi = 1

and at least one of the βi = 0 such that

N+2∑

i=1

βiyi = x.

We may assume without loss of generality that

N+2∑

i=1

βif (yi) >

N+2∑

i=1

βif (yi) , (2.36)

otherwise choosing γi = βi we would immediately obtain (2.35). We then let

J :=

i ∈ 1, · · · , N + 2 : βi − βi < 0

.

Observe that J = ∅, since otherwise βi ≥ βi ≥ 0 for every 1 ≤ i ≤ N + 2 andsince at least one of the βi = 0, we would have a contradiction with

∑N+2i=1 βi =∑N+2

i=1 βi = 1 and the fact that βi > 0 for every i. We then define

λ := mini∈Jβi/( βi − βi )

and we have clearly λ > 0. Finally let

γi := βi + λ(βi − βi

), 1 ≤ i ≤ N + 2.

We therefore have

γi ≥ 0 ,N+2∑

i=1

γi = 1, at least one of the γi = 0

Convex functions 55

and from (2.36)

N+2∑

i=1

γif (yi) =

N+2∑

i=1

βif (yi) + λ

N+2∑

i=1

(βi − βi

)f (yi)

≤N+2∑

i=1

βif (yi) .

We have therefore obtained (2.35) and this concludes Step 3 and thus thetheorem.

We will see several examples of convex envelopes in Section 2.3.5, but beforethat we want to make more precise the connection between the convex hull ofa set and the convex envelope of its indicator function.

Proposition 2.36 Let E ⊂ RN and χE be the indicator function of E, namely

χE (x) =

0 if x ∈ E

+∞ if x /∈ E.

Then

CχE = χcoE .

Moreover, if

FE∞ :=

f : RN → R ∪ +∞ : f |E ≤ 0

,

FE :=f : RN → R : f |E ≤ 0

,

then

coE =x ∈ RN : f (x) ≤ 0, for every convex f ∈ FE

∞

,

coE =x ∈ RN : f (x) ≤ 0, for every convex f ∈ FE

.

Remark 2.37 Anticipating the results of Section 2.3.5, we also have

χ∗∗E = χco E . ♦

Proof. (i) Let us start by showing CχE = χco E . Since χco E ≤ χE and χco E

is convex, we deduce immediately that

χcoE ≤ CχE .

In order to show the reverse inequality, it is sufficient to show that

χco E (x) = 0 ⇒ CχE (x) = 0.


This follows from Theorem 2.13. Indeed if χco E (x) = 0, this means x ∈ coEand hence from Caratheodory theorem we can find xi ∈ E, λ ∈ ΛN+1 so that

x =

N+1∑

i=1

λixi .

We therefore obtainN+1∑

i=1

λiχE (xi) = 0

and hence, from Theorem 2.35, CχE (x) = 0, as wished.

(ii) Since we obviously have

f ∈ FE∞ ⇔ f ≤ χE ,

we deduce that any convex f ∈ FE∞ must be such that

f ≤ CχE = χco E

and hence

coE ⊂x ∈ RN : f (x) ≤ 0, for every convex f ∈ FE

∞

.

The reverse inclusion follows from the fact that χco E ∈ FE∞ and is convex.

(iii) The set

X :=x ∈ RN : f (x) ≤ 0, for every convex f ∈ FE

is clearly convex and closed, since any convex function in FE is continuous;furthermore E ⊂ X. We can therefore infer that

coE ⊂ X.

Since the distance function (see Example 2.23) dco E is convex and belongs toFE we deduce the reverse inclusion coE ⊃ X.

2.3.4 Lower semicontinuous envelope

An important concept that we will encounter again in all the processes of relax-ation is the following.

Definition 2.38 Let f : RN → R∪+∞ . The lower semicontinuous envelopeof f, denoted by f, is the largest lower semicontinuous function that is below f.

Remark 2.39 (i) In view of Theorem 2.26, we can write, for every x ∈ RN ,

f (x) = sup g (x) : g ≤ f and g lower semicontinuous .

Convex functions 57

(ii) Another way of rewriting the function f is

f (x) = infxν

lim infxν→x

f (xν) . ♦

We have the following easy result that we state without proof.

Proposition 2.40 Let E ⊂ RN and χE be its indicator function. Then

χE = χE .

2.3.5 Legendre transform and duality

We now introduce the notions of duality and dual maps, following Fenchel [276]and Moreau [452]. These notions play a central role in convex analysis.

Definition 2.41 Let 〈·; ·〉 denote the scalar product in RN and

f : RN → R ∪ +∞

with f ≡ +∞.

(i) The function f∗ : RN → R ∪ +∞ defined by

f∗ (x∗) := supx∈RN

〈x; x∗〉 − f (x)

is called the conjugate, or dual, function of f.

(ii) The function f∗∗ : RN → R ∪ ±∞ defined by

f∗∗ (x) := supx∗∈RN

〈x; x∗〉 − f∗ (x∗)

is called the biconjugate, or bidual, function of f.

Remark 2.42 (i) The notion of duality is closely related to the concept ofLegendre transform and we will, by abuse of language, often use both notionsas equivalent.

(ii) In general, the bidual f∗∗, as well as Cf, may take the value −∞ eventhough f is never −∞.

(iii) If f ≡ +∞, then f∗ ≡ −∞ and f∗∗ ≡ +∞. ♦

Before giving some examples, we state some important properties of thesefunctions, as established by Fenchel [276] and Moreau [452].

Theorem 2.43 Let f : RN → R∪ +∞ . Then the following statements hold.

(i) f∗ is convex and lower semicontinuous.

(ii) If f is convex and lower semicontinuous, then f∗ ≡ +∞.


(iii) The following inequalities hold:

f∗∗ ≤ Cf ≤ f.

Moreover, if Cf is lower semicontinuous and Cf > −∞, then

f∗∗ = Cf,

so that, in particular, if f is convex and lower semicontinuous, then

f∗∗ = Cf = f.

(iv) The identity f∗∗∗ = f∗ is always valid.

Remark 2.44 We therefore see that f∗∗ is at the same time the convex andthe lower semicontinuous envelope of the function f, while Cf (respectively f)is only the convex envelope (respectively the lower semicontinuous envelope)of f. ♦

We now discuss some examples.

Example 2.45 (i) We recall that the indicator function of a set E ⊂ RN isgiven by

χE =

0 if x ∈ E

+∞ if x /∈ E.

We then haveχ∗

E (x∗) = supx∈E

〈x; x∗〉 ,

which is nothing but the support function of E. Again applying the duality, weobtain

χ∗∗E (x) = χco E (x) and CχE (x) = χco E (x) ,

where coE (respectively coE) denotes the convex hull (respectively the closedconvex hull) of E. In particular, if E = (0, 1) ⊂ R, we get

χ(0,1) = Cχ(0,1) and χ[0,1] = χ∗∗(0,1) .

The second identity, CχE = χco E , has been shown in Proposition 2.36. Theidentity χ∗∗

E = χco E follows from the following observations.- χco E is convex and lower semicontinuous and hence χco E = χ∗∗

co E. We

therefore deduce, recalling the trivial fact χco E ≤ χE , that

χco E = χ∗∗co E

≤ χ∗∗E .

- Appealing to Proposition 2.40 we see that χco E = χco E and, since χ∗∗E is

lower semicontinuous and χ∗∗E ≤ CχE = χcoE , we deduce that

χ∗∗E ≤ χco E

as wished.

Convex functions 59

(ii) The difference between Cf and f∗∗ is even more striking if we consider

f (x) =

(x2 − 1

)−1if |x| < 1

+∞ otherwise.

Then f∗ ≡ +∞ and hence f∗∗ ≡ −∞, while

Cf (x) =

−∞ if |x| < 1

+∞ otherwise.

(iii) Define for x ∈ RN and 1 ≤ p ≤ ∞ the Holder norm

|x|p :=

⎧⎨⎩

[∑Ni=1 |xi|p

]1/p

if 1 ≤ p < ∞

max1≤i≤N |xi| if p = ∞.

For 1 < p <∞, let

f (x) :=1

p|x|pp .

Then, if p′ = p/ (p− 1) ,

f∗ (x∗) =1

p′|x∗|p

′

p′ .

(iv) Let A ∈ Rn2

(the set of n×n matrices identified with Rn2

) and f (A) =detA. Then

f∗ (A∗) ≡ +∞ and f∗∗ (A) = Cf (A) ≡ −∞. ♦

We now turn to the proof of Theorem 2.43.

Proof. (i) Since

x∗ → 〈x; x∗〉 − f (x)

is convex (in fact affine) and lower semicontinuous (in fact continuous) then f∗

is convex and lower semicontinuous.

(ii) Note first that if f ≡ +∞, then f∗ ≡ −∞ and the result is proved. Sowe may assume that there exists x0 ∈ dom f. We next let a0 < f (x0) and weapply Theorem 2.10 (iii) to

A = epi f and B = (x0, a0) .

We then obtain that there exists a hyperplane over RN × R defined by

〈(x, a) ; (x∗; a∗)〉 = 〈x; x∗〉+ aa∗ = α

which separates strictly A and B, i.e.

〈x0; x∗〉+ a0a

∗ < α < 〈x; x∗〉+ f (x) a∗, for every x ∈ RN . (2.37)


Taking x = x0 in (2.37) we immediately get

〈x0; x∗〉+ a0a

∗ < α < 〈x0; x∗〉+ f (x0) a∗

and hence a∗ > 0. We therefore deduce immediately from (2.37) that

〈 x;− 1

a∗x∗ 〉 − f (x) < − α

a∗ (2.38)

and thus taking the supremum in (2.38) we obtain the result, i.e. f∗ ≡ +∞.

(iii) We proceed in three steps.

Step 1. Observe first that f∗∗ is convex and lower semicontinuous and that,by definition, f (x) ≥ 〈x; x∗〉 − f∗ (x∗) , hence f∗∗ ≤ f . The first inequality,f∗∗ ≤ Cf ≤ f, follows then immediately.

Step 2. There is no loss of generality if we assume Cf = f. We next reducethe problem to the case where f ≥ 0. We may assume without loss of generalitythat f ≡ +∞. Choosing x∗ ∈ dom f∗, which is non-empty as seen in (ii), anddefining

g (x) := f (x) − 〈x; x∗〉+ f∗ (x∗)

we obtain that g ≥ 0, convex, lower semicontinuous and g ≡ +∞. Observe alsothat

g∗∗ (x) = f∗∗ (x)− 〈x; x∗〉+ f∗ (x∗) .

Therefore the result, f = f∗∗, will follow from the corresponding result for g.

Step 3. We may then assume that f ≥ 0 (and thus f∗∗ ≥ 0), convex, lowersemicontinuous and f ≡ +∞. In view of Step 1, we only need to show thatf∗∗ ≥ f. We proceed by contradiction and assume that there exists x0 ∈ RN

such that

0 ≤ f∗∗ (x0) < f (x0) . (2.39)

Applying Theorem 2.10 (iii) to

A = epi f and B = (x0, f∗∗ (x0))

we have that there exists a hyperplane 〈x; x∗〉+aa∗ = α which separates strictlyA and B, i.e.,

〈x; x∗〉+ aa∗ > α for every (x, a) ∈ epi f (2.40)

〈x0; x∗〉+ f∗∗ (x0) a∗ < α. (2.41)

Since in (2.40) x ∈ dom f, letting a → +∞ immediately yields a∗ ≥ 0. We thenlet ǫ > 0 and use the fact that f ≥ 0 and (2.40) to get

〈x; x∗〉+ f (x) (a∗ + ǫ) > α for every x ∈ dom f

Convex functions 61

and hence

〈 x;−x∗

a∗ + ǫ〉 − f (x) <

−α

a∗ + ǫfor every x ∈ dom f.

The last inequality implies that

f∗(− x∗

a∗ + ǫ) ≤ − α

a∗ + ǫ.

Using the definition of f∗∗ we therefore have

f∗∗ (x0) ≥ 〈 x0;−x∗

a∗ + ǫ〉 − f∗(− x∗

a∗ + ǫ) ≥ 〈 x0;

−x∗

a∗ + ǫ〉+ α

a∗ + ǫ.

Thus〈x0; x

∗〉+ f∗∗ (x0) (a∗ + ǫ) ≥ α.

Using the arbitrariness of ǫ and (2.41) we have a contradiction and this termi-nates Step 3.

(iv) We now want to show that f∗∗∗ = f∗. Since we always have f∗∗ ≤ f,we deduce that f∗∗∗ ≥ f∗. Furthermore from the definition of duality we havefor every x ∈ RN , x∗ ∈ RN ,

〈x; x∗〉 − f∗∗ (x) ≤ f∗ (x∗)

and hence, taking the supremum in the left hand side, we obtain f∗∗∗ ≤ f∗.

2.3.6 Subgradients and differentiabilityof convex functions

In this section we will always assume that

f : RN → R ∪ +∞

is convex and f ≡ +∞.

Definition 2.46 We say that x∗ ∈ RN is a subgradient of f at x if

f (z) ≥ f (x) + 〈x∗; z − x〉 , ∀ z ∈ RN .

The set of all subgradients of f at x is called the subdifferential of f at x andis denoted by ∂f (x) .

We now give an elementary example and show a simple characterization ofthe subdifferential.

Example 2.47 Let f : R → R be defined by

f (x) = |x| .


Then

∂f (x) =

⎧⎪⎪⎨⎪⎪⎩

1 if x > 0

[−1, 1] if x = 0

−1 if x < 0.

♦

Theorem 2.48 Let f : RN → R∪ +∞ be convex, lower semicontinuous andf ≡ +∞. The following conditions are then equivalent:

(i) x∗ ∈ ∂f (x) ;

(ii) 〈x∗; z〉 − f (z) achieves its maximum at z = x;

(iii) f∗ (x∗) + f (x) = 〈x∗; x〉 ;(iv) x ∈ ∂f∗ (x∗) .

Proof. (i) ⇔ (ii) We have that x∗ ∈ ∂f (x) is equivalent to

〈x∗; x〉 − f (x) ≥ 〈x∗; z〉 − f (z) , ∀ z ∈ RN (2.42)

and therefore is equivalent to the fact that 〈x∗; z〉− f (z) achieves its maximumat z = x.

(ii) ⇔ (iii) Using the definition of the conjugate function

f∗ (x∗) = supx∈RN

〈x∗; x〉 − f (x) ,

combined with (2.42) we have the equivalence.

(iii) ⇔ (iv) Appealing to Theorem 2.43 we find that (iii) is equivalent to

f∗ (x∗) + f∗∗ (x) = 〈x∗; x〉 .

Using then the equivalence (i) ⇔ (iii) applied to f∗∗ we get the result.

The notion of subdifferential is intimately related to the notion of directionalderivative, notion that we now define.

Definition 2.49 Let f : RN → R ∪ +∞ and let x ∈ dom f.

(i) The one sided directional derivative of f at x in the direction y is thelimit, if it exists,

f ′ (x, y) := limλ→0+

f (x + λy)− f (x)

λ.

(ii) The directional derivative of f at x in the direction y is f ′ (x, y) , providedboth f ′ (x, y) and f ′ (x,−y) exist and

f ′ (x,−y) = −f ′ (x, y) .

We now have the following theorem.

Convex functions 63

Theorem 2.50 Let f : RN → R∪ +∞ be convex and lower semicontinuous,f ≡ +∞ and x ∈ int (dom f) . The following conclusions then hold.

(i) f ′ (x, y) exists and

f ′ (x, y) ≡ infλ>0

f (x + λy)− f (x)

λ.

Moreover the function y → f ′ (x, y) is convex and

f ′ (x, ty) = tf ′ (x, y) for every t ≥ 0

f ′ (x, y) ≥ −f ′ (x,−y) for every y ∈ RN .

(ii) x∗ ∈ ∂f (x) if and only if

f ′ (x, y) ≥ 〈x∗; y〉 for every y ∈ RN .

(iii) ∂f (x) is non-empty, convex and compact. Moreover, f ′ (x, y) is finitefor every y ∈ RN .

(iv) The function y → f ′ (x, y) is lower semicontinuous, convex and

f ′ (x, y) = sup 〈x∗; y〉 : x∗ ∈ ∂f (x) .

(v) If f is differentiable at x, then

∂f (x) = ∇f (x)

andf (x) + f∗ (∇f (x)) = 〈x;∇f (x)〉 .

(vi) If f has a unique subgradient at x, then f is differentiable at x.

(vii) The set D where f is differentiable is dense in int (dom f) and itscomplement in int (dom f) has zero measure. Furthermore, the usual gradientmap ∇f : x → ∇f (x) is continuous on D.

Proof. (i) Let us first show that since f is convex, then the function

λ → f (x + λy)− f (x)

λ

is an increasing function of λ > 0. Observe that if λ ≥ μ > 0, we have

f (x + μy) = f(μ

λ(x + λy) +

λ− μ

λx)

≤ μ

λf (x + λy) +

λ− μ

λf (x)

which implies that

f (x + μy)− f (x)

μ≤ f (x + λy)− f (x)

λ


as claimed. It then follows that f ′ (x, y) exists and

f ′ (x, y) ≡ infλ>0

f (x + λy)− f (x)

λ.

The other properties follow in an immediate way from the convexity of f andfrom the above formula.

(ii) From the definition of ∂f (x) and from (i), we get, for every y ∈ RN andevery λ > 0,

x∗ ∈ ∂f (x) ⇔ f (x + λy)− f (x)

λ≥ 〈x∗; y〉 ⇔ f ′ (x, y) ≥ 〈x∗; y〉 ,

as wished.

(iii) Since f is lower semicontinuous and convex, it follows from Theorem2.26 that epi f ⊂ RN+1 is closed and convex. Since (x, f (x)) ∈ ∂ (epi f) , we maytherefore use Theorem 2.10 (ii) to get that there exist a∗ = (a∗

1, a∗2) ∈ RN × R,

a∗ = 0, and α ∈ R so that, for every (y, a) ∈ epi f,

〈x; a∗1〉+ f (x) a∗

2 = α ≤ 〈y; a∗1〉+ aa∗

2 (2.43)

Note next that a∗2 ≥ 0, since (x, f (x) + 1) ∈ epi f. Moreover a∗

2 = 0, otherwise〈y − x; a∗

1〉 ≥ 0 for every y in the neighborhood of x (since x ∈ int (dom f)) andthis would imply that a∗

1 = 0, as well as a∗2 = 0, which is absurd. Therefore

a∗2 > 0 and we deduce from (2.43) and from the fact that (y, f (y)) ∈ epi f,

〈y ; a∗1/a∗

2 〉+ f (y) ≥ 〈x ; a∗1/a∗

2 〉+ f (x) . (2.44)

Letting x∗ = −a∗1/a∗

2 in (2.44) we have that x∗ ∈ ∂f (x) and hence ∂f (x) = ∅.We next prove that ∂f (x) is convex and compact; since it is clearly closed

and convex, we only need to show that it is bounded. From Theorem 2.31 andfrom the fact that x ∈ int (dom f) , we deduce that there exists L = L (x) > 0so that for every y in a neighborhood of x

|f (y)− f (x)| ≤ L |y − x| . (2.45)

So let x∗ ∈ ∂f (x) and use the definition of the subgradient to write

f (y) ≥ f (x) + 〈x∗; y − x〉

and hence1

|y − x| 〈x∗; y − x〉 ≤ f (y)− f (x)

|y − x| ≤ L.

We therefore have

|x∗| = sup|z|=1

〈x∗; z〉 ≤ L

and hence ∂f (x) is bounded.

Convex functions 65

It remains to show that f ′ (x, y) is finite for every y ∈ RN . From (2.45) wehave that, for every y ∈ RN and every λ > 0 sufficiently small,

−L |y| ≤ f (x + λy)− f (x)

λ≤ L |y| .

Invoking (i) we find the result, namely that for every y ∈ RN the inequality

|f ′ (x, y)| ≤ L |y|

is valid.

(iv) We refer for a proof of (iv) to Theorem 23.4 in Rockafellar [514].

(v) Assume that f is differentiable at x, then

f ′ (x, y) = 〈∇f (x) ; y〉 .

Applying (ii) we get

〈∇f (x) ; y〉 ≥ 〈x∗; y〉 , for every y ∈ RN

and hence ∂f (x) = ∇f (x) . Moreover the identity

f (x) + f∗ (∇f (x)) = 〈x;∇f (x)〉

then follows from (iii) of Theorem 2.48.

(vi) For the converse part of (v) we refer to Theorem 25.1 in Rockafellar[514].

(vii) We will not prove this last fact and we refer to Theorem 25.5 inRockafellar [514].

We have as an immediate corollary the following.

Corollary 2.51 Let f : RN → R be convex. Then, for every x ∈ RN , thereexists x∗ ∈ ∂f (x) and thus

f (z) ≥ f (x) + 〈x∗; z − x〉 , ∀z ∈ RN .

Moreover, the following identity holds for every x ∈ RN

f (x) = sup g (x) : g ≤ f and g affine .

Proof. (i) Since f takes only finite values, then int (dom f) = RN and f iscontinuous. Thus Theorem 2.50 applies and we find x∗ ∈ ∂f (x) . The inequalityfollows then from Theorem 2.48.

(ii) Since for every x ∈ RN there exists x∗ ∈ ∂f (x) , we obtain from Theorem2.48 that

supz∈RN

〈x∗; z〉 − f (z) = 〈x∗; x〉 − f (x) = f∗ (x∗) .


We thus have, for every z ∈ RN ,

〈x∗; z〉 − f∗ (x∗) ≤ f (z)

〈x∗; x〉 − f∗ (x∗) = f (x)

which completes the proof.

We now give some classical criteria equivalent to the convexity.

Theorem 2.52 Let f : RN → R, f ∈ C1(RN)

and 〈.; .〉 denote the scalarproduct in RN .

Part 1. The following conditions are then equivalent:

(i) f is convex;

(ii) for every x, y ∈ RN ,

f (y) ≥ f (x) + 〈y − x;∇f (x)〉 ;

(iii) for every x, y ∈ RN ,

〈y − x;∇f (y)−∇f (x)〉 ≥ 0.

Part 2. If f ∈ C2(RN), then f is convex if and only if its Hessian, ∇2f, is

positive semi definite.

Proof. Part 1. (i) ⇒ (ii). Let λ > 0, we have from the convexity of f that

1

λ[f (x + λ (y − x))− f (x)] ≤ f (y)− f (x) .

Letting λ → 0, we have immediately (ii).

(ii) ⇒ (i). We have from the inequality (ii) that, for λ ∈ [0, 1] ,

f (x) ≥ f (λx + (1− λ) y) + 〈x− (λx + (1− λ) y) ;∇f (λx + (1− λ) y)〉f (y) ≥ f (λx + (1− λ) y) + 〈y − (λx + (1− λ) y) ;∇f (λx + (1− λ) y)〉 .

Multiplying the first equation by λ and the second by (1− λ) and addingthem, yields the convexity of f.

(ii) ⇒ (iii). Using the inequality (ii) we have

f (y) ≥ f (x) + 〈y − x;∇f (x)〉f (x) ≥ f (y) + 〈x− y;∇f (y)〉 .

Combining these two inequalities we have

〈y − x;∇f (y)〉 ≥ f (y)− f (x) ≥ 〈y − x;∇f (x)〉

and thus the result.

Convex functions 67

(iii) ⇒ (ii). Let λ ∈ [0, 1] and consider

φ (λ) := f (x + λ (y − x)) .

Observe that

φ′ (λ)− φ′ (0) = 〈y − x;∇f (x + λ (y − x))−∇f (x)〉

=1

λ[〈x + λ (y − x)− x;∇f (x + λ (y − x))−∇f (x)〉] ≥ 0

where we have used (iii). Therefore integrating the inequality we obtain

φ (λ) ≥ φ (0) + λφ′ (0)

and thus letting z = x + λ (y − x) , we have

f (z) ≥ f (x) + 〈z − x;∇f (x)〉 .

Part 2. The monotonicity of the gradient of convex functions is ensured by(iii), which in turn is classically equivalent for C2 functions to the fact that theHessian, ∇2f, is positive semi-definite.

We end the section with the following corollary.

Corollary 2.53 Let Q ∈ RN×N be a symmetric positive semi definite matrix.Then f : RN → R defined by

f (x) := (〈Qx; x〉)1/2

is convex.

Proof. As a consequence of Theorem 13.3 (see also Theorem 2 in Section 4.7of Bellman [74]) we can find U ∈ SO (N) and

Λ = diag(λ21, · · · , λ2

N ) ∈ RN×N

so that

UQU t = Λ.

Observe then that the function

g (x) := (〈Λx; x〉)1/2 = (∑N

i=1 λ2i x

2i )

1/2

is convex. Since

f (x) = g(Ux),

we get the claim.


2.3.7 Gauges and their polars

We now recall some facts about gauges and their polars.

Definition 2.54 (i) Let E ⊂ RN be a convex set. Then the gauge (sometimesalso called Minkowski function) associated to E is defined as

ρ (x) := inf λ ≥ 0 : x ∈ λE .

(ii) The polar of a gauge ρ is defined as

ρ0 (x∗) := infλ∗ ≥ 0 : 〈x∗; x〉 ≤ λ∗ρ (x) , ∀x ∈ RN

.

The main properties of gauges and polars are summarized in the followingproposition.

Proposition 2.55 Let E ⊂ RN be a compact and convex set with 0 ∈ intE.The following properties then hold.

(i) The gauge ρ associated to E is finite everywhere, convex and satisfies

(a) ρ (x) > 0, ∀x = 0

(b) ρ (tx) = tρ (x) , ∀x ∈ RN , ∀t > 0.

(ii) One has E =x ∈ RN : ρ (x) ≤ 1

.

(iii) Another characterization of ρ0 is given by

ρ0 (x∗) = supx =0〈x

∗; x〉ρ (x)

.

(iv) The following identity holds: ρ00 = ρ.

(v) Let x = 0 and x∗ ∈ ∂ρ (x) . Then

ρ0 (x∗) = 1.

Remark 2.56 (i) Note that if 0 /∈ intE, then, in general, ρ is not finite every-where. Similarly if E is unbounded, then we may have ρ (x) = 0 for somex = 0.

(ii) The notions of a gauge and its polar are aimed at generalizing Cauchy-Schwarz inequality; namely, we have that

〈x∗; x〉 ≤ ρ (x) ρ0 (x∗) ,

in a similar manner as

ρ∗ (x∗) = supx∈RN

〈x∗; x〉 − ρ (x)

Convex functions 69

is the best possible inequality of the form

〈x∗; x〉 ≤ ρ (x) + ρ∗ (x∗) .

(iii) Note that in general we do not have ρ (x) = ρ (−x) .

(iv) The typical examples are the ones involving Holder norms; namely if1 ≤ p ≤ ∞, if 1/p + 1/p′ = 1 and

ρ (x) = |x|p :=

⎧⎨⎩

[∑Ni=1 |xi|p

]1/p

if 1 ≤ p < ∞

max1≤i≤N |xi| if p = ∞.

thenE =

x ∈ RN : |x|p ≤ 1

and ρ0 (x∗) = |x∗|p′ .

(v) If we compare the definition of the polar of a gauge with the usual dualfunction, defined as

ρ∗ (x∗) = supx∈RN

〈x∗; x〉 − ρ (x) ,

we get, under the hypotheses of the proposition,

ρ∗ (x∗) =

0 if ρ0 (x∗) ≤ 1

+∞ otherwise.♦

Proof. The proposition easily follows from the definitions and we do notdiscuss the details; we only, for the sake of illustration, prove (v).

From Theorem 2.50, we have that ∂ρ (x) is non empty and therefore

ρ (y) ≥ ρ (x) + 〈x∗; y − x〉 , for every y ∈ RN . (2.46)

We first choose y = 0 and get from (i) and (iii) that

〈x∗; x〉ρ (x)

≥ 1 ⇒ ρ0 (x∗) ≥ 1.

Moreover choosing y = 2x, we obtain that

ρ (x)− 〈x∗; x〉 ≥ 0

and thus returning to (2.46), we deduce that

ρ (y)− 〈x∗; y〉 ≥ ρ (x) − 〈x∗; x〉 ≥ 0, for every y ∈ RN .

This implies that

〈x∗; y〉ρ (y)

≤ 1, ∀ y ∈ RN − 0 (iii)⇒ ρ0 (x∗) ≤ 1

as claimed.


2.3.8 Choquet function

Extreme points of a convex set can be characterized through the Choquet func-tion and for more details we refer to Choquet [151] (see also Pianigiani [495]).

Theorem 2.57 Let E ⊂ RN be a non-empty compact convex set and Eext bethe set of its extreme points. Then there exists ϕE : RN → R ∪ +∞ (calledthe Choquet function) a convex function, strictly convex on E, so that

Eext = x ∈ E : ϕE (x) = 0

ϕE (x) ≤ 0 ⇔ x ∈ E.


f (x) :=

− |x|2 if x ∈ E

+∞ otherwise

and

ϕE (x) :=

Cf (x)− f (x) if x ∈ E

+∞ otherwise.

Observe that ϕE : RN → R∪+∞ is convex, since, letting χE be the indicatorfunction of the set E, we have

ϕE (x) = Cf (x) + |x|2 + χE(x), ∀x ∈ RN .

Furthermore ϕE is strictly convex on E, since Cf is convex and −f (x) = |x|2is strictly convex. Moreover we obtain that

ϕE (x) ≤ 0 if x ∈ E.

Indeed the inequality is clear since on E the function f is finite and, by definition,Cf is always not larger than f . We now show that

ϕE (x) = 0 ⇔ x ∈ Eext .

Note that if x ∈ E, then, applying Theorem 2.13, we have

ϕE (x) = |x|2 + infxi∈E

−∑N+1

i=1 λi |xi|2 : x =∑N+1

i=1 λixi , λ ∈ ΛN+1

whereΛs := λ = (λ1, · · · , λs) : λi ≥ 0 and

∑si=1 λi = 1 .

- Therefore if x ∈ Eext , we deduce, by definition, that in the infimum theonly admissible xi are xi = x (or if xi = x then the corresponding λi = 0, whichin any case leads to the same result) and hence we have ϕE (x) = 0.

Convex functions 71

- We now show the reverse implication, namely

ϕE (x) = 0 ⇒ x ∈ Eext .

From the above representation formula we obtain, since ϕE (x) = 0 and x ∈ E,that

|x|2 = supxi∈E

∑N+1i=1 λi |xi|2 : x =

∑N+1i=1 λixi , λ ∈ ΛN+1

.

Combining the above with the convexity of the function x→ |x|2 we get that

|x|2 ≥∑N+1i=1 λi |xi|2 ≥

∣∣∣∑N+1

i=1 λixi

∣∣∣2

= |x|2 ;

the strict convexity of x → |x|2 implies then that xi = x (or if xi = x then thecorresponding λi = 0, which is then an irrelevant index). Thus x ∈ Eext .

Chapter 3

Lower semicontinuityand existence theorems

3.1 Introduction

In the present chapter, we deal with the minimization problem

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where

- Ω ⊂ Rn is an open set;

- u : Ω → RN and hence ∇u ∈ RN×n;

- f : Ω × RN × RN×n → R, f = f (x, u, ξ) , is a Caratheodory function(see below for a definition).

Most of the results are concerned with the scalar case N = 1 or n = 1,although, in some cases, it is convenient to consider the general case N, n ≥ 1.

The main result of this chapter, investigated in Section 3.2, deals with the(sequential) weak lower semicontinuity of I, meaning that

lim infν→∞

I (uν) ≥ I (u)

for every sequence uν u in W 1,p. We show, roughly speaking, that thefunctional I is (sequentially) weakly lower semicontinuous if and only if ξ →f (x, u, ξ) is convex (see Theorem 3.15 and Corollary 3.24).

Since the presence of the lower order terms (x, u) induces many technicaldifficulties, we first prove both the necessary and the sufficient parts when thefunction f depends only on the term ξ, i.e. f = f (ξ) .

74 Lower semicontinuity and existence theorems

In Section 3.3.1, we obtain as a direct consequence of the results of thepreceding section that I is (sequentially) weakly continuous, meaning that

limν→∞

I (uν) = I (u)

for every sequence uν u in W 1,p, if and only if ξ → f (x, u, ξ) is affine.

In Section 3.4, we apply the above mentioned results to prove the existenceof minimizers for problem (P ). We then derive the necessary condition thatshould satisfy any minimizer namely the Euler-Lagrange equation.

For further references on this chapter we recommend Ambrosio-Fusco-Pallara[25], Buttazzo [112], Buttazzo-Giaquinta-Hildebrandt [117], Cesari [143],Dacorogna [179], Giaquinta [307], Giusti [316] and Morrey [455].

3.2 Weak lower semicontinuity

We now recall the following definition.

Definition 3.1 Let p ≥ 1 and Ω, u, f be as above. We say that I is (sequen-tially) weakly lower semicontinuous in W 1,p

(Ω; RN

)if for every sequence uν

u in W 1,p, thenlim infν→∞

I (uν) ≥ I (u) .

If p = ∞, we say that I is (sequentially) weak ∗ lower semicontinuous in

W 1,∞ (Ω; RN)

if the same inequality holds for every sequence uν∗ u in W 1,∞.

Remark 3.2 (i) In the remaining part of the book we usually drop the word”sequentially” when referring to lower semicontinuity.

(ii) If Ω is bounded, it is clear that the first notion implies the second one,

since any sequence uν∗ u in W 1,∞ is such that uν u in W 1,p for every

p ≥ 1. ♦

3.2.1 Preliminaries

We start with some definitions.

Definition 3.3 Let Ω ⊂ Rn be an open set and let f : Ω × RN → R ∪ +∞be a Borel measurable function. Then f is said to be a normal integrand ifξ → f (x, ξ) is lower semicontinuous for almost every x ∈ Ω.

Remark 3.4 The fact that we require f to be Borel measurable is just to ensurethat if u : Ω → RN is a measurable function, then the function g : Ω → R∪+∞defined by

g (x) := f (x, u (x))

is measurable. This property is ensured if, for example, f is (globally, meaningas a function on Ω × RN ) lower semicontinuous. However it is not, in general,true if we only assume that ξ → f (x, ξ) is lower semicontinuous and x → f (x, ξ)is measurable. ♦

Weak lower semicontinuity 75

The most important example of normal integrands is the following (we donot fully prove this fact and we refer to Proposition VIII.1.1 in Ekeland-Temam[264], however we prove the most important part of it in the proposition below).

Definition 3.5 Let Ω ⊂ Rn be an open set and let f : Ω× RN → R ∪ +∞ .Then f is said to be a Caratheodory function if

(i) ξ → f (x, ξ) is continuous for almost every x ∈ Ω,

(ii) x→ f (x, ξ) is measurable for every ξ ∈ RN .

Remark 3.6 In most of the uses of the above notion, we will apply it to func-tions f : Ω × Rm × RM → R ∪ +∞ , f = f (x, u, ξ) . When we speak ofCaratheodory functions in this context, we will consider the variable ξ as play-ing the role of (u, ξ) and RN = Rm × RM . ♦

The first result concerns the composition of these functions with measurableones.

Proposition 3.7 Let Ω ⊂ Rn be an open set, f : Ω × RN → R ∪ +∞ be aCaratheodory function and u : Ω → RN be a measurable function. Then thefunction g : Ω → R ∪ +∞ defined by

g (x) := f (x, u (x))

is measurable.

Proof. We start by proving the result for simple functions of the form

u (x) =

m∑

i=1

αi1Ai (x)

where αi ∈ R, Ai are measurable disjoint sets whose union is Ω and 1Ai is thecharacteristic function of the set Ai namely

1Ai (x) :=

1 if x ∈ Ai

0 if x /∈ Ai.

Let a ∈ R and observe that

x ∈ Ω : g (x) < a =

m⋃

i=1

x ∈ Ai : f (x, αi) < a .

Since x → f (x, ξ) is measurable, we deduce that the set on the right hand sideis measurable and hence g is measurable.

Since any measurable function u is a limit of simple functions uν and ξ →f (x, ξ) is continuous, we deduce that for almost every x ∈ Ω, we have

g (x) = f (x, u (x)) = limν→∞

f (x, uν (x))

and thus g is measurable.


The next theorem is a generalization of the classical Lusin theorem toCaratheodory functions (for a proof, see, for example, Ambrosio-Fusco-Pallara[25], Cesari [143], Ekeland-Temam [264] or Giusti [316], and we here follow thislast proof).

Theorem 3.8 (Scorza-Dragoni theorem) Let Ω ⊂ Rn be bounded and mea-surable, S ⊂ RN be compact and f : Ω × RN → R ∪ +∞ be a Caratheodoryfunction. Then for every ǫ > 0 there exists a compact set Kǫ ⊂ Ω such thatmeas (Ω−Kǫ) ≤ ǫ and f restricted to Kǫ × S is continuous.

Proof. We first define for i ∈ N

ωi (x) := sup |f (x, u)− f (x, v)| : u, v ∈ S, |u− v| < 1/i .

By hypothesis we know that ωi → 0 a.e. in Ω and hence, appealing to Egorovtheorem, we deduce that for every ǫ > 0 we can find a compact set K1

ǫ ⊂ Ω sothat

ωi → 0 uniformly in K1ǫ and meas

(Ω−K1

ǫ

)≤ ǫ/2.

In other words we can find for every η > 0 and u ∈ S, δ1 = δ1 (u, η) > 0 so thatfor every x ∈ K1

ǫ and v ∈ S we have

|u− v| < δ1 ⇒ |f (x, u)− f (x, v)| < η/4. (3.1)

We next choose a sequence ui∞i=1 dense in S. Applying Lusin theorem, we canfind for every fixed i ∈ N a compact set Ki ⊂ Ω such that

x → f (x, ui) is continuous in Ki and meas (Ω−Ki) ≤ ǫ/2i+1.

Letting K2ǫ =

⋂Ki we have that, for all i ∈ N,

x→ f (x, ui) is continuous in K2ǫ and meas

(Ω−K2

ǫ

)≤ ǫ/2.

In other words we can find for every η > 0, x ∈ K2ǫ and ui, δ2 = δ2 (x, ui, η) > 0

so that for every y ∈ K2ǫ the following implication holds

|x− y| < δ2 ⇒ |f (x, ui)− f (y, ui)| < η/4. (3.2)

We finally let Kǫ = K1ǫ

⋂K2

ǫ . It remains to show that f restricted to Kǫ× S iscontinuous. So let η > 0, x ∈ Kǫ and u ∈ S. We first choose δ1 = δ1 (u, η) as in(3.1) and then ui so that

|u− ui| < δ1.

This choice implies that if u, v ∈ S are such that

|u− v| < δ1


and x, y ∈ Kǫ , then

|f (x, u)− f (x, ui)| , |f (y, u)− f (y, ui)| , |f (y, u)− f (y, v)| < η/4. (3.3)

We then find δ2 = δ2 (x, ui, η) as in (3.2) so that for every y ∈ K2ǫ

|x− y| < δ2 ⇒ |f (x, ui)− f (y, ui)| < η/4. (3.4)

Finally letδ = δ (x, u, η) := min δ1 (u, η) , δ2 (x, ui, η) .

Combining (3.3) and (3.4) we obtain that for every y ∈ Kǫ and v ∈ S

|x− y|+ |u− v| < δ ⇒ |f (x, u)− f (y, v)| < η.


We finally point out an important result that allows a passage from weak tostrong convergence (see, for example, Theorem 3.13 in Rudin [519] or TheoremV.1.2 in Yosida [605]). It is a direct consequence of Hahn-Banach theorem.

Theorem 3.9 (Mazur theorem) Let (X, ‖.‖) be a normed space and let

xν x in X.

Then there exists a sequence yµ∞µ=1 ⊂ co xν∞ν=1 such that

yµ → x in X.

More precisely, for every μ there exist an integer mµ and

αiµ > 0 with

mµ∑

i=1

αiµ = 1

such that

yµ =

mµ∑

i=1

αiµxi

and‖yµ − x‖ → 0 as μ→∞.

3.2.2 Some approximation lemmas

On several occasions, we will have to construct functions whose gradient essen-tially takes only two values. The scalar version of Lemma 3.11 will be usedin Section 3.2.4, while the vectorial version will be used in Theorem 5.3. InChapter 10, we will have some more results in the same spirit.

We start with the case where n = 1 and we recall that by Affpiec we meanthe set of piecewise affine functions (see Chapter 12 for details).


Lemma 3.10 (N ≥ n = 1) Let a < b, λ, μ ∈ RN , t ∈ [0, 1] ,

ξ = tλ + (1− t)μ

and uξ : R → RN defined byuξ (x) = ξx.

For every ǫ > 0, there exist u ∈ Affpiec

([a, b] ; RN

)and disjoint open sets

Iλ, Iµ ⊂ (a, b) such that

meas Iλ = t (b− a) , meas Iµ = (1− t) (b− a) ,

u (a) = uξ (a) , u (b) = uξ (b) ,

‖u− uξ‖L∞ ≤ ǫ,

u′ (x) =

λ if x ∈ Iλ

μ if x ∈ Iµ .

Proof. We easily reduce the problem to the case where a = 0 and b = 1. Wethen let ν ∈ N and we divide the interval (0, 1) in disjoint intervals of length2−ν . Each of these subintervals is then further subdivided into two disjointintervals of respective length t2−ν and (1− t) 2−ν . More precisely we let fors = 0, · · · , 2ν − 1, an integer,

| | | |

Figure 3.1: Function ϕ

Is,ν :=

(s

2ν,s + t

2ν

)and Js,ν :=

(s + t

2ν,s + 1

2ν

)

and we let

Iλ :=2ν−1⋃

s=0

Is,ν , Iµ :=2ν−1⋃

s=0

Js,ν .


If we next define ϕ : [0, 1]→ RN (see Figure 3.1) by

ϕ (x) :=

(1− t) (λ− μ)

(x− s

2ν

)if x ∈ Is,ν

−t (λ− μ)(x− s+1

2ν

)if x ∈ Js,ν

we have constructed a function ϕ ∈ W 1,∞0

((0, 1) ; RN

), in fact ϕ ∈ Affpiec(

[0, 1] ; RN), which satisfies

ϕ′ (x) =

(1− t) (λ− μ) if x ∈ Iλ

−t (λ− μ) if x ∈ Iµ .

Choosing ν so that|λ− μ| 2−ν ≤ ǫ

and setting u = uξ + ϕ we have indeed established the lemma.

We next consider the case with several variables.

Lemma 3.11 (N, n ≥ 1) Let Ω ⊂ Rn be an open set with finite measure. Lett ∈ [0, 1] and α, β ∈ RN×n with rank α− β = 1. Let uξ be such that

∇uξ(x) = ξ = tα + (1 − t)β, ∀x ∈ Ω.

Then, for every ǫ > 0, there exist u ∈ Affpiec

(Ω; RN


Ωα , Ωβ ⊂ Ω, so that⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

|meas Ωα − t meas Ω| , |meas Ωβ − (1− t)meas Ω| ≤ ǫ,

u ≡ uξ near ∂Ω, ‖u− uξ‖L∞ ≤ ǫ,

∇u(x) =

α in Ωα

β in Ωβ ,

dist (∇u(x), co α, β) ≤ ǫ a.e. in Ω,

where co α, β = [α, β] is the closed segment joining α to β.

Remark 3.12 If N = 1 or n = 1, then the hypothesis rank α− β = 1 is nota restriction. We can also note that, when n = 1, Lemma 3.10 gives a sharperresult than the present lemma. ♦Proof. We divide the proof into two steps.

Step 1. Let us first assume that the matrix has the form

α− β = a⊗ e1

where e1 = (1, 0, · · · , 0) ∈ Rn and a ∈ RN , or equivalently

α− β = (a, 0, · · · , 0) =

⎛⎜⎜⎜⎝

a1 0 · · · 0a2 0 · · · 0...

.... . .

...aN 0 · · · 0

⎞⎟⎟⎟⎠ ∈ RN×n.


We can express Ω as union of cubes with faces parallel to the coordinate axesand a set of small measure. Then, by posing u ≡ uξ on the set of small measure,and by homotheties and translations, we can reduce ourselves to work with Ωequal to the unit cube.

Let Ωǫ be an open set compactly contained in Ω and let η ∈ Affpiec

(Ω)

andL > 0 be such that

⎧⎪⎪⎪⎨⎪⎪⎪⎩

meas (Ω− Ωǫ) ≤ ǫ, supp η ⊂ Ω

0 ≤ η(x) ≤ 1, ∀x ∈ Ω

η(x) = 1, ∀x ∈ Ωǫ

|∇η(x)| ≤ Lǫ , a.e. x ∈ Ω− Ωǫ .

(3.5)

Let us define a function ϕ : [0, 1] → RN , as in Lemma 3.10 (where ξ = 0,λ = (1 − t)a and μ = −ta), i.e. we can find for every δ > 0, I, J disjoint opensets such that

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

I ∪ J = [0, 1], I ∩ J = ∅meas I = t, meas J = 1− t

ϕ (0) = ϕ (1) = 0, |ϕ(x1)| ≤ δ, ∀x1 ∈ (0, 1)

ϕ′(x1) =

(1− t)a if x1 ∈ I

−ta if x1 ∈ J.

We next let ψ : Rn → RN be such that

ψ (x) = ψ (x1, · · · , xn) := ϕ (x1)

which implies in particular that

∇ψ (x) = ϕ′(x1)⊗ e1 =

(1− t) (α− β) in I × Rn−1

−t (α− β) in J × Rn−1.

We then define u as a convex combination of ψ +uξ and uξ in the following way

u := η(ψ + uξ) + (1− η)uξ = ηψ + uξ.

Choosing δ > 0 sufficiently small, namely

δ := minǫ2, ǫ2

L,

we find that u satisfies the conclusions of the lemma, with

Ωα := x ∈ Ωǫ : x1 ∈ I and Ωβ := x ∈ Ωǫ : x1 ∈ J .

In fact u ≡ uξ near ∂Ω and we have for every x ∈ Ω

‖u− uξ‖L∞ ≤ ǫ.


Since in Ωǫ we have η ≡ 1 we deduce that

∇u = ∇ψ +∇uξ = ∇ψ + tα + (1 − t)β =

α in Ωα

β in Ωβ .

Finally it remains to show that

dist (∇u(x), co α, β) ≤ ǫ a.e. in Ω.

We have that

∇u = η∇ψ +∇uξ + ψ ⊗∇η

where, by definition of δ,

|ψ ⊗∇η| ≤ ǫ.

Since both ∇ψ +∇uξ (= α or β) and ∇uξ = tα + (1 − t)β belong to co α, βwe obtain that

η∇ψ +∇uξ = η (∇ψ +∇uξ) + (1− η)∇uξ ∈ co α, β ;

since the remaining term is arbitrarily small we deduce the result i.e.,

dist (∇u; co α, β) ≤ ǫ.

Step 2. Let us now assume that α − β is any matrix of rank one of RN×n

and therefore it can be written as α− β = a⊗ ν, namely

(α− β)ij = aiνj

for a certain a ∈ RN and ν ∈ Rn (ν not necessarily e1 as in Step 1). Replacinga by |ν| a we can assume that |ν| = 1. We can then find

R =(rij

)1≤i≤n

1≤j≤n∈ O (n) ⊂ Rn×n

(the set of orthogonal matrices, see Chapter 13) so that ν = e1R and hence

e1 = νRt. We then set Ω = RΩ and for 1 ≤ i ≤ N, 1 ≤ j ≤ n we let

αij =

n∑

k=1

αikrj

k and βij =

n∑

k=1

βikrj

k .

i.e.,

α = αRt and β = βRt.

We observe that by construction

α− β = a⊗ e1 .


Indeed, since e1 = νRt, we have

( α− β )ij =

n∑

k=1

aiνkrjk = ai

n∑

k=1

νkrjk = ai (e1)j .

We can therefore apply Step 1 to Ω and to uξ (y) = uξ (Rty) and find Ωα , Ωβ

and u ∈ Affpiec(Ω; RN ) with the claimed properties. By setting

u (x) = u(Rx), x ∈ Ω

Ωα = Rt Ωα , Ωβ = Rt Ωβ

we get the result, since

∇u (x) = ∇u(Rx)R.

3.2.3 Necessary condition: the case without lowerorder terms

We start with a simpler version of Theorem 3.15.

Theorem 3.13 Let Ω ⊂ Rn be an open set, f : RN×n → R be continuous and

I (u) :=

∫

Ω

f (∇u (x)) dx.

Assume that there exists u0 ∈W 1,∞ (Ω; RN)

such that

|I (u0)| <∞. (3.6)

If I is weak ∗ lower semicontinuous in W 1,∞ (Ω; RN), then the following two

results hold.

(i) For every bounded open set D ⊂ Rn, for every ξ0 ∈ RN×n and for everyϕ ∈ W 1,∞

0

(D; RN

),

1

measD

∫

D

f (ξ0 +∇ϕ (y)) dy ≥ f (ξ0) . (3.7)

(ii) If either N = 1 or n = 1, then f is convex.

Remark 3.14 (i) The condition (3.6) is automatically satisfied if Ω is bounded,just choose u0 ≡ 0.

(ii) In Chapter 5, we will call a function f satisfying the inequality in (i)quasiconvex. Note that the statement (i) is valid for any N, n ≥ 1, while (ii) isonly valid in the scalar case. ♦


Proof. (i) We divide the proof into two steps.

Step 1. In view of Lemma 3.17, we can assume that Ω is bounded. We alsonote that if the inequality (3.7) holds for one bounded open set D ⊂ Ω, it holdsfor any such bounded open set, see Proposition 5.11. We therefore only showthe result for a particular cube defined below.

Step 2. Let D be a cube, whose faces are parallel to the axes, contained inΩ and ξ0 ∈ RN×n. Let ϕ ∈ W 1,∞

0

(D; RN

)be extended by periodicity from D

to Rn, meaning that, if the edge length of D is d,

ϕ (x + dz) = ϕ (x) , for every x ∈ D and z ∈ Zn.

Let ν be an integer and define

ϕν (x) :=1

νϕ (νx) .

Since ϕ = 0 on ∂D, we have that

ϕν∗ 0 in W 1,∞

0

(D; RN

).

Defining u := uξ0 , where uξ0 (x) := ξ0x, and letting

uν (x) :=

⎧⎨⎩

uξ0 (x) if x ∈ Ω−D

uξ0 (x) +1

νϕ (νx) if x ∈ D

we have

uν∗ uξ0 in W 1,∞ (Ω; RN

).

Observe also that

I (uν) =

∫

Ω

f (∇uν (x)) dx =

∫

Ω−D

f (ξ0) dx +

∫

D

f (ξ0 +∇ϕ (νx)) dx

= f (ξ0)meas (Ω−D) +1

νn

∫

νD

f (ξ0 +∇ϕ (y)) dy

= f (ξ0)meas (Ω−D) +

∫

D

f (ξ0 +∇ϕ (y)) dy,

where we have used in the last equality the periodicity of ϕ. Taking the limitin the above identity and using the weak ∗ lower semicontinuity of I we haveindeed obtained

1

measD

∫

D

f (ξ0 +∇ϕ (y)) dy ≥ f (ξ0) .

(ii) We want to show that

f (λα + (1− λ) β) ≤ λf (α) + (1− λ) f (β) (3.8)


for every α, β ∈ RN×n and λ ∈ [0, 1] . Recall also that we are now assuming thateither N = 1 or n = 1.

Step 1. We first construct for every ǫ > 0, using Lemma 3.11 (writingϕǫ = u − uξ), a function ϕǫ ∈ Affpiec

(D; RN

)⊂ W 1,∞ (D; RN

)and disjoint

open sets Dα , Dβ ⊂ D, so that

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

|meas Dα − λmeas D| , |meas Dβ − (1− λ)meas D| ≤ ǫ,

ϕǫ ≡ 0 near ∂D, ‖ϕǫ‖L∞ ≤ ǫ, ‖∇ϕǫ‖L∞ ≤ γ,

∇ϕǫ(x) =

(1− λ) (α− β) in Dα

−λ (α− β) in Dβ

where γ > 0 is a constant independent of ǫ.

Step 2. We are now in a position to show (3.8), i.e. that f is convex. Inview of (i) we have

1

meas D

∫

D

f (λα + (1− λ)β +∇ϕǫ (x)) dx ≥ f (λα + (1− λ) β)

where ϕǫ ∈ W 1,∞0

(D; RN

)is as in Step 1. Evaluating the integral we find

∫

D

f (λα + (1− λ) β +∇ϕǫ (x)) dx =

∫

Dα

f (α) dx +

∫

Dβ

f (β) dx

+

∫

D−Dα∪Dβ

f (λα + (1− λ)β +∇ϕǫ (x)) dx.

Letting ǫ→ 0 we have indeed obtained (3.8) and thus the theorem.

3.2.4 Necessary condition: the general case

We now discuss the necessary condition in the general context.

Theorem 3.15 Let Ω ⊂ Rn be an open set, f : Ω × RN × RN×n → R be aCaratheodory function satisfying, for almost every x ∈ Ω and for every (u, ξ) ∈RN × RN×n,

|f (x, u, ξ)| ≤ a (x) + b (u, ξ) ,

where a, b ≥ 0, a ∈ L1 (Rn) and b ∈ C(RN × RN×n

). Let

I (u) = I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx,

and assume that there exists u0 ∈W 1,∞ (Ω; RN)

such that

|I (u0, Ω)| < ∞. (3.9)

If I is weak ∗ lower semicontinuous in W 1,∞ (Ω; RN)

and if either N = 1 orn = 1, then ξ → f (x, u, ξ) is convex for almost every x ∈ Ω and for everyu ∈ RN .


Remark 3.16 (i) The condition (3.9) is a restriction only if Ω is of infinitemeasure.

(ii) If Ω ⊂ Rn is bounded, then any sequence uν∗ u in W 1,∞ automatically

verifies uν u in W 1,p, p ≥ 1. Therefore the convexity of f is also necessaryfor W 1,p weak lower semicontinuity of I.

(iii) The theorem remains valid if the functional I is lower semicontinuousfor every sequence

uν∗ u in W 1,∞ (Ω; RN

)

and uν ∈ u + W 1,∞0

(Ω; RN

), since in the proof of Theorem 3.15 we use such

sequences.

(iv) A theorem of the above type has been proved under various kinds ofhypotheses. The first to notice the importance of convexity was Tonelli [579].Then important contributions were made by Berkowitz [79], [80], Buttazzo [112],Cacciopoli-Scorza-Dragoni [120], Cesari [139], [140], [142], [143], Ioffe [347],[348], [349], MacShane [409], Marcellini-Sbordone [428], Morrey [455] and Olech[481], [482]. ♦

The proof that we are now about to give is neither the most direct nor theeasiest one, but it has the advantage of giving an important result (see Lemma3.18) that is also valid in the vectorial case. We start with two lemmas that holdfor any n, N ≥ 1; only the last part of the proof of Theorem 3.15 will requireN = 1 or n = 1.

Lemma 3.17 Let Ω, f and I be as in the theorem. Let O ⊂ Ω be a boundedopen set and

I (u, O) :=

∫

O

f (x, u (x) ,∇u (x)) dx.

Let I be weak ∗ lower semicontinuous in W 1,∞ (Ω; RN). Let u ∈W 1,∞ (Ω; RN

)

and uν ⊂ W 1,∞ (O; RN)

be such that

uν∗ u in W 1,∞ (O; RN

).

Thenlim infν→∞

I (uν , O) ≥ I (u, O) .


Step 1. Let U ⊂⊂ O be open and


).

We claim that, up to the extraction of a subsequence that we do not relabel, wecan find a sequence vν ∈ u + W 1,∞

0

(U ; RN

)such that

vν∗ u in W 1,∞ (U ; RN

)

|I (vν , U)− I (uν , U)| ≤ 1

ν.


Let us construct such a sequence. Since uν∗ u in W 1,∞ (O; RN

), we can

assume, up to the extraction of a subsequence (that we do not relabel), that

‖uν − u‖L∞ ≤ 1/ν2. (3.10)

We next choose Uν ⊂⊂ U and η ∈ C∞0 (U)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

meas(U − Uν) ≤ 1/ν

0 ≤ η(x) ≤ 1, ∀x ∈ U

η(x) = 1, ∀x ∈ Uν

|∇η(x)| ≤ Lν, ∀x ∈ U − Uν .

where L > 0 is a constant.We then set

vν := (1− η)u + ηuν

Note that vν ∈ u + W 1,∞0

(U ; RN

)and

‖vν − u‖L∞ ≤ 1/ν2.

Let γ > 0 be such that

‖uν‖W 1,∞ , ‖u‖W 1,∞ ≤ γ

and observe that

‖vν‖W 1,∞ ≤ γ +L

ν,

since (3.10) holds and

∇vν = (1− η)∇u + η∇uν +∇η ⊗ (uν − u) .

We can then extract a subsequence, still denoted vν , with the claimed prop-erties.

Step 2. Let O ⊂ Ω,


)and u ∈ W 1,∞ (Ω; RN

).

Let U ⊂⊂ O and let us show that

lim infν→∞

I (uν , U) ≥ I (u, U) .

Replacing, if necessary, uν by a subsequence, we can use Step 1 to constructvν ∈ u + W 1,∞

0

(U ; RN

)such that

vν∗ u in W 1,∞ (U ; RN

)

|I (vν , U)− I (uν , U)| ≤ 1

ν. (3.11)


Next let θ ∈ C∞0 (Ω) be such that θ ≡ 1 in U and define

w := θu + (1− θ)u0 .

Observe that |I (w, Ω)| <∞, since (3.9) holds. Finally let

wν :=

vν in U

w in Ω− U

and note thatwν

∗ w in W 1,∞ (Ω; RN

).

Moreover, sinceI (vν , U) = I (wν , Ω)− I (w, Ω− U)

we get from (3.11) and the fact that I (u, Ω) is weak ∗ lower semicontinuous inW 1,∞ (Ω; RN

)

lim infν→∞

I (uν , U) = lim infν→∞

I (vν , U) = lim infν→∞

[I (wν , Ω)]− I (w, Ω− U)

≥ I (w, Ω)− I (w, Ω− U) = I (u, U)

which is the claim.

Step 3. Since O is bounded, we can choose a sequence of open sets Oh ⊂⊂O ⊂ Ω so that

meas (O −Oh)→ 0, as h →∞.

We claim that for every ǫ > 0, we can find h0 = h0 (ǫ) , independent of ν, sothat, for every h ≥ h0 ,

|I (uν , O) − I (uν , Oh)| ≤ ǫ and |I (u, O)− I (u, Oh)| ≤ ǫ. (3.12)

Indeed use the bound on f to write

|I (uν , O) − I (uν , Oh)| ≤∫

O−Oh

|f (x, uν (x) ,∇uν (x))| dx

≤∫

O−Oh

a (x) dx +

∫

O−Oh

b (uν (x) ,∇uν (x)) dx.

Since the sequence uν∗ u in W 1,∞ (O; RN

), we have, using the hypotheses on

a and b, the first claim in (3.12). The second one follows in the same way.We therefore have, using Step 2 and (3.12),

lim infν→∞

I (uν, O) ≥ −ǫ + lim infν→∞

I (uν , Oh) ≥ −ǫ + I (u, Oh)

≥ −2ǫ + I (u, O) .

Since ǫ is arbitrary, we have indeed proved the lemma.

We continue with the following.


Lemma 3.18 Let Ω, f and I be as in the theorem. Assume that I is weak ∗lower semicontinuous in W 1,∞ (Ω; RN

). Then

1

measD

∫

D

f (x0, u0, ξ0 +∇ϕ (y)) dy ≥ f (x0, u0, ξ0)

for every bounded open set D ⊂ Rn, for almost every x0 ∈ Ω, for every (u0, ξ0) ∈RN × RN×n and for every ϕ ∈W 1,∞

0

(D; RN

).

Remark 3.19 (i) Note once more that, contrary to Theorem 3.15, we do notassume that either N = 1 or n = 1. The lemma is therefore valid also in thevectorial case, i.e. N, n ≥ 1. In Chapter 5, we will call a function f satisfyingthe above inequality, quasiconvex.

(ii) The lemma remains valid if we assume the functional I to be lowersemicontinuous for every sequence


)

and uν ∈ u + W 1,∞0

(Ω; RN

), since in the proof of Lemma 3.18, we use such

sequences.

(iii) The above lemma is essentially due to Morrey [453], [455] and has beenrefined by Meyers [442] and Silvermann [537] for the case of continuous functionsf and by Acerbi-Fusco [3] for the case of Caratheodory ones. We follow thislast proof. ♦Proof. (Lemma 3.18). We start by observing that, in view of Lemma 3.17,there is no loss of generality in assuming that Ω is bounded. In fact this ispossible since all limit functions u, that we will consider in the proof, will beaffine and therefore can be extended from any O ⊂⊂ Ω as v ∈ W 1,∞ (Ω; RN

).

As in the proof of Theorem 3.13, it is sufficient to prove the lemma only forthe unit cube D. We proceed in three steps.

Step 1. We first fix the notations.

- Let ϕ ∈W 1,∞0

(D; RN

), (u0, ξ0) ∈ RN × RN×n be given and define

λ = λ (u0, ξ0, ϕ) := ‖ϕ‖W 1,∞(D;RN ) + |ξ0|+ supx,y∈Ω

|u0 + ξ0 (x− y)|.

- We then define

Sλ :=(u, ξ) ∈ RN × RN×n : |u|+ |ξ| ≤ λ

and we letγ := max b (u, ξ) : (u, ξ) ∈ Sλ .

- For any μ ∈ N, we can find, by Theorem 3.8, a compact set Kµ ⊂ Ω suchthat meas (Ω−Kµ) ≤ 1/μ and f restricted to Kµ × Sλ is continuous. We nextdefine, using Tietze extension theorem (see Rudin [518]), a continuous function

fµ : Rn × RN × RN×n → R


which coincide with f on Kµ × Sλ and such that

|fµ| ≤ max |f (x, u, ξ)| : (x, u, ξ) ∈ Kµ × Sλ .

We can also assume that for every ǫ > 0, we have

∫

Ω−Kµ

|fµ (x, u (x)∇u (x))| dx ≤ ǫ (3.13)

for any u ∈ W 1,∞ (Ω; RN). This is indeed possible by replacing, if necessary,

fµ by ηµfµ , where ηµ ∈ C0 (Ω) , 0 ≤ ηµ ≤ 1, ηµ ≡ 1 on Kµ and

∫

Ω−Kµ

ηµ (x) dx ≤ ǫ/ max|f (x, u, ξ)| : (x, u, ξ) ∈ Kµ × Sλ .

- We, in addition, can find for every ǫ > 0, μ ∈ N, so that for any μ ≥ μ thefollowing holds ∫

Ω−Kµ

[a (x) + γ] dx ≤ ǫ. (3.14)

- We next define Ω0 ⊂ Ω to be the set of points x ∈ Ω so that

x ∈⋃

µ∈N

Kµ

and x is a Lebesgue point of 1Kµ and a.1Ω−Kµ , for every μ ∈ N, where

1Kµ (x) :=

1 if x ∈ Kµ

0 if x ∈ Ω−Kµ .

Observe that meas (Ω− Ω0) = 0.

- From now on x0 will be a given element of Ω0 .

- For h an integer we let

Qh := x0 +1

hD = x ∈ Rn : (x0)i < xi < (x0)i + 1/h, i = 1, · · · , n .

We choose h sufficiently large so that Qh ⊂ Ω.

- Let ϕ ∈ W 1,∞0

(D; RN

)be fixed as above. Extend ϕ by periodicity from

D to Rn and define

ϕν,h (x) :=

⎧⎨⎩

1

νhϕ (νh (x− x0)) if x ∈ Qh

0 if x /∈ Qh .

(3.15)

Fixing h we clearly have

ϕν,h∗ 0 in W 1,∞ (Ω; RN

), as ν →∞.


Observe that if

u (x) := u0 + ξ0 (x− x0) and uν (x) := u (x) + ϕν,h (x) (3.16)

we get


)(3.17)

and, for almost every x ∈ Ω,

(u (x) ,∇u (x)) , (uν (x) ,∇uν (x)) ∈ Sλ . (3.18)

- We then split Qh into cubes Qνh,j of edge length 1/νh (see Figure 3.2) and

denote by xj , 0 ≤ j ≤ νn − 1, the corner of Qνh,j closest to x0 . Therefore

Qh =

νn−1⋃

j=0

Q νh,j and Qν

h,j = xj +1

νhD. (3.19)

× ×

××

Figure 3.2: Cubes Qνh,j and points xj , 0 ≤ j ≤ νn − 1

Step 2. We now consider

I (uν) =

∫

Ω

f (x, uν (x) ,∇uν (x)) dx

=

∫

Ω−Qh

f (x, u (x) ,∇u (x)) dx +

∫

Qh


=

∫

Ω−Qh

f (x, u (x) ,∇u (x)) dx +

∫

Qh

fµ (x, uν (x) ,∇uν (x)) dx

+

∫

Qh

[f (x, uν (x) ,∇uν (x))− fµ (x, uν (x) ,∇uν (x))]dx.


We can rewrite it as

I (uν) =

∫

Ω−Qh

f (x, u (x) ,∇u (x)) dx

+νn−1∑

j=0

∫

Qνh,j

fµ (x, uν (x) ,∇uν (x)) dx + Iν3

=

∫

Ω−Qh

f (x, u (x) ,∇u (x)) dx + Iν1 + Iν

2 + Iν3

(3.20)

where

Iν1 =

νn−1∑

j=0

∫

Qνh,j

fµ (xj , u (xj) ,∇uν (x)) dx

Iν2 =

νn−1∑

j=0

∫

Qνh,j

[fµ (x, uν (x) ,∇uν (x))− fµ (xj , u (xj) ,∇uν (x))] dx

Iν3 =

∫

Qh

[f (x, uν (x) ,∇uν (x))− fµ (x, uν (x) ,∇uν (x))] dx.

Let us estimate the last term

|Iν3 | ≤

∫

Qh

|f (x, uν (x) ,∇uν (x))− fµ (x, uν (x) ,∇uν (x))| dx

≤∫

Ω

|f (x, uν (x) ,∇uν (x))− fµ (x, uν (x) ,∇uν (x))| dx.

Using the definition of f, fµ and Kµ as well as (3.18) and our hypotheses wefind that

|Iν3 | ≤

∫

Ω−Kµ

|f (x, uν (x) ,∇uν (x))| dx +

∫

Ω−Kµ

|fµ (x, uν (x) ,∇uν (x))| dx

≤∫

Ω−Kµ

[a (x) + γ] dx +

∫

Ω−Kµ

|fµ (x, uν (x) ,∇uν (x))| dx.

This, combined with (3.13) and (3.14), finally yields that, for every ǫ > 0, ν ∈ Nand μ ≥ μ,

|Iν3 | ≤ 2ǫ. (3.21)

The uniform continuity of fµ on Qh × Sλ , (3.16), (3.17) and (3.18) lead, for hand μ fixed, to

limν→∞

Iν2 = 0. (3.22)


It then remains to estimate the first term in (3.20), i.e.

Iν1 =

νn−1∑

j=0

∫

Qνh,j

fµ (xj , u (xj) ,∇uν (x)) dx

=

νn−1∑

j=0

∫

xj+1

νh D

fµ (xj , u0 + ξ0 (xj − x0) , ξ0 +∇ϕ (νh (x− x0))) dx

=

νn−1∑

j=0

1

(νh)n

∫

D

fµ (xj , u0 + ξ0 (xj − x0) , ξ0 +∇ϕ (y + νh (xj − x0))) dy

where we have used (3.16), (3.19) and performed a change of variables y =νh (x− xj) . Using finally the periodicity of ϕ we find that for h and μ fixed

Iν1 =

νn−1∑

j=0

1

(νh)n

∫

D

fµ (xj , u0 + ξ0 (xj − x0) , ξ0 +∇ϕ (y)) dy.

We then immediately deduce that, for h and μ fixed,

limν→∞

Iν1 =

∫

Qh

∫

D

fµ (x, u0 + ξ0 (x− x0) , ξ0 +∇ϕ (y)) dydx

=

∫

Qh

∫

D

fµ (x, u (x) , ξ0 +∇ϕ (y)) dydx.

(3.23)

Collecting (3.20), (3.22) and (3.23) and using the weak ∗ lower semicontinuityof I we have

lim infν→∞

I (uν) =

∫

Ω−Qh

f (x, u (x) ,∇u (x)) dx

+

∫

Qh

∫

D

fµ (x, u (x) , ξ0 +∇ϕ (y)) dydx + lim infν→∞

Iν3

≥ I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx.

Hence letting μ →∞ and since ǫ is arbitrary, we find, using (3.21),

1

measQh

∫

Qh

∫

D

f (x, u0 + ξ0 (x− x0) , ξ0 +∇ϕ (y)) dydx

≥ 1

meas Qh

∫

Qh

f (x, u0 + ξ0 (x− x0) , ξ0) dx.

(3.24)

Step 3. Denote by

F (x) :=

∫

D

f (x, u0 + ξ0 (x− x0) , ξ0 +∇ϕ (y)) dy−f (x, u0 + ξ0 (x− x0) , ξ0) .


Therefore (3.24) is equivalent to

1

meas Qh

∫

Qh

F (x) dx ≥ 0. (3.25)

Since f is continuous on Kµ × Sλ , we deduce that

limh→∞

1

meas(Qh ∩Kµ )

∫

Qh∩Kµ

F (x) dx = F (x0) . (3.26)

We next write

1

meas Qh

∫

Qh∩Kµ

F =meas(Qh ∩Kµ )

measQh

1

meas(Qh ∩Kµ )

∫

Qh∩Kµ

F

=1

measQh

∫

Qh

1Kµ

1

meas(Qh ∩Kµ )

∫

Qh∩Kµ

F.

Recalling that x0 ∈ Kµ is a Lebesgue point of 1Kµ and that (3.26) holds, wefind, letting h →∞,

limh→∞

1

meas Qh

∫

Qh∩Kµ

F (x) dx = 1Kµ (x0)F (x0) = F (x0) . (3.27)

On the other hand, we have

| 1

measQh

∫

Qh−(Qh∩Kµ )

F | = | 1

measQh

∫

Qh

F (x) 1Ω−Kµ (x) dx |

≤ 2

meas Qh

∫

Qh

[a (x) + γ] 1Ω−Kµ (x) dx.

We therefore get

limh→∞

| 1

meas Qh

∫

Qh−(Qh∩Kµ )

Fdx | ≤ 2 [a (x0) + γ] 1Ω−Kµ (x0) = 0.

Combining the above estimate with (3.25) and (3.27), we have indeed obtainedthe claim, namely

F (x0) =

∫

D

f (x, u0, ξ0 +∇ϕ (y)) dy − f (x, u0, ξ0) ≥ 0.

This finishes the proof of the lemma.

We now continue with the proof of the main theorem.

Proof. (Theorem 3.15). We want to show that

f (x0, u0, λα + (1− λ)β) ≤ λf (x0, u0, α) + (1− λ) f (x0, u0, β) (3.28)

for almost every x0 ∈ Ω, every u0 ∈ RN , α, β ∈ RN×n and λ ∈ [0, 1] . Recallalso that we are now assuming that either N = 1 or n = 1.

We then proceed exactly as in the proof of (ii) of Theorem 3.13, using Lemma3.18, instead of (i) of Theorem 3.13.


3.2.5 Sufficient condition: a particular case

We start with a simpler version of the general theorem that we will prove below(see Theorem 3.23); this simpler result will be used in the general one.

Theorem 3.20 Let Ω be an open set of Rn and q ≥ 1. Let f : Ω × RM →R ∪ +∞ be a normal integrand satisfying

f (x, ξ) ≥ 〈a (x) ; ξ〉+ b (x)

for almost every x ∈ Ω, for every ξ ∈ RM , for some a ∈ Lq′ (Ω; RM

), 1/q +

1/q′ = 1, b ∈ L1 (Ω) , and where 〈.; .〉 denotes the scalar product in RM . Let

J (ξ) :=

∫

Ω

f (x, ξ (x)) dx.

Assume that ξ → f (x, ξ) is convex and that

ξν ξ in Lq(Ω; RM

).

Thenlim infν→∞

J (ξν) ≥ J (ξ) .

Remark 3.21 Since Caratheodory functions are normal integrands, the theo-rem applies also to such functions. ♦

We then have as a direct consequence the following corollary that applies tothe setting of the calculus of variations.

Corollary 3.22 Let p ≥ 1, Ω ⊂ Rn be an open set and let

f : Ω× RN×n → R ∪ +∞ , f = f (x, ξ) ,

be a Caratheodory function satisfying

f (x, ξ) ≥ 〈a (x) ; ξ〉+ b (x)

for almost every x ∈ Ω, for every ξ ∈ RN×n, for some a ∈ Lp′ (Ω; RN×n

),

1/p + 1/p′ = 1, b ∈ L1 (Ω) and where 〈.; .〉 denotes the scalar product inRN×n. Let

I (u) :=

∫

Ω

f (x,∇u (x)) dx.

Assume that ξ → f (x, ξ) is convex and that

uν u in W 1,p(Ω; RN

).

Thenlim infν→∞

I (uν) ≥ I (u) .


We now continue with the proof of Theorem 3.20.


Step 1. Observe first that if we let

h (x, ξ) := 〈a (x) ; ξ〉+ b (x)

we have, by definition of weak convergence, that

limν→∞

∫

Ω

h (x, ξν (x)) dx =

∫

Ω

h (x, ξ (x)) dx

since a ∈ Lq′

and b ∈ L1 and

ξν ξ in Lq.

Writingg (x, ξ) = f (x, ξ) − h (x, ξ)

we have that g ≥ 0 and therefore if we can show that

lim infν→∞

∫

Ω

g (x, ξν (x)) dx ≥∫

Ω

g (x, ξ (x)) dx

the theorem will follow. Therefore from now on we assume, without loss ofgenerality, that f ≥ 0.

Step 2. We next show that J is (strongly) lower semicontinuous. Assumethat

ξν → ξ in Lq

and extract a sequence that we still label ξν so that

ξν → ξ a.e..

Since f ≥ 0, we can apply Fatou lemma and obtain that

lim infν→∞

∫

Ω

f (x, ξν (x)) dx ≥∫

Ω

lim infν→∞

f (x, ξν (x)) dx.

Combining the above inequality with the lower semicontinuity of f, we have theclaim.

Step 3. We now have to pass from (strong) lower semicontinuity to weaklower semicontinuity. First let

L := lim infν→∞

J (ξν)

and observe that L ≥ 0, since f ≥ 0. We may also assume that L < +∞,otherwise the theorem is trivial. Restricting our attention, if necessary, to asubsequence, we may furthermore assume that

L = limν→∞

J (ξν)


and therefore for every ǫ > 0 we can find νǫ so that for every ν ≥ νǫ the followinginequality holds

J (ξν) ≤ L + ǫ. (3.29)

We next fix ǫ > 0 and apply Mazur Theorem (Theorem 3.9) to find a sequenceηµ∞µ=1 ⊂ co ξν∞ν=νǫ

as in the theorem and in particular so that

ηµ → ξ in Lq (3.30)

and such that for every μ there exist an integer mµ ≥ νǫ and αiµ > 0 with

Σmµ

i=νǫαi

µ = 1 and

ηµ =

mµ∑

i=νǫ

αiµξi .

Appealing to the convexity of ξ → f (x, ξ) and to (3.29) we deduce that

J (ηµ) ≤mµ∑

i=νǫ

αiµJ (ξi) ≤ L + ǫ.

The above inequality combined with (3.30) and Step 2, lead to

J (ξ) ≤ L + ǫ.

Since ǫ is arbitrary, we have the theorem.

3.2.6 Sufficient condition: the general case

We now have our main lower semicontinuity result.

Theorem 3.23 Let Ω be an open set of Rnand p, q ≥ 1. Let f : Ω×Rm×RM →R ∪ +∞ be a Caratheodory function satisfying

f (x, u, ξ) ≥ 〈a (x) ; ξ〉+ b (x) + c |u|p

for almost every x ∈ Ω, for every (u, ξ) ∈ Rm×RM , for some a ∈ Lq′ (Ω; RM

),

1/q + 1/q′ = 1, b ∈ L1 (Ω) , c ∈ R and where 〈.; .〉 denotes the scalar product inRM . Let

J (u, ξ) :=

∫

Ω

f (x, u (x) , ξ (x)) dx.

Assume that ξ → f (x, u, ξ) is convex and that

uν → u in Lp (Ω; Rm) and ξν ξ in Lq(Ω; RM

).

Then

lim infν→∞

J (uν , ξν) ≥ J(u, ξ).


Before making some remarks, we have the following corollary that appliesto the calculus of variations.

Corollary 3.24 Let p ≥ 1, Ω ⊂ Rn be a bounded open set with Lipschitz bound-ary and

f : Ω× RN × RN×n → R ∪ +∞ , f = f (x, u, ξ) ,

be a Caratheodory function satisfying

f (x, u, ξ) ≥ 〈a (x) ; ξ〉+ b (x) + c |u|r

for almost every x ∈ Ω, for every (u, ξ) ∈ RN × RN×n, for some a ∈ Lp′

(Ω; RN×n

), 1/p + 1/p′ = 1, b ∈ L1 (Ω) , c ∈ R, 1 ≤ r < np/ (n− p) if p <

n and 1 ≤ r < ∞ if p ≥ n and where 〈.; .〉 denotes the scalar product inRN×n. Let

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx.

Assume that ξ → f (x, u, ξ) is convex and that


).

Thenlim infν→∞

I (uν) ≥ I (u) .

Remark 3.25 (i) Summarizing the results of Theorem 3.15 and Corollary 3.24we find that a necessary and sufficient condition for I to be weakly lower semi-continuous in W 1,p is that ξ → f (x, u, ξ) be convex.

(ii) Of course both the theorem and the corollary remain valid if we replacerespectively weak convergence in Lq or W 1,p by weak* convergence in L∞ orW 1,∞. Therefore the convexity of ξ → f (x, u, ξ) implies that I is weak* lowersemicontinuous.

(iii) There are some advantages in proving Theorem 3.23 as stated and notrestricting the functional to the case of the calculus of variations; one of thereasons will be clearer in Part II (Theorem 8.16).

(iv) The hypotheses of the theorem are nearly optimal. As mentioned abovethis theorem has a long history and we quote here only a few of the con-tributors starting with Tonelli [579]. Important contributions follow from thework of Berkowitz [79], [80], Buttazzo [112], Cesari [139], [140], [142], [143],De Giorgi [239], De Giorgi-Buttazzo-Dal Maso [242], Eisen [259], Ekeland-Temam [264], Ioffe [348], [349], MacShane [409], Marcellini-Sbordone [428],Marcus-Mizel [430], [431], Morrey [455], Olech [481], [482], Rockafellar [515],Sbordone [523] and Serrin [532] ,[533]. This theorem has also been generalizedin many respects, and we refer to the bibliography for more details. ♦

We first prove Corollary 3.24.


Proof. It follows from Rellich theorem (see Theorem 12.12) that

uν → u in Lr(Ω; RN

).

We therefore can apply Theorem 3.23 with q = p and p = r and the corollaryfollows.

We now proceed with the proof of Theorem 3.23 and we follow here that ofDe Giorgi [239].

Proof. We decompose the proof into four steps.

Step 1. Replacing if necessary f by f where

f (x, u, ξ) := f (x, u, ξ)− 〈a (x) ; ξ〉 − b (x)− c |u|p

we may assume, without loss of generality, that

f (x, u, ξ) ≥ 0, (x, u, ξ) ∈ Ω× Rm × RM .

Indeed note that

N (u, ξ) :=

∫

Ω

[〈a (x) ; ξ (x)〉+ b (x) + c |u (x)|p] dx

is continuous with respect to the weak convergence of ξν ξ in Lq and strongconvergence of uν → u in Lp.

Step 2. Observe that if


J (uν , ξν)

then L ≥ 0, since f ≥ 0. We may also assume that L < +∞, otherwise thetheorem is trivial. Restricting our attention to a subsequence, if necessary, wemay furthermore consider that

L = limν→∞

J (uν , ξν) .

We next show that there is no loss of generality in assuming that Ω is bounded.To emphasize the dependence on the domain let us write

J (u, ξ, Ω) :=

∫

Ω

f (x, u (x) , ξ (x)) dx.

From the above consideration we have

L = limν→∞

J (uν , ξν , Ω) < +∞.

We next suppose that we have shown the desired lower semicontinuity result forany bounded open set Ωµ ⊂ Ω, meaning that

J(u, ξ,Ωµ

)≤ lim inf

ν→∞J (uν , ξν , Ωµ) .


Since f ≥ 0, we obtain that

J (uν , ξν , Ωµ) ≤ J (uν , ξν , Ω)

and henceJ(u, ξ,Ωµ

)≤ L.

Choosing then a sequence of increasing bounded open sets Ωµ ⊂ Ω so thatΩµ ր Ω and applying Lebesgue monotone convergence theorem, we get theresult.

Step 3. So from now on we assume that Ω is bounded, f ≥ 0 and

limν→∞

J (uν , ξν) = L < +∞.

(Note that in the present step we will not use either the convexity of ξ →f (x, u, ξ) or the fact that f ≥ 0.)

We next fix ǫ > 0 and we wish to show that there exists a measurable setΩǫ ⊂ Ω and a subsequence νj , with νj →∞, such that

meas (Ω− Ωǫ) < ǫ∫

Ωǫ

∣∣f(x, uνj (x) , ξνj (x))− f(x, u (x) , ξνj (x))∣∣ dx < ǫ meas Ω.

(3.31)

We now construct Ωǫ with the property (3.31). Note first that since uν → u inLp (Ω) and ξν ξ in Lq (Ω) , we have that for every ǫ > 0, there exists Mǫ > 0,which is independent of ν, such that if

K1ǫ,ν := x ∈ Ω : |u (x)| or |uν (x)| ≥Mǫ

K2ǫ,ν := x ∈ Ω : |ξν (x)| ≥Mǫ

thenmeasK1

ǫ,ν , measK2ǫ,ν <

ǫ

6

for every ν. Hence ifΩ1

ǫ,ν := Ω−(K1

ǫ,ν ∪K2ǫ,ν

)

thenmeas

(Ω− Ω1

ǫ,ν

)<

ǫ

3. (3.32)

Since f is a Caratheodory function, there exists (see Scorza-Dragoni theorem,Theorem 3.8) Ω2

ǫ,ν ⊂ Ω1ǫ,ν a compact set with

meas(Ω1

ǫ,ν − Ω2ǫ,ν

)<

ǫ

3(3.33)

and such that f restricted to Ω2ǫ,ν × Sǫ is continuous where

Sǫ :=(u, ξ) ∈ Rm × RM : |u| < Mǫ and |ξ| < Mǫ

.


We therefore have that there exists δ (ǫ) > 0 such that

|u− v| < δ (ǫ) ⇒ |f (x, u, ξ)− f (x, v, ξ)| < ǫ, (3.34)

for every x ∈ Ω2ǫ,ν , every |u| , |v| < Mǫ and |ξ| < Mǫ .

Having fixed δ (ǫ) in this way and using the fact that uν → u in Lp we canfind νǫ = νǫ,δ(ǫ) such that if

Ω3ǫ,ν := x ∈ Ω : |uν (x)− u (x)| < δ (ǫ)

thenmeas

(Ω− Ω3

ǫ,ν

)<

ǫ

3, for every ν ≥ νǫ . (3.35)

Therefore lettingΩǫ,ν := Ω2

ǫ,ν ∩ Ω3ǫ,ν

we have from (3.32), (3.33), (3.34) and (3.35)

meas (Ω− Ωǫ,ν) < ǫ∫

Ωǫ,ν

|f (x, u (x) , ξν (x))− f (x, uν (x) , ξν (x))| dx < ǫ measΩ(3.36)

for every ν ≥ νǫ. We now choose ǫj = ǫ/2j, j ∈ N. We therefore have that(3.36) holds with ǫ and νǫ replaced by ǫj, νǫj . We then choose any νj ≥ νǫj withlim νj = ∞ and we let

Ωǫ :=

∞⋂

j=1

Ωǫj ,νj .

We immediately deduce (3.31) and this concludes Step 3.

Step 4. We are finally in a position to show the theorem. Let

1Ωǫ (x) :=

1 if x ∈ Ωǫ

0 if x ∈ Ω− Ωǫ .

Letg (x, ξ) := 1Ωǫ (x) f (x, u (x) , ξ)

then g : Ω × RM → R ∪ +∞ is a Caratheodory function and ξ → g (x, ξ) isconvex for almost every x ∈ Ω. Applying Theorem 3.20 to

G (ξ) :=

∫

Ω

g (x, ξ (x)) dx

and to ξνj ξ in Lq(Ω; RM

)we get

lim infνj→∞

G( ξνj ) = lim infνj→∞

∫

Ω

1Ωǫ (x) f(x, u (x) , ξνj (x))dx

≥ G( ξ ) =

∫

Ω

1Ωǫ (x) f(x, u (x) , ξ (x)

)dx.

(3.37)

Weak continuity and invariant integrals 101

Therefore, using (3.31), we have, for νj sufficiently large, that

∫

Ωǫ

f(x, uνj (x) , ξνj (x))dx

≥∫

Ωǫ

f(x, u (x) , ξνj (x))dx

−∫

Ωǫ

| f(x, uνj (x) , ξνj (x))− f(x, u (x) , ξνj (x)) | dx

≥∫

Ωǫ

f(x, u (x) , ξνj (x))dx− ǫ measΩ.

Combining the above inequality and the fact that f ≥ 0, we find that

∫

Ω

f(x, uνj (x) , ξνj (x))dx ≥∫

Ωǫ

f(x, u (x) , ξνj (x))dx − ǫ measΩ

= G(ξνj )− ǫ measΩ.

Letting νj →∞ and using (3.37) we have

L = lim infνj→∞

∫

Ω

f(x, uνj (x) , ξνj (x))dx

≥∫

Ω

1Ωǫ (x) f(x, u (x) , ξ (x)

)dx− ǫ measΩ.

Letting ǫ → 0, using the fact that meas (Ω− Ωǫ) → 0 and Lebesgue monotoneconvergence theorem in the right hand side of the above inequality, we haveindeed obtained the theorem.

3.3 Weak continuity and invariant integrals

We show in this section that if

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx

then I is weakly continuous in W 1,p if and only if ξ → f (x, u, ξ) is affine. In thesecond part of this section we show that invariant integrals (i.e. integrals that areconstant whenever the boundary condition is fixed) can be fully characterizedas those that are in divergence form.

3.3.1 Weak continuity

Combining Theorem 3.15 and Corollary 3.24 we immediately have the following.

Theorem 3.26 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundary,f : Ω×RN ×RN×n → R be a Caratheodory function satisfying, for almost every


x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|f (x, u, ξ)| ≤ a (x) + b (u, ξ) ,


). Assume that either N = 1

or n = 1 and let

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx.

Then I is weak ∗ continuous in W 1,∞ (Ω; RN)

if and only if ξ → f (x, u, ξ)is affine, i.e. there exist Caratheodory functions g : Ω × RN → RN×n andh : Ω× RN → R such that

f (x, u, ξ) = 〈g (x, u) ; ξ〉+ h (x, u) ,

where 〈.; .〉 denotes the scalar product in RN×n.

Remark 3.27 (i) Similar results hold in W 1,p, provided one imposes somerestrictions on g and h, in particular, that g (x, u) ∈ Lp′

(Ω) whenever u ∈W 1,p (Ω) .

(ii) Note also that the result is strictly restricted to the scalar case. It isfalse in the vectorial case (N, n > 1). We will see in Chapter 5 that if N = nand

I (u) =

∫

Ω

det∇u (x) dx

then I is weakly continuous although f (ξ) = det ξ is not affine in ξ.

(iii) The necessary part of the theorem remains valid if the function I iscontinuous for every sequence


)

and uν ∈ u+W 1,∞0

(Ω; RN

), since the proof is a direct consequence of Theorem

3.15. ♦

Proof. The necessity follows immediately from Theorem 3.15 applied to f, Iand then to −f, −I and we find

f (x, u, ξ) = 〈g (x, u) ; ξ〉+ h (x, u) .

The fact that h is a Caratheodory function follows by setting ξ = 0 and usethe fact that (x, u) → f (x, u, 0) is itself a Caratheodory function. A similarargument applies to g.

The sufficiency is also obvious since, if uν∗ u, in W 1,∞, then uν → u in

L∞ and the conclusion follows from the fact that g and h are Caratheodoryfunctions.

Weak continuity and invariant integrals 103

3.3.2 Invariant integrals

We now turn our attention to invariant integrals, which are important in thefield theories of the calculus of variations. Following Caratheodory and Weyl[599], we give here a complete characterization of such integrals.

Theorem 3.28 Let Ω ⊂ Rn be a bounded connected open set sufficiently regularso that the divergence theorem holds. Let f : Ω × RN × RN×n → R be a C∞

function satisfying, for every (x, u, ξ) ∈ Ω× RN × RN×n,

|f (x, u, ξ)| ≤ a (x) + b (u, ξ)

where a, b ≥ 0, a ∈ C∞ (Rn) and b ∈ C∞ (RN × RN×n). Let N = 1 or n = 1

and

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx.

The following two conditions are then equivalent.

(i) I is invariant, meaning that

I (u) = constant, for every u ∈ u0 + W 1,∞0

(Ω; RN

).

(ii) There exist C∞ functions ϕ : Ω× RN → Rn and β : Ω → R such that

f (x, u, ξ) = 〈ϕu (x, u) ; ξ〉+ divx ϕ (x, u) + β (x)

for every (x, u, ξ) ∈ Ω× RN × RN×n; where 〈.; .〉 denotes the scalar product inRN×n and, writing ϕx = ∂ϕ/∂x and ϕu = ∂ϕ/∂u,

ϕu (x, u) =

(ϕ1

u, · · · , ϕnu

)if N = 1

(ϕu1 , · · · , ϕuN ) if n = 1;

divx ϕ (x, u) =

∑ni=1 ϕi

xi(x, u) if N = 1

ϕx (x, u) if n = 1.

In particular if ξ = ∇u, then

f (x, u,∇u) = div [ϕ (x, u (x))] + β (x) .

Remark 3.29 As mentioned above this result is strictly restricted to the scalarcase; for the vectorial case, see Chapter 5 and Ericksen [265], Rund [520] andSivaloganathan [541]. ♦

Proof. (ii) ⇒ (i) Let f be as above then

I (u) =

∫

Ω

β (x) dx +

∫

Ω

div [ϕ (x, u (x))] dx

and since u = u0 on ∂Ω, we have after an integration by parts that I is constant.


(i) ⇒ (ii) Following Theorem 3.26 we have that if I is constant, then it isweak ∗ continuous and therefore there exist g and h such that

f (x, u, ξ) = 〈g (x, u) ; ξ〉+ h (x, u) .

Note that since f ∈ C∞, then so are g and h. We now study separately thecases N = 1 and n = 1.

Case 1: N = 1. By hypothesis we have, denoting partial derivatives byindices as for example ∂u/∂xi = uxi ,

I (u) =

∫

Ω

[∑n

i=1 gi (x, u)uxi + h (x, u)]dx = constant.

Choosing u ∈ u0 + W 1,∞0 (Ω) and v ∈ C∞

0 (Ω) we have that

d

dǫI (u + ǫv)

∣∣∣∣ǫ=0

=

∫

Ω

[∑n

i=1

[gi

u (x, u)uxiv + gi (x, u) vxi

]+ hu (x, u) v ]dx

=

∫

Ω

∑ni=1[g

iu (x, u)uxi −

[gi (x, u)

]xi

] + hu (x, u) v dx

=

∫

Ω

[−∑ni=1 gi

xi(x, u) + hu (x, u) ]v dx ≡ 0.

Applying Theorem 3.40, we obtain that, for every x ∈ Ω

hu (x, u (x)) = divx g (x, u (x)) =

n∑

i=1

gixi

(x, u (x)) . (3.38)

Since this holds for every u ∈ u0 +W 1,∞0 (Ω) , we deduce that the identity holds

for every x ∈ Ω and u ∈ R, namely

hu (x, u) = divx g (x, u) =

n∑

i=1

gixi

(x, u) .

Let

ϕi (x1, · · · , xn, u) =

∫ u

0

gi (x1, · · · , xn, s) ds, i = 1, · · · , n.

We have that if ϕ =(ϕ1, · · · , ϕn

), then

hu (x, u) = divx (ϕu (x, u)) = [divx ϕ (x, u)]u

and thus h (x, u) = β (x) + divx ϕ (x, u)

g (x, u) = ϕu (x, u) .

This concludes the proof in the case N = 1.

Case 2: n = 1. We now have Ω = (a, b). The same argument leading to(3.38) gives, for every x ∈ (a, b) , every j = 1, · · · , N and every u ∈ u0 +

Existence theorems and Euler-Lagrange equations 105

W 1,∞0

((a, b) ; RN

),

huj (x, u (x))− gjx (x, u (x)) +

n∑

i=1

[ giuj (x, u (x))− gj

ui (x, u (x)) ](ui (x))′ = 0.

This implies that, for every (x, u) ∈ (a, b)× RN and every i, j = 1, · · · , N,

huj (x, u) = gjx (x, u) and gi

uj (x, u) = gjui (x, u) . (3.39)

Letting

ψ (x, u) :=

∫ x

a

h (s, u)ds

we find from the first set of equations in (3.39) that

(ψuj (x, u)− gj (x, u))x = 0

and thus there exists γj ∈ C∞ (RN)

such that

gj (x, u) = ψuj (x, u) + γj (u) .

From the second set of equations in (3.39) we find

γiuj (u) = γj

ui (u)

and hence there exists γ ∈ C∞ (RN)

such that

γj (u) = γuj (u) .

Setting

ϕ (x, u) := ψ (x, u) + γ (u) and β (x) ≡ 0

we have the claim.

3.4 Existence theorems and Euler-Lagrangeequations

In this section, we first show how to apply the above results to the existence ofminima. We also derive the Euler-Lagrange equations under various types ofconditions. We then mention some regularity results, but we omit their proofs.

3.4.1 Existence theorems

We are now in a position to show the existence of minimizers for our problem.


Theorem 3.30 Let Ω be a bounded open set of Rn with a Lipschitz boundary.Let f : Ω×RN ×RN×n → R∪ +∞ be a Caratheodory function satisfying thecoercivity condition

f (x, u, ξ) ≥ α1 |ξ|p + α2 |u|q + α3 (x) (3.40)

for almost every x ∈ Ω and for every (u, ξ) ∈ RN × RN×n and for some α3 ∈L1 (Ω) , α2 ∈ R, α1 > 0 and p > q ≥ 1. Assume that ξ → f (x, u, ξ) is convex.Let

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx.

Assume that I (u0) <∞, then

(P ) infI (u) : u ∈ u0 + W 1,p

0

(Ω; RN

)

attains its minimum.

Furthermore, if (u, ξ)→ f (x, u, ξ) is strictly convex for almost every x ∈ Ω,then the minimizer is unique.

Remark 3.31 (i) The theorem is also valid in the vectorial case N, n > 1, butit can be extended a great deal in this case; see Chapter 8.

(ii) Of course the theorem applies to the Dirichlet integral ; indeed we havethat

f (x, u, ξ) = f (ξ) =1

2|ξ|2

satisfies all the hypotheses of the theorem with p = 2. The natural generalizationof the preceding example is

f (x, u, ξ) =1

p|ξ|p + g (x, u)

where g is continuous and non-negative and p > 1.

(iii) Note that the minimal surface case where

f (x, u, ξ) =

√1 + |ξ|2

is not contained in the above theorem although f is convex, since the coer-civity condition holds only for p = 1 and then W 1,1 is not a reflexive space.For the treatment of this problem we refer to Almgren [17], [19], De Giorgi[237], [238], Dierkes-Hildebrandt-Kuster-Wohlrab [248], Ekeland-Temam [264],Federer [275], Giusti [315], Morrey [455], Nitsche [476] and the references quotedthere.

(iv) The hypothesis I (u0) < ∞ can be ensured if for example we impose agrowth condition on the function f, such as

f (x, u, ξ) ≤ β1 (x) + β2( |u|p∗

+ |ξ|p ),


where β1 ∈ L1 (Ω) , β2 > 0 and p∗ = np/n− p if 1 < p < n and no condition onp∗ if p ≥ n.

(v) Note that, in general, neither the convexity of f, nor the coercivitycondition (3.40) can be weakened. Examples of non-existence of solutions inthese cases occur already when n = N = 1 and are given in Chapter 4. ♦

Proof. Step 1 (Existence). Write

infI (u) : u ∈ u0 + W 1,p

0

(Ω; RN

)= m

and observe that since I (u0) < ∞, we have that m < +∞. Note also thatbecause of the lower bound on f, m > −∞. Let uν be a minimizing sequence,i.e. I (uν)→ m. We have from (3.40) that for ν sufficiently large

m + 1 ≥ I (uν) ≥ α1 ‖∇uν‖pLp − |α2| ‖uν‖q

Lq −∫

Ω

|α3 (x)| dx.

From now on we will denote by γk > 0 constants that are independent of ν.Since by Holder inequality we have

‖uν‖qLq =

∫

Ω

|uν |q ≤(∫

Ω

|uν |p)q/p (∫

Ω

dx

)(p−q)/p

= (meas Ω)(p−q)/p ‖uν‖q

Lp ,

we deduce that we can find constants γ1 and γ2 such that

m + 1 ≥ α1 ‖∇uν‖pLp − γ1 ‖uν‖q

Lp − γ2

≥ α1 ‖∇uν‖pLp − γ1 ‖uν‖q

W 1,p − γ2 .

Invoking Poincare inequality, we can find γ3 , γ4 , γ5 , so that

m + 1 ≥ γ3 ‖uν‖pW 1,p − γ4 ‖u0‖p

W 1,p − γ1 ‖uν‖qW 1,p − γ5

and hence, γ6 being a constant,

m + 1 ≥ γ3 ‖uν‖pW 1,p − γ1 ‖uν‖q

W 1,p − γ6 .

Since 1 ≤ q < p, we can find γ7 , γ8 so that

m + 1 ≥ γ7 ‖uν‖pW 1,p − γ8

which, combined with the fact that m < ∞, leads to the claim, namely

‖uν‖W 1,p ≤ γ9 .

We may therefore extract a subsequence, that we still denote uν , and findu ∈ u0 + W 1,p

0

(Ω; RN

)so that


).


Appealing to Corollary 3.24 we get

lim infν→∞

I (uν) ≥ I (u)

and hence u is a minimizer of (P ) .

Step 2 (Uniqueness). Assume that there exist u, v ∈ u0 + W 1,p0 (Ω) so that

I (u) = I (v) = m

and let us prove that this implies u = v. Denote by w = (u + v) /2 and observethat w ∈ u0 + W 1,p

0 (Ω) . The function (u, ξ)→ f (x, u, ξ) being convex, we caninfer that w is also a minimizer since

m ≤ I (w) ≤ 1

2I (u) +

1

2I (v) = m,

which readily implies that∫

Ω

[1

2f (x, u,∇u) +

1

2f (x, v,∇v)− f(x,

u + v

2,∇u +∇v

2)] dx = 0.

The convexity of (u, ξ)→ f (x, u, ξ) implies that the integrand is non negative,while the integral is 0. This is possible only if

1

2f (x, u,∇u) +

1

2f (x, v,∇v)− f(x,

u + v

2,∇u +∇v

2) = 0 a.e. in Ω.

We now use the strict convexity of (u, ξ)→ f (x, u, ξ) to obtain that u = v and∇u = ∇v a.e. in Ω, which implies the desired uniqueness, namely u = v a.e.in Ω.

3.4.2 Euler-Lagrange equations

We now compute I ′ (u) , the Gateaux derivative of

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx.

We first recall some notation. We will write, for f = f (x, u, ξ) ,

Duf = (fu1 , · · · , fuN ) ∈ RN , where fui = ∂f/∂ui, i = 1, · · · , N

Dξf =(fξi

α

)1≤i≤N

1≤α≤n∈ RN×n, where fξi

α= ∂f/∂ξi

α , i = 1, · · · , N, α = 1, · · · , n.

We have to consider several restrictions on the growth of f and its derivativesand we list them below.

Condition 3.32 (Growth condition on f) The function

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,


is a Caratheodory function satisfying, for almost every x ∈ Ω, for every (u, ξ) ∈RN × RN×n,

|f (x, u, ξ)| ≤ α1 (x) + β (|u|p + |ξ|p)

where α1 ∈ L1 (Ω) and β ≥ 0. ♦

With the use of Sobolev imbedding theorem, we can further improve thegrowth condition as follows (here, for simplicity, we consider the case of boundedΩ with a Lipschitz boundary).

- If p > n, then we assume that for every R > 0, there exist α1 ∈ L1 (Ω)and β = β (R) ≥ 0 such that, for almost every x ∈ Ω and for every (u, ξ) ∈BN

R × RN×n,|f (x, u, ξ)| ≤ α1 (x) + β |ξ|p

where BNR :=

u ∈ RN : |u| ≤ R

.

- If p = n, then there exist α1 ∈ L1 (Ω) and β ≥ 0 such that, for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|f (x, u, ξ)| ≤ α1 (x) + β (|u|q + |ξ|p)

and where q ≥ 1.

- If 1 ≤ p < n, then there exist α1 ∈ L1 (Ω) and β ≥ 0 such that, for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|f (x, u, ξ)| ≤ α1 (x) + β( |u|p∗

+ |ξ|p )

and where p∗ = np/ (n− p) .

We now turn our attention to the conditions on the derivatives.

Condition 3.33 (Growth condition (I)) The functions fui , fξiα

: Ω×RN ×RN×n → R are Caratheodory functions for every i = 1, · · · , N, α = 1, · · · , n.Moreover, they satisfy, for almost every x ∈ Ω and for every (u, ξ) ∈ RN×RN×n,

|Duf (x, u, ξ)| , |Dξf (x, u, ξ)| ≤ α1 (x) + β (|u|p + |ξ|p)

where α1 ∈ L1 (Ω) and β ≥ 0. ♦

As before, the condition can be improved (we consider here only the case ofbounded Ω with a Lipschitz boundary) as follows.

- If p > n, then we assume that for every R > 0, there exist α1 ∈ L1 (Ω)and β = β (R) ≥ 0 such that, for almost every x ∈ Ω and for every (u, ξ) ∈BN

R × RN×n,

|Duf (x, u, ξ)| , |Dξf (x, u, ξ)| ≤ α1 (x) + β |ξ|p ,

where BNR :=

u ∈ RN : |u| ≤ R

.


- If p = n, then there exist α1 ∈ L1 (Ω) and β ≥ 0 such that, for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|Duf (x, u, ξ)| , |Dξf (x, u, ξ)| ≤ α1 (x) + β (|u|q + |ξ|p)

and where q ≥ 1.

- If 1 ≤ p < n, then there exist α1 ∈ L1 (Ω) and β ≥ 0 such that, for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|Duf (x, u, ξ)| , |Dξf (x, u, ξ)| ≤ α1 (x) + β( |u|p∗

+ |ξ|p )

and where p∗ = np/ (n− p) .

Condition 3.34 (Growth condition (II)) The functions fui , fξiα

: Ω×RN×RN×n → R are Caratheodory functions for every i = 1, · · · , N, α = 1, · · · , n.Moreover for every R > 0, there exist α1 ∈ L1 (Ω) , α2 ∈ Lp/(p−1) (Ω) and β =β (R) ≥ 0 such that, for almost every x ∈ Ω and for every (u, ξ) ∈ BN

R ×RN×n,

|Duf (x, u, ξ)| ≤ α1 (x) + β |ξ|p ,

|Dξf (x, u, ξ)| ≤ α2 (x) + β |ξ|p−1,

where BNR :=

u ∈ RN : |u| ≤ R

. ♦

Condition 3.35 (Growth condition (III)) The functions fui , fξiα

: Ω×RN×RN×n → R are Caratheodory functions for every i = 1, · · · , N, α = 1, · · · , n.Moreover they satisfy, for almost every x ∈ Ω and for every (u, ξ) ∈ RN×RN×n,

|Duf (x, u, ξ)| ≤ α1 (x) + β( |u|p−1+ |ξ|p−1

),

|Dξf (x, u, ξ)| ≤ α2 (x) + β( |u|p−1+ |ξ|p−1

),

where α1 , α2 ∈ Lp/(p−1) (Ω) and β ≥ 0. ♦

As before, the condition can be improved (we consider here only the case ofbounded Ω with a Lipschitz boundary) as follows.

- If p > n, then we assume that for every R > 0, there exist α1 ∈ L1 (Ω) ,α2 ∈ Lp/(p−1) (Ω) and β = β (R) ≥ 0 such that, for almost every x ∈ Ω and forevery (u, ξ) ∈ BN

R × RN×n,

|Duf (x, u, ξ)| ≤ α1 (x) + β |ξ|p

|Dξf (x, u, ξ)| ≤ α2 (x) + β |ξ|p−1

where BNR :=

u ∈ RN : |u| ≤ R

.


- If p = n, then there exist α1 ∈ Ls (Ω) , α2 ∈ Lp/(p−1) (Ω) and β ≥ 0 suchthat, for almost every x ∈ Ω and for every (u, ξ) ∈ RN × RN×n,

|Duf (x, u, ξ)| ≤ α1 (x) + β (|u|r1 + |ξ|r2)

|Dξf (x, u, ξ)| ≤ α2 (x) + β( |u|q + |ξ|p−1)

where s > 1, q, r1 ≥ 1 and 1 ≤ r2 < p.

- If 1 ≤ p < n, then there exist α1 ∈ Lnp/(np−n+p) (Ω) , α2 ∈ Lp/(p−1) (Ω)and β ≥ 0 such that, for almost every x ∈ Ω and for every (u, ξ) ∈ RN ×RN×n,

|Duf (x, u, ξ)| ≤ α1 (x) + β (|u|r1 + |ξ|r2)

|Dξf (x, u, ξ)| ≤ α2 (x) + β( |u|q + |ξ|p−1)

where 1 ≤ q ≤ (np− n) / (n− p) , 1 ≤ r1 ≤ (np− n + p) / (n− p) and 1 ≤ r2 ≤(np− n + p) /n.

Remark 3.36 (i) The conditions are more and more restrictive in the sensethat

(III)⇒ (II)⇒ (I) .

For example if 1 ≤ p ≤ n then (III) is a stronger hypothesis than (II), since wehave only 1 ≤ r2 < p for the growth condition on Duf for (III) while r2 = p isallowed in (II).

Another example is the case where N = 1,

f (x, u, ξ) = f (u, ξ) = a (u) |ξ|2

where 0 < a1 ≤ a (u) , a′ (u) ≤ a2 <∞ and n ≥ p = 2. Then

|Duf (u, ξ)| ≤ a2 |ξ|2

and therefore (II) is satisfied while (III) is not.

(ii) Growth condition (III) is sometimes called controllable growth condi-tion and (II) natural growth condition, see Giaquinta [307], Giusti [316],Ladyzhenskaya-Uraltseva [388] and Morrey [455]. ♦

We now prove the main theorem of this section which gives the weak form ofthe Euler-Lagrange equation. It is only based on several applications of Holderinequality and Sobolev imbedding theorem.

Theorem 3.37 (Weak form of Euler-Lagrange equation) Let f be as inCondition 3.32 and for ϕ : Ω → RN let

L (u, ϕ) :=

∫

Ω

∑Ni=1 [

∑nα=1

∂f∂ξi

α(x, u,∇u) ∂ϕi

∂xα] + ∂f

∂ui (x, u,∇u)ϕi dx

=

∫

Ω

〈Dξf (x, u,∇u) ;∇ϕ〉 + 〈Duf (x, u,∇u) ; ϕ〉 dx.


Assume that u ∈ u0 + W 1,p0

(Ω; RN

)is a minimizer of (P ) , where

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

).

(I) If Growth condition (I) holds, then

(Ew) L (u, ϕ) = 0 for every ϕ ∈ C∞0

(Ω; RN

).

(II) If Growth condition (II) holds and in addition u ∈ L∞ (Ω; RN), then

(Ew) L (u, ϕ) = 0 for every ϕ ∈ W 1,p0

(Ω; RN

)∩ L∞ (Ω; RN

).

(III) If Growth condition (III) holds, then

(Ew) L (u, ϕ) = 0 for every ϕ ∈W 1,p0

(Ω; RN

).

Conversely, if u satisfies (Ew) and if (u, ξ) → f (x, u, ξ) is convex for almostevery x ∈ Ω, then u is a minimizer of (P ) .

Proof. Step 1. Note first that because of the growth condition on f itself wehave, for every ǫ ∈ R, and every ϕ ∈W 1,p, that I (u + ǫϕ) is well defined.

Since u is a minimizer of (P ) then

I (u + ǫϕ) ≥ I (u) ,

for every ϕ ∈ C∞0

(Ω; RN

)in (I), ϕ ∈ W 1,p

0

(Ω; RN

)∩ L∞ (Ω; RN

)in (II) and

ϕ ∈ W 1,p0

(Ω; RN

)in (III).

We thus have, if the limit exists, that (cf. below)

L (u, ϕ) = limǫ→0

I (u + ǫϕ)− I (u)

ǫ= 0

which leads to (Ew), as wished.

Indeed let us show that

L (u, ϕ) = limǫ→0

I (u + ǫϕ)− I (u)

ǫ. (3.41)

We first introduce the following notation

g (x, ǫ) :=

∫ 1

0

〈Duf (x, u + tǫϕ,∇u + tǫ∇ϕ) ; ϕ〉

= + 〈Dξf (x, u + tǫϕ,∇u + tǫ∇ϕ) ;∇ϕ〉 dt.

We therefore find that

I (u + ǫϕ)− I (u)

ǫ=

1

ǫ

∫

Ω

dx

∫ 1

0

d

dt[ f (x, u (x) + tǫϕ (x) ,∇u (x) + tǫ∇ϕ (x)) ] dt

=

∫

Ω

g (x, ǫ) dx.


If we can show that there exists γ ∈ L1 (Ω) such that for every ǫ sufficientlysmall

|g (x, ǫ)| ≤ γ (x) , a.e. x ∈ Ω (3.42)

we will have (3.41) by applying Lebesgue dominated convergence theorem.

So let us show (3.42). We have to consider the three cases.

Growth condition (I). We find, since ϕ ∈ C∞0

(Ω; RN

), that

|〈Duf (x, u + tǫϕ,∇u + tǫ∇ϕ) ; ϕ〉|≤ [α1 (x) + β (|u + tǫϕ|p + |∇u + tǫ∇ϕ|p)] |ϕ|

|〈Dξf (x, u + tǫϕ,∇u + tǫ∇ϕ) ;∇ϕ〉|≤ [α1 (x) + β( |u + tǫϕ|p + |∇u + tǫ∇ϕ|p )] |∇ϕ| .

Summing up the two inequalities and taking the supremum in (t, ǫ) ∈ [0, 1] ×[−1, 1] , we have indeed obtained (3.42).

Growth condition (II). Since u, ϕ ∈ L∞ (Ω; RN), we can find R > 0 so that,

for every (t, ǫ) ∈ [0, 1]× [−1, 1] ,

|u + tǫϕ| , |ϕ| ≤ R, a.e. x ∈ Ω.

We therefore find

|〈Duf (x, u + tǫϕ,∇u + tǫ∇ϕ) ; ϕ〉| ≤ [α1 (x) + β |∇u + tǫ∇ϕ|p] |ϕ||〈Dξf (x, u + tǫϕ,∇u + tǫ∇ϕ) ;∇ϕ〉| ≤ [ α2 (x) + β |∇u + tǫ∇ϕ|p−1

] |∇ϕ| .

Noting that, since u, ϕ ∈ W 1,p(Ω; RN

)∩ L∞ (Ω; RN

), we have by Holder

inequalityα1 |ϕ| , |∇u + tǫ∇ϕ|p |ϕ| ∈ L1 (Ω)

α2 |∇ϕ| , |∇u + tǫ∇ϕ|p−1 |∇ϕ| ∈ L1 (Ω) .


Growth condition (III). We find

|〈Duf (x, u + tǫϕ,∇u + tǫ∇ϕ) ; ϕ〉|≤ [ α1 (x) + β( |u + tǫϕ|p−1

+ |∇u + tǫ∇ϕ|p−1) ] |ϕ|

|〈Dξf (x, u + tǫϕ,∇u + tǫ∇ϕ) ;∇ϕ〉|≤ [ α2 (x) + β( |u + tǫϕ|p−1

+ |∇u + tǫ∇ϕ|p−1) ] |∇ϕ| .

Noting that, since u, ϕ ∈ W 1,p(Ω; RN

), we have by Holder inequality

α1 |ϕ| , |u + tǫϕ|p−1 |ϕ| , |∇u + tǫ∇ϕ|p−1 |ϕ| ∈ L1 (Ω)


α2 |∇ϕ| , |u + tǫϕ|p−1 |∇ϕ| , |∇u + tǫ∇ϕ|p−1 |∇ϕ| ∈ L1 (Ω) .


The theorem is therefore proved. The use of Sobolev imbedding theoremallows to improve the exponents, but the proof is straightforward and we do notdiscuss the details.

Step 2. It remains to show that, provided (u, ξ)→ f (x, u, ξ) is convex, thenany solution u of (Ew) is a minimizer of (P ) . From the hypotheses on f wededuce that, for almost every x ∈ Ω,

f(x, u,∇u) ≥ f(x, u,∇u) + 〈Dξf(x, u,∇u);∇(u− u)〉+ 〈Duf(x, u,∇u); u− u〉.

Therefore for any u ∈ u0 +W 1,p0

(Ω; RN

), we have after integration and appeal-

ing to (Ew) (since u− u ∈ W 1,p0

(Ω; RN

)) that

∫

Ω

f (x, u (x) ,∇u (x)) dx ≥∫

Ω

f (x, u (x) ,∇u (x)) dx

as claimed.

We get as a corollary the classical form of the following equation.

Corollary 3.38 (Euler-Lagrange equation) Let f : Ω × RN × RN×n → Rbe a C2 function. Assume that u ∈ C2

(Ω; RN

)is a minimizer of

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

).

Then u satisfies, for every x ∈ Ω,

(E)n∑

α=1

∂

∂xα[

∂f

∂ξiα

(x, u,∇u) ] =∂f

∂ui(x, u,∇u) , i = 1, · · · , N.

Remark 3.39 (i) Note that if n = 1, then (E) is reduced to a system ofordinary differential equations, namely

(E)d

dx(

∂f

∂ξi(x, u, u′) ) =

∂f

∂ui(x, u, u′) , i = 1, · · · , N.

If N = 1 it is reduced to a single partial differential equation, given by

(E)n∑

α=1

∂

∂xα(

∂f

∂ξα(x, u,∇u) ) =

∂f

∂u(x, u,∇u)

which can be rewritten as

(E) div (Dξf (x, u,∇u) ) = fu (x, u,∇u) ,

while if N, n > 1, (E) is a system of partial differential equations.


(ii) Note also that if N = 1 and if ξ → f (x, u, ξ) is convex then we musthave, provided f is C2,

n∑

i,j=1

∂2f

∂ξi∂ξjλiλj ≥ 0

for every λ ∈ Rn, which in the context of a single partial differential equationis the usual ellipticity condition. ♦

Proof. It follows at once as in the theorem that if ϕ ∈ C∞0

(Ω; RN

)then

∫

Ω

∑Ni=1 [

∑nα=1

∂f

∂ξiα

(x, u,∇u)∂ϕi

∂xα] +

∂f

∂ui(x, u,∇u)ϕi dx = 0.

Integrating by parts we get that∫

Ω

∑Ni=1

∑nα=1

∂

∂xα[

∂f

∂ξiα

(x, u,∇u) ]− ∂f

∂ui(x, u,∇u) ϕidx = 0.

Applying Theorem 3.40 to each component of ϕ, we get the result.

We have used in the proof of the corollary the following classical result.

Theorem 3.40 (Fundamental lemma of the calculus of variations) LetΩ ⊂ Rn be an open set and u ∈ L1

loc (Ω) be such that∫

Ω

u (x) ψ (x) dx = 0, ∀ψ ∈ C∞0 (Ω) . (3.43)

Then u = 0 almost everywhere in Ω.

Proof. We divide the proof into two steps.

Step 1. By approximation, in the uniform norm, of any function ψ ∈ C0 (Ω)by ϕ ∈ C∞

0 (Ω) it is sufficient to prove the theorem assuming that∫

Ω

u (x)ψ (x) dx = 0, ∀ψ ∈ C0 (Ω) . (3.44)

Step 2. It clearly suffices to show the result for u ∈ L1 (O) where O ⊂⊂ Ω.Fix ǫ > 0 and find v ∈ C∞

0 (O) ⊂ C∞0 (Ω) such that

‖u− v‖L1(O) ≤ ǫ. (3.45)

We then define

K := K+ ∪K− where K± := x ∈ O : ±v (x) ≥ ǫ .

Since v is continuous, we find that K+ and K− are compact, disjoint and com-pactly contained in O. Define next η ∈ C (K) by

η (x) :=

1 if x ∈ K+

−1 if x ∈ K− .


Extend then η as a function in C0 (O) ⊂ C0 (Ω) so that |η (x)| ≤ 1 for everyx ∈ O, this is possible by Tietze extension theorem (cf. Rudin [518]).

Applying (3.44) and (3.45) we deduce that

|∫

O

vη | ≤ |∫

O

(v − u) η +

∫

O

uη | ≤ ǫ ‖η‖L∞(O) = ǫ . (3.46)

Observe now that∫

O

|v| =

∫

O−K

|v|+∫

K

vη =

∫

O−K

[ |v| − vη ] +

∫

O

vη

≤ 2

∫

O−K

|v|+ ǫ ≤ ǫ [2 meas (O −K) + 1] ≤ ǫ [2 measO + 1] ,

where we have used (3.46) and the fact that |v| ≤ ǫ in O −K.We therefore deduce from (3.45) and from the above inequality that

‖u‖L1(O) ≤ ‖v‖L1(O) + ‖u− v‖L1(O) ≤ 2ǫ[measO + 1]

and since ǫ is arbitrary, we have indeed obtained the theorem.

3.4.3 Some regularity results

The question of knowing if the minimizers, that we found to exist in a Sobolevspace, are in fact more regular, is the 19th of the famous problems of Hilbertand there is an extensive literature on this subject and refer to Giaquinta [307],Giusti [316], Ladyzhenskaya-Uraltseva [388] and Morrey [455].

Here we just mention some results without proof. We just consider thescalar case n > N = 1. The case N ≥ n = 1 is simpler and will be dealt with inChapter 4. For the vectorial case N, n > 1, we refer to the bibliography. Thefollowing theorem is Theorem IX.1.1 in Giaquinta [307].

Theorem 3.41 Let

I (u) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx

where f : Ω× R× Rn → R is a Caratheodory function satisfying

α1 |ξ|p − β |u|q − γ (x) ≤ f (x, u, ξ) ≤ α2 |ξ|p + β |u|q + γ (x)

where γ ∈ Ls (Ω) with s > n/p, α2 ≥ α1 > 0, β ≥ 0 and either 1 ≤ p < n and1 ≤ q < np

n−p or p = n and q ≥ 1. Assume that u ∈ W 1,p (Ω) is such that

I (u) ≤ I (u + ϕ)

for every ϕ ∈W 1,p (Ω) with supp ϕ ⊂⊂ Ω. Then u is (locally) Holder continuous(in particular, u is locally bounded).


Remark 3.42 (i) When p > n, the above result is trivial by the Sobolev imbed-ding theorem. The case p = n, is simpler than when p < n, see Remark 6.2 inGiusti [316]. Note also that no convexity hypothesis is required on f.

(ii) When looking at the global problem

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0 (Ω)

it is possible to extend the above result up to the boundary. Namely (seeTheorem 7.8 in Giusti [316]) if Ω ⊂ Rn is a bounded open set with a Lipschitzboundary and u0 is Holder continuous in Ω, then any minimizer is also Holdercontinuous in Ω. ♦

The result of Theorem 3.41 can now be improved if one assumes some moreproperties on the function f. We will not prove the following theorem and werefer to the above mentioned books (see in particular Theorem 1.10.4 in Morrey[455]). The improvements with respect to Theorem 3.41 are obtained using theEuler-Lagrange equations. We again recall that we are considering here onlythe scalar case N = 1.

Theorem 3.43 Let Ω ⊂ Rn be a bounded open set and f ∈ C∞ (Ω× R× Rn) ,f = f (x, u, ξ) . Let fx = (fx1, · · · , fxn) , fξ = (fξ1 , · · · , fξn) and similarly forthe higher derivatives. Let f satisfy, for every (x, u, ξ) ∈ Ω×R×Rn and λ ∈ Rn,

(A)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

α1Vp − α2 ≤ f (x, u, ξ) ≤ α3V

p,

|fξ| , |fxξ| , |fu| , |fxu| ≤ α3Vp−1, |fuξ| , |fu u| ≤ α3V

p−2,

α4Vp−2 |λ|2 ≤

n∑i,j=1

fξiξj (x, u, ξ)λiλj ≤ α5Vp−2 |λ|2

where p ≥ 2, V 2 = 1 + u2 + |ξ|2 and αi > 0, i = 1, · · · , 5, are constants (if

f (x, u, ξ) = f (x, ξ) , then take V 2 = 1 + |ξ|2 in (A)).

Then any minimizer of

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0 (Ω)

is in C∞ (D) , for every D ⊂ D ⊂ Ω and is analytic if f is analytic.

Remark 3.44 (i) The above results are strictly restricted to the scalar case, i.e.u : Rn → RN with N = 1 and they are false if N, n > 1, see the bibliography.

(ii) Note that the last condition in (A) is a kind of uniform convexity of fwith respect to the variable ξ and it ensures the ellipticity of the Euler-Lagrangeequation. ♦

Chapter 4

The one dimensional case

4.1 Introduction

In the one dimensional case, the results of Chapter 3, can be improved a greatdeal. The classical methods of the calculus of variations give some importantqualitative properties. Moreover, the regularity results are at the same timeeasier to obtain and more general.

We recall that we are considering

(P ) inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

,

where f : (a, b)× R× R → R is a Caratheodory function, p ≥ 1 and

X :=u ∈ W 1,p (a, b) , u (a) = α, u (b) = β

.

Most of the results that are given in the present chapter immediately extend tothe case where u : [a, b]→ RN with N ≥ 1.

The chapter is organized as follows.

In Section 4.2, we restate, without proof, the general existence theoremobtained in Chapter 3. We then recall the weak form of the Euler-Lagrangeequation.

In Section 4.3, we briefly discuss some aspects of the classical Euler-Lagrangeequation

(E)d

dx[fξ (x, u (x) , u′ (x))] = fu (x, u (x) , u′ (x)) , x ∈ [a, b]

and its integrated form

d

dx[f (x, u (x) , u′ (x))− u′ (x) fξ (x, u (x) , u′ (x))] = fx (x, u (x) , u′ (x)) .

120 The one dimensional case

This rewriting of the equation turns out to be particularly useful when f doesnot depend explicitly on the variable x. Indeed we then have a first integral of(E) that is

f (u (x) , u′ (x))− u′ (x) fξ (u (x) , u′ (x)) = constant, ∀x ∈ [a, b] .

We then discuss some classical examples.

In Section 4.4, we study two important inequalities, namely Poincare-Wirtinger inequality and its generalization, Wirtinger inequality, which is equiv-alent to the isoperimetric inequality.

In Section 4.5, we present the Hamiltonian formulation of the problem.Roughly speaking, the idea is that the solutions of (E) are also solutions (andconversely) of

(H)

u′ (x) = Hv (x, u (x) , v (x))

v′ (x) = −Hu (x, u (x) , v (x))

where v (x) = fξ (x, u (x) , u′ (x)) and H is the Legendre transform of f, namely

H (x, u, v) := supξ∈R

v ξ − f (x, u, ξ) .

In classical mechanics, f is called the Lagrangian and H the Hamiltonian.

In Section 4.6, we prove some simple and general regularity results.

Finally, in Section 4.7, we conclude with some remarks on Lavrentiev phe-nomenon. This phenomenon shows that substituting the space of admissiblefunctions by a dense one may give different values for the infimum.

We refer for more developments to the following books: Akhiezer [8], Bliss[84], Bolza [90], Buttazzo-Giaquinta-Hildebrandt [117], Caratheodory [121],Cesari [143], Courant [163], Courant-Hilbert [164], Dacorogna [180], Gelfand-Fomin [304], Giaquinta-Hildebrandt [309], Hestenes [337], Pars [490], Rund[520], Troutman [581] or Weinstock [598]. Our presentation closely follows theone in [180].

4.2 An existence theorem

We recall, without proof, the main existence theorem of Chapter 3.

Theorem 4.1 Let a q ≥ 1 and α1 > 0, α2, α3 ∈ R, such that, for almostevery x ∈ (a, b) and every (u, ξ) ∈ R× R,

f (x, u, ξ) ≥ α1 |ξ|p + α2 |u|q + α3 .

An existence theorem 121

Let

(P ) m := inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

,

X :=u ∈ W 1,p (a, b) , u (a) = α, u (b) = β

.

Assume that m < ∞. Then there exists u ∈ X a minimizer of (P ).

Furthermore, if (u, ξ) → f (x, u, ξ) is strictly convex for almost every x ∈(a, b), then the minimizer is unique.

Remark 4.2 (i) It is easy to see that uniqueness holds under a slightly weakercondition, namely that (u, ξ)→ f (x, u, ξ) is convex and either u→ f (x, u, ξ) isstrictly convex or ξ → f (x, u, ξ) is strictly convex.

(ii) This theorem has a long history and we refer to Chapter 3 for details.♦

We now discuss several examples showing that the theorem is nearly optimal,but let us first start with the prototype of examples where the theorem applies.

Example 4.3 A typical example is

f (x, u, ξ) =1

p|ξ|p + g (x, u) ,

where g is a non-negative Caratheodory function and p > 1. ♦

We now discuss several counterexamples showing that neither the coercivitynor the convexity of f can be weakened.

Example 4.4 We start with an example where f is neither convex nor coercive.Consider f (ξ) = e−ξ2

and

(P ) infu∈X

I (u) =

∫ 1

0

f (u′ (x)) dx

= m

where X = W 1,10 (0, 1) =

u ∈ W 1,1 (0, 1) : u (0) = u (1) = 0

. We now show

that (P ) has no minimizer. Assume for a moment that m = 0. Then, clearly,no function u ∈ X can satisfy

∫ 1

0

e−(u′(x))2

dx = 0

and hence (P ) has no solution. Let us now show that m = 0. Let ν ∈ N anddefine

uν (x) := ν(x− 1/2)2 − ν

4

then uν ∈ X and

I (uν) =

∫ 1

0

e−4ν2(x−1/2)2 dx =1

2ν

∫ ν

−ν

e−y2

dy → 0 as ν →∞.

Thus m = 0, as claimed. ♦


Example 4.5 This example is of the area type but easier and it also shows thatall the hypotheses of the theorem are satisfied except the coercivity (H2) which istrue with p = 1. This weakening of (H2) leads to the following counterexample.

Let f (x, u, ξ) = f (u, ξ) =√

u2 + ξ2 and

(P ) inf

I (u) =

∫ 1

0

f (u (x) , u′ (x)) dx : u ∈ X

= m

where X =u ∈ W 1,1 (0, 1) : u (0) = 0, u (1) = 1

. Let us prove that (P ) has

no solution. We first show that m = 1 and start by observing that m ≥ 1 since

I (u) ≥∫ 1

0

|u′ (x)| dx ≥∫ 1

0

u′ (x) dx = u (1)− u (0) = 1 .

To establish that m = 1, we construct a minimizing sequence uν ∈ X (ν beingan integer) as follows:

uν (x) :=

0 if x ∈

[0, 1− 1

ν

]

1 + ν (x− 1) if x ∈(1− 1

ν , 1].

We therefore have m = 1 since

1 ≤ I (uν) =

∫ 1

1− 1ν

√(1 + ν (x− 1))

2+ ν2 dx

≤ 1

ν

√1 + ν2 → 1 as ν →∞.

Assume now, for the sake of contradiction, that there exists u ∈ X a minimizerof (P ). We should then have, as above,

1 = I (u) =

∫ 1

0

√u2 + u′2 dx ≥

∫ 1

0

|u′| dx

≥∫ 1

0

u′dx = u (1)− u (0) = 1.

This implies that u = 0 a.e. in (0, 1). Since elements of X are continuous, wehave that u ≡ 0 and this is incompatible with the boundary data. Thus (P )has no solution. ♦Example 4.6 (Weierstrass example) Let f (x, u, ξ) = f (x, ξ) = xξ2 and

(P ) inf

I (u) =

∫ 1

0

f (x, u′ (x)) dx : u ∈ X

= m,

where X =u ∈W 1,2 (0, 1) : u (0) = 1, u (1) = 0

. All the hypotheses of the

theorem are verified with the exception of (H2), which degenerates only at onepoint. This is enough to show that (P ) has no minimizer in X. Let us first showthat m = 0. Let ν ∈ N and consider the sequence

uν (x) :=

1 if x ∈

[0, 1

ν

]

− log xlog ν if x ∈

(1ν , 1].

An existence theorem 123

We easily have

I (uν) =1

log ν→ 0 as ν →∞,

and hence m = 0. Now assume, by absurd hypothesis, that (P ) has a solutionu ∈ X. We should then have I (u) = 0, but since the integrand is non-negativewe deduce that u′ = 0 a.e. in (0, 1) . Since elements of X are continuous, wehave that u is constant, and this is incompatible with the boundary data. Hence(P ) has no solution. ♦

Example 4.7 (Poincare-Wirtinger inequality) The present example showsthat we cannot allow, in general, that q = p in (H2). Let λ > π and

f (x, u, ξ) = f (u, ξ) =1

2

(ξ2 − λ2u2

)

and

(P ) inf

I (u) =

∫ 1

0

f (u (x) , u′ (x)) dx : u ∈ W 1,20 (0, 1)

= m.

Clearly m = −∞, since, letting uα (x) = α sin πx with α ∈ R, we have

I (uα) = α2

∫ 1

0

[π2 cos2 (πx)− λ2 sin2 (πx)

]dx → −∞ as α →∞.

This means that (P ) has no solution. ♦

Example 4.8 (Bolza example) We now show that, as a general rule, onecannot weaken (H1) either. One such example was already seen above, where

we had f (x, u, ξ) = f (ξ) = e−ξ2

(which satisfied neither (H1) nor (H2)). Let

f (x, u, ξ) = f (u, ξ) =(ξ2 − 1

)2+ u4

(P ) inf

I (u) =

∫ 1

0

f (u (x) , u′ (x)) dx : u ∈ W 1,40 (0, 1)

= m.

Assume for a moment that we already proved that m = 0 and let us show that(P ) has no solution, using an argument by contradiction. Let u ∈ W 1,4

0 (0, 1)be a minimizer of (P ) ; i.e. I (u) = 0. This implies that u = 0 and |u′| = 1 a.e.in (0, 1) . Since the elements of W 1,4 are continuous, we have that u ≡ 0 andhence u′ ≡ 0 which is clearly absurd.

So let us show that m = 0 by constructing an appropriate minimizingsequence. Let uν ∈ W 1,4

0 (ν ≥ 2 being an integer; see Figure 4.1) definedon each interval [k/ν, (k + 1) /ν] , 0 ≤ k ≤ ν − 1, as follows

uν (x) :=

x− k

ν if x ∈[

2k2ν , 2k+1

2ν

]

−x + k+1ν if x ∈

(2k+12ν , 2k+2

2ν

].


1

2ν

uν(x)

x1

2ν1

ν3

2ν2

ν· · · 1

| | |

Figure 4.1: Minimizing sequence for Bolza example

Observe that |u′ν | = 1 a.e. and |uν | ≤ 1/ (2ν) , therefore leading to the desired

convergence, namely

0 ≤ I (uν) ≤ 1

(2ν)4 → 0, as ν →∞. ♦

The following elementary example shows that in absence of strict convexityone cannot expect uniqueness of minimizers.

Example 4.9 Let X =u ∈ W 1,1 (0, 1) : u (0) = 0, u (1) = 1

and

(P ) inf

I (u) =

∫ 1

0

|u′ (x)| dx : u ∈ X

= m .

By Jensen inequality, we have m = 1. Clearly any function u ∈ X with u′ ≥ 0,is a minimizer of (P ) . ♦

We now give two examples showing that, in general, solutions of (P ) are notsmooth, even if the integrand is smooth.

Example 4.10 Let f (ξ) =(ξ2 − 1

)2

(P ) infu∈X

I (u) =

∫ 1

0

f (u′ (x)) dx

= m

where X = W 1,10 (0, 1) . Clearly

v (x) :=

x if x ∈ [0, 1/2]

1− x if x ∈ (1/2, 1]

is a solution since v is piecewise C1 and satisfies v (0) = v (1) = 0 and I (v) = 0;thus m = 0. This readily implies that (P ) has no C1 solution. Indeed I (u) = 0

The Euler-Lagrange equation 125

implies that |u′| = 1 almost everywhere and no function u ∈ C1 can satisfy|u′| = 1 (since by continuity of the derivative we should have either u′ = 1everywhere or u′ = −1 everywhere and this is incompatible with the boundarydata). ♦

Example 4.11 This time the integrand is convex in the variable ξ. Letf (u, ξ) = u2 (1− ξ)

2and

(P ) inf

I (u) =

∫ 1

−1

f (u (x) , u′ (x)) dx : u ∈ X

= m

where X =u ∈W 1,1 (−1, 1) : u (−1) = 0, u (1) = 1

. Observe that

u (x) :=

0 if x ∈ [−1, 0]

x if x ∈ (0, 1]

is a solution of (P ). However, it is easy to see that (P ) has no C1 minimizer,since m = 0 and no u ∈ C1 ([−1, 1]) can satisfy I (u) = 0. ♦

4.3 The Euler-Lagrange equation

4.3.1 The classical and the weak forms

We first recall Theorem 3.37 and Corollary 3.38 of Chapter 3 applied to thepresent context.

Theorem 4.12 Let a < b, p ≥ 1 and f, fu , fξ : (a, b) × R × R → R, f =f (x, u, ξ) , be Caratheodory functions, where fξ = ∂f/∂ξ and fu = ∂f/∂u.Assume that at least one of the following two hypotheses hold.

(H3) For every R > 0, there exist α1 ∈ L1 (a, b) , α2 ∈ Lp/(p−1) (a, b) andβ = β (R) such that for almost every x ∈ (a, b) and every (u, ξ) ∈ [−R, R]× R,the following inequalities hold

|f (x, u, ξ)| , |fu (x, u, ξ)| ≤ α1 (x) + β |ξ|p , |fξ (x, u, ξ)| ≤ α2 (x) + β |ξ|p−1.

(H3’) For every R > 0, there exist α1 ∈ L1 (a, b) and β = β (R) such that foralmost every x ∈ (a, b) and every (u, ξ) ∈ [−R, R]×R, the following inequalitieshold:

|f (x, u, ξ)| , |fu (x, u, ξ)| , |fξ (x, u, ξ)| ≤ α1 (x) + β |ξ|p .

Let u ∈ X be a minimizer of

(P ) inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

= m,

X :=u ∈ W 1,p (a, b) , u (a) = α, u (b) = β

.


If (H3) holds, then u satisfies the weak form of the Euler-Lagrange equation

(Ew)

∫ b

a

[fu (x, u, u′) ϕ + fξ (x, u, u′)ϕ′] dx = 0, ∀ϕ ∈W 1,p0 (a, b) .

If (H3’) holds, then u satisfies the even weaker form of the Euler-Lagrangeequation

(E′w)

∫ b

a

[fu (x, u, u′)ϕ + fξ (x, u, u′)ϕ′] dx = 0, ∀ϕ ∈ C∞0 (a, b) .

Moreover if f ∈ C2 ([a, b]× R× Rn) and u ∈ C2 ([a, b]) then u satisfies theEuler-Lagrange equation

(E)d

dx[fξ (x, u, u′)] = fu (x, u, u′) , ∀x ∈ [a, b] .

Conversely, if (u, ξ) → f (x, u, ξ) is convex for almost every x ∈ (a, b) and if uis a solution of either (Ew), (E′

w), or (E) then it is a minimizer of (P ).

Remark 4.13 (i) The hypothesis (H3) corresponds to Growth condition (II) or(III) in Theorem 3.37, while (H3’) is just Growth condition (I). They both ensurethat fuϕ, fξϕ

′ ∈ L1 (a, b) . We also recall that hypothesis (H3) implies (H3’).

(ii) In the statement of the theorem, we do not need hypothesis (H1) or (H2)of Theorem 4.1. Therefore we do not use the convexity of f (naturally for theconverse we need the convexity of f). However, we require that a minimizer of(P ) does exist. ♦

We now discuss some simple examples.

Example 4.14 Consider the case where

f (x, u, ξ) = f (ξ) =1

p|ξ|p + g (x, u) .

The equation (E) becomes

d

dx

[|u′|p−2

u′]

= gu (x, u) , in (a, b) . ♦

Example 4.15 Let f (x, u, ξ) = f (ξ) . Then the Euler-Lagrange equation is

d

dx[f ′ (u′)] = 0, i.e. f ′ (u′) = constant.

Note that

u (x) =β − α

b− a(x− a) + α (4.1)

is a solution of the equation and furthermore satisfies the boundary conditionsu (a) = α, u (b) = β. It is not, however, always a minimizer of (P ) , as was seenin Example 4.4.


If f is convex the above u is indeed a minimizer. This follows from thetheorem but it can be seen in a more elementary way (which is also valid evenif f ∈ C0 (R)). From Jensen inequality, it follows that for any u ∈ W 1,∞ (a, b)with u (a) = α, u (b) = β

1

b− a

∫ b

a

f (u′ (x)) dx ≥ f(1

b− a

∫ b

a

u′ (x) dx) = f(u (b)− u (a)

b− a)

= f(β − α

b− a) = f (u′ (x))

=1

b− a

∫ b

a

f (u′ (x)) dx,

which is the claim.

(ii) Let f (x, u, ξ) = f (x, ξ) . The Euler-Lagrange equation is then

d

dx[fξ (x, u′)] = 0, i.e. fξ (x, u′) = constant.

The equation is already harder to solve than the preceding one and, in general,it does not have as simple a solution as in (4.1), see Example 4.33. ♦

We continue with two classical examples.

Example 4.16 (Brachistochrone) The function under consideration is

f (u, ξ) =

√1 + ξ2

√u

and

(P ) infu∈X

I (u) =

∫ b

0

f (u (x) , u′ (x)) dx

= m

where

X :=u ∈W 1,1 (0, b) : u (0) = 0, u (b) = β and u (x) > 0, ∀x ∈ (0, b]

.

The Euler-Lagrange equation and its first integral (see Theorem 4.20 below) are

[u′

√u√

1 + u′ 2

]′= −

√1 + u′ 2

2√

u3,

√1 + u′ 2√

u− u′[

u′√

u√

1 + u′ 2 ] = constant.

This leads (μ being a positive constant) to

u(1 + u′ 2) = 2μ .


The solution is a cycloid and is given in implicit form by

u (x) = μ(1− cos θ−1 (x)

)

whereθ (t) = μ (t− sin t) .

Note that u (0) = 0. It therefore remains to choose μ so that u (b) = β. ♦Example 4.17 (Minimal surfaces of revolution) The function under con-

sideration is f (u, ξ) = 2πu√

1 + ξ2 and the minimization problem (which cor-responds to minimization of the area of a surface of revolution) is

(P ) infu∈X

I (u) =

∫ b

a

f (u (x) , u′ (x)) dx

= m,

where, for α, β > 0, we set

X :=u ∈W 1,1 (a, b) : u (a) = α, u (b) = β, u > 0

.

The Euler-Lagrange equation and its first integral (see Theorem 4.20 below) are[

u′u√1 + u′ 2

]′=√

1 + u′ 2 ⇔ u′′u = 1 + u′ 2,

u√

1 + u′ 2 − u′ u′u√1 + u′ 2 = λ = constant.

This leads to

u′ 2 =u2

λ2− 1.

The solutions, if they exist (this depends on a, b, α and β), are of the form (μbeing a constant)

u (x) = λ cosh(x

λ+ μ). ♦

It is clear that if the function (u, ξ) → f (x, u, ξ) is not convex for everyx ∈ [a, b] , then, in general, a solution u of the Euler-Lagrange equation (E) isnot a minimizer of (P ). However, an important part of the classical calculusof variations is devoted to the fields theories , which sometimes allows us inthe absence of the convexity of (u, ξ) → f (x, u, ξ) to prove that a solutionof the Euler-Lagrange equation is a minimizer of (P ). We do not discuss thisapproach here, but we try to explain the nature of the theory with a particularlysimple case that is given in the next theorem and that turns out to be useful inSection 4.4.1.

Theorem 4.18 Let f ∈ C2 ([a, b]× R× R). If there exists Φ ∈ C3 ([a, b]× R)with Φ (a, α) = Φ (b, β) such that

(u, ξ)→ f (x, u, ξ) is convex for every x ∈ [a, b] ,


wheref (x, u, ξ) := f (x, u, ξ) + Φu (x, u) ξ + Φx (x, u) ;

then any solution u of (E) is a minimizer of (P ) .

Remark 4.19 We should immediately point out that in order to have (u, ξ)→f (x, u, ξ) convex for every x ∈ [a, b] , we should, at least, have that ξ →f (x, u, ξ) is convex for every (x, u) ∈ [a, b]× R. If (u, ξ) → f (x, u, ξ) is alreadyconvex, then choose Φ ≡ 0 and apply Theorem 4.12. ♦Proof. Define

ϕ (x, u, ξ) := Φu (x, u) ξ + Φx (x, u) .

Observe that the two following identities (the first one uses that Φ (a, α) =Φ (b, β) and the second one is just straight differentiation)

∫ b

a

d

dx[Φ (x, u (x))] dx = Φ (b, β)− Φ (a, α) = 0

d

dx[ϕξ (x, u, u′)] = ϕu (x, u, u′) , x ∈ [a, b]

hold for any u ∈ X =u ∈ C1 ([a, b]) : u (a) = α, u (b) = β

. The first iden-

tity expresses that the integral is invariant, while the second one says thatϕ (x, u, u′) satisfies the Euler-Lagrange equation identically (it is then called anull Lagrangian).

With the help of the above observations we immediately obtain the result byapplying Theorem 4.12 to f . Indeed we have that (u, ξ)→ f (x, u, ξ) is convex ,

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx =

∫ b

a

f (x, u (x) , u′ (x)) dx

for every u ∈ X and any solution u of (E) satisfies also

( E )d

dx[fξ (x, u, u′)] = fu (x, u, u′) , x ∈ (a, b) .

This concludes the proof.

4.3.2 Second form of the Euler-Lagrange equation

The next theorem gives a different way of expressing the Euler-Lagrange equa-tion, this new equation is sometimes called the DuBois-Reymond equation. Itturns out to be an important help when f does not depend explicitly on x, asalready seen in some of the above examples.

Theorem 4.20 Let f ∈ C2 ([a, b]× R× R) , f = f (x, u, ξ) , and

(P ) infu∈X

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx

= m,


where X =u ∈ C1 ([a, b]) : u (a) = α, u (b) = β

. Let u ∈ X ∩ C2 ([a, b]) be a

minimizer of (P ) . Then, for every x ∈ [a, b] , the following equation holds:

d

dx[f (x, u (x) , u′ (x))− u′ (x) fξ (x, u (x) , u′ (x))] = fx (x, u (x) , u′ (x)) . (4.2)

Remark 4.21 The theorem is particularly interesting when f does not dependexplicitly on x, namely f = f (u, ξ) . We then have

f (u (x) , u′ (x))− u′ (x) fξ (u (x) , u′ (x)) = constant, x ∈ (a, b) . ♦Proof. We will give two different proofs of the theorem. The first one isvery elementary and uses the Euler-Lagrange equation. The second one is moreinvolved but has several advantages, notably that it can be derived under weakerregularity hypotheses on the minimizer u.

Proof 1. Observe first that for any u ∈ C2 ([a, b]) we have, by straightdifferentiation,

d

dx[f (x, u, u′)− u′fξ (x, u, u′)]

= fx (x, u, u′) + u′[fu (x, u, u′)− d

dx[fξ (x, u, u′)]] .

By Theorem 4.12 we know that any solution u of (P ) satisfies the Euler-Lagrangeequation

d

dx[fξ (x, u (x) , u′ (x))] = fu (x, u (x) , u′ (x))

hence combining the two identities we have the result.

Proof 2. We will use a technique known as variations of the independentvariables; the classical derivation of Euler-Lagrange equation can be seen as atechnique of variations of the dependent variables.

Let ǫ ∈ R, ϕ ∈ C∞0 (a, b) , λ = (2 ‖ϕ′‖L∞)

−1and

ξ (x, ǫ) = x + ǫλϕ (x) = y.

Observe that for |ǫ| ≤ 1, then ξ (., ǫ) : [a, b] → [a, b] is a diffeomorphism withξ (a, ǫ) = a, ξ (b, ǫ) = b and ξx (x, ǫ) > 0. Let η (., ǫ) : [a, b]→ [a, b] be its inverse,i.e.

ξ (η (y, ǫ) , ǫ) = y.

Since

ξx (η (y, ǫ) , ǫ) ηy (y, ǫ) = 1 and ξx (η (y, ǫ) , ǫ) ηǫ (y, ǫ) + ξǫ (η (y, ǫ) , ǫ) = 0,

we find (O (t) stands for a function f so that |f (t) /t| is bounded in a neighbor-hood of t = 0)

ηy (y, ǫ) = 1− ǫλϕ′ (y) + O(ǫ2)

ηǫ (y, ǫ) = −λϕ (y) + O (ǫ) .


Set for u a minimizer of (P )

uǫ (x) = u (ξ (x, ǫ)) .

Note that, performing also a change of variables y = ξ (x, ǫ) ,

I (uǫ) =

∫ b

a

f(x, uǫ (x) , (uǫ)

′(x))

dx

=

∫ b

a

f (x, u (ξ (x, ǫ)) , u′ (ξ (x, ǫ)) ξx (x, ǫ)) dx

=

∫ b

a

f (η (y, ǫ) , u (y) , u′ (y) /ηy (y, ǫ)) ηy (y, ǫ) dy .

Denoting by g (ǫ) the last integrand, we get

g′ (ǫ) = ηyǫf + [ fxηǫ −ηyǫ

η2y

u′fξ ]ηy

which leads tog′ (0) = λ [−fxϕ + (u′fξ − f)ϕ′] .

Since by hypothesis u is a minimizer of (P ) and uǫ ∈ X, we have I (uǫ) ≥ I (u)and hence

0 =d

dǫI (uǫ)

∣∣∣∣ǫ=0

= λ

∫ b

a

−fx (x, u (x) , u′ (x)) ϕ (x)

+ [u′ (x) fξ (x, u (x) , u′ (x))− f (x, u (x) , u′ (x))] ϕ′ (x) dx

= λ

∫ b

a

ϕ (x) −fx (x, u (x) , u′ (x))

+d

dx[−u′ (x) fξ (x, u (x) , u′ (x)) + f (x, u (x) , u′ (x))] dx .

Invoking Theorem 3.40, we have indeed obtained the claim.

One should note, as seen in the following example, that it might happenthat a solution of (4.2) is not necessarily a solution of the Euler-Lagrangeequation (E) .

Example 4.22 Let

f (x, u, ξ) = f (u, ξ) =1

2ξ2 − u.

The second form of the Euler-Lagrange equation is

0 =d

dx[f (u (x) , u′ (x))− u′ (x) fξ (u (x) , u′ (x))] = −u′ (x) [u′′ (x) + 1] ,

and it is satisfied by u ≡ 1. However, u ≡ 1 does not verify the Euler-Lagrangeequation, which is in the present case

u′′ (x) = −1. ♦


4.4 Some inequalities

4.4.1 Poincare-Wirtinger inequality

Theorem 4.23 (Poincare-Wirtinger inequality) For every u ∈W 1,20 (a, b),

the following inequality holds

∫ b

a

u′ 2 dx ≥ (π

b− a)2∫ b

a

u2 dx.

Proof. By a change of variable we immediately reduce the study to the casea = 0 and b = 1 and we therefore have to prove that

∫ 1

0

u′ 2 dx ≥ π2

∫ 1

0

u2 dx, for every u ∈W 1,20 (0, 1) .

We will in fact prove that, for every 0 ≤ λ < π,

∫ 1

0

u′ 2 dx ≥ λ2

∫ 1

0

u2 dx, for every u ∈W 1,20 (0, 1) .

An elementary passage to the limit leads to Poincare-Wirtinger inequality. Fora different proof of a slightly more general form of Poincare-Wirtinger inequalitysee Theorem 4.24.

We first let fλ (u, ξ) :=(ξ2 − λ2u2

)/2 and

Iλ (u) :=

∫ 1

0

fλ (u (x) , u′ (x)) dx.

We then apply Theorem 4.18, with

Φ (x, u) :=λ

2tan[λ(x − 1

2)]u2,

f (x, u, ξ) :=1

2ξ2 + λ tan[λ(x− 1

2)]uξ +

λ2

2tan2[λ(x − 1

2)]u2

and observe that Φ satisfies all the properties of Theorem 4.18. It is easy to seethat (u, ξ)→ f (x, u, ξ) is convex and therefore applying Theorem 4.18 we havethat, for every 0 ≤ λ < π,

Iλ (u) ≥ Iλ (0) , ∀u ∈ X,

which is the claim.

4.4.2 Wirtinger inequality

The Wirtinger inequality is a generalization of the Poincare-Wirtinger one. Itturns out to be equivalent to the isoperimetric inequality; this will be brieflydiscussed below.

Some inequalities 133

We first introduce the following notation, for any p ≥ 1,

W 1,pper (a, b) :=

u ∈W 1,p (a, b) : u (a) = u (b)

.

Theorem 4.24 (Wirtinger inequality) Let

X :=

u ∈W 1,2

per (a, b) :

∫ b

a

u (x) dx = 0

.

Then ∫ b

a

u′ 2 dx ≥ (2π

b− a)2∫ b

a

u2 dx, ∀u ∈ X .

Furthermore, equality holds if and only if there exist α, β ∈ R such that

u (x) = α cos2πx

b− a+ β sin

2πx

b− a.

Remark 4.25 (i) The inequality can also be generalized (see Croce-Dacorogna[168]) to

(∫ b

a

|u′|p dx

)1/p

≥ α (p, q, r)

(∫ b

a

|u′|q dx

)1/q

, ∀u ∈ X

for some appropriate α (p, q, r) (in particular, α (2, 2, 2) = 2π/ (b− a)) andwhere

X :=

u ∈ W 1,p

per (a, b) :

∫ b

a

|u (x)|r−2u (x) dx = 0

.

(ii) The above inequality is a generalization of Theorem 4.23, namely

∫ d

c

v′ 2 dx ≥ (π

d− c)2∫ d

c

v2 dx, ∀v ∈W 1,20 (c, d) .

The Poincare-Wirtinger inequality can be inferred from the theorem by settingb = d, a = 2c− d and

u (x) :=

v (x) if x ∈ (c, d)

−v (2c− x) if x ∈ (2c− d, c) .♦

Proof. By a change of variable, we immediately reduce the study to the casea = −1 and b = 1 and we therefore have to prove that if

X :=

u ∈W 1,2

per (−1, 1) :

∫ 1

−1

u (x) dx = 0

then ∫ 1

−1

u′ 2 dx ≥ π2

∫ 1

−1

u2 dx, ∀u ∈ X .


We give here two proofs.

Proof 1. The first proof is the classical one of Hurwitz. We divide it intothree steps.

Step 1. We start by proving the theorem under the further restriction thatu ∈ X ∩C2 [−1, 1] . We express u in Fourier series

u (x) =

∞∑

n=1

[an cosnπx + bn sin nπx] .

Note that there is no constant term since∫ 1

−1u (x) dx = 0. We know at the

same time that

u′ (x) = π∞∑

n=1

[−nan sin nπx + nbn cosnπx] .

We can now invoke Parseval formula to get∫ 1

−1

u2 dx =

∞∑

n=1

(a2

n + b2n

)and

∫ 1

−1

u′ 2 dx = π2∞∑

n=1

(a2

n + b2n

)n2.

The desired inequality follows then at once∫ 1

−1

u′ 2 dx ≥ π2

∫ 1

−1

u2 dx, ∀u ∈ X ∩C2.

Moreover equality holds if and only if an = bn = 0, for every n ≥ 2. This impliesthat equality holds if and only if u (x) = α cosπx + β sinπx, for any α, β ∈ R,as claimed.

Step 2. We now show that we can remove the restriction u ∈ X ∩C2 [−1, 1] .By the usual density argument we can find for every u ∈ X a sequence uν ∈X ∩C2 [−1, 1] so that

uν → u in W 1,2 (−1, 1) .

Therefore, for every ǫ > 0, we can find ν sufficiently large so that∫ 1

−1

u′ 2 dx ≥∫ 1

−1

u′ 2ν dx− ǫ and

∫ 1

−1

u2ν dx ≥

∫ 1

−1

u2 dx− ǫ.

Combining these inequalities with Step 1 we find∫ 1

−1

u′ 2 dx ≥ π2

∫ 1

−1

u2 dx−(π2 + 1

)ǫ.

Letting ǫ→ 0 we have indeed obtained the inequality.

Step 3. We still need to see that equality in X holds if and only if u (x) =α cosπx + β sin πx, for any α, β ∈ R. This has been proved in Step 1 only ifu ∈ X ∩ C2 [−1, 1] . Since the minimum in (P ) is attained by u ∈ X, we have,for any v ∈ X ∩ C∞

0 (−1, 1) and any ǫ ∈ R, that

I (u + ǫv) ≥ I (u) .

Some inequalities 135

Therefore the Euler-Lagrange equation is satisfied, namely∫ 1

−1

(u′v′ − π2uv

)dx = 0, ∀v ∈ X ∩ C∞

0 (−1, 1) . (4.3)

Let us transform it in a more classical way and choose a function f ∈ C∞0 (−1, 1)

with∫ 1

−1 f = 1 and let ϕ ∈ C∞0 (−1, 1) be arbitrary. Set

v (x) := ϕ (x)− (

∫ 1

−1

ϕdx)f (x) and λ := − 1

π2

∫ 1

−1

(u′f ′ − π2uf

)dx.

Observe that v ∈ X ∩C∞0 (−1, 1) . Use (4.3), the fact that

∫ 1

−1 f = 1,∫ 1

−1 v = 0and the definition of λ to get, for every ϕ ∈ C∞

0 (−1, 1) ,∫ [

u′ϕ′ − π2 (u− λ) ϕ]

=

∫[u′(v′ + f ′

∫ϕ)− π2u(v + f

∫ϕ)] + π2λ

∫ϕ

=

∫ (u′v′ − π2uv

)+ [

∫ϕ][π2λ +

∫ (u′f ′ − π2uf

)]

= 0.

The regularity of u (which is a minimizer of (P ) in X) then follows (as inTheorem 4.36) at once from the above equation. Since we know (from Step 1)that among smooth minimizers of (P ) the only ones are of the form u (x) =α cosπx + β sin πx, we have the result.

Proof 2. An alternative proof, due to H. Lewy (cf. Hardy-Littlewood-Polya[334], page 185), more in the spirit of Section 4.4.1, is now discussed. Let u ∈ Xwhere

X :=

u ∈ W 1,2 (−1, 1) : u (−1) = u (1) with

∫ 1

−1

u = 0

.

Definez (x) := u (x + 1)− u (x)

and note that z (−1) = −z (0) , since u (−1) = u (1). We deduce that we canfind α ∈ (−1, 0] so that z (α) = 0, which means that u (α + 1) = u (α) . Wedenote this common value by a, namely

a := u (α + 1) = u (α) .

Since u ∈ W 1,2 (−1, 1) it is easy to see that the function

v (x) := (u (x)− a)2cot (π (x− α))

vanishes at x = α and x = α + 1 (this follows from Holder inequality). Wetherefore have (recalling that u (−1) = u (1))

∫ 1

−1

u′ 2 − π2 (u− a)2 − (u′ − π (u− a) cotπ (x− α))2 dx

= π[(u (x) − a)

2cot (π (x− α))

]1−1

= 0.


Since∫ 1

−1 u = 0, we get from the above identity that

∫ 1

−1

(u′ 2 − π2u2

)dx = 2π2a2 +

∫ 1

−1

(u′ − π (u− a) cotπ (x− α))2dx

and hence Wirtinger inequality follows. Moreover we have equality in Wirtingerinequality if and only if a = 0 and

u′ = πu cotπ (x− α) ⇔ u = c sinπ (x− α)

where c is a constant.

We get the following as a direct consequence of the theorem.

Corollary 4.26 The following inequality holds

∫ 1

−1

(u′ 2 + v′ 2

)dx ≥ 2π

∫ 1

−1

uv′ dx, ∀u, v ∈W 1,2per (−1, 1) .

Furthermore equality holds if and only if

(u (x)− r1)2

+ (v (x)− r2)2

= r23 , ∀x ∈ [−1, 1]

where r1, r2, r3 ∈ R are constants.

Proof. We first observe that if we replace u by u − r1 and v by v − r2 theinequality remains unchanged, therefore we can assume that

∫ 1

−1

u dx =

∫ 1

−1

v dx = 0

and hence that

u, v ∈ X :=

u ∈W 1,2

per (−1, 1) :

∫ 1

−1

u (x) dx = 0

.

We write the inequality in the equivalent form

∫ 1

−1

(u′ 2 + v′ 2 − 2πuv′

)dx =

∫ 1

−1

(v′ − πu)2

dx +

∫ 1

−1

(u′ 2 − π2u2

)dx ≥ 0 .

From Theorem 4.24 we deduce that the second term in the above inequality isnon negative while the first one is trivially non negative; thus the inequality isestablished.

We now discuss the equality case. If equality holds we should have

v′ = πu and

∫ 1

−1

(u′ 2 − π2u2

)dx = 0

which implies from Theorem 4.24 that

u (x) = α cosπx + β sin πx and v (x) = α sinπx − β cosπx.

Hamiltonian formulation 137

Since we can replace u by u− r1 and v by v − r2, we have that

(u (x)− r1)2

+ (v (x)− r2)2

= r23 , ∀x ∈ [−1, 1]

as wished.

We now briefly discuss the implication of the Wirtinger inequality and itscorollary. It is easily shown that they are equivalent (see below) to the isoperi-metric inequality, which states that

[L (∂A)]2 − 4πM (A) ≥ 0,

where for A ⊂ R2 a bounded open set whose boundary, ∂A, is a sufficientlyregular simple closed curve, L (∂A) denotes the length of the boundary andM (A) the measure (the area) of A. Furthermore, equality holds if and only ifA is a disk (i.e., ∂A is a circle).

The sketch of the proof of the claim is as follows. We first parametrize theboundary ∂A by u, v ∈ W 1,2

per (−1, 1) , so that the length and area are given by

L (∂A) = L (u, v) =

∫ 1

−1

√u′ 2 + v′ 2 dx

M (A) = M (u, v) =1

2

∫ 1

−1

(uv′ − vu′) dx =

∫ 1

−1

uv′ dx.

We next reparametrize the curve by a multiple of its arc length so that

[L (u, v)]2

= 2

∫ 1

−1

(u′ 2 + v′ 2

)dx

and then use the corollary to get the result.

There are several articles and books devoted to the isoperimetric inequalityin any dimension, we recommend the review article of Osserman [487] and thebooks by Berger [77], Blaschke [83], Dacorogna [180], Federer [275], Hardy-Littlewood-Polya [334] (for the two dimensional case) and Webster [597].

4.5 Hamiltonian formulation

Recall that we are considering functions f : [a, b]× R× R → R, f = f (x, u, ξ) ,and

I (u) :=

∫ b

a

f (x, u (x) , u′ (x)) dx .

The Euler-Lagrange equation is

(E)d

dx[fξ (x, u, u′)] = fu (x, u, u′) , x ∈ [a, b] .

We have seen in the preceding sections that a minimizer of I, if it is sufficientlyregular, is also a solution of (E). The aim of this section is to show that, in


certain cases, solving (E) is equivalent to finding stationary points of a differentfunctional, namely

J (u, v) :=

∫ b

a

[u′ (x) v (x)−H (x, u (x) , v (x))] dx

whose Euler-Lagrange equations are

(H)

u′ (x) = Hv (x, u (x) , v (x))

v′ (x) = −Hu (x, u (x) , v (x)) .

The function H is called the Hamiltonian and is defined as the Legendre trans-form of f, namely


v ξ − f (x, u, ξ) .

Sometimes the system (H) is called the canonical form of the Euler-Lagrangeequation.

We start our analysis with a lemma.

Lemma 4.27 Let k ≥ 2, f ∈ Ck ([a, b]× R× R) , f = f (x, u, ξ) , such that

fξξ (x, u, ξ) > 0, for every (x, u, ξ) ∈ [a, b]× R× R, (4.4)

lim|ξ|→∞

f (x, u, ξ)

|ξ| = +∞, for every (x, u) ∈ [a, b]× R. (4.5)

LetH (x, u, v) := sup

ξ∈R

v ξ − f (x, u, ξ) . (4.6)

Then H ∈ Ck ([a, b]× R× R) and

Hx (x, u, v) = −fx (x, u, Hv (x, u, v)) , (4.7)

Hu (x, u, v) = −fu (x, u, Hv (x, u, v)) , (4.8)

H (x, u, v) = v Hv (x, u, v)− f (x, u, Hv (x, u, v)) , (4.9)

v = fξ (x, u, ξ) ⇔ ξ = Hv (x, u, v) . (4.10)

Remark 4.28 (i) The lemma remains partially true if we replace the hypothesis(4.4) by the weaker condition

ξ → f (x, u, ξ) is strictly convex.

In general, however the function H is only C1, as the following simple exampleshows

f (x, u, ξ) =1

4|ξ|4 and H (x, u, v) =

3

4|v|4/3

.

(See also Example 4.31.)


(ii) The lemma also remains valid if the hypothesis (4.5) does not hold butthen, in general, H is no longer finite everywhere as the following simple examplesuggests. Consider the strictly convex function

f (x, u, ξ) = f (ξ) =√

1 + ξ2

and observe that

H (v) =

−√

1− v2 if |v| ≤ 1

+∞ if |v| > 1.♦

Proof. We only discuss the case k = 2, the general one, k ≥ 2, being handledsimilarly. We divide the proof into several steps.

Step 1. Fix (x, u) ∈ [a, b] × R. From the definition of H and from (4.5) wededuce that there exists ξ = ξ (x, u, v) such that

H (x, u, v) = v ξ − f (x, u, ξ)

v = fξ (x, u, ξ) .(4.11)

Step 2. The function H is easily seen to be continuous. Indeed let (x, u, v),(x′, u′, v′) ∈ [a, b]× R× R, using (4.11) we find ξ = ξ (x, u, v) such that

H (x, u, v) = v ξ − f (x, u, ξ) .

Appealing to the definition of H we also have

H (x′, u′, v′) ≥ v′ ξ − f (x′, u′, ξ) .

Combining the two facts we get

H (x, u, v)−H (x′, u′, v′) ≤ (v − v′) ξ + f (x′, u′, ξ)− f (x, u, ξ) ,

since the reverse inequality is obtained similarly, we deduce the continuity of Hfrom the one of f.

Step 3. The inverse function theorem, the fact that f ∈ C2 and the inequality(4.4) imply that ξ ∈ C1. Let us however discuss it in details. First let us provethat ξ is continuous (in fact locally Lipschitz). Let R > 0 be fixed. From (4.5)we deduce that we can find R1 > 0 so that

|ξ (x, u, v)| ≤ R1 , for every x ∈ [a, b] , |u| , |v| ≤ R.

Since fξ is C1, we can find γ1 > 0 so that

|fξ (x, u, ξ)− fξ (x′, u′, ξ′)| ≤ γ1 (|x− x′|+ |u− u′|+ |ξ − ξ′|) (4.12)

for every x, x′ ∈ [a, b] , |u| , |u′| ≤ R, |ξ| , |ξ′| ≤ R1 .


From (4.4), we find that there exists γ2 > 0 so that

fξξ (x, u, ξ) ≥ γ2 , for every x ∈ [a, b] , |u| ≤ R, |ξ| ≤ R1

and we thus have, for every x ∈ [a, b] , |u| ≤ R, |ξ| , |ξ′| ≤ R1 ,

|fξ (x, u, ξ)− fξ (x, u, ξ′)| ≥ γ2 |ξ − ξ′| . (4.13)

Let x, x′ ∈ [a, b] , |u| , |u′| ≤ R, |v| , |v′| ≤ R. By definition of ξ we have

fξ (x, u, ξ (x, u, v)) = v and fξ (x′, u′, ξ (x′, u′, v′)) = v′,

which leads to

fξ (x, u, ξ (x′, u′, v′)) − fξ (x, u, ξ (x, u, v))

= fξ (x, u, ξ (x′, u′, v′))− fξ (x′, u′, ξ (x′, u′, v′)) + v′ − v

Combining this identity with (4.12) and (4.13) we get

γ2 |ξ (x, u, v)− ξ (x′, u′, v′)| ≤ γ1 (|x− x′|+ |u− u′|) + |v − v′|

which, indeed, establishes the continuity of ξ.

We now show that ξ is in fact C1. From the equation v = fξ (x, u, ξ) wededuce that ⎧

⎪⎨⎪⎩

fxξ (x, u, ξ) + fξξ (x, u, ξ) ξx = 0

fuξ (x, u, ξ) + fξξ (x, u, ξ) ξu = 0

fξξ (x, u, ξ) ξv = 1.

Since (4.4) holds and f ∈ C2, we deduce that ξ ∈ C1 ([a, b]× R× R) .

Step 4. We therefore have that the functions

(x, u, v) → ξ (x, u, v) , fx (x, u, ξ (x, u, v)) , fu (x, u, ξ (x, u, v))

are C1. We then immediately obtain (4.7), (4.8), and thus H ∈ C2. Indeed wehave, differentiating (4.11),

⎧⎪⎨⎪⎩

Hx = vξx − fx − fξ ξx = (v − fξ) ξx − fx = −fx

Hu = vξu − fu − fξ ξu = (v − fξ) ξu − fu = −fu

Hv = ξ + vξv − fξ ξv = (v − fξ) ξv + ξ = ξ

and in particular

ξ = Hv (x, u, v) .

This achieves the proof of the lemma.

The main theorem of the present section is the following.


Theorem 4.29 Let f and H be as in the above lemma. Let (u, v) ∈ C2 ([a, b])×C2 ([a, b]) satisfy for every x ∈ [a, b]

(H)

u′ (x) = Hv (x, u (x) , v (x))

v′ (x) = −Hu (x, u (x) , v (x)) .

Then u verifies

(E)d

dx[fξ (x, u (x) , u′ (x))] = fu (x, u (x) , u′ (x)) , ∀x ∈ [a, b] .

Conversely, if u ∈ C2 ([a, b]) satisfies (E) , then (u, v) are solutions of (H) where

v (x) = fξ (x, u (x) , u′ (x)) , ∀x ∈ [a, b] .

Proof. Part 1. Let (u, v) satisfy (H). Using (4.10) and (4.8) we get

u′ = Hv (x, u, v) ⇔ v = fξ (x, u, u′)

v′ = −Hu (x, u, v) = fu (x, u, u′)

and thus u satisfies (E).

Part 2. Conversely by (4.10) and since v = fξ (x, u, u′) we get the firstequation

u′ = Hv (x, u, v) .

Moreover since v = fξ (x, u, u′) and u satisfies (E), we have

v′ =d

dx[v] =

d

dx[fξ (x, u, u′)] = fu (x, u, u′) .

The second equation follows then from the combination of the above identityand (4.8).

Example 4.30 Let m > 0, g ∈ C1 ([a, b]) and

f (x, u, ξ) =m

2ξ2 − g (x)u.

The integral under consideration is

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx

and the associated Euler-Lagrange equation is

mu′′ (x) = −g (x) , x ∈ (a, b) .


The Hamiltonian is then

H (x, u, v) =v2

2m+ g (x) u

while the associated Hamiltonian system is

u′ (x) = v (x) /m

v′ (x) = −g (x) .♦

Example 4.31 We now generalize the preceding example. Let p > 1 andp′ = p/ (p− 1) ,

f (x, u, ξ) =1

p|ξ|p − g (x, u) and H (x, u, v) =

1

p′|v|p

′

+ g (x, u) .

The Euler-Lagrange equation and the associated Hamiltonian system are

d

dx

[|u′|p−2

u′]

= −gu (x, u)

and u′ = |v|p

′−2v

v′ = −gu (x, u) .♦

Example 4.32 Consider the simplest case, where f (x, u, ξ) = f (ξ) with f ′′ >0 (or more generally f is strictly convex) and lim|ξ|→∞ f (ξ) /ξ = +∞. TheEuler-Lagrange equation and its integrated form are

d

dx[f ′ (u′)] = 0 ⇒ f ′ (u′) = λ = constant.

The Hamiltonian is given by

H (v) = f∗ (v) = supξvξ − f (ξ) .

The associated Hamiltonian system is

u′ = f∗′ (v)

v′ = 0.

We find trivially that, denoting by λ and μ some constants, v′ = λ and hence

u (x) = f∗′ (λ) x + μ. ♦

Example 4.33 We now look for the slightly more involved case wheref (x, u, ξ) = f (x, ξ) with the appropriate hypotheses. The Euler-Lagrange equa-tion and its integrated form are

d

dx[fξ (x, u′)] = 0 ⇒ fξ (x, u′) = λ = constant.

Regularity 143

The Hamiltonian of f is given by

H (x, v) = supξvξ − f (x, ξ) .


u′ (x) = Hv (x, v (x))

v′ = 0.

The solution is then given by v = λ = constant and u′ (x) = Hv (x, λ) . ♦

Example 4.34 We next consider the more difficult case where f (x, u, ξ) =f (u, ξ) with the hypotheses of the theorem. The Euler-Lagrange equation andits integrated form are

d

dx[fξ (u, u′)] = fu (u, u′) ⇒ f (u, u′)− u′fξ (x, u′) = λ = constant.

The Hamiltonian of f is given by

H (u, v) = supξvξ − f (u, ξ) with v = fξ (u, ξ) .


u′ (x) = Hv (u (x) , v (x))

v′ (x) = −Hu (u (x) , v (x)) .

The Hamiltonian system has also a first integral given by

d

dx[H (u (x) , v (x))] = Hu (u, v)u′ + Hv (u, v) v′ ≡ 0.

In physical terms, we can say that if the Lagrangian f is independent of thevariable x (which here is the time), the Hamiltonian H is constant along thetrajectories. ♦

4.6 Regularity

Let us restate the problem. We consider

(P ) inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

= m,

where X :=u ∈W 1,p (a, b) : u (a) = α, u (b) = β

,

f : [a, b]× R× R → R, f = f (x, u, ξ) ,

is a Caratheodory function.

We have seen (see Theorems 4.1 and 4.12) that if f satisfies


(H1) ξ → f (x, u, ξ) is convex for almost every x ∈ (a, b) and every u ∈ R;

(H2) there exist p > q ≥ 1 and α1 > 0, α2 , α3 ∈ R such that, for almostevery x ∈ (a, b) and every (u, ξ) ∈ R× R,

f (x, u, ξ) ≥ α1 |ξ|p + α2 |u|q + α3 ,

then (P ) has a solution u ∈ X.

If, furthermore, f ∈ C1 ([a, b]× R× R) and verifies that

(H3’) for every R > 0, there exists α4 = α4 (R) such that, for every (x, u, ξ) ∈[a, b]× [−R, R]× R,

|f (x, u, ξ)| , |fu (x, u, ξ)| , |fξ (x, u, ξ)| ≤ α4 (1 + |ξ|p) ,

then any minimizer u ∈ X satisfies the weak form of the Euler-Lagrange equa-tion

(Ew)

∫ b

a

[fu (x, u, u′) v + fξ (x, u, u′) v′] dx = 0, ∀v ∈ C∞0 (a, b) .

We will show that under some strengthening of the hypotheses, we have that iff ∈ C∞, then u ∈ C∞. These results are, in part, also valid if u : [a, b] → RN

for N > 1.

We start with a lemma.

Lemma 4.35 Let f ∈ C1 ([a, b]× R× R) satisfy (H1), (H2) and (H3’). Thenany minimizer u ∈ W 1,p (a, b) of (P ) is in fact in W 1,∞ (a, b) and the Euler-Lagrange equation holds almost everywhere, i.e.

d

dx[fξ (x, u, u′)] = fu (x, u, u′) , a.e. x ∈ (a, b) .

Proof. We know from Theorem 4.12 that the following equation holds

(Ew)

∫ b

a

[fu (x, u, u′) v + fξ (x, u, u′) v′] dx = 0, ∀v ∈ C∞0 (a, b) . (4.14)

We then divide the proof into two steps.

Step 1. Define

ϕ (x) := fξ (x, u (x) , u′ (x)) and ψ (x) := fu (x, u (x) , u′ (x)) .

We easily see that ϕ ∈ W 1,1 (a, b) and that ϕ′ (x) = ψ (x) , for almost everyx ∈ (a, b) , which means that

d

dx[fξ (x, u, u′)] = fu (x, u, u′) , a.e. x ∈ (a, b) . (4.15)

Regularity 145

Indeed since u ∈ W 1,p (a, b) , and hence u ∈ L∞ (a, b) , we deduce from (H3’)that ψ ∈ L1 (a, b) . We also have from (4.14) that

∫ b

a

ψ (x) v (x) dx = −∫ b

a

ϕ (x) v′ (x) dx, ∀v ∈ C∞0 (a, b) .

Since ϕ ∈ L1 (a, b) (from (H3’)), we have by definition of weak derivatives theclaim, namely ϕ ∈W 1,1 (a, b) and ϕ′ = ψ a.e.

Step 2. Since ϕ ∈W 1,1 (a, b) , we have that ϕ ∈ C0 ([a, b]) which means thatthere exists a constant α5 > 0 so that

|ϕ (x)| = |fξ (x, u (x) , u′ (x))| ≤ α5 , ∀x ∈ [a, b] . (4.16)

Since u is bounded (and even continuous), let us say |u (x)| ≤ R for everyx ∈ [a, b], we have from (H1) that

f (x, u, 0) ≥ f (x, u, ξ)− ξfξ (x, u, ξ) , ∀ (x, u, ξ) ∈ [a, b]× [−R, R]× R.

Combining this inequality with (H2) we find that there exists α6 ∈ R such that,for every (x, u, ξ) ∈ [a, b]× [−R, R]× R,

ξfξ (x, u, ξ) ≥ f (x, u, ξ)− f (x, u, 0) ≥ α1 |ξ|p + α6 .

Using (4.16) and the above inequality we find

α1 |u′|p + α6 ≤ u′fξ (x, u, u′) ≤ |u′| |fξ (x, u, u′)| ≤ α5 |u′| , a.e. x ∈ (a, b)

which implies, since p > 1, that |u′| is uniformly bounded. Thus the lemma.

Theorem 4.36 Let f ∈ C∞ ([a, b]× R× R) satisfy (H2), (H3’) and

(H1’) fξξ (x, u, ξ) > 0, ∀ (x, u, ξ) ∈ [a, b]× R× R.

Then any minimizer of (P ) is in C∞ ([a, b]) .

Remark 4.37 (i) Note that (H1’) is more restrictive than (H1). This strongercondition is usually, but not always, as will be seen in Theorem 4.38, necessaryto get higher regularity.

(ii) The proof will show that if f ∈ Ck, k ≥ 2, then the minimizer is also Ck.

(iii) Of course, the convexity of f is essential for regularity, see Exam-ple 4.10. ♦


Step 1. We know from Lemma 4.35 that

x→ ϕ (x) := fξ (x, u (x) , u′ (x))


is in W 1,1 (a, b) and hence it is continuous. Appealing to Lemma 4.27 (and theremark following this lemma), we have that if


v ξ − f (x, u, ξ)

then H ∈ C∞ ([a, b]× R× R) and, for every x ∈ [a, b], we have

ϕ (x) = fξ (x, u (x) , u′ (x)) ⇔ u′ (x) = Hv (x, u (x) , ϕ (x)) .

Since Hv , u and ϕ are continuous, we infer that u′ is continuous and henceu ∈ C1 ([a, b]). We therefore deduce that x→ fu (x, u (x) , u′ (x)) is continuous,which combined with the fact that (cf. (4.15))

d

dx[ϕ (x)] = fu (x, u (x) , u′ (x)) , a.e. x ∈ (a, b)

(or equivalently, by Lemma 4.27, ϕ′ = −Hu (x, u, ϕ)) leads to ϕ ∈ C1 ([a, b]) .

Step 2. Returning to our Hamiltonian system

u′ (x) = Hv (x, u (x) , ϕ (x))

ϕ′ (x) = −Hu (x, u (x) , ϕ (x))

we can start our iteration. Indeed since H is C∞and u and ϕ are C1 we deducefrom our system that, in fact, u and ϕ are C2. Returning to the system we getthat u and ϕ are C3. Finally we get that u is C∞, as wished.

We next give an example where we can get further regularity without assum-ing the non-degeneracy condition fξξ > 0.

Theorem 4.38 Let g ∈ C1 ([a, b]× R) satisfy

(H2) there exist p > q ≥ 1 and α2, α3 ∈ R such that

g (x, u) ≥ α2 |u|q + α3, ∀ (x, u) ∈ [a, b]× R.

Let

f (x, u, ξ) =1

p|ξ|p + g (x, u) .

Then there exists u ∈ C1 ([a, b]) , with |u′|p−2u′ ∈ C1 ([a, b]) , a minimizer of

(P ) and the Euler-Lagrange equation holds everywhere, i.e.

d

dx[ |u′ (x)|p−2

u′ (x) ] = gu (x, u (x)) , ∀x ∈ [a, b] .

Moreover, if 1 < p ≤ 2, then u ∈ C2 ([a, b]) .

If, in addition, u→ g (x, u) is convex for every x ∈ [a, b], then the minimizeris unique.

Proof. The existence (and uniqueness, if g is convex) of a solution u ∈W 1,p (a, b) follows from Theorem 4.1. According to Lemma 4.35, we know that

Regularity 147

u ∈ W 1,∞ (a, b) and since x → gu (x, u (x)) is continuous, we have that theEuler-Lagrange equation holds everywhere, i.e.

d

dx[ |u′ (x)|p−2

u′ (x) ] = gu (x, u (x)) , x ∈ [a, b] .

We thus have that |u′|p−2u′ ∈ C1 ([a, b]) . Call v := |u′|p−2

u′. We may theninfer that

u′ = |v|2−pp−1 v.

Since the function t → |t|2−pp−1 t is continuous if p > 2 and C1 if 1 2q > 2 and

f (x, u, ξ) = f (u, ξ) =1

p|ξ|p +

λ

q|u|q , where λ =

qpq−1 (p− 1)

(p− q)q

u (x) =p− q

p|x|p/(p−q)

(note that if, for example, p = 6 and q = 2, then f ∈ C∞ (R2)).

(i) It is easy to see that u ∈ C1 ([−1, 1]) but u /∈ C2 ([−1, 1]) ; indeed, wehave

u′ = |x|p

p−q −2 x and u′′ =q

p− q|x|

2q−pp−q .

(ii) Since

|u′|p−2u′ = |x|

p(q−1)p−q x and |u|q−2

u = ((p− q) /p)q−1 |x|

p(q−1)p−q

we find for instance that if p(q−1)p−q = 4 (which is realized, for example, if p = 8 and

q = 10/3), then |u′|p−2u′, |u|q−2 u ∈ C∞ ([−1, 1]) , although u /∈ C2 ([−1, 1]) .

(iii) Let

(P ) infu∈W 1,p(−1,1)

I (u) =

∫ 1

−1

f (u (x) , u′ (x)) dx : u (−1) = u (1) =p− q

p

.

Since the function (u, ξ)→ f (u, ξ) is strictly convex and satisfies all the hypothe-ses of Theorems 4.1, 4.12 and 4.38, we have that (P ) has a unique minimizerand that it should be the solution of the Euler-Lagrange equation

(|u′|p−2

u′)′

= λ |u|q−2u.

A direct computation shows that, indeed, u is a solution of this equation andtherefore it is the unique minimizer of (P ) . ♦

Finally, we conclude this section by giving a partial regularity result (for aproof see Buttazzo-Giaquinta-Hildebrandt [117]).


Theorem 4.40 (Tonelli partial regularity theorem) Let f ∈ C∞ ([a, b]×R× R) satisfy

(H1’) fξξ (x, u, ξ) > 0, ∀ (x, u, ξ) ∈ [a, b]× R× R .

Let u ∈W 1,1 (a, b) be a minimizer of

(P ) inf

I (u) =

∫ b

a

f (x, u (x) , u′ (x)) dx : u ∈ X

= m

where X :=u ∈ W 1,1 (a, b) : u (a) = α, u (b) = β

. Then u has a classical

derivative (possibly infinite) at every point in [a, b] and u′ : [a, b] → R ∪ ±∞is continuous. Furthermore, the singular set

E := x ∈ [a, b] : |u′ (x)| =∞

is closed and has zero measure and u is C∞ outside E.

4.7 Lavrentiev phenomenon

We conclude this chapter by presenting an example of the so-called Lavrentievphenomenon. It illustrates that some of the hypotheses used in order to getexistence results, to derive Euler-Lagrange equations or to obtain regularityresults are optimal. In particular, a careful choice of the space of admissiblefunctions is necessary.

Theorem 4.41 (Mania example) Let f (x, u, ξ) :=(x− u3

)2ξ6 and

I (u) :=

∫ 1

0

f (x, u (x) , u′ (x)) dx.

LetW∞ :=

u ∈W 1,∞ (0, 1) : u (0) = 0, u (1) = 1

,

W1 :=u ∈W 1,1 (0, 1) : u (0) = 0, u (1) = 1

.

Theninf I (u) : u ∈ W∞ > inf I (u) : u ∈ W1 = 0.

Moreover, u (x) = x1/3 is a minimizer of I over W1 .

Remark 4.42 (i) The first observation of this phenomenon was due toLavrentiev [391]. The example presented here is essentially due to Mania [417];see also Ball-Mizel [64] and Cesari [143]. For more on this phenomenon we referto Belloni [75], Buttazzo-Mizel [119], Clarke-Vinter [159], Davie [222], Ferriero[278], Mizel [448] and Sarychev [522].

(ii) It is interesting to note that the usual finite element methods (by takingpiecewise affine functions, which are in W 1,∞) in numerical analysis will then

Lavrentiev phenomenon 149

not be able to detect the minimum of some integrals such as the one in thetheorem.

(iii) Note also that one can show (see Ball-Mizel [64]) a similar result to thatof the theorem with a function such as

f (x, u, ξ) =(x4 − u6

)|ξ|s + ǫ |ξ|2

ǫ > 0, s ≥ 27. This last example has the advantage of leading to a coerciveintegral in W 1,2, while this is not the case in the above theorem. ♦

Before proceeding with the proof, we establish a preliminary lemma.

Lemma 4.43 Let 0 < α < β < 1 and

Wαβ : = u ∈W 1,∞ (α, β) : u (α) = 14α1/3, u (β) = 1

2β1/3;14x1/3 ≤ u (x) ≤ 1

2x1/3, for every x ∈ [α, β].If f (x, u, ξ) =

(x− u3

)2ξ6 and

Iαβ (u) :=

∫ β

α

f (x, u (x) , u′ (x)) dx,

thenIαβ (u) ≥ c0

β

for every u ∈ Wαβ and for c0 = 72352−185−5.

Proof. Since u (x) ≤ 12x1/3 we have

1− u3

x≥ 1− 1

x

(x1/3

2

)3

=7

23, for every x ∈ [α, β] .

We thus obtain

Iαβ (u) =

∫ β

α

x2

(1− u3

x

)2

u′ 6 dx ≥ 72

26

∫ β

α

x2u′ 6 dx. (4.17)

We next lety := x3/5 and u (x) := u (y) = u(x3/5).

We immediately deduce that

u′ (x) = u′ (y)dy

dx=

3

5u′ (y)x−2/5 =

3

5u′ (y) y−2/3.

Returning to (4.17) we have

Iαβ (u) ≥ 72

26

∫ β3/5

α3/5

y10/3

(3

5u′ (y) y−2/3

)6(5

3y2/3

)dy

≥ 7235

2655

∫ β3/5

α3/5

(u′ (y))6dy.


Applying Jensen inequality to the right hand side we obtain

Iαβ (u) ≥ 7235

2655

(u(β3/5

)− u(α3/5

))6(β3/5 − α3/5

)5 =7235

2655

(12β1/3 − 1

4α1/3)6

(β3/5 − α3/5

)5

and thus

Iαβ (u) ≥ 7235

21255

β2(1− 1

2 (α/β)1/3)6

β3(1− (α/β)3/5

)5 . (4.18)

Observe finally that since 0 < α < β, then

(1− 1

2(α/β)1/3

)6

≥(

1

2

)6

and(1− (α/β)1/3

)−5

≥ 1. (4.19)

Combining (4.18) and (4.19) we have indeed obtained the lemma.

We now prove Theorem 4.41.


Step 1. We first prove that if u ∈ W∞ , then there exist 0 < α < β < 1 suchthat u ∈ Wαβ (Wαβ as in the lemma), namely

⎧⎪⎨⎪⎩

u (α) =1

4α1/3, u (β) =

1

2β1/3

1

4x1/3 ≤ u (x) ≤ 1

2x1/3, for every x ∈ [α, β] .

(4.20)

The existence of such α and β is easily seen (see Figure 4.2). Let

A :=

a ∈ (0, 1) : u (a) =

1

4a1/3

B :=

b ∈ (0, 1) : u (b) =

1

2b1/3

.

Since u is Lipschitz, u (0) = 0 and u (1) = 1, it follows that A = ∅ and B = ∅.Next choose

α := max a : a ∈ A and β := min b : b ∈ B and b > α .

It is then clear that α and β satisfy (4.20).

Step 2. We may therefore use the lemma to deduce that, for every u ∈ W∞ ,

I (u) =

∫ 1

0

(x− u3

)2u′ 6 dx ≥

∫ β

α

(x− u3

)2u′ 6 dx ≥ c0

β> c0 > 0

and thusinf I (u) : u ∈ W∞ ≥ c0 > 0.

Lavrentiev phenomenon 151

1

1

u(x)

x

Γ0 = (x, x1/3) : x ∈ [0, 1]

Γ1 = (x, 1

2x1/3) : x ∈ [0, 1]

Γ2 = (x, 1

4x1/3) : x ∈ [0, 1]

C = (x, u(x)) : x ∈ [0, 1]

α β

Figure 4.2: Function u versus Γ0 , Γ1 , Γ2

Step 3. The fact that u (x) = x1/3 is a minimizer of I over all u ∈ W1 istrivial and hence

inf I (u) : u ∈ W1 = 0.

This achieves the proof of the theorem.

Chapter 5

Polyconvex, quasiconvexand rank one convexfunctions

5.1 Introduction

We now turn our attention to the vectorial case. Recall that we are consideringintegrals of the form

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx

where

- Ω ⊂ Rn is an open set;

- u : Ω → RN and hence ∇u ∈ RN×n;

- f : Ω× RN × RN×n → R, f = f (x, u, ξ) , is a Caratheodory function.

While in Part I we were essentially concerned with the scalar case (N = 1or n = 1), we now deal with the vectorial case (N, n > 1). The convexity ofξ → f (x, u, ξ) played the central role in the scalar case (N = 1 or n = 1),see Chapter 3. In the vectorial case, it is still a sufficient condition to ensureweak lower semicontinuity of I in W 1,p

(Ω; RN

); it is, however, far from being

a necessary one. Such a condition is the so-called quasiconvexity introduced byMorrey. It turns out (see Chapter 8) that

f quasiconvex ⇔ I weakly lower semicontinuous.

Since the notion of quasiconvexity is not a pointwise condition, it is hard toverify if a given function f is quasiconvex. Therefore one is led to introducea slightly weaker condition, known as rank one convexity, that is equivalentto the ellipticity of the Euler-Lagrange system of equations associated to the

156 Polyconvex, quasiconvex and rank one convex functions

functional I. We also define a stronger condition, called polyconvexity, thatnaturally arises when we try to generalize the notions of duality for convexfunctions to the vectorial context. One can relate all these definitions throughthe following diagram


We should again emphasize that in the scalar case all these notions are equivalentto the usual convexity condition.

The definitions and main properties of these generalized notions of convexityare discussed in Section 5.2.

In Section 5.3, we give several examples. In particular we show that all thereverse implications are false.

Finally, in an appendix (Section 5.4), we gather certain elementary propertiesof determinants.

5.2 Definitions and main properties

5.2.1 Definitions and notations

Recall that, if ξ ∈ RN×n, we write

ξ =

⎛⎜⎝

ξ11 · · · ξ1

n...

. . ....

ξN1 · · · ξN

n

⎞⎟⎠ =

⎛⎜⎝

ξ1

...ξN

⎞⎟⎠ = (ξ1, · · · , ξn) =

(ξiα

)1≤i≤N

1≤α≤n.

In particular if u : Rn → RN we write

∇u =

⎛⎜⎜⎜⎜⎜⎝

∂u1

∂x1· · · ∂u1

∂xn...

. . ....

∂uN

∂x1· · · ∂uN

∂xn

⎞⎟⎟⎟⎟⎟⎠

.

We may now define all the notions introduced above.

Definition 5.1 (i) A function f : RN×n → R ∪ +∞ is said to be rank oneconvex if

f (λξ + (1− λ) η) ≤ λf (ξ) + (1− λ) f (η)


(ii) A Borel measurable and locally bounded function f : RN×n → R is saidto be quasiconvex if

f (ξ) ≤ 1

measD

∫

D

f (ξ +∇ϕ (x)) dx

Definitions and main properties 157


0

(D; RN

).

(iii) A function f : RN×n → R ∪ +∞ is said to be polyconvex if thereexists F : Rτ(n,N) → R ∪ +∞ convex, such that

f (ξ) = F (T (ξ)) ,

where T : RN×n → Rτ(n,N) is such that

T (ξ) := (ξ, adj2 ξ, · · · , adjn∧N ξ) .

In the preceding definition, adjs ξ stands for the matrix of all s × s minors ofthe matrix ξ ∈ RN×n, 2 ≤ s ≤ n ∧N = min n, N and

τ (n, N) :=

n∧N∑

s=1

σ (s) , where σ (s) :=(Ns

)(ns

)=

N !n!

(s!)2(N − s)! (n− s)!

.

(iv) A function f : Rm → R ∪ +∞ is said to be separately convex, orconvex in each variable, if the function

xi → f (x1, · · · , xi−1, xi, xi+1, · · · , xm) is convex for every i = 1, · · · , m,

for every fixed (x1, · · · , xi−1, xi+1, · · · , xm) ∈ Rm−1.

(v) A function f is called polyaffine, quasiaffine or rank one affine if f and−f are, respectively, polyconvex, quasiconvex or rank one convex.

Remark 5.2 (i) The concepts were introduced by Morrey [453], but the ter-minology is that of Ball [53]; note, however, that Ball calls quasiaffine functionsnull Lagrangians.

(ii) If we adopt the tensorial notation, the notion of rank one convexity canbe read as follows: the function ϕ : R → R ∪ +∞ , ϕ = ϕ (t) , defined by

ϕ (t) := f (ξ + ta⊗ b)

is convex for every ξ ∈ RN×n and for every a ∈ RN , b ∈ Rn, where we havedenoted by

a⊗ b =(aibα

)1≤i≤N

1≤α≤n.

(iii) It is easily seen that in the definition of quasiconvexity, one can replacethe set of test functions W 1,∞

0 by C∞0

(D; RN

).

(iv) We will see in Proposition 5.11 that if in the definition of quasiconvexitythe inequality holds for one bounded open set D, it holds for any such set.

(v) We did not give a definition of quasiconvex functions f that may takethe value +∞, contrary to polyconvexity and rank one convexity. There havebeen such definitions given, for example by Ball-Murat [65] and Dacorogna-Fusco [186] (see also Wagner [594]), in the case where f is allowed to take the


value +∞. However, although such definitions have been shown to be necessaryfor weak lower semicontinuity, it has not been proved that they were sufficientand this seems to be a difficult problem. The notion of quasiconvexity beinguseful only as an equivalent to weak lower semicontinuity we have disregardedthe extension to the case R∪+∞ ; while those of polyconvexity and rank oneconvexity will be shown to be useful.

(vi) We have gathered in Section 5.4 some elementary facts about determi-nants and adjs of matrices. Note that in the case N = n = 2, the notion ofpolyconvexity can be read as follows

σ (1) = 4, σ (2) = 1, τ (n, N) = τ (2, 2) = 5,

T (ξ) = (ξ,det ξ) , f (ξ) = F (ξ,det ξ) .

(vii) In the definition of polyconvexity of a given function f, the associatedfunction F (i.e. f (ξ) = F (T (ξ))) in general is not unique. For example, letN = n = 2,

ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

)

and

f (ξ) = |ξ|2 =(ξ11

)2+(ξ21

)2+(ξ12

)2+(ξ22

)2

=(ξ11 − ξ2

2

)2+(ξ12 + ξ2

1

)2+ 2 det ξ.

Let F1, F2 : R5 → R be defined by

F1 (ξ, a) := |ξ|2 and F2 (ξ, a) :=(ξ11 − ξ2

2

)2+(ξ12 + ξ2

1

)2+ 2a.

Then F1 and F2 are convex, F1 = F2 and

f (ξ) = F1 (T (ξ)) = F1 (ξ,det ξ) = F2 (T (ξ)) = F2 (ξ,det ξ) .

We will see, after Theorem 5.6, that using either Caratheodory theorem or theseparation theorem one can privilege one among the numerous functions F.

(viii) The notion of separate convexity does not play any direct role in thecalculus of variations. However it can serve as a model for better understandingof the more difficult notion of rank one convexity.

(ix) We will see (see Theorem 5.20) that the notions of polyaffine, quasiaffineor rank one affine are equivalent. Therefore the first and third concepts will notbe used anymore. ♦

5.2.2 Main properties

In Section 5.3, we give several examples of polyconvex, quasiconvex and rankone convex functions, but before that we show the relationship between thesenotions. The following result is essentially due to Morrey [453], [455].


Theorem 5.3 (i) Let f : RN×n → R. Then


If f : RN×n → R ∪ +∞ , then

f convex ⇒ f polyconvex ⇒ f rank one convex.

(ii) If N = 1 or n = 1, then all these notions are equivalent.

(iii) If f ∈ C2(RN×n

), then rank one convexity is equivalent to Legendre-

Hadamard condition (or ellipticity condition)

N∑

i,j=1

n∑

α,β=1

∂2f (ξ)

∂ξiα∂ξj

β

λiλjμαμβ ≥ 0

for every λ ∈ RN , μ ∈ Rn, ξ =(ξiα

)1≤i≤N


(iv) If f : RN×n → R is convex, polyconvex, quasiconvex or rank one convex,then f is locally Lipschitz.

Remark 5.4 (i) We will show later that all the counter implications are false.

- The fact thatf polyconvex ⇒ f convex

is elementary. For example, when N = n = 2, the function

f (ξ) := det ξ

is polyconvex but not convex.

- We will see several examples (with N, n ≥ 2), notably in Sections 5.3.2,5.3.8 and 5.3.9, of quasiconvex functions that are not polyconvex so that wehave

f quasiconvex ⇒ f polyconvex.

However, there are no elementary examples of this fact.

- The result that

f rank one convex ⇒ f quasiconvex

is the fundamental example of Sverak (see Section 5.3.7), which is valid forn ≥ 2 and N ≥ 3. However it is still an open problem to know whether frank one convex implies f quasiconvex, when N = 2 (so, in particular, the caseN = n = 2 is open).

(ii) The Legendre-Hadamard condition is the usual inequality required forthe Euler-Lagrange system of equations and is known in this case as ellipticity(see Agmon-Douglis-Nirenberg [7]).


(iii) It is straightforward to see that

f rank one convex ⇒ f separately convex.

However, the reverse implication is false, as the following example shows. LetN = n = 2 and

f (ξ) := ξ11ξ1

2 .

This function is clearly separately convex but not rank one convex. ♦

Before proceeding with the proof of the theorem, we give a lemma involvingsome elementary properties of the determinants.

Lemma 5.5 Let ξ ∈ RN×n and T (ξ) be defined as above.

(i) For every ξ, η ∈ RN×n with rank ξ − η ≤ 1 and for every λ ∈ [0, 1] ,the following identity holds:

T (λξ + (1− λ) η) = λT (ξ) + (1− λ) T (η) .

(ii) For every D ⊂ Rn a bounded open set, ξ ∈ RN×n, ϕ ∈ W 1,∞0

(D; RN

),

the following result is valid:

T (ξ) =1

meas D

∫

D

T (ξ +∇ϕ (x)) dx.

Proof. The proof is elementary and can be found in Proposition 5.65 andTheorem 8.35. We give here, for the sake of illustration, the proof in the caseN = n = 2. We then have

ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

)

andT (ξ) = (ξ,det ξ) =

(ξ11 , ξ1

2 , ξ21 , ξ2

2 , ξ11ξ2

2 − ξ12ξ2

1

).

(i) Since rankξ − η ≤ 1, there exist a, b ∈ R2 such that

η = ξ + a⊗ b =

(ξ11 + a1b1 ξ1

2 + a1b2

ξ21 + a2b1 ξ2

2 + a2b2

).

We therefore get that

det (λξ + (1− λ) η) = det (ξ + (1− λ) a⊗ b)

= λdet ξ + (1− λ) det η.

We then deduce that, whenever rank ξ − η ≤ 1,

T (λξ + (1− λ) η) = (λξ + (1− λ) η, det (λξ + (1− λ) η))

= λT (ξ) + (1− λ)T (η) .


(ii) The proof is similar to the preceding one. Note first that if ϕ ∈C2(D; R2

), then

det∇ϕ =∂ϕ1

∂x1

∂ϕ2

∂x2− ∂ϕ1

∂x2

∂ϕ2

∂x1=

∂

∂x1(ϕ1 ∂ϕ2

∂x2)− ∂

∂x2(ϕ1 ∂ϕ2

∂x1).

Integrating by part the above identity, we have that, if ϕ ∈ C20

(D; R2

), then

det ξ measD =

∫

D

[det ξ + ξ11

∂ϕ2

∂x2+ ξ2

2

∂ϕ1

∂x1− ξ1

2

∂ϕ2

∂x1− ξ2

1

∂ϕ1

∂x2+ det∇ϕ]dx

=

∫

D

det (ξ +∇ϕ (x)) dx.

By density, the above identity holds also if ϕ ∈ W 1,∞0

(D; R2

). We then deduce

that for every ϕ ∈ W 1,∞0

(D; R2

), we must have

T (ξ) measD = (

∫

D

(ξ +∇ϕ (x)) dx,

∫

D

det (ξ +∇ϕ (x)) dx)

=

∫

D

T (ξ +∇ϕ (x)) dx.

This concludes the proof of the lemma.

We may now proceed with the proof of Theorem 5.3.

Proof. Part 1 : f convex⇒ f polyconvex. This implication is trivial.

Part 2 : f polyconvex⇒ f quasiconvex. Since f is polyconvex, there existsF : Rτ(n,N) → R convex, such that

f (ξ) = F (T (ξ)) .

Using Lemma 5.5 and Jensen inequality we obtain

1

measD

∫

D

f (ξ +∇ϕ (x)) dx =1

meas D

∫

D

F (T (ξ +∇ϕ (x))) dx

≥ F (1

measD

∫

D

T (ξ +∇ϕ (x)) dx) = F (T (ξ)) = f (ξ) ,


0

(D; RN

). The inequality is precisely the definition of quasiconvexity.

Part 3 : f quasiconvex ⇒ f rank one convex. The proof is similar to thatof Theorem 3.13 of Chapter 3. Recall that we want to show that

f (λξ + (1− λ) η) ≤ λf (ξ) + (1− λ) f (η)

for every λ ∈ [0, 1] , ξ, η ∈ RN×n with rank ξ − η ≤ 1. To achieve this goalwe let ǫ > 0 and we apply Lemma 3.11. We therefore find disjoint open sets


Dξ , Dη ⊂ D and ϕ ∈W 1,∞0

(D; RN

)such that

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

|meas Dξ − λmeas D| ≤ ǫ, |measDη − (1− λ) measD| ≤ ǫ

∇ϕ (x) =

(1− λ) (ξ − η) if x ∈ Dξ

−λ (ξ − η) if x ∈ Dη

‖∇ϕ‖L∞ ≤ γ

where γ = γ (ξ, η, D) is a constant independent of ǫ. We may then use thequasiconvexity of f to get

∫

D

f (λξ + (1− λ) η +∇ϕ (x)) dx

=

∫

Dξ

f (ξ) dx +

∫

Dη

f (η) dx +

∫

D−(Dξ∪Dη)

f (λξ + (1− λ) η +∇ϕ (x)) dx

≥ f (λξ + (1− λ) η)measD.

Using the properties of the function ϕ and the fact that ǫ is arbitrary, we haveindeed obtained that f is rank one convex.

Part 4. If we now consider the case where f : RN×n → R ∪ +∞ , thefirst implication: f convex ⇒ f polyconvex is still trivial. The implicationf polyconvex ⇒ f rank one convex is also elementary if we use Lemma 5.5.Indeed since f is polyconvex, there exists F : Rτ(n,N) → R ∪ +∞ convex sothat

f (ξ) = F (T (ξ)) .

Let λ ∈ [0, 1] , ξ, η ∈ RN×n with rank ξ − η ≤ 1, then, using Lemma 5.5,we get

f (λξ + (1− λ) η) = F (T (λξ + (1− λ) η)) = F (λT (ξ) + (1− λ) T (η))

≤ λF (T (ξ)) + (1− λ) F (T (η)) = λf (ξ) + (1− λ) f (η)

which is precisely the rank one convexity of f.

(ii) The second statement of the theorem, asserting that if N = 1 or n = 1,then all the notions are equivalent, is trivial.

(iii) We now assume that f is C2 and rank one convex, that is

ϕ (t) := f (ξ + tλ⊗ μ)

is convex in t ∈ R for every ξ ∈ RN×n and for every λ ∈ RN , μ ∈ Rn. Since ϕ isalso C2, we obtain immediately Legendre-Hadamard condition, by computingϕ′′ (t) and using the convexity of ϕ.

(iv) The last part of Theorem 5.3 is an immediate consequence of Theorem2.31 of Chapter 2, since a rank one convex function is evidently separatelyconvex.


5.2.3 Further properties of polyconvex functions

We now give different characterizations of polyconvex functions that are basedon Caratheodory theorem and separation theorems. The next result is due toDacorogna [177] and [179], following earlier results of Ball [53].

We first recall the notation that for any integer I

ΛI := λ = (λ1, · · · , λI) : λi ≥ 0 and∑I

i=1 λi = 1.

Theorem 5.6 Part 1. Let f : RN×n → R ∪ +∞ , then the following twostatements are equivalent:

(i) f is polyconvex;

(ii) the next two properties hold:

• there exists a convex function c : Rτ → R ∪ +∞ , where τ = τ (n, N) ,such that

f (ξ) ≥ c (T (ξ)) for every ξ ∈ RN×n; (5.1)

• for every ξi ∈ RN×n, λ ∈ Λτ+1 , satisfying∑τ+1

i=1 λiT (ξi) = T (∑τ+1

i=1 λiξi ), (5.2)

thenf(∑τ+1

i=1 λiξi ) ≤∑τ+1i=1 λif (ξi) . (5.3)

Part 2. If (ii) is satisfied and if F : Rτ → R ∪ +∞ is defined by

F (X) := inf∑τ+1i=1 λif (ξi) : λ ∈ Λτ+1 ,

∑τ+1i=1 λiT (ξi) = X , (5.4)

then F is convex and

f (ξ) = F (T (ξ)) for every ξ ∈ RN×n. (5.5)

Moreover, for every X ∈ Rτ ,

F (X) = supG (X) : G : Rτ → R ∪ +∞ convex

and f (ξ) = G (T (ξ)) , ∀ ξ ∈ RN×n .Part 3. Let f : RN×n → R, i.e. f takes only finite values. Then the following

conditions are equivalent:

(i) f is polyconvex;

(iii) for every ξ ∈ RN×n, there exists β = β (ξ) ∈ Rτ such that

f (η) ≥ f (ξ) + 〈β (ξ) ; T (η)− T (ξ)〉 (5.6)

for every η ∈ RN×n and where 〈·; ·〉 denotes the scalar product in Rτ .

Part 4. If (iii) is satisfied, then the function

h (X) := supξ∈RN×n

〈β (ξ) ; X − T (ξ)〉+ f (ξ) (5.7)

is convex, takes only finite values and satisfies

f (ξ) = h (T (ξ)) for every ξ ∈ RN×n. (5.8)


Example 5.7 Let N = n = 2. Then (5.3) and (5.2) become

f(∑6

i=1 λiξi ) ≤∑6i=1 λif (ξi) ,

∑6i=1 λi det (ξi) = det(

∑6i=1 λiξi )

and (5.6) is read

f (η) ≥ f (ξ) + 〈γ (ξ) ; η − ξ〉+ δ (ξ) (det η − det ξ)

where γ (ξ) ∈ R2×2 and δ (ξ) ∈ R. ♦

Remark 5.8 (i) The above theorem is a direct adaptation of Caratheodorytheorem and the separation theorems for polyconvex functions.

(ii) The condition (5.1) in the theorem implies that F defined in (5.4) doesnot take the value −∞.

(iii) The theorem is important for the following reasons.

- It gives an intrinsic definition of polyconvexity, in the sense that it is notgiven in terms of convexity properties of an associated function F.

- As already mentioned in the definition of the polyconvexity of a givenfunction f, the associated convex function F is not unique. Equation (5.4)allows us to privilege one such function F. A similar remark can be done using(5.7), as was also observed by Kohn and Strang [373], [374].

- If f : RN×n → R (i.e. f takes only finite values), then F defined by (5.4)also takes finite values.

(iv) In view of the above remark we can conclude that if f takes only finitevalues then (i), (ii) and (iii) of Theorem 5.6 are equivalent.

(v) Some other properties of polyconvex functions in the cases N = n = 2or N = n = 3 are given by Aubert [39]. ♦

Proof. We follow here the proof of Dacorogna [177], [179], inspired by earlierconsiderations by Ball [53], which were based on results of Busemann-Ewald-Shephard [110] and Busemann-Shephard [111].

Parts 1 and 2. (i) ⇒ (ii). Since f is polyconvex, there exists F : Rτ →R ∪ +∞ , τ = τ (n, N) , convex such that

f (ξ) = F (T (ξ)) . (5.9)

The existence of a function c is trivial, just choose c = F. The convexity of Fcoupled with (5.2) gives immediately (5.3).

(ii) ⇒ (i). Assume that (5.3) holds for every (λi, ξi) , 1 ≤ i ≤ τ + 1,satisfying (5.2). We wish to show that there exists F : Rτ(n,N) → R ∪ +∞convex satisfying (5.9). Let I ≥ τ + 1 (τ = τ (n, N)) be an integer and forX ∈ Rτ define

FI (X) := inf∑Ii=1 λif (ξi) : λ ∈ ΛI ,

∑Ii=1 λiT (ξi) = X . (5.10)


We will show that FI satisfies (5.9) and that one can choose I = τ + 1, withoutloss of generality, establishing hence (5.4). The proof is divided into four steps.

Step 1. We first show that FI is well defined.

Step 2. We next prove that I can be taken to be τ +1 in (5.10) without lossof generality and we therefore denote FI by F (satisfying then (5.4)).

Step 3. We then show that F is convex.

Step 4. We finally establish that F satisfies (5.5).

We now proceed with the details of these four steps.

Step 1. Let us start by showing that FI is well defined. To do this we mustsee that given X ∈ Rτ(n,N) and I ≥ τ + 1, then there exist λ ∈ ΛI and ξi suchthat

∑λiT (ξi) = X. In view of Caratheodory theorem, this is equivalent to

showing thatcoT

(RN×n

)= Rτ(n,N), (5.11)

where coM denotes the convex hull of M and

T(RN×n

)=X ∈ Rτ(n,N) : there exists ξ ∈ RN×n with T (ξ) = X

.

In order to establish (5.11), we proceed by contradiction. Assume that

co(T(RN×n

))= Rτ .

Then from the separation theorems (see Corollary 2.11), there exist 0 = α ∈Rτ , β ∈ R, such that

co(T(RN×n

))⊂ V := X ∈ Rτ : 〈α; X〉 ≤ β (5.12)

where 〈·; ·〉 denotes the scalar product in Rτ , τ = τ (n, N) . Recall from thedefinition of polyconvexity that

τ (n, N) =n∧N∑

s=1

σ (s)

where σ (s) =(Ns

)(ns

). We then let for X ∈ Rτ(n,N)

X = (X1, X2, · · · , Xn∧N) ∈ Rσ(1) × Rσ(2) × · · · × Rσ(n∧N) = Rτ(n,N)

and similarly for α ∈ Rτ . We may then write

〈α; X〉 =

n∧N∑

s=1

〈αs; Xs〉 .

Since α = 0, there exists t ∈ 1, · · · , n ∧N such that αt = 0 while αs = 0if s < t (if α1 = 0, then take t = 1). We now show that (5.12) leads to acontradiction and therefore (5.11) holds. Let ξ ∈ RN×n and therefore

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) ∈ T(RN×n

)⊂ coT

(RN×n

).


We choose ξ ∈ RN×n such that

〈α; T (ξ)〉 = 〈αt; adjt ξ〉 = 0.

This is possible by choosing (N − t) lines of ξ to be zero vectors of Rn andchoosing the other t lines of ξ so that 〈αt; adjt ξ〉 is non-zero.

Let λ ∈ R be arbitrary and multiply any of the t non zero lines of ξby λ. Denote the obtained matrix by η. We then have T (η) ∈ T

(RN×n

)

⊂ co T(RN×n

)and

〈α; T (η)〉 = 〈αt; adjt η〉 = λ 〈αt; adjt ξ〉 = λ 〈α; T (ξ)〉 .

Using (5.12), we deduce that T (ξ) , T (η) ∈ V, i.e.

〈α; T (ξ)〉 ≤ β

〈α; T (η)〉 = λ 〈α; T (ξ)〉 ≤ β.

The arbitrariness of λ and the fact that 〈α; T (ξ)〉 = 0 lead immediately to acontradiction. This completes Step 1.

Step 2. We now want to show that in (5.10) we can take I = τ + 1. This isdone as in Theorem 2.13.

So let X ∈ Rτ , ξi ∈ RN×n and λ ∈ ΛI be such that

X =

I∑

i=1

λiT (ξi) .

We first prove that there is no loss of generality if we choose I = τ + 2. Define

T (epi f) := (T (ξ) , a) ∈ Rτ × R : f (ξ) ≤ a ⊂ Rτ+1.

We then trivially have that (T (ξi) , f (ξi)) ∈ T (epi f) and if λ ∈ ΛI , we get

(X,∑I

i=1 λif (ξi)) =∑I

i=1 λi (T (ξi) , f (ξi)) ∈ co T (epi f) .

Using Caratheodory theorem, we find that in (5.10) we can take I = τ + 2. Itnow remains to reduce I from τ + 2 to τ + 1 and this is done as in Theorem2.35. We show that given X, T (ξi) ∈ Rτ , 1 ≤ i ≤ τ +2, f : RN×n → R∪+∞and α ∈ Λτ+2 with

τ+2∑

i=1

αiT (ξi) = X, (5.13)

then there exist β ∈ Λτ+2 such that at least one of the βi = 0 (meaning, uponrelabeling, that β ∈ Λτ+1) and

τ+2∑

i=1

βif (ξi) ≤τ+2∑

i=1

αif (ξi) withτ+2∑

i=1

βiT (ξi) = X. (5.14)


It is clear that (5.14) will imply Step 2. Assume all αi > 0 in (5.13) and (5.14),otherwise (5.14) would be trivial. Since from (5.13), we have

X ∈ co T (ξ1) , · · · , T (ξτ+2) ⊂ Rτ ,

it results, from Caratheodory theorem, that there exists α ∈ Λτ+2 with at leastone of the αi = 0 such that

τ+2∑

i=1

αiT (ξi) = X.

We may assume without loss of generality that

τ+2∑

i=1

αif (ξi) >τ+2∑

i=1

αif (ξi) , (5.15)

otherwise choosing βi = αi we would immediately obtain (5.14). We then let

J := i ∈ 1, · · · , τ + 2 : αi − αi < 0 .

Observe that J = ∅, since otherwise αi ≥ αi ≥ 0 for every 1 ≤ i ≤ τ + 2 andsince at least one of the αi = 0, we would have a contradiction with

∑τ+2i=1 αi =∑τ+2

i=1 αi = 1 and the fact that αi > 0 for every i. We then define

λ := mini∈J αi

αi − αi

and we have clearly λ > 0. Finally let

βi := αi + λ (αi − αi) , 1 ≤ i ≤ τ + 2.

We therefore have

βi ≥ 0,

τ+2∑

i=1

βi = 1, at least one of the βi = 0,

τ+2∑

i=1

βiT (ξi) = X

and from (5.15)

τ+2∑

i=1

βif (ξi) =

τ+2∑

i=1

αif (ξi) + λ

τ+2∑

i=1

(αi − αi) f (ξi)

≤τ+2∑

i=1

αif (ξi) .

We have therefore obtained (5.14) and this concludes Step 2. Since I can betaken to be τ + 1, we will then denote FI by F (i.e. (5.10) can be replacedby (5.4)).


Step 3. We now show F is convex. Let λ ∈ [0, 1] , X, Y ∈ Rτ . We want toprove that

λF (X) + (1− λ) F (Y ) ≥ F (λX (1− λ) Y ) .

Fix ǫ > 0. From (5.4) we deduce that there exist λ, μ ∈ Λτ+1 and ξi, ηi ∈ RN×n

such that

λF (X) + (1− λ)F (Y ) + ǫ ≥ λ

τ+1∑

i=1

λif (ξi) + (1− λ)

τ+1∑

i=1

μif (ηi) , (5.16)

withτ+1∑

i=1

λiT (ξi) = X,

τ+1∑

i=1

μiT (ηi) = Y. (5.17)

For 1 ≤ i ≤ τ + 1, let

λi = λλi Ci = ξi

λi+τ+1 = (1− λ)μi Ci+τ+1 = ηi .

Then (5.16) and (5.17) can be rewritten as

λF (X) + (1− λ) F (Y ) + ǫ ≥2τ+2∑

i=1

λif (Ci) (5.18)

with λ ∈ Λ2τ+2 and

2τ+2∑

i=1

λiT (Ci) = λX + (1− λ) Y. (5.19)

Taking the infimum in the right hand side of (5.18) over all λi , Ci satisfying(5.19), using (5.10) and Step 2 we have

λF (X) + (1− λ)F (Y ) + ǫ ≥ F (λX + (1− λ)Y ) ;

ǫ > 0 being arbitrary, we have indeed established the convexity of F.

Step 4. It now remains to show (5.5), i.e.

f (ξ) = F (T (ξ))

where F satisfies (5.4), namely

F (X) = inf∑τ+1i=1 λif (ξi) :

∑τ+1i=1 λiT (ξi) = X .

We have just shown that F is convex. Choosing X = T (ξ) we have from(5.3), (5.2) and (5.11) that the infimum in (5.4) is attained precisely by f (ξ) ,


hence (5.5) holds. The fact that F is the supremum over all convex functionsG satisfying

f (ξ) = G (T (ξ)) for every ξ ∈ RN×n,

follows at once from (5.4). This concludes Part 2.

Parts 3 and 4. (i) ⇒ (iii). Since f is polyconvex and finite we may useParts 1 and 2 to find F : Rτ → R convex and finite satisfying (see (5.4))

f (ξ) = F (T (ξ))

F (X) := inf∑τ+1i=1 λif (ξi) :

∑τ+1i=1 λiT (ξi) = X .

Since F is convex and finite, it is continuous and therefore (see Corollary 2.51of Chapter 2), for each X ∈ Rτ , there exists γ (X) ∈ Rτ such that

F (Y ) ≥ F (X) + 〈γ (X) ; Y −X〉

for all Y ∈ Rτ . Choosing Y = T (η) , X = T (ξ) , β (ξ) = γ (T (ξ)) , we get (5.6),namely

f (η) ≥ f (ξ) + 〈β (ξ) ; T (η)− T (ξ)〉 .

(iii)⇒ (i). We define h as in (5.7), namely

h (X) := supξ∈RN×n

〈β (ξ) ; X − T (ξ)〉+ f (ξ) .

The function h, being a supremum of affine functions, is convex. If X = T (η)then (5.6) ensures that the supremum in (5.7) is attained by f (η) and thereforewe have

f (η) = h (T (η))

as claimed. Moreover, h takes only finite values, since by Part 2 we have h ≤ F,where F is as in (5.4).

We now obtain as a corollary that a polyconvex function with subquadraticgrowth must be convex. This is in striking contrast with quasiconvex and rankone convex functions as was established by Sverak [549] (see Theorem 5.54)and later by Gangbo [300] in an indirect way; see also Section 5.3.10. We alsoprove that a polyconvex function cannot have an arbitrary bound from below,contrary to quasiconvex and rank one convex functions (see Section 5.3.8).

Corollary 5.9 Let f : RN×n → R be polyconvex.

(i) If there exist α ≥ 0 and 0 ≤ p < 2 such that

f (ξ) ≤ α (1 + |ξ|p) for every ξ ∈ RN×n,

then f is convex.

(ii) There exists γ ≥ 0 such that

f (ξ) ≥ −γ (1 + |ξ|n∧N ) for every ξ ∈ RN×n.


Proof. (i) Since f is polyconvex and finite, we can find, for every ξ ∈ RN×n,according to Theorem 5.6 (iii), β = β (ξ) ∈ Rτ such that

f (η) ≥ f (ξ) + 〈β (ξ) ; T (η)− T (ξ)〉 , for every η ∈ RN×n. (5.20)

Using the growth condition on f, we find that

f (ξ) + 〈β (ξ) ; T (η)− T (ξ)〉 ≤ f (η) ≤ α (1 + |η|p) , for every η ∈ RN×n.(5.21)

We can also rewrite it as

f (ξ)+〈β (ξ) ; T (η)− T (ξ)〉 = f (ξ)+〈β1 (ξ) ; η − ξ〉+n∧N∑

s=2

〈βs (ξ) ; adjs η − adjs ξ〉

and hence, for every η ∈ RN×n,

g (ξ) + 〈β1 (ξ) ; η〉+n∧N∑

s=2

〈βs (ξ) ; adjs η〉 ≤ α (1 + |η|p) (5.22)

where

g (ξ) := f (ξ)− 〈β1 (ξ) ; ξ〉 −n∧N∑

s=2

〈βs (ξ) ; adjs ξ〉 .

Replacing η by tη, with t ∈ R, in (5.22) we get

g (ξ) + t 〈β1 (ξ) ; η〉+

n∧N∑

s=2

ts 〈βs (ξ) ; adjs η〉 ≤ α (1 + |t|p |η|p) .

Letting t → ∞, using the fact that η is arbitrary and p < 2, we obtain thatβs (ξ) = 0 for every s = 2, · · · , n∧N. Returning to (5.21) we find that, for everyξ ∈ RN×n,

f (ξ) + 〈β1 (ξ) ; η − ξ〉 ≤ f (η) , for every η ∈ RN×n

which implies that f is convex. Indeed we have that, for λ ∈ [0, 1] ,

f (ξ) ≥ f (λξ + (1− λ) η) + 〈ξ − (λξ + (1− λ) η) ; β1 (λξ + (1− λ) η)〉f (η) ≥ f (λξ + (1− λ) η) + 〈η − (λξ + (1− λ) η) ; β1 (λξ + (1− λ) η)〉 .

Multiplying the first equation by λ and the second by (1− λ) and adding themwe obtain the convexity of f.

(ii) The second part of the corollary follows at once from (5.20). Moreprecisely, we have from (5.20) that, for every ξ ∈ RN×n,

f (ξ) ≥ f (0) + 〈β (0) ; T (ξ)〉 ≥ −γ (1 + |ξ|n∧N)

for an appropriate γ = γ (f (0) , β (0)) .


Another direct consequence of Theorem 5.6 is that we can easily construct(see Dacorogna [177]) rank one convex functions that are not polyconvex. Wewill see more sophisticated examples in the next sections.

Let N = n = 2, ξ1, ξ2, ξ3 ∈ R2×2 and λ1, λ2, λ3 ∈ (0, 1) be such that

λ1 + λ2 + λ3 = 1,

∑3i=1 λi det ξi = det(

∑3i=1 λiξi)

det (ξ1 − ξ2) = 0, det (ξ1 − ξ3) = 0, det (ξ2 − ξ3) = 0.

For example we can choose λ1 = λ2 = λ3 = 1/3 and

ξ1 =

(1 02 0

), ξ2 =

(0 10 1

), ξ3 =

(−1 −10 0

).

We then define f : R2×2 → R ∪ +∞ as

f (ξ) :=

0 if ξ = ξ1, ξ2, ξ3

+∞ otherwise.

Proposition 5.10 f is rank one convex but not polyconvex.

Proof. Part 1. To show that f is rank one convex, we have to prove that

f (λξ + (1− λ) η) ≤ λf (ξ) + (1− λ) f (η) (5.23)

for every λ ∈ [0, 1] and every ξ, η ∈ R2×2 such that rank ξ − η ≤ 1. Threecases can happen.

Case 1. ξ = ξi or η = ξi for every i = 1, 2, 3, then f (ξ) = +∞ or f (η) = +∞and therefore (5.23) is trivially satisfied.

Case 2. ξ = ξi and η = ξj with i = j. This case is impossible, since byconstruction rankξi − ξj = 2 if i = j.

Case 3. ξ = η = ξi , then (5.23) is trivially satisfied.

Part 2. It now remains to show that f is not polyconvex. We proceed bycontradiction. If f were polyconvex, we should have, using Theorem 5.6 andthe construction of (λi, ξi)1≤i≤3 , that

f(∑3

i=1 λiξi ) ≤∑3i=1 λif (ξi) .

This is however impossible since the left hand side takes the value +∞ whilethe right hand side is 0.

5.2.4 Further properties of quasiconvex functions

We first show that if in the definition of quasiconvexity the inequality holds forone bounded open set, it holds for any such set.


Proposition 5.11 Let f : RN×n → R be Borel measurable and locally bounded.Let D ⊂ Rn be a bounded open set and let the inequality

f (ξ)measD ≤∫

D

f (ξ +∇ϕ (x)) dx (5.24)

hold for every ξ ∈ RN×n and for every ϕ ∈W 1,∞0

(D; RN

). Then the inequality

f (ξ)meas E ≤∫

E

f (ξ +∇ψ (x)) dx (5.25)

holds for every bounded open set E ⊂ Rn, for every ξ ∈ RN×n and for everyψ ∈W 1,∞

0

(E; RN

).

Proof. We wish to show (5.25) assuming that (5.24) holds. So let ψ ∈W 1,∞

0

(E; RN

)be given and choose first a > 0 sufficiently large so that

E ⊂ Qa := (−a, a)n

Define next

v (x) :=

ψ (x) if x ∈ E

0 if x ∈ Qa − E

so that v ∈W 1,∞0

(Qa; RN

).

Let then x0 ∈ D and choose ν sufficiently large so that

x0 +1

νQa = x0 +

(−a

ν,a

ν

)n

⊂ D.

Define next

ϕ (x) :=

1ν v (ν (x− x0)) if x ∈ x0 + 1

ν Qa

0 if x ∈ D − [x0 + 1ν Qa].

Observe that ϕ ∈W 1,∞0

(D; RN

)and

∫

D

f (ξ +∇ϕ (x)) dx

= f (ξ)meas(D − [x0 +1

νQa ]) +

∫

[x0+1ν Qa ]

f (ξ +∇v (ν (x− x0))) dx

= f (ξ) [meas(D)− measQa

νn] +

1

νn

∫

Qa

f (ξ +∇v (y)) dy

= f (ξ) [meas(D)− measQa

νn+

meas(Qa − E)

νn] +

1

νn

∫

E

f (ξ +∇ψ (y)) dy.

Appealing to (5.24), we deduce that

f (ξ)meas(D) ≤ f (ξ) [meas(D)− measE

νn] +

1

νn

∫

E

f (ξ +∇ψ (y)) dy


which is equivalent to the claim, namely (5.25).

In some examples (such as Sverak example in Section 5.3.7), it might bemore convenient to replace the set of test functions W 1,∞

0 by the set of periodicfunctions.

Notation 5.12 For D := (0, 1)n

, we let

W 1,∞per

(D; RN

):=u ∈ W 1,∞ (Rn; RN

): u (x + ei) = u (x) , i = 1, · · · , n

where e1, · · · , en is the standard orthonormal basis. ♦

We therefore have the following.

Proposition 5.13 Let f : RN×n → R be Borel measurable and locally bounded.The following two statements are then equivalent:

(i) f is quasiconvex;

(ii) for D = (0, 1)n

, the inequality

f (ξ) ≤∫

D

f (ξ +∇ψ (x)) dx (5.26)

holds for every ξ ∈ RN×n and for every ψ ∈W 1,∞per

(D; RN

).

Proof. (ii) ⇒ (i). This follows at once from Proposition 5.11 and the factthat

W 1,∞0

(D; RN

)⊂ W 1,∞

per

(D; RN

).

(i) ⇒ (ii). Let ψ ∈ W 1,∞per

(D; RN

)and observe first that if ν ∈ N and if

ψν (x) :=1

νψ (νx)

then, from the periodicity of ψ, we get∫

D

f (ξ +∇ψν (x)) dx =1

νn

∫

νD

f (ξ +∇ψ (y)) dy =

∫

D

f (ξ +∇ψ (x)) dx.

(5.27)Choose then ην ∈ C∞

0 (D) such that 0 ≤ ην ≤ 1 in D,

ην ≡ 1 on Dν :=

(1

ν, 1− 1

ν

)n

and ‖∇ην‖L∞ ≤ c1ν

where c1 > 0 is a constant independent of ν.Let then

ϕν (x) := ην (x) ψν (x)

and observe that ϕν ∈W 1,∞0

(D; RN

)and

‖∇ϕν −∇ψν‖L∞ = ‖(ην − 1)∇ψν +∇ην ⊗ ψν‖L∞

≤ c2 ‖ψ‖W 1,∞


where c2 > 0 is a constant, independent of ν. Since the function f is locallybounded we can find c3 > 0, independent of ν, so that

‖f (ξ +∇ψν)− f (ξ +∇ϕν)‖L∞ ≤ c3 .

Appealing to the quasiconvexity of f, to (5.27) and to the preceding observa-tions, we find

∫

D

f (ξ +∇ψ (x)) dx =

∫

D

f (ξ +∇ϕν (x)) dx

+

∫

D

[f (ξ +∇ψν (x))− f (ξ +∇ϕν (x))] dx

=

∫

D

f (ξ +∇ϕν (x)) dx

+

∫

D−Dν

[f (ξ +∇ψν (x))− f (ξ +∇ϕν (x))] dx

≥ f (ξ)− c3 meas (D −Dν) .

Letting ν →∞ we have indeed obtained (5.26), as wished.

5.2.5 Further properties of rank one convex functions

There is no known equivalent to Theorem 5.6 for rank one convex functions.We, nevertheless, give here a characterization of rank one convex functions thatis in the same spirit as Part 1 of Theorem 5.6, but much weaker. It will turnout to be useful in Chapter 6.

To characterize rank one convex functions, we give a property of matricesξi ∈ RN×n that will play the same role as (5.2) of Theorem 5.6 for polyconvexfunctions. We follow here the presentation of Dacorogna [176] and [179].

We also recall that for any integer I

ΛI := λ = (λ1, · · · , λI) : λi ≥ 0 and∑I

i=1 λi = 1.

Definition 5.14 Let I be an integer and λ ∈ ΛI . Let ξi ∈ RN×n, 1 ≤ i ≤ I.We say that (λi, ξi)1≤i≤I satisfy (HI) if

(i) when I = 2, then rank ξ1 − ξ2 ≤ 1;

(ii) when I > 2, then, up to a permutation, rankξ1 − ξ2 ≤ 1 and if, forevery 2 ≤ i ≤ I − 1, we define

⎧⎪⎨⎪⎩

μ1 = λ1 + λ2 η1 =λ1ξ1 + λ2ξ2

λ1 + λ2

μi = λi+1 ηi = ξi+1

then (μi, ηi)1≤i≤I−1 satisfy (HI−1) .


Example 5.15 (a) When I = 2, λ ∈ Λ2 , then (λ1, ξ1) , (λ2, ξ2) satisfy (H2) ifand only if

rankξ1 − ξ2 ≤ 1.

(b) When I = 3, λ ∈ Λ3 , then (λi, ξi)1≤i≤3 satisfy (H3) if, up to a permu-tation, ⎧

⎨⎩

rank ξ1 − ξ2 ≤ 1

rankξ3 −λ1ξ1 + λ2ξ2

λ1 + λ2 ≤ 1.

(c) When I = 4, λ ∈ Λ4 , then (λi, ξi)1≤i≤4 satisfy (H4) if, up to a permu-tation, either one of the conditions

⎧⎪⎪⎨⎪⎪⎩

rank ξ1 − ξ2 ≤ 1, rankξ3 −λ1ξ1 + λ2ξ2

λ1 + λ2 ≤ 1

rankξ4 −λ1ξ1 + λ2ξ2 + λ3ξ3

λ1 + λ2 + λ3 ≤ 1

or ⎧⎨⎩

rank ξ1 − ξ2 ≤ 1, rank ξ3 − ξ4 ≤ 1

rankλ1ξ1 + λ2ξ2

λ1 + λ2− λ3ξ3 + λ4ξ4

λ3 + λ4 ≤ 1

holds. ♦

Proposition 5.16 Let f : RN×n → R ∪ +∞ , then the following two condi-tions are equivalent.

(i) f is rank one convex.

(ii) The expression

f(∑I

i=1 λiξi) ≤∑I

i=1 λif (ξi) (5.28)

holds whenever (λi, ξi)1≤i≤I satisfy (HI) .

Proof. (ii)⇒ (i). This is trivial since it suffices to choose I = 2 in (5.28).

(i) ⇒ (ii). We establish (5.28) by induction. By definition of rank oneconvexity, (5.28) holds for I = 2; assume therefore that the proposition is truefor I − 1. Observe that

I∑

i=1

λif (ξi) = (λ1 + λ2) (λ1

λ1 + λ2f (ξ1) +

λ2

λ1 + λ2f (ξ2)) +

I∑

i=3

λif (ξi) .

If we now use the rank one convexity of f and the hypothesis (HI) we get

(λ1 + λ2) f(λ1ξ1 + λ2ξ2

λ1 + λ2) +

I∑

i=3

λif (ξi) ≤I∑

i=1

λif (ξi) .

Using again the rank one convexity of f, hypothesis (HI) and the hypothesis ofinduction, we have indeed established (5.28).


The above result is much weaker than Theorem 5.6 in the sense that onecannot fix an upper bound on I. Two simple examples show that the situationis intrinsically more complicated for rank one convex functions.. The first onehas been established in Dacorogna [176], [179].

Example 5.17 Let N = n = 2,

A =

(0 00 0

), B =

(1 01 0

), C =

(0 −2

1/2 0

), D =

(−1/4 4

0 4

),

and λ1 = λ2 = λ3 = λ4 = λ5 = 1/5

ξ1 = A, ξ2 = B, ξ3 = C, ξ4 = D, ξ5 = A.

It is then easy to see that (λi, ξi)1≤i≤5 satisfy (H5) since

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

det (ξ1 − ξ2) = 0

detξ3 −λ1ξ1 + λ2ξ2

λ1 + λ2 = 0

detξ4 −λ1ξ1 + λ2ξ2 + λ3ξ3

λ1 + λ2 + λ3 = 0

detξ5 −λ1ξ1 + λ2ξ2 + λ3ξ3 + λ4ξ4

λ1 + λ2 + λ3 + λ4 = 0.

However, if we combine together ξ1 and ξ5 and if we consider

μ1 = λ1 + λ5 = 2/5, μ2 = μ3 = μ4 = 1/5

η1 = A, η2 = B, η3 = C, η4 = D

then it is easy to see that (μi, ηi)1≤i≤4 do not satisfy (H4) . In other words, if we

use Proposition 5.16, we have the surprising result that if f : R2×2 → R∪+∞is rank one convex then

f(2

5A +

1

5B +

1

5C +

1

5D) ≤ 2

5f (A) +

1

5f (B) +

1

5f (C) +

1

5f (D)

i.e.f(∑4

i=1 μiηi) ≤∑4

i=1 μif (ηi) (5.29)

even though (μi, ηi)1≤i≤4 do not satisfy (H4) . In order to show (5.28), we haveto write the inequality (with (λi, ξi)1≤i≤5) as

f(1

5A +

1

5B +

1

5C +

1

5D +

1

5A)

≤ 1

5f (A) +

1

5f (B) +

1

5f (C) +

1

5f (D) +

1

5f (A) . ♦

The next example is even more striking and has been given by CasadioTarabusi [127]. A similar example has also been found by Aumann-Hart [50]and Tartar [571].


Example 5.18 Let N = n = 2 and (see Figure 5.1)

⎧⎪⎪⎨⎪⎪⎩

ξ1 =

(−1 00 0

), ξ2 =

(1 00 −1

), ξ3 =

(2 00 1

), ξ4 =

(0 00 2

)

λ1 =8

15, λ2 =

4

15, λ3 =

2

15, λ4 =

1

15.

Observe that λ ∈ Λ4 and

×

×

×

×

Figure 5.1: The matrices ξ1, ξ2, ξ3, ξ4

rank ξi − ξj = 2, if i = j.

Let⎧⎪⎪⎪⎨⎪⎪⎪⎩

η1 = ξ1 , η2 = ξ2 , η3 = ξ3 , η4 = ξ4 , η5 = 0 =

(0 00 0

)=

4∑

i=1

λiξi

μ1 =8

16, μ2 =

4

16, μ3 =

2

16, μ4 =

1

16, μ5 =

1

16.

Observe that (μi, ηi)1≤i≤5 satisfy (H5) , since

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

det (η4 − η5) = 0

detη3 −μ4η4 + μ5η5

μ4 + μ5 = 0

detη2 −μ3η3 + μ4η4 + μ5η5

μ3 + μ4 + μ5 = 0

detη1 −μ2η2 + μ3η3 + μ4η4 + μ5η5

μ2 + μ3 + μ4 + μ5 = 0.


Therefore, using Proposition 5.16, we obtain for every f : R2×2 → R

f (0) = f(∑5

i=1 μiηi) ≤∑5

i=1 μif (ηi) ;

which means that

16f (0) ≤ 8f (ξ1) + 4f (ξ2) + 2f (ξ3) + f (ξ4) + f (η5) . (5.30)

Noting that η5 = 0 and dividing the above inequality by 15, we have that

f (0) = f(∑4

i=1 λiξi) ≤∑4

i=1 λif (ξi) . (5.31)

We have therefore obtained the inequality (5.31) of rank one convexity eventhough none of the ξi − ξj differs by rank one. ♦

Remark 5.19 An interesting point should be emphasized if one compares thetwo examples, namely the inequalities (5.29) and (5.31) of rank one convexity.The first one deals with any rank one convex function f : R2×2 → R ∪ +∞ ,while in the second one we have to restrict our analysis to functions f : R2×2 →R (i.e. that are finite everywhere), since we subtract f (0) from both sides inthe inequality (5.30).

Indeed, the inequality (5.31) does not hold if we allow the function f to takethe value +∞ as the following example shows. Let

f (ξ) = χξ1, ξ2, ξ3, ξ4 (ξ) =

0 if ξ ∈ ξ1, ξ2, ξ3, ξ4

+∞ otherwise.

This function is clearly rank one convex, since rank ξi − ξj = 2 for i = j.Therefore ∑4

i=1 λif (ξi) = 0 < f(∑4

i=1 λiξi) = f (0) = +∞. ♦

5.3 Examples

We have seen in Section 5.2 the definitions and the relations between the notionsof convexity, polyconvexity, quasiconvexity and rank one convexity. We nowdiscuss several examples, the most important being the following.

i) We start in Section 5.3.1 with the complete characterization of the quasi-affine functions (i.e. the functions f such that f and −f are quasiconvex) byshowing that they are linear combinations of minors of the matrix ∇u.

ii) In Section 5.3.2 we study the case of quadratic functions f. The mainresult being that rank one convexity and quasiconvexity are equivalent. Notethat the quadratic case is important in the sense that it leads to associatedlinear Euler-Lagrange equations. Therefore, in the linear case, the ellipticityof the Euler-Lagrange equations corresponds exactly to the quasiconvexity ofthe integrand and thus, anticipating the results of Chapter 8, to the weak lowersemicontinuity of the associated variational problem.

Examples 179

iii) The third important result is considered in Sections 5.3.3 and 5.3.4. Westudy functions invariant under rotations, notably those depending on singularvalues. We characterize their convexity and polyconvexity.

iv) In Section 5.3.7, we present the celebrated example of Sverak that pro-vides, in dimensions N ≥ 3 and n ≥ 2, an example of a rank one convex functionthat is not quasiconvex.

v) In Section 5.3.8, we consider the example of Alibert-Dacorogna-Marcellini,which is valid when N = n = 2. It characterizes for a homogeneous polynomialof degree four the different notions of convexity encountered in Section 5.2.

5.3.1 Quasiaffine functions

We start with a result established by Ball [53], that is an extension of resultsof Edelen [255], Ericksen [265] and Rund [520]. It characterizes completelythe quasiaffine functions (see also Anderson-Duchamp [27], Ball-Curie-Olver[59], Sivaloganathan [541] and Vasilenko [588]). We follow here the proof ofDacorogna [179].

Theorem 5.20 Let f : RN×n → R. The following conditions are equivalent.

(i) f is quasiaffine.

(ii) f is rank one affine, meaning that f and −f are rank one convex, i.e.

f (λξ + (1− λ) η) = λf (ξ) + (1− λ) f (η)


(ii’) The function f ∈ C1 and for every ξ ∈ RN×n, a ∈ RN , b ∈ Rn,

f (ξ + a⊗ b) = f (ξ) + 〈∇f (ξ) ; a⊗ b〉 ,

where 〈·; ·〉 denotes the scalar product in RN×n.

(iii) f is polyaffine, i.e. f and −f are polyconvex.

(iii’) There exists β ∈ Rτ(n,N) such that

f (ξ) = f (0) + 〈β; T (ξ)〉

for every ξ ∈ RN×n and where 〈·; ·〉 denotes the scalar product in Rτ(n,N) andT is as in Definition 5.1.

Example 5.21 (i) If N = n = 2, then the theorem asserts that the only quasi-affine functions are of the type

f (ξ) = f (0) + 〈β; ξ〉+ γ det ξ.

In particular the only fully non-linear quasiaffine function is det ξ.

(ii) More generally if n, N > 1, then the only non-linear quasiaffine functionsare linear combinations of the s × s minors of the matrix ξ ∈ RN×n, where2 ≤ s ≤ n ∧N = min n, N . ♦


Before proceeding with the proof of the theorem, we mention two corollaries.The first one is a straightforward combination of Theorems 5.20 and 8.35.

Corollary 5.22 Let Ω ⊂ Rn be a bounded open set and f : RN×n → R bequasiaffine. Let v ∈ u + W 1,p

0 (Ω), with p ≥ n ∧N, then

∫

Ω

f (∇u (x)) dx =

∫

Ω

f (∇v (x)) dx.

The second one was established by Dacorogna-Ribeiro [212] and we will useit in Theorems 6.24 and 7.47.

Corollary 5.23 Let f : RN×n → R be quasiaffine.

(i) If f is locally constant, then it is constant.

(ii) If f has a local extremum, then it is constant.

Proof. (Corollary 5.23). (i) We show that if f is locally constant arounda point ξ ∈ RN×n then f is constant everywhere, establishing the result. Soassume that there exists ǫ > 0 such that

f(ξ + v) = f(ξ), ∀ v ∈ RN×n with∣∣vi

j

∣∣ ≤ ǫ (5.32)

and let us show that

f(ξ + w) = f(ξ), ∀ w ∈ RN×n. (5.33)

The procedure consists in working component by component. We start to showthat for every w1

1 ∈ R and∣∣vi

j

∣∣ ≤ ǫ we have (denoting bye1, · · · , eN

and

e1, · · · , en the standard basis of RN and Rn respectively)

f(ξ + w11e

1 ⊗ e1 +∑

(i,j) =(1,1) vije

i ⊗ ej) = f(ξ + w11e

1 ⊗ e1) = f(ξ). (5.34)

Indeed if∣∣w1

1

∣∣ ≤ ǫ this is nothing else than (5.32) so we may assume that∣∣w1

1

∣∣ > ǫand use the fact that f is quasiaffine, to deduce that

f(ξ +ǫw1

1

|w11|

e1 ⊗ e1 +∑

(i,j) =(1,1) vije

i ⊗ ej)

=ǫ

|w11|

f(ξ + w11e

1 ⊗ e1 +∑

(i,j) =(1,1) vije

i ⊗ ej)

+(1− ǫ

|w11|

)f(ξ +∑

(i,j) =(1,1) vije

i ⊗ ej).

Therefore appealing to (5.32) and to the preceding identity we have indeedestablished (5.34). Proceeding iteratively in a similar manner with the othercomponents (w1

2, w13, · · · ) we have indeed obtained (5.33) and thus the proof of

(i) is complete.

Examples 181

(ii) We now show that if ξ is a local extremum point of f, then f is constantin a neighborhood of ξ and thus applying (i) we have the result.

Assume that ξ is a local minimum point of f (the case of a local maximizerbeing handled similarly). We therefore have that there exists ǫ > 0 so that

f(ξ) ≤ f(ξ + v), for every v ∈ RN×n so that∣∣vi

j

∣∣ ≤ ǫ. (5.35)

Let us show that this implies that

f(ξ) = f(ξ + v), for every v ∈ RN×n so that∣∣vi

j

∣∣ ≤ ǫ. (5.36)

We write

v =

N∑

i=1

n∑

j=1

vije

i ⊗ ej

and observe that, since f is quasiaffine,

f(ξ) =1

2f(ξ + v1

1e1 ⊗ e1) +

1

2f(ξ − v1

1e1 ⊗ e1)

and since (5.35) is satisfied we deduce that

f(ξ ± v11e1 ⊗ e1) = f(ξ),

∣∣v11

∣∣ ≤ ǫ. (5.37)

We next write, using again the fact that f is quasiaffine,

f(ξ + v11e1⊗ e1) =

1

2f(ξ + v1

1e1⊗ e1 + v1

2e1⊗ e2)+1

2f(ξ + v1

1e1⊗ e1− v1

2e1⊗ e2)

and since (5.35) and (5.37) hold, we deduce that

f(ξ + v11e1 ⊗ e1 ± v1

2e1 ⊗ e2) = f(ξ + v11e

1 ⊗ e1) = f(ξ),∣∣v1

1

∣∣ ,∣∣v1

2

∣∣ ≤ ǫ.

Iterating the procedure we have indeed established (5.36). Appealing to (i), wehave therefore proved the corollary.

We should mention that some of the results of Theorem 5.20 will be provedin a more straightforward way in Sections 5.4 and 8.5. Indeed, the implication(iii’) ⇒ (ii) can also be found in Proposition 5.65, while the implication (iii’)⇒ (i) is also established in Theorem 8.35.

We now turn to the proof of Theorem 5.20.

Proof. (i)⇒ (ii). This implication follows immediately from Theorem 5.3.

(ii’)⇒ (ii). This case is trivial.

(ii)⇒ (ii’). We fix ξ ∈ RN×n, a ∈ RN , b ∈ Rn and let for t ∈ [0, 1]

ϕ (t) := f (ξ + ta⊗ b) .

Since f is rank one affine then ϕ is affine and thus ϕ ∈ C1 and

ϕ (t) = ϕ (0) + tϕ′ (0) .


Since ϕ ∈ C1, then, obviously, f ∈ C1 and the result immediately follows fromthe above identity.

(iii’)⇒ (iii). This implication follows from the definition of polyconvexity.

(iii)⇒ (i). The result follows from Theorem 5.3.

(ii’)⇒ (iii’). This is the only non trivial implication. So recall that

ξ =

⎛⎜⎜⎜⎜⎝

ξ11 · · · ξ1

n

.... . .

...

ξN1 · · · ξN

n

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

ξ1

...

ξN

⎞⎟⎟⎠ = (ξ1, · · · , ξn) .

Assume also that f is such that

f (ξ + a⊗ b)− f (ξ) = 〈∇f (ξ) ; a⊗ b〉 , (5.38)

for every ξ ∈ RN×n, a ∈ RN , b ∈ Rn. We wish to show that there existsβ ∈ Rτ(n,N) such that

f (ξ)− f (0) = 〈β; T (ξ)〉 , for every ξ ∈ RN×n. (5.39)

In the sequel we assume that n ≥ N, otherwise we reverse the roles of n and N.We then proceed by induction on N.

Step 1. N = 1. Since N = 1, (5.38) can be read as

f (ξ + η)− f (ξ) = 〈∇f (ξ) ; η〉

for every ξ, η ∈ Rn. It is then trivial to see that the above identity implies thatf is affine and therefore if we choose β = ∇f (0) , we have immediately (5.39).

Step 2. N = 2. This step is unnecessary but we prove it for the sake ofillustration. Let

ξ =

(ξ11 · · · ξ1

n

ξ21 · · · ξ2

n

)=

(ξ1

ξ2

)= (ξ1, · · · , ξn)

and for a ∈ R2, b ∈ Rn

a⊗ b =

(a1b

a2b

)=

(a1b1 · · · a1bn

a2b1 · · · a2bn

).

We want to show that if f is rank one affine, i.e.

f (ξ + a⊗ b)− f (ξ) = 〈∇f (ξ) ; a⊗ b〉

then there exists β ∈ Rτ(n,2) such that

f (ξ) = f (0) + 〈β; T (ξ)〉

Examples 183

where

T (ξ) = (ξ, adj2 ξ) ∈ R2×n × R

( n2

)

= Rτ(n,2).

For the notations concerning adj2 ξ, see Section 5.4. But note that, up to asign and the ordering, an element of the matrix adj2 ξ is essentially det (ξk, ξl) ,1 ≤ k < l ≤ n. We then fix ξ2 and choose a = e1 = (1, 0) in (5.38) and define

g(ξ1)

:= f

(ξ1

ξ2

).

Thus the function

t→ g(ξ1 + tb

)= f

(ξ1 + tb

ξ2

)

is affine and we may then use Step 1 to find γ = γ(ξ2)∈ Rn such that

g(ξ1)

= g (0) +⟨γ(ξ2); ξ1⟩

= f

(0

ξ2

)+⟨γ(ξ2); ξ1⟩.

Repeating the argument when ξ1 = 0 for f(

0ξ2

), we have

f

(0

ξ2

)= f (0) +

⟨β2; ξ2

⟩.

Combining the above two identities, we obtain

f

(ξ1

ξ2

)= f (0) +

⟨β2; ξ2

⟩+⟨γ(ξ2); ξ1⟩. (5.40)

Since f is rank one affine, it is affine (when ξ1 is fixed) with respect to ξ2

and therefore γ(ξ2)

=(γ1

(ξ2), · · · , γn

(ξ2))

is affine and hence there exist

β1 =(β1

1 , · · · , β1n

)∈ Rn, δ1, · · · , δn ∈ Rn such that

γl

(ξ2)

= β1l +⟨δl; ξ

2⟩, l = 1, · · · , n.

Returning to (5.40), we therefore get

f

(ξ1

ξ2

)= f (0) +

⟨β1; ξ1

⟩+⟨β2; ξ2

⟩+

n∑

l=1

ξ1l

⟨δl; ξ

2⟩

or in other words

f

(ξ1

ξ2

)= f (0) +

⟨β1; ξ1

⟩+⟨β2; ξ2

⟩+

n∑

l=1

n∑

α=1

δlαξ1l ξ2

α . (5.41)

Since f is rank one affine we have from (5.41) that if

h (ξ) :=

n∑

l=1

n∑

α=1

δlαξ1l ξ2

α


then h is rank one affine and therefore using Lemma 5.24 we must have

δlα = −δαl .

Thus there exists ǫ ∈ R

(n2

)such that

h (ξ) =∑

1≤l<α≤n

δlα

(ξ1l ξ2

α − ξ1αξ2

l

)= 〈ǫ; adj2 ξ〉 .

Combining (5.41) with the above identity, we deduce (5.39) and this concludesStep 2.

Step N. We now proceed with the general case. Assume that we have provedthe theorem for every l < N. Fixing ξ2, · · · , ξN and using the fact that f is rankone affine, then f is affine in ξ1, for ξ2, · · · , ξN fixed. Therefore there exist

ψ

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ = (ψ1, · · · , ψn) ∈ Rn and χ

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ ∈ R,

such that

f (ξ) = 〈ψ

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ ; ξ1

⟩+ χ

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ . (5.42)

Using the hypothesis of induction and proceeding as in Step 2 we find that

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

χ

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ = f (0) +

⟨β0; T

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ 〉

ψl

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ = βl + 〈γl; T

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ 〉 , l = 1, · · · , n

(5.43)

for some β0, γ1, · · · , γn ∈ Rτ(n,N−1) and β1 = (β1, · · · , βn) ∈ Rn. Combining(5.42) and (5.43) we have that

f (ξ) = f (0) +⟨β0; T

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ 〉+

⟨β1; ξ1

⟩+

n∑

l=1

ξ1l 〈γl; T

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ 〉

Examples 185

which can be rewritten as

f (ξ) = f (0) +⟨β0; T

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠ 〉+

⟨β1; ξ1

⟩

+

N−1∑

s=1

n∑

l=1

(ns

)∑

α=1

(N−1

s

)∑

i=1

γislαξ1

l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

.

(5.44)

Letting

h (ξ) :=

N−1∑

s=1

hs (ξ) where hs (ξ) :=

n∑

l=1

(ns

)∑

α=1

(N−1

s

)∑

i=1

γislαξ1

l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

we deduce from the fact that f is rank one affine and from (5.44) that h is rankone affine. Since h is rank one affine, we deduce that so is hs . Indeed let us firstshow this for h1 . Write

h1 (ξ) =

N∑

i=2

hi1 (ξ) where hi

1 (ξ) :=

n∑

l=1

n∑

α=1

γi1lαξ1

l ξiα .

By first choosing ξ3 = · · · = ξN = 0, we obtain that h21 is rank one affine (since

then h = h21); iterating this process we find that all the hi

1 are rank one affineand thus h1 is rank one affine. We then infer that so is h− h1 . With the samereasoning, we get that all the hs , 1 ≤ s ≤ N − 1, are rank one affine.

We may then use Lemma 5.24 to deduce that there exist

δjsβ ∈ R, 1 ≤ s ≤ N − 1, 1 ≤ β ≤

(n

s+1

), 1 ≤ j ≤

(N

s+1

)

such that

h (ξ) =

N−1∑

s=1

(n

s+1

)∑

β=1

(N

s+1

)∑

j=1

δjsβ

(adjs+1 ξ

)jβ

.

Combining (5.44) and the above identity, we have indeed found β ∈ Rτ(n,N)

such that

f (ξ) = f (0) + 〈β; T (ξ)〉 ,

which is the claimed result.

In the above proof we have used the following lemma.


Lemma 5.24 Let n ≥ N and ξ ∈ RN×n,

ξ =

⎛⎜⎜⎜⎜⎝

ξ11 · · · ξ1

n

.... . .

...

ξN1 · · · ξN

n

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

ξ1

...

ξN

⎞⎟⎟⎠ = (ξ1, · · · , ξn) .

For 1 ≤ s ≤ N − 1, let

g (ξ) :=

n∑

l=1

(ns

)∑

α=1

(N−1

s

)∑

i=1

γilαξ1

l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

.

If g is rank one affine, meaning that

g (ξ + a⊗ b) = g (ξ) + 〈∇g (ξ) ; a⊗ b〉 ,

then there exist δjβ ∈ R, 1 ≤ β ≤

(n

s+1

), 1 ≤ j ≤

(N

s+1

)such that

g (ξ) =

(n

s+1

)∑

β=1

(N

s+1

)∑

j=1

δjβ

⎛⎜⎝adjs+1

⎛⎜⎝

ξ1

...ξN

⎞⎟⎠

⎞⎟⎠

j

β

=⟨δ; adjs+1 ξ

⟩.

Proof. Part 1. We start, for the sake of illustration, with the case N = 2,therefore s = 1 and

g (ξ) =

n∑

l=1

n∑

α=1

γlαξ1l ξ2

α .

Since g is rank one affine and quadratic then

d2

dt2g (ξ + ta⊗ b) = g (a⊗ b) =

n∑

l,α=1

γlαa1a2blbα = 0,

for every a =(a1, a2

)∈ R2, b = (b1, · · · , bn) ∈ Rn. We therefore immediately

deduce that γlα = −γαl and hence

g (ξ) =∑

1≤l<α≤n

γlα

(ξ1l ξ2

α − ξ1αξ2

l

)=

∑

1≤l<α≤n

γlα det

(ξ1l ξ1

α

ξ2l ξ2

α

)

=

( n2

)

∑

β=1

δβ (adj2 ξ)β = 〈δ; adj2 ξ〉 ,

since adj2 ξ is a vector of R( n

2

)composed of elements of the form det (ξl, ξα) ,

1 ≤ l < α ≤ n and therefore δβ is essentially γlα with the appropriate sign.

Examples 187

Part 2. We now proceed with the general case. Let

g (ξ) =

(N−1

s

)∑

i=1

gi (ξ) where gi (ξ) :=

n∑

l=1

(ns

)∑

α=1

γilαξ1

l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

.

As in the theorem, it is easy to see that g is rank one affine if and only if gi isrank one affine. Therefore it is enough to prove, the stronger version, that for

every i, 1 ≤ i ≤(

N−1s

)there exists j, 1 ≤ j ≤

(N

s+1

), and δj

β ∈ R, so that if

gi (ξ) :=

n∑

l=1

( ns

)

∑

α=1

γilαξ1

l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

is rank one affine, then

gi (ξ) =

( ns+1

)

∑

β=1

δjβ

⎛⎜⎝adjs+1

⎛⎜⎝

ξ1

...ξN

⎞⎟⎠

⎞⎟⎠

j

β

.

It is clear that the above identities imply the lemma. We should draw theattention that all the δj

β corresponding to

⎛⎜⎝adjs+1

⎛⎜⎝

ξ1

...ξN

⎞⎟⎠

⎞⎟⎠

j

β

which do not contain the row ξ1 are chosen to be 0.

For notational convenience, we show the above result only when i =(

N−1s

),

the general case being handled similarly. So let i =(

N−1s

), which corresponds

to j =(

Ns+1

)and we therefore have

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξN

⎞⎟⎠

⎞⎟⎠

i

α

= (−1)i+1

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξs+1

⎞⎟⎠

⎞⎟⎠

α

, 1 ≤ α ≤(

ns

).

We also, from now on, drop the indices i and j and write, to simplify thenotations, γi

lα = (−1)i+1 γlα in this case. We therefore have to show that if

g (ξ) :=n∑

l=1

( ns

)

∑

α=1

γlαξ1l

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξs+1

⎞⎟⎠

⎞⎟⎠

α


is rank one affine then there exists δβ ∈ R, 1 ≤ β ≤(

ns+1

), such that

g (ξ) =

(n

s+1

)∑

β=1

δβ

⎛⎜⎜⎜⎝adjs+1

⎛⎜⎜⎜⎝

ξ1

ξ2

...ξs+1

⎞⎟⎟⎟⎠

⎞⎟⎟⎟⎠

β

. (5.45)

Recall that for given α, 1 ≤ α ≤(

ns

), there exists a unique s-tuple

(λ1, λ2, · · · , λs) with 1 ≤ λ1 < λ2 < · · · < λs ≤ n, such that

⎛⎜⎝adjs

⎛⎜⎝

ξ2

...ξs+1

⎞⎟⎠

⎞⎟⎠

α

= (−1)1+α

det

⎛⎜⎝

ξ2λ1

· · · ξ2λs

.... . .

...ξs+1λ1

· · · ξs+1λs

⎞⎟⎠ . (5.46)

We now fix an arbitrary (s + 1)-tuple (λ1, · · · , λs+1) , where 1 ≤ λ1 < · · · <λs+1 ≤ n and we denote by β the associate integer (as in (5.46)), more precisely

⎛⎜⎝adjs

⎛⎜⎝

ξ1

...ξs+1

⎞⎟⎠

⎞⎟⎠

β

= (−1)1+β

det

⎛⎜⎝

ξ1λ1

· · · ξ1λs+1

.... . .

...ξs+1λ1

· · · ξs+1λs+1

⎞⎟⎠ .

Note that there are(

ns+1

)such (s + 1)-tuple. Denote by α1 the integer cor-

responding (as in (5.46)) to the s-tuple (λ1, · · · , λs) , by αk the integer corre-sponding to the s-tuple (λ1, · · · , λk−1, λk+1, · · · ., λs+1) , 2 ≤ k ≤ s and by αs+1

the integer corresponding to the s-tuple (λ2, · · · , λs+1) . Finally let

Xβ (ξ) :=

n∑

l1=1

(−1)1+α1 γl1α1ξ

1l1

det

⎛⎜⎝

ξ2λ1

· · · ξ2λs

.... . .

...ξs+1λ1

· · · ξs+1λs

⎞⎟⎠

+

s∑

k=2

n∑

lk=1

(−1)1+αk γlkαk

ξ1lk

det

⎛⎜⎝

ξ2λ1

· · · ξ2λk−1

ξ2λk+1

· · · ξ2λs+1

.... . .

......

. . ....

ξs+1λ1

· · · ξs+1λk−1

ξs+1λk+1

· · · ξs+1λs+1

⎞⎟⎠

+

n∑

ls+1=1

(−1)1+αs+1 γls+1αs+1ξ

1ls+1

det

⎛⎜⎝

ξ2λ2

· · · ξ2λs+1

.... . .

...ξs+1λ2

· · · ξs+1λs+1

⎞⎟⎠ .

(5.47)

Examples 189

We then obviously have that

g (ξ) =

( ns+1

)

∑

β=1

Xβ (ξ) .

Since g is rank one affine, then so is Xβ . Therefore in order to show (5.45) it is

then sufficient to find δβ ∈ R, 1 ≤ β ≤(

ns+1

)such that

Xβ (ξ) = δβ det

⎛⎜⎜⎜⎜⎝

ξ1λ1

· · · ξ1λs+1

.... . .

...

ξs+1λ1

· · · ξs+1λs+1

⎞⎟⎟⎟⎟⎠

. (5.48)

To deduce the claim we will use the fact that the function t → Xβ (ξ + ta⊗ b)is affine for every ξ ∈ RN×n, a ∈ RN , b ∈ Rn. We will always choose

a1 = a2 = 1 and a3 = · · · = aN = 0

and we will make several different choices of ξ ∈ RN×n and b ∈ Rn.

1) We first choose ξλ1 = ξλs+1 , meaning that

ξλ1 =

⎛⎜⎜⎝

ξ2λ1

...

ξs+1λ1

⎞⎟⎟⎠ = ξλs+1 =

⎛⎜⎜⎝

ξ2λs+1

...

ξs+1λs+1

⎞⎟⎟⎠ . (5.49)

For such a choice of ξ, we have

Xβ (ξ) =n∑

l1=1

(−1)1+α1 γl1α1ξ1l1

det

⎛⎜⎝

ξ2λ1

· · · ξ2λs

.... . .

...ξs+1λ1

· · · ξs+1λs

⎞⎟⎠

+

n∑

ls+1=1

(−1)1+αs+1 γls+1αs+1ξ

1ls+1

det

⎛⎜⎝

ξ2λ2

· · · ξ2λs+1

.... . .

...ξs+1λ2

· · · ξs+1λs+1

⎞⎟⎠ .

We then let

bl = 0 if l = λ2, · · · , λs . (5.50)

Using the fact that the function t → Xβ (ξ + ta⊗ b) is affine, we deduce thatthe coefficient of the term in t2 must be 0 for every above choices of ξ and b.


We thus obtain that

[

n∑

l1=1

(−1)1+α1 γl1α1bl1bλ1 +

n∑

ls+1=1

(−1)1+αs+1 (−1)s+1 γls+1αs+1bls+1bλs+1 ]

det

⎛⎜⎝

ξ3λ2

· · · ξ3λs

.... . .

...ξs+1λ2

· · · ξs+1λs

⎞⎟⎠ = 0.

Since ξ ∈ RN×n and b ∈ Rn are arbitrary, letting aside (5.49) and (5.50), wefind that

γl1α1 = 0 if l1 = λs+1 and γls+1αs+1 = 0 if ls+1 = λ1

(−1)1+αs+1

γλ1αs+1 = (−1)s+1+α1 γλs+1α1 .

(5.51)

2) We proceed in a similar manner with the other coefficients, namely welet, if 2 ≤ k ≤ s,

ξλk= ξλs+1 and bl = 0 if l = λ1, · · · , λk−1, λk+1, · · · , λs . (5.52)

We then use the fact that the function t → Xβ (ξ + ta⊗ b) is affine and thusthe coefficient of the term in t2 must be 0 for every ξ and b as in (5.52). Wetherefore get that

[

n∑

l1=1

(−1)1+α1 γl1α1bl1 (−1)

k+1bλk

+

n∑

lk=1

(−1)1+αk γlkαk

blk (−1)s+1

bλs+1 ]

det

⎛⎜⎝

ξ3λ1

· · · ξ3λk−1

ξ3λk+1

· · · ξ3λs

.... . .

......

. . ....

ξs+1λ1

· · · ξs+1λk−1

ξs+1λk+1

· · · ξs+1λs

⎞⎟⎠ = 0.

As above we can then deduce that, for every 2 ≤ k ≤ s,

γl1α1 = 0 if l1 = λs+1 and γlkαk

= 0 if lk = λk

(−1)1+αk γλkαk= (−1)s+k+α1 γλs+1α1 .

(5.53)

Examples 191

Combining (5.47), (5.51) and (5.53), we have

Xβ (ξ) = (−1)1+α1 γλs+1α1ξ

1λs+1

det

⎛⎜⎝

ξ2λ1

· · · ξ2λs

.... . .

...ξs+1λ1

· · · ξs+1λs

⎞⎟⎠

+

s∑

k=2

(−1)1+α1 (−1)

s+k+1γλs+1α1ξ

1λk

det

⎛⎜⎝

ξ2λ1

· · · ξ2λk−1

ξ2λk+1

· · · ξ2λs+1

.... . .

......

. . ....

ξs+1λ1

· · · ξs+1λk−1

ξs+1λk+1

· · · ξs+1λs+1

⎞⎟⎠

+ (−1)s(−1)

1+α1 γλs+1α1ξ1λ1

det

⎛⎜⎝

ξ2λ2

· · · ξ2λs+1

.... . .

...ξs+1λ2

· · · ξs+1λs+1

⎞⎟⎠ .

Letting, in the above computation,

δβ := (−1)s+1+α1 γλs+1α1

we have indeed obtained (5.48). This completes the proof of the lemma.

5.3.2 Quadratic case

We now turn our attention to the case where f is quadratic. This case is ofparticular interest since the associated Euler-Lagrange equations are linear. Ithas therefore received much attention. Let us first mention the theorem.

Theorem 5.25 Let M be a symmetric matrix in R(N×n)×(N×n). Let

f (ξ) := 〈Mξ; ξ〉 ,

where ξ ∈ RN×n and 〈·; ·〉 denotes the scalar product in RN×n. The followingstatements then hold.

(i) f is rank one convex if and only if f is quasiconvex.

(ii) If N = 2 or n = 2, then

f polyconvex ⇔ f quasiconvex ⇔ f rank one convex.

(iii) If N, n ≥ 3, then in general

f rank one convex f polyconvex.

Remark 5.26 (i) The proof of (i) of Theorem 5.25 was given by Van Hove[585], [586], although it was implicitly known earlier.


(ii) The second part of the theorem has received considerable attention. Thequestion was raised in 1937 by Bliss and received a progressive answer throughthe works of Albert [9], Hestenes-MacShane [338], MacShane [411], Marcellini[422], Reid [506], Serre [530] and Terpstra [575]. The proof of (ii) of Theorem5.25 relies on an algebraic lemma whose importance is summarized in Uhlig[582].

(iii) A counterexample to the third part of the theorem was given by Terpstra[575] and later by Serre [530] (see also Ball [56]).

(iv) Note also that even if N = n = 2 and f is quadratic, then in general

f polyconvex f convex,

as the trivial example f (ξ) = det ξ shows. ♦

Before proceeding with the proof of the theorem we mention two simple factsthat are summarized in the next lemmas.

Lemma 5.27 Let M be a symmetric matrix in R(N×n)×(N×n) and let

f (ξ) := 〈Mξ; ξ〉 .

Then the following results hold.

(i) f is convex if and only if

f (ξ) ≥ 0

for every ξ ∈ RN×n.

(ii) f is polyconvex if and only if there exists α ∈ Rσ(2) such that

f (ξ) ≥ 〈α; adj2 ξ〉

for every ξ ∈ RN×n and where 〈·; ·〉 denotes the scalar product in Rσ(2) and

σ (2) =(

N2

) (n2

).

(iii) f is quasiconvex if and only if

∫

D

f (∇ϕ (x)) dx ≥ 0

for every bounded open set D ⊂ Rn and for every ϕ ∈ W 1,∞0

(D; RN

).

(iv) f is rank one convex if and only if

f (a⊗ b) ≥ 0

for every a ∈ RN , b ∈ Rn.

Examples 193

Proof. (Lemma 5.27). Parts (i), (iii) and (iv) are trivial. The fact that

f (ξ) ≥ 〈α; adj2 ξ〉 (5.54)

implies that f is polyconvex follows immediately from the following observation.Let

g (ξ) := f (ξ)− 〈α; adj2 ξ〉

then by (5.54) and (i) of the lemma, we deduce that g is convex. Thus f (ξ) =g (ξ) + 〈α; adj2 ξ〉 is polyconvex.

Assume now that f is polyconvex. We wish to show that (5.54) holds forsome α ∈ Rσ(2). Using Theorem 5.6, bearing in mind that f (0) = 0, we findthat there exists β = (βσ(1), βσ(2), · · · , βσ(n∧N) ) ∈ Rτ(n,N) such that

f (ξ) ≥ 〈β; T (ξ)〉 =n∧N∑

s=1

〈 βσ(s); adjs ξ 〉 .

Multiplying ξ by ǫ > 0, we get

f (ǫξ) = ǫ2f (ξ) ≥ ǫ 〈 βσ(1); ξ 〉+ ǫ2 〈 βσ(2); adj2 ξ 〉+ O(ǫ3). (5.55)

Dividing by ǫ and letting ǫ→ 0, we obtain

〈 βσ(1); ξ 〉 ≤ 0

for every ξ ∈ RN×n, thus βσ(1) = 0. Returning to (5.55), dividing by ǫ2 andletting ǫ→ 0 we have indeed obtained (5.54) with α = βσ(2) .

The second important point that we wish to mention is the following lemmaconcerning Fourier transforms for which the proof is straightforward.

Lemma 5.28 Let Ω ⊂ Rn be a bounded open set. Let ϕ ∈ W 1,∞0

(Ω; RN

)be

extended by ϕ ≡ 0 outside of Ω. Define for ξ ∈ Rn

ϕα (ξ) :=

∫

Rn

ϕα (x) e−2πi〈ξ;x〉dx, 1 ≤ α ≤ N.

Then∇ϕ = 2πi

(ϕαξj

)1≤α≤N

1≤j≤n= 2πi ϕ⊗ ξ,

in particular rankRe(∇ϕ), rankIm(∇ϕ) ≤ 1.

Remark 5.29 Lemma 5.28 explains in a way other than that of Theorem 5.3why matrices of rank one play such an important role in quasiconvex analysis.♦

We now proceed with the proof of Theorem 5.25.

Proof. (i) Recall thatf (ξ) = 〈Mξ; ξ〉 .


Theorem 5.3 implies that if f is quasiconvex then f is rank one convex. We nowprove the converse. By Lemma 5.27 we have to show that

∫

Ω

〈M∇ϕ (x) ;∇ϕ (x)〉 dx ≥ 0 (5.56)

for every bounded open set Ω, for every ϕ ∈ W 1,∞0

(Ω; RN

)(we will set ϕ ≡ 0

outside of Ω), knowing that

f (a⊗ b) = 〈Ma⊗ b; a⊗ b〉 ≥ 0. (5.57)

We then use Plancherel formula (we write ξ for the complex conjugate of ξ) toget

∫

Ω

〈M∇ϕ (x) ;∇ϕ (x)〉 dx =

∫

Rn

〈M∇ϕ (x) ;∇ϕ (x)〉 dx

=

∫

Rn

〈 M∇ϕ (ξ) ; ∇ϕ (ξ) 〉 dξ.

(5.58)

Using Lemma 5.28 and (5.57) in (5.58), we obtain (5.56).

(ii) We do not prove this result and we refer to the above bibliography.

(iii) We now want to show that if N = n = 3, then there exists f rank oneconvex which is not polyconvex. We give here an example due to Serre [530].Let

ξ =

⎛⎜⎜⎜⎝

ξ11 ξ1

2 ξ13

ξ21 ξ2

2 ξ23

ξ31 ξ3

2 ξ33

⎞⎟⎟⎟⎠

and letf (ξ) :=

(ξ11 − ξ3

2 − ξ23

)2+(ξ12 − ξ3

1 + ξ13

)2

+(ξ21 − ξ3

1 − ξ13

)2+(ξ22

)2+(ξ33

)2.

We divide the proof into two steps.

Step 1. We first show that there exists ǫ > 0 such that

f (a⊗ b)− ǫ |a⊗ b|2 ≥ 0 (5.59)

for every a, b ∈ R3 and where |ξ|2 := 〈ξ; ξ〉 denotes the Euclidean norm. Lemma5.27 will then ensure that

g (ξ) = f (ξ)− ǫ |ξ|2 (5.60)

is rank one convex. In Step 2 we then prove that this g is not polyconvex andthis will end the proof of the theorem. We first let

ǫ0 := inff (a⊗ b) : a, b ∈ R3, |a⊗ b| = 1

. (5.61)

Examples 195

Then, since f ≥ 0, we have ǫ0 ≥ 0. In order to prove (5.59) it is sufficient toprove that ǫ0 > 0. We proceed by contradiction and assume that ǫ0 = 0. Observethat in (5.61) the minimum is attained and therefore there exist a, b ∈ R3 suchthat

f (a⊗ b) = ǫ0 = 0 and |a⊗ b| = 1. (5.62)

Recall that

a⊗ b =

⎛⎜⎜⎜⎝

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

⎞⎟⎟⎟⎠ ,

therefore the first equation of (5.62) becomes

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

a1b1 = a2b3 + a3b2

a1b2 = a3b1 − a1b3

a2b1 = a1b3 + a3b1

a2b2 = 0

a3b3 = 0.

(5.63)

We then show that (5.63) is in contradiction with the fact that |a⊗ b| = 1. Todo so, we carefully examine (5.63) and separate the discussion in several cases.

Case 1. a2 = a3 = 0 (cf. the two last equations of (5.63)), then (5.63)becomes ⎧

⎪⎨⎪⎩

a2 = a3 = 0

a1b1 = a1b3 = 0

a1b2 = −a1b3 .

(5.64)

Case 1a. a1 = 0, therefore a1 = a2 = a3 = 0 and hence |a⊗ b| = 0,contradiction.

Case 1b. b1 = 0, hence from (5.64), a1b3 = 0 and thus a1b2 = 0. We thenalso conclude that |a⊗ b| = 0 and this is a contradiction.

Case 2. a2 = b3 = 0 (cf. the two last equations of (5.63)), then (5.63)becomes ⎧

⎪⎪⎪⎨⎪⎪⎪⎩

a2 = b3 = 0

a1b1 = a3b2

a1b2 = a3b1

a3b1 = 0.

Case 2a. a3 = 0, then a1b1 = a1b2 = 0 and therefore |a⊗ b| = 0, contradic-tion.

Case 2b. b1 = 0, then a3b2 = a1b2 = 0 and therefore |a⊗ b| = 0, contradic-tion.

Similarly for the case a3 = b2 = 0 and b2 = b3 = 0. Thus ǫ0 > 0 and henceStep 1, i.e. g defined by (5.60), is rank one convex for every 0 < ǫ ≤ ǫ0 .


Step 2. We now show that g is not polyconvex. In view of Lemma 5.27 it issufficient to show that for every α ∈ R3×3, there exists ξ ∈ R3×3 such that

g (ξ) + 〈α; adj2 ξ〉 < 0.

We prove that the above inequality holds for matrices ξ of the following form

ξ :=

⎛⎝

b + d c− a ac + a 0 b

c d 0

⎞⎠ .

For such matrices we have f (ξ) = 0 and therefore

g (ξ) = −ǫ |ξ|2

= −ǫ[ (b + d)2

+ (c− a)2

+ a2 + (c + a)2

+ b2 + c2 + d2 ]

and

adj2 ξ =

⎛⎜⎜⎜⎝

−bd bc cd + ad

ad −ac −(bd + d2 − c2 + ac

)

bc− ab ac + a2 − b2 − bd a2 − c2

⎞⎟⎟⎟⎠ .

Therefore

〈α; adj2 ξ〉 = −α1bd + α2bc + α3 (cd + ad)

+α4ad− α5ac− α6

(bd + d2 − c2 + ac

)

+α7 (bc− ab) + α8

(ac + a2 − b2 − bd

)+ α9

(a2 − c2

).

As in Step 1 we consider several cases.

Case 1. If α8 > 0, then take a = c = d = 0 and b = 0, to get

g (ξ) + 〈α; adj2 ξ〉 = −ǫ |ξ|2 + 〈α; adj2 ξ〉= −ǫ

(2b2)− α8b

2 < 0.

Case 2. If α6 > 0, then take a = b = c = 0 and d = 0, to get

g (ξ) + 〈α; adj2 ξ〉 = −ǫ(2d2)− α6d

2 < 0.

We therefore can assume that α8 ≤ 0 and α6 ≤ 0.

Case 3. If α9−α6 > 0 (α8 ≤ 0, α6 ≤ 0) , then take a = b = d = 0 and c = 0to get

g (ξ) + 〈α; adj2 ξ〉 = −ǫ(3c2)

+ (α6 − α9) c2 < 0.

We therefore assume α8 ≤ 0, α6 ≤ 0 and α9 − α6 ≤ 0. From these threeinequalities we deduce that α8 + α9 ≤ 0, and then taking b = c = d = 0 anda = 0, we get

g (ξ) + 〈α; adj2 ξ〉 = −ǫ(3a2)

+ (α8 + α9) a2 < 0.

And this concludes the proof of the theorem.

Examples 197

5.3.3 Convexity of SO (n) × SO (n) and O (N) × O (n)invariant functions

We now discuss the different notions of convexity for functions having somesymmetries and follow the presentation of Dacorogna-Marechal [204].

Let f : RN×n → R ∪ +∞ and let Γ1 ⊂ RN×N be a subgroup of GL (N)(the set of invertible matrices) and Γ2 ⊂ Rn×n be a subgroup of GL (n) . Assumethat f is Γ1 × Γ2-invariant, meaning that

f (UξV ) = f (ξ) , ∀U ∈ Γ1 , ∀V ∈ Γ2 .

We will be concerned with groups Γ that are either O (n) (the set of orthogonalmatrices) or SO (n) (the set of special orthogonal matrices); see Chapter 13 forprecise definitions.

We start with some notation and we refer to Chapter 13 for more details.In the whole of this section, we assume that N ≥ n, but all the results can becarried in a straightforward way to the case where N ≤ n.

Notation 5.30 (i) Let N ≥ n and ξ ∈ RN×n. The singular values of ξ,denoted by

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) ,

are defined to be the square root of the eigenvalues of the symmetric and positivesemidefinite matrix ξtξ ∈ Rn×n. A similar definition holds when N ≤ n. We let

λ (ξ) = (λ1 (ξ) , · · · , λn (ξ)) .

(ii) When N = n, we denote by

0 ≤ μ1 (ξ) ≤ · · · ≤ μn (ξ) ,

the signed singular values of ξ ∈ Rn×n; they are defined as

μ1 (ξ) = λ1 (ξ) sign (det ξ) and μj (ξ) = λj (ξ) , j = 2, · · · , n.

We let

μ (ξ) = (μ1 (ξ) , · · · , μn (ξ)) .

(iii) We denote, for every integer m ≥ 1 :

- Π (m) the subgroup of O (m) that consists of the matrices having exactlyone nonzero entry per row and per column, moreover each entry belongs to−1, 1;

- Πe (m) the subgroup of Π (m) that consists of the matrices having an evennumber of entries equal to −1;

- S (m) the subgroup of Πe (m) of all permutation matrices.


We therefore have

S (m) ⊂ Πe (m) ⊂ Π(m) ⊂ O (m) ⊂ GL (m) .

(iv) We let RN×nd be the subspace of RN×n consisting of diagonal matrices,

meaning thatξ ∈ RN×n

d ⇒ ξij = 0 if i = j.

(v) For a vector x = (x1, · · · , xn) ∈ Rn, we denote by diagN×n (when N = n

we simply write diag) the matrix ξ ∈ RN×nd such that

ξii = xi . ♦

We start with some simple observations. The first proposition is an immedi-ate consequence of the singular values decomposition theorem (see Theorem 13.3).

Proposition 5.31 (i) Let f : Rn×n → R∪+∞ . Then f is SO (n)×SO (n)-invariant if and only if f satisfies

f = f diag μ,

andg := f diag

is then the unique Πe (n)-invariant function such that f = g μ.

(ii) Let f : RN×n → R ∪ +∞ , where N ≥ n. Then f is O(N) × O (n)-invariant if and only if f satisfies

f = f diagN×n λ,

andg := f diagN×n

is then the unique Π(n)-invariant function such that f = g λ.

It is clear that, if N = n, the notions of O (N)×O (n) , SO (N)×O (n) andO (N) × SO (n)-invariance coincide but differ from that of SO (N) × SO (n)-invariance. However, if N = n, all four notions coincide as we now show.

Proposition 5.32 Let f : RN×n → R∪+∞ , where N > n. Then the follow-ing are equivalent:

(i) f is O(N)×O (n)-invariant;

(ii) f is SO (N)× SO (n)-invariant.

Proof. Obviously, we need only prove that (ii) implies (i). We will see that,if f is SO (N)× SO (n)-invariant, then

f = f diagN×n λ. (5.65)

Examples 199

The conclusion will then follow from Proposition 5.31.

Let ξ ∈ RN×n. By the singular values decomposition theorem (Theorem13.3), there exist U ∈ O(N), V ∈ O (n) such that

ξ = UΛV t, where Λ := diagN×n(λ1 (ξ) , · · · , λn (ξ)).

So we have to consider several cases. First of all let us introduce the followingnotation. If m ≥ 1 is an integer, we let

Hm := diag(−1, 1, · · · , 1) ∈ Rm×m and Km := diag(1, · · · , 1,−1) ∈ Rm×m.

- If U ∈ SO (N) and V ∈ SO (n) , then, from (ii) the conclusion follows,namely

f(ξ) = f(Λ) = (f diagN×n λ)(ξ).

- If U ∈ O(N)−SO (N) and V ∈ O (n)−SO (n) , we may write Λ = HNΛHn ,so that

UΛV t = (UHN )Λ(V Hn)t

with UHN ∈ SO (N) and V Hn ∈ SO (n) . Thus (5.65) holds by (ii).

- If U ∈ O(N)− SO (N) and V ∈ SO (n) , we may write Λ = KNΛ, so that

UΛV t = (UKN )ΛV t

with UKN ∈ SO (N) . Equation (5.65) then follows from (ii).

- If U ∈ SO (N) and V ∈ O (n) − SO (n) , we may write Λ = HNKNΛHn ,so that

UΛV t = (UHNKN )Λ(V Hn)t,

with UHNKN ∈ SO (N) and V Hn ∈ SO (n) . Thus (5.65) holds.

We have therefore shown the claim, namely that f = f diagN×n λ.

The main result concerns the convexity of such functions.

Theorem 5.33 (A) Let f : Rn×n → R∪ +∞ be SO (n)× SO (n)-invariant,f ≡ +∞, and let g : Rn → R ∪ +∞ be the unique Πe (n)-invariant functionsuch that

f = g μ.

Then the following are equivalent:

(i) f is lower semicontinuous and convex;

(ii) the restriction of f to Rn×nd , the subspace of Rn×n of diagonal matrices,

is lower semicontinuous and convex;

(iii) g is lower semicontinuous and convex.


(B) Let N > n, let f : RN×n → R ∪ +∞ be SO (N) × SO (n)-invariantor, equivalently, O(N)×O (n)-invariant, f ≡ +∞, and let g : Rn → R∪ +∞to be the unique Π(n)-invariant function such that

f = g λ.



(ii) the restriction of f to RN×nd , the subspace of RN×n of diagonal matrices,

is lower semicontinuous and convex;


Remark 5.34 (i) We discuss now the history of this theorem first in the casewhere N = n and in the O(n) × O (n)-invariant case. The result was estab-lished by Ball [53], Hill [341] and Thompson-Freede [577]; see also Dacorogna-Marcellini [202] and Le Dret [397]. In elasticity, an O(n) × O (n)-invariantfunction is called isotropic.

(ii) The case N = n and SO (n)× SO (n)-invariant, was first established byDacorogna-Koshigoe [192] in the case n = 2, and later by Vincent [589] whenn ≥ 3, as a consequence of the convexity theorem of Kostant [377]. A differentproof, inspired by Rosakis [516] and based on the notion of signed singularvalues and a generalized Von Neumann inequality (see Theorem 13.10), wasgiven by Dacorogna-Marechal [204]. In this last paper, the case N = n was alsohandled. ♦Proof. (A) The fact that (i) implies (ii) is clear. The fact that (ii) implies(iii) results immediately from the equality g = f diag . Finally, suppose that(iii) holds. Then g∗∗ = g, and Theorem 6.17 (i) implies that

f∗∗ = g∗∗ μ = g μ = f,

which shows that f is lower semicontinuous and convex.

(B) The fact that (i) implies (ii) is clear. The fact that (ii) implies (iii)results immediately from the equality g = f diagN×n . Finally, suppose that(iii) holds. Theorem 6.17 (ii) then implies that

f∗∗ = g∗∗ λ = g λ = f,

which shows that f is lower semicontinuous and convex.

In the case of O (n) × O (n)-invariant functions, the analogous statementcan be derived in several ways from the above results and we do not discuss thedetails.

Corollary 5.35 Let f : Rn×n → R∪+∞ be O (n)×O (n)-invariant, f ≡ +∞,and let g : Rn → R ∪ +∞ be the unique Π(n)-invariant function such that

f = g λ.

Examples 201



(ii) the restriction of f to Rn×nd is lower semicontinuous and convex;


Remark 5.36 As a convex Π (n)-invariant function, the function g appearingin Theorem 5.33 (B) or in Corollary 5.35 must be such that each function

xk → g(x1, · · · , xn), k = 1, · · · , n

is non-decreasing on R+ . We now prove this only when k = 1, the other casesbeing handled similarly. As a matter of fact, for all x = (x1, · · · , xn) ∈ Rn withx1 ≥ 0,

g(0, x2, · · · , xn) ≤ 1

2g(−x1, x2, · · · , xn) +

1

2g(x1, x2, · · · , xn) = g(x),

and if z > 0, we see, using the above inequality, that

g(x) ≤ x1

x1 + zg(x1 + z, x2, · · · , xn) +

z

x1 + zg(0, x2, · · · , xn)

≤ x1

x1 + zg(x1 + z, x2, · · · , xn) +

z

x1 + zg(x1 + z, x2, · · · , xn)

= g(x1 + z, x2, · · · , xn).

Thus x1 → g(x1, · · · , xn) is non-decreasing on R+ . ♦

We now give a simple corollary, which follows from Theorem 5.33 and in amore direct way from Theorem 13.10. It will be used in Theorems 5.39, 5.43and 7.43.

Corollary 5.37 Let ξ ∈ Rn×n and

0 ≤ b1 ≤ · · · ≤ bn .

The functions

fν (ξ) =

n∑

i=ν

biλi (ξ)

are convex for every ν = 1, · · · , n.

If |b1| ≤ b2 ≤ · · · ≤ bn , then the following functions are also convex

gν (ξ) =n∑

i=ν

biμi (ξ) , ν = 1, · · · , n.


5.3.4 Polyconvexity and rank one convexity of SO (n) ×

SO (n) and O (N) × O (n) invariant functions

We now discuss the polyconvexity and rank one convexity of functions havingthe symmetries considered in the previous section. We first discuss the caseof a O (N) × O (n)-invariant function and then the SO (2) × SO (2)-invariantcase. We also assume, as in the previous section, that N ≥ n, but all the resultsimmediately extend to the case where N ≤ n.

We start with some notation.

Notation 5.38 Let N ≥ n.

(i) We letRn

+ := x ∈ Rn : xi ≥ 0, i = 1, · · · , n ,

Kn+ := x ∈ Rn : 0 ≤ x1 ≤ · · · ≤ xn .

In particular, when n = 1, K+ = R+ .

(ii) For X ∈ R(

Ns

)×(ns

), 1 ≤ s ≤ n − 1, we denote by Λs (X) ∈ K

(ns

)+ its

singular values. In particular, when s = 1, we have

Λ1 (ξ) = (λ1 (ξ) , · · · , λn (ξ)) .

In the notation of Section 5.3.3 we have Λ1 (ξ) = λ (ξ) .

(iii) For every x ∈ Kn+ , we adopt the following notation.

- If s = 2, we let

adj2 x ∈ K

(n2

)+

the vector in R(n2

)composed of every xixj with i < j rearranged in an increasing

way (for example if n = 3 then adj2 x = (x1x2, x1x3, x2x3)). Note that, unlessn = 2, 3, the ordering of adj2 x depends on x itself. For example, if n = 4, thenfor some x we can have

adj2 x = (x1x2, x1x3, x1x4, x2x3, x2x4, x3x4)

and for others

adj2 x = (x1x2, x1x3, x2x3, x1x4, x2x4, x3x4) .

- Similarly, if 2 < s < n, we let

adjs x ∈ K

(ns

)+

to be the vector in R(ns

)composed of every xi1 · · ·xis , i1 < · · · < is rearranged

in an increasing way.

Examples 203

- Finally, when s = n, we denote by either of the following symbols

adjn x = detx =

n∏

i=1

xi .

Note that with these notations we have for every ξ ∈ RN×n and every1 ≤ s ≤ n that

Λs (adjs ξ) = adjs Λ1 (ξ) . ♦

The next theorem is stated, for the convenience of the reader, first whenN = n = 2, then when N = n = 3 and finally in the general case N ≥ n.

Theorem 5.39 Let N ≥ n,

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) ,

be the singular values of ξ ∈ RN×n. Let f : RN×n → R and g : Rn+ → R be such

thatf (ξ) = g (λ1 (ξ) , · · · , λn (ξ)) .

(i) Let N = n = 2. Assume that there exists

G : R2+ × R+ → R, G = G (x, δ) = G (x1, x2, δ) ,

convex, non-decreasing in each variable, symmetric with respect to the first twovariables, meaning that

G (x2, x1, δ) = G (x1, x2, δ) ,

and such thatg (x1, x2) = G (x1, x2, x1x2) ,

then f is polyconvex.

(ii) Let N = n = 3. Assume that there exists

G : R3+ × R3

+ × R+ → R

G = G (x, y, δ) = G (x1, x2, x3, y1, y2, y3, δ)

convex, non-decreasing in each variable and symmetric in the variables x and yseparately, meaning that for every permutation P and P ′ of three elements

G (Px, P ′y, δ) = G (x, y, δ) ,

and such that

g (x1, x2, x3) = G (x1, x2, x3, x1x2, x1x3, x2x3, x1x2x3) .


Then f is polyconvex.

(iii) General case: N ≥ n. Assume that there exists

G : Rn+ × R

(n2

)+ × · · · × R

(n

n−1

)+ × R+ → R

G = G (z) = G(z1, z2, · · · , zn−1, zn

)

convex, non-decreasing in each variable and symmetric in each of the variableszi separately, i.e., for every permutation Pi of

(ni

)elements

G(P1Λ

1, P2Λ2, · · · , Pn−1Λ

n−1, Λn)

= G(Λ1, Λ2, · · · , Λn−1, Λn

)

and such that

g (x) = G(x, adj2 x, · · · , adjn−1 x, adjn x

).

Then f is polyconvex.

Remark 5.40 (i) The above result is due to Ball [53] when N = n = 2 andN = n = 3 and to Dacorogna-Marcellini [202] when N = n. Here we follow thislast proof. A different approach, more in the spirit of Section 5.3.3, has beengiven by Dacorogna-Marechal [205]. One can also consult Mielke [443].

(ii) The above sufficient condition is in some sense also necessary, once wehave taken care of the appropriate symmetries implied by the fact that f dependsonly on singular values. For example, since the function f does not see changesof signs of the determinant, then G should not see it either (and the functionF, defined in the proof, as well). This will be achieved in Theorem 5.43 whenN = n = 2. ♦

Proof. We first proceed, just for the sake of better understanding the proof,with the case N = n = 2.

Case: N = n = 2. We divide the proof into two steps.

Step 1. We start with the following preliminary observation. Since G isconvex over R2

+ × R+ we have (cf. Corollary 2.51)

G (x, δ) = supb0, b2 ∈ Rb1 ∈ R2

b0 + 〈b1; x〉+ b2δ :

b0 + 〈b1; y〉+ b2ǫ ≤ G (y, ǫ) , ∀ (y, ǫ) ∈ R2+ × R+

.

It is easy to see (cf. below) that since x ∈ K2+ and δ ≥ 0 and since G is non

decreasing in each variable and symmetric in the x variable, there is no loss ofgenerality in considering the supremum only on b2 ≥ 0 and b1 ∈ K2

+ . Hence,

Examples 205

for every (x, δ) ∈ K2+ × R+ , we have

G (x, δ) = supb0 ∈ Rb2 ≥ 0

b1 ∈ K2+

b0 + 〈b1; x〉 + b2δ :

b0 + 〈b1; y〉+ b2ǫ ≤ G (y, ǫ) , ∀ (y, ǫ) ∈ K2+ × R+

.

Let us now prove that we can indeed restrict the supremum to (b1, b2) ∈ K2+ ×

R+ . DefineL (b0, b1, b2, x, δ) := b0 + 〈b1; x〉+ b2δ.

1) Assume first that we have b2 < 0 and

L (b0, b1, b2, y, ǫ) ≤ G (y, ǫ) , ∀ (y, ǫ) ∈ K2+ × R+

and let us show that we can increase the value by considering b2 = 0. Indeed,since δ ≥ 0, we surely have

L (b0, b1, b2, x, δ) ≤ L (b0, b1, 0, x, δ)

and moreover, since G is non decreasing in the variable ǫ,

L (b0, b1, 0, y, ǫ) = L (b0, b1, b2, y, 0) ≤ G (y, 0)

≤ G (y, ǫ) , ∀ (y, ǫ) ∈ K2+ × R+ .

We have therefore shown that the supremum can be restricted to b2 ≥ 0.

2) A completely analogous argument shows that we can also restrict ourattention to b1 ∈ R2

+ . Once this is achieved, we can further consider only b1 ∈K2

+ , since x itself belongs to K2+ and G is symmetric with respect to the two

first variables.

Step 2. Let F : R2×2 × R → R be defined by

F (ξ, δ) := G(Λ1 (ξ) , |δ|

)= G (λ1 (ξ) , λ2 (ξ) , |δ|) .

Observe that

F (ξ,det ξ) = G (λ1 (ξ) , λ2 (ξ) , λ1 (ξ) λ2 (ξ))

= g (λ1 (ξ) , λ2 (ξ)) = f (ξ) .

Hence if we prove that F is convex, we will have established that f is polyconvex.We have by Step 1 that, for every (x, δ) ∈ K2

+ × R+ ,

G (x, δ) = supb0 ∈ Rb2 ≥ 0

b1 ∈ K2+

b0 + 〈b1; x〉 + b2δ :

b0 + 〈b1; y〉+ b2ǫ ≤ G (y, ǫ) , ∀ (y, ǫ) ∈ K2+ × R+

.


Since for every y ∈ K2+ , we can find η ∈ R2×2 so that

Λ1 (η) = y

(just choose η = diag (y1, y2)), we deduce that

F (ξ, δ) = supb0 ∈ R, b2 ≥ 0

b1 ∈ K2+

⎧⎪⎨⎪⎩

b0 +⟨b1; Λ

1 (ξ)⟩

+ b2 |δ| :b0 +

⟨b1; Λ

1 (η)⟩

+ b2 |ǫ| ≤ F (η, ǫ) ,

∀ (η, ǫ) ∈ R2×2 × R

⎫⎪⎬⎪⎭

.

Since the function (η, ǫ)→ b0 +⟨b1; Λ

1 (η)⟩+ b2 |ǫ| is convex (by Corollary 5.37

and since b2 ≥ 0 and b1 ∈ K2+), we deduce that F is convex. The proof, in the

case N = n = 2, is therefore complete.

General case: N ≥ n. Recall first the notations of Sections 5.2 and 5.4. Let

τ (n, N) :=n∑

s=1

(Ns

)(ns

)

and T : RN×n → Rτ(n,N) be such that

T (ξ) := (ξ, adj2 ξ, · · · , adjn ξ)

where

Rτ(n,N) := RN×n × R(N2

)×(n2

)× · · · × R

(N

n−1

)×(

nn−1

)× R(Nn

).

For X =(X1, X2, · · · , Xn−1, Xn

)∈ Rτ(n,N) we denote by

Λ (X) :=(Λ1(X1), Λ2(X2), · · · , Λn−1

(Xn−1

), Λn (Xn)

)∈ K

θ(n)+

where

Kθ(n)+ := Kn

+ ×K(n2)

+ × · · · ×K( n

n−1)+ ×K+ .

Finally define F : Rτ(n,N) → R by

F (X) := G (Λ (X)) .

Observe that, for ξ ∈ RN×n,

F (T (ξ)) = G (Λ (T (ξ)))

= G(Λ1 (ξ) , Λ2 (adj2 ξ) , · · · , Λn−1

(adjn−1 ξ

), Λn (adjn ξ)

)

= G(Λ1 (ξ) , adj2 Λ1 (ξ) , · · · , adjn−1 Λ1 (ξ) , adjn Λ1 (ξ)

)

= g(Λ1 (ξ)

)= g (λ1 (ξ) , · · · , λn (ξ)) = f (ξ) .

Hence to prove the polyconvexity of f it remains only to prove the convexity of

F. We then use the convexity of G to deduce, for every z =(z1, · · · , zn

)∈ K

θ(n)+ ,

Examples 207

that

G (z) = sup

b0, bν∈R(n

ν)

b0 +

∑nν=1 〈bν ; zν〉 :

b0 +∑n

ν=1 〈bν ; yν〉 ≤ G (y) , ∀y ∈ Rθ(n)+

.

The facts that G is non decreasing in each variable and symmetric in each of

the variables but the last one, that zν ∈ K(n

ν)+ , for every ν = 1, · · · , n, allow (as

in Step 1 of the case where N = n = 2) to restrict the above supremum to

G (z) = sup

b0∈R, bν∈K(n

ν)+

b0 +

∑nν=1 〈bν ; zν〉 :

b0 +∑n

ν=1 〈bν ; yν〉 ≤ G (y) , ∀y ∈ Kθ(n)+

.

Since for every yν ∈ K(n

ν)+ and every ν = 1, · · · , n, we can find ην ∈ R(N

ν )×(nν)

so thatΛν (ην) = yν

(just choose ην a diagonal matrix with the appropriate entries), we obtain thatfor every X =

(X1, · · · , , Xn

)∈ Rτ(n,N),

F (X) = G (Λ (X))

= sup

b0∈R, bν∈K(n

ν)+

b0 +

∑nν=1 〈bν ; Λν (Xν)〉 :

b0 +∑n

ν=1 〈bν ; Λν (ην)〉 ≤ F (η) , ∀η ∈ Rτ(n,N)

Observe that since bν ∈ K(n

ν)+ for ν = 1, · · · , n, we have that the function

η =(η1, · · · , ηn

)∈ Rτ(n,N) → b0 +

n∑

ν=1

〈bν ; Λν (ην)〉

is convex (cf. Corollary 5.37) and hence F is convex. Thus the function f ispolyconvex and this achieves the proof of the theorem.

The next example will turn out, in the subsequent chapters, to be useful.

Example 5.41 Let ξ ∈ Rn×n, then the functions

fν (ξ) :=n∏

i=ν

λi (ξ)

are polyconvex for every ν = 1, · · · , n. The proof follows from the theorem, butit can be seen in a more straightforward way from the following argument. For1 ≤ s ≤ n, the function

X ∈ R(ns

)×(ns

)→ λ(n

s

) (X)

is convex, according to Corollary 5.37. Hence the function

ξ → λ(ns

) (adjs ξ)


is polyconvex. Since

λ(ns)

(adjs ξ) =

n∏

i=n−s+1

λi (ξ) ,

we have the claim. ♦

We now turn our attention to the SO (2) × SO (2)-invariant case and givehere a theorem due to Dacorogna-Koshigoe [192], which shows, in particular,that at least when N = n = 2, the sufficient condition of Theorem 5.39 is alsonecessary. We here follow the proof of Dacorogna-Marechal [205]; but let usfirst introduce the following definition of polyconvexity for vectors.

Definition 5.42 A function g : R2 → R ∪ +∞ is said to be polyconvex ifthere exists G : R3 → R ∪ +∞ convex such that

g (x1, x2) = G (x1, x2, x1x2) .

There is of course a similar definition for polyconvex functions over Rn (fordetails see [205]), but we will not need this extension here.

In the next theorem we use the notations of Section 5.3.3.

Theorem 5.43 Let f : R2×2 → R be SO (2)×SO (2)-invariant and let g : R2 →R be the unique Πe (2)-invariant function such that

f = g μ.

The following statements are all equivalent.

(i) f is polyconvex.

(ii) g is polyconvex.

(iii) For every (ai, bi) ∈ R2, ti ≥ 0, i = 1, 2, 3, 4 with

∑4i=1 ti = 1 and

∑4i=1 tiaibi = (

∑4i=1 tiai )(

∑4i=1 tibi )

the following inequality holds

g(∑4

i=1 ti (ai, bi)) ≤∑4

i=1 tig (ai, bi) .

In particular, if G : R3 → R is defined by

G (a, b, δ) := inf

∑4i=1 tig (ai, bi) :

∑4i=1 ti (ai, bi, aibi) = (a, b, δ) and

∑4i=1 ti = 1

,

then G is well defined. Moreover if g satisfies the above condition, then G isconvex and

g (a, b) = G (a, b, ab)

Examples 209

for every (a, b) ∈ R2.

(iv) For every (a, b) ∈ R2, there exists β = β (a, b) ∈ R3 such that

g (x, y) ≥ g (a, b) + 〈β (a, b) ; (x, y, xy)− (a, b, ab)〉

for every (x, y) ∈ R2 and where 〈·; ·〉 denotes the scalar product in R3.

Remark 5.44 (i) The equivalence between (i) and (ii) can be restated as:

f |R

2×2d

is polyconvex ⇔ f is polyconvex,

where R2×2d is the subspace of diagonal matrices of R2×2 and f |

R2×2d

is the

restriction of f to this subspace.

(ii) The same result holds if f : R2×2 → R is O (2) × O (2)-invariant andg : R2 → R is the unique Π (2)-invariant function such that

f = g λ.

(iii) The result can be, in part, extended to the case where f : R2×2 →R ∪ +∞ , see Dacorogna-Marechal [205] for details.

(iv) We recall that when we say that a function g : R2 → R is Πe (2)-invariantwe mean that, for every x1, x2 ∈ R,

g (x1, x2) = g (x2, x1) = g (−x1,−x2) = g (−x2,−x1) . ♦

Proof. The equivalence between (ii), (iii) and (iv) is proved in exactly thesame way as the one of Theorem 5.6 and we will therefore omit the proof.

(i) ⇒ (ii). Since f is polyconvex, we can find a convex function

F : R2×2 × R → R

so that

f (ξ) = F (ξ,det ξ) .

Let (x1, x2, δ) ∈ R3 and let

G (x1, x2, δ) := F (ξ, δ)

where ξ = diag (x1, x2) ∈ R2×2. Observe that G : R3 → R is convex and, sinceg is Πe (2)-invariant, we have

g (x1, x2) = G (x1, x2, x1x2) .

Thus g is polyconvex.

(ii) ⇒ (i). We divide the proof into two steps.


Step 1. Since g is polyconvex, we can find G : R3 → R convex such that

g (x1, x2) = G (x1, x2, x1x2) .

In general the function (x1, x2)→ G (x1, x2, δ) is not Πe (2)-invariant, althoughg is. To remedy to this difficulty, we let H : R3 → R be defined by

H (x1, x2, δ) :=1

4[G (x1, x2, δ) + G (x2, x1, δ) + G (−x1,−x2, δ)

+G (−x2,−x1, δ)] .

The function H is convex and furthermore (x1, x2) → H (x1, x2, δ) is Πe (2)-invariant. Moreover, since g is Πe (2)-invariant we also have

g (x1, x2) = H (x1, x2, x1x2) .

We then define, for ξ ∈ R2×2,

F (ξ, δ) := H (μ1 (ξ) , μ2 (ξ) , δ) .

Since we clearly havef (ξ) = F (ξ,det ξ) ,

we will deduce the claim, namely that f is polyconvex, once we will have shownthat F : R2×2 × R → R is convex.

This is done in a completely analogous manner to the one of Theorem 5.39.Indeed since H is convex over R3 we have (cf. Corollary 2.51)

H (x1, x2, δ) = supb0,b1,b2,b3∈R

⎧⎪⎨⎪⎩

b0 + b1x1 + b2x2 + b3δ :

b0 + b1y1 + b2y2 + b3ǫ ≤ H (y1, y2, ǫ) ,

∀ (y1, y2, ǫ) ∈ R3

⎫⎪⎬⎪⎭

.

It is easy to see (cf. Step 2 below) that, if |x1| ≤ x2 , we have

H (x1, x2, δ) = supb0, b3 ∈ R|b1| ≤ b2

⎧⎪⎨⎪⎩

b0 + b1x1 + b2x2 + b3δ :

b0 + b1y1 + b2y2 + b3ǫ ≤ H (y1, y2, ǫ) ,

for every |y1| ≤ y2 and ǫ ∈ R

⎫⎪⎬⎪⎭

.

(5.66)since (x1, x2)→ H (x1, x2, δ) is Πe (2)-invariant.

Since for every |y1| ≤ y2 , we can find η ∈ R2×2 so that

μ1 (η) = y1 and μ2 (η) = y2

(just choose η = diag (y1, y2)), we deduce that

F (ξ, δ) = supb0, b3 ∈ R|b1| ≤ b2

⎧⎪⎨⎪⎩

b0 + b1μ1 (ξ) + b2μ2 (ξ) + b3δ :

b0 + b1μ1 (η) + b2μ2 (η) + b3ǫ ≤ F (η, ǫ) ,

∀ (η, ǫ) ∈ R2×2 × R

⎫⎪⎬⎪⎭

.

Examples 211

Since |b1| ≤ b2 , we find that the function

(η, ǫ)→ b0 + b1μ1 (η) + b2μ2 (η) + b3ǫ

is convex (by Corollary 5.37) and we thus deduce that F is convex. The proofis therefore complete.

Step 2. Let us now prove that (5.66) holds. So let, for |x1| ≤ x2 andb0, b1, b2, b3, δ ∈ R,

L (b1, b2, x1, x2, δ) := b0 + b1x1 + b2x2 + b3δ

(we do not denote in L the dependence on b0, b3, since they will not change inthe following computations) be such that

L (b1, b2, y1, y2, ǫ) ≤ H (y1, y2, ǫ) , ∀ (y1, y2, ǫ) ∈ R3. (5.67)

The claim (5.66) will follow, if we can find |c1| ≤ c2 so that

L (b1, b2, x1, x2, δ) ≤ L (c1, c2, x1, x2, δ) (5.68)

whileL (c1, c2, y1, y2, ǫ) ≤ H (y1, y2, ǫ) , ∀ (y1, y2, ǫ) ∈ R3. (5.69)

This is done as follows. Let

σ (b1, b2) :=

⎧⎪⎨⎪⎩

1 if b1b2 > 0

0 if b1b2 = 0

−1 if b1b2 < 0.

Let τ be a permutation of 1, 2 such that

∣∣bτ(1)

∣∣ ≤∣∣bτ(2)

∣∣

andc1 := σ (b1, b2)

∣∣bτ(1)

∣∣ and c2 :=∣∣bτ(2)

∣∣ .

According to Proposition 13.9, the inequality (5.68) is satisfied. Observe that,for every y1, y2 ∈ R,

c1y1 + c2y2 =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

b1y1 + b2y2 if b2 ≥ |b1|−b1y1 − b2y2 if − b2 ≥ |b1|b2y1 + b1y2 if b1 ≥ |b2|−b2y1 − b1y2 if − b1 ≥ |b2| .

This implies that

L (c1, c2, y1, y2, ǫ) ≤ maxL (b1, b2, y1, y2, ǫ) , L (b1, b2,−y1,−y2, ǫ) ,

L (b1, b2, y2, y1, ǫ) , L (b1, b2,−y2,−y1, ǫ).


Since (5.67) holds and (x1, x2) → H (x1, x2, δ) is Πe (2)-invariant, we get (5.69)and hence the claim (5.66) is established.

Having discussed the convexity and the polyconvexity of SO (2)×SO (2) orO (N)× O (n)-invariant functions, one would be tempted to think that similarresults exist for rank one and quasiconvex functions. This is not the case as wasfirst observed by Dacorogna-Koshigoe [192] (see Example 5.45) for rank oneconvex functions. Later Muller [463] showed the same result for quasiconvexfunctions.

Example 5.45 The examples are based on computations of Dacorogna-Douchet-Gangbo-Rappaz in [185]. In both examples, N = n = 2 and b ≥ 0.

(i) Let α > 2 +√

2 and

fα,b (ξ) = |ξ|2α − 2α−1b |det ξ|α .

(ii) Let α > (9 + 5√

5 )/4 and

fα,b (ξ) = |ξ|2α( |ξ|2 − 2b det ξ ).

Note that both functions are SO (2) × SO (2)-invariant. In both cases, thereexist b2 < b1 (for the precise values of b1 , b2 see [185]) such that

fα,b is rank one convex ⇔ b ≤ b2 ,

fα,b|R2×2d

is rank one convex ⇔ b ≤ b1 . ♦

We finally conclude this section by mentioning other results on rank oneconvexity of O (n)×O (n)-invariant functions. As seen in Proposition 5.31, anysuch function is necessarily of the form

f (ξ) = g (λ1 (ξ) , · · · , λn (ξ))

where 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) are the singular values of the matrix ξ ∈ Rn×n.Assuming that the function f is twice differentiable, it is therefore natural toask conditions on the derivatives of g that ensure the rank one convexity ofthe function f. This was achieved by Knowles-Sternberg [371] when n = 2 andthen in various different ways by Aubert [41], Aubert-Tahraoui [48], Ball [55],Dacorogna-Marcellini [202] and Davies [223]. When n = 3, Aubert-Tahraouiin [47] gave also some necessary conditions and, although in a slightly differentcontext, necessary and sufficient conditions were derived by Simpson-Spector[540] (see also Zee-Sternberg [613]). In the case of general n, certain resultsexist but are less explicit; see Dacorogna [182] and Silhavy [536].

5.3.5 Functions depending on a quasiaffine function

The following theorem was established in Dacorogna [173].

Examples 213

Theorem 5.46 Let f : RN×n → R, Φ : RN×n → R be quasiaffine but notidentically constant and g : R → R be such that

f (ξ) = g (Φ (ξ))

(in particular, if N = n, one can take Φ (ξ) = det ξ). Then

f polyconvex ⇔ f quasiconvex ⇔ f rank one convex ⇔ g convex.

Proof. The implications

g convex ⇒ f polyconvex ⇒ f quasiconvex ⇒ f rank one convex

follow immediately from Theorem 5.3. It therefore remains to show that

f rank one convex ⇒ g convex.

We want to prove that for t ∈ (0, 1) , α, β ∈ R, then

g (tα + (1− t)β) ≤ tg (α) + (1− t) g (β)

provided f is rank one convex. Following Theorem 5.20, we have that

Φ (ξ) = a0 + 〈a; T (ξ)〉 = a0 + 〈a1; ξ〉+

n∧N∑

j=2

⟨aj ; adjj ξ

⟩,

where a0 ∈ R, a1 ∈ RN×n and aj ∈ Rσ(j) where σ (j) =(Nj

)(nj

). Since Φ is not

identically constant, then at least one of the aj , 1 ≤ j ≤ n∧N is not zero. Lets be such that as = 0 but as−1 = as−2 = · · · = a1 = 0 (if a1 = 0, we then takes = 1). Since as = 0

(∈ Rσ(s)

)we have that at least one of the components of

as = (a1s, · · · , a

σ(s)s ) is non-zero. For notational convenience, we take a

σ(s)s = 0.


First choose η ∈ RN×n in the following way

η =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

η11 · · · η1

s η1s+1 · · · η1

n...

. . ....

.... . .

...ηs1 · · · ηs

s ηss+1 · · · ηs

n

ηs+11 · · · ηs+1

s ηs+1s+1 · · · ηs+1

n...

. . ....

.... . .

...ηN1 · · · ηN

s ηNs+1 · · · ηN

n

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

α− a0

aσ(s)s

· · · 0 0 · · · 0

.... . .

......

. . ....

0 · · · 1 0 · · · 00 · · · 0 0 · · · 0...

. . ....

.... . .

...0 · · · 0 0 · · · 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

More precisely, we take all components to be zero but the following ones:

η11 =

α− a0

aσ(s)s

, ηii = 1 for 2 ≤ i ≤ s.

We next choose λ ∈ RN×n in exactly the same manner except that we replace

the first component by (β − a0 )/aσ(s)s . We then immediately have

Φ (η) = α, Φ (λ) = β

rank η − λ ≤ 1

since aj = 0 if j < s,

adjs η = (0, · · · , 0,α− a0

aσ(s)s

) and adjs λ = (0, · · · , 0,β − a0

aσ(s)s

)

and adjj η = adjj λ = 0 if j ≥ s + 1.

We also clearly have from Theorem 5.20 that

Φ (tη + (1− t)λ) = tα + (1− t)β.

Using the rank one convexity of f and the above construction we get

g (tα + (1− t)β) = g (Φ (tη + (1− t)λ)) = f (tη + (1− t)λ)

≤ tf (η) + (1− t) f (λ) = tg (Φ (η)) + (1− t) g (Φ (λ))

= tg (α) + (1− t) g (β)

which is the desired result.

Examples 215

5.3.6 The area type case

The next result is due to Morrey [453], but we follow a different proof, estab-lished in Dacorogna [171].

Theorem 5.47 Let N = n + 1 and for ξ ∈ R (n+1)×n let

adjn ξ = (det ξ 1,− det ξ 2, · · · , (−1)k+1 det ξ k, · · · , (−1)n+2 det ξ n+1),

where ξ k is the n × n matrix obtained from ξ by suppressing the k th row. Letf : R(n+1)×n → R and g : Rn+1 → R be such that


Then

f polyconvex ⇔ f quasiconvex ⇔ f rank one convex ⇔ g convex.

Remark 5.48 It is clear that if u : Rn → Rn+1, then adjn∇u represents thenormal to the surface

u (x) : x ∈ Rn .

In the case n = 2, u (x1, x2) =(u1, u2, u3

)we have

adj2∇u =

⎛⎜⎜⎝

∂u2

∂x1

∂u3

∂x2− ∂u2

∂x2

∂u3

∂x1

∂u3

∂x1

∂u1

∂x2− ∂u1

∂x1

∂u3

∂x2

∂u1

∂x1

∂u2

∂x2− ∂u1

∂x2

∂u2

∂x1

⎞⎟⎟⎠ . ♦

Before proceeding with the proof of the theorem, we mention an algebraiclemma, stronger than needed, that will be fully used in Section 6.6.4. We willprove the lemma, established in Dacorogna [171], after the proof of Theorem5.47.

Lemma 5.49 Let 0 < t < 1, a, b ∈ Rn+1 and ξ ∈ R(n+1)×n be such that

adjn ξ = ta + (1− t) b = 0.

Then there exist α, β ∈ R(n+1)×n such that⎧⎪⎨⎪⎩

ξ = tα + (1− t)β

adjn α = a, adjn β = b

rank α− β ≤ 1.

Proof. (Theorem 5.47). The implications

g convex ⇒ f polyconvex ⇒ f quasiconvex ⇒ f rank one convex


follow immediately from Theorem 5.3.It therefore remains to show that

f rank one convex ⇒ g convex.

We let t ∈ (0, 1) , a, b ∈ Rn+1 and we wish to show that

g (ta + (1− t) b) ≤ tg (a) + (1− t) g (b) (5.70)

provided f is rank one convex and f (ξ) = g (adjn ξ) . We divide the proof intotwo cases.

Case 1 : ta + (1− t) b = 0. We let

c := ta + (1− t) b =(c1, · · · , cn+1

)∈ Rn+1.

Since c = 0, we assume, for notational convenience, that c1 = 0 (the generalcase is handled similarly). We then let

ξ :=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

ξ11 ξ1

2 · · · ξ1n

ξ21 ξ2

2 · · · ξ2n

ξ31 ξ3

2 · · · ξ3n

......

. . ....

ξn+11 ξn+1

2 · · · ξn+1n

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−c2 −c3

c1· · · −cn+1

c1

c1 0 · · · 0

0 1 · · · 0...

.... . .

...0 0 · · · 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

It is then easy to see that

adjn ξ = c = ta + (1− t) b = 0.

We may now apply Lemma 5.49 to get α, β ∈ R(n+1)×n such that⎧⎪⎨⎪⎩

ξ = tα + (1− t)β


rank α− β ≤ 1.

Returning to (5.70), using the rank one convexity of f , we obtain

g (ta + (1− t) b) = g (adjn ξ) = f (ξ) = f (tα + (1− t)β)

≤ tf (α) + (1− t) f (β) = tg (a) + (1− t) g (b) ,

which is precisely the result.

Case 2 : ta + (1− t) b = 0. Observe first that the rank one convexity of fimplies that f is continuous (cf. Theorem 5.3), thus from f (ξ) = g (adjn ξ) wededuce that g is continuous. Therefore using Case 1 for a = a + (ǫ, 0, · · · , 0)

and b = b+(ǫ, 0, · · · , 0) where ǫ > 0 is arbitrary, we deduce (5.70) by continuityof g.

Examples 217

We now conclude this section by proving Lemma 5.49.

Proof. We give here a different proof than the one in Dacorogna [171] or[179]. We decompose the proof into two steps.

Step 1. We start by assuming that ξ ∈ R(n+1)×n has the following specialform

ξ = diag(n+1)×n (x1, · · · , xn) =

⎛⎜⎜⎜⎝

x1 · · · 0...

. . ....

0 · · · xn

0 · · · 0

⎞⎟⎟⎟⎠

with x1, · · · , xn ∈ R, all different from 0, and thus

adjn ξ =

⎛⎜⎜⎜⎝

0...0

(−1)n

x1 · · ·xn

⎞⎟⎟⎟⎠ = ta + (1− t) b = 0.

We next observe that for every λ ∈ Rn+1 and μ ∈ Rn we have

adjn (ξ + λ⊗ μ) = adjn ξ +⟨adjn−1 ξ; λ⊗ μ

⟩

where

⟨adjn−1 ξ; λ⊗ μ

⟩= (−1)n

⎛⎜⎜⎜⎜⎜⎝

−λn+1μ1

∏j =1 xj

...

−λn+1μn

∏j =n xj∑n

s=1 [ λsμs

∏j =s xj ]

⎞⎟⎟⎟⎟⎟⎠

.

We then search for α, β ∈ R(n+1)×n of the form

α = ξ + (1− t)λ⊗ μ

β = ξ − tλ⊗ μ

where λ ∈ Rn+1 and μ ∈ Rn are to be determined. We therefore immediatelydeduce that

ξ = tα + (1− t) β and rank α− β ≤ 1.

We next observe that

adjn α = adjn ξ + (1− t)⟨adjn−1 ξ; λ⊗ μ

⟩

adjn β = adjn ξ − t⟨adjn−1 ξ; λ⊗ μ

⟩.

Thus the equations adjn α = a and adjn β = b reduce to the single system ofequations ⟨

adjn−1 ξ; λ⊗ μ⟩

= a− b := c (5.71)

that we solve by considering two cases.


Case 1 : c1 = · · · = cn = 0. We then choose

λ1 = 1, λ2 = · · · = λn+1 = 0, μ2 = · · · = μn = 0

and

μ1 = (−1)n cn+1

∏nj=2 xj

so as to satisfy (5.71).

Case 2 : there exists k ∈ 1, · · · , n with ck = 0. Equation (5.71) is thensatisfied if we choose

μi = (−1)n+1 ci

∏j =i xj

, i = 1, · · · , n

and λi = 0 whenever i = k, n + 1, λn+1 = 1 and

λk = (−1)n cn+1

μk

∏j =k xj

=−cn+1

ck.

Step 2. We now reduce the general case ξ ∈ R(n+1)×n to the special form ofthe previous step by using the singular values decomposition theorem (cf. The-orem 13.3). We can indeed find R ∈ O (n + 1) , Q ∈ SO (n) and x1, · · · , xn ∈ Rso that

ξ := RξQ =

⎛⎜⎜⎜⎝

x1 · · · 0...

. . ....

0 · · · xn

0 · · · 0

⎞⎟⎟⎟⎠ .

Using Proposition 5.66, and noting that adjn Q = det Q = 1, we find that

adjn ξ = adjn R adjn ξ = 0.

Observing that adjn R ∈ O (n + 1) (by Proposition 5.66), we set

a := adjn R a and b := adjn R b

and we can find, from Step 1, α, β ∈ R(n+1)×n such that

⎧⎪⎨⎪⎩

ξ = tα + (1− t) β


rankα− β ≤ 1.

Setting

α := Rt α Qt and β := Rt β Qt

we have indeed obtained the claim of the lemma.

Examples 219

5.3.7 The example of Sverak

We now turn to an example of a rank one convex function that is not quasicon-vex. This fundamental result was obtained by Sverak [551] when N ≥ 3 andn ≥ 2 and we follow his presentation here. The question of extending Sverakexample to the case where n ≥ N = 2 is still open.

Theorem 5.50 Let N ≥ 3 and n ≥ 2. Then there exists f : RN×n → R rankone convex but not quasiconvex.

Proof. The proof is divided into four steps.

Step 1. Assume that we have already constructed a rank one convex functiong : R3×2 → R, that is not quasiconvex. In particular (appealing to Proposition

5.13), there exists η ∈ R3×2 and ψ ∈ W 1,∞per

(D2; R3

), where D2 = (0, 1)

2such

that ∫

D2

g (η +∇ψ (x)) dx < g (η) .

Then define π : RN×n → R3×2 to be

π (ξ) =

⎛⎜⎜⎝

ξ11 ξ1

2

ξ21 ξ2

2

ξ31 ξ3

2

⎞⎟⎟⎠ , for ξ ∈ RN×n.

Finally, letf (ξ) = g (π (ξ)) .

This function is clearly rank one convex, since g is. It is also not quasiconvex,since choosing any ξ ∈ RN×n so that π (ξ) = η, Dn = (0, 1)

nand

ϕi (x1, · · · , xn) :=

ψi (x1, x2) if i = 1, 2, 3

0 if not

we get that ϕ ∈W 1,∞per

(Dn; RN

)and

∫

Dn

f (ξ +∇ϕ (x)) dx < f (ξ) .

Step 2. In view of Step 1, it is therefore sufficient to prove the theorem forfunctions f : R3×2 → R. We first let

L := ξ ∈ R3×2 : ξ =

⎛⎝

x 00 yz z

⎞⎠ where x, y, z ∈ R

and P : R3×2 → L be defined by

P (ξ) :=

⎛⎜⎜⎝

ξ11 0

0 ξ22(

ξ31 + ξ3

2

)/2

(ξ31 + ξ3

2

)/2

⎞⎟⎟⎠ .


We next let g : L → R be defined by

g

⎛⎝

x 00 yz z

⎞⎠ = −xyz.

Finally, for ǫ, γ ≥ 0 let the function fǫ,γ : R3×2 → R be such that

fǫ,γ (ξ) := g (P (ξ)) + ǫ |ξ|2 + ǫ |ξ|4 + γ |ξ − P (ξ)|2 .

We claim that we can find ǫ and γ so that fǫ,γ is rank one convex (see Step 4)but not quasiconvex (see Step 3), giving the desired claim.

Step 3. Choose ξ = 0 and

ϕ (x1, x2) =1

2π

⎛⎜⎝

sin 2πx1

sin 2πx2

sin 2π (x1 + x2)

⎞⎟⎠ .

Observe that ϕ ∈ W 1,∞per

(D; R3

), where D = (0, 1)

2and ∇ϕ ∈ L (hence

P (∇ϕ) = ∇ϕ). Moreover,

∫

D

g (∇ϕ (x)) dx = −∫ 1

0

∫ 1

0

(cos 2πx1)2(cos 2πx2)

2dx1dx2 < 0.

Therefore (see Proposition 5.13), for every ǫ ≥ 0 sufficiently small and for everyγ ≥ 0, the function fǫ,γ is not quasiconvex.

Step 4. We now show that for every ǫ > 0, we can find γ = γ (ǫ) > 0 sothat fǫ,γ is rank one convex. This is equivalent to showing that the Legendre-Hadamard condition is satisfied, namely

Lf (ξ, η) :=d2

dt2[fǫ,γ (ξ + tη)]

∣∣∣∣t=0

≥ 0, ∀ξ, η ∈ R3×2 with rank η = 1.

(5.72)Letting

Lg (ξ, η) :=d2

dt2[g (P (ξ + tη))]

∣∣∣∣t=0

we find

Lf (ξ, η) = Lg (ξ, η) + 2ǫ |η|2 + 4ǫ |ξ|2 |η|2 + 8ǫ (〈ξ; η〉)2 + 2γ |η − P (η)|2 .

We show (5.72) in two substeps.

Step 4’. Observe that since g is a homogeneous polynomial of degree three,we can find c > 0 such that

Lg (ξ, η) ≥ −c |ξ| |η|2 .

Examples 221

We therefore deduce that

Lf (ξ, η) ≥ (−c + 4ǫ |ξ|) |ξ| |η|2

and thus (5.72) holds for every η ∈ R3×2 (independently of the fact thatrank η = 1) and for every ξ ∈ R3×2 that satisfies

|ξ| ≥ c

4ǫ.

Step 4”. It therefore remains to show (5.72) in the compact set

K :=(ξ, η) ∈ R3×2 × R3×2 : |ξ| ≤ c

4ǫ, |η| = 1, rank η = 1

in view of Step 4’ and of the fact that Lf (ξ, η) is homogeneous of degree two inthe variable η.

Moreover, we also find that

Lf (ξ, η) ≥ H (ξ, η, γ) := Lg (ξ, η) + 2ǫ |η|2 + 2γ |η − P (η)|2

and therefore (5.72) will follow if we can show that for every ǫ > 0 we can findγ = γ (ǫ) so that H ≥ 0 on K.

Assume, for the sake of contradiction, that this is not the case. We can thenfind γν →∞, (ξν , ην) ∈ K so that

Lg (ξν , ην) + 2ǫ ≤ Lg (ξν , ην) + 2ǫ + 2γν |ην − P (ην)|2 < 0.

Since K is compact, we have up to a subsequence (still labeled (ξν , ην)) that

(ξν , ην)→ (ξ, η) ∈ K, Lg (ξ, η) + 2ǫ ≤ 0 and P (η) = η.

However we have ǫ > 0 and, by construction,

Lg (ξ, η) ≡ 0, ∀ ξ, η ∈ R3×2 with P (η) = η and rankη = 1.

This leads to the desired contradiction and therefore the theorem holds.

5.3.8 The example of Alibert-Dacorogna-Marcellini

We now turn our attention to an example where N = n = 2. It involves ahomogeneous polynomial of degree four. We characterize, with the help of onesingle real parameter, the different notions of convexity.

Theorem 5.51 Let γ ∈ R and let fγ : R2×2 → R be defined as

fγ (ξ) = |ξ|2 ( |ξ|2 − 2γ det ξ ).


Then

fγ is convex ⇔ |γ| ≤ γc =2

3

√2,

fγ is polyconvex ⇔ |γ| ≤ γp = 1,

fγ is quasiconvex ⇔ |γ| ≤ γq and γq > 1,

fγ is rank one convex ⇔ |γ| ≤ γr =2√3

.

We now make some comments about this theorem.

(i) The last result and the fact that if fγ is polyconvex, then |γ| ≤ 1, wereestablished by Dacorogna-Marcellini [193]. All the other results were first provedin Alibert-Dacorogna [14]. The most interesting fact is the third one.

(ii) The example also provides a quasiconvex function that is not polyconvex(such an example was already seen in Theorem 5.25 when N, n ≥ 3; see alsowhen n = N = 2, Theorem 5.54 and Sverak [552]).

(iii) The problem of knowing if γq = 2/√

3 is still open. If this is not thecase (meaning that γq < 2/

√3), then this would provide a rank one convex

function that is not quasiconvex, giving a final answer to this long standingquestion. However many numerical evidences tend to indicate that γq = 2/

√3,

see Dacorogna-Douchet-Gangbo-Rappaz [185], Dacorogna-Haeberly [191] andGremaud [321].

(iv) The polyconvexity of the function

f1 (ξ) = |ξ|2 ( |ξ|2 − 2 det ξ )

has, since the work of Alibert-Dacorogna [14], been reproved notably by Iwaniec-Lutoborski [353]. Hartwig [335] also proved this fact exhibiting a convex func-tion F : R2×2 × R → R, namely

F (ξ, δ) =

[ |ξ|2 + 2 det ξ − 2δ ][ |ξ|2 + 2 det ξ − 4δ ] if |ξ|2 + 2 det ξ ≥ 4δ

0 otherwise,

so thatf1 (ξ) = F (ξ,det ξ) .

We now proceed with the proof of the theorem. But before that we wantto observe that in all four statements we can restrict our attention to the casewhere γ ≥ 0. Indeed, observe first that

fγ (Qξ) = f−γ (ξ) for every ξ ∈ R2×2 and Q ∈ O (2) with detQ = −1.

This easily implies that fγ is convex (respectively polyconvex, quasiconvex, rankone convex) if and only if f−γ is convex (respectively polyconvex, quasiconvex,rank one convex). Hence, we may assume throughout, without loss of generality,that γ ≥ 0.

Examples 223

We first start with the statement on the convexity of fγ .

Proof. (Theorem 5.51: Convexity). We have to show that

fγ is convex ⇔ γ ≤ γc =2

3

√2.

This result was first proved by Alibert-Dacorogna, but we give here the proofbased on Dacorogna-Marechal [204].

According to Theorem 5.33, it is sufficient to verify the claim only on diag-onal matrices. So let

g (x, y) :=(x2 + y2

) [(x2 + y2

)− 2γxy

].

The Hessian of g is therefore given by

∇2g (x, y) =

(4(x2 + y2

)+ 8x2 − 12γxy 8xy − 6γ

(x2 + y2

)

8xy − 6γ(x2 + y2

)4(x2 + y2

)+ 8y2 − 12γxy

).

Setting

x = r cosθ

2, y = r sin

θ

2

we find that

∇2g (x, y) = 2r2

(4 + 2 cos θ − 3γ sin θ 2 sin θ − 3γ

2 sin θ − 3γ 4− 2 cos θ − 3γ sin θ

).

The function g is therefore convex if and only if the trace and the determinantof ∇2g are non negative. This is true if and only if

4− 3γ sin θ ≥ 0,

12− 9γ2 − 12γ sin θ + 9γ2 sin2 θ ≥ 0.

Step 1 : (⇐). We first consider the case where γ ≤ γc = 2√

2/3. Thisimmediately implies that the first inequality holds. Since the discriminant ofthe polynomial (in sin θ) that appears in the second inequality is given by

∆ = 36γ2(9γ2 − 8

)≤ 0,

we have indeed obtained the claim.

Step 2 : (⇒). We now show that if fγ is convex, then γ ≤ γc . We prove theresult by contradiction and write for a certain t > 1

γ = tγc =2

3

√2 t.

The polynomial that appears in the second inequality is then transformed into

12− 8t2 − 8√

2 t sin θ + 8t2 sin2 θ.


Observe that the minimum of this polynomial (in sin θ) is attained at

sin θ =1√2 t

and its value is then8(1− t2

)< 0.

This is the desired contradiction.

We now discuss the rank one convexity of fγ .

Proof. (Theorem 5.51: Rank one convexity). We have to show that

fγ is rank one convex ⇔ 0 ≤ γ ≤ γr = 2/√

3

and this was established first by Dacorogna-Marcellini [193].

We start with some notations and with the computation of the secondvariation.

Notation. To every ξ ∈ R2×2 we associate ξ ∈ R2×2 in the following way

ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

), ξ =

(ξ22 −ξ2

1

−ξ12 ξ1

1

).

We immediately observe that

⎧⎪⎨⎪⎩

|ξ| = | ξ | , det ξ = det ξ

〈 ξ; η 〉 = 〈 ξ; η 〉 , 〈 ξ; ξ 〉 = 2 det ξ

det (ξ + η) = det ξ + 〈 ξ; η 〉+ det η

where 〈·; ·〉 denotes the scalar product in R2×2. We also have that

∂

∂ξij

(det ξ) = ξ ij , i.e. ∇ (det ξ) = ξ.

(Note that in the notations of Section 5.4 ξ = adj1 ξ).It will be convenient to decompose any matrix in its ”conformal” and ”anti-

conformal” parts, which are given by

ξ+ :=1

2(ξ + ξ ), ξ− :=

1

2(ξ − ξ ).

We find the following relations. For ξ, η ∈ R2×2 we have

2 det ξ+ =∣∣ξ+∣∣2 and 2 det ξ− = −

∣∣ξ−∣∣2

|ξ|2 =∣∣ξ+∣∣2 +

∣∣ξ−∣∣2 and 2 det ξ =

∣∣ξ+∣∣2 −

∣∣ξ−∣∣2 = 2 det ξ+ + 2 det ξ−

〈 ξ; η 〉 = 〈 ξ+; η+ 〉+ 〈 ξ−; η− 〉 and 〈 ξ+; η− 〉 = 〈 ξ−; η+ 〉 = 0

Examples 225

|ξ|2 − 2 det ξ = 2∣∣ξ−∣∣2 and |ξ|2 + 2 det ξ = 2

∣∣ξ+∣∣2 .

Second variation. We next compute the second variation of fγ

ψγ (ξ, η) :=

2∑

i,j=1

2∑

α,β=1

∂2fγ

∂ξiα∂ξj

β

ηiαηj

β .

We first calculate ∇f and find

∂fγ

∂ξiα

= 4 |ξ|2 ξiα − 4γ (det ξ) ξi

α − 2γ |ξ|2 ξ iα .

We then deduce that, for 1 ≤ i, j, α, β ≤ 2,

∂2fγ

∂ξiα∂ξj

β

= 8ξiαξj

β + 4 |ξ|2 δijδαβ − 4γξiαξ j

β

−4γ (det ξ) δijδαβ − 4γξ iαξj

β − 2γ |ξ|2 δ ij δαβ ,

where

δij =

1 if i = j

0 if i = j, δ ij =

(−1)

jif i = j

0 otherwise

and similarly for δαβ and δαβ . We therefore have that, if

ψγ (ξ, η) =2∑

i,j,α,β=1

∂2fγ

∂ξiα∂ξj

β

ηiαηj

β ,

then

ψγ (ξ, η) = 8 (〈ξ; η〉)2 + 4 |ξ|2 |η|2 − 8γ 〈ξ; η〉〈 ξ; η〉−4γ |η|2 det ξ − 4γ |ξ|2 det η.

(5.73)

In terms of the above decomposition we have

14ψγ (ξ, η) = 2 (1− γ) 〈 ξ+; η+ 〉2 + 4 〈 ξ+; η+ 〉〈 ξ−; η− 〉

+2 (1 + γ) 〈 ξ−; η− 〉2 + (1− γ) |ξ+|2 |η+|2

+ |ξ+|2 |η−|2 + |ξ−|2 |η+|2 + (1 + γ) |ξ−|2 |η−|2 .

(5.74)

Step 1 : (⇐). We first show that if γ ≤ 2/√

3, then fγ is rank one convex.This is equivalent to showing (see Theorem 5.3) that the Legendre-Hadamardcondition holds, i.e.,

ψγ (ξ, η) ≥ 0, for every ξ, η ∈ R2×2 with det η = 0. (5.75)


Using (5.74) and the fact that det η = 0 if and only if |η+|2 = |η−|2 , weimmediately obtain

14ψγ (ξ, η) = [ (4− 3γ) 〈 ξ+; η+ 〉2 + 4 〈 ξ+; η+ 〉〈 ξ−; η− 〉

+ (4 + 3γ) 〈 ξ−; η− 〉2 ]

+[ (2− γ) (|ξ+|2 |η+|2 − 〈 ξ+; η+ 〉2)+ (2 + γ) (|ξ−|2 |η−|2 − 〈 ξ−; η− 〉2) ].

Since γ ≤ 2/√

3 ≤ 2, we deduce that the term in the second bracket is non-negative. The discriminant of the term in the first bracket is

∆ = 4 [4− (4− 3γ) (4 + 3γ)]

and is non-positive if γ ≤ 2/√

3. Therefore

ψγ (ξ, η) ≥ 0, for every γ ≤ 2√3,

as claimed and the proof of Step 1 is complete.

Step 2 : (⇒). We now prove that if fγ is rank one convex, then γ ≤ 2/√

3.In order to show the result, we prove that if γ > 2/

√3, then fγ is not rank one

convex, which is equivalent (see (5.75)) to showing that there exist ξγ , ηγ ∈ R2×2

with det ηγ = 0 such that ψγ (ξγ , ηγ) < 0. This is easily done. Choose

ξγ =

(a 00 1

), ηγ =

(1 00 0

)

with a defined below. A direct computation gives

1

4ψγ (ξγ , ηγ) = 3a2 − 3γa + 1.

If the discriminant ∆ = 9γ2 − 12 is positive, and this happens if γ > 2/√

3, wecan then choose a so that ψγ (ξγ , ηγ) < 0, as wished.

This concludes the study of the rank one convexity of the function fγ .

We next turn our attention to the polyconvexity of fγ .

Proof. (Theorem 5.51: Polyconvexity). We have to prove that

fγ is polyconvex ⇔ 0 ≤ γ ≤ γp = 1.

Step 1 : (⇒). We first show that if fγ is polyconvex, then 0 ≤ γ ≤ 1. UsingCorollary 5.9, we can find c ≥ 0 such that

fγ (ξ) ≥ −c(1 + |ξ|2) for every ξ ∈ R2×2.

In particular the inequality holds for

ξ = t I, t ∈ R.

Examples 227

We therefore find that

fγ (ξ) = 4 (1− γ) t4 ≥ −c(1 + 2t2).

Dividing both sides by t4 and letting t→∞, we find that

1− γ ≥ 0,

as wished.

Step 2: (⇐). We start with a preliminary step.

Step 2’. We show that if fγ is polyconvex, then fβ is polyconvex for every0 ≤ β ≤ γ. We have to prove, according to Theorem 5.6, that

fβ (ξ) ≤6∑

i=1

λifβ (ξi)

whenever ξ, ξi ∈ R2×2, λ ∈ Λ6 , satisfy

ξ =

6∑

i=1

λiξi ,

6∑

i=1

λi det ξi = det ξ.


Case 1. Assume that

6∑

i=1

λi |ξi|2 det ξi ≤ |ξ|2 det ξ.

Then the claim is trivial since, recalling that β ≥ 0 and observing that thefunction ξ → |ξ|4 is convex,

fβ (ξ) = |ξ|4 − 2β |ξ|2 det ξ ≤6∑

i=1

λi[ |ξi|4 − 2β |ξi|2 det ξi ] =

6∑

i=1

λifβ (ξi) .

Case 2. Assume now that

6∑

i=1

λi |ξi|2 det ξi ≥ |ξ|2 det ξ.

Then the claim follows from the observation

fβ (ξ) = fγ (ξ)− 2 (β − γ) |ξ|2 det ξ,

from the hypothesis 0 ≤ β ≤ γ and from the polyconvexity of fγ .

This achieves the proof of Step 2’.

Step 2”. It therefore remains to show that

f1 (ξ) = |ξ|2 ( |ξ|2 − 2 det ξ )


is polyconvex and the proof will be complete. As we already mentioned, thereare three proofs of the preceding fact: the original one of Alibert-Dacorogna,the one of Hartwig and that of Iwaniec-Lutoborski, which is in the same spiritas the one of Alibert-Dacorogna but slightly simpler, and we will follow herethis last one. We will show that, for every ξ, η ∈ R2×2,

f1 (η) ≥ f1 (ξ) + 4(|ξ|2 − det ξ) 〈ξ; η − ξ〉 − 2 |ξ|2 [det η − det ξ] .

This last inequality, combined with Theorem 5.6, gives that f1 is polyconvex.

In order to show the inequality, it is sufficient (see Theorem 5.43 and theremark following it) to verify it on diagonal matrices, so we will set

ξ = diag (a, b) and η = diag (x, y) .

We therefore have to prove that

(x− y)2 (

x2 + y2)≥ (a− b)

2 (a2 + b2

)

+ 4(a2 + b2 − ab

)[a (x− a) + b (y − b)]

− 2(a2 + b2

)(xy − ab) .

This can be rewritten, setting X = x− a and Y = y − b, as

αX2 − 2βXY + γY 2 ≥ 0 (5.76)

where

α = (x− y + a)2

+ a2 + (a− b)2

β = (a− b) (x− y + a− b)

γ = (x− y − b)2+ b2 + (a− b)

2.

The inequality (5.76), and thus the polyconvexity of f1 , follows from the factthat α, γ ≥ 0 and from

∆ = αγ − β2

= [ a2 + b2 − (x− y) (a− b) ]2

+ (x− y + a− b)2[ (x− y)

2+ (a− b)

2]

≥ 0.

This concludes the claim for the polyconvexity.

We finally show the statement on quasiconvexity. It is clearly the mostdifficult to prove and we will first start with the following result, proved byAlibert-Dacorogna [14], which is a consequence of regularity results for Laplaceequation. We will use it twice: once when ξ = 0 and p = 4, in the proof ofTheorem 5.51, and the second time when ξ = 0 and 1 0 such that

∫

Ω

[ |∇ϕ (x)|2 ± 2 det (∇ϕ (x)) ]p/2dx ≥ ǫ

∫

Ω

|∇ϕ (x)|p dx (5.77)

for every ϕ ∈ W 1,∞0

(Ω; R2

).

Moreover, when p = 4, the inequality∫

Ω

[ |ξ +∇ϕ (x)|2 ± 2 det (ξ +∇ϕ (x)) ]2dx

≥ (|ξ|2 ± 2 det ξ)2 meas Ω + ǫ

∫

Ω

|∇ϕ (x)|4 dx

(5.78)

holds for every ξ ∈ R2×2 and every ϕ ∈W 1,∞0

(Ω; R2

).

The result (5.77) is clearly non-trivial, except when p = 2 (in this case wecan take ǫ = 1 and equality, instead of inequality, holds). Observe also thatthe inequality (5.77) shows that the functional on the left-hand side of (5.77) iscoercive in W 1,p

0

(Ω, R2

), even though the integrand is not coercive (not even

up to a quasiaffine function, which here can be at most quadratic).

Proof. (Theorem 5.52). We prove (5.77) and (5.78) only for the minus sign,the proof being identical for the plus sign. For this purpose we adapt an ideaof Sverak [552].

Step 1. We first prove the result for ξ = 0 and 1 < p <∞. We start with analgebraic relation. We clearly have that there exists a constant α = α (p) suchthat for every ξ ∈ R2×2

[|ξ|2 − 2 det ξ

]p/2

=[(

ξ11 − ξ2

2

)2+(ξ12 + ξ2

1

)2]p/2

≥ α[∣∣ξ1

1 − ξ22

∣∣p +∣∣ξ1

2 + ξ21

∣∣p].

We now turn to the claim and note that it is sufficient to prove the claim forϕ =

(ϕ1, ϕ2

)∈ C∞

0

(Ω, R2

), the general result being obtained by density. We

also extend the function outside Ω by setting ϕ ≡ 0 there. Then denoting∂ϕj/∂xi by ∂iϕ

j , i, j ∈ 1, 2, we have from the above algebraic relation∫

Ω

[ |∇ϕ (x)|2 − 2 det (∇ϕ (x)) ]p/2dx

≥ α

∫

Ω

[∣∣∂1ϕ

1 (x)− ∂2ϕ2 (x)

∣∣p +∣∣∂2ϕ

1 (x) + ∂1ϕ2 (x)

∣∣p ]dx.

The classical regularity results for Cauchy-Riemann equations (see, for example,Proposition 4 on page 60 in Stein [543]) leads to the existence of a constant β > 0such that

‖∇ϕ‖pLp ≤ β

∫

Ω

[∣∣∂1ϕ

1 (x)− ∂2ϕ2 (x)

∣∣p +∣∣∂2ϕ

1 (x) + ∂1ϕ2 (x)

∣∣p ]dx.


Choosing ǫ ≤ α/β, we have (5.77).

Step 2. We now prove the general case, where ξ is not necessarily 0 butp = 4. We start with the following algebraic observation

[ 〈 ξ − ξ; η 〉 ]2 ≤ | ξ − ξ |2 |η|2 = 2[ |ξ|2 − 2 det ξ ] |η|2 . (5.79)

We next compute

[ |ξ +∇ϕ|2 − 2 det (ξ +∇ϕ) ]2

= [ |ξ|2 − 2 det ξ + 2 〈 ξ − ξ;∇ϕ 〉+ |∇ϕ|2 − 2 det (∇ϕ) ]2

= [ |ξ|2 − 2 det ξ ]2 + 4[ 〈 ξ − ξ;∇ϕ 〉 ]2 + [ |∇ϕ|2 − 2 det (∇ϕ) ]2

+4[ |ξ|2 − 2 det ξ ][ 〈 ξ − ξ;∇ϕ 〉 − det (∇ϕ) ]

+2[ |ξ|2 − 2 det ξ ] |∇ϕ|2 + 4[ |∇ϕ|2 − 2 det (∇ϕ) ] 〈 ξ − ξ;∇ϕ 〉 .

Using (5.79), we obtain

[ |ξ +∇ϕ|2 − 2 det (ξ +∇ϕ) ]2

≥ [ |ξ|2 − 2 det ξ ]2 + 5[ 〈 ξ − ξ;∇ϕ 〉 ]2 + [ |∇ϕ|2 − 2 det (∇ϕ) ]2

+4[ |ξ|2 − 2 det ξ ][ 〈 ξ − ξ;∇ϕ 〉 − det (∇ϕ) ]

+4[ |∇ϕ|2 − 2 det (∇ϕ) ] 〈 ξ − ξ;∇ϕ 〉 .

Noticing that

0 ≤ 5[ 〈 ξ − ξ;∇ϕ 〉 ]2

+4[ |∇ϕ|2 − 2 det (∇ϕ) ] 〈 ξ − ξ;∇ϕ 〉+4

5[ |∇ϕ|2 − 2 det (∇ϕ) ]2

we deduce that

[ |ξ +∇ϕ|2 − 2 det (ξ +∇ϕ) ]2 ≥ [ |ξ|2 − 2 det ξ ]2 + 15 [ |∇ϕ|2 − 2 det (∇ϕ) ]2

+4[ |ξ|2 − 2 det ξ ][ 〈 ξ − ξ;∇ϕ 〉 − det (∇ϕ) ].

We then integrate the above inequality, bearing in mind that ϕ = 0 on ∂Ω, andwe find∫

Ω

[ |ξ +∇ϕ|2 − 2 det (ξ +∇ϕ) ]2dx ≥ [ |ξ|2 − 2 det ξ ]2 measΩ

+1

5

∫

Ω

[ |∇ϕ|2 − 2 det (∇ϕ) ]2dx.

Using Step 1, with p = 4, we find that∫

Ω

[ |ξ +∇ϕ|2−2 det (ξ +∇ϕ) ]2dx ≥ [ |ξ|2−2 det ξ ]2 measΩ +α

5β

∫

Ω

|∇ϕ|4 dx.

Choosing ǫ = α/5β, we have indeed established (5.78) and thus the theorem isproved.

Examples 231

We now continue with the proof of the main theorem.

Proof. (Theorem 5.51: Quasiconvexity). We have to establish that

fγ is quasiconvex ⇔ γ ≤ γq and γq > 1.

In the first step, we prove the existence of a γq with the above property; this isthe easy part of the proof. The difficult part, which will be dealt with in Step2, is to show that γq > 1.

Step 1 : Existence of γq . We start by showing that if fγ is quasiconvex, thenfβ is quasiconvex for every 0 ≤ β ≤ γ. Let

Iγ (ξ, ϕ) :=

∫

Ω

[fγ (ξ +∇ϕ (x))− fγ (ξ)] dx

for every ξ ∈ R2×2 and every ϕ ∈ W 1,∞0

(Ω; R2

). We have to show that

Iγ (ξ, ϕ) ≥ 0 implies Iβ (ξ, ϕ) ≥ 0. We have to deal with two cases.Case 1. If

∫

Ω

[ |ξ +∇ϕ (x)|2 det (ξ +∇ϕ (x))− |ξ|2 det ξ ]dx ≤ 0,

then the claim is trivial using the convexity of ξ → |ξ|4 and the fact that β ≥ 0.Case 2. If

∫

Ω

[ |ξ +∇ϕ (x)|2 det (ξ +∇ϕ (x))− |ξ|2 det ξ ]dx ≥ 0,

we observe that

Iβ (ξ, ϕ) − Iγ (ξ, ϕ)

= 2 (γ − β)

∫

Ω

[ |ξ +∇ϕ (x)|2 det (ξ +∇ϕ (x))− |ξ|2 det ξ ]dx ≥ 0,

as wished.We may now define γq by taking the largest γ such that fγ is quasiconvex.

It exists because of the preceding observation and from the fact that

1 = γp ≤ γq ≤ γr =2√3

and this completes Step 1.

Step 2 : γq > 1. We therefore have to show that there exists α > 0 smallenough, so that if γ = 1+α, then fγ is quasiconvex. We start with a preliminaryresult.

Step 2’. We prove the quasiconvexity of fγ at 0 for γ = 1 + α with α > 0small enough. We have to prove that

∫

Ω

fγ (∇ϕ (x)) dx ≥ 0


for every ϕ ∈ W 1,∞0

(Ω; R2

)and for some α > 0. Observe first the following

algebraic inequality (we use the fact that |ξ|2 ≥ 2 det ξ), valid for any ξ ∈ R2×2,

fγ (ξ) = |ξ|4 − 2 (1 + α) |ξ|2 det ξ

=1

2[ |ξ|4 − 4 |ξ|2 det ξ + 4 (det ξ)

2]

+1

2[ |ξ|4 − 4 (det ξ)

2]− 2α |ξ|2 det ξ

≥ 1

2[ |ξ|2 − 2 det ξ ]2 − α |ξ|4 .

We then integrate and use Theorem 5.52 to get

∫

Ω

fγ (∇ϕ (x)) dx ≥ (ǫ− α)

∫

Ω

|∇ϕ (x)|4 dx. (5.80)

Choosing 0 ≤ α ≤ ǫ, we have indeed obtained the result.

Step 2”. We now proceed with the general case. We already know thatγq ≥ γp = 1, so we will assume throughout this step that γ ≥ 1 and we will setα = γ − 1.

Expanding fγ , keeping in mind its special structure, we find

fγ (ξ + η) = fγ (ξ) + 〈 ∇fγ (ξ) ; η 〉+1

2〈 ∇2fγ (ξ) η; η 〉

+ 〈 ∇fγ (η) ; ξ 〉+ fγ (η) .

Recall that 〈 ∇2fγ (ξ) η; η 〉 is given by (5.73). We rewrite this as

fγ (ξ + η)− fγ (ξ) = Aγ (ξ, η) + Bγ (ξ, η) + Cγ (ξ, η) + Dγ (η) + Eγ (η) (5.81)

where

Aγ (ξ, η) := 〈 ∇fγ (ξ) ; η 〉 − 2γ |ξ|2 det η

Bγ (ξ, η) := 12 〈 ∇2fγ (ξ) η; η 〉+ 2γ |ξ|2 det η

= 4 (〈ξ; η〉)2 + 2 |ξ|2 |η|2 − 4γ 〈ξ; η〉〈 ξ; η〉 − 2γ |η|2 det ξ

Cγ (ξ, η) := 〈 ∇fγ (η) ; ξ 〉= 4 〈 ξ; η 〉 |η|2 − 4γ 〈 ξ; η 〉det η − 2γ 〈 ξ; η 〉 |η|2

Dγ (η) := (1− ǫ) f1 (η) +ǫ2

2|η|4

Eγ (η) := ǫf1 (η)− 2 (γ − 1) |η|2 det η − ǫ2

2 |η|4

≥ ǫf1 (η)− (α + ǫ2

2 ) |η|4 .

Observe that

Dγ (η) + Eγ (η) = fγ (η) .

Examples 233

From Step 2’ (applying (5.80) with γ = 1 and hence α = 0), we have that, forevery ϕ ∈W 1,∞

0

(Ω; R2

),

∫

Ω

Eγ (∇ϕ (x)) dx ≥ [ ǫ2 − (α +ǫ2

2) ]

∫

Ω

|∇ϕ (x)|4 dx

which for α > 0 sufficiently small with respect to ǫ2 leads to

∫

Ω

Eγ (∇ϕ (x)) dx ≥ 0. (5.82)

We also have that for ǫ > 0 and α > 0 even smaller (see Lemma 5.53)

σǫ,α (ξ, η) = Bγ (ξ, η) + Cγ (ξ, η) + Dγ (η) ≥ 0 (5.83)

for every ξ, η ∈ R2×2.We are now in a position to conclude by combining (5.81), (5.82) and (5.83).

We therefore have, for every ξ ∈ R2×2, ϕ ∈ W 1,∞0

(Ω; R2

),

∫

Ω

[fγ (ξ +∇ϕ (x))− fγ (ξ)]dx ≥∫

Ω

Aγ (ξ,∇ϕ (x)) dx = 0.

This is the desired claim.

The above proof relied on the following algebraic lemma.

Lemma 5.53 Let

σǫ,α (ξ, η) = Bγ (ξ, η) + Cγ (ξ, η) + Dγ (η)

where γ = 1 + α and

Bγ (ξ, η) = 4 (〈ξ; η〉)2 + 2 |ξ|2 |η|2 − 4γ 〈ξ; η〉〈 ξ; η〉 − 2γ |η|2 det ξ

Cγ (ξ, η) = 4 〈 ξ; η 〉 |η|2 − 4γ 〈 ξ; η 〉det η − 2γ 〈 ξ; η 〉 |η|2

Dγ (η) = (1− ǫ) [ |η|4 − 2 |η|2 det η ] +ǫ2

2|η|4 .

For every ǫ > 0 sufficiently small, there exists α0 = α0 (ǫ) > 0 such that if0 ≤ α ≤ α0 , then

σǫ,α (ξ, η) ≥ 0, for every ξ, η ∈ R2×2.

Proof. The idea of the proof is to show that, for every ǫ > 0 sufficiently small,there exists α0 = α0 (ǫ) > 0 such that if 0 ≤ α ≤ α0 , then

ξ → σǫ,α (ξ, η)


is a strictly convex polynomial of degree two for every η ∈ R2×2. In Step 2we prove that by choosing both ǫ sufficiently small and α0 (ǫ) even smaller(uniformly with respect to η), then

σǫ,α

(ξ, η)≥ 0

at the unique minimum point ξ = ξ (η) .

Step 1. We first show that for α = γ − 1 > 0 sufficiently small

Bγ (ξ, η) ≥ 11− 9γ2

6|ξ|2 |η|2 for every ξ, η ∈ R2×2. (5.84)

The case ξ = 0 or η = 0 being trivial, we can assume because of the homogeneityof Bγ that

|ξ| = |η| = 1.

Moreover, since QξR = QξR for every Q, R ∈ SO (2) , we have

Bγ (ξ, QηR) = Bγ

(QtξRt, η

)

and thus it is enough to prove (5.84) for matrices ξ and η of the form (accordingto Theorem 13.3)

ξ =

(cos θ cosA sin A cosB

sin A sin B sin θ cosA

)and η =

(cosϕ 0

0 sin ϕ

).

We therefore find

Bγ (ξ, η) = 2 + γ sin (2B) sin2 A +[4 cos2 (θ − ϕ)− 4γ cos (θ − ϕ) sin (θ + ϕ)− γ sin (2θ)

]cos2 A.

Since sin (2B) ≥ −1, we find that

Bγ (ξ, η) ≥ 2− γ+[γ + 4 cos2 (θ − ϕ)− 4γ cos (θ − ϕ) sin (θ + ϕ) − γ sin (2θ)

]cos2 A.

Since γ > 1 is sufficiently close to 1 and we want to minimize Bγ (ξ, η) , we haveto choose cos2 A = 1. We can thus write

Bγ (ξ, η) ≥ 2 + 4 cos2 (θ − ϕ)− 4γ cos (θ − ϕ) sin (θ + ϕ)− γ sin (2θ)

or, writing a = 2θ and b = 2ϕ,

Bγ (ξ, η) ≥ g (a, b) := 4 + 2 cos (a− b)− 3γ sin a− 2γ sin b. (5.85)

We easily have that

∇g (a, b) = 0 ⇔ cos b = −3

2cos a =

1

γsin (a− b) . (5.86)

Examples 235

We can next write that

g (a, b) ≥ min g (a, b) : ∇g (a, b) = 0 (5.87)

and therefore two cases can happen.

Case 1: cos a = cos b = sin (a− b) = 0. At such a point (recalling that γ issufficiently close to 1) we have

g (a, b) ≥ 6− 5γ. (5.88)

Case 2: cos a = 0 and cos b = 0. From (5.86), we find

cos b = −3

2cos a and sin b =

3

2(γ − sin a) .

We hence deduce that

4

9=

4

9cos2 b +

4

9sin2 b = γ2 + 1− 2γ sin a.

Therefore at such a point (a, b) we have

g (a, b) = 4 + 2 cos a cos b + 2 sin a sin b− 3γ sin a− 2γ sin b

= 1− 3γ2 + 3γ sin a =11− 9γ2

6.

Combining (5.85), (5.87), (5.88) and the above identity, we have indeed obtained(5.84).

Step 2. We now prove that by choosing both ǫ sufficiently small and α0 (ǫ)even smaller (uniformly with respect to η), then

σǫ,α (ξ, η) ≥ 0 for every ξ, η ∈ R2×2.

We start by observing that

σǫ,α (ξ, 0) = 0 for every ξ ∈ R2×2.

So from now on we will assume that η = 0 and is fixed. From Step 1, we seethat the function

ξ → σǫ,α (ξ, η)

has a unique minimum, which satisfies

∇ξσǫ,α (ξ, η) = 0;

i.e.

4 |η|2 η − 4γ (det η) η − 2γ |η|2 η + 4 |η|2 ξ − 2γ |η|2 ξ

+8 〈 ξ; η 〉 η − 4γ 〈 ξ; η 〉 η − 4γ 〈 ξ; η 〉 η = 0.(5.89)


We now multiply (5.89) first by ξ, then by η and finally by η to get

2 〈 ξ; η 〉 (|η|2 − γ det η)− γ 〈 ξ; η 〉 |η|2

= 2 |ξ|2 |η|2 − 2γ |η|2 det ξ + 4 (〈 ξ; η 〉)2 − 4γ 〈 ξ; η 〉〈 ξ; η 〉+4 〈 ξ; η 〉 |η|2 − 4γ 〈 ξ; η 〉det η − 2γ 〈 ξ; η 〉 |η|2

(5.90)

−γ 〈 ξ; η 〉 |η|2

= − 23 |η|

4+ 4

3γ |η|2 det η − 2 〈 ξ; η 〉 |η|2 + 43γ 〈 ξ; η 〉det η

(5.91)

2 〈 ξ; η 〉 (|η|2 − 2γ det η)

= 〈 ξ; η 〉 (3γ |η|2 − 8 det η) + γ |η|4 − 4 |η|2 det η + 4γ (det η)2 .(5.92)

We next combine (5.89) to (5.92) to show that σǫ,α ≥ 0 at a stationary pointprovided α = γ − 1 and ǫ are small enough. Combining (5.91) and (5.92), so as

to eliminate 〈 ξ; η 〉 , we find that

〈 ξ; η 〉 [ 3(4− 3γ2

)|η|4 − 8γ |η|2 det η + 16γ2 (det η)

2]

= − |η|2 [(4− 3γ2

)|η|4 − 4γ |η|2 det η + 4γ2 (det η)

2].

(5.93)

We now use (5.90), (5.91) and (5.93) to compute σǫ,α at the minimum point.First appeal to (5.90) to obtain

σǫ,α = 2 〈 ξ; η 〉 (|η|2 − γ det η)− γ 〈 ξ; η 〉 |η|2

+(1− ǫ +ǫ2

2) |η|4 − 2 (1− ǫ) |η|2 det η.

Replacing the second term, with the help of (5.91), we find

σǫ,α = − 23γ 〈 ξ; η 〉det η + (

1

3− ǫ +

ǫ2

2) |η|4 − 2

(1− ǫ− 2

3γ)|η|2 det η.

Inserting (5.93) in the above identity, we obtain

3σǫ,α

|η|2[ 3(4− 3γ2

)|η|4 − 8γ |η|2 det η + 16γ2 (det η)2 ]

= [ (1− 3ǫ + 3ǫ2

2 ) |η|2 − 2 (3− 3ǫ− 2γ) det η ]

×[ 3(4− 3γ2

)|η|4 − 8γ |η|2 det η + 16γ2 (det η)

2]

+2γ det η[(4− 3γ2

)|η|4 − 4γ |η|2 det η + 4γ2 (det η)

2].

Setting

t = |η| and δ = 2 det (η/ |η|) ( ⇒ |δ| ≤ 1),

Examples 237

we get

3σǫ,α

t4[3(4− 3γ2

)− 4γδ + 4γ2δ2

]

=[(

1− 3ǫ + 32ǫ2)− (3− 3ǫ− 2γ) δ

] [3(4− 3γ2

)− 4γδ + 4γ2δ2

]

+γδ[(4− 3γ2

)− 2γδ + γ2δ2 ].

Letting α = γ − 1 ≥ 0 and using the fact that |δ| ≤ 1, we find the followingthree estimates for α small enough

[ 3(4− 3γ2

)− 4γδ + 4γ2δ2 ]

= [ 3(1− 6α− 3α2

)− 4 (1 + α) δ + 4 (1 + α)

2δ2 ]

≤ [ 3− 4δ + 4δ2 ] + 1 ≤ 12

[ (1 − 3ǫ + 32ǫ2)− (3− 3ǫ− 2γ) δ ] [ 3

(4− 3γ2

)− 4γδ + 4γ2δ2 ]

= 32ǫ2[ 2 + (1− 2δ)

2]

+ (1− 3ǫ) (1− δ)[3− 4δ + 4δ2

]+ Oδ (α)

γδ [(4− 3γ2

)− 2γδ + γ2δ2 ]

= (1 + α) δ[(1− 6α− 3α2

)− 2 (1 + α) δ + (1 + α)

2δ2 ]

= δ (1− δ)2+ Oδ (α)

where Oδ (α) stands for a term that goes to 0 as α tends to 0 uniformly for|δ| ≤ 1.

Combining these three estimates, we find for ǫ sufficiently small, since |δ| ≤ 1,

36σǫ,α

t4≥ 3ǫ2 +3 (1− δ)

[1− δ + δ2 − ǫ

(3− 4δ + 4δ2

)]+Oδ (α) ≥ 3ǫ2 +Oδ (α) .

Choosing α << ǫ (recalling that ǫ is small), we get the result; i.e.

σǫ,α (ξ, η) ≥ 0, for every ξ, η ∈ R2×2.

This concludes the proof of the lemma.

5.3.9 Quasiconvex functions with subquadratic growth.

We have seen in Corollary 5.9 that a polyconvex function having a subquadraticgrowth, must be convex. This, however, is not the case for quasiconvex and rankone convex functions. We now give such an example, following Sverak [549] (forthe case p = 1, see Theorem 5.55).

Theorem 5.54 Let 1 < p < 2. Then there exists a function f : R2×2 → Rquasiconvex, non-convex and satisfying

0 ≤ f (ξ) ≤ γ (1 + |ξ|p) , ∀ ξ ∈ R2×2

and where γ is a positive constant.


Proof. We start with the following easily established algebraic inequality validfor any ξ ∈ R2×2

min|ξ − I|2 , |ξ + I|2 ≥ 1

2[ |ξ|2 − 2 det ξ ] ≥ 0. (5.94)

We next defineg (ξ) := min |ξ − I|p , |ξ + I|p .

Anticipating on the definition and properties of the quasiconvex envelope givenin Chapter 6 (see Theorem 6.9), we let

f := Qg

and we claim that f has all the desired properties. By definition it is quasiconvexand satisfies the growth condition, we therefore only need to show that it is notconvex. This will be proved, once shown that

f (0) = Qg (0) > 0, (5.95)

since clearlyCg (0) = 0

where Cg denotes the convex envelope of g.

Assume for the sake of contradiction that

Qg (0) = 0

and use Theorem 6.9 to find a sequence ϕν ∈ W 1,∞0

(D; R2

), here D ⊂ R2 is a

bounded open set with measD = 1, such that

0 = Qg (0) ≥ −1

ν+

∫

D

g (∇ϕν (x)) dx. (5.96)

Invoking (5.94), we can deduce from the above inequality that

1

ν≥ 2−p/2

∫

D

[|∇ϕν (x)|2 − 2 det (∇ϕν (x))

]p/2

dx.

The estimate of Theorem 5.52 then implies that

ϕν → 0 in W 1,p(D; R2

).

This therefore leads to

limν→∞

∫

D

g (∇ϕν (x)) dx = 2p/2,

contradicting (5.96). We have therefore proved (5.95) and the theorem follows.

Examples 239

5.3.10 The case of homogeneous functions of degree one

We would now like to discuss the convexity properties of homogeneous functionsof degree one, f : R2×2 → R and we have the following theorem.

Theorem 5.55 Let f : R2×2 → R be positively homogeneous of degree one,namely

f (tξ) = tf (ξ) for every t ≥ 0 and every ξ ∈ R2×2. (5.97)

The following three properties hold.

(i) f is polyconvex if and only if it is convex.

(ii) If f is SO (2)× SO (2)-invariant, in the sense that

f (ξ) = f (QξR) for every Q, R ∈ SO (2) ,

then f is rank one convex if and only if it is convex.

(iii) The function

f (ξ) =

⎧⎨⎩

7 |ξ|+ 3(ξ11)

2+2ξ1

1ξ22+3(ξ2

2)2+4ξ1

2ξ21

|ξ| if ξ = 0

0 if ξ = 0

is rank one convex but not convex.

Remark 5.56 (i) The first statement follows at once from Corollary 5.9.

(ii) The second assertion has been proved by Dacorogna [181] and the lastone is a particular case of the study undertaken by Dacorogna-Haeberly [190].

(iii) Muller [461] (see also Zhang [618]) produced, in an indirect way similarto that of Theorem 5.54, an example of a quasiconvex function satisfying (5.97)and that is not convex.

(iv) It is not presently known if the function given in (iii) of the theorem isquasiconvex. Numerical evidences given in Dacorogna-Haeberly [191] tend toindicate that it is quasiconvex. ♦

Before proceeding with the proof we need the following elementary lemmaestablished in Dacorogna [181], for a different proof see Dacorogna-Marechal[206]. The lemma is false if either the function is not everywhere finite or indimensions 3 and higher, see [206] for details. Note that in dimension 4, thefunction given in Theorem 5.55, being rank one convex, is separately convexbut not convex.

Lemma 5.57 Let g : R2 → R be positively homogeneous of degree one andseparately convex (meaning that x → g (x, y) and y → g (x, y) are both convex).Then g is convex.

Proof. (Lemma 5.57). Since g is homogeneous of degree one, it is clear thatg is convex if and only if

g (x1 + x2, y1 + y2) ≤ g (x1, y1) + g (x2, y2) . (5.98)



Case 1: x1x2 ≥ 0 or y1y2 ≥ 0. Since the hypothesis x1x2 ≥ 0 is handled sim-ilarly to y1y2 ≥ 0, we will assume that this last one holds. Since g is separatelyconvex it is continuous (cf. Theorem 2.31) and hence it is enough to prove theresult for y1y2 > 0. Observe then that

σ :=y1 + y2

|y1 + y2|=

y1

|y1|=

y2

|y2|∈ ±1 .

We therefore have, using the convexity of g with respect to the first variable,

g (x1 + x2, y1 + y2) = |y1 + y2| g(|y1|

|y1 + y2|x1

|y1|+

|y2||y1 + y2|

x2

|y2|, σ)

≤ |y1| g(x1

|y1|, σ) + |y2| g(

x2

|y2|, σ) = g (x1, y1) + g (x2, y2)

as wished.

Case 2: x1x2 < 0 and y1y2 < 0. This case is more involved than the previousone and we divide the proof into two steps.

Step 1. We first show that

g (x1 + x2, 0) ≤ g (x1, y) + g (x2,−y) , ∀y ∈ R. (5.99)

Since x1x2 < 0, we have either

x1 (x1 + x2) ≥ 0 or x2 (x1 + x2) ≥ 0.

Without loss of generality (otherwise exchange the roles of (x1, y) with that of(x2,−y)), we will assume that

x1 (x1 + x2) ≥ 0. (5.100)

We then choose ǫ > 0 sufficiently small and let

a :=x1 + (1− 2ǫ)x2

(1− ǫ)and μ :=

1− 2ǫ

1− ǫ.

Observe that ⎧⎪⎨⎪⎩

−2ǫμ + 2 (1− μ) (1− 2ǫ) = 0

μ (x1 + x2) + 2 (1− μ)x1 = a

2ǫx2 + (1− ǫ) a = x1 + x2 .

Appealing to Case 1, since (−y) .0 ≥ 0, we find

g (x1 + x2,−2ǫy) = g (2ǫx2 + (1− ǫ) a, 2ǫ (−y) + (1− ǫ) 0)

≤ 2ǫg (x2,−y) + (1− ǫ) g (a, 0) .

Since (5.100) holds, we also have from Case 1

g (a, 0) = g (μ (x1 + x2) + 2 (1− μ)x1, μ (−2ǫy) + 2 (1− μ) (1− 2ǫ) y)

≤ μg (x1 + x2,−2ǫy) + (1− μ) g (2x1, 2 (1− 2ǫ) y)

= μg (x1 + x2,−2ǫy) + 2 (1− μ) g (x1, (1− 2ǫ) y) .

Examples 241

Combining the last two inequalities, we find

g (x1 + x2,−2ǫy) ≤ 2ǫg (x2,−y) + (1− 2ǫ) g (x1 + x2,−2ǫy)

+2ǫg (x1, (1− 2ǫ) y)

or, in other words,

2ǫg (x1 + x2,−2ǫy) ≤ 2ǫg (x2,−y) + 2ǫg (x1, (1− 2ǫ) y) .

Dividing by 2ǫ and letting ǫ tend to 0, using the continuity of g, we have indeedobtained (5.99).

Step 2. We now prove (5.98). Observe that the hypothesis y1y2 < 0 implies

y1 + y2

y1≥ 0 or

y1 + y2

y2≥ 0.

We will assume that the first possibility happens, the second one being handledsimilarly.

We can therefore write,

g (x1 + x2, y1 + y2) = g(y1 + y2

y1x1 + x2 −

y2

y1x1,

y1 + y2

y1y1 + 0).

Since (y1 + y2) · 0 ≥ 0, we can apply Case 1 and get

g (x1 + x2, y1 + y2) ≤y1 + y2

y1g (x1, y1) + g(x2 −

y2

y1x1, 0). (5.101)

We also have, invoking Step 1,

g(x2 −y2

y1x1, 0) ≤ g (x2, y2) + g(−y2

y1x1,−

y2

y1y1)

= g (x2, y2)−y2

y1g (x1, y1) .

Combining the above inequality and (5.101), we obtain (5.98) and thus thelemma.

We now proceed with the proof of the theorem.

Proof. (Theorem 5.55). (i) As already mentioned the proof of the first partimmediately follows from Corollary 5.9.

(ii) The implication f convex ⇒ f rank one convex, being always true, weneed only prove the reverse one. According to Theorem 5.33, it is sufficient toshow that f is convex on diagonal matrices. Therefore let

g (x1, x2) := f

(x1 00 x2

)

and observe that the rank one convexity of f implies the separate convexity ofg. Lemma 5.57 gives immediately the claim.


(iii) We first discuss the fact that f is non convex. We let

ξ =1√2

(1 00 1

)and η =

1√2

(0 1−1 0

)

and for t ∈ R, we define

t→ ϕ (t) := f (ξ + tη) = 5(1 + t2

)1/2+ 6(1 + t2

)−1/2.

A direct computation shows that

ϕ′′ (t) =(17t2 − 1

) (1 + t2

)−5/2

and hence ϕ′′ (0) = −1 < 0, which implies that f is non convex.

It therefore remains only to show that f is rank one convex. We divide theproof of this fact into three steps.

Step 1. The rank one convexity of f is equivalent to showing that for everyfixed ξ ∈ R2×2, a, b ∈ R2 the function

t→ ϕξ,a,b (t) := f (ξ + ta⊗ b)

is convex in t ∈ R.Since f (ξ) ≥ 0, we have that if there exists α ∈ R such that

ξ = α a⊗ b,

thenf (ξ + ta⊗ b) = f ((α + t) a⊗ b) = |α + t| f (a⊗ b)

and thus ϕξ,a,b is convex in t. From now on we may therefore assume that ξ isnot parallel to a⊗b. The function ϕξ,a,b is then twice continuously differentiableand its convexity is therefore equivalent to the Legendre-Hadamard condition,namely

〈 ∇2f (ξ) a⊗ b; a⊗ b 〉 ≥ 0 (5.102)

for every ξ ∈ R2×2, a, b ∈ R2 with ξ not parallel to a⊗ b.

Step 2. We now compute the Hessian of f. It will be more convenient, inthe present analysis, to identify R2×2 with R4 and, therefore, a matrix ξ will bewritten as a vector (ξ1, ξ2, ξ3, ξ4) . We then let

〈ξ; η〉 =

4∑

i=1

ξiηi, |ξ|2 = 〈ξ; ξ〉 , det ξ = ξ1ξ4 − ξ2ξ3 .

Letting

M =

⎛⎜⎜⎝

9 0 0 10 6 2 00 2 6 01 0 0 9

⎞⎟⎟⎠

Examples 243

we can rewrite f, when ξ = 0, as

f (ξ) = |ξ|+ 〈 Mξ; ξ 〉|ξ| .

Computing the Hessian of f, when ξ = 0, we first find, for α = 1, 2, 3, 4, that

∂f (ξ)

∂ξα=

ξα

|ξ| +

2 |ξ| (Mξ)α − 〈Mξ; ξ〉 ξα

|ξ||ξ|2

=ξα

|ξ| +2 |ξ|2 (Mξ)α − 〈Mξ; ξ〉 ξα

|ξ|3

and thus

∂2f (ξ)

∂ξα∂ξβ=

δαβ

|ξ| −ξαξβ

|ξ|3+

1

|ξ|6−3 |ξ| ξβ [ 2 |ξ|2 (Mξ)α − 〈Mξ; ξ〉 ξα ]

+ [ 4 (Mξ)α ξβ + 2 |ξ|2 Mαβ − 〈Mξ; ξ〉 δαβ− 2 (Mξ)β ξα ] |ξ|3,

where δαβ is the Kronecker symbol.Since the quadratic form 〈 ∇2f (ξ)λ; λ 〉 is homogeneous of degree −1 in ξ

and 2 in λ, we only need to consider the case where |ξ| = |λ| = 1. We hence getthat

4∑

α,β=1

∂2f (ξ)

∂ξα∂ξβλαλβ = 1− (〈ξ; λ〉)2 − 4 〈Mξ; λ〉〈ξ, λ〉+ 2 〈Mλ; λ〉

− 〈Mξ; ξ〉+ 3 〈Mξ; ξ〉 (〈ξ; λ〉)2 .

We can still transform this expression into a more amenable one, by choosing avector η ∈ R4 and θ ∈ R so that

λ = ξ cos θ + η sin θ, with |η| = 1 and 〈ξ; η〉 = 0.

We therefore obtain that

〈ξ; λ〉 = cos θ, 〈Mξ; λ〉 = 〈Mξ; ξ〉 cos θ + 〈Mξ; η〉 sin θ

〈Mλ; λ〉 = 〈Mξ; ξ〉 cos2 θ + 2 〈Mξ; η〉 cos θ sin θ + 〈Mη; η〉 sin2 θ.

Returning to the quadratic form we therefore find that

〈 ∇2f (ξ)λ; λ 〉 = [1 + 2 〈Mη; η〉 − 〈Mξ; ξ〉] sin2 θ.

Hence (5.102) is equivalent to showing that

1 + 2 〈Mη; η〉 − 〈Mξ; ξ〉 ≥ 0 (5.103)


for every ξ, η ∈ R4 and θ ∈ R satisfying

|ξ| = |η| = 1, 〈ξ; η〉 = 0 and det (ξ cos θ + η sin θ) = 0. (5.104)

Step 3. It therefore remains to show (5.103) whenever (5.104) holds. Westart by observing that the matrix M has eigenvalues

μ1 = 4 ≤ μ2 = μ3 = 8 ≤ μ4 = 10

and corresponding orthonormal eigenvectors

ϕ1 = 1√2

(0, 1,−1, 0) ϕ2 = 1√2

(0, 1, 1, 0)

ϕ3 = 1√2

(1, 0, 0,−1) ϕ4 = 1√2

(1, 0, 0, 1) .

Note thatdetϕ1 = detϕ4 = − detϕ2 = − detϕ3 = 1

2 .

Expanding the vectors ξ, η ∈ R4 in this basis we have

ξ =

4∑

i=1

ξiϕi , η =

4∑

i=1

ηiϕi ,

and from now on ξi and ηi will always denote the components of ξ and η in thisnew basis and in particular we find that

det ξ =1

2(ξ2

1 + ξ24 − ξ2

2 − ξ23).

Moreover, (5.103) is equivalent to showing that

2 〈Mη, η〉 − 〈Mξ, ξ〉 =

4∑

i=1

μi

(2η2

i − ξ2i

)≥ −1. (5.105)

Moreover, (5.104) can then be rewritten as

|ξ|2 = ξ21 + ξ2

2 + ξ23 + ξ2

4 = |η|2 = η21 + η2

2 + η23 + η2

4 = 1,

〈ξ; η〉 = 0 ⇔ ξ1η1 + ξ4η4 = − (ξ2η2 + ξ3η3) ,

with

det (ξ cos θ + η sin θ) = 0

⇔(ξ21 + ξ2

4 − ξ22 − ξ2

3

)cos2 θ +

(η21 + η2

4 − η22 − η2

3

)sin2 θ

+2 (ξ1η1 + ξ4η4 − ξ2η2 − ξ3η3) cos θ sin θ = 0.

We now argue by contradiction and assume that (5.105) does not hold, meaningthat we can find ξ, η ∈ R4 and θ ∈ R as above and so that

4∑

i=1

μi

(2η2

i − ξ2i

)< −1.

Examples 245

Observing that

2μ1 |η|2 −μ3 + μ4

2|ξ|2 = −1,

we can rewrite the above inequality as

12η24 + 5ξ2

1 + 8(η22 + η2

3

)< ξ2

4 − ξ22 − ξ2

3 . (5.106)

Similarly, writing

4∑

i=1

μi

(2η2

i − ξ2i

)< −1 < 2 = (μ1 + μ2) |η|2 − μ4 |ξ|2

we find that

8η24 + 6ξ2

1 + 2(ξ22 + ξ2

3

)< 4(η21 − η2

2 − η23

). (5.107)

From (5.106) and (5.107), we deduce that

8(η22 + η2

3

)< ξ2

4 − ξ22 − ξ2

3 and1

2

(ξ22 + ξ2

3

)< η2

1 − η22 − η2

3 .

Inserting these inequalities in the identity det (ξ cos θ + η sin θ) = 0 and alsousing the fact that ξ1η1 + ξ4η4 = − (ξ2η2 + ξ3η3) leads to the desired contradic-tion, namely

0 =(ξ21 + ξ2

4 − ξ22 − ξ2

3

)cos2 θ +

(η21 + η2

4 − η22 − η2

3

)sin2 θ

+2 (ξ1η1 + ξ4η4 − ξ2η2 − ξ3η3) cos θ sin θ

>[ξ21 + 8

(η22 + η2

3

)]cos2 θ + [ η2

4 +1

2

(ξ22 + ξ2

3

)] sin2 θ

−4 (ξ2η2 + ξ3η3) cos θ sin θ

≥ 8(η22 + η2

3

)cos2 θ +

1

2

(ξ22 + ξ2

3

)sin2 θ − 4 |(ξ2η2 + ξ3η3) cos θ sin θ|

≥ 1

2

[4 |cos θ|

√η22 + η2

3 − |sin θ|√

ξ22 + ξ2

3

]2≥ 0.


5.3.11 Some more examples

We now give some more examples.

Theorem 5.58 Let f : RN×n → R and let |.| denote the Euclidean norm,namely, for ξ ∈ RN×n, we let

|ξ| :=(∑n

α=1

∑Ni=1

(ξiα

)2)1/2

.


(i) Let g : R+ → R be such that

f (ξ) = g (|ξ|) .

Then

f convex ⇔ f polyconvex ⇔ f quasiconvex ⇔ f rank one convex

⇔ g convex and g (0) = inf g (x) : x ≥ 0 .

(ii) Let N = n, 1 ≤ α < 2n, h : R → R be such that

f (ξ) = |ξ|α + h (det ξ) .

Then

f polyconvex ⇔ f quasiconvex ⇔ f rank one convex ⇔ h convex.

(iii) Let N = n, p > 0, 1 ≤ s ≤ n− 1 and

f (ξ) =

⎧⎨⎩

(|adjs ξ|n/s

det ξ

)p

if det ξ > 0

+∞ otherwise.

Thenf polyconvex ⇔ f rank one convex ⇔ p ≥ s

n− s.

Remark 5.59 (i) The result (i) was established by Dacorogna [176].

(ii) Case (ii) was proved by Ball-Murat [65]. Note that the hypothesis α < 2ncannot be dropped in general. Indeed, if n = 2 and α = 4, then

f (ξ) = |ξ|4 − 2 (det ξ)2

is even convex.

(iii) Case (iii) is interesting in elasticity for slightly compressible materialsand was established by Charrier-Dacorogna-Hanouzet-Laborde [144]. It wasthen generalized by Dacorogna-Marechal [206]. ♦

Proof. (i) Let ξ ∈ RN×n and

f (ξ) = g (|ξ|) .

In view of Theorem 5.3, it remains to show that

f rank one convex ⇒ g convex and g (0) = inf g (x) : x ≥ 0

which will be proved in Step 1 and

g convex and g (0) = inf g (x) : x ≥ 0 ⇒ f convex

Examples 247

which we will establish in Step 2.

Step 1. Let x > 0 and define ξ ∈ RN×n to be such that

ξ11 = x and ξi

j = 0 if (i, j) = (1, 1) .

We then deduce that

g (0) = f(ξ − ξ

2) ≤ 1

2f (ξ) +

1

2f (−ξ) = g (x)

as wished.

Let us now show that g is convex. Let λ ∈ [0, 1] , α, β ≥ 0. Define ξ, η ∈ RN×n

byξ11 = α, η1

1 = β and ξij = ηi

j = 0 if (i, j) = (1, 1) .

Observing that rank ξ − η ≤ 1 and using the rank one convexity of f we get

g (λα + (1− λ)β) = f (λξ + (1− λ) η)

≤ λf (ξ) + (1− λ) f (η) = λg (|α|) + (1− λ) g (|β|)= λg (α) + (1− λ) g (β)

which is indeed the claimed convexity inequality.

Step 2. Note that since g is convex and

g (0) = inf g (x) : x ≥ 0 ,

then g is non decreasing on R+ .We now want to show that g convex ⇒ f convex. This is immediate since

f (λξ + (1− λ) η) = g (|λξ + (1− λ) η|) ≤ g (λ |ξ|+ (1− λ) |η|)≤ λg (|ξ|) + (1− λ) g (|η|) = λf (ξ) + (1− λ) f (η)

and this achieves the proof of the third part of the theorem.

(ii) Let n = N, ξ ∈ Rn×n, 1 ≤ α < 2n and

f (ξ) = |ξ|α + h (det ξ) .

It follows from Theorem 5.3 that it only remains to prove that

f rank one convex ⇒ h convex.

Let λ ∈ (0, 1) , a, b ∈ R, we want to show that

h (λa + (1− λ) b) ≤ λh (a) + (1− λ) h (b) . (5.108)

We will assume, with no loss of generality, that a = b and a = 0. Let ǫ = 0 withǫ (b− a) > 0 and

ξ := diag(aǫ

b− a,

(b− a

ǫ

) 1n−1

, · · · ,(

b− a

ǫ

) 1n−1

) ∈ Rn×n.


It is then easy to see that, letting e1 = (1, 0, · · · , 0) ∈ Rn,

det ξ = a, det (ξ + ǫ e1 ⊗ e1) = b

det (ξ + (1− λ) ǫ e1 ⊗ e1) = λa + (1− λ) b.

Since f is rank one convex, we have

|ξ + (1− λ) ǫ e1 ⊗ e1|α + h (λa + (1− λ) b)

= f (λξ + (1− λ) (ξ + ǫ e1 ⊗ e1))

≤ λf (ξ) + (1− λ) f (ξ + ǫ e1 ⊗ e1)

= λ |ξ|α + (1− λ) |ξ + ǫ e1 ⊗ e1|α + λh (a) + (1− λ) h (b) .

(5.109)

Observe that

λ |ξ|α + (1− λ) |ξ + ǫ (e1 ⊗ e1)|α − |ξ + (1− λ) ǫ (e1 ⊗ e1)|α

= λ[ (aǫ

b− a)2 + (n− 1) (

b− a

ǫ)

2n−1 ]α/2

+ (1− λ) [ (aǫ

b− a+ ǫ)2 + (n− 1) (

b− a

ǫ)

2n−1 ]α/2

−[ (aǫ

b− a+ (1− λ) ǫ)2 + (n− 1) (

b− a

ǫ)

2n−1 ]α/2

= O(ǫ2n−αn−1 )

where O (t) stands for a term that goes to 0 as t→ 0. It is clear that if 1 ≤ α <2n, then the right hand side in the above identity tends to zero as ǫ→ 0. Thuscombining (5.109) and the above identity, as ǫ → 0, we have indeed obtained(5.108), i.e. that h is convex.

(iii) We decompose the proof into two steps.

Step 1 : p ≥ s

n− s⇒ f polyconvex. Define first h : R× R → R by

h (x, δ) :=

xnp/sδ−p if x, δ > 0

+∞ otherwise.

It is then easy to see that h is convex if and only if p ≥ s

n− s. We then let, for

1 ≤ s ≤ n− 1, F : R(ns

)×(ns

)× R → R be defined by

F (η, δ) := h (|η| , δ) .

Then from the convexity of h and from the fact that x → h (x, δ) is non decreas-ing in R+ , we deduce that F is convex. Observing that

f (ξ) = F (adjs ξ,det ξ)

Appendix: some basic properties of determinants 249

we immediately obtain the polyconvexity of f from the fact that p ≥ s

n− s.

Step 2 : f rank one convex ⇒ p ≥ s

n− s. Let ξ ∈ Rn×n, a, b ∈ Rn be such

thatdet (ξ + ta⊗ b) > 0, for every t > 0.

Then the rank one convexity of f implies that

t→ ϕ (t) := f (ξ + ta⊗ b) =

(|adjs (ξ + ta⊗ b)|n/s

det (ξ + ta⊗ b)

)p

is convex. We next simplify the notations by letting λ1, · · · , λ5 be such that

|adjs (ξ + ta⊗ b)|2 = λ2

1t2 + λ2t + λ2

3

det (ξ + ta⊗ b) = λ4t + λ5 .

Such λ1, · · · , λ5 exist since

t→ adjs (ξ + ta⊗ b) and t→ det (ξ + ta⊗ b)

are linear functions (cf. Proposition 5.65). Combining the above notation withthe definition of ϕ, we find

ϕ (t) =(λ2

1t2 + λ2t + λ2

3

)np2s (λ4t + λ5)

−p.

After an elementary computation we obtain

ϕ′′ (t) =(λ2

1t2 + λ2t + λ2

3

)np2s −2

(λ4t + λ5)−p−2

×[ λ41λ

24t

4 p

s2(n− s)2 (p− s

n− s) + O

(t3)].

Since ϕ is convex for t > 0 we must have p ≥ s

n− s.

5.4 Appendix: some basic propertiesof determinants

In the whole of Chapter 5, we have seen the importance of determinants inquasiconvex analysis. We gather in this appendix some well known algebraicproperties of determinants. In the first part, we carefully introduce the notationfor the minors adjs ξ of a given matrix ξ.

We first introduce some notation. Let n ∈ N (the set of positive integers)and let 1 ≤ s ≤ n. We define

Ins := (α1, · · · , αs) ∈ Ns : 1 ≤ α1 < α2 < · · · < αs ≤ n .


We call the elements of Ins increasing s-tuples. The number of elements of In

s isthen

card Ins =

(ns

)=

n!

s! (n− s)!.

We next endow Ins with the following ordering relation:

α = (α1, · · · , αs) ≻ (β1, · · · , βs) = β

if and only ifαk < βk ,

where k is the largest integer less than or equal to s such that αk = βk andαl = βl for every l > k. (This is the inverse of the lexicographical order whenread backward.)

Example 5.60 (i) n = 4, s = 2. Then

(1, 2) ≻ (1, 3) ≻ (2, 3) ≻ (1, 4) ≻ (2, 4) ≻ (3, 4) .

(ii) n = 5, s = 3. Then

(1, 2, 3) ≻ (1, 2, 4) ≻ (1, 3, 4) ≻ (2, 3, 4) ≻ (1, 2, 5)

≻ (1, 3, 5) ≻ (2, 3, 5) ≻ (1, 4, 5) ≻ (2, 4, 5) ≻ (3, 4, 5) .

(iii) s = n− 1. Then

(1, · · · , n− 1) ≻ · · · ≻ (1, · · · , k − 1, k + 1, · · · , n) ≻ · · · ≻ (2, · · · , n) . ♦We then define the map ϕn

s

ϕns : 1, 2, 3, · · · ,

(ns

) → In

s

as the only bijection that respects the order defined above.

Example 5.61 (i) n = 4, s = 2. Then

ϕ42 (1) = (3, 4) , ϕ4

2 (2) = (2, 4) , ϕ42 (3) = (1, 4) ,

ϕ42 (4) = (2, 3) , ϕ4

2 (5) = (1, 3) , ϕ42 (6) = (1, 2) .

(ii) s = n− 1. Then

ϕnn−1 (1) = (2, · · · , n)

ϕnn−1 (k) = (1, · · · , k − 1, k + 1, · · · , n)

ϕnn−1 (n) = (1, · · · , n− 1) . ♦

We are now in a position to define, for a given matrix ξ ∈ RN×n, the adjugatematrix of order s, 1 ≤ s ≤ n ∧N = min n, N ,

adjs ξ ∈ R(Ns

)×(ns

).


Let ξ ∈ RN×n be such that

ξ =

⎛⎜⎜⎜⎜⎝

ξ11 · · · ξ1

n

.... . .

...

ξN1 · · · ξN

n

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

ξ1

...

ξN

⎞⎟⎟⎠ = (ξ1, · · · , ξn) .

We define adjs ξ to be the following matrix in R(Ns

)×(ns

):

adjs ξ =

⎛⎜⎜⎜⎜⎜⎜⎝

(adjs ξ)11 · · · (adjs ξ)

1

(ns)

.... . .

...

(adjs ξ)(N

s )1 · · · (adjs ξ)

(Ns )

(ns)

⎞⎟⎟⎟⎟⎟⎟⎠∈ R(Ns

)×(ns

)

=

⎛⎜⎜⎜⎝

(adjs ξ)1

...

(adjs ξ)(Ns )

⎞⎟⎟⎟⎠ =

((adjs ξ)1 , · · · , (adjs ξ)(n

s)

),

where

(adjs ξ)iα = (−1)i+α det

⎛⎜⎜⎜⎜⎝

ξi1α1

· · · ξi1αs

.... . .

...

ξisα1

· · · ξisαs

⎞⎟⎟⎟⎟⎠

and (i1, · · · , is) , (α1, · · · , αs) are the s-tuples corresponding to i and α by thebijections ϕN

s and ϕns , meaning that

ϕNs (i) = (i1, · · · , is) and ϕn

s (α) = (α1, · · · , αs) .

Notation 5.62 We sometimes, as in examples (iv) and (vii) below, denote by

ξ i1,··· ,ikα1,··· ,αl

the (N − k)×(n− l) matrix obtained from ξ ∈ RN×n by suppressing the k rowsi1, · · · , ik and the l columns α1, · · · , αl . ♦

Example 5.63 (i) N = n = 2, s = 1. Let

ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

).

ThenIns = IN

s = 1, 2


and the bijection ϕ21 : 1, 2 → 2, 1 . Hence

adj1 ξ =

((adj1 ξ)11 (adj1 ξ)12

(adj1 ξ)21 (adj1 ξ)22

)=

(ξ22 −ξ2

1

−ξ12 ξ1

1

).

(note that adj1 ξ is exactly ξ defined in Theorem 5.51 above).

(ii) N = n = s = 2. Then

Ins = IN

s = (1, 2)

and ϕ22 (1) = (1, 2) . Hence

adj2 ξ = det

(ξ11 ξ1

2

ξ21 ξ2

2

)= det ξ.

(iii) N = 3, s = n = 2. Then

Ins = I2

2 = (1, 2)

and ϕ22 (1) = (1, 2) , while

INs = I3

2 = (1, 2) ; (1, 3) ; (2; 3)

and ϕ32 (1) = (2, 3) , ϕ3

2 (2) = (1, 3) , ϕ32 (3) = (1, 2) . Therefore, if

ξ =

⎛⎜⎜⎝

ξ11 ξ1

2

ξ21 ξ2

2

ξ31 ξ3

2

⎞⎟⎟⎠ =

⎛⎜⎝

ξ1

ξ2

ξ3

⎞⎟⎠ = (ξ1, ξ2) ,

then

adj2 ξ =

⎛⎜⎝

(adj2 ξ)11

(adj2 ξ)21

(adj2 ξ)31

⎞⎟⎠ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

det

(ξ21 ξ2

2

ξ31 ξ3

2

)

− det

(ξ11 ξ1

2

ξ31 ξ3

2

)

det

(ξ11 ξ1

2

ξ21 ξ2

2

)

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

(iv) N = n + 1, s = n. We let

ξ =

⎛⎜⎜⎜⎜⎝

ξ11 · · · ξ1

n

.... . .

...

ξn+11 · · · ξn+1

n

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

ξ1

...

ξn+1

⎞⎟⎟⎠ = (ξ1, · · · , ξn) .


Then

adjn ξ =

⎛⎜⎜⎝

(adjn ξ)11...

(adjn ξ)n+11

⎞⎟⎟⎠ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

det

⎛⎜⎝

ξ21 · · · ξ2

n...

. . ....

ξn+11 · · · ξn+1

n

⎞⎟⎠

...

(−1)n+2 det

⎛⎜⎝

ξ11 · · · ξ1

n...

. . ....

ξn1 · · · ξn

n

⎞⎟⎠

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

det ξ 1

...

(−1)n+2 det ξ n+1

⎞⎟⎟⎠

where ξ k denotes the n× n matrix obtained by suppressing the k th row in thematrix ξ.

(v) N = n = s = 3. Then I33 = (1, 2, 3) and therefore

adj3 ξ = det ξ.

(vi) N = n = 3, s = 2. Then

adj2 ξ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

det

(ξ22 ξ2

3

ξ32 ξ3

3

)− det

(ξ21 ξ2

3

ξ31 ξ3

3

)det

(ξ21 ξ2

2

ξ31 ξ3

2

)

− det

(ξ12 ξ1

3

ξ32 ξ3

3

)det

(ξ11 ξ1

3

ξ31 ξ3

3

)− det

(ξ11 ξ1

2

ξ31 ξ3

2

)

det

(ξ12 ξ1

3

ξ22 ξ2

3

)− det

(ξ11 ξ1

3

ξ21 ξ2

3

)det

(ξ11 ξ1

2

ξ21 ξ2

2

)

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

The above expression is the usual transpose of the matrix of cofactors.

(vii) N = n and s = n− 1. Then

adjn−1 ξ ∈ Rn×n

and (adjn−1 ξ

)iα

= (−1)i+α det(ξ iα)

where ξ iα is the (n− 1)×(n− 1) matrix obtained from ξ ∈ Rn×n by suppressing

the i th row and the α th column. ♦


Remark 5.64 Note that one can write the rows of adjs ξ as

(adjs ξ)i= (−1)

i+1adjs

⎛⎜⎜⎝

ξi1

...

ξis

⎞⎟⎟⎠ , 1 ≤ i ≤

(Ns

),

where (i1, · · · , is) = ϕNs (i) is the s-tuple associated to the integer i. So, in

particular,

(adjs ξ)1 = adjs

⎛⎜⎜⎜⎜⎜⎜⎜⎝

ξN−s+1

ξN−s+2

...

ξN−1

ξN

⎞⎟⎟⎟⎟⎟⎟⎟⎠

, · · · , (adjs ξ)

(Ns

)= (−1)

(Ns

)+1 adjs

⎛⎜⎜⎜⎜⎜⎜⎜⎝

ξ1

ξ2

...

ξs−1

ξs

⎞⎟⎟⎟⎟⎟⎟⎟⎠

.

A similar remark applies to the columns of adjs ξ. ♦

We now give some elementary properties of determinants.

Proposition 5.65 Let ξ ∈ RN×n.

(i) If N = n, then, for every ξ ∈ Rn×n,

⟨ξµ;(adjn−1 ξ

)ν⟩=⟨ξµ;(adjn−1 ξ

)ν

⟩= δµν det ξ, μ, ν = 1, 2, · · · , n,

where 〈·; ·〉 denotes the scalar product in Rn and δµν denotes the Kroneckersymbol.

(ii) If N = n, then, for every ξ ∈ Rn×n,

ξ(adjn−1 ξ

)t= det ξ · I

where I is the identity matrix in Rn×n and ξt denotes the transpose of the matrixξ. In particular if det ξ = 0, then

ξ−1 =1

det ξ

(adjn−1 ξ

)t.

(iii) If N = n + 1, then, for every ξ ∈ R(n+1)×n,

〈ξν ; adjn ξ〉 = 0, ν = 1, · · · , n,

where 〈·; ·〉 denotes the scalar product in Rn+1.

(iv) If N = n− 1, then, for every ξ ∈ R(n−1)×n,

⟨ξν ; adjn−1 ξ

⟩= 0, ν = 1, · · · , n− 1,

where 〈·; ·〉 denotes the scalar product in Rn.


(v) If N = n, then, for every ξ ∈ Rn×n,

∂

∂ξiα

(det ξ) =(adjn−1 ξ

)iα

, 1 ≤ i, α ≤ n = N.

(vi) Denote

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) ∈ Rτ(n,N)

where n ∧N = min n, N and

τ (n, N) =

n∧N∑

s=1

σ (s) =

n∧N∑

s=1

(ns

)(Ns

).

Let a ∈ Rn, b ∈ RN . Define

a⊗ b =(aibα

)1≤i≤N


Let t ∈ [0, 1] , then, for every ξ ∈ RN×n,

T (ξ + (1− t) a⊗ b) = tT (ξ) + (1− t)T (ξ + a⊗ b) .

Proof. (i) The case μ = ν is just the way a determinant is computed, byexpanding it along the ν th row or the ν th column. When μ = ν, then both⟨ξµ;(adjn−1 ξ

)ν⟩and

⟨ξµ;(adjn−1 ξ

)ν

⟩are again determinants of n×n matrices,

but the first matrix has twice the row ξµ and the second has twice the columnξµ . Thus both determinants are equal to 0, as claimed.

(ii) This follows at once from (i).

(iii) Let N = n + 1 and ν ∈ 1, · · · , n . We have to show that

〈ξν ; adjn ξ〉 = 0.

Define the matrix η = [ξν ; ξ] ∈ R(n+1)×(n+1) (recall that ξ ∈ R(n+1)×n). Thenη1 = ην+1 and therefore det η = 0. Using (i), we obtain

0 = det η = 〈η1; (adjn η)1〉 = 〈ξν ; adjn ξ〉 .

(iv) This is established exactly as (iii).

(v) This is a direct consequence of (i).

(vi) We divide the proof into three steps.

Step 1. The result is equivalent to

adjs (ξ + (1− t) a⊗ b) = t adjs ξ + (1− t) adjs (ξ + a⊗ b)

for every 1 ≤ s ≤ n ∧N. In terms of components this is equivalent to

(adjs (ξ + (1− t) a⊗ b))iα

= t (adjs ξ)iα + (1− t) (adjs (ξ + a⊗ b))

iα ,

(5.110)


1 ≤ i ≤(Ns

), 1 ≤ α ≤

(ns

). Recall that

(adjs ξ)iα = (−1)

i+αdet

⎛⎜⎜⎜⎜⎝

ξi1α1

· · · ξi1αs

.... . .

...

ξisα1

· · · ξisαs

⎞⎟⎟⎟⎟⎠

.

By abuse of notation, let

ξ =

⎛⎜⎜⎜⎜⎝

ξi1α1

· · · ξi1αs

.... . .

...

ξisα1

· · · ξisαs

⎞⎟⎟⎟⎟⎠

, a⊗ b =

⎛⎜⎜⎜⎜⎝

ai1bα1 · · · ai1bαs

.... . .

...

aisbα1 · · · aisbαs

⎞⎟⎟⎟⎟⎠

.

Therefore (5.110) is equivalent to showing that, for every ξ ∈ Rs×s, a, b ∈ Rs,t ∈ [0, 1] ,

det (ξ + (1− t) a⊗ b) = t det ξ + (1− t) det (ξ + a⊗ b) . (5.111)

This is a standard property of determinants that we prove in the two stepsbelow.

Step 2. We start by proving (5.111) when

a = b = e1 = e1 = (1, 0, · · · , 0) ∈ Rs.

Note that, for every x ∈ R, we have

(ξ + xe1 ⊗ e1

)1= ξ1 + xe1 and

(adjs−1

(ξ + xe1 ⊗ e1

))1=(adjs−1 ξ

)1.

The first identity is obvious and the second one follows since the components of(adjs−1 ξ

)1are given by determinants where the first row of ξ does not appear.

We can therefore apply (i) to find

det(ξ + (1− t) e1 ⊗ e1

)

= 〈(ξ + (1− t) e1 ⊗ e1

)1;(adjs−1

(ξ + (1− t) e1 ⊗ e1

))1 〉= 〈 ξ1 + (1− t) e1;

(adjs−1 ξ

)1 〉= t 〈 ξ1;

(adjs−1 ξ

)1 〉+ (1− t) 〈 ξ1 + e1;(adjs−1 ξ

)1 〉= t 〈 ξ1;

(adjs−1 ξ

)1 〉+ (1− t) 〈 ξ1 + e1;

(adjs−1

(ξ + e1 ⊗ e1

))1 〉= t det ξ + (1− t) det

(ξ + e1 ⊗ e1

)

which is the claim of Step 2.


Step 3. The general statement (5.111) follows at once from Step 2 andTheorem 13.3. Indeed, we can find R, Q ∈ O (s) such that

R(e1 ⊗ e1

)Q = a⊗ b.

We therefore find, using Step 2,

det (ξ + (1− t) a⊗ b) = det(R(RtξQt + (1− t) e1 ⊗ e1

)Q)

= detR det(RtξQt + (1− t) e1 ⊗ e1

)detQ

= t detR det(RtξQt

)detQ

+ (1− t) det R det(RtξQt + e1 ⊗ e1

)detQ

= t det ξ + (1− t) det(ξ + R

(e1 ⊗ e1

)Q)

= t det ξ + (1− t) det (ξ + a⊗ b)

which is the claim.

We also have the following useful result (see Buttazzo-Dacorogna-Gangbo[113] and Dacorogna-Marechal [205]).

Proposition 5.66 (i) Let ξ ∈ RN×n, η ∈ Rn×m and

1 ≤ s ≤ N ∧ n ∧m := min N, n, m .

Thenadjs (ξη) = adjs ξ adjs η.

(ii) Let ξ ∈ RN×n and 1 ≤ s ≤ N ∧ n, then

adjs(ξt)

= (adjs ξ)t.

(iii) If N = n and R ∈ O (n) (respectively R ∈ SO (n)), then

adjs R ∈ O((

ns

))(respectively adjs R ∈ SO

((ns

))).

(iv) If N = n and ξ ∈ Rn×n is invertible, then adjs ξ ∈ R(ns)×(n

s) is invertibleand

(adjs ξ)−1

= adjs(ξ−1).

(v) If N = n and if R ∈ SO (n) , then

adjn−1 R = R.

Proof. (i) We have to prove that

(adjs (ξη))ij = (adjs ξ adjs η)i

j


for every 1 ≤ i ≤(Ns

), 1 ≤ j ≤

(ms

). To simplify the notation, we will write

α := ϕNs , β := ϕn

s , γ := ϕms .

Let the s-tuples corresponding to i and j (and later k) be given by

α (i) = (i1, · · · , is) , β (k) = (k1, · · · , ks) , γ (j) = (j1, · · · , js) .

For a matrix θ ∈ RN×m, we let

θα(i)γ(j) :=

⎛⎜⎜⎜⎜⎝

θi1j1

· · · θi1js

.... . .

...

θis

j1· · · θis

js

⎞⎟⎟⎟⎟⎠∈ Rs×s

and, for 1 ≤ ν ≤ m,

( θα(i)γ(j) )ν :=

⎛⎜⎜⎝

θi1ν

...

θisν

⎞⎟⎟⎠ ∈ Rs.

For 1 ≤ p, q ≤ s, we have that

( ( ξηα(i)γ(j) )q

p = ( ξη )iq

jp=

n∑

ν=1

ξiqν ην

jp.

In other words, the p th column vector of the matrix is given by

( (ξη)α(i)γ(j) )p =

⎛⎜⎜⎜⎝

( ( ξη )α(i)γ(j) )1p

...

( ( ξη )α(i)γ(j) )s

p

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎝

∑nν=1 ξi1

ν ηνjp

...∑nν=1 ξis

ν ηνjp

⎞⎟⎟⎠

=

n∑

ν=1

ηνjp

⎛⎜⎜⎝

ξi1ν

...

ξisν

⎞⎟⎟⎠ =

n∑

ν=1

ηνjp

( ξα(i) )ν .

We therefore have, by definition of adjs , that

(adjs (ξη))ij

= (−1)i+j

det( (ξη)α(i)γ(j) )

= (−1)i+j

det( ( (ξη)α(i)γ(j) )1 , · · · , ( (ξη)

α(i)γ(j) )s )

= (−1)i+j

det(∑n

ν=1 ηνj1( ξα(i) )ν , · · · ,

∑nν=1 ην

js( ξα(i) )ν )

= (−1)i+j det(∑n

ν1=1 ην1

j1( ξα(i) )ν1 , · · · ,

∑nνs=1 ηνs

js( ξα(i) )νs )

= (−1)i+j

n∑

ν1,··· ,νs=1

ην1

j1· · · ηνs

jsdet( ( ξα(i) )ν1 , · · · , ( ξα(i) )νs ).


Now, if νp = νq for two distinct integers p, q ∈ 1, · · · , s , we clearly have

det( ( ξα(i) )ν1 , · · · , ( ξα(i) )νs ) = 0.

Thus, writing Fn,s for all s-tuples (ν1, · · · , νs) in 1, · · · , ns such that the νp

are pairwise distinct, we find

(adjs (ξη))ij = (−1)

i+j∑

(ν1,··· ,νs)∈Fn,s

ην1

j1· · · ηνs

jsdet( ( ξα(i) )ν1 , · · · , ( ξα(i) )νs ).

(5.112)On the other hand we can write

(adjs ξ adjs η)ij =

(ns

)∑

k=1

(adjs ξ)ik (adjs η)

kj

=

(ns

)∑

k=1

(−1)i+k

det( ξα(i)β(k) ) (−1)

k+jdet( η

β(k)γ(j) )

= (−1)i+j

(ns

)∑

k=1

det( ξα(i)β(k)η

β(k)γ(j) ).

Since, for 1 ≤ p, q, r ≤ s,

( ξα(i)β(k) )q

p = ξiq

kpand ( η

β(k)γ(j) )p

r = ηkp

jr

we find

( ξα(i)β(k)η

β(k)γ(j) )q

r =

s∑

p=1

ξiq

kpη

kp

jr.

Phrased differently, we have that the r-th column vector of the matrix is given by

(ξ

α(i)β(k)η

β(k)γ(j)

)r

=

⎛⎜⎜⎜⎝

( ξα(i)β(k)η

β(k)γ(j) )1r

...

( ξα(i)β(k)η

β(k)γ(j) )s

r

⎞⎟⎟⎟⎠ =

⎛⎜⎜⎜⎝

∑sp=1 ξi1

kpη

kp

jr

...∑s

p=1 ξis

kpη

kp

jr

⎞⎟⎟⎟⎠

=

s∑

p=1

ηkp

jr

⎛⎜⎜⎝

ξi1kp

...

ξis

kp

⎞⎟⎟⎠ =

s∑

p=1

ηkp

jr( ξα(i) )kp .


We thus deduce that

(adjs ξ adjs η)ij

= (−1)i+j

(ns

)∑k=1

det( ( ξα(i)β(k)η

β(k)γ(j) )1 , · · · , ( ξ

α(i)β(k)η

β(k)γ(j) )s )

= (−1)i+j

(ns

)∑k=1

det(s∑

p=1η

kp

j1( ξα(i) )kp , · · · ,

s∑p=1

ηkp

js( ξα(i) )kp )

= (−1)i+j

(ns

)∑k=1

det(s∑

p1=1η

kp1

j1( ξα(i) )kp1

, · · · ,s∑

ps=1η

kps

js( ξα(i) )kps

)

= (−1)i+j

(ns

)∑k=1

s∑p1,··· ,ps=1

ηkp1

j1· · · ηkps

jsdet( ( ξα(i) )kp1

, · · · , ( ξα(i) )kps).

If (p1, · · · , ps) ∈ 1, · · · , ssis not a permutation of (1, · · · , s) , then

det( ( ξα(i) )kp1, · · · , ( ξα(i) )kps

) = 0.

Letting

νr := kpr , r = 1, · · · , s,

we note that, when (p1, · · · , ps) ∈ 1, · · · , ssis a permutation of (1, · · · , s) and

k ∈1, · · · ,

(ns

), then (ν1, · · · , νs) ∈ Fn,s , the set of s-tuples (ν1, · · · , νs) in

1, · · · , nssuch that the νp are pairwise distinct. We therefore get that

(adjs ξ adjs η)ij

= (−1)i+j

∑

(ν1,··· ,νs)∈Fn,s

ην1

j1· · · ηνs

jsdet( ( ξα(i) )ν1 , · · · , ( ξα(i) )νs ).

The above identity and (5.112) imply the result.

(ii) As above, let

α := ϕNs , β := ϕn

s .

We clearly have, for 1 ≤ i ≤(Ns

)and 1 ≤ j ≤

(ns

), that

(ξt)α(i)

β(j)=(ξ

β(j)α(i)

)t


since, for α (i) = (i1, · · · , is) and β (j) = (j1, · · · , js) , we can write

(ξt)α(i)

β(j)=

⎛⎜⎜⎜⎜⎜⎝

(ξt)i1j1

· · · (ξt)i1js

.... . .

...

(ξt)is

j1· · · (ξt)

is

js

⎞⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎝

ξj1i1

· · · ξjs

i1

.... . .

...

ξj1is

· · · ξjs

is

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎝

ξj1i1

· · · ξj1is

.... . .

...

ξjs

i1· · · ξjs

is

⎞⎟⎟⎟⎟⎠

t

=(ξ

β(j)α(i)

)t

.

We can therefore deduce that(adjs

(ξt))i

j= (−1)i+j det(

(ξt)α(i)

β(j)) = (−1)i+j det( ( ξ

β(j)α(i) )t )

= (−1)i+j

det( ξβ(j)α(i) ) = (adjs ξ)

ji

which is statement (ii).

(iii) From (i) and (ii) we immediately deduce the claim for R ∈ O (n) , since

adjs R (adjs R)t

= adjs R adjs Rt = adjs(RRt

)

= adjs In = I(ns

)

where for any integer m we have let Im to be the identity matrix in Rm×m.

We now discuss the case where R ∈ SO (n) . We already know that

adjs R ∈ O((

ns

)).

It therefore remains to prove that

det (adjs R) = 1.

We observe that SO (n) is a connected manifold, meaning that, for every R ∈SO (n) , there exists a continuous function

θ : [0, 1]→ SO (n) , θ (0) = In , θ (1) = R.

We then define, for t ∈ [0, 1] , the function

f (t) := det (adjs θ (t)) .

We observe that since any Q ∈ SO (n) ⊂ O (n) has

det (adjs Q) ∈ ±1 ,


then the function f takes only values in ±1 . Since it is a continuous function,as a composition of three continuous functions, and since f (0) = 1, we deducethat f (1) = 1, which is the assertion.

(iv) This follows from (i) exactly as above. Indeed

adjs ξ adjs(ξ−1)

= adjs In = I(ns

) .

(v) From (ii) of Proposition 5.65, we have, since R ∈ SO (n) ,

R(adjn−1 R

)t= In

and thus the claim.

We now want to write, for every ξ, η ∈ Rn×n, det (ξ + η) . To this aim let usintroduce the following notations.

- Let N1,··· ,n be the set of couples (I, J) , each of them ordered, so that

I ∪ J = 1, · · · , n , I ∩ J = ∅.

- For all (I, J) ∈ N1,··· ,n and all matrices ξ, η ∈ Rn×n, we denote by

( ξI , ηJ ) ∈ Rn×n

the n × n matrix whose row of index k is ξk if k ∈ I or ηk if k ∈ J. So, forexample, if n = 3, I = 1, 3 , J = 2 , then

( ξI , ηJ ) =

⎛⎜⎝

ξ1

η2

ξ3

⎞⎟⎠ .

Proposition 5.67 Let ξ, η ∈ Rn×n, then

det (ξ + η) =∑

(I,J)∈N1,··· ,n

det( ξI , ηJ ).

Proof. Let us first examine the case n = 2, where we trivially have

det (ξ + η) = det( ξ1, ξ2 ) + det( ξ1, η2 ) + det( η1, ξ2 ) + det( η1, η2 ).

The general case easily follows if we write the determinant as a multilinear form;namely, for ξ ∈ Rn×n, we write

det ξ = ξ1 ∧ · · · ∧ ξn.


The claim follows by induction, since

det (ξ + η) = ξ1 ∧(ξ2 + η2

)∧ · · · ∧ (ξn + ηn) + η1 ∧

(ξ2 + η2

)∧ · · · ∧ (ξn + ηn)

=∑

(I,J)∈N2,··· ,n

det( ξ1, ξI , ηJ ) +∑

(I,J)∈N2,··· ,n

det( η1, ξI , ηJ )

=∑

(I,J)∈N1,··· ,n

det( ξI , ηJ ).

This finishes the proof of the proposition.

Chapter 6

Polyconvex, quasiconvexand rank one convexenvelopes

6.1 Introduction

We now proceed with the characterization of the convex Cf, polyconvex Pf,quasiconvex Qf and rank one convex envelope Rf, which are, respectively,defined as the largest convex, polyconvex, quasiconvex and rank one convexfunction below f. In other words, we have, for every ξ ∈ RN×n,

Cf (ξ) = sup g (ξ) : g ≤ f and g convex ,

Pf (ξ) = sup g (ξ) : g ≤ f and g polyconvex ,

Qf (ξ) = sup g (ξ) : g ≤ f and g quasiconvex ,

Rf (ξ) = sup g (ξ) : g ≤ f and g rank one convex .

The first notion has already been encountered in Chapter 2, where wegave two different characterizations of Cf. The first one, in Section 2.3.3, viaCaratheodory theorem and the second one, in Section 2.3.5, via duality and theseparation theorems.

In view of Theorem 5.3, we have

Cf ≤ Pf ≤ Qf ≤ Rf ≤ f.

In Section 6.2, we start with the polyconvex envelope Pf, which is the mostsimilar to the convex envelope Cf. We always recall, without proofs, what hasalready been said about Cf in Chapter 2 to show the resemblance between thetwo envelopes.

266 Polyconvex, quasiconvex and rank one convex envelopes

In Section 6.3, we give a representation formula for the quasiconvex envelope,inspired by Caratheodory theorem.

In Section 6.4, we discuss a representation formula for Rf, also in the spiritof Caratheodory theorem.

In Section 6.5, we present a result that in some cases can simplify the com-putations of the different envelopes.

In Section 6.6, we discuss several examples, relevant for applications, whereone can compute these envelopes.

6.2 The polyconvex envelope

6.2.1 Duality for polyconvex functions

We first recall (see Section 2.3.5) some facts about duality in convex analysis.

Definition 6.1 Let f : RN×n → R∪+∞ with f ≡ +∞ and let f∗ : RN×n →R ∪ +∞ be defined as

f∗ (ξ∗) := supξ∈RN×n

〈ξ; ξ∗〉 − f (ξ) ,

where 〈·; ·〉 denotes the scalar product in RN×n, and let

f∗∗ := (f∗)∗ ,

or in other words

f∗∗ (ξ) = supξ∗∈RN×n

〈ξ; ξ∗〉 − f∗ (ξ∗) .

Remark 6.2 As seen in Chapter 2, we always have that(i) f∗∗∗ = f∗;

(ii) f∗∗ is convex and lower semicontinuous and therefore

f∗∗ ≤ Cf ≤ f.

Moreover, if Cf is lower semicontinuous and Cf > −∞, then

f∗∗ = Cf. ♦

We now proceed in an analogous way for polyconvex functions and followhere the idea of Kohn and Strang [373], [374] as presented in Dacorogna [179].We also adopt the notation of Chapter 5.

Definition 6.3 Let f : RN×n → R ∪ +∞ with f ≡ +∞.

(i) One defines the polyconvex conjugate function of f as

fp : Rτ(n,N) → R ∪ +∞ ,

The polyconvex envelope 267

where τ (n, N) =∑n∧N

s=1

(Ns

)(ns

), by

fp (X∗) := supξ∈RN×n

〈T (ξ) ; X∗〉 − f (ξ) ,

where 〈·; ·〉 denotes the scalar product in Rτ(n,N) and T is as in Definition 5.1.

(ii) Let (fp)∗ : Rτ(n,N) → R ∪ ±∞ be defined by

(fp)∗(X) := sup

X∗∈Rτ(n,N)

〈X ; X∗〉 − fp (X∗) .

Finally, let ξ ∈ RN×n and define the polyconvex biconjugate function of f as

fpp (ξ) := (fp)∗ (T (ξ)) .

Remark 6.4 (i) It is clear that if N = 1 or n = 1

fp = f∗ and fpp = f∗∗.

(ii) It is also simple to see that

fppp = fp

and that fpp is polyconvex, lower semicontinuous and less than f and therefore

fpp ≤ Pf ≤ f.

(iii) If N = n = 2, then τ (n, N) = 5 and we can write

fp (ξ∗, δ∗) = supξ∈R2×2

〈ξ; ξ∗〉+ δ∗ det ξ − f (ξ) ,

where ξ∗ ∈ R2×2, δ∗ ∈ R and 〈·; ·〉 denotes the scalar product in R2×2. Similarly

(fp)∗(ξ, δ) = sup

ξ∗∈R2×2

δ∗∈R

〈ξ; ξ∗〉+ δ∗δ − fp (ξ∗, δ∗)

and thereforefpp (ξ) = (fp)

∗(ξ,det ξ) . ♦

We now give a simple example where one can explicitly computef∗, f∗∗, fp, fpp.

Example 6.5 Let N = n = 2 and

f (ξ) = det ξ.

(i) It is easy to show that

f∗ (ξ∗) = supξ∈R2×2

〈ξ; ξ∗〉 − det ξ ≡ +∞


and thereforef∗∗ (ξ) ≡ −∞.

(ii) Similarly,

fp (ξ∗, δ∗) = supξ∈R2×2

〈ξ; ξ∗〉+ (δ∗ − 1) det ξ

=

0 if δ∗ = 1 and ξ∗ = 0

+∞ elsewhere.

We therefore obtain

(fp)∗ (ξ, δ) = supξ∗∈R2×2

δ∗∈R

〈ξ; ξ∗〉+ δ∗δ − fp (ξ∗, δ∗) = δ

and hencefpp (ξ) = det ξ. ♦

We now have the following result, which was already proved in Theorem 2.43for the convex case.

Theorem 6.6 Let f : RN×n → R (i.e. f is finite) and let f∗∗ and fpp bedefined as in the preceding section.

Part 1. If there exists g : RN×n → R convex such that

f (ξ) ≥ g (ξ) for every ξ ∈ RN×n,

thenf∗∗ = Cf. (6.1)

Part 2. If there exists g : RN×n → R polyconvex such that

f (ξ) ≥ g (ξ) for every ξ ∈ RN×n,

thenfpp = Pf. (6.2)

Remark 6.7 (i) One can also rewrite (6.1) in the following way (if f takes onlyfinite values):

Cf (ξ) = sup g (ξ) : g ≤ f and g affine .

Similarly, for (6.2),

Pf (ξ) = sup g (ξ) : g ≤ f and g quasiaffine .

(ii) It is also interesting to note that (6.1) and (6.2) do not hold if f isallowed to take the value +∞. For example, if N = n = 1 and

f (ξ) = χ(0,1) (ξ) =

0 if ξ ∈ (0, 1)

+∞ elsewhere.

then f = Cf and f∗∗ = χ[0,1] . ♦

The polyconvex envelope 269

Proof. As already pointed out, we always have fpp ≤ Pf. We now wish toprove the reverse inequality. We divide the proof into two steps.

Step 1. We first show that if f is polyconvex and finite then

fpp = f. (6.3)

From Theorem 5.6 of Chapter 5, we have that there exists F : Rτ → R, τ =τ (n, N) , convex and finite such that

f (ξ) = F (T (ξ)) .

It is obvious from the definition that

F ∗ (X∗) = supX∈Rτ

〈X ; X∗〉 − F (X)

≥ supξ∈RN×n

〈T (ξ) ; X∗〉 − F (T (ξ)) = fp (X∗)

and thereforeF ∗∗ (X) ≤ (fp)∗ (X) .

However since F is convex and finite, we have

f (ξ) = F (T (ξ)) = F ∗∗ (T (ξ)) ≤ (fp)∗(T (ξ)) = fpp (ξ) .

Since the reverse inequality is trivial, we have indeed established (6.3).

Step 2. Applying Step 1 to Pf which is polyconvex and finite (since f ≥Pf ≥ g = Pg) we get

Pf = (Pf)pp

.

We thus deducePf = (Pf)

pp ≤ fpp ≤ f

and the result follows.

6.2.2 Another representation formula

We start by recalling the notation, valid for an integer s,

Λs := λ = (λ1, · · · , λs) : λi ≥ 0 and∑s

i=1 λi = 1 .

In Theorem 2.35, we proved that for f : RN×n → R ∪ +∞

Cf (ξ) = inf∑nN+1

i=1 λif (ξi) : λ ∈ ΛnN+1 ,∑nN+1

i=1 λiξi = ξ

.

We now discuss an analogous formula for the polyconvex envelope, that wasfirst proved by Dacorogna in [176], see also [177], [179].


Theorem 6.8 Let f : RN×n → R ∪ +∞ . Let g : RN×n → R ∪ +∞ bepolyconvex and such that

f (ξ) ≥ g (ξ) for every ξ ∈ RN×n.

Then the following formula holds, for every ξ ∈ RN×n,

Pf (ξ) = inf∑τ(n,N)+1

i=1 λif (ξi) : λ ∈ Λτ(n,N)+1 ,∑τ(n,N)+1

i=1 λiT (ξi) = T (ξ)

.

Proof. When there is no ambiguity, we write τ for τ (n, N) . We first define

P ′f (ξ) := inf∑I

i=1 λif (ξi) : λ ∈ ΛI , I ≥ τ + 1,∑I


.

(6.4)We decompose the proof into three steps.

Step 1. We first show that P ′f is polyconvex. In view of Theorem 5.6, toshow polyconvexity of P ′f, it is sufficient to see that

∑τ+1ν=1 λνP ′f (ην) ≥ P ′f(

∑τ+1ν=1 λνην) (6.5)

whenever λ ∈ Λτ+1 and

∑τ+1ν=1 λνT (ην) = T (

∑τ+1ν=1 λνην).

Fix ǫ > 0. From (6.4), we have that there exist, for every 1 ≤ ν ≤ τ + 1,

Iν ≥ τ + 1, αν ∈ ΛIν and ξνi ∈ RN×n

such that ⎧⎪⎪⎪⎨⎪⎪⎪⎩

ǫ + P ′f (ην) ≥Iν∑

i=1

ανi f (ξν

i ) , 1 ≤ ν ≤ τ + 1

Iν∑i=1

ανi T (ξν

i ) = T (ην) , 1 ≤ ν ≤ τ + 1.

Relabeling ανi and ξν

i as

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

βi = λ1α1i Xi = ξ1

i 1 ≤ i ≤ I1

βI1+i = λ2α2i XI1+i = ξ2

i 1 ≤ i ≤ I2

......

...

βI1+···+Iτ +i = λτ+1ατ+1i XI1+···+Iτ+i = ξτ+1

i 1 ≤ i ≤ Iτ+1

we get that β ∈ ΛI1+···+Iτ+1 ,

ǫ +τ+1∑ν=1

λνP ′f (ην) ≥I1+···+Iτ+1∑

i=1

βif (Xi) (6.6)

The quasiconvex envelope 271

and

I1+···+Iτ+1∑i=1

βiT (Xi) =τ+1∑ν=1

λνT (ην) = T (τ+1∑ν=1

λνην)

= T (I1+···+Iτ+1∑

i=1

βiXi).

Using (6.4) in the right hand side of the inequality in (6.6) and the fact thatǫ is arbitrary, we have indeed obtained (6.5) and therefore shown that P ′f ispolyconvex.

Step 2. We next want to prove that P ′f = Pf. We have, using Theorem 5.6and Step 1, that if h ≤ f is any polyconvex function with h ≥ g, then P ′h = hand hence

h = P ′h ≤ P ′f ≤ f.

Thus P ′f ≥ Pf, and since P ′f is polyconvex, we have indeed P ′f = Pf.

Step 3. It now remains to show that in (6.4) we can choose I = τ + 1.The proof is almost identical to that of Step 2 of Theorem 5.6 and we will notreproduce it here.

6.3 The quasiconvex envelope

The following formula was established by Dacorogna in [172], see also [176], [177].

Theorem 6.9 (Dacorogna formula) Let f : RN×n → R be locally boundedand Borel measurable. Let g : RN×n → R be quasiconvex and such that


Then, for every ξ ∈ RN×n,

Qf (ξ) = inf

1

measD

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈W 1,∞0

(D; RN

)(6.7)

where D ⊂ Rn is a bounded open set. In particular, the infimum in the formulais independent of the choice of D.

Proof. We first set the notation. We have

Qf (ξ) := sup h (ξ) : h ≤ f and h quasiconvex .

We next call for D ⊂ Rn a bounded open set

Qf (ξ) := inf

1

meas D

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈W 1,∞0

(D; RN

)


the expression given in the right hand side of (6.7) and we finally let

Q′f (ξ) := inf

1

measD

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈ A (D)

(6.8)

where (see Chapter 12)

A (D) :=ϕ ∈ W 1,∞

0

(D; RN

)∩Affpiec

(D; RN

): supp ϕ ⊂ D

.

Note that since f ≥ g, then Qf, Q′f, Qf > −∞.

The aim is to show that the definitions of Qf and Q′f are independent ofthe choice of the set D and that the three definitions coincide, namely

Qf = Q′f = Qf.

To achieve this goal we divide the proof into five steps.

Step 1. We first show that the definition of Q′f is independent of the choiceof D.

Step 2. We then establish that∫

D

Q′f (ξ +∇ψ (x)) dx ≥ Q′f (ξ)measD

for every ξ ∈ RN×n and for every ψ ∈ A (D) .

Step 3. The previous step easily shows that Q′f is quasiconvex.

Step 4. We then deduce that Q′f = Qf.

Step 5. We finally establish (6.7).

We now discuss the details of the different steps.

Step 1. For D ⊂ Rn a bounded open set, let

Q′fD (ξ) := inf 1

measD

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈ A (D). (6.9)

We wish to show that given two such sets D and E, then

Q′fD = Q′fE .

This will, a posteriori, justify the notation Q′f instead of Q′fD in (6.8). Theproof is in the same spirit as that of Proposition 5.11.

(1) We first prove this result for sets D and E which are dilated and trans-lated one from each other, namely

E = x0 + λD,

where x0 ∈ Rn and λ > 0. This is straightforward, if we set for any ϕ ∈ A (D)

ϕλ (y) := λϕ(y − x0

λ), y ∈ E


and observing that ϕλ ∈ A (E) with

1

measD

∫

D

f (ξ +∇ϕ (x)) dx =1

measE

∫

E

f (ξ +∇ϕλ (x)) dx.

(2) We now discuss the case of open sets D and E with

meas (∂D) = 0.

The idea is to approximate the set E by dilation and translation of D. Moreprecisely, for every ǫ > 0 we can find (see Lemma 5.3 in Giusti [316]) Iǫ aninteger and disjoint open sets Di ⊂ E that are homothetic to D so that

meas(E −⋃Iǫ

i=1 Di) ≤ ǫ.

We next use (6.9) and the previous observation that Q′fD = Q′fDi to findϕǫ

i ∈ A (Di) such that

∫

Di

f (ξ +∇ϕǫi (x)) dx ≤ (ǫ + Q′fD (ξ)) measDi .

Define then ϕǫ ∈ A (E) by

ϕǫ (x) =

⎧⎪⎨⎪⎩

ϕǫi (x) if x ∈ Di

0 if x ∈ E −Iǫ⋃

i=1

Di .

By the definition of Q′fE we have

Q′fE (ξ)measE ≤∫

E

f (ξ +∇ϕǫ (x)) dx

≤Iǫ∑

i=1

∫

Di

f (ξ +∇ϕǫi (x)) dx + f (ξ)meas(E −⋃Iǫ

i=1 Di)

≤ (ǫ + Q′fD (ξ))meas(⋃Iǫ

i=1 Di) + ǫf (ξ) .

Since ǫ is arbitrary, we have indeed shown that

meas (∂D) = 0 ⇒ Q′fE ≤ Q′fD . (6.10)

A similar argument establishes the reverse inequality, namely

meas (∂E) = 0 ⇒ Q′fD ≤ Q′fE , (6.11)

in particular

meas (∂E) = meas (∂D) = 0 ⇒ Q′fD = Q′fE . (6.12)


(3) We finally consider general open sets D and E. By definition, we canfind, for every ǫ > 0, ϕ ∈ A (D) such that

Q′fD (ξ)measD ≥ −ǫ +

∫

D

f (ξ +∇ϕ (x)) dx. (6.13)

Since supp ϕ ⊂ D, we can find an open set A, with meas (∂A) = 0, such that

supp ϕ ⊂ A ⊂ D.

Returning to (6.13) and using the definition of Q′fA and the fact that ϕ ∈ A (A) ,we find

Q′fD (ξ)meas A + Q′fD (ξ) meas (D − A)

≥ −ǫ +

∫

A

f (ξ +∇ϕ (x)) dx + f (ξ)meas (D −A)

≥ −ǫ + Q′fA (ξ)measA + f (ξ)meas (D −A) .

We have thus obtained, since f ≥ Q′fD , that

[ Q′fD (ξ)−Q′fA (ξ) ] measA ≥ −ǫ + [f (ξ)−Q′fD (ξ)]meas (D −A)

≥ −ǫ.

Using (6.12), we find that the above inequality holds for any set A withmeas (∂A) = 0 and thus, since ǫ is arbitrary, we have obtained that Q′fD ≥Q′fA . Since meas (∂A) = 0, we have, appealing to (6.11) and to the precedinginequality, that

Q′fA = Q′fD . (6.14)

The same reasoning on E gives that there exists an open set B, with meas (∂B) =0, such that

Q′fE = Q′fB .

Combining the above equality, (6.14) and (6.12), we have

Q′fE = Q′fB = Q′fA = Q′fD

as wished.

Step 2. We now want to show that∫

D

Q′f (ξ +∇ψ (x)) dx ≥ Q′f (ξ)measD (6.15)

for every ξ ∈ RN×n and ψ ∈ A (D) . Note that the inequality (6.15) ensures, upto a density argument (see Step 3), that Q′f is quasiconvex. Since ψ ∈ A (D) ,there exist disjoint open sets Di ⊂ D with

meas(D −⋃∞i=1 Di) = 0


and ηi ∈ RN×n such that

∇ψ (x) = ηi , x ∈ Di .

Since f is locally bounded and g ≤ Q′f ≤ f, we can find γ = γ (ξ, ψ) > 0 suchthat

|Q′f (ξ +∇ψ (x))| , |f (ξ +∇ψ (x))| ≤ γ, a.e. x ∈ D.

We therefore have, for every ǫ > 0, that there exists an integer Iǫ = Iǫ (ǫ, ξ, ψ)such that∫

D−∪Iǫi=1Di

|Q′f (ξ +∇ψ (x))| dx,

∫

D−∪Iǫi=1Di

|f (ξ +∇ψ (x))| dx ≤ ǫ (6.16)

and thus

∫

D

Q′f (ξ +∇ψ (x)) dx ≥Iǫ∑

i=1

Q′f (ξ + ηi)measDi − ǫ. (6.17)

Fixing ǫ > 0 and using (6.9) we have that there exists ϕǫi ∈ A (Di) , 1 ≤ i ≤ Iǫ ,

such that

Q′f (ξ + ηi) ≥1

measDi

∫

Di

f (ξ + ηi +∇ϕǫi (x)) dx− ǫ. (6.18)

Let χ ∈ A (D) be defined by

χ (x) :=

⎧⎪⎨⎪⎩

ψ (x) + ϕǫi (x) if x ∈ Di , i = 1, · · · , Iǫ

ψ (x) if x ∈ D −Iǫ⋃

i=1

Di .

We therefore have

∫

D

Q′f (ξ +∇ψ (x)) dx ≥Iǫ∑

i=1

∫

Di

f (ξ +∇χ (x)) dx− ǫ (1 + measD)

≥∫

D

f (ξ +∇χ (x)) dx− ǫ (2 + measD)

≥ Q′f (ξ)measD − ǫ (2 + measD)

where we have used in the first line (6.16), (6.17) and (6.18), in the second line(6.16) and in the last line (6.9). Letting ǫ → 0, we have obtained (6.15).

Step 3. We next want to prove that Q′f is quasiconvex. It will be sufficientto show that (6.15) implies that Q′f is continuous and therefore combining(6.15), the continuity of Q′f, the fact that A (D) is dense in W 1,∞

0

(D; RN

)in

any W 1,p norm, 1 ≤ p < ∞, (see Theorem 12.15) and Lebesgue dominatedconvergence theorem, we will have that Q′f is quasiconvex.


In order to show that Q′f is continuous we prove that Q′f is rank one convex.The continuity will then follow from standard properties of convex functions (seeTheorem 2.31).

Let t ∈ [0, 1] , α, β ∈ RN×n be such that rank α− β = 1. We wish to showthat Q′f is rank one convex, meaning that

t Q′f (α) + (1− t)Q′f (β) ≥ Q′f (tα + (1− t)β) . (6.19)

We can then find from Lemma 3.11 that, for every ǫ > 0, there exist k =k (α, β) > 0, u ∈ A (D) and disjoint open sets Dα, Dβ ⊂ D, so that

⎧⎪⎨⎪⎩

|meas Dα − t meas D| , |meas Dβ − (1− t)measD| ≤ ǫ,

∇u(x) =

(1− t) (α− β) in Dα

−t (α− β) in Dβ

, ‖∇u‖L∞ ≤ k.

We then use (6.15) to obtain∫

D

Q′f (tα + (1− t)β +∇u (x)) dx ≥ Q′f (tα + (1− t)β) measD.

Letting ǫ → 0, we have indeed obtained (6.19) and thus the continuity of Q′fand consequently the quasiconvexity of Q′f.

Step 4. We now show that Q′f = Qf. Observe that if h is quasiconvex then,trivially, Q′h = h. Therefore let h ≤ f be quasiconvex, we then deduce that

h = Q′h ≤ Q′f ≤ f.

Hence Q′f ≥ Qf and since Q′f itself is quasiconvex, according to Step 3, wehave indeed established that Q′f = Qf.

Step 5. It remains to establish (6.7), i.e. if

Qf (ξ) = inf

1

measD

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈W 1,∞0

(D; RN

)

then Qf = Qf. From Step 4 we also have

Qf (ξ) = inf

1

meas D

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈ A (D)

.

Since A (D) ⊂ W 1,∞0

(D; RN

), we deduce that

Qf (ξ) ≥ Qf (ξ) . (6.20)

Since Qf is quasiconvex, we immediately obtain that

Q (Qf) = Qf.

Therefore, combining (6.20) and the above identity we have

Qf ≥ Qf ≥ Q (Qf) = Qf.

This indeed establishes the result.

The rank one convex envelope 277

6.4 The rank one convex envelope

Recall first that, for any integer I, we let

ΛI :=λ = (λ1, · · · , λI) : λi ≥ 0 and

∑Ii=1 λi = 1

.

Theorem 6.10 Let f : RN×n → R ∪ +∞ . Let g : RN×n → R ∪ +∞ berank one convex and such that


Part 1. For every ξ ∈ RN×n,

Rf (ξ) = inf∑I

i=1 λif (ξi) : λ ∈ ΛI ,∑I

i=1 λiξi = ξ, (λi, ξi) satisfy (HI)

where (HI) is as in Definition 5.14.

Part 2. Let R0f := f and for k ∈ N define inductively

Rk+1f (ξ) := inf

λRkf (ξ1) + (1− λ) Rkf (ξ2) :

λξ1 + (1− λ) ξ2 = ξ with rank ξ1 − ξ2 ≤ 1

.

ThenRf = lim

k→∞Rkf = inf

k∈N

Rkf.

Remark 6.11 Part 1 was established by Dacorogna in [176], see also [177],[179]. Part 2 was proved in Kohn-Strang [373], [374]. The two approaches arevery similar and both formulas present a serious defect in the sense that in thefirst one we cannot prescribe a priori the value of the integer I, while in thesecond one we cannot prescribe that the limit Rf is attained after a given finitenumber of steps. Therefore such formulas are useful for computing Rf onlywhen there is a hint on the number of steps required in order to get Rf. ♦

Proof. Part 1. We first define

R′f (ξ) := inf∑I



.

Note that, since f ≥ g, then R′f > −∞ (cf. Proposition 5.16).

We decompose the proof into three steps.

Step 1. We start with a preliminary step where we want to show that if

λ ∈ ΛI and (λi, ξi)1≤i≤I satisfy (HI)

μ ∈ ΛJ and (μj , ηj)1≤j≤J satisfy (HJ )

and ifrank∑I

i=1 λiξi −∑J

j=1 μjηj ≤ 1 (6.21)


then, for every t ∈ [0, 1] , we have((tλi, ξi)1≤i≤I , ((1− t)μj , ηj)1≤j≤J

)satisfy (HI+J) . (6.22)

To show (6.22) we proceed by induction over I + J.The case I + J = 2 is trivial since this implies that I = J = 1 and therefore

(6.22) is equivalent, by definition, to (6.21).Assume therefore that (6.22) has been established for I + J − 1. We may

also assume, without loss of generality, that I ≥ 2. Since (λi, ξi) satisfy (HI) wehave, up to a permutation, that

rank ξ1 − ξ2 ≤ 1 (6.23)

and if ⎧⎨⎩

λ1 = λ1 + λ2, ξ1 =λ1ξ1 + λ2ξ2

λ1 + λ2

λi = λi+1, ξi = ξi+1, i ≥ 2

then (λi, ξi

)1≤i≤I−1

satisfy (HI−1) .

Note that (6.21) implies then that

rank∑I−1i=1 λiξi −

∑Jj=1 μjηj ≤ 1.

The hypothesis of induction therefore ensures that((

tλi , ξi

)1≤i≤I−1

, ((1− t)μj , ηj)1≤j≤J

)satisfy (HI+J−1) .

Coupling the last statement and (6.23), we have indeed obtained (6.22).

Step 2. We now show that R′f is rank one convex, i.e.

tR′f (ξ) + (1− t)R′f (η) ≥ R′f (tξ + (1− t) η)

for every ξ, η ∈ RN×n such that

rank ξ − η ≤ 1.

Fix ǫ > 0 and use the definition of R′f to get⎧⎪⎨⎪⎩

ǫ + R′f (ξ) ≥I∑

i=1

λif (ξi) ,I∑

i=1

λiξi = ξ,

λ ∈ ΛI , (λi, ξi)1≤i≤I satisfy (HI)

and ⎧⎪⎨⎪⎩

ǫ + R′f (η) ≥J∑

j=1

μjf (ηj) ,J∑

j=1

μjηj = η,

μ ∈ ΛJ , (μj , ηj)1≤j≤J satisfy (HJ) .

The rank one convex envelope 279

Combining the above two inequalities with Step 1 we get

ǫ + tR′f (ξ) + (1− t)R′f (η) ≥I∑

i=1

tλif (ξi) +

J∑

j=1

(1− t)μjf (ηj)

where⎧⎪⎪⎨⎪⎪⎩

I∑i=1

tλiξi +J∑

j=1

(1− t)μjηj = tξ + (1− t) η

((tλi, ξi)1≤i≤I , ((1− t)μj , ηj)1≤j≤J

)satisfy (HI+J) .

Using the definition of R′f and the fact that ǫ is arbitrary, we have indeedobtained that R′f is rank one convex.

Step 3. We may now conclude. Note first that if h (g ≤ h) is rank oneconvex, then, by Proposition 5.16, we have R′h = h. Finally let h ≤ f be rankone convex and observe that

h = R′h ≤ R′f ≤ f

and thus R′f ≥ Rf. Since R′f is rank one convex we have indeed the result,namely R′f = Rf.

Part 2. We start by observing that

g ≤ Rk+1f ≤ Rkf

and therefore the following quantities are well defined

R′f := limk→∞

Rkf = infk∈N

Rkf.

To prove the result of Part 2, we first note that if f is rank one convex, thentrivially

R′f = f.

This implies that R′ (Rf) = Rf and hence

Rf = R′ (Rf) ≤ R′f ≤ f. (6.24)

We next prove that R′f is rank one convex, which combined with the inequality(6.24) leads to the claim, namely Rf = R′f.

It therefore remains to show that for every t ∈ [0, 1] and every ξ, η ∈ RN×n

with rank ξ − η ≤ 1 we have

tR′f (ξ) + (1− t)R′f (η) ≥ R′f (tξ + (1− t) η) .

By definition we have that for every ǫ > 0, we can find i, j ∈ N, such that (weassume, without loss of generality, that i ≤ j)

R′f (ξ) ≥ −ǫ + Rif (ξ) ≥ −ǫ + Rjf (ξ) and R′f (η) ≥ −ǫ + Rjf (η) .


We thus find that

tR′f (ξ) + (1− t)R′f (η) ≥ −ǫ + tRjf (ξ) + (1− t)Rjf (η)

≥ −ǫ + Rj+1f (tξ + (1− t) η)

≥ −ǫ + R′f (tξ + (1− t) η) .

The claim follows by letting ǫ→ 0.

6.5 Some more properties of the envelopes

6.5.1 Envelopes and sums of functions

We now give a result that allows us to separate in some cases the computationof the different envelopes.

Theorem 6.12 Let u = u (x, t) : Rn × Rm → RN ,

∇u = (∇xu,∇tu) ∈ RN×n × RN×m.

Let ξ = (α, β) ∈ RN×n × RN×m = RN×(n+m) and

f (ξ) = g (α) + h (β)

where g and h satisfy the hypotheses of Theorem 2.35 for the convex case (andfor the polyconvex, quasiconvex and rank one convex cases, the hypotheses ofTheorems 6.8, 6.9 and 6.10, respectively). Then

Cf = Cg + Ch

Pf = Pg + Ph

Qf = Qg + Qh

Rf = Rg + Rh.

Remark 6.13 This result is standard for Cf and for the other cases it hasbeen established by Dacorogna [175], [179]. ♦

Proof. All three formulas are easily proved and we do so only for Qf and Rf(the formula for Pf is proved in exactly the same way).

Formula for Rf. It is clear that if α ∈ RN×n and β ∈ RN×m, then

Rg (α) + Rh (β) ≤ Rf (α, β) . (6.25)

It therefore only remains to prove the reverse inequality. For this we first provethat

Rf (α, β) ≤ Rg (α) + h (β) . (6.26)

Some more properties of the envelopes 281

Fix ǫ ≥ 0, then by Theorem 6.10, there exist (λi, αi)1≤i≤I such that

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

λ ∈ ΛI ,I∑

i=1

λiαi = α

αi ∈ RN×n with (λi, αi)1≤i≤I satisfying (HI)

ǫ + Rg (α) ≥I∑

i=1

λig (αi) .

It is clear that if β ∈ RN×m, then (λi, (αi, β))1≤i≤I satisfy (HI) and therefore

⎧⎪⎪⎪⎨⎪⎪⎪⎩

I∑i=1

λi (αi, β) = (α, β) , (λi, (αi, β))1≤i≤I satisfy (HI)

ǫ + Rg (α) + h (β) ≥I∑

i=1

λi (g (αi) + h (β)) .

Again using Theorem 6.10 and the fact that ǫ is arbitrary, we obtain (6.26). Asimilar argument shows that

Rf (α, β) ≤ g (α) + Rh (β) .

We now combine Theorem 6.10, (6.26) and the above inequality to get

Rf = R (Rf) ≤ R (Rg + h) ≤ Rg + Rh

which, combined with (6.25), is the claimed result.

Formula for Qf. We establish the present formula similarly as the one forRf. We first prove that Qf ≤ Qg + h, then, in exactly the same way, thatQf ≤ g + Qh and conclude as above. Therefore we only show that

Qf (α, β) ≤ Qg (α) + h (β) . (6.27)

From Theorem 6.9, we have that if D ⊂ Rn and Ω ⊂ Rm are unit cubes, then

Qf (α, β)= infϕ∈W 1,∞

0 (D×Ω;RN )∫

D

∫

Ω

[g (α +∇xϕ (x, t)) + h (β +∇tϕ (x, t))] dxdt.

Let ǫ > 0 be fixed, then Theorem 6.9 implies that there exists σ ∈W 1,∞0

(D; RN

)

such that ∫

D

g (α +∇xσ (x)) dx ≤ ǫ + Qg (α) .

On extending σ by periodicity from D to Rn, we trivially have that, for ν ∈ N,∫

D

g (α +∇xσ (νx)) dx ≤ ǫ + Qg (α) . (6.28)

Let Ων ⊂ Ω be a cube with the same centre as Ω and such that

dist (∂Ω; Ων) =1

ν.


We then define ψ ∈ W 1,∞0 (Ω) , 0 ≤ ψ (t) ≤ 1, ‖gradψ‖L∞ ≤ Lν, for a certain

constant L, and such that

ψ (t) =

1 if t ∈ Ων ⊂ Ω ⊂ Rm

0 if t ∈ ∂Ω.

and choose

ϕ (x, t) :=1

νσ (νx) ψ (t) .


(D × Ω; RN

). Using the formula for Qf (α, β) we get

Qf (α, β) ≤∫

D

∫

Ω

[ g (α + ψ (t)∇xσ (νx)) + h(β +1

νσ (νx)⊗ gradψ (t)) ]dxdt

(6.29)where σ (νx) ⊗ gradψ (t) denotes the tensorial product in RN×(n+m). We nextuse (6.28) to get, recalling that measΩ = measD = 1,

∫

D

∫

Ω

g (α + ψ (t)∇xσ (νx)) dxdt

=

∫

D

∫

Ων

g (α +∇xσ (νx)) dxdt +

∫

D

∫

Ω−Ων

g (α + ψ (t)∇xσ (νx)) dxdt

≤ [ǫ + Qg (α)] measΩν + meas (Ω− Ων) sup |g (α + ψ (t)∇xσ (νx))| .

Similarly we have∫

D

∫

Ω

h(β +1

νσ (νx) ⊗ gradψ (t))dxdt

≤ h (β)measΩν +

∫

D

∫

Ω−Ων

h(β +1

νσ (νx) ⊗ gradψ (t))dxdt

≤ h (β)measΩν + meas (Ω− Ων) sup| h(β +1

νσ (νx)⊗ gradψ (t)) |.

Combining (6.29) and the above two inequalities, letting ν → +∞ and using thefact that ǫ is arbitrary, we have indeed established (6.27) and thus the result.

6.5.2 Envelopes and invariances

We now see that, in some cases, the different envelopes inherit the invariance ofthe function.

Theorem 6.14 Let f : RN×n → R ∪ +∞ satisfy the hypotheses of Theorem2.35 for the convex case (and for the polyconvex, quasiconvex and rank oneconvex cases, the hypotheses of Theorems 6.8, 6.9 and 6.10 respectively). LetΓ1 ⊂ RN×N be a subgroup of GL (N) and Γ2 ⊂ Rn×n be a subgroup of GL (n) .Assume that f is Γ1 × Γ2 -invariant, meaning that

f (UξV ) = f (ξ) , ∀U ∈ Γ1 , ∀V ∈ Γ2 .

Then Cf, Pf, Qf, Rf are Γ1 × Γ2 -invariant.

Some more properties of the envelopes 283

Proof. We first recall that for any integer s, we denote by

Λs := λ = (λ1, · · · , λs) : λi ≥ 0 and∑s

i=1 λi = 1 .

(i) We first prove the result for Cf. We use Theorem 2.35 to write

Cf (ξ) = inf∑nN+1

i=1 λif (ξi) : λ ∈ ΛnN+1 ,∑nN+1

i=1 λiξi = ξ

.

Since matrices in Γ1 and Γ2 are invertible, we have, for every U ∈ Γ1 , V ∈ Γ2 ,

nN+1∑

i=1

λiξi = ξ ⇔nN+1∑

i=1

λiUξiV = UξV

and since, moreover,

nN+1∑

i=1

λif (ξi) =nN+1∑

i=1

λif (UξiV )

we find immediately

Cf (UξV ) = Cf (ξ) .

(ii) We now discuss the case of Pf. We first invoke Theorem 6.8 that gives

Pf(ξ) = inf∑τ(n,N)+1

i=1 λif(ξi) : λ ∈ Λτ(n,N)+1,∑τ(n,N)+1


.

Since matrices in Γ1 and Γ2 are invertible and appealing to Proposition 5.66 (i),we find that

τ(n,N)+1∑

i=1

λiT (ξi) = T (ξ) ⇔τ(n,N)+1∑

i=1

λiT (UξiV ) = T (UξV ) .

Furthermore the invariance of f, leads immediately to

Pf (UξV ) = Pf (ξ) .

(iii) We next turn our attention to Qf. We first let D ⊂ Rn be a boundedopen set, U ∈ Γ1 and V ∈ Γ2 and we note that

ϕ ∈ W 1,∞0

(D; RN

)⇔ ψ ∈W 1,∞

0

(V D; RN

)

where

ϕ (x) = Uψ (V x)

which implies that

∇ϕ (x) = U∇ψ (V x)V.


The resultQf (UξV ) = Qf (ξ)

then follows from Theorem 6.9, which states that

Qf (ξ) = inf

1

meas D

∫

D

f (ξ +∇ϕ (x)) dx : ϕ ∈W 1,∞0

(D; RN

).

(iv) We finally prove the result for Rf. We first note that the invertibilityof matrices in Γ1 and Γ2 leads to

λξ1 + (1− λ) ξ2 = ξ ⇔ λUξ1V + (1− λ) Uξ2V = UξV

andrankξ1 − ξ2 ≤ 1 ⇔ rank Uξ1V − Uξ2V ≤ 1.

Theorem 6.10 Part 2 then implies that

Rif (UξV ) = Rif (ξ) , ∀i ∈ N

and thusRf (UξV ) = Rf (ξ) .


Note that if Γ1 is a subgroup of GL (n) , then the set

Γt1 :=

M t : M ∈ Γ1

is also a subgroup of GL (n) . We now have the following elementary proposition.

Proposition 6.15 Let f : RN×n → R ∪ +∞ , f ≡ +∞, let Γ1 be a sub-group of GL (N) , and let Γ2 be a subgroup of GL (n) . Consider the followingstatements:

(i) f is Γ1 × Γt2 -invariant;

(ii) f∗ is Γt1 × Γ2 -invariant.

Then (i) implies (ii), and the converse is true if f is lower semicontinuous andconvex.

Proof. Suppose that f is Γ1 × Γt2 -invariant, and let U ∈ Γ1 and V ∈ Γ2 . For

every ξ, X ∈ RN×n, we have⟨U tξV ; X

⟩= trace

(U tξV Xt

)= trace

(ξV XtU t

)=⟨ξ; UXV t

⟩.

Thus

f∗(U tξV ) = sup⟨

U tξV ; X⟩− f (X) : X ∈ RN×n

= sup⟨

ξ; UXV t⟩− f(UXV t

): X ∈ RN×n

= sup〈ξ; Y 〉 − f (Y ) : Y ∈ RN×n

Examples 285

since X → UXV t is bijective. Therefore, f∗(U tξV ) = f∗(ξ), so that f∗ isΓt

1×Γ2 -invariant. If f is lower semicontinuous and convex, the converse followsdually, since f∗∗ = f in this case.

6.6 Examples

We now turn our attention to some examples where one can explicitly computethe different envelopes. Usually one is interested (see Chapter 9) in computingQf (Cf in the scalar case), which is in general a difficult problem. One way ofdoing so is to compute Pf and Rf and then show that they are equal, whichis, of course, not always true. The examples should be compared with those ofSection 5.3.

6.6.1 Duality for SO (n) × SO (n) and O (N) × O (n)invariant functions

The results of the present section and Section 6.6.2 are closely related. In thissection, we discuss the duality and polyconvex duality aspects of the problem.

We adopt the notation of Section 5.3.3 and Chapter 13. In particular wealways assume that N ≥ n (a completely analogous treatment holds when N <n). For ξ ∈ RN×n, we denote its singular values by

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ)

and we let

λ (ξ) = (λ1 (ξ) , · · · , λn (ξ)) .

When N = n, we denote by

0 ≤ μ1 (ξ) ≤ · · · ≤ μn (ξ)

the signed singular values of ξ ∈ Rn×n and we let

μ (ξ) = (μ1 (ξ) , · · · , μn (ξ)) .

We now see how the duality in convex analysis is carried for the SO (n) ×SO (n)-invariant (respectively O(N)×O (n)-invariant) functions and we followthe presentation of Dacorogna-Marechal [204].

Theorem 6.16 (i) Let f : Rn×n → R ∪ +∞ be SO (n) × SO (n)-invariant,f ≡ +∞, and let g : Rn → R ∪ +∞ be the unique Πe (n)-invariant functionsuch that f = g μ. Then

f∗ = g∗ μ.


(ii) Let N ≥ n, let f : RN×n → R ∪ +∞ be O(N) × O (n)-invariant,f ≡ +∞, and let g : Rn → R ∪ +∞ be the unique Π(n)-invariant functionsuch that f = g λ. Then

f∗ = g∗ λ.

Proof. (i) We start by noticing that

f∗(ξ) = supX∈Rn×n

〈ξ; X〉 − f(X)

= supX∈Rn×n

〈ξ; X〉 − g(μ (X))

= supX∈Rn×n

supQ,R∈SO(n)

⟨ξ; QXRt

⟩− g(μ

(QXRt

)).

We obtain that, for every Q, R ∈ SO (n) ,

⟨ξ; QXRt

⟩= trace

(ξtQXRt

)= trace

(QXRtξt

)and μ

(QXRt

)= μ (X)

and therefore the inner supremum is, from Theorem 13.10 (i), equal to

n∑

k=1

μk (ξ) μk (X)− g(μ1 (X) , · · · , μn (X)).

Furthermore, μ (X) runs over

Kn := x = (x1, · · · , xn) ∈ Rn : |x1| ≤ x2 ≤ · · · ≤ xn ,

as X runs over Rn×n. Therefore,

f∗(ξ) = supx∈Kn

〈μ (ξ) ; x〉 − g(x) . (6.30)

On the other hand, let x′ ∈ Rn and first find M ∈ Πe (n) such that

x := Mx′ ∈ Kn.

Next apply Proposition 13.9 and the invariance of g under the action of Πe (n)to get, for every y ∈ Kn,

g(x′) = g(x) and 〈y; x′〉 ≤ 〈y; x〉 .

We therefore have that, for every y ∈ Kn,

g∗(y) := supx∈Rn

〈y; x〉 − g(x) = supx∈Kn

〈y; x〉 − g(x). (6.31)

The result follows from (6.30) and (6.31).

Examples 287

(ii) We first observe that

f∗(ξ) = supX∈RN×n

〈ξ; X〉 − f(X)

= supX∈RN×n

supQ∈O(N)R∈O(n)

⟨ξ; QXRt

⟩− f(QXRt)

= supX∈RN×n

supQ∈O(N)R∈O(n)

⟨ξ; QXRt

⟩− f(X)

By Theorem 13.10 (ii), we get

supQ∈O(N)R∈O(n)

⟨ξ; QXRt

⟩= sup

Q∈O(N)R∈O(n)

trace(QXRtξt)

=

n∑

k=1

λk (ξ)λk (X) .

Furthermore, λ (X) runs over

Kn+ := x = (x1, · · · , xn) ∈ Rn : 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn ,

as X runs over RN×n. We therefore deduce the following identity

f∗(ξ) = supx∈Kn

+

〈λ (ξ) ; x〉 − g(x) . (6.32)

On the other hand, let x′ ∈ Rn and first find M ∈ Π(n) such that

x := Mx′ ∈ Kn+ .

Next apply Proposition 13.9 and the invariance of g under the action of Π (n)to get, for every y ∈ Kn

+ ,

g(x′) = g(x) and 〈y; x′〉 ≤ 〈y; x〉 .

We thus deduce that, for every y ∈ Kn+ ,

g∗(y) := supx∈Rn

〈y; x〉 − g(x) = supx∈Kn

+

〈y; x〉 − g(x). (6.33)

The result follows from (6.32) and (6.33).

We get the following as an immediate corollary.

Theorem 6.17 (i) Let f : Rn×n → R ∪ +∞ be SO (n) × SO (n)-invariant,f ≡ +∞, and let g := f diag . Then

f∗∗ = g∗∗ μ.

Furthermore, let Cf and Cg denote the convex envelopes of f and g, respectively.Assume that the relationships Cf = f∗∗ and Cg = g∗∗ hold, which happensnotably when f and g are finite and bounded below by a convex function. Then

Cf = Cg μ.


(ii) Let N ≥ n and let f : RN×n → R ∪ +∞ be O(N) × O (n)-invariantand f ≡ +∞. Let g := f diagN×n . Then

f∗∗ = g∗∗ λ.

Assume that the relationships Cf = f∗∗ and Cg = g∗∗ hold, which happensnotably when f and g are finite and bounded below by a convex function. Then

Cf = Cg λ.

Proof. It is an immediate consequence of Theorem 6.16.

The corollary allows an important simplification in the computation of theconvex envelopes of SO (n)×SO (n)-invariant (or O(N)×O (n)-invariant) func-tions f. This is true if either one proceeds by duality or by Caratheodory the-orem. In this last case, for example, while to compute Cf we normally needNn + 1 matrices, for Cg it is sufficient to take n + 1 diagonal matrices.

We now turn to similar results for the polyconvex envelope and have thefollowing result due to Dacorogna-Marechal [205] (see also Section 6.6.2). Butwe first need the following definition (compare with Definition 6.3).

Definition 6.18 Let g : R2 → R ∪ +∞ with g ≡ +∞. We define

gp : R3 → R ∪ +∞

asgp (x∗

1, x∗2, δ

∗) := sup(x1,x2)∈R2

x1x∗1 + x2x

∗2 + x1x2δ

∗ − g (x1, x2) .

Similarly, we let gpp : R2 → R ∪ ±∞ be defined as

gpp (x1, x2) := (gp)∗(x1, x2, x1x2)

where

(gp)∗(x1, x2, δ) := sup

(x∗1,x∗

2,δ∗)∈R3

x1x∗1 + x2x

∗2 + δδ∗ − gp (x∗

1, x∗2, δ

∗) .

Theorem 6.19 Let f : R2×2 → R ∪ +∞ be SO (2)× SO (2)-invariant, f ≡+∞ and g : R2 → R ∪ +∞ be the unique Πe (2)-invariant function such thatf = g μ. The following two properties then hold.

(i) For every X∗ = (ξ∗, δ∗) ∈ R2×2 × R,

fp (X∗) = gp (μ (ξ∗) , δ∗) .

(ii) For every ξ ∈ R2×2,

fpp (ξ) = gpp (μ (ξ)) ,

Examples 289

or differently expressed

fpp = gpp μ.

Furthermore, if f : R2×2 → R and is bounded below by a polyconvex function,then Pf = fpp.

Proof. (i) By definition of fp, we have

fp (X∗) = fp (ξ∗, δ∗) = supξ∈R2×2

〈ξ; ξ∗〉+ δ∗ det ξ − f (ξ)

= supξ∈R2×2

supQ,R∈SO(2)

⟨QξRt; ξ∗

⟩+ δ∗ det(QξRt)− gμ(QξRt)) .

We then invoke Theorem 13.10 and the SO (2) × SO (2) invariance of det andμ, to get

fp (X∗) = supξ∈R2×2

∑2j=1 μj (ξ) μj (ξ∗) + δ∗ det ξ − g (μ (ξ))

= supξ∈R2×2

∑2j=1 μj (ξ) μj (ξ∗) + δ∗μ1 (ξ)μ2 (ξ)− g (μ (ξ))

= sup0≤|x1|≤x2

x1μ1 (ξ∗) + x2μ2 (ξ∗) + x1x2δ∗ − g (x1, x2) .

Using the definition of gp, we have

gp (μ (ξ∗) , δ∗) := sup(x1,x2)∈R2

x1μ1 (ξ∗) + x2μ2 (ξ∗) + x1x2δ∗ − g (x1, x2) .

Since 0 ≤ |μ1 (ξ∗)| ≤ μ2 (ξ∗) , we deduce that

sup(x1,x2)∈R2

x1μ1 (ξ∗) + x2μ2 (ξ∗) = sup0≤|x1|≤x2

x1μ1 (ξ∗) + x2μ2 (ξ∗)

and therefore, since the functions (x1, x2)→ x1x2 and g are Πe (2)-invariant,

gp (μ (ξ∗) , δ∗) = sup0≤|x1|≤x2

x1μ1 (ξ∗) + x2μ2 (ξ∗) + x1x2δ∗ − g (x1, x2) .

The claim has therefore been validated.

Let us show, for further reference, that the function

x∗ = (x∗1, x

∗2)→ gp (x∗, δ∗)

is Πe (2)-invariant, for every δ∗ ∈ R. Recall that

Πe (2) :=

±(

1 00 1

), ±(

0 11 0

).


We therefore have, using the Πe (2)-invariance of g, that, for every M ∈ Πe (2) ,

gp (Mx∗, δ∗) = supx∈R2

〈x; Mx∗〉+ x1x2δ∗ − g (x)

= supx∈R2

⟨M tx; x∗⟩+

(M tx

)1

(M tx

)2δ∗ − g

(M tx

)

= supy∈R2

〈y; x∗〉+ y1y2δ∗ − g (y)

= gp (x∗, δ∗) .

(ii) The proof is almost identical to that of (i). We have, using the definition,(i) and Theorem 13.10,

fpp (ξ) = (fp)∗(ξ,det ξ) = sup

ξ∗∈R2×2

δ∗∈R

〈ξ; ξ∗〉+ δ∗ det ξ − fp (ξ∗, δ∗)

= supξ∗∈R2×2

δ∗∈R

〈ξ; ξ∗〉+ δ∗ det ξ − gp (μ (ξ∗) , δ∗)

= supξ∗∈R2×2

δ∗∈R

supQ,R∈SO(2)

⟨ξ; Qξ∗Rt

⟩+ δ∗ det ξ − gp

(μ(Qξ∗Rt

), δ∗)

= supξ∗∈R2×2

δ∗∈R

∑2j=1 μj (ξ)μj (ξ∗) + δ∗μ1 (ξ)μ2 (ξ)− gp (μ (ξ∗) , δ∗)

= sup0≤|x∗

1 |≤x∗2

δ∗∈R

x∗1μ1 (ξ) + x∗

2μ2 (ξ) + δ∗μ1 (ξ)μ2 (ξ)− gp (x∗1, x

∗2, δ

∗) .

On the other hand

gpp (x1, x2) = (gp)∗ (x1, x2, x1x2)

= sup(x∗

1,x∗2,δ∗)∈R3

x1x∗1 + x2x

∗2 + x1x2δ

∗ − gp (x∗1, x

∗2, δ

∗)

and thus

gpp(μ(ξ)) = sup(x∗

1 ,x∗2,δ∗)∈R3

x∗1μ1(ξ) + x∗

2μ2(ξ) + δ∗μ1(ξ)μ2(ξ)− gp (x∗1, x

∗2, δ

∗) .

Since (cf. (i)) the function

(x∗1, x

∗2) → gp (x∗

1, x∗2, δ

∗)

is Πe (2)-invariant, for every δ∗ ∈ R and 0 ≤ |μ1 (ξ)| ≤ μ2 (ξ) , we obtain

sup(x∗

1,x∗2)∈R2

x∗1μ1 (ξ) + x∗

2μ2 (ξ) = sup0≤|x∗

1|≤x∗2

x∗1μ1 (ξ) + x∗

2μ2 (ξ)

and therefore

gpp (μ (ξ)) = sup0≤|x∗

1 |≤x∗2

δ∗∈R

x∗1μ1 (ξ) + x∗

2μ2 (ξ) + δ∗μ1 (ξ)μ2 (ξ)− gp (x∗1, x

∗2, δ

∗) .

The statement (ii) is thus established.

(iii) The fact that Pf = fpp was already discussed in Theorem 6.6.

Examples 291

6.6.2 The case of singular values

As already said, the results of the present section are closely related to those ofthe previous one, but the emphasis is now on Caratheodory type representationformulas.

We first start with an immediate corollary of Theorem 6.14. We will thensee how to improve the theorem in some special cases.

Theorem 6.20 Let f : RN×n → R ∪ +∞ . Let g : RN×n → R ∪ +∞ berespectively convex, polyconvex, quasiconvex (in this case, f, g : RN×n → R),rank one convex and such that


Assume that f is O (N)×O (n)-invariant, meaning that

f (RξQ) = f (ξ) , ∀R ∈ O (N) , ∀Q ∈ O (n) .

Then Cf, Pf, Qf, Rf are respectively O (N)×O (n)-invariant.

Remark 6.21 When N = n, a completely analogous result is valid for SO (n)×SO (n)-invariant functions. ♦

We next continue our study of the envelopes of SO (n) × SO (n)-invariantfunctions. We recall that we denote the singular values of ξ ∈ Rn×n by 0 ≤λ1(ξ) ≤ · · · ≤ λn(ξ). We consider functions of the form

f(ξ) = g(λ1(ξ), · · · , λn(ξ), det ξ). (6.34)

The first immediate consequence of Theorem 6.20 is that Cf, Pf, Qf, Rf are ofthe same form. In general, it is difficult to determine these envelopes in functionof some convex envelopes of g. Furthermore, in such a general context, theenvelopes might all be different. We now examine some examples and counterexamples and recall the notation that

Kn+ := x = (x1, · · · , xn) ∈ Rn : 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn .

Theorem 6.22 Let f : Rn×n → R.

(i) Let g : Kn−1+ → R be upper semicontinuous at (0, x2, · · · , xn−1) and

inf g > −∞. If

f(ξ) = g(λ1(ξ), · · · , λn−1(ξ)),

then

Cf(ξ) = Pf(ξ) = Qf(ξ) = Rf(ξ) = inf g.


(ii) There exists a function g : R → R, namely

g (t) =

1 + t2 if t = 0

0 if t = 0

such that iff(ξ) = g(λn(ξ)),

thenCf < Pf.

(iii) Let g : Kn−2+ → R be upper semicontinuous at (0, x3, · · · , xn−1) and

inf g > −∞. Let h : R → R and a, b ∈ R be such that

h (δ) ≥ aδ + b, ∀ δ ∈ R.

Iff(ξ) = g(λ2(ξ), · · · , λn−1(ξ)) + h (det ξ)

thenPf(ξ) = Qf(ξ) = Rf(ξ) = inf g + Ch(det ξ).

(iv) Let f : R2×2 → R be defined by

f(ξ) = |λ1(ξ)− 1|+ |det ξ| .

Then there exists ξ ∈ R2×2 such that

Pf(ξ) = |det ξ| .

Remark 6.23 (i) The first two statements were established in Buttazzo-Dacorogna-Gangbo [113], while the last two were proved in Dacorogna-Pisante-Ribeiro [211].

(ii) Assertion (iii) of the theorem is valid for more general functions of theform

f(ξ) = g(λ2(ξ), · · · , λn−1(ξ), det ξ).

(iii) It is interesting to note that (ii) and (iv) of the present theorem showthat results (i) and (iii) are not true if some dependence on λ1 or λn is allowed.♦

Proof. (i) The inequalities

inf g ≤ Cf(ξ) ≤ Pf(ξ) ≤ Qf(ξ) ≤ Rf(ξ),

being obvious, we only need to prove that

Rf(ξ) ≤ inf g, ∀ ξ ∈ Rn×n. (6.35)

Examples 293

To show this inequality, we proceed in two steps.Step 1. Let x ∈ Kn−1

+ with x1 > 0 and denote

Ex := η ∈ Rn×n : λi(η) = xi, i = 1, · · · , n− 1.

Using the results and the notations of Chapter 7, we find

RcoEx = Rn×n, (6.36)

where Rco Ex denotes the rank one convex hull of Ex . Indeed, let ξ ∈ Rn×n

and choose xn ≥ xn−1 so large that

n∏

i=ν

λi (ξ) ≤n∏

i=ν

xi, ν = 1, · · · , n

and apply Theorem 7.43 to get that

ξ ∈ Rco[Ex ∩ η ∈ Rn×n : λn(η) = xn

]⊂ RcoEx

and hence (6.36) is proved.

Step 2. Let x ∈ Kn−1+ with x1 > 0 and let us show that

Rf(ξ) ≤ g (x) , ∀ξ ∈ Rn×n. (6.37)

Define a function F : Rn×n → R by

F (η) := Rf(η)− g(x).

Observe that F is rank one convex and that

F (η) = Rf(η)− f (η) , ∀η ∈ Ex .

Therefore F |Ex≤ 0 and hence, by Definition 7.25 and Theorem 7.28, we deduce

thatF |Rco Ex

≤ 0

which combined with (6.36) leads to the claimed inequality (6.37).To prove the desired inequality (6.35), it remains to approximate any x ∈

Kn−1+ by xν ∈ Kn−1

+ with xν1 > 0 and apply (6.37) with xν and use the upper

semicontinuity of g at 0 to get (6.35).

(ii) It is sufficient to restrict our attention to the case n = 2. Since ξ → λ2(ξ)is a norm, we have (see Theorem 6.30) that

Cf (ξ) = Cg(λ2(ξ)),

where

Cg (t) =

1 + t2 if |t| ≥ 1

2 |t| if |t| < 1.


In view of Theorem 6.19, it is sufficient, for computing Pf, to only consider diag-onal matrices and therefore, by abuse of notation, we write a diagonal matrixξ ∈ R2×2 whose entries are x and y as ξ = (x, y) .

In order to prove the desired inequality, namely Cf < Pf, we show that if

0 < a < 2(√

2− 1) < 1

thenCf (a, a) = 2a < Pf (a, a) . (6.38)

Applying Theorem 6.19, we get

Pf (a, a) = sup|x|≤yδ∈R

a (x + y) + δa2 − fp (x, y, δ)

, (6.39)

wherefp (x, y, δ) = sup

|α|≤β

αx + βy + δαβ − g (β) .

We next compute, for y ≥ 0,

fp (y, y,−1) = sup|α|≤β

(α + β) y − αβ − g (β)

= max0, sup|α|≤β

(α + β) y − αβ − 1− β2

.

It is easy to see that the last supremum is attained for α = β = y/2 and henceif we denote

[x]+ =

x if x ≥ 0

0 otherwise

we find

fp (y, y,−1) =

[y2

2− 1

]

+

.

Returning to (6.39), we find that

Pf (a, a) ≥ supy≥0

2ay − a2 − fp (y, y,−1)

≥ 2a

√2− a2.

The last inequality and the fact that 0 < a < 2(√

2− 1)

immediately give (6.38).

(iii) We proceed in a way very similar to (i). The inequalities

inf g + Ch(det ξ) ≤ Pf(ξ) ≤ Qf(ξ) ≤ Rf(ξ)

being obvious, we only show

Rf(ξ) ≤ inf g + Ch(det ξ), ∀ξ ∈ Rn×n. (6.40)

Examples 295

To establish this inequality, we proceed in two steps. Recall that

Kn−2+ := x = (x2, · · · , xn−1) ∈ Rn−2 : 0 ≤ x2 ≤ · · · ≤ xn−1.

Step 1. Let c ∈ R and x ∈ Kn−2+ with x2 > 0. Denote

Ex,c := ξ ∈ Rn×n : λi(ξ) = xi, i = 2, · · · , n− 1, det ξ = c.

Using the results and the notations of Chapter 7, we find

Rco Ex,c = ξ ∈ Rn×n : det ξ = c. (6.41)

Indeed, let ξ ∈ Rn×n with det ξ = c, choose xn ≥ xn−1 so large that

n∏

i=ν

λi (ξ) ≤n∏

i=ν

xi, ν = 2, · · · , n

and apply Theorem 7.43 to get that

ξ ∈ Rco[Ex,c ∩ η ∈ Rn×n : λn(η) = xn

]⊂ RcoEx,c

and hence (6.41) is proved.

Step 2. Let x ∈ Kn−2+ with x2 > 0 and let us show that

Rf(ξ) ≤ g (x) + Ch(det ξ), ∀ ξ ∈ Rn×n, (6.42)

which will follow if we can prove that, for every ξ ∈ Rn×n,

Rf(ξ) ≤ g (x) + h(det ξ). (6.43)

In fact, if we get (6.43), then the rank one convex envelope of each memberpreserves the inequality and since the rank one convex envelope of h(det ξ) isCh(det ξ), we get (6.42).

So let ξ be any matrix in Rn×n with c := det ξ and let us show (6.43). Tothis aim, we define a function Fξ : Rn×n → R such that

Fξ(η) := Rf(η)− g(x)− h(det ξ).

Observe that Fξ is rank one convex and that

Fξ(η) = Rf(η)− f (η) , ∀ η ∈ Ex,c .

Therefore Fξ|Ex,c≤ 0 and hence, by Definition 7.25 and Theorem 7.28, we

deduce thatFξ|Rco Ex,c

≤ 0,

which means (see (6.41)) that, for every η ∈ Rn×n with det η = det ξ = c,

Fξ(η) = Rf(η)− g(x)− h(det ξ) ≤ 0.


In particular the above inequality holds for ξ, which is exactly (6.43).

To prove the final inequality (6.40) it remains to approximate any x ∈ Kn−2+

by xν ∈ Kn−2+ with xν

2 > 0 and apply (6.42) with xν and use the upper semi-continuity of g at 0 to get (6.40).

(iv) Let us suppose for the sake of contradiction that Pf(ξ) = |det ξ| forevery ξ ∈ R2×2. Then, for ξ such that λ1(ξ) = 0, we get

Pf(ξ) = |det ξ| = λ1(ξ)λ2(ξ) = 0.

From the representation formula for the polyconvex envelope (see Theorem 6.8),

we therefore get that there exist ξνi ∈ R2×2, tνi ∈ [0, 1] and

∑6i=1 tνi = 1 such

that

limν→∞

6∑

i=1

tνi f(ξνi ) = 0 with

6∑

i=1

tνi (ξνi , det ξν

i ) = (ξ,det ξ).

In particular, tνi |λ1(ξνi )− 1| → 0 and tνi |det ξν

i | → 0, i = 1, · · · , 6. Up to a

subsequence, tνi → ti ∈ [0, 1] with∑6

i=1 ti = 1. So, there is some j such thattj = 0 and thus for this j we have

⎧⎨⎩

∣∣λ1(ξνj )− 1

∣∣ = 1tνj

tνj∣∣λ1(ξ

νj )− 1

∣∣→ 0∣∣det ξν

j

∣∣ = 1tνj

tνj∣∣det ξν

j

∣∣→ 0.

The first condition implies that λ1(ξνj ) → 1, which contradicts the second one,

since then we would have∣∣det ξν

j

∣∣ ≥ (λ1(ξνj ))2 → 1.

6.6.3 Functions depending on a quasiaffine function

The following theorem, established by Dacorogna [176], [179], should be relatedto Theorem 5.46.

Theorem 6.24 Let g : R → R be such that there exist a1, a2 ∈ R with

g (δ) ≥ a1δ + a2, ∀ δ ∈ R.

Let f : RN×n → R, Φ : RN×n → R be quasiaffine not identically constant and

f (ξ) = g (Φ (ξ)) .

ThenPf = Qf = Rf = Cg Φ

and in generalPf > Cf.

Before proceeding with the proof, we establish a preliminary lemma.

Examples 297

Lemma 6.25 Let Φ : RN×n → R be quasiaffine and not identically constant.Let ξ ∈ RN×n be such that

∇Φ (ξ) =

(∂Φ

∂ξij

(ξ)

)1≤i≤N

1≤j≤n

= 0.

Let β, γ ∈ R and λ ∈ [0, 1] be such that

Φ (ξ) = λβ + (1− λ) γ.

Then there exist B, C ∈ RN×n such that

⎧⎪⎨⎪⎩

ξ = λB + (1− λ)C

Φ (B) = β, Φ (C) = γ

rank B − C ≤ 1.

Proof. Since ∇Φ (ξ) = 0, we can find a ∈ RN , b ∈ Rn such that

〈∇Φ (ξ) ; a⊗ b〉 = γ − β

where 〈·; ·〉 denotes the scalar product in RN×n and a⊗b =(aibj

)1≤i≤N

1≤j≤n. Define

then B := ξ − (1− λ) a⊗ b

C := ξ + λa⊗ b.

In order to obtain the lemma it is therefore sufficient to show that Φ (B) = βand Φ (C) = γ. Since Φ is quasiaffine we have (see Theorem 5.20)

Φ (B) = Φ (ξ − (1− λ) a⊗ b) = Φ (ξ)− (1− λ) 〈∇Φ (ξ) ; a⊗ b〉 = β

Φ (C) = Φ (ξ + λa⊗ b) = Φ (ξ) + λ 〈∇Φ (ξ) ; a⊗ b〉 = γ

which is indeed the claim.


Proof. It is easy to see that

Rf ≥ Qf ≥ Pf ≥ Cg Φ.

It therefore remains to show that for every ξ ∈ RN×n

Rf (ξ) ≤ Cg (Φ (ξ)) . (6.44)

Case 1: ∇Φ (ξ) = 0. Fix ǫ > 0; from Theorem 2.35 we have that there existβ, γ ∈ R, λ ∈ [0, 1] such that

λg (β) + (1− λ) g (γ) ≤ Cg (Φ (ξ)) + ǫ

λβ + (1− λ) γ = Φ (ξ) .


Using Lemma 6.25, we have that there exist B, C ∈ RN×n satisfying the con-clusions of the lemma. Using Theorem 6.10, we obtain that

Rf (ξ) ≤ λf (B) + (1− λ) f (C) = λg (β) + (1− λ) g (γ) ≤ Cg (Φ (ξ)) + ǫ.

Since ǫ is arbitrary we have indeed obtained (6.44).

Case 2: ∇Φ (ξ) = 0. Since Rf and Cg are continuous and Φ is not identicallyconstant we have (see Theorem 5.20 and Corollary 5.23) that, for every ǫ > 0,there exists η ∈ RN×n sufficiently close to ξ such that ∇Φ (η) = 0,

Cg (Φ (η)) ≤ Cg (Φ (ξ)) + ǫ and Rf (ξ) ≤ Rf (η) + ǫ.

Applying Case 1 to η we find that

Rf (ξ) ≤ Rf (η) + ǫ = Cg (Φ (η)) + ǫ ≤ Cg (Φ (ξ)) + 2ǫ.

Since ǫ is arbitrary, we have indeed obtained (6.44).

It therefore remains to show that, in general, Pf > Cf. Choosing, for exam-ple, N = n and

f (ξ) = (det ξ)2

we have immediately

Rf (ξ) = Qf (ξ) = Pf (ξ) = Cg (det ξ) = f (ξ) > Cf (ξ) ≡ 0.

The identity Cf (ξ) ≡ 0 is a consequence of the fact that

0 ≤ Cf (ξ) ≤ infλ (detα)

2+ (1− λ) (det β)

2: λα + (1− λ) β = ξ

and that the infimum on the right hand side is exactly zero.


The next theorem, established in Dacorogna [176], [179], should be comparedwith Theorem 5.47.

Recall first the notation (see Example 5.63) that for ξ ∈ R(n+1)×n we let

adjn ξ =(det ξ 1,− det ξ 2, · · · , (−1)

k+1det ξ k, · · · , (−1)

n+2det ξ n+1

)

where ξ k is the n× n matrix obtained from ξ by suppressing the k th row.

Theorem 6.26 Let g : Rn+1 → R be such that there exist a ∈ Rn+1 and b ∈ Rwith

g (δ) ≥ 〈a; δ〉+ b, ∀ δ ∈ Rn+1.

Examples 299

Let f : R(n+1)×n → R be such that


ThenPf = Qf = Rf = Cg adjn

and in generalPf > Cf.

Before proceeding with the proof, we establish a preliminary lemma that isan extension of Lemma 5.49.

Lemma 6.27 Let N = n + 1, ξ ∈ R(n+1)×n and adjn ξ = 0. Let I ∈ N, λi ≥ 0

with∑I

i=1 λi = 1, βi ∈ Rn+1 such that

adjn ξ =

I∑

i=1

λiβi .

Then there exist ξi ∈ R(n+1)×n such that (see Definition 5.14)⎧⎪⎨⎪⎩

ξ =

I∑

i=1

λiξi , (λi, ξi)1≤i≤I satisfy (HI) ,

adjn ξi = βi , 1 ≤ i ≤ I.

Proof. We proceed by induction on I. The case I = 2 is precisely Lemma5.49. Assume therefore that the lemma has been proved up to the order (I − 1)and we wish to prove that it holds for I. We let

γ =1

1− λ1

I∑

i=2

λiβi .

We may assume, upon a possible relabelling, that γ = 0. Observe also that wehave

adjn ξ = λ1β1 + (1− λ1) γ.

We now apply Lemma 5.49 to β1 and γ to get ξ1, η ∈ R(n+1)×n such that

ξ = λ1ξ1 + (1− λ1) η, rankξ1 − η ≤ 1

adjn ξ1 = β1, adjn η = γ.(6.45)

We may then use the hypothesis of induction to get that there exist ξi ∈R(n+1)×n such that

⎧⎪⎪⎨⎪⎪⎩

η =

I∑

i=2

λi

1− λ1ξi ,

(λi

1− λ1, ξi

)

2≤i≤I

satisfy (HI−1)

adjn ξi = βi , 2 ≤ i ≤ I.

(6.46)


Collecting (6.45) and (6.46) we have indeed obtained the lemma.


Proof. We trivially have that Rf ≥ Qf ≥ Pf ≥ Cg adjn and we thereforeonly need to show that, for every ξ ∈ R(n+1)×n,

Rf (ξ) ≤ Cg (adjn ξ) . (6.47)

Case 1 : adjn ξ = 0. Fix ǫ > 0, from Theorem 2.35 we have that there existβi ∈ Rn+1, λi ≥ 0 with

∑n+2i=1 λi = 1 such that

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

n+2∑

i=1

λig (βi) ≤ Cg (adjn ξ) + ǫ

n+2∑

i=1

λiβi = adjn ξ.

We then use Lemma 6.27 to get ξi satisfying the conclusions of the lemma.Appealing to Theorem 6.10, we have that

Rf (ξ) ≤n+2∑

i=1

λif (ξi) =

n+2∑

i=1

λig (βi) ≤ Cg (adjn ξ) + ǫ.

The inequality (6.47) follows by letting ǫ → 0.

Case 2 : adjn ξ = 0. Using the continuity of Rf and Cg and an argumentsimilar to that in the proof of Theorem 6.24, we immediately get (6.47).

In order to show that, in general, Pf > Cf, we choose

f (ξ) = |adjn ξ|2

and we get

Rf (ξ) = Qf (ξ) = Pf (ξ) = Cg (adjn ξ) = f (ξ) > Cf (ξ) ≡ 0.


6.6.5 The Kohn-Strang example

We now turn our attention to an important example in optimal design. Thefollowing is a result of Kohn and Strang [373], [374] and we only slightly modifytheir proof.

Theorem 6.28 Let N ≥ n = 2 (or n ≥ N = 2) and, for ξ ∈ RN×2,

f (ξ) =

1 + |ξ|2 if ξ = 0

0 if ξ = 0,

Examples 301

where |·| denotes the Euclidean norm. If θ : R+ → R+ is such that

θ (t) :=

1 + t2 if t ≥ 1

2t if 0 ≤ t < 1

then

Cf (ξ) = θ (|ξ|)

and

Pf (ξ) = Qf (ξ) = Rf (ξ) = θ((|ξ|2 + 2 |adj2 ξ|)1/2)− 2 |adj2 ξ| .

Proof. We first observe that the representation formula for Cf follows at oncefrom Theorem 6.30 below. We therefore only prove the formula for Pf, Qf andRf and to do so we divide the proof into two steps.

Step 1. Let h : RN×2 → R be defined by

h (ξ) := θ((|ξ|2 + 2 |adj2 ξ|)1/2)− 2 |adj2 ξ|

and let us first show that it is polyconvex. This will be achieved if we can finda convex function

H : RN×2 × R(N2 ) → R, H = H (ξ, A) ,

such that

h (ξ) = H (ξ, adj2 ξ) . (6.48)

Let us start with some observations.

(1) Note first that

|ξ|2 − 2 |adj2 ξ| ≥ 0.

Indeed, since both

ξ → |ξ|2 and ξ → |adj2 ξ|

are O (N) × O (2)-invariant, it is enough to check the inequality on diagonalmatrices of the form

ξ =

⎛⎜⎜⎜⎝

x 00 y...

...0 0

⎞⎟⎟⎟⎠ ∈ RN×2,

in which case it is a trivial inequality, since then

|ξ|2 = x2 + y2 and |adj2 ξ| = |xy| .


(2) Observe next that, if |α| = 1 and appealing to the above inequality, wehave

|ξ|2 + 2 〈α; adj2 ξ〉 ≥ |ξ|2 − 2 |adj2 ξ| ≥ 0,

where 〈.; .〉 denotes the scalar product in R(N2 ). The function

ξ → (|ξ|2 + 2 〈α; adj2 ξ〉)1/2

is therefore convex (see Corollary 2.53) and thus, since θ is convex and increas-ing, we get that

ξ → θ((|ξ|2 + 2 〈α; adj2 ξ〉)1/2)

is also convex.

(3) We then define, for α ∈ R(N2 ) with |α| = 1, a family of convex (because

of the above considerations) functions

Hα : RN×2 × R(N2 ) → R, Hα = Hα (ξ, A) ,

byHα (ξ, A) := θ((|ξ|2 + 2 〈α; adj2 ξ〉)1/2)− 2 〈α; A〉 .

(4) Finally, we let

H (ξ, A) := sup Hα (ξ, A) : |α| = 1 .

It is clearly a convex function and it therefore only remains to show (6.48). Wehave to consider two cases. But before that, we should observe that the caseadj2 ξ = 0 is straightforward, since then

Hα (ξ, 0) = H (ξ, 0) = θ(|ξ|) = h (ξ) , ∀α ∈ R(N2 ).

So, from now on, we will assume that adj2 ξ = 0 (and hence ξ = 0).

Case 1 : |ξ|2 + 2 |adj2 ξ| ≥ 1. Observe that (recalling that θ (t) ≤ 1 + t2 forevery t ∈ R+)

Hα (ξ, adj2 ξ) ≤ 1 + |ξ|2 = h (ξ) , ∀α ∈ R(N2 ) with |α| = 1.

Moreover, by choosing

α :=adj2 ξ

|adj2 ξ| ,

we have for such α that

Hα (ξ, adj2 ξ) = 1 + |ξ|2 = h (ξ) .

Thus (6.48) holds.

Examples 303

Case 2 : |ξ|2 + 2 |adj2 ξ| < 1. We then have (recall that adj2 ξ = 0)

Hα (ξ, adj2 ξ) = 2(|ξ|2 + 2 〈α; adj2 ξ〉)1/2 − 2 〈α; adj2 ξ〉

and we have to prove that, for every α ∈ R(N2 ) with |α| = 1, the supremum in

α is exactly

h (ξ) = 2(|ξ|2 + 2 |adj2 ξ|)1/2 − 2 |adj2 ξ| .

Denoting the Lagrange multiplier by λ, we find that the stationary points, on|α| = 1, of

α→ Hα (ξ, adj2 ξ)− λ(|α|2 − 1)

satisfy

[(|ξ|2 + 2 〈α; adj2 ξ〉)−1/2 − 1] adj2 ξ = λα.

Multiplying this equation first by α, bearing in mind that |α| = 1, then byadj2 ξ, we find

λ = [(|ξ|2 + 2 〈α; adj2 ξ〉)−1/2 − 1] 〈α; adj2 ξ〉 and α = ± adj2 ξ

|adj2 ξ| .

It is easy to see (see below) that the plus sign corresponds to the maximum andthe minus sign to the minimum. If this is the case, we have indeed establishedthat

H (ξ, adj2 ξ) = 2(|ξ|2 + 2 |adj2 ξ|)1/2 − 2 |adj2 ξ| = h (ξ)

as wished. So it only remains to show that

(|ξ|2 + 2 |adj2 ξ|)1/2 − |adj2 ξ| ≥ (|ξ|2 − 2 |adj2 ξ|)1/2 + |adj2 ξ|

whenever |ξ|2 + 2 |adj2 ξ| < 1. This is equivalent, to showing that

(r + 2s)1/2 − s ≥ (r − 2s)

1/2+ s ⇔ s2 − r + 1 ≥ 0

whenever 0 ≤ 2s ≤ r ≤ r + 2s < 1, and this is straightforward.

Step 2. We now prove that

Pf = Qf = Rf = h.

In view of the general results and those of Step 1, we have

h ≤ Pf ≤ Qf ≤ Rf ≤ f.

We therefore only have to prove that

Rf ≤ h. (6.49)


Appealing to Theorem 6.10, we find

Rf (ξ) = inf∑I



(6.50)where

ΛI =

λ = (λ1, · · · , λI) : λi ≥ 0 and∑I

i=1 λi = 1

.

Since h, f and hence Pf, Qf, Rf (see Theorem 6.20) are O (N) × O (2)-invariant, we can restrict, as in Step 1, our attention to matrices of the form,0 ≤ x ≤ y,

ξ =

⎛⎜⎜⎜⎝

x 00 y...

...0 0

⎞⎟⎟⎟⎠ .

We also let

g (x, y) :=

1 + x2 + y2 if (x, y) = 0

0 if (x, y) = 0

so that for such a ξ we have

f (ξ) = g (x, y) .

Because of the special form of ξ above, it is clear that for proving (6.49) we canrestrict our attention to the case n = N = 2 and then infer in a straightforwardway the general case N ≥ n = 2 (or n ≥ N = 2).

Observe that if we let, for α, β ∈ (0, 1] and α + β ≤ 1,

ξ1 =

(0 00 0

), ξ2 =

((1− β)x/α 0

0 0

), ξ3 =

(x 00 y/β

)

then, writing λ1 = 1− α− β, λ2 = α, λ3 = β, we find

λ1ξ1 + λ2ξ2 + λ3ξ3 = ξ =

(x 00 y

)

and

det [ ξ1 − ξ2 ] = det[λ1ξ1 + λ2ξ2

λ1 + λ2− ξ3 ] = 0.

We therefore obtain

λ1f (ξ1) + λ2f (ξ2) + λ3f (ξ3) = αg((1− β)x/α, 0) + βg(x, y/β).

Returning to (6.50) with the above ξi we find that

Rf (ξ) ≤ G (x, y) := infα,β∈(0,1]α+β≤1

αg ((1− β)x/α, 0) + βg (x, y/β) .

Examples 305

It whence remains to show that

G (x, y) ≤ h (x, y) :=

1 + x2 + y2 if |x|+ |y| ≥ 1

2 (|x|+ |y| − |xy|) if |x|+ |y| < 1.(6.51)

and the proof of (6.49) will be complete.By choosing α + β = 1 and then letting β → 1 in the definition of G, it is

clear that we always have

G (x, y) ≤ 1 + x2 + y2.

Since2 (|x|+ |y| − |xy|) ≤ 1 + x2 + y2,

we only need to show (6.51), when |x|+ |y| < 1.The case (x, y) = (0, 0) being trivial, we also assume that (x, y) = (0, 0) .

Since the function

α ∈ (0, 1]→ αg ((1 − β)x/α, 0) + βg (x, y/β)

is convex (this is obvious if β = 1 or x = 0 and is also clear otherwise since thenthe function equals α(1 + (1 − β)2x2/α2) + βg (x, y/β)), we see that it attainsits minimum (noticing that |x| ≤ |x|+ |y| < 1) at

α = |x| (1− β) .

We thus deduce that

G (x, y) = infβ∈(0,1]

2 |x| (1− β) + β[ 1 + x2 +y2

β2]

= 2 |x|+ infβ∈(0,1]

β (1− |x|)2 +y2

β

= 2 (|x|+ |y| − |xy|)

as wished.

6.6.6 The Saint Venant-Kirchhoff energy function

We now discuss an important function for nonlinear elasticity, namely the SaintVenant-Kirchhoff energy function. Upon rescaling, the function under consid-eration is, ν ∈ (0, 1/2) being a parameter,

f(ξ) =∣∣ξξt − I

∣∣2 +ν

1− 2ν(|ξ|2 − n)2,

or, in terms of the singular values 0 ≤ λ1(ξ) ≤ · · · ≤ λn(ξ) of ξ ∈ Rn×n,

f(ξ) =∑n

i=1

(λ2

i (ξ)− 1)2

+ν

1− 2ν

(∑ni=1 λ2

i (ξ)− n)2

.


It is clearly not a quasiconvex function and we therefore compute its quasiconvexenvelope. This was achieved by Le Dret-Raoult [399], [400] when n = 3 and wenow prove their result when n = 2.

Theorem 6.29 Let f : R2×2 → R be defined by


∣∣2 +ν

1− 2ν(|ξ|2 − 2)2

=(λ2

1(ξ)− 1)2

+(λ2

2(ξ)− 1)2

+ν

1− 2ν

(λ2

1(ξ) + λ22(ξ)− 2

)2.

Let

g (ξ) :=

⎧⎪⎪⎨⎪⎪⎩

f (ξ) if ξ /∈ D1 ∪D2

11−ν

(λ2

2(ξ)− 1)2

if ξ ∈ D2

0 if ξ ∈ D1

where

D1 =ξ ∈ R2×2 : (1− ν) [λ1(ξ)]

2+ ν [λ2(ξ)]

2< 1 and λ2(ξ) < 1

=ξ ∈ R2×2 : λ1(ξ) ≤ λ2(ξ) < 1

D2 =ξ ∈ R2×2 : (1− ν) [λ1(ξ)]

2+ ν [λ2(ξ)]

2< 1 and λ2(ξ) ≥ 1

.

Then

Cf (ξ) = Pf (ξ) = Qf (ξ) = Rf (ξ) = g (ξ) .

Proof. In view of the general results, we have

0 ≤ Cf ≤ Pf ≤ Qf ≤ Rf ≤ f.

We therefore only need to show that

Rf (ξ) = Cf (ξ) = g (ξ) , ∀ ξ ∈ R2×2. (6.52)

Since f and hence Cf, Pf, Qf, Rf (see Theorem 6.20) are SO (2) × SO (2)-invariant, we can restrict our attention to matrices of the form

ξ =

(x 00 y

),

where |x| ≤ y.

Recall that, from Theorem 6.10, we have

Rf (ξ) = inf∑I



.

(6.53)

Examples 307

Before proceeding further, it is convenient to introduce two new functionsdefined on R2 by

ψ (x, y) := 11−ν

(y2 − 1

)2

+ 1(1−ν)(1−2ν)

[(1− ν)

(x2 − 1

)+ ν(y2 − 1

)]2

ϕ (x, y) := 11−ν

[y2 − 1

]2+

+ 1(1−ν)(1−2ν)

[(1− ν)

(x2 − 1

)+ ν(y2 − 1

)]2+

where for z ∈ R we let

[z]2+ =

z2 if z ≥ 0

0 otherwise.

For every |x| ≤ y, a simple calculation leads to

f

(x 00 y

)= ψ (x, y) and g

(x 00 y

)= ϕ (x, y) .

Moreover, the function ϕ is clearly convex and thus, appealing to Theorem 5.33(A), we find that g is convex.

We may now proceed with the proof of the identity (6.52) and we divide itinto the study of three cases.

Case 1: ξ ∈ D1 . We first prove that, for |x| ≤ 1, we have

Rf

(x 00 ±1

)= 0. (6.54)

Indeed let λ = (1 + x) /2 and

ξ1 =

(1 00 ±1

), ξ2 =

(−1 00 ±1

)

which implies that

ξ = λξ1 + (1− λ) ξ2 and det (ξ1 − ξ2) = 0.

We therefore have from (6.53) that

0 ≤ Cf

(x 00 ±1

)≤ Rf

(x 00 ±1

)

≤ λf

(1 00 ±1

)+ (1− λ) f

(−1 00 ±1

)= 0

as wished.We now consider the general case ξ ∈ D1 and, recalling that |x| ≤ y ≤ 1, we

let λ = (1 + y) /2 and

ξ1 =

(x 00 1

), ξ2 =

(x 00 −1

),


which implies that

ξ = λξ1 + (1− λ) ξ2 and det (ξ1 − ξ2) = 0.

We therefore have from (6.53) and (6.54) that

ϕ (x, y) = g (ξ) = 0 ≤ Cf

(x 00 y

)≤ Rf

(x 00 y

)

≤ λRf

(x 00 1

)+ (1− λ)Rf

(x 00 −1

)= 0

and hence (6.52) holds and the discussion of Case 1 is complete.

Case 2: ξ ∈ D2 . We can then infer that for ξ ∈ D2 we have

ϕ (x, y) = g (ξ) ≤ Cf (ξ) ≤ Rf (ξ) ≤ f (ξ) . (6.55)

Since for ξ ∈ D2 we have

(1− ν) x2 + νy2 < 1 ≤ y

we can find |x| < x1 < 1 so that

(1− ν)x21 + νy2 = 1.

We hence get that

f

(±x1 00 y

)= ψ (x1, y) =

1

1− ν

(y2 − 1

)2

+1

(1− ν) (1− 2ν)

[(1− ν)

(x2

1 − 1)

+ ν(y2 − 1

)]2

=1

1− ν

[y2 − 1

]2= ϕ (x, y) .

Next define λ = (x + x1) /2x1 and

ξ1 =

(x1 00 y

), ξ2 =

(−x1 00 y

)

which implies that

ξ = λξ1 + (1− λ) ξ2 and det (ξ1 − ξ2) = 0.

We therefore have from (6.53) and (6.55) that

ϕ (x, y) = g (ξ) ≤ Cf (ξ) ≤ Rf (ξ)

≤ λf

(x1 00 y

)+ (1− λ) f

(−x1 00 y

)= ϕ (x, y) .

Therefore (6.52) holds also in Case 2.

Examples 309

Case 3: ξ /∈ D1 ∪D2 . This case is trivial, since we then have

f (ξ) = ϕ (x, y) = g (ξ) ≤ Cf (ξ) ≤ Rf (ξ) ≤ f (ξ)

and thus (6.52) is satisfied.

6.6.7 The case of a norm

Theorem 6.30 Let f : RN×n → R and g : R+ → R with

g (0) = inf g (x) : x ≥ 0

be such thatf (ξ) = g (|ξ|) ,

where |·| denotes any norm on RN×n. Then, for every ξ ∈ RN×n,

Cf (ξ) = Cg (|ξ|) .

If |·|2 denotes the Euclidean norm, namely

|ξ|2 := (∑N

i=1

∑nj=1

(ξij

)2)1/2,

then, in general,Pf (ξ) > Cf (ξ) .

If, however, there exists a ≥ 0 such that

g (a) = g (0) and Cg (x) = g (x) for every x ≥ a

thenRf (ξ) = Qf (ξ) = Pf (ξ) = Cf (ξ) = Cg (|ξ|2) . (6.56)

Remark 6.31 The result of this theorem is surprising when compared withTheorem 5.58, which would have suggested that

Rf = Qf = Pf = Cf.

We have already seen in Theorem 6.28 that, in general, Pf > Cf. A simplerexample is given below. What is even more striking is that there are examplesof functions as in the theorem where

Qf > Pf,

as shown by Gangbo [300]. ♦

We now give two examples that illustrate the theorem.


Example 6.32 (i) The first one shows that

Rf = Qf = Pf = Cf = Cg.

Letf (ξ) = g (|ξ|2) = ( |ξ|22 − 1 )2.

It follows from the theorem that

Rf (ξ) = Qf (ξ) = Pf (ξ) = Cf (ξ) = Cg (|ξ|2)

where

Cg (x) =

(x2 − 1

)2if |x| ≥ 1

0 if |x| < 1.

(ii) The second one is an example where strict inequality holds between Pfand Cf. Let N = n = 2 and g : R+ → R be continuous and such that

⎧⎪⎨⎪⎩

g (0) = inf g (x) : x ≥ 0g (x) ≥ a |x|α , a > 0 and α > 2

Cg strictly increasing and Cg ≡ g

and, for ξ ∈ R2×2, we letf (ξ) = g (|ξ|2) .

One can choose for example

g (x) =

⎧⎪⎪⎨⎪⎪⎩

x if x ∈ [0, 2]

−x + 4 if x ∈ [2, 3]

x3 − 26 if x ≥ 3.

Then

Cg (x) =

x/3 if x ∈ [0, 3]

x3 − 26 if x ≥ 3.

It can easily be seen that, for such g, we have Pf > Cf = Cg and we refer toDacorogna [176], [179] for a complete discussion of this example. ♦

We may now proceed with the proof of the theorem.

Proof. We decompose the proof into three steps.

Step 1. We first show that Cf = Cg. Observe first that one always hasCg ≤ Cf. We wish to show the reverse inequality. Let ǫ > 0 be fixed. Then,from Theorem 2.35, we get that there exist λ ∈ [0, 1] , b, c ∈ R+ , such that

ǫ + Cg (|ξ|) ≥ λg (b) + (1− λ) g (c)

|ξ| = λb + (1− λ) c.

Examples 311

Then choose (the case ξ = 0 is trivial)

β =bξ

|ξ| , γ =cξ

|ξ| .

We therefore get

ǫ + Cg (|ξ|) ≥ λf (β) + (1− λ) f (γ) ≥ Cf (ξ)

ξ = λβ + (1− λ) γ.

Since ǫ is arbitrary, we have indeed obtained the claimed result.

Step 2. As already noted the inequality Pf > Cf has been seen either inExample 6.32 (ii) or in Theorem 6.28.

Step 3. It now remains to show (6.56), i.e. that if there exists a ≥ 0 suchthat

Cg (x) =

g (x) if x ≥ a

g (0) = g (a) if x < a,(6.57)

then Rf = Qf = Pf = Cf = Cg (note that the functions g considered inExample 6.32 (ii) or in Theorem 6.28 do not satisfy (6.57)). It is obvious thatCg (|ξ|2) ≤ Rf (ξ) , therefore we only need to show that for every ξ ∈ RN×n

Rf (ξ) ≤ Cg (|ξ|2) . (6.58)

Note also that if |ξ|2 ≥ a, then (6.58) is trivially satisfied. Therefore we onlyneed to consider the case where 0 < |ξ|2 < a. From (6.57), we then obtain

Cg (|ξ|2) = g (a) .

Let ξ =(ξij

)1≤i≤N

1≤j≤n(we may assume for notational convenience that ξ1

1 = 0, the

general case being handled similarly) and

2λ := 1 +

∣∣ξ11

∣∣(a2 − |ξ|22 + (ξ1

1)2)1/2.

Then 12 < λ < 1. Let

η =(ηi

j

)1≤i≤N

1≤j≤nwith η1

1 =2ξ1

1

1− 2λ, ηi

j = 0 otherwise.

Finally let

β := ξ − (1− λ) η and γ := ξ + λη.

It is then easy to see that

ξ = λβ + (1− λ) γ

|β|2 = |γ|2 = a, rankβ − γ ≤ 1.


From Theorem 6.10 we find that

Rf (ξ) ≤ λf (β) + (1− λ) f (γ) = λg (a) + (1− λ) g (a) = Cg (|ξ|2) ,

which is precisely (6.58).

Chapter 7

Polyconvex, quasiconvexand rank one convex sets

7.1 Introduction

We now discuss the notions of polyconvex, quasiconvex and rank one convexsets. Contrary to the usual presentation of classical convex analysis, where thenotion of a convex set is defined prior to that of a convex function; this is not thecase for the generalized notions of convexity. This is of course due to historicalreasons. The notions of polyconvex, quasiconvex and rank one convex functionswere introduced, as already said, by Morrey in 1952, although the terminologyis that of Ball [53]. It was not until the systematic studies of partial differen-tial equations and inclusions by Dacorogna-Marcellini and Muller-Sverak, ini-tiated in 1996 and discussed in Chapter 10, that the equivalent definitions forsets became an important issue. Moreover these notions were essentially seenthrough the different generalized convex hulls, leading somehow to terminologiesthat do not exactly cover the same concepts.

We here try to imitate as much as possible the classical approach of convexanalysis in the present context. Throughout the two first sections, we follow thepresentation of Dacorogna-Ribeiro [213], following earlier results of Dacorogna-Marcellini [202].

In Section 7.2, we define the notions of polyconvex, quasiconvex and rankone convex sets. The first and third ones are straightforward and equivalent,as they should be, to the polyconvexity and rank one convexity of the indicatorfunction. The second one is more delicate. Indeed one would have liked to defineit as equivalent to the quasiconvexity of the indicator function, but quasiconvexfunctions allowed to take the value +∞ are, as we have already seen, poorlyunderstood. We give a definition of quasiconvex set that is compatible withmany of the desired properties that should have such a definition. Notably we

314 Polyconvex, quasiconvex and rank one convex sets

have that, for a set E ⊂ RN×n,

E convex⇒ E polyconvex⇒ E quasiconvex⇒ E rank one convex

and all counter implications turn out to be false whenever N, n ≥ 2. This lastresult is better than the corresponding one for functions, since we have examplesof rank one convex functions that are not quasiconvex (see Section 5.3.7) onlywhen n ≥ 2 and N ≥ 3.

We then prove separation and Caratheodory type theorems for polyconvexsets.

In Section 7.3, we give the definitions of polyconvex, quasiconvex and rankone convex hulls of a given set E denoted respectively PcoE, QcoE and RcoE.They are, as they should be, the smallest polyconvex, quasiconvex and rankone convex sets, respectively, that contain E. As we already alluded to, thesedefinitions are not exactly the same for all authors working in the field; however,ours is clearly the closest to classical convex analysis.

We also show that if we let

FE∞ :=

f : RN×n → R ∪ +∞ : f |E ≤ 0

,

FE :=f : RN×n → R : f |E ≤ 0

,

then

PcoE =ξ ∈ RN×n : f (ξ) ≤ 0 for every polyconvex f ∈ FE

∞

,

RcoE =ξ ∈ RN×n : f (ξ) ≤ 0 for every rank one convex f ∈ FE

∞

,

as for the convex case where

co E =ξ ∈ RN×n : f (ξ) ≤ 0 for every convex f ∈ FE

∞

.

In the convex case, we also have the representation formula for the closure ofthe convex hull as


.

However, the representation of the closure of the hulls analogous to the above isnot true for general sets. We discuss this question in details introducing threemore types of hulls, namely

Pcof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every polyconvex f ∈ FE

Qcof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every quasiconvex f ∈ FE

Rcof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every rank one convex f ∈ FE

.

Polyconvex, quasiconvex and rank one convex sets 315

It turns out that, in general,

PcoE ⊂=

Pcof E, QcoE ⊂=

Qcof E and RcoE ⊂=

Rcof E.

However, if E is compact, then, as for the convex case,

PcoE = Pcof E.

We end the section by introducing the notion of extreme points in these gener-alized senses and establish Minkowski type theorems. We moreover define, asfor the convex case, the gauge of a polyconvex set and the Choquet function ofa polyconvex or a rank one convex set.

In Section 7.4, we consider several sets that are defined in terms of singularvalues, for example sets of the form

E =ξ ∈ Rn×n : λi (ξ) = γi, i = 1, · · · , n

,

where 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) are the singular values of the matrix ξ and0 < γ1 ≤ · · · ≤ γn are given real numbers.

We also study sets such as

E = SO(2)A ∪ SO(2)B,

where A, B ∈ R2×2 are given matrices.In all these cases, we characterize their convex, polyconvex, quasiconvex and

rank one convex hulls.

7.2 Polyconvex, quasiconvex and rank one

convex sets

7.2.1 Definitions and main properties

We first introduce some notation that is used throughout this chapter.

Notation 7.1 - For s ∈ N, let

Λs := λ = (λ1, · · · , λs) : λi ≥ 0,∑s

i=1 λi = 1 .

- Recall that T : RN×n → Rτ(n,N) is such that

T (ξ) := (ξ, adj2 ξ, · · · , adjn∧N ξ) ,

where adjs ξ stands for the matrix of all s× s minors of the matrix ξ ∈ RN×n,2 ≤ s ≤ n ∧N = min n, N , and

τ (n, N) :=n∧N∑

s=1

σ (s) , where σ (s) :=(Ns

)(ns

).


- D stands for the unit cube (0, 1)n of Rn.

- e1, · · · , en is the standard orthonormal basis of Rn.

- W 1,∞per (D; RN ) denotes, as usual, the space of periodic functions in

W 1,∞(D; RN ), meaning that

u(x) = u(x + ei) for every x ∈ D and i = 1, · · · , n.

- Wper denotes the subspace of functions in W 1,∞per (D; RN ) and whose gradi-

ents take only a finite number of values. ♦

We start by giving the generalized definitions of convexity for sets.

Definition 7.2 (i) We say that E ⊂ Rm is convex if for every λ ∈ [0, 1] andξ, η ∈ E, then

λξ + (1− λ)η ∈ E.

(ii) We say that E ⊂ RN×n is polyconvex if there exists a convex set K ⊂Rτ(n,N) such that

π(K ∩ T (RN×n)) = E,

where π (see below) denotes the orthogonal projection of (the first N × n com-ponents of) Rτ(n,N) in RN×n. Equivalently, E is polyconvex if there exists aconvex set K ⊂ Rτ(n,N) such that

ξ ∈ RN×n : T (ξ) ∈ K

= E.

(iii) We say that E ⊂ RN×n is quasiconvex if we have

ξ +∇ϕ(x)R ∈ E, a.e. x ∈ D,

for some R ∈ O(n) and some ϕ ∈ Wper

⇒ ξ ∈ E.

(iv) We say that E ⊂ RN×n is rank one convex if for every λ ∈ [0, 1] andξ, η ∈ E such that rank ξ − η = 1, then

λξ + (1− λ)η ∈ E.

(v) We say that E ⊂ Rm is separately convex if for every λ ∈ [0, 1] andξ, η ∈ E such that ξ − η = sei , for some s ∈ R and i ∈ 1, · · · , m, then

λξ + (1− λ)η ∈ E.

Remark 7.3 (i) The operator π introduced in the above definition is moreprecisely defined as follows. If

X = (X1, · · · , Xτ(n,N)) ⇒ π(X) = (X1, · · · , XN×n),


this implies thatπ(T (ξ)) = ξ, ∀ ξ ∈ RN×n.

In particular, if N = n = 2 and X = (ξ, δ) ∈ R2×2 × R, then π(X) = ξ.

(ii) The definitions of convex, rank one convex and separately convex setsare standard.

(iii) The usual way to define polyconvexity, is with the condition (ii) inTheorem 7.4 below. However, the two conditions turn out to be equivalent.With our definition, given in Dacorogna-Ribeiro [213], we get some coherencewith the analogous notion for functions.

We note that one could think that if a set E is polyconvex, then T (E) isconvex. This, however, is not true. Consider, for example, the polyconvex setE = I, ξ, where I is the identity matrix and ξ = diag(2, 0). Then

T (E) = (I, 1), (ξ, 0),

which is not convex.

(iv) The best definition for quasiconvex sets is unclear. Several definitionshave already been considered (see Dacorogna-Marcellini [202], Muller [462],Zhang [616]). The one we propose here, following Dacorogna-Ribeiro [213], isconsistent with known properties for functions and have most properties whichare desirable (see Theorem 7.7 below).

(v) One can replace Wper by W 1,∞per (D; RN ) in the definition of quasiconvex

sets and keep the main result (Theorem 7.7) still valid. However the definitiongiven above is more convenient for Theorem 7.16. ♦

We first give an equivalent condition for polyconvexity.

Theorem 7.4 Let E ⊂ RN×n. The following conditions are equivalent.(i) E is polyconvex.

(ii) ∑Ii=1 λiT (ξi) = T (

∑Ii=1 λiξi)

ξi ∈ E, (λ1, · · · , λI) ∈ ΛI

⇒∑I

i=1 λiξi ∈ E.

Moreover one can take I = τ (n, N) + 1.

(iii) Denoting by coT (E) the convex hull of T (E), then

E = π(co T (E) ∩ T (RN×n))

or equivalentlyE = ξ ∈ RN×n : T (ξ) ∈ coT (E).

Proof. (i) ⇒ (ii). Suppose

∑Ii=1 λiT (ξi) = T (

∑Ii=1 λiξi), (7.1)


for some ξi ∈ E and (λ1, · · · , λI) ∈ ΛI . By hypothesis, ξi ∈ π(K ∩ T (RN×n))for some convex set K ⊂ Rτ(n,N) and so T (ξi) ∈ K. Therefore

I∑

i=1

λiT (ξi) ∈ coK = K

and, by (7.1), we conclude that∑I

i=1 λiξi ∈ E.The fact that we can take I = τ (n, N) + 1 in (ii) is a consequence of

Caratheodory theorem (see Theorem 5.6).

(ii) ⇒ (iii). We have to see that E = π(co T (E) ∩ T (RN×n)). Evidently Eis contained in the set on the right hand side. For the reverse inclusion, considerξ ∈ π(co T (E) ∩ T (RN×n)). So, T (ξ) ∈ co T (E) and we can write

T (ξ) =

I∑

i=1

λiT (ξi)

for some ξi ∈ E and (λ1, · · · , λI) ∈ ΛI . We then use (ii) to get that ξ ∈ E, aswished.

(iii) ⇒ (i). This is immediate.

The next result, whose proof is straightforward, shows the relation betweenthe notions of convexity for sets and the corresponding notions for functions.

Proposition 7.5 Let E ⊂ RN×n and χE denote the indicator function of E :

χE (ξ) =

0 if ξ ∈ E

+∞ if ξ /∈ E.

Then E is, respectively, convex, polyconvex, rank one convex or separately con-vex if and only if χE is, respectively, convex, polyconvex, rank one convex orseparately convex.

Remark 7.6 One would have liked to have the same result for quasiconvex setsbut, as already discussed, quasiconvex functions taking the value +∞ are notconsidered here. ♦

The convexity conditions are related in the following way (see [213]).

Theorem 7.7 Let E ⊂ RN×n. The following implications then hold

E convex ⇒ E polyconvex ⇒ E quasiconvex

⇒ E rank one convex ⇒ E separately convex.

All counter implications are false as soon as N, n ≥ 2.

Remark 7.8 We will see (see Proposition 7.24) that, as for the convex case:E, respectively, polyconvex, quasiconvex, rank one convex or separately conveximplies that intE is also, respectively, polyconvex, quasiconvex, rank one convexor separately convex. However, this is not true anymore for E. Indeed we willgive (see Proposition 7.24) an example of a bounded polyconvex set E ⊂ R2×2

with E not even separately convex. ♦


Proof. Part 1. We only prove the implications related to the notion ofquasiconvexity since the others are trivial.

(i) We prove that if E is polyconvex then E is quasiconvex. Assume that

ξ +∇ϕ(x)R ∈ E, a.e. x ∈ D

for some R ∈ O(n) and ϕ ∈ Wper . We can write

∇ϕ(x)R ∈ η1, · · · , ηk, a.e. x ∈ D

for some ηi such that ξ + ηi ∈ E, i = 1, · · · , k. Defining

λi := measx ∈ D : ∇ϕ(x)R = ηi,

we have λi ≥ 0,∑k

i=1 λi = 1. Since ϕ is periodic and the functions adjs arequasiaffine (s = 1, · · · , N ∧ n) we have

T (ξ) =

∫

D

T (ξ +∇ϕ(x)R) dx =

k∑

i=1

λiT (ξ + ηi).

Using the polyconvexity of the set E and Theorem 7.4, we obtain that ξ ∈ E.

(ii) We now prove that if a set E is quasiconvex then it is rank one convex.Let ξ, η ∈ E be such that rank ξ − η = 1 and λ ∈ (0, 1). We have to show thatλξ + (1− λ)η ∈ E. To achieve this, it is enough to find R ∈ O(n) and ϕ ∈ Wper

such thatλξ + (1 − λ)η +∇ϕ(x)R ∈ ξ, η, a.e. x ∈ D

or equivalently

∇ϕ(x)R ∈ (1− λ)(ξ − η),−λ(ξ − η), a.e. x ∈ D.

The result then follows from the quasiconvexity of E. The construction of suchϕ is standard for relaxation theorems (see, for a more sophisticated version,Lemma 3.11) and we now give the proof. Since rank ξ − η = 1, we can write

ξ − η = a⊗ ν

with a ∈ RN and ν a unit vector in Rn. Choose R ∈ O(n) an orthogonaltransformation such that e1R = ν. Let h : R → R be periodic, of period 1, andsuch that

h(s) :=

(1− λ)s if 0 ≤ s ≤ λ

−λ (s− 1) if λ < s ≤ 1

and define ϕ ∈ Wper as

ϕ(x) := h(x1)a ⇒ ∇ϕ (x) = h′(x1)a⊗ e1 ⇒ ∇ϕ (x) R = h′(x1)a⊗ ν.


It clearly satisfies the claim and this finishes the proof.

Part 2. We next see that the reverse implications are, in general, not true.

(i) Polyconvexity convexity. Consider, for example, the set E = ξ, η ⊂R2×2, where ξ = diag(1, 0) and η = diag(0, 1). The set E is polyconvex but notconvex.

(ii) Quasiconvexity polyconvexity. Consider the matrices (cf. Proposition5.10)

ξ1 =

(1 02 0

), ξ2 =

(0 10 1

), ξ3 =

(−1 −10 0

)

and

η =

(0 0

2/3 1/3

)=

1

3ξ1 +

1

3ξ2 +

1

3ξ3 .

We have

T (η) =1

3T (ξ1) +

1

3T (ξ2) +

1

3T (ξ3).

The set E = ξ1, ξ2, ξ3 is not a polyconvex set since η /∈ E. However, it isquasiconvex. Suppose ξ +∇ϕR ∈ E for some ϕ ∈ Wper where R ∈ O(2). Sincerank ξi − ξj = 2 for i = j, we have from Theorem 7.11 (with m = 3) thatthere exists ξi ∈ E such that

ξ +∇ϕ (x) R = ξi , a.e. x ∈ D.

Using then the periodicity of ϕ, we find

ξ =

∫

D

(ξ +∇ϕ(x)R) dx = ξi

and thus ξ = ξi ∈ E. We then conclude that E is quasiconvex.

(iii) Rank one convexity quasiconvexity. We should again draw the atten-tion to the fact that our result is better for sets than for functions. We provethis assertion in two steps.

Step 1. From Theorem 7.12, we can find ξ1, · · · , ξm ⊂ R2×2 with

rank ξi − ξj = 2, ∀ i = j,

ξ0 /∈ ξ1, · · · , ξm and u ∈ uξ0 + W 1,∞0 (D; R2) such that

∇u (x) ∈ ξ1, · · · , ξm , a.e. x ∈ D

where uξ0 (x) = ξ0x.

Step 2. Let E = ξ1, · · · , ξm . Since there are no rank one connectionsbetween the matrices ξi , the set E is rank one convex. We now see that E isnot quasiconvex. Let u be as in Step 1 and write

ϕ := u− uξ0 .


We therefore have ∇u(x) = ξ0 +∇ϕ(x) ∈ E, a.e. in D, with

ϕ ∈W 1,∞0 (D; R2) ∩Wper

but, by construction, ξ0 /∈ E, and thus E is not quasiconvex.

(iv) Separate convexity rank one convexity. Indeed, the set E = ξ, η ⊂R2×2, where

ξ =

(2 20 0

), η =

(1 11 1

)

is separately convex but not rank one convex.

7.2.2 Separation theorems for polyconvex sets

We next deal with the problem of separating polyconvex sets generalizing inthis way known results in the convex context (see Theorem 2.10 and Corollary2.11).

Theorem 7.9 Let E be a polyconvex set of RN×n.

(i) If η /∈ E or η ∈ ∂E, then there exists β ∈ Rτ(n,N), β = 0, such that

〈β; T (η)− T (ξ)〉 ≤ 0, ∀ ξ ∈ E.

(ii) If E is compact and η /∈ E, then there exists β ∈ Rτ(n,N), β = 0, such that

〈β; T (η)〉 < infξ∈E〈β; T (ξ)〉.

Proof. (i) Since E is polyconvex, if η /∈ E then (see Theorem 7.4 (iii))T (η) /∈ coT (E); in the case η ∈ ∂E then we get T (η) ∈ ∂ coT (E). In both cases,using the separation theorem for convex sets (see Theorem 2.10 and Corollary2.11) we obtain the existence of β = 0 satisfying

〈β; T (η)−X)〉 ≤ 0, ∀X ∈ coT (E),

and, in particular, for X ∈ T (E) as desired.

(ii) Since E is compact, then T (E) is compact and so is coT (E) (see The-orem 2.14 (i)). We may then apply Theorem 2.10 (iii) to get the result.

As a consequence of the previous separation theorem we have the character-ization of a polyconvex set given in the following result. This is an extension ofTheorem 2.10 (iv) for convex sets that ensures that a closed convex set is theintersection of the closed half spaces containing the set.

Theorem 7.10 A compact set E ⊂ RN×n is polyconvex if and only if

E =ξ ∈ RN×n : f (ξ) ≤ 0 for every quasiaffine f ∈ FE

,


where

FE :=f : RN×n → R : f |E ≤ 0

.

Proof. Call

X :=ξ ∈ RN×n : f (ξ) ≤ 0, for every quasiaffine f ∈ FE

.

(⇒) The fact that E ⊂ X is obvious, so let us show the reverse inclusion. Letξ0 ∈ X and let us prove that ξ0 ∈ E. If this was not the case, then, fromTheorem 7.9 (ii), there exist β ∈ Rτ(n,N) − 0 and ǫ > 0 such that

〈β; T (ξ0)〉 < −ǫ + infξ∈E〈β; T (ξ)〉.

Define then

f(ξ) := −〈β; T (ξ)〉+ ǫ + 〈β; T (ξ0)〉.

Observe that f is quasiaffine and f ∈ FE , although f(ξ0) = ǫ > 0. Thereforeξ0 /∈ X, which is a contradiction.

(⇐) Since X is clearly polyconvex and E = X, we have the claim.

7.2.3 Appendix: functions with finitely many gradients

We here gather some results, without proofs, that are used throughout thechapter.

Theorem 7.11 Let n, N ≥ 2, D = (0, 1)n ⊂ Rn and

E = ξ1, · · · , ξm ⊂ RN×n with rank ξi − ξj ≥ 2, ∀ i = j.

Let u ∈W 1,∞(D; RN ) be such that

∇u (x) ∈ E, a.e. x ∈ D.

If m ≤ 4, then there exists ξi ∈ E such that

∇u (x) = ξi , a.e. x ∈ D.

The theorem was established by Ball-James [60] when m = 2, by Sverak[550], [552] and Zhang [619] when m = 3 and by Chlebik-Kirchheim [150] whenm = 4.

The result is false as soon as m ≥ 5 (and n = N = 2), as was shown byKirchheim-Preiss [365]. In the same spirit, we now quote a theorem of Kirchheim[364].

The different types of convex hulls 323

Theorem 7.12 Let n, N ≥ 2 and D = (0, 1)n ⊂ Rn. Then there exist

ξ1, · · · , ξm ⊂ RN×n with rank ξi − ξj = n ∧N, ∀ i = j,

ξ0 /∈ ξ1, · · · , ξm and u ∈ uξ0 + W 1,∞0 (D; RN ) such that

∇u (x) ∈ ξ1, · · · , ξm , a.e. x ∈ D

where uξ0 (x) = ξ0x.

7.3 The different types of convex hulls

7.3.1 The different convex hulls

Having defined the generalized notions of convexity, we are now in a position tointroduce the concept of generalized convex hulls. We follow the same procedureas in the classical convex case.

Definition 7.13 The polyconvex, quasiconvex, rank one convex and sepa-rately convex hulls of a set E ⊂ RN×n are, respectively, the smallest polyconvex,quasiconvex, rank one convex and separately convex sets containing E and arerespectively denoted by PcoE, QcoE, RcoE and ScoE.

From the discussion made in Section 7.2.1, the following inclusions hold:

E ⊂ Sco E ⊂ RcoE ⊂ QcoE ⊂ PcoE ⊂ coE.

As we note below (see Remark 7.26), there are some authors who haveadopted other definitions for the rank one convex hull, but this one is moreconsistent with the convex case.

In the following (see Dacorogna-Marcellini [202] and Dacorogna-Ribeiro[213]) we give some representations of the hulls defined above. We start bygiving two characterizations of the polyconvex hull of a set. The second one isa consequence of Caratheodory theorem and is equivalent to that obtained inthe convex case.

Theorem 7.14 Let E ⊂ RN×n. Then the following two representation formulashold (recall that τ = τ (n, N) ; see Notation 7.1):

PcoE = π(co T (E) ∩ T (RN×n)),

PcoE =

ξ ∈ RN×n : T (ξ) =∑τ+1

i=1 λiT (ξi), ξi ∈ E, λ ∈ Λτ+1

.

In particular, if E is compact, then PcoE is also compact and if E is open, thenPcoE is also open.


Proof. (i) We prove the first representation of PcoE. It is clear, from Theorem7.4 (iii), that PcoE ⊂ π(co T (E)∩ T (RN×n)). For the other inclusion, we startby noting that, since PcoE is polyconvex, by definition,

PcoE = π(K ∩ T (RN×n))

for some convex set K ⊂ Rτ(n,N). Since E ⊂ PcoE, K must contain T (E) and,consequently, must contain coT (E), thus the desired inclusion follows.

(ii) Let

Y :=ξ ∈ RN×n : T (ξ) =

∑τ+1i=1 λiT (ξi), ξi ∈ E, λ ∈ Λτ+1

.

Let ξ ∈ Y, then there exist ξi ∈ E ⊂ PcoE and λ ∈ Λτ+1 such that

T (ξ) =τ+1∑

i=1

λiT (ξi).

We therefore deduce, from Theorem 7.4 (ii), that ξ ∈ PcoE.

The reverse inclusion follows from the fact that E ⊂ Y and that Y is easilyseen to be polyconvex (as in Theorems 5.6 and 6.8) and thus PcoE ⊂ Y.

(iii) Let E be compact, then PcoE is trivially bounded, so let us show thatit is also closed. Then let ξν ∈ PcoE with ξν → ξ. By the first representationof PcoE, T (ξν) ∈ coT (E), which is a compact set since T (E) is compact andby Theorem 2.14 (i). Then T (ξ) = limT (ξν) ∈ coT (E) and thus ξ ∈ PcoE aswished.

(iv) Assume (see below) that we have shown that for every ξ, ξi ∈ RN×n

and λ ∈ Λτ+1 , such that

T (ξ) =τ+1∑

i=1

λiT (ξi),

then

T (ξ + η) =

τ+1∑

i=1

λiT (ξi + η), ∀ η ∈ RN×n. (7.2)

From this and (ii), it easily follows that PcoE is open if E is open.

So it remains to show (7.2). Since T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) , (7.2) isproved if we can show that, for every 1 ≤ s ≤ n ∧N,

adjr ξ =τ+1∑

i=1

λi adjr ξi , ∀ 1 ≤ r ≤ s ⇒ adjs (ξ + η) =τ+1∑

i=1

λi adjs (ξi + η) .

(7.3)We prove the claim by induction on s. The result is trivially true when s = 1,so assume that it has been proved up to s − 1 and let us show it for s. Sinceadjs consists of s× s determinants of the matrix ξ ∈ RN×n, we find that (7.3)


is proved if we can show that, for α, αi, β ∈ Rs×s and λ ∈ Λτ+1 , then

adjr α =

τ+1∑

i=1

λi adjr αi , ∀ 1 ≤ r ≤ s ⇒ det (α + β) =

τ+1∑

i=1

λi det (αi + β) .

(7.4)We use Proposition 5.67 and the hypothesis in (7.4) to get the claim, namely

τ+1∑

i=1

λi det (αi + β) =

τ+1∑

i=1

λi

∑

(I,J)∈N1,··· ,s

det(αIi , βJ )

=

τ+1∑

i=1

λi detαi +

τ+1∑

i=1

λi

∑

(I,J)∈N1,··· ,s

J =∅

det(αIi , βJ )

= detα +∑

(I,J)∈N1,··· ,s

J =∅

det(αI , βJ ) = det (α + β) .

This proves the theorem.

We now give a different representation of the polyconvex hull using theseparation results of Section 7.2.2.

Theorem 7.15 Let E ⊂ RN×n be such that PcoE is compact. Then

PcoE =ξ ∈ RN×n : f (ξ) ≤ 0 for every quasiaffine f ∈ FE

,

where

FE :=f : RN×n → R : f |E ≤ 0

.

Proof. Let

X :=ξ ∈ RN×n : f (ξ) ≤ 0, for every quasiaffine f ∈ FE

The set X is clearly polyconvex and contains E, thus PcoE ⊂ X.

On the other hand, since PcoE is polyconvex and compact then, by Theorem7.10, we have

PcoE =ξ ∈ RN×n : f (ξ) ≤ 0, for every quasiaffine f ∈ FPcoE

.

Since FPcoE ⊂ FE , we get X ⊂ PcoE, as claimed.

We next give a representation for the quasiconvex hull similar to the sec-ond representation formula of Theorem 7.14. This representation is, however,weaker than the one obtained in the polyconvex case since we cannot obtain therepresentation formula in a prescribed finite number of steps.


Theorem 7.16 Let E ⊂ RN×n. Let Q0 co E = E and define by induction thesets

Qi+1 coE =

ξ ∈ RN×n :

∃ R ∈ O(n), ϕ ∈ Wper such that

ξ +∇ϕ(x)R ∈ Qi coE, a.e. x ∈ D

, i ≥ 0.

Then QcoE = ∪i∈NQi co E. In particular, if E is open, then QcoE is also open.

Proof. (i) We first show that

⋃

i∈N

Qi coE ⊂ QcoE.

It is sufficient to show that Qi coE ⊂ QcoE, for every i. We proceed by induc-tion; the result is, by definition, true for i = 0. Since QcoE is quasiconvex, wehave, by definition of quasiconvex sets and by induction, that if Qi coE ⊂ QcoE,then Qi+1 coE ⊂ QcoE. This proves the claim.

The reverse inclusion follows at once from the fact that ∪i∈NQi coE is, aswe now see, a quasiconvex set containing E. Let R ∈ O(n), ϕ ∈ Wper andξ +∇ϕ(x)R ∈ ∪i∈NQi coE, a.e. x ∈ D. One has

∇ϕ(x)R ∈ η1, · · · , ηk a.e. x ∈ D,

with

measx ∈ D : ∇ϕ(x)R = ηj > 0, j = 1, · · · , k.

Moreover, ξ + ηj ∈ Qα(j)

co E for some α(j) ∈ N. Let s = maxα(1), · · · , α(k).Since Qi coE ⊂ Qi+1 co E, we have, for all j = 1, · · · , k, ξ + ηj ∈ Qs coE. Thusξ +∇ϕ(x)R ∈ Qs coE and, by definition, we get

ξ ∈ Qs+1 coE ⊂⋃

i∈N

Qi coE;

thus the quasiconvexity of ∪i∈NQi coE has been proved.

(ii) Since E is open, one easily gets, using an induction argument, that eachQi coE is open. By the preceding representation of QcoE it follows that thisset is also open.

The analogous representation for the rank one convex hull of a set is givenin the result below (see also Dacorogna-Marcellini [202]).

Theorem 7.17 Let E ⊂ RN×n. Let R0 co E = E and define by induction thesets

Ri+1 coE =

ξ ∈ RN×n :

ξ = λξ1 + (1− λ) ξ2 , λ ∈ [0, 1],

ξ1, ξ2 ∈ Ri coE, rankξ1 − ξ2 ≤ 1

, i ≥ 0.

Then RcoE = ∪i∈NRi coE. In particular, if E is open, then Rco E is also open.


Remark 7.18 (i) A similar construction and results can be obtained for ScoE.

(ii) In general it is not true that rank one convex hulls or separately convexhulls of compact sets are compact (see Aumann-Hart [50] and Kolar [376]). ♦

Proof. (i) A straightforward induction leads to Ri coE ⊂ RcoE and thus⋃Ri co E ⊂ RcoE. We now show the reverse inclusion. Observe that, by defi-

nition,

E ⊂⋃

Ri coE.

If we can show that⋃

Ri coE is rank one convex, we will have the claim, namely

RcoE ⊂⋃

Ri coE.

So let us show that⋃

Ri coE is rank one convex. Let λ ∈ [0, 1] and

ξ, η ∈⋃

Ri co E with rank ξ − η = 1

and let us prove that

λξ + (1− λ) η ∈⋃

Ri co E.

By definition there exist i, j ∈ N, for notational convenience assume that i ≥ j,such that

ξ ∈ Ri coE, η ∈ Rj coE ⊂ Ri coE.

We therefore deduce the result, namely

λξ + (1− λ) η ∈ Ri+1 coE ⊂⋃

Ri coE.

(ii) It is easy to see, by an induction argument that every Ri coE is open,provided E is open; thus RcoE is open.

We now consider representations of the convex hulls through functions as wecan get in the convex case.

Notation 7.19 Given a set E ⊂ RN×n, we consider the sets of functions

FE∞ :=

f : RN×n → R ∪ +∞ : f |E ≤ 0

,

FE :=f : RN×n → R : f |E ≤ 0

. ♦

With the above notation, one has (see Proposition 2.36), for E ⊂ RN×n,


∞

, (7.5)


, (7.6)

where coE denotes the closure of the convex hull of E.


Representations analogous to (7.5) are obtained in the theorem below for thepolyconvex, rank one convex and separately convex cases. However, (7.6) canonly be generalized to the polyconvex case if the sets are compact (see Theorem7.28). When dealing with the other notions of convexity, (7.6) is not true, evenif compact sets are considered.

Theorem 7.20 Let E ⊂ RN×n, then


∞

,


∞

.

Remark 7.21 A similar result for separately convex hulls can be proved,namely

ScoE =ξ ∈ RN×n : f (ξ) ≤ 0 for every separately convex f ∈ FE

∞

. ♦

Proof. We prove the first identity, the other one being handled analogously.Let us call X the set on the right hand side. Evidently X is a polyconvex setcontaining E and thus PcoE ⊂ X.

Consider now ξ ∈ X. Since χPco E is a polyconvex function belonging to FE∞ ,

one has χPcoE(ξ) ≤ 0 and consequently ξ ∈ PcoE, thus obtaining the otherinclusion.

We now come to a simple but important result (see Dacorogna-Marcellini[202]). It shows that our definitions of polyconvex and rank one convex hullsare consistent with the notions of polyconvex and rank one convex envelopesdefined in Chapter 6.

Proposition 7.22 Let E ⊂ RN×n and χE be its indicator function. Then

PχE = χPcoE and RχE = χRco E

where PχE and RχE are, respectively, the polyconvex and rank one convexenvelopes of χE .

Remark 7.23 A similar result holds for separately convex hulls. ♦

Proof. (i) Since χPcoE ≤ χE and χPcoE is polyconvex, we get that χPcoE ≤PχE ; so it remains to show the reverse inequality. From Theorem 6.8, we have

PχE (ξ) = inf∑τ+1

i=1 λiχE (ξi) : λ ∈ Λτ+1 ,∑τ+1


.

Note that for every ξ ∈ PcoE (or equivalently χPco E (ξ) = 0), we have fromTheorem 7.14 that there exist ξi ∈ E, λ ∈ Λτ+1 , such that

τ+1∑

i=1

λiT (ξi) = T (ξ) .


Therefore PχE (ξ) = 0 and thus PχE ≤ χPco E .

(ii) We first recall the construction of the rank one convex envelope of agiven function f : RN×n → R∪ +∞ (cf. Theorem 6.10). Define by inductionR0f = f and

Ri+1f (ξ) := inf

λRif (ξ1) + (1− λ) Rif (ξ2) :

λξ1 + (1− λ) ξ2 = ξ with rankξ1 − ξ2 ≤ 1

.

We then get that the rank one convex envelope of f is given by

Rf(ξ) = infi∈N

Rif(ξ).

We apply this result to χE , the indicator function of E. We observe that byinduction

RiχE = χRi coE

and thus, invoking Theorem 7.17,

RχE = χ⋃ Ri co E = χRco E . (7.7)

The proposition has therefore been proved.

We next show, as already mentioned in Remark 7.8, that the interior of ageneralized convex set keeps the convexity (in the generalized sense), but that,contrary to the classical convex case (see Proposition 2.4), this is not true forits closure.

Proposition 7.24 (i) Let E ⊂ RN×n be, respectively, a polyconvex, quasicon-vex, rank one convex or separately convex set. Then intE is also, respectively,polyconvex, quasiconvex, rank one convex or separately convex.

(ii) There is E ⊂ R2×2 a polyconvex and bounded set such that E is notseparately convex.

Proof. (i) We present the proof in the context of polyconvexity. For the otherconvexities, the proof is analogous. It is sufficient to prove that Pco(intE) =intE. The non-trivial inclusion is Pco(int E) ⊂ intE. Since E is polyconvex,evidently

Pco(intE) ⊂ PcoE = E. (7.8)

On the other hand, intE is open and thus (see Theorem 7.14) Pco(intE) is alsoopen. From (7.8), the desired inclusion then follows.

(ii) We define (see Figure 7.1)

E =

±(

1 00 x

): 0 < x < 1

.

It is a bounded set and E is not separately convex. In fact, let

ξ1 = diag(1, 0) and ξ2 = diag(−1, 0).


E

E−1

−1

1

1

Figure 7.1: The set E

One has ξ1, ξ2 ∈ E, but λξ1 + (1− λ)ξ2 /∈ E for any 0 < λ < 1.

We now show that E is polyconvex. Let ξ1, · · · , ξ6 ∈ E and suppose

T (ξ) =6∑

i=1

λiT (ξi) for some λ = (λ1, · · · , λ6) ∈ Λ6 . (7.9)

We have to see that ξ ∈ E. We can write 1, · · · , 6 = I+ ∪ I− for some I+ andI− such that

ξi =

(1 00 xi

)if i ∈ I+ and ξi =

(−1 00 −xi

)if i ∈ I− ,

where 0 < xi < 1, i = 1, · · · , 6. In any case, det ξi = xi .

If I+ = ∅ or I− = ∅, then it is clear that ξ ∈ E. We will see that the othercase, namely I+ = ∅ and I− = ∅, is not an admissible one. In fact, from (7.9),we can write

ξ =

( ∑i∈I+

λi −∑

i∈I−λi 0

0∑

i∈I+λixi −

∑i∈I−

λixi

)=

(α 0

0 β

)

and

det ξ = αβ =

6∑

i=1

λixi .

Then |α| <∑6i=1 λi = 1, |β| <∑6

i=1 λixi and thus |αβ| <∑6i=1 λixi , which is

a contradiction.


7.3.2 The different convex finite hulls

We next introduce some new sets that will allow a better understanding of theclosure of the different hulls. Recall first that

FE :=f : RN×n → R : f |E ≤ 0

.

Definition 7.25 For a set E ⊂ RN×n, let

cof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every convex f ∈ FE

,

Pcof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every polyconvex f ∈ FE

,


,

Rcof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every rank one convex f ∈ FE

,

Scof E :=ξ ∈ RN×n : f (ξ) ≤ 0 for every separately convex f ∈ FE

.

We call them, respectively, the convex finite, polyconvex finite, rank one convexfinite, quasiconvex finite and separately convex finite hulls of E.

Remark 7.26 (i) We recall that (see Proposition 2.36)

cof E = co E.

(ii) The above sets are all closed because any separately convex functiontaking only finite values is continuous. Besides, they are, respectively, (accord-ing to our definitions) convex, polyconvex, quasiconvex, rank one convex andseparately convex.

(iii) Some authors (see, for example, Muller-Sverak [465], Sverak [554], Zhang[616]), when dealing with quasiconvexity and rank one convexity, have adoptedthe above definitions for the hull of a set (in the generalized sense). They call alaminate convex hull what we have called Rco E. ♦

As in Theorem 7.14, the following proposition can easily be shown.

Proposition 7.27 Let E ⊂ RN×n, then

Pcof E = π(cof T (E) ∩ T (RN×n)).

Proof. Start by observing that if f : RN×n → R and F : Rτ(n,N) → R aresuch that

f (η) = F (T (η)) , ∀ η ∈ RN×n

then

f ∈ FE ⇔ F ∈ FT (E).


Call then

X := π(cof T (E) ∩ T (RN×n)) = ξ ∈ RN×n : T (ξ) ∈ cof T (E).

(i) Let us first show that X ⊂ Pcof E. So let f ∈ FE be any polyconvexfunction and ξ ∈ X. By definition of polyconvexity, we can find a convex functionF : Rτ(n,N) → R such that

f (η) = F (T (η)) , ∀ η ∈ RN×n.

Moreover F ∈ FT (E) and therefore we find that

f (ξ) = F (T (ξ)) ≤ 0

which is the claim, namely ξ ∈ Pcof E.

(ii) Let F : Rτ(n,N) → R be convex and such that F ∈ FT (E) and letξ ∈ Pcof E. Define

f (η) := F (T (η)) , ∀ η ∈ RN×n

and observe that f is polyconvex and f ∈ FE . We therefore find

f (ξ) = F (T (ξ)) ≤ 0

which means that ξ ∈ X, as wished.

We next see the relations between the closures of the convex hulls and thesets introduced in the above definition. Recall that we let PcoE, QcoE, RcoEand Sco E denote the closure of, respectively, the polyconvex, quasiconvex, rankone convex and separately convex hulls of E.

Theorem 7.28 Let E ⊂ RN×n, then

E ⊂ Scof E ⊂ Rcof E ⊂ Qcof E ⊂ Pcof E ⊂ cof E

and moreover

PcoE ⊂ Pcof E, QcoE ⊂ Qcof E, Rco E ⊂ Rcof E, ScoE ⊂ Scof E.

In general, the four inclusions are strict. However, if E is compact, then

PcoE = PcoE = Pcof E.

Remark 7.29 (i) We should also draw attention to the fact (see Proposition7.24) that in general the sets

PcoE, QcoE, Rco E, Sco E

are not even separately convex. Hence, in particular, PcoE = Pcof E unless Eis compact.


(ii) Let us emphasize, once more, that all the above analysis shows thatthere are examples of compact quasiconvex (respectively, rank one convex andseparately convex) sets E such that

E ⊂=

Qcof E, E ⊂=

Rcof E, E ⊂=

Scof E

contrary, by definition, to the following:

E = QcoE, E = Rco E, E = Sco E. ♦

Proof. (i) The inclusions

E ⊂ Scof E ⊂ Rcof E ⊂ Qcof E ⊂ Pcof E ⊂ cof E.

are obvious.

(ii) The inclusions

PcoE ⊂ Pcof E, QcoE ⊂ Qcof E, Rco E ⊂ Rcof E, ScoE ⊂ Scof E.

are also easy, since all the sets in the right hand side of the inclusions are closed,contain E and are, respectively, polyconvex, quasiconvex, rank one convex andseparately convex.

(iii) Let us show that the first inclusion (PcoE ⊂ Pcof E) is strict. Thisfollows (cf. Proposition 7.24) from the fact that there are polyconvex sets whoseclosure is not polyconvex though Pcof E is always a polyconvex set.

(iv) We now deal with the fact that the last three inclusions are strict. Weuse Example 5.18 which will give at once

QcoE ⊂=

Qcof E, Rco E ⊂=

Rcof E and ScoE ⊂=

Scof E.

Consider the set

E := ξ1, ξ2, ξ3, ξ4 ⊂ R2×2

where

ξ1 = diag(−1, 0), ξ2 = diag(1,−1), ξ3 = diag(2, 1), ξ4 = diag(0, 2)

and let us show that E is quasiconvex (and hence rank one and separatelyconvex); this will imply that

E = QcoE = Rco E = Sco E.

Suppose that ξ +∇ϕR ∈ E for some ϕ ∈ Wper and R ∈ O(2). Since

rankξi − ξj = 2 for i = j,


we have from Theorem 7.11 (with m = 4) that there exists ξi ∈ E such that

ξ +∇ϕ (x) R = ξi , a.e. x ∈ D.

Using then the periodicity of ϕ, we find

ξ =

∫

D

(ξ +∇ϕ(x)R) dx = ξi

and thus ξ = ξi ∈ E. We then conclude that E is quasiconvex.

However, any separately convex function f ∈ FE and consequently any rankone convex or quasiconvex function in FE is such that f(0) ≤ 0, according to(5.31) in Example 5.18 (observing that separately convex functions are rank oneconvex when restricted to diagonal matrices). Thus 0 ∈ Scof E, but 0 /∈ QcoE.

(v) If we assume E to be compact, we then have, as we now see,

PcoE = PcoE = Pcof E.

We have already shown in (ii) that

PcoE ⊂ Pcof E

By Theorem 7.14, in this case, PcoE is compact and then PcoE = PcoE. Wetherefore have combining this identity and (ii) that

PcoE = PcoE ⊂ Pcof E.

Let us now show the reverse inclusion: Pcof E ⊂ PcoE. We start noting that,since E is compact, T (E) is compact and thus co T (E) is also compact (cf.Theorem 2.14 (i)). Note also that the function

η → f (η) := dist(T (η), coT (E))

is polyconvex and f ∈ FE . Therefore if ξ ∈ Pcof E we get

dist(T (ξ), coT (E)) = 0.

Since coT (E) is closed, we deduce that T (ξ) ∈ coT (E) and thus, ξ ∈ PcoE.

Gathering all the results, we can write

ScoE ⊂ RcoE ⊂ QcoE ⊂ PcoE ⊂ coE = cof E

and also

Scof E ⊂ Rcof E ⊂ Qcof E ⊂ Pcof E ⊂ coE = cof E.


Moreover, the same example and arguments used in the proof of Theorem7.28 (see also Proposition 7.24) show that, in general,

Scof E Rco E, Rcof E QcoE and Qcof E PcoE.

However, if E is compact one has Qcof E ⊂ PcoE.

We draw attention to the fact that several characterizations of the sets inDefinition 7.25 have been used in the literature according to the specific needs ofeach situation. These sets can be written in terms of measures (see Kirchheim[365], Muller [462]) or using the distance function (see Zhang [617]): if E ⊂RN×n is compact, then

Qcof E =ξ ∈ RN×n : Qdist(ξ, E) = 0

,

where Q dist(·, E) is the quasiconvex envelope of the function dist(·, E).

7.3.3 Extreme points and Minkowski type theorem forpolyconvex, quasiconvex and rank one convex sets

An important tool in convex analysis is the notion of extreme point. In astraightforward manner, we can define it for generalized convex sets as follows(see Dacorogna-Marcellini [202] and Dacorogna-Ribeiro [213]).

Definition 7.30 (i) If E ⊂ Rm is convex, ξ ∈ E is said to be an extreme pointof E in the convex sense if

ξ = λξ1 + (1 − λ)ξ2

λ ∈ (0, 1), ξ1, ξ2 ∈ E

⇒ ξ1 = ξ2 = ξ.

For an arbitrary set E ⊂ Rm, the set of extreme points in the convex sense ofcoE is denoted Ec

ext (in Chapter 2, since there was no ambiguity, we have justwritten for this set Eext).

(ii) If E ⊂ RN×n is polyconvex, ξ ∈ E is said to be an extreme point of Ein the polyconvex sense if

T (ξ) =∑I

i=1 λiT (ξi), I ∈ N

(λ1, · · · , λI) ∈ ΛI , λi > 0, ξi ∈ E

⇒ ξi = ξ, i = 1, · · · , I.

For an arbitrary set E ⊂ RN×n, the set of extreme points in the polyconvexsense of PcoE is denoted Ep

ext .

(iii) If E ⊂ RN×n is quasiconvex, ξ ∈ E is said to be an extreme point of Ein the quasiconvex sense if

ξ +∇ϕ(x)R ∈ E a.e. x ∈ D,

D = (0, 1)n, R ∈ O(n), ϕ ∈ Wper

⇒ ∇ϕ = 0 a.e. in D.

For an arbitrary set E ⊂ RN×n, the set of extreme points in the quasiconvexsense of QcoE is denoted Eq

ext .


(iv) If E ⊂ RN×n is rank one convex, ξ ∈ E is said to be an extreme pointof E in the rank one convex sense if

ξ = λξ1 + (1 − λ)ξ2

λ ∈ (0, 1), ξ1, ξ2 ∈ E, rank ξ1 − ξ2 ≤ 1

⇒ ξ1 = ξ2 = ξ.

For an arbitrary set E ⊂ RN×n, the set of extreme points in the rank one convexsense of RcoE is denoted Er

ext .

(v) If E ⊂ Rm is separately convex, ξ ∈ E is said to be an extreme point ofE in the separately convex sense if

ξ = λξ1 + (1− λ)ξ2

λ ∈ (0, 1), ξ1, ξ2 ∈ E, ξ1 − ξ2 = s ei,

with s ∈ R and ei a vector of the canonical basis of Rm

⎫⎪⎬⎪⎭⇒ ξ1 = ξ2 = ξ.

For an arbitrary set E ⊂ Rm, the set of extreme points in the separately convexsense of ScoE is denoted Es

ext .

We first see the relations between the sets of extreme points for the differentnotions of convexity.

Proposition 7.31 Let E ⊂ RN×n. Then

Ecext ⊂ Ep

ext ⊂ Eqext ⊂ Er

ext ⊂ Esext ⊂ E.

Proof. (i) Let us first show that any of these sets are in E. We do it, forexample, with Er

ext , the others being handled similarly. So let ξ ∈ Erext ⊂ RcoE

and thus, by Theorem 7.17, ξ ∈ Ri+1 coE for a certain i ∈ N. This means that

ξ ∈

ξ ∈ RN×n :ξ = λξ1 + (1− λ) ξ2 , λ ∈ [0, 1],

ξ1, ξ2 ∈ Ri coE ⊂ RcoE, rankξ1 − ξ2 ≤ 1

and thus, since ξ ∈ Erext , we deduce that in fact ξ ∈ Ri coE. Iterating the

procedure, we find that ξ ∈ R0 co E = E, as claimed.

(ii) The non-trivial inclusions are those related to Eqext , the set of extreme

points of QcoE, but it can be obtained with the same arguments used in theproof of Theorem 7.7, Part 1, and we just do it for

Eqext ⊂ Er

ext ;

the other one being handled similarly. Let ξ ∈ Eqext ⊂ E, we have to show that

ξ = λξ1 + (1 − λ)ξ2

λ ∈ (0, 1), ξ1, ξ2 ∈ Rco E, rank ξ1 − ξ2 ≤ 1

⇒ ξ1 = ξ2 = ξ.

So let λ ∈ (0, 1), ξ1, ξ2 ∈ RcoE, rank ξ1 − ξ2 ≤ 1 and find then, as in theproof of Theorem 7.7, R ∈ O(n) and ϕ ∈ Wper such that

∇ϕ(x)R ∈ (1− λ)(ξ1 − ξ2),−λ(ξ1 − ξ2) a.e. x ∈ D.


Since ξ ∈ Eqext , we then get, from the above construction, that

ξ +∇ϕ(x)R ∈ RcoE ⊂ QcoE a.e. x ∈ D,

R ∈ O(n), ϕ ∈ Wper

⇒ ∇ϕ ≡ 0;

thus ξ1 = ξ2 = ξ and hence ξ ∈ Erext , as wished.

We now give two examples (see [202]) showing that, in general, the inclusionsare strict. The first will show that a point can be extreme in the separatelyconvex sense but not in the usual sense (i.e., the convex sense). The secondone will provide an extreme point in the rank one convex sense but not in thepolyconvex sense.

Example 7.32 Let E =(x, y) ∈ R2 : 0 ≤ x, y ≤ x + y ≤ 1

(see Figure 7.2),

1

1

y

x

Figure 7.2: The set E

which is convex (and thus separately convex). Then any element of the linex + y = 1 is an extreme point of E in the separately convex sense but on thisline only (1, 0) and (0, 1) are extreme points (in the convex sense) of E. ♦

Example 7.33 (see Proposition 5.10). Let

ξ1 =

(1 02 0

), ξ2 =

(0 10 1

), ξ3 =

(−1 −10 0

), η =

(0 0

2/3 1/3

).

Let

E = ξ1, ξ2, ξ3, η and F = ξ1, ξ2, ξ3 .

Note that

η =1

3(ξ1 + ξ2 + ξ3) ,

det η =1

3(det ξ1 + det ξ2 + det ξ3)


and hence η is not an extreme point in the polyconvex sense of PcoE (thusPcoE = PcoF ). Moreover, since

det (ξi − ξj) = 0, ∀i = j and det (ξi − η) = 0, ∀i,

we deduce that η is an extreme point in the rank one convex sense of RcoE = E(thus RcoE = E ⊃ RcoF = F ). What is, however, more interesting is that ηis also an extreme point in the rank one convex sense of PcoE. Indeed one caneasily show that

PcoE =

ξ ∈ R2×2 : ξ = αξ1 + βξ2 + (1− α− β) ξ3

0 ≤ α, β ≤ α + β ≤ 1 and β = α + 12 ± 1

2

√12α2 − 4α + 1

.

We now prove that η is an extreme point in the rank one convex sense of PcoE.Assume that there exist 0 < t < 1, ξ, ξ′ ∈ PcoE, with rank ξ − ξ′ ≤ 1 suchthat

η = tξ + (1− t) ξ′.

We wish to show that η = ξ = ξ′. So let

ξ = αξ1 + βξ2 + (1− α− β) ξ3 =

(2α + β − 1 2β + α− 1

2α β

),

ξ′ = γξ1 + δξ2 + (1− γ − δ) ξ3 =

(2γ + δ − 1 2δ + γ − 1

2γ δ

),

with 0 ≤ α, β ≤ α + β ≤ 1, 0 ≤ γ, δ ≤ γ + δ ≤ 1 and

β = α +1

2± 1

2

√12α2 − 4α + 1 ,

δ = γ +1

2± 1

2

√12γ2 − 4γ + 1 .

From η = tξ + (1− t) ξ′, we deduce that

tα + (1− t) γ = tβ + (1− t) δ =1

3. (7.10)

Furthermore, from det (ξ − ξ′) = 0 we find

3 (α− γ)2

= [(α− γ)− (β − δ)]2

=1

4

[√12α2 − 4α + 1±

√12γ2 − 4γ + 1

]2.

This identity then implies that α = γ, which coupled with (7.10) leads to α =γ = β = δ = 1/3. Thus η = ξ = ξ′ as claimed. ♦

Minkowski theorem (see Theorem 2.20) ensures that the convex hull of acompact set coincides with the convex hull of its extreme points. We next dealwith the generalization of this result to the other convexities. We start with thepolyconvex case (see also Dacorogna-Tanteri [215]).


Theorem 7.34 Let E ⊂ RN×n be a compact set. Then

PcoE = PcoEpext .

Proof. The inclusion PcoEpext ⊂ PcoE is trivial, since Ep

ext ⊂ PcoE. Wethus show the reverse inclusion. We start noting that

PcoE = π(co T (E) ∩ T (RN×n))

PcoEpext = π(co T (Ep

ext) ∩ T (RN×n)).

Let ξ ∈ PcoE and let us prove that ξ ∈ PcoEpext . By the above characterization

of PcoE we have T (ξ) ∈ coT (E). Moreover, by the classical Minkowski theorem(cf. Theorem 2.20), and using the fact that T (E) is compact, we have

coT (E) = co(T (E)cext),

where T (E)cext is the set of extreme points of coT (E) (in the convex sense).

We next prove that

T (E)cext ⊂ T (Ep

ext),

which will finish the proof.Let then X ∈ T (E)c

ext . In particular, X ∈ T (E) and we can write X = T (η)with η ∈ E. It suffices then to see that η ∈ Ep

ext . Suppose that

T (η) =

I∑

i=1

λiT (ηi)

for some (λ1, · · · , λI) ∈ ΛI , λi > 0, ηi ∈ PcoE. Observing that, since ηi ∈PcoE then T (ηi) ∈ coT (E), it immediately follows, from the fact that T (η)is an extreme point of coT (E), that ηi = η for every i, that is to say η is anextreme point of PcoE. The proof is finished.

As remarked in Kirchheim [365], the above result is not true for quasiconvex,rank one convex or separately convex hulls. In fact, Example 5.18 considered inthe proof of Theorem 7.28 shows that, in general,

QcoEqext = QcoE, Rco Er

ext = Rco E and Sco Esext = Sco E

as we now prove it.

Example 7.35 We consider a set of diagonal matrices that we identify with ele-ments of R2. In particular, rank one convexity and separate convexity coincide.Let

E = E1 ∪ E2 ∪ E3 ∪E4 ∪ E5 ,


whereE1 = (x, y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,

E2 = (x, 1) ∈ R2 : 1 ≤ x ≤ 2, E3 = (0, y) ∈ R2 : 1 ≤ y ≤ 2,E4 = (x, 0) ∈ R2 : −1 ≤ x ≤ 0, E5 = (1, y) ∈ R2 : −1 ≤ y ≤ 0.

Note that E is a compact rank one convex set and

Eqext ⊂ Er

ext = ξ1, ξ2, ξ3, ξ4,

whereξ1 = (−1, 0), ξ2 = (1,−1), ξ3 = (2, 1), ξ4 = (0, 2).

Thus, since there are no rank one connections between the elements ξi ,

QcoEqext = Eq

ext and Rco Erext = Er

ext .

However, Eqext ⊂ Er

ext E = RcoE ⊂ QcoE. ♦

We now prove a weaker result than Theorem 7.34 but that is valid in thequasiconvex, rank one convex and separately convex cases. We follow the proofof Matousek-Plechac [439], which can also be adapted to the quasiconvex case(see also Zhang [617] for a different proof in the quasiconvex case).

Theorem 7.36 Let E ⊂ RN×n be a bounded set and Eqfext , Erf

ext , Esfext denote,

respectively, the set of extreme points of Qcof E (in the quasiconvex sense), theset of extreme points of Rcof E (in the rank one convex sense) and the set ofextreme points of Scof E (in the separately convex sense). Then

Qcof E = Qcof Eqfext , Rcof E = Rcof Erf

ext and Scof E = Scof Esfext.

Proof. We divide the proof into two steps. The first is common to the threeconvexities and we present it in the context of quasiconvexity. In the secondstep, we consider separately the quasiconvex and the rank one convex cases (thelatter being analogous to the separately convex case, which we will not consider

explicitly). In all that follows we will denote by Eqf

ext the closure of Eqfext .

Step 1. We remark that, for any set K ⊂ RN×n, since Qcof is automat-

ically closed, Qcof K = Qcof K. Thus, it is enough to prove that Qcof E =

Qcof Eqf

ext . The inclusion Qcof Eqf

ext ⊂ Qcof E is trivial. It remains to verifythe reverse inclusion. We argue by contradiction.

Suppose there is some η ∈ Qcof E − Qcof Eqf

ext . Then, by definition, there

exists a quasiconvex function f : RN×n → R with f ∈ FEqfext such that f(η) > 0.

Now let

M := maxQcof E

f and A := ξ ∈ Qcof E : f(ξ) = M.


The set A is non empty and compact (since Qcof E is compact and f is a

continuous function). Thus, considering RN×n with the lexicographic order(the elements of RN×n being seen as vectors), one can consider the maximum

element of A, say ξ0 . We have ξ0 /∈ Eqfext , which follows from

0 < f(η) ≤ maxQcof E

f = M = f(ξ0).

As we will see in Step 2, this leads to the existence of an element in A greaterthan ξ0 for the lexicographic order, which is the desired contradiction.

Step 2. Quasiconvex case. Since ξ0 ∈ Qcof E − Eqfext , there are R ∈ O(n)

and ϕ ∈ Wper such that

ξ0 +∇ϕ(x)R ∈ Qcof E, a.e. x ∈ D, with measx ∈ D : ∇ϕ(x) = 0 > 0.

We can write

∇ϕ(x)R ∈ ξ1, · · · , ξk and λi = measx ∈ D : ∇ϕ(x)R = ξi > 0.

Since ξ0 + ξi ∈ Qcof E, we have f(ξ0 + ξi) ≤ M. Consequently, by the quasi-convexity of f we get

M = f(ξ0) ≤∫

D

f(ξ0 +∇ϕ(x)R) dx =

k∑

i=1

λif(ξ0 + ξi) ≤M

implying f(ξ0 + ξi) = M for every i = 1, · · · , k, that is ξ0 + ξi ∈ A. Finally,from the fact that ∇ϕ ≡ 0 and

0 =

∫

D

∇ϕ(x)R dx =k∑

i=1

λiξi

we conclude that among the elements ξ0 + ξi there must be at least one that isgreater than ξ0 (in the lexicographic order), which contradicts the fact that ξ0

is the maximum element of A.

Rank one convex case. We recall that in this case the function f ∈ FErfext is a

rank one convex function. Since ξ0 ∈ Rcof E −Erfext , there are η1, η2 ∈ Rcof E,

with rank η1 − η2 ≤ 1 such that

ξ0 = λη1 + (1 − λ)η2 and ξ0 = η1 , ξ0 = η2 .

As in the quasiconvex case we get

f(η1) = f(η2) = M

and from ξ0 = λη1 + (1− λ)η2 it follows that η1 or η2 must be greater than ξ0 ,which is a contradiction.


7.3.4 Gauges for polyconvex sets

We define, as in convex analysis (see Section 2.3.7), the gauge of a polyconvexset. We follow here the presentation of Dacorogna-Tanteri [215].

Theorem 7.37 Let E ⊂ RN×n be a non-empty polyconvex set and let

χE(ξ) :=

0 if ξ ∈ E

+∞ if ξ /∈ E

be its indicator function. Let Hp : Rτ(n,N) → R ∪ +∞ be defined as

Hp (X∗) := supξ∈E

〈T (ξ) ; X∗〉 .

The following statements then hold.

(i) Hp is lower semicontinuous, convex and positively homogeneous of degreeone.

(ii) If E is compact and if

(Hp)∗

: Rτ(n,N) → R ∪ +∞

is the conjugate function of Hp meaning that

(Hp)∗(X) = sup

X∗∈Rτ(n,N)

〈X ; X∗〉 −Hp (X∗) ,

then

χE(ξ) = (Hp)∗ (T (ξ)) ,

E =ξ ∈ RN×n : (Hp)

∗(T (ξ)) ≤ 0

.

(iii) If 0 ∈ E, then

Hp (X∗) ≥ Hp (0) = 0 for every X∗ ∈ Rτ(n,N);

and if E is compact, then Hp takes only finite values.

(iv) If 0 ∈ intE and if E is compact, then

Hp (X∗) = 0 ⇔ X∗ = 0;

and in this case

E =ξ ∈ RN×n : (Hp)0 (T (ξ)) ≤ 1

,

where (Hp)0

(called the gauge in the polyconvex sense of E) is the polar of Hp,namely

(Hp)0 (X) := supX∗ =0

〈X ; X∗〉Hp (X∗)

.


Remark 7.38 (i) When N = n = 2, we have that Hp : R2×2×R→ R∪+∞is given by

Hp (ξ∗, δ∗) = supξ∈E

〈ξ; ξ∗〉+ δ∗ det ξ

andE =

ξ ∈ R2×2 : (Hp)

∗(ξ,det ξ) ≤ 0

.

(ii) Note that (Hp)0 is positively homogeneous of degree one but, of course,

this is not the case for the function ξ → (Hp)0 (T (ξ)) .

(iii) In the notation of Section 6.2.1, we have

Hp = (χE)p. ♦

Example 7.39 For ξ ∈ R2×2, let 0 ≤ λ1 (ξ) ≤ λ2 (ξ) denote its singular values,0 < a1 ≤ a2 and

E =ξ ∈ R2×2 : λ2 (ξ) ≤ a2, λ1 (ξ)λ2 (ξ) ≤ a1a2

,

which is a polyconvex set (see Theorem 7.43). Then

(Hp)0(ξ∗, δ∗) = max

λ2 (ξ∗)

a2,|δ∗|a1a2

is a gauge for E. ♦

Proof. (i) Since E is non-empty then Hp > −∞. Hp being the supremumof affine functions, is convex and lower semicontinuous. The fact that Hp ispositively homogeneous of degree one is easy.

(ii) Since E is compact, the function

χco T (E) : Rτ(n,N) → R ∪ +∞

is convex and lower semicontinuous. Moreover since E is polyconvex, we have,according to Theorem 7.4, that

E = ξ ∈ RN×n : T (ξ) ∈ coT (E)

and thusχE (ξ) = χco T (E) (T (ξ)) .

We then proceed as in the proof of Theorem 6.6 to deduce that

Hp (X∗) = χpE (X∗)

χE (ξ) = χppE (ξ) := (Hp)

∗(T (ξ))

hence the result.


(iii) This is obvious.

(iv) We now show that if 0 ∈ intE and if E is compact then

Hp (X∗) = 0 ⇔ X∗ = 0.

The implication (⇐) follows from (iii) and we therefore discuss only the reverseone. Let ξ ∈ RN×n be an arbitrary point, ξ = 0. Since 0 ∈ intE, we deducethat for every ǫ sufficiently small then ǫ ξ/ |ξ| ∈ E and therefore

0 = Hp (X∗) ≥ 〈 T ( ǫ ξ/ |ξ| ) ; X∗ 〉 . (7.11)

Since ξ ∈ RN×n is arbitrary, the above inequality implies that X∗ = 0, asclaimed. We prove this last fact only when N = n = 2, the general case beingproved similarly. The inequality (7.11) reads then (writing X∗ = (ξ∗, δ∗))

0 = Hp (X∗) ≥ ǫ

|ξ| 〈ξ; ξ∗〉+ ǫ2

det ξ

|ξ|2δ∗, ∀ ξ ∈ R2×2, ξ = 0.

We therefore get, using the fact that ǫ is arbitrary,〈ξ; ξ∗〉 = 0, ∀ ξ ∈ R2×2

δ∗ det ξ ≤ 0, ∀ ξ ∈ R2×2

hence (ξ∗, δ∗) = (0, 0) .The last identity

E =ξ ∈ RN×n : (Hp)

0(T (ξ)) ≤ 1

immediately follows from (ii).

7.3.5 Choquet functions for polyconvex and rank oneconvex sets

We finally define some functions that characterize the extreme points first inthe polyconvex and then in the rank one convex sense. In the convex case thisis known as the Choquet function (see Section 2.3.8). The first theorem wasestablished by Dacorogna-Tanteri [215].

Theorem 7.40 Let E ⊂ RN×n be a non-empty compact polyconvex set andEp

ext be its extreme points in the polyconvex sense. Then there exists a polyconvexfunction

ϕEp : RN×n → R ∪ +∞

such thatEp

ext =ξ ∈ E : ϕE

p (ξ) = 0

,

ϕEp (ξ) ≤ 0 ⇔ ξ ∈ E.



f (ξ) :=

− |ξ|2 if ξ ∈ E

+∞ otherwise.

Note thatf (ξ) ≥ − sup|ξ|2 : ξ ∈ E > −∞.

The Choquet function for polyconvex sets is then defined as

ϕEp (ξ) :=

Pf (ξ)− f (ξ) if ξ ∈ E

+∞ otherwise.

Note that, letting χE be the indicator function of the set E,

ϕEp (ξ) = Pf (ξ) + |ξ|2 + χE(ξ), ∀ ξ ∈ RN×n.

The function ϕEp : RN×n → R ∪ +∞ is therefore polyconvex and

ϕEp (ξ) ≤ 0, if ξ ∈ E and ϕE

p (ξ) = 0, if and only if ξ ∈ Epext .

Indeed the inequality is clear since in E the function f is finite and, by definition,Pf is always not larger than f. We now show that

ϕEp (ξ) = 0 ⇔ ξ ∈ Ep

ext .

Using Theorem 6.8, we find that if ξ ∈ E, then, letting τ := τ (n, N) ,

ϕEp (ξ) = |ξ|2 + inf

ξi∈E−∑τ+1

i=1 ti |ξi|2 : T (ξ) =∑τ+1

i=1 tiT (ξi) , t ∈ Λτ+1

where for s ∈ N, we have

Λs := λ = (λ1, · · · , λs) : λi ≥ 0,∑s

i=1 λi = 1 .

Therefore if ξ ∈ Epext , we deduce, by definition, that in the infimum the only

admissible ξi are ξi = ξ; and hence we have ϕEp (ξ) = 0.

We now show the reverse implication, namely

ϕEp (ξ) = 0 ⇒ ξ ∈ Ep

ext .

From the above representation formula we obtain, since ϕEp (ξ) = 0 and ξ ∈ E,

that|ξ|2 = sup

ξi∈E∑τ+1

i=1 ti |ξi|2 : T (ξ) =∑τ+1

i=1 tiT (ξi) , t ∈ Λτ+1.

Combining the above with the convexity of the function ξ → |ξ|2 we get that

|ξ|2 ≥∑τ+1i=1 ti |ξi|2 ≥ |

∑τ+1i=1 tiξi |2 = |ξ|2 ;


the strict convexity of ξ → |ξ|2 implies then that ξi = ξ. Thus ξ ∈ Epext .

A similar construction can be done for rank one convexity, as was achievedby Ribeiro [512].

Theorem 7.41 Let E ⊂ RN×n be a non-empty compact rank one convex setand Er

ext be its extreme points in the rank one convex sense. Then there existsϕE

r : RN×n → R ∪ +∞ a rank one convex function such that

Erext =

ξ ∈ E : ϕE

r (ξ) = 0

,

ϕEr (ξ) ≤ 0 ⇔ ξ ∈ E.


f (ξ) :=

− |ξ|2 if ξ ∈ E

+∞ otherwise

and the Choquet function for rank one convex sets is defined as

ϕEr (ξ) :=

Rf (ξ)− f (ξ) if ξ ∈ E

+∞ otherwise.

Observe that ϕEr : RN×n → R ∪ +∞ is rank one convex, since

ϕEr (ξ) = Rf (ξ) + |ξ|2 + χE(ξ), ∀ ξ ∈ RN×n.

We now prove that

ϕEr (ξ) ≤ 0, if ξ ∈ E and ϕE

r (ξ) = 0, if and only if ξ ∈ Erext .

Indeed the inequality is clear since in E the function f is finite and, by definition,Rf is always not larger than f. We now show that

ϕEr (ξ) = 0 ⇔ ξ ∈ Er

ext .

Recall first that (cf. Theorem 6.10) we have

Rf = limk→∞

Rkf = infk∈N

Rkf

where R0f := f and Rkf is inductively given by

Rk+1f (ξ) := inf

λRkf (ξ1) + (1− λ) Rkf (ξ2) :

λξ1 + (1− λ) ξ2 = ξ with rank ξ1 − ξ2 ≤ 1

.

Let us show that for ξ ∈ Erext we have

Rkf (ξ) = f (ξ) , for every k ∈ N (7.12)

Examples 347

(and hence Rf (ξ) = f (ξ)). We proceed by induction. The result is true fork = 0 and let us assume that it is valid for k and show the claim for k +1. SinceRkf ≤ f, we have that Rkf (ξ) is finite and hence, since ξ ∈ Er

ext , we get thatin the inf the only acceptable ξ1, ξ2 are ξ1 = ξ2 = ξ and thus

Rk+1f (ξ) = Rkf (ξ) .

The induction procedure therefore leads to (7.12). We have therefore shownthat

ξ ∈ Erext ⇒ ϕE

r (ξ) = 0.

Let us now prove the reverse implication

ϕEr (ξ) = 0 ⇒ ξ ∈ Er

ext .

From the above representation formula we obtain, since ϕEr (ξ) = 0 and ξ ∈ E,

thatRf (ξ) = Rkf (ξ) = f (ξ) , for every k ∈ N.

Let ξ ∈ E, λ ∈ (0, 1), ξ1, ξ2 ∈ E with rank ξ1 − ξ2 ≤ 1 and such that

ξ = λξ1 + (1 − λ)ξ2

we have to show that ξ1 = ξ2 = ξ.From the fact that R1f (ξ) = f (ξ) we get

|ξ|2 = supηi∈E

λ |η1|2 + (1− λ) |η2|2 :

λη1 + (1− λ) η2 = ξ with rank η1 − η2 ≤ 1

.

Combining the above with the convexity of the function ξ → |ξ|2 we get that

|ξ|2 ≥ λ |ξ1|2 + (1− λ) |ξ2|2 ≥ | λξ1 + (1− λ)ξ2 |2 = |ξ|2 ;

the strict convexity of ξ → |ξ|2 implies then that ξi = ξ. Thus ξ ∈ Erext .

7.4 Examples

We now discuss several examples that should be related to those of Sections 6.6,10.3 and 11.5.

In many instances, we will use below the following elementary lemma or asimilar argument to that in the proof.

Lemma 7.42 Let X ⊂ RN×n be compact and E be rank one convex. Then

∂X ⊂ E ⇒ X ⊂ E.


Proof. Let ξ ∈ X. If ξ ∈ ∂X, then nothing is to be proved; so we assume thatξ ∈ intX. Let η ∈ RN×n be any matrix of rank one and set for t ∈ R,

ξt := ξ + tη.

Since X is compact, we can find t1 < 0 < t2 so that

ξt1 , ξt2 ∈ ∂X ⊂ E and ξ =t2

t2 − t1ξt1 +

−t1t2 − t1

ξt2 .

Since E is rank one convex,

ξt1 , ξt2 ∈ E and rank ξt1 − ξt2 = 1

we have that ξ ∈ E.


One of the most general examples of such hulls concerns sets that involve singularvalues. Let us first recall (see Chapter 13) that the singular values of a givenmatrix ξ ∈ Rn×n, denoted by 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) , are the eigenvalues of

(ξξt)1/2

.

Our result (see Dacorogna-Tanteri [214], [215] and also [202] for the first twocases and Dacorogna-Ribeiro [212] for the third case) is the following.

Theorem 7.43 Let 0 < γ1 ≤ · · · ≤ γn .

Part 1. If

E =ξ ∈ Rn×n : λi (ξ) = γi , i = 1, · · · , n

,

then

coE =ξ ∈ Rn×n :

∑ni=ν λi (ξ) ≤∑n

i=ν γi , ν = 1, · · · , n

PcoE = QcoE = RcoE =ξ ∈ Rn×n :


i=ν γi , ν = 1, · · · , n

.

Part 2. If α = 0,n∏

i=1

γi = |α|

and

Eα =ξ ∈ Rn×n : λi (ξ) = γi , i = 1, · · · , n, det ξ = α

then

PcoEα = QcoEα = RcoEα

=ξ ∈ Rn×n :


i=ν γi , ν = 2, · · · , n, det ξ = α

.

Examples 349

Furthermore,

intRco Eα =ξ ∈ Rn×n :

∏ni=ν λi(ξ) <

∏ni=ν γi , ν = 2, · · · , n, det ξ = α

where the interior is to be understood relative to the manifold det ξ = α .

Part 3. Let α ≤ β. If either α = 0 or β = 0,

γ2

n∏

i=2

γi ≥ max |α| , |β|

and

Eα,β =ξ ∈ Rn×n : λi (ξ) = γi , i = 2, · · · , n, det ξ ∈ α, β

,

then

PcoEα,β = QcoEα,β = RcoEα,β

=ξ ∈ Rn×n :

∏ni=ν λi(ξ) ≤

∏ni=ν γi , ν = 2, · · · , n, det ξ ∈ [α, β]

.

Moreover, if α < β, then

intRcoEα,β =ξ ∈ Rn×n :

∏ni=ν λi(ξ) <

∏ni=ν γi, ν = 2, · · · , n, det ξ ∈ (α, β)

.

Before proceeding with the proof, let us make some comments. Since |det ξ| =∏ni=1 λi (ξ) and the singular values are ordered as 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) , we

should, and did, respectively impose in the second and third cases that

n∏

i=1

γi = |α|

γ2

n∏

i=2

γi ≥ max |α| , |β| .

Proof. First note that the third case contains the other ones as particularcases. Indeed the first one is deduced from the last one by setting β = −α and

γ1 = β [∏n

i=2 γi ]−1,

while the second one is obtained by setting β = α and

γ1 = |α| [∏n

i=2 γi ]−1

in the third case.

We therefore divide the proof into four parts: the first dealing with the rep-resentation of coE, the second with the formulas of the polyconvex, quasiconvexand rank one convex hulls of Eα,β (and thus of E and Eα) and the third andfourth with the representations of intRcoEα,β and intRcoEα , respectively.


(i) Representation of coE. Let

X :=ξ ∈ Rn×n :

∑ni=ν λi(ξ) ≤

∑ni=ν γi , ν = 1 , . . . , n

.

Step 1: coE ⊂ X. The inclusion is easy, since E ⊂ X and the functionsξ →∑n

i=ν λi(ξ) are convex (see Corollary 5.37), we find that X is convex andthus the inclusion coE ⊂ X.

Step 2: X ⊂ co E. Let ξ ∈ X. Since the functions ξ → λi(ξ) are invariant byorthogonal transformations, we can assume (see Theorem 13.3), without loss ofgenerality, that

ξ = diag (x1, · · · , xn) =

⎛⎜⎝

x1 · · · 0...

. . ....

0 · · · xn

⎞⎟⎠

with 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn and

n∑

i=ν

xi ≤n∑

i=ν

γi , ν = 1, · · · , n.

We proceed by induction. The case n = 1 is easy. Indeed let ξ = x1 ∈ Xmeaning that 0 ≤ x1 ≤ γ1. We then write

x1 = tγ1 + (1 − t)(−γ1) with t =x1 + γ1

2γ1

and we can thus deduce that ξ ∈ coE, as claimed.

We now assume that the result has been proved up to the order n − 1 andprove the claim for n. We divide the study into two cases.

Case 1:∑n

i=ν xi =∑n

i=ν γi for a certain ν ∈ 2, . . ., n . Observe that wecan apply the hypothesis of induction to

x1, · · · , xν−1 and γ1, · · · , γν−1

and toxν , · · · , xn and γν , · · · , γn .

Indeed for the second one this follows from the hypotheses

n∑

i=ν

xi ≤n∑

i=ν

γi , ν = ν, · · · , n

while for the first one we have, for ν = 1, · · · , ν − 1,

ν−1∑

i=ν

xi =

n∑

i=ν

xi −n∑

i=ν

xi =

n∑

i=ν

xi −n∑

i=ν

γi

≤n∑

i=ν

γi −n∑

i=ν

γi =

ν−1∑

i=ν

γi .

Examples 351

We can therefore deduce, by hypothesis of induction, that ξ ∈ coE.

Case 2:∑n

i=ν xi <∑n

i=ν γi for every ν ∈ 2, · · · , n . We then let

L :=

η ∈ Rn×n :

∑ni=ν λi(η) ≤∑n

i=ν γi , ν = 2 , · · · , nand

∑ni=1 λi(η) =

∑ni=1 xi

.

Observe that L ⊂ X is compact and that ξ ∈ relint (L) (where relint (L) standsfor the relative interior of L), since

n∑

i=ν

λi(ξ) =

n∑

i=ν

xi <

n∑

i=ν

γi and

n∑

i=1

λi(ξ) =

n∑

i=1

xi .

Note also that by Case 1 we have ∂L ⊂ coE. We therefore let, for t ∈ R,

ξt := diag (x1, · · · , xn−2, xn−1 − t, xn + t) .

Observe that by compactness of L and since ξ = ξ0 ∈ relint (L) we can find (asin Lemma 7.42) t1 < 0 < t2 such that

ξt1 , ξt2 ∈ ∂L ⊂ coE and ξ =t2

t2 − t1ξt1 +

−t1t2 − t1

ξt2 .

We have therefore obtained that ξ ∈ coE and hence the claimed result X ⊂coE.

(ii) Formula for PcoEα,β , QcoEα,β and RcoEα,β . We let

Y :=ξ ∈ Rn×n :


∏ni=ν γi , ν = 2, · · · , n, det ξ ∈ [α, β]

and we wish to show that

Y = PcoEα,β = QcoEα,β = Rco Eα,β .

Since we always have Rco Eα,β ⊂ QcoEα,β ⊂ PcoEα,β , it is sufficient to showthat PcoEα,β ⊂ Y and then Y ⊂ Rco Eα,β .

Step 1: PcoEα,β ⊂ Y. This is the easy inclusion. Indeed observe thatEα,β ⊂ Y and that the functions

ξ → ± det ξ, ξ →n∏

i=ν

λi(ξ), ν = 2, · · · , n,

are polyconvex (see Example 5.41). We therefore have that the set Y is poly-convex and thus the desired inclusion.

Step 2: Y ⊂ RcoEα,β . Since the set Y is compact (the function ξ → λn(ξ)being a norm), it is enough (see Lemma 7.42) to show that ∂Y ⊂ Rco Eα,β . Sowe let ξ ∈ ∂Y and we wish to prove that ξ ∈ RcoEα,β . Note that

∂Y = Yα ∪ Yβ ∪ Y2 ∪ · · · ∪ Yn


whereYα := ξ ∈ Y : det ξ = α , Yβ := ξ ∈ Y : det ξ = β ,

Yν := ξ ∈ Y :∏n

i=ν λi(ξ) =∏n

i=ν γi , ν = 2, · · · , n.

Since all the functions involved in the definition of Y are right and left SO(n)invariant, there is no loss of generality (see Theorem 13.3) in assuming that ξis diagonal

ξ = diag(x1, x2, · · · , xn),

with 0 ≤ |x1| ≤ x2 ≤ · · · ≤ xn . We therefore have

λ1(ξ) = |x1| , λi(ξ) = xi , i = 2, · · · , n.

We now proceed by induction on the dimension n; when n = 1 the result istrivial. Several possibilities can then happen, bearing in mind that ξ ∈ ∂Y.

Case 1: ξ ∈ Yν for a certain ν = 2, · · · , n, meaning that

n∏

i=ν

xi =

n∏

i=ν

γi .

We write ξ ∈ Rn×n as two blocks, one in R(ν−1)×(ν−1) and one inR(n−ν+1)×(n−ν+1) in the following way:

ξ = diag(ξν−1, ξn−ν+1)

where

ξν−1 = diag(x1, · · · , xν−1) and ξn−ν+1 = diag(xν , · · · , xn).

We then apply the hypothesis of induction on ξν−1 and ξn−ν+1 (we will checkthat we can do so below) and we deduce that ξ ∈ RcoEα,β . Let us now see thatwe can apply the hypothesis of induction first for ξν−1 . We have (when ν = 2

or ν = n, terms such as∏ν−1

i=2 or∏n

i=ν+1 should be replaced by 1)

γ2

∏ν−1i=2 γi = γ2

∏ni=2 γi (

∏ni=ν γi)

−1 ≥ max |α|γν · · · γn

,|β|

γν · · ·γn,

det ξν−1 =∏ν−1

i=1 xi =∏n

i=1 xi (∏n

i=ν xi)−1

=∏n

i=1 xi (∏n

i=ν γi)−1

= det ξ (∏n

i=ν γi)−1 ∈

[α

γν · · · γn,

β

γν · · ·γn

],

∏ν−1i=ν λi(ξν−1) =

∏ni=ν xi (

∏ni=ν xi)

−1=∏n

i=ν xi (∏n

i=ν γi)−1

≤ ∏ν−1i=ν γi , ν = 2, · · · , ν − 1

and thus the result.

Examples 353

Similarly for ξn−ν+1 since (here the roles for both α and β are played by∏ni=νγi)

γν+1

n∏

i=ν+1

γi ≥n∏

i=ν

γi ,

det ξn−ν+1 =

n∏

i=ν

xi =

n∏

i=ν

γi ,

n−ν+1∏

i=ν−ν+1

λi(ξn−ν+1) =n∏

i=ν

xi ≤n∏

i=ν

γi , ν = ν, · · · , n

we have the claim.

Case 2: ξ ∈ Yα (and similarly for the case ξ ∈ Yβ). We can also assume thatξ /∈ Yν , ν = 2, · · · , n, otherwise we apply Case 1. We therefore have

ξ ∈ intYα =η ∈ Rn×n : det η = α,

∏ni=ν λi(η) <

∏ni=ν γi , ν = 2, · · · , n

.

This is clearly an open set (relative to the manifold det η = α). Recall that

ξ = diag(x1, · · · , xn) =

⎛⎜⎝

x1 · · · 0...

. . ....

0 · · · xn

⎞⎟⎠ .

We then set for t ∈ R

ξt :=

⎛⎜⎜⎜⎝

x1 · · · 0 0...

. . ....

...0 · · · xn−1 t0 · · · 0 xn

⎞⎟⎟⎟⎠

and observe that det ξt = det ξ = α. Since intYα is bounded we can find (as inLemma 7.42) t1 < 0 < t2 so that ξt1 , ξt2 ∈ ∂Yα which means that ξti ∈ Yνi , i =1, 2, for a certain νi = 2, · · · , n and therefore, by Case 1, ξti ∈ Rco Eα,β andthus, since rankξt1 − ξt2 = 1 and

ξ =t2

t2 − t1ξt1 +

−t1t2 − t1

ξt2 ,

we deduce that ξ ∈ Rco Eα,β as wished.

(iii) Representation formula for intRcoEα,β . Let

Z :=ξ ∈ Rn×n :

∏ni=ν λi(ξ) <

∏ni=ν γi , ν = 2, · · · , n, det ξ ∈ (α, β)

.

We wish to show that intRco Eα,β = Z.

The inclusion Z ⊂ intRcoEα,β is clear, since by continuity Z is open andby the representation formula for RcoEα,β we have Z ⊂ RcoEα,β .


We now prove the reverse inclusion intRcoEα,β ⊂ Z. So let ξ ∈ intRcoEα,β .We can find R , Q ∈ SO (n) (see Theorem 13.3) so that

ξ = R diag(±λ1(ξ), · · · , λn(ξ))Q.

Since ξ ∈ intRcoEα,β , we can find ǫ sufficiently small so that B2ǫ(ξ) ⊂ RcoEα,β

(where B2ǫ(ξ) denotes the ball centered at ξ and of radius 2ǫ).


Case 1: λν (ξ) = 0 for every ν ∈ 1, · · · , n . Define

η+ := R diag (±λ1(ξ), · · · , λn−1(ξ), λn(ξ) + ǫ)Q

η− := R diag(±λ1(ξ)− ǫ, λ2(ξ), · · · , λn(ξ))Q.

Since |η± − ξ| = ǫ < 2ǫ, then η± ∈ Rco Eα,β , meaning that

det η± ∈ [α, β] ,

n∏

i=ν

λi(η±) ≤n∏

i=ν

γi , ν = 2, · · · , n.

This clearly implies that

det ξ ∈ (α, β) ,

n∏

i=ν

λi(ξ) <

n∏

i=ν

γi , ν = 2, · · · , n

which just means that ξ ∈ Z, as wished.

Case 2: λν(ξ) = 0 and λν+1(ξ) > 0 for a certain ν ∈ 1, · · · , n (if ν = n,this means that ξ = 0). Letting δ = ǫ/

√ν + 1, we define

η± := R diag (±δ, δ, · · · , δ, λν+1(ξ), · · · , λn−1(ξ), λn(ξ) + δ) Q.

We therefore have |η± − ξ| = ǫ < 2ǫ and thus η± ∈ RcoEα,β , meaning that

det η± ∈ [α, β] ,n∏

i=ν

λi(η±) ≤n∏

i=ν

γi , ν = 2, · · · , n.

This obviously implies that

det ξ ∈ (α, β) ,

n∏

i=ν

λi(ξ) <

n∏

i=ν

γi , ν = 2, · · · , n

and hence ξ ∈ Z, as wished.

(iv) Representation formula for intRcoEα . This is proved exactly as above.We assume that α > 0, the case α < 0 being handled similarly. We let

K :=ξ ∈ Rn×n :

∏ni=ν λi(ξ) <

∏ni=ν γi , ν = 2, · · · , n, det ξ = α

.

Examples 355

As above the inclusion K ⊂ intRco Eα is obvious. Let us show the second oneand therefore let ξ ∈ intRco Eα and find R , Q ∈ SO (n) such that

ξ = R diag(λ1(ξ), · · · , λn(ξ))Q.

Since ξ ∈ intRco Eα , we can find ǫ sufficiently small so that Bǫ(ξ) ⊂ RcoEα

(where Bǫ(ξ) denotes the ball, restricted to the manifold det ξ = α, centered atξ and of radius ǫ). Define next, for δ > 0 sufficiently small,

η := R diag(λ1(ξ)

1 + δ, · · · , λn−1(ξ)

1 + δ, (1 + δ)n−1 λn(ξ))Q,

so that η ∈ Bǫ(ξ) ⊂ RcoEα and thus

det η = α,n∏

i=ν

λi(η) ≤n∏

i=ν

γi , ν = 2, · · · , n.

This clearly shows that ξ ∈ K, as wished.

7.4.2 The case of potential wells

We now give a representation formula for Rco E where

E := SO(2)A ∪ SO(2)B

and detA, det B > 0.Up to rotation and dilation, we can assume without loss of generality that

A =

(a1 00 a2

)and B =

(b1 00 b2

)

with 0 < a1, a2, b1, b2 and b1a1≤ b2

a2; and we assume throughout this section that

A and B have this particular form.

We denote the elements of SO(2) by Rθ , i.e.,

Rθ =

(cos θ sin θ− sin θ cos θ

).

The following result was established by Sverak [554].

Theorem 7.44 LetE := SO(2)A ∪ SO(2)B,

then

coE =

ξ ∈ R2×2 :

ξ = αRaA + βRbB, Ra , Rb ∈ SO(2),

0 ≤ α, β, α + β ≤ 1

.

Furthermore, if det (RθA−B) = 0 for a certain Rθ ∈ SO(2), the followingresults hold.


Case 1. If detB = detA > 0, then

PcoE = QcoE = RcoE

=

ξ ∈ R2×2 :


0 ≤ α, β, α + β ≤ 1 and det ξ = detA = detB

.

Case 2. If detB > detA > 0, then

PcoE = QcoE = RcoE

=

ξ ∈ R2×2 :


0 ≤ α ≤ detB−det ξdetB−det A , 0 ≤ β ≤ det ξ−det A

det B−detA

.

Moreover, in this last case, the interior of Rco E is given by the same formulawith strict inequalities on the right hand side.

Remark 7.45 (i) If the wells are not rank one connected meaning that thereexists no Rθ ∈ SO(2) such that det (RθA−B) = 0, then it will be obvious fromthe proof that in this case E = Rco E. This connection of the wells for A, B asabove is equivalent to

b1

a1≤ 1 ≤ b2

a2.

(ii) When 0 < b1 < a1 ≤ a2 < b2 , then (see Corollary 8.3 in Dacorogna-Marcellini [202]) matrices of the form

Aδ =

(a1 − δ 0

0 a2 + Tδ

)

are in intRco E for every δ > 0 sufficiently small and for T satisfying

a2 (b2 − a2) (a1 + b1)

a1 (b2 + a2) (a1 − b1)< T <

b1

(b22 − a2

2

)

b2 (a21 − b2

1).

In a similar manner, for appropriate S > 0, matrices of the form

Bδ =

(b1 + Sδ 0

0 b2 − δ

)

are in intRco E for every δ > 0 sufficiently small. ♦

Proof. We start with the following obvious observation. For every α, β ≥ 0,Ra, Rb ∈ SO(2), there exist γ ≥ 0 and Rc ∈ SO(2) such that

αRa + βRb = γRc with γ ≤ α + β. (7.13)

Indeed, just choose

γ2 = (α cos a + β cos b)2 + (α sin a + β sin b)2

Examples 357

and

cos c =α cos a + β cos b

γ, sin c =

α sin a + β sin b

γ.

Part 1. Formula for coE. We let

X :=ξ ∈ R2×2 : ξ = αRaA + βRbB, Ra , Rb ∈ SO(2), 0 ≤ α, β, α + β ≤ 1

.

We will prove that X = co E in two steps.

Step 1. We first show that coE ⊂ X. Since E ⊂ X, we will have the claimedinclusion if we can show that X is convex. So let ξ1, ξ2 ∈ X and t ∈ [0, 1] , then

tξ1 + (1− t) ξ2 = t (α1Ra1A + β1Rb1B) + (1− t) (α2Ra2A + β2Rb2B) .

Using (7.13), we obtain that

tξ1 + (1− t) ξ2 = αRaA + βRbB,

0 ≤ α ≤ tα1 + (1− t)α2 , 0 ≤ β ≤ tβ1 + (1− t)β2 .

Henceα + β ≤ t (α1 + β1) + (1− t) (α2 + β2) ≤ 1

and thus X is convex.

Step 2. We now prove that X ⊂ co E. So let ξ ∈ X then

ξ = αRaA + βRbB

=1 + α + β

2

[2α

1 + α + βRaA +

1 + β − α

1 + α + βRbB

]+

1− α− β

2Rb+πB.

Note thatRaA, RbB, Rb+πB ∈ E

and hence ξ ∈ co E. This achieves the proof of this part.

Part 2. Formula for PcoE, QcoE and RcoE. We first observe that up torotations and dilations (replacing E by EA−1 and B by BA−1) we can furtherrestrict ourselves to considering

A = I =

(1 00 1

)and B =

(λ 00 μ

), with λ ≥ μ ≥ 0.

The fact that the wells are rank one connected implies that λ ≥ 1 ≥ μ; however,if we want the problem to be non-trivial, we also assume that λ > μ.

We let

Y :=

ξ ∈ R2×2 :

ξ = αRa + βRbB, Ra , Rb ∈ SO(2),

0 ≤ α, β, α + β ≤ 1 and det ξ = 1


and

Z :=

ξ ∈ R2×2 :


0 ≤ α ≤ det B−det ξdetB−1 , 0 ≤ β ≤ det ξ−1

detB−1

.

We have to prove that when detB = 1, then

Y = PcoE = QcoE = Rco E

and similarly when detB = 1, then

Z = PcoE = QcoE = RcoE.

Since we always have Rco E ⊂ QcoE ⊂ PcoE, we only need to show thatY ⊂ Rco E and PcoE ⊂ Y and similarly for Z. This is achieved through thefollowing steps.

Step1: PcoE ⊂ Y and PcoE ⊂ Z. We clearly have respectively E ⊂ Y andE ⊂ Z; so if we can show that both sets are polyconvex, we will have the desiredinclusions. Let ξi ∈ Y (respectively Z), t ∈ Λ6 be such that

ξ :=∑6

i=1 tiξi and∑6

i=1 ti det ξi = det(∑6

i=1 tiξi),

whereΛ6 :=

λ = (λ1, · · · , λ6) : λi ≥ 0,

∑6i=1 λi = 1

.

We have to show that ξ ∈ Y (respectively Z). We therefore have by iterating(7.13) that

ξ =∑6

i=1 tiξi =∑6

i=1 tiαiRai + (∑6

i=1 tiβiRbi)B

= αRa + βRbB

with

0 ≤ α ≤6∑

i=1

tiαi and 0 ≤ β ≤6∑

i=1

tiβi .

Using the fact that

det ξ =

6∑

i=1

ti det ξi

we deduce that ξ ∈ Y (respectively Z).

Step 2. We next establish a decomposition of matrices that keeps the deter-minant fixed and allows movements in rank one directions. Namely let

ξ = αRθ + βB = α


)+ β

(λ 00 μ

)

and assume that det ξ > 0. We can then find s and ϕ such that

ξ =√

det ξ

(cosϕ sin ϕ− sinϕ cosϕ

)+ s

(1 + sin ϕ − cosϕ

cosϕ −1 + sin ϕ

). (7.14)

Examples 359

Indeed we have to solve

α cos θ + βλ =√

det ξ cosϕ + s (1 + sinϕ)

α cos θ + βμ =√

det ξ cosϕ + s (−1 + sin ϕ)

α sin θ =√

det ξ sin ϕ− s cosϕ.

We thus choose

s =β

2(λ− μ)

(note that if β > 0 then s > 0 since λ > μ) and then solve

s sin ϕ +√

det ξ cosϕ = α cos θ + β2 (λ + μ)√

det ξ sin ϕ− s cosϕ = α sin θ.(7.15)

Observe that this system is indeed solvable since taking the square of each sideof each equation, summing them and using the fact that

det ξ = α2 + β2λμ + αβ (λ + μ) cos θ

we get that they are compatible. This therefore leads to

(det ξ + s2

)sin ϕ = αs cos θ +

βs

2(λ + μ) + α

√det ξ sin θ

(det ξ + s2

)cosϕ = α

√det ξ cos θ +

β

2

√det ξ (λ + μ)− αs sin θ.

A similar decomposition can be made for

ξ = αI + βRθB =

(α + βλ cos θ βμ sin θ−βλ sin θ α + βμ cos θ

). (7.16)

Step 3: Y ⊂ Rco E. Let us first observe that if

Y1 :=

ξ ∈ R2×2 :

ξ = αRa + (1− α)RbB, Ra , Rb ∈ SO(2),

0 ≤ α ≤ 1 and det ξ = 1

,

then Y1 = R1 coE, where we recall that

R1 coE =

ξ ∈ R2×2 :

ξ = tξ1 + (1− t) ξ2

ξ1, ξ2 ∈ E, det (ξ1 − ξ2) = 0

=

ξ ∈ R2×2 :

ξ = tRa + (1− t)RbB, Ra , Rb ∈ SO(2),

0 ≤ t ≤ 1 and det (Ra −RbB) = 0

.

It is clear that the two sets are equal since no nontrivial rank one connectioncan be achieved in any of the two wells and

det (RaI −RbB) = 0 ⇔ det ξ = t det I + (1− t) detB = 1.


We next show that Y ⊂ R2 coE and thus the claim Y ⊂ RcoE. We first shouldobserve that on the manifold det ξ = 1 we have that ∂Y = R1 co E. So letξ ∈ Y, then, in view of the previous observation, we can assume that ξ ∈ intY(understood as relative to the manifold det ξ = 1); otherwise the result is alreadyproved. Furthermore up to a rotation we can assume that ξ1

2 + ξ21 = 0. Upon

multiplication, if necessary, by −I, we can therefore assume, using (7.14), thatthere exist α, β > 0, ϕ and s > 0 such that

ξ = αRθ + βB = α


)+ β

(λ 00 μ

)

=


)+ s


cosϕ −1 + sin ϕ

).

Set

ξt =


)+ t


cosϕ −1 + sin ϕ

).

Observe that ξs = ξ, det ξt ≡ 1 and ξ0 ∈ SO (2) . Note also that since ξ ∈ intYand Y is compact we can find t > s such that ξt ∈ ∂Y. Therefore

ξs = ξ =(1− s

t

)ξ0 +

s

tξt .

Since ξ0 ∈ E, ξt ∈ R1 coE and det (ξ0 − ξt) = 0, we deduce that ξ ∈ R2 coE,which is the desired result.

Step 4: Z ⊂ Rco E. We show exactly as in Step 3 that

R1 co E =

ξ ∈ R2×2 :

ξ = αRa + (1− α)RbB,

Ra , Rb ∈ SO(2), α = det B−det ξdetB−1

.

Let ξ ∈ Z, since Z is compact, we have that any line containing ξ will intersect∂Z and thus we can write

⎧⎪⎨⎪⎩

ξ = tξ1 + (1− t) ξ2

t ∈ [0, 1] , ξ1, ξ2 ∈ ∂Z

det (ξ1 − ξ2) = 0.

It is therefore sufficient to prove that ∂Z ⊂ RcoE. We can write

∂Z = Z1 ∪ Z2 ,

where

Z1 : =

ξ ∈ R2×2 :


α = det B−det ξdetB−1 , 0 ≤ β ≤ det ξ−1

det B−1

,

Z2 : =

ξ ∈ R2×2 :


0 ≤ α ≤ detB−det ξdetB−1 , β = det ξ−1

det B−1

.

Examples 361

We only prove that Z2 ⊂ RcoE (the other inclusion being handled similarlybut using (7.16)). Note that (on the manifold β = det ξ−1

detB−1 ) ∂Z2 = R1 coE. Solet ξ ∈ Z2 ; in view of the preceding observation we can therefore assume thatξ ∈ intZ2 (relative to the manifold β = det ξ−1

det B−1 ). Up to a rotation, we canassume as in Step 3 (using (7.14)) that

ξ = αRθ + βB =√

det ξ

(cosϕ sin ϕ− sin ϕ cosϕ

)+ s

(1 + sinϕ − cosϕ

cosϕ −1 + sin ϕ

),

where

β =det ξ − 1

detB − 1.

If we then denote

ξt =√

det ξ


)+ t


cosϕ −1 + sinϕ

)

we have that

det ξt ≡ det ξ, ξs = ξ and det (ξt1 − ξt2) = 0, ∀ t1, t2 ∈ R.

Since Z2 is compact and ξ ∈ intZ2 , we have that there exist (as in Lemma 7.42)t1 < s < t2 such that ξt1 , ξt2 ∈ ∂Z2 = R1 coE. We can hence write

ξ = ξs =t2 − s

t2 − t1ξt1 +

s− t1t2 − t1

ξt2

det (ξt1 − ξt2) = 0, ξt1 , ξt2 ∈ R1 coE.

Thus ξ ∈ RcoE and the proof is complete.

Part 3. Formula for intRco E. It is clear by continuity that if for ξ ∈ RcoEstrict inequalities hold then ξ ∈ intRcoE. We now show the converse. Assumefor the sake of contradiction that ξ ∈ intRco E and that one of the inequality isactually an equality. Without loss of generality we assume that it is the secondone, more precisely


0 ≤ α ≤ det B − det ξ

detB − 1, β =

det ξ − 1

det B − 1.

Since ξ ∈ intRco E, we deduce that for t small enough

ξt = ξ + tRθ ∈ Rco E, ∀Rθ ∈ SO (2)

and hence by the representation formula for RcoE we deduce that

β =det ξ − 1

detB − 1≤ det ξt − 1

detB − 1⇒ det ξ ≤ det ξt , ∀ t small enough.


Therefore we get

〈ξ; Rθ〉 = 0, ∀Rθ ∈ SO (2) ⇒ ξ =

(σ ττ −σ

)

and hence det ξ ≤ 0, which is in contradiction with the fact that det ξ ≥ 1.

7.4.3 The case of a quasiaffine function

We need, prior to the main theorem, an elementary lemma, but we postponeits proof to the end of the present subsection. It will be used to assert thatcondition (7.17) below can be fulfilled by some ci

j > 0 and will also be used inTheorem 10.29.

Lemma 7.46 Let Φ : RN×n → R be a non-constant quasiaffine function andM, m > 0. Then there exist ci

j > m, i = 1, · · · , N, j = 1, · · · , n such that

inf|Φ(ξ)| :∣∣ξi

j

∣∣ = cij > M.

We can now state the main theorem, established by Dacorogna-Ribeiro [212].

Theorem 7.47 Let Φ : RN×n → R be a non-constant quasiaffine function,α < β, ci

j > 0 ,satisfying

inf|Φ(ξ)| :∣∣ξi

j

∣∣ = cij > max|α| , |β|. (7.17)

Let

E :=ξ ∈ RN×n : Φ(ξ) ∈ α, β,

∣∣ξij

∣∣ ≤ cij , i = 1, · · · , N, j = 1, · · · , n

.

Then

RcoE =ξ ∈ RN×n : Φ(ξ) ∈ [α, β],

∣∣ξij

∣∣ ≤ cij , i = 1, · · · , N, j = 1, · · · , n

,

intRcoE =ξ ∈ RN×n : Φ(ξ) ∈ (α, β),

∣∣ξij

∣∣ < cij , i = 1, · · · , N, j = 1, · · · , n

.

Proof. Part 1. We let

X :=ξ ∈ RN×n : Φ(ξ) ∈ [α, β],

∣∣ξij

∣∣ ≤ cij , i = 1, · · · , N, j = 1, · · · , n

and we show that X = Rco E. The inclusion Rco E ⊂ X follows from thecombination of the facts that E ⊂ X and that the set X is rank one convex(the functions Φ, −Φ and |·| being rank one convex).

We therefore have to show only that X ⊂ Rco E. So we let ξ ∈ X and wecan assume that α < Φ(ξ) < β otherwise the result is trivial. We observe that

Examples 363

(7.17) implies that for every ξ ∈ X there exists (i, j) so that∣∣ξi

j

∣∣ < cij . So let

for t ∈ Rξt := ξ + tei ⊗ ej

where ei (respectively ej) is the i th (respectively j th) vector of the canonicalbasis of RN (respectively Rn). Observe that by compactness there exist t1 <0 < t2 so that ξtν ∈ ∂X, ν = 1, 2 which implies that either

Φ(ξtν ) ∈ α, β or∣∣(ξtν )i

j

∣∣ = cij , ν = 1, 2.

If the first possibility happens then we are done, if however the second caseholds then we restart the process with a different (i, j), since it is not possibleby (7.17) that

∣∣(ξtν )ij

∣∣ = cij for every (i, j).

Part 2. We now define

Y :=ξ ∈ RN×n : Φ(ξ) ∈ (α, β),

∣∣ξij

∣∣ < cij , i = 1, · · · , N, j = 1, · · · , n

and observe that since Y ⊂ Rco E and Y is open, then Y ⊂ intRco E. So let usshow the reverse inclusion and choose ξ ∈ intRco E. Clearly such a ξ must have∣∣ξi

j

∣∣ < cij . Corollary 5.23 shows also that ξ should be so that α < Φ(ξ) < β.

These observations imply the result.

We now prove Lemma 7.46.

Proof. Since Φ is quasiaffine, we can write

Φ(ξ) = Φ(0) +∑

1≤q≤N∧n

∑

1≤i1<···<iq≤N1≤j1<···<jq≤n

μi1···iq

j1···jqdet

⎛⎜⎜⎝

ξi1j1

· · · ξi1jq

.... . .

...

ξiq

j1· · · ξ

iq

jq

⎞⎟⎟⎠ .

Since Φ is not constant we can find 1 ≤ s ≤ N ∧ n, 1 ≤ i1 < · · · < is ≤ N and1 ≤ j1 < · · · < js ≤ n so that

μi1···is

j1···js= 0 and μ

i1···iq

j1···jq= 0, ∀ q > s.

Assume, for notational convenience (the general case being handled similarly),that

μ1···s1···s = 0. (7.18)

Let us define the set

Θ =θ ∈ RN×n : θi

j ∈ ±1

and the product A⊙B ∈ RN×n, for two given matrices A, B ∈ RN×n, as

(A⊙B)ij := Ai

j ·Bij .


We want to find a matrix C ∈ RN×n such that its entries satisfy cij > m and

ξ := C ⊙ θ, θ ∈ Θ ⇒ |Φ(ξ)| > M.

In fact we prove that the matrix can be chosen of the form C = τA where τ > 0and for t > 0

Aij :=

t if 1 ≤ i = j ≤ s,

1 if i = j or if i = j ≥ s + 1.

We observe that

Φ(ξ) = Φ(C ⊙ θ)

= Φ(0)+∑

1≤q≤s

τq∑

1≤i1<···<iq≤N1≤j1<···<jq≤n

μi1···iq

j1···jqdet

⎛⎜⎜⎝

Ai1j1

θi1j1

· · · Ai1jq

θi1jq

.... . .

...

Aiq

j1θ

iq

j1· · · A

iq

jqθ

iq

jq

⎞⎟⎟⎠

and that for τ and t sufficiently large it is possible to find γ > 0 so that

|Φ(ξ)| ≥ γτsts.

So choosing τ and t sufficiently large we have indeed found cij > m and |Φ(ξ)| >

M as wished.

7.4.4 A problem of optimal design

Recall that the set of 2× 2 symmetric matrices is denoted by R2×2s and that to

every ξ ∈ R2×2 we associate ξ ∈ R2×2 in the following way

ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

), ξ =

(ξ22 −ξ2

1

−ξ12 ξ1

1

).

Our algebraic result, due to Dacorogna-Tanteri [215], is as follows.

Theorem 7.48 Let

E :=ξ ∈ R2×2

s : trace ξ ∈ 0, 1 , det ξ ≥ 0

,

thenRcoE = coE =

ξ ∈ R2×2

s : 0 ≤ trace ξ ≤ 1, det ξ ≥ 0

,

intRcoE =ξ ∈ R2×2

s : 0 < trace ξ < 1, det ξ > 0

,

where the interior is understood as relative to R2×2s .

Remark 7.49 Note that it is slightly surprising that the rank one convex hullis in fact convex since the function ξ → det ξ is not convex. ♦

Examples 365

Proof. We call

X :=ξ ∈ R2×2

s : 0 ≤ trace ξ ≤ 1, det ξ ≥ 0

,

Y :=ξ ∈ R2×2

s : 0 < trace ξ < 1, det ξ > 0

.

Step 1. We first prove that

RcoE ⊂ coE ⊂ X.

The first inclusion always holds and the second one follows from the fact thatE ⊂ X and that X is convex. Indeed, let ξ, η ∈ X, 0 ≤ t ≤ 1; we wish to showthat tξ + (1− t) η ∈ X.

- It is clear that the first inequality in the definition of X holds since

ξ → trace ξ

is linear.

- We now show the second one. Observe first that since det ξ = ξ11ξ2

2−(ξ12

)2,

det η = η11η

22−(η12

)2 ≥ 0 and trace ξ, trace η ≥ 0, then ξ11 , ξ2

2 , η11 , η2

2 ≥ 0 and wetherefore have (assume below that ξ1

1 , η11 > 0 otherwise, under our assumptions,

the inequality below is trivial)

〈 ξ; η 〉 = ξ11η2

2 + η11ξ2

2 − 2ξ12η1

2

≥ ξ11

(η12

)2

η11

+ η11

(ξ12

)2

ξ11

− 2ξ12η1

2 =

(ξ11η1

2 − η11ξ

12

)2

ξ11η1

1

≥ 0.

We therefore deduce that

det (tξ + (1− t) η) = t2 det ξ + t (1− t) 〈 ξ; η 〉+ (1− t)2 det η ≥ 0.

Step 2. We now show that

X ⊂ RcoE.

Since X is compact, it is enough to prove (see Lemma 7.42) that ∂X ⊂ RcoE.However, it is easy to see that

∂X = E ∪ξ ∈ R2×2

s : 0 < trace ξ < 1, det ξ = 0

and therefore the proof will be complete once we show that the second set inthe right hand side is contained in Rco E. Assume that ξ is such that 0 < t =trace ξ < 1 and det ξ = 0. We can then write

ξ =

(x ±

√x (t− x)

±√

x (t− x) t− x

)= tξ1 + (1− t) ξ2

= t

(α ±

√α (1− α)

±√

α (1− α) 1− α

)+ (1− t)

(0 0

0 0

)


where x = tα. The result follows from the facts that ξ1, ξ2 ∈ E anddet (ξ1 − ξ2) = 0.

Step 3. Let us now show that Y = intRcoE. The inclusion Y ⊂ intRcoEfollows from the fact that Y is open and from the obvious inclusion Y ⊂ X =RcoE. We therefore only need to prove that intRcoE ⊂ Y. So let ξ ∈ intRcoE,we consider two cases (the second one will be shown to be impossible).

Case 1: ξ22 > 0. Find ǫ sufficiently small so that Bǫ(ξ) ⊂ RcoE (where Bǫ(ξ)

denotes the ball, restricted to R2×2s , centered at ξ and of radius ǫ). Define

η± :=

(ξ11 ± δ ξ1

2

ξ12 ξ2

2

),

where δ ∈ (0, ǫ) is chosen sufficiently small so that η± ∈ Bǫ(ξ) ⊂ Rco E = X.We thus have

0 ≤ trace η± ≤ 1 and det η± ≥ 0.

This immediately leads to

0 < trace ξ < 1 and det ξ > 0

which is the claim, namely ξ ∈ Y.

Case 2: ξ22 = 0. Since ξ ∈ intRco E ⊂ RcoE, we deduce that ξ1

2 = 0 andhence

ξ =

(ξ11 0

0 0

).

However, such a ξ cannot be in intRcoE since

(ξ11 δ

δ 0

)/∈ RcoE, ∀ δ = 0.


Chapter 8

Lower semi continuityand existence theoremsin the vectorial case

8.1 Introduction

We now consider the minimization problem

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where

- Ω ⊂ Rn is a bounded open set,

- u : Ω → RN and hence ∇u ∈ RN×n,

- u0 ∈W 1,p(Ω; RN

)is a given map,

- f : Ω× RN × RN×n → R, f = f (x, u, ξ) , is a Caratheodory function.

In Section 8.2, we obtain the main result of this chapter, showing that theintegral I is (sequentially) weakly lower semicontinuous, namely

lim infν→∞

I (uν) ≥ I (u)

for every sequence uν u in W 1,p if and only if

ξ → f (x, u, ξ) is quasiconvex.

For the clarity of exposition, we prove the result several times. First when thereis no dependence on lower order terms, meaning that f = f (ξ) . Then in thegeneral case, f = f (x, u, ξ) , first with p = ∞ and then when 1 ≤ p < ∞, whichis the hardest case.

368 Lower semi continuity and existence theorems in the vectorial case

In Section 8.3, we characterize completely the functions f that lead to inte-grals I that are weakly continuous, meaning that I and −I are weakly lowersemicontinuous. These turn out to be the quasiaffine functions. In particular,when N = n = 2, we have that if

uν u in W 1,p(Ω; R2

), p > 2,

thendet∇uν det∇u in Lp/2 (Ω) .

In Section 8.4, we see how to apply the above results to the existence ofminimizers for the above problem (P ).

In Section 8.5, we gather some important properties of the Jacobian deter-minants that we will use throughout the present chapter.

8.2 Weak lower semicontinuity

8.2.1 Necessary condition

The main theorem of this paragraph has already been proved (see Lemma 3.18of Chapter 3) and we now restate it.

Theorem 8.1 Let Ω be an open set of Rn, u : Ω ⊂ Rn → RN and

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function satisfying, for almost every x ∈ Ω and for every(u, ξ) ∈ RN × RN×n,

|f (x, u, ξ)| ≤ a (x) + b (u, ξ) ,


). Finally, let

I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx

and assume that there exists u0 ∈W 1,∞ (Ω; RN)

such that

|I (u0, Ω)| < ∞. (8.1)

If I is weak ∗ lower semicontinuous in W 1,∞ (Ω, RN), meaning that

lim infuν

∗u

(uν , Ω) ≥ I (u, Ω) ,

then ξ → f (x, u, ξ) is quasiconvex, i.e.

1

measD

∫

D

f (x0, u0, ξ0 +∇ϕ (x)) dx ≥ f (x0, u0, ξ0)

for almost every x0 ∈ Ω, every (u0, ξ0) ∈ RN × RN×n, for every bounded openset D ⊂ Rn and for every ϕ ∈W 1,∞

0 (D; Rn) .


Remark 8.2 It is clear that if I is weakly lower semicontinuous in W 1,p, thenI is weak ∗ lower semicontinuous in W 1,∞ and therefore the quasiconvexity off is also necessary for the weak lower semicontinuity of I in W 1,p. ♦

8.2.2 Lower semicontinuity for quasiconvex functionswithout lower order terms

We now turn our attention to the sufficiency of the quasiconvexity to obtain weaklower semicontinuity in W 1,p. We consider here only the case where f = f (ξ) ,the general case f = f (x, u, ξ) will be discussed in the next sections.

We first introduce a growth condition that should satisfy the function f.

Definition 8.3 Let f : RN×n → R and 1 ≤ p ≤ ∞. Then f is said to satisfygrowth condition (Cp) if

(1) when p = ∞

(C∞) |f (ξ)| ≤ η (|ξ|) for every ξ ∈ RN×n,

where η is a continuous and increasing function;

(2) when 1 < p <∞

(Cp) − α (1 + |ξ|q) ≤ f (ξ) ≤ α (1 + |ξ|p) for every ξ ∈ RN×n,

where α ≥ 0, 1 ≤ q < p;

(3) when p = 1

(C1) |f (ξ)| ≤ α (1 + |ξ|) for every ξ ∈ RN×n,

where α ≥ 0.

We may now state the theorem.

Theorem 8.4 Let f : RN×n → R be quasiconvex and satisfying growth condi-tion (Cp). Let Ω ⊂ Rn be a bounded open set and

I (u, Ω) :=

∫

Ω

f (∇u (x)) dx.

Then I is weakly lower semicontinuous in W 1,p(Ω; RN

)(weak ∗ lower semicon-

tinuous if p = ∞), i.e.

lim infuνu

I (uν , Ω) ≥ I (u, Ω) .

Remark 8.5 (i) The above theorem is essentially due to Morrey [453], [455]and has been refined, notably by Meyers [442], Acerbi-Fusco [3] and Marcellini[423] and we will follow this last proof.


(ii) If f is convex instead of quasiconvex, then there exists ξ∗ ∈ RN×n suchthat

f (0) + 〈ξ∗; ξ〉 ≤ f (ξ) (8.2)

for every ξ ∈ RN×n. Therefore, in the convex case, we impose on f only theabove natural growth condition. As seen in Chapter 5 (in particular, see Sec-tion 5.3.8), there is no known equivalent to (8.2) for quasiconvex functions andtherefore one needs to impose conditions of the type (Cp) below and above.

(iii) One should also note that the condition (Cp) if 1 < p < +∞ is optimalin the sense that one cannot allow the lower bound in (Cp) to be of the form−α (1 + |ξ|p) with the same p as in the upper bound α (1 + |ξ|p) , as for the casep = 1. This is seen in the example below.

(iv) If f ≥ 0, the result remains valid even if Ω is unbounded (this is doneas in the proof of Theorem 3.23 or Remark 8.12 (iv)). ♦

Example 8.6 We give here an example that is essentially due to Tartar (seeBall-Murat [65]). Let N = n = p = 2 and

f (∇u) = det∇u,

then (C2) is satisfied only if q = p = 2. We show that if 0 < a < 1, Ω = (0, a)2

and

uν (x, y) =1√ν

(1− y)ν(sin νx, cos νx) ,

then

uν 0 = u in W 1,2(Ω; R2

),

while

lim supν→∞

∫

Ω

f (∇uν) dxdy < 0 =

∫

Ω

f (∇u) dxdy.

The fact that uν → 0 in L∞ (Ω; R2)

is obvious, while the convergence

∇uν 0 = ∇u in L2(Ω; R2×2

)

is easily obtained from the following computations. We have

∇uν =

( √ν (1− y)

νcos νx −√ν (1− y)

ν−1sin νx

−√ν (1− y)νsin νx −√ν (1− y)

ν−1cos νx

)

and therefore

‖∇uν‖2L2 =

∫ a

0

∫ a

0

ν[ (1− y)2ν + (1− y)2ν−2 ]dxdy

= aν[1

2ν + 1+

1

2ν − 1− (1− a)

2ν+1

2ν + 1− (1− a)

2ν−1

2ν − 1] < 2a


if ν ≥ 1. We thus deduce that, up to a subsequence (although in the presentcase, we do not need to restrict to a subsequence), ∇uν 0 in L2

(Ω; R2×2

).

However,

∫

Ω

f (∇uν (x, y)) dxdy =

∫ a

0

∫ a

0

(−ν (1− y)2ν−1

)dxdy

= −νa[1

2ν− (1− a)

2ν

2ν]

and therefore the lower semicontinuity inequality does not hold since

lim supν→∞

∫

Ω

f (∇uν) = −a

2<

∫

Ω

f (∇u) = 0. ♦

We now continue with the proof of Theorem 8.4. But since the proof is long,we prefer to prove the main step in a separate lemma.

Lemma 8.7 Let D ⊂ Rn be a cube parallel to the axes, ξ ∈ RN×n and f :RN×n → R be quasiconvex and satisfying growth condition (Cp). Let 1 ≤ p ≤ ∞and

vν 0 in W 1,p(D; RN

)

(vν∗ 0 in W 1,∞ (D; RN

)if p =∞). Then

lim infν→∞

∫

D

f (ξ +∇vν (x)) dx ≥ f (ξ)measD. (8.3)

Proof. (Lemma 8.7). To infer (8.3) from the quasiconvexity of f, the onlyproblem is to change vν slightly in order to have vν = 0 on ∂D. This is classicalin the calculus of variations, see Chapter 5, Acerbi-Fusco [3], Marcellini [423],Meyers [442], Morrey [455], [453] and others. However since the cases p = ∞and p = 1 are simpler and more natural, we will start with those. The moresophisticated case 1 0 andwe claim that we can find δ = δ (ǫ) > 0 (independent of ν) so that, for anymeasurable set E ⊂ Rn, we have

meas E ≤ δ ⇒∫

E

|∇vν (x)| dx ≤ ǫ. (8.4)

This is clear from the equiintegrability of the sequence ∇vν when p =∞ andp = 1.

We next construct an open set Dǫ ⊂ D and a function ηǫ ∈ C∞0 (D) such

that 0 ≤ ηǫ ≤ 1 in D and

meas[ D −Dǫ ] ≤ δ, ηǫ ≡ 1 on Dǫ and |∇ηǫ| ≤ c1/δ,


where c1 > 0 is a constant.We finally define a sequence of functions uν ∈ W 1,p

0

(D; RN

)by

uν (x) := ηǫ (x) vν (x) .

We then claim that we can find a constant c2 > 0 so that

lim supν→∞

∣∣∣∣∫

D−Dǫ

[ f (ξ +∇vν (x))− f (ξ +∇uν (x)) ]dx

∣∣∣∣ ≤ c2ǫ. (8.5)

Assume for a moment that we have proved this last inequality and let us con-clude the proof of the lemma. Since f is quasiconvex, uν ∈ W 1,p

0

(D; RN

), (8.5)

holds and∫

D

f (ξ +∇vν (x)) dx =

∫

D

f (ξ +∇uν (x)) dx

+

∫

D−Dǫ

[ f (ξ +∇vν (x))− f (ξ +∇uν (x)) ]dx

we obtain that

lim infν→∞

∫

D

f (ξ +∇vν (x)) dx ≥ f (ξ)measD − c2ǫ.

Letting ǫ→ 0, we have indeed obtained the lemma.It therefore remains to show (8.5). We separate the discussion in two cases.

Case 1: p = ∞. Since

∇uν = ηǫ∇vν +∇ηǫ ⊗ vν ,

‖∇vν‖L∞ is uniformly bounded and vν → 0 uniformly, we clearly have (8.5).

Case 2: p = 1. Use Proposition 2.32 to deduce that there exists a constantc3 > 0 such that

|f (ξ +∇vν (x))− f (ξ +∇uν (x))| ≤ c3 |∇vν (x) −∇uν (x)|= c3 |(1− ηǫ)∇vν −∇ηǫ ⊗ vν |≤ c3 |∇vν |+ c3 |∇ηǫ ⊗ vν |≤ c3 |∇vν |+

c3c1

δ|vν | .

Appealing to (8.4) and to the fact that vν → 0 in L1, we have indeed obtained(8.5).

Step 2: 1 < p < ∞. The above procedure is too simple in the presentcontext, because, contrary to the cases p = ∞ and p = 1, we now lack theequiintegrability of the sequence. In order to fix the boundary datum, we proceedas in Marcellini [423]. Let D0 ⊂⊂ D be a cube and let

R :=1

2dist

(D0, ∂D

). (8.6)


D0

D

2R

Figure 8.1: The sets D0, Dk, D

Let K be an integer and let D0 ⊂ Dk ⊂ D (see Figure 8.1), 1 ≤ k ≤ K, be suchthat

dist(D0; ∂Dk

)=

k

KR, 1 ≤ k ≤ K.

We then choose ϕk ∈ C∞ (D) , 1 ≤ k ≤ K, such that

0 ≤ ϕk ≤ 1,∣∣∇ϕk

∣∣ ≤ aK

R, ϕk (x) =

1 if x ∈ Dk−1

0 if x ∈ D −Dk,

where a > 0 is a constant. Let

vkν = ϕkvν ,

then vkν ∈W 1,p

0

(D; RN

). We may therefore use the quasiconvexity of f to get

∫

D

f (ξ) dx ≤∫

D

f(ξ +∇vk

ν (x))dx

=

∫

D−Dk

f (ξ) dx +

∫

Dk−Dk−1

f(ξ +∇vk

ν (x))dx

+

∫

Dk−1

f (ξ +∇vν (x)) dx.

We then deduce that

∫

Dk

f (ξ) dx ≤∫

Dk−Dk−1

f(ξ +∇vk

ν (x))dx +

∫

Dk−1

f (ξ +∇vν (x)) dx.


We may also rewrite the above inequality in the following way:

∫

Dk

f (ξ) dx ≤∫

D

f (ξ +∇vν (x)) dx−∫

D−Dk−1

f (ξ +∇vν (x)) dx

+

∫

Dk−Dk−1

f(ξ +∇vk

ν (x))dx

=

∫

D

f (ξ +∇vν (x)) dx + A1 + A2 .

(8.7)We now estimate A1 and A2 .

Estimation of A1 , where

A1 := −∫

D−Dk−1

f (ξ +∇vν (x)) dx.

We want to show that by choosing R sufficiently small (see (8.6)) we have

A1 ≤ ǫ (8.8)

uniformly in ν. Use (Cp) to get, α′ being a constant,

A1 ≤ α

∫

D−Dk−1

(1 + |ξ +∇vν |q) dx

≤ α′∫

D−D0

(1 + |ξ|q + |∇vν |q) dx.

Since q < p, we use Holder inequality to obtain

∫

D−D0

|∇vν |q dx ≤(∫

D−D0

|∇vν |p dx

)q/p

meas(D −D0

)(p−q)/p

and therefore, by choosing R sufficiently small, we have (8.8).

Estimation of A2 , where

A2 :=

∫

Dk−Dk−1

f(ξ +∇vk

ν (x))dx.

We have, using (Cp) and denoting by αi ≥ 0 constants that are independent ofK and ν,

A2 ≤ α

∫

Dk−Dk−1

( 1 +∣∣ξ +∇vk

ν

∣∣p )dx

≤ α1

∫

Dk−Dk−1

( 1 + |ξ|p +∣∣ϕk∇vν +∇ϕk ⊗ vν

∣∣p )dx

≤ α2

∫

Dk−Dk−1

( 1 + |ξ|p + |∇vν |p + (aK/R)p |vν |p )dx

where we have used the definition of ϕk in the last inequality.


Returning to (8.7), using (8.8) and the above estimate of A2 , we find, if wesum the left and right hand sides of (8.7) from k = 1 to K,

K

∫

D

f (ξ +∇vν (x)) dx− f (ξ)

K∑

k=1

meas Dk

≥ −Kǫ− α2

∫

DK−D0

(1 + |ξ|p + |∇vν |p + (aK/R)p |vν |p) dx.

Dividing the above inequality by K and letting ν → +∞ we get (recalling thatvν 0 in W 1,p

(D; RN

))

lim infν→∞

∫

D

f (ξ +∇vν (x)) dx− f (ξ)

K

K∑

k=1

measDk ≥ −ǫ− α3

K(8.9)

where α3 is a constant. Noting that

measD0 ≤ 1

K

K∑

k=1

meas Dk ≤ measD,

letting K → ∞ and taking into account the fact that ǫ and D0 are arbitrary(see (8.6)), we have indeed obtained from (8.9) that

lim infν→∞

∫

D

f (ξ +∇vν (x)) dx ≥ f (ξ) measD

which is the desired result.

We are now in a position to conclude the proof of Theorem 8.4.

Proof. We start by approximating Ω by a union of cubes parallel to theaxes in Rn and take the average of ∇u over each of these cubes (so that ∇uis constant on each cube). More precisely, we let δ > 0 and h be an integer.We approximate Ω by a union of cubes Ds parallel to the axes and whose edgelength is 1/h. We denote this union by Hh . We then choose h large enough sothat

meas (Ω−Hh) ≤ δ where Hh :=⋃

Ds .

We then take the average of ∇u over each of the Ds , namely

ξs :=1

measDs

∫

Ds

∇u (x) dx.

Now fix ǫ > 0. Choosing δ smaller if necessary, we have

(∑s

∫

Ds

|∇u (x)− ξs|p dx

)1/p

< ǫ. (8.10)

We recall thatuν u in W 1,p

(Ω; RN

)


(uν∗ u if p = +∞). We then consider

I (uν ; Ω)− I (u; Ω) =

∫

Ω

[ f (∇uν (x))− f (∇u (x)) ]dx

= J1 + J2 + J3 + J4 ,

(8.11)

where

J1 :=

∫

Ω−Hh

[ f (∇uν (x))− f (∇u (x)) ]dx,

J2 :=∑

s

∫

Ds

[ f (∇u + (∇uν −∇u))− f (ξs + (∇uν −∇u)) ]dx,

J3 :=∑

s

∫

Ds

[ f (ξs + (∇uν −∇u))− f (ξs) ]dx,

J4 :=∑

s

∫

Ds

[ f (ξs)− f (∇u) ]dx.

The difficult term to estimate is J3 and this has already been done in Lemma 8.7.We now estimate J1, J2 and J4 . In the sequel, we denote by αi ≥ 0 constantsthat are independent of δ and ν,

Estimation of J1 . We want to prove that if δ is chosen small enough, then

J1 ≥ −α1ǫ (8.12)

uniformly in ν.

Case 1. If p = +∞, then (8.12) is trivial since ‖∇uν‖L∞ is uniformlybounded and since (C∞) holds.

Case 2. If 1 < p < ∞, then use (Cp) to get

J1 ≥ −∫

Ω−Hh

[ α (1 + |∇uν |q) + f (∇u) ]dx

= −∫

Ω−Hh

(α + f (∇u))− α

∫

Ω−Hh

|∇uν |q .

Using Holder inequality and the fact that q < p we get

J1 ≥ −∫

Ω−Hh

(α + f (∇u))− α

(∫

Ω−Hh

|∇uν |p)q/p

meas (Ω−Hh)(p−q)/p .

Choosing δ small enough, we get (8.12).

Case 3. If p = 1, then by (C1) we have

J1 ≥ −∫

Ω−Hh

[ α (1 + |∇uν |) + f (∇u) ]dx.

Since ∇uν ∇u in L1, we may use the equiintegrability of ∇uν to get (8.12)immediately.


Estimation of J2 . Using Proposition 2.32, if 1 ≤ p < +∞, we find that thereexists a constant β > 0 such that

|J2| ≤ β∑

s

∫

Ds

( 1 + |∇uν |p−1+ |∇uν + ξs −∇u|p−1

) |∇u− ξs| dx.

Using Holder inequality, (8.10) and the fact that ‖∇uν‖Lp is uniformly bounded,we deduce that, for δ small enough,

|J2| ≤ α2ǫ (8.13)

uniformly in ν. If p = +∞, (8.13) is obtained in the same way using (C∞) andthe fact that ‖∇uν‖L∞ is uniformly bounded.

Estimation of J4 . This estimate is very similar but simpler than that of J2

and we skip the details. We also find that, for δ small enough,

|J4| ≤ α4ǫ. (8.14)

We now return to (8.11) gathering (8.12), (8.13) and (8.14). We thereforehave, for δ small enough,

I (uν; Ω)− I (u; Ω) ≥ − (α1 + α2 + α4) ǫ

+∑

s

∫

Ds

[ f (ξs +∇uν −∇u)− f (ξs) ]dx.

Taking the limit as ν →∞, we get

lim infν→∞

I (uν; Ω)− I (u; Ω)

≥ − (α1 + α2 + α4) ǫ +∑

s

lim infν→∞

∫

Ds

[ f (ξs + (∇uν −∇u))− f (ξs) ]dx.

(8.15)We now invoke Lemma 8.7 to get

lim infν→∞

∫

Ds

f (ξs + (∇uν −∇u)) dx ≥∫

Ds

f (ξs) dx.

Then combining (8.15), the above inequality and the fact that ǫ is arbitrary, wehave indeed obtained that I is weakly lower semicontinuous.

8.2.3 Lower semicontinuity for general quasiconvexfunctions for p =∞

We now turn our attention to general integrands of the type

I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx.

We start with the easiest case, where p = ∞. The strategy is to freeze the lowerorder terms and then apply Theorem 8.4.


Theorem 8.8 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundaryand let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function such that ξ → f (x, u, ξ) is quasiconvex, i.e.

∫

D

f (x0, u0, ξ0 +∇ϕ (x)) dx ≥ f (x0, u0, ξ0)measD

for every bounded open set D ⊂ Rn, for almost every x0 ∈ Ω, every (u0, ξ0) ∈RN × RN×n and for every ϕ ∈W 1,∞

0

(D; RN

).

Let

I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx

Assume that f satisfies

(C∞) |f (x, u, ξ)| ≤ β (x) + α (|u| , |ξ|)

where α, β ≥ 0, β ∈ L1 (Ω) and α is a continuous and increasing (in eachargument) function; then I is (sequentially) weak * lower semicontinuous inW 1,∞ (Ω; RN

).

Remark 8.9 (i) The theorem is due to Morrey [453], [455], under further hy-potheses, and has been refined by Meyers [442], Acerbi-Fusco [3] and Marcellini[423]. We follow here the proof in [3].

(ii) The result remains valid for any open set Ω (not necessarily bounded) ifwe assume that f ≥ 0; see Remark 8.12 (iv) for details. ♦


Step 1. We first prove that we can restrict our attention to sets Ω thatare finite unions of disjoint open cubes parallel to the axes and to functions fsatisfying

(C′∞) 0 ≤ f (x, u, ξ) ≤ β (x) + α (|u| , |ξ|) .

Indeed, since uν∗ u in W 1,∞, we can find γ ≥ 0 such that

‖uν‖W 1,∞ , ‖u‖W 1,∞ ≤ γ.

Then let

g (x, u, ξ) := f (x, u, ξ) + β (x) + k

where k = α (γ, γ) (α as in (C∞)). By hypothesis, we deduce that for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n with max |u| , |ξ| ≤ γ, then

g (x, u, ξ) ≥ 0.


We next choose Ωµ ⊂⊂ Ω to be a union of cubes parallel to the axes, and weobtain

∫

Ω

f (x, uν ,∇uν) dx =

∫

Ω

g (x, uν ,∇uν) dx−∫

Ω

[ β (x) + k ]dx

≥∫

Ωµ


Ω

[ β (x) + k ]dx.

We next apply the lower semicontinuity to Ωµ and to g ≥ 0 to get

lim infν→∞

∫

Ω

f (x, uν ,∇uν) dx ≥ lim infν→∞

∫

Ωµ


Ω

[ β (x) + k ]dx

≥∫

Ωµ

g (x, u,∇u)dx−∫

Ω

[ β (x) + k ]dx.

Then choosing a sequence of increasing Ωµ ⊂ Ω so that Ωµ ր Ω and applyingLebesgue monotone convergence theorem, we get the result.

From now on, we therefore assume, by working on each cube, that Ω itselfis a cube parallel to the axes and that f satisfies (C′

∞) .

Step 2. We now proceed in a very similar way to that of Theorem 8.4. Welet

‖uν‖W 1,∞ , ‖u‖W 1,∞ ≤ γ,

k := α (γ, γ) .

Let ǫ > 0, we can then find M = M (ǫ) so that if

Eǫ := x ∈ Ω : β (x) ≤ M

then

meas (Ω− Eǫ) <ǫ

2k,

∫

Ω−Eǫ

β (x) dx <ǫ

2

and, in particular,

M meas (Ω− Eǫ) <ǫ

2.

Appealing to Theorem 3.8, we can find a compact set Kǫ ⊂ Ω with

meas (Ω−Kǫ) ≤ ǫ/ (M + k)

and such that f : Kǫ × S → R is continuous, where

S :=(u, ξ) ∈ RN × RN×n : |u|+ |ξ| ≤ γ

.

We next decompose Ω, which is a cube parallel to the axes, in a finite unionof cubes Ds parallel to the axes and whose edge length is 1/h. We denote this


union by Hh . Choosing h proportional to the edge length of the cube Ω, wehave

meas (Ω−Hh) = 0, where Hh :=⋃

Ds .

We then take the average of u over each of the Ds , i.e.

us :=1

measDs

∫

Ds

u (x) dx

and we fix xs ∈ Ds ∩Kǫ ∩ Eǫ , if it is non-empty.We next make estimates of the two quantities

A :=

∫

Ω

f (x, u,∇u) dx, Aν :=

∫

Ω

f (x, uν ,∇uν) dx.

Estimation of A. We observe that

A1 :=

∫

Ω−Eǫ

f (x, u,∇u) dx ≤∫

Ω−Eǫ

[ α (γ, γ) + β (x) ]dx ≤ ǫ

A2 :=

∫

Eǫ−(Kǫ∩Eǫ)

f (x, u,∇u) dx ≤∫

Eǫ−(Kǫ∩Eǫ)

[ α (γ, γ) + β (x) ]dx

≤ [ k + M ] meas (Eǫ − (Kǫ ∩Eǫ)) ≤ [ k + M ] meas (Ω−Kǫ) ≤ ǫ.

Furthermore, since when letting h → ∞ we find that xs → x and us → u andsince f is uniformly continuous over Kǫ×S, we deduce that we can find h largeenough so that

A3 :=∑

s

∫

Ds∩Kǫ∩Eǫ

[ f (x, u,∇u)− f (xs, us,∇u) ]dx ≤ ǫ.

Combining these estimates, we have

A =

∫

Ω

f (x, u,∇u) dx =

∫

Eǫ

f (x, u,∇u) dx + A1

=

∫

Kǫ∩Eǫ

f (x, u,∇u) dx + A1 + A2

=∑

s

∫

Ds∩Kǫ∩Eǫ

f (xs, us,∇u) dx + A1 + A2 + A3

which leads, since f ≥ 0, to

A =

∫

Ω

f (x, u,∇u) dx ≤∑

s

∫

Ds

f (xs, us,∇u) dx + 3ǫ. (8.16)

Estimation of Aν . We observe that by choosing ν sufficiently large we have,since f is uniformly continuous over Kǫ × S and uν → u uniformly, that

A1ν :=

∫

Kǫ∩Eǫ

[ f (x, uν ,∇uν)− f (x, u,∇uν) ]dx ≥ −ǫ


and similarly, by choosing h sufficiently large, we get

A2ν :=

∑s

∫

Ds∩Kǫ∩Eǫ

[ f (x, u,∇uν)− f (xs, us,∇uν) ]dx ≥ −ǫ.

We next see that

A3ν := −

∑s

∫

Ds−(Ds∩Kǫ∩Eǫ)

f (xs, us,∇uν) dx

≥ −∑

s

∫

Ds−(Ds∩Kǫ∩Eǫ)

[ α (γ, γ) + β (xs) ]dx

≥ −[ k + M ] meas[ Ω− (Kǫ ∩Eǫ) ]

≥ −[ k + M ] meas[ Ω−Kǫ ]− [ k + M ] meas[ Ω− Eǫ ]≥ −2ǫ.

We now combine all the inequalities and the fact that f ≥ 0 to have

Aν =

∫

Ω

f (x, uν ,∇uν) dx ≥∫

Kǫ∩Eǫ

f (x, uν ,∇uν) dx

=

∫

Kǫ∩Eǫ

f (x, u,∇uν) dx + A1ν

=∑

s

∫

Ds∩Kǫ∩Eǫ

f (xs, us,∇uν) dx + A1ν + A2

ν

=∑

s

∫

Ds

f (xs, us,∇uν) dx + A1ν + A2

ν + A3ν

≥∑

s

∫

Ds

f (xs, us,∇uν) dx− 4ǫ.

Moreover, appealing to Theorem 8.4, we find that

lim infν→∞

∫

Ω

f (x, uν ,∇uν) dx ≥∑

s

∫

Ds

f (xs, us,∇u) dx− 4ǫ.

Combining the above estimate and (8.16), we find

lim infν→∞

∫

Ω


Ω

f (x, u,∇u) dx− 7ǫ.

Since ǫ is arbitrary, we have the claim.

8.2.4 Lower semicontinuity for general quasiconvexfunctions for 1 ≤ p < ∞

We now state the main theorem, first introducing a growth condition that shouldsatisfy the function f.

Definition 8.10 Let 1 ≤ p < ∞, Ω ⊂ Rn be an open set and let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,


be a Caratheodory function. The function f is said to satisfy growth condition(Cp) if for almost every x ∈ Ω and for every (u, ξ) ∈ RN×RN×n the inequalities

(Cp) − α (|u|r + |ξ|q)− β (x) ≤ f (x, u, ξ) ≤ g (x, u) (1 + |ξ|p)

hold, where α, β, g ≥ 0, β ∈ L1 (Ω) , 1 ≤ q < p, 1 ≤ r < np/ (n− p) if p < nand 1 ≤ r < ∞ if p ≥ n and

g : Ω× RN → R+ , g = g (x, u) ,

is a Caratheodory function. In the case p = 1 we assume that

(C1) |f (x, u, ξ)| ≤ α (1 + |ξ|)

where α ≥ 0.

We may now state the main theorem.

Theorem 8.11 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundaryand let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function such that ξ → f (x, u, ξ) is quasiconvex, i.e.

∫

D


for every bounded open set D ⊂ Rn, for almost every x0 ∈ Ω, every (u0, ξ0) ∈RN × RN×n and for every ϕ ∈W 1,∞

0

(D; RN

).

Let 1 ≤ p <∞ and assume that f satisfies (Cp) . Let

I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx,

then I is (sequentially) weakly lower semicontinuous in W 1,p(Ω; RN

).

Remark 8.12 (i) The above theorem is due to Morrey [453], [455], under fur-ther hypotheses, and has been refined by Meyers [442] and several authors sincethen, notably by Acerbi-Fusco [3] and Marcellini [423]. We will follow here theproof given in Ansini-Dacorogna [29], which is a combination of that of [3],[423] and a lemma on equiintegrability of Fonseca-Muller-Pedregal [288] andKristensen [379], see also Pedregal [492].

(ii) If one assumes that f is quasiconvex and satisfies

|f (x, u, ξ)| ≤ α (1 + |u|p + |ξ|p)

then Theorem 8.11, as well as Theorem 8.4, is proved in a much simpler way ifone wants to show that I is weakly lower semicontinuous in W 1,p+ǫ, where ǫ > 0,


instead of W 1,p. This observation is useful since for minimization problems it isoften possible to see that some minimizing sequences are bounded uniformly inW 1,p+ǫ instead of W 1,p (see, for example, Ekeland-Temam [264], Chapters IXand X, or Marcellini-Sbordone [428], [429]).

(iii) The condition (Cp) is optimal, in the sense that we cannot allow eitherp = q (when p > 1) according to Example 8.6 or r = p∗ = np/ (n− p) (when1 ≤ p < n) as the simple example given below shows. However when p = 1, wecan easily replace the hypothesis by

(C1) |f (x, u, ξ)| ≤ β (x) + α (|u|r + |ξ|)

where α, β ≥ 0, β ∈ L1 (Ω) and 1 ≤ r < n/ (n− 1) .

(iv) The result remains valid for any open set Ω (neither necessarily boundednor with a Lipschitz boundary) if we assume that f ≥ 0; see the proof belowfor details. ♦

We first prove Remark 8.12 (iv).

Proof. We here prove that we can restrict our attention to bounded Ω withsmooth boundary if f ≥ 0. Indeed choose Ωµ ⊂⊂ Ω with Lipschitz boundary(in particular Ωµ is bounded) and apply the lower semicontinuity to Ωµ to get

lim infν→∞

∫

Ω

f (x, uν ,∇uν) dx ≥ lim infν→∞

∫

Ωµ


Ωµ

f (x, u,∇u) dx.

Choosing then a sequence of increasing bounded open sets with smooth bound-ary Ωµ ⊂ Ω so that Ωµ ր Ω and applying Lebesgue monotone convergencetheorem, we get the result.

We now give an example showing that we cannot allow r = p∗ = np/ (n− p)in the theorem.

Example 8.13 Let 1 ≤ p < n and find a sequence uν such that

uν 0 in W 1,p (Ω) with uν → 0 in Lp∗

(Ω) ;

more precisely, we assume that

b := limν→∞

∫

Ω

|uν (x)|p∗

dx > 0.

We then let

a := lim infν→∞

∫

Ω

|∇uν (x)|p dx

and we define

I (u, Ω) =

∫

Ω

[ |∇u (x)|p − a + 1

b|u (x)|p

∗

]dx.

We clearly havelim infν→∞

I (uν , Ω) = −1 < I (0, Ω) = 0. ♦


Before proving the main theorem, we start with a particular case.

Lemma 8.14 Let 1 ≤ p < ∞, Ω ⊂ Rn be an open cube and let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function satisfying for almost every x ∈ Ω and for every(u, ξ) ∈ RN × RN×n

(C′

p

)− α |ξ|q ≤ f (x, u, ξ) ≤ α (1 + |ξ|p) ,

where α > 0, 1 ≤ q 1 and q = 1 if p = 1. Assume, in addition, thatξ → f (x, u, ξ) is quasiconvex, meaning that

∫

D


for every bounded open set D ⊂ Rn, almost every x0 ∈ Ω, every (u0, ξ0) ∈RN × RN×n and every ϕ ∈W 1,∞

0

(D; RN

).

Let


)with |∇uν |p equiintegrable

and let

I (u, Ω) :=

∫

Ω

f (x, u (x) ,∇u (x)) dx,

then

lim infν→∞

I (uν , Ω) ≥ I (u, Ω) .

It is clear that when p = 1, the equiintegrability hypothesis is not a restric-tion.

Proof. (Lemma 8.14). The strategy is first to freeze the lower order terms (asin Step 3 of Theorem 3.23) and then to apply Theorem 8.4 to get the result.

Step 1. We fix ǫ > 0 and we wish to show that there exists a measurable setΩǫ ⊂ Ω and a subsequence νj , with νj →∞, such that

meas (Ω− Ωǫ) < ǫ∫

Ωǫ

∣∣f(x, uνj (x) ,∇uνj (x)

)− f(x, u (x) ,∇uνj (x)

)∣∣ dx < ǫ measΩ.

(8.17)Indeed we first construct a set Ωǫ,ν ⊂ Ω in the following way.

Since uν → u in Lp (Ω) and ∇uν ∇u in Lp (Ω) , we have that, for everyǫ > 0, there exists Mǫ ≥ 1, which is independent of ν, such that if

K1ǫ,ν := x ∈ Ω : |u (x)|p or |uν (x)|p ≥ Mǫ


and

K2ǫ,ν := x ∈ Ω : |∇u (x)|p or |∇uν (x)|p ≥Mǫ ,

then

measK1ǫ,ν , measK2

ǫ,ν <ǫ

6

for every ν. Hence, if

Ω1ǫ,ν := Ω−

(K1

ǫ,ν ∪K2ǫ,ν

),

then

meas(Ω− Ω1

ǫ,ν

)<

ǫ

3. (8.18)

Since f is a Caratheodory function, there exists (see Scorza-Dragoni theorem,Theorem 3.8) Ω2

ǫ,ν ⊂ Ω1ǫ,ν a compact set with

meas(Ω1

ǫ,ν − Ω2ǫ,ν

)<

ǫ

3(8.19)

and such that f restricted to Ω2ǫ,ν × Sǫ is continuous where

Sǫ :=(u, ξ) ∈ RN × RN×n : |u|p < 2pMǫ and |ξ|p < Mǫ

.

We therefore have that there exists δ (ǫ) > 0 such that

|x− y|+ |u− v| < δ (ǫ) ⇒ |f (x, u, ξ)− f (y, v, ξ)| < ǫ (8.20)

for every x, y ∈ Ω2ǫ,ν , every |u|p , |v|p < 2pMǫ and |ξ|p < Mǫ .

Having fixed δ (ǫ) in this way and using the fact that uν → u in Lp (Ω) , wecan find νǫ = νǫ,δ(ǫ) such that if

Ω3ǫ,ν := x ∈ Ω : |uν (x)− u (x)| < δ (ǫ) ,

then

meas(Ω− Ω3

ǫ,ν

)<

ǫ

3for every ν ≥ νǫ. (8.21)

Therefore, letting

Ωǫ,ν := Ω2ǫ,ν ∩ Ω3

ǫ,ν ,

we have from (8.18), (8.19) and (8.21) that

meas (Ω− Ωǫ,ν) < ǫ (8.22)

and from (8.20) that

∫

Ωǫ,ν

|f (x, u (x) ,∇uν (x))− f (x, uν (x) ,∇uν (x))| dx < ǫ measΩ (8.23)


for every ν ≥ νǫ . We now choose ǫj = ǫ/2j, j ∈ N. We therefore have that(8.23) holds with ǫ and νǫ replaced by ǫj , νǫj . We then choose any νj ≥ νǫj withlim νj = ∞ and we let

Ωǫ :=

∞⋂

j=1

Ωǫj ,νj .

We therefore immediately have (8.17), as wished. From now on, in order not toburden the notations, we will still denote the subsequence

uνj

by uν .

Step 2. We next use the equiintegrability of the sequence |∇uν |p and thefact that u ∈W 1,p to get that there exists a non negative and increasing functionη such that η (t) → 0 as t→ 0, so that, for every measurable set A ⊂ Ω,

∫

A

[ 1 + max|∇u (x)|p , |∇uν (x)|p ]dx ≤ η (meas A) . (8.24)

Combining (8.17) and(C′

p

), we get, noting that |ξ|q ≤ 1 + |ξ|p ,

∫

Ω

|f (x, u (x) ,∇uν (x))− f (x, uν (x) ,∇uν (x))| dx

≤ ǫ measΩ + 2α

∫

Ω−Ωǫ

[ 1 + |∇uν (x)|p ]dx

and thus, appealing to (8.24) and to the fact that meas (Ω− Ωǫ) < ǫ, we inferthat∫

Ω

|f (x, u (x) ,∇uν (x))− f (x, uν (x) ,∇uν (x))| dx ≤ ǫ measΩ + 2αη (ǫ) .

(8.25)Step 3. (1) We then decompose the cube Ω into a finite union of cubes Ds

of edge length 1/h. We denote this union by Hh . Choosing h proportional tothe edge length of the cube Ω, we have

meas (Ω−Hh) = 0 where Hh :=⋃

Ds .

(2) We then define

us :=1

measDs

∫

Ds

u (x) dx and uh (x) :=∑

sus1Ds (x) ,

where

1Ds (x) :=

1 if x ∈ Ds

0 if x ∈ Ω−Ds .

Since uh → u in Lp (Ω) as h → ∞, by choosing h sufficiently large, we canassume that there exists Ωδ = Ωδ(ǫ) ⊂ Ω such that

meas(Ω− Ωδ) < ǫ and |u (x)− us| < δ (ǫ) /2, ∀x ∈ Ωδ ∩Ds .


Moreover for a fixed xs ∈ Ωǫ ∩Ds , we have

|x− xs| < δ (ǫ) /2, ∀x ∈ Ωǫ ∩Ds .

Combining the two estimates, we get

|x− xs|+ |u (x)− us| < δ (ǫ) , x ∈ Ωǫ ∩ Ωδ ∩Ds .

Note that, since |u (x)|p < Mǫ in Ωǫ and since we can always assume thatδp ≤ Mǫ , we get, for every x ∈ Ωǫ ∩ Ωδ ∩Ds ,

|us|p ≤ 2p−1(|u (x)|p + |u (x)− us|p) ≤ 2p−1(Mǫ + |δ|p) ≤ 2pMǫ .

We therefore get, from (8.20) and the fact that |u (x)|p , |∇uν (x)|p < Mǫ in Ωǫ ,

∫

Ωǫ∩Ωδ∩Ds

|f (x, u (x) ,∇uν (x))−f (xs, us,∇uν (x))| dx<ǫ meas(Ωǫ∩ Ωδ∩Ds

).

Summing up the last inequality we get

∑s

∫

Ωǫ∩Ωδ∩Ds

|f (x, u (x) ,∇uν (x))− f (xs, us,∇uν (x))| dx < ǫ measΩ.

From(C′

p

), with the observation that |ξ|q ≤ 1 + |ξ|p , and the above inequality,

we deduce that

∑s

∫

Ds

|f (x, u (x) ,∇uν (x))− f (xs, us,∇uν (x))| dx

≤ ǫ measΩ + 2α

∫

Ω−(Ωǫ∩Ωδ)

[ 1 + |∇uν (x)|p ]dx.

We thus get, from the fact that meas (Ω− Ωǫ) < ǫ, meas(Ω− Ωδ

)< ǫ and from

(8.24),

∑s

∫

Ds

|f (x, u (x) ,∇uν (x))− f (xs, us,∇uν (x))| dx ≤ ǫ measΩ + 2αη (2ǫ) .

(8.26)(3) For a similar reason we also have

∑s

∫

Ds

|f (x, u (x) ,∇u (x))− f (xs, us,∇u (x))| dx < ǫ measΩ + 2αη (2ǫ) .

(8.27)Step 4. We now gather all these inequalities. From (8.25), we have

I (uν , Ω) =

∫

Ω


≥∫

Ω

f (x, u (x) ,∇uν (x)) dx− ǫ measΩ− 2αη (2ǫ) .


We therefore get from (8.26)

I (uν , Ω) ≥∑

s

∫

Ds

f (xs, us,∇uν (x)) dx− 2ǫ measΩ− 4αη (2ǫ) .

We may now apply Theorem 8.4 to

ξ → f (xs, us, ξ)

and we hence find

lim infν→∞

I (uν , Ω) ≥∑

s

∫

Ds

f (xs, us,∇u (x)) dx− 2ǫ measΩ− 4αη (2ǫ) .

Thus, using (8.27), we find

lim infν→∞

I (uν , Ω) ≥∫

Ω

f (x, u (x) ,∇u (x)) dx− 3ǫ measΩ− 6αη (2ǫ) .

Since ǫ is arbitrary, we have the claim.

We now continue with the proof of Theorem 8.11.


Step 1. We first show that we can restrict ourselves to sets Ω that are afinite union of disjoint open cubes and to functions f satisfying

−α |ξ|q ≤ f (x, u, ξ) ≤ g (x, u) (1 + |ξ|p) , (8.28)

with 1 ≤ q 1 and q = 1 if p = 1.

This is already so when p = q = 1 and we therefore consider only the casep > 1. We start by defining

h (x, u, ξ) := f (x, u, ξ) + α |u|r + β (x) ,

which satisfies (8.28), with a different g than the one in (Cp) . By Rellich theo-rem, we have

uν → u in Lr(Ω; RN

).

We now use the equiintegrability in W 1,q ∩ Lr (since 1 ≤ q 1 andq = 1 if p = 1) of the sequence uν to get, for every ǫ > 0, δ = δ (ǫ) so that

meas A ≤ δ ⇒ 0 ≤∫

A

[ α (|uν (x)|r + |∇uν (x)|q) + β (x) ]dx ≤ ǫ.

We next choose Ωµ ⊂ Ω to be a finite union of cubes and μ sufficiently large sothat

meas (Ω− Ωµ) ≤ δ.


We thus obtain, from (Cp) and the equiintegrability, that

∫

Ω


≥∫

Ωµ

f (x, uν (x) ,∇uν (x)) dx−∫

Ω−Ωµ

[ α (|uν |r + |∇uν (x)|q) + β (x) ]dx

≥∫

Ωµ

h (x, uν (x) ,∇uν (x)) dx−∫

Ωµ

[ α |uν|r + β (x) ]dx− ǫ.

Applying the lower semicontinuity for functions h satisfying (8.28), we can write

lim infν→∞

∫

Ω


Ωµ

h (x, u,∇u) dx−∫

Ωµ

[ α |u|r + β (x) ]dx− ǫ

=

∫

Ωµ

f (x, u,∇u) dx− ǫ.

Letting μ→∞ and recalling that ǫ > 0 is arbitrary, we indeed have the claim.

Step 2. We next see that we can further assume that f verifies for almostevery x ∈ Ω and for every (u, ξ) ∈ RN × RN×n the following condition

−α |ξ|q ≤ f (x, u, ξ) ≤ α (1 + |ξ|p) (8.29)

where α > 0. This is already so when p = q = 1 and we hence discuss only thecase p > 1. First note that we can restrict our attention to f ≥ 0. Indeed since1 ≤ q 0, k = k (ǫ) > 0 so that

ǫ |ξ|p + k ≥ α |ξ|q , ∀ξ ∈ RN×n

and thus

fǫ (x, u, ξ) := f (x, u, ξ) + α |u|r + β (x) + ǫ |ξ|p + k ≥ 0.

Note that fǫ ≥ 0 and satisfies (Cp) , of course with a different function g. Apply-ing the lower semicontinuity for the non-negative function fǫ ≥ 0 satisfying (Cp)and letting ǫ → 0, we easily get the claim. So we may now assume that whenp > 1

0 ≤ f (x, u, ξ) ≤ g (x, u) (1 + |ξ|p) .

It therefore remains to show that we can replace g (x, u) by α so as to have(8.29). Indeed, define, for every integer i, a sequence of non-increasing contin-uous functions

hi : R+ → [0, 1]

such that

hi (s) : =

1 if 0 ≤ s ≤ i− 1

0 if s ≥ i


and let

ϕi (x, u) : =

hi (|u|) if g (x, u) ≤ i

ihi(|u|)g(x,u) if g (x, u) > i.

Then define

fi (x, u, ξ) := ϕi (x, u) f (x, u, ξ)

and observe that it is a non negative Caratheodory function, quasiconvex in thevariable ξ and that it satisfies (8.29) with α = i. Moreover,

limi→∞

fi (x, u, ξ) = supi

fi (x, u, ξ) = f (x, u, ξ) .

So, assume that we have proved the theorem for integrands fi satisfying (8.29)and let us prove the result for f. We in fact have

lim infν→∞

∫

Ω

f (x, uν (x) ,∇uν (x)) dx ≥ lim infν→∞

∫

Ω

fi (x, uν ,∇uν) dx

≥∫

Ω

fi (x, u,∇u)dx.

Taking the supremum over i on the right hand side, we have indeed obtainedthe claim.

Step 3. So, from now on, we will assume that Ω is a finite union of cubes andf satisfies (8.29). By working on each cube separately, we can even assume thatΩ itself is a cube. By restricting our attention to a subsequence, still denoteduν , we can assume that


I (uν, Ω) = limν→∞

I (uν , Ω) .

We now conclude the proof of the theorem. Applying Lemma 8.15, we can find asubsequence uµ and vµ ∈ W 1,p

(Ω; RN

)such that |∇vµ|p is equiintegrable,

vµ u in W 1,p(Ω; RN

)

and

limµ→∞

measΩµ = 0,

where

Ωµ := x ∈ Ω : uµ (x) = vµ (x) ∪ x ∈ Ω : ∇uµ (x) = ∇vµ (x).

Using the fact that f satisfies (8.29), we can write

I (uµ, Ω) = I (uµ, Ω− Ωµ) + I (uµ, Ωµ) ≥ I (uµ, Ω− Ωµ)− α

∫

Ωµ

|∇uµ (x)|q dx.


Since vµ = uµ in Ω − Ωµ , we find from the above inequality and from (8.29)that

I (uµ, Ω) ≥ I (vµ, Ω− Ωµ)− α

∫

Ωµ

|∇uµ (x)|q dx

≥ I (vµ, Ω)− α

∫

Ωµ

(1 + |∇uµ (x)|q + |∇vµ (x)|p) dx.

We then apply Lemma 8.14 (note that (8.29) is(C′

p

)of Lemma 8.14) to

the sequence vµ, using the equiintegrability of the sequences |∇vµ|p and|∇uµ|q (since 1 ≤ q 1 and q = 1 if p = 1) and the fact thatlimµ→∞ measΩµ = 0, to obtain the result, namely

L = limµ→∞

I (uµ, Ω) ≥ lim infµ→∞

I (vµ, Ω) ≥ I (u, Ω) .


One important step in the proof of Theorem 8.11 was to replace the originalsequence by an equiintegrable sequence, using the following result of Fonseca-Muller-Pedregal [288] (based on ideas contained in [3]) and Kristensen [379].

Lemma 8.15 Let 1 < p < ∞, Ω ⊂ Rn be a bounded open set with Lipschitzboundary and


).

Then there exists a subsequence uµ and vµ ∈W 1,p(Ω; RN

)such that |∇vµ|p

is equiintegrable,vµ u in W 1,p

(Ω; RN

)

andlim

µ→∞measΩµ = 0

where

Ωµ := x ∈ Ω : uµ (x) = vµ (x) ∪ x ∈ Ω : ∇uµ (x) = ∇vµ (x).

8.2.5 Lower semicontinuity for polyconvex functions

We now discuss the case of polyconvex functions. At first glance, it may seemthat, since polyconvex functions are quasiconvex, nothing new is to be proved.However we can now consider functions that have no upper bound and in par-ticular functions that may take the value +∞.

Theorem 8.16 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundary,p > n ∧N,

F : Ω× RN × Rτ(n,N) → R ∪ +∞ , F = F (x, u, X) ,


be a Caratheodory function that is such that for almost every x ∈ Ω and forevery (u, X) ∈ RN × Rτ(n,N)

X → F (x, u, X) is convex,

F (x, u, X) ≥ 〈a (x) ; X〉+ b (x) + c |u|r ,

where a ∈ Lp′ (Ω; Rτ(n,N)

), 1/p + 1/p′ = 1, b ∈ L1 (Ω) , 1 ≤ r < np/ (n− p) if

p < n and 1 ≤ r < ∞ if p ≥ n and c ∈ R. Then

I (u, Ω) :=

∫

Ω

F (x, u (x) , T (∇u (x))) dx

is (sequentially) weakly lower semicontinuous in W 1,p(Ω; RN

).

Remark 8.17 The restriction p > n ∧N can be slightly relaxed if we replacethe convergence


)

by uν → u in Lp

(Ω; RN

)

T (∇uν) T (∇u) in L1(Ω; Rτ(n,N)

).

We will see, in Theorem 8.20 and in Theorem 8.31, how this can be obtainedwith some p ≤ n ∧N. ♦

Proof. Anticipating the results of Theorem 8.20, we find that since p > n∧N,then

T (∇uν) T (∇u) in Lp/n∧N(Ω; Rτ(n,N)

).

We are therefore in a position to apply Theorem 3.23 and hence the resultfollows at once.

The above proof relied heavily on the weak continuity of determinants. Onecan however improve the result in the following context. As a simplification, weassume that N = n and that the function F ≥ 0 and does not depend on lowerorder terms.

Assume that p > n − 1 (note that in the theorem we need p > n) andu, uν ∈ W 1,n(Ω; Rn). Then

I(u, Ω) :=

∫

Ω

F (T (∇u(x)))dx

is weakly lower semicontinuous in W 1,p(Ω; RN

).

The above result was proved by Dacorogna-Marcellini [194] when p > n− 1.Maly [415] then gave a counterexample proving that the result does not holdif p < n− 1. Later, Celada-Dal Maso [128] and Fusco-Hutchinson [296] showedthat Dacorogna-Marcellini result holds if p = n− 1.

Weak Continuity 393

8.3 Weak Continuity

We now turn our attention to results on weak continuity of nonlinear functions.

Let f : RN×n → R be continuous. We show that

f (∇uν) f (∇u) in D′ (Ω)

for every sequence uν u in W 1,p(Ω; RN

)if and only if f is quasiaffine (i.e.

from Theorem 5.20, f is a linear combination of minors of the matrix ∇u).

Plainly, the existence of nonlinear weakly continuous functions is purely dueto the vectorial nature of the problem, since if N = 1 (or n = 1), the only minorsof the matrix ∇u are just the linear terms ∂u/∂xi , 1 ≤ i ≤ n (or if n = 1, thelinear terms dui/dx, 1 ≤ i ≤ N).

It is also clear that Theorem 8.1 and Theorem 8.4 applied to f, I and −f,−I, added to the fact that the domain Ω is arbitrary, immediately give theweak continuity if p is large enough. We use Theorem 8.1 for the necessarycondition; however, for reasons explained below, we do not use Theorem 8.4for the sufficiency result and we give a new proof of the weak continuity of theminors.

The results of this section are essentially due to Reshetnyak [509], [510] andBall [51], [53]. Considerations on weak continuity have been developed in a moregeneral context, called compensated compactness, by Murat and Tartar [469],[470], [471], [568] (for a presentation of this theory, see also Dacorogna [173]).

Before starting our analysis, we recall the following definition.

Definition 8.18 Let Ω ⊂ Rn be an open set and fν , f ∈ L1loc (Ω) . We say that

fν converges to f in the sense of distributions and we write

fν f in D′ (Ω)

if ∫

Ω

fν (x) ϕ (x) dx →∫

Ω

f (x)ϕ (x) dx

for every ϕ ∈ D (Ω) (the set of C∞ functions with compact support).

8.3.1 Necessary condition

Theorem 8.19 Let 1 ≤ p ≤ ∞, let Ω ⊂ Rn be a bounded open set and letf : RN×n → R be continuous. If, for every sequence uν u in W 1,p

(Ω; RN

)

(uν∗ u if p = ∞),

f (∇uν) f (∇u) in D′ (Ω) (8.30)

then f is quasiaffine, i.e. there exist α ∈ R, β ∈ Rτ(n,N) such that

f (ξ) = α + 〈β; T (ξ)〉 (8.31)


for every ξ ∈ RN×n, where n ∧N = min n, N and

⎧⎪⎨⎪⎩

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ)

τ (n, N) =

n∧N∑

s=1

σ (s) , σ (s) =(Ns

)(ns

)

and 〈·; ·〉 denotes the scalar product in Rτ(n,N).

Proof. Let ϕ ∈ D (Ω) and let

I (u, Ω) :=

∫

Ω

ϕ (x) f (∇u (x)) dx

then (8.30) is equivalent to

limν→∞

I (uν, Ω) = I (u, Ω) .

We may therefore apply Theorem 8.1 to I and −I and get that f and −f arequasiconvex, i.e. f is quasiaffine. Theorem 5.20 implies then (8.31) and thetheorem follows.

8.3.2 Sufficient condition

For the clarity of the exposition, we always give the results for the cases N =n = 2, N = n = 3 and then N = n, before giving the general result whenN, n ≥ 2.

We also recall some notations and elementary properties of determinantsand adjugate matrices (for more details, see Sections 5.4 and 8.5).

- For N = n = 3, we denote

adj2∇u =((adj2∇u)

iα

)1≤i≤3

1≤α≤3,

where

(adj2∇u)iα = (−1)

i+α ∂(uj, uk

)

∂ (xβ , xγ)

= (−1)i+α

[∂uj

∂xβ

∂uk

∂xγ− ∂uj

∂xγ

∂uk

∂xβ],

where j < k with j, k = i and β < γ with β, γ = α.

We also have

det∇u =3∑

i=1

∂u1

∂xi(adj2∇u)1i .

Weak Continuity 395

- When N = n, we recall that

det∇u =∂(u1, · · · , un

)

∂ (x1, · · · , xn)

=

n∑

α=1

(−1)α+1 ∂u1

∂xα

∂(u2, · · · , un

)

∂ (x1, · · · , xα−1, xα+1, · · · , xn).

- For N, n ≥ 2 and 2 ≤ s ≤ n ∧N = min n, N , we have

adjs∇u =((adjs∇u)

iα

)1≤i≤(Ns )

1≤α≤(ns)

,

where

(adjs∇u)iα = (−1)

i+α ∂(ui1 , · · · , uis

)

∂ (xα1 , · · · , xαs).

For the precise relation between i, i1, · · · , is and α, α1, · · · , αs see Section 5.4.

We now give the main theorem, which shows that these functions are actuallyweakly continuous.

Theorem 8.20 Let Ω ⊂ Rn be a bounded open set, 1 2, thendet∇uν det∇u in Lp/2 (Ω) .

Part 2. Let N = n = 3. If p ≥ 2, then

adj2∇uν adj2∇u in D′ (Ω; R9)

and if p > 2, then

adj2∇uν adj2∇u in Lp/2(Ω; R9

).

If p ≥ 3, thendet∇uν det∇u in D′ (Ω)

and if p > 3, thendet∇uν det∇u in Lp/3 (Ω) .

Part 3. Let N = n and p ≥ n. Then

det∇uν det∇u in D′ (Ω)


and if p > n, thendet∇uν det∇u in Lp/n (Ω) .

Part 4. Let N, n ≥ 2, 2 ≤ s ≤ n ∧N = min n, N and p ≥ s. Then

adjs∇uν adjs∇u in D′(Ω; Rσ(s)

),

where

σ (s) =(Ns

)(ns

)=

N !n!

(s!)2 (N − s)! (n− s)!.

Furthermore, if p > s, then

adjs∇uν adjs∇u in Lp/s(Ω; Rσ(s)

).

Part 5. Let N, n ≥ 2, 2 ≤ s ≤ n ∧N and assume that

adjs−1∇uν adjs−1∇u in Lr(Ω; Rσ(s−1)

),

where r > 1 with1

p+

1

r≤ 1. Then

adjs∇uν adjs∇u in D′(Ω; Rσ(s)

).

Remark 8.21 (i) Let N = n = 2. Note that if p = 2 and if we know, inaddition, that

det∇uν f in L1 (Ω) ,

then the uniqueness of the limit in D′ (Ω) ensures that f = det∇u.

(ii) Let N = n = 2. If p > 2, the statement in Part 1 results immediatelyfrom Theorem 8.4, since, trivially,

−(1 + |∇u|2

)≤ − |∇u|2 ≤ det∇u ≤ |∇u|2 ≤ 1 + |∇u|p .

(iii) Let N = n = 2. If p = 2, Theorem 8.4 cannot be applied, as seen inRemark 8.5 (iii), there are examples (see Example 8.6) of sequences uν uin W 1,2

(Ω; R2

)such that det∇uν det∇u in L1 (Ω) . Theorem 8.20 ensures,

however, that det∇uν det∇u in D′ (Ω) .

(iv) All the above remarks can be made for the general case N, n ≥ 2.

(v) If p = +∞, replace everywhere weak convergence by weak * convergencein the appropriate space (L∞ or W 1,∞).

(vi) For some extensions of the theorem, we refer to Muller [458], [459]. ♦

The main tool in proving Theorem 8.20 is the observation that any minor of∇u can be expressed as a divergence of a vector field and we are led to introducethe following operators.

Weak Continuity 397

Definition 8.22 Let Ω ⊂ Rn be an open set and let u ∈ C2(Ω; RN

).

(i) For N = n = 2, define

Det∇u :=∂

∂x1(u1 ∂u2

∂x2)− ∂

∂x2(u1 ∂u2

∂x1).

(ii) For N = n = 3, define (see the form of adj2∇u given above)

Adj2∇u :=((Adj2∇u)

iα

)1≤i≤3

1≤α≤3,

where

(Adj2∇u)iα := (−1)i+α [

∂

∂xβ(uj ∂uk

∂xγ)− ∂

∂xγ(uj ∂uk

∂xβ) ]

where j < k with j, k = i and β < γ with β, γ = α.

Similarly, we let

Det∇u :=

3∑

i=1

∂

∂xi

(u1 (adj2∇u)

1i

).

(iii) When N = n, we let

Det∇u :=n∑

α=1

(−1)α+1 ∂

∂xα(u1 ∂

(u2, · · · , un

)

∂ (x1, · · · , xα−1, xα+1, · · · , xn)).

(iv) Let N, n ≥ 2 and 2 ≤ s ≤ n∧N = min n, N . We define (see the formof adjs∇u given above)

Adjs∇u :=((Adjs∇u)

iα

)1≤i≤(Ns )

1≤α≤(ns)

,

where

(Adjs∇u)iα := (−1)i+α

s∑

t=1

(−1)t+1 ∂

∂xαt

(ui1∂(ui2 , · · · , uis

)

∂(xα1 , · · · , xαt−1 , xαt+1 , · · · , xαs

) ).

Remark 8.23 (i) In the case N = n = 2, we could have written as well

Det∇u :=∂

∂x2(u2 ∂u1

∂x1)− ∂

∂x1(u2 ∂u1

∂x2),

but this does not change anything in the analysis below. A similar remarkapplies to all other cases.

(ii) One should observe that if N = n = 3, Ball [53] defines

Det∇u :=3∑

i=1

∂

∂xi

(u1 (Adj2∇u)1i

),


which does not correspond to our definition (ii) (note the change from adj toAdj above). The two definitions need not be the same if u ∈ W 1,p

(Ω; R3

)and

p < 2, as suggested by the following lemma. ♦

We now see how to relate these operators Det and Adjs to the algebraicdefinitions of det and adjs .

Lemma 8.24 Let Ω ⊂ Rn be a bounded open set and let u ∈ W 1,p(Ω; RN

),

1 < p < ∞.

Part 1: N = n = 2. If p ≥ 43 , then Det∇u ∈ D′ (Ω) . Furthermore, if p ≥ 2,

thenDet∇u = det∇u in D′ (Ω) .

In particular, if u ∈ C2(Ω; R2

), then the above identity holds in the usual sense.

Part 2: N = n = 3.

(i) If p ≥ 32 , then Adj2∇u ∈ D′ (Ω; R9

)and if p ≥ 2, then

Adj2∇u = adj2∇u in D′ (Ω; R9).

In particular, if u ∈ C2(Ω; R3

), then the above identity holds in the usual sense.

(ii) If p ≥ 94 , then Det∇u ∈ D′ (Ω) . Moreover, if p ≥ 3, then

Det∇u = det∇u in D′ (Ω)

and, in particular, if u ∈ C2(Ω; R3

), then the identity holds in the usual sense.

Part 3: N = n. If p ≥ n2

n+1 , then Det∇u ∈ D′ (Ω) . Furthermore, if p ≥ n,then

Det∇u = det∇u in D′ (Ω) .

In particular, if u ∈ C2 (Ω; Rn) , then the above equality holds in the usual sense.

Part 4: N, n ≥ 2 and 2 ≤ s ≤ n ∧ N = min n, N . If p ≥ snn+1 , then

Adjs∇u ∈ D′ (Ω; Rσ(s)), where σ (s) =

(Ns

)(ns

). Moreover, if p ≥ s, then

Adjs∇u = adjs∇u in D′(Ω; Rσ(s)

)

and thus the identity holds in the usual sense if u ∈ C2(Ω; RN

).

Part 5: N, n ≥ 2 and 2 ≤ s ≤ n ∧N. Assume that, for r > 1,

adjs−1∇u ∈ Lr(Ω; Rσ(s−1)

).

(i) If 1p + 1

r ≤ 1 + 1n , then Adjs∇u ∈ D′ (Ω; Rσ(s)

).

(ii) If 1p + 1

r ≤ 1, then


).

Weak Continuity 399

Remark 8.25 (i) Let N = n = 2 and 43 ≤ p < 2. If det∇u is defined in the

usual way, namely

det∇u =∂u1

∂x1

∂u2

∂x2− ∂u2

∂x1

∂u1

∂x2,

then det∇u is not necessarily a distribution, while Det∇u is. In fact, Det∇uis the (unique) extension, as a distribution, by continuity of det∇u, when 4

3 ≤p < 2. Note also that if p < 4

3 , then even Det∇u need not be a distribution(see Example 8.28).

(ii) Similar remarks apply to the general case N, n ≥ 1. ♦

Proof. (Lemma 8.24). We prove Part 1 for illustration and then prove Parts4 and 5.

Part 1: N = n = 2. If p ≥ 43 and since u ∈ W 1,p

(Ω; R2

), we have by the

Sobolev imbedding theorem that u ∈ L4loc

(Ω; R2

). Using Holder inequality, we

deduce that

u1 ∂u2

∂x2, u1 ∂u2

∂x1∈ L1

loc (Ω)

and thus Det∇u ∈ D′ (Ω) .Moreover, if p ≥ 2, then det∇u ∈ L1 (Ω) . Observe that if u ∈ C2

(Ω; R2

),

then

det∇u =∂u1

∂x1

∂u2

∂x2− ∂u2

∂x1

∂u1

∂x2

=∂

∂x1(u1 ∂u2

∂x2)− ∂

∂x2(u1 ∂u2

∂x1) = Det∇u.

Therefore, multiplying the above identity by ϕ ∈ D (Ω) and integrating by parts,we find

∫

Ω

det∇u · ϕdx = −∫

Ω

(u1 ∂u2

∂x2

∂ϕ

∂x1− u1 ∂u2

∂x1

∂ϕ

∂x2)dx.

Since C2(Ω; R2

)is dense in W 1,2

loc

(Ω; R2

), then the above equality holds for

every u ∈W 1,2; and this concludes Part 1.

Part 4: N, n ≥ 2. We first consider the case where p ≥ snn+1 . Since

u ∈ W 1,p(Ω; RN

), we have by the Sobolev imbedding theorem that u ∈

Lsn/(n+1−s)loc

(Ω; RN

). Using Holder inequality, we also have that

∂(ui2 , · · · , uis

)

∂(xα1 , · · · , xαt−1 , xαt+1 , · · · , xαs

) ∈ Lsn/(n+1)(s−1) (Ω)

and hence

ui1∂(ui2 , · · · , uis

)

∂(xα1 , · · · , xαt−1 , xαt+1 , · · · , xαs

) ∈ L1loc (Ω) .

Therefore, using the definition of Adjs , we get that Adjs∇u ∈ D′(Ω; R

σ(s))

.


We now discuss the case p ≥ s and then adjs∇u ∈ L1(Ω; R

σ(s))

. Observe

that if u ∈ C2(Ω; RN

), then (cf. Theorem 8.33)

Adjs∇u = adjs∇u.

Multiplying the above equality by ϕ ∈ D (Ω) and integrating by parts the lefthand side, we find that

∫

Ω

((adjs∇u)

iα

)ϕdx

=

s∑

t=1

(−1)t∫

Ω

ui1∂(ui2 , · · · , uis

)

∂(xα1 , · · · , xαt−1 , xαt+1 , · · · , xαs

) ∂ϕ

∂xαt

dx.

Since C2(Ω; RN

)is dense in W 1,s

loc

(Ω; RN

), we deduce that the equality holds

for every u ∈W 1,s and this concludes Part 4 of the lemma.

Part 5: N, n ≥ 2. The case p ≥ n being easier, we assume that p <n. Using Sobolev imbedding theorem, we have, since u ∈ W 1,p, that u ∈L

np/(n−p)loc

(Ω; RN

). We now combine the definition of Adjs , adjs and the fact

that adjs−1∇u ∈ Lr with Holder inequality (recalling that 1p + 1

r − 1n ≤ 1) to

deduce thatui1(adjs−1∇u

)iα∈ L1

loc (Ω) .

We therefore have Adjs∇u ∈ D′ (Ω; Rσ(s)).

Furthermore, since 1p + 1

r ≤ 1 and adjs−1∇u ∈ Lr, we have that adjs∇u ∈L1. Since, using Theorem 8.33, the identity

Adjs∇u = adjs∇u

holds for every u ∈ C2, we deduce, by density, that


).

This concludes the proof of the Lemma.

We are now in a position to show Theorem 8.20.

Proof. We prove Part 1 for the sake of illustration, then Parts 4 and 5.

Part 1. Let N = n = 2 and p ≥ 2. Let ϕ ∈ D (Ω) , then by Lemma 8.24 wehave ∫

Ω

det∇uν · ϕdx = −∫

Ω

(u1ν

∂u2ν

∂x2

∂ϕ

∂x1− u1

ν

∂u2ν

∂x1

∂ϕ

∂x2)dx.

Since uν u in W 1,p(Ω; R2

), uν → u in Lq

loc

(Ω; R2

)with q < ∞, and therefore

(u1ν

∂u2ν

∂x2, u1

ν

∂u2ν

∂x1) (u1 ∂u2

∂x2, u1 ∂u2

∂x1) in L1

loc

(Ω; R2

).

Weak Continuity 401

We therefore deduce that∫

Ω

det∇uν · ϕdx →∫

Ω

det∇u · ϕdx.

Part 4. Let N, n ≥ 2, 2 ≤ s ≤ n ∧N and p ≥ s. In order to show the theoremit is sufficient to establish that for every ϕ ∈ D (Ω) we have

∫

Ω

∂(ui1

ν , · · · , uisν

)

∂ (xα1 , · · · , xαs)ϕdx →

∫

Ω

∂(ui1 , · · · , uis

)

∂ (xα1 , · · · , xαs)ϕdx. (8.32)

To show (8.32), we proceed by induction on s. Suppose that the theorem hasbeen established up to the order s − 1 (the case s = 2 has been dealt with inPart 1). Since uν u in W 1,p

(Ω; RN

)and p ≥ s we deduce that

uν → u in Lqloc

(Ω; RN

)with 1 ≤ q <

ns

n− s. (8.33)

By hypothesis of induction, denoting

(x1, · · · , xt, · · · , xs) = (x1, · · · , xt−1, xt+1, · · · , xs) ,

we have

∂(ui2

ν , · · · , uisν

)

∂ (xα1 , · · · , xαt , · · · , xαs)

∂(ui2 , · · · , uis

)

∂ (xα1 , · · · , xαt , · · · , xαs)in D′ (Ω)

for every 1 ≤ t ≤ s. Since the above (s− 1) minor is in Lp/(s−1) (Ω) and p ≥ s,we get that the above convergence is actually in Lp/(s−1) (Ω) . Combining (8.33)with the above convergence, we obtain that, for 1 ≤ r < n/ (n− 1) ,

ui1ν

∂(ui2

ν , · · · , uisν

)

∂ (xα1 , · · · , xαt , · · · , xαs) ui1

∂(ui2 , · · · , uis

)

∂ (xα1 , · · · , xαt , · · · , xαs)in Lr

loc (Ω)

for every 1 ≤ t ≤ s. We finally combine this convergence result with Lemma8.24 to get, for every ϕ ∈ D (Ω) ,

∫

Ω

∂(ui1

ν , · · · , uisν

)

∂ (xα1 , · · · , xαs)ϕdx

=

s∑

t=1

(−1)t∫

Ω

ui1ν

∂(ui2

ν , · · · , uisν

)

∂ (xα1 , · · · , xαt , · · · , xαs)

∂ϕ

∂xαt

dx

→s∑

t=1

(−1)t∫

Ω

ui1∂(ui2 , · · · , uis

)

∂ (xα1 , · · · , xαt , · · · , xαs)

∂ϕ

∂xαt

dx.

Using again Lemma 8.24 on the right hand side, we obtain (8.32) and thusPart 4.

Part 5. Let N, n ≥ 2, 2 ≤ s ≤ n ∧ N and 1p + 1

r ≤ 1. We then have

adjs∇u ∈ L1(Ω; Rσ(s)

). Proceeding exactly as in Part 4, we obtain the result

and hence the theorem.


It is clear that, from Theorem 8.20 and Lemma 8.24 (with the same notationsas in the lemma), we immediately get (see Ball [51], [53]) the following.

Corollary 8.26 Let Ω ⊂ Rn be a bounded open set and let


).

Part 1. Let N = n and p > n2

n+1 , then

Det∇uν Det∇u in D′ (Ω) .

Part 2. Let N, n ≥ 2, 2 ≤ s ≤ n ∧N and p > snn+1 , then

Adjs∇uν Adjs∇u in D′(Ω; Rσ(s)

).

Remark 8.27 (i) The proof of Corollary 8.26 is almost identical to that ofTheorem 8.20 using Lemma 8.24.

(ii) The result of the corollary is false if, for example for Part 1, p ≤ n2

n+1 ,see Example 8.28. ♦

Example 8.28 (Dacorogna-Murat [208]). The result of Part 1 of the corollary

is false if p ≤ n2

n+1 . When p = n2

n+1 , the result remains partially true (see

Theorem 1 in [208]) in the sense that, up to a subsequence,

Det∇uν Det∇u + μ in D′ (Ω)

for a certain μ ∈ D′ (Ω) , with, in general, μ = 0. We now give an example whereweak continuity does not hold. We construct a sequence uν ⊂ W 1,∞ (Ω; Rn)such that

uν u ≡ 0 in W 1,p (Ω; Rn) and Det∇uν Det∇u ≡ 0 in D′ (Ω) .

Note that since uν , u ∈W 1,∞ (Ω; Rn) , then

Det∇u = det∇u = 0 and Det∇uν = det∇uν in D′ (Ω) .

We first let x = (x1, · · · , xn) , r2 = x21 + · · ·+ x2

n and

Ω = x ∈ Rn : r < 1 .

We then consider the sequence

uν (x) := fν (r)

⎛⎜⎜⎜⎜⎝

1

x2/r

...

xn/r

⎞⎟⎟⎟⎟⎠

,

Existence theorems 403

where for p ≥ 1 and for ν ≥ 4, we let

fν (r) :=

⎧⎪⎪⎨⎪⎪⎩

νn/pr if r ∈[0, 1

ν

)

νn/p(

2ν − r

)if r ∈

[1ν , 2

ν

)

0 if r ∈[

2ν , 1).

A direct computation shows that the sequence uν has all the above propertiesand in particular that

det∇uν (x) = f ′ν (r)

(fν (r)

r

)n−1x1

r.

We now claim that we can find ϕ ∈ D (Ω) and a = 0 such that

limν→∞

∫

Ω

det∇uν (x) ϕ (x) dx→

a if p = n2

n+1

∞ if 1 ≤ p < n2

n+1 .

Indeed chooseϕ (x) = −x1ρ (x)

where ρ ∈ D (Ω) and ρ (x) ≡ 1 if |x| < 1/2. We therefore find, since det∇uν (x) =0 if |x| > 1/2, that

∫

Ω

det∇uν (x) ϕ (x) dx = −∫

|x|<1/2

det∇uν (x) x1dx

= −∫

|x|<1/2

f ′ν (r)

(fν (r)

r

)n−1x2

1

rdx

= −σn

n

∫ 1/2

0

rf ′ν (r) (fν (r))

n−1dr

=2σn

n2 (n + 1)ν

n2

p −(n+1),

where σn is the area of the unit sphere in Rn. The result follows at once. ♦

8.4 Existence theorems

We now collect the results of Sections 8.2 and 8.3 to get existence theorems inthe classical way. There will be two results:

(1) one involving quasiconvex functions that are finite everywhere;

(2) one using polyconvex functions that are allowed to take the value +∞in a subset of RN×n.

8.4.1 Existence theorem for quasiconvex functions

We first combine the lower semicontinuity result obtained in Theorem 8.11 witha coercivity condition to get the first existence theorem.


Theorem 8.29 Let p > 1, Ω ⊂ Rn be a bounded open set with a Lipschitzboundary. Let

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function satisfying for almost every x ∈ Ω, for every (u, ξ) ∈RN × RN×n

ξ → f (x, u, ξ) is quasiconvex,

α1 |ξ|p + β1 |u|q + γ1 (x) ≤ f (x, u, ξ) ≤ α2 |ξ|p + β2 |u|r + γ2 (x) , (8.34)

where α2 ≥ α1 > 0, β1 ∈ R, β2 ≥ 0, γ1, γ2 ∈ L1 (Ω) , p > q ≥ 1 and 1 ≤ r ≤np/ (n− p) if p < n and 1 ≤ r < ∞ if p ≥ n. Let

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

then (P ) admits at least one solution.

Remark 8.30 The above theorem is due to Acerbi-Fusco [3] and Marcellini[423], improving earlier results by Morrey [453], [455] and Meyers [442]. ♦

Proof. Observe first that inf (P ) is finite, since, for example, I (u0) < +∞,by the growth condition (8.34). So let uν be a minimizing sequence, i.e.

I (uν)→ inf (P ) .

Proceeding exactly as in Theorem 3.30, we obtain that ‖uν‖W 1,p is uniformlybounded. Since p > 1, we then deduce that, up to the extraction of a subse-quence still labeled uν ,


).

Using Theorem 8.11, we immediately get that

I (u) = inf (P ) ,

which is the claim.

8.4.2 Existence theorem for polyconvex functions

We now give a theorem that is applicable to functions in a smaller class thanthe previous one from the point of view of convexity (since f polyconvex ⇒ fquasiconvex) but in a larger class from the point of view of growth and coercivityconditions. More precisely, the previous theorem excludes two important cases:

(1) functions f allowed to take the value +∞,

(2) functions f of the type (if, for example, N = n = 2)

f (ξ) = |ξ|2 + |det ξ|2 .


These two cases are important for applications. For example, the first oneis useful when one deals with minimization problems with constraints, as is thecase, for example, in elasticity where a natural constraint is det ξ > 0.

Since polyconvex functions are defined through a convex function F, thetheorem will be stated in terms of the function F.

Theorem 8.31 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundary,p > n ∧N,

F : Ω× RN × Rτ(n,N) → R ∪ +∞ , F = F (x, u, X) ,

be a Caratheodory function which is such that for almost every x ∈ Ω, for every(u, X) ∈ RN × Rτ(n,N)

X → F (x, u, X) is convex,

F (x, u, X) ≥ a (x) + b1 |X1|p , (8.35)

where X = (X1, X2, · · · , Xn∧N) ∈ Rτ(n,N), a ∈ L1 (Ω) and b1 > 0. Let

(P ) inf

I (u) =

∫

Ω

F (x, u (x) , T (∇u (x))) dx : u ∈ u0 + W 1,p0

(Ω; RN

)= m.

Assume thatI (u0) < +∞. (8.36)

Then (P ) admits at least one solution.

Remark 8.32 (i) The above theorem is due to Ball [51], [53] and has beenapplied to find minima in nonlinear elasticity.

(ii) The hypothesis (8.36) is important to ensure that m < +∞. A way ofsatisfying (8.36) would be to impose a growth condition of the same type asthe coercivity condition (8.35), as was done in Theorem 8.29 and then u0 wouldtrivially satisfy (8.36).

(iii) The coercivity (8.35) ensures, for some appropriate sequence, that


)

and thus, since p > n ∧N, we have


).

However, since in the proof we only need this last convergence, any coercivitycondition which ensures it is enough. For example (see Step 2 in the proof ofthe theorem for details), one could consider

F (x, u, X) ≥ a (x) +n∧N∑

s=1

bs |Xs|ps , (8.37)


where X = (X1, X2, · · · , Xn∧N) , a ∈ L1 (Ω) , bs > 0 and p1 ≥ 2, ps ≥ p1

p1−1 if2 ≤ s < n ∧N and pn∧N > 1.

For example, if N = n = 2, then (8.37) is read, writing X = (ξ, δ) ∈ R2×2×R,

F (x, u, ξ, δ) ≥ α (x) + β1 |ξ|p1 + β2 |δ|p2

with p1 ≥ 2 and p2 > 1. Therefore f (ξ) = |ξ|2 + (det ξ)2

satisfies (8.37). ♦

Proof. (Theorem 8.31). We divide the proof into three steps.

Step 1. Let uν be a minimizing sequence for (P ) . Then by (8.35) and (8.36),we have ∫

Ω

a (x) dx + b1

∫

Ω

|∇uν (x)|p dx ≤ m + 1 < +∞.

Using Poincare inequality we find that there exists a constant γ so that

‖uν‖W 1,p ≤ γ

and therefore, up to the extraction of a subsequence still denoted uν , we have


).

Since p > n ∧N, we deduce from Theorem 8.20 that

uν → u in Lp

(Ω; RN

)


).

(8.38)

Step 2. We now show that (8.38) holds for p = p1 when we replace thecoercivity condition (8.35) by (8.37). By the same argument as in Step 1, wefind that there exists u ∈ W 1,p1

(Ω; RN

)and ξs ∈ Lps

(Ω; Rσ(s)

)so that, up to

the extraction of a subsequence,

uν u in W 1,p1(Ω; RN

)(8.39)

adjs∇uν ξs in Lps

(Ω; Rσ(s)

), s = 2, · · · , n ∧N. (8.40)

We now show that these two convergences imply ξs = adjs∇u and thus

adjs∇uν adjs∇u in L1(Ω; Rσ(s)

). (8.41)

This implies (8.38). We prove (8.41) by induction on s.If s = 2, then Theorem 8.20 combined with (8.39), (8.40) and the fact that

p1 ≥ 2 give immediately (8.41).Assume that we have proved (8.41) up to s− 1 ≤ (N ∧ n)− 2. We therefore

haveadjs−1∇uν ξs−1 = adjs−1∇u in Lps−1 .

Appendix: some properties of Jacobians 407

Since 1/p1 + 1/ps ≤ 1, we have immediately (8.41) by Part 5 of Theorem 8.20,combined with (8.40), and Step 2 is therefore complete.

Step 3. In view of Step 1 or Step 2, we are now in a position of applyingTheorem 8.16 to F. We therefore obtain that

lim infν→∞

I (uν) = lim infν→∞

∫

Ω

F (x, uν (x) , T (∇uν (x))) dx

≥∫

Ω

F (x, u (x) , T (∇u (x))) dx = I (u)

and thus u is a minimizer for (P ).

8.5 Appendix: some properties of Jacobians

We now let u : Rn → RN (hence ∇u ∈ RN×n) be a C2 function and studythe analytic properties of adjs∇u. We first introduce the notation for 2 ≤ s ≤n ∧N, 1 ≤ i1 < · · · < is ≤ n and 1 ≤ α1 < · · · < αs ≤ N,

∂(ui1 , · · · , uis

)

∂ (xα1 , · · · , xαs):= det

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

∂ui1

∂xα1

· · · ∂ui1

∂xαs

.... . .

...

∂uis

∂xα1

· · · ∂uis

∂xαs

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

For 1 ≤ t ≤ s, we also let

(x1, · · · , xt, · · · , xs) = (x1, · · · , xt−1, xt+1, · · · , xs) .

The main result is that Jacobians may be written in divergence form.

Theorem 8.33 Let u ∈ C2(Rn; RN

). Let 2 ≤ s ≤ n∧N, 1 ≤ i1 < · · · < is ≤ n

and 1 ≤ α1 < · · · < αs ≤ N. Then

s∑

t=1

(−1)t+1 ∂

∂xαt

[∂(ui2 , · · · , uis

)

∂ (xα1 , · · · , xαt , · · · , xαs)] = 0 (8.42)

and

∂(ui1 , · · · , uis

)

∂ (xα1 , · · · , xαs)=

s∑

t=1

(−1)t+1 ∂

∂xαt

[ ui1∂(ui2 , · · · , uis

)

∂ (xα1 , · · · , xαt , · · · , xαs)]. (8.43)

Remark 8.34 (i) Let N = n = 2, u (x1, x2) =(u1, u2

). Then (8.42) just

expresses the fact that

curl(gradu1

)= curl

(gradu2

)= 0.


Equation (8.43) is then

det∇u =∂

∂x1(u1 ∂u2

∂x2)− ∂

∂x2(u1 ∂u2

∂x1)

=∂

∂x2(u2 ∂u1

∂x1)− ∂

∂x1(u2 ∂u1

∂x2).

(ii) It is obvious that in (8.42) and (8.43) one can interchange the role of ui1

with any uit . ♦

Proof. Note that (8.43) is a direct consequence of (8.42). To show (8.42) weproceed by induction. To simplify notation, we take (i1, · · · , is) = (1, · · · , s) and(α1, · · · , αs) = (1, · · · , s) . Assume that (8.42) and (8.43) have been establishedup to the order (s− 1) , the case s = 1 being trivial. Using the hypothesis ofinduction we have

∂(u2, · · · , us

)

∂ (x1, · · · , xt, · · · , xs)

=

t−1∑

α=1

(−1)α+1 ∂

∂xα(u2

∂(u3, · · · , us

)

∂ (x1, · · · , xα, · · · , xt, · · · , xs))

+s∑

α=t+1

(−1)α ∂

∂xα(u2

∂(u3, · · · , us

)

∂ (x1, · · · , xt, · · · , xα, · · · , xs))

where we have denoted, as above,

(x1, · · · , xt, · · · , xs) = (x1, · · · , xt−1, xt+1, · · · , xs)

and similarly for (x1, · · · , xt, · · · , xα, · · · , xs) . Returning to (8.42) and using theabove identity we have

s∑

t=1

(−1)t+1 ∂

∂xt[

∂(u2, · · · , us

)

∂ (x1, · · · , xt, · · · , xs)]

=

s∑

t=1

[

t−1∑

α=1

(−1)α+t ∂2

∂xt∂xα(u2

∂(u3, · · · , us

)

∂ (x1, · · · , xα, · · · , xt, · · · , xs))

+s∑

α=t+1

(−1)α+t+1 ∂2

∂xt∂xα(u2

∂(u3, · · · , us

)

∂ (x1, · · · , xt, · · · , xα, · · · , xs)) ].

Now observe that for any r < β

∂2

∂xr∂xβ(u2 ∂

(u3, · · · , us

)

∂ (x1, · · · , xr, · · · , xβ , · · · , xs))[ (−1)

r+β+ (−1)

r+β+1] ≡ 0.

We therefore have, by combining the last two identities, established (8.42).

Theorem 8.33 allows us to prove now Lemma 5.5 and at the same time togeneralize it.


Theorem 8.35 Let Ω ⊂ Rn be a bounded open set and let

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) .

Then ∫

Ω

T (ξ +∇ϕ (x)) dx = T (ξ) ·measΩ (8.44)

for every ξ ∈ RN×n and for every ϕ ∈W 1,∞0

(Ω; RN

).

(ii) Let u ∈ v + W 1,p0 (Ω; RN ), with p ≥ n ∧N. Then∫

Ω

T (∇u (x)) dx =

∫

Ω

T (∇v (x)) dx.

Observe that when N = n = 2, the theorem reads as

T (ξ) = (ξ,det ξ)

and hence (8.44) becomes

T (ξ) ·meas Ω = (ξ,det ξ) ·measΩ

=

∫

Ω

(ξ +∇ϕ (x) , det (ξ +∇ϕ (x))) dx.

We now proceed with the proof of Theorem 8.35.

Proof. It is clear that (ii) is more general than (i); however, for the sake ofexposition, we still establish (i).

(i) Step 1. We first prove the result when ϕ ∈ C∞0

(Ω; RN

). In order to

establish (8.44), we only need to show that∫

Ω

adjs (ξ +∇ϕ (x)) dx = adjs ξ ·meas Ω (8.45)

for every 1 ≤ s ≤ n ∧ N, for every ξ ∈ RN×n and for every ϕ ∈ C∞0

(Ω; RN

).

Recall that for 1 ≤ i ≤(Ns

), 1 ≤ α ≤

(ns

), we have

(adjs ξ)iα = (−1)i+α det

⎛⎜⎜⎜⎜⎝

ξi1α1

· · · ξi1αs

.... . .

...

ξisα1

· · · ξisαs

⎞⎟⎟⎟⎟⎠

.

By abuse of notation, let

ξ =

⎛⎜⎜⎜⎜⎝

ξi1α1

· · · ξi1αs

.... . .

...

ξisα1

· · · ξisαs

⎞⎟⎟⎟⎟⎠

, ∇ϕ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

∂ϕi1

∂xα1

· · · ∂ϕi1

∂xαs

.... . .

...

∂ϕis

∂xα1

· · · ∂ϕis

∂xαs

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

.


Therefore (8.44) or (8.45) is equivalent to showing that∫

Ω

det (ξ +∇ϕ (x)) dx = det ξ ·measΩ (8.46)

for every bounded domain Ω ⊂ Rs, for every ξ ∈ Rs×s and for every ϕ ∈C∞

0 (Ω; Rs) . To show (8.46), we proceed by induction on s. The result is trivialif s = 1. Assume therefore that (8.46) has been established up to the order s−1.Using Proposition 5.65, we have that

det (ξ +∇ϕ)

= 〈 (ξ +∇ϕ)1;(adjs−1 (ξ +∇ϕ)

)1 〉= 〈 ξ1;

(adjs−1 (ξ +∇ϕ)

)1 〉+ 〈 (∇ϕ)1 ;(adjs−1 (ξ +∇ϕ)

)1 〉

=

s∑

α=1

[ ξ1α


)1α

+∂ϕ1

∂xα


)1α

].

Integrating the above identity, using the hypothesis of induction on the firstpart, an integration by part in the second term and (8.42), we have indeedobtained (8.46) and thus the theorem.

Step 2. From Step 1, we know that (8.44) holds for ϕ ∈ C∞0

(Ω; RN

). By an

elementary density argument we have the claim.

(ii) As in Step 1 of (i), it is clearly enough to prove the result when N = nand for ∫

Ω

det∇u (x) dx =

∫

Ω

det∇v (x) dx (8.47)

for every u ∈ v + W 1,n0 (Ω; Rn). By density, it will be sufficient to prove the

identity for u and v of the form

u = v + w

with v ∈ C∞ (Ω; Rn) ∩W 1,n(Ω; Rn) and w ∈ C∞0 (Ω; Rn) .

1) As usual we start with the case n = 2 to illustrate the purpose. We notethat

det∇u = det∇v + det(∇v1,∇w2

)+ det

(∇w1,∇v2

)+ det∇w

= det∇v +∂

∂x2(w2 ∂v1

∂x1)− ∂

∂x1(w2 ∂v1

∂x2)

+∂

∂x1(w1 ∂v2

∂x2)− ∂

∂x2(w1 ∂v2

∂x1) + det∇w.

Integrating both sides, we have, since w ∈ C∞0

(Ω; R2

), the identity (8.47).

2) We now proceed with the general case. We appeal to Proposition 5.67 towrite

det∇u = det∇v +∑

(I,J)∈N1,··· ,n

J =∅

det(∇vI ,∇wJ

).


Clearly, if we can show that, for every (I, J) ∈ N1,··· ,n with J = ∅, we have

∫

Ω

det(∇vI (x) ,∇wJ (x)

)dx = 0, (8.48)

then the result (8.47) will follow.

Since J = ∅, we can choose j ∈ J and use (8.43) in Theorem 8.33 to obtain

det(∇vI ,∇wJ

)=

n∑

t=1

(−1)t+1 ∂

∂xt[ wj ∂

(vI , wJ−j)

∂ (x1, · · · , xt, · · · , xn)].

Integrating this last identity, bearing in mind that w ∈ C∞0 (Ω; Rn) , we have

indeed obtained (8.48).

Chapter 9

Relaxation theorems

9.1 Introduction

In Chapters 3 and 8 we have seen that in order to get existence theorems for

(P ) inf

I (u) =

∫

Ω

f (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

the convexity (or quasiconvexity in the vectorial case) of f, with respect to thelast variable, plays a central role. In this chapter, we study the case where ffails to be convex (quasiconvex in the vectorial case).

It is then natural to replace the problem (P ) by the so called relaxed problem

(QP ) inf

I (u) =

∫

Ω

Qf (x, u (x) ,∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where Qf is the quasiconvex envelope of f (with respect to the last variable∇u). We show that, even though the original f is not quasiconvex (not convexin the scalar case) and therefore in general the infimum of (P ) is not attained,one has

inf (P ) = inf (QP ) ,

and with some extra coercivity condition, the infimum of (QP ) is attained. Moreprecisely, if u is a solution of (QP ), then there exists a minimizing sequence uνof (P ) such that

uν u in W 1,p

0

(Ω; RN

)

I (uν)→ I (u) = inf (P ) = inf (QP ) .

In other words, even if (P ) has no solution in W 1,p(Ω; RN

), one can consider

the solutions of (QP ) as generalized solutions of (P ) , in the sense of weakconvergence.

In the case N = n = 1, this result was proved by L.C. Young [606], [608]and then generalized by others to the scalar case, N = 1 or n = 1, notably

416 Relaxation theorems

by Berliocchi-Lasry [81], Ekeland [262], [264], Ioffe-Tihomirov [351], MacShane[412], [413] and Marcellini-Sbordone [427], [428]. Note that in this context

Qf = Cf = f∗∗,

where Cf is the usual convex envelope of f (with respect to the last variable).

The result for the vectorial case (i.e. N, n > 1; recall also that, in general,we now have Qf > Cf) was established by Dacorogna in [172], when there isno lower order terms. Following a different approach, it was later also provedby Acerbi-Fusco [3].

9.2 Relaxation Theorems

9.2.1 The case without lower order terms

We now turn our attention to the relaxation theorem when the integrand dependsonly on the higher order terms. We recall our minimization problem

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

),

where 1 ≤ p ≤ ∞.

We define the relaxed problem associated to (P ) to be

(QP ) inf

I (u) =

∫

Ω

Qf (∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

).

Theorem 9.1 (Relaxation theorem) Let Ω ⊂ Rn be a bounded open set.Let f : RN×n → R be Borel measurable satisfying, for 1 ≤ p <∞,

g (ξ) ≤ f (ξ) and |g (ξ)| , |f (ξ)| ≤ α1 (1 + |ξ|p) for every ξ ∈ RN×n, (9.1)

where g : RN×n → R is quasiconvex and α1 > 0 is a constant, while for p = ∞it is assumed that f is locally bounded and bounded below by g.

LetQf (ξ) := sup g (ξ) : g ≤ f and g quasiconvex

be the quasiconvex envelope of f.

Part 1. Theninf (P ) = inf(QP ).

More precisely, for every p ≤ q ≤ ∞ and u ∈ W 1,q(Ω; RN

), there exists a

sequence uν∞ν=1 ⊂ u + W 1,q0

(Ω; RN

)such that

uν → u in Lq(Ω; RN

)as ν →∞,

Relaxation Theorems 417

∫

Ω

f (∇uν (x)) dx →∫

Ω

Qf (∇u (x)) dx as ν →∞.

Part 2. Let α2 > 0, I ≥ 1 be an integer, ∞ > p ≥ pi > 1, i = 1, · · · , I andΦi : RN×n → R, i = 1, · · · , I, be quasiaffine functions satisfying

maxi=1,··· ,I

|Φi (ξ)|pi ≤ α2 (1 + |ξ|p) for every ξ ∈ RN×n.

Assume, in addition to the above hypothesis and those of Part 1, that there exista, c ∈ R, di ≥ bi > 0 such that

(C) a +I∑

i=1

bi |Φi (ξ)|pi ≤ f (ξ) ≤ c +I∑

i=1

di |Φi (ξ)|pi

for every ξ ∈ RN×n. Then, in addition to the conclusions of Part 1, the followingholds

Φi (∇uν) Φi (∇u) in Lpi (Ω) , i = 1, · · · , I, as ν →∞.

Before making some remarks, we give two significant examples of functionssatisfying (C).

Example 9.2 (i) The case where f satisfies a condition of the type

a + b |ξ|p ≤ f (ξ) ≤ c + d |ξ|p

is a particular case of (C). It suffices to choose I = N × n, pi = p > 1,

di = d ≥ bi = b > 0, i = 1, · · · , I and for ξ =(ξij

)1≤i≤N

1≤j≤n,

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Φ1 (ξ) = ξ11 , · · · , Φn (ξ) = ξ1

n

Φn+1 (ξ) = ξ21 , · · · , Φ2n (ξ) = ξ2

n

...

Φ(N−1)n+1 (ξ) = ξN1 , · · · , ΦNn (ξ) = ξN

n ,

which are all quasiaffine. In this case, the theorem implies that


)as ν →∞.

(ii) If N = n, p > n and

a + b |det ξ|p/n ≤ f (ξ) ≤ c + d |det ξ|p/n,

then choose in (C) I = 1, p1 = p/n and Φ1 (ξ) = det ξ which is quasiaffine. Wetherefore also have

det∇uν det∇u in Lp/n (Ω) as ν →∞,


and if f (ξ) = F (det ξ) , we then find

∫

Ω

F (det∇uν (x)) dx →∫

Ω

CF (det∇u (x)) dx,

since, by Theorem 6.24, Qf = CF = F ∗∗. ♦

We should also note that, in general, the sequence uν , which convergesin Lq, does not converge in any Sobolev space, as the following simple exampleshows.

Example 9.3 Consider n = N = 1, f (ξ) = e−|ξ| and

(P ) inf

I (u) =

∫ 1

0

f (u′ (x)) dx : u ∈W 1,10 (0, 1)

= m.

Since Cf = Qf ≡ 0, we have from the relaxation theorem that

inf (P ) = inf(QP ) = 0.

However, any sequence uν∞ν=1 ⊂W 1,10 (0, 1) such that

∫ 1

0

f (u′ν (x)) dx →

∫ 1

0

Qf (u′ (x)) dx = 0 as ν →∞

also satisfies by Jensen inequality

0 ≤ e−∫

1

0|u′

ν |dx ≤∫ 1

0

e−|u′ν | dx =

∫ 1

0

f (u′ν (x)) dx → 0

and therefore must satisfy ∫ 1

0

|u′ν | dx →∞.

Thus it cannot converge weakly to any u in W 1,1, though there exists a sequencesatisfying

uν → u in L∞ (0, 1) as ν →∞. ♦

We now make some further remarks.

Remark 9.4 (i) The history of the theorem has already been discussed. Thisapproach of relaxing non-convex (non-quasiconvex in the vectorial case) prob-lems is not the only one. There is a closely related idea due to L.C. Young[606], [608] (and in fact prior to the one presented here); see also MacShane[412], [413] that, instead of replacing f by Qf, enlarges the space of admissiblefunctions from Sobolev spaces to spaces of parametrized measures (called gen-eralized curve by L.C. Young and nowadays called Young measure). This ideaof L.C. Young has been very fruitful in the calculus of variations as well as inoptimal control theory and in partial differential equations following the workof Tartar [568]; see also Dacorogna [173] and Pedregal [492].


(ii) Theorem 9.1 implies in particular that if (QP ) has a solution u, thenthere exists a minimizing sequence uν for (P ) satisfying the conclusions ofthe theorem. Conversely, since every minimizing sequence uν for (P ) is alsoa minimizing sequence of (QP ), then, up to the extraction of a subsequence,uν converges weakly to a solution u of (QP ) (provided it exists). One shouldunderstand, in this sense, that solutions of (QP ) are generalized solutions of(P ) .

(iii) Note also that in the scalar case (QP ) and (P ∗∗) are the same problemswhere

(P ∗∗) inf

I∗∗ (u) =

∫

Ω

f∗∗ (∇u (x)) dx : u ∈ u0 + W 1,p0

(Ω; RN

).

However, this is not true in the vectorial case, one has in general

inf (P ) = inf (QP ) > inf (P ∗∗) .

For example, if N = n ≥ 2 and f (ξ) = (det ξ)2, then

f (ξ) = Qf (ξ) = (det ξ)2

> Cf (ξ) = f∗∗ (ξ) ≡ 0

(see Theorem 6.24) and therefore, using Jensen inequality and Theorem 8.35,we have, if det∇u0 > 0,

inf (P ) = inf (QP ) ≥ measΩ (1

measΩ

∫

Ω

det∇u0 (x) dx )2

> 0 = (inf P ∗∗) .

One can prove, for some u0 and Ω, (see Corollary 14.9) that

inf (P ) = inf (QP ) = (1

measΩ

∫

Ω

det∇u0 (x) dx )2 measΩ

and the infimum of (QP ) is attained.

(iv) The above Part 2 of the theorem does not apply to area type problems(see Section 5.3.6), since the growth condition (C) holds in this case with pi = 1.However, the minimal surface problem in parametric form can be handled in asimilar way; see Dacorogna [171].

(v) With the notation

I (u) =

∫

Ω

f (∇u (x)) dx, I (u) =

∫

Ω

Qf (∇u (x)) dx,

we find that the theorem, under the hypotheses of Example 9.2 (i), implies that

I (u) = infuν

lim inf [ I (uν) : uν − u 0 in W 1,p

0

(Ω; RN

)]

.

(vi) Recently Anza Hafsa-Mandallena [32], [33], [34], [35] have extendedthe above theorem so as to take into account some constraints of nonlinearelasticity. ♦


We finally proceed with the proof of the theorem.


Step 1. We first show Part 1 when Ω = D = (0, 1)n

and u is affine on D,meaning that there exists ξ ∈ RN×n such that

∇u (x) = ξ for every x ∈ D.

Now use Theorem 6.9 to find ϕν ∈ W 1,∞0

(D; RN

)with the property that

∫

D

f (ξ +∇ϕν (x)) dx ≥ Qf (ξ) ≥ −1

ν+

∫

D

f (ξ +∇ϕν (x)) dx. (9.2)

Let s (ν) be an integer such that s (ν) →∞ as ν →∞ and satisfying

ν ‖ϕν‖L∞ ≤ s (ν) .

Extend ϕν by periodicity, in each variable, from D to Rn and let

ψν (x) :=1

s (ν)ϕν (s (ν) x) .

Note that ψν ∈W 1,∞0

(D; RN

). Finally, define

uν := u + ψν .

Note that, using the periodicity of ϕν , we have

∫

D

f (ξ +∇ψν (x)) dx =1

sn

∫

sD

f (ξ +∇ϕν (y)) dy =

∫

D

f (ξ +∇ϕν (y)) dy.

(9.3)Combining the above identity with (9.2), we have indeed shown that uν has allthe properties stated in Part 1, more precisely, uν ∈ u + W 1,∞

0

(Ω; RN

)and

uν → u in L∞ (Ω; RN)

as ν →∞,

0 ≤∫

Ω

[ f (∇uν (x))−Qf (∇u (x)) ]dx → 0 as ν →∞.

Step 2. We now show Part 2, still under the same restrictions as in Step 1.Since the following reasoning applies to each i = 1, · · · , I, we may assume, fornotational convenience, that I = 1 and we therefore write Φi = Φ and pi = p.We thus have to show that

Φ (ξ +∇ψν) Φ (ξ) in Lp (D) as ν →∞;

in other words, we have to prove that for every w ∈ Lp′

(D)

limν→∞

∫

D

[ Φ (ξ +∇ψν (x))− Φ (ξ) ]w (x) dx = 0. (9.4)


From the hypothesis (C), we have that there exists a constant γ, independentof ν, such that ∫

D

|Φ (ξ +∇ψν (x))|p dx ≤ γ. (9.5)

Since p > 1 and (9.5) holds, it is sufficient in order to show (9.4) to prove thatif

E = x0 + δD ⊂ D,

where x0 ∈ D, δ > 0, then∫

E

[ Φ (ξ +∇ψν (x))− Φ (ξ) ]dx → 0 as ν →∞. (9.6)

We will assume that x0 = 0, since by periodicity we can always get back to thiscase. We therefore have, letting s = s (ν) ,

∫

E

Φ (ξ +∇ψν (x)) dx

=

∫

E

Φ (ξ +∇ϕν (sx)) dx =1

sn

∫

sE

Φ (ξ +∇ϕν (y)) dy

=1

sn

∫

[sδ]D

Φ (ξ +∇ϕν (y)) dy +1

sn

∫

sE−[sδ]D


= ([sδ] /s)n∫

D

Φ (ξ +∇ϕν (y)) dy +1

sn

∫

sE−[sδ]D


where [sδ] denotes the integer part of sδ and where we have used in the aboveidentity the periodicity of ϕν (in a similar way as in (9.3)). Since ϕν = 0 on ∂Dand Φ is quasiaffine, we have∫

E

Φ (ξ +∇ψν (x)) dx = ([sδ] /s)n

Φ (ξ) +1

sn

∫

sE−[sδ]D

Φ (ξ +∇ϕν (y)) dy.

(9.7)Let us now estimate the last term in the above identity. Observe first that ify ∈ sE − [sδ]D, then there exists i ∈ 1, · · · , n such that

[sδ] ≤ yi ≤ sδ and 0 ≤ yi ≤ sδ, for every i ∈ 1, · · · , n.

We then find, using once more the periodicity of ϕν ,

∫

sE−[sδ]D

|Φ (ξ +∇ϕν (y))| dy ≤ n

∫ sδ

[sδ]

∫ sδ

0

· · ·∫ sδ

0

|Φ (ξ +∇ϕν (y))| dy1 · · · dyn

≤ n

∫ [sδ]+1

[sδ]

∫ [sδ]+1

0

· · ·∫ [sδ]+1

0

|Φ (ξ +∇ϕν (y))|

dy1 · · · dyn

≤ n([sδ] + 1)n−1

∫

D

|Φ (ξ +∇ϕν (y))| dy.


Hence, in view of (9.5), we have

| 1

sn

∫

sE−[sδ]D

Φ (ξ +∇ϕν (y)) dy | ≤ γ1

s

where γ1 denotes a constant. Therefore returning to (9.7), we deduce

|∫

E

[ Φ (ξ +∇ψν (x))− Φ (ξ) ]dx | ≤ (δn − [sδ]n

/sn) |Φ (ξ)|+ γ1

s.

Letting ν → ∞ and thus s → ∞, we have indeed obtained (9.6) and henceStep 2.

Step 3. We finally remove the assumptions of Steps 1 and 2, namely thatΩ = D = (0, 1)n and u is affine on D. First observe that Steps 1 and 2 remainunchanged if the unit cube is replaced by any cube with faces parallel to theaxes.

We start with an approximation of the given function u. Let ǫ > 0 bearbitrary, we can then find disjoint open cubes with faces parallel to the axesΩ1, · · · , Ωk ⊂ Ω, ξ1, · · · , ξk ∈ RN×n, γ independent of ǫ and v ∈ u+W 1,q

0

(Ω; RN

)

such that ⎧⎪⎨⎪⎩

meas[Ω− ∪k

j=1Ωj

]≤ ǫ

‖u‖W 1,q , ‖v‖W 1,q ≤ γ, ‖u− v‖W 1,q ≤ ǫ

∇v (x) = ξj if x ∈ Ωj

(9.8)

(if q = ∞, we only have ‖u− v‖W 1,r ≤ ǫ for every 1 ≤ r < ∞). We can thenfind a non-negative increasing function η satisfying η (t) → 0 as t→ 0 and suchthat, using the continuity of Qf and the growth condition on f,

∫

Ω

|Qf (∇u (x))−Qf (∇v (x))| dx ≤ η (ǫ) , (9.9)

0 ≤∫

Ω−∪kj=1Ωj

[ f (∇v (x))−Qf (∇v (x)) ]dx ≤ η (ǫ) . (9.10)

Indeed, let us discuss the case 1 ≤ p < ∞, the case p = ∞ being easier. Recallthat (see Theorem 5.3) any quasiconvex function is locally Lipschitz continuous.Since it also satisfies (9.1), we can find β > 0 (see Proposition 2.32) such that

|Qf (∇u)−Qf (∇v)| ≤ β( 1 + |∇u|p−1+ |∇v|p−1

) |∇u −∇v| .

Using Holder inequality, we obtain∫

Ω

|Qf (∇u)−Qf (∇v)| dx

≤ β

[∫

Ω

[ ( 1 + |∇u|p−1+ |∇v|p−1

) ]p

p−1

]p−1p[∫

Ω

|∇u−∇v|p] 1

p

and (9.9) therefore follows from (9.8). The inequality (9.10) follows from (9.8),since f (∇v) , Qf (∇v) ∈ L1.


Make the construction of Step 1 on every Ωj and find

uj,ν ∈ v + W 1,∞0

(Ωj ; RN

).

Then define

uν (x) =

uj,ν (x) if x ∈ Ωj , j = 1, · · · , kv (x) if x ∈ Ω− ∪k

j=1Ωj .

We get that uν ∈ u + W 1,q0

(Ω; RN

)and


), as ν →∞.

Let ǫ > 0 be fixed, then, for ν sufficiently large, we have from Step 1 that

0 ≤∫

∪kj=1Ωj

[ f (∇uν (x))−Qf (∇v (x)) ]dx ≤ ǫ meas[∪k

j=1Ωj

].

Using (9.10), we get

0 ≤∫

Ω−∪kj=1Ωj

[ f (∇uν (x))−Qf (∇v (x)) ]dx

=

∫

Ω−∪kj=1Ωj

[ f (∇v (x))−Qf (∇v (x)) ]dx ≤ η (ǫ) .

In other words, combining these inequalities, we have proved that

0 ≤∫

Ω

[ f (∇uν (x))−Qf (∇v (x)) ]dx ≤ η (ǫ) + ǫ measΩ.

Invoking (9.9), we find

|∫

Ω

[ f (∇uν (x))−Qf (∇u (x)) ]dx | ≤ 2η (ǫ) + ǫ measΩ.

Letting ǫ → 0 (and thus ν →∞), we have indeed obtained Part 1 of the theorem.

Let us now show Part 2. Let w ∈ Lp′i (Ω) , we then have

∫

Ω

[ Φi (∇uν)− Φi (∇u) ]w =

∫

Ω

[ Φi (∇uν)− Φi (∇v) ]w

+

∫

Ω

[ Φi (∇v)− Φi (∇u) ]w

=

k∑

j=1

∫

Ωj

[ Φi (∇uj,ν)− Φi (∇v) ]w

+

∫

Ω

[ Φi (∇v)− Φi (∇u) ]w.

Passing to the limit, appealing to Step 2, as ν →∞ on every Ωj , we get

limν→∞

∫

Ω

[ Φi (∇uν)− Φi (∇u) ]w =

∫

Ω

[ Φi (∇v)− Φi (∇u) ]w.


Next, using (9.8) and the fact that ǫ is arbitrary, we have indeed obtained Part2 of the theorem.

9.2.2 The general case

We now generalize the result of Theorem 9.1 to integrands that depend not onlyon ∇u but also on x and u.

We start with some general considerations on quasiconvex envelopes. Thefollowing proposition was established, in the scalar case N = 1, by Marcellini-Sbordone [428], following earlier work of Ekeland-Temam [264]. The vectorialversion is, essentially, in Acerbi-Fusco [3] and Marcellini [423].

Proposition 9.5 Let D ⊂ Rn be a bounded open set and

f : Rn × RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function. Assume that there exist a Caratheodory function

g : Rn × RN × RN×n → R, g = g (x, u, ξ) ,

quasiconvex in the last variable (i.e. ξ → g (x, u, ξ) is quasiconvex for almostevery x ∈ Rn and every u ∈ RN ), β ≥ 0, β ∈ L1 (Rn) and α, a continuous andincreasing (in each argument) function, satisfying

g (x, u, ξ) ≤ f (x, u, ξ) , |f (x, u, ξ)| ≤ β (x) + α (|u| , |ξ|)

for almost every x ∈ Rn and for every (u, ξ) ∈ RN × RN×n.

For almost every x ∈ Rn and for every (u, ξ) ∈ RN × RN×n, let

Qf (x, u, ξ) := inf

1

measD

∫

D

f (x, u, ξ +∇ϕ (y)) dy : ϕ ∈W 1,∞0

(D; RN

)

and, for r > 0, set

Qfr(x, u, ξ) := inf

1

measD

∫

D

f(x, u, ξ +∇ϕ(y))dy :ϕ ∈W 1,∞

0 (D; RN )

‖ξ +∇ϕ‖L∞ ≤ r

.

Part 1. The function Qf : Rn ×RN ×RN×n → R is measurable in x, uppersemicontinuous in u and quasiconvex in ξ (and hence continuous). Moreover,the definition of Qf is independent of the choice of the set D.

Part 2. The function Qfr : Rn × RN × RN×n → R is a Caratheodoryfunction on |ξ| < r, the definition of Qfr is independent of the choice of D and,furthermore,

Qf = limr→∞

Qfr .

Part 3. Moreover, Qf is a Caratheodory function provided f satisfies anyof the following conditions:


(a) f (x, u, ξ) = f1 (x, u) f2 (x, ξ) + f3 (x, u) with f1 (x, u) ≥ 0;

(b) f is continuous in u, uniformly with respect to ξ;

(c) there exist p > 1, γ1 > 0, γ2 ∈ L1loc (Rn) and θ : Rn × R → R a non-

negative Caratheodory function, increasing in the last argument, with θ (x, 0) =0 such that

γ2 (x) + γ1 |ξ|p ≤ f (x, u, ξ) ,

|f (x, u, ξ)− f (x, v, ξ)| ≤ θ (x, |u− v|) (1 + |ξ|p) ,

for almost every x ∈ Rn, every u, v ∈ RN and every ξ ∈ RN×n.

It should be emphasized that, in general, the function Qf is not aCaratheodory function, even in the scalar case N = 1 or n = 1, where Qfcoincides with the usual convex envelope Cf = f∗∗, as the following example,given by Marcellini-Sbordone [428], shows.

Example 9.6 Let N = n = 1 and consider the function

f (u, ξ) = (|ξ|+ 1)|u|

.

An easy computation (recall that here Qf = Cf = f∗∗ and that the convexenvelope is understood as the envelope only with respect to the variable ξ) givesthat

f∗∗ (u, ξ) =

f (u, ξ) if |u| ≥ 1

1 if |u| < 1.

Clearly the function u → f∗∗ (u, ξ) is not continuous, but only upper semicon-tinuous. ♦

Proof. Part 1. The measurability in x follows from Part 2 (ii) and (iii). Thequasiconvexity in ξ, as well as the independence of the definition on the set D,follows from Theorem 6.9. The only thing that remains to be proved is theupper semicontinuity in u.

Without loss of generality, we may assume that β and f are defined for every(x, u, ξ) ∈ Rn × RN × RN×n and that the inequalities assumed in the theoremhold everywhere. Moreover, choose x ∈ Rn so that the function

(u, ξ)→ f (x, u, ξ)

is continuous.So let uν , u ∈ RN be such that

uν → u as ν →∞

(assume without loss of generality that |uν | ≤ |u|+1) and let ǫ > 0 be arbitrary.By definition, we have (letting D be the unit cube of Rn) that there exists


ϕ ∈ W 1,∞0

(D; RN

)such that

Qf (x, u, ξ) ≥ −ǫ +

∫

D

f (x, u, ξ +∇ϕ (y)) dy

≥ −ǫ−∫

D

λν (y) dy +

∫

D

f (x, uν , ξ +∇ϕ (y)) dy

≥ −ǫ−∫

D

λν (y) dy + Qf (x, uν , ξ)

where

λν (y) := |f (x, u, ξ +∇ϕ (y))− f (x, uν , ξ +∇ϕ (y))| .

Observe that, by hypothesis, λν (y)→ 0 a.e. y ∈ D and

0 ≤ λν (y) ≤ 2 [β (x) + α (|u|+ 1, |ξ|+ ‖∇ϕ‖L∞) ] .

We may therefore conclude that

Qf (x, u, ξ) ≥ −ǫ− limν→∞

∫

D

λν (y) dy + lim supν→∞

Qf (x, uν , ξ)

≥ −ǫ + lim supν→∞

Qf (x, uν , ξ) .

Since ǫ > 0 is arbitrary, we have the claim.

Part 2. (i) The fact that the definition of Qfr is independent of the choiceof the set D is shown exactly as in Theorem 6.9.

(ii) We next prove that

Qf = limr→∞

Qfr .

Observe first that trivially

r ≥ s > 0 ⇒ Qf ≤ Qfr ≤ Qfs .

From the definition of Qf (choosing D to be the unit cube of Rn) we can find,for every ν ∈ N, ϕν ∈ W 1,∞

0

(D; RN

)such that

Qf (x, u, ξ) ≥ −1

ν+

∫

D

f (x, u, ξ +∇ϕν (y)) dy.

Denote

r (ν) := ‖ξ +∇ϕν‖L∞ .

From the definition of Qfr , we find

∫

D

f (x, u, ξ +∇ϕν (y)) dy ≥ Qfr(ν) (x, u, ξ) .


Combining the three inequalities, we get

Qfr(ν) (x, u, ξ) ≥ Qf (x, u, ξ) ≥ −1

ν+ Qfr(ν) (x, u, ξ) .

Letting ν →∞, we have the claim.

(iii) We now establish that, for every (u, ξ) ∈ RN × RN×n, the function

x → hr (x) := Qfr (x, u, ξ)

is measurable. Choose r > 0 sufficiently large so that |u| , |ξ| < r and let

Br := (u, ξ) ∈ RN × RN×n : |u| , |ξ| ≤ r.

It follows from Scorza-Dragoni theorem (Theorem 3.8) that, for every boundedopen set Ω ⊂ Rn and every ǫ > 0, there exists a compact set Kǫ ⊂ Ω such that

meas (Ω−Kǫ) ≤ ǫ and f |Kǫ×Bris continuous.

Therefore hr|Kǫis upper semicontinuous. Since ǫ > 0 is arbitrary we deduce

that hr is measurable in Ω and thus, since Ω is arbitrary, we have the claim.

(iv) Let us now show that Qfr is a Caratheodory function on |ξ| < r, wherer > 0 is given. We already discussed the measurability in x, so we now considerthe continuity in (u, ξ) . Let ǫ > 0 and (u, ξ) ∈ RN × RN×n, with |ξ| < r. Let

θ :=r − |ξ|

2> 0.

We have to prove that, for almost every x ∈ Rn, we can find λ = λ (ǫ, x, u, ξ, r) >0 such that

|u− v|+ |ξ − η| ≤ λ ⇒ |Qfr (x, u, ξ)−Qfr (x, v, η)| ≤ ǫ. (9.11)

(1) Since f is a Caratheodory function, we have, for almost every x ∈ Rn,that there exists δ = δ (ǫ, x, u, r) such that if

|A| , |B| ≤ r and |v| , |w| ≤ |u|+ 1

then|v − w|+ |A−B| ≤ δ ⇒ |f (x, v, A)− f (x, w, B)| ≤ ǫ/2.

(2) From the definition of Qfr (choosing D to be the unit cube of Rn), wecan find, for every ν ∈ N, ϕν ∈W 1,∞

0

(D; RN

)such that

‖B +∇ϕν‖L∞ ≤ r

Qfr (x, w, B) ≥ −1

ν+

∫

D

f (x, w, B +∇ϕν (y)) dy.


(3) We now let

t := δ/2r , λ := tθ .

Note that λ < δ (since θ ≤ r/2). Choosing δ smaller if necessary, we can assumethat δ < 2r, and hence t ∈ (0, 1) and 0 < λ < θ.

(4) From now on we assume that

|A−B| ≤ λ and |A| , |B| ≤ r − θ.

(5) Defining, for t as above,

ψν (y) := (1− t)ϕν (y)

we have ψν ∈W 1,∞0

(D; RN

)and, using (2) and (4),

‖B +∇ψν‖L∞ ≤ t |B|+(1− t) ‖B +∇ϕν‖L∞ ≤ t (r − θ)+(1− t) r = r−tθ ≤ r.

Furthermore, using the above inequality and the definition of t and λ, we find,recalling that |A−B| ≤ λ and |A| , |B| ≤ r − θ,

‖A +∇ψν‖L∞ ≤ |A−B|+ ‖B +∇ψν‖L∞ ≤ λ + r − tθ = r.

Finally, noting that

‖∇ϕν‖L∞ ≤ |B|+ ‖B +∇ϕν‖L∞ ≤ 2r

we have

‖(B +∇ψν)− (B +∇ϕν)‖L∞ = t ‖∇ϕν‖L∞ ≤ 2rt = δ.

We now combine (1), (2) and (5) to get, for |w| ≤ |u|+ 1 and |B| ≤ r − θ,

Qfr (x, w, B) ≥ −1

ν+

∫

D

f (x, w, B +∇ϕν (y)) dy

≥ −1

ν+

∫

D

f (x, w, B +∇ψν (y)) dy

−∫

D

|f (x, w, B +∇ϕν (y)− f (x, w, B +∇ψν (y)))| dy

≥ −1

ν− ǫ

2+

∫

D

f (x, w, B +∇ψν (y)) dy.

Again using (5) we find, from the definition of Qfr ,

Qfr (x, v, A) ≤∫

D

f (x, v, A +∇ψν (y)) dy.


Therefore, if |A−B| ≤ λ, |A| , |B| ≤ r − θ and |v| , |w| ≤ |u| + 1, the twoinequalities lead to

Qfr (x, v, A)−Qfr (x, w, B)

≤ 1

ν+

ǫ

2+

∫

D

[f (x, v, A +∇ψν (y))− f (x, w, B +∇ψν (y))] dy.

We have therefore obtained, from (1) and the above inequality, that if |A| , |B| ≤r − θ, |v| , |w| ≤ |u|+ 1 and

|v − w|+ |A−B| ≤ λ

then

Qfr (x, v, A)−Qfr (x, w, B) ≤ 1

ν+ ǫ.

Letting ν →∞, we have found that

Qfr (x, v, A)−Qfr (x, w, B) ≤ ǫ.

Since the inequality Qfr (x, w, B) − Qfr (x, v, A) ≤ ǫ can be obtained in thesame way, we have indeed obtained (9.11).

Part 3. (a) The first statement results from the observation that

Qf (x, u, ξ) = f1 (x, u)Qf2 (x, ξ) + f3 (x, u)

and from the fact that Qf2 : Rn × RN×n → R is, according to Part 1, aCaratheodory function.

(b) Let ǫ > 0 and (u, ξ) ∈ RN × RN×n. We have to find, for almost everyx ∈ Rn, δ = δ (ǫ, x, u, ξ) > 0 such that

|u− v|+ |ξ − η| ≤ δ ⇒ |Qf (x, u, ξ)−Qf (x, v, η)| ≤ ǫ. (9.12)

(1) Since f is continuous in u uniformly with respect to the last variable, wecan find, for almost every x ∈ Rn and for every A ∈ RN×n, δ1 = δ1 (ǫ, x, u) >0 such that

|u− v| ≤ δ1 ⇒ |f (x, u, A)− f (x, v, A)| ≤ ǫ/4.

Using the definition of Qf (choosing D to be the unit cube of Rn), we haveϕǫ ∈W 1,∞

0

(D; RN

)such that

Qf (x, v, η) ≥ − ǫ

4+

∫

D

f (x, v, η +∇ϕǫ (y)) dy

≥ − ǫ

4−∫

D

λ (y)dy +

∫

D

f (x, u, η +∇ϕǫ (y)) dy,

whereλ (y) := |f (x, u, η +∇ϕǫ (y))− f (x, v, η +∇ϕǫ (y))| .


We therefore get from the definition of Qf (x, u, η) that, for |u− v| ≤ δ1 ,

Qf (x, u, η)−Qf (x, v, η) ≤ ǫ/2.

Since the opposite inequality is obtained in a similar manner, we get

|u− v| ≤ δ1 ⇒ |Qf (x, u, η)−Qf (x, v, η)| ≤ ǫ/2.

(2) The function Qf being continuous in ξ, by Part 1, we can find, for almostevery x ∈ Rn, δ2 = δ2 (ǫ, x, u, ξ) > 0 such that

|ξ − η| ≤ δ2 ⇒ |Qf (x, u, η)−Qf (x, u, ξ)| ≤ ǫ/2.

Letting δ = min δ1 , δ2 and combining the two inequalities, we have indeedobtained (9.12).

(c) The argument is very similar to the one in (b).

(1) We start by observing that, because of the coercivity condition on f, wehave, for almost every x ∈ Rn and every (v, η) ∈ RN ×RN×n with |v| , |η| ≤ R,that, if ϕǫ ∈W 1,∞

0

(D; RN

)is such that

Qf (x, v, η) ≥ −ǫ +

∫

D

f (x, v, η +∇ϕǫ (y)) dy,

then there exists a constant γ = γ (x, R) , independent of ǫ, such that

‖η +∇ϕǫ‖pLp ≤ γ.

(2) By hypothesis we have, for almost every x ∈ Rn and for every A ∈ RN×n,that

|f (x, u, A)− f (x, v, A)| ≤ θ (x, |u− v|) (1 + |A|p) .

Choose δ1 = δ1 (ǫ, x) such that

|u− v| ≤ δ1 ⇒ θ (x, |u− v|) ≤ ǫ.

Using the definition of Qf (choosing D to be the unit cube of Rn), we can findϕǫ ∈W 1,∞

0

(D; RN

)such that

Qf (x, v, η) ≥ −ǫ +

∫

D

f (x, v, η +∇ϕǫ (y)) dy

≥ −ǫ−∫

D

λ (y)dy +

∫

D

f (x, u, η +∇ϕǫ (y)) dy,

where

λ (y) := |f (x, u, η +∇ϕǫ (y))− f (x, v, η +∇ϕǫ (y))| .


We therefore get from the definition of Qf (x, u, η) that, for |u− v| ≤ δ1 ,

Qf (x, u, η)−Qf (x, v, η) ≤ (2 + ‖η +∇ϕǫ‖pLp) ǫ.

Invoking the constant γ in (1), we get that

Qf (x, u, η)−Qf (x, v, η) ≤ (2 + γ) ǫ.

Since the opposite inequality is obtained in a similar manner, we get

|u− v| ≤ δ1 ⇒ |Qf (x, u, η)−Qf (x, v, η)| ≤ (2 + γ) ǫ.

(3) The function Qf being continuous in ξ by Part 1, we can find, for almostevery x ∈ Rn, δ2 = δ2 (ǫ, x, u, ξ) > 0 such that

|ξ − η| ≤ δ2 ⇒ |Qf (x, u, η)−Qf (x, u, ξ)| ≤ ǫ.

Letting δ = min δ1 , δ2 and combining the two inequalities, we have indeedobtained

|u− v|+ |ξ − η| ≤ δ ⇒ |Qf (x, u, ξ)−Qf (x, v, η)| ≤ (3 + γ) ǫ.

This concludes the proof of the proposition.

We are now in a position to state the main theorem, but prior to that weexpress the growth condition that should satisfy the function.

Definition 9.7 Let 1 ≤ p ≤ ∞ and

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

be a Caratheodory function. We say that f satisfies growth condition (Gp) ifthere exists a Caratheodory function

g : Ω× RN × RN×n → R, g = g (x, u, ξ) ,

quasiconvex in the last variable (i.e. ξ → g (x, u, ξ) is quasiconvex for almostevery x ∈ Ω and every u ∈ RN ) and

g (x, u, ξ) ≤ f (x, u, ξ)

for almost every x ∈ Ω and for every (u, ξ) ∈ RN × RN×n.

Moreover the following inequalities hold for almost every x ∈ Ω and for every(u, ξ) ∈ RN × RN×n.

(i) When 1 ≤ p < ∞

(Gp) |g (x, u, ξ)| , |f (x, u, ξ)| ≤ α (1 + |u|p + |ξ|p) ,


where α ≥ 0 is a constant.

(ii) If p = ∞, it verifies

(G∞) |g (x, u, ξ)| , |f (x, u, ξ)| ≤ β (x) + α (|u| , |ξ|) ,

where α, β ≥ 0, β ∈ L1 (Ω) and α is a continuous and increasing (in eachargument) function.

Theorem 9.8 Let 1 ≤ p ≤ ∞, Ω ⊂ Rn be a bounded open set and

f : Ω× RN × RN×n → R, f = f (x, u, ξ) ,

a Caratheodory function satisfying growth condition (Gp) (see Definition 9.7).For almost every x ∈ Ω and for every (u, ξ) ∈ RN × RN×n, let

Qf (x, u, ξ) := inf

1

measD

∫

D

f (x, u, ξ +∇ϕ (y)) dy : ϕ ∈W 1,∞0

(D; RN

)

(D ⊂ Rn being a bounded open set), which is the quasiconvex envelope (withrespect to the last variable) of f. Assume that

Qf : Ω× RN × RN×n → R

is a Caratheodory function.

Part 1. Let p ≤ q ≤ ∞ and u ∈ W 1,q(Ω; RN

), then there exists a sequence

uν∞ν=1 ⊂ u + W 1,q0

(Ω; RN

)such that


)as ν →∞,

∫

Ω

f (x, uν (x) ,∇uν (x)) dx→∫

Ω

Qf (x, u (x) ,∇u (x)) dx as ν →∞.

Part 2. Assume, in addition to the hypotheses of Part 1, that 1 ≤ p < ∞and there exist α2 > 0 and α3 ∈ R such that, for almost every x ∈ Ω and forevery (u, ξ) ∈ RN × RN×n,

f (x, u, ξ) ≥ α2 |ξ|p + α3 . (9.13)

Then, in addition to the conclusions of Part 1, the following holds:


)as ν →∞.

Remark 9.9 (i) In the scalar case N = 1 or n = 1, we recall that Qf is nothingelse than Cf = f∗∗, which is the convex envelope of f with respect to the lastvariable.

(ii) Note that both hypotheses of Parts 1 and 2 of the present theorem arestronger than the corresponding ones in Theorem 9.1.


- Indeed, in Part 1 we have required continuity of f in the variable ξ. As itwill be seen in the proof, this is not necessary (upon some modification of theproof), since the main requirement of continuity (with respect to ξ) is on Qf,which is automatically continuous.

- In Part 2, we could also impose, instead of (9.13), a coercivity condition ofthe type (C) of Theorem 9.1, getting the corresponding conclusion. ♦

We now proceed with the proof of the theorem.

Proof. We divide the proof into three steps (the first two corresponding toPart 1 and the other to Part 2).

Step 1. We start by proving the theorem when p = ∞ and u ∈ C∞ (Ω; RN)

(this last restriction will be removed at the end of Step 1).

(1) Let r > 0 so that‖u‖W 1,∞ < r/2.

Let Qfr be as in Proposition 9.5, namely

Qfr(x, u, ξ) := inf

1

measD

∫

D

f(x, u, ξ +∇ϕ(y))dy :ϕ ∈W 1,∞

0

(D; RN

)

‖ξ +∇ϕ‖L∞ ≤ r

.

Observe that Qfr also satisfies (G∞) .

(2) Let η be a non-negative increasing function satisfying η (t)→ 0 as t→ 0and such that, for every measurable set A ⊂ Ω,

0 ≤∫

A

β (x) dx ≤ η (measA) ,

∫

A

|g (x, u (x) ,∇u (x))| dx,

∫

A

|Qfr (x, u (x) ,∇u (x))| dx ≤ η (measA) .

(3) Let ǫ > 0, we can then find M = M (ǫ) such that if

Eǫ := x ∈ Ω : β (x) ≤ M ,

then, letting k = α (2r, 2r) ,

meas (Ω− Eǫ) ≤ min ǫ, ǫ/2k

and, in particular,M meas (Ω− Eǫ) ≤ η (ǫ) .

(4) Appealing to Theorem 3.8, we can find a compact set Kǫ ⊂ Ω with

meas (Ω−Kǫ) ≤ min ǫ, ǫ/ (M + 2k)

and such that f : Kǫ×S2r → R and Qfr : Kǫ×Sr/2 → R are continuous, where

Sr :=(u, ξ) ∈ RN × RN×n : |u| , |ξ| ≤ r

.


We can therefore find δ = δ (ǫ) such that, if x, y ∈ Kǫ , (u, ξ) , (v, η) ∈ S2r , then

|x− y|+ |u− v|+ |ξ − η| ≤ δ ⇒ |f (x, u, ξ)− f (y, v, η)| ≤ ǫ/ measΩ

and, for x, y ∈ Kǫ , (u, ξ) , (v, η) ∈ Sr/2 , we also have

|x− y|+ |u− v|+ |ξ − η| ≤ δ ⇒ |Qfr (x, u, ξ)−Qfr (y, v, η)| ≤ ǫ/ measΩ.

(5) We then let h > 0 be small and decompose Ω in a finite union of disjointopen sets Ωs so that

meas(Ω−⋃Ss=1 Ωs ) = 0 and measΩs ≤ h.

We then fix xs ∈ Ωs ∩Kǫ ∩ Eǫ , whenever this set is non-empty, and define

(us, ξs) :=1

measΩs

∫

Ωs

(u (x) ,∇u (x)) dx.

Note that|us| , |ξs| < r/2

and, by choosing h sufficiently small, we can assume (recall that u ∈ C∞) that,for every x ∈ Ωs ,

|x− xs|+ |u (x) − us|+ |∇u (x)− ξs| ≤ δ/2.

(6) This has a direct consequence that

|∫

Ω

Qfr (x, u (x) ,∇u (x)) dx−∑Ss=1 Qfr (xs, us, ξs)measΩs | ≤ 3ǫ + 3η (ǫ) .

Indeed, letting

λs (x) := |Qfr (x, u (x) ,∇u (x))−Qfr (xs, us, ξs)| ,

we have, using (4) and (5), that

S∑

s=1

∫

Ωs∩Kǫ∩Eǫ

λs (x) dx ≤ ǫ.

Furthermore, since Qfr also satisfies (G∞) and xs ∈ Eǫ , we have

S∑

s=1

∫

Ωs−(Kǫ∩Eǫ)

λs (x) dx

≤S∑

s=1

[ β (xs) + 2α (r/2, r/2) ] meas [ Ωs − (Kǫ ∩ Eǫ) ] +

∫

Ω−(Kǫ∩Eǫ)

β (x) dx

≤ [ M + 2k ] [meas (Ω− Eǫ) + meas (Ω−Kǫ)] + 2η (ǫ) ≤ 2ǫ + 3η (ǫ) .


Combining the two estimates, we have, as wished,

S∑

s=1

∫

Ωs

λs (x) dx ≤ 3ǫ + 3η (ǫ) .

(7) Using the definition of Qfr and the same argument as in Theorem 9.1,we can find for every s a function ϕs

ν ∈W 1,∞0

(Ωs; RN

)with the properties that

ϕsν → 0 in L∞ (Ωs; RN

)as ν →∞, ‖ξs +∇ϕs

ν‖L∞ ≤ r

0 ≤∫

Ωs

[f (xs, us, ξs +∇ϕsν (x))−Qfr (xs, us, ξs)] dx ≤ ǫ measΩs

measΩ.

Letting

ϕν (x) := ϕsν (x) , for x ∈ Ωs

we have constructed a function ϕν ∈W 1,∞0

(Ω; RN

)satisfying

ϕν → 0 in L∞ (Ω; RN)

as ν →∞, sup1≤s≤S

‖ξs +∇ϕν‖L∞ ≤ r,

0 ≤S∑

s=1

∫

Ωs

[f (xs, us, ξs +∇ϕν (x))−Qfr (xs, us, ξs)]dx ≤ ǫ.

(8) We now estimate, as in (6) above,

S∑

s=1

∫

Ωs

|f (xs, us, ξs +∇ϕν (x))− f (x, u (x) + ϕν (x) ,∇u (x) +∇ϕν (x))| dx.

We denote

λsν (x) := |f (xs, us, ξs +∇ϕν (x))− f (x, u (x) + ϕν (x) ,∇u (x) +∇ϕν (x))| .

We then choose ν sufficiently large so that ‖ϕν‖L∞ ≤ δ/2 and hence, using (4)and (5),

S∑

s=1

∫

Ωs∩Kǫ∩Eǫ

λsν (x) dx ≤ ǫ.

Furthermore, since f satisfies (G∞) and xs ∈ Eǫ , we have

S∑

s=1

∫

Ωs−(Kǫ∩Eǫ)

λsν (x) dx

≤S∑

s=1

[ β (xs) + 2α (2r, 2r)] meas [Ωs − (Kǫ ∩ Eǫ)] +

∫

Ω−(Kǫ∩Eǫ)

β (x) dx

≤ [M + 2k] [meas (Ω− Eǫ) + meas (Ω−Kǫ)] + 2η (ǫ) ≤ 2ǫ + 3η (ǫ) .


Combining the two estimates, we have

S∑

s=1

∫

Ωs

λsν (x) dx ≤ 3ǫ + 3η (ǫ) .

We finally collect the estimates (6), (7) and (8) to get that, writing uν :=u+ϕν , for every ‖u‖W 1,∞ < r/2, there exists νǫ,r = ν (ǫ, r) such that for ν ≥ νǫ,r

|∫

Ω

Qfr (x, u (x) ,∇u (x)) dx−∫

Ω

f (x, uν (x) ,∇uν (x)) dx | ≤ 7ǫ + 6η (ǫ) .

(9.14)Using Proposition 9.5 and Lebesgue dominated convergence theorem, we canfind r = r (ǫ) so that

∫

Ω

|Qfr (x, u (x) ,∇u (x))−Qf (x, u (x) ,∇u (x))| dx ≤ ǫ.

Combining (9.14) and the above inequality, we find that for every ǫ > 0 we canfind νǫ = ν (ǫ, r (ǫ)) such that for ν ≥ νǫ

|∫

Ω

Qf (x, u (x) ,∇u (x)) dx−∫

Ω

f (x, uν (x) ,∇uν (x)) dx | ≤ 8ǫ + 6η (ǫ) .

Letting ǫ → 0, we have indeed obtained the result for any function u ∈C∞ (Ω; RN

).

In order to have the claim for u ∈W 1,∞ (Ω; RN)

we proceed as follows. Letǫ > 0 be arbitrary, we can then find an open set O ⊂ Ω with smooth boundary,γ > 0 independent of ǫ and v ∈ u + W 1,∞

0

(Ω; RN

)such that

meas

[Ω−O

]≤ ǫ, v ∈ C∞ (O; RN

)

‖u‖W 1,∞(Ω) , ‖v‖W 1,∞(Ω) ≤ γ, ‖u− v‖W 1,1(Ω) ≤ ǫ.

We can therefore find a non-negative increasing function η satisfying η (t) → 0as t→ 0 and so that

∫

Ω

|Qf (x, u (x) ,∇u (x))−Qf (x, v (x) ,∇v (x))| dx ≤ η (ǫ)

∫

Ω−O

|Qf (x, v (x) ,∇v (x))− f (x, v (x) ,∇v (x))| dx ≤ η (ǫ) .

We then apply the above construction to v and O to get uν ∈ v+W 1,∞0

(O; RN

)

such that

|∫

O

Qf (x, v (x) ,∇v (x)) dx−∫

O

f (x, uν (x) ,∇uν (x)) dx | ≤ ǫ.

Letting uν = v in Ω − O and combining the three estimates, we have indeedobtained the claim for p = ∞.

Step 2. We now discuss the case 1 ≤ p < ∞. We first approximate u ∈W 1,q

(Ω; RN

). Indeed, let ǫ > 0 be arbitrary, we can then find an open set O ⊂ Ω


with a smooth boundary, γ > 0 independent of ǫ and v ∈ u+W 1,q0

(Ω; RN

)such

that meas

[Ω−O

]≤ ǫ, v ∈ W 1,∞ (O; RN

)

‖u‖W 1,q(Ω) , ‖v‖W 1,q(Ω) ≤ γ, ‖u− v‖W 1,p(Ω) ≤ ǫ.

Since Qf satisfies (Gp) , we can find a non-negative increasing function η satis-fying η (t)→ 0 as t→ 0 and so that

∫

Ω

|Qf (x, u (x) ,∇u (x))−Qf (x, v (x) ,∇v (x))| dx ≤ η (ǫ)

∫

Ω−O

|Qf (x, v (x) ,∇v (x))− f (x, v (x) ,∇v (x))| dx ≤ η (ǫ) .

Apply Step 1 to find uν∞ν=1 ⊂ v + W 1,∞0

(O; RN

)such that

uν → v in L∞ (O; RN)

as ν →∞,

∫

O

f (x, uν (x) ,∇uν (x)) dx →∫

O

Qf (x, v (x) ,∇v (x)) dx as ν →∞.

Letting uν ≡ v in Ω − O, we have indeed, combining the two above estimates,established Part 1 of the theorem.

Step 3. We now conclude with Part 2. The coercivity condition (9.13) andPart 1 imply that, up to the extraction of a subsequence (still denoted uν),we have


)as ν →∞,

as wished.

Chapter 10

Implicit partial differentialequations

10.1 Introduction

In this chapter, we discuss the existence of solutions u ∈ W 1,∞ (Ω; RN)

for theDirichlet problem involving differential inclusions of the form

∇u (x) ∈ E a.e. x ∈ Ω

u (x) = ϕ (x) x ∈ ∂Ω,

where ϕ ∈ W 1,∞ (Ω; RN)

is a given map and E ⊂ RN×n is a given set. Closelyrelated is the implicit partial differential equation

F (∇u (x)) = 0 a.e. x ∈ Ω

u (x) = ϕ (x) x ∈ ∂Ω,

where F : RN×n → R is a given function. It suffices to set

E :=ξ ∈ RN×n : F (ξ) = 0

.

The results obtained here, combined with the relaxation theorems of Chapter9, will lead in Chapter 11 to proving the existence of minimizers in problems ofthe calculus of variations without appealing to lower semicontinuity theorems.

In the scalar case (n = 1 or N = 1), a sufficient condition, for finding asolution of our problem, is

∇ϕ (x) ∈ E ∪ int co E a.e. in Ω,

where int co E stands for the interior of the convex hull of E. This fact wasobserved by several authors with different proofs and different levels of general-ity, notably by Bressan-Flores [103], Cellina [134], Dacorogna-Marcellini [196],

440 Implicit partial differential equations

[198], [202], De Blasi-Pianigiani [230] or Friesecke [291]. It should be noted thatthis sufficient condition is, in some sense, a necessary one (see Theorem 10.24).

When turning to the vectorial case (n, N ≥ 2), the problem becomes consid-erably harder and no result with such a degree of elegancy is available. The firstgeneral results were obtained by Dacorogna-Marcellini (see the bibliography, inparticular [202]) and by Muller-Sverak [464] with the help of the method ofconvex integration of Gromov.

The presentation follows Dacorogna-Marcellini [202], to which we constantlyrefer. In particular, we do not discuss the case with lower order terms, meaningequations of the form

F (x, u (x) ,∇u (x)) = 0 a.e. x ∈ Ω,

which is considered in [202]. However, we mention several new results, whichwe discuss in detail.

The chapter is organized as follows. In Section 10.2, we present an abstractexistence theorem based on Baire category theorem. In Section 10.3, we give sev-eral examples, notably one involving singular values and one concerning poten-tial wells, where the abstract theorem applies.

10.2 Existence theorems

10.2.1 An abstract theorem

We start by recalling the notation for various convex hulls of sets (seeSection 7.3).

Notation 10.1 For E ⊂ RN×n, we let

FE∞ :=

f : RN×n → R ∪ +∞ : f |E ≤ 0

,

FE :=f : RN×n → R : f |E ≤ 0

.

We then have, respectively, that the convex, polyconvex and rank one convexhulls of E satisfy (see Theorem 7.20)

coE =ξ ∈ RN×n : f (ξ) ≤ 0 for every convex f ∈ FE

∞

,


∞

,


∞

,

while the finite quasiconvex hull of E is defined as


. ♦

The following definition was introduced by Dacorogna-Marcellini in [201](see also [202]) and it is the key condition to get the existence of solutions.


Definition 10.2 (Relaxation property) Let E, K ⊂ RN×n. We say that Khas the relaxation property with respect to E if for every bounded open setΩ ⊂ Rn, for every affine map uξ satisfying

∇uξ (x) = ξ ∈ K,

there exists a sequence uν ⊂ Affpiec

(Ω; RN

)with the following properties

uν ∈ uξ + W 1,∞0

(Ω; RN

), ∇uν (x) ∈ E ∪K, a.e. in Ω,

uν∗ uξ in W 1,∞,

∫

Ω

dist (∇uν (x) ; E) dx→ 0 as ν →∞.

Remark 10.3 (i) With the exception of the condition ∇uν (x) ∈ E ∪K, whosestatus as a necessary condition is unclear, the relaxation property is obviouslya natural condition for solving differential inclusions of the form u ∈ uξ +

W 1,∞0

(Ω; RN

)and

∇u (x) ∈ E

for every ξ ∈ K. Indeed it then just states that there is an approximate solutionof the problem under consideration.

(ii) It is interesting to note that in the scalar case (n = 1 or N = 1),K = int coE has the relaxation property with respect to E.

(iii) In the vectorial case, we have that, if K is bounded and has the relax-ation property with respect to E, then necessarily

K ⊂ Qcof E.

Indeed first recall that the definition of quasiconvexity implies that, for everyquasiconvex f ∈ FE ,

f (ξ)measΩ ≤∫

Ω

f (∇uν (x)) dx.

Combining this last result with the fact that ∇uν is uniformly bounded, thefact that any quasiconvex function is continuous and the last property in thedefinition of the relaxation property, we get the inclusion K ⊂ Qcof E. ♦

We now give the main abstract theorem.

Theorem 10.4 Let Ω ⊂ Rn be a bounded open set. Let E, K ⊂ RN×n be suchthat E is compact and K is bounded. Assume that K has the relaxation propertywith respect to E. Let ϕ ∈ Affpiec

(Ω; RN

)be such that

∇ϕ (x) ∈ E ∪K a.e. in Ω.

Then there exists (a dense set of) u ∈ ϕ + W 1,∞0

(Ω; RN

)such that

∇u (x) ∈ E a.e. in Ω.


Remark 10.5 (i) We will see in Section 10.3.1 that, in the scalar case, thelargest such K is

K = E ∪ int coE.

(ii) Although we will not discuss the details, we can consider (using the resultsin Chapter 10 in [202]) the more general boundary datum ϕ, if we make thefollowing extra hypotheses:

- in the scalar case (see Corollary 10.11 in [202]), if K is open, ϕ can evenbe taken in W 1,∞ (Ω; RN

)and considering the previous remark, we should have

(see Theorem 10.18 below) that

∇ϕ (x) ∈ E ∪ int co E;

- in the vectorial case, if the set K is open, ϕ can be taken in C1piec

(Ω; RN

)

(see Corollary 10.15 or Theorem 10.16 in [202]) with ∇ϕ (x) ∈ E ∪K; while ifK is open and convex, ϕ can be taken in W 1,∞ (Ω; RN

)provided

∇ϕ (x) ∈ C a.e. in Ω,

where C ⊂ K is compact (see Corollary 10.21 in [202]).

(iii) The present theorem was first proved by Dacorogna-Marcellini in [201](see also Theorem 6.3 in [202]) under the further hypothesis that

E =ξ ∈ RN×n : Fi (ξ) = 0, i = 1, 2, · · · , I

,

where Fi : RN×n → R, i = 1, 2, · · · , I, are quasiconvex. This hypothesis waslater removed by Sychev in [559] (see also Muller-Sychev [468]). Kirchheim in[364] pointed out that using a classical result (see Theorem 10.15 below), theproof of Dacorogna-Marcellini was still valid without the extra hypothesis onE. Kirchheim’s idea, combined with the proof of [202], is used below, followingDacorogna-Pisante [210].

(iv) The theorem can be extended to the case with higher derivatives or lowerorder terms, see Dacorogna-Marcellini [202], Dacorogna-Pisante [210], Muller-Sychev [468] and also Theorem 10.9 below. ♦

Proof. By working on each piece where ϕ is affine, we can assume, with-out loss of generality, that ϕ itself is affine. We next let V be the closure inL∞ (Ω; RN

)of

V :=u ∈ Affpiec

(Ω; RN

): u ∈ ϕ + W 1,∞

0

(Ω; RN

)and ∇u (x) ∈ E ∪K

.

V is non empty since ϕ ∈ V. Let, for k ∈ N,

V k := int

u ∈ V :

∫

Ω

dist (∇u (x) ; E) dx ≤ 1

k


where ”int” stands for the interior of the set.

We claim that V k is, in addition to be open, dense in the complete metricspace V . Postponing the proof of this fact for the end of the proof, we concludeby Baire category theorem that

∞⋂

k=1

V k ⊂u ∈ V : dist (∇u (x) ; E) = 0, a.e. in Ω

⊂ V

is dense, and hence non empty, in V . The result then follows, since E is compact.

We finally show that V k is dense in V . So let u ∈ V and ǫ > 0 be arbitrary.We wish to find v ∈ V k so that

‖u− v‖L∞ ≤ ǫ.

We recall (cf. Section 10.2.3) that

ω∇(u) := limδ→0

supϕ,ψ∈Bδ(u)

‖∇ϕ−∇ψ‖L1

whereBδ (u) :=

v ∈ V : ‖u− v‖L∞ < δ

.

- We start by finding α ∈ V a point of continuity of the operator ∇ (inparticular ω∇ (α) = 0) so that

‖u− α‖L∞ ≤ ǫ/3.

This is always possible by virtue of Theorem 10.15 and Proposition 10.17.

- We next approximate α ∈ V by β ∈ V so that, using Proposition 10.14,

‖β − α‖L∞ ≤ ǫ/3 and ω∇ (β) < 1/4k .

- Finally we use the relaxation property on every piece where ∇β is constantand we then construct v ∈ V, by patching all the pieces together, such that

‖β − v‖L∞ ≤ ǫ/3, ω∇ (v) < 1/2k and

∫

Ω

dist (∇v (x) ; E) dx < 1/2k.

Moreover, since ω∇ (v) < 1/2k, we can find δ = δ (k, v) > 0 so that

‖v − ψ‖L∞ ≤ δ ⇒ ‖∇v −∇ψ‖L1 ≤ 1/2k

and hence∫

Ω

dist (∇ψ (x) ; E) dx ≤∫

Ω

dist (∇v (x) ; E) dx + ‖∇ψ −∇v‖L1 < 1/k

for every ψ ∈ Bδ (v) ; which implies that v ∈ V k.

Combining these three facts we have indeed obtained the desired densityresult.


10.2.2 A sufficient condition for the relaxation property

We now give a sufficient condition that ensures the relaxation property. In con-crete examples, this condition is usually much easier to check than the relaxationproperty. It resembles the so called in approximation in the convex integrationmethod of Gromov as revisited by Muller and Sverak. We start with a definition.

Definition 10.6 (Approximation property) Let d be an integer and

E ⊂ K ⊂M ⊂ Rd.

We say that (E, K,M) (when M = Rd we simply write (E, K)) has the approx-imation property if there exists a family of sets Eδ and K (Eδ) , δ > 0, suchthat

(1) Eδ ⊂ K (Eδ) ⊂⊂ intM K for every δ > 0 (where intM stands for theinterior relative to M and A ⊂⊂ B means that A ⊂ B and is compact);

(2) for every ǫ > 0 there exists δ0 = δ0 (ǫ) > 0 such that dist(η; E) ≤ ǫ forevery η ∈ Eδ and δ ∈ (0, δ0];

(3) if η ∈ intM K, then η ∈ K (Eδ) for every δ > 0 sufficiently small.

Before proceeding further, we first recall the notation (see Chapter 12) forthe higher derivatives. The aim is to write in a simple way the matrix ∇mu ofall partial derivatives of order m of a map u : Rn → RN .

Notation 10.7 (i) Let N, n, m ≥ 1 be integers. We denote by RN×nm

s the setof matrices

A =(Ai

j1···jm

)1≤i≤N

1≤j1,··· ,jm≤n∈ RN×nm

such that for every permutation σ of j1, · · · , jm we have

Aiσ(j1···jm) = Ai

j1···jm.

The number of different entries (because of the different symmetries) is

N ×(n+m−1

m

).

- When m = 1, we have RN×ns = RN×n.

- When N = 1 and m = 2, we get Rn2

s = Rn×ns , i.e. the set of symmetric

matrices.

(ii) Let u : Rn → RN . We therefore have

∇mu =

(∂mui

∂xj1 · · · ∂xjm

)1≤i≤N

1≤j1,··· ,jm≤n

∈ RN×nm

s .

- When m = 1, this is the usual gradient map.


- When N = 1 and n = m = 2, we have

∇2u =

(∂2u/∂x2

1 ∂2u/∂x1∂x2

∂2u/∂x1∂x2 ∂2u/∂x22

)∈ R2×2

s .

(iii) Given α ∈ Rn denote by α⊗m = α⊗ α · · · ⊗ α (m times), it is a matrixin Rnm

s . Therefore a generic matrix of rank one in RN×nm

s is of the form

β ⊗ α⊗m =(βiαj1 · · ·αjm

)1≤i≤N

1≤j1,··· ,jm≤n,

where β ∈ RN and α ∈ Rn.- When m = 1, this is the usual tensorial product; i.e.

β ⊗ α =(βiαj

)1≤i≤N

1≤j≤n.

- When N = 1 and n = m = 2, we have

α⊗2 = α⊗ α =

((α1 )2 α1α2

α1α2 (α2 )2

)∈ R2×2

s . ♦

The next result gives the appropriate generalization to higher derivatives ofLemma 3.11. Its proof is very similar and we will not reproduce it here; we referto Lemma 6.8 in Dacorogna-Marcellini [202] for details.

Lemma 10.8 Let Ω ⊂ Rn be an open set with finite measure. Let t ∈ [0, 1] andA, B ∈ RN×nm

s with rankA−B = 1. Let ϕ be such that

∇mϕ(x) = tA + (1− t)B, ∀x ∈ Ω.

Then, for every ǫ > 0, there exist u ∈ Affmpiec

(Ω; RN


ΩA , ΩB ⊂ Ω, such that⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

|measΩA − t measΩ| , |meas ΩB − (1− t)meas Ω| ≤ ǫ

u ≡ ϕ near ∂Ω, ‖u− ϕ‖W m−1,∞ ≤ ǫ

∇mu(x) =

A in ΩA

B in ΩB

dist (∇mu(x), co A, B) ≤ ǫ a.e. in Ω.

We now give the following theorem (see Theorem 6.14 in [202] and for aslightly more flexible one, see Theorem 6.15 in [202]), which will be used in animportant way in the examples.

Theorem 10.9 Let m, N, n ∈ N and

E ⊂M = RN×nm

s ⊂ RN×nm

with E compact. Assume that there exist Eδ such that Eδ , K (Eδ) = RcoEδ

satisfy (1), (2) and (3) in the definition of the approximation property withK = intM Rco E. Then K has the relaxation property with respect to E.


Remark 10.10 (i) We recall that Rco E stands for the rank one convex hullof E (see Section 7.3.1). Note also that M is convex and thus rank one convex.Consequently, if E ⊂M, then RcoE ⊂M.

(ii) The theorem contains, in particular, the case where m = 1 and hence

M = RN×n.

The theorem will always be applied when m = 1 in the examples, but in Section10.3.5 where N = 1 and n = m = 2. ♦

Proof. We wish to show that for every bounded open set Ω ⊂ Rn, for everyaffine map uξ ∈ Affpiec(Ω; RN×nm−1

) satisfying

∇uξ (x) = ξ ∈ K,

there exists a sequence uν ⊂ Affpiec(Ω; RN×nm−1

) such that

uν ∈ uξ + W 1,∞0 (Ω; RN×nm−1

), ∇uν (x) ∈ E ∪K, a.e. in Ω


∫

Ω

dist (∇uν (x) ; E) dx→ 0 as ν →∞.

Since ξ ∈ intM RcoE, we have, by (3) in the definition of the approximationproperty, that, for any δ > 0 sufficiently small,

ξ ∈ RcoEδ .

We can therefore find (cf. Theorem 7.17) an integer I so that

ξ ∈ RI coEδ .

We then proceed by induction on I.

Case: I = 1. We can thus find t ∈ [0, 1] and

A, B ∈ Eδ with rank A−B = 1

so that

ξ = tA + (1 − t)B.

We then use Lemma 10.8 (with ϕ such that ∇mϕ = ξ) to get that there existsa sequence

vν ⊂ Affmpiec

(Ω; RN

)

such that, if we write

uν = ∇m−1vν ∈ Affpiec(Ω; RN×nm−1

)


then (letting Ω := ΩA ∪ ΩB and ǫ = 1/ν in the conclusion of the lemma)

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

0 ≤ measΩ−meas Ω ≤ 1/ν

uν ∈ uξ + W 1,∞0 (Ω; RN×nm−1

)

‖uν − uξ‖L∞ ≤ 1/ν

∇uν(x) ∈ Eδ , a.e. in Ω

dist (∇uν(x), R1 coEδ) ≤ 1/ν, a.e. in Ω.

Since ∇uν ∈ M, RcoEδ ⊂⊂ K (by (1) in the definition of the approximationproperty), K is a bounded set and (2) in the definition of the approximationproperty holds, we have the claim.

Case: I > 1. We now assume that the result has been proved for I − 1 andlet us show the claim for I. We can thus find t ∈ [0, 1] and

A, B ∈ RI−1 coEδ with rank A−B = 1

so thatξ = tA + (1 − t)B.

Appealing to Lemma 10.8 (with ϕ such that ∇mϕ = ξ), we get that there existsa sequence

ψν ⊂ Affmpiec

(Ω; RN

)

such that, if we write

ϕν = ∇m−1ψν ∈ Affpiec(Ω; RN×nm−1

)

then ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

|meas (ΩA ∪ ΩB)−measΩ| ≤ 1/ν

ϕν ∈ uξ + W 1,∞0 (Ω; RN×nm−1

)

‖ϕν − uξ‖L∞ ≤ 1/ν

∇ϕν(x) =

A in ΩA

B in ΩB

dist (∇ϕν(x), RI coEδ) ≤ 1/ν, a.e. in Ω.

We now use the hypothesis of induction on ΩA , ΩB and A, B ∈ RI−1 co Eδ . Wetherefore find sequences

αµ ⊂ Affpiec(ΩA; RN×nm−1

), βµ ⊂ Affpiec(ΩB; RN×nm−1

)

such that

αµ ∈ ϕν + W 1,∞0 (ΩA; RN×nm−1

), ∇αµ (x) ∈ E ∪K, a.e. in Ω

αµ∗ ϕν in W 1,∞,

∫

ΩA

dist (∇αµ (x) ; E) dx→ 0 as μ →∞


and similarly for the sequence βµ .We next write, taking a diagonal sequence,

uν(x) =

⎧⎪⎪⎨⎪⎪⎩

ϕν(x) in Ω− (ΩA ∪ ΩB)

αν(x) in ΩA

βν(x) in ΩB

and use the facts that ∇uν ∈M, Rco Eδ ⊂⊂ K, K is a bounded set and (2) inthe definition of the approximation property holds, to get the claim.

Finally we discuss an example showing that the approximation property isnot always fulfilled. We consider the case N = n = 2 and denote by R2×2

d theset of 2× 2 diagonal matrices, we write any such matrix as a vector of R2.

Example 10.11 Let E = ξ1, ξ2, ξ3, ξ4, ξ5, ξ6 ⊂ M = R2×2d (see Figure 10.1)

be defined by

ξ1 = (1, 0) , ξ2 = (1,−1) , ξ3 = (0,−1) , ξ4 = (−1, 0) , ξ5 = (−1, 1) , ξ6 = (0, 1) .

It is easy to find that

ξ4

ξ5

ξ6

ξ3

ξ2

ξ1

×

××

×

× ×

Figure 10.1: The set E and RcoE

RcoE = ξ : ξ = (x, y) ∈ [0, 1]× [−1, 0]∪ ξ : ξ = (x, y) ∈ [−1, 0]× [0, 1]

and its interior (relative to R2×2d ) is given by

intM Rco E = ξ : ξ = (x, y) ∈ (0, 1)× (−1, 0)∪ ξ : ξ = (x, y) ∈ (−1, 0)× (0, 1) .


However, there is no way of finding a set Eδ satisfying the approximation prop-erty with K (Eδ) = Rco Eδ . In fact, condition (3) in the definition of the approx-imation property will be violated. ♦

10.2.3 Appendix: Baire one functions

In this appendix, we recall some well known facts about so called Baire onefunctions (see, for example, Oxtoby [488] or Yosida [605]). We start with thefollowing definitions.

Definition 10.12 Let X, Y be metric spaces and f : X → Y. We define theoscillation of f at x0 ∈ X as

ωf(x0) := limδ→0

supx,y∈BX(x0,δ)

dY (f(y), f(x)) ,

where BX(x0, δ) := x ∈ X : dX(x, x0) < δ and dX , dY are the metrics on thespaces X and Y, respectively.

Definition 10.13 A function f is said to be a Baire one function (or a functionof first class) if it can be represented as the pointwise limit of an everywhereconvergent sequence of continuous functions.

In the next proposition, we recall some elementary properties of the oscilla-tion function ωf .

Proposition 10.14 Let X, Y be metric spaces and f : X → Y.

(i) f is continuous at x0 ∈ X if and only if ωf (x0) = 0.

(ii) The set Ωǫf := x ∈ X : ωf(x) < ǫ is an open set in X.

Using the notion of oscillation and Proposition 10.14, we can write the setDf of all points at which a given function f is discontinuous as an Fσ set asfollows

Df =

∞⋃

ν=1

x ∈ X : ωf (x) ≥ 1/ν . (10.1)

We therefore have the following theorem, due to Baire, for Baire one functions.For the convenience of the reader, we give a proof of this theorem (see alsoTheorem 7.3 in Oxtoby [488] or Yosida [605] page 12).

Theorem 10.15 Let X, Y be metric spaces, with X complete, and f : X → Y.If f is a Baire one function, then Df is a set of first category.

Proof. Using the representation (10.1) of Df , it suffices to show that, forevery ǫ > 0, the set F := x ∈ X : ωf (x) ≥ 5ǫ is nowhere dense.

Let f(x) := limν→∞ fν(x), with fν continuous and define the sets

Eν :=⋂

i,j ≥ν

x ∈ X : dY (fi(x), fj(x)) ≤ ǫ , ∀ ν ∈ N.


Then Eν is closed in X, by continuity of fν , and Eν ⊂ Eν+1 . Moreover,⋃ν∈N

Eν = X, since for every x ∈ X the sequence fν(x) is convergent andthus a Cauchy sequence in Y.

Consider any closed set with non-empty interior I ⊂ X. Since I =⋃

(Eν∩I),the sets Eν ∩ I cannot all be nowhere dense. Indeed (see Yosida [605] page 12)in this case the complement of I in X, Ic, should be a dense set as a complementof a set of first category by Baire category theorem and this is a contradictionwith the fact that I has a non-empty interior. Hence for some positive integerν, Eν ∩ I contains an open subset J, by definition (see Yosida [605] page 11) ofa nowhere dense set.

We have dY (fj(x), fi(x)) ≤ ǫ for all x ∈ J and for all i, j ≥ ν. Putting j = νand letting i tend to ∞, we find that dY (fν(x), f(x)) ≤ ǫ for all x ∈ J. By thecontinuity of fν , for any x0 ∈ J, there exists a neighborhood I(x0) ⊂ J suchthat dY (fν(x), fν(x0)) ≤ ǫ for all x ∈ I(x0) and hence

dY (f(x), fν(x0)) ≤ 2ǫ, ∀x ∈ I(x0).

Therefore

dY (f(x), f(y)) ≤ dY (f(x), fν(x0)) + dY (f(y), fν (x0)) ≤ 4ǫ, ∀x, y ∈ I(x0),

then ωf (x0) ≤ 4ǫ, and so no point of J belongs to F. Thus, for every closed setI with non-empty interior there is an open set J ⊂ I ∩ F c. This shows that Fis nowhere dense and therefore Df is of first category.

Remark 10.16 From Theorem 10.15 and the Baire category theorem, it followsin particular that the set of points of continuity of a Baire one function from acomplete metric space X to any metric space Y (i.e. the set Dc

f complement ofDf ) is a dense Gδ set. Indeed, for any ǫ > 0, the set

Ωǫf := x ∈ X : ωf (x) < ǫ

is open and dense in X. ♦

In the proof of our main theorem, we have used Theorem 10.15 applied to thefollowing, quite surprising, special case of a Baire one function. This result wasobserved by Kirchheim in [364], [365] for complete sets of Lipschitz functionsand the same argument in fact gives the result for general complete subsets ofW 1,∞(Ω) functions.

Proposition 10.17 Let Ω ⊂ Rn be a bounded open set and let V ⊂ W 1,∞(Ω)∩W 1,∞

loc (Rn) be a non-empty complete space with respect to the L∞ metric. Thenthe gradient operator ∇ : V → Lp(Ω; Rn) is a Baire one function for any 1 ≤p <∞.

Proof. For h = 0, we let

∇h = (∇h1 , · · · ,∇h

n ) : V → Lp(Ω; Rn)

Examples 451

be defined, for every u ∈ V and x ∈ Ω, by

∇hi u (x) :=

u(x+hei)−u(x)

h if dist(x, Ωc) > |h|0 elsewhere

for i = 1, · · · , n, where e1, · · · , en stand for the vectors from the Euclidean basis.The claim follows once we have proved that for any fixed h the operator ∇h

is continuous and that, for any sequence h → 0,

limh→0

‖ ∇hi u−∇iu ‖Lp(Ω) = 0

for any i = 1, · · · , n, u ∈ V.The continuity of ∇h follows easily by observing that for every i = 1, · · · , n,

ǫ > 0 and u, v ∈ V we have that

‖ ∇hi u−∇h

i v ‖Lp(Ω) =1

|h|

(∫

Ωh

|u(x)− v(x) + v (x + hei)− u (x + hei)|p dx

) 1p

≤ 2(measΩ)1p

|h| ‖u− v‖L∞(Ω) ,

where Ωh := x ∈ Ω : dist(x, Ωc) > |h|.For the second claim we start observing that for any h and for any u ∈ V

we have

‖ ∇hi u ‖L∞(Ω) ≤

1

h‖ u (x + hei)− u (x) ‖L∞(Ωh) ≤ ‖ ∇iu ‖L∞(Ω) < +∞.

Moreover by Rademacher theorem (see Theorems 6.2.1 and 6.2.2 in Evans-Gariepy [273]), for any sequence h → 0,

limh→0

∇hi u(x) = ∇iu(x), a.e. x ∈ Ω.

The result follows by Lebesgue dominated convergence theorem.

10.3 Examples

We now give several examples of existence theorems that follow from the abstractones.

10.3.1 The scalar case

The first one concerns the scalar case, where we can even get sharper results(see Bressan-Flores [103], Cellina [134], Dacorogna-Marcellini [196], [198], [202],De Blasi-Pianigiani [230] or Friesecke [291]).

Theorem 10.18 Let Ω ⊂ Rn be a bounded open set and E ⊂ Rn. Let ϕ ∈Affpiec

(Ω)

satisfy∇ϕ (x) ∈ E ∪ int coE a.e. x ∈ Ω (10.2)


(where int coE stands for the interior of the convex hull of E). Then thereexists u ∈ ϕ + W 1,∞

0 (Ω) such that

∇u (x) ∈ E a.e. x ∈ Ω. (10.3)

Remark 10.19 (i) The theorem easily follows from the abstract Theorems 10.9and 10.4, but we prefer to give here a different proof, based on the method ofpyramids introduced by Cellina [134] and Friesecke [291]. Below we follow theproof of Lemma 2.11 in Dacorogna-Marcellini [202], showing, in particular, thatu ∈ Affpiec

(Ω).

(ii) The theorem can also be proved (see Theorem 2.10 in Dacorogna-Marcellini [202]) if ϕ ∈W 1,∞ (Ω) satisfy

∇ϕ (x) ∈ E ∪ int co E a.e. x ∈ Ω. ♦

Proof. Step 1. By working on each piece, where ϕ is affine, we can assumethat in fact ϕ is affine and therefore

∇ϕ = ξ0 ∈ int coE, ∀x ∈ Ω,

the case ξ0 ∈ E being trivial. Invoking Corollary 2.16, we can find

ξ1, ξ2, · · · , ξm ∈ E, m ≥ n + 1,

such that ξ1 − ξ0, ξ2 − ξ0, · · · , ξm − ξ0 spans the whole of Rn,

ξ0 ∈ int co ξ1, ξ2, · · · , ξm

and there exist si > 0, i = 1, 2, · · · , m, with∑m

i=1 si = 1 such that

m∑

i=1

si(ξi − ξ0) = 0. (10.4)

Step 2. Let x0 ∈ Ω and define for r > 0 the function

vr,x0 (x) := r + mini=1,··· ,m

〈ξi − ξ0; x− x0〉

which we call a ”pyramid”. Let

G (r, x0) := x ∈ Rn : vr,x0 (x) > 0

and observe (see Step 3) that this set is bounded. Finally, let

u (x) := ϕ (x) + vr,x0 (x) .

Then u ∈ ϕ + W 1,∞0 (G (r, x0)) , with u ∈ Affpiec(G (r, x0) ), and

∇u (x) ∈ ξ1, · · · , ξm ⊂ E a.e. x ∈ G (r, x0) .

Examples 453

Covering Ω by dilation and translation of sets of the form G (r, x0) and appealingto Vitali covering theorem (see Corollary 10.6 in [202]), we have the result.

Step 3. It remains to prove that G (r, x0) is bounded. Let us assume, for thesake of contradiction, that for some x0 ∈ Rn and r > 0 the set G (r, x0) is notbounded. Then there exists a sequence xk ∈ Rn, k ∈ N, such that

limk→∞

|xk| = +∞ and r + 〈ξi − ξ0; xk − x0〉 ≥ 0, ∀ i = 1, 2, · · · , m, ∀ k ∈ N.

Letyk :=

xk

|xk|.

Then, up to the extraction of a subsequence that we still denote by yk , we havethat yk → y0 for some y0 ∈ Rn with |y0| = 1. Passing to the limit as k → +∞on both sides of the inequality, we obtain

r

|xk|+ 〈 ξi − ξ0; yk −

x0

|xk|〉 ≥ 0, ∀ i = 1, 2, · · · , m, ∀ k ∈ N,

and we get 〈ξi − ξ0; y0〉 ≥ 0 for every i = 1, 2, · · · , m. By using (10.4), we obtain

0 = 〈0; y0〉 =m∑

i=1

si 〈ξi − ξ0; y0〉 .

Sincesi > 0, 〈ξi − ξ0; y0〉 ≥ 0, ∀ i = 1, 2, · · · , m,

we deduce that〈ξi − ξ0; y0〉 = 0, ∀ i = 1, 2, · · · , m.

Recall that the set ξ1 − ξ0, ξ2 − ξ0, · · · , ξm − ξ0 spans the whole of Rn. There-fore there exist ci ∈ R, i = 1, 2, · · · , m, such that y0 =

∑mi=1 ci(ξi − ξ0). Com-

bining all the above identities, we obtain the desired contradiction, namely

1 = |y0|2 = 〈y0; y0〉 =

m∑

i=1

ci 〈ξi − ξ0; y0〉 = 0.

This finishes the proof.

We have as immediate corollaries the following.

Corollary 10.20 Let Ω ⊂ Rn be a bounded open set and F : Rn → R becontinuous and such that lim|ξ|→∞ F (ξ) = +∞. Let ϕ ∈ Affpiec

(Ω)

be suchthat

F (∇ϕ (x)) ≤ 0 a.e. x ∈ Ω.

Then there exists u ∈ ϕ + W 1,∞0 (Ω) such that

F (∇u (x)) = 0 a.e. x ∈ Ω.


Remark 10.21 As in Remark 10.19 (ii), the corollary still holds for ϕ ∈W 1,∞ (Ω) . ♦

Proof. Let

E = ξ ∈ Rn : F (ξ) = 0 , K = ξ ∈ Rn : F (ξ) ≤ 0

and observe that, under our hypotheses,

K ⊂ E ∪ int coE.

Indeed let ξ ∈ K and observe that if F (ξ) = 0, then the inclusion is triviallytrue; we therefore assume that

F (ξ) < 0.

We then let e1 = (1, 0, · · · , 0) ∈ Rn, t ∈ R and

ξt := ξ + te1 .

Since F is continuous and lim|ξ|→∞ F (ξ) = +∞, we can find t− < 0 < t+ suchthat

F (ξt) < 0, ∀t ∈ (t− , t+) and F (ξt±) = 0, i.e. ξt± ∈ E.

We can therefore write

ξ =t+

t+ − t−ξ− +

−t−t+ − t−

ξ+ ∈ coE.

Since F (ξ) < 0, it is easy to see that, in fact, ξ ∈ int coE. Thus K ⊂ E∪int coE.

We are therefore in a position to apply Theorem 10.18 to get the result.

The above corollary generalizes in a straightforward way to the vectorialcase.

Corollary 10.22 Let Ω ⊂ Rn be a bounded open set and F : RN×n → R becontinuous and such that lim|ξ|→∞ F (ξ) = +∞. Let ϕ ∈ Affpiec

(Ω; RN

)be such

thatF (∇ϕ (x)) ≤ 0 a.e. x ∈ Ω.

Then there exists u ∈ ϕ + W 1,∞0

(Ω; RN

)such that

F (∇u (x)) = 0 a.e. x ∈ Ω.

Proof. By working on each piece where ϕ is affine, we can assume that it isaffine and thus ∇ϕ = ξ0 ∈ RN×n. We can also assume that ξ0 = 0, otherwisemake a translation. Set for ξ ∈ RN×n

ξ = ( ξ1, · · · , ξN ), where ξi ∈ Rn

Examples 455

andG(ξ1)

:= F(ξ1, 0, · · · , 0

).

Observe that G satisfies all the hypotheses of Corollary 10.20 and

G (0) = F (0) ≤ 0;

therefore there exists u1 ∈W 1,∞0 (Ω) such that

G(∇u1 (x)

)= 0, a.e. x ∈ Ω.

Settingu (x) :=

(u1 (x) , 0, · · · , 0

),

we have indeed obtained the claim.

We now conclude with an approximation result of Muller-Sychev [468] (com-pare it with Lemma 3.11).

Corollary 10.23 Let n ≥ 2, N ≥ 1, Ω ⊂ Rn be a bounded open set. Lett ∈ [0, 1] and A, B ∈ RN×n such that

A−B = a⊗ b

with a ∈ RN and b ∈ Rn. Let b3, · · · , bk ∈ Rn, k ≥ n + 2, such that

0 ∈ int cob,−b, b3, · · · , bk. (10.5)

Let ϕ be an affine map such that

∇ϕ(x) = ξ0 = tA + (1 − t)B, x ∈ Ω

(i.e. A = ξ0 + (1− t) a⊗ b and B = ξ0 − ta⊗ b). Then, for every ǫ > 0, thereexist u ∈ Affpiec

(Ω; RN

)and disjoint open sets ΩA , ΩB ⊂ Ω, such that

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

|meas ΩA − t meas Ω| , |meas ΩB − (1− t)meas Ω| ≤ ǫ

u(x) = ϕ(x), x ∈ ∂Ω and |u(x)− ϕ(x)| ≤ ǫ, x ∈ Ω

∇u(x) =

A in ΩA

B in ΩB

∇u(x) ∈ ξ0 + (1− t) a⊗ b,−ta⊗ b, a⊗ b3, · · · , a⊗ bk a.e. in Ω.

Proof. We follow here the proof of Kirchheim [365]. We divide the proof intothree steps.

Step 1. There is no loss of generality if we assume the two next hypotheses.

1) ξ0 = 0, by settingu := u + ϕ


and solving the problem for u replaced by u and ϕ by 0.

2) N = 1, by lettingu (x) := v (x) a

where we have to find v ∈ Affpiec

(Ω)

and disjoint open sets ΩA, ΩB ⊂ Ω, suchthat

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

|measΩA − t measΩ| , |measΩB − (1− t)meas Ω| ≤ ǫ

v(x) = 0, x ∈ ∂Ω and |v(x)| ≤ ǫ, x ∈ Ω

∇v(x) =

(1− t) b in ΩA

−tb in ΩB

∇v(x) ∈ (1− t) b,−tb, b3, · · · , bk, a.e. in Ω.

(10.6)

Step 2. We now prove (10.6), without the conclusion |v(x)| ≤ ǫ, for Ω of theform

G :=

x ∈ Rn : 1 + min

i=1,··· ,k〈bi; x〉 > 0

where b1 = b and b2 = −b. It is easily proved (cf. Step 3 in the proof of Theorem10.18) that under our hypothesis (10.5) the set G is bounded.

Define next a non negative periodic function, of period 1, h : R → R suchthat h (0) = 0 and

h′ (τ) =

(1− t) if τ ∈ (0, t)

−t if τ ∈ (t, 1) .

Finally let, for ν ∈ N,

v (x) := min1 + mini=3,··· ,k

〈bi; x〉 ,1

νh (ν 〈b; x〉).

We claim that, by choosing ν sufficiently large, the function v has all the desiredproperties in (10.6), with the exception of |v(x)| ≤ ǫ.

Note that, since h ≥ 0 and by definition of G,

v (x) ≥ 0, x ∈ G.

Since at a boundary point of G, either the first term in the minimum in v (x)vanishes or |〈b; x〉| = 1 (recall that b1 = b = −b2), which implies

h (ν 〈b; x〉) = 0;

thusv (x) = 0, x ∈ ∂G.

It is also clear that

∇v(x) ∈ (1− t) b,−tb, b3, · · · , bk, a.e. in G.

Examples 457

So it remains to show that, by choosing ν sufficiently large, the sets

ΩA := x ∈ G : ∇v(x) = (1− t) b

ΩB := x ∈ G : ∇v(x) = −tb

satisfy the desired estimates on their measures. Indeed note that, for everys ∈ [0, 1] , we have (1− s)G ⊂ G, and

1 + mini=3,··· ,k

〈bi; x〉 ≥ s, ∀x ∈ (1− s)G.

Therefore by choosing ν sufficiently large, ν > 1/s, we have that

v (x) =1

νh (ν 〈b; x〉) , ∀x ∈ (1− s)G.

Therefore

measΩA ≥ (1− s)n t measG and measΩB ≥ (1− s)n (1− t)measG.

Thus choosing s > 0 sufficiently small so that

ǫ = [1− (1− s)n

] measG

we get

t measG−measΩA ≤ tǫ and (1− t)meas G−measΩB ≤ (1− t) ǫ (10.7)

which gives one set of inequalities for the measures in (10.6). To prove thesecond ones, we proceed by contradiction and assume, for example, that

|measΩA − t measG| > ǫ.

Since the first inequality in (10.7) holds, this implies that

measΩA − t measG > ǫ.

Combining this inequality with the second one in (10.7), we get

0 ≥ meas ΩA + measΩB −measG > ǫ− (1− t) ǫ = tǫ

which is the desired contradiction. Therefore the inequalities for the measuresin (10.6) have been proved.

Step 3. The conclusion for general Ω follows by using Vitali covering theorem(cf. Corollary 10.6 in [202]), covering Ω by dilations and translations of the aboveset G. More precisely if x0 ∈ Ω and r > 0, we consider sets of the form

G (r, x0) :=

x ∈ Rn : r + min

i=1,··· ,k〈bi; x− x0〉 > 0

.


By choosing r > 0 sufficiently small, we can also ensure that |v(x)| ≤ ǫ. Thisconcludes the proof.

We finally show that (10.2) is in fact also a necessary condition, at leastwhen ϕ is affine; for the general case, see Section 2.4 in Dacorogna-Marcellini[202]. For the affine case, the result is in some of the above mentioned articles(notably in Cellina [134] or Friesecke [291]), but we follow here Bandyopadhyay-Barroso-Dacorogna-Matias [68].

Theorem 10.24 Let Ω ⊂ Rn be a bounded open set, E ⊂ Rn, ξ0 ∈ Rn andu ∈ uξ0 + W 1,∞

0 (Ω) (uξ0 being such that ∇uξ0 = ξ0) such that

∇u (x) ∈ E a.e. x ∈ Ω

then

ξ0 ∈ E ∪ int co E.

Proof. Assume that ξ0 /∈ E, otherwise nothing is to be proved. It is easy tosee (see Proposition 2.36) that, by Jensen inequality and since ∇u (x) ∈ E,

ξ0 =1

meas Ω

∫

Ω

∇u (x) dx ∈ co E.

Let us show that we cannot have ξ0 ∈ ∂(coE

). If we can prove this, we will

deduce that ξ0 ∈ int coE. Since int coE = int coE (cf. Theorem 2.6) we willhave the result.

If ξ0 ∈ ∂(coE

), we find from the separation theorem (see Corollary 2.11)

that there exists α ∈ Rn, α = 0, such that

〈α; z − ξ0〉 ≥ 0, ∀z ∈ co E.

We therefore have that

〈α;∇u (x)− ξ0〉 ≥ 0, a.e. x ∈ Ω.

Recalling that u ∈ uξ0 + W 1,∞0 (Ω) , we find that

∫

Ω

〈α;∇u (x)− ξ0〉 dx = 0,

which coupled with the above inequality leads to

〈α;∇u (x)− ξ0〉 = 0, a.e. x ∈ Ω.

Applying Lemma 11.17, we get that u ≡ uξ0 and hence ξ0 ∈ E, a contradictionwith the hypothesis made at the beginning of the proof. Therefore ξ0 /∈ ∂

(coE

)

as claimed and hence the theorem is proved.

Examples 459


The next example, studied by Dacorogna-Ribeiro [212], deals with the singularvalues case that we encountered in Section 7.4.1.

Theorem 10.25 (Singular values) Let Ω ⊂ Rn be a bounded open set, α < βand 0 < γ2 ≤ · · · ≤ γn be such that

max |α| , |β| < γ2

n∏

i=2

γi .

Let ϕ ∈ Affpiec

(Ω; Rn

)be such that, for almost every x ∈ Ω,

α < det∇ϕ(x) < β and

n∏

i=ν

λi(∇ϕ(x)) <

n∏

i=ν

γi , ν = 2, · · · , n.

Then there exists u ∈ ϕ + W 1,∞0 (Ω; Rn) such that, for almost every x ∈ Ω,

det∇u (x) ∈ α, β and λν(∇u (x)) = γν , ν = 2, · · · , n.

Remark 10.26 (i) If α = −β < 0 and if we set

γ1 = β [∏n

i=2 γi ]−1,

we recover the result of Dacorogna-Marcellini [202], namely that if

n∏

i=ν

λi(∇ϕ(x)) <

n∏

i=ν

γi , ν = 1, · · · , n,

then there exists u ∈ ϕ + W 1,∞0 (Ω; Rn) such that

λν(∇u) = γν , ν = 1, · · · , n, a.e. in Ω.

(ii) If α = β = 0, it can also be proved, as in Dacorogna-Tanteri [215], that if

det∇ϕ(x) = α,n∏

i=ν

λi(∇ϕ(x)) <n∏

i=ν

γi , ν = 2, · · · , n,

then there exists u ∈ ϕ + W 1,∞0 (Ω; Rn) such that

λν(∇u) = γν , ν = 2, · · · , n and det∇u = α, a.e. in Ω. ♦

Proof. We let

E :=ξ ∈ Rn×n : det ξ ∈ α, β , λi(ξ) = γi , i = 2, · · · , n

and recall that from Theorem 7.43 we have

Rco E =ξ ∈ Rn×n : det ξ ∈ [α, β] ,


∏ni=ν γi , ν = 2, · · · , n


and a similar formula for intRco E. Since ϕ ∈ Affpiec

(Ω; Rn

)and ∇ϕ ∈

intRco E, we only need, in order to apply Theorems 10.9 and 10.4 to getthe result, to verify that E and Rco E have the approximation property (cf.Definition 10.6).

For δ > 0 such that γ2 − δ > 0 and α + δ < β − δ, let

Eδ :=ξ ∈ Rn×n : det ξ ∈ α + δ, β − δ , λi(ξ) = γi − δ, i = 2, · · · , n

.

For a sufficiently small δ we have

(γ2 − δ)

n∏

i=2

(γi − δ) ≥ max |α + δ| , |β − δ|

and thus Theorem 7.43 ensures that

Rco Eδ =

ξ ∈ Rn×n :

det ξ ∈ [α + δ, β − δ] ,∏ni=ν λi(ξ) ≤

∏ni=ν (γi − δ), ν = 2, · · · , n

.

We have to verify the three conditions of Definition 10.6. The first one isobvious. We next verify the second condition. Since η ∈ Eδ , we assume thatdet η = α + δ, the case det η = β − δ being handled in an analogous way. Theset Eδ being left and right SO(n) invariant, we can assume that

η = diag(α + δ

(γ2 − δ) · · · (γn − δ), γ2 − δ, · · · , γn − δ).

If we let

ξ = diag(α

γ2 · · · γn, γ2, · · · , γn)

we have ξ ∈ E and

dist(η; E) ≤ max | α + δ

(γ2 − δ) · · · (γn − δ)− α

γ2 · · ·γn| , δ → 0, as δ → 0.

The second condition of Definition 10.6 then follows.

The third condition of the approximation property follows from the conti-nuity of the functions involved in the definition of Rco Eδ .

We also find the following immediate corollary.

Corollary 10.27 Let Ω ⊂ Rn be a bounded open set and ϕ ∈ Affpiec

(Ω; Rn

)

be such that

|det∇ϕ(x)| < 1 a.e. x ∈ Ω.

Then there exists u ∈ ϕ + W 1,∞0 (Ω; Rn) such that, for almost every x ∈ Ω,

|det∇u (x)| = 1 a.e. x ∈ Ω.

Examples 461


With the help of Theorem 7.44 and the abstract results of the present chapter,we can prove the following existence theorem. The result was proved by Muller-Sverak [464] using the method of convex integration of Gromov [324] and byDacorogna-Marcellini in [196] and [202].

Theorem 10.28 Let Ω ⊂ R2 be a bounded open set,

A =

(a1 0

0 a2

)and B =

(b1 0

0 b2

),

where 0 < b1 < a1 ≤ a2 < b2 and a1a2 < b1b2 . Let

E := SO(2)A ∪ SO(2)B

and ξ0 ∈ intRcoE. Denote by uξ0 an affine map such that ∇uξ0 = ξ0 . Then

there exists u ∈ uξ0 + W 1,∞0

(Ω; R2

)such that

∇u (x) ∈ E a.e. in Ω.

Before proceeding with the proof, let us make some comments on the hypothe-ses (see Section 7.4.2).

(i) The hypothesis a1a2 < b1b2 guarantees that detA = detB. The case ofequality can also be handled but requires a special treatment (see Dacorogna-Tanteri [215] and Muller-Sverak [466]).

(ii) Up to rotations, we can always assume that the matrices A and B arediagonal.

(iii) The hypothesis 0 < b1 < a1 ≤ a2 < b2 ensures that there exists R ∈SO(2) such that

det (RA−B) = 0

and guarantees also that intRco E = ∅.We now proceed with the proof of the theorem.

Proof. We recall that (see Theorem 7.44)

RcoE =

ξ ∈ R2×2 :

ξ = αRaA + βRbB, Ra, Rb ∈ SO(2),

0 ≤ α ≤ det B−det ξdetB−detA , 0 ≤ β ≤ det ξ−detA

det B−detA

and its interior is given by the same formula with strict inequalities on the righthand side.

Moreover,

Aδ =

(a1 − δ 0

0 a2 + Tδ

), Bδ =

(b1 + Sδ 0

0 b2 − δ

)


both belong to intRco E, for appropriate S, T > 0 and for every δ > 0 sufficientlysmall.

Then write

Eδ := SO(2)Aδ ∪ SO(2)Bδ.

Note that

RcoEδ ⊂⊂ intRco E,

which is (1), with K (Eδ) = RcoEδ , in the definition of the approximationproperty (see Definition 10.6). Since the properties (2) and (3) in Definition10.6 are also true, we can apply Theorems 10.9 and 10.4 to get the theorem.

10.3.4 The case of a quasiaffine function

The next theorem will use Theorem 7.47.

Theorem 10.29 Let Ω ⊂ Rn be a bounded open set, α < β, Φ : RN×n → R anon-constant quasiaffine function and ϕ ∈ Affpiec

(Ω; RN

)such that, for almost

every x ∈ Ω,

α < Φ(∇ϕ(x)) < β.

Then there exists u ∈ ϕ + W 1,∞0 (Ω; RN ) satisfying

Φ(∇u) ∈ α, β a.e. in Ω.

Proof. By working on each piece where ϕ is affine, we can assume that ϕ isaffine. By Lemma 7.46, we can find constants ci

j such that∣∣∂ϕi(x)/∂xj

∣∣ < cij

and

infi=1,··· ,Nj=1,··· ,n

|Φ(ξ)| :

∣∣ξij

∣∣ = cij

> max|α| , |β|. (10.8)

We then define

E :=ξ ∈ RN×n : Φ(ξ) ∈ α, β,

∣∣ξij

∣∣ ≤ cij , i = 1, · · · , N, j = 1, · · · , n

.

As before we only need to verify that the sets E and Rco E have the approxi-mation property. Let, for δ > 0 sufficiently small,

Eδ :=

ξ ∈ RN×n :

Φ(ξ) ∈ α + δ, β − δ,∣∣ξij

∣∣ ≤ cij − δ, i = 1, · · · , N, j = 1, · · · , n

.

We first observe that, by continuity, it follows from (10.8) that

infi=1,··· ,Nj=1,··· ,n

|Φ(ξ)| :

∣∣ξij

∣∣ = cij − δ

> max|α + δ| , |β − δ|.

Examples 463

We can then apply Theorem 7.47 to find

Rco Eδ =

ξ ∈ RN×n :

Φ(ξ) ∈ [α + δ, β − δ],∣∣ξij

∣∣ ≤ cij − δ, i = 1, · · · , N, j = 1, · · · , n

.

It immediately follows that the first and third conditions of Definition 10.6 areverified. It therefore remains to check the second one.

We proceed by contradiction and assume that there exist ǫ > 0 and asequence ην ∈ E1/ν with dist(ην , E) > ǫ. As

∣∣(ην)ij

∣∣ ≤ cij we can extract a

convergent subsequence, still denoted ην , and η ∈ E so that ην → η, whichcontradicts the fact that dist(ην , E) > ǫ.

We can therefore invoke Theorems 10.9 and 10.4 to conclude the proof.


We now turn our attention to the problem already considered in Sections 6.6.5and 7.4.4. The present result will be fully used in Theorem 11.35.

Theorem 10.30 Let Ω ⊂ R2 be a bounded open set and

E :=ξ ∈ R2×2

s : det ξ ≥ 0 and trace ξ ∈ 0, 1

,

where R2×2s denotes the set of 2× 2 symmetric matrices. Let ξ0 ∈ R2×2

s be suchthat

det ξ0 > 0 and 0 < trace ξ0 < 1

and denote by uξ0 an affine map such that ∇uξ0 = ξ0 . Then there exists u ∈uξ0 + W 1,∞

0

(Ω; R2

)such that

∇u (x) ∈ E a.e. in Ω.

Proof. According to Theorem 7.48, we have that

ξ0 ∈ intRco E

and that Rco E = co E. So let δ ∈ (0, 1) and let

Eδ := δξ0 + (1− δ)E.

It is easy to see that

RcoEδ = δξ0 + (1− δ) coE ⊂⊂ intRcoE

which is (1), with K (Eδ) = RcoEδ , in the definition of the approximationproperty (see Definition 10.6). Since the properties (2) and (3) in Definition10.6 are also true, we can apply Theorems 10.9 and 10.4 to get the result.

Chapter 11

Existence of minima fornon-quasiconvex integrands

11.1 Introduction

In this chapter, we discuss the existence of minimizers for the problem

(P ) inf

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

,

where

- Ω ⊂ Rn is a bounded open set,

- u : Ω → RN and ∇u =(

∂ui

∂xj

)∈ RN×n,

- f : RN×n → R is lower semicontinuous, locally bounded and non-negative,

- uξ0 is a given affine map (i.e. ∇uξ0 = ξ0 , where ξ0 ∈ RN×n is a fixedmatrix).

If the function f is quasiconvex (recall that in the scalar case n = 1 or N = 1,quasiconvexity and ordinary convexity are equivalent), then the problem (P ) ,trivially, has uξ0 as a minimizer.

The aim of the present chapter is to study the case where f fails to bequasiconvex. The general rule is that the problem has no solution, as alreadyseen even in the simplest case N = n = 1 in Chapter 4. However, there arestill many instances where solutions do exist, although the direct methods donot apply. We now explain how to deal with such problems. The first step isto apply the relaxation theorem (see Chapter 9). It has as a direct consequence(see Theorem 11.1) that (P ) has a solution u ∈ uξ0 + W 1,∞

0 (Ω; RN ) if andonly if

f (∇u (x)) = Qf (∇u (x)) a.e. x ∈ Ω,

466 Existence of minima for non-quasiconvex integrands

∫

Ω

Qf (∇u (x)) dx = Qf (ξ0) measΩ,

where Qf is the quasiconvex envelope (see Section 6.3) of f, namely

Qf (ξ) := sup g (ξ) : g ≤ f and g quasiconvex .

The problem is then to discuss the existence or non-existence of a u satisfyingthe two equations. The two equations are not really of the same nature. Thefirst one is an implicit partial differential equation of the type studied in Chapter10. The second one is more geometric in nature and has to do with some”quasiaffinity” of the quasiconvex envelope Qf.

The scalar case (n = 1 or N = 1) has been intensively studied bymany authors including: Aubert-Tahraoui [42], [43], [46], Bauman-Phillips [72],Buttazzo-Ferone-Kawohl [116], Celada-Perrotta [130], [131], Cellina [133], [134],Cellina-Colombo [135], Cesari [141], [143], Cutri [169], Dacorogna [179], Ekeland[262], Friesecke [291], Fusco-Marcellini-Ornelas [297], Giachetti-Schianchi [306],Klotzler [369], Marcellini [419], [420], [426], Mascolo [433], Mascolo-Schianchi[437], [438], Monteiro Marques-Ornelas [451], Ornelas [485], Raymond [502],[503], [504], Sychev [558], Tahraoui [564], [565], Treu [580] and Zagatti[610], [611].

The vectorial case has been investigated for some special examplesnotably by Allaire-Francfort [15], Cellina-Zagatti [138], Dacorogna-Ribeiro[212], Dacorogna-Tanteri [215], Mascolo-Schianchi [436], Muller-Sverak [464],Raymond [505] and Zagatti [612]. A more systematic study was achieved byDacorogna-Marcellini in [195], [202], [203] and Dacorogna-Pisante-Ribeiro [211].We will closely follow the survey article of Dacorogna [184], which is based on[195] and [211].

We have, throughout this chapter, made two important restrictions:

- f does not depend on lower order terms, i.e. f (x, u, ξ) = f (ξ) ;

- the boundary datum u0 is affine, i.e. there exists ξ0 ∈ RN×n such that

∇u0 = ξ0 .

In the above literature, some authors have considered either of the two moregeneral cases. The results are then much less general and essentially apply onlyto the scalar case.

We now briefly describe the content of the chapter. We start by makingsome abstract considerations on sufficient (in Section 11.2) and necessary (inSection 11.3) conditions. We then apply these abstract results first to the scalarcase (see Section 11.4), getting general existence theorems, particularly in thecase of single integrals (i.e. n = 1). In the vectorial case, we investigate severalexamples that are relevant for applications and that we have already encounteredin the previous chapters.

Sufficient conditions 467

11.2 Sufficient conditions

With the help of the relaxation theorem (see Theorem 9.1) and Theorem 10.4,we are now in a position to discuss some existence results for the problem(P ) . The following theorem (see [195]) is elementary and gives a necessary andsufficient condition for the existence of minima. It will be crucial in several ofour arguments.

Theorem 11.1 Let Ω ⊂ Rn be a bounded open set, f : RN×n → R a lowersemicontinuous, locally bounded and non-negative function, ξ0 ∈ RN×n and uξ0

be such that ∇uξ0 = ξ0 . The problem (P ) has a solution if and only if there

exists u ∈ uξ0 + W 1,∞0 (Ω; RN ) such that

f (∇u (x)) = Qf (∇u (x)) a.e. x ∈ Ω, (11.1)∫

Ω

Qf (∇u (x)) dx = Qf (ξ0) measΩ. (11.2)

Proof. By the relaxation theorem (see Theorem 9.1) and since uξ0 is affine,we have

inf (P ) = inf (QP ) = Qf (ξ0) measΩ.

Moreover, since we always have f ≥ Qf and we have a solution of (11.1) satis-fying (11.2), we get that u is a solution of (P ). The fact that (11.1) and (11.2)are necessary for the existence of a minimum for (P ) follows in the same way.

The previous theorem explains why the set

K :=ξ ∈ RN×n : Qf (ξ) < f (ξ)

plays a central role in the existence theorems that follow. In order to ensure(11.1), we will have to consider differential inclusions of the form studied inChapter 10, namely: find u ∈ uξ0 + W 1,∞

0 (Ω; RN ) such that

∇u (x) ∈ ∂K a.e. x ∈ Ω.

In order to deal with the second condition (11.2), we will have to impose somehypotheses of the type ”Qf is quasiaffine on K”.

The main abstract theorem is the following.

Theorem 11.2 Let Ω ⊂ Rn be a bounded open set, ξ0 ∈ RN×n, f : RN×n → Ra lower semicontinuous, locally bounded and non-negative function and let

K :=η ∈ RN×n : Qf (η) < f (η)

.

Assume that there exists K0 ⊂ K such that

• ξ0 ∈ K0 ,

• K0 is bounded and has the relaxation property (see Definition 10.2) withrespect to K0 ∩ ∂K,


• Qf is quasiaffine on K0 .

Let uξ0 (x) = ξ0x. Then the problem

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

has a solution u ∈ uξ0 + W 1,∞0 (Ω; RN ).

Remark 11.3 (i) Although this theorem applies to functions f that take onlyfinite values, it can sometimes be extended to functions f : RN×n → R∪+∞.

(ii) Of course, if ξ0 /∈ K, then uξ0 is a minimizer of (P ) .

(iii) The last hypothesis in the theorem means that∫

Ω

Qf (ξ +∇ϕ (x)) dx = Qf (ξ) measΩ

for every ξ ∈ K0 and every ϕ ∈ W 1,∞0 (Ω; RN ) with

ξ +∇ϕ (x) ∈ K0 a.e. in Ω. ♦Proof. Since ξ0 ∈ K0 and K0 is bounded and has the relaxation propertywith respect to K0 ∩ ∂K, we can find, appealing to Theorem 10.4, a map u ∈uξ0 + W 1,∞

0 (Ω; RN ) satisfying

∇u ∈ K0 ∩ ∂K, a.e. in Ω,

which means that (11.1) of Theorem 11.1 is satisfied. Moreover, since Qf isquasiaffine on K0 , we have that (11.2) of Theorem 11.1 holds and thus theclaim.

The second hypothesis in the theorem is clearly the most difficult to ver-ify; nevertheless, there are some cases when it is automatically satisfied. Forexample, if K is bounded, we can take K0 = K (see Corollary 11.8).

We will see that, in many applications, the set K turns out to be unboundedand in order to apply Theorem 11.2 we need to find some weaker conditions onK that guarantee the existence of a subset K0 of K satisfying the requestedproperties. With this aim in mind, we give the following notation and defini-tions.

Notation 11.4 Let K ⊂ RN×n be open and λ ∈ RN×n.

(i) For ξ ∈ K, we denote by LK(ξ, λ) the largest segment of the form[ξ + tλ, ξ + sλ] , t < 0 < s, such that (ξ + tλ, ξ + sλ) ⊂ K.

(ii) If LK(ξ, λ) is bounded, we denote by t− (ξ) < 0 < t+ (ξ) the elementssuch that LK(ξ, λ) = [ξ + t−λ, ξ + t+λ] . They therefore satisfy

ξ + t±λ ∈ ∂K and ξ + tλ ∈ K, ∀t ∈ (t−, t+).

(iii) If H ⊂ K, we let

LK(H, λ) :=⋃

ξ∈H

LK(ξ, λ). ♦


Definition 11.5 Let K ⊂ RN×n be open, ξ0 ∈ K and λ ∈ RN×n.

(i) We say that K is bounded at ξ0 in the direction λ if LK(ξ0, λ) is bounded.

(ii) We say that K is stably bounded at ξ0 in the rank one direction λ = α⊗β(with α ∈ RN and β ∈ Rn) if there exists ǫ > 0 such that LK(ξ0 + α⊗Bǫ, λ) isbounded, where we have denoted

ξ0 + α⊗Bǫ :=ξ ∈ RN×n : ξ = ξ0 + α⊗ b with |b| < ǫ

.

Clearly a bounded open set K is bounded at every point ξ ∈ K in anydirection λ and consequently it is also stably bounded.

We now give an example of a globally unbounded set that is bounded incertain directions.

Example 11.6 Let N = n = 2, ξ0 ∈ R2×2, α, β ∈ R such that α < det ξ0 < βand

K =ξ ∈ R2×2 : α < det ξ < β

.

The set K is clearly unbounded.

(i) If ξ0 = I, then K is bounded, and even stably bounded, at ξ0 , in adirection of rank one, for example with

λ =

(1 00 0

)or λ =

(0 00 1

).

(ii) However, if ξ0 = 0, then K is unbounded in any rank one direction butis bounded in any rank two direction. ♦

In the following result, we deal with sets K that are bounded in a rank onedirection only. This corollary says, roughly speaking, that if K is bounded atξ0 in a rank one direction λ and this boundedness (in the same direction) ispreserved under small perturbations of ξ0 along rank one λ-compatible direc-tions, then we can ensure the relaxation property required in the main existencetheorem.

Corollary 11.7 Let Ω ⊂ Rn be a bounded open set, f : RN×n → R a lowersemicontinuous, locally bounded and non-negative function and let ξ0 ∈ K where

K :=ξ ∈ RN×n : Qf (ξ) < f (ξ)

.

If there exists a rank one direction λ ∈ RN×n (meaning that λ = α ⊗ β withα ∈ RN and β ∈ Rn) such that

(i) K is stably bounded at ξ0 in the direction λ,

(ii) Qf is quasiaffine on the set LK(ξ0 + α⊗Bǫ, λ),

then the problem

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )



The following corollary is strictly contained in the previous one but, since ittakes a much simpler form, we state it now.

Corollary 11.8 Let Ω ⊂ Rn be a bounded open set, f : RN×n → R a lowersemicontinuous, locally bounded and non-negative function and let ξ0 ∈ K where

K :=ξ ∈ RN×n : Qf (ξ) < f (ξ)

.

If the connected component of K containing ξ0 is bounded and if Qf is quasi-affine on this connected component, then the problem

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )


We now proceed with the proof of Corollary 11.7.


Step 1. Assume that |β| = 1, otherwise replace it by β/ |β| , and let βk ∈ Rn,k ≥ n + 2, with |βk| = 1, be such that

0 ∈ H := int coβ,−β, β3, · · · , βk ⊂ B1 (0) := x ∈ Rn : |x| < 1 .

Since K is stably bounded at ξ0 , we can find ǫ > 0 so that LK(ξ0 + α⊗Bǫ, λ)is bounded. Define then

K0 := (ξ0 + α⊗ ǫH) ∪[∂K ∩ LK(ξ0 + α⊗ ǫH, λ)

].

We therefore have that ξ0 ∈ K0 and, by hypothesis, that K0 is bounded, since

K0 ⊂ K0 ⊂ LK(ξ0 + α⊗Bǫ, λ).

We furthermore have

K0 ∩ ∂K = ∂K ∩ LK(ξ0 + α⊗ ǫH, λ).

In order to deduce the corollary from Theorem 11.2, we only need to show thatK0 has the relaxation property (cf. Definition 10.2) with respect to K0 ∩ ∂K.This is achieved in the next step.

Step 2. We now prove that K0 has the relaxation property with respect toK0 ∩ ∂K. Let ξ ∈ K0 and let us find a sequence uν ∈ Affpiec

(Ω; RN

)so that

uν ∈ uξ + W 1,∞0

(Ω; RN

), ∇uν (x) ∈

(K0 ∩ ∂K

)∪K0 , a.e. in Ω


∫

Ω

dist(∇uν (x) ; K0 ∩ ∂K

)dx → 0 as ν →∞.

(11.3)

If ξ ∈ ∂K ∩ LK(ξ0 + α ⊗ ǫH, λ), nothing is to be proved; so we assume thatξ ∈ ξ0 + α⊗ ǫH. By hypothesis (i), we can find t− (ξ) < 0 < t+ (ξ) so that

ξ± := ξ + t±λ ∈ ∂K and ξ + tλ ∈ K, ∀ t ∈ (t−, t+)


and hence ξ± ∈ K0 ∩ ∂K. We moreover have that

ξ =−t−

t+ − t−ξ+ +

t+t+ − t−

ξ− with ξ± ∈ K0 ∩ ∂K. (11.4)

Furthermore, since ξ ∈ ξ0 + α⊗ ǫH, we can find γ ∈ ǫH such that

ξ = ξ0 + α⊗ γ.

The set H being open we have that Bδ (γ) ⊂ ǫH, for every sufficiently smallδ > 0. Moreover since for every δ > 0, we have

0 ∈ δH = int co±δβ, δβ3, · · · , δβk

and since for every sufficiently small δ > 0, we have

±δβ ∈ co± (t+ − t−)β ⊂ co± (t+ − t−)β, δβ3, · · · , δβk,

we get that

0 ∈ δH = int co±δβ, δβ3, · · · , δβk ⊂ int co± (t+ − t−)β, δβ3, · · · , δβk.

We are therefore in a position to apply Corollary 10.23 to

a = α, b = (t+ − t−)β, bj = δβj for j = 3, · · · , k, t =−t−

t+ − t−,

A = ξ+ = ξ +t+

t+ − t−α⊗ (t+ − t−)β = ξ + (1− t) a⊗ b,

B = ξ− = ξ +t−

t+ − t−α⊗ (t+ − t−)β = ξ − ta⊗ b

and find uδ ∈ Affpiec

(Ω; RN

), disjoint open sets Ω+ , Ω− ⊂ Ω, such that

⎧⎪⎪⎪⎨⎪⎪⎪⎩

0 ≤ measΩ−meas (Ω+ ∪ Ω−) ≤ δ

uδ(x) = uξ(x), x ∈ ∂Ω and |uδ(x)− uξ(x)| ≤ δ, x ∈ Ω

∇uδ(x) = ξ± a.e. in Ω±∇uδ(x) ∈ ξ + t+α⊗ β, t−α⊗ β, α⊗ δβ3, · · · , α⊗ δβk, a.e. in Ω.

(11.5)Since ξ± ∈ K0 ∩ ∂K and

ξ+α⊗δβj ∈ ξ+α⊗δH = ξ0+α⊗(γ + δH

)⊂ ξ0+α⊗ǫH ⊂ K0 for j = 3, · · · , k,

we deduce, by choosing δ = 1/ν as ν → ∞ in (11.5), the relaxation property(11.4). This achieves the proof of Step 2 and thus of the corollary.

We finally want to point out that, as a particular case of Corollary 11.7, wefind the existence theorem (Theorem 3.1) proved by Dacorogna-Marcellini in[195].


11.3 Necessary conditions

Recall that we are considering the minimization problem

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

,

where Ω is a bounded open set of Rn, uξ0 is affine (i.e. ∇uξ0 = ξ0) andf : RN×n → R is a lower semicontinuous, locally bounded and non-negativefunction. In order to avoid the trivial case, we always assume that

Qf (ξ0) < f (ξ0) .

Most non-existence results for problem (P ) follow by showing that the relaxedproblem (QP ) has a unique solution, namely uξ0 , which by hypothesis is not asolution of (P ). This approach was strongly used in Marcellini [420], Dacorogna-Marcellini [195] and Dacorogna-Pisante-Ribeiro [211]. We should point out thatwe will give an example (see Proposition 11.38 in Section 11.5.5) related to areatype integrands, where non-existence occurs, while the relaxed problem hasinfinitely many solutions, none of them being a solution of (P ).

The right notion in order to have uniqueness of the relaxed problem is thefollowing.

Definition 11.9 A quasiconvex function f : RN×n → R is said to be strictlyquasiconvex at ξ0 ∈ RN×n if for some bounded open set U ⊂ Rn the equality

∫

U

f (ξ0 +∇ϕ (x)) dx = f (ξ0) measU

holds for some ϕ ∈ W 1,∞0

(U ; RN

), then necessarily ϕ ≡ 0.

We should observe that as in Proposition 5.11 the notion of strict quasi-convexity is independent of the choice of the set U, more precisely we have thefollowing.

Proposition 11.10 If a function f : RN×n → R is strictly quasiconvex atξ0 ∈ RN×n for one bounded open set U ⊂ Rn, it is so for any such set.

Proof. Let V ⊂ Rn be a bounded open set and ψ ∈ W 1,∞0

(V ; RN

)be such

that ∫

V

f (ξ0 +∇ψ (x)) dx = f (ξ0) measV (11.6)

and let us conclude that we necessarily have ψ ≡ 0.Choose first a > 0 sufficiently large so that

V ⊂ Qa = (−a, a)n

and then define

v (x) :=

ψ (x) if x ∈ V

0 if x ∈ Qa − V

Necessary conditions 473

so that v ∈W 1,∞0

(Qa; RN

).

Let then x0 ∈ U and choose ν sufficiently large so that

x0 +1

νQa = x0 +

(−a

ν,a

ν

)n

⊂ U.

Define next

ϕ (x) :=

1ν v (ν (x− x0)) if x ∈ x0 + 1

ν Qa

0 if x ∈ U −[x0 + 1

ν Qa

].


(U ; RN

)and

∫

U

f (ξ0 +∇ϕ (x)) dx = f (ξ0) meas(U − [x0 +1

νQa])

+

∫

x0+1ν Qa

f (ξ0 +∇v (ν (x− x0))) dx

= f (ξ0) [measU − measQa

νn] +

1

νn

∫

Qa

f (ξ0 +∇v (y)) dy

= f (ξ0) [measU − measQa

νn+

meas(Qa − V )

νn]

+1

νn

∫

V

f (ξ0 +∇ψ (y)) dy.

Appealing to (11.6), we deduce that∫

U

f (ξ0 +∇ϕ (x)) dx = f (ξ0) measU.

Since f is strictly quasiconvex at ξ0 ∈ RN×n for the set U, we deduce that ϕ ≡ 0,which in turn implies that

v (y) ≡ 0, for every y ∈ Qa .

This finally implies that ψ ≡ 0 as claimed.

We will see below some sufficient conditions that can ensure strict quasicon-vexity, but let us start with the elementary following non-existence theorem.

Theorem 11.11 Let f : RN×n → R be lower semicontinuous, locally boundedand non-negative, ξ0 ∈ RN×n with Qf (ξ0) < f (ξ0) and Qf be strictly quasicon-vex at ξ0 . Then the relaxed problem (QP ) has a unique solution, namely uξ0 ,while (P ) has no solution.

Proof. The fact that (QP ) has only one solution follows by definition ofthe strict quasiconvexity of Qf and Proposition 11.10. Assume for the sake ofcontradiction that (P ) has a solution u ∈ uξ0 + W 1,∞

0 (Ω; RN ). We should havefrom Theorem 11.1 that (writing u (x) = ξ0x + ϕ (x))

f (ξ0 +∇ϕ (x)) = Qf (ξ0 +∇ϕ (x)) , a.e. x ∈ Ω


∫

Ω

Qf (ξ0 +∇ϕ (x)) dx = Qf (ξ0) measΩ.

Since Qf is strictly quasiconvex at ξ0 , we deduce from the last identity thatϕ ≡ 0. Hence we have, from the first identity, that Qf (ξ0) = f (ξ0) , which is incontradiction with the hypothesis.

We now want to give some criteria that can ensure the strict quasiconvexityof a given function. The first one was introduced by Dacorogna-Marcellini in[195].

Definition 11.12 A convex function f : RN×n → R is said to be strictlyconvex at ξ0 ∈ RN×n in at least N directions if there exist αi ∈ Rn, αi = 0 forevery i = 1, · · · , N, such that: if for some η ∈ RN×n the identity

1

2f (ξ0 + η) +

1

2f (ξ0) = f(ξ0 +

1

2η)

holds, then necessarily⟨αi; ηi

⟩= 0, i = 1, · · · , N.

In order to better understand the generalization of this notion to polyconvexfunctions (see Proposition 11.18), it might be enlightening to state the definitionin the following way.

Proposition 11.13 Let f : RN×n → R be a convex function and, for ξ ∈RN×n, denote by ∂f (ξ) the subdifferential of f at ξ. The following two conditionsare then equivalent:

(i) f is strictly convex at ξ0 ∈ RN×n in at least N directions;

(ii) there exist αi ∈ Rn with αi = 0 for every i = 1, · · · , N, so that whenever

f (ξ0 + η)− f (ξ0)− 〈λ; η〉 = 0

for some η ∈ RN×n and for some λ ∈ ∂f (ξ0) , then⟨αi; ηi

⟩= 0, i = 1, · · · , N.

Proof. Step 1. We start with a preliminary observation that if

1

2f (ξ0 + η) +

1

2f (ξ0) = f(ξ0 +

1

2η), (11.7)

then, for every t ∈ [0, 1] , we have

tf (ξ0 + η) + (1− t) f (ξ0) = f (ξ0 + tη) . (11.8)

Let us show this under the assumption that t > 1/2 (the case t < 1/2 is handledsimilarly). We can therefore find α ∈ (0, 1) such that

1

2= αt + (1− α) 0 = αt.


From the convexity of f and by hypothesis, we obtain

1

2f (ξ0 + η) +

1

2f (ξ0) = f(ξ0 +

1

2η) ≤ αf (ξ0 + tη) + (1− α) f (ξ0) .

Assume, for the sake of contradiction, that

f (ξ0 + tη) < tf (ξ0 + η) + (1− t) f (ξ0) .

Then combine this inequality with the previous one to get

1

2f (ξ0 + η) +

1

2f (ξ0) < α [tf (ξ0 + η) + (1− t) f (ξ0)] + (1− α) f (ξ0)

=1

2f (ξ0 + η) +

1

2f (ξ0) ,

which is clearly a contradiction. Therefore the convexity of f and the abovecontradiction imply (11.8) and also that

f ′ (ξ0, η) := limt→0+

f (ξ0 + tη)− f (ξ0)

t= f (ξ0 + η)− f (ξ0) .

Applying Theorem 2.50, we get that there exists λ ∈ ∂f (ξ0) such thatf (ξ0 + η)− f (ξ0) = 〈λ; η〉 and hence, from (11.8), we find

f (ξ0 + tη)− f (ξ0)− t 〈λ; η〉 = 0, ∀ t ∈ [0, 1] . (11.9)

We have therefore proved that (11.7) implies (11.9). Since the converse is obvi-ously true, we conclude that they are equivalent.

Step 2. Let us show the equivalence of the two conditions.

(i) ⇒ (ii). We first observe that, for any μ ∈ RN×n, we have

12f (ξ0 + η) + 1

2f (ξ0)− f(ξ0 + 1

2η)

= 12 [f (ξ0 + η)− f (ξ0)− 〈μ; η〉]− [ f

(ξ0 + 1

2η)− f (ξ0)− 1

2 〈μ; η〉 ].(11.10)

Assume that, for λ ∈ ∂f (ξ0) , we have

f (ξ0 + η)− f (ξ0)− 〈λ; η〉 = 0.

From (11.10) applied to μ = λ, from the definition of ∂f (ξ0) and from theconvexity of f, we get

0 ≤ 1

2f (ξ0 + η) +

1

2f (ξ0)− f(ξ0 +

1

2η)

= −[f(ξ0 +1

2η)− f (ξ0)−

1

2〈λ; η〉] ≤ 0.

Using the above identity, we then are in the framework of (i) and we deducethat

⟨αi; ηi

⟩= 0, i = 1, · · · , N, and thus (ii).


(ii) ⇒ (i). Assume now that we have (11.7), namely

1

2f (ξ0 + η) +

1

2f (ξ0)− f(ξ0 +

1

2η) = 0

which, by Step 1, implies that there exists λ ∈ ∂f (ξ0) such that

f (ξ0 + tη)− f (ξ0)− t 〈λ; η〉 = 0, ∀ t ∈ [0, 1] .

We are therefore choosing t = 1 in the framework of (ii) and we get⟨αi; ηi

⟩=

0, i = 1, · · · , N, as wished.

Of course any strictly convex function is strictly convex in at least N direc-tions, but the above condition is much weaker. For example, in the scalar case,N = 1, it is enough that the function is not affine in a neighborhood of ξ0 toguarantee the condition (see below).

We now have the following result established by Dacorogna-Marcellini in[195].

Proposition 11.14 If a convex function f : RN×n → R is strictly convex atξ0 ∈ RN×n in at least N directions, then it is strictly quasiconvex at ξ0 .

Theorem 11.11, combined with the above proposition, immediately gives asharp result for the scalar case, namely the following corollary.

Corollary 11.15 Let f : Rn → R be lower semicontinuous, locally bounded andnon-negative, ξ0 ∈ Rn with Cf (ξ0) < f (ξ0) and Cf not affine in a neighborhoodof ξ0 . Then (P ) has no solution.

Remark 11.16 In the scalar case this result has been obtained by severalauthors, in particular Cellina [133], Friesecke [291] and Dacorogna-Marcellini[195]. It also gives (see Theorem 11.26), combined with the result of the pre-ceding section, that, provided some appropriate boundedness is assumed, a nec-essary and sufficient condition for the existence of minima for (P ) is that f beaffine on the connected component of ξ : Cf (ξ) < f (ξ) that contains ξ0 . ♦

Before proceeding with the proof of Proposition 11.14, we need the followingelementary lemma.

Lemma 11.17 Let Ω be a bounded open set of Rn and ϕ ∈ W 1,∞0 (Ω; RN ) be

such that ⟨αi;∇ϕi (x)

⟩= 0 a.e. x ∈ Ω, i = 1, · · · , N,

for some αi = 0, i = 1, · · · , N, then ϕ ≡ 0.

Proof. (Lemma 11.17). Working component by component, we can assumethat N = 1 and therefore we will drop the indices. So let ϕ ∈ W 1,∞

0 (Ω) satisfy,for some α ∈ Rn, α = 0,

〈α;∇ϕ (x)〉 = 0 a.e. x ∈ Ω.


We then choose α2, · · · , αn ∈ Rn such that α, α2, · · · , αn generate a basis ofRn. Let a > 0 and for m an integer

Qma := (−a, a)

m.

Let x ∈ Ω and let a > 0 and t > 0 be sufficiently small so that

x + τα + τ2α2 + · · ·+ τnαn ∈ Ω, for every τ ∈ (0, t) and (τ2, · · · , τn) ∈ Qn−1a .

Observe that if ϕ ∈ C10 (Ω), then

∫

Qn−1a

[ϕ (x + tα + τ2α2 + · · ·+ τnαn)− ϕ (x + τ2α2 + · · ·+ τnαn)] dτ2 · · · dτn

=

∫

Qn−1a

∫ t

0

d

dτ[ϕ (x + τα + τ2α2 + · · ·+ τnαn)] dτdτ2 · · ·dτn

=

∫

Qn−1a

∫ t

0

〈∇ϕ (x + τα + τ2α2 + · · ·+ τnαn) ; α〉 dτdτ2 · · ·dτn .

By a standard regularization procedure, the above identity also holds for anyϕ ∈ W 1,∞

0 (Ω). Since 〈α;∇ϕ〉 = 0, we deduce that∫

Qn−1a

[ϕ (x + tα + τ2α2 +· · ·+ τnαn)− ϕ (x + τ2α2 +· · ·+ τnαn)] dτ2· · ·dτn = 0.

Since ϕ is continuous, we deduce, by dividing by the measure of Qn−1a and

letting a → 0, that, for every t sufficiently small so that x + tα ∈ Ω,

ϕ (x + tα) = ϕ (x) .

Choosing t so that

x + τα ∈ Ω, ∀ τ ∈ [0, t) and x + tα ∈ ∂Ω,

we obtain the claim, namely

ϕ (x) = 0, ∀x ∈ Ω.

This achieves the proof of the lemma.

Proof. (Proposition 11.14). Assume that for a certain bounded open setU ⊂ Rn and for some ϕ ∈W 1,∞

0

(U ; RN

)we have

∫

U


and let us show that ϕ ≡ 0.Since f is convex and the above identity holds, we find

f (ξ0) meas U =

∫

U

[1

2f (ξ0) +

1

2f (ξ0 +∇ϕ (x)) ]dx

≥∫

U

f(ξ0 +1

2∇ϕ (x))dx ≥ f (ξ0) measU,


which implies that∫

U

[1

2f (ξ0) +

1

2f (ξ0 +∇ϕ (x))− f(ξ0 +

1

2∇ϕ (x))]dx = 0.

The convexity of f implies then that, for almost every x in U, we have

1

2f (ξ0) +

1

2f (ξ0 +∇ϕ (x))− f(ξ0 +

1

2∇ϕ (x)) = 0.

The strict convexity in at least N directions leads to⟨αi;∇ϕi (x)

⟩= 0, a.e. x ∈ Ω, i = 1, · · · , N.

Lemma 11.17 gives the claim.

We now generalize Proposition 11.14. Since the notation in the next resultis involved, we first write the proposition when N = n = 2. We also adopt thenotation of Definition 5.1.

Proposition 11.18 Let f : RN×n → R be polyconvex, ξ0 ∈ RN×n and λ =λ (ξ0) ∈ Rτ(n,N) such that

f (ξ0 + η)− f (ξ0)− 〈λ; T (ξ0 + η)− T (ξ0)〉 ≥ 0 for every η ∈ RN×n.

(i) Let N = n = 2 and assume that there exist α1,1, α1,2, α2,2 ∈ R2, α1,1 =0, α2,2 = 0, β ∈ R, so that if for some η ∈ R2×2 the equality

f (ξ0 + η)− f (ξ0)− 〈λ; T (ξ0 + η)− T (ξ0)〉 = 0

holds, then necessarily⟨α2,2; η2

⟩= 0 and

⟨α1,1; η1

⟩+⟨α1,2; η2

⟩+ β det η = 0.

Then f is strictly quasiconvex at ξ0 .

(ii) Let N, n ≥ 2 and assume that there exist, for every ν = 1, · · · , N,

αν,ν , αν,ν+1, · · · , αν,N ∈ Rn, αν,ν = 0,

βν,s ∈ R

(N−ν+1

s

)×

(ns

), 2 ≤ s ≤ n ∧ (N − ν + 1) ,

such that if for some η ∈ RN×n the equality

f (ξ0 + η)− f (ξ0)− 〈λ; T (ξ0 + η)− T (ξ0)〉 = 0

holds, then necessarily

N∑

s=ν

〈αν,s; ηs〉+n∧(N−ν+1)∑

s=2

⟨βν,s; adjs

(ην , · · · , ηN

)⟩= 0, ν = 1, · · · , N.

Then f is strictly quasiconvex at ξ0 .


Remark 11.19 (i) The existence of a λ as in the hypotheses of the propositionis automatically guaranteed by the polyconvexity of f (see Theorem 5.6, whichcorresponds in the case of a convex function to an element of ∂f (ξ0)).

(ii) We have adopted the convention that if l > k > 0 are integers, then

k∑

l

= 0. ♦

Example 11.20 Let N = n = 2 and consider the function

f (η) =(η22

)2+(η11 + det η

)2.

This function is trivially polyconvex and according to the proposition it is alsostrictly quasiconvex at ξ0 = 0 (choose λ = 0 ∈ R5, α2,2 = (0, 1) , α1,2 = (0, 0) ,α1,1 = (1, 0) , β = 1). ♦

Proof. We prove the proposition only in the case N = n = 2, the general casebeing handled similarly.

Assume that for a certain bounded open set U ⊂ R2 and for some ϕ ∈W 1,∞

0

(U ; R2

)we have

∫

U


and let us prove that ϕ ≡ 0. This is equivalent, for every μ ∈ Rτ(2,2), to

∫

U

[ f (ξ0 +∇ϕ (x))− f (ξ0)− 〈μ; T (ξ0 +∇ϕ (x))− T (ξ0)〉 ]dx = 0.

Choosing μ = λ (λ as in the statement of the proposition) in the previousequation and using the polyconvexity of the function f, we get

f (ξ0 +∇ϕ (x))− f (ξ0)− 〈λ; T (ξ0 +∇ϕ (x))− T (ξ0)〉 = 0, a.e. x ∈ Ω.

We hence infer that, for almost every x ∈ Ω, we have

⟨α2,2;∇ϕ2

⟩= 0 and

⟨α1,1;∇ϕ1

⟩+⟨α1,2;∇ϕ2

⟩+ β det∇ϕ = 0.

Lemma 11.17, applied to the first equation, implies that ϕ2 ≡ 0. Using thisresult in the second equation we get

⟨α1,1;∇ϕ1

⟩= 0

and hence, appealing once more to the lemma, we have the claim, namelyϕ1 ≡ 0.

Summarizing the results of Theorem 11.11, Proposition 11.14 and Proposi-tion 11.18, we get the following corollary.


Corollary 11.21 Let f : RN×n → R be lower semicontinuous, locally boundedand non-negative and ξ0 ∈ RN×n with

Qf (ξ0) < f (ξ0) .

If either one of the two conditions

(i) Qf (ξ0) = Cf (ξ0) and Cf is strictly convex at ξ0 in at least N directions;

(ii) Qf (ξ0) = Pf (ξ0) and Pf is strictly polyconvex at ξ0 (in the sense of

Proposition 11.18);

holds, then (QP ) has a unique solution, namely uξ0 , while (P ) has no solution.

Proof. The proof is almost identical under both hypotheses and so we establishthe corollary only in the first case. The result follows from Theorem 11.11 if wecan show that Qf is strictly quasiconvex at ξ0 . So assume that

∫

Ω

Qf (ξ0 +∇ϕ (x)) dx = Qf (ξ0) measΩ

for some ϕ ∈ W 1,∞0 (Ω; RN ) and let us prove that ϕ ≡ 0. Using Jensen inequality

combined with the hypothesis Qf (ξ0) = Cf (ξ0) and the fact that Qf ≥ Cf,we find that the above identity implies

∫

Ω

Cf (ξ0 +∇ϕ (x)) dx = Cf (ξ0) measΩ.

The hypotheses on Cf and Proposition 11.14 imply that ϕ ≡ 0, as wished.

We now conclude this section with a different necessary condition that isbased on Caratheodory theorem (see Theorem 2.13).

Recall first that, for any integer s, we let

Λs := λ = (λ1, · · · , λs) : λi ≥ 0 and∑s

i=1 λi = 1 .

Theorem 11.22 Let f : RN×n → R be lower semicontinuous, locally boundedand non-negative. If (P ) has a solution u ∈ uξ0 + W 1,∞

0

(Ω; RN

), then there

exist μ ∈ ΛNn+1 and ξν ∈ RN×n, |ξν | ≤ ‖u‖W 1,∞ , 1 ≤ ν ≤ Nn + 1 such that

Qf (ξ0) ≥Nn+1∑

ν=1

μνf (ξν) and ξ0 =

Nn+1∑

ν=1

μνξν .

Moreover, if either n = 1 or N = 1, the inequality becomes an equality, namely

Cf (ξ0) =

Nn+1∑

ν=1

μνf (ξν) and ξ0 =

Nn+1∑

ν=1

μνξν .


Remark 11.23 The theorem is just a curiosity in the vectorial case n, N > 1.However, in the scalar case n > N = 1, under some extra hypotheses (seeTheorem 11.26), one of them being

ξ0 ∈ int co ξ1, · · · , ξn+1 ,

it turns out that the necessary condition is also sufficient. But it is in the caseN ≥ n = 1 that it is particularly interesting since then this condition is alsosufficient (see Theorem 11.24). ♦

Proof. We decompose the proof into three steps.

Step 1. Let u ∈ uξ0 +W 1,∞0

(Ω; RN

)be a solution of (P ). It should therefore

satisfy∫

Ω

f (∇u (x)) dx = inf (P ) = inf (QP ) = Qf (ξ0)meas Ω. (11.11)

Let r = ‖u‖W 1,∞ and use the fact that f is locally bounded to find R = R (r)such that

0 ≤ f (∇u (x)) ≤ R a.e. x ∈ Ω.

Denote

Kr :=(ξ, y) ∈ RN×n × R : |ξ| ≤ r and |y| ≤ R

,

epi f :=(ξ, y) ∈ RN×n × R : f (ξ) ≤ y

,

E := epi f ∩Kr .

Note that since f is lower semicontinuous then epi f is closed and hence E iscompact. Therefore its convex hull co E is also compact.

Observe that, for almost every x ∈ Ω, we have

(∇u (x) , f (∇u (x))) ∈ E

and thus by Jensen inequality and (11.11) we deduce that

(ξ0, Qf (ξ0)) =1

measΩ

∫

Ω

(∇u (x) , f (∇u (x))) dx ∈ coE.

Appealing to Caratheodory theorem, we can find λ ∈ ΛNn+2 , (ξi, yi) ∈ E,1 ≤ i ≤ Nn + 2 (in particular, f (ξi) ≤ yi) such that

Qf (ξ0) =

Nn+2∑

i=1

λiyi ≥Nn+2∑

i=1

λif (ξi) and ξ0 =

Nn+2∑

i=1

λiξi .

(Note, in passing, that if f is continuous, we can replace epi f in the aboveargument by

graphf :=(ξ, y) ∈ RN×n × R : f (ξ) = y

,


therefore obtaining equality instead of inequality in the above statement.)

Step 2. To obtain Part 1 of the theorem it therefore remains to show thatone can take only (Nn + 1) elements. This is a classical procedure in convexanalysis and we have encountered it in Theorem 2.35. The result is equivalentto showing that there exist μi , 1 ≤ i ≤ Nn + 2, such that

⎧⎪⎪⎪⎨⎪⎪⎪⎩

μi ≥ 0,Nn+2∑

i=1

μi = 1, at least one of the μi = 0

Nn+2∑i=1

μif (ξi) ≤Nn+2∑i=1

λif (ξi) , ξ0 =Nn+2∑i=1

μiξi .

(11.12)

meaning in fact that μ ∈ ΛNn+1 as wished.Assume that λi > 0, 1 ≤ i ≤ Nn + 2; otherwise nothing is to be proved.

Observe first that ξ0 ∈ co ξ1, · · · , ξNn+2 ⊂ RN×n. Thus it follows fromCaratheodory theorem that there exist ν ∈ ΛNn+2 with at least one of theνi = 0 (i.e. ν ∈ ΛNn+1) such that

ξ0 =Nn+2∑

i=1

νiξi .

Assume, without loss of generality, that

Nn+2∑

i=1

νif (ξi) >

Nn+2∑

i=1

λif (ξi) ; (11.13)

otherwise, choosing μi = νi we would immediately have (11.12). Let

J := i ∈ 1, · · · , Nn + 2 : λi − νi < 0 .

Observe that J = ∅, since otherwise λi ≥ νi ≥ 0 for every i and since at leastone of the νi = 0, we would have a contradiction with

∑νi =

∑λi = 1 and

λi > 0 for every i. We then define

γ := mini∈J λi

νi − λi.

We clearly have that γ > 0. Finally, let

μi = λi + γ (λi − νi) , 1 ≤ i ≤ Nn + 2.

We immediately get that

μi ≥ 0,

Nn+2∑

i=1

μi = 1, at least one of the μi = 0. (11.14)

From (11.13), we obtain∑Nn+2

i=1 μif (ξi) =∑Nn+2

i=1 λif (ξi) + γ(∑Nn+2

i=1 λif (ξi)−∑Nn+2

i=1 νif (ξi))

≤∑Nn+2i=1 λif (ξi) .

The scalar case 483

The combination of the above with (11.14) (assuming for the sake of notationsthat μNn+2 = 0) immediately gives

Qf (ξ0) ≥Nn+1∑

i=1

μif (ξi) and ξ0 =

Nn+1∑

i=1

μiξi .

Step 3. The result for the scalar case follows from the fact that Qf (ξ0) =Cf (ξ0) and from Theorem 2.35.

11.4 The scalar case

We now see how to apply the above abstract considerations to the case whereeither n = 1 or N = 1. We recall that

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0

(Ω; RN

).

We first treat the more elementary case where n = 1 and then the case N = 1.

11.4.1 The case of single integrals

In this very elementary case, we can get much simpler and sharper results.

Theorem 11.24 Let N ≥ 1 and f : RN → R be non-negative, locally boundedand lower semicontinuous. Let a < b, α, β ∈ RN and

(P ) inf

I (u) =

∫ b

a

f (u′ (x)) dx : u ∈ X

,

where

X :=u ∈W 1,∞ ((a, b) ; RN

): u (a) = α, u (b) = β

.

The following two statements are then equivalent:

(i) problem (P ) has a minimizer;

(ii) there exist λν ≥ 0 with∑N+1

ν=1 λν = 1, γν ∈ RN , 1 ≤ ν ≤ N + 1 suchthat

Cf(β − α

b− a) =

N+1∑

ν=1

λνf (γν) andβ − α

b− a=

N+1∑

ν=1

λνγν , (11.15)

where Cf is the convex envelope of f.

Furthermore, if (11.15) is satisfied and if

Ip := [ a + (b− a)∑p−1

ν=1 λν , a + (b− a)∑p

ν=1 λν ], 1 ≤ p ≤ N + 1,


then

u (x) = γp (x− a) + (b− a)

p∑

ν=1

λν (γν − γp) + α, x ∈ Ip , 1 ≤ p ≤ N + 1,

is a solution of (P ).

Remark 11.25 (i) The sufficiency of (11.15) is implicitly or explicitly proved inthe papers mentioned in the introduction of the present chapter. The necessityis less known but is also implicit in the literature. The theorem as stated canbe found in Dacorogna [179].

(ii) Recall that by Caratheodory theorem (see Theorem 2.35) we always have

Cf ((β − α) / (b− a)) = inf∑N+1ν=1 λνf (γν) :

∑N+1ν=1 λνγν =

β − α

b− a. (11.16)

Therefore (11.15) states that a necessary and sufficient condition for existenceof solutions is that the infimum in (11.16) be attained. Note also that if f isconvex or f coercive (in the sense that f (ξ) ≥ a |ξ|p + b with p > 1, a > 0),then the infimum in (11.16) is always attained.

(iii) Therefore if f (x, u, ξ) = f (ξ) , counterexamples to existence must benon-convex and non-coercive; see Example 4.4, where

(P ) inf

I (u) =

∫ 1

0

e−(u′(x))2

dx : u ∈ W 1,∞0 (0, 1)

(i.e. f (ξ) = e−ξ2

), then Cf (ξ) ≡ 0 and therefore by the relaxation theorem

inf (P ) = inf (QP ) = 0.

However, it is obvious that I (u) = 0 for every u ∈ W 1,∞0 (0, 1) and hence the

infimum of (P ) is not attained.

(iv) A similar proof to that of Theorem 11.24 (see for example Marcellini[419]) shows that a sufficient condition to ensure existence of minima to

(P ) inf

I (u) =

∫ b

a

f (x, u′ (x)) dx : u ∈ X

is (11.15), where λν and γν are then measurable functions. Of course, if fdepends explicitly on u, the example of Bolza (see Example 4.8) shows that thetheorem is then false. ♦Proof. (Theorem 11.24). It is easy to see that we can reduce our study to thecase where

a = 0, b = 1 and α = 0.

Sufficient condition. The sufficiency part is elementary. Let

(QP ) inf

I (u) =

∫ 1

0

Cf (u′ (x)) dx : u ∈ X

The scalar case 485

where now

X :=u ∈ W 1,∞ ((0, 1) ; RN

): u (0) = 0, u (1) = β

.

Then u (x) = βx is trivially a solution of (QP ) and therefore

inf (QP ) = Cf (β) .

Let now u be as in the statement of the theorem. Observe first that u ∈W 1,∞ ((0, 1) ; RN

)and u (0) = 0, u (1) = β. We now compute

I (u) =

∫ 1

0

f (u′ (x)) dx =

N+1∑

p=1

∫

Ip

f (u′ (x)) dx =

N+1∑

p=1

f (γp)meas Ip

=

N+1∑

p=1

λpf (γp) = Cf (β) = inf (QP ) ≤ inf (P ) .

Necessary condition. This has already been proved in Theorem 11.22.

11.4.2 The case of multiple integrals

We now discuss the case n > N = 1. This is of course a more difficult case thanthe preceding one and no such simple result like Theorem 11.24 is available.However we immediately have from Sections 11.2 and 11.3 (Theorem 10.18 andCorollary 11.15) the theorem stated below. For some historical comments onthis theorem, see the remark following Corollary 11.15.

But let us first recall the problem and the notation. We have

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω)

,

where Ω is a bounded open set of Rn, uξ0 is affine (i.e. ∇uξ0 = ξ0) and f :Rn → R is a lower semicontinuous, locally bounded and non-negative function.Let

Cf (ξ) := sup g (ξ) : g ≤ f and g convex .

In order to avoid the trivial situation, we assume that

Cf (ξ0) < f (ξ0) .

We next set

K := ξ ∈ Rn : Cf (ξ) < f (ξ)

and we assume that it is connected, otherwise we replace it by its connectedcomponent that contains ξ0 .


Theorem 11.26 Necessary condition. If (P ) has a minimizer, then Cf isaffine in a neighborhood of ξ0 .

Sufficient condition. If there exists E ⊂ ∂K such that ξ0 ∈ int coE andCf |E∪ξ0 is affine, then (P ) has a solution.

Remark 11.27 (i) By Cf |E∪ξ0 affine we mean that there exist α ∈ Rn, β ∈R such that

Cf (ξ) = 〈α; ξ〉+ β for every ξ ∈ E ∪ ξ0 .

Usually one proves that Cf is affine on the whole of co E.

(ii) The theorem applies, of course, to the case where E = ∂K and Cf isaffine on the whole of K (since K is open and ξ0 ∈ K ⊂ int coK). However, inmany simple examples such as the one given below, it is not realistic to assumethat E = ∂K. ♦

Proof. The necessary part is just Corollary 11.15. We therefore discuss onlythe sufficient part. We use Theorem 10.18 to find u ∈ uξ0 +W 1,∞

0 (Ω) such that

∇u (x) ∈ E ⊂ ∂K, a.e. x ∈ Ω

and hencef (∇u (x)) = Cf (∇u (x)) , a.e. x ∈ Ω.

Then use the fact that Cf |E∪ξ0 is affine to deduce that

∫

Ω

Cf (∇u (x)) dx = Cf (ξ0) measΩ.

The conclusion then follows from Theorem 11.1.

We would now like to give two simple examples.

Example 11.28 Let N = 1, n = 2, Ω = (0, 1)2, u0 ≡ 0, a ≥ 0, ξ = (ξ1, ξ2)

andf (ξ) = ((ξ1)

2 − 1)2 + ((ξ2)2 − a2)2.

We find that

Cf (ξ) =[(ξ1)

2 − 1]2+

+[(ξ2)

2 − a2]2+

where

[x]+ =

x if x ≥ 0

0 if x < 0.

We therefore have that

K =ξ ∈ R2 : |ξ1| < 1 or |ξ2| < a

and note that it is unbounded and that Cf is not affine on the whole of K.

The vectorial case 487

Let us discuss the two different cases.

Case 1: a = 0. Then clearly Cf is not affine in the neighborhood of ξ0 = 0,since it is strictly convex in the direction e2 = (0, 1) . Hence (P ) has no solution.

Case 2: a > 0. We let

E :=ξ ∈ R2 : |ξ1| = 1 and |ξ2| = a

⊂ ∂K.

Note that ξ0 = 0 ∈ int coE and Cf |co E ≡ 0 is affine. Therefore the theoremapplies and we obtain that (P ) has a solution. ♦Example 11.29 (see Marcellini [420] and Dacorogna-Marcellini [195]). Letn ≥ 2 and

f (∇u) = g (|∇u|) ,

where g : R → R is lower semicontinuous, locally bounded and non-negativewith

g (0) = inf g (t) : t ≥ 0 .

Theorem 6.30 implies that Cf = Cg. Let

S := t ≥ 0 : Cg (t) < g (t)

K := ξ ∈ Rn : Cf (ξ) < f (ξ) = ξ ∈ Rn : |ξ| ∈ S .

Assume that ξ0 ∈ K and that S is connected, otherwise replace it by its con-nected component containing |ξ0| .

We then have to consider two cases.

Case 1: Cg is strictly increasing at |ξ0| . Then clearly Cf is not affine in anyneighborhood of ξ0 and hence (P ) has no solution.

Case 2: Cg is constant on S. Assume that S is bounded, this can be guar-anteed if, for example,

limt→+∞

g(t)

t= +∞.

So let |ξ0| ∈ S = (α, β) and choose in the sufficient part of the theorem

E := ξ ∈ Rn : |ξ| = β

and apply the theorem to find a minimizer for (P ). ♦

11.5 The vectorial case

We now consider several examples of the form studied in the previous sections,namely

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

,


where Ω is a bounded open set of Rn, uξ0 is affine (i.e. ∇uξ0 = ξ0) andf : RN×n → R is a lower semicontinuous, locally bounded and non-negativefunction.

All the cases have already been encountered on several occasions.

(1) We consider in Section 11.5.1 (see also Sections 6.6.2 and 10.3.2) the casewhere N = n and

f(ξ) = g(λ2(ξ), · · · , λn−1(ξ), det ξ),

where 0 ≤ λ1(ξ) ≤ · · · ≤ λn(ξ) are the singular values of ξ ∈ Rn×n.

(2) In Section 11.5.2 (see also Sections 6.6.3 and 10.3.4), we deal with thecase

f (ξ) = g (Φ (ξ))

where Φ : RN×n → R is quasiaffine (so in particular we can have, when N = n,

Φ(ξ) = det ξ, as in the previous case).

(3) We next discuss in Section 11.5.3 (see also Section 6.6.6) the SaintVenant-Kirchhoff energy functional. Up to rescaling, the function under consid-eration is (here N = n and ν ∈ (0, 1/2) is a parameter)


∣∣2 +ν

1− 2ν( |ξ|2 − n )2

or in terms of the singular values 0 ≤ λ1(ξ) ≤ · · · ≤ λn(ξ) of ξ ∈ Rn×n

f(ξ) =∑n

i=1

(λ2

i − 1)2

+ν

1− 2ν

(∑ni=1 λ2

i − n)2

.

(4) In Section 11.5.4 (see also Sections 6.6.5 and 10.3.5), we consider a prob-lem of optimal design where N = n = 2 and

f (ξ) =

1 + |ξ|2 if ξ = 0

0 if ξ = 0.

(5) In Section 11.5.5 (see also Section 6.6.4), we deal with the area type case,namely when N = n + 1 and f(ξ) = g(adjn ξ).

(6) Finally, in Section 11.5.6 (see also Sections 7.4.2 and 10.3.3), we discussthe case of potential wells.


In this section, we let N = n and we denote by λ1(ξ), · · · , λn(ξ) the singularvalues of ξ ∈ Rn×n with 0 ≤ λ1(ξ) ≤ · · · ≤ λn(ξ) and by Kn−2

+ the set

Kn−2+ := x = (x2, · · · , xn−1) ∈ Rn−2 : 0 ≤ x2 ≤ · · · ≤ xn−1,


which is the natural set where to consider (λ2(ξ), · · · , λn−1(ξ)) for ξ ∈ Rn×n.

The following theorem has been established by Dacorogna-Pisante-Ribeiro[211].

Theorem 11.30 Let

f(ξ) = g(λ2(ξ), · · · , λn−1(ξ)) + h(det ξ),

where g : Kn−2+ → R is upper semi continuous and verifies

inf g = g(m2, · · · , mn−1), with 0 < m2 ≤ · · · ≤ mn−1

and h : R → R is a lower semicontinuous, locally bounded and non-negativefunction such that

lim|t|→+∞

h(t)

|t| = +∞. (11.17)

Let ξ0 ∈ Rn×n be such that

Ch(det ξ0) < h(det ξ0). (11.18)

Then

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; Rn)

has a solution.

Remark 11.31 It can be shown (see the proof for details) that the condition(11.18) is not needed and that the conclusion is valid for every ξ0 ∈ Rn×n. ♦

Proof. We note that, by Theorem 6.22, Qf(ξ) = inf g + Ch(det ξ). Letting

K :=ξ ∈ RN×n : Qf (ξ) < f (ξ)

we see that

K = L1 ∪ L2

where

L1 :=ξ ∈ Rn×n : Ch(det ξ) < h(det ξ)

L2 :=ξ ∈ Rn×n : Ch(det ξ) = h(det ξ), inf g < g(λ2(ξ), · · · , λn−1(ξ))

.

Clearly, if ξ0 /∈ K then uξ0 is a solution of (P ). Let us suppose that ξ0 ∈ K.Our hypothesis (11.18) ensures that ξ0 ∈ L1 . The case ξ0 ∈ L2 is more delicateand can be handled as in Dacorogna-Pisante-Ribeiro [211] (cf. also [184]), butwe do not discuss here the details.


We first observe that hypothesis (11.17) allows us to write S as an, at mostdenumerable, union of intervals, namely

S := t ∈ R : Ch(t) < h(t) =⋃

j∈N

(αj , βj),

Ch being affine in each interval (αj , βj); thus Qf is quasiaffine on each connectedcomponent of L1 and

L1 = ξ ∈ Rn×n : det ξ ∈ ∪j∈N(αj , βj).

Let (αj , βj) be an interval as above such that

det ξ0 ∈ (αj , βj).

We then choose mn ≥ mn−1 sufficiently large so that

n∏

i=ν

λi(ξ0) <n∏

i=ν

mi , ν = 2, · · · , n and max |αj | , |βj | < m2

n∏

i=2

mi .

We are then in a position to apply Theorem 10.25 to find u ∈ uξ0 +W 1,∞0 (Ω; Rn)

so that, for almost every x ∈ Ω,

det∇u (x) ∈ αj, βj, λν(∇u (x)) = mν , ν = 2, · · · , n.

Since Qf is quasiaffine on the connected component of L1 containing ξ0 , we canapply Theorem 11.1 to get the result.

11.5.2 The case of quasiaffine functions

We next study the minimization problem

(P ) inf

∫

Ω

g(Φ(∇u(x))) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

,

where Ω is a bounded open set of Rn, ∇uξ0 = ξ0 and

- g : R → R is a lower semicontinuous, locally bounded and non-negativefunction,

- Φ : RN×n → R is quasiaffine and non-constant.

We recall that in particular we can have, when N = n, Φ(ξ) = det ξ.

The relaxed problem is then

(QP ) inf

∫

Ω

Cg(Φ(∇u(x))) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

,

where Cg is the convex envelope of g (see Theorem 6.24).

The existence result is the following.


Theorem 11.32 Let Ω ⊂ Rn be a bounded open set, g : R → R a lowersemicontinuous, locally bounded and non-negative function such that

lim|t|→+∞

g(t)

|t| = +∞ (11.19)

and uξ0 (x) = ξ0x with ξ0 ∈ RN×n. Then there exists u ∈ uξ0 + W 1,∞0 (Ω; RN )

solution of

(P ) inf

∫

Ω

g(Φ(∇u(x))) dx : u ∈ uξ0 + W 1,∞0 (Ω; RN )

.

Remark 11.33 This result was first established by Mascolo-Schianchi [436]and later by Dacorogna-Marcellini [195] for the case of the determinant. Thegeneral case is due to Cellina-Zagatti [138] and later to Dacorogna-Ribeiro [212].Here we see that it can be obtained as a particular case of Theorem 11.1. ♦

Proof. We first let

S := t ∈ R : Cg(t) < g(t).

From the hypothesis on g we can write

S =⋃

j∈N

(αj , βj)

with Cg affine in each interval (αj , βj).

Case 1: Φ(ξ0) /∈ S. Then uξ0 is a solution of (P ).

Case 2: Φ(ξ0) ∈ (αj , βj) ⊂ S for some αj and βj . We apply Theorem 10.29

to find u ∈ uξ0 + W 1,∞0 (Ω; RN ) satisfying

Φ(∇u) ∈ αj , βj, a.e. in Ω.

Note also that Qf = Cg Φ is quasiaffine on the connected component of

K :=ξ ∈ RN×n : Qf(ξ) < f (ξ)

containing ξ0 . Invoking then Theorem 11.1, we have the claim.

The problem under consideration is sufficiently flexible that we could alsoproceed as in Dacorogna-Marcellini [195], using Corollary 11.7. Indeed if∇Φ(ξ0) = 0 (in the case Φ(ξ) = det ξ this means that rank ξ0 ≥ n − 1), wecan apply the corollary, since the connected component of K containing ξ0 isbounded, in the neighborhood of ξ0 , in a direction of rank one. We do notdiscuss the details of this different approach.


11.5.3 The Saint Venant-Kirchhoff energy

We recall that the Saint Venant-Kirchhoff function is given by


∣∣2 +ν

1− 2ν( |ξ|2 − n )2

where ν ∈ (0, 1/2) is a parameter. We here discuss only the case n = 2 and werecall (see Theorem 6.29) that

Qf (ξ) = Cf (ξ) ,

where

Qf (ξ) :=

⎧⎪⎪⎨⎪⎪⎩

f (ξ) if ξ /∈ D1 ∪D2

11−ν ((λ2)

2 − 1)2 if ξ ∈ D2

0 if ξ ∈ D1

where

D1 =ξ ∈ R2×2 : (1− ν) [λ1(ξ)]

2+ ν [λ2(ξ)]

2< 1 and λ2(ξ) < 1

=ξ ∈ R2×2 : λ1(ξ) ≤ λ2(ξ) < 1

,

D2 =ξ ∈ R2×2 : (1− ν) [λ1(ξ)]

2+ ν [λ2(ξ)]

2< 1 and λ2(ξ) ≥ 1

.

The existence theorem, which was first studied in Dacorogna-Marcellini[195], is then the following.

Theorem 11.34 Let Ω ⊂ R2 be a bounded open set, f : R2×2 → R be as above,ξ0 ∈ R2×2 and

(P ) inf

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; R2)

.

The following statements then hold.

(i) If ξ0 ∈ D1 or ξ0 /∈ D1 ∪D2 then (P ) has a solution.

(ii) If ξ0 ∈ intD2 then (P ) has no solution.

Proof. (i) The case where ξ0 /∈ D1∪D2 corresponds to the trivial case, whereQf = f.

The case ξ0 ∈ D1 was not settled in [195] and can be treated as follows.From Theorem 10.25, we find u ∈ uξ0 + W 1,∞

0 (Ω; R2) such that

λ1(∇u) = λ2(∇u) = 1, a.e. in Ω.

Note that Qf is quasiaffine on D1 (in fact Qf (ξ) ≡ 0) and therefore we canapply Theorem 11.1, to find that u is indeed a minimizer of (P ).

(ii) It was shown in [195] that if ξ0 ∈ intD2 then the function Qf isstrictly quasiconvex at ξ0 and therefore (P ) has no solution. We refer for detailsto [195].



We now consider the case, studied by many authors following the pioneeringwork of Kohn-Strang [374], where

(P ) inf

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; R2)

,

Ω is a bounded open set of R2, ∇uξ0 = ξ0 and

f (ξ) =

1 + |ξ|2 if ξ = 0

0 if ξ = 0.

We have seen in Theorem 6.28 that the quasiconvex envelope is then

Qf (ξ) =

1 + |ξ|2 if |ξ|2 + 2 |det ξ| ≥ 1

2(|ξ|2 + 2 |det ξ|)1/2 − 2 |det ξ| if |ξ|2 + 2 |det ξ| < 1.

The existence of minimizers for problem (P ) was then established by Dacorogna-Marcellini in [195] and [202], namely the following.

Theorem 11.35 Let Ω ⊂ R2, f : R2×2 → R be as above and ξ0 ∈ R2×2. Thena necessary and sufficient condition for (P ) to have a solution is that one of thefollowing conditions hold:

(i) ξ0 = 0 or |ξ0|2 + 2 |det ξ0| ≥ 1, (i.e. f (ξ0) = Qf (ξ0))

(ii) det ξ0 = 0.

Proof. We refer for the necessary part to [195]. Observe that if ξ0 satisfy (i),we are in the trivial situation; so we assume from now on that

|ξ0|2 + 2 |det ξ0| < 1 and det ξ0 = 0.

Since f is O (2) × O (2)-invariant and det ξ0 = 0, we can assume, without lossof generality, that ξ0 ∈ K0 where (denoting by R2×2

s the set of 2× 2 symmetricmatrices)

K0 :=ξ ∈ R2×2

s : det ξ > 0 and trace ξ ∈ (0, 1)

.

Using Theorem 10.30, we can find, letting

E =ξ ∈ R2×2

s : det ξ ≥ 0 and trace ξ ∈ 0, 1

= 0 ∪ξ ∈ R2×2

s : det ξ ≥ 0 and trace ξ = 1

u ∈ uξ0 + W 1,∞0

(Ω; R2

)such that

∇u (x) ∈ E, a.e. in Ω.


This last condition means that

f (∇u (x)) = Qf (∇u (x)) , a.e. x ∈ Ω.

Since Qf is quasiaffine on K0 (Qf (ξ) = 2 trace ξ − 2 det ξ), we have that

∫

Ω

Qf (∇u (x)) dx = Qf (ξ0) measΩ.

Theorem 11.1 implies that u is a minimizer of (P ).


Following Dacorogna-Pisante-Ribeiro [211], we now deal with the case whereN = n + 1 and

f(ξ) = g(adjn ξ).

The minimization problem is then

(P ) inf

∫

Ω

g(adjn(∇u(x))) dx : u ∈ uξ0 + W 1,∞0 (Ω; Rn+1)

,

where Ω is a bounded open set of Rn, ∇uξ0 = ξ0 and g : Rn+1 → R is anon-negative, lower semicontinuous and locally bounded non-convex function.

From Theorem 6.26, we have

Qf(ξ) = Cg(adjn ξ).

We next setS := y ∈ Rn+1 : Cg(y) < g(y)

and assume, in order to avoid the trivial situation, that adjn ξ0 ∈ S. We alsoassume that S is connected, otherwise we replace it by its connected componentthat contains adjn ξ0 .

Observe that

K := ξ ∈ R(n+1)×n : Qf(ξ) < f(ξ) =ξ ∈ R(n+1)×n : adjn ξ ∈ S

.

Theorem 11.36 If S is bounded, Cg is affine in S and rank ξ0 ≥ n− 1, then(P ) has a solution.

Remark 11.37 The fact that Cg be affine in S is not a necessary conditionfor existence of minima, as seen in Proposition 11.38. ♦

Proof. The result follows if we choose a convenient rank one direction λ =α⊗ β ∈ R(n+1)×n satisfying the hypothesis of Corollary 11.7. We remark that,since we suppose Cg affine in S, Qf is quasiaffine in LK(ξ0 + α ⊗ Bǫ, λ) (cf.


Notation 11.4 and Definition 11.5) independently of the choice of λ. So we onlyhave to prove that K is stably bounded at ξ0 in a direction λ = α⊗ β.

Firstly we observe that we can find (cf. Theorem 13.3) P ∈ O (n + 1) ,Q ∈ SO (n) and 0 ≤ λ1 ≤ · · · ≤ λn , so that

ξ0 = PΛQ, where Λ = diag(n+1)×n (λ1, · · · , λn) ;

in particular when n = 2 we have

Λ =

⎛⎜⎜⎝

λ1 0

0 λ2

0 0

⎞⎟⎟⎠ .

Since rank ξ0 ≥ n− 1 we have that λ2 > 0. We also note that

adjn ξ0 = adjn P . adjn Λ and adjn Λ =

⎛⎜⎜⎜⎜⎝

0

...

0

(−1)n λ1 · · ·λn

⎞⎟⎟⎟⎟⎠

.

Without loss of generality we assume ξ0 = Λ. We then choose λ = α ⊗ βwhere α = (1, 0, · · · , 0) ∈ Rn+1 and β = (1, 0, · · · , 0) ∈ Rn. We will see thatLK(ξ0 +α⊗Bǫ, λ) is bounded for every small ǫ > 0. Let η ∈ LK(ξ0 +α⊗Bǫ, λ)then we can write η = ξ0 +α⊗γǫ + tλ for some γǫ ∈ Bǫ and t ∈ R. By definitionof LK(ξ0 + α⊗Bǫ, λ) we have adjn η ∈ S. Since S is bounded and

|adjn η| =∣∣λ1 + γ1

ǫ + t∣∣λ2 · · ·λn

it follows, using the fact that rank ξ0 ≥ n− 1, that |t| is bounded by a constantdepending on S, ξ0 and ǫ. Consequently |η| ≤ |ξ0|+ |α⊗ γǫ|+ |t| |λ| is boundedfor any fixed positive ǫ and we get the result.

As already alluded in Section 11.3, we now obtain a result of non-existencealthough the integrand of the relaxed problem is not strictly quasiconvex. Weconsider the case where N = 3, n = 2 and f : R3×2 → R is given by

f (ξ) = g (adj2 ξ)

where g : R3 → R is defined by, letting ν = (ν1, ν2, ν3) ,

g (ν) = ((ν1)2 − 4)2 + (ν2)

2 + (ν3)2.

We therefore get Qf (ξ) = Cg (adj2 ξ) and

Cg (ν) =[(ν1)

2 − 4]2+

+ (ν2)2 + (ν3)

2,


where

[x]+ =

x if x ≥ 0

0 if x < 0.

We choose the boundary datum

uξ0 (x) =

⎛⎜⎝

u1ξ0

(x) = α1x1 + α2x2

u2ξ0

(x) = 0

u3ξ0

(x) = 0

⎞⎟⎠

and hence

∇uξ0 (x) = ξ0 =

⎛⎝

α1 α2

0 00 0

⎞⎠ , adj2∇uξ0 (x) = adj2 ξ0 =

⎛⎝

000

⎞⎠ .

The problem is then

(P ) inf

I (u) =

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; R3)

.

Note also that Qf (ξ0) = 0 < f (ξ0) = 16.In terms of the previous notation, we have

S = y ∈ R3 : Cg(y) < g(y) = y = (y1, y2, y3) ∈ R3 : |y1| < 2,

K = ξ ∈ R3×2 : Qf(ξ) < f(ξ) =ξ ∈ R3×2 : adj2 ξ ∈ S

and we observe that Cg is not affine on S, which in turn implies that Qf is notquasiaffine on K.

The following result shows that the hypothesis of strict quasiconvexity ofQf is not necessary for non-existence.

Proposition 11.38 (P ) has a solution if and only if uξ0 ≡ 0. Moreover, Qf isnot strictly quasiconvex at any ξ0 ∈ R3×2 of the form

ξ0 =

⎛⎝

α1 α2

0 00 0

⎞⎠ .

Proof. Step 1. We first show that if (P ) has a solution then uξ0 ≡ 0. If

u ∈ uξ0 + W 1,∞0 (Ω; R3) is a solution of (P ) we necessarily have, denoting by

ν (ξ) = adj2 ξ,|ν1 (∇u)| = 2, ν2 (∇u) = ν3 (∇u) = 0,

sinceQf (∇uξ0) = Cg (adj2∇uξ0) = Cg (0) = 0.


The three equations read as

⎧⎪⎨⎪⎩

∣∣u2x1

u3x2− u2

x2u3

x1

∣∣ = 2

u1x1

u3x2− u1

x2u3

x1= 0

u1x1

u2x2− u1

x2u2

x1= 0.

(11.20)

Multiplying the second equation of (11.20) first by u2x1

, then by u2x2

, using thethird equation of (11.20), we get

0 = u2x1

u1x1

u3x2− u2

x1u1

x2u3

x1= u2

x1u1

x1u3

x2− u1

x1u2

x2u3

x1= u1

x1

(u2

x1u3

x2− u2

x2u3

x1

)

0 = u2x2

u1x1

u3x2− u2

x2u1

x2u3

x1= u2

x1u1

x2u3

x2− u2

x2u1

x2u3

x1= u1

x2

(u2

x1u3

x2− u2

x2u3

x1

).

Combining these last equations with the first one of (11.20), we find

u1x1

= u1x2

= 0, a.e.

We therefore find that any solution of (P ) should have ∇u1 = 0 a.e. and henceu1 ≡ constant on each connected component of Ω. Since u1 agrees with u1

ξ0on

the boundary of Ω, we deduce that u1ξ0≡ 0 and thus uξ0 ≡ 0, as claimed.

Step 2. We next show that if uξ0 ≡ 0, then (P ) has a solution. It suffices tochoose u1 ≡ 0 and to solve

∣∣u2x1

u3x2− u2

x2u3

x1

∣∣ = 2 a.e. in Ω

u2 = u3 = 0 on ∂Ω.

This is possible by virtue of Corollary 10.27.

Step 3. We finally prove that Qf is not strictly quasiconvex at any ξ0 ∈ R3×2

of the form given in the statement of the proposition. Indeed let 0 < R1 <R2 < R and denote by BR the ball centered at 0 and of radius R. Chooseλ, μ ∈ C∞ (BR) such that

1) λ = 0 on ∂BR and λ ≡ 1 on BR2 .

2) μ ≡ 0 on BR −BR2 , μ ≡ 1 on BR1 and

∣∣μ2 + μ (x1μx1 + x2μx2)∣∣ < 2, for every x ∈ BR .

This last condition (which is a restriction only in BR2 −BR1) is easily ensuredby choosing appropriately R1 , R2 and R.

We then choose u (x) = uξ0 (x) + ϕ (x) where

ϕ1 (x) = −λ (x)u1ξ0

(x) , ϕ2 (x) = μ (x) x1 and ϕ3 (x) = μ (x) x2 .

We therefore have that ϕ ∈ W 1,∞0 (BR; R3), adj2∇u ≡ 0 on BR − BR2 , while

on BR2 we have

adj2∇u =(μ2 + μ (x1μx1 + x2μx2) , 0, 0

).


We have thus obtained that Cg (adj2∇u) ≡ 0 and hence

Qf (ξ0 +∇ϕ) ≡ Qf (ξ0) = 0.

This implies that (QP ) has infinitely many solutions. However since ϕ does notvanish identically, we deduce that Qf is not strictly quasiconvex at any ξ0 ofthe given form.


The general problem of potential wells has been intensively studied by manyauthors in conjunction with crystallographic models involving fine microstruc-tures. The reference paper on the subject is Ball and James [60]. It has sincethen been studied by many authors including Bhattacharya-Firoozye-James-Kohn, Dacorogna-Marcellini, De Simone-Dolzmann, Dolzmann-Muller, Erick-sen, Firoozye-Kohn, Fonseca-Tartar, Kinderlehrer-Pedregal, Kohn, Luskin,Muller-Sverak, Pipkin and Sverak and we refer to [202] for full bibliographicreferences.

In mathematical terms the problem of potential wells can be described asfollows. Find a minimizer of the problem

(P ) inf

∫

Ω

f (∇u (x)) dx : u ∈ uξ0 + W 1,∞0 (Ω; Rn)

,

where Ω ⊂ Rn is a bounded open set, uξ0 is an affine map with ∇uξ0 = ξ0 andf : Rn×n → R+ is such that

f (ξ) = 0 ⇔ ξ ∈ E :=m⋃

i=1

SO (n)Ai .

The m wells are SO (n)Ai , 1 ≤ i ≤ m (and SO (n) denotes the set of matricesU such that U tU = UU t = I and detU = 1).

The interesting case is when

ξ0 ∈ intRco E

and we have, since Rco E ⊂ QcoE ⊂ Qcof E (see Theorem 7.28), that

Qf (ξ0) = 0.

Therefore, by the relaxation theorem, we have

inf (P ) = inf (QP ) = 0.

The existence of minimizers, since Qf is affine on RcoE (indeed Qf ≡ 0), for(P ) is then reduced to finding a function u ∈ uξ0 + W 1,∞

0 (Ω; Rn) such that

∇u (x) ∈ E =m⋃

i=1

SO (n)Ai .


The problem is relatively well understood only in the cases of two wells, i.e.m = 2, and in dimension n = 2. It is this case that we briefly discuss now. Wetherefore now have A, B ∈ R2×2 and we assume that

A =

(a1 0

0 a2

)and B =

(b1 0

0 b2

),

where 0 < b1 < a1 ≤ a2 < b2 and a1a2 < b1b2 . We want to find u ∈ uξ0 +

W 1,∞0

(Ω; R2

), where Ω ⊂ R2 is a bounded open set, satisfying

∇u(x) ∈ SO(2)A ∪ SO(2)B a.e. in Ω.

The first important result is to identify the set where the gradient of the bound-ary datum, ξ0, should lie. We have seen in Theorem 7.44 that

Rco E =

ξ ∈ R2×2 :


0 ≤ α ≤ det B−det ξdet B−detA , 0 ≤ β ≤ det ξ−det A

detB−detA

,

while the interior is given by the same formulas with strict inequalities on theright hand side.

We therefore have the following (see Theorem 10.28).

Theorem 11.39 Let Ω ⊂ R2 be a bounded open set and

ξ0 ∈ intRcoE.

Then there exists u ∈ uξ0 + W 1,∞0 (Ω; R2) such that

∇u(x) ∈ E = SO(2)A ∪ SO(2)B a.e. in Ω

and therefore (P ) has a solution.

As already discussed in Section 10.3.3, the case where detA = detB > 0can also be handled (see Muller-Sverak [466] and also Dacorogna-Tanteri [215]),using the representation formula of Sverak [554] (see Theorem 7.44), namely

Rco E =

ξ ∈ R2×2 :


0 ≤ α, β, α + β ≤ 1 and det ξ = detA = detB

.

Chapter 12

Function spaces

12.1 Introduction

We have gathered in this chapter the notation and the most important resultson different function spaces that we have used or will use throughout the book.We precisely fix the notations and state the theorems. But we provide almostno proof, since the results are standard.

12.2 Main notation

We first recall the usual notation for derivatives.

(i) If u : Rn → R, u = u (x1, · · · , xn) , we denote partial derivatives by eitherof the following ways

Dju = uxj =∂u

∂xj

and

∇u = gradu = (∂u

∂x1, · · · , ∂u

∂xn) = (ux1, · · · , uxn) ∈ Rn.

(ii) For maps u : Rn → RN , we write u =(u1, · · · , uN

)and

∇u =

(∂ui

∂xj

)1≤i≤N

1≤j≤n

∈ RN×n.

(iii) For higher derivatives, we proceed as follows. Let m ≥ 1 be an integer;an element of

Am := a = (a1, · · · , an) ∈ Nn :∑n

j=1 aj = m

504 Function spaces

is called a multi-index of order m. We also write for such elements

|a| =n∑

j=1

aj = m.

For a ∈ Am , we write

Dau = Da11 · · ·Dan

n u =∂|a| u

∂xa11 · · ·∂xan

n.

- Let N, n, m ≥ 1 be integers. For u : Rn → RN we write

∇mu =

(∂mui

∂xj1 · · · ∂xjm

)1≤i≤N

1≤j1,··· ,jm≤n

∈ RN×nm

s .

(The index s here stands for all the natural symmetries implied by the inter-change of the order of differentiation.) When m = 1, we have

RN×ns = RN×n,

while if N = 1 and m = 2, we obtain

Rn2

s = Rn×ns

(i.e., the usual set of symmetric matrices).

- We also let∇[m]u = (u,∇u, · · · ,∇mu)

stand for the matrix of all partial derivatives of u up to the order m. Note that

∇[m−1]u ∈ RN×Ms = RN × RN×n × RN×n2

s × · · · × RN×n(m−1)

s ,

where

M := 1 + n + · · ·+ n(m−1) =nm − 1

n− 1.

Hence∇[m]u = (∇[m−1]u,∇mu ) ∈ RN×M

s × RN×nm

s .

We next define some function spaces.

Definition 12.1 Let Ω ⊂ Rn be an open set.

(i) C0 (Ω) = C (Ω) is the set of continuous functions u : Ω → R.

(ii) C0(Ω)

= C(Ω)

is the set of continuous functions u : Ω → R, which can

be continuously extended to Ω. The norm over C(Ω)

is given by

‖u‖C0 = supx∈Ω

|u (x)| .

Main notation 505

(iii) The support of a function u : Ω → R is defined as

supp u := x ∈ Ω : u (x) = 0.

(iv) C0 (Ω) := u ∈ C (Ω) : suppu ⊂ Ω is compact .

We now proceed similarly for the spaces involving derivatives.

Definition 12.2 Let n, m ∈ N and Ω ⊂ Rn be an open set.

(i) The set of functions u : Ω → R that have all partial derivatives, Dau,a ∈ Ak , 0 ≤ k ≤ m, continuous is denoted by Cm (Ω) .

(ii) Cm(Ω)

is the set of Cm (Ω) functions whose derivatives up to the order

m can be extended continuously to Ω. It is equipped with the following norm

‖u‖Cm = max0≤|a|≤m

supx∈Ω

|Dau (x)| .

(iii) Affm(Ω)

stands for the set of polynomials of degree m; in particular,

if u ∈ Affm(Ω), there exists ξ ∈ Rnm

s such that

∇mu (x) = ξ for every x ∈ Ω.

Most of the time when m = 1, we let Aff(Ω)

instead of Aff1(Ω).

(iv) Cm0 (Ω) := Cm (Ω) ∩ C0 (Ω) .

(v) C∞ (Ω) :=∞⋂

m=0Cm (Ω) , C∞ (Ω

):=

∞⋂m=0

Cm(Ω).

(vi) C∞0 (Ω) = D (Ω) := C∞ (Ω) ∩ C0 (Ω) .

(vii) When dealing with maps, u : Ω → RN , we accordingly writeCm(Ω; RN

), Cm

(Ω; RN

)or Affm

(Ω; RN

)and similarly for the other nota-

tions.

We often have to consider the above spaces as split in several pieces and wetherefore have the following definitions.

Definition 12.3 Let n, m ∈ N and Ω ⊂ Rn be an open set.

(i) A function u ∈ Cmpiec

(Ω)

if u ∈ Cm−1(Ω)

and ∇mu is piecewise con-tinuous, meaning that there exists a partition of Ω into a countable union ofdisjoint open sets Ωk ⊂ Ω for every k ∈ N, more precisely

Ωh ∩ Ωk = ∅, ∀h, k ∈ N, h = k, and meas(Ω−⋃k∈N

Ωk

)= 0

and so that ∇mu ∈ C(Ωk; Rnm

s

)for every k ∈ N.

(ii) Affmpiec

(Ω)

stands for the subset of Cmpiec

(Ω)

so that ∇mu is piecewise

(in the above sense) constant. Most of the time when m = 1, we let Affpiec

(Ω)

instead of Aff1piec

(Ω).

(iii) Similarly for maps, u : Ω → RN , we accordingly write Cmpiec

(Ω; RN

)or

Affmpiec

(Ω; RN

).

506 Function spaces

On several occasions we used the following definition.

Definition 12.4 Let Ω ⊂ Rn be an open set and uν : Ω → R be a sequenceof measurable functions. We say that uν is equiintegrable, if there exists anincreasing function

η : R+ → R+

with η (t) → 0 as t→ 0, so that

∫

A

|uν (x)| dx ≤ η (meas A) ,

for every measurable set A ⊂ Ω.

We recall (see Dunford-Pettis theorem) that if the sequence uν convergesweakly in L1, then it is equiintegrable.

12.3 Some properties of Holder spaces

We recall here some basic properties of Holder spaces. We use as referenceson this part: Adams [5], Dacorogna [180], Gilbarg and Trudinger [313] orHormander [343].

Definition 12.5 Let D ⊂ Rn, u : D → R and 0 < α ≤ 1. We let

[u]α,D := supx,y∈D

x =y

|u (x)− u (y)||x− y|α

.

Let Ω ⊂ Rn be open and m ≥ 0 be an integer. We define the different spaces ofHolder continuous functions in the following way.

(i) C0,α (Ω) is the set of u ∈ C (Ω) such that

[u]α,K = supx,y∈K

x =y

|u (x)− u (y)||x− y|α

< ∞

for every compact set K ⊂ Ω.

(ii) C0,α(Ω)

is the set of functions u ∈ C(Ω)

such that

[u]α,Ω <∞ .

It is equipped with the norm

‖u‖C0,α(Ω) := ‖u‖C0(Ω) + [u]α,Ω .

If there is no ambiguity, we drop the dependence on the set Ω and write simply

‖u‖C0,α := ‖u‖C0 + [u]α .

Some properties of Holder spaces 507

(iii) Cm,α (Ω) is the set of u ∈ Cm (Ω) such that

[Dau]α,K < ∞

for every compact set K ⊂ Ω and every a ∈ Am .

(iv) Cm,α(Ω)

is the set of functions u ∈ Cm(Ω)

such that

[Dau]α,Ω <∞

for every multi-index a ∈ Am . We equip Cm,α(Ω)

with the following norm:

‖u‖Cm,α := ‖u‖Cm + maxa∈Am

[Dau]α .

(v) For maps, u : Ω → RN , we write Cm,α(Ω; RN

).

Remark 12.6 (i) Cm,α(Ω)

with its norm ‖·‖Cm,α is a Banach space.

(ii) By abuse of notation we write Cm (Ω) = Cm,0 (Ω) , or, in other words,the set of continuous functions is identified with the set of Holder continuousfunctions with exponent 0.

(iii) Similarly, when α = 1, we see that C0,1(Ω)

is in fact the set of Lipschitzcontinuous functions, namely the set of functions u such that there exists aconstant γ > 0 such that

|u (x) − u (y)| ≤ γ |x− y| , ∀x, y ∈ Ω.

The best such constant is γ = [u]C0,1 . ♦

We now list some important properties of Holder spaces.

Proposition 12.7 Let Ω ⊂ Rn be a bounded open set with a Lipschitz boundary,m ≥ 0 an integer and 0 ≤ α ≤ 1. The following properties then hold.

(i) If u, v ∈ Cm,α(Ω), then uv ∈ Cm,α

(Ω). More precisely, if u, v ∈ Cm,α,

then there exists a constant γ > 0 such that

‖uv‖Cm,α ≤ γ (‖u‖Cm,α ‖v‖Cm + ‖u‖Cm ‖v‖Cm,α) ≤ 2γ ‖u‖Cm,α ‖v‖Cm,α .

Moreover, if Ω is convex or its boundary is Cm,α, then

‖uv‖Cm,α ≤ γ (‖u‖Cm,α ‖v‖C0 + ‖u‖C0 ‖v‖Cm,α) .

(ii) Let O ⊂ RNbe open bounded and with a Lipschitz boundary, m ≥ 1,v ∈ Cm,α

(Ω; O

)and u ∈ Cm,α

(O). Then

u v ∈ Cm,α(Ω).

(iii) If 0 ≤ α ≤ β ≤ 1, then

Cm(Ω)⊃ Cm,α

(Ω)⊃ Cm,β

(Ω)⊃ Cm,1

(Ω)⊃ Cm+1

(Ω).

508 Function spaces

(iv) Let m ≥ l be non-negative integers and 0 < β ≤ α < 1 be such that

l + β < m + α.

Then, for every ǫ > 0 and every u ∈ Cm,α(Ω), there exists v ∈ C∞ (Ω

)such

that‖u− v‖Cl,β ≤ ǫ.

We conclude with the following lower semicontinuity result.

Proposition 12.8 Let m ≥ 1 be an integer and 0 < α < 1. Let Ω ⊂ Rn be abounded open set with a Lipschitz boundary. Let r > 0 and

Cr :=u ∈ Cm,α

(Ω)

: ‖u‖Cm,α ≤ r

.

Let uν ⊂ Cr be a sequence such that

uν → u in L∞ (Ω) as ν →∞,

then u ∈ Cr and‖u‖Cm,α ≤ lim inf

ν→∞‖uν‖Cm,α .


Step 1. We recall that

‖u‖Cm,α = ‖u‖Cm + maxa∈Am

[Dau]α

and observe that since (see Section 12.4 for the definition and properties ofSobolev spaces)

‖u‖Cm = ‖u‖W m,∞

we can deduce that, up to the extraction of a subsequence still labeled uν ,there exists v ∈Wm,∞ (Ω) so that

uν∗ v in Wm,∞ (Ω) as ν →∞. (12.1)

By uniqueness of the limit we can identify u and v. We therefore have that

uν → u in Wm−1,∞ (Ω) as ν →∞, (12.2)

u ∈ Wm,∞ (Ω) and‖u‖W m,∞ ≤ lim inf

ν→∞‖uν‖W m,∞ .

Step 2. We prove the claim only for the case m = 1. The general case followsin a similar manner, since we have (12.2). We already know that u ∈ W 1,∞ (Ω)and

‖u‖W 1,∞ ≤ lim infν→∞

‖uν‖W 1,∞ (12.3)

Some properties of Sobolev spaces 509

so it remains to show that u ∈ C1,α(Ω)

and that

[Diu]α ≤ lim infν→∞

[Diuν ]α , i = 1, · · · , n. (12.4)

The combination of (12.3) and (12.4) gives the proposition.Since u ∈ W 1,∞ (Ω) and ∂Ω is regular, we have that u is Lipschitz and there-

fore by Rademacher theorem we have that u is almost everywhere differentiable(see for example Evans [272]). So let x, y ∈ Ω be points of differentiability of u,i.e.

Diu (x) = limh→0

u (x + hei)− u (x)

h, Diu (y) = lim

h→0

u (y + hei)− u (y)

h. (12.5)

Observe next that

| u(x+hei)−u(x)h − u(y+hei)−u(y)

h | ≤ 4h ‖uν − u‖L∞

+ | uν(x+hei)−uν(x)h − uν(y+hei)−uν(y)

h | .(12.6)

Since uν ∈ C1,α(Ω), we can find θν

x and θνy so that

|θνx − x| ,

∣∣θνy − y

∣∣ ≤ h

and

Diuν (θνx) =

uν (x + hei)− uν (x)

h, Diuν

(θν

y

)=

uν (y + hei)− uν (y)

h.

This leads to

| uν (x + hei)− uν (x)

h− uν (y + hei)− uν (y)

h| ≤ [Diuν]α

∣∣θνx − θν

y

∣∣α

≤ [Diuν]α (2h + |x− y|)α.

Combining this last inequality with (12.6), we get

| u (x + hei)− u (x)

h− u (y + hei)− u (y)

h| ≤ 4

h‖uν − u‖L∞

+ [Diuν ]α (2h + |x− y|)α .

Letting first ν →∞ and then h → 0 and appealing to (12.5), we have obtainedthat, for almost every x, y ∈ Ω,

|Diu (x) −Diu (y)| ≤ lim infν→∞

[Diuν ]α |x− y|α .

This easily leads to the conclusion of Step 2 and thus the proposition is proved.

12.4 Some properties of Sobolev spaces

For more details concerning Sobolev spaces, we refer to Adams [5], Brezis [105],Dacorogna [180], Dacorogna-Marcellini [202], Ekeland-Temam [264], Evans[272], Gilbarg and Trudinger [313], Giusti [316], Kufner-John-Fucik [384],Ladyzhenskaya-Uraltseva [388] or Morrey [455].

510 Function spaces

12.4.1 Definitions and notations

We first recall the definition of Sobolev spaces.

Definition 12.9 Let Ω ⊂ Rn be an open set and 1 ≤ p ≤ ∞.

(i) We let W 1,p (Ω) be the set of functions u : Ω → R, u ∈ Lp (Ω) , whoseweak partial derivatives uxi ∈ Lp (Ω) for every i = 1, · · · , n. We endow thisspace with the following norm

‖u‖W 1,p := (‖u‖pLp + ‖∇u‖p

Lp)1/p

if 1 ≤ p < ∞,

‖u‖W 1,∞ := max ‖u‖L∞ , ‖∇u‖L∞ if p = ∞.

(ii) If 1 ≤ p < ∞, the set W 1,p0 (Ω) is defined as the closure of C∞

0 (Ω)functions in W 1,p (Ω) .

(iii) We also write u ∈ u0 + W 1,p0 (Ω) , meaning that u, u0 ∈ W 1,p (Ω) and

u− u0 ∈W 1,p0 (Ω) .

(iv) We let W 1,∞0 (Ω) := W 1,∞ (Ω) ∩W 1,1

0 (Ω) .

(v) Analogously, we define the Sobolev spaces with higher derivatives as fol-lows. If m > 0 is an integer, we let (by abuse of notation, we will writeW 0,p := Lp) Wm,p (Ω) be the set of functions u : Ω → R whose weak par-tial derivatives Dau ∈ Lp (Ω) for every multi-index a ∈ Ak , 0 ≤ k ≤ m. Thenorm is then

‖u‖W m,p :=

⎧⎪⎨⎪⎩

(∑0≤|a|≤m ‖Dau‖p

Lp

)1/p

if 1 ≤ p < ∞

max0≤|a|≤m

(‖Dau‖L∞) if p = ∞.

(vi) If 1 ≤ p <∞, Wm,p0 (Ω) denotes the closure of C∞

0 (Ω) in Wm,p (Ω) and

Wm,∞0 (Ω) := Wm,∞ (Ω) ∩Wm,1

0 (Ω) .

(vii) For maps u : Ω → RN , u =(u1, · · · , uN

), we say that u ∈

W 1,p(Ω; RN

)if ui ∈W 1,p (Ω) for every i = 1, · · · , N. Similar definitions apply

to W 1,p0

(Ω; RN

), Wm,p

(Ω; RN

)or Wm,p

0

(Ω; RN

).

12.4.2 Imbeddings and compact imbeddings

We recall here the Sobolev and the Rellich-Kondrachov theorems. We start withthe definition of Lipschitz boundary of a given set.

Definition 12.10 Let Ω ⊂ Rn be open and bounded. We say that Ω is abounded open set with a Lipschitz boundary if for every x ∈ ∂Ω, there exista neighborhood U ⊂ Rn of x and a one-to-one and onto map H : Q → U, where(see Figure 12.1)

Q := x ∈ Rn : |xj | < 1, j = 1, 2, · · · , n ,


H ∈ C0,1(Q), H−1 ∈ C0,1

(U), H (Q+) = U ∩ Ω, H (Q0) = U ∩ ∂Ω,

with Q+ := x ∈ Q : xn > 0 and Q0 := x ∈ Q : xn = 0 .

Similarly, we say that Ω has a Cm,α boundary, if the above H ∈ Cm,α aswell as H−1 ∈ Cm,α.

Figure 12.1: Regularity of the boundary

Theorem 12.11 (Sobolev imbedding theorem) Let Ω ⊂ Rn be a boundedopen set with a Lipschitz boundary.

Case 1. If 1 ≤ p < n, then

W 1,p (Ω) ⊂ Lq (Ω)

for every q ∈ [1, p∗] , where p∗ is the Sobolev exponent defined by

1

p∗=

1

p− 1

n, i.e. p∗ =

np

n− p.

More precisely, for every q ∈ [1, p∗] there exists c = c (Ω, p, q) such that

‖u‖Lq ≤ c ‖u‖W 1,p .

Case 2. If p = n, then

W 1,n (Ω) ⊂ Lq (Ω) for every q ∈ [1,∞) .

More precisely, for every q ∈ [1,∞) there exists c = c (Ω, p, q) such that

‖u‖Lq ≤ c ‖u‖W 1,n .

512 Function spaces

Case 3. If p > n, then

W 1,p (Ω) ⊂ C0,α(Ω)

for every α ∈ [0, 1− n/p] .

In particular, there exists a constant c = c (Ω, p) such that

‖u‖L∞ ≤ c ‖u‖W 1,p .

We continue with compact imbeddings.

Theorem 12.12 (Rellich-Kondrachov theorem) Let Ω ⊂ Rn be a boundedopen set with a Lipschitz boundary.

Case 1. If 1 ≤ p < n, then the imbedding of W 1,p in Lq is compact for everyq ∈ [1, p∗) . This means that any bounded set of W 1,p is precompact (i.e., itsclosure is compact) in Lq for every 1 ≤ q < p∗ (the result is false if q = p∗).

Case 2. If p = n, then the imbedding of W 1,n in Lq is compact for everyq ∈ [1,∞) .

Case 3. If p > n, then the imbedding of W 1,p in C0,α(Ω)

is compact forevery 0 ≤ α < 1− n/p.

In particular, in all cases (i.e., 1 ≤ p ≤ ∞), the imbedding of W 1,p (Ω) in Lp (Ω)is compact.

Remark 12.13 (i) Similar results for W k,p spaces exist, but we will not needthem.

(ii) If in both theorems we replace the spaces W 1,p by W 1,p0 , then no regu-

larity on the boundary of the domain is required. ♦

12.4.3 Approximation by smooth and piecewise affinefunctions

We now turn our attention to density results for Sobolev functions. The firstresult concerns the approximation by smooth functions.

Theorem 12.14 Let Ω ⊂ Rn be a bounded open set and 1 ≤ p < ∞. Thenthe space C∞ (Ω)∩W 1,p (Ω) is dense in W 1,p (Ω) . Moreover, if Ω is a boundeddomain with Lipschitz boundary, then C∞ (Ω

)is also dense in W 1,p (Ω) .

The second result deals with approximation by piecewise affine functions.

Theorem 12.15 Let Ω be a bounded open set of Rn, 1 ≤ p ≤ ∞ and u ∈W 1,p

0 (Ω) . Then, for every ǫ > 0, there exists uǫ ∈ Affpiec

(Ω)

such that

⎧⎪⎨⎪⎩

uǫ ∈W 1,p0 (Ω)

‖uǫ‖W 1,p(Ω) ≤ ‖u‖W 1,p(Ω) + ǫ

‖uǫ − u‖W 1,p(Ω) ≤ ǫ.

(If p =∞, we only have ‖uǫ − u‖W 1,q(Ω) ≤ ǫ for any q < ∞.)


More sophisticated approximation results exist (see Corollaries 10.11, 10.13,10.14, 10.15 or 10.21 in Dacorogna-Marcellini [202]), such as the next one.

Theorem 12.16 Let Ω be an open set of Rn. Let A, B be disjoint sets of Rn,with A open and B possibly empty. Let u ∈W 1,∞(Ω) such that

∇u(x) ∈ A ∪B a.e. x ∈ Ω.

Then, for every ǫ > 0, there exists a function v ∈ W 1,∞(Ω) and an open setΩ′ ⊂ Ω (Ω′ = Ω if B = ∅) such that

⎧⎪⎪⎪⎨⎪⎪⎪⎩

v ∈ Affpiec (Ω′) , v = u on ∂Ω,

‖v − u‖L∞(Ω) < ǫ,

∇v(x) ∈ A a.e. x ∈ Ω′,

∇v(x) = ∇u(x) ∈ B a.e. x ∈ Ω− Ω′.

Chapter 13

Singular values

13.1 Introduction

In this chapter, we refer to the following books: Bellman [74], Ciarlet [153],Dacorogna-Marcellini [202], Horn-Johnson [345] and [346], Marshall-Olkin [432]and Serre [531].

13.2 Definition and basic properties

We collect here the definition and some properties of singular values of matrices(see Ciarlet [153] Theorems 1.2.1 and 1.2.2; Section 7.3 in Horn-Johnson [345]and Section 3.1 in [346]). First we recall our notation for matrices ξ ∈ RN×n,which we write as

ξ =

⎛⎜⎝

ξ11 · · · ξ1

n...

. . ....

ξN1 · · · ξN

n

⎞⎟⎠ =

⎛⎜⎝

ξ1

...ξN

⎞⎟⎠ = (ξ1, · · · , ξn) .

We start with the following well known notation (in the sequel n and N denotepositive integers).

Definition 13.1 (i) We denote by GL (n) the set of invertible matrices inRn×n.

(ii) The set of orthogonal matrices is denoted by O (n) . It is the set ofmatrices R ∈ Rn×n such that

RRt = I,

where I denotes the identity matrix in Rn×n.

(iii) The set of special orthogonal matrices, denoted by SO (n) , is the subsetof O (n) where the matrices satisfy

detR = 1.

516 Singular values

(iv) We denote by RN×nd the set of diagonal matrices, meaning that ξ ∈

RN×nd if and only if

ξij = 0 whenever i = j.

In particular, if N ≥ n (and similarly if N < n), such a matrix is written as

ξ = diagN×n

(ξ11 , · · · , ξn

n

)

and when N = n, we simply let

ξ = diag(ξ11 , · · · , ξn

n

).

We now give the definition of the singular values.

Definition 13.2 (i) Let N ≤ n and ξ ∈ RN×n. The singular values of ξ,denoted by

0 ≤ λ1 (ξ) ≤ · · · ≤ λN (ξ) ,

are defined to be the square root of the eigenvalues of the symmetric and positivesemidefinite matrix ξξt ∈ RN×N .

(ii) Let N ≥ n and ξ ∈ RN×n. The singular values of ξ, denoted by

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) ,

are defined to be the square root of the eigenvalues of the symmetric and positivesemidefinite matrix ξtξ ∈ Rn×n.

The following theorem is the standard decomposition theorem (see Theorem7.3.5 in [345] or Theorem 3.1.1 in [346], for example).

Theorem 13.3 (Singular values decomposition theorem) (i) Let N ≤ n,ξ ∈ RN×n and 0 ≤ λ1 (ξ) ≤ · · · ≤ λN (ξ) be its singular values. Then there existsR ∈ O (N) such that, δij denoting the Kronecker symbol,

Rξ = ξ =

⎛⎜⎜⎝

ξ 1

...

ξ N

⎞⎟⎟⎠ , with 〈 ξ i; ξ j 〉 = | ξ i | | ξ j | δij , λi (ξ) = | ξ i | .

Furthermore, there exists Q ∈ O (n) such that

(RξQ)ij =

0 if i = j

λi (ξ) if i = j

or equivalently RξQ ∈ RN×nd and

RξQ = diagN×n (λ1 (ξ) , · · · , λN (ξ)) .

Definition and basic properties 517

(ii) Let N ≥ n, ξ ∈ RN×n and 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) be its singularvalues. Then there exists Q ∈ O (n) such that, δij denoting the Kroneckersymbol,

ξQ = ξ =(ξ1, · · · , ξn

), with 〈 ξi; ξj 〉 = | ξi | | ξj | δij , λi (ξ) = | ξi | .

Moreover, there exists R ∈ O (N) such that

(RξQ)ij =

0 if i = j

λi (ξ) if i = j

or equivalently RξQ ∈ RN×nd and

RξQ = diagN×n (λ1 (ξ) , · · · , λn (ξ)) .

We gather below some elementary facts about the singular values, that canbe deduced easily from the above theorem (see Dacorogna-Marcellini [202] page171).

Proposition 13.4 Let 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) be the singular values of thematrix ξ ∈ Rn×n. Then

|ξ|2 =

n∑

i,j=1

( ξij )2 =

n∑

i=1

(λi (ξ))2,

|adjs ξ|2 =∑

i1<···<is

(λi1 (ξ))2 · · · (λis (ξ))

2,

|det ξ| =

n∏

i=1

λi (ξ) ,

where adjs ξ ∈ R(ns)×(n

s) denotes the matrix obtained by forming all the s × sminors, 2 ≤ s ≤ n− 1, of the matrix ξ. (See Section 5.4 for the notation.)

In particular, if n = 3, then

|ξ|2 =

3∑

i,j=1

(ξij

)2= (λ1 (ξ))

2+ (λ2 (ξ))

2+ (λ3 (ξ))

2,

|adj2 ξ|2 = (λ1 (ξ)λ2 (ξ))2

+ (λ1 (ξ)λ3 (ξ))2+ (λ2 (ξ) λ3 (ξ))

2,

|det ξ| = λ1 (ξ)λ2 (ξ)λ3 (ξ) .


Step 1. We recall some elementary facts about adjs ξ (see Proposition 5.66).Let ξ, η ∈ Rn×n and 1 ≤ s ≤ n (by abuse of notation, we write ξ = adj1 ξ anddet ξ = adjn ξ). Then

adjs (ξη) = adjs ξ adjs η, adjs(ξt)

= (adjs ξ)t .

518 Singular values

This implies that if U ∈ O (n) then

adjs U ∈ O((

ns

)).

Indeed, if for N an integer we denote by IN the identity matrix in RN×N , wehave

(adjs U) (adjs U)t

= adjs(UU t

)= adjs (In) = I(n

s)

(adjs U)t(adjs U) = adjs

(U tU

)= adjs (In) = I(n

s).

Step 2. Observe first that by Step 1 and Theorem 13.3, we can find, forevery ξ ∈ Rn×n, Q, R ∈ O (n) such that

RξQ = Λ := diag (λ1 (ξ) , · · · , λn (ξ))

adjs R adjs ξ adjs Q = adjs Λ.

Note next that if ζ ∈ RN×N and if we denote by

|ζ|2 =

N∑

i,j=1

(ζij

)2

then it is well known (see, for example, Theorem 1.4.4 in [153]) that

|ζU | = |Uζ| = |ζ| for every U ∈ O (N) .

Combining the two above observations, the second one applied to ζ = adjs ξ,we immediately deduce the proposition.

Remark 13.5 From the above proof, it is clear that if ξ ∈ Rn×n has singularvalues 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) , and if 1 ≤ s ≤ n− 1, then adjs ξ has singularvalues λi1 (ξ) · · ·λis (ξ) , where 1 ≤ i1 < · · · < is ≤ n. ♦

When n = 2, a simple formula for the singular values can be obtained fromProposition 13.4.

Proposition 13.6 Let ξ ∈ R2×2 and 0 ≤ λ1 (ξ) ≤ λ2 (ξ) be its singular values.Then

λ1 (ξ) =1

2[

√|ξ|2 + 2 |det ξ| −

√|ξ|2 − 2 |det ξ| ],

λ2 (ξ) =1

2[

√|ξ|2 + 2 |det ξ|+

√|ξ|2 − 2 |det ξ| ].

In dimension two there is also a standard way of decomposing matrices thatis useful for computing singular values (see Alibert-Dacorogna [14] and Section5.3.8), namely the following.

Signed singular values and von Neumann type inequalities 519

Remark 13.7 For ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

)define

ξ =

(ξ22 −ξ2

1

−ξ12 ξ1

1

)

and

ξ+ =1

2( ξ + ξ ), ξ− =

1

2( ξ − ξ ).

Then the following properties hold (where we denote the scalar product in R2×2

by 〈.; .〉)

ξ = ξ+ + ξ−, ξ = ξ+ − ξ−

2 det ξ+ =∣∣ξ+∣∣2 , 2 det ξ− = −

∣∣ξ−∣∣2

|ξ|2 =∣∣ξ+∣∣2 +

∣∣ξ−∣∣2

2 det ξ =∣∣ξ+∣∣2 −

∣∣ξ−∣∣2 = 〈 ξ; ξ 〉 = 2 det ξ+ + 2 det ξ−

∣∣ξ+∣∣2 =

1

2[ |ξ|2 + 2 det ξ ],

∣∣ξ−∣∣2 =

1

2[ |ξ|2 − 2 det ξ ].

In particular, from the last identities and the above proposition, we deduce that

λ1 (ξ) =1√2

∣∣∣∣ξ+∣∣−∣∣ξ−∣∣∣∣ ,

λ2 (ξ) =1√2

(∣∣ξ+∣∣+∣∣ξ−∣∣) . ♦

13.3 Signed singular values and von Neumann

type inequalities

We now define the concept of signed singular values, valid only when N = n.Given ξ ∈ Rn×n, we denote by

0 ≤ |μ1 (ξ)| ≤ · · · ≤ μn (ξ)

the signed singular values; they are defined by

μ1 (ξ) = λ1 (ξ) sign (det ξ) and μj (ξ) = λj (ξ) , j = 2, · · · , n,

where 0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) are the singular values of the matrix ξ ∈ Rn×n.

From Theorem 13.3, we immediately deduce that, for every ξ ∈ Rn×n, thereexist Q, R ∈ SO (n) so that

RξQ = diag (μ1 (ξ) , · · · , μn (ξ)) .

520 Singular values

In the sequel we write

λ (ξ) = (λ1 (ξ) , · · · , λn (ξ)) and μ (ξ) = (μ1 (ξ) , · · · , μn (ξ)).

When we consider matrices ξ ∈ RN×n, we always assume that N ≥ n, the caseN < n being immediately inferred from the previous one by transposition.

This section is devoted to von Neumann type inequalities (see Theorem 13.10below). We follow here the approach of Dacorogna-Marechal [204], inspired byRosakis [516]. It combines a variational argument and the resolution of somediscrete optimization problem. The main advantage of our proof is that we getthe classical von Neumann inequality as a by-product, while Rosakis uses it inhis proof. We need the following technical result.

Lemma 13.8 (i) Let D ∈ Rn×n be diagonal, with diagonal entries whose abso-lute values are pairwise distinct. If M ∈ Rn×n is such that both MD and DMare symmetric, then M is diagonal.

(ii) Let D ∈ RN×n be diagonal (N > n), with nonzero diagonal entrieswhose absolute values are pairwise distinct. If M ∈ Rn×N is such that both MDand DM are symmetric, then M is diagonal.

Proof. (i) Let D = diag(d1, · · · , dn). Assuming that MD and DM aresymmetric, we have

MD2 = DM tD = D2M

where D2 is diagonal and has pairwise distinct diagonal entries. We thereforehave for every i, j ∈ 1, · · · , n

(MD2

)ij

= mij d2

j and(D2M

)ij

= mij d2

i .

If i = j, then d2i = d2

j and thus mij = 0.

(ii) Let us write

D =

[∆

Z

]with ∆ = diag(d1, · · · , dn) and Z = 0 ∈ R(N−n)×n.

Assuming that MD and DM are symmetric, we get

MDDt = DtM tDt = DtDM.

Writing

M = [M1 , M2] with M1 ∈ Rn×n and M2 ∈ Rn×(N−n)

we find that the above equation implies that

M1∆2 = ∆2M1 and ∆2M2 = 0.


Part (i) then shows that M1 is diagonal and, since ∆2 is diagonal with nonzeroentries, we get M2 = 0. The proof is therefore complete.

The following proposition may be regarded as a primary version of our gen-eralized von Neumann inequality for diagonal matrices.

Proposition 13.9 Let a1, · · · , an, b1, · · · , bn ∈ R and τ be a permutation of1, · · · , n such that

|b1| ≤ b2 ≤ · · · ≤ bn and∣∣aτ(1)

∣∣ ≤ · · · ≤∣∣aτ(n)

∣∣ .

(i) If∏n

j=1 aj ≥ 0, then

n∑

j=1

ajbj ≤∣∣aτ(1)

∣∣ b1 +n∑

j=2

∣∣aτ(j)

∣∣ bj .

(ii) If∏n

j=1 aj < 0, then

n∑

j=1

ajbj ≤ −∣∣aτ(1)

∣∣ b1 +

n∑

j=2

∣∣aτ(j)

∣∣ bj .

In other words, if b belongs to the set

Kn := x = (x1, · · · , xn) ∈ Rn : |x1| ≤ x2 ≤ · · · ≤ xn ,

then, letting Πe (n) be as in Notation 5.30,

maxM∈Πe(n)

〈Ma; b〉 = 〈μ (diag a) ; b〉 .

Proof. Step 1 : n = 2. It says that, if |b1| ≤ b2 and if τ is a permutation of1, 2 such that |aτ(1)| ≤ |aτ(2)|, then

(i’) a1a2 ≥ 0 implies

a1b1 + a2b2 ≤ |aτ(1)|b1 + |aτ(2)|b2 ,

(ii’) a1a2 < 0 implies

a1b1 + a2b2 ≤ −|aτ(1)|b1 + |aτ(2)|b2 .

Let us prove the claim.

Case 1: a1a2 ≥ 0. Note that

(|aτ(2)|−a2)b2+(|aτ(1)|−a1)b1 =

(|a2| − a2)b2 + (|a1| − a1)b1 if |a2| ≥ |a1|(|a1| − a2)b2 + (|a2| − a1)b1 if |a2| ≤ |a1|.

Since |b1| ≤ b2 , a direct inspection shows that both quantities are non-negativeand the claim (i’) follows.

522 Singular values

Case 2: a1a2 < 0. Observe that

(|aτ(2)|−a2)b2−(|aτ(1)|+a1)b1 =

(|a2| − a2)b2 − (|a1|+ a1)b1 if |a2| ≥ |a1|(|a1| − a2)b2 − (|a2|+ a1)b1 if |a2| ≤ |a1|.

The fact that |b1| ≤ b2 easily leads to the claim (ii’).

Step 2 : n > 2. We will use the rules of Step 1 to prove the result in thegeneral case. The given permutation τ will be decomposed as a well chosenproduct of transpositions, each of them giving rise to an inequality via (i’) or(ii’). For example, assuming that |ak| ≥ |ak+1| for some k, we can write, ifakak+1 ≥ 0,

a1b1 + · · ·+ akbk + ak+1bk+1 + · · ·+ anbn

≤ a1b1 + · · ·+ |ak+1|bk + |ak|bk+1 + · · ·+ anbn (13.1)

or, if akak+1 < 0,

a1b1 + · · ·+ akbk + ak+1bk+1 + · · ·+ anbn

≤ a1b1 + · · · − |ak+1|bk + |ak|bk+1 + · · ·+ anbn. (13.2)

Since the bk will keep the same place throughout, we will symbolize inequalitiessuch as (13.1) and (13.2) by

(a1, · · · , ak, ak+1, · · · , an) → (a1, · · · , |ak+1|, |ak|, · · · , an), (13.3)

(a1, · · · , ak, ak+1, · · · , an) → (a1, · · · ,−|ak+1|, |ak|, · · · , an), (13.4)

respectively.

Case 1: b1 ≥ 0. We have to consider two subcases.

Case 1.1:∏n

j=1 aj ≥ 0. Clearly,

(a1, · · · , an) → (|a1|, · · · , |an|).

Now, |aτ(n)| can migrate rightward by means of a transposition of type (13.3).Thus

(|a1|, · · · , |an|)→ (|a1|, · · · , |aτ(n)−1|, |aτ(n)+1|, · · · , |an−1|, |aτ(n)|).

Repeating this process, with |aτ(n−1)|, |aτ(n−2)| and so on, gives rise to thedesired inequality.

Case 1.2:∏n

j=1 aj < 0. In this case, we decide to replace all but one of thenegative aj by their absolute values: for example, if ak is negative,

(a1, · · · , an) → (|a1|, · · · , |ak−1|,−|ak|, |ak+1|, · · · , |an|).

Now we let |aτ(n)| migrate rightward using either a transposition of type (13.3)or a transposition of type (13.4) according to the signs of the elements under


consideration. Each transposition leaves one negative element. Repeating thisprocess, with |aτ(n−1)|, |aτ(n−2)| and so on, eventually sorts the |aj | accordingto τ and gives rise to

(|a1|, · · · , |ak−1|,−|ak|, |ak+1|, · · · , |an|)→ (|aτ(1)|, |aτ(2)|, · · · ,−|aτ(l)|, · · · , |aτ(n−1)|, |aτ(n)|).

Finally, it is clear that the minus sign is allowed to migrate leftward, since allelements are now sorted increasingly. Therefore,

(|aτ(1)|, |aτ(2)|, · · · ,−|aτ(l)|, · · · , |aτ(n−1)|, |aτ(n)|)→ (−|aτ(1)|, |aτ(2)|, · · · , |aτ(n)|)

and we are done.

Case 2: b1 < 0. This is easily obtained from the above strategy by observingthat

a1b1 + a2b2 + · · ·+ anbn = (−a1)(−b1) + a2b2 + · · ·+ anbn .

This achieves the proof of the proposition.

We are now ready to prove the main theorem of this section.

Theorem 13.10 (i) Let ξ, η ∈ Rn×n. Then

maxQ,R∈SO(n)

trace(QξRtηt) =

n∑

j=1

μj (ξ)μj (η)

and consequently


j=1

μj (ξ) μj (η) .

(ii) Let ξ, η ∈ RN×n, where N ≥ n. Then

maxQ∈O(N)R∈O(n)

trace(QξRtηt) =

n∑

j=1

λj (ξ) λj (η)

and consequently


j=1

λj (ξ) λj (η) .

Remark 13.11 The set of all transformations ξ → UξV t with U, V ∈ SO (n) ,endowed with the composition, is obviously a group that is isomorphic to theproduct group SO (n)×SO (n) . By abuse of notation, we may denote this groupby SO (n)× SO (n) . It results from Theorems 13.3 and 13.10 that the system

(Rn×n, SO (n)× SO (n) , diag μ)

524 Singular values

satisfies

(i) diag μ is SO (n)× SO (n)-invariant;

(ii) for all ξ ∈ Rn×n, there exists (U, V ) ∈ SO (n) × SO (n) such that ξ =U diag (μ (ξ)) V t;

(iii) for all ξ, η ∈ Rn×n, trace(ξηt) ≤ trace(diag (μ (ξ)) diag (μ (η))).

According to the terminology of Lewis [401],

(Rn×n, SO (n)× SO (n) , diag μ)

is a normal decomposition system. Our preceding results also show that, simi-larly,

(RN×n, O(N)×O (n) , diagN×n λ)

is a normal decomposition system. ♦

Proof. (i) As already said, the beginning of our proof follows the one ofRosakis [516]. Observe first that we can assume that η satisfies

η = diag (μ1 (η) , · · · , μn (η)) . (13.5)

As a matter of fact, suppose that the result is proved in this case. Let ζ beany element of Rn×n, and let U, V ∈ SO (n) be such that (cf. Theorem 13.3)ζ = UMV t, with M := diag (μ1 (ζ) , · · · , μn (ζ)) . For all Q, R ∈ SO (n) ,

trace(QξRtζt) = trace(QξRtV MU t) = trace((U tQ

)ξ(RtV

)M).

Since U tSO (n) = SO (n)V = SO (n) , we see that

maxQ,R∈SO(n)

trace(QξRtζt) = maxQ1,R1∈SO(n)

trace(Q1ξRt1M)

=

n∑

j=1

μj (ξ)μj (M)

=

n∑

j=1

μj (ξ)μj (ζ) ,

where the second equality results from the fact that M satisfies (13.5).

Notice that we can also assume, in addition to (13.5), that η satisfies|μ1 (η) | < μ2 (η) < · · · < μn (η) , since a continuity argument will then allow toextend the result to the case of wide inequalities.

Since SO (n)×SO (n) is compact and the function (Q, R)→ trace(QξRtηt)is continuous, there exist Q0, R0 ∈ SO (n) such that

trace(Q0ξRt0η

t) = maxQ,R∈SO(n)

trace(QξRtηt). (13.6)


We now prove that Q0 and R0 must be such that Q0ξRt0 is diagonal. Let A and

B be skew-symmetric matrices, that is, At = −A and Bt = −B. For all s ∈ R,let

Q(s) := esAQ0 and R(s) := esBR0.

Clearly, Q(s) and R(s) are in SO (n) , and the function

ϕ(s) := trace(Q (s) ξR (s)t ηt)

is differentiable. The optimality condition (13.6) implies that s = 0 maximizesϕ. Consequently,

0 = ϕ′(0) = trace(AQ0ξRt0η

t) + trace(Q0ξRt0B

tηt).

We have therefore shown that, for all skew-symmetric matrices A and B,

trace(AQ0ξRt0η

t) = 〈 A;(Q0ξR

t0η

t)t 〉 = 0,

trace(ηtQ0ξRt0B

t) =⟨ηtQ0ξR

t0; B⟩

= 0.

Recall that Rn×n is the orthogonal direct sum of Rn×ns and Rn×n

as , the subspacesof symmetric and skew-symmetric matrices, respectively. Therefore, the aboveconditions tell us that Q0ξR

t0η

t and ηtQ0ξRt0 must be symmetric. Lemma 13.8

(i) then implies that Q0ξRt0 is diagonal. We have thus shown so far that

maxQ,R∈SO(n)

trace(QξRtηt) = trace(Q0ξRt0η

t),

where Q0, R0 ∈ SO (n) are such that Q0ξRt0 is diagonal. It remains to see that

Q0 and R0 are such that

Q0ξRt0 = diag (μ1 (ξ) , · · · , μn (ξ)) .

But this is an immediate consequence of Proposition 13.9.

(ii) The case where N = n, which results immediately from Part (i), corre-sponds to Von Neumann inequality itself. Thus, let us assume that N > n. Theargument is analogous to that of Part (i), so we merely outline the main steps.We can assume that η satisfies

η = diagN×n (λ1 (η) , · · · , λn (η)) , (13.7)

with 0 < λ1 (η) < · · · < λn (η), the case of non strict inequalities being deducedby a passage to the limit. The compactness of O(N)×O (n) and the continuityof the function (Q, R)→ trace(QξRtηt) imply the existence of Q0 ∈ O(N) andR0 ∈ O (n) such that

trace(Q0ξRt0η

t) = maxQ∈O(N)R∈O(n)

trace(QξRtηt). (13.8)

526 Singular values

The same variational argument as in Part (i), together with Lemma 13.8 (ii),shows that Q0 and R0 must be such that Q0ξR

t0 is diagonal. Finally, it is clear

(similarly to Proposition 13.9) that, among all diagonal (N×n)-matrices ξ′ withprescribed singular values λ1 (ξ) , · · · , λn (ξ) , the matrix

diagN×n(λ1 (ξ) , · · · , λn (ξ))

maximizes trace(ξηt). Thus we must have

Q0ξRt0 = diagN×n(λ1 (ξ) , · · · , λn (ξ)),

and the result follows.

Observe that, in the square case,

− trace(ξηt) = trace(−ξηt) ≤n∑

j=1

λj (−ξ)λj (η) =

n∑

j=1

λj (ξ)λj (η) ,

so that∣∣trace(ξηt)

∣∣ ≤n∑

j=1

λj (ξ)λj (η)

for all ξ, η ∈ Rn×n. This is the classical von Neumann inequality (see vonNeumann [591], Mirsky [447] or Section 7.4 in Horn-Johnson [345]).

It is worth noticing that the analogous inequality for signed singular valuesholds as well if n is even.

Corollary 13.12 Let ξ, η ∈ Rn×n. If n is even, then

| trace(ξηt)| ≤n∑

j=1

μj (ξ) μj (η) . (13.9)

If n is odd, (13.9) is false in general.

Proof. If n is even, then det(−ξ) = det ξ and μj (−ξ) = μj (ξ) for allj = 1, · · · , n. Since

trace(−ξηt) = − trace(ξηt),

we conclude that both trace(ξηt) and − trace(ξηt) are not larger than∑nj=1 μj (ξ) μj (η) .If n is odd, counterexamples are easy to construct. For example, if n = 3,

let ξ := diag (−1, 1, 1) and η := diag (1,−1,−1) . Then

trace(ξηt) = −3 and

3∑

j=1

μj (ξ)μj (η) = 1.

This finishes the proof.

The next result, which we do not use, relates the eigenvalues and the singularvalues of a given matrix (see Theorem 3.3.2 in Horn-Johnson [346]). It shouldbe compared, at least formally, to Theorem 7.43 in Chapter 7 (see also Buliga[108]).


Theorem 13.13 (Weyl theorem) Let ξ ∈ Rn×n and denote by 0 ≤ λ1 (ξ) ≤· · · ≤ λn (ξ) its singular values and by σ1 (ξ) , · · · , σn (ξ) its eigenvalues, whichare complex in general, ordered by their modulus (0 ≤ |σ1 (ξ)| ≤ · · · ≤ |σn (ξ)|).Then the following result holds

n∏

i=ν

|σi (ξ)| ≤n∏

i=ν

λi (ξ) , ν = 2, · · · , n,

n∏

i=1

|σi (ξ)| =n∏

i=1

λi (ξ) = |det ξ| .

Chapter 14

Some underdeterminedpartial differential equations

14.1 Introduction

In this chapter, we consider Dirichlet problems associated with some underde-termined equations, that are important in geometry as well as in applications,notably in fluid mechanics and elasticity.

We start in Section 14.2.2 by studying the equation div u = f, while inSection 14.2.3 we investigate the system curl u = f in dimension 3 (a similarresult holds in any dimension, see below). We solve both problems in Holderspaces; of course similar results exist (see the bibliography below) in Lp spaces.

Both problems are part of a more general program, that of solving the Dirich-let problem associated with du = f, where u is a k form and d stands for theexterior derivative. This problem is fundamental in differential geometry andalgebraic topology. The literature on the subject is vast and we refer for furtherreferences to Dacorogna [183], where the problem is solved in Holder spacesCm,α and to Schwarz [528], where the problem is studied in Lp spaces.

In Section 14.3 we discuss a Dirichlet problem associated with the non-linearequation det∇u = f. Note that, in terms of fluid mechanics, this last equationis the Lagrangian version of the equation div u = f. The nonlinear problem issolved in Holder spaces.

14.2 The equations div u = f and curl u = f

14.2.1 A preliminary lemma

We start with this elementary lemma, whose proof can be found in Dacorogna-Moser [207]. This lemma is used to fix the boundary data. We denote by‖·‖Cm,α the Cm,α norm (see Section 12.3 for details).

530 Some underdetermined partial differential equations

Lemma 14.1 Let m ≥ 1 be an integer and 0 < α < 1. Let Ω ⊂ Rn be abounded connected open set with a Cm+2,α boundary consisting of finitely manyconnected components (ν denotes the outward unit normal). Let c ∈ Cm,α

(Ω),

then there exists b ∈ Cm+1,α(Ω)

satisfying

grad b = c ν on ∂Ω.

Furthermore, there exists K = K (α, m, Ω) > 0 such that

‖b‖Cm+1,α ≤ K ‖c‖Cm,α .

Proof. If one is not interested in the sharp regularity result, a solution of theproblem is given by

b (x) := −c (x) ζ (dist (x, ∂Ω)) ,

where dist (x, ∂Ω) stands for the distance from x to the boundary and ζ is asmooth function such that ζ (0) = 0, ζ′ (0) = 1 and ζ ≡ 0 outside a smallneighborhood of 0.

To construct a smoother solution we proceed as follows. First find aCm+1,α

(Ω)

solution of (see Gilbarg-Trudinger [313] or Ladyzhenskaya-Uraltseva[388])

∆d =

∫∂Ω

c dσ/ measΩ in Ω

∂d/∂ν = c on ∂Ω.

Moreover there exists K = K (α, m, Ω) > 0 such that

‖d‖Cm+1,α ≤ K ‖c‖Cm,α . (14.1)

We then let χ, ζ ∈ C∞ (R) be such that χ, ζ ≡ 0 outside a small neighborhoodof 0 and

χ (0) = 1, ζ (0) = 0, χ′ (0) = 0, ζ′ (0) = 1.

Defineb (x) := d (x) − χ (dist (x, ∂Ω)) d (ψ (x)) (14.2)

andψ (x) := x− ζ(dist (x, ∂Ω)) grad (dist (x, ∂Ω)) .

It remains to check that b has the claimed property. Indeed, if x ∈ ∂Ω (notethat ψ (x) = x on ∂Ω), then

grad b (x) = gradd (x)− gradd (ψ (x))∇ψ (x)

= gradd (x)− gradd (x) [I − grad (dist (x, ∂Ω))⊗ grad (dist (x, ∂Ω))]

= gradd (x) [ν ⊗ ν] =∂d

∂νν

= c ν.

The equations div u = f and curlu = f 531

From (14.1), (14.2) and the fact that the distance function is (near the bound-ary) as smooth as the boundary itself provided the boundary is at least C2 (see,for example, Gilbarg-Trudinger [313]), we deduce that

‖b‖Cm+1,α ≤ K ‖c‖Cm,α

as wished.

Once the functions χ and ζ are fixed, the above construction has defined abounded linear operator

A : Cm,α(Ω)→ Cm+1,α

(Ω)

that to every c ∈ Cm,α(Ω)

associates a unique b ∈ Cm+1,α(Ω)

such that

grad b = c ν on ∂Ω.

14.2.2 The case div u = f

Theorem 14.2 Let m ≥ 0 be an integer and 0 < α < 1. Let Ω ⊂ Rn bea bounded connected open set with a Cm+3,α boundary consisting of finitelymany connected components (ν denotes the outward unit normal). The followingconditions are then equivalent.

(i) f ∈ Cm,α(Ω)

satisfies

∫

Ω

f (x) dx = 0.

(ii) There exists u ∈ Cm+1,α(Ω; Rn

)verifying

div u = f in Ω

u = 0 on ∂Ω,(14.3)

where div u =∑n

i=1∂ui

∂xi. Furthermore, there exists K = K (α, m, Ω) > 0 such

that‖u‖Cm+1,α ≤ K ‖f‖Cm,α .

Remark 14.3 (i) If the set Ω is disconnected, then the result holds true if thecompatibility condition is understood on each connected component.

(ii) This problem has been investigated by several authors, in particularBogovski [89], Borchers-Sohr [92], Dacorogna [183], Dacorogna-Moser [207],Dautray-Lions [221], Galdi [298], Girault-Raviart [314], Kapitanskii-Pileckas[359], Ladyzhenskaya [386], Ladyzhenskaya-Solonnikov [387], Necas [473], Tartar[567] and Von Wahl [592], [593]. We follow here the presentation of Dacorogna[183], which is, however, similar to many of the above mentioned articles.


(iii) Similar type of results hold in Lp, 1 < p < ∞, see the above bibliography.However, the result is false if p = 1 or p = ∞ and it is also false in C0,α

when α = 0 or α = 1, see Bourgain-Brezis [97], Dacorogna-Fusco-Tartar [187],McMullen [408] and Preiss [500].

(iv) In fact, the proof of the theorem shows that if

X :=f ∈ Cm,α

(Ω)

:∫

Ω f (x) dx = 0

,

Y :=u ∈ Cm+1,α

(Ω; Rn

): u = 0 on ∂Ω

,

then we can construct a bounded linear operator L : X → Y which associatesto every f ∈ X, a unique u = Lf ∈ Y satisfying (14.3). ♦

Proof. (ii)⇒ (i). This implication is just the divergence theorem.

(i)⇒ (ii). We split the proof into two steps.

Step 1. We first find a ∈ Cm+2,α (see Gilbarg-Trudinger [313] orLadyzhenskaya-Uraltseva [388]) satisfying

∆a = f in Ω

∂a/∂ν = 0 on ∂Ω.

Moreover there exists K = K (α, m, Ω) > 0 such that

‖a‖Cm+2,α ≤ K ‖f‖Cm,α . (14.4)

Step 2. We then write

u = curl∗ b + grada (14.5)

where b = (bij)1≤i<j≤n ∈ Rn(n−1)/2,

curl∗ b := ((curl∗ b)1 , · · · , (curl∗ b)n)

and

(curl∗ b)i :=i−1∑

j=1

∂bji

∂xj−

n∑

j=i+1

∂bij

∂xj.

Since div curl∗ b = 0 it remains to find b ∈ Cm+2,α such that

curl∗ b = − grada on ∂Ω.

An easy computation (using the fact that ∂a/∂ν = 0) shows that a solution ofthis problem is given by

grad bij = (∂a

∂xiνj −

∂a

∂xjνi )ν on ∂Ω

The equations div u = f and curlu = f 533

whose solvability is ensured by Lemma 14.1 and moreover there exists K =K (α, m, Ω) > 0 such that

‖b‖Cm+2,α ≤ K ‖a‖Cm+2,α . (14.6)

The combination of (14.4), (14.5) and (14.6) leads to the proof of the theorem.As in the proof of Lemma 14.1, we have also proved Remark 14.3 (iv).

In order to clarify the link with differential forms, we rewrite the proof inthis terminology (see Dacorogna [183] for details). We consider u as a 1 formand therefore the problem we want to solve is

δu = f in Ω

u = 0 on ∂Ω

where δ is the codifferential. We therefore write

u = da + δb

(where a is a 0 form and b is a 2 form). This leads to

f = δu = δda = ∆a

since δδb = 0, ∆a = δda + dδa and δa = 0, a being a 0 form. (The fact that∆a = δda makes easier the case of 1 forms u in comparison with k forms k ≥ 2).

We also observe that (for the exact definition of dν and δν , see [183])

δν (da) := 〈grada; ν〉 =∂a

∂ν

which leads to our choice in Step 1.

Now in order to have u = 0 on the boundary it remains to solve (cf. Step 2)

δb = −da on ∂Ω.

The idea is then to find a solution, via Lemma 14.1, of

grad bij = − [dνda]ij ν = (∂a

∂xiνj −

∂a

∂xjνi )ν on ∂Ω

and then to check that such b satisfies δb = −da on ∂Ω.

14.2.3 The case curl u = f

The problem under investigation is important in fluid mechanics and hasbeen considered by Borchers-Sohr [92], Dacorogna [183], Dautray-Lions [221],Griesinger [322] and Von Wahl [592], [593]. In this section, we follow theapproach of Dacorogna [183], which is inspired by that of Von Wahl [592], [593].


Theorem 14.4 Let m ≥ 1 be an integer and 0 < α < 1. Let Ω ⊂ R3 bea bounded convex set with a Cm+3,α boundary and ν denote the outward unitnormal. The following conditions are then equivalent.

(i) f ∈ Cm,α(Ω; R3

)verifies

div f = 0 in Ω and 〈f ; ν〉 = 0 on ∂Ω.

(ii) There exists u ∈ Cm+1,α(Ω; R3

)satisfying

curlu = f in Ω

u = 0 on ∂Ω,

where if u =(u1, u2, u3

), then

curlu =

(∂u3

∂x2− ∂u2

∂x3,∂u1

∂x3− ∂u3

∂x1,∂u2

∂x1− ∂u1

∂x2

).

Proof. (ii)⇒ (i) The fact that div f = 0 is obvious. We now show that〈f ; ν〉 = 0 on ∂Ω. For this purpose, we let ψ ∈ C2

(Ω)

be an arbitrary function.The integration by parts formula and the facts that u = 0 on ∂Ω, curlu = fand div f = 0 lead to

∫

Ω

〈gradψ; f〉 dx =

∫

Ω

〈gradψ; curlu〉 dx = 0,

∫

Ω

〈gradψ; f〉dx =

∫

∂Ω

ψ 〈f ; ν〉 dσ.

Combining these two equations and the fact that ψ is arbitrary, we have indeedobtained that 〈f ; ν〉 = 0 on ∂Ω.

(i)⇒ (ii) We divide the proof into two steps.

Step 1. We first find w ∈ Cm+1,α that solves the system (denoting thevectorial product by w ∧ ν)

⎧⎪⎪⎨⎪⎪⎩

curlw = f in Ω

div w = 0 in Ω

w ∧ ν = 0 on ∂Ω.

This is possible, using a result for the existence part due to Kress [378] (see alsoDuff-Spencer [253] and Morrey [455] Sections 7.7 and 7.8). The regularity thenfollows from standard arguments (see Morrey [455]).

In terms of the notations of differential forms, we are in fact solving (con-sidering w as a 1 form and f as a 2 form)

⎧⎪⎪⎨⎪⎪⎩

dw = f in Ω

δw = 0 in Ω

dνw = 0 on ∂Ω,

The equation det∇u = f 535

where for f = (f12, f13, f23) we let

f = (f23,−f13, f12) .

The compatibility conditions for solving this problem are exactly

d f = div f = 0 in Ω and dν f = 0⇔ 〈f ; ν〉 = 0 on ∂Ω.

Step 2. A solution of our problem is then given by

u = w + gradv,

where v ∈ Cm+2,α solves on ∂Ω

gradv = −w.

Indeed, this is possible by Lemma 14.1 and by the fact that w ∧ ν = 0.

We conclude this section by discussing the case where Ω is not necessarilyconvex. We assume that Ω ⊂ R3 is a bounded connected set with smoothboundary (ν then denotes the outward unit normal) consisting of finitely manyconnected components.

Denote

D2 (Ω) :=

ψ ∈ C0

(Ω; R3

)∩ C1

(Ω; R3

):

curlψ = 0, div ψ = 0 in Ω and 〈ψ; ν〉 = 0 on ∂Ω

,

which is the set of 2 harmonic fields with a Dirichlet boundary condition. IfΩ is convex or more generally contractible, we have D2 (Ω) = 0 . In general,however, this is not the case and the dimension of D2 (Ω) is then related to theBetti numbers of Ω (see Duff-Spencer [253] and Kress [378]).

Theorem 14.4 remains valid for such general sets if we add the followingnecessary condition ∫

Ω

〈f ; ψ〉dx = 0, ∀ψ ∈ D2 (Ω) .

14.3 The equation det ∇u = f

14.3.1 The main theorem and some corollaries

We now state the main result of this section and we follow the presentation ofDacorogna-Moser [207]. We start first with the following notation.

Notation 14.5 Let Ω, O ⊂ Rn be bounded open sets, m ≥ 1 (includingm = ∞) be an integer and 0 < α ≤ 1. We denote by Diffm

(Ω; O

)(respec-

tively Diffm,α(Ω; O

)) the set of diffeomorphisms u : Ω → O such that u ∈

Cm(Ω; Rn

)(respectively Cm,α

(Ω; Rn

)) and u−1 ∈ Cm

(O; Rn

)(respectively

Cm,α(O; Rn

)). When Ω = O, we simply write Diffm

(Ω)

(respectively

Diffm,α(Ω)). ♦


Theorem 14.6 (Dacorogna-Moser theorem) Let m ≥ 0 be an integer and0 < α < 1. Let Ω ⊂ Rn be a bounded connected open set with a Cm+3,α boundaryconsisting of finitely many connected components. Let f ∈ Cm,α

(Ω), f > 0 in

Ω and ∫

Ω

f (x) dx = meas Ω. (14.7)


satisfying

det∇u (x) = f (x) x ∈ Ω

u (x) = x x ∈ ∂Ω.(14.8)

Conversely, if there exists u ∈ Diffm+1,α(Ω)

satisfying (14.8), then f ∈Cm,α

(Ω)

and (14.7) holds.

Remark 14.7 (i) The theorem is due to Dacorogna-Moser [207]. It finds itsorigins in Moser [457], who proved the result for manifolds without bound-ary and in the C∞ case. His result was improved notably by Banyaga [69],Reimann [507] and Zehnder [614]. Independently, Tartar [569] and Dacorogna[170] proved a similar result for the case where Ω is the unit ball of R2 andR3. Some counter examples or extensions of the above theorem can be found inBurago-Kleiner [109], MacMullen [408], Riviere-Ye [513] and Ye [604].

(ii) Of course no uniqueness is to be expected. For example, if n = 2, Ω isthe unit disk and f ≡ 1. Indeed, writing u in polar coordinates,

u (x) = u (x1, x2) =(r cos

(θ + 2kπr2

), r sin

(θ + 2kπr2

)),

with r ∈ [0, 1] , k ∈ Z (the set of integers); we find that u satisfies (14.8) forevery k. ♦

We first mention the following immediate corollary.

Corollary 14.8 Let m ≥ 0 be an integer and 0 < α < 1. Let Ω ⊂ Rn be abounded connected open set with a Cm+3,α boundary consisting of finitely manyconnected components. Let f, g ∈ Cm,α

(Ω), f, g > 0 in Ω and

∫

Ω

f (x) dx =

∫

Ω

g (x) dx.


satisfying

g (u (x)) det∇u (x) = f (x) x ∈ Ω

u (x) = x x ∈ ∂Ω.


Proof. (Corollary 14.8). It suffices to set u = v−1 w where (see Theorem14.6) v and w satisfy

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

det∇w (x) =f (x) measΩ∫

Ω f (x) dxx ∈ Ω

det∇v (x) =g (x)measΩ∫

Ωg (x) dx

x ∈ Ω

w (x) = v (x) = x x ∈ ∂Ω.

This concludes the proof of the corollary.

The theorem has also as a direct consequence the following result.

Corollary 14.9 Let m ≥ 1 be an integer and 0 < α < 1. Let Ω, O ⊂ Rn bebounded connected open sets with a Cm+3,α boundary consisting of finitely manyconnected components. Let u0 ∈ Diffm,α

(Ω; O

)such that

det∇u0 > 0 in Ω.

Let g : R → R be convex and

(P ) inf

I (u) =

∫

Ω

g (det∇u (x)) dx : u ∈ u0 + W 1,∞0 (Ω; Rn)

.

Then there exists a minimizer u ∈ X of (P ) and moreover u ∈ Diffm,α(Ω; O

).

Remark 14.10 (i) This problem was first considered in Dacorogna [170]. Itshould be pointed out that, although the function

f (ξ) := g (det ξ)

is quasiconvex (and even polyconvex), since g is convex, the direct methods ofChapter 8 do not apply, because we lack the appropriate coercivity hypothesis.

(ii) The non-convex case has been considered by Mascolo-Schianchi [437],see also Theorem 11.32. ♦

Proof. We let f : O → R be defined by

f (y) :=det∇u−1

0 (y)

measΩ

∫

Ω

det∇u0 (z)dz.

Note that f > 0 in O and∫

O

f (y) dy = measO.

We may therefore apply Theorem 14.6 to find v ∈ Diffm,α(O)

satisfying

det∇v (y) = f (y) y ∈ O

v (y) = y y ∈ ∂O.


Settingu = v u0

it is easy to see that

det∇u (x) =

∫Ω det∇u0 (z)dz/ measΩ x ∈ Ω

u (x) = u0 (x) x ∈ ∂Ω.(14.9)

We now claim that u is indeed a minimizer of (P ) . Take any u ∈ u0 +W 1,∞

0 (Ω; Rn) and apply first Jensen inequality, then the fact that u = u0 on∂Ω (combined with Theorem 8.35) and finally (14.9) to get

I (u) =

∫

Ω

g (det∇u (x)) dx ≥ measΩ g(1

measΩ

∫

Ω

det∇u (x) dx)

= measΩ g(1

measΩ

∫

Ω

det∇u0 (x) dx)

= measΩ g (det∇u (x)) =

∫

Ω

g (det∇u (x)) dx = I (u)

as wished.

We now describe roughly the idea of the proof of the theorem. We giveseveral ways of constructing solutions of (14.8). All of them require as a firststep to solve the linearized problem (setting u (x) = x + v (x))

div v = f − 1 in Ω

v = 0 on ∂Ω.(14.10)

This was already achieved in Theorem 14.2. Although the solution of thisproblem is clearly not unique, our construction provides a well defined solution(see Remark 14.3 (iv)).

We now present the different ways of solving the nonlinear problem (14.8);for a still different approach, see Dacorogna [170] or Dacorogna-Moser [207].

- In Section 14.3.2 (see Lemma 14.11), we find a Cm,α solution by a defor-mation argument, i.e. by solving the ordinary differential equations

⎧⎨⎩

d

dtΦt (x) =

v (Φt (x))

t + (1− t) f (Φt (x))

Φ0 (x) = x,

where v is as in (14.10). Standard properties of ordinary differential equationsgive that u (x) = Φ1 (x) is a solution of (14.8), but it is only in Cm,α and notin Cm+1,α as wished.

- In Section 14.3.3 (see Lemma 14.12), using (14.10) and a smallness assump-tion on the C0,β norm, 0 < β ≤ α < 1, of f − 1, we obtain a Cm+1,α solutionby linearizing the equation around the identity.


- Finally, in Section 14.3.4, we give two proofs of the theorem, obtaining theclaimed regularity conclusion and removing the smallness assumption on f − 1.This is achieved in two different ways, one as a combination of Lemmas 14.11and 14.12 and the other by several iterations of Lemma 14.12.

14.3.2 A deformation argument

We turn our attention to proving Theorem 14.6 with a weaker regularity thanstated in the conclusion of the theorem. We follow here the original proof ofMoser [457] and Dacorogna-Moser [207].

Lemma 14.11 Let m ≥ 1 be a integer and 0 < α < 1. Let Ω ⊂ Rn be abounded connected open set with a Cm+3,α boundary consisting of finitely manyconnected components. Let f ∈ Cm,α

(Ω), f > 0 in Ω and

∫

Ω

f (x) dx = measΩ.

Then there exists u ∈ Diffm,α(Ω)

satisfying

det∇u (x) = f (x) x ∈ Ω

u (x) = x x ∈ ∂Ω.(14.11)


Step 1. For t ∈ [0, 1] and z ∈ Ω, let

vt (z) :=v (z)

t + (1− t) f (z), (14.12)

where v ∈ Cm+1,α(Ω; Rn

)(but vt ∈ Cm,α

(Ω; Rn

)) satisfies

div v = f − 1 in Ω

v = 0 on ∂Ω.(14.13)

(Such a v exists by Theorem 14.2.)We then define Φt (x) : [0, 1]× Ω → Rn as the solution of

⎧⎨⎩

d

dt[Φt (x)] = vt (Φt (x)) , t > 0

Φ0 (x) = x.(14.14)

First note that Φt ∈ Cm,α(Ω; Rn

)for every t and that Φt is uniquely defined on

[0, 1] ; moreover, Φt is, by construction, a diffeomorphism. Observe also that,for every t ∈ [0, 1] , we have

Φt (x) ≡ x if x ∈ ∂Ω.


This follows from the observation that if x ∈ ∂Ω, then x is a solution of (14.14),since v = 0 on ∂Ω; the uniqueness then implies that Φt (x) ≡ x for every x ∈ ∂Ω.

We now show that u (x) := Φ1 (x) is a solution of (14.11). The boundarycondition has already been verified, so we only need to check that det∇Φ1 (x) =f (x) . To prove this, we let

h (t, x) := [det∇Φt (x)] . [t + (1− t) f (Φt (x))] . (14.15)

If we show (see Step 2) that

∂

∂th (t, x) ≡ 0 (14.16)

we will have the result from the fact that h (1, x) = h (0, x) .

Step 2. We therefore only need to show (14.16). Let A be an n× n matrix,then it is a well known fact (see Coddington-Levinson [160], page 28) that if ψsatisfies ψ′ (t) = A (t) ψ (t) , then

(detψ)′

= trace (A) . detψ,

where trace (A) stands for the trace of A. We therefore get that

∂

∂t[det∇Φt (x)] = det∇Φt (x) . div vt (Φt (x)) . (14.17)

We now differentiate (14.15) to get

∂

∂th (t, x) =

∂

∂t[det∇Φt] . [t + (1− t) f (Φt)]

+ [det∇Φt] [ 1− f (Φt) + (1− t) 〈 ∇f (Φt) ;d

dtΦt 〉 ].

Using (14.14) and (14.17), we obtain

∂

∂th (t, x) = [det∇Φt] [(t + (1− t) f (Φt)) div vt (Φt)

+ (1− t) 〈∇f (Φt) ; vt (Φt)〉+ (1− f (Φt))] .

Using the definition of vt (see (14.12)), we deduce that

div v (y) = (t + (1− t) f (y)) div vt (y) + (1− t) 〈∇f (y) ; vt (y)〉 .

Combining the two identities, we have

∂

∂th (t, x) = [det∇Φt] . [div v (Φt) + (1− f (Φt))] .

The definition of v (see (14.13)) immediately gives (14.16) and thus the lemma.


14.3.3 A proof under a smallness assumption

We now prove Theorem 14.6 under a smallness assumption on the C0,β normof f − 1.

Lemma 14.12 Let Ω, m, α and f ∈ Cm,α(Ω)

be as in Theorem 14.6. Let 0 <β ≤ α < 1. Then there exists ǫ = ǫ (α, β, m,Ω) > 0 such that if ‖f − 1‖C0,β ≤ ǫ,then there exists u ∈ Diffm+1,α

(Ω)

such that

det∇u (x) = f (x) x ∈ Ω

u (x) = x x ∈ ∂Ω.(14.18)

Remark 14.13 (i) A similar result can be found in Zehnder [614].

(ii) We use below some elementary properties of Holder continuous functionsthat are gathered in Section 12.3. ♦


Step1. We start by defining two constants K1, K2 as follows.

(i) Let

X :=b ∈ Cm,α

(Ω)

:∫

Ωb (x) dx = 0

,

Y :=a ∈ Cm+1,α

(Ω; Rn

): a = 0 on ∂Ω

.

As seen in Theorem 14.2, we can then define a bounded linear operator L : X →Y that associates to every b ∈ X a unique a ∈ Y such that

div a = b in Ω

a = 0 on ∂Ω.

Furthermore, there exists K1 > 0 such that

‖Lb‖C1,β ≤ K1 ‖b‖C0,β (14.19)

‖Lb‖Cm+1,α ≤ K1 ‖b‖Cm,α . (14.20)

(ii) For ξ any n× n matrix, let

Q (ξ) := det (I + ξ)− 1− trace (ξ) , (14.21)

where I stands for the identity matrix and trace (ξ) for the trace of ξ. Notethat Q is a sum of monomials of degree t, 2 ≤ t ≤ n. We can therefore find (seeProposition 12.7) K2 > 0 such that if v, w ∈ Cm+1,α with ‖v‖C1,β , ‖w‖C1,β ≤ 1,then

‖Q (∇v)−Q (∇w)‖C0,β ≤ K2 (‖v‖C1,β + ‖w‖C1,β ) ‖v − w‖C1,β

‖Q (∇v)‖Cm,α ≤ K2 ‖v‖C1 ‖v‖Cm+1,α .(14.22)


Step 2. In order to solve (14.18), we set v (x) = u (x) − x and we rewrite itas

div v = f − 1−Q (∇v) in Ω

v = 0 on ∂Ω.(14.23)

If we set

N (v) := f − 1−Q (∇v) ,

then (14.23) is satisfied for any v ∈ Y with

v = LN (v) . (14.24)

Note first that the equation is well defined (i.e. N : Y → X), since if v = 0 on∂Ω then

∫Ω N (v (x)) dx = 0. Indeed, from (14.21) we have that

∫

Ω

N (v (x)) dx =

∫

Ω

[f (x)− 1−Q (∇v (x))] dx

=

∫

Ω

[f (x) + div v (x)− det (I +∇v (x))] dx;

since v = 0 on ∂Ω and∫Ω f = measΩ, it follows immediately (see Theorem

8.35)) that the right hand side of the above identity is 0.

We now solve (14.24) by the contraction principle. We first let for r > 0

Br :=

u ∈ Cm+1,α

(Ω; Rn

):

u = 0 on ∂Ω, ‖u‖C1,β ≤ r,

‖u‖Cm+1,α ≤ 2K1 ‖f − 1‖Cm,α

.

We endow Br with the C1,β norm. We observe that Br is complete (see Propo-sition 12.8) and we will show that by choosing ‖f − 1‖C0,β and r small enough,then LN : Br → Br is a contraction mapping. The contraction principle willthen immediately lead to a solution v ∈ Br and hence in Cm+1,α of (14.24).

Indeed, let

‖f − 1‖C0,β ≤ min

1

8K21K2

,1

2K1

(14.25)

r := 2K1 ‖f − 1‖C0,β . (14.26)

If v, w ∈ Br (note that by construction r ≤ 1), we then have

‖LN (v)− LN (w)‖C1,β ≤ 1

2‖v − w‖C1,β (14.27)

‖LN (v)‖C1,β ≤ r, ‖LN (v)‖Cm+1,α ≤ 2K1 ‖f − 1‖Cm,α . (14.28)


The inequality (14.27) follows from (14.19), (14.22), (14.25) and (14.26) through

‖LN (v)− LN (w)‖C1,β ≤ K1 ‖N (v)−N (w)‖C0,β

= K1 ‖Q (∇v)−Q (∇w)‖C0,β

≤ K1K2 (‖v‖C1,β + ‖w‖C1,β ) ‖v − w‖C1,β

≤ 2rK1K2 ‖v − w‖C1,β

= 4K21K2 ‖f − 1‖C0,β ‖v − w‖C1,β

≤ 1

2‖v − w‖C1,β .

To obtain the first inequality in (14.28) we observe that

‖LN (0)‖C1,β ≤ K1 ‖N (0)‖C0,β = K1 ‖f − 1‖C0,β =r

2

and hence combining (14.27) with the above inequality, we have immediatelythe first inequality in (14.28). To obtain the second one, we just have to observethat

‖LN (v)‖Cm+1,α ≤ K1 ‖N (v)‖Cm,α ≤ K1 ‖f − 1‖Cm,α + K1 ‖Q (∇v)‖Cm,α

(14.29)and use the second inequality in (14.22) to get

‖Q (∇v)‖Cm,α ≤ K2 ‖v‖C1 ‖v‖Cm+1,α ≤ K2 ‖v‖C1,β ‖v‖Cm+1,α

≤ 2K1K2 ‖f − 1‖C0,β ‖v‖Cm+1,α

where we have used in the third inequality the fact that

‖v‖C1,β ≤ r = 2K1 ‖f − 1‖C0,β .

The above inequality combined with (14.25) gives

‖Q (∇v)‖Cm,α ≤ 1

4K1‖v‖Cm+1,α .

Combining this last inequality, (14.29) and the fact that v ∈ Br , we deducethat

‖LN (v)‖Cm+1,α ≤ 2K1 ‖f − 1‖Cm,α .

Thus the contraction principle gives immediately the existence of a Cm+1,α

solution.It now remains to show that u (x) = v (x) + x is a diffeomorphism. This is

a consequence of the fact that det∇u = f > 0 and u (x) = x on ∂Ω (see, forexample, Corollary 2 on page 79 in Meisters-Olech [441]).

14.3.4 Two proofs of the main theorem

We may now turn to the first proof of Theorem 14.6.


Proof. The fact that if there exists u ∈ Diffm+1,α(Ω)

satisfying (14.8), then

f ∈ Cm,α(Ω)

and (14.7) holds, is straightforward (cf. Theorem 8.35), using thefact that if u (x) = x on ∂Ω, then

∫

Ω

det∇u (x) dx = measΩ

and hence the claim.

We now prove the converse and we divide the proof into two steps.

Step 1. Let us first show that, if f ∈ Cm,α(Ω), f > 0 in Ω and if 0 < β <

α < 1, we can then find, for every ǫ > 0, a function g ∈ C∞ (Ω), g > 0 in Ω

such that ∥∥∥∥f

g− 1

∥∥∥∥C0,β

≤ ǫ and

∫

Ω

f (x)

g (x)dx = measΩ. (14.30)

1) We first start by observing that if h ∈ C0,β(Ω)

and

h (x) ≥ h > 0, for every x ∈ Ω

then ∥∥∥∥1

h

∥∥∥∥C0,β

≤ 1

h+

1

h2 [h]β ≤

1

h2 ‖h‖C0,β . (14.31)

2) Let

f :=1

2inff (x) : x ∈ Ω

.

From Proposition 12.7 (v), we can find, for every δ > 0 sufficiently small, afunction fδ ∈ C∞ (Ω

)such that

‖f − fδ‖C0,β ≤ δ and fδ (x) ≥ f > 0, for every x ∈ Ω. (14.32)

Note that from Proposition 12.7 (i), (14.31) and (14.32), we have

∥∥∥∥f

fδ− 1

∥∥∥∥C0,β

≤ 2C ‖f − fδ‖C0,β

∥∥∥∥1

fδ

∥∥∥∥C0,β

≤ 2C

f2 δ ‖fδ‖C0,β

≤ δ′ :=2C

f2 δ [ ‖f‖C0,β + δ ] .

3) We next set

t :=1

meas Ω

∫

Ω

f (x)

fδ (x)dx

and observe that |t− 1| ≤ δ′. Defining

g := tfδ

and choosing δ and thus δ′ small enough, we have indeed shown that g satisfies(14.30).


Step 2. Choose now ǫ as in Lemma 14.12 and apply Step 1 to find g ∈C∞ (Ω

), g > 0 in Ω, satisfying (14.30). We then define b ∈ Diffm+1,α

(Ω)

to bea solution, which exists by (14.30) and Lemma 14.12, of

⎧⎪⎨⎪⎩

det∇b (x) =f (x)

g (x)x ∈ Ω

b (x) = x x ∈ ∂Ω.

We further let a ∈ Diffm+1,α(Ω)

to be a solution of

det∇a (y) = g

(b−1 (y)

)y ∈ Ω

a (y) = y y ∈ ∂Ω.

Such a solution exists by Lemma 14.11 since g b−1 ∈ Cm+1,α(Ω)

(cf. Propo-sition 12.7 (ii)) and

∫

Ω

g(b−1 (y)

)dy =

∫

Ω

g (x) det∇b (x) dx =

∫

Ω

f (x) dx = measΩ.

Finally observe that the function u = a b has all the claimed properties.

We conclude with a second proof of Theorem 14.6 that does not use the flow,as in Lemma 14.11, but only appeals to Lemma 14.12, with β = α (see Ye [604]for a similar procedure in Sobolev spaces).

Proof. We proceed in two steps.

Step 1. We start by defining for s = 0, · · · , N + 1, N an integer,

fs := (1− s

N + 1)f +

s

N + 1∈ Cm,α

(Ω).

(1) Note that f0 = f and fN+1 ≡ 1. Recall that∫

Ω

f (x) dx = meas Ω, (14.33)

which implies in particular that min f ≤ 1 ≤ max f . We moreover have,for every x ∈ Ω,

0 < f := minx∈Ω

f (x) ≤ min f (x) , 1 ≤ fs (x) ≤ max f (x) , 1 ≤ ‖f‖C0 .

(14.34)(2) We also have the following estimates for every s = 0, · · · , N + 1 :

‖fs‖C0 ≤ ‖f‖C0 and [fs]α = (1− s

N + 1) [f ]α ≤ [f ]α

thus‖fs‖C0,α ≤ ‖f‖C0,α

and ∥∥∥∥1

fs

∥∥∥∥C0,α

≤ 1

f+

1

f2 [fs]α ≤

1

f2 (‖f‖C0 + [fs]α) ≤ 1

f2 ‖f‖C0,α .


We also clearly have

‖fs+1 − fs‖C0,α ≤‖f − 1‖C0,α

N + 1, s = 0, · · · , N.

(3) We next set

ts :=1

meas Ω

∫

Ω

f (x)

fs (x)dx, s = 0, · · · , N + 1

and observe that t0 = tN+1 = 1. Moreover, from (14.33) and (14.34), we obtain

1

‖f‖C0

≤ ts ≤1

f, s = 0, · · · , N + 1

as well as, for s = 0, · · · , N + 1,

|ts − ts+1| ≤ 1

measΩ

∫

Ω

f (x)

∣∣∣∣1

fs (x)− 1

fs+1 (x)

∣∣∣∣ dx

≤∥∥∥∥

1

fs− 1

fs+1

∥∥∥∥C0

≤ ‖f − 1‖C0,α

f2(N + 1)

.

Defininggs := tsfs , s = 0, · · · , N + 1

we find gs ∈ Cm,α(Ω),

f

‖f‖C0

≤ gs ≤‖f‖C0

f, s = 0, · · · , N + 1

and ∫

Ω

f (x)

gs (x)dx = measΩ, s = 0, · · · , N + 1.

Finally note that, from Proposition 12.7 (i) and the above estimates, we get∥∥∥∥

gs

gs+1− 1

∥∥∥∥C0,α

=

∥∥∥∥tsfs

ts+1fs+1− 1

∥∥∥∥C0,α

=

∥∥∥∥ts(fs − fs+1) + (ts − ts+1)fs+1

ts+1fs+1

∥∥∥∥C0,α

≤ 2C

ts+1

∥∥∥∥1

fs+1

∥∥∥∥C0,α

[ ts‖fs+1 − fs‖C0,α + |ts − ts+1|‖fs+1‖C0,α ]

≤ 2C ‖f‖C0

f2 ‖f‖C0,α [

1

f

‖f − 1‖C0,α

N + 1+‖f − 1‖C0,α

f2(N + 1)

‖f‖C0,α ].

Setting

γ := 2C‖f‖C0

f4 ‖f‖C0,α ‖f − 1‖C0,α [ f + ‖f‖C0,α ]

we get that ∥∥∥∥gs

gs+1− 1

∥∥∥∥C0,α

≤ γ

N + 1.


(4) We therefore choose N sufficiently large so that, for ǫ as in Lemma 14.12,we have g0 = f, gN+1 ≡ 1 and, for every s = 0, · · · , N + 1,

∥∥∥∥gs

gs+1− 1

∥∥∥∥C0,α

≤ ǫ and

∫

Ω

f (x)

gs (x)dx = measΩ.

Step 2. We then set

u0 (x) := x and g0 = f

and define inductively, with the help of Lemma 14.12, us+1 ∈ Diffm+1,α(Ω),

s = 0, · · · , N, satisfying

⎧⎨⎩

det∇us+1 (x) =gs(u−1

s (x))gs+1(u −1

s (x))x ∈ Ω

us+1 (x) = x x ∈ ∂Ω,

(14.35)

whereus := us · · · u1 u0 .

If such a us exists, then a straightforward induction procedure shows that

det∇us+1 (x) =f (x)

gs+1 (x), s = 0, · · · , N (14.36)

and hence uN+1 is the claimed solution of the theorem.It therefore remains to show that there exists us ∈ Diffm+1,α

(Ω)

satisfying(14.35) and we proceed by induction. Indeed if s = 0, we have

g0

g1∈ Cm,α

(Ω),

g0

g1> 0,

∥∥∥∥g0

g1− 1

∥∥∥∥C0,α

≤ ǫ

and ∫

Ω

g0 (x)

g1 (x)dx = measΩ.

So we may apply Lemma 14.12 to get u1 ∈ Diffm+1,α(Ω).

Assume now that we have proved the result up to s and let us prove it fors + 1. We clearly have

gs

(u−1

s

)

gs+1

(u−1

s

) ∈ Cm,α(Ω),

gs

gs+1> 0,

∥∥∥∥gs

gs+1− 1

∥∥∥∥C0,α

≤ ǫ.

Moreover we see, from (14.36), that

∫

Ω

gs

(u−1

s (x))

gs+1

(u−1

s (x))dx =

∫

Ω

gs (y)

gs+1 (y)det∇us (y) dy =

∫

Ω

f (y)

gs+1 (y)dy = measΩ.

So we may again apply Lemma 14.12 to find, as wished, us+1 ∈ Diffm+1,α(Ω)

satisfying (14.35). This concludes the induction procedure and thus the proofof the theorem.

Chapter 15

Extension of Lipschitzfunctions on Banach spaces

15.1 Introduction

In this chapter, we deal with the extension of Lipschitz maps and we follow thepresentation of Dacorogna-Gangbo [188].

We consider two Banach spaces (E, ‖.‖E) and (F, ‖.‖F ). We ask when anymap u : D ⊂ E → F satisfying

‖u (x) − u (y)‖F ≤ ‖x− y‖E , x, y ∈ D (15.1)

can be extended to the whole of E so as to preserve the inequality.

This is by now a classical problem and we revisit this question in the followingsections. In applications to the calculus of variations, we often use this type ofextension for E = Rn and F = RN .

15.2 Preliminaries and notation

Throughout this chapter (E, ‖.‖E) and (F, ‖.‖F ) are normed spaces and even inmost cases Banach spaces. We denote by SE the unit sphere in E, namely the

set of x ∈ E such that ‖x‖E = 1. The convex hull of SE is the closed ball BE

of interior BE .

Definition 15.1 (i) We say that u : E → F is a contraction on D ⊂ E or uis 1–Lipschitz on D if

‖u(x)− u(y)‖F ≤ ‖x− y‖E for every x, y ∈ D.

In this case, we write that u ∈ Lip1(D, F ).

(ii) When u ∈ Lip1(E, F ), we simply say that u is a contraction.

550 Extension of Lipschitz functions on Banach spaces

Definition 15.2 We say that [E; F ] has the extension property for contractionson D if every u ∈ Lip1(D, F ) has an extension u ∈ Lip1(E, F ). If [E; F ] has theextension property for contractions for every D ⊂ E, we simply say that [E; F ]has the extension property for contractions.

Many extension theorems of Lipschitz maps can be derived from a principledue to Minty [445] that we state in Theorem 15.3. It gives a sufficient conditionfor extending Lipschitz maps from sets of cardinality k into sets of cardinalityk + 1.

We recall the following notation. When k is an integer, we let

Λk := λ = (λ1, · · · , λk) ∈ [0, 1]k :∑k

i=1 λi = 1.

We next need the function F : Λk×Λk → R (for more general types of functionsF, see [188]) defined by

F (λ, μ) :=∑k

i=1 λi [ ‖ yi −∑k

j=1 μjyj ‖pF − ‖xi − x‖p

E ],

where x, x1, · · · , xk ∈ E, y1, · · · , yk ∈ F are kept fixed and p ≥ 1.

Theorem 15.3 (Minty theorem) Assume that k+1 points x, x1, · · · , xk ∈ Eand k points y1, · · · , yk ∈ F are given and are such that

F (λ, λ) ≤ 0 (15.2)

for every λ ∈ Λk . Then there exists y ∈ coy1, · · · , yk such that

‖yi − y‖F ≤ ‖xi − x‖E

for every i = 1, · · · , k.

Proof. Clearly, λ → F (λ, μ) is concave (in fact affine) for every μ andμ→ F (λ, μ) is convex for every λ. Since Λk is a convex compact set, the minimaxtheorem holds (see Zeidler [615] III page 458) and there exists (λ, μ) ∈ Λk ×Λk

such thatminµ∈Λk

maxλ∈Λk

F (λ, μ) = F (λ, μ) = maxλ∈Λk

minµ∈Λk

F (λ, μ). (15.3)

One can readily conclude from (15.3) that (λ, μ) is a saddle point in the sensethat

F (λ, μ) ≤ F (λ, μ) ≤ F (λ, μ) (15.4)

for every λ, μ ∈ Λk . Setting μ = λ in (15.4) and using (15.2), we obtain that

F (λ, μ) ≤ F (λ, μ) ≤ F (λ, λ) ≤ 0

for every λ ∈ Λk . We set y =∑k

j=1 μjyj and choose λi ∈ Λk such that λij = 0

for j = i and λii = 1. Note that F (λi, μ) ≤ 0 is equivalent to

‖yi − y‖F ≤ ‖xi − x‖E

which is the claim.

Norms induced by an inner product 551

15.3 Norms induced by an inner product

We start by collecting some well known facts about inner product spaces. Onecan consult, as a general reference, Amir [26]; Lemmas 15.7 and 15.9 will beexplicitly used in the proofs of the next section.

Definition 15.4 An ellipse centered at 0 in Rn is a set

Σα := x ∈ Rn :∑n

i=1 α2i x

2i = 1,

where α = (α1, · · · , αn) ∈ (0, +∞)n. We refer to the convex hull of Σα as theregion enclosed by Σα and we denote it by Bα.

The next lemma is due to Lowner in an apparently unpublished work.

Lemma 15.5 (Lowner theorem) Assume that n ≥ 2 and that E = Rn.Then there exist a unique ellipse Σmax of maximal volume inscribed in SE anda unique ellipse of minimal volume Σmin circumscribed about SE . Furthermore,both Σmax ∩ SE and Σmin ∩ SE contain at least 2n distinct points.

Proof. Existence of ellipses of maximal volume. If Σα is inscribed in BE ,then

n∑

i=1

α2i x

2i ≥ ‖x‖2E (15.5)

for every x ∈ Rn. Assume that, for some ǫ > 0, we have

ǫ ≤ vol(Bα) =ωn∏αi

, (15.6)

where ωn is the volume of the unit Euclidean ball. The set of α such that αi > 0

×

×

×

×

×

× ×

×

Figure 15.1: Ellipses of maximal and minimal volume

and (15.5)-(15.6) hold is a compact subset Kǫ ⊂ Rn. Every maximizing sequenceof the set of ellipses inscribed in BE , of maximal volume, has its accumulationpoints in Kǫ for some small ǫ > 0. This shows that there exists an ellipse Σmax

inscribed in SE and of maximal volume (see Figure 15.1). Similarly, one obtainsan ellipse Σmin circumscribed about SE and of minimal volume.


Uniqueness of ellipses of maximal volume. Assume that Σa, Σc are twoellipses inscribed in SE and of maximal volume. By an affine transformation,we may assume that c = (1, · · · , 1) so that the volumes of these two regions are

ωn = vol(Bc) = vol(Ba) = vol(Bc) /n∏

i=1

ai .

We therefore deduce that∏n

i=1 ai = 1.

Let ‖.‖0E be the polar conjugate of ‖.‖E defined by

‖z‖0E := supx〈x; z〉 : ‖x‖E ≤ 1.

Denote (see Section 2.3.7) by ρΣa (respectively ρΣc) the gauge associated to Ba

(respectively Bc) and let ρ0Σa (respectively ρ0

Σc) be its polar; we more preciselyhave

ρΣa (z) =(∑n

i=1 a2i z

2i

)1/2and ρ0

Σa(z) =(∑n

i=1 z2i /a2

i

)1/2

ρΣc(z) = ρ0Σc(z) =

(∑ni=1 z2

i

)1/2.

Since Σa, Σc are inscribed in SE , we have that ‖.‖E ≤ ρΣa , ρΣc and so ρ0Σa , ρ0

Σc ≤‖.‖0E . Hence,

[ρ0(z)

]2:=

1

2

∑ni=1(1 + 1/a2

i )z2i =

1

2

(∑ni=1 z2

i /a2i +∑n

i=1 z2i

)

=1

2(ρ0

Σa(z)2 + ρ0Σc(z)2) ≤ (‖z‖0E )2

holds for every z ∈ Rn. The previous inequality yields that ρ2 ≥ ‖.‖2E , whichmeans that

n∑

i=1

2

1 + 1/a2i

x2i ≥ ‖x‖2E for every x ∈ Rn. (15.7)

Letting b2i = 2

1+1/a2i

, b = (b1, · · · , bn) , we find from (15.7) that Σb is inscribed

in SE.We now show that Σa and Σc coincide and we proceed by contradiction

assuming that they are distinct. Then, ai = 1 for at least one i = 1, · · ·n. Thevolume of the region enclosed by Σb is (recalling that

∏ni=1 ai = 1)

vol(Bb) = ωn(∏n

i=1(1 + 1/a2i )/2)1/2

= ωn(∏n

i=1 [ (1− 1/ai )2/2 + 1/ai ])1/2 > ωn(∏n

i=1 1/ai)1/2 = ωn.

This contradicts the maximality of the volume of Σc. Thus, Σc = Σa and so wehave a unique ellipse of maximal volume in SE . Replacing ρΣ and ‖.‖E by theirpolar conjugates, we conclude that Σmin is also unique.

Intersection of the maximal ellipse with SE . As before, we assume that Σmax

= Σα where α = (1, · · · , 1). Since Σmax and SE are compact sets, they have


a non-empty intersection; otherwise the maximality of Σmax would be contra-dicted. By symmetry there are therefore at least two points in SE ∩Σmax . Letus show that if we have 2s points in SE ∩ Σmax , 1 ≤ s < n, then in fact wehave at least 2 (s + 1) points in the intersection, therefore showing the claim.Up to a rotation, we may assume that the points ±p1, · · · ,±ps ∈ SE ∩Σmax layin the subspace generated by the first s elements e1, · · · , es of the standardEuclidean basis, which means that for every j = 1, · · · , s, we have

pji = 0 for every i ≥ s + 1.

For ǫ ∈ (0, 1), define

αǫ := ((1 − ǫ)−1, · · · , (1− ǫ)−1, (1− ǫ)s, 1, · · · , 1).

Since Σmax is unique and

vol(Σαǫ) = ωn = vol(Σmax),

we conclude that Σαǫ is not inscribed in SE . Consequently, there exists pǫ =

(pǫ1, · · · , pǫ

n) ∈ BE

, which is in Bαǫ , the region enclosed by Σαǫ , and hence wehave

‖pǫ‖E > 1 (15.8)

and

1 ≥ ρ2Σαǫ (pǫ) = ρ2

Σmax(pǫ) + ((1 − ǫ)−2−1)

s∑

i=1

(pǫi)

2 − (1− (1− ǫ)2s

)(pǫ

s+1

)2.

(15.9)

Because pǫ ∈ Σmax ⊂ BE

, (15.9) implies that

((1 − ǫ)−2−1)s∑

i=1

(pǫi)

2 ≤ (1− (1− ǫ)2s)(pǫ

s+1

)2.

Dividing both sides of the previous inequality by ǫ, we get

2− ǫ

(1− ǫ)2

s∑

i=1

(pǫi)

2 ≤ 1− (1− ǫ)2s

ǫ

(pǫ

s+1

)2. (15.10)

Let pǫν∞ν=1 be a subsequence of pǫ0<ǫ<1 converging, as ǫν → 0, to somep ∈ E. We use (15.8)-(15.10) to obtain that

ρΣmax(p) ≤ 1 ≤ ‖p‖E and

s∑

i=1

p2i ≤ s p2

s+1 . (15.11)

The first two inequalities in (15.11) and the fact that ρΣmax ≥ ‖.‖E yield thatp ∈ SE ∩ Σmax. The last inequality in (15.11) gives that p /∈ span e1, · · · , es(in particular, p = ±p1, · · · ,±ps) and thus by symmetry ±p ∈ SE ∩Σmax . This


proves that SE ∩ Σmax has at least 2 (s + 1) distinct points, if s < n. Iteratingthe process, we have indeed shown that SE ∩ Σmax has at least 2n distinctpoints. Existence of at least 2n distinct points in SE ∩ Σmin is obtained in asimilar manner.

In [357], Jordan and von Neumann gave a condition that characterizes anorm induced by an inner product.

Lemma 15.6 (Jordan-von Neumann theorem) Assume that dimE ≥ 2.Then, the norm ‖.‖E is induced by an inner product if and only if the parallel-ogram rule holds for every x, y ∈ E, namely

2(‖x‖2E + ‖y‖2E) = ‖x + y‖2E + ‖x− y‖2E . (15.12)

Proof. The fact that every norm induced by an inner product satisfies (15.12)can be checked by direct computation. Conversely, if (15.12) holds, one defines

〈x; y〉 :=‖x + y‖2E − ‖x− y‖2E

4

and check that, for every x, y ∈ E, we have

〈x; y〉 = 〈y; x〉 , 〈x; x〉 = ‖x‖2E , 〈x; 0〉 = 0, 〈−x; y〉 = −〈x; y〉

Direct computations give that if x, y, z ∈ E then

〈x + y; z〉+ 〈x− y; z〉 = 2 〈x; z〉 . (15.13)

In particular, if we set first x = y, then x = x + y and y = x− y in (15.13), weobtain that

〈2x; z〉 = 2 〈x; z〉 , 〈x + y; z〉 = 〈x; z〉+ 〈y; z〉 .

By induction, if m is an integer, we get

〈mx; z〉 = m 〈x; z〉 and 〈 x

m; z 〉 =

1

m〈x; z〉 .

We conclude that〈 m

nx; z 〉 =

m

n〈x; z〉

for every m, n integers. By continuity of ‖.‖E we conclude that 〈tx; z〉 = t 〈x; z〉for every t ∈ R. Thus, 〈·; ·〉 is an inner product that induces ‖.‖E .

The following lemma, which is a corollary of Lemma 15.5, is directly usedin the proof of Theorem 15.12.

Lemma 15.7 Assume that dimE ≥ 2. If ‖.‖E is not induced by an inner prod-uct, then there exist x, y, X, Y ∈ SE such that

‖x + y‖2E + ‖x− y‖2E < 4 < ‖X + Y ‖2E + ‖X − Y ‖2E .


Proof. As usual, it is enough to establish the result for E = R2. Let us showthe first inequality, the second one being obtained dually by replacing Σmax byΣmin . Since Σmax is inscribed in SE , we have

‖z‖E ≤ ρΣmax (z) for every z ∈ E,

where ρΣmax is the gauge associated to Σmax . It is also clear that we cannothave (see below)

‖x + y‖E = ρΣmax (x + y) (15.14)

for every x, y ∈ Σmax ∩ SE. Therefore choose x, y ∈ Σmax ∩ SE such that

‖x + y‖E < ρΣmax (x + y) .

Since we always have ‖x− y‖E ≤ ρΣmax (x− y) and ρΣmax satisfies the parallel-ogram rule, we have indeed established the claimed inequality.

We now show, by contradiction, that (15.14) does not hold. Up to an affinetransformation, we may assume that Σmax is the Euclidean disk:

Σmax = (x1, x2) ∈ R2 : x21 + x2

2 = 1.

By Lemma 15.5, Σmax ∩ SE contains at least four distinct points p11, p

12, p

13, p

14

(ordered in the clockwise direction, in particular p13 = −p1

1 and p14 = −p1

2) andwe denote by F1 := p1

1 , p12 , p1

3 , p14. Note that

ρΣmax(p1i+1 − p1

i ) ≤ π

(with the convention that p15 = p1

1) for every point in F1 .We next use (15.14) for every x, y ∈ F1 to obtain a family F2 ⊂ Σmax ∩ SE

of eight distinct points that contains F1 (see Figure 15.2). More precisely, weset

p21 = p1

1 , p23 = p1

2 , p25 = p1

3 , p27 = p1

4 ,

p22 =

p11 + p1

2

ρΣmax(p11 + p1

2), p2

4 =p12 + p1

3

ρΣmax(p12 + p1

3),

p26 =

p13 + p1

4

ρΣmax(p13 + p1

4), p2

8 =p14 + p1

1

ρΣmax(p14 + p1

1).

We clearly have that ρΣmax(p2i+1−p2

i ) ≤ π/2 (with the convention that p29 = p2

1).We iterate this process to inductively obtain families

Fn ⊂ Fn+1 ⊂ Σmax ∩ SE

such that Fn = pni 2

n+1

i=1 has 2n+1 distinct points and

ρΣmax(pni+1 − pn

i ) ≤ π/2n−1


p1

1= p2

1

p28

p27

= p14

p26

p2

5= p1

3

p24

p12

= p23

p2

2×

Figure 15.2: The points pij

(with the convention that pn2n+1+1 = pn

1 ). This gives that⋃∞

n=1 Fn is dense in

Σmax and in SE . Consequently, Σmax = SE and thus ‖.‖E is induced by aninner product, which is the desired contradiction.

We immediately obtain as a corollary the following result established by Day[224], which is a refinement of the theorem of Jordan-von Neumann.

Corollary 15.8 Assume that dim E ≥ 2. The norm ‖.‖E is induced by an innerproduct if and only if for every x, y ∈ SE the following identity holds:

‖x + y‖2E + ‖x− y‖2E = 4. (15.15)

Proof. The fact that (15.15) is a necessary condition for ‖.‖E to be inducedby an inner product is a by-product of the parallelogram rule (15.12) provedin Lemma 15.6. Conversely, we proceed by contradiction and assume that thenorm ‖.‖E is not induced by an inner product. By Lemma 15.7, we have that(15.15) does not hold and thus the claim.

We conclude with Nordlander inequality [477].

Lemma 15.9 (Nordlander inequality) Assume that dim E ≥ 2 and that0 < t < 1. Then

inf‖x + y‖E : (x, y) ∈ St ≤ 2√

1− t2 ≤ sup‖x + y‖E : (x, y) ∈ St, (15.16)

whereSt :=

(x, y) ∈ SE × SE : ‖x− y‖E = 2t

.

Proof. Observe that it suffices to prove that (15.16) holds on a subspaceof E of dimension 2. For that we may assume without loss of generality thatdimE = 2.


Since SE is the boundary of a convex set, we can find

s → u(s) = (u1(s), u2(s))

a Lipschitz parametrization of SE (in the counterclockwise direction).

We next consider, for every s, the set u(s) + 2tSE . It intersects SE at twodistinct points and we let v(s) denote the one on the ”left” of u(s). Thus (seeFigure 15.3)

s→ v(s) = (v1(s), v2(s))

is another Lipschitz parametrization in the counterclockwise direction of SE .

×

u(s)

×

×

u(s) + 2tSE

SE

v(s)

Figure 15.3: The point v (s)

By Green formula, we get

area(BE) =

∫

SE

u1du2 =

∫

SE

v1dv2. (15.17)

Let Et be the region enclosed by the curve s → (u(s) + v(s))/2 . The curve

Ct : s → (u(s)− v(s))/2

is a closed curve contained in tSE . Hence, it coincides with tSE and so theregion enclosed by Ct is tBE . We use again Green formula and (15.17) to obtainthat

area(tBE) =

∫

SE

(u1 − v1)

2d(u2 − v2)

2=

1

2area(BE)− 1

4

∫

SE

(v1du2 + u1dv2)

and

area(Et) =

∫

SE

(u1 + v1)

2d(u2 + v2)

2=

1

2area(BE) +

1

4

∫

SE

(v1du2 + u1dv2).

We add up both sides of the above identities to conclude that

area(Et) = (1− t2) area(BE).


This last identity implies that Et neither strictly contains nor is strictly con-tained in the ball of radius

√1− t2 as asserted either on the left-hand side or

the right-hand side of (15.16).

15.4 Extension from a general subset of E to E

We now present the main results of this chapter. We discuss some necessary andsufficient conditions on the spaces E and F , which in most of our analysis areBanach spaces, ensuring that [E; F ] has the extension property for contractions.

The earliest result in this direction is the celebrated MacShane lemma [410](see also Whitney [600]) asserting that if dim F = 1, then [E; F ] has the exten-sion property for contractions for any E. It turns out that this is also true forany F if dim E = 1.

At the same time, Kirszbraun [368] proved that if E and F are both finitedimensional spaces whose norms are induced by a scalar product, then [E; F ]has the extension property for contractions. This result, known as Kirszbrauntheorem, has been proved, and at the same time extended to Hilbert spaces, inseveral different ways, notably by Valentine [583], [584], Grunbaum [325], Minty[445] and others; one could also consult books such as those of Federer [275] orSchwartz [527].

When turning to necessary conditions, it was established by Schonbeck [524]that if dimE, dimF ≥ 2 and if the unit sphere SF of F is strictly convex(see below for a precise definition), then [E; F ] has the extension property forcontractions if and only if both E and F are Hilbert spaces. It can also beshown that [E; F ] has the extension property for contractions if and only iffor every set D ⊂ D′ of respective cardinality 3, 4, every map u ∈ Lip1(D, F )admits an extension u ∈ Lip1(D

′, F ). When E = F, one can prove some strongerresults, see De Figueiredo-Karlovitz [235], [236], Edelstein-Thompson [256] andSchonbeck [525].

It is one of our goals to give a still different, and somehow more elementaryand more self contained, proof of the result of Schonbeck (see Theorem 15.12).The approach used to obtain this result involves the smallest norm above ‖.‖E

that is induced by an inner product. This norm is precisely the gauge ρΣEmax

of the ellipse ΣEmax of maximal volume, inscribed in SE . Similarly, one also

considers the largest norm below ‖.‖E that is induced by an inner product.This norm turns out to be the gauge ρΣE

minof the ellipse of minimal volume,

circumscribed about SE . One seeks conditions under which

ρΣEmax

= ‖.‖E = ρΣEmin

and ρΣFmax

= ‖.‖F = ρΣFmin

.

We start with a definition that is used in the main theorem.

Extension from a general subset of E to E 559

Definition 15.10 The unit sphere SF is said to be strictly convex if it has noflat part, meaning that

‖(1− t)x + ty‖F < (1 − t) ‖x‖F + t ‖y‖F = 1

for every t ∈ (0, 1) and every x, y ∈ SF such that x = y.

Let us recall that, for 1 ≤ p ≤ ∞, the Holder norms |x|p over Rn are definedas

|x|p :=

[∑n

i=1 |xi|p ]1/p if 1 ≤ p < ∞max1≤i≤n |xi| if p = ∞.

When n ≥ 2, the unit sphere for | · |p is strictly convex if and only if 1 < p < ∞.

We can now state our main theorems.

Theorem 15.11 (i) Let (E, ‖.‖E) be a normed space. Then [E; R] has theextension property for contractions.

(ii) Let (F, ‖.‖F ) be a Banach space. Then [R; F ] has the extension propertyfor contractions.

We next turn our attention to the case where both E and F have dimensionat least 2 and we give a theorem that characterizes the Banach spaces for which[E, F ] has the extension property for contractions.

Theorem 15.12 Assume that (E, ‖.‖E) and (F, ‖.‖F ) are Banach spaces suchthat dimE, dim F ≥ 2 and that the unit sphere in F is strictly convex. Assumealso that every closed set D ⊂ E contains a countable set Dc ⊂ D whose closureis D. Then, the following three properties are equivalent:

(i) ‖.‖E and ‖.‖F are induced by an inner product;

(ii) [E; F ] has the extension property for contractions;

(iii) for every x ∈ E and every S := x1, x2, x3 ⊂ E, every u ∈ Lip1(S, F )has an extension u ∈ Lip1(S ∪ x, F ).

Remark 15.13 (i) We should point out that if S consists of only two pointsx, y ∈ E, x = y, then the extension to any third point is always possible. Indeedassume that

‖u (x)− u (y)‖F ≤ ‖x− y‖E .

Then let z ∈ E and define

t := min1,‖z − y‖E

‖x− y‖E

and u (z) := tu (x) + (1− t)u (y) .

It is immediate to check that

‖u (x)− u (z)‖F ≤ ‖x− z‖E and ‖u (z)− u (y)‖F ≤ ‖z − y‖E

as wished.


(ii) Interestingly enough, if one drops the assumption that SF is strictlyconvex, the extension property for contractions may hold for [E; F ] even if noneof the norm is induced by an inner product. This happens, for example, in thefollowing cases.

- If F = R2 (or Rn, n ≥ 2) and ‖.‖F = | · |∞ , MacShane lemma (Theorem15.11) applied to each component of a vector valued map ensures that [E; F ]has the extension property for contractions for every normed space E.

- If F = R2 with ‖.‖F = | · |1 , then [E; F ] has the extension property forcontractions for any normed space E. This follows from the simple observationthat if

R = 1/2

(1 −11 1

),

then |Ry|1 = |y|∞ for any y ∈ R2. This, together with the above argument forthe | · |∞ norm, gives that [E; R2] has the extension property for contractionsfor any normed space E.

(iii) Proceeding by contradiction in the proof that (iii) ⇒ (i), we find S :=x1, x2, x3, x ∈ (x1, x2) and u ∈ Lip1(S, F ) such that there is no extensionu ∈ Lip1(S ∪ x, F ). A continuity argument can show that there is also noextension u ∈ Lip1(S ∪ xδ, F ) where for δ > 0 small enough

xδ = x + δ (x3 − x) .

Observe that therefore xδ ∈ int cox1, x2, x3. ♦

In the proof of Theorem 15.12, we need the following lemma.

Lemma 15.14 Assume that dim E, dimF ≥ 2 and that at least one of thesenorms is not induced by an inner product. Then there exist y1, y2 ∈ F andx1, x2 ∈ E so that

‖x1‖E = ‖x2‖E = ‖y1‖F = ‖y2‖F = 1 and ‖y1 ± y2‖F < ‖x1 ± x2‖E .

Proof. It is enough to prove the lemma when dimE = dimF = 2. Weassume that ‖.‖F is not induced by a scalar product; a similar argument holdsif ‖.‖E is not induced by a scalar product. By Lemma 15.7, we can thereforefind y1, y2 ∈ R2 so that

‖y1‖F = ‖y2‖F = 1 and ‖y1 − y2‖2F + ‖y1 + y2‖2F < 4.

Let

s =1

2‖y1 − y2‖F

and use the triangle inequality to see that 0 < s < 1. We therefore have

‖y1 + y2‖F < 2√

1− s2.


We next choose t ∈ (s, 1) so that

‖y1 + y2‖F < 2√

1− t2 < 2√

1− s2.

We then apply Nordlander inequality (15.16) to get that there exist x1, x2 ∈ R2

so that

‖x1‖E = ‖x2‖E = 1 and ‖x1 − x2‖E = 2t, ‖x1 + x2‖E ≥ 2√

1− t2.

Combining all these results we have indeed found y1, y2 ∈ F and x1, x2 ∈ Esatisfying

‖y1‖F = ‖y2‖F = ‖x1‖E = ‖x2‖E = 1,

‖y1− y2‖F = 2s < 2t = ‖x1− x2‖E and ‖y1 + y2‖F < 2√

1− t2 ≤ ‖x1 + x2‖E ,

as claimed in the lemma.

It is interesting to see how to construct elements satisfying the conclusionsof Lemma 15.14 in the case of Holder norms.

Example 15.15 Assume that E = F = R2, ‖.‖F = | · |q and ‖.‖E = | · |p ,where 1 < p, q < ∞. Denote also by p′ and q′ the conjugate exponents of p andq. We then have the following cases.

Case 1. If q > p, we set x1 = y1 = (0, 1), x2 = y2 = (1, 0) and observe that

|y1 − y2|q = |y1 + y2|q = 21/q < |x1 − x2|p = |x1 + x2|p = 21/p.

Case 2. If p > q, we set x1 = 2−1/p (1, 1) , x2 = 2−1/p (1,−1) , y1 =2−1/q (1, 1) , y2 = 2−1/q (1,−1) and observe that

|y1 − y2|q = |y1 + y2|q = 21/q′

< |x1 − x2|p = |x1 + x2|p = 21/p′

.

Case 3. We assume here that p = q.

(i) If q > p′, we set x1 = 2−1/p (1, 1) , x2 = 2−1/p (1,−1) , y1 = (1, 0),y2 = (0, 1) and observe that

|y1 − y2|q = |y1 + y2|q = 21/q < |x1 − x2|p = |x1 + x2|p = 21/p′

.

(ii) If q < p′, we let x1 = (1, 0), x2 = (0, 1), y1 = 2−1/q (1, 1) , y2 =2−1/q (1,−1) to obtain that

|y1 − y2|q = |y1 + y2|q = 21/q′

< |x1 − x2|p = |x1 + x2|p = 21/p. ♦

We can now proceed with the proofs of the theorems stated above.

Proof. (Theorem 15.11). (i) In fact, the arguments used in the proof of thispart of the theorem are still valid in metric spaces. The fact that [E, R] has theextension property for contractions is, as already discussed, MacShane lemma.


We recall that if D ⊂ E and u ∈ Lip1(D, R) then both of the functions beloware extensions of u that belong to Lip1(E, R) :

u+(x) := infy∈D

u(y) + ‖x− y‖E , u−(x) := supy∈D

u(y)− ‖x− y‖E .

Furthermore, if u ∈ Lip1(E, R) is another extension of u, then u− ≤ u ≤ u+.

(ii) We now check that [R, F ] has the extension property for contractions.So we assume that we have D ⊂ R and u : D → F satisfying

‖u(x)− u(y)‖F ≤ |x− y| for every x, y ∈ D.

We wish to show that we can find u : R → F, an extension of u, satisfying

‖u(x)− u(y)‖F ≤ |x− y| for every x, y ∈ R.

We proceed in two steps.

Step 1. If D is not closed, we extend u to D by continuity. More precisely, letx ∈ D and xν ∈ D converging to x. Observe that u(xν) is a Cauchy sequence,since

‖u(xν)− u(xµ)‖F ≤ |xν − xµ| .

It therefore converges to an element of F, independent of the choice of thesequence, denoted by u(x). With this definition, we clearly deduce that

‖u(x) − u(y)‖F ≤ |x− y| for every x, y ∈ D.

Step 2. From now on we assume that D is closed. Let

α := inf x : x ∈ D and β := sup x : x ∈ D .

Thenint coD = (α, β) .

For x ∈ R, we define

x+ := inf y : y ∈ D and y ≥ x and x− := sup y : y ∈ D and y ≤ x .

Since D is closed, if x ∈ int co D, we deduce that x± ∈ D. Moreover if x ∈ D,we have that x± = x, while if x ∈ int coD but x /∈ D, we find x− < x < x+.If α < x < β, then −∞ < x− ≤ x ≤ x+ < +∞ and therefore there exists aunique t = t (x) ∈ [0, 1] such that

x = tx+ + (1− t)x−.

We are now in a position to define u : R → F through

u (x) :=

⎧⎪⎪⎨⎪⎪⎩

u (α) if x ≤ α

tu (x+) + (1− t)u (x−) if α < x < β

u (β) if x ≥ β.


In the above definition, it is understood that if α = −∞ (respectively β = +∞),then the first (respectively the third) possibility does not happen. Furthermore,since when x ∈ D we have that x± = x, we deduce that u is indeed an extensionof u. The fact that u ∈ Lip1(R, F ) is easily checked once we note that

∥∥u(x)− u(x+)∥∥

F≤ x+ − x and

∥∥u(x)− u(x−)∥∥

F≤ x− x−

if α < x < β.

We continue with the proof of Theorem 15.12.

Proof. (i) ⇒ (ii). When E and F are finite dimensional spaces, the fact that(i) implies (ii) is Kirszbraun theorem and we present here the proof of Minty[445].

In view of Zorn lemma, it is sufficient to prove that [E; F ] has the exten-sion property for contractions for finitely many points (for more details, seeDacorogna-Gangbo [188]).

So assume that we are given x1, · · · , xk ∈ E, y1, · · · , yk ∈ F such that

||yi − yj ||F ≤ ||xi − xj ||E , i, j = 1, · · · , k (15.18)

and let us show that for any x ∈ E, we can find y ∈ F such that

||yi − y||F ≤ ||xi − x||E , i = 1, · · · , k. (15.19)

In order to obtain this result, we first check that the condition (15.2) with p = 2is satisfied. Theorem 15.3 then implies the claim (15.19).

We therefore have to prove that, for every λ ∈ Λk ,

F (λ, λ) :=∑k

i=1 λi||yi −∑k

j=1 λjyj||2F − ∑ki=1 λi||xi − x||2E ≤ 0, (15.20)

whereΛk := λ = (λ1, · · · , λk) ∈ [0, 1]k :

∑ki=1 λi = 1.

Note that since the norm is induced by an inner product, the identity

∑ki,j=1 λiλj ||yi − yj ||2F = 2

∑ki=1 λi||yi −

∑kj=1 λjyj ||2F (15.21)

holds for every λ ∈ Λk . Similarly, the inequality

∑ki,j=1 λiλj ||xi − xj ||2E ≤ 2

∑ki=1 λi||xi − x||2E (15.22)

holds for every x ∈ E and every λ ∈ Λk . In fact, the right-hand side of (15.22)is minimized by the average value

x :=∑k

i=1 λixi .

We combine (15.18), (15.21) and (15.22) to conclude that (15.20) holds.

(ii) ⇒ (iii). This implication is obvious.


(iii) ⇒ (i). We proceed by contradiction, assuming that either ‖.‖E or ‖.‖F

is not induced by an inner product. We will construct

u : S := x1, x2, x3 ⊂ E → u (x1) = y1, u (x2) = y2, u (x3) = y3 ⊂ F

so that u ∈ Lip1(S, F ), but there is no extension u ∈ Lip1(S ∪ x = 0, F ).

We proceed in two steps.

Step 1. From Lemma 15.14, there exist y1, y3 ∈ F and x1, x3 ∈ E such that

‖y1‖F = ‖y3‖F = ‖x1‖E = ‖x3‖E = 1 and ‖y1 ± y3‖F < ‖x1 ± x3‖E .

We can therefore find ǫ > 0 sufficiently small so that if

y3 = (1 + ǫ)y3

we still have‖y1 ± y3‖F ≤ ‖x1 ± x3‖E .

Letting y2 = −y1 and x2 = −x1 we find that

‖y1‖F = ‖y2‖F = 1, ‖y3‖F = 1 + ǫ, ‖x1‖E = ‖x2‖E = ‖x3‖E = 1,

‖y1 − y2‖F = ‖2y1‖F = 2 = ‖2x1‖E = ‖x1 − x2‖E ,

‖y1 − y3‖F ≤ ‖x1 − x3‖E ,

‖y2 − y3‖F = ‖y1 + y3‖F ≤ ‖x1 + x3‖E = ‖x2 − x3‖E .

Hence u ∈ Lip1(S, F ), meaning that

‖yi − yj‖F ≤ ‖xi − xj‖E , ∀ i, j = 1, 2, 3. (15.23)

Step 2. The claim that there is no extension u ∈ Lip1(S ∪ x = 0, F ) willfollow if we can show that no y ∈ F can verify

‖y − yj‖F ≤ ‖xj‖E = 1, ∀ j = 1, 2, 3,

which is equivalent to showing that

A := y ∈ F : ‖y − yj‖F ≤ 1, ∀ j = 1, 2, 3 = ∅.

To prove this, we only need to show that

B := y ∈ F : ‖y − y1‖F , ‖y − y2‖F = ‖y + y1‖F ≤ 1 = 0

and use that ‖y3‖F = 1 + ǫ to obtain the claim. If y ∈ B, we obtain

1 = ‖y1‖F = ‖ 1

2(y1 − y) +

1

2(y1 + y) ‖F ≤

1

2‖y1 − y‖F +

1

2‖y1 + y‖F ≤ 1

Extension from a convex subset of E to E 565

and consequently

‖y1‖F =1

2‖y1 − y‖F +

1

2‖y1 + y‖F = 1.

Since y ∈ B, we get that

‖y1‖F = ‖y1 − y‖F = ‖y1 + y‖F = 1.

Since the unit sphere SF is strictly convex, we obtain

y1 − y = y1 + y ⇒ y = 0

as wished.

15.5 Extension from a convex subset of E to E

In many applications it is important to know if for every closed convex setΩ ⊂ E, every 1–Lipschitz map u : Ω → F admits a 1–Lipschitz extension overE. These questions have been investigated by De Figueiredo-Karlovitz in [234],[235], [236] in the case where E = F and ‖.‖E = ‖.‖F . The general case, whichstill remains open, is apparently closely related to whether or not projections onconvex sets are contractions. In this section, we address the extension propertyfor contractions for convex sets in simple cases where E is a Hilbert space.

Throughout this subsection, we assume that E is a reflexive Banach space(mostly a Hilbert space) and that Ω ⊂ E is a closed convex set. We will specifyit when we need to impose that ∂Ω, the boundary of Ω, is strictly convex. Thismeans that (1 − t)x + ty ∈ intΩ whenever t ∈ (0, 1) and x, y ∈ ∂Ω, x = y.Here, intΩ denotes the interior of Ω. The following result should be related toTheorem 2.9.

Lemma 15.16 (i) For every x ∈ E, there exists x∞ ∈ Ω minimizing

z → ‖x− z‖E

over Ω. Moreover, if x /∈ intΩ, then x∞ ∈ ∂Ω.

(ii) If in addition either SE is strictly convex or ∂Ω is strictly convex, thenx∞ is uniquely determined. In that case, the map

x → pΩ(x) := x∞

is well defined and is referred to as the projection map onto Ω.

Proof. (i) Let x ∈ E and let xν∞ν=1 ⊂ Ω be such that

limν→+∞

‖x− xν‖E = infz∈Ω

‖x− z‖E . (15.24)


The set xν∞ν=1 , being bounded, is weakly precompact and so has a subse-quence that we still label xν∞ν=1 , converging weakly to some x∞ ∈ Ω. Since‖.‖E is convex, we conclude that ‖.‖E is weakly lower semicontinuous and hence

‖x− x∞‖E ≤ limν→+∞

‖x− xν‖E .

This, together with (15.24), yields that x∞ is a minimizer of ‖x− z‖E over Ω.Let us show that if x /∈ intΩ, then x∞ ∈ ∂Ω. By contradiction, if x∞ ∈ intΩ,

we would have for t ∈ (0, 1) small enough that

xt = (1− t)x∞ + tx ∈ Ω

and thus‖x− xt‖E = (1− t) ‖x− x∞‖E < ‖x− x∞‖E

contradicting the definition of x∞ .

(ii) Let x /∈ Ω and x∞, x∞ ∈ Ω be two minimizers of ‖x− z‖E over Ω. Sincex∞, x∞ ∈ ∂Ω, we find that

x0 :=x∞ + x∞

2∈ Ω

is another minimizer of ‖x− z‖E . Assume for the sake of contradiction thatx∞ = x∞ . If ∂Ω is strictly convex, then x0 ∈ ∂Ω, which yields a contradiction.On the other hand, if SE is strictly convex, we have from the fact that

r := ‖x− x∞‖E = ‖x− x∞‖E > 0

that ‖x− x0‖E < r, which also yields a contradiction. This proves that theminimizer of ‖x− z‖E over Ω is unique.

Lemma 15.17 If E is a Hilbert space, then pΩ : E → E is a contraction.

Proof. Every Hilbert space is reflexive. Furthermore, the parallelogram rule(15.12) gives that SE is strictly convex. Hence, by Lemma 15.16, pΩ is welldefined.

Since for every x ∈ E, t ∈ [0, 1] and z ∈ Ω, we have

‖x− pΩ(x)‖2E ≤ g (t) := ‖x− [(1− t) pΩ(x) + tz]‖2E

we find, since g′ (0) ≥ 0, that pΩ(x) satisfies

〈x− pΩ(x); z − pΩ(x)〉 ≤ 0 for every z ∈ Ω. (15.25)

If x1, x2 ∈ E, we use (15.25), once with z = pΩ(x2) and once with z = pΩ(x1),to obtain that

〈x1 − pΩ(x1); pΩ(x2)− pΩ(x1)〉 ≤ 0 and 〈x2 − pΩ(x2); pΩ(x1)− pΩ(x2)〉 ≤ 0.

Extension from a convex subset of E to E 567

Adding up these two inequalities yields that

‖pΩ(x1)− pΩ(x2)‖2E ≤ 〈pΩ(x1)− pΩ(x2); x1 − x2〉 .

This, together with Cauchy-Schwarz inequality, leads to

‖pΩ(x1)− pΩ(x2)‖E ≤ ‖x1 − x2‖E ,

which is the claim.

Corollary 15.18 Assume that E is a Hilbert space and F is a normed space.Then every contraction u : Ω ⊂ E → F has an extension u : E → F that is stilla contraction.

Proof. By Lemma 15.17, pΩ is a contraction and thus the map

u := u pΩ

is a contraction as a composition of two contractions.

Remark 15.19 Let E be a finite dimensional normed space (not necessarilyinduced by a scalar product) and consider the radial map

x → pE(x) :=x

max1, ‖x‖E.

(i) In [234], under the assumption that dimE ≥ 3, De Figueiredo-Karlovitz,proved that: pE ∈ Lip1(E, E) if and only if ‖.‖E is induced by an inner product.

(ii) As is well known, we next verify that pE satisfies

‖x− pE(x)‖E ≤ ‖x− z‖E for every z ∈ BE

. (15.26)

Since the result is trivial if x ∈ BE

, we assume that x ∈ E \ BE

. We then letρ = ‖.‖E and observe that it trivially is the gauge of BE . Let ρ0 be its polar;then, according to Proposition 2.55,

p ∈ ∂ρ(x) ⇒ ρ0 (p) = 1,

where ∂ρ(x) denotes the subdifferential of ρ at x. So let p ∈ ∂ρ(x) and z ∈ BE

;we then have

‖x− z‖E ≥ ‖x‖E − 〈p; z〉 ≥ ‖x‖E − ρ0 (p) ‖z‖E ≥ ‖x‖E − 1 = ‖x− pE(x)‖E

as claimed in (15.26). ♦

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Notation

General notation

- For a given set E ⊂ RN , E, respectively ∂E, intE and Ec stand for theclosure, respectively the boundary, the interior and the complement of E.

- The sum of two sets E, F ⊂ RN is denoted by E + F ; see Section 2.2.2.

- Bǫ (x) :=y ∈ RN : |y − x| < ǫ

and Bǫ (x) :=

y ∈ RN : |y − x| ≤ ǫ

.

- 〈.; .〉 stands for the scalar product in RN .

- The following is used throughout:

Λs := λ = (λ1, · · · , λs) : λi ≥ 0 and∑s

i=1 λi = 1 ;

see Section 2.2.3.

- For a set E ⊂ RN , we denote by χE the indicator function and by 1E thecharacteristic function of E (see Sections 2.3.1 and 3.2.6) and they are given by

χE (x) :=

0 if x ∈ E

+∞ if x /∈ Eand 1E (x) :=

1 if x ∈ E

0 if x /∈ E .

- The domain of a function f is defined as

dom f :=x ∈ RN : f (x) < +∞

.

- The support of a function f is denoted by supp f ; see Section 12.2.

- Weak and weak∗ convergence are denoted by and∗ .

- For integers 1 ≤ s ≤ n, we let

(ns

)=

n!

s! (n− s)!.

Convex analysis

- For a given convex set E ⊂ RN , the relative interior and the relativeboundary of E are denoted respectively by ri E and rbdE; see Section 2.2.1.

612 NOTATION

- The projection on a convex set E is denoted by pE ; see Section 2.2.2.

- The convex hull of a set E ⊂ RN is denoted by co E; see Section 2.2.3.

- The set of extreme points of a convex set E is denoted by Eext ; see Section2.2.4.

- The gauge and the support function of a convex set E are denoted respec-tively by ρE and χ∗

E ; see Section 2.3.1.

- The epigraph and the level set of a function f are denoted respectively byepi f and levelα f ; see Section 2.3.1.

- The convex envelope of a function f is denoted by Cf ; see Section 2.3.3.

- The setsFE

∞ :=f : RN → R ∪ +∞ : f |E ≤ 0

FE :=f : RN → R : f |E ≤ 0

,

are defined in Section 2.3.3.

- The dual and bidual of a function f are respectively denoted by f∗ andf∗∗; see Section 2.3.5.

- The subdifferential ∂f (x) and the directional derivative f ′ (x, y) are definedin Section 2.3.6.

- The polar of a gauge ρ is denoted by ρ0; see Section 2.3.7.

Determinants and singular values

- RN×n stands for the set of N × n real matrices ξ,

ξ =

⎛⎜⎜⎜⎜⎝

ξ11 · · · ξ1

n

.... . .

...

ξN1 · · · ξN

n

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

ξ1

...

ξN

⎞⎟⎟⎠ = (ξ1, · · · , ξn) .

For such a matrix, ξt ∈ Rn×N denotes the transpose and, if N = n, its trace isdenoted by trace (ξ) .

- Rn×ns is the set of n× n symmetric matrices; see Section 7.4.4.

- RN×nd is the subspace of RN×n consisting of diagonal matrices. When

n ≤ N (and similarly when n > N) and x = (x1, · · · , xn) ∈ Rn, we let

diagN×n (x1, · · · , xn)

denote the diagonal matrix whose entries are the xi (when N = n, we simplywrite diag (x1, · · · , xn)); see Section 5.3.3.

- For a ∈ RN , b ∈ Rn, we let a⊗ b ∈ RN×n be defined (see Section 5.2.1) by

a⊗ b =(aibα

)1≤i≤N

1≤α≤n.

NOTATION 613

- We denote

n ∧N = min n, N , σ (s) =(ns

)(Ns

)

and

τ (n, N) =

n∧N∑

s=1

σ (s) =

n∧N∑

s=1

(ns

)(Ns

);

see Section 5.4.

- For ξ ∈ RN×n and 1 ≤ s ≤ n∧N, adjs ξ ∈ R(Ns )×(n

s) stands for the adjugatematrix and

T (ξ) = (ξ, adj2 ξ, · · · , adjn∧N ξ) ∈ Rτ(n,N);

see Section 5.4.

- For matices ξ =

(ξ11 ξ1

2

ξ21 ξ2

2

)∈ R2×2, the notations

ξ =

(ξ22 −ξ2

1

−ξ12 ξ1

1

), ξ+ =

1

2( ξ + ξ ) and ξ− =

1

2( ξ − ξ )

are used; see Remark 13.7.

- Let N ≥ n (similarly if N < n) and ξ ∈ RN×n. The singular values of ξare denoted

0 ≤ λ1 (ξ) ≤ · · · ≤ λn (ξ) ;

see Section 13.2.

- The signed singular values of ξ ∈ Rn×n are denoted by

0 ≤ |μ1 (ξ)| ≤ · · · ≤ μn (ξ) ;

see Section 13.3.

- GL (n) , Π(n) , Πe (n) , S (n) , stand for some subsets of Rn×n matrices;see Section 5.3.3.

- O (n) and SO (n) denote respectively the sets of orthogonal and specialorthogonal matrices; see Section 13.2.

- For u : Rn → RN (hence ∇u ∈ RN×n), we denote (see Section 8.5), for2 ≤ s ≤ n ∧N, 1 ≤ i1 < · · · < is ≤ n and 1 ≤ α1 < · · · < αs ≤ N,

∂(ui1 , · · · , uis

)

∂ (xα1 , · · · , xαs):= det

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

∂ui1

∂xα1

· · · ∂ui1

∂xαs

.... . .

...

∂uis

∂xα1

· · · ∂uis

∂xαs

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

614 NOTATION

Quasiconvex analysis

- Hypothesis (HI) ; see Section 5.2.5.

- The polyconvex, quasiconvex and rank one convex envelopes of a functionf are respectively denoted Pf, Qf and Rf ; see Sections 6.2, 6.3 and 6.4.

- The polyconvex conjugate and biconjugate of a function f are respectivelydenoted fp and fpp; see Section 6.2.

- The polyconvex, quasiconvex, rank one convex and separately convex hullsof a set E ⊂ RN×n are respectively denoted by

PcoE, QcoE, RcoE and ScoE;

see Section 7.3.1.

- The convex, polyconvex, quasiconvex, rank one convex and separately con-vex finite hulls of a set E ⊂ RN×n are respectively denoted by

cof E, Pcof E, Qcof E, Rcof E and Scof E;

see Section 7.3.2.

- The extreme points in the convex, polyconvex, quasiconvex, rank one con-vex and separately convex senses are respectively denoted by Ec

ext (also denotedEext in Chapter 2), Ep

ext , Eqext , Er

ext and Esext ; see Section 7.3.3.

Function spaces

Let m ≥ 0 be an integer, 1 ≤ p ≤ ∞ and 0 < α ≤ 1.

- Cm, Cm0 , C∞, C∞

0 , Cmpiec , Affm, Affpiec , Affm

piec ; see Section 12.2.

- Cm,α denote Holder spaces; see Section 12.3.

- Wm,p, Wm,p0 , denote Sobolev spaces; see Section 12.4.

Index

Adjugate matrix, 250, 517Affine hull

definition, 32Affine set, 32

definition, 32Alibert-Dacorogna-Marcellini example,

179, 221Anticonformal part of a matrix

definition, 224Approximation property, 445, 446, 448,

460, 462, 463definition, 444

Baire category theorem, 440, 443,450

Baire one function, 449, 450definition, 449

Betti numbers, 535Biconjugate function

definition, 57Bidual function

definition, 57Bolza example, 123, 484Brachistochrone, 2, 127

Canonical form, 138Caratheodory function

definition, 75Caratheodory theorem, 38, 39, 41, 52,

54, 56, 158, 163–167, 265, 288,318, 323, 480–482, 484

Cauchy-Riemann equations, 229Cauchy-Schwarz inequality, 36, 68, 567Characteristic function

definition, 75Choquet function

definition, 70

Choquet function for polyconvex setsdefinition, 345

Choquet function for rank one convexsets

definition, 346Conformal part of a matrix

definition, 224Conjugate function, 62, 342

definition, 57Contraction map

definition, 549Controllable growth condition, 111Convergence in the sense of distribu-

tionsdefinition, 393

Convexfunction, 207

Convex envelope, 55, 58, 238, 265, 287,416, 425, 432, 483, 490

definition, 52Convex function, 44–47, 51, 53, 57, 60–

63, 65, 66, 68, 70, 73, 82, 84,94, 96, 97, 100, 106, 112, 114,115, 120, 121, 125–129, 132,144, 146, 164, 168, 169, 179,192, 193, 199–201, 203–207,209–211, 213, 215, 222, 239,241, 246, 266, 318, 345, 347,350, 370, 405

definition, 44Convex hull, 55, 58, 165, 317, 327, 338,

364, 439, 440, 452, 551definition, 38

Convex integration, 461Convex set, 32, 33, 35–37, 42, 44, 45,

64, 68, 70, 318, 320, 335, 337,357, 442, 550

616 INDEX

definition, 32Cycloid, 128

Dacorogna formula, 271Dacorogna-Moser theorem, 536Directional derivative

definition, 62Dirichlet integral, 2, 6, 106Distance function, 56, 531

definition, 45Divergence theorem, 532Domain of a function

definition, 45Dual function, 69

definition, 57DuBois-Reymond equation, 129Dunford-Pettis theorem, 506

Egorov theorem, 76Ellipticity condition, 115, 117, 155, 159Epigraph

definition, 45Equiintegrability, 371, 376, 384, 386,

389, 391definition, 506

Euler-Lagrange equation, 3, 111, 117,119, 126–131, 137, 138, 141–144, 146–148, 155, 159, 178,191

Extension property for contractions,559, 560, 562

definition, 550Extreme point, 42, 43, 70

definition, 42Extreme point in the convex sense, 337

definition, 335Extreme point in the polyconvex sense,

337, 338, 344definition, 335

Extreme point in the quasiconvex sense,340

definition, 335Extreme point in the rank one convex

sense, 338, 340, 346definition, 336

Extreme point in the separately convexsense, 337, 340

definition, 336

Fatou lemma, 95Fermat principle, 2Field theory, 128Fourier series, 134Fourier transform, 193Fundamental lemma of the calculus of

variations, 115

Gateaux derivative, 108Gauge, 68, 342, 552, 555, 558, 567

definition, 45, 68Gauge in the polyconvex sense, 343

definition, 342Generalized von Neumann inequality,

520, 521, 523Green formula, 557

Hahn-Banach theorem, 34, 77Hamiltonian, 120, 138, 142, 143Hamiltonian system, 142, 143, 146Harmonic field, 535Hessian, 66Hoelder continuous function, 506, 507,

541Hoelder inequality, 107, 111, 113, 135,

374, 376, 399, 400, 422Hoelder norm, 59, 69, 559, 561Hoelder space, 529

definition, 506Hyperplane

definition, 32

Indicator function, 45, 55, 57, 58, 70,318, 328, 329, 342, 345

definition, 44Invariant integral, 103, 129Isoperimetric inequality, 120, 132, 137Isotropic function

definition, 200

Jensen inequality, 46, 124, 127, 150,161, 418, 419, 458, 481, 538

INDEX 617

Jordan-von Neumann theorem,554, 556

Kirszbraun theorem, 558, 563Kohn-Strang example, 300Kostant theorem, 200Krein-Milman theorem, 42

Lagrangian, 8, 120, 143Laminate convex hull, 331Laplace equation, 7, 228Lavrentiev phenomenon, 120,

148Lebesgue dominated convergence

theorem, 451Lebesgue monotone convergence

theorem, 99, 101, 379, 383Legendre transform, 120, 138

definition, 57Legendre-Hadamard condition, 159,

162, 220, 225, 242Level set

definition, 45Lipschitz boundary

definition, 510Lipschitz continuous function, 507Loewner theorem, 551Lower semicontinuous envelope, 58

definition, 56Lower semicontinuous function, 46

definition, 45Lusin theorem, 76

MacShane lemma, 558, 560, 561Mania example, 148Matrix of cofactors, 253Mazur theorem, 77, 96Minimal surface

non parametric form, 2, 106of revolution, 128parametric form, 2, 13, 419

Minimax theorem, 550Minkowski function

definition, 68Minkowski theorem, 42, 338, 339Minty theorem, 550

Natural growth condition, 111Newton problem, 2Nordlander inequality, 556, 561Normal integrand, 94

definition, 74Null Lagrangian, 129, 157

One sided directional derivativedefinition, 62

Orthogonal matrixdefinition, 515

Oscillation of a functiondefinition, 449

Parallelogram rule, 554, 556, 566Parseval formula, 134Piecewise continuous function

definition, 505Plancherel formula, 194Poincare inequality, 107, 406Poincare-Wirtinger inequality, 120,

123, 132, 133Polar function, 68, 552, 567

definition, 68Polyaffine function, 158, 179

definition, 157Polyconvex biconjugate function

definition, 267Polyconvex conjugate function

definition, 266Polyconvex envelope, 269, 288, 328

definition, 265Polyconvex finite hull

definition, 331Polyconvex function, 158, 159, 161–

164, 169, 171, 174, 179, 182,192–194, 196, 203–210, 213,215, 222, 226, 228, 237, 239,246, 248, 266–271, 289, 291,301, 318, 344, 351, 391, 403,404, 474, 478, 479

definition, 157Polyconvex hull, 328, 332, 349, 440

definition, 323Polyconvex set, 317–321, 325, 329, 333,

335, 342–344definition, 316

618 INDEX

Potential wells, 355, 440, 461, 488, 498Projection map, 35, 565Proper separating hyperplane

definition, 35Pyramid, 452

Quasiaffine function, 157, 158, 178–181,212, 213, 229, 296, 297, 322,362, 368, 393, 394, 417, 421,462, 467–470, 488, 490–492,494, 496

definition, 157Quasiconvex envelope, 238, 415, 416,

432, 466, 493definition, 265

Quasiconvex finite hull, 440definition, 331

Quasiconvex function, 155, 157–159,161, 162, 169, 171, 173, 178,191, 192, 194, 212, 213, 215,219, 222, 231, 237, 239, 246,271, 272, 274–276, 291, 318,367–373, 377, 378, 381, 382,384, 390, 391, 394, 403, 404,415, 424, 441, 442, 465, 472

definition, 156Quasiconvex hull, 325, 332, 339, 349

definition, 323Quasiconvex set, 319, 326, 329, 333,

335definition, 316

Rademacher theorem, 451, 509Rank one affine function, 158, 179, 181–

189definition, 157

Rank one convex envelope, 295, 328,329

definition, 265Rank one convex finite hull

definition, 331Rank one convex function, 155, 157,

159–162, 169, 171, 174–176,178, 179, 191, 192, 194, 195,212–216, 219, 220, 222, 225,

226, 239, 241, 242, 246–249,276–279, 291, 293, 318, 346

definition, 156Rank one convex hull, 293, 326, 328,

332, 339, 349, 364, 440, 446definition, 323

Rank one convex set, 318, 319, 329,333, 336, 340, 346, 347, 362,446

definition, 316Relative boundary of a convex set

definition, 32Relative interior of a convex set

definition, 32Relaxation property, 441, 443–445,

467–471definition, 441

Relaxation theorem, 319, 416, 432, 439,465, 467, 498

Relaxed problem, 415, 473, 490, 495definition, 416

Rellich theorem, 98, 388, 512

Saint Venant-Kirchhoff energyfunction, 305, 488, 492

Scorza-Dragoni theorem, 76, 99, 385,427

Separately convex finite hulldefinition, 331

Separately convex function, 47, 48, 51,158, 160, 162, 239, 318

definition, 47, 157Separately convex hull, 327, 332, 339

definition, 323Separately convex set, 318, 329, 333,

336, 337definition, 316

Separating hyperplanedefinition, 35

Separation theorem, 36, 158, 163–165,265, 321, 458

Set of permutation matricesdefinition, 197

Signed singular values, 197, 200, 285definition, 519

Simple function, 75

INDEX 619

Singular values, 197, 202, 203, 212, 285,291, 348, 440, 459, 488, 517,518, 526

definition, 516Singular values decomposition theorem,

198, 199, 218, 516Sobolev exponent, 511Sobolev imbedding theorem, 109, 111,

114, 117, 399, 400, 511Sobolev space

definition, 510Special orthogonal matrix

definition, 515Stationary point, 138Strictly convex function, 35, 36, 70,

106, 121, 139, 142, 147, 346,347, 559

definition, 44Strictly convex function in at least N

directions, 474, 476, 480definition, 474

Strictly quasiconvex function, 472–474,476, 478, 492, 495, 496, 498

definition, 472Strictly separating hyperplane, 35Subdifferential, 62, 567

definition, 61Subgradient, 61, 63, 64

definition, 61

Support function, 58definition, 45

Sverak example, 159, 173, 179, 219

Tietze extension theorem, 88, 116Tonelli partial regularity theorem, 148

Variations of independent variables,130

Vitali covering theorem, 453, 457Von Neumann inequality, 200, 520, 525,

526

Weak * lower semicontinuity, 82–85, 87,88, 92, 97, 368, 378

definition, 74Weak continuity, 368, 392, 393Weak form of Euler-Lagrange equation,

119, 125, 126Weak lower semicontinuity, 85, 95, 97,

367, 369, 377, 392definition, 74

Weierstrass example, 122Weyl theorem, 527Wirtinger inequality, 120, 132, 133, 137

Young measure, 418

Zorn lemma, 563

61. Sattinger/Weaver: Lie Groups and Algebras with

Applications to Physics, Geometry, and Mechanics.

62. LaSalle: The Stability and Control of Discrete

Processes.

63. Grasman: Asymptotic Methods of Relaxation

Oscillations and Applications.

64. Hsu: Cell-to-Cell Mapping: A Method of Global

Analysis for Nonlinear Systems.

65. Rand/Armbruster: Perturbation Methods,

Bifurcation Theory and Computer Algebra.

66. Hlavácek/Haslinger/Necasl/Lovísek: Solution

of Variational Inequalities in Mechanics.

67. Cercignani: The Boltzmann Equation and Its

Application.

68. Temam: Infi nite Dimensional Dynamical

Systems in Mechanics and Physics, 2nd ed.

69. Golubitsky/Stewart/Schaeffer: Singularities and

Groups in Bifurcation Theory, Vol. II.

70. Constantin/Foias/Nicolaenko/Temam: Integral

Manifolds and Inertial Manifolds for Dissipative

Partial Differential Equations.

71. Catlin: Estimation, Control, and The Discrete

Kalman Filter.

72. Lochak/Meunier: Multiphase Averaging for

Classical Systems.

73. Wiggins: Global Bifurcations and Chaos.

74. Mawhin/Willem: Critical Point Theory and

Hamiltonian Systems.

75. Abraham/Marsden/Ratiu: Manifolds, Tensor

Analysis, and Applications, 2nd ed.

76. Lagerstrom: Matched Asymptotic Expansions:

Ideas and Techniques.

77. Aldous: Probability Approximations via the

Poisson Clumping Heuristic.

78. Dacorogna: Direct Methods in the Calculus of

Variations.

79. Hernández-Lerma: Adaptive Markov Processes.

80. Lawden: Elliptic Functions and Applications.

81. Bluman/Kumei: Symmetries and Differential

Equations.

82. Kress: Linear Integral Equations, 2nd ed.

83. Bebernes/Eberly: Mathematical Problems from

Combustion Theory.

84. Joseph: Fluid Dynamics of Viscoelastic Fluids.

85. Yang: Wave Packets and Their Bifurcations in

Geophysical Fluid Dynamics.

86. Dendrinos/Sonis: Chaos and Socio-Spatial

Dynamics.

87. Weder: Spectral and Scattering Theory for wave

Propagation in Perturbed Stratifi ed Media.

88. Bogaevski/Povzner: Algebraic Methods in

Nonlinear Perturbation Theory.

89. O’Malley: Singular Perturbation Methods for

Ordinary Differential Equations.

90. Meyer/Hall: Introduction to Hamiltonian

Dynamical Systems and the N-body Problem.

91. Straughan: The Energy Method, Stability, and

Nonlinear Convection, 2nd ed.

92. Naber: The Geometry of Minkowski Spacetime.

93. Colton/Kress: Inverse Acoustic and

Electromagnetic Scattering Theory, 2nd ed.

94. Hoppensteadt: Analysis and Simulation

of Chaotic Systems.

95. Hackbusch: Iterative Solution of Large Sparse

Systems of Equations.

96. Marchioro/Pulvirenti: Mathematical Theory

of Incompressible Nonviscous Fluids.

97. Lasota/Mackey: Chaos, Fractals, and Noise:

Stochastic Aspects of Dynamics, 2nd ed.

98. de Boor/Höllig/Riemenschneider: Box Splines.

99. Hale/Lunel: Introduction to Functional


100. Sirovich (ed): Trends and Perspectives in

Applied Mathematics.

101. Nusse/Yorke: Dynamics: Numerical

Explorations, 2nd ed.

102. Chossat/Iooss: The Couette-Taylor Problem.

103. Chorin: Vorticity and Turbulence.

104. Farkas: Periodic Motions.

105. Wiggins: Normally Hyperbolic Invariant

Manifolds in Dynamical Systems.

106. Cercignani/Ilner/Pulvirenti: The Mathematical

Theory of Dilute Gases.

107. Antman: Nonlinear Problems of Elasticity, 2nd ed.

108. Zeidler: Applied Functional Analysis:

Applications to Mathematical Physics.

109. Zeidler: Applied Functional Analysis: Main

Principles and Their Applications.

110. Diekman/van Gils/Verduyn Lunel/Walther:

Delay Equations: Functional-, Complex-,

and Nonlinear Analysis.

111. Visintin: Differential Models of Hysteresis.

112. Kuznetsov: Elements of Applied Bifurcation

Theory, 3rd ed.

113. Hislop/Sigal: Introduction to Spectral Theory.

114. Kevorkian/Cole: Multiple Scale and Singular

Perturbation Methods.

115. Taylor: Partial Differential Equations I, Basic

Theory.

116. Taylor: Partial Differential Equations II,

Qualitative Studies of Linear Equations.

117. Taylor: Partial Differential Equations III,

Nonlinear Equations.

118. Godlewski/Raviart: Numerical Approximation of

Hyperbolic Systems of Conservation Laws.

119. Wu: Theory and Applications of Partial

Functional Differential Equations.

120. Kirsch: An Introduction to the Mathematical

Theory of Inverse Problems.

121. Brokate/Sprekels: Hysteresis and Phase Transitions.

122. Gliklikh: Global Analysis in Mathematical

Physics: Geometric and Stochastic Methods.


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123. Khoi Le/Schmitt: Global Bifurcation in

Variational Inequalities: Applications to

Obstacle and Unilateral Problems.

124. Polak: Optimization: Algorithms and

Consistent Approximations.

125. Arnold/Khesin: Topological Methods in

Hydrodynamics.

126. Hoppensteadt/Izhikevich: Weakly Connected

Neural Networks.

127. Isakov: Inverse Problems for Partial Differential

Equations, 2nd ed.

128. Li/Wiggins: Invariant Manifolds and

Fibrations for Perturbed Nonlinear Schrödinger

Equations.

129. Müller: Analysis of Spherical Symmetries in

Euclidean Spaces.

130. Feintuch: Robust Control Theory in Hilbert Space.

131. Ericksen: Introduction to the Thermodynamics of

Solids, Revised Edition.

132. Ihlenburg: Finite Element Analysis of Acoustic

Scattering.

133. Vorovich: Nonlinear Theory of Shallow Shells.

134. Vein/Dale: Determinants and

Their Applications in Mathematical Physics.

135. Drew/Passman: Theory of Multicomponent

Fluids.

136. Cioranescu/Saint Jean Paulin: Homogenization of

Reticulated Structures.

137. Gurtin: Confi gurational Forces as Basic

Concepts of Continuum Physics.

138. Haller: Chaos Near Resonance.

139. Sulem/Sulem: The Nonlinear Schrödinger

Equation: Self-Focusing and Wave Collapse.

140. Cherkaev: Variational Methods for Structural

Optimization.

141. Naber: Topology, Geometry, and Gauge Fields:

Interactions.

142. Schmid/Henningson: Stability and Transition in

Shear Flows.

143. Sell/You: Dynamics of Evolutionary Equations.

144. Nédélec: Acoustic and Electromagnetic

Equations: Integral Representations for

Harmonic Problems.

145. Newton: The N-Vortex Problem: Analytical

Techniques.

146. Allaire: Shape Optimization

by the Homogenization Method.

147. Aubert/Kornprobst: Mathematical Problems in

Image Processing: Partial Differential

Equations and the Calculus of Variations.

148. Peyret: Spectral Methods for Incompressible

Viscous Flow.

149. Ikeda/Murota: Imperfect Bifurcation

in Structures and Materials.

150. Skorokhod/Hoppensteadt/Salehi: Random

Perturbation Methods with Applications

in Science and Engineering.

151. Bensoussan/Frehse: Regularity Results for

Nonlinear Elliptic Systems and Applications.

152. Holden/Risebro: Front Tracking for Hyperbolic

Conservation Laws.

153. Osher/Fedkiw: Level Set Methods and Dynamic

Implicit Surfaces.

154. Bluman/Anco: Symmetries and Integration

Methods for Differential Equations.

155. Chalmond: Modeling and Inverse Problems in

Image Analysis.

156. Kielhöfer: Bifurcation Theory: An Introduction

with Applications to PDEs.

157. Kaczynski/Mischaikow/Mrozek: Computational

Homology.

158. Oertel: Prandtl–Essentials of Fluid Mechanics,

10th Revised Edition.

159. Ern: Theory and Practice of Finite Elements.

160. Kaipio: Statistical and Computational Inverse

Problems.

161. Ting: Viscous Vortical Flows II.

162. Ammari/Kang: Polarization and Moment

Tensors: With Applications to Inverse

Problems and Effective Medium Theory.


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Smooth Dynamical Systems.

163. di Bernardo, M., Budd, C.J. (et al.): Piece-wise