Introduction to Harmonic Analysis (Student Mathematical Library) 147047199X, 9781470471996

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Introduction to Harmonic Analysis

S T U D E N T M AT H E M AT I C A L L I B R A R Y IAS/PARK CITY MATHEMATICAL SUBSERIES Volume 105

Introduction to Harmonic Analysis Ricardo A. Sáenz

American Mathematical Society Institute for Advanced Study

EDITORIAL COMMITTEE John McCleary Rosa C. Orellana (Chair)

Paul Pollack Kavita Ramanan

2020 Mathematics Subject Classification. Primary 31B05, 31B10, 31B25, 42A16, 42A20, 42B10, 42B25, 28A20, 28A80.

For additional information and updates on this book, visit www.ams.org/bookpages/stml-105

Library of Congress Cataloging-in-Publication Data Cataloging-in-Publication Data has been applied for by the AMS. See http://www.loc.gov/publish/cip/. DOI: https://doi.org/10.1090/stml/105

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established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

28 27 26 25 24 23

A Irene

Contents

IAS/Park City Mathematics Institute Preface Chapter 1.

xi xiii

Motivation and preliminaries

1

§1.1. The heat equation in equilibrium

1

§1.2. Holomorphic functions

3

§1.3. Know thy calculus

4

§1.4. The Dirichlet principle

9

Exercises Chapter 2.

11 Basic properties

13

§2.1. The mean value property

13

§2.2. The maximum principle

20

§2.3. Poisson kernel and Poisson integrals in the ball

24

§2.4. Isolated singularities

29

Exercises

31

Notes

34

Chapter 3.

Fourier series

35

§3.1. Separation of variables

35

§3.2. Fourier series

38 vii

viii

Contents

§3.3. Abel means and Poisson integrals

43

§3.4. Absolute convergence

48

§3.5. Fejér’s theorem

54

§3.6. Mean-square convergence

59

§3.7. Convergence for continuous functions

62

Exercises

68

Notes

70

Chapter 4.

Poisson kernel in the half-space

71

§4.1. The Poisson kernel in the half-space

71

§4.2. Poisson integrals in the half-space

74

§4.3. Boundary limits

77

Exercises

80

Notes

82

Chapter 5.

Measure theory in Euclidean space

83

§5.1. The need for an integration theory

83

§5.2. Outer measure in Euclidean space

85

§5.3. Measurable sets and measure

89

§5.4. Measurable functions

96

Exercises

102

Notes

103

Chapter 6.

Lebesgue integral and Lebesgue spaces

105

§6.1. Integration of measurable functions

105

§6.2. Fubini’s theorem

120 1

124

2

§6.4. The Lebesgue space 𝐿

130

Exercises

134

Notes

135

§6.3. The Lebesgue space 𝐿

Chapter 7.

Maximal functions

137

§7.1. Indefinite integrals and averages

137

§7.2. The Hardy–Littlewood maximal function

138

Contents

ix

§7.3. The Lebesgue differentiation theorem

143

§7.4. Boundary limits of harmonic functions

145

Exercises

148

Notes

150

Chapter 8.

Fourier transform

151

§8.1. Integrable functions

151

§8.2. The Fourier inversion formula

156

§8.3. Mean-square convergence

159

Exercises

164

Notes

165

Chapter 9.

Hilbert transform

167

§9.1. The conjugate function

167

§9.2. Mean-square convergence

168

§9.3. The Hilbert transform of integrable functions

172

§9.4. Convergence in measure

179

Exercises

181

Notes

183

Chapter 10.

Mathematics of fractals

185

§10.1. Hausdorff dimension

185

§10.2. Self-similar sets

191

Exercises

201

Notes

202

Chapter 11.

The Laplacian on the Sierpiński gasket

203

§11.1. Discrete energy on the interval

203

§11.2. Harmonic structure on the Sierpiński gasket

207

§11.3. The Laplacian on the Sierpiński gasket

212

Exercises

219

Notes

220

Chapter 12.

Eigenfunctions of the Laplacian

§12.1. Discrete eigenfunctions on the interval

223 224

x

Contents §12.2. Discrete eigenfunctions on the Sierpiński gasket

227

§12.3. Dirichlet eigenfunctions

233

Exercises

241

Notes

242

Chapter 13.

Harmonic functions on post-critically finite sets

243

§13.1. Post-critically finite sets

243

§13.2. Harmonic structures and discrete energy

245

§13.3. Discrete Laplacians

250

§13.4. The Laplacian on a PCF set

255

Exercises

257

Notes

258

Appendix A. Some results from real analysis

259

§A.1. The real line

259

§A.2. Topology

261

§A.3. Riemann integration

262

§A.4. The Euclidean space

265

§A.5. Complete metric spaces

267

Acknowledgments

271

Bibliography

273

Index

277

IAS/Park City Mathematics Institute

The IAS/Park City Mathematics Institute (PCMI) was founded in 1991 as part of the “Regional Geometry Institute” initiative of the National Science Foundation. In mid-1993 the program found an institutional home at the Institute for Advanced Study (IAS) in Princeton, New Jersey. The annual PCMI summer programs take place in Park City, Utah. Each year’s PCMI summer program is an intensive three-week session where several different activities take place in parallel. These include individual programs for researchers, graduate students, undergraduate faculty, undergraduate students, and K-12 teachers, as well as a workshop devoted to issues surrounding equity in the mathematics classroom. Over 300 people are in attendance each year. A main goal of PCMI is to make all participants aware of the broad spectrum of mathematical activities and to promote interactions between these groups, often leading to new collaboration and new mentoring arrangements. Each summer a different research topic is chosen as the focus of the Research Program and Graduate Summer School. The Undergraduate Program typically focuses on closely related material. Lecture notes from the Graduate Summer School are published each year in the IAS/Park City Mathematics Series. Course material for the Undergraduate Program at PCMI is published intermittently under the IAS/Park City Mathematical Subseries in the Student Mathematical Library. We are very

xi

xii

IAS/Park City Mathematics Institute

pleased to make available to a wide audience the expanded versions of these undergraduate lectures, which we believe should be of great value to students everywhere. Rafe Mazzeo, Series Editor March 2023

Preface

This text grew out of the lecture notes for the course Introduction to Harmonic Analysis given at the Undergraduate Summer School during the 2018 Park City Mathematics Institute (IAS/PCMI). The twelve-hour course contained the basic properties of harmonic functions and the explicit solutions to the Laplace equation in special cases. It also contained a study of the behavior of harmonic functions at the boundary of domains, and introduced the Hilbert transform. The last lectures were dedicated to the construction of harmonic functions in fractals. The lectures were then extended to a full semester course at the University of Colima, aimed to junior and senior undergraduate mathematics majors. Besides a more detailed discussion on the previous topics, the course also included a discussion on Fourier series and their convergence, an introduction to Lebesgue measure and integration, the Hardy–Littlewood maximal function, the Fourier transform and a more extended study of analysis on fractals, in particular the Laplacian on the Sierpiński gasket and the construction of its eigenfunctions. The purpose of both the minicourse at IAS/PCMI and the course at Colima is to introduce the modern ideas and problems of harmonic analysis to undergraduate students, from the point of view of harmonic functions. Most books on harmonic and Fourier analysis are too advanced to be appropriate at the undergraduate level, and undergradute textbooks, as those by Thomas William Körner [K8̈8] and Elias M. Stein and Rami Shakarchi [SS03], focus on Fourier series and their convergence, rather

xiii

xiv

Preface

than on harmonic functions and their behavior at boundary domains. So this text starts with a motivation to study harmonic functions and the Dirichlet problem. The first chapter discusses the solution to the heat equation in equilibrium, the real and imaginary parts of holomorphic functions, and the minimizing functions of energy, all of which are harmonic functions. This book is intended for junior and senior undergraduates with a basic knowledge of real analysis. It requires familiarity with the properties of complete metric and normed spaces, uniform convergence and density. The needed results are reviewed in Appendix A. It does not require knowledge of measure theory, as the text includes two chapters on measure theory, Lebesgue integration, and approximation theorems. It does requires knowledge of linear algebra, in particular familiarity with vector spaces, subspaces, linear operators and orthogonality. Complex analysis is not required but it is recommended, as a couple of calculations of the Fourier transform of some functions are easily done using line integrals and the residue theorem. The text can be roughly partitioned in three parts. The first part, Chapters 1–4, discusses the basic properties of harmonic functions and the problem of their behavior at the boundary of their domains. Basic properties as the mean value property, the maximum principle or the classification of singularities are discussed in Chapter 2. Explicit solutions to the Dirichlet problem, in terms of Poisson integrals, are discussed for the ball (Chapter 2) and the half-space (Chapter 4). We also discuss the problem of the behavior at the boundary of these domains in those chapters, for the case of continuous boundary values. In Chapter 3, we discuss the solution of the Dirichlet problem in the disk using Fourier series, so in this chapter we also discuss the problem of their convergence. In particular, we discuss Abel means, Fejér’s theorem and mean-square convergence. We end Chapter 3 with the construction of an example of a continuous function with divergent Fourier series at a point. The second part, Chapters 5–9, discusses the problem of the behavior of harmonic functions at the boundary of their domains, in particular the half-space, for noncontinuous functions. This requires the use of measure theory, and thus is developed in Chapters 5 and 6. This is not intended to be a comprehensive course in measure theory, but a brief introduction to the main ideas of Lebesgue integration and the basic

Preface

xv

properties of the Lebesgue spaces 𝐿1 and 𝐿2 . In Chapter 7 we discuss the Hardy–Littlewood maximal function and the problem of almost everywhere convergence of Poisson integrals of integrable functions at the boundary of the half-space. In Chapter 9 we introduce the Hilbert transform, which describes the limits at the boundary of the conjugate function to the Poisson integral. We discuss the 𝐿1 theory of the Hilbert transform and the concept of operators of weak type. In order to study the 𝐿2 theory of the Hilbert transform, we introduce the Fourier transform in the previous Chapter 8, where we discuss its basic properties, as the Riemann–Lebesgue lemma for integrable functions and the Plancherel theory of square integrable functions. In the third part, Chapters 10–13, we discuss the theory of harmonic functions on fractals. We start with the fundamental ideas of self-similarity and Hausdorff dimension in Chapter 10, and then proceed to the study of harmonic analysis, the Laplacian and its eigenfunctions, on the Sierpiński gasket, in Chapters 11 and 12. We discuss in these chapters the construction of a harmonic structure and the harmonic functions by interpolation. We describe the construction of the Laplacian and an algorithm to construct its eigenfunctions. In particular, we explicitly describe the Dirichlet eigenfunctions on the Sierpiński gasket. We discuss in Chapter 13 harmonic functions on more general self-similar sets. This text can be used in an introductory course on Harmonic Analysis in several ways. In Colima, semesters are sixteen weeks long, so one has enough time to cover almost all of the material,1 but for shorter semesters we can choose accordingly to the interests and previous knowledge of the audience. Chapters 1–9 provide an introduction to classical harmonic analysis, and students who already took a course on measure theory may skip Chapters 5 and 6. For a course on analysis on fractals, one may choose Chapters 1–3 and then move on to Chapters 10–13, as the first chapters serve as motivation for the study of harmonic functions and eigenfunctions of the Laplacian. Each chapter has a list of exercises and bibliographic and historical notes. Ricardo A. Sáenz Colima, Mexico, October 2022

1 The Fall 2020 course was transmitted online and is available at the page https://www. facebook.com/HarmonicAnalysis

Chapter 1

Motivation and preliminaries

1.1. The heat equation in equilibrium In this chapter we discuss a number of motivations for the study of harmonic functions, with examples taken from physics to complex analysis. We start in this section with a deduction of the heat equation in equilibrium, using the original argument given by Joseph Fourier in his seminal work Analytical Theory of Heat [Fou55]. Consider the propagation of heat through a solid in space. For example, you can consider a potato in the oven, receiving heat in part of its peel. If you wait sufficiently long, the temperature inside the potato will be in equilibrium; though not necessarily constant in its interior, it will not depend on time. Let 𝑄 be a small cube inside this solid, which we describe with edges parallel to the axes in ℝ3 . Suppose two of its opposite vertices are given by (𝑥0 , 𝑦0 , 𝑧0 ) and (𝑥0 +𝜀, 𝑦0 +𝜀, 𝑧0 +𝜀) for some small 𝜀 > 0, and we consider the propagation of heat in 𝑄, with temperature function 𝑢(𝑥, 𝑦, 𝑧, 𝑡). Since we are assuming the system is in equilibrium, the temperature does not depend on time, so it is then a function 𝑢(𝑥, 𝑦, 𝑧) in 𝑄. We also assume 𝑢 is a smooth function in a neighborhood of 𝑄 (that is, an open set that contains 𝑄).

1

2

1. Motivation and preliminaries

Figure 1.1. The small cube 𝑄, with heat propagating in the 𝑥 direction.

By Newton’s law of heat flow, the amount of heat that enters through the side 𝑥 = 𝑥0 of 𝑄 (the left side in Figure 1.1) is proportional to the change in temperature, from hotter to colder, in the 𝑥 direction on this side, so it is given by 𝜕𝑢 −𝐾𝜀2 (𝑥0 , 𝑦0 , 𝑧0 ), 𝜕𝑥 where 𝜀2 is the surface area of the left side and the proportionality constant 𝐾 > 0, which depends only on the material of the solid, is called the conductivity constant. The amount of heat that exits through the side 𝑥 = 𝑥0 + 𝜀 of 𝑄 is then given by 𝜕𝑢 (𝑥 + 𝜀, 𝑦0 , 𝑧0 ). 𝜕𝑥 0 The quantity of heat accumulated in 𝑄 as a consequence of propagation in the 𝑥 direction is the difference between these two quantities, −𝐾𝜀2

𝜕𝑢 𝜕𝑢 (𝑥0 , 𝑦0 , 𝑧0 ) − ( − 𝐾𝜀2 (𝑥0 + 𝜀, 𝑦0 , 𝑧0 )) 𝜕𝑥 𝜕𝑥 𝜕𝑢 2 𝜕𝑢 (𝑥 , 𝑦 , 𝑧 )). = 𝐾𝜀 ( (𝑥0 + 𝜀, 𝑦0 , 𝑧0 ) − 𝜕𝑥 𝜕𝑥 0 0 0 By the mean value theorem, there exists 0 < 𝛿 < 𝜀 such that − 𝐾𝜀2

𝜕𝑢 𝜕𝑢 𝜕2 𝑢 (𝑥0 + 𝜀, 𝑦0 , 𝑧0 ) − (𝑥0 , 𝑦0 , 𝑧0 ) = 𝜀 2 (𝑥0 + 𝛿, 𝑦0 , 𝑧0 ), 𝜕𝑥 𝜕𝑥 𝜕𝑥 so the propagation of heat through 𝑄 in the 𝑥 direction is then 𝐾𝜀3

𝜕2 𝑢 (𝑥 + 𝛿, 𝑦0 , 𝑧0 ). 𝜕𝑥2 0

1.2. Holomorphic functions

3

Similarly, there exist 0 < 𝜂, 𝜃 < 𝜀 so that the propagation of heat through 𝑄 in the 𝑦 and 𝑧 directions is given by 𝐾𝜀3

𝜕2 𝑢 (𝑥 , 𝑦 + 𝜂, 𝑧0 ) 𝜕𝑦2 0 0

and

𝐾𝜀3

𝜕2 𝑢 (𝑥 , 𝑦 , 𝑧 + 𝜃), 𝜕𝑧2 0 0 0

respectively, and the total propagation is then given by 𝐾𝜀3 (

𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢 (𝑥 + 𝛿, 𝑦 , 𝑧 ) + (𝑥 , 𝑦 + 𝜂, 𝑧 ) + (𝑥 , 𝑦 , 𝑧 + 𝜃)). 0 0 0 0 0 0 𝜕𝑥2 𝜕𝑦2 𝜕𝑧2 0 0 0

Since the system is in equilibrium, the total propagation must be equal to 0. As we are assuming that 𝑢 is a smooth function, its partial derivatives are continuous, so we obtain, as 𝜀 → 0, the equation 𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢 + + 2 =0 𝜕𝑥2 𝜕𝑦2 𝜕𝑧

(1.1) at the point (𝑥0 , 𝑦0 , 𝑧0 ).

Equation (1.1) is called the Laplace equation. We can also write it as Δ𝑢 = 0, where the differential operator Δ is given by Δ𝑢 =

𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢 + + 2. 𝜕𝑥2 𝜕𝑦2 𝜕𝑧

Δ𝑢 is called the Laplacian of 𝑢. The solutions of equation (1.1) are called harmonic functions.

1.2. Holomorphic functions In this section we observe that harmonic functions also appear in complex analysis. Recall that 𝑓 is holomorphic (or analytic) in an open set 𝐷 ⊂ ℂ if, for each 𝑧 ∈ 𝐷, its derivative (1.2)

𝑓(𝑧 + ℎ) − 𝑓(𝑧) ℎ ℎ→0

𝑓′ (𝑧) = lim

exists. If we write the holomorphic function 𝑓 as 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦), where 𝑧 = 𝑥+𝑖𝑦 and 𝑢 and 𝑣 are its real and imaginary parts, respectively, then 𝑢 and 𝑣 satisfy the Cauchy–Riemann equations (1.3)

𝜕𝑢 𝜕𝑣 = 𝜕𝑥 𝜕𝑦

and

𝜕𝑣 𝜕𝑢 =− . 𝜕𝑥 𝜕𝑦

4

1. Motivation and preliminaries

These equations follow directly from the differentiability of 𝑓. Indeed, if we take the limit in (1.2) by approaching ℎ → 0 with real numbers, we obtain 𝜕𝑢 𝜕𝑣 𝑓′ (𝑧) = (𝑥, 𝑦) + 𝑖 (𝑥, 𝑦). 𝜕𝑥 𝜕𝑥 Meanwhile, if we approach ℎ → 0 with purely imaginary numbers, we get 1 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢 𝑓′ (𝑧) = (𝑥, 𝑦) + (𝑥, 𝑦) = (𝑥, 𝑦) − 𝑖 (𝑥, 𝑦). 𝑖 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦 As these two expressions for 𝑓′ (𝑧) must be equal, we obtain (1.3). Assuming 𝑢 and 𝑣 are smooth functions,1 we can differentiate the Cauchy–Riemann equations and get 𝜕2 𝑢 𝜕2 𝑣 = 𝜕𝑥𝜕𝑦 𝜕𝑥2

and

𝜕2 𝑣 𝜕2 𝑢 = − 2. 𝜕𝑦𝜕𝑥 𝜕𝑦

All mixed derivatives are continuous, so they must be equal and thus 𝜕2 𝑢 𝜕2 𝑢 = − . 𝜕𝑥2 𝜕𝑦2 Therefore 𝑢 is a harmonic function. We can similarly verify that 𝑣 is harmonic. As 𝑣 is the imaginary part of a holomorphic function of which 𝑢 is the real part, we say that the function 𝑣 is a conjugate harmonic function to 𝑢. Note that conjugate harmonic functions are not unique, because adding any constant to 𝑣 will give another conjugate harmonic function. Also, observe that −𝑢 is the conjugate harmonic function to 𝑣. Under appropiate conditions on the set 𝐷, one can prove that every harmonic 𝑢 has a conjugate harmonic function 𝑣. See Exercise (3) for the special case of the complex plane 𝐷 = ℂ. We will dedicate Chapter 9 to the study of the properties of conjugate harmonic functions in the upper half-plane.

1.3. Know thy calculus Before moving on, let’s dedicate a section to set the notation used in this text, and review some of the results from advanced calculus that we’ll need later on. This will just be a quick summary of these results, so we 1 It is a fact, proven in any basic complex analysis text (see [Gam01], for example), that both ᵆ and 𝑣 are smooth functions whenever 𝑓 is holomorphic.

1.3. Know thy calculus

5

invite the reader to consult advanced calculus texts, such as [Fle77] or [Spi65] for the details and proofs. We denote the 𝑑-dimensional Euclidean space by ℝ𝑑 . Thus ℝ𝑑 = {𝑥 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑑 ) ∶ 𝑥𝑖 ∈ ℝ}. We will usually denote the points in the plane ℝ2 and in the space ℝ3 by (𝑥, 𝑦) and (𝑥, 𝑦, 𝑧), respectively. We denote the Euclidean norm of a vector 𝑥 ∈ ℝ𝑑 by |𝑥|. Thus |𝑥| = √𝑥12 + 𝑥22 + . . . + 𝑥𝑑2 . For 𝑥 ∈ ℝ𝑑 , 𝑥′ is the point in ℝ𝑑−1 formed by the first 𝑑 − 1 coordinates of 𝑥. We can thus write 𝑥 = (𝑥′ , 𝑥𝑑 ). If we need to explicitly distinguish the last coordinate, then we refer to ℝ𝑑+1 = {(𝑥, 𝑡) ∶ 𝑥 ∈ ℝ𝑑 , 𝑡 ∈ ℝ}. We denote by ℝ𝑑+1 the upper half-space of points (𝑥, 𝑡) ∈ ℝ𝑑+1 with + 𝑡 > 0. The open ball of radius 𝑟 > 0 centered at 𝑥0 is given by 𝐵𝑟 (𝑥0 ) = {𝑥 ∈ ℝ𝑑 ∶ |𝑥 − 𝑥0 | < 𝑟}. If 𝑥0 = 0, we simply denote it by 𝐵𝑟 . If, in addition, 𝑟 = 1, we denote it by 𝔹. The sphere of radius 𝑟 > 0 centered at 𝑥0 is given by 𝑆𝑟 (𝑥0 ) = {𝑥 ∈ ℝ𝑑 ∶ |𝑥 − 𝑥0 | = 𝑟}, and we denote it by 𝑆𝑟 if 𝑥0 = 0, and by 𝕊 if we also have 𝑟 = 1. For an open set Ω ∈ ℝ𝑑 , 𝐶(Ω) is the space of continuous functions ̄ the space of continuous functions on its closure. We in Ω, and 𝐶(Ω) 𝑘 denote by 𝐶 (Ω) the space of 𝑘-continuously differentiable functions in Ω, and by 𝐶 ∞ (Ω) the space of smooth functions. Note that 𝐶 ∞ (Ω) =

⋂

𝐶 𝑘 (Ω).

𝑘≥1

We denote by 𝐶𝑐∞ (Ω) the space of smooth functions with compact support in Ω. That is, 𝑓 ∈ 𝐶𝑐∞ (Ω) if 𝑓 is smooth in Ω and there exists a compact subset 𝐾 ⊂ Ω such that 𝑓(𝑥) = 0 for all 𝑥 ∉ 𝐾. In particular, we say that 𝑓 is zero “close to the boundary.”

6

1. Motivation and preliminaries

We denote the partial derivative of 𝑓 with respect to 𝑥𝑖 either by 𝜕𝑓 , by 𝜕𝑥𝑖 𝑓 or simply by 𝜕𝑖 𝑓, if there is no confusion. The gradient of a 𝜕𝑥𝑖 function 𝑓 is given by the vector ∇𝑓 = (𝜕1 𝑓, 𝜕2 𝑓, . . . , 𝜕𝑑 𝑓). Note that its norm is given by |∇𝑓| = (|𝜕1 𝑓|2 + |𝜕2 𝑓|2 + . . . + |𝜕𝑑 𝑓|2 )

1/2

.

If 𝛼 = (𝛼1 , 𝛼2 , . . . , 𝛼𝑑 ) is a multi-index, where each 𝛼𝑖 ∈ ℕ, we define 𝑥𝛼 as the monomial 𝛼 𝛼 𝛼 𝑥 𝛼 = 𝑥 1 1 𝑥2 2 ⋯ 𝑥 𝑑 𝑑 and 𝜕𝛼 𝑓 as the higher order derivative 𝛼

𝛼

𝛼

𝜕𝛼 𝑓 = 𝜕1 1 𝜕2 2 ⋯ 𝜕𝑑 𝑑 𝑓. The order of the multi-index 𝛼 is given by |𝛼| = 𝛼1 + 𝛼2 + . . . + 𝛼𝑑 . A hypersurface in ℝ𝑑 is a differentiable manifold 𝑆 of dimension 𝑑 − 1. Locally, for each 𝑥0 ∈ 𝑆, there exists an open set 𝑈 that contains 𝑥0 such that 𝑈 ∩ 𝑆 is the solution set to the equation (1.4)

𝜙(𝑥) = 0,

for some continuously differentiable function 𝜙 in 𝑈 with ∇𝜙 ≠ 0. By the implicit function theorem, and relabeling the coordinates if needed, we can assume 𝜙 is of the form 𝜙(𝑥) = 𝑥𝑑 − 𝜓(𝑥′ ), and thus 𝑈 ∩ 𝑆 is given by (1.5)

𝑥𝑑 = 𝜓(𝑥′ ).

We say that 𝑆 is a 𝐶 𝑘 -hypersurface if the function 𝜙 above is in 𝐶 𝑘 (𝑈) (and hence 𝜓 is a 𝐶 𝑘 function in its domain). A domain in ℝ𝑑 is an open and connected subset Ω ⊂ ℝ𝑑 . The domain Ω is a 𝐶 𝑘 -domain if its boundary 𝜕Ω is a 𝐶 𝑘 -hypersurface. If Ω is a 𝐶 1 -domain and 𝑥0 ∈ 𝜕Ω, then the normal vector at 𝑥0 is the unit vector 𝜈(𝑥0 ) orthogonal to the hypersuface 𝜕Ω pointing outwards of Ω (Figure 1.2). Thus, ∇𝜙 𝜈=± , |∇𝜙| where 𝜙 is a function that describes 𝜕Ω locally near 𝑥0 , as in (1.4).

1.3. Know thy calculus

7

Figure 1.2. The normal vector 𝜈 at a point in the boundary of Ω.

Example 1.6. The open ball 𝐵𝑅 (𝑥0 ) of radius 𝑅 and centered at 𝑥0 ∈ ℝ𝑑 is a 𝐶 1 -domain (in fact, a 𝐶 𝑘 -domain for every 𝑘), with boundary equal to the sphere 𝑆 𝑅 (𝑥0 ). Note that 𝕊 is the solution set to the equation 𝑥12 + 𝑥22 + . . . + 𝑥𝑑2 = 1. Hence, for 𝑥 ∈ 𝕊, 𝜈(𝑥) = 𝑥. The surface measure on a hypersurface 𝑆 is denoted by 𝑑𝜎. Locally, if 𝜓 is as in (1.5), we have that 𝑑𝜎 = √1 + |∇𝜓|2 𝑑𝑥′ . 1.7. One can integrate over ℝ𝑑 (or over a subset with rotational symmetry) by using spherical coordinates. If we write a point 𝑥 ≠ 0 in ℝ𝑑 as 𝑥 = 𝑟𝜉, where 𝑟 = |𝑥| > 0 and 𝜉 = 𝑥/|𝑥| ∈ 𝕊, then ∞

∫ 𝑓(𝑥)𝑑𝑥 = ∫ ∫ 𝑓(𝑟𝜉)𝑑𝜎(𝜉)𝑟𝑑−1 𝑑𝑟. ℝ𝑑

0

𝕊

1.8. The area of the unit sphere in ℝ𝑑 is denoted by 𝜔𝑑 . Thus, 𝜔𝑑 = ∫ 𝑑𝜎. 𝕊

We leave it as an exercise (Exercise (7)) to prove that 𝜔𝑑 =

2𝜋𝑑/2 , Γ(𝑑/2)

8

1. Motivation and preliminaries

where Γ(𝑠) is the gamma function given by ∞

Γ(𝑠) = ∫ 𝑒−𝑡 𝑡𝑠−1 𝑑𝑡, 0

for every 𝑠 > 0. We will also make use of Theorem 1.9. Theorem 1.9 (Divergence theorem). Let Ω ⊂ ℝ𝑑 be a bounded 𝐶 1 domain and 𝐹 a continuously differentiable vector field defined in a neighborhood of Ω.̄ Then ∫ ∇ ⋅ 𝐹𝑑𝑥 = ∫ 𝐹 ⋅ 𝜈𝑑𝜎.

(1.10)

Ω

𝜕Ω

If 𝐹 = (𝐹 1 , 𝐹 2 , . . . , 𝐹 𝑑 ), then ∇ ⋅ 𝐹 = 𝜕1 𝐹 1 + 𝜕2 𝐹 2 + . . . + 𝜕 𝑑 𝐹 𝑑 is called the divergence of 𝐹, and is also denoted by div 𝐹. Note that the divergence theorem is a multi-dimensional version of the fundamental theorem of calculus. Indeed, in the line ℝ, Ω is just an open interval, say Ω = (𝑎, 𝑏), the normal vector at its boundary is given by 𝜈(𝑎) = −1, 𝜈(𝑏) = 1, and ∇ ⋅ 𝐹 is the derivative of 𝐹, so (1.10) is 𝑏

∫ 𝐹 ′ (𝑥)𝑑𝑥 = −𝐹(𝑎) + 𝐹(𝑏). 𝑎

1.11. Taking 𝐹 = (0, 0, . . . , 𝑢𝑣, . . . , 0), where the nonzero component is the 𝑖th term, we obtain the formula for integration by parts: ∫ Ω

𝜕𝑢 𝜕𝑣 𝑣𝑑𝑥 = ∫ 𝑢𝑣𝜈 𝑖 𝑑𝜎 − ∫ 𝑢 𝑑𝑥. 𝜕𝑥𝑖 𝜕𝑥 𝑖 𝜕Ω Ω

Theorem 1.12 (Green’s identities). Let Ω be a bounded 𝐶 1 -domain in ℝ𝑑 . (1) If 𝑢 is continuously differentiable and 𝑣 is twice continuously differentiable in a neighborhood of Ω,̄ then (1.13)

∫ (𝑢Δ𝑣 + ∇𝑢 ⋅ ∇𝑣)𝑑𝑥 = ∫ 𝑢𝜕𝜈 𝑣𝑑𝜎, Ω

𝜕Ω

where 𝜕𝜈 𝑣 = ∇𝑣 ⋅ 𝜈 is the normal derivative of 𝑣 at the boundary of Ω.

1.4. The Dirichlet principle

9

(2) If 𝑢 and 𝑣 are twice continuously differentiable in a neighborhood of Ω,̄ then (1.14)

∫ (𝑢Δ𝑣 − 𝑣Δ𝑢)𝑑𝑥 = ∫ (𝑢𝜕𝜈 𝑣 − 𝑣𝜕𝜈 𝑢)𝑑𝜎. Ω

𝜕Ω

Theorem 1.12 follows almost immediately from the divergence theorem (or 1.11), and we leave it as an exercise (Exercise (9)).

1.4. The Dirichlet principle The Green identities provide us with another motivation for the study of harmonic functions: they are minimizers of energy. Let Ω be a 𝐶 1 domain. We define the energy form on Ω as the bilinear form (1.15)

ℰ(𝑢, 𝑣) = ∫ ∇𝑢 ⋅ ∇𝑣𝑑𝑥, Ω

for smooth functions 𝑢 and 𝑣 in a neighborhood of Ω̄ (we denote the ̄ The energy of the function 𝑢, denoted space of such functions as 𝐶 ∞ (Ω)). simply as ℰ(𝑢), is then given by ℰ(𝑢) = ℰ(𝑢, 𝑢) = ∫ |∇𝑢|2 𝑑𝑥. Ω

We now consider the following question: can we find the function 𝑢 that minimizes ℰ(𝑢), given its values at the boundary 𝜕Ω of Ω? That ̄ such that 𝑢|𝜕Ω = 𝑓 is, given a function 𝑓 defined on 𝜕Ω, find 𝑢 ∈ 𝐶 ∞ (Ω) and ̄ and 𝑣|𝜕Ω = 𝑓}. ℰ(𝑢) = min{ℰ(𝑣) ∶ 𝑣 ∈ 𝐶 ∞ (Ω) It is clear that we cannot expect the above problem to always have a ̄ 𝑓 cannot be arbitrary because solution 𝑢. First, as we require 𝑢 ∈ 𝐶 ∞ (Ω), it must be the restriction of such a function to 𝜕Ω. Moreover, although it is true that the set ̄ and 𝑣|𝜕Ω = 𝑓} {ℰ(𝑣) ∶ 𝑣 ∈ 𝐶 ∞ (Ω) is bounded from below, because ℰ(𝑣) ≥ 0 for any smooth function 𝑣, it is not clear whether it has a minimum or not. However, we have the following fact: in the case when ℰ takes its minimum at 𝑢, then 𝑢 is a harmonic function in Ω, that is a function that satisfies the equation Δ𝑢 = 0

10

1. Motivation and preliminaries

in Ω. To prove this, suppose ℰ takes its minimum at 𝑢 for given values at 𝜕Ω. Now, for any function 𝑣 ∈ 𝐶𝑐∞ (Ω) and any 𝑡 ∈ ℝ, the function 𝑢 + 𝑡𝑣 is smooth and has the same values as 𝑢 at the boundary. Since ℰ(𝑢) is minimal, we have that ℰ(𝑢) ≤ ℰ(𝑢 + 𝑡𝑣). Hence, as a function of 𝑡, the function 𝐼(𝑡) = ℰ(𝑢+𝑡𝑣) takes its minimum value at 𝑡 = 0, and thus 𝐼 ′ (0) = 0.

(1.16) Now

𝐼(𝑡) = ℰ(𝑢 + 𝑡𝑣) = ∫ ∇(𝑢 + 𝑡𝑣) ⋅ ∇(𝑢 + 𝑡𝑣)𝑑𝑥 Ω

= ∫ |∇(𝑢)|2 𝑑𝑥 + 2𝑡 ∫ ∇𝑢 ⋅ ∇𝑣𝑑𝑥 + 𝑡2 ∫ |∇𝑣|2 𝑑𝑥, Ω

Ω

Ω

so 𝐼 ′ (𝑡) = 2 ∫ ∇𝑢 ⋅ ∇𝑣𝑑𝑥 + 2𝑡 ∫ |∇𝑣|2 𝑑𝑥 Ω

Ω

and (1.16) implies ∫ ∇𝑢 ⋅ ∇𝑣𝑑𝑥 = 0. Ω

By the Green identity (1.13), we thus have (1.17)

∫ 𝑣Δ𝑢𝑑𝑥 = ∫ 𝑣𝜕𝜈 𝑢𝑑𝜎 = 0, Ω

𝜕Ω

because 𝑣 is zero near the boundary. Moreover, since (1.17) holds for every 𝑣 ∈ 𝐶𝑐∞ (Ω), we conclude Δ𝑢 = 0, and thus 𝑢 is harmonic. We have left many open questions in the discussion above. We have mentioned that we cannot expect to have a minimizer 𝑢 of the energy form satisfying that 𝑢|𝜕Ω = 𝑓 for any function 𝑓. However, as a minimizer is a harmonic function, this leads to the following problem: given a domain Ω and a function 𝑓 defined on 𝜕Ω, find a harmonic function 𝑢 in Ω such that it is equal to 𝑓 on the boundary. This is known as the Dirichlet problem, in honor of the french mathematician Lejeune Dirichlet. It opens a handful of questions, such as the following: • For which domains Ω can we solve the Dirichlet problem? For which functions on its boundary does the solution exist?

Exercises

11

• If 𝑢 is a harmonic function in Ω, what can we say about its behavior at the boundary of Ω? Does it extend continuously to 𝜕Ω? Throughout this text we will be discussing results related to the previous questions. In particular, we will focus our attention to harmonic functions in the domains Ω = 𝔹, the unit ball, and Ω = ℝ𝑑+1 + , the upper half-space, and the behavior of such harmonic functions at the boundaries of their domains. The fact that the minimizers of the energy form ℰ are harmonic functions is called the Dirichlet principle. The first to give it this name was Bernard Riemann in [Rie51], who used this fact to prove the result in complex analysis that we now know as the Riemann mapping theorem. See, for example, [Ull08] for a study of the Riemann mapping theorem and its relation to the Dirichlet problem.

Exercises (1) Let 𝑅 be a rotation in the plane. (a) Consider the change of variables (𝜉, 𝜂) = 𝑅(𝑥, 𝑦). Then 𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢 + = + . 𝜕𝜂2 𝜕𝑥2 𝜕𝑦2 𝜕𝜉2 (b) If 𝑢 is harmonic, then 𝑢 ∘ 𝑅 is also harmonic. (2) Let (𝑟, 𝜃) be the polar coordinates of the plane. Then Δ𝑢 =

𝜕2 𝑢 1 𝜕𝑢 1 𝜕2 𝑢 + + . 𝑟 𝜕𝑟 𝑟2 𝜕𝜃2 𝜕𝑟2

(3) Let 𝑢 be a harmonic function in ℝ2 . Then there exists a conjugate harmonic function 𝑣 to 𝑢. (Hint: Consider a line integral of the 1form 𝜕𝑢 𝜕𝑢 − 𝑑𝑥 + 𝑑𝑦.) 𝜕𝑦 𝜕𝑥 (4) If 𝑣 1 and 𝑣 2 are conjugate to 𝑢 in the plane, then 𝑣 1 − 𝑣 2 is constant. (5) (a) If 0 is conjugate to 𝑢 in the plane, then 𝑢 is constant. (b) If 𝑓 is holomorphic in ℂ and real valued, then 𝑓 is constant. (6) Let Γ(𝑠) be the gamma function.

12

1. Motivation and preliminaries (a) Integrate by parts to verify the identity Γ(𝑠 + 1) = 𝑠Γ(𝑠). (b) For every 𝑛 ∈ ℤ+ , Γ(𝑛) = (𝑛 − 1)!.

(7) (a) Use polar coordinates to verify the identity 2

∫ 𝑒−𝜋|𝑥| 𝑑𝑥 = 1. ℝ2

(b) For every dimension 𝑑, 2

∫ 𝑒−𝜋|𝑥| 𝑑𝑥 = 1. ℝ𝑑

(c) Use spherical coordinates to verify 2𝜋𝑑/2 . Γ(𝑑/2) (8) Use integration in spherical coordinates, fact 1.7, to prove that the volume of the unit ball 𝔹 is given by 𝜔𝑑 =

∫ 𝑑𝑥 = 𝔹

𝜔𝑑 . 𝑑

(9) Prove Theorem 1.12. (10) Consider the unit interval and, for smooth functions in [0, 1], define the form 1

ℰ(𝑓) = ∫ 𝑓′ (𝑥)2 𝑑𝑥. 0

(a) The minimizers of this form, given the values of 𝑓 at 𝑥 = 0 and 𝑥 = 1, are the linear functions 𝑓(𝑥) = 𝑎𝑥 + 𝑏. (b) If ℰ(𝑓) is a minimum, then ℰ(𝑓) = (𝑓(1) − 𝑓(0))2 .

Chapter 2

Basic properties

2.1. The mean value property Let Ω ⊂ ℝ𝑑 be an open set. As discussed in Chapter 1, we say that a twice differentiable function 𝑢 is harmonic in Ω if it satisfies Δ𝑢 = 𝜕12 𝑢 + 𝜕22 𝑢 + . . . + 𝜕𝑑2 𝑢 = 0 in Ω. Example 2.1. Any linear function 𝑢 = 𝑎1 𝑥1 + 𝑎2 𝑥2 + . . . + 𝑎𝑑 𝑥𝑑 is harmonic in ℝ𝑑 , as all of its second derivatives are zero. Observe that, in the case 𝑑 = 1, the linear functions 𝑢(𝑥) = 𝑎𝑥 + 𝑏 are precisely the functions that satisfy 𝑢″ (𝑥) = 0, so the only harmonic functions in ℝ (or in any interval in the real line) are the linear functions. Example 2.2. The quadratic polynomial 𝑢(𝑥, 𝑦) = 𝑥2 − 𝑦2 is harmonic in ℝ2 , as its second derivatives are equal to 𝜕12 𝑢 = 2 and 𝜕22 𝑢 = −2. Note that 𝑢 is the real part of the holomorphic function 𝑓(𝑧) = 𝑧2 . As observed in Section 1.2, the real and imaginary parts of an analytic function are harmoinc. The imaginary part of 𝑧2 , and thus a conjugate harmonic to 𝑢, is 𝑣(𝑥, 𝑦) = 2𝑥𝑦. Similarly, the functions 𝑢(𝑥, 𝑦) = ℜ((𝑥 + 𝑖𝑦)𝑛 )

and

𝑣(𝑥, 𝑦) = ℑ((𝑥 + 𝑖𝑦)𝑛 ),

the real and imaginary parts of 𝑧𝑛 , are harmonic in ℝ2 for each 𝑛 ∈ ℕ. 13

14

2. Basic properties

Example 2.3. The function 𝑢(𝑥, 𝑦) = sin 𝑥 sinh 𝑦 is harmonic in ℝ2 , because and 𝜕22 𝑢 = 𝑢. 𝜕12 𝑢 = −𝑢 Note that 𝑢 is the imaginary part of the holomorphic function − cos 𝑧. It is easy to see that the harmonic functions in any open set in ℝ𝑑 form a vector space because, if 𝑢 and 𝑣 are harmonic, then any linear combination 𝛼𝑢 + 𝛽𝑣 of 𝑢 and 𝑣 is also a harmonic function. Moreover, the space of harmonic functions in ℝ𝑑 is invariant under translations and orthogonal transformations (see Exercises (1) and (2)). We observed in Example 2.1 that the linear functions are harmonic functions and, in fact they are the only harmonic functions in ℝ. We now make a rather immediate observation: if 𝑢 is linear, say in the interval [𝑎, 𝑏], then its value at the midpoint (𝑎 + 𝑏)/2 is the average of its values at 𝑎 and 𝑏, 𝑎+𝑏 𝑢(𝑎) + 𝑢(𝑏) 𝑢( . )= 2 2 It turns out that this is true in every dimension. Theorem 2.4 (Mean value property). Let 𝑢 be a harmonic function in a neighborhhod of the closed ball 𝐵𝑟̄ (𝑥0 ). Then (2.5)

𝑢(𝑥0 ) =

1 ∫ 𝑢(𝜉)𝑑𝜎(𝜉). |𝑆𝑟 (𝑥0 )| 𝑆 (𝑥 ) 𝑟

0

In other words, the average of the values of 𝑢 over any sphere around 𝑥0 is equal to 𝑢(𝑥0 ). In the identity (2.5), 𝑑𝜎 is the surface measure on the sphere 𝑆𝑟 (𝑥0 ), and |𝑆𝑟 (𝑥0 )| is its surface area, |𝑆𝑟 (𝑥0 )| = 𝜔𝑑 𝑟𝑑−1 . With an appropiate change of variables, we can also write (2.5) as (2.6)

𝑢(𝑥0 ) =

1 ∫ 𝑢(𝑥0 + 𝑟𝜉)𝑑𝜎(𝜉), 𝜔𝑑 𝕊

where 𝕊 is the unit sphere around the origin.

2.1. The mean value property

15

2.7. Integrating over the ball 𝐵𝑟 (𝑥0 ) in spherical coordinates, it follows from (2.5) that, under the same assumptions of Theorem 2.4, 𝑢(𝑥0 ) =

1 ∫ |𝐵𝑟 (𝑥0 )| 𝐵

𝑢(𝑥)𝑑𝑥 =

𝑟 (𝑥0 )

𝑑 ∫ 𝑢(𝑥0 + 𝑟𝑥)𝑑𝑥, 𝜔𝑑 𝔹

where 𝔹 is the unit ball centered at the origin. We leave this as an exercise (Exercise (3)). Again, this identity is immediate in ℝ (Exercise (4)). Proof of Theorem 2.4. We only need to prove the case 𝑑 ≥ 2, by the observations made before the statement of the theorem. By translating by 𝑥0 , we can assume 𝑥0 = 0 (see Exercise (1)). For 0 < 𝜀 < 𝑟, let Ω = 𝐵𝑟 ⧵ 𝐵𝜀̄ , where 𝐵𝑟 = 𝐵𝑟 (0) and 𝐵𝜀 = 𝐵𝜀 (0), as in Figure 2.1. Define

Figure 2.1. The domain Ω = 𝐵𝑟 ⧵ 𝐵𝜀̄ . On a point in 𝑆 𝑟 , the normal vector 𝜈 points away from the origin, while on a point in 𝑆 𝜀 points towards the origin.

the function 𝑣 in ℝ𝑑 ⧵ {0} by log |𝑥| 𝑣(𝑥) = { 2−𝑑 |𝑥|

𝑑=2

𝑑 ≥ 3. 𝑥 Note that, for any 𝑥 ∈ Ω, ∇𝑣(𝑥) = 𝑐 𝑑 𝑑 , where 𝑐 𝑑 = 1 if 𝑑 = 2 and |𝑥| 𝑐 𝑑 = 2 − 𝑑 if 𝑑 ≥ 3. Also, 𝜕Ω = 𝑆𝑟 − 𝑆𝜀 , where we have written 𝑆𝑟 and 𝑆𝜀 for 𝑆𝑟 (0) and 𝑆𝜀 (0), respectively (as oriented manifolds,1 see Figure 2.1), 1 We are only interested, at this moment, in the fact that the normal vectors on −𝑆𝑒 point opposite to those at 𝑆𝑟 .

16

2. Basic properties

and

𝑥 𝜈(𝑥) = { 𝑟 𝑥 − 𝜀

on 𝑆𝑟 on − 𝑆𝑒 .

Thus

𝑐𝑑 on 𝑆𝑟 𝑑−1 𝑟 𝜕𝜈 𝑣 = { 𝑐 𝑑 − 𝑑−1 on − 𝑆𝑒 . 𝜀 We can also verify explicitly that Δ𝑣 = 0 (Exercise (5)), and hence ∫ (𝑢Δ𝑣 − 𝑣Δ𝑢)𝑑𝑥 = 0. Ω

Applying Green’s identity (1.14), and the previous explicit calculations, we obtain 0 = ∫ (𝑢𝜕𝜈 𝑣 − 𝑣𝜕𝜈 𝑢)𝑑𝜎 𝜕Ω

= ∫ (𝑢 𝑆𝑟

𝑐𝑑 𝑐𝑑 − 𝜐𝑑 (𝑟)𝜕𝜈 𝑢)𝑑𝜎 − ∫ (𝑢 𝑑−1 − 𝜐𝑑 (𝜀)𝜕𝜈 𝑢)𝑑𝜎, 𝑟𝑑−1 𝑒 𝑆𝑒

where log 𝑠 𝑑 = 2 𝜐𝑑 (𝑠) = { 1 𝑑 ≥ 3, 𝑠𝑑−2 for 𝑠 = 𝑑 or 𝑠 = 𝜀, which are constant over 𝑆𝑟 and 𝑆𝜀 . Thus, since the surface integral of 𝜕𝜈 𝑢 over a sphere is zero (Exercise (6)), we obtain 𝑐𝑑 𝑐𝑑 ∫ 𝑢𝑑𝜎 = 𝑑−1 ∫ 𝑢𝑑𝜎 𝑟𝑑−1 𝑆𝑟 𝜀 𝑆𝜀 for any 𝜀 > 0. Since 𝑢 is continuous we obtain, taking 𝜀 → 0 (see Exercise (7)), 1 ∫ 𝑢𝑑𝜎 = 𝑢(0). 𝜔𝑑 𝑟𝑑−1 𝑆𝑟 □ If a continuous function 𝑢 in ℝ satisfies the mean value property, that is, 𝑢(𝑥) + 𝑢(𝑦) 𝑥+𝑦 𝑢( )= 2 2

2.1. The mean value property

17

for all 𝑥, 𝑦 ∈ ℝ, then 𝑢 must be a linear function. Indeed, let 𝑎 = 𝑢(1) − 𝑢(0) and 𝑏 = 𝑢(0), so we have 𝑢(1) = 𝑎 + 𝑏 and 𝑢(0) = 𝑏. Since 𝑢(1) = (𝑢(0) + 𝑢(2))/2, we see that 𝑢(2) = 2𝑢(1) − 𝑢(0) = 2𝑎 + 𝑏, and we can verify, inductively, that (2.8)

𝑢(𝑛) = 𝑎𝑛 + 𝑏

for every 𝑛 ∈ ℕ. We can similarly prove that (2.8) holds for negative integers 𝑛. Now, for every 𝑛 ∈ ℤ, 𝑢(𝑛) + 𝑢(𝑛 + 1) 𝑎𝑛 + 𝑏 + 𝑎(𝑛 + 1) + 𝑏 2𝑛 + 1 = )= 2 2 2 2𝑛 + 1 = 𝑎( ) + 𝑏, 2 and similarly for every number of the form 𝑘/2𝑛 , for every 𝑘 ∈ ℤ and every 𝑛 ∈ ℕ. Since such numbers are dense in ℝ and 𝑢 is continuous, we conclude that 𝑢(𝑥) = 𝑎𝑥 + 𝑏 𝑢(

for every 𝑥 ∈ ℝ. As we have stated above, the linear functions are the harmonic functions in ℝ so, therefore, the continuous functions that satisfy the mean value property are precisely the harmonic functions. This is also true in higher dimensions. Theorem 2.9 (Converse to the mean value property). Let Ω ⊂ ℝ𝑑 be open and 𝑢 a continuous function on Ω that satisfies that, whenever 𝐵𝑟̄ (𝑥) ⊂ Ω, (2.10)

𝑢(𝑥) =

1 ∫ 𝑢(𝑥 + 𝑟𝜉)𝑑𝜎(𝜉). 𝜔𝑑 𝕊

Then 𝑢 ∈ 𝐶 ∞ (Ω) and 𝑢 is harmonic in Ω. As harmonic functions must be twice differentiable, we must prove that a function 𝑢 that satisfies (2.10) is at least twice differentiable before proving that Δ𝑢 = 0 in Ω. However, the conclusion of Theorem 2.9 is much stronger: 𝑢 is actually an infinitely differentiable function. We thus conclude Corollary 2.11, which follows by applying Theorems 2.4 and 2.9. Corollary 2.11. If 𝑢 is harmonic in an open set Ω, then it is infinitely differentiable in Ω.

18

2. Basic properties In order to prove Theorem 2.9, we will make use of Lemma 2.12.

Lemma 2.12. There exists a smooth radial function 𝜙 on ℝ𝑑 such that it is supported in 𝔹 and ∫ 𝜙 = 1. Proof. Consider the function 𝜓 on ℝ given by 1

𝑒 (4𝑡−1)(2𝑡−1) 𝜓(𝑥) = { 0

1/4 < 𝑡 < 1/2 otherwise.

It is a standard calculus exercise to verify that 𝜓 is a smooth function in ℝ, supported in [1/4, 1/2]. Indeed, it is infinitely flat at the points 1/4 and 1/2 (see Figure 2.2). Now, we define on ℝ𝑑 the function 𝜙(𝑥) = 𝑐𝜓(|𝑥|),

1

1

4

2

Figure 2.2. The cut-off function 𝜓(𝑡). Note that it is supported in [1/4, 1/2], and infinitely flat at the points 1/4 and 1/2.

where 𝑐 is such that ∫ 𝜙(𝑥)𝑑𝑥 = 1. ℝ𝕕

Such 𝑐 exists because 𝜓 is nonnegative, and thus ∫ℝ𝕕 𝜓(|𝑥|)𝑑𝑥 > 0. Now, 𝜓 is 𝐶 ∞ and supported away from zero, and hence 𝜙 ∈ 𝐶 ∞ (ℝ𝑑 ), because

2.1. The mean value property

19

𝑥 ↦ |𝑥| is smooth away from zero. Finally, as 𝜓(𝑡) = 0 unless 1/4 < 𝑡 < 1/2, then 𝜙(𝑥) = 0 unless 1/4 < |𝑥| < 1/2, and thus supp 𝜙 ⊂ 𝔹. □ Proof of Theorem 2.9. Let 𝑢 be a continuous function that satisfies ̄ (𝑥0 ) ⊂ Ω. (2.10) for every 𝐵𝑟̄ (𝑥) ⊂ Ω. Let 𝑥0 ∈ Ω, and 𝜀 > 0 such that 𝐵2𝜀 ̃ Let 𝜙(𝑥) = 𝜙(|𝑥|) as in Lemma 2.12, and define the function 𝜙𝜀 (𝑥) = 𝜀−𝑑 𝜙(𝜀−1 𝑥). Note that 𝜙𝜀 ∈ 𝐶 ∞ (ℝ𝑑 ), it is supported in 𝐵𝜀 (0), and ∫ 𝜙𝜀 (𝑥)𝑑𝑥 = 1. ℝ𝕕

In particular, for any 𝑥 ∈ 𝐵𝜀 (𝑥0 ), the function 𝑦 ↦ 𝜙𝜀 (𝑥 −𝑦) is supported in 𝐵2𝜀 (𝑥0 ) ⊂ Ω. Hence, we observe that, for 𝑥 ∈ 𝐵𝜀 (𝑥0 ), ∫ 𝑢(𝑦)𝜙𝜀 (𝑥 − 𝑦)𝑑𝑦 = ∫ 𝑢(𝑥 − 𝑦)𝜙𝜀 (𝑦)𝑑𝑦 = ∫ ℝ𝕕

ℝ𝕕 𝜀

𝑢(𝑥 − 𝑦)𝜙𝜀 (𝑦)𝑑𝑦

𝐵𝜀 (0)

̃ −1 𝑟)𝑟𝑑−1 𝑑𝑟. = ∫ ∫ 𝑢(𝑥 − 𝑟𝜉)𝑑𝜎(𝜉) ⋅ 𝜀−𝑑 𝜙(𝜀 0

𝕊

Using (2.10) we obtain 𝜀

̃ −1 𝑟)𝑟𝑑−1 𝑑𝑟 ∫ 𝑢(𝑦)𝜙𝜀 (𝑥 − 𝑦)𝑑𝑦 = 𝑢(𝑥) ⋅ 𝜔𝑑 ∫ 𝜀−𝑑 𝜙(𝜀 ℝ𝕕

0 𝜀

̃ −1 𝑟)𝑑𝜎(𝜉)𝑟𝑑−1 𝑑𝑟 = 𝑢(𝑥) ∫ ∫ 𝜀−𝑑 𝜙(𝜀 0

𝕊

= 𝑢(𝑥) ∫

𝜙𝜀 (𝑦)𝑑𝑦 = 𝑢(𝑥).

𝐵𝜀 (0)

Note that the function 𝑢(𝑦)𝜙𝜀 (𝑥 − 𝑦) is 𝐶 ∞ in 𝑥 and continuous with compact support in 𝑦, so we can differentiate under the integral the function 𝑥 ↦ ∫ 𝑢(𝑦)𝜙𝜀 (𝑥 − 𝑦)𝑑𝑦 ℝ𝕕

as many times as we want, and thus we conclude 𝑢 ∈ 𝐶 ∞ (𝐵𝜀 (𝑥0 )). In particular, Δ𝑢 is a continuous function in 𝐵𝜀 (𝑥0 ). Now, for any 𝑥 ∈ 𝐵𝜀 (𝑥0 ) and 0 < 𝑟 < 𝜀 such that 𝐵𝑟̄ (𝑥) ⊂ 𝐵𝜀 (𝑥0 ), 𝑢(𝑥) =

1 ∫ 𝑢(𝑥 + 𝑟𝜉)𝑑𝜎(𝜉), 𝜔𝑑 𝕊

20

2. Basic properties

so if we differentiate with respect to 𝑟 we obtain 0= =

𝑑 ∫ 𝑢(𝑥 + 𝑟𝜉)𝑑𝜎(𝜉) = ∫ ∇𝑢(𝑥 + 𝑟𝜉) ⋅ 𝜉𝑑𝜎(𝜉) 𝑑𝑟 𝕊 𝕊 1 𝑟𝑑−1

∫

𝜕𝜈 𝑢𝑑𝜎 =

𝑆𝑟 (𝑥)

1 𝑟𝑑−1

∫

Δ𝑢𝑑𝑥.

𝐵𝑟 (𝑥)

In the last equality we have used Green’s identity (1.14) with 𝑣 = −1. Hence, the integral of Δ𝑢 over any ball in 𝐵𝜀 (𝑥0 ) is zero. By the continuity of Δ𝑢, Δ𝑢 = 0 in 𝐵𝜀 (𝑥0 ). Since 𝑥0 ∈ Ω is arbitrary, we conclude 𝑢 is harmonic in Ω.

□

2.2. The maximum principle From the mean value property 2.4 we obtain another basic property of harmonic functions, the maximum principle. Corollary 2.13 (Maximum principle). If Ω ⊂ ℝ𝑑 is a domain and 𝑢 is harmonic in Ω, then 𝑢 does not take a maximum nor a minumum in Ω, unless 𝑢 is constant. This is easy to see in the case 𝑑 = 1, where harmonic functions coincide with linear functions: if 𝑢 is linear in the interval (𝑎, 𝑏), then it clearly does not take neither a maximum or a minimum, because 𝑢 is either strictly increasing or strictly decreasing, unless it is constant. Proof of Corollary 2.13. Suppose that 𝑢 takes its maximum 𝑀 at some 𝑥0 ∈ Ω, so 𝑢(𝑥0 ) = 𝑀. Let 𝑈 = {𝑥 ∈ Ω ∶ 𝑢(𝑥) = 𝑀} be the set of points in Ω where 𝑢 takes the value 𝑀. Note that 𝑈 ≠ ∅ because 𝑥0 ∈ 𝑈. We prove that 𝑈 = Ω. First, 𝑈 is closed in Ω because 𝑈 = 𝑢−1 ({𝑀}) and 𝑢 is continuous, so it is the pre-image of a closed set under a continuous function.2 Now let 𝑥 ∈ 𝑈. Since Ω is open, there exists 𝑟 > 0 such that 𝐵𝑟̄ (𝑥) ⊂ Ω. By the mean value property, 𝑢(𝑥) =

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

𝑢(𝑦)𝑑𝑦.

𝑟 (𝑥)

2

See Sections A.2 and A.4 for a summary of results from topology in Euclidean spaces.

2.2. The maximum principle

21

As we are assuming 𝑥 ∈ 𝑈, this integral must equal 𝑀. Now 𝑢(𝑦) ≤ 𝑀 for all 𝑦 ∈ Ω, because 𝑀 is the maximum of 𝑢. Hence, if at some 𝑦 ∈ 𝐵𝑟 (𝑥) we had 𝑢(𝑦) < 𝑀, this integral would be smaller than 𝑀, because 𝑢 is continuous. Thus 𝑢(𝑦) = 𝑀 for all 𝑦 ∈ 𝐵𝑟 (𝑥). Therefore 𝐵𝑟 (𝑥) ⊂ 𝑈, and 𝑈 is also open in Ω. Since Ω is connected, we conclude 𝑈 = Ω, and therefore 𝑢 is the constant function 𝑢(𝑥) = 𝑀. By taking −𝑢, we also see that 𝑢 takes its minimum in Ω only if it is constant. □ The maximum principle implies that if Ω is bounded, so Ω̄ is compact, and 𝑢 is harmonic in Ω and continuous on Ω,̄ then 𝑢 takes its maximum (and its minimum) at the boundary of Ω (Exercise (8)). Again, this is clear in the case of a linear function in a closed interval. The maximum principle also implies uniqueness of harmonic functions in a bounded domain Ω, given their values on the boundary. See Exercise (9) for details. The maximum principle states that harmonic functions, unless they are constant, do not take their maxima nor minima in their domains. If the domain Ω is bounded, a nonconstant harmonic function in Ω may be bounded, of course, and in that case it may be possible to extend it to the boundary of Ω, and hence its extrema would be achieved in 𝜕Ω. Even if the domain Ω is unbounded, we may have bounded harmonic functions, as we will see later on. However, in the case where the domain of the harmonic function is all of the Euclidean space, we have the following result. Theorem 2.14 (Liouville). If 𝑢 is harmonic and bounded in ℝ𝑑 , then it is constant. Proof. Suppose 𝑢 is harmonic and |𝑢(𝑥)| ≤ 𝑀 for all 𝑥 ∈ ℝ𝑑 . We prove that 𝑢(𝑥) = 𝑢(0) for all 𝑥 ∈ ℝ𝑑 . Fix 𝑥 ∈ ℝ𝑑 and let 𝑅 > |𝑥|. By the mean value property in balls, 2.7, we have that 𝑢(𝑥) − 𝑢(0) =

1 ∫ |𝐵𝑅 (𝑥)| 𝐵

𝑅 (𝑥)

𝑢(𝑦)𝑑𝑦 −

1 ∫ 𝑢(𝑦)𝑑𝑦, |𝐵𝑅 | 𝐵 𝑅

22

2. Basic properties

where 𝐵𝑅 = 𝐵𝑅 (0). Since |𝐵𝑅 (𝑥)| = |𝐵𝑅 | = 𝜔𝑑 𝑅𝑑 /𝑑, we can write this difference as 𝑑 𝑢(𝑥) − 𝑢(0) = 𝑢(𝑦)𝑑𝑦 − ∫ 𝑢(𝑦)𝑑𝑦). (∫ 𝜔𝑑 𝑅𝑑 𝐵𝑅 (𝑥) 𝐵𝑅 Now, if 𝐴 is the annulus 𝐴 = {𝑦 ∈ ℝ𝑑 ∶ 𝑅 − |𝑥| ≤ |𝑦| ≤ 𝑅 + |𝑥|}, we see that the symmetric difference of the balls 𝐵𝑅 (𝑥) and 𝐵𝑅 satisfies 𝐵𝑅 (𝑥) △ 𝐵𝑅 ⊂ 𝐴. (See Figure 2.3) Indeed, if 𝑦 ∈ 𝐵𝑅 (𝑥) ⧵ 𝐵𝑅 , then |𝑦 − 𝑥| < 𝑅 and |𝑦| ≥

Figure 2.3. The annulus 𝐴 containing the symmetric difference of the balls 𝐵𝑅 (𝑥) and 𝐵𝑅 .

𝑅 ≥ 𝑅 − |𝑥|, and further |𝑦| ≤ |𝑦 − 𝑥| + |𝑥| < 𝑅 + |𝑥|; similarly, if 𝑦 ∈ 𝐵𝑅 ⧵ 𝐵𝑅 (𝑥), then |𝑦| < 𝑅 ≤ 𝑅 + |𝑥| and |𝑦 − 𝑥| ≥ 𝑅, so |𝑦| ≥ |𝑦 − 𝑥| − |𝑥| ≥ 𝑅 − |𝑥|. If we integrate using spherical coordinates, we obtain 𝑅+|𝑥|

𝑑 𝑑 ∫ |𝑢(𝑦)|𝑑𝑦 ≤ ∫∫ 𝑀𝑟𝑑−1 𝑑𝑟𝑑𝜎 |𝑢(𝑥) − 𝑢(0)| ≤ 𝜔 𝑑 𝑅𝑑 𝐴 𝜔𝑑 𝑅𝑑 𝕊 𝑅−|𝑥| =

(𝑅 + |𝑥|)𝑑 − (𝑅 − |𝑥|)𝑑 𝐶 𝑀 𝑑 ⋅ 𝑀𝜔 ⋅ ≤ 𝑥 , 𝑑 𝑑 𝑅 𝜔 𝑑 𝑅𝑑

2.2. The maximum principle

23

where we have used the fact that |𝑢(𝑦)| ≤ 𝑀 for all 𝑦 ∈ ℝ𝑑 , and the constant 𝐶𝑥 only depends on 𝑑 and |𝑥|. As 𝑥 is fixed and 𝑅 is arbitrary, we conclude that |𝑢(𝑥) − 𝑢(0)| = 0, and therefore 𝑢(𝑥) = 𝑢(0). □ We have seen above that, if 𝑓(𝑧) is holomorphic, then it is real and imaginary parts are harmonic functions. Therefore, Theorem 2.14 implies that, if 𝑓 is a holomorphic function in ℂ (such a function is called an entire function) and is bounded, then 𝑓 must be constant (this is also know as Liouville’s theorem). This fact provides a proof for the fundamental theorem of algebra: If 𝑝(𝑧) is polynomial over ℂ of degree at least 1, then it has a root in ℂ. Indeed, if 𝑝(𝑧) is a polynomial over ℂ with no roots, then 1/𝑝(𝑧) is an entire bounded function, and thus constant, so 𝑝(𝑧) is a constant polynomial. See Exercise (10) for the details. We can refine Theorem 2.14 to obtain the same conclusion even when 𝑢 is only bounded from below or from above. Theorem 2.15. If 𝑢 is harmonic and nonnegative in ℝ𝑑 , then it is constant. Proof. The proof of Theorem 2.15 follows similarly as the proof of Theorem 2.14, but we now have to be more careful when estimating the difference 𝑢(𝑥) − 𝑢(0) =

𝑑 𝑢(𝑦)𝑑𝑦 − ∫ 𝑢(𝑦)𝑑𝑦), (∫ 𝜔𝑑 𝑅𝑑 𝐵𝑅 (𝑥) 𝐵𝑅

for 𝑥 ∈ ℝ𝑑 and 𝑅 > |𝑥|. This time, we use the fact that 𝑢(𝑦) ≥ 0 to observe that, if again 𝐴 is the annulus 𝐴 = {𝑦 ∈ ℝ𝑑 ∶ 𝑅 − |𝑥| ≤ |𝑦| ≤ 𝑅 + |𝑥|}, then |𝑢(𝑥) − 𝑢(0)| ≤ =

𝑑 ∫ 𝑢(𝑦)𝑑𝑦 𝜔 𝑑 𝑅𝑑 𝐴 𝑑 𝑢(𝑦)𝑑𝑦 − ∫ 𝑢(𝑦)𝑑𝑦) (∫ 𝜔𝑑 𝑅𝑑 𝐵𝑅+|𝑥| 𝐵𝑅−|𝑥|

where 𝐵𝑅+|𝑥| and 𝐵𝑅−|𝑥| are the balls of radii 𝑅 + |𝑥| and 𝑅 − |𝑥| centered at 0, respectively. We use again the mean value property and, as before,

24

2. Basic properties

for some constant 𝐶𝑥 that depends only on 𝑑 and |𝑥|, |𝑢(𝑥) − 𝑢(0)| ≤

𝑑 |𝑢(0) − |𝐵𝑅−|𝑥| |𝑢(0)) (|𝐵 𝜔𝑑 𝑅𝑑 𝑅+|𝑥|

(𝑅 + |𝑥|)𝑑 − (𝑅 − |𝑥|)𝑑 ⋅ 𝑢(0) 𝑅𝑑 𝐶 𝑢(0) ≤ 𝑥 . 𝑅 We can now conclude again that 𝑢(𝑥) = 𝑢(0). =

□

It is clear that we obtain the same conclusion of Theorem 2.15 whenever 𝑢 is a harmonic function in ℝ𝑑 and there exists a constant 𝛼 ∈ ℝ such that either 𝑢(𝑥) ≥ 𝛼 for all 𝑥 ∈ ℝ𝑑 , or 𝑢(𝑥) ≤ 𝛼 for all 𝑥 ∈ ℝ𝑑 .

2.3. Poisson kernel and Poisson integrals in the ball We have seen that the value of a harmonic function at the center of a sphere is equal to the average of its values over the sphere. One can ask naturally if the value at any other point in the interior of the sphere is similarly determined by the values over the sphere, and if this value corresponds to a perhaps weighted average over such values. This is certainly the case for a linear function in a close interval [𝑎, 𝑏]: if 𝑡 ∈ (𝑎, 𝑏) and 𝑢 is linear, then 𝑏−𝑡 𝑡−𝑎 𝑢(𝑎) + 𝑢(𝑏), 𝑏−𝑎 𝑏−𝑎 which is a convex combination of 𝑢(𝑎) and 𝑢(𝑏). 𝑢(𝑡) =

We will prove that this is true for harmonic functions in ℝ𝑑 , for 𝑑 ≥ 2, as well. The weight function is called the Poisson kernel. For 𝑥 ∈ 𝔹 and 𝜉 ∈ 𝕊, we define (2.16)

𝑃(𝑥, 𝜉) =

1 1 − |𝑥|2 . 𝜔𝑑 |𝑥 − 𝜉|𝑑

The Poisson kernel satisfies the following facts. 2.17. 𝑃(𝑥, 𝜉) > 0 for any 𝑥 ∈ 𝔹 and 𝜉 ∈ 𝕊, which is easily seen from (2.16) because |𝑥| < 1. 2.18. For each fixed 𝜉 ∈ 𝕊, the function 𝑥 ↦ 𝑃(𝑥, 𝜉) is harmonic in 𝔹. This is followed by explicit differentiation (Exercise (13)).

2.3. Poisson kernel and Poisson integrals in the ball

25

2.19. For each fixed 𝑥 ∈ 𝔹, ∫ 𝑃(𝑥, 𝜉)𝑑𝜎(𝜉) = 1.

(2.20)

𝕊

This is clear if 𝑥 = 0, since (2.21)

𝑃(0, 𝜉) =

1 𝜔𝑑

for any 𝜉 ∈ 𝕊, and thus ∫ 𝑃(0, 𝜉)𝑑𝜎(𝜉) = ∫ 𝕊

𝕊

1 𝑑𝜎(𝜉) = 1. 𝜔𝑑

For 𝑥 ≠ 0, 𝑥/|𝑥| ∈ 𝕊 and, since 𝑃(⋅, 𝑥/|𝑥|) is harmonic in 𝔹, the mean value property implies (2.22)

𝑃(0,

𝑥 1 𝑥 ∫ 𝑃(|𝑥|𝜉, )= )𝑑𝜎(𝜉), |𝑥| 𝜔𝑑 𝕊 |𝑥|

because the integral on the right side is the average over the sphere centered at the origin of radius |𝑥| < 1, contained in the ball 𝔹. By the identity ||𝑥|𝜉 − 𝑥 | = |𝑥 − 𝜉|, | |𝑥| | known as the symmetry lemma (Exercise (14)), we have that 2

𝑃(|𝑥|𝜉,

1 1 − ||𝑥|𝜉 | 1 1 − |𝑥|2 𝑥 = = 𝑃(𝑥, 𝜉), )= |𝑥| 𝜔𝑑 | 𝜔𝑑 |𝑥 − 𝜉|𝑑 𝑥 |𝑑 ||𝑥|𝜉 − |𝑥| |

and thus 𝑃(0,

𝑥 1 ∫ 𝑃(𝑥, 𝜉)𝑑𝜎(𝜉). )= |𝑥| 𝜔𝑑 𝕊

The identity (2.20) follows using (2.21). Observe that, if 𝑑 = 1, we simply have 𝜔1 = 2 and hence 𝑃(𝑥, −1) =

1 1 − |𝑥|2 1 = (1 − 𝑥) 2 |𝑥 − (−1)| 2

and 𝑃(𝑥, 1) =

1 1 − |𝑥|2 1 = (1 + 𝑥) 2 |𝑥 − 1| 2

26

2. Basic properties

for every 𝑥 ∈ (−1, 1), since the boundary of the unit interval (−1, 1) is the set {−1, 1} of two points. Note that each function 𝑥 ↦ 𝑃(𝑥, ±1) is linear, so it is harmonic in (−1, 1), and (2.20) is just 𝑃(𝑥, −1) + 𝑃(𝑥, 1) = 1. Example 2.23. In the case when 𝑑 = 2, we can write the Poisson kernel 𝑃(𝑥, 𝜉) in polar coordinates. Indeed, if 𝑥 = 𝑟𝑒𝑖𝜃 for some 0 ≤ 𝑟 < 1 and 𝜉 = 𝑒𝑖𝜏 , then 𝑃(𝑟𝑒𝑖𝜃 , 𝑒𝑖𝜏 ) =

1 1 − 𝑟2 1 1 − 𝑟2 = ⋅ . 2𝜋 |𝑟𝑒𝑖𝜃 − 𝑒𝑖𝜏 |2 2𝜋 1 + 2𝑟 cos(𝜏 − 𝜃) + 𝑟2

Note that (2.20) is now the identity 2𝜋

1 ∫ 2𝜋 0

1 − 𝑟2 𝑑𝜏 = 1. 1 + 2𝑟 cos(𝜏 − 𝜃) + 𝑟2

The following fact states that, as we approach a point in the boundary, the weight of the Poisson kernel concentrates on that point. 2.24. For any 𝜁 ∈ 𝕊 and 𝜂 > 0, ∫

𝑃(𝑥, 𝜉)𝑑𝜎(𝜉) → 0

|𝜉−𝜁|≥𝜂

as 𝑥 → 𝜁, where the integral is taken over the subset of 𝕊 of points 𝜉 ∈ 𝕊 that satisfy |𝜉 − 𝜁| ≥ 𝜂. To verify this limit observe that, if |𝜉 − 𝜁| ≥ 𝜂 and |𝑥 − 𝜁| < 𝜂/2, then 𝜂 𝜂 |𝑥 − 𝜉| = |𝜉 − 𝜁 + 𝜁 − 𝑥| ≥ |𝜉 − 𝜁| − |𝜁 − 𝑥| > 𝜂 − = , 2 2 so we have 1 1 − |𝑥|2 1 1 − |𝑥|2 𝑃(𝑥, 𝜉) = ≤ . 𝜔𝑑 |𝑥 − 𝜉|𝑑 𝜔𝑑 (𝜂/2)𝑑 Therefore, if |𝑥 − 𝜁| < 𝜂/2, ∫

𝑃(𝑥, 𝜉)𝑑𝜎(𝜉) ≤ ∫

|𝜉−𝜁|≥𝜂

|𝜉−𝜁|≥𝜂

1 1 − |𝑥|2 2 𝑑 𝑑𝜎(𝜉) ≤ ( ) (1 − |𝑥|2 ), 𝑑 𝜔𝑑 (𝜂/2) 𝜂

and hence ∫

𝑃(𝑥, 𝜉)𝑑𝜎(𝜉) → 0

|𝜉−𝜁|≥𝜂

as 𝑥 → 𝜁, because |𝜁| = 1 and thus |𝑥| → 1.

2.3. Poisson kernel and Poisson integrals in the ball

27

The facts 2.17, 2.20 and 2.24 make the family {𝜉 ↦ 𝑃(𝑥, 𝜉) ∶ 𝑥 ∈ 𝔹} of functions on 𝕊 resemble a family of good kernels as 𝑥 → 𝜁 ∈ 𝕊, as defined in [SS03]. We will study such families later in this text. Let 𝑓 ∈ 𝐶(𝕊). The Poisson integral of 𝑓 is given by (2.25)

𝒫𝑓(𝑥) = ∫ 𝑃(𝑥, 𝜉)𝑓(𝜉)𝑑𝜎(𝜉), 𝕊

for each 𝑥 ∈ 𝔹. The function 𝑢(𝑥) = 𝒫𝑓(𝑥) defined by the Poisson integral of 𝑓 is well defined in 𝔹 for any continuous function 𝑓 on 𝕊. This follows because 𝑃(𝑥, 𝜉) is continuous as well in 𝕊. In fact, it is not required for 𝑓 to be continuous on 𝕊 for the integral in (2.25) to be defined. It is sufficient for 𝑓 to be Riemann-integrable on 𝕊. 2.26. The Poisson integral 𝑢(𝑥) of 𝑓 ∈ 𝐶(𝕊) is harmonic in 𝔹. This is followed by differentiating inside the integral (2.25), and using the fact that 𝑃(𝑥, 𝜉) is harmonic in 𝑥. (This is true for a Riemann-integrable function 𝑓 on 𝕊, as well; see Exercise (15).) In the case 𝑑 = 1, the Poisson integral of 𝑓 ∶ {−1, 1} → ℝ is the sum 𝑢(𝑥) = 𝑃(𝑥, −1)𝑓(−1) + 𝑃(𝑥, 1)𝑓(1) 1 1 = (1 − 𝑥)𝑓(−1) + (1 + 𝑥)𝑓(1). 2 2 This is a linear combination of linear functions, so it is linear and clearly harmonic in (−1, 1). Note that 𝑢(𝑥) → 𝑓(±1) as 𝑥 → ±1. The Poisson integral solves the Dirichlet problem for the ball: given 𝑓 ∈ 𝐶(𝕊), find a function 𝑢 on 𝔹̄ such that it is harmonic in the interior and coincides with 𝑓 on the boundary, that is (2.27)

Δ𝑢 = 0 { 𝑢=𝑓

in 𝔹 on 𝕊.

We prove Theorem 2.28. Theorem 2.28. Let 𝑓 ∈ 𝐶(𝕊) and 𝑢 = 𝒫𝑓 its Poisson integral. Then 𝑢 is harmonic in 𝔹, extends continuously to 𝔹̄ and 𝑢|𝕊 = 𝑓.

28

2. Basic properties

Proof. From fact 2.26, we know that 𝑢 is harmonic in 𝔹. It is thus sufficient to prove that, for each 𝜁 ∈ 𝕊, 𝑢(𝑥) → 𝑓(𝜁) as 𝑥 → 𝜁. Since 𝑓 is continuous on the compact set 𝕊, it is bounded.3 Let 𝑀 > 0 be such that |𝑓(𝜉)| ≤ 𝑀 for all 𝜉 ∈ 𝕊. Given 𝜀 > 0, we can choose 𝜂 > 0 such that, if |𝜉 − 𝜁| < 𝜂, then 𝜀 |𝑓(𝜉) − 𝑓(𝜁)| < . 2 We write, using identity (2.20), |𝑢(𝑥) − 𝑓(𝜁)| = || ∫ 𝑃(𝑥, 𝜉)𝑓(𝜉)𝑑𝜎(𝜉) − 𝑓(𝜁) ∫ 𝑃(𝑥, 𝜉)𝑑𝜎(𝜉)|| 𝕊

𝕊

≤ ∫ 𝑃(𝑥, 𝜉)|𝑓(𝜉) − 𝑓(𝜁)|𝑑𝜎(𝜉) 𝕊

=∫

+∫

|𝜉−𝜁| 0, we define the function in 𝔹̄ ∗ 𝑣 𝜀 (𝑥) = 𝑢(𝑥) − 𝒫(𝑢|𝕊 )(𝑥) + 𝜀(|𝑥|2−𝑑 − 1).

Exercises

31

As |𝑥|2−𝑑 is harmonic in ℝ𝑑 ⧵ {0}, 𝑣 𝜀 is harmonic in 𝔹∗ and, if we set 𝒫(𝑢|𝕊 )(𝜉) = 𝑢(𝜉) if 𝜉 ∈ 𝕊, 𝑣 𝜀 is continuous on 𝔹̄ ∗ . We observe that, for 𝜉 ∈ 𝕊, 𝑣 𝜀 (𝜉) = 0. Moreover, as 𝑥 → 0, we have 𝑣 𝜀 (𝑥) → ∞ because 𝑢 is bounded near 0. Thus, by the maximum principle, 𝑣 𝜀 (𝑥) > 0 for all 𝑥 ∈ 𝔹∗ , because otherwise 𝑢 would take a negative minimum in 𝔹∗ , and that is not possible by the maximum principle. Since 𝜀 > 0 is arbitrary, we obtain that 𝑢(𝑥) ≥ 𝒫(𝑢|𝕊 )(𝑥) for all 𝑥 ∈ 𝔹∗ . If we repeat the argument for −𝑢, we obtain 𝑢(𝑥) ≤ 𝒫(𝑢|𝕊 )(𝑥) for all 𝑥 ∈ 𝔹∗ . Thus 𝑢(𝑥) = 𝒫(𝑢|𝕊 )(𝑥) in 𝔹∗ . Therefore, the Poisson integral 𝒫(𝑢|𝕊 )(𝑥) is the harmonic extension of 𝑢 to all of 𝔹. □ Note that, in the proof of Theorem 2.33, we are extending 𝑢 to 𝑥 = 0 by its average over 𝕊. It is actually not necessary to assume that 𝑢 is bounded near 𝑥0 to conclude that 𝑥0 is a removable singularity. See Exercise (21).

Exercises (1) A translation in ℝ𝑑 is a map 𝑇 ∶ ℝ𝑑 → ℝ𝑑 of the form 𝑇(𝑥) = 𝑥 + ℎ, for some ℎ ∈ ℝ𝑑 . (a) If 𝑇 is a translation, Δ(𝑢 ∘ 𝑇) = (Δ𝑢) ∘ 𝑇. (b) If 𝑢 is harmonic in ℝ𝑑 and 𝑇 is a translation, then 𝑢 ∘ 𝑇 is also harmonic in ℝ𝑑 . (2) An orthogonal transformation in ℝ𝑑 is a map 𝑃 ∶ ℝ𝑑 → ℝ𝑑 of the form 𝑃(𝑥) = 𝐴𝑥, for some orthogonal 𝑛 × 𝑛 matrix 𝐴, that is, 𝐴 satisfies that 𝐴𝐴𝑡 = 𝐼𝑛 , where 𝐴𝑡 is the transpose of 𝐴 and 𝐼𝑛 is the 𝑛 × 𝑛 identity matrix. (a) If 𝑃 is an orthogonal transformation, then Δ(𝑢 ∘ 𝑃) = (Δ𝑢) ∘ 𝑃. (b) If 𝑢 is harmonic in ℝ𝑑 and 𝑃 is orthogonal, then 𝑢 ∘ 𝑃 is also harmonic in ℝ𝑑 . (3) Prove fact 2.7.

32

2. Basic properties

(4) Let 𝑢(𝑥) = 𝑎𝑥 + 𝑏. Then 𝑥 +𝑟

0 1 ∫ 𝑢(𝑥0 ) = 𝑢(𝑥)𝑑𝑥. 2𝑟 𝑥 −𝑟 0

(5) Prove that the function 𝑣 in the proof of Theorem 2.4 is harmonic. (6) Suppose 𝑢 is harmonic in a neighborhood of Ω,̄ where Ω is a 𝐶 1 domain. Then ∫ 𝜕𝜈 𝑢 𝑑𝜎 = 0. 𝜕Ω

(7) Let 𝑓 be Riemann-integrable on the rectangle 𝑅, and continuous at the interior point 𝑥0 ∈ 𝑅. As 𝜀 → 0, 1 ∫ 𝑓 → 𝑓(𝑥0 ); and (a) |𝐵𝜀 (𝑥0 )| 𝐵 (𝑥 ) 𝜀 0 1 ∫ 𝑓 → 𝑓(𝑥0 ). (b) |𝑆𝜀 (𝑥0 )| 𝑆 (𝑥 ) 𝜀

0

(8) If Ω ⊂ ℝ𝑑 is a bounded domain and 𝑢 is harmonic in Ω and continuous on Ω,̄ then 𝑢 takes its maximum and its minimum on 𝜕Ω. (9) Let Ω ⊂ ℝ𝑑 be a bounded domain, 𝑢 and 𝑣 harmonic in Ω and continuous on Ω.̄ If 𝑢 = 𝑣 on 𝜕Ω, then 𝑢 = 𝑣 in Ω. (10) The following exercises provide the details of the proof of the fundamental theorem of algebra. (a) If 𝑓 is a holomorphic function without zeroes in its domain Ω, then 1/𝑓 is holomorphic in Ω. (b) If 𝑝(𝑧) is a polynomial over ℂ, then either 𝑝(𝑧) is constant or |𝑝(𝑧)| → ∞ as |𝑧| → ∞. (c) If 𝑝(𝑧) is polynomial over ℂ with no roots, then 1/𝑝(𝑧) is an entire bounded function. (11) If 𝑓 is an entire function and its real part is nonnegative, then 𝑓 is constant. (12) If 𝑢 is a radial harmonic function in 𝔹, then it is constant. (13) 𝑥 ↦ 𝑃(𝑥, 𝜉) is harmonic in 𝔹, for each 𝜉 ∈ 𝕊. (Hint: Write 𝑃(𝑥, 𝜉) = 2 −𝑑 𝜔−1 and use the identity Δ(𝑢𝑣) = (Δ𝑢)𝑣 + 2∇𝑢 ⋅ 𝑑 (1 − |𝑥| )|𝑥 − 𝜉| ∇𝑣 + 𝑢Δ𝑣.) (14) Symmetry Lemma: If 𝑥 ∈ 𝔹 and 𝜉 ∈ 𝕊, then ||𝑥|𝜉 − 𝑥 | = |𝑥 − 𝜉|. | |𝑥| |

Exercises

33

See Figure 2.4.

Figure 2.4. If 𝑥 ∈ 𝔹 and 𝜉 ∈ 𝕊, the distance between the points 𝑥/|𝑥| and |𝑥|𝜉 is the same as the distance between 𝑥 and 𝜉, as stated by the symmetry lemma.

(15) Let 𝑓 be Riemann-integrable on 𝕊. Then its Poisson integral 𝑢 is harmonic in 𝔹. (16) Hopf lemma: If 𝑢 is a nonconstant harmonic function in 𝔹, is continuous on 𝔹,̄ and attains its maximum at 𝜁 ∈ 𝕊, then there exists 𝑐 > 0 such that 𝑢(𝜁) − 𝑢(𝑟𝜁) > 𝑐(1 − 𝑟), for any 0 < 𝑟 < 1. (17) Harnack inequality: If 𝑢 is a harmonic function in 𝔹, is continuous on 𝔹,̄ and is positive, then 1 − |𝑥| 1 + |𝑥| 𝑢(0) ≤ 𝑢(𝑥) ≤ 𝑢(0) 𝑑−1 (1 + |𝑥|) (1 − |𝑥|)𝑑−1 for all 𝑥 ∈ 𝔹. (18) If 𝑢 is harmonic in Ω and 𝐵𝑟̄ (𝑥0 ) ⊂ Ω, then the values of 𝑢 in 𝐵𝑟 (𝑥0 ) are determined by its values on 𝑆𝑟 (𝑥0 ). (19) Let 𝑢𝑛 be a sequence of harmonic functions in Ω such that 𝑢𝑛 ⇉ 𝑢 on any compact 𝐾 ⊂ Ω. Then 𝑢 is harmonic in Ω. (20) Prove Theorem 2.33 for 𝑑 = 2.

34

2. Basic properties

(21) Let 𝑢 be harmonic in a domain in ℝ𝑑 with an isolated singularity at 𝑥0 . If 𝑑 = 2 and lim 𝑢(𝑥) log |𝑥 − 𝑥0 | = 0,

𝑥→𝑥0

or 𝑑 > 2 and lim 𝑢(𝑥)|𝑥 − 𝑥0 |𝑑−2 = 0,

𝑥→𝑥0

then 𝑥0 is a removable singularity.

Notes The results of this chapter are basic classical results, proven in every text in harmonic functions and partial differential equations. A classical reference for the theory of harmonic functions is [Kel67]. Theorem 2.4 is a result by Gauss [Gau40]. The proof presented here is the most popular and can be found, for instance, in [ABR01] or in [Fol95]. It can also be proven by differentiating the integral over a sphere with respect to its radius, as in [Eva10] or in [MS13]. The proof of Theorem 2.9 is also in [Fol95]. The proof of Theorem 2.14 is an elaboration of the proof by Nelson [Nel61]. Siméon Denis Poisson developed explicit expressions to solutions to the Laplace equation in terms of integrals over the sphere in [Poi20], and thus the Poisson kernel and integral are named after him. Theorem 2.33 is a result by Riemann [Rie51]. The proof presented here can be found in [ABR01].

Chapter 3

Fourier series

3.1. Separation of variables In Chapter 2 we solved the Dirichlet problem (2.27) using the Poisson integral, which provides an explicit integral form of the solution from its values on the boundary. We now attempt to solve the problem by decomposing the function 𝑓 on 𝕊 in fundamental pieces, which is the original Fourier approach in [Fou55]. In this chapter we consider the Dirichlet problem in the unit disk 𝔻 in the plane ℝ2 . In order to find such fundamental pieces, we first recall that the Laplacian in polar coordinates (𝑟, 𝜃) is given by Δ𝑢 =

𝜕2 𝑢 1 𝜕𝑢 1 𝜕2 𝑢 + + 2 2. 2 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑟

(See Exercise (2) of Chapter 1.) We now search for solutions of the form 𝑢(𝑟, 𝜃) = 𝑣(𝑟)𝜙(𝜃), where 𝑣(𝑟) is a twice differentiable function defined for 0 ≤ 𝑟 < 1 and 𝜙(𝜃) is a twice differentiable periodic function defined on ℝ, with period 2𝜋, as the pair (𝑟, 𝜃) denotes a point in the disk. Thus we want to solve the equation 1 1 Δ𝑢 = 𝑣″ (𝑟)𝜙(𝜃) + 𝑣′ (𝑟)𝜙(𝜃) + 2 𝑣(𝑟)𝜙″ (𝜃) = 0, 𝑟 𝑟 35

36

3. Fourier series

which we can rewrite, when 𝑣(𝑟) ≠ 0 and 𝜙(𝜃) ≠ 0, as 𝜙″ (𝜃) 𝑟2 𝑣″ (𝑟) + 𝑟𝑣′ (𝑟) =− . 𝑣(𝑟) 𝜙(𝜃) The left hand side of this equation does not depend on 𝜃, and the right hand side does not depend on 𝑟, so we conclude that both sides are equal to a constant, say, 𝜆 ∈ ℝ. Hence we obtain the equations (3.1)

𝑟2 𝑣″ (𝑟) + 𝑟𝑣′ (𝑟) = 𝜆𝑣(𝑟)

and 𝜙″ (𝜃) = −𝜆𝜙(𝜃),

(3.2)

subject to the constrains stated above. Equation (3.2) has periodic solutions 𝜙(𝜃) = cos(𝑛𝜃)

and

𝜙(𝜃) = sin(𝑛𝜃),

2

with period 2𝜋, when 𝜆 = 𝑛 and 𝑛 ∈ ℕ. We have a pair of linearly independent solutions for each natural number 𝑛 ≥ 1. For 𝑛 = 0, we have the linearly independent solutions 𝜙(𝜃) = 1 and 𝜙(𝜃) = 𝜃, but the latter is not periodic. Now, with 𝜆 = 𝑛2 , equation (3.1) is 𝑟2 𝑣″ (𝑟) + 𝑟𝑣′ (𝑟) − 𝑛2 𝑣(𝑟) = 0 and has linearly independent solutions 𝑣(𝑟) = 𝑟𝑛

and

𝑣(𝑟) = 𝑟−𝑛

for each 𝑛 ≥ 1, though only the former is well defined on [0, 1). Again, if 𝑛 = 0, we have the solutions 𝑣(𝑟) = 1 and 𝑣(𝑟) = log 𝑟, but only the former is defined on [0, 1). We thus obtain, for each 𝑛 ∈ ℕ, the harmonic functions 𝑢𝑛 (𝑟, 𝜃) = 𝑟𝑛 (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)), where 𝑎𝑛 , 𝑏𝑛 ∈ ℝ (note that 𝑢0 is the constant function 𝑢0 (𝑟, 𝜃) = 𝑎0 ). Any linear combination of such functions, 𝑁

(3.3)

𝑢(𝑟, 𝜃) = ∑ 𝑟𝑛 (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)), 𝑛=0

3.1. Separation of variables

37

is harmonic in the disk, and its limit at the boundary 𝑟 → 1 is the trigonometric polynomial 𝑁

(3.4)

𝑝(𝜃) = ∑ (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)). 𝑛=0

In fact, this limit is uniform as 𝑟 → 1 on 𝕊. We have thus solved the problem Δ𝑢 = 0 in 𝔻 { 𝑢=𝑝 on 𝕊, where 𝑝 is the trigonometric polynomial (3.4). It is clear that the trigonometric polynomial (3.4) can be seen either as a function on 𝕊 or as a periodic function on ℝ, through the map 𝜃 ↦ (cos 𝜃, sin 𝜃). In general, that is true for any 2𝜋-periodic function 𝑓 on ℝ: the function 𝐹(cos 𝜃, sin 𝜃) = 𝑓(𝜃) is a well-defined function on 𝕊. We note that 𝑓 is continuous on ℝ if and only if 𝐹 is continuous on 𝕊, so we can identify the space 𝐶(𝕊) of continuous functions on 𝕊 with the subspace of 𝐶(ℝ) of periodic functions with period 2𝜋, or with the subspace of 𝐶([0, 2𝜋]) of functions satisfying 𝑓(0) = 𝑓(2𝜋). Similarly, Riemann-integrable functions on 𝕊 can be identified with Riemann-integrable functions 𝑓 on [0, 2𝜋] satisfying 𝑓(0) = 𝑓(2𝜋), as well as with 2𝜋-periodic functions on ℝ that are Riemann-integrable on each closed interval. We will interchangeably use the terms “function on 𝕊”, “function on [0, 2𝜋]” and “2𝜋-periodic function on ℝ” throughout this chapter. If we recall de Moivre’s formula for complex numbers, (cos 𝜃 + 𝑖 sin 𝜃)𝑛 = cos(𝑛𝜃) + 𝑖 sin(𝑛𝜃), we see that (3.4) is indeed a polynomial in cos 𝜃 and sin 𝜃 (that is why we call it a trigonometric polynomial), and thus a polynomial in the coordinate functions on 𝕊. We can ask whether we can use the linear combinations (3.3) to solve the Dirichlet problem (3.5)

Δ𝑢 = 0 { 𝑢=𝑓

in 𝔻 on 𝕊,

given any continuous function 𝑓 on the boundary 𝕊, and not only a trigonometric polynomial. More precisely, we have:

38

3. Fourier series Question 1: For 𝑓 ∈ 𝐶(𝕊), is there a sequence of trigonometric polynomials 𝑝𝑁 that converge to 𝑓 (pointwise or uniformly) such that the corresponding solutions 𝑢𝑁 as in (3.3) converge to a solution of the Dirichlet problem (3.5)? Question 2: For 𝑓 ∈ 𝐶(𝕊), is there a series ∞

∑ (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)) 𝑛=0

such that the series ∞

𝑢(𝑟, 𝜃) = ∑ 𝑟𝑛 (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)) 𝑛=0

converges to a solution of the Dirichlet problem (3.5)? Note that an affirmative answer to Question 2 would not necessarily give a positive answer to Question 1, as the partial sums of the series might not satisfy the requirement of the trigonometric polynomials 𝑝𝑁 and the functions 𝑢𝑁 . Note also that a sequence 𝑝𝑁 of trigonometric polynomials converging uniformly to 𝑓 is by all means not unique, and thus neither the sequence 𝑢𝑁 . These questions also make sense if 𝑓 is only Riemann-integrable, except for the fact that we cannot expect uniform convergence anymore because the uniform limit of continuous functions is continuous. In that case, we would restrict to convergence at certain points of 𝕊.

3.2. Fourier series Consider the problem of writing a function 𝑓 on 𝕊 as a series ∞

(3.6)

∑ (𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃) 𝑛=0

for 𝜃 ∈ [0, 2𝜋], where we agree that 𝑏0 = 0. By Euler’s formula 𝑒𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥, we can write, for 𝑛 ≥ 1, 𝑎𝑛 cos 𝑛𝜃 + 𝑏𝑛 sin 𝑛𝜃 = 𝑐𝑛 𝑒𝑖𝑛𝜃 + 𝑐−𝑛 𝑒−𝑖𝑛𝜃 , where 𝑐 ±𝑛 =

𝑎𝑛 ∓ 𝑖𝑏𝑛 . 2

3.2. Fourier series

39

If we set 𝑐 0 = 𝑎0 , the series (3.6) can be written as ∞

(3.7)

∞

𝑐 0 + ∑ (𝑐𝑛 𝑒𝑖𝑛𝜃 + 𝑐−𝑛 𝑒−𝑖𝑛𝜃 ) = ∑ 𝑐𝑛 𝑒𝑖𝑛𝜃 , 𝑛=1

𝑛=−∞

where the double infinite series on the right of (3.7) is understood as the limit of the partial sums 𝑁

∑ 𝑐𝑛 𝑒𝑖𝑛𝜃 𝑛=−𝑁

as 𝑁 → ∞. Suppose the series (3.7) converges uniformly to 𝑓, so 𝑁

𝑠𝑁 (𝜃) = ∑ 𝑐𝑛 𝑒𝑖𝑛𝜃 ⇉ 𝑓(𝜃). 𝑛=−𝑁

Thus, for any 𝑛 ∈ ℤ (see Section A.3), 2𝜋

2𝜋

∫

𝑠𝑁 (𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃 → ∫

0

0

𝑓(𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃.

Now for any 𝑚 ∈ ℤ, 2𝜋

(3.8)

∫

𝑒𝑖𝑚𝜃 𝑒−𝑖𝑛𝜃 𝑑𝜃 = {

0

2𝜋 𝑚 = 𝑛 0

𝑚≠0

(Exercise (3)) and thus, for 𝑁 ≥ 𝑛, 2𝜋

∫

𝑠𝑁 (𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃 = 2𝜋𝑐𝑛 .

0

Therefore, if the series (3.7) converges uniformly to 𝑓, the coefficients 𝑐𝑛 are given by 𝑐𝑛 =

1 ∫ 2𝜋 0

2𝜋

𝑓(𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃

for each 𝑛 ∈ ℤ. For any Riemann-integrable function 𝑓 on 𝕊, we define its Fourier coefficients, for 𝑛 ∈ ℤ, by 2𝜋

(3.9)

̂ = 1 ∫ 𝑓(𝑛) 2𝜋 0

𝑓(𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃.

40

3. Fourier series

The series 𝑖𝑛𝜃 ̂ ∑ 𝑓(𝑛)𝑒

(3.10)

𝑛∈ℤ

is called the Fourier series of 𝑓. Note that, since a Riemann-integrable function 𝑓 is bounded, its Fourier coefficients (3.9) are bounded. Indeed, they are bounded by any bound for 𝑓 because, if |𝑓(𝜃)| ≤ 𝑀 for all 𝜃, then 1 ̂ ∫ |𝑓(𝑛)| ≤ || 2𝜋 0

(3.11)

2𝜋

2𝜋

𝑓(𝜃)𝑒

−𝑖𝑛𝜃

1 ∫ 𝑑𝜃|| ≤ 2𝜋 0

|𝑓(𝜃)|𝑑𝜃 ≤ 𝑀.

We also observe that in the definition (3.9) of the Fourier coefficients of 𝑓, as 𝑓 is periodic, we can choose any interval of length 2𝜋 in the integration, as convenient (see Exercise (4)). Example 3.12 (The sawtooth function). Consider the periodic function 𝑓, with period 2𝜋, given in [−𝜋, 𝜋) by 𝑓(𝜃) = 𝜃 (Figure 3.1). Its Fourier π

-2 π

-π

2π

π

3π

-π

Figure 3.1. The sawtooth function given by 𝑓(𝜃) = 𝜃, −𝜋 ≤ 𝜃 < 𝜋. Note that 𝑓 is discontinuous at 𝜋.

coefficients are given, for 𝑛 = 0, by 𝜋

𝜋

̂ = 1 ∫ 𝑓(𝜃)𝑑𝜃 = 1 ∫ 𝜃𝑑𝜃 = 0, 𝑓(0) 2𝜋 −𝜋 2𝜋 −𝜋 and, for 𝑛 ≠ 0, by 𝜋

𝜋

𝑛 ̂ = 1 ∫ 𝑓(𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃 = 1 ∫ 𝜃𝑒−𝑖𝑛𝜃 𝑑𝜃 = 𝑖(−1) , 𝑓(𝑛) 2𝜋 −𝜋 2𝜋 −𝜋 𝑛

3.2. Fourier series

41

and thus the Fourier series of 𝑓 is given by ∞

(−1)𝑛 𝑖𝑛𝜃 (−1)𝑛 𝑒 = −2 ∑ sin 𝑛𝜃. 𝑛 𝑛 𝑛≠0 𝑛=1

𝑖∑

Using Dirichlet’s test,1 one can prove that this series converges for every 𝜃 (Exercise (5)). At 𝜃 = 0, the series clearly converges to 0, and at 𝜃 = 𝜋/2, the series ∞

∞

(−1)𝑛 𝑛𝜋 (−1)𝑘 1 1 1 sin =2∑ = 2(1 − + − + ⋯ ) 𝑛 2 2𝑘 + 1 3 5 7 𝑛=1 𝑘=0

−2 ∑

converges to 𝜋/2, as seen in a calculus course. In both cases, the series at each 𝜃 converges to 𝑓(𝜃). However, at 𝜃 = 𝜋, the series converges to 0, even though 𝑓(𝜋) = −𝜋. Note that 𝑓 is discontinuous at 𝜋. Example 3.13 (The sharkteeth function). We now consider the periodic function 𝑔 given in [−𝜋, 𝜋) by 𝑔(𝜃) = |𝜃| (Figure 3.2). This time the π

-2 π

-π

π

2π

3π

Figure 3.2. The sharkteeth function given by 𝑔(𝜃) = |𝜃|, −𝜋 ≤ 𝜃 < 𝜋. Note that 𝑔 is continuous at every point.

function 𝑔 is continuous at every point. Its Fourier coefficients are given by 𝜋 𝑛=0 ⎧ ⎪2 even 𝑛 ≠ 0 𝑔(𝑛) ̂ = 0 ⎨ 2 ⎪− ⎩ 𝜋𝑛2 odd 𝑛 (Exercise (6)), and thus its Fourier series is given by ∞

𝑒𝑖𝑛𝜃 cos(2𝑘 + 1)𝜃 𝜋 2 𝜋 4 ∑ 2 = − ∑ − . 2 𝜋 odd 𝑛 𝑛 2 𝜋 𝑘=0 (2𝑘 + 1)2 1

Theorem A.5 in Section A.1.

42

3. Fourier series

We see that, this time, the Fourier series of 𝑔 converges uniformly on [−𝜋, 𝜋), by the Weierstrass 𝑀-test,2 as the coefficients decrease as 1/𝑛2 . For example, for 𝜃 = 𝜋/2, each 𝜋 cos(2𝑘 + 1) = 0, 2 so the series converges to 𝜋/2, the value of 𝑔 at 𝜋/2. We can ask then if the series converges to 𝑔(𝜃) at any other 𝜃. Example 3.14 (The flounces function). Consider now the periodic function ℎ given in [−𝜋, 𝜋) by ℎ(𝜃) = (𝜋2 −𝜃2 )2 , shown in Figure 3.3. ℎ is not

-π

π

Figure 3.3. The flounces function given by ℎ(𝜃) = (𝜋2 − 𝜃2 )2 , −𝜋 ≤ 𝜃 < 𝜋. Note that ℎ is continuous and differentiable at every point.

only continuous, but also differentiable at everypoint, with ℎ′ (𝑘𝜋) = 0 for every 𝑘 ∈ ℤ. Its Fourier coefficients (Exercise (7)) are given by 8 4 𝜋 15 ̂ ℎ(𝑛) = { (−1)𝑛 −24 4 𝑛 so its Fourier series is then

𝑛=0 𝑛 ≠ 0, ∞

8 4 (−1)𝑛 𝑖𝑛𝜃 8 4 (−1)𝑛 𝑒 = cos 𝑛𝜃. 𝜋 − 24 ∑ 𝜋 − 48 ∑ 4 15 15 𝑛 𝑛4 𝑛≠0 𝑛=1 As in example 3.13 above, the Fourier series converges absolutely and uniformly, though this time it is not clear what the limit of the series is at any point. However, we observe that the series converges more rapidly than in the previous example, and thus one questions if the regularity of ℎ—the fact that it is not only continuous but also differentiable—has any effect in the decrease of the coefficients. We have now another set of questions. 2

Theorem A.6 in Section A.1.

3.3. Abel means and Poisson integrals

43

Question 3: Does the Fourier series (3.10) converge? In what sense? Pointwise, uniformly? In any other sense? Does the regularity of the function have any effect on the convergence? Question 4: If the Fourier series (3.10) of 𝑓 converges, does it converge to 𝑓? It turns out that Question 4 can be answered thanks to our previous analysis of Poisson integrals.

3.3. Abel means and Poisson integrals ∞

Consider a series ∑𝑛=0 𝑎𝑛 . The Abel means of ∑ 𝑎𝑛 are given by ∞

𝐴𝑟 = ∑ 𝑎𝑛 𝑟𝑛 ,

(3.15)

𝑛=0

for 0 < 𝑟 < 1. We say that the series ∑ 𝑎𝑛 is Abel-summable to 𝑠 if lim 𝐴𝑟 = 𝑠. 𝑟→1

All convergent series are also Abel-summable, a result know as Abel’s theorem. Theorem 3.16. Suppose that the series ∑ 𝑎𝑛 converges to 𝑠. Then ∑ 𝑎𝑛 is Abel-summable to 𝑠. Proof. Let 𝑠𝑛 = 𝑎0 + 𝑎1 + . . . 𝑎𝑛 be the 𝑛th partial sum of the series, so we have 𝑠𝑛 → 𝑠. In particular, 𝑠𝑛 is bounded, so there exists 𝑀 > 0 such that |𝑠𝑛 | ≤ 𝑀 for all 𝑛. The boundedness of 𝑠𝑛 implies the convergence of ∑ 𝑠𝑛 𝑟𝑛 , so for each 0 ≤ 𝑟 < 1, using the fact that 𝑎𝑛 = 𝑠𝑛 − 𝑠𝑛−1 for each 𝑛 ≥ 1, ∞

∞

𝐴𝑟 = ∑ 𝑎𝑛 𝑟𝑛 = 𝑎0 + ∑ (𝑠𝑛 − 𝑠𝑛−1 )𝑟𝑛 𝑛=0

𝑛=1 ∞

∞

= 𝑠0 + ∑ 𝑠𝑛 𝑟𝑛 − ∑ 𝑠𝑛−1 𝑟𝑛 𝑛=1 ∞

𝑛=1 ∞

∞

= ∑ 𝑠𝑛 𝑟𝑛 − ∑ 𝑠𝑛 𝑟𝑛+1 = (1 − 𝑟) ∑ 𝑠𝑛 𝑟𝑛 . 𝑛=0

𝑛=0

𝑛=0

44

3. Fourier series Given 𝜀 > 0 there exists 𝑁 such that |𝑠𝑛 − 𝑠|
0 we have |𝑓′ (𝜏)| ≤ 𝑀, so (3.38)

|𝑔ℎ (𝜃)| ≤ 2𝑀ℎ

52

3. Fourier series

for every 𝜃. Now, 2𝜋

𝑔 ˆ ℎ (𝑛) =

1 ∫ (𝑓(𝜃 + ℎ) − 𝑓(𝜃 − ℎ))𝑒−𝑖𝑛𝜃 𝑑𝜃 2𝜋 0 2𝜋

2𝜋

=

1 ∫ 2𝜋 0

𝑓(𝜃 + ℎ)𝑒−𝑖𝑛𝜃 𝑑𝜃 −

1 ∫ 2𝜋 0

=

1 ∫ 2𝜋 0

𝑓(𝜃)𝑒−𝑖𝑛(𝜃−ℎ) 𝑑𝜃 −

1 ∫ 2𝜋 0

=

𝑒𝑖𝑛ℎ − 𝑒−𝑖𝑛ℎ ∫ 2𝜋 0

2𝜋

𝑓(𝜃 − ℎ)𝑒−𝑖𝑛𝜃 𝑑𝜃

2𝜋

𝑓(𝜃)𝑒−𝑖𝑛(𝜃+ℎ) 𝑑𝜃

2𝜋

̂ 𝑓(𝜃)𝑒−𝑖𝑛𝜃 𝑑𝜃 = 2𝑖 sin 𝑛ℎ𝑓(𝑛),

so we have ̂ 2. ∑ |ˆ 𝑔ℎ (𝑛)|2 = 4 ∑ | sin 𝑛ℎ|2 |𝑓(𝑛)|

(3.39)

𝑛∈ℤ

𝑛∈ℤ

Lemma 3.36, together with equations (3.38) and (3.39), implies ̂ 2≤ ∑ | sin 𝑛ℎ|2 |𝑓(𝑛)| 𝑛∈ℤ

1 1 ∫ ⋅ 4 2𝜋 0

2𝜋

|𝑔ℎ (𝜃)|2 𝑑𝜃 ≤

(2𝑀ℎ)2 ⋅ 2𝜋 = 𝑀 2 ℎ2 . 8𝜋

In particular, for each 𝑝 ≥ 1, ̂ 2 ≤ 𝑀 2 ℎ2 . | sin 𝑛ℎ|2 |𝑓(𝑛)|

∑

(3.40)

2𝑝−1 ≤|𝑛| 0 such that, if |𝜏| < 𝜂, then 𝜀 |𝑓(𝜃 − 𝜏) − 𝑓(𝜃)| < . 2 Now, as 𝑓 is Riemann-integrable, it is bounded, so there exists 𝑀 > 0 such that |𝑓(𝜏)| ≤ 𝑀 for all 𝜏. By (3.48), there exists 𝐾 such that, if 𝑁 ≥ 𝐾, 𝜀 | 1 ∫ 𝐹𝑁 (𝜏)𝑑𝜏|| < . | 2𝜋 4𝑀 𝜂≤|𝜏|≤𝜋 Therefore, for 𝑁 ≥ 𝐾, using (3.47), 𝜋

1 ∫ 𝐹𝑁 (𝜏)𝑓(𝜃 − 𝜏)𝑑𝜏 − 𝑓(𝜃)|| |𝜎𝑁 (𝜃) − 𝑓(𝜃)| = || 2𝜋 −𝜋

𝜋

𝜋

= ||

1 1 ∫ 𝐹 (𝜏)𝑓(𝜃 − 𝜏)𝑑𝜏 − 𝑓(𝜃) ⋅ ∫ 𝐹 (𝜏)𝑑𝜏|| 2𝜋 −𝜋 𝑁 2𝜋 −𝜋 𝑁

= ||

1 ∫ 𝐹 (𝜏)(𝑓(𝜃 − 𝜏) − 𝑓(𝜃))𝑑𝜏|| 2𝜋 −𝜋 𝑁

𝜋

≤

1 ∫ 𝐹 (𝜏)|𝑓(𝜃 − 𝜏) − 𝑓(𝜃)|𝑑𝜏 2𝜋 |𝜏| 0, ∫

|𝐾𝑁 (𝜃)|𝑑𝜃 → 0

𝜂≤|𝜃|≤𝜋

as 𝑁 → ∞. It is not hard to verify that the proof of Theorem 3.45 applies to any family of good kernels, so, if a function is continuous at 𝜃, then 𝜋

1 ∫ 𝐾 (𝜃 − 𝜏)𝑓(𝜏)𝑑𝜏 → 𝑓(𝜃) 2𝜋 −𝜋 𝑁 as 𝑁 → ∞. The fact that the Cesàro sums of a convergent series converge to the same limit implies the same results of Corollary 3.21. However, we additionally have the following stronger result. Corollary 3.49. The space of trigonometric polynomials is dense in 𝐶(𝕊). In other words, if 𝑓 ∈ 𝐶(𝕊) and 𝜀 > 0, there exists a trigonometric polynomial 𝑝 such that |𝑓(𝜃) − 𝑝(𝜃)| < 𝜀 for every 𝜃. Proof. By Fejér’s theorem, the Cesàro sums 𝜎𝑁 of the Fourier series of 𝑓 converge uniformly to 𝑓. In other words, given 𝜀 > 0, there exists 𝑁 such that |𝑓(𝜃) − 𝜎𝑁 (𝜃)| < 𝜀

3.6. Mean-square convergence

59

for all 𝜃. The results follows from the fact that 𝜎𝑁 is a trigonometric polynomial. □ Corollary 3.49 gives an affirmative answer to Question 1 above.

3.6. Mean-square convergence We now come back to the orthogonality of the Fourier expansions to prove Theorem 3.50. Theorem 3.50. If 𝑓 is a 2𝜋-periodic Riemann-integrable function and, for each 𝑁 ∈ ℕ, 𝑁 𝑖𝑛𝜃 ̂ 𝑠𝑁 (𝜃) = ∑ 𝑓(𝑛)𝑒 , 𝑛=−𝑁

then (3.51)

lim ‖𝑓 − 𝑠𝑁 ‖ = 0.

𝑁→∞

In other words, the Fourier series of 𝑓 𝑖𝑛𝜃 ̂ ∑ 𝑓(𝑛)𝑒 𝑛∈ℤ

converges to 𝑓 in the sense, 2𝜋

𝑁

1 ∫ 2𝜋 0

2

𝑖𝑛𝜃 | |𝑓(𝜃) − ∑ 𝑓(𝑛)𝑒 ̂ | | 𝑑𝜃 → 0 𝑛=−𝑁

as 𝑁 → ∞. This is called mean-square convergence. Proof. As 𝑓 is Riemann-integrable, it is bounded, so there exists 𝑀 > 0 such that |𝑓(𝜃)| ≤ 𝑀 for all 𝜃. Now, given 𝜀 > 0, by Theorem A.10 (Section A.3) we can choose a continuous function 𝑔 on 𝕊 such that |𝑔(𝜃)| ≤ 𝑀 and 2𝜋

∫

|𝑓(𝜃) − 𝑔(𝜃)|𝑑𝜃
0 small, let 𝑔𝑁 be the continuous function on [−𝜋, 𝜋] that is equal to 𝜙𝑁 (𝜃) in each interval [𝜃𝑘 + 𝛿, 𝜃𝑘+1 − 𝛿], and is equal to the linear function from 𝜙𝑁 (𝜃𝑘 − 𝛿) to 𝜙𝑁 (𝜃𝑘 + 𝛿) around each 𝜃𝑘 (see Figure 3.4).4 We now observe that 𝜃𝑘 =

Figure 3.4. The functions 𝐷𝑁 , 𝜙𝑁 and 𝑔𝑁 for 𝑁 = 5. 𝜋

∫ |𝜙𝑁 (𝜃) − 𝑔𝑁 (𝜃)|𝑑𝜃 = 2𝑁𝛿, −𝜋

so, using the fact that |𝐷𝑁 (𝜃)| ≤ 2𝑁 + 1, we obtain 𝜋

1 1 1 ∫ |𝐷 (𝜃)||𝜙𝑁 (𝜃) − 𝑔𝑁 (𝜃)|𝑑𝜃 ≤ (2𝑁 + 1)2𝑁𝛿 < , 2𝜋 −𝜋 𝑁 2𝜋 4 if we choose 𝛿< 4

Cf. the proof of Theorem A.10.

𝜋 . 4𝑁(2𝑁 + 1)

3.7. Convergence for continuous functions

65

Hence 𝜋

1 ∫ 𝐷 (𝜃)𝑔𝑁 (𝜃)𝑑𝜃 2𝜋 −𝜋 𝑁 𝜋

𝜋

1 1 ∫ 𝐷 (𝜃)𝜙𝑁 (𝜃)𝑑𝜃 − ∫ |𝐷 (𝜃)||𝜙𝑁 (𝜃) − 𝑔𝑁 (𝜃)|𝑑𝜃 ≥ 2𝜋 −𝜋 𝑁 2𝜋 −𝜋 𝑁 4 ≥ 2 log 𝑁. 𝜋 Set 𝑁 large enough so that 4 log 𝑁 ≥ 𝑀 + 1. 𝜋2 By Corollary 3.49, there exists a trigonometric polynomial 𝑝(𝜃) such that |𝑔𝑁 (𝜃) − 𝑝(𝜃)| < Hence

1 . 2𝑁 + 1

𝜋

1 ∫ |𝐷 (𝜃)||𝑔𝑁 (𝜃) − 𝑝(𝜃)|𝑑𝜃 < 1 2𝜋 −𝜋 𝑁 and, as above 𝜋

1 ∫ 𝐷 (𝜃)𝑝(𝜃)𝑑𝜃 2𝜋 −𝜋 𝑁 𝜋

≥

𝜋

1 1 ∫ 𝐷 (𝜃)𝑔𝑁 (𝜃)𝑑𝜃 − ∫ |𝐷 (𝜃)||𝑔𝑁 (𝜃) − 𝑝(𝜃)|𝑑𝜃 2𝜋 −𝜋 𝑁 2𝜋 −𝜋 𝑁

≥ 𝑀. Since |𝑔𝑁 (𝜃)| ≤ 1, we clearly have |𝑝(𝜃)| ≤ 2.

□

Proof of Theorem 3.56. For each 𝑘 ≥ 0, by Lemma 3.57 we can find a trigonometric polynomial 𝑝 𝑘 and an integer 𝑁 𝑘 such that |𝑝 𝑘 (𝜃)| ≤ 2 and 𝜋

(3.58)

1 ∫ 𝐷 (𝜃)𝑝 𝑘 (𝜃)𝑑𝜃 ≥ 2𝑘 . 2𝜋 −𝜋 𝑁𝑘

Let 𝑑𝑘 be a sequence of integers such that 𝑑𝑘+1 > 3𝑑𝑘 and such that 𝑑𝑘 is at least as large as the degree of 𝑝 𝑘 and as 𝑁 𝑘 , so that we can write 𝑑𝑘

(3.59)

𝑝 𝑘 (𝜃) = ∑ 𝑝 ˆ𝑘 (𝑛)𝑒𝑖𝑛𝜃 , 𝑛=−𝑑𝑘

66

3. Fourier series

and estimate (3.58) implies 𝑁𝑘

∑ 𝑝 ˆ𝑘 (𝑛) ≥ 2𝑘 .

(3.60)

𝑛=−𝑁𝑘

Now define 𝑞𝑘 (𝜃) = 𝑒2𝑖𝑑𝑘 𝜃 𝑝 𝑘 (𝜃), so by (3.59) we have 3𝑑𝑘

𝑑𝑘

ˆ𝑘 (𝑛)𝑒 𝑞𝑘 (𝜃) = ∑ 𝑝

𝑖(𝑛+2𝑑𝑘 )𝜃

𝑛=−𝑑𝑘

= ∑ 𝑝 ˆ𝑘 (𝑛 − 2𝑑𝑘 )𝑒𝑖𝑛𝜃 , 𝑛=𝑑𝑘

so (3.60) is now 2𝑑𝑘 +𝑁𝑘

𝑞ˆ𝑘 (𝑛) ≥ 2𝑘 .

∑

(3.61)

𝑛=2𝑑𝑘 −𝑁𝑘

Note that |𝑞𝑘 (𝜃)| ≤ 2 and, for each 𝑛, 𝑞ˆ𝑘 (𝑛) ≠ 0 for at most one 𝑘 because 𝑑𝑘 > 3𝑑𝑘−1 . We can now define ∞

𝑓(𝜃) = ∑ 2−𝑘 𝑞𝑘 (𝜃). 𝑘=0

By the Weierstrass 𝑀-test, the series above converges uniformly and, since each 𝑞𝑘 is continuous (it’s a trigonometric polynomial), then 𝑓 is continuous. Moreover, by the observations above and the uniform con̂ = 2−𝑘 𝑞ˆ𝑘 (𝑛) for 𝑑𝑘 ≤ 𝑛 ≤ vergence of the series, for each 𝑛 we have 𝑓(𝑛) 3𝑑𝑘 . Therefore, if 𝑠𝑁 (0) is the 𝑁th partial sum of the Fourier series of 𝑓 at 0, 2𝑑𝑘 +𝑁𝑘

𝑠2𝑑𝑘 +𝑁𝑘 (0) − 𝑠2𝑑𝑘 −𝑁𝑘 −1 (0) = 2−𝑘

∑

𝑞ˆ𝑘 (𝑛) ≥ 1,

𝑛=2𝑑𝑘 −𝑁𝑘

by (3.61). Therefore, the sequence 𝑠𝑁 (0) is not a Cauchy sequence, and cannot converge. □ By an appropiate translation of 𝑓, we can verify the existence of a continuous function with divergent Fourier series at any 𝜃0 ∈ 𝕊. By adding such functions, we conclude that there exist continuous functions with divergent Fourier series at any finite, or even countable infinite, number of points (Exercise (18)). Theorem 3.56 implies that being continuous is not enough for a function to have convergent Fourier series. However, if the function is “regular enough”, then we can guarantee its convergence, as in Theorem

3.7. Convergence for continuous functions

67

3.32, where we have 𝑓 ∈ 𝐶 1 (𝕊). Moreover, it is enough for a function to be differentiable at a point to conclude that its Fourier series converges at that point. Theorem 3.62. Let 𝑓 be Riemann-integrable on 𝕊 and differentiable at 𝜃0 . Then its Fourier series converges at 𝜃0 . Proof. Let 𝑓 be differentiable at 𝜃0 . We want to prove that 𝜋

(3.63)

𝑠𝑁 (𝜃0 ) − 𝑓(𝜃0 ) =

1 ∫ 𝐷 (𝜏)(𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ))𝑑𝜏 2𝜋 −𝜋 𝑁

converges to 0 as 𝑁 → ∞. Since 𝐷𝑁 (𝜏) =

sin 𝑁𝜏 cos 𝜏/2 + cos 𝑁𝜏 sin 𝜏/2 sin(𝑁 + 1/2)𝜏 = , sin 𝜏/2 sin 𝜏/2

we can write (3.63) as 𝜋

(3.64)

𝑠𝑁 (𝜃0 ) − 𝑓(𝜃0 ) =

1 sin 𝑁𝜏 cos 𝜏/2 ∫ (𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ))𝑑𝜏 2𝜋 −𝜋 sin 𝜏/2 𝜋

+

1 ∫ cos 𝑁𝜏(𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ))𝑑𝜏. 2𝜋 −𝜋

Note that the second integral in (3.64) is the Fourier coefficient 𝑎𝑁 of the Riemann-integrable function 𝜏 ↦ 𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ), and thus converges to 0 as 𝑁 → ∞, by the Riemann–Lebesgue lemma 3.43. The first integral in (3.64) is the Fourier coefficient 𝑏𝑁 of the function 𝜏 ↦ cos 𝜏/2

𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ) , sin 𝜏/2

which is also Riemann-integrable as the limit lim

𝜏→0

𝑓(𝜃0 − 𝜏) − 𝑓(𝜃0 ) sin 𝜏/2

exists because 𝑓 is differentiable at 𝜃0 . Thus, it also converges to 0 when 𝑁 → ∞. □

68

3. Fourier series

Exercises (1) Calculate the solutions of equations (3.1) and (3.2) when 𝜆 < 0. (2) If the sequences 𝑎𝑛 and 𝑏𝑛 are bounded, then ∞

𝑢(𝑟, 𝜃) = ∑ 𝑟𝑛 (𝑎𝑛 cos(𝑛𝜃) + 𝑏𝑛 sin(𝑛𝜃)) 𝑛=0

is harmonic in 𝔻. (3) For 𝑚, 𝑛 ∈ ℤ, 2𝜋

∫ 0

2𝜋 𝑚 = 𝑛 𝑒𝑖𝑚𝜃 𝑒−𝑖𝑛𝜃 𝑑𝜃 = { 0 𝑚 ≠ 𝑛.

(4) If 𝑓 is Riemann-integrable and periodic with period 𝑇, then 𝑎+𝑇

∫

𝑇

𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥

𝑎

0

for any 𝑎 ∈ ℝ. (5) Use Dirichlet’s test to prove that the Fourier series of Example 3.12 converges for every 𝜃. (6) Complete the details of Example 3.13. (7) Complete the details of Example 3.14. (8) If ∑ 𝑎𝑛 is Abel-summable to 𝑠 and 𝑛𝑎𝑛 → 0, then ∑ 𝑎𝑛 converges to 𝑠. (9) Use Example 3.14 to obtain the identity ∞

1 𝜋4 = . 4 90 𝑛 𝑛=1 ∑

(10) Suppose 𝑓 is a Riemann-integrable function that has left and right limits at 𝜃0 , say lim 𝑓(𝜃) = 𝑓(𝜃0 −)

𝜃→𝜃0−

and

lim 𝑓(𝜃) = 𝑓(𝜃0 +),

𝜃→𝜃0+

then 𝑓(𝜃0 −) + 𝑓(𝜃0 +) , 2 as 𝑟 → 1, where 𝒜𝑟 𝑓(𝜃) are the Abel means of its Fourier series. 𝒜𝑟 𝑓(𝜃0 ) →

Exercises

69

(11) The following exercises prove inequality (3.35) for the inner product (3.33) and the partial sums (3.34). All functions are Riemannintegrable periodic functions with period 2𝜋. (a) For a function 𝑓, if 𝑠𝑁 is given by (3.34) and 𝑝 ∈ 𝒯𝑁 , then 𝑓 −𝑠𝑁 and 𝑝 are orthogonal, ⟨𝑓 − 𝑠𝑁 , 𝑝⟩ = 0. (b) (Pythagoras’s theorem) If 𝑓, 𝑔 are orthogonal, then ‖𝑓 + 𝑔‖2 = ‖𝑓‖2 + ‖𝑔‖2 . (c) Conclude (3.35) by noting that 𝑓 − 𝑝 = 𝑓 − 𝑠𝑁 + 𝑠𝑛 − 𝑝, and that 𝑓 − 𝑠𝑁 is orthogonal to 𝑠𝑁 − 𝑝. (12) Let ℛ(𝕊) be the space of Riemann-integrable functions on 𝕊, and let 𝑅 = ℛ(𝕊)/ ∼, where ∼ is the equivalence relation so that 𝑓 ∼ 𝑔 if, and only if, 𝑓(𝑥) = 𝑔(𝑥) for all 𝑥 except at a set of measure 0. (a) The subspace of continuous functions 𝐶(𝕊) identifies with itself in 𝑅. (b) The bilinear form ⟨𝑓, 𝑔⟩ =

1 ∫ 2𝜋 0

2𝜋

𝑓(𝜃)𝑔(𝜃)𝑑𝜃

induces an inner product on 𝑅. (c) The quadratic form ‖𝑓‖ = √⟨𝑓, 𝑓⟩ induces a norm in 𝑅. (13) We say that 𝑓 is Hölder continuous with exponent 𝛼, and write 𝑓 ∈ 𝐶 𝛼 (𝕊), for some 0 < 𝛼 ≤ 1, if there exists 𝑀 > 0 such that |𝑓(𝑥) − 𝑓(𝑦)| ≤ 𝑀|𝑥 − 𝑦|𝛼 . Prove Bernstein’s theorem: If 𝑓 ∈ 𝐶 𝛼 (𝕊) for some 𝛼 > 1/2, then ̂ < ∞. (Hint: Proceed as in the proof of Theorem 3.32). ∑ |𝑓(𝑛)| (14) If ∑ 𝑎𝑛 is Cesàro-summable to 𝑠, then is it Abel-summable to 𝑠. (15) For any 𝜏 ∈ ℝ and 𝑁 ≥ 1, 𝑁−1

𝑛

1 1 sin2 (𝑁𝜏/2) ∑ ∑ 𝑒𝑖𝑘𝜏 = . 𝑁 𝑛=0 𝑘=−𝑛 𝑁 sin2 (𝜏/2) (16) Suppose 𝑓 is a Riemann-integrable function that has left and right limits at 𝜃0 , say lim 𝑓(𝜃) = 𝑓(𝜃0 −)

𝜃→𝜃0−

and

lim 𝑓(𝜃) = 𝑓(𝜃0 +),

𝜃→𝜃0+

70

3. Fourier series then

𝑓(𝜃0 −) + 𝑓(𝜃0 +) , 2 where 𝜎𝑁 (𝜃) are the Cesàro sums of its Fourier series. 𝜎𝑁 (𝜃0 ) →

(17) The Dirichlet kernel 𝑁

𝐷𝑁 (𝜃) = ∑ 𝑒𝑖𝑛𝜃 𝑛=−𝑁

is given explicitly by sin(𝑁 + 1/2)𝜃 . sin 𝜃/2 (18) (a) For each 𝜃 ∈ 𝕊, there exists 𝑓 ∈ 𝐶(𝕊) with divergent Fourier series at 𝜃. (b) If 𝜃1 , 𝜃2 , . . . , 𝜃𝑘 ∈ 𝕊, there exists 𝑓 ∈ 𝐶(𝕊) with divergent Fourier series at 𝜃1 , 𝜃2 , . . . , 𝜃𝑘 . (c) If 𝐴 ⊂ 𝕊 is countable, there exists 𝑓 ∈ 𝐶(𝕊) with divergent Fourier series at each point in 𝐴. 𝐷𝑁 (𝜃) =

Notes The idea of solving Laplace’s equation, and the heat equation, by decomposing its solutions in trigonometric series was introduced by Fourier in [Fou55]. Parseval’s identity was stated by Marc-Antoine Parseval in [PdC06], and it was proven by Michel Plancherel in [Pla10]. The Riemann– Lebesgue’s lemma was first proven by Riemann for the case of Riemannintegrable functions, and then proven for general measurable functions by Lebesgue in [Leb03]. Theorem 3.45 was proven by Lipót Fejér in [Fej00]. An extensive discussion of summability methods can be found in [Zyg02]. The proof of Theorem 3.62 is taken from [SS03]. Bernstein’s theorem was proven in [Ber14]. The steps to prove it in Exercise (13) are also taken from [SS03], which are elaborated from its proof in [Zyg02]. The first to provide an explicit example of a continuous function with divergent Fourier series was Paul du Bois-Reymond in [du 76]. The construction discussed here follows [K8̈8]. Another construction can be found in [SS03].

Chapter 4

Poisson kernel in the half-space

4.1. The Poisson kernel in the half-space We now study harmonic functions in the upper half-space ℝ𝑑+1 = {(𝑥, 𝑡) ∶ 𝑥 ∈ ℝ𝑑 , 𝑡 > 0}. + As in the case of the ball in ℝ𝑑 , we will study explicit formulas for harmonic functions in ℝ𝑑+1 + , given their values at its boundary ℝ𝑑 × {0} = {(𝑥, 0) ∶ 𝑥 ∈ ℝ𝑑 } Note that, while the ball is a bounded set, and hence its boundary is compact, the upper half-space is unbounded, and so is its boundary ℝ𝑑 × {0}. This will force us to be more careful when defining objects analogous to the Poisson integrals studied before. The Poisson kernel for the upper half-space is given by the function (4.1)

𝑃𝑡 (𝑥) =

𝜔𝑑+1

(|𝑥|2

2𝑡 , + 𝑡2 )(𝑑+1)/2

𝑑

defined for 𝑥 ∈ ℝ and 𝑡 > 0. Compare the function 𝑃𝑡 (𝑥) with the Poisson kernel for the ball, given by 1 − |𝑥|2 𝑃(𝑥, 𝜉) = 𝜔𝑑 |𝑥 − 𝜉|𝑑 71

72

4. Poisson kernel in the half-space

for 𝑥 ∈ 𝔹 and 𝜉 ∈ 𝕊. Both functions have a multiple of the distance to the boundary (2𝑡 and (1 + |𝑥|)(1 − |𝑥|), respectively), and both have the distance to a boundary point powered to the dimension: in the case of the upper half-space, this is the distance between (𝑥, 𝑡) to the origin, so it is |(𝑥, 𝑡)|𝑑+1 , and in the case of the ball, the distance between 𝑥 and 𝜉, for each 𝜉 ∈ 𝕊, so it is |𝑥 − 𝜉|𝑑 . Example 4.2. For 𝑑 = 1, since 𝜔2 = 2𝜋, the circunference of the unit circle, the Poisson kernel for the upper half-plane is given by 𝑃𝑡 (𝑥) =

1 𝑡 . 2 𝜋 𝑥 + 𝑡2

𝑃𝑡 (𝑥) is an even function, and since we can write 𝑃𝑡 (𝑥) =

1 1 1 ⋅ , 𝑡 𝜋 (𝑥/𝑡)2 + 1

we see that it has a bump at the origin that increases as 𝑡 → 0, as seen in Figure 4.1.

Figure 4.1. The Poisson kernel for the upper half-plane. Note that it is an even function, with a bump at the origin.

4.3. The function (𝑥, 𝑡) ↦ 𝑃𝑡 (𝑥) is harmonic in ℝ𝑑+1 + . This can be verified explicitly by differentiating (Exercise (1)). 4.4. For each 𝑡 > 0, 𝑃𝑡 (𝑥) =

𝑥 1 𝑃1 ( ). 𝑑 𝑡 𝑡

4.1. The Poisson kernel in the half-space

73

Indeed, we have 2𝑡 2𝑡 = 𝜔𝑑+1 (|𝑥|2 + 𝑡2 )(𝑑+1)/2 𝜔𝑑+1 𝑡𝑑+1 (|𝑥/𝑡|2 + 1)(𝑑+1)/2 1 2 𝑥 1 = 𝑑 = 𝑑 𝑃1 ( ). (𝑑+1)/2 2 𝑡 𝑡 𝜔𝑑+1 (|𝑥/𝑡| + 1) 𝑡

𝑃𝑡 (𝑥) =

Thus, 𝑃𝑡 is the dilation of 𝑃1 , which motivates the notation of 𝑡 as a subindex parameter, rather than as another variable of 𝑃. For any 𝑡 > 0, ∫ 𝑃𝑡 (𝑥)𝑑𝑥 = 1.

(4.5)

ℝ𝑑

The integral in (4.5) must be understood in the improper sense, i.e. ∫ 𝑃𝑡 (𝑥)𝑑𝑥 = lim ∫ 𝑁→∞

ℝ𝑑

𝑃𝑡 (𝑥)𝑑𝑥.

|𝑥|≤𝑁

The fact the this limit exists for 𝑡 > 0 follows from the estimate 2𝑡 2𝑡 ≤ , 𝑃𝑡 (𝑥) = (𝑑+1)/2 2 2 𝜔𝑑+1 |𝑥|𝑑+1 𝜔𝑑+1 (|𝑥| + 𝑡 ) so, for 𝑁 ≥ 𝑀, using spherical coordinates we obtain |∫ |

𝑃𝑡 (𝑥)𝑑𝑥 − ∫

|𝑥|≤𝑁

|𝑥|≤𝑀

𝑃𝑡 (𝑥)𝑑𝑥|| ≤ ≤

𝑑𝑥 2𝑡 ∫ 𝜔𝑑+1 𝑀 0, ∫

𝑃𝑡 (𝑥)𝑑𝑥 → 0

|𝑥|≥𝛿

as 𝑡 → 0. Thus, the weight of the Poisson kernel gets concentrated at the origin as 𝑡 → 0. To verify this limit, observe first that, for any 𝑡 > 0, 𝑃𝑡 (𝑥) ≤

2𝑡 , 𝜔𝑑+1 |𝑥|𝑑+1

74

4. Poisson kernel in the half-space

and thus, using spherical coordinates as above ∞

∫

𝑃𝑡 (𝑥)𝑑𝑥 ≤ 𝑐𝑡 ∫

|𝑥|≥𝛿

|𝑥|≥𝛿

1 𝑡 𝑑𝑥 = 𝑐𝑡 ∫ ∫ 𝑑+1 𝑑𝜎𝑟𝑑−1 𝑑𝑟 = 𝑐′ , 𝛿 |𝑥|𝑑+1 𝑟 𝛿 𝕊

′

where 𝑐, 𝑐 are the same constants as above, and all integrals at infinity are improper integrals. Therefore, the integral clearly goes to 0 as 𝑡 → 0, for each 𝛿 > 0.

4.2. Poisson integrals in the half-space In this section we study the Poisson integrals in the half-space. Again, as in the previous section, we understand all integrals in ℝ𝑑 in the improper sense. For this, let ℛ(ℝ𝑑 ) be the space of bounded, locally Riemannintegrable functions 𝑓 on ℝ𝑑 (𝑓 is Riemann-integrable on any rectangle in ℝ𝑑 ) such that the sequence ∫

(4.7)

|𝑓(𝑥)|𝑑𝑥

|𝑥|≤𝑁

is bounded. For 𝑓 ∈ ℛ(ℝ𝑑 ), we define its improper integral as ∫ 𝑓(𝑥)𝑑𝑥 = lim ∫

(4.8)

𝑁→∞

ℝ𝑑

𝑓(𝑥)𝑑𝑥.

|𝑥|≤𝑁

To see that this limit exists, note that, since the sequence (4.7) is increasing and bounded, it converges, so for any 𝜀 > 0 there exists 𝐾 such that, if 𝑁 ≥ 𝑀 ≥ 𝐾, then ∫ 𝑀≤|𝑥|≤𝑁

|𝑓(𝑥)|𝑑𝑥 = || ∫

|𝑓(𝑥)|𝑑𝑥 − ∫

|𝑥|≤𝑀

|𝑥|≤𝑁

|𝑓(𝑥)|𝑑𝑥|| < 𝜀.

Thus, for 𝑁 ≥ 𝑀 ≥ 𝐾, |∫ |

𝑓(𝑥)𝑑𝑥 − ∫

|𝑥|≤𝑀

|𝑥|≤𝑁

𝑓(𝑥)𝑑𝑥|| ≤ ∫

|𝑓(𝑥)|𝑑𝑥 < 𝜀,

𝑀≤|𝑥|≤𝑁

and therefore the limit in (4.8) exists. Improper integrals on ℝ𝑑 are translation invariant, that is, (4.9)

∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥 − ℎ)𝑑𝑥, ℝ𝑑

ℝ𝑑

4.2. Poisson integrals in the half-space

75

for any 𝑓 ∈ ℛ(ℝ𝑑 ) and any ℎ ∈ ℝ𝑑 . They also satisfy, for any 𝑟 > 0, ∫ 𝑓(𝑥)𝑑𝑥 =

(4.10)

ℝ𝑑

𝑥 1 ∫ 𝑓( )𝑑𝑥. 𝑟𝑑 ℝ𝑑 𝑟

These two properties follow from the boundedness of the sequence (4.7), and are left as an exercise (Exercise (3)). For a bounded locally Riemann-integrable function 𝑓 in ℝ𝑑 , we define its Poisson integral to be the function 𝑢(𝑥, 𝑡) given by 𝑢(𝑥, 𝑡) = 𝒫𝑡 𝑓(𝑥) = ∫ 𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦,

(4.11)

ℝ𝑑

for each (𝑥, 𝑡) ∈ ℝ𝑑+1 + . The integral in (4.11) is well defined because 𝑓 is bounded and thus 𝑦 ↦ 𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦) is in ℛ(ℝ𝑑 ). Note that we can also write the Poisson integral as 𝑢(𝑥, 𝑡) = ∫ 𝑃𝑡 (𝑦)𝑓(𝑥 − 𝑦)𝑑𝑦, ℝ𝑑 1

using (4.9).

The Poisson integral 𝑢(𝑥, 𝑡) of a bounded function 𝑓 is a bounded continuous function. Indeed, if |𝑓(𝑥)| ≤ 𝑀, |𝑢(𝑥, 𝑡)| ≤ ∫ 𝑃𝑡 (𝑥 − 𝑦)|𝑓(𝑦)|𝑑𝑦 ≤ 𝑀 ∫ 𝑃𝑡 (𝑥 − 𝑦)𝑑𝑦 = 𝑀, ℝ𝑑

ℝ𝑑

by (4.5) and (4.9). The continuity follows from the continuity of 𝑃𝑡 (𝑥), its integrability, and the boundedness of 𝑓. Note that, for |𝑧| > 2|𝑥|, we have |𝑥 − 𝑧| > |𝑧|/2 and thus 𝑃𝑡 (𝑥 − 𝑧) ≤

2𝑡 2𝑑+2 𝑡 < . 𝜔𝑑+1 |𝑥 − 𝑧|𝑑+1 𝜔𝑑+1 |𝑧|𝑑+1

Hence, if 𝑁 > 2|𝑥|, 2𝑑+2 𝜔𝑑 𝑡𝑀 𝑃𝑡 (𝑥 − 𝑧)𝑓(𝑧)𝑑𝑧|| ≤ . 𝜔𝑑+1 𝑁 |𝑧|>𝑁

|∫ | 1

The operator 𝑓 ↦ 𝒫𝑡 𝑓 is also called the Poisson semigroup operator. It satisfies 𝒫𝑡 𝒫𝑠 𝑓 = 𝒫𝑡+𝑠 𝑓 𝑑

for any 𝑓 ∈ ℛ(ℝ ) and 𝑡, 𝑠 > 0. This follows from the identity ∫ 𝑃𝑡 (𝑥 − 𝑧)𝑃𝑠 (𝑧 − 𝑦)𝑑𝑧 = 𝑃𝑡+𝑠 (𝑥 − 𝑦), ℝ𝑑

which will be proved later using the Fourier transform (Chapter 10).

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4. Poisson kernel in the half-space

Given (𝑥, 𝑡) ∈ ℝ𝑑+1 and 𝜀 > 0, choose 𝑁 > 2(|𝑥| + 1) such that + 2𝑑+2 𝜔𝑑 (𝑡 + 1)𝑀 𝜀 < . 𝜔𝑑+1 𝑁 3 Now, choose 𝛿 > 0 such that 𝛿 < 1 and, if (|𝑥 − 𝑦|2 + (𝑡 − 𝑠)2 )1/2 < 𝛿, then 𝑑𝜀 |𝑃𝑡 (𝑥 − 𝑧) − 𝑃𝑠 (𝑦 − 𝑧)| < 3𝜔𝑑 𝑀𝑁 𝑑 for any |𝑧| ≤ 𝑁. Such 𝛿 exists since the closed ball of radius 𝑁 + 1 is compact and, hence, 𝑃𝑡 (𝑥 − 𝑧) is uniformly continuous. Therefore, using the fact that |𝑥 − 𝑦| < 1 and |𝑡 − 𝑠| < 1 if (|𝑥 − 𝑦|2 + (𝑡 − 𝑠)2 )1/2 < 𝛿, so that we have 𝑁 > 2|𝑦| and 𝑠 < 𝑡 + 1, | | | ∫ 𝑃𝑡 (𝑥 − 𝑧)𝑓(𝑧)𝑑𝑧 − ∫ 𝑃𝑠 (𝑦 − 𝑧)𝑓(𝑧)𝑑𝑧| ℝ𝑑

ℝ𝑑

≤ ∫ |𝑃𝑡 (𝑥 − 𝑧) − 𝑃𝑠 (𝑦 − 𝑧)||𝑓(𝑧)|𝑑𝑧 ℝ𝑑

2𝑑+2 𝜔𝑑 𝑡𝑀 2𝑑+2 𝜔𝑑 𝑠𝑀 𝑑𝜀 𝑀𝑑𝑧 + + 𝑑 𝜔𝑑+1 𝑁 𝜔𝑑+1 𝑁 |𝑧|≤𝑁 3𝜔 𝑑 𝑀𝑁 𝜀 𝜀 𝜀 < + + = 𝜀. 3 3 3 ≤∫

As above, we can also verify that 𝑢(𝑥, 𝑡) is differentiable, using the fact that 𝑃𝑡 (𝑥) is continuously differentiable and in ℛ(ℝ𝑑 ) for each 𝑡. Indeed, the partial derivatives of 𝑃𝑡 (𝑥) are given by 2(𝑑 + 1)𝑡𝑥𝑗 𝜕𝑃𝑡 =− 𝜕𝑥𝑗 𝜔𝑑+1 (|𝑥|2 + 𝑡2 )(𝑑+3)/2 for 𝑗 = 1, . . . , 𝑑, and 2(|𝑥|2 − 𝑑𝑡2 ) 𝜕𝑃𝑡 , = 𝜕𝑡 𝜔𝑑+1 (|𝑥|2 + 𝑡2 )(𝑑+3)/2 and thus, as above, for |𝑧| > 2|𝑥|, | 𝜕𝑃𝑡 (𝑥 − 𝑧)| ≤ 𝐶𝑡 | 𝜕𝑥 | |𝑧|𝑑+2 𝑗

and

| 𝜕𝑃𝑡 (𝑥 − 𝑧)| ≤ 𝐶 , | 𝜕𝑡 | |𝑧|𝑑+1

4.3. Boundary limits

77

for some constant 𝐶 that depends only on 𝑑. Moreover, note that for ℎ ∈ ℝ, 𝑢(𝑥 + ℎ𝑒𝑗 , 𝑡) − 𝑢(𝑥, 𝑡) 𝜕𝑃𝑡 −∫ (𝑥 − 𝑧)𝑓(𝑧)𝑑𝑧 ℎ 𝜕𝑥 𝑗 ℝ𝕕 =∫ ( ℝ𝑑

𝑃𝑡 (𝑥 + ℎ𝑒𝑗 − 𝑧) − 𝑃𝑡 (𝑥 − 𝑧) 𝜕𝑃 − 𝑡 (𝑥 − 𝑧))𝑓(𝑧)𝑑𝑧 ℎ 𝜕𝑥𝑗 =∫ ( ℝ𝑑

𝜕𝑃𝑡 𝜕𝑃 (𝑦 − 𝑧) − 𝑡 (𝑥 − 𝑧))𝑓(𝑧)𝑑𝑧, 𝜕𝑥𝑗 𝜕𝑥𝑗

𝑑

where 𝑦 ∈ ℝ is of the form 𝑦 = 𝑥 + 𝑠𝑒𝑗 where 𝑠 is a number between 0 and ℎ that depends on 𝑥, 𝑡 and 𝑧. Since 𝜕𝑃𝑡 𝜕𝑃 (𝑦 − 𝑧) → 𝑡 (𝑥 − 𝑧) 𝜕𝑥𝑗 𝜕𝑥𝑗 as ℎ → 0 for each 𝑧, we can proceed as above and conclude that 𝜕𝑃𝑡 𝜕𝑢 =∫ (𝑥 − 𝑧)𝑓(𝑧)𝑑𝑧, 𝜕𝑥𝑗 𝜕𝑥 𝑗 ℝ𝕕 so we can “differentiate under the integral”. Similarly for the partial derivative of 𝑢 with respect to 𝑡. Hence, we have the following result. Proposition 4.12. If 𝑓 is a bounded locally Riemann-integrable function in ℝ𝑑 , its Poisson integral 𝑢(𝑥, 𝑡) is harmonic in ℝ𝑑+1 + . Proof. Similarly, as in the above cases, we can differentiate inside the Poisson integral to verify that 𝑑

𝜕2 𝑃𝑡 𝜕2 𝑃𝑡 (𝑥 − 𝑧) + (𝑥 − 𝑧))𝑓(𝑧)𝑑𝑧. 2 𝜕𝑡2 𝑗=1 𝜕𝑥𝑗

Δ𝑢(𝑥, 𝑡) = ∫ ( ∑ ℝ𝕕

The theorem then follows from the fact that (𝑥, 𝑡) ↦ 𝑃𝑡 (𝑥) is harmonic. We leave the details as an exercise (Exercise (4)). □

4.3. Boundary limits We now study the behavior of a Poisson integrals 𝑢(𝑥, 𝑡) as the point (𝑥, 𝑡) approaches a boundary point. We start with Theorem 4.13, analogous to Theorem 2.28.

78

4. Poisson kernel in the half-space

Theorem 4.13. Let 𝑓 be a bounded locally Riemann-integrable function in ℝ𝑑 , continuous at 𝑥0 ∈ ℝ𝑑 , and let 𝑢(𝑥, 𝑡) be its Poisson integral. Then lim

(𝑥,𝑡)→(𝑥0 ,0)

𝑢(𝑥, 𝑡) = 𝑓(𝑥0 ).

In particular, if 𝑓 is continuous at every point of ℝ𝑑 , its Poisson integral 𝑢(𝑥, 𝑡) in ℝ𝑑+1 extends continuously to the boundary ℝ𝑑 × {0}, with + the value 𝑓(𝑥) at each (𝑥, 0). Unsurprinsingly, the proof is very similar to the proofs of Theorems 2.28 and 3.45. Proof. Given 𝜀 > 0, since 𝑓 is continuous at 𝑥0 there exists 𝜂 > 0 such that, if |𝑥 − 𝑥0 | < 𝜂, then |𝑓(𝑥) − 𝑓(𝑥0 )| < 𝜀/2. Now, write 𝑢(𝑥, 𝑡) − 𝑓(𝑥0 ) = ∫ 𝑃𝑡 (𝑧)𝑓(𝑥 − 𝑧)𝑑𝑧 − 𝑓(𝑥0 ) ℝ𝑑

= ∫ 𝑃𝑡 (𝑧)(𝑓(𝑥 − 𝑧) − 𝑓(𝑥0 ))𝑑𝑧, ℝ𝑑

where we have used (4.5). Now, if |𝑥 − 𝑥0 | < 𝜂/2 and |𝑧| < 𝜂/2, |𝑥 − 𝑧 − 𝑥0 | < 𝜂 and thus |∫ |

|𝑧| 0, ∫

|Φ𝑡 (𝑥)|𝑑𝑥 → 0 as 𝑡 → 0.

|𝑥|≥𝛿

These operators are called convolution operators, and are denoted by Φ𝑡 ∗ 𝑓. Note that the Poisson integral of 𝑓 is the convolution 𝑃𝑡 ∗ 𝑓. The collection of functions {Φ𝑡 }𝑡>0 form a family of good kernels. See Exercise (9).

80

4. Poisson kernel in the half-space

Exercises (1) The function (𝑥, 𝑡) ↦ 𝑃𝑡 (𝑥) is harmonic in ℝ𝑑+1 + . (2) For any dimension 𝑑 ≥ 1, ∫ ℝ𝑑

(|𝑥|2

𝑑𝑥 𝜋(𝑑+1)/2 = , (𝑑+1)/2 Γ((𝑑 + 1)/2) + 1)

and verify (4.5). (Hint: Use spherical coordinates and the identity ∞

∫ 𝑡𝛼 𝑒−𝑡𝑠 0

Γ(𝛼) 𝑑𝑡 = 𝛼 𝑡 𝑠

for any 𝛼, 𝑠 > 0.) (3) Let 𝑓 ∈ ℛ(ℝ𝑑 ). (a) For any ℎ ∈ ℝ𝑑 , ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥 − ℎ)𝑑𝑥. ℝ𝑑

ℝ𝑑

(b) For any 𝑟 > 0, ∫ 𝑓(𝑥)𝑑𝑥 = ℝ𝑑

1 𝑥 ∫ 𝑓( )𝑑𝑥. 𝑟𝑑 ℝ𝑑 𝑟

(Hint: The integrals ∫|𝑥|≤𝑁 𝑓(𝑥)𝑑𝑥 form a Cauchy sequence.) (4) Complete the details of the proof of Proposition 4.12. (5) If 𝑓 ∈ 𝐶0 (ℝ𝑑 ), then 𝑢(𝑥, 𝑡) → 𝑓(𝑥) as 𝑡 → 0, uniformly in 𝑥 ∈ ℝ𝑑 . (6) Let 𝑓 ∈ 𝐶𝑐 (ℝ𝑑 ). Then ∫ |𝑓(𝑥 − ℎ) − 𝑓(𝑥)|𝑑𝑥 → 0 ℝ𝑑 𝑑

as ℎ → 0 in ℝ . (Hint: 𝑓 is uniformly continuous on its compact support.) (7) Prove the following version of Fubini’s theorem: Let 𝑓(𝑥, 𝑦) ∈ 𝐶(ℝ𝑑+𝑑 ) such that (a) there exists 𝐴 > 0 such that, for all 𝑦 ∈ ℝ𝑑 , |𝑓(𝑥, 𝑦)| ≤

𝐴 ; (|𝑥|𝑑+1 + 1)

Exercises

81

(b) there exists a compact 𝐾 ⊂ ℝ𝑑 such that, for all 𝑥 ∈ ℝ𝑑 , 𝑓(𝑥, ⋅) is supported in 𝐾. Then 𝑓(𝑥, 𝑦) is integrable in ℝ2𝑑 and ∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∫ ( ∫ 𝑓(𝑥, 𝑦)𝑑𝑥)𝑑𝑦 = ∫ ( ∫ 𝑓(𝑥, 𝑦)𝑑𝑦)𝑑𝑥. ℝ2𝑑

ℝ𝑑

ℝ𝑑

ℝ𝑑

ℝ𝑑

(8) If 𝑓 ∈ 𝐶𝑐 (ℝ𝑑 ), ∫ |𝑢(𝑥, 𝑡) − 𝑓(𝑥)|𝑑𝑥 → 0 ℝ𝑑

as 𝑡 → 0. (Hint: Use Exercises (6) and (7).) (9) Let {𝐾𝑡 }𝑡>0 be a family of functions in ℛ(ℝ𝑑 ). We say that it is a family of good kernels if • ∫ 𝐾𝑡 (𝑥)𝑑𝑥 = 1 for all 𝑡 > 0; ℝ𝑑

• there exists 𝑀 > 0 such that ∫ |𝐾𝑡 (𝑥)|𝑑𝑥 ≤ 𝑀 for all 𝑡 > 0; ℝ𝑑

and • for 𝛿 > 0, ∫

|𝐾𝑡 (𝑥)|𝑑𝑥 → 0 as 𝑡 → 0.

|𝑥|≥𝛿 𝑑

(a) If Φ ∈ ℛ(ℝ ) and ∫ Φ = 1, then its dilations {Φ𝑡 }𝑡>0 form a family of good kernels. (b) If {𝐾𝑡 }𝑡>0 is a family of good kernels and 𝑓 ∈ 𝐶(ℝ𝑑 ) is bounded, then, for each 𝑥 ∈ ℝ𝑑 , lim

(𝑦,𝑡)→(𝑥,0)

𝐾𝑡 ∗ 𝑓(𝑦) = 𝑓(𝑥).

(c) If {𝐾𝑡 }𝑡>0 is a family of good kernels and 𝑓 ∈ 𝐶0 (ℝ𝑑 ), then 𝐾𝑡 ∗ 𝑓 ⇉ 𝑓 as 𝑡 → 0. (10) (Principle of subordination) Let 𝐻𝑡 (𝑥) be the heat kernel, 𝐻𝑡 (𝑥) =

1 2 𝑒−|𝑥| /4𝑡 . (4𝜋𝑡)𝑑/2

Then 𝑃𝑡 (𝑥) =

𝑡 2√𝜋

∞ 2 /4𝑠

∫ 𝑒−𝑡 0

𝐻𝑠 (𝑥)

𝑑𝑠 . 𝑠3/2

(Hint: use the identity of the gamma function as in Exercise (2).)

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4. Poisson kernel in the half-space

Notes The results of this chapter are classical, and can be found in the references cited previously, as [ABR01]. An extended treatment on Poisson integrals can be found in [Ste70].

Chapter 5

Measure theory in Euclidean space

5.1. The need for an integration theory We have seen in Chapter 4 that, when dealing with Poisson integrals in the upper half-space, that improper integrals add a further difficulty when dealing with boundary limits. However, the main problem with Riemann integration is the fact that the Riemann integral is not compatible with pointwise limits. Indeed, consider the set 𝑄 = [0, 1] ∩ ℚ of rational numbers in the interval [0, 1]. This set is countable, so we can write 𝑄 = {𝑞𝑛 ∶ 𝑛 ∈ ℕ}. Now, for each 𝑛, let 𝑓𝑛 be the function on [0, 1] given by 1 𝑥 = 𝑞𝑗 for some 𝑗 ≤ 𝑛 𝑓𝑛 (𝑥) = { 0 otherwise. Then, we see that 𝑓𝑛 (𝑥) → 𝑓(𝑥) for every 𝑥 ∈ [0, 1], where 𝑓 is the function 1 𝑓(𝑥) = { 0

𝑥∈𝑄 𝑥 ∉ 𝑄.

However, while each 𝑓𝑛 is Riemann-integrable on [0, 1], 𝑓 is not. Given any partition 𝒫 = {𝑥0 = 0 < 𝑥1 < 𝑥2 < . . . < 𝑡𝑛 = 1} 83

84

5. Measure theory in Euclidean space

of [0, 1], we see that 𝑀𝑗 = sup{𝑓(𝑥) ∶ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} = 1 for all 𝑗 = 1, . . . , 𝑛, because every interval in ℝ contains a rational number, so the upper Riemann sum of 𝑓 is 𝑛

𝑛

𝑈(𝑓, 𝒫) = ∑ 𝑀𝑗 (𝑥𝑗 − 𝑥𝑗−1 ) = ∑ (𝑥𝑗 − 𝑥𝑗−1 ) = 1. 𝑗=1

𝑗=1

On the other hand, 𝑚𝑗 = inf{𝑓(𝑥) ∶ 𝑥 ∈ [𝑥𝑗−1 , 𝑥𝑗 ]} = 0 for all 𝑗 = 1, . . . , 𝑛, because every interval in ℝ also contains an irrational number, so the lower Riemann sum is 𝑛

𝐿(𝑓, 𝒫) = ∑ 𝑚𝑗 (𝑥𝑗 − 𝑥𝑗−1 ) = 0. 𝑗=1

As it is impossible to find a partition to get 𝑈(𝑓, 𝒫) − 𝐿(𝑓, 𝒫) < 1, we conclude 𝑓 is not Riemann integrable. We now ask if it is possible to extend the definition of the integral to satisfy the following requirements: (1) every Riemann-integrable function is also integrable in the new definition; (2) if each 𝑓𝑛 is integrable and 𝑓𝑛 → 𝑓 pointwise, then 𝑓 is also integrable; (3) under appropiate conditions, if 𝑓𝑛 → 𝑓 pointwise, then we have ∫ 𝑓𝑛 → ∫ 𝑓 in the extended definition. Property (1) is necessary because we want to extend the definition of the integral, so that the results on integration we already have must still hold in the new definition. The “appropiate conditions” above should not be too restrictive, of course. For example, the Riemann integral is consistent with uniform convergence, which is an hypothesis too strong to be useful in most situations. The purpose of this chapter is to give an introduction to measure and integration theory. Lebesgue’s theory provides an integral that satisfies the above conditions, and will allow us to both extend the results of Chapter 4 and deal with further related problems.

5.2. Outer measure in Euclidean space

85

5.2. Outer measure in Euclidean space The first ingredient in Lebesgue’s theory is the idea of measure of a set. Recall that if 𝑅 is a rectangle in ℝ𝑑 , say 𝑅 = 𝐼 1 × 𝐼2 × ⋯ × 𝐼 𝑑 , where each 𝐼𝑗 is a bounded interval, then its volume is given by vol(𝑅) = |𝐼1 | ⋯ |𝐼𝑑 |, the product of the lengths of the intervals 𝐼1 , . . . , 𝐼𝑑 . A cube 𝑄 is a rectangle with all intervals 𝐼𝑗 above of the same length, say 𝐿, and in that case vol(𝑄) = 𝐿𝑑 . 𝑄 is a dyadic cube if 𝐿 = 2𝑚 for some 𝑚 ∈ ℤ, and the limits of each 𝐼𝑗 are of the form 𝑘 ⋅ 2𝑚 for some 𝑘 ∈ ℤ. The intervals that define the rectangle 𝑅 may be open, closed, or neither. Note that 𝑅 is an open set if all the intervals 𝐼𝑗 are open, and 𝑅 is closed if all of them are closed. For any rectangle 𝑅, its interior 𝑅0 and its closure 𝑅̄ satisfy vol(𝑅0 ) = vol(𝑅)̄ = vol(𝑅). For any 𝜀 > 0, we can find an open rectangle 𝑆 ⊃ 𝑅 such that vol(𝑆) < vol(𝑅) + 𝜀, by widening (if necessary) each interval 𝐼𝑗 , and a closed rectangle 𝑇 ⊂ 𝑅 such that vol(𝑇) > vol(𝑅) − 𝜀, by shrinking (if necessary) each interval 𝐼𝑗 (Exercise (1)). In order to extend the concept of volume to general sets 𝐴 ⊂ ℝ𝑑 , we first define the outer measure of 𝐴 by |𝐴|∗ = inf { ∑ vol(𝑅𝑗 ) ∶ each 𝑅𝑗 is a rectangle and 𝐴 ⊂ 𝑗≥1

⋃

𝑅𝑗 }.

𝑗≥1

Thus, |𝐴|∗ is the infimum of the set of all possible sums of volumes of rectangles covering 𝐴, where each cover may have a finite or a countable infinite number of rectangles. If no such sum is finite, then we say |𝐴|∗ = ∞. Thus, for each 𝜀 > 0, there exists a cover of rectangles 𝑅1 , 𝑅2 , . . . for 𝐴 such that ∑ vol(𝑅𝑗 ) < |𝐴|∗ + 𝜀, 𝑗≥1

86

5. Measure theory in Euclidean space

and, by the observations above, we can choose all of these rectangles to be open, or all of them to be closed.1 The outer measure is also usually called exterior measure. 5.1. If 𝐴 = ∅, then clearly |𝐴|∗ = 0. Also if 𝐴 contains only one point, or if 𝐴 is finite. In fact, if 𝐴 is countable, |𝐴|∗ = 0 (Exercise (2)). As seen in Appendix A, if |𝐴|∗ = 0, we say that 𝐴 has measure zero, or that it’s a measure zero set. One can show that the countable union of measure zero sets is also of measure zero (Exercise (2)). The Cantor ternary set is also a set of measure zero (Exercise (3)). 5.2. If 𝑅 is a closed rectangle, then |𝑅|∗ = vol(𝑅). That is, its outer measure coincides with its volume. Indeed, first note that |𝑅|∗ ≤ vol(𝑅), because 𝑅 is a cover for itself. For the reverse inequality, suppose 𝜀 > 0 is given and 𝑅1 , 𝑅2 , . . . are open rectangles that cover 𝑅 such that ∑ vol(𝑅𝑗 ) < |𝑅|∗ + 𝜀. 𝑗≥1

Since 𝑅 is closed, it is compact, and thus we can assume we have a finite number of the 𝑅𝑗 , say, we have 𝑁 of them. Now, the edges of these rectangles can be extended to form a grid of subrectangles 𝑆 𝑖 of 𝑅, as in

Figure 5.1. Given a cover of rectangles for a rectangle, the edges can be extended to form a grid of subrectangles.

Figure 5.1, each one of them also a subrectangle of the 𝑅𝑗 , such that 𝑅=

⋃

𝑆𝑖

𝑖

1

We use the convention that 𝑥 < ∞ for all 𝑥 ∈ ℝ, and later on we will also use ∞ + ∞ = ∞.

5.2. Outer measure in Euclidean space

87

and the 𝑆 𝑖 are almost disjoint, that is, 𝑆 𝑖 ∩ 𝑆 𝑘 is either empty or just part of their boundary. Therefore 𝑁

vol(𝑅) = ∑ vol(𝑆 𝑖 ) ≤ ∑ vol(𝑅𝑗 ) < |𝑅|∗ + 𝜀 𝑖

𝑗=1

(Exercise (4)) and, since 𝜀 > 0 is arbitrary, vol(𝑅) ≤ |𝑅|∗ . 5.3. If 𝑅 is any rectangle, then we also have that |𝑅|∗ = vol(𝑅). Indeed, |𝑅|∗ ≤ vol(𝑅) because again 𝑅 forms a cover for itself. Also, for any closed rectangle 𝑆 ⊂ 𝑅, |𝑆|∗ ≤ |𝑅|∗ , because any cover of rectangles for 𝑅 also a cover for 𝑆. Now, |𝑆|∗ = vol(𝑆) and, for any 𝜀 > 0, we can choose 𝑆 such that vol(𝑆) > vol(𝑅) − 𝜀. Thus vol(𝑅) − 𝜀 < |𝑅|∗ , so we again have vol(𝑅) ≤ |𝑅|∗ because we can make 𝜀 arbitrarily small. 5.4. The fact for rectangles used above is true for general subsets of ℝ𝑑 : if 𝐴 ⊂ 𝐵, then |𝐴|∗ ≤ |𝐵|∗ , as any cover of rectangles for 𝐵 is also a cover for 𝐴. This property is called monotonicity. 5.5. If 𝐴 = ⋃𝑗≥1 𝐴𝑗 , then |𝐴|∗ ≤ ∑ |𝐴𝑗 |∗ . 𝑗≥1 𝑗

Indeed, let 𝜀 > 0 be given and, for each 𝐴𝑗 , let {𝑅𝑖 } be a cover of rectangles for 𝐴𝑗 such that 𝜀 𝑗 ∑ vol(𝑅𝑖 ) < |𝐴𝑗 |∗ + 𝑗 . 2 𝑖≥1 𝑗

Thus ⋃𝑖,𝑗≥1 {𝑅𝑖 } is a cover of rectangles for 𝐴, and 𝑗

|𝐴|∗ ≤ ∑ vol(𝑅𝑖 ) < ∑ (|𝐴𝑗 |∗ + 𝑖,𝑗≥1

𝑗≥1

𝜀 ) = ∑ |𝐴𝑗 |∗ + 𝜀. 2𝑗 𝑗≥1

We obtain the result because 𝜀 > 0 is arbitrary. Property 5.5 is called countable subadditivity. A natural question to ask is whether we have equality in 5.5 when the 𝐴𝑗 are disjoint. In general we don’t,2 but we have the following special cases. 2

This is beyond the goal of this text, but you can find an example in [Fol99, Section 1.1].

88

5. Measure theory in Euclidean space

5.6. If 𝐴, 𝐵 ⊂ ℝ𝑑 and dist(𝐴, 𝐵) > 0, then |𝐴 ∪ 𝐵|∗ = |𝐴|∗ + |𝐵|∗ . Note that 𝐴 and 𝐵 are not only disjoint, but separated by a positive distance from each other, that is, there is some 𝛿 > 0 such that, for any 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵, then |𝑥 − 𝑦| > 𝛿. By 5.5 we have |𝐴 ∪ 𝐵|∗ ≤ |𝐴|∗ + |𝐵|∗ , so we have to prove the reverse inequality. Let 𝜀 > 0 be given and {𝑅𝑗 } be a cover of rectangles for 𝐴 ∪ 𝐵 such that ∑ vol(𝑅𝑗 ) < |𝐴 ∪ 𝐵|∗ + 𝜀 𝑗

and, by subdividing them if necessary, the diameter of each 𝑅𝑗 is smaller than 𝛿. Thus, if 𝑅𝑗 ∩ 𝐴 ≠ ∅, then 𝑅𝑗 ∩ 𝐵 = ∅, and vice versa. Hence, if 𝔍𝐴 = {𝑗 ∶ 𝑅𝑗 ∩ 𝐴 ≠ ∅} and 𝔍𝐵 = {𝑗 ∶ 𝑅𝑗 ∩ 𝐵 ≠ ∅}, then 𝔍𝐴 ∩ 𝔍𝐵 = ∅ and each of {𝑅𝑗 }𝑗∈𝔍𝐴 and {𝑅𝑗 }𝑗∈𝔍𝐵 is a cover for 𝐴 and 𝐵, respectively. Thus |𝐴|∗ + |𝐵|∗ ≤ ∑ vol(𝑅𝑗 ) + ∑ vol(𝑅𝑗 ) ≤ ∑ vol(𝑅𝑗 ) < |𝐴 ∪ 𝐵|∗ + 𝜀, 𝑗∈𝔍𝐴

𝑗∈𝔍𝐵

𝑗

and the result follows because 𝜀 > 0 is arbitrary. Note that, inductively, we can extend 5.6 for any finite number of sets at positive distance. 5.7. If 𝐴 = ⋃𝑗 𝑄𝑗 where the 𝑄𝑗 are almost disjoint cubes, then |𝐴|∗ = ∑ |𝑄|∗ = ∑ vol(𝑄). 𝑗

𝑗

To verify this, let 𝜀 > 0 be given and consider strictly thinner cubes 𝑄̃𝑗 ⊂ 𝑄𝑗 such that 𝜀 vol(𝑄̃𝑗 ) > vol(𝑄𝑗 ) − 𝑗 . 2 With strictly thinner we mean that the closure of each 𝑄̃𝑗 is contained in the interior of 𝑄𝑗 . Thus, for any 𝑖 ≠ 𝑗, dist(𝑄̃ 𝑖 , 𝑄̃𝑗 ) > 0 and, by 5.6, for each 𝑁 we have 𝑁

𝑁

𝑁

𝑁

𝜀 | | | ⋃ 𝑄̃𝑗 |∗ = ∑ |𝑄̃𝑗 |∗ > ∑ (|𝑄𝑗 |∗ − 2𝑗 ) > ∑ |𝑄𝑗 |∗ − 𝜀. 𝑗=1 𝑗=1 𝑗=1 𝑗=1

5.3. Measurable sets and measure

89

𝑁

Since 𝐴 ⊃ ⋃𝑗=1 𝑄̃𝑗 , we have, by 5.4, 𝑁

|𝐴|∗ > ∑ |𝑄𝑗 |∗ − 𝜀. 𝑗=1

As 𝑁 is arbitrary, we have |𝐴|∗ ≥ ∑ |𝑄𝑗 |∗ − 𝜀, 𝑗

and we obtain the result because 𝜀 > 0 is also arbitrary. Fact 5.7 is particularly useful when 𝐴 is an open set, since every nonempty open set in ℝ𝑑 is an almost disjoint union of closed cubes which can even be chosen to be dyadic (Exercise (5)). As we have mentioned above, we don’t have equality in 5.5, even if we have a finite number of sets. Thus, we need to restrict ourselves to those sets on which the equality would be true.

5.3. Measurable sets and measure We say that 𝐴 ⊂ ℝ𝑑 is measurable if, for each 𝐵 ⊂ ℝ𝑑 , |𝐵|∗ = |𝐵 ∩ 𝐴|∗ + |𝐵 ⧵ 𝐴|∗ .

(5.8)

Note that 𝐵 ∩ 𝐴 and 𝐵 ⧵ 𝐴 are disjoint and 𝐵 = (𝐵 ∩ 𝐴) ∪ (𝐵 ⧵ 𝐴), so 𝐴 is measurable when it splits any other set 𝐵 in parts whose outer measures add to the outer measure of 𝐵. Recall that we always have, by 5.5, the inequality |𝐵|∗ ≤ |𝐵 ∩ 𝐴|∗ + |𝐵 ⧵ 𝐴|∗ , so 𝐴 is measurable if the reverse inequality |𝐵|∗ ≥ |𝐵 ∩ 𝐴|∗ + |𝐵 ⧵ 𝐴|∗ is true for all 𝐵 ⊂ ℝ𝑑 . 5.9. ∅ and ℝ𝑑 are measurable. This follows because 𝐵∩∅=∅

and

𝐵 ⧵ ∅ = 𝐵,

𝐵 ∩ ℝ𝑑 = 𝐵

and

𝐵 ⧵ ℝ𝑑 = ∅.

as well as 5.10. If 𝐴 is measurable, then its complement ℝ𝑑 ⧵𝐴 is measurable. This also follows immediately because 𝐵 ∩ (ℝ𝑑 ⧵ 𝐴) = 𝐵 ⧵ 𝐴

and

𝐵 ⧵ (ℝ𝑑 ⧵ 𝐴) = 𝐵 ∩ 𝐴.

90

5. Measure theory in Euclidean space

5.11. If 𝐴 and 𝐵 are measurable, then 𝐴 ∪ 𝐵 is measurable. To prove this note that, for any 𝐶 ⊂ ℝ𝑑 , |𝐶|∗ = |𝐶 ∩ 𝐴|∗ + |𝐶 ⧵ 𝐴|∗ = |𝐶 ∩ 𝐴 ∩ 𝐵|∗ + |(𝐶 ∩ 𝐴) ⧵ 𝐵|∗ + |(𝐶 ⧵ 𝐴) ∩ 𝐵|∗ + |(𝐶 ⧵ 𝐴) ⧵ 𝐵|∗ , using (5.8) for each of 𝐴 and 𝐵, which are both measurable. Now 𝐴 ∪ 𝐵 = (𝐴 ∩ 𝐵) ∪ (𝐴 ⧵ 𝐵) ∪ (𝐵 ⧵ 𝐴), 𝐶 ∩ (𝐴 ⧵ 𝐵) = (𝐶 ∩ 𝐴) ⧵ 𝐵,

and

𝐶 ∩ (𝐵 ⧵ 𝐴) = (𝐶 ⧵ 𝐴) ∩ 𝐵,

so we have 𝐶 ∩ (𝐴 ∪ 𝐵) = (𝐶 ∩ 𝐴 ∩ 𝐵) ∪ (𝐶 ∩ (𝐴 ⧵ 𝐵)) ∪ (𝐶 ∩ (𝐵 ⧵ 𝐴)) = (𝐶 ∩ 𝐴 ∩ 𝐵) ∪ ((𝐶 ∩ 𝐴) ⧵ 𝐵) ∪ ((𝐶 ⧵ 𝐴) ∩ 𝐵), and thus |𝐶 ∩ 𝐴 ∩ 𝐵|∗ + |(𝐶 ∩ 𝐴) ⧵ 𝐵|∗ + |(𝐶 ⧵ 𝐴) ∩ 𝐵|∗ ≥ |𝐶 ∩ (𝐴 ∪ 𝐵)|∗ . Also (𝐶 ⧵ 𝐴) ⧵ 𝐵 = 𝐶 ⧵ (𝐴 ∪ 𝐵), and hence |𝐶|∗ ≥ |𝐶 ∩ (𝐴 ∪ 𝐵)|∗ + |𝐶 ⧵ (𝐴 ∪ 𝐵)|∗ , so 𝐴 ∪ 𝐵 is measurable. You can visualize the decomposition of 𝐶 used in the proof above with the Venn diagram shown in Figure 5.2. Inductively, for any 𝑁, the union of 𝑁 measurable sets 𝐴1 , . . . , 𝐴𝑁 is measurable. 5.12. If 𝐴1 , 𝐴2 , . . . are measurable, then their union 𝐴 = ⋃𝑗 𝐴𝑗 is also measurable. To prove this, let 𝐵1 = 𝐴1 and, for each 𝑗 ≥ 2, define 𝐵𝑗 = 𝑗−1

𝑁

𝑁

𝐴𝑗 ⧵ ⋃𝑖=1 𝐴𝑖 . Thus the 𝐵𝑗 are disjoint, ⋃𝑗=1 𝐵𝑗 = ⋃𝑗=1 𝐴𝑗 for all 𝑁 and 𝑁

⋃𝑗 𝐵𝑗 = 𝐴. Also, each ⋃𝑗=1 𝐵𝑗 is measurable. Now, for 𝐶 ⊂ ℝ𝑑 and any 𝑁, 𝑁

|𝐶 ∩

⋃ 𝑗=1

𝑁

𝐵𝑗 |∗ = |(𝐶 ∩

⋃

𝑁

𝐵𝑗 ) ∩ 𝐵𝑁 |∗ + |(𝐶 ∩

𝑗=1

𝑗=1

𝑁−1

= |𝐶 ∩ 𝐵𝑁 |∗ + |𝐶 ∩

⋃

⋃ 𝑗=1

𝐵𝑗 |∗ ,

𝐵𝑗 ) ⧵ 𝐵𝑁 |∗

5.3. Measurable sets and measure

91

Figure 5.2. Venn diagram of the decomposition of 𝐶 ∩ (𝐴 ∪ 𝐵) as (𝐶 ∩ 𝐴 ∩ 𝐵) ∪ ((𝐶 ∩ 𝐴) ⧵ 𝐵) ∪ ((𝐶 ⧵ 𝐴) ∩ 𝐵).

and inductively 𝑁

|𝐶 ∩

⋃

𝑁

𝐵𝑗 |∗ = ∑ |𝐶 ∩ 𝐵𝑗 |∗ . 𝑗=1

𝑗=1

Thus 𝑁

𝑁

|𝐶|∗ = |𝐶 ∩

⋃ 𝑗=1

𝐵𝑗 |∗ + |𝐶 ⧵

⋃

𝑁

𝐵𝑗 |∗ = ∑ |𝐶 ∩ 𝐵𝑗 |∗ + |𝐶 ⧵

𝑗=1

𝑗=1

𝑁

⋃

𝐵𝑗 |∗

𝑗=1

𝑁

≥ ∑ |𝐶 ∩ 𝐵𝑗 |∗ + |𝐶 ⧵ 𝐴|∗ , 𝑗=1 𝑁

because ⋃𝑗=1 𝐵𝑗 ⊂ 𝐴. As 𝑁 is arbitrary, we obtain |𝐶|∗ ≥ ∑ |𝐶 ∩ 𝐵𝑗 |∗ + |𝐶 ⧵ 𝐴|∗ ≥ || (𝐶 ∩ 𝐵𝑗 )|| + |𝐶 ⧵ 𝐴|∗ ⋃ ∗ 𝑗≥1

= |𝐶 ∩ 𝐴| + |𝐶 ⧵ 𝐴|∗ , and we conclude 𝐴 is measurable.

𝑗≥1

92

5. Measure theory in Euclidean space

Note that we can conclude, by 5.10 and 5.12, that the intersection of a countable number of measurable sets is measurable, as we can write ⋂

𝐴𝑗 = ℝ𝑑 ⧵

𝑗

⋃

(ℝ𝑑 ⧵ 𝐴𝑗 ).

𝑗

5.13. A closed cube 𝑄 is measurable. Let 𝐴 ⊂ ℝ𝑑 , and we want to prove |𝐴|∗ ≥ |𝐴 ∩ 𝑄|∗ + |𝐴 ⧵ 𝑄|∗ . This inequality is obvious if |𝐴|∗ = ∞, so we assume |𝐴|∗ < ∞. For each 𝑗, define 𝑄𝑗 = {𝑥 ∈ ℝ𝑑 ∶ dist(𝑥, 𝑄)
0 be given. (1) There exists an open set 𝑈 ⊃ 𝐴 such that |𝑈 ⧵ 𝐴| < 𝜀. (2) There exists a closed set 𝐸 ⊂ 𝐴 such that |𝐴 ⧵ 𝐸| < 𝜀. (3) If |𝐴| < ∞, there exists a compact set 𝐾 ⊂ 𝐴 such that |𝐴⧵𝐾| < 𝜀. (4) If |𝐴| < ∞, there exist finitely many closed cubes 𝑄1 , 𝑄2 , . . ., 𝑄𝑁 𝑁 such that, if 𝐹 = ⋃𝑗=1 𝑄𝑗 , then |𝐴△𝐹| < 𝜀. The symbol △ above denotes the symmetric difference of the sets, 𝐴△𝐹 = (𝐴 ⧵ 𝐹) ∪ (𝐹 ⧵ 𝐴). 5.18. Moreover, statement (1) of Corollary 5.17 is equivalent to the statement that 𝐴 is measurable. Indeed, suppose 𝐴 ⊂ ℝ𝑑 such that, for any 𝜀 > 0, there exists an open 𝑈 ⊃ 𝐴 such that |𝑈 ⧵ 𝐴|∗ < 𝜀.

96

5. Measure theory in Euclidean space

Let 𝐵 ⊂ ℝ𝑑 . Given 𝜀 > 0, choose 𝑈 ⊃ 𝐴 such that |𝑈 ⧵ 𝐴|∗ < 𝜀. Now 𝐵 ∩ 𝐴 ⊂ 𝐵 ∩ 𝑈 and 𝐵 ⧵ 𝐴 ⊂ (𝐵 ⧵ 𝑈) ∪ (𝑈 ⧵ 𝐴), so |𝐵 ∩ 𝐴|∗ + |𝐵 ⧵ 𝐴|∗ ≤ |𝐵 ∩ 𝑈|∗ + |𝐵 ⧵ 𝑈|∗ + |𝑈 ⧵ 𝐴|∗ < |𝐵|∗ + 𝜀, because 𝑈 is measurable. Since 𝜀 > 0 is arbitrary, |𝐵 ∩ 𝐴|∗ + |𝐵 ⧵ 𝐴|∗ ≤ |𝐵|∗ , so 𝐴 is measurable.3 5.19. Lebesgue measure is translation and dilation invariant: For 𝐴 ⊂ ℝ𝑑 , 𝑥0 ∈ ℝ𝑑 and 𝛿 > 0, we define 𝑥0 + 𝐴 = {𝑥0 + 𝑥 ∶ 𝑥 ∈ 𝐴}

and

𝛿𝐴 = {𝛿𝑥 ∶ 𝑥 ∈ 𝐴}.

These sets are measurable, and we have |𝑥0 + 𝐴| = |𝐴| and |𝛿𝐴| = 𝛿𝑑 |𝐴| (Exercises (12) and (13)).

5.4. Measurable functions In this section we discuss the functions that are to be integrated with respect to the Lebesgue measure discussed in Section 5.3. Such functions are called measurable. For convenience, we will allow functions to have infinite values at some points, so we consider the extended real line [−∞, ∞] with the conventions 𝑥+∞=∞+𝑥=∞

for all 𝑥 ∈ ℝ,

𝑥 − ∞ = −∞ + 𝑥 = −∞

for all 𝑥 ∈ ℝ,

𝑥⋅∞=∞⋅𝑥=∞

for all 𝑥 ∈ ℝ,

𝑥 > 0,

𝑥 ⋅ ∞ = ∞ ⋅ 𝑥 = −∞

for all 𝑥 ∈ ℝ,

𝑥 < 0,

and the corresponding products with −∞. We also agree that ∞+∞=∞

and

− ∞ − ∞ = −∞,

or ±∞ ⋅ ∞ = ±∞, but we do not define ∞ − ∞ nor 0 ⋅ ∞. We also say that −∞ < 𝑥 and 𝑥 < ∞ for all 𝑥 ∈ ℝ. Let 𝑓 ∶ ℝ𝑑 → [−∞, ∞] be an extended real valued function. We say that 𝑓 is measurable if, for all 𝑎 ∈ ℝ, the set 𝑓−1 ([−∞, 𝑎)) = {𝑥 ∈ ℝ𝑑 ∶ 𝑓(𝑥) < 𝑎} 3

Statement (1) is indeed used as the definition of a measurable set in some texts, as [SS05].

5.4. Measurable functions

97

is measurable. The definition of a measurable function is equivalent to saying that 𝑓−1 ([−∞, 𝑎]) is measurable for all 𝑎 ∈ ℝ. Indeed, since 1 1 [−∞, 𝑎] = [ − ∞, 𝑎 + ) and [−∞, 𝑎) = [ − ∞, 𝑎 − ], ⋃ ⋂ 𝑗 𝑗 𝑗

𝑗

−1

we see that the sets 𝑓 ([−∞, 𝑎)) are measurable for all 𝑎 ∈ ℝ if and only if the sets 𝑓−1 ([−∞, 𝑎]) are measurable for all 𝑎 ∈ ℝ. We could also have used the sets 𝑓−1 ((𝑎, ∞]) or 𝑓−1 ((𝑎, ∞]), or even the pre-images of bounded intervals 𝑓−1 ((𝑎, 𝑏]), 𝑓−1 ((𝑎, 𝑏)), etc. together with 𝑓−1 ({∞}) and 𝑓−1 ({−∞}), to define measurability. We usually denote the set 𝑓−1 ([−∞, 𝑎)) by {𝑓 < 𝑎}, and the set 𝑓 ([−∞, 𝑎]) by {𝑓 ≤ 𝑎}. Similarly, 𝑓−1 ((𝑎, ∞]) by {𝑓 > 𝑎}, 𝑓−1 ([𝑎, ∞]) by {𝑓 ≥ 𝑎}, etc. −1

The fact that 𝑓 is measurable is also equivalent to the fact that 𝑓−1 (𝑈) is measurable for any open 𝑈 ⊂ ℝ, or 𝑓−1 (𝐸) is measurable for any closed 𝐸 ⊂ ℝ (Exercise (14)). The previous equivalence implies that all continuous functions 𝑓 are measurable. Indeed, since every open set is measurable and, for any open 𝑈 ⊂ ℝ, 𝑓−1 (𝑈) is open, then 𝑓 is measurable, as stated above. The most important property of measurability of functions is its stability under pointwise limits. This is a consequence of the following result. Theorem 5.20. Let 𝑓1 , 𝑓2 , . . . be measurable extended real valued functions. Then the functions sup 𝑓𝑛 ,

inf 𝑓𝑛 ,

lim sup 𝑓𝑛 ,

and

lim inf 𝑓𝑛

are measurable. In the theorem, sup 𝑓𝑛 denotes the function sup 𝑓𝑛 (𝑥) = sup{𝑓𝑛 (𝑥) ∶ 𝑛 ≥ 1}, which we define as ∞ in the case when the set {𝑓𝑛 (𝑥) ∶ 𝑛 ≥ 1} is not bounded. Similarly for inf 𝑓𝑛 . lim sup 𝑓𝑛 and lim inf 𝑓𝑛 are defined by lim sup 𝑓𝑛 (𝑥) = inf sup 𝑓𝑘 (𝑥) = inf { sup{𝑓𝑘 (𝑥) ∶ 𝑘 ≥ 𝑛} ∶ 𝑛 ≥ 1}, 𝑛≥1 𝑘≥𝑛

lim inf 𝑓𝑛 (𝑥) = sup inf 𝑓𝑘 (𝑥) = sup { inf{𝑓𝑘 (𝑥) ∶ 𝑘 ≥ 𝑛} ∶ 𝑛 ≥ 1}. 𝑛≥1 𝑘≥𝑛

98

5. Measure theory in Euclidean space

Proof. To see that sup 𝑓𝑛 is measurable, observe that {sup 𝑓𝑛 > 𝑎} =

⋃

{𝑓𝑛 > 𝑎}

𝑛

and each set {𝑓𝑛 > 𝑎} is measurable. Similarly, inf 𝑓𝑛 is measurable because {inf 𝑓𝑛 < 𝑎} = {𝑓 < 𝑎} ⋃ 𝑛 𝑛

and also each {𝑓𝑛 < 𝑎} is measurable. Now, lim sup 𝑓𝑛 is measurable because, for each 𝑛, sup𝑘≥𝑛 𝑓𝑘 is measurable, and therefore inf𝑛≥1 sup𝑘≥𝑛 𝑓𝑘 is measurable. Similarly for lim inf 𝑓𝑛 . □ Corollary 5.21. If 𝑓1 , 𝑓2 , . . . is a sequence of extended real valued measurable functions and 𝑓𝑛 → 𝑓 pointwise, then 𝑓 is measurable. □

Proof. If 𝑓𝑛 → 𝑓, then lim sup 𝑓𝑛 = lim inf 𝑓𝑛 = 𝑓. 𝑑

5.22. If 𝑓 is measurable and 𝑔(𝑥) = 𝑓(𝑥) for all 𝑥 ∈ ℝ except at a set of measure zero, then 𝑔 is also measurable. Indeed, for each 𝑎 ∈ ℝ, whenever 𝑔(𝑥) < 𝑎 we have either 𝑓(𝑥) < 𝑎 or 𝑓(𝑥) ≥ 𝑎, so {𝑔 < 𝑎} is contained in the union of {𝑓 < 𝑎} and {𝑓 ≥ 𝑎} ∩ {𝑔 < 𝑎}. However, if 𝑓(𝑥) < 𝑎, it could happen that 𝑔(𝑥) ≥ 𝑎, so we need to remove the set {𝑓 < 𝑎} ∩ {𝑔 ≥ 𝑎}. Thus, we have {𝑔 < 𝑎} = {𝑓 < 𝑎} ∪ ({𝑓 ≥ 𝑎} ∩ {𝑔 < 𝑎}) ⧵ ({𝑓 < 𝑎} ∩ {𝑔 ≥ 𝑎}). As the two sets {𝑓 ≥ 𝑎} ∩ {𝑔 < 𝑎} and {𝑓 < 𝑎} ∩ {𝑔 ≥ 𝑎} are contained in the set where 𝑓(𝑥) ≠ 𝑔(𝑥), they are sets of measure zero and thus measurable, so we conclude {𝑔 < 𝑎} is measurable. If 𝑔(𝑥) = 𝑓(𝑥) for all 𝑥 ∈ ℝ𝑑 except at a set of measure zero, we say that 𝑔 = 𝑓 almost everywhere, or at almost every point, and we denote it by a.e. In general, we say that a property 𝑃(𝑥) holds a.e. if the set where 𝑃(𝑥) is false is of measure zero. We can then refine Corollary 5.21 as the following statement. Corollary 5.23. If 𝑓1 , 𝑓2 , . . . is a sequence of extended real valued measurable functions and 𝑓𝑛 → 𝑓 a.e., then 𝑓 is measurable. 5.24. If 𝑓 is measurable and 𝑘 ∈ ℤ+ , then (𝑓)𝑘 is measurable. This follows because, if 𝑘 is odd, {(𝑓)𝑘 < 𝑎} = {𝑓 < 𝑎1/𝑘 }

5.4. Measurable functions

99

and, if 𝑘 is even, {(𝑓)𝑘 < 𝑎} = {−𝑎1/𝑘 < 𝑓 < 𝑎1/𝑘 } if 𝑎 > 0, or empty if 𝑎 ≤ 0, and both of the sets on right side are measurable because 𝑓 is measurable. 5.25. If 𝑓 and 𝑔 are measurable and real valued, then 𝑓 + 𝑔 and 𝑓𝑔 are measurable. For the sum, observe that (5.26)

{𝑓 + 𝑔 < 𝑎} =

⋃

({𝑓 < 𝑎 − 𝑟} ∩ {𝑔 < 𝑟})

𝑟∈ℚ

and that each set {𝑓 < 𝑎 − 𝑟} ∩ {𝑔 < 𝑟} is measurable. To prove the identity (5.26), first note that, if 𝑓(𝑥) < 𝑎 − 𝑟 and 𝑔(𝑥) < 𝑟, we clearly have 𝑓(𝑥) + 𝑔(𝑥) < 𝑎, so the union of the right side of (5.26) is contained in the left side. For the reverse inclusion, assume 𝑓(𝑥)+𝑔(𝑥) < 𝑎, so 𝑔(𝑥) < 𝑎−𝑓(𝑥). Let 𝑟 ∈ ℚ such that 𝑔(𝑥) < 𝑟 < 𝑎−𝑓(𝑥). Hence 𝑔(𝑥) < 𝑟 and 𝑓(𝑥) < 𝑎−𝑟, so 𝑥 ∈ {𝑓 < 𝑎 − 𝑟} ∩ {𝑔 < 𝑟}, which implies (5.26). Now, the multiplication 𝑓𝑔 is a measurable function because we can write 1 𝑓𝑔 = ((𝑓 + 𝑔)2 − (𝑓 − 𝑔)2 ), 4 and each of the functions (𝑓 ± 𝑔)2 is measurable by the previous results. 5.27. If 𝐴 is measurable set, its characteristic function 1 𝜒𝐴 (𝑥) = { 0

𝑥∈𝐴 𝑥 ∉ 𝐴,

is a measurable function. Indeed, for each 𝑎 ∈ ℝ, 𝑑

⎧ℝ {𝜒𝐴 < 𝑎} = ℝ𝑑 ⧵ 𝐴 ⎨ ⎩∅

𝑎>1 0 0}.

𝑘=0

We have thus partitioned the set where 𝑓 is positive in the 22𝑛 + 1 sets 2𝑛 𝐴0𝑛 , 𝐴1𝑛 , . . . , 𝐴2𝑛 −1 and 𝐵𝑛 . Now define the simple function 22𝑛 −1

𝜙𝑛 = ∑ 𝑘=0

𝑘 𝜒 𝑘 + 2𝑛 𝜒𝐵𝑛 . 2𝑛 𝐴𝑛

Note that 0 ≤ 𝜙𝑛 ≤ 2𝑛 , and 𝜙𝑛 splits the values of 𝑓, up to 2𝑛 , in small jumps of size 2−𝑛 (see Figure 5.4). The 𝜙𝑛 clearly satisfy 𝜙𝑛 ≤ 𝜙𝑛+1 and

2n

Figure 5.4. The approximation of a function with simple functions.

𝜙𝑛 ↗ 𝑓. Moreover, given 𝑀 > 0, if 𝐴 = {𝑥 ∈ ℝ𝑑 ∶ 𝑓(𝑥) ≤ 𝑀}, then, for 𝑛 such that 2𝑛 > 𝑀, |𝑓(𝑥) − 𝜙𝑛 (𝑥)| < 2−𝑛 □

for all 𝑥 ∈ 𝐴. Therefore 𝜙𝑛 ⇉ 𝑓 on 𝐴. f (x)

f + (x)

f - (x)

Figure 5.5. The positive and negative parts of a function 𝑓(𝑥).

5.30. We can apply Theorem 5.29 to approximate a general measurable function 𝑓 with with simple functions 𝜙𝑛 such that |𝜙𝑛 | ≤ |𝜙𝑛+1 |, |𝜙𝑛 | ≤ |𝑓| and 𝜙𝑛 → 𝑓, and the convergence to be uniform on any set where 𝑓

102

5. Measure theory in Euclidean space

is bounded. We just need to write 𝑓 = 𝑓+ − 𝑓− , where 𝑓+ and 𝑓− are the positive and negative parts of 𝑓, respectively, given by 𝑓+ = 𝑓 ⋅ 𝜒{𝑓≥0}

𝑓− = −𝑓 ⋅ 𝜒{𝑓≤0} .

and

(See Figure 5.5.) We leave the details as an exercise (Exercise (16)). Note that we also have |𝑓| = 𝑓+ + 𝑓− .

Exercises (1) Let 𝑅 be a rectangle and 𝜀 > 0. Then there exist an open rectangle 𝑆 ⊃ 𝑅 and a closed rectangle 𝑇 ⊂ 𝑅 such that vol(𝑆) < vol(𝑅) + 𝜀

and

vol(𝑇) > vol(𝑅) − 𝜀.

(2) (a) If 𝐴 is countable, then |𝐴|∗ = 0. (b) If 𝐴1 , 𝐴2 , . . . are sets of measure zero, then their union has measure zero. (3) Let 𝑋 be the Cantor ternary set 𝐶, constructed by the removal of middle third intervals starting from 𝐶0 = [0, 1]. Then 𝐶 is of measure zero. (4) Let 𝑅 = [𝑎1 , 𝑏1 ] × [𝑎2 , 𝑏2 ] × ⋯ × [𝑎𝑑 , 𝑏𝑑 ] be a rectangle and, for each 𝑖 = 1, 2, . . . , 𝑑, 𝑗

𝑗

𝑗

𝑗

𝒫𝑗 = {𝑥0 = 𝑎𝑗 < 𝑥1 < 𝑥2 < . . . < 𝑥𝑁𝑗 = 𝑏𝑗 } a partition of [𝑎𝑗 , 𝑏𝑗 ]. If 𝒮 is the grid of subrectangles 𝑅 of the form 𝑗

𝑗

𝑆 = 𝐼1 × 𝐼2 × ⋯ × 𝐼𝑑 , where each 𝐼𝑗 = [𝑥𝑖−1 , 𝑥𝑖 ], then vol(𝑅) = ∑ vol(𝑆). 𝑆∈𝒮

(5) If 𝑈 ⊂ ℝ𝑑 is open and nonempty, then 𝑈 = ⋃𝑗 𝑄𝑗 where 𝑄𝑗 are dyadic almost disjoint closed cubes. For the proof, follow the next steps: (a) For 𝑚 ≥ 0, let 𝔔𝑚 be the collection of closed dyadic cubes 𝑄 with sides of length 2−𝑚 such that 𝑄 ⊂ 𝑈. (b) Show that, for 𝑚 ≤ 𝑛, 𝑄 ∈ 𝔔𝑚 and 𝑄′ ∈ 𝔔𝑛 , then either 𝑄 ⊃ 𝑄′ or they are almost disjoint.

Notes

103

(c) Define 𝔍0 = 𝔔0 and, for each 𝑚 > 0, let 𝔍𝑚 be collection of 𝑄 ∈ 𝔔𝑚 such that 𝑄 is almost disjoint to every 𝑄′ ∈ 𝔔𝑘 , for 𝑘 = 0, . . . , 𝑚 − 1. (d) Let 𝔍 = ∪𝑚≥0 𝔍𝑚 . (e) Prove that 𝑈= 𝑄. ⋃ 𝑄∈𝔍

𝑑

(6) For any 𝐴 ⊂ ℝ , |𝐴|∗ = inf{|𝑈|∗ ∶ 𝑈 is open and 𝐴 ⊂ 𝑈}. (7) If 𝐴1 , 𝐴2 , . . . are measurable, then ⋂𝑗 𝐴𝑗 is measurable. (8) A set of measure zero is measurable. (9) Let 𝐴 ⊂ ℝ𝑑 . The following are equivalent. (a) 𝐴 is measurable. (b) 𝐴 = 𝑃 ⧵ 𝑀, where 𝑃 is a 𝐺 𝛿 set and |𝑀| = 0. (c) 𝐴 = 𝑄 ∪ 𝑁, where 𝑄 is an 𝐹𝜍 set and |𝑁| = 0. (10) Part (2) of Corollary 5.16 is false if all 𝐴𝑗 have infinite measure. (11) Prove Corollary 5.17. (12) For 𝐴 ⊂ ℝ𝑑 measurable and 𝑥0 ∈ ℝ𝑑 , |𝑥0 + 𝐴| = |𝐴|. (13) Let 𝐴 ⊂ ℝ𝑑 be a measurable set. (a) For 𝛿 > 0, |𝛿𝐴| = 𝛿𝑑 |𝐴|. (b) For a 𝑑-tuple 𝛿 ̄ = (𝛿1 , . . . , 𝛿 𝑑 ) with each 𝛿𝑗 > 0, 𝑗 = 1, . . . , 𝑑, define ̄ = {(𝛿1 𝑥1 , . . . , 𝛿 𝑑 𝑥𝑑 ) ∶ (𝑥1 , . . . , 𝑥𝑑 ) ∈ 𝐴}. 𝛿𝐴 ̄ = 𝛿1 ⋯ 𝛿 𝑑 |𝐴|. Then |𝛿𝐴| (14) Let 𝑓 ∶ ℝ𝑑 → [−∞, ∞] be an extended valued function. Then the following are equivalent. (a) 𝑓 is measurable. (b) For any open 𝑈 ⊂ ℝ, 𝑓−1 (𝑈) is measurable. (c) For any closed 𝐸 ⊂ ℝ, 𝑓−1 (𝐸) is measurable. (15) If 𝑓 ∶ ℝ → [−∞, ∞] is monotone, then it is measurable. (16) Write the details for Fact 5.30.

Notes Measure theory has a long history with a long list of motivations, as discussed in detail in the texts [Bre07] and [Bre08] by David M. Bressoud.

104

5. Measure theory in Euclidean space

Henri Lebesgue’s contribution was to consider countable covers for sets, as in the definition of outer measure presented here, and the approximation of a function by decomposition of its image, as it was done in 5.29. Lebesgue’s theory is discussed in his papers [Leb98], [Leb99a], and [Leb99b], as well as his text [Leb04]. More detailed introductions to measure theory can be found in [Fol99] and [SS05].

Chapter 6

Lebesgue integral and Lebesgue spaces

6.1. Integration of measurable functions We are now ready to define the Lebesgue integral of a measurable function, and we first consider nonnegative functions. By Theorem 5.29, we can approximate any nonnegative measurable function by simple functions. Thus, we start by defining the integral of a nonnegative simple function. 6.1.1. Nonnegative simple functions. Let 𝜙 be a nonnegative simple function on ℝ𝑑 and ∑𝑗 𝑎𝑗 𝜒𝐴𝑗 its reduced form, so each 𝑎𝑗 > 0 and 𝐴𝑖 ∩ 𝐴𝑗 = ∅ if 𝑖 ≠ 𝑗. We define the integral of 𝜙 by (6.1)

∫ 𝜙 = ∑ 𝑎𝑗 |𝐴𝑗 |. 𝑗

The integral of 𝜙 is also written ∫ 𝜙(𝑥)𝑑𝑥 if we need to explicitly make reference to its argument 𝑥. We require no special conditions on the sets 𝐴𝑗 that define the simple function 𝜙, except to be measurable. Thus, the integral (6.1) may 105

106

6. Lebesgue integral and Lebesgue spaces

be infinite, for example, if some |𝐴𝑗 | = ∞. Note that assuming 𝑎𝑗 > 0 avoids the conflicting operations ∞ − ∞ or 0 ⋅ ∞ in (6.1). 6.2. For any 𝑐 > 0, ∫ 𝑐𝜙 = 𝑐 ∫ 𝜙. This follows clearly from the fact that 𝑐 ∑ 𝑎𝑗 𝜒𝐴𝑗 = ∑ 𝑐𝑎𝑗 𝜒𝐴𝑗 , 𝑗

𝑗

if 𝑐 > 0. 6.3. For nonnegative simple functions 𝜙 and 𝜓, ∫(𝜙 + 𝜓) = ∫ 𝜙 + ∫ 𝜓. Indeed, if 𝜙 = ∑𝑗 𝑎𝑗 𝜒𝐴𝑗 and 𝜓 = ∑𝑘 𝑏𝑘 𝜒𝐵𝑘 , then we can write the reduced form of 𝜙 + 𝜓 as 𝜙 + 𝜓 = ∑ 𝑐 𝑖 𝜒𝐶𝑖 = ∑(𝑐′𝑖 + 𝑐″𝑖 )𝜒𝐶𝑖 , 𝑖

𝑖

where each 𝐴𝑗 =

𝐶𝑖 ,

⋃

𝐵𝑘 =

𝑎 𝑐′𝑖 = { 𝑗 0

if 𝐶𝑖 ⊂ 𝐴𝑗 otherwise

⋃

𝐶𝑖 ,

𝑏𝑘

if 𝐶𝑖 ⊂ 𝐵𝑘

0

otherwise.

𝐶𝑖 ∩𝐵𝑘 ≠∅

𝐶𝑖 ∩𝐴𝑗 ≠∅

and

𝑐″𝑖 = {

Note that each 𝐶𝑖 is either of the form 𝐴𝑗 ∩ 𝐵𝑘 for some 𝑗, 𝑘, of the form 𝐴𝑗 ⧵ ⋃𝑘 𝐵𝑘 for some 𝑗, or 𝐵𝑘 ⧵ ⋃𝑗 𝐴𝑗 for some 𝑘. Thus ∫(𝜙 + 𝜓) = ∑ 𝑐 𝑖 |𝐶𝑖 | = ∑(𝑐′𝑖 + 𝑐″𝑖 )|𝐶𝑖 | = ∑ 𝑐′𝑖 |𝐶𝑖 | + ∑ 𝑐″𝑖 |𝐶𝑖 | 𝑖

=∑

𝑖;𝑐′𝑖 ≠0

𝑖

∑

𝑐′𝑖 |𝐶𝑖 | + ∑

𝑗 𝐶𝑖 ∩𝐴𝑗 ≠∅

𝑐″𝑖 |𝐶𝑖 |

∑

𝑘 𝐶𝑖 ∩𝐵𝑘 ≠∅

= ∑ 𝑎𝑗

∑

𝑗

𝐶𝑖 ∩𝐴𝑗 ≠∅

|𝐶𝑖 | + ∑ 𝑏𝑘 𝑘

∑

|𝐶𝑖 |

𝐶𝑖 ∩𝐵𝑘 ≠∅

= ∑ 𝑎𝑗 |𝐴𝑗 | + ∑ 𝑏𝑘 |𝐵𝑘 | = ∫ 𝜙 + ∫ 𝜓, 𝑗

𝑘

because the sets 𝐶𝑖 are pairwise disjoint and measurable.

𝑖;𝑐″ 𝑖 ≠0

6.1. Integration of measurable functions

107

6.4. If 𝜙 ≥ 0 is simple, 𝑥0 ∈ ℝ𝑑 and 𝜓 is the translation of 𝜙 by 𝑥0 , 𝜓(𝑥) = 𝜙(𝑥 − 𝑥0 ), then ∫ 𝜓 = ∫ 𝜙. This follows from the fact that |𝑥0 + 𝐴| = |𝐴| for any measurable set 𝐴 ⊂ ℝ𝑑 (Exercise (12), Chapter 5), because this implies that ∫ 𝜒𝐴 (𝑥 − 𝑥0 )𝑑𝑥 = ∫ 𝜒𝑥0 +𝐴 = |𝑥0 + 𝐴| = |𝐴| = ∫ 𝜒𝐴 , and thus the result follows for any linear combination of 𝜒𝐴 , by 6.2 and 6.3. 6.5. If 𝜙 ≥ 0 is simple, 𝛿 > 0 and 𝜓 is the dilation of 𝜙 by 𝛿, 𝜓(𝑥) = 𝛿−𝑑 𝜙(𝑥/𝛿), then ∫ 𝜓 = ∫ 𝜙. This follows from the fact that |𝛿𝐴| = 𝛿𝑑 |𝐴| for any measurable set 𝐴 ⊂ ℝ𝑑 (Exercise (13), Chapter 5), because this implies that, using 6.2, ∫ 𝛿−𝑑 𝜒𝐴 (𝑥/𝛿)𝑑𝑥 = 𝛿−𝑑 ∫ 𝜒𝛿𝐴 = 𝛿−𝑑 |𝛿𝐴| = 𝛿−𝑑 ⋅ 𝛿𝑑 |𝐴| = ∫ 𝜒𝐴 , and thus, as above, the result follows for any linear combination of 𝜒𝐴 . 6.6. If 𝜙 ≤ 𝜓, ∫ 𝜙 ≤ ∫ 𝜓. Using the same decomposition as in 6.3, we have ∫ 𝜙 = ∑ 𝑎𝑗 |𝐴𝑗 | = ∑ 𝑐′𝑖 |𝐶𝑖 | ≤ ∑ 𝑐″𝑖 |𝐶𝑖 | = ∑ 𝑏𝑘 |𝐵𝑘 | = ∫ 𝜓, 𝑗

since each

𝑐′𝑖

≤

𝑖

𝑖

𝑘

𝑐″𝑖 .

If 𝐴 ⊂ ℝ𝑑 is measurable, we define the integral of 𝜙 over 𝐴 as ∫ 𝜙 = ∫ 𝜙𝜒𝐴 .

(6.7)

𝐴

Note that, if 𝜙 = ∑𝑗 𝑎𝑗 𝜒𝐴𝑗 , then ∫ 𝜙 = ∑ 𝑎𝑗 |𝐴𝑗 ∩ 𝐴|. 𝐴

𝑗

108

6. Lebesgue integral and Lebesgue spaces

6.8. If 𝐴, 𝐵 ⊂ ℝ𝑑 are measurable and 𝐴 ∩ 𝐵 = ∅, then ∫

𝜙 = ∫ 𝜙 + ∫ 𝜙.

𝐴∪𝐵

𝐴

𝐵

Follows from the fact that, since 𝐴 ∩ 𝐵 = ∅, 𝜒𝐶∩(𝐴∪𝐵) = 𝜒𝐶∩𝐴 + 𝜒𝐶∩𝐵 for any set 𝐶. 6.1.2. Nonnegative measurable functions. For any measurable extended real valued function 𝑓 ≥ 0, we can approximate 𝑓 pointwise with simple functions 𝜙, by Theorem 5.29. This allows us to define the integral of 𝑓 by (6.9)

∫ 𝑓 = sup { ∫ 𝜙 ∶ 𝜙 is simple and 0 ≤ 𝜙 ≤ 𝑓}.

Again, it is possible to have ∫ 𝑓 = ∞. The integral of 𝑓 is also usually denoted by ∫ 𝑓, ℝ𝕕

if one wants to make explicit the fact that 𝑓 is a function on ℝ𝑑 , or ∫ 𝑓(𝑥)𝑑𝑥, ℝ𝕕

making the argument 𝑥 explicit. As before, ∫𝐴 𝑓 means ∫ 𝑓 = ∫ 𝑓𝜒𝐴 , 𝐴 𝑑

for any measurable 𝐴 ⊂ ℝ . The definition of the integral implies the following properties, which follow from their versions for simple functions. 6.10. For any measurable function 𝑓 ≥ 0 and any 𝑐 > 0, ∫ 𝑐𝑓 = 𝑐 ∫ 𝑓. This follows directly from 6.2 and the definition (6.9).

6.1. Integration of measurable functions

109

6.11. The translation and dilation invariance of the integral for simple functions, 6.4 and 6.5, also imply the invariance of the integral for measurable 𝑓 ≥ 0, by (6.9). That is, if 𝑓 ≥ 0 is measurable, 𝑥0 ∈ ℝ𝑑 and 𝛿 > 0, then ∫ 𝑓(𝑥 − 𝑥0 )𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 and ∫ 𝛿−𝑑 𝑓(𝑥/𝛿)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥. 6.12. If 𝑓, 𝑔 are measurable and 0 ≤ 𝑓 ≤ 𝑔, then ∫ 𝑓 ≤ ∫ 𝑔. Again, this follows from the definition (6.9), and the fact that, if 𝜙 is a simple function that satisfies 0 ≤ 𝜙 ≤ 𝑓, then 0 ≤ 𝜙 ≤ 𝑔 and thus { ∫ 𝜙 ∶ 𝜙 is simple and 0 ≤ 𝜙 ≤ 𝑓} ⊂ { ∫ 𝜙 ∶ 𝜙 is simple and 0 ≤ 𝜙 ≤ 𝑔}. The supremum of the set on the left side cannot be larger than that of the set on the right side. We are ready to state and prove the first of the Lebesgue integration theorems, called the monotone convergence theorem. Theorem 6.13 (Monotone convergence). If 𝑓𝑛 ≥ 0 are measurable and 𝑓𝑛 ↗ 𝑓, then ∫ 𝑓𝑛 → ∫ 𝑓. Proof. The limit 𝑓 is measurable by Theorem 5.20. By 6.12, the sequence ∫ 𝑓𝑛 is increasing, so the limit exists (possibly ∞). Also, by 6.12, ∫ 𝑓𝑛 ≤ ∫ 𝑓 for all 𝑛, so lim ∫ 𝑓𝑛 ≤ ∫ 𝑓. To prove the reverse inequality, we verify that the limit of the integrals is at least as large as the integral of any simple function 𝜙 that

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6. Lebesgue integral and Lebesgue spaces

satisfies 0 ≤ 𝜙 ≤ 𝑓, so this limit is an upper bound for the set of all such integrals. By the definition (6.9) of the integral, this will imply lim ∫ 𝑓𝑛 ≥ ∫ 𝑓. So let 𝜙 be a simple function such that 0 ≤ 𝜙 ≤ 𝑓, and let 0 < 𝜀 < 1. Define 𝐸𝑛 = {𝑥 ∶ 𝑓𝑛 (𝑥) ≥ (1 − 𝜀)𝜙(𝑥)}. Since 𝑓𝑛 is increasing, the sequence 𝐸𝑛 of sets is increasing because, if 𝑥 ∈ 𝐸𝑛 , then 𝑓𝑛+1 (𝑥) ≥ 𝑓𝑛 (𝑥) ≥ (1 − 𝜀)𝜙(𝑥) and thus 𝑥 ∈ 𝐸𝑛+1 . We also have ⋃𝑛 𝐸𝑛 = ℝ𝑑 because 𝑓𝑛 ↗ 𝑓 and 𝑓 ≥ 𝜙. Suppose 𝜙 = ∑ 𝑎𝑗 𝜒𝐴𝑗 . Then, for each 𝑛, ∫ 𝑓𝑛 ≥ ∫ 𝑓𝑛 𝜒𝐸𝑛 ≥ (1 − 𝜀) ∫ 𝜙𝜒𝐸𝑛 = (1 − 𝜀) ∑ 𝑎𝑗 |𝐴𝑗 ∩ 𝐸𝑛 |. 𝑗

For each 𝑗, the sequence of sets 𝐴𝑗 ∩ 𝐸𝑛 is increasing in 𝑛 and ⋃

𝐴𝑗 ∩ 𝐸𝑛 = 𝐴𝑗 ,

𝑛

so we have, by the monotone continuity of the measure (Corollary 5.16), |𝐴𝑗 ∩ 𝐸𝑛 | → |𝐴𝑗 | and thus lim ∫ 𝑓𝑛 ≥ (1 − 𝜀) ∫ 𝜙. Since 0 < 𝜀 < 1 is arbitrary, we obtain lim ∫ 𝑓𝑛 ≥ ∫ 𝜙, □

as desired.

The monotone convergence theorem can be used to prove the linear and analytic properties of the Lebesgue integral, as in the following two facts. 6.14. If 𝑓, 𝑔 ≥ are measurable, then ∫(𝑓 + 𝑔) = ∫ 𝑓 + ∫ 𝑔.

6.1. Integration of measurable functions

111

This is the analogue of fact 6.3 for nonnegative functions. For the proof, let 𝜙𝑛 and 𝜓𝑛 be sequences of simple functions with 0 ≤ 𝜙𝑛 ≤ 𝑓 and 0 ≤ 𝜓𝑛 ≤ 𝑔 such that 𝜙𝑛 ↗ 𝑓 and 𝜓𝑛 ↗ 𝑔, which exist by Theorem 5.29. Thus 𝜙𝑛 + 𝜓𝑛 ↗ 𝑓 + 𝑔 and, by the monotone convergence theorem, ∫(𝑓 + 𝑔) = lim ∫(𝜙𝑛 + 𝜓𝑛 ) = lim ∫ 𝜙𝑛 + lim ∫ 𝜓𝑛 = ∫ 𝑓 + ∫ 𝑔. Inductively, we can extend 6.14 to any finite number of nonnegative measurable functions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 , so 𝑁

(6.15)

𝑁

∫ ∑ 𝑓𝑗 = ∑ ∫ 𝑓𝑗 . 𝑗=1

𝑗=1

The monotone convergence theorem also implies, in fact, that the integral of a series of nonnegative functions is equal to the series of the integrals. 6.16. If 𝑓𝑛 is a sequence of nonnegative measurable functions and 𝑓 = ∑𝑛 𝑓𝑛 , then ∫ 𝑓 = ∑ ∫ 𝑓𝑛 . 𝑛

Indeed, the partial sums 𝑠𝑁 =

𝑁 ∑𝑛=1 𝑓𝑛

satisfy

𝑁

∫ 𝑠𝑁 = ∑ ∫ 𝑓𝑛 , 𝑛=1

by (6.15). Since each 𝑓𝑛 is nonnegative, 𝑠𝑁 ↗ 𝑓, so by the monotone convergence theorem we obtain 𝑁

∑ ∫ 𝑓𝑛 → ∫ 𝑓. 𝑛=1

The monotone convergence theorem is true even if we only assume 𝑓𝑛 ↗ 𝑓 a.e. (Exercise (4)). If the sequence 𝑓𝑛 is not monotone, nor converges to 𝑓, we can still say something about the sequence of integrals. The following result, our second of the Lebesgue theorems, is known as Fatou’s lemma.

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6. Lebesgue integral and Lebesgue spaces

Theorem 6.17 (Fatou’s lemma). For any sequence 𝑓𝑛 of nonnegative measurable functions, ∫ lim inf 𝑓𝑛 ≤ lim inf ∫ 𝑓𝑛 . Proof. For each 𝑛, consider the function inf𝑘≥𝑛 𝑓𝑘 given by inf 𝑓𝑘 (𝑥) = inf{𝑓𝑘 (𝑥) ∶ 𝑘 ≥ 𝑛},

𝑘≥𝑛

the largest lower bound of the set of values 𝑓𝑘 (𝑥) for 𝑘 ≥ 𝑛. Thus, for each 𝑗 ≥ 𝑛, we have inf𝑘≥𝑛 𝑓𝑘 ≤ 𝑓𝑗 and thus ∫ inf 𝑓𝑘 ≤ ∫ 𝑓𝑗 ,

(6.18)

𝑘≥𝑛

by 6.12. Now, since 𝑗 ≥ 𝑛 is arbitrary, the number on the left of (6.18) is then a lower bound for the set { ∫ 𝑓𝑘 ∶ 𝑘 ≥ 𝑛}, so we have that ∫ inf 𝑓𝑘 ≤ inf ∫ 𝑓𝑘 .

(6.19)

𝑘≥𝑛

𝑘≥𝑛

Since inf𝑘≥𝑛 𝑓𝑘 ↗ lim inf 𝑓𝑛 as 𝑛 → ∞, the monotone convergence theorem implies ∫ lim inf 𝑓𝑛 = lim ∫ inf 𝑓𝑘 . 𝑛

𝑘≥𝑛

Therefore, by (6.19), ∫ lim inf 𝑓𝑛 ≤ lim inf ∫ 𝑓𝑘 = lim inf ∫ 𝑓𝑛 . 𝑛 𝑘≥𝑛

□ 6.20. In the case when 𝑓𝑛 → 𝑓, Fatou’s lemma implies that ∫ 𝑓 ≤ lim inf ∫ 𝑓𝑛 . Thus, the integral of the limit of a sequence of functions can never be larger that the limit of the integrals, if it exists.

6.1. Integration of measurable functions

113

Example 6.21. In general, we don’t have equality in 6.20. Consider the sequence 𝑓𝑛 = 𝜒(𝑛,𝑛+1] . Clearly 𝑓𝑛 → 0, but each ∫ 𝑓𝑛 = 1, so ∫ 𝑓𝑛 → 1 > 0. 6.1.3. Extended real valued functions. Now we consider the integral for general measurable extended real valued functions 𝑓. We say the 𝑓 is integrable if ∫ 𝑓+ < ∞

∫ 𝑓− < ∞,

and

where 𝑓+ and 𝑓− are the positive and negative parts of 𝑓, defined above by 𝑓+ = 𝑓 ⋅ 𝜒{𝑓≥0} and 𝑓− = −𝑓 ⋅ 𝜒{𝑓≤0} . + − See Figure 5.5. As |𝑓| = 𝑓 + 𝑓 , the integrability of 𝑓 is equivalent to ∫ |𝑓| < ∞. Note that 𝑓 = 𝑓+ − 𝑓− . Thus, if 𝑓 is integrable, its integral is defined by ∫ 𝑓 = ∫ 𝑓+ − ∫ 𝑓− . Example 6.22. Let 𝑓 ∶ ℝ → ℝ be given by sin 𝑥 𝑥>0 𝑓(𝑥) = { 𝑥 0 𝑥 ≤ 0. 𝑓 is not integrable. We have that (2𝑘+1)𝜋

∞

∫ 𝑓+ = ∑ ∫ 𝑘=0 2𝑘𝜋

sin 𝑥 𝑑𝑥 𝑥 (2𝑘+1)𝜋

∞

1 ∫ (2𝑘 + 1)𝜋 2𝑘𝜋 𝑘=0

≥ ∑

sin 𝑥𝑑𝑥 = ∞,

because each integral in the series is equal to 2.1 1

Recall, however, that the improper integral of 𝑓 indeed exists: ∞

∫ 0

𝑅

sin 𝑥 𝜋 sin 𝑥 𝑑𝑥 = lim ∫ 𝑑𝑥 = . 𝑥 𝑥 2 𝑅→∞ 0

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6. Lebesgue integral and Lebesgue spaces

It is not hard to verify that the integral of extended real valued functions, by the definition above, satisfies the linear properties ∫ 𝑐𝑓 = 𝑐 ∫ 𝑓

(6.23) and (6.24)

∫ 𝑓 + 𝑔 = ∫ 𝑓 + ∫ 𝑔.

Indeed, for the proof of (6.23), consider first the case 𝑐 > 0. We have 𝑐𝑓 = (𝑐𝑓)+ − (𝑐𝑓)− = 𝑐𝑓+ − 𝑐𝑓− , so ∫ 𝑐𝑓 = ∫ 𝑐𝑓+ − ∫ 𝑐𝑓− = 𝑐 ∫ 𝑓+ − 𝑐 ∫ 𝑓− = 𝑐 inf 𝑓, where we have used 6.10. The case 𝑐 = 0 is trivial, as we have both sides of (6.23) equal to 0. If 𝑐 < 0, we write 𝑐𝑓 = (−𝑐)(−𝑓) and thus ∫ 𝑐𝑓 = ∫(−𝑐)(−𝑓) = (−𝑐) ∫(−𝑓) = −𝑐( ∫ 𝑓− − ∫ 𝑓+ ) = 𝑐 ∫ 𝑓, because (−𝑓)+ = 𝑓− and (−𝑓)− = 𝑓+ . Since we are allowing the functions to be extended real valued, the undefined operation 0⋅∞ will occur in the case 𝑐 = 0 at the points where 𝑓 is equal to ∞. However this may only occur in a set of measure zero (Exercise (3)) for an integrable function 𝑓, so the scalar multiplication 𝑐𝑓 is zero a.e. and ∫ 𝑐𝑓 = 0 (Exercise (1)). To prove (6.24), first note that, in the case when the measurable functions 𝑓, 𝑔 satisfy 𝑓, 𝑔 ≥ 0, 𝑓 − 𝑔 ≥ 0 and ∫ 𝑔 < ∞, then ∫(𝑓 − 𝑔) = ∫ 𝑓 − ∫ 𝑔. Indeed, by 6.14, we have ∫ 𝑓 = ∫(𝑓 − 𝑔 + 𝑔) = ∫(𝑓 − 𝑔) + ∫ 𝑔, and we can substract ∫ 𝑔 from both sides because ∫ 𝑔 < ∞. Now, to verify (6.24), for extended real valued functions 𝑓, 𝑔 we have (𝑓 + 𝑔)+ = 𝑓+ 𝜒𝐴 − 𝑓− 𝜒𝐴 + 𝑔+ 𝜒𝐴 − 𝑔− 𝜒𝐴 ,

6.1. Integration of measurable functions

115

where 𝐴 = {𝑥 ∶ 𝑓(𝑥) + 𝑔(𝑥) ≥ 0}, and similarly for (𝑓 + 𝑔)− (with opposite signs). Thus, by the previous observation and 6.14, ∫(𝑓 + 𝑔)+ = ∫(𝑓+ 𝜒𝐴 − 𝑓− 𝜒𝐴 + 𝑔+ 𝜒𝐴 − 𝑔− 𝜒𝐴 ) = ∫(𝑓+ 𝜒𝐴 + 𝑔+ 𝜒𝐴 ) − ∫(𝑓− 𝜒𝐴 + 𝑔− 𝜒𝐴 ) = ∫ 𝑓+ 𝜒𝐴 + ∫ 𝑔+ 𝜒𝐴 − ∫ 𝑓− 𝜒𝐴 − ∫ 𝑔− 𝜒𝐴 = ∫ 𝑓𝜒𝐴 + ∫ 𝑔𝜒𝐴 . Combining with the corresponding expresion for ∫(𝑓 + 𝑔)− we obtain (6.24). We leave the rest of the details as an exercise (Exercise (7)). Again, as we are allowing the functions to be extended valued, we may have the operation ∞ − ∞. As before, this may only occur in a set of measure zero, which doesn’t modify the integrals (Exercise (2)). 6.1.4. Complex valued functions. Now, let 𝑓 ∶ ℝ𝑑 → ℂ. We say that 𝑓 is integrable if both its real and imaginary parts are measurable and ∫ |𝑓| < ∞. In this case, the integral of 𝑓 is defined by ∫ 𝑓 = ∫ ℜ𝑓 + 𝑖 ∫ ℑ𝑓, where ℜ𝑓 and ℑ𝑓 are the real and imaginary parts of 𝑓, respectively. Note that the complex valued function 𝑓 is integrable if and only if its real and imaginary parts are integrable. As above, the integral of 𝑓 is denoted by ∫ 𝑓 ℝ𝕕

when we want to make the Euclidean space ℝ𝑑 explicit, or by ∫ 𝑓(𝑥)𝑑𝑥, ℝ𝕕

if we need to show explicitly the variable 𝑥 of 𝑓. We denote the set of complex valued integrable functions by 𝐿1 (ℝ𝑑 ). We sometimes denote it simply by 𝐿1 , if there is no confusion. It is not hard to see that 𝐿1 is a complex vector space and 𝑓 ↦ ∫ 𝑓 is a linear

116

6. Lebesgue integral and Lebesgue spaces

functional on 𝐿1 (Exercise (8)). This time we don’t have to worry about operations with infinity, but one still needs to verify carefully the identity ∫ 𝛼𝑓 = 𝛼 ∫ 𝑓 for complex scalars 𝛼. 6.25. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), 𝑥0 ∈ ℝ𝑑 and 𝛿 > 0, let 𝑔(𝑥) = 𝑓(𝑥 − 𝑥0 ) be the translation of 𝑓 by 𝑥0 , and ℎ(𝑥) = 𝛿−𝑑 𝑓(𝑥/𝛿) the dilation of 𝑓 by 𝛿. Then, by applying 6.11 to (ℜ𝑓)± , (ℑ𝑓)± we obtain ∫ 𝑔 = ∫ ℎ = ∫ 𝑓. 6.26. For 𝑓 ∈ 𝐿1 (ℝ𝑑 ), | ∫ 𝑓| ≤ ∫ |𝑓|. | | This inequality is obvious if ∫ 𝑓 = 0. If 𝑓 is real valued, | ∫ 𝑓| = | ∫ 𝑓+ − ∫ 𝑓− | ≤ ∫ 𝑓+ + ∫ 𝑓− = ∫ |𝑓|. | | | | If 𝑓 is complex valued and ∫ 𝑓 ≠ 0, write ∫ 𝑓 = 𝑟𝑒𝑖𝜃 , its polar form. Then | ∫ 𝑓| = 𝑟 = 𝑒−𝑖𝜃 ∫ 𝑓 = ∫ 𝑒−𝑖𝜃 𝑓, | | so ∫ 𝑒−𝑖𝜃 𝑓 is a real number, and hence ∫ 𝑒−𝑖𝜃 𝑓 = ∫ ℜ(𝑒−𝑖𝜃 𝑓) + 𝑖 ∫ ℑ(𝑒−𝑖𝜃 𝑓) = ∫ ℜ(𝑒−𝑖𝜃 𝑓). Therefore, by 6.12, | ∫ 𝑓| = ∫ 𝑒−𝑖𝜃 𝑓 = ∫ ℜ(𝑒−𝑖𝜃 𝑓) | | ≤ ∫ |ℜ(𝑒−𝑖𝜃 𝑓)| ≤ ∫ |𝑒−𝑖𝜃 𝑓| = ∫ |𝑓|. For a measurable 𝐴 ⊂ ℝ𝑑 , we say that 𝑓 is integrable on 𝐴 if ∫ |𝑓| = ∫ |𝑓|𝜒𝐴 < ∞. 𝐴

If 𝑓 is integrable on 𝐴, we have ∫ 𝑓 = ∫ 𝑓𝜒𝐴 . 𝐴

6.1. Integration of measurable functions

117

We denote the set of integrable functions on 𝐴 by 𝐿1 (𝐴). Recall that we say that 𝑓 = 𝑔 almost everywhere (and we write a.e.) if the set {𝑥 ∶ 𝑓(𝑥) ≠ 𝑔(𝑥)} has measure zero. Proposition 6.27. Let 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ). The following are equivalent. (1) 𝑓 = 𝑔 a.e. (2) ∫ |𝑓 − 𝑔| = 0. (3) For all measurable 𝐴 ⊂ ℝ𝑑 , ∫ 𝑓 = ∫ 𝑔. 𝐴

𝐴

Proof. We prove the implications (1) ⇒ (2) ⇒ (3) ⇒ (1). (1) ⇒ (2): If 𝑓 = 𝑔 a.e., then |𝑓 − 𝑔| = 0 a.e. Hence ∫ |𝑓 − 𝑔| = 0. (2) ⇒ (3): For any measurable 𝐴 ⊂ ℝ𝑑 , by 6.26, | ∫ 𝑓 − ∫ 𝑔| = | ∫(𝑓 − 𝑔)𝜒 | ≤ ∫ |𝑓 − 𝑔|𝜒 ≤ ∫ |𝑓 − 𝑔| = 0. 𝐴| 𝐴 | | | 𝐴

𝐴

(3) ⇒ (1): We prove the contrapositive, so assume 𝑓 ≠ 𝑔 in a set of positive measure. Then at least one of the functions ℜ(𝑓 − 𝑔)± or ℑ(𝑓 − 𝑔)± is positive in a set of positive measure. Assume 𝐴 = {𝑥 ∈ ℝ𝑑 ∶ ℜ(𝑓 − 𝑔)+ > 0} has positive measure. Note that, if 𝑥 ∈ 𝐴, ℜ(𝑓 − 𝑔)− (𝑥) = 0. Thus ∫ 𝑓 − ∫ 𝑔 = ∫ (𝑓 − 𝑔) = ∫ ℜ(𝑓 − 𝑔) + 𝑖 ∫ ℑ(𝑓 − 𝑔) 𝐴

𝐴

𝐴

𝐴

𝐴

and ∫ ℜ(𝑓 − 𝑔) = ∫ ℜ(𝑓 − 𝑔)+ > 0, 𝐴

so ∫𝐴 𝑓 ≠ ∫𝐴 𝑔.

𝐴

□

We are ready for our third Lebesgue convergence theorem for integrals, known as the dominated convergence theorem.

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6. Lebesgue integral and Lebesgue spaces

Theorem 6.28 (Dominated convergence). Let 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) such that 𝑓𝑛 → 𝑓 and there exists 𝑔 ∈ 𝐿1 (ℝ𝑑 ) with |𝑓𝑛 | ≤ 𝑔 for all 𝑛. Then 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and ∫ 𝑓𝑛 → ∫ 𝑓. Proof. 𝑓 is measurable by Corollary 5.21 and is integrable by 6.12, because |𝑓| ≤ 𝑔. By taking real and imaginary parts, we can assume all functions are real valued. Since |𝑓𝑛 | ≤ 𝑔, we have 𝑔 ± 𝑓𝑛 ≥ 0 for all 𝑛, and we can apply Fatou’s lemma (Theorem 6.17) to the sequences of nonnegative functions 𝑔 ± 𝑓𝑛 . Hence we have ∫(𝑔 + 𝑓) ≤ lim inf ∫(𝑔 + 𝑓𝑛 ) = ∫ 𝑔 + lim inf ∫ 𝑓𝑛 and ∫(𝑔 − 𝑓) ≤ lim inf ∫(𝑔 − 𝑓𝑛 ) = ∫ 𝑔 − lim sup ∫ 𝑓𝑛 . As ∫ 𝑔 < ∞ and ∫(𝑔 ± 𝑓) = ∫ 𝑔 ± ∫ 𝑓, we have lim sup ∫ 𝑓𝑛 ≤ ∫ 𝑓 ≤ lim inf ∫ 𝑓𝑛 , □

and therefore ∫ 𝑓𝑛 → ∫ 𝑓.

As in the case of the monotone convergence theorem, Theorem 6.28 also holds if we only assume 𝑓𝑛 → 𝑓 a.e. As a corollary, we can extend 6.16 to complex valued integrable functions. Corollary 6.29. Let 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) such that ∑𝑛 ∫ |𝑓𝑛 | < ∞. Then the series ∑𝑛 𝑓𝑛 converges almost everywhere to an integrable function 𝑓 and ∫ 𝑓 = ∑ ∫ 𝑓𝑛 . 𝑛

Proof. Consider the function 𝑔 = ∑𝑛 |𝑓𝑛 |. Then 𝑔 is the limit of an increasing sequence, because each |𝑓𝑛 | ≥ 0. By the monotone convergence

6.1. Integration of measurable functions

119

theorem we have ∫ 𝑔 = ∑ ∫ |𝑓𝑛 | < ∞, 𝑛 1

𝑑

so 𝑔 ∈ 𝐿 (ℝ ). In particular, 𝑔 is finite almost everywhere (Exercise (3)) and hence ∑𝑛 𝑓𝑛 converges almost everywhere. Also, for every 𝑁, 𝑁

𝑁

𝑛=1

𝑛=1

| ∑ 𝑓 | ≤ ∑ |𝑓 | ≤ 𝑔. 𝑛| 𝑛 | By the dominated convergence theorem, if ∑𝑛 𝑓𝑛 → 𝑓, we have that 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and ∫ 𝑓 = ∑ ∫ 𝑓𝑛 . 𝑛

□ 6.30. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and, for each 𝑛 ≥ 1, 𝑓𝑛 = 𝑓𝜒𝐵𝑛 , where 𝐵𝑛 is the ball around the origin of radius 𝑛, then clearly |𝑓𝑛 | ≤ |𝑓|, 𝑓𝑛 → 𝑓 and thus, by the dominated convergence theorem, ∫ 𝑓𝑛 → ∫ 𝑓. This can be written as ∫ 𝑓 → ∫ 𝑓, 𝐵𝑛

ℝ𝕕

or, explicitly, lim ∫

𝑛→∞

|𝑥|≤𝑛

𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥, ℝ𝕕

a fact that will later be useful to calculate integrals explicitly. There is nothing special about the balls 𝐵𝑛 : we can use any increasing sequence of measurable sets 𝐴𝑛 such that ⋃𝑛 𝐴𝑛 = ℝ𝑑 and obtain the same result. Corollary 6.31 states the conditions for the continuity and differentiability of integrals, also implied by the dominated convergence theorem. We leave its proof as an exercise (Exercise (9)). Corollary 6.31. Let 𝑓(𝑥, 𝑡) be a function on ℝ𝑑 × [𝑎, 𝑏] such that 𝑓(⋅, 𝑡) ∈ 𝐿1 (ℝ𝑑 ) for each 𝑡 ∈ [𝑎, 𝑏]. (1) If 𝑓(𝑥, ⋅) is continuous on [𝑎, 𝑏] for each 𝑥 ∈ ℝ𝑑 and there exists 𝑔 ∈ 𝐿1 (ℝ𝑑 ) such that |𝑓(𝑥, 𝑡)| ≤ 𝑔(𝑥) for all 𝑥 ∈ ℝ𝑑 and 𝑡 ∈ [𝑎, 𝑏], then 𝑡 ↦ ∫ℝ𝕕 𝑓(𝑥, 𝑡)𝑑𝑥 is continuous on [𝑎, 𝑏].

120

6. Lebesgue integral and Lebesgue spaces (2) If 𝑓(𝑥, ⋅) is differentiable in (𝑎, 𝑏) for each 𝑥 ∈ ℝ𝑑 and there exists ℎ ∈ 𝐿1 (ℝ𝑑 ) such that | 𝑓(𝑥, 𝑡) − 𝑓(𝑥, 𝑠) | ≤ ℎ(𝑥) | | 𝑡−𝑠 for all 𝑥 ∈ ℝ𝑑 and 𝑡, 𝑠 ∈ (𝑎, 𝑏), 𝑡 ≠ 𝑠, then 𝑡 ↦ ∫ℝ𝕕 𝑓(𝑥, 𝑡)𝑑𝑥 is differentiable in (𝑎, 𝑏) and 𝑑 𝜕 ∫ 𝑓(𝑥, 𝑡)𝑑𝑥 = ∫ 𝑓(𝑥, 𝑡)𝑑𝑥. 𝑑𝑡 ℝ𝕕 𝜕𝑡 ℝ𝕕

6.2. Fubini’s theorem In this section we prove Fubini’s theorem on iterated integrals. For 𝑑1 , 𝑑2 ∈ ℤ+ , we write the elements of the space ℝ𝑑1 +𝑑2 as (𝑥, 𝑦), where 𝑥 ∈ ℝ𝑑1 and 𝑦 ∈ ℝ𝑑2 . Now consider a measurable set 𝐴 ⊂ ℝ𝑑1 +𝑑2 . For each 𝑥 ∈ ℝ𝑑1 , the 𝑥-section of 𝐴 is defined as 𝐴𝑥 = {𝑦 ∈ ℝ𝑑2 ∶ (𝑥, 𝑦) ∈ 𝐴}, while the 𝑦-section of 𝐴 is defined by 𝐴 𝑦 = {𝑥 ∈ ℝ𝑑1 ∶ (𝑥, 𝑦) ∈ 𝐴}. 𝐴𝑥 and 𝐴 𝑦 are the projections of the cross sections of 𝐴 onto ℝ𝑑2 and ℝ𝑑1 , respectively (see Figure 6.1). It may happen that 𝐴𝑥 or 𝐴 𝑦 are nonmeasurable sets. However, we have the following result. Lemma 6.32. The sets 𝐴𝑥 and 𝐴 𝑦 are measurable for a.e. 𝑥 ∈ ℝ𝑑1 and a.e. 𝑦 ∈ ℝ𝑑2 , respectively, the functions 𝑥 ↦ |𝐴𝑥 | and 𝑦 ↦ |𝐴 𝑦 | are measurable, and |𝐴| = ∫ |𝐴𝑥 |𝑑𝑥 = ∫ |𝐴 𝑦 |𝑑𝑦.

(6.33)

ℝ𝑑 2

ℝ𝑑 1

Proof. The lemma follows immediately if 𝐴 = 𝑅 × 𝑆, where 𝑅 ⊂ ℝ𝑑1 and 𝑆 ⊂ ℝ𝑑2 are rectangles, because then 𝑥∈𝑅 𝑥∉𝑅

and

𝑅 𝐴𝑦 = { ∅

|𝐴𝑥 | = |𝑆|𝜒𝑅 (𝑥)

and

|𝐴 𝑦 | = |𝑅|𝜒𝑆 (𝑦),

𝐴𝑥 = {

𝑆 ∅

𝑦∈𝑆 𝑦 ∉ 𝑆,

so

6.2. Fubini’s theorem

121

Figure 6.1. The 𝑥-section and the 𝑦-section of 𝐴. Note that they are the projections of the cross sections of 𝐴 onto ℝ𝑑2 and ℝ𝑑1 .

and |𝐴| = |𝑅| ⋅ |𝑆| = ∫ |𝑆|𝜒𝑅 (𝑥)𝑑𝑥 = ∫ |𝑅|𝜒𝑆 (𝑦)𝑑𝑦. ℝ𝑑2

ℝ𝑑 1

If 𝐴 is a finite almost disjoint union of closed rectangles, then each 𝐴𝑥 and 𝐴 𝑦 is a finite disjoint union of closed rectangles (except for a finite number of points 𝑥 and 𝑦 corresponding to intersecting boundaries), so they are measurable, and (6.33) follows by the linearity of the integral. 𝑁 Indeed, if 𝐴 = ⋃𝑗=1 𝑅𝑗 where the 𝑅𝑗 are almost disjoint, then 𝐴𝑥 = 𝑁

𝑁

⋃𝑗=1 (𝑅𝑗 )𝑥 and 𝐴 𝑦 = ⋃𝑗=1 (𝑅𝑗 )𝑦 are also almost disjoint unions, so 𝑁

|𝐴𝑥 | = ∑ |(𝑅𝑗 )𝑥 | 𝑗=1

𝑁

and

|𝐴 𝑦 | = ∑ |(𝑅𝑗 )𝑦 | 𝑗=1

122

6. Lebesgue integral and Lebesgue spaces

except at most a finite number of 𝑥 and 𝑦. Hence, 𝑁

𝑁

|𝐴| = ∑ |𝑅𝑗 | = ∑ ∫ |(𝑅𝑗 )𝑥 |𝑑𝑥 𝑗=1 ℝ𝑑1

𝑗=1 𝑁

=∫

∑ |(𝑅𝑗 )𝑥 |𝑑𝑥 = ∫ |𝐴𝑥 |𝑑𝑥,

ℝ𝑑1 𝑗=1

ℝ𝑑1

and similarly for the integrals of the 𝑦-sections. By Exercise (5) of Chapter 5, if 𝐴 is open then it is the countable union of almost disjoint closed cubes, so the lemma follows by the monotone convergence theorem. Indeed, if we write 𝐴 = ⋃𝑗 𝑅𝑗 , where the 𝑅𝑗 are almost disjoint closed rectangles, then we also have 𝐴 = ⋃𝑛 𝐴𝑛 , where the 𝐴𝑛 is an increasing sequence of sets which are a finite union of almost disjoint rectangles, so by monotone continuity |𝐴| = lim |𝐴𝑛 | = lim ∫ |(𝐴𝑛 )𝑥 |𝑑𝑥 = ∫ |𝐴𝑥 |𝑑𝑥 𝑛

𝑛

ℝ𝑑 1

ℝ𝑑1

as |(𝐴𝑛 )𝑥 | ↗ |𝐴𝑥 | for each 𝑥. The result for the 𝑦-sections follows in the same way. If 𝐴 is a bounded 𝐺 𝛿 set, then 𝐴 = ⋂𝑛 𝑈𝑛 , where 𝑈𝑛 is a decreasing sequence of bounded open sets. Then 𝐴𝑥 =

⋂

(𝑈𝑛 )𝑥

and

𝑛

𝐴𝑦 =

⋂

(𝑈𝑛 )𝑦

𝑛

𝑦

are measurable. Since |(𝑈1 )𝑥 |, |(𝑈1 ) | < ∞ because 𝑈1 is bounded, the sequences |(𝑈𝑛 )𝑥 | and |(𝑈𝑛 )𝑦 | converge to |𝐴𝑥 | and |𝐴 𝑦 |, respectively, satisfy that |(𝑈𝑛 )𝑥 | ≤ |(𝑈1 )𝑥 |, |(𝑈𝑛 )𝑦 | ≤ |(𝑈1 )𝑦 |, and ∫ |(𝑈1 )𝑥 |𝑑𝑥 = ∫ |(𝑈1 )𝑦 |𝑑𝑦 = |𝑈1 | < ∞, ℝ𝑑 2

ℝ𝑑1

𝑦

so |(𝑈1 )𝑥 | and |(𝑈1 ) | are integrable. Thus (6.33) follows by the dominated convergence theorem, following similar lines as in the previous cases. For a nonbounded 𝐺 𝛿 set 𝐴, we can write 𝐴 = ⋃𝑛 (𝐴 ∩ 𝐵𝑛 ), where each 𝐵𝑛 is the open ball of radius 𝑛 around the origin, and thus 𝐴 is the increasing union of bounded 𝐺 𝛿 sets, so the lemma again follows by the monotone convergence theorem.

6.2. Fubini’s theorem

123

Let 𝐴 be a measure zero set. By Corollary 5.17(1), for each 𝑛 there exists an open set 𝑈𝑛 ⊃ 𝐴 such that |𝑈𝑛 | < 1/𝑛. If 𝑈 = ⋂𝑛 𝑈𝑛 , then 𝑈 is a 𝐺 𝛿 set of measure zero, 𝑈𝑥 and 𝑈 𝑦 are measurable, and ∫ |𝑈𝑥 |𝑑𝑥 = ∫ |𝑈 𝑦 |𝑑𝑦 = 0. ℝ𝑑 2

ℝ𝑑1

Thus |𝑈𝑥 | = 0 for a.e. 𝑥 ∈ ℝ𝑑1 and |𝑈 𝑦 | = 0 for a.e. 𝑦 ∈ ℝ𝑑2 . Since 𝐴 ⊂ 𝑈, each 𝐴𝑥 ⊂ 𝑈𝑥 and 𝐴 𝑦 ⊂ 𝑈 𝑦 , so 𝐴𝑥 is a measure zero set for a.e. 𝑥 ∈ ℝ𝑑1 and 𝐴 𝑦 is a measure zero set for a.e. 𝑦 ∈ ℝ𝑑2 , and thus measurable. Therefore (6.33) is true for the measure zero set 𝐴. For a general measurable set 𝐴, we can use Corollary 5.17(1) as above to write 𝐴 = 𝑈 ⧵ 𝑁, where 𝑈 is a 𝐺 𝛿 set and 𝑁 is a measure zero set. Since 𝐴𝑥 = 𝑈𝑥 ⧵ 𝑁𝑥 and 𝐴 𝑦 = 𝑈 𝑦 ⧵ 𝑁 𝑦 , 𝐴𝑥 and 𝐴 𝑦 are measurable for a.e. 𝑥 ∈ ℝ𝑑1 and for a.e. 𝑦 ∈ ℝ𝑑2 , respectively, and (6.33) follows by the previous cases. □ Indeed, the monotone convergence theorem is the protagonist in the proof of Lemma 6.32. Do not be surprised, as it will be in many of the results further in this book. It is, of course, in the proof of Theorem 6.34. Given a function 𝑓 ∶ ℝ𝑑1 +𝑑2 → ℂ we define, for each 𝑥 ∈ ℝ𝑑1 , the function 𝑓𝑥 on ℝ𝑑2 by 𝑓𝑥 (𝑦) = 𝑓(𝑥, 𝑦) and, for each 𝑦 ∈ ℝ𝑑2 , the function 𝑓𝑦 on ℝ𝑑1 by 𝑓𝑦 (𝑥) = 𝑓(𝑥, 𝑦). Theorem 6.34 (Fubini). If 𝑓 ∈ 𝐿1 (ℝ𝑑1 +𝑑2 ), for a.e. 𝑥 ∈ ℝ𝑑1 and 𝑦 ∈ ℝ𝑑2 we have 𝑓𝑥 ∈ 𝐿1 (ℝ𝑑2 ) and 𝑓𝑦 ∈ 𝐿1 (ℝ𝑑1 ) and ∫

(6.35)

ℝ𝑑1 +𝑑2

𝑓 = ∫ ( ∫ 𝑓𝑥 )𝑑𝑥 = ∫ ( ∫ 𝑓𝑦 )𝑑𝑦. ℝ𝑑1

ℝ𝑑 2

ℝ𝑑2

ℝ𝑑 1

Proof. By taking (ℜ𝑓)± and (ℑ𝑓)± , we can assume 𝑓 ≥ 0. If 𝑓 = 𝜒𝐴 for some measurable set 𝐴 ⊂ ℝ𝑑1 +𝑑2 , the identity (6.35) is the same as (6.33). By linearity of the integral, (6.35) follows if 𝑓 is a simple function. For general 𝑓 ≥ 0, take a sequence of nonnegative simple functions 𝜙𝑛 ↗ 𝑓, as in Theorem 5.29. Then (𝜙𝑛 )𝑥 ↗ 𝑓𝑥 and (𝜙𝑛 )𝑦 ↗ 𝑓𝑦 as well, for any 𝑥 ∈ ℝ𝑑1 and 𝑦 ∈ ℝ𝑑2 . Thus 𝑓𝑥 and 𝑓𝑦 are measurable for a.e. 𝑥 ∈ ℝ𝑑1 , 𝑦 ∈ ℝ𝑑2 and by the monotone convergence theorem we have the two limits ∫ (𝜙𝑛 )𝑥 ↗ ∫ 𝑓𝑥 ℝ𝑑 2

ℝ𝑑2

and

∫ (𝜙𝑛 )𝑦 ↗ ∫ 𝑓𝑦 , ℝ𝑑1

ℝ𝑑1

124

6. Lebesgue integral and Lebesgue spaces

and a second application of the monotone convergence theorem gives us ∫ ( ∫ (𝜙𝑛 )𝑥 )𝑑𝑥 → ∫ ( ∫ 𝑓𝑥 )𝑑𝑥 ℝ𝑑2

ℝ𝑑 1

ℝ𝑑 1

and

ℝ𝑑2

∫ ( ∫ (𝜙𝑛 )𝑦 )𝑑𝑦 → ∫ ( ∫ 𝑓𝑦 )𝑑𝑦. ℝ𝑑 2

ℝ𝑑2

ℝ𝑑 1

ℝ𝑑1

Equation (6.35) follows because, once more, the monotone convergence theorem implies ∫ 𝜙𝑛 → ∫ 𝑓 on ℝ𝑑1 +𝑑2 . Since we are assuming ∫ 𝑓 < ∞, we see that all integrals in (6.35) are finite, so the functions 𝑥 ↦ ∫ 𝑓𝑥 and 𝑦 ↦ ∫ 𝑓𝑦 are finite almost everywhere, so 𝑓𝑥 and 𝑓𝑦 are integrable for a.e. 𝑥 ∈ ℝ𝑑1 and 𝑦 ∈ ℝ𝑑2 . □ We usually write (6.35) as 𝑓 = ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥 = ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦,

∫ ℝ𝑑1 +𝑑2

ℝ𝑑1

ℝ𝑑 2

ℝ𝑑2

ℝ𝑑1

and we call the second and third integrals as the iterated integrals of ∫ 𝑓. From the proof of Fubini’s theorem, we can see that (6.35) is true if we only assume 𝑓 ≥ 0, as the integrability of 𝑓 was only used to conclude the integrability of 𝑓𝑥 and 𝑓𝑦 , needed to extend the result to general integrable functions. The result for the case of nonnegative functions is known as Tonelli’s theorem. We usually use Tonelli’s theorem to verify the integrability of 𝑓, as we can estimate ∫ |𝑓| using its iterated integrals. 6.36. An immediate consequence of Fubini’s theorem is following inequality, known as Minkowski’s inequality, ∫ || ∫ 𝑓(𝑥, 𝑦)𝑑𝑦||𝑑𝑥 ≤ ∫ ∫ |𝑓(𝑥, 𝑦)|𝑑𝑥𝑑𝑦. ℝ𝑑 1

ℝ𝑑 2

ℝ𝑑2

ℝ𝑑 1

6.3. The Lebesgue space 𝐿1 We observed above that 𝐿1 (ℝ𝑑 ) is a vector space under the usual operations (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥)

and

(𝛼𝑓)(𝑥) = 𝛼 𝑓(𝑥),

which follow from the basic properties of the Lebesgue integral. Now we can define ‖𝑓‖𝐿1 = ∫ |𝑓|,

6.3. The Lebesgue space 𝐿1

125

and see that it satisfies: (1) ‖𝑓‖𝐿1 ≥ 0 for all 𝑓 ∈ 𝐿1 (ℝ𝑑 ); (2) ‖𝛼𝑓‖𝐿1 = |𝛼| ‖𝑓‖𝐿1 for all 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝛼 ∈ ℂ; and (3) ‖𝑓 + 𝑔‖𝐿1 ≤ ‖𝑓‖𝐿1 + ‖𝑔‖𝐿1 , which also follow from the basic properties of the integral. For instance, inequality (3), called the triangle inequality, follows from 6.12, ‖𝑓 + 𝑔‖𝐿1 = ∫ |𝑓 + 𝑔| ≤ ∫(|𝑓| + |𝑔|) = ∫ |𝑓| + ∫ |𝑔| = ‖𝑓‖𝐿1 + ‖𝑔‖𝐿1 , because |𝑓(𝑥) + 𝑔(𝑥)| ≤ |𝑓(𝑥)| + |𝑔(𝑥)| for all 𝑥. So we can almost say that ‖ ⋅ ‖𝐿1 defines a norm, except for the fact that the condition “‖𝑓‖𝐿1 = 0 if and only if 𝑓 = 0” is not true. For instance, take 𝐴 = ℚ (or any other subset of ℝ of measure zero) and 𝑓 = 𝜒𝐴 . Then 𝑓 ≠ 0, but ‖𝑓‖𝐿1 = 0. However, by Proposition 6.27, 𝑓 = 𝑔 a.e. if and only if ∫ |𝑓−𝑔| = 0, so 𝑓 = 𝑔 a.e. if and only if ‖𝑓 − 𝑔‖𝐿1 = 0. Thus, if we define the equivalence relation 𝑓 ∼ 𝑔 if and only if 𝑓 = 𝑔 a.e., we see that ‖ ⋅ ‖𝐿1 defines a norm on the space ℒ = 𝐿1 (ℝ𝑑 )/ ∼ of equivalence classes. Indeed, first note that the sum and scalar multiplication are well defined on ℒ, because if 𝑓1 ∼ 𝑓2 and 𝑔1 ∼ 𝑔2 , then {𝑥 ∈ ℝ𝑑 ∶ 𝑓1 (𝑥) + 𝑔1 (𝑥) ≠ 𝑓2 (𝑥) + 𝑔2 (𝑥)} ⊂ {𝑥 ∈ ℝ𝑑 ∶ 𝑓1 (𝑥) ≠ 𝑓2 (𝑥)} ∪ {𝑥 ∈ ℝ𝑑 ∶ 𝑔1 (𝑥) ≠ 𝑔2 (𝑥)}, so 𝑓1 + 𝑔1 = 𝑓2 + 𝑔2 a.e., and similarly for the scalar multiplication. For ‖ ⋅ ‖𝐿1 , note that if 𝑓 ∼ 𝑔 then |𝑓| = |𝑔| a.e., so ∫ |𝑓| = ∫ |𝑔| and hence ‖𝑓‖𝐿1 = ‖𝑔‖𝐿1 . Thus, ‖ ⋅ ‖𝐿1 is well defined on ℒ, and it is a norm, by (1), (2) and (3). Therefore, ℒ is a normed space. As there is no reason for confusion, we will also denote the space ℒ of equivalence classes of integrable functions by 𝐿1 (ℝ𝑑 ). 6.37. The norm ‖ ⋅ ‖𝐿1 is invariant under translations and dilations, which follows immediately by 6.25. Theorem 6.38. 𝐿1 (ℝ𝑑 ) is a complete normed space.

126

6. Lebesgue integral and Lebesgue spaces Any norm ‖ ⋅ ‖ on a vector space 𝑋 induces the metric 𝑑(𝑓, 𝑔) = ‖𝑓 − 𝑔‖,

so it makes 𝑋 a metric space. Recall that a complete metric space is a metric space where its Cauchy sequences converge. If 𝑋 is a complete normed vector space, we call it a Banach space. See Appendix A.5. Proof of Theorem 6.38. We use Theorem A.13: assume that we have a sequence 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) such that ∑𝑛 ‖𝑓𝑛 ‖𝐿1 < ∞, and we have to prove that the series ∑𝑛 𝑓𝑛 converges in 𝐿1 (ℝ𝑑 ), that is, there exists 𝑓 ∈ 𝐿1 (ℝ𝑑 ) such that 𝑁

|| ∑ 𝑓 − 𝑓|| → 0 𝑛 || ||𝐿1 𝑛=1

as 𝑁 → ∞. By Corollary 6.29, there exists 𝑓 ∈ 𝐿1 (ℝ𝑑 ) such that ∑𝑛 𝑓𝑛 = 𝑓 a.e. If we set 𝑔 = ∑𝑛 |𝑓𝑛 | as in the proof of 6.29, then 𝑁

| ∑ 𝑓 − 𝑓| ≤ 2𝑔 𝑛 | | 𝑛=1

for all 𝑁. Since 𝑁

| ∑ 𝑓 − 𝑓| → 0 𝑛 | |

a.e.,

𝑛=1

the dominated convergence theorem implies that 𝑁

∫ || ∑ 𝑓𝑛 − 𝑓|| → 0. 𝑛=1

□ 6.39. Simple functions are dense in 𝐿1 (ℝ𝑑 ). This follows by 5.30, which states that we can find simple functions 𝜙𝑛 such that |𝜙𝑛 | ≤ |𝜙𝑛+1 |, |𝜙𝑛 | ≤ |𝑓| and 𝜙𝑛 → 𝑓, and the dominated convergence theorem. Note that, since each 𝜙𝑛 is integrable, if 𝜙𝑛 = ∑ 𝑎𝑗 𝜒𝐴𝑗 , 𝑗

is the reduced form of 𝜙𝑛 , then the sets 𝐴𝑗 have finite measure.

6.3. The Lebesgue space 𝐿1

127

6.40. The space 𝐶𝑐 (ℝ𝑑 ) of continuous functions of compact support is also dense in 𝐿1 (ℝ𝑑 ). By 6.30, we can approximate any 𝑓 ∈ 𝐿1 (ℝ𝑑 ) with integrable functions with compact support, and thus we can assume that the simple functions in 6.39 also have compact support, so each 𝐴𝑗 in their reduced form is bounded. Now, given 𝜀 > 0, we can find a compact set 𝐹𝑗 ⊂ 𝐴𝑗 and a bounded open set 𝑈 𝑗 ⊃ 𝐴𝑗 such that |𝐴𝑗 ⧵𝐹𝑗 |, |𝑈 𝑗 ⧵𝐴𝑗 | < 𝜀, and by Theorem A.11 in Appendix A there exists a continuous function 𝑓𝑗 , with 0 ≤ 𝑓𝑗 ≤ 1, such that 𝑓𝑗 is supported in 𝑈 𝑗 and 𝑓 = 1 on 𝐹𝑗 . Thus ∫ |𝜒𝐴𝑗 − 𝑓𝑗 | ≤ |𝑈 𝑗 ⧵ 𝐴𝑗 | < 2𝜀, so ∫ || ∑ 𝑎𝑗 𝜒𝐴𝑗 − ∑ 𝑎𝑗 𝑓𝑗 || ≤ 2𝜀 ∑ |𝑎𝑗 |, 𝑗

𝑗

𝑗

which implies the result because 𝜀 > 0 is arbitrary. Note that, explicitly, 6.39 and 6.40 imply that, for every 𝜀 > 0, we can find a simple 𝜙 and 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ) such that ‖𝑓 − 𝜙‖𝐿1 < 𝜀

and

‖𝑓 − 𝑔‖𝐿1 < 𝜀.

6.41. In particular, 6.40 implies that translations are continuous under the norm in 𝐿1 , in the following sense. Let 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and, for each 𝑦 ∈ ℝ𝑑 , let 𝑓𝑦 (𝑥) = 𝑓(𝑥 − 𝑦) be the translation of 𝑓 by 𝑦. Then lim ‖𝑓𝑦 − 𝑓‖𝐿1 = 0,

𝑦→0

so 𝑓𝑦 → 𝑓 in 𝐿1 as 𝑦 → 0. Note first that the result is true for 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ), because every continuous function of compact support is uniformly continuous. Indeed, asume 𝑔 is supported in the ball 𝐵𝑟 . Given 𝜀 > 0, there exists 𝛿 > 0 such that 𝛿 < 1 and, if |𝑦| < 𝛿 then 𝜀 |𝑔(𝑥 − 𝑦) − 𝑔(𝑥)| < . |𝐵𝑟+1 | Hence, if |𝑦| < 𝛿, ‖𝑔𝑦 − 𝑔‖𝐿1 = ∫ 𝐵𝑟+1 𝑦

|𝑔(𝑥 − 𝑦) − 𝑔(𝑥)|𝑑𝑥 ≤ ∫ 𝐵𝑟+1

𝜀 𝑑𝑥 = 𝜀. |𝐵𝑟+1 |

1

Thus 𝑔 → 𝑔 in 𝐿 as 𝑦 → 0. For any other 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and given 𝜀 > 0, choose 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ) such that ‖𝑓 − 𝑔‖𝐿1 < 𝜀. Thus, by the triangle inequality and the invariance of the norm under translations, ‖𝑓𝑦 − 𝑓‖𝐿1 ≤ ‖𝑓𝑦 − 𝑔𝑦 ‖𝐿1 + ‖𝑔𝑦 − 𝑔‖𝐿1 + ‖𝑔 − 𝑓‖𝐿1 < 2𝜀 + ‖𝑔𝑦 − 𝑔‖𝐿1 .

128

6. Lebesgue integral and Lebesgue spaces

Since ‖𝑔𝑦 − 𝑔‖𝐿1 → 0 and 𝜀 > 0 is arbitrary, we conclude that 𝑓𝑦 → 𝑓 in 𝐿1 as 𝑦 → 0. 6.42. For 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), their convolution is defined by 𝑓 ∗ 𝑔(𝑥) = ∫ 𝑓(𝑥 − 𝑦)𝑔(𝑦)𝑑𝑦. ℝ𝕕

By Minkowski’s inequality and the translation invariance of the integral, ∫ |𝑓 ∗ 𝑔(𝑥)|𝑑𝑥 = ∫ || ∫ 𝑓(𝑥 − 𝑦)𝑔(𝑦)𝑑𝑦||𝑑𝑥 ℝ𝕕

ℝ𝕕

ℝ𝕕

≤ ∫ ∫ |𝑓(𝑥 − 𝑦)𝑔(𝑦)|𝑑𝑥𝑑𝑦 ℝ𝕕

ℝ𝕕

= ∫ |𝑔(𝑦)| ∫ |𝑓(𝑥 − 𝑦)|𝑑𝑥𝑑𝑦 ℝ𝕕

ℝ𝕕

= ∫ |𝑔(𝑦)| ∫ |𝑓(𝑥)|𝑑𝑥𝑑𝑦 = ‖𝑓‖𝐿1 ‖𝑔‖𝐿1 . ℝ𝕕

ℝ𝕕

Therefore 𝑓 ∗ 𝑔 ∈ 𝐿1 (ℝ𝑑 ) and its norm satisfies (6.43)

‖𝑓 ∗ 𝑔‖𝐿1 ≤ ‖𝑓‖𝐿1 ‖𝑔‖𝐿1 .

Note that we have 𝑓 ∗ 𝑔 = 𝑔 ∗ 𝑓 for any 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ) (Exercise (10)). 6.44. Recall, as an example, the convolution operators 4.14 defined by 𝑓 ↦ Φ𝑡 ∗ 𝑓, where the dilations Φ𝑡 (𝑥) =

1 𝑥 Φ( ), 𝑡 𝑡𝑑

form a family of good kernels. Note that this is true for any Φ ∈ 𝐿1 (ℝ𝑑 ) with ∫ Φ = 1, because we don’t use the continuity of Φ in Exercise (9) of Chapter 4. Also (6.43) and 6.37 imply ‖Φ𝑡 ∗ 𝑓‖𝐿1 ≤ ‖Φ𝑡 ‖𝐿1 ‖𝑓‖𝐿1 = ‖Φ‖𝐿1 ‖𝑓‖𝐿1 , so the convolutions Φ𝑡 ∗ 𝑓 are uniformly bounded in 𝐿1 .

6.3. The Lebesgue space 𝐿1

129

Now, by Minkwoski’s inequality and invariance under dilations, ‖Φ𝑡 ∗ 𝑓 − 𝑓‖𝐿1 = ∫ ||𝑓 ∗ Φ𝑡 (𝑥) − 𝑓(𝑥) ∫ Φ𝑡 ||𝑑𝑥 ℝ𝕕

ℝ𝕕

= ∫ || ∫ (𝑓(𝑥 − 𝑦) − 𝑓(𝑥))Φ𝑡 (𝑦)𝑑𝑦||𝑑𝑥 ℝ𝕕

ℝ𝕕

≤ ∫ ∫ |𝑓(𝑥 − 𝑦) − 𝑓(𝑥)| |Φ𝑡 (𝑦)|𝑑𝑥𝑑𝑦 ℝ𝕕

ℝ𝕕

= ∫ ∫ |𝑓(𝑥 − 𝑡𝑦) − 𝑓(𝑥)| |Φ(𝑦)|𝑑𝑥𝑑𝑦 ℝ𝕕

ℝ𝕕

= ∫ ‖𝑓𝑡𝑦 − 𝑓‖𝐿1 |Φ(𝑦)|𝑑𝑦. ℝ𝕕

Now, by 6.41, ‖𝑓𝑡𝑦 − 𝑓‖𝐿1 → 0 for each 𝑦 ∈ ℝ𝑑 as 𝑡 → 0. Since ‖𝑓𝑡𝑦 − 𝑓‖𝐿1 ≤ 2‖𝑓‖𝐿1 and Φ is integrable, the dominated convergence theorem implies that ‖Φ𝑡 ∗ 𝑓 − 𝑓‖𝐿1 → 0

as 𝑡 → 0.

6.45. As a consequence of 6.44, we also see that the space 𝐶𝑐∞ (ℝ𝑑 ) of smooth functions of compact support are dense in 𝐿1 (ℝ𝑑 ). Indeed, by Corollary 6.31, if 𝜙 ∈ 𝐶𝑐∞ (ℝ𝑑 ) and 𝑓 ∈ 𝐿1 (ℝ𝑑 ), 𝜙∗𝑓 is a smooth function because we can differentiate 𝜙 ∗ 𝑓(𝑥) inside the integral. Now, if 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ), then 𝜙𝑡 ∗ 𝑔 has compact support for any 𝑡 > 0, and 𝜙𝑡 ∗ 𝑔 → 𝑔 in 𝐿1 as 𝑡 → 0, if ∫ 𝜙 = 1. Therefore, given 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝜀 > 0, we can choose 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ) such that ‖𝑔−𝑓‖𝐿1 < 𝜀/2, and for a fixed 𝜙 ∈ 𝐶𝑐∞ (ℝ𝑑 ) with ∫ 𝜙 = 1 we can choose 𝑡 > 0 such that ‖𝜙𝑡 ∗ 𝑔 − 𝑔‖𝐿1 < 𝜀/2, and hence ‖𝜙𝑡 ∗ 𝑔 − 𝑓‖𝐿1 ≤ ‖𝜙𝑡 ∗ 𝑔 − 𝑔‖𝐿1 + ‖𝑔 − 𝑓‖𝐿1 < 𝜀. 6.46. Observe that, if {𝐾𝑡 } is a family of good kernels, 𝑓 ∈ 𝐿1 (ℝ𝑑 ) is bounded and continuous at 𝑥, then 𝐾𝑡 ∗ 𝑓(𝑥) → 𝑓(𝑥) as 𝑡 → 0, as the same proof as in the ones the case of the Poisson integral in Chapter 4 will apply to this case.

130

6. Lebesgue integral and Lebesgue spaces

6.4. The Lebesgue space 𝐿2 We now define 𝐿2 (ℝ𝑑 ) as the set of square integrable measurable functions on ℝ𝑑 , that is, functions that satisfy ∫ |𝑓|2 < ∞. As in the case of 𝐿1 (ℝ𝑑 ), we in fact identify 𝐿2 (ℝ𝑑 ) to be the set of equivalence classes with respect to the relation 𝑓 ∼ 𝑔 if and only if 𝑓(𝑥) = 𝑔(𝑥) for almost every 𝑥. 𝐿2 (ℝ𝑑 ) is a vector space (Exercise (11)) with inner product (6.47)

⟨𝑓, 𝑔⟩ = ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥. ℝ𝕕

The integral in (6.47) converges, as Cauchy’s inequality |𝑓(𝑥)𝑔(𝑥)| ≤

1 (|𝑓(𝑥)|2 + |𝑔(𝑥)|2 ) 2

implies 𝑓𝑔̄ is integrable whenever |𝑓|2 and |𝑔|2 are integrable. The inner product (6.47) induces the 𝐿2 norm ‖𝑓‖𝐿2 =

∫ |𝑓|2 . √

Recall the Cauchy–Schwarz inequality | ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥| ≤ ‖𝑓‖ 2 ‖𝑔‖ 2 , 𝐿 𝐿 | | which is the main ingredient when verifying that ‖ ⋅ ‖𝐿2 satisfies the triangle inequality. The 𝐿2 norm is also invariant by translations, as the 𝐿1 norm above, although it is not invariant by dilations. Theorem 6.48. 𝐿2 (ℝ𝑑 ) is a complete inner product space. As in the proof of Theorem 6.38, the proof of Theorem 6.48 consists on verifying that every absolutely convergent series of functions in 𝐿2 (ℝ𝑑 ) converges in 𝐿2 , and we leave the proof as an exercise (Exercise (12)). Moreover, we can also prove that the simple functions, as well as the space 𝐶𝑐 (ℝ𝑑 ) of continuous functions of compact support, are dense in

6.4. The Lebesgue space 𝐿2

131

𝐿2 (ℝ𝑑 ), with similar proofs to the 𝐿1 (ℝ𝑑 ) case. The density of 𝐶𝑐 (ℝ𝑑 ) in 𝐿2 (ℝ𝑑 ) also implies the continuity of translations in the 𝐿2 norm, that is (6.49)

‖𝑓𝑦 − 𝑓‖𝐿2 → 0

as 𝑦 → 0.

To discuss the analogous results to convolutions in 𝐿2 (ℝ𝑑 ), we need the following result. Theorem 6.50. Let 𝑓 be a measurable function such that, for any 𝑔 ∈ 𝐿2 (ℝ𝑑 ), 𝑓𝑔 ∈ 𝐿1 (ℝ𝑑 ) and 𝑀 = sup {|| ∫ 𝑓𝑔|| ∶ 𝑔 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑔‖𝐿2 = 1} < ∞. Then 𝑓 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑓‖𝐿2 = 𝑀. Theorem 6.50 can be seen as the converse to the Cauchy–Schwarz inequality. Proof. The result is trivial if 𝑓 = 0 a.e., so we assume it is not. Let 𝜙𝑛 be a sequence of simple funtions such that |𝜙𝑛 | ≤ |𝜙𝑛+1 |, |𝜙𝑛 | ≤ |𝑓| and 𝜙𝑛 → 𝑓, as in 5.30. If we set 𝜓𝑛 = 𝜙𝑛 𝜒𝐵𝑛 , then 𝜓𝑛 → 𝑓 with |𝜓𝑛 | ↗ |𝑓| and each 𝜓𝑛 is supported in the ball 𝐵𝑛 , so 𝜓𝑛 ∈ 𝐿2 (ℝ𝑑 ). Since we are assuming 𝑓 is not zero almost everywhere, we can also assume each ‖𝜓𝑛 ‖𝐿2 > 0. We write 𝑓 in polar coordinates as 𝑓(𝑥) = |𝑓(𝑥)|𝑒𝑖𝜃(𝑥) , and define 𝑔𝑛 =

|𝜓𝑛 |𝑒−𝑖𝜃 . ‖𝜓𝑛 ‖𝐿2

The sequence 𝑔𝑛 has the following properties. • 𝑔𝑛 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑔𝑛 ‖𝐿2 = 1. Indeed, ∫ |𝑔𝑛 |2 =

1 ∫ |𝜓𝑛 |2 = 1. ‖𝜓𝑛 ‖2𝐿2

• ∫ |𝜓𝑛 | |𝑔𝑛 | = ‖𝜓𝑛 ‖𝐿2 . This follows directly by the calculation ∫ |𝜓𝑛 | |𝑔𝑛 | = ∫ |𝜓𝑛 |

|𝜓𝑛 | 1 ∫ |𝜓𝑛 |2 = ‖𝜓𝑛 ‖𝐿2 . = ‖𝜓𝑛 ‖𝐿2 ‖𝜓𝑛 ‖𝐿2

132

6. Lebesgue integral and Lebesgue spaces • ∫ 𝑓𝑔𝑛 = ∫ |𝑓| |𝑔𝑛 |. It follows from the fact that 𝑓𝑒−𝑖𝜃 = |𝑓|, because 𝑓𝑔𝑛 = 𝑓

|𝜓𝑛 |𝑒−𝑖𝜃 |𝜓𝑛 | = |𝑓| = |𝑓| |𝑔𝑛 |. ‖𝜓‖𝐿2 ‖𝜓‖𝐿2

Thus, by Fatou’s lemma and the above properties ‖𝑓‖𝐿2 ≤ lim inf ‖𝜓𝑛 ‖𝐿2 = lim inf ∫ |𝜓𝑛 | |𝑔𝑛 | ≤ lim inf ∫ |𝑓| |𝑔𝑛 | = lim inf ∫ 𝑓𝑔𝑛 ≤ 𝑀.

Hence 𝑓 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑓‖𝐿2 ≤ 𝑀. By the Cauchy–Schwarz inequality, for any 𝑔 ∈ 𝐿2 (ℝ𝑑 ) such that ‖𝑔‖𝐿2 = 1, | ∫ 𝑓𝑔| ≤ ‖𝑓‖ 2 ‖𝑔‖ 2 = ‖𝑓‖ 2 . 𝐿 𝐿 𝐿 | | □

Therefore 𝑀 ≤ ‖𝑓‖𝐿2 , as required.

6.51. As a consequence of Theorem 6.50, we have the Minkowski’s inequality for 𝐿2 , 2

√

∫ || ∫ 𝑓(𝑥, 𝑦)𝑑𝑦|| 𝑑𝑥 ≤ ∫ ℝ𝑑 1

ℝ𝑑2

ℝ𝑑2

∫ |𝑓(𝑥, 𝑦)|2 𝑑𝑥𝑑𝑦. √ ℝ𝑑1

Note that, if we write 𝑓𝑦 = 𝑓(𝑥, 𝑦), this can be written as || ∫ 𝑓 𝑑𝑦|| ≤ ∫ ‖𝑓 ‖ 2 𝑑𝑦, 𝑦 𝑦 𝐿 || ||𝐿2 so it can be seen as an extension of the triangle inequality for integrals of functions. This follows by Theorem 6.50 because, for any 𝑔 ∈ 𝐿2 (ℝ𝑑1 ) such that ‖𝑔‖𝐿2 = 1, using Tonelli’s theorem and the Cauchy–Schwarz

6.4. The Lebesgue space 𝐿2

133

inequality, ∫ || ∫ 𝑓(𝑥, 𝑦)𝑑𝑦|| |𝑔(𝑥)|𝑑𝑥 ≤ ∫ ∫ |𝑓(𝑥, 𝑦)|𝑑𝑦 |𝑔(𝑥)|𝑑𝑥 ℝ𝑑 1

ℝ𝑑 2

ℝ𝑑 1

ℝ𝑑 2

= ∫ ∫ |𝑓(𝑥, 𝑦)| |𝑔(𝑥)|𝑑𝑥𝑑𝑦 ℝ𝑑2

≤∫ ℝ𝑑 2

=∫ ℝ𝑑2

ℝ𝑑 1

√

∫ |𝑓(𝑥, 𝑦)|2 𝑑𝑥 ⋅ ‖𝑔‖𝐿2 𝑑𝑦 ℝ𝑑 1

∫ |𝑓(𝑥, 𝑦)|2 𝑑𝑥𝑑𝑦. √ ℝ𝑑1

Therefore, if the last integral is finite, 𝑥 ↦ ( ∫ 𝑓(𝑥, 𝑦)𝑑𝑦)𝑔(𝑥) ℝ𝑑2

is integrable and satisfies | ∫ ( ∫ 𝑓(𝑥, 𝑦)𝑑𝑦)𝑔(𝑥)𝑑𝑥| ≤ ∫ | | ℝ𝑑 1

ℝ𝑑 2

ℝ𝑑2

∫ |𝑓(𝑥, 𝑦)|2 𝑑𝑥𝑑𝑦, √ ℝ𝑑 1

so we obtain the result. 6.52. We can obtain the following inequality for convolutions. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑔 ∈ 𝐿2 (ℝ𝑑 ), then 𝑓 ∗ 𝑔 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑓 ∗ 𝑔‖𝐿2 ≤ ‖𝑓‖𝐿1 ‖𝑔‖𝐿2 . This follows by Minkowski’s inequality because 2

∫ |𝑓 ∗ 𝑔(𝑥)|2 𝑑𝑥 = ∫ || ∫ 𝑓(𝑦)𝑔(𝑥 − 𝑦)𝑑𝑦|| 𝑑𝑥 √ √ ≤∫

∫ |𝑓(𝑦)|2 |𝑔(𝑥 − 𝑦)|2 𝑑𝑥𝑑𝑦 √

∫ |𝑔(𝑥 − 𝑦)|2 𝑑𝑥𝑑𝑦 √ = ‖𝑓‖𝐿1 ‖𝑔‖𝐿2 , = ∫ |𝑓(𝑦)|

where have used the invariance of the 𝐿2 norm under translations.

134

6. Lebesgue integral and Lebesgue spaces

We can also prove analogous result in 𝐿2 to the convergence of convolution operators with good kernels discussed in 6.44, in 𝐿1 . We leave it as an exercise.

Exercises (1) Let 𝑓 ≥ 0 be measurable. Then ∫ 𝑓 = 0 if and only if 𝑓 = 0 a.e. (2) Let 𝑓 ≥ 0 be measurable and 𝐴 a set of measure 0. Then ∫𝐴 𝑓 = 0. (3) Let 𝑓 ≥ 0 be measurable. If ∫ 𝑓 < ∞, then 𝑓 < ∞ a.e. (4) If 𝑓𝑛 ≥ 0 are measurable and 𝑓𝑛 ↗ 𝑓 a.e., then ∫ 𝑓𝑛 → ∫ 𝑓. (5) If 𝑓𝑛 ≥ 0 are measurable, 𝑓𝑛 → 𝑓 and ∫ 𝑓 = lim ∫ 𝑓𝑛 < ∞, then ∫ 𝑓𝑛 → ∫ 𝑓 𝐴

𝐴

for all measurable 𝐴 ⊂ ℝ𝑑 . The statement is false if ∫ 𝑓 = lim ∫ 𝑓𝑛 = ∞. (6) Fatou’s lemma implies the monotone convergence theorem. (7) Complete the details of the proof of (6.24). (8) The set 𝐿1 (ℝ𝑑 ) of integrable functions is a vector space with the usual pointwise operations, and 𝑓 ↦ ∫ 𝑓 defines a linear functional on 𝐿1 (ℝ𝑑 ). (9) Prove Corollary 6.31. (10) If 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), then 𝑓 ∗ 𝑔 = 𝑔 ∗ 𝑓. (11) The set of equivalence clases 𝐿2 (ℝ𝑑 ) of square integrable functions forms a vector space, and (6.47) defines an inner product. (12) Prove Theorem 6.48. (13) Let Φ ∈ 𝐿1 (ℝ𝑑 ) with ∫ Φ = 1. If {Φ𝑡 }𝑡>0 are the dilations of Φ, then lim ‖Φ𝑡 ∗ 𝑓 − 𝑓‖𝐿2 = 0 𝑡→0

for any 𝑓 ∈ 𝐿2 (ℝ𝑑 ).

Notes

135

Notes Lebesgue developed what is now known as the Lebesgue integral in his papers cited in Chapter 5, as well as [Leb01a], [Leb01b], [Leb01c], his thesis [Leb02] and his text [Leb04], where he also proved the dominated convergence theorem for the special case when the functions are dominated on [0, 1] by a constant. The general case was proved in [Leb10]. Fatou’s lemma was stated and proved by Pierre Fatou in [Fat06]. Beppo Levi stated and proved the monotone convergence theorem in [Lev06]. The spaces 𝐿𝑝 (for 1 ≤ 𝑝 ≤ ∞, although we only discussed 𝑝 = 1, 2) were introduced by Frigyes Riesz in [Rie10], where he used the letter “𝐿” to denote them in honor to Lebesgue. Hermann Minkowski proved the inequality now known as Minkowski’s inequality for sums and series in [Min96], and the integral version is due to Riesz [Rie13].

Chapter 7

Maximal functions

7.1. Indefinite integrals and averages A well known result from calculus states that, if 𝑓 is Riemann-integrable on the interval [𝑎, 𝑏], then 1 ∫ ℎ→0 ℎ 𝑥

𝑥+ℎ

lim

𝑓 = 𝑓(𝑥)

at each point 𝑥 ∈ (𝑎, 𝑏) where 𝑓 is continuous. This says that, if 𝑥

𝐹(𝑥) = ∫ 𝑓 𝑎

is the indefinite integral of 𝑓, then 𝐹 is differentiable at each point 𝑥 where 𝑓 is continuous, and 𝐹 ′ (𝑥) = 𝑓(𝑥). This result is true if 𝑓 is Lebesgue integrable, of course, and, in fact, can be extended to functions on ℝ𝑑 if we replace difference quotients by averages. Proposition 7.1. Let 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and continuous at 𝑥 ∈ ℝ𝑑 . Then 1 ∫ |𝐵 𝑟→0 𝑟 (𝑥)| 𝐵

lim

𝑓 = 𝑓(𝑥).

𝑟 (𝑥)

Proof. Given 𝜀 > 0, let 𝛿 > 0 such that, if |𝑦 − 𝑥| < 𝛿, then |𝑓(𝑦) − 𝑓(𝑥)| < 𝜀. 137

138

7. Maximal functions

Then, if 𝑟 < 𝛿, | 1 ∫ | |𝐵 (𝑥)| 𝑟

𝐵𝑟 (𝑥)

𝑓 − 𝑓(𝑥)|| ≤

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

|𝑓(𝑦) − 𝑓(𝑥)|𝑑𝑦


0

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

𝑟 (𝑥)

|𝑓|,

7.2. The Hardy–Littlewood maximal function

139

for each 𝑥 ∈ ℝ𝑑 . Note that we are taking the supremum of the averages of |𝑓| over all balls centered at 𝑥, of all radii. If the set of averages is not bounded, then 𝑀𝑓(𝑥) = ∞. If 𝑓 is bounded, say, |𝑓(𝑥)| ≤ 𝐴 for all 𝑥 ∈ ℝ𝑑 (or almost everywhere), then 𝑀𝑓(𝑥) is finite at every 𝑥, of course, and |𝑀𝑓(𝑥)| ≤ 𝐴. Example 7.3. A priori, nothing guarantees that 𝑀𝑓 is finite in a positive measure set, or anywhere (see Exercises (2) and (3)). It is easy to see that, in general, 𝑀𝑓 ∉ 𝐿1 (ℝ𝑑 ), even if 𝑓 ∈ 𝐿1 (ℝ𝑑 ). For example, consider 𝑓 = 𝜒[0,1] on ℝ. For 𝑥 > 1, 𝑀𝑓(𝑥) ≥

1 ∫ 2𝑥 0

2𝑥

𝜒[0,1] =

1 , 2𝑥

and thus 𝑀𝑓 ∉ 𝐿1 (ℝ). However, Hardy and Littlewood proved that, indeed, 𝑀𝑓 is finite almost everywhere if 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and, even though it might not be integrable, Example 7.3 is essentially the worst case scenario. Theorem 7.4 (Hardy–Littlewood). There exists 𝐴 > 0 such that, for any 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝛼 > 0, |{𝑥 ∈ ℝ𝑑 ∶ 𝑀𝑓(𝑥) > 𝛼}| ≤

𝐴 ‖𝑓‖𝐿1 . 𝛼

We say then that the maximal function is an operator of weak type (1, 1) . We first observe that Theorem 7.4 states that the situation of Example 7.3 is pretty much the “worst case” we would get for 𝑀𝑓, as the function 1/𝑥 is the natural example of a measurable function 𝑓 that satisfies 𝐴 |{𝑥 ∈ ℝ ∶ |𝑓(𝑥)| > 𝛼}| = 𝛼 for some constant 𝐴. Indeed, we clearly have 1 2 |{𝑥 ∈ ℝ ∶ | | > 𝛼}| = 𝑥 𝛼 for any 𝛼 > 0, as the set on the left hand side is the interval (−𝛼, 𝛼).

140

7. Maximal functions In fact, any integrable function satisfies the above inequality, as |{𝑥 ∈ ℝ𝑑 ∶ |𝑓(𝑥)| > 𝛼}| = ∫

1𝑑𝑥

{𝑥∈ℝ𝑑 ∶|𝑓(𝑥)|>𝛼}

≤∫ {𝑥∈ℝ𝑑 ∶|𝑓(𝑥)|>𝛼}

|𝑓(𝑥)| 𝑑𝑥 𝛼

1 (7.5) ≤ ‖𝑓‖𝐿1 . 𝛼 The inequality (7.5) is called Chebyshev’s inequality. In the proof of Theorem 7.4, we will use Lemma 7.6, due to Vitali. Lemma 7.6 (Vitali). Let 𝐵1 , 𝐵2 , . . . , 𝐵𝑁 be a finite collection of balls in ℝ𝑑 . Then there exist disjoint 𝐵𝑖1 , 𝐵𝑖2 , . . . , 𝐵𝑖𝑘 among them such that 𝑁

𝑘

𝑑 | | | ⋃ 𝐵𝑗 | ≤ 3 ∑ |𝐵𝑖𝑗 |. 𝑗=1

𝑗=1

In other words, from any finite collection of balls we can obtain a disjoint subcollection such that, even as its union might be a smaller set than the union of the original balls, the volume of the new union is at least as large as a fixed proportion of the original total volume. Proof. Given 𝐵1 , 𝐵2 , . . . , 𝐵𝑁 , choose a ball of maximal radius, say 𝐵𝑖1 . Once 𝐵𝑖1 , 𝐵𝑖2 , . . . , 𝐵𝑖𝑗 are chosen, choose 𝐵𝑖𝑗+1 as a ball of maximal radius among all remaining balls disjoint to 𝐵𝑖1 , 𝐵𝑖2 , . . . , 𝐵𝑖𝑗 , until exhausting the collection. For each 𝐵𝑗 in the collection, there exists 𝑖𝑙 such that • 𝐵𝑖𝑙 intersects 𝐵𝑗 ; and • the radius of 𝐵𝑖𝑙 is at least as large as the radius of 𝐵𝑗 . Otherwise, 𝐵𝑗 would have been chosen in the construction. Thus, if 3𝐵𝑖𝑙 is the ball with the same center as 𝐵𝑖𝑙 and three times its radius, then 𝐵𝑗 ⊂ 3𝐵𝑖𝑙 (see Figure 7.1) and, thus, 𝑁

⋃ 𝑗=1

𝑘

𝐵𝑗 ⊂

⋃

3𝐵𝑖𝑗 .

𝑗=1

The lemma follows from the fact that |3𝐵𝑖𝑗 | = 3𝑑 |𝐵𝑖𝑗 |.

□

7.2. The Hardy–Littlewood maximal function

141

Figure 7.1. The ball 3𝐵𝑖𝑙 with the same center as 𝐵𝑖𝑙 and three times its radius contains all balls intersecting it with smaller or equal radii.

Proof of Theorem 7.4. As the measure of any measurable set is the supremum of the measures of its compact subsets, we let 𝐾 ⊂ {𝑥 ∈ ℝ𝑑 ∶ 𝑀𝑓(𝑥) > 𝛼} be compact. Hence, it is enough to estimate the measure of 𝐾. Now, by definition, for each 𝑥 ∈ 𝐾 there exists a ball 𝐵𝑥 centered at 𝑥 such that 1 ∫ |𝑓| > 𝛼. |𝐵𝑥 | 𝐵 𝑥

The balls 𝐵𝑥 , for 𝑥 ∈ 𝐾, cover the compact set 𝐾, so there exists a finite collection 𝐵1 , 𝐵2 , . . . , 𝐵𝑁 of them such that 𝑁

𝐾⊂

⋃

𝐵𝑗 .

𝑗=1

By Lemma 7.6, there exist disjoint 𝐵𝑖1 , 𝐵𝑖2 , . . . , 𝐵𝑖𝑘 among them such that 𝑁

𝑘

𝑑 | | | ⋃ 𝐵𝑗 | ≤ 3 ∑ |𝐵𝑖𝑗 |. 𝑗=1

𝑗=1

142

7. Maximal functions

Thus 𝑁

𝑘

|𝐾| ≤ || 𝐵 | ≤ 3𝑑 ∑ |𝐵𝑖𝑗 |. ⋃ 𝑗| 𝑗=1

𝑗=1

Now, for each of those balls, 1 ∫ |𝑓| > 𝛼, |𝐵𝑖𝑗 | 𝐵 𝑖𝑗

so |𝐵𝑖𝑗 |
𝛼, for any fixed positive 𝛼, is finite. 7.7. We can also consider, for 𝑓 ∈ 𝐿1loc (ℝ𝑑 ), the maximal function 1 ̃ ∫ |𝑓| ∶ 𝐵 is a ball with 𝑥 ∈ 𝐵}. 𝑀𝑓(𝑥) = sup { |𝐵| 𝐵 ̃ is the supremum of the averages of |𝑓| over all balls that That is, 𝑀𝑓 contain 𝑥, and not only those centered at 𝑥. If 𝑥 ∈ 𝐵 and 𝐵 has radius 𝑟, then 𝐵 ⊂ 𝐵2𝑟 (𝑥) and |𝐵2𝑟 (𝑥)| = 2𝑑 |𝐵|, and thus 1 2𝑑 ∫ |𝑓| ≤ ∫ |𝐵| 𝐵 |𝐵2𝑟 (𝑥)| 𝐵

|𝑓|,

2𝑟 (𝑥)

̃ ̃ also satisfies the concluand hence 𝑀𝑓(𝑥) ≤ 2𝑑 𝑀𝑓(𝑥). Therefore, 𝑀𝑓 sion of the Hardy–Littlewood maximal theorem.

7.3. The Lebesgue differentiation theorem

143

7.3. The Lebesgue differentiation theorem We are now ready to discuss the question posed in Section 7.1, on whether lim 𝑟→0

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

𝑓

𝑟 (𝑥)

exists for any 𝑥 ∈ ℝ𝑑 , if 𝑓 ∈ 𝐿1 (ℝ𝑑 ). In fact, it turns out that the limit exists almost everywhere, as stated by Lebesgue’s theorem. Theorem 7.8 (Lebesgue). If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then, for almost every 𝑥 ∈ ℝ𝑑 , lim 𝑟→0

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

𝑓 = 𝑓(𝑥).

𝑟 (𝑥)

Lebesgue’s theorem 7.8 states that the limit not only exists, but that it is actually equal to 𝑓 almost everywhere. This might seem remarkable when compared to Proposition 7.1, because 𝑓 could be discontinuous everywhere.1 Proof of Theorem 7.8. Let 𝐹 ⊂ ℝ𝑑 be the set where either the limit does not exist or is not equal to 𝑓. Thus, 𝐹 = {𝑥 ∈ ℝ𝑑 ∶ lim sup |𝐼𝑟 𝑓(𝑥) − 𝑓(𝑥)| > 0}. 𝑟→0

where we have written 𝐼𝑟 𝑓(𝑥) for the average of 𝑓 on the ball of radius 𝑟 around 𝑥, 1 ∫ 𝐼𝑟 𝑓(𝑥) = 𝑓. |𝐵𝑟 (𝑥)| 𝐵 (𝑥) 𝑟

We want to prove that |𝐹| = 0. As 𝐹 = ⋃𝑛 𝐹1/𝑛 , where 𝐹𝛼 = {𝑥 ∈ ℝ𝑑 ∶ lim sup |𝐼𝑟 𝑓(𝑥) − 𝑓(𝑥)| > 𝛼}, 𝑟→0

it is sufficient to prove that |𝐹𝛼 | = 0 for any 𝛼 > 0. Given 𝜀 > 0, let 𝑔 ∈ 𝐶𝑐 (ℝ𝑑 ) such that ‖𝑓 − 𝑔‖𝐿1 < 𝜀. By Proposition 7.1, lim 𝐼𝑟 𝑔(𝑥) = 𝑔(𝑥) 𝑟→0

1 Recall, however, that Riemann-integrable functions are indeed continuous almost everywhere (Theorem A.9).

144

7. Maximal functions

for every 𝑥 ∈ ℝ𝑑 , and thus lim sup|𝐼𝑟 𝑓(𝑥) − 𝑓(𝑥)| 𝑟→0

= lim sup |𝐼𝑟 (𝑓 − 𝑔)(𝑥) + 𝐼𝑟 𝑔(𝑥) − 𝑔(𝑥) + 𝑔(𝑥) − 𝑓(𝑥)| 𝑟→0

= lim sup |𝐼𝑟 (𝑓 − 𝑔)(𝑥)| + |𝑔(𝑥) − 𝑓(𝑥)| 𝑟→0

≤ 𝑀(𝑓 − 𝑔)(𝑥) + |𝑔(𝑥) − 𝑓(𝑥)|, where, in the last inequality, we have used the fact |𝐼𝑟 (𝑓 − 𝑔)(𝑥)| ≤

1 ∫ |𝐵𝑟 (𝑥)| 𝐵

|𝑓 − 𝑔| ≤ 𝑀(𝑓 − 𝑔)(𝑥).

𝑟 (𝑥)

Then |𝐹𝛼 | ≤ |{𝑥 ∶ 𝑀(𝑓 − 𝑔)(𝑥) >

𝛼 𝛼 }| + |{𝑥 ∶ |𝑔(𝑥) − 𝑓(𝑥)| > }|. 2 2

The first term is estimated by the Hardy–Littlewood theorem, |{𝑥 ∶ 𝑀(𝑓 − 𝑔)(𝑥) >

𝛼 𝐴 2𝐴 }| ≤ ‖𝑓 − 𝑔‖𝐿1 < 𝜀, 2 𝛼/2 𝛼

and the second term by Chebyshev’s inequality (7.5), |{𝑥 ∶ |𝑔(𝑥) − 𝑓(𝑥)| >

𝛼 2 1 }| ≤ ‖𝑔 − 𝑓‖𝐿1 < 𝜀 2 𝛼/2 𝛼

Therefore 2(𝐴 + 1) 𝜀, 𝛼 and the theorem follows because 𝜀 > 0 is arbitrary. |𝐹𝛼 |
0 such that, for every 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and (𝑥, 𝑡) ∈ ℝ𝑑+1 + , |𝑢(𝑥, 𝑡)| ≤ 𝐴𝑀𝑓(𝑥). Hence Poisson integrals are uniformly estimated from above by the maximal function. Proof. We estimate the Poisson integral 𝑢(𝑥, 𝑡) by integrating over dyadic annuli around the point 𝑥. Indeed, since 𝑓 ∈ 𝐿1 (ℝ𝑑 ), for each (𝑥, 𝑡) ∈ ℝ𝑑+1 we can write + ∞

𝑢(𝑥, 𝑡) = ∫

𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦 + ∑ ∫

𝐵𝑡 (𝑥)

𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦,

𝑘=0 𝐴𝑘 (𝑥)

where each 𝐴𝑘 (𝑥) is the annulus 𝐴𝑘 (𝑥) = {𝑦 ∈ ℝ𝑑 ∶ 2𝑘 𝑡 ≤ |𝑥 − 𝑦| < 2𝑘+1 𝑡}. Now 𝑃𝑡 (𝑥 − 𝑦) =

2 2𝑡 ≤ , (𝑑+1)/2 2 2 𝜔𝑑+1 𝑡𝑑 𝜔𝑑+1 (|𝑥 − 𝑦| + 𝑡 )

so the first integral in the sum above is estimated by |∫ |

𝐵𝑡 (𝑥)

𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦|| ≤

2 𝜔𝑑+1 𝑡𝑑

∫

|𝑓(𝑦)|𝑑𝑦 ≤ 𝑐𝑀𝑓(𝑥),

𝐵𝑡 (𝑥)

where 𝑐 > 0 is a constant that depends only on the dimension 𝑑. Similarly, if |𝑥 − 𝑦| ≥ 2𝑘 𝑡, we have 𝑃𝑡 (𝑥 − 𝑦) ≤

2𝑡 2 ≤ , 𝜔𝑑+1 |𝑥 − 𝑦|𝑑+1 𝜔𝑑+1 2𝑘(𝑑+1) 𝑡𝑑

146

7. Maximal functions

and thus |∫ |

𝐴𝑘 (𝑥)

𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦|| ≤

2 ∫ 𝜔𝑑+1 2𝑘(𝑑+1) 𝑡𝑑 𝐵 𝑘+1 2

|𝑓(𝑦)|𝑑𝑦 (𝑥) 𝑡

𝑐′ ≤ 𝑘 𝑀𝑓(𝑥), 2 ′ where 𝑐 > 0 is another constant that depends only on 𝑑. Therefore ∞

𝑐′ 𝑀𝑓(𝑥) = 𝐴𝑀𝑓(𝑥), 2𝑘 𝑘=0

|𝑢(𝑥, 𝑡)| ≤ 𝑐𝑀𝑓(𝑥) + ∑

□

where 𝐴 = 𝑐 + 2𝑐′ .

We can now state the following result on pointwise boundary limits of Poisson integrals. Theorem 7.12. Lef 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑢(𝑥, 𝑡) its Poisson integral. Then lim 𝑢(𝑥, 𝑡) = 𝑓(𝑥) 𝑡→0

𝑑

for almost every 𝑥 ∈ ℝ . The proof of Theorem 7.12 follows as the proof of Theorem 7.8, by showing that the set {𝑥 ∈ ℝ𝑑 ∶ lim sup |𝑢(𝑥, 𝑡) − 𝑓(𝑥)| > 𝛼} 𝑡→0

has measure 0 for any 𝛼 > 0, by first approximating with a compactly supported continuous function and then comparing with the maximal function, using Proposition 7.11. We leave the details as an exercise (Exercise (5)). If we compare Theorem 7.12 with Theorem 4.13, we see that this time we are only approaching the boundary point (𝑥, 0) vertically, as we only consider 𝑢(𝑥, 𝑡) as 𝑡 > 0, while in Theorem 4.13 we approach (𝑥, 0) from any direction in the upper half-space. We can extend Theorem 7.12 if we consider nontangential limits. For 𝑥 ∈ ℝ𝑑 , the cone of aperture 𝜃 > 0 over 𝑥 is defined as the set Γ𝜃 (𝑥) = {(𝑦, 𝑡) ∈ ℝ𝑑+1 ∶ |𝑥 − 𝑦| < 𝜃𝑡}. + See Figure 7.2. Thus, we consider the limit of 𝑢(𝑦, 𝑡) when we approach (𝑥, 0) within this cone. We first prove Theorem 7.13.

7.4. Boundary limits of harmonic functions

147

Figure 7.2. The cone Γ𝜃 (𝑥) over the point 𝑥 ∈ ℝ𝑑 .

Theorem 7.13. For any 𝜃 > 0, there exists 𝐴𝜃 > 0 such that, for any 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑥 ∈ ℝ𝑑 , |𝑢(𝑦, 𝑡)| ≤ 𝐴𝜃 𝑀𝑓(𝑥) for all (𝑦, 𝑡) ∈ Γ𝜃 (𝑥). Proof. By Proposition 7.11, the theorem follows once we prove that there exists 𝑐 𝜃 > 0 such that, for all 𝑧 ∈ ℝ𝑑 , (7.14)

𝑃𝑡 (𝑦 − 𝑧) ≤ 𝑐 𝜃 𝑃𝑡 (𝑥 − 𝑧),

whenever (𝑦, 𝑡) ∈ Γ𝜃 (𝑥). For this, fix 𝑥, 𝑧 ∈ ℝ𝑑 and (𝑦, 𝑡) ∈ Γ𝜃 (𝑥). If |𝑥 − 𝑧| ≥ 2|𝑥 − 𝑦|, then 1 1 |𝑦 − 𝑧| = |𝑥 − 𝑧 + 𝑦 − 𝑥| ≥ |𝑥 − 𝑧| − |𝑦 − 𝑥| ≥ |𝑥 − 𝑧| − |𝑥 − 𝑧| = |𝑥 − 𝑧|, 2 2 and hence 1 1 |𝑦 − 𝑧|2 + 𝑡2 ≥ |𝑥 − 𝑧|2 + 𝑡2 ≥ (|𝑥 − 𝑧|2 + 𝑡2 ), 4 4 so 2𝑡 𝑃𝑡 (𝑦 − 𝑧) = 𝜔𝑑+1 (|𝑦 − 𝑧|2 + 𝑡2 )(𝑑+1)/2 2𝑡 = 2𝑑+1 𝑃𝑡 (𝑥 − 𝑧). ≤ (𝑑+1)/2 2 2 𝜔𝑑+1 ((|𝑥 − 𝑧| + 𝑡 )/4)

148

7. Maximal functions

If |𝑥 − 𝑧| < 2|𝑥 − 𝑦|, we have |𝑥 − 𝑧| < 2𝜃𝑡 beacuse (𝑦, 𝑡) ∈ Γ𝜃 (𝑥), and thus 1 1 1 1 |𝑦 − 𝑧|2 + 𝑡2 ≥ 𝑡2 = 𝑡2 + 𝑡2 ≥ 2 |𝑥 − 𝑧|2 + 𝑡2 ≥ 𝑐(|𝑥 − 𝑧|2 + 𝑡2 ), 2 2 2 8𝜃 where 𝑐 = min{1/8𝜃2 , 1/2}, and thus, as above, 𝑃𝑡 (𝑦 − 𝑧) ≤

1

𝑃𝑡 (𝑥 𝑐(𝑑+1)/2

− 𝑧).

We obtain the inequality (7.14) with 𝑐 𝜃 = max{2𝑑+1 , 1/𝑐(𝑑+1)/2 }.

□

We then have the following result on the existence of nontangential limits. Corollary 7.15. Lef 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑢(𝑥, 𝑡) its Poisson integral. Then, for any 𝜃 > 0, lim 𝑢(𝑥, 𝑡) = 𝑓(𝑥) (𝑦,𝑡)→(𝑥,0) (𝑦,𝑡)∈Γ𝜃 (𝑥)

for almost every 𝑥 ∈ ℝ𝑑 . We leave its proof as an exercise (Exercise (10)).

Exercises (1) The set 𝐿1loc (ℝ𝑑 ) of locally integrable functions is a vector space. (2) If 𝑓 ∈ 𝐿1loc (ℝ𝑑 ), then 𝑀𝑓 might be infinite at every point. (3) If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then 𝑀𝑓 might be infinite at some points. (4) Define on ℝ the function 1 𝑓(𝑥) = { |𝑥|(log |𝑥|)2 0

if |𝑥| ≤ 1/2 otherwise.

Then 𝑓 ∈ 𝐿1 (ℝ), but 𝑀𝑓 ∉ 𝐿1loc (ℝ). (5) Prove Theorem 7.12. (6) We say that {𝐾𝑡 }𝑡>0 is a family of better kernels if it satisfies • ∫ 𝐾𝑡 (𝑥)𝑑𝑥 = 1 for all 𝑡 > 0; ℝ𝑑

Exercises

149

• there exists 𝐴 > 0 such that |𝐾𝑡 (𝑥)| ≤ 𝑡 > 0; and • there exists 𝐴′ > 0 such that |𝐾𝑡 (𝑥)| ≤

𝐴 for all 𝑥 ∈ ℝ𝑑 and 𝑡𝑑 𝐴′ 𝑡 for all 𝑥 ∈ ℝ𝑑 |𝑥|𝑑+1

and 𝑡 > 0. (a) If Φ ∈ 𝐿1 (ℝ𝑑 ), ∫ Φ = 1 and |Φ(𝑥)| ≤ 𝐴/(1 + |𝑥|)𝑑+1 , then its dilations {Φ𝑡 }𝑡>0 form a family of better kernels. (b) If {𝐾𝑡 }𝑡>0 is a family of better kernels, then it is a family of good kernels. (c) If {𝐾𝑡 }𝑡>0 is a family of better kernels, then there exists a constant 𝑐 > 0 such that, if 𝑓 ∈ 𝐿1 (ℝ𝑑 ), |𝐾𝑡 ∗ 𝑓(𝑥)| ≤ 𝑐𝑀𝑓(𝑥) for all 𝑥 ∈ ℝ𝑑 and 𝑡 > 0. (d) If {𝐾𝑡 }𝑡>0 is a family of better kernels and 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then lim 𝐾𝑡 ∗ 𝑓(𝑥) = 𝑓(𝑥) 𝑡→0

for almost every 𝑥 ∈ ℝ𝑑 . (7) The results of the previous exercise are still true if we change the third hypothesis by |𝐾𝑡 (𝑥)| ≤

𝐴′ 𝑡𝜀 , |𝑥|𝑑+𝜀

for some 𝜀 > 0. (8) If {𝐾𝑡 }𝑡>0 is a family of better kernels and 𝑓 ∈ 𝐿1 (ℝ𝑑 ) is continuous at 𝑥, then lim 𝐾𝑡 ∗ 𝑓(𝑥) = 𝑓(𝑥). 𝑡→0

(Hint: Write 𝑓 = 𝑓 ⋅ 𝜒𝐵𝛿 (𝑥) + 𝑓 ⋅ (1 − 𝜒𝐵𝛿 (𝑥) ), where 𝐵𝛿 (𝑥) is a ball on which 𝑓 is bounded.) (9) Let {𝐾𝑡 }𝑡>0 be a family that satisfies the hypothesis of a family of better kernels except that, instead of the first hypothesis in Exercise (6), it satisfies, for some 𝜆 ∈ ℂ, ∫ 𝐾𝑡 (𝑥)𝑑𝑥 = 𝜆 ℝ𝑑

for all 𝑡 > 0. Then lim 𝐾𝑡 ∗ 𝑓(𝑥) = 𝜆𝑓(𝑥) 𝑡→0

150

7. Maximal functions for almost every 𝑥 ∈ ℝ𝑑 .

(10) Prove Corollary 7.15. (11) Let {𝐾𝑡 }𝑡>0 be a family of better kernels and 𝜃 > 0. (a) There exists a constant 𝑐 𝜃 > 0 such that, for 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑥 ∈ ℝ𝑑 , |𝐾𝑡 ∗ 𝑓(𝑦)| ≤ 𝑐 𝜃 𝑀𝑓(𝑥) for all (𝑦, 𝑡) ∈ Γ𝜃 (𝑥). (b) If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then lim

(𝑦,𝑡)→(𝑥,0) (𝑦,𝑡)∈Γ𝜃 (𝑥)

𝐾𝑡 ∗ 𝑓(𝑦) = 𝑓(𝑥)

for almost every 𝑥 ∈ ℝ𝑑 .

Notes The maximal function was introduced by Hardy and Littlewood in [HL30], where they proved Theorem 7.4 for the 1-dimensional case. They also proved Theorem 7.11 in the same paper, for Poisson integrals in the circle. Lemma 7.6 appeared in the paper [Vit08] by Giuseppe Vitali, in the 1-dimensional case. The 𝑑-dimensional case, and the proof presented here, is due to Stephan Banach [Ban24]. Lebesgue’s differentiation theorem 7.8 appeared in the one variable case in his book [Leb04]. A proof using Vitali’s lemma appeared in [Leb10], and the proof using the Hardy–Littlewood maximal theorem is due to Riesz [Rie32]. Our discussion on nontangential limits can be found in [Ste70].

Chapter 8

Fourier transform

8.1. Integrable functions In this chapter we discuss the representation of a function on ℝ𝑑 in terms of its Fourier transform, analogously to the Fourier series representation of a function in a circle discussed in previous chapters. The Fourier transform of a function 𝑓 ∈ 𝐿1 (ℝ𝑑 ) is defined as ̂ = ∫ 𝑓(𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥, 𝑓(𝜉)

(8.1)

ℝ𝑑

for each 𝜉 ∈ ℝ𝑑 . The integral above converges since 𝑓, and thus the function 𝑥 ↦ 𝑓(𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 , is integrable for each 𝜉. 8.2. For any 𝑓 ∈ 𝐿1 (ℝ𝑑 ), 𝑓 ̂ is a bounded continuous function. Indeed, ̂ |𝑓(𝜉)| ≤ ∫ |𝑓(𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 |𝑑𝑥 = ‖𝑓‖𝐿1 , ℝ𝑑

and, if 𝜉 → 𝜉0 , then ̂ − 𝑓(𝜉 ̂ 0 )| ≤ ∫ |𝑓(𝑥)| |𝑒−2𝜋𝑖𝑥⋅𝜉 − 𝑒−2𝜋𝑖𝑥⋅𝜉0 |𝑑𝑥 → 0, |𝑓(𝜉) ℝ𝑑

by the dominated convergence theorem. 8.3. The Fourier transform defines on 𝐿1 (ℝ𝑑 ) a linear operator, i.e. 151

152

8. Fourier transform ˆ ̂ • for each 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝛼 ∈ ℂ, 𝛼𝑓(𝜉) = 𝛼𝑓(𝜉); and 1 𝑑 ̂ + 𝑔(𝜉). • for 𝑓, 𝑔 ∈ 𝐿 (ℝ ), 𝑓ˆ + 𝑔(𝜉) = 𝑓(𝜉) ̂

By 8.2, this operator continuously maps 𝐿1 (ℝ𝑑 ) into the space 𝐶𝐵 (ℝ𝑑 ) of bounded continuous functions with the uniform norm. Example 8.4. If 𝑓 = 𝜒[−1,1] in ℝ, then, for 𝜉 ≠ 0, 1

2𝜋𝑖𝜉 sin(2𝜋𝜉) − 𝑒−2𝜋𝑖𝜉 ̂ = ∫ 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 = 𝑒 𝑓(𝜉) = . 2𝜋𝑖𝜉 𝜋𝜉 −1

In particular, 𝑓 ̂ ∉ 𝐿1 (ℝ𝑑 ). Although the Fourier transform of a function 𝑓 ∈ 𝐿1 (ℝ𝑑 ) is not necessarily an integrable function, as in Example 8.4, we observe that, in this ̂ example, 𝑓(𝜉) → 0 as |𝜉| → ∞. This property is true for all functions 𝑓 ∈ 𝐿1 (ℝ𝑑 ). ̂ Proposition 8.5 (Riemann–Lebesgue lemma). If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then 𝑓(𝜉) → 0 as |𝜉| → ∞. Proof. One can check explicitly that, if 𝑅 is a rectangle in ℝ𝑑 , then 𝜒 ˆ𝑅 (𝜉) → 0

as |𝜉| → ∞.

(Exercise (2)). If 𝐴 ⊂ ℝ𝑑 is a finite measure set, by Corollary 5.17(4), for any 𝜀 > 0 there exist cubes 𝑄1 , . . . , 𝑄𝑛 such that |𝐴△ ⋃ 𝑄𝑗 | < 𝜀, so ‖𝜒𝐴 − 𝜒⋃ 𝑄𝑗 ‖𝐿1 < 𝜀 and thus |ˆ 𝜒𝐴 (𝜉) − 𝜒ˆ ⋃ 𝑄𝑗 (𝜉)| < 𝜀. Since we can write 𝑁

𝜒⋃ 𝑄𝑗 = ∑ 𝜒𝑅𝑗 , 𝑗=1

where each 𝑅𝑗 is a rectangle, we have that 𝜒ˆ ⋃ 𝑄𝑗 (𝜉) → 0 as |𝜉| → ∞, and thus 𝜒 ˆ 𝐴 (𝜉) → 0 as |𝜉| → ∞ beacuse 𝜀 > 0 above is arbitrary. We ̂ thus have that, for a simple 𝜙 ∈ 𝐿1 (ℝ𝑑 ), 𝜙(𝜉) → 0 as |𝜉| → ∞, because an integrable simple function is the linear combination of characteristic functions of finite measure sets. We thus get the proposition for every 𝑓 ∈ 𝐿1 (ℝ𝑑 ), because the simple functions are dense in 𝐿1 (ℝ𝑑 ). □ The Riemann–Lebesgue lemma implies that the Fourier transform defines a linear operator on 𝐿1 (ℝ𝑑 ) into 𝐶0 (ℝ𝑑 ).

8.1. Integrable functions

153

8.6. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ), the Fourier transform of its dilation 𝑓𝑡 (𝑥) = 𝑡−𝑑 𝑓(𝑥/𝑡), for 𝑡 > 0, is given by ̂ 𝑓ˆ𝑡 (𝜉) = 𝑓(𝑡𝜉). This follows from the dilation property of Lebesgue measure. Indeed, 1 𝑓ˆ𝑡 (𝜉) = ∫ 𝑓𝑡 (𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = 𝑑 ∫ 𝑓(𝑥/𝑡)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 𝑡 ℝ𝑑 ℝ𝑑 ̂ = ∫ 𝑓(𝑦)𝑒−2𝜋𝑖𝑡𝑦⋅𝜉 𝑑𝑦 = 𝑓(𝑡𝜉), ℝ𝑑

where we have applied the dilation 𝑥 ↦ 𝑡𝑦. 8.7. The translation invariance of Lebesgue measure on ℝ𝑑 further implies, for 𝑓 ∈ 𝐿1 (ℝ𝑑 ), • if 𝑓ℎ (𝑥) = 𝑓(𝑥 − ℎ) for some ℎ ∈ ℝ𝑑 , then ˆℎ (𝜉) = 𝑒−2𝜋𝑖𝑥⋅ℎ 𝑓(𝜉); ̂ 𝑓 and • if 𝑔(𝑥) = 𝑒2𝜋𝑖𝑥⋅ℎ 𝑓(𝑥) for some ℎ ∈ ℝ𝑑 , then ̂ − ℎ). 𝑔(𝜉) ̂ = 𝑓(𝜉 We leave the proof of these properties as an exercise (Exercise (3)). 2

Example 8.8. Consider the function 𝐺(𝑥) = 𝑒−𝜋|𝑥| in ℝ𝑑 . Then 2 ̂ 𝐺(𝜉) = 𝑒−𝜋|𝜉| = 𝐺(𝜉),

so 𝐺 is equal to its own Fourier transform. To prove this by Fubini’s theorem it is enough to show that ∞ 2

∫ 𝑒−𝜋𝑥 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 = 𝑒−𝜋𝜉

2

−∞ 2

2

2

2

for each 𝜉 ∈ ℝ, because 𝑒−𝜋|𝑥| = 𝑒−𝜋𝑥1 𝑒−𝜋𝑥2 ⋯ 𝑒−𝜋𝑥𝑑 . Since ∞

𝑁 2

2

∫ 𝑒−𝜋𝑥 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 = lim ∫ 𝑒−𝜋𝑥 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 𝑁→∞

−∞

−𝑁 𝑁

2

2

= 𝑒−𝜋𝜉 lim ∫ 𝑒−𝜋(𝑥+𝑖𝜉) 𝑑𝑥, 𝑁→∞

−𝑁

it remains to verify that 𝑁

𝑁 2

2

lim ∫ 𝑒−𝜋(𝑥+𝑖𝜉) 𝑑𝑥 = lim ∫ 𝑒−𝜋𝑥 𝑑𝑥,

𝑁→∞

−𝑁

𝑁→∞

−𝑁

154

8. Fourier transform

as we know that the integral on the right hand side is equal to 1. The above identity is achieved by considering the contour integral 2

∫ 𝑒−𝜋𝑧 𝑑𝑧 = 0 𝛾

over the rectangle 𝛾 with vertices 𝑁, 𝑁 + 𝑖𝜉, −𝑁 + 𝑖𝜉 and −𝑁. See Figure 8.1. We leave the details as an exercise (Exercise (4)).

Figure 8.1. The contour 𝛾 to obtain the Fourier transform of the function 𝐺(𝑥).

8.9. The function 𝐺(𝑥) of Example 8.8 is called the Gaussian kernel, and is related to the heat kernel introduced in Exercise (10) of Chapter 4, 𝐻𝑡 (𝑥) =

1 2 𝑒−|𝑥| /4𝑡 . 𝑑/2 (4𝜋𝑡)

Note that 𝐻𝑡 correspond to the dilation 𝐺√4𝜋𝑡 (𝑥) of the Gaussian. By the dilation property 8.6 of the Fourier transform we have that ˆ𝑡 (𝜉) = 𝑒−4𝜋2 𝑡|𝜉|2 . 𝐻 Example 8.10. Consider now, for a fixed 𝑡 > 0, the Poisson kernel 𝑃𝑡 (𝑥). Then ˆ𝑡 (𝜉) = 𝑒−2𝜋𝑡|𝜉| . 𝑃 To show this, recall that 𝑃𝑡 is the dilation of 𝑃1 , so it is sufficient to prove that 𝑃ˆ1 (𝜉) = 𝑒−2𝜋|𝜉| . We calculate Γ((𝑑 + 1)/2) 𝑒−2𝜋𝑖𝑥⋅𝜉 ∫ 𝑃ˆ1 (𝜉) = ∫ 𝑃1 (𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = 𝑑𝑥, (𝑑+1)/2 (𝑑+1)/2 2 𝜋 ℝ𝑑 ℝ𝑑 (|𝑥| + 1)

8.1. Integrable functions

155

where we have used the fact 2𝜋(𝑑+1)/2 . Γ((𝑑 + 1)/2)

𝜔𝑑+1 = Now, by the identity

∞

1 1 𝑑𝑠 2 ∫ 𝑠(𝑑+1)/2 𝑒−(|𝑥| +1)𝑠 , = (𝑑+1)/2 2 𝑠 Γ((𝑑 + 1)/2) 0 (|𝑥| + 1) (see Exercise (2) of Chapter 4) and Fubini’s theorem we obtain 𝑃ˆ1 (𝑥) = =

∞

1 𝜋(𝑑+1)/2

∫ 𝑒−2𝜋𝑖𝑥⋅𝜉 ( ∫ 𝑠(𝑑+1)/2 𝑒−(|𝑥| ℝ𝑑 ∞

2 +1)𝑠

0

𝑑𝑠 )𝑑𝑥 𝑠

1 2 ∫ 𝑠(𝑑−1)/2 𝑒−𝑠 ( ∫ 𝑒−2𝜋𝑖𝑥⋅𝜉 𝑒−|𝑥| 𝑠 𝑑𝑥)𝑑𝑠. 𝜋(𝑑+1)/2 0 𝑑 ℝ

From 8.8 and the dilation 𝑥 ↦ (√𝜋/𝑠)𝑦, the integral inside is 2

∫ 𝑒−2𝜋𝑖𝑥⋅𝜉 𝑒−|𝑥| 𝑠 𝑑𝑥 = ℝ𝑑

and hence

∞

1 ∫ 𝑃ˆ1 (𝑥) = √𝜋 0

1 √𝑠

𝜋𝑑/2 −𝜋2 |𝜉|2 /𝑠 𝑒 , 𝑠𝑑/2

𝑒−𝑠 𝑒−𝜋

2 |𝜉|2 /𝑠

𝑑𝑠.

We calculate the last integral in two steps: (I) For any 𝑢 > 0, ∞

∞

𝑒−2𝜋𝑖ᵆ𝑣 1 1 ∫ ∫ 𝑑𝑣 = 𝜋 −∞ 1 + 𝑣2 √𝜋 0

1 √𝑠

𝑒−𝑠 𝑒−𝜋

2 ᵆ2 /𝑠

(II) For any 𝑢 > 0, ∞

1 𝑒−2𝜋𝑖ᵆ𝑣 ∫ 𝑑𝑣 = 𝑒−2𝜋ᵆ . 𝜋 −∞ 1 + 𝑣2 Step (I) follows from the identities ∞

1 2 = ∫ 𝑒−(1+𝑣 )𝑠 𝑑𝑠 1 + 𝑣2 0 and

∞ 2

∫ 𝑒−2𝜋𝑖ᵆ𝑣 𝑒−𝑣 𝑠 𝑑𝑣 = −∞

𝜋 −𝜋2 ᵆ2 /𝑠 𝑒 . √𝑠

𝑑𝑠;

156

8. Fourier transform

The first is a straightforward calculation, while the second follows from Example 8.8 in the 1-dimensional case and the dilation property 8.6 of the Fourier transform. For Step (II), we calculate the contour integral ∫ 𝛾

𝑒−2𝜋𝑖ᵆ𝑧 𝑑𝑧 1 + 𝑧2

over the lower semicircle around the origin of radius 𝑁, and we let 𝑁 → ∞. See Figure 8.2. We leave the details of Steps (I) and (II) as an exercise (Exercise (5)).

Figure 8.2. The contour 𝛾 to obtain the Fourier transform of the Poisson kernel 𝑃1 (𝑥). Note that the function 𝑧 ↦ 𝑒−2𝜋𝑖ᵆ𝑧 /(1 + 𝑧2 ) has a pole at −𝑖.

8.2. The Fourier inversion formula A natural question to ask is whether one can recover a function from its Fourier transform. Recall that the expansion formula 𝑖𝑛𝜃 ̂ 𝑓(𝜃) = ∑ 𝑓(𝑛)𝑒 𝑛∈ℤ

of a function on the circle in terms of its Fourier series holds for appropiate continuous functions. In fact, it holds whenever the series ̂ ∑ |𝑓(𝑛)| 𝑛∈ℤ

converges, so the Fourier series of 𝑓 is absolutely convergent.

8.2. The Fourier inversion formula

157

Example 8.8, as well as the convergence of the Fourier series above, suggests that, if 𝑓 ̂ is integrable, then we should have the following Fourier inversion formula, 2𝜋𝑖𝑥⋅𝜉 ̂ 𝑓(𝑥) = ∫ 𝑓(𝜉)𝑒 𝑑𝜉.

(8.11)

ℝ𝑑

It is indeed the case. Theorem 8.12. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ) as well as 𝑓 ̂ ∈ 𝐿1 (ℝ𝑑 ), then the Fourier inversion formula (8.11) holds for almost every 𝑥 ∈ ℝ𝑑 . In the proof of Theorem 8.12 we will use Lemma 8.13. Lemma 8.13. If 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), then ∫ 𝑓𝑔̂ = ∫ 𝑓𝑔.̂ ℝ𝑑

ℝ𝑑

Proof. We first observe that both integrals converge because 𝑓 and 𝑔 are integrable, and both 𝑓 ̂ and 𝑔̂ are continuous and bounded. Also, if we define 𝐹(𝑥, 𝑦) = 𝑓(𝑥)𝑔(𝑦)𝑒−2𝜋𝑖𝑥⋅𝑦 𝑑 𝑑 on ℝ × ℝ , then 𝐹 is integrable, because ∫

|𝐹| ≤ ∫ |𝑓(𝑥)|𝑑𝑥 ∫ |𝑔(𝑦)|𝑑𝑦 < ∞.

ℝ𝑑 ×ℝ𝑑

ℝ𝑑

ℝ𝑑

Thus, by Fubini’s theorem, ∫ ℝ𝑑 ×ℝ𝑑

𝐹 = ∫ ( ∫ 𝐹(𝑥, 𝑦)𝑑𝑥)𝑑𝑦 ℝ𝑑

ℝ𝑑

= ∫ ( ∫ 𝑓(𝑥)𝑒−2𝜋𝑖𝑥⋅𝑦 𝑑𝑥)𝑔(𝑦)𝑑𝑦 ℝ𝑑

ℝ𝑑

̂ = ∫ 𝑓(𝑦)𝑔(𝑦)𝑑𝑦, ℝ𝑑

and ∫ ℝ𝑑 ×ℝ𝑑

𝐹 = ∫ ( ∫ 𝐹(𝑥, 𝑦)𝑑𝑦)𝑑𝑥 ℝ𝑑

ℝ𝑑

= ∫ 𝑓(𝑥)( ∫ 𝑔(𝑦)𝑒−2𝜋𝑖𝑥⋅𝑦 𝑑𝑥)𝑑𝑥 ℝ𝑑

ℝ𝑑

= ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥. ̂ ℝ𝑑

158

8. Fourier transform □

Proof of Theorem 8.12. For each 𝑥 ∈ ℝ𝑑 and 𝑡 > 0, consider the function 2 𝑔(𝜉) = 𝑒−𝜋𝑡|𝜉| 𝑒2𝜋𝑖𝑥⋅𝜉 . By 8.6, 8.7 and 8.8, its Fourier transform is the Gaussian kernel 𝑔(𝑦) ̂ =

1 𝑡𝑑/2

2

𝑒−𝜋|𝑥−𝑦| /𝑡 .

The collection {𝐾𝑡 }𝑡>0 given by 1 −𝜋|𝑥|2 /𝑡 𝑒 𝑡𝑑/2 is a collection of better kernels (see Exercise (6) of Chapter 7; see also Exercise (6) of this chapter), because it is the collection of dilations of the Gaussian kernel seen in 8.8. By Lemma 8.13 with 𝑓 and 𝑔, we obtain 𝐾𝑡 (𝑥) =

(8.14)

−𝜋𝑡|𝜉|2 2𝜋𝑖𝑥⋅𝜉 ̂ ∫ 𝑓(𝜉)𝑒 𝑒 𝑑𝜉 = ∫ 𝑓(𝑦)𝐾𝑡 (𝑥 − 𝑦)𝑑𝑦, ℝ𝑑

ℝ𝑑

where we have again used 8.7. Since 𝑓 ̂ ∈ 𝐿1 (ℝ𝑑 ), the left side of (8.14) converges to 2𝜋𝑖𝑥⋅𝜉 ̂ ∫ 𝑓(𝜉)𝑒 𝑑𝜉 ℝ𝑑

as 𝑡 → 0, by the dominated convergence theorem. Since the collection {𝐾𝑡 }𝑡>0 is a collection of better kernels, ∫ 𝑓(𝑦)𝐾𝑡 (𝑥 − 𝑦)𝑑𝑦 = 𝑓 ∗ 𝐾𝑡 (𝑥) → 𝑓(𝑥) ℝ𝑑

for almost every 𝑥 ∈ ℝ𝑑 , by Exercise (6) of Chapter 7.

□

8.15. Note that, in particular, if 𝑓 and 𝑓 ̂ are integrable and 𝑓 is continuous at 𝑥 ∈ ℝ𝑑 , then 2𝜋𝑖𝑥⋅𝜉 ̂ 𝑓(𝑥) = ∫ 𝑓(𝜉)𝑒 𝑑𝜉. ℝ𝑑

(Exercise (8) of Chapter 7.) Example 8.16. Since the Fourier transform of the Poisson kernel 𝑃𝑡 (𝑥) is given by ˆ𝑡 (𝜉) = 𝑒−2𝜋𝑡|𝜉| , 𝑃

8.3. Mean-square convergence

159

ˆ𝑡 is integrable, we have that by 8.10, and clearly 𝑃 ∫ 𝑒−2𝜋𝑡|𝜉| 𝑒2𝜋𝑖𝑥⋅𝜉 𝑑𝜉 = 𝑃𝑡 (𝑥), ℝ𝑑

for every 𝑥 ∈ ℝ𝑑 . Moreover, since both 𝑃𝑡 (𝑥) = 𝑃𝑡 (−𝑥) and 𝑒−2𝜋𝑡|𝑥| = 𝑒−2𝜋𝑡|−𝑥| , we see that the Fourier transform of 𝑒−2𝜋𝑡|𝑥| is 𝑃𝑡 (𝜉), as well. Theorem 8.12 leads to the question of whether the Fourier transform of a given function is integrable. In general, we have seen that the Fourier transform is continuous and has limit 0 at infinity but, in general, it’s not a function in 𝐿1 (ℝ𝑑 ), as the simple example of a characteristic function shows (Example 8.4). However, an application of the limit results above and the monotone convergence theorem give us the following test. ̂ Proposition 8.17. Let 𝑓 ∈ 𝐿1 (ℝ𝑑 ) be continuous at 0. If 𝑓(𝜉) ≥ 0 for 𝑑 1 𝑑 ̂ every 𝜉 ∈ ℝ , then 𝑓 ∈ 𝐿 (ℝ ). Proof. From equation (8.14) (which is true for all 𝑓 ∈ 𝐿1 (ℝ𝑑 )) and the hypothesis that 𝑓 is continuous at 0, we have that −𝜋𝑡|𝜉|2 ̂ lim ∫ 𝑓(𝜉)𝑒 𝑑𝜉 = 𝑓(0). 𝑡→0

ℝ𝑑

̂ ≥ 0, we clearly have that Also, since 𝑓(𝜉) −𝜋𝑡|𝜉|2 ̂ ̂ 𝑓(𝜉)𝑒 ↗ 𝑓(𝜉)

as 𝑡 → 0 so, by the monotone convergence theorem, −𝜋𝑡|𝜉|2 ̂ ̂ ∫ 𝑓(𝜉)𝑑𝜉 = lim ∫ 𝑓(𝜉)𝑒 𝑑𝜉. ℝ𝑑

𝑡→0

ℝ𝑑

Therefore, 𝑓 ̂ ∈ 𝐿1 (ℝ𝑑 ) and, in fact, ‖𝑓‖̂ 𝐿1 = 𝑓(0).

□

8.3. Mean-square convergence The Fourier transform is only defined for 𝑓 ∈ 𝐿1 (ℝ𝑑 ), as we require that the integral (8.1) converges. However, it is possible to extend the definition of the Fourier transform to functions in 𝐿2 (ℝ𝑑 ), and even obtain its inverse.

160

8. Fourier transform

For this, we will use the fact that the set 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) is dense in 𝐿 (ℝ𝑑 ): for any 𝑓 ∈ 𝐿2 (ℝ𝑑 ), there exists a sequence 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) that converges to 𝑓 in 𝐿2 (ℝ𝑑 ), that is 2

∫ |𝑓 − 𝑓𝑛 |2 → 0. ℝ𝑑

This follows from the fact that both 𝐿1 (ℝ𝑑 ) and 𝐿2 (ℝ𝑑 ) contain the space 𝐶𝑐 (ℝ𝑑 ) of continuous functions of compact support, and that 𝐶𝑐 (ℝ𝑑 ) is dense in both 𝐿1 (ℝ𝑑 ) and 𝐿2 (ℝ𝑑 ), as we discussed in Chapter 6. We have Theorem 8.18. Theorem 8.18. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ), then ∫ |𝑓|2 = ∫ |𝑓|̂ 2 . ℝ𝑑

ℝ𝑑

That is, if ⟨⋅, ⋅⟩ is the inner product in 𝐿2 (ℝ𝑑 ), then the operator 𝑓 ↦ 𝑓̂ is an isometry on the subset 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) of 𝐿2 (ℝ𝑑 ), so ‖𝑓‖𝐿2 = ‖𝑓‖̂ 𝐿2 . By the polarization identity, we also have ⟨𝑓, 𝑔⟩ = ⟨𝑓,̂ 𝑔⟩̂

(8.19)

for any 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 )∩𝐿2 (ℝ𝑑 ) (Exercise (7)). See Section A.5 for a review of results of inner product spaces. Proof. For the proof of Theorem 8.18, we will use the identity ˆ ̂ 𝑔(𝜉), 𝑓 ∗ 𝑔(𝜉) = 𝑓(𝜉) ̂ for any 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), which follows from Fubini’s theorem applied to the function 𝐹(𝑥, 𝑦) = 𝑓(𝑥 − 𝑦)𝑔(𝑦)𝑒−2𝜋𝑖𝑥⋅𝜉 (Exercise (8)). In particular, if we take 𝑔(𝑥) = 𝑓(−𝑥) and ℎ = 𝑓 ∗ 𝑔, then ̂ 𝑔(𝜉) ̂ 2, ̂ = 𝑓(𝜉) ℎ(𝜉) ̂ = |𝑓(𝜉)| because ̂ 𝑔(𝜉) ̂ = ∫ 𝑓(−𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = ∫ 𝑓(𝑥)𝑒2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = 𝑓(𝜉). ℝ𝑑

ℝ𝑑

8.3. Mean-square convergence

161

Now, ℎ is a continuous function because, by the Cauchy–Schwarz inequality, |ℎ(𝑥) − ℎ(𝑦)| = |𝑓 ∗ 𝑔(𝑥) − 𝑓 ∗ 𝑔(𝑦)| = || ∫ (𝑓(𝑥 − 𝑧) − 𝑓(𝑦 − 𝑧))𝑔(𝑧)𝑑𝑧|| ℝ𝑑

≤ ‖𝑓(𝑥 − ⋅) − 𝑓(𝑦 − ⋅)‖𝐿2 ‖𝑔‖𝐿2 goes to 0 as 𝑦 → 𝑥, by the continuity of translations in the 𝐿2 norm (6.49). Then, ℎ is continuous at 0 and ℎ ̂ ≥ 0, so by Proposition 8.17 we have that ℎ ̂ ∈ 𝐿1 (ℝ𝑑 ) and ℎ(0) = ∫ ℎ.̂ ℝ𝑑

Therefore ∫ |𝑓|̂ 2 = ∫ ℎ ̂ = ℎ(0) = 𝑓 ∗ 𝑔(0) = ∫ 𝑓(−𝑥)𝑔(𝑥)𝑑𝑥 ℝ𝕕

ℝ𝑑

ℝ𝕕

= ∫ 𝑓(−𝑥)𝑓(−𝑥)𝑑𝑥 = ∫ |𝑓|2 . ℝ𝕕

ℝ𝕕

□ Theorem 8.18 allows us to extend the definition of the Fourier transform to any function in 𝐿2 (ℝ𝑑 ). Indeed, if 𝑓 ∈ 𝐿2 (ℝ𝑑 ), as we have seen above, there exists a sequence 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) such that 𝑓𝑛 → 𝑓 in 𝐿2 (ℝ𝑑 ). Now the sequence of Fourier transforms 𝑓𝑛̂ is a Cauchy sequence in 𝐿2 (ℝ𝑑 ), as ‖𝑓𝑛̂ − 𝑓𝑚̂ ‖𝐿2 = ‖𝑓𝑛 − 𝑓𝑚 ‖𝐿2 and the fact that 𝑓𝑛 converges in 𝐿2 (ℝ𝑑 ). 𝐿2 (ℝ𝑑 ) is complete, so 𝑓𝑛̂ converges in 𝐿2 (ℝ𝑑 ) and we can define ℱ𝑓 = lim 𝑓𝑛̂ .

(8.20)

The operator ℱ ∶ 𝐿2 (ℝ𝑑 ) → 𝐿2 (ℝ𝑑 ) is well defined because, if 𝑓𝑛 → 𝑓 and 𝑔𝑛 → 𝑓 in 𝐿2 (ℝ𝑑 ), then ‖𝑓𝑛̂ − 𝑔𝑛̂ ‖𝐿2 = ‖𝑓𝑛 − 𝑔𝑛 ‖𝐿2 → 0, so lim 𝑓𝑛̂ = lim 𝑔𝑛̂ 2

𝑑

in 𝐿 (ℝ ). It is also an isometry, as clearly (8.21)

‖ℱ𝑓‖𝐿2 = ‖𝑓‖𝐿2

162

8. Fourier transform

for all 𝑓 ∈ 𝐿2 (ℝ𝑑 ). Theorem 8.22. The operator ℱ ∶ 𝐿2 (ℝ𝑑 ) → 𝐿2 (ℝ𝑑 ) is a unitary operator in 𝐿2 (ℝ𝑑 ) with inverse ℱ −1 𝑔(𝑥) = ℱ𝑔(−𝑥). A unitary operator is a surjective isometry. Theorem 8.22 is commonly known as Plancherel’s theorem. Proof. We first observe that ℱ has closed range. If ℱ𝑓𝑛 → 𝑔 in 𝐿2 (ℝ𝑑 ), then ‖𝑓𝑛 − 𝑓𝑚 ‖𝐿2 = ‖ℱ𝑓𝑛 − ℱ𝑓𝑚 ‖𝐿2 because ℱ is an isometry. Thus 𝑓𝑛 is Cauchy in 𝐿2 (ℝ𝑑 ), so it converges, say 𝑓𝑛 → 𝑓 in 𝐿2 (ℝ𝑑 ). Hence ℱ𝑓𝑛 → ℱ𝑓, and 𝑔 = ℱ𝑓. Let ℳ be the range of ℱ. If ℳ ≠ 𝐿2 (ℝ𝑑 ), the orthogonal complement to ℳ would be nontrivial (see Appendix A.5), so there would be a nonzero 𝑔 ∈ 𝐿2 (ℝ𝑑 ) so that ⟨ℱ𝑓, 𝑔⟩ = 0 2

𝑑

for every 𝑓 ∈ 𝐿 (ℝ ), which implies ∫(ℱ𝑓)𝑔 = 0 for all 𝑓 ∈ 𝐿2 (ℝ𝑑 ). Now, the density of 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) and Lemma 8.13 imply that ∫ ℱ𝑓 𝑔 = ∫ 𝑓 ℱ𝑔 ℝ𝑑

ℝ𝑑

(Exercise (10)), so ∫ 𝑓 ℱ𝑔 = 0 ℝ𝑑 2

𝑑

for every 𝑓 ∈ 𝐿 (ℝ ), and thus, choosing 𝑓 = ℱ𝑔, we obtain ∫ |ℱ𝑔|2 = 0. ℝ𝑑

But then ‖𝑔‖𝐿2 = ‖ℱ𝑔‖𝐿2 = 0, contradicting the fact that 𝑔 ≠ 0 in 𝐿2 (ℝ𝑑 ). Thus ℳ = 𝐿2 (ℝ𝑑 ) and ℱ is surjective. In order to show that ℱ −1 𝑔(𝑥) = ℱ𝑔(−𝑥), we first note that, for any 𝑓 ∈ 𝐿2 (ℝ𝑑 ), ̄ ℱ𝑓(𝜉) = ℱ 𝑓(−𝜉)

8.3. Mean-square convergence

163

̄ = 𝑓(𝑥). Indeed, if 𝑓𝑛 ∈ 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) and where 𝑓 ̄ is the function 𝑓(𝑥) 2 𝑑 𝑓𝑛 → 𝑓 in 𝐿 (ℝ ), then 𝑓𝑛̂ (𝜉) = ∫ 𝑓𝑛 (𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = ∫ 𝑓𝑛 (𝑥)𝑒2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 = 𝑓𝑛̂̄ (−𝜉), ℝ𝕕

ℝ𝕕

̄ so 𝑓𝑛̂ (𝜉) → ℱ 𝑓(−𝜉) in 𝐿2 (ℝ𝑑 ). Then, for any 𝑓 ∈ 𝐿2 (ℝ𝑑 ), if we set ℎ(𝑥) = ℱ𝑔(−𝑥), ̄ ⟨ℱℎ, 𝑓⟩ = ∫ ℱℎ(𝑥)𝑓(𝑥)𝑑𝑥 = ∫ ℎ(𝑥)ℱ 𝑓(𝑥)𝑑𝑥 ℝ𝕕

ℝ𝕕

̄ ̄ = ∫ ℱ𝑔(𝑥)ℱ 𝑓(−𝑥)𝑑𝑥 = ∫ ℱ𝑔(−𝑥)ℱ 𝑓(𝑥)𝑑𝑥 ℝ𝕕

ℝ𝕕

= ∫ ℱ𝑔(𝑥)ℱ𝑓(𝑥)𝑑𝑥 = ⟨ℱ𝑔, ℱ𝑓⟩ = ⟨𝑔, 𝑓⟩, ℝ𝕕

where in the last identiy we have used the fact that ℱ is an isometry. Since 𝑓 ∈ 𝐿2 (ℝ𝑑 ) is arbitrary, we have shown that ℎ = ℱ −1 𝑔. □ For a function 𝑓 ∈ 𝐿2 (ℝ𝑑 ), we will write its Fourier transform ℱ𝑓 simply as 𝑓.̂ 8.23. If 𝑓 ∈ 𝐿2 (ℝ𝑑 ), then 𝑓𝑁 = 𝑓𝜒𝐵𝑁 ∈ 𝐿1 (ℝ𝑑 ), where 𝜒𝐵𝑁 is the characteristic function of the ball 𝐵𝑁 of radius 𝑁 around the origin. Indeed, ∫ |𝑓𝑁 | = ∫ |𝑓|𝜒𝐵𝑁 ≤ ‖𝑓‖𝐿2 ⋅ √|𝐵𝑁 | < ∞ ℝ𝕕

ℝ𝕕

for each 𝑁. Moreover, 𝑓𝑁 → 𝑓 in 𝐿2 (ℝ𝑑 ) as 𝑁 → ∞, because ‖𝑓 − 𝑓𝑁 ‖2𝐿2 = ∫

|𝑓(𝑥)|2 𝑑𝑥 → 0.

|𝑥|≥𝑁

Thus ̂ = lim 𝑓𝑁̂ (𝜉) = lim ∫ 𝑓(𝑥)𝑒−2𝜋𝑖𝑥⋅𝜉 𝑑𝑥 𝑓(𝜉) 𝑁→∞

2

𝑁→∞

𝐵𝑁

𝑑

in 𝐿 (ℝ ) (i.e. the limit is taken in the 𝐿2 sense). By Theorem 8.22, 2𝜋𝑖𝑥⋅𝜉 ̂ 𝑑𝜉 𝑓(𝑥) = lim ∫ 𝑓(𝜉)𝑒 𝑁→∞

2

𝐵𝑁

𝑑

in 𝐿 (ℝ ), that is 2

2𝜋𝑖𝑥⋅𝜉 ̂ ∫ ||𝑓(𝑥) − ∫ 𝑓(𝜉)𝑒 𝑑𝜉 || 𝑑𝑥 → 0 ℝ𝕕

𝐵𝑁

164

8. Fourier transform

as 𝑁 → ∞. Hence we have an analog of Theorem 3.50, on the meansquare convergence of Fourier series, for the Fourier transform. 8.24. If 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑔 ∈ 𝐿2 (ℝ𝑑 ), we have seen above that their convolution 𝑓 ∗ 𝑔 ∈ 𝐿2 (ℝ𝑑 ) and ‖𝑓 ∗ 𝑔‖𝐿2 ≤ ‖𝑓‖𝐿1 ‖𝑔‖𝐿2 . Thus, if 𝑔𝑛 ∈ 𝐿1 (ℝ𝑑 ) ∩ 𝐿2 (ℝ𝑑 ) and 𝑔𝑛 → 𝑔 in 𝐿2 (ℝ𝑑 ), then ‖𝑓 ∗ 𝑔𝑛 − 𝑓 ∗ 𝑔‖𝐿2 ≤ ‖𝑓‖𝐿1 ‖𝑔𝑛 − 𝑔‖𝐿2 , so 𝑓 ∗ 𝑔𝑛 → 𝑓 ∗ 𝑔 in 𝐿2 (ℝ𝑑 ). Note that we also have 𝑓 ∗ 𝑔𝑛 ∈ 𝐿1 (ℝ𝑑 ), and thus 𝑓ˆ ∗ 𝑔𝑛 = 𝑓𝑔̂ 𝑛̂ for each 𝑛. Therefore we obtain ˆ 𝑓 ∗ 𝑔 = 𝑓𝑔̂ ̂ for any 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑔 ∈ 𝐿2 (ℝ𝑑 ). Example 8.25. Let 𝑓 ∈ 𝐿2 (ℝ𝑑 ) and let 𝒫𝑡 𝑓(𝑥) be its Poisson integral. Then, since 𝒫𝑡 𝑓 = 𝑃𝑡 ∗ 𝑓, we have −2𝜋𝑡|𝜉| ̂ ̂ ˆ 𝑓(𝜉). 𝒫ˆ 𝑡 𝑓(𝜉) = 𝑃𝑡 (𝜉)𝑓(𝜉) = 𝑒

Exercises (1) The Fourier transform is a linear operator: (a) If 𝛼 ∈ ℂ and 𝑓 ∈ 𝐿1 (ℝ𝑑 ), then ˆ ̂ 𝛼𝑓(𝜉) = 𝛼𝑓(𝜉). (b) If 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), then ̂ + 𝑔(𝜉). 𝑓ˆ + 𝑔(𝜉) = 𝑓(𝜉) ̂ (2) If 𝑅 is a rectangle in ℝ𝑑 , then 𝜒 ˆ𝑅 (𝜉) → 0 as |𝜉| → ∞. (3) Let 𝑓 ∈ 𝐿1 (ℝ𝑑 ). (a) If 𝑓ℎ (𝑥) = 𝑓(𝑥 + ℎ) is the translation of 𝑓 with ℎ ∈ ℝ𝑑 , then ˆℎ (𝜉) = 𝑒2𝜋𝑖ℎ⋅𝜉 𝑓(𝜉). ̂ 𝑓 (b) If 𝑔(𝑥) = 𝑒2𝜋𝑖𝑥⋅ℎ with ℎ ∈ ℝ𝑑 , then ̂ − ℎ). 𝑔(𝜉) ̂ = 𝑓(𝜉

Notes

165

(4) For any 𝜉 ∈ ℝ, 𝑁

𝑁 2

2

lim ∫ 𝑒−𝜋(𝑥+𝑖𝜉) 𝑑𝑥 = lim ∫ 𝑒−𝜋𝑥 𝑑𝑥 = 1.

𝑁→∞

𝑁→∞

−𝑁

−𝑁 2

(Hint: Consider the contour integral ∫𝛾 𝑒−𝜋𝑧 𝑑𝑧 = 0 over the rectangle 𝛾 with vertices 𝑁, 𝑁 + 𝑖𝜉, −𝑁 + 𝑖𝜉 and −𝑁.) (5) Let 𝑢 > 0. Then ∞ −2𝜋𝑖ᵆ𝑣 ∞ 1 𝑒 1 1 −𝑠 −𝜋2 ᵆ2 /𝑠 ∫ ∫ (a) 𝑑𝑣 = 𝑒 𝑒 𝑑𝑠; 𝜋 −∞ 1 + 𝑣2 √𝜋 0 √𝑠 ∞ −2𝜋𝑖ᵆ𝑣 1 𝑒 ∫ (b) 𝑑𝑣 = 𝑒−2𝜋ᵆ . 𝜋 −∞ 1 + 𝑣2 2

(6) If Φ(𝑥) = 𝑒−𝜋|𝑥| , the collection {Φ𝑡 (𝑥)}𝑡>0 of its dilations is a collection of better kernels (see Exercise (6) of Chapter 9). (7) Use the polarization identity for complex inner products spaces to show 8.19. (8) If 𝑓, 𝑔 ∈ 𝐿1 (ℝ𝑑 ), then ˆ ̂ 𝑔(𝜉). 𝑓 ∗ 𝑔(𝜉) = 𝑓(𝜉) ̂ (Hint: Apply Fubini’s theorem to the function 𝐹(𝑥, 𝑦) = 𝑓(𝑥 − 𝑦)𝑔(𝑦)𝑒−2𝜋𝑖𝑥⋅𝜉 on ℝ𝑑 × ℝ𝑑 .) (9) Let 𝑓 ∈ 𝐿1 (ℝ𝑑 ) and 𝑅 a rotation on ℝ𝑑 . (a) If 𝑔 = 𝑓 ∘ 𝑅, ̂ 𝑔(𝜉) ̂ = 𝑓(𝑅𝜉). (b) If 𝑓 is a radial function, then 𝑓 ̂ is also radial. (10) For every 𝑓, 𝑔 ∈ 𝐿2 (ℝ𝑑 ), ∫ ℱ𝑓𝑔 = ∫ 𝑓ℱ𝑔. ℝ𝕕

ℝ𝕕

Notes The results in this chapter, and a deeper study of the Fourier transform, can be found in [SW71]. As we mentioned above, Theorem 8.22 is usually known as Plancherel’s theorem, and the collective results of Section 8.3 are commonly described as Plancherel’s theory due to Michel Plancherel’s work in [Pla10].

Chapter 9

Hilbert transform

9.1. The conjugate function Let 𝑓 ∈ 𝐿1 (ℝ) and 𝑢(𝑥, 𝑡) be its Poisson integral, 𝑢(𝑥, 𝑡) = 𝒫𝑡 𝑓(𝑥) = ∫ 𝑃𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦, ℝ

where 𝑃𝑡 is the Poisson kernel on ℝ, 𝑃𝑡 (𝑥) =

1 𝑡 . 2 𝜋 𝑥 + 𝑡2

We have seen above, in 4.3, that 𝑢(𝑥, 𝑡) is harmonic in the upper half plane ℝ2+ . Consider now the function (9.1)

𝑄𝑡 (𝑥) =

1 𝑥 . 𝜋 𝑥2 + 𝑡 2

The function 𝑄𝑡 is not integrable, as it only decays as 1/|𝑥| when |𝑥| → ∞. However, it is bounded for each 𝑡 > 0, so we can calculate (9.2)

𝑣(𝑥, 𝑡) = ∫ 𝑄𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦 ℝ

for any 𝑓 ∈ 𝐿1 (ℝ). 𝑣(𝑥, 𝑡) is a conjugate harmonic function to 𝑢(𝑥, 𝑡) in ℝ2+ , so the function 𝑓(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) + 𝑖𝑣(𝑥, 𝑡) 167

168

9. Hilbert transform

is holomorphic in ℝ2+ . To prove this, it is sufficient to verify explicitly that (𝑥, 𝑡) ↦ 𝑄𝑡 (𝑥) is harmonic, and so is 𝑣(𝑥, 𝑡), in ℝ2+ , and that 𝑢 and 𝑣 satisfy the Cauchy-Riemann equations (Exercise (1)). In fact, we note that 𝑖 𝑃𝑡 (𝑥) + 𝑖𝑄𝑡 (𝑥) = , 𝜋(𝑥 + 𝑖𝑡) so 𝑃𝑦 (𝑥) and 𝑄𝑦 (𝑥), for 𝑦 > 0, are the real and imaginary parts of 𝑖/𝜋𝑧, where 𝑧 = 𝑥 + 𝑖𝑦. We also observe that 𝑄𝑡 (𝑥) =

1 1 𝑥/𝑡 1 𝑥 ( ) = 𝑄1 ( ), 𝑡 𝜋 (𝑥/𝑡)2 + 1 𝑡 𝑡

so {𝑄𝑡 }𝑡>0 is the family of dilations of 𝑄1 . However, since 𝑄1 ∉ 𝐿1 (ℝ), {𝑄𝑡 }𝑡>0 is not a family of good kernels, so we cannot yet determine the existence of (9.3)

lim ∫ 𝑄𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦. 𝑡→0

ℝ

Note that the limit (9.3) represents the limit at the boundary of the conjugate harmonic function 𝑣 to the Poisson integral of 𝑓. We have the following questions. Question 1: If 𝑓 ∈ 𝐿1 (ℝ), does the limit (9.3) exist for any 𝑥 ∈ ℝ? Question 2: Under what conditions on 𝑓 can we guarantee, and in what sense, that the limit (9.3) exists?

9.2. Mean-square convergence We first consider Question 2 by asking if the limit (9.3) exists in the mean-square sense, that is, in 𝐿2 (ℝ). The first problem we encounter is that, since 𝑄𝑡 is not integrable, we cannot guarantee that the convolution 𝑄𝑡 ∗ 𝑓 exists if 𝑓 ∈ 𝐿2 (ℝ). However, since indeed 𝑄𝑡 ∈ 𝐿2 (ℝ), the convolution 𝑄𝑡 ∗ 𝑓 ∈ 𝐿2 (ℝ) if 𝑓 ∈ 𝐿1 (ℝ) and, in particular, if 𝑓 ∈ 𝐿1 (ℝ) ∩ 𝐿2 (ℝ). Thus, before we consider the limit as 𝑡 → 0, we have to ask Question 3. Question 3: Can we extend the operator 𝑓 ↦ 𝑄𝑡 ∗ 𝑓 from 𝐿1 (ℝ) ∩ 𝐿2 (ℝ) to 𝐿2 (ℝ), for any 𝑡 > 0? In that case, does the limit (9.3) exist in the 𝐿2 sense?

9.2. Mean-square convergence

169

To answer Question 3 we need to find out if the family of operators 𝑓 ↦ 𝑄𝑡 ∗ 𝑓 is uniformly bounded in 𝐿2 (ℝ), that is, if there exists a constant 𝐴 > 0, independent of 𝑡, such that ‖𝑄𝑡 ∗ 𝑓‖𝐿2 ≤ 𝐴‖𝑓‖𝐿2 1

2

for every 𝑓 ∈ 𝐿 (ℝ) ∩ 𝐿 (ℝ) and all 𝑡 > 0. This will allow us to extend 𝑄𝑡 ∗ 𝑓 to all of 𝐿2 (ℝ), as we did in the case of the Fourier transform in the previous chapter, for each 𝑡 > 0, and then take the limit as 𝑡 → 0. Lemma 9.4. For any 𝑡 > 0, the Fourier transform of 𝑄𝑡 in 𝐿2 (ℝ) is given by ˆ𝑡 (𝜉) = −𝑖 sgn(𝜉)𝑒−2𝜋𝑡|𝜉| , where sgn(𝜉) is the sign function the function 𝑄 of 𝜉. Proof. As we noted in 8.23, the Fourier transform of the 𝐿2 function 𝑄𝑡 is given by 𝑁

𝑁

𝑥 ˆ𝑡 (𝜉) = lim ∫ 𝑄𝑡 (𝑥)𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 = 1 lim ∫ 𝑄 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥. 2 + 𝑡2 𝜋 𝑥 𝑁→∞ 𝑁→∞ −𝑁 −𝑁 As 𝑄𝑡 is odd, the above limit is 0 if 𝜉 = 0. For 𝜉 ≠ 0, the limit on the right can be obtained by using the residue theorem [Gam01]. Indeed, if 𝜉 < 0, consider the contour 𝛾 given by the upper semicircle of radius 𝑁

Figure 9.1. The contour 𝛾 in the proof of Lemma 9.4. For 𝑁 sufficiently large, the pole 𝑖𝑡 of the function 𝑓(𝑧) is inside 𝛾.

with center at the origin (as in Figure 9.1), and 1 𝑧 𝑓(𝑧) = 𝑒−2𝜋𝑖𝑧𝜉 . 𝜋 𝑧 2 + 𝑡2

170

9. Hilbert transform

Then, by the residue theorem, ∫ 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖 Res𝑧=𝑖𝑡 𝑓(𝑧) = 𝑖𝑒2𝜋𝑡𝜉 . 𝛾

We also have ∫ 𝑓(𝑧)𝑑𝑧 = 𝛾 𝑁

∫ −𝑁

𝜋

1 𝑥 1 𝑁𝑒𝑖𝜃 𝑖𝜃 𝑒−2𝜋𝑖𝑥𝜉 𝑑𝑥 + ∫ 𝑒−2𝜋𝑖𝑁𝑒 𝜉 𝑖𝑁𝑒𝑖𝜃 𝑑𝜃, 2 2 𝑖𝜃 )2 + 𝑡2 𝜋𝑥 +𝑡 𝜋 (𝑁𝑒 0

where the integrals on the right hand side correspond to the integrals over the segment [−𝑁, 𝑁] and the semicircle 𝑁𝑒𝑖𝜃 , 0 ≤ 𝜃 ≤ 𝜋, of 𝛾, respectively. The first integral, as 𝑁 → ∞, gives our desired limit, while the second satisfies 𝜋

𝜋

𝑁𝑒𝑖𝜃 𝑖𝜃 𝑖𝜃 |∫ 1 𝑒−2𝜋𝑖𝑁𝑒 𝜉 𝑖𝑁𝑒𝑖𝜃 𝑑𝜃|| ≤ 𝐶 ∫ |𝑒−2𝜋𝑖𝑁𝑒 𝜉 |𝑑𝜃 | 𝑖𝜃 )2 + 𝑡2 𝜋 (𝑁𝑒 0 0 𝜋

= 𝐶 ∫ 𝑒2𝜋𝑁𝜉 sin 𝜃 𝑑𝜃 0 𝜋/2

≤ 2𝐶 ∫

𝑒−4𝑁|𝜉|𝜃 𝑑𝜃 → 0,

0

where we have used the fact that 𝜉 < 0 and the estimate 2 sin 𝜃 ≥ 𝜃 𝜋 on [0, 𝜋/2], along with the symmetry of sin 𝜃 around 𝜃 = 𝜋/2. Thus, we obtain ˆ𝑡 (𝜉) = 𝑖𝑒2𝜋𝑡𝜉 𝑄 for 𝜉 < 0. We leave as an exercise the details for 𝜉 > 0 (Exercise (2)), where we obtain ˆ𝑡 (𝜉) = −𝑖𝑒−2𝜋𝑡𝜉 . 𝑄 Combining the above we conclude ˆ𝑡 (𝜉) = −𝑖 sgn(𝜉)𝑒−2𝜋𝑡|𝜉| . 𝑄 □ 1

2

Corollary 9.5. If 𝑓 ∈ 𝐿 (ℝ) ∩ 𝐿 (ℝ), then, for any 𝑡 > 0, ‖𝑄𝑡 ∗ 𝑓‖𝐿2 ≤ ‖𝑓‖𝐿2 .

9.2. Mean-square convergence

171

Proof. The Fourier transform is an isometry in 𝐿2 (ℝ), so, using Lemma 9.4, −2𝜋𝑡|𝜉| ̂ ˆ ‖𝑄𝑡 ∗ 𝑓‖𝐿2 = ‖𝑄 𝑓‖𝐿2 ≤ ‖𝑓‖̂ 𝐿2 = ‖𝑓‖𝐿2 . 𝑡 ∗ 𝑓‖𝐿2 = ‖ − 𝑖 sgn(𝜉)𝑒

□ Corollary 9.5 allows us to define the convolution 𝑄𝑡 ∗ 𝑓 for any 𝑓 ∈ 𝐿2 (ℝ), and we can do it in two different ways. The first way is analogous to the way we extended the Fourier transform to 𝐿2 (ℝ𝑑 ): using the fact that 𝐿1 (ℝ) ∩ 𝐿2 (ℝ) is dense in 𝐿2 (ℝ), we can find, for any 𝑓 ∈ 𝐿2 (ℝ), a sequence 𝑓𝑛 ∈ 𝐿1 (ℝ) ∩ 𝐿2 (ℝ) such that 𝑓𝑛 → 𝑓 in 𝐿2 (ℝ), and then define 𝑄𝑡 ∗ 𝑓 as the limit in 𝐿2 (ℝ) of the sequence 𝑄𝑡 ∗ 𝑓𝑛 . This limit exists because 𝑄𝑡 ∗ 𝑓𝑛 is a Cauchy sequence in 𝐿2 (ℝ), by Corollary 9.5, and 𝐿2 (ℝ) is complete. The other way is through the Fourier transform of 𝑄𝑡 ∗ 𝑓. Since the Fourier transform is a unitary operator, by Theorem 8.18, we can define, for 𝑓 ∈ 𝐿2 (ℝ), 𝑄𝑡 ∗ 𝑓 ∈ 𝐿2 (ℝ) as the function whose Fourier transform is equal to (9.6)

−2𝜋𝑡|𝜉| ̂ ˆ 𝑄 𝑓(𝜉). 𝑡 ∗ 𝑓(𝜉) = −𝑖 sgn(𝜉)𝑒

That is, we define, for 𝑓 ∈ 𝐿2 (ℝ), 𝑄𝑡 ∗ 𝑓 as the inverse Fourier transform of (9.6). Moreover, it is clear that, as 𝑡 → 0, the limit ˆ lim 𝑄 𝑡 ∗ 𝑓(𝜉) 𝑡→0

̂ exists for every 𝜉 ∈ ℝ, and is equal to −𝑖 sgn(𝜉)𝑓(𝜉). This limit also converges in 𝐿2 (ℝ) because 2

̂ − ( − 𝑖 sgn(𝜉)𝑓(𝜉)) ̂ | 𝑑𝜉 ∫ | − 𝑖 sgn(𝜉)𝑒−2𝜋𝑡|𝜉| 𝑓(𝜉) ℝ 2

̂ | 𝑑𝜉 → 0, = ∫ |(𝑒−2𝜋𝑡|𝜉| − 1)𝑓(𝜉) ℝ

by the dominated convergence theorem. We can therefore make Definition 9.7. Definition 9.7. Let 𝑓 ∈ 𝐿2 (ℝ). The Hilbert transform of 𝑓 is the function 𝐻𝑓 ∈ 𝐿2 (ℝ) equal to the inverse Fourier transform of ˆ ̂ 𝐻𝑓(𝜉) = −𝑖 sgn(𝜉)𝑓(𝜉).

172

9. Hilbert transform

9.8. The Hilbert transform 𝐻 is an isometry in 𝐿2 (ℝ), because ̂ ̂ | − 𝑖 sgn(𝜉)𝑓(𝜉)| = |𝑓(𝜉)| and hence ˆ 𝐿2 (ℝ) = ‖𝑓‖̂ 𝐿2 (ℝ) = ‖𝑓‖𝐿2 (ℝ) . ‖𝐻𝑓‖𝐿2 (ℝ) = ‖𝐻𝑓‖ With the observations above, and Definition 9.7, we have answered Questions 2 and 3 of this and Section 9.1. However, to answer Question 1, we need to understand the local nature of the convolution 𝑄𝑡 ∗ 𝑓 as 𝑡 → 0, which we discuss in Section 9.3.

9.3. The Hilbert transform of integrable functions In this section we study the Hilbert transform of integrable functions. As noted above, the kernels 𝑄𝑡 are not integrable, and thus we cannot discuss the behavior of 𝑄𝑡 ∗ 𝑓 as 𝑡 → 0 pointwise, nor in 𝐿1 (ℝ), directly. However, we have Lemma 9.9. Lemma 9.9. For 𝑓 ∈ 𝐿1 (ℝ), lim ( ∫ 𝑓(𝑥 − 𝑦) 𝑡→0

ℝ

𝑓(𝑥 − 𝑦) 𝑦 𝑑𝑦 − ∫ 𝑑𝑦) = 0 𝑦 𝑦2 + 𝑡 2 |𝑦|≥𝑡

for almost every 𝑥 ∈ ℝ. Proof. Define

𝑦 1 − 2+1 𝑦 𝑦 Φ(𝑦) = { 𝑦 𝑦2 + 1

Since

|𝑦| ≥ 1 |𝑦| < 1.

𝑦 1 1 − =− 2 +1 𝑦 𝑦(𝑦 + 1) for |𝑦| > 1, Φ is integrable. Moreover, the collection of its dilations 𝑦 1 − |𝑦| ≥ 𝑡 2 + 𝑡2 1 𝑦 𝑦 𝑦 Φ𝑡 (𝑦) = Φ( ) = { 𝑦 𝑡 𝑡 |𝑦| < 𝑡 𝑦2 + 𝑡 2 Φ(𝑦) =

𝑦2

satisfies (1) ∫ Φ𝑡 = 0 for all 𝑡 > 0; ℝ

9.3. The Hilbert transform of integrable functions

173

1 for all 𝑡 > 0 and 𝑦 ∈ ℝ; and 𝑡 𝑡 (3) |Φ𝑡 (𝑦)| ≤ 2 for all 𝑡 > 0 and 𝑦 ∈ ℝ, 𝑦 ≠ 0. 𝑦 (2) |Φ𝑡 (𝑦)| ≤

The first one follows because Φ is odd, and the other two from the explicit form of Φ𝑡 (𝑦) (Exercise (4)). Then ∫ 𝑓(𝑥 − 𝑦) ℝ

𝑓(𝑥 − 𝑦) 𝑦 𝑑𝑦 − ∫ 𝑑𝑦 = Φ𝑡 ∗ 𝑓(𝑥) 𝑦 𝑦2 + 𝑡 2 |𝑦|≥𝑡

and {Φ𝑡 }𝑡>0 is a collection of better kernels, as introduced in Exercises (6) and (9) of Chapter 7, with constant 𝜆 = 0 in the latter. Therefore, lim Φ𝑡 ∗ 𝑓(𝑥) = 0 𝑡→0

□

for almost every 𝑥 ∈ ℝ. Lemma 9.9 implies that, for almost every 𝑥 ∈ ℝ, the limit lim ∫ 𝑓(𝑥 − 𝑦) 𝑡→0

ℝ

𝑦 𝑑𝑦 𝑦2 + 𝑡 2

exists if and only if lim ∫ 𝑡→0

|𝑦|≥𝑡

𝑓(𝑥 − 𝑦) 𝑑𝑦 𝑦

exists. Note that the latter limit exists at 𝑥 ∈ ℝ if 𝑓 ∈ 𝐿1 (ℝ) and 𝑥 is not in the support of 𝑓, because lim 𝑡→0

𝑓(𝑥 − 𝑦) 𝑓(𝑦) 𝑓(𝑦) 1 1 1 ∫ 𝑑𝑦 = lim ∫ 𝑑𝑦 = ∫ 𝑑𝑦, 𝜋 |𝑦|≥𝑡 𝑦 𝑥 − 𝑦 𝜋 𝑥 −𝑦 𝑡→0 𝜋 |𝑥−𝑦|≥𝑡 ℝ

as there exist some 𝛿 > 0 such that, if |𝑥 − 𝑦| < 𝛿, then 𝑦 ∉ supp 𝑓, so 𝑓(𝑦) = 0. The limit also exists if 𝑓 is differentiable at 𝑥, because in that case the integrals 𝑓(𝑥 − 𝑦) ∫ 𝑑𝑦 𝑦 𝑡≤|𝑦| 0, ‖𝐻𝑓‖𝐿1 ≤ 𝐴‖𝑓‖𝐿1 𝐶𝑐∞ (ℝ).

for every 𝑓 ∈ However, this is far from true, because 𝐻𝑓 might not even be integrable. Example 9.11. Let 𝑓 ∈ 𝐶𝑐∞ (ℝ) such that 𝑓 ≥ 0, 𝑓(𝑥) = 1 if 0 ≤ 𝑥 ≤ 1, and 𝑓(𝑥) = 0 if 𝑥 ≥ 2 or 𝑥 ≤ −1 (as in Figure 9.2). Then, for 𝑥 > 2,

Figure 9.2. A function 𝑓 ∈ 𝐶𝑐∞ (ℝ) such that 𝑓 ≥ 0, 𝑓(𝑥) = 1 if 0 ≤ 𝑥 ≤ 1, and 𝑓(𝑥) = 0 if 𝑥 ≥ 2 or 𝑥 ≤ −1. 2

𝐻𝑓(𝑥) =

1

𝑓(𝑦) 1 1 1 1 ∫ 𝑑𝑦 ≥ ∫ 𝑑𝑦 ≥ , 𝜋 −1 𝑥 − 𝑦 𝜋 0 𝑥−𝑦 𝜋𝑥

so 𝐻𝑓 is not an integrable function. We see then that the Hilbert transform cannot be bounded in 𝐿1 (ℝ). However we have Theorem 9.12. Theorem 9.12. There exists a constant 𝐴 > 0 such that, for any 𝑓 ∈ 𝐶𝑐∞ (ℝ) and 𝛼 > 0, 𝐴 |{𝑥 ∈ ℝ ∶ |𝐻𝑓(𝑥)| > 𝛼}| ≤ ‖𝑓‖𝐿1 . 𝛼 As in the case of the maximal function, the Hilbert transform is of weak type (1, 1) (although, so far, it is only defined on 𝐶𝑐∞ (ℝ)).

9.3. The Hilbert transform of integrable functions

175

Proof. Fix 𝛼 > 0. We will write the function 𝑓 as 𝑔 + 𝑏, the sum of a “good” and a “bad” part. For this, we construct a collection ℐ of dyadic intervals (intervals of the form [𝑘 ⋅ 2𝑛 , (𝑘 + 1) ⋅ 2𝑛 ], with 𝑘, 𝑛 ∈ ℤ) in the following way. Let 𝑁 be large enough so that 1 ∫|𝑓| ≤ 𝛼 |𝐼| 𝐼 for every dyadic interval 𝐼 of length |𝐼| = 2𝑁 . Such 𝑁 exists because 𝑓 is integrable. Now, subdivide each 𝐼 in two subintervals 𝐼 ′ of half the length of 𝐼. For each one, we have either 1 ∫ |𝑓| ≤ 𝛼 |𝐼 ′ | 𝐼 ′

1 ∫ |𝑓| > 𝛼. |𝐼 ′ | 𝐼 ′

or

In the second case, we add 𝐼 ′ to the collection ℐ. Note that we have 𝛼
𝛼/2}| ≤ ∫ ℝ

|𝐻𝑔|2 4 20 ≤ 2 ⋅ 5𝛼‖𝑓‖𝐿1 = ‖𝑓‖𝐿1 . 𝛼 𝛼 (𝛼/2)2

Define 𝑏 = 𝑓 − 𝑔. Since |{𝑥 ∈ ℝ ∶ |𝐻𝑓(𝑥)| > 𝛼}| ≤ |{𝑥 ∈ ℝ ∶ |𝐻𝑔(𝑥)| > 𝛼/2}| + |{𝑥 ∈ ℝ ∶ |𝐻𝑏(𝑥)| > 𝛼/2}|, it remains to estimate |{𝑥 ∈ ℝ ∶ |𝐻𝑏(𝑥)| > 𝛼/2}|. 𝑏 is what we refer as the bad part of 𝑓, because we have no control over the size of |𝑏(𝑥)|. However, 𝑏 = 0 outside of Ω and, for each 𝐼 ∈ ℐ, ∫𝑏 = ∫𝑓 − ∫𝑔 = ∫𝑓 − ∫( 𝐼

𝐼

𝐼

𝐼

𝐼

1 ∫𝑓) = 0. |𝐼| 𝐼

Write 𝑏 = ∑ 𝑏𝐼 , 𝐼∈ℐ

where each 𝑏𝐼 = 𝑏 ⋅ 𝜒𝐼 , for 𝐼 ∈ ℐ. Then 𝐻𝑏 = ∑ 𝐻𝑏𝐼 . 𝐼∈ℐ

Note that each 𝐻𝑏𝐼 is defined almost everywhere, since 𝑏𝐼 is either 0, outside of 𝐼, or 1 ∫𝑓 𝑏𝐼 (𝑥) = 𝑓(𝑥) − |𝐼| 𝐼 in the interior of 𝐼, so it is differentiable, as we are assuming that 𝑓 is differentiable.

9.3. The Hilbert transform of integrable functions

177

Write 3𝐼 for the interval with the same center as 𝐼, but 3 times its length (see Figure 9.3).

Figure 9.3. The interval 3𝐼 has the same center as 𝐼 and 3 times its length.

If 𝑥 ∉ 3𝐼, and 𝑦0 is the center of 𝐼,

𝐻𝑏𝐼 (𝑥) =

𝑏 (𝑦) 𝑏 (𝑦) 1 1 ∫ 𝐼 𝑑𝑦 − ∫ 𝐼 𝑑𝑦 𝜋 𝐼 𝑥−𝑦 𝜋 𝐼 𝑥 − 𝑦0

=

1 1 1 ∫𝑏 (𝑦)( − )𝑑𝑦 𝜋 𝐼 𝐼 𝑥 − 𝑦 𝑥 − 𝑦0

=

𝑦 − 𝑦0 1 ∫𝑏 (𝑦) 𝑑𝑦, 𝜋 𝐼 𝐼 (𝑥 − 𝑦)(𝑥 − 𝑦0 )

where, in the first equality, we have used the fact that ∫𝐼 𝑏𝐼 = 0. As 𝑥 ∉ 3𝐼 and 𝑦 ∈ 𝐼 in the integral above, we see that

|𝑦 − 𝑦0 | ≤

1 1 |𝐼| ≤ |𝑥 − 𝑦|, 2 2

so |𝑥 − 𝑦0 | ≥ |𝑥 − 𝑦| − |𝑦 − 𝑦0 | ≥

1 |𝑥 − 𝑦|. 2

Thus

|𝐻𝑏𝐼 (𝑥)| ≤

|𝑦 − 𝑦0 | |𝑏 (𝑦)| 2 1 ∫|𝑏 (𝑦)| 𝑑𝑦 ≤ |𝐼| ∫ 𝐼 2 𝑑𝑦, 𝜋 𝐼 𝐼 𝜋 |𝑥 − 𝑦|2 |𝑥 − 𝑦| 𝐼

178

9. Hilbert transform

and therefore ∫

|𝐻𝑏𝐼 (𝑥)|𝑑𝑥 ≤

ℝ⧵3𝐼

|𝑏 (𝑦)| 1 |𝐼| ∫ ∫ 𝐼 2 𝑑𝑦𝑑𝑥 𝜋 |𝑥 − 𝑦| ℝ⧵3𝐼 𝐼

≤

1 1 |𝐼| ∫|𝑏𝐼 (𝑦)| ∫ 𝑑𝑥𝑑𝑦 𝜋 |𝑥 − 𝑦|2 𝐼 |𝑥−𝑦|≥|𝐼|

≤

2 ∫|𝑏 (𝑦)|𝑑𝑦 𝜋 𝐼 𝐼

≤

1 2 | ∫ 𝑓(𝑦) − ∫𝑓|𝑑𝑦 𝜋 𝐼| |𝐼| 𝐼 |

≤

2 1 ∫( ∫|𝑓|)) ( ∫|𝑓| + 𝜋 𝐼 |𝐼| 𝐼 𝐼

=

4 8 ∫|𝑓| ≤ 𝛼|𝐼|, 𝜋 𝐼 𝜋

because 𝐼 ∈ ℐ. Hence, if we define Ω∗ =

⋃

3𝐼,

𝐼∈ℐ

we obtain ∫

|𝐻𝑏(𝑥)|𝑑𝑥 ≤ ∑ ∫

ℝ⧵Ω∗

|𝐻𝑏𝐼 (𝑥)|𝑑𝑥 ≤

𝐼∈ℐ ℝ⧵3𝐼

=

8𝛼 ∑ |𝐼| 𝜋 𝐼∈ℐ

8𝛼 8 |Ω| ≤ ‖𝑓‖𝐿1 . 𝜋 𝜋

Thus, by Chebyshev’s inequality, |{𝑥 ∈ ℝ ⧵ Ω∗ ∶ |𝐻𝑏(𝑥)| > 𝛼/2}| ≤

16/𝜋 2 ∫ |𝐻𝑏(𝑥)|𝑑𝑥 ≤ ‖𝑓‖𝐿1 . 𝛼 ℝ⧵Ω∗ 𝛼

It just remains to estimate |{𝑥 ∈ Ω∗ ∶ |𝐻𝑏(𝑥)| > 𝛼/2}| ≤ |Ω∗ | ≤ ∑ |3𝐼| = 3|Ω| ≤ 𝐼∈ℐ

3 ‖𝑓‖𝐿1 . 𝛼 □

Note that, from the proof above, we can take the constant 𝐴 in Theorem 9.12 as 16 𝐴 = 20 + + 3 < 29. 𝜋

9.4. Convergence in measure

179

9.4. Convergence in measure Theorem 9.12 doesn’t guarantee that the limit lim 𝑡→0

𝑓(𝑥 − 𝑦) 1 ∫ 𝑑𝑦 𝜋 |𝑦|≥𝑡 𝑦

converges for a function in 𝐿1 . However, the Hilbert transform can be defined “in measure”. We say that the sequence 𝑓𝑛 converges in measure to 𝑓 if, for every 𝜀 > 0, |{𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛 (𝑥) − 𝑓(𝑥)| ≥ 𝜀}| → 0 as 𝑛 → ∞. The function 𝑓 is unique almost everywhere (Exercise (6)). If a sequence 𝑓𝑛 → 𝑓 in 𝐿1 , then it converges in measure to 𝑓, but almost everywhere convergence does not imply convergence in measure (Exercise (7)). A sequence 𝑓𝑛 of measurable functions on ℝ𝑑 is Cauchy in measure if, for every 𝜀 > 0, |{𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| ≥ 𝜀}| → 0 as 𝑛, 𝑚 → ∞. If 𝑓𝑛 converges in measure to 𝑓, then 𝑓𝑛 is Cauchy in measure because {𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| ≥ 𝜀} ⊂ {𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛 (𝑥) − 𝑓(𝑥)| ≥ 𝜀/2} ∪ {𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑚 (𝑥) − 𝑓(𝑥)| ≥ 𝜀/2}, by the triangle inequality. If 𝑓𝑛 is Cauchy in measure, there exists a sequence 𝑛𝑘 such that |{𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛𝑘 (𝑥) − 𝑓𝑛𝑘+1 (𝑥)| ≥

1 1 }| < 𝑘 . 𝑘 2 2

If we define 𝐴𝑘 =

⋃

{𝑥 ∈ ℝ𝑑 ∶ |𝑓𝑛𝑗 (𝑥) − 𝑓𝑛𝑗+1 (𝑥)| ≥

𝑗≥𝑘

then ∞

1 1 = 𝑘−1 . 𝑗 2 2 𝑗=𝑘

|𝐴𝑘 | ≤ ∑

1 }, 2𝑗

180

9. Hilbert transform

In particular, |𝐴𝑘 | → 0 and | ⋂𝑘 𝐴𝑘 | = 0. If 𝑥 ∉ 𝐴𝑘 and 𝑖 ≥ 𝑗 ≥ 𝑘, 𝑖

𝑖

(9.13)

|𝑓𝑛𝑗 (𝑥) − 𝑓𝑛𝑖 (𝑥)| ≤ ∑ |𝑓𝑛𝑙 (𝑥) − 𝑓𝑛𝑙+1 (𝑥)| < ∑ 𝑙=𝑗

𝑙=𝑗

1 1 < 𝑗−1 . 𝑙 2 2

Thus the sequence 𝑓𝑛𝑗 (𝑥) is Cauchy for each 𝑥 ∉ 𝐴𝑘 , and thus converges. We can then define 𝑓(𝑥) = lim 𝑓𝑛𝑗 (𝑥) for each 𝑥∈

(ℝ𝑑 ⧵ 𝐴𝑘 ) = ℝ𝑑 ⧵ 𝐴 . ⋃ ⋂ 𝑘 𝑘

𝑘

Therefore the subsequence 𝑓𝑛𝑘 → 𝑓 almost everywhere. By (9.13), |𝑓𝑛𝑗 (𝑥) − 𝑓(𝑥)|
0, |{𝑥 ∈ ℝ ∶ |𝐻(𝑓𝑛 − 𝑓𝑚 )(𝑥)| > 𝜀}| ≤

𝐴 ‖𝑓 − 𝑓𝑚 ‖𝐿1 → 0, 𝜀 𝑛

so the sequence 𝐻𝑓𝑛 is Cauchy in measure. Therefore, there exists a function, which we may call 𝐻𝑓, such that 𝐻𝑓𝑛 → 𝐻𝑓 in measure. As we have seen before, 𝐻𝑓 is well defined and there exists a subsequence 𝑓𝑛𝑘 such that 𝐻𝑓𝑛𝑘 → 𝐻𝑓 almost everywhere (Exercise (8)). Theorem 9.12 implies that 𝐻𝑓 thus defined is finite almost everywhere.

Exercises

181

Exercises (1) (a) The function 𝑞(𝑥, 𝑡) = 𝑄𝑡 (𝑥) =

1 𝑥 𝜋 𝑥2 + 𝑡 2

is harmonic in ℝ2+ . (b) If 𝑓 ∈ 𝐿1 (ℝ), the function 𝑣(𝑥, 𝑡) = ∫ 𝑄𝑡 (𝑥 − 𝑦)𝑓(𝑦)𝑑𝑦 ℝ

is harmonic in ℝ2+ . (c) If 𝑓 ∈ 𝐿1 (ℝ), the functions 𝑣(𝑥, 𝑡) and 𝑢(𝑥, 𝑡), the Poisson integral of 𝑓, are conjugate harmonic. (2) Let 𝛾 be the lower semicircle of radius 𝑁 around the origin, and 𝜉 > 0. Then ∫ 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖 Res𝑧=−𝑖𝑡 𝑓(𝑧) = −𝑖𝑒−2𝜋𝑡𝜉 , 𝛾

where 𝑓(𝑧) is the function defined in the proof of Lemma 9.4. (3) For 𝑓 ∈ 𝐶𝑐 (ℝ), consider the Cauchy integral 𝐹(𝑧) =

𝑓(𝑡) 1 ∫ 𝑑𝑡, 𝑖𝜋 ℝ 𝑡 − 𝑧

for 𝑧 = 𝑥 + 𝑖𝑦 ∈ ℝ2+ . (a) There exists a constant 𝐴 > 0 such that |𝐹(𝑧)| ≤ 𝐴/|𝑧|. (b) ∫ 𝐹(𝑥)2 𝑑𝑥 = 0. ℝ

(c) Calculate ℜ(𝐹(𝑥)2 ) and conclude ‖𝐻𝑓‖𝐿2 = ‖𝑓‖𝐿2 . (4) Let 𝑦 1 − |𝑦| ≥ 1 2+1 𝑦 𝑦 Φ(𝑦) = { 𝑦 |𝑦| < 1. 𝑦2 + 1 Then its collection {Φ𝑡 }𝑡>0 of dilations satisfies (a) ∫ Φ𝑡 = 0 for all 𝑡 > 0; ℝ

(b) |Φ𝑡 (𝑦)| ≤

1 for all 𝑡 > 0 and 𝑦 ∈ ℝ; and 𝑡

182

9. Hilbert transform

𝑡 for all 𝑡 > 0 and 𝑦 ∈ ℝ, 𝑦 ≠ 0. 𝑦2 (5) If 𝑓 ∈ 𝐿1 (ℝ) is differentiable at 𝑥 ∈ ℝ, then the limit (c) |Φ𝑡 (𝑦)| ≤

lim ∫ 𝑡→0

|𝑦|≥𝑡

𝑓(𝑥 − 𝑦) 𝑑𝑦 𝑦

exists. (Hint: Use the identity, for any 𝛿𝑛 > 0, ∫ 𝑡≤|𝑦| 0 such that ∫ |𝑇𝑓|𝑞 ≤ 𝐴 ∫ |𝑓|𝑞 ℝ

ℝ

Notes

183 for some 𝑞 > 1.

Notes The study of the conjugate function started in the work of several authors in the early 20th century, using complex variable methods (as in Exercise (3)). See the references in [Zyg02] for a list of such authors. The decomposition method in the proof of Theorem 9.12 is due to Alberto P. Calderón and Antoni Zygmund [CZ52], where they generalize to other kernels (see Exercises (9), (10) and (11)) and other dimensions. See [Ste70] for a further discussion of these methods, and a more extensive study of singular integrals.

Chapter 10

Mathematics of fractals

10.1. Hausdorff dimension The purpose of this chapter is to introduce the basic ideas in the study of fractals. We start with the Hausdorff dimension, which provides a means to quantify the complexity of a fractal set. For a set 𝐴 ⊂ ℝ𝑑 , we denote its diameter by diam 𝐴, diam 𝐴 = sup{|𝑥 − 𝑦| ∶ 𝑥, 𝑦 ∈ 𝐴}, the supremum over all distances between points in 𝐴. We observe, for example, that diam 𝐵𝑟 (𝑥) = 2𝑟, and that, if 𝑥 ∈ 𝐴 and 𝑟 = diam 𝐴, 𝐴 ⊂ 𝐵𝑟̄ (𝑥). For 𝛿 > 0, a 𝛿-cover for 𝐴 ⊂ ℝ𝑑 is a collection 𝑈1 , 𝑈2 , . . . of subsets of ℝ𝑑 such that diam 𝑈 𝑗 ≤ 𝛿,

𝑗 = 1, 2, . . . ,

and

𝐴⊂

⋃

𝑈𝑗.

𝑗

The collection {𝑈 𝑗 } may be finite or infinite (as long as it is countable), and the sets 𝑈 𝑗 are arbitrary, as long as they cover 𝐴 and have diameter not larger than 𝛿. 185

186

10. Mathematics of fractals For 𝑠 ≥ 0, consider the number, for each 𝐴 ⊂ ℝ𝑑 ,

(10.1)

𝐻𝛿𝑠 (𝐴) = inf { ∑(diam 𝑈 𝑗 )𝑠 ∶ {𝑈 𝑗 } is a 𝛿-cover for 𝐴}. 𝑗

In other words, we take all 𝛿-covers {𝑈 𝑗 } for 𝐴, calculate for each one ∑𝑗 (diam 𝑈 𝑗 )𝑠 , which is clearly nonnegative, and then take the infimum of all such sums. It is possible that all of them are divergent, and in that case we have 𝐻𝛿𝑠 (𝐴) = ∞. It is also possible that 𝐻𝛿𝑠 (𝐴) = 0. For example, if 𝐴 = ∅, or if 𝐴 is finite. Indeed, if 𝐴 = {𝑥1 , 𝑥2 , . . . , 𝑥𝑘 } ⊂ ℝ𝑑 , consider the 𝛿-cover for 𝐴 given by the balls 𝐵𝜀 (𝑥1 ), 𝐵𝜀 (𝑥2 ), . . . , 𝐵𝜀 (𝑥𝑘 ), with 𝜀 > 0. Then, for 𝑠 > 0, 𝑘

∑ (diam 𝐵𝜀 (𝑥𝑗 ))𝑠 = (2𝜀)𝑠 𝑘, 𝑗=1

and thus 𝐻𝛿𝑠 (𝐴) ≤ (2𝜀)𝑠 𝑘. As 𝜀 > 0 is arbitrary, we have 𝐻𝛿𝑠 (𝐴) = 0. It is not hard to see that, if 𝐴 ⊂ 𝐵, then 𝐻𝛿𝑠 (𝐴) ≤ 𝐻𝛿𝑠 (𝐵), and, if 𝐴 = ⋃𝑗 𝐴𝑗 , then 𝐻𝛿𝑠 (𝐴) ≤ ∑ 𝐻𝛿𝑠 (𝐴𝑗 ). 𝑗

See Exercise (1). These are the same properties which are also satisfied by the outer measure | ⋅ |∗ . Now, observe that, if 𝛿 < 𝜂, then every 𝛿-cover for a set 𝐴 is also an 𝜂-cover for 𝐴. Hence, by the definition (10.1), we have 𝐻𝛿𝑠 (𝐴) ≥ 𝐻𝜂𝑠 (𝐴). We see that 𝐻𝛿𝑠 (𝐴) increases as 𝛿 decreases, so the limit exists as 𝛿 → 0 (it might be infinite). We thus define (10.2)

ℋ 𝑠 (𝐴) = lim 𝐻𝛿𝑠 (𝐴) = sup{𝐻𝛿𝑠 (𝐴) ∶ 𝛿 > 0}. 𝛿→0

ℋ (𝐴) is called the Hausdorff measure with exponent 𝑠 of 𝐴. As 𝐻𝛿𝑠 , the Hausdorff measure also satisfies the properties of the outer measure | ⋅ |∗ (Exercise (2)). 𝑠

If {𝑈 𝑗 } is a 𝛿-cover for 𝐴 ⊂ ℝ𝑑 , and 𝑠 > 𝑡 ≥ 0, ∑(diam 𝑈 𝑗 )𝑠 = ∑(diam 𝑈 𝑗 )𝑠−𝑡 (diam 𝑈 𝑗 )𝑡 ≤ 𝛿𝑠−𝑡 ∑(diam 𝑈 𝑗 )𝑡 , 𝑗

𝑗

𝑗

10.1. Hausdorff dimension

187

and, by the definition (10.1), (10.3)

𝐻𝛿𝑠 (𝐴) ≤ 𝛿𝑠−𝑡 𝐻𝛿𝑡 (𝐴).

We thus have Theorem 10.4. Theorem 10.4. Let 𝐴 ⊂ ℝ𝑑 and 𝑠 > 𝑡 ≥ 0. (1) If ℋ 𝑡 (𝐴) < ∞, then ℋ 𝑠 (𝐴) = 0. (2) If ℋ 𝑠 (𝐴) > 0, then ℋ 𝑡 (𝐴) = ∞. Proof. To prove (1), assume that ℋ 𝑡 (𝐴) < ∞. Thus 𝐻𝛿𝑡 (𝐴) is bounded in 𝛿 and, by (10.3), we obtain ℋ 𝑠 (𝐴) = lim 𝐻𝛿𝑠 (𝐴) = 0. 𝛿→0

(2) is the contrapositive of (1).

□

Theorem 10.4 allows us to conclude that there exists some number 𝐷 such that ℋ 𝑠 (𝐴) = ∞ if 𝑠 < 𝐷, and ℋ 𝑠 (𝐴) = 0 if 𝑠 > 𝐷 (see Figure H s (A) ∞

D Figure 10.1. The number 𝐷, the Hausdorff dimension of 𝐴, satisfies that ℋ 𝑠 (𝐴) = ∞ if 𝑠 < 𝐷, and ℋ 𝑠 (𝐴) = 0 if 𝑠 > 𝐷.

10.1). Indeed, we see that 𝐷 = inf{𝑠 ∶ ℋ 𝑠 (𝐴) = 0}.

s

188

10. Mathematics of fractals

The value of ℋ 𝐷 (𝐴) may be zero or infinity, or even a number 0 < ℋ 𝐷 (𝐴) < ∞. 𝐷 is called the Hausdorff dimension of 𝐴, and is denoted by dim 𝐴. Example 10.5. For 𝑠 = 0, ℋ 0 (𝐴) = #𝐴, the number of elements of 𝐴, if 𝐴 is finite, and thus dim 𝐴 = 0. If 𝐴 is infinite, ℋ 0 (𝐴) = ∞. However, if 𝐴 is countably infinite, its Hausdorff dimension is also zero (Exercise (3)). 10.6. We observe that, if 𝐴 ⊂ ℝ, ℋ 1 (𝐴) is the outer Lebesgue measure of 𝐴, |𝐴|∗ . Thus, for any interval [𝑎, 𝑏], 𝑏 > 𝑎, ℋ 1 ([𝑎, 𝑏]) = 𝑏 − 𝑎, and thus intervals have Hausdorff dimension 1. 10.7. If 𝑠 > 𝑑 and 𝐴 ⊂ ℝ𝑑 , ℋ 𝑠 (𝐴) = 0. Indeed, consider first a unit cube 𝑄 in ℝ𝑑 and, given 𝛿 > 0, subdivide it in 2𝑑𝑁 subcubes 𝑄𝑗 of sides 2−𝑁 < 𝛿/𝑑. Each subcube has diameter diam(𝑄𝑗 ) = 2−𝑁 √𝑑 < 𝛿, and hence {𝑄𝑗 } is a 𝛿-cover for 𝑄. Thus ∑(diam 𝑄𝑗 )𝑠 = (2−𝑁 √𝑑)𝑠 ⋅ 2𝑑𝑁 = 𝑑 𝑠/2 2(𝑑−𝑠)𝑁 . 𝑗

As 𝑁 is an arbitrary positive integer (as long as 2−𝑁 < 𝛿/𝑑) and 𝑠 > 𝑑, then 𝐻𝛿𝑠 (𝑄) = 0. Hence ℋ 𝑠 (𝑄) = 0, and therefore ℋ 𝑠 (𝐴) = 0 for any 𝐴 ⊂ ℝ𝑑 , because any 𝐴 is contained in the union of countably many unit cubes. Therefore dim 𝐴 ≤ 𝑑 for any 𝐴 ⊂ ℝ𝑑 . Example 10.8. Consider the Cantor ternary set, constructed by the removal of middle intervals starting from 𝐶0 = [0, 1]. Thus 𝐶1 = [0, 1/3] ∪ [2/3, 1], 𝐶2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1],

...

as seen in Figure 10.2. The Cantor set is

∞

𝐶=

⋂

𝐶𝑛 .

𝑛=0

The Cantor set is compact, perfect (does not have any isolated points), uncountable and of measure 0. Thus ℋ 1 (𝐶) = 0. We show that its

10.1. Hausdorff dimension

189

Figure 10.2. The construction of the Cantor ternary set.

Hausdorff dimension is 𝐷=

log 2 ≈ 0.63. log 3

To see that dim 𝐶 ≤ 𝐷, we prove ℋ 𝐷 (𝐶) < ∞. Given 𝛿 > 0, let 𝑛 so that 3−𝑛 < 𝛿. Thus, 𝐶𝑛 is the union of 2𝑛 intervals 𝐼𝑗 of length 3−𝑛 < 𝛿, so they form a 𝛿-cover for 𝐶. Hence 2𝑛

𝐻𝛿𝐷 (𝐶)

≤ ∑ (diam 𝐼𝑗 )𝐷 = (3−𝑛 )𝐷 2𝑛 = ( 𝑗=1

2 𝑛 ) = 1, 3𝐷

as 𝐷 is the number such that 3𝐷 = 2. Since 𝛿 > 0 is arbitrary, we have that ℋ 𝐷 (𝐶) ≤ 1, and therefore dim 𝐶 ≤ 𝐷. To show that dim 𝐶 ≥ 𝐷 (and thus conclude they are equal), we verify that ℋ 𝐷 (𝐶) > 0. In fact, we prove that ℋ 𝐷 (𝐶) ≥ 1, and we will do it by contradiction. Assume ℋ 𝐷 (𝐶) < 1, so there exist intervals 𝐼𝑗 so that 𝐶⊂

⋃

𝐼𝑗

and

∑(diam 𝐼𝑗 )𝐷 < 1. 𝑗

𝑗

By widening the intervals a little bit so that their sum is still smaller than 1, we can assume all 𝐼𝑗 are open, and thus, by the compactness of 𝐶, we can choose a finite number of them that still cover 𝐶. Hence we have 𝑁

(10.9)

𝐶⊂

⋃ 𝑗=1

𝑁

𝐼𝑗

and

∑ (diam 𝐼𝑗 )𝐷 < 1. 𝑗=1

Now, again by compactness of 𝐶, we can find 𝑀 large enough so that each interval of 𝐶𝑀 is completely contained in one of the 𝐼𝑗 . We can now shorten each 𝐼𝑗 to a closed interval in such a way that its extreme points coincide with extreme points of the intervals of 𝐶𝑀 , and they still cover

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10. Mathematics of fractals

𝐶. That is, if 𝐼𝑗 contains the intervals 𝐷1 , 𝐷2 , . . . , 𝐷𝑘 in 𝐶𝑀 , we replace 𝐼𝑗 with the smallest closed interval that contains 𝐷1 ∪ 𝐷2 ∪ . . . ∪ 𝐷𝑘 . Fix 𝐼𝑗 . Let 𝐸0 be the gap of largest length between the 𝐷𝑖 . By the construction of 𝐶𝑀 , such gap must exist. We write 𝐼𝑗 ⧵𝐸0 as the union 𝐸1 ∪ 𝐸2 of closed intervals, as in Figure 10.3. We see that, by the construction, diam 𝐸1 , diam 𝐸2 ≤ diam 𝐸0 .

Figure 10.3. The largest gap inside 𝐼𝑗 , for the special case when 𝐼𝑗 contains the intervals 𝐷1 , 𝐷2 , 𝐷3 . By the construction of 𝐶 𝑀 , such largest gap must exist and is unique.

Thus diam 𝐸0 ≥

diam 𝐸1 + diam 𝐸2 2

and diam 𝐼𝑗 = diam 𝐸0 + diam 𝐸1 + diam 𝐸2 3 3 1 1 ≥ diam 𝐸1 + diam 𝐸2 = 3( diam 𝐸1 + diam 𝐸2 ). 2 2 2 2 Now, since 𝐷 < 1, the function 𝑥 ↦ 𝑥𝐷 is concave (Exercise (6)), and thus 𝐷 1 1 (diam 𝐼𝑗 )𝐷 ≥ 3𝐷 ( diam 𝐸1 + diam 𝐸2 ) 2 2 1 1 ≥ 3𝐷 ( (diam 𝐸1 )𝐷 + (diam 𝐸2 )𝐷 ) 2 2 = (diam 𝐸1 )𝐷 + (diam 𝐸2 )𝐷 ,

where we have again used the fact that 3𝐷 = 2. Therefore, we can replace 𝐼𝑗 with the intervals 𝐸1 and 𝐸2 , and the new set of intervals still satisfies (10.9). We can continue replacing the intervals 𝐼𝑗 until all gaps are removed, so we arrive to the cover formed precisely by the 2𝑀 intervals of 𝐶𝑀 ,

10.2. Self-similar sets

191

which of course satisfies 2𝑀

∑ (diam 𝐼𝑗 )𝐷 = 2𝑀 (3−𝑀 )𝐷 = 1, 𝑗=1

a contradiction with (10.9).

10.2. Self-similar sets Cantor’s example shows us the difficulty of calculating the Hausdorff dimension of a given set 𝐴 ⊂ ℝ𝑑 . However, under certain conditions, it is possible to calculate the Hausdorff dimension a self-similar set. A function 𝐹 ∶ ℝ𝑑 → ℝ𝑑 is called a contraction if there exists a constant 𝛼 < 1 such that |𝐹(𝑥) − 𝐹(𝑦)| ≤ 𝛼|𝑥 − 𝑦| for all 𝑥, 𝑦 ∈ ℝ𝑑 . The number 𝛼 is called the contraction constant of 𝐹. A nonempty compact set 𝐾 ⊂ ℝ𝑑 is self-similar if there exist contractions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 ∶ ℝ𝑑 → ℝ𝑑 such that (10.10)

𝐾 = 𝑓1 (𝐾) ∪ 𝑓2 (𝐾) ∪ . . . ∪ 𝑓𝑁 (𝐾).

One can show that 𝐾 is determined by the contractions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 , that is, there exists only one nonempty compact set 𝐾 that satisfies (10.10) [YHK97]. Let 𝛼1 , 𝛼2 , . . . , 𝛼𝑁 be the contraction constants of the functions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 in (10.10) and let 𝐷 be the unique positive number such that (10.11)

𝐷 𝐷 𝛼𝐷 1 + 𝛼2 + . . . + 𝛼𝑁 = 1.

Such number exists because the function 𝑥 ↦ 𝛼𝑥1 + 𝛼𝑥2 + . . . + 𝛼𝑥𝑁 is strictly decreasing on [0, ∞), equal to 𝑁 at 𝑥 = 0 and goes to 0 as 𝑥 → ∞. We have the following result. Proposition 10.12. If 𝐾 is the self-similar set determined by (10.10) and 𝐷 satisfies (10.11), then dim 𝐾 ≤ 𝐷. Proof. We prove that ℋ 𝐷 (𝐾) < ∞. For this, we will use the following notation. For a finite sequence 𝑤 = 𝑤 1 𝑤 2 ⋯ 𝑤 𝑚 of length 𝑚 (we will call such sequences words), where each 𝑤𝑗 ∈ {1, 2, . . . , 𝑁}, let 𝑓𝑤 denote

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10. Mathematics of fractals

the composition 𝑓𝑤 = 𝑓𝑤1 ∘ 𝑓𝑤2 ∘ ⋯ ∘ 𝑓𝑤𝑚 , and 𝐾𝑤 the set 𝑓𝑤 (𝐾). Each 𝐾𝑤 is called a cell of level 𝑚, and, by (10.10), 𝐾=

⋃

𝐾𝑤 ,

𝑤∈𝑊𝑚

where 𝑊𝑚 is the set of words of length 𝑚. Also, since each 𝑓𝑗 is a contraction with constant 𝛼𝑗 , we have diam 𝐾𝑤 ≤ 𝛼𝑤1 𝛼𝑤2 ⋯ 𝛼𝑤𝑚 diam 𝐾, and thus diam 𝐾𝑤 ≤ 𝐴𝑚 diam 𝐾, where 𝐴 = max{𝛼1 , 𝛼2 , . . . , 𝛼𝑁 }. Since 𝐴 < 1, given 𝛿 > 0 we can choose 𝑚 large enough so that the collection of cells of level 𝑚, {𝐾𝑤 }𝑤∈𝑊𝑚 , is a 𝛿-cover for 𝐾. Also, 𝐷 𝐷 𝐷 ∑ (diam𝐾𝑤 )𝐷 ≤ ∑ 𝛼𝐷 𝑤1 𝛼𝑤2 ⋯ 𝛼𝑤𝑚 (diam 𝐾) 𝑤∈𝑊𝑚

𝑤∈𝑊𝑚 𝑁

𝑁

𝑁

𝐷 𝐷 𝐷 = ( ∑ 𝛼𝐷 𝑤1 )( ∑ 𝛼𝑤2 ) ⋯ ( ∑ 𝛼𝑤𝑚 )(diam 𝐾) 𝑤1 =1

𝑤2 =1

𝑤𝑚 =1

𝐷

= (diam 𝐾) , by (10.11). Hence 𝐻𝛿𝐷 (𝐾) ≤ (diam 𝐾)𝐷 and, as 𝛿 > 0 is arbitrary, we have that ℋ 𝐷 (𝐾) ≤ (diam 𝐾)𝐷 < ∞. Therefore dim 𝐾 ≤ 𝐷. □ Consider the Cantor set 𝐶 discussed in Example 10.8. 𝐶 is selfsimilar with respect to the contractions 𝑓1 , 𝑓2 ∶ ℝ → ℝ given by 𝑓1 (𝑥) =

1 𝑥 3

and

Thus, in this case, 𝛼1 = 𝛼2 =

𝑓2 (𝑥) =

1 2 𝑥+ . 3 3

1 and the number 𝐷 that satisfies 3

1 𝐷 1 𝐷 ( ) +( ) =1 3 3 is precisely 𝐷=

log 2 , log 3

which we proved to be the Hausdorff dimension of 𝐶. However, (10.11) does not guarantee that 𝐷 is the Hausdorff dimension of 𝐾 for any self-similar set, as Example 10.13 shows.

10.2. Self-similar sets

193

Example 10.13. Consider the functions 𝑔1 , 𝑔2 ∶ ℝ → ℝ given by 2 2 1 𝑥 and 𝑔2 (𝑥) = 𝑥 + . 3 3 3 𝑔1 and 𝑔2 are contractions and, if 𝐼 = [0, 1], then 𝑔1 (𝑥) =

𝐼 = 𝑔1 (𝐼) ∪ 𝑔2 (𝐼), so 𝐼 is self-similar with respect to 𝑔1 and 𝑔2 . Both functions have contraction constant 2/3, and the number 𝐷 that satisfies 2 𝐷 2 𝐷 ( ) +( ) =1 3 3 is

log 2 . log 3/2 Note that 𝐷 > 1, so it is not equal to dim 𝐼. 𝐷=

The problem with Example 10.13 is the overlap of the images 𝑔1 (𝐼) = [0, 2/3]

and

𝑔2 (𝐼) = [1/3, 1]

of 𝐼 under the contractions 𝑔1 and 𝑔2 , so we need conditions to avoid this problem. We say that a function 𝑓 ∶ ℝ𝑑 → ℝ𝑑 is a similitude if there exists a constant 𝛼 such that |𝑓(𝑥) − 𝑓(𝑦)| = 𝛼|𝑥 − 𝑦|. It can be verified that, if 𝑓 is a similitude, then it’s a composition of a dilation, a rotation and a translation (Exercise (7)). Thus, for any ball 𝐵𝑟 (𝑥) ⊂ ℝ𝑑 , (10.14)

𝑓(𝐵𝑟 (𝑥)) = 𝐵𝛼𝑟 (𝑓(𝑥)).

Let 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 be contractive similitudes. They satisfy the open set condition if there exists a bounded open set 𝑈 ⊂ ℝ𝑑 such that (1) 𝑓𝑗 (𝑈) ⊂ 𝑈 for each 𝑗 = 1, 2, . . . , 𝑁; (2) 𝑓𝑖 (𝑈) ∩ 𝑓𝑗 (𝑈) = ∅, for 𝑖, 𝑗 = 1, 2, . . . , 𝑁 such that 𝑖 ≠ 𝑗. One can verify that, since 𝑓1 (𝑈) ∪ 𝑓2 (𝑈) ∪ . . . ∪ 𝑓𝑁 (𝑈) ⊂ 𝑈, then 𝐾 ⊂ 𝑈,̄ where 𝑈̄ is the closure of 𝑈 (Exercise (8)).

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10. Mathematics of fractals

Condition (2) guarantees that there are no interior overlaps between the images 𝑓𝑗 (𝐾) of the contractions. Theorem 10.15. Let 𝑓𝑗 ∶ ℝ𝑑 → ℝ𝑑 , 𝑗 = 1, 2, . . . , 𝑁, be contractive similitudes with constants 𝛼𝑗 < 1, respectively, that satisfy the open set condition. If 𝐾 is the self-similar set with respect to the 𝑓𝑗 , then dim 𝐾 = 𝐷, where 𝐷 is the unique solution to (10.11). Proof. We already know that dim 𝐾 ≤ 𝐷, by Proposition 10.12. To prove that dim 𝐾 ≥ 𝐷, we verify that ℋ 𝐷 (𝐾) > 0. Let 𝑈 ⊂ ℝ𝑑 be a bounded open set as in the open set condition, and let 𝛼, 𝛽 > 0 and 𝑥0 , 𝑦0 ∈ ℝ𝑑 such that (10.16)

𝐵𝛼̄ (𝑥0 ) ⊂ 𝑈 ⊂ 𝐵𝛽 (𝑦0 ).

Since the 𝑓𝑗 are similitudes, for each word 𝑤 ∈ 𝑊𝑚 we have, by (10.14), (10.17)

𝐵𝛼̄ 𝑤 𝛼 (𝑓𝑤 (𝑥0 )) ⊂ 𝑈𝑤 ⊂ 𝐵𝛼𝑤 𝛽 (𝑓𝑤 (𝑦0 )),

where 𝛼𝑤 = 𝛼𝑤1 𝛼𝑤2 ⋯ 𝛼𝑤𝑚 and 𝑈𝑤 = 𝑓𝑤 (𝑈). Let 𝛾 = min{𝛼1 , 𝛼2 , . . . , 𝛼𝑁 } and 𝜏=(

𝛼𝛾 𝑑 ) . 2𝛽 + 1

We will prove that ℋ 𝐷 (𝐾) ≥ 𝜏. In order to arrive to a contradiction, assume ℋ 𝐷 (𝐾) < 𝜏. Let 𝛿 > 0 be small enough (𝛿 < 𝜏 is sufficient) and {𝑉 𝑗 } a 𝛿-cover for 𝐾 such that (10.18)

∑ diam(𝑉 𝑗 )𝐷 < 𝜏. 𝑗

By widening the 𝑉 𝑗 up we can assume they are open, and so we can also assume they are finite because 𝐾 is compact. For each 𝑗, consider the set of words Λ𝑗 = {𝑤 ∶ 𝛼𝑤 ≤ diam 𝑉 𝑗 < 𝛼𝑤′ } where, if 𝑤 = 𝑤 1 𝑤 2 . . . 𝑤 𝑚 , then 𝑤′ = 𝑤 1 𝑤 2 . . . 𝑤 𝑚−1 . That is, Λ𝑗 is the set of words 𝑤 of sequences 𝑤 1 𝑤 2 . . . 𝑤 𝑚 which make the product 𝛼𝑤1 𝛼𝑤2 ⋯ 𝛼𝑤𝑚 precisely smaller or equal to diam 𝑉 𝑗 , and larger if we miss the last factor 𝛼𝑤𝑚 . The words in Λ𝑗 satisfy the following two properties.

10.2. Self-similar sets

195

• If 𝑢, 𝑤 ∈ Λ𝑗 are different, then the letters of 𝑢 cannot be the first letters of 𝑤, nor the other way around. This guarantees that, by the open set condition, 𝑈ᵆ ∩ 𝑈𝑤 = ∅

if 𝑢, 𝑤 ∈ Λ𝑗 , 𝑢 ≠ 𝑤.

• If 𝑤 ∈ Λ𝑗 , 𝛾 diam 𝑉 𝑗 < 𝛼𝑤 ≤ diam 𝑉 𝑗 , so all the products 𝛼𝑤 , for 𝑤 ∈ Λ𝑗 , are essentially of the same size. Let 𝑥 ∈ 𝑉 𝑗 . Then 𝑉 𝑗 ⊂ 𝐵diam 𝑉𝑗 (𝑥). Using (10.17), we have 𝑈𝑤 ⊂ 𝐵𝛼𝑤 𝛽 (𝑓𝑤 (𝑦0 )) ⊂ 𝐵2𝛼𝑤 𝛽+diam 𝑉𝑗 (𝑥) ̄ ∩ 𝑉 𝑗 ≠ ∅. As 𝛼𝑤 𝛽 ≤ 𝛽 diam 𝑉 𝑗 , we have for each 𝑤 ∈ Λ𝑗 such that 𝑈𝑤 (10.19)

𝑈𝑤 ∈ 𝐵(2𝛽+1) diam 𝑉𝑗 (𝑥).

Thus, the ball 𝐵(2𝛽+1) diam 𝑉𝑗 (𝑥) contains all sets 𝑈𝑤 such that 𝑤 ∈ Λ𝑗 ̄ ∩ 𝑉 𝑗 ≠ ∅. Moreover, each set 𝑈𝑤 contains the ball 𝐵𝛼 𝛼 (𝑓𝑤 (𝑥0 )), and 𝑈𝑤 𝑤 by (10.17), and all of them are disjoint, so the ball 𝐵(2𝛽+1) diam 𝑉𝑗 (𝑥) contains, say, 𝑝 balls of radius 𝛼𝑤 𝛼 = 𝛼𝑤′ 𝛼𝑤𝑚 𝛼 > 𝛾𝛼 diam 𝑉 𝑗 , where ̄ ∩ 𝑉 𝑗 ≠ ∅}. 𝑝 = #{𝑤 ∈ Λ𝑗 ∶ 𝑈𝑤 Since the measure |𝐵𝑟 (𝑥)| of a ball of radius 𝑟 is given by 𝜔𝑑 𝑟𝑑 /𝑑, we have the inequality 𝑑

𝑑

((2𝛽 + 1) diam 𝑉 𝑗 ) ≥ 𝑝(𝛾𝛼 diam 𝑉 𝑗 ) , from the fact that we have a ball of radius (2𝛽 + 1) diam 𝑉 𝑗 that contains 𝑝 disjoint balls of radius 𝛾𝛼 diam 𝑉 𝑗 . Therefore 𝑝≤(

2𝛽 + 1 𝑑 1 ) = . 𝛾𝛼 𝜏

Now, for each 𝑚 and 𝑗, we consider the sum 𝐴𝑚 (𝑗) =

∑ 𝑤=𝑤1 𝑤2 . . .𝑤𝑚 𝑈̄ 𝑤 ∩𝑉𝑗 ≠0

𝛼𝐷 𝑤.

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10. Mathematics of fractals

𝐴𝑚 (𝑗) is decreasing in 𝑚 because 𝑈𝑤𝑖 ⊂ 𝑈𝑤 for each 𝑖 = 1, 2, . . . , 𝑁, and 𝑁 ∑𝑖=1 𝛼𝐷 𝑖 = 1. Hence, if 𝐿 = max{𝑚 ∶ 𝑤 1 𝑤 2 . . . 𝑤 𝑚 ∈ Λ𝑗 }, we have, as each 𝛼𝑖 < 1, ∑

𝐴𝐿 (𝑗) ≤

𝛼𝐷 𝑤 ≤

𝑤∈Λ𝑗 𝑈̄ 𝑤 ∩𝑉𝑗 ≠∅

∑

(diam 𝑉 𝑗 )𝐷 = 𝑝(diam 𝑉 𝑗 )𝐷

𝑤∈Λ𝑗 𝑈̄ 𝑤 ∩𝑉𝑗 ≠∅

1 (diam 𝑉 𝑗 )𝐷 . 𝜏 However, as 𝐾 ⊂ 𝑈,̄ 𝐾 ⊂ ⋃𝑗 𝑉 𝑗 , we conclude ≤

∑

1=

𝑤=𝑤1 𝑤2 . . .𝑤𝐿 𝑈̄ 𝑤 ∩𝐾≠∅


0. (a) If 𝐴 ⊂ 𝐵, then 𝐻𝛿𝑠 (𝐴) ≤ 𝐻𝛿𝑠 (𝐵). (b) If 𝐴 = ⋃𝑗 𝐴𝑗 , then 𝐻𝛿𝑠 (𝐴) ≤ ∑ 𝐻𝛿𝑠 (𝐴𝑗 ). 𝑗

(2) Let 𝑠 ≥ 0. (a) If 𝐴 ⊂ 𝐵, then ℋ 𝑠 (𝐴) ≤ ℋ 𝑠 (𝐵). (b) If 𝐴 = ⋃𝑗 𝐴𝑗 , then ℋ 𝑠 (𝐴) ≤ ∑ ℋ 𝑠 (𝐴𝑗 ). 𝑗 𝑠

(c) If dist(𝐴, 𝐵) > 0, then ℋ (𝐴 ∪ 𝐵) = ℋ 𝑠 (𝐴) + ℋ 𝑠 (𝐵). (3) If 𝐴 is countable, then dim 𝐴 = 0. (4) There exist constants 𝑐, 𝐶 > 0 such that, for any measurable 𝐴 ⊂ ℝ𝑑 , 𝑐|𝐴| ≤ ℋ 𝑑 (𝐴) ≤ 𝐶|𝐴|, and thus ℋ 𝑑 is comparable to Lebesgue measure on ℝ𝑑 . (5) If 𝐴 ⊂ ℝ𝑑 is open, then dim 𝐴 = 𝑑. (6) If 0 < 𝑝 < 1, the function 𝑥 ↦ 𝑥𝑝 is concave, that is, for 𝑥, 𝑦 > 0 and 0 ≤ 𝑡 ≤ 1, (𝑡𝑥 + (1 − 𝑡)𝑦)𝑝 ≥ 𝑡𝑥𝑝 + (1 − 𝑡)𝑦𝑝 . (7) Let 𝑓 ∶ ℝ𝑑 → ℝ𝑑 be a similitude with coefficient 𝛼 > 0: for every 𝑥, 𝑦 ∈ ℝ𝑑 , |𝑓(𝑥) − 𝑓(𝑦)| = 𝛼|𝑥 − 𝑦| 1 Let 𝑔(𝑥) = (𝑓(𝑥) − 𝑓(0)). 𝛼 (a) For all 𝑥, 𝑦 ∈ ℝ𝑑 , 𝑔(𝑥) ⋅ 𝑔(𝑦) = 𝑥 ⋅ 𝑦. (b) If 𝑒 1 , 𝑒 2 , . . . , 𝑒 𝑑 is the standard basis for ℝ𝑑 , then 𝑔(𝑒 1 ), 𝑔(𝑒 2 ), . . ., 𝑔(𝑒 𝑑 ) is an orthonormal basis for ℝ𝑑 . (c) For 𝑥, 𝑦 ∈ ℝ𝑑 and 𝑗 = 1, 2, . . . , 𝑑, 𝑔(𝑥 + 𝑦) ⋅ 𝑔(𝑒𝑗 ) = (𝑔(𝑥) + 𝑔(𝑦)) ⋅ 𝑔(𝑒𝑗 ). (d) For 𝑥 ∈ ℝ𝑑 , 𝜆 ∈ ℝ and 𝑗 = 1, 2, . . . , 𝑑, 𝑔(𝜆𝑥) ⋅ 𝑔(𝑒𝑗 ) = 𝜆𝑔(𝑥) ⋅ 𝑔(𝑒𝑗 ).

202

10. Mathematics of fractals (e) 𝑔 is an orthogonal linear transformation. (f) Conclude that every similitude in ℝ𝑑 is of the form 𝑓(𝑥) = 𝛼𝑀𝑥 + 𝑥0 , where 𝛼 > 0, 𝑀 ∈ 𝑂(𝑑) and 𝑥0 ∈ ℝ𝑑 .

(8) Let 𝐾 be the self-similar set with respect to the contractions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 ∶ ℝ𝑑 → ℝ𝑑 , and suppose 𝐴 ⊂ ℝ𝑑 is nonempty and satisfies 𝑓1 (𝐴) ∪ 𝑓2 (𝐴) ∪ . . . ∪ 𝑓𝑁 (𝐴) ⊂ 𝐴. Then 𝐾 ⊂ 𝐴,̄ where 𝐴̄ is the closure of 𝐴. (Hint: Prove that, if 𝐴𝜀 = {𝑥 ∈ ℝ𝑑 ∶ there exists 𝑦 ∈ 𝐴 such that |𝑥 − 𝑦| < 𝜀}, then 𝐾 ⊂ 𝐴𝜀 for any 𝜀 > 0.) (9) Describe explicitly the convex hull of the Hata tree set of Example 10.22, and prove that its interior is an open set under which the contractions that define the Hata set satisfy the open set conditions. (10) Describe explicitly an open set, under which the open set condition is satisfied, for the golden fractal of Example 10.23. (11) The polynomial 𝑥3 + 2𝑥 − 1 has only one real root, and it belongs to the interval (0, 𝛾), where 𝛾 = 1/𝜑 is the reciprocal of the golden ratio.

Notes The Hausdorff dimension was introduced by Felix Hausdorff in [Hau18]. Theorem 10.15 was proven by John Hutchinson in [Hut81]. It is sometimes known as Moran’s theorem due to Patrick Alfred Pierce Moran’s related work in [Mor46]. The proof presented here is from [YHK97]. The Sierpiński gasket was studied by Wacław Sierpiński in [Sie15]. The snowflake set is presented in [Kig01]. The Hata set was introduced by Masayoshi Hata in [Hat85], and the golden fractal by Marc Frantz in [Fra09].

Chapter 11

The Laplacian on the Sierpiński gasket

In the previous chapters, we studied harmonic funcions on regions given by open subsets of the Euclidean space ℝ𝑑 . However, in the study of diffusion in disordered media, fractals sets, as the ones we studied in Chapter 10, are better suited as models to those systems. The purpose of this chapter is to construct a Laplacian—and study the harmonic functions, on the Sierpiński gasket—the set 𝑆 described in Example 10.20. We will do this by working on discrete approximations to 𝑆. To motivate our methods, we start by discussing a discrete approach to harmonic analysis on the interval [0, 1].

11.1. Discrete energy on the interval We start with the following model. Consider a spring placed between two nodes at points (0, 𝑎) and (1, 𝑏) in the plane, as in Figure 11.1 (left). If we assume that it is in equilibrium when 𝑎 = 𝑏, then its energy is given by a constant multiple of (11.1)

ℰ0 = (𝑎 − 𝑏)2 .

Now, assume we put a node at the point (1/2, 𝑥), as in Figure 11.1 (right). We now have a pair of springs whose total energy is given by (11.2)

ℰ1 = 𝛼((𝑎 − 𝑥)2 + (𝑥 − 𝑏)2 ), 203

204

11. The Laplacian on the Sierpiński gasket

Figure 11.1. The spring on the left has energy ℰ0 = (𝑎 − 𝑏)2 , while the pair of springs on the right have total energy ℰ1 = 𝛼((𝑎 − 𝑥)2 + (𝑥 − 𝑏)2 ).

for an appropriate constant 𝛼. It is not hard to see that the value of 𝑥 that minimizes (11.2) is 𝑎+𝑏 𝑥∗ = , 2 and at this value the energy ℰ1 in (11.2) is equal to 𝛼 ℰ1 = . 2 Note that if the node is placed at (1/2, 𝑥∗ ), then it is the midpoint of the line segment from (0, 𝑎) to (1, 𝑏). Hence the total energy of the springs must be the same as ℰ0 , and thus we must choose 𝛼 = 2. If we continue subdividing dyadically the interval [0, 1], we obtain a partition 1 2 3 𝑉𝑚 = {0, 𝑚 , 𝑚 , 𝑚 , . . . , 1} 2 2 2 for each 𝑚 ≥ 1. Given a real valued function 𝑢 on 𝑉𝑚 , we define its energy ℰ𝑚 (𝑢) as the quadratic form 2𝑚

(11.3)

ℰ𝑚 (𝑢) = 2

𝑚

∑ (𝑢( 𝑘=1

𝑘−1 𝑘 2 − 𝑢( ) )) . 2𝑚 2𝑚

If 𝑢(0) = 𝑎, 𝑢(1) = 𝑏, then ℰ0 (𝑢) = (𝑎 − 𝑏)2 , and, as we observed above, min{ℰ1 (𝑢) ∶ 𝑢(0) = 𝑎, 𝑢(1) = 𝑏} = ℰ0 (𝑢) = (𝑎 − 𝑏)2 , and is attained when 𝑢(1/2) = (𝑎 + 𝑏)/2. Inductively, we can verify that min{ℰ𝑚 (𝑢) ∶ 𝑢(0) = 𝑎, 𝑢(1) = 𝑏} = ℰ0 (𝑢) for each 𝑚 ≥ 1, and that is attained at the function 𝑢 that satisfies 1 𝑘−1 2𝑘 − 1 𝑘 (11.4) 𝑢( 𝑚 ) = (𝑢( 𝑚−1 ) + 𝑢( 𝑚−1 )) 2 2 2 2

11.1. Discrete energy on the interval

205

for each 𝑚 ≥ 1 and each 1 ≤ 𝑘 ≤ 2𝑚−1 , and thus can be constructed inductively by the algorithm (11.4). In fact, (11.4) describes the dyadic points of the line segment from (0, 𝑢(0)) to (1, 𝑢(1)). By the mean value theorem, if 𝑢 is a differentiable function in [0, 1], for each 1 ≤ 𝑘 ≤ 2𝑚 there is some 𝑡 𝑘 ∈ [(𝑘 − 1)/2𝑚 , 𝑘/2𝑚 ] such that 1 𝑘−1 𝑘 ) − 𝑢( 𝑚 ) = 𝑢′ (𝑡 𝑘 ) ⋅ 𝑚 , 2𝑚 2 2 and hence (11.3) can be written as 𝑢(

2𝑚

2𝑚

𝑢′ (𝑡 ) 2 1 ℰ𝑚 (𝑢) = 2 ∑ ( 𝑚𝑘 ) = ∑ 𝑢′ (𝑡 𝑘 )2 ⋅ 𝑚 , 2 2 𝑘=1 𝑘=1 𝑚

which, if 𝑢′ is Riemann-integrable on [0, 1], is a Riemann sum of the integral 1

ℰ(𝑢) = ∫ 𝑢′ (𝑡)2 𝑑𝑡,

(11.5)

0

the energy of the function 𝑢 on [0, 1]. We note that the minimizers of this energy are the linear functions, which are the harmonic functions in [0, 1], and the continuous limit of algorithm (11.4). Moreover, we can observe that, if we polarize the quadratic form ℰ𝑚 , we see that the bilinear form 2𝑚

ℰ𝑚 (𝑢, 𝑣) = 2

𝑚

∑ (𝑢( 𝑘=1

𝑘−1 𝑘 𝑘−1 𝑘 ) − 𝑢( 𝑚 ))(𝑣( 𝑚 ) − 𝑣( 𝑚 )) 2𝑚 2 2 2

converges, for 𝑢, 𝑣 differentiable in [0, 1] and 𝑢′ , 𝑣′ Riemann-integrable, to 1

ℰ(𝑢, 𝑣) = ∫ 𝑢′ (𝑡)𝑣′ (𝑡)𝑑𝑡, 0

the energy form on the interval studied in Section 1.4. If, say, 𝑢 ∈ 𝐶 2 ([0, 1]), and 𝑣 is zero at the boundary points 0 and 1, then we can integrate by parts to obtain 1

ℰ(𝑢, 𝑣) = − ∫ 𝑢″ (𝑡)𝑣(𝑡)𝑑𝑡.

(11.6)

0 ″

Note that 𝑢 is the Laplacian of 𝑢 for a one-variable function, and that 𝑢 is harmonic (𝑢″ (𝑡) = 0 for all 𝑡 ∈ [0, 1]) if and only if it is a linear function.

206

11. The Laplacian on the Sierpiński gasket We can also obtain 𝑢″ (𝑡) by a discrete limit. Note that we can rewrite 2𝑚

ℰ𝑚 (𝑢, 𝑣) = − ∑ 𝐻𝑚 𝑢( 𝑘=0

𝑘 𝑘 )𝑣( 𝑚 ), 2𝑚 2

where 𝐻𝑚 𝑢(

𝑘 1 𝑘−1 𝑘+1 𝑘 ) = −𝑚 (𝑢( 𝑚 ) + 𝑢( 𝑚 ) − 2𝑢( 𝑚 )) 𝑚 2 2 2 2 2

if 1 ≤ 𝑘 ≤ 2𝑚 − 1, 1 1 (𝑢( 𝑚 ) − 𝑢(0)), 2−𝑚 2

𝐻𝑚 𝑢(0) = and

2𝑚 − 1 ) − 𝑢(1)). 2𝑚 If, for 1 ≤ 𝑘 ≤ 2𝑚 − 1, we define the piecewise linear function 𝑣 by 𝐻𝑚 𝑢(1) =

𝑣(

1

2−𝑚

(𝑢(

1 if 𝑗 = 𝑘 𝑗 )={ 𝑚 2 0 otherwise,

then

𝑘 ). 2𝑚 In fact, one can verify explicitly (Exercise (1)) that ℰ𝑚 (𝑢, 𝑣) = −𝐻𝑚 𝑢( 1

∫ 𝑢″ (𝑡)𝑣(𝑡)𝑑𝑡 = 𝐻𝑚 𝑢( 0

𝑘 ). 2𝑚

2

As 𝑢 ∈ 𝐶 ([0, 1]), given 𝜀 > 0 we can choose 𝑚 large enough so that |𝑢″ (𝑥) − 𝑢″ (𝑦)| < 𝜀 for |𝑥 − 𝑦| ≤ 2−𝑚 . Thus 1

1

1

| ∫ 𝑢″ (𝑡)𝑣(𝑡)𝑑𝑡 − 𝑢″ ( 𝑘 ) ∫ 𝑣(𝑡)𝑑𝑡| ≤ ∫ |𝑢″ (𝑡) − 𝑢″ ( 𝑘 )|𝑣(𝑡)𝑑𝑡 | | | 2𝑚 2𝑚 | 0

0

0

1

< 𝜀 ∫ 𝑣(𝑡)𝑑𝑡. 0

Since

1

∫ 𝑣(𝑡)𝑑𝑡 = 0

1 , 2𝑚

11.2. Harmonic structure on the Sierpiński gasket

207

we have that, for any dyadic point 𝑥0 in [0, 1], 1 𝐻𝑚 𝑢(𝑥0 ) = 𝑢″ (𝑥0 ). lim 𝑚→∞ 2−𝑚 The previous limit is also a known result from calculus. We call the 𝐻𝑚 the sequence of discrete Laplacians on the interval [0, 1].

11.2. Harmonic structure on the Sierpiński gasket We now proceed to define a Laplacian on the Sierpiński gasket 𝑆 introduced in Example 10.20. Recall that 𝑆 is the self-similar set in the plane that satisfies 𝑆 = 𝑓1 (𝑆) ∪ 𝑓2 (𝑆) ∪ 𝑓3 (𝑆), where the 𝑓𝑖 ∶ ℝ2 → ℝ2 are the contractions 1 𝑓𝑖 (𝑥) = (𝑥 + 𝑝 𝑖 ), 2 and the points 𝑝1 = (1/2, √3/2), 𝑝2 = (0, 0) and 𝑝3 = (1, 0) are the vertices of an equilateral triangle (Figure 11.2). As in the case of the interval, we will do this by constructing a sequence of quadratic forms

Figure 11.2. The Sierpiński gasket.

on approximating points to 𝑆, whose limit will define a quadratic form on functions on 𝑆. This quadratic form willl induce a Laplacian on the set 𝑆, which will also be a limit of discrete difference operators. We start by considering three springs on the sides of an equilateral triangle, with nodes at each of its vertices, at heights 𝑎, 𝑏 and 𝑐, as in

208

11. The Laplacian on the Sierpiński gasket

Figure 11.3. Three springs with nodes at heights 𝑎, 𝑏 and 𝑐.

Figure 11.3. The energy of this system of springs is now defined to be ℰ0 = (𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2 . Suppose that we want to add nodes at their middle points, with heights 𝑥, 𝑦, 𝑧, as in Figure 11.4. The energy is now given by ℰ1 = 𝛼((𝑎 − 𝑧)2 + (𝑧 − 𝑦)2 + (𝑦 − 𝑎)2 + (𝑧 − 𝑏)2 + (𝑏 − 𝑥)2 + (𝑥 − 𝑧)2 + (𝑦 − 𝑥)2 + (𝑥 − 𝑐)2 + (𝑐 − 𝑦)2 ), where the constant 𝛼 is chosen so that, if 𝑥, 𝑦 and 𝑧 minimize ℰ1 , then this minimum is equal to ℰ0 . Thus, we can calculate (Exercise (3)) that the minimizing values are (11.7)

𝑥∗ =

𝑎 + 2𝑏 + 2𝑐 , 5

𝑦∗ =

2𝑎 + 𝑏 + 2𝑐 , 5

𝑧∗ =

2𝑎 + 2𝑏 + 𝑐 , 5

and that

5 . 3 We observe that the values (11.7) are averages of the values at the vertices, with weights 2 to 1 depending on whether the corresponding vertex is adyacent or opposite from each middle point node. 𝛼=

Let 𝑉0 = {𝑝1 , 𝑝2 , 𝑝3 } and define, for each 𝑚 ≥ 1, the set 𝑉𝑚 = 𝑓1 (𝑉𝑚−1 ) ∪ 𝑓2 (𝑉𝑚−1 ) ∪ 𝑓3 (𝑉𝑚−1 ). Note that each 𝑥 ∈ 𝑉𝑚 is of the form 𝑓𝑤 (𝑝 𝑖 ), where, as in Chapter 10, 𝑤 = 𝑤 1 𝑤 2 . . . 𝑤 𝑚 ∈ 𝑊𝑚 , where 𝑊𝑚 is the set of words of length 𝑚 (with

11.2. Harmonic structure on the Sierpiński gasket

209

Figure 11.4. We add three more nodes, at the middle points of each previous spring, with heights 𝑥, 𝑦, 𝑧 that minimize the energy ℰ1 .

letters 𝑤𝑗 = 1, 2, 3), and 𝑓𝑤 = 𝑓𝑤1 ∘ 𝑓𝑤2 ∘ ⋯ ∘ 𝑓𝑤𝑚 .

Figure 11.5. The sets 𝑉 0 , 𝑉 1 and 𝑉 2 . The edges of the graphs join adjacent vertices at each level.

We thus have 𝑉𝑚 =

⋃

𝑓𝑤 (𝑉0 ).

𝑤∈𝑊𝑚

We say that two vertices 𝑥, 𝑦 ∈ 𝑉𝑚 are adjacent, or neighbors, and write 𝑥 ∼𝑚 𝑦, if there exists a word 𝑤 ∈ 𝑊𝑚 such that 𝑥 = 𝑓𝑤 (𝑝 𝑖 ) and 𝑦 = 𝑓𝑤 (𝑝𝑗 ), for some 𝑖, 𝑗 = 1, 2, 3. Figure 11.5 shows 𝑉0 , 𝑉1 and 𝑉2 , with edges joining neighboring vertices. For a function 𝑢 defined on 𝑉𝑚 , define the quadratic form 2 5 𝑚 (11.8) ℰ𝑚 (𝑢) = ( ) ∑ (𝑢(𝑥) − 𝑢(𝑦)) . 3 𝑥∼ 𝑦 𝑚

210

11. The Laplacian on the Sierpiński gasket

The quadratic form ℰ𝑚 defined in (11.8) is positive semidefinite, and from its definition on can verify that ℰ𝑚 (𝑢) = 0 if and only if 𝑢 is constant on 𝑉𝑚 . ℰ𝑚 is called the discrete energy of level 𝑚 on the Sierpiński gasket. If 𝑢 is a function on 𝑉𝑚 , then, for each 𝑗 = 1, 2, 3, 𝑢 ∘ 𝑓𝑗 is a function on 𝑉𝑚−1 . From (11.8) we see that 3

(11.9)

ℰ𝑚 (𝑢) =

5 ∑ℰ (𝑢 ∘ 𝑓𝑗 ). 3 𝑗=1 𝑚−1

Thus, from the previous discussion, if 𝑣 is a function on 𝑉𝑚−1 , then (11.10) min{ℰ𝑚 (𝑢) ∶ functions 𝑢 on 𝑉𝑚 such that 𝑢|𝑉𝑚−1 = 𝑣} = ℰ𝑚−1 (𝑣), where the minimizing function 𝑢 is calculated using (11.7). The sequence ℰ𝑚 is called a harmonic structure on 𝑆. Using (11.10), one can verify by induction that, given a function 𝜌 on 𝑉0 , we have that min{ℰ𝑚 (𝑢) ∶ functions 𝑢 on 𝑉𝑚 such that 𝑢|𝑉0 = 𝜌} = ℰ0 (𝜌). Again, the minimizer function can be calculated recursively using the algorithm (11.7). The points in 𝑉0 are called the boundary of 𝑆, and the minimizer function 𝑢 is called harmonic with boundary values 𝜌. Figure 11.6 shows the harmonic function with boundary values given by 𝜌(𝑝1 ) = 1 and 𝜌(𝑝2 ) = 𝜌(𝑝3 ) = 0. If 𝑢 is harmonic, we can show that there exists 𝐴 > 0 such that, for 𝑚 ≥ 0, 3 𝑚 (11.11) |𝑢(𝑥) − 𝑢(𝑦)| ≤ 𝐴( ) 5 whenever 𝑥 ∼𝑚 𝑦. Indeed, let 𝐴 = max{|𝑢(𝑝1 ) − 𝑢(𝑝2 )|, |𝑢(𝑝2 ) − 𝑢(𝑝3 )|, |𝑢(𝑝3 ) − 𝑢(𝑝1 )|}. Then clearly (11.11) is satisfied at 𝑚 = 0, and suppose it is true at 𝑚 − 1, for some 𝑚 ≥ 1. If 𝑥 ∼𝑚 𝑦, then either 𝑥, 𝑦 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 or, say, 𝑥 ∈ 𝑉𝑚−1 and 𝑦 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 . Let 𝑤 ∈ 𝑊𝑚−1 , and 𝑥𝑖 = 𝑓𝑤 (𝑝 𝑖 ), for 𝑖 = 1, 2, 3. If 𝑦 𝑖 is the point in 𝑉𝑚 ⧵ 𝑉𝑚−1 in the cell bounded by the points 𝑥1 , 𝑥2 , 𝑥3 , opposite to 𝑥𝑖 , then, by (11.7), 𝑢(𝑦 𝑖 ) =

𝑢(𝑥𝑖 ) + 2𝑢(𝑥𝑗 ) + 2𝑢(𝑥𝑘 ) , 5

11.2. Harmonic structure on the Sierpiński gasket 3

i 3

i

2

4

0 1

1 2

0 0 1 2

1

Figure 11.6. Harmonic function with boundary values 𝜌(𝑝 1 ) = 1 and 𝜌(𝑝 2 ) = 𝜌(𝑝 3 ) = 0.

where 𝑖, 𝑗, 𝑘 are the three different numbers 1, 2, 3. Thus |𝑢(𝑦 𝑖 ) − 𝑢(𝑦𝑗 )| 𝑢(𝑥𝑖 ) + 2𝑢(𝑥𝑗 ) + 2𝑢(𝑥𝑘 ) 𝑢(𝑥𝑗 ) + 2𝑢(𝑥𝑖 ) + 2𝑢(𝑥𝑘 ) | − | 5 5 |𝑢(𝑥𝑖 ) − 𝑢(𝑥𝑗 )| 1 3 𝑚 3 𝑚−1 = ≤ 𝐴( ) , ≤ ⋅ 𝐴( ) 5 5 5 5

= ||

by the induction hypothesis, and 𝑢(𝑥𝑗 ) + 2𝑢(𝑥𝑖 ) + 2𝑢(𝑥𝑘 ) | | 5 |𝑢(𝑥𝑖 ) − 𝑢(𝑥𝑗 )| 2|𝑢(𝑥𝑖 ) − 𝑢(𝑥𝑘 )| ≤ + 5 5 𝑚−1 1 3 2 3 𝑚−1 3 𝑚 ≤ ⋅ 𝐴( ) + ⋅ 𝐴( ) = 𝐴( ) . 5 5 5 5 5

|𝑢(𝑥𝑖 ) − 𝑢(𝑦𝑗 )| = ||𝑢(𝑥𝑖 ) −

Thus (11.11) holds for every 𝑚 ≥ 0. If 𝑉∗ =

⋃

𝑚≥0

𝑉𝑚 ,

211

212

11. The Laplacian on the Sierpiński gasket

we can use estimate (11.11) to show that 𝑢 is a uniformly continuous function on 𝑉∗ (Exercise (4)), which, since 𝑉∗ is dense in 𝑆 (Exercise (5)), can be extended to a continuous function on 𝑆 (Exercise (6)).

11.3. The Laplacian on the Sierpiński gasket Let 𝑢 ∈ 𝐶(𝑆), a continuous function on 𝑆, and for each 𝑚 ≥ 0, let 𝑢𝑚 = 𝑢|𝑉𝑚 , its restriction to 𝑉𝑚 . By (11.10), the sequence ℰ𝑚 (𝑢𝑚 ) is increasing. Let ℱ = {𝑢 ∈ 𝐶(𝑆) ∶ ℰ𝑚 (𝑢𝑚 ) is bounded}. Then, for each 𝑢 ∈ ℱ, the limit ℰ(𝑢) = lim ℰ𝑚 (𝑢𝑚 ) = sup ℰ𝑚 (𝑢𝑚 ) 𝑚→∞

𝑚≥0

exists. We call ℰ(𝑢) the energy of 𝑢 on 𝑆. Example 11.12. If 𝑢 is constant, then ℰ𝑚 (𝑢𝑚 ) = 0 for all 𝑚, and thus 𝑢 ∈ ℱ and ℰ(𝑢) = 0. In fact, since ℰ(𝑢) = sup ℰ𝑚 (𝑢𝑚 ), 𝑚≥0

then ℰ(𝑢) = 0 only if 𝑢 is a constant function. Example 11.13. If 𝑢 is the continuous extension of a harmonic function, then ℰ𝑚 (𝑢𝑚 ) = ℰ0 (𝑢0 ) for all 𝑚, and thus ℰ(𝑢) = ℰ0 (𝑢0 ). Moreover, since ℰ(𝑢) = sup ℰ𝑚 (𝑢𝑚 ), 𝑚≥0

then 𝑢 minimizes ℰ among all functions in ℱ with boundary values 𝑢0 . We thus also call 𝑢 harmonic in 𝑆. We say that 𝑢 ∈ 𝐶(𝑆) is 𝑚-harmonic if, for each 𝑤 ∈ 𝑊𝑚 , the function 𝑢 ∘ 𝑓𝑤 is harmonic. Equivalently, 𝑢 is the continuous extension of the function constructed using (11.7) starting from a given function on 𝑉𝑚 . Thus, for each 𝑛 ≥ 𝑚, ℰ𝑛 (𝑢𝑛 ) = ℰ𝑚 (𝑢𝑚 ), and thus 𝑢 ∈ ℱ and ℰ(𝑢) = ℰ𝑚 (𝑢𝑚 ). Example 11.14. Let 𝑚 ≥ 0 and 𝑥 ∈ 𝑉𝑚 . Consider the 𝑚-harmonic function 𝜓𝑥,𝑚 (if 𝑚 = 0 we just say harmonic) constructed from the function on 𝑉𝑚 given by 1 𝑦=𝑥 𝜒(𝑦) = { 0 𝑦 ≠ 𝑥.

11.3. The Laplacian on the Sierpiński gasket

213

𝜓𝑥,𝑚 is called the 𝑚-harmonic spline at 𝑥. Note that, for any 𝑚-harmonic function 𝑢, we have 𝑢 = ∑ 𝑢(𝑥)𝜓𝑥,𝑚 . 𝑥∈𝑉𝑚

In particular, if 𝑚 = 0, the splines of Example 11.14 correspond to the harmonic functions with boundary values 1, 0, 0, as the one in Figure 11.6. If we denote 𝜓𝑝𝑖 ,0 simply by 𝜓𝑖 , we see that 𝑢 = 𝑢(𝑝1 )𝜓1 + 𝑢(𝑝2 )𝜓2 + 𝑢(𝑝3 )𝜓3 , and thus 𝜓1 , 𝜓2 and 𝜓3 form a basis for the (3-dimensional) space of harmonic functions on 𝑆. 2𝜋 Note that, by the symmetry of 𝑆 and (11.7), each 𝜓𝑖 is a rotation 3 of each other. Also, any other 𝜓𝑥,𝑚 is either of the form 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1

or

−1 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1 + 𝜒𝑆𝑤̄ ⋅ 𝜓𝑗 ∘ 𝑓𝑤 ̄ ,

where 𝑤, 𝑤̄ ∈ 𝑊𝑚 , depending on whether 𝑥 ∈ 𝑉0 or 𝑥 ∈ 𝑉𝑚 ⧵𝑉0 (Exercise (8)). We want to obtain the analogous result of integration by parts in the interval, as in (11.6), for the Sierpiński gasket. This will give us a way to define the Laplacian on 𝑆. For this, we first need to define an integral of functions on 𝑆. The usual Lebesgue integral that we have been using in the previous chapters is not useful for us here, since 𝑆 is a set of measure zero in the plane. So we have to start by defining a measure on subsets of 𝑆. Recall that, for any word 𝑤 ∈ 𝑊𝑚 , we define the cell 𝑆𝑤 = 𝑓𝑤 (𝑆). For any given cell, we define 1 , 3𝑚 where 𝑚 is the length of the word 𝑤. We now define, for 𝐴 ⊂ 𝑆, (11.15)

(11.16)

𝜇(𝑆𝑤 ) =

𝜇(𝐴) = inf { ∑ 𝜇(𝑇𝑗 ) ∶ 𝑇𝑗 are cells and 𝐴 ⊂ 𝑗

⋃

𝑇𝑗 }.

𝑗

As in the case of the Hausdorff measure that we discussed in Chapter 10, 𝜇 satisfies the properties of the Lebesgue outer measure. Proposition 11.17. Let 𝜇 be defined on subsets of 𝑆 as in (11.16). (1) If 𝐴 ⊂ 𝐵, then 𝜇(𝐴) ≤ 𝜇(𝐵).

214

11. The Laplacian on the Sierpiński gasket (2) If 𝐴 = ⋃𝑗 𝐴𝑗 , then 𝜇(𝐴) ≤ ∑𝑗 𝜇(𝐴𝑗 ). (3) If dist(𝐴, 𝐵) > 0, 𝜇(𝐴 ∪ 𝐵) = 𝜇(𝐴) + 𝜇(𝐵).

We leave the proof of Proposition 11.17 as an exercise (Exercise (9)). We can then define a set 𝐴 ⊂ 𝑆 to be measurable if, for any 𝐵 ⊂ 𝑆, 𝜇(𝐵) = 𝜇(𝐵 ∩ 𝐴) + 𝜇(𝐵 ⧵ 𝐴). As in the case of the Lebesgue measure, part (3) of Proposition 11.17 implies that every open subset of 𝑆 is measurable (and hence every closed subset ). Also, every open 𝑈 ⊂ 𝑆 can be written as an almost disjoint union ⋃𝑗 𝑇𝑗 of cells, so 𝜇(𝑈) = ∑ 𝜇(𝑇𝑗 ). 𝑗

We, in fact, can prove more. Proposition 11.18. (1) If 𝐴1 , 𝐴2 , . . . are disjoint measurable subsets of 𝑆, then 𝐴 = ⋃𝑗 𝐴𝑗 is measurable and 𝜇(𝐴) = ∑ 𝜇(𝐴𝑗 ). 𝑗

(2) If 𝐴 ⊂ 𝑆 is measurable, its complement 𝑆 ⧵ 𝐴 is also measurable and 𝜇(𝑆 ⧵ 𝐴) = 1 − 𝜇(𝐴). The proofs of these statements are similar to their analogous results on Lebesgue measure and we again leave them as an exercise (Exercise (9)). We can now proceed to define the integral in an analogous way, first for characteristic functions of measurable sets, ∫ 𝜒𝐴 𝑑𝜇 = 𝜇(𝐴), 𝑆

then for simple functions 𝜙 = ∑𝑗 𝑐𝑗 𝜇(𝐴𝑗 ), ∫ 𝜙𝑑𝜇 = ∑ 𝑐𝑗 𝜇(𝐴𝑗 ), 𝑆

𝑗

11.3. The Laplacian on the Sierpiński gasket

215

and then by approximating any measurable 𝑓 ≥ 0 with simple functions ∫ 𝑓𝑑𝜇 = sup { ∫ 𝜙𝑑𝜇 ∶ 𝜙 is simple and 0 ≤ 𝜙 ≤ 𝑓}. 𝑆

𝑆

However, while the Lebesgue measure is traslation invariant and dilations by a positive 𝛿 induce a factor of 𝛿𝑑 , where 𝑑 is the dimension of the space, in this case we have the following properties. 11.19. If 𝜎 is a symmetry of 𝑆, say, a rotation or a reflection over an axis passing through one of its vertices 𝑝 𝑖 , then ∫ 𝑓 ∘ 𝜎𝑑𝜇 = ∫ 𝑓𝑑𝜇. 𝑆

𝑆

To prove this fact, observe that it is enough to prove it for characteristic functions 𝜒𝐴 , and in that case ∫ 𝜒𝐴 ∘ 𝜎𝑑𝜇 = ∫ 𝜒𝜍−1 (𝐴) 𝑑𝜇 = 𝜇(𝜎−1 (𝐴)), 𝑆

𝑆

so we just need to verify that 𝜇 is invariant under the symmetries of 𝑆. This follows from the observation that, for any cell 𝑆𝑤 and any symmetry 𝜎 of 𝑆, then 𝜎(𝑆𝑤 ) is also a cell of level 𝑚, and thus have the same measure. Indeed, if 𝜎 is the reflection over the axis passing through 𝑝 𝑖 , so it keeps 𝑝 𝑖 fixed and switches 𝑝𝑗 and 𝑝 𝑘 , then, for any 𝑤 ∈ 𝑊𝑚 , 𝜎(𝑆𝑤 ) = 𝑆𝑤̄ , where 𝑤̄ ∈ 𝑊𝑚 is the word obtained from 𝑤 by switching the letters 𝑗 and 𝑘. If 𝜎 is the rotation 𝑝1 ↦ 𝑝2 , 𝑝2 ↦ 𝑝3 and 𝑝3 ↦ 𝑝1 , then 𝜎(𝑆𝑤 ) = 𝑆𝑤̄ where 𝑤̄ is obtained by replacing any letter 𝑖 of 𝑤 by the letter 𝑖 + 1 (mod 3). 11.20. For any 𝑤 ∈ 𝑊𝑚 , ∫ 𝑓𝑑𝜇 = 𝑆𝑤

1 ∫ 𝑓 ∘ 𝑓𝑤 𝑑𝜇. 3𝑚 𝑆

In order to verify this for characteristic functions, we have to verify that 1 𝜇(𝑓−1 (𝐴 ∩ 𝑆𝑤 )), 3𝑚 𝑤 which follows from the fact that, if the 𝑇𝑗 are cells, 𝜇(𝐴 ∩ 𝑆𝑤 ) =

𝑓𝑤−1 (𝐴 ∩ 𝑆𝑤 ) ⊂

⋃ 𝑗

𝑇𝑗

if and only if

𝐴 ∩ 𝑆𝑤 ⊂

⋃ 𝑗

𝑓𝑤 (𝑇𝑗 ),

216

11. The Laplacian on the Sierpiński gasket

and 𝜇(𝑓𝑤 (𝑇𝑗 )) =

1 𝜇(𝑇𝑗 ) 3𝑚

for each cell 𝑇𝑗 . Example 11.21. If 𝑓 is a harmonic function on 𝑆, then 𝑓 = 𝑓(𝑝1 )𝜓1 + 𝑓(𝑝2 )𝜓2 + 𝑓(𝑝3 )𝜓3 , where each 𝜓𝑖 is the harmonic function with boundary values 𝜓𝑖 (𝑝 𝑖 ) = 1 and 𝜓𝑖 (𝑝𝑗 ) = 0, for 𝑗 ≠ 𝑖. As we have seen above, each 𝜓𝑖 is a rotation of the others. Since 𝜓1 + 𝜓2 + 𝜓3 = 1 and 𝜇(𝑆) = 1, we have that ∫ 𝜓1 𝑑𝜇 = ∫ 𝜓2 𝑑𝜇 = ∫ 𝜓3 𝑑𝜇 = 𝑆

𝑆

Therefore ∫ 𝑓𝑑𝜇 = 𝑆

𝑆

1 . 3

𝑓(𝑝1 ) + 𝑓(𝑝2 ) + 𝑓(𝑝3 ) . 3

Example 11.22. Let 𝜓𝑥,𝑚 be the 𝑚-harmonic spline at 𝑥. As we have seen above, if 𝑥 ∈ 𝑉0 , 𝜓𝑥,𝑚 = 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1 for some 𝑖 and 𝑤 ∈ 𝑊𝑚 , and thus 1 1 ∫ 𝜓𝑥,𝑚 𝑑𝜇 = ∫ 𝜓𝑖 ∘ 𝑓𝑤−1 𝑑𝜇 = 𝑚 ∫ 𝜓𝑖 𝑑𝜇 = 𝑚+1 . 3 𝑆 3 𝑆 𝑆 𝑤

−1 If 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 , then 𝜓𝑥,𝑚 = 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1 + 𝜒𝑆𝑤̄ ⋅ 𝜓𝑗 ∘ 𝑓𝑤 ̄ for some 𝑖, 𝑗 and 𝑤, 𝑤̄ ∈ 𝑊𝑚 , and thus −1 ∫ 𝜓𝑥,𝑚 = ∫ 𝜓𝑖 ∘ 𝑓𝑤−1 𝑑𝜇 + ∫ 𝜓𝑗 ∘ 𝑓𝑤 ̄ 𝑑𝜇 𝑆

𝑆𝑤

𝑆𝑤̄

1 1 = 𝑚 ∫ 𝜓𝑖 𝑑𝜇 + 𝑚 ∫ 𝜓𝑗 𝑑𝜇 3 𝑆 3 𝑆 2 = 𝑚+1 . 3 We are ready to define the Laplacian. First, consider the polarization ℰ(𝑢, 𝑣) of the energy, defined for 𝑢, 𝑣 ∈ ℱ. We can see that ℰ(𝑢, 𝑣) = lim ℰ𝑚 (𝑢, 𝑣), where in the right-hand side we also denote by 𝑢 and 𝑣 their restrictions to 𝑉𝑚 , and, for functions 𝑉𝑚 , ℰ𝑚 (𝑢, 𝑣) is the polarization of the quadratic form ℰ𝑚 , 5 𝑚 ℰ𝑚 (𝑢, 𝑣) = ( ) ∑ (𝑢(𝑥) − 𝑢(𝑦))(𝑣(𝑥) − 𝑣(𝑦)). 3 𝑥∼ 𝑦 𝑚

11.3. The Laplacian on the Sierpiński gasket

217

In fact, one can show (Exercise (10)) that 5 𝑚 (11.23) ℰ𝑚 (𝑢, 𝑣) = −( ) ∑ Δ𝑚 𝑢(𝑥) 𝑣(𝑥), 3 𝑥∈𝑉 𝑚

where Δ𝑚 is the difference operator (11.24)

Δ𝑚 𝑢(𝑥) = ∑ (𝑢(𝑦) − 𝑢(𝑥)). 𝑦∼𝑚 𝑥

Note that this sum has two terms if 𝑥 ∈ 𝑉0 , or four terms if 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 . Δ𝑚 is called the discrete Laplacian at level 𝑚. We say that 𝑢 ∈ dom Δ if 𝑢 ∈ ℱ and there exists 𝑓 ∈ 𝐶(𝑆) such that, for all 𝑣 ∈ ℱ with 𝑣|𝑉0 = 0, (11.25)

ℰ(𝑢, 𝑣) = − ∫ 𝑓𝑣𝑑𝜇. 𝑆

We write 𝑓 = Δ𝑢, and we call it the Laplacian of 𝑢. 11.26. If 𝑢 is harmonic, for each 𝑚 ≥ 1 and all 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 , we have that Δ𝑚 𝑢(𝑥) = 0 (Exercise (11)). Thus, if 𝑣 ∈ ℱ with 𝑣|𝑉0 = 0, ℰ𝑚 (𝑢, 𝑣) = 0 for all 𝑚, and thus ℰ(𝑢, 𝑣) = 0. We thus conclude that, if 𝑢 is harmonic, 𝑢 ∈ dom Δ and Δ𝑢 = 0. 11.27. If 𝑢 is 𝑚-harmonic, Δ𝑛 𝑢(𝑥) = 0 for all 𝑛 > 𝑚 and 𝑥 ∈ 𝑉𝑛 ⧵ 𝑉𝑚 . Thus, for any other 𝑣 ∈ ℱ, ℰ(𝑢, 𝑣) = ℰ𝑚 (𝑢, 𝑣). However, 𝑢 ∉ dom Δ, because Δ𝑢(𝑥) cannont be defined for 𝑥 ∈ 𝑉𝑚 . Fix 𝑥0 ∈ 𝑉∗ ⧵ 𝑉0 , and let 𝑚 be so 𝑥0 ∈ 𝑉𝑚 . If 𝑣 = 𝜓𝑥0 ,𝑚 , then, as we have seen before, 𝑣 ∈ ℱ, and of course 𝑣|𝑉0 = 0, because 𝑥0 ∉ 𝑉0 . Let 𝑢 ∈ dom Δ. We first note that, since 𝑣 is 𝑚-harmonic, 5 𝑚 ℰ(𝑢, 𝑣) = ℰ𝑚 (𝑢, 𝑣) = −( ) ∑ Δ𝑚 𝑢(𝑥) 𝑣(𝑥) 3 𝑥∈𝑉 𝑚

5 𝑚 = −( ) Δ𝑚 𝑢(𝑥0 ), 3 because 𝑣(𝑥0 ) = 1 and 𝑣(𝑥) = 0 for any other 𝑥 ∈ 𝑉𝑚 . Also ∫ Δ𝑢(𝑥)𝑣(𝑥)𝑑𝜇(𝑥) − Δ𝑢(𝑥0 ) ∫ 𝑣(𝑥)𝑑𝜇(𝑥) 𝑆

𝑆

= ∫ (Δ𝑢(𝑥) − Δ𝑢(𝑥0 ))𝑣(𝑥)𝑑𝜇(𝑥). 𝑆

218

11. The Laplacian on the Sierpiński gasket

Given 𝜀 > 0, we can choose 𝑚 large enough so that |Δ𝑢(𝑥) − Δ𝑢(𝑥0 )| < 𝜀 for every 𝑥 ∈ supp 𝑣, because Δ𝑢 is continuous. Thus, for such 𝑚, and using our result in Example 11.22, | ∫ (Δ𝑢(𝑥) − Δ𝑢(𝑥 ))𝑣(𝑥)𝑑𝜇(𝑥)| < 𝜀 ∫ 𝑣𝑑𝜇 = 2 𝜀. 0 | | 3𝑚+1 𝑆 𝑆 Putting the previous identities and inequalities together we obtain 𝑚 |( 5 ) Δ 𝑢(𝑥 ) − 2 Δ𝑢(𝑥 )| < 2 𝜀, 𝑚 0 0 | | 3 3𝑚+1 3𝑚+1

and thus

(11.28)

| 3 5𝑚 Δ 𝑢(𝑥 ) − Δ𝑢(𝑥 )| < 𝜀. 𝑚 0 0 | |2

We have therefore proven Theorem 11.29, that states that the Laplacian is a limit of normalized difference operators. Theorem 11.29. If 𝑢 ∈ dom Δ and 𝑥0 ∈ 𝑉∗ ⧵ 𝑉0 , then

Δ𝑢(𝑥0 ) = lim

𝑚→∞

3 𝑚 5 Δ𝑚 𝑢(𝑥0 ). 2

We can also prove that the limit in Theorem 11.28 is uniform, in the sense that, given 𝜀 > 0, we can choose 𝑚 such that (11.29) is true independently of the particular choice of 𝑥0 ∈ 𝑉𝑚 . Moreover, the existence of this uniform limit also implies that 𝑢 ∈ dom Δ. We have left these facts as exercises (Exercises (12) and (13)).

Exercises

219

Exercises (1) Let 𝑥0 ∈ (0, 1) and ℎ > 0 so that (𝑥0 − ℎ, 𝑥0 + ℎ) ⊂ (0, 1). Let 𝑣 be the piecewise linear function given on [0, 1] by ⎧0 ⎪ ⎪ 𝑡 − 𝑥0 + ℎ 𝑣(𝑡) = ℎ ⎨ 𝑥0 + ℎ − 𝑡 ⎪ ℎ ⎪ ⎩0

𝑡 < 𝑥0 − ℎ 𝑥0 − ℎ ≤ 𝑡 < 𝑥 0 𝑥0 ≤ 𝑡 < 𝑥 0 + ℎ 𝑥0 + ℎ ≤ 𝑡.

Then, for 𝑢 ∈ 𝐶 2 ([0, 1]), 1

∫ 𝑢″ (𝑡)𝑣(𝑡)𝑑𝑡 = 0

𝑢(𝑥0 − ℎ) + 𝑢(𝑥0 + ℎ) − 2𝑢(𝑥0 ) . ℎ

(2) For functions 𝑢 and 𝑣 as in the previous exercise, 1

1 ∫ 𝑢″ (𝑡)𝑣(𝑡)𝑑𝑡 = 𝑢″ (𝑥0 ). ℎ→0 ℎ 0 lim

(3) Let 𝑓(𝑥, 𝑦, 𝑧) =(𝑎 − 𝑧)2 + (𝑧 − 𝑦)2 + (𝑦 − 𝑎)2 + (𝑧 − 𝑏)2 + (𝑏 − 𝑥)2 + (𝑥 − 𝑧)2 + (𝑦 − 𝑥)2 + (𝑥 − 𝑐)2 + (𝑐 − 𝑦)2 . Then 𝑓 takes its minimum at 𝑥∗ =

𝑎 + 2𝑏 + 2𝑐 , 5

𝑦∗ =

2𝑎 + 𝑏 + 2𝑐 , 5

𝑓(𝑥∗ , 𝑦∗ , 𝑧∗ ) =

3 ((𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2 ). 5

𝑧∗ =

2𝑎 + 2𝑏 + 𝑐 , 5

and

(4) If 𝑢 is a harmonic function, then it is uniformly continuous on 𝑉∗ , the set of all vertices in 𝑆. (5) The union 𝑉∗ of all vertices in 𝑆 is dense in 𝑆. (6) If 𝑢 is a harmonic function, then it can be extended to a continuous function on 𝑆. (Hint: Use Exercises (4) and (5).)

220

11. The Laplacian on the Sierpiński gasket

(7) If 𝑢 is the continuous extension of a harmonic function in 𝑆, then 𝑢 is a Hölder continuous function with exponent 𝛼=

log(5/3) . log 2

(8) Let 𝑚 ≥ 1 and 𝑥 ∈ 𝑉𝑚 . (a) If 𝑥 ∈ 𝑉0 , 𝜓𝑥,𝑚 = 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1 for some 𝑖 = 1, 2, 3, and 𝑤 is the word 𝑤 = 𝑖𝑖 ⋯ 𝑖 ∈ 𝑊𝑚 . (b) If 𝑥 ∉ 𝑉0 , there exist 𝑖, 𝑗 = 1, 2, 3 and 𝑤, 𝑤̄ ∈ 𝑊𝑚 such that −1 𝜓𝑥,𝑚 = 𝜒𝑆𝑤 ⋅ 𝜓𝑖 ∘ 𝑓𝑤−1 + 𝜒𝑆𝑤̄ ⋅ 𝜓𝑗 ∘ 𝑓𝑤 ̄ .

(9) (a) Prove Proposition 11.17. (b) Prove Propostion 11.18. (10) For functions 𝑢, 𝑣 on 𝑉𝑚 , ∑ (𝑢(𝑥) − 𝑢(𝑦))(𝑣(𝑥) − 𝑣(𝑦)) = − ∑ Δ𝑚 𝑢(𝑥) 𝑣(𝑥), 𝑥∼𝑚 𝑦

𝑥∈𝑉𝑚

where Δ𝑚 is the difference operator Δ𝑚 𝑢(𝑥) = ∑ (𝑢(𝑦) − 𝑢(𝑥)). 𝑦∼𝑚 𝑥

(11) If 𝑢 is harmonic, then, for each 𝑚 ≥ 1 and 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 , Δ𝑚 𝑢(𝑥) = 0. (12) Let 𝑢 ∈ dom Δ. Then, for any 𝜀 > 0, there exists 𝑁 such that, for any 𝑚 ≥ 𝑁, | 3 5𝑚 Δ 𝑢(𝑥) − Δ𝑢(𝑥)| < 𝜀 𝑚 |2 | for any 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 . (13) Let 𝑢 ∈ ℱ and suppose there exists 𝑓 ∈ 𝐶(𝑆) such that, for any 𝜀 > 0, we can find 𝑁 such that, for any 𝑚 ≥ 𝑁, | 3 5𝑚 Δ 𝑢(𝑥) − 𝑓(𝑥)| < 𝜀 𝑚 | |2 for any 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 . Then 𝑢 ∈ dom Δ and 𝑓 = Δ𝑢.

Notes The calculus on the Sierpiński gasket developed in this chapter was introduced by Jun Kigami in [Kig89], where the motivation from the discretization of the interval is also presented. The idea of using springs to define discrete energy was also used in [YHK97]. The discrete energy can also be obtained from the theory of electrical networks, as in

Notes

221

[Str06] or in [Kig01]. The Laplacian on the Sierpiński gasket can also be introduced through Brownian motion, as in [Bar98].

Chapter 12

Eigenfunctions of the Laplacian

In Chapter 3, we studied the decomposition of functions in the interval in terms of trigonometric functions, as a means to solve the Dirichlet problem in the disk. The trigonometric functions sine and cosine satisfy the equations 𝑑2 sin 𝑥 = − sin 𝑥 𝑑𝑥2

and

𝑑2 cos 𝑥 = − cos 𝑥, 𝑑𝑥2

which imply that the trigonometric functions sin 𝑘𝑥 and cos 𝑘𝑥 are all eigenfunctions of the operator 𝑑 2 /𝑑𝑥2 , as they satisfy 𝑑2 𝜙(𝑥) = −𝜆𝜙(𝑥) 𝑑𝑥2 for some nonnegative 𝜆 ∈ ℝ. In this chapter we study the eigenfunctions and eigenvalues of the Laplacian on the Sierpiński gasket, defined in Chapter 11. That is, we will study the solutions to the equation Δ𝑢(𝑥) = −𝜆𝑢(𝑥), for some scalar 𝜆 and nonzero 𝑢 ∈ dom Δ, and each 𝑥 ∈ 𝑆 ⧵ 𝑉0 . The first natural question is whether such eigenfunctions can be constructed through an interpolation algorithm, as is the case of the harmonic functions, studied above. Though there is no a priori reason for the existence 223

224

12. Eigenfunctions of the Laplacian

of such algorithm, we first note that, in the case of the interval [0, 1], it indeed exists.

12.1. Discrete eigenfunctions on the interval We first observe that if 𝜙 is an eigenfunction of the Laplacian Δ = 𝑑 2 /𝑑𝑥2 on the interval, then, for a sufficiently large 𝑚, 𝜙|𝑉𝑚 is an eigenfunction of the difference operator 1 1 ) + 𝑢(𝑥 + 𝑚 ) − 2𝑢(𝑥), 2𝑚 2 where 𝑢 is a function on the dyadic partition 𝑉𝑚 of level 𝑚, and 𝑥 is of the form 𝑘/2𝑚 , 1 ≤ 𝑘 ≤ 2𝑚 − 1. Indeed, the eigenfunctions of Δ are linear combinations of functions of the form 𝜙(𝑥) = 𝑒𝜔𝑥 , since Δ𝑚 𝑢(𝑥) = 𝑢(𝑥 −

𝑑 2 𝜔𝑥 𝑒 = 𝜔2 𝑒𝜔𝑥 . 𝑑𝑥2 Now 1

1

Δ𝑚 𝜙(𝑥) = 𝑒𝜔(𝑥− 2𝑚 ) + 𝑒𝜔(𝑥+ 2𝑚 ) − 2𝑒𝜔𝑥 = −𝜆𝑚 𝑒𝜔𝑥 , where 𝜔

𝜔

𝜔

−𝜆𝑚 = 𝑒− 2𝑚 + 𝑒 2𝑚 − 2 = (𝑒 2𝑚+1 − 𝑒

−

𝜔 2𝑚+1

)2 = 4 sinh

2

𝜔 2𝑚+1

.

Thus, the restriction of 𝜙 to 𝑉𝑚 is an eigenfunction of the difference operator Δ𝑚 with respect to the eigenvalue −𝜆𝑚 . Moreover, note that, as we have seen in Chapter 11, 4𝑚 Δ𝑚 𝜙(𝑥) → 𝜙″ (𝑥) = 𝜔2 𝜙(𝑥) as 𝑚 → ∞, and 𝜔 → 𝜔2 , 2𝑚+1 so the discrete eigenvalues, when properly normalized, converge to the true eigenvalues of Δ. 2

−4𝑚 𝜆𝑚 = (2𝑚+1 )2 sinh

Note that it may happen that 𝜙|𝑉𝑚 ≡ 0 for some 𝑚: for instance, if 𝜙(𝑥) = sin(2𝑛 𝜋𝑥), then 𝜙|𝑉𝑚 ≡ 0 for every 𝑚 ≤ 𝑛. Since an eigenfunction 𝜙 is not identically zero, 𝜙|𝑉𝑚 ≢ 0 for sufficiently large 𝑚, so the argument above applies.

12.1. Discrete eigenfunctions on the interval

225

Conversely, we can construct the eigenfunctions of Δ by extending discrete eigenfunctions of the operators Δ𝑚 . Suppose we have a function 𝑢 on 𝑉𝑚−1 which is an eigenfunction of Δ𝑚−1 with respect to the eigenvalue −𝜆𝑚−1 , that is Δ𝑚−1 𝑢(𝑥) = −𝜆𝑚−1 𝑢(𝑥) for each 𝑥 ∈ 𝑉𝑚−1 ⧵ 𝑉0 . We want to extend 𝑢 to a function on 𝑉𝑚 , that we also denote by 𝑢, that satisfies Δ𝑚 𝑢(𝑥) = −𝜆𝑚 𝑢(𝑥) for all 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 , for some 𝜆𝑚 . In particular, for 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 of the form (2𝑘 + 1)/2𝑚 , we want Δ𝑚 𝑢(

2𝑘 + 1 2𝑘 + 1 ) = −𝜆𝑚 𝑢( 𝑚 ), 2𝑚 2

and thus 2𝑘 2𝑘 + 2 2𝑘 + 1 2𝑘 + 1 ) + 𝑢( 𝑚 ) − 2𝑢( 𝑚 ) = −𝜆𝑚 𝑢( 𝑚 ), 2𝑚 2 2 2 so 𝑢 satisfies, for each 𝑘, 2𝑘 + 1 𝑘 𝑘+1 (12.1) (2 − 𝜆𝑚 )𝑢( 𝑚 ) = 𝑢( 𝑚−1 ) + 𝑢( 𝑚−1 ). 2 2 2 Note that 𝑘/2𝑚−1 , (𝑘 + 1)/2𝑚−1 ∈ 𝑉𝑚−1 so, provided 𝜆𝑚 ≠ 2, we have the extension algorithm 𝑢(

(12.2)

𝑘 𝑘+1 𝑢( 𝑚−1 ) + 𝑢( 𝑚−1 ) 2𝑘 + 1 2 2 . 𝑢( 𝑚 ) = 2 2 − 𝜆𝑚

The equation (12.2) tells us how to extend 𝑢 from 𝑉𝑚−1 to 𝑉𝑚 , if we want 𝑢 to be an eigenfunction of Δ𝑚 with respect to −𝜆𝑚 . Now, the eigenfunction equation must also be satisfied at points in 𝑉𝑚−1 ⧵ 𝑉0 , so we also want 𝑘 𝑘 Δ𝑚 𝑢( 𝑚−1 ) = −𝜆𝑚 𝑢( 𝑚−1 ) 2 2 for 1 ≤ 𝑘 ≤ 2𝑚−1 − 1, which gives the equation (12.3)

(2 − 𝜆𝑚 )𝑢(

𝑘 2𝑘 − 1 2𝑘 + 1 ) = 𝑢( 𝑚 ) + 𝑢( 𝑚 ). 2 2 2𝑚−1

By (12.2), 𝑘−1 𝑘 𝑢( 𝑚−1 ) + 𝑢( 𝑚−1 ) 2𝑘 − 1 2 2 𝑢( 𝑚 ) = , 2 2 − 𝜆𝑚

226

12. Eigenfunctions of the Laplacian

and substituting in (12.3) we obtain 𝑘−1 𝑘+1 𝑘 ) + 𝑢( 𝑚−1 ) = (2 − 4𝜆𝑚 + 𝜆2𝑚 )𝑢( 𝑚−1 ). 2𝑚−1 2 2 Since 𝑢 is assumed to be an eigenfunction of Δ𝑚−1 with respect to the eigenvalue −𝜆𝑚−1 , we also have 𝑢(

𝑘−1 𝑘+1 𝑘 ) + 𝑢( 𝑚−1 ) = (2 − 𝜆𝑚−1 )𝑢( 𝑚−1 ), 2𝑚−1 2 2 so we obtain the equation 𝑢(

2 − 𝜆𝑚−1 = 2 − 4𝜆𝑚 + 𝜆2𝑚 , which can be written as (12.4)

𝜆𝑚−1 = 𝜆𝑚 (4 − 𝜆𝑚 ).

We have proven that we can always extend an eigenfunction of Δ𝑚−1 with respect to the eigenvalue −𝜆𝑚−1 to an eigenfunction of Δ𝑚 with respect to −𝜆𝑚 , provided 𝜆𝑚 ≠ 2 and the condition (12.4) holds. Together with (12.2), this provides an algorithm to construct all eigenfunctions of Δ. Indeed, since the roots of (12.4) are 𝜆𝑚 = 2 ± √4 − 𝜆𝑚−1 , we can start from an eigenfunction of Δ𝑚0 with respect to 𝜆𝑚0 , for some 𝑚0 , and extend to each 𝑚 > 𝑚0 choosing 𝜆𝑚 as any of these roots. Note that, if 𝑤 = 2−√4 − 𝑧, using the principal branch of √4 − 𝑧 on ℂ⧵[4, ∞), then 𝑧 𝑤 = + 𝑂(|𝑧|2 ), 4 and we can conclude that 4𝑚 𝜆𝑚 converges if we choose the minus sign for all but finitely many 𝑚 (Exercise (1)). Note that, if 𝜆𝑚0 ≠ 2, the starting eigenfunction on 𝑉𝑚0 will be an extension of its restriction to 𝑉𝑚0 −1 , with respect to the eigenvalue −𝜆𝑚0 −1 given by (12.4). If 𝜆𝑚0 = 2, then, by (12.1) and (12.3), 𝑢( and

𝑘 𝑘+1 ) + 𝑢( 𝑚 −1 ) = 0, 2𝑚0 −1 2 0

0 ≤ 𝑘 ≤ 2𝑚0 −1 − 1,

2𝑘 − 1 2𝑘 + 1 1 ≤ 𝑘 ≤ 2𝑚0 −1 − 1, ) + 𝑢( 𝑚0 ) = 0, 2 𝑚0 2 so we either have 𝑢|𝑉𝑚 −1 ≡ 0 and 𝑢|𝑉𝑚 ⧵𝑉𝑚 −1 equal to an alternating 0 0 0 sequence of ±1, or the other way around, as is shown in Figure 12.1. 𝑢(

12.2. Discrete eigenfunctions on the Sierpiński gasket

227

Their limit corresponds to the eigenfunctions of the form sin(𝑛𝜋𝑥) and

Figure 12.1. Eigenfunctions of Δ3 with respect to the eigenvalue 𝜆3 = 2. Note that they correspond to the restrictions to 𝑉 3 of the functions sin(4𝜋𝑥) (top) and cos(4𝜋𝑥) (bottom).

cos(𝑛𝜋𝑥), respectively.

12.2. Discrete eigenfunctions on the Sierpiński gasket We now study the eigenfunctions and eigenvalues of the Laplacian on the Sierpiński gasket 𝑆 and, in particular, the possiblity of constructing them through an interpolation process as in the case of the interval. This time we don’t have any explicit formulae nor identities for the eigenfunctions, so we cannot prove directly that their restrictions to each 𝑉𝑚 are discrete eigenfunctions of the difference operators Δ𝑚 given by (11.24). However, in this section we show how to extend an eigenfunction of Δ𝑚−1 on 𝑉𝑚−1 to the next level, provided certain conditions are satisfied, and we discuss whether this process generates all eigenfunctions of the Laplacian Δ on 𝑆. Assume 𝑢 is an eigenfunction of Δ𝑚−1 on 𝑉𝑚−1 with respect to the eigenvalue −𝜆𝑚−1 , so it satisfies the equation (12.5)

Δ𝑚−1 𝑢(𝑥) = −𝜆𝑚−1 𝑢(𝑥)

for every 𝑥 ∈ 𝑉𝑚−1 ⧵𝑉0 . We want to extend to a function on 𝑉𝑚 , which we also denote by 𝑢, which will be an eigenfunction of Δ𝑚 with eigenvalue −𝜆𝑚 . Let 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝑉𝑚−1 be the boundary points of a cell 𝑆𝑤 , for some

228

12. Eigenfunctions of the Laplacian

𝑤 ∈ 𝑊𝑚−1 . Let 𝑦1 , 𝑦2 , 𝑦3 ∈ 𝑉𝑚 ∩ 𝑆𝑤 be the points in 𝑉𝑚 ⧵ 𝑉𝑚−1 inside the cell, where each 𝑦 𝑖 is opposite to each 𝑥𝑖 , as in Figure 12.2.

Figure 12.2. The vertices in the cell 𝑆 𝑤 , for 𝑤 ∈ 𝑊𝑚 . 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝑉𝑚−1 , while 𝑦 1 , 𝑦 2 , 𝑦 3 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 .

We want to construct 𝑢 so that, for each 𝑦 𝑖 , Δ𝑚 𝑢(𝑦 𝑖 ) = −𝜆𝑚 𝑢(𝑦 𝑖 ). By (11.24), this gives us the three equations (12.6a)

𝑢(𝑥2 ) + 𝑢(𝑥3 ) + 𝑢(𝑦2 ) + 𝑢(𝑦3 ) = (4 − 𝜆𝑚 )𝑢(𝑦1 ),

(12.6b)

𝑢(𝑥1 ) + 𝑢(𝑥3 ) + 𝑢(𝑦1 ) + 𝑢(𝑦3 ) = (4 − 𝜆𝑚 )𝑢(𝑦2 ),

(12.6c)

𝑢(𝑥1 ) + 𝑢(𝑥2 ) + 𝑢(𝑦1 ) + 𝑢(𝑦2 ) = (4 − 𝜆𝑚 )𝑢(𝑦3 ),

which can be added to obtain (12.7)

(2 − 𝜆𝑚 )(𝑢(𝑦1 ) + 𝑢(𝑦2 ) + 𝑢(𝑦3 )) = 2(𝑢(𝑥1 ) + 𝑢(𝑥2 ) + 𝑢(𝑥3 )).

If 𝜆𝑚 ≠ 2, we obtain the relation 𝑢(𝑦1 ) + 𝑢(𝑦2 ) + 𝑢(𝑦3 ) =

2(𝑢(𝑥1 ) + 𝑢(𝑥2 ) + 𝑢(𝑥3 )) , 2 − 𝜆𝑚

which can be used, by adding 𝑢(𝑦 𝑖 ) to its corresponding equation in (12.6a)–(12.6c), to obtain (12.8)

2𝑢(𝑥𝑖 ) + (4 − 𝜆𝑚 )(𝑢(𝑥𝑗 ) + 𝑢(𝑥𝑘 )) = (5 − 𝜆𝑚 )𝑢(𝑦 𝑖 ), 2 − 𝜆𝑚

12.2. Discrete eigenfunctions on the Sierpiński gasket

229

where 𝑖, 𝑗, 𝑘 are the three distinct numbers 1, 2, 3. Thus, if 𝜆𝑚 ≠ 5, we have the extension algorithm (12.9)

𝑢(𝑦 𝑖 ) =

2𝑢(𝑥𝑖 ) + (4 − 𝜆𝑚 )(𝑢(𝑥𝑗 ) + 𝑢(𝑥𝑘 )) . (2 − 𝜆𝑚 )(5 − 𝜆𝑚 )

Note that (12.9) reduces to the extension algorithm (11.7) for harmonic functions if 𝜆𝑚 = 0. The resulting function 𝑢 on 𝑉𝑚 will be an eigenfunction of Δ𝑚 if it also satifies Δ𝑚 𝑢(𝑥) = −𝜆𝑚 𝑢(𝑥) for each 𝑉𝑚−1 ⧵ 𝑉0 . Suppose that 𝑥1 above is not in 𝑉0 . Then 𝑥1 belongs to two cells in level 𝑚 − 1, as in

Figure 12.3. If 𝑥1 ∈ 𝑉𝑚−1 ⧵ 𝑉 0 , then it belongs to two cells of level 𝑚 − 1, with vertices 𝑥2 , 𝑥3 , 𝑥4 , 𝑥5 ∈ 𝑉𝑚−1 and 𝑦 1 , 𝑦 2 , 𝑦 3 , 𝑦 4 , 𝑦5 , 𝑦 6 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 .

Figure 12.3. Thus, Δ𝑚 𝑢(𝑥1 ) = −𝜆𝑚 𝑢(𝑥1 ) implies the equation 𝑢(𝑦2 ) + 𝑢(𝑦3 ) + 𝑢(𝑦4 ) + 𝑢(𝑦5 ) = (4 − 𝜆𝑚 )𝑢(𝑥1 ).

230

12. Eigenfunctions of the Laplacian

As each 𝑢(𝑦 𝑖 ), 𝑖 = 2, 3, 4, 5, satisfies (12.9), we see that (12.10)

𝑢(𝑦2 ) + 𝑢(𝑦3 ) + 𝑢(𝑦4 ) + 𝑢(𝑦5 ) =

4(4 − 𝜆𝑚 )𝑢(𝑥1 ) + (6 − 𝜆𝑚 )(𝑢(𝑥2 ) + 𝑢(𝑥3 ) + 𝑢(𝑥4 ) + 𝑢(𝑥5 )) . (2 − 𝜆𝑚 )(5 − 𝜆𝑚 )

Now, 𝑢(𝑥1 ) also satisfies Δ𝑚−1 𝑢(𝑥1 ) = −𝜆𝑚−1 𝑢(𝑥1 ), so we have 𝑢(𝑥2 ) + 𝑢(𝑥3 ) + 𝑢(𝑥4 ) + 𝑢(𝑥5 ) = (4 − 𝜆𝑚−1 )𝑢(𝑥1 ), which substituting in (12.10) gives us 𝑢(𝑦2 ) + 𝑢(𝑦3 ) + 𝑢(𝑦4 ) + 𝑢(𝑦5 ) =

4(4 − 𝜆𝑚 )𝑢(𝑥1 ) + (6 − 𝜆𝑚 )(4 − 𝜆𝑚−1 )𝑢(𝑥1 ) , (2 − 𝜆𝑚 )(5 − 𝜆𝑚 )

so 4(4 − 𝜆𝑚 )𝑢(𝑥1 ) + (6 − 𝜆𝑚 )(4 − 𝜆𝑚−1 )𝑢(𝑥1 ) = (4 − 𝜆𝑚 )𝑢(𝑥1 ). (2 − 𝜆𝑚 )(5 − 𝜆𝑚 ) It is not hard to see that this must hold for every 𝑥 ∈ 𝑉𝑚−1 ⧵ 𝑉0 , and, since 𝑢 is an eigenfunction of Δ𝑚−1 , it cannot be zero at all points, we obtain the relation 4(4 − 𝜆𝑚 ) + (6 − 𝜆𝑚 )(4 − 𝜆𝑚−1 ) = (4 − 𝜆𝑚 )(2 − 𝜆𝑚 )(5 − 𝜆𝑚 ), which we can rewrite as (6 − 𝜆𝑚 )𝜆𝑚−1 = (6 − 𝜆𝑚 )𝜆𝑚 (5 − 𝜆𝑚 ). If 𝜆𝑚 ≠ 6, we obtain the condition (12.11)

𝜆𝑚−1 = 𝜆𝑚 (5 − 𝜆𝑚 ).

We summarize the previous analysis in Proposition 12.12. Proposition 12.12. Suppose that 𝜆𝑚 ≠ 2, 5, 6, and that we have the condition (12.11). (1) If 𝑢 is an eigenfunction of Δ𝑚−1 on 𝑉𝑚−1 with respect to the eigenvalue −𝜆𝑚−1 and is extended to 𝑉𝑚 by (12.9), then we obtain an eigenfunction of Δ𝑚 with respect to the eigenvalue −𝜆𝑚 . (2) If 𝑢 is an eigenfunction of Δ𝑚 on 𝑉𝑚 with respect to the eigenvalue −𝜆𝑚 , then 𝑢|𝑉𝑚−1 is an eigenfunction of Δ𝑚−1 with respect to −𝜆𝑚−1 .

12.2. Discrete eigenfunctions on the Sierpiński gasket

231

The discrete process described in Proposition 12.12 can be used to construct eigenfunctions of the Laplacian Δ on 𝑆. Indeed, note that the roots of (12.11) are given by 𝜆𝑚 =

5 ± √25 − 4𝜆𝑚−1 . 2

Using the principal branch of √5 − 𝑧 on ℂ ⧵ [5, ∞), we observe that, if 𝑤=

5 − √25 − 4𝑧 , 2

then 𝑤=

𝑧 + 𝑂(|𝑧|2 ) 5

as 𝑧 → 0, and thus 3 𝑚 5 𝜆𝑚 2 converges if we start from some |𝜆𝑚0 | < 5 and choose the minus sign for all but infinitely many 𝑚 (Exercise (3)). Moreover, the functions 𝑢 defined by Proposition 12.12 converge to a uniformly continuous function on 𝑉∗ (Exercise (4)), which satisfies (12.13)

3 𝑚 3 5 Δ𝑚 𝑢(𝑥) = − 5𝑚 𝜆𝑚 𝑢(𝑥) 2 2 3 for all 𝑥 ∈ 𝑉∗ ⧵ 𝑉0 . Thus, if 5𝑚 𝜆𝑚 → 𝜆, we can continuously extend 𝑢 2 to a function on 𝑆 in dom Δ that satisfies Δ𝑢 = −𝜆𝑢, and thus is an eigenfunction with respect to the eigenvalue −𝜆. We described how to construct an eigenfunction of Δ through this interpolation process, so we now dicuss whether we can obtain all eigenfunctions of Δ in this way. First, we deal with the question of whether we can obtain any eigenvalue −𝜆 as a limit (12.13) starting from a proper 𝜆 𝑚0 . An eigenfunction 𝑢 that satisfies 𝑢|𝑉0 = 0 is called a Dirichlet eigenfunction, and we say that an eigenvalue is a Dirichlet eigenvalue if it has a corresponding Dirichlet eigenfunction. Note that, if −𝜆 is an eigenvalue that is not a Dirichlet eigenvalue, its corresponding eigenspace is at most 3-dimensional, because #𝑉0 = 3

232

12. Eigenfunctions of the Laplacian

and otherwise we could obtain a linear combination of 4 or more of its eigenfunctions to obtain 0 on 𝑉0 . Moreover, for any 𝜆1 ≠ 2, (12.9) defines an eigenfunction of Δ1 for any choice of 𝑢(𝑝1 ), 𝑢(𝑝2 ) and 𝑢(𝑝3 ), so it defines a 3-dimensional space of eigenfunctions and thus, by taking the minus sign for every 𝑚 in the roots of (12.11), defines a 3-dimensional space of eigenfunctions with respect to each eigenvalue −𝜆, if 𝜆 is a limit (12.13). To see which points in ℂ are such limits, define 5 − √25 − 4𝑧 2 and, for |𝑧| < 2 (so to avoid 𝑧 = 2), 𝜓(𝑧) =

Ψ(𝑧) = lim 5𝑚 𝜓𝑚 (𝑧), 𝑚→∞

𝑚

where 𝜓 is the composition of 𝜓 with itself 𝑚 times. Ψ is well defined in this way (Exercise (3)), and by the above discussion, for any |𝜆1 | < 2, if 3 𝜆 = 5Ψ(𝜆1 ), 2 then −𝜆 is an eigenvalue of Δ. The convergence of the limit that defines Ψ is uniform in compact subsets of 𝐵2 (0), so Ψ is holomorphic in 𝐵2 (0). Moreover, since 𝜓′ (0) = 1/5, we see that each 5𝑚 𝜓𝑚 (𝑧) has derivative 1 at 0, so Ψ′ (0) = 1. By the open mapping theorem of complex analysis (see, for instance, [Ull08, Theorem 5.7]), we see that Ψ(𝐵2 (0)) contains a ball 𝐵𝑟 (0) for some 𝑟 > 0. Therefore, given any non-Dirichlet eigenvalue in the disc of radius 15𝑟/2, this eigenvalue and its corresponding eigenfunctions are the result of the discrete algorithm starting from some |𝜆1 | < 2. Now, for a larger non-Dirichlet eigenvalue −𝜆, we choose 𝑚0 large enough so that |𝜆| < 3 ⋅ 5𝑚0 𝑟/2. Thus 3 𝜆 = 5𝑚0 Ψ(𝜆𝑚0 ) 2 for some 𝜆𝑚0 ∈ 𝐵2 (0), so we need to make sure that we can construct a 3dimensional eigenspace of Δ𝑚0 with respect to −𝜆𝑚0 . However, observe that the system of #(𝑉𝑚0 ⧵ 𝑉0 ) linear equations Δ𝑚0 𝑢(𝑥) = −𝜆𝑚0 𝑢(𝑥), can be written as 𝒟𝑢(𝑥) = 𝑏𝑥 ,

12.3. Dirichlet eigenfunctions

233

where 𝒟 is the linear operator given, for each 𝑥 ∈ 𝑉𝑚0 ⧵ 𝑉0 , by 𝒟𝑢(𝑥) = ∑ 𝑢(𝑦) − (4 − 𝜆𝑚0 )𝑢(𝑥), 𝑦∼𝑚 𝑥 𝑦∉𝑉0

and −𝑢(𝑝 𝑖 ) 𝑏𝑥 = { 0

𝑥 ∼𝑚 𝑝 𝑖 𝑥 is not a neighbor of a point 𝑝 𝑖 ∈ 𝑉0 .

Thus, either we have a unique solution for each choice of 𝑢(𝑝1 ), 𝑢(𝑝2 ), 𝑢(𝑝3 ), so we have a 3-dimensional space of eigenfunctions with respect to −𝜆𝑚0 , or −𝜆𝑚0 is a Dirichlet eigenvalue, because in that case we would have solutions when all 𝑏𝑥 = 0, and thus −𝜆 would be a Dirichlet eigenvalue. We have proven that any eigenfunction with respect to a non-Dirichlet eigenvalue can be constructed through the discrete process. The Dirichlet eigenfunctions will be discussed in Section 12.3.

12.3. Dirichlet eigenfunctions In this section we discuss the construction of the Dirichlet eigenfunctions of Δ on the Sierpiński gasket 𝑆. For this, we will analyze more carefully the discrete process, keeping track on the number of linearly independent eigenfunctions constructed, in order to conclude that we have constructed all of them. We first construct all discrete Dirichlet eigenfunctions of the operators Δ𝑚 , 𝑚 ≥ 1. We start with the Dirichlet eigenfunctions of Δ1 on 𝑉1 , so we need to solve the three equations Δ1 𝑢(𝑥) = −𝜆1 𝑢(𝑥), one for each of the three points 𝑥 ∈ 𝑉1 ⧵ 𝑉0 , with the Dirichlet condition 𝑢(𝑝1 ) = 𝑢(𝑝2 ) = 𝑢(𝑝3 ) = 0. By Exercise (5), we have three linearly independent solutions shown in Figure 12.4, with corresponding eigenvalues −2 and −5, the latter with multiplicity 2. Note that these are two of the forbidden eigenvalues of Proposition 12.12, which makes sense as their restrictions to 𝑉0 are not eigenfunctions of Δ0 . Also, observe that one of the Dirichlet eigenfunctions with 𝜆1 = 5 is the rotation by 2𝜋/3 of the other, and that we have

234

12. Eigenfunctions of the Laplacian

Figure 12.4. The Dirichlet eigenfunctions of Δ1 on 𝑉 1 , corresponding to 𝜆1 = 2, 5 and 5, respectively. For simplicity, we only show the values of the eigenfunction at the vertices, and we don’t show the ones that have value zero.

a third eigenfunction corresponding to the next rotation. However, the latter is not linearly independent of the other two. If we remove the Dirichlet condition, we have two more linearly independent solutions for 𝜆1 = 2 and one more for 𝜆1 = 5, shown in Figure 12.5. Note that, for the case 𝜆1 = 2,

Figure 12.5. Eigenfunctions of Δ1 on 𝑉 1 corresponding to 𝜆1 = 2, 2 and 5, respectively. The third rotation of the first two of them is also an eigenfunction corresponding to 𝜆1 = 2, but is not linearly independent.

𝑢(𝑝1 ) + 𝑢(𝑝2 ) + 𝑢(𝑝3 ) = 0, which is a necessary condition given by (12.7). For 𝜆1 = 5, we have 𝑢(𝑝1 ) = 𝑢(𝑝2 ) = 𝑢(𝑝3 ), which follows from (12.8) (Exercise (6)). We now move on to the eigenvalues of Δ2 on 𝑉2 . As #(𝑉2 ⧵ 𝑉0 ) = 12, we need to construct twelve linearly independent Dirichlet eigenfunctions. Although we could just solve explicitly the system of equations (12.14)

Δ2 𝑢(𝑥) = −𝜆2 𝑢(𝑥)

12.3. Dirichlet eigenfunctions

235

for 𝑥 ∈ 𝑉2 ⧵ 𝑉0 , with 𝑢(𝑝 𝑖 ) = 0, we will construct them in such a way that we can generalize for any 𝑚 ≥ 2. First, six of the Dirichlet eigenfunctions correspond to the extensions of the eigenfunctions of Δ1 shown in Figure 12.4, with eigenvalues −𝜆2 where 5 ± √25 − 4𝜆1 𝜆2 = . 2 Thus, we have six eigenfunctions with 𝜆2 given by 5 + √17 5 − √17 5 + √5 5 − √5 , , , , 2 2 2 2 the last two with multiplicity 2. The other six Dirichlet eigenfunctions are not extensions from any eigenfunction on 𝑉1 , so they must correspond to any of the forbidden values 𝜆2 = 2, 5 or 6; otherwise, their restrictions would be Dirichlet eigenfunctions on 𝑉1 , by Proposition 12.12. For 𝜆2 = 2, note that the restriction of such eigenfunction to each cell 𝑆 𝑖 ∩ 𝑉2 must correspond to one of eigenfunctions with 𝜆1 = 2 shown in Figures 12.4 or 12.5, since the same difference equation must be satisfied. However, since (12.14) must also be satisfied in the points 𝑉1 ⧵ 𝑉0 , this impose an extra restriction of how the 𝑉1 -eigenfunctions are pasted into 𝑉2 . In fact, one sees that it is impossible to do this because, when one starts at one of the corners, the next one is either determined by this condition if one chooses to start with the eigenfunction of Figure 12.4, and then it is impossible to past the third one, or it is already impossible to paste the second one if we choose one of those shown in Figure 12.5. See Figure 12.6. Therefore, there are no Dirichlet eigenfunctions of Δ2 with respect to the eigenvalue −2. For 𝜆2 = 5, we now see that it is possible to paste two consecutive eigenfunctions shown in Figure 12.4, and their rotations, so we obtain the three linearly independent Dirichlet eigenfunctions shown in Figure 12.7. For 𝜆2 = 6, we first consider the eigenfunctions of Δ1 with 𝜆1 = 6. These are shown in Figure 12.8. These are not Dirichlet eigenfunctions, of course, but they can be pasted consecutively (as in the case of 𝜆2 = 5) to form Dirichlet eigenvalues of Δ2 , as is shown in Figure 12.9. Note that not only each is a rotation of the other, but each one is “centered” at a point in 𝑉1 , the point where two of the eigenfunctions from Figure 12.8

236

12. Eigenfunctions of the Laplacian

Figure 12.6. Failed attempts to construct a Dirichlet eigenfunction on 𝑉 2 with 𝜆2 = 2.

Figure 12.7. Three linearly independent Dirichlet eigenfunctions with 𝜆2 = 5, constructed by pasting two consecutive eigenfunctions on 𝑉 1 .

Figure 12.8. The three linearly independent eigenfunctions Δ1 with 𝜆1 = 6. Note that each is a rotation of the other.

where pasted. These form the last three linearly independent Dirichlet eigenfunctions of Δ2 on 𝑉2 . We can now generalize these constructions for all 𝑚 ≥ 3. Since #(𝑉𝑚 ⧵ 𝑉0 ) =

3𝑚+1 − 3 2

12.3. Dirichlet eigenfunctions

237

Figure 12.9. Three linearly independent Dirichlet eigenfunctions of Δ2 on 𝑉 2 with 𝜆2 = 6. Note that each is centered at a point in 𝑉 1 .

(Exercise (8)), there must be (3𝑚+1 − 3)/2 linearly independent Dirichlet eigenfunctions of Δ𝑚 . First, note that, as in the case for 𝑚 = 2, there are no Dirichlet eigenfunctions with 𝜆𝑚 = 2, for 𝑚 ≥ 3. The argument is the same as above, and we leave it as an exercise (Exercise (7)). Also, note that if 𝜆𝑚−1 = 6, then 5 ± √25 − 4𝜆𝑚−1 5±1 = = 2, 3, 2 2 so we cannot extend such Dirichlet eigenfunctions of Δ𝑚−1 using the negative sign above. Thus, in order to count the eigenfunctions of Δ𝑚 that are extensions from 𝑉𝑚−1 , we first need to count the discrete Dirichlet eigenfunctions with respect to the eigenvalue −6. These are constructed as in the case 𝑚 = 2 above. Indeed, for each 𝑉𝑚 , we can paste two of the eigenfunctions in Figure 12.8 centered at each point 𝑥 ∈ 𝑉𝑚−1 , as in Figure 12.9. Hence, we have (3𝑚 − 3)/2 such eigenfunctions. On 𝑉𝑚−1 , we have then (3𝑚−1 − 3)/2 Dirichlet eigenfunctions with 𝜆𝑚−1 = 6, so they extend to Dirichlet eigenfunctions of Δ𝑚 with 𝜆𝑚 = 3. The number of Dirichlet eigenfunctions with 𝜆𝑚−1 ≠ 6 is then 3𝑚 − 3 3𝑚−1 − 3 − = 3𝑚−1 , 2 2 and they extend to 2 ⋅ 3𝑚−1 eigenfunctions of Δ𝑚 with 𝜆𝑚 =

5 ± √25 − 4𝜆𝑚−1 . 2

Thus, the number of Dirichlet eigenfunctions of Δ𝑚 extended from 𝑉𝑚−1 is (3𝑚−1 − 3)/2 + 2 ⋅ 3𝑚−1 = (5 ⋅ 3𝑚−1 − 3)/2 which, together with the

238

12. Eigenfunctions of the Laplacian

ones with 𝜆𝑚 = 6 that we already constructed, make a total of 8 ⋅ 3𝑚−1 − 6 5 ⋅ 3𝑚−1 − 3 3𝑚 − 3 + = . = 4 ⋅ 3𝑚−1 − 3. 2 2 2 Thus, we need to construct 3𝑚+1 − 3 3𝑚−1 + 3 − (4 ⋅ 3𝑚−1 − 3) = 2 2 more eigenfunctions. They will correspond to 𝜆𝑚 = 5. To construct these, first note that we can paste together a chain of eigenfunctions with 𝜆1 = 5 (as the last two in Figure 12.4 and their rotations) in each of the cells 𝑆𝑤 , for 𝑤 ∈ 𝑊𝑚−1 , surrounding each downward triangular cycle formed with points in 𝑉 𝑘 ⧵ 𝑉 𝑘−1 , for 𝑘 = 1, 2, . . . , 𝑚 − 1. We show this for the largest and one of the next largest downward triangular cycles for 𝑉3 in Figure 12.10. As #𝑉 𝑘 ⧵ 𝑉 𝑘−1 = 3𝑘 ,

Figure 12.10. Two of the Dirichlet eigenfunctions of Δ3 with 𝜆3 = 5, chained around triangles with vertices in 𝑉 1 ⧵ 𝑉 0 and 𝑉 2 ⧵ 𝑉 1 , respectively.

we have that the number of downward triangular cycles with vertices in 𝑉 𝑘 ⧵ 𝑉 𝑘−1 is 3𝑘−1 , and thus the total number of such cycles is 3𝑚−1 − 1 . 2 We obtain two more linearly independent eigenfunctions by pasting a chain if eigenfunctions along the edge from vertex 𝑝1 to 𝑝2 , and another one from 𝑝2 to 𝑝3 . We thus have a total of (3𝑚−1 + 3)/2 eigenfunctions with 𝜆𝑚 = 5, as required. 1 + 3 + 32 + 3𝑚−2 =

We have thus constructed all Dirichlet eigenfunctions of the discrete Laplacian Δ𝑚 on 𝑉𝑚 . As in Section 12.2, starting from each eigenfunction on 𝑉𝑚0 with 𝜆𝑚0 , the sequence obtained from discrete converges to

12.3. Dirichlet eigenfunctions

239

a uniformly continuous function on 𝑉∗ , which extends continuously to a Dirichlet eigenfunction 𝑢 of Δ on 𝑆 with respect to the eigenvalue −𝜆, where 3 𝜆 = lim 5𝑚 𝜆𝑚 . 2 𝑚→∞ However, recall that, for each 𝑚 > 𝑚0 , 5 + 𝜖𝑚 √25 − 4𝜆𝑚−1 , 2 where all but infinitely many 𝜖𝑚 = −1. This means that, for each finite sequence 𝜖𝑚0 +1 , 𝜖𝑚0 +2 , . . . , 𝜖𝑀 of ±1, we obtain a distinct eigenvalue −𝜆 with a distinct eigenfunction 𝑢 when choosing 𝜖𝑚 = −1 for all 𝑚 > 𝑀. Therefore, we obtain a sequence of Dirichlet eigenvalues and eigenfunctions from each starting 𝜆𝑚0 . 𝜆𝑚 =

In Figure 12.11 we show the Dirichlet eigenfunctions that we ob-

Figure 12.11. Dirichlet eigenfunctions of Δ on 𝑆 with respect to the eigenvalues −𝜆, with 𝜆 ≈ 16.816, 𝜆 ≈ 240.1686 and 𝜆 ≈ 920.6197, respectively, obtained from 𝜆1 = 2 when choosing 𝜖𝑚 = −1 for all 𝑚 ≥ 2 in the first case, 𝜖2 = 1 and 𝜖𝑚 = −1 for all 𝑚 ≥ 3 in the second, and 𝜖2 = 𝜖3 = 1 and 𝜖𝑚 = −1 for all 𝑚 ≥ 4 in the third case.

tain when we start from the Dirichlet eigenfunction with respect to the eigenvalue −2 shown in Figure 12.4, where 𝑢(𝑥) = 1 for 𝑥 ∈ 𝑉1 ⧵ 𝑉0 . We obtain the first eigenfunction shown when we choose 𝜖𝑚 = −1 for all 𝑚 ≥ 2. The first terms of the sequence 𝜆𝑚 in this case are 5 − √17 5 − √15 + 2√17 5 − √15 + 2√15 + 2√17 , , , ... , 2 2 2 and one can approximate its normalized limit by 2,

3 lim 5𝑚 𝜆𝑚 ≈ 16.816. 2 𝑚→∞

240

12. Eigenfunctions of the Laplacian

We obtain the second eigenfunction when choosing 𝜖2 = 1 and 𝜖𝑚 = −1 for all 𝑚 ≥ 3, so the first terms of the sequence 𝜆𝑚 are 5 + √17 5 − √15 − 2√17 5 − √15 + 2√15 − 2√17 , , , ... , 2 2 2 and its limit 3 lim 5𝑚 𝜆𝑚 ≈ 240.1686. 2 𝑚→∞ We obtain the third eigenfunction in Figure 12.11 when choosing 𝜖2 = 𝜖3 = 1 and 𝜖𝑚 = −1 for all 𝑚 ≥ 4, and in this case 3 lim 5𝑚 𝜆𝑚 ≈ 920.6197. 2 𝑚→∞ In Figure 12.12 we show the eigenfunctions obtained starting with 𝜆1 = 2,

Figure 12.12. Dirichlet eigenfunctions of Δ on 𝑆 with respect to the eigenvalues −𝜆, with 𝜆 ≈ 55.8858, 𝜆 ≈ 172.3645 and 𝜆 ≈ 1032.0357.

5, corresponding to the second eigenfunction in Figure 12.4, and the same choice of starting sequences 𝜖2 , 𝜖3 as above. In this case we obtain the limits 𝜆 ≈ 55.8858, 𝜆 ≈ 172.3645, 𝜆 ≈ 1032.0357, respectively. 12.15. The argument in Section 12.2 can also be applied here to conclude that, if −𝜆 is a Dirichlet eigenvalue of Δ on 𝑆, then 𝜆 is the limit of (3/2)5𝑚 𝜆𝑚 , starting from 𝜆1 = 2, 𝜆𝑚0 = 5 or 𝜆𝑚0 = 6, from an appropiately chosen 𝑚0 and a particular choice of 𝜖𝑚 , 𝑚 > 𝑚0 (Exercise (5)). Our construction above produces all Dirichlet eigenfunctions with respect to a Dirichlet eigenvalue −𝜆. 12.16. If −𝜆 is a Dirichlet eigenvalue, so is −5𝑚 𝜆, for each 𝑚 ≥ 1. Indeed, if 𝜆 is the discrete limit starting from 𝜆𝑚0 , then 𝜆 ̄ = 5𝑚 𝜆 is the limit ̄ +𝑚 = 𝜆𝑚 . starting from 𝜆𝑚 0 0

Exercises

241

It is also possible to prove that we can construct all eigenfunctions 𝑢 with respect to a Dirichlet eigenvalue through this discrete algorithm, even those where 𝑢|𝑉0 ≠ 0. The argument involves Neumann derivatives at the vertices, which we won’t discuss here. The reader can find the argument in [Str06].

Exercises (1) Let 𝜙(𝑧) = 2 − √4 − 𝑧, for |𝑧| < 4, using the principal branch of √4 − 𝑧 on ℂ ⧵ [4, ∞). 1 (a) 𝜙(𝑧) = 𝑧 + 𝑂(|𝑧|2 ) as 𝑧 → 0. 4 (b) Given |𝜆1 | < 4, the sequence defined by 𝜆𝑚 = 𝜙(𝜆𝑚−1 ) and 𝑧𝑚 = 4𝑚 𝜆𝑚 for 𝑚 ≥ 1 satisfies 𝑧𝑚 − 𝑧𝑚−1 = 𝑂(2−𝑚 ). (c) 𝑧𝑚 is Cauchy and hence converges. (2) Let 𝜆𝑚 be the sequence defined in Exercise (1) with 𝜆1 = 2. Then 4𝑚 𝜆𝑚 → 𝜋2 . 5 − √25 − 4𝑧 25 , for |𝑧| < , with the principal branch of 2 4 √5 − 𝑧 on ℂ ⧵ [5, ∞). 1 (a) 𝜓(𝑧) = 𝑧 + 𝑂(|𝑧|2 ) as 𝑧 → 0. 5 25 (b) Given |𝜆1 | < , the sequence defined by 𝜆𝑚 = 𝜓(𝜆𝑚−1 ) and 4 𝑧𝑚 = 5𝑚 𝜆𝑚 for 𝑚 ≥ 1 satisfies

(3) Let 𝜓(𝑧) =

𝑧𝑚 − 𝑧𝑚−1 = 𝑂(5−𝑚 ). (c) 𝑧𝑚 is Cauchy and hence converges. (4) The sequence functions resulting from the discrete process on the Sierpiński gasket constructs a uniformly continuous function on 𝑉∗ . (5) Calculate the eigenvalues and eigenvectors of the matrix 4 −1 −1 (−1 4 −1) . −1 −1 4

242

12. Eigenfunctions of the Laplacian

(6) If 𝜆𝑚 = 5, the equation (12.8) implies that 𝑢(𝑥1 ) = 𝑢(𝑥2 ) = 𝑢(𝑥3 ). (7) For 𝑚 ≥ 2, Δ𝑚 does not have any Dirichlet eigenfunctions on 𝑉2 with respect to the eigenvalue −2. (8) (a) For any 𝑚 ≥ 0, #𝑉𝑚 =

3𝑚+1 + 3 . 2

(b) For any 𝑚 ≥ 1, 3𝑚+1 − 3 . 2 (9) If −𝜆 is a Dirichlet eigenvalue of Δ on 𝑆, then there exists 𝑚0 and a sequence 𝜖𝑚 such that 3 𝜆 = lim 5𝑚 𝜆𝑚 , 2 𝑚→∞ with 𝜆𝑚0 = 2, 5 or 6. #(𝑉𝑚 ⧵ 𝑉0 ) =

(10) Let 𝑢 be the Dirichlet eigenfunction obtained by discrete with 𝜆1 = 2, and choosing 𝜖𝑚 = −1 for all 𝑚 ≥ 2. This is the first eigenfunction shown in Figure 12.11. Then 𝑢(𝑥) > 0 for all 𝑥 ∈ 𝑆 ⧵ 𝑉0 .

Notes The algorithm described in this chapter for the construction of the eigenfunctions, as well as the relation for the discrete eigenvalues between levels, is known as the spectral decimation method and is due to Tadashi Shima [Shi91], who based his work from the results in [Ram84]. Our discussion follows the analysis presented in [Str06].

Chapter 13

Harmonic functions on post-critically finite sets

In this chapter we discuss how to construct the Laplacian on other fractal sets, analogous to the construction of the Laplacian on the Sierpiński gasket.

13.1. Post-critically finite sets Let 𝐾 ⊂ ℝ𝑑 be a self-similar set with respect to the contractions 𝑓1 , 𝑓2 , . . . , 𝑓𝑁 . Recall that 𝐾 is nonempty, compact, and satisfies 𝐾 = 𝑓1 (𝐾) ∪ 𝑓2 (𝐾) ∪ . . . ∪ 𝑓𝑁 (𝐾). We assume that the contractions 𝑓𝑗 are restricted to 𝐾, so we see them as functions 𝑓𝑗 ∶ 𝐾 → 𝐾. We use the notation of Chapter 10, so 𝑊𝑚 , for 𝑚 ≥ 1, is the set of words 𝑤 = 𝑤 1 𝑤 2 . . . 𝑤 𝑚 of length 𝑚 with each 𝑤 𝑖 ∈ {1, 2, . . . , 𝑁}. We define 𝑓𝑤 = 𝑓𝑤1 ∘ 𝑓𝑤2 ∘ ⋯ ∘ 𝑓𝑤𝑚 and 𝐾𝑤 = 𝑓𝑤 (𝐾). For 𝑚 = 0 we define 𝑊0 = {∅}, where ∅ is the empty word, 𝑓∅ is the identity function, and 𝐾∅ = 𝐾. Each 𝐾𝑤 , with 𝑤 ∈ 𝑊𝑚 , is called a cell of level 𝑚. We have that 𝐾 = ⋃𝑤∈𝑊 𝑓𝑤 (𝐾) where, if 𝛼𝑗 𝑚 is the contraction constant of 𝑓𝑗 , diam(𝐾𝑤 ) ≤ 𝛼𝑤 diam(𝐾), 243

244

13. Harmonic functions on post-critically finite sets

with 𝛼𝑤 = 𝛼𝑤1 𝛼𝑤2 ⋯ 𝛼𝑤𝑚 . Since each 𝛼𝑗 < 1, then diam(𝐾𝑤 ) → 0 if 𝑚 → ∞. Thus, for each sequence 𝜔 = 𝑤 1 , 𝑤 2 , . . ., if we denote 𝜔𝑚 = 𝑤 1 𝑤 2 . . . 𝑤 𝑚 , then ⋂𝑚≥1 𝐾𝜔𝑚 contains exactly one point of 𝐾. Conversely, if 𝑥 ∈ 𝐾, there exists a sequence 𝜔 such that (13.1)

⋂

𝐾𝜔𝑚 = {𝑥}.

𝑚≥1

For a self-similar set 𝐾, define the set (13.2)

𝒞=

⋃

𝐾𝑖 ∩ 𝐾𝑗 .

𝑖≠𝑗

Then 𝒞 is the set of all overlaps between the images 𝐾𝑗 of 𝐾 under the contractions. 𝒞 is called the critical set. We assume 𝒞 ≠ ∅, and define the post-critical set of 𝐾 by (13.3)

𝑉0 =

⋃ ⋃

𝑓𝑤−1 (𝒞),

𝑚≥1 𝑤∈𝑊𝑚

Observe that 𝑉0 is the union of all possible pre-images, under any number of iterations of the 𝑓𝑗 , of the critical set. We say that 𝐾 is a post-critically finite set, denoted as PCF set, if 𝑉0 is finite. Example 13.4. Consider the Sierpiński gasket 𝑆 discussed in the previous chapters. Its critical set 𝒞 consists of the three middle points of its sides, which clearly are the images of the triangle vertices 𝑝1 , 𝑝2 , 𝑝3 (see Figure 11.2) under the contractions 𝑓1 , 𝑓2 , 𝑓3 . Thus 𝑝1 , 𝑝2 , 𝑝3 ∈ 𝑉0 . Now, since 𝑝1 ∉ 𝑆 2 nor 𝑝1 ∉ 𝑆 3 , and 𝑝1 is the fixed point of 𝑓1 , the further preimages of 𝑝1 contain only the point 𝑝1 . Similarly for 𝑝2 and 𝑝3 , which are the fixed points of 𝑓2 and 𝑓3 , respectively, we have that 𝑉0 = {𝑝1 , 𝑝2 , 𝑝3 }. Therefore 𝑆 is a PCF set. Example 13.5. The interval 𝐼 = [0, 1], seen as a self-similar set with contractions 𝑓1 (𝑥) = 𝑥/2 and 𝑓2 (𝑥) = 𝑥/2+1/2, is also a PCF set. Indeed, 𝒞 = {1/2} and 𝑉0 = {0, 1}. Note that, in both of the previous examples, 𝑉0 is known as the boundary of 𝑆 and 𝐼, respectively. In general, 𝑉0 is called the boundary of the PCF set 𝐾. Example 13.6. The Hata tree set 𝐾 of Example 10.22, with contractions 2 1 𝑓1 (𝑧) = 𝑐𝑧,̄ 𝑓2 (𝑧) = 𝑧 ̄ + , 3 3

13.2. Harmonic structures and discrete energy

245

√3 1 where 𝑐 = + 𝑖. Its critical set consists only of the point 1/3 (see 2 6 Figure 10.6), which is given by 1 = 𝑓1 (𝑐) = 𝑓2 (0). 3 Thus 𝑐, 0 ∈ 𝑉0 . Now 𝑐 = 𝑓1 (1), so 1 ∈ 𝑉0 . Since 0, 𝑐 ∈ 𝐾1 ⧵ 𝐾2 and 1 ∈ 𝐾2 ⧵ 𝐾1 , and 0 and 1 are the fixed points of 𝑓1 and 𝑓2 , respectively, we conclude that 𝑉0 = {𝑐, 0, 1}. For each 𝑚 ≥ 1, we define 𝑉𝑚 = 𝑓1 (𝑉𝑚−1 ) ∪ 𝑓2 (𝑉𝑚−1 ) ∪ . . . ∪ 𝑓𝑁 (𝑉𝑚−1 ). 𝑉𝑚 is the set of vertices of level 𝑚. We see that 𝑉𝑚 =

⋃

𝑓𝑤 (𝑉0 )

𝑤∈𝑊𝑚

and, by (13.3), each 𝑉𝑚−1 ⊂ 𝑉𝑚 . The set of all vertices is denoted by 𝑉∗ , so 𝑉∗ =

⋃

𝑉𝑚 .

𝑚≥0

By (13.1), 𝑉∗ is dense in 𝐾 (Exercise (1)). If we denote 𝑉0 = {𝑝1 , 𝑝2 , . . . , 𝑝𝑀 }, where 𝑀 = #𝑉0 , then each 𝑥 ∈ 𝑉𝑚 is of the form 𝑥 = 𝑓𝑤 (𝑝 𝑖 ) for some 𝑤 ∈ 𝑊𝑚 and 𝑗 = 1, 2, . . . , 𝑁. We say that two vertices 𝑥, 𝑦 ∈ 𝑉𝑚 are adjacent, or neighbors, and write 𝑥 ∼𝑚 𝑦, if there exists 𝑤 ∈ 𝑊𝑚 such that 𝑥 = 𝑓𝑤 (𝑝 𝑖 ) and 𝑦 = 𝑓𝑤 (𝑝𝑗 ), for some 𝑖, 𝑗. That is, 𝑥, 𝑦 ∈ 𝑓𝑤 (𝑉0 ), so they belong to the same cell 𝐾𝑤 .

13.2. Harmonic structures and discrete energy For a nonempty finite set 𝑉, let 𝑙(𝑉) be the vector space of real valued functions on 𝑉. Thus, if #𝑉 = 𝑀, then 𝑙(𝑉) ≅ ℝ𝑀 . 𝑙(𝑉) has the inner product ⟨𝑢, 𝑣⟩ = ∑ 𝑢(𝑥)𝑣(𝑥), 𝑥∈𝑉

so ⟨⋅, ⋅⟩ is essentially the dot product in ℝ𝑀 .

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13. Harmonic functions on post-critically finite sets

The standard basis of 𝑙(𝑉) is given by the functions 𝜒{𝑥} , 𝑥 ∈ 𝑉. A linear operator 𝑇 on 𝑙(𝑉) can be identified with the 𝑀 ×𝑀 matrix which, with respect to the standard basis, has entries 𝑇𝑥𝑦 = ⟨𝑇𝜒{𝑥} , 𝜒{𝑦} ⟩. Any bilinear form 𝐵 on 𝑙(𝑉) is induced by an operator 𝑇 via 𝐵(𝑢, 𝑣) = −⟨𝑇𝑢, 𝑣⟩. We use the negative sign for technical reasons that will be clear below. If 𝐵 is symmetric, that is 𝐵(𝑢, 𝑣) = 𝐵(𝑣, 𝑢) for any 𝑢, 𝑣 ∈ 𝑙(𝑉), then 𝑇 is symmetric. If 𝐵 is symmetric, we denote 𝐵(𝑢, 𝑢) simply by 𝐵(𝑢), so 𝑢 ↦ 𝐵(𝑢) is a quadratic form on 𝑙(𝑉). The symmetric bilinear form ℰ is a Dirichet form if it satisfies (1) ℰ(𝑢) ≥ 0 for all 𝑢 ∈ 𝑙(𝑉); (2) ℰ(𝑢) = 0 if and only if 𝑢 is constant; and (3) ℰ(𝑢)̄ ≤ ℰ(𝑢) for every 𝑢 ∈ 𝑙(𝑉), where 𝑢(𝑥) < 0 ⎧0 𝑢(𝑥) ̄ = 𝑢(𝑥) 0 ≤ 𝑢(𝑥) ≤ 1 ⎨ 𝑢(𝑥) > 1 ⎩1 is the cut of 𝑢 by [0, 1]. 13.7. Any bilinear form ℰ given by ℰ(𝑢, 𝑣) = ∑ 𝑐𝑥𝑦 (𝑢(𝑥) − 𝑢(𝑦))(𝑣(𝑥) − 𝑣(𝑦)), 𝑥,𝑦∈𝑉

with 𝑐𝑥𝑦 ≥ 0, is a Dirichlet form if and only if, for each 𝑥0 , 𝑦0 ∈ 𝑉, we can find 𝑥1 , 𝑥2 , . . . , 𝑥𝑘 = 𝑦0 such that 𝑐𝑥𝑗−1 𝑥𝑗 > 0. We just need to observe that ℰ(𝑢) = ∑ 𝑐𝑥𝑦 (𝑢(𝑥) − 𝑢(𝑦))2 𝑥,𝑦∈𝑉

is a nonnegative linear combination of the squares (𝑢(𝑥) − 𝑢(𝑦))2 , so (1) and (3) are clearly satisfied, and that ℰ(𝑢) = 0 when, for each pair 𝑥, 𝑦 ∈ 𝑉, 𝑐𝑥𝑦 = 0 or 𝑢(𝑥) = 𝑢(𝑦), so the condition implies that this can only happen if 𝑢 is constant. Note that ℰ(𝑢, 𝑣) is the polarization of ℰ(𝑢).

13.2. Harmonic structures and discrete energy

247

Let 𝐾 be a PCF set and consider the sequence 𝑉𝑚 of vertices of level 𝑚, defined for 𝑚 ≥ 0 as above. For each 𝑚 ≥ 0, let ℰ𝑚 be a Dirichlet form on 𝑙(𝑉𝑚 ). We say that the sequence ℰ𝑚 is a harmonic structure on 𝐾 if is satisfies: • (self-similarity) there exist numbers 0 < 𝑟1 , 𝑟2 , . . . , 𝑟𝑁 < 1 such that 𝑁

(13.8)

1 ℰ𝑚−1 (𝑢 ∘ 𝑓𝑗 , 𝑣 ∘ 𝑓𝑗 ) 𝑟 𝑗=1 𝑗

ℰ𝑚 (𝑢, 𝑣) = ∑

for all 𝑢, 𝑣 ∈ (𝑉𝑚 ) and 𝑚 ≥ 1; and • (compatibility) for each 𝑚 ≥ 1 and 𝑣 ∈ 𝑙(𝑉𝑚−1 ), (13.9)

ℰ𝑚−1 (𝑣) = min{ℰ𝑚 (𝑢) ∶ 𝑢 ∈ 𝑙(𝑉𝑚 ), 𝑢|𝑉𝑚−1 = 𝑣}.

We call each Dirichlet form ℰ𝑚 on 𝑙(𝑉𝑚 ) the energy on 𝑉𝑚 . For each 𝑢 ∈ 𝑙(𝑉𝑚 ) and 𝑗 = 1, 2, . . . , 𝑁, 𝑢 ∘ 𝑓𝑗 defines a function on 𝑉𝑚−1 , so selfsimilarity of the harmonic structure ℰ𝑚 means that the energy on each level can be distributed over each cell 𝐾𝑗 , weighted by the 1/𝑟𝑗 (although they are not really weights). We can—inductively—verify that the selfsimilarity property (13.8) implies that (13.10)

ℰ𝑚 (𝑢, 𝑣) = ∑ 𝑤∈𝑊𝑚

1 ℰ (𝑢 ∘ 𝑓𝑤 , 𝑣 ∘ 𝑓𝑤 ), 𝑟𝑤 0

for any 𝑢, 𝑣 ∈ 𝑙(𝑉𝑚 ), where 𝑟𝑤 = 𝑟𝑤1 𝑟𝑤2 ⋯ 𝑟𝑤𝑚 . Hence, the harmonic structure is determined by the initial energy ℰ0 on 𝑉0 and the numbers 𝑟1 , 𝑟2 , . . . , 𝑟𝑁 . The compatibility property (13.9) means that the energy is preserved between levels. Inductively, using and (13.10) we can see that, given 𝑣 ∈ 𝑙(𝑉0 ), (13.11)

ℰ0 (𝑣) = min{ℰ𝑚 (𝑢) ∶ 𝑢 ∈ 𝑙(𝑉𝑚 ), 𝑢|𝑉0 = 𝑣}

for every 𝑚 ≥ 1. We can then prove by induction that, given ℰ0 and 𝑟1 , 𝑟2 , . . . , 𝑟𝑁 , if ℰ𝑚 is given by (13.10) and (13.11) is satisfied with 𝑚 = 1, then (13.11) is satisfied for every 𝑚 ≥ 1 (Exercise (4)). If 𝑢 ∈ 𝑙(𝑉𝑚 ) satifies (13.11), we say that 𝑢 is harmonic on 𝑉𝑚 . A harmonic function is a function 𝑢 on 𝑉∗ such that each 𝑢|𝑉𝑚 is harmonic on 𝑉𝑚 , for every 𝑚.

248

13. Harmonic functions on post-critically finite sets

Example 13.12. Consider the interval [0, 1] with contractions as above. The vertices 𝑉𝑚 of level 𝑚 correspond to the dyadic partition 1 2 2𝑚 − 1 , , . . . , , 1}. 2𝑚 2𝑚 2𝑚 The sequence of energies defined in (11.3), 𝑉𝑚 = {0,

2𝑚

ℰ𝑚 (𝑢) = 2

𝑚

∑ (𝑢( 𝑘=1

𝑘−1 𝑘 2 ) − 𝑢( 𝑚 )) , 2𝑚 2

is a harmonic structure on 𝐼, as discussed in Section 11.1. Harmonic functions are restrictions to 𝑉∗ of linear functions. Example 13.13. Similarly, the sequence of energies on the vertices of the Sierpiński gasket 𝑆 discussed in Section 11.2, 2 5 𝑚 ∑ (𝑢(𝑥) − 𝑢(𝑦)) , ℰ𝑚 (𝑢) = ( ) 3 𝑥,𝑦∈𝑉 𝑥∼𝑦

𝑚

is a harmonic structure on 𝑆. Harmonic functions can be constructed by interpolation, as discussed in that section. Example 13.14. Consider now the Hata tree set 𝐾 described in Example 13.6. We have that 𝑉0 = {𝑐, 0, 1}. For ℎ > 0, we consider the Dirichlet form on 𝑙(𝑉0 ) given by (13.15)

ℰ0 (𝑢) = (𝑢(0) − 𝑢(1))2 + ℎ(𝑢(0) − 𝑢(𝑐))2 .

ℰ0 is a Dirichlet form, as discussed in 13.7. For 𝑟1 , 𝑟2 > 0, we have, for 𝑢 ∈ 𝑙(𝑉1 ), ℰ1 (𝑢) =

1 ((𝑢(0) − 𝑢(𝑐))2 + ℎ(𝑢(0) − 𝑢(1/3))2 ) 𝑟1 1 + ((𝑢(1/3) − 𝑢(1))2 + ℎ(𝑢(1/3) − 𝑢(𝑑))2 ), 𝑟2

where 𝑑 = 𝑓2 (𝑐) (see Figure 13.1). Given 𝑢(𝑐) = 𝛼, 𝑢(0) = 𝛽 and 𝑢(1) = 𝛾, we want to find 𝑢(1/3) = 𝑥̄ and 𝑢(𝑑) = 𝑦 ̄ such that the quadratic function 1 1 𝑓(𝑥, 𝑦) = ((𝛼 − 𝛽)2 + ℎ(𝛽 − 𝑥)2 ) + ((𝑥 − 𝛾)2 + ℎ(𝑥 − 𝑦)2 ) 𝑟1 𝑟2 ̄ and 𝑟1 , 𝑟2 such that 0 < 𝑟1 , 𝑟2 < 1 and takes its minimum at (𝑥,̄ 𝑦), 𝑓(𝑥,̄ 𝑦)̄ = ℎ(𝛼 − 𝛽)2 + (𝛽 − 𝛾)2 .

13.2. Harmonic structures and discrete energy

249

c

1/ 3

0

1 d

Figure 13.1. The set 𝑉 1 of vertices of level 1 of the Hata tree set.

By elementary calculus 𝑓 takes its minimum at 𝑥̄ = 𝑦 ̄ =

(13.16)

ℎ𝑟2 𝛽 + 𝑟1 𝛾 ℎ𝑟2 + 𝑟1

and 𝑓(𝑥,̄ 𝑦)̄ =

1 ℎ (𝛼 − 𝛽)2 + (𝛽 − 𝛾)2 . 𝑟1 ℎ𝑟2 + 𝑟1

Thus 1 =ℎ 𝑟1

and

ℎ = 1, ℎ𝑟2 + 𝑟1

so 1 1 and 𝑟2 = 1 − 2 = 1 − 𝑟12 . ℎ ℎ Note that 0 < 𝑟1 , 𝑟2 < 1 if and only if ℎ > 1. Thus, we have a family of harmonic structures on the Hata set, one for each number ℎ > 1. (13.17)

𝑟1 =

By (13.16), a harmonic function 𝑢 on 𝑉1 satisfies 1 1 )𝑢(0) + 2 𝑢(1). 2 ℎ ℎ Note that (13.18) is a convex combination of 𝑢(0) and 𝑢(1), and that it does not depend on the value of 𝑢(𝑐). By the observations above, (13.18) provides an algorithm to construct a harmonic function on 𝑉∗ . Figure 13.2 shows harmonic functions with ℎ = 2 and boundary values 𝑢 = 𝜒{1} and 𝑢 = 𝜒{𝑐} . (13.18)

𝑢(1/3) = 𝑢(𝑑) = (1 −

250

13. Harmonic functions on post-critically finite sets

Figure 13.2. Harmonic functions on the Hata tree set with ℎ = 2 and boundary values 𝑢(𝑐) = 𝑢(0) = 0 and 𝑢(1) = 1 on the left, and 𝑢(𝑐) = 1 and 𝑢(0) = 𝑢(1) = 0 on the right.

13.3. Discrete Laplacians Let 𝑉 be a finte set and 𝐻 a symmetric operator on 𝑙(𝑉). 𝐻 is a Laplacian on 𝑉 if (a) 𝐻 is nonpositive definite; (b) 𝐻𝑢 = 0 if and only if 𝑢 is a constant; and (c) 𝐻𝑥𝑦 ≥ 0 for all 𝑥, 𝑦 ∈ 𝑉, 𝑥 ≠ 𝑦. Let ℰ be the bilinear form induced by 𝐻, so ℰ(𝑢, 𝑣) = −⟨𝐻𝑢, 𝑣⟩. It is clear that ℰ(𝑢) ≥ 0 for all 𝑢 if and only if 𝐻 is nonpositive definite, by definition. By the spectral theorem, ℰ(𝑢) = 0 if and only if 𝑢 is a constant is equivalent to saying that 𝐻𝑢 = 0 if and only if 𝑢 is a constant (Exercise (6)). We have Proposition 13.19. Proposition 13.19. The bilinear form ℰ induced by 𝐻 is a Dirichlet form if and only if 𝐻 is a Laplacian on 𝑉.

13.3. Discrete Laplacians

251

Proof. We have seen that (1) and (2) of the definition of a Dirichlet form are equivalent to (a) and (b) of the definition of a Laplacian. Now, ℰ(𝑢, 𝑣) = −⟨𝐻𝑢, 𝑣⟩ = − ∑ 𝐻𝑢(𝑥)𝑣(𝑥) 𝑥∈𝑉

= − ∑ 𝐻𝑥𝑦 𝑢(𝑦)𝑣(𝑥) 𝑥,𝑦∈𝑉

=

1 ∑ 𝐻 (𝑢(𝑥) − 𝑢(𝑦))(𝑣(𝑥) − 𝑣(𝑦)), 2 𝑥,𝑦∈𝑉 𝑥𝑦

using the fact that ∑ 𝐻𝑥𝑦 = 0 𝑦∈𝑉

for each 𝑥, because 𝐻 is zero on a constant. Hence, if 𝐻𝑥𝑦 ≥ 0 for all pairs 𝑥 ≠ 𝑦, ℰ is a Dirichlet form by 13.7. Suppose some 𝐻𝑥𝑦 < 0. By multiplying by a positive constant, we can assume 𝐻𝑥𝑦 = −1. Let 𝑡 < 0 and define 𝑢 on 𝑉 by setting 𝑢(𝑥) = 1, 𝑢(𝑦) = 𝑡 and 𝑢(𝑧) = 0 for all 𝑧 ∈ 𝑉, 𝑧 ≠ 𝑥, 𝑦. Then ℰ(𝑢) =

1 ∑ 𝐻 (𝑢(𝑧) − 𝑢(𝑤))2 = −(1 − 𝑡)2 + 𝐴𝑡2 + 𝐵, 2 𝑧,𝑤∈𝑉 𝑧𝑤

for some 𝐴, 𝐵 ≥ 0. Since 𝑡 < 0, the cut 𝑢̄ of 𝑢 by [0, 1] is 𝑢(𝑥) = 1 and 𝑢(𝑧) = 0 for all 𝑧 ≠ 𝑥, so ℰ(𝑢)̄ = −1 + 𝐵. Thus ℰ(𝑢) − ℰ(𝑢)̄ = 2𝑡 + (𝐴 − 1)𝑡2 < 0 if 𝐴 ≤ 1 or 0>𝑡>−

2 , 𝐴−1

so ℰ doesn’t satisfy (3) of the definition of Dirichlet form.

□

Hence, if ℰ𝑚 is a harmonic structure on the PCF set 𝐾, then we have a sequence of Laplacians 𝐻𝑚 on 𝑉𝑚 , where each 𝐻𝑚 induces ℰ𝑚 . The self-similarity condition (13.8) gives a relation between each 𝐻𝑚−1 and 𝐻𝑚 , which then defines an extension of 𝐻𝑚−1 on 𝑉𝑚−1 to 𝑉𝑚 . We leave this as an exercise (Exercise (9)). To write the compatiblity condition (13.9) in terms of Laplacians, we make the following observations. Let 𝑉 a finite set and 𝑈 ⊂ 𝑉 a proper

252

13. Harmonic functions on post-critically finite sets

subset. If 𝐻 is a Laplacian on 𝑉, we write the matrix of 𝐻, which we also denote by 𝐻, in blocks 𝑇 𝐻=( 𝐽

(13.20)

𝐽𝑡 ), 𝑋

where 𝑇 ∶ 𝑙(𝑈) → 𝑙(𝑈), 𝐽 ∶ 𝑙(𝑈) → 𝑙(𝑉 ⧵𝑈) and 𝑋 ∶ 𝑙(𝑉 ⧵𝑈) → 𝑙(𝑉 ⧵𝑈). Hence, if we write 𝑢 ∈ 𝑙(𝑉) as 𝑢 𝑢 = ( 0) , 𝑢1 where 𝑢0 = 𝑢|𝑈 and 𝑢1 = 𝑢|𝑉 ⧵𝑈 , then (13.21)

𝐻𝑢 = (

𝑇𝑢0 + 𝐽 𝑡 𝑢1 ). 𝐽𝑢0 + 𝑋𝑢1

𝑋 is invertible. Indeed, for 𝑣 ∈ 𝑙(𝑉 ⧵ 𝑈), consider its extension 𝑢 ∈ 𝑙(𝑉) with 𝑢|𝑈 = 0 and 𝑢|𝑉 ⧵𝑈 = 𝑣. Then, if ℰ is the Dirichlet form induced by 𝐻, by (13.21) we have ℰ(𝑢) = −⟨𝐻𝑢, 𝑢⟩ = −⟨ (

𝐽𝑡𝑣 0 ) , ( ) ⟩ = −⟨𝑋𝑣, 𝑣⟩. 𝑋𝑣 𝑣

Thus 𝑋𝑣 = 0 implies ℰ(𝑢) = 0, so 𝑢 must be constant and therefore 𝑣 = 0. Note that this also implies that 𝑋 is negative definite. We can thus write ℰ(𝑢) = −⟨𝑇𝑢0 + 𝐽 𝑡 𝑢1 , 𝑢0 ⟩ − ⟨𝐽𝑢0 + 𝑋𝑢1 , 𝑢1 ⟩ = ⟨(𝑇 − 𝐽 𝑡 𝑋 −1 𝐽)𝑢0 , 𝑢0 ⟩ − ⟨𝑋(𝑢1 + 𝑋 −1 𝐽𝑢0 ), 𝑢1 + 𝑋 −1 𝐽𝑢0 ⟩ Given 𝑢0 on 𝑈, we see that ℰ(𝑢) is minimal when ⟨𝑋(𝑢1 + 𝑋 −1 𝐽𝑢0 ), 𝑢1 + 𝑋 −1 𝐽𝑢0 ⟩ = 0, because 𝑋 is negative definite. Thus we require (13.22)

𝑢1 + 𝑋 −1 𝐽𝑢0 = 0

and ℰ(𝑢) = ℰ′ (𝑢0 ), where ℰ′ is the quadratic form induced by 𝑇 − 𝐽 𝑡 𝑋 −1 𝐽. Equations (13.21) and (13.22) imply that a minimizer function 𝑢 satifies (13.23)

𝐻𝑢|𝑉 ⧵𝑈 = 0.

13.3. Discrete Laplacians

253

13.24. If 𝑢 ∈ 𝑙(𝑉) satisfies (13.23), then, for all 𝑥 ∈ 𝑉, min 𝑢(𝑦) ≤ 𝑢(𝑥) ≤ max 𝑢(𝑦). 𝑦∈𝑈

𝑦∈𝑈

If we have any of the above equalities for some 𝑥 ∈ 𝑉 ⧵ 𝑈, then 𝑢 is constant. Thus, a minimizer function satisfies the maximum principle. Indeed, for any 𝑥 ∈ 𝑉, 𝐻𝑢(𝑥) = ∑ 𝐻𝑥𝑦 𝑢(𝑦) = ∑ 𝐻𝑥𝑦 (𝑢(𝑦) − 𝑢(𝑥)), 𝑦∈𝑉

𝑦∈𝑉

because ∑𝑦∈𝑉 𝐻𝑥𝑦 = 0 for all 𝑥. Thus, if 𝑥 ∈ 𝑉 ⧵ 𝑈, 𝐻𝑢(𝑥) = 0 and thus (13.25)

∑ 𝐻𝑥𝑦 (𝑢(𝑦) − 𝑢(𝑥)) = 0. 𝑦∈𝑉

If 𝑢 takes its maximum at 𝑥 ∈ 𝑉 ⧵ 𝑈, (13.25) implies that 𝑢(𝑦) = 𝑢(𝑥) whenever 𝐻𝑥𝑦 > 0, because each 𝐻𝑥𝑦 ≥ 0 and 𝑢(𝑦) − 𝑢(𝑥) ≤ 0, so 𝑢 = 𝑢(𝑥) on the set of points 𝑦 such that 𝐻𝑥𝑦 > 0. By 13.7, for any 𝑦 ∈ 𝑉 there exists a sequence 𝑦0 = 𝑥, 𝑦1 , 𝑦2 , . . ., 𝑦𝑛 = 𝑦 such that 𝐻𝑦𝑖−1 𝑦𝑖 > 0, so recursively the above argument shows that 𝑢(𝑦) = 𝑢(𝑥), and 𝑢 is a constant. If we apply the previous analysis with 𝑉 = 𝑉𝑚 and 𝑈 = 𝑉𝑚−1 , we see that the compatibility condition (13.9) is equivalent to (13.26)

𝐻𝑚−1 = 𝑇 − 𝐽 𝑡 𝑋 −1 𝐽,

where 𝑇, 𝐽 and 𝑋 are the blocks of 𝐻𝑚 as in (13.20). Note that the minimizer function 𝑢 ∈ 𝑙(𝑉𝑚 ), for a given 𝑣 ∈ 𝑙(𝑉𝑚−1 ), is given by extending (13.27)

𝑢|𝑉𝑚 ⧵𝑉𝑚−1 = −𝑋 −1 𝐽𝑣.

By (13.21), 𝑢 is a minimizer for a given 𝑣 ∈ 𝑙(𝑉𝑚−1 ) if (13.28)

𝐻𝑚 𝑢|𝑉𝑚 ⧵𝑉𝑚−1 = 0.

By (13.11) and (13.28), 𝑢 is harmonic if (13.29)

𝐻𝑚 (𝑢|𝑉𝑚 )(𝑥) = 0

for all 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 . Since (13.29) is a linear equation, the set of harmonic functions is a vector space. If #𝑉0 = 𝑀, this vector space is 𝑀-dimensional. Moreover, (13.27) provides an algorithm to construct a harmonic function by interpolation to each level 𝑚 from level 𝑚 − 1.

254

13. Harmonic functions on post-critically finite sets

13.30. As a consequence of (13.28) and 13.24, we have that harmonic functions satisfy the maximum principle: if 𝑢 is harmonic, then min 𝑢 ≤ 𝑢(𝑥) ≤ max 𝑢 𝑉0

𝑉0

for all 𝑥 ∈ 𝑉∗ , and we have any of the equalities for some 𝑥 ∈ 𝑉∗ ⧵ 𝑉0 only if 𝑢 is a constant. 13.31. If 𝑢 is harmonic, then it is uniformly continuous in 𝑉∗ , so it has a unique continuous extension to 𝐾. To see this, choose first 𝑚 large enough so that, if 𝑥 ∼𝑚 𝑦, at most one of them belong to 𝑉0 . If 𝑢 is harmonic and not a constant, then (13.32) max{|𝑢(𝑥) − 𝑢(𝑦)| ∶ 𝑥 ∼𝑚 𝑦} < max{|𝑣(𝑥) − 𝑣(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉0 }, where 𝑣 = 𝑢|𝑉0 is the boundary value of 𝑢. By iterating (13.27), for each 𝑤 ∈ 𝑊𝑚 there exists an operator 𝑃𝑤 ∶ 𝑙(𝑉0 ) → 𝑙(𝑓𝑤 (𝑉0 )) such that 𝑢|𝑓𝑤 (𝑉0 ) = 𝑃𝑤 𝑣. Note that, if 𝑣(𝑥) ̃ = 𝑣(𝑥) −

1 ∑ 𝑣(𝑦), 𝑀 𝑦∈𝑉 0

then max{|𝑣(𝑥) − 𝑣(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉0 } = max{|𝑣(𝑥) ̃ − 𝑣(𝑦)| ̃ ∶ 𝑥, 𝑦 ∈ 𝑉0 }, and ∑𝑥∈𝑉 𝑣(𝑥) ̃ = 0. By Exercise (12) and (13.32), there exists 𝛾𝑤 < 1 0 such that max{|𝑃𝑤 𝑣(𝑥) − 𝑃𝑤 𝑣(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉0 } ≤ 𝛾𝑤 max{|𝑣(𝑥) − 𝑣(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉0 }, and thus max{|𝑢(𝑥) − 𝑢(𝑦)| ∶ 𝑥 ∼𝑚 𝑦} ≤ 𝛾 max{|𝑣(𝑥) − 𝑣(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉0 }, where 𝛾 = max{𝛾𝑤 ∶ 𝑤 ∈ 𝑊𝑚 } < 1. Thus, for any 𝑘 ≥ 1, if 𝑥 ∼𝑘𝑚 𝑦 then |𝑢(𝑥) − 𝑢(𝑦)| ≤ 𝐴𝛾𝑘 , which implies that 𝑢 is uniformly continuous because 𝐴𝛾𝑘 → 0. We say that 𝑢 ∈ 𝐶(𝐾) is harmonic if 𝑢|𝑉∗ is a harmonic function.

13.4. The Laplacian on a PCF set

255

13.4. The Laplacian on a PCF set A Laplacian on a PCF set can be constructed in the same way that we constructed a Laplacian on the Sierpiński gasket in Section 11.3. For a given 𝑢 ∈ 𝐶(𝐾), the sequence ℰ𝑚 (𝑢), where we are denoting the restriction of 𝑢 to 𝑉𝑚 simply by 𝑢, is increasing, so we define ℱ = {𝑢 ∈ 𝐶(𝐾) ∶ ℰ𝑚 (𝑢) is bounded} and define the bilinear form ℰ on ℱ by ℰ(𝑢, 𝑣) = lim ℰ𝑚 (𝑢, 𝑣). We also call ℰ(𝑢) the energy of 𝑢 on 𝐾. As in the case of the Sierpiński gasket, ℱ contains constant and harmonic functions, as well as 𝑚-harmonic functions, where a function 𝑢 ∈ 𝐶(𝐾) is called 𝑚-harmonic if 𝑢 ∘ 𝑓𝑤 is harmonic for each 𝑤 ∈ 𝑊𝑚 . Let 𝑑 > 0 be the unique number such that 𝑑 𝑟1𝑑 + 𝑟2𝑑 + . . . + 𝑟𝑁 = 1.

(13.33)

For each cell 𝐾𝑤 , we define 𝜇(𝐾𝑤 ) = 𝑟𝑤𝑑 . By (13.33) we have that ∑ 𝜇(𝐾𝑤 ) = 1 𝑤∈𝑊𝑚

and, as in the case of the Sierpiński gasket, if we define for each 𝐴 ⊂ 𝐾 𝜇(𝐴) = inf { ∑ 𝜇(𝑇𝑗 ) ∶ 𝑇𝑗 are cells and 𝐴 ⊂ 𝑗

⋃

𝑇𝑗 },

𝑗

the 𝜇 satisfies the analogous to Propositions 11.17 and 11.18 in Chapter 11. We thus obtain a self-similar measure on 𝐾, and hence an integral on 𝐾 with respect to the measure 𝜇. As above, we say that 𝑢 ∈ dom Δ if 𝑢 ∈ ℱ and there exists 𝑓 ∈ 𝐶(𝐾) such that, for all 𝑣 ∈ ℱ with 𝑣|𝑉0 = 0, (13.34)

ℰ(𝑢, 𝑣) = − ∫ 𝑓𝑣𝑑𝜇. 𝐾

We write 𝑓 = Δ𝑢, and we call it the Laplacian of 𝑢. If 𝑢 is harmonic, for each 𝑚 ≥ 1 we have 𝐻𝑚 𝑢(𝑥) = 0 for all 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 , and thus ℰ(𝑢, 𝑣) = ℰ0 (𝑢, 𝑣)

256

13. Harmonic functions on post-critically finite sets

for any 𝑣 ∈ ℱ. Hence, if further 𝑣|𝑉0 = 0, we have ℰ(𝑢, 𝑣) = 0 and therefore Δ𝑢(𝑥) = 0 for all 𝑥 ∈ 𝐾 ⧵ 𝑉0 . For 𝑥0 ∈ 𝑉∗ ⧵ 𝑉0 , let 𝑚 large enough so that 𝑥0 ∈ 𝑉𝑚 and let 𝜓𝑥0 ,𝑚 be the 𝑚-harmonic function such that 𝜓𝑥0 ,𝑚 |𝑉𝑚 = 𝜒{𝑥0 } . As in the case of the Sierpiński gasket, we call 𝜓𝑥0 ,𝑚 the 𝑚-harmonic spline on 𝑥0 . Write 𝜇𝑥0 ,𝑚 = ∫ 𝜓𝑥0 ,𝑚 𝑑𝜇. 𝐾

Thus, if 𝑢 ∈ dom Δ, for large 𝑚 we have ∫ Δ𝑢 𝜓𝑥0 ,𝑚 𝑑𝜇 ≈ Δ𝑢(𝑥0 )𝜇𝑥0 ,𝑚 , 𝐾

because Δ𝑢 is continuous. Also ℰ(𝑢, 𝜓𝑥0 ,𝑚 ) = ℰ𝑚 (𝑢, 𝜓𝑥0 ,𝑚 ) = − ∑ 𝐻𝑚 𝑢(𝑥) 𝜓𝑥0 ,𝑚 (𝑥) = −𝐻𝑚 𝑢(𝑥0 ), 𝑥∈𝑉𝑚

so we have that Δ𝑢(𝑥0 ) ≈

1 𝐻 (𝑥 ). 𝜇𝑥0 ,𝑚 𝑚 0

We can make this argument precise to prove Theorem 13.35, the generalization of Theorem 11.29 to PCF sets. Theorem 13.35. Let 𝑢 ∈ ℱ. 𝑢 ∈ dom Δ if and only if there exists 𝑓 ∈ 𝐶(𝐾) such that 1 lim max {|| 𝐻𝑚 (𝑥) − 𝑓(𝑥)|| ∶ 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉0 } = 0. 𝜇

𝑚→∞

𝑥,𝑚

In such case, 𝑓 = Δ𝑢. We leave the details as an exercise (Exercise (13)).

Exercises

257

Exercises (1) 𝑉∗ is dense in 𝐾. (2) Let 𝑆 be the Sierpiński gasket, 𝑚 ≥ 1, and 𝑥 ∈ 𝑉𝑚 ⧵ 𝑉𝑚−1 . Then there exists a unique 𝑤 ∈ 𝑊𝑚−1 and 𝑖, 𝑗 = 1, 2, 3 such that 𝑥 = 𝑓𝑤𝑖 (𝑝𝑗 ) = 𝑓𝑤𝑗 (𝑝 𝑖 ). (3) Let 𝑆 be the Sierpiński gasket, 𝑚 ≥ 1, 0 ≤ 𝑛 < 𝑚, and 𝑥 ∈ 𝑉𝑚 such that 𝑥 ∈ 𝑉𝑛+1 ⧵ 𝑉𝑛 . Then there exists a unique 𝑤 ∈ 𝑊𝑛 and 𝑖, 𝑗 = 1, 2, 3 such that 𝑥 = 𝑓𝑤𝑖𝑗. . .𝑗 (𝑝𝑗 ) = 𝑓𝑤𝑗𝑖. . .𝑖 (𝑝 𝑖 ), where the words 𝑤𝑖𝑗 . . . 𝑗 and 𝑤𝑗𝑖 . . . 𝑖 of length 𝑚 have 𝑚 − 1 − 𝑛 repeated 𝑗’s and 𝑖’s, respectively. (4) Let ℰ0 be a Dirichlet form on 𝑙(𝑉0 ) and 0 < 𝑟1 , 𝑟2 , . . . , 𝑟𝑁 < 1. If ℰ𝑚 is given by (13.10) and (13.11) is satisfied with 𝑚 = 1, then (13.11) is satisfied for every 𝑚 ≥ 1. (5) Calculate all the harmonic structures of the interval, self-similar with the contractions 𝑓1 (𝑥) = 𝑥/2 and 𝑓2 (𝑥) = 𝑥/2 + 1/2 on the real line. (6) Let 𝑉 be a finite set, 𝐻 a symmetric nonpositive definite operator on 𝑙(𝑉), and ℰ(𝑢, 𝑣) − ⟨𝐻𝑢, 𝑣⟩ the bilinear form induced by 𝐻. The following are equivalent. (a) ℰ(𝑢) = 0 if and only if 𝑢 is a constant. (b) 𝐻𝑢 = 0 if and only if 𝑢 is a constant. (Hint: Use the spectral theorem.) (7) Calculate the initial Laplacian 𝐻0 of the harmonic structure of the Sierpiński gasket. (8) Calculate the Laplacian 𝐻0 of the harmonic structure of the Hata tree set of Example 13.14. (9) For each 𝑗 = 1, 2, . . . , 𝑁, let 𝑅𝑗 ∶ 𝑙(𝑉𝑚 ) → 𝑙(𝑉𝑚−1 ) be given by 𝑅𝑗 𝑢 = 𝑢 ∘ 𝑓𝑗 . Then the self-similarity condition (13.8) is equivalent to 𝑁

1 𝑡 𝑅𝑗 𝐻𝑚−1 𝑅𝑗 . 𝑟 𝑗=1 𝑗

𝐻𝑚 = ∑

258

13. Harmonic functions on post-critically finite sets

(10) For each 𝑤 ∈ 𝑊𝑚 , let 𝑅𝑤 ∶ 𝑙(𝑉𝑚 ) → 𝑙(𝑉0 ) be given by 𝑅𝑤 𝑢 = 𝑢 ∘ 𝑓𝑤 . Then (13.10) is equivalent to 1 𝑡 𝐻𝑚 = ∑ 𝑅 𝑤 𝐻0 𝑅 𝑤 . 𝑟 𝑤 𝑤∈𝑊 𝑚

(11) Calculate the Laplacian 𝐻1 of the harmonic structure of the Hata tree set and verify (13.26) explicitly. (12) (a) Let 𝑉 a finite set and ℒ be the set of equivalence classes of 𝑙(𝑉) modulo constants, that is, under the equivalence relation 𝑢 ∼ 𝑣 if and only if 𝑢 − 𝑣 is a constant. Then ℒ is finite dimensional, ‖𝑢‖ = max{|𝑢(𝑥) − 𝑢(𝑦)| ∶ 𝑥, 𝑦 ∈ 𝑉} and is a norm on ℒ. (b) The set {𝑢 ∈ 𝑙(𝑉) ∶ ∑𝑥∈𝑉 𝑢(𝑥) = 0, ‖𝑢‖ ≤ 1} is compact. (13) Prove Theorem 13.35.

Notes The development of the Laplacian on post-critically finite sets was done by Kigami in [Kig93], where he proves that harmonic functions satisfy the linear equations involving discrete Laplacians, and hence form a finite dimensional space. The number 𝑑 defined by equation (13.33) is equal to the Hausdorff dimension of 𝐾 with respect to the effective resistence metric, which was developed in [Kig94]. If 𝐵𝜀 (𝑥) is a ball in 𝐾 with respect to this metric, then we can prove that 𝜇(𝐵𝜀 (𝑥)) ∼ 𝜀𝑑 for small 𝜀, where 𝜇 is the measure defined in this chapter, and thus 𝜇 satisfies the analogous result to the Hardy–Littlewood Theorem 7.4 on a PCF set [Sáe12]. Further results on the analysis on PCF sets can be found in the text [Kig01].

Appendix A

Some results from real analysis

In this appendix we review some basic results from real analysis. With a few exceptions that may be useful for the discussion in the text, we won’t include the proofs of the results listed here. All can be found in standard introductory texts as [Gau09] or [BS92] for the case of the real line, [Spi65] or [Fle77] for Euclidean space, and [Fol99] or [MH93] for metric and Banach spaces.

A.1. The real line The set of real numbers ℝ is a complete ordered field. It is unique up to isomorphism. With complete we mean that it satisfies the least upper bound axiom: Axiom (Completeness). If 𝐴 ⊂ ℝ is nonempty and bounded, then it has a least upper bound. A sequence 𝑥𝑛 in ℝ converges to 𝑥, and we write 𝑥𝑛 → 𝑥, if, given any 𝜀 > 0, there exists 𝑁 such that 𝑛 ≥ 𝑁 implies |𝑥𝑛 − 𝑥| < 𝜀. Any convergent sequence is a Cauchy sequence: given 𝜀 > 0, there exists 𝑁 such that 𝑛, 𝑚 ≥ 𝑁 implies |𝑥𝑛 − 𝑥𝑚 | < 𝜀. The converse is also true. Theorem A.1. A sequence in ℝ converges if and only if it is a Cauchy sequence. 259

260

A. Some results from real analysis

In fact, Theorem A.1 is equivalent to the completeness axiom. It is clear that convergent sequences are bounded. The converse is not true, but we have the following fact. Theorem A.2 (Bolzano–Weierstrass). If 𝑥𝑛 is bounded, then it has a convergent subsequence. It turns out that the Bolzano–Weierstrass theorem is also equivalent to the completeness axiom. A function 𝑓 ∶ 𝐴 → ℝ, where 𝐴 ⊂ ℝ, is continuous at 𝑥0 ∈ 𝐴 if given 𝜀 > 0, there exists 𝛿 > 0 such that |𝑥 − 𝑥0 | < 𝛿 implies |𝑓(𝑥) − 𝑓(𝑥0 )| < 𝜀, whenever 𝑥 ∈ 𝐴. 𝑓 is continuous at 𝑥0 if, and only if, for every sequence 𝑥𝑛 in 𝐴 that converges to 𝑥0 , then we have 𝑓(𝑥𝑛 ) → 𝑓(𝑥0 ). Sums and products of continuous functions are continuous. Quotients are also continuous, provided they are well defined. 𝑓 is continuous on 𝐴 if it is continuous at each point of 𝐴. 𝑓 is uniformly continuous on 𝐴 if, for each 𝜀 > 0, there exist 𝛿 > 0 such that |𝑥 − 𝑦| < 𝛿, for any 𝑥, 𝑦 ∈ 𝐴, implies |𝑓(𝑥) − 𝑓(𝑦)| < 𝜀. If 𝑓 is uniformly continuous and 𝑥𝑛 is a Cauchy sequence in 𝐴, then 𝑓(𝑥𝑛 ) is also a Cauchy sequence. This has the following consequence. A limit point of 𝐴 is a point 𝑥 ∈ ℝ such that there exists a sequence 𝑥𝑛 in 𝐴 such that 𝑥𝑛 → 𝑥. 𝑥 may or may not be in 𝐴. Theorem A.3. If 𝑓 is uniformly continuous on 𝐴 and 𝑥 is a limit point of 𝐴, then 𝑓 can extended to a continuous function on 𝐴 ∪ {𝑥}. Given a sequence 𝑥𝑛 in 𝐴 that converges to 𝑥, the extension of 𝑓 to 𝑥 is well-defined by the limit of 𝑓(𝑥𝑛 ), which converges because it is a Cauchy sequence. The union 𝐴̄ of 𝐴 and its limit points is called the closure of 𝐴. Thus, by Theorem A.3, if 𝑓 is uniformly continuous on 𝐴, then it can be extended continuously to its closure 𝐴.̄ Let 𝑓𝑛 ∶ 𝐴 → ℝ be a sequence of functions on 𝐴. We say that 𝑓𝑛 converges pointwise to 𝑓 ∶ 𝐴 → ℝ if, for each 𝑥 ∈ 𝐴, 𝑓𝑛 (𝑥) → 𝑓(𝑥). We usually just write to 𝑓𝑛 → 𝑓 to denote that 𝑓𝑛 converges pointwise to 𝑓. We say that 𝑓𝑛 converges uniformly to 𝑓 if, given 𝜀 > 0, there exists 𝑁 such that 𝑛 ≥ 𝑁 implies |𝑓𝑛 (𝑥) − 𝑓(𝑥)| < 𝜀 for all 𝑥 ∈ 𝐴. In this case we write 𝑓𝑛 ⇉ 𝑓. Theorem A.4. If each 𝑓𝑛 is continuous and 𝑓𝑛 ⇉ 𝑓, then 𝑓 is continuous.

A.2. Topology

261

A series ∑ 𝑎𝑛 converges to 𝑠 if the sequence 𝑠𝑛 of partial sums, 𝑠𝑛 = 𝑎1 + 𝑎2 + . . . + 𝑎𝑛 , converges to 𝑠. By Theorem A.1, if ∑ |𝑎𝑛 | converges, then ∑ 𝑎𝑛 converges. If ∑ |𝑎𝑛 | converges, we write ∑ |𝑎𝑛 | < ∞ and say that ∑ 𝑎𝑛 converges absolutely. Thus, any absolutely convergent series converges. Not every convergent series is absolutely convergent; we say that such a series converges conditionally. A test for conditional convergence is the following: Theorem A.5 (Dirichlet’s test). Let 𝑎𝑛 , 𝑏𝑛 be sequences in ℝ satisfying the following. (1) Each 𝑎𝑛 > 0, 𝑎𝑛+1 ≤ 𝑎𝑛 and 𝑎𝑛 → 0 (we write 𝑎𝑛 ↘ 0). (2) The sequence 𝑠𝑛 = 𝑏1 + 𝑏2 + . . . 𝑏𝑛 is bounded. Then the series ∑ 𝑎𝑛 𝑏𝑛 converges. A series ∑ 𝑓𝑛 of functions converges pointwise or uniformly if the corresponding sequence of partial sums 𝑠𝑛 (𝑥) = 𝑓1 (𝑥) + . . . + 𝑓𝑛 (𝑥) converges pointwise or uniformly, respectively. Thus, if the series ∑ 𝑓𝑛 converges uniformly to 𝑓 and each 𝑓𝑛 is continuous, then 𝑓 is also continuous. We have the following test for uniform convergence of a series. Theorem A.6 (Weierstrass 𝑀-test). Let 𝑓𝑛 be a sequence of functions on 𝐴 satisfying the following: (1) Each 𝑓𝑛 is bounded, with |𝑓𝑛 (𝑥)| ≤ 𝑀𝑛 for all 𝑥 ∈ 𝐴. (2) ∑ 𝑀𝑛 < ∞. Then the series ∑ 𝑓𝑛 converges uniformly.

A.2. Topology A set 𝐴 ⊂ ℝ is closed if it contains its limit points. Equivalently, 𝐴 is closed if 𝐴̄ = 𝐴. 𝐴 is open if, for each 𝑥 ∈ 𝐴, there exists 𝜀 > 0 such that (𝑥 − 𝜀, 𝑥 + 𝜀) ⊂ 𝐴. A set is open if, and only if, it’s complement is closed. Any union of open sets in ℝ is open, and any intersection of closed sets is closed. Finite intersections of open sets are open, while finite unions of closed sets are closed. A countable intersection of open sets is called a 𝐺 𝛿 set, and a countable union of closed set is called an 𝐹𝜍 set.

262

A. Some results from real analysis

A set 𝐵 ⊂ 𝐴 is said to be open in 𝐴 is there exists an open set 𝑈 in ℝ such that 𝐵 = 𝐴 ∩ 𝑈. Similarly, 𝐵 is closed in 𝐴 if there exists a closed set 𝐸 in ℝ such that 𝐵 = 𝐴 ∩ 𝐸. 𝐵 is open in 𝐴 if, and only if, 𝐴 ⧵ 𝐵 is closed in 𝐴. If 𝑓 ∶ 𝐴 → ℝ is a function, then 𝑓 is continuous on 𝐴 if, and only if, for any open set 𝑈 in ℝ its preimage 𝑓−1 (𝑈) is open in 𝐴, and equivalently, if for any closed set 𝐸 in ℝ its preimage 𝑓−1 (𝐸) is closed in 𝐴. By the Bolzano–Weierstrass theorem, if 𝐴 is closed and bounded, then any sequence in 𝐴 has a convergent subsequence, and its limit is also in 𝐴 (we say it converges in 𝐴). In fact, if any sequence in 𝐴 has a subsequence that converges in 𝐴, then 𝐴 must be closed and bounded. As a consequence, if 𝑓 is continuous on 𝐴 and 𝐴 is closed and bounded, then its image 𝐴 is also closed and bounded. An open cover for a set 𝐴 ⊂ ℝ is a collection {𝑈𝛼 } of open sets such that 𝐴 ⊂ ⋃𝛼 𝑈𝛼 . We say that the open cover {𝑈𝛼 } for 𝐴 has a finite subcover if we can choose a finite subcollection 𝑈𝛼1 , 𝑈𝛼2 , . . . , 𝑈𝛼𝑘 that also covers 𝐴. We have the following result. Theorem A.7 (Heine–Borel). Let 𝐴 ⊂ ℝ. The following are equivalent. (1) 𝐴 is closed and bounded. (2) Every open cover for 𝐴 has a finite subcover. A set that satisfies the statements of the Heine–Borel theorem is called compact. We also have the following results for continuous functions on compact sets. Theorem A.8. Let 𝐴 ⊂ ℝ be compact and 𝑓 ∶ 𝐴 → ℝ continuous on 𝐴. Then: (1) 𝑓 is uniformly continuous. (2) 𝑓(𝐴) is compact. In particular, if 𝐴 is compact and 𝑓 is continuous on 𝐴, then 𝑓 takes its maximum and its minimum value in 𝐴.

A.3. Riemann integration Let 𝑓 ∶ [𝑎, 𝑏] → ℝ be bounded and 𝒫 = {𝑥0 = 𝑎 < 𝑥1 < . . . < 𝑥𝑛 = 𝑏} a partition of [𝑎, 𝑏]. The lower and upper sums of 𝑓 with respect to 𝒫 are

A.3. Riemann integration

263

given by 𝑛

𝑛

𝐿(𝑓, 𝒫) = ∑ 𝑚𝑖 (𝑥𝑖 − 𝑥𝑖−1 ) and

𝑈(𝑓, 𝒫) = ∑ 𝑀𝑖 (𝑥𝑖 − 𝑥𝑖−1 ),

𝑖=1

𝑖=1

respectively, where 𝑚𝑖 = inf{𝑓(𝑥) ∶ 𝑥 ∈ [𝑥𝑖−1 , 𝑥𝑖 ]} and 𝑀𝑖 = sup{𝑓(𝑥) ∶ 𝑥 ∈ [𝑥𝑖−1 , 𝑥𝑖 ]}. We say that 𝑓 is Riemann-integrable on [𝑎, 𝑏] if, for any 𝜀 > 0, there exists a partition 𝒫 such that 𝑈(𝑓, 𝒫) − 𝐿(𝑓, 𝒫) < 𝜀. In such case, the unique number 𝐼 that satisfies 𝐿(𝑓, 𝒫) < 𝐼 < 𝑈(𝑓𝒫) for all partitions is called the integral of 𝑓, and is denoted by 𝑏

𝑏

∫ 𝑓(𝑥)𝑑𝑥, 𝑎

∫ 𝑓, or simply ∫ 𝑓. 𝑎

Linear combinations and products of Riemann-integrable functions are Riemann-integrable, and we have 𝑛

𝑛

∫ ∑ 𝑓𝑘 = ∑ ∫ 𝑓𝑘 . 𝑘=1

𝑘=1

Also, if 𝑓 ≤ 𝑔 on [𝑎, 𝑏], then ∫ 𝑓 ≤ ∫ 𝑔. If 𝑓 is Riemann-integrable, then so is |𝑓| and | ∫ 𝑓| ≤ ∫ |𝑓|. | | We also have that, for any 𝑐 ∈ (𝑎, 𝑏), 𝑏

𝑐

𝑏

∫ 𝑓 = ∫ 𝑓 + ∫ 𝑓. 𝑎

𝑎

𝑐

Continuous and monotone functions are Riemann-integrable, as well as piecewise continuous or piecewise monotone functions. A set 𝐴 ⊂ ℝ is of measure zero if for each 𝜀 > 0 there exists intervals {𝐼𝑛 } such that 𝐴 ⊂ ⋃𝑛 𝐼𝑛 and ∑ |𝐼𝑛 | < 𝜀, where |𝐼𝑛 | denotes the length ot 𝐼𝑛 . Finite and countable infinite sets are of measure zero. There exist also uncountable sets of measure zero, as the Cantor set (see Chapter 5). We have the following criterion for Riemann integrability. Theorem A.9. A bounded function 𝑓 ∶ [𝑎, 𝑏] → ℝ is Riemann-integrable if, and only if, the set where 𝑓 is not continuous is of measure zero.

264

A. Some results from real analysis

A Riemann-integrable function can be approximated by continuous functions, in the following sense. Theorem A.10. Let 𝑓 ∶ [𝑎, 𝑏] → ℝ be Riemann-integrable, and let 𝑀 such that |𝑓(𝑥)| ≤ 𝑀 for all 𝑥 ∈ [𝑎, 𝑏]. For any 𝜀 > 0, there exists a continuous function 𝑔 on [𝑎, 𝑏] such that |𝑔(𝑥)| ≤ 𝑀

and

∫ |𝑓 − 𝑔| < 𝜀.

We include the proof of this theorem here, as its ideas are useful for the discussion in the text (see Chapter 3). Proof. Given 𝜀 > 0, let 𝒫 = {𝑥0 = 𝑎 < 𝑥1 < . . . < 𝑥𝑛 = 𝑏} be a partition so that 𝑈(𝑓, 𝒫) − 𝐿(𝑓, 𝒫) < 𝜀/2. In particular, |𝑈(𝑓, 𝒫) − ∫ 𝑓| < 𝜀 . | | 2 Note that 𝑈(𝑓, 𝒫) is the integral of the function ℎ on [𝑎, 𝑏] given by ℎ(𝑥) = 𝑀𝑖 , if 𝑥 ∈ [𝑥𝑖−1 , 𝑥𝑖 ), which satisfies ℎ ≥ 𝑓. Hence |𝑈(𝑓, 𝒫) − ∫ 𝑓| = ∫ |ℎ − 𝑓|. | | Let 𝛿 > 0 so that

𝜀 . 4𝑀𝑛 Let 𝑔 be the continuous function on [𝑎, 𝑏] such that it is equal to 𝑀𝑖 on each interval [𝑥𝑖−1 + 𝛿, 𝑥𝑖 − 𝛿], linear from 𝑀𝑖−1 to 𝑀𝑖 on each [𝑥𝑖−1 − 𝛿, 𝑥𝑖−1 +𝛿], 𝑖 = 2, . . . , 𝑛, linear from 𝑓(𝑎) to 𝑀1 on [𝑥0 , 𝑥0 +𝛿], and linear from 𝑀𝑛 to 𝑓(𝑏) on [𝑥𝑛 −𝛿, 𝑥𝑛 ]. As each |𝑀𝑖 | ≤ 𝑀, we have that |𝑔| ≤ 𝑀. Also 𝛿
0, there exists a partition 𝒫 such that 𝑈(𝑓, 𝒫) − 𝐿(𝑓, 𝒫) < 𝜀. If 𝑓 is Riemann integrable, its integral is denoted by ∫ 𝑓(𝑥)𝑑𝑥, 𝑅

∫ 𝑓, or simply ∫ 𝑓. 𝑅

As in the case of the Riemann integral of single variable functions on an interval, linear combinations and products of Riemann-integrable func𝑛 𝑛 tions are Riemann-integrable, we have ∫ ∑𝑘=1 𝑓𝑘 = ∑𝑘=1 ∫ 𝑓𝑘 , and, if 𝑓 ≤ 𝑔 on 𝑅, then ∫ 𝑓 ≤ ∫ 𝑔. If 𝑓 is Riemann-integrable on 𝑅, then so is |𝑓| and | ∫ 𝑓| ≤ ∫ |𝑓|. We also have that, for any partition 𝒫 of 𝑅 with subrectangles, each 𝑓|𝑆 is Riemann-integrable for each 𝑆 ∈ 𝒫 and ∫ 𝑓 = ∑ ∫ 𝑓. 𝑅

𝑆∈𝒫 𝑆

As above, a set 𝐴 ⊂ ℝ𝑑 is of measure zero if for each 𝜀 > 0 there exist rectangles {𝑅𝑛 } such that 𝐴 ⊂ ⋃𝑛 𝑅𝑛 and ∑ vol(𝑅𝑛 ) < 𝜀. The analog

A.5. Complete metric spaces

267

of Theorem A.9 is true for multivariable functions: a bounded function 𝑓 ∶ 𝑅 → ℝ is Riemann-integrable if and only if the set where 𝑓 is not continuous is of measure zero. If we write 𝑅 = 𝑆 × 𝑇, where 𝑆 and 𝑇 are rectangles in ℝ𝑙 and ℝ𝑘 , respectively, with 𝑑 = 𝑙 + 𝑘, we have the identity, for a continuous 𝑓 ∶ 𝑅 → ℝ, (A.12)

∫ 𝑓 = ∫ ( ∫ 𝑓𝑥 (𝑦)𝑑𝑦)𝑑𝑥, 𝑅

𝑆

𝑇

where, for each 𝑥 ∈ 𝑆 and 𝑦 ∈ 𝑇, 𝑓𝑥 (𝑦) = 𝑓(𝑥, 𝑦). The assumption that 𝑓 is continuous is needed because 𝑓𝑥 might not be a Riemann-integrable function on 𝑇 for all 𝑥 ∈ 𝑆, given a Riemann-integrable function 𝑓 on 𝑅. See [Spi65, Theorem 3.10] for a precise version of (A.12) for general Riemann-integrable functions. If 𝑅 ⊂ ℝ𝑑 is a closed rectangle, 𝐴 ⊂ 𝑅 and 𝑓 ∶ 𝑅 → ℝ is Riemannintegrable, we define ∫ 𝑓 = ∫ 𝑓 ⋅ 𝜒𝐴 , 𝐴

𝑅

where 𝜒𝐴 is the characteristic function of the set 𝐴, provided 𝜒𝐴 is Riemann integrable, which occurs when the set of boundary points1 of 𝐴 is of measure zero.

A.5. Complete metric spaces The Euclidean space ℝ𝑑 with the Euclidean distance induced by the norm 𝑥 ↦ |𝑥| is an example of a metric space. A metric space is a set provided with a metric 𝑑 ∶ 𝑋 × 𝑋 → [0, ∞) that satisfies, for all 𝑥, 𝑦, 𝑧 ∈ 𝑋, (1) 𝑑(𝑥, 𝑦) = 0 if and only if 𝑥 = 𝑦, (2) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥), and (3) 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑧) + 𝑑(𝑧, 𝑦). A sequence 𝑥𝑛 in a metric space 𝑋 converges to 𝑥 if, for any 𝜀 > 0, there exists 𝑁 such that 𝑑(𝑥𝑛 , 𝑥) < 𝜀. A convergent sequence 𝑥𝑛 is clearly a Cauchy sequence: given 𝜀 > 0, there exists 𝑁 such that 𝑑(𝑥𝑛 , 𝑥𝑚 ) < 𝜀 for any 𝑚, 𝑛 ≥ 𝑁. 1

A point 𝑥 is in the boundary of 𝐴 if, for all 𝜀 > 0, 𝐴 ∩ 𝐵𝜀 (𝑥) ≠ ∅ and 𝐵𝜀 (𝑥) ⧵ 𝐴 ≠ ∅.

268

A. Some results from real analysis

In general, though, not every Cauchy sequence converges in a metric space. A complete metric space is a metric space in which every Cauchy sequence converges. By Theorem A.1, the Euclidean space is complete. If 𝑋 is a real or complex vector space, ‖ ⋅ ‖ ∶ 𝑋 → [0, ∞) is a norm if it satisfies, for any vectors 𝑥, 𝑦 ∈ 𝑋, (1) ‖𝑥‖ = 0 if and only if 𝑥 = 0, (2) ‖𝜆𝑥‖ = |𝜆|‖𝑥‖ for any scalar 𝜆, and (3) ‖𝑥 + 𝑦‖ ≤ ‖𝑥‖ + ‖𝑦‖. The properties above imply that, if ‖ ⋅ ‖ is a norm on 𝑋, then 𝑑(𝑥, 𝑦) = ‖𝑥 − 𝑦‖ is a metric, and thus 𝑋 is metric space. If 𝑋 is a complete normed vector space, we call it a Banach space. A series ∑ 𝑥𝑛 in the normed space 𝑋 converges to 𝑥 if the sequence 𝑛 of partial sums 𝑠𝑛 = ∑𝑘=1 𝑥𝑘 converges to 𝑥. Equivalently, 𝑛

|| ∑ 𝑥 − 𝑥|| → 0. 𝑘 || || 𝑘=1

A series converges absolutely if the series ∑ ‖𝑥𝑛 ‖ converges. If 𝑋 is a Banach space and the series ∑ 𝑥𝑛 converges absolutely, then it converges. It turns out that this fact is equivalent to the completeness of 𝑋, as stated in Theorem A.13 due to Banach. Theorem A.13. The normed vector space 𝑋 is a Banach space if and only if every absolutely convergent series converges in 𝑋. An inner product space in the vector space 𝑋 is a scalar form ⟨⋅, ⋅⟩ on 𝑋 × 𝑋 that satisfies (1) ⟨𝑥, 𝑥⟩ ≥ 0, and ⟨𝑥, 𝑥⟩ = 0 if and only if 𝑥 = 0, (2) ⟨𝑥, 𝑦⟩ = ⟨𝑦, 𝑥⟩ for any 𝑥, 𝑦 ∈ 𝑋, and (3) ⟨𝜆𝑥 + 𝜇𝑦, 𝑧⟩ = 𝜆⟨𝑥, 𝑧⟩ + 𝜇⟨𝑦, 𝑧⟩ for any scalars 𝜆, 𝜇 and 𝑥, 𝑦, 𝑧 ∈ 𝑋. Two vectors 𝑥, 𝑦 ∈ 𝑋 are called orthogonal, and we write 𝑥 ⟂ 𝑦, if ⟨𝑥, 𝑦⟩ = 0. If 𝑆 ⊂ 𝑋, 𝑥 is orthogonal to 𝑆, and we write 𝑥 ⟂ 𝑆, if

A.5. Complete metric spaces

269

𝑥 is orthogonal to every vector in 𝑆. The orthogonal complement of 𝑆, denoted as 𝑆⟂ , is given by all the orthogonal vectors to 𝑆. 𝑆 ⟂ is always a vector subspace of 𝑋. If 𝑋 is a real vector space, then an inner product is a positive symmetric bilinear form. An inner product induces the norm ‖𝑥‖ = √⟨𝑥, 𝑥⟩. The norm satisfies the parallelogram identity ‖𝑥 + 𝑦‖2 + ‖𝑥 − 𝑦‖2 = 2‖𝑥‖2 + 2‖𝑦‖2 , and the polarization identity, given by 1 ⟨𝑥, 𝑦⟩ = (‖𝑥 + 𝑦‖2 − ‖𝑥 − 𝑦‖2 ) 4 in the real case, and by 1 ⟨𝑥, 𝑦⟩ = (‖𝑥 + 𝑦‖2 − ‖𝑥 − 𝑦‖2 + 𝑖‖𝑥 + 𝑖𝑦‖2 − 𝑖‖𝑥 − 𝑖𝑦‖2 ) 4 in the complex case. If 𝑋 is a complete inner product space, then it is called a Hilbert space. Note that a Hilbert space is a Banach space whose norm is induce by an inner product. If 𝑋 is a Hilbert space and 𝑌 is a closed vector subspace, then, for each 𝑥 ∈ 𝑋, there is 𝑦0 ∈ 𝑌 closest to 𝑥, that is, ‖𝑦0 − 𝑥‖ ≤ ‖𝑦 − 𝑥‖ for all 𝑦 ∈ 𝑌 . 𝑦0 is called the orthogonal projection of 𝑥 in 𝑌 , and satisfies 𝑥 −𝑦0 ⟂ 𝑌 . As a consequence, if 𝑌 is a proper closed subspace of 𝑋, then 𝑌 ⟂ is nontrivial.

Acknowledgments

I would like to thank all the students who have taken this course, both in IAS/PCMI and Colima, who provided the motivation to work on this text. Their questions and commentary to the manuscript were very valuable during its writing. I would also like to thank the reviewers of the first version of the manuscript, as their observations lead to several improvements to this book. I finally want to thank Eriko Hironaka, Ina Mette, Marcia C. Almeida, John F. Brady Jr., and Abigail Lawson at the AMS, who took care of all the logistics in the publication of this book.

271

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Index

𝐹𝜍 set, 261 𝐺 𝛿 set, 261 Abel means, 43 summable, 43, 55 theorem, 43 adyacent vertex, 209, 245 almost everywhere, 98 average, 54, 137, 143 Banach space, 126, 268 theorem, 268 Bernstein’s theorem, 69 Bessel’s inequality, 51 better kernels, 148, 158, 165, 173 Bolzano–Weierstrass theorem, 260 boundary, 210, 244 Cantor set, 188 Cauchy sequence, 259, 267 cell, 192, 243 Cesàro summable, 55 summation, 54 sums, 54 Chebyshev’s inequality, 140 closed set, 261 closure, 260 compact, 262

complete, 259, 268 connected set, 266 continuous, 260 uniformly, 260 contraction, 191 contraction constant, 191 convergence absolute, 46, 48, 261 conditional, 261 in measure, 179 mean-square, 59, 164, 168 nontangential, 146 pointwise, 260 radial, 29 uniform, 260 convolution, 128 cover 𝛿-cover, 185 critical set, 244 cube, 85 dyadic, 85 strictly thinner, 88 diameter, 185 Dirichlet eigenfunction, 231, 233 eigenvalue, 231 form, 246 principle, 11 problem, 10, 27

277

278 test, 261 discrete energy, 204, 210 discrete Laplacian, 207, 217 domain, 6 dominated convergence theorem, 118 dyadic decomposition, 51 eigenfunction, 223 eigenvalue, 223 energy form, 9, 212, 216, 247, 255 minimizers, 9 extended real line, 96 Fatou’s lemma, 111 Fejér kernel, 58 theorem, 55 flounces function, 42 Fourier coefficients, 39 inversion formula, 157 series, 40 transform, 151 Fubini’s theorem, 123

Index Hilbert space, 269 transform, 171, 173 Hopf lemma, 33 integrable function, 113, 115 locally integrable, 138 Riemann-integrable, 263, 266 square integrable, 130 integral complex valued function, 115 extended real valued function, 113 improper, 74 nonnegative function, 108 nonnegative simple function, 105 on the Sierpiński gasket, 214 Riemann, 263, 266 self-similarity, 215 symmetry, 215 isometry, 160, 172

Gaussian kernel, 154, 158 golden fractal, 199 good kernels, 27, 58, 79, 81

Laplace equation, 3 Laplacian, 3, 217, 250, 255 polar coordinates, 11, 35 Lebesgue differentiation theorem, 143 measure, 94 space, 124, 130 Lipschitz function, 53

Hölder continuous, 53, 69 Hardy-Littlewood maximal function, 138 theorem, 139 harmonic 𝑚-harmonic function, 212, 217 conjugate, 4, 167 function, 3, 13, 210, 212, 217, 247, 254 harmonic structure, 210, 247 compatibility, 247 self-similarity, 247 Harnack inequality, 33 Hata tree set, 197, 244, 248 Hausdorff dimension, 185, 188 measure, 186 heat kernel, 81, 154 Heine–Borel theorem, 262

maximal function, 138 uncentered, 142 maximum princple, 20, 253, 254 measurable 𝐹𝜍 set, 94 𝐺 𝛿 set, 94 closed cube, 92 closed set, 93, 214 complement, 89, 214 countable intersection, 92 countable union, 90, 214 function, 96 open set, 93, 214 pointwise limit, 97 set, 89, 214 union, 90 measure countable additivity, 94, 214 countable subadditivity, 87

Index exterior, 86 invariance, 96 Lebesgue, 94 monotone continuity, 94 monotonicity, 87 on the Sierpiński gasket, 213 outer, 85, 186 regularity, 95 zero, 86, 263, 266 metric space, 267 Minkowski’s inequality, 124, 132 monotone convergence theorem, 109 negative part, 102 neighbor, 209, 245 nontangential limit, 146 norm, 268 open cover, 262 open set, 261 open set condition, 193 orthogonal transformation, 31 orthogonality, 49, 59, 268 Parseval’s identity, 60 PCF set, 244 Plancherel’s theorem, 162 Poisson integral, 27, 75, 164, 167 kernel, 24, 71, 154, 158, 167 polarization identity, 160 positive part, 102 post-critical set, 244 post-critically finite set, 244 punctured ball, 30 Riemann–Lebesgue lemma, 53, 152 sawtooth function, 40, 46 section of a function, 123 of a set, 120 self-similar, 191, 243 sharkteeth function, 41, 46 Sierpiński gasket, 196, 207, 244 similitude, 193 simple function, 99 approximation, 100 reduced form, 100

279 singularity isolated, 29 removable, 30 snowflake set, 197 spherical coordinates, 7 spline, 213, 256 integral, 216 step function, 100 subordination, 81 symmetry lemma, 25, 32 Tonelli’s theorem, 124 translation, 31 trigonometric polynomial, 37, 58 unitary operator, 162 upper half-space, 5, 71 Vitalli’s covering lemma, 140 volume, 85 weak type, 139, 174, 182 Weierstrass 𝑀-test, 261 word, 191, 208, 243

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