THEORY OF HP SPACES
This is Volume 38 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Editors: PAULA. SMITHAND SAMUEL EILENBERG A complete list of titles in this series appears at the end of this volume
THEORY OF HpSPACES
Peter L. Duren Department of Mathematics University of Michigan Ann Arbor, Michigan
Academic Press New York and London 1970
COPYRIGHT 0 1970, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED NO PART OF THIS BOOK MAY BE REPRODUCED I N ANY FORM, BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS.
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LIBRARY OF CONGRESS CATALOG CARD NUMBER : 74-1 17092
PRINTED IN THE UNITED STATES OF AMERICA
TO M Y FATHER William L. Duren
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CONTENTS
Preface, xi 1. HARMONIC AND SUBHARMONIC FUNCTIONS
1 .I. Harmonic Functions, 1 1.2. Boundary Behavior of Poisson-Stieltjes Integrals, 4 1.3. Subharmonic Functions, 7 1.4. Hardy's Convexity Theorem, 8 1.5. Subordination, 10 1.6. Maximal Theorems, 1 1 Exercises, 13 2. BASIC STRUCTURE OF
HP
FUNCTIONS
2.1. Boundary Values, 15 2.2. Zeros, 18 2.3. Mean Convergence to Boundary Values, 20 2.4. Canonical Factorization, 23 2.5.The Class N + , 25 2.6. Harmonic Majorants, 28 Exercises, 29 3. APPLICATIONS
3.1. Poisson Integrals and H', 33 3.2. Description of the Boundary Functions, 35 3.3. Cauchy and Cauchy-Stieltjes Integrals, 39 3.4. Analytic Functions Continuous in [ z ]5 1, 42 3.5. Applications to Conformal Mapping, 43 3.6. Inequalities of FejBr-Riesz, Hilbert, and Hardy, 46 3.7. Schlicht Functions, 49 Exercises, 57 4. CONJUGATE FUNCTIONS
4.1.Theorem of M. Riesz, 53 4.2. Kolmogorov's Theorem, 56 vii
viii
CONTENTS
4.3. 4.4. 4.5. 4.6.
Zygmund's Theorem, 58 Trigonometric Series, 61 The Conjugate of an h' Function, 63 The Case p < 1 : A Counterexample, 65 Exercises, 67
5. M E A N GROWTH AND SMOOTHNESS
5.1. 5.2. 5.3. 5.4. 5.5. 5.6.
Smoothness Classes, 71 Smoothness of the Boundary Function, 74 Growth of a Function and its Derivative, 79 More on Conjugate Functions, 82 Comparative Growth of Means, 84 Functions with H p Derivative, 88 Exercises, 90
6. TAYLOR COEFFICIENTS
6.1. 6.2. 6.3. 6.4.
Hausdorff-Young Inequalities, 93 Theorem of Hardy and Littlewood, 95 1, 98 The Case p I Multipliers, 99 Exercises, 706
7.
HP
AS A LINEAR SPACE
7.1. 7.2. 7.3. 7.4. 7.5. 7.6.
Quotient Spaces and Annihilators, 110 Representation of Linear Functionals, 112 Beurling's Approximation Theorem, 113 Linear Functionals on H p ,0 < p < 1, 115 Failure of the Hahn-Banach Theorem, 118 Extreme Points, 123 Exercises, 126
8. EXTREMAL PROBLEMS
8.1. 8.2. 8.3. 8.4. 8.5.
The Extremal Problem and its Dual, 129 Uniqueness of Solutions, 132 Counterexamples in the Case p = 1, 134 Rational Kernels, 136 Examples, 139 Exercises, 143
9. INTERPOLATION THEORY
9.1. Universal Interpolation Sequences, 147 9.2. Proof of the Main Theorem, 149
CONTENTS
9.3. The Proof for p < 1,153 9.4. Uniformly Separated Sequences, 154 9.5. A Theorem of Carleson, 156 Exercises, 164 10. H P SPACES OVER GENERAL DOMAINS
10.1. 10.2. 10.3. 10.4. 10.5.
Simply Connected Domains, 167 Jordan Domains with Rectifiable Boundary, 169 Smirnov Domains, 173 Domains not of Smirnov Type, 176 Multiply Connected Domains, 179 Exercises, 183
11. H P SPACES OVER A HALF-PLANE
11.1. 11.2. 11.3. 11.4. 11.5.
Subharmonic Functions, 188 Boundary Behavior, 189 Canonical Factorization, 192 Cauchy Integrals, 194 Fourier Transforms, 195 Exercises, 197
12. THE CORONA THEOREM
12.1. 12.2. 12.3. 12.4. 12.5.
Maximal Ideals, 201 Interpolation and the Corona Theorem, 203 Harmonic Measures, 207 Construction of the Contour 21 1 Arclength of I?, 21 5 Exercises, 2 18
r,
Appendix A . Rademacher Functions, 221 Appendix
B. Maximal Theorems, 231
References, 237 Author index, 253 Subject index, 256
ix
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PREFACE
The theory of H P spaces has its origins in discoveries made forty or fifty years ago by such mathematicians as G. H. Hardy, J. E. Littlewood, I. I. Privalov, F. and M. Riesz, V. Smirnov, and G. Szego. Most of this early work is concerned with the properties of individual functions of class H P , and is classical in spirit. In recent years, the development of functional analysis has stimulated new interest in the H P classes as linear spaces. This point of view has suggested a variety of natural problems and has provided new methods of attack, leading to important advances in the theory. This book is an account of both aspects of the subject, the classical and the modern. It is intended to provide a convenient source for the older parts of the theory (the work of Hardy and Littlewood, for example), as well as to give a self-contained exposition of more recent developments such as Beurling’s theorem on invariant subspaces, the Macintyre-RogosinskiShapiro-Havinson theory of extremal problems, interpolation theory, the dual space structure of H P with p < 1, H P spaces over general domains, and Carleson’s proof of the corona theorem. Some of the older results are proved by modern methods. In fact, the dominant theme of the book is the interplay of “ hard” and “ soft” analysis, the blending of classical and modern techniques and viewpoints. The book should prove useful both to the research worker and to the graduate student or mathematician who is approaching the subject for the first time. The only prerequisites are an elementary working knowledge of real and complex analysis, including Lebesgue integration and the elements of functional analysis. For example, the books (cited in the bibliography) of Ahlfors or Titchmarsh, Natanson or Royden, and Goffman and Pedrick are more than adequate background. Occasionally, particularly in the last few chapters, some more advanced results enter into the discussion, and appropriate references are given. But the book is essentially self-contained, and it can serve as a textbook for a course at the second- or third-year graduate level. In fact, the book has evolved from lectures which I gave in such a course at the University of Michigan in 1964 and again in 1966. With the student in mind, I have tried to keep things at an elementary level wherever possible. xi
PREFACE
xii
On the other hand, some sections of the book (for example, parts of Chapters 4,6,7,9, 10, and 12) are rather specialized and are directed primarily to research workers. Many of these topics appear for the first time in book form. In particular, the last chapter, which gives a complete proof of the corona theorem, is “ for adults only.” Each chapter contains a list of exercises. Some of them are straightforward, others are more challenging, and a few are quite difficult. Those in the last category are usually accompanied by references to the literature. Many of the exercises point out directions in which the theory can be extended and applied. Further indications of this type, as well as historical remarks and references, appear in the Notes at the end of each chapter. Two appendices are included to develop background material which the average mathematician cannot be expected to know. The chapters need not be read in sequence. For example, Chapters 8 and 9 depend only upon the first three chapters (with some deletions possible) and upon the first two sections of Chapter 7. Chapter 12 can be read immediately after Chapters 8 and 9. The coverage is reasonably complete, but some topics which might have been included are mentioned only in the Notes, or not at all. Inevitably, my own interests have influenced the selection of material. I wish to express my sincere appreciation to the many friends, students, and colleagues who offered valuable advice or criticized earlier versions of the manuscript. I am especially indebted to J. Caughran, W. L. Duren, F. W. Gehring, W. K. Hayman, J. Hesse, H. J. Landau, A. Macdonald,’ B. Muckenhoupt, P. Rosenthal, W. Rudin, J. V. Ryff, D. Sarason, H. S. Shapiro, A. L. Shields, B. A. Taylor, G. D. Taylor, G. Weiss and A. Zygmund. I am very thankful to my wife Gay, who accurately prepared the bibliography and proofread the entire book. Renate McLaughlin’s help with the proofreading was also host valuable. In addition, I am grateful to the Alfred P. Sloan Foundation for support during the academic year 19641965, when I wrote the first coherent draft of the book. I had the good fortune to spend this year at Imperial College, University of London and at the Centre d’Orsay, UniversitC de Paris. The scope of the book was broadened as a result of my mathematical experiences at both of these institutions. In 1968-1969, while at the Institute for Advanced Study on sabbatical leave from the University of Michigan, I added major sections and made final revisions. I am grateful to the National Science Foundation for partial support during this period. Peter L. Duren
THEORY OF HP SPACES
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HARMONIC AND SUBHARMONIC FUNCTIONS
CHAPTER 1
This chapter begins with the classical representation theorems for certain classes of harmonic functions in the unit disk, together with some basic results on boundary behavior. After this comes a brief discussion of subharmonic functions. Both topics are fundamental to the theory of H P spaces. In particular, subharmonic functions provide a strikingly simple approach to Hardy’s convexity theorem and to Littlewood’s subordination theorem, as shown in Sections 1.4 and 1.5. Finally, the Hardy-Littlewood maximal theorem (proved in Appendix B) is applied to establish an important maximal theorem for analytic functions. 1.1. HARMONIC FUNCTIONS
Many problems of analysis center upon analytic functions with restricted growth near the boundary. For functions analytic in a disk, the integral means
2
1 HARMONIC A N D SUBHARMONIC FUNCTIONS
provide one measure of growth and lead to a particularly rich theory with broad applications. A functionf(z) analytic in the unit disk JzI < 1 is said to be of class H P (0 < p Ia)if M p ( r , f )remains bounded as r -+ 1. Thus H" is the class of bounded analytic functions in the disk, while H 2is the class of power series a,z" with la,12 < 00. It is convenient also to introduce the analogous classes of harmonic functions. A real-valued function u(z) harmonic in IzI < 1 is said to be of class hp (0 < p 5 a)if Mp(r,u) is bounded. Since
c
c
ap5 (a + b)P5 2P(ap+ b p ) ,
a 2 0, b 2 0,
for 0 < p < 00, an analytic function belongs to H P if and only if its real and imaginary parts are both in hP. The same inequality shows that H P and hP are linear spaces. Finally, it is evident that H P 3 H 4 if 0 < p < q I 00, and likewise for the hP spaces. Any real-valued function u(z) harmonic in IzI < 1 and continuous in 121 I 1 can be recovered from its boundary function by the Poisson integral
1 u(z) = u(reie)= 2n
2n
P(r, 0 - t)u(e") d t ,
0
(1)
where
P(r, 0) =
I-r2 1 - 2r cos 8 + r2
is the Poisson kernel. Now replace u(e") in the integral (1) by an arbitrary continuous function q(t) with q(0) = ~ ( 2 7 ~The ) . resulting function u(z) is still harmonic in IzI <1, continuous in IzI 1, and has boundary values u(e") = q(t). Generalizing this idea, one is led to the notion of a PoissonStieltjes integral. This is a function of the form
1 2n
u(z> = u(reie>= -
1
2=
P(r, e - t ) dp(t),
0
(2)
where p(t) is of bounded variation on [0,2x]. Again, each such function is harmonic in IzI < 1. THEOREM 1.I.ThefollowingthreeclassesoffunctionsinIzI
< 1areidentical:
(i) Poisson-Stieltjes integrals; (ii) differences of two positive harmonic functions; (iii) h'. The proof is based on the HeIly selection theorem, which we now state for the convenience of the reader. (For a proof, see Natanson [l] or Widder [I]. Also, see Notes.)
1.1 HARMONIC FUNCTIONS
3
L E M M A (Helly selection theorem). Let (p,(t)} be a uniformly bounded sequence of functions of uniformly bounded variation over a finite interval [a, b]. Then some subsequence ( p J t ) } converges everywhere in [a, b] to a function p(t) of bounded variation, and for every continuous function cp(t),
PROOF OF THEOREM 1.1. (i)* (ii). Expressing p(t) as the difference of two bounded nondecreasing functions, we see that every Poisson-Stieltjes integral is the difference of two positive harmonic functions. (ii) * (iii). Suppose u(z) = ul(z) - u2(z), where u1 and u2 are positive harmonic functions. Then
lu(reie)l d8 4
2n
2n
0
0
j ul(reiB)d8 +
= 2nL-u,(O) so that u E h'. (iii) 3 (i). Given u E h', define
u2(reie)dB
+ UZ(O)l,
Then pr(0) =0, and for 0 = to < t , < ... < t, = 2n,
Hence the functions p r ( t ) are of uniformly bounded variation. By the Helly selection theorem, there is a sequence {r,} tending to 1 for which prn(t)-+ p(t), a function of bounded variation in 0 I t 5271. Thus
8 - t)u(r, ei') d t = lim u(r, z ) = u(z). n+
(Here, as always, z
=
a l
reie.)
As a corollary to the proof, we see that every positive harmonic function in the unit disk can be represented as a Poisson-Stieltjes integral with respect to a nondecreasing function p(t). This is usually called the Herglotz representation. The function p(t) of bounded variation corresponding to a given u E h' is
4 1 HARMONIC AND SUBHARMONIC FUNCTIONS
essentially unique. Indeed, if P(r, 8 - t ) dp(t) = 0, analytic completion gives
where y is a real constant. Since
we conclude that joz'ein' dp(t) = 0,
IZ = 0,
5 1, 5 2 , .
Since the characteristic function of any interval can be approximated in I,' by a continuous periodic function, hence by a trigonometric polynomial, this shows that the measure of each interval is zero. Thus dpis the zero measure. 1.2. BOUNDARY BEHAVIOR OF POISSON-STIELTJES
INTEGRALS
If u(z) is the Poisson integral of an integrable function p(t), then for any point t = 8, where p is continuous, u(z) + cp(f3,) as z +eioO.This can be generalized to Poisson-Stieltjes integrals : u(z) +p'(8,) wherever p is continuously differentiable. Actually, it is enough that p be differentiable ; or, slightly more generally, that the symmetric derivative
exist, as the following theorem shows. THEOREM 1.2. Let u(z) be a Poisson-Stieltjes integral of the form (2), where p is of bounded variation. If the symmetric derivative Dp(8,) exists at a point B0, then the radial limit Iim,.+[ u(reieD)exists and has the value DP(80)-
PROOF. We may assume 8, = 0. Set A =Dp(O), and write 1
rlt
1 -2n
1
"
-z
1
[ p ( t ) - At] - P ( r , t ) d t . Kt
1.2 BOUNDARY BEHAVIOR
OF POISSON-STIELTJES INTEGRALS 5
The integrated term tends to zero as r + 1. For 0 < 6 I It1 l n,
Hence for each fixed 6 > 0, u(r) - A - I, + 0, where
Given E >0, choose 6 > O so small that
Then
for r sufficiently near 1, as an integration by parts shows. Thus u(r) + A as r + 1, and the proof is complete. Since a function of bounded variation is differentiable almost everywhere, we obtain two important corollaries. COROLLARY 1.
Each function u E h' has a radial limit almost everywhere.
COROLLARY 2. If u is the Poisson integral of a function p EL', then u(reie)4 q(8) almost everywhere.
By a refinement of the proof it is even possible to show that u(z) tends to Dp(Bo) along any path not tangent to the unit circle. However, we shall arrive at this result (almost everywhere) by an indirect route. For the present, we content ourselves with showing that a bounded analytic function has such a nontangential limit almost everywhere. For 0 < CI < n/2, construct the sector with vertex eie, of angle 2a,symmetric with respect to the ray from the origin through eie. Draw the two segments from the origin perpendicular to the boundaries of this sector, and let SJe) denote the " kite-shaped " region so constructed (see Fig. 1).
6 1
HARMONIC AND SUBHARMONIC FUNCTIONS
Figure 1 T H E O R E M 1.3. If f E H", the radial limit limp+,f ( r e i e ) exists almost everywhere. Furthermore, if Oo is a value for which the radial limit exists, then f ( z ) tends to the same limit as z +eieDinside any region S,(Oo), a < 4 2 .
PROOF. The existence almost everywhere of a radial limit follows from Corollary 1 to Theorem 1.2, since h" c h'. To discuss the angular limit, it is convenient to deal instead with a bounded analytic function f (z) in the disk lz - 1 I < 1, having a limit L as z + 0 along the positive real axis. Let f,(z) = f ( z / n ) , n = 1, 2, . . . . The functions L(z) are uniformly bounded, so they form a normal family (see Ahlfors [ 2 ] , Chap. 5). This implies that a subsequence tends to an analytic function F(z) uniformly in each closed subdomain of the disk, hence in the region
larg zl 5 a < 4 2 ;
(cos a)/2 s ( z (s cos a.
(3)
(The ray arg z = a has a segment of length 2 cos c1 in common with the disk ( z - 1I < 1.) But for all real z in the interval 0 < z < 2, f,(z) + L. It follows that F(z) E L, and that fn(z)+ L uniformly in the region (3). This implies that f ( z )+ L as z + 0 inside the sector (argzI 5 a, which proves the theorem. The function f is said to have a nontangential limit L a t eieoif f ( z ) -,L a s eieo inside . each region S,(O,), a
1.3 SUBHARMONIC FUNCTIONS
7
1.3. SUBHARMONIC FUNCTIONS
A domain is an open connected set in the complex plane. Let D be a bounded domain. The boundary 8D of D is defined to be the closure D minus D. A real-valued function g(z) continuous in D is said to be subharmonic if it has the following property. For each domain B with B c D,and for each function U(z) harmonic in B, continuous in B, such that g(z) 5 U(z) on aB, the inequality g(z) < U(z) holds throughout B. In particular, if there is a function U(z) harmonic in B with boundary values g(z), then g(z) < U(z) in B. Subharmonic functions are also characterized by the " local sub-mean-value property," which is often easier to work with. THEOREM 1.4. Necessary and sufficient that a continuous function g(z) be subharmonic in D is that for each zo E D there exist po > 0 such that the disk Iz - zol < po is in D and
for every p < po . PROOF. The necessity is easy. Let ) z - zo) < p be in D,and let U(z) be the function harmonic in this disk and equal to g(z) on Iz - zoI = p. Then
To prove the sufficiency, suppose there exists a domain B with B c D and a harmonic function U(z) such that g(z) IU(z) on aB, yet g(z) > U(z) somewhere in B. Let E be the set of points in B at which h(z) = g(z) - U(z) attains its maximum m > 0. Then E c B, because h(z) 0 on aB. Since E is a closed set, some point zo E E has no circular neighborhood entirely contained in E. Hence there exists a sequence {p,} tending to zero such that the disk Iz - zoI
< m = h(zo) = g(zo) - U(zo). But this contradicts property (4), so the proof is complete. ;EXAMPLE 1. If f ( z ) is analytic in a domain D and p > 0, then g(z) = If(z)lpis subharmonic in D.To see this, we need only verify (4). If f ( z o ) = 0,
8 1
HARMONIC AND SUBHARMONIC FUNCTIONS
there is nothing to prove. Iff(zo) # 0, then some branch of [f(z)lP is analytic in a neighborhood of z o , so that
for sufficiently small p. From this (4) easily follows. EXAMPLE 2. If u(z) is harmonic in D and p 2 1, then lu(z)lp is subharmonic in D.Again, we have only to check (4). This is obvious if p = 1. If p > 1, we apply Holder's inequality to the mean vaIue relation and' obtain
where l/p + l/q = 1. This implies (4). EXAMPLE 3.
Let log+ x=
(F
x21 01x<1.
xy
Then g(z) = log' If(z)l is subharmonic if f ( z ) is analytic. Indeed, (4) is trivially true if If(zo)l I 1. If If(zo)I > 1, then this inequality persists in some neighborhood Iz - zoI s p . Thus log'If(z)l coincides with the harmonic function loglf(z)l in Iz - zoI 5 p. 1.4. HARDY'S CONVEXITY THEOREM
The class H P was introduced in Section 1.1 as the set of all functions f ( z ) analytic in IzI < 1 for which the means MP(r,f)are bounded. It is natural to ask how these means may behave as functions of r, for an arbitrary analytic function$ The case p = 2 is especially simple. If f ( z ) = a, z" is analytic in IzI < I , then by Parseval's relation W
M z Z ( r , f )=
1 lan12r2n. n=O
This shows that M 2 ( r , f ) increases with r, and that f~ H 2 if and only if lu,12 < co. Likewise, it follows from the maximum modulus principle that Mm(r,f) increases with r. The situation is more complicated for other values of p , but M P ( r , f )is always a nondecreasing function. In fact, much more is true.
1.4 HARDY'S CONVEXITY THEOREM
THEOREM 1.5 (Hardy's convexity theorem).
9
Let f ( z ) be analytic in
< 1, and let 0 < p I00. Then
IzI
(i) M p ( r , f )is a nondecreasing function of r ; (ii) log M , ( r , f ) is a convex function of log r. To say that log M,(r,f) is a convex function of log r means that if log r
=a
log r1 + (1 - a) log r2
(0 < rl < r2 < 1 ; 0 < a < l),
then log M,(r,f)< a log M p ( r l , f )+ (1 - a) log Mp(r2,f>, or M p ( r , f )I CMp(rl,f ) l a [ M p ( h> f)l' - a . For p = co, this is the well-known Hadamard " three-circles " theorem (see Titchmarsh [l],p. 172), which is an easy consequence of the maximum modulus principle. We shall deduce Hardy's theorem from a general theorem on subharmonic functions. THEOREM 1.6.
Let g(z) be subharmonic in IzI < 1, and let
Then m(r) is nondecreasing and is a convex function of log r. PROOF. Choose 0 5 r1 < r2 < 1, and let U(z) be the function harmonic in JzI< r2 , continuous in Iz] 5 r 2 , and equal to g ( z ) on IzI = r2 . Then g ( z ) I
U(z)in IzI I r 2 , so
To prove the convexity, let 0 < r1 < r2 < 1, and let U ( z ) be the function harmonic in the annulus r l < Iz( < r2 and agreeing with g ( z ) on both boundary circles. Then
10 1 HARMONIC AND SUBHARMONIC FUNCTIONS
with equality at the end points. On the other hand, by Green's theorem,
where a p n denotes the normal derivative, ds = r do is the element of arclength, and c is a constant. The right-hand side of ( 5 ) therefore has the form a log r b, which shows that m(r) is a convex function of log r.
+
REMARK. The proof shows that m(r) is a convex function of log r if g(z) is subharmonic only in an annulus p 1 < ( z (< p 2 . PROOF OF HARDY'S THEOREM. (i) As pointed out in Section 1.3, g(z) = If(z)lp is subharmonic iff(z) is analytic, so Theorem 1.6 applies. In fact, the same is true of lu(z)["if u(z) is harmonic and p 2 1, so MJr, u ) is also nondecreasing. (ii) The argument of Section 1.3 shows that g(z) = Izl'llf(z)lp is subharmonic in 0 < IzI < 1 if p >O, /z is any real number, and f(z) is analytic. Thus, by the remark following Theorem 1.6, r'lMpP(r,f)is a convex function of log r. Given 0 < rl < r2 < 1, let A < 0 be chosen so that
ri"M,P(ri,
f) = r2"M,P(r2
9
f) = K ,
say. Let r = rla ri-' (0 < CI < 1). Then
r 'MM,P(r,f) I K = KaK1-" = (r,a~M,P(rl,f))Q{r21MpP(r2,f)>'-" = r " ~ p P ( r l , f ) ) a w p P ( i "J2> > -la ,
which completes the proof. 1.5. SUBORDINATION
Let F ( z ) be analytic and univalent in IzI < 1, with F(0) = 0. Letf(z) be analytic in Jz]< 1, withf(0) = 0, and suppose the range offis contained in that of F . Then w(z) = F-l(f(z)) is well-defined and analytic in IzI < 1, o(0) = 0, and lo(z)l 5 1. By Schwarz's lemma, then, lo(z)l < IzI. This implies, in particular, that the image under f ( z ) =F(o(z)) of each disk IzI Ir < 1 is contained in the image of the same disk under F(z). More generally, a functionf(z) analytic in IzI < 1 is said to be subordinate to an analytic function F(z) (writtenf< F ) iff(z) =F(w(z))for some function w(z) analytic in IzI < 1, satisfying lo(z)l IIzI. F ( z ) need not be univalent. The following result has many applications. THEOREM 1.7 (Littlewood's subordination theorem). Let f ( z ) and F(z) be analytic in IzI < 1, and su p p o sefi F . Then M p ( r , f )IMp(r,F ) , 0 < p Ico.
1.6 MAXIMAL T H E O R E M S
11
PROOF.As in the proof of Hardy’s theorem, we shall deduce this from a more general result concerning subharmonic functions. Let G(z) be subharmonic in ] z ]< 1, and let g(z) = G(w(z)), where o ( z ) is analytic in Iz] < 1 and Iw(z)] I Jz].Then
I.
2rr
2n
g(reie) dd I G(reiO)do. 0
To prove (6), from which the theorem easily follows, let U ( z ) be the function harmonic in IzJ< r and equal to G(z) on JzI= r. Then G(z) 5 U(z) in IzI 5 r, so that g(z) Iu(z) = U(w(z))on IzI = r . Hence
1 g(reie)d0 5 271
1
2r
u(r.eie)d0 = u(0)
0
1 U(reie)d0 271 1
= U(0) = -
2rr
1
j G(reie)do, 271
=-
0
2*
0
which proves (6). 1.6. MAXIMAL THEOREMS
For
E
Lp =LP(O, 271), let
p 5 co, an easy If u(r, 0) is the Poisson integral of a function ip E Lp, 1 I calculation based on Holder’s inequality shows that u E hP and M,(r, u ) 5 I(q/(p. There is, however, a much deeper theorem which extends this result in a striking and useful way. THEOREM 1.8 (Hardy-Littlewood). cp E Lp, 1 < p i co, and let
Let u(r, 0) be the Poisson integral of
Then U E Lp, and there is a constant A , depending only on p such that
/I UII,
5
~,/lip/l,.
PROOF. The proof depends upon a theorem on “maximal rearrangements” of functions, which is discussed in Appendix B. Extend 40 periodically with period 271, and let
12
1 HARMONIC AND SUBHARMONIC FUNCTIONS
By Theorem B.3, Jl E Lp and
IlJa,5 CPll'pllP,
1 < P Ice.
U(Q)is a measurable function, since it is the upper envelope of the functions lu(r, 0)l as r runs through the sequence of rational numbers in the interval (0, 1). Hence it will suffice t o show that
lu(r, 0)l i 2 A ( Q ) ,
0I r < 1.
(7)
This can be seen by writing the Poisson formula in the form
and integrating by parts. For fixed 8, let O(t) = j'cp(0
+ u ) du.
0
Then
From this (7) follows, since IO(t)/tl< J l ( 0 ) and
2r
< 1.
It should be observed that (7) remains true in the case p = 1, although the theorem is then false. (See Exercise 6.) However, the corresponding theorem for analytic functions is true for allp in the range 0 < p 5 co. THEOREM 1.9 (Hardy-Littlewood).
Let f~ H P , 0 < p 5 a,and k t
F(e) = suplf(re")I. r<S
Then F E Lp and
IIFll, 5 Bpllfll, , where Bp depends only on p and
Ilfll,
= lim r +S
Mp(r,f).
PROOF. For each fixed R, 0 < R < 1, the function g(z) = l f ( R ~ ) 1 ~is' ~ subharmonic in IzI < 1, so it follows from Theorem 1.8 that
q e ) = supjg(reiB)lE L~ rcS
NOTES
13
and IIG112
AZ~Z(L9).
In other words, IIFRllp
5 A;'PMp(R,f),
where ~ ~ ( =d supIf(re")>l. ) r i R
Since FR(t?) increases to F(6) as R 1, the proof is completed by appeal to the Lebesgue monotone convergence theorem. --f
EXERCISES
1. Show that (1 -z)-' is in H P for everyp <1, but is not in H1. 2. Show that iff(z) is analytic in ( z (< 1 and its range is contained in a sector of angle CI (0 < CI I 2 n ) , thenfe H P for every p < nja. 3. Show that the function
belongs to H' if c > 1. 4. Show directly (i.e., by refining the proof of Theorem 1.2) that a PoissonStieltjes integral has a nontangential limit almost everywhere. 5. Use Littlewood's subordination theorem to show that M p ( r , f ) is a nondecreasing function of r. 6. Use the modified Poisson kernel
u(r, 0) =
R 2 - r2 RZ - 2Rrcos 6 + r 2 '
R > 1,
to show that Theorem 1.8 is false for p = 1. 7. Show that if f(z) and g(z) are subharmonic, then so is max{f(z), g(z)}. NOTES
In the language of functional analysis, the Helly selection theorem simply asserts the weak-star sequential compactness of the unit sphere of the dual space of the space C of continuous functions over the interval [a, 61. That is, if $n E C* and supll+,II < co,then some subsequence {+nk} converges pointwise to $ E C*.
14 1 HARMONIC AND SUBHARMONIC FUNCTIONS
The paper of Fatou [l] is one of the earliest studies of the boundary behavior of analytic and harmonic functions in the disk. Fatou showed, among other things, that a Poisson-Stieltjes integral has a nontangential limit almost everywhere. The existence of such a limit for a bounded analytic function is known as “Fatou’s theorem.” Theorem 1.1 goes back at least to Plessner [l], although Herglotz [I] had earlier found the representation for positive harmonic functions. Hardy’s convexity theorem is in his paper [2], which is now regarded as the historical starting point of the theory of H P spaces. (Thus the “ H ” in “ H p ” is for Hardy.) The elegant proof of Hardy’s theorem via subharmonic functions is due to F. Riesz [2]. A. E. Taylor [l] found another proof using Banach space methods. Littlewood’s subordination theorem is in his paper [l]; again F. Riesz [ 5 ] vastly simplified the proof using subharmonic functions. For an extension, see Gabriel [1,2]. See also Ryff [l]. For a thorough discussion of subharmonic functions, see Rado [I]. The Hardy-Littlewood maximal theorems are in their paper [4]. They also proved more general theorems in which the supremum is taken not over a radial segment, but over a “ kite-shaped region” S,(fl) with vertex at a boundary point.
BASIC STRUCTURE OF H p FUNCTIONS
CHAPTER 2
In this chapter we consider, among other things, the boundary behavior and the zeros of H p functions. As may be expected, the growth restriction on an H p function imposes a restriction on the density of its zeros. This leads to the formation of " Blaschke products " and the canonical factorization theorems, which play a vital role in later work. Another important result is the mean convergence of an H p function to its boundary function. Two new classes of functions, N and N', arise in the course of the discussion, and their properties are explored. Finally, an equivalent definition of H pis given in terms of harmonic majorants. 2.1. BOUNDARY VALUES
We have seen that a harmonic function of class h' must have a radial limit in almost every direction. The same is therefore true for analytic functions of class H'. The theorem remains valid for H P functions with p < I, despite the fact (to be noted in Chapter 4) that the corresponding statement for harmonic functions is false. Surprisingly, the generalization to p < 1 is obtained most simply by dealing with an even wider class of functions. This is the Nevanlinna
16 2
BASIC STRUCTURE OF HD FUNCTIONS
class N , also known as the class of functions of bounded characteristic. A functionf(z) analytic in Iz[ < 1 is said to be of class N if the integrals
are bounded for r < 1. Since log'lfl is subharmonic wheneverfis analytic, these integrals always increase with r . It is clear that N contains H P for every p > 0. The following theorem is basic to the further development of the theory. THEOREM 2.1 (F. and I?. Nevanlinna). A function analytic in the unit disk belongs to the class N if and only if it is the quotient of two bounded analytic functions.
PROOF. Suppose first that f ( z ) = q ( z ) / $ ( z ) , where q and $ are analytic and bounded in IzI < 1. There is no loss of generality in assuming [q(z)l < 1, l$(z)l I 1, and $(O) # 0. Then
But by Jensen's formula (see Ahlfors [2], p. 206), logl$(reie)l dB
= logl$(O)l
r log -,
+ IZ,I<~
IznI
where zn are the zeros of $. This shows that 1logl$l increases with r, s o f ~N . Conversely, let f ( z ) f 0 be of class N . Let f have a zero of multiplicity m 2 0 at the origin, so that z-"f(z) -+ a # 0 as z -+0. Let zn be the other zeros of f, repeated according to multiplicity and arranged so that 0 < [zll I < 1. Iff@) # 0 on the circle IzI = p < 1, the function [ z 2 (5
is analytic in (zI I p, and Re{F(z)} = log)f(z)l on 1zI. completion of the Poisson formula,
= p.
Hence, by analytic
This is sometimes called the Poisson-Jensen formula. After exponentiation, it takes the form f ( z ) = q,(z)/$,(z), where
2.1 BOUNDARY VALUES
17
Now choose a sequence {Pk} increasing to 1, such thatf(z) # 0 on the circles
IzI = P k . Let k')('
= q)Pk(Pk'>;
yk(z> = $pk(Pk
'1'
Thenf(p,z) = 4t)k(Z)/yk(Z)for Iz( < 1. But the functions akand y, are analytic in the unit disk, and I@k(z)l < 1, IYk(Z)I I 1 there. Hence {ak} and {yk} are normal families, and there is a sequence (ki}such that 4t)k,(Z) q ( z ) and Y&) .+ $(z) uniformly in each disk IzI 5 R < 1. The functions cp and $ are analytic in the unit disk, and lq(z)l 5 1, l$(z)l 5 1. According to the definition of the fact that j l o g + ) f ) is bounded gives a uniform estimate IYk(0)(2 6 > 0, so $(z) $ 0. T h u s f = q/$,and the proof is complete. --f
The importance of this representation theorem is that it allows properties of functions in N to be deduced from the corresponding properties of bounded analytic functions. The boundary behavior, for example, can now be discussed. THEOREM 2.2. For each function f~ N , the nontangential limit f(eio) exists almost everywhere, and logl,f(eiO)lis integrable unless f ( z ) = 0. If f E H P for some p > 0, then,f(eie) E Lp. PROOF. Assuming f ( z ) $0, represent it in the f x m q(z)/$(z), where lq(z)l I 1 and 1$(z)1 I 1. Since cp and $ are bounded analytic functions, they have nontangential limits q(eie)and $(do) almost everywhere, by Theorem 1.3. Appealing to Fatou's lemma, we find
~ o z ~ ~ l o g ~ q (dd e i eI ) (!iminf(jr-rl
lozn: loglq(ueio)(d o ) .
But j loglq(reie)l dd increases with r, by Jensen's theorem. Hence log/q(e")I E L', and similarly for $. In particular, $(eie) cannot vanish on a set of positive measure. The radial limit f(eie) therefore exists almost everywhere, and logIf(eie)i E L'. Another application of Fatou's lemma shows that f(eie) E Lp iff E Hp. The theorem says that i f f € N and iff(eie) = 0 on a set of positive measure, thenf(z) = 0. In other words, a function of class N is uniquely determined by its boundary values on any set of positive measure.
18 2 BASIC STRUCTURE OF H 9 FUNCTIONS
It is evident from the representation f = q/$ that J log-If(reie)I dt' is bounded i f f € N . (Here log- x = max{ -log x, O}.) Hence f E N if and only if J Iloglf(reie)ll dt' is bounded. 2.2. ZEROS
The zeros of an analytic function cannot cluster inside the domain of analyticity unless the function vanishes identically. More is true if the function satisfies a growth restriction: the zeros must tend "rapidly " to the boundary. The following theorem illustrates this principle. THEOREM 2.3. Let f ( z ) $ 0 be analytic in I zI < 1, and let zl, z 2 , .. . be its zeros, repeated according to multiplicity, in Iz( < 1. Then J;' loglf(reie)l dt' is bounded if and only if (1 - lz,,l) < co. PROOF. Letfhave a zero of multiplicity ?n 2 0 at the origin, so thatf(z) = urn* * * , c t # 0. Call the other zeros a,,, and let them be ordered so that 0 < lall I la,] I * . By Jensen's formula,
+
-
This increases monotonically with r ; suppose it is bounded from above by some constant C. Then, fixing an integer N , we have for all r > la,l
Letting r tend to 1, we find
n N
0 < lale-c I Iu,,~, n= 1
so that X(1 - la,l) < a.To prove the converse, we have only to note that Jensen's theorem implies
COROLLARY.
The zeros { z n } of a function of class N must satisfy
C (1 - lznl) < One might well expect a stronger growth condition such asfE H p to imply a correspondingly stronger statement about the rate at which the zeros
2.2 ZEROS
19
approach the boundary. Even iff E H", however, nothing more can be said about its zeros. In fact, given an arbitrary sequence {a,) of numbers in the unit disk such that C (1 - la,l) < co, one can produce a bounded analytic function whose zeros are precisely {a,). It is natural to construct such a function as an infinite product. The factors will be essentially the functions (z- a,)/(l - Gz), which map the unit disk onto itself and vanish at a, . THEOREM 2.4. Let a,, a 2 , < 1 and 0 < la,] I la,! I
. . . be a sequence of complex numbers such that
IF=, (1 - la,!) <
00.
Then the infinite product
converges uniformly in each disk (zI 2 R < 1. Each a, is a zero of B(z), with multiplicity equal to the number of times it occurs in the sequence; and B(z) has no other zeros in IzI < 1. Finally, IB(z)l < 1 in IzI < 1, and IB(eie)(= 1 a.e. PROOF.
In the disk ( z (I R ,
1
Ian1 a n - z (an + lanlz>(l- Ian11 a, 1 - G Z / = a,(l- a , z )
2(1- lanl)
/'1-R
2
'
Since (1 - la,l) < co,it follows that the infinite product converges uniformly in each disk IzI I R < 1, and therefore represents a function B(z) analytic in JzJ < 1. Furthermore, each a, is a zero of B(z) with the correct multiplicity, and B(z) # 0 otherwise. That IB(z)l < 1 in Iz] < 1 is obvious since this is true for the partial products. The radial limit B(eie) therefore exists almost everywhere, and IB(eie))I 1. It remains to show that IB(eie)I= 1 a.e. But for any f E H", the Lebesgue bounded convergence theorem and the monotonicity of M,(r,f ) give 22
If(reie)l d e I Jo If(eit>ldt. Now apply this to the function f = B/E, ,where
Since lB,(eie)l = 1, we have
1
B( reie) B,(reie)
1
- do 5 1:'IB(eit)1 dt.
20
2 BASIC STRUCTURE OF HP FUNCTIONS
But Bn(z)+B(z) uniformly on IzI =r, so
Because IB(eie)I5 1 a.e., this shows IB(eie)[= 1 a.e. A function of the form
is called a Bluschkeproduct. Here m is a nonnegative integer and (1 - lu,,l) < co.The set (a,} may be finite, or even empty. If (a,} is empty, it is understood that B(z) =zm. 2.3. MEAN CONVERGENCE TO BOUNDARY VALUES
We have seen that every Hp function f(reie)converges almost everywhere to an Lp boundary function f(eie).For the further development of the theory, it is important to know that f(reie) always tends to f ( e i e )in the sense of the Lp mean. The following factorization theorem will enter into the proof and will become a standard tool in later chapters. THEOREM 2.5 (F. Riesz). Every functionf(z) $ 0 of class H p ( p > 0 ) can be factored in the form f ( z ) =B(z)g(z),where B(z) is a Blaschke product and g(z) is an H p function which does not vanish in Izl < 1. Similarly, eachfe N has a factorization f = Bg, where g is a nonvanishing function of class N .
PROOF. We may suppose thatf’(z) has infinitely many zeros, since otherwise the theorem is trivial. Let
denote the partial Blaschke product, and let g,(z) = f(z)/B,,(z).For fixed n and E > 0, JB,(z)l > 1 - E for lzl sufficiently close to I. Therefore,
for r sufficiently large; hence, by monotonicity, for all r . Letting E - 0 , we find ~oz~~gn(reie)~P dtJ I M
2.3 MEAN CONVERGENCE TO BOUNDARY VALUES
21
for all r < 1 and all n. By Theorem 2.4, however, g,,(z)tends to g(z) =f(z)/B(z) uniformly on each circle IzI = R < 1. Thus g is in H p , and it has no zeros. (The proof for f~ N is similar.) We are now ready for the mean convergence theorem. THEOREM 2.6.
I f f € H p (0 < p < co), then
!$ 1, If(reie)IPdl3 2n
2n
=
0
If(eie))lpdo
and
lolf(reie)-f(eie)lP do 2n
lim
= 0.
(3)
PROOF. First let us prove (3) for p = 2. If .f(z) =
a,, z" is in H 2 , then
r-tl
(un12< co. But by Fatou's lemma,
/:ffjj-(reie)-f(eie)12 d o I lim inf P-'1
l:nlf(de) -f(peie)>i2d6
a,
= 27c
C1 lunI2(1 - rn)2,
n=
which tends to zero as r -+ 1. This proves (3), and hence (2), in the case p = 2. Iff€ H p (0 < p < co), we use the factorization f =Bg given in Theorem 2.5. Since [g(~)]~'~ E H 2 , it follows from what we have just proved that
I. If(reie)>lp lo 2n
2n
do I
lg(reie)lPdo-+
1
0
loIf(eie)>lpdo. 2rr
2K
lg(eie)lPdo =
This together with Fatou's lemma proves (2). The following lemma can now be applied to deduce (3) from (2). L E M M A 1.
Let i2 be a measurable subset of the real line, and let < co; n = I, 2, . . . . As n -+ co, suppose cp,,(x)-+ q(x) a.e.
qnE Lp(i2), 0
on R and
Then
22
2 BASIC STRUCTURE OF
PROOF.
j E IqIP.Let
H P
FUNCTIONS
For a measurable set E c R, let Jn(E)= j E lqnlPand J(E) = = R - E. Then
J(E) I Iim inf Jn(E) I lim sup J,(E) n+w
n+m
s lim Jn(sZ) - lim inf Jn@) I J(R) - J(E) = J(E). n+w
n +m
This shows that Jn(E)-+ J(E) for each E c R. Given E > 0, choose a set F c R of finite measure such that J(F) < E . Choose 6 > 0 so that J(Q) < E for every set Q c F of measure m(Q) < 6. By Egorov's theorem, there exists a set Q c F with m(Q) < 6, such that (P,,(x) -+ q ( x ) uniformly on E = F - Q . Thus
_< 2 ' { J n ( O
+ J(O + Jn(Q) + J(Q>> + JE Itp, - ( P I P
< (2j' * 6 + 1 ) ~ for n sufficiently large, since Jn(E)+J(F) and Jn(Q) -+ J(Q). This proves the lemma, and Theorem 2.6 follows. COROLLARY.
I f f € H p for some p > 0, then lim j;*llog+lf(re")I
- log' If(eie)l I do = 0.
r+ 1
The corollary is an immediate consequence of Theorem 2.6 and the following lemma. LEMMA 2.
For a 2 0, b 2 0, and for 0 < p I 1, IlOg+ u - log+ bl I (l/pl) la - blP.
PROOF.
It is enough to assume 1 I b c a. The result then follows from the
inequality l o g x s (l/p) ( x - I)",
x 2 1,
by setting x = a/b. The latter inequality can be proved by noting that the difference of the two expressions has a positive derivative and vanishes at x = 1.
It is tempting to think that the corollary holds generally for a llfe N . This is false; the function f(z) = exp((1 + z)/(l - z)} is a counterexample.
2.4 CANONICAL FACTORIZATION
23
There is also a " short " proof of the mean convergence theorem (Theorem 2.6) which does not use the Riesz factorization theorem but appeals instead to the Hardy-Littlewood maximal theorem (Theorem 1.9). I f f E Hp, then $(reie) +f(eie) almost everywhere,andf(eie) E E. But by the maximal theorem, If(reie)l 5 F(O), where F ELp. Hence Theorem 2.6 follows at once from the Lebesgue dominated convergence theorem! It should be observed at this point that H p is a normed linear space if 1 ~p I 03. The norm is defined as the Lp norm of the boundary function. Thusif1 5 p < 00,
while
2.4. CAN0 N ICAL FACT0 R lZATl0 N
The Riesz factorization (Theorem 2.5) can be refined to produce a canonical factorization which is of supreme importance both for the theory of H p spaces and for its applications. This refinement rests upon the following inequality. THEOREM 2.7. Iff E Hp, p
>o, then
1 257 loglf(reie)I I - P(r, 8 - t ) log)f(e")l d t . 2n Jo PROOF. After factoring out the Blaschke product, whose presence would only strengthen the inequality, we may assume f(z) # 0 in IzI < 1. Then loglf(z)l is harmonic in Iz( < 1, and
1 loglf(preie)l = 2n
2n
j0P(r, 8 - t ) Ioglf(peir)) dt,
r < p < I.
By the corollary to Theorem 2.6, 2a
lim P+l
joP(r, 8 - t ) log+(f(peir)ld t =
2n
0
P(r, 8 - t ) log+If(e")( dt.
On the other hand, by Fatou's lemma, 2%
lim P-rl
f, P(r, O - t ) log-lf(peir))d t 2
Subtraction gives the desired result.
2n
0
P(r, 8 - t) log-lf(eif)l d t .
24 2 BASIC STRUCTURE OF
H n FUNCTIONS
+
The theorem is false for the class N , as the example exp((1 z)/(l - z)} again shows. The reciprocal of this function reveals, incidentally, that strict inequality may occur even if f(z) is bounded and has no zeros. Returning to the problem of factorization, let f(z) $ 0 be of class H P for some p > 0. According to Theorem 2.2, f ( e i e )E Lp and log)f(eie)lE L1. Consider the analytic function
Letf(z) = B(z)g(z) as in Theorem 2.5; thus g(z) # 0 and lg(eiQ)I= If(eie)I a.e. By Theorem 2.7, Ig(z)I
O < I S(z)l I 1;
IS(eiQ)l= 1 a.e.;
S(O) > 0.
This shows that -loglS(z)l is a positive harmonic function which vanishes almost everywhere on the boundary. Thus by the Herglotz representation and Theorem 1.2, -loglS(z)l can be represented as a Poisson-Stieltjes integral with respect to a bounded nondecreasing function ,u(t), and p ' ( t ) = 0 a.e. Since S(0) > 0, analytic completion gives
Putting everything together, we have the factorizationf(z)
= eiyB(z)S(z)F(z).
We now introduce some terminology. An outer,function for the class H p is a function of the form
where y is a real number, $ ( t ) 2 0, log $(t) E L', and $ ( t ) E Lp.Thus (4) is an outer function. An inner function is any functionf(z) analytic in IzI < 1, having the properties If(z)l 5 1 and lf(eio)l = 1 a.e. We have shown that every inner finction has a factorization eiYB(z)S(z),where B(z) is a Blaschke product and S(z) is a function of the the form (5), p ( t ) being a bounded nondecreasing singular function (p'(t) = 0 a.e.). Such a function S(z) is called a singular inner function. THEOREM 2.8 (Canonical factorization theorem). Every functionf(z) $ 0 of class H p ( p > 0) has a unique factorization of the formf(z) =B(z)S(z)F(z), where B(z) is a Blaschke product, S(z) is a singular inner function, and F(z)
2.5 THE CLASS N +
25
is an outer function for the class Hp (with $ ( t ) = l.f(e")l). Conversely, every such product B(z)S(z)F(z) belongs to Hp. PROOF. We have already shown that everyf E HPcan be factored as claimed, and the uniqueness is obvious. To prove the converse, it suffices to show that an outer function (6) must belong to HP.Applying the arithmetic-geometric mean inequality (see Exercise 2), we find
Thus
/:'IF(reie)Ip d6 I /oz'[$(t)]p d t . There is a similar factorization forfE N. A function F(z) of the form (6), where $(t) 2 0 and log $ ( t ) E I,', will be called an outerfunction for the class N . (Note that the condition $ E Lp has been dropped.) THEOREM 2.9.
Every functionf(z) $ 0 of class N can be expressed in the
form f(z )
= B(z)CS,(z)/S,(z)lF(z),
(7)
where B(z) is a Blaschke product, S,(z) and S,(z) are singular inner functions, and F(z) is an outer function for the class N (with $ ( t ) = If(e")l). Conversely, every function of the form (7) belongs to N. PROOF. Letf(z) =eiYB(z)g(z),where g E N , g(z) # 0 in IzI < 1, and g(0) > 0. Since loglg(z)l E h', it has a representation as a Poisson-Stieltjes integral
with respect to a function v(t) of bounded variation. Analytic compIetion and separation of v ( t ) into its absoluteIy continuous and singular components gives the desired representation. The converse follows directly from the fact that every Poisson-Stieltjes integral is of class h'. 2.5.THE CLASS N'
The two preceding theorems point out the sharp structural difference between functions in the classes H P and N . In factoring functions of class N , it is necessary not only to enlarge the class of admissible outer functions, but also to replace the singular factor by a quotient of two singular inner functions.
26
2 BASIC STRUCTURE OF HD FUNCTIONS
+
This allows, for instance, our "pathological" example exp((1 z)/(l - z)}, which we now recognize as the reciprocal of a singular inner function. It is useful to distinguish the class N + of all functionsf€ N for which Sz(z) = 1. That is,,f E N + if it has the form f = BSF, where B is a Blaschke product, S is a singular inner function, and F is an outer function for the class N . In a sense, N + is the natural limit of H p as p -+0. The proper inclusions H p c N + c N are obvious. THEOREM 2.10.
A function f E N belongs to the class N + if and only if
PROOF. Suppose first that f E N + , so that f =BSF. Then, in view of (4),
1 log+lf(re")l I log+(F(rei8)1I 2n
2z
P(r, 0 - t ) log+lf(e")l dt.
0
Hence 2n
277
lim Jo log+lf(rei8)1d0 I r+l
log+If(e")l d t . 0
Fatou's lemma gives the reverse inequality. It is more difficult to prove the sufficiency of the condition (8). We first observe that for an arbitrary Blaschke product B(z), lim r+ 1
2"
loglB(reie)l d0
= 0.
(9)
This is obvious if B(z) has only a finite number of factors, so we may assume
By Jensen's theorem,
Holding N fixed, we then have for all r > laN\
= Nlog
r+
m
C n=N+l
logla,l.
2.5
Consequently,
Jazz
m
C n=N+
2i-c
loglanl Ilim r+l
1
THE CLASS N+
27
loglB(rei"')l d8 I0,
and (9) follows by letting N -+ 00. Continuing the proof of the theorem, let the given function f E N be expressed in the form f(z) =B(z)g(z), where g E Nand g(z) # 0 in 1zI < 1. Since 1Oglm)l + log+ldz)l
los+lf(z>l Ilog+Ig(z)l,
combination of (8) and (9) gives r+l
j0
1 log+lg(eie)l de. 2n
277
lim
log+lg(reie)l
=
0
(10)
Since loglg(z)l E hl, it has a representation 1 loglg(z)l = 2.n
s,
2n
P(r, 8 - t > dv(0
(11)
with respect to a function v(t) of bounded variation. Recalling the proof of Theorem 1.1, we see that v can be chosen to have the form e
v(8)
=
lim n-m
S, loglg(r,ei')l dt,
0 I8 5271,
where {r,} is an appropriate sequence increasing to 1. On the other hand, by Fatou's lemma,
1: log+lg(eif)ld t 2 lim inf J: log+Ig(rnei')] d t n-tm
=
v+(e),
(12)
say. If there were strict inequality for some 8, then a similar application of Fatou's lemma in [O, 2711 and addition of the two results would give a contradiction to (10). Equality therefore holds in (12) for all 8, which shows that v+(O) is absolutely continuous. On the other hand,
v-(o) = v+(e) - v(8) = lim inf n-rm
:1
log-Ig(rn eif)Id t
is nondecreasing. In view of (1 l), this shows that g = SG, where S is a singular inner function and G is an outer function. Hence f E N + , which was to be shown. The following useful result is an easy consequence of the factorization theorems.
28 2 BASIC STRUCTURE O F
THEOREM 2.11.
H p FUNCTIONS
I f f € N + andf(eio) E Lp for somep > 0, then f E H P .
The a priori assumption that f E N + cannot be relaxed. The reciprocal of any (nontrivial) singular inner function is bounded on [zI = 1, but is not of class N'. 2.6. HARMONIC MAJORANTS
We noted in Section 1.3 that iff(z) is analytic in a domain D, then I f ( z ) l pis subharmonic in D. This means that in each disk contained in D, If(z)lPis dominated by a harmonic function, the Poisson integral of its boundary function. However, there may not be a single harmonic function which dominates If(z)lpthroughout D. In general, a function g(z) is said to have a harmonic majorant in D if there is a function U(z)harmonic in D such that g(z) _< U(z) for all z in D. If g is continuous and has a harmonic majorant, it is obviously subharmonic; but the converse is false. THEOR'EM 2.12. I f f ( z ) is analytic in ( z (t l , then and only if If(z)lphas a harmonic majorant in ( z (< 1. PROOF.
f E
H p (0 < p < co) i f
If U(z) is a harmonic majorant of If(z)Ip,then by the mean value
theorem
M P k ,f )
I: U(O>l
Conversely, if f E H p , it follows from Theorem 2.7 and the arithmeticgeometric mean inequality (Exercise 2) that
I
1 2n 271 Jo P(r, 6 - t ) l f ( e i ' ) [ dt. p
In other words, If(z)lpis dominated by the Poisson integral U(z)of its boundary function. It is easy to see, in fact, that this is the least harmonic majorant of If(z)lp. That is, if V ( z ) is any other harmonic majorant, then U(z) 5 V ( z )for all z, /zl < 1. Indeed, for any p < 1.
1
271 Jo
1 2n P(r, 6 - t)lf(pei')lp d t I 2n Jo P ( r , 0 - t)V(pe
2rr
= V(pz).
As p + 1, this shows U(z) 5 V(z).
dt
EXERCISES
29
COROLLARY. If f E H p (0 < p < a) and if cp is analytic and satisfies Ip(z)l < 1 in IzJ< 1, then g(z) =f(cp(z)) E H p and
PROOF. Let U ( z ) be the Poisson integral of If(e”)l”. Then If(z)lpI U(z), so that U(p(z)) is a harmonic majorant of [g(z)Ip. Hence the mean value theorem gives
(1 sinceP(r, 0) I
+ r)/(l - r ) .
EXE R Cl SES
I
1. Prove in detail that , f ’ N ~ if and only if Iloglf(re”)) do is bounded. 2. (Arithmetic-geometric mean inequality.) Letf(x) 2 0 be integrable with respect to a nonnegative measure dp of unit total mass. Let %(f) = Sf(.) dp(x) be the “arithmetic mean” off, and let S ( f )= exp{%(logf)) be the “geometric mean.” Prove that S ( f )_< 9l(f), with equality if and only if f ( x ) is constant a.e. (Hint: Note that log t 5 t - 1, t 2 0, and set t = f ( x ) / % ( f ) . This proof is due to F. Riesz [7].) 3. Show that the classical inequality (a1az *
’*
a”)”” < (l/n)
(a1 f a2
+
’
.
*
-f
an)?
ak > 0,
is a special case of the general arithmetic-geometric mean inequality. 4. (Jensen’s inequality) Let cp(zr) be a convex function in an interval a < u < b, and suppose a < f ( x ) < b. As in Exercise 2, let %(f) denote the arithmetric mean o f f with respect to a unit measure dp. Prove Jensen’s inequality cp(%(f)) < %?l(cp(f)),and deduce the general arithmetic-geometric mean inequality. (Hint: Let a = % ( f ) and let 1 be the slope of a supporting line through (a, cp(a)). Show that
cp(u) - cp(a) 2 A(u - a),
a < u
This proof is due to Zygmund [4].) Interpret Jensen’s inequality for the case in which the unit measure dp is concentrated at a finite number of points. 5. Show that if $(u) is a nondecreasing convex function and g(z) is subharmonic, then $(g(z)) is subharmonic.
30
2 BASIC STRUCTURE OF
HP
FUNCTIONS
6. Show that a function f(z) analytic in IzI < 1 is a Blaschke product (aside from a constant factor of modulus one) if and only if lim ~,"llogjl(reie)Il de = 0. r-11
7. Show that iff(z) is an inner function and q(z) is a conformal mapping of the unit disk onto itself, thenf(q(z)) and q(f(z)) are inner functions. 8. Letf(z) be an inner function, and suppose lcll <1. Prove that for all M outside a set of planar measure zero, the function f(z> 1 - Ef(z) is a Blaschke product! (Frostman [l].) 9. Prove the corollary to Theorem 2.6 under the weaker hypothesis that fEN+. 10. Show that for functions f analytic in the unit disk, f E N if and only if log+If(z)I has a harmonic majorant.
NOTES
Fatou [ l ] showed that eachfE H" has a nontangential limit f(eie) almost everywhere, and that f(eie) cannot vanish on an arc 8, I 8 I 8, + 6 unless f(z) = 0. F. and M, Riesz [l] improved this to H 1 and showed thatf(z) = 0 if its boundary function vanishes on a set of positive measure. Later, Szego [l] showed that if f~ H 2 and f(z) f 0, then loglf(eie)IE L'. After F. Riesz [3] proved his factorization theorem (Theorem 2.5), he was able to show that each f € H P (p > 0) has a radial limit almost everywhere, and that log1f(eie)l E L' unless f(z) = 0. He also proved the mean convergence of f(reie) to f(eie) (Theorem 2.6). Lemma 1, which simplifies the proof of this theorem, is also due to F. Riesz [6], at least for p 2 1. The proof given in the text is in Littlewood's book [5]. A somewhat different approach to the structure of HP functions is in the paper of M. and G. Weiss [I]. For further information on the boundary behavior of H p functions, see Tanaka [l]. Theorem 2.1 is due to F. and R. Nevanlinna [l]; it unifies the presentation of the H p theory. A paper of Ostrowski [I] is also relevant here. Blaschke [I] introduced " Blaschke products " and proved Theorem 2.4. Regarding matters of priority, however, see Landau [3]. The canonical factorization theorems (Theorems 2.8 and 2.9) are due to Smirnov [2], who also noted a weaker form of Theorem 2.1 1 : i f f € H p andf(eie) E L4 for some q > p , then
NOTES
31
f~ Hq. Beurling [l] coined the terms “inner function” and “ outer function.” Smirnov [3] introduced the class N + (called “ D ” in his paper and in subsequent Soviet literature) and cited Theorem 2.10, which he attributed to “Madame P. Kotchine”. More about the class N + may be found in Privalov’s book [4]. The condition (9) actually characterizes Blaschke products among all analytic functions with If(z)l I 1 (see Exercise 6). This theorem is due to M. Riesz, but was first published in Frostman [l]. (See also Zygmund [4], Vol. I, p. 281.) Theorem 2.12 was mentioned in passing by Smirnov ([3], p. 341). There is a large literature on the boundary behavior of Blaschkeproducts; see, for example, Frostman [2], Cargo [l, 21, and Somadasa [l]. Paley and Zygmund [2] have shown that if the hypothesisfE N is “ slightly relaxed” (which they interpret in two different ways), thenf(z) need not have a radial limit on any set of positive measure.
This Page Intentionally Left Blank
APPLICATIONS
CHAPTER 3
We now turn to some applications of the H P theory to such diverse topics as measure theory (the F. and M. Riesz theorem), Cauchy and Poisson integrals, and conformal mapping. Some of the results obtained will be useful in the later development of the H P theory. Further applications will appear in subsequent chapters.
3.1. POISSON INTEGRALS AND HI
In Chapter 1, the harmonic functions u E hl were characterized as PoissonStieltjes integrals. We shall now show that if the harmonic conjugate of u also belongs to h', then the representing function p ( t ) is absolutely continuous. This result will have a number of interesting consequences. The harmonic conjugate of u is any function o such that f ( z ) = " harmonic conjugate even though it is determined only up to an additive constant. Later it will be convenient to normalize o by the requirement ~ ( 0 = ) 0. NOTE.
u(z)
+ iv(z)is analytic in the disk. We shall speak of '' the
34 3 APPLICATIONS
THEOREM 3.1.
A functionf(z) analytic in lzl < 1 is representable in the
form
' if and only iff as the Poisson integral of a function cp E L cp(t) =f(eif) a.e.
E H1. In
this case,
PROOF. If an analytic functionf(z) has the form (l), then
J;zlf(rei@)l dB 5 d'lcp(t)l d t ,
so that f~ H'. Conversely, suppose f~ H I , and write 1 @(z)= 2.n
joP(r, 8 - t)f(e") d t . 2n
Foranyfixedp,O
But, by Theorem 2.6, If(pe") -f(eif)l dt +. 0 as p + 1, so f@z) + @(z). Hence @(z)=f(z), and the theorem is proved. COROLLARY 1. Let u E hl, so that it is the Poisson-Stieltjes integral of a function p ( t ) of bounded variation. If the conjugate harmonic function z1 E h', then p ( t ) is absolutely continuous. COROLLARY 2. A functionf(z) analytic in lzl < 1 is the Poisson integral of a function cp E Lp (1 5 p 5 00) if and only i f f e H P . COROLLARY 3.
then S,(z)/S,(z)
Letf(z) =B(z)[s,(z)/s,(z)]~(z) EN. If log[f(z)/B(z)] fH1,
= 1.
Corollary 2 follows from Theorem 2.11. We shall prove in Chapter 4 that the harmonic conjugate of any function u E hP (1 < p < 00) is also of class hP.This cannot be true for p = 1, as Corollary 1 shows, even if u is a positive harmonic function. In fact, the Poisson kernel P(r, 0 ) itself is a counterexample. The following weaker theorem, however, remains true. THEOREM 3.2. Every analytic functionf(z)with positive real part in IzI < 1 is of class HPfor all p < 1.
3.2 DESCRIPTION OF THE BOUNDARY FUNCTIONS
35
PROOF. Without loss of generality, suppose f ( 0 ) = 1. The range of f is contained in the right half-plane, so f is subordinate to
where P(r, 6) is the Poisson kernel and Q(r’ e,
2r sin 8 = 1 - 2r cos e + r2
is the conjugate Poisson kernel. It follows from Littlewood’s subordination theorem (Theorem 1.7) that
for anyp < 1, since (1 COROLLARY.
+ z)/(l - z) is in HPand P(1, 0) = 0 for 0 # 0.
If u E hl, then its harmonic conjugate belongs to hP for all
p < 1.
This follows from the representation of h’ functions as the difference of two positive harmonic functions (Theorem 1.1). In Chapter 4 the corollary will be reproved in slightly stronger form (Kolmogorov’s theorem). 3.2. DESCRIPTION OF THE BOUNDARY FUNCTIONS
We have seen that each function f E HP (0 < p 5 00) has a nontangential limit f ( e i e ) at almost every boundary point. It is of interest to characterize these boundary functions in a simple way. Let Z pdenote the set of boundary functionsf(eie) of functions f E H P . As usual, two functions are identified if they coincide almost everywhere, so the elements of Z pare actually equivalence classes. We know that Z pc Lp, and in fact that Z pis a vector subspace of LP: it is closed under addition and scalar multiplication. We shall see presently that Z pis also topologically closed. For E Lp,let
It is convenient to retain the norm notation even for p < 1, when 11 I l p is not a genuine norm. The notation llfll, for f E H P will refer to the boundary function. Thus
36 3
APPLICATIONS
LEMMA.
Iff. H P (0 < p < a),then
PROOF. If p = 1, the estimate follows easily from the Poisson integral representation of H1 functions (Theorem 3.1). For p # 1, we may use the Riesz factorization (Theorem 2.5) of a function f~ H P in the form f = Bg, where B is a Blaschke product and g is a nonvanishing H P function. Since If(z)l I Ig(z)l and I l f l l , = llgllp, it suEces t o prove the lemma for nonvanishing H P functions. But if g E H P and g(z) # 0, some branch of [g(z)Ip is single-valued and. belongs to H'. Thus the desired result follows from the H 1 case.
We are now ready to characterize functions of the form
%p.
It is clear that SPcontains all
where the a, are complex constants. These functions will be called polynomials in eis. They are dense in SP, as we shall now show. T H E O R E M 3.3.
For 0 < p < 03,
is the Lp closure of the set of poly-
nomials in eie. PROOF.
For 0 < p < 1, let fp(z)=f@z). Given f E H P and
E
> 0, choose
p < 1 such that
llfp --flip < &/2. (This is possible, by Theorem 2.6.) Now let S,(z) denote the nth partial sum of the Taylor series o f , f a t the origin. Since S,(z)-tf(z) uniformly on the circle IzI = p ,
Il$lp
-fpllp <42
for n sufficiently large. Thus, using Minkowski's inequality in the case p 2 1, we find
lls"p-flip< E . This shows that the boundary functionf(eie) belongs to the Lp closure of the polynomials in eie. For p < 1, the inequality (a
+ b)PI 2j'(aP + bP),
gives the same result.
a > 0, b > 0,
3.2 DESCRIPTION OF THE BOUNDARY FUNCTIONS
37
To complete the proof, we have to show that Zpis closed. But suppose that a sequence {f,(eie)}of Zpfunctions converges in Lp mean to p(Q)E Lp.Then by the lemma, {f,(z)} is uniformly bounded in each disk IzI 5 R < 1. Thus {Ji}is a normal family, so a diagonalization argument produces a subsequence {f,,(z)> which converges uniformly in each disk IzI I R < 1 to an analytic function f(z). It is clear that f~ H P . We wish to show that p(Q) = f ( e i e )a.e. But given E > 0, choose N such that l f , -fmllp < E for n, m 2 N . Then for rn>Nandr
f -f m )
= lim Mp(r, f n , k+m
-f m )
lim SUP I1fn, k-rm
- f m I/p < 8.
Letting r -+ 1, we find l\f-fmllp < E for all m 2 N ; thus (p(e)= f ( e i e )a.e.
\lf-hllp-+
0, and
The same argument, in simpler form, shows that Zm is closed. COROLLARY 1.
If 1 I p I co, H P is a Banach space.
If p < 1, then 11 ( j p is not a true norm and in fact the space H P is not normable. However, the expression
4f, g) = llf- 911; defines a metric on H P if p < 1. This can be verified using the inequality (a
+ b)PI ap + bP,
a > 0, b > 0,
which is valid for 0 < p < 1. (See Section 4.2.) Furthermore, the theorem shows that H P is complete under the topology induced by this metric. COROLLARY 2.
If 0 < p < 1, H P is a complete metric space.
Theorem 3.3 is false for p = co,since Xmcontains functions which do not coincide almost everywhere with continuous functions. One example is the singular inner function exp{(z -t- l)/(z - l)}, whose boundary function is exp{ - i cot(Q/2)},8 # 0. There is another approach, however, which leads to a description of ~p for 1 Ip 5 00. The Fourier coeficients of a function p E L1are the numbers
It is important to note that the Taylor coefficients of a function f E H P (1 Ip 5 co) coincide with the Fourier coefficients of its boundary function.
The following theorem expresses this more precisely.
38
3 APPLICATIONS
c:=o
THEOREM 3.4. Let f ( z ) = anz" belong to H', and let {cn> be the Fourier coefficients of its boundary function f(e"). Then c,, = a,, for n 2 0, and c,, = 0 for n < 0. Furthermore, Z p(1 5 p 5 co) is exactly the class of L p functions whose Fourier coefficients vanish for all n < 0. PROOF.
The Taylor coefficients offcan be expressed in the form a,, = -
r-"e-'"f(re'') d t ,
2n
O
0
Thus Ir"an - cnl 5
Ilf, -fill -+O
as r 1, by Theorem 2.6. [Here, as above, f,(z) =f(rz).] This shows that c,, = a, for n 2 0. Similarly, c, = 0 for n < 0. Suppose now that q E Lp (1 I p 5 00) and that its Fourier coefficients c, vanish for all n < 0. Let --f
be the Poisson integral of q. Since the Poisson kernel has the expansion m
P(r, t ) = 1
+ C rn(ein'+ n=l
it follows that
c c,z", oc
f(z)=
I1 =
Iz( < 1.
0
This shows that f(z) is analytic in IzI < 1 ; hence f~ H P , by Corollary 2 to Theorem 3.1. But ~ ( t=)f ( e i ' ) a.e., so q E SP, and the proof is complete. According to the Riemann-Lebesgue lemma, the Fourier coefficients of an integrable function tend to zero as n -+ t- 00. Thus we have the following corollary. COROLLARY.
Iff(z)
=
a,z" E H ' , then a, + 0 as n + 00.
From this point on we will usually suppress the notation A?' for the space of boundary functions. It is convenient, and custonary in the literature, to identify Z pwith HP,and thus to regard H P as a subspace of Lp.
3.3 CAUCHY AND CAUCHY-STIELTJES INTEGRALS
39
3.3. CAUCHY A N D CAUCHY-STIELTJES INTEGRALS
For each complex-valued function q ( i )integrable over the unit circle Ill = 1, the Cauchy integral dt represents a pair of functions: one analytic in IzI < 1, the other in JzJ> 1. More generally, the same is true for a Cauchy-Stieltjes integral
with respect to a complex-valued function p ( t ) of bounded variation. THEOREM 3.5. If a function F(z) can be represented in I zI < 1 as a CauchyStieltjes integral, then it belongs to H P for all p < 1.
PROOF. It suffices to assume p ( t ) is real and nondecreasing. Then Re{F(z)} > 0, since
1 - r cos(t - 0)
and the result follows from Theorem 3.2. The converse is false: a function can be in H P for all p < 1, yet not be representable as a Cauchy-Stieltjes integral. (See Exercise 2.) We shall also see that a Cauchy-Stieltjes integral, or even a Cauchy integral, can easily fail to be of class H’. A Cauchy-Stieltjes integral is related in a simple way to the PoissonStieltjes integral formed with respect to the same function p ( t ) :
In view of Theorem 3.5 and Theorem 2.2, the radial limit lim F(reie) r+l-
exists almost everywhere (taken from inside the unit circle). But as r -+ 1, the right-hand side of (2) tends to p’(0) a.e., by Theorem 1.2. Thus the exterior radial limit also exists a.e., and lim F(re”) - lim F(re”) = ~ ’ ( 8 ) a.e. r+l-
r+l+
40 3 APPLICATIONS THEOREM 3.6. Every function f e H 1 can be expressed as the Cauchy integral of its boundary function. In fact,
Each of these integrals vanishes identically in IzI > 1. PROOF. It suffices to establish the result f o r k = 0; the rest will then follow by the familiar procedure of " differentiation under the integral sign." (See, for example, Ahlfors [2],p. 121.) Let F(z) be the Cauchy integral off(eif), and let {P,(z)} be a sequence of polynomials such that Ilf- P,ll --f 0. (See Theorem 3.3.) Writing
we find that F(z) = 0 in IzI > 1. Hence it follows from the relation (2) with dp(t) =f(eif) dt, and from Theorem 3.1, that
1 F(z) = 2.n
Jo
2rr
P ( r , 0 - t)f(e") d t = f ( z ) ,
IzI < 1.
(3)
This completes the proof. The relation (2) can also be used to prove the following theorem. THEOREM 3.7. For a complex-valued function p ( t ) of bounded variation, the following three statements are equivalent :
(0
%'eint d p ( t ) = 0,
n
= 1, 2,
...
(ii) The Cauchy-Stieltjes integral
vanishes identically in JzI> 1. (iii) The Poisson-Stieltjes integral
is analytic in Izl < 1. PROOF. (i) e (ii). This follows by expanding the kernel ei'(ei' - z)-' in powers of l/z and integrating term by term.
3.3. CAUCHY A N 0 CAUCHY-STIELTJES INTEGRALS
41
(ii) (iii). Since the Cauchy-Stieltjes integral F(z) is always analytic in < 1, this follows at once from (2). (iii)=.(ii).The Cauchy-Stieltjes integral F ( z ) is analytic in IzI > 1. If the Poisson-Stieltjes integral f(z) is analytic in IzI < 1, then (2) shows that F(5) is also analytic in IzI > 1. Thus by the Cauchy-Riemann equations, F(z) is identically constant in IzI > 1. Since F(z) + 0 as z -,03, the constant must be zero. Hence F ( z ) = 0 in Izl > 1. IzI
A complex-valued function p ( t ) of bounded variation over [0,271] will be called normalizedif p ( t ) = ,u(t + ) for 0 < t < 271 ;that is, if p is continuous from the right. We can now use the preceding results to establish an important measure-theoretic theorem. THEOREM 3.8 (F. and M . Riesz). Let p ( t ) be a normalized complex-valued function of bounded variation on [0,271], with the property S,Z'ein'
d p ( t ) = 0,
n = 1,2, . . . .
Then p ( t ) is absolutely continuous. PROOF. By Theorem 3.7, the hypothesis implies the corresponding PoissonStieltjes integral f(z) is analytic in IzI < 1 ; hence f E H'. But according to Theorem 3.1, every H' function can be represented in the form 1
f(z)=
2.n
..2n
J
P ( r , 8 - t>f(e")d t . 0
Since the Poisson representation is unique (Section 1.2), dp(t) = f ( e i ' ) dt. The essential step in the proof was the use of Theorem 3.1, the representation of H' functions by Poisson integrals. In fact, this result is in a sense equivalent to the theorem of F. and M. Riesz. Below is another variant which is sometimes useful. The proof is left as an exercise. THEOREM 3.9. If a functionf(2) analytic in IzI < 1 can be represented in any one of the following four ways: (i) as a Cauchy-Stieltjes integral with F(z) = 0 in IzI > 1; (ii) as a Cauchy integral with F ( z ) = 0 in IzI > 1 ; (iii) as a Poisson-Stieltjes integral; (iv) as a Poisson integral, then it can be represented in each of the other three ways. The class of functions so representable is H'.
42
3 APPLICATIONS
3.4. ANALYTIC FUNCTIONS CONTINUOUS IN
(21
51
In general, a continuous function of bounded variation need not be absolutely continuous. For example, the familiar Lebesgue function over the Cantor set is monotonic, continuous, and purely singular. For the boundary values of functions analytic in the disk, however, continuity is equivalent to absolute continuity. In fact, a somewhat stronger statement can be made, as follows. THEOREM 3.10. I f f € H' and its boundary functionf(e") is equal almost everywhere to a function of bounded variation, then f ( z ) is continuous in jz( 5 1 andf(eio) is absolutely continuous. PROOF. Suppose f(eiO) = p(0) a.e., where p is a normalized function of bounded variation and p(0) = 4271). By Theorem 3.4,
jozze'nep(0)dt? =
IOZR ei"y(eLe)
d8 = 0,
n = 1, 2, . . .
Integrating by parts, we find ~ ~ n e dp(8) i n e = 0,
n = 1,2, . . . .
By the F. and M. Riesz theorem, then, p(0) is absolutely continuous. On the other hand,f(z) can be represented as a Poisson integral of p(t?), by Theorem 3.1 and the fact that f ( e i O )= p(d) a.e. Thus the radial limit ,f(eie) exists everywhere, coincides with p(Q), and is absolutely continuous. We shall now show that the functions just considered-analytic functions with absolutely continuous boundary values-are characterized by the simple condition f ' E H I . The expressionf'(e'') then can have two possible meanings. It may indicate the radial limit off'(z); or it may indicate the derivative with respect t o 8 of the boundary function, apart from a factor ie". It is a remarkable fact that the two are essentially the same. THEOREM 3.11. A functionf(z) analytic in jz/ < 1 is continuous in jzj 5 1 and absolutely continuous on IzJ= 1 if and only i f f ' E H1. Iff' E H1, then
d -f(eie) dO
= ieielimf'(reie)
a.e.
r-rl
(4)
PROOF. If f(z) is analytic in (z( <1 and continuous in fzi < 1, it is the Poisson integral of its boundary function:
f(z)
=
2n1 I.Z*P(r, B - t)f(e")
dt.
3.5 APPLICATIONS TO CONFORMAL MAPPING
43
Differentiate with respect to 8:
and integrate by parts [using the absolute continuity off(e")] to obtain
Thus izf'(z) E H ' , which impliesf' E H1. Conversely, i f f ' E H1, then izf'(z) can be represented in the form (5) as the Poisson integral of its boundary function. Here it is understood that f'(ei') denotes lim, f'(re"). Since the function --t
e
ieitf'(ei') dt
g(O) = 0
is absolutely continuous in [0,271], and g(0) = g(2n) = 0, integration by parts in ( 5 ) gives
Thus
Since C(r) is the difference of two (complex-valued) harmonic functions, it is itself harmonic : C"(r) + (l/r)C'(r) = 0. Thus C(r) = a log r + b, where a and b are constants. To ensure continuity at the origin, a must be zero, Hencef(z) is the Poisson integral of the continuous function [g(t) + b], and is therefore continuous in IzI I 1 and has boundary valuesf(e") = g(e) + b. The relation (4) now follows from the definition of do). 3.5. APPLICATIONS TO CONFORMAL MAPPING
The theorems just given, together with a few general facts about HP functions, can be applied to obtain some rather deep results in the theory of conformal mapping. A Jordan curve (or a simple closed curve) C is the image of a continuous complex-valued function w = w ( t ) (0 I t 5 271) such that w(0) = w(271) and
44 3 APPLICATIONS
w(tl) # w(t2) for 0 I t , < t2 < 271. The curve C is said to be rectijiable if w(t) is of bounded variation. Its length Lis then defined as the total variation of w(t): n
where the supremum is taken over all finite partitions 0 = to < t , < -..< tn = 271 of [0,271]. It is easily seen that Ldepends only on the curve C, and is invariant under a change of parameter. If w(t) is absolutely continuous, the length of the arc of C corresponding to an arbitrary interval a 5 t < b is given by
1Iw'(t)l dt. -b
a
(A proof may be found in Natanson [I], Vol. 11, p. 227.) It follows that any Lebesgue measurable set E c [0,271] has an image on C of measure
Indeed, the formula is obviously valid for open sets E, hence for G, sets (countable intersections of open sets). Therefore, since an arbitrary measurable set E is contained in a G, set with the same measure, the formula holds generally. A famous theorem of Carathkodory asserts that every conformal mapping w = f ( z ) of 1z/ < 1 onto the interior of a Jordan curve C has a one-one continuous extension to IzI I 1. In particular, w = f ( e i 8 )is a parametrization of C. Applying Theorems 3.10 and 3.11, we therefore obtain the following important result. THEOREM 3.12. Let f ( z ) map IzI < 1 conformally onto the interior of a Jordan curve C. Then C is rectifiable if and only iff' E H'.
This theorem is highly plausible, perhaps even geometrically. It says that the lengths
"
obvious," when viewed
2n
L, = r
joff'(reie)ldB
of the images of the circles IzI = r are bounded if and only if the boundary has finite length. In the presence of a rectifiable boundary, this result puts the general HP theory at our disposal. The derivative of the mapping function has an angular
3.5 APPLICATIONS TO CONFORMAL MAPPING
45
limitf’(e’’) almost everywhere on the boundary, and loglf’(ei6)I is integrable. The function J”(e”) is related as in (4) to the derivative of the absolutely continuous boundary function f(eie). A measurable set E on the unit circle is carried onto a subset of C with measure
Consequently, a subset of IzI = 1 has measure zero if and only if its image on C has measure zero. In other words, the boundary sets of measure zero are preserved under the conformal mapping. This remains true (in view of the Riemann mapping theorem) for a conformal mapping between any two Jordan domains with rectifiable boundaries. More can be said. If f ( z ) is a conformal mapping of IzI < 1 onto the interior of a rectifiable Jordan curve C , then its continuous extension to IzI I 1 is conformal at almost every boundary point. To be precise, let y be a continuous curve in IzI < 1 which terminates at a point zo = eieo in a well-defined direction not tangent to the unit circle; i.e., the limit of arg{z - z,) is to exist as z -+ zo along y, and is not to equal 0, k 71/2. The image of y is then a curve L‘ inside C, terminating atf(zo). Since (d/d8)(f(ei8)} exists a.e., C has a tangent direction at almost every point. We assert that for almost every 6, , the angle between r and the tangent t o C atJ’(z,) exists and is equal t o the angle between y and the tangent t o the unit circle at z,. In other words, the mapping preserves angles at almost every boundary point. We have to show that
ZEY
= 8,
+ n/2 - lim arg(z - z,}. Z‘ZO
=SY
In view of relation (4), it is enough to prove
Z€Y
wherever the angular limitf‘fz,) off’(z) exists; hence almost everywhere. But
the integration being performed along the segment joining zo and z. Iff’(z,) exists, the right-hand side of (7) approachesf’b,) as z + zo along y, which was assumed to be a nontangential path. This proves (6).
46 3 APPLICATIONS
3.6. INEQUALITIES OF FEJER-RIESZ, HILBERT, AND HARDY
We shall now discuss some interesting inequalities which will have applications in later chapters. THEOREM 3.13 (FejBr-Riesz inequality). I f f € H P (0 < p < a),then the x 5 1 converges, and integral of If(x)lp along the segment - 1 I
The constant
+ is best possible.
PROOF. Consider first the case p = 2, and suppose for the moment that f ( z ) is real on the real axis. By Cauchy’s theorem with a semi-circular path,
j : r [ f ( x ) ] 2 dx
+ i /OR [f(reie)]’eied 6 = 0
for each r < 1. It follows that
Adding a similar inequality for the lower semicircle, we find 2R
2
f [ f ( x ) l 2 dx I/:Rlf(reie)12 d8 If lf(eie)I2 do. -r
0
The desired inequality is now obtained by letting r tend to 1. More generally, we may express an arbitrary H 2 functionf(z) in the form m
m
m
where g and h are in H 2 and are real on the real axis. By what we have just proved,
But the last integral must vanish, since the integrand is an odd function of 8. Hence inequality (8) is established in the case p = 2. To deduce the result for general p , we use a familiar device. Iff(z) belongs
3.6 FEJCR-RIESZ.
HILBERT, AND HARDY INEQUALITIES
47
to H P and vanishes nowhere in IzI < 1, then [ f ( ~ ) ] " ' ~E H 2 , and (8) follows from the special case p = 2 already proved. If f ( z ) has zeros, we factor out the Blaschke product B(z) to obtain a nonvanishing function g(z) = f(z)/B(z) which is again of class H P . Thus
j-If(x)lP dx 5 J- Ig(x)lp dx 5 t':1 lg(eie)IPdB = + /:'lf(eio))lp do. 1
1
1
1
In view of Theorem 3.12, there is an interesting geometric application. COROLLARY. If the unit disk lzl < 1 is mapped conformally onto the interior of a rectifiable Jordan curve C , the image of any diameter has length at most half the length of C.
+
We may now finish the proof of the theorem by showing that is the best possible constant. Let w = q(z) map IzI < 1 conformally onto the interior of the rectangle with vertices f1 f iE, the diameter - 1 5 z 5 1 corresponding to the real segment - 1 I w 5 1. The ratio of the length of this segment to the perimeter of the rectangle is 2[4(1 + &)]-I, which tends to 4 as E --+ 0. The functionf(z) = [q'(z)]l'" E H P therefore shows that the constant cannot be improved. Two further inequalities, named after Hilbert and Hardy, lie in the same circle of ideas. We shall prove them in generalized form, with a view to later applications. For a complex vector x = (xo,x l , . . . , xN),let N
c
1 / ~ / 1 2 = n = O 1Xnl2. THEOREM 3.14.
Let $ E Lw and
1 2n
An=-!
2n
e-'"'$(t) d t ,
0
Let
Then
PROOF.
Let P(t) =
x,e-'"'. Then
n = 0, 1,2, . . . .
(9)
48 3 APPLICATIONS
COROLLARY (Hilbert's inequality).
PROOF.
Choose $ ( t ) = ie-ir(n - t),so that A, = (n + I)-' and /[$llm = n.
THEOREM 3.15.
f(z)=
c
U,Z"
E
Let A n 2 0 be given by (9) for some $ € L r n . Then if
HI, m
PROOF. Every f E H 1 has a factorization f = gh, where g and h are H 2 functions with 1)g)122 = JJhJ122 = 1) f )I1. Indeed, f = Bip, where B is a Blaschke product and cp is a nonvanishing H1 function (Theorem 2.5). Let g = Brp1l2 and h = @ I 2 . Now let g(z) = 1 bnznand h(z) = cnzn;then
c
n
Hence
by Theorem 3.14. COROLLARY (Hardy's inequality).
If f ( z ) =
c anz" f?'then , E
3.7 SCHLICHT FUNCTIONS
49
Hardy's inequality is proved with the same choice of $,I that gave Hilbert's inequality. One interesting consequence should be mentioned. Suppose f ( z ) = a,z" is analytic in Iz( < 1. If (a,! < 03, then f has a continuous extension to Izl II , but the converse is false (see Exercise 7). Hardy's inequality shows, however, that iff' E H' (or equivalently, in light of Theorem 3.11, iffis continuous in [zI I 1 and absolutely continuous on (zI = l), then (a,,]< co. In particular, la,l < 03 iffis a conformal mapping of the unit disk onto a Jordan domain with rectifiable boundary.
c
c
c
C
3.7. SCHLICHT FUNCTIONS
A function analytic in a domain is said to be schlicht (or uniualent) if it does not take any value twice; that is, iff(z,) # f ( z 2 )whenever 2, # z2 . Our main aim in this section is to prove that every schlicht function in the unit disk is of class H P for all p < +. This will show, for instance, that a conformal mapping of the unit disk onto an arbitrary simply connected domain, however pathological the boundary, automatically has a finite radial limit in almost every direction. The Koebe function
which maps (zI < I onto the full plane slit along the negative real axis from -$ to co, shows that a schlicht function need not belong to H1I2. We say t h a t f e Y iffis analytic and schlicht in (zI < 1 and is normalized so that f ( 0 ) = 0 and f ' ( 0 ) = 1. The first of the following lemmas is very well known, and the proof is omitted here. (See, for example, Hayman [l], p. 4.) L E M M A 1.
IffE9,then
L E M M A 2. Let C, and C, be twice continuously differentiable Jordan curves in the plane, surrounding the origin, with C , in the interior of C, . Then
where (r, 9) are polar coordinates and both curves are traversed in the positive direction.
50 3
APPLICATIONS
PROOF. Let B be the ring domain between C,and C 2 ,let C be its boundary, and let ( x , y ) denote rectangular coordinates. Applying Green's theorem and the Cauchy-Riemann equations
a(i0g -=ax
r)
ae. ay
>
a(iog r ) - - ae -aY ax
we find
This proves the lemma. We are now ready to state the main theorem. THEOREM 3.16.
Iff(z) is analytic and schlicht in lzl < 1, t h e n f s H P for
allp < 1/2. PROOF. Without loss = 0. Now set
of generality, we may assumefs 9, so thatf(z) # 0
unless z
f(reie)= Re'@
(r > 0)
and apply the Cauchy-Riemann equation
to obtain
where Tris the image underfof the circle lzl = r. Thus if M(r) = M , ( r , f ) is the maximum of If(z)l on IzI = r, then for each E > 0 the circle JwI= M(r) + E surrounds Tr,and it follows from Lemma 2 that
EXERCISES
Now let E
-+0, integrate
51
from 0 to r, and apply Lemma 1 :
1
5 p l o r p - l ( l - r)-’p dr c co
if 0 < p < 3.This concludes the proof. As a function of class H P ( p < +), each schlicht function f has a canonical factorization f(4 = B(z)WF(z) (see Theorem 2.8). The Blaschke product B(z) obviously has at most one factor. Less obvious is the fact thatfcan have no singular part. THEOREM 3.17.
factor S(z) = 1.
Iff(z) is analytic and schlicht in JzI < 1, then its singular
PROOF. Iff does not vanish in the open disk, then llfis analytic and schlicht, so l i f ~H P for all p < 4.Hence by the uniqueness of the canonical factorization of a function of class N (Theorem 2.9), S(z) = 1. Iff([) = 0 for some [, l[l < 1, then the function
By Lemma 1, then, belongs to 9’.
+
Ig(z)/zl 2 (1 r > - 2 > 4. From this it follows that B i f ~H ” , where z - i
B(z) = -, 1- ( z In particular, .f cannot have a (nontrivial) singular factor. EXERCISES
1. Prove: If f(z) is analytic and Re{f(z)} > 0 in IzI < 1, thenf is an outer function. (Hint: Show thatf,(z) = E + f ( z ) is outer and let E -+ 0.) 2. Showthat 1 1 f(z>= 1 - z log(-) 1 - z belongs to H P for all p < 1, but has unbounded Taylor coefficients. In particular, show thatf(z) is not a Cauchy-Stieltjes integral, so that the converse to Theorem 3.4 is false.
52 3 APPLICATIONS
3. Prove Theorem 3.9. 4. Prove the integral analogue of Hilbert’s inequality:
iffand g are in L2(0,co). 5. It is a natural conjecture that i f f € N andf(eis) E L’, then / ~ “ e i n ~ ( ede i ~ )= 0,
n = 1,2, . .. .
Show this is false. 6. Construct an analytic functionf(z) = a,z” such that la,l < co, but f ’ 4 H’. 7. Give an example of a function f(z) = a,z” analytic in Izl < 1 and 1, such that C la,l = 03. (Suggestion: Let f map the unit continuous in IzI I disk conformally onto a Jordan domain constructed in such a way that the z < 1 corresponds to a curve of infinite length.) radius 0 I
c
c
1
NOTES
Theorems 3.1 and 3.8 are essentially due to F. and M. Riesz [I]. They also showed that the coefficients of an H’ function tend to zero. Theorem 3.4 and the falsity of its converse (Exercise 2) may be found in Smirnov [2]. Theorem 3.9 is due to Fichtenholz [l]. Privalov’s book [4] discusses these results. Helson [l] has given a “soft” proof of the F. and M. Riesz theorem, making no use of complex function theory. Rudin [6] obtained a generalization of this theorem in which 1eintdp = 0 only outside a certain “thin ” set of positive integers. Theorem 3.11 is due to Privalov [l, 21, as are most of the results in Section 3.5. Theorems 3.10 and 3.12 appear in a paper of Smirnov [4]. The theorem of CarathCodory mentioned before Theorem 3.12 may be found in Goluzin [3], Chap 11, Sec. 3, or in Zygmund [4], Chap. VII, Sec. 10. Golubev [l J gave the first example of a conformal mapping of the unit disk onto a Jordan domain which carries a boundary set of measure zero onto a set of positive measure on the (nonrectifiable) boundary of the image domain. Theorem 3.13 is in a paper of FejCr and Riesz [I]. Hardy [l] had essentially proved it for p = 2. Many proofs of Hilbert’s inequality have been given; see Hardy, Littlewood, and Pblya [l]. “Hardy’s inequality” seems to have appeared first in a paper of Hardy and Littlewood [I]. Theorem 3.16 is essentially in a paper of Prawitz [I]; see also Goluzin [3], Chap. IV, Sec. 6. Another proof is in Littlewood [5]. Theorem 3.17 is due to Lohwater and Ryan [l].
CONJUGATE FUNCTIONS
CHAPTER 4
If a harmonic function has a certain property, must the same be true of its conjugate ? Questions of this kind have been widely investigated, both for their own interest and for their importance in applications. In this chapter and the next, we consider growth and smoothness properties of functions harmonic in the unit disk. Generally speaking, a harmonic function and its conjugate behave alike, but there are some rather surprising exceptions. 4.1. THEOREM OF M. RlESZ
Given a real-valued function u(z) harmonic in /zI < 1, let v(z) be its harmonic conjugate, normalized so that v(0) = 0. Thusf(z) = u(z) + iv(z) is analytic in (z(< 1, andf(0) is real. Iff(z) = C c,z" and c, = a, - ib,, then co
u(z) = a.
+ 11P(a, cos n e + b, sin no), n=
m
v(z) =
n=l
r"( - b, cos no + a, sin no).
54
4 CONJUGATE FUNCTIONS
By the Parseval relation,
Hence Mh.3 v> I Mz(r, u), (2) with equality if and only if a, = 0. It was the discovery of M. Riesz that something similar is true for every p in the range 1 < p < 00. THEOREM 4.1 (M.Riesz). If u E hP for somep, 1 < p < 00, then its harmonic conjugate u is also of class hP. Furthermore, there is a constant A , , depending only on p , such that
M,(r, v) I A,M,(r, u),
0I r < 1,
(3)
for all u E hP. PROOF. First let us observe that only the case 1 < p I 2 needs to be considered. Indeed, if the theorem holds for some index p (1 < p < co), then it is also true for the "conjugate index" q defined by l/p + l/q = 1, and in fact with A , = A , . To show this, let N
g ( z ) = g(reie)= a ,
+n = 1 F(a, cos no + b, sin no)
be an arbitrary harmonic polynomial, and let N
h(z) =
C rn(- b, cos no + a, sin no) n=1
be the conjugate function, normalized by h(0) = 0. Thus g(z) + ih(z) is an ordinary algebraic polynomial; and for fixed p (0 < p < l), the function
m)= C 4 P 4 + ~v(Pz)lCg(z)+ W Z ) l is analytic in lz] I 1. The mean value theorem applied to I m ( F ( z ) } gives
+
lozn[u(peio)h(eie) u(peie)g(eie)]dB = 2n[u(O)h(O)+ v(O)g(O)] = 0.
Assuming the theorem has been proved for the index p , it follows that
4.1 THEOREM
OF M. RlESZ 55
Now take the supremum of 1(1/2n)jvg(over all g with llgllp I 1, and use the fact that trigonometric polynomials are dense in Lp.The result is
Thus we may assume 1 < p I 2. The next step is to apply Green’s theorem in the form
where V2 denotes the Laplacian. We suppose for the moment that u(z) > 0, and compute
By Green’s theorem, then,
Integrating from 0 to
Y,
and recalling that v(0) = 0, we find
If(reie))lPd0 - 2n[u(O)lp5
p-1
(jznlu(reiR)Ip d6 - 2n[u(O)Ip 0
Thus
if u(z) > 0. For general u E hP,fix Y < 1 and set ~ ~ ( =6 max{u(reie), ) O} ;
Then
~ ~ ( =6 maxi ) - u(reie), O}.
56
4 CONJUGATE FUNCTIONS
For 0 Ip < 1, define
Let u j be the harmonic conjugate of uj . Then
u(pre") = ul(peie)- u2(peie); +reie) = ul(peie) - vz(peiO). Since u1 and u2 are positive harmonic functions, it follows from what has been proved that
Letting p + 1 and using relations (4) and (5), one now obtains (3) with
The factor 2 can be removed by a more refined argument (see Exercise 5). As pointed out iil Section 3.1, Riesz's theorem breaks down for p = 1. The Poisson kernel P(r, 0) E hl, but its analytic completion (1 + z)/(l - z ) $ H I . The next two sections will be concerned with weaker theorems which may be said to replace Riesz's theorem in the case p = 1. Let us note that Riesz's theorem also fails in the case p = co. Consider, for example, the function
f(z) = u ( z ) + iv(z)
=i
l+z log -, 1-2
which maps IzI < 1 conformally onto the vertical strip -n/2 < u < n/2. Here u is bounded, but u is not. On the other hand, Riesz's theorem guarantees that f E H P for all p < co,a fact not easy to verify by direct calculation. 4.2. KOLMOGOROV'S THEOREM
Although the harmonic conjugate of an 12' function u(z) need not be in h', it does belong to hP for all p < 1. We have already proved this (Corollary to Theorem 3.2) by appeal to Littlewood's subordination theorem. We now give an independent proof of a slightly stronger result. The following lemma will be needed.
4.2 KOLMOGOROVS THEOREM
LEMMA.
For arbitrary positive numbers a and b,
ap + bP,
PROOF.
57
O
It is enough to find the supremum of
for x 2 1. Differentiation gives
g’(x) = p ( l
+ x)p-l(l + x p ) - 2 ( 1
- xp-’).
Thus g(x) is decreasing in the interval 1 I x < 00 if p > 1, increasing if 0 < p < 1. Since g(1) = 2p-1 and g(x) -+ I as x + co,the lemma is proved. For p > 1, the result also follows easily from the fact that xp is a convex function. THEOREM 4.2 (Kolmogorov). If u E h‘, then its conjugate o E hP for all p < 1. Furthermore, there is a constant B, , depending only on p , such that
M&r, v) 4 B,M,(r, u),
0I Y < 1,
for all u E h’. PROOF.
Assume first u(z) > 0, and set f(z)
= u(z)
+ io(z) = Re“,
I@\ < n/2.
Then F(z) = [f(z)]” = RPeiPQis analytic in IzI < 1. By the mean value theorem,
Hence
1
2.n
s,
2*
RP cos pa do = [ M l ( r , u)]”.
In view of the inequalities Iv(z)l < R and 0 < cos pn/2 < cos p@, it follows that
M ~ ( Yv), < (sec p 7 ~ / 2 > ~ ’ pu) ~~(r, forpositive u. The constant is in fact best possible for u(z) > 0 (see Exercise 3).
58 4 CONJUGATE FUNCTIONS
The extension to general ti E hl is similar to that in the proof of Riesz’s theorem. Using the same notation, we find by the lemma that
4.3. ZYGMUND’S THEOREM
Since u E h’ is not enough to ensure v E hl, but the stronger hypothesis u E hP (for some p > 1) is sufficient, it is natural to ask for the “minimal” growth restriction on u which will imply u E h’. Such a condition is the boundedness of
We shall denote by h log’ h the class of harmonic functions u(z) for which these integrals are bounded. Clearly, hP c h log’ h for all p > 1. THEOREM 4.3 (Zygrnund).
If u E h log’ h, then its conjugate u is of class
h’, and 2R
2n
j
Jo Io(reie)l d9 I lu(reie)l log’ lu(reie)I d9 0
+ Gze,
0 I r < 1.
PROOF. The proof is similar to that of Riesz’s theorem. Assuming first u(z) 2 e , one finds
V2{u1og u> = lf’I2/u. V2{lfl> = lf’I2/ I f l , Thus V2 { I f 1 } I V2 { u log u } , and an application of Green’s theorem gives, as before, 2n
2a
Jo lf(reie)l dB - 27cu(O) I u(reie)log u(reie)d9 - 27-40) log u(0). 0
4.3 ZYGMUND'S THEOREM
From this we conclude [since u(0) 2 el
provided u(z) 2 e. For general u(z), fix r < 1 and let
u,(e)= max{u(rei6), e}, u,(d) = max{-u(reie), U3(d)= u(reie) - [u,(o) - u,(d)].
e},
Thus lU3(d)l Ie. For 0 5 p < 1, let uj(peie)=
1
1
2n
0
P(p, 8 - t ) U j ( t )dt,
j = 1, 2, 3,
and let v j be the respective harmonic conjugates. Then
u(preie) = ul(peie)- u2(peiB)+ u3(peie), v(preie)= vl(peis)- v,(peie) + u3(peie).
By what we have already proved, Ml(P-2 u) s M 1 h V l ) + M l ( P , vz> + M d P , u3) 1 2 2n I -C luj(pei6)llog' luj(peie)l dd
1
2nj=1 0
+ M l ( p , v3).
On the other hand, applying the Schwarz inequality and (2), we find M1(P, v3) 5 MZ(P, 0 3 ) I M2(P, u3) 5 e.
Now let p tend to 1, with the result 27cM1(r,v ) <
2
j=1
2n
IUj(d)l log'lUj(d)l do 0
+ 271..
To complete the proof, consider the subsets of [O, 2x1 E , = (0 :u(reie)2 e ) and E , = { d : u(reie) 5 - e l .
Then
2n
I
jo lu(reie)llog+)u(rei6)1dd + 47re.
59
60
4 CONJUGATE FUNCTIONS
Zygmund's theorem is, in a sense, best possible: the growth restriction it imposes on u(z) cannot be weakened. The following is a partial converse. THEOREM 4.4. If both u and its conjugate v belong to h', and if u(z) > C for some constant C, then u E h log' h. PROOF. We may assume u(z) > 1, since addition of a constant to u does not affect u. Set
f ( z ) = U(Z)
+ ic.(z) = ReiQ,
I@l < 71/2.
The function f ( z ) logf(z) is analytic in IzI < 1. Applying the mean value theorem to its real part, we find
I', R cos CD log R
1
2n
2%
do =
Thus
I',
@Rsin @ dO + 2nu(O) log u(0).
0
2n
u(reiQ)log u(reie)dB 5
s
2n
u log R dO
0
COROLLARY. Let q(x) 2 0 be a convex nondecreasing function of x 2 0, such that q ( x ) = o(x log x) as x --* 00. Then there exists a harmonic function u(z), with conjugate ~ ( z )such , that
is bounded for r < 1, while u # h'. PROOF. Choose an integrable function U ( t ) 2 0 on [0, 2x1 such that cp(U(i(t))is integrable but U ( i ) logfU(t) is not. (We will show in a moment that this is possible.) Let u(z) be the Poisson integral of U(t). Then u(z) > 0, and an application of Jensen's inequality (Chapter 2, Exercise 4) shows that Sq(u(re'"))dtl is bounded. By the preceding theorem, u ~ h 'would imply u E h logt h, which is not the case (by Fatou's lemma). Thus u $ h'. It remains to show that a function U(t)can be constructed with the required properties. Suppose, more generally, that q(x) and $(x) are any continuous, nonnegative, nondecreasing functions (convex or not), defined on 0 I x < KJ, such that $(x) + co and cp(x)/$(x) -+ 0 as x + co. Choose increasing numbers
4.4 TRIGONOMETRIC SERIES
x,(n = 1,2, . . .) such that $(xn)= 2". Then 2-"(p(x,) Ink} for which
--f
61
0 ; select a subsequence
Now define to = 0 and t k = t k- l + 2-nk, k 2 1. Notice that 0 < t , < t , < * . * 5 1. Let T = lim t k . Finally, let U ( t ) be the step function with the value xnkin the interval t k - l 5 t < t , and with U ( t ) = 0 in T 5 t 2n. Observe that
C 2-"kcp(~,k)+ ( m
=
2 -~ T)cp(O)< a.
k= 1
On the other hand, $(U(t)) is not integrable because 2-""$(x,,)
= 1.
4.4. TRIGONOMETRIC SERIES
The preceding results can also be expressed in the language of trigonometric series. We shall now briefly indicate the connection between the two theories. A formal trigonometric series
- + C (a, cos nO + b, sin no)
a0
2
n=l
(6)
is called a Fourier series if there exists an integrable (real-valued) function q(t)such that 1 2R a, = - cos nt q(t) d t , n = 0,1,2, . . . ;
/
n o 1 2n b, = sin nt ( p ( t ) d t , n o
n = 1,2, ... .
(7)
It is called a Fourier-Stieltjes series if there is a function p ( t ) of bounded variation such that (7) holds with q(t)dt replaced by d,u(t). The formal series 00
(- b, cos nO n=l
+ a, sin no)
(8)
is called the conjugate trigonometric series of (6). The use of the word conjugate is easily justified. Assuming that (6) is the Fourier series of q(O), consider the Poisson integral u(z) =
2721
1
2*
0
P( r , 0 - t ) q ( t ) dt.
62
4 CONJUGATE FUNCTIONS
Let 1 2neif+z f(z) = u(z) + iu(2) = 4o(t) dt 2n J0 eif- z
(9)
be its analytic completion, normalized as usual by v(0) = 0. The expansion
and the relations (7) now give {(z) =
a0
a,
+nC= c,z",
C,
= a, - ib, ,
1
Taking real and imaginary parts, we find a0 2
u(z) = -
a,
+nC= 1 ?(a,
cos ne + b, sin ne),
co
v(z) = C r"( - b, cos n8 + a, sin no). n=l
These series are convergent for every r < 1. On the other hand u(reio)--f (p(8) a.e.,
by Theorem 1.2. In the language of summability theory, we may therefore say that the series (6) is Abel summable to q ( 0 ) for almost every 8. It is known, however, that if cp is merely integrable, the series (6) need not converge anywhere in the usual sense. A similar statement may be made about the conjugate series (8). If (6) is a Fourier series, or even a Fourier-Stieltjes series, then u E h'. Hence, by Kolmogorov's theorem, v E hP for all p < 1, and (by Theorem 2.2) u(z) has a radial limit a.e. The function ~ ( 8=) lim v(reie) r+1
is called the conjugate function of (p(8). Thus the conjugate series of the Fourier series of q is Abel summable almost everywhere to @(d), and 4 E Lp for a l l p < 1. Must 4 be integrable? If so, then v E h1 (Theorem 2.1 l), and v(z) is the Poisson integral of 4; thus (8) is actually the Fourier series of 4. Conversely, if (8) is the Fourier series of some integrable function, that function must be 4. Thus 4 E L1if and only if the conjugate series (8) is itself a Fourier series, or if and only if v(z) E hl. If 4 is not integrable, then the conjugate series is not even
4.5 THE CONJUGATE OF AN h' FUNCTION
63
a Fourier-Stieltjes series (by Theorem 1.1). An example constructed in Section 4.5 will show that this can actually happen. If cp is more than integrable, a correspondingly stronger statement can be made about $5. Suppose, for example, that cp E Lpfor somep, 1 < p < co.Then u E hP, so u E hP by Riesz's theorem, and @ E Lp.In fact, there is a constant A , suchthat ll$511p I A,((cp(l,.Ifcp ~Llog+L(notationobvious), t h e n u c h log'h, and Zygmund's theorem tells us that u E hl; hence $5 E L'. It was in terms of cp and @ that these theorems were originally formulated. There is another representation of @(O) which is more direct and which is convenient for some purposes. If we take the imaginary part of (9) and let r -+ 1 under the integral sign, we find formally
Unfortunately, this last integral need not converge, because of the singularity at t = 8. The formula is correct, however, if the integral is interpreted in the Cauchy principal-value sense. That is, after changing variables and extending cp periodically,
1
1 " @(O) = lim - cot(t/2)[cp(O - t ) - q(0 + t ) ] d t a.e. &+0271&
We omit the proof, since we shall make no use of this formula. Proofs may be found in most books on trigonometric series; for example, in Bary [l], Vol. 2, pp. 60-62. 4.5. THE CONJUGATE OF A N h'
FUNCTION
If u(z) and its conjugate u(z) both belong to h', then u(z) is a Poisson integral (Theorem 3.1); that is, the function p(t) in its Poisson-Stieltjes representation must be absolutely continuous. The converse is false. Even if u(z) is a Poisson integral, v(z) need not be of class hl. We are about to present a counterexample. In view of Hardy's inequality (Corollary to Theorem 3.15), the function X f ( 2 )
c -$HI. , log n z"
=
=2
However, its real part belongs to hl, and is in fact a Poisson integral. To see this, it is enough to observe that
:=
,,=2
log n
64 4 CONJUGATE FUNCTIONS
is the Fourier series of an integrable function. But this is a particular case of the following more general proposition.
If the real sequence { a o ,a,, . . .} is convex and a, + 0 as then the series
THEOREM 4.5.
n
--f
03,
a0
-
2
+ C a,, cos nt
(10)
n=l
converges for all t # 0 in [ - T C , T C to ] a nonnegative integrable function q(t), and (10) is the Fourier series of cp(t). PROOF. Let A an = a,,,, - a, and A’ a, = A(A a,,). The convexity of {a,} means that A’ a,, 2 0; thus {A a,,) is an increasing sequence. Since {a,,} converges, A an+ 0; hence A a,, I 0. Since a,, -+ 0, it follows that a,, 2 0. Let S N ( t )denote the Nth partial sum of (10). After two summations by parts, we find N
Sdt)=
c (A’ %J(n + 1)K,,(t)
n=O
-(A a,+,)(N
+ l ) K v ( t ) + a,+,~,(O,
(11)
where
1 N sin(N + +)t + cos nt = 2 n=l 2 sin t/2
C
~,(t) =-
and
are the Dirichlet and the FejCr kernel, respectively. For each t # 0, the sequences { D N ( t ) }and {NK,-,(t)) are bounded; hence the last two terms in (1 1) tend to zero. Consequently,
Because (n + I)K,,(t) is uniformly bounded in each set 0 < 6 5 It] 5 n, the series (12) converges uniformly there, and so represents a function q ( t ) continuous for t # 0. Since the terms of the series are all nonnegative, so is q(t). By the Lebesgue monotone convergence theorem, m
q(t)cos mt dt
=
1(A’ n=o m
= 7c n=m
a,,)(n + 1)
jrrK,,(t)cos mt dt -rr
(A’ a,,)(n - m + 1),
172
= 0, 1, 2,
. .. .
4.6 THE CASE p < 1 : A COUNTEREXAMPLE
65
This last series may be evaluated through summation by parts:
c (A2 a,)(n
k
k
n=m
-m
+ 1) = - 1 A a, + ( k - m + 1) A a k + l = + ( k - m + 1) A n=m
Urn
ak+l
Uk+l.
As k -+ co, the right-hand side approaches a,, , and so
1
1 " -
71
m = 0 , 1, 2 , . . . .
q ( t ) cos mt dt = a,,
-n
[In particular, q ( t )is integrable.] Thus (10) is a Fourier series, which was to be proved. The fact that k A ak -+ 0 follows from a classical theorem of Abel : Zf { b k }is a monotonically decreasing sequence and 6 , converges, then kb, + 0. For a proof, observe that 2"
4.6. THE CASE p
<1:
A COUNTEREXAMPLE
For 1 < p < co, the class hP is preserved under conjugation. This is false for h', but it is " almost true " in the sense that the conjugates of h' functions belong to hPfor all p < I. If the hypothesis is further weakened by requiring only that u E hp for some p < 1, hardly a trace of the theorem remains. The conjugate function v may not belong to h4 for any positive q. In fact, we are about to construct an analytic functionf(z) = u(z) + iv(z) such that u E hPfor all p < 1, yet v $ hP for any p > 0. The example is
(13) We are going to show that for every choice of the signs E, ,u E hPfor all p < 1 ; while for "almost every" sequence of signs,f(z) has a radial limit on no set of positive measure. In particular, some choice of the F , gives a functionf(z) which is not even of class N , but whose real part belongs to hPfor all p < 1. LEMMA.
For each p > 3,
66
4 CONJUGATE FUNCTIONS
PROOF.
Since sin x 2 (2/n)xfor 0 I x 5 $2,
+ 4r sin2 812 2 (1 - r)' + (4r/n2)Q2.
1 - 2r cos 8 + r 2 = (1 - r)'
Hence, for r 2 4, the integral does not exceed
.L
dtl
[ ( I - r)2
dt
m
1
+ 2n-2e21p < ( 1 - r l z p - 1
S , [ I + 2n-2tZIP'
The last integral is convergent because p > 3. In showing u E hP for all p < 1, we may suppose p > 4. With z = rei8, we begin by computing
Rn(l - R:) cos 2"8 Re( 1 - z ~ " ' ~ )1 - 2R: cos 2""8 + R:' Z2"
Rn = rZn
From this we find, using the lemma,
I
2%
m
1
lu(reiB)IPd8 I RflP(l- R,2)p n= 1
dtl ( 1 - 2Rf12 cos 8
W
1RZ(1 - R n ) l - P ,
I A C RZ(1 - R, )l - p _< 2 1 - p A n= 1
n= 1
where A is a constant. But R,P(~ - Q - P since 1 - r m
= (1 - r)(l
+r +
1- rp 1-r
+ R,4)p
m
< $ " ~ 2 n ( l - ~ )-( lr ) l - P , a * -
__-p
+ rrn-') < m(1 - r). Furthermore,
+ 0(1 - r )
(r + l ) ,
by the binomial expansion of r p = [ l - ( 1 - r)]". We will have shown u E hP,then, if we prove m
C 2nar2"= O((1 - r)-"), n= 1
CI
> 0.
(14)
It is equivalent to estimate the integral
Since log r < r - 1, the second integral is majorized by Jlme-tl-r)u
u a - I du = ( 1
- r)-"
m
e-"o'-l 1-r
which proves (14). Thus u E hP for all p < 1.
do,
EXERCISES
67
To prove the nonexistence of radial limits we appeal to the theory of Rademacher functions, as developed in Appendix A, and to Theorem A.4 in particular. For the present application we have
Thus a,
do
C 1gn(Z)12 =n C= N Rn2(1- 2R2 cos 2"+'6' + R:)-' n=N
as r .+ 1. Since the other hypotheses of Theorem A.4 are obviously satisfied, we conclude that for almost every choice of signs, the analytic function (1 3) has a radial limit almost nowhere, which is what we wanted to prove. EXERCISES
1. Show that if p E Lp (1
< p < co) and
then its " analytic projection " f ( z ) = , anzn belongs to H P , and Ilffl, 5 A,llpJlp for some constant A , independent of p. 2. Show that the Poisson kernel P(r, 6) does not belong to hPfor any p > 1. 3. Show that the constant B, = ( s e ~ p n / 2 ) lis/ ~best possible for the special case of Kolmogorov's theorem in which u(z) > 0. (Hint: Try the Poisson kernel.) 4. Use Green's theorem to prove Kolmogorov's theorem with B, = 21/P-1(1 - p ) - l / P .
+
5. Letf(z) = u(z) iv(z) be analytic in (21 < 1, and suppose u E hPfor some p, 1 < p 2 2. For fixed E > 0, set
+
G(z) = { [ u ( z ) ] ~ E } " ~ ;
H(z) = {lf(~)1~
Show that
V 2 N I -V2G. P-1
+ E}'".
68 4 CONJUGATE FUNCTIONS
Apply Green’s theorem, integrate, and let E + 0 to obtain
thus proving the theorem of M. Riesz with a smaller constant. (This idea is due to W. K. Hayman. It is an open problem to find the best possible constant, even for u(z) > 0. The above constant reduces to J?for p = 2, while the best possible constant in this case is 1.) 6. For the example of Section 4.6, show that there is a choice of signs such that
NOTES
Theorem 4.1, or an equivalent form of it, is in the paper of M. Riesz [I]. The proof based on Green’s theorem, as presented in the text, is due to P. Stein [l]. This approach has the advantage of leading to a relatively good value of the constant A,, . Calder6n [I] has given still another proof; see also Zygmund [4], Chap. VII. Shortly after Kolmogorov [l] proved Theorem 4.2, Littlewood [2] suggested the proof using his subordination theorem. Hardy [3] then discovered the elementary argument given in the text as a second proof. Theorem 4.3 and the converse results (Theorem 4.4 and its corollary) are due to Zygmund [1, 41. Another proof is in a paper of Littlewood [3]. Theorem 4.5 and similar results may be found in Zyginund [4], Chap. V. Phenomena of the type illustrated in Sec. 4.6 are discussed in the paper of Hardy and Littlewood [6]. Previously, Littlewood [l] had constructed a harmonic function which belongs to hPfor allp < 1, yet has a radial limit almost nowhere. The example (13) given in Sec. 4.6 is essentially due to Hardy and Littlewood [6], who showed by a highly nonelementary argument that a function similar to this (but with E, E 1) fails t o have a radial limit on some set of positive measure. Paley and Zygmund [2] introduced Rademacher functions and “ constructed ” an example of the type (13) for which f $ H Pfor all p < 1. The argument given in the text is considerably simpler than theirs. Hardy and Littlewood [6] also proved that if u E hP for some p I 1, then its conjugate satisfies
For p = 1, +,+, . . . , they showed by an elementary example that this estimate is best possible. Whether or not it can be improved for other values of p
NOTES
69
remains an open question, although Swinnerton-Dyer [l] has shown that it cannot be improved to
(Unfortunately, Hardy and Littlewood stated the positive result incorrectly in the introduction to their paper, and Swinnerton-Dyer reproduced this error.) Gwilliam [I] simplified some of the work of Hardy and Littlewood in this area.
This Page Intentionally Left Blank
MEAN GROWTH AND SMOOTHNESS
CHAPTER 5
If f ( z ) is analytic in the unit disk, there is a very close relation between the mean growth of the derivative f ' ( z ) and the smoothness of the boundary function f (eie).This principle takes several precise forms, as we shall see in the present chapter. Other topics to be discussed are the relations between the growth of Mp(r,f ) and MJr, f '), between the growth of M J r , f ) and Mq(r,.f), and between the growth and smoothness of a harmonic function and its conjugate. Some of these results will be applied repeatedly in later chapters. We shall begin by discussing several ways to measure smoothness and exploring the connections between them. 5.1. SMOOTHNESS CLASSES
Let cp(x)be a complex-valued function defined on - co < x < co and periodic with period 2n. The modulus of continuity of cp is the function
4 9 = W ( t ; cp> =
SUP lcp(x) - d Y ) l .
Ix--Yl
Thus cp is continuous if and only if o(t) 0 as t --f
4
0. We say that cp is of class
72 5
MEAN GROWTH AND SMOOTHNESS
A, (0 < a 5 1) if o(t)= O(t7 as 1 -+ 0. Alternatively, A, is the class of functions which satisfy a Lipschitz condition of order a :
IcpW - cp(Y)l
5 A I(x - Al".
The definition is of no interest for a > 1, since A, would then contain only the constant functions. It is clear that AD c A, if a < p. A continuous function q ( x ) is said to be of class A* if there is a constant A such that Iq(x
+ h) - 2cp(x) + cp(x - h)l I Ah
(1)
for all x and for all h > 0. The condition (1) alone does not imply continuity; indeed, it is well known that there exist nonmeasurable functions such that
cp(x + v) = c p ( 4
+ cp(y>.
For any a < 1, the proper inclusions A, c A* c A, hold. In fact, every cp E A, has modulus of continuity o(t)= O(t log lit), as we shall see later. (See also Exercise 14.) It often turns out that A,, not A,, is the " natural limit " of A, as a 3 1. Another class, apparently larger than A,, could be defined by requiring that the left-hand side of (1) be O(ha). However, this class is actually the same as A, if a < 1. For cp E Lp = Lp(O,2n), 1 5 p < co, the function o,(t>= % i t ; cp) =
SUP O
~JoZZIcp(x+ h ) - cp(x)J Pdr)
1i P
is called the integral modulus of continuity of order p. For every Lp function cp, w,(t) + 0 as t -+ 0, by a theorem of F. Riesz (see, for example, Titchmarsh [l], p. 397). If wp(t)= @la), 0 < a I 1, we say that cp belongs to the class AaP.Because the Lp means increase with p , it is evident that A 2 c AaPif p < q. If cp is continuous, w,(t) -+ w(t) a sp + 03. The classes A/ may therefore be viewed as generalizations of A,. Obviously, A, c Aap. I n the proof of Theorem 5.4 we shall need the following result, which is of some interest in itself. L E M M A 1 (Hardy-Littlewood). If cp is of bounded variation over [o, 2711, then cp E A l l . Conversely, every function cp E A l l is equal almost everywhere to a function of bounded variation. PROOF. Suppose first that cp(x) is of bounded variation, and let V(x) be the total variation of cp on [0, x]. Then for small h > 0,
5.1 SMOOTHNESS CLASSES
The converse is more difficult. Assuming y
E
73
A l l , let
ja
F(x)=
y(t)di
and y J x ) = n[F(x
+ l/n) - F ( x ) ] .
Then y,(x) is absolutely continuous and yn(x)-+ y(x) a.e. as n -+ co. For h > 0 we have y,(x Since y
E
+ h ) - Y,(x) =
h
[y (x
y1.r
A l l , it follows that
Jbzn/Y.(x + h ) - cP,,(x>/dx 5
0
J:
dt
+ 1+ t )
s:'
Iy(x
-
y(x
+ t ) ] dt.
+ -I + t ) - y(x + l ) l dx n
5 Ch, where the constant C is independent of n. Thus, by Fatou's lemma
1 J r \ y d ( x ) \dx 5 lim inf h-0
2n j0
\y,(x + h ) - (pn(x)I dx 5 C.
Now let 0 = xo < x1 < . * * < x, = 2n be an arbitrary finite partition of the interval [0,2n]. Bearing in mind that yn(x) is absolutely continuous, we have m
.2z
But yn(x) y(x) if x is not in a certain set E of measure zero. Hence for any partition such that xkC$ E, --f
It can now be shown from (21, by imitating the classical argument (see, for example, Rudin [ 7 ] )that y ( x ) coincides on the complement of E with the difference of two monotonic functions. Consequently, y ( x ) has a natural extension to [0, 2x1, and the extended function satisfies (2) for all partitions. This proves that q(x) coincides almost everywhere with a function of bounded variation.
74 5 MEAN GROWTH AND SMOOTHNESS
5.2. S M O O T H N E S S O F THE BOUNDARY FUNCTION
Generally speaking, it is reasonable to expect an analytic function to be smooth on the boundary if its derivative grows slowly, and conversely. For the unit disk, this principle can be expressed i n surprisingly precise form, as follows. T H E O R E M 5.1 (Hardy-Littlewood). Let f ( z ) be a function analytic in IzI < 1. Then f ( z ) is continuous in IzI 2 1 and f ( e i e )E A, (0 < a I l), if and only if
PROOF. First let us dispose of the case a = 1. By Theorem 3.1 1, the Lipschitz condition on the boundary implies that f'E H 1 and that f ' ( e i e )E L". Hencef' E H ", by Theorem 2.11. Conversely, iff' E H", another application of Theorem 3.11 shows that f ( z ) is continuous in IzI I 1, f ( e i e )is absolutely continuous, and (d/dO)f(eie)E L". Integration of the derivative therefore gives f ( e i e )E A l . Now let f ( z ) be continuous in IzI 5 1, and suppose f ( e i e )E A,, 0 < u < 1. By the Cauchy formula,
Thus
Using the Lipschitz condition and the relation
1 - 2r cos t
+ r2 = (1
-
r)'
+ 4r sin2 42 -> (1 - r>2 + 2,4rt2 n
0I tI n,we find
The assumption u < 1 assures the convergence of the integral u"du
lom 7 which arises after the substitution u Conversely, suppose
=
t/(l - r).
5.2 SMOOTHNESS
OF
THE BOUNDARY FUNCTION
75
Then the radial limit f ( e i o )= f(0)
+ lim R-1
exists for every 0. Furthermore, f~ H " ,
I
f'(re'') dr
SO
f ( z ) is the Poisson integral of
f(e'@).Hence the continuity of f ( e i o )would imply the continuity of f ( z ) in Iz[ 5 1. We shall prove the continuity by showing f ( e i s )E A,. For this purpose, choose 0 and q with 0 < cp - 0 < 1. Fix p , 0 < p < 1, and let be the contour consisting of the radial segment from eis to pei8, the arc of the circle [ z [= p from pe" to pe", and the radial segment from peiV to eiq.(See
Fig. 2.) Then
Figure 2
f(e'")
r:
- f ( e i s )= f'(5) d5.
Breaking up the integral into its three components, we find [ f ( e i q )- f(e'@)I5
If'("')[
+ lp[ f ' ( r e i q ) [dr + JoVlf'(peif)[d t 1
dr
76 5 MEAN GROWTH AND SMOOTHNESS
With the choice p
1 - (cp - e), this gives
=
and the proof is complete. The method may be used to prove other theorems of the same type. We mention one example, leaving the proof as an exercise. THEOREM 5.2.
Letf(z) be analytic in jzI < 1, and suppose f'(z)
=0
i
log-
l!?
Thenf(z) is continuous in IzI I 1, and f ( e i B )has modulus of continuity o(t) =
o(t log
i).
According to Theorem 5.1,f' E H" if and only iff(eiB) E Al. If the growth condition is slightly weakened to f"(z) = O((1 - r ) - ' ) , Theorem 5.2 says that f(eis) must still be "almost" of class A, : w ( t ) = O(t log lit). This degree of smoothness, however, does not ensure thatf"(z) = O((1 - r)-'). The necessary and sufficient condition is that the boundary function belong to the class A, introduced in Section 5.1. The precise result is as follows. THEOREM 5.3 (Zygrnund). Let f(z) be analytic in IzI < 1. Then f ( z ) is continuous in IzI I 1 andf(eie) E A, if and only if
f"(z)
1Y).
=0 -
il
(4)
COROLLARY. If f(z) is analytic in Iz/ < 1, continuous in (zI 5 1 , and f ( e i B E) A, thenf(e") has modulus of continuity o(t)= O(t log lit).
PROOF OF THEOREM.
lff(z) is continuous in IzI I 1, it can be represented
as a Poisson integral: f(z)
1 Z J" ~ ( r ,- t)f(ei'>d t , 2n - n
=-
e
z
= re''.
Since the second derivative P8@(r,0) is an even function of B and PBo(r,t ) d t Jo
= Po(I*,
n) - Po(r, 0) = 0,
(5)
5.2 SMOOTHNESS OF THE BOUNDARY FUNCTION
77
it follows that 1 " f o O ( z )= 2 7 ~/oPOO(r, t ) { f ( e " e ~ '-~ )2f(eiB)
+ f(e'('-'))} d t .
The hypothesis f ( e i o )E A, therefore implies
since P,,(v, 0) 2 0 for 0 I 0I 71 and r 2 2 -$, as a calculation shows. On the other hand, ( 5 ) and the boundedness of f ( e i ' ) easily show
Thus
The proof of the converse is more difficult. In view of Theorem 5.1, the condition (4) implies f ( z ) is continuous in IzI 5 1 ; it must be shown that f ( e i e )E A,. For 0 < h < 1, let us use the notation G(0) = G(O Ah2 G(8) = G(O Ah
+ h) - G(O), + 2h) - 2G(d + h) + G(8).
We are required to show that A h 2 f ( e i B= ) O(h), uniformly in 0, as /I strategy is to write
-+0.
Our
+
Ah2f(eioj= A h 2 { f ( e i e) f(peis)>> Ah2f(pe")
(6) (0 < p < l), to set p = 1 - h, and to show that as h + 0, each of the two terms in (6) is uniformly O(h). The identity
+ e2io / (1 - r)f"(reioj d r I
f(eio> - j ( p e i 9 )
= (1
-
p)e'y'(pei8)
(7)
P
is easily verified through integration by parts. Now set p = 1 - / I . Under the hypothesis (4), the integral in (7) is then uniformly O(h). Thus
A h 2 { f ( e i e) f ( p e i o ) }= h Ah2{eiBf'(peie)}+ O(h).
(8)
But Ah{eiBf'(peio)}= Ah{e'8}f'(pei(B+'')) + eis Ah{f'(peie)} =0
(h log :I) -
+ ipeiO
l
ei(e+Of"(pei(Q+r)
1 d t = 0(1)7
uniformly in 0. Hence the expression (8) is uniformly O(h).
(9)
78
5 MEAN GROWTH AND SMOOTHNESS
Finally,
Analyzing the integrand as in (9), we see that it is dominated by
[ ;
Ct log - + - , where Cis independent of 8. Hence A: f (peie)is uniformly O(h),and the proof is complete. The next theorem may be viewed as the Lp analogue of Theorem 5.1. It is included here for the sake of completeness; we shall make no use of it in this book. THEOREM 5.4 (Hardy-Littlewood).
f E H P and f ( e i e )E AaP(1
5p
Let f ( z ) be analytic in (zI < 1. Then
< 03 ; 0 < a 5 1) if and only if 1
- r)l -a)*
f') = '('1
~ p ( r >
(10)
PROOF. We first show that the condition (10) for any G! > 0 impliesf€ HP. It is convenient to assume f ( 0 ) = 0, so that
f ( r e i e ) = f f ' ( p e i e ) e i ed p . 0
Applying the continuous form of Minkowski's inequality, we find
hence f E H P . Now let us deduce from (10) that f ( e i o )E AnP.Let 0 < p < Y < 1 and 0 < cp - 8 < 2n. Then f ( r e i q ) - f(re")
=
1rf ' ( Q d i ,
where the contour goes radially from reie to pe", along radially to reiq. It follows that If(reiq) - f ( r e i e ) JI J'rlf'(seie)Ids
[zI
= p to peiq, then
+ flf'(se''P)Id s
P
rlf
' P
+
e
'(pe")I dt.
5.3 GROWTH OF FUNCTION A N D DERIVATIVE
79
Now choose h, 0 < h < 4,and let cp = 8 + h. Applying Minkowski's inequality in both discrete and continuous form, we obtain
[J;
f(re'@+k)) - f(reie)IPdB)'" I 2 J i M p ( s ,f') ds + hM,(p, f').
(11)
The parameters r and p are still free. Assuming r > 1 - h, set p = r - h. Using the hypothesis (lo), we conclude that for all r > 1 - h, the left-hand side of (1 1) is no larger than Ch", where the constant C is independent of r. Letting r 1, it follows that f (8') E AmP. To prove the converse, we begin with the estimate (3) for [ f ' ( z ) ] .If f ( e i e )E Amp,Minkowski's inequality gives --f
Hence, as in the proof of Theorem 5.1, we obtain (10) for CI < 1 . If ct = 1 , it must be shown that f E H P and f ( e i R ) E Alp imply f'E H P , 1 I p < 00. Consider first the case p = 1. Then f E H' and by Lemma 1, f ( e i @is ) equivalent to a function of bounded variation. Invoking Theorems 3.10 and 3.11, we conclude that f ( e i e ) is absolutely continuous, ,f' E H ' , and the radial limit f '(e") is obtained by differentiation off(e'@).If it is also known thatf(e") E Alp for some p > 1, then
It then follows by use of Fatou's lemma that the H' function f '(2) has an Lp boundary function. Therefore f'E H p , by Theorem 2.1 1. 5.3.GROWTH OF A FUNCTION AND ITS DERIVATIVE
Our next object is to explore the relation between the mean growth of an analytic function and that of its derivative. We begin with a lemma which is especially useful in dealing with the case p < 1. L E M M A 2. Any function f E H P (0 < p 5 co)can be expressed in the form f ( z ) = f l ( z ) +f2(z), wheref, and f2 are nonvanishing H P functions such that llLlllP 211fllp , = 1, 2. PROOF. If f ( z ) # 0, we may take fi = f 2 = +J I f f has zeros, we apply Theorem 2.5 to write f ( z ) = B(z)g(z), where B is a Blaschke product and g is an H p function with no zeros. Thus
f(z)= C W - llg(4
+ g(4.
80 5 MEAN GROWTH AND SMOOTHNESS
THEOREM 5.5.
Suppose 0 < p I 00 and fl > 0, and let f ( z ) be analytic in
JzI < 1. Then
1
if and only if
PROOF. First let I - < p I 00, and assume that (12) holds. By the Cauchy formula,
where p
= t(l
+ r ) . Minkowski's
inequality (in continuous form) then gives
Conversely, if (13) holds, we apply Minkowski's inequality to the relation
and obtain
We remark that the proof could also have been based on Jensen's inequality. Suppose now that 0 < p < 1 and that
Under the preliminary assumption that f ( z ) # 0 in Iz/ < 1, the function F(z) = [ f ( z ) l Pis analytic and
5.3 GROWTH OF FUNCTION AND DERIVATIVE
With p
= $(I
81
+ r ) , this implies
On the other hand, f’(2) =
$ [F(z>]””-’F’
(2>;
thus Holder’s inequality gives
Iff(z) has zeros, we fix p (0 < p < 1) and appeal to the lemma to write
f(P4
=
fi(4 +f m ,
where fiand fiare nonvanishing H P functions such that
and the constant C is independent of p . Hence, by the result just obtained,
Since P P I f ’ ( P 4 l ” 5 If1’(z>lP+ If2’(Z>lP, it follows that
where K is independent of p . This easily gives (13). The proof that (13) implies (12) when 0 < p < 1 is much more difficult, and is omitted here (see Notes). The theorem says, roughly, thatf’ grows faster thanfby a factor (I - r>-’. The same phenomenon is reflected in the following more delicate result, which will be needed later in this chapter. THEOREM 5.6. Let f ( z > be analytic in IzI < 1. Then for 1 < p < 00, 1 < a < co,and - 1 < b < cx), 1
j,(l - r ) b { M p ( rf>>” , dr I C(jol(l - r)a+b{Mp(r,f’)>”d r + lf(O)l“), where C is a constant independent of$
82
5 MEAN GROWTH A N D SMOOTHNESS
PROOF.
First supposefis analytic in the closed unit disk, and integrate by
parts:
But
and a calculation based on the Cauchy-Riemann equations gives
Thus it follows from Holder's inequality that
j
a
1
f>>"-l%(r, f ' ) .
{filp(r>f>>" 5 4 M p ( r ,
Consequently, the integral on the right-hand side of (14) is dominated by (1 - J - ) ~ + ' ( M ~f)>"-'M,(r, (Y,
f') d r
where Holder's inequality has been used again. This easily gives the desired inequality. Iff(z) is not analytic in IzI 5 1, replace it byf(pz), where 0 < p < 1, and let p tend to 1 to obtain the result by means of the Lebesgue monotone convergence theorem. 5.4. MORE ON CONJUGATE FUNCTIONS
We saw in Chapter 4 that the class hPis preserved under harmonic conjugation if 1 < p < 00, but not if 0 < p I 1 or if p = 00. Nevertheless, the symmetry is restored in these latter cases if instead of the boundedness of the means one considers their order of growth. We shall confine attention to 1 Ip I co. The theorem remains true for 0 < p < 1, but the proof is much more difficult.
5.4 MORE O N CONJUGATE FUNCTIONS
THEOREM 5.7. Letf(z) = u(z)
83
+ b(z) be analytic in IzI < 1, and suppose
Then
PROOF. For 1 < p < 00, we could apply the theorem of M. Riesz, but we shall give a proof which makes no appeal to this deeper result. Let p = +(I + r ) , and express f ( z ) by the Poisson formula:
1 f(z) =
-+ z u(pe") dt + iy. -p--z
2npeif
Then
By Minkowski's inequality,
Theorem 5.5 now gives the desired conclusion. Generally speaking, then, a harmonic function and its conjugate have the same rate of growth. We are now in a position to show that they also have the same degree of smoothness on the boundary. THEOREM 5.8. Let f ( z ) = u(z) + iu(z) be analytic in Iz( < 1, and suppose u(z) is continuous in IzI I 1. If u(eie)E A, (a < l), then u(z) is continuous in IzI 5 1 and u(eie)E A,. If u(eie)E A * , then u(eig)E A*.
PROOF. If we represent f ( z ) as a Poisson integral of u(e") and follow the proof of Theorem 5.1, we see that the weaker condition u(eie) E A, still implies f'(z) = O((1 - ry-'). Thusf(e'@)E A,, by Theorem 5.1. To show that A* is also preserved under conjugation, we may express u(z) as the Poisson integral of u(e") and follow the proof of Theorem 5.3 to see that u,,(z) = O((1 - r ) - l ) .
84 5
MEAN GROWTH AND SMOOTHNESS
Theorem 5.7 then implies vee(z)= 0((1 - v ) - l ) , so thatf”(z) Hence f ( e i e )E A,, by Theorem 5.3.
= O((1
- I)-’).
COROLLARY. Every function q(x) of class A* has modulus of continuity o(t)= O(t log lit).
PROOF. Let u(z) be the Poisson integral of the given function q E A*, and let f (z) = u(z) + iv(z) be its analytic completion. Then f ( z ) is continuous in IzJ5 1, and j(ei6) E A,. The result therefore follows from the special case already noted as a corollary to Theorem 5.3. 5.5. COMPARATIVE GROWTH O F MEANS
Iff(z) E H P(0 < p
p . This theorem has a number of interesting applications. The proof will make use of the following lemma, which can be established by essentially the same argument used to prove the lemma in Section 4.6. LEMMA 3. If a
> 1 and p
= +(I
+ r ) , then
THEOREM 5.9 (Hardy-Littlewood). Let f ( z ) be analytic in 1zI
< 1, and
suppose
j q r , f) 5 Then there is a constant K
If (p
p = 0 (i.e.,
C 3
= K(p, p)
o
p20.
(15)
independent o f j s u c h that
if , f H ~P ) , then Mq(r,f)
= o((1
+ l/p - l/q) is best possible in all cases.
- r)l/q-l/p). The exponent
PROOF. Let us first observe that it suffices to consider the case q = 00. Indeed, suppose that (16) has been proved for q = co, and suppose for convenience that K 2 1. Then for p < q < 00,
5.5 COMPARATIVE GROWTH OF MEANS
85
where
A = P P h + (P + l/P)(l
-P / d = P
+ l/P -
Similar remarks apply to the " o " part of the theorem. Hence we may confine attention to the case q = co. Supposenowthat 1 < p < m,andletp'be theconjugateindex: I/p+ I/$= 1. By the Cauchy formula,
Applying Holder's inequality and the hypothesis (1 5), we find
We now set p = +(I + r ) and use Lemma 3 to obtain (I 6) for q = co.The proof forp = 1 is similar but easier. Turning now to the " o " assertion, let f E H p , 1 5 p < co. Given E > 0, there exists 6 > 0 such that 6
If(ei(e+t))lP d t < E~
for all 8.
1-6
On the other hand, the Cau&y formula gives
For each fixed 6, the last two integrals are obviously bounded as r -+ 1. If p > 1, the integral over ( - 6, 6) is dominated by
where the lemma of Section 4.6 has been used. Similarly for p = 1. Hence f ( z ) = o((1 - r)-'lP), which is what we wanted to show. It remains to deal with the case p < 1. If f ( z ) # 0 in 1zI < 1, the function F(z) = [f(z)lP is analytic, and (15) gives f P
Thus, by what we have already proved,
86 5
MEAN GROWTH AND SMOOTHNESS
which is the desired result. The same argument showsf@) = o((1 - r)-'lp) for nonvanishing H p functions. Iff(z) has zeros, we fix p < 1 and use Lemma 2 to write
f
@z>=f l ( 4 +fZ(4,
wheref, and f z do not vanish and
Sincefn(z) # 0, it follows that
The constant K is independent of p. Finally, i f f E H P , the general result f ( z ) = o((1 - r)-l'p) follows directly from Lemma 2. The functionf(z) = (1 - z ) - ~ ,for suitable y > 0, shows that the exponent (p + l/p - l/q) cannot be improved. We leave the details to the reader. The " o " assertion can also be reduced to the special case p = 2, which is easily proved as follows. Letf(z) = anznbe in H Z ,and fix N so that
1
Then
I O(1)
which shows thatf(z)
+ &(I- r)-''',
= o((1
- r)-'I2).
It is natural to ask whether Theorem 5.9 has a converse. In particular, if 3 00 sufficiently slowly, must f be in H P ?Unfortunately, no such thing can be true, as the following theorem shows.
p < q and M,(r, f )
THEOREM 5.10. Let $(r) be continuous and increasing on 0 5 r < 1, with $(O) = 1 and $(r) 00 as r 1. Then there is a function f ( z ) analytic in IzI c 1, satisfying If(z)l I $(lzI), yet having a radial limit on no set of positive measure. In particular, f (z) is not of class H p for any p > 0. --f
--f
PROOF.The proof rests upon the following elementary observation.
5.5 COMPARATIVE GROWTH OF MEANS
87
LEMMA 4. Given a function $(r) as in Theorem 5.10, there is a sequence of integers 0 < n1 < n2 < . * such that m
1 rnks $(r),
o I r < 1.
h= 1
Momentarily taking the lemma for granted, let
Then If(z)[ I $(r), and it follows from Theorem A.5 (in Appendix A) that for almost every choice of signs { E , ) , ~ ( Z ) has a radial limit almost nowhere. This proves the theorem. PROOF OF LEMMA. Let ro = 0, and let r l , r 2 , . . . be determined by $(rh) = k + 1. Then 0 < rl < r2 < . * < 1. Choose n, such that r;' I +. Generally, select nh > nk-l such that r,"kI 2 - k . Given r in [O, I), let s be the index for which r s - l <_ r < r , . Then
-
m
2
c r?
s- 1
a,
rnk
2
h= 1
a,
=-CrF
+ 1 r:
h= 1
h=s
h= 1
m
< (s - 1)
+ C r?
_< ( s -
1)
+ 2-'+'
k=s
Is = $(rs-l) I $(r).
Although Theorem 5.9 is in a sense best possible, it can be slightly sharpened in one direction to produce a very useful result. THEOREM 5.11 (Hardy-Littiewood). = l/p - l/q, then
If 0 < p < q _<
CO,
f~ HP,A 2 p ,
and CI
1
j o ( l - r)Aa-1{A4q(r, f ) } Adr < 00. PROOF.
It suffices to take A = p , since by Theorem 5.9,
rd
Jol(l - r)Ia-'{Mq(r, f)}A dr ICA--p (1 - r ) p a - l { M q ( r ,f ) > p dr.
Next, suppose the theorem were proved for A = p = 2. Then if , f H* ~ (0 < p < CO) andf(z) # 0 in \zl < 1 , g(z) =
Cf(W2 EH2,
88 5
MEAN GROWTH A N D SMOOTHNESS
and the desired result (with I = p ) would follow. The general theorem could then be deduced by writing f E H P as the sum of two non-vanishing HP functions, as in Lemma 2. Hence the theorem reduces to the special case A = p = 2. To prove this, suppose first that 2 < q < 03, and let m v
f(z) =
1 a,z"EZ n=O
Then by Theorem 5.6, J:(1
i
(1 - r ) 2 - z / q { M q ( rf,' ) } 2 d r .
- r ) - 2 / q { M , ( r f)}' , dr 5 C
(17) But it follows from Theorem 5.9 that
M q ( r , f') 5 K ( l - r)l/q-l/ZM2 (r1IZ,f 0, so the second integral in (17) is dominated by a constant multiple of m
Jol(l- r){M2(r1/', f')}' d r
= n= 1
Finally, the result for q = co is easily deduced from the case q < co by Theorem 5.9. 5.6. FUNCTIONS WITH HP DERIVATIVE
According to Theorem 3.11, every function which has an H' derivative is continuous in the closed disk and absolutely continuous on the boundary. In particular, f' E H' implies f E H". This latter result has an interesting generalization. THEOREM 5.12 (Hardy-Littlewood).
f~ H4, where q = p/(l
If f ' E H P for some p < 1, then of p , the index q is best possible.
- p). For each value
For convenience, assume f ( 0 ) = 0. Case I : 3 5 p 5 1. Since q 2 1, Minkowski's inequality may be applied (as in the proof of Theorem 5.5) to give PROOF.
5.6 FUNCTIONS WITH
HP
DERIVATIVE
89
But iff' E H P and q = p / ( l - p ) , it follows from Theorem 5.11 that j;M&,
f') ds < 00.
Case IZ:p < +. Fix a in the range
1/2p < a < l / p ;
then CI > 1 . In view of Lemma 2, we may assume that f ' ( 2 ) has no zeros in /zI < I. Letting g(z) = [f'(z)]''", we have If(reio)i I flg(seie)Ia ds I [G(Q)]"-'h(r, O), 0
where
G(8) = sup Ig(rei8)l O
and
h(r7 8) = jrlg(seio)l ds. 0
Hence
If(re")lq 5 [ G ( Q ) ] ( a - 1 ) 4 [ h8)Iq. (~, Now set
and let
p'
= p/(B -
1) be the conjugate index. By Holder's inequality,
since (a - I)@ = ap. By the Hardy-Littlewood maximal theorem (Theorem 1.9), G E Lap.On the other hand, aP qj3' = > 1; 1- ap thus Minkowski's inequality gives
which remains bounded as r + 1 , by Theorem 5.1 1. Thus f E H4.
90 5
MEAN GROWTH AND SMOOTHNESS
To show that q = p/(1 - p ) cannot be replaced by any larger index, one need only consider f"(z) = (1 - z)e-lLp for small E > 0. EXERCISES
1. Let f ( z ) =
a, zn be analytic in
IzI < 1. Show that if
N
2n lanl = O ( N ~ - ' ) ,
o < a I I, then f ( z ) is continuous in IzJI 1 and f ( e i e )E A, . Show further that the converse is true if a < 1 and a, > 0, n = 1, 2, . .. . 2. Letf(z) = EF=oanznbe analytic in Iz( < 1. Show that if n=l
N
C n Ian1 = O(log N ) , n= 1 thenf(z) is continuous in JzI 4 1 andf(e") has modulus of continuity w(t) = O(t log l / t ) . 3. Show that if N
thenf(z) is continuous in Izl I1 andf(eie) E A*. Show that the converse is t r u e i f a , > O , n = 1,2, ... . 4. Show that iff@) is continuous in (21 < 1 andf(e") E A,, 0 < a < 1, then
If f ( e i e )E A,, show that N
= o(N);
n3 n= 1
hence
(Duren, Shapiro, and Shields [l]). 5. Show that if m is an integer greater than 1, and if 0 < a < 1, the function a,
f(z) = k= 1
is continuous in Iz( I 1 and f(eie) E A,.
m-kazmk
NOTES
91
6. Show that the first two estimates in Exercise 4 are best possible in the sense that " 0 " cannot be replaced by " 0.'' (The third estimate is also best possible; see Duren, Shapiro, and Shields [l].) 7. Show that if f(z) is analytic in (z(< 1, continuous in lzl I 1, and f(ei') E A, for some a > 4, then C;!o (a,(< 03. Show further that the index 3 is best possible: ihe series la,l may diverge if onlyf(e'e) E Al,z (Theorem of S. Bernstein). 8. Show that iff(z) is continuous in JzJI I andf(e'e) E A,, 0 < a I 1, then lanlY< 00 for all y > 2/(2a + 1). (Hint: Show that nB ]a,(' < 00 for all
p < 2a.)
9. Show that iff' E H P , 1 < p Ico, thenf(z) is continuous in Iz( 5 1 and f ( e i e )E A,, a = 1 - l/p. 10. Show that if q(t) is continuous and
lq(t
+ h) - 2&) + q(t - h)l I Ah",
0 < a < I,
then q E A,. (Suggestion: Imitate the proof of Theorem 5.1.) 11. For 1 5 p co, show that the class Aup contains only the constant functions if a > 1. 12. Letf(z) = u(z) + iv(z)be analytic in Izl < 1. Suppose u E hP(1 < p < 03) and u(eie) E A$ (0 < a < 1). Show that v(e") E A .: In other words, show that the class AaPis self-conjugate. 13. Prove that i f f € H P (1 < p < co) and f ( e i e )E AaP (l/p < a I I), then f ( z ) is continuous in (zJI 1 and f ( e i e )E A,, 6 = a - I/p. 14. Show that the function
-=
p(x) = x log l/x,
x > 0,
belongs to A* but not to A,. 15. Prove Theorem 5.2. 16. Using the example f ( z ) = (1 - z)-?, show that the exponent (p + l/p - I/q) in Theorem 5.9 is best possible. NOTES
Almost everything in this chapter is due to Hardy and Littlewood. Lemma 1 is in their paper [2]. Theorems 5.1, 5.4-5.6, 5.9, and 5.11 occur in their paper [5]. The proofs of Theorems 5.5 and 5.7 for the case p < I may be found in their paper [q.The simple but enormously useful remark that every HP function is the sum of two nonvanishing ones (Lemma 2) can be traced to their
92 5
MEAN GROWTH AND SMOOTHNESS
paper [l]. Further results and generalizations are in their papers [5] and [S]. The “0” growth condition given in Theorem 5.9 is best possible; see G. D. Taylor [l] and Duren and Taylor [I]. Flett [2] has recently based a proof of Theorem 5.11 on the Marcinkiewicz interpolation theorem. Zygmund [2,4] introduced the class A, and proved Theorem 5.3 in somewhat different form. The self-conjugacy of A, (Theorem 5.8) goes back to Privalov [l]. Zygmund [2] showed that A, is self-conjugate. For a direct proof that every A, function has modulus of continuity which is O(t log 1i t ) , see Zygmund [4], VOl. I, p. 44. The converse to Theorem 5.12 is totally false. The “ Bloch-Nevanlinna conjecture ” asserted that iff E N , then f ’ E N , but this has been disproved in many ways. In fact, there exist functions f E H“ continuous in the closed disk, such that f ’ ( z ) has a radial limit almost nowhere. See, for example, Lohwater, Piranian, and Rudin [I] and Duren [4]. Hayman [3] has recently shown that f ’ E N does not imply f E N , an unexpected result in view of Theorem 5.12. Caughran [2] has obtained results comparing the canonical factorizations off and f ’ in case f’ E H ’ .
TAYLOR COEFFICIENTS
CHAPTER 6
C
If a function f ( z ) = a,, z” belongs to a certain H P space, what can be said about its Taylor coefficients a,? Clearly, one can hope to describe only the “eventual behavior” of {a,} as n --f co, since any finite number of coefficients can be changed arbitrarily without upsetting the fact thatfis in H P . It is also interesting to ask how an H P function can be recognized by the behavior of its Taylor coefficients. Ideally, one would like to find a condition on the a, which is both necessary and sufficient for f to be in H P . For p = 2, of course, the problem is completely solved: f E H 2 if and only if la,I2 < 00. But the general situation is much more complicated, and no complete answer is available. If 1 < p < 03, the problem is equivalent to that of describing the Fourier coefficients of Lp functions, as the M. Riesz theorem shows. This chapter contains some scattered information about the coefficients of H P functions. Curiously, the results are most complete in the case 0 < p < 1.
1
6.1. HAUSDORFF-YOUNG
lNEQUALlTlES
Any available information about the Fourier coefficients of Lp functions can be applied, in particular, to the Taylor coefficients ofHP functions. In fact, the
94
6 TAYLOR COEFFICIENTS
two sets of coefficient sequences are essentially the same if 1 < p < co. The Hausdorff-Young theorem states that if cp(x) E Lp = Lp[O,2711, 1 < p 5 2, then its sequence {c,} of Fourier coefficients is in tq(l/p l/q = 1) and
+
Conversely, every t psequence {c,} of complex numbers (I < p 5 2) is the sequence of Fourier coefficients of some cp E L4 (l/p + l/q = l), and
This result may also be expressed in H P language, as follows. THEOREM 6.1.
If W
f(z) =
C a,z" E H P n=O
(1 I p I 2),
then {a,} E L4 (l/p + l/q = 1) and II{an>llq5
Ilfllp.
Conversely, if {a,} is any tPsequence of complex numbers (1 < p < 2 ) , then f ( z ) = C a,z" is in H q (l/p + l/q = I) and
Ilfllq 5
II{an>Ilp*
(2)
PROOF. The first statement follows easily from the Hausdorff-Young theorem. Indeed, i f f € H P , then the radial limitf(eie) is in Lp and {a,} is its 2), sequence of Fourier coefficients. On the other hand, if {a,} E tp(1 5 p I then f~ H 2 , and the numbers a, are the Fourier coefficients of f(eie). The ~ Hausdorff-Young theorem then tells us thatf(eie) E Lq,which implies j - Hq, and (2) follows.
Neither part of the theorem remains valid if p > 2. In fact, the hypothesis tha t f e H p for somep > 2 implies nothing more about the la,l than {a,} E t2. And if the condition {a,} E t 2is weakened to {a,} E t pfor somep > 2, nothing reasonable can be said about f(z). These claims are justified by Theorem A S of Appendix A. According to this theorem, the mere assumption that {a,} E t 2 implies that, for almost every choice of signs { E , } , the function E,U,Z" is in H P for allp < 03. On the other hand, if {a,} # d2, almost every choice of signs E, a, zn having a radial limit almost nowhere. produces a function 1
6.2 THEOREM
OF
HARDY AND LITTLEWOOD
95
6.2. THEOREM OF HARDY AND LITTLEWOOD
The following theorem provides further information about the coefficients of HP functions. If
THEOREM 6.2 (Hardy-Littlewood).
c an m
f(2)=
2“ E HP,
0 < p 5 2,
n=O
where C, depends only on p. PROOF. We suppose 1 Ip I 2, postponing the case 0 < p < 1 to Section 6.4. I f p = 1, the theorem reduces to Hardy’s inequality (Corollary to Theorem 3.15). If p = 2, it follows from the Parseval relation. A proof for intermediate values of p can be based on the Marcinkiewicz interpolation theorem, but the argument given here will be self-confained. Let p be the measure defined on the set of integers by
n = 0, f l , k2,. . . ,
p(n) = (In1 + 1)-2, If g E L2 and
-c m
g(t)
b, einr,
n=-m
let
g(n) = (In\ + l)bn,
n
= 0,
+ I , t-2, . . . .
For s > 0, let E,
Then
=
{n : Is”(n)l > s}.
96 6 TAYLOR COEFFICIENTS
Now write
where
Then
where A , is a constant depending only on p . Similarly, by (4), m
I$s(n)lP~(n)= n=-m
-J
m
s p dB(s) = p JmsP-’p(s) ds 0
Combining these two estimates with (6), we have
if g EL’.
0
6.2 THEOREM OF HARDY AND LITTLEWOOD
97
I f f € H (1 < p < 2), the desired estimate ( 3 ) now follows after approximating the boundary function f ( e " ) by
applying (7) to g M , and letting M
-+ 00.
As with the Hausdorff-Young theorem, the converse to Theorem 6.2 is false if p < 2. However, the converse is true for indices larger than 2, and can be deduced from Theorem 6.2 by a duality argument. The exact statement is as follows. THEOREM 6.3. Let {a,}
be a sequence of complex numbers such that n4-21an(4
C;=o
m
n=O
where Cq depends only on q. PROOF. Let p
= q/(q -
l), and let n
G(eie)=
ckeike k= -n
be an arbitrary trigonometric polynomial with llGllP I 1. By the M. Riesz theorem, its " analytic projection " n
g(eie)=
C ckeike k=O
Then
98
6 TAYLOR COEFFICIENTS
Taking the supremum over all G with IIGII,
< 1, we have
and the result follows by letting n -+ 00. 6.3. THE CASE p I1
As pointed out in the corollary to Theorem 3.4, the Taylor coefficients of an H1 function must tend to zero, by the Riemann-Lebesgue lemma. The simple example (1 - z)-' shows this is false for H P functions withp < 1, but a sharp asymptotic estimate for the coefficients can be given as follows. THEOREM 6.4.
If
" f(z)=
O
CU,Z"EHP, n=O
then a, = o ( n l i P - 1 )
and Ian/ICn'iP-'IIfIlp.
(9) Furthermore, the estimate (8) is best possible for each p : given any positive sequence (6,) tending to zero, there exists f E H P such that a,
z O(6, n ' / p - ' ) .
PROOF. We have already observed that a, = o(1) iff E H I . This statement cannot be improved, even iff 6 H". Indeed, given a sequence (6,) tending to zero, choose integers 0 < nl < n2 < * such that 6,, < 2 - k . Define
a, =
(,"
dm
if n = n k , otherwise.
Then C la,[ < 00, so f(z) = a,z" is continuous in IzI < 1, yet a, # O(6,). Now suppose f E H P ,p < 1. The coefficient a, has the representation
Hence
6.4 MULTIPLIERS
99
But by Theorem 5.9, M,(r,f)
= o((1
- r)""p).
With the choice r = 1 - l/n, (8) follows. Theorem 5.9 also gives (9) in the same way. Functions of the form (1 - z ) - ~show the exponent in (8) cannot be reduced. The proof that the estimate is best possible will be postponed to Section 6.4. 6.4. MULTIPLIERS
A complex sequence (A,,} is said to be a multiplier of H P into the sequence space t4if (Anan}E t4whenever a,,z"E H P . Similarly, {A,,} is a multiplier of H P into H 4 if a,,znE H P implies A,, a,,z" E H4. One way to give information about the coefficients of H P functions is to identify the multipliers of H P into various spaces. Results of this type have a number of interesting consequences. We shall content ourselves with describing the multipliers of H p (0 < p < 1) into t4( p 5 q I a)and of H' into t' and into 8' (alias H'). Some preliminary remarks are in order. First of all, even though H p and C p are not Banach spaces if p < 1, they are complete topological vector spaces whose topologies are induced by the translation invariant metrics II f - g1lPpand
2
2
IIx - vII,"
2
=
2 Ixn - VnIp,
respectively (see Section 3.2). In other words, H P and t pare F-spaces, in the terminology introduced by Banach [ 11. The closed graph theorem therefore applies to mappins between these spaces (see Banach [l] or Dunford and Schwartz [l], Chap. 11). A sequence {A,,} may be regarded as inducing a linear operator
A : 2 a,,z" -P {A,,a,,} from H P to C4. The domain of A is the linear manifold
9(A)= { f E H P : A ( f ) E t 4 } . Ifp < 00,9(A) is dense in H P ,since it contains all polynomials. Furthermore, it is clear that 9(A) = H P if and only if {A,} is a multiplier of H P into P. The closed graph theorem therefore tells us that A is a bounded operator from H p to t4if {A,,} is a multiplier. We have only to check that A is closed. To show A is a closed operator, suppose fk
9(A),
h+ f
E HP,
and
gk
= Auk)+
g
E
t4.
100
6 TAYLOR COEFFICIENTS
Let fk(z)
=
U?)Z",
f ( z )=
u,z",
and g
= {b,}.
Then by Theorem 6.4, aLk)+.a, , n = 0, 1, . . . . Since also A, akk)+ b, , this shows that A, a, = b, , n = 0 , 1, . . . . In other words, f E 9 ( A ) and A(f) = g, which proves that A is closed. THEOREM 6.5. Suppose 0 < p S 1. Then {A,} is a multiplier of HPinto em
if and only if
A, = O(n'-'/P).
(10)
1
PROOF. If f ( z ) = a,z" is in H P and (10) holds, then {A, a,} E L", by Theorem 6.4. (In fact, A, a, --f 0.) Conversely, if {A,} is a multiplier, then by the closed graph theorem (as discussed above), A is a bounded operator from H P to em. In other words,
SUP IAflafll5 CllfIlP>
.fE
HP.
n
Now let
-
g ( z ) = (1 - z ) - l - l / p =
(11)
c bnz"7
where b, Bn'lP; and choose f ( z ) = g(rz) for fixed r < 1 . Then by (1 1) and the lemma in Section 4.6, IA,In'/pr" I~ ( -1r>-'. The choice r = 1 - l/n now gives (10).
As a corollary, we can now show that the estimate (8) in Theorem 6.4 is best possible. COROLLARY. If {d,} is a sequence of positive numbers such that a, = O(d,) for every function a,z" in H P ( p 5 1), then there is an E > 0 such that
d , n l - l / P 2 E,
n = 1,2,. .. .
PROOF. If a, = O(d,) for every f E H P , then {1/d,} multiplies H P into P . Thus l / d , = O(nl-'/P), as claimed. THEOREM 6.6. Suppose 0 < p < 1 andp I q < 03. Then {A,) is a multiplier of H p into tqif and only if N
1 nq/pIA,(q= O(Nq). n= 1
6.4 MULTIPLIERS
101
Before passing to the proof, let us apply the result to establish the HardyLittlewood theorem (Theorem 6.2) for p < 1. COROLLARY.
a,z" E H P (0 < p < l), then
If
m
PROOF. Let q = p , and observe that A,, = n 1 - 2 i pthen satisfies (12). Hence { n 1 - 2 i P } is a multiplier of H P into tp.The closed graph theorem now gives the
inequality (3) stated in Theorem 6.2. The following lemma will be needed in the proof of Theorem 6.6. LEMMA.
Suppose b, 2 0 and 0 < p < CI. Then N
2nab, = O(NB) n=l
if and only if a,
C bn = O(NB-"). n=N PROOF.
Letting M Then
-+
Let (13) hold and let Sn =
c;=lkabk.Then
co,we obtain (14). Conversely, assume (14) and let R,, = Ckm,, bk.
c nab,
N
N
n= 1
= n= 1
[nu - (n - l)*]Rn- N"RN+1 I CNB.
PROOF OF THEOREM 6.6.
By the lemma, the condition (12) implies
Assume without loss of generality that An 2 0 and and
,I2
= 1.
Let s1 = 0
102
6 TAYLOR COEFFICIENTS
where y = q ( l / p - 1). Note that sn increases to 1 as n + co.By Theorem 5.11, f E HP (0 < p < 1) implies ~ ~ ' ( 1r > - ' - l M l q ( r , j ) dr < oo,
Since M , ( r , f ) 2 rnlu,,l,wheref(z) 03
>
2 anzn,it follows that
5 r+'(l - r ) Y - 1 M 1 4 ( r , fdr) c j (1 r ) y - l r n qdr
n= 1
s,,
m
2
=
p 5 q < co.
S"
+1
-
n=l 1
S ,
m
by the definition of s, . But by (15),
which shows that (s,)"q
2 (I - C/nyq-+ e-Q > 0.
Since these factors (sJ4 are eventually bounded away from zero, we have shown that {An} is a multiplier of H P into l4if it satisfies (12). Conversely, if {A,,} is a multiplier, then by the closed graph theorem,
Choosingf(z)
= g(rz) as
in the proof of Theorem 6.5, we now find
Hence
and (12) follows upon setting r = 1 - l/N. Note that the argument shows (12) is necessary even if p 2 1 or q < p . The condition (12) also characterizes the multipliers of H' into P, if 2 5 q < 00. We give the proof first in the case q = 2.
6.4 MULTIPLIERS
THEOREM 6.7.
103
The sequence {An} is a multiplier of H’ into H 2 if and only
if N
C n2jAnI2= o(N’). n= 1 PROOF. The necessity of (16) was observed in the proof of Theorem 6.6. In proving the sufficiency, it is enough to consider functions without zeros, since by the lemma in Section 5.3, every H’ function is the sum of two nonvanishing H1functions. Thus suppose f ( z ) = C anz” is an H’ function which does not vanish in IzI < 1, and writef= (p2, where q(z) = bnznis in H 2 . Then
bkbn-k.
an = k=O
It is to be shown that (16) and 1 lbn\’ < co imply of generality, suppose An 2 0 and b, 2 0. Then
1 [Anan)’< co.Without loss
where [ x ] denotes the greatest integer S x . Now apply the Cauchy-Schwarz inequality : n
m
n
2 b 2 = C k = [Cn / 2 ] b;=Cpn, k=[n/2]
an214Cb,2 k=O
say. It follows that N
N
C An2an25 cnC= 1An2pn. n= 1
But a summation by parts gives
where
and A x , = x , , , ~- xn . The term N - ’ S , p , is bounded, by (16) and the assumption 1bk2< 00, so it remains only to show that m
n21A(n-2fin)I< 00.
n= 1
But = (n
+ 1)-2 Apn + pnA(n-2).
(17)
104
6 TAYLOR COEFFICIENTS
Since C bk2< co, it is clear that
and m
“ 1
1 n2P,,lA(n-’)l
S 2C
n= 1
m
n=l
2k+l
< 2 C bk2 k=l
2
n=k
-
* 1
k=[n/2]
b:
1 m - < 6 C b: < CO. n k=l
This establishes (17) and completes the proof of the theorem. As a first illustration, let An = n-’I2. Then (16) is satisfied, so
for everyfc H‘. However, this follows from Hardy’s inequality and the fact that an-+ 0. There is another application which is more interesting. Let n l , n 2 , . . . be a lacunary sequence of integers in the sense that nk+llnk
2
Q > 1.
Let An={
1 if n = fzk 0 otherwise.
We thus obtain PALEY’S THEOREM.
If
then for every lacunary sequence {n,},
6.4 MULTIPLIERS
105
As a final corollary of Theorem 6.7, we now show that, more generally,
{A,} is a multiplier of H' into C4 (2 Iq < co)if (and only if) it satisfies (12). Let p,
=
(An14'2, and observe that (12) gives a condition equivalent to N
n 2 p 2 = O(N2). n=l
Thus {p,} multiplies H' into f 2 , which implies (since the coefficients of an H' function are bounded) that {A,} multiplies H' into d4. The multipliers from H1 to f 1 are more difficult to describe. Hardy's inequality shows that the sequence {(n + 1)-'} is one example. It is possible to characterize all the multipliers, but only by a condition difficult to verify in most situations. Thus the following theorem, interesting though it may be, is really more a translation of the problem than a solution. THEOREM 6.8. The sequence (A,} is a multiplier of H' into f' if and only if there is a function II/ EL" such that
1 IAnl = 2.n
1
2=
0
e-i"'$(t) dt,
n = 0, 1, 2, . . . .
PROOF. The sufficiency of the condition (18) was established in Theorem 3.15. Conversely, suppose {A,} is a multiplier from H1 to 1'. Then by the closed graph theorem, the mapping
is a bounded operator from H1 to f' : m
m
is a bounded linear functional on H I . By the Hahn-Banach theorem, can be extended t o a bounded linear functional @ on L1. Thus by the Riesz representation theorem, there exists $ EL" with O(f)
1
=
2.n
1
2n
0
f(ei')9(t> d t ,
f~ I,'.
106 6 TAYLOR COEFFICIENTS
Choosing the H’ functionf(z)
= z” (n = 0,
1,2, ...), we have
which becomes (1 8) after conjugation. EXERCISES
1. Show that Theorem 6.2 is best possible if 0 < p < 1 : For each positive sequence {k,} increasing to infinity, there exists a, Z” E H P with
C knnP-2la,,lP=
CQ.
’
(This is also true if 1 Ip 5 2; see Duren and Taylor [l].) 2. Show that Theorem 6.2 is false for p > 2. 3. Show that the converse to Theorem 6.2 is false (0 < p I2). 4. Show that Paley’s theorem does not generalize to Fourier series: There existsf€ L’ with Fourier series C c,eins, for which Ic2r;I2= 00. (Suggestion: Try C (log n)-’ cos no, a > 0. See Theorem 4.5.) 5. Show thatf(z) may be analytic in IzI < 1 and continuous in (zI 1, yet f ’(z)have a radiallimit almost nowhere. (Thisdisproves the “ Bloch-Nevanlinna conjecture ” that f E N implies f ‘ E N . See Duren [4].) 6. A functionf(2) analytic in IzI < 1 is said to have finite Dirichlet integral (YED ) if
Show that f’E H’ implies f E D , but that f need not be in D iff’ E H P for all p < 1. 7. Show that D is contained in H P for all p < 00, but D is not contained in H”. NOTES
For the Hausdorff-Young theorem, see Zygmund [4]. Hardy and Littlewood [ 11 gave a long and difficult proof of Theorem 6.2 in their original paper. The idea to prove it for 0 < p < I via Theorem 5.11 is due to Flett [2]. The fact that a, = ~ ( n ” ~ - ’for ) f E H P ( p < 1) is due to Hardy and Littlewood [5], although in the Soviet literature it is often ascribed to G. A. Fridman, who rediscovered it in 1949. Evgrafov [l] gave a direct construction to show that
NOTES
107
this estimate is best possible; a simpler argument was given by Duren and Taylor [I]. Theorem 6.5 (and its corollary) and Theorem 6.6 are due to Duren and Shields [I, 21. Hardy and Littlewood [7, 81 gave a sufficient condition for multipliers from H P to H q (1 p 5 2 5 q < co) which includes the “ i f ” part of Theorem 6.7; the proof in the text is theirs. They also stated without proof a sufficient condition for multipliers from H P to H 4 (0 < p < 1 5 q < a).A proof appears in Duren and Shields [2]. The closed graph theorem has often been applied to multiplier problems in various spaces. J. H. Wells pointed out to the author that the necessity of (16) could be proved by this method. Paley [ I ] proved Paley’s theorem ” and inspired the original Hardy-Littlewood work on multipliers. A converse to Paley’s theorem, which Rudin [4] proved by explicit construction, is a simple consequence of the necessity of the multiplier condition (16); see Stein and Zygmund [l]. Hedlund [l] recently gave a sufficient condition for multipliers from H P (1 < p < 2) to H 2 which extends that of Hardy and Littlewood. The condition (12) is necessary but not sufficient if p = 1 and q < 2; see Duren and Shields [2]. E. M. Stein [ I ] has given a sufficient condition for multipliers from H P to H P (0 < p < I ) . For further information on multipliers, see Kaczmarz [ 11, Caveny [ 1, 21, Wells [I], Duren and Shields [ I , 21, and Duren [3]. The coefficients of inner functions have been studied by Newman and Shapiro [l]. They show that no inner function except a finite Blaschke product can have coefficients o(l in), although the coefficients of an infinite Blaschke product can be O(l/n). The coefficients of a singular inner function cannot be ~ ( n - ~ l but ~ ) they , can be O ( n - 3 1 4 ) . Information is also available on the coefficients of functions of bounded characteristic. Mergelyan has shown that iff(z) = a, 2’’ E N, then “
lim sup n-’i2 logla,( 5 C, n+ m
where C can be given explicitly in terms off. (See Privalov [4].) Cantor [ I ] recently proved that iff. Nand A , denotes the matrix (aiT,),0 I i, j I n, then 0 as n -+ co. (det A,\”” --f
This Page Intentionally Left Blank
H P A S A LINEAR SPACE
CHAPTER 7
In order to solve problems concerning H P functions, it is often advantageous to view H P as a linear space and to use the methods of functional analysis. We have already seen in Chapter 6 that the closed graph theorem is an effective tool for describing coefficient multipliers. The same tool will be used in Chapter 9 to discuss interpolation problems. In the present chapter, we shall study the linear space structure of H P in some detail. One major objective is to represent the continuous linear functionals on H P both for p 2 1 and in the more interesting case p < I. The results for y 2 1 help to solve an approximation problem (Section 7.3) and prepare the ground for a full discussion of extremal and interpolation problems in Chapters 8 and 9. The dual space structure of H P with y < 1 is applied to demonstrate the failure of the HahnBanach theorem in a non-locally convex space with " reasonable " properties. This chapter concludes with a description of the extreme points of the unit sphere in H', a topic also of interest in the general context of linear space theory.
110 7
Hp
AS A LINEAR SPACE
7.1. QUOTIENT SPACES AND ANNIHILATORS
Let us begin by recalling a few general Banach space concepts. Let X be a Banach space, and let S be a (closed) subspace. A coset of X modulo S is a subset 5 = x + S consisting of all elements of the form x + y , where x is some fixed member of X and y ranges over S. Two cosets are either identical or disjoint.The quotient space X / S has as its elements all distinct cosets of X modulo S. With the natural definitions of addition and scalar multiplication, X / S is a linear space. To be specific, (XI
and
+ S ) + (x2 + S ) = ( X I + x2) + S a(x + S ) = ax + S.
The zero element of X / S is the coset S. Finally, the norm of a coset 4 is defined by 11511 = inf IIX + Yll.
=x
+S
YES
It is easy to check that this is a genuine norm. Since S is closed, 11511 = 0 implies 5 = S. The relations 11511 2 0, lia511 = la1 11511, and lltI t211 I II t1ll + I1 t2II are obvious. Under the given norm, X / S is complete, and therefore is itself a Banach space. To see this, let (4,) be a Cauchy sequence of cosets. Choose a sequence such that of integers 0 < n, < n2 <
+
k
Iltnk- 5,k+,11 I 2 - k , Then choose xk
E
t,
=
1,2, .. . .
(k = 1, 2, . . .) such that
With this construction, { x k ) is a Cauchy sequence, so xk tends to a limit x E X . Let ( = x + S. Then &, + 5, because (by definition of the norm) lltnk
- 411 5
llXk
-xll.
Finally, since (5,) is a Cauchy sequence, this implies 5, + 5. The annihilator of the subspace S is the set S' of all linear functionals 4 E X * such that $ ( x ) = 0 for all x E S. It can be easily verified that S' is a subspace of X * . The following results play an essential role in the theory of extremal problems (Chapter 8). THEOREM 7.1. The quotient space X*/S' is isometrically isomorphic to S*. Furthermore, for each fixed 4 E X * ,
sup
x e s , Ilxll
where
"
I$(X>l
=
min 114 + $11,
)€Sl
min " indicates that the infimum is attained.
7.1 QUOTIENT SPACES AND ANNIHILATORS
111
THEOREM 7.2. The space ( X / S ) * is isometrically isomorphic to S'. Furthermore, for each fixed x E X ,
max @ESI.
where
"
II$llSl
I*(x>l = inf
IIX
+ Yll,
Y ES
max' ' indicates that the supremum is attained.
PROOF OF THEOREM 7.1. For each fixed
E
s*,the class of all extensions
(b E X * is a coset in X*/S'. It is clear that this correspondence between S*
and X*/S' is an isomorphism. It is also an isometry. In fact, 11$11 s 11qb11 for every extension 4 ; and, by the Hahn-Banach theorem, there is at least one extension for which il*il = 11411. In other words, for the coset of extensions of $, the infimum defining the norm is attained and is equal to 11$11. PROOF OF THEOREM 7.2. In terms of a given a linear functional (b in S' unambiguously by
E
(x/S)*, one can define
+(x) = CD(x + S). Conversely, given 4 E S', this relation defines a linear functional CD on X / S . The correspondence is clearly an isomorphism. We assert that the boundedness of either functional implies that of the other, and in fact IIdll = ll@ll. This follows from the relations
I4(x)l = I@(X + S)l 5 ll@Il IIX + SII 5 IIQII IIXII ; .Y E s. lW + S)l = Iqb(4l = I+(x + Y>l Ilqbll Ilx + YII, Hence S' and ( X / S ) * are isometrically isomorphic. To prove the rest of the theorem, observe first that for any $ E S' with 11(b11 I I, and for any x E X and y E S ,
I+(x)l =
+ .Y)l llx + Yll.
Thus
On the other hand, given x E X , a corollary of the Hahn-Banach theorem (see Dunford and Schwartz [l], p. 65) shows the existence of @ E ( X / S ) * such that l@(x
+ S)l = Ilx + Sl/
and
ll@l\ = 1.
Now let $ E S' correspond to CD as above, so that (b(x)= CD(x + S) and (1 (b /I = (1 CD (1 = 1. Equality therefore holds in (l), and the supremum is attained.
112 7 H" AS A LINEAR SPACE
7.2. REPRESENTATION OF LINEAR FUNCTIONALS
As we saw in Chapter 3, H P is a Banach space if 1 IP I 03, with norm l l f l l =Mp(l,f).ThepolynomialsaredenseinHPifO
In particular, if each f E H P is identified with its boundary function, H P can be regarded as a subspace of Lp, 0 < p 2 00. According to the Riesz representation theorem, every bounded linear functional 4 on L p (1 s p < 00) has a unique representation 1
2n
4 ~ =)g j0 f(eie>g(eie)do,
g E~4,
(2)
where l/p + l / q = 1. In fact, llq!ll = / / g / I 4and , (Lp)*is isometrically isomorphic to E. Since H P is a subspace of Lp,then, Theorem 7.1 can be used to describe (HP)* if the annihilator of H P in (Lp)* can be determined. But if g E L4 annihilates every H P function, then surely
n = 0, 1, 2, . . . .
/.ne"eg(e") d6' = 0,
Therefare g(eie) is the boundary function of some g(z) E H 4 , and g(0) = 0. Call this class of functions H04. Conversely, if g E H04, it is clear that ~ 0 2 > ( e i e ) g ( ed~o~= ) o
for every f E H P . Hence Ho4 is the annihilator of H P , and it follows from Theorem 7.1 that (HP)* is isometrically isomorphic to L4/H04.Actually, we may as well replace E/HO4by L?/H4, since the correspondence < - e i e 5 between cosets of the two spaces is an isometric isomorphism. I t is even possible to give a canonical representation of the bounded linear functionals on H P . Any 4 E (ElP)* can be extended (by the Hahn-Banach theorem) to a functional on Lp,and hence may be represented in the form (2) for some g E L4. This representation is certainly not unique. Two functions g1 and g 2 belonging to the same coset of E/HO4,so that
-C m
gl(eie)- g2(eie)
c,
cine,
n= 1
obviously represent the same functional Q, on H P . Functions in different cosets, however, generate different functionals. Thus forp > 1, the representation becomes unique if we distinguish in each coset that function g for which
7.3 B E U R L I N G ' S A P P R O X I M A T I O N T H E O R E M
Equivalently, there is a unique function g
E H 4 for
113
which
for all f E H P , 1 < p < co. Since 1 < q < 00, the M. Riesz theorem (Theorem 4.1) guarantees that the '' analytic projection " g of the original L? function is in H 4 (see Chapter 4, Exercise 1). In summary: THEOREM 7.3. For 1 5 p < a, the space (Hp)* is isometrically isomorphic to L4/Hq, where I/p f l/q = 1. Furthermore, if 1 < y < co, each 4 E (HP)* is representable in the form ( 3 ) by a unique function g E H 4 , while each 4 E ( H I ) * can be represented in the form ( 3 ) by some g E L". ___
We might have chosen to put g(e-") instead o f g ( e i 8 )in (3). This would have the advantage of setting up an isomorphism between (HP)* and H 4 . But in either case the correspondence need not be an isometry. In fact, ll4li is equal to the norm of the coset determined by g(eiB)(or by g(e-")), so that only the inequality
lldll S IMI
I ApII4lI,
1 < p < a,
is true generally. The right-hand inequality comes from the M. Riesz theorem; = 2 it is easy to see that 11 4 Ij =
A, is a constant independent of 4. In the case p Ilgll. 7.3. BEURLING'S APPROXI MATlON THEOREM
The representation of linear fmctionals can be applied to prove an interesting theorem on polynomial approximation in H P , 1 I p < co. Recall that a function f E H P has a canonical factorization f ( 4 =B(z)WF(zf,
where B ( z ) is a Blaschke product, S(z) is a singular inner function generated by a nondecreasing singular function p ( t ) , and F ( z ) is an outer function. The inner function f o ( z ) = B(z)S(z) is called the inner factor of f ( z ) . Let go(z) be another inner function, with an associated singular function v(t). fo is said to be a divisor of g o if go(z)/fo(z)is an inner function. This is clearly the case if and only if every zero offo(z) in IzI < 1 is also a zero of go(z) (with the same or higher multiplicity) and [v(t) - p ( t ) ] is nondecreasing.
114 7 H P AS A LINEAR SPACE
For fixed f E H P , let P [ f ] denote the subspace generated by the functions z"f(z), n = 0, 1, 2, . . . . Thus S [ f ] consists of all H P functions which can be approximated by polynomial multiples of J: The problem is to characterize P [ f ] . It is already known that S[l] = H P , since the polynomials are dense in H P . More generally, 9 " f ] can be described as follows. THEOREM 7.4 (Beurling). Let f and g be H P functions (1 ( p < m), not identically zero, with inner factorsf, and g o , respectively. Then g E P [f] if and only iff, is a divisor of g o . COROLLARY 1.
P [ f ] = P [ f O ] If . 9"fJ = 9 Q o ] ,thenf, = g o .
PROOF OF THEOREM. Under the assumption thatf, divides g o , it is to be shown that g E S [ f ] .Since
it suffices to prove that Y [ F ] = H P for every outer function F(z). If S [ F ] is not the whole space, there is a nontrivial bounded linear functional which annihilates - it. That is, there exists a function h(ei8)E E (l/p + l/q = 1) such that h(ei8)I$ Ho4and
/;ei"'F(eie)h(e'3
do = 0,
n = 0, 1, 2, .. . .
__
Thus F(eie)h(eie)= k(eie),where k(z) E H,'. Since k(e")/F(eie)E E , it follows from Theorem 2.11 that - k(ei8) h(eie)= E HO4,
F( ele)
a contradiction. Conversely, if g E P [ f ] it is obvious that g must vanish in ( z [< I wherever f does, since norm convergence implies pointwise convergence. Thus h(z) = go(z)/fo(z)is analytic. It has to be proved that h is an inner function. But since g E 9 " f ] , there is a sequence of polynomials (P,} for which l]P,f-- 911 0. Division byf,(eie) then shows that h(ei8)G(eis)is an Lp function approximable in norm by H P functions. Thus hG E H P , which implies that h E H p . This shows (by Theorem 2.1 1) that h is an inner function, completing the proof. --f
COROLLARY 2.
fo - H P .
Let 0 < p < co. Then for any inner functionf, , S [ f O ]=
This is true even for 0 < p < 1, since the proof used only the fact that HP is the Lp closure of the polynomials.
7.4 LINEAR FUNCTIONALS ON
HP,
0
< p < 1 115
7.4. LINEAR FUNCTIONALS ON H9, 0 < p < 1
Although H P is not normable in the case p < 1, its bounded linear functionals can still be defined in the usual manner. Thus a linear functional 4 on H P is said to be bounded if
It is easy to verify that a linear functional is bounded if and only if it is continuous. It can also be checked that the bounded linear functionals on H P form a Banach space under the given norm. We observed in Section 6.4 that H P is an F-space even if p < 1. Hence the principle of uniform boundedness (Banach-Steinhaus theorem) still applies : every pointwise bounded sequence of bounded linear functionals on H P is uniformly bounded. (See Dunford and Schwartz [l], Chapter 11.) We are about to derive a complete representation of the bounded linear functionals on H P ,p < 1. First, however, we need to introduce some notation. Let A denote the class of functions analytic in IzI < 1 and continuous in JzI5 1. It is convenient to writefe A, to indicate thatfE A and its boundary function f ( e " ) belongs to the Lipschitz class A,, 0 < a 5 1. Similarly, f E A, will mean that f~ A and f ( e i o )E A, . (See Section 5.1 .) THEOREM 7.5. To each bounded linear functional there corresponds a unique function g E A such that
4
on H P , 0 < p < 1,
+
If (n 1)-l < p <TI-'(n = 1,2, ...), then g("-') € A , , where a = l/p - n. Conversely, for any g with g("-l) E A,, the limit (4) exists for all YE HP and defines a bounded linear functional. If p = (n + I)-', then 9'" - ') €A* ; and conversely, any g with g("-') E A* defines through (4) a bounded linear functional on H P . PROOF.
Given
4 E (HP)*: let bk = 4(zk),k = 0, 1, .. . . Then Ibkl 5 11$11 IIZkll = 11411,
so the function m
g(Z) =
bkZk k=O
is well defined and analytic in IzI < 1. Suppose now that a,
f ( z )=
CU k=O
~ E ZH P ~.
116 7
HP
AS A LINEAR SPACE
For fixed p , 0 < p < 1, let fp(z) = f ( p z ) . Since f p is the uniform limit on IzI = 1 of the partial sums of its power series, and since 4 is continuous, it follows that N
= lim $( N+m
Butf, +fin H P norm as p
m
akpkzk)
=
k=O
1
akbkpk
k=O
-+ 1, so on
To deduce (4) from (5), it would suffice to show g E H I . But for fixed [, ICl < ],let m
1
f(z) = (1 - [z)-' =
Then by
[kzk.
k=O
(9,
Hence that is, g E H". This proves (4), with g E H". Now suppose (n 1)-' < p < n-', and let
+
F(z) =
n ! Z"
I l l < 1.
(1 - [Z)"+"
In view of (4)and the Cauchy formula, we have
4 ( F ) = g'"'(i>. It now follows from the lemma in Section 4.6 that
ls("'(5)l
5
11411 llFllp=
- l m l ' p - n - l 1,
so that g("-') € A a with a = l/p - n, by Theorem 5.1. If p = (n + a similar argument shows g'"'l'(() = O((1 - Ill)-'), which implies g("-') E A,, by Theorem 5.3. To prove the converse, we first suppose (n + l)-' < p < n-' and that g ( z ) = bkzkis given with g("-l)€ A cr ,a = I/p - n. It is to be shown that for every f ( z ) = a, zk in HP, the function
1
has a limit as r --* 1, and that
7.4 LINEAR FUNCTIONALS ON HB, 0 < p i 1
We shall prove the existence of the limit by showing that
jO1l*V)l d r < 00. Setting h(z) = z"-lg(z) and
where fo(z) =f (z), we have the relation
which can be checked by comparing power series. By Theorem 5.12, v
f, E HPu,
=
1, 2,
. . . , n,
where p o = p and Pv Pv+l = 1 - Pv Simple induction shows that
P
pv =
Hence p
G9 <**.
v = 1 , 2 ,..., n.
< 1 < p n < co.
On the other hand, the fact that g("-') EA, gives
by Theorem 5.1. It therefore follows from (7) that
rnI$'(r2)I 5 ~ ( 1 -ry-lMl(r,fn-l),
a = l/p
But according to Theorem 5.11, IOi(1- #Ml(LLn-l) d r < a, where
p=--
1
Pn- 1 This proves (6) for (n
+ 1)-'
2=a-1.
< p < n-l.
- n.
117
118 7
HP
AS A LINEAR SPACE
+
If p = (n 1)-' and gCn-')E A*, (6) can be established by a similar argument. We omit the details, since they involve the theory of fractional integration, and a development of this background material would carry us too far afield. A full account may be found in Duren, Romberg, and Shields [l]. To complete the proof, we now show that if g ( z ) = bkzkis any function such that m
exists for everyf(z) =
akzkin H P , then 4 E (Hp)*. For fixed Y < 1, let m
dr(f By Theorem 6.5,
=
2
a k bk
k=O
4r E (HP)*for each Y.
But for each fixedfE HP,
sup14 r ( f > l < 00. r
Thus by the principle of uniform boundedness, SUPlI4,Il < CQ, r
which implies 4 E (I€')*. This concludes the proof. We remark that the principle of uniform boundedness need not have been used. The proofs of the various auxiliary theorems actually give norm estimates which could have been carried through the proof to establish the boundedness of 4 directly. The function g ( z ) = (1 - [z)-', [[I < 1, provides an interesting example. Here the integral in (4) reduces simply to f ( r [ ) , so that 4 ( f ) =f((').Point evaluation is therefore a bounded linear functional on H P ,p < 1. (This can also be proved directly; see the lemma in Section 3.2.) In particular, there are enough linear functionals on H P to distinguish elements of the space. This contrasts sharply with the situation for Lp(p < l), where there are no bounded linear functionals at all except the zero functional. 7.5. FAILURE OF THE HAHN-BANACH THEOREM
We have just seen that H P has enough continuous linear functionals to distinguish elements of the space, even if p < 1. On the other hand, we shall now show that there are not always enough functionals to separate points from subspaces. That is, given a proper subspace M of H P ( p < 1) and an H p function f I$ M , there may not be any functional 4 E (HP)*such that $ ( M ) = 0
7.5. FAILURE OF THE HAHN-BANACH THEOREM
119
and &J ' ) # 0. To show this, we shall construct a proper subspace of H P which is annihilated by no functional 4 E (HP)* except the zero functional. Before turning to the actual construction, we have to develop some background material which is of interest itself. Let
be a singular inner function; and let o(t;p) denote the modulus of continuity of p. LEMMA 1. If w ( t ; p) = O(t log l / t ) , then there exist positive constants a and C such that
(S(z)(2 C(l
- r)n,
(z(< 1.
But another in gration by parts shows this last integral is less than
k,
+ J;"P(r,
t ) log -1 d t t
I k, +logI k,
1
1 + log +-
= (k,
+ 2) + 3 log -.I -1r
1 log - d t
120 7 HW AS A LINEAR SPACE
Hence
which proves the lemma. We now introduce the space B, (a > 0) of functions f (2) = in IzI < 1, such that
a, z" analytic
m
IIfII,"
=
C ( n + 1)-aIan12 < a* n=O
B, is a Hilbert space with inner product m
(f, 9 ) =
1( n + l ) - a a n b n
n=O
where g(z) = bnz". Obviously, the polynomials are dense in B, , for each a > 0. A straightforward calculation shows that
5
c2
llf lla2,
(8)
for some positive constants C, and C2 depending only on a. A function f E B, will be called an a-outer function if the set of all polynomial multiples Pf is dense in B E .Clearly, f is an a-outer function if and only if 1 belongs to the B, closure of the polynomial multiples off. L E M M A 2.
If w ( t ; p) = O(t log lit), then S is an a-outer function, for
every a > 0. PROOF. By Lemma 1 and (8), S - l i r n ~ Bfor , some positive integer m. Given G > 0, it follows from (8) that
IIPS'I" - 1 1, ICljP - S - l / m l j a < E
for a suitable polynomial P. This shows that S""' is an a-outer function. To conclude that S is an a-outer function, we need only observe that iff E H" and g E B, are a-outer functions, so is their product f g . Indeed, by (S), IlPQfS
- 111,s llPf(QS - 1>11, + lip?-111, 5 CllWll I1 Qg - 1 1, + Ilpf - 1 1.,
Choose P to make the second term small; then choose Q to make the first term small. This shows thatfg is an a-outer function, and the proof is complete.
7.5 FAILURE O F T H E HAHN-BANACH THEOREM
121
We now consider the Hilbert space H 2 and denote by M I the orthogonal complement of a subspace M of H 2 . For fixed f E H", fHpwill denote the subspace of H P (0 < p I a)consisting of all multiples fg, g E H P . THEOREM 7.6. Let S be a singular inner function such that w ( t ; p) = O(t log lit). Then (SH')' contains no nonnull function g(z) = b,z" such that
c
c nYlbn12 < co, m
y > 0.
n=i
(9)
PROOF. Let S(z) = a,z". Suppose g E H 2 is orthogonal to S H 2 . Then g is orthogonal to zkS(z),k = 0, 1, . . . ; so
(10) c, = (n
If (9) is satisfied, then
+ 1)%,
c m
h(z) =
C,Z"
E
By;
n=O
while the conditions (10) are equivalent to
2 n=k
an-kCn
(n
+ 1)' -0,
k=0,1)
....
This says that h is orthogonal to z k S ( z )(k = 0, 1, . . .) in the space B y . But by Lemma 2, the functions zkS(z) span By.Hence h = 0, which implies g = 0. One final lemma is needed. LEMMA 3. If g ( z ) =
b,z" € A afor some a > 0, then
PROOF. By Theorem 5.1, [g'(z)[ I C(1 - ry-'
Thus
122 7
HP
AS A LINEAR SPACE
Setting r = 1 - l / N , N
= 2,
3, . . . , we find
A summation by parts now gives
if y < 2u. We are now ready to give the construction promised at the beginning of this section. THEOREM 7.7.
Let S be a singular inner function with o(t;p) = O(t log l/t).
Then for each p (0 < p < 11, the only bounded linear functional on H P which annihilates the subspace SHP is the zero functional. PROOF. Suppose
4 E (HP)* annihilates
S H P . Then, in particular,
n = 0, 1, . . .
1
where g(z) = bnz" E A, for some a > 0 (see Theorem 7.5). This implies that g, regarded as an element of H 2 , is orthogonal to the subspace SH2. Therefore, by Theorem 7.6, m
C1 nYlb,J2= co
for each y > 0,
fl=
unless g
= 0.
But this contradicts Lemma 3, since g E A,.
We remark that S H P is a proper subspace unless S = 1. COROLLARY I . If S is a singular inner function as described in the theorem, the quotient space H P / ( S H P )has no continuous linear functionals except the zero functional, for each p < 1.
7.6 EXTREME POINTS
123
COROLLARY 2. Ifp < I , there is a subspace M of H P and a bounded linear functional on M which cannot be extended to a bounded linear functional on HP.
The proofs are left as exercises.
7.6. EXTREME POINTS
A set S in a linear space X is said to be convex if whenever x1 and x2 are in S, every proper convex combination ax,+(l-a)x,,
O
is also in S. An element x E S is called an extremepoint of S if it is not a proper convex combination of any two distinct points in S. There is considerable interest in finding the extreme points of a convex set, especially in view of the Krein-Milman theorem, which states that a compact convex set in a locally convex topological vector space is the closed convex hull of its extreme points. In particular, such a set has extreme points. In any Banach space X , the problem arises to describe the extreme points of the unit sphere
s = {x E x : llxll Il}. Every extreme point obviously lies on the boundary of S; that is, it has unit norm. It is useful to observe that if a point x with ilxl[ = 1 is not an extreme point, it can be represented as the midpoint of the segment joining two distinct points ( x + y ) and (x - y ) in S. Furthermore, Ilx + YII
= IIX - YII =
1,
(11)
since 1 = llxll = Hllx + Yll
+ Ilx - Yll}
1.
Hence a point x with llxll = 1 is an extreme point if and only if (11) implies y = 0. If X is a Hilbert space, it is not hard to show that every boundary point of S is an extreme point. More generally, the same is true if X is a uniformry convex space; that is, if for each E > 0 there exists a 6 > 0 such that llxll = Ilyll = 1 and I/x - y/I > E imply Il(x y)/2II < 1 - 6. If 1 < p < co, the space Lp, and therefore H P , is uniformly convex, so every boundary point is an extreme point of the unit sphere. On the other hand, it is not hard to show that the unit sphere in L1 has no extreme points at all. What is the situation for H' ?
+
124 7 H 9 AS A LINEAR SPACE
THEOREM 7.8 (deLeeuw-Rudin). A function f E H 1 with llfll = 1 is an extreme point of the unit sphere in H1 if and only iff is an outer function.
PROOF. Suppose first that f is an outer function and lifl] = 1. To show f is an extreme point, suppose g E H 1 and llf+911 = Ilf-gll = 1.
(12)
Let h =g/$ Then h E H' and
J~"{~I+ h(eio)l+ 11 - h(ei6)l- 2)jj(eie)l
= 0.
The integrand is nonnegative, since the sum of the distances from h(ei8) to 1 and (- 1) is not less than 2. Sincef(eie) does not vanish on any set of positive measure, it follows that -1 5 h(e") 5 1
a.e.
(13)
Thus, in view of the Poisson representation for H 1 functions (Theorem 3.11, h(z) is real everywhere in ( z I < 1, hence is identically constant. But (12) and the fact that Iifll = 1 then give
11
+ hl = 11 - h(,
so h = 0. This shows g(z) 5 0, proving thatfis an extreme point. Conversely, l e t f = I F be an H' function with nonconstant inner factor I and llfll = 1. To show that f is not an extreme point, we shall construct a gj( = l l f - 911 = 1. This will be achieved function g # 0 in H' such that / I f + if h = g/fhas the properties (13) and
g = +f(J
+ l/J).
(15)
Then g E H',and (13) is obviously satisfied. For (14) also to hold, we have only to choose a such that /oz'
Re{ei"l(eie)}~f(eie)~ dO = 0.
This is possible, since the integral is a real-valued continuous function of a which either vanishes identically or changes sign in the interval 0 5 M < 71. Hence the proof is complete.
7.6 EXTREME POINTS
125
= 1, and f is not an extreme point of S , COROLLARY. (i) I f f € H', then there are two extreme pointsf, and f 2 such that f = 3(fl + f 2 ) . (ii) If [If11 < 1, thenfis a convex combination of two extreme points.
PROOF. (i) If llfli = 1 and f = IF is not an outer function, define g by (15) and (16). Then I l f f 911 = 1, so it will be enough to show thatf+ g are outer functions. But we claim that f f Lg are outer functions whenever 1 2 1. Indeed,
f+
f
;Ig = - [2J f L ( J 2
25
=
+ -A2 e-'"F(I
+ l)]
f eiPJ)(l
+ e-'PJ),
where 1 cos /3 = 1. The last two factors are outer functions, because they have positive real part (see Exercise 1, Chapter 3). Thusff Ag is outer, since it is the product of outer functions. (ii) If lifli < 1 andfis not an outer function, let g be as above and choose A, > 1 and ;I2 > 1 such that
Ilf+M I = llf- &gll = 1. by continuity, since Ilfll < 1. Then ( f + L l g ) and (f- L,g)
This is possible, are extreme points, as shown in (i). I f f i s outer, the functions +fiiifii are extreme points. The next theorem identifies the extreme points of the unit sphere in H". By way of motivation, let us first note that f E H" is an extreme point if llfll = 1 and / f ( e i o ) l= 1 on some set E of positive measure. Indeed, if g E H" and Ilf+ gjl = / I f - 911 = I , theng(eiB)= 0 a.e. on E, which implies g = 0. THEOREM 7.9. A functionfE H" with Ilfll unit sphere in H" if and only if
=:-! PROOF.
=
1 is an extreme point of the
log(1 - If(e"))l) dd = - co.
Let g E H" such that
Ilf+
911 = Ilf-sll = 1.
(17)
126 7
HP
AS A LINEAR SPACE
Then
ls(z>I2I 1 - If(z)lZ I 2 ( 1 - If(z>l), and it follows from (17) that
which implies g = 0. Hence f is an extreme point. Conversely, if the integral (17) converges, let
Then Ig(z)l I 1 and lg(eiB)I I 1 - If(eie)>l
Thus
\If+
gll I 1 and
Ilf-
a.e.
gI( I 1, sofis not an extreme point.
EXERClS ES
1. Show that for each fixed i, < 1, and for each positive integer n, $ ( f ) = f ( " ) ( i )is a bounded linear functional on H P , p < 1. 2. Prove Corollary 1 to Theorem 7.7. 3. Prove Corollary 2 to Theorem 7.7. 4. Show that every Hilbert space is uniformly convex. 5. Show that in a uniformly convex space, every boundary point of the unit sphere is an extreme point. 6. Show thatfE L" with ilfli
=
1 is an extreme point of the unit sphere in
L" if and only if lf(ei@)l= 1 a.e. 7. Show that the unit sphere in L1 has no extreme points. NOTES
A. E. Taylor [2,3] was among the first to study H P (1 4p I 00) as a Banach space. He represented the linear functionals (Theorem 7.3) in his paper [3]. Beurling proved Theorem 7.4 for H 2 in his fundamental paper [I]. The proof given in the text is essentially his, suitably extended to 1 < p < 00. Actually, Beurling showed that every subspace of H 2 invariant under multiplication by z has the form S [ f O ]for some (unique) inner functionf, ; and he described
NOTES
127
the lattice structure of these invariant subspaces. Equivalently, he described the invariant subspaces of the shift operator on f 2 . Subsequently, a large literature has evolved on invariant subspaces. References up to 1964 may be found in Helson [3]. Gamelin [l] has extended Beurling's theory to H P with O
T(f)(4 = aC~'(411'"f(4(4>, where a is some complex number of modulus one and cp is some conformal mapping of the unit disk onto itself. The Hilbert space H 2 of course has many other isometries. Forelli also describes the isometries of H P into itself for p # 2. These are more complicated.
This Page Intentionally Left Blank
EXTREMAL PROBLEMS
CHAPTER 8
We are now prepared to discuss the theory of extremal problems in H P spaces. It turns out that extremal problems come in pairs, and that even a partial solution to one problem may help solve the other. We first discuss the existence and uniqueness of solutions to the two types of problems, and establish an important " duality relation. " Then we find the qualitative form of the solutions in the case of a rational kernel, which is general enough to include most of the interesting examples. These results offer a systematic and practical method for solving various extremal problems. At the end of the chapter, a few examples are discussed to illustrate the technique.
8.1. THE EXTREMAL PROBLEM AND ITS DUAL
The most general bounded linear functional on H p (1 I p < a)can be expressed in the form
130
8 EXTREMAL PROBLEMS
where k(eie)E Lq, l / p + l/q = 1. The functionals of greatest interest are generated by kernels k(eie)which are boundary values of rational functions. Some examples are given below.
(ii)
k ( z ) = n ! ( z - P>-"-l,
< 1;
+(f)
=f("'(P).
cimLq,,oajthe typical extremal problem is to find
wheref(z) = For fixed k
zj.
E
11411 =
SUP IEHP,
Iml.
IlfllpSl
In addition to finding the value of the supremum, several questions are to be considered. Is the supremum attained, and if so, is the extremal function unique ? What are the extremal functions ? A function h E Lq is said to be equivalent to the given kernel k (written h k) if h and k belong to the same coset of E/H4; that is, if h - k E H 4 . Thus h and k determine the same functional on H P if and only if h k . An application of Theorem 7.1 gives at once the important relation
-
-
sup
_f_
f e H P , IlfllP5l
2n
1s
1z1=1
I
f ( z ) k ( z )d z = min Ilk
- gll,
(3)
geHq
for 1 < p < a.This is called the duaIity relation. It connects the original extremal problem with what is caled the dual extremalprobfem: to find the function g E H 4 which is closest to the given kernel k E E (or, equivalently, to find the function h k of minimal norm). According to Theorem 7.1, the minimum is actually attained; that is, the dual extremal problem always has a solution if 1 < q 00. The argument breaks down for p = co (q = l), since the conjugate space of L" is larger than L'. However, the dual extremal problem still has a solution in this case, and the duality relation remains true. To prove this, let C be the subspace of L" consisting of all continuous periodic functions, and let P be the subspace of C generated by 1, eie, eZie,. . . The idea is to view as a functional on C . By Theorem 7.1,
-
.
8.1 THE EXTREMAL PROBLEM AND ITS DUAL
131
where 'P indicates the subspace of C* which annihilates P. It is easy to describe P'-.According to the Riesz representation theorem, each $ E C* has the form
for some function p(0) of bounded variation. But $ E'P implies $(e'"') = 0, n = 0,1,2, . . . . Hence it follows from the theorem of F. and M. Riesz that
dp(0) = g(eie)eiedo,
g E H1,
so that
Conversely, every $ of this form annihilates P.Thus the relation (4) takes the form
I W ) l = min Ilk + 9111. gsH'
SUP fEP?Ilfllmsl
(5)
In particular, the dual extremal problem has a solution in the case q = 1. On the other hand, for every f~ H" with llfll I1 and for every g E H', it is trivially true that I4(f)l IIlk g1I1. Hence
+
I W ) l 5 J e H mS,Ullfll P I9Kf)l Igmin Ilk + 911. S1 E H ~
SUP f e P , IlfIls1
In view of (5), then, equality must hold throughout, and the duality relation is true even for p = 03. Let us now return to the original extremal problem (2) and show that a solution always exists for 1 < p I03. In other words, for each L4 function k(eie),the supremum in (3) is attained. The idea of the proof is to regard
as a bounded linear functional on L4 generated by a fixed function f ( e i e ) which belongs (more generally) to Lp.Then 11$11 = Ilfll. For 1 Iq < 03, each $ E (Lq)*has the form (6) for somefe Lp; and those $ which annihilate H 4 are precisely the functionals generated by functions YE H P . This implies that the extremal problem in question has a solution, since SUP
lW>l
r l r ~ ( H V11rlrIls1 ,
is attained for each fixed k E L4,by Theorem 7.2.
132 8
EXTREMAL PROBLEMS
The proof fails in the case p = 1, and for a very good reason: the extremal problem (2) need not have a solution in H'. A counterexample will be given in Section 8.3. A solution does exist in H' if the given kernel k(eie)is continuous. To prove this, consider again the space C of continuous periodic functions and the subspace P which is the uniform closure of the polynomials in eie. Each function f~ L1 defines, by (6), a functional $ E C" with ll$ll = l l f l l . The argument given above, using the F. and M. Riesz theorem, shows that the annihilator of P consists of all )I E C* of the form (6) withfE H'. Theorem 7.2 may therefore be invoked t o conclude that the supremum in (3) is attained, which was to be shown. Observe that periodicity of the kernel is not really necessary; continuity on the interval [O, 2x1 is sufficient. The conclusion obviously may be extended t o any kernel which is the sum of a continuous function and an H" function. 8.2. UNIQUENESS OF SOLUTIONS
-
It will be convenient to reserve the term extrema1,function to indicate a solution to the original extremal problem (2). A function K(eie) k(eiB)for which
will be called an extremal kernel. Having established the existence of extremal functions and extremal kernels, it is natural to ask about uniqueness. Since 14(ei"f)l = I4(f)l for every real a, some normalization clearly must be imposed before an extremal function can be unique. It is convenient to single out the extremal functions for which $(f)= ll$ll; these will be called normalized extremal functions. To avoid trivial complications, it is always assumed that 4 is not the zero functional; i.e., k H 4 . The main results on existence and uniqueness can be stated as follows. THEOREM 8.1 (Main existence and uniqueness theorem). For each p (1 s p 5 a)and for each function k(eio)E E (l/p l/q = 1) with k H4,the duality relation Id(S)I = inf Ilk - 911, SUP
+
J-EHP,
IlflIpCl
gEH4
holds, where d(f) is defined by (1). If p > 1, there is a unique extremal function f for which 4 ( f ) > 0. If p = 1 and k(eiB)is continuous, at least one extremal function exists. If p > 1 (q < a),the dual extremal problem has a unique solution. If p = 1 (q = a),the dual extremal problem has at least one solution; it is unique if an extremal function exists.
8.2 UNIQUENESS
OF SOLUTIONS
133
The existence of solutions was established in Section 8.1 ; only the uniqueness assertions remain to be proved. The argument is based upon the following simple observation. LEMMA. In order that a function F E H P , with llFll = 1 and @ ( F ) > 0, be an extremal function and that K ( K k ) be an extremal kernel, it is necessary and sufficient that N
eiBF(eio)K(eiB) 20
a.e.
(7)
and that
IK(eie)l = I l ~ l l IF(eiB)IP= I
if p = l ;
a.e.
l ~ l-qlK(eiB)lq l
IF(eio)l = 1
a.e.
if 1 < p < co; if p = co.
a.e. on (6 : K(eiB)# O>
(8)
PROOF OF L E M M A . It is clear from the duality relation that F is a normalized extremal function and K an extremal kernel if and only if,@(F)= IlKIl. The lemma then merely expresses the conditions for equality in Holder's inequality. In the condition for p = 1, the fact is used that F(eie) cannot vanish on a set of positive measure without vanishing identically. PROOF O F THEOREM. Let F and K be a normalized extremal function and an extremal kernel, respectively. Since k E H 4 , K(eLe)must be different from zero on a set E of positive measure. Through relations (7) and (8), K determines both sgn F(eiB)and IF(eiB)Ialmost everywhere on E, provided 1 < p < co. This means that if 1 < p < co, all normalized extremal functions coincide on some set of positive measure, hence coincide almost everywhere. In other words, F is unique. For p = 1, the argument shows at least that sgn F(eio)is determined almost everywhere, if any extremal functions F exist. To prove the uniqueness of K , let F be a fixed normalized extremal function (uniqueness is not used), K any extremal kernel. According to (7),
Re{ieiBF(eiB)K(eie)} =0 Let G = k
a.e.
- K , so that G E H q ; and let h(z) = izF(z)G(z).Then Re{h(eie)} = Re{ieiBF(eiB)k(ei~)} a.e.
But since h E H1 and h(0) = 0, the knowledge of Re{h(eie)} completely determines h(z), through the Poisson formula. This shows that G, and hence K , is uniquely determined. The proof is valid in the case p = 1 (q = co) if an extremal function F exists.
134 8 EXTREMAL PROBLEMS
8.3. COUNTEREXAMPLES IN THE CASE p = 1
The existence and uniqueness of extremal functions and extremal kernels has now been established for 1 < p < CQ. The situation is entirely different, however, in the case p = 1. An extremal function need not exist in H i , and if it does, it need not be unique up to normalization. At least one extremal kernel must exist, as shown above, but it is not in general unique. Counterexamples will now be given t o justify the three negative statements. (i) To show that an extremal function may fail to exist, consider the example 1
-1
The existence of the integral (in the principal value sense) can be proved by applying Cauchy's theorem with a contour consisting of the segment -r I x I r and the semicircle z = rei0,0 5 8 I 7c. Sincef E H I , it follows that lim r-tl
so that
J: rf(x> dx = - i j;f(eie)eie d e ,
4(f)has the usual form (1) with a kernel k(eie)=
{- 01,,
-7c<e
o5eI.n. An equivalent kernel is K(z) = k(z) + 5; hence ll$l/ 5 l(KII = +. This is also a consequence of the FejCr-Riesz inequality (Theorem 3.13). In fact, the example used to prove the sharpness of the FejCr-Riesz inequality, a conformal mapping of Iz( < 1 onto an elongated rectangle, shows that ll4Il = $. Thus K is an extremal kernel. If there exists an extremal function F , normalized so that $(F) = +,then by (7) eieF(eie)K(eio) 20
a.e.
Therefore, F(eiO)=
e-ie)F(eie)l, { -e-ie)F(ei8)1,
<e <0 O
-7c
Since F E H ' , it follows that
0=
1'
ei(n+
i)eF(p)
--x
0
=
J- zeinelF(eie)IdQ - lffein01F(eie)\do, 0
n = 0, 1, 2, . . . .
8.3 COUNTEREXAMPLES IN THE CASE p = 1
135
Consequently,
1- einelF(e")I do 0
n
=
rein0lF(eiB)Ido,
n
= 0,
& 1, +2, . .. .
0
In other words, the two functions
and
have identical Fourier coefficients. This implies $l(e) which is obviously impossible.
=
$,(e)
a.e., so F(z) = 0,
(ii) The elementary example
shows that a normalized extremal function need not be unique. Here it is obvious that \l4\\= 1, and thatf(z) = z is an extremal function. But a simple calculation shows that for every complex constant a,
is also extremal. (iii) A more elaborate example is needed to demonstrate the nonuniqueness of an extremal kernel. In view of the main theorem, this will automatically be another case in whish no extremal function exists. Let
and let 4 ( f ) be the functional with kernel k , defined by (1). Obviously, 11411 5 1. To show that 11411 = 1, consider the functiong(z) which maps 17.1 < 1 conformally onto the rectangle with vertices & 1 Jr i E , in such a way that g(1) = - 1 - k , g(i) = 1, and g(- 1) = -1 + k.Let eia(0 < CI < n/2) be the
136 8 EXTREMAL PROBLEMS
point carried into (1 - k).Then g(e’(”-a))= 1 + ie, by symmetry. It is geometrically clear that the tangent vector
while eieg’(eie)2 0 for
CI
< 0 < n - c(. Thus
On the other hand, ll9’lIl =
4(1
+ E)
7
Therefore,
which implies I1411 = 1. It now follows from the duality theorem that k is itself an extremal kernel, since llkll, = 1. The object is to produce another extremal kernel; that is, another function K k with 11KlIm= 1. Let K = k + h, where h(z) maps J z J< 1 conformally onto the half-disk J w J< 1, Im(w> > 0, with h(1) = - 1, h(i) = 0, h( - 1) = 1. Obviously, K k . But
-
N
-z< 0 <0; 0I 0 42;
Ih(eie)l= 1,
s
1,
71/2 < 0 s x.
Hence l(KI(, = 1. 8.4. RATIONAL KERNELS
If the kernel k(eis)is the restriction to the unit circle of a rational function k(z), it is possible to describe the extremal functions and extremal kernels more or less explicitly. This is of some interest, because most of the functionals which arise in practice have rational kernels. Some examples were given at the beginning of the chapter. Let k ( z ) be analytic in IzI < 1 except for poles at PI, P 2 , . . . ,pH (IPJ < I), each pole being repeated according to multiplicity. Suppose k(z) is continuous elsewhere in J z J_< I. Then by Theorem 8.1, a normalized extremal function F
8.4 RATIONAL K E R N E L S
137
exists for eachp, 1 5 p 5 03. F is unique for 1 < p 5 03, but not necessarily for p = 1. An extremal kernel K exists for each p, 1 5 p 5 co, and is unique (even for p = 1, since F exists). Consider the function R(z) = zF(z)K(z), which is analytic in IzI < 1 except perhaps for poles at the points P k . According to the lemma of Section 8.2, R(eie)2 0 a.e., because F and K are extremal. Furthermore, since lim
r
r-rl a
IR(reie)- R(eie)l de = 0
for any arc (a, b) on IzJ = 1, the classical technique of Schwarz reflection may be applied. (See, for instance, Nehari [ 5 ] , pp. 183-185.) The conclusion is that R(z) has a continuation onto the extended - z-plane as a function analytic except perhaps for poles at pl, . . . , fin ; lip1, . . . , lip,. Hence R is a rational function. If R(a) = 0 for IaJ < 1, then l/C is a zero of the same multiplicity. Furthermore, any zero on the boundary IzJ= 1 must be of even multiplicity, since R(z) 2 0 there. Thus there are points a,,. . . , a,, lai/ 5 1, in terms of which R has the structure
By the reflection property, the order of the zero (or pole) of R at the origin must be the same as the order of the zero (or pole) at infinity. If p of the cli and v of the pi are equal to zero, this statement is expressed by the equation p - - v + l =(2n-v)-(2m-p)-11.
Hence m = n - 1. It follows, incidentally, that C > 0, since R(z) 2 0 on 1 2 1 = 1. Some of the ai which are zeros of F may possibly coincide with numbers pi.However, K has a pole at each p i , and in fact its principal part at each pole must coincide with that of k. Let the ai be renumbered, if necessary, so that a,, . . . , a, are the zeros of K and a,,,, . . . , a, those of F in IzI < 1. Then [ail = 1 for i = c + 1, . . . ,y1 - 1. The functions
and F(2) = F ( z )
f J
i=s+1
1-Gz z - a,
___
are analytic and nonvanishing in IzI < 1. F E H P , while R(z) is continuous in /zJ 1. Furthermore, P and I? are outer functions. To prove this, it suffices
138 8 EXTREMAL PROBLEMS
to show that the product B(z) = F(z)l?(z) has no singular factor. In view of the structure of R(z), however, i?(zj is a rational function without zeros or poles in ( z (I1, except perhaps for zeros on J z J= 1. Thus the problem reduces to showing that ( z - a ) is an outer function if jell = 1. But this follows from the fact that ( z - a)-’ E HP for p < 1. Consequently,
loglF(e”jl d t ] ,
(9)
and similarly for I?(z). The relations (8) between F and K are yet to be used. Suppose first that 1 < p < m.Then JF(z)Ip= IIK/J-4JK(z)J4 a.e. on
J z J= 1.
But JF(z)K(z)l = R(z) on (21 = 1, so
1F(z)l = IJKlJ-1’”[R(z)]1’p
a.e. on IzJ = 1. Combination of this expression with (9) gives
where A is a complex constant. Since zF(z)K(z)= R(z), it follows that
Here the fact has been used that
The structural formulas (10) and (1 1) remain valid in the cases p = 1 and p = co,with the understanding that 1/00 = 0. This may be verified by similar calculations based upon the appropriate relations (8). F o r p = co,the extremal function takes the simple form
The practical value of the formulas ( 1 0) and ( 1 1 ) will in general depend upon the actual calculation of the parameters a i. This can be a very difficult problem.
8.5 EXAMPLES
139
Suppose, however, that an integer 0 5 n - 1 and numbers B and a l , a 2 , . . . , an-,canbefoundwithlail < l f o r l 5 is aandlaJ = l f o r o + 1 5 i l n - 1, such that for some s 5 CJ the function K given by the formula (1 1) is equivalent to the original kernel k. Then K is the extremal kernel and, with an appropriate choice of the constant A , F defined by (10) is a normalized extremal function (unique i f p > 1). This follows from the lemma of Section 8.2. The problem therefore reduces to an interpolation problem: to find numbers B and x i , subject to the above restrictions, such that K has the same principal part as k at each of the poles p i . This problem is guaranteed to have a unique solution if 1 < p 2 00, since there is a unique kernel K of the form (1 1) and a unique normalized extremal function F with the structure (10). The interpolation problem has at least one solution in the case p = 1. Here it is a question of finding numbers B and C L ~ , . . . ,CL, with ]ctiI < 1 and sI n - 1, such that
has the given principal parts at the poles pi.This modified extremal problem will have a unique solution, because there is only one extremal kernel. If it turns out that s = n - 1, the extremal function F is uniquely determined by (lo), with the first product missing. Otherwise (if 0 5 s < n - l), the remaining m i can be chosen arbitrarily in the disk la1 _< 1. If lail < 1 for s + 1 5 i I 0 and ]ail = 1 for 0 + 1 5 i I n - 1, the function (10) (withp = 1) is necessarily an extremal function, and every extremal function is obtained in this manner. This characterizes the family of extremal functions in the indeterminate case. 8.5. EXAMPLES
(i) A coeficient problem and its dual. Given complex constants . . . , cn (c, # 0), let it be required to find
co, cl,
where f ( z ) = Cj"=oajz'. This coefficient problem was posed and elegantly solved by E. Landau in the special case p = co and co = c1 = ... = c, = 1. From the present point of view, it is an extremal problem with the rational kernel n
k(z)=
C cjz-j-'. j=O
The dual extremal problem is
140 0 EXTREMAL PROBLEMS
which is equivalent to the problem
In other words, the dual extremal problem is to find inf (lhl(among all H 4 friimctions of the foim
The latter is a " minimal interpolation " problem at the orjgin. According to the general theory, there is always a unique solution (1 I qI GO), and the value of the minimum is equal to the supremum in (13). The problem (13) also has a unique normalized solution if 1 < p _< 03, and at least one solution ifp 1. In fact, these solutions take the forms (11) and (lo), respectively. The parameters B and ui must be determined so that [ail _< 1, lai/ < 1 for 1 < i _< s, and
Assumingp > 1 and recalling that cn # 0, let
, . . . , A,, can be expressed in terms of for small Iz]. The numbers l o A,, co, c1, . . . , c,. lf P,,(z) = 1, l , z -t- * * * + z"
+
does not vanish in jzI < 1, then n
-
P,(z) = A, n(1- E i Z ) , i= 1
lail
I 1,
and n
[Pn(z)]2'*= c,
rl[(1 - ?&z)2'q i=l
is the required solution to (14). Thus
n n
K(z) = cn
(1 - Gz)2'qz-"-'
i= 1
is the extremal kernel, and the extremal function has the form
8.5 EXAMPLES
141
where a,, . , . , a, are the reflections of the zeros of P,(z), if any, which lie outside the unit circle. The functions K and F are not so easily found if P,,(z)vanishes in IzI < 1. Various criteria are available, however, for concluding tbat P,(z)has no zeros inside the unit circle. (See M. Marden [I].) For example, the classical Enestrom-Kakeya theorem asserts that this is the case if
A,
2
A, 2 - * .
2
A, > 0.
If
2, > A, > . . . > An > 0, all the zeros of P,(z) lie strictly outside the unit circle. (ii) Minimal interpolation. Let distinct points z , , . . . , z,, be given in IzI < 1. It is required to find an H P function (I I p 5 co) of minimal norm which takes prescribed values f ( z j )= w j ,
j = I,.
.., n,
at the given points. The problem could be generalized by prescribing the values of the function and its first few derivatives at the points z j , but the more special problem will serve to illustrate the main ideas. It is not immediately clear that a solution exists at all. However, the problem can be cast in the form of a dual extremal problem with rational kernel, which will ensure the existence and uniqueness of a solution. Let z-zi
B(z) =
and
-
i=,
1- F z
Bj(z)=
11 ~.z - z i
i+jl-
Define
Then h(z) is analytic in IzJI 1 and h(zj) = w j , .j H P function taking the values w j at z j is
=
1, 2, ~.. , n. The general
f(4 = h(z) - B(zlg(z), where g is an arbitrary H Pfunction. The problem of minimizing fore equivalent to the problem of finding minlljc - g/lp, gENP
where
llfli is there-
142 8 EXTREMAL PROBLEMS
This is a dual extremal problem with the rational kernel k. The associated functional on H 4 ( l / p + l/q = 1) is
The duality relation (3) therefore gives the curious result
(iii) Extremal problems in H 2 . If the kernel is rational and p = q = 2, the extremal kernel (1 1) takes the form S
K(z)= B
(2
- mi)
i= 1
n- 1
n
i=s+ I
i=l
IT (1 - .iz ) n:
(2
- pi)- 1 .
The parameters must be chosen so that this function K is equivalent to the given kernel k. But the '' natural kernel " k*, defined as the sum of the principal parts of k at each of the poles pi, has precisely the form (16): n
ITI -
k*(z) = Q ( z )
(2
Pi)-
9
i=
where Q is a polynomial of degree (n - 1) or less. Hence the natural kernel is always the extremal kernel in the H 2 case. Knowing the extremal kernel, it is easy to find the extremal function F as given by (lo), without actually determining the roots ai. Indeed, in view of (12)9
n( 1 - % z ) n - ai) n (1 - CZ). F(z) n( 1 =C
(Z
i= 1
Hence
n- 1
a
S
Zn-'Q(l/Z)
i=s+ 1
i=a+l
- n
= Azn-lQ(l/Z)
-Biz)-',
i= 1
(iv) The coe$cient problem in H I . Let us return now to the coefficient problem (13) and its dual, the minimal interpolation problem at the origin. Let mp be the value of the maximum in (13). Obviously,
m2 = {lco12
+ Ic1I2+ + IC~}'~. *
It is also possible to find m,, at least up to an algebraic computation. Any H 1 functionf(z) = anz" can be expressed as the product of two H 2 functions g and h with llflll = 119112 llhllz *
EXERCISES
143
If m
m
then
Thus
c
c
n
n
Ckak=
n Cj+kXjyk,
i=O k=O
k=O
and the problem reduces to finding
+ + lx,I2 and ( A j k )is the Hankel matrix
where llx1I2 = lxOl2
The maximum of this symmetric bi-quadratic form can be characterized as the greatest absolute value of the eigenvalues of the matrix ( A j k ) . By the duality relation, this is also the norm of the minimal interpolating function. EXERCISES
1
1. Show that if f ( z ) = ukzkE HP (1 < p I co),then lan(I I/ f ( I p , with equality if and only iff(z) = Az". 2. Show that iff(z) = ukzkE H ' , then lanl _< I/ f /II, with equality if and only iff has the form
c
n
f ( z )= A
fl( z - aJ(1 i=
KZ),
1
where the aiare arbitrary complex constants. 3. For distinct points z1 = 0, z 2 , . . . ,z, in IzI < 1, and for 0 < p i co, find the H Pfunction of minimum norm such that f ( 0 ) = 1;
What is the minimum norm?
f(zJ
=*
* *
= f ( z , ) = 0.
144 8 EXTREMAL PROBLEMS
4. Show that iff
E H P (0
If(4l I ( 1 - l~12)-1'pllfllp~ I4 < 1. Show that the inequality is sharp for each fixed z.
5. For fixed z, IzI < 1, and for each positive integer n, find the maximum of If(")(z)l over all H 2 functions f with llfl12 2 1. Determine the extremal functions. 6. ForfE H 2 and for (zl/< 1, Iz2(< 1, and z1 # z 2 , prove
and show that the constant is best possible. 7. For each fixed z, Iz( < 1, and for each odd integer n = 2m + 1, prove the sharp inequality
and identify the extremal functions. (Hint: Show that the natural kernel is extremal.) (SzBsz [2]; Macintyre and Rogosinski [2].) 8. Let M,, denote the maximum of la, + a, + ... + a,,j among all H" functions f(z) = C akzkwith I l f l I , I 1. Show that
Hence M ,
N
I/. log n as n --+ a.(Hint: Observe that
(Landau [1,21.)
9. Find the distance from H' to the function k(z) = (1 - z-"-l)(l - z)-';
that is, find the minimum of Jjk-fill 10. Show that forf(z)
=
for allfe H I .
akzkE H ' ,
max l a o + a l + . . . + a , l
IIf II 1 5 1
+
1 (n 1)n ==-sec-. 2 2n+3
(Hint: Observe that the eigenvalue problem X , - ~ + X , - ~ + ~~ -~* + X , , = ? ~ . X ~ ,
k = = Q , I.,. . . 17,
has the solution
1 2
+
(n l)n* 2n+3'
A = - sec ____
xk = sin-
(k 4- l ) n 2n -t3 '
I<- 0, 1 ,
I
.
.
,
..>
(Egervdry [I].) NOTES
The theory of extremal problems in H p spaces has evolved from a long history of scattered work on special examples. Landau [ 1,2] solved thecoefficient problem (i) in Section 8.5 in the case p = GO and c, = c1 = . . . = L, = 1. Szzisz [3, 41 discussed the more general problem in 15". CarathCodory and FejCr [2] proposed the "minimal interpolation " problem to minimize Ilfil among all f~ H" whose Taylor coefficients a,, a , , . . . , a, at the origin are prescribed. They proved the existence and uniqueness of a wlution, and they gave an algebraic method for computing the minimum. Gronwall [l ] simplified this work. F. Riesz [l] discussed the same problem in N',proved the existelice and uniqueness of a solution, and characterized it using a variational method. Fejtr [ l ] solved the coefficient problem (i) of Section 8.5 in N ' , as well as its dual, the Garathtodory-Fejer problem, essentially by the method described in (iv) of Section 8.5. Egervgry [I] obtained a completely explicit solution i i i the case co = c1 = = c, = 1. Kakeya [I, 21 considered the more general interpolation problem (ii) of Section 8.5 and found the form of the solution. In the H" case, this problem was also discussed by Pick [ I , 21, Sclur [l], and R. Nevanlinna [I]. Geronimus [ I , 2, 31 treated some problems similar to those of Caratheodory-FejCr and Riesz. Doob [1] considered the problem of approximating an arbitrary rational kernel by an HI fimction, proved the existence and uniqueness of a solution, and found its qiiditative form. Goluzin [I, 21 discussed a wide variety of extremal problems. The first systematic attempt to unify the theory and to extend it to H P (1 5 p 5 co) was the paper of Macintyre and Rogosinski [I]. They formulated the duality relation in full generality, proved the existence and uniqueness of extremal functions and extremal kernels, described their form in the case of a rational kernel, and applied the theory to obtain explicit solutions to a great many extremal problems. Nevertheless, the theory achieved complete unity and elegance only when Havinson [1, 21 and (independently) Rogosinski and Shapiro [I] introduced 3 .
146
8 EXTAEMAL PROBLEMS
the methods of functional analysis. A. L. Shields discusses this development in a historical note following his translation of Havinson’s paper [2]. Subsequent refinements by H. S. Shapiro [l], Havinson [3], Bonsall [I], and others have led to the main existence and uniqueness theorem as stated and proved in Sections 8.1 and 8.2. The three counterexamples given in Section 8.3 are essentially due to Rogosinski and Shapiro [l]; the third is given in a simplified form suggested by Caughran [l]. Extremal problems for H P spaces over multiply connected domains (see Chapter 10) have been studied by Ahlfors [l], Garabedian [l], Nehari [2,3,4], Rudin [I, 21, Havinson [3], Penez [I], Lax [l], Tumarkin and Havinson [8], and others. Akutowicz and Carleson [l] have studied the analytic continuation properties of extremal interpolatory functions.
INTERPOLATION THEORY
CHAPTER 9
Let zl, z2 , . . . be distinct points in the open unit disk, and let ivl, w 2 ,. . . be arbitrary complex numbers. The general interpolation problem is to characterize the pairs of sequences {zk} and { w k } for which there exists a function J’E H P with f (zk) = wk , k = 1,2, . . . , and to find all of the interpolating functions. But in such general form the problem is difficult, and has not yet been fully solved (see Notes). In this chapter we confine attention to the more modest problem of describing the “ universal interpolation sequences.” These are the sequences {zk} which admit an H“ interpolation for every bounded sequence ( w k ) . This restricted problem has a natural generalization to H P (0 < p < 00) which we also discuss. The chapter concludes with a theorem of Carleson which generalizes a special result used in proving the main interpolation theorem. Carleson’s theorem plays an important part in the proof of the corona theorem (Chapter 12), and it has other applications. 9.1. UNIVERSAL INTERPOLATION SEQUENCES
Consider the problem of finding the most general function f E H P such that = W k , k = 1, 2, . . . . If a particular solutionfcan be found, it is an easy
f(Zk)
148
9 INTERPOL4TION THEORY
matter to describe the general solution. Indeed, the interpolation f is unique if and only if 1 (1 - lzkl) = co,since the difference of two interpolations would vanish at the points z,. If (1 - lzkl)< co, then the general solution is f + B g , where B is the Blaschke product with zeros (2,) and g is an arbitrary HPfunction. In discussing the existence of an H P interpolation, it is helpful to adopt a more abstract point of view. Fixing p (0 < p 5 co) and a sequence {z,}, consider the linear operator T which assigns the sequence ( f ( z k ) } to each f~ H P . The range of Tis clearly the set of all sequences {w,}for which interpolation is possible. The problem is then to characterize the range T(HP) for each sequence (2,). It is obvious that T ( H “ ) c em for every sequence (2,); and it follows from Theorem 5.9 that if Iz,~ + 1, then lim W,(I - Izkl)’’P = 0 k+ m
for each {w,} E T(HP), 0 < p < 00. It is of interest to identify the sequences {z,} for which T(H“) = em. These are called universal interpolation sequences, since to every {w,} E em there corresponds a functionfE H“ withf(z,) = w,. It is intuitively clear that { z k }cannot be a universal interpolation sequence if the points are “ too close together,” because it would then be impossible to interpolate a “ highly oscillatory” sequence {w,} by a function with “reasonably small ” derivative. This can be made precise. A sequence {zk}is said to be uniformly separated if there is a number 6 > 0 such that
The convergence of the Blaschke product implies, in particular, that C (1 - lzkl) < co. It turns out that {z,} is a universal interpolation sequence if and only if it is uniformly separated. This surprisingly sharp result can be extended to an arbitrary H P space. For this purpose it is convenient to consider a kind of weighted interpolation, Given a sequence (zk}, let T, be the linear operator on H P (0 < p 5’00) defined by In particular, T, = T. We noted above that if p < co, T, maps HP into the space of sequences which tend to zero, provided only that ( z k } has no limit point in the open disk. On the other hand, T, need not map HP into P, even if (1 - lzkl)
9.2 PROOF OF THE MAIN THEOREM
149
For 0 < p i 00, T,(HP)= L p
THEOREM 9.1 ( M a i n interpolation theorem).
if and only if { z k }is uniformly separated. 9.2. PROOF OF THE MAIN THEOREM
Let us first assume 1 I p i co, deferring the case 0 < p < 1 to Section 9.3. It is obvious that for every sequence { z k } , T = T, is a bounded operator from H“ to L“. More generally, if { z k }is such that T,(HP)c P,then T, is a bounded operator from H P into P.This will follow from the closed graph theorem if it can be shown that T, is closed. But suppose f,EHP,
L+f,
and Tp(f,)
= {(l
- Izk12)”pfn(zk)}
-+
Lp*
’v
Thenf,(z,) + f ( z k ) for each k,so T,(f)= w.This proves T, is a closed operator. Suppose now that T,(HP)= P,and let
N P = (fE H” : f ( Z k ) = 0, k
=
1, 2,
. . .)
denote the kernel of T, . The quotient space H P / N Pis a Banach space under the usual norm. (See Section 7.1.) Since T,(HP) = L”, T, induces a one-one bounded linear operator 9,from H P / N p onto P.By the open mapping theorem, 9,’ is bounded. In other words, corresponding to each w = { w k } E P,there is a function f E H P with
k = 1, 2, .. .
(1 - I z k 1 2 ) ’ / ” f ( z k ) = wk,
(1)
and
llfll 5 M I l M 4 , where M is a constant depending only on p . In particular, for each positive integer k there is a functionf, E H P with llfkll I M and .fiCzj>
=
((1 -
,
j =k j#k.
For n > k, let
Then F , k E H P and IlFnklI = llfkll 5 M. On the other hand, for any F follows from Theorem 5.9 that IF(z>l i CIIFIl(1 - 142)-1/p,
E H P , it
150 9 INTERPOLATION THEORY
where C depends only on p . Consequently,
ICllFnkI/ 5 C M .
This shows that {zk} is uniformly separated if Tp(Hp)= eP. The converse is more difficult. The plan is to prove it first in the special case p = 2, then to apply this to establish the general result. Two lemmas will be needed. LEMMA I.If
{zk}is a uniformly separated sequence, then
where A is a constant independent of k. PROOF.
A simple calculation gives
Thus
This gives the desired result with A = 1 - 2 log 6. LEMMA 2.
Let ajk ( j , k = 1, 2,
...,n)
be complex numbers such that
akj = aiikand n
C lajkl I M , j=1 Then for any numbers xl,
...,xn,
k = 1,2,
... , n.
9.2 PROOF OF THE MAIN THEOREM
151
PROOF. Write lajk x j
%I = ( l a j k ( 1 / 2 1 x j ~ ” a j k 1 1 / ’ ~ ~ ~ ~ )
and apply the Cauchy-Schwarz inequality twice. Since the calculation is straightforward, the details will be omitted. We return now to the proof of the main interpolation theorem. Forfand g in H’, let
It will be convenient also to use the notation
bnk
= Bnk(2k)*
Given w = ( w k } E 8‘ and a uniformly separated sequence show that a functionfe H 2 exists such that Tz(f)= w.Let and let
Then and
{zk},
we have to
152
9 INTERPOLATION THEORY
it follows that I(gnj
7
gnk)l
5 2(1 - lZjl2)(l - lzk12)11 - z j % l - 2 *
By Lemma 1, this implies n
C
I(qnj3
j=1
grid/ 5 2 A .
Consequently, in view of (2) and Lemma 2,
llf,l12
5 2A 8-411Jvl127
where the fact that { z k } is uniformly separated ( l b n k l 2 8) has been used. It follows that the functions f n form a normal family, so that a subsequence tends uniformly in each disk JzI 5 R < 1 to a functionfE H 2 for which T2(f)= w
and
llfll 5 CII4I,
(3)
where C = (2AI1l2 It remains to show that T,(f) E L2 for every f E H 2 . Let i v = { w k } be an arbitrary t2sequence, and let /z,(z) be the H 2 function of minimal norm such that 2 -112 h n ( z k ) = (1 - lzkl ) 7 k = 1, 2, * . 12. 9
We have just proved the existence of a function f E H 2 satisfying the conditions (3). Clearly, then,
I l ~ f l l5l cll~4. Appealing to relation (15) of Chapter 8, we therefore see that each f E H 2 satisfies
Taking the supremum over all iv E L2 with )/ivJI= 1, we conclude that
1. In particular, T 2 ( f )E t2for every f~ H 2 . since Ib,,k/I It is now an easy step to show that T p ( f )E L P for everyfE H P , 0 < p < a. Simply write
f(4=~(z)cs(~)12’p> where B is a Blaschke product and g is a nonvanishing H 2 function. Then m
co
1( l - IZk12>/.f(zk)lp k= 1
5 h=l
(l - lzk1z)>1g(zh)12
5 C211gl12= c
~II~II~.
9.3 THE PROOF F O R p < 1 153
That is,
ll~p(f)ll5 C2’pllfll,
.fE
HP.
(4)
Finally, it has to be shown that Tp(HP)3 C p (I I p I co) if {zk) is uniformly separated. Given {wk}E Gp, let g,(z) be the N P function of minimal norm which satisfies k = 1, 2,. . . > N. g r t ( Z k ) = (1 - ] z k \ 2 ) - 1 ’ p l r k , Appealing again to relation (15) of Chapter 8, we have
for somefE H q with IlfIl
5 6-
=
1, where q is the coiljugate index. Thus, by (4),
* c2’qw / /
,
Trivial modifications are required in the cases p = I and p = co, but the final result is the same. The functions g,(z) therefore constitute a normal family, and a subsequence converges to a function g E HP for which Tp(g)= IV. Since w was arbitrary, this proves Tp(HP)3 P,1 I p 5 a. 9.3. T H E PROOF FOR p < 1
We have already shown, even for p < I , that Tp(HP)c l pif {zk} is uniformly separated. On the other hand, the proof that Tp(HP)2 C p for 1 I p 5 co was based on the duality relation, which has no counterpart for p < 1. Nevertheless, a direct and surprisingly simple construction is available in this case. Let ( z k )be uniformly separated, and let w = ( w k ) E CP, 0 < p < 1. we shall construct a function J E H P such that T,(f) = IY. This for arbitrary I V E C p will prove Tp(HP)3 Cp. Let
bk(Z) =
m
IZ.1
z.- z
JJ J J j=1 j #k
zj 1- z j z
and gk(Z)= (1
Since Ibk(zk)l 2 6 > 0, the series
-kz)-2’p.
154 9 INTERPOLATION THEORY
converges uniformly in each disk Iz( I R c 1. Hence f is analytic in IzI < 1. Furthermore, m
proving that f E f I P . It is easy to see that Tp(f) = w. This shows that Tp(Hp)= C p if (zh) is uniformly separated. The converse can be proved by virtually the same argument used in the case 1 5 p 5 co. Of course, H P and C p are no longer Banach spaces if p < 1, but they are F-spaces (see Section 6.4) under the translation invariant metrics
and m
d(x, y ) =
Ixh h= 1
- yhIp,
respectively. Hence the closed graph and open mapping theorems are still available. It has to be checked that the quotient space H P / N Pis again an F-space, but this presents no difficulty. In fact, if Xis any F-space and S is a closed subspace, X / S is an F-space under the obvious metric. 9.4. UNIFORMLY SEPARATED SEQUENCES
The condition for uniform separation is difficult both to understand intuitively and to verify in practice. In fact, the very existence of uniformly separated sequences is not obvious. It is important to find a more accessible characterization. This is a purely geometric problem which will be discussed independently of the interpolation theory. LEMMA.
PROOF,
It is enough to assume a = a > 0. Then
--
9.4 UNIFORMLY SEPARATED SEQUENCES
155
where
The problem is to find the minimum of A+x B+x
d x )=-
for - 1 Ix I1. But in this interval q’(x) =
B-A
( B + x)’ > O’ ~
since 1 IA
which proves the lemma. THEOREM 9.2. If there is a constant c
1 - 1z,+11 I c(1 -
IZ”l),
< 1 such that n
=
1, 2 , . ..)
(5 )
then {z,} is uniformly separated. The condition (5) is also necessary if 0 5 z 1 < z 2 < ..*. PROOF.
The condition ( 5 ) implies 1 - lZjl I C j P k ( l - l z k l ) , j > k ;
In particular,
k = 1, 2, .. .
.
2 (1 - Izjl) < 00. It follows from (6) that f0r.j > k lzjl
- lzkl
2 (1 - d-’)>(1 - lzkl)
and 1 - IZjZkl = 1 - ] Z j l
+ IZjl(l
- I Z J ) 5 (1 + cj-”>(r - IZfJ).
Hence, by the lemma,
For j < k this inequality takes the form 2-
j < k.
(6)
156
9 INTERPOLATION THEORY
Consequently,
which shows that {zk} is uniformly separated. Now suppose 0 < zI < z2< .* * and k = l , 2 ,....
26>0, Then
z,+~- z, 2 6(1 - Z,Z,+~),
n
=
1, 2 , . .. ,
so that
Thus {z,} satisfies ( 5 ) , as claimed. Sequences (z,} which satisfy the condition (5) are called exponentiul sequences. 9.5. A THEOREM OF CARLESON
Part of the proof of the main interpolation theorem consisted in showing that if {z,} is uniformly separated a n d f c H P (0 < y < a), then
In other words, if
1-1 is
the discrete measure on IzI < 1 defined by
p(z,)
=
1 - Iz,~
.
n = 1, 2, . . ,
then
I,
zl
If(z>lPM z ) < 0-3
for eachfc H P , and the injection mapping from H P to the space LPp(with the obvious definition) is bounded. I t is natural to ask what other measures p have this property. One trivial example is the ordinary Lebesgue measxe on IzJ< 1. As it turns out, the exact class of admissible measures can be characterized very neatly. A finite
9.5 A THEOREM OF CARLESON
157
measure p on IzI < 1 will be called a Carleson measure if there is a constant A such that p(S) 2 Ah for every set S of the form S={z=re“:l - h s r < l ; 8 , ~ ~ 1 6 ~ + h ) . The theorem may now be stated as follows.
(7)
THEOREM 9.3 (Carleson). Let p be a finite measure on Iz( < 1, and suppose 0 < p < 03. Then in order that these exist a constant C such that
it is necessary and sufficient that p be a Carleson measure. PROOF OF NECESSITY. It is easy to see that the validity of (8) for any given p is equivalent to its validity for p = 2. (Simply factor out Blaschke products and use the standard argument.) Suppose, then, that (8) holds with p = 2. Let zo = pein # 0 be an arbitrary point in the open disk, and consider the H 2 function
g(2) = (1 - z,z)-l,
whose norm is
llsllz = (1 - P 21-112
*
Let h = 1 - p, and let S be the set of points z = reiesuch that p
If (8) holds with p
sr<1 = 2,
and
CI
- h/2 5 0s ct + h/2.
then
1
lg(z)12 d p ( z ) I C2(1 - p
y
s C2h-I
But a simple geometric argument shows that for z E S, I(l/p)eia- z~ I I(l/p)e’“ - pei(a+h12) I
1
= - [l
P
- 2p2 cos(h/2) + ~ ~ 1 ’ ’ ~ ;
(9)
158
9 INTERPOLATION THEORY
Combining this with (8), we find
which was to be shown. PROOF OF SUFFICIENCY. This is much more difficult. We first give an outline of the argument, which is essentially based on the Marcinkiewicz interpolation theorem (see Zygmund [4], Chap. XII), although the special case actually needed is relatively easy, and will be proved independently. The first step is to show that (8) will follow if it can be shown that a certain sublinear operator T is of “ type ” (2,2). The second and most difficult step is to show that T is of “weak type ” (1, I). Here the argument relies upon a certain covering lemma, and it is essential that ,u be a Carleson measure. The third step, which could be avoided by direct appeal to the Marcinkiewicz theorem, is to deduce that Tis of type (2,2); it is trivially of type (GO,a).
First Step. For eachf E H 2 we have
If(z)l
1 5
1 P(r, t - e)l.f(e’‘)l 2n
dt.
0
Therefore, to show that (8) holds for a given measure p, it will be sufficient to prove that
if u(z) is the Poisson integral of an L2 function q(t) 2 0. With each point z = reie # 0 in the open disk we associate the boundary arc
I, = { e i t : e - +(I - r ) I tI e + +(I - r ) } . Choose 0 = 0 if z = 0. (See Figure 3.) It will be convenient to identify the arc I, with the corresponding segment on the real line (choosing 0 I 8 t271). Given an integrable function q(t)2 0, periodic with period 271, define
where the supremum is taken over all intervals I containing I,, of length 111 < 1. It is evident that @(z)is continuous in 0 < IzI < 1. We shall now prove that
44 I 1 6 n Z C W+ llqll11,
I4 < 1,
(11)
9.5
A THEOREM
OF CARLESON
159
Figure 3
where
is the Poisson integral of cp. Later, in the third step of the proof, we will show that (10) holds with u replaced by 4; this together with (11) will prove (10). To prove (Il), fix the point z = re", let t, = 2"-'(1 - r), and define the intervals 0, =
[--t,) t,],
n = 0, 1, .. . ) N ,
where N is the largest integer n such that t, < 3.Now define the sets Go=wo;
G,=w,-w,-l,
n=l,
GN+1 = [-n, n] - w N . Then 1
cJ c
N+1
P(r, t ) q ( t + 6 ) d t 2nn=0 cn 1 N+1 _< P(r, t,-l 2n ,=O
u(z)=-
..., N ;
160
9 INTERPOLATION THEORY
where tLl
= 0.
But from the inequality
one finds
p(r, t,-
1)
1671' 4"(1 - r)'
+ir
I
It follows that for r 2 +,
c
8n N + l u(z) I - 4-" 1 - r ,,=o
J
cp(t
+ 0) d t
G,
N
by the definitions of i j and N . If r < 3,we have trivially 2 " u(.) i q(t 71
J
+ 0) d t = 4/lcpl/,.
--K
This proves (1 1).
Second Step. The operator T : cp + i j given by the relation
assigns to every periodic q E 15' a function @ continuous in 0 < IzI < 1. We are going to show that if ,u is a Carleson measure, then T is of iveak type (1, 1):
A'%) cs-'llcpll,,
(12)
where E,
= {Z
: @(z) > S } ,
s
> 0.
For the proof, we define for each E > 0 the sets A," = [ z :
I'
Iq(t)l dt > S(E
+ lIzI))
1,
and B," = ( z : I, c I,,, for some Observe that the sets B," expand as
E -+
E, =
0 and
u B,". EZO
iv E
A,").
9.5 A THEOREM
OF
CARLESON
161
Hence p ( E J = lim p ( B i ) . &+O
If z,
E A,"
and the arcs Iznare disjoint, then s
c
(6
II
+ II,,,I>
1
Izn
icp(t)l d t 5 2nllpll1.
(14)
In particular, there can be at most a j n i t e number of points z, in A," whose associated arcs IZnare disjoint. The following lemma is now needed. Covering lemma. Let A be a nonempty set in (zI< 1 which contains no infinite sequence of points whose associated arcs 1," are pairwise disjoint. Then there exist a finite number of points zl, . . . ,zm in A such that the arcs I,, are disjoint and m
where J, is the arc of length 511,l whose center coincides with that of I,. Before proving the lemma, let us use it to complete the proof of (12). If E, is empty, there is nothing to prove. Otherwise, A," is nonempty for sufficiently small E , so that
u {z m
A," c
: I , c J,,>,
n=l
where z,
E A,"
and the arcs I,, are disjoint. It follows easily that m
B," = (j { z : I,
= J,J.
,= 1
But since p is a Carleson measure, we have
p ( { z : I , c J,,}) 5 5AlIZnl, Hence, by (15) and (14),
~1
= 1, . .
. , m.
m
In view of (13), this completes the proof of (12). Proof of covering lemma. Let p1 = inf{lzj : z
and choose z1 E A with Iz, 1 < f(l zl, . . . , z, - let pn = inf{lzl : z
EA
EA},
+ pl). If 0 E A , let z1 = 0. Having chosen
and I, n Izj = 0, j = 1, . . . , n - 1).
162 9 INTERPOLATION THEORY
Then choose z, E A with I,, n Z,, = 0 for j = 1,. .., n - 1 and lz,l < +(I p,). By hypothesis, this process must terminate after the choice of a finite number m of the points z,. Now for each z E A we have I, n I," # 0 for some n (n = 1, . .., m) ; suppose n is the smallest such index. Then IzI 2 p, , which gives 1 - [ z [I1 - p , < 2(1 - [z,l).
+
But this together with the fact that 1, meets Zznimplies I, c Jz, ,which proves the lemma.
Third Step. It now remains to make use of (12) to show that T is of type (2,2) :
4
14(41'
442) 5
cp EL2.
Kllcpl12Z,
(17)
21<1
For s > 0, let cp = $,
+ x,, where
*'
wherever ]cp[ > s/2C; otherwise.
icp = 10
Here C is the constant in (12); we assume C 2 1. Then
4(z) 5 3,(4 + l?s(z> 5 3,w +
2c S
9
from which it follows that
In particular, p(E,) 5 p(F,). Letting
-1
~ ( s )= p(E,),
1/41' d p =
we therefore have by (12)
lo m
00
S'
da(s) 5 2
0
SN(S)
ds
But with the notation 0s
=
it : Idt)l > s/2C),
we have
2cldr)I
= 1;711cp(t)lJ
0
27I
ds d t = 2c
jb I 4 w d 4
which proves (17). This completes the proof of the theorem.
9.5 A THEOREM OF CARLESON
163
REMARK. For later reference (in Chapter 12), we note that if 11 is a Carleson measure with constant A 2 1, then the constant C in (8) may be taken to be 4(80)4A2. This can be verified by a careful examination of the proof. Finally, it should be observed that the argument, with appropriate modifications, actually proves the following more general theorem. THEOREM 9.4. Let p be a finite measure on 121 < 1, and suppose 0 < p I q < 00. Then in order that there exist a constant C such that
for allfE H P , it is necessary and sufficient. that p(S) IAhq/Pfor every set S of the form (7). PROOF. The general theorem can be reduced as before to the case p = 2 I q < co. The necessity can then be proved as it was in the case q = p . To prove the sufficiency, one again uses the Marcinkiewicz interpolation theorem (in more general form). What now has to be shown is that under the hypothesis p(S) I Ahq, the operator T : cp -P 4 is of weak type (1, q) for q 2 1: PL(Q 5
c~-qll~l114,4 2 1,
(18)
for all sets E,defined as before. (Here we have replaced q/2 by q, for notational convenience.) The proof of (18) is the same as for q = 1, except that (16) must be replaced by
s c(27C)4s-q
cpIl14.
Theorem 9.4 can be applied to prove the Hardy-Littlewood theorem (Theorem 5.1 1) that f~ H P implies
Another application is a generalization of the FejCr-Riesz theorem (aside from the value of the constant):
-=
for allfE HP, 0
00.
164 9 INTERPOLATION THEORY
EXERCISES
1. Show that a necessary and sufficient condition for T, to map H P into OG) is that the measure ki defined by ~ ( ( z ,= ) 1 - 1zkI2be a Carleson measure. 2. Show that if zk = 1 then
L'J' (0 < p 5
rn
for somefE H P (0 < p < co), even though (1 - lzkl) < 00. 3. Show that if { z k }is an exponential sequence, then its associated measure p (as defined in Exercise 1) is a Carleson measure.
-
4. Construct a sequence { z k } with 0 < z1 < z2 < * . which is not exponential, but whose associated measure p is a Carleson measure. 5. Construct a sequence ( z k } such that T, maps H P into Cp, but not onto. 6. Provide the details for the two applications of Theorem 9.4 mentioned at the end of Section 9.5. 7. Use Theorem 9.4 to prove thatf' E H112impliesf€ H1, a special case of Theorem 5.12. NOTES
The problem of describing the universal interpolation sequences for H" was proposed by R. C. Buck. After partial solutions by Hayman [2] and by Newman [l], Carleson [l] proved Theorem 9.1 in the case p = co. Shortly thereafter, Shapiro and Shields [l] gave a simpler proof (which we have essentially followed) and generalized the theorem to 1 5 p 5 03. The extension to 0 < p < 1 is due to K a b a h [2, 51. Hayman actually constructed an interpolating function under hypotheses a little stronger than uniform separation. P. Beurling later proved the existence of an absolutely convergent interpolation series whenever { z k } is uniformly separated; see Carleson [3]. Theorem 9.2 was found independently by Hayman [2], Kab& [I], and Newman [l]. Newman's paper contains a further result of this type. Newman also showed, in an unpublished 1961 manuscript, that {z,} is uniformly separated if and only if the Blaschke product B(z) with zeros z, has the property that for sufficiently small E > 0, the level set IB(z)[= E decomposes into a system of disjoint curves, each surrounding exactly one point z,. This justifies the term " uniformly separated ". The result recently reappeared in a paper of Hoffman [2].
NOTES
165
Nevanlinna and Pick solved the general problem of characterizing the pairs of sequences { z k }and { w k } for which there exists f~H" with f ( z k )= w k , k = 1,2, . . . . An account of this work is in the paper of R. Nevanlinna [2]. Krein and Rechtman [I] gave another solution based on a certain moment theorem. More recently, Sz-Nagy and Koranyi [I, 21 based a proof on the spectral theorem for a unitary operator in Hilbert space. Carleson [I, 21 proved Theorem 9.3 en route to the corona theorem (see Chapter 12). The idea of basing the proof on the Marcinkiewicz interpolation theorem is due to Hormander [I], although the proof that Tis of weak type (1, 1) is similar to Carleson's original argument. The generalization to Theorem 9.4 is in Duren [2]. For a statement and proof of the Marcinkiewicz interpolation theorem, see Zygmund [4].
This Page Intentionally Left Blank
Hp SPACES OVER GENERAL DOMAINS
CHAPTER 10
Up to this point we havedealt almost exclusivelywith functions analytic in the unit disk. Our objective is now to extend the H P theory to other domains. Even in the case of a simply connected domain, there are two “natural” generalizations of the space H P which need not coincide. We begin by presenting these two definitions and exploring the connection between them. Specializing to the case of a Jordan domain with rectifiable boundary, we then discuss boundary behavior and introduce the concept of a Smirnov domain, which we investigate in some detail. We conclude by indicating some extensions of the theory to multiply connected domains. The discussion of H P spaces over a half plane is reserved for the next chapter. 10.1. SIMPLY CONNECTED DOMAINS
Let D be an arbitrary simply connected domain with at least two boundary points. There is no difficulty in defining the space H“(D) of bounded analytic functions in D ; it is a Banach space under the norm
llfll = SUP If(z)l* Z S D
168 10 HP SPACES OVER GENERAL DOMAINS
If 0 < p < co,there are at least two alternatives. One can require the boundedness of the integrals of I f / ” over certain curves tending to the boundary, or one can demand that Ifl” have a harmonic majorant (see Section 2.6). The latter definition is simpler, and will be considered first. A functionfanalytic in D is said to belong to the class HP(D) if the subharmonic function If(z)lp has a harmonic majorant in D. The norm can then be defined as I l f l I = [U(Z,)]~’~,where zo is some fixed point in D and u is the least harmonic majorant of I f l ” . It is easy to see that the space H P ( D )is conformally invariant. That is, i f f € H P ( D ) and if z = q(w) is a conformal mapping of a domain D* onto D, thenf(q(w)) E HP(D*).Furthermore, if the norm in HP(D*) is defined in terms of the point w o = q-’(z0), this correspondence f -f 0 q~ is an isometric isomorphism. A function f analytic in D is said to be of class EP(D) if there exists a sequence of rectifiable Jordan curves C,, C, , . .. in D, tending to the boundary inthesensethat C,eventuallysurroundseachcompactsubdomainof D,suchthat
It is not immediately clear that E P ( D )reduces to the usual H P space when D is the unit disk. As it turns out, however, it suffices to consider curves C, which are level curves of an arbitrary conformal mapping of the unit disk onto D. THEOREM 10.1. Let q(w) map JwI< 1 conformally onto D, and let r, be the image under q of the circle IwI = r. Then for each functionfF Ep(D),
PROOF. Let zo = q(O), and suppose ~ ’ ( 0> ) 0. LetfE EP(D), and let {C,} be a corresponding sequence of curves. Let D, denote the interior of C,. Without loss of generality, suppose zo belongs to all of the domains D, . Let qn(w) be the conformal mapping of [wI< 1 onto D,, , normalized by Pn’(0) > 0. ~ n ( 0= ) zo > Then 9,‘E H I , and f ( q n ( w ) )is continuous in ( w [i 1. By the CarathCodory convergence theorem, q, tends to q uniformly in each disk IwI I R < 1. Thus
10.2 J O R D A N DOMAINS WITH RECTIFIABLE BOUNDARY
COROLLARY.
169
The following are equivalent:
( 0 f ( Z ) E E P ( D ); (ii) F(w)=f(cp(w)) [cp'(~)]''~ E H P for some conformal mapping ~ ( wof) Iwl < 1 onto D ; (iii) F(w) E H P for all such mappings cp. This last result points out the difference between HP(B)and EP(D).In fact, E H P , while f E E P ( D )if and only if
f E H P ( D )if and only iff(cp(w))
f(cp("~Ccp'(~~l''" E HP. The two spaces therefore coincide if Icp'(w)l is bounded away from 0 and a. This will be the case, for example, if D is the interior of an analytic Jordan curve. (See, however, Exercise 3.) It is a surprising fact that the obvious sufficient condition for the equivalence of H P ( D )and E P ( D )is also necessary. THEOREM 10.2. Let cp(w) be an arbitrary conformal mapping of IwI < 1 onto the domain D.Then HP(D)= E P ( D )if and only if there are positive constants a and b such that
a I\cp'(w)\ I b,
w E D.
(1)
PROOF. It should be observed that (1) is really a property of D, not of q. We have already shown that (1) implies the equivalence of the two spaces. Conversely, suppose H P ( D )= EP(D).This means that for functionsf analytic in D, f ( c p ( W ) ) E H P*f(cp"~cp'(~~l''"E H P . The choice f ( z ) = 1 therefore shows that cp' E H I . On the other hand, iff is chosen so thatf"p(~))[q'(w)]''~ = 1, it follows that l/cp' E H'. Furthermore, since the modulus of an H P function can be prescribed arbitrarily on the boundary, subject only to the conditions f ( e " ) E Lp and loglf(e")l E L', the hypothesis implies that g E L1* cp'g E L'.
+
(Write g = g1 g 2 , where log)g,l E L' and g 2 EL"'.) But the operator T : g + c p ' g from L' to L1 is closed, hence bounded, by the closed graph E L". Since cp' E H', it follows from Theorem theorem. In other words, cp'"") , is half of (1). A similar argument shows that 2.11 that ~ ' E H " which l/q' E H". 10.2. JORDAN DOMAINS WITH RECTIFIABLE BOUNDARY
Suppose now that D is a domain bounded by a rectifiable Jordan curve C. Let ~ ( wmap ) IwI < 1 conformally onto D, and let $(z) be the inverse mapping.
170 10
H P
SPACES OVER GENERAL DOMAINS
Since C is rectifiable, it makes sense to speak of a tangential direction almost everywhere. THEOREM 10.3. Each function f of class E P ( D ) or H P ( D ) has a nontangential limit almost everywhere on C, which cannot vanish on a set of positive measure unless f(z) G 0. Furthermore,
If(z)lp Idzl < 03
if f
E E”(D),
JC
and Jc
If(z>IPl$’(z)l Idzl < co
if f E Hp(D).
PROOF. I f f € E P ( D ) ,then
F(w) = f(cp(”cp’(w)l’’P E HP. As observed in Section 3.5, cp preserves sets of measure zero on the boundary, and it preserves angles at almost every boundary point. But F(w) has a nontangential limit, and so does [cp’(w)]’’”,since cp‘ E H’. Hence f(q(bv)), and therefore f ( z ) , has a nontangential limit almost everywhere. If f ( z ) = 0 on a boundary set of positive measure, the same is true for F(w); thus F(1c) = 0, andf(z) = 0. The integrability of If(z)lpfollows from the fact that (F(w)lP is integrable over I wI = 1. The discussion is similar for f~ HP(D).
One may now ask under what conditions an analytic function f’ can be recovered from its boundary values by the Cauchy integral formula. In the case of the unit disk, we know (Theorem 3.6) this is true if and only i f f € H’. It will be convenient to use the notation L p ( C )for the class of measurable functions g on C such that Ig(z)Ip is integrable with respect to arclength. THEOREM 10.4.
EachfE E’(D) has a Cauchy representation
and the integral vanishes for all z outside C. Conversely, if g E L’(C) and /$“g(z)
dz = 0,
n = 0, 1,2, . . . ,
then
and g coincides almost everywhere on C with the nontangential limit off:
(3)
10.2 JORDAN DOMAINS WITH RECTIFIABLE BOUNDARY
171
REMARK. It should be noted that the condition (3) is equivalent to the identical vanishing of the integral (4) outside C, as a series expansion of the Cauchy kernel shows. PROOF OF THEOREM.
Iff’€ E’(D), then
so that 1 F(w)=-,j 2nz
F ( w ) do > o-w ~
/wl=l
IWI
< 1.
(5)
On the other hand, for fixed w,the function R(w) = is analytic in
cp‘(w)
1
-wcp(w>- rp(w)
w
101 < 1 and continuous in IwI I 1. Thus F(w)R(w)E H ’ ,
Adding this to
and
(9,we obtain F(w) dw q(w> - cp(w)’
which is equivalent to (2). To prove that the integral vanishes if z is outside C, we need only note that
since F E H’. Conversely, given g E L’(C) satisfying (3), the relation (5) allows us to write the function (4) in the form
where G(w) =g(cp(w))cp’(w). But I?($([)) is analytic in D and continuous in B, so by Walsh’s theorem it can be approximated uniformly in D by a polynomial. It now follows from (3) that the second integral in (6) is equal to zero. Hence F(w) is represented as a Cauchy integral which vanishes outside the unit circle (as another application of Walsh’s theorem shows). But this implies F E H’, or f~ E’(D). The uniqueness of the Cauchy representation
172 10
H p SPACES OVER GENERAL DOMAINS
(with the integral vanishing identically outside C ) is obvious. This completes the proof. We now recall that in the unit disk the space H' coincides not only with the class of Cauchy integrals, but also with the class of analytic Poisson integrals (Theorem 3.1). For a general Jordan domain with rectifiable boundary, the Poisson formula would generalize to Green's f u r m i h
where G ( [ ;z ) is Green's function of D with pole at z and ajan indicates the exterior normal derivative. We recall that Green's function is the unique function of the form G(i; 2)
= log15 -
ZI + A([),
zE
D,
where h([) is harmonic in D and continuous in the closure B, and G ( i ;z ) = 0 for on C. By a Green integral over C we mean a function of the form (7), withf(4') replaced by an arbitrary function k ( i ) for which the integral exists. Just as E 1 ( D ) coincides with the set of Cauchy integrals, it turns out that H ' ( D ) is the class of analytic Green integrals over the boundary.
c
THEOREM 10.5. EachfE H 1 ( D ) satisfies Green's formula (7). Conversely, every Green integral over C which is analytic in D is of class H 1 ( D ) . PROOF. Let i= q ( w ) map IwI < 1 onto = cp(0). The Green's function of D is
let z
D, let w then
= $([)
be its inverse, and
G ( i ; z ) = hMOl, and a straightforward calculation shows that
ButfE H 1 ( D ) impliesf(q(w))
E
H', and it follows that
which is equivalent to (7). Conversely, iff is the Green integral of k , then
will serve as a harmonic majorant of If(z)l.
10.3 S M I R N O V DOMAINS
173
10.3. SMIRNOV DOMAINS
Again let D be a Jordan domain with rectifiable boundary C, let z = q ( w ) map D onto Iwl < 1, and let w = $(z) be the inverse mapping. We have seen that everyfE EP(D)has a boundary function of class Lp(C). By analogy with the result for the unit disk, one might expect this set of boundary functions to coincide with the L pclosure of the polynomials in z. However, this turns out to be true only for a certain subclass of domains which we are about to describe. Since q‘ is an H’ fiinction with no zeros, it has a canonical factorization of the form q ’ ( 4= S ( W ) W W ) , where S is a singular inner function and CD is an outer function. D is called a Smirnov domain if S(w) 3 1 ; that is, if q’ is purely outer. It is easy to check that this is a property only of the domain D, and is independent of the choice of mapping function. Indeed, any other mapping q, has the form ql(w) = q(A(w)), where ilis some linear fractional mapping of the unit disk onto itself. Thus q l y W ) = qylb(w))ayw) = s(a(w))~(a(w))a’(w). But if S(w) is a nontrivial inner function, so is S(A(w)). (See Exercise 7, Chapter 2.) A function g e L P ( C )will be said to belong to the L p ( C ) closure of the polynomials if there is a sequence {qn(z)) of polynomials such that lim
JJdZ)
n+m
- 4n(z)IP Id4 = 0.
It is convenient to identify E p ( D ) with its set of boundary functions. Thus EP(D)is a closed subspace of Lp(C)which contains all polynomials, hence also their closure. THEOREM 10.6. Let D be a Jordan domain with rectifiable boundary C, and let 1 p < co. Then Ep(D)coincides with the L p ( C )closure of the polynomials if and only if D is a Smirnov domain.
PROOF.
Iff€ EP(D),then
F(w) =f(~(”~’(w)l’’p
E
HP.
If D is a Smirnov domain, then [q‘]’’pis outer, and by Beurling’s theorem (Theorem 7.4), there is a polynomial Q(w) such that IlP -
Q[~‘l’’pll <4 2 .
(8)
174 10
HP
SPACES OVER GENERAL DOMAINS
On the other hand, Q($(z)) is analytic in D and continuous in D, so by Walsh’s theorem (see Notes) there exists a polynomial q(z) such that
I Q($(z>)- 4 Z ) I < (@)L-’/‘
on C,
where L is the length of C. This implies
II[Q
-4
0
cplC~’l’~pll <~ / 2 .
(9)
Combination of (8) and (9) gives
which was to be shown. Conversely, if every f E EP(D)can be approximated by polynomials, a similar argument shows that the polynomial multiples of [ y ‘ ] l ‘ pare dense in H P ,so [4”]‘’p is outer, by Beurling’s theorem. Hence cp‘ is outer, and D is a Smirnov domain. (See Exercise 1.) If D is not a Smirnov domain, the problem arises to characterize the L p ( C )closure of the polynomials. This can be done in terms of the class N which was discussed in Section 2.5. +
THEOREM 10.7. For 1 < p < co, a function f E E P ( D ) is in the L p ( C ) closure of the polynomials if and only iff (cp(w)) E N + for some mapping cp of Iw[< 1 onto D (and therefore for all such mappings).
PROOF. LetfE E P ( D ) ,so that
F(w) =f (cpO(w>)ccp‘(w)ll’p E H P . Thusf(y(w)) E N in any case, since it is the quotient of two H P functions. Let S(w) be the singular factor of cp’(w). By appeal to Beurling’s theorem in its full strength, the proof of Theorem 10.3 may be adapted to show that f belongs to the L p ( C ) closure of the polynomials if and only if [S(w)]”” divides the inner factor of F(w). This is clearly equivalent to saying that f ( V ( 4 ) EN+. A simple sufficient condition for D to be a Smirnov domain is that log y’(w) E Hi.
This follows from Corollary 3 to Theorem 3.1. In fact, it is enough that arg ~ ’ ( wE) h’,
10.3 SMIRNOV DOMAINS
175
since log(qo'(w)l E h' whenever D has rectifiable boundary. This follows from the mean value theorem for harmonic functions and the fact that
In particular, D is a Smirnov domain if arg cp'(w) is bounded either from above or from below. Geometrically, this means that the local rotation of the mapping is bounded ; loosely speaking, the boundary curve cannot spiral too much. These considerations show that a domain is of Smirnov type if it is starlike (or even " close-to-convex "), or if it has analytic boundary. Smirnov domains also arise in the study of polynomial expansions of analytic functions. Our next aim is to generalize the simple fact that H 2 of the unit disk is the class of power series with square-summable coefficients. Associated with the rectifiable Jordan curve C is a unique sequence of polynomials po(z),p l ( z ) , . . . such that p n ( z ) = C,,Zn
+ C n , n - l Z " - l + * . . + C"05
C",
> 0,
and
where L is the length of C. These are called the Szego polynomials of C ; they can be constructed by orthonormalization of the sequence { z " } . A full discussion of these orthogonal polynomials and their remarkable connection with conformal mapping is beyond the scope of this book. If C is the unit circle, the nth Szego polynomial reduces to z", n = 0, 1, . . . . THEOREM 10.8.
If D is a Smirnov domain, every function f E E 2 ( D )has a
unique expansion
Furthermore, every series of the form (10) converges uniformly in each closed subdomain of D t o a function f E E 2 ( D ) . PROOF. First let (a,} be an arbitrary square-summable complex sequence. In the space L2(C),the functions n sn(z)
=
k=O
ak P k ( Z )
form a Cauchy sequence, so there is a functionfE L 2 ( C )such that
176 10 H p SPACES OVER GENERAL DOMAINS
Since f is in the L 2 ( C )closure of the polynomials, it is the boundary function of somefE E2(D). By Theorem 10.4, the Cauchy formula
is valid. Using the Schwarz inequality, we now see that s,(z) + f ( z ) uniformly in each closed subdomain of D. It should be observed that, for this half of the theorem, D need not be a Smirnov domain. (See Exercise 4, however.) Conversely, supposefE E 2 ( D ) , and let
be the “ Fourier coefficients ” of J: Among all polynomials of degree n, the “ Fourier polynomial ” sn(z) approximates f most closely in the L 2 ( C )sense. Thus it follows from Theorem 10.6 that (1 1) holds if D is a Smirnov domain. Applying the Schwarz inequality to the formula (12), we conclude as before that sn(z) + f ( z ) in D. By Parseval’s relation,
In particular, this proves the uniqueness of the coefficients. 10.4. DOMAINS NOT OF SMIRNOV TYPE
The notion of a Smirnov domain has been seen to play a decisive role in the theory of approximation and polynomial expansion. The question arises whether there actually exist Jordan domains with rectifiable boundary which are not of Smirnov type. It turns out that non-Smirnov domains do exist, but they are extremely pathological. They can be constructed by an elaborate geometric process due to Keldysh and Lavrentiev. In this section we shall outline a different approach to the problem which shows its close relation to a certain “ real-variables ” question. The discussion is based on a remarkably simple criterion for univalence, involving the Schwarzian derivative
THEOREM (Nehari; Ahlfors-Weill).
Let f ( z ) be analytic in Iz[ < 1, and
suppose I{f(z), z>I I k(1 - r2)-’,
r
= tzt,
(13)
10.4 DOMAINS NOT OF SMIRNOV TYPE
177
where k < 2. Thenf(z) maps IzI < 1 conformally onto a Jordan domain (on the Riemann sphere). The proof is beyond the scope of this book. (See Notes.) For a function p ( t ) of bounded variation over [0, 2n], it is convenient to adopt the normalization 0 < t 5 2n. p(t) = p(t-), For t E [0, 2n], define
and let v(t) be extended periodically to - co < t < co. Thus v(t) is continuous on (- co, co) if and only if p(t) is continuous on [O, 2n]. Let us say that ,u E A, if v E A,, in the sense of Chapter 5 . THEOREM 10.9,
Let p(t) be a normalized real-valued function of bounded
variation, and let J
A t ) = P,(t>
+ J0 w ( 7 ) dz
be its canonical decomposition into singular and absolutely continuous parts. Let
Then there exists a constant a > 0 such that exp{ -aF(z)} is the derivative of a fuiictionf(z) which maps Iz1 < 1 conformally onto a Jordan domain, if and only if p E A,. The boundary of this domain is rectifiable if and only if pL,(t)is nondecreasing and exp{ - 2naw(t)}is integrable. REMARK. This theorem shows, in particular, that the construction of a Jordan domain with rectifiable boundary, the derivative of whose mapping function is a singular inner function alone (as in the Keldysh-Lavrentiev example), is equivalent to the problem of constructing a singular, nondecreasing, bounded function p ( t ) of class A,. This latter construction can be carried out directly. (See Notes.) PROOF OF THEOREM.
Set
178 10
H p SPACES OVER GENERAL DOMAINS
so that ”2
- {f(z), z } = aF”(2)
+ 7[F’(z)I2. U
Let us first observe that there exists a number a > 0 such thatf(z) maps ( z [< 1 conformally onto a Jordan domain, if and only if
:r).
F’(z) = 0 (1
Indeed, if (1 5) holds, then F”(z) = O((1 - r)-’), and the inequality (1 3) can be achieved by a suitably small choice of a. Conversely, iff(.) is univalent in ( z (< 1, it must satisfy the elementary inequality
(See, e.g., Hayman [l], p. 5 or Nehari [ 5 ] , p. 216.) This implies (15). The next step is t o show that (15) holds if and only if p E A,. Let
where v is defined by (14). Integrating by parts, we find
Thus F’(z) = ig’(z)+ izg”(z). This shows that (15) is equivalent to the condition
But according to Theorem 5.3, (16) holds if and only if
gEA
and g(eiB) E A,.
(17)
We claim that g has these properties if and only if v E A,. Indeed, suppose g(z) = u(z) + iv(z) satisfies (17). Since u(z) is the Poisson integral of v,
u(eie)=v(O) wherever v is continuous. Hence, by the normalization of p, v is continuous everywhere, and is of class A*. Conversely, if v E A,, then u(z) is continuous in [z[s 1 and u(e“) E A,. Since A+ is preserved under conjugation (Theorem 5.8), u(z) has the same properties, so g(z) satisfies (17).
10.5 MULTIPLY CONNECTED DOMAINS
179
It remains to discuss the rectifiability of the boundary. Here we suppose f ( z ) maps IzI < 1 conformally onto a Jordan domain and has a derivative of the form
where a > 0, p,(t) is a singular function of bounded variation, and w(t) is integrable. The canonical factorization theorem (Theorem 2.9) makes it clear that f‘E N and that loglf’(eit)( = - 2naw(t)
a.e.
It now follows from the factorization theorem for H’ functions (Theorem 2.8) that f’E H’ if and only if ~ 1 is, nondecreasing ~ and If’(e’‘)l = exp{ -2naw(t)} E L ‘ .
Sincef‘ E H’ is equivalent to the rectifiability of the boundary (Theorem 3.12), the proof is complete. 10.5. MULTIPLY CONNECTED DOMAINS
The definitions of the classes HP(D)and Ep(D),given in Section 10.1 for a simply connected domain, are easily extended to the case in which D is multiply connected. Suppose, then, that D is an arbitrary domain in the complex plane. For 0 < p < 00, let HP(D) again be the space of analytic functionsfsuch that If(z)[” has a harmonic majorant in D. Fixing an arbitrary point zo E D, let
llfll
= [u(zo)l‘~”,
f~ HP(D),
(18)
where u is the least harmonic majorant of Ifl”. As we shall see shortly, this is a genuine norm if p 2 1. It is not difficult t o show that different points of reference zo induce equivalent norms (see Exercise 2). If D is simply connected (and has at least two boundary points), we know that f E HP(D)if and only if f(z)[$’(~)]l/~ E EP(D),where $ is any conformal mapping of D onto the unit disk. The following theorem may be viewed as a generalization of this result to multiply connected domains. THEOREM 10.10. Fix a point zo E D, letf(z) be analytic in D , and let A be any subdomain of D, containing z o , whose boundary r consists of a finite
180 10 H P SPACES OVER GENERAL DOMAINS
number of continuously differentiable Jordan curves in D. ThenfE H P ( D )if and only if there exists a constant M , independent of A, such that
where G(5; zo) is the Green’s function of A with pole at zo . PROOF. If Iflp has a harmonic majorant u in D, the integral (19) is always less than or equal to u(zo). Conversely, suppose (19) holds and let ( A k } be an expanding sequence of domains of the type described in the theorem, whose union is D.Let rkdenote the boundary of A , , and let Gk(c;z ) be the Green’s function with pole at z E Ak . Then the function
is harmonic in A , . Since I f l p is subharmonic, U k ( z ) 5 uk+l(z) for all z E A k ; while (19) implies that vk(zo)<M for all k . By Harnack’s principle (see Ahlfors [2], p. 236), the sequence { U k } therefore converges to a function u harmonic in D which is clearly a majorant (in fact, the least harmonic majorant) of Iflp. As a corollary to the proof, we note that iff E HP(D),then
where uk is defined by (20) and [[f[i by (18). In particular, the limit is independent of the sequence ( A k } .From (21) and Minkowski’s inequality, it follows that I( [( is a genuine norm if p 2 1. The space HP(D)can be identified in a natural way with a certain subspace of HP over the unit disk. To develop this correspondence, we first discuss a basic result which is essentially the uniformization theorem for planar domains. (See Goluzin [3], Chap. VI, Section 1 ; and Ahlfors and Sario [l].) If D has at least three boundary points, there exists a function q ( w ) analytic and locally univalent in IwI < 1, whose range is precisely D and which is invariant under a group 9 of linear fractional mappings of the unit disk onto itself: 9 E 9. c p M W ) ) = cp(w), Furthermore, if zo is an arbitrary point in D, cp may be chosen so that q(0)= z o and p’(0) > 0; and these conditions determine cp uniquely. In other words, the pair (Iwl < 1, p) is the universal covering surface of D,and 9 is the group of cover trartsformations, or the azrtomorphic group of D. If q(w,) = q ( w 2 ) , there is some g E 9 such that g(w,) = w 2 .
10.5 MULTIPLY CONNECTED DOMAINS
THEOREM 10.11.
181
The mapping
f ( z > F(w) =f ( q ( 4 ) is an isometric isomorphism of HP(D)onto the subspace of HP invariant under 9. +
REMARK. If D is simply connected, then 9 is trivial, and the theorem reduces to an observation already made in Section 10.1. PROOF OF THEOREM. The function F associated with a givenfis clearly invariant under 9.If 1f (z)\” has a harmonic majorant u(z), then u(q(w)) is a harmonic majorant of F(w). Conversely, let F be an arbitrary H P function invariant under 9,and let U(w) be the least harmonic majorant of IF(w)lp.In fact, U is simply the Poisson integral of (FIPover the unit circle, so (IF((= [U(O)]’/”. Since F is invariant under 9,U(g(w)) is also the least harmonic majorant of JF(w)IP, and we have U(g(w)) = U(w) for all g E 9.In other words, U is invariant under 9.This means that if q(w,) = q(w,), then U(w,)= U(w,),since there is some g E 9 for which g(w,) = w, . Hence the function u(z) = U(q-’(z)) is well defined and is the least harmonic majorant of If(z)lP, wheref(z) = F ( q - ’ ( z ) ) . In particular,
llfll = Cu(z0)l”P = CU(0>l1’P = IlFll, so the correspondence between f and F is an isometry. If D is,finitely connected, certain questions about functions in HP(D)can be reduced to the simply connected case by means of a decomposition theorem. For simplicity, we assume that no component of the complement of D reduces to a point, so that (by successive applications of the Riemann mapping theorem) D is conformally equivalent to a domain bounded by a finite number of analytic Jordan curves. THEOREM 10.12. Let D be a finitely connected domain whose boundary consists of disjoint Jordan curves C,, C, , . . . , C,,. Let Dk be the domain with boundary Ck which contains D ( k = 1, 2, . . . ,n). Then every f E HP(D)can be represented in the form
f ( z >=fi(z) + fZ(z) + * * * where fk E HP(Dk),k = 1, 2, . . . , n.
+ fn(z),
PROOF. Since HP(D) is conformally invariant, we may suppose that C, is a circle which surrounds C , , . . . , C,,. Let r,, T, , . . . , r,,be disjoint rectifiable Jordan curves in D which are homologous to C , , Cz ,. . . ,C,,,respectively.
182 10
HP
SPACES OVER GENERAL DOMAINS
+ + ... +f,, where
Then f =f, f i
is analytic in Dk (after the obvious analytic continuation). To show that f k E HP(Dk),we need only deal with the case k = 1, since any of the curves C, can be made to play the role of C, by a suitable conformal mapping. Let R c D be an annulus with outer boundary C, and inner boundary r, a circle in D. By hypothesis, If I p has a harmonic majorant u in D. Thus, since f , , . . . ,f , are bounded in R, there is a constant a such that
+
z E R. If,(z)l” I u(z) a, But u = u, + u2 , where u1 is harmonic inside C, and u2 is harmonic outside and on r. In particular, u2 is bounded in R, so Ifl(Z>l” I u,(z) + b, z E R, (22) for some constant b. But (ul + b) is harmonic and If,l” is subharmonic throughout D,, so the inequality (22) holds throughout D,. Hencef, E HP(D,), and the proof is complete.
If the boundary curves C, C2 , . . ., C,, are rectifiable, it follows at once ) a nonfrom the decomposition theorem that a function f E H H P ( D has tangential limit at almost every boundary point, and the boundary function cannot vanish on a set of positive measure unless f = 0. The classes Ep(D)can also be considered in multiply connected domains. A function f analytic in D is said to belong to Ep(D)if there is a sequence {Av} of domains whose boundaries {r,} consist of a finite number of rectifiable Jordan curves, such that A” eventually contains each compact subset of D, the lengths of the I’,are bounded, and
1 f(z)l” ldzl < a.
lim sup v’m
y
Suppose now that D is finitely connected, and that its boundary C consists of rectifiable Jordan curves C,, C, , . . . , C, . If y1 = 1, we know by Theorem 10.1 that the boundedness of the lengths of the rvis a superfluous requirement in the definition of Ep(D);but whether this condition can be dispensed with in the multiply connected case is still unknown. For domains D of this type, it is easy to prove a decomposition theorem analogous to Theorem 10.12 for functions f E Ep(D). It then follows that every such f has a nontangential limit almost everywhere on C, and that (if p 2 1)f can be recovered from its boundary function by a Cauchy integral over C. Also, EP(D)= HP(D)if all the boundary curves are analytic.
EXERCISES
183
If D is a finitely connected domain, none of whose boundary components consists of a single point, there is a characterization of Ep(D)which generalizes a result proved earlier for simply connected domains. Let w = $(z) be any conformal mapping of D onto a domain bounded by analytic Jordan curves, and let z = cp(w) be the inverse mapping. Iff(z) is analytic in D , thenfE EP(D) if and only if If(cp(w))lp Icp’(w)l has a harmonic majorant in G = $(D). In particular, if arg(cp’(w)) is single-valued in G, one can say that f E Ep(D)if and only iff(cp(~)>[cp’(w)]’~~ E HP(G).The proof is omitted, since it is similar to the proof of Theorem 10.10. EXERCISES
1. Let D be a Jordan domain with rectifiable boundary and let cp map
IwI < 1 conformally onto D . For 0 < p < co, show that every F E H P has the form F(w) =f(40(w))[cp’(w)11’” for some f~ EP(D). 2. Let D be an arbitrary domain, let z1 and z2 be any points in D , and let
llflll = cu(zl>ll/p?
l l f l l z = cu(z2>1’/p
be the corresponding norms on HP(D).Prove the existence of positive constants A and B such that AIIfIIl 5
I I f I I z 5 B Ilflli,
fE
HP(D).
(Hint: Choose a simply connected subdomain containing z1 and z 2 , map it onto the unit disk, and apply the Poisson formula.) 3. Let D be the interior of a Jordan curve which is analytic except at one point, where it has a corner with interior angle a. For 0 < p < co, show that EP(D)5 HP(D)if 0 < a < n;while HP(D)5 Ep(D)if n < a < 27c. 4. Let D be a Jordan domain with rectifiable boundary C, not a Smirnov domain. Show that f E E 2(D)has an expansion f(z> = C anPn(z),
C Ian12 < 00,
in terms of the Szego polynomials p n , if and only iffcan be approximated in the L2(C) sense by polynomials. Hence show that the first statement in Theorem 10.8 is false if D is not a Smirnov domain. 5. Let D be a Jordan domain with rectifiable boundary C, and let w = $(z) map D onto IwI < 1. Show that $’ E E 1 ( D ) .Show, however, that $’ $ E P ( D ) for a nyp > 1 if D is not a Smirnov domain.
184 10 HP
SPACES OVER GENERAL DOMAINS
6. Let D be a Jordan domain with rectifiable boundary C. Show that if f E E P ( D )and its boundary function is in L4(C)for some q > p, it need not follow that f E E4(D). Show, however, that the conclusion is true if D is a Smirnov domain. (Hint: See Exercise 5 and Theorem 2.1 1.) 7. Let D be a Jordan domain with rectifiable boundary C. According to Theorem 10.4, the set of boundary functions o f f € E 1 ( D )is precisely the class of functions f E L 1 ( C )such that JCznf(z)d z
= 0,
n = 0, 1, . . . .
Show that the analogous statement for 1 < p 5 domain, but need not be true otherwise.
GO
is true if D is a Smirnov
8. Show that if q maps IwI < 1 onto a Jordan domain D with rectifiable boundary and log(cp’(e”)(E L log’L, then D is a Smirnov domain. Thus D is a Smirnov domain if lq’1 is not “too small” on the boundary. NOTES
For a Jordan domain with rectifiable boundary, the spaces E p ( D )were first considered by Smirnov [3], who defined them in terms of level curves. The equivalence with the apparently more general definition (Theorem 10.1) was proved by Keldysh and Lavrentiev [ I ] ; see Privalov [4]. Smirnov also introduced the definition of H P ( D ) and proved Theorems 10.4 and 10.5. The recent book of Smirnov and Lebedev [I] discusses these matters. Theorem 10.2 is due to Tumarkin and Havinson [3]. Privalov [4] gives further information on integrals of Cauchy-Stieltjes type and generalizations of the F. and M. Riesz theorem. A discussion of the Carathtodory convergence theorem and Walsh’s theorem can be found in Goluzin [3]. Smirnov gave a “Hilbert space ” proof of Theorem 10.6 in the case p = 2 (see also Goluzin [3]). Since our proof is based on Beurling’s theorem, it is valid even for p < I (see Gamelin [I]). The general result (for 0 < y < a)was stated by Keldysh [l]. Theorem 10.7 is due to Tumarkin [2], with a different proof; it is also valid f o r p < 1 . Theorem 10.8 goes back to Smirnov [I, 31: For further information on the Szego polynomials, the reader is referred to Szego [2, 31. Smirnov apparently tried without success to prove that every Jordan domain with rectifiable boundary is a Smirnov domain (see Smirnov [3], p. 353), but several years later Keldysh and Lavrentiev [l] produced a counterexample. Their construction was simplified in the book of Privalov [4], but the technical details remain formidable. The relatively simple approach described in Section 10.4, is due to Duren, Shapiro, and Shields [l]. Nehari [I] obtained
NOTES
185
a Schwarzian derivative criterion for univalence in the open disk, the condition (13) with k replaced by 2 . Ahlfors and Weill [l] sharpened it to the form given. See also Duren [5]. Constructions of a singular nondecreasing function of class A, have been carried out by Piranian [I], Kahane [I], and H. S. Shapiro [4]. For sufficient conditions that a domain be of Smirnov type, see Privalov [4], Tumarkin [3], and H. S. Shapiro [3]. Rudin [I, 21 developed the theory of H P ( D )in the multiply connected case and proved Theorems 10.10, 10.11, and 10.12. For the decomposition theorem in EP(D),see Tumarkin and Havinson [4]. They also found [2] the characterization of Ep(D)mentioned at the end of Section 10.5. Their survey paper [7] gives a clear account of the theory and contains further references. For a discussion of H P spaces over an annulus, see Kas’yanyuk [l, 21, Sarason [l], and Coifman and Weiss [I]. There is a large and rapidly growing literature on H P spaces over Riemann surfaces. Some of the relevant papers are Parreau [I 1, Royden [I], Voichick [l, 2, 31, Voichick and Zalcman [I], Gamelin and Voichick [l], Forelli [3,4], Fisher [l], Heins [l], and Earle and Marden [ I , 21. The notes of Heins [2] survey certain aspects of this theory. Rudin [9] describes some generalizations of the H P theory to several complex variables, where the domain is a polydisk. Some of the one-variable theory extends t o higher dimensions, but there are many counterexamples. Recently, various aspects of the H P theory have been generalized to the abstract setting of function algebras. See, for example, the surveys of Gamelin and Lumer [I], Lumer [l], and Gamelin [2], which give further references.
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H p SPACES OVER A HALF-PLANE
CHAPTER 11
This chapter deals with functions analytic in the upper half-plane D
= {Z = x
+ iy : y > O}.
It turns out that the spaces H P ( D )and E P ( D ) ,as defined in Chapter 10, do not coincide in this case, and in fact E P ( D )is properly contained in H P ( D ) .It is natural also to consider the space 5 j p (0 < p < GO) of functionsfanalytic in D, such that If(x + iy)lp is integrable for each y > 0 and
is bounded, 0 < y < GO. fjmwill denote the space of bounded analytic functions in D. Eventually it will turn out that 5 j p = EP(D),but the general theory of Chapter 10 is not entirely applicable because the boundary of D is not rectifiable. It is possible to develop the theory of bPby mapping the half-plane onto the unit disk, but this approach runs into difficulties because the lines y = y o are mapped onto circles tangent to the unit circle. Mainly to deal with this problem, or rather to avoid it, we begin with some lemmas on subharmonic
188 11
Hp
SPACES OVER A HALF-PLANE
functions in D. We then discuss boundary behavior, factorization, and integral representations of s j p functions, basing most of the proofs on known properties of H P functions in the disk. The chapter concludes with the Paley-Wiener theorem, in which a Fourier transform plays the role of the Taylor coefficients. 11.1. SUBHARMONIC FUNCTIONS L E M M A 1.
If g ( z ) 2 0 is subharmonic in the upper half-plane D and
+ iy) dx 5 M ,
/:mg(x
y > 0,
then
PROOF. Fix zo = xo
+ iyo ( y o > 0), and map D onto the unit disk by w
- zo z--f;; 2
=; -,
Then
G(w)= g ( ' O - z o 1-w
")
is subharmonic in lwI < 1, so by the mean value theorem, PC
4
where w = li + iu. Letting p
-+
1, we therefore have
L E M M A 2. If a subharmonic function g ( z ) satisfies the hypotheses of Lemma I , then it has a harmonic majorant in D.
11 2 BOUNDARY BEHAVIOR
PROOF.
189
Map D onto lwl < 1 by
w
i(1
z-i
+ w)
z = q(w) = ~. 1-w
= -.
= I)(.)
z+i'
The line y = b then corresponds to the circle c,, with center b(l + b)-' and radius R = (1 + b)-'. By Lemma 1, C(w)= g((p(w)) is bounded inside cb, so it has a least harmonic majorant U,(w) there. If a < b, it is clear that U,(w) 2 Ub(w) for each w inside Cb. Thus by Harnack's principle (see Ahlfors [2], p. 236) and a diagonalization argument, lim U,(w)
=
( w I < 1;
U(w),
a+O
and U(w) is a harmonic majorant of G(w) unless U(w) = GO. Hence u(z) = U($(z)) is a harmonic majorant of g(z) if U(w) GO. To show that U(w) f co,let rb be a circle concentric with C, and having radius p < R. Let V,(w) be the Poisson integral of C over rb. Then, in particular,
+
But as p -+ R, vb(w) -+ ub(w) inside convergence theorem that
5
cb;
l+b 7 /-,g(x
so it follows from (2) and the bounded
+ ib) dx 5 (1 +7cb)M
As b + 0, this shows U(0) 5 M / x . Thus U(w)
+
00, and
the proof is complete.
11.2. BOUNDARY BEHAVIOR
The following theorem is an immediate consequence of Lemma 2. THEOREM 11.1. COROLLARY.
If 0 < p < GO and f E g p , thenfE H P ( D ) .
I f , f E !$', then the boundary function
f ( x ) = limf(x Y+O
+ iy)
190 11
HP
SPACES OVER A HALF-PLANE
exists almost everywhere, f
E Lp,and
PROOF OF COROLLARY. As in the proof of Theorem 10.3, the existence (more generally) of a nontangential limit f (x) = limz+xf ( z ) follows from the fact that F(w) = f ( q ( w ) ) is in H P , where q is the mapping (1). Fatou’s lemma shows If(x)lp is integrable over ( - co, co). Finally, (3) follows from the fact (Theorem 2.2) that
It is also true thatf(x + iy) tends to f ( x ) in the Lp mean. Before showing this, it is convenient to prove a Poisson integral representation for functions and a factorization theorem analogous to that of F. Riesz (Theorem 2.5). THEOREM 11.2.
I f f € sjp, 1 I p I co, then
Conversely, if h E L P (1 5 p I a) and f(z)
is analytic in D,then f
1
=n1 -“m ( X -
E 5”
t y + y h ( t ) dt
and its boundary functionf(x)
= h(x)
a.e.
PROOF. Since F(w) =f ( q ( w ) ) is in H P (1 i p I co), it has a Poisson representation (Theorem 3.1)
1 F(w) = 27c
s
2~ 0
+w
eie
Re(---]F(eiB) e” - w
do.
This gives (4) after a change of variable and a straightforward calculation. The converse is obtained from Jensen’s inequality. COROLLARY.
Iff E sjp, 1 I p < co,then W
lim y-0
SO0
-m
If(x
+ iy)Ip dx = j- m lf(x)lp dx.
(5)
11.2 BOUNDARY BEHAVIOR
191
PROOF. Applying Jensen’s inequality to (4), we have
Jy
+ iy)Ip dx
If(x
m
m
If(x>lp dx,
Y > 0.
(6)
This together with Fatou’s lemma gives the result. THEOREM 11.3. I f f € 5jP(0 < p I co) and f ( z ) $0, then f ( z ) = b(z)g(z), where g is a nonvanishing !+’ function with Ig(x)l = If(x)l a.e., and
is a Blaschke product for the upper half-plane. Here m is a nonnegative integer and z, are the zeros (z, # i ) off in D,finite or infinite in number. Furthermore,
PROOF. According to Theorem 2.5,
. f ( d w ) ) = B(w)G(w),
where B is a Blaschke product in the disk and G E H P has no zeros. If we define b(z) = B($(z)j, the expression (7) follows from the corresponding formula for B ; and (8) is equivalent to (1 - I$(z,)l) < co. It remains to show that the nonvanishing function g(z) = G($(z)) is in JjP. But by Theorem 11.2, or rather by its proof, [g(z)lPis the Poisson integral of its boundary function; hence g p E &’. COROLLARY 1.
Iff€
aP,0 < p < co, then (5) holds. \
PROOF. Since
J;mls(x
g p E &I,
+ iY)lP dx 4
1; I& m
+ iy)IPdx
+
1; Is(x>l”dx 1;
If(x)l” dx
=
m
m
.
COROLLARY 2. If f e & p , 0 < p < co, then f ( z ) -+0 as z -+ co within each half-plane y 2 6 > 0.
PROOF. Since If(z)lpI Ig(z)lp,it is enough to prove this for functions. But each f~ 5$’ has a Poisson representation of the form (4). Given E > 0, choose T large enough so that
192 11
H P SPACES
OVER A HALF-PLANE
Then (4) gives
It is now a short step to the theorem on mean convergence. THEOREM 11.4.
Iff€
SSP (0 < p
< a),then
PROOF. Apply Corollary 1 above and Lemma 1 in Section 2.3.
A further application of the factorization theorem shows that 9Xp(y,f ) is a nonincreasing function of y i f f € S j p . This is expressed by the following theorem. THEOREM 11.5.
I f f € S j p (0 < p < a)and 0 < y1 < y 2 , then
~ ~ , 0 1f1> >2 q J ( Y 2 f 1. 2
+
fl
PROOF. Let f i ( z ) = f ( z iyl). Then fi E S j p , so it has the factorization = b,g, as in Theorem 11.3. But since glPE $ j l , an application of (6) gives mpoi2- Y l , . f l ) I
5qY2
- y,, Sl)
‘JJZ,(O, Sl>=
qo>f,),
which proves the theorem. The hypothesis f
E
$ j p is essential, as the example
f ( z ) = e-iz(i
+ z)-’/P
shows. Here
rnn,P(Y,f )
= neY(y
+ 1)
- >
which increases to infinity with y. 11.3. CANONICAL FACTORIZATION
The factorizationf= bg given jn Theorem 11.3 can be refined as it was in the case of the disk (Theorem 2.8) to produce a canonical factorization for $ j p functions. The space H P ( D )will be considered first.
11.3 CANONICAL FACTORIZATION
THEOREM 11.6.
Each function f
E H P ( D ) ,0
193
< p < co,has a unique factor-
ization of the form f (z) = e'"'b(z)~(z)G(z),
where
CI
(9)
2 0, b(z) is a Blaschke product of the form (7),
1+tz
s ( z ) = exp(i
t - z dV(t)\
for some nondecreasing function v(t) of bounded variation with v'(t) = 0 a.e., and a!
~ ( z=) eiY exp
+
(1 t z ) log w ( t ) ( t - z)(l t 2 )
+
for some real number y and some measurable function w ( t ) 2 0 with
Conversely, iff has the form (9), where CI 2 0, b(z) is an arbitrary Blaschke product, and the functions v(t) and w(t) have the properties indicated, then j-€ HP(D). PROOF.
Iff
E H P ( D ) ,then
by Theorem 2.8,
F(w) =f(cp(wN
=
B(w)S(w)@(w),
where B(w) is a Blaschke product for the disk,
p(e) being a bounded nondecreasing singular function, and 2 R eie
+
loglF(eie)I d e ) . As in the proof of Theorem 11.3, b(z) = B(rc/(z)) is a Blaschke product of the form (7). With w = $(z) and eie = $(t), a calculation gives
eie + w 1 + tz cis - w i(t - z) and 2dt do= 1 t2'
+
194 11 H p SPACES OVER A HALF-PLANE
Thus G(z) = a($(.)) has the form (1 l), with w(t) = If(t)l. The properties (12) of o follow from the properties loglF(eie)IE L1 and F(eie)E Lp.Finally, taking into account the possible jumps of p at 0 and at 2n, we have log S($(z))
=
i
s
1+tz
-d v ( t ) + iaz,
a t - z
where v ( t ) = p(arg{$(t)}) and a = p(O+) - p(0)
+ p(271) - p(2n-)
2 0.
Conversely, iff is an arbitrary function of the form (9), then
by the geometric-arithmetic mean inequality. Thus If(z)lp has a harmonic majorant in D. THEOREM 11.7. Each functionye Jjp, 0 < p < co, has a unique factorization of the form (9), with the factors defined as in Theorem 11.6 except that the second condition in (12) is replaced by w E Lp. Conversely, each product of such factors belongs to !$'. PROOF. Since , f !$ ~ ' implies f E H P ( D ) and f ( t ) E Lp, the first statement follows from Theorem 11.6. The converse is proved by integrating (13) with respect to x.
11.4. CAUCHY INTEGRALS
We now wish to show that every function in Jjp (1 Ip < 00) can be recovered from its boundary function by a Cauchy integral. It would be possible to prove this by mapping the half-plane onto the disk, as we did in Section 10.2 for the case of a domain bounded by a rectifiable Jordan curve. However, it is much easier to base the discussion on the identity
and essentially to follow the argument used for the disk in Section 3.3.
11.5 FOURIER TRANSFORMS
THEOREM 11.8.
Iff
E
195
S j p (1 I p < co), then
and the integral vanishes for all y < 0. Conversely, if h E Lp (1 Ip < co) and dt
eo,
y
then for y > 0 this integral represents a function f tion f ( x ) = h(x) a.e. PROOF.
Iff
E
S j p whose boundary func-
5', the Cauchy integral
is analytic in both of the half-planes y > 0 and y < 0. According to (14), it is related to the Poisson integral by the identity
Therefore, in view of Theorem 11.2, y > 0. F(Z) = F(z) - f (z), In particular, F(?) is analytic for y > 0, so F(z) must be identically constant in the lower half-plane. But since F(z) -+ 0 as z -+ 00, the constant is zero. Thus F(z) =f ( z ) in y > 0 and F(z) = 0 in y < 0, which was to be shown. The
converse follows immediately from Theorem 11.2. 11.5. FOURIER TRANSFORMS
We come now to the Paley-Wiener theorem, a half-plane analogue of the fact that H 2 is the class of powcr series a, Z" with la,J2< 00. For functions analytic in the upper half-plane, the Fourier integral
c
c
W
f(z> =
joeiZ'F(t) d t
(15)
plays the role of a power series. Before stating the Paley-Wiener theorem, we recall a few facts about Fourier transforms of L2 functions. I f f € L2,its Fourier transform is defined as R
f ( x ) = 1.i.m. R-tw
j e-'"'f(t) d t , 2n -R
196 1 1
H P SPACES OVER A HALF-PLANE
where “1.i.m.” stands for “limit in mean” in the L2 sense. It is a theorem of Plancherel that ?exists, lifi122= 2n I l f 1Iz2, and R
f(t) = 1.i.m. R+m
1- eixtf^(x)
dx.
R
If g is another L2 function with Fourier transform @, the PIunchereI formula is (16)
-m
THEOREM 11.9 (Paley-Wiener). A functionf(z) belongs to Sj2 if and only if it has the form (15) for some F E L2.
PROOF. Iff has the form (15) with F E L2, it is analytic in the upper halfplane, as an application of Morera’s theorem shows. For fixed y > 0, the functionf,(x) = f ( x + iy) is the inverse Fourier transform of
Hence
Jymlf(x + i y ) I 2 d x = 271
lorn
e-2y‘lF(t)12 d t
.m
S 2n
j,
IF(t)I2 dt < co,
s2.
showing that f E Conversely, each f in 5j2 is the Cauchy integral of its boundary function, by Theorem 11.8 : f(z)
1 2ni
= -Jm
-m
f(t> dt, t-z
y >O.
But
where o(5)= eiz6 for 5 2 0 and a(() formula combined with (17) gives
=0
which proves the Paley-Wiener theorem.
for
< < 0. Thus
the Plancherel
EXERCISES
COROLLARY.
tion, thenf((5)
197
I f f € Bz and f is the Fourier transform of its boundary funcfor almost all < 0.
<
=0
PROOF. By Theorem 11.8, the Cauchy integral (17) vanishes for all y < 0. But if y < 0,
1 27ci(z - t )
1 =-J’ 271
0
e-it< i z y
e
dS,
-00
and we find as before that 0
J’-
e i z y ( ( )d< E 0,
y < 0.
m
In particular,
which provesf((5)
=0
for almost all
5 < 0.
The argument can be generalized to give a similar representation for 5jp functions, 1 < p < 2. We shall content ourselves with a discussion of B‘. If f~L1, its Fourier transform
.f(x)
1 m eCix‘f(t)d t 27c - 0 0
=-
J’
is continuous on - co < x < co, andf(x) -,0 as x -+ rt co. If also g E L1,the formula (16) is a simple consequence of Fubini’s theorem. Thus the proof of the Paley-Wiener theorem can be adapted to obtain the following result. THEOREM 11.10. Ifj’E &’ a ndpis the = 0 for all 5 0 and
function, then?(<)
Fourier transform of the boundary
5
EXERCISES
1. Show that , f ~ if and only if f(q(w))[q‘(~>]’’~ E H P , 0 < p < 03. Hence show that 3jp = E P ( D ) ,where D is the upper half-plane. [Suggestion: Use the canonical factorization theorems (Theorems 11.6 and 11.7).] 2. Show that E P ( D )is properly contained in H p ( D )if D is the upper halfplane.
198 11
H P
SPACES OVER A HALF-PLANE
3. Give an example of a function f(z) which is analytic in a half-plane y > -6 (6 > 0), with f ( x ) E I.’, but which is not the Cauchy integral off(x). 4. Let p(t) be a complex-valued function of bounded variation over (- co, a), such that /:meixt
dp(t) = o
for all x > 0.
Show that dp is absolutely continuous with respect to Lebesgue measure.
5. Prove the half-plane analogue of Hardy’s inequality (Section 3.6): If f~ 6’and .f is the Fourier transform of its boundary function, then
(Hille and Tamarkin [3]. See Exercise 4 of Chapter 3.)
6. For f E S j p , 0 < p < co, prove
(This analogue of the FejCr-Riesz theorem is due to M. Riesz [l].) 7. Show that if b(z) is a Blaschke product in the upper half-plane, then
lim y’o
+ 1- loglb(x 1+ x 2
iy)J
dx
= 0.
m
Conversely, show that if f ( z ) is analytic in y > 0, If(z)I < 1, and lim y’o
/-
m
loglf(x + iY)l dx 1+X2
= 0,
thenf(z) = ei(Y+az)b(z), where y is a real number, a 2 0, and b(z) is a Blaschke product (Akutowicz [l]). NOTES
Most of the results in Section 11.2 are due to Hille and Tamarkin [3], who considered only 1 I p < co. They proved the key result &” c H P ( D )using a lemma of Gabriel [ 2 ] on subharmonic functions. But in order to apply Gabriel’s lemma, it must first be shown that a functionfe BP tends to a limit as z -+ co within each half-plane y 2 6 > 0. Hille and Tamarkin were able to show this only after a difficult argument proving the Poisson representation
NOTES
199
(Theorem 11.2) from first principles. Kawata [l] extended the Hille-Tamarkin results to 0 < p < 1. The relatively simple approach via harmonic majorants, as presented in the text, is due to Krylov [l]. Krylov also obtained the canonical factorization theorems of Section 11.3. Theorem 11.9 is in the book of Paley and Wiener [11. For proofs of the Plancherel theorems and other information about Fourier transforms, see Goldberg [l]. Theorem 11.10 is due to Hille and Tamarkin [l]; see their papers [2, 31 for further results. Kawata [l] proved theorems on the growth of 9Jlp(y,f) analogous to those of Hardy and Littlewood for the disk.
This Page Intentionally Left Blank
THE CORONA THEOREM
CHAPTER 1 2
The purpose of this final chapter is to give a self-contained proof of the “ corona theorem,” which concerns the maximal ideal space of the Banach algebra H “ . After describing the result in its abstract form, we show how it reduces to a certain “concrete theorem. Here the discussion must presuppose an elementary acquaintance with the theory of Banach algebras. However, the proof of the reduced theorem (which occupies most of the chapter) uses purely classical methods, and makes no further reference to Banach algebras. ”
12.1. MAXIMAL IDEALS
Let A be a commutative Banach algebra with unit, and let A2 be its maximal ideal space, endowed with the Gelfand topology. In other words, the basic neighborhoods of a point M* E 4 have the form % ! = (hf E &:if l f k ( M ) - Ak(hf*)l < &, k = 1, . . .,/?>, where E > 0, the xk are arbitrary elements of A , and A, is the Gelfand transform of xk. That is, A ( M ) = CpM(x),where (pM is the multiplicative linear functional with kernel M . Now let
yk = xk - Ak(M*)e,
202
1 2 THE
CORONA THEOREM
where e is the unit element of A . Then y , the equivalent form
E
M" (since j,(M*) = 0), and
% = { M E 4 : ljk(M)I < E ,
k
=
takes
1, . . . , ti}.
It is well known that A is a compact Hausdorff space under the Gelfand topology. Associated with each fixed point [, < 1, the Banach algebra H" has the maximal ideal
[cl
M,
=
{YE H" :f([)=O}.
The problem arises to describe the closure of these ideals M, in the maximal ideal space 4 of H " , under the Gelfand topology. Are there points in &' which are outside this closure? To put the question in more picturesque language, does the unit disk have a "corona"? As it turns out, the answer is negative. CORONA THEOREM. The maximal ideals M , , maximal ideal space of H".
< 1, are dense in the
The corona theorem is a direct consequence of the following purely function-theoretic result. THEOREM 12.1 (Reduced corona theorem).
Let,f,,f2, . . . , f, be functions
in H" such that
+ I f Z ( Z ) I + ... + Ifn(4l 2 6, Izl < 1, for some 6 > 0. Then there are functions gl, g 2 , . . . , g, in H" fi(z)g,(Z) +fz(z)gZ(z)+ . . . +fn(z)gn(z) 1. Ifl(4I
such that
To derive the corona theorem from Theorem 12.1, suppose the maximal ideals M , are not dense in A. Then some M* E A has a neighborhood of the form % = { M E Jt' : Ifk(M)I < E,
k
=
1, . . . , n } ,
fk E M",
which contains no ideal M , . In other words, to each point [ (/[I < 1) there corresponds an integer k ( k = 1, . . . , 17) such that
Ifk(0I = Ifk(M<)I2 8 . In particular, If,([)l + ... +\)[(,fI 2 &forall [, < 1. But by Theorem 12.1, this implies the existence of gl, . . . , g, in H" such that figl + * - . + f , g , = e .
12.2 INTERPOLATION AND THE CORONA THEOREM
203
Sincef, E M Y ( k = 1, . . . , n ) , it then follows that e E M * , which is impossible. Thus the ideals M , are dense in A?,as the corona theorem asserts. The rest of the chapter will be devoted t o the proof of Theorem 12.1. 12.2. INTERPOLATION AND THE CORONA THEOREM
The most difficult step in the proof of the corona theorem is to show that the zeros of an arbitrary finite Blaschke product can be surrounded by a contour which is not too long,” and on which the Blaschke product i s neither “too large” nor “too small”. That the contour is not too long will mean that the arclength measure it induces on the unit disk is a Carleson measure, in the sense of Section 9.5. The precise statement is as follows. “
L E M M A 1 (Carleson’s lemma). There exist absolute constants (0 < K < 1) and C > 0 for which the following is true. Corresponding to each E (0 < E 5 $) and to each finite Blaschke product IC
there is a rectifiable contour r such that (i) r has winding number I about each point a,; (ii) E I IB(z)I I cK for all z E r; (iii) the measure p defined on (zI < 1 by letting p ( E ) be the arclength of that part of which lies in the set E, has the property p ( S ) 5 CE-2h
for each set S o f the form
S = { z = r e i 0 :1 - h < r <
1; O , ~ O I ~ , + / I } .
Since the proof of this lemma is long and technical, we shall defer it to the end of the chapter (Sections 12.3-12.5). It seems advantageous first to motivate the lemma by showing how it leads to a proof of the (reduced) corona theorem, by way of the following result on interpolation. L E M M A 2. Let B(z) be a finite Blaschke product of the form (1) with distinct zeros a, (v = 1, . . . , s). For 6 < 3,let E(z) be analytic in the (possibly disconnected) set { z : lB(z)l < S} and satisfy IF(z)I < 1 there. Then there existsf€ H“ withf(a,,) = F(a,), v = 1, . . . , s; and i l f l l _< 6-“, where a is an absolute constant.
204
12 THE CORONA T H E O R E M
PROOF. Obviously, there are many functions f E H“ with f ( a v )= F(av), v = 1, . .., s. Choose such an interpolating function f with minimal norm. Then by the duality relation [see Section 8.5, example (ii)],
for some g E H’ of norm llgl!l= 1. With K as in Carleson’s lemma, choose such that E~ = 612, and let T’be the corresponding contour. Then we have
E
But the arclength measure ,LL induced by r is a Carleson measure [Lemma 1, (iii)], so by Carleson’s theorem (Theorem 9.3 and the remark following the proof), there is an absolute constant c such that
!lfll, where
CI
CE-’(Igj!1
= CE-’
I 6-“,
is an absolute constant. This proves Lemma 2.
In using Lemma 2 to prove the corona theorem, one must approximate certain functions by finite Blaschke products. The following lemma is slightly stronger than what is actually needed. Letf(z) be analytic in the open unit disk D and continuous in I 1 on Iz] = 1 and let E be the subset of / z J= 1 on which If(z)l < 1. Suppose E is nonempty. Then there exists a sequence (B,(z)) of finite Blaschke products with simple zeros, such that IB,(z)I + If(z)[ uniformly in each closed subset of ( D - E ) , and B,(z) + f ( z ) uniformly in each closed subset of D. L E M M A 3.
D. Suppose 0 < If(z)l
PROOF. Let S be an arbitrary closed subset of (6 - E). For 0 < p < 1, let fJz) = f ( p z ) , and let ED be the subset of Iz[ = 1 for which If,(z)[ < 1. Then fp(z)+f(z) uniformly in 6 as p -+ 1, and for p sufficiently near 1, S c (D-Ep). Hence it will suffice to prove the lemma under the assumption thatfis analytic in 6, and therefore has at most a finite number of zeros in D.Then, since it is clear that a finite Blaschke product can be approximated by one with simple zeros, uniformly in D, it is enough to suppose that f does not vanish in Assuming for convenience that f (0) > 0, we then have
a.
12.2 INTERPOLATION A N D THE CORONA THEOREM
205
Now let wk =eZnik/"be the nth roots of unity (k = 1, . . ., n), and let
Elementary considerations show thatf,(z) +f(z) uniformly in S. Let &k
= -
n1 loglf(ok)l
>
so that 1 0 5 &k 5 - -log n
= an,
say, where p is the minimum of If(z)l on IzI 6, < 5, let 1 - p: = 2~~
=
1. Choosing n so large that
and ak = p k w,;
and define
Note that (ak(= I if ck = 0, so that the corresponding factor in B,(z) is trivial. A calculation gives n
2 log(B,(z)l
=
-2(1 - 1212)
1 F k ( l - akz1-2 + O(S,2), k= 1
uniformly in S. From this it follows that loglBn(z)l = loglf,(z)I
+ 0(6l>>
uniformly in S. Hence log(B,(z)I -log(f(z)(, which implies (Bn(z)l-,l f ( z ) [ , uniformly in S. Since B,(O) > 0, it also follows (by analytic completion of the Poisson formula) that &(z) + f ( z ) uniformly in each disk (zI 5 ro < 1. Using Lemmas 2 and 3, we can now carry out the proof of the corona theorem. In fact, the argument will give Theorem 12.1 in the following sharper form. THEOREM 12.2.
Letfi,. . . ,f,be H" functions with
l i f k l l I 1 ( k = 1 , . . . ,n)
and Ifl(Z)I
+ ... + If"(Z)I
2 6,
IZI
< 1,
(2)
where 0 < 6 < +.Then there exist functions gl, . .. , gnin H" such that
f,(z)g,(z) + ... +f,(z)gn(z) = 1, and /(gkl/ 5 f i - p n , where
I4 < 1,
Pn is a constant depending only on n.
(3)
206
1 2 T H E CORONA THEOREM
PROOF. The argument will proceed by induction on n. The theorem is trivial for n = 1. Suppose it has been proved for all collections of (n - 1) functions, and letf,, ..., f , satisfy the given hypotheses. Suppose first thatf, is a finite Blaschke product:
with distinct a,, 0 5 la,( < 1. Then
where the Djare the simply connected components of S, . Since Ifl(Z>l
+ . * . + lfn-I(Z)l
2 8/27
z E s,>
and since the statement of the theorem is conformally invariant, it follows from the inductive hypothesis that in each domain Dj there exist (n - 1) bounded analytic functions Gjk with
and n- 1
c
fk(Z)Gjk(Z)
=
k=l
for all z in Dj.By Lemma 2,there exist functionsg,, .. . ,g , - l in H" such that gk(%)
= Gjk(av),
Dj
3
j =
*.
* 3
m,
and
Now define
This function is analytic in [ z (< 1, since
The relation (3) is automatically satisfied, and for all z $ S,,
12.3 HARMONIC MEASURES
207
where y, depends only on n. By the maximum modulus theorem, the same estimate holds throughout IzJ < 1. This concludes the proof for the case in whichf, is a finite Blaschke product with simple zeros. Continuing the inductive argument, suppose next that, more generally, f,(z) is analytic in (zI 5 1 and f , ( z ) # 0 on IzI = 1. Let Fn(z) be analytic and nonvanishing in IzJ < 1, continuous in IzI i 1, and satisfy
(Fn(z)l= min{lf,(z)]-', 3/6}
on
( z J= 1.
Then On= F,f, has the property l(Dn(z)l5 1 in JzI I 1, and l@,,(z)J = 1 on JzJ= 1 except on the set where Ifn(z)I < 6/3. Thus by Lemma 3, there is a sequence {B,(z)} of finite Blaschke products with simple zeros, such that B,,(z) + @,(z)in Iz( < 1 as m a3 ; and --f
lBrn(z>l2 I@n(z)l - 6/25
m 2 mo
9
(4)
uniformly on the set where Ifn(z)\ 2 Sj2. Since IF,,(z)l 2 1 in Jzl i 1, it now follows from (2) and (4) that Ifi(z>I + ... + Ifn-1(z)l
+ IBrn(2)l 2 6/2,
IzI < 1, m 2 mo.
Thus by what has already been proved, there are H" functions glm, . . . , z m,) such that
g,lrn(m
fi(z)g,rn(z)
+ .. + f n -
,(z)gn- l,m(z) -t Bm(z)gnm(z) = 1
in (z(< 1 and
such that for a sufficiently large choice of /I,. Now choose a sequence {mi} gkrni(z) -gk(z) uniformly in each disk Iz] 5 Y,, < 1. Then& E H", llgklli 6-'., and (3) holds. Finally, iff, is an arbitrary H" function, we may choose p < 1 such that ,f,(z) # 0 on IzI = p. Then by what has just been proved, there are functions 9:"' E H" with lig?)Il I and f,(pz)g'P)(z)
+
*
. . + f,(pz)g$)(z)
E
1.
Now let p -+ 1 through a suitable sequence, so that gp) tends pointwise to k = 1, . . . , n. This concludes the proof.
g k E H",
12.3. HARMONIC MEASURES
In order to complete the proof of the corona theorem, it now remains only to verify Carleson's lemma. This we shall do in several stages. In the present section, we digress to establish a general result ("Hall's lemma") on the
208 12 THE
CORONA THEOREM
harmonic measures of certain sets, and to apply it to obtain a special estimate which will enter into the argument at a later stage (Section 12.5). Hall’s lemma also has more direct applications to function theory, but these lie outside the scope of the present discussion. Let E be a closed subset of the right half-plane
H
=
{z : Re{z} > O}.
Suppose E does not divide the plane, and that the Dirichlet problem is solvable for D = H - E. Let dE denote the boundary of E, and let E*
=
{ i ( z (: z
E}
E
be the circular projection of E onto the positive imaginary axis. Let w(z) be the harmonic measure of E with respect to D. In other words, w(z) is the bounded harmonic function in D for which w(iy) = 0 (- GO < y < GO) and w(z) = 1 for z E aE. Finally, let
1
w”(z) = -
j xz +x( dty - t)” ’
z = x + iy,
E*
be the harmonic measure of E* with respect to H . L E M M A 4 (Hall’s Lemma).
w(x
For (x+iy)
E
D,
+ iy) 2 fo*(x - ilyl).
(5)
PROOF. Suppose first that E consists of a finite number of radial segments
k = 1, 2, . . . , n ; /8,1 < 7112, I I’ 5 b k } , with the intervals (ak, bk) disjoint. Let {reiek
ak
denote the Green’s function of H , and consider the function
where ds is the element of arclength on E. We claim that
o*(x) I U(x),
x > 0;
and
U(z) < +,
Re{z} > 0.
12.3 HARMONIC MEASlJRES
209
To prove (6), observe that on the semicircle = p, Re{<} 2 0, the function t-'C(x, [) attains its minimum for 5 = 0; hence 1
- G(x, [)
5
2x
l i l = P.
2 ___ x2 P2'
+
But this gives 1 U(x)Z-j" 7L E*
x
___
x2
+p
d p = W*(X).
The proof of (7) is somewhat more difficult. Fix z = x + iy, and let M ( p ) be the maximum of <-'G(z, <) over the part of the circle I[ - zI = p where Re{<) 2 0. Since on this circle
is a decreasing function of value o f t . Thus
4 , the maximum occurs for the smallest possible
Now let $ ( p ) denote the total length of the part of E which lies in the disk - zI < p . Since $(p) I2 p and M ( p ) is a decreasing function, integration by parts gives
From (6) and (7), it is easy to deduce (5). Indeed, it follows from (7) that
3w(z) - U(2) 2 0 on the boundary of D,so by the maximum principle, the same is true in D. This and (6) show that the function q ( z ) = 3w(z) - w*(z)
is non-negative on the positive real axis; while cp(iy) = 0 for y < 0, and q(z) 2 5 for z E E. By the maximum principle, then, o ( x + iy) 2 +w*(x+ iy),
x > 0, y < 0.
210 1 2 THE
CORONA THEOREM
Hence by symmetry, w(x
+ iy) 2 +w*(x - iy),
x > 0, y > 0.
This proves the lemma for the special case in which E is a union of radial segments with nonoverlapping projections. For a general compact set E, choose E > 0 and consider the set
s, = { z : W(.)
> 1-E}.
Clearly, aE c S,. Choose a set E" c S, which consists of a finite number of radial segments with nonoverlapping projections, for which E"* = E*. (To see that this is possible, cover E by open disks in S, and apply the Heine-Bore1 theorem.) Let &(z) be the harmonic measure of E". By what has just been proved, &(x iy) 2 +*(x - i lyl),
+
since E"* = E*. But the function [w(z) - &(z)]vanishes on the imaginary axis, is 2 0 on aE, and is 2 - E wherever it is defined on E". Thus by the maximum principle, w(x
+ iy) + E 2 6(x + iy) 2 +w*(x - I iyl)
+
for (x iy) E D.Now let E -+ 0, and the lemma is proved for compact sets E. Finally, suppose E is closed but unbounded. Let E, be the intersection of E with the disk Iz1 5 r, let E," be its circular projection, and let w,(z) and w,*(z) denote the respective harmonic measures. Then w(x
+ iy) 2 w,(x + iy) 2 $w,*(x - ilyl)
for each point (x + iy) E D.But it is clear from the integral representations that w,*(z) + w*(z) pointwise as r co. This completes the proof. --f
Hall's lemma will now be applied to obtain a special result needed in the proof of Carleson's lemma. Let R be the annulus p < IzI < 1, and let El be a closed subset of R which does not divide the plane. Let wl(z) be the harmonic measure of El with respect to ( R - El), and let El*
=
{eie : reie E E l }
be the radial projection of El onto the outer boundary of R. For fixed B < n/ llog P I , let Fl* be the part of El* such that 161 5 p llog P I . Then the total length IFl*/ of Fl* can be estimated as follows. LEMMA 5.
1 f p 3 6 E,,
12.4 CONSTRUCTION O F THE CONTOUR
PROOF.
r
211
The multiple-valued function
maps R onto the right half-plane Re{(} > 0. Let z valued) inverse, and let
E
=
{i: cp(i)
= q([) denote
the (single-
E El}.
Then E is a closed (unbounded) subset of H , and Hall's lemma may be = applied. The harmonic measure of E with respect to D = H - E is o([) wl((p(5)). Since cp maps the circular projection E* of E onto El*, the harmonic measure of E* with respect to H is w * ( i ) = ol*(cp(i)), where wl*(z) is the harmonic measure of El* with respect to R. Thus Hall's lemma gives Wl(cpP(t
+ i d ) 2 3Wl*(cp(5
- ilrll)),
5 + iy E D.
Choosing y 2 0, we obtain in particular o ~ ( z )2 +01*(~/2),
zE
(R - E l ) ; IzI
2 p"'.
(8)
On the other hand, if F* is the image of F,* under the restriction of cp-l to 161 < rc, we have
2 2110g pl-'[~osh(rc~/2>1-~ IF1*[ I
This together with (8) proves Lemma 5. 12.4. CONSTRUCTION OF THE CONTOUR
r
We are now ready to prove Carleson's lemma (Lemma l), the key to the proof of the corona theorem. Let B(z) be a finite Blaschke product, suppose 0 < E < 1, and consider the sets d(E) =
{ z : lB(z)l 4 E l ,
9?(E) = ( 2
For n
= 0,
1,
: IZI 4 1, lB(z)l 2 E } .
... and k = 1, 2, . . ., 2"+', let
Rnk= {reie : 1 - 2-" I r 2 1 - 2-"-l,
(k - 1 ) 2 3 I 6 4 k2-"rc}.
212
1 2 THE CORONA T H E O R E M
Let N be a positive integer (to be chosen later), and subdivide Rnkinto 22N parts by means of the radial lines
6 = ( k - 1 +j2-N)2-"n,
j = 1, . .., 2N- 1 ,
and the circular arcs = 1 - 2-n
+j 2 - N - " - 1
,
j=1,
...) 2 N - l .
Denote these parts (boundaries included) by Rnk(i),i = 1 , . . . , 22N.The actual numbering scheme is not important. The sets Rnk(i)will be called the blocks of the rectangle R,, . Now fix E 5 4 and let d obe the union of all the blocks Rnk(i)which meet &(E). Thus d ( ~ c) do. LEMMA 6.
If 2 - N < 618, t h e n d o c d ( 2 ~ ) .
PROOF. If zo €.do, then zo is in some block Rnk(i)which contains a point z1 E ~ ( E ) Then .
lB(z,)I 5 I&,>
- %)I
+ 8.
But it follows easily from the Cauchy formula that IB'(z)l I(1 - - 1 ~ 1 ~ )so- ~ ,
5 (1
+ 2 ~ ) 2 - ~ - " 2<"8.2-N 5 E.
Thus IB(zo)l< 28, which shows that zo E d ( 2 ~ )This . proves Lemma 6. From now on, let N be the smallest integer such that 2-N I 4 8 . (In partiSince E I 4 and K < 5, cular, N 2 5.) For fixed IC (0 < IC < +), let go= it is easily seen that 8 > 2 ~Hence . by Lemma 6 , d 0 n a0= @. For later use (in Section 12.51, we now record a lemma on the harmonic measure o ( z ; d o ) o f d o with respect to the unit disk. Roughly speaking, it is an estimate on the distance from d oto go,which increases as IC decreases. LEMMA 7. PROOF.
But for z
For
ZE
go,
@(z;d o )2 2K.
By Lemma 6,
E
go, loglB(z)l 2 K log E 2 2 K log ( 2 4 ,
since E 5 $. This gives the desired result.
12.4 CONSTRUCTION OF THE CONTOUR
r
213
Before turning directly to the proof of Carleson's lemma, it will be helpful to introduce some further notation. If n 2 1 and Rnk(i)[resp., Rnk]has the form
{re'@: a I r I b, c I 0 I d } , let vnk(i) [resp., vnk]consist of the circular arc {r = a, c 5 0 5 d } plus the two radial segments {a 5 r < 1, 0 = c, d } . Let Snk(i) [resp., S,k] be the set
{re'@: a I r < 1, c I 0 5 d). Finally, let Gnk(i)be the union of v n k ( i ) and the boundaries of all the blocks RmI(jc ) Snk(i) such that m < n + N . This set Gnk(i)will be known as the grating of Rnk(i).(See Figure 4,where N = 2 for the sake of the illustration.)
Figure 4 The grating GIz(i) of a block R l z ( i ) (shaded).
In terms of the Set do,we shall define a set Erik associated with Rnk,as follows. A block Rml(j ) c S n k n d owill be called a leading block of Snkif it is not contained in Smr1,(j') for any other block RmrIr(j') c Snk n do.Then Enkis defined as the union of v,k and the gratings G m l ( j )of all the leading blocks Rml(j)of S n k . There can be at most a finite number of gratings in Enk, since Bo contains some annular region ro < JzII1. Conceivably, E,,kmay consist of Vnkalone.
214 1 2 THE CORONA THEOREM
Within each set s , k we now construct a set Fnkcalled itsjinal set. A rectangle RmIc s n k will be called a leading rectangle of s,k if R,, meets W,but is not contained in S,,,, for any other rectangle RrnrlI c snkwhich meets 93,.Thus the leading rectangles of snk are defined in terms of W,just as the leading blocks are defined in terms of d o .Although s,k may have no leading blocks, it does have finitely many leading rectangles, since W,contains the annulus ro < IzI I 1. The jinal set Fnkof s,k is now defined as the union of the sets Em,associated with the leading rectangles R,, of snk. We are now equipped to describe a set A composed of certain radial segments and circular arcs in the open unit disk. The set required for Carleson's lemma will be an appropriate subset of A. It will be shown in Section 12.5 that the arclength measure induced by A (hence that induced by r) satisfies the metrical condition (iii) of Carleson's lemma. Let A, be the union of the boundaries of the blocks ROk(i),k = 1, 2; i = 1, . .., 2',. The sets S l k (k = 1, 2, 3, 4) will be known as residual sets of the zeroth generation. Let At be the union of the final sets F l k , k = 1, 2, 3, 4. Adjacent to each grating Gnk(i)in Al, and within the region snk(i) is a well determined set S,,,, I ,called a residual set of thejirst generation. Specifically, if S,k(i)
=
{reie: a I r < 1, c I eI c + 2-n-N7c},
then the corresponding residual set is S,,,,,
=
{reie:1 - 2-"-" I r < 1; c I eI c + 2-n-N71}.
Now let A2 be the union of the final sets Fn+N,I of all the residual sets S,,, I of the first generation. Adjacent to each grating in A2 is a residual set of the second generation; let A3 be the union of the final sets of these residual sets. The process continues until, after finitely many steps, a set AM is constructed which contains no gratings. Let
A
= A0
v A1 v
. * *
v AM.
It follows from the construction that s , k n d o= @ for each leading rectangle Rnkof a residual set of ( M - 1)st generation. Thus A surrounds d oand separates it from the boundary of the disk. We claim that A separates d ofrom W,, in the sense that each continuous curve in IzI < 1 joining a point z1 E d o to a point z2 e g o must meet A (perhaps at z1 or z2). It is clear that z1 and z 2 cannot belong to the same block Rnk(i),since d on W,= @. If 1z21 I+ then A,, separates z2 from zl. If 1z21 > +, let S be the residual set of highest generation which contains z 2 . Then z 2 E snk for some leading rectangle Rnkof S. If z16 Sn,, then vnkc A separates z1 from z 2 . If z1 E s n k , then z1 E S m l ( j )for some leading block
12.5 ARCLENGTH OF
r
215
R J j ) of s n k . But then the grating G m l ( j belongs ) to A and separates z1from z 2 , since z2 belongs to no residual set of generation higher than that of S. Thus A separates d ofrom go.Now let A be the set formed from A by deleting all interior points of d oand of go.It is clear that A still separates d ofrom go.Let R be the union of all the components of the complement of A which meet d o and , let r be the boundary of 0.Then &(E) c d oc R; and gon R = /a, since A separates d ofrom go.Consequently, E
I ~ B ( z )I( E~
if
ZE
r.
Finally, let r have the orientation compatible with the counter-clockwise orientation of the blocks &k(i). Then I? is a contour having winding number 1 about each point in R, and in particular, about each zero of the Blaschke product B(z). Hence r has properties (i) and (ii) of Carleson's lemma. We shall show in Section 12.5 that it also has property (iii). 12.5. ARCLENGTH OF
r
To complete the proof of Carleson's lemma, it remains to show that the arclength measure p induced by r is a Carleson measure. More specifically, it has to be shown that there is an absolute constant C such that p ( S ) ICE-2h for each set S of the form S = {reio: 1 - h I r < 1;
eo I e I eo + h } .
We shall prove this (if K is sufficiently small) for the arclength measure corresponding to the larger set A. In fact, the reader may find it convenient to identify p with this larger measure. It is clearly sufficient to show that p(&)
I cE-22-n
(9)
for all the sets Snk.Observe that p(RoK)I 2 N <~1 6 n ~ - ' < 1 6 ~ ~ - ~k ,= 1,2,
since ~ / < 8 2-N+1.(Recall that N is the smallest integer with 2-N Hence we need only prove (9) for n 2 1 .
Let Gnk(i) be an arbitrary grating, subtending an angle a = at the origin. Then
L E M M A 8. 2 - n - N 7~
48.)
p(G,k(i)) IbE-2a,
where b is an absolute constant.
216 1 2 T H E CORONA THEOREM
PROOF. Directly from the definition, one sees that the radial segments in G,,(i) have total length less than ( N + 4)2-"-', while the circular arcs have total length less than
( N i - 1)2-"n.
Since N 2 5, these two estimates give p(Gnk(i))< 5N2-" < bs-'a,
where b = 256. Next we consider rectangles which contain points of the set .go.The following lemma is the crucial step in the proof of (9). Recall that the parameter ti (0 < K <+) is still at our disposal. LEMMA 9. Suppose ti I K~ = Q[cosh (n2/2)]-', let R,, (n 2 1) be a rectangle which meets go,and let the leading blocks of S,,, subtend angles cq,a2,. . . , a, at the origin. Then c('
+
c(2
f
'
. . + G!"
I 42,
where a = 2-"n is the angle which Rnksubtends at the origin. PROOF. We shall apply Lemma 5. Choose
z o = p 1 i 3 e i e ~ E R,, n go. Let L,, be the union of the leading blocks of S,,,, and let L:k be the radial projection of L,, onto the unit circle. Note that
L,,cR= ( z : p < ( z I < l } , since (1 - 2-"-1)3 < 1 - 2-". Note alse that L:k has total length
IL,*,l
= 011
+ + * . . + a, a2
*
Let ol(z) be the harmonic measure of L,, with respect to R,and let ol*(z) be the harmonic measure of L:, with respect to R. Let w ( z ; L n k )denote the harmonic measure of L,, with respect to the unit'disk. Since L,, c d o ,it follows from the maximum principle and Lemma 7 that o1(zo) 5 o ( z 0 ; L,,)
Io(z0 ;d o ) I 2lC.
(10)
On the other hand, the sector 10 - OoI I p [log p ( contains S,,, (and hence Lnk)if p llog pI 2 2-%. Since 2-" 5 IIOg(1 - 2-")( I /log p ( I3(10g(l - 2-")[ _< 2'-"7~,
12.5 ARCLENGTH OF I?
this will be the case if and (10) give
p = 71.
217
L,k, since d‘, n W,= 0, so Lemma 5
But z,
I G l 5 l h PI c o s h Z ( ~ p / 2 > ~ , ( z o > < 2’-”71ti cosh2(n2/2)I a12 if
ti
5 tco. This proves Lemma 9.
The inequality (9) can now be proved for an arbitrary residual set. As above, E* will denote the radial projection of a set E onto the boundary of the unit disk. We assume henceforth that ti I t c 0 . L E M M A 10. Let S be a residual set of the mth generation, and let T be the union of the (m + 1)st generation residual sets contained in S. Then
p(S n Am+1)I Cc-’IS*I
(1 1)
and
IT*( I+ (SYI.
(12)
PROOF. Let R,, be a leading rectangle of S, and let the leading blocks (if any) of S,, subtend angles a l , a’, . . . , a, at the origin. The set Enkassociated with R,, is the union of v,k and the gratings of these leading blocks. Thus by Lemmas 8 and 9,
+
+
p(E,,,) 5 p(V,,,) be-’(x1 .. . < 3 ~ ~ l E z k+I $ b f 2 IE,*kl.
+ a,)
Summing over all the leading rectangles of S, we obtain (11). A similar application of Lemma 9 gives (12). COROLLARY. There is an absolute constant C such that p(S) I C&-’1S*I for all residual sets S.
PROOF. Let S be a residual set of mth generation. Then successive applications of Lemma 10 show that P(S n A m +
j )
< 2-jt1C&-’ IS*/,
j = 1 , 2 , . ..,M
- m.
The corollary then follows by addition over j . It is now a simple matter to prove (9) for an arbitrary set S,,, . Indeed, let c1, a 2 , . . . , a,,be the maximal residual sets contained in s,k. That is, aj is a
218
12 THE CORONA THEOREM
residual set in S,, , not contained in any lower-generation residual set in S,, By the construction of A, s , k fl
A
C
c,, U 0 1 U
' * '
.
U 0,,
where c,k is the union of the gratings G,,(z), i = 1, . .. ,22N.Thus by Lemma 8 and the corollary to Lemma 10,
+ P(Ol> -I-* * ' + P(>,. 5 2Nb~-22-"-Nn C E - ' ( ~ O , * (
P(!jn!f) 5 P(G,L)
+
+
+ I.,*/)
I c , E - 2 IS,*,l. This establishes the required inequality (9), and the proof of Carleson's lemma is complete. This also finishes the proof of the corona theorem. EXERCISES
1. Let A be the Banach algebra of functionsf(z) analytic in IzI < 1 and continuous in IzI I 1, with the uniform norm. Show that iffi, . .. ,A, are functions in A with no common zero in IzI I 1, then there exist g l , ...,g, in A such that f1(z)g,(z) + . * +f,(z)g,(z) = 1, 121 5 1. 2. Prove the following " zero-one " interpolation theorem. Let {a,} and (b,} be sequences of complex numbers in IzI < 1 such that (1 - la,[) < co and (1 - lb,l) < co, and let A(z) and B(z) denote the respective Blaschke products. Then there existsf€ H" such thatf(a,) = 0 andf(b,) = 1 (n = 1,2, . ..) if and only if there is a number 6 > 0 such that
1441+ lWl 2 6,
14 < 1.
Under this condition, f c a n be chosen with llfll I C, a constant depending only on 6 (Carleson [2]). 3. Use the zero-one interpolation theorem (Exercise 2) to prove the main interpolation theorem (Theorem 9.1) for H" : if {z,} is uniformly separated and (w,} E d", then there exists f~ H" such that f(z,) = w,, n = 1, 2, . . . . (Carleson [a].) NOTES
The corona theorem was conjectured by S. Kakutani as early as 1941. Carleson [2]gave a proof in 1961, basing part of the argument on unpublished work of D. J. Newman. In particular, the deduction of Theorem 12.1 from
NOTES
219
Carleson's lemma (as presented in Section 12.2) is essentially due to Newman. Hormander 12, 31 recently found a somewhat different approach which uses techniques borrowed from the theory of partial differential equations. Ha!l's lemma and some of its function-theoretic applications may be found in Hall [l]. The book of Gelfand, Raikov, and Shilov [I] is a good source for the basic theory of Banach algebras. The works of Wermer [l], Browder [l], and Gamelin [2] survey the theory of function algebras. Further information about the maximal ideal space of H" is in the papers of Kakutani 111, Newman [2], Schark [l], Kerr-Lawson [I], and Hoffman [2].
This Page Intentionally Left Blank
RADEMACHER FUNCTIONS
The Rademacher functions cpl(t),
cp2(t),
APPENDIX A
. .. are
cp,(t) = sgn{sin (2"nt)},
0 s t 5 1.
Thus, for example, 1, o < t < *
0, t
= 0, f, 1.
In general, cp,(t) vanishes at all multiples of 2-" and takes the values + I elsewhere. Let R denote the set of all dyadic rationals in the interval [0, 11; that is, numbers of the form m2-" (rn = 0 , 1, . . . ,2"; n = 1,2, . . .). The set R is countable, and so has measure zero. Each number t E [0, 11 - R has a unique binary expansion t = O.b,b, b3 . ..,
b,
=0
or
1.
(1)
It is easy to see from the definition of the Rademacher functions that cp,(t) = 1 if 6, = 0, while cp,(t) = - 1 if b, = 1. The number t then determines a sequence of " signs " +_ 1 : (2) cpl(0, %(O, cp3(t), . * * ;
222 APPENDIX A, RADEMACHER FUNCTIONS
and different t’s generate different sign sequences. Since t .$ R,the sequence (2) cannot be eventually constant; it must assume each of the values + 1 and - 1 infinitely often. Furthermore, every sequence of signs { E , } (E, = k 1) not eventually constant is representable in the form (2) by a unique t E [O, 1) - R. We have only to set b, = (1 - cn)/2 and let the expansion (1) determine t. In short, the collection of all sign sequences { c n } not eventually constant is in one-one correspondence with the set [0, I] - R.It is now natural to define the measure of a given collection of sequences { E , } as the Lebesgue measure of the corresponding subset of [0, I], provided this set is measurable. The set of all eventually constant sequences ( E , } is assigned measure zero. It becomes meaningful now to speak of “almost every sequence of signs.” The Rademacher functions form an orthonormal system over the interval co, 11:
More generally, if n, < n2 <
< nk ,
fdqfll(t)qII,(t) ’ ’ ’ q , k ( t > d t = *’
(3)
To see this, observe that on each interval ( j / 2 n k - 1(, j + 1)/2”k-1), the product q,,(t) qnk-,(t)is constant, while qnk(t) takes the values + 1 and - 1 equally often. e . 0
Let a,, a 2 , ... be complex numbers such
THEOREM A . l (Rademacher).
that
c
la,J2 < co. Then the series m
converges almost everywhere. Equivalently, the theorem says that for any square-summable sequence +a, converges for almost every choice of signs.
{a,,}, the series
PROOF O F THEOREM. Let s,(t) = CE=, akqk(t). By the Riesz-Fischer theorem, there is an L’ function @(t) such that
lo I@(t)- s,(t)I2 d t 1
lim n4 m
= 0.
In particular, @(t)is integrable and (by the Schwarz inequality)
1
B
dt = n-r w
a
@(t)d t ,
0 I a < /3 5 1.
APPENDIX A
223
fi
For almost every t E [0, I], the indefinite integral @(u) du has a derivative equal to @(t); let to 4 R be such a point. For each integer m, to is contained in a unique interval (a,, p,) of the form (j/2,, ( j 1)/2"). ,On the interval (a,, p,), pk(t)is constant if k I m, but takes the values f1 equally often if k > m. Thus
+
/ar[sn(t)
for n > m.
- s,(t)] dt = 0
In view of (5), we conclude that
s"
m
0 = lim n-tw
dt
um
=
1 ~
Pm
[ @ ( t ) - s,(t)]
dt.
Urn
Hence, because s,(t) is constant on (a,, s,(to)
=I
Bm
- s,(t)]
[s,(t)
- am
fm@(t)
pm),
dt -+
as m -+ co.
@(to)
em
Thus ( s n ( t ) } converges almost everywhere. COROLLARY.
If ($,(x)} is an orthonormal system in L2[u,b], and if
< co,then for almost every choice of signs { e n ] , the series
JU,,~~
converges almost everywhere. PROOF. By the Lebesgue monotone convergence theorem, the series lan everywhere. Thus, for almost every
c,"= $,,(x)]' converges almost x [a, b], the series E
converges for almost every % E[0, I]. Let E be the set of all points ( t , x) at which (6) converges, and let r(t,x) be the characteristic function of E. We have just observed that for almost every x , r(t,x ) = 1 for almost all t. Fubini's theorem therefore gives
la b
(b-U) =
1
r(t,X) dt dx = /
1
b
O
a
r(t,x) dx dt.
If the corollary were false, however, the right-hand side would be less than (b - a).
224 APPENDIX A, RADEMACHER FUNCTIONS
THEOREM A.2 (Khinchin‘s inequality). Suppose as in Theorem A.1 that 00, and let @(t)be the sum (4). Then @ €LPIO,11 for everyp < 00,
candlunI2<
I f p = 2m is an even integer, the constant ( p / 2 PROOF.First let p
= 2m,
+
can be replaced by m1I2.
and let the coefficients aj be real. Then
where the sum is extended over all systems of nonnegative integers vj such that v1 * * * + v, = 2m. In view of the orthogonality property (3), however, all the terms vanish except those for which all vj are even integers (zero included). Thus the sum is equal to
+
On the other hand,
But the ratio of the respective coefficients is ( 2 m ) !k , ! k 2 ! * . - k,! <-( 2 m ) ! < mrn. m ! ( 2 k l ) ! ( 2 k 2 )...( ! 2k,)! - m ! 2 ” I 2 - k ; the inequality involving mm may be [If k > 0, k ! / ( 2 k ) !s (k + verified by induction.] Therefore,
if the u j are real. In the complex case, let uj = aj + iPj and let s, = u,
+ iu, = C aj ‘ p j + i C
Pj
yj.
By Minkowski’s inequality,
I m j=l
which is equivalent to (8).
aj2
+ m f1 j=
c laj12, n
pj2
=m
j= 1
APPENDIX A
225
If 2m - 2 < p < 2m, it follows that
Letting n -+ 00 here and in (8), we obtain (7) and its sharpened form for the case p = 2m. The argument also shows that s, + (D in the L p mean for every p < co. THEOREM A.3 (Khinchin-Kolmogorov).
If
= co, then the series
a, qn(t)diverges almost everywhere. PROOF, Suppose, in fact, that the partial sums sn(t) are bounded on a set E of measure [El > 0. Then
Is,(t) - sm(t)lI C,
1 I m < n ; t E E,
where C is a constant. It follows that
= IEI
k
lakI2
k=m+ 1
+
f
j , k=m+ 1 j
a j G
/
E
(9) qj(t)qk(t)
dt.
On the other hand, according to (3), the doubly indexed system of functions q j ( t ) ( P k ( t ) (1 - < j < k < 00) is orthonormal over [0, 11. We may regard J E q j ( t ) q k ( t ) dt as the “Fourier coefficient ” of r(t), the characteristic function of the set E. By Bessel’s inequality, then,
Therefore, for m sufficiently large,
It follows from (10) and the Schwarz inequality that the absolute value of the last term in (9) is no greater than
226 APPENDIX A. RADEMACHER FUNCTIONS
Introducing this estimate into (9), we find
The next theorem serves a special purpose in Section 4.6 and has other applications. THEOREM A.4. Let g,(z), g2(z), . . . be complex-valued functions each of which is continuous in IzI 5 I except perhaps at a finite number of points on IzI = 1. Suppose
,
Jg,(z)l < co in 1zI < 1, the convergence being uniform in each disk IzI I p < 1 ; and (ii) for each N , CFZN 1gn(reie)>12 co uniformly in 0 as r -+ 1. (i)
--f
Then, for almost every choice of signs
{gn},
the function
has a radial limit almost nowhere. (That is, the 0’s for which limr+l G(reie) exists constitute a set of measure zero.) PROOF. Because of hypothesis (i), the function
m,0, t> = c cpn(t)gn(rei8) m
n= 1
is well defined for all r E [0, I), 0 E [0,271], and t E [0, 11. It is continuous in r and 0. Let E = ((8, t) :lim F(r, 8, t ) does not exist.} r+l
We are going to show that for every 8 E [0,271], the “0-section” be = {t : (0, t ) E S}
has measure Idel = 1. It will then follow from Fubini’s theorem, as in the proof of the Corollary to Theorem A.1, that for almost every t E [0, 11, the set E‘
=
{e : (0, t ) E 6)
has measure 271, as the theorem asserts.
APPENDIX A
227
Suppose, then, that lSsl< 1 for some fixed 8. Then the complementary set has measure
80
a = I&,]
I
=1
- 1601 > 0,
and F(r, 8, t) has a (finite) radial limit for all t E 8 0 . Because of the continuity of the functions gn(z),we may conclude that m
FN(r,8, t ) =
C qn(t)gn(re") n=N
( N = 1,2, . . .)
has a radial limit for all t in a set A of measure a, obtained from Reby the deletion of a countable set. Now, for K = 1, 2, ...,let BK = { t E A : IF(r, 8, t)l I K for all r < I}.
Then B, c B, c . - -and &=, BK = A , so that lBKl -+ IAl = a. Choose K = K , so large that IBK,I 2 $a, and let A , = BK1.Then IF(r, 8, t)l IK , for all t E A , and r < 1. Similarly, there are a set Az c Al and a constant K, such that ] A z ]2 +a and IF,(r, 8, t)l IK, if t E Az and r < 1. Proceeding inductively, we may construct a sequence of sets
AxAl xAzx.-*xAN3*-. and a sequence of constants { K N }such that
IAN1 2 (a/2)(1
+2-N)
and
lFN(Y,8,t)l
tEAN; r < l .
Finally, let a,
B= n A N . N=l
Then IB( 2 a/2 > 0,and
We are now in a position to imitate the proof of Theorem A.3. First fix N so large that
m
228
APPENDIX A. RADEMACHER FUNCTIONS
In view of (I 1), an argument entirely similar to the proof of Theorem A.3 now gives
or m
But tAs contradicts hypothesis (ii). Hence Idel = 1 for every ”, and the proof is complete. THEOREM A.5.
Let a, , a,,
... be complex numbers such that
’ ~1. lim sup ~ q l l = n+
(i) If
C
m
< co, then for almost every choice of signs {q,}, m
f(z)=
C
E,u,z“EHP
for all p < 00.
n=O
1
(ii) If la,I2 = 00, then for almost every choice of signs { E , } , ~ ( z )has a radial limit almost nowhere. PROOF.
Part (ii) is an immediate corollary of Theorem A.4, with g,(z)
=
anz”.We have only to observe that m
02
m
n=N
n=N
n=N
c 1anznl2= C lan12r2n-+ c lanlZ= co
as r -+ 1. (i) If
C lanlZ< co,then W
g ( z , t> =
C qn(t)anzn E H Z
n= 1
for every t, so (by Theorem 2.1 1) it need only be shown that for almost every t E 10, I], the radial limit G(8, t ) = lim g(re”, t ) r -*1
belongs to LP, as a function of 0, for all p < co. By Khinchin’s inequality (Theorem A.2),
APPENDIX
A 229
On the other hand, Theorem A . l implies that for each fixed 8, the limit (12) exists for almost every t. By Fatou's lemma, then,
Thus IG(8, t)lP is integrable over the rectangle, and by Fubini's theorem, G(8, t ) E Lp for almost every t. To complete the proof, choose a sequence p1,p2,...-)a, andlet E, = (t :G(0, t ) 4 Ek};
E
=
{ t : G(8, t) $ Lp for some p < co}.
Then lEkl = 0 for each k , so that E = UF'
Ek is also of measure zero.
Even if log n < co,it can happen that E, a, Z" $ H of signs. However, the slightly stronger condition
for every choice
C [a,(2(logn)lfd< co
(6 > 0 ) forces 1E, a,, z" to be continuous in JzII 1 for almost every choice of signs. For the proofs we must refer the reader to the literature (see Notes). NOTES
More information on Rademacher functions and related questions can be found in the books of Alexits [l], Kaczmarz and Steinhaus [I], and Zygmund [4]. Paley and Zygmund [l, 21 used Rademacher functions to prove various theorems in function theory. Proofs of the assertions in the last paragraph (above) may be found in Paley and Zygmund [l]. Theorem A.5 is due to Littlewood [4].
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MAXIMAL THEOREMS
APPENDIX B
The Hardy-Littlewood “ maximal theorem ” becomes clearer if it is stated first in discrete form. Let al, u 2 , . . . , a, be given nonnegative numbers, and let b,, b,, . . . , b, be the same numbers rearranged in nonincreasing order. For each fixed k (k = 1, . . . , n), let u k = max lsjck
1 k-j
+ 1 2ai i=j
be the optimal average of successive ui terminating with a,; and let
be the corresponding quantities for the bi. Let s(x) be any nondecreasing function defined for all x 2 0. Then, as Hardy and Littlewood showed, k= 1
k= 1
Hardy and Littlewood interpreted the a, as cricket scores and s as the batsman’s “ satisfaction function.” The theorem then says that the batsman’s
232
APPENDIX 6, MAXIMAL THEOREMS
total satisfaction is maximized if he plays a given collection of innings in decreasing order. In order to state a continuous version of this theorem, it is necessary to define the “ rearrangement ” of a function. Let f(x) be nonnegative and integrable over a finite interval [0, a ] , and let p ( y ) be the measure of the set in whichf(x) > y . Note that p(y) is nonincreasing. Two functionsf,(x) and f2(x) are said to be equimeasurable if they give rise to the same function p(y). It is then clear from the definition of Lebesgue integral that “
”
J)dX)
dx
dx.
= f0k x )
If p ( y ) is associated as above with f(x), its inverse function f*(x) = p-’(x), normalized so that f*(x) =f*(x +), is called the decreasing rearrangement of f(x). It is easy to see that f*(x) and f(x) are equimeasurable. Finally, let A(x,
1
”
5 ) = A ( x , 5 ; f)= 2Jt f(t)
dt,
0
5: < X ;
and let O(x) = @(x; f) = sup A(x,
5:; f).
0564x
The maximal theorem may now be stated as follows. THEOREM B . l (Hardy-Littlewood maximal theorem).
If s(y) is any non-
decreasing function defined for y 2 0, then Jis(O(x;
f))d x
Hardy and Littlewood used a limiting process to base a proof on the discrete form of the theorem. Shortly afterwards, F. Riesz showed that in fact O(x;f*), O*(x;f) I
0 < x Ia,
(1)
which immediately implies the Hardy-Littlewood result. Riesz’s proof of (1) makes use of his “rising sun lemma,” which may be stated as follows. LEMMA. Let g(x) be continuous on the interval [0, a], and let E be the set of all x in (0, a ) for which there exists 5 in [0, x) with g(5) < g(x). Then E is an open set: E = U(ak, bk), where the intervals (ak, bk) are disjoint; and &k) dbk).
PROOF OF LEMMA. Since the inequalities 5 < x and g(5)
APPENDIX B
233
choosex, E ( a k ,bk)and consider the point x 1 where g(x) attains its minimum in [O, xO]. Then x, $ ( a k ,x,], since these points are in E. Thus x1 E [0, a k ] . But since uk $ E, g(ak)5 g(x) for all x E [0, ak). Hence x, = a,, and in particular, g(u,) 5 g(xo). Now let xo -+ b, to conclude that g(a,) 5 g(bk). PROOF OF (I). For fixed y o 2 0, apply the lemma to the function
dx) =
Jkr) d t - Yo x. 0
E is then the set of all x for which there exists 5 in [O, x) with A(x, 5 ; f ) > y o . In other words, E = {x : O(x;f) >yo). (2) On the other hand, the condition g(ak)I g(bk) says that A(bk, a k ; f )2 y o . Thus
where m ( E ) denotes the measure of E. Now let
All this is for arbitrary y o 2 0. Given xo E (0, a], choose y o = O * ( x o ; f ) . Then by (2) and the definitioii of O*, m ( E ) = xo. Hence (4) gives @(xo;f*) 2 Y o = @*(xo;f), which proves (1). The following application of the maximal theorem will serve to illustrate its usefulness. THEOREM B.2. If f ( x ) belongs to Lp (1 c p < co) over some interval [O, a ] , then the maximal function O(x) = O(x; I f / ) is also in Lp,and
234 APPENDIX 6, MAXIMAL THEOREMS
PROOF. Apply the maximal theorem with s(y) = y p , note that O ( x ; f * )= A(x, O ; f * ) , and use the following inequality. L E M M A (Hardy's inequality).
If 1 < p < co,g(x) is in Lp over (0, co),and
then G E Lp and
PROOF OF LEMMA.
Fix a > 0. Since
the continuous form of Minkowski's inequality gives
Now let a -+ co to obtain Hardy's inequality. The next theorem is essentially a restatement of Theorem B.2 in a form convenient for certain applications (see Section 1.6). THEOREM B.3. Letf(x) be periodic with period 2n, and suppose f E Lp = Lp(O,2n), I < p < co.Then
If jo T
F ( x ) = sup O<(T(sn
is also in Lp,and [IFlip PROOF.
f(x
+ t ) dt
I
C p ~ ~ fwhere l ~ p ,Cpis a constant depending only on p .
It is clearly enough to assumef(x) 2 0 and to consider
A 1- T f ( x + t ) d t O<Tin 0
F l ( x ) = sup
APPENDIX B
235
This function is easily compared with the maximal function O(x) = O ( x ; f ) associated withf. For x 2 rc, it follows from the definitions that Fl(x) 2 O(x). For 0 I x < rc, the periodicity offmay be used to show that F,(x)5 O ( x + 271). Hence by Theorem B.2,
as desired. The constant Cpcan be taken as 4””p(p - I)-’. NOTES
Maximal theorems are the invention of Hardy and Littlewood [4], who proved Theorem B.l and pointed out its applications to function theory. The simple proof given above is due to F. Riesz [8]. See also Hardy, Littlewood, and P6lya [I]. Riesz had introduced the “rising sun lemma” as a tool in differentiation theory (see, for example, Boas [2]).Hardy’s inequality and related results are in Hardy, Littlewood, and P6lya [l].
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Voichick, M., and Zalcman, L. [l] Inner and outer functions on Riemann surfaces. Proc. Amer. Math. Soc. 16 (1965), 1 200-1 204. Walters, S. S. [I] The space H p with O < p < 1 . Pvoc. Amer. Math. Soc. 1 (1950), 80e805. [2] Remarks on the space H p .Pacific J . Math. 1 (1951), 455-471. Weiss, G . [l] An interpolation theorem for sublinear operators on Hp spaces. Proe. Amer. Math. SOC.8 (1957), 92-99. Weiss, M. and G. [l] A derivation of the main results of the theory of H p spaces. Reu. Un. Mat. Argentina 20(1962), 63-71. Wells, J. H. [l] Some results concerning multipliers of H p . J . London Math. Soc. 2 (1970). Wermer, J. [l] Banach algebras and analytic functions. Advances in Math. 1 (1961), fasc. 1, 51-102. Widder, D. V. [l] “The Laplace Transform.” Princeton Univ. Press, Princeton, New Jersey, 1941. Zygmund, A. [l] Sur les fonctions conjugu6es. Fund. Math. 13 (1929), 284-303; corr. 18 (1932), 312. [2] Smooth functions. Duke Math. J. 12 (1945), 47-76. [3] On the preservation of classes of functions. J . Math. Mech. 8 (1959), 889-895. [4] “Trigonometric Series.” Cambridge Univ. Press, London and New York, 1959.
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AUTHOR INDEX
Numbers in italics show the page on which the complete reference is listed
A Ahlfors, L. V., 6 , 16740,146, 180,1859 189, 237 Akutowicz, E. J., 146, 198, 237 Alexits, G . , 229, 237
C Calderbn, A. P., 68, 238 Cantor, D. G., 107, 238 CarathCodory, C., 145, 238 Cargo, G. T., 31, 238 Carleson, L., 146,164,165,218,237,238 Caughran, J. G., 92, 146, 238 Caveny, J., 107, 239 Coifman, R., 185, 239 Collingwood, E. F., 239
B D Banach, S., 99, 237 Bary, N. K., 63, 238 Beurling, A., 31, 126, 238 Blaschke, W., 30, 238 Boas, R. P., 235, 238 Bonsall, F. F., 146, 238 Browder, A., 219, 238
Day, M. M., 127, 239 Doob, J. L., 145, 239 Douglas, R. G., 239 Dunford, N., 99, 111, 115,239 Duren, P. L., 90-92, 106, 107, 118, 127, 165, 184, 185 239
254
AUTHOR INDEX
E Earle, C. J., 185, 239 Egerviry, E., 145,239 Epstein, B., 240 Evgrafov, M. A,, 106, 145, 240 F Fatou, P., 14, 240 Fejer, L., 52, 145, 238, 240 Fichtenholz, G., 52, 240 Fisher, S. D., 185, 240 Flett, T. M., 92, 106, 240 Forelli, F., 127, 185, 240 Frostman, O., 31, 240
K Kabaila, V., 164,243 Kaczmarz, S., 107, 229, 243 Kahane, J.-P., 185, 243 Kakeya, S., 145, 244 Kakutani, S., 219, 244 Kas’yanyuk, S. A., 185, 244 Kawata, T., 199, 244 Keldysh, M., 184, 244 Keldysh, M. V., 184, 244 Kerr-Lawson, A., 219, 244 Kolmogorov, A., 68, 244 Koranyi, A., 165, 249 Kothe, G., 127, 244 Krein, M., 165, 244 Krylov, V. I., 199, 244
G 1
Gabriel, R. M., 14, 198, 240 Gamelin, T. W., 127, 184, 185, 219, 240,241 Garabedian, P. R., 146, 241 Gaudry, G. I., 241 Gelfand, 1. M., 219, 241 Geronimus, Ja. L., 145, 241 Gleason, A. M., 241 Goffrnan, C., 241 Goldberg, R. R., 199, 241 Golubev, V., 52, 241 Goluzin, G. M., 52, 145, 180, 184, 241 Gronwall, T. H., 145, 241 Gwilliam, A. E., 69, 241 H Hall, T., 219, 241 Hardy, G. H., 14, 52, 68, 91, 92, 106, 107, 235, 242 Havin, V. P., 242 Havinson, S. Ja., 145,146,184, 185,242,250 Hayrnan, W. K., 49,92, 164, 178, 242 Hedlund, J. H., 107,242 Heins, M., 185, 243 Helson, H., 52, 127, 243 Herglotz, G., 14, 243 Hille, E., 198, 199, 243 Hoffman, K., 164, 219, 243 Hormander, L., 165, 219, 243
Landau, E., 30,145,244 Landsberg, M., 127, 245 Lavrentiev, M. A., 184, 244 Lax, P. D., 146, 245 Lebedev, N. A., 184, 249 deleeuw, K., 127, 245 Littlewood, J. E., 14, 30, 52, 68, 91, 92, 106, 107,229,235,242,245 Livingston, A. E., 127,245 Lohwater, A. J., 52,92,239,245 Lumer, G., 185,240,245 M Macintyre, A. J., 144, 145,245 Marden, A., 185, 239 Marden, M., 141, 245 Minker, J., 240 N
Nagasawa, M., 127,245 Natanson, I. P., 2, 44, 245 Nehari, Z., 137, 146, 178, 184, 185, 245 Neuwirth, J., 246 Nevanlinna, F., 30, 246 Nevanlinna, R., 30, 145, 165, 246
AUTHOR INDEX
Newman, D. J., 107, 127, 164, 219, 246 Noshiro, K., 246 0
Ostrowski, A., 30,246
P Paley, R. E. A. C., 31,68,107,199,229,246 Parreau, M., 185, 246 Pedrick, G., 241 Penez, J., 146, 246 Pick, G., 145, 246 Piranian, G., 92, 185, 245, 247 Plessner, A., 14, 247 Pblya, G., 52, 235, 242 Porcelli, P., 247 Prawitz, H., 52, 247 Privalov, I. I., 31, 52, 92, 107, 184, 185, 247
R Rado, T., 14,247 Raikov, D., 219, 241 Rechtrnan, P., 165, 244 Riesz, F., 14, 29, 30, 52, 145, 235, 240, 247 Riesz, M., 30, 52, 68, 198,247 Rogosinski, W., 144-146,247 Rogosinski, W. W., 144-146, 245, 247 Romberg, B. W., 118,127,239,247 Royden, H. L., 185, 245 Rudin, W., 52, 73, 92, 107, 127, 146, 185, 239,245,248 Ryan, F. B., 52, 245 Ryff, J. V., 14, 248
S Sarason, D., 185, 248 Sario, L., 180, 237 Schark, 1. J., 219, 248 Schur, I., 145, 248 Schwartz, J. T., 99, 111, 115, 239 Shapiro, H. S., 90, 91, 107, 127, 145, 146, 164, 184, 185, 239, 246-248
255
Shapiro, J., 127, 248 Shields,A.L.,90,91, 107,118, 127, 164,184, 239,248 Shilov, G., 219,241 Smirnov, V. I., 30,31, 52,184,248,249 Somadasa, H., 31, 249 Stein, E. M., 107, 249 Stein, P., 68, 249 Steinhaus, H., 229, 243 Sunouchi, G., 249 ~
~
~F’268, 249 ~
Szego, G., 30, 184, 249 Sz.-Nagy, B., 165, 249
T Tamarkin, J. D., 198, 199, 243 Tanaka, C . , 30, 249 Taylor, A. E.3 14, 126, 250 Taylor, G. D., 92, 106, 107, 239, 250 Titchmarsh, E. C., 9, 72, 250 Turnarkin, G. C., 146, 184, 185, 250
V
Voichick, M., 127, 185, 241, 250, 251
W Walters, S. S., 127, 251 Weill, G., 185, 237 Weiss, G., 30, 185, 239, 249, 251 Weiss, M., 30, 251 Wells, J. H., 107, 251 Wermer, J., 127, 219, 245, 251 Whitney, H., 241 Widder, D. V., 2, 251 Wiener, N., 199, 246
Z Zalcman, L., 185, 251 Zygrnund, A., 29, 31, 52, 68, 92, 106, 107, 158, 165, 229, 246, 249, 251
~
SUBJECT INDEX
A
Abel summability, 62 Absolute continuity, 41, 42 Annihilator, 110 Arithmetic-geometric mean inequality, 29 Automorphic group, 180
B Banach algebra, 201 Banach space, 37 Beurling’s theorem, 114, 173 Blaschke product, 20, 191, 204 Block, 212 Boundary function, 6, 17, 35, 74, 170, 189 Bounded characteristic, 16 Bounded variation, 2, 39, 72
C
Carathkodory convergence theorem, 168, 184 Carathkodory’s theorem, 44 Carleson measure, 157 Carleson’s lemma, 203 Carleson’s theorem, 157 Cauchy integral, 39, 170, 194 Cauchy-Stieltjes integral, 39 Class A,, 72 Class A,, 72 Class N , 16 Class N + , 26 Coefficients, 93, 139, 142 Conformal mapping, 43 Conjugate function, 33, 53, 62 Conjugate harmonic function, 33, 53, 83 Conjugate index, 54 Conjugate Poisson kernel, 35 Conjugate trigonometric series, 61 Canonical factorization, 24, 25, 192
SUBJECT INDEX
Corona theorem, 202 Cover transformations, 180 Covering lemma, 161
D Divisor, 113 Dual extremal problem, 130 Dual space, 112, 115 Duality relation, 130
E Equimeasurable functions, 232 Exponential sequence, 156 Extremal function, 132 Extremal kernel, 132 Extremal problem, 129 Extreme point, 123
F F-space, 99, 115, 154 Fatou’s theorem, 14 Fejer-Riesz inequality, 46 Final set, 214 Fourier coefficients, 37 Fourier series, 61 Fourier-Stieltjes series, 61 Fourier transform, 195
G Gelfand topology, 201 Gelfand transform, 201 Grating, 213 Green integral, 172 Green’s function, 172 Green’s theorem, 55
257
on coefficients of H p functions, 95, 98 on comparative mean growth, 84, 87 on integrals of H p functions, 88 on multipliers, 103, 107 on rearrangements, 232 Hardy’s convexity theorem, 9 Hardy’s inequality, 48, 234 Harmonic conjugate, 33, 53, 83 Harmonic majorant, 28, 168, 179, 188 Harmonic measure, 208 Hausdorff-Young inequalities, 93 HeIIy selection theorem, 3 Herglotz representation, 3 Hilbert’s inequality, 48
1
Inner factor, 113 Inner function, 24 Integral modulus of continuity, 72 Interpolation, 140, 141, 147, 203 Invariant subspace, 126
J
Jensen’s inequality, 29 Jensen’s theorem, 16 Jordan curve, 43 Jordan domain. 169
K Keldysh-Lavrentiev example, 176, 184 Khinchin-Kolmogorov theorem, 225 Khinchin’s inequality, 224 Kite-shaped region, 5 Kolmogorov’s theorem, 57
L H Hall’s lemma, 208 Hardy-Littlewood theorem on boundary smoothness, 74, 78
Lacunary sequence, 104 Leading block, 213 Leading rectangle, 214 Least harmonic majorant, 28, 168, 179, 189
258 SUBJECT INDEX
Linear functionals on H p 1 c p < co, 112 0 < p < 1, 115 Lipschitz classes, 72 Littlewood‘s subordination theorem, 10
M Marcinkiewicz interpolation theorem, 95, 158, 163, 165 Maximal ideal, 201 Maximal theorems, 11, 23, 89, 231 Mean convergence, 21, 192 Mean growth, 74 Modulus of continuity, 71 Multipliers, 99 Multiply connected domains, 179 N
Natural kernel, 142 Nevanlinna class, 16 Nevanlinna theorem, 16 Nontangential limit, 6, 17 Normal family, 6 Normalized extremal function, 132
0 Orthogonal polynomials, 175, 184 Orthonormal system, 175, 222, 223 Outer function, 24, 25
P Paley-Wiener theorem, 196 Paley’s theorem, 104 Plancherel formula, 196 Poisson kernel, 2 Poisson integral, 2, 41, 190, 195 Poisson-Jensen formula, 16 Poisson-Stieltjes integral, 2, 40
Q Quotient space, 110
R Rademacher functions, 67, 87, 94, 221 Rademacher’s theorem, 222 Rational kernels, 136 Rearrangement, 232 Rectangle, 212 Rectifiable curve, 44 Residual set, 214 Riesz factorization theorem, 20 Riesz (F. and M.) theorem, 41 Riesz (M.) theorem, 54 Rising sun lemma, 232
S Schlicht functions, 49 Schwarzian derivative, 176 Singular inner function, 24 Smirnov domain, 173 Smoothness classes, 71 Subharmonic function, 7, 188 Subordinate function, 10 Symmetric derivative, 4 Szego polynomials, 175, 184 T Taylor coefficients, 93, 139, 142 Trigonometric series, 61
U Uniformly convex space, 123 Uniformly separated sequence, 148, 154 Universal covering surface, 180 Universal interpolation sequence, 148
w Walsh’s theorem, 174, 184 Weak type, 158, 160, 163, 165
Z Zeros, 18, 191 Zygmund’s theorem on boundary smoothness, 76 on conjugate functions, 58
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Pure and Applied Mathematics A Series of Monographs and Textbooks Editors Paul A. Smith and Samuel Eilenberg
Columbia University, New York
1: ARNOLD SOMMERFELD. Partial Differential Equations in Physics. 1949 (Lectures on Theoretical Physics, Volume V I )
2 : REINHOLD BAER.Linear Algebra and Projective Geometry. 1952 3 : HERBERTBUSEMANN A N D PAULKELLY.Projective Geometry and Projective Metrics. 1953 4: STEFAN BERCMAN A N D M. SCHIFFER. Kernel Functions and Elliptic Differential Equations in Mathematical Physics. 1953 5: RALPHPHILIP BOAS,JR. Entire Functions. 1954 6: HERBERT BUSEMANN. The Geometry of Geodesics. 1955 7 : CLAUDECHEVALLEY. Fundamental Concepts of Algebra. 1956 8: SZE-TSEN Hu. Homotopy Theory. 1959 9: A. M. OSTROWSKI. Solution of Equations and Systems of Equations. Second Edition. 1966 ~. on Analysis. Volume I, Foundations of Modern Analy10: J. D I E U W N NTreatise sis, enlarged and corrected printing, 1969. Volume 11, 1970. 11 : S. I. GOLDBERC. Curvature and Homology. 1962. 12 : SICURDUR HELGASON. Differential Geometry and Symmetric Spaces. 1962 13 : T. H. HILDEBRANDT. Introduction to the Theory of Integration. 1963. 14 : SHREERAM ABIIYANKAR. Local Analytic Geometry. 1964 15: RICHARD L. BISHOPA N D RICHARD J. CRITTENDEN. Geometry of Manifolds. 1964 16: STEVEN A. GAAL.Point Set Topology. 1964 17: BARRYMITCHELL.Theory of Categories. 1965 18: ANTHONY P. MORSE.A Theory of Sets. 1965
Pure and Applied Mathematics A Series of Monographs and Textbooks
19 : GUSTAVECHOQUET. Topology. 1966 20: 2. I. BOREVICH A K D I. R. SHAFAREVICH. Number Theory. 1966 21 : Josh LUIS MASSERA A N D J U A NJORGE SCHAFFER. Linear Differential Equations and Function Spaces. 1966 22 : RICHARD D. SCHAFER. An Introduction to Nonassociative Algebras. 1966 23: MARTINEICHLER. Introduction to the Theory of Algebraic Numbers and Functions. 1966 24 : SHREERAM ABHYANKAR. Resolution of Singularities of Embedded Algebraic Surfaces. 1966 25 : FRANCOIS TREVES. Topological Vector Spaces, Distributions, and Kernels. 1967 26: PETER D. LAXand RALPHS. PHILLIPS. Scattering Theory. 1967 27: OYSTEINORE.The Four Color Problem. 1967 28 : MAURICE HEINS.Complex Function Theory. 1968 29 : R. M. BLUMENTHAL A N D R. K. GETOOR. Markov Processes and Potential Theory. 1968 30 : L. J. MORDELL. Diophantine Equations. 1969 31 : J. BARKLEYROSSER.Simplified Independence Proofs : Boolean Valued Models of Set Theory. 1969 32: WILLIAMF. DONOGHUE, JR. Distributions and Fourier Transforms. 1969 33: MARSTONMORSEA N D STEWART S. CAIRNS.Critical Point Theory in Global Analysis and Differential Topology. 1969 34: EDWIN WEISS.Cohomology of Groups. 1969 35: HANSFREUDENTHAL AND H. DE VRIES. Linear Lie Groups. 1969 36 : LASZLO FUCHS. Infinite Abelian Groups : Volume I. 1970 37: KEIO NAGAMI.Dimension Theory. 1970 38: PETER L. DUREN.Theory of HP Spaces. 1970
In firebaration EDUARD PRUGOVECKI. Quantum Mechanics in Hilbert Space. BODOPAREICIS. Categories and Functors : An Introduction
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