Lectures by: Monica Visanjosephbreen/Lectures_in_Harmonic_Analysis.pdf1.3 Complex Interpolation An important tool in harmonic analysis is interpolation. Broadly speaking, interpolation

Lectures in Harmonic Analysis

Joseph Breen

Lectures by: Monica Visan

Last updated: June 29, 2017

Department of MathematicsUniversity of California, Los Angeles

Contents

Preface 3

1 Preliminaries 41.1 Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Complex Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Some General Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4.1 Integrating |x|−α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 Dyadic Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 The Fourier Transform 102.1 The Fourier Transform on Rd . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 The Fourier Transform on Td. . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Lorentz Spaces and Real Interpolation 193.1 Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Duality in Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.3 The Marcinkiewicz Interpolation Theorem . . . . . . . . . . . . . . . . . . . 26

4 Maximal Functions 294.1 The Hardy-Littlewood Maximal Function . . . . . . . . . . . . . . . . . . . . 294.2 Ap Weights and the Weighted Maximal Inequality . . . . . . . . . . . . . . . 304.3 The Vector-Valued Maximal Inequality . . . . . . . . . . . . . . . . . . . . . . 36

5 Sobolev Inequalities 415.1 The Hardy-Littlewood-Sobolev Inequality . . . . . . . . . . . . . . . . . . . . 415.2 The Sobolev Embedding Theorem . . . . . . . . . . . . . . . . . . . . . . . . 445.3 The Gagliardo-Nirenberg Inequality . . . . . . . . . . . . . . . . . . . . . . . 47

6 Fourier Multipliers 496.1 Calderon-Zygmund Convolution Kernels . . . . . . . . . . . . . . . . . . . . 496.2 The Hilbert Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566.3 The Mikhlin Multiplier Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 57

7 Littlewood-Paley Theory 617.1 Littlewood-Paley Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.2 The Littlewood-Paley Square Function . . . . . . . . . . . . . . . . . . . . . . 687.3 Applications to Fractional Derivatives . . . . . . . . . . . . . . . . . . . . . . 71

8 Oscillatory Integrals 798.1 Oscillatory Integrals of the First Kind, d = 1 . . . . . . . . . . . . . . . . . . . 808.2 The Morse Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 878.3 Oscillatory Integrals of the First Kind, d > 1 . . . . . . . . . . . . . . . . . . . 89

1

8.4 The Fourier Transform of a Surface Measure . . . . . . . . . . . . . . . . . . 92

9 Dispersive Partial Differential Equations 949.1 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 949.2 The Linear Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . 95

9.2.1 The Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . 959.2.2 Dispersive Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 969.2.3 Strichartz Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

10 Restriction Theory 9710.1 The Restriction Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9710.2 The Tomas-Stein Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10010.3 Restriction Theory and Strichartz Estimates . . . . . . . . . . . . . . . . . . . 103

11 Rearrangement Theory 10811.1 Definitions and Basic Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 10811.2 The Riesz Rearrangement Inequality, d = 1 . . . . . . . . . . . . . . . . . . . 11411.3 The Riesz Rearrangement Inequality, d > 1 . . . . . . . . . . . . . . . . . . . 11711.4 The Polya-Szego inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12211.5 The Energy of the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . 125

12 Compactness in Lp Spaces 12812.1 The Riesz Compactness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 12812.2 The Rellich-Kondrachov Theorem . . . . . . . . . . . . . . . . . . . . . . . . 13112.3 The Strauss Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13412.4 The Refined Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

13 The Sharp Gagliardo-Nirenberg Inequality 13713.1 The Focusing Cubic NLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

14 The Sharp Sobolev Embedding Theorem 138

15 Concentration Compactness in Partial Differential Equations 139

References 139

2

Preface

**Eventually I’ll probably write a better introduction, this is a placeholder for now**

The following text covers material from Monica Visan’s Math 247A and Math 247Bclasses, offered at UCLA in the winter and spring quarters of 2017. The content of eachchapter is my own adaptation of lecture notes taken throughout the two quarters.

Chapter 1 contains a selection of background material which was not covered in lec-tures. The rest of the chapters roughly follow the chronological order of lectures giventhroughout the winter and spring, with some slight rearranging of my own choice. Forexample, basic results about the Fourier transform on the torus are included in Chapter2, despite being the subject of lectures in the spring. A strictly chronological (and incom-plete) version of these notes can be found on my personal web page at http://www.math.ucla.edu/˜josephbreen/.

Finally, any mistakes are likely of my own doing and not Professor Visan’s. Commentsand corrections are welcome.

3

http://www.math.ucla.edu/~josephbreen/

http://www.math.ucla.edu/~josephbreen/

Chapter 1

Preliminaries

The material covered in these lecture notes is presented at the level of an upper divisiongraduate course in harmonic analysis. As such, we assume that the reader is comfortablewith real and complex analysis at a sophisticated level. However, for convenience andcompleteness we recall some of the basic facts and inequalities that are used throughoutthe text. We also establish conventions with notation, and discuss a few select topics thatthe reader may be unfamiliar with (for example, the Riesz-Thorin interpolation theorem).Any standard textbook in real analysis or harmonic analysis is a suitable reference for thismaterial, for example, [3], [6], and [8].

1.1 Notation and Conventions

If x ≤ Cy for some constant C, we say x . y. If x . y and y . x, then x ∼ y. If the implicitconstant depends on additional data, this is manifested as a subscript in the inequalitysign. For example, if A denotes the ball of radius r > 0 in Rd and | · | denotes Rd-Lebesguemeasure, we have |A| ∼d rd. However, the dependencies of implicit constants are usuallyclear from context, or stated otherwise.

FINISH

1.2 Lp Spaces

ADD

Proposition 1.1 (Young’s Convolution inequality). For 1 ≤ p, q, r ≤ ∞,

‖f ∗ g‖Lr(Rd) . ‖f‖Lp(Rd) ‖g‖Lq(Rd)

whenever 1 + 1r = 1

p + 1q .

Young’s convolution inequality may be proved directly, or by using the Riesz-Thorininterpolation theorem (see Section 1.3). For details on both methods, refer to [3].

It is convenient to record a few special cases of Young’s inequality that arise frequently.

Corollary 1.2 (Young’s Convolution inequality, special cases). The following convolution in-equalities hold:

1. For any 1 ≤ p ≤ ∞, ‖f ∗ g‖L∞ . ‖f‖Lp ‖g‖Lp′ . In particular, ‖f ∗ g‖L∞ . ‖f‖L2 ‖g‖L2

and ‖f ∗ g‖L∞ . ‖f‖L1 ‖g‖L∞ .

2. For any 1 ≤ p ≤ ∞, ‖f ∗ g‖Lp . ‖f‖L1 ‖g‖Lp .

4

1.3 Complex Interpolation

An important tool in harmonic analysis is interpolation. Broadly speaking, interpolationconsiders the following question: given estimates of some kind on two different spaces,say Lp0 and Lp1 , what can be said about the corresponding estimate on Lp for values of pbetween p0 and p1?

There are a number of theorems which answer this kind of question. In Chapter 3, wewill prove the Marcinkiewicz interpolation theorem using real analytic methods. Here, weprove a different interpolation theorem using complex analytic methods. Both theoremshave their own advantages and disadvantages, and will be used frequently.

Our treatment of the Riesz-Thorin interpolation theorem follows [8]. The main state-ment is as follows.

Theorem 1.3 (Riesz-Thorin interpolation). Suppose that T is a bounded linear map from Lp0 +Lp1 → Lq0 + Lq1 and that

‖Tf‖Lq0 ≤M0 ‖f‖Lp0 and ‖Tf‖Lq1 ≤M1 ‖f‖Lp1

for some 1 ≤ p0, p1, q0, q1 ≤ ∞. Then

‖Tf‖Lqt ≤M1−t0 M t

1 ‖f‖Lpt

where 1pt

= 1−tp0

+ tp1

and 1qt

= 1−tq0

+ tq1

, for any 0 ≤ t ≤ 1.

As noted above, the proof of this theorem relies on complex analysis. Specifically, weneed the following lemma.

Lemma 1.4 (Three Lines lemma). Suppose that Φ(z) is holomorphic in S = {0 < Re z < 1}and continuous and bounded on S. Let

M0 := supy∈R|Φ(iy)| and M1 := sup

y∈R|Φ(1 + iy)|.

Thensupy∈R|Φ(t+ iy)| ≤M1−t

0 M t1

for any 0 ≤ t ≤ 1.

Proof. We can assume without loss of generality that Φ is nonconstant, otherwise the con-clusion is trivial.

We first prove a special case of the lemma. Suppose that M0 = M1 = 1 and thatsup0≤x≤1 |Φ(x + iy)| → 0 as |y| → ∞. In this case, φ has a global (on S) supremum ofM > 0. Let {zn} be a sequence such that |Φ(zn)| → M . Because sup0≤x≤1 |Φ(x + iy)| → 0as |y| → ∞, the sequence {zn} is bounded, hence there is a point z∞ ∈ S such that asubsequence of {zn} converges to z∞. By continuity, |Φ(z∞)| = M . Since Φ is nonconstant,by the maximum modulus principle, z∞ is on the boundary of S. This means that |Φ(z)| isglobally bounded by 1. Since M0 = M1 = 1, this proves the special case.

Next, we remove the decay assumption and only suppose thatM0 = M1 = 1. For ε > 0,define

Φε(z) := Φ(z)eε(z2−1).

First observe that |Φε(z)| ≤ 1 if Re z = 0 or Re z = 1. Indeed, note that for y ∈ R we have

|Φε(iy)| = |Φ(iy)|∣∣∣eε((iy)2−1)

∣∣∣ ≤ ∣∣∣e−ε(y2+1)∣∣∣ ≤ 1

and

|Φε(1 + iy)| = |Φ(1 + iy)|∣∣∣eε((1+iy)2−1)

∣∣∣ ≤ ∣∣∣eε(1+2iy−y2−1)∣∣∣ ≤ ∣∣e2iyε

∣∣ ∣∣∣e−εy2∣∣∣ ≤ 1.

5

Next, we claim that Φε satisfies the decay condition from the previous case. Indeed, notethat

|Φε(x+ iy)| = |Φ(x+ iy)|∣∣∣eε((x+iy)2−1)

∣∣∣ = |Φ(x+ iy)|∣∣∣eε(x2−y2−1+2xyi)

∣∣∣≤ |Φ(x+ iy)|

∣∣∣eε(x2−y2−1)∣∣∣ .

Because 0 ≤ x ≤ 1 and because |Φ| is bounded, as |y| → ∞, |Φε(x + iy)| → 0 uniformly inx. By the previous case, |Φε(z)| ≤ 1 uniformly in S. Since this holds for all ε > 0, lettingε→ 0 gives the desired result for Φ.

Finally we assume that M0 and M1 are arbitrary positive numbers. Define φ(z) :=M z−1

0 M−z1 φ(z). Note that if Re z = 0, then

|Φ(z)| ≤ |M iy−10 M−iy1 |M0 ≤ |M iy

0 | · |M−iy1 | ≤ 1

and if Re z = 1, then

|Φ(z)| ≤ |M iy0 M

−1−iy1 |M1 ≤ |M iy

0 | · |M−iy1 | ≤ 1.

The claim then follows by applying the previous case.

Proof of Theorem 1.3. By scaling appropriately, we may assume without loss of generalitythat ‖f‖Lpt = 1.

First, consider the case when f is a simple function. By duality,

‖Tf‖Lqt = sup

∣∣∣∣∫ Tf(x) g(x) dx

∣∣∣∣where the supremum is taken over simple functions g with ‖g‖

Lq′t

= 1. Fix such a g. First,consider the case pt <∞ and qt > 1. Let

γ(z) := pt

(1− zp0

+z

p1

)and δ(z) := q′t

(1− zq′0

+z

q′1

).

Definefz := |f |γ(z) f

|f |and gz := |g|δ(z) g

|g|.

Note that γ(t) = 1. Hence, ft = f . Also, if Re z = 0, then

‖fz‖p0

Lp0 =

∫ ∣∣∣ |f |ptγ(z)∣∣∣ dx =

∫ ∣∣∣∣ |f |pt−z(pt+ p0p1

) ∣∣∣∣ dx =

∫|f |pt dx = ‖f‖ptLpt = 1.

A similar computation gives ‖fz‖Lp1 = 1 if Re z = 1. Also, ‖gz‖Lq′0 = 1 if Re z = 0 and‖gz‖Lq′1 = 1 if Re z = 1.

Next, define

Φ(z) :=

∫Tfz(x) gz(x) dx.

As f and g are simple, we may write f =∑

k akχEk where the sets Ek are disjoint and offinite measure, and likewise g =

∑j bjχFj . Then

fz =∑k

|ak|γ(z) ak|ak|

χEk and gz =∑j

|bj |δ(z)bj|bj |

χFj .

Also with this notation we have

Φ(z) =∑j,k

|ak|γ(z)|bj |δ(z)ak|ak|

bj|bj |

∫T (χEk)χFj dx.

6

From this it is evident that Φ(z) is holomorphic in S = {0 < Re z < 1} and continuous andbounded on S. Moreover, if Re z = 0, then Holder’s inequality gives

|Φ(z)| ≤∫|Tfz(x) gz(x)| dx ≤ ‖Tfz‖Lq0 ‖gz‖Lq′0 ≤M0 ‖f‖Lp0 = M0.

Similarly, if Re z = 1, |Φ(z)| ≤M1. Therefore, by the Three Lines lemma, if Re z = t then

|Φ(z)| ≤M1−t0 M t

1.

As Φ(t) =∫Tf(x) g(x) dx, this is the desired result.

The passage from simple functions to general functions is a straightforward limitingargument, and the remaining cases pt = 1 and qt = ∞ are handled similarly. For moredetails, consult [8].

It is convenient to state a few special cases of Riesz-Thorin interpolation. These occurfrequently, and also provide more concrete examples of interpolation at work.

Corollary 1.5 (Riesz-Thorin interpolation, special cases).

1. Fix 1 ≤ p0 < p1 ≤ ∞. Suppose that T is bounded as a linear map from Lp0 → Lp0 and alsofrom Lp1 → Lp1 . Then T is bounded from Lp → Lp for all p0 ≤ p ≤ p1.

2. Suppose that T is bounded as a linear map from L2 → L2 and also from L1 → L∞. Then Tis bounded from Lp → Lp

′ for all 1 ≤ p ≤ 2.

Proof.

1. If T is bounded from Lp0 → Lp0 and Lp1 → Lp1 , then Riesz-Thorin implies that T isbounded from Lpt → Lqt for all 0 ≤ t ≤ 1 with

1

pt=

1− tp0

+t

p1and

1

qt=

1− tp0

+t

p1.

Thus, pt = qt. Moreover, as t ranges from 0 to 1, pt ranges from p0 to p1. Therefore, Tis bounded from Lp → Lp for all p0 ≤ p ≤ p1.

2. Riesz-Thorin gives boundedness of T : Lpt → Lqt for 0 ≤ t ≤ 1 with

1

pt=

1− t1

+t

2and

1

qt=

1− t∞

+t

2.

Thus, 1pt

= 1− t2 and 1

qt= t

2 so that qt = p′t. Moreover, pt = 22−t , so as t ranges from 0

to 1, pt ranges from 1 to 2.

The first substantial application of interpolation is found in found in Chapter 2, wherewe show that the Fourier transform is well-defined on Lp spaces for 1 ≤ p ≤ 2. As a finalremark, the Riesz-Thorin interpolation theorem is an interpolation result based on complexanalytic tools, namely, the maximum principle for holomorphic functions. In Chapter 3,we prove a different interpolation theorem — the Marcinkiewicz interpolation theorem —based on real analytic techniques.

1.4 Some General Principles

Throughout the rest of this text, there are a number of straightforward principles that wewill employ over and over again, often without explicit comment. For convenience, weinclude a short discussion of these recurring principles here.

7

1.4.1 Integrating |x|−α

The first concern integration of functions of the form |x|−α. In words, integrating |x|−α overa region of Rd increases the exponent by d. Explicitly, we have the following proposition.

Proposition 1.6. Fix R > 0. Then

1. The integral∫|x|>R |x|

−α dx is finite when α > d, in which case∫|x|>R

|x|−α dx ∼ R−α+d.

2. The integral∫|x|<R |x|

−α dx is finite when α ≤ d, in which case∫|x|<R

|x|−α dx ∼ R−α+d.

Consequently, |x|−α /∈ Lp(Rd) for 1 ≤ p <∞, for any choice of α.

Proof. Fix R > 0. Set r = |x|. Then dx = rd−1 dr. Thus,∫|x|>R

1

|x|αdx =

∫r>R

1

rαrd−1 dr =

∫ ∞r=R

r−α+d−1 dr ∼ r−α+d

∣∣∣∣∣∞

R

.

This limit exists precisely when −α+ d < 0, i.e., when α > d, in which case∫|x|>R

1

|x|αdx ∼ R−α+d.

Similarly, ∫|x|<R

1

|x|αdx ∼ r−α+d

∣∣∣∣∣R

0

.

This limit exists precisely when −α+ d ≥ 0, i.e., α ≤ d, in which case∫|x|<R

1

|x|αdx ∼ R−α+d.

1.4.2 Dyadic Sums

Another computational tool ubiquitous in harmonic analysis is dyadic sums; that is, geo-metric series with r = 1

2 . The following proposition is easy, but worth recording.

Proposition 1.7. Let 2Z = { 2n : n ∈ Z } be the set of dyadic numbers. Then∑N∈2Z;N≥N0

1

N= 2 · 1

N0

and ∑N∈2Z;N≤N0

N = 2 ·N0

8

Proof. Fix a dyadic number N0 = 2n0 ∈ 2Z. Then

∑N∈2Z;N≥N0

1

N=

∑n∈Z;n≥n0

2−n =∞∑

n=n0

(1

2

)n=

(12

)n0

1− 12

= 2 · 2−n0 = 2 · 1

N0.

Essentially the same computation gives∑N∈2Z;N≤N0

N = 2 ·N0.

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Chapter 2

The Fourier Transform

Our study of harmonic analysis naturally begins with the Fourier transform. Historically,the Fourier transform was first introduced by Joseph Fourier in his study of the heat equa-tion. In this context, Fourier showed how a periodic function could be decomposed into asum of sines and cosines which represent the frequencies of the function. From a modernpoint of view, the Fourier transform is a transformation which accepts a function and re-turns a new function, defined via the frequency data of the original function. As a bridgebetween the physical domain and the frequency domain, the Fourier transform is the maintool of use in harmonic analysis.

The basic theory of the Fourier transform is standard, and there are a wealth of refer-ences (for example, [3], [6], and [7]) for the following material.

2.1 The Fourier Transform on Rd

We begin by discussing the Fourier transform of complex-valued functions on the Eu-clidean space Rd. The Fourier transform is defined as an integral transformation of a func-tion; as such, it is natural to first consider integrable functions.

Definition 2.1. The Fourier transform of a function f ∈ L1(Rd) is given by

F(f)(ξ) := f(ξ) := (f)ˆ(ξ) :=

∫Rde−2πix·ξf(x) dx.

There are a number of conventions for placement of constants when defining the Fouriertransform. For example, another common definition of the Fourier transform is

f(ξ) =1

(2π)d2

∫Rde−ix·ξf(x) dx.

These choices clearly do not alter the theory in any meaningful way; they are simply a mat-ter of notational preference. In this text, we will use either convention when convenient. Inthis section, we will use the former.

We have two immediate goals. One is to understand the basic properties of the Fouriertransform and how it interacts with operations such as differentiation, convolution, etc.The other goal is to understand which classes of functions the Fourier transform can bereasonably defined on. As and aid towards both of these goals, we introduce (or recall)a suitably nice class of functions. Here, we use the following multi-index notation: forα = (α1, . . . , αd) ∈ Nd, define

|α| := α1 + · · ·αd; xα := xα11 · · ·x

αdd ; Dα :=

∂|α|

∂xα11 · · · ∂x

αdd

.

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Definition 2.2. A C∞-function f : Rd → C is Schwartz if xαDβf ∈ L∞x (Rd) for everymultiindex α, β ∈ Nd. The collection of Schwartz functions, Schwartz space, is denotedS(Rd).

In words, a function is Schwartz if it is smooth, and if all of its derivatives decay fasterthan any polynomial. An example of a Schwartz function is e−|x|

2.

The set of Schwartz functions forms a Frechet space, i.e., a locally convex space (a vectorspace endowed with a family of seminorms {ρα} that separates points: if ρα(f) = 0 for allα, then f = 0) which is metrizable and complete. In the case of S(Rd), the collection ofseminorms is given by {ρα,β}α,β∈Nd , where

ρα,β(f) :=∥∥∥xαDβf

∥∥∥L∞x (Rd)

.

The Frechet space structure of S(Rd) is not our main concern; rather, density in Lp spaces(indeed, note that C∞c (Rd) ⊆ S(Rd)) and its interaction with the Fourier transform are ofmore importance.

Here, we collect a number of basic and important properties of the Fourier transformof Schwartz functions.

Proposition 2.3. Let f ∈ S(Rd). Then:

1. If g(x) = f(x− y), then g(ξ) = e−2πiy·ξ f(ξ).

2. If g(x) = e2πix·ηf(x), then g(ξ) = f(ξ − η).

3. If g(x) := f(Tx) for T ∈ GL(Rd), then g(ξ) = | detT |−1f(T−tξ). In particular, if T isa rotation or reflection and f(Tx) = f(x) (i.e. f is rotation or reflection invariant), theng(ξ) = f(ξ).

4. If g(x) := f(x), then g(ξ) = f(−ξ).

5. If g(x) := Dαf(x), then g(ξ) = (2πiξ)αf(ξ).

6. If g(x) := xαf(x), then g(ξ) =(i

2π

)|α|Dαξ f(ξ).

7. If g(x) := (k ∗ f)(x) for k ∈ L1(Rd), then g(ξ) = k(ξ) · f(ξ).

8. We have the basic estimate∥∥∥f∥∥∥

L∞ξ

≤ ‖f‖L1x.

Proof.

1. Making the change of variables z = x− y, we have

g(ξ) =

∫e−2πix·ξf(x− y) dx =

∫e−2πi(z+y)·ξf(z) dz = e−2πiy

∫e−2πiz·ξf(z) dz

= e−2πiyf(ξ).

2. Computing directly gives:

g(ξ) =

∫e−2πix·ξe2πix·ηf(x) dx =

∫e−2πix·(ξ−η)f(x) dx = f(ξ − η).

3. For T ∈ GL(Rd), make the change of variables y = Tx. Then

g(ξ) =

∫e−2πix·ξf(Tx) dx =

∫e−2πi(T−1y)·ξf(y)|detT |−1 dy

= |detT |−1

∫e−2πiy·(T−tξ)f(y) dy

= |detT |−1f(T−tξ).

11

If T is a rotation or reflection (or more generally, T is orthogonal), then T tT = TT t =I and detT = 1. In this case, g(ξ) = |detT |−1f(T−tξ) = f(ξ).

4. Computing directly gives:

g(ξ) =

∫e−2πix·ξf(x) dx =

∫e−2πix·ξf(x) dx =

∫e−2πix·(−ξ)f(x) dx = f(−ξ).

5. Integrating by parts, we have:

g(ξ) =

∫e−2πix·ξDαf(x) dx = (−1)|α|

∫Dαe−2πix·ξf(x) dx

= (−1)|α|∫

(−2πiξ)αe−2πix·ξf(x) dx = (2πiξ)α∫e−2πix·ξf(x) dx

= (2πξ)αf(ξ).

6. Note that Dαxe−2πix·ξ = (−2πix)αe−2πix·ξ, so that xαe−2πix·ξ. Thus, integrating by

parts again gives:

g(ξ) =

∫e−2πix·ξxαf(x) dx =

(i

2π

)|α|f(ξ).

7. Computing, we have

g(ξ) =

∫e−2πix·ξ(k ∗ f)(x) dx =

∫e−2πix·ξ

∫k(x− y)f(y) dy dx

=

∫ ∫e−2πix·ξk(x− y)f(y) dx dy.

We make a change of variables z = x− y in the inner integral to get

g(ξ) =

∫ ∫e−2πi(z+y)·ξk(z)f(y) dz dy =

∫e−2πiz·ξk(z) dz ·

∫e−2πiy·ξf(y) dy

= k(ξ) · f(ξ).

8. This follows from the triangle inequality for integrals:∣∣∣∣∫ e−2πix·ξf(x) dx

∣∣∣∣ ≤ |∫ |e−2πix·ξ||f(x)| dx =

∫|f(x)| dx.

From this proposition, it becomes immediately clear why the Fourier transform is sucha powerful computational tool. For example, properties 5 and 7 above describe how theFourier transform turns complicated function operations like differentiation and convolu-tion into multiplication. This becomes incredibly useful when studying partial differentialequations, for instance. Also note that properties 1,2,3,4,7, and 8 extend immediately tofunctions in L1(Rd).

Proposition 2.4. If f ∈ S(Rd), then f ∈ S(Rd). Moreover, if fn → f ∈ S(Rd), then fn → f ∈S(Rd).

12

Proof. Suppose that f ∈ S(Rd). Note that

|ξαDβ f(ξ)| ∼ |ξαxβf(ξ)| ∼ |Dαxβf(ξ)| =∣∣∣∣∫ e−2πix·ξDα(xβf(x)) dx

∣∣∣∣ .Because f is Schwartz, Dα(xβf(x)) ∈ L1(Rd). Therefore,∥∥∥ξαDβ f(ξ)

∥∥∥L∞ξ

.∥∥∥Dα(xβf(x))

∥∥∥L1<∞

so that f ∈ S(Rd). It is clear that {fn} → f ∈ S(Rd) if and only if {fn} → f ∈ S(Rd) byproperties 5 and 6 of the previous proposition.

Corollary 2.5 (Riemann-Lebesgue Lemma). If f ∈ L1(Rd), then f is uniformly continuous andvanishes at∞.

Proof. Let C0(Rd) denote the space of continuous functions which vanish at∞.Let {fn} be a sequence of Schwartz functions with fn → f in L1(Rd). Since∥∥∥fn − f∥∥∥

L∞≤ ‖fn − f‖L1

it follows that fn → f in L∞(Rd). As f ∈ S(Rd), f ∈ C0(Rd). Since C0(Rd) is closed underuniform convergence, we are done.

Aside from simple properties and estimates of the Fourier transform and its interactionwith various function operations, we have not computed any Fourier transforms of actualfunctions.

Lemma 2.6. Let A be a real, symmetric, positive definite d× d matrix. Then∫e−x·Axe−2πix·ξ dx = π

d2 (detA)−

12 e−π

2ξ·A−1ξ.

Proof. SinceA is real, symmetric, and positive definite, it is diagonalizable. Explicitly, thereis an orthogonal matrix O and a diagonal matrix D = diag(λ1, . . . , λd), λj > 0, so thatA = OTDO. Let y = Ox and η = Oξ. Then

x ·Ax = x ·OTDOx = Ox ·DOx = y ·Dy =d∑j=1

λjy2j

andx · ξ = OT y ·OT η = y · η.

Thus, ∫e−x·Axe−2πix·ξ dx =

∫e−∑dj=1 λjy

2j e−2πiy·η dy =

d∏j=1

∫Re−λjy

2j−2πiyjηj dyj .

Computing each of these one-variable integrals, we have:∫Re−λy

2−2πiyη dy =

∫Re−λ(y−

πiηλ )

2−π2η2

λ dy = e−π2η2

λ

∫Re−λy

2dy = e−

π2η2

λ λ−12π

12 .

So ∫e−x·Axe−2πix·ξ dx =

d∏j=1

e−π2η2

jλj λ

− 12

j π12 = π

d2 (λ1 · · ·λd)−

12 e−∑dj=1

π2

λjη2j

= πd2 (detA)−

12 e−π

2η·D−1η.

The result follows from the fact that η ·D−1η = Oξ ·D−1Oξ = ξ ·A−1ξ.

13

Corollary 2.7. The function e−π|x|2 is an eigenvalue of the Fourier transform with eigenvalue 1.

Explicitly, e−π|x|2 = e−π|ξ|2 .

Proof. This follows from the previous lemma with A = πId. Indeed,∫e−π|x|

2e−2πix·ξ dx = π

d2

(πd)− 1

2e−π

2ξ· 1πξ = e−π|ξ|

2.

Above, we computed that the Fourier transform of a Schwartz function is again Schwartz.In fact, the Fourier transform is an isometry on Schwartz space and extends to a unitaryoperator on L2. The next few results summarize these facts.

Theorem 2.8 (Fourier Inversion). If f ∈ S(Rd), then (f )(x) = f(−x). Equivalently,

f(x) = (f)ˆ(−x) =: (f)ˇ(x) =: F−1(f)(x).

Proof. For ε > 0, let

Iε(x) :=

∫e−πε

2|ξ|2e2πix·ξ f(ξ) dξ.

By the dominated convergence theorem, as ε→ 0, Iε(x)→∫e2πix·ξ f(ξ) dξ. Thus, to prove

the theorem, it suffices to show that Iε(x)→ f(x) as ε→ 0. We have

Iε(x) =

∫e−πε

2|ξ|2e2πix·ξ f(ξ) dξ =

∫e−πε

2|ξ|2e2πix·ξ∫e−2πiy·ξf(y) dy dξ

=

∫f(y)

∫e−πε

2|ξ|2e−2πi(y−x)·ξ dξ dy.

Note that∫e−πε

2|ξ|2e−2πi(y−x)·ξ dξ =(e−πε

2|ξ|2)

(y − x) = πd2

((πε2)d

)− 12e−π

2(y−x)· 1πε2

(y−x)

= ε−de−π|y−x|2

ε2 .

Consequently,

Iε(x) =

∫f(y) ε−de−π

|y−x|2

ε2 dy = (f ∗ φε)(x)

where φε(x) = ε−dφ(xε

)with φ(x) = e−π|x|

2. It is a standard result that {φε} is an approxi-

mation to the identity. Thus, as ε→ 0,

Iε(x) = (f ∗ φε)(x)→ f(x).

Lemma 2.9. If f, g ∈ S(Rd), then∫f(ξ)g(ξ) dξ =

∫f(x)g(x) dx. In particular,

∫f(ξ)g(ξ) dx =∫

f(x)g(x) dx, so that∥∥∥f∥∥∥

L2= ‖f‖L2 . Hence, the Fourier transform is an isometry on S(Rd).

Proof. Computing, we have∫f(ξ)g(ξ) dξ =

∫ ∫e−2πix·ξf(x) dx g(ξ) dξ =

∫f(x)

∫e−2πix·ξg(ξ) dξ dx

=

∫f(x)g(x) dx.

For the “in particular” statement, let h = g. Then

h(x) = ˆg = (g)(−x) = g(x).

14

Theorem 2.10 (Plancharel). The Fourier transform extends from an operator on S(Rd) to a uni-tary operator on L2(Rd).

Proof. Fix f ∈ L2(Rd). Let {fn} ⊆ S(Rd) such that fnL2

−→ f . As the Fourier transform is anisometry on S(Rd), we have

∥∥∥fn − fm∥∥∥L2

= ‖fn − fm‖L2 . This shows that {fn} is Cauchy

in L2. Let f := limn→∞ fn, the limit being taken in the L2-sense.We claim that f is well-defined. Let {fn} and {gn} be two sequences of Schwartz func-

tions such that fn, gnL2

−→ f . Define

hn :=

{fk if n = 2k − 1gk if n = 2k

.

Then hnL2

−→ f . By the argument above, {hn} converges in L2, which implies by the unique-ness of limits that limn→∞ fn = limn→∞ gn.

Next we claim that f ∈ L2(Rd), then∥∥∥f∥∥∥

L2= ‖f‖L2 , so the Fourier transform is an

isometry on L2(Rd). Because the norm function is continuous in the L2 topology, we have∥∥∥f∥∥∥L2

= limn→∞

∥∥∥fn∥∥∥L2

= limn→∞

‖fn‖L2 =∥∥∥ limn→∞

fn

∥∥∥L2

= ‖f‖L2 .

Before completing the proof, we remark that in infinite dimensions, an isometry is not nec-essarily a unitary operator. For example, let T : `2(N) → `2(N) be the right-shift operatorgiven by

T (a0, a1, a2, . . . ) := (0, a0, a1, . . . ).

ThenT ∗(a0, a1, a2, . . . ) := (a1, a2, a3, . . . ).

Clearly T is an isometry, but TT ∗ 6= I .However, to prove that the isometry F is unitary, it suffices to prove that F is surjective.

We claim that imF is closed in L2. From this claim, since S(Rd) ⊆ imF and S(Rd) is densein L2(Rd), the proof is complete.

To demonstrate this claim, let g ∈ imF . Then there is a sequence {fn} of L2 functions

so that fnL2

−→ g. As the Fourier transform is an isometry on L2, this implies that {fn}converges in L2. Let f := limn→∞ fn. Then

∥∥∥fn − f∥∥∥L2

= ‖fn − f‖L2 → 0 so that g = fn.

To summarize our progress thus far, we have defined the Fourier transform on L1(Rd),investigated its properties on Schwartz functions, and used the class of Schwartz functionsto define the Fourier transform on L2(Rd). Next, we used Riesz-Thorin interpolation (seeChapter 1) to define the Fourier transform on Lp(Rd) for 1 ≤ p ≤ 2. Even more, we willshow that the Fourier transform cannot be defined as a bounded operator on Lp(Rd) spacesfor p > 2.

Theorem 2.11 (Hausdorff-Young). If f ∈ S(Rd), then∥∥∥f∥∥∥Lp′

. ‖f‖Lp

for 1 ≤ p ≤ 2, where 1p + 1

p′ = 1.

Proof. We have the estimates∥∥∥f∥∥∥

L∞≤ ‖f‖L1 and

∥∥∥f∥∥∥L2

= ‖f‖L2 . Applying the Riesz-Thorin interpolation theorem with p0 = 1, p1 = 2, q0 = ∞, and q1 = 2, it follows that theFourier transform is bounded from Lpθ → Lqθ , where

1

pθ=

1− θ1

+θ

2and

1

qθ=θ

2

15

so that 1pθ

= 1 − θ2 , hence 1

pθ+ 1

qθ= 1. Since pθ = 2

2−θ , as θ ∈ [0, 1] we have pθ ∈ [1, 2] asdesired.

Conversely, we have the following. This is one of our first examples of the power ofscaling arguments.

Theorem 2.12. If∥∥∥f∥∥∥

Lq. ‖f‖Lp for all f ∈ S(Rd) for some 1 ≤ p, q ≤ ∞, then 1 ≤ p ≤ 2 and

q = p′.

Proof. Fix f 6= 0 ∈ S(Rd). For λ > 0, let fλ(x) := f(x/λ). Then

fλ(ξ) =

∫e−2πix·ξf

(xλ

)dx = λdf(λξ).

We have∥∥∥fλ∥∥∥

Lq. ‖fλ‖Lp , which implies

λd · λ−dq

∥∥∥f∥∥∥Lq

. λdp ‖f‖Lp .

Thus, λd · λ−dq . λ

dp , so that λ

dq′ . λ

dp and hence λ

1q′ . λ

1p for all 0 < λ < ∞. Letting

λ→∞ gives 1q′ ≤

1p , and letting λ→ 0 gives 1

q′ ≥1p . Thus, so q = p′.

Next, we show 1 ≤ p ≤ 2. It suffices to show p ≤ p′. Let ϕ ∈ C∞c (Rd) with suppϕ ⊆B(0, 1/2). Let ϕk(x) = e−2πix·λke1ϕ(x − ke1). Using properties 1 and 2 of the Fouriertransform, it follows that

ϕk(ξ) = e−2πiξ·ke1ϕ(ξ − λke1).

Let f =∑N

k=1 ϕk. Since the supports of ϕk are disjoint, it follows that ‖f‖Lp ∼ N1p . Next,

we compute, expanding out∥∥∥f∥∥∥

Lp′as follows:∥∥∥∥∥

N∑k=1

e−2πiξ·ke1ϕ(ξ − λke1)χB(λke1,λ/2)(ξ) +N∑k=1

e−2πiξ·ke1ϕ(ξ − λke1)χCB(λke1,λ/2)(ξ)

∥∥∥∥∥Lp′

≥

∥∥∥∥∥N∑k=1

e−2πiξ·ke1ϕ(ξ − λke1)χB(λke1,λ/2)(ξ)

∥∥∥∥∥Lp′

−N∑k=1

∥∥∥e−2πiξ·ke1ϕ(ξ − λke1)χCB(λke1,λ/2)(ξ)∥∥∥Lp′

& N1p′ −N

∥∥∥ϕ(ξ)χCB(0,λ/2)(ξ)∥∥∥Lp′

.

Because ϕ is Schwartz,

∥∥∥ϕ(ξ)χCB(0,λ/2)(ξ)∥∥∥Lp′

.

(∫|ξ|>λ/2

1

|ξ|2ddξ

) 1p′

. λ− dp′

which→ 0 as λ → ∞. Thus,∥∥∥f∥∥∥

Lp′. ‖f‖Lp if and only if N

1p′ . N

1p for all N , which is

true if and only if p ≤ p′.

2.2 The Fourier Transform on Td.

The primary focus of this text is harmonic analysis on Euclidean space, and consequentlythe Fourier transform on Rd is the most important tool for our purposes. One can alsodefine the Fourier transform on the torus Td := Rd/Zd, equivalently, for periodic functions.In the interest of studying dispersive partial differential equations on the torus (see Chapter9), we include a brief discussion of the Fourier transform in this setting.

16

Definition 2.13. For aC∞-function f : Td → C, we define its Fourier transform f : Zd → Cby

f(k) =

∫Tde−2πik·xf(x) dx.

Then f(k) = 〈f, ek〉where ek(x) := e2πik·x.

The characters ek are orthonormal, since

〈ek, el〉 =

∫Tde2πi(k−l)·x = δkl.

We quickly establish some familiar properties of the Fourier transform.

Proposition 2.14 (Bessel’s Inequality). For f ∈ C∞(Td),∑k∈Zd| 〈f, ek〉 |2 ≤ ‖f‖2L2(Td) .

Proof. Let S ⊆ Zd be a finite set. Then

0 ≤

∥∥∥∥∥f −∑k∈S〈f, ek〉 ek

∥∥∥∥∥2

L2(Td)

=

⟨f −

∑k∈S〈f, ek〉 ek, f −

∑l∈S〈f, el〉 el

⟩

= ‖f‖2L2(Td) − 2∑k∈S〈f, ek〉〈ek, f〉+

⟨∑k∈S〈f, ek〉 ek,

∑l∈S〈f, el〉 el

⟩= ‖f‖2L2(Td) − 2

∑k∈S| 〈f, ek〉 |2 +

∑| 〈f, ek〉 |2

where the last equality follows from orthogonality of the ek’s. Simplifying and rearranginggives ∑

k∈S| 〈f, ek〉 |2 ≤ ‖f‖2L2(Rd)

for every finite set S, hence the desired inequality.

This fact shows that if f ∈ C∞(Td), then∑

k∈Zd 〈f, ek〉 ek ∈ L2(Td). Unsurprisingly,these two objects can be identified.

Proposition 2.15 (Fourier inversion). If f ∈ C∞(Td), then

f =∑k∈Zd〈f, ek〉 ek =

∑k∈Zd

f(k)e2πik·x.

Proof. Suppose for the sake of contradiction that f 6=∑

k∈Zd 〈f, ek〉 ek. Then by Stone-Weierstrass, there is a trigonometric polynomial g such that⟨

f −∑k∈Zd〈f, ek〉 ek, g

⟩6= 0.

For any character el, ⟨f −

∑k∈Zd〈f, ek〉 ek, el

⟩= 〈f, el〉 − 〈f, el〉 = 0.

This is a contradiction.

17

The following result is unique to the periodic case.

Lemma 2.16 (Poisson summation). Let ϕ ∈ S(R). Then∑n∈Z

ϕ(x+ n) =∑n∈Z

ϕ(n)e2πinx

for all x ∈ R. In particular,∑

n∈Z ϕ(n) =∑

n∈Z ϕ(n).

Proof. Define F1(x) =∑

n∈Z ϕ(x + n) and F2(x) =∑

n∈Z ϕ(n)e2πinx. Note that both F1

and F2 are 1-periodic in x. Also, since ϕ is Schwartz, each sum converges absolutely in nand converges uniformly in x on compact sets. Therefore, F1 and F2 are both continuousfunctions on T. As such, to give the desired equality it suffices to prove that F1 and F2 havethe same Fourier coefficients. The definition of F2 immediately gives F2(k) = ϕ(k) for allk ∈ Z.

Next, we compute:

F1(k) =∑n∈Z

∫ 1

0ϕ(x+ n)e−2πikx dx =

∑n∈Z

∫ n+1

nϕ(y)e−2πik(y−n) dy

=∑n∈Z

∫ n+1

nϕ(y)e−2πiky dy

=

∫Rϕ(y)e−2πiky dy

= ϕ(k).

The in particular statement follows by considering x = 0.

18

Chapter 3

Lorentz Spaces and Real Interpolation

In this chapter, we introduce a new class of functions space, namely, Lorentz spaces. Thesespaces measure the size of functions in a more refined manner than the classical Lp spaces.After developing some of the basic theory of Lorentz spaces, we then state and prove theMarcinkiewicz interpolation theorem in its most general setting.

Some references for Lorentz spaces are [5] and [6].

3.1 Lorentz Spaces

Before defining Lorentz spaces in their full generality, we first consider a simpler and morefamiliar class of functions.

Definition 3.1. For 1 ≤ p ≤ ∞ and f : Rd → C measurable, define

‖f‖∗Lpweak := supλ>0

λ∣∣∣{x ∈ Rd : |f(x)| > λ

}∣∣∣ 1p. (3.1)

The weak-Lp space, written Lpweak(Rd), is the family of measurable functions f for which

‖f‖∗Lpweak is finite.

The quantity defined in (3.1) contains an asterisk as a superscript because is not a norm.However, it is a quasinorm. Recall that a quasinorm satisfies the same properties as a norm,except that the triangle inequality is replaced by the quasi triangle inequality: ‖f + g‖ ≤C (‖f‖+ ‖g‖) for some universal constant C > 0.

To clarify the definition of the weak-Lp (quasi)norm, consider a function f ∈ Lp(Rd).Using the fundamental theorem of calculus, we can write

‖f‖pLp(Rd)

=

∫Rd|f(x)|p dx =

∫Rd

∫ |f(x)|

0pλp−1 dλ dx.

By Tonelli’s theorem,

‖f‖pLp(Rd)

=

∫ ∞0

pλp−1

∫{x∈Rd : |f(x)|>λ}

dx dλ =

∫ ∞0

pλp−1∣∣∣{x ∈ Rd : |f(x)| > λ

}∣∣∣ dλ= p

∫ ∞0

(λ∣∣∣{x ∈ Rd : |f(x)| > λ

}∣∣∣ 1p

)pdλ

λ.

So ‖f‖Lp(Rd) = p1p

∥∥∥λ ∣∣{x ∈ Rd : |f(x)| > λ}∣∣ 1

p

∥∥∥Lp((0,∞), dλ

λ ). Using the convention p

1∞ =

1, we can then write

‖f‖∗Lpweak(Rd) = p1∞

∥∥∥∥λ ∣∣∣{x ∈ Rd : |f(x)| > λ}∣∣∣ 1

p

∥∥∥∥L∞((0,∞), dλ

λ ).

In this way, weak-Lp spaces are a clear generalization of Lp spaces.

19

Example 3.2. Let f(x) = |x|−dp . Then f ∈ Lpweak(R

d), but f /∈ Lp(Rd).To see this, first note that∫

Rd|f(x)|p dx =

∫Rd

1

|x|ddx ∼

∫ ∞0

1

rdrd−1 dr ∼

∫ ∞0

1

rdr.

This latter integral does not converge, hence the Lp-norm of f is not finite, so f /∈ Lp(Rd).On the other hand,

∣∣∣{x ∈ Rd : |f(x)| > λ}∣∣∣ 1

p=

∣∣∣∣∣{x ∈ Rd :

1

|x|dp

> λ

}∣∣∣∣∣1p

=

∣∣∣∣∣{x ∈ Rd : |x| <

(1

λ

) pd

}∣∣∣∣∣1p

.

The set being measured is a ball of radius (1/λ)p/d. The volume of a such a ball scalesaccording to

((1/λ)p/d

)d= (1/λ)p. Taking pth roots as above then gives∣∣∣{x ∈ Rd : |f(x)| > λ

}∣∣∣ 1p.

1

λ.

Hence,

‖f‖∗Lpweak(Rd) . supλ>0

λ1

λ= 1 <∞

so that f ∈ Lpweak(Rd).

Lorentz spaces are an even further generalization of Lp and weak-Lp spaces.

Definition 3.3. For 1 ≤ p ≤ ∞ and 1 ≤ q ≤ ∞, the Lorentz space Lp,q(Rd) is the space ofmeasurable functions f : Rd → C for which the quantity

‖f‖∗Lp,q(Rd) := p1q

∥∥∥∥λ ∣∣∣{x ∈ Rd : |f(x)| > λ}∣∣∣ 1

p

∥∥∥∥Lq((0,∞), dλ

λ )(3.2)

is finite.

By our previous computation, Lp(Rd) = Lp,p(Rd), and Lpweak(Rd) = Lp,∞(Rd). Also, as

with the weak-Lp norm, the Lp,q-norm defined in (3.2) is not actually a norm.

Lemma 3.4. The quantity ‖·‖∗Lp,q(Rd) is a quasinorm.

Proof. First, note that if ‖f‖∗Lp,q(Rd) = 0, then∥∥∥∥λ ∣∣∣{x ∈ Rd : |f(x)| > λ}∣∣∣ 1

p

∥∥∥∥Lq((0,∞), dλ

λ )= 0

which implies that, for almost all λ > 0,∣∣{x ∈ Rd : |f(x)| > λ

}∣∣ = 0. It follows thatf(x) = 0 almost everywhere. Thus, f = 0.

Next, let a ∈ C. Then

‖af‖∗Lp,q(Rd) = p1q

∥∥∥∥λ ∣∣∣{x ∈ Rd : |af(x)| > λ}∣∣∣ 1

p

∥∥∥∥Lq((0,∞), dλ

λ )

= |a|p1q

∥∥∥∥∥λa∣∣∣∣{x ∈ Rd : |f(x)| > λ

a

}∣∣∣∣ 1p

∥∥∥∥∥Lq((0,∞), dλ

λ )

.

By making the change of variables η = λ/a, we have ‖af‖∗Lp,q(Rd) = |a| ‖f‖∗Lp,q(Rd).

20

It remains to prove that quasi-triangle inequality. To do this, we invoke the fact thatfor 0 < α < 1, the map x 7→ xα is concave. It is a standard fact that concave functions aresubadditive, so that (x+ y)α ≤ xα + yα. With this in mind, we compute:

‖f + g‖∗Lp,q(Rd) = p1q

∥∥∥∥λ ∣∣∣{x ∈ Rd : |f(x) + g(x)| > λ}∣∣∣ 1

p

∥∥∥∥Lq((0,∞), dλ

λ ).

Observe that{x ∈ Rd : |f(x) + g(x)| > λ

}⊆{x ∈ Rd : |f(x)| > λ

2

}∪{x ∈ Rd : |g(x)| > λ

2

}.

Thus,

‖f + g‖∗Lp,q(Rd) ≤ p1q

∥∥∥∥∥λ(∣∣∣∣{x : |f(x)| > λ

2

}∣∣∣∣+

∣∣∣∣{x : |g(x)| > λ

2

}∣∣∣∣) 1p

∥∥∥∥∥Lq((0,∞), dλ

λ )

.

By subbaditivity of the concave map x 7→ x1p , we have

‖f + g‖∗Lp,q(Rd) ≤ p1q

∥∥∥∥∥λ(∣∣∣∣{x : |f(x)| > λ

2

}∣∣∣∣ 1p

+

∣∣∣∣{x : |g(x)| > λ

2

}∣∣∣∣ 1p

)∥∥∥∥∥Lq((0,∞), dλ

λ )

.

Distributing the λ, factoring out a 2, and appliyng the usual Lq-Minkowski inequality thenyields:

‖f + g‖∗Lp,q(Rd) ≤ 2p1q

∥∥∥∥∥λ2∣∣∣∣{x : |f(x)| > λ

2

}∣∣∣∣ 1p

+λ

2

∣∣∣∣{x : |g(x)| > λ

2

}∣∣∣∣ 1p

∥∥∥∥∥Lq((0,∞), dλ

λ )

≤ 2p1q

∥∥∥∥∥λ2∣∣∣∣{x : |f(x)| > λ

2

}∣∣∣∣ 1p

∥∥∥∥∥Lq((0,∞), dλ

λ )

+

∥∥∥∥∥λ2∣∣∣∣{x : |g(x)| > λ

2

}∣∣∣∣ 1p

∥∥∥∥∥Lq((0,∞), dλ

λ )

= 2

(‖f‖∗Lp,q(Rd) + ‖g‖∗Lp,q(Rd)

).

For 1 < p < ∞ and 1 ≤ q ≤ ∞, we will show that the quasinorm ‖·‖∗Lp,q(Rd) is infact equivalent to a norm. For p = 1, q 6= 1, the quasinorm is not equivalent to a norm.However, in this case, there does exist a metric that generates the same topology. Thus, ineither of these cases, Lp,q(Rd) is a complete metric space.

As another quick remark, we note that if |g| ≤ |f |, then for any λ > 0,{x ∈ Rd : |g(x)| > λ

}⊆{x ∈ Rd : |f(x)| > λ

}.

This implies that ‖g‖∗Lp,q(Rd) ≤ ‖f‖∗Lp,q(Rd).

It is natural to ask whether different Lorentz spaces have any simple embedding prop-erties. The next result gives a partial answer to this question.

Proposition 3.5. Let 1 ≤ p <∞, 1 ≤ q ≤ ∞, and f ∈ Lp,q(Rd). Write f =∑

m∈Z fm, where

fm(x) := f(x)χ{x∈Rd : 2m≤|f(x)|≤2m+1 }.

Then‖f‖∗Lp,q(Rd) ∼p,q

∥∥∥‖fm‖Lp(Rd)

∥∥∥`qm(Z)

.

In particular, Lp,q1(Rd) ⊆ Lp,q2(Rd) whenever q1 ≤ q2.

21

Proof. By the preceding remark, note that∥∥∥∥∥∑m∈Z

2m χ{x : 2m≤|f |≤2m+1 }

∥∥∥∥∥∗

Lp,q(Rd)

≤

∥∥∥∥∥∑m∈Z

fm

∥∥∥∥∥∗

Lp,q(Rd)

≤

∥∥∥∥∥∑m∈Z

2m+1 χ{x : 2m≤|f |≤2m+1 }

∥∥∥∥∥∗

Lp,q(Rd)

.

Therefore, it suffices to prove the proposition for a function of the form f(x) =∑

m∈Z 2mχEmwhere {Em} is a pairwise disjoint collection of sets.

In this case, we need to show that ‖f‖∗Lp,q(Rd) ∼p,q∥∥∥2m|Em|

1p

∥∥∥`qm(Z)

. We compute:(‖f‖∗Lp,q(Rd)

)q= p

∫ ∞0

λq | {x : |f(x)| > λ } |qpdλ

λ

= p∑m∈Z

∫ 2m

2m+1

λq | {x : |f(x)| > λ } |qpdλ

λ.

Next, observe that for λ ∈ [2m+1, 2m), {x : |f(x)| > λ } =⋃n≥mEM . As the Em’s are

disjoint, we then have

(‖f‖∗Lp,q(Rd)

)q= p

∑m∈Z

∫ 2m

2m+1

λq

∑n≥m|Em|

qp

dλ

λ.

This removes the dependence on λ in the set inside the integral. Because

p

∫ 2m

2m+1

λqdλ

λ=p

q(2mq − 2(m+1)q) =

p

q(1− 2q) 2mq

it follows that

(‖f‖∗Lp,q(Rd)

)q∼p,q

∑m∈Z

2mq

∑n≥m|En|

qp

=

∥∥∥∥∥∥∥2m

∑n≥m|Em|

1p

∥∥∥∥∥∥∥q

`qm(Z)

.

Thus,

‖f‖∗Lp,q(Rd) ∼p,q

∥∥∥∥∥∥∥2m

∑n≥m|En|

1p

∥∥∥∥∥∥∥`qm(Z)

&p,q

∥∥∥2m (|Em|)1p

∥∥∥`qm(Z)

.

This gives half of the ∼p,q relation that we need.To get the other inequality, we compute as follows. First, invoking concavity of frac-

tional powers as before,

‖f‖∗Lp,q(Rd) ∼p,q

∥∥∥∥∥∥∥2m

∑n≥m|En|

1p

∥∥∥∥∥∥∥`qm(Z)

≤

∥∥∥∥∥∥2m∑n≥m|En|

1p

∥∥∥∥∥∥`qm(Z)

.

Next, making the change of variables n = m+ k,

=

∥∥∥∥∥∥2m∑k≥0

|Em+k|1p

∥∥∥∥∥∥`qm(Z)

=

∥∥∥∥∥∥∑k≥0

2−k2m+k|Em+k|1p

∥∥∥∥∥∥`qm(Z)

≤∑k≥0

2−k∥∥∥2m+k|Em+k|

1p

∥∥∥`qm(Z)

=∑k≥0

2−k∥∥∥2m|Em|

1p

∥∥∥`qm(Z)

=∥∥∥2m|Em|

1p

∥∥∥`qm(Z)

22

so that ‖f‖∗Lp,q(Rd) .p,q

∥∥∥2m|Em|1p

∥∥∥`qm(Z)

. Therefore,

‖f‖∗Lp,q(Rd) ∼p,q∥∥∥2m|Em|

1p

∥∥∥`qm(Z)

as desired.The “in particular” statement follows from the fact that if q1 ≤ q2, then `q1(Z) ⊆ `q2(Z).

As noted in the proof of the above proposition, the monotonicty of the quasinormallows for the following reduction in many computations involving Lorentz spaces. Iff ∈ Lp,q(Rd) is real-valued and nonnegative, then we can assume without loss of general-ity that f =

∑m∈Z 2mχEm , where the sets Em are pairwise disjoint. In this case,

‖f‖∗Lp,q(Rd) =∥∥∥2m|Em|

1p

∥∥∥`qm(Z)

.

We can further decompose a general function into its real and imaginary parts, and theninto their corresponding positive and negative parts, to apply this reduction.

3.2 Duality in Lorentz Spaces

One of the most important principles in Lp space theory is that of duality. The analogousfact for Lorentz spaces is given by the following proposition.

Proposition 3.6. Let 1 < p < ∞ and 1 ≤ q ≤ ∞, and let p′ and q′ denote the respective Holderconjugates. Then

‖f‖∗Lp,q(Rd) ∼p,q sup‖g‖∗

Lp′,q′ (Rd)

≤1

∣∣∣∣∫Rdf(x)g(x) dx

∣∣∣∣ . (3.3)

Proof. As the quasinorm is positively homogeneous, we can scale the function f appro-priately and assume without loss of generality that ‖f‖∗Lp,q(Rd) = 1. Also, as remarkedpreviously, we can assume without loss of generality that f and g are real-valued and non-negative, hence we can take f =

∑n∈Z 2nχFn for disjoint sets Fn and g =

∑m∈Z 2mχEm for

disjoint sets Em.In this case, we have

1 =(‖f‖∗Lp,q(Rd)

)q=∥∥∥2n|Fn|

1p

∥∥∥q`qn(Z)

=∑n∈Z

2nq|Fn|qp .

We decompose the above sum as follows. Let 2Z denote the set of dyadic numbers. Then

∑n∈Z

2nq|Fn|qp =

∑N∈2Z

∑n:|Fn|∼N

2nqNqp =

∑N∈2Z

Nqp

∑n:|Fn|∼N

2nq ∼∑N∈2Z

Nqp

∑n:|Fn|∼N

2n

q

.

The latter equivalence is a straightforward exercise in dealing with dyadic sums, togetherwith the fact that the `q-norm is bounded by the ell1-norm. Thus,

1 ∼∑N∈2Z

∑n:|Fn|∼N

2nN1p

q

.

Similarly, since ‖g‖∗Lp′,q′ (Rd)

≤ 1, the same computation gives

∑M∈2Z

∑m:|Em|∼M

2m|Em|1p′

q′

. 1.

23

With these computations and our reductions in mind, we demonstrate the equivalence in(3.3). We first show that the right hand side is bounded by the left.∫

f(x)g(x) dx =∑n,m∈Z

∫2n2mχFn∩Em(x) dx =

∑n,m∈Z

2n2m|Fn ∩ Em|

.∑

N,M∈2Z

∑n:|Fn|∼N

2n∑

m:|Em|∼M

2m min{N,M}.

By our above computations, we eventually want to introduce the quantitiesN1p and |Em|

1p′ .

We force them in the sum as follows:

.∑

N,M∈2Z

∑n:|Fn|∼N

2nN1p

∑m:|Em|∼M

2m|Em|1p′ min

{N

N1pM

1p′,

M

N1pM

1p′

}

=∑

N,M∈2Z

∑n:|Fn|∼N

2n|Fn|1p

∑m:|Em|∼M

2m|Em|1p′ min

{(N

M

) 1p′

,

(M

N

) 1p

}.

Next, we use Holder’s inequality on the above sum, viewing the n-sum as living in `qN andthe m-sum as living in `q

′

M . This yields

.

∑N,M∈2Z

∑n:|Fn|∼N

2n|Fn|1p

q

min

{(N

M

) 1p′

,

(M

N

) 1p

}1q

·

∑N,M∈2Z

∑m:|Em|∼M

2M |Em|1p′

q′

min

{(N

M

) 1p′

,

(M

N

) 1p

}1q′

.

In the first term, freeze N and sum over M . Because p > 1 and hence p′ <∞, we have

∑M∈2Z

min

{(N

M

) 1p′

,

(M

N

) 1p

}=∑M≤N

(M

N

) 1p

+∑M>N

(N

M

) 1p′

. 1 + 1 . 1.

Similarly, in the second term we freeze M and sum over N . This gives

∫f(x)g(x) dx .

∑N∈2Z

∑n:|Fn|∼N

2n|Fn|1p

q1q

·

∑M∈2Z

∑m:|Em|∼M

2M |Em|1p′

q′

1q′

. 1.

This gives one inequality.Next, we consider the converse inequality. Again we can take f to be of the form∑n 2nχFn with ‖f‖∗Lp,q(Rd) ∼ 1. Let

g =∑n

(2n|Fn|

1p

)q−1|Fn|

− 1p′ χFn =

∑n

2n(q−1)|Fn|qp−1χFn .

We make two claims, from which the desired inequality clearly follows:

1.∫f(x)g(x) dx & 1;

2. ‖g‖∗Lp′,q′ (Rd)

. 1.

24

The first claim follows from∫f(x)g(x) dx =

∫ ∑n

2n(q−1)2n|Fn|qp−1χFn =

∑n

2nq|Fn|qp

=∥∥∥2n|Fn|

1p

∥∥∥q`qn(Z)

∼ 1.

To see the second claim, writeg ∼

∑N∈2Z

Nχ⋃n∈Sn Fn

where Sn ={n : 2n(q−1)|Fn|

qp−1 ∼ N

}. Note that the Sn’s are disjoint sets. Then

(‖g‖∗

Lp′,q′ (Rd)

)q′∼∑N∈2Z

N q′

(∑n∈Sn

|Fn|

) q′p′

∼∑N∈2Z

N q′

(∑n∈Sn

(N2−n(q−1)

) pq−p

) q′p′

∼∑N∈2Z

N q′∑n∈Sn

Npq−p ·

q′p′ 2−nq p−1

q−p

where in the last equivalence we have used the fact about dyadic sums and `q norms fromabove and the fact that q′ = q

q−1 . It remains to compute and simplify all of the Holderexponents. Doing this gives(

‖g‖∗Lp′,q′ (Rd)

)q′∼∑n∈Z

(2n(q−1)|Fn|

q−pp

) qq−p

2−nq p−q

q−p

∼∑n∈Z|Fn|

qp 2nq

. 1.

This proves both claims, the the converse inequality, and hence completes the proof.

The right hand side of (3.3) defines a norm which is equivalent to the quasinorm Lorentzquasinorm. With respect to this norm, Lp,q(Rd) for 1 < p < ∞, 1 ≤ q ≤ ∞, is a Banachspace. The proof of completeness is the same as the proof of completeness in Lp spaces.The dual space in this case is Lp

′,q′(Rd).As remarked above, if p = 1 and q 6= 1, there is no norm which is equivalent to the

quasinorm. The following example demonstrates this explicitly.

Example 3.7. Consider the case p = 1, q = ∞, d = 1. Let f(x) =∑N

n=11

|x−n| . We

saw in a previous example that∥∥∥ 1|x−n|

∥∥∥∗L1,∞(R)

=∥∥∥ 1|x−n|

∥∥∥∗L1weak(R)

. 1. This implies that∑Nn=1

∥∥∥ 1|x−n|

∥∥∥∗L1,∞(R)

. N .

We claim that∥∥∥∑N

n=11

|x−n|

∥∥∥∗L1,∞(R)

& N logN . We have

∥∥∥∥∥N∑n=1

1

|x− n|

∥∥∥∥∥∗

L1,∞(R)

= supλ>0

λ

∣∣∣∣∣{x ∈ Rd :

N∑n=1

1

|x− n|> λ

}∣∣∣∣∣ .25

Fix x ∈ [0, N ]. Note that, for n ≥ x, |x−n| < n, so that 1|x−n| >

1n . For n < x, we can consider

each finite number and rearrange them so that their sum is comparable to∑

n<x1n . Thus,

for x ∈ [0, N ] we have f(x) &∑N

n=11n ∼ logN . Choose λ = C logN for some appropriate

constant C, we then have{x ∈ Rd :

∑Nn=1

1|x−n| > C logN

}⊇ [0, N ]. Therefore,

‖f‖L1,∞(R) ≥ C logN

∣∣∣∣∣{x ∈ Rd :

N∑n=1

1

|x− n|> C logN

}∣∣∣∣∣ ≥ CN logN

as claimed.Now, suppose that the quasinorm was equivalent to a norm. Then by the triangle in-

equality,

N logN .

∥∥∥∥∥N∑n=1

1

|x− n|

∥∥∥∥∥∗

L1,∞(R)

.N∑n=1

∥∥∥∥ 1

|x− n|

∥∥∥∥∗L1,∞(R)

. N,

which is a contradiction.

3.3 The Marcinkiewicz Interpolation Theorem

Finally, we arrive at the Marcinkiewicz interpolation theorem. Before stating and provingthe theorem, we recall the notion of sublinearity and introduce some language that will beused throughout the remainder of this text.

Definition 3.8. A mapping T on a class of measurable functions is sublinear if |T (cf)| ≤|c||T (f)| and |T (f + g)| ≤ |T (f)|+ |T (g)| for all c ∈ C and f, g in the support of T .

Any linear operator is obviously sublinear. A more interesting class of examples, andperhaps the primary motivation for defining sublinearity, is the following.

Example 3.9. Given a family of linear mappings {Tt}t∈I , the mapping defined by

T (f)(x) := ‖Tt(f)(x)‖Lqt

is sublinear. When q =∞, operators of this form are called maximal functions. When q = 2,the term square function is used.

Definition 3.10. Let 1 ≤ p, q ≤ ∞. A mapping of functions T is of (strong) type (p, q) if‖Tf‖Lq(Rd) .T ‖f‖Lp(Rd). That is, T is of type (p, q) if it is bounded as a mapping fromLp(Rd) → Lq(Rd). Similarly, for q < ∞, T is of weak type (p, q) if ‖Tf‖∗Lq,∞(Rd) .T

‖f‖Lp(Rd), and is of restricted weak type (p, q) if ‖TχF ‖∗Lq,∞(Rd) .T |F |1p for all finite mea-

sure sets F .

Note that type (p, q) implies weak type (p, q), which implies restricted weak type (p, q).An alternative characterization of restricted weak type operators is given by the followingproposition.

Proposition 3.11. A mapping T is of restricted weak type (p, q) if and only if∫|TχF ||χE | dx . |E|

1p |F |

1q

for all finite measure sets E,F .

26

Proof. First, suppose that T is of restricted type (p, q). Then ‖TχF ‖∗Lq,∞(Rd) .T |F |1p . By the

duality of Lorentz quasinorms, this implies∫|TχF ||χE | dx . ‖TχF ‖∗Lq,∞(Rd) ‖χE‖

∗Lq′,1(Rd)

.

Note that (‖χE‖∗Lq′,1(Rd)

)q′= q′

∫ 1

0λq′ |{x : χE(x) > λ }| dλ

λ∼ |E|.

Hence, ∫|TχF ||χE | dx . |E|

1p |F |

1q .

Conversely, suppose that the above inequality holds for all finite measure sets E and F .We again invoke duality to write

‖TχF ‖∗Lq,∞(Rd) ∼ sup‖g‖∗

Lq′,1≤1

∣∣∣∣∫ TχF g dx

∣∣∣∣ . sup‖g‖∗

Lq′,1≤1

∫|TχF ||g| dx.

Consider g :=∑

m 2mEm for an appropriate disjoint collection of sets Em. Then∫|TχF ||g| dx .

∑m

2m|F |1p |E|

1q′ . |F |

1p ‖g‖∗

Lq′,1(Rd)

. |F |1p

as desired.

The formulation of the Marcinkiewicz interpolation theorem that we state below is dueto Hunt (see [5]). The classical statement of the theorem is given as Corollary 3.14 below.

Theorem 3.12 (Marcinkiewicz interpolation). Fix 1 ≤ p1, p2, q1, q2 ≤ ∞ such that p1 6= p2

and q1 6= q2. Let T be a sublinear operator of restricted weak type (p1, q1) and of restricted weaktype (p2, q2). Then for any 1 ≤ r ≤ ∞ and 0 < θ < 1,

‖Tf‖∗Lqθ,r . ‖f‖∗Lpθ,r

where 1pθ

= θp1

+ 1−θp2

and 1qθ

= θq1

+ 1−θq2

.

Before proving the theorem, we make a few remarks. First, if pθ ≤ qθ, taking r = qθgives

‖Tf‖Lqθ . ‖f‖∗Lpθ,qθ . ‖f‖Lpθso that T is of strong type (pθ, qθ). The requirement that pθ ≤ qθ is essential for the strongtype conclusion, as evidenced by the following example.

Example 3.13. Define T (f)(x) := f(x)

|x|12

. Then T is bounded as an operator from Lp(R) →

L2pp+2

,∞(R) for any 2 ≤ p ≤ ∞, but is not bounded as an operator from Lp(R)→ L

2pp+2 (R).

To prove this, we invoke Holder’s inequality in Lorentz spaces: for 1 ≤ p1, p2, p < ∞and 1 ≤ q1, q2, q ≤ ∞ satisfying 1

p = 1p1

+ 1p2

and 1q = 1

q1+ 1

q2, we have ‖fg‖∗Lp,q .

‖f‖∗Lp1,q1 ‖g‖∗Lp2,q2 . By this inequality, we have:

‖Tf‖∗L

2pp+2 ,∞(R)

. ‖f‖∗Lp,∞(R)

∥∥∥|x|− 12

∥∥∥∗L2,∞(R)

. ‖f‖Lp(R) .

27

On the other hand, consider the function f(x) = |x|−1p

∣∣∣log(|x|+ 1

|x|

)∣∣∣− p+22p . We first claim

that f ∈ Lp(R). Indeed, we can write

‖f‖pLp(R) =

∫1

|x|

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣− p+22

dx

= 2

(∫ 2

0

1

|x|

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣− p+22

dx+

∫ ∞2

1

|x|

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣− p+22

dx

).

Consider the second integral. Since p ≥ 2, we have

∫ ∞2

1

|x|

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣− p+22

dx ≤∫ ∞

2

1

|x|

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣−2

dx ≤∫ ∞

2

1

x(log x)2dx

=

∫ ∞log 2

1

u2du

<∞.

By symmetry, the singularity at 0 behaves similarly. Hence, f ∈ Lp(R). However, it is notdifficult to show that the quantity

‖Tf‖2pp+2

L2pp+2

=

∫|x|−

2p+2

∣∣∣∣log

(|x|+ 1

|x|

)∣∣∣∣−1

dx

is not finite.

We now prove Theorem 3.12.

Proof.

Corollary 3.14 (Marcinkiewicz interpolation).

28

Chapter 4

Maximal Functions

The notion of averaging is one which is ubiquitous in analysis. For example, convolutioncan be viewed as a sort of averaging against a given function. Convolving a function witha smooth function then smooths the original function, a fact which is immensely useful.

In this chapter, we study averaging in more depth via maximal functions. We beginby recalling the classical Hardy-Littlewood maximal inequality, and spend the rest of thechapter developing generalizations.

4.1 The Hardy-Littlewood Maximal Function

The reader may already be familiar with the Hardy-Littlewood maximal function, definedas

M(f)(x) := supr>0

1

|Br(x)|

∫Br(x)

|f(y)| dy. (4.1)

The quantity 1|Br(x)|

∫Br(x) |f(y)| dy represents the average value of the function on a ball

centered at x. Thus, the maximal function measures the largest possible average values forf on various balls.

The following theorem is standard.

Theorem 4.1 (Hardy-Littlewood maximal inequality). Let M(f) denote the Hardy-Littlewoodmaximal function as in (4.1). Then

1. For f ∈ Lp(Rd), 1 ≤ p ≤ ∞, M(f) is finite almost everywhere.

2. The operator M is of weak type (1, 1), and of strong type (p, p) for 1 < p ≤ ∞.

Later in this chapter, we will prove a generalization of Theorem 4.1, and so we defer a proofuntil then.

Unraveling statement 2 of Theorem 4.1, we see that M being of weak type (1, 1) means

‖Mf‖∗L1,∞(Rd) . ‖f‖L1(Rd)

and hence| {x : (Mf)(x) > λ } | ≤ C

λ‖f‖L1(Rd)

for all λ > 0. This is the usual formulation of the Hardy-Littlewood maximal inequality.We remark that the p > 1 condition in statement 2 is necessary, because M is not of strongtype (1, 1). To see this, let φ ∈ C∞c (Rd) with suppφ ⊆ B1/2(0). Fix |x| > 1. If r < |x| − 1/2,then

∫Br(x) φ(y) dy = 0. If r > |x|+ 1/2, then

1

|Br(x)|

∫Br(x)

φ(y) dy =1

|Br(x)|

∫B|x|+1/2(x)

φ(y) dy ≤ 1

|B|x|+1/2(x)|

∫B|x|+1/2(x)

φ(y) dy.

29

Thus,

(Mφ)(x) = sup|x|−1/2≤r≤|x|+1/2

1

|Br(x)|

∫Br(x)

φ(y) dy &1

|x|d.

But 1|x|d is not in L1(Rd).

We close this section with a classic application of the Hardy-Littlewood maximal in-equality.

Theorem 4.2 (Lebesgue differentiation). Let f ∈ L1loc(Rd). Then

limr→0

1

|Br(x)|

∫Br(x)

f(y) dy = f(x)

for almost every x ∈ Rd.

Proof. ADD IN

More details on the classical Hardy-Littlewood maximal function can be found in [9].

4.2 Ap Weights and the Weighted Maximal Inequality

The main theorem of this section, which is a generalization of Theorem 4.1, is the following.

Theorem 4.3 (Weighted maximal inequality). Let ω : Rd → [0,∞) be locally integrable. Asso-ciate to ω the measure defined by ω(E) :=

∫E ω(y) dy. Then

M : L1(M(ω) dx)→ L1,∞(ω dx)

andM : Lp(M(ω) dx)→ Lp(ω dx)

for 1 λ }) . 1

λ

∫Rd|f(x)| (Mω)(x) dx

and ∫Rd|(Mf)(x)|pω(x) dx .

∫Rd|f(x)|p (Mω)(x) dx.

Note that if ω ≡ 1, then M(ω) ≡ 1, and we recover the classical Hardy-Littlewoodmaximal inequality in Theorem 4.1.

Before proving the weighted maximal inequality, we establish some preliminary factson Ap weights.

Definition 4.4. A function ω satisfies theA1 condition, written ω ∈ A1, ifM(ω) . ω almosteverywhere.

Note that if ω ∈ A1, then Theorem 4.3 gives M : L1(ω dx) → L1,∞(ω dx) and M :Lp(ω dx)→ Lp(ω dx) for 1 < p ≤ ∞.

Lemma 4.5. The following are equivalent:

1. ω ∈ A1;

2. ω(B)|B| . ω(x) for almost all x ∈ B and all balls B;

3. 1|B|∫B f(y) dy . 1

ω(B)

∫B f(y)ω(y) dx for all f ≥ 0 and all balls B.

30

Proof. We first show (1)⇒ (2). Fix a ball B0 of radius r0 and let x ∈ B0. Then

ω(x) & (Mω)(x) = supr>0

1

|Br(x)|

∫Br(x)

ω(y) dy ≥ 1

|B2r0(x)|

∫B2r0 (x)

ω(y) dy.

BecauseB2r0(x) ⊇ B0 (drawing a picture makes this clearer) and because |B2r0(x)| .d |B0|,

ω(x) &1

|B0|

∫B0

ω(y) dy =ω(B0)

|B0|

as desired.The implication (2) ⇒ (1) follows immediately from the definition of the maximal

function.Next, we show (2)⇒ (3). Fix a ball B and f ≥ 0. Then

1

ω(B)

∫Bf(y)ω(y) dy &

1

ω(B)

∫Bf(y)

ω(B)

|B|dy =

1

|B|

∫Bf(y) dy.

Finally (3) ⇒ (2). Fix a ball B and let x ∈ B be a Lebesgue point of B (i.e., a point forwhich the conclusion of the Lebesgue differentiation theorem holds). Choose r << 1 sothat Br(x) ⊆ B. Let f = χBr(x). Then by (3),

1

|B|

∫BχBr(x)(y) dy .

1

ω(B)

∫BχBr(x)(y)ω(y) dx ⇒ |Br(x)|

|B|.

1

ω(B)

∫Br(x)

ω(y) dy.

Since x is a Lebesgue point,

ω(B)

|B|.

1

|Br(x)|

∫Br(x)

ω(y) dy → ω(x)

as r → 0.

Definition 4.6. A function ω : Rd → [0,∞) satisfies the Ap condition, written ω ∈ Ap, if

supB

ω(B)

|B|

(1

|B|

∫Bω(y)

− p′p dy

) pp′

. 1,

where the supremum is taken over all open balls in Rd.

We note for convenience that the above condition is equivalent to

supB

ω(B)

|B|p

(∫Bω(y)

− p′p dy

) pp′

. 1.

Also, if 1 < p < q <∞, then Ap ⊆ Aq. This follows from Holder’s inequality.The importance of the class ofAp weights is the following theorem, which characterizes

when the maximal function is a bounded operator on a weighted Lp space.

Theorem 4.7. Fix 1 < p <∞. Then M : Lp(ω dx)→ Lp(ω dx) if and only if ω ∈ Ap.

Proof. Here, we only explicitly prove one direction (⇒) of this statement. For the otherdirection, we will provide an outline of the argument and leave the details as an exercisefor the reader.

Suppose that M : Lp(ω dx) → Lp(ω dx). Fix a ball B and let f = (ω + ε)− p′p χB . For

x ∈ B,

(Mf)(x) = supr>0

1

|B(x, r)|

∫B(x,r)

(ω(y) + ε)− p′p χB(y) dy ≥ 1

|2B|

∫B

(ω(y) + ε)− p′p dy

&1

|B|

∫B

(ω(y) + ε)− p′p dy =: λ.

31

Here our definition of λ includes any implicit constants from the above inequality. Then

ω(B) ≤ ω({

x : (Mf)(x) >λ

2

}).

1

λp

∫|f(y)|pω(y) dy

where the final inequality follows from the fact that M : Lp(M(ω) dx) → Lp(ω dx), andhence

M : Lp(M(ω) dx)→ Lp,∞(ω dx).

Plugging in our definition of λ gives

ω(B) . |B|p(∫

B(ω(y) + ε)

− p′p dy

)−p ∫B

(ω(y) + ε)−p′ω(y) dy

. |B|p(∫

B(ω(y) + ε)

− p′p dy

)−p ∫B

(ω(y) + ε)−p′+1 dy

= |B|p(∫

B(ω(y) + ε)

− p′p dy

)−p ∫B

(ω(y) + ε)− p′p dy

= |B|p(∫

B(ω(y) + ε)

− p′p dy

)−(p−1)

.

Thus,ω(B)

|B|p

(∫B

(ω(y) + ε)− p′p dy

) pp′

. 1.

Letting ε→ 0 gives ω ∈ Ap.

Now, we sketch the proof of the converse direction. Suppose that ω ∈ Ap. Define thefollowing weighted maximal function:

(Mωf)(x) := supr>0

1

ω(B(x, r))

∫B(x,r)

|f(y)|ω(y) dy.

It can be shown (via techniques from this section) that Mω : L1(ω dx) → L1,∞(ω dx), andmoreover (Mf)p .Mω(fp).

Fix f ∈ Lp(ω dx). Then(‖Mf‖∗Lp,∞(ω dx)

)p= sup

λ>0λpω ({x : (Mf)(x) > λ }) = sup

λ>0λpω ({x : (Mf)p(x) > λp })

= supη>0

η ω ({x : (Mf)p(x) > η }) .

Since (Mf)p(x) .Mω(|f |p)(x),

ω ({x : (Mf)p(x) > η }) . ω ({x : Mω(|f |p)(x) > η }) .

Thus, (‖Mf‖∗Lp,∞(ω dx

)p. sup

η>0η ω ({x : Mω(|f |p)(x) > η }) = ‖Mω(|f |p)‖L1,∞(ω dx) .

Since Mω : L1(ω dx)→ L1,∞(ω dx),

‖Mω(|f |p)‖L1,∞(ω dx) . ‖|f |p‖L1(ω dx) =

∫|f(x)|pω(x) dx = ‖f‖pLp(ω dx) .

Therefore,‖Mf‖∗Lp,∞(ω dx . ‖f‖Lp(ω dx)

so that M : Lp(ω dx)→ Lp,∞(ω dx).

32

Theorem 4.8. Fix 1 ≤ p ≤ ∞, and let dµ be a nonnegative Borel measure. If M : Lp(dµ) →Lp,∞(dµ), then dµ = ω dx for some ω ∈ Ap.

Proof. If we can prove that dµ is absolutely continuous, then in light of the previous proof,we are done.

Decompose dµ = ω(x) dx + dν where dν is the singular part of dµ, i.e., there existsa compact set K with |K| = 0 but ν(K) > 0. Define Un = {x : d(x,K) < 1/n } andfn = χUn\K . As Un \K ⊇ Un+1 \K, and

⋂(Un \K) = ∅, it follows that fn → 0 pointwise.

We claim that dµ is finite on compact sets. To see this, pick a measurable set E with0 < µ(E) < ∞; this possible since µ is nontrivial. We may assume that E is compact byinner regularity. Then

(MχE)(x) = supr>0

1

|B(x, r)|

∫B(x,r)

χE(y) dy.

Let r = d(x,E) + diamE. Then

(MχE)(x) &|E|rd.

Since E is compact, if we restrict x to a compact set then d(x,E) is bounded below andabove. Thus, MχE &E,F 1 uniformly for x ∈ F with F compact. So suppose for the sakeof contradiction that there is a compact set F with µ(F ) =∞. Then

∞ = µ(F ) ≤ µ ({x : MχE(x) &E,F 1 }) .E,F

∫Edµ = µ(E) <∞

which is a contradiction.With this claim proven, next, note that

∫|fn|p dµ → 0 by the dominated convergence

theorem. Fix x ∈ K. Then since B(x, 1/n) ⊆ Un and since |K| = 0,

(Mf)(x) = supr>0

1

|B(x, r)|

∫B(x,r)

χUn\K(y) dy ≥ 1

|B(x, 1/n)|

∫B(x,1/n)

χUn\K(y) dy

=1

|B(x, 1/n)|

∫KC

χB(x,1/n)(y) dy

=1

Rd

∫KC

χB(x,1/n)(y) dy

= 1.

So x ∈ {x : (Mfn)(x) > 1/2 }. Thus, K ⊆ {x : (Mfn)(x) > 1/2 }. So

0 < ν(K) < µ(K) ≤ µ ({x : (Mfn)(x) > 1/2 }) .∫|fn|p dµ→ 0

which is a contradiction. Thus, dµ = ω(x) dx, and by the previous theorem, ω ∈ Ap.

Lemma 4.9. We have ω ∈ Ap if and only if(1

|B|

∫Bf(y) dy

)p.

1

ω(B)

∫Bf(y)pω(y) dy

uniformly for all f ≥ 0 and all balls B.

Proof. First, suppose that ω ∈ Ap. Then

supB

ω(B)

|B|p

(∫Bw(y)

− p′p

) pp′

. 1.

33

Then, by Holder,

1

|B|p

(∫Bf(y) dy

)p=

1

|B|p

(∫Bf(y)ω(y)

1pω(y)

− 1p dy

)p.

1

|B|p

∫B|f(y)|pω(y) dy

(∫Bω(y)

− p′p

) pp′

.1

|B|p

∫B|f(y)|pω(y) dy

|B|p

ω(B)

=1

ω(B)

∫Bf(y)pω(y) dy.

To see the converse direction, suppose that(1

|B|

∫Bf(y) dy

)p.

1

ω(B)

∫Bf(y)pω(y) dy

uniformly for all f ≥ 0 and all balls B. Let f = (ω + ε)− p′p . Then

1

|B|p

(∫B

(ω(y) + ε)− p′p dy

)p.

1

ω(B)

∫B

(ω(y) + ε)−p′ω(y) dy

≤ 1

ω(B)

∫B

(ω(y) + ε)−p′+1 dy

=1

ω(B)

∫B

(ω(y) + ε)p′p dy.

Soω(B)

|B|p

(∫B

(ω(y) + ε)− p′p dy

)p−1

. 1.

Letting ε→ 0 gives the result.

The final step before proving the weighted maximal inequality is a version of a coveringlemma, due to Vitali.

Lemma 4.10 (Vitali). Let F be a finite collection of open balls in Rd. Then there exists a subcollec-tion S of F such that distinct balls in S are disjoint, and

⋃B∈F ⊆

⋃B∈S 3B.

Proof. Run the following algorithm.

1. Set S := ∅.

2. Choose a ball in F of largest radius, and add it to S.

3. Discard all of the balls in F which intersect balls in S.

4. If all balls in F are removed, stop. Otherwise, return to step 2.

This algorithm terminates because at least one ball in F is discarded at every step. Byconstruction, distinct balls in S are disjoint. Finally, if B is a ball in F which is not in S, itnecessarily intersects some ball B′ ∈ S of larger radius. By the triangle inequality (draw apicture), 3B′ contains B. Thus,

⋃B∈F ⊆

⋃B∈S 3B.

At last, we prove the weighted maximal inequality.

34

Proof of Theorem 4.3. First, we claim that M : L∞(M(ω) dx) → L∞(ω dx). To see this, letf ∈ L∞(M(ω) dx). Then

‖Mf‖L∞(ω dx) = infω(E)=0

supx∈EC

|(Mf)(x)| ≤ inf|E|=0

supx∈EC

|(Mf)(x)|

because |E| = 0 implies ω(E) = 0, asω(E) =∫E w(y) dy. We also have the trivial estimate

that ‖Mf(x)‖L∞(dx) ≤ ‖f‖L∞(dx), which follows from moving absolute value signs insidethe integral definition of Mf . Then

‖Mf‖L∞(ω dx) ≤ inf|E|=0

supx∈EC

|f(x)|.

Note that, unless ω ≡ 0, Mω(x) > 0 for all x. Thus,

‖Mf‖L∞(ω dx) ≤ inf|E|=0

supx∈EC

|f(x)| = inf(Mω)(E)=0

supx∈EC

|f(x)| = ‖f‖L∞(Mω dx)

as desired.Since M is bounded from L∞(M(ω) dx) → L∞(ω dx), by the Marcinkiewicz interpo-

lation theorem, it suffices to prove that M is bounded from L1(M(ω) dx) → L1,∞(ω dx).Indeed, set p1 = 1, q1 = 1 and p2 =∞, q2 =∞, and for θ ∈ (0, 1), define

1

pθ=

θ

p1+

1− θp2

= θ ⇒ pθ =1

θ

and1

qθ=

θ

q1+

1− θq2

= θ ⇒ qθ =1

θ.

For r = qθ = pθ = 1θ , the interpolation theorem gives boundedness of M as an operator

from L1θ (M(ω) dx)→ L

1θ (ω dx). As 0 < θ < 1, 1 < 1

θ <∞, hence the theorem is proved.So it remains to prove M : L1(M(ω) dx) → L1,∞(ω dx). For f ∈ L1(M(ω) dx), we need

to show that ‖Mf‖∗L1,∞(ω dx) . ‖f‖L1(M(ω) dx), i.e.,

supλ>0

λω ({x : |Mf(x)| > λ }) .∫|f(x)|M(ω)(x) dx.

Fix λ > 0 and consider {x : |Mf(x)| > λ }. Let K ⊆ {x : |Mf(x)| > λ } be compact. Forx ∈ K, Mf(x) > λ, so by definition of the maximal function there exists rx > 0 so that

1

|Br(x)|

∫Brx (x)

|f(y)| dy > λ.

Then K ⊆⋃x∈K Brx(x), and compactness then gives a finite subcover of such balls. By

Vitali’s covering lemma, there exists a subcollection S of these balls such that distinct ballsin S are disjoint, and K ⊆

⋃B∈S 3B. Then ω(K) ≤

∑B∈S ω(3B).

Consider a ball Bj ∈ S with radius rj . For y ∈ Bj , note that B4rj (y) ⊇ 3Bj . Thus,

ω(3Bj) =

∫3Bj

ω(x) dx ≤∫B4rj

(y)ω(x) dx ≤ (Mω)(y) |B4rj (y)| = (Mω)(y) 4d |Bj |.

Therefore,

ω(3Bj) ·∫Bj

|f(y)| dy ≤ 4d|Bj |∫Bj

(Mω)(y)|f(y)| dy

so thatω(3Bj) ·

1

|Bj |

∫Bj

|f(y)| dy ≤ 4d∫Bj

(Mω)(y)|f(y)| dy.

35

But 1|Bj |

∫Bj|f(y)| dy > λ, so

ω(3Bj) .d1

λ

∫Bj

(Mω)(y)|f(y)| dy.

Since the balls Bj are all disjoint, we then have

ω(K) ≤∑Bj∈S

ω(3Bj) .1

λ

∫Rd

(Mω)(y)|f(y)| dy =1

λ‖f‖L1(M(ω) dx) .

As this holds for all compact K ⊆ {x : |Mf(x)| > λ }, by the inner regularity of the mea-sure ω, we are done.

4.3 The Vector-Valued Maximal Inequality

Our first application of the weighted maximal inequality is a similar estimate for vector-valued maximal functions.

Definition 4.11. For f : Rd → `2(N) given by f(x) = {fn(x)}n≥1, define

‖f‖Lp :=∥∥∥∥{fn(x)}`2n

∥∥∥∥Lpx.

The vector-valued maximal function M is given by

M(f)(x) := ‖{Mfn(x)}‖`2n .

Note that Mf : Rd → [0,∞], so that Mf is not itself vector-valued. Rather, the phrase“vector-valued” refers only to the input functions f .

The main estimate on M is the following vector-valued version of the classical Hardy-Littlewood maximal inequality.

Theorem 4.12 (Vector-valued maximal inequality).

1. The operator M is of weak type (1, 1), that is,∥∥Mf∥∥∗L1,∞ . ‖f‖L1

for f : Rd → `2(N);

2. M is of strong type (p, p) for all 1 2 will follow from the weighted maximalinequality. Having established the case p = 2, by the Marcinkiewicz Interpolation theoremit will suffice to prove the weak type estimate for p = 1 to prove the entire theorem. Toprove this case we employ a Calderon-Zygmund decomposition technique.

We adopt the convention that for f : Rd → `2(N), |f(x)| := ‖{fn(x)}‖`2n .

Lemma 4.13 (Calderon-Zygmund decomposition). Given f ∈ L1(Rd) (possibly vector valued)and λ > 0, there is a decomposition f = g + b such that:

1. |g(x)| ≤ λ for almost all x ∈ Rd;

36

2. b = fχQk , where {Qk} is a collection of cubes whose interiors are disjoint and such that

λ <1

|Qk|

∫Qk

|b(y)| dy ≤ 2dλ.

In this decomposition, g represents the good part of f , i.e., the part which is essentiallyuniformly bounded by λ, and b represents the bad part. The requirement of b in the decom-position says that the average of b on any given cube is on the order of λ.

Proof. Decompose Rd into dyadic cubes of the form:

Qk = [2nk1, 2n(k1 + 1)× · · · × [2nkd, 2

n(kd + 1))

where the diameter of each cube is chosen to be large enough so that

1

Qk

∫Qk

|f(y)| dy ≤ λ.

Because∫Rd |f(y)| dy <∞, this is certainly possible.

Run the following algorithm. Fix a cube Q from the above decomposition. Divide Qinto 2d equal sized cubes. Let Q′ denote one of these smaller cubes.

If Q satisfies1

Q′

∫Q′|f(y)| dy > λ,

then stop and add Q′ to the collection of cubes which define the support of b. Note that, forsuch a cube,

λ <1

|Q′|

∫Q′|f(y)| dy ≤ 2d

|Q|

∫Q|f(y)| dy ≤ 2dλ

as required by the definition of b.If Q satisfies

1

Q′

∫Q′|f(y)| dy ≤ λ,

then subdivide Q′ as we did with Q and continue until, if ever, we land in the previouscase.

Having run this algorithm, let b = fχ⋃Qk , where {Qk} is the collection of cubes fromthe above algorithm. Set g = f − b. By construction, the cubes Qk have disjoint interiors.

It only remains to check that |g| ≤ λ almost everywhere. If x /∈⋃Qk, then by our

application there is a sequence of cubes, each containing x, with diameters shrinking to 0,such that the average value of |f | on these cubes is all ≤ λ. By the Lebesgue differentiationtheorem, |g| ≤ λ almost everywhere.

Now, we prove the vector-valued maximal inequality.

Proof Theorem 4.12. First consider the case p = 2. For a vector-valued f = {fn} ∈ L2(Rd),we need to show that

∥∥Mf∥∥L2 . ‖f‖L2 . By definition,

∥∥Mf∥∥2

L2 =

∫Rd‖{Mfn(x)}‖2`2n dx =

∫Rd

∑n≥1

|Mfn(x)|2 dx.

Since all the quantities in question are nonnegative, we can interchange the integral andsummation (by Tonelli’s theorem) to get∥∥Mf

∥∥2

L2 =∑n≥1

∫Rd|Mfn(x)|2 dx.

37

Previously, we proved that M is of strong type (2, 2). So

∥∥Mf∥∥2

L2 =∑n≥1

∫Rd|Mfn(x)|2 dx .

∑n≥1

∫Rd|fn(x)|2 dx =

∫Rd

∑n≥1

|fn(x)|2 dx = ‖f‖2L2

as desired.Next, suppose that 2 < p <∞. We wish to show that

∥∥Mf∥∥Lp

. ‖f‖Lp . Observe that

∥∥Mf∥∥Lp

=∥∥∥‖Mfn(x)‖`2n

∥∥∥Lpx

=∥∥∥(‖Mfn(x)‖`2n)2

∥∥∥ 12

Lp/2x

so that∥∥Mf

∥∥2

Lp=∥∥(Mf(x))2

∥∥Lp/2

. By the duality characterization of norms,

∥∥Mf∥∥2

Lp=∥∥(Mf(x))2

∥∥Lp/2

= sup‖ω‖

L(p/2)′=1

∫(Mf(x))2 ω(x) dx

= sup‖ω‖

L(p/2)′=1

∫ ∑n≥1

|Mfn(x)|2 ω(x) dx

= sup‖ω‖

L(p/2)′=1

∑n≥1

∫|Mfn(x)|2 ω(x) dx.

For ω ≥ 0, the weighted maximal inequality that we proved earlier tells us that M :L2(M(ω) dx) → L2(ω dx). Thus, supposing without loss of generality that the supremumin the above expression is taken over ω ≥ 0, we have

∥∥Mf∥∥2

Lp= sup‖ω‖

L(p/2)′=1

∑n≥1

∫|Mfn(x)|2 ω(x) dx . sup

‖ω‖L(p/2)′=1

∑n≥1

∫|fn(x)|2 (Mω)(x) dx

= sup‖ω‖

L(p/2)′=1

∫ ∑n≥1

|fn(x)|2 (Mω)(x) dx.

Applying Holder’s inequality,

∥∥Mf∥∥2

Lp. sup‖ω‖

L(p/2)′=1

∥∥∥∥∥∥∑n≥1

|fn|2∥∥∥∥∥∥L(p/2)

‖Mω‖L(p/2)′ .

By the standard Hardy-Little maximal inequality theorem, ‖Mω‖L(p/2)′ . ‖ω‖L(p/2)′ = 1.Finally, note that ∥∥∥∥∥∥

∑n≥1

|fn|2∥∥∥∥∥∥L(p/2)

=∥∥∥‖fn‖2`2n∥∥∥L(p/2)

= ‖f‖2Lp

so that∥∥Mf

∥∥Lp

. ‖f‖Lp as desired.As noted above, it remains to prove the weak type (1, 1) claim. Explicitly, given f =

{fn} ∈ L1(Rd), we need to show that

supλ>0

λ∣∣{x : (Mf)(x) > λ

}∣∣ . ‖f‖L1 .

Using the Calderon-Zygmund decomposition, write f = g + b with |g| ≤ λ almost every-where and b = fχ⋃Qk where Qk are cubes with disjoint interiors satisfying

1

|Qk|

∫Qk

|b(y)| dy ∼ λ.

38

Because M is a sublinear operator,

{x : (Mf)(x) > λ

}⊆{x : (Mg)(x) >

λ

2

}∪{x : (Mb)(x) >

λ

2

}.

Consider the first set. We have already shown that M is of strong type (2, 2), so in particularit is of weak type (2, 2). Thus,∣∣∣∣{x : (Mg)(x) >

λ

2

}∣∣∣∣ . 1

λ‖g‖2L2 =

1

λ2

∥∥|g|2∥∥L1 ≤

1

λ2‖λ|g|‖L1 =

1

λ‖g‖L1 ≤

1

λ‖f‖L1 .

Therefore, it remains to prove the same type of estimate for the set{x : (Mb)(x) > λ

2

}.

Let 2Qk denote the cube with same center as Qk and twice the side length. Then, be-cause the interior of the Qk’s are disjoint, we have:∣∣∣⋃ 2Qk

∣∣∣ ≤ 2d∑k

|Qk| .∑k

1

λ

∫Qk

|b(y)| dy .1

λ

∫Rd|b(y)| dy

≤ 1

λ‖f‖L1 .

With this estimate, it remains to show that∣∣∣∣{x /∈⋃ 2Qk : (Mb)(x) >λ

2

}∣∣∣∣ . 1

λ‖f‖L1 .

Define bavgn (x) =∑

k χQk(x) 1|Qk|

∫Qk|bn(y)| dy. If x ∈ Qk, then

‖bavgn (x)‖`2n =1

|Qk|

∥∥∥∥∫Qk

|bn(y)| dy∥∥∥∥`2n

≤ 1

|Qk|

∫Qk

‖bn(y)‖`2n dy =1

|Qk|

∫Qk

|b(y)| dy . λ

invoking Minkowski’s integral inequality to move the `2n norm inside the integral. Thus,

‖bavg‖L1 =∥∥∥‖bavgn (x)‖`2n

∥∥∥L1x

. λ∣∣∣⋃Qk

∣∣∣ . λ · 1

λ‖f‖L1 = ‖f‖L1 .

Now, fix x /∈⋃

2Qk. Then

Mbn(x) = supr>0

1

|B(x, r)|

∫B(x,r)

|bn(y)| dy = supr>0

1

|B(x, r)|∑k

∫B(x,r)∩Qk

|bn(y)| dy.

Suppose that x /∈⋃

2Qk but that B(x, r) ∩ Qk 6= ∅. Let l be the side length of Qk. Thennecessarily r ≥ l/2. Note that the diameter of Qk is

√dl ≤ 2r

√d. This implies that

B(x, r(1 + 2√d)) ⊇ Qk. Thus,

Mbn(x) .d supr>0

1

|B(x, r(1 + 2√d)|

∑k

∫Qk

|bn(y)| dy

= supr>0

1

|B(x, r(1 + 2√d)|

∑k

(∫B(x,r(1+2

√d))χQk(z) dz

)(∫Qk

|bn(y)| dy)

= supr>0

1

|B(x, r(1 + 2√d)|

∫B(x,r(1+2

√d))

∑k

χQk(z) · 1

|Qk|

∫Qk

|bn(y)| dy dz

= Mbavgn (x).

39

It follows that Mb . Mbavg. So we again apply the (2, 2) estimate of M to get∣∣∣∣{x /∈⋃ 2Qk : (Mb)(x) >λ

2

}∣∣∣∣ ≤ ∣∣∣{x /∈⋃ 2Qk : (Mbavg)(x) & λ}∣∣∣ . 1

λ2‖bavg‖2L2

=1

λ2

∥∥∥‖bavgn (x)‖`2n∥∥∥2

L2x

=1

λ2

∥∥∥‖bavgn (x)‖2`2n∥∥∥L1x

.1

λ

∥∥∥‖bavgn (x)‖`2n∥∥∥L1x

=1

λ‖bavg‖L1

.1

λ‖f‖L1 .

40

Chapter 5

Sobolev Inequalities

This chapter contains a brief foray into the world of so-called Sobolev inequalities. A typicalinequality of this type bounds a certain norm of a function by a norm of a derivative of thefunction. In vague terms, this kind of estimate bounds the average gradient (or energy) ofa function below by its overall spread. This turns out to be a useful fact.

The inequalities we prove in this chapter will appear again throughout the rest ofthe book in different contexts. For example. we will provide an alternate proof for theGagliardo-Nirenberg inequality after we develop the machinery of Littlewood-Paley the-ory. Also, we will discuss the optimal constants of the Gagliardo-Nirenberg and Sobolevembedding inequalities in the future.

5.1 The Hardy-Littlewood-Sobolev Inequality

As a precursor to the Sobolev inequalities we will prove, we begin with an important con-volution estimate called the Hardy-Littlewood-Sobolev inequality. We actually prove twosuch inequalities in this section, the latter a generalization of the first.

There are many ways to prove the following theorem. The approach we take is due toHedberg [4] and is amenable to proving certain inverse inequalities.

Theorem 5.1 (Hardy-Littlewood-Sobolev I). Let f ∈ S(Rd). Then∥∥∥∥f ∗ 1

|x|α

∥∥∥∥Lr

. ‖f‖Lp

whenever 1 + 1r = 1

p + αd for 1 < p < r <∞ and 0 < α < d.

Proof. We remark that requiring f ∈ S(Rd) ensures that f ∈ Lp(Rd) for all necessary p.Decompose the convolution as follows:(

f ∗ 1

|x|α

)(x) =

∫f(y)

|x− y|αdy =

∫|x−y|≤R

f(y)

|x− y|αdy +

∫|x−y|>R

f(y)

|x− y|αdy

for some R > 0. We will estimate each integral and then optimize in R.Consider the first integral.∣∣∣∣∣∫|x−y|≤R

f(y)

|x− y|αdy

∣∣∣∣∣ ≤∫|x−y|≤R

|f(y)||x− y|α

dy ≤∑

r∈2Z;r≤R

∫r<|x−y|<2r

|f(y)||x− y|α

dy

.∑

r∈2Z;r≤R

r−α∫|x−y|<2r

|f(y)| dy

.∑

r∈2Z;r≤R

rd−α

|B(x, 2r)|

∫|x−y|<2r

|f(y)| dy.

41

The quantity 1|B(x,2r)|

∫|x−y|<2r |f(y)| dy is bounded by Mf(x), so∣∣∣∣∣∫|x−y|≤R

f(y)

|x− y|αdy

∣∣∣∣∣ ≤Mf(x)∑

r∈2Z;r≤R

rd−α.

Because d − α > 0 and because the r’s are dyadic numbers, the sum is summable and isbounded (up to a constant) by the largest term. Thus,∣∣∣∣∣

∫|x−y|≤R

f(y)

|x− y|αdy

∣∣∣∣∣ ≤ Rd−αMf(x).

Now we consider the second integral. We have∫|x−y|≤R

f(y)

|x− y|αdy =

(f ∗(

1

|x|αχ{|x|>R}

))(x).

By Young’s convolution inequality (or more directly, Holder’s inequality),

∥∥∥∥f ∗ ( 1

|x|αχ{|x|>R}

)∥∥∥∥L∞

. ‖f‖Lp∥∥∥∥ 1

|x|αχ{|x|>R}

∥∥∥∥Lp′

= ‖f‖Lp

(∫|x|>R

1

|x|αp′dx

) 1p′

.

If αp′ > d, then the integral quantity is finite, and in particular . Rd−αp′p′ . Since

α

d+

1

p> 1 ⇒ α

d> 1− 1

p=

1

p′

this is indeed the case. Thus,∣∣∣∣∣∫|x−y|≤R

f(y)

|x− y|αdy

∣∣∣∣∣ ≤∥∥∥∥f ∗ ( 1

|x|αχ{|x|>R}

)∥∥∥∥L∞

. ‖f‖Lp Rdp′−α.

We optimize our choice of R by requiring the two estimates to be comparable, so that

Rd−αMf(x) ∼ Rdp′−α ‖f‖p. Choose R so that R ∼

(‖f‖LpMf(x)

) pd .

With this choice,∣∣∣∣f ∗ 1

|x|α

∣∣∣∣ (x) .

(( ‖f‖pMf(x)

) pd

)d−αMf(x) = ‖f‖

d−αdp

Lp (Mf(x))1− pd

(d−α).

Simplifying the exponents,

1− p

d(d− α) = p

(1

p− 1 +

α

d

)=p

r.

So ∣∣∣∣f ∗ 1

|x|α

∣∣∣∣ (x) . ‖f‖1−pr

Lp (Mf(x))pr .

Taking the Lr norm and using the fact that M is of type (p, p) then gives∥∥∥∥f ∗ 1

|x|α

∥∥∥∥Lr

. ‖f‖1−pr

Lp

∥∥∥(Mf(x))pr

∥∥∥Lr

= ‖f‖1−pr

Lp ‖Mf‖prLp . ‖f‖

1− pr

Lp ‖f‖prLp = ‖f‖Lp

as desired.

42

It is tempting to view the Hardy-Littlewood-Sobolev inequality as having more or lessthe same content as Young’s convolution inequality, as the statements of the estimates aresimilar. However, the fact that |x|−α is not in any Lp spaces makes Hardy-Littlewood-Sobolev much more subtle.

Note, however, that the function |x|−α lives in the weak space Ldα,∞(Rd). This suggests

the following generalization of the previous theorem.

Theorem 5.2 (Hardy-Littlewood-Sobolev II). For 1 < p < r <∞ and 1 < q <∞, we have

‖f ∗ g‖Lr . ‖f‖Lp ‖g‖∗Lq,∞

whenever 1 + 1r = 1

p + 1q .

Proof. We begin with a few reductions. By rescaling, we can assume without loss of gener-ality that ‖f‖Lp = ‖g‖∗Lq,∞ = 1. Furthermore, note that for a fixed such g it suffices to provethat the operator f 7→ f ∗ g is of strong type (p, r). In fact, by the Marcinkiewicz interpo-lation theorem, it suffices to prove that this operator is of weak type (p, r). This is becausethe condition 1 0. We need to show that ‖f ∗ g‖∗Lr,∞ . 1, i.e., |{x : |f ∗ g|(x) > λ }| . λ−r.As in the previous proof, we decompose g = gχ|g|≤R + gχ|g|>R := g1 + g2. Then

|{x : |f ∗ g|(x) > λ }| ≤∣∣∣∣{x : |f ∗ g1|(x) >

λ

2

}∣∣∣∣+

∣∣∣∣{x : |f ∗ g2|(x) >λ

2

}∣∣∣∣ .We will show that the first quantity on the right is 0 for an appropriately chosen R. Intu-itively, since g1 is bounded by R, the convolution of f with g cannot be too large. Comput-ing,

‖f ∗ g1‖L∞ . ‖f‖Lp ‖g1‖Lp′ = ‖g1‖Lp′ .(∫ ∞

0αp′ |{x : |g1(x)| > α }| dα

α

) 1p′

=

(∫ R

0αp′ |{x : |g1(x)| > α }| dα

α

) 1p′

where we have used the layer-cake representation for the Lp′

norm and (see Chapter 3) andthe fact that |g1| ≤ R.

Continuing,

‖f ∗ g1‖L∞ .

(∫ R

0supα>0

[αq |{x : |g1(x)| > α }|] αp′−q dαα

) 1p′

=

(supα>0

[αq |{x : |g1(x)| > α }|]) 1p′(∫ R

0αp′−q dα

α

) 1p′

.

The quantity on the left is precisely (‖g‖∗Lq,∞)1p′ , which we are assuming is 1. The integral

on the right is integrable if p′ − q > 0. By assumption,

1− 1

p=

1

q− 1

r⇒ 1

p′<

1

q⇒ p′ > q.

So the integral is finite, and in particular,(∫ R

0 αp′−q dα

α

) 1p′ . R

p′−qp′ . Putting this all together

gives

‖f ∗ g1‖L∞ . Rp′−qp′ .

43

Recall that we are estimating∣∣{x : |f ∗ g1|(x) > λ

2

}∣∣. Thus, if we choose R small enough(dependent on λ, say R = cλ for some sufficiently small constant c, then ‖f ∗ g1‖L∞ ≤ λ/2,and so ∣∣∣∣{x : |f ∗ g1|(x) >

λ

2

}∣∣∣∣ = 0.

Thus, it remains to estimate∣∣{x : |f ∗ g2|(x) > λ

2

}∣∣. By Chebychev’s inequality,∣∣∣∣{x : |f ∗ g2|(x) >λ

2

}∣∣∣∣ . λ−p ‖f ∗ g2‖pLp . λ−p (‖f‖Lp ‖g2‖L1)p

= λ−p(∫ ∞

0α |{x : |g2(x)| > α }| dα

α

)p.

We consider the integral separately.∫ ∞0

α |{x : |g2(x)| > α }| dαα

=

∫ R

0α |{x : |g2(x)| > α }| dα

α+

∫ ∞R

α |{x : |g2(x)| > α }| dαα

≤ R · |{x : |g(x)| > R }|+ supα>0

αq |{x : |g(x)| > α }|∫ ∞R

α1−q dα

α.

As before, supα>0 αq |{x : |g(x)| > α }| = ‖g‖∗Lq,∞ = 1. Furthermore, since q > 1, we have∫∞

R α1−q dαα . R1−q. So∫ ∞

0α |{x : |g2(x)| > α }| dα

α. Rq · |{x : |g(x)| > R }| R1−q +R1−q.

But since Rq · |{x : |g(x)| > R }| ≤ ‖g‖∗Lq,∞ = 1, we then have∫ ∞0

α |{x : |g2(x)| > α }| dαα

. R1−q.

Plugging this into our original estimate and using R1− qp′ = cλ,

|{x : |f ∗ g2| > λ/2 }| . λ−pRp(1−q) . λ−pRp(1−q) . λ−pλ

(p′p′−q

)p(1−q)

. λ−r.

5.2 The Sobolev Embedding Theorem

As a consequence of the Hardy-Littlewood Sobolev inequalities, we prove a version of theSobolev embedding theorem.

Towards this goal, we begin with a computation. We wish to (formally) take the Fouriertransform of functions of the form 1

|x|d−α . As this function is not in Lp(Rd), we consider the

Fourier transform in the sense of distributions. Recall that if T ∈ S ′(Rd) is a tempereddistribution, its Fourier transform is defined by T (f) := T (f) for f ∈ S(Rd).

Proposition 5.3. Let 0 < α < d. Then 1|x|d−α ∼

1|x|α , in the sense of tempered distributions.

Explicitly, (π−

d−α2 Γ

(d− α

2

)1

|x|d−α

)= π−

α2 Γ(α

2

) 1

|x|α. (5.1)

44

Proof. First, observe that∫ ∞0

e−πt|x|2td−α

2dt

t=

∫ ∞0

e−u(

u

π|x|2

) d−α2 du

u= π−

d−α2 Γ

(d− α

2

)1

|x|d−α.

So for f ∈ S(Rd),(π−

d−α2 Γ

(d− α

2

)1

|x|d−α

)(f) =

(π−

d−α2 Γ

(d− α

2

)1

|x|d−α

)(f)

=

∫Rdπ−

d−α2 Γ

(d− α

2

)1

|x|d−αf(x) dx.

Our first observation, together with the definition of the Fourier transform, gives

=

∫Rd

∫ ∞0

∫Rde−πt|x|

2td−α

2dt

t

∫Rde−2πix·yf(y) dy dx

=

∫Rd

∫ ∞0

(∫Rde−πt|x|

2e−2πix·y dx

)t−

d−α2dt

tf(y) dy.

The quantity in the parentheses is precisely the Fourier transform of e−π|x|2. Recall that(

e−x·Ax)

= πd2 (detA)−

12 e−π

2ξ·A−1ξ.

Thus, ∫Rde−πt|x|

2e−2πix·y dx = π

d2 (πt)−

d2 e−π

2y· 1πty = t

d2 e−

π|y|2t .

Continuing the computation,

=

∫Rd

∫ ∞0

e−π|y|2t t−

α2dt

tf(y) dy =

∫Rd

∫ ∞0

td2 e−

π|y|2t t−

d−α2dt

tf(y) dy

=

∫Rd

∫ ∞0

(π|y|2

u

)−α2

e−udu

uf(y) dy

=

∫Rd

∫ ∞0

π−α2 |y|−αu

α2 e−u

du

uf(y) dy

=

∫Rdπ−

α2 Γ(α

2

)|y|−α f(y) dy

=(π−

α2 Γ(α

2

)|y|−α

)(f).

This proves (5.1) in the sense of tempered distributions.

The are various forms of the Sobolev embedding theorem. The one we consider hereinvolves a fractional derivative operator |∇|s, which generalizes the usual notion of deriva-tive. The operator |∇|s is defined via its action in the Fourier domain.

Definition 5.4. Fix s > −d. For f ∈ S(Rd), define |∇|s via

(|∇|sf)ˆ(ξ) := (2π|ξ|)sf(ξ),

where this equality is understood in the sense of tempered distributions.

We require s > −d so that the quantity (2π|ξ|)s makes sense as a distribution on Rd. As(Dαf)ˆ(ξ) = (2πiξ)αf(ξ), it is clear why the operator |∇|s generalizes the usual notion of aderivative. It is natural to wonder how estimates on |∇|s compare to estimates on the usualderivative operator ∇. It turns out to be a fact that ‖|∇|f‖Lp . ‖∇f‖Lp . We will prove thisin the future when we discuss Calderon-Zygmund convolution kernels.

45

Theorem 5.5 (Sobolev embedding). Fix 1 0. Let f ∈ S(Rd) such that |∇|sf ∈Lp(Rd).1 Then

‖f‖Lq . ‖|∇|sf‖Lp

provided 1p = 1

q + sd .

Proof. Recall that S(Rd) is dense in Lq′. Thus, by duality of Lp-norms and Plancharel’s

theorem,

‖f‖Lq = supg∈S(Rd):‖g‖

Lq′=1

〈f, g〉 = supg∈S:‖g‖

Lq′=1

⟨f , g⟩

= supg∈S:‖g‖

Lq′=1

⟨(2π|ξ|)sf(ξ), (2π|ξ|)−sg(ξ)

⟩.

By our previous comments, (2π|ξ|)sf(ξ) ∈ S ′(Rd). We would like for (2π|ξ|)−sg(ξ) to be aSchwartz function but if g does not vanish near the origin then the singularity at ξ = 0 mayprevent this from being true. Thus, we consider a slightly smaller space of functions.

We claim that the family

F :={g ∈ S(Rd) : g vanishes on a neighborhood of ξ = 0

}is dense in Lp for 1 < p < ∞. Obviously it suffices to show that F is dense in S. Fixg ∈ S(Rd). Let ϕ be a smooth bump function on Rd such that ϕ(ξ) = 1 for |ξ| ≤ 1 andϕ(ξ) = 0 for |ξ| ≥ 2. Define

gε(ξ) := g(ξ)

(1− ϕ

(|ξ|ε

)).

Note that

g(ξ)

(1− ϕ

(|ξ|ε

))= g(ξ)− g(ξ)ϕ

(|ξ|ε

)=

(g(x)− g ∗

(ϕ

(|ξ|ε

))(x)

).

Hence

‖g − gε‖Lp =

∥∥∥∥∥g ∗(ϕ

(|ξ|ε

))∥∥∥∥∥Lp

=∥∥∥g ∗ εdϕ(ε|x|)

∥∥∥Lp

. εd ‖g‖L1 ε− dp ‖ϕ‖Lp .

Since g and ϕ are fixed and because p > 1, as ε→ 0, the above quantity→ 0.This estimate clearly fails for p = 1. Intuitively, F is not dense in L1 because g(0) is the

total integral over Rd of g. Functions with mean zero are certainly not dense in L1.For 1 < p <∞, we have a new dense subset F . So

‖f‖Lq = supg∈F :‖g‖

Lq′=1

⟨(2π|ξ|)sf(ξ), (2π|ξ|)−sg(ξ)

⟩.

The functions (2π|ξ|)−sg(ξ) are certainly Schwartz, since g is a Schwartz function whichvanishes in a neighborhood of 0.

Applying Plancharel again gives

‖f‖Lq = supg∈F :‖g‖

Lq′=1

⟨|∇|sf, |∇|−sg

⟩. sup

g∈F :‖g‖Lq′=1‖|∇|sf‖Lp

∥∥|∇|−sg∥∥Lp′

.

We then have (|∇|−sg

)(x) =

((2π|ξ|)−sg(ξ)

)(x).

1In the sense of distributions, i.e., |∇|sf is given by integration against a function in Lp.

46

Now, recall that if T is a tempered distribution and g is a Schwartz function, then (T ∗ g) =T · g. Since (2π|ξ|)−s is a tempered distribution and g is Schwartz, it follows that

(|∇|−sg

)(x) =

((((2π|ξ|)−s)ˇ∗ g

))(x).

Our computation from (5.1) tells us that (2π|ξ|)−s) ∼d,s |x|s−d. Therefore,

|∇|−sg ∼d,s |x|s−d ∗ g.

By the first Hardy-Littlewood-Sobolev inequality,∥∥∥∥g ∗ 1

|x|d−s

∥∥∥∥Lp′

. ‖g‖Lq′

provided 1 + 1p′ = 1

q′ + d−sd . But this is true precisely when

1 + 1− 1

p= 1− 1

q+ 1− s

d⇐⇒ −1

p= −1

q− s

d⇐⇒ 1

p=

1

q+s

d

which holds by assumption.Therefore,

‖f‖Lq . supg∈F :‖g‖

Lq′=1‖|∇|sf‖Lp

∥∥|∇|−sg∥∥Lp′

. supg∈F :‖g‖

Lq′=1‖|∇|sf‖Lp ‖g‖Lq′ = ‖|∇|sf‖Lp

as desired.

It will be useful to record the special case of Sobolev embedding when p = 2 and s = 1.

Corollary 5.6 (Sobolev embedding, p = 2). Fix d ≥ 3. Suppose that f ∈ S(Rd) with |∇|f ∈L2(Rd). Then

‖f‖L

2dd−2

. ‖|∇|f‖L2 .

Proof. By the Sobolev embedding theorem, ‖f‖Lq . ‖|∇|f‖L2 provided 12 + 1

q + 1d . This

means1

q=d− 2

2d

and so q = 2dd−2 as desired.

Using the yet-unproven fact that ‖|∇|f‖L2 . ‖∇f‖L2 , this special case of the Sobolev

embedding theorem gives the inclusion H1(Rd) ⊆ L2dd−2 (Rd).

5.3 The Gagliardo-Nirenberg Inequality

In this section, we use the Sobolev embedding theorem to prove a version of the Gagliardo-Nirenberg inequality. As remarked at the beginning of this chapter, this theorem will berestated and revisited in different contexts after developing more sophisticated tools.

The following estimate is stated for d ≥ 3, but in fact it holds for d = 1 and d = 2, where0 < p <∞. The case d = 1 is an easy calculus exercise, while the d = 2 case is more subtle.

Theorem 5.7 (Gagliardo-Nirenberg). Fix d ≥ 3 and f ∈ S(Rd). Then for all 0 < p < 4d−2 ,

‖f‖p+2p+2 . ‖f‖p+2− pd

22 ‖|∇|f‖

pd2

2 .

47

Proof. For convenience, we recall the log-convexity of Lp norms: for any 0 < θ < 1 and1r = 1−θ

p + θq , we have ‖f‖r ≤ ‖f‖

1−θp ‖f‖θq .

Let 0 0, θ > 0. Because 0 0,

1− θA

+θ

B=

4 + 2p− pd2A(p+ 2)

+pd

2B(p+ 2)=

1

p+ 2

(4 + 2p− pd

2A+pd

2B

).

With log-convexity of Lp norms in mind, we want to chooseA andB so that 4+2p−pd2A + pd

2B =1. Towards our end goal, set A = 2. Then

4 + 2p− pd4

+pd

2B= 1 ⇒ 1 +

p

2− pd

4+pd

2B= 1 ⇒ 1

2− d

4+

d

2B= 0.

Solving for B yields

d

2B=d

4− 1

2=d− 2

4⇒ 2B

d=

4

d− 2⇒ B =

2d

d− 2.

These computations demonstrate that

1

p+ 2=

1− θ2

+θ2dd−2

.

Thus,‖f‖p+2 ≤ ‖f‖

1−θ2 ‖f‖θ2d

d−2.

Next, by the p = 2 Sobolev embedding theorem, ‖f‖ 2dd−2

. ‖|∇|f‖2. So

‖f‖p+2 . ‖f‖1−θ2 ‖|∇|f‖θ2 = ‖f‖1− pd

2(p+2)

2 ‖|∇|f‖pd

2(p+2)

2

as desired.

48

Chapter 6

Fourier Multipliers

In Chapter 5, we introduced the operator |∇|s. This was our first experience with a nontriv-ial Fourier multiplier operator, i.e., an operator defined by inverting a multiplication operatorin the Fourier domain. The operator |∇|s is a Fourier multiplier with symbol (2π|ξ|)s. Wesaw other simple examples of when first defining the Fourier transform. For example,the regular partial derivative operator ∂xk is a Fourier multiplier with symbol 2πiξk; thetranslation operator f(·) 7→ f(· − y) is a Fourier multiplier with symbol e2πiy·ξ, etc. Moregenerally, the Fourier multiplier with symbol m(ξ) is the operator

f(x) 7→(m(ξ)f(ξ)

)(x) =

∫e2πix·ξm(ξ)f(ξ) dξ.

In this chapter, we study the Lp-boundedness of various Fourier multipliers.

6.1 Calderon-Zygmund Convolution Kernels

When the operator |∇|s was introduced in Chapter 5, we remarked that the estimate ‖|∇|f‖Lp .‖∇f‖Lp holds for Schwartz functions f . The usefulness of this estimate is apparent fromthe fact that |∇|s is a nonlocal operator, in that it was defined by multiplication in the Fourierdomain and hence by convolution in the physical domain. On the other hand,∇ is a purelylocal operator. Recall that |∇|f(ξ) = 2π|ξ|f(ξ). Write

2π|ξ| = 2π|ξ|2

|ξ|= 2π

∑ ξ2j

|ξ|=∑−i ξj|ξ|· 2πiξj =

∑mj(ξ)2πiξj

wheremj(ξ) := −i ξj|ξ| are called Riesz multipliers. To show the desired estimate ‖|∇|f‖Lp .‖∇f‖Lp , it is enough to prove that the Riesz multipliers are bounded on Lp, that is, the op-erators Rj defined via Rjf(ξ) := mj(ξ)f(ξ), or equivalently Rjf = mj ∗ f , are bounded onLp. Indeed, |∇|f(ξ) = 2π|ξ|f(ξ) =

∑mj(ξ)2πiξj f(ξ)

so that |∇|f =∑Rj(∂jf). By the triangle inequality and the equivalence of norms on finite

dimensional spaces,

‖|∇|f‖Lp ≤∑‖Rj(∂jf)‖Lp .

∑‖∂jf‖Lp . ‖∇f‖Lp .

To prove that this is the case, we will prove Lp-boundedness for a more general class ofobjects. In particular, by passing to distributions we can view any Fourier multiplier as aconvolution kernel, i.e., an operator defined by f 7→ K ∗ f for some distribution K. Thisfollows immediately from the fact that the (inverse) Fourier transform turns multiplicationinto convolution. In this section, we consider a special kind of convolution kernel.

49

Definition 6.1. A function K : Rd \ {0} → C is a Calderon-Zygmund convolution kernelif it satisfies:

1. |K(x)| . 1|x|d uniformly in x 6= 0;

2.∫R1≤|x|≤R2

K(x) dx = 0 for all 0 < R1 < R2 <∞;

3.∫|x|≥2|y| |K(x)−K(x+ y)| dx . 1 uniformly in y ∈ Rd.

The first condition says that K is allowed to have a singularity at 0, as long as thesingularity is not too bad. The second condition says that K satisfies a kind of cancellationproperty; note that if K is an odd function, this condition is satisfied. The third conditionis a smoothness requirement.

As a first remark, we claim that if |∇K|(x) . 1|x|1+d , thenK satisfies condition 3. Indeed,

by the fundamental theorem of calculus,

K(x)−K(x+ y) = −∫ 1

0y · ∇K(x+ θy) dθ.

So ∫|x|≥2|y|

|K(x)−K(x+ y)| dx ≤∫|x|≥2|y|

∫ 1

0|y| · 1

|x+ θy|1+ddθ dx.

Since |x| ≥ 2|y|, |x+ θy| & |x|. Thus,∫|x|≥2|y|

|K(x)−K(x+ y)| dx .∫|x|≥2|y|

∫ 1

0|y| · 1

|x|1+ddθ dx =

∫|x|≥2|y|

|y| · 1

|x|1+ddx

. |y| · 1

|y|= 1.

Therefore, K satisfies the third as claimed.Relevant to our motivating discussion, we claim that the kernel of the Riesz multiplier

defined above is a Calderon-Zygmund convolution kernels. To see this, first note that

Kj := mj =

(−i ξj|ξ|

)=

(− 1

2π· 2πiξj|ξ|

).

In the Sobolev embedding theorem section, we computed the inverse Fourier transform offunctions of the form 1

|ξ|s . Recalling this computation (see Equation (5.1)) we have

Kj(x) ∼d −1

2π∂j

(1

|x|d−1

)∼d

xj|x|d+1

since ∂j |x| = xj|x| . Clearly, |Kj(x)| .d

1|x|d and

∫R1≤|x|≤R2

Kj(x) dx = 0 (since x 7→ xj is odd).

Finally, we wish to show |∇Kj | . 1|x|1+d to verify the third condition. Note that

∂kKj(x) ∼ ∂kxj |x|−(d+1) = δjk|x|−(d+1) − (d+ 1)|x|−(d+1)−1xk|x|−1

=δjk|x|d+1

− (d− 1)xk|x|d+3

,

so certainly |∇Kj | . 1|x|1+d . Thus, Kj is a Calderon-Zygmund convolution kernel.

The first result on Calderon-Zygmund convolution kernels is anL2 → L2 type estimate.

Theorem 6.2. Suppose K is a Calderon-Zygmund convolution kernel. For ε > 0, let Kε :=K · χε≤|x|≤1/ε. Then ‖Kε ∗ f‖L2 . ‖f‖L2 uniformly in ε > 0. Moreover, K ∗ f = limε→0Kε ∗ fand the operator K ∗ f extends as a bounded operator from S(Rd) to L2(Rd).

50

Proof. First, we will show that Kε is a Calderon-Zygmund convolution kernel. Because Kε

is just a restriction of K, conditions 1 and 2 are immediate. We only need to verify that∫|x|≥2|y|Kε(x)−Kε(x+ y)| dy ≤ 1 uniformly in y ∈ Rd and ε > 0.

To do this, consider the following three subregions of |x| ≥ 2|y|:

A = {|x| ≥ 2|y|, ε ≤ |x|, |x+ y| ≤ 1/ε}B = {|x| ≥ 2|y|, ε ≤ |x| ≤ 1/ε, |x+ y| < ε or |x+ y| > 1/ε}C = {|x| ≥ 2|y|, ε ≤ |x+ y| ≤ 1/ε, |x| < ε or |x| > 1/ε}.

Then ∫|x|≥2|y|

|Kε(x)−Kε(x+ y)| dy ≤∫A|K(x)−K(x+ y)| dy

+

∫B|K(x)| dy +

∫C|K(x+ y)| dy.

The integral over A is . 1 because K is a Calderon-Zygmund convolution kernel.Consider the integral over B. Assume first that |x + y| < ε. Then |x| ≤ |x + y| + |y| <

ε+ |x|2 , hence |x| < 2ε. The contribution of this part of the integral is bounded by∫

ε≤|x|≤2ε|K(x)| dx ≤

∫ε≤|x|≤2ε

1

|x|ddx . 1.

If |x + y| > 1/ε, then |x| ≥ |x + y| − |y| > 1/ε − |x|/2, hence |x| ≥ 23ε . The contribution of

this part of the integral is bounded by∫23ε≤|x|≤ 1

ε

|K(x)| dx ≤∫

23ε≤|x|≤ 1

ε

1

|x|ddx . 1.

Thus,∫B |K(x)| dy . 1.

Bounding the integral over C is essentially the same argument, paired with a changeof variables. Thus,

∫C |K(x + y)| dy . 1. This verifies the third condition, hence Kε is a

Calderon-Zygmund convolution kernel.Next, we want to show ‖Kε ∗ f‖L2 . ‖f‖L2 uniformly in ε > 0. By Plancharel, we have

‖Kε ∗ f‖L2 =∥∥∥Kε ∗ f

∥∥∥L2

=∥∥∥Kε · f

∥∥∥L2≤∥∥∥Kε

∥∥∥L∞

∥∥∥f∥∥∥L2

∥∥∥Kε

∥∥∥L∞‖f‖L2 .

Thus, it suffices to prove that∥∥∥Kε

∥∥∥L∞

. 1 uniformly in ε.We will decompose the Fourier transform of Kε into an integral without any oscillation

and an integral with oscillation. Explicitly,

Kε(ξ) =

∫e−2πix·ξKε(x) dx =

∫|x|≤1/|ξ|

e−2πix·ξKε(x) dx+

∫|x|≥1/|ξ|

e−2πix·ξKε(x) dx.

In the first integral, |x · ξ| ≤ |x||ξ| ≤ 1, thus the term e−2πix·ξ is restricted in its oscilla-tion. Using the cancellation property of Calderon-Zygmund convolution kernels, we have∫|x|≤1/|ξ|Kε(x) dx =

∫ε≤|x|≤1/|ξ|K(x) dx = 0. Thus,∣∣∣∣∣

∫|x|≤1/|ξ|

e−2πix·ξKε(x) dx

∣∣∣∣∣ =

∣∣∣∣∣∫|x|≤1/|ξ|

(e−2πix·ξ − 1

)Kε(x) dx

∣∣∣∣∣≤∫|x|≤1/|ξ|

|e−2πix·ξ − 1||Kε(x)| dx.

51

Because Kε is a Calderon-Zygmund convolution kernel, |Kε(x)| . |x|−d. The quantity|e−2πx·ξ − 1| is the distance from the point e−2πix·ξ of the unit circle to 1. This is boundedby the arc-length of that segment on the unit circle corresponds to the phase, which is|x · ξ| ≤ |x||ξ|. Therefore,∣∣∣∣∣

∫|x|≤1/|ξ|


∣∣∣∣∣ ≤∫ε≤|x|≤1/|ξ|

|x||ξ||x|−d dx . |ξ| · 1

|ξ|= 1.

Next, consider the second integral. Here we invoke the smoothness condition of Calderon-Zygmund convolution kernels. Write

1 =1− eiπ

2=

1− e2πi ξ·ξ2|ξ|2

2.

Then ∫|x|>1/|ξ|

e−2πix·ξKε(x) dx =

∫|x|>1/|ξ|

e−2πix·ξ

1− e2πi ξ·ξ2|ξ|2

2

Kε(x) dx

=1

2

∫|x|>1/|ξ|

(e−2πix·ξ − e−2πiξ·

(x− ξ

2|ξ|2

))Kε(x) dx

=1

2

∫|x|>1/|ξ|


− 1

2

∫∣∣∣x+ ξ

2|ξ|2

∣∣∣>1/|ξ|e−2πix·ξKε

(x+

ξ

2|ξ|2

)dx

We want to combine these integrals, but have to account for the shifted circles which definethe domains; it helps to draw a picture here. We have:

=1

2

∫|x|>1/|ξ|

e−2πix·ξ(Kε(x)−Kε

(x+

ξ

2|ξ|2

))dx

+1

2

∫Ae−2πix·ξKε

(x+

ξ

2|ξ|2

)dx− 1

2

∫Be−2πix·ξKε

(x+

ξ

2|ξ|2

)dx

where

A =

{x : |x| ≥ 1

|ξ|≥∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ } ,B =

{x : |x| ≤ 1

|ξ|≤∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ } .Estimate the first of these three integrals by employing the smoothness condition of theCalderon-Zygmund convolution kernels:∣∣∣∣∣12

∫|x|>1/|ξ|

e−2πix·ξ(Kε(x)−Kε

(x+

ξ

2|ξ|2

))dx

∣∣∣∣∣.∫|x|>2

∣∣∣ ξ

2|ξ|2

∣∣∣∣∣∣∣Kε(x)−Kε

(x+

ξ

2|ξ|2

)∣∣∣∣ dx . 1.

Next, consider the integral over A. By the reverse triangle inequality,

|x| ≥ 1

|ξ|≥∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ ⇒∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ ≥ |x| − 1

2|ξ|≥ 1

|ξ|.

52

Thus, ∣∣∣∣12∫Ae−2πix·ξKε

(x+

ξ

2|ξ|2

)dx

∣∣∣∣ . ∫ 12|ξ|≤

∣∣∣x+ ξ

2|ξ|2

∣∣∣≤ 1|ξ|

∣∣∣∣Kε

(x+

ξ

2|ξ|2

)∣∣∣∣ dx.∫

12|ξ|≤|x|≤

1|ξ|

1

|x|ddx

. 1

where we have made the obvious change of variables and invoked the first property ofCalderon-Zygmund convolution kernels. Use a similar trick for bounding the integral overB, noting that

|x| ≤ 1

|ξ|≤∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ ⇒∣∣∣∣x+

ξ

2|ξ|2

∣∣∣∣ ≤ |x|+ 1

2|ξ|≤ 3

2|ξ|.

Combining all of this gives∣∣∣∫|x|>1/|ξ| e

−2πix·ξKε(x) dx∣∣∣ . 1, and hence |Kε(ξ)| . 1 uni-

formly in ε and ξ, thus,∥∥∥Kε(ξ)

∥∥∥L∞

. 1 uniformly in ε. Therefore, ‖Kε ∗ f‖L2 . ‖f‖L2 .

Next, we wish to show that Kε ∗ f converges in L2 as ε → 0. It suffices to show that{Kε ∗ f} is Cauchy. So fix 0 < ε2 < ε1 and fix f ∈ S(Rd). Then

(Kε1 ∗ f −Kε2 ∗ f)(x) =

∫ε1≤|y|≤1/ε1

K(y)f(x− y) dy −∫ε2≤|y|≤1/ε2

K(y)f(x− y) dy

= −∫ε2≤|y|≤ε1

K(y)f(x− y) dy −∫

1/ε1≤|y|≤1/ε2

K(y)f(x− y) dy.

Consider the second integral.∥∥∥∥∥∫

1/ε1≤|y|≤1/ε2

K(y)f(x− y) dy

∥∥∥∥∥L2

.∥∥K · χ1/ε1≤|y|≤1/ε2

∥∥L2 ‖f‖L1

.f

(∫1/ε1≤|y|≤1/ε2

1

|x|2ddx

) 12

. εd21

which→ 0 as ε1 → 0. For the first integral, the cancellation property of K gives∫ε2≤|y|≤ε1

K(y)f(x− y) dy =

∫ε2≤|y|≤ε1

K(y)(f(x− y)− f(x)) dy

=

∫ε2≤|y|≤ε1

K(y) y

∫ 1

0∇f(x− θy) dθ dy.

Thus, ∥∥∥∥∥∫ε2≤|y|≤ε1

K(y)f(x− y) dy

∥∥∥∥∥L2x

.∫ 1

0

∫ε2≤|y|≤ε1

1

|y|d|y| ‖∇f(x− θy)‖L2

xdy dθ

.f

∫ε2≤|y|≤ε1

1

|y|d−1dy ≤ ε1

→ 0

as ε1 → 0.

53

This shows that for a Schwartz function f , {Kε ∗ f} is Cauchy. By density, {Kε ∗ f} isCauchy for any f ∈ L2. To see this argument explicitly, fix δ > 0 and let g ∈ S(Rd) be suchthat ‖f − g‖L2 < δ. Then by the usual triangle inequality argument,

‖Kε1 ∗ f −Kε2 ∗ f‖L2 ≤ ‖Kε1 ∗ g −Kε2 ∗ g‖L2 + ‖Kε1 ∗ (f − g)‖L2 + ‖Kε2 ∗ (f − g)‖L2 .

The first quantity → 0 as ε1, ε2 → 0 since g is Schwartz. We have already seen thatCalderon-Zygmund convolution kernels are of type (2, 2), thus,

‖Kε1 ∗ (f − g)‖L2 . ‖f − g‖L2 ≤ δ

and likewise for Kε2 . Therefore, taking ε1, ε2 → 0 and then taking δ → 0 gives the desiredresult.

Therefore, we can define K ∗ f := L2 − limε→0Kε ∗ f . This operator extends to L2 byuniform boundedness of the Kε operators.

This theorem says that Calderon-Zygmund convolution kernels are operators of type(2, 2). Using this fact together with interpolation gives a wider range of estimates.

Theorem 6.3. Suppose that K is a Calderon-Zygmund convolution kernel. For ε > 0, let Kε :=K · χε≤|x|≤1/ε. Then

1. K is of weak-type (1, 1), i.e.,∣∣∣{x ∈ Rd : |Kε ∗ f | > λ}∣∣∣ . 1

λ‖f‖L1(Rd)

uniformly in λ > 0 and ε > 0.

2. Kε is of (strong) type (p, p) for 1 0.

Moreover, for 1 < p <∞, K ∗ f := Lp − limε→0Kε ∗ f extends as a bounded map on S(Rd) to abounded map on Lp(Rd).

Proof. Assume first that we have proven 1, thatKε is of weak-type (1, 1). SinceKε is of type(2, 2) by the previous theorem, it follows from the Marcinkiewicz interpolation theoremthat Kε is of type (p, p) for 1 < p < 2. We then use a duality argument to achieve theestimate for 2 < p <∞. Explicitly, for p in this range,

‖Kε ∗ f‖Lp = sup‖g‖

Lp′=1〈Kε ∗ f, g〉 = sup

‖g‖Lp′=1

∫ ∫Kε(x− y)f(y) g(x) dy dx

= sup‖g‖

Lp′=1

∫ ∫(Kε)R(y − x)g(x) dx f(y) dy

= sup‖g‖

Lp′=1

⟨f, (Kε)R ∗ g

⟩where (Kε)R(x) = Kε(−x) is the reflection of Kε. By Holder’s inequality,

‖Kε ∗ f‖Lp ≤ sup‖g‖

Lp′=1‖f‖Lp

∥∥∥(Kε)R ∗ g∥∥∥Lp′

.

Note that (Kε)R is also a Calderon-Zygmund convolution kernel. Since 1 < p′ < 2, we thenhave

∥∥∥(Kε)R ∗ g∥∥∥Lp′

. ‖g‖Lp′ = 1, so that ‖Kε ∗ f‖Lp . ‖f‖Lp as desired.

54

For the “moreover” statement, we wish to show that ‖Kε1 ∗ f −Kε2 ∗ f‖Lp → 0 asε1, ε2 → 0 as in the previous theorem. Though the same proof technique as before willwork, we present an alternative argument here. For 1 < p < 2, choose 1 < q < p andcompute, using the log-concavity of Lp-norms:

‖Kε1 ∗ f −Kε2 ∗ f‖Lp . ‖Kε1 ∗ f −Kε2 ∗ f‖θLq ‖Kε1 ∗ f −Kε2 ∗ f‖

1−θL2

where 1p = θ

q + 1−θ2 . We have ‖Kε1 ∗ f −Kε2 ∗ f‖

θLq . ‖f‖

θLq and ‖Kε1 ∗ f −Kε2 ∗ f‖

1−θL2 →

0 from previous results. For 2 < p <∞, choose p < q <∞ and argue similarly.Thus, to prove the entire theorem it only remains to show 1. Let f ∈ L1(Rd). We

perform a modified version of the Calderon-Zygmund decomposition introduced earlier.For a fixed λ > 0, write f = g + b where:

- |g| . λ a.e.;

- b is supported on a union of cubes {Qk} whose interiors are disjoint and |⋃Qk| .

1λ

∫|f(y)| dy;

- b |Qk= f |Qk − 1|Qk|

∫Qkf(y) dy, so that consequently

∫Qkb(y) dy = 0;

- 1|Qk|

∫Qk|b(y)| dy . λ.

Then

|{x : |Kε ∗ f | > λ }| ≤∣∣∣∣{x : |Kε ∗ g| >

λ

2

}∣∣∣∣+

∣∣∣∣{x : |Kε ∗ b| >λ

2

}∣∣∣∣ .We estimate the first quantity on the right using the (2, 2) estimate for Kε. Using Cheby-chev’s inequality and the uniform bound on g, we have:∣∣∣∣{x : |Kε ∗ g| >

λ

2

}∣∣∣∣ . λ−2 ‖Kε ∗ g‖2L2

. λ−2 ‖g‖2L2 = λ−2∥∥|g|2∥∥

L1 . λ−1 ‖g‖L1 ≤1

λ‖f‖L1 .

To estimate the second quantity, we first remove a set of comparable size to 1λ ‖f‖L1 . Let

Q∗k be the cube with same center, call it xk, as Qk but with side lengths 2√d`(Qk). (To

draw a picture, drawQk and circumscribe a circle; draw a circle with twice the radius, thencircumscribe a cube around this circle.) Then∣∣∣⋃Q∗k

∣∣∣ ≤∑ |Qk| ≤ (2√d)d∑|Qk| .d

1

λ‖f‖L1 .

Thus, it suffices to estimate the quantity∣∣{x /∈ ⋃Q∗k : |Kε ∗ b| > λ

2

}∣∣. By Chebychev, wehave ∣∣∣∣{x /∈⋃Q∗k : |Kε ∗ b| >

λ

2

}∣∣∣∣ . 1

λ‖Kε ∗ b‖L1((

⋃Q∗k)C) .

Writing out the definition of the convolution above gives

(Kε ∗ b)(x) =

∫Kε(x− y)b(y) dy =

∑∫Qk

Kε(x− y)b(y) dy

=∑∫

Qk

(Kε(x− y)−Kε(x− xk)) b(y) dy

55

where the last equality follows from the fact that b has mean 0. So

‖Kε ∗ b‖L1((⋃Q∗k)C) =

∫(⋃Q∗k)C)

|Kε ∗ b|(x) dx

≤∑∫

(Q∗k)C

∫Qk

|Kε(x− y)−Kε(x− xk)| |b(y)| dy dx

=∑∫

Qk

|b(y)|∫

(Q∗k)C|Kε(x− y)−Kε(x− xk)| dx dy.

Making the change of variables x− xk 7→ x yields

‖Kε ∗ b‖L1((⋃Q∗k)C) =

∑∫Qk

|b(y)|∫

(Q∗k)C−xk|Kε(x+ xk − y)−Kε(x)| dx dy

where (Q∗k)C−xk denotes the translation of (Q∗k)

C by the point xk. Note that if x ∈ (Q∗k)C−

xk, then |x| > 2√d`(Qk). Since

√d`(Qk) is the diameter of Qk and y ∈ Qk, this implies that

|x| ≥ 2|y−xk|. By the the smoothness property of Calderon-Zygmund convolution kernels,∫(Q∗k)C−xk

|Kε(x+ xk − y)−Kε(x)| dx . 1.

Thus,

‖Kε ∗ b‖L1((⋃Q∗k)C) .

∑∫Qk

|b(y)| dy .∫|f(y)| dy

and we are done.

In the proof of this theorem, having already shown (2, 2) boundedness of K, the only prop-erty of Calderon-Zygmund convolution kernels that we explicitly used was the smoothnesscondition. Interpolation was enough to get boundedness for other p values. This gives usa more general fact.

Proposition 6.4. If K is any convolution kernel (not necessarily Calderon-Zygmund) which is oftype (2, 2) and satisfies

∫|x|≥2|y| |K(x + y) − K(x)| dx . 1 uniformly in y, then K extends to a

bounded map on Lp(Rd) for 1 < p <∞.

6.2 The Hilbert Transform

As a first application of Lp-boundedness of Calderon-Zygmund convolution kernels, wediscuss the Hilbert transform. The Hilbert transform is the marquee example of a singularintegral operator, and is important in fields such as signal processing.

Definition 6.5. The Hilbert transform is the convolution operator corresponding to thekernel K : R \ {0} → R given by K(x) = 1

πx . Explicitly, the Hilbert transform of a functionf : R→ R is:

(Hf)(x) := limε→0

∫ε≤|y|≤ 1

ε

1

πyf(x− y) dy.

A priori, it is not clear that the Hilbert transform is well-defined for even nice functionsf . However, we claim that K is a Calderon-Zygmund convolution kernel. The estimate|K(x)| . 1

|x| is immediate from the definition, and the fact that K satisfies the cancellationproperty over annuli in R follows from the fact that 1

x is an odd function. Checking thesmoothness condition, we have∫

|x|≥2|y||K(x+ y)−K(x)| dx =

1

π

∫|x|≥2|y|

∣∣∣∣ 1

x+ y− 1

x

∣∣∣∣ dx =1

π

∫|x|≥2|y|

|y||x+ y||x|

dx.

56

Since |x| ≥ 2|y|, |x+ y| ≥ |x| − |y| & |x|, so that∫|x|≥2|y|

|K(x+ y)−K(x)| dx .∫|x|≥|y|

|y||x|2

dx . 1.

Thus, K is a Calderon-Zygmund convolution kernel. By the results from the previoussection, this proves the following.

Proposition 6.6. The Hilbert transform is a bounded operator from Lp(R) → Lp(R) for 1 < p <∞.

It is worth nothing that the strong-type estimates do indeed fail for p = 1 and p = ∞.For example, let f = χ[a,b]. Then f ∈ L1(R) and f ∈ L∞(R), and

(Hf)(x) =1

πlimε→0

∫ε≤|y|≤ 1

ε

1

yχ[a,b](x− y) dy.

Consider the case where x− a, x+ b > 0. Then for ε sufficiently small,

(Hf)(x) =1

π

∫ x−b

x−a

1

ydy =

1

πlog

∣∣∣∣x− ax− b

∣∣∣∣for almost all x, which is neither in L1(R) nor L∞(R). The cases x − a, x − b < 0 andx− a < 0 < x− b are handled similarly.

It can be shown that the Hilbert transform is a Fourier multiplier with symbol−i sgn(ξ).That is, for f ∈ S(R),

Hf(ξ) = −i sgn(ξ)f(ξ).

As such, the Riesz multipliers Rjf(ξ) = −i ξj|ξ| f(ξ) defined earlier in this chapter can beviewed as a higher dimensional analogue of the Hilbert transform.

6.3 The Mikhlin Multiplier Theorem

In this section we prove a multiplier theorem which gives Lp-boundedness of a large classof multipliers. This particular theorem, and its proof technique, offers a segue into theLittlewood-Paley theory developed in the next chapter.

The main result is the following.

Theorem 6.7 (Mikhlin multiplier theorem). Let m : Rd \ {0} → C satisfy∣∣Dαξm(ξ)

∣∣ . 1

|ξ||α|(6.1)

uniformly in ξ for all 0 ≤ |α| ≤⌈d+1

2

⌉. Then f 7→ (m · f ) = m ∗ f is bounded on Lp(Rd) for

1 < p <∞.

To prove this theorem, it will useful to localize at different frequencies in the Fourierdomain. With this in mind, we introduce a particular family of bump functions. Let ϕ ∈C∞(Rd) be a smooth bump function satisfying

ϕ(x) =

{1 |x| ≤ 10 |x| ≥ 11

10

.

Let ψ(x) = ϕ(x)− ϕ(2x). Then

ψ(x) =

0 |x| ≤ 1

21 11

20 ≤ |x| ≤ 10 |x| ≥ 11

10

.

57

For N ∈ 2Z, define ψN (x) = ψ(xN

). These annular bump functions which are ∼ 1 on a

scale of N will be used in the proof of the Theorem 6.7, and are central to Littlewood-Paleytheory.

Proof of Theorem 6.7. For p = 2, the desired conclusion follows from Plancharel. Indeed,

‖m ∗ f‖L2x

=∥∥∥m · f∥∥∥

L2ξ

≤ ‖m‖L∞ξ∥∥∥f∥∥∥

L2ξ

.

By choosing α = 0 in (6.1) it follows that m is uniformly bounded, so ‖m‖L∞ξ < ∞. Thus,

‖m ∗ f‖L2x.∥∥∥f∥∥∥

L2ξ

.

With this (2, 2) estimate, by Proposition 6.4, to prove boundedness on Lp for 1 < p <∞it suffices to prove that K = m satisfies the smoothness condition of Calderon-Zygmundconvolution kernels. Before doing this with the hypotheses of the theorem, we prove aslightly easier case. In particular, suppose that (6.1) holds for all |α| ≤ d+ 2 rather than all|α| ≤

⌈d+1

2

⌉. In this case, we will show that |∇K(x)| . 1

|x|d+1 . By previous remarks, thisimplies that K satisfies the desired smoothness condition. Towards this goal, note that∑

N∈2Z

ψN (x) = 1

for almost all x ∈ Rd. Write

m(ξ) =∑N∈2Z

m(ξ)ψN (ξ) =:∑N∈2Z

mN (ξ).

Then |∇K(x)| ≤∑

N∈2Z |∇mN (x)|. By the properties of the Fourier transform,

‖xα∇mN (x)‖L∞x .∥∥Dα

ξ ξmN (ξ)∥∥L1ξ

.

By the product rule,

Dαξ (ξmN (ξ)) = Dα

ξ (ξm(ξ)ψN (ξ)) =∑

α1+α2=α

Cα1,α2 Dα1ξ (ξm(ξ))Dα2

ξ (ψN (ξ)).

By assumption, |Dα1ξ m(ξ)| . |ξ|−|α1|. Therefore, |Dα1

ξ (ξm(ξ))| . |ξ|1−|α1|, again by theproduct rule. Next, by the chain rule,

|Dα2ξ ψN (ξ)| =

∣∣∣∣Dα2ξ

(ψ

(ξ

N

))∣∣∣∣ = N−|α2|∣∣∣∣(Dα2

ξ ψ)

(ξ

N

)∣∣∣∣ .Thus,

|Dαξ ξmN (ξ)| .

∑α1+α2=α

|ξ|1−|α1|N−|α2|∣∣∣∣(Dα2

ξ ψ)

(ξ

N

)∣∣∣∣ .Since ψN and all of its derivatives are supported on an annulus of radius comparable to N ,

‖xα∇mN (x)‖L∞x .∥∥Dα

ξ ξmN (ξ)∥∥L1ξ

.∑

α1+α2=α

∫|ξ|∼N

|ξ|1−|α1|N−|α2| dξ

.∑

α1+α2=α

N1+d−|α1|N−|α2|

. N1+d−|α|.

58

By choosing α = 0 and |α| = d+ 2, it then follows that

|∇mN (x)| . min

{Nd+1,

1

N |x|d+2

}so that

|∇K(x)| ≤∑N∈2Z

|∇mN (x)| .∑N∈2Z

min

{Nd+1,

1

N |x|d+2

}.

Splitting this sum over small and large frequencies, chosen appropriately,

|∇K(x)| .∑N≤ 1

|x|

Nd+1 +∑N> 1

|x|

1

N |x|d+2.

Since both of these sums are over dyadic numbers, the first sum is bounded (up to a con-stant) by its largest term and the second sum is bounded (up to a constant) by its smallestterm. Thus,

|∇K(x)| . 1

|x|d+1+

11|x| |x|d+2

.1

|x|d+1.

By our initial remark, we are done.Next, we return to the original assumption that |α| ≤

⌈d+1

2

⌉. The proof is similar, except

that we perform the computation in L2(Rd) instead of L∞(Rd). Using Plancharel and thenproceeding like before, we have:

‖(−2πix)αmN (x)‖L2x

=∥∥Dα

ξmN (ξ)∥∥L2ξ

.∑

α1+α2=α

∥∥∥∥Dα1ξ m(ξ) ·Dα2

ξ

(ψ

(ξ

N

))∥∥∥∥L2ξ

.∑

α1+α2=α

∥∥∥∥ 1

|ξ||α1|·N−|α2|

∥∥∥∥L2ξ(|ξ|∼N)

=∑

α1+α2=α

N−|α2|

(∫|ξ|∼N

|ξ|−2|α1| dξ

) 12

.∑

α1+α2=α

N−|α2|(N−2|α1|+d

) 12

= N−|α|+d2 .

In particular, by choosing α = 0 we have ‖mN (x)‖L2x. N

d2 . Thus, for a fixed A > 0,

Holder’s inequality gives∫|x|≤A

|mN (x)| dx . ‖mN (x)‖L2

∥∥χ|x|≤A∥∥L2 . Nd2A

d2

and ∫|x|>A

|mN (x)| dx .∥∥∥|x||α|mN (x)

∥∥∥L2

∥∥∥|x|−|α|χ|x|>A∥∥∥L2.

By the above computation,∥∥|x||α|mN (x)

∥∥L2 . N−|α|+

d2 . Computing the other norm,

∥∥∥|x|−|α|χ|x|>A∥∥∥L2

=

(∫|x|>A

|x|−2|α| dx

) 12

. A−|α|+d2

provided |α| ≤⌈d+1

2

⌉. Combining these two estimates gives∫

|x|>A|mN (x)| dx . (NA)−d

d+12 e+ d

2 .

59

In particular, choosing A = 1/N , we have∫|mN (x)| dx . 1

uniformly in N ∈ 2Z. Essentially the same computation gives∫|∇mN (x)| dx . N

where the implicit constant is independent of N . The only difference in the calculation isthat we begin by estimating

∥∥(−2πix)α ∂∂xmN (x)

∥∥L2x∼∥∥∥Dα

ξ ξmN (ξ)∥∥∥L2ξ

and consequently

pick up an extra power of N throughout. Thus,∫|x|≥2|y|

|K(x+ y)−K(x)| dx ≤∑N∈2Z

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx

=∑N≤ 1

|y|

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx

+∑N> 1

|y|

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx.

In the first sum, over low frequencies, we can use the fundamental theorem of calculus toget ∑

N≤ 1|y|

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx ≤∑N≤ 1

|y|

∫|x|≥2|y|

|y|∫ 1

0|∇mN (x+ θy)| dθ dx.

Using Fubini’s theorem and integrating over Rd instead of |x| ≥ 2|y| gives∑N≤ 1

|y|

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx ≤∑N≤ 1

|y|

|y| ‖∇mN‖L1 .∑N≤ 1

|y|

|y|N . 1

where the last inequality follows from the usual dyadic sum argument. For the secondsum, we have the crude estimate∑

N> 1|y|

∫|x|≥2|y|

|mN (x+ y)− mN (x)| dx ≤∑N> 1

|y|

∫|x|≥2|y|

|mN (x+ y)|+ |mN (x)| dx

.∑N> 1

|y|

∫|x|≥|y|

|mN (x)| dx

.∑N> 1

|y|

(N |y|)−dd+1

2 e+ d2 .

Here, we have used our previous estimate with A = |y|. Since −⌈d+1

2

⌉+ d

2 < 0 and thesum ranges over dyadic numbers greater than 1

|y| , we have

∑N> 1

|y|

(N |y|)−dd+1

2 e+ d2 .

(1

|y||y|)−d d+1

2 e+ d2

= 1.

Therefore, ∫|x|≥2|y|

|K(x+ y)−K(x)| dx . 1 + 1 . 1

and we are done.

60

Chapter 7

Littlewood-Paley Theory

In this chapter, we develop the theory of Littlewood-Paley projections and discuss a few oftheir immediate applications. More applications will become apparent in various contextsthroughout the rest of this text.

The philosophy behind Littlewood-Paley theory is to decompose functions by localiz-ing at different frequencies in the Fourier domain. As we shall see, this turns out to bea powerful idea. One source of motivation is the desire for a notion of orthogonality inLp spaces for p 6= 2. Indeed, suppose that we can write f ∈ L2 as f =

∑j fj , where the

functions f have disjoint support in the frequency domain. Then by Plancharel,

‖f‖2L2 =

∥∥∥∥∥∥∑j

fj

∥∥∥∥∥∥2

L2

=

∥∥∥∥∥∥∑j

fj

∥∥∥∥∥∥2

L2

=∑j

∥∥∥fj∥∥∥2

L2=∑j

‖fj‖2L2 .

The calculation shows that the Fourier transform has some nice orthogonality propertiesin L2. Unfortunately, this is not strictly true in other Lp spaces. Littlewood-Paley theorygives a way of understanding the orthogonality properties of the Fourier transform in suchspaces.

Some references for the following material are [2] and [3].

7.1 Littlewood-Paley Projections

We begin by defining the Littlewood-Paley projection operators. Throughout the rest ofthis chapter, we will use the following bump functions: let ϕ be a smooth function (whichwe view on the frequency domain) such that

ϕ(ξ) =

{1 |ξ| ≤ 10 |ξ| ≥ 11

10

.

Let ψ(ξ) = ϕ(ξ)− ϕ(2ξ). Then

ψ(ξ) =

0 |ξ| ≤ 1

21 11

20 ≤ |ξ| ≤ 10 |ξ| ≥ 11

10

.

For N ∈ 2Z, define ψN (ξ) := ψ(ξN

).

Definition 7.1. The Littlewood-Paley projection onto frequencies |ξ| ∼ N , denoted PN ,is defined in the Fourier domain for Schwartz functions f via

PNf(ξ) = ψN (ξ)f(ξ).

Equivalently,PNf = f ∗Ndψ(N ·).

61

In words, the operator PN localizes the function f to frequencies of order N . Note thatPN is not a true projection operator, despite its name, since P 2

N 6= PN .Littlewood-Paley projections can naturally be defined on wider ranges of frequencies.

Definition 7.2. The Littlewood-Paley projection onto low frequencies, denoted P≤N , isdefined in the Fourier domain for Schwartz functions f via

P≤Nf(ξ) = ϕ

(ξ

N

)f(ξ).

Equivalently,P≤Nf = f ∗Ndϕ(N ·).

The Littlewood-Paley projection onto high frequencies is P>N := I − P≤N , and theLittlewood-Paley projection onto medium frequencies is PM≤≤N :=

∑M≤K≤N PK .

We often use the shorthand notation fN , f≤N , f>N , and fM≤≤N for PNf , P≤Nf , P>Nf ,and PM≤≤Nf , respectively.

The following proposition contains basic properties of the Littlewood-Paley projectionsthat will be used frequently.

Proposition 7.3. Fix N ∈ 2Z.

1. The operators PN and P≤N are bounded on Lp for 1 ≤ p ≤ ∞, i.e., ‖fN‖Lp + ‖f≤N‖Lp .‖f‖Lp .

2. We have the pointwise estimate |fN (x)|+ |f≤N (x)| . (Mf)(x).

3. For f ∈ Lp with 1 < p <∞,∑

N∈2Z PNfLp−→ f .

4. (Bernstein inequality I) ‖fN‖Lq . Ndp− dq ‖fN‖Lp for 1 ≤ p ≤ q ≤ ∞.

5. (Bernstein inequality II) ‖|∇|sfN‖Lp ∼ N s ‖fN‖Lp for 1 ≤ p ≤ ∞ and s ∈ R.

Proof.

1. We compute:

‖fN‖Lp =∥∥∥f ∗Ndψ(N ·)

∥∥∥Lp

. ‖f‖Lp∥∥∥Ndψ(N ·)

∥∥∥L1.

By construction, Ndψ(N ·) is L1-normalized, so that∥∥Ndψ(N ·)

∥∥L1 =

∥∥ψ∥∥L1 . Thus,

‖fN‖Lp . ‖f‖Lp∥∥ψ∥∥

L1 .

The same computation works for ‖f≤N‖Lp .

2. We show the estimate for |fN (x)|, and as before, the same proof works for |f≤N (x)|.We have

fN (x) =(f ∗Ndψ(N ·)

)(x) = Nd

∫f(x− y)ψ(Ny) dy.

Because ψ is Schwartz, ψ is Schwartz, hence is bounded and decays up to any order.Thus,

|fN (x)| . Nd

∫|f(x− y)|〈N |y|〉100d

dy

62

where we use the Japanese angle bracket notation 〈x〉 := (1 + |x|2)12 . When |y| is small,

we exploit the boundedness of the angle bracket, and when |y| is large we exploit itsdecay. Explicitly,

|fN (x)| . Nd

∫|y|≤ 1

N

|f(x− y)| dy +Nd

∫|y|> 1

N

|f(x− y)|(N |y|)100d

dy.

Because |B(0, 1/N)| ∼ N−d so that 1/|B(0, 1/N)| ∼ Nd, we have by definition of themaximal function

Nd

∫|y|≤ 1

N

|f(x− y)| dy . (Mf)(x).

To estimate the second integral, we sum over dyadic annuli.

Nd

∫|y|> 1

N

|f(x− y)|(N |y|)100d

dy . Nd∑

M∈2Z;M> 1N

∫M≤|y|≤2M

|f(x− y)|(NM)100d

dy

. Nd∑

M∈2Z;M> 1N

(NM)−100d Md

|B(0, 2M)|

∫B(0,2M)

|f(x− y)| dy

.∑

M∈2Z;M> 1N

(NM)−99d(Mf)(x).

By the usual dyadic sum argument,∑M∈2Z;M> 1

N

(NM)−99d(Mf)(x) .

(1

MM

)−99d

(Mf)(x) = (Mf)(x)

and we are done.

3. By the density of S(Rd) ⊆ Lp(Rd) and by property (1), it suffices by the usual approx-imation argument to prove the claim for Schwartz functions. Fix f ∈ S(Rd).

For p = 2, by Plancharel we have

‖f − fN≤≤M‖L2 =

∥∥∥∥∥∥f1−

∑N≤K≤M

ψK

∥∥∥∥∥∥L2

≤

∥∥∥∥∥f ∑K<N

ψK

∥∥∥∥∥L2

+

∥∥∥∥∥f ∑K>M

ψK

∥∥∥∥∥L2

.

Since∑

K<N ψK ≤ χB(0,N) and∑

K>M ψK ≤ χ|ξ|≥M , by the dominated convergencetheorem, the left hands side tends to 0 as N → 0, and the right hand side tends to 0as M →∞.

For 1 < p < 2, we use interpolation. Choose θ so that 1p = θ

1 + 1−θ2 . Then

‖f − fN≤≤M‖Lp ≤ ‖f − fN≤≤M‖θL1 ‖f − fN≤≤M‖1−θL2 .

By (1),‖f − fN≤≤M‖θL1 ≤

(‖f‖L1 + ‖fN≤≤M‖L1

)θ. ‖f‖θL1 .

Our previous computation shows that ‖f − fN≤≤M‖1−θL2 → 0 and thus

‖f − fN≤≤M‖Lp → 0.

For 2 < p <∞, the same trick works:

‖f − fN≤≤M‖Lp ≤ ‖f − fN≤≤M‖1− 2

p

L1 ‖f − fN≤≤M‖2p

L2 .

The left hand side is bounded by (1), and the right hand side → 0 by the previouscalculation.

63

4. By Young’s convolution inequality, for r satisfying 1p + 1

r = 1q + 1,

‖fN‖Lq =∥∥∥f ∗Ndψ(N ·)

∥∥∥Lq

. ‖f‖Lp∥∥∥Ndψ(N ·)

∥∥∥Lr

= ‖f‖Lp Nd− d

r

∥∥ψ∥∥Lr

. Ndp− dq ‖f‖Lp .

This is great but this isn’t the estimate we need. To recover fN on the right, we usefattened Littlewood-Paley projections PN := PN

2≤≤2N . Note that PNPN = PN . Also,

we have

PNf(ξ) =

∑N2≤K≤2N

ψK

(ξ)f(ξ)

so that

PNf = f ∗

∑N2≤K≤2N

ψK

ˇ

= f ∗∑

N2≤K≤2N

Kdψ(K·) ∼ f ∗Nd∑

N2≤K≤2N

ψ(K·)

= f ∗Nd ˇψ(N ·)

for an appropriately defined ψ. 1 Thus,

‖fN‖Lq =∥∥∥PNfN∥∥∥

Lq∼∥∥∥fN ∗Nd ˇ

ψ(N ·)∥∥∥Lq

. ‖fN‖Lp∥∥∥Nd ˇ

ψ(N ·)∥∥∥Lr

= ‖fN‖Lp Nd− d

r

∥∥∥ ˇψ∥∥∥Lr

. Ndp− dq ‖fN‖Lp .

5. Fix s ∈ R. By definition, we have:

|∇|sfN (ξ) ∼ |ξ|sψN (ξ)f(ξ) = N s

[(|ξ|N

)sψ

(ξ

N

)]f(ξ).

Since the support of ψ is localized away from the origin, ρ(ξ) := |ξ|sψ(ξ) ∈ C∞c (Rd \{0}) for any value of s. So |∇|sfN (ξ) ∼ N sρ

(ξN

)f(ξ), hence

|∇|sfN = N s[f ∗Ndρ(N ·)

].

Thus,

‖|∇|sfN‖Lp . N s ‖f‖Lp∥∥∥Ndρ(N ·)

∥∥∥L1

= N s ‖f‖Lp ‖ρ‖L1 . N s ‖f‖Lp .

Via the same fattened Littlewood-Paley projections technique, we get the estimate‖|∇|sfN‖Lp . N s ‖fN‖Lp .

To get the reverse inequality, observe that

fN (ξ) = |ξ|−s|ξ|sfN (ξ) ∼ (|∇|−s|∇|sfN )ˆ(ξ).

Thus, applying the inequality we already have to |∇|−s|∇|sfN gives

‖fN‖Lp ∼∥∥|∇|−s|∇|sfN∥∥Lp . N−s ‖|∇|sfN‖Lp

so that N s ‖fN‖Lp . ‖|∇|sfN‖Lp as desired.

1Note that, since we are summing over dyadic numbers,∑

N2≤K≤2N ψ(K·) = ψ

(N2·)

+ ψ (N ·) + ψ (2N ·).

64

We remark that (3) does not hold for p = 1 or p = ∞. Note that∫fN (x) dx = fN (0) =

0 for any N ∈ 2Z, so∫fN≤≤M (x) dx = 0. But

∑PNf

L1

−→ f implies that their meansconverge. The claim fails for p = 1 by choosing an f ∈ L1 with

∫f(x) dx 6= 0. To see why

the claim fails for p = ∞, note that fN≤≤M ∈ C∞. Since convergence in L∞ is uniformconvergence, lim

∑PNf is continuous.

We also remark that the same proof technique in (4) gives the estimate ‖f≤N‖Lq .

Ndp− dq ‖f≤N‖Lp .

Though (3) does not hold for p = 1, we do have the following result.

Proposition 7.4. Let f ∈ L1(Rd) with∫Rd f(x) dx = 0. Then

∑PNf

L1

−→ f .

Proof. First, we claim that f can be approximated by a smooth function with compact sup-port and mean 0. Indeed, fix ε > 0. Then since f ∈ L1, there exists an R > 0 such that∫|x|>R |f(x)| dx < ε. Then∫

f(x)χB(0,R)(x) +O(ε)χR≤|x|≤2R(x) dx = 0

for some appropriately chosen constant O(ε). Convolving with a smooth approximationto the identity the desired approximation.

Thus, we can assume that f ∈ C∞c (Rd) and∫f(x) dx = 0. By the triangle inequality,∥∥∥∥∥∥f −

∑N≤K≤M

PNf

∥∥∥∥∥∥L1

≤

∥∥∥∥∥∑K<N

PKf

∥∥∥∥∥L1

+∑K>M

‖PKf‖L1 .

First, consider the high frequencies. By Bernstein’s inequality,∑K>M

‖PKf‖L1 .∑K>M

K−1 ‖|∇|fK‖L1 .

Since f ∈ C∞c (Rd) and the sum is a geometric sum over dyadic numbers,∑K>M

‖PKf‖L1 .f M−1

which → 0 as M → ∞. Next, consider the low frequencies. We exploit the fact thatconvolving with a mean 0 function is like taking a derivative, in the following sense: wehave

(P≤f)(x) =

∫f(y)Ndϕ(N(x− y)) dy = Nd

∫f(y) [ϕ(N(x− y))− ϕ(Nx)] dy

since f has mean 0. By the fundamental theorem of calculus,

ϕ(N(x− y))− ϕ(Nx) = Ny

∫ 1

0∇ϕ(Nx− θNy) dθ.

Using the fact that the support of f is contained in B(0, R), along with the fact that ϕ isSchwartz and hence decays as fast as we need,

‖P≤Nf‖L1 . Nd+1

∫ ∫|y||f(y)|

∫ 1

0|∇ϕ(Nx− θNy)| dθ dy dx

. Nd+1R

∫ ∫|f(y)| 1

〈Nx〉100ddy dx.

65

Here we have also used the fact that NR << 1 for N << 1, so we don’t need to include itin the Japanese angle bracket. So

‖P≤Nf‖L1 . Nd+1R ‖f‖L1

∫1

〈Nx〉100ddx . Nd+1R ‖f‖L1 N

−d

which→ 0 as N → 0.

Broadly speaking, low frequencies of a function contribute to smoothness, whereashigh frequencies contribute to irregularity. This is made somewhat precise by the followingproposition.

Proposition 7.5. Let f ∈ L∞(Rd) and fix 0 < α < 1. Then f is α-Holder continuous if and onlyif ‖P≥Nf‖L∞ . N−α for all N ≥ 1.

Proof. Suppose first that f is α-Holder continuous. Then for all x, y ∈ Rd with y 6= 0,|f(x− y)− f(x)| . |y|α. Fix N ∈ 2Z such that N ≥ 1. By definition of the Littlewood-Paleyprojections,

|P≥Nf(x)| =

∣∣∣∣∣∣∑K≥N

PKf(x)

∣∣∣∣∣∣ ≤∑K≥N

|PKf(x)| =∑K≥N

∣∣∣(f ∗Kdψ(K·)(x)∣∣∣

=∑K≥N

∣∣∣∣Kd

∫Rdf(x− y)ψ(Ky) dy

∣∣∣∣ .Observe that

∫Rd ψ(y) dy = ˆψ(0) = ψ(0) = 0. Thus, ψ has mean zero, hence

Kd

∫Rdf(x)ψ(Ny) dy = 0.

Inserting this into the above expression gives

|P≥Nf(x)| ≤∑K≥N

∣∣∣∣Kd

∫Rdf(x− y)ψ(Ky) dy −Kd

∫Rdf(x)ψ(Ky) dy

∣∣∣∣=∑K≥N

∣∣∣∣Kd

∫Rd

[f(x− y)− f(x)] ψ(Ky) dy

∣∣∣∣≤∑K≥N

Kd

∫Rd|f(x− y)− f(x)||ψ(Ky)| dy.

Since f is α-Holder continuous, we have

|P≥Nf(x)| .∑K≥N

Kd

∫Rd|y|α|ψ(Ky)| dy.

Making the change of variables z = Ky gives dy = K−d dz, hence

|P≥Nf(x)| .∑K≥N

Kd

∫Rd

∣∣∣ zK

∣∣∣α |ψ(z)|K−d dz =∑K≥N

K−α∫Rd|z|α |ψ(z)| dz.

Since ψ is a smooth function with compact support on an annulus of radius ∼ 1, ψ isSchwartz, and hence ψ is Schwartz. Thus, ψ decays up to any polynomial order, and|z|α|ψ(z)| is bounded. Therefore,∫

Rd|z|α |ψ(z)| dz .

∫|z|≤1

|z|α |ψ(z)| dz +

∫|z|>1

|z|α 1

|z|100ddz . 1 + 1 = 1.

66

This gives|P≥Nf(x)| .

∑K≥N

K−α . N−α

This bound is uniform in x. Thus, ‖P≥Nf‖L∞ . N−α for all dyadic N ≥ 1.

Next, we show the converse statement.Suppose that ‖P≥Nf‖L∞ . N−α for all dyadic N ≥ 1. We need to show that |f(x−y)−

f(x)| . |y|α for all x ∈ Rd and y 6= 0 ∈ Rd. Because f ∈ L∞(Rd) and hence |f(x − y) −f(x)| ≤ 2 ‖f‖L∞ . 1, it suffices to consider 0 < |y| < 1.

So fix x, y ∈ Rd with 0 < |y| < 1. Decompose |f(x− y)− f(x)| as

|f(x− y)− f(x)| ≤ |f<1(x− y)− f<1(x)|+ |f≥1(x− y)− f≥1(x)|.

We estimate these terms separately. First, consider the projection of f onto frequencies< 1.By definition of the Littlewood-Paley projection P<1,

|f<1(x− y)− f<1(x)| = |(f ∗ ϕ)(x− y)− (f ∗ ϕ)(x)|

=

∣∣∣∣∫ f(x− y − z)ϕ(z) dz −∫f(x− z)ϕ(z) dz

∣∣∣∣ .Changing variables in the first integral via y + z 7→ z gives

|f<1(x− y)− f<1(x)| =∣∣∣∣∫ f(x− z)ϕ(z − y) dz −

∫f(x− z)ϕ(z) dz

∣∣∣∣≤∫|f(x− z)| |ϕ(z − y)− ϕ(z)| dz.

By the fundamental theorem of calculus,

|f<1(x− y)− f<1(x)| ≤∫|f(x− z)||y|

∫ 1

0|∇ϕ(z − θy)| dθ dz

≤ ‖f‖∞ |y|∫ ∫ 1

0|∇ϕ(z − θy)| dθ dz.

Because ϕ is Schwartz, ∇ϕ is Schwartz, hence is bounded and decays up to any order.Consequently, ∫ ∫ 1

0|∇ϕ(z − θy)| dθ dz .

∫1

〈z〉100ddz . 1.

Thus, |f<1(x− y)− f<1(x)| . |y|. Since |y| < 1, |y|α ≥ |y|, so |f<1(x− y)− f<1(x)| . |y|α.Next we consider |f≥1(x− y)− f≥1(x)|. For any N ≥ 1, by assumption,

|f≥N (x− y)− f≥N (x)| ≤ |f≥N (x− y)|+ |f≥N (x)| ≤ 2 ‖P≥Nf‖L∞ . N−α. (7.1)

On the other hand, for any K ≥ 1 we compute as above with fattened Littlewood-Paleyprojections to get

|fK(x− y)− fK(x)| = |PKfK(x− y)− PKfK(x)|

=∣∣∣(fK ∗Kd ˇ

ψ(K·))(x− y)− (fK ∗Kd ˇψ(K·))(x)

∣∣∣=

∣∣∣∣Kd

∫RdfK(x− y − z) ˇ

ψ(Kz) dz −Kd

∫RdfK(x− z) ˇ

ψ(Kz) dz

∣∣∣∣=

∣∣∣∣Kd

∫RdfK(x− z) ˇ

ψ(K(z − y)) dz −Kd

∫RdfK(x− z) ˇ

ψ(Kz) dz

∣∣∣∣=

∣∣∣∣Kd

∫Rd

[ˇψ(K(z − y))− ˇ

ψ(Kz)]fK(x− z) dz

∣∣∣∣≤ Kd ‖fK‖L∞

∫Rd

∣∣∣ ˇψ(Kz −Ky)− ˇψ(Kz)

∣∣∣ dz.67

Changing variables via Kz 7→ z and hence dz 7→ K−d dz gives

|fK(x− y)− fK(x)| ≤ ‖fK‖L∞∫Rd

∣∣∣ ˇψ(z −Ky)− ˇψ(z)

∣∣∣ dz.Since fK = f≥K − f≥2K ,

‖fK‖L∞ ≤ ‖f≥K‖L∞ + ‖f≥2K‖L∞ . K−α + (2K)−α . K−α.

Then by this fact and by the fundamental theorem of calculus,

|fK(x− y)− fK(x)| ≤ K−αK|y|∫ ∫ 1

0

∣∣∣∇ ˇψ(z − θNy

∣∣∣ dθ dz.Since ψ is Schwartz, the same argument as above gives

∫ ∫ 10

∣∣∣∇ ˇψ(z − θNy

∣∣∣ dθ dz . 1. Thus,for any K ≥ 1 we have the estimate

|fK(x− y)− fK(x)| . K1−α|y|. (7.2)

So for any dyadic N ≥ 1, by (7.1) and (7.2),

|f≥1(x− y)− f≥1(x)| ≤∑

1≤K<N|fK(x− y)− fK(x)| + |f≥N (x− y)− f≥N (x)|

.∑

1≤K<NK1−α|y| + N−α.

Since the sum is over dyadic K,∑

1≤K<N K1−α|y| . N1−α|y|. So for any dyadic N ≥ 1,

|f≥1(x− y)− f≥1(x)| . N1−α|y|+N−α.

Choose N ≥ 1 such that1

|y|≤ N ≤ 2

1

|y|.

Since |y| < 1, this is certainly possible. Then N ∼ |y|−1, and thus

|f≥1(x− y)− f≥1(x)| . (|y|−1)1−α|y|+ (|y|−1)−α = |y|α−1|y|+ |y|α

. |y|α.

Putting the two estimates together yields

|f(x− y)− f(x)| ≤ |f<1(x− y)− f<1(x)|+ |f≥1(x− y)− f≥1(x)| . |y|α + |y|α

. |y|α

so that f is α-Holder continuous as desired.

7.2 The Littlewood-Paley Square Function

In this section, we introduce the Littlewood-Paley square function, and the key fact that itsLp-norm is comparable to the Lp-norm of the input function.

Definition 7.6. For f ∈ S(Rd), the Littlewood-Paley square function of f is given by

(Sf)(x) =

∑N∈2Z

|fN (x)|2 1

2

.

68

To better estimate the size of the square function, we borrow an inequality from proba-bility theory.

Lemma 7.7 (Kinchin’s Inequality). Let Xn be independent identically distributed random vari-ables such that Xn = ±1 with equal probability. Then for any 0 λ) dλ

λ.

We have

P(∣∣∣∑ cnXN

∣∣∣ > λ)≤ P

(∑cnXN > λ

)+ P

(∑cnXN < −λ

)= 2P

(∑cnXN > λ

)because the Xn are i.i.d with Xn = ±1 with equal probability. Next, recall the exponentialChebychev’s inequality: for a random variable X and t > 0,

P (X ≥ λ) ≤ e−tλ E(etX).

Thus, for t > 0,

P(∣∣∣∑ cnXN

∣∣∣ > λ)≤ 2e−λt E

(et∑cnXn

).

Because the Xn are independent, this gives

P(∣∣∣∑ cnXN

∣∣∣ > λ)≤ 2e−λt

∏E(etcnXn

)= 2e−λt

∏(1

2etcn +

1

2e−tcn

)= 2e−λt

∏cosh(tcn).

Recall that cosh(x) ≤ ex2

2 ; one quick way to see this is by comparing Taylor series. So

P(∣∣∣∑ cnXN

∣∣∣ > λ)≤ 2e−λt

∏e

(tcn)2

2 = 2e−λtet2

2

∑c2n .

Choosing t such that λt = t2∑c2n, hence t = λ∑

c2n, gives

P(∣∣∣∑ cnXN

∣∣∣ > λ)≤ 2e−λte

λt2 = 2e−

λt2 = 2e

− λ2

2∑c2n .

So

E∣∣∣∑ cnXN

∣∣∣p .p

∫ ∞0

λpe− λ2

2∑c2ndλ

λ.

Making the changes of variables z = λ√∑c2n

yields

E∣∣∣∑ cnXN

∣∣∣p . (∑ c2n

) p2

∫ ∞0

zpe−z2

2dz

z.(∑

c2n

) p2.

This gives the . direction of the statement.Next, we need the & direction. We first consider the case 1 ≤ p < ∞. Note that∑|cn|2 = E |

∑cnXn|2. This is because the Xn are independent, and so

E(XnXm) = E(Xn)E(Xm) = 0.

69

Then by Holder, and by the above inequality,

∑|cn|2 ≤

(E∣∣∣∑ cnXn

∣∣∣p) 1p

(E∣∣∣∑ cnXn

∣∣∣p′) 1p′

.(E∣∣∣∑ cnXn

∣∣∣p) 1p(∑

c2n

) 12.

This gives (∑c2n

) p2.(E∣∣∣∑ cnXn

∣∣∣p) 1p

as desired.For 0 < p < 1, we use Cauchy-Schwarz:∑

|cn|2 = E∣∣∣∑ cnXn

∣∣∣2 = E(∣∣∣∑ cnXn

∣∣∣ p2 ∣∣∣∑ cnXn

∣∣∣2− p2)≤(E∣∣∣∑ cnXn

∣∣∣p) 12

(E∣∣∣∑ cnXn

∣∣∣4−p) 12

.

Again by the first demonstrated inequality we have∑|cn|2 .

(E∣∣∣∑ cnXn

∣∣∣p) 12(∑

|cn|2) 4−p

4 ⇒(∑

|cn|2) p

4.(E∣∣∣∑ cnXn

∣∣∣p) 12.

Raising both sides to the power 2p gives the desired result.

The main result of this section is the following theorem.

Theorem 7.8 (Square function estimate). For 1 < p <∞, we have ‖Sf‖Lp ∼ ‖f‖Lp .

Proof. We will first show ‖Sf‖Lp . ‖f‖Lp . As a remark, the proof of this direction does notrely on the specific multiplier ψ defining P1, so this inequality holds in more generality. Inparticular, ψ can be replaced by any element of C∞c (Rd \ {0}).

Let m(ξ) :=∑

N∈2Z ψN (ξ)XN , where the XN are i.i.d random variables such XN = ±1with equal probability. Then ‖m‖L∞ . 1, because for any given ξ, only finitely many of thesummands give any nonzero contribution due to the compact supports of the φN . Also,

|Dαξm(ξ)| ≤

∑N∈2Z

N−|α|∣∣∣∣Dα

ξ ψ

(ξ

N

)∣∣∣∣ .Because ψN is smooth and has support |ξ| ∼ N , and because only finitely summandscontribute,

|Dαξm(ξ)| . |ξ|−|α|.

As this holds for any α ∈ Nd, the Mikhlin multiplier theorem gives ‖m ∗ f‖Lp . ‖f‖Lp forall 1 < p <∞. Note that m ∗ f =

∑N∈2Z fNXN . So by Kinchin’s inequality,

(Sf)(x) =

∑N∈2Z

|fN (x)|2 1

2

∼

E

∣∣∣∣∣∣∑N∈2Z

fN (x)XN

∣∣∣∣∣∣p

1p

= (E |m ∗ f |p)1p .

Using Fubini’s theorem,

‖Sf‖pLp ∼∫

E |m ∗ f |p (x) dx . E ‖m ∗ f‖pLp

Since the expectation of a constant is itself,

‖Sf‖pLp . E ‖m ∗ f‖pLp . E ‖f‖pLp = ‖f‖pLp .

70

Next, we show ‖f‖Lp . ‖Sf‖Lp using duality and the fattened Littlewood-Paley projec-tions. Recall that PNPN = PN . We have

‖f‖Lp = sup‖g‖

Lp′=1

〈f, g〉 = sup‖g‖

Lp′=1

⟨∑PNf, g

⟩= sup‖g‖

Lp′=1

⟨∑PNPNf, g

⟩= sup‖g‖

Lp′=1

∑⟨PNf, PNg

⟩since the operators PN are self-adjoint. By Cauchy-Schwarz and then Holder,

‖f‖Lp ≤ sup‖g‖

Lp′=1

⟨(∑|PNf |2

) 12,(∑

|PNg|2) 1

2

⟩≤ sup‖g‖

Lp′=1

‖Sf‖Lp∥∥∥∥(∑ |PNg|2

) 12

∥∥∥∥Lp′

.

By the remark at the beginning of the proof,∥∥∥∥(∑ |PNg|2) 1

2

∥∥∥∥Lp′

. ‖g‖Lp′ . Thus,

‖f‖Lp . ‖Sf‖Lp .

7.3 Applications to Fractional Derivatives

One of the first properties of Littlewood-Paley projections that we proved was Bernstein’sinequality, which says ‖|∇|sfN‖Lp ∼ N s ‖fN‖Lp . This should not be surprising, as differ-entiation of a function amount to multiplication by ξ in the frequency domain, and thefunction fN is localized to frequencies |ξ| ∼ N . Because the Littlewood-Paley projectionsinteract with differentiation so nicely, they naturally have applications in partial differen-tial equations. In this section, we prove a product rule and chain rule for the fractionalderivative operator |∇|s.

First, we provide a square function estimate involving the fractional derivative. Thiscaptures the notion that the derivative |∇|s can be expressed as a linear combination ofLittlewood-Paley projection operators.

Proposition 7.9. Let 1 0,

‖|∇|sf‖Lp ∼

∥∥∥∥∥∥∥∑N∈2Z

|N sf≥N |2 1

2

∥∥∥∥∥∥∥Lp

.

Proof.

1. First, consider the & inequality. We have∥∥∥∥∥∥∥∑N∈2Z

|N sfN |2 1

2

∥∥∥∥∥∥∥Lp

=

∥∥∥∥∥∥∥∑N∈2Z

|N s|∇|−s|∇|sfN |2 1

2

∥∥∥∥∥∥∥Lp

.

71

Also, |∇|sfN = |∇|sPNf = PN (|∇|sf), as these operators are given by multiplicationon the Fourier side. Thus,∥∥∥∥∥∥∥

∑N∈2Z

|N sfN |2 1

2

∥∥∥∥∥∥∥Lp

=

∥∥∥∥∥∥∥∑N∈2Z

|N s|∇|−sPN (|∇|sf)|2 1

2

∥∥∥∥∥∥∥Lp

.

Recall that in the proof of the Littlewood-Paley square function estimate from, weproved one direction in a greater generality; we showed that ‖Sf‖Lp . ‖f‖Lp whereS is defined via any ψ ∈ C∞c (Rd \ {0}). In particular, this holds for ψ(ξ) := |ξ|−sψ(ξ),where ψ is the usual Littlewood-Paley ψ. Because ψ ∈ C∞c (Rd \ {0}), ψ ∈ C∞c (Rd \{0}). Defining ψN (ξ) := N s|ξ|−sψ(ξ/N), the general square function estimate gives∥∥∥∥∥∥∥

∑N∈2Z

|N s|∇|−sPN (|∇|sf)|2 1

2

∥∥∥∥∥∥∥Lp

. ‖|∇|sf‖Lp .

Next, we show ., using duality and the fattened Littlewood-Paley projections.

‖|∇|sf‖Lp = sup‖g‖

Lp′=1〈|∇|sf, g〉 = sup

‖g‖Lp′=1

⟨∑PN (|∇|sf), g

⟩= sup‖g‖

Lp′=1

⟨∑|∇|sPNPNf, g

⟩= sup‖g‖

Lp′=1

∑⟨PNf, |∇|sPNg

⟩= sup‖g‖

Lp′=1

∑⟨N sPNf,N

−s|∇|sPNg⟩.

Applying Cauchy-Schwarz and then Holder,

‖|∇|sf‖Lp ≤ sup‖g‖

Lp′=1

⟨(∑|N2fN |2

) 12,(∑

|N−s|∇|sPNg|2) 1

2

⟩

≤ sup‖g‖

Lp′=1

∥∥∥∥(∑ |N sfN |2) 1

2

∥∥∥∥Lp

∥∥∥∥(∑ |N−s|∇|sPNg|2) 1

2

∥∥∥∥Lp′

.

By the same remark from above,∥∥∥∥(∑ |N−s|∇|sPNg|2) 1

2

∥∥∥∥Lp′

. ‖g‖Lp′ .

Thus, ‖|∇|sf‖Lp .∥∥∥(∑ |N sfN |2

) 12

∥∥∥Lp

.

2. Note that PN = P≥N − P≥2N . Thus,

‖|∇|sf‖Lp ∼∥∥∥∥(∑ |N sfN |2

) 12

∥∥∥∥Lp

≤∥∥∥∥(∑ |N sf≥N |2

) 12

∥∥∥∥Lp

+

∥∥∥∥(∑ |N sf≥2N |2) 1

2

∥∥∥∥Lp

where we have invoked the `2 triangle inequality followed by the Lp triangle inequal-ity. But up to a factor of 2,∥∥∥∥(∑ |N sf≥2N |2

) 12

∥∥∥∥Lp

.

∥∥∥∥(∑ |N sf≥N |2) 1

2

∥∥∥∥Lp.

72

So

‖|∇|sf‖Lp .∥∥∥∥(∑ |N sf≥N |2

) 12

∥∥∥∥Lp.

Next, we show &. We have∑|N sf≥N |2 =

∑N2s|f≥N |2 ≤

∑N∈2Z

N2s∑K≥N

|fK |2

≤∑N∈2Z

N2s

∑N1≥N

|fN1 |

∑N2≥N

|fN2 |

.

Multiplying the latter sums out, rearranging, and picking up a factor of 2 from sym-metry gives ∑

|N sf≥N |2 ≤ 2∑N∈2Z

N2s∑

N≤N1≤N2

|fN1 ||fN2 |.

Cleverly inserting constants yields∑|N sf≥N |2 ≤ 2

∑N∈2Z

N2s∑

N≤N1≤N2

1

N s1N

s2

|N s1fN1 ||N s

2fN2 |

= 2∑

N,N1,N2;N≤N1≤N2

N2s

N s1N

s2

|N s1fN1 ||N s

2fN2 |.

Freeze N1 and N2 and consider the sum over N . This is a dyadic sum, and by theusual argument we can bound this by plugging in the largest term, which is N1.Thus, ∑

|N sf≥N |2 .∑

N1≤N2

(N1

N2

)s|N s

1fN1 ||N s2fN2 |.

This is summable in both N1 and N2, since s > 0. By Schur’s test, it follows that

∑|N sf≥N |2 .

∑N2

|N s2fN2 |2

12∑

N1

|N s1fN1 |2

12

=∑|N sfN |2.

Taking the square root of both sides and then the Lp norm gives∥∥∥∥(∑ |N sf≥N |2) 1

2

∥∥∥∥Lp

.

∥∥∥∥(∑ |N sfN |2) 1

2

∥∥∥∥Lp.

Applying the result from the previous part gives the desired result.

Using these estimates, we now prove the product rule and chain rule for the operator|∇|s, both results due to Christ and Weinstein [1]. As |∇|s is a non-local operator, theserules are given as Lp norm estimates, rather than as pointwise facts.

Theorem 7.10 (Fractional product rule). Let 1 0. Then

‖|∇|s(fg)‖Lp . ‖|∇|sf‖Lp1 ‖g‖Lp2 + ‖|∇|sg‖Lq1 ‖f‖Lq2

where 1p = 1

p1+ 1

p2= 1

q1+ 1

q2.

73

Proof. The previous proposition gives ‖|∇|s(fg)‖Lp ∼∥∥∥(∑N2s|PN (fg)|2

) 12

∥∥∥Lp

. We per-form a paraproduct decomposition on fg:

fg = f≥N8g + f<N

8g = f≥N

8g + f<N

8g≥N

8+ f<N

8g<N

8.

ThenPN (fg) = PN (f≥N

8g) + PN (f<N

8g≥N

8) + PN (f<N

8g<N

8).

Note that f<N8

and g<N8

have frequency supports bounded by 2N8 = N4 . Thus, the maxi-

mum frequency attained by f<N8g<N

8is bounded by N

4 + N4 = N

2 . Thus, PN (f<N8g<N

8) = 0.

Thus, it remains to consider the first two terms.Using the fact that the Littlewood-Paley projections are bounded by the maximal func-

tion,|PN (fg)| .M(f≥N

8g) +M(f<N

8g≥N

8) .M(f≥N

8g) +M

[(Mf) g≥N

8

].

Therefore, the `2 and Lp triangle inequalities yield∥∥∥∥(∑N2s|PN (fg)|2) 1

2

∥∥∥∥Lp

.

∥∥∥∥∥(∑∣∣∣M(N sf≥N

8g)∣∣∣2) 1

2

∥∥∥∥∥Lp

+

∥∥∥∥∥(∑∣∣∣M [

(Mf)N sg≥N8

]∣∣∣2) 12

∥∥∥∥∥Lp

.

Next, apply the vector-valued Hardy-Littlewood maximal inequality to both norms:∥∥∥∥(∑N2s|PN (fg)|2) 1

2

∥∥∥∥Lp

.

∥∥∥∥∥(∑∣∣∣N sf≥N

8g∣∣∣2) 1

2

∥∥∥∥∥Lp

+

∥∥∥∥∥(∑∣∣∣(Mf)N sg≥N

8

∣∣∣2) 12

∥∥∥∥∥Lp

=

∥∥∥∥∥|g|(∑∣∣∣N sf≥N

8

∣∣∣2) 12

∥∥∥∥∥Lp

+

∥∥∥∥∥|Mf |(∑∣∣∣N sg≥N

8

∣∣∣2) 12

∥∥∥∥∥Lp

.

Applying Holder’s inequality,∥∥∥∥(∑N2s|PN (fg)|2) 1

2

∥∥∥∥Lp

. ‖g‖Lp2

∥∥∥∥∥(∑∣∣∣N sf≥N

8

∣∣∣2) 12

∥∥∥∥∥Lp1

+ ‖Mf‖Lq2

∥∥∥∥∥(∑∣∣∣N sg≥N

8

∣∣∣2) 12

∥∥∥∥∥Lq1

.

Since the maximal function is of type (q2, q2), ‖Mf‖Lq2 . ‖f‖Lq2 . Applying part 2 of theprevious proposition to the p1 and q1 norms yields

‖|∇|s(fg)‖Lp ∼∥∥∥∥(∑N2s|PN (fg)|2

) 12

∥∥∥∥Lp

. ‖g‖Lp2 ‖|∇|sf‖Lp1 + ‖f‖Lq2 ‖|∇|

sg‖Lq1 .

Theorem 7.11 (Fractional chain rule). Suppose F : C→ C satisfies

|F (u)− F (v)| . |u− v| · |G(u)−G(v)|

for all functions u, v : Rd → C and for some function G : C → [0,∞). Then for 1 < p < ∞ and0 < s < 1,

‖|∇|sF (u)‖Lp . ‖|∇|su‖Lp1 ‖G(u)‖Lp2

where 1p = 1

p1+ 1

p2and 1 < p2 ≤ ∞.

74

Proof. As before, the previous proposition yields

‖|∇|sF (u)‖Lp ∼∥∥∥∥(∑N2s|PNF (u)|2

) 12

∥∥∥∥Lp.

Observe that the function ψ is a mean zero function. Indeed,∫Rdψ(x) dx = ˆψ(0) = ψ(0) = 0.

We use this and the fact that convolving with a mean zero function is like differentiation asfollows:

PNF (u)(x) = (F (u) ∗Ndψ(N ·))(x) = Nd

∫NyF (u(x− y)) dy

= Nd

∫ψ(Ny) [F (u(x− y))− F (u(x))] dy

and so

|PNF (u)(x)| . Nd

∫|ψ(Ny)| |u(x− y)− u(x)| |G(u(x− y))−G(u(x)| dy.

Decomposing |u(x− y)− u(x)| based on where we expect cancellation, we have

|u(x− y)− u(x)| ≤ |u>N (x− y)|+ |u>N (x)|+∑K≤N

|uK(x− y)− uK(x)|.

We proceed by considering each of these terms separately. First, we claim that

|uK(x− y)− uK(x)| . K|y| |(MuK)(x− y) + (MuK)(x)| .

Indeed, if K|y| & 1, then

|uK(x− y)− uK(x)| ≤ |uK(x− y)|+ |uK(x)| . |(MuK)(x− y) + (MuK)(x)|. K|y| |(MuK)(x− y) + (MuK)(x)| .

If K|y| << 1, then

uK(x− y)− uK(x) = PK(uK(x− y)− uK(x))

= Kd

∫ˇψ(Ky) [uK(x− y − z)− uK(x− z)] dz

=

∫Kd[

ˇψ(K(z − y))− ˇ

ψ(Kz)]uK(x− z) dz

via the change of variables y + z 7→ z. By the fundamental theorem of calculus,

|uK(x− y)− uK(x)| ≤∫Kd

∫ 1

0K|y|

∣∣∇ψ(Kz − θKy)∣∣ dθ |uK(x− z)| dz.

Because ˇψ is Schwartz and thus is bounded and decays as quickly as we need, we have

|uK(x− y)− uK(x)| . K|y|∫Kd 1

〈Kz〉100d|uK(x− z)| dz.

In previous arguments, by considering z < 1/K and z > 1/K separately, we have seen that∫Kd 1

〈Kz〉100d|uK(x− z)| dz . (MuK)(x).

75

Our claim then follows from this fact.Next, we consider the contribution of |u>N (x− y)| to |PNF (u)(x)|. The contribution is

bounded by ∫Nd |ψ(Ny)| |u>N (x− y)| [G(u)(x− y) +G(u)(x)] dy.

By the same trick as above, use the quickly decaying and bounded nature of ψ to estimatethis integral by

M (u>N ·G(u)) (x)+G(u)(x)·M(u>N )(x) .M (u>N ·G(u)) (x)+M(G(u))(x)·M(u>N )(x).

Similarly, the contribution of |u>N (x)| to |PNF (u)(x)| is bounded by∫Nd |ψ(Ny)| |u>N (x)| [G(u)(x− y) +G(u)(x)] dy

. |u>N (x)| ·M(G(u))(x) + |u>N (x)| ·G(u)(x)

.M(u>N )(x) ·M(G(u))(x).

Thus, the contribution of |u>N (x− y)|+ |u>N (x)| to |PNF (u)(x)| is bounded by

M (u>N ·G(u)) (x) +M(G(u))(x) ·M(u>N )(x).

Estimate the contribution of M (u>N ·G(u)) (x) to ‖|∇|sF (u)‖Lp as follows:∥∥∥∥∥∥∥∑N∈2Z

N2s |M(u>NG(u))|2 1

2

∥∥∥∥∥∥∥Lp

=

∥∥∥∥∥∥∥∑N∈2Z

|M(N su>NG(u))|2 1

2

∥∥∥∥∥∥∥Lp

.

∥∥∥∥∥∥∥∑N∈2Z

|N su>NG(u)|2 1

2

∥∥∥∥∥∥∥Lp

using the vector-valued Hardy-Littlewood maximal inequality. Next, pull the G(u) out ofthe sum and apply Holder’s inequality followed by our derivative estimate from earlier:

‖G(u)‖Lp2

∥∥∥∥∥∥∥∑N∈2Z

|N su>N |2 1

2

∥∥∥∥∥∥∥Lp1

. ‖G(u)‖Lp2 ‖|∇|su‖Lp1 .

The contribution of M(G(u))(x) · M(u>N )(x) to ‖|∇|sF (u)‖Lp is almost the exact samecalculation.

Thus, it only remains to bound the contribution of the∑

K≤N |uK(x−y)−uK(x)| term.

76

The contribution to |PNF (u)| is∑K≤N

∫Nd|ψ(Ny)||uK(x− y)− uK(x)||G(u)(x− y)−G(u)(x)| dy

.∑K≤N

∫Nd|ψ(Ny)|K|y| |(MuK)(x− y) + (MuK)(x)| |G(u)(x− y)−G(u)(x)| dy

=∑K≤N

K

N

∫Nd|ψ(Ny)|N |y| |(MuK)(x− y) + (MuK)(x)| |G(u)(x− y)−G(u)(x)| dy

.∑K≤N

K

N[M(M(uK)G(u))(x) +M(M(uK))(x)G(u)(x)

+ (M(uK)M(G(u)))(x) + (M(uK)G(u))(x)]

.∑K≤N

K

N[M(M(uK)G(u))(x) +M(M(uK))(x)G(u)(x)]

=:∑K≤N

K

NCK .

Towards the goal of estimating the contribution of this term to ‖|∇|sF (u)‖Lp , we first com-pute:

∑N

N2s

∣∣∣∣∣∣∑K≤N

K

NCK

∣∣∣∣∣∣2

=∑N

∑K≤N

K

NCK

∑L≤N

L

NCL

≤ 2

∑N

N2s∑

K≤L≤N

KL

N2CKCL

where the factor of 2 comes from the symmetry of the summations. Inserting Ks and Ls

terms and then summing in N gives

= 2∑

N,K,L;K≤L≤NN2(s−1)K1−sL1−s(KsCK)(LsCL)

.∑K≤L

L2(s−1)K1−sL1−s(KsCK)(LsCL)

=∑K≤L

(K

L

)1−s(KsCK)(LsCL).

This is summable in both K and L. Hence, by Schur’s test,

∑N

N2s

∣∣∣∣∣∣∑K≤N

K

NCK

∣∣∣∣∣∣2

.

(∑K

(KsCK)2

) 12(∑

L

(LsCL)2

) 12

=∑K

K2sC2K .

Using this, we can estimate the contribution to ‖|∇|sF (u)‖Lp by∥∥∥∥∥∥(∑

K

K2s |M(M(uK)G(u))|2) 1

2

∥∥∥∥∥∥Lp

+

∥∥∥∥∥∥(∑

K

K2s |M(M(uK))M(G(u))|2) 1

2

∥∥∥∥∥∥Lp

.

The rest of the computation uses the vector valued maximal inequality in exactly the samemanner as above. To avoid unnecessary repetition, the details are left to the reader.

77

An example of such a function F which appears in applications in partial differentialequations is F (u) = |u|pu. Indeed, by the fundamental theorem of calculus,

|F (u)− F (v)| ≤ |u− v| · ||u|p + |v|p| .

Nonlinearities in equations such as the Schrodinger equation often take this form.Also, this fractional product rule is useful for the case s > 1, as we can typically write

derivative operators as an integer part plus a fractional part 0 < s < 1.

78

Chapter 8

Oscillatory Integrals

In this chapter, we develop the basic tools to study oscillatory integrals. The fundamentalexample of an oscillatory integral is the Fourier transform:∫

Rde−2πix·ξf(x) dx.

The exponential term in the above integral has unit modulus, but nontrivial phase oscilla-tion. The most naive way to estimate the size of a particular Fourier transform is to use theintegral triangle inequality:∣∣∣∣∫

Rde−2πix·ξf(x) dx

∣∣∣∣ ≤ ∫Rd|e−2πix·ξf(x)| dx =

∫Rd|f(x)| dx.

Though useful in some contexts, this estimate is crude because it ignores any oscillationcoming from the phase e−2πix·ξ. For example, if the function f is not integrable, estimatingin this way is useless. Oftentimes, taking advantage of phase oscillation and cancellationis a more effective way to bound the size of an integral like the Fourier transform.

More generally, we define two different notions of oscillatory integrals.

Definition 8.1. An oscillatory integral of the first kind is an integral of the form

I(λ) =

∫Rdeiλφ(x) ψ(x) dx

where φ : Rd → R, ψ : Rd → C, and λ > 0.

The main goal of this chapter is to understand the asymptotic behavior of these inte-grals as λ→∞. We will eventually see applications of this to various PDEs.

Oscillatory integrals of the first kind will be our only focus, but we provide the follow-ing definition for the sake of completeness.

Definition 8.2. An oscillatory integral of the second kind is an operator of the form

(Tλf)(x) =

∫Rdeiλφ(x,y)K(x, y) f(y) dy

where φ : Rd × Rd → R, K : Rd × Rd → C, f : Rd → C, and λ > 0.

In this case, it is desirable to understand the asymptotic behavior of ‖Tλ‖op as λ → ∞.We refer to [10] for more details on oscillatory integrals of the second kind, as well as theother material in this chapter.

79

8.1 Oscillatory Integrals of the First Kind, d = 1

We begin by studying oscillatory integrals of the first kind in one dimension. In this setting,the theory is completely understood and admits a simpler presentation. There are threemain techniques presented in this section: nonstationary phase, the van der Corput lemma,and stationary phase.

We begin with nonstationary phase. Roughly, the principle of nonstationary phase saysthat if the phase eiλφ(x) is nonstationary in the sense that φ′(x) 6= 0, then the oscillatoryintegral I(λ) is rapidly decaying. In other words, integrating a wildly varying phase leadsto cancellation, and hence decay. This should make sense from the perspective of addingsinusoidal waves: if the waves have similar phases, their sum amplifies, whereas summingwaves with different phases leads to cancellation in amplitude.

More directly, the principle of nonstationary phase uses the compact support of ψ to-gether with integration by parts to achieve decay up to any order.

Proposition 8.3 (Nonstationary phase). Let φ : R → R and ψ : R → C be smooth with ψcompactly supported in (a, b). Assume φ′(x) 6= 0 for all x ∈ [a, b]. Then∣∣∣∣∫ b

aeiλφ(x)ψ(x) dx

∣∣∣∣ .N λ−N (8.1)

for any N ≥ 0. Here the implicit constant does not depend on a and b.

Proof. As remarked, the proof of this proposition is simply repeated applications of inte-gration by parts. The details are as follows.

Because φ′(x) 6= 0, we have eiλφ(x) = 1iλφ′(x)

ddxe

iλφ(x). Let D := 1iλφ′(x)

ddx . Then eiλφ(x) =

DN(eiλφ(x)

)for any N ≥ 0. To integrate by parts with D, we need to compute its adjoint.

For any smooth f , normal integration by parts gives∫RDf(x)ψ(x) dx =

∫R

1

iλφ′(x)

df

dx(x)ψ(x) dx = −

∫Rf(x)

d

dx

(1

iλφ′(x)ψ(x)

)dx.

Here, the boundary terms vanish because of the compact support of ψ.Thus, the adjoint of D is given by

tDf(x) := − d

dx

[1

iλφ′(x)f(x)

].

We have ∫ b

aeiλφ(x)ψ(x) dx =

∫ b

aDN (eiλφ(x))ψ(x) dx =

∫ b

aeiλφ(x) (tD)Nψ(x) dx

=

∫ b

aeiλφ(x)

[− d

dx

1

iλφ′(x)

]N(ψ)(x) dx.

Therefore, ∣∣∣∣∫ b

aeiλφ(x)ψ(x) dx

∣∣∣∣ ≤ λ−N ∫ b

a

∣∣∣∣∣[d

dx

1

φ′(x)

]N(ψ)(x)

∣∣∣∣∣ dx.Inside this integral, the derivative can hit either the φ′(x) in the denominator or the ψ(x).So ∫ b

a

∣∣∣∣∣[d

dx

1

φ′(x)

]N(ψ)(x)

∣∣∣∣∣ dxis some finite constant which depends on N + 1 derivatives of φ(x) and N derivatives ofψ(x). Thus, ∣∣∣∣∫ b

aeiλφ(x)ψ(x) dx

∣∣∣∣ .N λ−N

80

for any N ≥ 0, where the implicit constant depends on N + 1 derivatives of φ(x) and Nderivatives of ψ(x).

If ψ does not have compact support, the best decay in λ we can expect is λ−1. Forexample, when ψ(x) = 1 and φ(x) = x,∣∣∣∣∫ b

aeiλx dx

∣∣∣∣ =

∣∣∣∣eiλb − eiλaiλ

∣∣∣∣ . λ−1.

Summarizing our analysis so far, we have the following informal heuristics:{control of φ′

}⇒ I(λ) decays like λ−1

whereas control of φ′

&compact support of ψ

⇒ I(λ) decays to any order.

The first heuristic is suggested by the preceding example, and the second heuristic is theprinciple of nonstationary phase. Here, “control” means bounding away from 0. As φdetermines the phase oscillation of I(λ), these ideas suggest that a lot of oscillation leadsto a lot of cancellation, hence decay.

The next principle is a generalization of the first heuristic, which analyses the decay ofI(λ) if we only have control of the k-th derivative of φ, rather than φ′.

Lemma 8.4 (van der Corput). Let φ : R→ R be smooth. Fix k ≥ 1, and assume that |φ(k)(x)| ≥1. If k = 1, assume further that φ′ is monotone. Then∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ .k λ− 1k (8.2)

where the implicit constant does not depend on a or b.

Proof. We proceed by induction on k.Consider the k = 1 case. Because φ′ 6= 0,∫ b

aeiλφ(x) dx =

∫ b

a

1

iλφ(x)

d

dx

[eiλφ

′(x)]dx.

Integrating by parts (with boundary terms!) gives∫ b

aeiλφ(x) dx = −

∫ b

a

d

dx

[1

iλφ′(x)

]eiλφ(x) dx +

eiλφ(b)

iλφ′(b)− eiλφ(a)

iλφ′(a)

=1

λ

[eiφ(b)

iλφ′(b)− eiλφ(a)

iφ′(a)

]− 1

iλ

∫ b

aeiλφ(x) d

dx

[1

φ′(x)

]dx.

Thus, ∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ . 1

λ+

1

λ

∫ b

a

∣∣∣∣ ddx 1

φ′(x)

∣∣∣∣ dx.Because φ′ is monotone, d

dx1

φ′(x) = ±∣∣∣ ddx 1

φ′(x)

∣∣∣. So we can estimate further by∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ . 1

λ+

1

λ

∫ b

a

∣∣∣∣ ddx 1

φ′(x)

∣∣∣∣ dx=

1

λ+

1

λ

∣∣∣∣∫ b

a

d

dx

1

φ′(x)dx

∣∣∣∣=

1

λ+

1

λ

∣∣∣∣ 1

φ′(b)− 1

φ′(a)

∣∣∣∣.

1

λ.

81

This establishes the base case k = 1.Next, assume by induction that the desired estimate holds for some k ≥ 1. Suppose

that |φ(k+1)(x)| ≥ 1. By replacing φ with −φ if necessary, we may assume without loss ofgenerality that φ(k+1)(x) ≥ 1. Then φ(k) is strictly increasing, and consequently there is atmost one point c ∈ [a, b] such that φ(k)(c) = 0.Case 1. First, consider the case where there is such a c. Since φ(k+1) ≥ 1, |φ(k)(x)| ≥ δ for allx of distance at least δ from c, i.e., for all x ∈ [a, b] \ (c − δ, c + δ). Close to c, we have thetrivial estimate ∣∣∣∣∣

∫[a,b]∩(c−δ,c+δ)

eiλφ(x) dx

∣∣∣∣∣ ≤ 2δ.

Away from c, we estimate as follows. Because |φ(k)(x)| ≥ δ on the complement of (c −δ, c + δ), we can rescale φ to use the inductive hypothesis. Let φδ(x) := φ(δ−

1kx). Then

φ(k)δ (x) = δ−1φ(k)(δ−

1kx) and hence |φ(k)

δ (x)| ≥ 1 outside of (c− δ, c+ δ). Thus,1

∣∣∣∣∫ c−δ

aeiλφ(x) dx

∣∣∣∣ =

∣∣∣∣∣∣δ− 1k

∫ δ−1k (c−δ)

δ−1k a

eiλφ(δ−1k y) dy

∣∣∣∣∣∣ =

∣∣∣∣∣∣δ− 1k

∫ δ−1k (c−δ)

δ−1k a

eiλφδ(y) dy

∣∣∣∣∣∣. δ−

1kλ−

1k .

The same computation gives∣∣∣∫ bc+δ eiλφ(x) dx

∣∣∣ . δ−1kλ−

1k . Combining these estimates with

the first trivial estimate yields ∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ . δ + δ−1kλ−

1k

for any δ > 0. We optimize the choice of δ by requiring δ = δ−1kλ−

1k , and thus δ = λ−

1k+1 .

This gives ∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ .k λ− 1k+1

as desired.Case 2. Next, suppose that no such c exists, i.e., φ(k)(x) 6= 0 for all x ∈ [a, b]. It is eitherthe case that φ(k)(a) > 0 or φ(k)(a) < 0. Suppose first that φ(k)(a) > 0. Arguing as before,because φ(k+1)(x) ≥ 1 we know that φ(k)(x) ≥ δ for all x ∈ (a+ δ, b]. Then∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ ≤ ∣∣∣∣∫ a+δ

aeiλφ(x) dx

∣∣∣∣+

∣∣∣∣∫ b

a+δeiλφ(x) dx

∣∣∣∣ .The first quantity is trivially bounded by δ. In the second integral, we rescale by δ and usethe inductive hypothesis exactly as before to get

∣∣∣∫ ba+δ eiλφ(x) dx

∣∣∣ . δ−1kλ−

1k . Optimizing in

δ by choosing δ = λ−1k+1 gives ∣∣∣∣∫ b

aeiλφ(x) dx

∣∣∣∣ .k λ− 1k+1

as desired.Suppose now that φ(k)(a) < 0. Then φ(k)(b) < 0. The argument from here is analogous,

this time decomposing [a, b] into [a, b− δ] and (b− δ, b].

1The following integral is understood to be 0 if c− δ ≤ a.

82

If k = 1, then the monotonicity assumption on φ′ is in fact needed. Note that∣∣∣∣∫ b

aeiφ(x) dx

∣∣∣∣ =

∣∣∣∣∫ b

acos(φ(x)) + i sin(φ(x)) dx

∣∣∣∣ ≥ ∣∣∣∣∫ b

asin(φ(x)) dx

∣∣∣∣ .It is possible to construct a function φ such that φ′(x) is large when sin(φ(x)) is small, andφ′(x) is small when sin(φ(x)) is large. Consequently,

| {x : sinφ(x) } | << | {x : sinφ(x) > 0 } |

and ∣∣∣∣∫ b

asin(φ(x)) dx

∣∣∣∣→∞as |a|, |b| → ∞. We leave the details to the reader.

Finally, we come to the principle of stationary phase. By the principle of nonstationaryphase, I(λ) decays as rapidly as we like when φ′ 6= 0. Thus, it remains to consider whathappens near critical points of φ, i.e., where the phase is not oscillating rapidly.

The following proposition gives a complete description of the asymptotic behavior ofI(λ) near a well-behaved critical point.

Proposition 8.5 (Stationary phase). Let φ : R → R be smooth, and assume that φ has a nonde-generate critical point2 at x0. If ψ is smooth and supported in a sufficiently small3 neighborhoodneighborhood of x0, then

I(λ) =

∫Reiλφ(x)ψ(x) dx

satisfies

I(λ) =[√

2πi∣∣φ′′(x0)

∣∣− 12 ei

π4

sgnφ′′(x0)eiλφ(x0)ψ(x0)]λ−

12 + O

(λ−

32

)(8.3)

as λ→∞.

This statement of the principle of stationary phase gives an explicit constant in (8.3)that will prove to be useful. Before providing the proof of this fact, we can easily derive thegeneral behavior if we ignore the explicit constant.

In particular, let a ∈ C∞c (R) be a smooth bump function satisfying a(x) = 1 for |x| ≤ 1and a(x) = 0 for |x| ≥ 2. We use this bump function to localize close to and far away fromx0 as follows:

|I(λ)| ≤∣∣∣∣∫

Reiλφ(x)ψ(x) a

(λ

12 (x− x0)

)dx

∣∣∣∣+

∣∣∣∣∫Reiλφ(x)ψ(x)

[1− a

(λ

12 (x− x0)

)]dx

∣∣∣∣ .Consider the first integral. The support of the rescaled a function is λ−

12 , so ignoring any

oscillation and using the triangle inequality gives∣∣∣∣∫Reiλφ(x)ψ(x) a

(λ

12 (x− x0)

)dx

∣∣∣∣ . λ−12 .

Consider the second integral. As ψ(x) has compact support and 1 − a(λ

12 (x− x0)

)has

support away from x0, φ′ 6= 0 on the support of ψ(x)[1− a

(λ

12 (x− x0)

)]. We would like

to use the principle of nonstationary phase to estimate this term, but because the function

2That is, φ′(x0) = 0 and φ′′(x0) 6= 0.3The precise nature of sufficiently small will be clear from the proof. Primarily, the support of ψ must not

contain any other critical points of φ.

83

ψ(x)[1− a

(λ

12 (x− x0)

)]has support which scales with λ, the implicit constant in (8.1)

depends on λ. However, as a is rescaled by a factor of λ12 , every iteration of integration by

parts in the proof of nonstationary phase picks up a a factor of λ−1 · λ12 = λ−

12 . Thus,∣∣∣∣∫

Reiλφ(x)ψ(x)

[1− a

(λ

12 (x− x0)

)]dx

∣∣∣∣ . λ−N2

for any order N . We can certainly choose N large enough so that this term is O(λ−

32

).

Thus,|I(λ)| . λ−

12 +O

(λ−

32

)as desired.

Now we provide the full proof, keeping track of the explicit constants.

Proof of Proposition 8.5. Because φ′(x0) = 0 and φ′′(x0) = 0, φ is well-approximated by itsorder-2 Taylor expansion near x0. Expanding gives us

φ(x) = φ(x0) + (x− x0)φ′(x0) +1

2(x− x0)2φ′′(x0) +O(|x− x0|3)

= φ(x0) + φ′′(x0)(x− x0)2 (1 + η(x))

where η : R→ R is smooth and η(x) = O(|x− x0|).Let U be an open neighborhood of x0, chosen sufficiently small enough so that |η(x)| <

1 for all x ∈ U , and φ(x) 6= 0 for all x 6= x0 ∈ U . Suppose that the support of ψ is containedin U .

Change variables via y = (x − x0)√

1 + η(x). This is a diffeomorphism from U to asmall neighborhood of y = 0. Then

I(λ) =

∫Reiλ[φ(x0)+ 1

2φ′′(x0)(x−x0)2(1+η(x))]ψ(x) dx

= eiλφ(x0)

∫Rei

12φ′′(x0)λy2

ψ(y) dy

where ψ ∈ C∞c (R) is the appropriately transformed function under the indicated diffeo-morphism. Note that ψ has support contained the the small neighborhood surroundingy = 0. Let λ := 1

2φ′′(x0)λ and let γ ∈ C∞c (R) be a function which is identically 1 on the

support of ψ. Then

I(λ) = eiλφ(x0)

∫Reiλy

2ψ(y) dy

= eiλφ(x0)

∫Reiλy

2e−y

2[ey

2ψ(y)

]γ(y) dy.

Taylor expand ey2ψ(y) by

ey2ψ(y) =

N∑j=0

ajyj + yN+1RN (y) =: P (N)

where RN is smooth and bounded. With this, we decompose I(λ) as follows:

I(λ) = eiλφ(x0)N∑j=0

aj

∫Reiλy

2e−y

2yj dy

+ eiλφ(x0)

∫Reiλy

2e−y

2yN+1RN (y)γ(y) dy

+ eiλφ(x0)

∫Reiλy

2e−y

2P (y) [γ(y)− 1] dy.

84

Let (1), (2), and (3) denote the first, second, and third lines of this decomposition, respec-tively.

Consider (3) first. Note that e−y2P (y) [γ(y)− 1] is supported outside of a neighborhood

of y = 0. As (y2)′ = 2y 6= 0 away from 0, by the principle of nonstationary phase we have|(3)| . |λ|−m for any order m.

Next, we estimate (2). Let a ∈ C∞c (R) be a smooth bump function satisfying a(x) = 1for |x| ≤ 1 and a(x) = 0 for |x| ≥ 2. For any ε > 0, we have a decomposition

(2) = eiλφ(x0)

∫Reiλy

2e−y

2yN+1RN (y)γ(y)ϕ(y/ε) dy

+ eiλφ(x0)

∫Reiλy

2e−y

2yN+1RN (y)γ(y) [1− ϕ(y/ε)] dy.

Label the first line of this decomposition (2i), and the second line (2ii). We will estimateeach term separately and then optimize in ε. Because the support of ϕ(y/ε) is small (of size∼ ε, when ε is small we do not expect much oscillation from the phase in this integral. Soto estimate (2i) we simply use the triangle inequality, boundedness of RN (y) and e−y

2, and

the support of ϕ(y/ε) to get

|(2i)| .∫R

∣∣yN+1ϕ(y/ε) dy∣∣ . εN+1ε = εN+2.

To estimate (2ii) we use the nonstationary oscillation of the phase eiλy2. Because we need

to keep track of the ε’s, we will integrate by parts directly instead of applying our previousnonstationary phase result. To simplify notation, let X(y) := e−y

2RN (y)γ(y). Define the

differential operator D by

D :=1

2iλy

d

dy

so that Deiλy2

= eiλy2. Then the adjoint is given by

(tDf)(y) = − d

dy

(1

2iλyf(y)

).

Thus,

(2ii) = eiλφ(x0)

∫Reiλy

2e−y

2yN+1RN (y)γ(y) [1− ϕ(y/ε)] dy

= eiλφ(x0)

∫RDm[eiλy

2] yN+1X(y) [1− ϕ(y/ε)] dy

= eiλφ(x0)

∫Reiλy

2(tD)m

{yN+1X(y) [1− ϕ(y/ε)]

}dy.

We carefully analyze the (tD)m{yN+1X(y) [1− ϕ(y/ε)]

}term. For clarity, consider a single

application of tD:

tD{yN+1X(y) [1− ϕ(y/ε)]

}= − d

dy

[1

2iλy

{yN+1X(y) [1− ϕ(y/ε)]

}].

By applying tD to{yN+1X(y) [1− ϕ(y/ε)]

}m times, we pick up m copies of y in the

denominator. Moreover, for each application of tD, the derivative can either hit the yin the denominator, the yN+1 term, X(y), or 1 − ϕ(y/ε). Let k denote the number oftimes the derivative hits the y denominator, let α1 denote the number of times the deriva-tive hits the yN+1, and let α2 the number of times the derivative hits 1 − ϕ(y/ε). As

85

X(y) = e−y2RN (y)γ(y), all of its derivatives are bounded and hence uninteresting. Ap-

plying the integral triangle inequality to (2ii) and summing over all possible combinationsof derivations described above, we have

|(2ii)| . 1

|λ|m

m∑k=0

∑α1+α2≤m−k

∫|y|≥ε

|y|N+1−α1ε−α2

|y|m+kdy.

The ≤ sign in the second sum is to account for the fact that we are ignoring the derivativesof X(y). The region of integration is due to the support of 1 − ϕ(y/ε). Continuing theestimate,

|(2ii)| . |λ|−mm∑k=0

∑α1+α2≤m−k

ε−α2

∫|y|≥ε

|y|N+1−α1−(m+k) dy.

The most singular integral in this sum corresponds to α1 = 0 and k = 0. This partic-ular integral is finite provided N + 1 − m < −1, i.e., N + 2 < m, and is bounded byεN+2−(α1+α2)−(m+k). So, as long as m is chosen so that N + 2 < m, we have

|(2ii)| . |λ|−mm∑k=0

∑α1+α2≤m−k

εN+2−(α1+α2)−(m+k).

Because ε is small, the summand is majorized when α1 + α2 = m− k. Thus,

|(2ii)| . |λ|−mm∑k=0

∑α1+α2≤m−k

εN+2−(m−k)−(m+k) = |λ|−mm∑k=0

∑α1+α2≤m−k

εN+2−2m

. |λ|−mεN+2−2m.

We now optimize in ε. Choose ε so that the (2i) and (2ii) estimates are equal:

εN+2 = |λ|−mεN+2−2m ⇒ ε = |λ|−12 .

Then|(2)| . |λ|−

N+22 .

This estimate is acceptable for N ≥ 1.It remains to estimate (1):

(1) = eiλφ(x0)N∑j=0

aj

∫Reiλy

2e−y

2yj dy.

Note that a0 = ψ(0) = ψ(x0). For j ≥ 0,∫Reiλy

2e−y

2yj dy =

∫Re−(1−iλ)y2

yj dy.

Let z = (1− iλ)12 . Then∫

Re−(1−iλ)y2

yj dy = (1− iλ)−j2− 1

2

∫(1−iλ)

12 Re−z

2zj dz

where we have chosen a branch of the function z−j2− 1

2 with cut along the negative real axis.Using the decay of e−z

2to shift the contour, we have∫Reiλy

2e−y

2yj dy = (1− iλ)−

j2− 1

2

∫Re−y

2yj dy

86

When j is odd, the function e−y2yj is odd, and consequently the above integral is 0. When

j ≥ 2 is even, the asymptotic contribution of the above integral is |λ|−j+1

2 . As j ≥ 2, thisquantity is O(λ−

32 ) and hence is acceptable. It remains to consider the contribution of the

j = 0 term. Write (1 − iλ) = reiθ. Then r =√

1 + λ2 and tan θ = −λ1 = −λ. Recall that∫

R e−y2

dy =√π. Thus, the contribution of the j = 0 term to (1) is therefore

eiλφ(x0)a0(1− iλ)−12√π = eiλφ(x0)ψ(x0)r−

12 e−i

θ2 .

As λ → ∞, θ → −π2 , and as λ → −∞, θ → π

2 . Recall that λ = 12φ′′(x0)λ. Thus, as λ → ∞

we have e−iθ2 → ei

π4

sgnφ′′(x0). Thus, the contribution to (1) is

eiλφ(x0)ψ(x0)√π

[2

λ|φ′′(x0)|

] 12

eiπ4

sgnφ′′(x0).

Careful inspection shows that this is precisely the first term of (8.3), which completes theproof.

8.2 The Morse Lemma

The remainder of this chapter is dedicated to proving the principles of nonstationary andstationary phase in higher dimensions. In the proof of stationary phase, it will be useful tochange variables in a particularly nice way, as described by the following lemma.

Lemma 8.6 (Morse). Suppose that φ : Rd → R is smooth with a nondegenerate critical point atx0 ∈ Rd. There is a change of coordinates x 7→ y(x) such that y(x0) = 0, ∂y∂x(x0) = Id, and

φ(x)− φ(x0) =1

2λ1y

21 + · · · 1

2λdy

2d,

where λ1, . . . , λd are the eigenvalues of D2φ(x0).

We have dedicated an entire section to the Morse lemma, as its proof is technical anddoes not strictly involve any harmonic analysis or techniques from oscillatory integrals.The higher dimensional analogues of nonstationary and stationary phase are contained inthe next section, and the reader may be inclined to skip ahead.

Proof. Because D2φ(x0) is symmetric, we may change coordinates if necessary and assumewithout loss of generality that D2φ(x0) = diag (λ1, . . . , λd). By Taylor expansion and inte-gration by parts, we may write

φ(x) = φ(x0) + (x− x0)∇φ(x0) +

∫ 1

0(1− t) d

2

dt2[φ(x0 + t(x− x0))] dt

= φ(x0) +

∫ 1

0(1− t)

d∑i,j=1

(x− x0)i(x− x0)j∂2φ

∂xi ∂xj(x0 + t(x− x0)) dt.

So φ(x)− φ(x0) =∑d

i,j=1(x− x0)i(x− x0)jµij(x), where

µij(x) =

∫ 1

0(1− t) ∂2φ

∂xi ∂xj(x0 + t(x− x0)) dt.

Since φ is smooth, µij is smooth. Also, µij = µji and

µij(x0) =1

2

∂2φ

∂xi ∂xj(x0).

87

From here, we argue by induction. Suppose that we have found a change of variablesx 7→ y(x) such that y(x0) = 0, ∂y∂x(x0) = I , and

φ(x)− φ(x0) =1

2λ1y

21 + · · · 1

2λr−1y

2r−1 +

d∑i,j≥r

yiyjµij(y).

For the base case, we take r = 1, y = x− x0, and µij(y) = µij(x).Next, we perform some computations assuming the above decomposition. First,

∂2

∂xi ∂xj

(1

2λky

2k

) ∣∣∣∣∣x=x0

= λk∂yk∂xi

∣∣∣∣∣x0

· ∂yk∂xj

∣∣∣∣∣x0

+ λk∂2yk∂xi ∂xj

∣∣∣∣∣x0

· yk

∣∣∣∣∣x0

.

The second summand is 0 because yk(x0) = 0. Because ∂y∂x(x0) = I by assumption, we are

left with∂2

∂xi ∂xj

(1

2λky

2k

) ∣∣∣∣∣x=x0

= λkδijδik.

As D2φ(x0) = diag (λ1, . . . , λd), we get

D2

d∑i,j≥r

yiyjµij(y)

∣∣∣∣∣x=x0

= diag (λr, . . . , λd).

Note that

∂2

∂xl ∂xk

d∑i,j≥r

yiyjµij(y)

∣∣∣∣∣x=x0

=∑i,j≥r

(∂yi∂xl

∣∣∣∣∣x0

· ∂yj∂xk

∣∣∣∣∣x0

+∂yi∂xk

∣∣∣∣∣x0

· ∂yj∂xl

∣∣∣∣∣x0

)µij(x = x0)

=∑i,j≥r

(δilδjk + δikδjl)µij(0)

= 2mlk(0).

This implies thatµij(0) = diag (λr, . . . , λd).

Next, perform the change of variables y 7→ y′(y) defined byy′j := yj for j 6= r;

y′r :=√

µrr(y)λr/2

(yr +

∑j≥r+1

µjr(y)µrr(y)yj

) .

We need to show that this diffeomorphism satisfies y′(x0) = 0, ∂y′

∂x (x = x0) = I , and

φ(x)− φ(x0) =1

2λ1(y′1)2 + · · · 1

2λr(y

′r)

2 +

d∑i,j≥r+1

y′iy′jµ′ij(y).

for some functions µ′ij(y). It is clear that y′(x0) = 0, since yk(x0) = 0 for all k. next, we have

∂y′r∂xi

∣∣∣∣∣x0

=

√µrr(y(x0))

λr/2

δir +∑j≥r+1

δij · 0

= δir

88

and so ∂y′

∂x (x = x0) = I . Finally,

d∑i,j≥r

yiyjµij(y)− 1

2λr(y

′r)

2

=d∑

i,j≥ryiyjµij(y)− mrr(y)

y2r + 2yr

∑j≥r+1

µjr(y)

µrr(y)yj +

∑j≥r+1

µjr(y)

µrr(y)yj

2=

∑i,j≥j+1

yiyj

[µij(y)− µir(y)µjr(y)

µrr(y)

].

Setting µ′ij = µij(y)− µir(y)µjr(y)µrr(y) gives the desired conclusion.

8.3 Oscillatory Integrals of the First Kind, d > 1

In this section we consider the asymptotics of oscillatory integrals on Rd. We begin withthe principle of nonstationary phase. The statement and proof is essentially the same as inthe one-dimensional case.

Proposition 8.7 (Nonstationary phase). Let φ : Rd → R and ψ : Rd → C be smooth. Assumethat ψ has compact support, and that∇φ 6= 0 on the support of ψ. Then∣∣∣∣∫

Rdeiλφ(x)ψ(x) dx

∣∣∣∣ . λ−N

for any N ≥ 0.

Proof. As in the one-dimensional case, we simply use integration by parts to exploit thephase oscillation.

Because∇φ 6= 0 on the support of ψ, we have

eiλφ(x) =∇φ(x)

iλ|∇φ(x)|2∇eiλφ(x).

Then

I(λ) =

∫Rdeiλφ(x)ψ(x) dx =

∫Rd

∇φ(x)

iλ|∇φ(x)|2∇eiλφ(x)ψ(x) dx

= − 1

iλ

∫Rdeiλφ(x)∇

[∇φ(x)

iλ|∇φ(x)|2ψ(x)

]dx.

Thus, |I(λ)| . λ−1, where the implicit constant depends on two derivatives of φ and onederivative of ψ. Iterating integration by parts as before gives decay of |I(λ)| to any order.

Next, we consider the principle of stationary phase. The one-dimensional principle ofstationary phase described how I(λ) decays like λ−

12 near a nondegenerate critical point of

the phase. In the higher-dimensional case, the decay is like becomes λ−d2 .

The proof is structurally the same as in the one-dimensional case, with some slightlymore involved computations.

Proposition 8.8 (Stationary phase). Let φ : Rd → R and ψ : Rd → C be smooth. Assume thatφ has a nondegenerate critical point at x0 ∈ Rd, and that ψ is supported in a sufficiently smallneighborhood of x0. Then I(λ) =

∫Rd e

iλφ(x)ψ(x) dx satisfies

I(λ) =

eiλφ(x0)ψ(x0)(2πi)d2

d∏j=1

µ− 1

2j

λ− d2 + O(λ−

d2− 1

2

)(8.4)

89

as λ→∞, where µ1, . . . , µd are the eigenvalues of D2φ(x0).

Proof. By the Morse lemma, there is a change of variables x 7→ y(x) such that y(x0) = 0,∂y∂x(x0) = Id, and

φ(x)− φ(x0) =1

2µ1y

21 + · · ·+ 1

2µdy

2d.

With this change of variables, we have

I(λ) = eiλφ(x0)

∫eiλ

∑dj=1

12µjy

2j ψ(y) dy

where ψ(y) absorbs the Jacobian from the change of variables. As in the proof of the one-dimensional case, let γ ∈ C∞c (Rd) be identically 1 on the support of ψ. Then

I(λ) = eiλφ(x0)

∫eiλ

∑dj=1

12µjy

2j e−|y|

2[e|y|

2ψ(y)

]γ(y) dy.

By Taylor expansion, we can write

e|y|2ψ(y) =

∑|α|≤N

cαyα +

∑|β|=N+1

yβRβ(y) =: P (y) +∑

|β|=N+1

yβRβ(y).

As in the one-dimensional case, we then have a decomposition

I(λ) = eiλφ(x0)∑|α|≤N

cα

∫eiλ

∑dj=1

12µjy

2j e−|y|

2yα dy

+ eiλφ(x0)∑

|β|=N+1

∫eiλ

∑dj=1

12µjy

2j e−|y|

2yβRβ(y)γ(y) dy

+ eiλφ(x0)

∫eiλ

∑dj=1

12µjy

2j e−|y|

2P (y) (γ(y)− 1) dy.

Let (1), (2), and (3) denote the first, second, and third lines of this decomposition, respec-tively.

Estimating (1).For |α| ≤ N , we have∫

eiλ∑dj=1

12µjy

2j e−|y|

2yα dy =

d∏j=1

∫eiλµ2

j2y2j e−y

2j yαjj dyj

=d∏j=1

(1−

iλµ2j

2

)− 12−αj2 ∫

e−y2j yαjj dyj .

For |α| ≥ 2, this computation gives∣∣∣∣∫ eiλ∑dj=1

12µjy

2j e−|y|

2yα dy

∣∣∣∣ = O(|λ|−

d2− |α|

2

)= O

(λ−

d2−1)

as λ → ∞. When |α| = 1, the contribution is 0 because∫e−y

2j yαjj dyj = 0 when αj is odd.

When α = 0, write 1− iλµ2j

2 = rjeiθj where rj =

√1 +

λ2µ2j

4 and tan θj = −λµj2 . As λ →∞,

θj → −π2 sgnµj . Thus,

d∏j=1

r− 1

2j e−i

θj2 →

(λ

2

)− d2

d∏j=1

eiπ4

sgnµj (π|µj |)−12

90

as λ → ∞. Finally, observe that c0 = ψ(x0). Rewriting the above constant and combiningthe estimates thus far gives

(1) =

eiλφ(x0)ψ(x0)(2πi)d2

d∏j=1

µ− 1

2j

λ− d2 + O(λ−

d2− 1

2

).

Estimating (2).Let a ∈ C∞c (Rd) satisfy a(x) = 1 for |x| ≤ 1 and a(x) = 0 for |x| ≥ 2. Let Xβ(y) =

e−|y|2Rβ(y)γ(y). For ε > 0 (to be determined later), we have

(2) = eiλφ(x0)∑

|β|=N+1

∫eiλ

∑dj=1

12µjy

2j yβXβ(y)a(y/ε) dy

+ eiλφ(x0)∑

|β|=N+1

∫eiλ

∑dj=1

12µjy

2j yβXβ(y) [1− a(y/ε)] dy.

Let (2i) and (2ii) denote the first and second lines of this decomposition, respectively. Asin the one-dimensional case, we do not expect any cancellation in (2i) because of the smallsupport of a(·/ε), thus we estimate (2i) crudely by

|(2i)| .∫|y|≤2ε

|y|N+1 dy . εN+1+d.

Next we consider (2ii). We will effectively use nonstationary phase on this integral, but wehave to be explicit with the integration by parts because we are optimizing our choice of ε.Let

Γj =

{x ∈ Rd : xj >

|x|2

d

}.

Then Rd =⋃dj=1 Γj . Fatten the regions as follows:

Γj =

{x ∈ Rd : xj >

|x|2

2d

}.

We will define a particular partition of unity subordinate to this collection of sets. Letϕj : Sd−1 ⊆ Rd → R be functions with support in Γj ∩ Sd−1 satisfying ϕj(x) = 1 forx ∈ Γj ∩Sd−1. Set ϕj =

ϕj∑dj=1 ϕj

. Finally, extend ϕj to all of Rd by setting ϕj(x) = ϕj(x/|x|).

Next, let Q(y) =∑d

j=112µjy

2j . Then ∇Q(y) = (µ1y1, . . . , µdyd), and hence Q has one

critical point at y = 0. We have

(2ii) = eiλφ(x0)d∑j=1

∑|β|=N+1

∫eiλQ(y)yβXβ(y) [1− a(y/ε)]ϕj(y) dy.

Note thateiλQ(y) =

1

iλµjyj

∂

∂yjeiλQ(y).

The computation from here is similar to the one-dimensional case. Integrating by parts mtimes via ∂

∂yjin each set Γj , and using the fact that yj ≥ |y|

2

d in Γj , we compute as follows.Let k denote the number of times the adjoint derivative hits the yi in the denominator ofthe above fraction. Then

|(2ii)| . 1

λm

m∑k=0

∑α1+α2≤m−k

∫|y|≥ε

|y|N+1−|α1|

|y|m+kε−|α2| dy.

91

This integral is finite provided m+ k −N − 1 + |α1| > d, in which case

|(2ii)| . 1

λm

m∑k=0

∑α1+α2≤m−k

εN+1+d−(|α1|+|α2|)−(m−k)

.1

λmεN+1+d−2m.

To optimize our choice in ε > 0 between (2i) and (2ii), we require

εN+1+d =1

λmεN+1+d−2m

and thus ε = λ−12 . With this choice,

|(2)| . λ−d2−N+1

2 ,

which is acceptable for N ≥ 1.

Estimating (3).Finally, we estimate (3). As γ(y)−1 is supported away from the origin, and because we

don’t need to worry about optimizing between different estimates, we can blindly applythe principle of nonstationary phase to get |(3)| . λ−m for any order m, which is obviouslyacceptable. This completes the proof.

8.4 The Fourier Transform of a Surface Measure

We end this chapter by applying the theory of oscillatory integrals to the surface measureon the unit sphere Sd−1 ⊆ Rd. This will have applications when we study dispersive PDEsin Chapter 9, and also when we study restriction theory in Chapter 10.

Let dσ denote the surface measure supported on Sd−1 ⊆ Rd. Define the inverse Fouriertransform of the measure dσ to be the function on Rd given by

dσ(x) :=1

(2π)d2

∫Sd−1

eix·ξ dσ(ξ).

This function satisfies the following decay property.

Lemma 8.9. If dσ is the surface measure on Sd−1 ⊆ Rd, then |dσ(x)| . 〈x〉−d−1

2 .

Proof. First, observe that

dσ(0) =

∫Sd−1

dσ(ξ) ∼ 1.

Thus, to prove the lemma it suffices to prove |dσ(x)| . |x|−d−1

2 .As dσ is rotation invariant, dσ is also rotation invariant. In particular,

dσ(x) = dσ((0, . . . , 0, |x|))

∼∫Sd−1

ei|x|ξd dσ(ξ).

Let θ denote the angle between ξ and the north pole (0, . . . , 0, 1). Changing variables viaspherical coordinates gives

dσ(x) ∼∫ π

0ei|x| cos θ sind−2(θ) dθ.

92

Note that ddθe

i|x| cos θ = ei|x| cos θ(−i|x| sin θ). Thus, if d ≥ 4 we can integrate by parts to get

dσ(x) ∼ −∫ π

0

d

dθ

[ei|x| cos θ

] 1

i|x| sin θsind−2(θ) dθ

= − 1

i|x|

∫ π

0

d

dθ

[ei|x| cos θ

]sind−3(θ) dθ

= − 1

i|x|ei|x| cos θ sind−3(θ)

∣∣∣∣∣π

0

+d− 3

i|x|

∫ π

0ei|x| cos θ sind−4(θ) cos θ dθ.

The boundary terms are 0, since sin 0 = sinπ = 0. So if d ≥ 4, one iteration of integrationby parts gives

dσ(x) ∼ |x|−1

∫ π

0ei|x| cos θ sind−4(θ) cos θ dθ.

In words, one iteration of integration by parts picked up a power of |x|−1, decreased thepower of sin θ in the integral by 2, and picked up a power of cos θ.

If d is even, we can integrate by parts d−22 times to eliminate the original d− 2 copies of

sin θ. Along the way we pick up more copies of sin θ and cos θ from differentiating productsof sin θ and cos θ, but the elimination of the original d− 2 copies of sin θ ensures

dσ(x) ∼ |x|−d−2

2

∫ π

0ei|x| cos θP (θ) dθ

where P (θ) is a trigonometric polynomial containing at least one term without a sin θ.Similarly, if d is odd we can integrate by parts d−3

2 times to get

dσ(x) ∼ |x|−d−3

2

∫ π

0ei|x| cos θ sin θQ(θ) dθ

for some trigonometric polynomial Q(θ). Integrating by parts one more time gives thedesired estimate:

dσ(x) ∼ |x|−d−3

2 |x|−1

∫ π

0

d

dθ

[ei|x| cos θ

]Q(θ) dθ

∼ |x|−d−1

2 .

Thus, it remains to consider when d is even. Let a1 ∈ C∞c (−ε, ε), a2 ∈ C∞c (ε/2, π − ε/2),and a3 ∈ C∞c (π − ε, π + ε) satisfy a1 + a2 + a3 = 1 on (0, π). In the support of a2, the phaseφ(θ) = cos θ has no critical points, so nonstationary phase gives∣∣∣∣∫ π

0ei|x| cos θP (θ)a2(θ) dθ

∣∣∣∣ . |x|−1.

On the support of a1, we have∫ π

0ei|x| cos θP (θ)a2(θ) dθ =

∫ ε

0ei|x| cos θP (θ)a2(θ) dθ.

The phase cos θ has one nondegenerate critical point at 0. The principle of stationary phasegives a contribution of O(|x|−

12 ), and similarly on the support of a3. Thus,

dσ(x) . |x|−d−2

2 |x|−12 = |x|−

d−12

as desired.The lower dimensional cases are straightforward exercises.

93

Chapter 9

Dispersive Partial DifferentialEquations

ADD INTRODUCTIONThe most important examples of dispersive partial differential equations are the nonlin-

ear Schrodinger equationi∂tu+ ∆u = F (u,Du) (9.1)

and the Korteweg-de Vries (KdV) equation

ut + uxxx + γuux = 0. (9.2)

In (9.1), u : Rdx×Rt → C or u : Tdx×Rt → C, and in (9.2), u : Rx×Rt → R or u : Tx×Rt → R.These equations arise in a number of physical contexts and are an active area of research.

FINISH

9.1 Dispersion

What does it mean for a partial differential equation to be dispersive? Informally, an equa-tion is dispersive if its solutions are waves whose spatial profiles spread out over time, if noboundary conditions are imposed. This heuristic is nice, but is not mathematically rigorousin any way. The goal of this section is to give a (mostly) formal definition of dispersion.

A slightly more tractable way of identifying a dispersive equation is via the frequencysupport of its fundamental solution. Consider the general linear evolution equation

∂tu = iP (D)u. (9.3)

Here, u : Rdx × Rt → C is a complex-valued function of space and time, and P (D) isa linear differential operator with Fourier symbol 2πϕ(ξ). Taking the space-time Fouriertransformation of (9.3) gives 2πiτ u(ξ, τ) = 2πiϕ(ξ)u(ξ, τ). Rearranging and dividing by2πi gives

[τ − ϕ(ξ)] u(ξ, τ) = 0. (9.4)

This equation tells us that the space-time Fourier transform of a solution of (9.3) is sup-ported on the surface Σ := { (ξ, τ) : τ = ϕ(ξ) }. We may identify an evolution equation ofthe form (9.3) as dispersive if the surface Σ is curved.1 Indeed, the surface corresponding tothe Schrodinger equation (9.1) is a parabaloid, and the surface corresponding to the KdVequation (9.2) is a cubic curve. An example of an equation which is not dispersive in thissense is the transport equation

ut − cux = 0. (9.5)1Here we are being intentionally vague about curvature to avoid any unnecessary excursions into differen-

tial geometry.

94

The corresponding surface is the hyperplane {τ = cξ}, which is decidedly not curved.To be more formal, consider the evolution equation{

∂tu = Luu(x, 0) = u0(x)

. (9.6)

Here, u : Rdx×Rt → V , where V is either Cd or Rd, and L is a skew-adjoint constant-coefficientlinear differential operator, that is,∫

Rd〈Lu(x), v(x)〉V dx = −

∫Rd〈u(x), Lv(x)〉V dx

for all functions u and v. With D = 1i∇x, we may write

Lu =∑|α|≤k

cα∂αxu := ih(D)u,

where h : Rd → EndV is the polynomial

h(ξ) ∼π∑|α|≤k

i|α|−1cαξα.

We refer to the symbol h as the dispersion relation of (9.6). For the linear Schrodinger equa-tion, the dispersion relation is h(ξ) = −|ξ|2, and for the Airy equation (the linear part of theKdV equation) the dispersion relation is h(ξ) = ξ3. For a fixed v ∈ Rd, the general transportequation

∂tu = −v · ∇xu (9.7)

has dispersion relation h(ξ) = −v · ξ.By considering these dispersion relations, we are irrevocably led to the following defi-

nition.

Definition 9.1. An evolution equation of the form (9.6) is dispersive if ∇h is not constant.

9.2 The Linear Schrodinger Equation

A necessary prerequisite for understanding the NLS equation (9.1) or the KdV equation(9.2) is to develop a solid understanding of their linear counterparts. Thus, we begin withthe linear Schrodinger equation, given by{

iut + ∆2 u

u(x, 0) = u0(x). (9.8)

Here, u : Rdx × Rt → C and u0 : Rdx → C. For now, we will assume that u0 ∈ S(Rd).

9.2.1 The Fundamental Solution

Our first order of business is to compute a concrete representation of a solution to thelinear Schrodinger equation. To solve (9.8), we first apply the spatial Fourier transform.2

As ∆ = (iξ) · (iξ) = −|ξ|2, this gives{iut(ξ) = |ξ|2

2 u(ξ)u(0, ξ) = u0(ξ)

2In this chapter we will adopt the convention that f(ξ) = 1

(2π)d2

∫Rd e

−ix·ξf(x) dx.

95

This is an ODE with respect to t, and can be solved by separation of variables. Indeed, thesolution is given on the Fourier side by

u(t, ξ) = e−i|ξ|2

2t u0(ξ). (9.9)

Inverting the Fourier transform, we have

u(t, x) =1

(2π)d2

∫Rdeix·ξ−i

|ξ|22t u0(ξ) dξ

=1

(2π)d2

∫Rdeit

(xt·ξ− |ξ|

2

2

)u0(ξ) dξ.

9.2.2 Dispersive Estimates

9.2.3 Strichartz Estimates

96

Chapter 10

Restriction Theory

In this chapter, we study one of the most important unsolved problems in harmonic anal-ysis: the restriction problem. Vaguely, the restriction problem asks when one can mean-ingfully restrict the Fourier transform of a function to a subspace of Rd. While this is aninteresting question in harmonic analysis on its own, we will see that the restriction prob-lem has close ties with dispersive partial differential equations. There are also connectionsto other major problems in harmonic analysis, such as the Kakeya conjecture, though wewill only focus on the dispersive PDE connection in this chapter. For more information onrestriction theory and its connections to other parts of harmonic analysis, refer to [10] and[11].

We begin by formalizing the statement of the restriction problem.

10.1 The Restriction Conjecture

A simple motivator for the restriction problem stems from the following observation. Fixd ≥ 2. In Chapter 2 we defined the Fourier transform for functions in Lp(Rd) for 1 ≤ p ≤2. If f ∈ L1(Rd), then by the Riemann-Lebesgue lemma we know that f ∈ C0(Rd). Inparticular, we can meaningfully restrict f to a set of a measure 0 in Rd, as f is continuous.On the other hand, if f ∈ L2(Rd), we only know that f ∈ L2(Rd). As an arbitrary L2

function can be redefined on sets of measure zero, we cannot restrict f to a sets of measure0.

What about functions f ∈ Lp(Rd) for 1 < p < 2? By the above discussion, restrictionto measure zero sets is completely well-defined for p = 1 and entirely not well-definedfor p = 2. Ideally, we would like a critical value between 1 and 2 for which restriction iswell-defined for p smaller than this critical value, but this is unfortunately not the case. Itturns out that there are functions that belong to Lp(Rd) for all p > 1 but for which f blowsup on a hyperplane.

Example 10.1. Let ψ : Rd−1 → C be a smooth bump function. Define f : Rd → C by

f(x1, . . . , xd) =ψ(x2, . . . , xd)

1 + |x1|.

Observe that∫Rd|f(x)|p dx =

∫Rd−1

ψ(x2, . . . , xd) dx2 . . . dxd ·∫

1

(1 + |x1|)pdx1 <∞

97

for any p > 1, hence f ∈ Lp(Rd) for all p > 1. The Fourier transform of f is

f(ξ) = (2π)−d2

∫e−ix·ξ

ψ(x2, . . . , xd)

1 + |x1|dx

= (2π)−12 ψ(ξ2, . . . , ξd)

∫e−ix1ξ1

1 + |x1|dx1.

On the hyperplane {ξ1 = 0}, the integral∫e−ix1ξ1

1+|x1| dx1 =∫

11+|x1| dx1 is divergent. Thus, the

restriction of f to this set is not well-defined.

Stein discovered (see ??) that if a surface S of measure 0 has “sufficient curvature,” thenone may restrict f to S when f ∈ Lp(Rd) for certain values of p. As such, we can informallystate the question of restriction of the Fourier transform as follows: for which zero-measuresets S and which 1 < p < 2 can one meaningfully restrict f to S if f ∈ Lp(Rd)?

There are three surfaces which have received a lot of attention:

- The sphere, Ssphere ={ξ ∈ Rd : |ξ| = 1

}.

- The parabaloid,1 Sparab ={ξ ∈ Rd : ξd = 1

2 |ξ|2}

.

- The cone, Scone ={ξ ∈ Rd : ξd = |ξ|

}.

These three surfaces are endowed with the following canonical measures:

- The sphere is given the usual surface measure dσ.

- The parabaloid is given the measure dσ defined via∫Sparab

f(ξ) dσ(ξ) :=

∫Rd−1

f

(ξ,

1

2|ξ|2)dξ.

- The cone is given the measure dσ defined via∫Scone

f(ξ) dσ(ξ) :=

∫Rd−1

f(ξ, |ξ|

) dξ|ξ|.

With these surfaces and measures, we can give a quantitative formulation of the restrictionproblem. Explicitly, we seek an estimate of the form∥∥∥f |S∥∥∥

Lq(S,dσ).p,q,S ‖f‖Lp(Rd) (10.1)

We shall refer to an estimate of the form (10.1) by R(p → q). For now, our focus will be onthe case S = Ssphere.

First, a few remarks.

1. If p = 1, then R(p→ q) holds for all 1 ≤ q ≤ ∞. Indeed, by Holder’s inequality,∥∥∥f |S∥∥∥Lq(S,dσ)

. σ(S)1q

∥∥∥f∥∥∥L∞(S,dσ)

. ‖f‖L1(Rd)

because S has finite measure.

2. If p = 2, then by our initial discussion R(p→ q) fails for every 1 ≤ q ≤ ∞.

1Here we are using the notation ξ = (x1, . . . , xd−1).

98

3. If R(p → q) holds, then R(p → q) holds for all p ≤ p and q ≤ q. Indeed, by Holder’sinequality, ∥∥∥f |S∥∥∥

Lq(S,dσ). σ(S)

1q− 1q

∥∥∥f |S∥∥∥Lq(S,dσ)

.

From here, let ϕ be a function such that ϕ is compactly supported and ϕ = 1 on a ballcontaining S. Then∥∥∥f |S∥∥∥

Lq(S,dσ).∥∥∥f |S ·ϕ∥∥∥

Lq(S,dσ).∥∥∥f ∗ ϕ∥∥∥

Lq(S,dσ). ‖f ∗ ϕ‖Lp(Rd)

where the last inequality holds because R(p → q) holds. Next, we apply Young’sinequality with r chosen so that 1 + 1

p = 1p + 1

r . Note that, for such an r, we have1r = 1 + p−p

pp ≤ 1 so that this choice is possible. Then∥∥∥f |S∥∥∥Lq(S,dσ)

. ‖f‖Lp(Rd) ‖ϕ‖Lr(Rd) . ‖f‖Lp(Rd)

as desired.

Because of this, when studying restriction problems we adopt a philosophy of maxi-mizing p and q for which R(p→ q) holds.

In modern papers dealing with the restriction problem, the formulation is often pre-sented using the adjoint of the expression R(p → q). In particular, let R : Lp(Rd) →Lq(S, dσ) be the restriction operator given by Rf = f |S . Consider the adjoint operatorR∗ : Lq

′(S, dσ)→ Lp

′(Rd). For a Schwartz function f , we have

〈Rf, g〉L2(S,dσ) =

∫S

(Rf)(ξ)g(ξ) dσ(ξ) = (2π)−d2

∫S

∫Rde−ix·ξf(x) dx g(ξ) dσ(ξ)

=

∫Rdf(x)(2π)−

d2

∫Seix·ξg(ξ) dσ(ξ) dx

=

∫Rdf(x) (g dσ)ˇdx

=⟨f, (g dσ)

⟩L2(S,dσ)

.

This implies that R∗g = (g dσ). Thus, R(p → q) holds if and only if R∗(q′ → p′) holds, i.e.,if there is an estimate of the form∥∥(g dσ)

∥∥Lp′ (Rd)

.p,q,S ‖g‖Lq′ (S,dσ) . (10.2)

We begin by considering necessary conditions for R∗(q′ → p′) to hold. It is natural to firstanalyze what happens when g ≡ 1. By Lemma 8.9, |(dσ)| . 〈x〉−

d−12 . Thus, (dσ) ∈ Lp

′

precisely when d−12 p′ > d, which is equivalent to p < 2d

d+1 . So for the function g ≡ 1,R∗(q′ → p′) holds when

p <2d

d+ 1.

This provides one necessary condition. Another condition comes from the following ex-ample.

Example 10.2 (Knapp). Let κ ⊆ S be a cap on the sphere centered at the point ξ0 =(0, . . . , 0,−1) of horizontal radius 1/R for some R >> 1. Near ξ0, we can parametrizethe cap κ and Taylor expand via

ξd = −√

1− |ξ|2 = −(

1− |ξ|2

2+O(|ξ|4)

)= −1 +O(R−2)

99

because |ξ| ≤ 1/R on κ. This means that the ξd-height of the cap is on the order of R−2, i.e.,the cap κ is contained in a cylinder D of radius 1/R and thickness ∼ 1/R2. Let T denotethe “dual” cylinder to D, i.e., the cylinder centered at 0 with radius R and height ∼ R2.

Let g = χκ. Then‖g‖Lq′ (S,dσ) = σ(κ)

1q′ . (R−1)

d−1q′ .

Also,

(g dσ)ˇ(x) = (2π)−d2

∫κeix·ξ dσ(ξ) = (2π)−

d2 e−ixd

∫κeix(ξ−ξ0) dσ(ξ).

Note that, for ξ ∈ κ,

|(ξ − ξ0)i| .{

1/R 1 ≤ i ≤ d− 11/R2 i = d

and for x ∈ T ,

|xi| .{

R 1 ≤ i ≤ d− 1R2 i = d

.

Thus, for most x ∈ T , we have

|(g dσ)ˇ(x)| & σ(κ) ∼ R−(d−1).

So ∥∥(g dσ)∥∥Lp′ (Rd)

&∥∥(g dσ)

∥∥Lp′ (T )

& R−(d−1)|T |1p′ ∼ R−(d−1)R

d−1+2p′ = R−(d−1)R

d+1p′ .

Therefore, if R∗(q′ → p′) holds, then R−(d−1)Rd+1p′ . R

− d−1q′ . Letting R → ∞ gives the

conditiond+ 1

p′≤ d− 1

q.

From these two examples, two necessary scaling conditions for R∗(q′ → p′) to hold are1 ≤ 2d

d+1 and d+1p′ ≤

d−1q . One might wonder whether such conditions are also sufficient.

Formally, we pose the following conjecture.

Conjecture 10.3 (Restriction conjecture for the sphere). Let d ≥ 2, and let S = Sd−1 ⊆ Rd bethe unit sphere endowed with the canonical surface measure. Then∥∥∥f |S∥∥∥

Lq(S,dσ).p,q ‖f‖Lp(Rd) ,

or equivalently ∥∥(g dσ)∥∥Lp′ (Rd)

.p,q ‖g‖Lq′ (S,dσ) ,

whenever 1 ≤ p ≤ 2dd+1 and 1 ≤ q ≤ d−1

d+1p′.

Zygmund proved this conjecture for d = 2 (see ??); for all other d, this remains an openproblem.

10.2 The Tomas-Stein Inequality

Though the restriction conjecture for the sphere is open for d > 2, we have the followingresult, due to Tomas and Stein.

100

Theorem 10.4 (Tomas-Stein). Let d ≥ 2, and let S = Sd−1 ⊆ Rd be the unit sphere endowedwith the canonical surface measure. Then∥∥∥f |S∥∥∥

L2(S,dσ).p ‖f‖Lp(Rd)

whenever 1 ≤ p ≤ 2(d+1)d+3 .

When q = 2, the scaling conditions in Conjecture 10.3 become 1 ≤ p ≤ 2dd+1 and p′ ≥

2(d+1)d−1 , which is equivalent to p ≤ 2(d+1)

d−3 . Moreover, for d > 1, 2(d+1)d+3 < 2d

d+1 . Thus, theTomas-Stein inequality is indeed the full restriction conjecture for the sphere in the specialcase q = 2.

Proof. For the sake of instruction, we begin with a proof attempt which will not prove thefull statement of Theorem 10.4.

Attempt 1: The first approach relies on using the decay condition |σ(x)| . 〈x〉−d−1

2 . Wewish to show that the operator R : Lp(Rd) → L2(S, dσ) is bounded, which is equivalentto showing boundedness of R∗ : L2(S, dσ) → Lp

′(Rd), which is equivalent to showing

boundedness of R∗R : Lp(Rd)→ Lp′(Rd).

Let f be a Schwartz function. By definition,

(R∗Rf)(x) = (Rf dσ)ˇ(x) = (2π)−d2

∫Seix·ξ(Rf)(ξ) dσ(ξ) = (2π)−

d2

∫Seix·ξ f(ξ) dσ(ξ)

= (2π)−d2 (2π)−

d2

∫Seix·ξ

∫Rde−iy·ξf(y) dy dσ(ξ)

= (2π)−d2

∫Rdf(y)σ(x− y) dy.

Hence, R∗Rf = f ∗ σ. Next, because |σ(x)| . 〈x〉−d−1

2 , we have σ ∈ L2dd−1

,∞(Rd). By theHardy-Littlewood-Sobolev inequality,

‖R∗Rf‖Lp′ (Rd) . ‖f‖Lp(Rd) ‖σ‖L

2dd−1

,∞(Rd)

. ‖f‖Lp

provided that 1 + 1p′ = 1

p + d−12d . This is equivalent to 2

p = d−12d , hence p′ = 4d

d−1 , hencep = 4d

3d+1 . So R(p → 2) holds for all 1 ≤ p ≤ 4d3d+1 . But a quick calculation shows that

4d3d+1 <

2(d+1)d+3 , so we do not have the full scaling range in the statement of the theorem.

Attempt 2: To remedy this, we use both the decay of |σ(x)| and the oscillation. We have

ϕ(x) +∑N>1

ψN (x) = 1

where ϕ and ψN are the usual Littlewood projections, and N is a dyadic number. So wehave a decomposition

f ∗ σ = f ∗ (ϕσ) +∑N>1

f ∗ (ψN σ).

By the triangle inequality,

‖R∗Rf‖Lp′ (Rd) ≤ ‖f ∗ (ϕσ)‖Lp′ (Rd) +∑N>1

‖f ∗ (ψN σ)‖Lp′ (Rd) .

Consider the first term. Note that 1 + 1p′ = 1

p + 1r when r = p′

2 . Because p < 2, p′ > 2, hence

r = p′

2 is a sensible choice. So by Young’s convolution inequality, we have

‖f ∗ (ϕσ)‖Lp′ (Rd) . ‖f‖Lp(Rd) ‖ϕσ‖Lp′2 (Rd)

.

101

As ϕ is a bounded function with compact support, ‖ϕσ‖Lp′2 (Rd)

<∞. Thus,

‖f ∗ (ϕσ)‖Lp′ (Rd) . ‖f‖Lp(Rd) .

Next, we consider the sum. In order to retain summability, we seek an estimate of the form

‖f ∗ (ψN σ)‖Lp′ (Rd) . N−ε(p) ‖f‖Lp(Rd)

for some ε(p) > 0. The strategy is to use real interpolation between a L1 → L∞ andL2 → L2 type estimate.

First, because ψN is supported on an annulus of radius ∼ N , the decay of |σ| gives

‖f ∗ (ψN σ)‖L∞(Rd) . ‖f‖L1(Rd) ‖ψN σ‖L∞(Rd) . N−d−1

2 ‖f‖L1(Rd) .

For the L2 → L2 type estimate, we use Plancharel to get

‖f ∗ (ψN σ)‖L2(Rd) =∥∥∥f · (ψN σ)

∥∥∥L2(Rd)

≤∥∥∥f∥∥∥

L2(Rd)

∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

= ‖f‖L2(Rd)

∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

.

Expanding the convolution,

(ψN ∗ dσ)(x) =

∫SψN (x− y)dσ(y) = Nd

∫Sψ(N(x− y)) dσ(y).

So for any m, since ψ is Schwartz,

|(ψN ∗ dσ)(x)| . Nd

∫S

1

〈N(x− y)〉mdσ(y)

= Nd

∫|x−y|≤1/N ; y∈S

dσ(y) +Nd∑M≤N

∫|x−y|∼1/M ; y∈S

(M

N

)mdσ(y).

The first integral∫|x−y|≤1/N ; y∈S dσ(y) is the measure of a cap on the sphere of radius . 1/N ,

hence∫|x−y|≤1/N ; y∈S dσ(y) . (1/N)d−1. Similarly,

∫|x−y|∼1/M ; y∈S dσ(y) is the measure of

two caps on the sphere with radius . 1/M , so in total we have

|(ψN ∗ dσ)(x)| . Nd

(1

N

)d−1

+Nd∑M≤N

(M

N

)m( 1

M

)d−1

= N +N∑M≤N

(M

N

)m−(d−1)

.

The sum in question is finite if m > d − 1. For such a choice of m, we then have |(ψN ∗dσ)(x)| . N , hence

∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

. N . Therefore, ‖f ∗ (ψN σ)‖L2(Rd) . N ‖f‖L2(Rd).

Now we interpolate between the estimates ‖f ∗ (ψN σ)‖L∞(Rd) . N−d−1

2 ‖f‖L1(Rd) and‖f ∗ (ψN σ)‖L2(Rd) . N ‖f‖L2(Rd). Note that

1

p′=

2/p′

2+

1− 2/p′

∞.

Thus,

‖f ∗ (ψN σ)‖Lp′ (Rd) . ‖f ∗ (ψN σ)‖2p′

L2(Rd)‖f ∗ (ψN σ)‖

1− 2p′

L∞(Rd)

. N2p′N

− d−12

(1− 2

p′

)‖f‖

2p′

L2(Rd)‖f‖

1− 2p′

L1(Rd)

. N− d−1

2+ d+1

p′ ‖f‖Lp(Rd) .

102

Set ε(p) = −d−12 + d+1

p′ . Then ε(p) > 0 precisely when p′ > 2(d+1)d−1 , or equivalently p < 2(d+1)

d+3 .

This gives the desired estimate in all cases, save for the endpoint p = 2(d+1)d+3 .

Attempt 3: The above argument, which proves the theorem except for the endpoint case,is due to Tomas. The loss of the endpoint is essentially due to the use of the triangle in-equality in the dyadic sum. To recover the endpoint case, Stein used complex interpolationwith estimates of the form∥∥∥∥∥∑

N>1

Nd−1

2+itf ∗ (ψN σ)

∥∥∥∥∥L∞(Rd)

. ‖f‖L1(Rd)∥∥∥∥∥∑N>1

N−1+itf ∗ (ψN σ)

∥∥∥∥∥L2(Rd)

. ‖f‖L2(Rd) .

We will not consider the details of the argument here; for reference, see [10].

10.3 Restriction Theory and Strichartz Estimates

As alluded to in the introduction of this chapter, the restriction of the Fourier transformis deeply connected to dispersive partial differential equations. In this section, we willdescribe this connection by studying an analogue of Theorem 10.4 for the paraboloid, andshowing its equivalence to a symmetric Strichartz inequality.

In this setting, because Strichartz estimates involve space and time, it will be convenientto increase the dimension of our restriction estimates from d to d+1. As such, we now seekan estimate of the form ∥∥(g dσ)

∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ) (10.3)

where dσ is the canonical measure on the paraboloid. This is dual formulation of the end-point case for the Tomas-Stein inequality on the paraboloid.

For the reader’s convenience, we recall some estimates from Chapter 9. Recall that thefree propagator eit

∆2 is defined for Schwartz functions by

eit∆2 f = F−1

ξ

(e−it

|ξ|22 f(ξ)

)where u0(x) = u(0, x). By interpolating the estimates∥∥∥eit∆

2 f∥∥∥L2x

= ‖f‖L2x

and∥∥∥eit∆

2 f∥∥∥L∞x

. |t|−d2 ‖f‖L1

x

we obtained (see Theorem ??) the dispersive estimate∥∥∥eit∆2 f∥∥∥Lpx

. |t|−d(

12− 1p

)‖f‖

Lp′x

(10.4)

for 2 ≤ p ≤ ∞. From this, we derived (see Theorem ??) the following inequality.

Theorem 10.5 (Strichartz inequality). Let f ∈ S(Rdx). Then∥∥∥eit∆2 f∥∥∥LqtL

rx

. ‖f‖L2x

for 2 ≤ q, r ≤ ∞ satisfying 2q + d

r = d2 and (d, q, r) 6= (2, 2,∞).

103

When q = r, the above inequality is known as a symmetric Strichartz inequality. In thiscase, the scaling relation gives

2

q+d

q=d

2⇒ q =

2(d+ 2)

d.

As 2(d+2)d > 2, this proves:

Theorem 10.6 (Symmetric Strichartz inequality). Let f ∈ S(Rdx). Then∥∥∥eit∆2 f∥∥∥L

2(d+2)d

t,x

. ‖f‖L2x.

We will now prove the following result.

Proposition 10.7. The Tomas-Stein inequality for the paraboloid is equivalent to the symmetricStrichartz inequality; that is, the estimate∥∥(g dσ)

∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ)

is equivalent to the estimate ∥∥∥eit∆2 f∥∥∥L

2(d+2)d

t,x

. ‖f‖L2x.

Proof. Let u0 ∈ S(Rd) and ϕ ∈ S(R × Rd). We adopt the notation ϕ = Ft,xϕ to denote thespace-time Fourier transform of ϕ. Let ψ = F−1

x ϕ.We wish to calculate the space-time Fourier transform of eit

∆2 u0, in the distributional

sense. We have⟨eit

∆2 u0, ϕ

⟩t,x

=⟨eit

∆2 u0, ϕ

⟩t,x

= (2π)−d2

⟨eit

∆2 u0,

∫Rde−ix·ξψ(t, ξ) dξ

⟩t,x

= (2π)−d2

∫Rd

∫R

∫Rd

(eit

∆2 u0

)(x)e−ix·ξψ(t, ξ) dx dt dξ

=

∫Rd

∫Re−it

|ξ|22 u0(ξ)ψ(t, ξ) dt dξ

= (2π)12

∫Rdu0(ξ) [Ftψ(ξ)]

(|ξ|2

2

)dξ.

Observe that [Ftψ(ξ)](|ξ|22

)= ϕ

(|ξ|22 , ξ

). Thus,⟨

eit∆2 u0, ϕ

⟩t,x

= (2π)12 〈u0 dσparab, ϕ〉t,x .

So, in the distributional sense, and up to a factor of (2π)12 ,

eit∆2 u0 = u0 dσparab ⇒ eit

∆2 u0 = Ft,x (u0 dσparab) .

Let g(ξ, |ξ|

2

2

)= u0(ξ). Then

‖g‖L2(Sparab,dσ) = ‖u0‖L2 = ‖u0‖L2

and (g dσparab)ˇ = eit

∆2 u0. Thus, the estimate

∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x

. ‖u0‖L2 yields the Tomas-

Stein estimate ∥∥(g dσ)∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ) .

The converse implication is essentially the same.

104

This proves that the endpoint Tomas-Stein inequality holds for the paraboloid by wayof the symmetric Strichartz estimate, which by our remark earlier in this chapter provesthe full Tomas-Stein inequality for the paraboloid. However, it is of interest to provethat the Tomas-Stein inequality on the sphere also implies the Tomas-Stein estimate onthe paraboloid, without appealing directly to Strichartz estimates.

Proposition 10.8. The Tomas-Stein estimate on the sphere implies the Tomas-Stein estimate on theparaboloid.

Before proving this proposition, we explicitly compute the surface area measure on thesphere.

Lemma 10.9. The canonical surface measure on the unit sphere Sd ⊆ Rd+1 is given explicitly by

dσSd =1√

1− |ξ|2dξ.

Proof. Let φ : Rd → Rd+1 be a coordinate parametrization of the sphere given by φ(ξ) =(ξ,√

1− |ξ|2). Then

dσSd =√

det [(∇φ)T (∇φ)] dξ.

It remains to compute√

det [(∇φ)T (∇φ)].We have

∇φ(ξ) =

1 0 · · · 00 1 · · · 0...

......

...0 0 · · · 1−ξ1√1−|ξ|2

−ξ2√1−|ξ|2

· · · −ξd√1−|ξ|2

.

Thus,

(∇φ(ξ))T (∇φ(ξ)) =

[δij +

ξiξj1− |ξ|2

]1≤i,j≤d

= I +|ξ|2

1− |ξ|2Pξ

where Pξ =[ξiξj|ξ|2

]is the projection matrix onto the ξ direction.

Consequently,

(∇φ(ξ))T (∇φ(ξ))ξ =

(I +

|ξ|2

1− |ξ|2Pξ

)ξ = ξ +

|ξ|2

1− |ξ|2ξ =

(1

1− |ξ|2

)ξ

so that (∇φ)T (∇φ) has an eigenvector ξ with eigenvalue 11−|ξ|2 . Next, observe that if x is a

vector orthogonal to ξ, then (I +

|ξ|2

1− |ξ|2Pξ

)x = x

so that (∇φ)T (∇φ) has a d− 1-dimensional eigenspace corresponding to the eigenvalue 1.Thus, √

det [(∇φ)T (∇φ)] =1√

1− |ξ|2

as desired.

With this, we can prove Proposition 10.8.

105

Proof. Assume the Tomas-Stein estimate on the sphere, that is,∥∥(g dσSd)∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sd,dσSd

) .

Because Tomas-Stein on the paraboloid is equivalent to the symmetric Strichartz estimate,our goal is to derive an estimate of the form∥∥∥eit∆

2 u0

∥∥∥L

2(d+2)d

t,x (R×Rd)

. ‖u0‖L2x

where we can take u0 ∈ C∞c (Rd) by density.Fix λ >> 1. Define a function gλ on the sphere Sd as follows:{

gλ(ξ,√

1− |ξ|2) = λd2 u0(λξ)

√1− |ξ|2

gλ(ξ,−√

1− |ξ|2) = 0.

Because u0 ∈ C∞c (Rd), for λ large, gλ is supported on a small ξ-region, i.e., a small caparound the north pole. Using the surface area lemma, we have

‖gλ‖2L2(Sd,dσSd

) = λd∫|u0(λξ)|2(1− |ξ|2)

1√1− |ξ|2

dξ

= λd∫|u0(λξ)|2

√1− |ξ|2 dξ

=

∫|u0(ξ)|2

√1− |ξ|

2

λdξ.

So for very large λ we have ‖gλ‖2L2(Sd,dσSd

) ∼ ‖u0‖L2ξ

= ‖u0‖L2x.

Next, we compute (gλ dσSd ):

(gλ dσSd)ˇ(t, x) = (2π)−

d+12

∫eitω+ix·ξλ

d2 u0(λξ)

√1− |ξ|2 1√

1− |ξ|2dξ δ(ω =

√1− |ξ|2)

= (2π)−d+1

2

∫eit√

1−|ξ|2+ix·ξλd2 u0(λξ) dξ

= (2π)−d+1

2 λ−d2

∫eit

√1− |ξ|

2

λ2 +i xλ·ξu0(ξ) dξ.

By Taylor expansion, we have√

1− |ξ|2

λ2 = 1− |ξ|2

2λ2 +O(|ξ|4λ4

). Also note that

(2π)−d+1

2 λ−d2

∫eit−

it|ξ|2

2λ2 +i xλ·ξu0(ξ) dξ = (2π)−

d+12 λ−

d2 eit ·

[ei

tλ2

∆2 u0

] (xλ

).

Thus, we can write

(gλ dσSd)ˇ(t, x) = (2π)−

d+12 λ−

d2 eit ·

[ei

tλ2

∆2 u0

] (xλ

)+ (2π)−

d+12 λ−

d2

∫eit−

it|ξ|2

2λ2 +i xλ·ξu0(ξ)

eit(√

1− |ξ|2

λ2 −1+|ξ|2

2λ2

)− 1

dξ.By the Taylor expansion comment and by the usual arc-length type estimate, for λ large wehave ∣∣∣∣∣∣∣e

it

(√1− |ξ|

2

λ2 −1+|ξ|2

2λ2

)− 1

∣∣∣∣∣∣∣ . |t||ξ|4

λ4.

106

Consequently,∣∣∣∣∣∣∣(2π)−d+1

2 λ−d2

∫eit−

it|ξ|2

2λ2 +i xλ·ξu0(ξ)

eit(√

1− |ξ|2

λ2 −1+|ξ|2

2λ2

)− 1

dξ∣∣∣∣∣∣∣

. λ−d2−4|t|

∫|u0(ξ)||ξ|4 dξ . λ−

d2−4|t|

where the last inequality follows from the fact that u0 ∈ C∞c (Rd). Thus,∥∥(gλ dσSd)∥∥L

2(d+2)d (Rd+1)

&∥∥(gλ dσSd)

∥∥L

2(d+2)d (|t|≤λ2(1+ε),|x|≤λ1+ε)

& λ−d2

∥∥∥[ei tλ2∆2 u0

] (xλ

)∥∥∥L

2(d+2)d (|t|≤λ2(1+ε),|x|≤λ1+ε)

− λ−d2−4λ2(1+ε)λ

(2(1+ε)+d(1+ε)) d2(d+2)

where the final power of λ comes from considering the volume of the set. Changing vari-ables and simplifying the power of λ in the error term then gives∥∥(gλ dσSd)

∥∥L

2(d+2)d (Rd+1)

& λ−d2λ

(2+d) d2(d+2)

∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d (|t|≤λ2ε,|x|≤|λ|ε)

− λ−2+2ε+ dε2 .

Thus, ∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x (|t|≤λ2ε,|x|≤|λ|ε)+O

(λ−2+ d+4

2ε). ‖u0‖L2

x.

As λ→∞, for ε << 1 we then get∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x (Rd+1). ‖u0‖L2

x

as desired.

In the end, via either method, we have proven the Tomas-Stein inequality on the para-baloid.

Theorem 10.10 (Tomas-Stein). Let d ≥ d, and let Sparab ⊆ Rd+1 be the paraboloid endowed withthe canonical surface measure. Then∥∥(g dσ)

∥∥Lp′ (Rd+1)

. ‖g‖L2(Sparab,dσ)

whenever p′ ≥ 2(d+2)d .

107

Chapter 11

Rearrangement Theory

Our next topic of study is the theory of rearrangements. Rearrangement inequalities haveapplications to problems involving the minimization of energy functionals and other vari-ational problems, and can be used in showing the existence and uniqueness of minimizers.Later in this chapter, we will see an explicit application of this type to the energy func-tional of the hydrogen atom. There are also connections to classical problems such as theisoperimetric inequality.

A good reference for the following material is Chapter 3 of [6].

11.1 Definitions and Basic Estimates

We begin by recalling the layer-cake decomposition of a Borel measurable function f : Rd →[0,∞); namely, the identity

f(x) =

∫ ∞0

χ{f>λ}(x) dλ.

This equality follows from the observation that

χ{f>λ}(x) =

{1 f(x) > λ0 f(x) ≤ λ =

{1 λ < f(x)0 λ ≥ f(x)

= χ[0,f(x))(λ)

and so ∫ ∞0

χ{f>λ}(x) dλ =

∫ ∞0

χ[0,f(x))(λ) dλ =

∫ f(x)

0dλ = f(x).

This decomposition will frequently be used in the proofs of the inequalities to come. Withthis in mind, we begin discussing rearrangements.

Definition 11.1. If A ⊆ Rd is Borel measurable, its symmetric rearrangement is the set

A∗ ={x ∈ Rd : |x| < r

}where r is chosen so that |A∗| = |A|.

Definition 11.2. Let f : Rd → C be a Borel measurable function which vanishes at infinity,that is, |{|f | > λ}| < ∞ for all λ > 0. The symmetric decreasing rearrangement of f isgiven by

f∗(x) =

∫ ∞0

χ{|f |>λ}∗(x) dλ.

The intuition and geometry behind the rearrangement of a set A is clear: A∗ is simplythe open ball centered at the origin with the same measure as A. To give intuition for

108

the symmetric rearrangement of a function, we compute a simple example. Suppose thatf : R→ C is given by f(x) = χ[−2,−1](x) + χ[1,2](x). Fix λ > 0. Then

{|f | > λ}∗ =

{([−2,−1] ∪ [1, 2])∗ 0 < λ < 1

∅∗ λ ≥ 1=

{(−1, 1) 0 < λ < 1∅ λ ≥ 1

.

Therefore,

f∗(x) =

∫ 1

0χ{|f |>λ}∗(x) dλ =

∫ 1

0χ(−1,1)(x) dλ =

{1 x ∈ (−1, 1)0 x /∈ (−1, 1)

= χ(−1,1)(x).

For a slightly more interesting example, suppose that g : R → C is given by g(x) =χ[−2,−1](x) + 2χ[1,2](x). Fix λ > 0. Then

{|g| > λ}∗ =

([−2,−1] ∪ [1, 2])∗ 0 < λ < 1

([1, 2])∗ 1 ≤ λ < 2∅∗ λ ≥ 2

=

(−1, 1) 0 < λ < 1(−1

2 ,12

)1 ≤ λ < 2

∅ λ ≥ 2.

Therefore,

g∗(x) =

∫ 1

0χ(−1,1)(x) dλ+

∫ 2

1χ(− 1

2, 12)(x) dλ = χ(−1,1)(x) + χ(− 1

2, 12)(x).

Roughly, the symmetric decreasing rearrangement of a function takes the mass under thegraph of the function and rearranges it to return a function which is radially symmetricand decreasing with the same total mass.

These examples suggest the following basic properties of symmetric decreasing rear-rangements.

Proposition 11.3. If f : Rd → C is a Borel measurable function which vanishes at infinity, thenf∗ is nonnegative, radially symmetric, radially non-increasing, and lower-semicontinuous.

Proof. Nonnegativity is clear. Radial symmetry of f∗ follows from the fact that the set{|f | > λ}∗ is radially symmetric. The fact that f∗ is radially decreasing follows from theobservation that if λ > µ, then {|f | > λ}∗ ⊆ {|f | > µ}∗.

To prove lower-semicontinuity, recall that f∗ is lower-semicontinuous if and only if{f∗ > λ} is open for all λ. We have{

x ∈ Rd : f∗(x) > λ}

=

{x ∈ Rd :

∫ ∞0

χ{|f |>t}∗(x) dt > λ

}=⋃t>λ

{|f | > t}∗.

Thus, {f∗ > λ} is open.

Next, observe that the rearrangement function satisfies the following monotonicityproperty.

Proposition 11.4. If f, g : Rd → C are Borel measurable functions which vanishes at infinity, andif |f | ≤ |g|, then f∗ ≤ g∗.

Proof. If |f | ≤ |g|, then {|f | > λ} ⊆ {|g| > λ}. Thus, {|f | > λ}∗ ⊆ {|g| > λ}∗, and the claimfollows.

Next, we make an observation that will be used frequently. By definition,

f∗(x) =

∫ ∞0

χ{|f |>λ}∗(x) dλ.

On the other hand, we have the layer-cake decomposition of f∗, which gives

f∗(x) =

∫ ∞0

χ{f∗>λ}(x) dλ.

This suggests the equality {|f | > λ}∗ = {f∗ > λ}. Indeed, this is the case.

109

Proposition 11.5. If f : Rd → C is a Borel measurable function which vanishes at infinity, then

{|f | > λ}∗ = {f∗ > λ}.

Proof. Note that

|{|f | > t}∗| = |{|f | > t}| =∫Rdχ{|f |>t}(x) dx.

As t→ λ from above, by the dominated convergence theorem, it then follows that

|{|f | > t}∗| →∫Rdχ{|f |>λ}(x) dx = |{|f | > λ}| = |{|f | > λ}∗|.

Because {f∗ > λ} =⋃t>λ{|f | > t}∗, we have thus shown |{f∗ > λ}| = |{|f | > λ}∗|.

Because f∗ is radially symmetric and lower-semicontinuous, {f∗ > λ} is an open ballcentered at the origin. By definition of a set rearrangement, {|f | > λ}∗ is also an open ballcentered at the origin. As two open balls of equal measure with same center, {|f | > λ}∗ ={f∗ > λ}.

Corollary 11.6. If f : Rd → C is a Borel measurable function which vanishes at infinity, then

‖f‖Lp(Rd) = ‖f∗‖Lp(Rd)

for 1 ≤ p ≤ ∞.

Proof. This follows from

‖f‖pLp(Rd)

= p

∫ ∞0

λp−1|{|f | > λ}| dλ

when 1 ≤ p <∞. The case p =∞ is similar.

In preparation for our first major result, we make a few remarks. We say that a functionf : Rd → [0,∞) is strictly symmetric decreasing if f is radial and if |x| > |y| implies thatf(x) < f(y). Note that this implies f(x) > 0 for all x ∈ Rd, and furthermore f∗ = f .

Theorem 11.7 (Hardy-Littlewood). Let f, g : Rd → [0,∞) be vanishing at infinity. Then∫f(x)g(x) dx ≤

∫f∗(x)g∗(x) dx.

Moreover, if f is strictly symmetric decreasing, then equality holds if and only if g = g∗ almosteverywhere.

Proof. First, we demonstrate the inequality. We compute, using a layer-cake decomposi-tion: ∫

f(x)g(x) dx =

∫Rd

∫ ∞0

χ{f>λ}(x) dλ

∫ ∞0

χ{g>µ} dµ dx

=

∫ ∞0

∫ ∞0|{f > λ} ∩ {g > µ}| dλ dµ.

Suppose without loss of generality that |{f > λ}| ≤ |{g > µ}|. Then {f > λ}∗ ⊆ {g > µ}∗.So

|{f > λ} ∩ {g > µ}| ≤ |{f > λ}| = |{f > λ}∗| = |{f > λ}∗ ∩ {g > µ}∗|

=

∫χ{f>λ}∗(x)χ{g>µ}∗(x) dx

110

and hence∫f(x)g(x) dx ≤

∫Rd

∫ ∞0

χ{f>λ}∗(x) dλ

∫ ∞0

χ{g>µ}∗(x) dµ dx =

∫f∗(x)g∗(x) dx.

Next, we consider the case of equality. Suppose that f is strictly symmetric decreasing andthat

∫f(x)g(x) dx =

∫f∗(x)g∗(x) dx. Then

∫f(x)g(x) dx =

∫f(x)g∗(x) dx, hence∫ ∞

0

∫Rdf(x)χ{g>µ}(x) dx dµ =

∫ ∞0

∫Rdf(x)χ{g>µ}∗(x) dx dµ.

Because ∫Rdf(x)χ{g>µ}(x) dx ≤

∫Rdf(x)χ{g>µ}∗(x) dx

by the demonstrated inequality, it follows that∫Rdf(x)χ{g>µ}(x) dx =

∫Rdf(x)χ{g>µ}∗(x) dx

for almost every µ.We claim that |{g > µ}∆{g > µ}∗| = 0. To see this, first observe that all open balls

centered at 0 occur as super-level sets of f . That is, {f > λ} = B(0, r(λ)) where λ 7→ r(λ) isnon-increasing and r(λ)→∞ as λ→ 0, since f is strictly symmetric decreasing. Note thatr(λ) is continuous. If it were not, then by monotonicity there would be a jump discontinuityat some point λ0, so that

limε→0|{f > λ0 + ε}| − |{f > λ0 − ε}| < 0

and thuslimε→0|{λ0 − ε ≤ f < λ0 + ε}| > 0.

This implies that |{f = λ0}| > 0, which is a contradiction, as f is strictly symmetric de-creasing. Now, by the above equality, we have∫ ∞

0

∫χ{f>λ}(x)χ{g>µ}(x) dx dλ =

∫ ∞0

∫χ{f>λ}(x)χ{g∗>µ}(x) dx dλ.

If A ⊆ Rd is a Borel measurable function, then the function

FA(λ) :=

∫χ{f>λ}(x)χA(x) dx = |A ∩B(0, r(λ))|

is continuous. By our demonstrated inequality,

F{g>µ}(λ) ≤∫χ{f>λ}(x)χ{g∗>µ}(x) dx = F{g∗>µ}(λ).

The assumed equality implies∫ ∞0

F{g>µ}(λ) dλ =

∫ ∞0

F{g∗>µ}(λ) dλ

and thus F{g>µ}(λ) = F{g∗>µ}(λ) for almost every λ. Continuity of these functions givesequality for all λ > 0. That is,

|B(0, r(λ)) ∩ {g > µ}| = |B(0, r(λ)) ∩ {g∗ > µ}|

for all λ > 0. Because r(λ) is continuous and r(λ)→∞ as λ→ 0,

|B(0, r) ∩ {g > µ}| = |B(0, r) ∩ {g∗ > µ}|

for all r > 0. From this, it follows that |{g > µ}∆{g > µ}∗| = 0 for almost every µ asdesired. From this, we have g = g∗ almost everywhere.

111

A quick corollary of this inequality is that the symmetric decreasing rearrangementcompresses the L2 distance between functions.

Corollary 11.8. Let f, g ∈ L2(Rd → [0,∞)). Then ‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 .

Proof. We have

‖f∗ − g∗‖2L2 =

∫(f∗(x)− g∗(x))2 dx = ‖f∗‖2L2 + ‖g∗‖2L2 − 2

∫f∗(x)g∗(x) dx

≤ ‖f‖2L2 + ‖g‖2L2 − 2

∫f(x)g(x) dx

= ‖f − g‖2L2 .

This estimate holds for more general Lp spaces. In fact, we have the following moregeneral theorem.

Theorem 11.9. Let φ : R→ [0,∞) be convex with φ(0) = 0, and suppose that f, g : Rd → [0,∞)are vanishing at infinity. Then∫

φ(f∗ − g∗)(x) dx ≤∫φ(f − g)(x) dx.

Moreover, if φ is strictly convex and f is strictly symmetric decreasing, then equality holds if andonly if g = g∗ almost everywhere.

Proof. Write φ = φ+ + φ−, where

φ+(t) =

{φ(t) t ≥ 0

0 t < 0; φ−(t) =

{0 t ≥ 0φ(t) t < 0

.

Both φ+ and φ− are convex, and it suffices to prove the inequality for only φ+.Recall that convex functions are Lipschitz on compact domains, and that if c < a < x <

y < b < d, we have

φ+(a)− φ+(c)

a− c≤ φ+(y)− φ+(x)

y − x≤ φ+(d)− φ+(b)

d− b.

Thus, convex functions are absolutely continuous, hence differentiable almost everywhere,and so φ+(t) =

∫ t0 φ′+(s) ds. Moreover, φ′+ is a nondecreasing function, and strictly increas-

ing if φ is strictly convex. Thus, φ′+ has countably many (jump) discontinuities, and so wemay modify φ′+ on a countable set to ensure that φ′+ is left-continuous.

So,

φ+(f(x)− g(x)) =

∫ f(x)−g(x)

0φ′+(t) dt =

∫ f(x)

g(x)φ′+(f(x)− µ) dµ

=

∫ ∞0

φ′+(f(x)− µ)χ{g≤µ}(x) dµ

where the last equality follows from the fact that φ′+(t) = 0 for t < 0. Thus,∫φ+(f(x)− g(x)) dx =

∫ ∞0

∫φ′+(f(x)− µ)χ{g≤µ}(x) dx dµ.

We now make two claims, which are left as exercises to the reader.

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1. We have the inequality∫f(x)χ{g≤µ}(x) dx ≥

∫f∗(x)χ{g∗≤µ}(x) dx.

2. If F is nondecreasing, left continuous, and satisfies F (0) = 0, then (F ◦ f)∗ = F (f∗).

Using these two facts, we have∫φ+(f(x)− g(x)) dx ≥

∫ ∞0

∫[φ′+(f(x)− µ)]∗χ{g∗≤µ}(x) dx dµ

=

∫ ∞0

∫φ′+(f∗(x)− µ)χ{g∗≤µ}(x) dx dµ

where we have adopted the obvious convention that (f − µ)∗ := f∗ − µ to account for thefact that f − µ does not vanish at∞. Unraveling the above integral, it follows that∫

φ+(f(x)− g(x)) dx ≥∫φ+(f∗(x)− g∗(x)) dx,

which is the desired inequality.Next, we consider the equality case. Assume that φ is strictly convex, f is strictly sym-

metric decreasing, and that∫φ+(f(x)−g(x)) dx =

∫φ+(f∗(x)−g∗(x)) dx. Then for almost

every µ, ∫φ′+(f(x)− µ)χ{g≤µ}(x) dx =

∫φ′+(f(x)− µ)χ{g∗≤µ}(x) dx

because f = f∗. Let h(x) := φ′+(f(x) − µ). Because f is strictly symmetric decreasing andφ′+ is strictly increasing, h is strictly symmetric decreasing on the set {f(x) > µ}. Rewritingthe above equality with a layer-cake decomposition then gives∫ ∞

0

∫χ{h>λ}(x)χ{g≤µ}(x) dx dλ =

∫ ∞0

∫χ{h>λ}(x)χ{g∗≤µ}(x) dx dλ.

Define

F{g≤µ}(λ) :=

∫χ{h>λ}(x)χ{g≤µ}(x) dx;

F{g∗≤µ}(λ) :=

∫χ{h>λ}(x)χ{g∗≤µ}(x) dx.

By claim 1, F{g≤µ}(λ) ≥ F{g∗≤µ}(λ). Since∫∞

0 F{g≤µ}(λ) dλ =∫∞

0 F{g∗≤µ}(λ) dλ, we thenhave F{g≤µ}(λ) = F{g∗≤µ}(λ) for almost all λ > 0. Since these functions are continuous inλ, equality in fact holds for all λ > 0. So for all λ > 0,

|{h > λ} ∩ {g ≤ µ}| = |{h > λ} ∩ {g∗ ≤ µ}|

and so

|{f > r}∩{g ≤ µ}| = |{f > r}∩{g∗ ≤ µ}| ⇒ |{f > r}∩{g > µ}| = |{f > r}∩{g∗ > µ}|

for all r > µ, since h is strictly symmetric decreasing. Working through the exact sameargument with φ− in place of φ+ gives the above equality for all r < µ. As before, itfollows that

|{g > µ}∆{g∗ > µ}| = 0

and consequently g = g∗ almost everywhere.

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11.2 The Riesz Rearrangement Inequality, d = 1

Next, we consider a classical rearrangement inequality due to Riesz. The motivation forstudying inequalities of the kind to follow originates with Poincare and his work on theshapes of fluid bodies in equilibrium. The resulting variational problem deals with inte-grals of the form ∫∫

f(x)h(y)|x− y|−1 dx dy.

More generally, variational problems involving symmetric decreasing functions h(x − y)

such as the heat kernel (4πt)11 e−

|x−y|2t are of interest.

The following inequality holds more generally in Euclidean space of any dimension,but because of the technical nature of the proof, we begin by considering the one dimen-sional case.

Theorem 11.10 (Riesz rearrangement inequality, d = 1). Let f, g, h : R→ [0,∞) be vanishingat infinity. Then

I(f, g, h) :=

∫∫f(x)g(x− y)h(y) dx dy ≤

∫∫f∗(x)g∗(x− y)h∗(y) dx dy = I(f∗, g∗, h∗).

Moreover, if g is strictly symmetric decreasing, then equality holds if and only if there exists x0 ∈ Rsuch that f(x) = f∗(x− x0) and h(x) = h∗(x− x0) for almost all x ∈ R.

Proof. Using a layer-cake decomposition on f, g and h, we have

I(f, g, h) =

∫ ∞0

∫ ∞0

∫ ∞0

∫∫χ{f>λ}(x)χ{g>µ}(x− y)χ{h>ν}(y) dx dy dλ dµ dν.

Thus, it suffices to prove the inequality for f = χA, g = χB , and h = χC for finite-measuresets A,B,C. Approximating these sets from above by open sets An, Bn, and Cn, we haveI(χAn , χBn , χCn) → I(χA, χB, χC) and I(χA∗n , χB∗n , χC∗n) → I(χA∗ , χB∗ , χC∗) by the dom-inated convergence theorem, hence we may reduce further and assume without loss ofgenerality that A,B, and C are open finite-measure sets.

Because A is an open subset of R, we can write A =⋃j≥1 Ij where Ij are open, disjoint

intervals such that |Ij | ≥ |Ij+1|. Let Am :=⋃mj=1 Ij . Let Bm and Cm denote the analogous

decompositions forB and C. By the monotone convergence theorem, I(χAm , χBm , χCm)→I(χA, χB, χC) and I(χA∗m , χB∗m , χC∗m) → I(χA∗ , χB∗ , χC∗). Consequently, we may assumewithout loss of generality that A,B,C are each finite unions of disjoint open intervals.

Write

χA(x) =

J∑j=1

fj(x− aj)

χB(x) =K∑k=1

gk(x− bk)

χC(x) =L∑l=1

hl(x− cl)

where fj , gk, hl are characteristic functions of open intervals which are centered at 0. Thenwe have

I(χA, χB, χC) =

J∑j=1

K∑k=1

L∑l=1

∫∫fj(x− aj)gk(x− y − bk)hl(y − cl) dx dy.

114

Observe that (fj(x− aj))∗ = f∗j . To demonstrate the desired inequality, our goal is to takeeach of the characteristic functions in the integral and shift them towards 0. For t ∈ [0, 1]we define

Ijkl(t) :=

∫∫fj(x− taj)gk(x− y − tbk)hl(y − tcl) dx dy.

We claim that this function is non-increasing with respect to t. To see this, change variablesvia u = x− taj and u− v = x− y − tbk, so that y = v + t(aj − bk). Then

Ijkl(t) =

∫∫fj(u)gk(u− v)hl(v + t(aj − bk − cl)) du dv.

Let wjk(v) =∫fj(u)gk(u − v) du. Because fj and gk are both characteristic functions of

intervals centered at 0, this is a symmetric decreasing function in v. Thus

Ijkl(t) =

∫wjk(v)hl(v + t(aj − bk − cl)) dv

is non-increasing in t by the same reasoning.Next, define

It(χA, χB, χC) :=

J∑j=1

K∑k=1

L∑l=1

∫∫fj(x− taj)gk(x− y − tbk)hl(y − tcl) dx dy.

As t→ 0, the intervals corresponding to the functions fj(x−taj), gk(x−y−tbk), and hl(y−tcl) shift towards the origin. As t descends from 1 to 0, when two intervals correspondingto the same set A, B, or C, touch, redefine the fj ’s, gk’s, and hl’s accordingly. Carrying thisout for finitely many times (because there are finitely many intervals comprising A,B, andC), we end up with A∗,B∗, and C∗ intervals centered at 0. Moreover, since the functionsIjkl(t) are non-increasing,

I(χA, χB, χC) = I1(χA, χB, χC) ≤ I0(χA, χB, χC).

By construction, I0(χA, χB, χC) = I(χA∗ , χB∗ , χC∗). Thus, we have demonstrated the in-equality

I(χA, χB, χC) ≤ I(χA∗ , χB∗ , χC∗)

as desired.Next, we consider the case of equality. Assume that g is strictly symmetric decreasing

and that I(f, g, h) = I(f∗, g∗, h∗). Applying a layer-cake decomposition to f and h yieldsthe equality∫ ∞

0

∫ ∞0

∫∫χ{f>λ}(x)g(x− y)χ{h>µ}(y)dxdy dλ dµ

=

∫ ∞0

∫ ∞0

∫∫χ{f∗>λ}(x)g(x− y)χ{h∗>µ}(y)dxdy dλ dµ

since g = g∗. Consider the inner dx dy integrals. By the demonstrated inequality,∫∫χ{f>λ}(x)g(x− y)χ{h>µ}(y) dx dy ≤

∫∫χ{f∗>λ}(x)g(x− y)χ{h∗>µ}(y) dx dy.

Hence, equality of the above integrals implies that∫∫χ{f>λ}(x)g(x− y)χ{h>µ}(y) dx dy =

∫∫χ{f∗>λ}(x)g(x− y)χ{h∗>µ}(y) dx dy

for almost all λ, µ > 0. Fix λ and µ for which this holds.

115

Set A = {f > λ} and B = {h > µ}. We wish to show that there exists x0 ∈ R such thatf(x) = f∗(x − x0) and h(x) = h∗(x − x0) for almost all x ∈ R. That is, we wish to showthat A and B are intervals (up to a zero-measure set) centered at the same point, and thatthis central point does not depend on λ or µ. Fixing λ and letting µ vary (and vice versa)shows that the central point will not depend on λ or µ.

Performing a layer-cake decomposition on g in the above equality yields∫ ∞0

∫∫χ{f>λ}(x)χ{g>σ}(x− y)χ{h>µ}(y) dx dy dσ

=

∫ ∞0

∫∫χ{f∗>λ}(x)χ{g>σ}(x− y)χ{h∗>µ}(y) dx dy dσ.

Thus, applying our demonstrated inequality to the dx dy integrals as before, it follows that∫∫χ{f>λ}(x)χ{g>σ}(x− y)χ{h>µ}(y) dx dy

=

∫∫χ{f∗>λ}(x)χ{g>σ}(x− y)χ{h∗>µ}(y) dx dy

for almost all σ > 0. But

σ 7→∫∫

χ{f>λ}(x)χ{g>σ}(x− y)χ{h>µ}(y) dx dy;

σ 7→∫∫

χ{f∗>λ}(x)χ{g>σ}(x− y)χ{h∗>µ}(y) dx dy

are continuous functions in σ, since g is strictly symmetric decreasing. Therefore, equalityholds for all σ. The fact that g is strictly symmetric decreasing then gives the equality∫∫

χA(x)χ(− r2, r2

)(x− y)χB(y) dx dy =

∫∫χA∗(x)χ(− r

2, r2

)(x− y)χB∗(y) dx dy

for all r > 0, with A and B as defined above.We claim thatA andB must be intervals. To see this, pick r > |A|+ |B|. With this choice

of r, ∫∫χA∗(x)χ(− r

2, r2

)(x− y)χB∗(y) dx dy = |A| · |B|.

Then for all r > |A|+ |B|, we must have

|A| · |B| =∫∫

χA(x)χ(− r2, r2

)(x− y)χB(y) dx dy

=

∫χ(− r

2, r2

)(x)

∫χA(x+ y)χB(y) dy dx.

By this equality, the continuous function w(x) :=∫χA(x + y)χB(y) dy must have support

in the interval(− r

2 ,r2

). Let IA be the smallest interval such that |IA ∩A| = |A|, let IB be the

smallest interval such that |IB ∩ B| = |B|, and let IAB be the smallest interval containingthe support of w.

We leave it as an exercise to the reader to show that |IAB| = |IA|+ |IB|. With this claim,we then have |IA| + |IB| < r for all r > |A| + |B|. Taking the infimum over such r gives|IA| + |IB| ≤ |A| + |B|. Because IA and IB were chosen to be intervals that essentiallycontain A and B respectively, it follows that IA = A and IB = B, up to sets of measurezero, so that A and B are intervals.

Thus∫∫χA(x)χ(− r

2, r2


∫∫χA∗(x)χ(− r

2, r2

)(x− y)χB∗(y) dx dy

116

for all r > 0, where A and B are intervals. Letting r → 0 yields∫χA(x)χB(x) dx =

∫χA∗(x)χB∗(x) dx.

From here we consider two cases. If |A| = |B|, then∫χA∗(x)χB∗(x) dx = |A|. But if A and

B are not centered at the same point, then∫χA(x)χB(x) dx < |A|, which is a contradiction.

Thus, A and B are intervals centered at the same point, as desired. If |A| > |B|, let r =|A| − |B|. Then∫∫

χA∗(x)χ(− r2, r2

)(x− y)χB∗(y) dx dy =

∫B∗

∫A∗∩(y− r2 ,y+ r

2)dx dy = r|B|.

On the other hand, if A and B are not centered at the same point, then∫∫χA(x)χ(− r

2, r2


∫B

∫A∩(y− r2 ,y+ r

2)dx dy < r|B|.

This is a contradiction, thus A and B are centered at the same point. This completes theproof.

11.3 The Riesz Rearrangement Inequality, d > 1

In this section, we prove a generalization of the previous theorem for Euclidean space Rd.In order to do this, it is necessary another notion of symmetrizing sets and functions.

Definition 11.11. Let A ⊆ Rd be a Borel measurable set of finite measure. Let e ∈ Sd−1 ⊆Rd. The Steiner symmetrization of A along e is the set A∗e which is obtained from A bysymmetrizing along lines parallel to e.

Definition 11.12. Let f : Rd → [0,∞) be Borel measurable and vanishing at infinity. Fore ∈ Sd−1 ⊆ Rd, let ρ : Rd → Rd be the rotation taking e to e1. Write ρf(x) := f(ρ−1(x)),and let (ρf)∗1 be the one-dimensional symmetric decreasing rearrangement of ρf along x1,keeping x2, . . . , xd fixed. The Steiner symmetrization of f along e is f∗e := ρ−1(ρf)∗1.

To clarify the definition of the Steiner symmetrization of a function, observe that f∗1

can be written as follows:

f∗1(x1, x2, . . . , xd) = (x1 7→ f(x1, x2, . . . , xd))∗ .

Also note that these quantities are well-defined by the usual considerations of productmeasures and measurability of slices.

Theorem 11.13 (Riesz rearrangement inequality). Let f, g, h : Rd → [0,∞) be vanishing atinfinity. Then

I(f, g, h) :=

∫∫f(x)g(x− y)h(y) dx dy ≤

∫∫f∗(x)g∗(x− y)h∗(y) dx dy = I(f∗, g∗, h∗).

Moreover, if g is strictly symmetric decreasing, then equality holds if and only if there exists x0 ∈ Rdsuch that f(x) = f∗(x− x0) and g(x) = g∗(x− x0) for almost all x ∈ Rd.

Proof. The proof proceeds as follows. We consider the inequality when d = 2 separately,then prove the general inequality by induction. Finally, we consider the case of equality.

The case d = 2.

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Suppose that d = 2. By the usual layer-cake decomposition argument that we havegiven multiple times, it suffices to prove the inequality when f = χA, g = χB , and h =χC for finite measure sets A,B,C ⊆ R2. To simplify notation, we write I(A,B,C) :=I(χA, χB, χC). First, observe that I(A,B,C) ≤ I(A∗e, B∗e, C∗e) for any e ∈ S1. This followsfrom Fubini’s theorem and the one-dimensional Riesz rearrangement inequality, since inthis case Steiner symmetrization is essentially symmetric rearrangement one direction.

Next, we define sets Ak, Bk, and Ck for k ≥ 1 by more or less iterating Steiner sym-metrization in various directions, with the goal of showing Ak → A∗, Bk → B∗, andCk → C∗. So let α be an irrational multiple of 2π and let Rα denote the rotation aroundthe origin by α. Let X and Y denote the Steiner symmetrizations along the x and y axes,respectively. Note then that Xf = f∗1 and Y f = Rπ/2

(R−π/2f

)∗1. For k ≥ 1, define

Ak = (Y XRα)k(A)

Bk = (Y XRα)k(B)

Ck = (Y XRα)k(C).

In words, we obtain Ak by rotating A, symmetrizing in the X direction and then sym-metrizing in the Y direction, and then iterating this process k times. Because these opera-tions are invariant under measure, |Ak| = A. Moreover,Ak has double-reflection symmetryover both the x and y axes, and hence we can write

Ak ={

(x, y) ∈ R2 : |y| < ωAk (|x|)}

where ωAk : [0,∞) → [0,∞) is non-increasing, symmetric, and lower-semicontinuous. Theanalogous statements obviously hold for B and C.

Our goal is to show that χAkL2

−→ χA∗ , and similarly for B and C. To see why this issufficient, first observe that

|I(A∗, B∗, C∗)− I(Ak, Bk, Ck)|

≤∣∣∣∣∫∫ (χA∗ − χAk)(x)χB∗(x− y)χC∗(y) dx dy

∣∣∣∣+

∣∣∣∣∫∫ χAk(x)(χB∗ − χBk)(x− y)χC∗(y) dx dy

∣∣∣∣+

∣∣∣∣∫∫ χAk(x)χBk(x− y)(χC∗ − χCk)(y) dx dy

∣∣∣∣≤ ‖χA∗ − χAk‖L2 ‖χB∗ ∗ χC∗‖L2 + ‖χB∗ − χBk‖L2 ‖χAk ∗ χC∗‖L2

+ ‖χC∗ − χCk‖ ‖χAk ∗ χBk‖L2

≤ ‖χA∗ − χAk‖L2 ‖χB∗‖L1 ‖χC∗‖L2 + ‖χB∗ − χBk‖L2 ‖χAk‖L2 ‖χC∗‖L1

+ ‖χC∗ − χCk‖ ‖χAk‖L1 ‖χBk‖L1 .

Because A, B, and C are finite measure sets, if we can show that χAk → χA∗ and likewisefor B and C, then this shows that I(Ak, Bk, Ck)→ I(A∗, B∗, C∗). Next, observe that

I(Ak+1, Bk+1, Ck+1) = I(Y XRαAk, Y XRαBk, Y XRαCk)

≥ I(XRαAk, XRαBk, XRαCk)

≥ I(RαAk, RαBk, RαCk)

= I(Ak, Bk, Ck).

The first and second inequalities follow from the one-dimensional Riesz rearrangementinequality, and the last equality is because the quantity I(f, g, h) is invariant under rotation,

118

via the obvious change of variables. Hence, I(Ak, Bk, Ck) is an increasing sequence whichconverges to I(A∗, B∗, C∗). In particular,

I(A,B,C) = I(A0, B0, C0) ≤ I(A∗, B∗, C∗)

as desired.With this goal in mind, we make the following observations. Suppose that F and G

are finite measure sets. Then ‖RαχF −RαχG‖L2 = ‖χF − χG‖L2 , and by the inequality‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 we also have

‖XχF −XχG‖L2 ≤ ‖χF − χG‖L2

‖Y χF − Y χG‖L2 ≤ ‖χF − χG‖L2 .

Consequently, we may assume without loss of generality that A,B, and C are boundedsets. Indeed, for every ε > 0 there exists an r > 0 and a set A′ ⊆ Br(0) such that‖χA − χA′‖L2 < ε. Let A′k = (Y XRα)k(A′). Then by the above inequalities we have∥∥∥χAk − χA′k∥∥∥L2

≤ ‖χA − χA′‖L2 < ε.

Thus, it is enough to consider convergence of the finite measure sets A′k, and similarly forB and C.

From now on we suppose that A,B, and C are bounded. This implies that the func-tions ωAk , ω

Bk , and ωCk are uniformly bounded. For now we focus onA, as the arguments for

B and C are exactly the same. Using a diagonalization argument and passing to a subse-quence l(k), we see that {ωAl(k)} converges at every single rational number. By monotonicity,{ωAl(k)} converges everywhere except at countably many points, at most. This implies that

χAl(k)→ χA almost everywhere for some set A. Necessarily, A has double reflection sym-

metry across the x and y axes. To conclude that A = A∗, which then concludes the proof ofthe d = 2 inequality, it suffices to show that |A| is a ball, since |A| = limk→∞ |Al(k)| = |A|.

We will accomplish this by essentially showing thatRαA = A, and using that fact that αis an irrational multiple of 2π and thus A has rotational symmetry on a dense set of angles.To do this, we take advantage of the equality case in the ‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 , whichpresumes f to be strictly symmetric decreasing. Let

γ(x, y) = e−|x|2−|y|2

be a Gaussian on R2, which is indeed strictly symmetric decreasing. Let ak = ‖γ − χAk‖L2 .Note that Rαγ = γ, Xγ = γ, and Y γ = γ. So

ak+1 =∥∥γ − χAk+1

∥∥L2 = ‖Y XRαγ − Y XRαχAk‖L2 ≤ ‖γ − χAk‖L2

= ak.

Thus, {ak} is a decreasing sequence. Because ak ≥ 0, a := limk→∞ ak = limk→∞ al(k) exists.Moreover, by the dominated convergence theorem, a =

∥∥γ − χA∥∥L2 . Next, note that∣∣al(k)+1 −∥∥γ − Y XRαχA∥∥L2

∣∣ =∣∣∣∥∥∥γ − χAl(k)+1

∥∥∥L2−∥∥γ − Y XRαχA∥∥L2

∣∣∣≤∥∥∥χAl(k)+1

− Y XRαχA∥∥∥L2

=∥∥∥Y XRαχAl(k)

− Y XRαχA∥∥∥L2

≤∥∥∥χAl(k)

− χA∥∥∥L2

119

which→ 0 as k →∞. On the other hand,∣∣al(k)+1 −∥∥γ − Y XRαχA∥∥L2

∣∣→ ∣∣a− ∥∥γ − Y XRαχA∥∥L2

∣∣as k →∞. Thus, a =

∥∥γ − Y XRαχA∥∥L2 . But we have

a =∥∥γ − Y XRαχA∥∥L2 =

∥∥Y γ − Y XRαχA∥∥L2 ≤∥∥γ −XRαχA∥∥L2

=∥∥Xγ −XRαχA∥∥L2 ≤

∥∥γ −RαχA∥∥L2

=∥∥γ − χA∥∥L2

= a.

Therefore, we must have∥∥γ −RαχA∥∥L2 =

∥∥γ −XRαχA∥∥L2 , i.e.,∫∫|γ −RαχA|

2 dx dy =

∫∫|γ −XRαχA|

2 dx dy.

Because∫|γ − RαχA|

2 dx ≤∫|γ −XRαχA|

2 dx, it follows that∫|γ − RαχA|

2 dx =∫|γ −

XRαχA|2 dx for almost every y. Thus, for almost every y,XRαχA = RαχA for almost every

x. By Fubini’s theorem, XRαχA = RαχA almost everywhere in R2. The same argumentshows that Y XRαχA = XRαχA almost everywhere in R2, and hence Y XRαχA = RαχAalmost everywhere in R2. Thus, RαχA has double reflection symmetry about both axes. IfP denotes reflection across the y-axis, then

RαχA = PRαχA = R−αPχA = R−αχA

which implies that χA = R2αχA. Because α, and hence 2α, is an irrational multiple of 2π,the set {n(2α) mod 2π} is dense in [0, 2π). Thus, the function

µ(θ) =∥∥χA −Rθ∥∥

has dense set of zeroes. To conclude that A is a circle, and thus conclude the proof of theRiesz rearrangement inequality in the d = 2 case, it suffices to prove that µ is continuousin θ. By writing and expanding µ2 as an inner product, it suffices to show continuity of themap

θ 7→∫∫

χA ·RθχA dx dy.

To see this, let fn ∈ C∞ such that fn → χA. Then∣∣∣∣∫∫ χA ·RθχA dx dy −∫∫

fn ·RθχA dx dy∣∣∣∣ ≤ ∥∥χA − fn∥∥L2 · |A|

12 → 0

uniformly in θ. Because θ 7→∫∫

fn · RθχA dx dy is continuous, it then follows that θ 7→∫∫χA ·RθχA dx dy is a uniform limit of continuous functions, hence continuous. Thus, µ is

continuous, and so A is a ball and therefore A = A∗.

The case d > 2.Let e ∈ Sd−1 ⊆ Rd. Recall that the Steiner symmetrization of f : Rd → [0,∞) along e

is f∗e(x) = (f ◦ ρ−1)∗1(ρx), where ρ is a rotation defined by ρe = e1. Define the Schwartzsymmetrization along directions perpendicular to e by

f∗e⊥

(x) := (f ◦ ρ−1)∗e⊥1 (ρx),

where ∗e⊥1 indicates symmetrization in the variables x2, . . . , xd, keeping x1 fixed.As before, to prove the d > 2 case it suffices to consider f = χA, g = χB , and h =

χC for finite and bounded measure sets A,B,C. Let R be a rotation such that Red =

120

ed−1. Let Y denote the Steiner symmetrization along ed, and let X denote the Schwartzsymmetrizations perpendicular to ed. Let

Ak = (Y XR)kA

Bk = (Y XR)kB

Ck = (Y XR)kC.

Write x = (x′, xd) where x′ = (x1, . . . , xd−1). By induction, we have∫∫f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx

′ dy′

≤∫∫

(Xf)(x′, xd)(Xg)(x′ − y′, xd − yd)(Xh)(y′, yd) dx′ dy′.

In particular, I(Ak+1, Bk+1, Ck+1) ≥ I(Ak, Bk, Ck), so by the same argument as before it

suffices to prove χAkL2

−→ χA∗ , and likewise for B and C.By construction, we can write

Ak ={x ∈ Rd : |xd| < ωAk (|x′|)

}where ωAk is a symmetric and non-increasing function. The ωAk ’s are uniformly boundedby boundedness of A. Exactly as before, we may pass to a subsequence to get χAl(k)

→ χAalmost everywhere. As each Al(k) is rotationally symmetric around the xd axis, the limitingset A is also rotationally symmetric around the xd axis. Also as before, define the Gaussianγ and set ak = ‖γ − χAk‖L2 . Running through the same computation gives Y XRA = RA

almost everywhere. Hence, RA is rotationally symmetric along the xd axis, which impliesthat A is rotationally symmetric around the xd axis and the xd−1 axis. We will show thatthis implies that A is a ball. Then because |A| = |A| = |A∗| it will follow that A = A∗ almosteverywhere, as desired.

Let ϕε be a radial smooth approximation to the identity and set χε(x) = (ϕε ∗ χA)(x).Then χε and χA have the same symmetry properties. As such, we can write

χε(x1, . . . , xd) = u(√x2

1 + · · ·+ x2d−1, xd) = v(

√x2

1 + · · ·+ x2d−2 + x2

d, xd−1)

for smooth functions u and v of two variables. Writing ρ2 = x21 + · · · + x2

d−1, we have

u(√ρ2 + x2

d−1, xd) = v(√ρ2 + x2

d, xd−1) for all xd−1, xd ∈ R and for all ρ > 0. In particular,setting xd = 0 we have

u(√ρ2 + x2

d−1, 0) = v(√ρ2, xd−1)

for all xd−1 ∈ R and for all ρ > 0. Choosing ρ2 = x21 + · · ·+x2

d−2 +x2d gives χε(x) = u(|x|, 0),

hence χε is spherically symmetric and so A is a ball.This completes the proof of the inequality for d > 2.

The case of equality.Finally, we consider the equality statement. Suppose that g is strictly symmetric de-

creasing and that equality holds. As before, we assume that f = χA and h = χB for finitemeasure sets A and B. Using the (x′, xd) notation from above, we have∫

R

∫R

∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′ dxd dyd

=

∫R

∫R

∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′ dxd dyd.

121

By the inequality in the statement of the theorem,∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′

≤∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′.

Hence, the assumed equality gives∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′

=

∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′

for almost all (xd, yd).By induction we get setsAxd andByd corresponding to the functions f(·, xd) and h(·, yd)

which are balls in Rd−1 centered at the same point, by the equality case of the one dimen-sional Riesz rearrangement inequality. Also, by the same argument as before, this centralpoint is independent of xd and yd. Thus, A and B are rotationally symmetric with respectto a line L1 which is parallel to the xd axis. Similarly, A and B must also be rotationallysymmetric with respect to another line L2 parallel to the ed−1 axis.

We claim that L1∩L2 6= ∅. In d = 2, this is obvious, but the existence of skew-symmetriclines in higher dimensions makes the d ≥ 3 case more complicated. This can be seen via aping pong argument:1 if L1 and L2 do not intersect, then by rotating around L2 180 degreesit follows that the sets are rotationally symmetric about L1 and and an affine shift of L1.By rotating around this shift of L1, the sets are then also rotationally symmetric around anaffine shift of L2. Continuing this process, we contradict the finite measure of A and B.

Thus, A and B are rotationally symmetric about two axes. As we argued previously, itfollows that A and B are balls.

11.4 The Polya-Szego inequality

Our first application of the Riesz rearrangement inequality is a special case of the Polya-Szego inequality. We begin with a lemma.

Lemma 11.14. Let f : Rd → C be such that f ∈ L1loc and ∇f ∈ L1

loc. Then

∇|f |(x) =

{(u∇u+v∇v)(x)

|f(x)| if f = u+ iv 6= 0

0 if f = u+ iv = 0

in the sense of distributions. In particular, |∇|f || ≤ |∇f | almost everywhere.

Proof. First we demonstrate the “in particular” statement, assuming the formula for ∇|f |.For f(x) 6= 0 (up to a choice of representative), we have by Cauchy-Schwarz

|∇|f(x)||2 =

[u2|∇u|2 + v2|∇v|2 + 2uv∇u∇v

](x)

|f(x)|2

≤[u2|∇u|2 + v2|∇v|2 + u2|∇v|2 + v2|∇u|2

](x)

|f(x)|2

=(u2 + v2)(|∇u|2 + |∇v|2)

|f |2(x)

= |∇f(x)|2

1Drawing a picture here is helpful.

122

as desired.To prove the formula in the statement, let ψ ∈ C∞c . We wish to show∫

∇ψ |f | dx = −∫{f 6=0}

ψu∇u+ v∇v|f |

dx.

The first issue we need to deal with is moving the derivative onto |f |. To do this, for δ > 0consider

√δ2 + |f |2. Note that for small δ,

√δ2 + |f |2 ≤ 1 + |f |, so by the dominated

convergence theorem ∫∇ψ

√δ2 + |f |2 dx→

∫∇ψ |f | dx

as δ → 0. Next, let ϕε be an approximation to the identity, uε = u ∗ ϕε, and vε = v ∗ ϕε.Then uε → u and vε → v almost everywhere and in L1

loc. Also, ∇(uε + ivε) → ∇f almosteverywhere and in L1

loc. Note that∫∇ψ

(√δ2 + u2 + v2 −

√δ2 + u2

ε + v2ε

)dx

=

∫∇ψ (u− uε)(u+ uε) + (v − vε)(v + vε)√

δ2 + u2 + v2 +√δ2 + u2

ε + v2ε

dx

so that∣∣∣∣∫ ∇ψ (√δ2 + u2 + v2 −√δ2 + u2

ε + v2ε

)dx

∣∣∣∣≤∫|∇ψ| |u− uε|(|u|+ |uε|) + |v − vε|(|v|+ |vε|)√

δ2 + u2 + v2 +√δ2 + u2

ε + v2ε

dx

≤ 2

∫|∇ψ| (|u− uε|+ |v − vε|) dx.

Because uε → u and vε → v in L1loc, since ψ has compact support, it follows that∫

∇ψ√δ2 + u2

ε + v2ε dx→

∫∇ψ√δ2 + u2 + v2 dx

as ε→ 0. Integrating by parts, we then have∫∇ψ√δ2 + u2 + v2 dx = lim

ε→0

∫∇ψ√δ2 + u2

ε + v2ε dx

= − limε→0

∫ψuε∇uε + vε∇vε√δ2 + u2

ε + v2ε

dx

= − limε→0

∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx

− limε→0

∫ψuε(∇uε −∇u) + vε(∇vε −∇v)√

δ2 + u2ε + v2

ε

dx.

Consider the second integral. We have∣∣∣∣∣∫ψuε(∇uε −∇u) + vε(∇vε −∇v)√

δ2 + u2ε + v2

ε

dx

∣∣∣∣∣ ≤∫|ψ| (|∇uε −∇u|+ |∇vε −∇v|) dx

Again by L1loc convergence and compact support of ψ, this integral converges to 0 as ε→ 0.

In the first integral, observe that∣∣∣∣∣∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx

∣∣∣∣∣ ≤∫|ψ| (|∇u|+ |∇v|) dx <∞.

123

Thus, by the dominated convergence theorem,∫∇ψ√δ2 + u2 + v2 dx = − lim

ε→0

∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx

= −∫ψ

u∇u+ v∇v√δ2 + u2 + v2

dx.

Letting δ → 0 gives ∫∇ψ|f | dx = −

∫ψu∇u+ v∇v|f |

dx

as desired.

Theorem 11.15 (Polya-Szego). Let f : Rd → C be vanishing at infinity such that f ∈ L1loc and

∇f ∈ L2. Then ‖∇f∗‖L2 ≤ ‖∇f‖L2 .

Proof. First, note that ∇f ∈ L2 implies ∇f ∈ L1loc, hence by the previous lemma we have

|∇|f || ≤ |∇f |. Consequently, it suffices to consider functions f satisfying f ≥ 0.Next, we make a further reduction. For ε > 0, define

fε := min

{1

ε, [f − ε]+

}=

0 f ≤ ε

f − ε ε ≤ f ≤ ε+ ε−1

1ε f ≥ ε+ ε−1

= ϕε ◦ f

where

ϕε(x) =

0 x ≤ ε

x− ε ε ≤ x ≤ ε+ ε−1

1ε x ≥ ε+ ε−1

.

Note that ∫|fε| dx =

1

ε

∣∣{f ≥ ε+ ε−1}∣∣+

∫{ε≤f≤ε+ε−1}

|f − ε| dx

≤ 1

ε

∣∣{f ≥ ε+ ε−1}∣∣+

1

ε|{f ≥ ε}| .

Because f vanishes at infinity, the above super-level sets have finite measure, hence∫|fε| dx

is finite and so fε ∈ L1. Similarly, it can be shown that fε ∈ L2. Also note that

∇fε = ∇fχ{ε≤f≤ε+ε−1}L2

−→ ∇f

as ε→ 0 by the monotone convergence theorem.Next, recall that because ϕε is symmetric and non-decreasing, (fε)

∗ = (ϕε ◦ f)∗ =

ϕε ◦ f∗ = (f∗)ε. Thus, we have ∇(fε)∗ = ∇(f∗)ε

L2

−→ ∇f∗. This implies that it suffices toconsider f ∈ L1 ∩ L2 and∇f ∈ L2.

By Plancharel and the monotone convergence theorem, we have∫|∇f |2 dx =

∫|ξ|2|f(ξ)|2 dξ = lim

t→0

∫1− e−t|ξ|2

t|f(ξ)|2 dξ.

Because f ∈ L1 and f ≥ 0, we can distribute and use Fubini to get∫|∇f |2 dx = lim

t→0

1

t

[‖f‖2L2 − (2π)−d

∫e−t|ξ|

2

∫e−ix·ξf(x) dx

∫eiy·ξf(y) dy dξ

]= lim

t→0

1

t

[‖f‖2L2 − (2π)−d

∫∫f(x)f(y)

∫e−t|ξ|

2−i(x−y)·ξ dξ dx dy

]= lim

t→0

1

t

[‖f‖2L2 − (2π)−d

∫∫f(x)f(y)

∫e−t(ξ+

i(x−y)2t

)2

e−|x−y|2

4t dξ dx dy

]

124

where we are using the convention that for z ∈ Cd, z2 =∑d

j=1 z2j . Changing variables and

integrating the resulting ξ-Gaussian then gives∫|∇f |2 dx = lim

t→0

1

t

[‖f‖2L2 − (2π)−dt−

d2π

d2

∫∫f(x) e−

|x−y|24t f(y) dx dy

].

By the Riesz rearrangement inequality, and by equi-measurability of f∗ with f ,∫|∇f |2 dx ≥ lim

t→0

1

t

[‖f∗‖2L2 − (2π)−dt−

d2π

d2

∫∫f∗(x) e−

|x−y|24t f∗(y) dx dy

].

Beginning with the expression on the right and undoing all of the previous calculations,we recover the inequality ‖∇f‖2L2 ≥ ‖∇f∗‖2L2 .

We make passing observation that the heat kernel (4πt)−d2 e−

|x−y|24t appears in the above

proof. This use of the heat kernel is possible because the stated estimate is in L2. A gener-alization of the Polya-Szego inequality holds for other values of p, but the above methoddoes not give this result.

11.5 The Energy of the Hydrogen Atom

As an application of the Polya-Szego inequality, we consider the energy functional

E(f) :=

∫R3

|∇f(x)|2 − |f(x)|2

|x|dx

for f ∈ H1(R3). Such a functional can be taken to represent the energy of a hydrogen atom.We will demonstrate that

inf‖f‖L2=1

E(f)

is achieved by a symmetric decreasing function. Via a scaling argument, we show that theinfimum is negative. Let ϕ ∈ C∞c (R3) with ‖ϕ‖L2 = 1. For λ > 0, let ϕλ(x) = λ−

32ϕ(xλ

).

Then ‖ϕλ‖L2 = ‖ϕ‖2 = 1, and

E(ϕλ) = λ−2λ−3

∫ ∣∣∣(∇ϕ)(xλ

)∣∣∣2 dx− λ−3

∫ ∣∣ϕ (xλ)∣∣2|x|

dx

= λ−2 ‖∇ϕ‖2L2 − λ−1

∫|ϕ(x)|2

|x|dx

Because ‖∇ϕ‖2L2 and |ϕ(x)|2|x| dx are fixed numbers, for large λ the above quantity is negative.

Hence, Emin := inf‖f‖2=1E(f) < 0. To show that the problem is well-posed, we also mustprove that the infimum is not identically−∞. To do this, we first recall Hardy’s inequality:

Proposition 11.16 (Hardy’s Inequality). Let f ∈ S(Rd) and 0 ≤ s < d. Then∥∥∥∥f(x)

|x|s

∥∥∥∥Lp

. ‖|∇|sf‖Lp

for all 1 < p < ds .

By this inequality, we have∥∥∥∥∥f(x)

|x|12

∥∥∥∥∥L2

.∥∥∥|∇| 12 f(x)

∥∥∥L2

=∥∥∥|ξ| 12 f(ξ)

∥∥∥L2

=∥∥∥|ξ| 12 |f(ξ)|

12 |f(ξ)|

12

∥∥∥L2

≤∥∥∥|ξ| 12 |f(ξ)|

12

∥∥∥L4

∥∥∥|f(ξ)|12

∥∥∥L4

= ‖∇f‖12

L2 ‖f‖12

L2 .

125

So for ‖f‖L2 = 1, E(f) ≥ ‖∇f‖2L2 − C ‖∇f‖L2 where the constant C comes from Hardy’sinequality. So Emin ≥ infx>0 x

2 − Cx > −∞ as desired.Next, we show that the infimum is achieved. Let fn ∈ H1(R3) be a decreasing minimiz-

ing sequence, that is, ‖fn‖L2 = 1 for all n and E(fn) → Emin. First, we claim that we canfind a minimizing sequence of radially symmetric decreasing functions. Indeed, considerf∗n. We have ‖f∗n‖L2 = ‖fn‖L2 = 1, and by the Polya-Szego inequality, ‖∇f∗n‖L2 ≤ ‖∇fn‖L2 .By the Hardy-Littlewood rearrangement estimate,∫|fn(x)|2

|x|dx =

⟨1

|x|, |fn(x)|2

⟩≤⟨

1

|x|, (|fn(x)|2)∗

⟩=

⟨1

|x|, |f∗n(x)|2

⟩=

∫|f∗n(x)|2

|x|dx.

Hence, E(f∗n) ≤ E(fn), which implies that f∗n is also a minimizing sequence. Consequently,we may assume from now on that each fn is radially symmetric and decreasing.

Next, we claim that the sequence {fn} is bounded inH1(R3). If it were not, then because‖fn‖L2 = 1, the quantities ‖∇f‖L2 are unbounded. But since E(f) ≥ ‖∇f‖2L2 − C ‖∇f‖L2 ,this implies that lim supE(fn) = ∞. This contradicts the assumption that E(fn) decreasesto Emin < 0. Hence, {fn} is indeed bounded in H1(R3). Because H1(R3) can be reformu-lated on the Fourier side as an L2-space with a weighted measure, and bounded sets in L2

spaces are weakly compact, it follows that (passing to a subsequence) fn ⇀ f weakly inH1(R3).

We need to demonstrate that the weak limit f is an optimizer. First, we will show that

E(f) ≤ lim inf E(fn) = Emin

as desired. By weak lower semi-continuity of the L2 norm, ‖∇f‖L2 ≤ lim inf ‖∇fn‖L2 .This is great, but it doesn’t help in demonstrating the energy inequality because it tellsus nothing about the potential energy term −

∫ |f(x)|2|x| dx. To remedy this, we invoke the

following fact.

Lemma 11.17 (Strauss). The space H1(R3) embeds compactly into Lp(R3) for 2 < p < 6.

The proof of this lemma is found in Chapter 12, after discussing some compactnessresults in Lp spaces. Here we only remark that the upper bound 6 for p comes from thescaling condition of the Sobolev embedding theorem.

In particular, this lemma implies that because fn ⇀ f weakly in H1(R3), then fn → fstrongly in Lp(R3) for 2 < p < 6. With this in mind, we estimate as follows:∣∣∣∣∫ |fn(x)|2

|x|dx−

∫|f(x)|2

|x|dx

∣∣∣∣ ≤ ∣∣∣∣∫ (|fn| − |f |)(|fn|+ |f |)|x|

dx

∣∣∣∣≤∫|fn − f |(|fn|+ |f |)

|x|dx.

In order to use the fact that fn → f strongly in certainLp spaces, we want to apply Holder’sinequality. But since 1

|x| is only in weak Lp spaces, we split up the integral:∣∣∣∣∫ |fn(x)|2

|x|dx−

∫|f(x)|2

|x|dx

∣∣∣∣≤∫|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx+

∫|x|≥1

|fn − f |(|fn|+ |f |)|x|

dx.

In the first integral, 1|x| ∈ L

p(B(0, 1) ⊆ R3) for p < 3. Applying Holder’s inequality twicegives ∫

|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx ≤∥∥∥∥ 1

|x|

∥∥∥∥L2(B(0,1))

‖|fn − f |(|fn|+ |f |)‖L2(B(0,1))

≤∥∥∥∥ 1

|x|

∥∥∥∥L2(B(0,1))

‖fn − f‖L4 (‖fn‖L4 + ‖f‖L4) .

126

The first norm is finite by the above comment. We claim that the quantities ‖fn‖L4 + ‖f‖L4

are bounded in n. To see this, recall the Gagliardo-Nirenberg inequality:

Proposition 11.18 (Gagliardo-Nirenberg inequality). Fix d ≥ 1 and 0 < p < ∞ for d = 1, 2,or 0 < p < 4

d−2 for d ≥ 3. Then for all f ∈ S(Rd),

‖f‖p+2Lp+2 . ‖f‖p+2− pd

2

L2 ‖∇f‖pd2

L2 .

A proof of this inequality is also contained in Chapter 12. When p = 2 and d = 3,the inequality above takes the form ‖f‖4L4 . ‖f‖L2 ‖∇f‖3L2 . Because {fn} is a boundedsequence inH1(R3), it follows that ‖fn‖L4 +‖f‖L4 is bounded. Then because ‖fn − f‖L4 →0, ∫

|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx→ 0.

Similarly,∫|x|≥1

|fn − f |(|fn|+ |f |)|x|

dx ≤∥∥∥∥ 1

|x|

∥∥∥∥L4(|x|≥1)

‖fn − f‖L4 (‖fn‖L2 + ‖f‖L2) . ‖fn − f‖L4 .

Thus, ∫|fn − f |(|fn|+ |f |)

|x|dx→ 0.

From this and the fact that ‖∇f‖L2 ≤ lim inf ‖∇fn‖L2 it follows that

E(f) ≤ lim inf E(fn) = Emin

as desired.In fact, a stronger statement is true. All of the optimizers of the energy functional E(f)

above are radially symmetric; in particular, any optimizer satisfies f = eiθf∗. We sketch theargument here. If f is an optimizer, then as seen above it follows that f∗ is also an optimizer.From E(f) = E(f∗) and Polya-Szego we deduce the equalities ‖∇f‖L2 = ‖∇f∗‖L2 and∥∥∥∥f(x)

|x|12

∥∥∥∥L2

=

∥∥∥∥f∗(x)

|x|12

∥∥∥∥L2

. By the equality case of our first rearrangement inequality, |f | = f∗

almost everywhere. Unfortunately this does not rule out the possibility that f itself is notradial. However, we then have

‖∇f‖L2 = ‖∇f∗‖L2 = ‖∇|f |‖L2 .

Having previously shown |∇|f || ≤ |∇f | almost everywhere, it follows that |∇|f || = |∇f |almost everywhere. Then if f = u + iv, we have u∇v = v∇u almost everywhere. Writingf = reiθ where r and θ are functions of x, we then have r∇θ = 0 almost everywhere.Because f is not identically 0, θ is constant as desired.

127

Chapter 12

Compactness in Lp Spaces

The analysis of the energy of the hydrogen atom in Chapter 11 is a model problem inthe calculus of variations. Given a functional on some spaces of functions, we wish todetermine its optimimum value, either a supremum or infimum, and moreover show thatthis optimum is achieved. By definition of the supremum and infimum, we can alwaysfind a sequence of functions whose functional values tend towards the optimum. Fromhere, we would ideally invoke some sort of compactness of the function space to extracta limiting function which optimizes the functional. However, as we know from previousmathematical experiences, compactness in infinite dimensional spaces, and in particularfunction spaces, is far from ideal. For example, it is well-known that a closed and boundedset in an infinite dimensional Banach space is not compact in the norm-topology. We can,however, achieve varying degrees of compactness by considering weaker topologies. Alsowell-known is the the following result, which we have invoked throughout the text already.

Theorem 12.1 (Banach-Alaoglu). Let X be a Banach space. A closed and bounded set in X∗ iscompact in the weak-* topology.

If X is reflexive, it immediately follows that a closed and bounded set in X is weaklycompact. In the context of Lp-spaces (where we also have separability), this is manifestedby the following theorem.

Theorem 12.2 (Weak compactness). If {fn} is a bounded sequence in Lp(Rd) for 1 < p < ∞,then, passing to a subsequence, fn ⇀ f for some f ∈ Lp(Rd).

Oftentimes, weak compactness is not enough to solve optimization problems. For ex-ample, if fn is a translation of a bump function whose profile marches off to infinity, thenfn ⇀ 0. As such, we must seek compactness with more care.

The next four chapters of this text may be taken a single unit; a careful study of com-pactness in various function spaces and its consequences in harmonic analysis and partialdifferential equations. In the present chapter, we begin by developing a foundation ofstandard compactness results, and in Chapters 13, 14, and 15 we study the delicate no-tion of concentration compactness, which measures the defects of compactness in variouscontexts.

12.1 The Riesz Compactness Theorem

The first theorem we prove addresses the question of precompactness in the norm topology.For continuous functions, the answer is given by the well-known Arzela-Ascoli theorem.The main compactness tool used in this chapter is an analogous result for Lp spaces.

Theorem 12.3 (Riesz compactness). Fix 1 ≤ p < ∞. A family F ⊆ Lp(Rd) is precompact inLp(Rd) if and only if it satisfies the following three properties:

128

1. (Boundedness) There exists A > 0 such that ‖f‖Lp ≤ A for all f ∈ F .

2. (Equicontinuity) For every ε > 0 there exists a δ > 0 such that∫|f(x+ y)− f(x)|p dx < ε

for all |y| < δ and for all f ∈ F .

3. (Tightness) For every ε > 0, there exists an R > 0 such that∫|x|≥R |f(x)|p dx < ε for all

f ∈ F .

Proof. The forward direction is fairly straightforward. Suppose that F is precompact inLp(Rd). Then for every ε > 0, there exists f1, . . . , fn ∈ F such that F ⊆

⋃nj=1B(fj ,

ε10).

Because any function f ∈ F lies in one of these balls, ‖f‖Lp ≤ε10 + maxj ‖fj‖Lp . Thus, F is

bounded. Next, for |y| < δ = δ(ε) << 1 sufficiently small, we have

‖f(·+ y)− f‖Lp ≤ 2ε

10+ max

j‖fj(·+ y)− fj‖Lp < ε,

so that F is equicontinuous. Finally,

‖f‖Lp(|x|≥R) ≤ε

10+ max

j‖fj‖Lp(|x|≥R) < ε

for R sufficiently large. Hence, F is tight.The other direction is more involved. Suppose that F is bounded, equicontinuous, and

tight. Fix ε > 0. We need to show the existence of f1, . . . , fn ∈ F so that F ⊆⋃nj=1B(fj , ε).

We will approximate F by a continuous family of functions and then apply Arzela-Ascoli. Let ϕ ∈ C∞c be a radial bump function satisfying

∫ϕdx = 1 and

ϕ(x) =

{cd |x| ≤ 1

20 |x| > 1

for some dimensional constant cd. For R > 0 and f ∈ F , define

fR(x) = ϕ( xR

)∫f(x− x

R

)ϕ(y) dy = ϕ

( xR

)∫f(y)Rdϕ(R(x− y)) dy.

The passage from f to fR “fuzzes” f at a scale 1/R. Define FR = { fR : f ∈ F }. BecausefR is defined as convolution with a smooth function multiplied by a compactly supportedfunction, FR is a continuous family of functions supported on B(0, R). Also,

‖f − fR‖Lp =∥∥∥(1− ϕ

( xR

))f(x)

∥∥∥Lp

+

∥∥∥∥ϕ( xR)(f(x)−

∫f(x− x

R

)ϕ(y) dy

)∥∥∥∥Lp.

Because F is tight, we can estimate the first norm by∥∥∥(1− ϕ( xR

))f(x)

∥∥∥Lp

= ‖f‖Lp(|x|>R) <ε

10

for R = R(ε) >> 1 sufficiently large. For the second norm, we use the fact that ϕ has unitmass. We have∥∥∥∥ϕ( xR)

(f(x)−

∫f(x− x

R

)ϕ(y) dy

)∥∥∥∥Lp≤∥∥∥∥∫ (f(x)− f

( xR

))ϕ(y) dx

∥∥∥∥Lpx

≤∫|ϕ(y)|

∥∥∥f(x)− f( xR

)∥∥∥Lpx

dy

≤ ε

10

for R sufficiently large, by equicontinuity of F . Thus, ‖f − fR‖Lp <ε5 for R sufficiently

large.

129

Next, we claim thatFR is uniformly bounded and equicontinuous. To see boundedness,by Young’s inequality and a change of variables we have

‖fR‖L∞ .∥∥∥f (x− y

R

)∥∥∥Lpy‖ϕ‖Lp′ . R

dp ‖f‖Lp . R

dpA.

To see equicontinuity, write

|fR(x+ y)− fR(x)| ≤∣∣∣∣[ϕ(x+ y

R

)− ϕ

( xR

)]∫f(x+ y − z

R

)ϕ(z) dz

∣∣∣∣+

∣∣∣∣ϕ( xR)∫ [

f(x+ y − z

R

)− f

(x− z

R

)]ϕ(z) dz

∣∣∣∣.

∣∣∣∣ϕ(x+ y

R

)− ϕ

( xR

)∣∣∣∣R dp ‖f‖Lp ‖ϕ‖Lp′

+∥∥∥f (x+ y − z

R

)− f

(x− z

R

)∥∥∥Lpz‖ϕ‖Lp′

.

∣∣∣∣ϕ(x+ y

R

)− ϕ

( xR

)∣∣∣∣R dpA+R

dp ‖f(·+ y)− f(·)‖Lp .

Because ϕ is smooth and because F is tight, for |y| < δ = δ(ε,R) with δ sufficiently small,|fR(x+ y)− fR(x)| < ε uniformly in f .

Therefore, as FR is a continuous family of uniformly bounded and equicontinuouscompactly supported functions, it follows by the Arzela-Ascoli theorem that FR is pre-compact in the L∞ topology. So for every ε > 0 there exists f1, . . . , fn ∈ F such thatFR ⊆

⋃nj=1BL∞ ((fj)R, ε).

Fix f ∈ F . There exists a j such that fR ∈ BL∞((fj)R, ε). By the triangle inequality, wehave

‖f − fj‖Lp ≤ ‖f − fR‖Lp + ‖fj − (fj)R‖Lp + ‖fR − (fj)R‖Lp

<ε

5+ε

5+ cd ‖fj − (fj)R‖L∞ R

dp .

Because ‖fj − (fj)R‖L∞ < ε, by choosing ε sufficiently small it follows that ‖f − fj‖Lp < ε,and hence F ⊆

⋃nj=1B(fj , ε) as desired.

Our first corollary of the Riesz compactness theorem is an alternative characterizationof precompact families when p = 2. In particular, in this case we can replace the equiconti-nuity condition by tightness in the Fourier domain.

Corollary 12.4. A family F ⊆ L2(Rd) is precompact if and only if it satisfies the following twoproperties:

1. There exists A > 0 such that ‖f‖L2 ≤ A for all f ∈ F .

2. For every ε > 0, there exists R > 0 such that∫|x|≥R

|f(x)|2 dx+

∫|ξ|≥R

|f(ξ)|2 dξ < ε

for all f ∈ F .

Proof. The forward direction is almost exactly the same as in the proof of the Riesz com-pactness theorem; tightness in the Fourier domain (i.e. the estimate

∫|ξ|≥R |f(ξ)|2 dξ < ε)

comes from the fact that the Fourier transform is an isometry on L2(Rd).Conversely, suppose the two conditions hold. By the Reisz compactness theorem, it

suffices to verify that F is equicontinuous.

130

Fix ε > 0. Using the fact that the Fourier transform is an isometry on L2(Rd), for anyR > 0 we can write

‖f(x+ y)− f(x)‖L2x

=∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2ξ

≤∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2(|ξ|≥R)

+∥∥∥f(ξ)

[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

.

When |ξ| is large, there is a lot of oscillation coming from the eiy·ξ phase. Thus, we simplyuse the triangle quality and tightness of f to estimate the first term:∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2(|ξ|≥R)

≤ 2∥∥∥f∥∥∥

L2(|ξ|≥R)<ε

2

for R sufficiently large. For the second term, we have∥∥∥f(ξ)[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

≤∥∥∥f · |y| · |ξ|∥∥∥

L2(|ξ|≤R)≤ |y|RA.

Then∥∥∥f(ξ)

[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

< ε/2 for |y| < δ where δ(R, ε) is sufficiently small. For

such a choice of δ, ‖f(x+ y)− f(x)‖L2x< ε uniformly in f and hence F is equicontinuous.

12.2 The Rellich-Kondrachov Theorem

In this section, we consider the issue of compactness in the Sobolev space H1(Rd) and thehomogeneous Sobolev space H1(Rd). For convenience, we recall their definitions here.

Definition 12.5.

1. The Hilbert space H1(Rd) is the completion of S(Rd) under the norm ‖f‖H1 :=‖f‖L2 + ‖∇f‖L2 .

2. The Hilbert space H1(Rd) is the completion of S(Rd) under the norm ‖f‖H1 :=‖f‖L2 + ‖∇f‖L2 .

In words, functions inH1(Rd) are L2-functions whose (distributional) derivative is alsoin L2(Rd). For a function in H1(Rd) we only assume that the gradient is in L2(Rd); thefunction itself may not be in L2(Rd).

The two most important estimates in the study of the Sobolev spaces H1(Rd) andH1(Rd) are the Gagliardo-Nirenberg inequality and the p = 2 Sobolev embedding theo-rem, respectively. Both results have already been stated and proven in this text, but werecall them here for convenience.

Proposition 12.6 (Gagliardo-Nirenberg). Fix d ≥ 1. Let p satisfy 2 ≤ p <∞ if d = 1 or d = 2,and 2 ≤ p < 2d

d−2 if d ≥ 3. Then for f ∈ H1(Rd),

‖f‖Lp . ‖f‖2d−p(d−2)

2p

L2 ‖∇f‖d(p−2)

2p

L2 .

Proposition 12.7 (Sobolev embedding, p = 2). Fix d ≥ 3. Suppose that f ∈ S(Rd) with∇f ∈ L2(Rd). Then

‖f‖L

2dd−2

. ‖∇f‖L2 .

From Gagliardo-Nirenberg we have the relationship ‖f‖p . ‖f‖H1 , and from Sobolevembedding we have ‖f‖

L2dd−2

. ‖f‖H1 . As promised in Chapter 5, we provide a slick proofof Proposition 12.6 using Littlewood-Paley theory, which we previously did not have atour disposal.

131

Proof of Proposition 12.6. It suffices to consider Schwartz functions f .Because the Littlewood-Paley projections of f converge to f in Lp for the prescribed

values of p, we have‖f‖Lp ≤

∑N∈2Z

‖fN‖Lp .

By Bernstein’s first inequality,

‖fN‖Lp . Nd2− dp ‖fN‖L2 . N

d2− dp ‖f‖L2 .

On the other hand, by Bernstein’s second inequality we have

‖fN‖Lp . Nd2− dpN−1 ‖∇fN‖L2 . N

d2− dp−1 ‖∇f‖L2 .

Together, this implies that

‖f‖Lp .∑N∈2Z

min{N

d2− dp ‖f‖L2 , N

d2− dp−1 ‖∇f‖L2

}.

Because the Gagliardo-Nirenberg inequality is trivial when p = 2, we may assume thatp > 2. Consequently, d2 −

dp > 0. Also, d2 −

dp − 1 = d−2

2 −dp < 0. This latter statement is

clear when d = 1 or d = 2, and when d ≥ 3 this follows from the fact that p < 2dd−2 . Using

the positivity and negativity of the above exponents, we split the sum into two summableseries to get the desired inequality. Explicitly,

‖f‖Lp .∑

N.‖∇f‖

L2‖f‖

L2

Nd2− dp ‖f‖L2 +

∑N&

‖∇f‖L2

‖f‖L2

Nd2− dp−1 ‖∇f‖L2

. ‖∇f‖d2− dp

L2 ‖f‖1− d

2+ dp

L2 + ‖∇f‖d2− dp

L2 ‖f‖1+ d

p− d

2

L2

. ‖f‖2d−p(d−2)

2p

L2 ‖∇f‖d(p−2)

2p

L2 .

For an H1-bounded family of functions, the Reisz compactness theorem leads to thefollowing important result.

Theorem 12.8 (Rellich-Kondrachov, H1(Rd)). Let χR be a cutoff function on BR(0) ⊆ Rd,smooth or not. Then the family F := {χRf : ‖f‖H1 ≤ 1 } is compact in Lp(Rd) for{

2 ≤ p <∞ d = 1, 2

2 ≤ p < 2dd−2 d ≥ 3

.

Proof. LetB1 denote the unit ball inH1(Rd). We begin with the observation thatF is closed.To see this, let fn ∈ B1 be a sequence of functions such that χRfn converges in Lp. Becausefn ∈ B1, the sequence fn is bounded in L2. Passing to a subsequence, we have fn ⇀ fweakly in L2. This implies that χRfn ⇀ χRf weakly in L2. As weak limits are unique, itfollows that χRfn → χrf in Lp and hence F is closed. Thus, to prove the theorem it sufficesto prove that F is precompact.

We will invoke the Reisz compactness theorem. As such, we need to demonstrateboundedness, equicontinuity, and tightness. By Gagliardo-Nirenberg, ‖χRf‖Lp ≤ ‖f‖Lp .‖f‖H1 and hence the family F is bounded. Tightness of F is immediate, as each functionin F has compact support.

132

It remains to show equicontinuity. Fix ε > 0. We have

‖χRf(x+ y)− χRf(x)‖Lpx≤ ‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx + ‖χR(x)[f(x+ y)− f(x)]‖Lpx .

To estimate the first term, we want to use Holder, the Lp continuity property of translationson χR to get smallness, and then Gagliardo-Nirenberg on the f term to get uniformity inf . However, because translation of χR is not continuous in the L∞-norm, we have to becareful. We have

‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx ≤ ‖χR(x+ y)− χR(x)‖Lrx ‖f(x+ y)‖Lqx

for r and q satisfying

p < q <

{∞ d = 1, 22dd−2 d ≥ 3

;1

p=

1

r+

1

q.

By Gagliardo-Nirenberg, ‖f(x+ y)‖Lqx is uniformly bounded in f . For |y| < δ(ε) << 1where δ is sufficiently small, we have ‖χR(x+ y)− χR(x)‖Lrx < ε.

To estimate the second term, we simply ignore the χR and use Gagliardo-Nirenberg:

‖χR(x)[f(x+ y)− f(x)]‖Lpx ≤ ‖f(x+ y)− f(x)‖Lpx. ‖f(x+ y)− f(x)‖θL2

x‖∇x[f(x+ y)− f(x)]‖1−θL2

x

where θ is the correct exponent dictated by Gagliardo-Nirenberg. By the triangle inequality,

‖∇x[f(x+ y)− f(x)]‖1−θL2x

. ‖∇f‖1−θL2

and hence is uniformly bounded in f . By the fundamental theorem of calculus,

‖f(x+ y)− f(x)‖θL2x. (|y| ‖∇f‖L2)θ.

Thus, for |y| sufficiently small, ‖χR(x)[f(x+ y)− f(x)]‖Lpx < ε.These two estimates give equicontinuity of F . By the Riesz compactness theorem, the

proof is complete.

The same result holds for H1(Rd). The proof is for the most part the same, with theSobolev embedding theorem replacing the Gagliardo-Nirenberg inequality as the main es-timation tool.

Theorem 12.9 (Rellich-Kondrachov, H1(Rd)). Fix d ≥ 3 and 2 ≤ p < 2dd−2 . Let χR be a cutoff

function on BR(0) ⊆ Rd, smooth or not. Then the family F :={χRf : ‖f‖H1 ≤ 1

}is compact

in Lp(Rd).

Proof. The proof of closedness of F is the same, this time using the fact that the sequenceis bounded in L

2dd−2 (Rd) (by Sobolev embedding) rather than L2(Rd).

To use the Riesz compactness theorem, we need to show boundedness, tightness, andequicontinuity. As before, tightness is immediate by compact support. To see bounded-ness, we first apply Holder, and then Sobolev embedding:

‖χRf‖Lp ≤ ‖χR‖Lα ‖f‖L

2dd−2

. ‖f‖H1 . 1

where 1α = 1

p −1

2d/(d−2) .

133

To prove equicontinuity, write

‖χRf(x+ y)− χRf(x)‖Lpx≤ ‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx + ‖χR(x)[f(x+ y)− f(x)]‖Lpx

as before. Estimating the first term, we have

‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx ≤ ‖χR(x+ y)− χR(x)‖Lαx ‖f(x+ y)‖L

2dd−2x

. ‖χR(x+ y)− χR(x)‖Lαx .

Continuity of translations in Lα(Rd) gives equicontinuity in this term. To estimate thesecond term, write

‖χR(x)[f(x+ y)− f(x)]‖Lpx ≤ ‖f(x+ y)− f(x)‖Lpx≤ ‖f(x+ y)− f(x)‖1−θL2

x‖f(x+ y)− f(x)‖θ

L2dd−2x

where θ = d(

12 −

12d/(d−2)

). By Sobolev embedding, the latter term is uniformly bounded.

The first term we estimate by first using the fundamental theorem of calculus:

‖f(x+ y)− f(x)‖1−θL2x

. (|y| ‖∇f‖L2)1−θ . |y|1−θ

which gives equicontinuity..

12.3 The Strauss Lemma

In Chapter 11, we studied the energy functional of the hydrogen atom. A crucial step inour analysis was the Strauss lemma, which we did not prove. In this section, we provide aproof.

First, we have the following theorem, which describes the decay of radial functions inH1(Rd).

Proposition 12.10 (Radial Sobolev embedding). Fix d ≥ 2 and 1 ≤ p < ∞. For a radialfunction f ∈ H1(Rd) ∩ Lp(Rd), we have

r2(d−1)p+2 |f(r)| . ‖f‖

p2+p

Lp ‖∇f‖2

2+p

L2

for almost all r > 0.

Proof. It suffices to prove the inequality for Schwartz functions. Indeed, S(Rd) is densein H1(Rd) ∩ Lp(Rd), and by passing to a subsequence we can ensure almost everywhereconvergence. As |∇|f || ≤ |∇f | almost everywhere, we can further assume that f ≥ 0.

By the fundamental theorem of calculus we have

rd−1f(r)1+ p2 =

∣∣∣∣rd−1

∫ ∞r

f ′(σ)f(σ)p2 dσ

∣∣∣∣ . ∫ ∞r

σd−1|f ′(σ)|f(σ)p2 dσ

.

(∫ ∞r

σd−1|f ′(σ)|2 dσ) 1

2(∫ ∞

rσd−1|f(σ)|p dσ

) 12

. ‖∇f‖L2 ‖f‖p2Lp .

The final inequality here comes from the fact that f is radial and noting that σd−1 dσ de-scribes a polar coordinate transformation. Raising both sides of this inequality to the power

2p+2 completes the proof.

134

Observe that as p increases in the above inequality, the decay condition weakens. Inparticular, the case p = 2 gives particularly good decay. As H1(Rd) ∩ L2(Rd) = H1(Rd),this proves the following.

Corollary 12.11. If f ∈ H1rad(Rd) for d ≥ 2, then

rd−1

2 |f(r)| . ‖f‖12

L2 ‖∇f‖12

L2

for almost all r > 0.

Finally, we prove the Strauss lemma.

Lemma 12.12 (Strauss). Fix d ≥ 2 and 2 < p < ∞ if d = 2 or 2 < p < 2dd−2 if d ≥ 3. Then

H1rad(Rd) embeds compactly into Lp(Rd).

Proof. By normalizing, it suffices to prove that the unit ball B1 ⊆ H1rad is precompact in

Lp(Rd). Naturally, we will apply the Riesz compactness theorem.The arguments for boundedness and equicontinuity are similar to those used in the

proof of the Rellich-Kondrachov theorem. For f ∈ B1, Gagliardo-Nirenberg gives

‖f‖Lp . ‖f‖θL2 ‖∇f‖1−θL2 . 1

for some the appropriate exponent θ. Thus,B1 isLp-bounded. Equicontinuity follows fromGagliardo-Nirenberg and the fundamental theorem of calculus:

‖f(x+ y)− f(x)‖Lpy . ‖f(x+ y)− f(x)‖θL2 ‖∇x(f(x+ y)− f(x))‖1−θL2

. (|y| ‖∇f‖L2)θ ‖∇f‖1−θL2

. |y|θ.

It remains to show tightness. We wish to use Corollary 12.11. If we try to use the theoremdirectly, we get ∫

|x|≥R|f(x)|p dx .

∫|x|≥R

r−d−1

2p dx =

∫ ∞R

r−d−1

2prd−1 dr

. Rd−d−1

2p

provided that d − d−12 p < 0, hence p > 2d

d−1 . But the conclusion holds for p > 2, so this isnot quite strong enough. Instead, we first use the fact that f ∈ L2(Rd). Writing∫

|x|≥R|f(x)|p dx =

∫|x|≥R

|f(x)|2 |f(x)|p−2 dx

and using the radial Sobolev embedding theorem on |f(x)|p−2 rather than |f(x)|p gives∫|x|≥R

|f(x)|p dx . ‖f‖2L2 R− d−1

2(p−2) ‖f‖2L2 .

This can be made uniformly small for p > 2 by choosing R sufficiently large.By the Riesz compactness theorem, the proof is complete.

The endpoint cases in the Strauss lemma, namely p = 2 and p = 2dd−2 when d ≥ 3, do

not hold. That is, H1rad(Rd) does not embed compactly into L2(Rd) or L

2dd−2 (Rd). To see that

H1rad(Rd) does not embed compactly into L2(Rd), we choose a radial bump function and

135

rescale on a large scale. Explicitly, let ϕ ∈ C∞c be radial, and define ϕλ(x) := λ−d2ϕ(xλ).

Then ‖ϕλ‖L2 = ‖ϕ‖L2 . Also,

‖∇ϕλ‖L2 = λ−d2λ−1λ

d2 ‖∇ϕ‖L2 = λ−1 ‖∇ϕ‖L2

which is . 1 for λ >> 1, hence ϕλ ∈ H1(Rd). Note that

| 〈ϕλ, ψ〉 | . λ−d2 ‖ϕ‖L∞ ‖ψ‖L1

and so ϕ ⇀ 0as λ → ∞. However, as the L2-norm of ϕλ is constant in λ, the ϕλ’s cannotconverge strongly to 0 in L2(Rd). The argument for showing thatH1

rad(Rd) does not embed

compactly intoL2dd−2 (Rd) is similar; here, we rescale to small scales. Letϕλ(x) = λ−

d−22 ϕ(xλ).

Then ‖ϕλ‖L

2dd−2

= ‖ϕ‖L

2dd−2

, and in fact ‖ϕλ‖H1 = ‖ϕ‖H1 . Also,

‖ϕλ‖L2 = λ−d−2

2 λd2 ‖ϕ‖L2 = λ ‖ϕ‖L2

which is . 1 for λ << 1. We have ϕλ ⇀ 0 as λ → 0, but ϕλ cannot converge to 0 inL

2dd−2 (Rd).

12.4 The Refined Fatou’s Lemma

The last result that we include in this chapter is an improvement on Fatou’s lemma. Itwill be important for establishing so-called asymptotic decoupling in Lp(Rd) when we studyconcentration compactness in Chapters 13 and 14.

Lemma 12.13 (Brezis-Lieb). Fix 1 ≤ p <∞, and let fn be an Lp-bounded sequence of functions.Suppose that fn → f almost everywhere. Then∫

| |fn|p − |f − fn|p − |f |p | dx = 0.

Proof. First, recall that for any real numbers a and b,∣∣∣ |a+ b|p − |a|p∣∣∣ ≤ ε|a|p + Cε|b|p (12.1)

for any ε > 0, where Cε is some constant dependent on ε and p. Let

gεn :=(∣∣∣ |fn|p − |fn − f |p − |f |p ∣∣∣ − ε|fn − f |p)+

.

Because fn → f almost everywhere, gεn → 0 almost everywhere. By the triangle inequalityand (12.1) with a = fn − f and b = f ,

gεn ≤(∣∣∣ |fn|p − |fn − f |p∣∣∣ + |f |p − ε|fn − f |p

)+

≤ (Cε|f |p + |f |p)+

= (Cε + 1) |f |p.

As f ∈ Lp (the sequence fn is Lp-bounded!), |f |p is integrable and so by the Dominatedconvergence theorem

∫gεn dx→ 0. Thus,∫

| |fn|p − |f − fn|p − |f |p | dx

≤∫ (∣∣∣ |fn|p − |fn − f |p − |f |p ∣∣∣ − ε|fn − f |p)+

dx+

∫ε|fn − f |p dx

=

∫gεn dx+ ε

∫|fn − f |p dx.

For any ε > 0, the first quantity goes to 0. As fn is Lp-bounded by assumption, the secondterm goes to 0 uniformly in ε. This completes the proof.

136

Chapter 13

The Sharp Gagliardo-NirenbergInequality

As alluded to in the introduction to Chapter 12, in some sense, the next three chapters are insome sense devoted to the notion of concentration compactness. Broadly, concentration com-pactness studies the defects of compactness, and describes how to deal with these defects.This will be described formally in great deal in the future.

Although concentration compactness is the underlying theme of the next three chap-ters, it is contextualized by various sharp inequalities. Indeed, our primary use for theprinciple of concentration compactness will be to determine optimal constants for theGagliardo-Nirenberg inequality, the Sobolev embedding theorem, and the Strichartz in-equality.

In this chapter, our focus is the Gagliardo-Nirenberg inequality. For good measure, werecall the inequality here.

Theorem 13.1 (Gagliardo-Nirenberg). Fix d ≥ 1. Let p satisfy 2 ≤ p < ∞ if d = 1 or d = 2,and 2 ≤ p < 2d

d−2 if d ≥ 3. Then for f ∈ H1(Rd),

‖f‖Lp . ‖f‖2d−p(d−2)

2p

L2 ‖∇f‖d(p−2)

2p

L2 .

We will prove the sharp version of inequality first without using concentration com-pactness, and then spend the rest of the chapter developing the principle of concentra-tion compactness to provide and alternate proof. Chapters 14 and 15 develop analogousconcentration compactness principles for the Sobolev embedding theorem and Strichartzinequality, respectively.

13.1 The Focusing Cubic NLS

Before we determine the optimal constant in the Gagliardo-Nirenberg inequality, we feel itnecessary to provide a suitable source of motivation. After all, we have spent the entiretyof this text generously employing estimates involving ., with no regard for explicit con-stants. Unsurprisingly, the optimal constant in Theorem 13.1 is important in the study ofcertain nonlinear Schrodinger equations. In particular, we will consider the focusing cubicnonlinear Schrodinger equation in d = 2, given by{

i∂tu+ ∆u = −|u|2uu(0) = u0

. (13.1)

Here, t ∈ R and x ∈ R2, and for now we take the initial data u0 to be a complex-valuedSchwartz function. This is one of the most commonly studied nonlinear Schrodinger equa-tions.

137

Chapter 14

The Sharp Sobolev EmbeddingTheorem

138

Chapter 15

Concentration Compactness in PartialDifferential Equations

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References

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Documents

Lectures by: Monica Visanjosephbreen/Lectures_in_Harmonic_Analysis.pdf1.3 Complex Interpolation An important tool in harmonic analysis is interpolation. Broadly speaking, interpolation