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

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

9 Dispersive Partial Differential Equations 949.1 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 949.2 The Linear Schrödinger 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.

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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 =1p +

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‖L2and ‖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

t1 ‖f‖Lpt

where 1pt =1−tp0

+ tp1 and1qt

= 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 S̄. Let

M0 := supy∈R|Φ(iy)| and M1 := sup

y∈R|Φ(1 + iy)|.

Thensupy∈R|Φ(t+ iy)| ≤M1−t0 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 S̄) 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∞ ∈ S̄ 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 S̄. 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 S̄. 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 φ̃(z) :=M z−10 M

−z1 φ(z). Note that if Re z = 0, then

|Φ̃(z)| ≤ |M iy−10 M−iy1 |M0 ≤ |M

iy0 | · |M

−iy1 | ≤ 1

and if Re z = 1, then

|Φ̃(z)| ≤ |M iy0 M−1−iy1 |M1 ≤ |M

iy0 | · |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 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‖p0Lp0 =∫ ∣∣∣ |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 S̄. 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 Mt1.

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

1qt

= t2 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.

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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

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|

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

)n01− 12

= 2 · 2−n0 = 2 · 1N0

.

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) := f̂(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

f̂(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 · · ·xαdd ; D

α :=∂|α|

∂xα11 · · · ∂xαdd

.

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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 ĝ(ξ) = e−2πiy·ξ f̂(ξ).

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

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

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

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

6. If g(x) := xαf(x), then ĝ(ξ) =(i

2π

)|α|Dαξ f̂(ξ).

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

8. We have the basic estimate∥∥∥f̂∥∥∥

L∞ξ

≤ ‖f‖L1x .

Proof.

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

ĝ(ξ) =

∫e−2πix·ξf(x− y) dx =

∫e−2πi(z+y)·ξf(z) dz = e−2πiy

∫e−2πiz·ξf(z) dz

= e−2πiyf̂(ξ).

2. Computing directly gives:

ĝ(ξ) =

∫e−2πix·ξe2πix·ηf(x) dx =

∫e−2πix·(ξ−η)f(x) dx = f̂(ξ − η).

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

ĝ(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 |−1f̂(T−tξ).

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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, ĝ(ξ) = |detT |−1f̂(T−tξ) = f̂(ξ).

4. Computing directly gives:

ĝ(ξ) =

∫e−2πix·ξf(x) dx =

∫e−2πix·ξf(x) dx =

∫e−2πix·(−ξ)f(x) dx = f̂(−ξ).

5. Integrating by parts, we have:

ĝ(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)αf̂(ξ).

6. Note that Dαxe−2πix·ξ = (−2πix)αe−2πix·ξ, so that xαe−2πix·ξ. Thus, integrating byparts again gives:

ĝ(ξ) =

∫e−2πix·ξxαf(x) dx =

(i

2π

)|α|f̂(ξ).

7. Computing, we have

ĝ(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

ĝ(ξ) =

∫ ∫e−2πi(z+y)·ξk(z)f(y) dz dy =

∫e−2πiz·ξk(z) dz ·

∫e−2πiy·ξf(y) dy

= k̂(ξ) · f̂(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 f̂ ∈ S(Rd). Moreover, if fn → f ∈ S(Rd), then f̂n → f̂ ∈S(Rd).

12

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

|ξαDβ f̂(ξ)| ∼ |ξαx̂βf(ξ)| ∼ |D̂αxβf(ξ)| =∣∣∣∣∫ e−2πix·ξDα(xβf(x)) dx∣∣∣∣ .

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

.∥∥∥Dα(xβf(x))∥∥∥

L1 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−π2η2jλj 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, ê−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 (f̂ )̂(x) = f(−x). Equivalently,

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

Proof. For ε > 0, let

Iε(x) :=

∫e−πε

2|ξ|2e2πix·ξ f̂(ξ) dξ.

By the dominated convergence theorem, as ε→ 0, Iε(x)→∫e2πix·ξ f̂(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 f̂(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∫f̂(ξ)g(ξ) dξ =

∫f(x)ĝ(x) dx. In particular,

∫f̂(ξ)ĝ(ξ) dx =∫

f(x)g(x) dx, so that∥∥∥f̂∥∥∥

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

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

∫ ∫e−2πix·ξf(x) dx g(ξ) dξ =

∫f(x)

∫e−2πix·ξg(ξ) dξ dx

=

∫f(x)ĝ(x) dx.

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

ĥ(x) = ˆ̂g = (ĝ)̂(−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 an

isometry on S(Rd), we have∥∥∥f̂n − f̂m∥∥∥

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

in L2. Let f̂ := limn→∞ f̂n, the limit being taken in the L2-sense.We claim that f̂ 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, {ĥn} converges in L2, which implies by the unique-

ness of limits that limn→∞ f̂n = limn→∞ ĝn.Next we claim that f ∈ L2(Rd), then

∥∥∥f̂∥∥∥L2

= ‖f‖L2 , so the Fourier transform is anisometry on L2(Rd). Because the norm function is continuous in the L2 topology, we have∥∥∥f̂∥∥∥

L2= lim

n→∞

∥∥∥f̂n∥∥∥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 functionsso that f̂n

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

∥∥∥f̂n − f̂∥∥∥L2

= ‖fn − f‖L2 → 0 so that g = f̂n.

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∥∥∥f̂∥∥∥Lp′

. ‖f‖Lp

for 1 ≤ p ≤ 2, where 1p +1p′ = 1.

Proof. We have the estimates∥∥∥f̂∥∥∥

L∞≤ ‖f‖L1 and

∥∥∥f̂∥∥∥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

1pθ

+ 1qθ = 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∥∥∥f̂∥∥∥

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

f̂λ(ξ) =

∫e−2πix·ξf

(xλ

)dx = λdf̂(λξ).

We have∥∥∥f̂λ∥∥∥

Lq. ‖fλ‖Lp , which implies

λd · λ−dq

∥∥∥f̂∥∥∥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

1q′ ≥

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

ϕ̂k(ξ) = e−2πiξ·ke1ϕ̂(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∥∥∥f̂∥∥∥

Lp′as follows:∥∥∥∥∥

N∑k=1

e−2πiξ·ke1ϕ̂(ξ − λke1)χB(λke1,λ/2)(ξ) +N∑k=1

e−2πiξ·ke1ϕ̂(ξ − λke1)χCB(λke1,λ/2)(ξ)

∥∥∥∥∥Lp′

≥

∥∥∥∥∥N∑k=1

e−2πiξ·ke1ϕ̂(ξ − λke1)χB(λke1,λ/2)(ξ)

∥∥∥∥∥Lp′

−N∑k=1

∥∥∥e−2πiξ·ke1ϕ̂(ξ − λke1)χCB(λke1,λ/2)(ξ)∥∥∥Lp′& N

1p′ −N

∥∥∥ϕ̂(ξ)χCB(0,λ/2)(ξ)∥∥∥Lp′

.

Because ϕ̂ is Schwartz,

∥∥∥ϕ̂(ξ)χCB(0,λ/2)(ξ)∥∥∥Lp′

.

(∫|ξ|>λ/2

1

|ξ|2ddξ

) 1p′

. λ−dp′

which→ 0 as λ → ∞. Thus,∥∥∥f̂∥∥∥

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 f̂ : Zd → Cby

f̂(k) =

∫Tde−2πik·xf(x) dx.

Then f̂(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

f̂(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

ϕ̂(n)e2πinx

for all x ∈ R. In particular,∑

n∈Z ϕ(n) =∑

n∈Z ϕ̂(n).

Proof. Define F1(x) =∑

n∈Z ϕ(x + n) and F2(x) =∑

n∈Z ϕ̂(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 F̂2(k) = ϕ̂(k) for allk ∈ Z.

Next, we compute:

F̂1(k) =∑n∈Z

∫ 10ϕ(x+ n)e−2πikx dx =

∑n∈Z

∫ n+1n

ϕ(y)e−2πik(y−n) dy

=∑n∈Z

∫ n+1n

ϕ(y)e−2πiky dy

=

∫Rϕ(y)e−2πiky dy

= ϕ̂(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>0λ∣∣∣{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)|0

pλp−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)p dλ

λ.

So ‖f‖Lp(Rd) = p1p

∥∥∥λ ∣∣{x ∈ Rd : |f(x)| > λ}∣∣ 1p∥∥∥Lp((0,∞), dλλ )

. Using the convention p1∞ =

1, we can then write

‖f‖∗Lpweak(Rd) = p1∞

∥∥∥∥λ ∣∣∣{x ∈ Rd : |f(x)| > λ}∣∣∣ 1p∥∥∥∥L∞((0,∞), dλλ )

.

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}∣∣∣ 1p = ∣∣∣∣∣{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>0λ

1

λ= 1 λ}∣∣∣ 1p∥∥∥∥Lq((0,∞), dλλ )

(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}∣∣∣ 1p∥∥∥∥Lq((0,∞), dλλ )

= 0

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

f(x) = 0 almost everywhere. Thus, f = 0.Next, let a ∈ C. Then

‖af‖∗Lp,q(Rd) = p1q

∥∥∥∥λ ∣∣∣{x ∈ Rd : |af(x)| > λ}∣∣∣ 1p∥∥∥∥Lq((0,∞), dλλ )

= |a|p1q

∥∥∥∥∥λa∣∣∣∣{x ∈ Rd : |f(x)| > λa

}∣∣∣∣ 1p∥∥∥∥∥Lq((0,∞), dλλ )

.

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}∣∣∣ 1p∥∥∥∥Lq((0,∞), dλλ )

.

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)| > λ2

}∣∣∣∣+ ∣∣∣∣{x : |g(x)| > λ2}∣∣∣∣) 1p

∥∥∥∥∥Lq((0,∞), dλλ )

.

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

‖f + g‖∗Lp,q(Rd) ≤ p1q

∥∥∥∥∥λ(∣∣∣∣{x : |f(x)| > λ2

}∣∣∣∣ 1p + ∣∣∣∣{x : |g(x)| > λ2}∣∣∣∣ 1p

)∥∥∥∥∥Lq((0,∞), dλλ )

.

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)| > λ2

}∣∣∣∣ 1p + λ2∣∣∣∣{x : |g(x)| > λ2

}∣∣∣∣ 1p∥∥∥∥∥Lq((0,∞), dλλ )

≤ 2p1q

∥∥∥∥∥λ2∣∣∣∣{x : |f(x)| > λ2

}∣∣∣∣ 1p∥∥∥∥∥Lq((0,∞), dλλ )

+

∥∥∥∥∥λ2∣∣∣∣{x : |g(x)| > λ2

}∣∣∣∣ 1p∥∥∥∥∥Lq((0,∞), dλλ )

= 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

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 2mχEm

where {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

∫ 2m2m+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

∫ 2m2m+1

λq

∑n≥m|Em|

qp

dλ

λ.

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

p

∫ 2m2m+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 2

mχEm , 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 2

nχFn 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{(NM

) 1p′

,

(M

N

) 1p

}1q

·

∑N,M∈2Z

∑m:|Em|∼M

2M |Em|1p′

q′ min{(NM

) 1p′

,

(M

N

) 1p

}1q′

.

In the first term, freeze N and sum over M . Because p > 1 and hence p′ 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 2

nχFn 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′ = qq−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∑N

n=1

∥∥∥ 1|x−n|∥∥∥∗L1,∞(R) . N .We claim that

∥∥∥∑Nn=1 1|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 C logN

}⊇ [0, N ]. Therefore,

‖f‖L1,∞(R) ≥ C logN

∣∣∣∣∣{x ∈ Rd :

N∑n=1

1

|x− n|> C logN

}∣∣∣∣∣ ≥ CN logNas 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′

∫ 10λq′ |{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= p2and 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 and1qθ

= 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‖Lpθ

so 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 1p =

1p1

+ 1p2 and1q =

1q1

+ 1q2 , 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 claimthat f ∈ Lp(R). Indeed, we can write

‖f‖pLp(R) =∫

1

|x|

∣∣∣∣log(|x|+ 1|x|)∣∣∣∣− p+22 dx

= 2

(∫ 20

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

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 ballcentered 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 < p ≤ ∞. Explicitly,

ω ({x : |(Mω)(x)| > λ }) . 1λ

∫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 .

1ω(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

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) 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

≤ 1ω(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|=0supx∈EC

|f(x)|.

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

‖Mf‖L∞(ω dx) ≤ inf|E|=0supx∈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 operatorfrom L

1θ (M(ω) dx)→ L

1θ (ω dx). As 0 < θ < 1, 1 < 1θ 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 = 1λ‖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 M̄ is given by

M̄(f)(x) := ‖{Mfn(x)}‖`2n .

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

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

Theorem 4.12 (Vector-valued maximal inequality).

1. The operator M̄ is of weak type (1, 1), that is,∥∥M̄f∥∥∗L1,∞

. ‖f‖L1

for f : Rd → `2(N);

2. M̄ is of strong type (p, p) for all 1 < p 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, 2

n(k1 + 1)× · · · × [2nkd, 2n(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 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 ≤ 2

d

|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

∥∥M̄f∥∥L2

. ‖f‖L2 . By definition,∥∥M̄f∥∥2L2

=

∫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∥∥M̄f∥∥2

L2=∑n≥1

∫Rd|Mfn(x)|2 dx.

37

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

∥∥M̄f∥∥2L2

=∑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 0

λ∣∣{x : (M̄f)(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 M̄ is a sublinear operator,

{x : (M̄f)(x) > λ

}⊆{x : (M̄g)(x) >

λ

2

}∪{x : (M̄b)(x) >

λ

2

}.

Consider the first set. We have already shown that M̄ is of strong type (2, 2), so in particularit is of weak type (2, 2). Thus,∣∣∣∣{x : (M̄g)(x) > λ2

}∣∣∣∣ . 1λ ‖g‖2L2 = 1λ2 ∥∥|g|2∥∥L1 ≤ 1λ2 ‖λ|g|‖L1 = 1λ ‖g‖L1 ≤ 1λ ‖f‖L1 .Therefore, it remains to prove the same type of estimate for the set

{x : (M̄b)(x) > λ2

}.

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 . 1λ

∫Rd|b(y)| dy

≤ 1λ‖f‖L1 .

With this estimate, it remains to show that∣∣∣∣{x /∈⋃ 2Qk : (M̄b)(x) > λ2}∣∣∣∣ . 1λ ‖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 · 1λ‖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 M̄b . M̄bavg. So we again apply the (2, 2) estimate of M̄ to get∣∣∣∣{x /∈⋃ 2Qk : (M̄b)(x) > λ2}∣∣∣∣ ≤ ∣∣∣{x /∈⋃ 2Qk : (M̄bavg)(x) & λ}∣∣∣ . 1λ2 ‖bavg‖2L2

=1

λ2

∥∥∥‖bavgn (x)‖`2n∥∥∥2L2x = 1λ2∥∥∥‖bavgn (x)‖2`2n∥∥∥L1x

.1

λ

∥∥∥‖bavgn (x)‖`2n∥∥∥L1x = 1λ ‖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 =1p +

αd for 1 < p < r 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

The quantity 1|B(x,2r)|∫|x−y| 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− p

d(d−α).

Simplifying the exponents,

1− pd

(d− α) = p(

1

p− 1 + α

d

)=p

r.

So ∣∣∣∣f ∗ 1|x|α∣∣∣∣ (x) . ‖f‖1− prLp (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 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) > λ2

}∣∣∣∣+ ∣∣∣∣{x : |f ∗ g2|(x) > λ2}∣∣∣∣ .

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′

=

(∫ R0αp′ |{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 dα

α

) 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− 1p

=1

q− 1r

⇒ 1p′<

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) > λ2 }∣∣. 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) > λ2

}∣∣∣∣ = 0.Thus, it remains to estimate

∣∣{x : |f ∗ g2|(x) > λ2 }∣∣. By Chebychev’s inequality,∣∣∣∣{x : |f ∗ g2|(x) > λ2}∣∣∣∣ . 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 }|dα

α

=

∫ R0α |{x : |g2(x)| > α }|

dα

α+

∫ ∞R

α |{x : |g2(x)| > α }|dα

α

≤ R · |{x : |g(x)| > R }|+ supα>0

αq |{x : |g(x)| > α }|∫ ∞R

α1−qdα

α.

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

R α1−q dα

α . 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 }|dα

α. 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 L

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

Proposition 5.3. Let 0 < α < d. Then 1̂|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

)(f̂)

=

∫Rdπ−

d−α2 Γ

(d− α

2

)1

|x|d−αf̂(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 he