The E. M. Stein Lectures on Hardy Spaces 3031219511, 9783031219511

Table of contents :
Preface
Contents
1 Introductory Material
1.1 Various Maximal Functions
1.2 Nontangential Convergence
1.3 Unrestricted Convergence
1.4 The Area Integral
1.5 Generalizations of R+ and Hp(R+)
1.6 Relationships Among Domains
2 More on Hardy Spaces
2.1 Hardy Spaces and Maximal Functions
2.2 More Maximal Functions
2.3 Real Variable Hp
2.4 Some Thoughts on Summability
3 Background on Hp Spaces
3.1 Where Did Hp Spaces Get Started?
3.2 Hardy Spaces in C1
3.3 The Hardy–Littlewood Maximal Function
3.4 The Poisson Kernel and Fourier Inversion
4 Hardy Spaces on D
4.1 The Role of the Hilbert Transform
4.2 Blaschke Products
4.3 Passage from D to R2+
5 Hardy Spaces on Rn
5.1 The Poisson Kernel on the Ball
5.2 The Poisson Kernel on the Upper Halfspace
5.3 Cauchy–Riemann Systems
5.4 A Characterization of Hp, p > 1
5.5 The Area Integral
5.6 Applications of the Maximal Function Characterization
5.7 H1(Rn) and Duality with Respect to BMO
6 Developments Since 1974
6.1 The Atomic Theory
6.2 The Local Theory of Hardy Spaces
6.3 The Work of Chang/Krantz/Stein on Hardy Spaces for Elliptic Boundary Value Problems
6.4 Multi-Parameter Harmonic Analysis
6.5 The T1 Theorem of David/Journé
6.6 Contributions of Tom Wolff
6.7 Wavelets
6.7.1 Localization in the Time and Space Variables
6.7.2 Building a Custom Fourier Analysis
6.7.3 The Haar Basis
Axioms for a Multi-Resolution Analysis (MRA)
7 Concluding Remarks
References
Index

Citation preview

Lecture Notes in Mathematics  2326

Steven G. Krantz

The E. M. Stein Lectures on Hardy Spaces

Lecture Notes in Mathematics Volume 2326

Editor-in-Chief Jean-Michel Morel, CMLA, ENS, Cachan, France Bernard Teissier, IMJ-PRG, Paris, France Series Editors Karin Baur, University of Leeds, Leeds, UK Michel Brion, UGA, Grenoble, France Annette Huber, Albert Ludwig University, Freiburg, Germany Davar Khoshnevisan, The University of Utah, Salt Lake City, UT, USA Ioannis Kontoyiannis, University of Cambridge, Cambridge, UK Angela Kunoth, University of Cologne, Cologne, Germany Ariane Mézard, IMJ-PRG, Paris, France Mark Podolskij, University of Luxembourg, Esch-sur-Alzette, Luxembourg Mark Policott, Mathematics Institute, University of Warwick, Coventry, UK Sylvia Serfaty, NYU Courant, New York, NY, USA László Székelyhidi , Institute of Mathematics, Leipzig University, Leipzig, Germany Gabriele Vezzosi, UniFI, Florence, Italy Anna Wienhard, Ruprecht Karl University, Heidelberg, Germany

This series reports on new developments in all areas of mathematics and their applications - quickly, informally and at a high level. Mathematical texts analysing new developments in modelling and numerical simulation are welcome. The type of material considered for publication includes: 1. Research monographs 2. Lectures on a new field or presentations of a new angle in a classical field 3. Summer schools and intensive courses on topics of current research. Texts which are out of print but still in demand may also be considered if they fall within these categories. The timeliness of a manuscript is sometimes more important than its form, which may be preliminary or tentative. Titles from this series are indexed by Scopus, Web of Science, Mathematical Reviews, and zbMATH.

Steven G. Krantz

The E. M. Stein Lectures on Hardy Spaces

Steven G. Krantz Department of Mathematics Washington University St. Louis, MO, USA

ISSN 0075-8434 ISSN 1617-9692 (electronic) Lecture Notes in Mathematics ISBN 978-3-031-21951-1 ISBN 978-3-031-21952-8 (eBook) https://doi.org/10.1007/978-3-031-21952-8 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To the memory of E. M. Stein, for his teaching and his friendship.

Preface

Elias M. Stein was one of the pre-eminent harmonic analysts of the twentieth century. He directed more than 50 Ph.D. students at Princeton University, and many of them are quite distinguished mathematicians in their own right. So his influence continues on into the twenty-first century. One of Stein’s seminal contributions to modern mathematical analysis is the real variable theory of Hardy spaces. Hardy spaces were first developed by F. Riesz and M. Riesz in the early part of the twentieth century. It was in fact F. Riesz who, in 1923, named these spaces in honor of G. H. Hardy. These were spaces of holomorphic functions on the unit disc in the complex plane. The Hardy spaces, or H p , are important because of their structural properties and also because of their behavior under certain important mappings. People long suspected that, lurking in the background, there is a real-variable theory of Hardy spaces—one that rejects the complex variable context in which they were originally formulated. A real variable theory would allow us to focus on the essential structure and mapping properties of these spaces. The pioneering work along these lines was done by E. M. Stein and Guido L. Weiss in 1960. There they realized that the right way to formulate a real-variable H p space was as the gradients of harmonic functions. A number of important calculations and insights appear in their Acta Mathematica paper. A 1971 paper by Burkholder et al. [BGS] provides a glimpse of what is possible in this direction. This paper depends decisively on probabilistic methods. The paper that broke the subject wide open was the 1972 paper of C. Fefferman and E. M. Stein. This paper is one of the most highly cited in modern mathematics. It inspired a flood of work in the 1970s and on into the 1980s, and continues to have a strong influence today. A major AMS Summer Workshop was held in Williamstown, Massachusetts in 1978 to celebrate and study this paper of Fefferman/Stein and its consequences. The present book is based on a year-long course that Stein taught at Princeton University in 1973–1974. He in fact taught the course at the request of Robert Fefferman and this author. It was a huge and protracted effort for him to produce this course on the spot, and the results were stunning. vii

viii

Preface

The course that Stein taught was wide-ranging, deep, and insightful. In the characteristic Stein manner, he not only stated theorems and proved them but also worked examples, formulated conjectures, and handed out open problems. Every lecture was an object lesson and a valuable commodity. This author wrote up very careful notes of the Stein course on Hardy spaces. In fact he wrote up notes while he was sitting in the course in Fine Hall. But then, a year or two later, he rewrote and polished the notes a second time. The book that we are presenting here is a formal development of those notes. The purpose now is to share with the mathematical world the perspective of Stein on this subject area that he invented. It is a glowing look back at one of the milestones of modern mathematics. It is a tribute to E. M. Stein. The content of E. M. Stein’s 1973–1974 lecture course, “Real Variable Hardy Spaces,” has been reproduced by Steven G. Krantz with permission from the Stein estate. Of course all errors and mis-steps contained herein are the responsibility of the author. We look forward to hearing from readers as the book is read and appreciated. St. Louis, Mo, USA

Steven G. Krantz

Contents

1 Introductory Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Various Maximal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Nontangential Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Unrestricted Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The Area Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Generalizations of R+ and H p (R+ ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Relationships Among Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 6 16 19 22 33

2

More on Hardy Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Hardy Spaces and Maximal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 More Maximal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Real Variable H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Some Thoughts on Summability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

35 35 43 51 59

3

Background on H p Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Where Did H p Spaces Get Started? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Hardy Spaces in C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 The Hardy–Littlewood Maximal Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Poisson Kernel and Fourier Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73 73 76 86 93

4

Hardy Spaces on D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 The Role of the Hilbert Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Blaschke Products. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Passage from D to R2+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

101 101 120 129

5

Hardy Spaces on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 The Poisson Kernel on the Ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Poisson Kernel on the Upper Halfspace . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Cauchy–Riemann Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 A Characterization of H p , p > 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Area Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Applications of the Maximal Function Characterization . . . . . . . . . . . . . . 5.7 H 1 (Rn ) and Duality with Respect to BMO . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 137 140 149 159 163 179 206 ix

x

6

7

Contents

Developments Since 1974. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 The Atomic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Local Theory of Hardy Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Work of Chang/Krantz/Stein on Hardy Spaces for Elliptic Boundary Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Multi-Parameter Harmonic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The T 1 Theorem of David/Journé . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Contributions of Tom Wolff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.1 Localization in the Time and Space Variables. . . . . . . . . . . . . . . . . 6.7.2 Building a Custom Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.3 The Haar Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

227 227 228 231 232 233 233 234 234 235 237

Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

Chapter 1

Introductory Material

1.1 Various Maximal Functions Stein began this first lecture by giving us three references: [RUD1, STW1, ZYG]. We begin by considering the space (R2+ )n ≡ R+ . Thus z ∈ R+ ⇒ z = x + iy , x ∈ Rn , y ∈ Rn . The notation y > 0 means yj > 0, j = 1, . . . , n. Notice that R+ ⊆ Cn canonically. Now R+ has two boundaries: (i) The topological boundary = {(x1 + iy1 , . . . , xn + iyn ) : yj = 0, some j } (ii) The distinguished boundary = {(x1 + iy1 , . . . , xn + iyn ) : yj = 0 for all j } The distinguished boundary is homeomorphic to Rn and is equal to the Šilov boundary. Definition 1.1.1 Let u(z1 , z2 , . . . , zn ) be a function of n variables, each zj ∈ C. Then u is multiply harmonic or n-harmonic if it is C ∞ and harmonic in each variable. Lemma 1.1.2 Suppose that u : R+ → C is multiply harmonic, bounded, and  ≡ 0. Then u ≡ 0. continuous on R+ . Furthermore, assume that u disting. bdry.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Krantz, The E. M. Stein Lectures on Hardy Spaces, Lecture Notes in Mathematics 2326, https://doi.org/10.1007/978-3-031-21952-8_1

1

2

1 Introductory Material

Proof Suppose without loss of generality that n = 2. Fix z2 ∈ R2+ and consider u( · , z2 ) on R2+ sup |u(z1 , z2 )| = sup |u(x1 , z2 )| x1 ∈R

z1

by the maximum modulus principle. But for each x1 , |u(x1 , · )| takes its maximum when z2 is on ∂R2+ , that is when Imz2 = 0. Therefore sup |u(z1 , z2 )| ≤

sup

|u(z1 , z2 )| = 0 .

(z1 ,z2 )∈dist. bdry. bR+



That completes the proof. Corollary 1.1.3 We have that sup |u(z)| = z

sup

|u(z)| .

z∈dist. bdry. bR+

Let us now define the Poisson kernel for R+ by Py (x) = Py1 (x1 ) · · · · · Pyn (xn ) , where here Pyj (xj ) ≡

yj 1 · 2 . π xj + yj2

Then P has the following properties:  (1) dist. bdry. R+ =Rn Py (x) dx = 1 , each y > 0. This is proved trivially from the classical case. (2) Let U 0 be any open set. Then  cU

Py (x), dx → 0 as y → 0 .

This is proved from the classical case or by inspection. (3) If f ∈ L∞ (Rn ) and u(x, y) ≡ Py ∗ f , then u is n-harmonic. Lemma 1.1.4 If f is continuous and bounded on the distinguished boundary of R+ and if u(x, y) = (multiple)P I f ,

1.1 Various Maximal Functions

3

then (a) u is multiply harmonic. (b) u is bounded and uL∞ ≤ f L∞ . (c) uis continuous on R+ .  (d) u = f. dist. bdry. R+

Proof These are straightforward from (1), (2), and (3) above or check the classical 

case. Problem 1.1.5 On R2+ , u is harmonic if and only if u ≡ 0. This is in turn true if and only if y 2 u ≡ 0. It happens that y 2  is the only second order differential operator that is invariant under the linear fractional transformations. The analogue for this operator on R2+ × R2+ is y12 1 + y22 2 . This operator degenerates on ∂R+ and does so strongly on the distinguished boundary. A theorem of Furstenberg says that if u : R+ → C, u bounded, and (y12 1 + 2 y2 2 )u = 0, then 1 u ≡ 2 u ≡ 0. His proof is rather deep. The problem is to give an elementary proof. It is worth noting that log(y1 /y2 ) is annihilated by y12 1 + y22 2 but is not multiply harmonic (it is also not bounded). Definition 1.1.6 Let f ∈ Lp (Rn ), 1 ≤ p ≤ ∞. We define the strong maximal function 1 MS f (x) ≡ sup R m(R)





|f (x − y)| dy , y∈R

where R ranges over rectangles of the form R = {y : |yj | ≤ aj } . Let M be the maximal function defined in the same way except that R ranges only over cubes {y : |yj | ≤ a}. Clearly Mf (x) ≤ MS f (x). Remark 1.1.7 In the past, one has considered Rn to be ∂Rn+1 . We now think of 2 )n (where db stands for distinguished boundary). The two theories are Rn as db(R+ completely different. In the latter case, many expected things are false. Recall that (i) Mf (x)Lp ≤ Ap f Lp , , 1 < p ≤ ∞. (ii) m{x : Mf (x) > α} ≤ (A/α)f L1 , 0 < α < ∞. It follows easily from (ii) that (iii) If f is supported on a cube Q0 , then

4

1 Introductory Material

   

1/p  |Mf |p dx  ≤ Af L1 , 0 < p < 1 .

Q0

Under the hypotheses of (iii), we also have (iv) 



|f |(1 + log+ |f |) dx

|Mf | dx ≤ A Q0

Q0

and more generally (v) 

|Mf |(1 + log+ Mf )r dx ≤ A



|f |(1 + log+ |f |)r+1 dx .

Q0

For the strong maximal function, one has the following theorem: Theorem We have (1) MS f Lp ≤ Bp f Lp , 1 < p ≤ ∞ . One proves this result by iteration of the estimate for M on R1 . Here Bp ∼ (Ap )n . Recall that Ap ∼ 1/(p − 1) as p → 1. (2) There exists f ∈ L1 such that MS f = +∞ everywhere. (3) If f is supported in a cube Q0 and 

|f |(1 + log+ |f |)n−1 dx < ∞ ,

Q0

then MS f < ∞ a.e. in Q0 and MS f ∈ Lp (Q0 ) for p < 1. This follows from the iteration argument indicated  in (1). (4) There exists an f such that f (1 + log+ |f |)n−1− dx < ∞ but MS f = ∞ everywhere. The positive results in this theorem are mainly due to Jessen, Marcinkiewicz, and Zygmund during the period 1930–1934 (cf. Fundamenta Mathematicae). Some of the other results are due to Saks. Corollary 1.1.8 If f ∈ Lp , 1 < p ≤ ∞, or f ∈ L(log L)n−1 (Q0 ), then 1 diam R→0 m(R)

 f (x − t) dt → f (x) a.e. x ∈ Q0 .

lim

R

This follows by standard arguments. We now indicate how some of the counterexamples indicated in the theorem are obtained.

1.1 Various Maximal Functions

5

(2) If a weak type inequality held for all functions, then it would hold for all measures, in particular for the Dirac measure. Now, on R2 , MS (δ)(x) =

1 . |x1 ||x2 |

So  {x : MS (δ)(x) > α} = x :

   1 1 > α = x : |x1 ||x2 | < . |x1 ||x2 | α

One can easily compute that this region has infinite area. Remark 1.1.9 Note, by comparison, that {x : Mδ(x) > α} = which is OK.

  x : |x|2 < α1 ,

Next, we construct a positive, finite measure dμ which is supported in the unit square and such that MS (dμ) = +∞ a.e. (for a reference, see [STE4]). The main lemma is the following, which is a Borel–Cantelli type of result. Lemma 1.1.10 Suppose that E1 , E2 , . . . is an infinite collection of subsets of the unit square Q0 such that

m(Ei ) = ∞ .

Then there exist sets Fn which are translates of the En , each n, so that almost every point of Q0 is contained in infinitely many Fn s. Equivalently, ⎧ ⎨

⎛

m Q0 ∩ ⎝ ⎩

∞ ∞  

k=1 j =k

⎞⎫ ⎬ Fj ⎠ = m(Q0 ) = 1 . ⎭

Now, for α > 1, we have 

 1 m[Q0 ∩ {x : MS δ(x) > α}] = m x : |x1 x2 | < , |xj | < 1 α  1  1/α 1 1 dx 2 dx + 4 = 2· 0 1/α x1 α   log α 1 + ≈C α α ≈C Let

log α . α

6

1 Introductory Material

En = {x ∈ Qn : MS δ(x) > n log2 n} . Then m(En ) ∼

log[n log2 n] n log2 n

∼

1 . n log n

So

m(En ) = +∞ .

Let Fn be the sets given by the lemma, each Fn a translate of En . Let μn be the corresponding translate of δ. Finally, let dμ =



1 dμn . n(log n)1+

This is a finite, positive measure. But MS (dμ)(x) ≥ >

1 MS (dμn ) n(log n)1+ n log2 n n(log n)1+

= (log n)1− , each n , for x ∈ Fn . Since almost every point of Q0 is in infinitely many of the Fn s, this shows that MS (dμ) = +∞ a.e.

1.2 Nontangential Convergence Let f ∈ Lp (RN ) and let u(x, y) be the multiple Poisson integral of f . We define the analogue of the nontangential maximal function for the product setting. Let y = (y1 , y2 , . . . , yn ) ∈ Rn , y > 0. Let j = {zj ∈ R1 : |xj | < yj } . Set  = 1 × · · · × n = {z : |xj | < yj , j = 1, . . . , n} . This is the product cone. Define

1.2 Nontangential Convergence

7

MU f (x) ≡ sup |u(z)| . z∈(x)

Lemma 1.2.1 We have (i) MU f (x) ≤ MS f (x). (ii) If f ≥ 0, then MS f (x) ≤ C · MU f (x) . Proof Suppose without loss of generality that n = 2. Assume that f ≥ 0. Then u(x, y) = =

1 π2

 R2

∞ k,j =0

+

≤

0≤|t2 |≤y2 /2 2k−1 y1 ≤|t1 |≤2k y1

|t1 |∼2k y1 |t2 |∼2j y2



∞ k,j =0

 +



k,j =0

0≤|t1 |≤y1 /2 2j −1 y2 ≤|t2 |≤2j y2



∞

∞

j =0

2k−1 y1 ≤|t1 |≤2k y1 2j −1 y2 ≤|t2 |≤2j y2

k=0

≤

y1 y2 f (x1 − t1 , x2 − t2 ) dt1 dt2 (y12 + t12 )(y22 + t22 )   ∞ +

|t1 |≤2k y1 |t2 |≤2j y2

0≤|t2 |≤y2 /2 0≤|t1 |≤y1 /2

1 −2k −2j 2 2 f (x1 − t1 , x2 − t2 ) dt1 dt2 y1 y2

plus similar terms

1 · 2−(j +k) f (x1 − t1 , x2 − t2 ) dt1 dt2 . (2k y1 )(2j y2 )

Now fix x 0 = (x10 , x20 ) and suppose that (x, y) ∈ x 0 . Make the change of variable tj → xj − xj0 + sj . Then, since |xj − xj0 | ≤ yj , we have that the last displayed line is ≤

∞ k,j =0

≤

∞

 |s1 |≤C2k y1 |s2 |≤C2j y2

1 (2k y1 )(2j y2 )

2−(j +k) f (x1 − s1 , x2 − s2 ) ds1 ds2

2−(j +k) MS (f )(x 0 )

k,j =0

≤ CMS f (x 0 ) .

This proves assertion (i). Assertion (ii) is now clear since the (0, 0) term in the sum decomposition of u(x, y) dominates MS f . 

8

1 Introductory Material

We can define a more general product cone by α (x) = {z : |xj − x j | < αj yj , j = 1, . . . , n} , where α = (α1 , . . . , αn ) > 0. Definition 1.2.2 We say that u has an unrestricted limit  at x if lim u(z) = 

z∈α (x) z→x

for all α > 0 .

Corollary of the Preceding Lemma Suppose that f ∈ Lp (Rn ) and that u is the multiple Poisson integral of f . Then (1) If 1 < p ≤ ∞, then u has unrestricted limits at almost every x ∈ Rn and the limit function is f . (2) Assertion (1) fails in general for p = 1. (3) If f is continuous and bounded, then lim u(x + iy) = f (x) as y → 0 uniformly on compact subsets of x ∈ Rn . Proof Assume that f is bounded and continuous on Rn . Let  u(x + iy) =

Py (t)f (x − t) dt .

Hence  Py (t)[f (x − t) − f (x)] dt

u(x + iy) − f (x) =  =

 |t|≤η

+

|t|>η

≡ T 1 + T2 . Now T1 may be made < /2 by choosing η so small that |f (x −t)−f (x)| < /2 for all x ∈ K compact—so |t| < η. Once η is fixed, letting y → 0 makes T2 vanish.  We formalize this last observation: for η > 0 fixed, |t|≥η Py (t) dt → 0 as y → 0 since P is a product of one-dimensional kernels. Remark 1.2.3 This is about localization. If f ∈ L∞ and f vanishes in a neighborhood of x, then u(x + iy) → 0 as y → 0. This follows from the above proof. From this, we can apply results for Lp , p < ∞, to an f ∈ L∞ by first multiplying f by a ϕ ∈ Cc∞ . One has ϕf ∈ Lp for all 1 ≤ p ≤ ∞. Example 1.2.4 (Localization Fails in General for Unrestricted Convergence) Precisely, on R2 , with p < ∞, there exists f ∈ Lp (R2 ) with f ≡ 0 in a neighborhood of 0, but u = P I (f ) does not converge at 0 unrestrictedly.

1.2 Nontangential Convergence

9

Fig. 1.1 The function f1

In fact, let f (x1 , x2 ) = f1 (x1 ) · f2 (x2 ), where the fi are compactly supported, fi ≥ 0. Suppose that f2 (x2 ) = 0 for all x2 in a neighborhood of 0. We will choose f1 to be a function that looks like Fig. 1.1, where f1 blows up at 0 very slowly. Then the multiple Poisson integral of f is u(x + iy) = u1 (x1 + iy1 ) · u2 (x2 + iy2 ) and u(iy) = u1 (iy1 ) · u2 (iy2 ). For any y2 > 0, u2 (y2 ) > 0 by the maximum principle and u2 (y2 ) → 0 as y2 → 0. But u1 (iy1 ) → +∞ as y1 → 0. We need only to choose f1 so that it blows up at 0 but is locally integrable. Since we are considering unrestricted convergence, we let y1 → 0 so much faster than y2 that lim u1 (y1 ) · u2 (y2 ) → ∞. Remark 1.2.5 Peaking for the product kernel is on a big subvariety instead of on the diagonal as in R1 . In order for localization to work in the product case, we would have to require that f vanishes in a neighborhood of the whole variety. Theorem 1.2.6 (Fatou’s Theorem) Suppose that u is multiply harmonic in (R2+ )n ⊆ Cn and that u is bounded. Then unrestricted limits exist almost everywhere and u is the multiple Poisson integral of some bounded function f . Also, if 1 < p < ∞, then the condition  sup

y>0 Rn

|u(x + iy)|p dx < ∞

implies that u has unrestricted limits almost everywhere and the limiting function f is in Lp . Remark 1.2.7 For p = 1, these results do not hold in general. Proof of the Theorem Imitation of the one-dimensional case.



Definition 1.2.8 Let MR f (x) ≡ sup |u(z)| , zj = xj + iyj , c1 ≤ z∈(x)

y1 ≤ c2 , k = 1, 2, . . . , n . y2

Lemma 1.2.9 We have: (1) It is the case that MR f (x) ≤ CMf (x). That is to say, there exists a function f so that the left side is ∞ and the right side is < ∞ at a point. (2) If f ≥ 0, then Mf (x) ≤ CMR f (x).

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Proof Note that (2) is clear by mimicking the proof of (ii) in Lemma 1.2.1. To see (1), let f = f1 (x1 ) · f2 (x2 ) with fi ≥ 0 for i = 1, 2. Assume that the fi are compactly supported and integrable. Choose f2 so that f2 (x2 ) = |x2 | δ near x2 = 0. Choose f1 so that 0 f1 (x1 ) dx1 → 0 as δ → 0 very slowly, say f1 (x1 ) = |x1 |−1 · (log |x1 |)−3/2 . Now 1 Mf (0) = sup 2 4δ δ

 |x1 |≤δ |x2 |≤δ

f1 (x1 ) · f2 (x2 ) dx1 dx2 < ∞ .

But  lim

δ→0



C ≥ δ

Pδ (t1 ) · Pδ (t2 ) · f1 (t1 ) · f2 (t2 ) dt1 dt2 

δ

   f1 (t1 ) dt1 · Cδ

0

≥ C| log δ|−1/2 · = C| log δ|

−1/2

δ

≥ C| log δ|

−1/2

= C| log δ|

δ





∞

∞ δ

1

f (t2 ) dt2 t22 



f2 (t2 ) dt2 t22

f2 (t2 ) dt2 + t22



∞ 1

f2 (t2 ) dt2 t22



· | log δ|

1/2

→ ∞ as δ → 0 .

Theorem 1.2.10 Suppose that f ∈ Lp (Rn ), 1 ≤ p ≤ ∞. Then MR f (x) < ∞ almost everywhere. In fact, m{x : MR f (x) > α} ≤

A f L1 α

and MR f Lp ≤ Ap f Lp , 1 < p ≤ ∞ , Ap ∼

1 as p → 1 . p−1

If instead f ∈ L1 , then the restricted limits of the Poisson integral exist almost everywhere and equal f .

1.2 Nontangential Convergence

11

Proof First Idea:

Define, in R2 ,

Rk,j = {(x1 , x2 ) : |x1 | ≤ 2k , |x2 | ≤ 2j } , k = 0, 1, 2, . . . , and j = 0, 1, 2, . . . . [As usual, we are assuming that n = 2.] Let 1 m(yR k,j ) y>0



Mk,j f (x) = sup

|f (x − t)| dt . yRk,j

Note that m(Rk,j ) = 2k+1 2j +1 and m(yRk,j ) = y 2 2k+1 2j +1 . It follows that Mk,j = Mk  ,j  if k − j = k  − j  . It is classical (cf. [STE1]) that m{x : Mk,j f (x) > α} ≤

C f L1 α

with C independent of k and j . We may let C = (3)n = 32 = 9. Second Idea: If f1 , f2 , . . . are uniformly  of weak type 1, i.e., m{x : |fk (x)| > α} ≤ α1 , then what can be said about ∞ k=1 ak fk ? [That is to say, for what choice  1 1 ?] We now give a sufficient condition: weak type 1 on R of ak is ak |x−x k|  1/2  ak ≤ 1. Then f = Lemma 1.2.11 Suppose that ak fk satisfies m{x : |f (x)| > α} ≤ 1/α with fk as above.   1/2 1/2 Proof Write f (x) = ak fk (x) = ak (ak fk (x)). We claim that {x : |f (x)| > α} ⊆

∞ 

1/2

{x : ak |fk (x)| > α} .

k=1

For if x ∈ ∪∞ {x : ak |fk (x)| > α}, then ak |fk (x)| ≤ α for all k, so  1/2 k=1 1/2 |f (x)| ≤ ak (ak fk (x)) ≤ α. This proves the inclusion. Thus 1/2

m{x : |f (x)| > α} ≤

1/2



 m{x :

1/2 ak |fk (x)|

> α} ≤

k

This completes the proof. It should be noted that the best possible condition on the ak is

1/2

ak α



≤

1 . α 

|ak | log |1/ak | ≤

1. Proof of Theorem 1.2.10 By the lemma and the First Idea, it will suffice to show that

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1 Introductory Material

MR f (x) ≤ C

∞

2−k−j Mk,j f (x) .

k,j =0

Let x = (x 1 , x 2 ). Let z = (z1 , z2 ) be the variable point, zj = xj + iyj . So we suppose that |xj − x j | < yj . Recall that Py (t) = Py1 (t1 ) · Py2 (t2 ) with Pyj (tj ) = (1/π )yj /(tj2 + yj2 ). There are two key estimates: (1) Since c1 ≤ y1 /y2 ≤ c2 , we have Py1 ( · ) ∼ Py2 ( · ). This is clear. (2) Py1 (t1 ) ∼ Py1 (t1 + h) provided that |h| ≤ cy1 . Let us prove this last result. Suppose without loss of generality that C > 1. If |t1 | ≤ 2C|y1 |, then 1 y1 32C 2 y1 1 y1 1 32C 2 y1 1 ≤ ≤ = . 2 2 2 2 2 2 π t1 + y1 π y1 π 32C y1 π 16C y12 + 16C 2 y12

(∗)

But |t1 + h|2 ≤ (3C|y1 |)2 ≤ 16C 2 |y1 |2 so that (∗) ≤

32C 2 y1 32C 2 y1 1 1 ≤ . π (t1 + h)2 + 16C 2 y12 π (t1 + h)2 + y12

If |t1 | ≥ 2C|y1 |, then |h| ≤ |t1 |/2, and hence 1 y1 y1 2y1 1 1 ≤ ≤ . π t12 + y12 π |t1 + h|2 + y12 π |t1 + h|2 + y12 Now       Py1 (t1 )Py2 (t2 )f (x1 − t1 , x2 − t2 ) dt1 dt2  . |u(x1 + iy1 , x2 + iy2 )| =  Using estimates (1) and (2) above, we majorize this last by

≤

Py (t1 )Py (t2 )|f (x 1 − t1 , x 2 − t2 )| dt1 dt2 + other leftover terms

2k−1 y≤t1 ≤2k y 2j −1 y≤t2 ≤2j y

≤

k,j

≤C

y −2 · 2−2k−2j



 |f (x 1 − t1 , x 2 − t2 )| dt1 dt2 yRkj

2−k−j Mk,j (f )(x)

1.2 Nontangential Convergence

13

as desired.



Problem 1.2.12 Is there a Fatou theorem for restricted convergence? More precisely, if u is multiply harmonic in (R2+ )n and bounded in the set where c1 ≤ y1 /2 ≤ c2 , then does u have boundary values in the restricted sense? We know that the global analogue for Theorem 1.2.6 is true.  Remark 1.2.13 Let ϕ ≥ 0, ϕ ∈ L1 (Rn ), ϕ dx = 1. Let ϕ (x) =  −n ϕ(x/). So  Rn ϕ (x) dx = 1 for all  > 0. We ask the following question: Is sup>0 (|f | ∗ ϕ ) of weak type 1 when f ∈ L1 ?   A sufficient condition is the following: define (x) ≡ sup|x  |≥|x| φ(x ). If Rn (x) dx < ∞, then the answer is “yes,” for then sup>0 |f | ∗ ϕ ≤ AMf , where A = .

Here is the difficulty: the ϕ which arises from restricted convergence is, in R2 , ϕ(x1 , x2 ) =

1 1 1 . 2 2 π 1 + x1 1 + x22

(x1 , x2 ) ≥

1 1 ·1· , π2 1 + |x|2

Hence

which is not integrable at ∞ in R2 . One can ask what is the right maximal function which dominates restricted convergence? Here is a possibility: Let : [−π, π ] → R+ . Then one can prove that the operator M (f )(x) ≡ sup h>0

1 h2



h π 0

−π

|f (x + reiθ )| (θ)r dθdr

satisfies M f Lp ≤ Ap f Lp  L1 (circle) , 1 < p ≤ ∞ . One uses the method of rotations on the one-dimensional result (see [STW1]). Here, as usual, Ap = 1/(p − 1) as p → 1.

An interesting special case of these ideas is when is a characteristic function. Then M amounts to a generalized Hardy–Littlewood maximal function taken over star-shaped balls. The ball is a union of radii through E where = χ E ; see Fig. 1.2. The point of this discussion is MR f (x) ≤ A M  (f )(x) ,

(∗)

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1 Introductory Material

Fig. 1.2 A picture of the ball

where  (θ ) = | sin 2θ |−2 and A ∼ 1/ as  → 0. We now prove inequality (∗). Observation If μ(x) ≥ 0 is radial and decreasing as a function of r = |x| on R2 ,  if R2 μ(x) dx < ∞, and if we let ϕ(r, θ ) = μ(r) · (θ ), then sup(|f | ∗ ϕ )(x) ≤ AμL1 M (f )(x) .

()

>0

This is immediate when μ(r) = αχ [0,r ] (r), some r0 > 0. 0 The general result follows by taking monotone limits of finite positive linear combinations of such functions. By the above remarks and the proof of Theorem 1.2.10, it suffices for us to show that if ϕ(x) =

1 1 · , 1 + x12 1 + x22

then ϕ(r, θ ) ≤

1 A| sin 2θ |−2 . (1 + r 2 )1+

For  R2

dx ≤ (1 + |x|2 )1+

 |x|≤2

≤ C+C

 + 

|x|>2

|x|>2

dx |x|2+2

C ≤ .  By (), this is enough. Now 1 (1 + x12 )(1 + x22 )

=

1 . 1 + r 2 + r 4 (sin θ cos θ )2

1.2 Nontangential Convergence

15

Since we wish to show that this is ≤

A| sin 2θ |−2 , (1 + r 2 )1+

we may as well suppose that r > 1, otherwise the result is trivial. So it suffices to show that (1 + r 2 )1+ ≤ A| sin 2θ |−2 1 + r 2 + r 4 | sin 2θ |2 or, since r > 1, (r 2 )1+ ≤ A| sin 2θ |−2 , r 2 + r 4 (sin 2θ )2 or, letting r 2 = s, s 1+ s

+ s 2 (sin 2θ )2

≤ A| sin 2θ |−2 .

With t = s/(sin 2θ ), this becomes essentially t 1+ ≤ C −2 . t + C2t 2 But if t ≤ 1/C 2 , then we have t 1+ t 1+ ≤ C −2 . ≤ 2 2 t t +C t If instead t > 1/C 2 , then t 1+ t 1+ 1 ≤ 2 2 ≤ 2 t −1 ≤ C −2 . 2 2 t +C t C t C That does it. Remark 1.2.14 Note that, by the method of rotations or by inspection, one can check that M  is of type (p, p), 1 < p ≤ ∞. The L1 result, unfortunately, does not follow in this fashion. But at least now we have a geometrical maximal function which dominates restricted convergence.

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1 Introductory Material

y2 A = {y1 ~ y2 }

y1

Fig. 1.3 Restricted convergence

y2 B = {(y1 , y )} 2

y1

Fig. 1.4 Unrestricted convergence

1.3 Unrestricted Convergence Now we discuss unrestricted convergence. We remark that when we consider holomorphic functions on R2+ , instead of just multiply harmonic functions, then the notions of restricted convergence and unrestricted convergence come closer together. Consider R2+ × R2+ . The restricted convergence situation is shown in Fig. 1.3. The unrestricted convergence situation is shown in Fig. 1.4. Now A can be mapped by a real linear transformation to B. This map can be extended to a complex linear map on Cn in the obvious way. Call the extended map L. If f is holomorphic, then so is f ◦ L. This fact enables one to pass from one theory to the other. But, if f is harmonic (or even multiply harmonic), then the same need not be true for f ◦ L. Definition 1.3.1 If u is multiply harmonic on R+ and x ∈ Rn , then u is said to be unrestrictedly bounded at x provided that there exist an α > 0 and h > 0 such that u is bounded on αh ≡ αh11 (x1 ) × αh22 (x2 ) × · · · × αhnn (xn ). [Note that there is no relationship among the yj s, hence the name.] Theorem 1.3.2 Let u be multiply harmonic in R+ A. Let E ⊆ Rn . If u is unrestrictedly bounded at each x ∈ E, then u has unrestricted limits at almost every point of E. Proof The proof of this result for Rn+ is well known, so we only indicate in detail where this proof differs from that one.

1.3 Unrestricted Convergence

17

We begin, as usual, by uniformizing the situation: namely, we must check that it suffices to assume that for some α > 0 and h > 0,  αh (x0 ) , E compact. |u(x, y)| ≤ 1 for all (x, y) ∈ x0 ∈E

The proof of this reduction relies on a point-of-density argument. However, in the present situation we must use the fact that almost every point of a set E ⊆ Rn is a point of strong density of E. That is to say, lim

diam R→0

m(E ∩ R) = 1, m(R)

where R is any rectangle centered at x with sides parallel to the axes. After uniformizing things, we suppose that, for E ⊆ R2 compact, |u| ≤ 1 in R ≡ ∪x0 ∈E α1 (x0 ). For each n ≥ 1, set  ϕn (x1 , x2 ) =

u(x1 + i/n, x2 + i/n) if (x1 + i/n, x2 + i/n) ∈ R 0 otherwise.

Let ϕn (z) = multiple PI[ϕn (x)]. Then, of course, |ϕn (x)| ≤ 1 for x ∈ R2 , n ≥ 1 . Define un (z) = u(x1 + i(y1 + 1/n), x2 + i(y2 + 1/n)), and finally let ψn (z) = un (z) − ϕn (z). All this notation is analogous to the proof for Rn+ . Now since ϕn L∞ ≤ 1, there is a subsequence ϕnk converging weak-∗ to some ϕ ∈ L∞ (R2 ). Let ϕ(z) = multiple PI(ϕ). Then of course unk (z) → u(z) and ϕnk (z) → ϕ(z), so ψnk (z) → u(z) − ϕ(z) ≡ ψ(z). Since ϕ is a multiple Poisson integral of an L∞ function, ϕ has unrestricted limits almost everywhere. So, in order to see that u has unrestricted limits almost everywhere, it behooves us to show that ψ(z) → 0 as z → x0 through α1 (x0 ) for almost every x0 ∈ E. As usual, in order to show this, it is enough to find a multiply harmonic h ≥ 0 in R+ such that, in R, |ψ(z)| ≤ h(z) and h has unrestricted limits  0 almost everywhere  on E. For the function h, we use h(z) = C y1 + y2 + MP [χ c E ] . Clearly h is multiply harmonic, h ≥ 0, and h has unrestricted limits 0 almost everywhere in E. We shall check that |ψn (z)| ≤ h(z) in R, finishing the proof. Of course, we must choose C appropriately. Now we need only to check our inequality on bR. More precisely, we must show that if z ∈ bR, then ψn (z) ≤ lim inf h(z ) . z →z z ∈R

()

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1 Introductory Material

Fig. 1.5 The set E

Fig. 1.6 The fact that Λ ⊆ cE

z

x

Consider Fig. 1.5. The boundary of R consists of four parts: (1) (2) (3) (4)

Either y1 or y2 equals h. 0 < y1 < h, 0 < y2 < h. y1 = 0 but 0 < y2 < h or vice versa. y1 = 0 and y2 = 0.

So we wish to check () in each of these four cases. For (1), just make C sufficiently large. For (4), note that when y1 = y2 , we have ψn = 0. To check (2) is more difficult. Let z = (x 1 + iy 1 , x 2 + iy 2 ) ∈ bR. We must check that |ψn (z)| ≤ h(z). Now let  = {(x1 , x2 ) : |xj − x j | < αy j }. We claim that  ⊆ c E. This is clear pictorially—see Fig. 1.6. Thus h(z) ≥ MP I (χ )  y1y2 χ (x 1 − t1 , x 2 − t2 ) = dt1 dt2 π2 (t12 + y 21 )(t22 + y 22 )  C ≥ χ (x 1 − t1 , x 2 − t2 ) dt1 dt2 y1y2 =

C m() y1y2

≥ C · ψn (z) . That proves (4). For (3), let z = (x 1 , x 2 + iy 2 ), 0 < y 2 . Set z = (x1 + iy1 , x2 , +iy2 ). Then MP I (χ c E )(z1 , z2 ) =

y2 π2



dt2 t22 + y  22

· y1 ·



χ c E (x1 − t1 , x2 − t2 ) t12 + y  21

dt1 ,

1.4 The Area Integral

19

so lim inf MP I (χ c E )(x 1 + iy1 , z2 ) y1 →0

1 ≥ 2 y2 π



χ c E (x 1 , x 2 − t2 ) t22 + y 22

dt2 ,

(∗)

by Fatou’s lemma, once we check that lim inf y1 



χ c E (x 1 − t1 , x 2 − t2 ) t12 + y  21

y1 →0

dt1 ≥ χ c E (x 1 , x 2 − t2 ) .

Now in case (x 1 , x 2 − t2 ) ∈ E, then the right side is 0 so the inequality is clear. But if (x 1 , x 2 − t2 ) ∈ c E, we have, since c E is open, that χ c E (x 1 − t1 , x 2 − t2 ) = 1 for t1 small. So the approximate identity property of the Poisson kernel shows that the left side is 1 (as is the right). Thus we have established (∗). But it is easy to see that the right side of (∗) is bounded from 0, independent of z, so we have established (), hence the theorem. 

Assertion: Somewhere in the above theory, the “distinguished boundary” is playing a role. As usual assume n = 2. Let z lie in the topological boundary of R. Furthermore, let z = (z1 , z2 ) = (x 1 + iy 1 , x 2 + iy 2 ). We say that z is a boundary point of type 1 if (x 1 + iy 1 − , x 2 + iy 2 ) ∈ R for all sufficiently small  > 0. Similarly, z is a boundary point of type 2 if (x 1 + iy 1 , x 2 + iy 2 − ) ∈ R for all sufficiently small  > 0. Every nontrivial boundary point is of either type 1 or type 2 or both (a trivial boundary point is one for which y 1 = y 2 = 0 or y 1 = y 2 = h). Let  = {z ∈ bR : z is of type 1 and of type 2}. We conjecture that  should play the role of the distinguished boundary for this local theory. It is unknown whether this can be made to work.

1.4 The Area Integral Let n = 2 and u(z1 , z2 ) be a multiply harmonic function in (R2+ )2 . Let   ∂f |∇j f | =  ∂x 2

j

2     +  ∂f   ∂y

j

2   , j = 1, 2 . 

Define ⎛ ⎜ S(u) = ⎝

 (x)

where (x) = 1 (x1 ) × 2 (x2 ).

⎞1/2 ⎟ |∇1 ∇2 u(z)| dxdy ⎠

,

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1 Introductory Material

One can check that if u = MP I (f ), f ∈ L2 (R2 ), then 

 R2

(S(u))2 dx1 dx2 = 4

R2

|f |2 dx1 dx2 .

To prove this, we can use Green’s theorem or we can recall that 1 (u2 ) = 2|∇1 u|2 and 1 2 (u2 ) = 4|∇1 ∇2 u|2 . Problem 1.4.1 Show that unrestricted convergence for a multiply harmonic function is locally almost everywhere equivalent with the finiteness of S(u). Definition 1.4.2 For 1 ≤ p < ∞, let  H (R+ ) = {f holomorphic on R+ : sup

p

p

y>0 R+

|f (x + iy)|p dx ≡ f H p < ∞} .

[Use the usual variant when p = ∞.] Theorem 1.4.3 Suppose that F ∈ H p (R+ ), p > 0. Then (1) F (x + iy) has unrestricted limits as y → 0 for almost every x ∈ Rn . (2) If F ∗ (x) ≡ supz∈(x) |F (z)|, then F ∗ Lp ≤ CF H p . (3) If p < ∞ and if F (x) = limy→0 F (x + iy), then  Rn

|F (x + iy) − F (x)|p dx → 0 as y → 0 (unrestrictedly).

Proof (In Case n = 2) First observe that if F (z1 , z2 ) is in H p (R+ ), then F ( · , z2 ) ∈ H p · ), each fixed z2 . Now fix a z2 and a y1 > 0. We claim that 

∞

−∞

p

|F (x1 + iy1 , z2 )|p dx1 ≤ Cy2 F H p (R+ ) .

To see this, note that by the subharmonicity of |F |p in each variable, F (z1 , z2 )|p ≤

C (y1 )2 (y2 )2





B(z1 ,y1 /2) B(z2 ,y2 /2)

Thus

|F (x1 +iy1 , x2 +iy2 )|p dx1 dy1 dx2 dy2 .

1.4 The Area Integral



∞

−∞

≤

≤

21

|F (x1 + iy1 , z2 )|p dx1

C y1 (y2 )2





|F (x1 + iy1 , x2 + iy2 )|p dx1 dy1 dx2 dy2

x1 ∈R x2 ∈R y1 /2≤y1 ≤3y1 /2 y2 /2≤y2 ≤3y2 /2

C p F H p . y2

This proves the claim. Now if F ∈ H p (R+ ) and z2 is fixed, then |F (z1 + i, z2 )|p/2 is subharmonic in z1 and has bounded L2 norms along the lines y1 = constant, as we just verified. We have in addition  ∞ p/2 |F (z1 + i, z2 + i)| ≤ Py1 (x1 − t1 )|F (t1 + i, z2 + i)|p/2 dt1 (a) −∞

by subharmonicity. A similar notion applied to |F (t1 + i, · )|p/2 shows that  |F (t1 + i, z2 + i)|

p/2

≤

∞ −∞

Py2 (x2 − t2 )|F (t1 + i, t2 + i)|p/2 dt2 .

(b)

Putting together (a) and (b) gives  |F (z1 +i, z2 +i)|p/2 ≤

Py1 (x1 −t1 )Py2 (x2 −t2 )|F (t1 +i, t2 +i)|p/2 dt1 dt2 . R2

(∗) Now the functions |F (t1 + i, t2 + i)|p/2 are uniformly bounded in L2 , so there exists a weak-∗ convergent subsequence |F (t1 +ik , t2 +ik )|p/2 with limit F . Also p/2 f L2 ≤ F H p . Taking the limit of (∗) as k → ∞ yields |F |p/2 (z) ≤ MP I (f )(z) whence (F ∗ )p/2 (x) ≤ sup |MP I (f )(z)| ≤ MS f (x) z∈(x)

by Lemma 1.2.1. Therefore F ∗ p ≤ Cp f p ≤ Cp F H p , p > 0 . p

p

p

[Note that we have arranged everything above to take place in L2 and H 2 so that all relevant operators are bounded.]

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Fig. 1.7 The cone 

Part (2) is thus proved. Part (1) now follows from the local Fatou theorem. Part (3) follows from (1) and dominated convergence. 

1.5 Generalizations of R+ and H p (R+ ) We now pass to generalizations of R+ and H p (R+ ) (Fig. 1.7). Definition 1.5.1 A set  ⊆ Rn is called an open cone if (1) (2) (3) (4)

x ∈ , y ∈  ⇒ x + y ∈ . x ∈ , r > 0 ⇒ rx ∈ . 0 ∈ .  is open.

Definition 1.5.2 The tube domain T over  is {z ∈ Cn : z = x + iy, y ∈ }. Definition 1.5.3 The cone  is said to be regular if  contains no lines. Example 1.5.4 In R1 , the only cones are {x : x > 0} and {x : x < 0}. In R2 , all cones are essentially of the form {z : 0 < arg z < θ } , θ ≤ π . The cone is regular if θ < π . In R3 , there are a great many different cones. Two examples are: (a) The first octant = {(y1 , y2 , y3 ) : yj > 0} (b) The forward light cone = {(y1 , y2 , y3 ) : y3 > 0, y12 + y22 < y32 } Examples (a) and (b) are “homogeneous” in the sense that, given two points of the cone, there exists a linear transformation under which the cone is invariant and one point is moved to the other. An example of a non-homogenous cone in R3 is readily seen to be the open, convex hull of four rays emanating from 0 and in general position. Definition 1.5.5 Let  ⊆ Rn be a cone. The dual cone  ∗ is defined to be  ∗ ≡ {y ∈ Rn : y · x ≥ 0, x ∈ }. It is a fact (see [STW1]) that  ∗ contains an interior point if and only if  is regular.

1.5 Generalizations of R+ and H p (R+ )

23

Definition 1.5.6 Define  H p (T ) = {F : F is holomorphic on T and sup

y∈ Rn

p

|F (x +iy)|p dx = F H p < ∞} .

The main theorem on H 2 (T ) is then the following. Theorem 1.5.7 Let  be a regular cone. Then if f ∈ L2 ( ∗ ), we see that  F (z) =

∗

f (t)e2π it·z dt

converges absolutely to a function which is in H 2 (T ). Every F ∈ H 2 (T ) arises in this fashion from a unique f and the correspondence f ↔ F is norm preserving, i.e., F H 2 (T ) = f L2 ( ∗ ) . More precisely, one can say that the integral converges absolutely and uniformly when y remains in a compact subset of . Also 

|F (x + iy)|2 dx = f L2 ( ∗ ) .

y∈

It is sometimes convenient to think of a partial ordering given by  as follows: We say that y > 0 if and only if y ∈  and y1 > y2 if and only if y1 − y2 ∈ .

Remark 1.5.8 Suppose that y0 is a fixed point in . Then there exists a δy0 > 0 such that e−2πy0 ·t ≤ e−(δy0 )|t| whenever t ∈  ∗ . Proof of the Theorem If y  is any unit vector in the y space, not necessarily in the cone, then y0 − y  ∈ ,  sufficiently small. So (y0 − y  ) · t ≥ 0 for all t ∈  ∗ , i.e., y0 · t ≥ +y  · t . By symmetry, y0 · t ≥ |y  · t| for all unit vectors y  or y0 · t ≥ |t| . This gives the result.

24

1 Introductory Material

Now clearly  ∗

e−2πy·t dt ≤

 ∗

e−2π δy |t| dt < ∞

for any y ∈ . By definition, moreover,  F (z) =

e2π ix·t e−2πy·t f (t) dt .

∗

By Schwarz’s inequality,  |F (z)| ≤

∗

e−4πy·t dt

1/2 · f L2 ( ∗ ) < ∞ .

To prove the assertion about uniform convergence, it suffices to consider convex hulls of finitely many points y 1 , y 2 , y 3 , . . . , y N .  N j aj = 1, Suppose that y is in the convex hull of {y j }N j =1 aj y , j =1 . So y = aj ≥ 0. We consider e2π iz·t f (t), Im z = y. Then     2π iz·t f (t) = e−2πy·t |f (t)| e = e−2π ≤

N



aj y j ·t

|f (t)|

aj e−2πy ·t |f (t)| . j

j =1

 The convergence of  ∗ on the left side hangs, of course, on that of the right side. Since every compact subset of  is contained in a finite union of such balls, we are done. Now, by Plancherel’s theorem, 

 Rn

|F (x + iy)|2 dx =  ≤

∗

∗

e−4πy·t |f (t)|2 dt for y ∈  |f (t)|2 dt .

Since  lim

y→0  ∗

e

−4πy·t

the asserted isometry is proved.

 |f (t)| dt = 2

∗

|f (t)|2 dt ,

1.5 Generalizations of R+ and H p (R+ )

25

To see that every F ∈ H 2 (T ) arises as the integral of an f ∈ L2 ( ∗ ), we need a lemma: Lemma 1.5.9 Suppose that F ∈ H 2 (T ). If K is relatively compact in , then F is bounded on Rn × K. Proof Fix K relatively compact in . Find an  > 0 so small that if y ∈ K and |y −y| < , then y ∈ . If z = x +iy, y ∈ K, let P (z) = {w ∈ Cn : |zj −wj | < }. Thus  1 |F (z)|2 ≤ |F (z )|2 dx  dy  ≤ CK F 2H 2 (T ) .  (π  2 )n P (z) Now suppose that F ∈ H 2 (T ), |F (x + iy)| ≤ Ay (1 + |x|)−n−1 , where the constant Ay is uniform for y in compact subsets of . Then y ∈  implies that  Rn

F (x + iy)e−2π ix·t dx ≡ fy (t)

is well defined. We assert that fy (t) = e−2πy·t f (t) gives a well-defined f (t) which meets our requirements. In order to see that this well defines f , it suffices to have fy (t) · e2πy·t = fy (t)e2π y·t , all y, y ∈  . Say that y = (y 1 , y 2 , . . . , y n ) and y = (y1 , y2 , . . . , yn ). We will now use Cauchy’s integral formula one variable at a time. Fix z2 , z3 , . . . , zn and look at F ( · , z2 , z3 , . . . , zn )e−2π iz·t . We wish to see that  F (x1 + iy1 , z2 , . . . , zn )e

−2π iz·t

 dx1 =

F (x1 + iy 1 , z2 , . . . , zn )e−2π iz·t dx1 ,

where z = (z1 , z2 , . . . , zn ) and  z = (x1 + iy 1 , z2 , . . . , zn ). Use the Cauchy Integral theorem on the indicated box (see Fig. 1.8) to verify this inequality. Now repeat this procedure in each variable. This defines f (t). Moreover, Plancherel’s formula shows that F and f have the desired relation. Now we may remove the growth restriction on F ; this is where we use Lemma 1.5.9. Look at      −(z12 +···+zn2 )  − xj2 + yj2 . =e e

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1 Introductory Material

Fig. 1.8 The Cauchy integral over a box in the complex plane

y

1

y1

If y is contained in a fixed compact set, then this is a rapidly decreasing function of x. n  2 Let F (z) = e− j =1 zj F (z) and fy (t) = F (x + iy)e−2π ix·t dx. If, in this last expression, we replace F by F and use the first part of the proof, we get  (f )y (t) =

F (x + iy)e2π ix·t dx,

and the expression on the right is independent of y. So we define  f (t) = e2πy·t (f )y (t) = e2πy·t

F (x + iy)e2π ix·t dx .

If we fix y (so that F ( · + iy) is bounded by the lemma), then we may let  → 0 to obtain the desired f . Let us summarize what we have accomplished. Given F ∈ H 2 (T ), we may write  F (x + iy) =

e2π ix·t e−2πy·t f (t) dt ,

where e−2πy·t f (t) is in L2 . We wish to see that f is supported in T ∗ . But  sup y>0

Rn

|F (x + iy)|2 dx < ∞

if and only if  sup y∈

Rn

e−4πy·t |f (t)|2 dt < ∞ .

Now if f has support not in , then −4πy · t will be positive there and the expression e−4πy·t can get very large. This will cause a contradiction. More precisely, let t ∈  ∗ and let y0 ∈  satisfy y0 · t < 0. There exists an  > 0 such that if t ∈ B(t, ), then y0 · t ≤ −δ < 0. Now let y = Ny0 , where N > 0 is large. Then

1.5 Generalizations of R+ and H p (R+ )

 sup

y∈ Rn

27

e−4πy·t |f (t)|2 dt < ∞ ⇒ sup N



e−4π Ny0 ·t |f (t)|2 dt < ∞

B(t,)



⇒ sup N

e4π N δ |f (t)|2 dt < ∞ B(t,)

⇒ f (t) ≡ 0 on B . So f is supported in  ∗ . This completes our rather long and chatty proof of Theorem 1.5.7.



Corollary 1.5.10 If F ∈ H 2 (T ), then limy→0 F (x +iy) = f (x) exists in L2 norm and F L2 = F H 2 (T ) = f L2 ( ∗ ) . 

Proof Use Plancherel’s theorem. Now we define a Cauchy kernel for T . Definition 1.5.11 With  as usual, let  e2π iz·t dt , z = x + iy , y ∈  . C(z) = ∗

This function C will be our Cauchy kernel. This definition is due to S. Bochner. Observe that C(z) is holomorphic on T . We have, with the notation Cy (x) = C(x + iy), y ∈ , that Cy ∈ L2 (Rn )



L∞ (Rn ),

and the norms are uniform when y is restricted to compact subsets of . Proposition 1.5.12 If F ∈ H 2 (T ), then  F (x + iy) =

Cy (t)F (x − t) dt .

If y1 and y2 are in , then  F (x + iy1 + iy2 ) =

Rn

Cy1 (t)F (x − t + iy2 ) dt .

Proof This is clear from what we have already done. For example, let  = {y ∈ 2 . Then Proposition 1.5.12 is well known. R1 : y > 0}. Consider T = R+



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1 Introductory Material

In many cases we can explicitly compute C. We define the Poisson kernel associated with the tube domain T by P(x, y) =

|Cy (x)|2 , y ∈ . C(2iy)

In many cases, P satisfies an appropriate second order differential equation, but in many other cases it does not. This definition of the Poisson kernel is due to Hua. From the properties of C, we see that Py ∈ L1 ∩ L∞ , each y ∈  . Also, by Plancherel,  Rn

|K(x + iy)|2 dx = K(2iy),

(∗)

so P (x, y) ≥ 0. Theorem 1.5.13 The Poisson kernel P satisfies the following properties:  (1) Rn Py (x) dx = 1. (2) If U is any open set containing 0, then  Py (x) dx = 1.

lim y→0 y∈

U

(3) If F ∈ H 2 (T ), then  F (x + iy) =

Rn

 Py (t)F (x − t) dt =

Rn

Py (x − t)F (t) dt.

(3’) If y1 , y2 ∈ , then  F (x + iy1 + iy2 ) =

Rn

Py1 (t)F (x − t + iy2 ) dt .

It is not generally the case that Py1 ∗Py2 = Py1 +y2 . It is worthwhile investigating in what sense this semigroup property holds. Proof of the Theorem Part (1) is clear from observation (∗). For part (2), we construct an F ∈ H 2 (T ) such that F is continuous on T , F (0) = 1, |F (x)| < 1 if x = 0, and |F (x)| → 0 as |x| → +∞. This is a “peak function” for T . The existence of such an F , plus the statement (3) of the theorem, will yield the result. For pick  > 0. Choose δ > 0 such that |F (x)| < 1 −  if |x| > δ and also such that B(0, δ) ⊆ U . Then

1.5 Generalizations of R+ and H p (R+ )

29

 1 = lim y∈ y→0



 |x|≤δ

F (x)P(x, y) dx +

F (x)P(x, y) dx

|x|>δ

.

Thus  1 ≤ lim y∈ y→0



 |x|≤δ

P(x, y) dx + (1 − )

  ≤ lim 1 −  y∈ y→0

|x|>δ

P(x, y) dx

 |x|>δ

P(x, y) dx

.

Therefore  lim y∈ y→0

|x|>δ

P(x, y) dx = 0 .

In conclusion,  1 = lim y∈ y→0

 Rn

P(x, y) dx = lim y∈ y→0

P(x, y) dx U

as desired. So we must construct such an F . In fact we let  e2π iz·t f (t) dt , F (x + iy) = ∗

 with f (t) ≥ 0, smooth, compactly supported in  ∗ , and such that f (t) dt = 1. Then all the assertions we have made about F are obvious except that |F | < 1 when x = 0. But  e2π ix·t f (t) dt . F (x) = ∗

This would be 1 if we were to take the absolute value of the integrand, but as it stands there will be cancellation. This proves (2). It now behooves us to prove (3) and (3’) without using (2). Let F ∈ H 2 (T ). Fix w ∈ T , w = u + iv, v ∈ , u ∈ Rn , and let G(z) = F (z)C(z − w) . From our observations about C, it follows that G ∈ H 2 . So, by the Cauchy formula,

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1 Introductory Material

 G(w) =

Rn

C(u − t + iv)F (t)C(t − u + iv) dt .

But, from the definition of C, C(x + iy) = C(−x + iy). So  F (w)C(2iw) = G(w) =

Rn

|C(u − t + iv)|2 F (t) dt

or  F (w) =

Rn

|C(u − t + iv)|2 F (t) dt = C(2iv)

 Rn

Pv (t)F (x − t) dt . 

The proof of (3’) is similar, using (3). Example 1.5.14 We actually present six examples, some in detail. 1. Let  = first octant in Rn = {y : y > 0}. Write Py (x) = Py1 (x1 ) · Py2 (x2 ) · · · · · Pyn (xn ) . 2. The Lorentz inner product is (x, y) = x1 y1 − x2 y2 − x3 y3 − · · · xn yn . So (y, y) = y12 −

n

yj2 .

j =2

Let  = {y : (y, y) > 0, y1 > 0} . This space is only interesting when n ≥ 3. For, when n = 2,  is simply a rotation of the first quadrant. We claim that  ∗ = . Proof of Claim Suppose that x ∈ Rn with x · y ≥ 0 for all y ∈ . Then x · (1, 0, . . . , 0) ≥ 0, so x1 ≥ 0. Suppose that yj , j = 2, 3, . . . , n, are real numbers satisfying y22 + y32 + · · · + yn2 = 1. Then (1, y2 , . . . , yn ) ∈  so that x · (1, y2 , . . . , yn ) ≥ 0 ,

1.5 Generalizations of R+ and H p (R+ )

31

that is, x1 −

n

xj yj ≥ 0 ,

j =2

that is, x1 ≥

n

xj yj .

j =2

Since the yj were arbitrary, it follows that ⎛ ⎝

n

⎞1/2 xj2 ⎠

≤ x1

j =2

or n

xj2 ≤ x12 .

j =2

That is to say, x ∈ . So  ∗ ⊆ . The reverse inclusion is trivial. If we define (z, w) = z1 w1 − z2 w 2 − · · · zn w n , z, w ∈ Cn , then one can compute that C(z) = cn (z, z)−n/2 ,

(∗)

where cn = (n/2)/(2π n/2 i n ). So Py (x) =

|cn |(y, y)n/2 |cn |(y, y)n/2 = . n |(x + iy, x + iy)| [((x, x) − (y, y))2 + 4(x, y)]n/2

When n is odd, there are subtleties lurking in this formula for C, but the function does not vanish on the cone, so we may define (z, z)−n/2 by choosing a branch. We remark that one computes (∗) by computing C at a point and then using the fact that  is a homogeneous space under the Lorentz group, where one can define the Lorentz group to be those linear transformations which respect ( · , · ). See [STE2], where this is carried out in a computation of the Bergman and Cauchy– Szeg˝o kernels.

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1 Introductory Material

For example, one can explicitly compute  C(i, 0, . . . , 0))

=

∗



e2π i·i·t dt

e−2π t dt

= 

 =

∞

m{(t2 , . . . , tn ) :

0

 = 0

n

tj2 ≤ t12 }e−2π t1 dt1

j =2 ∞

Vn−1 t1n−1 e−2π t1 dt1

= · · · = Vn−1 (n − 1)!(2π )−n+1



∞

e−2π t1 dt1

0

=

Vn−1 (n − 1)!(2π )−n ,

where Vn−1 is the volume of the unit ball in (n − 1)-space. So the forward light cone has the following interesting properties: (a) Self-duality (b) Homogeneity (c) Positivity under a Hermitian inner product 3. Now let n = [m(m + 1)]/2, m > 0, m ∈ Z. We may identify Rn with all m × m real symmetric matrices. Let Sm ⊆ E m be all positive definite real symmetric matrices. The automorphism group of Sm can be obtained from the general linear group on Em . Namely, if g ∈ GL(Em ), x ∈ Sm , let ρg x = gx t g. The tube over this cone is the “Siegel generalized upper half plane.” Note that in case m = n = 1, then TSm = R2+ . One can compute (n+1)/2    det y   Py (x) = constant ·  . 2 |det (x + iy)|  4. When n = m2 , we obtain a complex analogue of 3. Namely, we identify Rn with complex Hermitian matrices and Cn with m × m complex matrices. Then, of course, the cone Hm is the set of all m×m positive definite hermitian matrices. As in 3, the automorphism group on Hm can be obtained from GL(n, C). Namely, if g ∈ GL(n, C), define ρg x = gx t g for x ∈ Rn . 5. When n = 2m2 − m, one identifies Rn with the m × m quaternionic hermitian matrices and carries out the above program. 6. There is an analogue of 3, 4, and 5 for the Cayley numbers in case n = 16.

1.6 Relationships Among Domains

33

Theorem 1.5.15 (Vinberg, Koecher, and Rothaus) Every homogeneous self-dual cone is a product of the cases 1–6. One has a good chance of doing some function theory in the above setting, as one can compute the Cauchy and Poisson integrals explicitly, and one can use the symmetries and group actions. There exists a theory of “Jordan algebras,” which encompasses the above examples.

1.6 Relationships Among Domains We now make some remarks on the relationships among the domains we have studied. Every tube domain over a regular cone is biholomorphically equivalent to a bounded domain in Cn . One sees this as follows: after a linear transformation, surround the cone by the first octant. The first octant is of course a product of halfplanes which is biholomorphically equivalent to a product of discs. However, there exist bounded domains that are not biholomorphically equivalent to tubes over cones (e.g., the unit ball). Of course, we now have bounded domains



tube domains over regular cones

⊇

tube domains over homogeneous regular cones (there are uncountably many non-equivalent such domains—cf. Piatetski-Shapiro, Vinberg, etc.)

⊇

tubes over homogeneous, self-dual cones (there are finitely many of these in each dimension)

The latter collection coincides with the bounded symmetric domains of Cartan with distinguished boundary having 1/2 the real dimension of the domain. Recall that D ⊆ Cn is a bounded, symmetric domain of Cartan if (i) D is relatively compact in Cn . (ii) D is homogeneous in the sense that the group of holomorphic self-mappings of D is transitive. (iii) (symmetry) Given p ∈ D, there exists a mapping T : D → D holomorphic such that T 2 = id and T (p) = p for p an isolated fixed point. Examples of bounded symmetric domains of Cartan: • The unit ball in Cn . If p = 0 ∈ B, then let T : z → −z. Obtain T for any other point in the unit ball by conjugation by an appropriate biholomorphic map.

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1 Introductory Material

• On the Siegel domain, the transformation T for the point (i, 0, . . . , 0) is T : z → −1/z. It was a longstanding open problem whether the defining conditions (i) and (ii) for a bounded symmetric domain of Cartan implied condition (iii). It turns out that the answer is “no.” Also, it has been shown that tube domains over homogeneous cones do not coincide with tube domains over homogeneous, self-dual cones. We return now to tube domains over cones. Let  be a regular cone. We say that 0 is a proper subcone of  if 0 ⊆ Γ ∪ {0}. Definition 1.6.1 If F : T → C, x ∈ Rn , we say that the limit of F exists as z → x restrictedly if lim F (z)

z→x z=x+iy

exists, where |x − x| ≤ C|y|, y ∈ 0 , some 0 a regular subcone of , some C > 0. Theorem 1.6.2 Let  be a regular, convex cone, F ∈ H p (T ), 0 < p ≤ ∞. Then (1) lim  z→x F (z) = F (x) holds almost everywhere in the restricted sense. (2) Rn |F (x + iy)| − F (x)|p dx → 0 as y → 0 restrictedly, 0 < p < ∞. (3) For p ≥ 1, convergence works in the sense of (2) without restricting y to a subcone. Unsolved Problems (a) What happens to almost everywhere unrestricted convergence? (b) What happens to (3) when p < 1? The answers to these questions are unknown, even for the forward light cone. Proof of the Theorem Consider first the case when  is polygonal, i.e.,  is generated by finitely many lines through 0. We may even suppose that the lines are linearly independent. For any other polygonal cone is a union of such cones. There exists a non-singular linear transformation which takes our cone’s generators to the n unit coordinate vectors. The theorem is known on tubes over such cones. Now, if 0 is a proper subcone of , then one can find a polygonal cone between the two. Hence (1) and (2) follow. In the case of (3), F is a Poisson integral of an Lp function on the full cone. So the conclusion (3) follows from standard results about Poisson integrals. For details, see [STW1], p. 119.

Chapter 2

More on Hardy Spaces

2.1 Hardy Spaces and Maximal Functions Before proceeding we emphasize a few fundamental points. We have Theorem 2.1.1 Suppose that F ∈ H p (T ), p > 0, where  is the first octant in Rn . Then (1) F has a nontangential limit almost everywhere in each variable. In particular, lim y∈ F (x + iy) = F (x), for almost every x ∈ Rn . This is essentially y→0

unrestricted convergence (no subcone in y). (2) Rn |F (x + iy) − F (x)|p dx → 0 as y ∈  tends to 0. 

Proof See Stein and Weiss, page 115.

Remark 2.1.2 This result may fail for tubes over general cones. Unrestricted convergence of Poisson integrals may even fail for functions in Lp (Rn ), p > 1. For example, if  = Cn (the forward light cone in Rn ), n ≥ 3, then there exists an f ∈ Lp , p < ∞, such that, for almost every x,  lim sup u(x, y) = lim sup

Rn

P (x − t, y)f (t) dt = ∞ ,

as y ∈  tends to 0 unrestrictedly. However, if we use restricted convergence and if we use as a base for the cone one of the so-called domains of positivity described in the Vinberg–Koecher–Rothaus theorem, then for almost every x ∈ Rn , u(x, y) → f (x) restrictedly. For details, see [STW2]. We of course used the result of Theorem 2.1.1 to prove the last theorem in the last chapter.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Krantz, The E. M. Stein Lectures on Hardy Spaces, Lecture Notes in Mathematics 2326, https://doi.org/10.1007/978-3-031-21952-8_2

35

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2 More on Hardy Spaces

In what follows, we concentrate on circular cones, although the results go over to domains of positivity.  Theorem 2.1.3 Let  = Cn = {y ∈ Rn : y12 − nj=2 yj2 > 0}. (1) Let 0 ⊆  be a proper subcone, f ∈ Lp (Rn ), 1 ≤ p ≤ ∞. Let Py be the Poisson kernel for . Set u(x, y) = Py ∗ f and Mf (x) ≡ sup |u(x, y)| . |x−x|0

1 m(R)

 |f (x − t)| dt . R

The estimates that we get for the operator f → fR∗ are independent of the size and shape of R. In fact fR∗ L∞ ≤ f L∞ and m{x : fR∗ (x) > α} ≤

3n f L1 . α

These estimates hold uniformly when R is any compact, convex set. SECOND IDEA: (This is an improvement of a previous idea about summing weak type 1 functions.)

2.1 Hardy Spaces and Maximal Functions

37

Lemma 2.1.4 Suppose that fj (x) ≥ 0, j = 1, 2, . . . , satisfy m{x : fj (x) > α} ≤ 1/α, all α > 0, cj log(1/cj ) = K < ∞. We all j . Also suppose 1 > cj > 0, claim that f = cj fj is weak type 1 and m{x : |f (x)| > α} ≤

2K + 4 . α

Proof Start by supposing that K = 1. Fix α > 0. Let fj = fj1 + fj2 + fj3 for each j , where  fj1  fj2 =

=

fj when fj ≤ α/2 0 otherwise

fj when α/2 < fj ≤ α/(2cj ) 0 otherwise 

fj3 =

fj when fj > α/(2cj ) 0 otherwise

Write f = f 1 + f 2 + f 3 accordingly. Notice that m{x : f > 0} ≤ 3

≤ ≤



  α m x : fj (x) > 2cj

2cj α 2 . α

So 2 + m{x : f 1 + f 2 > α} α   α α 2 ≤ m x : f1 > + m x : f2 > + . 2 2 α

m{x : f (x) > α} ≤

But of course |f 1 | ≤



cj (α/2) ≤ α/2. Hence

 α = m(∅} = 0 . m x : f1 > 2 Finally,

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2 More on Hardy Spaces

  α 2 ≤ m x : f2 > f 2 dx 2 α by Chebyshev’s inequality. And this is  2 ≤ fj dx cj α α/2 r, then |x1 | ∼ |x1 | + r ∼ 2i

40

2 More on Hardy Spaces

2

2

2

2

2

j

j

j

( )

)

(

2

j

Fig. 2.2 The dotted “circle” is (n − 2)-dimensional

and if |x1 | ≤ r, then |x1 | ∼ |x1 | + r − (r − |x1 |) ∼ 2i − 2j ∼ 2i . Hence P(1,0,...,0) (x) ≥

C C ∼ (i+j )n . [(2i+j )2 + 4 · 22i ]n/2 2

This proves the claim. Thus u(x, (1, 0, . . . , 0)) = P(1,0,...,0) ∗ f ≤ C



2−n(i+j )

 f (x − t) dt . Sij

Clearly m(Si,j ) ∼ 2i(n−1)+j . So u(x, (1, 0, . . . , 0)) ≤ C



2−(i+(n−1)j )

1 m(Sij )

 f (x − t) dt .

()

Sij

Now unfortunately each Sij is shaped like a sleeve and there is no maximal theorem for sets of that shape. Hence we must write each Sij as a union of rectangles. Each rectangle will have dimensions 2i × 2j × · · · × 2j . If we write i − j = k, then it is easy to see that this will require essentially C1 · 2k(n−2) rectangles. See Fig. 2.2. Now let us consider a schematic of Sij . It is a sleeve-shaped section of Rn , such that the sleeve is modeled on a cone. Consider Fig. 2.3 Thus no. of rectangles × volume of rectangle ≈ 2(i−j )(n−2) × 2i+j (n−1) ≈ 2i(n−1)+j ≈ m(Sij ) .

2.1 Hardy Spaces and Maximal Functions

41

Fig. 2.3 The sleeve-shaped section of Rn

2

-k

2 - 2

-1

Now let , j be fixed, 0 ≤  ≤ C1 2k(n−2) , and let f∗ be the maximal function N ∗ corresponding to the th rectangle of Sij . Clearly fij∗ ≤ =1 f , where N ≈ C1 2k(n−2) and  1 ∗ fij = sup f (x + t) dt .  m(Sij ) Sij We will use the following version of Lemma 2.1.4: Lemma 2.1.5 Suppose the fj are as in Lemma 2.1.4. Then ⎧ ⎨

⎫ N ⎬ 2 log(N + 2) 1 , α > 0. m x: fj (x) > α ≤ ⎩ ⎭ N α j =1



Proof Trivial from Lemma 2.1.4. Thus m{x : fij∗ (x) > α} ≤

C log(N + 2) α

and, by (), a previous observation, and the lemma, this is C log(C1 2k(n−2) ) α C + Ck(n − 2) ≤ . α

≤

42

2 More on Hardy Spaces

Fig. 2.4 Conical approach

 −(i+(n−1)j ) ∗ 2 An application of Lemma 2.1.4 now shows that fij (x) is weak type 1. Standard interpolation arguments now give Lp estimates for p > 1. A simple dilation now gives sup u(x, (y, 0, . . . 0)) ≤ C



y∈R y>0

2−(i+(n−1)j ) fij∗ (x)

and the function on the right is weak type 1 or Lp according as f ∈ L1 or f ∈ Lp , p > 1. By standard arguments, this estimate gives all the positive results of the theorem when the subcone 0 is chosen to be the axis of . For general 0 the argument is the same, except that one must estimate P more carefully. Now we indicate how the negative results are proved. [For details and greater generality, see [STN]. Let f : Rn → C, u(x, y) = Py ∗ f . What happens when y → 0, but not through a proper subcone? Let dμy = Py (x) dx. If y0 ∈ b, y0 = 0, write y0 = sy0 , |y0 | = 1, s > 0. If we approach y0 through a conical neighborhood of ys as shown in Fig. 2.4, we claim that μy → the one-dimensional Poisson kernel . We now make more precise the sense in which this is true. Suppose that L is a directed line through 0 with e ∈ Rn the unit vector determining L. Suppose that dμ is a finite measure on R1 . We claim that we may concentrate dμ on L in such a way that this determines a singular measure on Rn . Namely, suppose that ϕ ∈ C0 (Rn ). Let  dμL (ϕ) =

 ϕ dμL ≡

∞

−∞

ϕ(es) dμ(s) .

n A measure  dν is said to be supported on L if, whenever ϕ ∈ C0 (R ), (supp ϕ)∩L = ∅, then ϕ dν = 0.

Proposition 2.1.6 We have that dν is supported on L if and only if dν = (dμ)L , some dμ on R1 .

2.2 More Maximal Functions

43

Proof Let dν be a measure on Rn . Define  ν(ξ ) =

Rn

e2π ix·ξ dν(x) .

Then dν is supported on L if and only if ν(ξ1 ) = ν(ξ2 ) whenever ξ1 − ξ2 ⊥ L. But then ν restricted to any line parallel to L is the same so ν can be thought of (via projection) as a function on the line. It follows that if we define dμ on R1 by (dμ) = ν (in the above sense), then dν = (dμ)L . With this definition of (dμ)L , we let dνy = Py (t) dt on R1 and (dηy )L be the concentration of dηy on a line L through 0 ∈ Rn (the x-space). Let μy be as in the discussion preceding the proposition. We wish to show that dμy → (dηS )L as y → y0 , where the choice of L depends on the initial choice of y0 . This will clearly show that unrestricted convergence cannot work, since an f ∈ Lp (Rn ) looks completely different in different directions. 

What the above argument purports to show, in an admittedly vague way, is that the problem of unrestricted convergence for Poisson integrals is equivalent to the problem of differentiation of the integral over long and skinny rectangles. Therefore the following considerations will complete our counterexample.

2.2 More Maximal Functions We define the hyper-maximal function MH by 1 MH f (x) = sup R∈R m(R)

 f (x, t) dt , R

where R = {all rectangles centered at 0} . Let S be a set. We say that a point x ∈ S ⊆ Rn is a point of hyperdensity if lim

diam ρ→0 ρ∈R

m(S ∩ ρ) = 1. m(ρ)

Theorem 2.2.1 We have: (1) For all p < ∞, there exists an f ∈ Lp (R2 ) such that MH f = ∞ almost everywhere. (2) There exists a set E, m(E) > 0, such that no point of E is a point of hyperdensity.

44

2 More on Hardy Spaces

Proof (1) For p < 2, simply let  f (x) =

|x|−1 if |x| < 1 0 if |x| ≥ 1 .

Then f ∈ Lp for all p < 2. Let x ∈ Rn . For rectangles ρ we choose long, skinny rectangles passing through x, 0 such that 1 ρ



1 f (x + t) dt ∼ 2N ρ



N −N

f (x + t) dt .

Since f restricted to a line through 0 is not in L1 (R1 ), this gives (1) for p < 2. The proof for p ≥ 2 we omit. The proof of (2) follows from a construction of Nikodym in Fundamenta Mathematicae 10(1927), 116–168. The observation that (2) follows from Nikodym’s work is contained in a remark of Zygmund at the end of that paper. Nikodym showed that There exists a set E ⊆ unit square ⊆ R2 , m(E) = 1, such that: At any point in the set there exists a ray (half-infinite line) such that the ray does not intersect E.

Now pick E0 ⊆ E closed, m(E0 ) > 0. One can easily see that 1 diam ρ→0 m(ρ)

 f (x + t) dt
E0

 =

y 1−n |∇u|2 dxdydx0

ψ(x0 , x, y) dx0 y 1−n |∇u|2 dxdy . E0

It will suffice to see that

βk (x0 )

5.5 The Area Integral

171

 ψ(x0 , x, y)dx0 ≥ Cy n E0

whenever (x, y) ∈ R. If (x, y) ∈ R, then there exists a z ∈ E such that |x − z| < αy, 0 < y < h. Thus 

 ψ(x0 , x, y) dx0 ≥ E0

E0 ∩{|x0 −z|

S(u)(x0 ) dx0 

≥

E

R

y|∇u|2 dxdy

 ≥

R

(Green)

≈

y|∇u|2 dxdy   C 

  ∂u2 2 ∂y y −u dσ  . ∂η B ∂η

(*)

Here B ≡ B0 ∪ B1 . Note that the points on B1 are bounded from {y = 0} hence the contribution to (∗) from B0 is bounded. Thus we have  ∞>

y B0

∂y ∂u2 − u2 dσ . ∂η ∂η

Now we claim that −

∂y ≥C>0 ∂η

on B0 . In fact ∂/∂η is the outward unit normal to the surface defined by ϕ (x, y) = αy − δ (x) ≡ 0. Thus ∂/∂η is given by     ∂ϕ ∂ϕ ∂δ ∂ϕ ∂δ , − = −α, ,..., ,..., ∂y ∂x1 ∂xn ∂x1 ∂xn normalize to have length 1. So

5 Hardy Spaces on Rn

172

∂y ≤ −α(α 2 + n2 )−1/2 ∂η or ∂y ≥ α(α 2 + n2 )−1/2 , ∂η

− as desired. So we know that



 C1

B0

y∂u2 ∂y − u2 ∂η ∂η

 dσ ≤ C2

or  C1

   y∂u2  ∂y 2  − u dσ ≤ C2 + C1    B0 ∂η  ∂η B0  ≤ C2 + C 1 y|u||∇u| dσ B0

 ≤ C2 + C3

B0

|u| dσ .

()

Let ' J =

(1/2 2

B0

u dσ

.

Then () just says that  C4 J2 = C4

B0

u2 dσ

≤ C2 + C3

 B0

|u| dσ '

 ≤ C2 + C3

B0

|u|

2

dσ1/2

·

B0

(1/2 dσ

≤ C2 + C5 J  , since B0 is rectifiable. It follows that J ≤ C6 , with C6 independent of . Thus  2 B0 u dσ ≤ C. This is what we set to prove a while back. Now we complete the proof of the converse.

5.5 The Area Integral

173

We wish to see that u is nontangentially bounded almost everywhere on E. We will find a V harmonic satisfying the following: (i) V = P I (f ) with f ∈ L2 (Rn ), f ≥ 0. (ii) |u(x, y)| ≤ C · V (x, y) + C1 for (x, y) ∈ R. By the local Fatou theorem, this will do the trick. We use what is at hand to construct V. For each  > 0, set  f (x) =

|u(x, δ (x)/α)| if (x, δ (x)/α) ∈ B0 0 otherwise .

Recall here that B0 = {y = δ (x)/α : 0 < y < h − }. Trivially, dσ ≥ dx so that 

 Rn

|f (x)|2 dx ≤

B0

|u(x)|2 dσ ≤ C

by the estimate we have proved above. Let v (x, y) = P I (f )(x). Our main claim is that, for appropriate C > 0, |u(x, y)| ≤ C(v (x, y) + 1) for (x, y) ∈ R .

(∗∗∗)

By the maximum principle, it suffices to check this inequality on ∂R . The noncritical part of ∂R (bounded away from {y = 0} and compact) is taken care of by making C large enough, in the C · 1 term. So we check (∗∗∗) on B0 . Now B0 ⊆ R = ∪x0 ∈E αh (x0 ) so that there exists a constant c with 1/2 > c > 0 ∗ such that B((x, y), cy) ⊆ ∪x0 ∈E αh∗ for all (x, y) ∈ B0 . Here α < α ∗ < β and h < h∗ < k fixed. Fix p1 = (x, y) ∈ B0 and let p2 be any other point of B = B((x, y), cy). Then |u(p1 ) − u(p2 )| ≤ |p1 − p2 | ·

|∇u(s, t)|

sup (s,t)∈p1 p2

≤ c·y·

sup y/2 α} , α > 0 . Then 

 Rn

∞

|f (x)|p dx =

(parts)



∞

α p dλ(α) = p ·

0

λ(α)α p−1 dα .

0

Lemma 5.5.7 Suppose that u(x, y) is continuous on Rn+1 + . Then, for γ > 0, we have m{x : u∗β1 (x) > γ } ≤ Cβ1 ,β2 m{x : uβ2 ∗ (x) > γ } independent of γ > 0, any β1 , β2 > 0. The essence of this lemma is that u∗β1 Lp ≈ u∗β2 Lp . Proof of the Lemma This is only interesting when β1 > β2 . Now fix γ temporarily. Let E = {u∗β2 > γ }. Let χE be the characteristic function of E. Set E ∗ = {x : MχE (x) > C} for a fixed constant C and M the Hardy–Littlewood maximal function. Of course m(E ∗ ) ≤ We claim that

5n · m(E) . C

{x : u∗β1 > γ } ⊆ E ∗

for the right choice of C. This is because u∗β1 (x) > γ means that there exists a p = (a, b) ∈ β1 (x) such that |u(p)| > γ . See Fig. 5.5. But then u∗β2 (t) > γ for all t in the upside down cone at (a, b), that is all t so that |t − a| < β2 b. But then m(B(a, β2 b) m(E ∩ B(x, |a − x| + β2 b) ≥ m(B(x, |a − x| + β2 b) m(B(x, |a − x| + β2 b)) ≥

(β2 b)n (β1 b + β2 b)n

≥

β2n . (β1 + β2 )n

5 Hardy Spaces on Rn

176

(a,b) (x)

upside down

1

2

(a,b)

x Fig. 5.5 The approach region β1 (x)

Thus if we let C=

β2n , (β1 + β2 )n

then we find that p ∈ {x : MχE (x) > C} so that m{x : u∗β1 > γ } ≤

5n · (β1 + β2 )n · m{x : u∗β2 (x) > γ } . β2n 

That proves the result. A reference for this work is Fefferman and Stein [FES].

Now we prove the theorem. We restrict attention to p < 2. The case p ≥ 2 is quite different, has been known longer, and can be handled with different techniques. It can also be handled by a more elaborate version of the same method, c.f. [STE1]. We will suppose that the situation has been uniformized in the following way: (1) u = P I (f ), f ∈ L2 . (2) The cone which defines u∗ is strictly larger than that which defines S(u). Say they are of apertures β > α. Taking inspiration from the lemma, we estimate that m{x : S(u) > λ} in terms of m{x : u∗ (x) > λ}. Let E = {x : u∗ (x) ≤ λ}

and

B = c E = {x : u∗ (x) > λ} .

Then E is closed and B is open. Let R = ∪x∈E α (x). As usual, we have    2 2 S (u)(x) dx ≤ y|∇u| dxdy = lim y|∇u|2 dxdy . E

R

→0

R

5.5 The Area Integral

177

Fig. 5.6 The boundary sets BE and BB

B

E Again we introduce the approximating regions so that it is legitimate to apply Green’s theorem. But  y|∇u|2 dxdy

(Green)

=

R

1 2

 ∂ R

y

∂u2 1 dσ − ∂η 2



∂y 2 u dσ . ∂ R ∂η

(∗)

Now since we have assumed that u = P I (f ), f ∈ L2 , we have supy |u(x, y)| ≤ AMf (x) ∈ L2 . In addition, since the kernel of y∇u is y · ∇Py , which is essentially an approximation to the identity, we have sup y|∇u| ∈ L2 .

()

y

Also y|∇u(x, y)| → 0 almost everywhere since this is easy to check for f ∈ Cc and () tells us that the associated maximal operator is bounded. Write ∂R = B = BE ∪ BB , where of course BE is the portion which lies over E and BB is the portion which lies over B. See Fig. 5.6. But over BB we know that |u| ≤ λ so |y∇u| ≤ Cλ. Thus the contribution to (∗) from BB is ≤ Cλ2 m(B). Therefore         ∂y 2 ∂u2     S (u)(x) dx ≤ Cλ m{x : u (x) > λ} +  y dσ  +  u dσ  .     E E ∂η ∂η   E B B



2

2

∗

Since y∇u → 0 as y → 0, the second term will die with . The last term is dominated by E |u∗ |2 dx. Altogether then, 

∗



S (u)(x) dx ≤ Cλ m{x : u (x) > λ} + 2

E

Clearly

2

E

|u∗ |2 dx .

5 Hardy Spaces on Rn

178



(u∗ (x))2 dx ≤ 2



λ

α · m{x : u∗ (x) > α} dα .

0

E

So 

S 2 (u)(x) dx ≤ Cλ2 · m{x : u∗ (x) > λ} + 2



λ

α · m{x : u∗ (x) > α} dα .

0

E

Now m ({x : |S(u)(x)| > λ} ∩ E) ≤



1 λ2

S 2 (u)(x) dx , E

so m{x : |S(u)(x)| > λ} ≤

1 λ2

 S 2 (u)(x) dx + m(c E) .

(∗∗)

E

But m(c E) = m(B) = m{u∗ > λ} . Altogether then, m{S(u) > λ} ≤ C · m{u∗ > λ} + Thus   S(u)p dx = p

∞

C λ2





α · m{u∗ > α} dα .

0

λp−1 · m{Su > λ} dλ

0

∞

≤ Cp

λ

λp−1 m{u∗ > λ} dλ + Cp

0 ∗



≤ Cu Lp + Cp 0

∞ ∞



∞



λ

λp−3

0

α · m{u∗ > α} dαdλ

0

λp−3 dλα · m{u∗ > α} dα .

α

For p < 2, the second integral in λ converges at ∞ so this is ≤ Cu∗ Lp +

Cp 2−p



∞

α p−1 m{u∗ > α} dα

0

≤ Cp u∗ Lp . This completes the proof.



5.6 Applications of the Maximal Function Characterization

179

Γα Γβ Γy x Fig. 5.7 Comparison of cones

We omit the proof of the converse, the removal of the condition u = P If , f ∈ L2 . The lemma clears up the cone condition. Note that the assumption u = P If was used to control u at ∞ so that we could apply Green’s theorem. Burkholder and Gundy have a proof of the above result in 1972 in Studia. We will only use this result for p ≤ 1.

5.6 Applications of the Maximal Function Characterization Suppose that u and v are conjugate harmonic functions on R2+ . Hence |∇u|2 = |∇v|2 . The above theorem enables us to pass from u to v or, more precisely, from u∗ to v ∗ . This is because u∗ Lp ≤ Cp SuLp = Cp SvLp ≤ Cp v ∗ Lp and conversely. In higher dimensions, we have u, v1 , . . . , vn with ∂vj ∂u . = ∂xj ∂y There is no simple relationship between grad u and grad vj . So we need an additional tool to relate u, vj via the area integral. See Fig. 5.7. Lemma 5.6.1 Let u be harmonic in Rn+1 + and bounded on some halfspace y ≥ y0 . Then  y α

 2     ∂y  dxdy ≤ Cα,β

1−n  ∂u 

β

y

  ∂u  ∂x

1−n 

j

2   dxdy . 

5 Hardy Spaces on Rn

180 Fig. 5.8 The vector ρ in the cone β

Γα ϱ

Γβ

[We assume here that ∂u/∂xj → 0 as y → ∞ in the cone β .] The motivation for the lemma is that, if we know the area integral for u, then we know ∂u/∂xj , hence we know ∂vj /∂y and, by the lemma, we therefore know ∂vj /∂xk . We note that, for n = 1, it suffices to let α = β. It is unknown whether this can be done for n > 1. Outline of the Proof of the Lemma Let ρ be a unit vector in α . We prove the lemma for each ρ, then integrate out over ρ. Refer to Fig. 5.8 as we proceed. So we will prove that  2     ∂u (ρs) s ds ≤ Cα,β  ∂x  j β

∞ 

 0

 2  ∂u  1−n   y dxdy .  ∂y 

As long as ρ ∈ α , then s and the y component of ρs are comparable. So, integrating both sides of the inequality over S n ∩ α , yields the result. Now fix ρ once and for all. Write ∂u(x, y) =− ∂xj



∞ y

∂ 2 u(x, τ ) dτ . ∂y∂xk

The integral converges since u is bounded. As usual, there exists C > 0 such that (x, τ ) ∈ α implies that B((x, τ ), Cτ ) ⊆ β . Call this ball Bτ . Clearly Bτ is contained in the strip Sτ = {(1 − C)τ ≤ y ≤ (1 + C)τ }. We claim that   2     ∂ u(x, τ ) 2  ∂u(x  , y  ) 2  dx  dy    ≤ Cτ −2 1    ∂y∂x  m(Bτ ) Bτ  ∂y k     ∂u(x  , y  ) 2  dx  dy  .  = Cτ −n−3   ∂y Bτ

(*)

This follows by convolving u with a radial bump function ϕτ such that ϕ ≥ 0,  supp ϕ ⊆ B(0, 1/2), ϕ = 1, ϕ ∈ C ∞ . Harmonicity implies that u ∗ ϕ = u and

5.6 Applications of the Maximal Function Characterization

181

differentiation under the integral sign yields (∗). Thus  2    ∂ u(x, τ ) 2   ≤ C · τ −n−3  ∂y∂x  k

Sτ ∩β

 2  ∂u    dx  dy  .  ∂y 

In conclusion,     ∂u   ≤C (ρs)  ∂x  j

∞ cs

τ (−n−3)/2 Iτ1/2 dτ ,

where  2  ∂u    dx  dy  .  ∂y 

 Iτ =

Sτ ∩β

We need a special case of Hardy’s inequality: ∞ (a) If (s) = s ϕ(τ ) dτ , then 

∞



∞

|(s)| s ds ≤ C 2

0

|ϕ(τ )|2 τ 3 dτ .

0

Proof We see that 

∞

|(s)| ≤

|ϕ(τ )| · τ · τ −1 dτ

s

so, by Schwarz, 

∞

|(s)|2 ≤ 



s

s ∞

≤ s

∞

|ϕ(τ )|2 τ 2 dτ ·

1 dτ τ2

1 |ϕ(τ )|2 τ 2 dτ · s

or 

∞



0



∞

|(s)|2 s ds ≤

dx 0

∞

|ϕ(τ )|2 τ 2 dτ

s

 =

|ϕ(τ )|2 τ 2 dτ 0≤s≤τ

=

1 2



0

∞

|ϕ(τ )|2 τ 3 dτ .

()

5 Hardy Spaces on Rn

182



That proves Hardy’s inequality. Applying this last result to () we get  0

 2    ∂u (ρs) s ds ≤ C  ∂x 

∞

j

∞

0

 =C

τ −n−3 · Iτ · τ 3 dτ

   ∂u   2  (x , y ) dx  dy  τ −n dτ   Sτ ∩β ∂y    ∂u   2  (x , y ) · |y  |1−n dx  dy  ,  ∂y 

∞

0

 ≤C β

as desired. We will have considerable use for the following fact: Proposition 5.6.2 If u ∈ H p (in the sense that (u, R 1 u, . . . , R n u) ∈ H p ), 0 < p < ∞, then u∗ ∈ Lp and u∗ Lp ≤ Ap uH p . Proof We do the proof for p > (n − 1)/n only. Note that the case p > 1 is obvious. Now we know that  (n−1)/(2n)   |uj (x, y + )|2   is subharmonic. Therefore !

|uj (x, y + )|2

"(n−1)/(2n)

 ≤

Py (x − t)

!

|uj (t, )|2

"(n−1)/(2n) dt

since  the left side is subharmonic and the right side harmonic. But the ( |uj (t, )|2 )(n−1)/(2n) are uniformly bounded in Lp/((n−1)/n) so there exists a weak-∗ subsequence converging to some h(x). If h(x, y) = P I h, then it follows that !

|uj (x, y)|2

"(n−1)/(2n)

≤ h(x, y)

and  sup

y>0 Rn

q

p

h(x, y)p/((n−1)/n) dx = hLq (Rn ) = uH p ,

where we let q = p/((n − 1)/n) > 1. So we have u∗ (x) ≤ h∗ (x)n/(n−1) ≤ C(Mh(x))n/(n−1) , where M is the Hardy–Littlewood maximal operator. So

5.6 Applications of the Maximal Function Characterization

 Rn

|u∗ (x)|p dx ≤ C

 Rn

183

 |Mh(x)|q dx ≤ C

Rn

p

hq dx = uH p , 

as desired.

Before continuing, we present an outline of where we are going and what the remainder of these notes will contain. I. Real Variable Theory: Let us consider H p (Rn ). Here we can characterize H p without appealing to any notion of conjugate harmonic function. II. Several Complex Variables: (a) The polydisc. One of the key features is the presence of a distinguished boundary, different from the topological boundary. (b) The unit ball in Cn . The boundary and the distinguished boundary coincide. The boundary is smooth and pseudoconvex. (c) The domains of Cartan. These have some of the characteristics of (a) and (b). The boundaries turn out to be “semi-distinguished.” Returning to the subject at hand, we proceed by giving a characterization of H p . We begin with H p (D). In what follows, T is a fixed triangle in D with vertex at 1 and Tθ is the triangle obtained by rotating T to have vertex at θ . Theorem 5.6.3 If F ∈ H (D) with F = u + iv, u, v real, v(0) = 0, and if u∗ (θ ) = supz∈Tθ |u(z)|, then F ∈ H p if and only if u∗ ∈ Lp , 0 < p < ∞, and in this case, F H p ≈ u∗ Lp . Remark 5.6.4 This theorem of Burkholder, Gundy, and Silverstein was originally proved by means of Brownian motion and represents the first theorem of classical analysis which was proved by these methods. In the classical setting, the area integral takes the form  S 2 (u)(θ ) =

|∇u|2 (z) dxdy . θ

Then the area theorem for harmonic functions in D is this: Theorem 5.6.5 Let u be harmonic in D, 0 < p < ∞. Then (1) If u∗ ∈ Lp , then S(u) ∈ Lp and Sup ≤ Cp u∗ p . (2) If u(0) = 0 and S(u) ∈ Lp , then u∗ ∈ Lp and u∗ p ≤ Ap Sup . Let us show how to obtain Theorem 5.6.3 from Theorem 5.6.5. Suppose that u ∈ H (D) and u∗ ∈ Lp . By the area theorem, S(u) ∈ Lp . Now the crucial point is that S(u) = S(v) since |∇u|2 = |∇v|2 . So S(v) ∈ Lp . By the area theorem, since v(0) = 0, v ∗ ∈ Lp and v ∗ Lp ≤ Ap u∗ Lp . (Here v is the conjugate of u.)

5 Hardy Spaces on Rn

184

So F = u + iv ∈ H p , F H p ≤ Ap u∗ Lp . So the central point is that the area integral is invariant under the Hilber transform. We already know that F ∗ Lp ≤ Ap F H p . This gives the converse. Now we have a characterization of H p independent of any notion of conjugate harmonic function. We shall seek another characterization of the following type: The class of distributions on the circle S which corresponds to the H p functions on D are those distributions whose maximal function associated with any regularization are in Lp . p Now we can characterize H p (Rn+1 + ) in the same way we did H (D). We have Theorem 5.6.6 (1) Let F = (u0 , u1 , . . . , un ), u = u0 , be a system of conjugate ∗ p harmonic functions in H p (Rn+1 + ), and p > (n − 1)/n. Then u ∈ L and ∗ u p ≤ Ap F p . (2) Conversely, suppose that u is harmonic on Rn+1 and u∗ ∈ Lp , p > (n − + so that F = 1)/n. Then there exist harmonic functions u1 , . . . , un in Rn+1 + n+1 p ∗ p p (u, u1 , . . . , un ) ∈ H (R+ ) and F H ≤ Cp u L . n+1 Recall the area theorem for Rn+1 + : If u ∈ h(R+ ), then

(1) For u∗ ∈ Lp , SuLp ≤ Ap u∗ Lp , 0 < p < ∞. (2) If u(x, y) → 0 as y → ∞ and S(u) ∈ Lp , then u∗ ∈ Lp and u∗ Lp ≤ Cp S(u)Lp . In order to prove this last theorem, we need the earlier lemma and also this coming lemma: Lemma h(Rn+1 + ) satisfies  5.6.7 (i)p Suppose that 0 < p < ∞ and u ∈ −n/p |u(x, y)| dx ≤ A for all y > 0. Then |u(x, y)| = O(y ). n R −β , some β ∈ R, then (ii) If u ∈ h(Rn+1 + ) and |u(x, y)| ≤ Ay    ∂ 2 α    u ≤ Ay −β−|α|   ∂x∂y  for any multi-index α. Now we will see how the preceding theorem follows from these two lemmas. One direction is clear, for F ∈ H p , F = (u0 , u1 , . . . , un ) implies that u∗ ≤ F ∗ ∈ Lp and u∗ Lp ≤ Ap F H p . Conversely, if u is harmonic, bounded near ∞, and u∗ ∈ Lp , p > (n − 1)/n, then by the second lemma, |u(x, y)| ≤ Ay −n/p and |∂u/∂xk | = O(y −n/p−1 ). If we let  ∞ ∂u uj (x, y) = − (x, y + s) ds , ∂xj 0 then it is easy to check that (u, u1 , . . . , un ) is a Cauchy–Riemann system which we call F . Now clearly |uj (x, y)| ≤ Ay −n/p and, since u∗ ∈ Lp , we get, by the area

5.6 Applications of the Maximal Function Characterization

185

theorem, that S(u) ∈ Lp . Let Sj2 (x)

   ∂u  =  ∂x

j

2  1−n  y dxdy . 

β (x)

Then Sj ≤ S implies Sj (u) ∈ Lp . Thus if we let Sy2 (uj )(x)

    ∂uj 2 1−n   = dxdy ,  ∂y  y β (x)

then Sy (uj ) = Sj (u) ∈ Lp . But, by the first lemma, it then follows that if we define  S (uj ) =

|∇uj |2 y 1−n dxdy ,

2

α (x)

then S(uj ) ∈ Lp and S(uj )p ≤ Ap u∗ Lp . By the area theorem, therefore, u∗j p ≤ Ap u∗ p so that F ∈ H p and F H p ≤ Ap u∗ p . We have already proved the first theorem, so now we prove the second. We will use the mean value property. We know that 



u(x, y) =

u(s, t) dsdt · B((x,y),y/2)

1 . m(B((x, y), y/2))

By Jensen’s inequality, |u(x, y)|p ≤ ≤

1 m(B) C

 |u(s, t)|p dsdt 

y n+1

B

|u(s, t)|p dsdt y/2≤t≤3y/2

≤ C  y −n or |u(x, y)| ≤ C  y −n/p . By a similar argument, if |u(x, y)| ≤ Cy −α , and D is a differential monomial of order k, ϕ a real bump function, ϕ ≥ 0, supp ϕ ⊆ B(0, 1/2), and ϕδ (x) = δ −n−1 ϕ(x/δ), then write u(x, y) = ϕy ∗ u(x, y) .

5 Hardy Spaces on Rn

186

Differentiation under the integral sign yields the desired result.



Now we come to a very tricky lemma of Hardy and Littlewood. Lemma 5.6.8 Fix 0 < p < ∞. If u is harmonic on a ball B(0, r) ⊆ Rn , and if u ∈ C(B), then there exists Cp > 0 so that |u(0)|p ≤

Cp m(B)

 |u(x)|p dx . B

Proof Assume that B is the unit ball in Rn . Further assume that p < 1, otherwise the assertion is trivial by convexity (with Cp = 1). Finally, assume that 1 m(B)

 |u|p = 1 . B

For 0 < r < 1, denote by  mp (r) =

S n−1

1/p |u(rξ )|p dσ (ξ ) , 0 < p < ∞.

Also set m∞ (r) = sup |u(rξ )| . ξ ∈S n−1

We may suppose that m∞ (r) ≥ 1, all 0 ≤ r ≤ 1, since in case m∞ (r) < 1 for some r then |u(0)| ≤ 1 so |u(0)|p ≤ 1 =

1 m(B)

 |u|p . B

Claim (a) There exists a θ , 0 < θ < 1, such that m1 (r) ≤ (mp (r))1−θ · (m∞ (r))θ . Proof of the Claim We have 

 S n−1

|u(rξ )| dσ (ξ ) =

S n−1

|u|p |u|1−p dσ (ξ )

 ≤

|u|p dσ · sup |u(rξ )|1−p ξ ∈S n−1

= (mp (r))p · (m∞ (r))1−p .

5.6 Applications of the Maximal Function Characterization

187

So θ = 1 − p will do. Let us restrict attention to 1/2 < r ≤ 1. Let 0 < ρ < r. Then m∞ (p) = sup |u(ψ)| ≤ Pρr (ξ )L∞ · m1 (r)L1 , ξ ∈S n−1

where P r is the Poisson kernel for B(0, r), which is ≤ C(1 − ρ/r)−n+1 · m1 (r) . We now let ρ = r a , a > 1, where the constant a will be chosen in a moment. So we have m∞ (r a ) ≤ C(1 − r a−1 )−n+1 · m1 (r) .

Claim (b) We have 

1 1/2

dr ≤ Ca,θ + (1 − θ ) r

log m∞ (r a )



1

log mp (r) 1/2

dr +θ r



1 1/2

log m∞ (r)

dr . r

Proof of Claim (b) We see that 

1

   r a  dr  log C + log 1 −  · |n − 1| + log m1 (r) r r 1/2  1 dr ≤ log C · log 2 + C  + log m1 (r) r 1/2  1  1 dr dr + . ≤ C  + (1 − θ ) log mp (r) θ log m∞ (r) r r 1/2 1/2

dr ≤ log m∞ (r ) r 1/2 a





1

This is our new claim. Thus 

1

1/2

log m∞ (r a )

dr ≤ C  + r



1 1/2

θ log m∞ (r)

dr r

()

since 

1

dr ≤K (1 − θ ) log mp (r) r 1/2



1

dr ≤ K + K (1 + mp (r)) r 1/2



|u|p ≤ K  . B

5 Hardy Spaces on Rn

188

By a change of variables in (), 1 a



1

dr ≤ C  + θ log m∞ (r) r 1/2



1

log m∞ (r)

1/2

dr . r

Now we merely choose a > 1 so that 1/a > θ = 1−p. Note that log m∞ (r) ≥ 0 since m∞ (r) ≥ 1 by assumption. So 

1 1/2

log m∞ (r)

dr ≤ C  . r

Thus, for some r, log m∞ (r) ≤ C  or m∞ (r) ≤ C ∗ ≡ exp(C  ) . But then |u(0)|p ≤ (C ∗ )p = (C ∗ )p ·

1 m(B)



B

|u(x)|p dx .

0

This completes the proof of the Hardy–Littlewood lemma. Now we make a sequence of remarks about



u∗ .

∗ p (1) Let 1 < u harmonic on Rn+1 + . Then u ∈ L if and only if  p < ∞ and p supy>0 Rn |u(x, y)| dx < ∞. (2) Let (n + 1)/n < p < ∞ and u harmonic. Then u∗ ∈ Lp if and only if there exists (u0 , u1 , . . . , un ) a Cauchy–Riemann system with u = u0 such that

 sup y>0

⎛ ⎝

n

⎞p/2 |uj (x, y)|2 ⎠

dx < ∞ .

j =0

(3) Let us push this one step further, to (n − 1)/(n + 1) < p < ∞. Suppose ∗ p that u is harmonic on Rn+1 + . Then u ∈ L if and only  if there exist uij (x, y), 0 ≤ i, j ≤ n, such that u(x) = u00 , uij = uj i , and i uii = 0. Moreover, if we define uij k =

∂ uij , ∂xk

5.6 Applications of the Maximal Function Characterization

189

then we require that the uij k be symmetric in i, j, k and have vanishing trace. (these are the substitute for the Cauchy–Riemann equations.) We also require that ⎛



⎝ Rn



⎞p/2 |uij k (x)|2 ⎠

dx < ∞ .

i,j,k

(4) There is an analogous definition, using tensors of rank k, for the case p > (n − 1)/(n − 1 + k). Note that the case k = 1 coincides with (2). (4’) Recall that, in the classical case, the square of the Hilbert transform is −I so polynomials in the Hilbert transform are quite simple. This is not the case on Rn+1 + . So a Cauchy–Riemann system is more complicated. + Now define, for u ∈ h(Rn+1 + ), u (x) = supy>0 |u(x, y)|. Clearly

u+ Lp ≤ u∗ Lp . However, we will prove that u∗ Lp ≤ Cp u+ Lp , 0 < p < ∞ . There is an example due to Bagemihl and Seidel which distinguishes radial and nontangential convergence. In fact they show that, given f ∈ C(D) and H ⊆ S of first category, with S = ∂D, there exists an F ∈ H (D) such that |F −f | → 0 along all radii through H. The case for nontangential convergence is quite different. In fact, if F ∈ H (D) and F has nontangential limit 0 on a set of positive measure, then F ≡ 0. Now we will prove that Cp [M(u+ (x))p/2 ]2/p ≥ u∗ (x) . In fact, with  fixed as usual, we have, for (z, y) ∈ (x), that |u(x, y)|p/2 ≤ Cp

1 m(B((x, y), y/2)



|u|p/2 dx  dy 

B((x,y),y/2)

by the Hardy–Littlewood lemma. And this last is 1 ≤ Cp m(B) ≤ Cp

1 y n+1



|u+ (x  )|p/2 dx  dy  B

·y·

 |x  −x|0 Rn |u(x, y)|p dx < ∞, 0 < p < ∞. Then limy→0 u(x, y) = f as a tempered distribution and u is uniquely determined by f . That is, there exists a unique f a tempered distribution so that  lim u(x, y)ϕ(x) dx = f (ϕ) for all ϕ ∈ S . y→0

Proof The lemma is well known for p ≥ 1. For p > 1, the convergence is in Lp and for p = 1 the convergence is in the weak-∗ topology of finite measures. We restrict attention to p ≤ 1. Let uδ (x, t) = u(x, t +δ), δ > 0, for (x, t) ∈ Rn+1 + . By the second lemma above, we know that uδ (x, t)∞ ≤ A · δ −n/p . This, combined with the pth power integrability hypothesis, implies that  Rn

|uδ (x, t)| dx ≤ Cδ for all δ ,

with Cδ independent of t. Thus uδ is the Poisson integral of a finite measure μδ , with μδ the weak-∗ limit of uδ (x, t) as t → 0. It follows by continuity of u that

5.6 Applications of the Maximal Function Characterization

191

μδ = uδ (x, 0) = u(x, δ) . Using the Fourier transform and corresponding identities for the Poisson kernel, we have uδ (ξ, t) = uδ (ξ, 0) · e−2π |ξ |t . That is to say, u(ξ, t + δ) = uδ (ξ, 0) · e−2π |ξ |t ≡ u0 (ξ )e−2π |ξ |(t+δ) .

(∗)

Trivially this definition of u0 (ξ ) is unambiguous in δ. Note that u0 is a continuous function. Moreover, we know that     −2π |ξ |t  |u(x, t)| dx u0 (ξ )e ≤ Rn

 ≤

Rn

|u(x, t)|p · t −(n/p)·(1−p) dx

≤ C · t n(1−1/p) . In conclusion, |u0 (ξ )| ≤ C · t n(1−1/p) · e2π |ξ |t for all t > 0 . Letting t = 1/|ξ | gives |u0 (ξ )| ≤ C|ξ |n(1/p−1) . Hence u0 is of polynomial growth. Let f = (u0 ) ˇ distributions. By (∗), for any ϕ ∈ S, 

 Rn

u(x, t)ϕ(x) dx =

Rn

=f

in the sense of tempered

u0 (ξ )e−2|ξ |π t ϕ(−ξ ) dξ

! " ˇ . e−2π |ξ |t · ϕ(−ξ )

Since ( −2π |ξ |t ϕ) ˇ → ϕ(ξ ) uniformly on compact sets as t → 0, we obtain  Rn

as desired.

u(x, t)ϕ(x, t) dx → f (ϕ) 

5 Hardy Spaces on Rn

192

Question Which tempered distributions arise as boundary values of H p functions? The answer is: Theorem 5.6.10 Suppose that f is a tempered distribution on Rn . Then the following are equivalent for 0 < p < ∞: (1) f = limy→0 u(x, y) in the distribution sense, where u is harmonic and u∗ ∈ Lp . p (2) For all ϕ ∈ S with ϕ (x) =  −n ϕ(x/), we have sup>0  (f ∗ ϕ )(x) ∈ L . −n (3) There exists a  ∈ S with  (x) =  (x/),  dx = 0, such that sup>0 (f ∗  ) ∈ Lp . The proof of this theorem is going to take some doing, so we begin with some remarks and discussion. We ask the reader to bear in mind that the next fifteen or more pages will be dealing with this theorem. Remark 5.6.11 Recall that a set B ⊆ S is bounded if B ⊆ {ϕ ∈ S; ϕα,β ≤ C for some C and for finitely many α, β} . Now part (2) of the theorem can be given a more precise statement. Fix 0 < p < ∞. Then f ∈ H p if and only if there exists a bounded set B ⊆ S such that sup sup |f ∗ ϕ | ∈ Lp .

ϕ∈B >0

We claim that the size of B required, that is the amount of control on the smoothness of elements of B, is   1 − 1 , 0 < p ≤ 1. (order of differentiability) > n · p This will be seen. This constant has already come up in an earlier lemma. Example 5.6.12 (1) The Poisson kernel for Rn+1 + is c . (1 + |x|2 )(n+1)/2 This is not a Schwartz function, but we know from the previous pages that this kernel will work. (2) The characteristic function of the unit ball will work for p ≥ 1, even though it has no smoothness. Note that in this case the corresponding maximal function is

5.6 Applications of the Maximal Function Characterization

  1 sup  n 

|y|≤

>0

193

  f (x − y) dy  .

This is dominated by the Hardy–Littlewood maximal function. So it is bounded on Lp = H p , p > 1. Will this work for p ≤ 1? For p < 1, no, since the boundary value of u ∈ H p , p < 1, is, in general, a distribution and we cannot talk about the absolute value of a distribution. What about p = 1? Well, the class of L1 functions such that sup >0

1 n

 |y|≤

|f (x − y)| dy ∈ L1

is locally the same as L log L, which is not the same as H 1 . These questions have close ties with the theory of singular integral operators and pseudo-differential operators. See [FES] and also [STE1]. Consider the Riesz transform (Rj f ) (ξ ) =

iξ f (ξ ) . |ξ |

It is the case that f ∈ H p implies Rj f ∈ H p , j = 1, . . . , n. Thus Pf ∈ H p for any polynomial P in the Rj . This is clear for p > 1, will follow from the duality of H 1 and BMO for p = 1, and follows from characterizations of the space H p for p < 1 in terms of maximal operators. Recall that the H p are defined in terms of the Riesz transforms. Now we prove the theorem. We show that, if we know (2) for all testing functions, then we know (2) for the Poisson kernel. One can do this by piecing together P by weighted Schwartz functions as in the Fefferman–Stein paper. But there is a simple trick which gives the result. 2 Let us assume (2). Choose the particular Schwartz function ϕ(x) = e−π |x| . Then 2 ϕ(ξ ) = e−π |ξ | . Now Py (ξ ) = e−2πy|ξ | and, just as (Py ∗ f ) (ξ ) = e−2π |ξ |y f (ξ ) , we also have (ϕ4πy ∗ f ) (ξ ) = e−4π

2 y|ξ |2

f (ξ ) .

(∗)

5 Hardy Spaces on Rn

194

Given a locally integrable f , if we let Wy f denote ϕ4πy ∗ f , then ∂Wy f (x) ∂ 2 Wy f (x) = . ∂y ∂xj2 n

j =1

This is the heat equation and ϕ4πy is the heat kernel. Now, assuming (2), we have supy>0 |Wy f | ∈ Lp . Now we claim that 

∞

Py (x) =

K(u)ϕπy 2 /u (x) du

0

for some function K ≥ 0 with

∞ 0

K(u) du = 1. We will use the classical formula

1 e−α = √ π



∞

0

e−u −α 2 /(4u) du √ e u

(see [STW1]). With α = 2π |ξ |y, this says e

−2π |ξ |y

1 =√ π

 0

∞

e−u −π 2 |ξ |2 y 2 /u du = √ e u



∞

0

1 e−u √ √ π u

 ϕ πy 2 /u (ξ ) du ,

that is to say,  Py (x) =

∞

K(u)ϕπy 2 /u (x) du

0

with 1 e−u K(u) = √ √ . π u From this identity it follows easily that  sup |f ∗ Py (x)| ≤

K(u) sup |ϕt ∗ f (x)| du

y>0

t>0

≤ sup |ϕt ∗ f (x)| ∈ Lp . t>0

Thus, if u = Py ∗ f , then we have u+ ∈ Lp so that u∗ ∈ Lp and therefore u ∈ H p . Thus we have learned that the theory of H p in terms of boundary values of the heat equation instead of Laplace’s equation is no different from the classical theory. Sometimes the heat kernel is nicer since the product of one-dimensional kernels gives the n-dimensional kernel. Now we show that (1) ⇒ (2). We remark that the Fefferman–Stein paper has a mistake in this segment of the proof.

5.6 Applications of the Maximal Function Characterization

195

Definition 5.6.13 Suppose that u(x, y) is defined on Rn+1 + and continuous. Let u+ (x) = sup |u(x, y)| , y>0

u∗ (x) = sup |u(x − z, y)| , |z| n/p, then u∗∗ r ∈L .

(b) 

p (u∗∗ r (x)) dx ≤ Cr,p

 Rn

(u∗ (x))p dx .

Proof Let Eα = {x : |u∗ (x)| > α}. Then, of course,  Rn

(u∗ (x))p dx = p



∞

α p−1 m(Eα ) dα .

()

0

Let Eα∗ = {x : u∗N (x) > α}. We wish to see that m(Eα∗ ) ≤ CN n m(Eα ) . Let χEα be the characteristic function of Eα . We claim that, for C small enough,   C Eα∗ ⊆ M(χEα )(x) > n . N Thus we will have

5 Hardy Spaces on Rn

196

x

x

Fig. 5.9 Comparison of x and x

m(Eα∗ )



C ≤ m M(χEα (x)) > n N

 ≤

m(Eα ) · N n . C

Then () will imply (a) of the lemma. So we prove the claim. In fact we prove the contrapositive: M(χEα )(x) ≤ N −n ⇒ u∗N (x) ≤ α . See Fig. 5.9. For C we choose C=

1 2



N N +1

n .

We claim now that, for each z with |z − x| < Ny, y > 0, there exists an x with |x − z| < y and u∗ (x) ≤ α. For if not then there exists a z with |x − z | < Ny and u∗ (x) > α for all x with |x −z | < y. This means that, if we look at B(x, (N +1)y), then M(χEα )(x) ≥

m[B(x, (N + 1)y) ∩ Eα ] m[B(x, (N + 1)y)]

≥

m[B(z , y)] Vn (N + 1)n y n

=

Vn y n Vn (N + 1)n y n

=

1 (N + 1)n

= 2CN −n > CN −n ,

5.6 Applications of the Maximal Function Characterization

197

which is a contradiction. This prove the claim. Now if u∗N (x) > α, then there exist (z, y) satisfying |x −z| < Ny and |u(z, y)| > α. Therefore, for all x satisfying |x − z| < y we have u∗ (x) > α. This contradicts the claim which we have just proved. So u∗N (x) ≤ α as desired. Now part (b) of the lemma follows from part (a). Namely,  sup |u(x − z, y)| · z,y

y y + |z|

r

)



y ≤ sup |u(x − z, y)| · sup y + |z| 0≤k 1. Problem Return to non-isotropic dilations. One can define BMO in this setting. Is BMO now the dual of some H 1 space? Can one isolate the right Riesz transforms? Can one show that f ∈ BMO if and only if f = f0 + R j fj , fj ∈ L∞ ? More specifically, can one do this theory on the Heisenberg group? Here the Heisenberg group (simplest case) can be thought of as ⎧⎛ ⎞⎫ ⎨ 1 x1 x3 ⎬ ⎝ 0 1 x2 ⎠ ⎩ ⎭ 0 0 1 under matrix multiplication. (See [FOS] and [KRA1] for more on the Heisenberg group.) And the appropriate dilations, which are automorphisms of the group, are ⎛

⎞ ⎛ ⎞ 1 x1 x3 1 δx1 δ 2 x3 αδ : ⎝ 0 1 x2 ⎠ −→ ⎝ 0 1 δx2 ⎠ . 0 0 1 0 0 1

Chapter 6

Developments Since 1974

Nearly fifty years have passed since the conclusion of Stein’s 1973–1974 course. Since those years marked the very beginning of the modern theory of real variable Hardy spaces, it is natural to expect that there have been many new developments. In the present chapter we shall briefly describe just some of the landmarks.

6.1 The Atomic Theory One of the most decisive new ideas that has come about since the conclusion of Stein’s class is the atomic decomposition for Hardy spaces. Based on a remark of C. Fefferman, the first paper in the subject was [COI]. This was quickly followed by [LAT]. The idea of atoms was brought into full bloom by the more substantial paper [COW]. We now give some definitions and state a few definitive results. A function a : RN → C is called a 1-atom if • a is supported in a ball B(P , r). • |a(x)| ≤ |B(P1 ,r)| .  • a(x) dx = 0. The basic theorem about these atoms is as follows: Theorem 6.1.1 A function f : RN → C is in H 1 if and only if f =



αj aj ,

j

where each aj is a 1-atom and {αj } ∈ 1 and f H p ≈ {αj }1 . © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Krantz, The E. M. Stein Lectures on Hardy Spaces, Lecture Notes in Mathematics 2326, https://doi.org/10.1007/978-3-031-21952-8_6

227

228

6 Developments Since 1974

It is easy to see that this theorem is easily generalized to spaces of homogeneous type (see [COW]). Matters are a bit more subtle for H p with 0 < p < 1. For such p we define a : RN → C to be a p-atom if and only if • a is supported in a ball B(P , r). • |a(x)| ≤ |B(P , r)|−1/p .  a(x)x α dx = 0 for any multi-index α satisfying |α| ≤ [N · (1/p − 1)]. • Here [ ] denotes the greatest integer function. The main theorem about H p for 0 < p < 1 is this: Theorem 6.1.2 Let 0 < p < 1. A function f : RN → C is in H p if and only if f =



αj aj ,

j

where each aj is a p-atom and {αj } ∈ p . The series is assumed to converge in the distribution sense. Furthermore, f H p ≈ {αj }p . It is easy to see that the definition of H p for p less than 1 but very near 1 makes sense on a space of homogeneous type—just because the mean value condition only involves orthogonality to constants. But the definition of H p for smaller p makes no sense on a general space of homogeneous type (although see [KRA2, KRA3] for some ideas about how to address this issue). The atomic theory of H p spaces is today very well developed. It should be noted, for instance, that singular integral operators do not map atoms to atoms. For this reason the theory of molecules was developed by M. Taibleson and others (see [TAW, CRTW]).

6.2 The Local Theory of Hardy Spaces In his thesis (see, for instance, [GOL]), David Goldberg developed a local theory of Hardy spaces. This theory was, in turn, based in part on the theory of atoms. The local Hardy spaces are denoted by hp (RN ). Among the many advantages of this local theory are (i) the Schwartz space is a dense subspace of hp , (ii) pseudodifferential operators of order 0 are bounded on hp , (iii) hp is closed under multiplication by smooth cutoff functions, and (iv) hp makes sense on manifolds. Let us now provide a few of the details (without proofs). Fix a dimension N ≥ 1. Define S = {(x1 , x2 , . . . , xN , y) ∈ RN +1 : 0 < y < 1} .

6.2 The Local Theory of Hardy Spaces

229

We shall refer to S as the strip in RN +1 . It is easy to see that the Poisson kernel Py for S exists (see [KRA4]). Now, if p > (N − 1)/N, then set hp ≡ {(u0 , u1 , u2 , . . . , uN ) :  ! "p/2 sup |u0 (x, y)|2 + |u1 (x, y)|2 + · · · + |uN (x, y)|2 dx < ∞} , 0