Elliptic PDE with almost-real coefficients

Elliptic Partial Differential Equations with

Almost-Real Coefficients

Ariel Barton

Author address:

Department of Mathematics, University of Chicago, 5734 S. Uni-versity Ave., Chicago, IL 60637

Current address: School of Mathematics, University of Minnesota, VincentHall, 206 Church St. SE, Minneapolis, MN 55455-0488

E-mail address: [email protected]

Contents

Chapter 1. Introduction 11.1. History 3

Chapter 2. Definitions and the Main Theorem 92.1. Geometric definitions 92.2. Definitions of function spaces 112.3. Layer potentials 132.4. The main theorem 142.5. Additional definitions 16

Chapter 3. Useful Theorems 213.1. Nontangential maximal functions 213.2. Bounds on solutions 233.3. Existence results 253.4. Preliminary uniqueness results 273.5. The Neumann and regularity problems in unusual domains 28

Chapter 4. The Fundamental Solution 334.1. A fundamental solution exists 334.2. Uniqueness of the fundamental solution 354.3. Symmetry of the fundamental solution 364.4. Conjugates to the fundamental solution 384.5. Calderon-Zygmund kernels 394.6. Analyticity 39

Chapter 5. Properties of Layer Potentials 435.1. Limits of layer potentials and the adjoint formulas 44

Chapter 6. Boundedness of Layer Potentials 496.1. Proof for a small Lipschitz constant: preliminary remarks 496.2. A B1 for the TB theorem 516.3. Weak boundedness of operators 546.4. The adjoint inequalities 566.5. Proof for a small Lipschitz constant: final remarks 636.6. Buildup to arbitrary special Lipschitz domains 646.7. Patching: special Lipschitz domains to bounded Lipschitz domains 67

Chapter 7. Invertibility of Layer Potentials and Other Properties 697.1. Nontangential maximal functions of layer potentials 697.2. Jump relations 737.3. Layer potentials on H1(∂V ) 76

v

vi CONTENTS

7.4. Invertibility of layer potentials on Lp(∂V ) 79

Chapter 8. Uniqueness of Solutions 838.1. Counterexamples to uniqueness 838.2. Uniqueness results 84

Chapter 9. Boundary Data in H1(∂V ) 899.1. Solutions with boundary data in H1 899.2. Invertibility of layer potentials on H1(∂V ) 95

Chapter 10. Concluding Remarks 9710.1. Converses 98

Bibliography 105

Abstract

In this monograph we investigate divergence-form elliptic partial differentialequations in two-dimensional Lipschitz domains whose coefficient matrices havesmall (but possibly nonzero) imaginary parts and depend only on one of the twocoordinates.

We show that for such operators, the Dirichlet problem with boundary datain Lq can be solved for q < ∞ large enough. We also show that the Neumannand regularity problems with boundary data in Lp can be solved for p > 1 smallenough, and provide an endpoint result at p = 1.

2010 Mathematics Subject Classification. Primary 35J25; Secondary 31A25.

vii

CHAPTER 1

Introduction

In this monograph, we consider solutions to boundary value problems for thesecond-order divergence form partial differential equation

divA(X)∇u(X) = 0.

The matrix of coefficients A is taken to be measurable; we do not assume that A isdifferentiable. Thus, the solutions u lie in the Sobolev space W 1,2

loc of functions withone weak derivative, and the equation divA∇u = 0 must be interpreted weakly. IfV is an open set, we say that divA∇u = 0 in V if

(1.1)

ˆV

∇η ·A∇u = 0 for all η ∈ C∞0 (V ).

We always assume that the coefficient matrix A is bounded and elliptic, thatis, there exist some constants Λ > λ > 0 such that

(1.2) λ|η|2 ≤ Re η ·A(X)η, |ξ ·A(X)η| ≤ Λ|η||ξ|for every X ∈ Rn and every ξ, η ∈ Cn. In this monograph, we will prove results inthe special case of two dimensions (V ⊂ R2), and of coefficients A(x, t) independentof one of the two coordinates. Under these conditions, solutions u are locally Holdercontinuous and their gradients are locally bounded.

We consider three boundary-value problems. If 1 < q < ∞, then we saythat the Dirichlet problem with boundary data in Lq(∂V ), or (D)Aq , holds in thedomain V with constant C, if for every f ∈ Lq(∂V ), there exists a unique function

u ∈W 1,2loc (V ) such that

(D)Aq

divA∇u = 0 in V,

u = f on ∂V,

‖Nu‖Lq(∂V ) ≤ C‖f‖Lq(∂V ).

The bottom two lines are interpreted as follows. If X ∈ ∂V , let

γV,a(X) = Y ∈ U : |X − Y | < (1 + a) dist(Y, ∂V )(1.3)

for some fixed positive number a. We say that u = f on ∂V if f is the nontangentiallimit of u; that is, limY→X, Y ∈γV,a(X) u(Y ) = f(X) for almost every X ∈ ∂V . Thenontangential maximal function Nu is defined by

Nu(X) = NU,au(X) = ess sup|u(Y )| : Y ∈ γU,a(X).(1.4)

If there exists a number C > 0 such that (D)Aq holds in V with constant C, we

simply say that (D)Aq holds in V .

Similarly, if 1 < p <∞, we say that the Neumann problem (N)Ap or regularity

problem (R)Ap holds in V with constant C if, for every g ∈ Lp(∂V )∩H1(∂V ), there

1

2 1. INTRODUCTION

is a unique function u such that

(N)Ap

divA∇u = 0 in V,

ν ·A∇u = g on ∂V,

‖N(∇u)‖Lp(∂V ) ≤ C‖g‖Lp(∂V )

or (R)Ap

divA∇u = 0 in V,

∂τu = g on ∂V,

‖N(∇u)‖Lp(∂V ) ≤ C‖g‖Lp(∂V ).

If V C is bounded we additionally require that lim|X|→∞ u(X) exist. Here H1(∂V )denotes the atomic Hardy space of harmonic analysis, and ν and τ are the unitoutward normal and unit tangent vectors to the domain V . We say that ν ·A∇u = gweakly if

(1.5)

ˆV

A∇u · ∇η =

ˆ∂V

gTr η dσ for all η ∈ C∞0 (V ).

If f is the nontangential limit of u, then ∂τu = g if g is the derivative of f along ∂V .This monograph has two main results. The first main result is that under

certain conditions, the boundary value problems above hold in V .

Theorem 1.6. Suppose that A0 and A are bounded, elliptic matrices definedon R2 which depend only on one of the two coordinates. Assume that A0 is real-valued; A may be complex-valued. Let V be a Lipschitz domain in R2 which hasconnected boundary.

Then there exist some constants ε > 0 and p0 > 1, depending only on V andA0, such that if ‖A−A0‖L∞ < ε, 1 < p ≤ p0, and 1/p+1/q = 1, then (D)Aq , (N)Ap ,

and (R)Ap hold in V .

Our second main result is an endpoint result for the Neumann and regularityproblems. We say that (N)A1 or (R)A1 holds in V with constant C if, for everyg ∈ H1(∂V ), there exists a unique function u defined in V such that

(N)A1

divA∇u = 0 in V,

ν ·A∇u = g on ∂V,

‖N(∇u)‖L1(∂V ) ≤ C‖g‖H1(∂V ),

or (R)A1

divA∇u = 0 in V,

∂τu = g on ∂V,

‖N(∇u)‖L1(∂V ) ≤ C‖g‖H1(∂V ).

That is, we consider only boundary data g in H1(∂V ) ( L1(∂V ).

Theorem 1.7. Let A0, A, and V be as in Theorem 1.6. There is some ε > 0depending only on V and A0 such that if ‖A − A0‖L∞ < ε, then (N)A1 and (R)A1hold in V .

Conversely, if V is simply connected, divA∇u = 0 in V and N(∇u) ∈ L1(∂V ),then the boundary values ν · A∇u and ∂τu exist and lie in H1(∂V ). Further-more, there is a constant C such that ‖ν · A∇u‖H1(∂V ) ≤ C‖N(∇u)‖L1(∂V ) and‖∂τu‖H1(∂V ) ≤ C‖N(∇u)‖L1(∂V ).

If A = A0 is real-valued, then the conclusions of Theorem 1.6 are known tohold; the conclusion regarding (D)A0

q was proven in [KKPT00] by Kenig, Koch,

Pipher and Toro, and the conclusions for (N)A0p and (R)A0

p were proven in [KR09]and [Rul07] by Kenig and Rule. The conclusions of Theorem 1.7 were proven in[DK87] by Dahlberg and Kenig in the case of harmonic functions; the conclusions(N)A1 and (R)A1 were proven in [KP93] by Kenig and Pipher under the conditionsthat A is real symmetric and (N)Ap and (R)Ap hold for some p > 1.

The organization of this monograph is as follows. The main results were statedabove in Theorem 1.6 and Theorem 1.7. We will conclude this chapter by reviewing

1.1. HISTORY 3

the history of boundary value problems in Lipschitz domains with Lp(∂V ) boundarydata. In Chapter 2, we define our terminology. The proof of our main results isby the classic method of layer potentials; for the reader’s convenience, we providean outline of this method in Section 2.4. In Chapter 3, we provide a number ofuseful theorems regarding nontangential maximal functions and solutions to partialdifferential equations.

We work out the details of the method of layer potentials in Chapters 4–7. InChapter 4, we construct a fundamental solution for the operator divA∇. In Chap-ter 5, we use this fundamental solution to construct layer potentials. In Chapter 6,we show that layer potentials are bounded operators. In Chapter 7, we prove someconsequences of boundedness, including a perturbative invertibility result. Theresults of these chapters let us prove that solutions to (D)Aq0 , (N)Ap0

and (R)Ap0exist.

We prove uniqueness of solutions in Chapter 8. In Chapter 9, we use existenceand uniqueness of solutions to (N)Ap0

and (R)Ap0to prove existence and uniqueness

of solutions to (N)A1 and (R)A1 . We may then interpolate to prove that (D)Aq , (N)Apand (R)Ap hold if 1 < p < p0.

Most of the results of Chapters 4–9 assume that the coefficient matrices A aresmooth. In Chapter 10, we pass from smooth coefficients to bounded measurablecoefficients. We also prove the converses mentioned in Theorem 1.7.

This monograph is a revision of my thesis written at the University of Chicago.My advisor was Carlos E. Kenig, to whom I would like to extend my grateful thanks;without his guidance and advice, the work here would not have been possible.

1.1. History

The study of second-order elliptic boundary value problems in Lipschitz do-mains has a long and rich history. The study began with harmonic functions, thatis, with solutions u to (1.1) where A ≡ I, the identity matrix. Many of the resultsof this study can be extended to more general problems under some conditions onthe coefficients A. This monograph concerns one particular such condition, namelythat the coefficients A be independent of some specified coordinate.

In this section, we begin by discussing some known results for harmonic func-tions. We then discuss how these results have been extended under various otherassumptions on A, before focusing on the study of coefficients independent of onecoordinate. We conclude this section by briefly reviewing the history of boundaryvalue problems with data in Hardy spaces.

In [Dah77] and [Dah79], Dahlberg showed that if V is a Lipschitz domain,then there is some ε > 0 depending on V such that if 2 − ε < q < ∞, then (D)Iqholds in V . This range is sharp in the sense that given q < 2, there is some Lipschitzdomain V such that (D)Iq does not hold in V ; see [FJL77].

In [JK81b], Jerison and Kenig showed that (N)Ip and (R)Ip hold in Lipschitzdomains in all dimensions provided p = 2. This was extended to the case 1 <p < 2 + ε, ε again depending on the domain, in [Ver84] (the regularity problem)and in [DK87] (the Neumann problem in three or more dimensions). It had beenobserved by Kenig and Fabes that (N)Ip holds, for p in this range, in Lipschitz

domains contained in R2. The same results hold if A is an arbitrary constant-coefficient elliptic matrix; in [She06] and [She07], Shen has proven similar resultsfor systems of constant-coefficient elliptic operators.

4 1. INTRODUCTION

One of the classic tools for studying harmonic functions in smooth domainsis the method of layer potentials. In [Ver84], Verchota showed that the solutionsto (D)Iq , (N)Ip and (R)Ip in a Lipschitz domain V could be constructed using layerpotentials; his construction used the celebrated result of Coifman, McIntosh andMeyer [CMM82] that the Cauchy integral on a Lipschitz curve is a bounded op-erator. The layer potential construction is very useful, as it is often easier to provetheorems about layer potentials than about arbitrary harmonic functions. Most ofthe theorems in this monograph are proven using layer potentials as well.

In order to pass to more general coefficients A, some conditions must be im-posed; solutions to (1.1) for arbitrary coefficients A can be very badly behaved. In[CFK81], Caffarelli, Fabes and Kenig provided an example of such behavior. Theyconstructed coefficient matrices A, defined in the unit ball in Rn, such that the L-harmonic measure associated to L = divA∇ is completely singular with respect toarc length on the unit sphere. Thus, (D)Aq does not hold in the unit ball for any1 < q <∞. The constructed matrices A, in addition to being bounded and elliptic,were real, symmetric, and continuous up to the boundary of the unit ball.

Some results can be proven under the assumption that A is more regular. In[JK81a], Jerison and Kenig showed that (D)Aq holds in Lipschitz domains V , for2 − ε < q < ∞, provided A is smooth. In [FJK84], Fabes, Jerison and Kenigsolved the Dirichlet problem for continuous real symmetric coefficients under someassumptions on their moduli of continuity. In [KP01], [DPP07], and [DR10],boundary value problems have been investigated for coefficients A that are regularin the sense that the gradient or oscillations of A satisfy a certain Carleson-measurebound.

Many results hold under other assumptions on A. One important case is thatof Carleson-measure perturbation. In [Dah86], Dahlberg showed that if A andA0 are real symmetric, (D)A0

q holds in a Lipschitz domain and A0 − A satisfies a

Carleson-measure estimate then (D)Aq must also hold in that domain.Weaker Carleson conditions were investigated for the Dirichlet problem by R.

Fefferman, Kenig and Pipher in [Fef89], [FKP91] and [Fef93], and Carleson-measure perturbations were investigated for the Neumann and regularity problemsby Kenig and Pipher in [KP93] and [KP95]. Recently in [AA11], [AR11] and[HM], Auscher, Axelsson, Hofmann and Mayboroda have investigated Carleson-measure perturbations for complex nonsymmetric coefficients.

We remark that the Neumann and regularity problems investigated in [KP93]are not precisely those of Theorem 1.6. Specifically, the nontangential maximalfunction N must be replaced by a modified nontangential maximal function N ; werefer the reader to [KP93] for a precise definition of N .

A third important field of investigation, which encompasses the results of thismonograph, is the study of coefficients independent of the radial coordinate or of aspecified Cartesian coordinate. Such coefficients have been used as the unperturbedmatrices A0 of Carleson-measure perturbation theory; see in particular [KP93].They are also brought to our attention by considering changes of variables. LetΩ ⊂ Rn+1 be the domain above the graph of a Lipschitz function ϕ : Rn 7→ R.The simple change of variables (x, t) 7→ (x, t−ϕ(x)) transforms Ω to the upper half-space. If u is harmonic in Ω, then the function v given by u(x, t) = v(x, t − ϕ(x))is a solution to an elliptic partial differential equation in the upper half-space.(See Figure 1 for an illustration of this change of variables in two dimensions.)

1.1. HISTORY 5

t = ϕ(x)

(x, t) 7→ (x, t− ϕ(x))

∆u = 0div

(1 −ϕ′(x)

−ϕ′(x) 1 + ϕ′(x)2

)∇v = 0

Figure 1. Change of variables to straighten the boundary of adomain in R2

The coefficients of this equation are real, symmetric, bounded, elliptic, and aret-independent, that is, do not depend on the t-coordinate. A similar change ofvariables, this one mapping starlike Lipschitz domains to the unit ball, generatesreal symmetric radially independent coefficients.

We now review the history associated to such coefficients.Soon after the publication of [JK81a], it was observed that the methods of that

paper only required smoothness in a direction transverse to the boundary. Thus,if V ⊂ Rn denotes a domain star-like with respect to the origin, then (D)Aq holdsfor 2− ε < q <∞ provided A is real, symmetric and smooth in the radial variable.In [KP93] Kenig and Pipher proved that (N)Ap and (R)Ap hold for such coefficientsprovided 1 < p ≤ 2 + ε.

These results use techniques, in particular the Rellich identity, that require thatthe coefficients be real and symmetric. Real nonsymmetric coefficients or complexcoefficients are less well understood in general, but a few other special cases havebeen studied.

In the special case where A is a block matrix independent of the nth coordinate,the conditions (D)A2 , (N)A2 and (R)A2 hold in the upper half-space Rn

+. Here A =(aij)ni,j=1

is a block matrix if ain = anj = 0 for i, j < n. The condition (D)A2 is a

consequence of the semigroup theory, while the conditions (N)A2 and (R)A2 followfrom Kato’s square root conjecture for elliptic operators (proven in [AHL+02]).See [Ken94, Remark 2.5.6] for a discussion of the Kato problem and its connectionto elliptic partial differential equations.

More results are known in the case of two dimensions. In [AT95], Auscher andTchamitchian studied two-dimensional block matrices (that is, diagonal matrices).They showed that for such coefficients, (D)Ap , (N)Ap and (R)Ap can be solved in theupper half-plane for any p with 1 < p <∞.

In [KKPT00], Kenig, Koch, Pipher and Toro proved that if V ⊂ R2 is aLipschitz domain, and A is real, t-independent but not necessarily symmetric, thenthere is some (possibly large) q0 < ∞ such that (D)Aq holds in V for every q0 <q < ∞. In [KR09] and [Rul07], Kenig and Rule showed that under the same

assumptions, if (D)Aq holds in V then so do (N)A/ detAp and (R)A

t

p ; thus, if A is real,

elliptic and t-independent then there is some p0 > 1 such that (N)Ap and (R)Ap holdin V for all 1 < p < p0.

These results concern t-independent coefficients in bounded Lipschitz domains;that is, the coefficients are constant along a direction not necessarily transverse tothe boundary. The authors observed that these results produce the optimal range

6 1. INTRODUCTION

of exponents, in the sense that for any given q < ∞ or p > 1, there is a realnonsymmetric coefficient matrix A such that (D)Aq , (N)Ap , and (R)Ap do not hold in

the upper half-plane R2+.

This monograph proves the same results under the assumption that A has asmall imaginary part. The proofs in [KKPT00] use harmonic measure techniquesand results concerning positive solutions to elliptic equations, such as Harnack’s in-equality and the comparison principle, which are unavailable if A is complex. Thus,our proofs must proceed by a different technique, the method of layer potentials;this technique is particularly suited to perturbative results.

Boundary value problems for t-independent coefficients have been investigatedin higher dimensions. Many of the techniques of [KKPT00] and [KR09] areunavailable in higher dimensions, and at present there is no known analogue to theirresults for real nonsymmetric coefficients. However, there are important knownresults, many involving L∞ perturbations.

In [FJK84], Fabes, Jerison and Kenig showed that if A0 is a constant (pos-sibly complex) matrix and ‖A − A0‖L∞ is small enough, for some t-independentmatrix A defined on Rn, then the Dirichlet problem (D)A2 can be solved in the up-per half-space Rn

+. In [AAA+11], a more general result was proven. The authors

showed that if solutions to (D)A02 , (N)A0

2 , and (R)A02 in the upper half-space can

be constructed using the method of layer potentials, and if ‖A − A0‖L∞ is smallenough, then solutions to (D)A2 , (N)A2 , and (R)A2 can be constructed in the sameway. The authors showed that constant or real symmetric matrices A0 satisfy theirassumptions; this included the first explicit proof that (D)A0

2 , (N)A02 , and (R)A0

2

can be solved for real symmetric matrices using the method of layer potentials.A different method has been used recently to analyze such problems. The

second-order differential equation divA∇u = 0 may be translated into a first-ordersystem and analyzed using semigroups. In [AAH08] this method was used toshow that (D)A2 , (N)A2 and (R)A2 can be solved in the upper half-plane providedA is a small, t-independent perturbation of a constant, real symmetric, or blockmatrix, without assumption on the layer potentials for A. In [AAM08], the method

was used to show that if (D)A02 , (N)A0

2 or (R)A02 holds in the upper half-plane,

then the corresponding problem holds for all A with ‖A − A0‖L∞ small enough;the only underlying assumption was t-independence of the coefficients A and A0.These methods apply to elliptic systems as well as elliptic equations. The results in[AA11] and [AR11] concerning Carleson-measure perturbations, mentioned above,were also obtained using this method.

The semigroup analysis of [AAH08], [AAM08], [AA11] and [AR11] relies onthe functional calculus of Hilbert space operators; at present these techniques havebeen used only for boundary data in the spaces L2(∂V ) or W 1,2(∂V ). [FJK84]and [AAA+11] also investigated boundary-value problems with boundary data inL2(∂V ) or W 1,2(∂V ).

The techniques of this monograph require only that (N)A0p and (R)A0

p hold forsome p > 1, not necessarily p = 2. However, we use many of the two-dimensionaltechniques of [KKPT00] and [KR09], and so our results cannot be easily gen-eralized to higher dimensions. In particular, we use a change of variables from[KKPT00]. The proofs in [KKPT00] and [KR09] used this change of variablesto transform their real coefficient matrices to upper triangular matrices. In Sec-tion 6.5 we use the same change of variables to transform A to a matrix whose real

1.1. HISTORY 7

part is upper triangular. We will use the fact that if ImA is small, then the trans-formed matrix is close (in L∞) to an upper triangular matrix. This requirement isthe reason why the matrix A0 of Theorem 1.6 must be real.

We now review the history of boundary value problems with data in the Hardyspace H1(∂V ). In [SW60], Stein and Weiss studied functions u harmonic in theupper half-space Rn+1

+ that satisfied N(∇u) ∈ Lp(∂Rn+1+ ) for some p > 0. In

[FS72], C. Fefferman and Stein defined Hp(Rn) to be the set of normal derivativeson ∂Rn+1

+ of such functions u. Thus, (N)I1 holds in Rn+1+ by definition.

There exist many equivalent characterizations of the space H1(Rn). One suchcharacterization of H1(Rn) (see [FS72]) is as the dual of the space BMO of func-tions of bounded mean oscillation. BMO and its dual may be easily generalizedto an arbitrary rectifiable curve (e.g., the boundary of a Lipschitz domain). Thisdefines the Hardy space H1(∂V ) of (N)A1 and (R)A1 above.

In [FK81], Fabes and Kenig showed that (N)I1 holds in all C1 domains V .In [DK87], Dahlberg and Kenig showed that (N)I1 and (R)I1 hold in all boundedLipschitz domains. Recall that by [JK81b], (N)I2 holds in Lipschitz domains. Itis possible to interpolate between H1 and Lp0 for any p0 > 1; Dahlberg and Kenigused their result and interpolation to show that (N)Ip holds in bounded Lipschitzdomains for all 1 < p ≤ 2. In [KP93], Kenig and Pipher showed that if A is realsymmetric, then (N)Ap0

and (R)Ap0imply (N)A1 and (R)A1 ; they used this result to

show that (N)Ap0and (R)Ap0

imply (N)Ap and (R)Ap for any 1 < p < p0.

In the present paper, we show that if A is t-independent and defined on R2,then (N)Ap0

and (R)Ap0imply (N)A1 and (R)A1 . As in the papers above, we may then

interpolate between (N)A1 and (N)Ap0, or (R)A1 and (R)Ap0

.

Other equivalent definitions of H1(Rn) may be generalized to yield other Hardyspaces. We mention one particular generalization, not considered in this paper butdirectly related to elliptic partial differential equations and studied extensively inthe literature. By [FS72], if f ∈ Hp(Rn) then f is the trace of a harmonic functionu with Nu ∈ Lp(∂Rn+1

+ ). Thus, H1A,D may be defined as

H1A,D(∂V ) = Tru : divA∇u = 0 in V, Nu ∈ L1(∂V ).

So H1I,D(∂Rn+1

+ ) = H1(Rn). In [AT95], it was shown that H1A,D(∂R2

+) = H1(R),provided the coefficients A are diagonal and t-independent. In other words, for suchcoefficients, the condition (D)A1 defined analogously to (N)A1 holds.

We remark that by interpolation, if H1A,D(∂V ) = H1(∂V ) and (D)Aq0 holds in

V then (D)Aq holds in V for any 1 < q < q0. Thus, H1A,D(∂V ) does not equal

H1(∂V ) even in the case where A ≡ I and V is a general Lipschitz domain. How-ever, the spaces H1

A,D are interesting in their own right, and have been studied

in many papers, including [FKN81], [JK82], [KP87], [AR03], [DY05], [HM09]and [HMM11].

CHAPTER 2

Definitions and the Main Theorem

In this chapter, we precisely define a number of geometric concepts, the functionspaces we intend to consider, and the layer potentials that solve certain boundary-value problems. Once we have done this, we will provide an outline of the proofof this monograph’s main results; we will conclude the chapter with some extradefinitions.

Recall from the introduction that A is an elliptic matrix if A is a matrix-valuedfunction defined on Rn that satisfies (1.2). In this monograph, we will restrict ourattention to n = 2 and to t-independent matrices; that is, we require A to be acomplex matrix-valued function defined on R2 which satisfies

(2.1) λ|η|2 ≤ Re η ·A(x, t)η, |ξ ·A(x, t)η| ≤ Λ|η||ξ|, A(x, t) = A(x, s)

for all x, t, s ∈ R and all η, ξ ∈ C2. We refer to the numbers λ and Λ of (1.2) or(2.1) as the ellipticity constants of A.

Throughout this monograph, the letter C will represent a positive constant,whose value may change from line to line. Unless otherwise specified, such con-stants are assumed to depend only on a few parameters. These parameters are theconstant a in the definition (1.3) of nontangential cone, the ellipticity constantsof any relevant coefficient matrices A, and the Lipschitz character of any relevantdomains. (See Definition 2.3 for a definition of Lipschitz character.)

We will use the symbol ≈ to indicate that two quantities are comparable up toa multiplicative constant; that is, a ≈ b if 1

C |b| ≤ |a| ≤ C|b|.

2.1. Geometric definitions

Let U ⊂ R2 be a domain. We define U+ = U , U− = UC = R2 \ U .If e ∈ C2 is a vector, we let a superscript ⊥ denote the perpendicular vector

e⊥ =

(0 1−1 0

)e.

If ∂U is rectifiable, we let σ denote surface measure on ∂U ; this is the only measurewe will use on ∂U . We let ν represent the unit outward normal to ∂U , and letτ = ν⊥ be the unit tangent vector to ∂U .

Recall the definition (1.3) of the nontangential cone γU,a(X) for X ∈ ∂U . Theexact value of a is usually unimportant; see Lemma 3.2. When no ambiguity willarise we suppress the subscripts U or a. We let γ±(X) = γU±(X).

We remark that if ∂U is bounded, X ∈ ∂U and Y ∈ R2, then dist(Y, ∂U) ≥|X−Y |−diam(∂U). Suppose that |X−Y | > (1+1/a) diam(U). Then dist(Y, ∂U) ≥|X − Y |/(1 + a), and so Y ∈ γ+(X) ∪ γ−(X). Put another way, if UC is boundedthen so is R2 \ γ(X) for any X ∈ ∂U .

9

10 2. DEFINITIONS AND THE MAIN THEOREM

Recall from (1.4) that if u is a function defined in U , the nontangential maximalfunction Nu of u is given by N(X) = NU,au(X) = ess sup|u(Y )| : Y ∈ γU,a(X).If ∂U is rectifiable and u is defined in U , we say that f is the nontangential limitof u if

(2.2) limη→0+

sup|u(Y )− f(X)| : Y ∈ γU,a(X), |X − Y | < η = 0

for a.e. X ∈ ∂U . We often write u = f on ∂U to indicate that f is the nontangentiallimit of u. If u = f on ∂U and f ∈W 1,1

loc (U), we will frequently write τ · ∇u for thetangential derivative ∂τf .

Suppose that the nontangential limit of u exists in Lp(∂U), and u ∈ W 1,p(U)and so the trace Tru exists. It was shown in [BLRR10, Section 5.4] that if Uis smooth then Tru is equal in Lp(∂U) to the nontangential limit of u. We cangeneralize this result to Lipschitz domains by changing variables to straighten theboundary in a small neighborhood.

In this monograph, we will work exclusively in Lipschitz domains, which aredefined as follows.

Definition 2.3. We say that the domain Ω is a special Lipschitz domain if,for some Lipschitz function ϕ and unit vector e,

Ω = X ∈ R2 : ϕ(X · e⊥) < X · e.We refer to M = ‖ϕ′‖L∞(R) as the Lipschitz constant of Ω.

Suppose V ⊂ R2 is a domain. We say that V is a Lipschitz domain if V is aspecial Lipschitz domain, or if ∂V may be covered by finitely many balls Bj suchthat V coincides with a special Lipschitz domain in each ball.

More precisely, let XjNj=1 ⊂ R2 be a set of points in the plane, let rjNj=1 be

a set of positive real numbers, let ejNj=1 be a set of unit vectors, and let ϕjNj=1

be a set of Lipschitz functions with ϕj(0) = 0 and max‖ϕ′j‖L∞(R)Nj=1 ≤M . Let

Ωj = X ∈ R2 : ϕj((X −Xj) · e⊥j ) < (X −Xj) · ej,Rj = X ∈ R2 : |(X −Xj) · e⊥j | < 2rj , |(X −Xj) · ej | < (2 + 2M)rj.

We say that V is a Lipschitz domain if either V is a special Lipschitz domain, or ifwe can find Xj ∈ V , rj , ej , ϕj such that

∂V ⊂N⋃

j=1

B(Xj , rj) and V ∩Rj = Ωj ∩Rj for each 1 ≤ j ≤ N .

If V is a special Lipschitz domain, let N = c0 = 1. Otherwise, let N be asabove, and let c0 = maxj rj/minj rj .

We refer to M , N , c0 as the Lipschitz constants or Lipschitz character of V .

We will reserve Ω for special Lipschitz domains, and V for general Lipschitzdomains. We require M , N , c0 < ∞. This means that every Lipschitz domainV which is not special has compact boundary, and so must be bounded or havebounded complement, and that every connected component of ∂V has surface mea-sure at least σ(∂V )/C. We will usually restrict our attention to Lipschitz domainswith connected boundary; that is, to special Lipschitz domains, simply connectedbounded domains, and domains with simply connected bounded complements.

2.2. DEFINITIONS OF FUNCTION SPACES 11

X

χ+(X, r)

χ−(X, r)r (1 + k1)r

∆(X, r)

e⊥j

ejQ(X, r)

Figure 1. Tents on the boundary of a Lipschitz domain

We remark that if V is a Lipschitz domain, X ∈ ∂V and r > 0, then

(2.4) σ(B(X, r) ∩ ∂V ) ≤ 2N√

1 +M2r.

That is, Lipschitz domains are Ahlfors regular.In analyzing functions in the upper half-plane R2

+, it is often useful to considerB(X, r) ∩ R2

+ for some X ∈ ∂R2+. If V is a Lipschitz domain, B(X, r) ∩ V may

be very badly behaved. We instead work with tents Q(X, r) defined as follows. IfX ∈ ∂V , then X ∈ Bj = B(Xj , rj) for one of the balls Bj of Definition 2.3. Letej , e⊥j , ϕj be the unit vectors and Lipschitz function associated with the specialLipschitz domain Ωj . Then for any 0 < r < rj ,

Q(X, r) = Y ∈ R2 : |(X − Y ) · e⊥j | < r,(2.5)

ϕj(Y · e⊥j ) < Y · ej < ϕj(Y · e⊥j ) + (1 +M)r.

(See Figure 1.) We let ∆(X, r) = ∂Q(X, r) ∩ ∂V . Then Q(X, r) is a simply con-nected, bounded Lipschitz domain whose Lipschitz constants depend only on M ,which contains V ∩ B(X, r) and is contained in B(X,Cr), and which satisfiesσ(∂Q(X, r)) ≤ Cr.

Let χ±(X, r) = X ± re⊥j + ϕj(X · e⊥j ± r)ej be the two endpoints of ∆(X, r).Then for a large enough (depending on M), ∂Q(X, r) \ ∂V ⊂ γa(χ+) ∪ γa(χ−).

It should be noted that Q(X, r) depends on our choice of Ωj , ej , and also thatif V is not a special Lipschitz domain, then Q(X, r) is defined only for r/σ(∂V )sufficiently small. These technicalities will not matter to our applications.

2.2. Definitions of function spaces

Throughout this monograph, we will reserve the letters p and q for the expo-nents of Lp-spaces. We will always let p and q be conjugate exponents given by1/p+1/q = 1; if multiple such exponents are needed, we will distinguish them withsubscripts or accents.


Recall that the norm in such spaces is defined by ‖f‖Lp(E) =(É|f |p dµ

)1/p.

If E ⊆ Rn is open, the measure µ is taken to be Lebesgue measure; if E ⊂ ∂U forsome domain U , the measure µ is taken to be the surface measure σ.

The inner product between Lp(E) and Lq(E) will be given by

〈G,F 〉 =

Ê

G(x)tF (x) dµ(x).

This is more convenient than the usual inner productÉGt(x)F (x) dµ(x). A super-

script of t will denote the transpose of a matrix or the adjoint of an operator withrespect to this inner product; so if P is an operator, then 〈F, PG〉 = 〈G,P tF 〉t.

If µ is a measure on a set E with µ(E) < ∞, and if f is a µ-measurablefunction defined on E, then we let

fflEf dµ = 1

µ(E)

Éf dµ be the average integral

of f over E.Recall the Hardy-Littlewood maximal function Mf(x) = supr>0

fflB(x,r)

|f | of

functions f defined on Rn. If U ⊂ R2 is a domain with rectifiable boundary, wemay generalize this maximal function to functions defined on ∂U by

Mf(X) = supr>0

B(X,r)∩∂U

|f | dσ.

We remark that if ∆ ⊂ ∂U is connected and ∂U is Ahlfors regular, thenffl

∆|f | dσ ≤

CMf(X) for any X ∈ ∆.The space H1(R), as defined in [SW60] and [FS72], has an atomic decom-

position. That is, if f is in H1(R), then f =∑k λkak, where λk ∈ C,

ák = 0,

‖ak‖L∞(R) ≤ 1/rk, and supp ak ⊂ B(xk, rk) for some xk ∈ R, rk > 0. Further-more,

∑k|λk| ≈ ‖f‖H1 . Functions a satisfying these conditions are called atoms.

See [Ste93, Section III.2] for a nice proof of this decomposition.We may extend the definition of H1 to H1(∂V ), where V is a Lipschitz domain.

We say that f ∈ H1(∂V ) if f =∑k λkak, where the λk are complex numbers, and´

∂Vak dσ = 0, supp ak ⊂ ∆k for some ∆k ⊂ ∂V connected, and ‖ak‖L∞(∂V ) ≤

1/σ(∆k). The norm is the smallest∑k|λk| among all such representations of f .

If V is a Lipschitz domain with connected boundary, then this is equivalent todefining H1 atoms to be functions a which satisfy

´∂V

a dσ = 0, supp a ⊂ B(X, r)∩∂V and ‖a‖L∞(∂V ) ≤ 1/r for some X ∈ ∂V and some r > 0.

If 1 < p < ∞, we let Lp0(∂V ) = H1(∂V ) ∩ Lp(∂V ), regarded as a subspaceof Lp(∂V ). If ∆ is a bounded connected set, g is supported on ∆,

´∆g = 0 and

1 < p ≤ ∞, then

(2.6) ‖g‖H1 ≤ C‖g‖Lp(∆)σ(∆)1/q.

See [Ste93, Section III.5.7] for a proof. Thus, if ∂V is bounded then Lp0(∂V ) ismerely the set of functions in Lp(∂V ) which integrate to zero on each connectedcomponent of ∂V . Conversely, if ∂V is unbounded and 1 < p < ∞ then Lp0(∂V )is dense in Lp(∂V ). We remark that Lp0(∂V ) is dense in H1(∂V ) for any Lipschitzdomain V .

We consider BMO(∂V ) to be the dual of H1(∂V ). This means that

‖f‖BMO(∂V ) = sup∆⊂∂V connected

1

σ(∆)

ˆ∆

|f −ffl

∆f | dσ.

2.3. LAYER POTENTIALS 13

2.3. Layer potentials

A method for constructing solutions to partial differential equations is themethod of layer potentials. In this section, we define the layer potentials appli-cable to our problems.

Lemma 2.7. Let A satisfy (2.1). Then, for each X ∈ R2, there is a functionΓX = ΓAX , unique up to an additive constant, such that for every Y ∈ R2,

|∇ΓX(Y )| ≤ C

|X − Y |

and for every η ∈ C∞0 (R2),

(2.8)

ˆR2

A(Y )∇ΓX(Y ) · ∇η(Y ) dy = −η(X).

We refer to this function as the fundamental solution for divA∇ with pole at X.This lemma will be proven in Chapter 4. By ∇ΓX(Y ) we mean the gradient in Y .We will sometimes wish to refer to the gradient in X; we will then write ∇XΓX(Y ).

If a function or operator is defined in terms of the coefficient matrix A, thena superscript of T will denote the corresponding function or operator defined in

terms of its transpose At. (So At = AT , and ΓTX(Y ) = ΓAT

X (Y ) is the fundamentalsolution for divAt∇ with pole at X.)

Let V be a Lipschitz domain. If f : ∂V 7→ C is a function, and X ∈ R2 \ ∂V ,we define the layer potentials by

Df(X) = DAV f(X) =

ˆ∂V

ν(Y ) ·AT (Y )∇ΓTX(Y )f(Y ) dσ(Y ),(2.9)

∇Sf(X) = ∇SAV f(X) =

ˆ∂V

∇XΓTX(Y )f(Y ) dσ(Y ).(2.10)

This defines Sf up to an additive constant on each connected component of R2 \∂V . These integrals converge under reasonable assumptions on f ; see Lemma 5.1.Under somewhat more restrictive assumptions, the integral

´∂V

ΓTX(Y )f(Y ) dσ(Y )converges for X /∈ ∂V ; in such cases, we let

Sf(X) =

ˆ∂V

ΓTX(Y )f(Y ) dσ(Y ).

If X ∈ ∂V , we define the boundary layer potentials K, L via

KAV f(X) = limZ→X, Z∈γ(X)

ˆ∂V

ν(Y ) ·AT (Y )∇ΓTZ(Y )f(Y ) dσ(Y ),(2.11)

K±f(X) = ±KAV±f(X) = limZ→X, Z∈γ±(X)

DAV f(X),(2.12)

Lf(X) = LAV f(X) = limZ→X, Z∈γ(X)

ˆ∂V

τ(Y ) · ∇ΓTZ(Y )f(Y ) dσ(Y ).(2.13)

When no confusion will arise we omit the subscripts and superscripts. If f ∈Lp(∂V ), 1 < p < ∞, then the limits above exist for a.e. X ∈ ∂V ; see Lemma 5.7and Corollary 7.3.


2.4. The main theorem

We are now in a position to outline the proofs of Theorem 1.6 and Theorem 1.7,the main results of this monograph. The proof will be by the classic method of layerpotentials; for the reader’s convenience, we provide an outline of this method. Theremainder of this monograph will be devoted to resolving the details.

Our results will build directly on two theorems, the first proven by Kenig, Koch,Pipher and Toro, and the second by Kenig and Rule.

Theorem 2.14 ([KKPT00]). Suppose that A0 : R2 7→ R2×2 is real-valued (butnot necessarily symmetric) and satisfies (2.1). Let V be a be a simply connectedLipschitz domain.

Then there is some (possibly large) number q0 < ∞, depending only on theconstants λ, Λ in (2.1) and the Lipschitz character of the domain V , such that ifq0 < q < ∞, then (D)A0

q holds in V with constant C(q) depending only on q andthe quantities mentioned above.

Theorem 2.15 ([KR09] and [Rul07]). Let A0, V be as in Theorem 2.14. Let

1/p + 1/q = 1. If (D)A0q holds in V with constant C(q), then (N)

A0/ detA0p and

(R)At0p hold in V with constant C(p), where C(p) depends only on p, λ, Λ, C(q) and

the Lipschitz character of V .

The theorem that we intend to prove is the following.

Theorem 2.16. Suppose that A0 and A satisfy (2.1). Assume that A0(x) isreal-valued; A(x) may complex-valued. Let V be a Lipschitz domain with connectedboundary.

Then there is some ε > 0, p0 > 1 depending only on λ, Λ and the Lipschitzcharacter of V , such that if ‖A − A0‖L∞ < ε and 1 < p ≤ p0, then (N)Ap , (R)Apand (D)Aq hold in V with constants depending only on p, λ, Λ and the Lipschitzcharacter of V .

Furthermore, (N)A1 and (R)A1 hold in V , again with constants depending onlyon λ, Λ and the Lipschitz character of V .

The converses mentioned in Theorem 1.7 will be proven in Section 10.1.

Outline of the proof. Recall that (D)Aq , (N)Ap and (R)Ap have two condi-tions, that solutions exist, and that solutions be unique.

We first establish the existence of solutions. Let V be a Lipschitz domain withconnected boundary. Choose some f : ∂V 7→ C in Lp(∂V ) or Lp0(∂V ) for some1 < p < ∞. In Lemma 5.1, we will show that if X ∈ R2 \ ∂V , then Df(X) and∇Sf(X) are well-defined complex numbers, and divA∇(Df) = 0, divA∇(Sf) = 0in R2 \ ∂V .

Our candidates for solutions to (D)Aq are the functions Df . Our candidates for

solutions to (N)Ap and (R)Ap are the functions Sf . We must establish bounds onN(Df) and N(∇Sf), and must show that the boundary values Df |∂Ω, ν · A∇Sfand τ · ∇Sf can be made to be any Lp(∂V ) functions we choose.

By definition, KA±f = Df |∂V± . In Lemma 5.8, we will show that

(LA)tf = τ · ∇STf |∂V , (KA±)tf = ∓ν ·AT∇STf |∂V∓in appropriate weak senses. Furthermore, in (5.5), we will show that if ∂V iscompact and f ∈ H1(∂V ), then lim|X|→∞ STf(X) = 0.

2.4. THE MAIN THEOREM 15

The first step is the boundedness of the operators K± and L on Lp(∂V ). InTheorem 6.1, working much as in [KR09], we will find an ε0 > 0 depending onlyon λ and Λ, such that if A is smooth, A satisfies (2.1), ‖ReA‖L∞ < ε0, and Ω isa special Lipschitz domain, then (KA±)t and (LA)t are bounded Lp(∂Ω) 7→ Lp(∂Ω)for 1 < p < ∞. In Theorem 6.27 we will show that these operators are boundedLp(∂V ) 7→ Lp(∂V ) for any Lipschitz domain V and any 1 < p <∞.

In Theorem 7.2, we will show that, if KA±, LA are bounded Lp(∂V ) 7→ Lp(∂V )then

‖N(Df)‖Lp(∂V ) ≤ Cp‖f‖Lp(∂V ), ‖N(∇ST f)‖Lp(∂V ) ≤ Cp‖f‖Lp(∂V )

for all 1 < p <∞. In Theorem 7.4, we will show that if f ∈ H1(∂V ), then

‖N(∇STf)‖L1(∂V ) ≤ C‖f‖H1(∂V ).

Using this fact, in Theorem 7.10 we will show that (KA±)t and (LA)t are boundedH1(∂V ) 7→ H1(∂V ) for any Lipschitz domain V .

Suppose that KA+ is invertible on Lq(∂V ) for some 1 < q <∞, and that (KA+)−1

has operator norm at most cq. Choose some g ∈ Lq(∂V ) and let u = D((KA+)−1g).Then

divA∇u = 0 in V, u|∂V = g, ‖Nu‖Lq ≤ Cq‖(KA+)−1g‖Lq ≤ Cqcq‖g‖Lq

and so u is a solution to (D)Aq .

Recall that for (N)Ap or (R)Ap to hold, we need only find solutions for boundary

data g ∈ Lp0(∂V ). If (KA−)t or (LA)t is bounded and invertible on H1(∂V ) or

Lp0(∂V ), then as before u = ST(((KA−)t)−1g) or u = ST(((LA)t)−1g) is a solution to

(N)AT

1 , (R)AT

1 , (N)AT

p , or (R)AT

p .

So we need only show that KA±, (KA±)t, (LA)t are invertible. If (KA±)t is boundedand invertible on a reflexive Banach space, then by elementary functional analysis,KA± is bounded and invertible on its dual space, and so we need only consider (KA±)t,

(LA)t. This is the classical method of layer potentials.In Theorem 7.11 and Corollary 7.16, we will show that there exists a p0 > 1

such that for every 1 0such that if ‖A0 −A‖L∞ < ε(p) and A is smooth and satisfies (2.1), then the layerpotentials (KA±)t and (LA)t are invertible on Lp0(∂V ).

The proof of Theorem 7.11 will rely on a number of facts. One is the bound-edness result mentioned above. A second is the invertibility of the operators (KI±)t

on Lp0(∂V ); this was proven by Verchota in [Ver84] provided ∂V is compact andconnected, and is straightforward to show if V = Ω is a special Lipschitz domain.

The third result is the fact that (N)AT0p and (R)

AT0p hold in V and V C provided p > 1

is small enough; this follows from Theorem 2.15 if V is special, and will be provenfrom Theorem 2.15 in Theorem 3.15 and Lemma 3.21 if V or V C is bounded.

The number ε(p) depends only on λ, Λ, p, M , N and c0. The number p0

depends only on λ, Λ, M , N and c0; thus, the same may be said of ε(p0).If ∂V is unbounded then Lp0(∂V ) is dense in Lp(∂V ), and so by duality KA± is

invertible on Lq(∂V ). Suppose ∂V is bounded. Then Lp0 and Lq0 are dual spaces, soKA± is invertible on Lq0(∂V ). We can take as our Dirichlet solutions D((KA±)−1(f −fV )) + fV for fV =

ffl∂V

f dσ.


Thus, if p > 1 is small enough, and if ‖A0 − A‖L∞ < ε(p), then solutions to(N)Ap , (R)Ap and (D)Aq exist. This result is summarized at the start of Chapter 8as Theorem 8.1.

It is straightforward (see Section 3.4) to show that if ∂V is compact, thensolutions to (N)Ap and (R)Ap are unique if 1 ≤ p < ∞. In Theorem 8.2, we will

show that if V = Ω is a special Lipschitz domain, and (N)Ap and (R)Ap hold with

uniform constants in the domains Q(X,R) of (2.5), then solutions to (N)Ap , (R)Apare unique in Ω. In Theorem 8.3, we will show that if V is any Lipschitz domain

and (R)AT

p holds in V , then solutions to (D)Ap in V are unique.Thus, if p > 1 is small enough, and if ‖A0 −A‖L∞ < ε(p), then the conditions

(N)Ap , (R)Ap and (D)Aq hold in V .We wish to remove the dependence of ε on p. We also wish to consider boundary

data in H1(∂V ). In Chapter 9, we will prove that (KA±)t and (LA)t are invertibleon H1(∂V ) provided ‖ImA‖L∞ < ε(1) for some ε(1) > 0 small. This will implythat solutions to (N)A1 and (R)A1 exist. Uniqueness of solutions to (N)A1 and (R)A1follows from results of Chapter 8.

Let ε = min(ε(1), ε(p0)). If ‖A0 − A‖L∞ < ε and A is smooth and satisfies(2.1), then the layer potentials (KA±)t and (LA)t are invertible on H1(∂V ) and onLp0

0 (∂V ). By [RS73], it is possible to interpolate from H1 to Lp; thus, if 1 < p < p0,then (KA±)t and (LA)t are bounded and invertible with bounded inverse on Lp0(∂V ).

Thus, if A is smooth, 1 < p ≤ p0 is, and ‖A−A0‖L∞ < ε, then (N)Ap , (R)Ap holdin V . We remark that ε and p0 depend only on λ, Λ and the Lipschitz constantsof V .

Finally, we pass to arbitrary (rough) A in Theorem 10.1.

2.5. Additional definitions

To prove the boundedness and invertibility of layer potentials, we will need anumber of auxiliary matrices, potentials and functions.

If A is a complex matrix that satisfies (2.1) and A0 is a real matrix that satisfies(2.1), define their components by

(2.17) A(X) =

(a11(X) a12(X)a21(X) a22(X)

), A0(X) =

(a0

11(X) a012(X)

a021(X) a0

22(X)

).

Let the matrix B0(X) be given by

(2.18) B0(X) = BA0 (X) =

(a11(X) a12(X)

0 1

).

In this monograph, the main interest of the matrix B0 is the fact (3.20) that ifdivA∇u = 0 in some open set and A is t-independent, then B0∇u is Holder contin-uous in that set. The transformation to first-order systems of [AAM08, Section 3]and [AA11, Proposition 4.1] used two auxiliary matrices A and A. Our matrix B0

is a special (two-dimensional) case of their A.We may now define a matrix-valued layer potential T with a Holder continuous

kernel.

KA(X,Y ) =(BT0 (Y )∇ΓTX(Y ) BT0 (Y )∇ΓTX(Y )

)t,(2.19)

T AV F (X) = limZ→X n.t.,Z∈V

ˆ∂V

KA(Z, Y )F (Y ) dσ(Y ).(2.20)

2.5. ADDITIONAL DEFINITIONS 17

If X /∈ ∂V , let

(2.21) RAV F (X) =

ˆ∂V

KA(X,Y )F (Y ) dσ(Y )

so that TV is the nontangential limit of RV , as K is the nontangential limit of D.We may recover the boundary layer potentials K, L from T as follows. Define

B1(X) = BA1 (X) =(BA

T

0 (X)t)−1 (

A(X)ν(X) τ(X)).(2.22)

If F =

(f1 f2

f3 f4

), then

RAV (B1F )(X) =

ˆ∂V

(∇ΓTX ∇ΓTX

)t (Aν τ

)F dσ(2.23)

=

(Df1(X)− S(∂τf3)(X) Df2(X)− S(∂τf4)(X)Df1(X)− S(∂τf3)(X) Df2(X)− S(∂τf4)(X)

)

and so

TV (B1F )(X) =

(Kf1(X) + Lf3(X) Kf2(X) + Lf4(X)Kf1(X) + Lf3(X) Kf2(X) + Lf4(X)

).(2.24)

Since B1 is bounded with a bounded inverse, TV is bounded Lp(∂V ) 7→ Lp(∂V ) ifand only if both KV and LV are bounded Lp(∂V ) 7→ Lp(∂V ). In Chapter 6, wewill establish the boundedness of TV using a T (B) theorem; B1 is further usefulbecause it provides one of the matrices B for the T (B) theorem.

Suppose that divA∇u = 0 in some domain U . The conjugate to u is a functionu which satisfies

(2.25)

(0 1−1 0

)∇u = A∇u.

In Lemma 3.16 we will show that u is well-defined up to an additive constant onany simply connected subset of U . It is easy to check that div A∇u = 0 in U , where

(2.26) A =1

detAAt.

We call A the conjugate matrix to A.The fundamental solution ΓX is a solution to divA∇ΓX = 0 in any domain

not containing X, and so its conjugate ΓX is a continuous function in any simplyconnected domain not containing X. We will use ΓX to construct variants on theusual layer potentials as follows:

KA(X,Y ) =(BT0 (Y )∇Y ΓY (X) BT0 (Y )∇Y ΓY (X)

)t,(2.27)

T AV F (X) = limZ→X n.t.,Z∈V

ˆ∂V

KA(Z, Y )F (Y ) dσ(Y ),(2.28)

RAV F (X) =

ˆ∂V

KA(X,Y )F (Y ) dσ(Y ).(2.29)

We will consider the case of special Lipschitz domains extensively. First, wewill need some terminology. Suppose that Ω = X ∈ R2 : ϕ(X · e⊥) < X · e. Ife1, e2 are the components of the vector e, then

e =

(e1

e2

), e⊥ =

(0 1−1 0

)e =

(e2

−e1

).


We define

ψ(x) = xe⊥ + ϕ(x)e ∈ ∂Ω(2.30)

ψ(x, h) = ψ(x) + he =

(xe2 + (ϕ(x) + h)e1

−xe1 + (ϕ(x) + h)e2

).(2.31)

Then ψ parametrizes Ω or ∂Ω in the obvious way; we use it to simplify our notation.If f is a function defined on ∂Ω, we will often use f(x) as shorthand for f(ψ(x)).

We have that Ω = ψ(x, h) : x ∈ R, h > 0 and that (x, t) = ψ(e2x−e1t, e1x+e2t− ϕ(e2x− e1t)).

We let the unit tangent and normal vectors to Ω be given by

τ(x) =1√

1 + ϕ′(x)2(e⊥ + ϕ′(x)e) =

1√1 + ϕ′(x)2

(e1ϕ′(x) + e2

e2ϕ′(x)− e1

),(2.32)

ν(x) =

(0 1−1 0

)τ(x) =

1√1 + ϕ′(x)2

(−e1 + e2ϕ

′(x)−e2 − e1ϕ

′(x)

).(2.33)

We provide variants on K, T , B1 to be used in the case of special Lipschitzdomains:

Kh(x, y) =(BT0 (ψ(y))∇ΓTψ(x,h)(ψ(y)) BT0 (ψ(y))∇ΓTψ(x,h)(ψ(y))

)t(2.34)

=

(∇ΓTψ(x,h)(ψ(y))t

∇ΓTψ(x,h)(ψ(y))t

)BT0 (ψ(y))t = KA(ψ(x, h), ψ(y))

Kh(x, y) = KA(ψ(x, h), ψ(y))(2.35)

〈G,T±F 〉 = limh→0±

ˆR2

G(x)tKh(x, y)F (y) dy dx(2.36)

〈G, T±F 〉 = limh→0±

ˆR2

G(x)tKh(x, y)F (y) dy dx(2.37)

B1(ψ(x, h)) =(BT0 (ψ(x, h))t

)−1√1 + ϕ′(x)2

(A(ψ(x, h))ν(x) τ(x)

)(2.38)

B1(x) = B1(ψ(x)) =√

1 + ϕ′(x)2BA1 (ψ(x))

We remark that TΩ±F (ψ(x)) = T±(√

1 + (ϕ′)2F ψ)(x).

Finally, we define slightly different forms T ′, T ′ of T and T . If X is a point in adomain V , then the integral

´∂V

K(X,Y )F (Y ) dσ(Y ) converges under reasonableassumptions on F . In defining boundary integrals, it is customary to fix the domainV and let the point X approach a point on the boundary; this is how T is defined.However, it is also possible to fix the point X and move the boundary ∂V to X; itis this alternative formulation that we use for T ′.

2.5. ADDITIONAL DEFINITIONS 19

T ′ and T ′ are given by the expressions

K ′h(x, y) =(BT0 (ψ(y, h))∇ΓTψ(x)(ψ(y, h)) BT0 (ψ(y, h))∇ΓTψ(x)(ψ(y, h))

)t(2.39)

=

(∇ΓTψ(x)(ψ(y, h))t

∇ΓTψ(x)(ψ(y, h))t

)BT0 (ψ(y, h))t = KA(ψ(x), ψ(y, h)),

K ′h(x, y) = KA(ψ(x), ψ(y, h)),(2.40)

〈G,T ′±F 〉 = limh→0±

ˆR2

G(x)tK ′h(x, y)F (y) dy dx,(2.41)

〈G, T ′±F 〉 = limh→0±

ˆR2

G(x)tK ′h(x, y)F (y) dy dx.(2.42)

We will show (Section 4.6) that if A − I ∈ C∞0 (R 7→ C2×2), then T± = T ′∓ and

T± = T ′∓ on C∞0 (R 7→ C2×2). These requirements will be dealt with in Section 6.6and Theorem 10.1.

CHAPTER 3

Useful Theorems

In this chapter we collect some lemmas that will be useful throughout thismonograph.

3.1. Nontangential maximal functions

We begin with some results concerning nontangential maximal functions.Let Y ∈ V . Suppose that NF ∈ Lp(∂V ). We can bound |F (Y )| as follows.

Observe that if Y ∈ γ(X) then NF (X) ≥ |F (Y )|. Therefore,

‖NF‖pLp(∂V ) ≥ˆX∈∂V :Y ∈γ(X)

NF (X)p dσ(X) ≥ |F (Y )|pσX : Y ∈ γ(X).

But Y ∈ γ(X) if and only if |X − Y | < (1 + a) dist(Y, ∂V ), so

X ∈ ∂V : Y ∈ γ(X) = ∂V ∩B(Y, (1 + a) dist(Y, ∂V ))

which either contains an entire boundary component of ∂V and so has measure atleast σ(∂V )/C, or is contained in a boundary component and has measure at least2adist(Y, ∂V ). See Figure 1). So

(3.1) |F (Y )| ≤ C‖NF‖Lp(∂V )

min(σ(∂V ),dist(Y, ∂V ))1/p.

We now show that the exact value of a in the definition of nontangential max-imal function is largely irrelevant as long as a > 0.

Lemma 3.2. Recall that

NaF (X) = sup|F (Y )| : Y ∈ V, |X − Y | ≤ (1 + a) dist(Y, ∂V ).

Y

dist(Y, ∂V )

a dist(Y, ∂V )

Figure 1. Points X ∈ ∂V such that Y ∈ γ(X)

21

22 3. USEFUL THEOREMS

Suppose that 0 < a < b and V is a Lipschitz domain. Then there is a constant Cdepending only on a, b and the Lipschitz constants of V such that for all 1 ≤ p ≤ ∞,

‖NbF‖Lp(∂V ) ≤ C‖NaF‖Lp(∂V ).

Proof. If V = R2+, then this lemma is proven in [FS72, Section 7, Lemma 1];

the proof may easily be extended to any special Lipschitz domain.If ∂V is compact, then we may reduce to the case of special Lipschitz domains.

Recall the points Xj , domains Ωj and balls Bj = B(Xj , rj) of Definition 2.3. LetEj = Y ∈ V : dist(Y, ∂V ∩Bj) < σ(∂V )/C, where C is a large number (dependingon a and b) to be chosen later.

Let F0 = F on V \ ∪jEj . If 1 ≤ j ≤ N then let Fj = F on Ej . Let Fj = 0elsewhere. Since ∂V ⊂ ∪jBj , we have that |F (Y )| = max|Fj(Y )| : 0 ≤ j ≤ N.So NbF (X) ≤ ∑N

j=0NbFj(X), and so to complete the proof we need only show

that ‖NV,bFj‖Lp(∂V ) ≤ C‖NV,aF‖Lp(∂V ) for all 0 ≤ j ≤ N .

First, note that |F0(Y )| ≤ C‖NaF‖Lp(∂V )σ(∂V )−1/p and so ‖NbF0‖Lp(∂V ) ≤C‖NaF‖Lp(∂V ).

Next, observe that if C is large enough, then Ej ∩ γa,Ωj (X) = Ej ∩ γa,V (X)

and Ej ∩γb,Ωj (X) = Ej ∩γb,V (X) for all X ∈ ∂V ∩B(X, 32rj), and γa,V (X)∩Ej =

γb,V (X) ∩ Ej = ∅ for all X ∈ ∂V \B(X, 32rj).

Then

‖NV,bFj‖Lp(∂V ) = ‖NΩj ,bFj‖Lp(∂V ) ≤ C‖NΩj ,aFj‖Lp(∂V )

= C‖NV,aFj‖Lp(∂V ) ≤ C‖NV,aF‖Lp(∂V )

as desired.

Lemma 3.3. Suppose that V is a Lipschitz domain, and that NF ∈ Lp(∂V ) forsome 1 ≤ p <∞.

If V is bounded or special, then there is a constant C, depending only on a andthe Lipschitz constants of V , such that ‖F‖L2p(V ) ≤ C‖NF‖Lp(∂V ).

If V C is bounded, then F = F1 + F2, where ‖F1‖L2p(V ) ≤ C‖NF‖Lp(∂V ) and

σ(∂V )1/p‖F2‖L∞(V ) ≤ C‖NF‖Lp(∂V ).Furthermore, if F = ∇u for some function u, then u ∈ L∞(V ∩ B(0, R)) for

any R > 0.

Proof. If NF ∈ Lp(∂V ), define

E(α) = X ∈ V : |F (X)| > α, e(α) = X ∈ ∂V : NF (X) > α.

Then αpσ(e(α)) < ‖NF‖pLp(∂V ) provided F is not identically zero. If ∂V is compact

let σ be the surface measure of the smallest connected component of ∂V , and letα0 = ‖NF‖Lp(∂V )σ

−1/p; otherwise let α0 = 0. If α ≥ α0 and α > 0 then there issome point in each connected component of ∂V not in e(α).

Choose some α with α ≥ α0 and α > 0, and let X ∈ E(α). Let X∗ ∈ ∂Vwith |X−X∗| = dist(X, ∂V ), and let ∆ ( ∂V be the connected component of e(α)containing X∗. Then X /∈ γ(Y ) for any Y ∈ ∂V \ e(α). So

dist(X, ∂V ) +1

2σ(∆) ≥ dist(X, ∂V \ e(α)) ≥ (1 + a) dist(X, ∂V )

3.2. BOUNDS ON SOLUTIONS 23

and so dist(X,∆) = dist(X, ∂V ) ≤ 12aσ(∆). So

E(α) ⊂⋃

∆⊂e(α)connected

⋃

X∗∈∆

B

(X∗,

1

2aσ(∆)

).

But if ∆ is a connected curve segment in the plane, then⋃X∗∈∆B (X∗, cσ(∆)) is

of size at most (1 + 2c)2σ(∆)2. So |E(α)| ≤ Cσ(e(α))2 for all α ≥ α0.Let F1 = F on E(α0), and let F1 = 0 otherwise. Let F2 = F − F1. If

V is a special Lipschitz domain then F1 = F and F2 = 0. If V C is boundedthen ‖F2‖L∞(V ) ≤ α0 ≤ C‖NF‖Lp(∂V )σ(∂V )−1/p. Finally, if V is bounded then

‖F2‖L2p(V ) ≤ α0|E(α0)|1/2p ≤ C‖NF‖Lp(∂V ). So in any case we need only provethat ‖F1‖L2p(V ) ≤ C‖NF‖Lp(∂V ).

Now,ˆV

|F1|2p =

ˆ ∞0

2pα2p−1|X :|F1(X)|> α| dα

≤ α2p0 |E(α0)|+ C

ˆ ∞α0

2pα2p−1 σ(e(α))2 dα

≤ α2p0 |E(α0)|+ C‖NF‖pLp(∂V )

ˆ ∞α0

pαp−1 σ(e(α)) dα ≤ C‖NF‖2pLp(∂V ).

We now must establish that if N(∇u) ∈ Lp(∂V ) then u ∈ L∞loc(V ). By (3.1), uis continuous on compact subsets of V because ∇u is bounded; we need only lookat a small neighborhood of the boundary, and so we need only consider V = Ω aspecial Lipschitz domain.

By Lemma 3.2, we may assume that a is large enough that N(∇u)(ψ(x)) <|∇u(ψ(x, t))| for all t > 0. For some X0 = ψ(x0) ∈ ∂Ω, N(∇u)(X0) is finite. Thenfor any t > 0, |u(ψ(x0, t))− u(X0)| ≤ tN(∇u)(X0) is finite.

Now, for any x ∈ R and any t > 0,

|u(ψ(x, t))− u(X0)| ≤ |u(ψ(x, t))− u(ψ(x0, t))|+ |u(ψ(x0, t))− u(ψ(x0))|

≤ tN(∇u)(X0) +

ˆ x

x0

|∇u(ψ(y, s))| dy

≤ tN(∇u)(X0) + |x− x0|1/q‖N(∇u)‖Lp(∂Ω)

and so u is bounded on compact sets.

3.2. Bounds on solutions

We now turn to solutions to elliptic partial differential equations. Let Br ⊂ R2

be a ball of radius r, Br/2 be the concentric ball of radius r/2. Suppose that Asatisfies (1.2). Then the following four useful lemmas hold.

Lemma 3.4 (The Caccioppoli inequality). Suppose that V is a Lipschitz do-main, and that divA∇u = 0 in V , ∇u ∈ L2(Br∩V ), and either u ≡ 0 or ν·A∇u = 0on ∂V ∩Br. Then there exists a constant C depending only on λ, Λ such thatˆ

V ∩Br/2|∇u|2 ≤ C

r2

ˆV ∩Br\Br/2

|u|2.


Lemma 3.5. For some C > 0 and p > 2, depending only on λ, Λ, we have thatif divA∇u = 0 in Br then

(1

r2

ˆBr/2

|∇u|p)1/p

≤ C(

1

r2

ˆBr

|∇u|2)1/2

.

Lemma 3.6. For all 1 ≤ p <∞, there is a constant C(p) depending only on λ,Λ, p, such that if divA∇u = 0 in Br then

supBr/2

|u| ≤ C(p)

(1

r2

ˆBr

|u|p)1/p

.

Lemma 3.7. For some C > 0 and some α > 0 depending only on λ, Λ, we havethat if divA∇u = 0 in Br then

supX,Y ∈Br/2

|u(X)− u(Y )| ≤ C |X − Y |α

rα

(1

r2

ˆBr

|∇u|2)1/2

.

The Caccioppoli inequality is well known and its proof is straightforward. Lem-ma 3.5 follows from the Caccioppoli inequality by [Gia83, Theorem 1.2, Chapter V]and preceding remarks.

Lemmas 3.6 and 3.7 hold in all dimensions under the additional assumptionthat A is real; these were first proven in [DG57], [Nas58] and [Mos61] for Asymmetric, and extended to nonsymmetric real equations in [Mor66].

If A is complex, then Lemmas 3.6 and 3.7 may not hold in higher dimen-sions; see [MNP91] and [Fre08] for specific counterexamples. However, if u solvesdivA∇u = 0 in a domain in R2, then Lemmas 3.6 and 3.7 follow from Lemma 3.5using the Poincare inequality and Morrey’s inequality.

Now, suppose that A(x, t) = A(x) is t-independent. We wish to control ∇upointwise. We first recall the following theorem from [AT95].

Lemma 3.8 ([AT95, Theoreme II.2]). If divA∇u = 0 in B(X, 2r) ⊂ R2, andA(x, t) = A(x) is t-independent, then

supY ∈B(X,r)

|∇u(Y )| ≤ C(

B(X,2r)

|∇u|2)1/2

.

By Lemmas 3.4, 3.6 and the Poincare inequality we may strengthen this con-dition to

supB(X,r/2)

|∇u| ≤ C(p)

( B(X,r)

|∇u|p)1/p

(3.9)

for any 1 ≤ p <∞.Recall that by Lemma 3.3, if u is a solution to (N)Ap or (R)Ap in a domain V ,

then ∇u lies in L2loc(V ). If V is bounded, this implies that ∇u ∈ L2(V ). We may

use (3.9) to extend this result to all Lipschitz domains with compact boundary.

Lemma 3.10. If V C is bounded, divA∇u = 0 in V , ∇u ∈ L2loc(V ) and u(X)

is bounded for all |X| sufficiently large, then ∇u ∈ L2(V ). Conversely, if ∂V isbounded, divA∇u = 0 in V , and ∇u ∈ L2(V ), then lim|X|→∞|u(X)| exists.

3.3. EXISTENCE RESULTS 25

Proof. Suppose that ∇u ∈ L2loc(V ) and u(X) is bounded for all |X| suffi-

ciently large. Let r be large enough that V C ⊂ B(0, r). By assumption, there issome constant U > 0 such that

´B(0,2r)∩V |∇u|2 ≤ U2 and such that if |X| > 2r

then |u(X)| 3r.If ρ ≥ r, let ηρ be a smooth cutoff function such that ηρ = 1 on B(0, ρ), ηρ = 0

outside B(0, 2ρ), with |∇η| ≤ C/ρ. Choose some R > r. ThenˆB(0,R)∩V

|∇u|2 ≤ˆB(0,2r)∩V

|∇u|2 + CRe

ˆV

(ηR − ηr)∇u ·A∇u

≤ U2 + CRe

ˆV

∇(uηR − uηr) ·A∇u

− CRe

ˆV

u∇(ηR − ηr) ·A∇u

≤ U2 +C

r

ˆB(0,2r)\B(0,r)

|u||∇u|+ C

R

ˆB(0,2R)\B(0,R)

|u||∇u|

since A is elliptic and divA∇u = 0. But each of those terms is at most CU2, andso by taking the limit as R→∞, we see that ∇u ∈ L2(V ).

Conversely, suppose ∇u ∈ L2(V ). Let Vj = V ∩ B(0, 2j+1) \ B(0, 2j), and let

Vj = Vj−1 ∪ Vj ∪ Vj+1. If ∇u ∈ L2(V ), then by (3.9)∞∑

j=0

2j supVj

|∇u|

2

≤∞∑

j=0

22j supVj

|∇u|2 ≤ C∞∑

j=0

ˆVj

|∇u|2 ≤ C‖∇u‖2L2(V ).

Thus, for any ε > 0, there is some R > 0 such that if |X|, |Y | > R, then thereis some path ω connecting X and Y such that

´ω|∇u| dσ < ε. So lim|X|→∞ u(X)

exists.

3.3. Existence results

The well-known Lax-Milgram lemma provides a way to construct solutions tothe Dirichlet and Neumann problems for relatively well-behaved boundary data.This method does not presume smoothness of coefficients. Solutions u constructedby this method do not satisfy the bounds on Nu and N(∇u) required by theformulations (D)Ap and (N)Ap ; instead their gradients ∇u lie in L2(V ). We providesuch constructions for the Dirichlet and Neumann problems in the plane.

The results we require are proven for real coefficients and special Lipschitzdomains in [KR09, Lemma 1.1 and 1.2], and we refer to that paper for someresults. However, we must work through the case of Lipschitz domains with compactboundary because we will need fairly precise control on the L2 norm of ∇u.

Recall that W 1,2(V ) is the Sobolev space of functions with one weak derivativein L2(V ), with ‖f‖2W 1,2(V ) = ‖f‖2L2(V ) + ‖∇f‖2L2(V ). We define the superspace

W 1,2(V ) as the space of all (equivalence classes modulo additive constants of)

functions in W 1,2loc (V ) such that the norm ‖f‖W 1,2(V ) = ‖∇f‖L2(V ) is finite.

Then solutions to the Dirichlet or Neumann problems exist in this space.

Lemma 3.11. If V is a Lipschitz domain and

f ∈W 1,2(∂V ) ∩ L7/6(∂V ) ∩ L17/6(∂V ),


then there is some function u ∈ W 1,2(V ) such that divA∇u = 0 in V , Tru = fand

‖u‖W 1,2(V ) ≤ C(‖f‖W 1,2(∂V ) + ‖f‖L7/6(∂V ) + ‖f‖L17/6(∂V )).

Lemma 3.12. If V is a Lipschitz domain and g ∈ H1(∂V ), then there is some

function u ∈ W 1,2(V ) such that ‖u‖W 1,2(V ) ≤ C‖g‖H1(∂V ) and such that ν ·A∇u =

g on ∂V in the weak sense.

Remark 3.13. Observe that if u ∈ L2(V ) for a bounded domain V , then by(1.5) ˆ

∂V

ν ·A∇u dσ = 0.

This is why we consider Neumann solutions with boundary data that integrates tozero on ∂V . It should be emphasized that if ∂V is not connected, then Lemma 3.12implies that solutions exist only for boundary data that integrates to zero on eachconnected component of ∂V . By adding multiples of the fundamental solution withpoles in various components of V C , we could circumvent this requirement; we leavethe details to the interested reader.

To prove Lemmas 3.11 and 3.12, we use a generalization of the Lax-Milgramlemma to the complex case. The Babuska-Lax-Milgram theorem [Bab71, Theo-rem 2.1] states that, if B is a bounded bilinear form on two complex Hilbert spacesH1 and H2, and if B is weakly coercive in the sense that

sup‖w‖1=1

|B(w, v)| ≥ λ‖v‖2, sup‖w‖2=1

|B(u,w)| ≥ λ‖u‖1

for every u ∈ H1, v ∈ H2, for some fixed λ > 0, then for every linear functional Tdefined on H2 there is a unique uT ∈ H1 such that B(uT , v) = T (v). Furthermore,‖uT ‖1 ≤ 1

λ‖T‖. (Here ‖·‖1, ‖·‖2, and ‖·‖ denote norm in H1, H2, and operatornorm H1 7→ H2, respectively.)

We apply this theorem to the bilinear form

B(ξ, η) =

ˆV

∇η ·A∇ξ

on H1 = H2 = W 1,2(V ) or a subspace; then B is clearly bounded and coercive.

Proof of Lemma 3.11. Choose some such function f . Suppose that thereexists a w ∈ W 1,2(V ) such that Trw = f . Then the map T defined by T (η) =

B(w, η) is a bounded linear functional on W 1,2(V ) (and hence on W 1,20 (V ), the

subspace of functions with trace zero). So there exists a v ∈ W 1,20 (V ) such that

T (η) = B(v, η) for every η ∈ W 1,20 (V ). Let u = w − v.

Then Tru = f , divA∇u = 0 in V in the weak sense, and

‖u‖W 1,2(V ) ≤ ‖w‖W 1,2(V ) + ‖v‖W 1,2(V ) ≤ ‖w‖W 1,2(V ) + C‖T‖ ≤ C‖w‖W 1,2(V ).

So we need only construct a w with Trw = f and ‖w‖W 1,2(V ) ≤ C9f9, where

9f9 = ‖f‖W 1,2(∂V ) + ‖f‖L7/6(∂V ) + ‖f‖L17/6(∂V ). If V is the upper half-plane,

then w is constructed in [KR09, Lemma 1.1]; by change of variables it exists foran arbitrary special Lipschitz domain Ω. In fact, for fixed X0 ∈ ∂Ω, this w alsosatisfies ˆ

Ω

|w(X)|21 + |X −X0|2

dX ≤ C9f92

3.4. PRELIMINARY UNIQUENESS RESULTS 27

and so if η is a smooth cutoff function supported in B(X0, R) with |∇η| ≤ C/R,we have that

‖∇(wη)‖L2(V ) = ‖η∇w + w∇η‖L2(V ) ≤ C9f9.Suppose that ∂V is bounded. Let

∑j ηj be a smooth partition of unity near

∂V , with ηj supported in B(Xj ,32rj), where the Xjs, rjs are as in Definition 2.3.

Let wj = fηj on ∂Ωj , w =∑j ηjwj where ηj = 1 on supp ηj and is supported in

B(Xj , 2rj). We may require |∇ηj | ≤ C/rj , |∇ηj | ≤ C/rj . Then Trw = f , and

‖∇w‖L2(V ) ≤N∑

j=1

‖∇(ηjwj)‖L2(V ) ≤N∑

j=1

C9fηj9.

We have that |∂τηj | ≤ C/rj ≤ C/σ(∂V ). So

9fηj9 = ‖fηj‖W 1,2(∂V ) + ‖fηj‖L7/6(∂V ) + ‖fηj‖L17/6(∂V )

≤ ‖∂τf‖L2 + ‖f∂τηj‖L2 + ‖f‖L2 + ‖f‖L7/6 + ‖f‖L17/6

≤ ‖∂τf‖L2 + C‖f‖L7/6 + C‖f‖L17/6 ≤ C9f9.

So ‖∇w‖L2(V ) ≤ C9f9, as desired.

Proof of Lemma 3.12. Pick some g ∈ H1(∂V ). Define the linear map T by

T (ξ) =

ˆ∂V

gTr ξ dσ.

If T is bounded, and if u ∈ W 1,2(V ) is such thatˆ∂V

gTr ξ dσ = B(u, ξ) =

ˆV

∇ξ ·A∇u

for all ξ ∈ W 1,2(V ), then divA∇u = 0, ν · A∇u = g in the weak sense, and‖u‖W 1,2(V ) ≤ 1

λ‖T‖.So we need only show that T is a bounded operator on W 1,2(V ).

It suffices to show that Tr is bounded from W 1,2(V ) to BMO(∂V ). That is, if∆ ⊂ ∂V is connected, then we must show that

ffl∆|Tr ξ−

ffl∆

Tr ξ| dσ ≤ C‖∇ξ‖L2(V ).In fact, we need only do this for σ(∆) ≤ σ(∂V )/C; so we may assume that ∆ ⊂B(Xj , 2rj) ∩ ∂V for one of the Xjs, rjs of Definition 2.3.

By the Poincare inequality,´Rj|ξ −

fflRjξ|2 ≤ Cr2

j

´Rj|∇ξ|2. We may assume

thatfflRjξ = 0. Multiplying ξ by a smooth cutoff function ηj as before, we see that

‖∇(ηjξ)‖L2(V ) ≤ ‖∇ξ‖L2(Rj) +C

rj‖ξ‖L2(Rj) ≤ C‖∇ξ‖L2(Rj).

So, taking ηj ≡ 1 on ∆, we need only show´

∆|ηjξ −

ffl∆ηjξ| dσ ≤ ‖∇(ηjξ)‖L2(Rj).

So we need only show that Tr : W 1,2(Ωj) 7→ BMO(∂Ωj) is bounded. This is donein the proof of [KR09, Lemma 1.2].

3.4. Preliminary uniqueness results

Our first uniqueness result is a simple corollary of the Caccioppoli inequality.

Lemma 3.14. Suppose that divA∇u = 0 in V for some Lipschitz domain V ,and that ∇u ∈ L2(V ). Assume that either ν · A∇u = 0 on ∂V or u ≡ 0 on ∂V .Then u is constant in V .


Proof. If V is bounded, this follows immediately from Lemma 3.4. Otherwiselet X ∈ ∂V and let R > 0 be large. If V C is bounded, let U(R) = B(X,R) ∩ V . IfV = Ω is special, let U(R) = Q(X,R). Let W (R) = U(2R) \ U(R).

Let η be a smooth, nonnegative cutoff function with η = 1 in U(R), η supportedin U(2R) with |∇η| < C/R. Then as in the proof of the Caccioppoli inequality,

λ

Û(R)

|∇u|2 ≤ Re

ˆV

η2∇u ·A∇u = −Re

ˆV

2ηu∇η ·A∇u

≤ C(

1

R2

ˆW (R)

|u|2)1/2(ˆ

W (R)

|∇u|2)1/2

.

If ν ·A∇u = 0 on ∂V , then we may assumefflW (R)

u = 0; by the Poincare inequality´W (R)

|u|2 ≤ CR2´W (R)

|∇u|2 and soÛ(R)

|∇u|2 ≤ CˆW (R)

|∇u|2.

Since ∇u ∈ L2(V ), this goes to zero as R → ∞ and so u is constant in V . Byboundedness of the trace map, the same argument applies if V = Ω is a specialLipschitz domain and u = 0 on ∂V .

If V C is bounded, then by Lemma 3.10, lim|X|→∞ u(X) exists. If R is largeenough, then for some constant C(u) independent of R,

Û(R)

|∇u|2 ≤ C(u)

(ˆW (R)

|∇u|2)1/2

.

Again, taking the limit as R→∞ yields that u is constant in V .

Suppose that N(∇u) ∈ Lp(∂V ) for some 1 ≤ p < ∞. If V is bounded, thenby Lemma 3.3, ∇u ∈ L2p(V ) ⊆ L2(∂V ). If V C is bounded, then by Lemmas 3.3and 3.10 we have that ∇u ∈ L2(V ) if in addition lim|X|→∞|u(X)| exists. Finally,

if p = 1 and V = Ω is special then ∇u ∈ L2(Ω).Thus, solutions to (N)Ap and (R)Ap are unique in domains with compact bound-

ary, and solutions to (N)A1 and (R)A1 are unique in arbitrary Lipschitz domains.Furthermore, if u is a solution to the Dirichlet or Neumann problem constructed

by Lemma 3.11 or Lemma 3.12, then ∇u ∈ L2(V ); thus, these solutions must equalthe solutions to (N)Ap or (R)Ap discussed above.

3.5. The Neumann and regularity problems in unusual domains

Many known results (and many of the theorems of this monograph) hold onlyin simply connected bounded or special Lipschitz domains. In this section, we showthat for the Neumann problem, we can pass from these domains to more generalLipschitz domains. We also show that certain Neumann and regularity problemsare equivalent; this implies in particular that we can solve the regularity problemin other domains.

We begin with the Neumann problem.

Theorem 3.15. Let V be a planar Lipschitz domain with compact boundary.We do not require that ∂V be connected. Let A satisfy (2.1), and suppose that1 < p ≤ 2.

3.5. THE NEUMANN AND REGULARITY PROBLEMS IN UNUSUAL DOMAINS 29

If (N)Ap holds with constant C(p) in the domains Q(X,R), for all X ∈ ∂V

and all R small enough that Q(X,R) exists, then (N)Ap holds in V with constantsdepending only on p, λ, Λ, C(p) and the Lipschitz constants of V .

Recall from (2.5) that Q(X,R) is a bounded, simply connected Lipschitz do-main whose Lipschitz character depends only on the Lipschitz constant M of V .Thus, if A is real, then by Theorem 2.15 the hypotheses of Theorem 3.15 hold forsome p > 1.

In this monograph, this theorem is mainly of interest if V C is bounded andsimply connected. Such domains are not simply connected but do have connectedboundary. As in Remark 3.13, if ∂V is not connected then (N)Ap holds only in thesense that we can find Neumann solutions for boundary data which integrate tozero on each connected component of ∂V .

Proof. By Lemma 3.14, solutions are unique. By Lemma 3.12, we knowthat a function u exists which satisfies divA∇u = 0 in V , ν · A∇u = g on ∂V ,such that ‖∇u‖L2(V ) ≤ C‖g‖H1(∂V ). We need only show that ‖N(∇u)‖Lp(∂V ) ≤C(p)‖g‖Lp(∂V ).

By (3.9), we have that |∇u(Y )| ≤ Cdist(Y,∂V )‖∇u‖L2(V ). But by (2.6),

C

dist(Y, ∂V )‖∇u‖L2(V ) ≤

C

dist(Y, ∂V )‖g‖H1(∂V ) ≤

C(p)‖g‖Lp(∂V )σ(∂V )1/q

dist(Y, ∂V ).

This lets us bound |∇u| pointwise. Furthermore, if V C is bounded, then by Lem-ma 3.10, lim|X|→∞ u(X) exists.

Define

N1F (X) = sup|F (Z)| : Z ∈ γ(X), dist(Z, ∂V ) < σ(∂V )/β,N2F (X) = sup|F (Z)| : Z ∈ γ(X), dist(Z, ∂V ) ≥ σ(∂V )/β,

for some constant β to be chosen later. Then NF (X) = max(N1F (X), N2F (X)),and so to bound N(∇u) we need only bound N1(∇u) and N2(∇u).

But by our previous remarks, N2(∇u)(X) ≤ C(p)β‖g‖Lp(∂V )σ(∂V )−1/p for anyX ∈ ∂V and so ‖N2(∇u)‖Lp(∂V ) ≤ C(p)β‖g‖Lp(∂V ).

We now consider N1. Define Xj , rj as in Definition 2.3. Let r be any numberwith 3

2rj < r < 2rj , and let Q(Xj , r) be as in (2.5). We have that Q(Xj , r) ⊂ V andB(Xj , rj)∩V ⊂ B(Xj , r)∩V ⊂ Q(Xj , r). Furthermore, we have that σ(∂V ) ≤ Crj .

If X ∈ B(Xj , rj) ∩ ∂V and β is small enough, then

Z ∈ γ(X) : dist(Z, ∂V ) < σ(∂V )/β ⊂ γQ(Xj ,r)(X)

and so N1(∇u)(X) ≤ NQ(Xj ,r)(∇u)(X).Thus ˆ

Bj∩∂VN1(∇u)p dσ ≤

ˆ∂Q(Xj ,r)

NQ(Xj ,r)(∇u)p dσ

for 32rj < r < 2rj . But (N)Ap holds in all the Q(Xj , r)s with constant at most C(p).

Then for each r,ˆBj∩∂V

N1(∇u)p dσ ≤ˆ∂Q(Xj ,r)

NQ(Xj ,r)(∇u)p dσ ≤ C(p)

ˆ∂Q(Xj ,r)

|ν ·A∇u|p dσ

≤ C(p)

ˆ∆(Xj ,r)

|g|p dσ dr + C(p)pˆ∂Q(Xj ,r)\∂V

|∇u|p dσ


Taking the average over r in ( 32rj , 2rj), we have thatˆ

Bj∩∂VN1(∇u)p dσ ≤ C(p)‖g‖pLp(∆(Xj ,2rj))

+C

rj

ˆ 2rj

3rj/2

ˆ∂Q(Xj ,r)\∂V

|∇u|p dσ dr

≤ C(p)‖g‖pLp(∆(Xj ,2rj))+C(p)

rj

ˆQ(Xj ,2rj)

|∇u|p

But if p ≤ 2, then by Holder’s inequality and our bound on ‖∇u‖L2(V ), thisis at most C(p)‖g‖pLp(∂V ). Since there are at most N such balls, we have that´∂V

N1(∇u)p dσ ≤ C(p)‖g‖pLp(∂V ), as desired. So (N)Ap holds in V for V any Lip-

schitz domain, with constants depending only on p, C(p), λ, Λ and the Lipschitzconstants of V .

We now show that certain Neumann and regularity problems are equivalent.Recall the conjugates to solutions of (2.25). Conjugates have been used extensivelyin the literature; they were used in [Pip97, Section 3], in [AT95] to prove Lem-ma 3.8, and in [KR09] and [Rul07] to prove Theorem 2.15. Conjugates can beconstructed even in the case of elliptic systems in two dimensions (see [AR11,Section 5]).

In this section, we will use conjugates to prove the desired equivalence. Wewill use conjugates in several other ways: we will use them to derive a regularityresult for the gradients of solutions, and in Chapter 6, we will use them to establishboundedness of layer potentials.

Lemma 3.16. Suppose that u satisfies divA∇u = 0 in some simply connecteddomain U ⊂ R2. Then there is a continuous function u defined in U , called theconjugate to u, unique up to an additive constant, such that

(3.17)

(0 1−1 0

)∇u = A∇u.

Furthermore, div A∇u = 0 in U .If U = V is a Lipschitz domain and N(∇u) ∈ L1

loc(∂V ), then

(3.18) τ · ∇u = ν ·A∇u on ∂V .

Observe that ˜u = −u up to an additive constant. Thus, (3.18) implies that if

V is simply connected, then (N)Ap holds in V if and only if (R)Ap holds in V .

Proof. If X0, X ∈ U , let ω ⊂ U be a path from X0 to X. Let τ be the unittangent vector to ω and let ν = −τ⊥ be the unit normal. The integralˆ

ω

ν(Z) ·A(Z)∇u(Z) dl(Z)

does not depend on choice of ω. Thus, if we choose some constant C and someX0 ∈ U , then

(3.19) u(X) = C +

ˆ X

X0


is well-defined.But sinceˆ X

X0

(0 −11 0

)A(Z)∇u(Z) · τ(Z) dl(Z) =

ˆ X

X0


3.5. THE NEUMANN AND REGULARITY PROBLEMS IN UNUSUAL DOMAINS 31

we readily derive (3.17). Since A = (1/detA)At, we have that A∇u =(−ut ux

)t

and so div A∇u = 0, as desired.We now establish (3.18). By Lemma 3.3, ∇u ∈ L2

loc(V ) and so ν · A∇u isdefined in the weak sense. We show that for a.e. X0 and X1 ∈ ∂V , if X0 and X1

are the endpoints of some connected set I ⊂ ∂V , then u(X0) and u(X1) exist, andfor the appropriate ordering of X0, X1,

u(X1)− u(X0) =

Î

ν ·A∇u dσ.

By (3.1), ∇u is bounded on compact subsets of V . Suppose that N(∇u)(Xi)is finite; this is true for a.e. Xi ∈ ∂V . Let ω be a path from X1 to X0 lyingentirely in V such that ω ∪ I forms the boundary of a simply connected boundedset W . By Lemma 3.2, we may assume that the number a in the definition (1.3)of nontangential cone is large. Thus, we may require that ω approach X0 and X1

through their nontangential cones. Since N(∇u)(Xi) is finite, this implies that ∇uis bounded uniformly on ω, and so u(X0), u(X1) exist.

By definition of u,

u(X0)− u(X1) =

ˆω

ν ·A∇u dσ.

But since W ⊂ V is bounded and simply connected, and ∇u ∈ L2loc(V ), we have

that Î

ν ·A∇u dσ +

ˆω

ν ·A∇u dσ =

ˆ∂W

ν ·A∇u dσ = 0

and so

u(X1)− u(X0) =

Î

ν ·A∇u dσ,as desired.

It should be emphasized that these conjugates can be constructed only in twodimensions. This construction is unlikely to be generalizable to higher dimensions.In [May10], the author shows that for every p < 2n

n+2 , there is some block matrix

A such that (N)Ap fails to hold in Rn+1+ . But the author also showed that (R)Ap

holds for all block matrices A provided max 2nn+4 , 1 < p ≤ 2; thus, passing from

regularity problems to Neumann problems is problematic in higher dimensions.Observe that by (2.25), ∂tu = a11∂xu+ a12∂tu, and so

B0∇u =

(∂tu∂tu

).

Thus by Lemma 3.7, if divA∇u = 0 in B(X, r) then

(3.20) |B0(Y )∇u(Y )−B0(Y ′)∇u(Y ′)| ≤ C|Y − Y ′|αr1+α

‖∇u‖L2(B(X,r))

for all Y , Y ′ ∈ B(X, r/2).

We have that (N)Ap is equivalent to (R)Ap in simply connected Lipschitz do-mains. We now extend this result to their complements.

Lemma 3.21. Let V be a planar Lipschitz domain with compact boundary. LetA satisfy (2.1). Suppose that u is a solution to (N)Ap in V .

Then u is well-defined up to an additive constant on any simply connected subsetof V . We may choose these additive constants such that u is continuous and solves


div A∇u = 0 in all of V . Furthermore, u is a solution to (R)Ap in V with the sameboundary data as u.

Finally, if (N)Ap holds in V and ∂V is connected, then (R)Ap holds in V .

Proof. Again define u by (3.19). To show that u(X) is well-defined up to anadditive constant, we need only show that

´ων · A∇u = 0 for all Jordan curves

ω ⊂ V .We may assume ω = ∂U for some simply connected bounded domain U . Be-

cause u is a solution to (N)Ap , we have that´ν · A∇u = 0 over every connected

component of ∂V . Soˆ∂U∪(U∩∂V )

ν ·A∇u =

ˆ∂(U∩V )

ν ·A∇u =

Û∩V

∇1 ·A∇u = 0

by the weak definition of ν · A∇u. But U ∩ ∂V is the union of one or more entirecomponents of ∂V ; therefore,ˆ

∂U

ν ·A∇u = −Û∩∂V

ν ·A∇u = 0.

So u is well-defined on V .Let g = ν · A∇u. By Lemma 3.16, div A∇u = 0 in V and τ · ∇u = g on ∂V .

By (2.25), N(∇u)(X) ≈ N(∇u)(X) and so ‖N(∇u)‖Lp ≤ C(p)‖g‖Lp(∂V ). If V C is

bounded, then by Lemmas 3.3 and 3.10 ∇u ∈ L2(V ) and thus ∇u ∈ L2(V ), and soagain by Lemma 3.10 lim|X|→∞ u(X) exists.

Thus, if (N)Ap holds in V , then solutions to (R)Ap exist in V . If ∂V is connected,

then solutions are unique up to additive constants by Lemma 3.14, and so (R)Apholds in V .

Thus if we can solve (N)Ap in all bounded, simply connected Lipschitz domains,

then by Theorem 3.15 and Lemma 3.21 we can solve (N)Ap in all Lipschitz domains

with compact boundary and (R)Ap in all Lipschitz domains with compact, connectedboundary. In particular, by Theorem 2.15 we can solve these problems if A is real.

The regularity problem is difficult to formulate in domains with disconnectedboundary. Suppose that we wish to solve the regularity problem with Dirichletboundary data f . If we require that solutions u satisfy u = f on ∂V , then it isobviously impossible to control ‖N(∇u)‖Lp(∂V ) by ‖∂τf‖Lp(∂V ); simply considerdata f which is constant on each connected component. Conversely, suppose werequire only that τ ·∇u = ∂τf on ∂V . Such solutions are not unique. A formulationof (R)Ap which avoids these problems is beyond the scope of this monograph.

CHAPTER 4

The Fundamental Solution

4.1. A fundamental solution exists

Recall that Lemma 2.7 states that if A satisfies (2.1), then there is a func-tion ΓAX(Y ), called the fundamental solution of the operator divA∇, such that|∇ΓX(Y )| ≤ C/|X−Y | for some constant C depending only on λ, Λ, and such that

ˆR2

A(Y )∇ΓAX(Y ) · ∇η(Y ) dY = −η(X)

for every η ∈ C∞0 (R2). This function ΓX is unique up to an additive constant. We

will let Γ = ΓA, Γ0 = ΓA0 , and ΓT = ΓAT

.This function is defined and constructed for real A in the appendix to [KN85],

and in 3 or more dimensions in [HK07]. For complex-valued A defined on R2, thefundamental solution is constructed in [AMT98, Theorem 3.16]; we review thisconstruction, paying special attention to bounds on the gradient ∇ΓX .

From [AT98, pp. 29–31], we know that there is a function Kt(X,Y ) such that,for all η ∈ C∞0 (R2),

ˆη(X)∂tKt(X,Y ) dX = −

Â(X)∇XK(X,Y ) · ∇η(X) dX.

Furthermore, there is some β, µ, C > 0 depending only on λ, Λ such that

|Kt(X,Y )| ≤ C

texp

−β|X − Y |

2

t

|Kt(X,Y )− Kt(X′, Y )| ≤ C

t

( |X −X ′|√t+ |X − Y |

)µexp

−β|X − Y |

2

t

|Kt(X,Y )− Kt(X,Y′)| ≤ C

t

( |Y − Y ′|√t+ |X − Y |

)µexp

−β|X − Y |

2

t

whenever |X − X ′|, |Y − Y ′| < 12 (√t + |X − Y |). This Kt is called the Schwartz

kernel of the operator e−tL, where L = −divA∇.Formally, we wish to construct ΓX = −L−1δX ; by the Laplace formula [AT98,

(60), p. 52], this is given by

ΓX = −L−1δX = −ˆ ∞

0

e−tLδX dt = −ˆ ∞

0

ˆKt(·, Z)δX(Z) dZ dt

= −ˆ ∞

0

Kt(·, X) dt.

33

34 4. THE FUNDAMENTAL SOLUTION

We study ΓX as follows. Let Jt(Y,X) = Kt(Y,X)−fflr≤|Z−X|≤2r

Kt(Z,X) dZ,

so that if r < |X − Y | < 2r,

|Jt(Y,X)| =∣∣∣∣Kt(Y,X)−

r≤|Z−X|≤2r

Kt(Z,X) dZ

∣∣∣∣

≤ C

t

( |X − Y |√t+ |X − Y |

)µexp

−β|X − Y |

2

t

.

So for fixed X and Y , t 7→ Jt(X,Y ) ∈ L1((0,∞)).From [AT98, p. 54], there is some ε, β, c > 0, such that

(ˆr≤|X−Y |≤2r

|∇Y Kt(Y,X)|2 dY)1/2

≤ c

t

(r2

t

)εe−βr

2/t.

Let fr(t) = ct

(r2

t

)εe−βr

2/t. Observe that fr(t) ∈ L1((0,∞)) and´∞

0fr(t) dt is

independent of r. By Holder’s inequality,

ˆr≤|X−Y |≤2r

∣∣∣∣ˆ ∞

0

∇Y Kt(Y,X) dt

∣∣∣∣2

dY

≤ˆr≤|X−Y |≤2r

ˆ ∞0

1

fr(t)|∇Y Kt(Y,X)|2 dt

ˆ ∞0

fr(t) dt dY

≤(ˆ ∞

0

fr(t) dt

)2

≤ C.

Thus,´∞

0∇Y Kt(Y,X) dt converges almost everywhere inB(X, 2r)\B(X, r). Define

ΓX(Y ) = C(r)−ˆ ∞

0

Jt(Y,X) dt

so

∇ΓX(Y ) = −ˆ ∞

0

∇Y Jt(Y,X) dt = −ˆ ∞

0

∇Y Kt(Y,X) dt

where C(r) is chosen such that the values of Γ on different annuli agree. ThenÂ∇ΓX · ∇η = −

Â(Y )

ˆ ∞0

∇Y Kt(Y,X) dt · ∇η(Y ) dY

= −ˆ ∞

0

Â(Y )∇Y Kt(Y,X) · ∇η(Y ) dY dt

=

ˆ ∞0

ˆ∂tKt(Y,X)η(Y ) dY dt

= limt→∞

ˆKt(Y,X)η(Y ) dY − lim

t→0+

ˆKt(Y,X)η(Y ) dY = 0− η(X)

whenever η ∈ C∞0 .So we have constructed a fundamental solution.Furthermore, we have the following bound on its gradient:ˆ

r≤|X−Y |≤2r

|∇ΓX(Y )|2 dY ≤ C.

4.2. UNIQUENESS OF THE FUNDAMENTAL SOLUTION 35

So by (3.9), if A satisfies (2.1) we have that

(4.1) |∇ΓX(Y )| ≤ C

|X − Y | .

This implies that |ΓX(Y )| ≤ C|log|X − Y || + C if we choose additive constantsappropriately.

4.2. Uniqueness of the fundamental solution

Let u = ΓX . Assume that |∇v(Y )| ≤ C(v)/|X − Y |, and thatÁ∇v · ∇η =

−η(X) for all compactly supported η ∈ W 1,2(R2). Let w = u − v. To prove thatthe fundamental solution is unique, we need only show that w is a constant.

We have that |∇w(Y )| ≤ C(v)/|X−Y | and so |w(Y )| ≤ C(v)(1+ |log|X−Y ||).Furthermore,

´R2 ∇η · A∇w = 0 for all smooth, compactly supported functions η.

If ∇w ∈ L2loc(R

2), then Lemma 3.7 will allow us to conclude that

|w(Y )− w(Z)| ≤ C|Y − Z|αr1+α

‖w‖L2(B(X,r)) ≤ C(v)|Y − Z|α log r

rα

for any r large enough. Letting r → ∞, we see that w is a constant; thus ΓX isunique up to an additive constant.

We need only show that ∇w ∈ L2loc(R

2). It suffices to show that ∇w ∈L2(B(X, 1) \ B(X, ε)) for all 0 < ε < 1, uniformly in ε. Let wε = w on B(X, ε)C ,wε constant on B(X, ε/2); by our condition on ∇v we may choose wε such that|∇wε(Y )| ≤ C(v)/ε. Thus, wε ∈ L2

loc(R2). Suppose η = 1 on B(X, 1), |∇η| ≤ C

and η is smooth and compactly supported. Then η2wε ∈ W 1,2(R2) is compactlysupported and is smooth near X. As in the standard proof of the Caccioppoliinequality,

0 =

ˆ∇(η2wε) ·A∇w

=

ˆη2∇wε ·A∇wε +

ˆη2∇wε ·A∇(wε − w) +

ˆ2ηwε∇η ·A∇w.

and soˆB(X,1)\B(X,ε)

|∇w|2 ≤ CRe

ˆη2∇wε ·A∇wε

= −CRe

ˆη2∇wε ·A∇(wε − w)− CRe

ˆ2ηwε∇η ·A∇w.

≤ CˆB(X,ε)

|∇wε| |∇(wε − w)|+ C

ˆsupp∇η

|w||∇w|

since w = wε outside B(X, ε). Because |∇wε| ≤ C(v)/ε and |∇w(Y )| ≤ C(v)/|X −Y |, the first integral is at most C(v). The second integral is clearly finite andindependent of ε, and so w ∈ L2

loc(R2), as desired.

We remark on a useful property of the fundamental solution. Let ξ ∈ R2 andlet Aξ(X) = A(X + ξ), ζ(X) = η(X − ξ). Then

−η(X) = −ζ(X + ξ) =

Â(Y )∇ΓAX+ξ(Y ) · ∇ζ(Y ) dY

=

Âξ(Y )∇ΓAX+ξ(Y + ξ) · ∇η(Y ) dY


and so by uniqueness of the fundamental solution,

(4.2) ∇ΓAX+ξ(Y + ξ) = ∇ΓAξX (Y ).

In particular, if A satisfies (2.1) and t is a real number then

∇ΓX+(0,t)(Y + (0, t)) = ∇ΓX(Y ).

4.3. Symmetry of the fundamental solution

Recall that ΓX(Y ) is defined only up to an additive constant C(X). We wishto show that there is some choice of additive constant such that, for all X and Y ,

(4.3) ΓTX(Y ) = ΓY (X).

This was shown for Γ constructed as in [KN85] (the real case) in [KR09,Lemma 2.7]; we generalize to the complex case as follows.

Let η ∈ C∞0 be 1 on a neighborhood of B(0, R). If R |X|, |Y |, then

ΓTY (X)− ΓX(Y ) =

ˆR2

∇(ηΓX) ·AT∇ΓTY −∇(ηΓTY )A∇ΓX

=

ˆB(0,R)

∇ΓX ·AT∇ΓTY −∇ΓTY ·A∇ΓX

+

ˆB(0,R)C

∇(ηΓX)AT∇ΓTY −∇(ηΓTY )A∇ΓX

= 0 +

ˆ∂B(0,R)C

ΓX ν ·AT∇ΓTY − ΓTY ν ·A∇ΓX dσ.

So we need only prove that the last integral converges to zero as R→∞. Thiswill be easier if we work with Ar and Γr instead of A, Γ, where Ar = I on B(0, 2r)C

and Ar = A on B(0, r).Consider only r > 2|X|+ 2|Y |. Then ΓrX is harmonic in B(0, 2r)C .Think of R2 as the complex plane, and let u(Z) = ΓrX(Z). There is some

bounded harmonic function w(Z) onB(0, 2r)C such that u(Z) = w(Z) on ∂B(0, 2r);let v(Z) = ΓrX(eZ)− w(eZ).

Then v is harmonic in a half-plane, and v = 0 on the boundary of that half-plane; by the reflection principle we may extend v to an entire function. Further-more, |v(Z)| ≤ C + CReZ, so v is linear.

So ΓrX(eZ) = w(eZ) + C1 + C2 ReZ; thus, ΓrX(Z) = w(Z) + C1 + C2 log|Z|for some bounded w and constants C1, C2. Using a test function which is 1 onB(0, 3r), we see that C2 = 1

2π . We thus have a standard normalization: we choose

additive constants such that ΓrX(Z) = w(Z) + ΓI(Z), where ΓI(Z) = 12π log|Z| and

lim|Z|→∞ w(Z) = 0.

Since w is bounded and harmonic on B(0, 2r)C , the function f(Z) = w(1/Z)is bounded and harmonic in a disk; thus, so are its partial derivatives. By ournormalization, f(0) = lim|Z|→∞ w(Z) = 0. Then |w(Z)| = |f(1/Z)| ≤ C(r)/|Z| on

B(0, 3r)C , and w′(Z) = −f ′(1/Z)/Z2, so |∇w(Z)| ≤ C(r)/|Z|2 on B(0, 3r)C .Let R > 3r. Then on ∂B(0, R),

|ΓrX ν · ∇Γr,TY − ΓIX ν · ∇ΓIY dσ| ≤ |ΓrX − ΓIX ||∇Γr,TY |+ ΓI |∇Γr,TY −∇ΓIY |

≤ C(r)

R

C

R+ C

logR

R

C

R.

4.3. SYMMETRY OF THE FUNDAMENTAL SOLUTION 37

So since Ar = Ar,T = I on B(0, 2r)C ,

|ΓrX(Y )− Γr,TY (X)| =∣∣∣∣ˆ∂B(0,R)

ν ·(

ΓrX∇Γr,TY − Γr,TY ∇ΓrX

)dσ

∣∣∣∣

=

∣∣∣∣ˆ∂B(0,R)

ΓrX ν · ∇Γr,TY − ΓI ν · ∇ΓI dσ

∣∣∣∣

+

∣∣∣∣ˆ∂B(0,R)

Γr,TY ν · ∇ΓrX − ΓI ν · ∇ΓI dσ

∣∣∣∣

≤ˆ∂B(0,R)

C(r) + C logR

R2dσ ≤ C(r)

1 + logR

R

which goes to 0 as R→∞. So Γr,TX (Y ) = ΓrY (X).In two dimensions, we do not have a natural normalization condition. (In

higher dimensions, we expect ΓX(Y ) to approach a constant as Y → ∞, and sowe can normalize by requiring this constant to equal zero.) Thus, limr→∞ ΓrX(Y )need not exist, and so (in particular) need not be ΓX(Y ); we will need to chooseour normalization carefully to recover ΓTX(Y ) = ΓY (X).

Let ur(Z) = ΓrY (Z) − ΓY (Z); we want to show that ∇ur → 0 as r → ∞.

As in Section 4.2 we have that ur ∈ W 1,2loc (R2) and divA∇ur = 0 in B(0, r). By

Lemma 3.4, the Poincare inequality and our bound on ∇ΓY , we have thatˆB(0,r/2)

|∇ur|2 ≤ CˆB(0,r)\B(0,r/2)

|∇ur|2 ≤ C

provided |Y | < r/4. By Lemma 3.5 and Holder’s inequality, for fixed ρ with 4|Y | ≤ρ ≤ r/4,

ˆB(0,ρ)

|∇ur|2 ≤ Cρ2−4/p

(ˆB(0,ρ)

|∇ur|p)2/p

≤ Cρ2−4/p

(ˆB(0,r/4)

|∇ur|p)2/p

≤ Cρ2−4/pr4/p−2

ˆB(0,r/4)

|∇ur|2 ≤ C(ρr

)2−4/p

for some p > 2. Thus, we have that ∇ur → 0 as r → ∞ in L2(B(0, ρ)) for anyfixed ρ > 0.

Using Lemma 3.7 and the Poincare inequality, this implies that

|ΓY (Z)− ΓrY (Z)− ΓY (Z0) + ΓrY (Z0)| ≤ C

rα|Z − Z0|α

for some α > 0, provided |Y |, |Z|, |Z0| < r/2.Fix some choice of Z0 and of ΓTZ0

. Normalize each ΓX such that ΓX(Z0) =

ΓTZ0(X). Then

ΓY (Z) = limr→∞

ΓrY (Z) + ΓY (Z0)− ΓrY (Z0) = limr→∞

Γr,TZ (Y ) + ΓTZ0(Y )− Γr,TZ0

(Y )

where the limit converges uniformly for Y and Z in compact sets. Thus, we maytake the gradient in Y of each side to see that

∇Y ΓY (Z) = limr→∞

∇Γr,TZ (Y ) +∇ΓTZ0(Y )−∇Γr,TZ0

(Y )

and so ∇Y ΓY (Z) = ∇ΓTZ(Y ) in L2(B(0, ρ)) for any ρ > 0. Thus, for any fixedZ, we must have that ΓY (Z) = ΓTZ(Y ) for all Y ∈ R2 \ Z, up to an additiveconstant; since ΓZ0

(Z) = ΓTZ(Z0) this constant must be zero, as desired.


4.4. Conjugates to the fundamental solution

Recall the conjugate of Lemma 3.16. Since divA∇ΓY = 0 in any domain notcontaining Y , we may define the conjugate ΓY to ΓY on any simply connecteddomain not containing Y . Note that

∇ΓY (X) =

(0 1−1 0

)A(X)∇ΓY (X)

is defined on R2 \ Y , even though ΓY itself is necessarily undefined on a ray.Since ΓY (X) solves an elliptic equation in both X and Y , using (3.9) and (4.1)

we have that

(4.4) |∂xj∂yiΓY (X)| ≤ C

|X − Y |

( B(X,|X−Y |/2)

|∇Y ΓY (Z)|2)1/2

≤ C

|X − Y |2 .

Now, since

ΓY (X) = ζ(Y ) +

ˆ X

X0

ν(Z) ·A(Z)∇ΓY (Z) dl(Z)

we have that

∇Y ΓY (X) = ∇ζ(Y ) +

ˆ X

X0

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z)

Since |∂xi∂zjΓY (Z)| ≤ C/|Y − Z|2, the limit

lim|X0|→∞

ˆ X

X0

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z)

exists, and so we may choose ζ(Y ) such that

(4.5) ∇Y ΓY (X) = lim|X0|→∞

ˆ X

X0

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z).

By letting X0 = X + r(X − Y ) and letting r →∞, we see that

(4.6) |∇Y ΓY (X)| ≤ C

|X − Y | .

We have that ΓY (X) solves an elliptic equation in X. We claim that it alsosolves an elliptic equation in Y . If η ∈ C∞0 (W ) for some bounded simply connecteddomain W with X /∈W , thenˆ

W

AT (Y )∇Y ΓY (X) · ∇η(Y ) dY

= lim|X0|→∞

ˆW

AT (Y )

ˆ X

X0

A(Z)ν(Z) · ∇Z(∇Y ΓY (Z)) dl(Z) · ∇η(Y ) dY

= lim|X0|→∞

ˆ X

X0

A(Z)ν(Z) · ∇ZˆW

AT (Y )∇ΓTZ(Y ) · ∇η(Y ) dY dl(Z)

= lim|X0|→∞

−ˆ X

X0

A(Z)ν(Z) · ∇Zη(Z) dl(Z) = 0.

Thus, if Y 6= X, then

(4.7) divY AT (Y )∇Y ΓY (X) = 0.

4.6. ANALYTICITY 39

4.5. Calderon-Zygmund kernels

Recall that we wish to analyze layer potentials as singular integral operators.Thus, we must construct Calderon-Zygmund kernels from ∇ΓX(Y ) and ∇X ΓX(Y ).

By (4.4), we have that for any 0 ≤ α ≤ 1 and any matrix B with |B(Y )| < C,if |X −X ′| ≤ |X − Y |/2 then

|B(Y )∇ΓTX(Y )−B(Y )∇ΓTX′(Y )| ≤ C |X −X′|α

|X − Y |1+α.

Apply (3.20) to u(Y ) = ΓTX(Y ). Recalling that

K(X,Y ) =(BT0 (Y )∇ΓTX(Y ) BT0 (Y )∇ΓTX(Y )

)t

we have that K(X,Y ) satisfies the Calderon-Zygmund kernel conditions

|K(X,Y )| ≤ C

|X − Y | ,(4.8)

|K(X,Y )−K(X ′, Y )| ≤ C |X −X′|α

|X − Y |1+α,

|K(X,Y )−K(X,Y ′)| ≤ C |Y − Y′|α

|X − Y |1+α,

provided |X −X ′|, |Y − Y ′| < |X−Y |2 .

We claim that K(X,Y ) =(BT0 (Y )∇Y ΓY (X) BT0 (Y )∇Y ΓY (X)

)tsatisfies the

same conditions. The bound on |K(X,Y )| follows directly from (4.6). Recall that

∇Y ΓY (X) = lim|X0|→∞

ˆ X

X0

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z).

Since |∇Z(∇Y ΓY (Z))| ≤ C/|Y −Z|2 and the integral is path-independent, we havethat

|∇Y ΓY (X)−∇Y ΓY (X ′)| ≤ˆ X′

X

C

|Y − Z|2 dl(Z) ≤ C |X −X′|

|Y −X|2whenever |X −X ′| ≤ |Y −X|/2.

Now, we wish to show that K is Cα in Y as well as X. But by (4.7), ΓY (X)solves an elliptic partial differential equation in Y away from X, so we may againapply (3.20).

So we have that, if |Y − Y ′|, |X −X ′| < 12 |Y −X|, then

|K(X,Y )| ≤ C

|Y −X| ,(4.9)

|K(X,Y )− K(X ′, Y )| ≤ C |X −X′|α

|Y −X|1+α

|K(X,Y )− K(X,Y ′)| ≤ C |Y − Y′|α

|Y −X|1+α.

4.6. Analyticity

We will eventually want to compare the fundamental solutions (and relatedoperators) for a real matrix A0 and a nearby complex matrix A. We can explorethis using analytic function theory. Let z 7→ Az be an analytic function from C to


L∞(R2 7→ C2×2). Assume that Az is uniformly elliptic in some neighborhood of 0,say B(0, 1).

The most useful example is to take Az = A0 + z(λ/2ε)(A−A0), where

ε = ‖A−A0‖L∞ = sup|ξ · (A(x)−A0(x))η| : x ∈ R, η, ξ ∈ C2, |η| = |ξ| = 1

.

Let Lz = divAz∇. Since we are working in R2, we know from [AT98] that

the operator e−tLz has a Schwartz kernel Kzt (X,Y ) = KAz

t (X,Y ). Furthermore,by ([AT98, p. 57]), the map A 7→ KA is analytic in the sense that, if z 7→ Az is ananalytic map C 7→ L∞(R2 7→ C2×2), then the map z 7→ KAz is also analytic.

Fix some Y . Recall that, as in Section 4.1,

∇ΓAzY (X) =

ˆ ∞0

∇Kzt (X,Y ) dt.

Since ∇Kzt ∈ L1

t , uniformly in z, we have that ∇ΓAzY (X) is analytic in z. Thislets us compute many useful inequalities.

If |z| < 12 and ω is an appropriately chosen simple closed curve lying in B(0, 1),

then

∇ΓAzY (X)−∇ΓA0

Y (X) =1

2πi

˛ω

∇ΓAζY (X)

(1

ζ − z −1

ζ

)dζ

=1

2πi

˛ω

∇ΓAζY (X)

(z

ζ(ζ − z)

)dζ

which has norm at most |z| C|X−Y | . Taking Az = A0 + z(λ/2ε)(A − A0) and then

applying this equation to z = 2ε/λ, we get that

(4.10) |∇ΓY (X)−∇Γ0Y (X)| < Cε

|X − Y | .

Similarly,

|∇ΓY (X)−∇Γ0Y (X)−∇ΓY (X ′) +∇Γ0

Y (X ′)| ≤ Cε|X −X ′|α|X − Y |1+α

,

|∇ΓY (X)−∇Γ0Y (X)−∇ΓY ′(X) +∇Γ0

Y ′(X)| ≤ Cε|Y − Y ′|α|X − Y |1+α

,

provided that |X −X ′|, |Y − Y ′| are less than 12 |X − Y |.

Suppose that J z is a Calderon-Zygmund operator whose kernel Kz(X,Y ) is an-alytic in z, and suppose that J z is uniformly bounded on Lp in some neighborhoodof z = 0. Then

J zf(X)− J 0f(X) =1

2πi

˛ω

J ζf(X)

(z

ζ(ζ − z)

)dζ

and so for |z| small enough,

(4.11) ‖J zf − J 0f‖Lp ≤1

2π

˛ω

‖J ζf‖Lp∣∣∣∣

z

ζ(ζ − z)

∣∣∣∣ dζ ≤ C|z| ‖f‖Lp .

We now assume that A is smooth. We define Az(X) = A(X) + z(A(X + ξ)−A(X)), so A1(X) = A(X + ξ). Then Az is uniformly elliptic for all |z| ‖A′‖L∞ |ξ| <λ/2, and so if |ξ| ≤ 2/‖A′‖L∞ , then

|∇ΓA1

X (Y )−∇ΓX(Y )| ≤ 1

|X − Y |C‖A′‖L∞ |ξ|.

4.6. ANALYTICITY 41

If A(x) is smooth in x and A = I for large x, then ‖A′‖L∞ is finite (if large).

Recall from (4.2) that ΓX+ξ(Y + ξ) = ΓA1

X (Y ). So if ξ ≤ 2/‖A′‖L∞ , then

(4.12) |∇ΓY+ξ(X + ξ)−∇ΓY (X)| ≤ C‖A′‖L∞|X − Y | |ξ|.

Similarly, if A is smooth and |ξ| is small, then

|∇(∇Y ΓY+ξ(X + ξ))−∇(∇Y ΓY (X))| ≤ C‖A′‖L∞|X − Y |2 |ξ|.

So

(4.13) |∇Y ΓY+ξ(X + ξ)−∇Y ΓY (X)|

= lim|X0|→∞

∣∣∣∣ˆ X

X0

ν(Z) ·A(Z + ξ)∇(∇Y ΓY+ξ(Z + ξ)) dl(Z)

−ˆ X

X0

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z)

∣∣∣∣

≤ C‖A′‖L∞|X − Y | |ξ|.

These equations are not useful in most circumstances, since we do not wantour estimates to depend on A′. However, we can use these formulas to compare theoperators T and T ′ of (2.41) and (2.42).

Recall that

T±F (x) = limh→0±

ˆR

Kh(x, y)F (y) dy,

T ′±F (x) = limh→0±

ˆR

K ′h(x, y)F (y) dy,

where

Kh(x, y) =



)BT0 (ψ(y))t,

K ′h(x, y) =

(∇ΓTψ(x)(ψ(y, h))t

∇ΓTψ(x)(ψ(y, h))t

)BT0 (ψ(y, h))t.

By (4.12) and (2.18), if h > 0, then |Kh(x, y)−K ′h(x, y)| ≤ C‖A′‖L∞ |h|/(|x−y|+ |h|), which has L2 norm C‖A′‖L∞

√|h|. Letting ThF (x) =

´RKh(x, y)F (y) dy,

and T ′hF (x) =´RK ′h(x, y)F (y) dy, we see that if F ∈ L2(R), then

|ThF (x)− T ′−hF (x)| ≤ C‖A′‖L∞‖F‖L2

√|h|.

Thus, if T±F (x) exists, then T∓F ′(x) must exist, and furthermore, T±F (x) =T ′∓F (x).

Similarly, if T±F (x) exists and F ∈ L2(R), then by (4.13) T ′∓F (x) exists and

T±F (x) = T ′∓F (x).

CHAPTER 5

Properties of Layer Potentials

Recall that we intend to construct solutions to (D)Aq , (N)Ap , (R)Ap as layer po-tentials. In this chapter, we establish some elementary properties of layer potentialsand formulas for their adjoints. Chapter 6 will be devoted to a proof that layerpotentials are bounded as operators on Lp(∂V ). Boundedness of layer potentialshas many interesting consequences, which will be explored in Chapter 7.

Recall the definitions (2.9) and (2.10) of the layer potentials

Df(X) =

ˆ∂V

ν ·AT (Y )∇ΓTX(Y )f(Y ) dσ(Y ),

∇Sf(X) =

ˆ∂V

∇XΓTX(Y )f(Y ) dσ(Y ).

We begin by listing some properties which are easy to check.

Lemma 5.1. Let f ∈ Lp(∂V ), 1 ≤ p < ∞. If X /∈ ∂V , then Df(X) and∇Sf(X) are well-defined (the integrals above converge) and are bounded on compactsubsets of R2 \ ∂V . Furthermore, if u = Df or u = Sf , then divA∇u = 0 inR2 \ ∂V .

If f ∈ H1(∂V ), or if ∂V is compact and f ∈ L1(∂V ), then the integral

(5.2)

ˆ∂V

ΓTX(Y )f(Y ) dσ(Y )

converges and is bounded on compact subsets of R2 \ ∂V .

If both integrals converge, then the gradient in X of´∂V

ΓTX(Y )f(Y ) dσ(Y ) isclearly ∇Sf(X); this gives us a direct formula for Sf(X).

Proof. Recall from (4.1) that |∇ΓTX(Y )| ≤ C/|X − Y |, and recall from (4.3)that ΓTX(Y ) = ΓY (X).

If 1 < p <∞ then

(5.3)

ˆ∂V \B(X,r)

rp−1

|Y −X|p dσ(Y ) ≤∞∑

j=0

rp−1

(2jr)pσ(B(X, r2j+1) ∩ ∂V ) ≤ Cp

p− 1.

In particular, if X0 /∈ ∂V then by (4.1)

(5.4)

ˆ∂V

|∇ΓX(Y )|p dσ(Y ) ≤ˆ∂V

1

|Y −X|p dσ(X) ≤ Cp

p− 1dist(X, ∂V )1−p

and so if f ∈ Lq(∂V ) and X /∈ ∂V , then Df(X), ∇Sf(X) are defined by convergentintegrals and are bounded on compact subsets of R2 \ ∂V .

By the bound on |∇ΓTX(Y )|, ΓX ∈ BMO(∂V ) for any X /∈ V , and if X /∈ ∂Vand ∂V is compact then ΓX is bounded on ∂V . Thus if f ∈ H1(∂V ), or if f ∈

43

44 5. PROPERTIES OF LAYER POTENTIALS

L1(∂V ) and ∂V is compact, then the integral (5.2) converges and is bounded oncompact subsets of R2 \ ∂V .

Finally, the equations divA∇Df = 0, divA∇Sf = 0 follow from (1.1), byusing (2.8) and (4.3) and interchanging the order of integration.

If ∂V is compact and f ∈ L1(∂V ), then by (5.4), lim|X|→∞Df(X) = 0 and forany fixed Y0,

(5.5) lim|X|→∞

Sf(X)− ΓTX(Y0)

ˆ∂V

f dσ = 0.

In particular, if f ∈ H1(∂V ) then lim|X|→∞ Sf = 0.

Let V be a bounded set. Choose some X ∈ R2 \ ∂V . Let η ∈ C∞0 (R2) withη ≡ 1 near V ; if X /∈ V , assume that η = 0 near X. Then by (1.5) and (2.8),

D1(X) =

ˆ∂V

ν ·AT∇ΓTX dσ =

ˆ∂V

η ν ·AT∇ΓTX dσ = −ˆV C∇η ·AT∇ΓTX(5.6)

= −ˆR2

∇η ·AT∇ΓTX = η(X)

and so D1 ≡ 1 on V and D1 ≡ 0 on V C . (If V C is bounded then D1 ≡ −1 on V C

and D1 ≡ 0 on V .)

5.1. Limits of layer potentials and the adjoint formulas

Recall (2.11) and (2.13): the boundary layer potentials K±, L are given by

K±f(X) = limZ→X,Z∈γ±(X)

Df(X)

= limZ→X,Z∈γ(X)

ˆ∂V

ν(Y ) ·AT (Y )∇ΓTZ(Y )f(Y ) dσ(Y ),

Lf(X) = limZ→X,Z∈γ(X)

ˆ∂V

τ(Y ) · ∇ΓTZ(Y )f(Y ) dσ(Y ).

In this section, we will show that under some conditions on f , the limitsK±f(X) and Lf(X) exist pointwise. We will also show that the operator adjointsKt± and Lt are given by

Kt±f = ∓ν ·AT∇STf |∂V∓ , Ltf = ∂τSTf.

The formula for Kt is well known; see, for example, [FJR78, Theorem 1.3 andTheorem 1.10], where it is proven in the case A ≡ I. The potentials L are ofinterest in this monograph principally because of the formula for Lt.

Lemma 5.7. If f ∈ Lp(∂V ) for some 1 ≤ p <∞, and if ∂τf ∈ L∞loc(∂V ), thenthe limits in the definitions of K = KV , L = LV exist at every point X ∈ ∂V .

If F ∈ Lp(∂V 7→ C2×2) and ∂τF ∈ L∞loc(∂V ), then the limits in the definitions

of T (B1F )(X) and T (B1F )(X) exist.

Proof. Let f ∈ Lp(∂V ). If f ∈ Lp(∂V ) and ∂τf ∈ L∞loc(∂V ), then F = Ifalso lies in Lp(∂V ) and ∂τF lies in L∞loc(∂V ). Furthermore, by (2.24), if T (B1F )(X)exists, then Kf(X) and Lf(X) exist. Thus, to complete the proof, we need only

consider T (B1F )(X) and T (B1F )(X).

5.1. LIMITS OF LAYER POTENTIALS AND THE ADJOINT FORMULAS 45

Recall that

RF (X) =

ˆ∂V

K(X,Y )F (Y ) dσ(Y ), RF (X) =

ˆ∂V

K(X,Y )F (Y ) dσ(Y ),

and that T F and T F are the nontangential limits of RF and RF . We then wantto show that if ∂τF ∈ L∞loc(∂V ), then the nontangential limits of R(B1F ), R(B1F )exist at X.

Suppose e, e are two small vectors such that X + e, X + e are both in V ,and that |e| < (1 + a) dist(X + e, ∂V ), |e| < (1 + a) dist(X + e, ∂V ). Then, ifρ > 2|e|+ 2|e| is small enough that the Q(X, ρ), ∆(X, ρ) of (2.5) are well-defined,then

|R(B1F )(X + e)−R(B1F )(X + e)|

=

∣∣∣∣ˆ∂V

(K(X + e, Y )−K(X + e, Y ))B1(Y )F (Y ) dσ(Y )

∣∣∣∣

≤∣∣∣∣ˆ

∆(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y )(F (Y )− F (X)) dσ(Y )

∣∣∣∣

+

∣∣∣∣F (X)

ˆ∆(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y )

∣∣∣∣

+

∣∣∣∣ˆ∂V \∆(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y )F (Y ) dσ

∣∣∣∣.

We may bound these integrals using (4.8). If V is a Lipschitz domain, byHolder’s inequality and (5.3) the third term is at mostˆ

∂V \∆(X,ρ)

C|e− e|α|Y −X|1+α

|F (Y )| dσ(Y ) ≤ ‖F‖Lp(∂V )C|e− e|αρα+1/p

.

To control the first term, recall that ∂τF ∈ L∞loc(∂V ), and so there is someC(X) such that, if |X − Y | is small, then |F (X) − F (Y )| < C(X)|X − Y |. So,provided ρ is small enough,∣∣∣∣ˆ

∆(X,ρ)

K(X + e, Y )B1(Y )(F (Y )− F (X)) dσ

∣∣∣∣ ≤ˆ

∆(X,ρ)

C(X)|X − Y ||X + e− Y | dσ(Y ).

Since |e| ≤ (1 + a) dist(X + e, ∂V ) ≤ (1 + a)|X + e− Y |, we have that |X − Y | ≤C|X + e− Y | and so

∣∣∣∣ˆ

∆(X,ρ)

K(X + e, Y )B1(Y )(F (Y )− F (X)) dσ(Y )

∣∣∣∣ ≤ C(X)ρ.

Finally, note that the middle term is equal to

F (X)

ˆQ(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y )−

F (X)

ˆQ(X,ρ)\∆(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y )

and that∣∣∣∣ˆQ(X,ρ)\∆(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y )

∣∣∣∣ ≤ C‖e− e‖α

ρα.


So

|RF (X + e)−RF (X + e)|

≤ C(X)ρ+ C|e− e|αρ1+α

‖F‖Lp(∂V ) + C|F (X)| ‖e− e‖αρα

+ |F (X)|∣∣∣∣ˆ∂Q(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y )

∣∣∣∣.

Similarly, using (4.9), we have that

|RF (X + e)− RF (X + e)|

≤ C(X)ρ+ C|e− e|αρ1+α

‖F‖Lp(∂V ) + C|F (X)| ‖e− e‖αρα

+ |F (X)|∣∣∣∣ˆ∂Q(X,ρ)

(K(X + e, Y )− K(X + e, Y ))B1(Y ) dσ(Y )

∣∣∣∣.

We may control the first three terms by first making ρ small and then making |e|and |e| small.

So we need only consider

ˆ∂Q(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y ),

ˆ∂Q(X,ρ)

(K(X + e, Y )− K(X + e, Y ))B1(Y ) dσ(Y ).

But

K(X + e, Y )B1(Y ) =

(ν(Y ) ·AT (Y )∇ΓTX+e(Y ) τ(Y ) · ∇ΓTX+e(Y )ν(Y ) ·AT (Y )∇ΓTX+e(Y ) τ(Y ) · ∇ΓTX+e(Y )

),

K(X + e, Y )B1(Y ) =

(ν(Y ) ·AT (Y )∇Y ΓY (X + e) τ(Y ) · ∇Y ΓY (X + e)

ν(Y ) ·AT (Y )∇Y ΓY (X + e) τ(Y ) · ∇Y ΓY (X + e)

).

Butˆ∂Q(X,ρ)

ν ·AT∇ΓTX+e dσ = 1,

ˆ∂Q(X,ρ)

τ · ∇ΓTX+e dσ = 0

by continuity of ΓTX and by (5.6); thus

ˆ∂Q(X,ρ)

(K(X + e, Y )−K(X + e, Y ))B1(Y ) dσ(Y ) = 0

and T F (X) exists.Recall from (4.5) that

∇Y ΓY (X + e)−∇Y ΓY (X + e) =

ˆ X+e

X+e

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z)

5.1. LIMITS OF LAYER POTENTIALS AND THE ADJOINT FORMULAS 47

Soˆ∂Q(X,ρ)

ν(Y ) ·AT (Y )(∇Y ΓY (X + e)−∇Y ΓY (X + e)) dσ(Y )

=

ˆ∂Q(X,ρ)

ν(Y ) ·AT (Y )

ˆ X+e

X+e

ν(Z) ·A(Z)∇(∇Y ΓY (Z)) dl(Z) dσ(Y )

=

ˆ X+e

X+e

ν(Z) ·A(Z)∇Zˆ∂Q(X,ρ)

ν(Y ) ·AT (Y )∇ZΓTZ(Y ) dσ(Y ) dl(Z)

=

ˆ X+e

X+e

ν(Z) ·A(Z)∇Z1 dl(Z) = 0.

The other term may be dealt with similarly; thus, T F (X) exists.

We now establish formulas for the adjoints of the layer potentials.

Lemma 5.8. If f ∈ Lp(∂V ), 1 < p <∞, then we have the following equationsfor the adjoints of K and L:

Kt±f = ∓ν ·AT∇STf |∂V∓ , Ltf(X) = ∂τSTf(Z).

In our equation for Kt±, ν is taken to be the outward unit normal to V∓, sothat we may easily use the weak definitionˆ

∂V

ηKt±f = ∓ˆV∓

∇η ·AT∇STf.

If we prefer to let ν = νV be the outward unit normal to V in both cases, thenKt±f = ν ·AT∇STf |∂V∓ .

Proof. Consider Kt+ first. Pick some η ∈ C∞0 (R2).

By definition of K, and since divAT∇ΓTZ = 0 in V− provided Z ∈ V , we havethatˆ

∂V

ηKt+f dσ =

ˆ∂V

K+η(Y )f(Y ) dσ(Y )

=

ˆ∂V

limZ→Y, Z∈γ+

ˆ∂V

ν ·AT∇ΓTZ(X) η(X) dσ(X) f(Y ) dσ(Y )

= −ˆ∂V

limZ→Y, Z∈γ+

ˆV−

∇η(X) ·AT∇ΓTZ(X) dX f(Y ) dσ(Y ).

If Z ∈ γ+(Y ) and X ∈ V−, then by considering the three cases |X−Y | ≤ 12 |Y −Z|,

|X − Y | ≥ 2|Y − Z|, and 12 |Y − Z| < |X − Y | < 2|Y − Z| separately, we see that

|X − Y | ≤ C|X − Z|. So by (4.1), we have that

|∇η(X) ·AT∇ΓTZ(X)| ≤ C(η)

|X − Y |on supp η, and so by the dominated convergence theoremˆ

∂V

ηKt+f dσ = −ˆ∂V

ˆV−

∇η(X) ·AT∇ΓTY (X) dX f(Y ) dσ(Y ).


Since η ∈ C∞0 , the inner integral is bounded uniformly in Y and is also at mostC‖∇η‖L1(R2)/ dist(Y, supp η); thus, if f ∈ Lp(∂V ) for some p <∞ then by Fubini’stheoremˆ

∂V

ηKt+f dσ = −ˆV−

ˆ∂V

∇η(X) ·AT (X)∇ΓTY (X) f(Y ) dσ(Y ) dX

= −ˆV−

∇η(X) ·AT (X)∇STf(X) dX

= −ˆ∂V−

η(X) ν(X) ·AT (X)∇STf(X) dσ(X)

as desired. Similarly, Kt−f = ν ·AT∇STf |∂V+.

Now we come to Lf . Let η ∈ C∞0 (R2) again. Thenˆ∂V

ηLtf dσ =

ˆ∂V

f(Y ) limZ→Y, Z∈γ(Y )

ˆ∂V

τ(X) · ∇ΓTZ(X) η(X) dσ(X) dσ(Y ).

Integrating by parts, we have thatˆ∂V

ηLtf dσ = −ˆ∂V

f(Y ) limZ→Y, Z∈γ(Y )

ˆ∂V

ΓTZ(X) τ(X) · ∇η(X) dσ(X) dσ(Y ).

As before, we may use the dominated convergence theorem and Fubini’s theoremto see thatˆ

∂V

ηLtf dσ = −ˆ∂V

τ(X) · ∇η(X)

ˆ∂V

ΓTY (X)f(Y ) dσ(Y ) dσ(X)

= −ˆ∂V

τ(X) · ∇η(X)STf(X) dσ(X)

=

ˆ∂V

η(X) τ(X) · ∇STf(X) dσ(X)

as desired.

CHAPTER 6

Boundedness of Layer Potentials

Theorem 6.1. Let A0, A be 2× 2 matrices defined on R2 which satisfy (2.1).Suppose that A0 is real-valued, and that A, A0 are smooth.

Then there exists an ε0 = ε0(λ,Λ) > 0 such that, if ‖A − A0‖L∞ ≤ ε0, then

the layer potentials T and T defined by (2.20) and (2.28) are bounded on Lp(∂V )for any Lipschitz domain V and any 1 < p < ∞, with bounds depending only onλ, Λ, p, and the Lipschitz constants of V .

If T is bounded L2 7→ L2, then by standard Calderon-Zygmund theory (see,for example, [Gra09, Theorem 8.2.1] or [Ste93, I.7]), T is bounded Lp 7→ Lp for1 < p <∞.

We will begin by proving a special case:

Theorem 6.2. Let Ω = X : ϕ(X · e⊥) < X · e be a special Lipschitz domain.Suppose that ϕ is smooth and compactly supported. Further assume that there existfunctions a0

12, a022, and a number R0 large, such that

A0(x) =

(1 a0

12(x)0 a0

22(x)

),(6.3)

A(x) = A0(x) = I for |x| > R0, and(6.4) R0

−R0

a21(y)

a11(y)dy = 0,

R0

−R0

1

a11(y)dy = 1.(6.5)

Then there is a δ0 = δ0(λ,Λ) > 0 and an ε0 = ε0(λ,Λ) > 0 such that if

‖ϕ′‖L∞ < δ0 and ‖A−A0‖L∞ < ε0, then T and T are bounded on L2(R).

The proof of Theorem 6.2 is involved; it will be the subject of Sections 6.1–6.5.In Section 6.6, we will use Theorem 6.2 to show that Theorem 6.1 holds in specialLipschitz domains. The main tool here is the buildup scheme of David. In Sec-tion 6.7, we will pass from special Lipschitz domains to arbitrary Lipschitz domains.Many of the ideas in the proof of Theorem 6.2 are taken from [KR09], where theyare used to prove boundedness of layer potentials for A real.

We let O(λ,Λ) denote a term which, while not a constant, may be boundedby a constant depending only on λ, Λ; for example, since we know |∇ΓX(Y )| ≤C/|X − Y |, we may write ∇ΓX(Y ) = O(λ,Λ)/|X − Y |.

6.1. Proof for a small Lipschitz constant: preliminary remarks

The proof of Theorem 6.2 will use the powerful T (B) theorem of David, Journe,and Semmes. In order to state this theorem, we review some standard notation.

49

50 6. BOUNDEDNESS OF LAYER POTENTIALS

Definition 6.6. A function F is called a normalized bump function if thereexists an x0 such that for any multiindex α with |α| ≤ 2,

|∂αF (x)| ≤ 1 and suppF ⊂ B(x0, 10).

For such a function F , let FR(x) = 1RF (x/R).

Let P be an operator that maps the space of Schwartz functions to its dual. Ifthere is a constant C such that, for any R > 0 and any normalized bump functionsF and G, the equation

|〈GR, PFR〉| ≤C

R

holds, then we say the operator P is weakly bounded.

Suppose that T is a linear operator with kernel K(x, y), and that B ∈ L∞(R).TB ∈ BMO is defined as follows. If M0 is a smooth H1 atom, so suppM0 ⊆[x0 −R, x0 +R], ‖M0‖L∞ ≤ 1

R , and´M0 = 0, then

〈M0, TB〉 = 〈M0, T (ηB)〉+

ˆR

ˆR

M0(x)(K(x, y)−K(x0, y)) dx (1− η(y))B(y) dy

whenever η ∈ C∞0 , 0 ≤ η ≤ 1, and η ≡ 1 on [x0 − 2R, x0 + 2R].We now state the T (B) theorem we intend to use.

Theorem 6.7 ([DJS85, p. 42]). Suppose that B1, B2 : R 7→ C2×2 are invert-ible matrices at all points, and suppose that ‖B1‖L∞ , ‖B2‖L∞ ≤ C1.

Assume that there exist nonnegative real smooth functions vi with supp vi ⊂[−1, 1],

´vi = 1, and ‖vi‖L∞ , ‖v′i‖L∞ ≤ C2, and such that for all x ∈ R and all

t > 0,

(6.8)

∣∣∣∣(ˆ

1

tvi

(x− yt

)Bi(y) dy

)−1∣∣∣∣ ≤ C3.

Suppose that T is an operator such that, whenever F , G ∈ C∞0 (R) have disjointsupport,

〈B2G,TB1F 〉 =

ˆR

ˆR

G(x)tB2(x)tK(x, y)B1(y)F (y) dy dx

for some K(x, y) which satisfies

|K(x, y)| ≤ C4

|x− y| ,(6.9)

|K(x, y)−K(x, y′)| ≤ C4|y − y′|αmin(|x− y|, |x− y′|)1+α

,

|K(x′, y)−K(x, y)| ≤ C4|x− x′|αmin(|x− y|, |x′ − y|)1+α

for some fixed C4, α > 0 and for all x, x′, y, y′ ∈ R.Suppose that the operator f 7→ Bt2T (B1f) is weakly bounded, and that the

constants in the definition of weak boundedness are no more than C5.Suppose finally that T (B1) and T t(B2) have BMO norm no more than C6.Then T has a continuous extension to L2, and its norm may be bounded by a

constant depending only on α, C1, C2, C3, C4, C5 and C6.

6.2. A B1 FOR THE TB THEOREM 51

We intend to apply Theorem 6.7 to the operators T and T of (2.36) and (2.37).In the following sections, we will prove results that allow us to do this. In Sec-tion 6.2, we will show that the matrix B1 of (2.38) is bounded, invertible andsatisfies (6.8). Observe that by (4.8) and (4.9), the kernel conditions (6.9) hold for

T and T . In Section 6.3, we will show that f 7→ Bt2T (B1f) and f 7→ Bt2T (B1f) areweakly bounded for any bounded matrix B2. As a corollary, ‖TB1‖BMO ≤ C and

‖TB1‖BMO ≤ C.The most involved step is to find matrices B2 that satisfy the conditions of

Theorem 6.7. This will be done in Section 6.4. In Section 6.5 we will complete theproof of Theorem 6.2.

We remark that if K satisfies (6.9), if M0 is a H1 atom supported in B(x0, R),and if η ≡ 1 on B(x0, 2R), then

(6.10)

∣∣∣∣ˆR

ˆR

M0(x)(K0(x, y)−K0(x0, y)) dx (1− η(y))B(y) dy

∣∣∣∣

≤ˆR

|M0(x)|ˆ|y−x0|>2R

C|x− x0|α|y − x0|1+α

dx ‖B‖L∞ dy ≤ C‖B‖L∞ .

6.2. A B1 for the TB theorem

Recall (2.38):

B1(y) =

(a11(ψ(y)) 0a21(ψ(y)) 1

)−1 (A(ψ(y))ν(y) τ(y)

)√1 + ϕ′(y)2.

In this section, we will show that if ‖ImA‖L∞ is small enough, then B1 satis-fies (6.8).

We first show that, for any interval I, Re detfflIB1 ≥ µ for some constant µ.

We adopt the convention that unless otherwise indicated, inside integrals Aand aij are to be evaluated at ψ(y). Then

I

B1(y) dy =

I

1

a11

(1 0−a21 a11

)((a11 a12

a21 a22

)ν(y) τ(y)

)√1 + ϕ′(y)2 dy

=

I

1

a11

((a11 a12

0 detA

)ν(y)

(1 0−a21 a11

)τ(y)

)√1 + ϕ′(y)2 dy

Recalling the definitions of ν(y) and τ(y), and letting ψ(x) =(ψ1(x) ψ2(x)

), we

have that I

B1(y) dy =

I

1

a11

((a11 a12

0 detA

)(ψ′2(y)−ψ′1(y)

) (1 0−a21 a11

)(ψ′1(y)ψ′2(y)

))dy

=

I

1

a11

(a11ψ

′2(y)− a12ψ

′1(y) ψ′1(y)

−detAψ′1(y) a11ψ′2(y)− a21ψ

′1(y)

)dy.

Let

α =

I

(ψ′2(y)− a12

2a11ψ′1(y)− a21

2a11ψ′1(y)

)dy, γ =

I

1

a11ψ′1(y) dy,

β =

I

(a12

2a11− a21

2a11

)ψ′1(y) dy, δ =

I

detA

a11ψ′1(y) dy.

So I

B1(y) dy =

(α− β γ−δ α+ β

)


and so

det

I

B1(y) dy = αα+ 2i Im(αβ)− ββ + γδ.

We bound β. First,

∣∣∣∣a21

a11− a12

a11

∣∣∣∣2

=a21a21

|a11|2+a12a12

|a11|2− a12a21

a211

− a12a21

a211

=|a12 + a21|2|a11|2

− a12a21

|a11|2− a12a21

|a11|2− a12a21

a211

− a12a21

a211

If A ∈ C2×2 satisfies (1.2), then, by either letting η be a coordinate vector, orletting η =

(√Re a22 − λ ζ

√Re a11 − λ

)for an appropriate ζ ∈ C with |ζ| = 1,

we have that

Re a11 ≥ λ, Re a22 ≥ λ, |a12 + a21| ≤ 2√

(Re a11 − λ)(Re a22 − λ).

So we have the bound∣∣∣∣a21

a11− a12

a11

∣∣∣∣2

≤ 4(Re a11 − λ)(Re a22 − λ)

|a11|2− 2 Re

a12a21

a211

− 2 Rea12a21

|a11|2

=4 Re a11 Re a22

|a11|2− 4λ

Re a22 + Re a11 − λ|a11|2

− 4 Re1

a11Re

a12a21

a11

= 4 Re1

a11Re

detA

a11− 4λ

(Re a22 + Re a11 − λ)

|a11|2

= 4 Re1

a11Re

detA

a11− 4λRe

1

a11− 4λ

Re a22 − λ|a11|2

Since Re(1/a11) ≥ λ/Λ2 and Re a22 − λ > 0 are nonnegative, this implies thatRe(detA/a11) ≥ λ.

Let I = (a, b). Because A(y, s) = A(y), we have that

α =ψ2(b)− ψ2(a)

b− a − 1

b− a

ˆ ψ1(b)

ψ1(a)

(a12(y)

2a11(y)+

a21(y)

2a11(y)

)dy,

β =

I

(a12

2a11− a21

2a11

)ψ′1(y) dy =

1

b− a

ˆ ψ1(b)

ψ1(a)

(a12(y)

2a11(y)− a21(y)

2a11(y)

)dy,

γ =

I

1

a11ψ′1(y) dy =

1

b− a

ˆ ψ1(b)

ψ1(a)

1

a11(y)dy,

δ =

I

detA

a11ψ′1(y) dy =

1

b− a

ˆ ψ1(b)

ψ1(a)

detA(y)

a11(y)dy.

So

|β| = 1

2

∣∣∣∣1

b− a

ˆ ψ1(b)

ψ1(a)

(a12(y)

a11(y)− a21(y)

a11(y)

)dy

∣∣∣∣

≤ 1

2√b− a

(ˆ ψ1(b)

ψ1(a)

∣∣∣∣a21(y)

a11(y)− a12(y)

a11(y)

∣∣∣∣2

1

Re(1/a11(y))dy

)1/2 ∣∣∣∣Re γ

∣∣∣∣1/2

.

6.2. A B1 FOR THE TB THEOREM 53

Applying the bound on a21(y)a11(y) −

a12(y)a11(y) we see that

|β| ≤ 1

2√b− a

∣∣∣∣ˆ ψ1(b)

ψ1(a)

(4 Re

detA(y)

a11(y)− 4λ

)∣∣∣∣1/2

|Re γ|1/2

=

∣∣∣∣Re δ − λψ1(b)− ψ1(a)

b− a

∣∣∣∣1/2

|Re γ|1/2

We want to bound∣∣det

fflIB1

∣∣ from below. Since Re γ and Re δ−λψ1(b)−ψ1(a)b−a

have the same sign, we have that

Re det

I

B1 = αα− ββ + Re γRe δ − Im γ Im δ

≥ αα−Re γ

(Re δ − λψ1(b)− ψ1(a)

b− a

)+ Re γRe δ − Im γ Im δ,

= αα+ λRe γψ1(b)− ψ1(a)

b− a − Im γ Im δ.

But

|α| ≥ |ψ2(b)− ψ2(a)|b− a − (Λ/λ)

|ψ1(b)− ψ1(a)|b− a , Re γ ≥ (λ/Λ2)

|ψ1(b)− ψ1(a)|b− a .

Thus,

Re det

I

B1 ≥ (1/C)|ψ(b)− ψ(a)|2

(b− a)2− Im γ Im δ ≥ 1/C − Im γ Im δ.

Since ‖A − A0‖L∞ is small, Im γ and Im δ are small; thus, there is a µ > 0 suchthat µ < Re det

fflIB1 for all intervals I.

But (6.8) requires a smooth average of B1.

Lemma 6.11. Suppose that B : R 7→ C2×2 satisfies |ξ · B(x)η| ≤ M |η||ξ| forall x ∈ R2, η, ξ ∈ C2. Assume further that for some number µ > 0,

(6.12)

∣∣∣∣det

I

B(x) dx

∣∣∣∣ ≥ µ

for all intervals I ⊂ R.Then there is a smooth real function v, with 0 ≤ v ≤ 1,

´v = 1, supp v ⊂

[−1, 1], and ‖v′‖ ≤ C(µ,M), such that if vt(x) = 1t v(xt

), then |detB ∗vt(x)| ≥ µ/2

for all t > 0 and all x ∈ R.

Since B ∗ vt(x) is invertible with bounded inverse if and only if |detB ∗ vt(x)|is bounded from below, this completes the proof that B1 satisfies (6.8).

Proof. Choose some v such that v ≡ 1 on(− 1

2 ,12

), supp v ⊂

[− 1

2 − ρ, 12 + ρ

],

and´v = 1, 0 ≤ v ≤ 1, and ‖v′‖L∞ ≤ 2

ρ for some positive real number ρ < 1/2 to

be determined. Then

vt ∗B(x) =

ˆvt(x− y)B(y) dy

=1

t

ˆ x+t/2

x−t/2B(y) dy +

ˆt/2<|y|<t/2+tρ

B(y)vt(x− y) dy


and so since |vt(x− y)| ≤ 1/t for all x, y,∣∣∣∣vt ∗B(x)−

x+t/2

x−t/2B(y) dy

∣∣∣∣ ≤ 2ρ‖B‖L∞ .

Therefore,|det vt ∗B(x)| ≥ µ− Cρ‖B‖2L∞ .

So if ρ < µ/CM2, then |det vt ∗B(x)| ≥ µ/2 for all x.

6.3. Weak boundedness of operators

For A real-valued, weak boundedness of T and T is shown in [KR09], in Lem-mas 4.3, 4.7, 4.8, and 4.10. We prove this for A complex-valued. In this section,we let S denote the class of Schwartz functions (not the single layer potential).

Lemma 6.13. The operators T± are continuous linear operators from B1S2×2

to (B2S2×2)′ for any bounded matrix B2. The map F 7→ Bt2 T±(B1F ) is weaklybounded; in fact, ‖T±(B1FR)‖L∞ ≤ C/R for any normalized bump function F .

Proof. We know from Lemma 5.7 that if f : ∂Ω 7→ C is well-behaved, thenT±(B1f)(x) is pointwise well-defined. Recall that

B1(y) = (BT0 (ψ(y))t)−1(A(ψ(y))ν(y) τ(y)

)√1 + ϕ′(y)2.

Choose some F , G ∈ S2×2 (that is, matrices whose components are all in S). Recallthat

〈B2G,T±(B1F )〉 = limh→0±

ˆR

ˆR

G(x)tB2(x)tKh(x, y)B1(y)F (y) dy dx

Now, the components of

Kh(x, y)B1(y) =



)(A(ψ(y))ν(y) τ(y)

)√1 + ϕ′(y)2

ared

dyΓTψ(x,h)(ψ(y)) and

d

dyΓTψ(x,h)(ψ(y)).

Let f be a component of F . We need only show that

limh→0+

ˆR

f(y)d

dyΓTψ(x,h)(ψ(y)) dy, lim

h→0+

ˆR

f(y)d

dyΓTψ(x,h)(ψ(y)) dy

are bounded by C‖F‖S , and that if F = FR for some normalized bump functionF , then those integrals are at most C/R.

For any number R > 0, we have that∣∣∣∣ˆR

f(y)d


∣∣∣∣

≤∣∣∣∣ˆ|x−y|>2R

f(y)d


∣∣∣∣+

∣∣∣∣f(x)

ˆ x+2R

x−2R

d


∣∣∣∣

+

∣∣∣∣ˆ x+2R

x−2R

(f(y)− f(x))d


∣∣∣∣.

By (4.1), the first integrand is at most CR |f(y)|, and so the first integral is at most

CR‖f‖L1 . The second integral is equal to ΓTψ(x,h)(ψ(x+ 2R))− ΓTψ(x,h)(ψ(x− 2R));

6.3. WEAK BOUNDEDNESS OF OPERATORS 55

by (4.1) this has norm at most C, and so the second term is at most C|f(x)|.Finally, the last integrand is at most

‖f ′‖L∞ |x− y|∣∣∣∣d

dyΓTψ(x,h)(ψ(y))

∣∣∣∣ ≤ C‖f ′‖L∞

and so∣∣∣∣ˆR

f(y)d


∣∣∣∣ ≤C

R‖f‖L1 + C‖f‖L∞ + CR‖f ′‖L∞ .

Picking R = 1, we see that this is bounded by the Schwartz norm of f ; letting Rbe arbitrary and letting F = FR where F is a normalized bump function, we seethis is at most C/R.

Similarly,

limh→0+

ˆR

d

dyΓTψ(x,h)(ψ(y))f(y) dy

exists and bounded by C/R provided F = FR for some normalized bump func-tion F .

Lemma 6.14. The operator T± is a continuous linear operator from B1S2×2 to

(B2S2×2)′ for any bounded B2. The map F 7→ Bt2 T±B1F is weakly bounded; in

fact, ‖T±(B1FR)‖L∞ ≤ C/R for any normalized bump function F .

Proof. Fix F , G ∈ S2×2. Again, we wish to show that 〈B2G, T (B1F )〉 existsand is bounded, and that if F = FR, G = GR for some normalized bump functionsF , G, then |〈B2G, T (B1F )〉| ≤ 1/R. As before, by Lemma 5.7 the limits exist.

Now,

〈B2G, T±(B1F )〉 = limh→0±

ˆR

G(x)B2(x)

ˆR

Kh(x, y)B1(y)F (y) dy dx.

But

Kh(x, y)B1(y) =

(∇Y Γψ(y)(ψ(x, h))t

∇Y Γψ(y)(ψ(x, h))t

)(A(ψ(y))ν(y) τ(y)

)√1 + ϕ′(y)2.

Observe that√

1 + ϕ′(y)2∇Y Γψ(y)(ψ(x, h)) · τ(y) = ddy Γψ(y)(ψ(x, h)), and so

we may deal with this component as in the proof of Lemma 6.13. We are left tryingto show that

limh→0±

ˆR

ν(y) ·AT (ψ(y))∇Y Γψ(y)(ψ(x, h))f(y)√

1 + ϕ′(y)2 dy

is bounded and converges uniformly in x, for any f a component of F .But this integral isˆ

∂Ω

ν(Y ) ·AT (Y )∇Y ΓY (ψ(x, h))f(ψ−1(Y )) dσ(Y ).

Let m ∈ C∞0 (R) with m ≡ 1 on (−R − R‖ϕ′‖L∞ , R + R‖ϕ′‖L∞), and 0 ≤m ≤ 1, suppm ⊂ (−CR,CR), and |m′| < C/R. Let u(ψ(y, t)) = f(y)m(t+ ϕ(y)),so u(ye⊥ + te) = f(y)m(t) and |∇u(ye⊥ + te)| ≤ |f(y)||m′(t)| + |f ′(y)||m(t)| ≤


C‖f ′‖L∞ + C‖f‖L∞/R. Then, provided ±h > 0, by (4.7) and (1.5) we have thatˆ∂Ω

ν(Y ) ·AT (Y )∇Y ΓY (ψ(x, h))f(ψ−1(Y )) dσ(Y )

= ∓ˆ

Ω∓

AT (Y )∇Y ΓY (ψ(x, h)) · ∇u(Y ) dY.

Therefore,∣∣∣∣ˆ∂Ω

ν(Y ) ·AT (Y )∇Y ΓY (ψ(x, h))f(ψ−1(Y ))dσ(Y )

∣∣∣∣

=

∣∣∣∣ˆ

Ω∓

AT (Y )∇Y ΓY (ψ(x, h)) · ∇u(Y ) dY

∣∣∣∣ ≤ˆ

Ω∓

C

|Y − ψ(x, h)| |∇u(Y )| dY.

Since |∇u| is bounded and in L1(R2), and since C/|Y−ψ(x, h)| ∈ L1(R2)+L∞(R2),

uniformly in h, this integral is also bounded. Thus, T is continuous on B1S.Furthermore, if f is a component of a normalized bump function, then |f | ≤

C/R, |f ′| ≤ C/R2, |supp f | ≤ CR, so |∇u| ≤ C/R2 and is supported in someball of radius CR. So if F = FR where F is a normalized bump function, then|T (B1F )(x)| ≤ C/R.

So F 7→ Bt2T (B1F ) is weakly bounded as well.

Corollary 6.15. We have that ‖T±(B1)‖BMO ≤ C and ‖T±(B1)‖BMO ≤ C.

Proof. Recall that if M0 is a smooth H1 atom and η ∈ C∞0 is 1 in a neigh-borhood of its support, then

〈M0, TB〉 = 〈M0, T (ηB)〉+ˆR

ˆR

M0(x)(K0(x, y)−K0(x0, y)) dx (1−η(y))B(y) dy.

By (6.10), we need only bound 〈M0, T (ηB1)〉 and 〈M0, T (ηB1)〉. In fact, we need

only bound ‖T (ηB1)‖L∞ and ‖T (ηB1)‖L∞ . If supp η ⊂ B(x0, R), then η = RFRfor some normalized bump function F , and so this follows immediately from Lem-ma 6.13 and Lemma 6.14.

6.4. The adjoint inequalities

In this section we will find bounded, invertible matrices Bk which satisfy (6.8),

and such that we can make some statement about ‖T tBk‖BMO and ‖T tBk‖BMO.These matrices will be needed to prove Theorem 6.2 from Theorem 6.7.

In order to prove these lemmas, we begin by investigating some integrals in-volving ΓX , ΓX for X /∈ ∂Ω.

If X ∈ Ω and f ∈ C∞0 (R2), then the divergence theorem and (2.8) tells us that

−f(X) =

ˆΩ

∇f ·A∇ΓX +

ˆΩC∇f ·A∇ΓX

=

ˆΩ

div(ΓXAT∇f)−

ˆΩ

ΓX divAT∇f +

ˆΩC∇f ·A∇ΓX

and so since divA∇ΓX = 0 in ΩC ,

(6.16) −f(X) +

ˆΩ

ΓX divAT∇f =

ˆ∂Ω

ΓX ν ·AT∇f dσ −ˆ∂Ω

f ν ·A∇ΓX dσ.

6.4. THE ADJOINT INEQUALITIES 57

By taking the gradient in X of (6.16), we have that

∇Xˆ

ΩCΓX(Y ) div(AT (Y )∇f(Y )) dY −∇f(X)

= ∇Xˆ∂Ω

ΓX(Y ) ν(Y ) ·AT (Y )∇f(Y ) dσ(Y )

−∇Xˆ∂Ω

f(Y ) ν(Y ) ·A(Y )∇ΓX(Y ) dσ(Y )

We may simplify this:

∇Xˆ∂Ω

f ν ·A∇ΓX dσ = ∇XˆR

f(ψ(y)) ν(y) · (A(ψ(y))∇ΓX(ψ(y))√

1 + ϕ′(y)2 dy

= ∇XˆR

f(ψ(y))τ(y) · ∇ΓX(ψ(y))√

1 + ϕ′(y)2 dy

= −ˆR

∇X ΓX(ψ(y))d

dyf(ψ(y)) dy

and

ˆ∂Ω

∇XΓX ν ·AT∇f dσ

=

ˆR

∇XΓX(ψ(y)) ν(y) ·AT (ψ(y))∇f(ψ(y))√

1 + ϕ′(y)2 dy.

So

(6.17)

ˆΩC∇XΓX(Y ) div(AT (Y )∇f(Y )) dY −∇f(X)

=

ˆR

∇XΓX(ψ(y)) ν(y) ·AT (ψ(y))∇f(ψ(y))√

1 + ϕ′(y)2 dy

+

ˆR

∇X ΓX(ψ(y))d

dyf(ψ(y)) dy.

We now consider functions f of a special form. Let f(y, s) = ρ(y, s)g(y, s).We require that there exist an R0 > 0 such that divAT (y)∇g(y, s) = 0 outside ofB(0, R0). We further require that |∇g(y, s)| be bounded and that g(0, 0) = 0, sothat |g(X)| ≤ ‖∇g‖L∞ |X|. We assume that ρ ≡ 1 on B(0, R), ρ ∈ C∞0 (B(0, R+1)),and |ρ′| < C, |ρ′′| < C.

Then

∇f(y, s) = ∇g(y, s)ρ(y, s) + g(y, s)∇ρ(y, s)

and so

AT (y)∇f(y, s) = ρ(y, s)AT (y)∇g(y, s) + g(y, s)AT (y)∇ρ(y, s).

Then

divAT (y)∇f(y, s) = ρ(y, s) divAT (y)∇g(y, s) +∇ρ(y, s) ·AT (y)∇g(y, s)

+∇g(y, s) ·AT (y)∇ρ(y, s) + g(y, s) divAT (y)∇ρ(y, s).


Then

ˆΩC∇XΓX divAT∇f =

ˆΩC∇XΓXρ divAT∇g +

ˆΩC∇XΓX ∇ρ ·AT∇g

+

ˆΩC∇XΓX ∇g ·AT∇ρ+

ˆΩC∇XΓXg divAT∇ρ.

Assume that |X| < R/2. The second and third integrands are zero away fromsupp∇ρ = B(0, R + 1) \ B(0, R), where they are at most C‖∇g‖L∞/R; thus, thesecond and third integrals are O(λ,Λ)‖∇g‖L∞ .

The first integral is O(λ,Λ)R0‖divAT∇g‖L∞ ; note that it is zero if divAT∇g ≡0.

So

ˆΩC

∂xiΓX(Y ) divAT (Y )∇f(Y ) dY

= O(λ,Λ)(‖∇g‖L∞ +R0‖divAT∇g‖L∞) +

ˆΩC

∂xiΓXg divAT∇ρ

= O(λ,Λ)(‖∇g‖L∞ +R0‖divAT∇g‖L∞)

+

ˆΩC

div(∂xiΓXgA

T∇ρ)−ˆ

ΩC∇ (∂xiΓXg) ·AT∇ρ.

The second integrand is zero away from supp∇ρ, where it is at most C‖∇g‖L∞/R,so as before the second integral is O(λ,Λ)‖∇g‖L∞ . By the divergence theorem thefirst integral is equal to

−ˆ∂Ω

ν ·(∂xiΓX(Y )g(Y )AT (Y )∇ρ(Y )

)dY

which is O(λ,Λ)‖∇g‖L∞ .Now,

ˆ∂Ω

∇XΓX(Y ) ν(Y ) ·AT (Y )∇f(Y ) dσ(Y )

=

ˆ∂Ω

∇XΓX(ρ ν ·AT∇g + g ν ·AT∇ρ

)dσ

=

ˆ∂Ω

∇XΓX ρ ν ·AT∇g dσ +O(λ,Λ)‖∇g‖L∞

and

ˆR

∇X ΓX(ψ(y))d

dyf(ψ(y)) dy

=

ˆR

∇X ΓX(ψ(y))

(ρ(ψ(y))

d

dyg(ψ(y)) + g(ψ(y))

d

dyρ(ψ(y))

)dy

=

ˆR

∇X ΓX(ψ(y))ρ(ψ(y))d

dyg(ψ(y)) dy +O(λ,Λ)‖∇g‖L∞


So, letting η(y) = ρ(ψ(y)), we have that by (6.17),

O(λ,Λ)(‖∇g‖L∞ +R0‖divAT∇g‖L∞)

=

ˆR

∇XΓX(ψ(y)) η(y) ν(y) ·AT (ψ(y))∇g(ψ(y))√

1 + ϕ′(y)2 dy

+

ˆR

∇X ΓX(ψ(y)) η(y)d

dyg(ψ(y)) dy.

Let

β1(y) = β1(y; g) =d

dyg(ψ(y)),(6.18)

β2(y) = β2(y; g) = −ν(y) ·AT (ψ(y))∇g(ψ(y))√

1 + ϕ′(y)2.

Then

(6.19) O(λ,Λ)(‖∇g‖L∞ +R0‖divAT∇g‖L∞)

= −ˆR

∇XΓX(ψ(y)) η(y)β2(y; g) dy +

ˆR

∇X ΓX(ψ(y)) η(y)β1(y; g) dy.

Recall that we seek matrices Bk that are bounded and satisfy (6.8) and such

that T t±Bk ∈ BMO or T t±Bk ∈ BMO. By (4.12) and Lemma 5.7, if A, ϕ and F

are smooth and F ∈ L2(R) then T±F (x) = T ′∓F (x) and T±F (x) = T ′∓F (x). Thus,

we may instead seek matrices Bk such that (T ′)tBk and (T ′)tBk lie in BMO.Observe that by (2.41) and (2.42) we have that

〈ηβI, T ′±M0〉 = limh→0±

ˆR2

η(y)β(y)

(∇ΓTψ(y)(ψ(x, h))t

∇ΓTψ(y)(ψ(x, h))t

)BT0 (ψ(x, h))tM0(x) dx dy,

〈ηβI, T ′±M0〉 = limh→0±

ˆR2

η(y)β(y)

(∇X Γψ(x,h)(ψ(y))t

∇X Γψ(x,h)(ψ(y))t

)BT0 (ψ(x, h))tM0(x) dx dy.

So by (6.19),

〈M0, (T′±)t(ηβ2I)〉 − 〈M0, (T

′±)t(ηβ1I)〉

= limh→0±

ˆR

M0(x)tBT0 (ψ(x, h))

ˆR

η(y)β2(y)

(∇ΓTψ(y)(ψ(x, h))t

∇ΓTψ(y)(ψ(x, h))t

)tdy dx

−ˆR

M0(x)tBT0 (ψ(x, h))

ˆR

η(y)β1(y)

(∇X Γψ(x,h)(ψ(y))t

∇X Γψ(x,h)(ψ(y))t

)tdy dx

= O(λ,Λ)(‖∇g‖L∞ +R0‖divAT∇g‖L∞)‖M0‖L1(R).

In combination with (6.10), this implies that if g((0, 0)) = 0, if ∇g is bounded, andif there exists an R0 > 0 such that divAT (y)∇g(y, s) = 0 outside of B(0, R0), then

‖(T ′±)t(β2I)‖BMO ≤ ‖(T ′±)t(β1I)‖BMO + C(‖∇g‖L∞ +R0‖divAT∇g‖L∞),

(6.20)

‖(T ′±)t(β1I)‖BMO ≤ ‖(T ′±)t(β2I)‖BMO + C(‖∇g‖L∞ +R0‖divAT∇g‖L∞).

(6.21)

We remark that if Bk(x) = β(x)I, and if 1/C ≤ Reβ(x) and |β(x)| ≤ C for allx ∈ R, then Bk is bounded, invertible and satisfies (6.8).


We now use (6.20) to find a matrix B3 such that T t(B3) ∈ BMO. We willultimately use this matrix B3 to establish that ‖T‖L2 7→L2 is finite.

Lemma 6.22. Suppose that A satisfies (6.3), (6.4) and (6.5), that A and ϕ aresmooth, and that for some (large) R0, A(y) = I and ϕ(y) = 0 for |y| > R0. Supposefurther that ‖ϕ′‖L∞ ≤ 1/2.

There exists a bounded matrix B3 such that

‖T t(B3)‖BMO ≤ C(λ,Λ, R0, ‖A′‖L∞ , ‖ϕ′′‖L∞)

Furthermore, if ‖ϕ′‖ is small enough, then B3 is invertible and satisfies (6.8).

Proof. First, we consider the case where e2 = 0 (that is, that Ω is the domainto the left or the right of a graph). Then either e1 = 1 or e1 = −1; for simplicitywe consider only the case where e1 = 1. Let ζ be a smooth cutoff function, withζ ≡ 1 on (−R0, R0), ζ ≡ 0 outside of (−2R0, 2R0), and |ζ ′| < C/R0, |ζ ′′| ≤ C/R2

0.Define

g(x, t) =

ˆ x

ζ(x)ϕ(−t)

1

a11(w)dw.

We have that

∇g(x, t) =

(1

a11(x) −ζ′(x)ϕ(−t)

a11(ζ(x)ϕ(−t))ζ(x)ϕ′(−t)

a11(ζ(x)ϕ(−t))

)

and so outside of (−2R0, R0) × (−2R0, R0), we have that ∇g(x, t) =(

1a11(x) 0

).

Therefore, |divAT∇g| ≤ C(‖ϕ′′‖L∞ + ‖A′‖L∞), and divAT (y)∇g(y, s) = 0 outside

of B(0, 2√

2R0).Observe that ∂Ω = (ϕ(−t), t) : t ∈ R and ψ(y) =

(ϕ(y) −y

). So g ≡ 0 on

∂Ω, and

ν(y) ·AT (ψ(y))∇g(ψ(y))√

1 + ϕ′(y)2 = − 1

a11(ϕ(y))

(1

ϕ′(y)

)·AT (ϕ(y))

(1

ϕ′(y)

).

By (6.18),

β1(y; g) =d

dyg(ψ(y)) = 0,

β2(y; g) =1

a11(ϕ(y))

(1

ϕ′(y)

)·AT (ϕ(y))

(1

ϕ′(y)

)

So by (6.20)

‖T t(B3)‖BMO ≤ C(λ,Λ, R0, ‖A′‖L∞ , ‖ϕ′′‖L∞)

where

B3(y) =1

a11(ϕ(y))

(1

ϕ′(y)

)·AT (ϕ(y))

(1

ϕ′(y)

)(1 00 1

)

is bounded, invertible, and if ‖ϕ′‖L∞ is small enough, then B3 satisfies (6.8).Now, consider the case where e2 6= 0. If |x| > R0/|e2|, then ϕ(x) = 0 and

ψ(x) = (xe2,−xe1) and |ψ1(x)| = |xe2| > R0, so A(ψ(x)) = I.Assume that R1 ≥ R0/|e2| is large enough (to be determined later). Choose

ζ ∈ C∞ such that ζ ≡ 1 on (−R1, R1), ζ ≡ 0 outside of (−2R1, 2R1), and |ζ ′| <


C/R1, |ζ ′′| < C/R21. Let

g(x, t) = e2t+ (1− ζ(t))

(ˆ x

−R0

e1 − e2a21(w)

a11(w)dw

)

+ ζ(t) (e1x− ϕ(e2x− e1t) + e1R0) .

If |x| ≥ R0, our a priori assumption´ R0

−R0

e1−e2a21

a11= 2R0e1 means that

g(x, t) = e2t+ e1x+ e1R0 − ζ(t)ϕ(e2x− e1t)

So if |y| ≥ R0/|e2| ≥ R0, then ϕ(y) = 0, and

g(ψ(y)) = g(ye2,−ye1) = e1R0 − ζ(−ye1)ϕ(y) = e1R0.

Conversely, we may take R1 large enough that if |y| < R0/|e2|, then |ψ(y)| <R1. So ζ(ψ2(y)) = 1, and so

g(ψ(y)) = g(ye2 + ϕ(y)e1,−ye1 + ϕ(y)e2)

= e2(−ye1 + ϕ(y)e2) + e1(ye2 + ϕ(y)e1)− ϕ(y) + e1R0 = e1R0.

So g is constant on ∂Ω.Now, consider

∇g(x, t) =

((1− ζ(t)) e1−e2a21(x)

a11(x) + e1ζ(t)− e2ζ(t)ϕ′(e2x− e1t)

e2 + e1ζ(t)ϕ′(e2x− e1t)

)

+ ζ ′(t)

(0

−´ x−R0

e1−e2a21(w)a11(w) dw + e1x− ϕ(e2x− e1t) + e1R0

).

If |t| > 2R1, or if |x|, |e1t− e2x| > R0, then

∇g(x, t) =

(e1−e2a21(x)

a11(x)

e2

)

and so divAT (x)∇g(x, t) = 0. So divAT∇g is zero outside a bounded set. Clearly,it is bounded in this set by a constant which depends only on A′, ϕ′′, R0, λ, Λ.

But ddy g(ψ(y)) = 0. By (6.18) and (6.20), if

B3(y) = β2(y; g)I = ν(y) ·AT (ψ(y))∇g(ψ(y))√

1 + ϕ′(y)2

(1 00 1

)

then

‖T t(B3)‖BMO ≤ C(λ,Λ, R0, 1/|e2|, ‖A′‖L∞ , ‖ϕ′′‖L∞).

But

∇g(ψ(y)) =√

1 + ϕ′(y)2 ν(y).

So if ‖ϕ′‖L∞ is small enough, then B3 is bounded, invertible and satisfies(6.8).

We have now found a matrix B3 such that ‖T t(B3)‖BMO is finite. However,‖T t(B3)‖BMO could potentially be very large. We would like to find a matrix B4

such that ‖T t(B4)‖BMO ≤ C. We will do this in a roundabout fashion.


Lemma 6.23. There exist bounded matrices B4, B5, B6 and B7 such that

‖(T ′)tB4‖BMO ≤ C + C‖(T ′)tB6‖BMO, ‖(T ′)tB5‖BMO ≤ C + C‖(T ′)tB7‖BMO

where the constant C depends only on the ellipticity constants λ and Λ and on‖ϕ′‖L∞ .

Furthermore, if ‖A − A0‖L∞ and ‖ϕ′‖L∞ are small enough, then B4 and B5

are invertible and satisfy (6.8), and

‖B6‖L∞ ≤ C(‖A−A0‖L∞ + ‖ϕ′‖L∞).

Proof. Let ξ, ζ be constants. Choose gξ,ζ(y, s) = ξs+mξ,ζ(y), where

mξ,ζ(y) =

ˆ y

0

ζ − ξa21(z)

a11(z)dz.

Then

∇gξ,ζ(y, s) =

(ζ−ξa21(y)a11(y)

ξ

)

and so

AT (y)∇gξ,ζ(y, s) =

(ζ

a12(y) ζ−ξa21(y)a11(y) + ξa22(y)

)=

(ζ

ζa12(y)+ξ detA(y)a11(y)

),

divAT∇gξ,ζ ≡ 0,

gξ,ζ(ψ(y)) = g(ye2 + ϕ(y)e1,−ye1 + ϕ(y)e2) = ξϕ(y)e2 − ξye1 +m(ye2 + ϕ(y)e1).

Let

β1(y, ζ, ξ) = β1(y; gξ,ζ) = ξϕ′(y)e2 − ξe1 +ζ − ξa21(ψ(y))

a11(ψ(y))(ϕ′(y)e1 + e2),

β2(y, ζ, ξ) = β2(y; gξ,ζ)

= ζ(e1 − ϕ′(y)e2) +ζa12(ψ(y)) + ξ detA(ψ(y))

a11(ψ(y))(ϕ′(y)e1 + e2)

Define β4, β5, β6, and β7 as follows:

β4(y) = β2(y, e1, e2) = e21 − ϕ′(y)e1e2 +

e1a12(y) + e2 detA(y)

a11(y)(ϕ′(y)e1 + e2),

β5(y) = β1

(y,

Λ4

λ3e2,−e1

)= (e1)2 − ϕ′(y)e1e2 +

Λ4

λ3 e2 + e1a21(y)

a11(y)(ϕ′(y)e1 + e2),

β6(y) = β1(y, e1, e2) = ϕ′(y)e22 − e2e1 +

e1 − e2a21(y)

a11(y)(ϕ′(y)e1 + e2)

β7(y) = β2

(y,

Λ4

λ3e2,−e1

)

= −Λ4

λ3ϕ′(y)(e2)2 +

Λ4

λ3e1e2 +

Λ4

λ3 e2a12(y)− e1 detA(y)

a11(y)(ϕ′(y)e1 + e2)

where y = ψ1(y).Then if Bi = Iβi, by (6.20) and (6.21),

‖(T ′)tB4‖BMO ≤ C + C‖(T ′)tB6‖BMO, ‖(T ′)tB5‖BMO ≤ C + C‖(T ′)tB7‖BMO.

Note the following:

6.5. PROOF FOR A SMALL LIPSCHITZ CONSTANT: FINAL REMARKS 63

• Re 1a11

= Re a11

|a11|2 ≥λ

Λ2 .

• If ‖ϕ′‖L∞ is small enough, then β5(y) is close to

(e1)2 +Λ4

λ3 (e2)2 + e1e2a21(y)

a11(y)

and so

Reβ5(y) ≥ 1

C

((e1)2 +

Λ2

λ2(e2)2 − |e1e2|

Λ

λ

)≥ 1

2C

since e21 + e2

2 = 1.• Similarly, if ‖ϕ′‖L∞ is small enough, then β4(y) is close to

1

a11(y)

(a11(y)(e1)2 + a12(y)(e1e2) + (e2)2 detA(y)

)

and if ‖A−A0‖ is small enough, so that a11−1, a21 are small, then β4(y)is close to

1

a11(y)

(e1 e2

)A(y)

(e1

e2

)

so Reβ4(y) ≥ λ2

Λ2 .• Finally,

|β6(y)| =∣∣∣∣ϕ′(y)e2

2 + ϕ′(y)e2

1 − e1e2a21(y)

a11(y)+e1e2(1− a11(y))− e2

2a21(y)

a11(y)

∣∣∣∣≤ C(‖1− a11‖L∞ + ‖a21‖L∞ + ‖ϕ′‖L∞)

≤ C(‖A−A0‖L∞ + ‖ϕ′‖L∞).

6.5. Proof for a small Lipschitz constant: final remarks

We recall the statement of Theorem 6.2.

Theorem. Suppose that ϕ is smooth and compactly supported. Suppose that A0

and A are smooth and satisfy (6.3), (6.4) and (6.5). Then there is a δ0 = δ0(λ,Λ) >0 and an ε0 = ε0(λ,Λ) > 0 such that if ‖ϕ′‖L∞ < δ0 and ‖A−A0‖L∞ < ε0, then T

and T are bounded on L2(R).

We remind the reader of Theorem 6.7; we wish to use it to prove Theorem 6.2.

Theorem ([DJS85, p. 42]). Suppose that B1, B2 : R 7→ C2×2 are invertibleat all points and are uniformly bounded. Assume that B1 and B2 satisfy (6.8).

Suppose that T is a linear operator with kernel K(x, y), and that K(x, y) sat-isfies (6.9).

Suppose that f 7→ Bt2T (B1f) is weakly bounded, and that T (B1) and T t(B2) liein BMO.

Then T has a continuous extension to L2, and its norm depends only on‖B1‖L∞ , ‖B2‖L∞ , ‖T (B1)‖BMO, ‖T t(B2)‖BMO, and the constants in (6.8), (6.9)and the definition of weak boundedness.

A converse to Theorem 6.7 is the statement that, for such an operator T ,

‖TB‖BMO ≤ C(C5 + ‖T‖L2 7→L2)‖B‖L∞ .This is a classic result of Calderon-Zygmund theory; see, for example, [Gra09,Theorem 8.2.7].


Proof of Theorem 6.2. From Section 6.2, we know that the matrix B1 of(2.38) is bounded, invertible, and satisfies (6.8). By Lemmas 6.13 and 6.14, f 7→Bt2T (B1f) and f 7→ Bt2T (B1f) are weakly bounded for any bounded matrix B2.

By Corollary 6.15, ‖TB1‖BMO ≤ C and ‖TB1‖BMO ≤ C.By Theorem 6.7, if B2 is bounded and satisfies (6.8), then

‖T‖L2 7→L2 ≤ C + C‖TB1‖BMO + C‖T tB2‖BMO ≤ C + C‖T tB2‖BMO,

‖T‖L2 7→L2 ≤ C + C‖TB1‖BMO + C‖T tB2‖BMO ≤ C + C‖T tB2‖BMO.

By Lemma 6.22, if A and ϕ satisfy the hypotheses of Theorem 6.2, and if‖ϕ′‖L∞ is small enough, then there exists a B3 such that ‖T t(B3)‖BMO is finite.So ‖T‖L2 7→L2 is finite. Unfortunately, our bound on ‖T t(B3)‖BMO depends onquantities such as ‖A′‖L∞ ; therefore, we will seek a better bound on ‖T‖L2 7→L2 .

By Lemma 6.23, there exist matrices B4, B5, B6 and B7 such that

‖T tB5‖BMO ≤ C + C‖T tB7‖BMO, ‖T tB4‖BMO ≤ C + C‖T tB6‖BMO,

where B4 and B5 satisfy (6.8) and B6 is small (depending on ‖A − A0‖L∞ and‖ϕ′‖L∞). This implies that

‖T‖L2 7→L2 ≤ C + C‖T tB4‖BMO ≤ C + C‖T tB6‖BMO ≤ C + C‖T‖L2 7→L2‖B6‖L∞

≤ C + C(C + ‖T tB5‖BMO

)‖B6‖L∞ ≤ C + C‖T tB7‖BMO‖B6‖L∞

≤ C + C‖T‖L2 7→L2‖B7‖L∞‖B6‖L∞Since ‖T‖L2 7→L2 is finite, if ‖B6‖L∞ is small enough then ‖T‖L2 7→L2 ≤ C.

Finally, T is also bounded on L2:

‖T‖L2 7→L2 ≤ C + C‖T tB5‖BMO ≤ C + C‖T‖L2 7→L2‖B7‖L∞ ≤ C.

6.6. Buildup to arbitrary special Lipschitz domains

In this section, we will prove that Theorem 6.1 holds in special Lipschitz do-

mains. Recall that if Ω is a special Lipschitz domain, then T or T is bounded onL2(∂Ω) if and only if T or T is bounded on L2(R). For simplicity, we work only

with T and T ; the proof for T and T is identical.

Theorem 6.24. Theorem 6.2 holds if we relax the conditions (6.4) and (6.5) onA and the requirement that ϕ ∈ C∞0 , and replace the requirement that ‖ϕ′‖L∞ < δ0with the requirement that ‖ϕ′−γ‖L∞ < δ0 for some γ ∈ R, and permit δ0 to dependon γ as well as λ, Λ.

Proof. Recall that T±F (x) = limh→0± ThF (x), where

ThF (x) =

ˆR

Kh(x, y)F (y) dy.

If F is well-behaved, we know that the limits exist for arbitrary A and ϕ; seeLemma 5.7.

Observe that if x /∈ suppF , then T±F (x) =´RK0(x, y)F (y) dy; thus, T± is a

Calderon-Zygmund operator with kernel K0. Define

ThF (x) =

ˆ|x−y|>h

K0(x, y)F (y) dy.

6.6. BUILDUP TO ARBITRARY SPECIAL LIPSCHITZ DOMAINS 65

It is well-known (see, for example, [Ste93, p. 34] or [Gra09, Theorem 8.2.3]) thatif T± is bounded on L2(R), then the operators Th are bounded on L2(R) uniformlyin h.

Observe that as in the proof of [DV90, Proposition 4.3],

|ThF (x)− ThF (x)| ≤∣∣∣∣ˆ|x−y|>h

(Kh(x, y)−K0(x, y))F (y) dy

∣∣∣∣

+

∣∣∣∣ˆ|x−y|≤h

Kh(x, y)F (y) dy

∣∣∣∣.

By (2.34) and (4.8), this is at most CMF (x), where M is the Hardy-Littlewoodmaximal operator. Since M is bounded on Lp for any 1 < p <∞, we have that ifT± is bounded on L2(R) then the operators Th are bounded on L2(R), uniformlyin h.

We first remove the requirements (6.4) and (6.5). Assume that A is smoothand satisfies (2.1). Pick some F ∈ L2(R). Observe that

|ThF (x)| ≤∣∣∣∣ˆ R

−RF (y)Kh(x, y), dy

∣∣∣∣+

ˆ|y|>R

|F (y)| C

|x− y|+ |h| dy.

Let µ = ‖F‖L2(R\(−R,R)). Then for all x with |x| < R/2 we have that

|ThF (x)| ≤ µ C√R

+ |Th(F1(−R,R))|

So by Fatou’s lemma and Theorem 6.2

‖T±F‖L2((−R/2,R/2)) ≤ Cµ+ lim infh→0±

‖Th(F1(−R,R))‖L2((−R/2,R/2)).(6.25)

Assume that R > 1, and let Aδ(x) = A(x) on (−R2, R2), Aδ(x) = I if |x| >2R2, such that Aδ is smooth and satisfies (2.1) and (6.5). Let Γδ = ΓAδ and defineKδh and T δh in the obvious way. For now, we continue to assume that ϕ ∈ C∞0 with‖ϕ′‖L∞ small.

As in Section 4.2, ΓX − ΓδX ∈ W 1,2loc (R2) and divA∇(ΓX − ΓδX) = 0 in the set

(−2R2, 2R2)×R. By (3.9), Lemma 3.4 and the bound |ΓX(Y )| ≤ C+C|log|X−Y ||,

|∇ΓX(Y )−∇ΓδX(Y )| < C logR

R3

provided X, Y ∈ B(0, R2/4).

Taking R large, we have that |Kδh(x, y)−Kh(x, y)| < C logR

R3 if |x|, |y| < R andh is small. Therefore, for such x,

‖Kδh(x, ·)−Kh(x, ·)‖L2((−R,R)) ≤

C logR

R5/2.

So if supp F ⊂ (−R,R), then

‖ThF − T δh F‖L2(−R/2,R/2) ≤ ‖F‖L2(−R,R)C logR

R2.

By Theorem 6.2, ‖T δh F‖L2(R) ≤ C‖F‖L2(R). So by (6.25),

‖T±F‖L2((−R/2,R/2)) ≤ Cµ+ C‖F‖L2(R) + C‖F‖L2(−R,R)C logR

R2.

As R → ∞, we have that µ → 0 and logR/R2 → 0, and so ‖T±F‖L2(R) ≤C‖F‖L2(R), as desired.


We next remove the requirement that ϕ ∈ C∞0 . We will again do this bycomparing T±F to a well-behaved operator Tδ. Assume that ‖ϕ′‖L∞ < δ0. Chooseϕδ compactly supported and smooth with ‖ϕδ − ϕ‖L∞ < δ on (−R,R). DefineT δ, Kδ

h in the obvious way. By the previous remarks, T δh is bounded on L2(R)uniformly in δ and h.

We have that

|Kh(x, y)−Kδh(x, y)| ≤ |ϕ(x)− ϕδ(x)|α + |ϕ(y)− ϕδ(y)|α

|x− y|1+α + |h|1+α

Thus, if F is supported in (−R,R), then

‖ThF‖L2((−R/2,R/2)) ≤ ‖ThF − T δh F‖L2((−R/2,R/2)) + ‖T δh F‖L2(R)

≤ C‖F‖L2

δα√R

hα+1/2+ C‖F‖L2(−R,R).

Letting δ → 0, we see that ‖Th(F1(−R,R))‖L2((−R/2,R/2)) ≤ C‖F‖L2(R). Applying(6.25) and letting R→∞, we again see that ‖TF‖L2(R) ≤ C‖F‖L2(R).

Finally, we relax from ‖ϕ′‖L∞ < δ0 to ‖ϕ′ − γ‖L∞ < δ0(γ) for some realnumber γ. Fix some choice of e, ϕ and γ. Then Ω = X ∈ R2 : ϕ(X ·e⊥) < X ·e.

Define e = e−γe⊥√1+γ2

. If ‖ϕ′ − γ‖L∞ is small enough, relative to |γ|, then there is

some function ϕ : R 7→ R such that

ϕ(X · e⊥) < X · e if and only if ϕ(X · e⊥) < X · e.Applying Theorem 6.2 to ϕ, we see that the layer potentials T± are bounded

L2 7→ L2. By definition, so are the potentials TΩ± . But TΩ± = TΩ± , and thereforeT± is bounded L2 7→ L2.

If 0 < δ0 ≤ k, define

Λk(δ0) =ϕ : for some constant γ ∈ (−k, k) we have ‖ϕ′ − γ‖L∞ < δ0

.

We have shown that, for every k > 0, there is some δ = δ(k, λ,Λ) such that T±is bounded on L2(R) for every ϕ ∈ Λk(δ).

We wish to remove the assumption that ϕ′ − γ must be small. This may bedone using the buildup scheme of David from [Dav84]; the proof of the followinglemma, in the case of bounded measurable non-symmetric matrices A, is carriedout in [KR09, Section 5].

Lemma 6.26. Suppose that for some fixed choice of e and e⊥, we have thatfor every k > 0 there is a δ0(k) > 0 such that T± is bounded on L2(R) for everyϕ ∈ Λk(δ0) with bounds depending only on λ, Λ and k.

Then T± is bounded on L2(R) for any Lipschitz function ϕ with bounds de-pending on λ, Λ and ‖ϕ′‖L∞ .

Finally, we deal with the assumption that A0 is upper triangular with a011 ≡ 1.

We use a change of variables (taken from [KKPT00, Lemma 3.47]).Consider the mapping J : (x, t) 7→ (f(x), t + g(x)) where λ < f ′ < Λ. Then

ΓX(Y ) = ΓJ(X)(J(Y )) is the fundamental solution with pole at X associated withthe elliptic matrix

A(y) =1

f ′(y)

(1 0

−g′(y) f ′(y)

)A(f(y))

(1 −g′(y)0 f ′(y)

).

6.7. PATCHING: SPECIAL LIPSCHITZ DOMAINS TO BOUNDED LIPSCHITZ DOMAINS 67

If we choose f ′(y) = 1a0

11(f(y))(or, put another way, choose f−1(y) =

´ y0a0

11),

then we will have a011 = 1, and if we choose g′(y) =

f ′(y)a021(f(y))

a011(f(y))

, then we will have

a021 = 0.

But if ϕ ∈ Λk(δ0) for δ0 sufficiently small (depending on λ, Λ and k), where ϕis the Lipschitz function in the definition of Ω, then

J(∂Ω) = (f(xe2 + ϕ(x)e1),−xe1 + ϕ(x)e2 + g(xe2 + ϕ(x)e1)) : x ∈ R .Recall e2

1 + e22 = 1. If ‖ϕ′‖L∞ and |e2| are small enough, then the function x 7→

−xe1 + ϕ(x)e2 + g(xe2 + ϕ(x)e1) is invertible on R, and both it and its inversehave bounded derivatives; thus J(Ω) is a special Lipschitz domain with coordinatevectors e =

(±1 0

)and e⊥ =

(0 ∓1

). If |e2| is not small but ‖ϕ′‖L∞ is, then

x 7→ f(xe2 + ϕ(x)e1) is invertible on R and both it and its inverse have boundedderivatives; thus, J(Ω) is a special Lipschitz domain with coordinate vectors e =(0 ±1

)and e⊥ =

(±1 0

).

In any case, if ‖ϕ′‖L∞ is small enough, then J(Ω) is a special Lipschitz domainand so T± is bounded on L2(∂R); thus, T± is bounded on L2(∂R), and so byLemma 6.26 we may build up to arbitrary special Lipschitz domains.

6.7. Patching: special Lipschitz domains to bounded Lipschitz domains

We now complete the proof of Theorem 6.1.

Theorem 6.27. If TΩ and TΩ are bounded Lp(∂Ω) 7→ Lp(∂Ω) for all specialLipschitz domains Ω, then for any Lipschitz domain V with compact boundary, the

operators TV and TV are bounded Lp(∂V ) 7→ Lp(∂V ), with bounds depending only

on λ, Λ, p, the Lipschitz constants of V , and the operator norms of the TΩs and TΩs.

As in Section 6.6, we work with T ; the proof for T is identical.

Proof. For any domains U , V and any function F defined on ∂U ∩ ∂V , if weextend F to ∂U and ∂V by zero, then TUF = TV F .

From Definition 2.3, we may partition ∂V as follows: there are N points Xj ∈∂V with associated numbers rj > 0, such that ∂V ⊂ ∪jB(Xj , rj) and B(Xj , 2rj)∩V = B(Xj , 2rj)∩Ωj for some special Lipschitz domains Ωj . Let

∑j ηj be a partition

of unity with supp ηj ⊂ ∂V ∩B(Xj , rj), and let Fj = Fηj .Then

‖TV F‖Lp(∂V ) ≤N∑

i=1

‖TV Fj‖Lp(∂V ).

But

‖TV Fj‖pLp(∂V ) = ‖TV Fj‖pLp(∂V ∩B(Xj ,2rj))+ ‖TV Fj‖pLp(∂V \B(Xj ,2rj))

and

‖TV Fj‖Lp(∂V ∩B(Xj ,2rj)) = ‖TΩjFj‖Lp(∂V ∩B(Xj ,2rj)) ≤ ‖TΩjFj‖Lp(∂Ωj)

≤ C‖Fj‖Lp(∂Ωj) = C‖Fj‖Lp(∂V ).

But if |Y −Xj | > 2rj , then by (4.8)

|T Fj(Y )| =∣∣∣∣ˆ∂V

K(Y,Z)Fj(Z) dσ(Z)

∣∣∣∣ ≤Cr

1/qj

|Xj − Y |‖Fj‖Lp(∂V )


and so

‖T Fj‖pLp(∂V ) ≤ˆ|Xj−Y |>2rj

|T Fj(Y )|p dσ + C‖Fj‖pLp(∂V ) ≤ Cp‖Fj‖pLp(∂V )

where Cp depends on the Lipschitz constants of V .Therefore,

‖TV F‖pLp(∂V ) ≤ CpN∑

i=1

‖TV Fj‖pLp(∂V ) ≤∑

j

Cp‖Fj‖pLp(∂V ) ≤ Cp‖F‖pLp(∂V ).

CHAPTER 7

Invertibility of Layer Potentials and OtherProperties

We have now established that if V is a Lipschitz domain, and if ‖ImA‖L∞ is

small enough, then the layer potentials TV and TV defined by (2.20) and (2.28) arebounded operators on Lp(∂V ) for all 1 < p < ∞. By (2.24), this implies that theoperators KV and LV are also bounded on Lp(∂V ) for all 1 < p < ∞. We nowexplore some consequences.

Recall that we seek to show that u = Df and u = Sg are solutions to (D)Aq ,

(N)Ap , or (R)Ap . In Section 7.1, we will show that layer potentials have the requirednontangential maximal estimates. In Section 7.2, we will prove some relationshipsbetween the operators KV+ , KV− , LV+ and LV− . In Section 7.3, we will show thatlayer potentials are bounded on the spaces H1(∂V ) as well. Finally, in Section 7.4we will use these results to show that (K±)t and Lt are invertible on Lp0(∂V ).

Many of the results of this chapter are known to hold for harmonic functionsor more general classes of solutions. See [Ken94, Theorem 2.2.13] for a summaryof some known results for harmonic functions. We verify that these results hold forarbitrary Lipschitz domains and for coefficients A that satisfy (2.1). The invert-ibility results of Section 7.4 are known for harmonic functions from [Ver84], andfor real symmetric coefficients from [AAA+11]; we establish invertibility underthe conditions of Theorem 2.16, using the results of [KR09] and [Rul07] for realcoefficients.

7.1. Nontangential maximal functions of layer potentials

In this section we establish bounds on the nontangential maximal functions oflayer potentials. Our main tool for controlling nontangential maximal functions isthe following generalization of [DV90, Proposition 4.3].

Lemma 7.1. Assume that for some α, β > 0, K(X,Y ) satisfies the Calderon-Zygmund kernel conditions

|K(X,Y )| ≤ β

|X − Y | ,

|K(X,Y )− K(X ′, Y )| ≤ β|X −X ′|αmin(|X − Y |, |X ′ − Y |)1+α

,

|K(X,Y )− K(X,Y ′)| ≤ β|Y − Y ′|αmin(|X − Y |, |X − Y ′|)1+α

.

Let

RF (X) =

ˆ∂V

K(X,Y )F (Y ) dσ(Y )

69

70 7. INVERTIBILITY OF LAYER POTENTIALS AND OTHER PROPERTIES

and let T F (X) be the nontangential limit of RF (X). Define the truncated operators

Th and truncated maximal operator T∗ by

ThF (X) =

ˆY ∈∂V, |X−Y |>ε

K(X,Y )F (Y ) dσ(Y ), T∗F (X) = suph>0|ThF (X)|.

Then for any X ∈ V , X∗ ∈ ∂V with |X −X∗| ≤ (1 + a) dist(X, ∂V ), we have that

|RF (X)| ≤ CβMF (X∗) + T∗F (X∗)

where C depends only on a and the Lipschitz constants of V .

Furthermore, if the operator T or the operators Th are bounded on L2(∂V ) withconstant β, then for any 1 0

B(X,r)∩∂U

|f | dσ.

Define

ThF (X∗) =

ˆ|Y−X∗|>h, Y ∈∂V

K(X∗, Y )F (Y ) dσ(Y ).

If h = dist(X, ∂V ), and if |X −X∗| < (1 + a)h, then

|RF (X)− ThF (X∗)| ≤∣∣∣∣ˆ|Y−X∗|>h

(K(X,Y )− K(X∗, Y ))F (Y ) dσ(Y )

∣∣∣∣

+

∣∣∣∣ˆ|Y−X∗|<h

K(X,Y )F (Y ) dσ(Y )

∣∣∣∣

≤ˆ|Y−X∗|>h

Cβhα|F (Y )||Y −X∗|1+α

dσ(Y )

+

ˆ|Y−X∗|<h

Cβ

h|F (Y )|dσ(Y )

≤ CβMF (X∗)

where MF is the maximal function and our constants depend on a and the Lipschitzconstants of V .

But |ThF (X)| ≤ T∗F (X), and so |RF (X)| ≤ CβMF (X∗) + T∗F (X∗).It is well known that M is bounded on Lp(R) for any 1 < p ≤ ∞. It is also well

known that if T is bounded on L2(R then T∗ is bounded on Lp(R) for 1 < p <∞.

Similarly, if the operators Th are uniformly bounded on L2(R) then so is T∗. See,for example, [Ste93, p. 13], [Ste93, p. 34] or [Gra09, Theorem 8.2.3], and [Gra09,Proposition 8.1.11].

It is straightforward to establish that these results still holds if V is a Lipschitzdomain. Thus,

‖N(RF )‖Lp(∂V ) ≤ Cβ‖MF‖Lp(∂V ) + ‖T∗F‖Lp(∂V ) ≤ Cβ‖F‖Lp(∂V )

as desired.

7.1. NONTANGENTIAL MAXIMAL FUNCTIONS OF LAYER POTENTIALS 71

Theorem 7.2. Let V be a Lipschitz domain. If TV is bounded from L2(∂V ) toitself, then for any 1 < q <∞, there is some Cq depending on q and the L2-boundof TV such that

‖NV±(Df)‖Lq ≤ Cq‖f‖Lq , ‖NV±(∇STg)‖Lp ≤ Cq‖g‖Lp .If TV (∂V ) is bounded on L2(∂V ), then there is a Cp depending on p and the L2-

bound of TV such that

‖NV±(∇DTf)‖Lp(∂V ) ≤ Cp‖∂τf‖Lp(∂V ).

Proof. By Lemma 7.1, we have that F 7→ N(RV F ) is bounded Lq(∂V ) 7→Lq(∂V ). By (2.23), this completes the proof for N(Df).

Let ThF (X) =´∂V \B(X,h)

K(X,Y )F (Y ) dσ(Y ). By Lemma 7.1, these opera-

tors are bounded on Lq(∂V ) uniformly in h.Then T thF (X) =

´∂V \B(X,h)

K(Y,X)tF (Y ) dσ(Y ) is also bounded on Lp(∂V ),

uniformly in h. So again by Lemma 7.1, the nontangential maximal function ofˆ∂V

K(Y,X)tF (Y ) dσ(Y ) =

ˆ∂V

BT0 (X)(∇XΓX(Y ) ∇XΓX(Y )

)F (Y ) dσ(Y )

lies in Lp(∂V ) whenever F ∈ Lp(∂V ). But B0(X) is bounded and invertible and

∇STg(X) =

ˆ∂V

∇XΓX(Y )g(Y ) dσ(Y )

and so choosing F = gI, we see that ‖N(∇STg))‖Lp(∂V ) ≤ C(p)‖g‖Lp(∂V ).Similarly,

∇DTf(X) = ∇Xˆ∂V

ν ·A∇ΓXf dσ = ∇Xˆ∂V

τ · ∇ΓXf dσ

= −∇Xˆ∂V

ΓX ∂τf dσ = −ˆ∂V

∇X ΓX ∂τf dσ

and

KA(Y,X)t = BT0 (X)(∇X ΓX(Y ) ∇X ΓX(Y )

)

and so the inequality ‖N(∇DTf)‖Lp(∂V ) ≤ C‖∂τf‖Lp(∂V ) follows from the L2-

boundedness of T .

Corollary 7.3. If the conditions of Theorem 7.2 hold, then the limits in thedefinition of Kf , Lf exist pointwise a.e. for f ∈ Lq(∂V ), even if f is not smooth.

Proof. We work with K only; the proof for L is identical. Let fn ∈ Lq(∂V )be smooth and such that ‖fn − f‖Lq(∂V ) ≤ 4−n. Then limY→X n.t.Dfn(Y ) existsfor each X ∈ ∂V . Since K is bounded on Lq(∂V ), limn→∞Kfn = Kf exists inLq(∂V ).

Let En = X ∈ ∂V : |Kf(X)−Kfn(X)| > 2−n orN(D(fn−f))(X) > 2−n. ByTheorem 7.2, ‖N(D(fn−f))‖Lq(∂V ) ≤ Cq‖f−fn‖Lq(∂V ) ≤ 4−nCq. By boundedness

of K, ‖Kf −Kfn‖Lq(∂V ) ≤ Cq4−n.

Thus, σ(En) ≤ C2−n, and so σ (∪∞m=nEm) ≤ C2−n. Therefore

E =

∞⋂

n=1

∞⋃

m=n

Em

has measure 0.


Suppose X ∈ ∂V , X /∈ E. So there is some N > 0 such that, if n > N , then

|Df(Y )−Kf(X)|≤ |Df(Y )−Dfn(Y )|+ |Dfn(Y )−Kfn(X)|+ |Kfn(X)−Kf(X)|≤ N(D(f − fn))(X) + |Dfn(Y )−Kfn(X)|+ |Kfn(X)−Kf(X)|≤ C2−n + |Dfn(Y )−Kfn(X)|.

So for every ε > 0, there is some n > N such that C2−n < ε/2, and some δ > 0such that |Dfn(Y ) − Kfn(X)| < ε/2 provided Y ∈ γ(X) and |X − Y | < δ; thus,|Df(Y ) − Kf(X)| < ε if |X − Y | < δ, and so the nontangential limit exists at X,as desired.

We now consider nontangential bounds for layer potentials withH1(∂V ) bound-ary data.

Theorem 7.4. If T TV is bounded on L2(∂V ), then there exists a β, C > 0depending on the L2-boundedness of T T such that if g is a H1 atom supported inB(X0, R), then ˆ

∂V

N(∇Sg)(X)(1 + |X −X0|/R)β dσ(X) ≤ C.

As an immediate consequence we have that ‖N(∇Sg)‖L1(∂V ) ≤ C‖g‖H1(∂V )

for all g ∈ H1(∂V ).

Proof. Suppose that g is a H1 atom supported in some connected set ∆ ⊂ ∂Vwith σ(∆) = R and X0 ∈ ∆. Then

‖g‖L∞(∂V ) ≤1

σ(∆)=

1

R,

and so ‖g‖L2 ≤ 1/√R.

Therefore, letting b > 1 be a constant, we have that by Theorem 7.2 andHolder’s inequality,

ˆB(X0,bR)∩∂V

|N(∇Sg)| ≤ CbR(

B(X0,2bR)∩∂V|N(∇Sg)|2

)1/2

≤ Cb√RC‖g‖L2 ≤ bC.

We need to bound´∂V \B(X0,bR)

|N(∇Sg)|(1 + |X −X0|/R)β dσ. If Y ∈ γ(X)

for some X ∈ ∂V , then either |X − Y | ≤ |X0 −X|/2 and so

|X0 − Y | ≥ |X0 −X|/2,or |X − Y | > |X0 −X|/2 and so

|X0 − Y | ≥ dist(Y, ∂V ) ≥ 1

1 + a|Y −X| > |X0 −X|

2 + 2a.

In any case,

N(∇Sg)(X) ≤ sup

|∇Sg(Y )| : Y ∈ V, |Y −X0| >

|X0 −X|2 + 2a

.

7.2. JUMP RELATIONS 73

If |Y −X0| > R, then

|∇Sg(Y )| ≤∣∣∣∣ˆB(X0,R)∩V

∇Y ΓY (Z)g(Z) dσ(Z)

∣∣∣∣

≤∣∣∣∣ˆB(X0,R)∩V

(∇Y ΓY (Z)−∇Y ΓY (X0)) g(Z) dσ(Z)

∣∣∣∣

≤ˆB(X0,R)∩V

|Z −X0|α(|Y −X0| −R)1+α

C

Rdσ(Z)

≤ C Rα

(|Y −X0| −R)1+α.

Therefore, if |X −X0| > (2 + 2a)R, then

N(∇Sg)(X) ≤ C Rα

(|X −X0| −R(2 + 2a))1+α.

and so if we choose b = 4 + 4a, thenˆ∂V \B(X0,bR)

N(∇Sg)(X)(1 + |X −X0|/R)α/2 dσ(X)

≤ˆ∂V \B(X0,bR)

CRα/2

|X −X0|1+α/2dσ(X) ≤ C.

This completes the proof.

7.2. Jump relations

In this section we will verify that the classical jump relation K+f − K−f = fholds for our operators, and will show that if f ∈ Lp(∂V ), 1 < p < ∞, then Sf iscontinuous on all of R2. If Sf is continuous, then the adjoint formula Ltf = ∂τSTfimplies that LV+

f = −LV−f . This means that we may take the limits in thedefinition of Lf(X) in either γ+(X) or in γ−(X).

Lemma 7.5. Suppose that u is defined on V , N(∇u) ∈ Lp(∂V ) for 1 < p ≤ ∞.Then there exists an extension of u to V which is Holder continuous on compactsubsets of V . If in particular u = Sf for some f ∈ Lp(∂V ), and TV is bounded onL2(∂V ), then u is Holder continuous on all compact subsets of R2.

Proof. If V = Ω is a special Lipschitz domain, N(∇u) ∈ Lp(∂Ω), and 0 ≤τ < t or 0 ≥ τ > t, then by (3.1)

|u(ψ(x, t))− u(ψ(x, τ))| ≤ˆ |t||τ ||∇u(ψ(x,±s))| ds ≤

ˆ |t||τ |

Cs−1/p‖N(∇u)‖Lp(∂Ω) ds.

But since t, τ have the same sign,

ˆ |t||τ |

s−1/p ds ≤ min

(ˆ |t|0

s−1/p ds, |t− τ ||τ |−1/p

)

= min

(q|t|1/q, |t− τ |1/q

( |t− τ ||τ |

)1/p)≤ C|t− τ |1/q

by considering the cases |t− τ | < |t|/2, |t− τ | ≥ |t|/2 separately.


If t 6= 0, then by Holder’s inequality

|u(ψ(x, t))− u(ψ(y, t))| ≤∣∣∣∣ˆ y

x

N(∇u)(ψ(z))√

1 + ϕ′(z)2 dz

∣∣∣∣

≤ C(p)|x− y|1/q‖N(∇u)‖Lp(∂Ω).

So if X, Y ∈ Ω, then

(7.6) |u(X)− u(Y )| ≤ C(p)|X − Y |1/q‖N(∇u)‖Lp(∂Ω).

Thus, u is Holder continuous on Ω.If V is a Lipschitz domain with compact boundary, then this result holds near

its boundary; by (3.1) ∇u is bounded away from ∂V , and so u is Holder continuouson all compact subsets of V .

By Theorem 7.2, if f ∈ Lp(∂V ), then N(∇Sf) ∈ Lp(∂V ) and so Sf extendscontinuously to each of V+ = V and V− = V C ; we need only show that the twoextensions agree.

Pick some X ∈ ∂V , t > 0 small. Let e be a vector such that X ± te ∈ γ±(X)for all sufficiently small positive t. Then

|Sf(X + te)− Sf(X − te)| =∣∣∣∣ˆ∂V

(ΓY (X + te)− ΓY (X − te)) f(Y ) dσ(Y )

∣∣∣∣

But

|ΓY (X + te)− ΓY (X − te)| ≤∣∣∣∣ˆ t

−t∇ΓY (X + re) dr

∣∣∣∣ ≤∣∣∣∣ˆ Ct

0

C√|X − Y |2 + r2

dr

∣∣∣∣

But that integral is at most Ct|X−Y | , and if |X − Y | < t, then

ˆ Ct

0

C√|X − Y |2 + r2

dr =

ˆ Ct/|X−Y |

0

C√1 + r2

dr ≤ C lnCt

|X − Y | .

So

|Sf(X + te)− Sf(X − te)|

=

∣∣∣∣ˆ∂V

(ΓY (X + te)− ΓY (X − te)) f(Y ) dσ(Y )

∣∣∣∣

≤ˆ|X−Y |>|t|

|f(Y )| C|t||X − Y | dσ(Y ) + C

ˆ|X−Y |<|t|

|f(Y )| ln Ct

|X − Y |dσ(Y ).

Applying Holder’s inequality to each integral, we see that

|Sf(X + te)− Sf(X − te)| ≤ C(p)t1/q‖f‖Lp(∂V )

provided 1 < p <∞.So u is continuous across the boundary, and so is continuous on R2.

Lemma 7.7. If f is a Lipschitz function defined on ∂V that lies in L1(∂V ),then K+f(X)−K−f(X) = f(X).

Since K± are bounded as operators on Lp(∂V ) and H1(∂V ), and integrableLipschitz functions are dense in those spaces, we have that K+−K− is the identityon those spaces as well.

7.2. JUMP RELATIONS 75

Proof. Let V ρ± = V±\B(X, ρ), Ψρ± = V± ∩ B(X, ρ). Fix some ρ > 0 small.

Let e be a vector such that X ± te ∈ γ±(X) for all sufficiently small positive t.Then, extending f in some reasonable fashion to R2, we have

K+f(X)−K−f(X) = limt→0+

ˆ∂V

ν ·AT∇ΓTX+tef dσ − limt→0+

ˆ∂V

ν ·AT∇ΓTX−tef dσ

= limt→0+

(ˆ∂V ρ+

ν ·AT∇ΓTX+tef dσ +

ˆ∂V ρ−


)

+ limt→0+

ˆ∂Ψρ+

ν ·AT∇ΓTX+tef dσ

+ limt→0+

ˆ∂Ψρ−

ν ·AT∇ΓTX−tef dσ.

But limt→0 ν(Y ) ·AT (Y )∇ΓTX+te(Y )− ν(Y ) ·AT (Y )∇ΓTX−te(Y ) = 0, uniformly for

Y ∈ ∂V ρ+ ∩ ∂V ρ−; thus, since f ∈ L1(∂V ), we have that

K+f(X)−K−f(X) = −ˆ∂B(X,ρ)

ν ·AT∇ΓTXf dσ + limt→0+

ˆ∂Ψρ+

ν ·AT∇ΓTX+tef dσ

+

ˆ∂Ψρ−


Recall (5.6): if U is a bounded domain, and X ∈ U , thenˆ∂U

ν ·AT∇ΓTX dσ = 1.

So

K+f(X)−K−f(X) = f(X)−ˆ∂B(X,ρ)

ν(Y ) ·AT (Y )∇ΓTX(Y )(f(Y )− f(X)) dσ

+ limt→0+

ˆ∂Ψρ+

ν(Y ) ·AT (Y )∇ΓTX+te(Y )(f(Y )− f(X)) dσ

+ limt→0+

ˆ∂Ψρ−

ν(Y ) ·AT (Y )∇ΓTX−te(Y )(f(Y )− f(X)) dσ.

The integrands are at most C‖f ′‖L∞ , and so the integrals are at most Cρ‖f ′‖L∞ .Taking the limit as ρ→ 0 yields the desired result.

Lemma 7.8. Let A0 satisfy (2.1) and let V be a Lipschitz domain. If (N)AT0p

and (R)AT0p hold in V and V C with constants cp, and if KA0

± , LA0 are bounded onLp(∂V ), then for all f ∈ Lp0(∂V ),

‖(KA0+ )tf‖Lp(∂V ) ≈ ‖(LA0)tf‖Lp(∂V ) ≈ ‖(KA0

− )tf‖Lp(∂V )

where the comparability constants depend on λ, Λ, and cp.

By Theorem 2.15, Theorem 3.15 and Lemma 3.21, the assumptions of Lem-ma 7.8 hold provided A0 is real. We would like to emphasize that Lemma 7.8 holdseven if A0 is complex-valued. We remark that for real symmetric coefficients, thecomparability ‖(KA0)tf‖Lp(∂V ) ≈ ‖(LA0)tf‖ is essentially the Rellich identity.


Proof. By definition of (N)AT0p , (R)

AT0p , there is some constant cp such that

‖N(∇u)‖Lp(∂V ) ≤ cp‖ν ·AT0∇u‖Lp(∂V ), and

‖N(∇u)‖Lp(∂V ) ≤ cp‖τ · ∇u‖Lp(∂V ).

provided u is a solution to (N)AT0p or (R)

AT0p . That is, these equations hold if

divAT0∇u = 0 in V , N(∇u) ∈ Lp(∂V ), and (if V C is bounded) lim|X|→∞ u(X)exists.

Since |ν · AT∇u| ≤ ΛN(∇u) and |τ · ∇u| ≤ N(∇u), we may reverse eitherinequality (up to a multiplicative constant) and so

‖ν ·AT0∇u‖Lp(∂V±) ≈ ‖τ · ∇u‖Lp(∂V±).

By Lemma 5.1, (2.24) and Theorem 7.2, and (5.5), if f ∈ Lp(∂V ) ∩H1(∂V ) =

Lp0(∂V ) then u = SAT0f is a solution to (N)A0p and (R)A0

p in both V+ and V−.

But by Lemma 7.5, SAT0f is continuous on R2, so |τ ·∇SAT0f | must be the sameon ∂V+ and ∂V−. So by Lemma 5.8,

‖(KA0+ )tf‖Lp(∂V ) ≈ ‖(LA0)tf‖Lp(∂V ) ≈ ‖(KA0

− )tf‖Lp(∂V ).

7.3. Layer potentials on H1(∂V )

In this section we show thatKt and Lt are bounded onH1(∂V ) for any Lipschitzdomain V . We remark that boundedness on H1 is a natural endpoint result ofthe Lp-theory of Calderon-Zygmund operators; we prove boundedness using Theo-rem 7.4 and the following lemma, which will also be useful in Chapter 9.

Lemma 7.9. Suppose that f ∈ L1(∂V ), V is a Lipschitz domain, and for some1 0 and some X0 ∈ ∂V ,ˆ∂V

f dσ = 0, ‖f‖Lp(∂V ) ≤cp

R1−1/p,

ˆ∂V

|f(X)|(1 + |X −X0|/R)α dσ(X) ≤ c1

Then f is in H1(∂V ) with H1 norm depending only on c1, cp, p, α and the Lipschitzconstants of V . (Specifically, not on R.)

Proof. Since we can parameterize ∂V by arc length, it suffices to prove thisin the case where ∂V = R and where X0 = 0. If g(x) = Rf(Rx), then g satisfiesthe conditions of the lemma with R = 1, and ‖g‖H1 = ‖f‖H1 ; so we may assumeR = 1.

Let Φ be a Schwartz function with´

Φ = 1. Recall from [FS72, Section V]that f ∈ H1(R) if ˆ

supt

∣∣∣∣ˆf(y)

1

tΦ

(x− yt

)dy

∣∣∣∣ dx

is finite, and that its H1 norm is comparable to the value of this integral (with com-parability constants depending only on Φ). Choose Φ nonnegative with Schwartznorm 1.

So the inner integral is at most CMf(x). Soˆ|x|<1

supt

∣∣∣∣ˆf(y)

1

tΦ

(x− yt

)dy

∣∣∣∣ dx ≤ Cˆ|x|<1

Mf(x) dx ≤ C‖Mf‖Lp

≤ C(p)‖f‖Lp ≤ C(p)cp.

7.3. LAYER POTENTIALS ON H1(∂V ) 77

For any x ∈ R, we have thatˆf(y)

1

tΦ

(x− yt

)dy =

ˆf(y)

1

t

(Φ

(x− yt

)− Φ

(xt

))dy.

Now, assume |x| > 1. Then

∣∣∣∣ˆ|y|<|x|/2

f(y)1

t

(Φ

(x− yt

)− Φ

(xt

))dy

∣∣∣∣ ≤ˆ|y|<|x|/2

|f(y)|1t

|y|t

C

(|x|/t)2dy

= C

ˆ|y|<|x|/2

|f(y)||y|α |y|1−α

|x|2 dy ≤ C

|x|1+α

ˆ|f(y)||y|α dy ≤ Cc1

|x|1+α

and∣∣∣∣ˆ

2|x|<|y|f(y)

1

t

(Φ

(x− yt

)− Φ

(xt

))dy

∣∣∣∣

≤ˆ

2|x|<|y||f(y)|1

t

(Ct

|y − x| +Ct

|x|

)dy ≤

ˆ2|x|<|y|

|f(y)||y|α C

|x|1+αdy ≤ Cc1

|x|1+α

and similarly∣∣∣∣ˆ|x|/2<|y|<2|x|

f(y)1

tΦ(xt

)dy

∣∣∣∣ ≤Cc1|x|1+α

.

But´|x|>1

1|x|1+α dx = 2

α . So to complete the proof, we need only bound

ˆ|x|>1

supt

∣∣∣∣ˆ|x|/2<|y|<2|x|

f(y)1

tΦ

(x− yt

)dy

∣∣∣∣ dx.

This is equal to

∞∑

k=1

ˆ2k−1<|x|<2k

supt

∣∣∣∣ˆ|x|/2<|y|<2|x|

f(y)1

tΦ

(x− yt

)dy

∣∣∣∣ dx

≤∞∑

k=1

ˆ2k−1<|x|<2k

supt

ˆ2k−2<|y|<2k+1

|f(y)|1tΦ

(x− yt

)dy dx

≤∞∑

k=1

ˆ2k−1<|x|<2k

CMfk(x) dx

where fk(x) = f(x) if 2k−2 < |x| < 2k+1 and fk(x) = 0 otherwise.Now, ˆ

2k−1<|x|<2kMfk(x) dx =

ˆ ∞0

λ(β) dβ

where λ(β) = |x : 2k−1 <|x|< 2k,Mfk(x) < β|. We have three upper bounds onλ(β):

• λ(β) ≤ 2k.

• Since f 7→Mf is weak (1, 1)-bounded, λ(β) ≤ C‖fk‖L1

β ≤ Cc12kαβ

.

• Since f 7→Mf is strong (p, p)-bounded, λ(β) ≤ C‖fk‖pLpβp ≤ Ccp

βp .


So

ˆ ∞0

λ(β) dβ =

ˆ 2−k−kα

0

λ(β) dβ +

ˆ 2kα/(p−1)

2−k−kαλ(β) dβ +

ˆ ∞2kα/(p−1)

λ(β) dβ

≤ˆ 2−k−kα

0

2k dβ +

ˆ 2kα/(p−1)

2−k−kα

Cc12kαβ

dβ +

ˆ ∞2kα/(p−1)

Ccpβp

dβ

=1

2kα+Cc12kα

ln(2k+kα+kα/(p−1)) +Ccpp− 1

2−kα

≤ C(p)(cp + c1)(1 + k)2−kα

Summing from k = 1 to infinity gives a finite number depending only on p and α,as desired.

Theorem 7.10. If V is a Lipschitz domain, and if Kt and Lt are bounded onLp(∂V ) for some 1 < p < ∞, then they are bounded H1(∂V ) 7→ H1(∂V ) as well,with bounds depending only on λ, Λ, p, the Lp-bound and the Lipschitz constantsof V .

Proof. Suppose that f is an atom in H1(∂V ). Then´f dσ = 0, supp f ⊂

B(X0, R), σ(supp f) < 2R, ‖f‖L∞(∂V ) ≤ 1/R for some X0 ∈ ∂V and some R > 0;without loss of generality we let X0 = 0.

Since Kt and Lt are bounded on Lp(∂V ),

‖Ktf‖Lp(∂V ), ‖Ltf‖Lp(∂V ) ≤ C(p)‖f‖Lp(∂V ) ≤ C(p)R1/p−1.

Next, by Theorem 7.4 and Lemma 5.8, there exists some α > 0 so thatˆ∂V

|Ktf(X)|(1 + |X|/R)α dσ(X) ≤ C,ˆ∂V

|Ltf(X)|(1 + |X|/R)α dσ(X) ≤ C.

We need only show that´Ktf dσ =

´Ltf dσ = 0. By Lemma 5.8,

ˆ∂V

LtV f(X) dσ(X) =

ˆ∂V

τ · ∇STf(X) dσ(X).

By Lemma 7.5, STf is continuous on R2; thus, if ∂V is bounded then this must bezero.

If V is bounded, then by (5.6) K1 = 1 and so´∂VKtf dσ =

´∂V

f dσ = 0 for

any f ∈ H1(∂V ). If V C is bounded, again by (5.6) K1 = 0 and so´∂VKtf dσ = 0

for any function f such that Ktf is well-defined.Finally, consider the case where V = Ω is a special Lipschitz domain. Then

STf(X) =

ˆ∂Ω

ΓX(Y )f(Y ) dσ(Y ) =

ˆ∂Ω

(ΓX(Y )− ΓX(0))f(Y ) dσ(Y )

and so if |X| is large enough, then

|STf(X)| ≤ˆ∂Ω

|ΓX(Y )− ΓX(0)||f(Y )| dσ(Y ) ≤ˆ

supp f

C|Y ||X|

1

Rdσ(Y ) ≤ CR

|X|

and so´∂ΩLtf(X) dσ(X) =

´∂Ωτ · ∇STf(X) dσ(X) = 0.

7.4. INVERTIBILITY OF LAYER POTENTIALS ON LP (∂V ) 79

Now, pick some r > 2R, and let η ∈ C∞0 (B(0, 2r)) with η ≡ 1 on B(0, r) and|∇η| < C/r. Thenˆ

∂Ω

η(X)Kt±f(X) dσ(X) = ∓ˆ∂Ω∓

η(X) ν ·AT∇STf(X) dσ(X)

= ∓ˆ

Ω∓

∇η(X) ·AT∇STf(X) dX

= ∓ˆ

Ω∓∩B(0,2r)\B(0,r)

∇η(X) ·AT∇STf(X) dX

We have that for |X| > 2R,

|∇STf(X)| ≤ˆ∂Ω

|∇ΓX(Y )−∇ΓX(0)||f(Y )| dσ(Y ) ≤ CRα

|X|1+α

and so ∣∣∣∣ˆ∂Ω

η(X)Kt±f(X) dσ(X)

∣∣∣∣ ≤ˆB(0,2r)\B(0,r)

CRα

r|X|1+αdX ≤ CRα

rα.

Since this goes to 0 as r →∞, we know´∂ΩKtf dσ = 0.

So, applying Lemma 7.9, we are done.

7.4. Invertibility of layer potentials on Lp(∂V )

Recall that in the proof of Theorem 2.16, we used boundedness and invertibilityof layer potentials. In Chapter 6 we proved boundedness; in this section we willprove invertibility.

Theorem 7.11. Let 1 0. Let V be a Lipschitz domainwith Lipschitz character M , N , c0. Let A0 satisfy (2.1). Assume that the followingconditions hold.

(7.12) There exists a cp such that, if ‖A − A0‖L∞ ≤ ε0 and A is smooth andsatisfies (2.1), then (KA±)t and (LA)t are bounded linear operators Lp(∂V ),with operator norms at most cp.

(7.13) The layer potentials (KA0± )t : Lp0(∂V ) 7→ Lp0(∂V ) are onto.

(7.14) (N)AT0p , (R)

AT0p hold in V = V+ and V C = V− with constants at most cp.

Then there is some ε > 0 such that, if ‖A − A0‖L∞ ≤ ε and A is smooth andsatisfies (2.1), then (KA±)t, (LA)t are one-to-one and onto on the space Lp0(∂V ).Furthermore, ε ≥ 1/C(λ,Λ,M,N, c0, cp, p, ε0), and

‖f‖Lp(∂V ) ≤ C(λ,Λ,M,N, c0, cp, p, ε0)‖(KA±)tf‖Lp(∂V ),

‖f‖Lp(∂V ) ≤ C(λ,Λ,M,N, c0, cp, p, ε0)‖(LA)tf‖Lp(∂V )

for all f ∈ Lp0(∂V ).

Notice that the ε produced by this theorem depends on p.Theorem 7.11 has three conditions. We will need Theorem 7.11 to prove that

(7.13) holds for real A0; thus, we prove Theorem 7.11 before we prove that we canuse it.

Proof. By (4.11) and (7.12), we have that

‖(KA±)t − (KA0± )t‖Lp 7→Lp , ‖(LA)t − (LA0)t‖Lp 7→Lp ≤ Cp‖A−A0‖L∞


for all smooth A sufficiently near A0. By Theorem 7.10, (KA±)t and (LA)t mapLp0(∂V ) = Lp(∂V ) ∩H1(∂V ) into itself.

From elementary functional analysis (see, for example, [RY00, Corollary 4.42]),we know that if G is an invertible linear operator defined on a Banach space, thenevery operator G′ with ‖G−G′‖ < ‖G−1‖−1 is also invertible. So to see that (KA±)t

and (LA)t are invertible, for ‖A − A0‖L∞ small enough, we need only show that

(KA0± )t, (LA0)t are invertible.

Let K0± = KA0

± , L0 = LA0 . We want to show that if (7.12), (7.13), and (7.14)hold, then (K0

±)t and (L0)t are invertible on Lp0(∂V ).

By Lemma 7.7, KA+ − KA− is the identity operator on Lp0(∂V ) for any ellipticmatrix A; so

‖f‖Lp(∂V ) = ‖(K0+)tf − (K0

−)tf‖Lp(∂V ) ≤ ‖(K0+)tf‖Lp(∂V ) + ‖(K0

−)tf‖Lp(∂V ).

By Lemma 7.8, if (7.12) and (7.14) hold, and if f ∈ Lp0(∂V ), then

‖(K0V )tf‖Lp(∂V ) ≈ ‖(L0

V )tf‖Lp(∂V ).

Finally, by Lemma 7.5, (L0V+

)t = −(L0V−

)t as operators defined on Lp0(∂V ).

Thus,

(7.15) ‖f‖Lp(∂V ) ≤ ‖(K0+)tf‖Lp(∂V ) + ‖(K0

−)tf‖Lp(∂V )

≤ Cp‖Lt0f‖Lp(∂V ) ≤ Cp‖(K0±)tf‖Lp(∂V ).

This implies the following:

• (L0)t and (K0±)t are one-to-one on Lp0(∂V ).

• If (L0)t and (K0±)t are onto, then their inverses have norms at most C(p).

So we need only show that (L0)t and (K0±)t are surjective Lp0(∂V ) 7→ Lp0(∂V ).

Condition (7.13) is simply that (K0±)t be surjective on Lp0(∂V ). Let f ∈ Lp0(∂V ).

We want to find a h ∈ Lp0(∂V ) such that Lth = ∂τSTh = f . We may assume thatV is either bounded or special.

Since (R)AT0p holds in V , there is some u with divAT0∇u = 0 in V , ∂τu = f on

∂V and ‖N(∇u)‖Lp(∂V ) ≤ C‖f‖Lp(∂V ).

Then there is some g ∈ Lp(∂V ) with g = ν · AT0∇u. If V is bounded, thenffl∂V

g dσ =´V∇1 · AT0∇u = 0, so g ∈ H1(∂V ). If V is special, then Lp0(∂V ) = is

dense in Lp(∂V ) for p > 1 and so (K0±)t is invertible on all of Lp(∂V ). In either

case, ((K0−)t)−1g exists.

Let h = ((K0−)t)−1g ∈ Lp(∂V ). Then by uniqueness of solutions to (N)

AT0p ,

ST0 h = u, and so (L0)th = f , as desired. Thus (L0)t as well as (K0±)t is invertible

on Lp0(∂V ).

Corollary 7.16. Suppose that A0 is real and V is a Lipschitz domain withconnected boundary. Then the conditions of Theorem 7.11 hold for some p > 1depending only on λ, Λ, M , N and c0.

Proof. Theorem 7.11 has three conditions. If V is a Lipschitz domain and A0

is real, then by Theorem 6.1, (7.12) holds.By [KR09] and [Rul07], if V is a bounded or special Lipschitz domain and A0

is real, then there exists some p = p0 > 1, depending only on λ, Λ and the Lipschitzconstants of V , such that (N)A0

p and (R)A0p hold in V with constants depending on

the same quantities. The complement of a special Lipschitz domain is also a special

7.4. INVERTIBILITY OF LAYER POTENTIALS ON LP (∂V ) 81

Lipschitz domain. By Theorem 3.15 and Lemma 3.21, if V is a Lipschitz domainand V C is bounded and connected, then (N)A0

p and (R)A0p hold in V . Thus if V is

a Lipschitz domain with connected boundary, then there is some p > 1 dependingonly on λ, Λ, M , N and c0 such that (7.14) holds.

By [Ver84, Theorem 4.2 and Corollary 4.4], if V is a bounded, simply connectedLipschitz domain and A ≡ I, then (7.13) holds. Let As = sI + (1− s)A0; then thematrices As are real-valued and satisfy (2.1), uniformly in s. By the above remarks,there is some p > 1 such that the matrices As also satisfy (7.12) and (7.14). LetKs = KAs . Then by Theorem 7.11, there is an ε > 0, independent of s, such thatif 0 ≤ s ≤ 1 and 0 ≤ r ≤ 1, and if |s − r| < ε and (Kr±)t is invertible, then so is

(Ks±)t. In particular, if (KI±)t is surjective on Lp0(∂V ), then so is (K0±)t.

Thus, if ∂V is bounded and connected then the conditions of Theorem 7.11hold.

It remains to show that (7.13) holds if V = Ω is a special Lipschitz domain. Bythe previous argument, we need only show that (KI±)t is invertible. In fact, since

(7.12) and (7.14) hold for real coefficients in special Lipschitz domains, if (KB)t isinvertible on Lp(∂Ω) ∩H1(∂Ω), for any real matrix B satisfying (2.1), then (KI)tis as well.

By symmetry of the Laplacian it suffices to prove invertibility for domains ofthe form (x, t) : t > ϕ(x) for some Lipschitz function ϕ.

Let H = (x, t) : t > 0 = R2+ be the upper half-plane. Then DIH is half of

the Poisson integral for the upper half-plane, so KIHf = 12f . Thus KIH, (KIH)t are

clearly onto.It is easy to check that ΓB(x,t)(y, s) = ΓI(x,t−ϕ(x))(y, s− ϕ(y)), where

B =

(1 ϕ′(x)

ϕ′(x) 1 + ϕ′(x)2

).

So if Y = (y, ϕ(y)) ∈ ∂Ω, then

ν(Y ) ·B(Y )∇ΓB(x,t+ϕ(x))(Y ) =

(0−1

)· ∇ΓI(x,t)(y, 0)

1√1 + ϕ′(y)2

and so for any sufficiently well-behaved function f ,

KBΩf(x, ϕ(x)) = limt→0+

ˆ∂Ω

ν(Y ) ·B(Y )∇ΓB(x,t+ϕ(x))(Y ) f(Y ) dσ(Y )

= limt→0+

ˆR

(0−1

)· ∇ΓI(x,t)(y, 0) f(y) dy = KIHf(x) =

1

2f(x).

Thus, KBΩ = 12 , and so (KBΩ )t = 1

2 is also invertible.

CHAPTER 8

Uniqueness of Solutions

We have established the following theorem.

Theorem 8.1. Fix some Λ > λ > 0 and some M > 0, N ≥ 1 and c0 ≥ 1.Suppose that V is a Lipschitz domain with Lipschitz constants at most M , N andc0, and that A is smooth and satisfies (2.1).

Then there exists a number p0 > 1, depending only on λ, Λ, M , N , and c0,such that, if 1 < p ≤ p0, then there exists an ε(p) = ε(λ,Λ,M,N, c0, p) such that,if ‖ImA‖L∞ < ε(p), then solutions to (D)Aq , (N)Ap , and (R)Ap exist in V .

Observe that ε(p0) may be said to depend only on λ, Λ, M , N , and c0. For anoutline of the proof of Theorem 8.1, see Section 2.4.

Recall that (D)Aq , (N)Ap , and (R)Ap are said to hold in the domain V if solutions

exist and are unique. From Section 3.4, we know that that solutions to (N)Ap and

(R)Ap are unique if ∂V is compact. Thus, we have shown that if ∂V is compact, if

p > 1 is small enough, and if ‖ImA‖L∞ < ε(p), then (N)Ap and (R)Ap hold in V .We will begin this chapter with some counterexamples to uniqueness. We will

then show that if p > 1 is small enough, then solutions to (N)Ap and (R)Ap in special

Lipschitz domains are unique. We will also show that solutions to (D)Aq in arbitraryLipschitz domains are unique for q <∞ large enough.

This will establish that if p > 1 is small enough, and if ‖ImA‖L∞ < ε(p), then(D)Aq , (N)Ap and (R)Ap hold.

8.1. Counterexamples to uniqueness

Suppose that divA∇u = 0 in some domain V . We would like to show that uis constant in V provided one of the following conditions holds.

• Nu ∈ Lq(∂V ) and u = 0 on ∂V .• N(∇u) ∈ Lp(∂V ) and ν ·A∇u = 0 on ∂V .• N(∇u) ∈ Lp(∂V ) and u = 0 on ∂V .

This would show that solutions to (D)Aq , (N)Ap , (R)Ap are unique. Unfortunately,this cannot be done for general p and Lipschitz domains V .

If V C is bounded, then the condition N(∇u) ∈ Lp(∂V ) is not enough to guar-antee uniqueness of solutions. This is elementary: consider the harmonic functionsu(x, t) = x+ x

x2+t2 or u(x, t) = log(x2 + t2) in R2 \B(0, 1). Imposing the additional

condition that lim|X|→∞ u(X) exists does ensure uniqueness; see Section 3.4.Special Lipschitz domains and the Dirichlet problem require more complicated

examples. In [KKPT00, Theorem 3.2.1], the authors constructed an example ofpoor regularity in the upper half-plane: if u(x, t) = Im(|x| + it)α, and if k =

83

84 8. UNIQUENESS OF SOLUTIONS

tan(π2 (1− α)

), then divA∇u = 0 in R2

+, where

A(x, t) =

(1 −kk 1

)if x < 0, A(x, t) =

(1 k−k 1

)if x > 0.

This u (with 0 < α < 1) was used in that paper to show that (D)Aq does not holdin the upper half-plane for q ≤ 1/α, and in the appendix to [KR09] to show that

(R)Ap , (N)Ap do not hold in the upper half-plane for p > 1/(1− α).We can use the same example to produce counterexamples to uniqueness. A

simple computation yields that |∇u(X)| = α|X|α−1. Let Ω be the domain above a

level set of u. Then u is a constant on ∂Ω, and so ν · A∇u = 0 on ∂Ω. It is easy tocheck that Ω is a special Lipschitz domain, and that if p > 1

1−α > 1, then NΩ(∇u),

NΩ(∇u) ∈ Lp(∂Ω).Thus, the condition N(∇u) ∈ Lp(∂Ω) is not enough to guarantee uniqueness in

special Lipschitz domains. However, we can recover uniqueness if p is small enough;see Theorem 8.2.

The same function u provides a counterexample to uniqueness of solutionsto the Dirichlet problem. Let V = B(0, 1) ∩ R2

+. Take −1 < α < 0, and let1 ≤ q < 1/|α| < q0 < ∞. Then Nu ∈ Lq(∂V ), Nu /∈ Lq0(∂V ). But u is boundedon ∂V . So by [KKPT00, Theorem 3.1], there is some v with divA∇v = 0 in V ,v = u on ∂V and Nv ∈ Lq0(∂V ) ⊂ Lq(∂V ). Then w = u − v = 0 on ∂V , andNw ∈ Lq(∂V ), but w 6= 0 in V .

So the condition Nw ∈ Lq(∂V ) is not enough to guarantee uniqueness, even inbounded domains. We can recover uniqueness if q is large enough; see Theorem 8.3.

8.2. Uniqueness results

In this section, we first show that solutions to (N)Ap and (R)Ap are unique inspecial Lipschitz domains for p ≥ 1 large enough. We next show that solutions to(D)Aq are unique in all Lipschitz domains for q <∞ small enough.

Recall that the solution to (D)Aq with boundary data f , constructed earlier in

this monograph, is D(K−1f), while the solution to (R)Ap with boundary data f is

S(((LT )t)−1∂τf) +ffl∂V

f dσ. It is not immediately obvious that these solutions areequal; the final theorem of this section will show that they are.

Theorem 8.2. Let Ω be a special Lipschitz domain. Suppose that divA∇u = 0in Ω and that N(∇u) ∈ L1(∂Ω) + Lp(∂Ω) for some 1 ≤ p <∞.

If ν ·A∇u = 0 and (N)Ap holds in Q(X,R) for all R > 0 and all X ∈ ∂Ω, then

u is constant in Ω. Similarly, if τ · ∇u = 0 on ∂Ω and (R)Ap holds in Q(X,R) forall R > 0 and all X ∈ ∂Ω, then u is constant in Ω.

The domains Q(X,R) are simply connected bounded Lipschitz domains. Thus,by Theorem 8.1, for any special Lipschitz domain Ω there is some p = p0 > 1 suchthat the conditions of Theorem 8.2 hold provided ‖ImA‖L∞ < ε(p0).

Proof. Since N(∇u) ∈ L1(∂Ω) + Lp(∂Ω), for any ε, R0 > 0, there must besome R > R0 such that N(∇u)(ψ(±R)) ≤ εR−1/p. Recall from Lemma 3.2 thatwe may take a large enough that ∂Q(0, R) ⊂ γa(χ+) ∪ γa(χ−).

Pick some R0, ε. Then

‖ν ·A∇u‖Lp(∂Q(0,R)) ≤ Cε or ‖τ · ∇u‖Lp(∂Q(0,R)) ≤ Cε

8.2. UNIQUENESS RESULTS 85

depending on whether ν · A∇u = 0 or τ · ∇u = 0 on ∂Ω. (If p = 1, then theH1(∂Q(0, R)) norm is at most Cε as well.)

Then ‖N(∇u)‖Lp(∂Q(0,R)) ≤ Cε. So by (3.1) |∇u(X)| ≤ Cεdist(X, ∂Ω)−1/p forall |X| ≤ R0/C. By taking the limits as R0 →∞ and ε→ 0, we see that ∇u ≡ 0,as desired.

We now prove uniqueness for the Dirichlet problem.

Theorem 8.3. Let V be a Lipschitz domain, and let 1 < q <∞. Assume that

solutions to (R)AT

p exist in V , where 1/p + 1/q = 1. Then if divA∇u = 0 in V ,Nu ∈ Lq(∂V ), and u ≡ 0 on ∂V , then u ≡ 0 in V .

Proof. As is common (see, for example, [Ken94, Theorem 1.4.4] for A real,and [AAA+11, Lemma 4.31] for p = q = 2), we prove this by using the Green’sfunction.

Fix any X ∈ V . By (5.4), ‖∇ΓTX‖Lp(∂V ) ≤ Cp dist(X, ∂V )1/p−1 for any p > 1.

If V is bounded or special, let ΦTX be a (R)AT

p -solution with boundary data ΓX , so

ΦTX = ΓTX on ∂V , ‖N(∇ΦTX)‖Lp(∂V ) ≤ Cp dist(X, ∂V )1/p−1. Let GTX = ΓTX − ΦTX .

If V C is bounded, choose some X0 in V C with dist(X0, ∂V ) ≥ σ(∂V )/C. Thenlet ΦTX be the regularity solution with boundary data ΓTX −ΓTX0

. Since ΦTX is a reg-

ularity solution, limY→∞ ΦTX(Y ) exists. Observe that lim|Y |→∞ ΓTX(Y )−ΓTX0(Y ) =

0; thus, lim|Y |→∞GTX(Y ) exists.Let η ∈ C∞0 (V ). Then uη is compactly supported with bounded gradient in V .

We may extend uη to R2 by zero. Then by (1.1) and (2.8),

u(X)η(X) =

ˆV

∇(uη) ·AT∇ΓTX =

ˆV

∇(uη) ·AT∇GTX

=

ˆV

u∇η ·AT∇GTX −ˆV

GTX∇η ·A∇u.

Suppose that Y ∈ γ(Q) and that 12 |X −Q| ≥ |Y −Q|. Then

|GTX(Y )| = |GTX(Y )−GTX(Q)| ≤ |ΓTX(Y )− ΓTX(Q)|+ˆ Y

Q

|∇ΦTX(Z)| dσ(Z)(8.4)

≤ C|Y −Q||X −Q| + C|Y −Q|N(∇ΦTX)(Q) .

Let Vδ = Y ∈ V : dist(Y, ∂V ) < δ. Then if δ is small enough,ˆVδ

|GTX |p ≤ Cδˆ∂V

sup|GTX(Y )| : Y ∈ γ(Q), |Y −Q| < Cδ dσ(Q)

≤ Cδˆ∂V

δpC

|X −Q|p + CδpN(∇ΦTX(Q))p dσ(Q)

≤ Cδp+1 dist(X, ∂V )1−p.

Also,ˆVδ

|∇GTX |p ≤ Cδˆ∂V

C

|X −Q|p +N(∇ΦTX(Q))p dσ(Q) ≤ Cδ dist(X, ∂V )1−p.

Let

Nδu(Q) = sup|u(Y )| : Y ∈ γ(Q), dist(Y, ∂V ) < δ.


By the dominated convergence theorem, Nδu → 0 in Lq(∂V ) as δ → 0. Further-more, ˆ

Vδ

|u|q ≤ Cδˆ∂V

Nδu(Q)q dσ(Q).

By Lemma 3.4 and (3.9), we also have thatˆVδ\Vδ/2

|∇u|q ≤ Cδ1−qˆ∂V

Nδu(Q)q dσ(Q).

Let δ < 12 dist(X, ∂V ) be small. Let ζ ≡ 1 on V \ Vδ, ζ ≡ 0 on Vδ/2, with

|∇ζ| ≤ C/δ.If V is bounded, let η = ζ. Then

∣∣∣∣ˆV

GTX∇η ·A∇u∣∣∣∣ ≤

C

δ

(ˆVδ

|GTX |p)1/p

(ˆVδ\Vδ/2

|∇u|q)1/q

≤ C dist(X, ∂V )−1/q‖Nδu‖Lq(∂V )

and∣∣∣∣ˆV

u∇η ·AT∇GTX∣∣∣∣ ≤

C

δ

(ˆVδ

|∇GTX |p)1/p(ˆ

Vδ

|u|q)1/q

≤ C dist(X, ∂V )−1/q‖Nδu‖Lq(∂V )

which goes to zero as δ → 0. Thus, u(X)η(X) = 0, and so u ≡ 0 in V .If V = Ω is a special Lipschitz domain, let θ ≡ 1 on Q(0, R), θ ≡ 0 on

Ω \Q(0, 2R), with |∇θ| ≤ C/R. Let W (R) = Q(0, 2R) \Q(0, R) \ Vδ. Then if R islarge enough,

|u(X)η(X)| ≤∣∣∣∣ˆVδ

u∇η ·AT∇GTX +

ˆVδ

GTX∇η ·A∇u∣∣∣∣

+

∣∣∣∣ˆW (R)

u∇η ·AT∇GTX +

ˆW (R)

GTX∇η ·A∇u∣∣∣∣.

We may bound the integrals over Vδ as for bounded domains. We have that∣∣∣∣ˆW (R)


C

R

ˆ 2R

R

ˆ∂Q(0,ρ)\∂Ω

|u| |∇ΦTX |+ |u| |∇ΓTX | dσ dρ,∣∣∣∣ˆW (R)

GTX∇η ·A∇u∣∣∣∣ ≤

C

R

ˆ 2R

R

ˆ∂Q(0,ρ)\∂Ω

|∇u(Y )| |GTX(Y )| dσ(Y ) dρ.

Let χ±(ρ) = χ±(0, ρ) be the endpoints of ∆(0, ρ). Recall that ∂Q(0, ρ) \ ∂Ω ⊂γa(χ+(ρ)) ∪ γa(χ−(ρ)). Assume that R > 2|X|. Then the integrands are at most

Nu(χ+(ρ))

(C

R+ CN(∇ΦTX)(χ+(ρ))

)+Nu(χ−(ρ))

(C

R+ CN(∇ΦTX)(χ−(ρ))

).

This is clear for the the first integral, and follows for the second integral from (8.4)and by applying Lemma 3.4 and (3.9) to u. So

|u(X)η(X)| ≤ C dist(X, ∂Ω)−1/q‖Nδu‖Lq(∂Ω)

+ C

ˆ∆(0,2R)\∆(0,R)

NuN(∇ΦTX) +C

RNudσ.

8.2. UNIQUENESS RESULTS 87

Since Nu ∈ Lq(∂Ω) and N(∇ΦTX) ∈ Lp(∂Ω), the integral goes to zero as R → ∞.Thus u(X) = 0.

Finally, suppose that V C is bounded. Let θ ≡ 1 on B(0, R), θ ≡ 0 outsideB(0, 2R), with |∇θ| ≤ C/R, for some R large. Let η(X) = ζ(X)θ(X) and let

W (R) = supp∇θ = B(0, 2R) \ B(0, R). Let W (R) = B(0, 3R) \ B(0, R/2). Asbefore, to show that u(X) = 0, we need only show that

∣∣∣∣ˆW (R)

u∇η ·AT∇GTX +

ˆW (R)

GTX∇η ·A∇u∣∣∣∣→ 0

as R→ 0.Recall that G∞ = lim|Y |→∞GTX(Y ) exists. Since Nu ∈ Lq(∂V ), u(Y ) is

bounded for |Y | large and so by Lemma 3.10 ∇u ∈ L2(R2 \ B(0, R)) for all Rlarge enough and u∞ = lim|Y |→∞ u(Y ) exists.

So by Lemma 3.4,

∣∣∣∣ˆW (R)


C

R

(ˆW (R)

|u|2)1/2(ˆ

W (R)

|∇GTX |2)1/2

≤ C(

W (R)

|u|2)1/2(

W (R)

|GTX −G∞|2)1/2

and

∣∣∣∣ˆW (R)

GTX∇η ·A∇u∣∣∣∣ ≤

C

R

(ˆW (R)

|∇u|2)1/2(ˆ

W (R)

|GTX |2)1/2

≤ C(

W (R)

|u− u∞|2)1/2(

W (R)

|GTX |2)1/2

Letting R→∞, we see that both terms go to zero, as desired.

Theorem 8.5. Suppose that V is a Lipschitz domain with connected boundary.Let f be defined on ∂V such that f ∈ Lq(∂V ), ∂τf ∈ Lp(∂V ), where 1 < q < ∞and 1 ≤ p < ∞ are such that solutions to (D)Aq , (R)Ap exist and are unique in V(and, if V = Ω is special, in all the Q(0, R)s).

If u is the solution to (D)Aq and v is the solution to (R)Ap with boundary data f ,then u ≡ v in V .

We do not require that p, q be conjugate.

Proof. If ∂V is compact then v is bounded in compact sets by Lemma 3.3. IfV C is bounded then lim|X|→∞ v(X) exists by definition of regularity solution, so vis bounded in V . But then Nv is bounded. Since ∂V is bounded, v is a solution to(D)Aq in V for any 1 < q <∞ and we need only apply the uniqueness assumption.

If V = Ω is a special Lipschitz domain, then for every R0 > 0, ε > 0, there issome R > R0 such that limZ→ψ(±R) n.t. u(Z) = u(ψ(±R)), Nu(ψ(±R)) < εR−1/q,

N(∇v)(±R) < εR−1/p.Define uR in Q(0, R) as follows: divA∇uR = 0 in Q(0, R), uR = u = v on

∂Q(0, R) ∩ ∂Ω, and on ∂Q(0, R) \ ∂Ω, uR = 0 except for the two segments of


length R1−1/q near ∂Ω, where uR is to decrease linearly from u(ψ(±R)) to zero.Then |∂τuR| < ε/R on ∂Q(0, R) \ ∂Ω, so if R is large enough then

‖uR‖Lq(∂Q(0,R)\∂Ω) ≤ 2εR−1/q2

< ε, ‖∂τuR‖Lp(∂Q(0,R)\∂Ω) ≤ 2εR−1/qp < ε.

Since Q(0, R) is bounded, uR is both a Dirichlet and regularity solution, so

‖NuR‖Lq(∂Q(0,R)) ≤ ‖uR‖Lq(∂Q(0,R)),

‖N(∇uR)‖Lp(∂Q(0,R)) ≤ ‖∂τuR‖Lp(∂Q(0,R)).

We claim that as R0 → ∞ and ε → 0, ∇uR(X) approaches both ∇u(X) and∇v(X) pointwise (not uniformly); this suffices to show that ∇u ≡ ∇v, and so u ≡ vup to an additive constant (which must be 0).

First,

‖u− uR‖Lq(∂Q(0,R)) = ‖u− uR‖Lq(∂Q(0,R)\∂Ω)

≤ ‖uR‖Lq(∂Q(0,R)\∂Ω) + ‖u‖Lq(∂Q(0,R)\∂Ω) ≤ Cε,and so ‖N(u− uR)‖Lq(∂Ω) ≤ Cε; therefore, if X ∈ Q(0, R/C) for C large enough,

|u(X)− uR(X)| ≤ Cεdist(X, ∂Q(0, R))−1/q = Cεdist(X, ∂Ω)−1/q.

Therefore, by Lemma 3.4 and (3.9),

|∇u(X)−∇uR(X)| ≤ Cεdist(X, ∂Ω)−1−1/q.

Next, if p > 1,

‖∂τv − ∂τuR‖Lp(∂Q(0,R)) = ‖∂τv − ∂τuR‖Lp(∂Q(0,R)\∂Ω)

≤ ‖∂τuR‖Lp(∂Q(0,R)\∂Ω) + ‖∂τv‖Lp(∂Q(0,R)\∂Ω) ≤ Cε.So by Theorem 8.2, ‖N(∇v −∇uR)‖Lq(∂Ω) ≤ Cε.

If p = 1 then ∂τv − ∂τuR is supported on ∂Q(0, R) \ ∂Ω. This set is of mea-sure CR, and on this set |∂τuR|, |∂τv| are each of size at most ε/R. So

‖N(∇uR −∇v)‖L1(∂Q(0,R)) ≤ C‖∂τuR − ∂τv‖H1(∂Q(0,R)) ≤ Cε.In either case, if X ∈ Q(0, R/C), then

|∇v(X)−∇uR(X)| ≤ Cεdist(X, ∂Q(0, R))−1/p = Cεdist(X, ∂Ω)−1/p.

Thus, by letting ε→ 0, we see that ∇u(X) ≡ ∇v(X), as desired.

CHAPTER 9

Boundary Data in H1(∂V )

We now consider the Neumann and regularity problems with boundary data inHardy spaces. The main results of this chapter are as follows.

Theorem 9.1. Suppose that A satisfies (2.1) and V is a Lipschitz domain. Leta be an atom of H1(∂V ), so ‖a‖L∞(∂V ) ≤ 1/r,

á = 0, and supp a ⊂ ∆ for some

connected set ∆ ⊂ ∂V with σ(∆) = r.Let divA∇u = 0 in V . Suppose that, for some 1 0

small enough that Q(X,R) exists, with constants at most cp,

or

• ν ·A∇u = a on ∂V and• (N)Ap and (D)Aq hold in V and Q(X,R), for all X ∈ ∂V and all R > 0

small enough that Q(X,R) exists, with constants at most cp.

If 0 < α < 1/q, then there exists a number C depending only on α, cp, theLipschitz constants of V and the ellipticity constants of A such that for any X0 ∈supp a,

(9.2)

ˆ∂V

N(∇u)(X)(1 + |X −X0|/r)α dσ(X) ≤ C.

The method of proof is essentially that of [DK90, Lemma 1.6].

Corollary 9.3. Suppose that A and V satisfy the conditions of Theorem 9.1.Suppose that for some p > 1, the layer potentials (KT±)t and (LT )t are invertible

on Lp0(∂V ). Then the layer potentials (KT±)t and (LT )t are invertible on H1(∂V )as well.

We have established that, if V is a Lipschitz domain with connected boundary,then there is some p = p0(λ,Λ,M,N, c0) > 1 and some ε = ε(p0) such that if‖ImA‖L∞ < ε then the conditions of Theorem 9.1 and Corollary 9.3 hold. As inSection 2.4, invertibility of layer potentials on H1(∂V ) together with the uniquenessresults of Section 3.4 imply that the two conditions (N)A1 and (R)A1 hold in V .

We will prove Theorem 9.1 in Section 9.1, and will prove Corollary 9.3 inSection 9.2.

9.1. Solutions with boundary data in H1

In this section we prove Theorem 9.1. We begin with some useful bounds onthe function u described in Theorem 9.1.

89

90 9. BOUNDARY DATA IN H1(∂V )

Lemma 9.4. Suppose that V , p, A, a, u satisfy the conditions of Theorem 9.1.If X ∈ V , then

|∇u(X)| ≤ Cr1/q

min(σ(∂V ),dist(X, ∂V ))1+1/q.(9.5)

If V = Ω is a special Lipschitz domain and |X −X0| > 4R > Cr thenˆQ(X,R)

|∇u|2 ≤ Cr2/q

R2/q.(9.6)

The constants depend only on p, cp, λ, Λ and the Lipschitz constants of V .

Proof. Theorem 9.1 has two cases. Consider the first case. Then τ · ∇u = aon ∂V , and (D)Aq and (R)Ap hold in V and the domains Q(X,R). Let a = ∂τf . Wemay assume that f is supported in a connected set of surface measure r.

By Theorem 8.5, u is also a solution to (D)Aq with boundary data f , so

‖Nu‖Lq ≤ cp‖f‖Lq ≤ Cr1/q. By (3.1),

|u(X)| ≤ Cr1/q

min(σ(∂V ),dist(X, ∂V ))1/q.

So (9.5) follows from (3.9) and Lemma 3.4. If V = Ω is a special Lipschitz domain,then by Lemma 3.4 ˆ

Q(X,R)

|∇u|2 ≤ C

R2

ˆQ(X,2R)

|u|2 ≤ Cr2/q

R2/q.

Now consider the second case, so that ν · A∇u = a, and (N)Ap and (D)Aq holdin V and the domains Q(X,R). By Lemma 3.21, the conjugate u is continuous

in V and (R)Ap holds in the same domains. So u satisfies (9.5) and (9.6), and since|∇u| ≈ |∇u|, we have that u satisfies (9.5) and (9.6).

We now prove (9.2). The main idea is to show that, for all 0 < h < 1,

(9.7)

ˆ∂V

N(∇u)(X)(1 + |X −X0|/r)α dσ

≤ Ch1/p

ˆ∂V

N(∇u)(X)(1 + |X −X0|/r)α dσ(X) +C

h1+1/q.

By choosing h small but positive, (9.7) immediately implies thatˆ∂V

N(∇u)(X)(1 + |X −X0|/r)α dσ ≤ C

provided the left-hand side is finite.We begin by looking at subsets of the boundary.

Lemma 9.8. Suppose that V , p, A, a, u satisfy the conditions of Theorem 9.1.Then there is some constant C depending only on p, cp, λ, Λ and the Lipschitzconstants of V , such that if Y ∈ ∂V , 0 < h and R is small enough that Q(Y, 2R)exists, and a = 0 on ∆(Y, 2R), then

ˆ∆(Y,R)

N(∇u) ≤ Ch1/p

ˆ∆(Y,2R)

N(∇u)(X) dσ(X) +Cr1/q

h1+1/qR1/q.

9.1. SOLUTIONS WITH BOUNDARY DATA IN H1 91

Proof. Pick some such Y and R. Define

γ1(X) = Y ∈ γ(X) : |Y −X| < R/4,γ2(X) = Y ∈ γ(X) : |Y −X| ≥ R/4,

Ni(∇u)(X) = sup |∇u(Y )| : Y ∈ γi(X),Qτ = Q(Y,R+ τR), ∆τ = ∆(Y,R+ τR),

and let N(∇u)(X) = NV (∇u)(X). We have thatˆ∆(Y,R)

N(∇u)(X) dσ(X) ≤ˆ

∆(Y,R)

N1(∇u)(X) +N2(∇u)(X) dσ(X).

If Z ∈ γ2(X,R), then dist(Z, ∂V ) ≥ 11+a |Z−X| ≥ R

4+4a . So by (9.5), N2(∇u)(X) ≤Cr1/qR−1−1/q.

Observe that if X ∈ ∆(Y,R) then γ1(X) ⊂ γQτ (X) for τ ≥ 12 . Assume that

ν · A∇u = a on ∂V ; the case τ · ∇u = a is similar. Then since (N)Ap holds inQ(Y,R+ τR), and ν ·A∇u = 0 on ∆(Y, 2R), if 1/2 ≤ τ ≤ 1 then

ˆ∆(Y,R)

N1(∇u) dσ ≤ˆ

∆(Y,R)

NQτ (∇u) dσ ≤ CR1/q

(ˆ∂Qτ

NQτ (∇u)p dσ

)1/p

≤ CR1/q

(ˆ∂Qτ

|ν ·A∇u|p dσ)1/p

.

Let χ±(t, τ) = ψ(R ± τR, t), and let χτ± = χ±(0, τ). Define I(h, τ) = χ±(t, τ) :0 < t < hR, so I(h, τ) ⊂ γa(χτ+) ∪ γa(χτ−) and σ(I(h, τ)) = 2hR. Then by (9.5)ˆ

∂Qτ

|ν ·A∇u|p dσ =

Î(h,τ)

|ν ·A∇u|p dσ +

ˆ∂Qτ\I(h,τ)\∂V

|ν ·A∇u|p dσ

≤ ChRN(∇u)(χτ−)p + ChRN(∇u)(χτ+)p +CRrp/q

(hR)p+p/q.

Taking the integral from τ = 1/2 to τ = 1, we get thatˆ

∆(Y,R)

N(∇u) ≤ˆ 1

1/2

CRh1/p(N(∇u)(χτ−) +N(∇u)(χτ+)

)dτ +

Cr1/q

h1+1/qR1/q

≤ Ch1/p

ˆ∆(Y,2R)

N(∇u)(X) dσ(X) +Cr1/q

h1+1/qR1/q.

We now consider subsets of the boundary that include supp a.

Lemma 9.9. Suppose that V , p, A, a, u satisfy the conditions of Theorem 9.1.Assume that α < 1/q = 1− 1/p, and that r is small enough that Q(X0, 4r) exists.

Then there is some constant C depending only on p, cp, λ, Λ and the Lipschitzconstants of V , such that if we define

I(R) =

ˆ∆(X0,R)

N(∇u)(X)(1 + |X −X0|/r)α dσ(X)

then for any h > 0 and any R small enough that Q(0, 2R) exists,

I(R) ≤ Ch1/pI(2R) +C

h2−1/p.


X0 Y j+

∆(X0, 2jρ) \∆(X0, 2

j−1ρ) ∆(Y j+, 2

j−1r)

−2jρ −2j−1ρ 0 2j−2ρ 2j−1ρ 2jρ

Figure 1. Connected subsets of the boundary

Proof. Let ρ = 2−j0R, where j0 is such that r ≤ ρ < 2r. Then supp a ⊂∆(X0, ρ), and

I(R) ≤ Cˆ

∆(X0,2ρ)

N(∇u) dσ + C

j0∑

j=2

2jαˆ

∆(X0,2jρ)\∆(X0,2j−1ρ)

N(∇u) dσ.

We can bound the first integral easily:ˆ

∆(X0,2ρ)

N(∇u)(X) dσ(X) ≤ Cρ1/q‖N(∇u)‖Lp(∂V ) ≤ Cr1/q‖a‖Lp(∂V ) ≤ C.

Now, ∆(X0, 2jρ) \ ∆(X0, 2

j−1ρ) is two connected sets, each of which can be

written as ∆(Y j±, 2j−2ρ) for some Y j±. Then a = 0 on ∆(Y j±, 2

j−1ρ). (See Figure 1.)So by Lemma 9.8,

ˆ∆(X0,2jρ)\∆(X0,2j−1ρ)

N(∇u) dσ

≤ Ch1/p

ˆ∆(X0,2j+1ρ)\∆(X0,2j−2ρ)

N(∇u) dσ +C

h1+1/q2j/q.

So

I(R) ≤ Cˆ

∆(X0,2ρ)

N(∇u) dσ + C

j0∑

j=2

2jαˆ

∆(X0,2jρ)\∆(X0,2j−1ρ)

N(∇u) dσ

≤ C + C

j0∑

j=2

2jαh1/p

ˆ∆(X0,2j+1ρ)\∆(X0,2j−2ρ)

N(∇u) dσ +C

h1+1/q

j0∑

j=2

2jα

2j/q.

If α < 1/q, then this is at most

I(R) ≤ C + Ch1/p

ˆ∆(X0,2j0+1ρ)

N(∇u)(1 + |X −X0|/r)α dσ +C

h1+1/q

and since 2j0ρ = R, this completes the proof.

Corollary 9.10. Theorem 9.1 holds if ∂V is bounded.

Proof. For any ∆ ⊂ ∂V , define I(∆) =´

∆N(∇u)(X)(1 + |X|/r)α dσ(X).

We wish to bound I(∂V ).

9.1. SOLUTIONS WITH BOUNDARY DATA IN H1 93

First,

I(∂V ) ≤ˆ∂V

N(∇u)

(σ(∂V )

r

)αdσ ≤

(σ(∂V )

r

)ασ(∂V )1/q

(ˆ∂V

N(∇u)p dσ

)1/p

≤ C(σ(∂V )

r

)ασ(∂V )1/q‖a‖Lp(∂V ) = C

(σ(∂V )

r

)α+1/q

.

This is finite for all r > 0. To complete the proof we need only consider the casewhere r = σ(supp a) is small compared to σ(∂V ).

By Definition 2.3, there exist at most C numbers Rj and points Xj ∈ ∂V suchthat Q(Xj , 2Rj) exists for all X ∈ ∂V , and

∂V ⊂C⋃

j=0

∆(Xj , Rj).

We may further assume that ∆(Xj , 2Rj)∩∆(X0, R0/2) = ∅ for all j ≥ 1, and thatRj ≥ σ(∂V )/C for all j. Finally, we may assume that 2r ≤ R0 and X0 ∈ supp a.

If X ∈ ∆(Xj , 2Rj) for some j ≥ 1, then |X −X0| ≈ σ(∂V ), so by Lemma 9.8,

I(∆(Xj , Rj)) ≤ Ch1/qI(∆(Xj , 2Rj)) +C

h1+1/q.

So by Lemma 9.9,

I(∂V ) ≤C∑

j=0

I(∆(Xj , Rj)) ≤C∑

j=0

Ch1/qI(∆(Xj , 2Rj)) +C

h1+1/q

≤ Ch1/qI(∂V ) +C

h1+1/q

and if h > 0 is small enough then I(∂V ) ≤ C.

Lemma 9.11. Theorem 9.1 holds if V = Ω is a special Lipschitz domain.

Proof. By Lemma 9.9, if we define

I(R) =

ˆ∆(X0,R)

N(∇u)(X)(1 + |X −X0|/r)α dσ(X)

then for any h > 0,

I(R) ≤ Ch1/pI(2R) +C

h2−1/p.

By taking the limit as R→∞, we recover (9.7). We need only show that I(∂Ω)is finite.

Recall that Ω = X ∈ R2 : ϕ(X ·e⊥) < X ·e; we begin by assuming A(x, t)−Iand ϕ have compact support. Let R be so large that B(0, R) contains supp a andxe⊥ + ϕ(x)e : ϕ 6= 0, and such that A(x) ≡ I for |x| > R. If e⊥ 6= (0,±1), wemay further assume that A(x) ≡ I on ∂Ω \B(0, R).

In Ω− \B(0, R), define

• u(xe⊥ − te) = u(xe⊥ + te), if u is a Neumann solution,• u(xe⊥ − te) = −u(xe⊥ + te), if u is a regularity solution.

Then ∇u(x, t) = ±E∇u(x,−t), where E is a constant matrix that representsreflection about the line xe⊥ : x ∈ R. Note that Et = E−1 = E.

Redefine A in ΩC such that A(xe⊥− te) = EA(xe⊥+ te)E outside of B(0, R).Figure 2 shows some lines on which A is constant after redefinition.


∂Ω

∂B(0, R)

e⊥e

Figure 2. Redefining A in ΩC

Then in Ω− \ B(0, R), it is straightforward to check that divA∇u = 0. Butoutside of B(0, R), where ∂Ω coincides with xe⊥ : x ∈ R, we have that ν+ ·A∇u|∂Ω+

= −ν− · A∇u|∂Ω− , where ν± = ∓e are the outward normal vectors toΩ±. (In the Neumann case, this is because both conormal derivatives are zero.)

Then if η ∈ C∞0 (R2 \B(0, R), thenˆ∇η ·A∇u =

ˆΩ+

∇η ·A∇u+

ˆΩ−

∇η ·A∇u

=

ˆ∂Ω+

η ν ·A∇u dσ +

ˆ∂Ω−

η ν ·A∇u dσ = 0

and so divA∇u = 0 in all of R2 \B(0, R).Now, if |X| > CR, thenA is independent of some direction in all ofB(X, |X|/4).

So

|∇u(X)| ≤ Cr1/q

|X|1+1/q;

if dist(X, ∂Ω) ≥ |X|/C this follows from (9.5), and if dist(X, ∂Ω) < |X|/C then by(3.9) and (9.6),

|∇u(X)| ≤ C(

B(X,|X|/C)

|∇u|2)1/2

≤ C

|X|

(ˆQ(X,|X|/C)

|∇u|2)1/2

≤ Cr1/q

|X|1+1/q.

Since N(∇u) ∈ Lp(∂Ω), we have thatˆ|X|<CR,X∈∂Ω

N(∇u)(X)(1 + |X|/r)α dσ(X)

is finite. But for large |X|, N(∇u)(X) ≤ C/|X|1+1/q; so if α < 1/q, thenˆ|X|≥CR,X∈∂Ω

N(∇u)(X)(1 + |X|/r)α dσ(X)

is finite as well and we are done.We now remove the assumption on ϕ and A. Assume ϕ(0) = 0, and let ϕR = ϕ

on (−R,R) and let ϕR = 0 outside of (−2R, 2R). Let ΩR = X ∈ R2 : ϕR(X ·e⊥) <X ·e. Let AR(x) = A for |x| < CR and AR(x) ≡ I for |x| > 2CR, C large enough.

9.2. INVERTIBILITY OF LAYER POTENTIALS ON H1(∂V ) 95

Let divAR∇uR = 0 in ΩR, ν · AR∇uR = a or τ · ∇uR = a on ∂Ω ∩ ∂ΩR, 0 on∂ΩR\∂Ω.

Suppose that |Y | is small compared to R, S with R < S. Then divA∇(uS −uR) = 0 in B(0, R) ∩ ΩR, and ν · A∇(uS − uR) = 0 or τ · ∇(uS − uR) = 0 onψ((−R,R)).

As in the proof of (9.6),ˆΩ∩B(0,R/2)

|∇uS −∇uR|2 ≤ Cr2/q

R2/q.

Thus, if |X| < R/4 then by (3.9),

|∇uS(X)−∇uR(X)| ≤ C r1/q

R1/q dist(X, ∂Ω).

Let Nδ,ρF (Q) = sup|F (Y )| : Y ∈ γ(Q), |Y − Q| > δ, |Y | < ρ. Defineu = limR→∞ uR. Then divA∇u = 0 in Ω, ν · A∇u = a or τ · ∇u = a on ∂Ω. Byconsidering Nδ,ρ(∇u−∇uR) for some R large (depending on δ and ρ) we see thatˆ

∂Ω

Nδ,ρ(∇u)(X)(1 + |X|/r)α dσ(X) ≤ C

for any ρ, δ > 0; thus´∂ΩN(∇u)(X)(1 + |X|/r)α dσ(X) ≤ C. By Lemma 3.14, we

have that u = u, and so Theorem 9.1 holds.

9.2. Invertibility of layer potentials on H1(∂V )

In this section we prove Corollary 9.3. This is an invertibility result, and soits proof is similar to the proof of Theorem 7.11. We begin by proving a lemmathat lets us compare the H1(∂V ) norms of (K±)tf and Ltf . This lemma serves thesame purpose as Lemma 7.8.

Lemma 9.12. Suppose that (K±)t and Lt are invertible on Lp0(∂V ) for some

1 < p < ∞, and that (N)AT

p and (R)AT

p hold in V±. Suppose further that theconclusions of Theorem 9.1 hold in V±.

Then for all f ∈ H1(∂V ),

‖(K+)tf‖H1(∂V ) ≈ ‖Ltf‖H1(∂V ) ≈ ‖(K−)tf‖H1(∂V )

where the comparability constants depend on λ, Λ, p, the Lipschitz constants of V ,and the invertibility constants of (K±)t, Lt.

Proof. Let f ∈ H1(∂V ). By Theorem 7.10, (K±)tf ∈ H1(∂V ), Ltf ∈H1(∂V ). We first show that ‖Ltf‖H1(∂V ) ≤ C(p)‖(K−)tf‖H1(∂V ). By the atomic

decomposition, (K−)tf =∑i λiai for some H1 atoms ai and constants λi with∑|λi| ≈ ‖(K−)tf‖H1(∂V ).

Let u = STf , so that (K−)tf = ν ·AT∇u. Then Ltf = τ · ∇u. We want to useLemma 7.9 to bound ‖τ · ∇u‖H1 .

Since (N)AT

p holds in V , we can write u =∑i λiui where ν · AT∇ui = ai,

‖N(∇ui)‖Lp(∂V ) ≤ C‖ai‖Lp(∂V ). If ai is supported in a set of surface measure ri

containing Xi, then ‖N(∇ui)‖Lp(∂V ) ≤ Cr1/p−1i . Furthermore, by assumptionˆ

∂V

N(∇ui)(X)(1 + |X −Xi|/ri)α dσ(X) ≤ C.

Since |τ · ∇ui| ≤ N(∇ui), these inequalities hold for τ · ∇ui as well.


If ∂V is compact, then´∂V

τ · ∇ui = 0 by continuity of ui. Otherwise, weare working in a special Lipschitz domain Ω. Recall the χ± of (2.5). SinceN(∇ui)(Y )(1 + |Y |/ri)α ∈ L1, for any ε > 0 and any R0 > 0 there must besome R > R0 with N(∇u)(χ±(Xi, R)) < ε/R1+α. So∣∣∣∣

ˆ∂Q(Xi,R)∩∂Ω

τ · ∇ui dσ∣∣∣∣ =

∣∣∣∣ˆ∂Q(Xi,R)\∂Ω

τ · ∇ui dσ∣∣∣∣ < Cε/Rα.

This may be made arbitrarily small by making ε small or R large; thus,´∂Ωτ ·

∇ui dσ = 0.So in either case, by Lemma 7.9, ‖τ · ∇ui‖H1(∂V ) ≤ C; therefore,

‖Ltf‖H1(∂V ) =∥∥∥τ ·

∑

i

λiui

∥∥∥H1≤ C(p)

∑

i

|λi| = C(p)‖Kt−f‖H1(∂V ).

Similarly, since (N)AT

p holds in V C , ‖Ltf‖H1(∂V ) ≤ C(p)‖(K+)tf‖H1(∂V ). We

want to show that ‖(K±)tf‖H1(∂V ) ≤ ‖Ltf‖H1(∂V ). We may say Ltf =∑i λiai;

since (R)AT

p holds in V±, we may let τ · ∇ui = ai. As before, we need only show

that´∂V

ν ·AT∇ui = 0 to establish ‖Ktf‖H1 ≤ C‖Ltf‖H1 . If V or V C is boundedthen this follows as in the proof of Theorem 7.10.

If V = Ω is a special Lipschitz domain, let η ∈ C∞0 (R2) with η ≡ 1 on Q(Xi, R),η ≡ 0 on Ω\Q(Xi, 2R), and |∇η| ≤ C/R. Then by the weak definition of ν ·AT0∇ui,∣∣∣∣

ˆ∂Ω

η ν ·AT∇ui dσ∣∣∣∣ =

∣∣∣∣ˆ

Ω

∇η ·AT∇ui dσ∣∣∣∣ ≤

C

R

ˆQ(Xi,2R)\Q(Xi,R)

|∇ui|

≤ C

R

ˆ 2R

R

ˆ∂Q(X,r)\∆(X,r)

|∇ui(Y )| dσ(Y ) dr

≤ Cˆ 2R

R

N(∇ui)(χ+(Xi, r)) +N(∇ui)(χ−(Xi, r)) dr.

Since N(∇ui) ∈ L1(∂Ω), this integral goes to zero as R →∞. This completes theproof.

Proof of Corollary 9.3. By Lemma 7.5, LtV+= −LtV− as an operator de-

fined on H1(∂V ). By Lemma 9.12, ‖KtV f‖H1(∂V ) ≈ ‖LtV f‖H1(∂V ). By Lemma 7.7,

K+ −K− is the identity operator on H1(∂V ). So

‖f‖H1(∂V ) = ‖Kt+f −Kt−f‖H1(∂V ) ≤ ‖Kt+f‖H1(∂V ) + ‖Kt−f‖H1(∂V )(9.13)

≤ C(p)‖Ltf‖H1(∂V ) ≤ C(p)‖Kt±f‖H1(∂V ).

So Lt and Kt± are one-to-one on H1(∂V ), and their inverses have norms at mostC(p).

Observe that Lp0(∂V ) is dense in H1(∂V ). Recall that Lt and Kt± are surjectiveon Lp0(∂V ). By (9.13) and by boundedness of Lt and Kt±, they must be surjectiveon H1(∂V ) as well.

CHAPTER 10

Concluding Remarks

We have now completed all the steps of the proof of Theorem 2.16 under theassumption that the coefficients A are smooth. In Theorem 10.1, we will removethe smoothness assumption. In Section 10.1, we will conclude this monograph byexploring some converses to Theorem 2.16.

Theorem 10.1. Let V ⊂ R2 be a Lipschitz domain with connected boundary.Suppose that Aj → A pointwise a.e. in V , where the Ajs satisfy (2.1) uniformly.

If (R)Aj1 holds in V uniformly in j, then if (D)

Ajq or (R)

Ajp holds in V , uniformly

in j, and their solutions are equal to the (R)A1 -solutions, then (D)Aq or (R)Ap holdsin V .

If (N)Aj1 holds in V uniformly in j, then if (N)

Ajp holds in V , uniformly in j,

and the (N)Ajp -solutions are equal to the (N)A1 -solutions, then (N)Ap holds in V .

We recover Theorem 2.16 by letting the matrices Aj be smooth. See Chapter 8for conditions under which (D)Aq -solutions or (R)Ap -solutions necessarily equal (R)A1 -solutions. The core of the proof comes from [Ken94, Section 1.10]; the technicalitiesarise in dealing with unbounded domains.

Proof. If we are establishing (D)Aq or (R)Ap , let f be compactly supported

on ∂V with f , ∂τf bounded. Let uj be the (R)Aj1 -solution with boundary data f ,

and let u be the solution to divA∇u = 0 in V , u = f on ∂V provided by Lem-ma 3.11. Then ∇u ∈ L2(V ), ∇uj ∈ L2(V ) by Lemmas 3.3 and 3.10.

If we are establishing (N)Ap , let f ∈ Lp(∂V ) ∩ H1(∂V ), let uj be the (N)Aj1 -

solution with boundary data f , and let u be the solution to divA∇u = 0 in V ,ν ·A∇u = f on ∂V provided by Lemma 3.12. Again ∇u ∈ L2(V ), ∇uj ∈ L2(V ).

It suffices to prove that Nu ∈ Lq(∂V ) or N(∇u) ∈ Lp(∂V ) with norm depend-ing only on ‖f‖Lq(∂V ) or ‖f‖Lp(∂V ).

Pick some R 0. Let W (R) = B(0, R)∩ V if V C (or V ) is bounded, W (R) =Q(0, R) if V = Ω is special. Let η = 1 in W (R), η = 0 in V \W (2R), |∇η| < C/R.Let vj = u− uj . Then by (1.1) or (1.5),

ˆV

η2|∇vj |2 ≤1

λRe

ˆV

η2∇vj ·Aj∇vj

=1

λRe

ˆV

∇(η2vj) · (Aj −A)∇u− 2

λRe

ˆV

η vj∇η ·Aj∇vj .

97

98 10. CONCLUDING REMARKS

Therefore, ˆV

η2|∇vj |2 ≤1

2

ˆV

η2|∇vj |2 + C

ˆV

|Aj −A|2|∇u|2

+C

R

ˆW (2R)\W (R)

|vj |(|∇vj |+ |∇u|).

Since ∇u and ∇uj lie in L2(V ), the left-hand integral is finite and we may hide thefirst term on the right-hand side. If V is bounded then the last integral is equalto zero. If V = Ω is special then by the Poincare inequality and the bounded-ness of the trace map, the last integral is controlled by ‖∇vj‖L2(W (2R)\W (R)) and

‖∇u‖L2(W (2R)\W (R)), which go to zero as R→∞. Finally, if V C is bounded, thenby Lemma 3.10 lim|X|→∞ vj(X) exists, and so again the last term goes to zero asR→∞.

In any case. ˆV

|∇(u− uj)|2 ≤ CˆV

|Aj −A|2|∇u|2.

Since ∇u ∈ L2(V ), by the dominated convergence theorem, this goes to zero asj →∞.

Let Nδ,RF (Q) = sup|F (Y )| : Y ∈ γ(Q), |Y − Q| > δ, |Y | < R. By (3.9), ifY ∈ γ(Q) then

|∇u(Y )|2 ≤ C B(Y,dist(Y,∂V )/2)

|∇u|2

≤ C B(Y,dist(Y,∂V )/2)

|∇u−∇uj |2 + C

B(Y,dist(Y,∂V )/2)

|∇uj |2

≤ C

dist(Y, ∂V )2‖∇(u− uj)‖2L2(V ) +N(∇uj)(Q).

So

‖Nδ,R(∇u)‖Lp(∂V ) ≤C

δ2R1/p‖∇(u− uj)‖2L2(V ) + ‖N(∇uj)‖Lp(∂V ).

Letting j →∞, and then letting δ → 0 and R→∞, we see that if (R)Ajp or (N)

Ajp

holds in V uniformly in j, then (R)Ap or (N)Ap holds in V as well.

We now turn to (D)Aq . By the Poincare inequality and the boundedness of thetrace map, if u−uj = 0 on ∂V , then ‖u−uj‖L2(W (R)) ≤ C(W (R))‖∇(u−uj)‖L2(V ).Thus, using Lemma 3.6 instead of (3.9),

‖Nδu‖Lq(∂V ∩B(0,R)) ≤C(W (R))

δ2R1/q‖∇(u− uj)‖2L2(V ) + ‖Nuj‖Lq(∂V ).

Again letting j →∞ and then letting δ → 0, R→∞ completes the proof.

10.1. Converses

We are interested in the converses to Theorem 2.16. Suppose that N(∇u) ∈Lp(∂V ), 1 ≤ p ≤ ∞. We want to show that ν · A∇u or τ · ∇u exist on ∂V and liein appropriate spaces.

It is elementary to show that the boundary values lie in Lp(∂V ). If N(∇u)is finite a.e., then the nontangential limit f of u exists a.e.; it is straightforwardto show that |∂τf | ≤ N(∇u). Similarly, since ∇u ∈ L2(V ∩ B(0, R)) for any

10.1. CONVERSES 99

R > 0 by Lemma 3.3, we have that ν · A∇u, as defined by (1.5), exists; by (1.5),|ν ·A∇u| ≤ ΛN(∇u).

However, Theorem 2.16 produces solutions u with N(∇u) ∈ L1(∂V ) only ifthe boundary data τ · ∇u or ν · A∇u lies in H1(∂V ) ( L1(∂V ). Thus, we wantto show that, if N(∇u) ∈ L1(∂V ), then ν · A∇u, ∂τu have H1 norms at mostC‖N(∇u)‖L1(∂V ).

This was shown in [DK87, Lemma 2.10] for harmonic functions in boundedLipschitz domains, using the duality of H1 and BMO and an extension theorem ofVaropoulos. Similar techniques work for more general solutions and in more generaldomains.

We begin with some results involving functions in BMO(∂V ).

Lemma 10.2. Suppose that V is a Lipschitz domain. Let f ∈ BMO(∂V ) besupported in some ∆(X,R) small enough that Q(X, 2R) exists. Then there existssome function F , supported in Q(X, 2R), such that F → f nontangentially a.e. in∂V , and such that

‖∇F‖C = supX0∈∂V,R>0

1

σ(B(X0, R) ∩ ∂V )

ˆB(X0,R)∩V

|∇F | ≤ C‖f‖BMO

where the constant C depends only on the Lipschitz constants of V .If ∂τf is bounded we may require ∇F bounded.

Remark 10.3. Let f ∈ BMO(∂V ) for some Lipschitz domain V with compactboundary. As in [FK81, Lemma 2.3], we may extend Lemma 10.2 to such functionsby decomposing f into pieces with small support.

Let the points Xj ∈ ∂V and the numbers rj be such that ∂V ⊂ ∪j∆(Xj , rj) andsuch that for each j, Q(Xj , 4rj) exists. Since V is a Lipschitz domain, this may bedone with at most CN points Xj . Let ηjCNj=1 be a set of smooth, compactly sup-ported functions such that

∑j ηj(X) = 1 for all X with dist(X, ∂V ) ≤ σ(∂V )/C,

|∇ηj | ≤ C/σ(∂V ) and supp ηj is contained in Q(Xj , 2rj).Let fj = fηj . If the connected component of ∂V containing Xj is ω, then

f =

CN∑

j=1

fj and ‖fj‖BMO(∂V ) ≤ C‖f‖BMO(∂V ) + C

∣∣∣∣ ω

f dσ

∣∣∣∣.

Before proving Lemma 10.2, we show that it implies our converses.

Theorem 10.4. Suppose that divA∇u = 0 in V for some Lipschitz domain V ,and assume N(∇u) ∈ L1(∂V ).

If V is simply connected, then there is a C depending only on Λ and the Lip-schitz constants of V such that

‖ν ·A∇u‖H1(∂V ) ≤ C‖N(∇u)‖L1(∂V )

in the sense that, if η is smooth and compactly supported in R2, then∣∣´ η ν ·A∇u∣∣ ≤

C‖N(∇u)‖L1(∂V )‖η‖BMO(∂V ).If V is not simply connected, then there is some fu, defined on ∂V and constant

on each connected component of ∂V , such that ν ·A∇u− fu is in H1(∂V ).

Proof. By the remarks at the start of this section, ν ·A∇u exists in the weaksense and lies in L1(∂V ). Let f ∈ C∞0 (∂V ). To show that ν ·A∇u− fu ∈ H1(∂V ),


it suffices to show that∣∣∣∣ˆ∂V

f(ν ·A∇u− fu) dσ

∣∣∣∣ ≤ C‖f‖BMO‖N(∇u)‖L1 .

But if F is compactly supported, ∇F ∈ L2 and TrF = f thenˆ∂V

f ν ·A∇u dσ =

ˆV

∇F ·A∇u.

If ∂V is bounded let fu =ffl∂ων · A∇u dσ on each connected component ω

of ∂V . If V is simply connected this implies fu ≡ 0. Thenˆω

f(ν ·A∇u− fu) dσ =

ˆω

f ν ·A∇u dσ −ˆω

f

ω

ν ·A∇u dσ

=

ˆω

(f −

fflωf)ν ·A∇u dσ

and so if ∂V is bounded we may assumefflωf dσ = 0. Thus, by Remark 10.3, we

need only consider functions f supported in ∆(X,R) for some X ∈ ∂V and someR small enough that Q(X, 2R) exists.

Since f is Lipschitz and supported in some Q(X,R), the F of Lemma 10.2 iscompactly supported and has bounded gradient, with ‖∇F‖C ≤ C‖f‖BMO. So weneed only show that ∣∣∣∣

ˆV

∇F ·A∇u∣∣∣∣ ≤ C‖∇F‖C‖N(∇u)‖L1 .

We must review some basic theorems about Carleson measures. Let G, H betwo functions. It is well known (see, for example, [Ste93, Section II.2]) thatˆ

R2+

|G||H| ≤ C‖G‖C‖NH‖L1(∂R2+).

This clearly extends by a change of variables to special Lipschitz domains. If V is aLipschitz domain with compact boundary, then the inequality holds if we integratenot over all of V but over the N domains Q(Xj , 2rj).

So ˆV

|∇F ||∇u| ≤ C‖f‖BMO‖N(∇u)‖L1

as desired.

We now move on to the regularity problem.

Theorem 10.5. Suppose that divA∇u = 0 in V and N(∇u) ∈ L1(∂V ) forsome Lipschitz domain V . Then g(X) = limZ→X n.t. u(Z) exists for almost everyX ∈ ∂V . Furthermore, ∂τg exists in the weak sense and satisfies

‖∂τg‖H1(∂V ) ≤ C‖N(∇u)‖L1(∂V ),

for some C depending only on λ, Λ and the Lipschitz constants of V .

Proof. Let f ∈ BMO(∂V ) have compact support. We seek to show that∣∣∣∣ˆ∂V

f∂τg dσ

∣∣∣∣ ≤ C‖f‖BMO(∂V )‖N(∇u)‖L1(∂V ).

If ∂V is bounded, then´ω∂τg dσ = 0 for all connected components of ∂V .

So we may assume thatfflωf dσ = 0, and so again we need only consider BMO

10.1. CONVERSES 101

functions f supported in ∆(X,R) for some R small enough that Q = Q(X, 2R)exists.

Then Q is a simply connected bounded Lipschitz domain. Recall the conjugateu of (2.25); we have that u exists in Q.

Since ˜u = −u up to an additive constant, (3.18) implies that τ ·∇u = −ν ·A∇u.So, letting F be the function produced by Lemma 10.2,

∣∣∣∣ˆ∂V

f τ · ∇u dσ∣∣∣∣ =

∣∣∣∣ˆ∂Q

f ν · A∇u dσ∣∣∣∣ =

∣∣∣∣ˆQ

∇F · A∇u∣∣∣∣

≤ C‖f‖BMO(∂V )‖NQ(1suppF∇u)‖L1(∂Q)

≤ C‖f‖BMO(∂V )‖N(∇u)‖L1(∂V )

since F is supported in Q(X, 2R).

We conclude this section by proving Lemma 10.2.

Proof of Lemma 10.2. We begin with the half-plane R2+. By [Var77] and

[Var78], if f is a compactly supported function on R = ∂R2+, with ‖f‖BMO(R) = 1,

then there exists an F ∈ C∞(R2+) such that limt→0 F (x, t) = f(x) for a.e. x ∈ R

and |∇F (x, t)| dx dt is a Carleson measure on R2+.

We will need a few nice properties of this function F . We claim that

(10.6)

∣∣∣∣F (x, t)− x+t

x−tf(y) dy

∣∣∣∣ ≤ C‖f‖BMO.

Without loss of generality take supp f ⊂ (0, 1). We review the construction of[Var77] and [Var78] as follows: if f ∈ BMO(R) and I ⊂ R is a dyadic interval,then there exists a family W = W (I) of dyadic intervals ω ⊂ I and a functionα : W 7→ C such that

• |α(ω)| ≤ C‖f‖BMO for all intervals ω ∈W (I),• ∑ω⊂I, ω∈W (I)|ω| ≤ C|I| for all intervals I ⊂ I, and

• f = b +fflIf +

∑ω∈W α(ω)1ω for some function b such that ‖b‖L∞(I) ≤

C‖f‖BMO.

If I ⊃ supp f , then∣∣fflIf∣∣ ≤ C‖f‖BMO, so we may redefine b to include this term.

Let W = W ((0, 1)). Define

F (x, t) =∑

ω∈Wαω1ω(x)1[0,σ(ω)](t) +

1

0

f.

Let ηt(y, s) = 1t2 η(y/t, s/t), where η is a smooth cutoff function supported in

B(0, 1/2). Let F (x, t) = F ∗ ηt(x, t).In [Var78], a smooth v is constructed with trace b such that ‖v‖L∞(R2

+) ≤C‖b‖L∞(R), |∇v(x, t)| ≤ ‖b‖L∞(R)/t and ‖ |∇v(x, t)| dx dt‖C ≤ C‖b‖L∞(R). Let

F = F +v. See [Var77] and [Var78] for details; here we only remark that, if I 3 x


is the dyadic interval with t < |I| ≤ 2t, then∣∣∣∣F (x, t)−

x+t

x−tf(y) dy

∣∣∣∣ ≤∣∣∣∣F (x, t)−

I

f(y)− b(y) dy

∣∣∣∣+ ‖b‖L∞(R) + C‖f‖BMO

=

∣∣∣∣∑

ω3x, |ω|>tα(ω)−

∑

ω∈Wα(ω)

|ω ∩ I||I|

∣∣∣∣+ C‖f‖BMO

=

∣∣∣∣∑

ω(Iα(ω)

|ω||I|

∣∣∣∣+ C‖f‖BMO ≤ C‖f‖BMO.

So since F is a convolution of F with a smooth cutoff, we have that∣∣∣∣F (x, t)−

x+t

x−tf(y) dy

∣∣∣∣ ≤ |v(x, t)|+ |F (x, t)− F (x, t)|+∣∣∣∣F (x, t)−

x+t

x−tf(y) dy

∣∣∣∣≤ C‖b‖L∞(R) + C‖f‖BMO ≤ C‖f‖BMO

and so (10.6) holds. Furthermore, observe that |∇F (x, t)| ≤ C/t.We claim that if f ′ is bounded, then we may require that ∇F is bounded. By

(10.6),

|f(x)− F (x, t)| ≤∣∣∣∣f(x)−

x+t

x−tf(y) dy

∣∣∣∣+

∣∣∣∣ x+t

x−tf(y) dy − F (x, t)

∣∣∣∣≤ t‖f ′‖L∞(R) + C‖f‖BMO.

Let η be a smooth cutoff function, η = 1 on (0, ρ), η(t) = 0 for t > 2ρ and|η′| < C/ρ. Then if G(x, t) = f(x)η(t) + F (x, t)(1 − η(t)) then |∇G(x, t)| dx dt isa Carleson measure, with Carleson norm at most ρ‖f ′‖+ ‖f‖BMO + ‖∇F (x, t)‖C .So by choosing ρ = ‖f‖BMO/‖f ′‖L∞(R) we see that we may replace F by G andtake ∇F bounded.

We now show that we may restrict the support of F by multiplying F by acutoff function.

Suppose f is supported in an interval (a, a + b). Let η be smooth, supportedin (a − c, a + b + c) × (0, b + c), with η(y, s) = 1 on (a, a + b) × (0, b), |η| ≤ 1 and|∇η| < 2/c everywhere. Let F be as above with F → f nontangentially. ThenFη → f nontangentially, and |∇(Fη)| ≤ |∇F |+ |F ||∇η|.

If (x, t) ∈ supp∇η, then either f ≡ 0 on (x, x + t) or (x − t, x), or supp f ⊂(x − t, x + t). In either case

∣∣ffl x+t

x−t f∣∣ ≤ C‖f‖BMO, so |F (x, t)| ≤ C‖f‖BMO. We

have that ‖∇F‖C ≤ C‖f‖BMO and ‖∇η‖C ≤ C, so ‖∇(Fη)‖C ≤ C‖f‖BMO.So we may assume that F is supported in a small neighborhood of supp f ×

(0, |supp f |). This completes the proof for the upper half-plane.We now pass to more general Lipschitz domains. If f ∈ BMO(∂Ω) for Ω

an arbitrary special Lipschitz domain, we may construct such an F by change ofvariables. Let g(x) = f(ψ(x)), and let G→ g with ‖∇G(x, t)‖C ≤ C‖g‖BMO. Let

F (ψ(x, t)) = G(x, t) and so F (xe⊥ + te) = G(x, t− ϕ(x)).

Then F → f and |∇F | ≤ C|∇G|, so

‖∇F (x, t)‖C ≤ C‖f‖BMO.

Finally, suppose that V is a Lipschitz domain with compact boundary, and thatf ∈ BMO(∂V ) is supported in some ∆(X,R). Simply construct F in the special

10.1. CONVERSES 103

Lipschitz domain Ωj of Definition 2.3; since suppF ⊂ Q(X, 2R) and Q(X, 2R) ⊂Ωj ∩V , we construct F in V by extending F by zero. This completes the proof.

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Elliptic PDE with almost-real coefficients