Introduction to Function Spaces on the Disk

Miroslav Pavlovi´c

Matematiˇckiinstitut SANU Beograd 2004 Miroslav Pavlovi´c Faculty of Mathematics Belgrade University 11001 Belgrade, p.p. 550 Serbia

Typeset by the author in LATEX. c M. Pavlovi´cand Matematiˇckiinstitut SANU. All rights reserved. No part of this publication may be reproduced, stored in a re- trieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior permission of the copyright owner. To Mirjana and Pavle Contents

Preface 6

1 Quasi-Banach spaces 7 1.1 Quasinorm and p-norm ...... 7 1.2 Linear operators ...... 10 1.3 Open mapping, closed graph ...... 12 1.4 F -spaces ...... 14 1.5 The spaces `p ...... 16

2 Interpolation and maximal functions 18 2.1 The Riesz/Thorin theorem ...... 18 2.2 Weak Lp-spaces and Marcinkiewicz’s theorem ...... 22 2.3 Maximal function and Lebesgue points ...... 25 2.4 The Rademacher functions ...... 28 2.5 Nikishin’s theorem ...... 30 2.6 Nikishin and Stein’s theorem ...... 34 2.7 Banach’s principle ...... 36

3 Poisson integral 38 3.1 Harmonic functions ...... 38 3.2 Borel measures and the space h1 ...... 42 3.3 Radial limits of the Poisson integral ...... 45 p p 3.4 The spaces h and L (T) ...... 49 3.5 The Littlewood/Paley theorem ...... 51 3.6 Harmonic Schwarz lemma ...... 53

4 Subharmonic functions 55 4.1 Basic properties ...... 55 4.2 Properties of the mean values ...... 60 4.3 Integral means of univalent functions ...... 62 4.4 The subordination principle ...... 64 4.5 The Riesz measure ...... 67 4.6 A Littlewood/Paley theorem ...... 70

5 Classical Hardy spaces 73 5.1 Basic properties ...... 73 5.2 The space H1 ...... 78 5.3 Blaschke product ...... 80 5.4 Inner and outer functions ...... 85 5.5 Composition with inner functions ...... 88

4 6 Conjugate functions 92 6.1 Harmonic conjugates ...... 92 6.2 Riesz projection theorem ...... 96 6.3 Applications of the projection theorem ...... 99 6.4 Aleksandrov’s theorem ...... 100 6.5 Strong convergence in H1 ...... 101 6.6 Quasiconformal harmonic homeomorphisms ...... 104 7 Maximal functions, interpolation, coefficients 110 7.1 Maximal theorems ...... 110 7.2 Maximal characterization of Hp ...... 114 7.3 “Smooth” Cesàromeans ...... 116 7.4 Interpolation of operators on Hardy spaces ...... 119 7.5 On the Hardy/Littlewood inequality ...... 123 7.6 On the dual of H1 ...... 127 8 Bergman spaces: Atomic decomposition 129 8.1 Bergman spaces ...... 129 8.2 Reproductive kernels ...... 130 8.3 The Coifman/Rochberg theorem ...... 133 8.4 Coefficients of vector-valued functions ...... 137 9 Subharmonic behavior 143 9.1 Subharmonic behavior and Bergman spaces ...... 143 9.2 The space hp, p < 1 ...... 147 9.3 Subharmonic behavior of smooth functions ...... 148 10 Lipschitz spaces 154 10.1 Lipschitz spaces of first order ...... 154 10.2 Lipschitz condition for the modulus ...... 158 10.3 Lipschitz spaces of higher order ...... 160 10.4 Growth of derivatives ...... 162 11 Lacunary series 170 11.1 Lacunary series in Hp ...... 170 11.2 Karamata’s theorem and Littlewood’s theorem ...... 172 11.3 Lacunary series in C[0, 1] ...... 176 11.4 Lp-integrability of lacunary series on (0, 1) ...... 178 Bibliography 182 Index 187

5 Preface

This text contains some facts, ideas, and techniques that can help or motivate the reader to read books and papers on various classes of functions on the disk and the circle. The reader will ﬁnd several well known, fundamental theorems as well as a number of the author’s results, and new proofs or extensions of known results. Most of assertions are proved, although sometimes in a rather concise way. A number of assertions are named by Exercise, while certain assertions are collected in Miscellaneous or Remarks; most of them can be treated by the reader as exercises. The reader is assumed to have good foundation in Lebesgue integration, complex analysis, functional analysis, and Fourier series, which means in particular that he/she had a good training through these areas. It is of some importance that the reader can accept the following: Throughout this text, constants are often given without computing their exact values. In the course of a proof, the value of a constant C may change from one occurrence to the next. Thus, the inequality 2C 6 C is true even if C > 0.

Acknowledgment I want to express my appreciation to those who pointed out to me several typos as well as suggestions for improvement. In particular, I want to mention the detailed comments from Professor Miroljub Jevtićand Professor MiloˇsArsenović. I also want to express my deep gratitude to Mathematical Institute of Serbian Academy of Arts and Sciences and to the Faculty of Economics, Finance and Ad- ministration, Belgrade, for financial support.

Cvetke and Belgrade, 20 March – 22 April 2003.

6 1 Quasi-Banach spaces

In this text we mention only two examples of locally convex spaces: h(Ω), the space of all complex-valued functions harmonic in Ω ⊂ C, and its subspace H(Ω) consisting of analytic functions. For our purposes, the class of locally bounded spaces is more important. By Kolmogorov’s theorem, the intersection of this class with the class of locally convex spaces consists precisely of normable spaces. The topology of a locally bounded space can be described by a “quasinorm”; conversely, a “quasinormable” space is locally bounded. In the class of quasi-Banach spaces, there hold the “basic principles of functional analysis.” A concise discussion of these principles is contained in Section 1.3 and, in the context of F -spaces, in 1.4. Some properties of `p are stated, without proofs, in Section 1.5.

1.1 Quasinorm and p-norm

Let X be a (complex) vector space. A functional k · k: X 7→ [0, ∞) is called a quasinorm if the following conditions hold:

kf + gk 6 K(kfk + kgk), (1.1) where K (> 1) is a constant independent of f, g ∈ X; and kfk > 0 (f 6= 0), kλfk = |λ| kfk (λ ∈ C). (1.2) The couple (X, k · k) is then called a quasinormed space. The standard example are Lebesgue spaces: if µ is a positive measure deﬁned on a sigma-algebra of subsets p p p of a set S, then the space L (µ) = L (S, µ) = L (S) (0 < p 6 ∞) consists of all measurable complex-valued functions f on S for which Z 1/p p kfk = kfkp = |f| dµ < ∞, S with the usual interpretation in the case p = ∞. When p < 1, this functional is not a norm but satisﬁes (1.1) with K = 21/p−1 and, moreover, p p p kf + gk 6 kfk + kgk . (1.3) A functional satisfying (1.3) and (1.2) is called a p-norm. From (1.3) it follows that p p p p kf1 + f2 + ··· + fnk 6 kf1k + kf2k + ··· + kfnk . A similar inequality holds in the general case although a quasinorm need not be a p-norm for any p.

7 8 1 Quasi-Banach spaces

1.1.1 Lemma If k · k is a quasinorm on X, then there exist constants p ∈ (0, 1) and C 6 4 such that

p p p p kf1 + f2 + ··· + fnk 6 C kf1k + kf2k + ··· + kfnk (1.4)

for every ﬁnite sequence f1, . . . , fn ∈ X.

From this one can deduce that X is “p-normable” for some p > 0.

1.1.2 Theorem (Aoki/Rolewicz) If k · k is a quasi-norm on X, then there is p > 0 and a p-norm ||| · ||| on X such that kfk/C 6 |||f||| 6 kfk, f ∈ X, where C is independent of f.

The p-norm is deﬁned by

n 1/p n n X p X o |||f||| = inf kfjk : f = fj , j=1 j=1

where the infimum is taken over all finite sequences {fj} ⊂ X. Proof of Lemma. Take p so that (2K)p = 2, where K is the constant from (1.1), and define the functional H on X in the following way: H(0) = 0 and p k k−1 p k H(f) = 2 if 2 6 kfk < 2 for some integer k. Since p p p kfk 6 H(f) 6 2kfk , (1.5) inequality (1.4) is a consequence of the inequality

p p p kf1 + ··· + fnk 6 2 H(f1) + ··· + H(fn) .

The latter holds for n = 1. If n > 2, we consider two cases. (i) Let the summands H(fj) be mutually distinct and arranged in decreasing order. Then we have

p 1−j p H(fj) 6 2 H(f1) (1 6 j 6 n). (1.6)

From (1.1) it follows that kf + gk 6 2K max{kfk, kgk}, whence, by (1.5),

j kf1 + ··· + fnk 6 max{(2K) H(fj) : 1 6 j 6 n}.

p p Because of (1.6) and the choice of p, it turns out that kf1 + ··· + fnk 6 2H(f1) , which implies the required inequality. (ii) Assume that the sequence H(fj) contains at least two equal elements; for m m−1 p p m example, let H(f1) = H(f2) = 2 . Then 2 6 kfk1, kfk2 < 2 . Since

p p p p p p m+1 kf1 + f2k 6 (2K) max{kf1k , kf2k } = 2 max{kf1k , kf2k } 6 2 , 1.1 Quasinorm and p-norm 9

p m+1 p p we have H(f1 +f2) 6 2 = H(f1) +H(f2) . This and the induction hypothesis imply

p p p k(f1 + f2) + ··· + fnk 6 2 H(f1 + f2) + ··· + H(fn) p p p 6 2 H(f1) + H(f2) + ··· + H(fn) p p p 6 2 kf1k + kf2k + ··· + kfnk . 2

The space X is endowed with the structure of a topological vector space by declaring “a neighborhood of zero” to mean “a set containing {f : kfk < 1/n} for some n = 1, 2,....”(∗) This topology is metrizable, according to the Aoki/Rolewicz theorem; namely, if a p-norm |||· ||| is equivalent to the original quasinorm, then the formula d(f, g) = |||f − g|||p deﬁnes a metric that induces the same topology. The space X need not be locally convex(†); it is locally bounded because the neighborhoods {f : kfk < 1/n} are bounded in the sense of theory of topological vector spaces. On the other hand, it is known that a locally bounded vector topology can be described by a quasinorm (cf. [87]).

1.1.3 Exercise On the space Lp(0, 1) (0 q > p. The same holds for the sequence space ` .

Quasi-Banach and p-Banach spaces A quasinormed space X is called a quasi- Banach space if it is complete, which means that a sequence {fn} ⊂ X is convergent if (and only if) kfm − fnk → 0 as m, n → ∞. If X is p-normed and complete, then X is said to be p-Banach.

1.1.4 Proposition Let X be p-normed. Then X is complete iﬀ convergence of the P p P P series kfnk implies convergence of fn. If X is complete and fn converges, P∞ p P∞ p then there holds the inequality n=1 fn 6 n=1 kfnk .

1.1.5 Proposition Let {fjk} (j, k > 1) be a double sequence in a p-Banach ∞ ∞ ∞ ∞ P p P P P P space X. If kfjkk < ∞, then the iterated series fjk and fjk j,k j=1 k=1 k=1 j=1 converge and have the same sum.

1.1.6 Exercise (Peck [81]) Let (X, k · k) be a complex p-normed space of dimension n < ∞. By a theorem of Carathe´odory, every point from the convex hull of the unit ball can be represented as a convex combination of 2n points from the ball (because the real dimension is 2n). This can be used to show that there exists a 1/p−1 1/p−1 norm k · kn on X such that kfkn 6 kfk 6 (2n) kfkn. Note that (2n) is not the best constant, at least for n = 1.

(∗)The “ball” {f : kfk < 1} need not be an open set. Therefore a quasinorm, in contrast to a p-norm, need not be continuous. (†)For example, the space Lp(0, 1), 0 < p < 1, is not locally convex. 10 1 Quasi-Banach spaces

1.2 Linear operators

In the class of quasinormed spaces, continuity and boundedness of linear operators are equivalent. In the space L(X,Y ), of continuous linear operators from X to Y , the quasinorm is deﬁned by kT k := sup kT fk. kfk61 The space L(X,Y ) is complete iﬀ so is Y . An operator T ∈ L(X,Y ) is said to be invertible if it is bijective and its inverse is continuous.

1.2.1 Proposition Let X be a quasi-Banach space and T ∈ L(X,X) such an operator that kI − T k < 1, where I is the identity operator. Then T is invertible −1 p p −1 and there holds the inequality kT k 6 C 1 − kI − T k , where C and p are the constants from Lemma 1.1.1.

P∞ k Proof. Consider the series k=0(I − T ) . From inequality (1.4), applied to the space L(X,X), we get n p n X k X pk (I − T ) C kI − T k . 6 k=m k=m Therefore the series converges; denote its sum by S. Then we have ST = TS = I p P∞ pk and kSk 6 C k=0 kI − T k , which was to be proved. 2 The following statement is important although its proof is very simple.

1.2.2 Theorem Let X and Y be quasi-Banach spaces and E a dense subset of X. Let Tn ∈ L(X,Y ) be a sequence such that supn kTnk < ∞. If the limit limn→∞ Tnf exists for all f ∈ E, then it exists for all f ∈ X and the operator T f := limn→∞ Tnf is linear and continuous.

1.2.3 Exercise Let T be a continuous linear operator from a quasi-normed space X to quasi-normed space Y , and let E be a subset of X such that the linear hull of E is dense in X. If Y0 is a closed subspace of Y such that T (E) ⊂ Y0, then T (X) ⊂ Y0.

q-Banach envelope In the general case, a quasi-Banach space is embedded into many q-Banach spaces; the “smallest” of them is called the q-Banach envelope of X. To be more precise, deﬁne the functional Nq (0 < q 6 1) on X in the following way: 1/q X q X Nq(f) = inf kfjk : fj = f , (1.7) j j

where the infimum is taken over the set of finite sequences {fj} ⊂ X. This functional is a “ q-seminorm”, i.e., satisfies the conditions

q q q {Nq(f + g)} 6 {Nq(f)} + {Nq(g)} ,Nq(λf) = |λ| Nq(f). 1.2 Linear operators 11

The set {f ∈ X : Nq(f) = 0} =: Ker Nq is a closed subspace of X. If Ker Nq = {0}, i.e., if Nq is a q-norm, then the “completion” of the space (X,Nq) is a q- Banach space and is called the q-Banach envelope of X; denote it by [X]q. According to the Aoki/Rolewicz theorem, always there exists a q such that X = p [X]q, with equivalent quasinorms. A simple but illustrative example is X = ` ; q then [X]q = ` (p < q 6 1) and the corresponding quasinorms are equal (see 1.2.8 and 1.2.6). It is much more diﬃcult to identify the envelops of the Hardy space Hp (see Theorem 8.3.5). The importance of the space [X]q lies in the fact that every operator from X to an arbitrary q-Banach space extends to an operator on [X]q; more precisely:

1.2.4 Proposition Let X possess the q-Banach envelope ( i.e., let Nq be a q- norm) and let Y be an arbitrary q-Banach space. If T ∈ L(X,Y ), then there exists a unique operator S ∈ L([X]q,Y ) such that Sf = T f for all f ∈ X.

The following fact is useful in identifying the envelope:

1.2.5 Proposition Let X be continuously embedded into a q-Banach space Y in P∞ such a way that every f ∈ Y can be represented as f = n=1 fn, fn ∈ X, with P∞ q q n=1 kfnkX 6 CkfkY , where C does not depend of f. Then Y = [X]q (with equivalent quasinorms).

Proof. The space X is a dense subset of Y . Since X is dense in [X]q, we see that it suﬃces to prove that the q-norms k · kY and Nq are equivalent on X. P Let f = fj, where {fj} is a ﬁnite sequence in X. Then

q X q q X q kfkY 6 kfjkY 6 C kfjkX .

Taking the inﬁmum over {fj} ⊂ X, we get kfkY 6 CNq(f). (Incidentally this shows that Nq is a q-norm.) P∞ To prove the reverse inequality, let f ∈ X. Then f = n=1 fn, where P∞ q q P∞ q q n=1 kfnkX 6 CkfkY . Since Nq(fn) 6 kfnkX , we get n=1 Nq(fn) 6 CkfkY . P q P q q Hence fn converges to f in [X]q to f, and Nq(f) 6 Nq(fn) 6 CkfkY , which completes the proof. 2

Miscellaneous

1.2.6 The functional Nq is a q-norm on X iﬀ there is a q-Banach space Y such that L(X,Y ) separates points in X. The latter means that for every f 6= 0 there is T ∈ L(X,Y ) with T f 6= 0.

1.2.7 The dual of a quasi-Banach space X is X∗ = L(X, C). If X∗ separates points in X, then the Banach envelope of X is equal to the completion of the ∗ normed space (X,N), where N(f) = sup{|Λf| :Λ ∈ X , kΛk 6 1}. 12 1 Quasi-Banach spaces

p 1.2.8 If X = L (0, 1), 0 q > p, then Nq(f) = 0 for all f ∈ X. This is connected with the relation L(X,Y ) = {0}, where Y is an arbitrary q- Banach space.

1.3 Open mapping, closed graph

Let X,Y be a pair of complete spaces such that X is a dense subset of Y , which means that each member of Y can be approximated by members of X. This does not imply that members of a ball K1 ⊂ Y can be approximated by members of any ﬁxed ball K2 ⊂ X, i.e., that K2 ⊃ K1.(K2 = the closure of K2 in the topology of Y .) Namely, as the following theorem states, if K2 ⊃ K1, then X = Y .

1.3.1 Theorem Let X and Y be quasi-Banach spaces. Let T ∈ L(X,Y ) be such that the closure of T (B), where B = {f ∈ X : kfk < 1}, contains a neighborhood of zero in Y . Then the mapping T : X 7→ Y is open and the operator Tb : X/ Ker T 7→ Y is invertible.

A mapping is open if it maps open sets onto open sets. If T ∈ L(X,Y ), then the operator Tb ∈ L(X/ Ker T,Y ) is deﬁned by Tb(f + Ker T ) = T f. The quasinorm in X/Z is deﬁned by kf + Zk = inf{kf − gk : g ∈ Z}. Proof. Because of the Aoki/Rolewicz theorem, we can suppose that X and Y are p-normed for some p < 1. Let δ > 0 and U = {f ∈ X : kfkp < δ}. From the p hypotheses of the theorem it follows that there are balls Un = {f ∈ X : kfk < δn} p and Vn = {g ∈ Y : kgk < εn}, n > 1, limn εn = 0, such that

Vn ⊂ T (Un), (1.8) ∞ X δn < δ. (1.9) n=1

We will prove that V1 ⊂ T (U); then it will be easy to complete the proof. Let g ∈ V1. It follows from (1.8) that there exists f1 ∈ U1 such that g−T f1 ∈ V2. Similarly, there is f2 ∈ U2 such that (g − T f1) − T f2 ∈ V3. Continuing in this way, Pn we get the sequence of relations g − k=1 T fk ∈ Vn+1, fk ∈ Uk. It follows that P∞ p P g = n=1 T fn. And since kfkk < δk, inequality (1.9) implies that the series k fk p P∞ p converges; denote its sum by f. Thus we have g = T f and kfk 6 n=1 kfnk < δ, which was to be proved. 2

The open mapping theorem

1.3.2 Theorem Let X and Y be complete spaces and T ∈ L(X,Y ). If T is onto, then T is open. In particular, T is invertible if it is onto and one-to-one.

Proof. Let U be the unit ball in X. Choose a zero-neighborhood W so that W − W ⊂ U. By the hypothesis, the space Y is the union of the sets T (nW ) 1.3 Open mapping, closed graph 13

(n > 1). By Baire’s category theorem, the closure at least of one of them has nonempty interior, which implies that T (W ) contains an open set V 6= ∅. Then V − V is a neighborhood of zero and there hold the inclusions

V − V ⊂ T (W ) − T (W ) ⊂ T (W ) − T (W ) ⊂ T (U).

Now the desired result follows from Theorem 1.3.1. 2

1.3.3 Exercise A subspace E of a quasi-Banach space X is said to have the Hahn/Banach extension property (HBEP) if each λ ∈ E∗ has an extension Λ ∈ X∗. If E has HBEP, then Λ can be chosen so that kΛkX∗ 6 CkλkE∗ , where C is independent of λ.

As a special case of the open mapping theorem we have:

1.3.4 Theorem (on equivalent norms) Let k · k1 and k · k2 be quasinorms on a a vector space X, and let kfk1 6 kfk2 for every f ∈ X. If X is complete with respect to both quasinorms, then there exists a constant C < ∞ such that kfk2 6 Ckfk1 for all f ∈ X.

The uniform boundedness principle

1.3.5 Theorem (Banach/Steinhauss) Let X and Y be quasi-Banach spaces, and let {As} ⊂ L(X,Y ) be a family of operators. If sups kAsfk < ∞, for all f ∈ X, then sups kAsk < ∞. In particular the limit of an everywhere convergent sequence of bounded operators is a bounded operator.

Proof. Let kfk2 = kfkX + sups kAsfkY (f ∈ X). From the hypotheses it follows that the functional k · k2 is a quasinorm on X. It is not hard to prove that the space (X, k · k2) is complete and therefore the conclusion follows from Theorem 1.3.4. 2

1.3.6 Corollary Let B : X×Y 7→ Z be a separately continuous bilinear operator, where X,Y,Z are quasi-Banach spaces. Then there is a constant C < ∞ such that kB(f, g)kZ 6 CkfkX kgkY for all f ∈ X, g ∈ Y . “Separately continuous” means that every operator of the form f 7→ B(f, g) (f ∈ X) or g 7→ B(f, g)(g ∈ Y ) is continuous.

Shauder basis

A sequence {en : n > 1} in a quasi-Banach space X is called a Shauder basis of X if to each f ∈ X there corresponds a unique scalar sequence {λn(f)} such that P∞ f = n=1 λn(f)en, the series converging in the topology of X.

1.3.7 Proposition If {en : n > 1} is a Shauder basis of X, then the functionals λn are continuous and the linear operators Sn : X 7→ X deﬁned by Snf = Pn k=1 λk(f)ek are uniformly bounded. 14 1 Quasi-Banach spaces

Proof. Let |||f||| = supn kSnfk. Since kf − Snfk → 0, we have kfk 6 K|||f|||, where K is the constant from (1.1), and therefore, by Theorem 1.3.4, it is enough ∞ to prove that X is complete with respect to the quasinorm |||· |||. Let {fj}j=1 be a Cauchy sequence in ||| · |||. This implies, because of the completeness of k · k, that there is a sequence gn such that

sup kSnfj − gnk → 0 as j → ∞, (1.10) n>1

∞ and that for every k the sequence {λk(fj)}j=1 converges; let γk = limj λk(fj). Since the functional λk is linear and the space Sn(X) is ﬁnite-dimensional, it follows that λk(gn) = limj λk(Snfj) = limj λk(fj) = γk for k 6 n, and λk(gn) = 0 for k > n. Pn Hence gn = k=1 γkek. On the other hand, (1.10) implies that {gn} converges in P∞ k · k to some g. Thus g = n=1 γnen, whence gn = Sng. Returning to (1.10) we see that |||fj − g||| → 0 as j → ∞, which was to be proved. 2

1.3.8 Exercise A sequence {en : n > 1} of nonzero vectors in a quasi-Banach space X is a Shauder basis of X if and only if the following conditions are satisﬁed: (a) The closed linear span of {en} is X; (b) There is a constant K such that Pm Pn k j=0 ajejk 6 Kk j=0 ajejk for all scalar sequences {aj} and m < n.

The closed graph theorem 1.3.9 Theorem Let T : X 7→ Y be a linear operator, where X and Y are complete spaces. Then T is continuous if the following condition is satisﬁed: For every sequence {fn} ⊂ X such that fn tends to 0 ∈ X and T fn tends to some g ∈ Y we have g = 0.

Proof. It follows from the hypotheses that X is complete with respect to the quasinorm kfk2 = kfkX + kT fkY so we can apply Theorem 1.3.4. 2

1.4 F -spaces

The closed graph theorem remains valid in a wider class of spaces, the so called F -spaces. By the term “F -norm” on a vector space X we mean a functional N : X 7→ [0, ∞) satisfying: (a) N(f) = 0 =⇒ f = 0; (b) N(f +g) 6 N(f)+N(g); (c) N(λf) 6 N(f) for |λ| 6 1, and lim N(λf) = 0. (1.11) λ→0 The formula d(f, g) = N(f − g) deﬁnes an invariant metric on X and the topology induced by this metric is vectorial, which means in particular that multiplication by scalars is continuous on C × X. In the case where the metric d is complete, the space X is called an F -space. A p-Banach space can be treated as an F -space by introducing the F -norm p N(f) = kfkX . 1.4 F -spaces 15

Besides, if X is a locally convex space whose topology is given by a sequence of seminorms pn (n = 1, 2,... ), then the formula

∞ −n X 2 pn(f) N(f) = 1 + p (f) n=1 n deﬁnes an F -norm on X that induces the same topology. As an example one can take the space h(D) consisting of all harmonic functions on the unit disk D ⊂ C as well as its analytic analogue H(D). These spaces are endowed with the topology of uniform convergence on compact subsets of D. This topology can be given by the sequence of norms p (f) = max |f(z)|, where r ∈ (0, 1) is an arbitrary n |z|6rn n sequence tending to 1. Concerning the requirement (1.11), which guarantees the continuity of scalar multiplication, it is useful to consider the case of the Nevanlinna class N (D). This class consists of the functions f ∈ H(D) for which 1 Z π N(f) := sup log 1 + |f(reiθ)| dθ < ∞. (1.12) 0

The functional N induces a complete invariant metric on the vector space N (D) but (1.11) is not satisﬁed. For example, 1 + z if f(z) = exp , then lim N(εf) = 1. 1 − z ε→0

The set on which (1.11) holds coincides with the Smirnov class N +(D) consisting iθ of those f ∈ N (D) for which the family θ 7→ log 1 + |f(re )| , 0 < r < 1, is uniformly integrable. This and other topological properties of the Nevanlinna class are discussed in [90]. The Nevanlinna theory is exposed in, e.g., [100, 18, 22]. In the class of F -spaces, there holds the open mapping theorem 1.3.2 as well (the formulation is the same). Theorem 1.3.1 is now stated as follows: Let X and Y be F -spaces and T ∈ L(X,Y ). If for each neighborhood U of 0 ∈ X the set T (U) contains a neighborhood of 0 ∈ Y , then T (X) = Y and T is open. The proof is identical to the proof of Theorem 1.3.1 up to the obvious changes of notation. The property (1.11) is used only in proving that T (X) = Y . As a special case of the open mapping theorem, we have the following generalization of Theorem 1.3.4, from which the closed graph theorem is obtained immediately.

1.4.1 Theorem Let N1 and N2 be F -norms on a vector space X, and let N1(f) 6 N2(f) for all f ∈ X. If X is complete with respect to both of them, then every sequence {fn} ⊂ X satisﬁes the condition: N1(fn) → 0 =⇒ N2(fn) → 0.

1.4.2 Exercise Let X and Y be quasi-Banach spaces “continuously” contained ∞ in H(D). A scalar sequence µ = {µn}0 is called a multiplier from X to Y if 16 1 Quasi-Banach spaces

P∞ n for every f ∈ X the series (µ ∗ f)(z) = n=0 µnfb(n)z converges in D and µ ∗ f belongs to Y . If µ is a multiplier, then there is a constant C < ∞ such that kµ ∗ fkY 6 CkfkX for all f ∈ X. In particular if Y = X and X contains all the polynomials, then the sequence µ is bounded.

1.4.3 Exercise Let Ts : X 7→ Y be a family of operators between F -spaces X and Y . If sups NY (Tsf/n) → 0, as n → ∞, for every f ∈ X, then the following holds: If NX (fn) → 0, where fn ∈ X, then sups NY (Tsfn) → 0.

1.5 The spaces `p

The simplest examples of inﬁnite-dimensional (quasi-)Banach spaces are `p (0 < ∞ p 6 ∞) and c0 = {a ∈ ` : lim an = 0}. They play an exceptional role in many areas of analysis and especially in geometry of Banach spaces; we refer the reader to [53]. Their importance for the theory of spaces of analytic functions lies in the following fact:

1.5.1 Theorem For every p ∈ (0, ∞) the Bergman space Ap,

p p A = {f ∈ L (D): f analytic in D}, D = {z : |z| < 1},

is isomorphic with `p.

In the case 1 1) and [97](p 6 1). The case of mixed norm spaces was considered in [97, 59]. p Here we list a few properties of ` and c0. A slight modiﬁcation of the proof of Theorem 1.3.1 yields the following:

1.5.2 Theorem If X and Y are p-Banach spaces and T ∈ L(X,Y ) is such that kT k 6 1 and T (BX ) ⊃ BY , where B indicates the open unit ball, then the spaces Y and X/ Ker T are isometrically isomorphic.

In particular:

1.5.3 Theorem Every separable p-Banach space (0 < p 6 1) is isometrically isomorphic to some quotient space of `p.

p ∞ P∞ Proof. Deﬁne the operator T : ` 7→ X, by T ({an}1 ) = n=1 anfn, where {fn} is a dense subset of the unit ball of X. 2

1.5.4 Exercise Let Y be a subspace of `p such that Lp(0, 1) is isometric to `p/Y . If a functional Λ ∈ (`p)∗ = `∞ vanishes on Y , then it vanishes everywhere on `p.

The proof of the following is much more delicate (cf. [35, 52]). 1.5 The spaces `p 17

p 1.5.5 Theorem Let X be either c0 or ` , 0 < p 6 ∞. Then: (a) Every closed subspace (of inﬁnite dimension) of X contains an isomorphic copy of X. And: (b) Every complemented subspace of X is isomorphic to X.

In the case 1 6 p < ∞ both assertions were proved Pe lczy´nski [82], while the case p < 1 was discussed by Stiles [96, 95]. Assertion (b) for p = ∞ was proved by Lindenstrauss [49]. The fact that the spaces `p and `q (p 6= q) are not isomorphic is contained in the following assertion of Pitt (p > 1, cf. [52, Theorem I.2.7]) and Stiles [96](p < 1, cf. [35, Proposition 2.9]):

1.5.6 Theorem Every bounded linear operator from `q to `p (0 < p < q < ∞) is p compact; the same is true for linear operators from c0 to ` . (Moreover, if p < q 6 1, then every operator from a q-Banach space to `p (p < q) is compact.) Consequently, p no space of the class ` , 0 < p < ∞, and c0 is isomorphic to a subspace of another member of this class.

p On the other hand, if 0 < q < p 6 2, then L (0, 1) is isometrically isomorphic to a subspace of Lq(0, 1) (see [52, Theorem II.3.4]). 2 Lebesgue spaces: Interpolation and maximal functions

Thorin’s proof of two variants of the Riesz/Thorin theorem is in Section 2.1; as an example, we prove the Hausdorﬀ/Young theorem. The simplest version of Marcinkiewicz’s interpolation theorem is proved in Section 2.2; as an example, we prove Paley’s theorem on Fourier coeﬃcients (Theorem 2.2.5). Section 2.3 contains the Hardy/Littlewood maximal theorem with application to Lebesgue points. In Section 2.4 we prove Khintchine’s inequality, which says that the subspace of Lp(0, 1), 0 < p < ∞, spanned by the Rademacher functions is isomorphic with `2. The rest of this chapter is devoted to the proof of Nikishin’s theorem. This theorem states, in particular, that if a bounded linear operator T maps Lp(T) into Lq(T), p where 0 < q < p 6 2, then actually T maps L (T) into the weak Lebesgue space Lp,∞(T r A), where T r A is of arbitrarily small measure (see Theorem 2.5.2). If in addition T “commutes with rotations”, then we can take A = ∅. Also, we prove the so called Banach’s principle and the theorem on a.e. convergence (Theorems 2.7.1 and 2.7.2).

2.1 The Riesz/Thorin theorem

The proof of various variants of the Riesz/Thorin (convexity) theorem can be found, e.g., in the books [7, Ch. IV §2] and [8].

The case of bilinear forms

Let γ = (γ1, . . . , γm) and δ = (δ1, . . . , δn) be sequences of positive real numbers. For a ∈ Cm, b ∈ Cn and p, q > 0 let

m 1/p n 1/q X p X p kakγ,p = |aj| γj , kbkδ,q = |bk| δk . j=1 k=1

2.1.1 Theorem Let 0 < p0, q0, p1, q1 6 ∞. Let

m,n X m n B(a, b) = Bjkajbk, a ∈ C , b ∈ C , j,k=1

18 2.1 The Riesz/Thorin theorem 19

where Bjk ∈ C, and suppose that

|B(a, b)| 6 M0kakγ,p0 kbkδ,q0 , |B(a, b)| 6 M1kakγ,p1 kbkδ,q1 for all a ∈ Cm, b ∈ Cn. If 1 1 − η η 1 1 − η η = + , = + and 0 < η < 1, p p0 p1 q q0 q1 then 1−η η |B(a, b)| 6 M0 M1 kakγ,pkbkδ,q. In other words, if

M(α, β) = sup{ |B(α, β)|: kakγ,1/α 6 1, kbkδ,1/β 6 1}, then the function log M(α, β) is convex in the quadrant α > 0, β > 0. Proof. (Throughout the proof we omit the indices γ, δ.) Let p, q < ∞, and

kakp = 1 and kbkq = 1. (2.1)

Deﬁne a(z) ∈ Cm and b(z) ∈ Cn by

p/p(z) i arg aj q/q(z) i arg bk a(z)j = |aj| e and b(z)k = |bk| e , where 1 1 − z z 1 1 − z z = + and = + . p(z) p0 p1 q(z) q0 q1 z−1 −z The function F (z) := M0 M1 B(a(z), b(z)) is entire, as a sum of exponential functions. We consider the restriction of F to the strip Π = {z : 0 6 Re z 6 1}. We have, for t ∈ R,

p/p0 q/q0 |a(it)j| = |aj| , |b(it)k| = |bk| ,

p/p1 q/q1 |a(1 + it)j| = |aj| , |b(1 + it)k| = |bk| , whence, in view of (2.1),

ka(it)kp0 = 1, kb(it)kq0 = 1,

ka(1 + it)kp1 = 1, kb(1 + it)kq1 = 1.

−1 It follows that |F (it)| = M0 |B(a(it), b(it))| 6 1 and similarly |F (1 + it)| 6 1 for every t ∈ R. Thus the function F is analytic and bounded on Π and |F | 6 1 on the 1−η η boundary of Π. It follows that |F | 6 1 on Π and in particular |B(a, b)| 6 M0 M1 , which completes the proof in the case p, q < ∞. The remaining case is similar. 2 20 2 Interpolation and maximal functions

The case of linear operators 2.1.2 Theorem Let (R, µ) and (S, ν) be sigma-ﬁnite measure spaces and let 1 6 p0, q0, p1, q1 6 ∞. Let T be a (complex-)linear operator deﬁned on µ-simple functions on R and taking values in the set of all complex ν-measurable functions,

and let kT fkq0 6 M0kfkp0 and kT fkq1 6 M1kfkp1 for all µ-simple functions f on R. If 1 1 − η η 1 1 − η η = + , = + and 0 < η < 1, p p0 p1 q q0 q1 1−η η then kT fkq 6 M0 M1 kfkp for all µ-simple functions f. Remark. Concerning the validity of this theorem in the entire ﬁrst quadrant, that is, for 0 < pk, qk 6 ∞, see [7, page 281]. R Proof of Theorem. We consider the bilinear form B(f, g) = S(T f)g dν, where f, g are simple functions on R,S, respectively, and use the formula

kT fkq = sup{ |B(f, g)|: kgkq0 = 1, g simple},

0 Pm Pn where 1/q + 1/q = 1. Let f = j=1 ajKj, g = k=1 bkHk, where Kj and Hk are sequences of “pairwise disjoint” characteristic functions. Then straightforward 0 0 calculation shows that we can apply Theorem 2.1.1 with the indices p0, q0, p1, q1, Z Z Z Bj,k = (TKj)Hk dν, γj = Kj dµ, δk = Hk dν. S R S The details are left to the reader. 2 The following form of the preceding theorem is perhaps more convenient in application.

2.1.3 Theorem Let (R, µ) and (S, ν) be sigma-ﬁnite measure spaces and let 1 6 p0, q0, p1, q1 6 ∞. Let T be a linear operator deﬁned on the complex space Lp0 (R, µ)+Lp1 (R, µ) and taking values in the set of all complex ν-measurable func- p0 p1 tions, and let kT fkq0 6 M0kfkp0 , kT fkq1 6 M1kfkp1 for all f ∈ L , f ∈ L , respectively. If 1 1 − η η 1 1 − η η = + , = + and 0 < η < 1, p p0 p1 q q0 q1 then T is a bounded operator from Lp(R, µ) into Lq(S, ν) and

1−η η kT fkq 6 M0 M1 kfkp.

Proof. We shall consider the case where the measure µ is ﬁnite and p0 < p1 0 6 ∞. Then Lp(µ) ⊂ Lp0 (µ). With the above notation, let g ∈ Lq (ν) be a simple p function and let f ∈ L (µ) be arbitrary. Choose a sequence fn of simple functions

on R such that kfn − fkp → 0. Then kfn − fkp0 → 0 and therefore Z T (fn − f)g dν → 0, S 2.1 The Riesz/Thorin theorem 21

p q p q0 because T : L 0 7→ L 0 is continuous on L 0 (µ) and g ∈ L 0 . Hence

Z 1−η η 0 T (f)g dν 6 M0 M1 kfkpkgkq . S

The result follows. 2

The case of real-linear operators

1−η η For the validity of the conclusion kT fkq 6 M0 M1 kfkp in the Riesz/Thorin theorem, it is essential that the operator T be complex-linear, i.e., that T (λf) = λT f for all complex scalars λ (see [7, Ch. 4, Example 1.3]). For real spaces, the conclusion of the theorem remains valid if pk 6 qk (k = 0, 1). Otherwise, we 1−η η have kT fkq 6 2M0 M1 kfkp. However, if T is a positive linear operator, then Theorem 2.1.3 remains valid for any pk, qk > 1.

The Hausdorﬀ/Young theorem

Let T denote the unit circle of the complex plane. For a function f ∈ L1(T), let fb(n) be the Fourier coeﬃcients of f,

1 Z π fb(n) = f(eiθ)e−inθ dθ. 2π −π

p 2.1.4 Theorem If f ∈ L (T), 1 6 p 6 2,

∞ 1/p0 X p0 |fb(n)| 6 kfkp , (2.2) n=−∞

0 p p where 1/p + 1/p = 1. In other words: If f ∈ L (T) and {bn} is a two-sided ` - P sequence, 1 6 p 6 2, then the series bnfb(n) is absolutely convergent and there holds the inequality ∞ X bnfb(n) 6 kfkp k{bn}kp . (2.3) n=−∞

Proof. The theorem is true for p = 1 because |fb(n)| 6 kfk1, and is true for p = 2 because of Parseval’s formula. Then the result is obtained by the Riesz/Thorin theorem. 2

∞ p0 2.1.5 Exercise If p > 2 and {bn}−∞ ∈ ` , then there exists a unique function 0 p P∞ p0 1/p g ∈ L (T) such that gb(n) = bn for all n and kgkp 6 n=−∞ |bn| . 22 2 Interpolation and maximal functions

2.2 Weak Lp-spaces and Marcinkiewicz’s theorem

The space Lp,∞ Let Ω be a measure space with a (positive) sigma-ﬁnite measure µ. The weak Lp space Lp,∞(µ), 0 λ} = µ{ω ∈ Ω: |f(ω)| > λ}. Chebyshev’s inequality, 1 Z µ(g, λ) 6 |g| dµ, λ Ω shows that Lp ⊂ Lp,∞, while the formula Z Z ∞ |g|q dµ = µ(g, λ) d(λq) (2.4) Ω 0 (proved by means of Fubini’s theorem) implies Lp,∞ ⊂ Lq for q < p, if µ is ﬁnite. The quantity k · kp,∞ is a norm for no p, but we have

max(1/p,1) kf + gkp,∞ 6 Cp (kfkp,∞ + kgkp,∞)(Cp = 2 ),

and hence k · kp,∞ is a (complete) quasinorm. It is interesting, however, that if p = 1, then the space need not be locally convex (if, for example, Ω = [0, 1] with Lebesgue measure), although it can be q-renormed for every q < 1. For p > 1 the space is locally convex, and for p < 1 it is p-convex, i.e., there is an equivalent p-norm on it.(∗)

2.2.1 Exercise There hold the inequalities

µ(f1 + f2, λ1 + λ2) 6 µ(f1, λ1) + µ(f2, λ2), µ(f1f2, λ1λ2) 6 µ(f1, λ1) + µ(f2, λ2).

Marcinkiewicz’s theorem Quasilinear operators Let T be an operator acting from a vector space X to the set of all nonnegative measurable functions deﬁned on a measure space (Ω, µ). Then T is called a quasilinear operator if there exists a constant K such that

T (f + g) 6 K(T f + T g)(f, g ∈ X). If K = 1, then T is said to be subadditive. If an operator S with values in the set of ﬁnite measurable functions on Ω is linear, then the operator T f = |Sf| is subadditive. (∗)For further information see Kalton [36]. See also [7] for the general theory of weak Lp spaces. 2.2 Weak Lp-spaces and Marcinkiewicz’s theorem 23

2.2.2 Theorem Let µ and σ be sigma-ﬁnite measures on Ω and S, respectively, p q let 0 < p < q 6 ∞ and let T be a quasilinear operator from L (σ) + L (σ) to the set of all nonnegative µ-measurable functions. Assume there exist constants C1 and C2, independent of f, such that

kT fkp,∞ 6 C1kfkp , (2.5) kT fkq,∞ 6 C2kfkq . (2.6) Then for every s ∈ (p, q) there exists a constant C independent of f such that

kT fks 6 Ckfks. (2.7)

In the case q = ∞ inequality (2.6) should be interpreted as kT fk∞ 6 C2kfk∞.

Weak type and strong type If T satisﬁes (2.5), i.e., if T maps Lp into Lp,∞ and is continuous at zero, then we say that T is of weak type (p, p); if (2.7) holds, then T is of strong type (s, s).

Proof of Theorem 2.2.2

We consider the case where K = C1 = C2 = 1 and q < ∞, leaving the remaining cases to the reader. We have to deduce the inequality Z Z s s |T f| dµ 6 C |f| dσ Ω S from two “weak” inequalities: Z 1 p µ(T f, λ) 6 p |f| dσ, (2.8) λ S Z 1 q µ(T f, λ) 6 q |f| dσ. (2.9) λ S

To show this we represent the function f in the form f = gλ + hλ, where ( f(ζ), if |f(ζ)| > λ, gλ(ζ) = 0, if |f(ζ)| < λ.

Since T f 6 T (gλ) + T (hλ), we have µ(T f, λ) 6 G(λ) + H(λ), where

G(λ) = µ(T gλ, λ/2) and H(λ) = µ(T hλ, λ/2).

It follows from (2.8) and (2.9) that Z Z p p p p G(λ) 6 (2/λ) |gλ| dσ = (2/λ) |f| dσ (2.10) S |f|>λ 24 2 Interpolation and maximal functions

and Z Z q q q q H(λ) 6 (2/λ) |hλ| dσ = (2/λ) |f| dσ. S |f|6λ Now we use the formula Z Z ∞ Z ∞ s s−1 s−1 |T f| dµ = s µ(T f, λ)λ dλ 6 s G(λ) + H(λ) λ dλ. Ω 0 0 Multiplying inequality (2.10) by sλs−1 and then integrating over λ ∈ (0, ∞) we get Z ∞ Z ∞ Z s−1 p s−p−1 p s G(λ)λ dλ 6 s2 λ |f| dσ dλ 0 0 |f|>λ Z Z |f| = s2p λs−p−1 dλ |f|p dσ S 0 s2p Z = |f|s dσ. s − p S The analogous inequality for H(λ) is proved in a similar way. 2

Marcinkiewicz’s theorem for L log+ L 2.2.3 Theorem Let µ and σ be ﬁnite measures on Ω and S, respectively, let 1 1 < q 6 ∞ and let T be a quasilinear operator from L (σ) to the set of all nonnegative µ-measurable functions. If T satisﬁes (2.5)(p = 1) and (2.6), then Z Z + T f dµ 6 K1 + K2 |f| log |f| dσ, (2.11) Ω S

where K1 and K2 are independent of f. The class of those σ-measurable functions f on S for which the integral on the right hand side of (2.11) is ﬁnite is denoted by L log+ L(S). The proof is similar to that of Theorem 2.2.2. Other variants of Marcinkiewicz’s theorem can be found in Zygmund [100, Ch. XII§4]; the proof of some of them is much more diﬃcult.

Paley’s theorem The implication

∞ p X p0 f ∈ L (T) =⇒ |fb(n)| < ∞ (1 < p < 2), n=−∞ which is a weak form of the Hausdorﬀ/Young theorem, was improved by Hardy and Littlewood; namely:

p P∞ p−2 ∗ p 2.2.4 Theorem If f ∈ L (T), 1 < p < 2, then n=0(n + 1) (cn) < ∞, where ∗ {cn} is the decreasing rearrangement of the sequence {fb(n)}. 2.3 Maximal function and Lebesgue points 25

An application of Marcinkiewicz’s theorem yields a more general result, due to Paley:

∞ 2.2.5 Theorem Let (Ω, µ) be a ﬁnite measure space and let {ϕn}1 be an or- 2 thonormal sequence in L (Ω, µ) such that supn kϕnk∞ < ∞. Then

∞ X p−2 p p p n |an| 6 Ckfkp, f ∈ L (Ω, µ), n=1 R where 1 < p < 2 and an = Ω fϕn dµ.

Proof. Let µ(Ω) = 1 and supn kϕnk∞ = K. Deﬁne the measure σ on N, the set −2 1 of positive integers, by σ({n}) = n . Deﬁne the operator T : L (Ω, µ) 7→ L0(N, σ) by (T f)(n) = nan. Bessel’s inequality implies that T is of strong type (2, 2). To prove that T is of weak type (1, 1) and therefore to conclude the proof (by Marcinkiewicz’s theorem), observe that |an| 6 Kkfk1. Hence, if kfk1 = 1, we have

X −2 σ{n: |T f(n)| > λ} 6 σ{n: Kn > λ} 6 n 6 CK min(1, 1/λ), n>λ/K which concludes the proof. 2

∞ 2.2.6 Exercise Let (Ω, µ) be a sigma-ﬁnite measure space and let {ϕn}1 be an 2 γ orthonormal sequence in L (Ω, µ) such that kϕnk 6 Mn , where M and γ are positive constants. Then for 1 < p < 2 there holds the inequality

∞ X (γ+1)(p−2) p p n |an| 6 Ckfkp . n=1

2.3 Maximal function and Lebesgue points

The maximal function of a 2π-periodic function φ ∈ L1(−π, π) is the (2π-periodic) function Mφ deﬁned as

Z θ+h 1 (Mφ)(θ) = sup φ(t) dt. (2.12) 0

The function Mφ is above semicontinuous (and, consequently, measurable) as the supremum of a family of continuous functions. The (sublinear) operator M taking φ to Mφ is called the maximal operator of Hardy and Littlewood. If g ∈ L1(T), then we deﬁne

(Mg)(eiθ) = (Mφ)(θ), where φ(θ) = g(eiθ). (2.13) 26 2 Interpolation and maximal functions

The main maximal theorem 2.3.1 Theorem (a) If φ is in L1(−π, π), then there exists an absolute constant C C such that {θ ∈ (−π, π): Mφ(θ) > λ} 6 λ kφk1. p p (b) If φ ∈ L (−π, π), p > 1, then Mφ ∈ L (−π, π) and kMφkp 6 Cpkφkp, where Cp depends only of p.

By | · · · | we denote the Lebesgue measure on the line. Proof. Assertion (b) is obtained from (a) by Marcinkiewicz’s theorem. To prove (a), let φ ∈ L1(−π, π), let E = {θ ∈ (−π, π): Mφ(θ) > 1} and let K be a compact subset of E. It suffices to find an absolute constant C such that |K| 6 Ckφk1. By the definition of Mφ and the compactness of E, there are S R intervals Ii (i = 1, . . . , n) such that Ii ⊂ (−2π, 2π), K ⊂ Ii and |Ii| |φ(t)| dt. 6 Ii Assume that the sequence |Ii| is decreasing. Let J1 = I1. Let J2 = Ik, where k is the smallest i for which Ii ∩ J1 = ∅. Then let J3 = Im, where m is the smallest i > k such that Ii ∩ (J1 ∪ J2) = ∅. Continuing in this way we find a sequence S S ∗ Jj ⊂ (−2π, 2π) of pairwise disjoint intervals such that Ii ⊂ Jj , where, for each ∗ ∗ j, Jj is the interval “concentric” with Jj and |Jj | = 3|Jj|. It follows that

X X Z (1/3)|K| 6 |Jj| 6 |φ(t)| dt, j j Jj

which gives the desired inequality with C = 6. 2

Lebesgue points The maximal theorem has many important applications. It is useful, for example, in proving almost everywhere convergence. Usually, we can easily prove a.e. convergence for a dense set of functions, and then use the maximal theorem to interchange the limits. Here we consider the existence of Lebesgue points. The Lebesgue point of a function φ is a point x ∈ R such that Z h 1 lim φ(t + x) − φ(x) dt = 0. h→0 2h −h The set of all Lebesgue points of f is called the Lebesgue set of f.

2.3.2 Theorem If a 2π-periodic function φ is integrable on (−π, π), then almost every point in R is a Lebesgue point for φ.

2.3.3 Corollary The inequality |φ(θ)| 6 (Mφ)(θ) holds almost everywhere. Proof of Theorem. The operator

Z h 1 T φ(x) = lim sup φ(t + x) − φ(x) dt h→0 2h −h 2.3 Maximal function and Lebesgue points 27

satisﬁes: (a) T (φ1 + φ2) 6 T φ1 + T φ2 ; (b) T φ 6 |φ| + Mφ ; (c) T g = 0 if g is continuous. Let φ ∈ L1(−π, π), λ > 0 and ε > 0. Choose a continuous function g so that kφ − gk1 < ε. From (a) we get T φ 6 T g + T (φ − g) = T (φ − g) and then, from (b), by Theorem 2.3.1 and Chebyshev’s inequality, we get

{θ : T (φ − g)(θ) > λ} 6 {θ : |φ − g|(θ) > λ/2} + {θ : M(φ − g)(θ) > λ/2} 4π 2C 2(2π + C)ε kφ − gk + kφ − gk . 6 λ 1 λ 1 6 λ

Thus {θ : T (φ − g)(θ) > λ} = 0, for every λ > 0, because ε is arbitrary. 2

Non-periodic case

n Let φ be a locally integrable function deﬁned on R , n > 1. The maximal function Mφ is deﬁned on Rn by Z 1 (Mφ)(z) = sup n φ(ξ) dVn(ξ), (2.14) r>0 r |ξ−z|

Density of rational functions in Lp

2.3.4 Theorem If D is a bounded subdomain of C, then the set of rational functions with simple poles is dense in Lp(D), for 0 < p < 2.

Observe that the function 1/(z − a), where a ∈ D, belongs to Lp(D) if and only if p < 2. q Proof. It suﬃces to consider the case 1 6 p < 2. Let ϕ ∈ L (D), 1/p+1/q = 1, and let Z ϕ(w)R(w) dA(w) = 0, where dA = dV2, D for every rational function R. Then Z ϕ(w) dA(w) = 0 D z − w for every z ∈ C, whence Z Z ϕ(w) dA(w) dz = 0, |z−z0|=r D z − w 28 2 Interpolation and maximal functions

where z0 ∈ D and r < dist(z0, ∂D). Here we can apply Fubini’s theorem because

Z ϕ(w) Z 1/p −p dA(w) 6 kϕkq |z − w| dA(w) 6 C, D z − w D

for |z − z0| = r, where C is independent of z. Hence, by Cauchy’s integral formula, Z ϕ(w) dA(w) = 0. |w−z0|

If z0 is a Lebesgue point of ϕ, then we obtain 1 Z 0 = lim ϕ(w) dA(w) = ϕ(z0), r→0 2 r |w−z0|

and this concludes the proof. 2

2.3.5 Exercise An interesting fact, observed in [10], can be deduced from Theo- rem 2.3.4 and Runge’s theorem. If f is a Lebesgue measurable function deﬁned on C, then there is a sequence Pn of (holomorphic) polynomials such that Pn → f a.e.

2.3.6 Exercise Let R denote the set of all rational functions. If p < 2 then the set R ∩ Lp(C) is dense in Lp(C).

2.4 The Rademacher functions

j The Rademacher functions rj(t) are deﬁned by rj(t) = sign sin(2 tπ)(j > 0, t ∈ R). For example, r0(t) = 1 for 0 < t < 1,  1, 0 < t < 1/2,  r1(t) = −1, 1/2 < t < 1, 0, t = 0, 1/2, 1,

2 and rn(t) ≡ rn−1(2t). These functions form an orthonormal sequence in L (0, 1) and therefore

Z 1 n 2 n X X 2 n ajrj(t) dt = |aj| , {ak}0 ⊂ C. (2.15)

0 j=0 j=0

This generalization of the parallelogram law can also be written as

n 2 n 1 X X X 2 a0 + εjak = |aj| , 2n n k=0 (εj )∈{−1,1} j=0 2.4 The Rademacher functions 29 or in a more symmetric form

n 2 n 1 X X X 2 εjak = |aj| . 2n+1 n+1 k=0 (εj )∈{−1,1} j=0

P∞ 2 P∞ It was proved by Rademacher that k=0 |ak| < ∞, then k=0 akrk(t) converges P∞ 2 P∞ a.e. On the other hand, if k=0 |ak| = ∞, then k=0 akrk(t) diverges a.e. (Khint- chine and Kolmogorov). The proof of these facts can be found in Duren [18] and, in a stronger form, in Zygmund [100].

Khintchine’s inequality

2.4.1 Theorem For every p ∈ (0, ∞) there are positive constants cp and Cp such that

n 1/2 Z 1 n p 1/p n 1/2 X 2 X X 2 cp |aj| ajrj(t) dt Cp |aj| (2.16) 6 6 j=0 0 j=0 j=0

n for every ﬁnite sequence {aj}1 of complex scalars.

Note that we can take cp = 1 for p 2, and Cp = 1 for p 2. > 6 n P Proof. Suppose we have proved the theorem for p > 2. Let φn(t) = ajrj(t). Then j=0 Z 1 Z 1 2 2 1/2 3/2 kφnk2 = |φn(t)| dt = |φn(t)| |φn(t)| dt 0 0 1/2 3/2 3/2 1/2 3/2 6 kφnk1 kφnk3 6 C3 kφnk1 kφnk2 ,

3 and hence kφnk2 6 C3 kφnk1, which proves the left-hand side inequality in (2.16) for p = 1. Using this we can prove that kφnk1 6 const kφnk1/2, and so on. To discuss the case p > 2, we can suppose that p is an even integer. Using the binomial formula we ﬁnd that there holds the inequality

p p |x + y| + |x − y| p/2 |x|2 + κ |y|2 , x, y ∈ , (2.17) 2 6 p C

1/2 where κp > 1 is a constant. We shall prove that kφnkp 6 κp kφnk2 by induction n on the length of {ak}0 . In the case n = 1 we have |a + a |p + |a − a |p 1/p kφ k = 0 1 0 1 κ1/2 kφ k , n p 2 6 p n 2 because of (2.17). Let n > 2. Then, as is easily veriﬁed, Z 1 n p Z 1 X 1 p p ajrj(t) dt = |a0 + ψ(t)| + |a0 − ψ(t)| dt, (2.18) 2 0 j=0 0 30 2 Interpolation and maximal functions

n n−1 P P where ψ(t) = akrk−1(t) = ak+1 rk(t). From this and (2.17) it follows that k=1 k=0 Z 1 n p Z 1 X 2 2p/2 ajrj(t) dt |ψ(t)| + κp |a0| dt. 6 0 j=0 0

2 2p/2 The last integral is 6 kψkp +κp |a0| , which can be seen by using the binomial formula, or by using Jensen’s inequality for the concave function x 7→ (x2/p +1)p/2, 1/2 x > 0. On the other hand, by induction hypothesis we have kψkp 6 κp kψk2, which implies Z 1 n p X 2 2p/2 ajrj(t) dt κp kψk + κp |a0| . 6 2 0 j=0 This completes the proof because n 1/2 2 21/2 1/2 X 2 κp kψk2 + κp |a0| = κp |aj| . 2 j=0

Miscellaneous

n−1 2.4.2 What is essential in the above proof is that rn is deﬁned by rn(t) = r(2 t), ∞ n > 1, where r ∈ L (R) is a 1-periodic function such that r(t + 1/2) ≡ −r(t). As an example we can take r(t) = ei 2πt to get Paley’s inequality; this inequality will be discussed in a diﬀerent context (see Theorems 11.1.1 and 11.1.3, page 170).

2.4.3 The Lp-closure of the linear span of the Rademacher functions is complemented in Lp, for 1 j>1 Ft = εjrj(t)fj, where εj = ±1. If X0 is a subset of X, then j>1

{t ∈ [0, 1]: Gt ∈ X0} = {t ∈ [0, 1]: Ft ∈ X0} .

2.5 Nikishin’s theorem

While Marcinkiewicz’s theorem enables us to deduce a strong inequality from two weak inequalities, by using Nikishin’s theorem we can obtain a weak inequality from a “very weak” one. Before stating this theorem we introduce some terminology.

The space L0 and sublinear operators

Let (Ω, µ) be a ﬁnite measure space. The vector space L0(µ) = L0(Ω, µ) of all ﬁnite measurable functions on Ω becomes an F -space when endowed with the F -norm Z |f| N0(f) = dµ. Ω 1 + |f| 2.5 Nikishin’s theorem 31

Convergence in L0(µ) is equivalent to convergence in measure, i.e., fn → 0 iﬀ limn µ(fn, λ) = 0 for every λ > 0. This can be deduced from the formula

Z ∞ µ(g, λ) N0(g) = 2 dλ, (2.19) 0 (1 + λ) which can easily be deduced from (2.4)(q = 1) by taking g = |f|/(1 + |f|). An operator T that maps a quasi-normed space X to the set of nonnegative measurable functions is said to be sublinear if (almost everywhere): (a) T (f + g) 6 T f + T g for f, g ∈ X; (b) T (λf) = |λ| T f for f ∈ X and λ ∈ C. If T f is ﬁnite a.e. for all f, then we can treat it as an operator from X to L0. Since (a) and (b) imply |T f − T g| 6 T (f − g), we see that continuity of T at the origin implies continuity of T on all of X.

Spaces of type p

A quasi-Banach space X is of type p (0 0 j for every ﬁnite sequence {fj} ⊂ X, where rj are the Rademacher functions; recall j that rj(t) = sign sin(2 tπ). Every p-Banach space (0 < p 6 1) satisﬁes (2.20) with K = 1. And by application of Khintchine’s inequality (2.16), one proves the following:

2.5.1 Theorem Lp has type min(p, 2) (p > 0).

From Khintchine’s inequality also follows that the notion of type p has no sense for p > 2 because if X = C, then the integral in (2.20) is “proportional” to the 2 ` -norm of the sequence {fj}.

Nikishin’s theorem

In order to state Nikishin’s theorem let TBf = χBT f, where χB is the characteristic function of B ⊂ Ω.

2.5.2 Theorem (Nikishin [66]) Let X be a quasi-Banach space of type p, 0 0 there exists a measurable set B ⊂ Ω such that µ(Ω r B) < ε and TB maps X to Lp,∞ and is continuous.

Lemmas For the proof we need two lemmas. The ﬁrst lemma provides a characterization of continuity of T by means of “weak” inequalities. 32 2 Interpolation and maximal functions

2.5.3 Lemma Let T :(X, k · k) 7→ L0(µ) be a sublinear operator on a quasi- Banach space X. Then T is continuous iﬀ there exists a decreasing function c(λ), λ > 0, such that limλ→∞ c(λ) = 0 and

µ(T f, λkfk) 6 c(λ)(λ > 0, f ∈ X). (2.21) Therefore, Nikishin’s theorem says that under some conditions inequality (2.21) can be improved so that c(λ) = C/λp outside a set of arbitrarily small measure. Proof. Let T be continuous. Then there exists a decreasing function ε(λ) deﬁned for λ > 0 such that ε(λ) → 0 (λ → ∞), and that there holds the implication

kfk = 1/λ =⇒ N0(T f) < ε(λ).

On the other hand, (2.19) implies µ(T f, 1) 6 2N0(T f) because the function µ(T f, λ) is decreasing. Thus if kfk = 1/λ, then µ(T f, λkfk) 6 2ε(λ) =: c(λ), and hence µ(T f, λkfk) = µ(T f/λkfk, 1) 6 c(λ) for every λ > 0. Conversely, if (2.21) is satisﬁed, let Z ∞ c(λ/δ) (δ) = 2 dλ. 0 (1 + λ) From (2.19) it follows that there holds the implication

kfk < δ =⇒ N0(T f) < (δ), which implies that T is continuous, because (δ) → 0 (δ → 0). 2

2.5.4 Lemma Let the hypotheses of Theorem 2.5.2 be satisﬁed. Then its conclusions hold if (and only if) there exists a decreasing function c(λ), λ > 0, such that limλ→∞ c(λ) = 0 and µ(sup T fj(ω), λ) 6 c(λ) (2.22) j P p for all sequences {fj} ⊂ X such that j kfjk 6 1. This lemma can also be interpreted in the following way: The conclusions of p Theorem 2.5.2 hold iﬀ the operator Sf = supj T fj maps the space ` (X) into L0(µ) p ∞ and is continuous. The space ` (X) consists of those sequences {fn}1 ⊂ X for which ∞ 1/p X p k{fn}k = kfnk < ∞. n=1 Proof. We will consider “if” part.(†) Let ε > 0 and choose R > 0 so that c(R) < ε, where c(λ) is the function from (2.22). We say that a measurable set F

(†)Concerning the opposite direction, which is not needed in our text, one could see Wojtaszczyk [98, Ch. III.H]; the proof of “if” part is essentially the same as in that book. 2.5 Nikishin’s theorem 33

satisﬁes condition (W ) if there exists f ∈ X, kfk 6 1, such that µ(F ) T f(ω)p > Rp (ω ∈ F ). (W ) If the collection, say F, of such sets is empty, then for ﬁxed λ > 0 and f ∈ X, p p kfk 6 1, we take F = {ω ∈ Ω: T f(ω) > λ} to get µ(F ) T f(ω) 6 R (ω ∈ F ), p p p whence µ(F )λ 6 R . i.e., µ{ω ∈ Ω: T f(ω) > λ} 6 (R/λ) , and this means that T maps X into Lp,∞ and is continuous. Therefore we can suppose that F is nonempty. Let (Fj) be a maximal collection of pairwise disjoint measurable sets 1/p satisfying (W ) and let fj ∈ X be the corresponding vectors. Putting cj = µ(Fj) P p S we have j kcjfjk 6 1 and supj T (cjfj)(ω) > R (ω ∈ Fj). From (2.22) it follows that µ(∪Fj) 6 c(R) < ε. Since the collection is maximal, there holds the S p inequality µ ω ∈ Ω r Fj : T f(ω) > λ 6 (R/λ) , for every λ > 0. This means p,∞ that the operator TB, where B = Ω r ∪Fj, maps X into L , which concludes the proof. 2

Proof of Theorem 2.5.2.

First observe that Lemma 2.5.4 remains true if we assume that the sequence fj is ﬁnite. Let fj ∈ X be such a sequence, and let X p kfjk 6 1. (2.23) j>1

According to Lemma 2.5.4, it is enough to ﬁnd a decreasing function c1(λ) deﬁned for λ > 0 such that limλ→∞ c1(λ) = 0 and

µ(Eλ) 6 c1(λ)(λ > 0), (2.24) P where Eλ = {ω ∈ Ω : maxj T fj λ}. Let gt = rj(t)fj (0 t 1), where rj > j>1 6 6 k are the Rademacher functions. Then 2rk(t)fk = gt + gt , where k gt = −r1(t)f1 − · · · − rk−1(t)fk−1 + rk(t)fk − .... k Therefore there holds the inequality 2T (fk) 6 T (gt) + T (gt ) almost everywhere on [0, 1]. Since k { t ∈ [0, 1] : T gt(ω) > η} = { t ∈ [0, 1] : T gt (ω) > η} for every η > 0 (see 2.4.4), we see that { t ∈ [0, 1] : T gt(ω) > T fk(ω) } > 1/2 for all ω ∈ Ω and k > 1. Thus we have {t ∈ [0, 1] : T gt(ω) > λ} > 1/2, ω ∈ Eλ, so it follows that Z

µ(Eλ)/2 6 {t ∈ [0, 1] : T gt(ω) > λ} dµ(ω). Eλ This inequality is suﬃcient to conclude the proof in the case when X is p-Banach, p < 1. Namely, replacing R by R and applying the formula Eλ Ω Z Z Z

{t ∈ [0, 1] : T gt(ω) > λ} dµ(ω) = dµ(ω) dt Ω Ω T gt(ω)>λ Z 1 Z Z 1 = dt dµ(ω) = µ{ω ∈ Ω: T gt(ω) > λ} dt, 0 T gt(ω)>λ 0 34 2 Interpolation and maximal functions

we get Z 1 µ(Eλ)/2 6 µ{ω ∈ Ω: T gt(ω) > λ} dt. 0 Now Lemma 2.5.3 and the inequality kgtk 6 1, which holds for every t, give µ(Eλ) 6 2c(λ/2), where c(λ) is the function from Lemma 2.5.3. So we have proved (2.24) in this special case. Let X be a space of type p. Then the inequality kgtk 6 1 can be false for some t, but from (2.23) and (2.20) it follows that

p {t : kgtk > λ} 6 K/λ . (2.25) In order to exploit this fact, we start from the inequality √ √ {t : T gt(ω) > λ} 6 {t : kgtk > λ} + {t : T gt > kgtk λ} . Hence, by integration over ω ∈ Ω and using (2.25) and Fubini’s theorem as above, we get Z

µ(Eλ)/2 6 {t : T gt(ω) > λ} dµ(ω) Ω Z 1 √ −p/2 6 Kλ µ(Ω) + µ ω : T gt(ω) > kgtk λ dt. 0 √ The last integral is 6 c( λ/2), because of Lemma 2.5.3, so we have (2.24) again. 2

2.6 Nikishin and Stein’s theorem

This theorem tells us that under additional, algebraic, conditions we can take B = Ω in Theorem 2.5.2. We restrict ourselves to the case of the multiplicative group T. 2.6.1 Theorem (Nikishin/Stein) Let X be a space of type p ∈ (0, 2], and let ζ 7→ fζ be a mapping of the unit circle T to the set of all isometric endomorphisms of X. If T : X 7→ L0(T) is sublinear, continuous and “commutes with rotations”, i.e., for every ζ ∈ T satisﬁes the condition (T fζ )(ω) = (T f)(ωζ), ω ∈ T a.e., then T (X) ⊂ Lp,∞(T) and T is continuous as an operator from X to Lp,∞(T).

p p In the case where X is L (T) or L (D), we put fζ (z) = f(ζz). Proof. Let dµ(ζ) = |dζ|/2π. Theorem 2.5.2 guarantees that there is a set B so p that µ(B) > 0 and µ Aλ ∩ B 6 C (kfk/λ) , where Aλ = {ω ∈ T : |(T f)(ω)| > λ}, and C is independent of f and λ. If we put fζ (ζ ∈ T) instead of f and apply the −1 p hypotheses of the theorem, we get µ (ζ Aλ) ∩ B 6 C (kfk/λ) . Integrating this with respect to dµ(ζ) and using the formula Z −1 µ (ζ Aλ) ∩ B dµ(ζ) = µ(Aλ)µ(B), (2.26) T 2.6 Nikishin and Stein’s theorem 35

C kfkp we get µ(A ) , which was to be proved. λ 6 µ(B) λ To verify formula (2.26), we write the left-hand side as Z Z Z −1 µ ζ A ∩ B dµ(ζ) = dµ(ζ) χζ−1A(ω) dµ(ω) T T B (χ is the characteristic function), and then apply Fubini’s theorem together with the relation χζ−1A(ω) = χω−1A(ζ). 2

A theorem on multipliers As a nice application of the Nikishin/Stein theorem and the Marcinkiewicz theorem we have the following.

1 2.6.2 Theorem Let T : L (T) 7→ L0(T) be a continuous linear operator that commutes with rotations. Then T is of weak type (1, 1) and of strong type (p, p) for every p ∈ (1, ∞), and there exists a bounded sequence mn (−∞ < n < ∞) such that ∞ iθ X inθ T f(e ) = mnfb(n)e (2.27) n=−∞ for every trigonometric polynomial f. Further, if f ∈ Lp, p > 1, then

Tc f(n) = mnfb(n) (2.28) for every integer n.

p Proof. Since L is of type p for 0 < p 6 2, Nikishin/Stein theorem tells us that the operator T with the above properties is of weak type (p, p) for p = 1 and p = 2. Hence, by Marcinkiewicz’s theorem, T is of strong type (p, p) for 1 < p < 2. To prove the rest, let g(w) = wn, w ∈ T, for a ﬁxed integer n. By the hypothesis, for every ζ ∈ T we have ζn(T g)(w) = (T g)(ζw) for a.e., w ∈ T. The function T g belongs to L1 because g ∈ L2 and T is of strong type (p, p) for p ∈ (1, 2). It follows that if φ ∈ L∞, then Z Z (T g)(w)φ(w) |dw| = ζ−n(T g)(ζw)φ(w) |dw| T T for every ζ ∈ T. Integrating this with respect to ζ and using Fubini’s theorem, we get

Z 1 Z Z (T g)(w)φ(w) |dw| = φ(w) |dw| ζ−n(T g)(ζw) |dζ| 2π T T T 1 Z Z = φ(w)wn |dw| ζ−n(T g)(ζ) |dζ|. 2π T T 36 2 Interpolation and maximal functions

Hence Z n −n n (T g)(w) = w ζ (T g)(ζ) |dζ| =: mnw for a.e. w ∈ T; T this proves formula (2.27). The validity of (2.28) can then be deduced from the Weierstrass theorem that the trigonometric polynomials are dense in Lp. It remains to prove that T is of strong type (q, q) for q > 2. By Marcinkiewicz’s theorem (or by the Riesz/Thorin theorem), we can assume that q > 2. Let f, g be trigonometric polynomials. Then, in view of (2.27), we have 1 Z π 1 Z π (T f)(eiθ)g(e−iθ) dθ = f(eiθ)(T g)(e−iθ) dθ. 2π −π 2π −π Using this and the fact that T is of strong type (p, p) for p = q/(q −1), we conclude that kT fkq 6 Ckfkq, (2.29) q where C is independent of f. Now let f ∈ L be arbitrary, and let fn be a sequence of trigonometric polynomials such that kfn − fkq → 0. The validity of (2.29) for trigonometric polynomials implies

kT fnkq 6 Cqkfkq. (2.30) 1 Since kfn −fk1 6 kfn −fkq and T is continuous from L to L0, we see that T fn → T f in measure; after extracting a subsequence we can assume that T fn → T f almost everywhere. Now Fatou’s lemma and (2.30) give kT fkq 6 Cqkfkq, which was to be proved. 2

2.7 Banach’s principle

The following fact, known as Banach’s principle, plays an important role in applications of Theorem 2.6.1 to maximal operators.

2.7.1 Theorem Let X be a quasi-Banach space, let Tn (n > 1) be a sequence of continuous linear operators from X to L0(Ω, µ), and let

Tmaxf(ω) := sup |Tnf(ω)| < ∞ n>1

for almost all ω ∈ Ω. Then the operator Tmax : X 7→ L0(Ω, µ) is continuous.

(‡) ∞ ∞ Proof. Let L0(` ) denote the set of all functions F = (f1, f2,... ):Ω 7→ ` with measurable coordinates. The following two facts, the proof of which is left to the reader, imply the validity of the theorem. (a) With the F -norm Z kF (ω)k ∞ dµ(ω), Ω 1 + kF (ω)k∞ (‡)See [7, theorem IV.5.7], where a proof is given based on Baire’s theorem. 2.7 Banach’s principle 37

∞ the set L0(` ) is an F -space. ∞ (b) The operator T g = (T1g, T2g, . . . ) maps X into L0(` ) and has closed graph. 2

Theorem on a.e. convergence 2.7.2 Theorem Suppose that the conditions of Theorem 2.7.1 are satisﬁed. If the limit lim Tnf(ω) := T f(ω) a.e. n→∞ exists and is ﬁnite for every f from a dense subset of X, then it exists for every f ∈ X, and T is continuous as an operator from X to L0.

Proof. Let X0 denote the dense subset. Consider the following sublinear operator on X :

Sf(ω) = lim sup |Tmf(ω) − Tnf(ω)| (ω ∈ Ω). m,n→∞

By Banach’s principle, this operator is continuous because Sf 6 2Tmaxf. By Lemma 2.5.3, we have

µ(Sf, ε) 6 c(ε/kfk)(ε > 0, f ∈ X), (2.31) where c(λ) → 0 as λ → ∞. On the other hand, since Sg = 0 for g ∈ X0, we have S(f) = S(f − g) for all f ∈ X, g ∈ X0. From this and (2.31) it follows that

µ(Sf, ε) 6 c(ε/kf − gkk)(ε > 0), where, for a ﬁxed f ∈ X, we have chosen a sequence gk ∈ X0 so that kf − gkk → 0 (k → ∞). Thus µ(Sf, ε) = 0 for every ε > 0. The result follows. 2 3 Poisson integral

P inθ Harmonic functions occur in Fourier analysis naturally. If cne is the 1 iθ P∞ |n| inθ Fourier series of a function φ ∈ L (T), then the function f(re ) = −∞ cnr e is harmonic in D. The function f = P [φ] is called the Poisson integral of φ. In this chapter we are mainly concerned with some classical results on radial limits of P [φ], or, what is the same, on the Abel summability of the Fourier series p of φ, where φ belongs to L (T), 1 6 p 6 ∞, or C(T). However, there are much stronger results in this direction, and the reader should consult Zygmund’s book. For example, in many cases we can replace Abel’s method of summability by the method (C, α), α > 0. But, in contrast to the case of (C, α), Abel’s method leads us to considering not only the radial but also the non-tangential limits of f(z). At the end we prove the Littlewood/Paley inequality and a variant of Schwarz lemma for harmonic functions (Theorems 3.5.1 and 3.6.1)

3.1 Harmonic functions

A complex-valued function f, deﬁned on a domain Ω ⊂ C, is said to be harmonic if it is of class C2 and ∆f ≡ 0 in Ω, where ∆ denotes the Laplacian, ∂2f ∂2f ∂2f ∆f = + = 4 (z = x + iy); ∂x2 ∂y2 ∂z∂z¯ here ∂f 1 ∂f ∂f ∂f 1 ∂f ∂f = ∂f(z) = − i , = ∂f¯ (z) = + i . ∂z 2 ∂x ∂y ∂z¯ 2 ∂x ∂y Thus f is harmonic iﬀ the function ∂f/∂z is analytic. This implies the following.

3.1.1 Theorem A harmonic function f deﬁned on a simply connected domain Ω can be represented in the form f(z) = h(z)+g(z), z ∈ Ω, where h and g are analytic and uniquely determined up to an additive constant; conversely, if f = h+g ¯, where h and g are analytic, then f is harmonic.

Using this theorem one can deduce various properties of harmonic functions from the corresponding properties of analytic functions. For instance, the composition of a harmonic function with an analytic function is harmonic.

Uniqueness theorem As a further consequence of Theorem 3.1.1 we have: If f is harmonic in a simply connected domain Ω and f = 0 in an open subset of Ω, then f = 0 in Ω.

38 3.1 Harmonic functions 39

Series expansion If f is harmonic in D = {z : |z| < R}, then there are unique iθ P∞ |k| ikθ complex numbers fb(k), −∞ < k < ∞, such that f(re ) = k=−∞ fb(k)r e , with the series converging uniformly and absolutely on every compact subset of D. Conversely, if such a series converges in D, then it converges uniformly on compact subsets of D and its sum is harmonic in D.

Heine/Borel property The set of all functions harmonic in a domain Ω and endowed with the topology of uniform convergence on compact subsets of Ω is denoted by h(Ω). By H(Ω) we denote the subspace of h(Ω) consisting of analytic functions. Both h(Ω) and H(Ω) are complete and have the Heine/Borel property. The latter means:

If a sequence fn ∈ h(Ω), resp. fn ∈ H(Ω), is uniformly bounded on compact subsets, then there is a subsequence tending uniformly on compact subsets to a harmonic, resp. analytic, function.

Mean value property A characteristic property of harmonic functions is the mean value property on circles: Z π 1 iθ f(a) = f(a + Re ) dθ, {z : |z − a| 6 R} ⊂ Ω. (3.1) 2π −π

Green’s formula If f is harmonic, then (3.1) can be deduced from, say, Green’s formula, a special case of which can be stated as follows: If F is a C2-function in D = {z : |z| < R}, then

d 1 Z π 1 Z F (reiθ) dθ = (∆F )(z) dm(z) (3.2) dr 2π −π 2πr |z|

Exercises

3.1.2 If f is harmonic in D and f(z) = g(|z|) in a “rectangle” r1 < |z| < r2, θ1 < arg z < θ2, then f = const in D. This can be shown by using the formula ∂2f ∂f ∂2f ∆f = + + . ∂r2 r ∂r r2 ∂θ2 40 3 Poisson integral

3.1.3 The mean value of f over a rectiﬁable Jordan curve Γ is deﬁned as

1 Z I(f, Γ) = f(z) |dz|, where |Γ| = length of Γ, |Γ| Γ provided the integral is somehow deﬁned. If [a, b] is the straight line joining a and b and f is analytic in a domain containing Γ, then

F (b) − F (a) I(f, [a, b]) = , b − a where F is a primitive function of f. Using this one can prove the following: Let Πn be a regular polygon centered at c with vertices a1, a2, . . . , an, n > 4. Let f be a function harmonic in a domain containing the curve Πn and its interior. Then n X I(f, Πn) = I(f, [c, ak]). k=1

In particular, I(f, Π4) = I(f, d1) + I(f, d2), where d1, d2 are the diagonals of the square. For the case n = 4 see [7, Ch. 5, Lemma 6.12].

3.1.4 (maximum modulus principle) If f ∈ h(Ω), where Ω is simply connected, and |f| = const in Ω, then f = const.

The Poisson integral of a continuous function Poisson kernel One of everywhere occurring harmonic functions is the Poisson kernel: 1 − |z|2 1 + z P (z) = = Re . |1 − z|2 1 − z There hold the formulas 1 − r2 P (r, θ) : = P (reiθ) = 1 + r2 − 2r cos θ ∞ ∞ (3.4) X X = r|n|einθ = 1 + 2 rn cos nθ, n=−∞ n=1

1 Z π P (r, t) > 0 and P (r, t) dt = 1. (3.5) 2π −π The Poisson kernel has the reproducing property:

3.1.5 Theorem If a function f continuous on D is harmonic in D, then Z π iθ 1 it iθ f(re ) = P (r, θ − t)f(e ) dt (re ∈ D). (3.6) 2π −π 3.1 Harmonic functions 41

Proof. The equality 1 Z f(0) = f(ζ) |dζ| ( = ∂ ) 2π T D T follows from the mean value property on the circles |z| = ρ < 1 and the continuity of f. Applying this equality to the harmonic function f ◦ ϕa, where a − z ϕ (z) = , |a| < 1, |z| 1, a 1 − az¯ 6 we get 1 Z f(a) = f(ϕ (ζ)) |dζ|. 2π a T Now introduce the substitution ζ = ϕa(w), i.e., w = ϕa(ζ). Since 1 − |a|2 |dζ| = |dw|, |1 − aw¯ |2 we get 1 Z f(a) = P (¯aw)f(w) |dw|, 2π T which is another form of (3.6). 2

The Poisson integral of a function The Poisson integral of a function φ ∈ L1(T) is the harmonic function P [φ] deﬁned by Z π iθ 1 it iθ P [φ](re ) = P (r, θ − t)φ(e ) dt (re ∈ D). 2π −π The proof of Theorem 3.1.5 shows that 1 Z P [φ](z) = φ(ϕ (w)) |dw| (z ∈ ). (3.7) 2π z D T The Poisson integral can be used to solve the Dirichlet problem for the disk:

3.1.6 Theorem If φ is a continuous function deﬁned on T, then φ has a unique continuous extension to D that is harmonic in D; this extension equals P [φ]. An immediate consequence is the well known Weierstrass theorem on approximation by trigonometric polynomials: The set of all trigonometric polynomials is dense in each of the spaces C(T), Lp(T) (0 < p < ∞). Proof of Theorem 3.1.6. The uniqueness follows from Theorem 3.1.5. Let f = P [φ]. From (3.7) and the continuity of the function φ, by using, for instance, the dominated convergence theorem, we get limD3 z→1 f(z) = φ(1), which proves that the extension is continuous at the point z = 1; and so on. 2 42 3 Poisson integral

The above proof, although very short, has a disadvantage in that it is based on very special properties of the Poisson kernel. The standard proof depends on (3.5) and the following: lim sup P (r, t) = 0 for all δ ∈ (0, π) (see [86, 100, 22, 46]). r→1 δ<|t|<π Theorems 3.1.5 and 3.1.6 yield the following.

3.1.7 Theorem The Poisson integral acts as an isometric isomorphism from C(T) onto hC(D) ⊂ L∞(D), the space of functions that are deﬁned and continuous on D and harmonic in D.

3.2 Borel measures and the space h1

The space M(T) Let M(T) denote the space of all complex Borel measures on the circle T; the norm is given by kµk = (1/2π)|µ|(T), where |µ| is the variation of µ. If dµ(eiθ) = φ(eiθ) dθ 1 with φ ∈ L (T), then kµk = kφk1, which means in particular that M(T) contains an isometric copy of L1(T). By one of Riesz’ representation theorems, M(T) is isometrically isomorphic to the dual of C(T) with respect to the bilinear form 1 Z (ϕ, µ) = ϕ(ζ¯) dµ(ζ), ϕ ∈ C( ), µ ∈ M( ). 2π T T T (See [86, Theorem 6.19] for a general representation theorem.)

The Poisson integral of a measure The Poisson integral of µ ∈ M(T) is defined by 1 Z P [µ](z) = P (zζ¯) dµ(ζ). (3.8) 2π T Again, if dµ(eiθ) = φ(eiθ) dθ with φ ∈ L1(T), then P [µ] = P [φ]. 3.2.1 Proposition The coefficients of the function P [µ] are equal to the corresponding coefficients of the measure µ, i.e., there holds Z π 1 −int it Pd[µ](n) = µb(n) := e dµ(e ). 2π −π Proof. The formula can be deduced from (3.8) by term-by-term integration of the series (3.4) with respect to µ; this is possible because the series (3.4) converges uniformly on the circles |z| = r < 1. 2

3.2.2 Theorem The Poisson integral acts as a continuous and injective linear operator from M(T) into h(D). Proof. To prove the injectivity assume P [µ] = 0. From Proposition 3.2.1 we get µb(n) = 0 for every n, and hence (φ, µ) = 0 for every trigonometric polynomial φ. And since trigonometric polynomials are dense in C(T) we have (φ, µ) = 0 for every φ ∈ C(T). Now the Riesz representation theorem gives µ = 0. 2 3.2 Borel measures and the space h1 43

The Riesz/Herglotz theorem

The space h1 consists of the functions f ∈ h(D) for which Z π 1 iθ kfk1 := sup |f(re )| dθ < ∞. r<1 2π −π

The Poisson integral of a measure µ ∈ M(T) belongs to h1 because

k P [µ] k1 6 kµk, (3.9) which is deduced from (3.5). The converse is true as well: every function belonging to h1 is equal to the Poisson integral of some (unique) measure. More precisely:

3.2.3 Theorem (Riesz/Herglotz) The Poisson integral acts as an isometric isomorphism from M(T) onto h1.

1 Proof. Let f ∈ h . Then the “sequence” fr (r → 1), where fr(ζ) = f(rζ) for ζ ∈ T, is bounded in L1(T). This means that the “sequence” of measures iθ iθ dµr(e ) = f(re ) dθ is bounded in M(T). Since M(T) is the dual of C(T), we can apply the Banach/Alaoglu theorem; there exists a sequence rn → 1 such that µrn converges weakly (star) to a measure µ ∈ M(T), which means that Z π Z π 1 it −it 1 −it it f(rne )g(e ) dt → g(e ) dµ(e ) 2π −π 2π −π for every g ∈ C(T). Take g(e−it) = P (ze−it), with ﬁxed z ∈ D. Then Z π 1 it −it f(rne )g(e ) dt = f(rnz) 2π −π because the function f(rnz) is harmonic in a neighborhood of the closed disk. From the last two relations we get f(z) = P [µ](z). And because of the weak convergence we have kµk 6 lim infn kµrn k = lim infn kfrn k1. Hence kµk 6 kfk1, which along with (3.9) implies the desired result. 2

3.2.4 Exercise The closure in h1 of the set of all harmonic polynomials is equal to {P [φ]: φ ∈ L1}.

Positive harmonic functions

A positive real function f ∈ h(D) belongs to h1 because 1 Z π |f(reit)| dt = f(0). 2π −π

From the proof of Theorem 3.2.3 we obtain the following variant of the Riesz/Her- glotz theorem. 44 3 Poisson integral

3.2.5 Theorem (a) If a function f ∈ h(D) is real-valued and positive, then f is equal to the Poisson integral of a finite positive measure. (b) A real harmonic function belongs to h1 iff it is equal to the difference of two positive harmonic functions.

The following fact is a consequence of Theorem 3.2.5(a) and the inequality 1 − r 1 + r P (r, θ) . 1 + r 6 6 1 − r

3.2.6 Theorem (Harnack’s inequality) If u is a function harmonic and positive in the disk |z| < R, then R − r R + r u(0) u(reiθ) u(0) (0 < r < R) R + r 6 6 R − r and |∇u(0)| 6 2u(0)/R. 3.2.7 Corollary If u is a positive harmonic function in a disk D, and K a compact subset of D, then u(z)/u(w) 6 C for z, w ∈ K, where C is a constant depending only on K.

3.2.8 Corollary (Hurwitz) If {fn} is a sequence of analytic functions with no zeros in D such that fn → f uniformly on compact subsets, then f has no zeros in D unless f vanishes identically.

Proof. Let D be a closed disk contained in D. Then there is a constant M such that supD |fn| < M for all n. Now we put un = log(M/|fn|) and apply Corollary 3.2.7. 2

3.2.9 Theorem (Harnack) Let {un} be an increasing sequence of real-valued harmonic functions deﬁned in a disk D ⊂ C. If there exists a point a ∈ D such that un(a) converges, then {un} converges uniformly on compact subsets of D. Proof. Let K ⊂ D be a compact set containing a. By Corollary 3.2.7 we have |um(z) − un(z)| 6 C|um(a) − un(a)|, where C is independent of m, n and z. 2

Poisson/Stieltjes integral Instead with measures, it is sometimes more convenient to work with functions of bounded variations. Let BV [a, b] denote the set of functions of bounded variation on [a, b]. The Poisson/Stieltjes integral of a function γ ∈ BV = BV [−π, π] is deﬁned as to be the harmonic function 1 Z π PS[γ](reiθ) = P (r, θ − t) dγ(t). (3.10) 2π −π In view to the well known connection between Borel measures and functions of bounded variation, the above assertions can be stated as follows. 3.3 Radial limits of the Poisson integral 45

3.2.10 Theorem A function f ∈ h(D) belongs to h1 iﬀ f is equal to the Pois- son/Stieltjes integral of a function γ ∈ BV . A positive harmonic function is equal to the Poisson/Stieltjes integral of an increasing function.

It is worthwhile to note that (3.10) can be rewritten in the form 1 Z π PS[γ](reiθ) = k · P (r, θ + π) + P 0(r, θ − t)γ(t) dt, (3.11) 2π −π γ(π) − γ(−π) where k = and 2π ∂P −2r sin t P 0(r, t) = = P (r, t), (3.12) ∂t 1 + r2 − 2r cos t or in the form ∂ PS[γ](reiθ) = k · P (r, θ + π) + P [γ](reiθ). (3.13) ∂θ

Exercises

3.2.11 Let f be analytic in D and Re f > 0. Then there exists a positive measure µ ∈ M(T) and a real constant c such that 1 Z π ζ + z f(z) = i Im f(0) + dµ(ζ)(z ∈ D). 2π −π ζ − z 3.2.12 Let u ∈ h(D) be a real-valued function such that |u(z)| 6 1 u D. Then 1 − |u(z)| |∇u(z)| 2 (z ∈ ). 6 1 − |z| D The assertion does not hold for complex-valued functions even if we replace “2” by an arbitrary constant.

3.2.13 Let {un} be a sequence of positive harmonic functions in a domain D ⊂ C. If it is not true that un → +∞ uniformly on compact subsets of D, then there exists a subsequence of {un} converging uniformly on compact subsets of D.

3.3 Radial limits of the Poisson integral

By the Jordan/Dirichlet test from Fourier analysis, the Fourier series of a 2π- periodic function γ of bounded variation on [−π, π] converges everywhere and its sum is equal to γ(θ + 0) + γ(θ − 0)/2 . This implies, via Abel’s theorem, that the iθ limit limr→1 P [γ](re ) exists everywhere and has the same value, so one can expect that ∂ lim P [γ](reiθ) = γ0(θ), r→1− ∂θ provided the derivative γ0 exists. It follows from (3.13) and the following assertion that this is true. 46 3 Poisson integral

3.3.1 Proposition If a function γ ∈ BV [−π, π] has the ﬁnite derivative at θ ∈ (−π, π), then lim PS[γ](reiθ) = γ0(θ). (3.14) r→1−

Proof. Let θ = 0 and γ0(θ) = 0. From (3.11) it follows that

Z δ 1 0 L := lim sup |PS[γ](r)| 6 lim sup |P (r, t)| · |γ(t)| dt r→1− r→1− 2π −δ

for every (small) δ > 0; this is so because, according to (3.12),

0 |P (r, t)| 6 C(δ)(1 − r)(|t| > δ),

whence Z lim sup |P 0(r, t)| · |γ(t)| dt = 0. r→1− |t|>δ

0 Now we apply the inequality |t P (r, t)| 6 2P (r, t), |t| 6 π, to obtain

1 Z δ γ(t)

L 6 lim sup P (r, t) dt. r→1− π −δ t

Finally, (3.14) follows from this and the hypothesis γ(t)/t → 0, t → 0. 2

3.3.2 (symmetric derivative) In Proposition 3.3.1, the ordinary derivative γ(θ + t) − γ(θ − t) γ0(θ) can be replaced by the symmetric derivative lim . t→0 2t

Fatou’s theorem Proposition 3.3.1 is of elementary character and only deeper results on diﬀerentia- bility of functions give deeper results on the existence of radial limits. Namely, by the Lebesgue theorem, the derivative γ0(θ) exists for almost all θ (see [86]), which along with Proposition 3.3.1 shows that the Poisson/Stieltjes integral has ﬁnite radial limits almost everywhere. Returning to the Poisson integral of a measure, we can state a more precise form of this result as follows.(∗)

3.3.3 Theorem (Fatou) If µ ∈ M(T), then P [µ] has radial limits almost everywhere. Besides lim P [µ](reiθ) = φ(eiθ) r→1− for almost all θ, where φ(eiθ) dθ is the absolutely continuous part of µ.

(∗)Rudin considers the case of “M-harmonic” functions on the complex ball, [84, Ch. V]. Much more information on the classical case can be found in Garnett [22, Ch. I] and, of course, Zygmund [100, Ch. III]. 3.3 Radial limits of the Poisson integral 47

Recall that the measure µ is uniquely represented as dµ(eiθ) = φ(eiθ) dθ + iθ dµs(e ), where φ is an integrable function and µs is a singular measure (the theorem of Lebesgue and Radon/Nikodym). As a special case of Theorem 3.3.3 we have:

3.3.4 Corollary Every bounded harmonic function on D has radial limits almost everywhere.

On the other hand, this special case is suﬃcient to prove the qualitative part of Fatou’s theorem, i.e., to prove the existence of the limits for h1-functions (see the proof of Corollary 3.3.9). Another special case:

3.3.5 Corollary A measure µ ∈ M( ) is singular iﬀ lim P [µ](reiθ) = 0 almost T − everywhere. r→1

3.3.6 Exercise If u is the Poisson integral of a singular measure, then Z 2π lim |u(reiθ)|p dθ = 0 for 0 < p < 1. r→1 0

Theorems of Kolmogorov and Carleson iθ P∞ |n| inθ Since P [φ](re ) = n=−∞ φb(n)r e , Theorem 3.3.3 shows that the Fourier series of an integrable function φ is almost everywhere summable by the Abel/Poisson method with the sum equal to φ. In connection with this, let us mention that Kolmogorov proved the existence of an everywhere divergent Fourier series (cf. [44, 100]). Carleson [12] proved that the Fourier series of an L2-function converges almost everywhere, which is generalized to the case p > 1 by Hunt [31].

Nontangential limits Slightly modifying the proof of Proposition 3.3.1 one can prove the “nontangential” variant of Fatou’s theorem:

3.3.7 Theorem If f ∈ h1, then for almost all ζ ∈ T there exists the limit ^ lim f(z) := lim f(z), z→ζ Uζ 3z→ζ where Uζ is the convex envelope of the union {z : |z| < ρ} ∪ {ζ}, and ρ < 1 is ﬁxed.

(0 < δ < 1). For a given ε > 0, take δ so that |γ(t)/t| < ε for |t| < δ, whence

Z δ Z δ 0 0 |P (r, θ − t)| · |γ(t)| dt 6 ε |t P (r, θ − t)| dt. −δ −δ

0 Finally, by using the property (3.15) we ﬁnd |t P (r, θ − t)| 6 const P (r, θ − t) iθ (re ∈ U1); thus

1 Z π L 6 Cε P (r, θ − t) dt = Cε, etc. 2 2π −π

Radial =⇒ nontangential

The above theorem can also be deduced from the following.

3.3.8 Theorem Suppose f is bounded and analytic in D, and ζ ∈ T. Then the existence of the radial limit of f at ζ implies the existence of the nontangential limit at ζ.

3.3.9 Corollary If f ∈ H(D) and u = Re f ∈ h1, then f has nontangential limits at almost every point ζ ∈ T.

Proof of Corollary. We can assume that u is positive. Then the function 1/(1 + f) is analytic and bounded in D. The result follows. 2

Proof of Theorem. Let ζ = 1 and limr→1 f(r) = 0. If f fails to have the limit 0 within Uζ , then there exist ε > 0 and sequences tn → 0, 0 < tn < 1, and wn, |wn| < ρ0 < 1, such that

|f(tnwn + 1 − tn)| > ε for all n. (3.16)

Consider the functions fn(w) = f(tnw + 1 − tn), w ∈ D. The sequence fn is uniformly bounded in D and therefore there exists a subsequence, denote it by fn, that converges to a function g ∈ H(D) uniformly on |w| < ρ0; this means that fn(wn)−g(wn) → 0. By the hypothesis, we have f(tnr +1−tn) → 0 for 0 < r < 1, which implies that g(r) = 0 for 0 < r < 1, whence g(w) = 0 for all w ∈ D. Thus fn(wn) → 0, which contradicts (3.16). 2

Lindel¨of’s theorem Theorem 3.3.8 is a special case of Lindel¨of’s theorem: Let f ∈ H∞(D) and let γ : [0, 1) 7→ D be a continuous curve such that γ(t) → 1 as t → 1. If there exists limt→1 f(γ(t)) = L, then f has nontangential limit L at the point 1. For the proof see [84, Theorem 8.4.1]. 3.4 The spaces hp and Lp(T) 49

3.4 The spaces hp and Lp(T) p The harmonic Hardy space h (0 < p 6 ∞) is deﬁned as p h = { f ∈ h(D): kfkp = sup Mp(r, f) < ∞ }, r<1 where Mp(r, f) is the integral mean of order p,

Z π 1/p 1 iθ p Mp(r, f) = |f(re )| dθ . 2π −π In the case p = ∞ the integral is to be interpreted as a supremum:

iθ M∞(r, f) = sup |f(re )|. θ∈[−π,π]

Therefore h∞ = h∞(D) is the subspace of L∞(D) spanned by harmonic functions. 2 P∞ 2 2|n| By Parseval’s formula we have M2 (r, f) = n=−∞ |fb(n)| r , and consequently ∞ 1/2 X 2 kfk2 = |fb(n)| . n=−∞

3.4.1 Proposition If p > 1, and f ∈ h(D), then Mp(r, f) increases with r. More- over, if p > 1, then Mp(r, f) is strictly increasing unless f = const.

It should be noted, however, that if p < 1 and f is positive, then Mp(r, f) is decreasing. Proof. The “increasing” property can be deduced from the subharmonicity of the function |f|p (see Theorem 4.2.1), or by application of Minkowski’s inequality (†) in continuous form (“norm of integral 6 integral of norm”) to the formula Z π iθ 1 f(λre ) = P (λ, t)fr(θ + t) dt (0 < λ < 1), 2π −π

it where fr(t) = f(re ). To prove the second assertion, let p > 1 and suppose that Mp(r1, f) = Mp(r2, f) for some 0 6 r1 < r2 < 1. Let Z π 1 it p p u(z) = |f(ze )| dt = Mp (|z|, f), |z| < r2. 2π −π

The function u is subharmonic in the disk |z| < r2 and attains its maximum at z = r1. It follows that u=const, by the maximum principle (see Theorem 4.1.7), whence 1 Z π |f(0)|p = |f(reit)|p dt, 0 < r < ρ. 2π −π

(†) The case p = ∞ is trivial, while in the case 1 6 p < ∞ we can apply Jensen’s inequality as well. 50 3 Poisson integral

Since 1 Z π p 1 Z π p it it p |f(0)| = f(re ) dt 6 |f(re )| dt 2π −π 2π −π (by Jensen’s inequality or by H¨older’s inequality), we see that

1 Z π p 1 Z π it it p f(re ) dt = |f(re )| dt, r < r2. 2π −π 2π −π

it This implies that f(re ) depends only on r, when r < r2, and therefore f(z) = const for |z| < 1; see 3.1.2. 2

Radial limits Since ∞ 1/2 X 2 kφk2 = |φb(n)| , n=−∞

where φ ∈ L2, as well as fb(n) = φb(n) provided f = P [φ], the Poisson integral acts as an isometric isomorphism from L2 onto h2. It is very important that this fact extends to the case 1 < p 6 ∞. On the other hand, as we have seen, the operator P : L1 7→ h1 is not onto.

p 3.4.2 Theorem The function f belongs to h (1 < p 6 ∞) iﬀ it is equal to the Poisson integral of some function φ ∈ Lp(T). And if f = P [φ], φ ∈ Lp(T), then

kfkp = kφkp (1 6 p 6 ∞),

Z π 1 iθ iθ p lim |f(re ) − φ(e )| dθ = 0 (1 6 p < ∞) (3.17) r→1− 2π −π and lim f(reiθ) = φ(eiθ) almost everywhere. r→1− The proof is very similar to the proof of Theorem3.2.3 and will be omitted here. The Poisson kernel shows that an h1-function need not be equal to the Poisson integral of the boundary function. However, (3.17) implies the following.

p iθ iθ 3.4.3 Corollary If f ∈ h , p > 1, then f = P [f∗], where f∗(e ) = lim f(re ). r→1−

The Poisson kernel also shows that boundedness of f∗ does not imply boundedness of f. However, if p > 1, we have the following.

iθ 3.4.4 Corollary Let a function f ∈ h(D) have radial limits f∗(e ) almost every- ∞ p ∞ where and f∗ ∈ L (T). If f ∈ h for some p > 1, then f ∈ h and kfk∞ = kf∗k∞. 3.5 The Littlewood/Paley theorem 51

Exercises

p q 3.4.5 [72] The Poisson kernel satisﬁes: Mp (r, P ) = Mq (r, P )(q = 1 − p) and

Mp(r, P ) (1 − r), for 0 1/2.

p 3.4.6 The inclusion h ⊂ h(D) (1 6 p 6 ∞) is compact, i.e., every closed ball of the space hp are compact in the topology of uniform convergence on compact subsets of D.

3.4.7 Let z ∈ D. The norm of the linear functional z 7→ f(z) on the space hp 2 −1/p (1 6 p < ∞) is equal to Kp(|z|)(1 − |z| ) , where Z π 1/q 1 it 2q−2 Kp(r) = |1 − re | dt (1/p + 1/q = 1). 2π −π 2 1/2 Observe that K2(r) = (1 + r ) .

3.5 The Littlewood/Paley theorem

Here we consider the simplest variant of the Littlewood/Paley theorem.

p 3.5.1 Theorem (a) If u ∈ h , 2 6 p < ∞, then Z K := |∇u(z)|p(1 − |z|)p−1 dA(z) < ∞ (3.18) D and there holds the inequality Z p p−1 p |∇u(z)| (1 − |z|) dA(z) 6 Cpkukp. D (b) If u is harmonic in D and satisﬁes condition (3.18) for some 1 < p < 2, then p p p u ∈ h and we have kukp 6 Cp (K + |u(0)| ).

The case p > 2 There are many proofs of (a). For example, it is possible to apply the Riesz/Thorin interpolation theorem; namely, the operators ∂u ∂u (S u)(z) = (1 − |z|) and (S u)(z) = (1 − |z|) 1 ∂z 2 ∂z¯ map hp into Lp(D, dA/(1 − |z|)) for p = ∞ and p = 2. An elementary but rather long proof, based on local estimates deduced from the Hardy/Stein identity, was found by Luecking [55]. We shall present a proof that is both elementary and short. We only need three simple lemmas on positive harmonic functions. 52 3 Poisson integral

2u(z) 3.5.2 Lemma If u is a positive harmonic function in , then |∇u(z)| . D 6 1 − |z|

Proof. Applying the inequality |∇u(0)| 6 2u(0) (Theorem 3.2.6) to the function w 7→ u(z + (1 − |z|)w), we get the result. 2

3.5.3 Lemma If u ∈ hp is a real-valued function and 1 < p < ∞, then there are p nonnegative functions h1 and h2 from h such that p p p u = h1 − h2, kukp = kh1kp + kh2kp.

Proof. Let g1(ζ) = max{u(ζ), 0} and g2(ζ) = max{−u(ζ), 0} for ζ ∈ T. Then, because of Theorem 3.4.2, the required conditions are satisﬁed by the functions h1 = P [g1] and h2 = P [g2]. 2

3.5.4 Lemma Let u > 0 belong to hp, 1 < p < ∞. Then p(p − 1) Z 1 kukp = |u(0)|p + up−2|∇u(z)|2 log dA. p 2 |z| D Proof. This is easily deduced from Green’s formula and the formula ∆(up) = p(p − 1)up−2|∇u|2. 2

Proof of Theorem 3.5.1(a). It is enough to consider real-valued functions. Because of Lemma 3.5.3, we can suppose that u is positive. Then Lemmas 3.5.4 and 3.5.2 give p2 − p Z kukp |∇u|222−p|∇u|p−2(1 − |z|)p−1 dA, p > 2 D which implies the desired conclusion.

The case 1 < p < 2

p We shall apply the method of “dualization.” Let Xp be the (real) subspace of h consisting of real-valued functions. Let Yp be the space of real-valued harmonic functions that satisfy (3.18) or, equivalently, Z 1 p−1 |∇u(z)|p log dA(z) < ∞; |z| D the norm is introduced in the obvious way. From the proof of (a) it follows that Xq ⊂ Yq (1/q+1/p = 1), the inclusion being continuous. From this we can conclude ∗ ∗ that Yp ⊂ Xp provided that we have the inclusions Yp ⊂ (Yq) and (Xq) ⊂ Xp with respect to the same bilinear form. The results of Section 3.4 can be formulated p p ∗ in terms of real L and h spaces. As a consequence we we get (Xq) = Xp with respect to the bilinear form Z π 1 iθ iθ (u, v)0 = u(e ) v(e ) dθ. 2π −π 3.6 Harmonic Schwarz lemma 53

∗ On the other hand, by Hölder’s inequality, we have Yp ⊂ (Yq) with respect to a different form, namely Z 1 (u, v) = u(0)v(0) + ∇u(z) · ∇v(z) log dA(z). 1 |z| D Fortunately, from Green’s formula and the equality ∆(uv) = 2∇u · ∇v it follows that (u, v)0 = (u, v)1 provided that u and v are harmonic in a neighborhood of the closed disk. Then, by Hölder’s inequality, we get

|(u, v)0| = |(u, v)1| 6 kukYp kvkYq .

Now we use assertion (a) to get |(u, v)0| 6 CkukYp kvkXq , where C depends only on q. And since sup{|(u, v)0| : kvkXq 6 1} = kukXp , we see that kukp 6 CkukYp , provided the function u is harmonic in a neighborhood of the closed disk. In the general case, we apply this inequality to the functions u(ρz), ρ → 1, and this completes the proof for 1 < p < 2.

3.6 Harmonic Schwarz lemma

If f ∈ h(D), |f| 6 1, and f(0) = 0, then there holds the sharp inequality 4 |f(z)| arctan |z| . 6 π See [6]; see also 4.4.6.

3.6.1 Theorem Let f be a complex-valued function harmonic in D such that |f(z)| 6 1 for z ∈ D. Then there holds the inequality 1 − |z|2 4 f(z) − f(0) arctan |z|. (3.19) 1 + |z|2 6 π

If equality occurs for some z ∈ D r {0}, then there are real constants α and β such that f(z) = eiακ(eiβz) for all z ∈ D, where 2 1 − z 2 2r sin θ κ(z) = arg = arctan (z = reiθ). π 1 + z π 1 − r2 Proof. We start from the formula 2 Z π 2 2 1 − r 1 1 − r 1 − r iθ f(r) − 2 f(0) = 2 − 2 f∗(e ) dθ 1 + r 2π −π 1 + r − 2r cos θ 1 + r 2 Z π (1 − r )2r 1 cos θ iθ = 2 2 f∗(e ) dθ. 1 + r 2π −π 1 + r − 2r cos θ From this and the hypothesis |f| 6 1 we get 1 − r2 (1 − r2)2r 1 Z π | cos θ| f(r) − f(0) dθ. 2 6 2 2 1 + r 1 + r 2π −π 1 + r − 2r cos θ 54 3 Poisson integral

So we have to compute the integral

1 Z π | cos θ| J = 2 dθ. 2π −π 1 + r − 2r cos θ We have 1 Z π/2 cos θ cos θ J = 2 + 2 dθ 2π −π/2 1 + r − 2r cos θ 1 + r + 2r cos θ

1 Z π/2 2(1 + r2) cos θ = 2 2 2 2 dθ 2π −π/2 (1 + r ) − 4r cos θ

1 Z π/2 2(1 + r2) cos θ = dθ 2 2 2 2 π 0 (1 − r ) + 4r sin θ Since Z 2(1 + r2) cos θ 1 + r2 2r sin θ dθ = arctan , (1 − r2)2 + 4r2 sin2 θ r(1 − r2) 1 − r2 we see that 1 + r2 2r 1 + r2 J = arctan = 2 arctan r. r(1 − r2) 1 − r2 r(1 − r2) Combining all these results we get (3.19) for z = r. If equality holds for a ﬁxed z = r > 0, then, as the above inequalities show, we iθ iα have |f∗| = 1 a.e., and f∗(e ) cos θ = e g(θ), where α is a constant and g > 0. It follows that ( −iα iθ 1, θ ∈ (−π/2, π/2), e f∗(e ) = −1, θ ∈ (π/2, 3π/2). Hence Z π iθ 1 iθ f(re ) = P (r, θ − t)f∗(e ) dt 2π −π

1 Z π/2 (1 − r2)4r cos(t − θ) = eiα dθ 2 2 2 2 2π −π/2 (1 − r ) + 4r sin θ

2 2r cos θ = eiα arctan = eiακ(ireiθ). 2 π 1 − r2 ¯ 3.6.2 With the hypotheses of Theorem 3.6.1 we have |∂f(0)|+|∂f(0)| 6 4/π with equality iﬀ f(z) ≡ eiακ(eiβz) for some α, β ∈ R. Therefore, if f is in addition real-valued, then |∇f(0)| 6 4/π. 4 Subharmonic functions

2 2 A real-valued C -function u is subharmonic iff ∆u > 0. If u is not of class C , then u is subharmonic iff it is the limit of a decreasing sequence of subharmonic functions of class C2. The importance of subharmonic functions for spaces of analytic and harmonic functions lies in the fact that if f is analytic (resp. harmonic), p then |f| is subharmonic for every p > 0 (resp. p > 1). This chapter contains concise proofs of the basic properties such as the maximum principle, local integrability, approximation by smooth functions, the subordination principle. The discussion of the integral means (Sections 4.2, 4.3) follows Hörmander’s book [30](∗) and in- cludes Prawitz’ theorem 4.3.1 and, as consequences, Koebe’s one-quarter theorem, and Bieberbach’s theorem. In Section 4.5 we prove a weak version of Riesz’ representation theorem. Section 4.6 is devoted to the proof of a Littlewood/Paley theorem for subharmonic functions.

4.1 Basic properties

A function u : D 7→ [−∞, ∞), where D is a subdomain of the complex plane, is said to be subharmonic if it is upper semicontinuous, i.e.,

u(a) > lim sup u(z) for all a ∈ D, (4.1) z→a and for every a ∈ D there exists R > 0 such that {z : |z − a| < R} ⊂ D and Z π 1 iθ u(a) 6 u(a + ρe ) dθ for every 0 < ρ < R. (4.2) 2π −π Upper semicontinuity implies boundedness from above, which guarantees the existence of the integral in (4.2). From (4.2) it follows that for every a ∈ D there exists a sequence zn → a such that u(a) 6 u(zn), which implies lim supz→a u(z) = u(a). In particular, u is continuous at a if u(a) = −∞. There are discontinuous subharmonic functions; e.g., the function

∞ X log |z − 2−n| u(z) = 2n n=1 is subharmonic in the entire plane and is discontinuous at zero. Because of the mean value property, every real-valued harmonic function is both subharmonic and superharmonic.

(∗)The only diﬀerence is in the proof of Prawitz’ theorem.

55 56 4 Subharmonic functions

A function u is said to be superharmonic if −u is subharmonic. In the case of C2-functions there is a simple criterion of subharmonicity deduced from Green’s formula: 2 A function u ∈ C (D) is subharmonic iﬀ ∆u > 0 in D. From this and the formula ∆(u ◦ ϕ)(z) = (∆u)(ϕ(z)) |ϕ0(z)|2, where ϕ is an analytic function, we get: The composition u ◦ ϕ is subharmonic if u is subharmonic and ϕ is analytic. In the general case this assertion can be reduced to the “smooth” case by ap- proximating an arbitrary subharmonic function by smooth ones (Theorem 4.1.15, later on). An important example of a subharmonic function taking the value −∞ is the function log |z − a|. More generally:

4.1.1 Theorem If f is analytic in D, then the function log |f| is subharmonic in D, and |f|p is subharmonic for every p > 0.

New examples can be produced by using the following assertions:

4.1.2 Theorem The sum and the maximum of a ﬁnite sequence of subharmonic functions are subharmonic functions. The same holds for the limit of a decreasing sequence of subharmonic functions.

4.1.3 Theorem Let φ be an increasing convex function that is deﬁned and continuous on an interval I ⊂ [−∞, +∞). If v is subharmonic and takes its values in I, then the function u = φ(v) is subharmonic. In particular, u is subharmonic in the following cases: p (a) u = |h| , where p > 1 and h is harmonic; p (b) u = v , p > 1, where v is subharmonic and nonnegative.

Exercises

Pn k 4.1.4 [59] If p > 0 and f(z) = k=m akz , then for 0 < p < ∞ we have

n m r Mp(1, f) 6 Mp(r, f) 6 r Mp(1, f) 0 < r < 1.

4.1.5 Let h 6≡ 0 be a real-valued harmonic function and 0 < p < 1. The function |h|p is subharmonic iﬀ h is constant. The function |h|p is superharmonic iﬀ h has no zeros.

4.1.6 Let u1, . . . , un be a sequence of nonnegative functions subharmonic in D. If p p 1/p p > 1, then the function u := (u1 + ··· + un) is subharmonic in D. If uk = |fk|, where fk are analytic, then u is subharmonic for every p > 0. 4.1 Basic properties 57

The maximum principle The simplest variant of the maximum principle says:

4.1.7 Theorem A nonconstant subharmonic function cannot attain its maximum inside the domain. In particular, a nonconstant harmonic function attains neither the maximum nor the minimum inside the domain.

Proof. Let u be subharmonic in a domain D and let M denote the set of points in D where u attains its maximum. Because of the semicontinuity, M is closed. Let us prove that M is open as well, which will imply M = D or M = ∅, so the proof will be ﬁnished. Let a ∈ M. Then (4.2) implies that, for R small enough and for all ρ < R, we have u(a) = u(a + ρζ) almost everywhere on the circle |ζ| = 1. From this and (4.1) it follows that u(a) 6 u(a + ρζ) everywhere; thus, u(a) = u(a + ρζ) everywhere, i.e., {z : |z − a| < R} ⊂ M. 2

4.1.8 Corollary If u is upper semicontinuous on D and subharmonic in D, then

max{u(z): z ∈ D} = max{u(ζ): ζ ∈ ∂D}.

4.1.9 Corollary Let D be a bounded domain, let u be a function subharmonic in D and upper semicontinuous on D, and let h be a real-valued function harmonic in D and continuous on D. If u 6 h on ∂D, then u 6 h in D; besides, u < h if u is not harmonic.

4.1.10 Corollary If a function is both subharmonic and superharmonic, then it is harmonic.

Proof. If u and −u are subharmonic in D, then u is continuous. Let K be a disk, relatively compact in D, and let h be harmonic and h = u on ∂K. Then u 6 h and −u 6 −h and K. 2 Corollary 4.1.9 gives a partial explanation of the term “subharmonic.” Moreover, we have:

4.1.11 Theorem Let u be an upper semicontinuous function in a domain D, with values in [−∞, ∞). Then u is subharmonic iﬀ for each disk K with K ⊂ D, and each function h continuous on K and harmonic in K, the condition u 6 h on ∂K implies u 6 h in K.

Proof. Necessity follows from Corollary 4.1.8 applied to the function u − h. Let D ⊂ D and let u be an upper semicontinuous function satisfying the above condition. Let φ be an arbitrary continuous function on ∂D such that φ > u on ∂D. Then the function 1 Z π h(reiθ) = P (r, θ − t)φ(eit) dt 2π −π 58 4 Subharmonic functions

equals the continuous extension of φ to D (Theorem 3.1.6). Hence u 6 h in D and in particular u(0) 6 h(0), i.e., Z π 1 it u(0) 6 φ(e ) dt. 2π −π

Being upper semicontinuous, the function u(eit) is the inﬁmum of a family of continuous functions; therefore,

Z π 1 it u(0) 6 u(e ) dt. 2π −π

Applying this inequality to the functions z 7→ u(a + ρz), we ﬁnd that there holds (4.2), and this was to be proved. 2

Local integrability If u is subharmonic in D, then

1 Z u(a) 6 2 u(z) dm(z) < ∞, (4.3) πρ |z−a|<ρ

whenever {z : |z − a| 6 r} ⊂ D, and therefore u is integrable in a neighborhood of any point at which u has a ﬁnite value. Moreover, u is integrable near an arbitrary point:

4.1.12 Theorem Every subharmonic function u 6≡ −∞ is locally integrable.

4.1.13 Remark It follows from Theorem 4.5.2 that |u|p is locally integrable for every p < ∞.

4.1.14 Corollary If u 6≡ −∞ is subharmonic, then the integral in (4.2) is ﬁnite.

Proof of Theorem 4.1.12. Let E be the set of those points in D in a neighborhood of which u is integrable. This set is nonempty because of (4.3). The deﬁnition implies that E is open. Let b belong to ∂E and let G ⊂ D be the disk of radius ε centered at b. Then G contains at least one point a such that |a − b| < ε/4 and u(a) > −∞. (Otherwise, b is in the interior of the complement of E.) The disk G0 = {|z − a| < ε/2} contains b and u is integrable on G0 because of (4.3). Hence, b ∈ E, which means that E is closed as well. 2

Approximation by smooth functions 4.1.15 Theorem Let u 6≡ −∞ be subharmonic in a domain D. Then there exists an increasing sequence of open sets Dn, whose union is D, and a decreasing ∞ sequence of subharmonic functions un ∈ C (Dn) tending to u. 4.1 Basic properties 59

∞ Proof. Let ω(z) = ω0(|z|) be a nonnegative “radial” function of class C (C) with compact support in D such that Z ω(w) dm(w) = 1. D For ε > 0 small enough consider the sets Dε = {z : dist(z, C r D) > ε} and the functions Z Z −2 uε(z) = ω(w)u(z + εw) dm(w) = ε ω (w − z)/ε u(w) dm(w), C C where u ≡ 0 outside of D. Then uε is ﬁnite (because of local integrability of u), ∞ subharmonic and of class C in Dε. From the formula Z 1 Z π it uε(z) = 2 rω0(r) dr u(z + rεe ) dt 0 −π and the inequality Z π Z π it it u(z + rεe ) dt 6 u(z + rδe ) dt, δ < ε, −π −π

(see Theorem 4.2.1 and its proof), it follows that u(z) 6 uδ(z) 6 uε(z), δ < ε, z ∈ Dε. And since u is bounded from above on compact subsets, we can apply the “limsup” variant of Fatou’s lemma; we get Z 1 Z π it lim sup uε(z) 6 2 rω0(r) dr lim sup u(z + rεe ) dt. ε→0 0 −π ε→0

Hence lim sup uε(z) 6 u(z), which completes the proof. 2 ε→0

Miscellaneous

4.1.16 Let u 6≡ −∞ be upper semicontinuous on D and subharmonic in D. Then Z π iθ 1 it u(re ) 6 P (r, θ − t)u(e ) dt (0 < r < 1). 2π −π 4.1.17 If a real-valued function h is semicontinuous and satisﬁes the condition 1 Z π h(a) = h(a + reiθ) dθ, 2π −π locally, then h is harmonic.

4.1.18 If an upper semicontinuous function u satisﬁes the condition 1 Z u(a) 6 2 u(z) dm(z), πr |z−a|

4.2 Properties of the mean values

Convexity and monotonicity By deﬁnition, a real function ϕ(r), r > 0, is convex of log r if the function x 7→ ϕ(ex) is convex. In other words, ϕ(r) is convex of log r if there holds the inequality

1−λ λ ϕ(r1 r2 ) 6 (1 − λ)ϕ(r1) + λϕ(r2), 0 < λ < 1.

2 00 0 If ϕ is of class C , then it is convex of log r iﬀ ϕ (r) + ϕ (r)/r > 0. 4.2.1 Theorem Let u be subharmonic in the disk |z| < R. Then the function 1 Z π I(r, u) = u(reiθ) dθ (0 < r < R) 2π −π is ﬁnite, increasing and convex of log r. The same holds for the function

it I∞(r, u) = max u(re ). 06t62π Proof. If u is subharmonic, then so is the function 1 Z π I(z) := u(zeiθ) dθ = I(|z|, u)(|z| < R). 2π −π

For fixed 0 < r1 < r2 define the harmonic function h(z) = a log |z| + b by h(rj) = I(rj). Since I(z) = h(z) on the boundary of the annulus r1 6 |z| 6 r2, we have I(z) 6 h(z) for r1 < |z| < r2. From this it follows that I(r, u) is convex of log r. That I(r, u) increases with r follows from the fact that the function ϕ(x) = I(r, ex) is convex and bounded for −∞ < x < log R, and this completes the proof in case of I(r, u). In case of I∞ the proof is similar; we define h by h(z) = I∞(rj) for |z| = rj. 2 The function I(r, u) need not be increasing if u is defined and subharmonic in an annulus; a simple example is the function u(z) = − log |z| which is subharmonic (and harmonic) in the annulus C r {0}. 4.2.2 Theorem Let u be subharmonic in the annulus ρ < |z| < R. Then the function 1 Z π I(r, u) = u(reiθ) dθ (ρ < r < R) 2π −π is finite and convex of log r.

Remark. The same holds for the function I∞(r, u). Proof. Since u is locally integrable and

Z Z r1 Z π u(z) dm(z) = r dr u(reiθ) dθ, r0<|z|

Logarithmic convexity A positive real function ϕ is said to be logarithmically convex if log ϕ is convex. A necessary and suﬃcient condition for ϕ to be logarithmically convex in (a, b) is that for every c ∈ R the function ecxφ(x) be convex in (a, b). We say that ϕ(r) is logarithmically convex of log r if ϕ(r) > 0 and log ϕ is convex of log r, which can 1−λ λ 1−λ λ be expressed as ϕ(r1 r2 ) 6 ϕ(r1) ϕ(r2) , 0 < λ < 1. If ϕ is continuous, then the validity of this for λ = 1/2 is suﬃcient for ϕ to be log-convex.

4.2.3 Theorem Let u be subharmonic in the annulus ρ < |z| < R, ρ > 0. Then the function I(r, eu) is logarithmically convex of log r in the interval ρ < r < R.

Proof. The function v(z) = ec log |z|+u(z) is subharmonic in the annulus ρ < |z| < R for every c > 0. Hence, the function I(ex, v) = ecxI(ex, eu) is convex for every c > 0. The result follows. 2

4.2.4 Corollary If f is a function analytic in the annulus ρ < |z| < R, then the function Z π 1/p 1 iθ p Mp(r, f) = |f(re )| dθ 2π −π is logarithmically convex of log r in the interval ρ < r < R, for every 0 < p 6 ∞. In the case p = ∞ this is Hadamard’s three circles theorem. The case p < ∞ was discussed by Hardy [25] in the ﬁrst paper from “theory of Hardy spaces.”

Miscellaneous 4.2.5 [30, Theorem 3.2.17] If u is subharmonic in the plane and satisﬁes the condition 1 Z π u(reiθ) dθ = o(log r)(r → ∞), 2π −π then u harmonic.

4.2.6 (Liouville’s theorem [30, Theorem 3.2.24]) If u is subharmonic in all of C and u(z) 6 o(log |z|)(z → ∞), then u is a constant.

4.2.7 If u > 0 is subharmonic in the annulus ρ < |z| < R and p > 1, then the p 1/p function Mp(r, u) = {I(r, u )} is convex of log r for ρ < r < R. 62 4 Subharmonic functions

4.3 Integral means of univalent functions

A function f deﬁned on D is said to be univalent if it is analytic and one-to-one. The leading example is the Koebe function f(z) = z/1 − z)2 mapping D to C slit from −1/4 to −∞ along the real axis.

Prawitz’ theorem

4.3.1 Theorem Let f : D 7→ C be a univalent function and f(0) = 0. Then for every p > 0 the function Z π 1 iθ −p Jp(r) = Jp(r, f) = |f(re )| dθ, 0 < r < 1, 2π −π is decreasing.

What is interesting here is that the function u = |f|−p is subharmonic in the annulus D r {0} but not in D, because u(0) = +∞. Also, the function −u is not subharmonic in D and therefore we cannot apply Theorem 4.2.1. Proof. We have Z 2π 0 iθ −p−2 iθ 0 iθ iθ 2πJ p(r) = −p |f(re )| Re{f(re )f (re )e } dθ 0 Z = −(p/r) Im |f(ζ)|−p−2f(ζ)f 0(ζ) dζ |ζ|=r Z = −(p/r) Im |w|−p−2w¯ dw (w = u + iv) Γr Z = −(p/r) |w|−p−2(u dv − v du), Γr

where Γr is the image under f of the circle |ζ| = r; the curve Γr is oriented positively. Now we apply Green’s formula to the domain Ωr,R bounded by Γr and the circle |w| = R, where R > max|z|=r |f(z)|. Since ∂(|w|−p−2u) ∂(−|w|−p−2v) − = −p|w|−p−2, ∂u ∂v we have Z Z ZZ |w|−p−2(u dv − v du) − |w|−p−2(u dv − v du) = −p |w|−p−2du dv |w|=R Γr Ωr,R The ﬁrst integral is equal to 2πR−p, and therefore ZZ 0 −p 2 −p−2 J p(r) = −(p/r)R − (p /2πr) |w| du dv. Ωr,R This concludes the proof. 2 4.3 Integral means of univalent functions 63

Distortion theorems

00 4.3.2 Theorem (Bieberbach) If f is a univalent function in D, then |f (0)| 6 4|f 0(0)|.

Proof. [30] We can assume that f(0) = 0 and f 0(0) = 1. Then f(z)−1/2 = z−1/21 − f 00(0)z/4 + z2h(z), where h is analytic in D. By Theorem 4.3.1, case p = 1, the function −1 00 2 2 4 2 J1(r) = r 1 + |f (0)/4| r + r M2 (r, h) is decreasing. Hence the function r−1 + |f 00(0)/4|2r is decreasing, and hence 00 |f (0)/4| 6 1. 2 Remark. The famous theorem of de Branges [17] states that if f is univalent in D, then f (n)(0) n|f 0(0)|, n 1. n! 6 > This was conjectured by Bieberbach.

4.3.3 Theorem (Koebe) If f is univalent in D, then f(D) contains the disk of radius |f 0(0)|/4 centered at f(0).

Proof. [30] Let f(0) = 0. If w is not in the range of f, then the function 00 0 g(z) = 1/(f(z) − w) is univalent in D and hence |g (0)| 6 4|g (0)|. It follows that 00 0 2 0 0 2 0 00 0 |f (0) + 2f (0) /w| 6 4|f (0)|, whence 2|f (0) /w| 6 4|f (0)| + |f (0)| 6 8|f (0)|. 0 Thus |w| > |f (0)/4|, and this concludes the proof. 2

4.3.4 Corollary If f is a conformal mapping of D ⊂ C onto G, then |f 0(z)| dist(f(z), ∂G) 4|f 0(z)|, z ∈ . 4 6 dist(z, ∂D) 6 D 4.3.5 Theorem If f is univalent in D, then 1 − r |f 0(z)| 1 + r , |z| = r. (1 + r)3 6 |f 0(0)| 6 (1 − r)3 z − w Proof. If we apply Bieberbach’s theorem to the function g(w) = f , 1 − zw¯ we get f 00(z) 2r2 r z − . f 0(z) 1 − r2 6 1 − r2 ∂ f 00 Since r log |f 0| = Re z we see that ∂r f 0 2r − 4 ∂ 2r + 4 log |f 0| . 1 − r2 6 ∂r 6 1 − r2 The desired result is obtained by integration. 2 64 4 Subharmonic functions

f(z) − f(0) |f 0(0)| 4.3.6 Corollary If f is univalent in , then . D z 6 (1 − |z|)2

f(z) − f(0) |f 0(0)| 4.3.7 Exercise If f is univalent in , then . D z > (1 + |z|)2 p p 4.3.8 Exercise If f is univalent in , then M(r)−p −J 0 (r) m(r)−p, D r 6 p 6 r where M(r) = max |f(z)|, m(r) = min |f(z)|. |z|=r |z|=r

Mean growth

If p > 0, then the function Ip(r) = Ip(r, f) = J−p(r, f) is increasing, and this fact does not depends on the hypothesis that f is univalent (Theorem 4.2.1). However, the proof of Theorem 4.3.1 gives additional information on Ip(r), namely: If f is univalent in D, f(0) = 0 and f 0(0) = 1, then

2 ZZ 0 p p p p−2 Ip(r) = R − |w| du dv. r 2πr Ωr,R

0 p This implies that Ip(r) 6 (p/r)M(r) . Combining this with Corollary 4.3.6 we get:

4.3.9 Theorem If f is univalent in D and 0 < p < 1/2, then Ip(r, f) is bounded, and 1 I (r, f) = O log , r → 1. 1/2 1 − r

4.4 The subordination principle

Let F be a univalent function deﬁned in D. A function f analytic in D is said to be subordinate to F if f(D) ⊂ F (D) and f(0) = F (0). In other words, f is subordinate to F if f(z) = F (ω(z)), where |ω(z)| 6 |z|, z ∈ D, and ω is analytic. In this form the notion of subordination is deﬁned for arbitrary functions. This notion is important because of the following theorem, known as Littlewood’s subordination principle.

4.4.1 Theorem If a function u is subordinate to a subharmonic function U, then

Z π Z π 1 iθ 1 iθ u(re ) dθ 6 U(re ) dθ (0 < r < 1). (4.4) 2π −π 2π −π

In the simplest case ω(z) = ρz, 0 < ρ < 1, this theorem reduces to Theo- rem 4.2.1. Proof. We can assume that U is continuous. Let h be a function harmonic in Dr = {|z| < r}, continuous on the closure and equal to U on the boundary. Then 4.4 The subordination principle 65

U 6 h on Dr and, hence, u(z) = U(ω(z)) 6 h(ω(z)) for |z| = r. It follows that Z π Z π Z π 1 iθ 1 iθ 1 iθ u(re ) dθ 6 h ω(re ) dθ = h ω(0) = h(0) = h(re ) dθ. 2π −π 2π −π 2π −π

This concludes the proof because h(reiθ) = U(reiθ). 2 For various applications of the subordination principle in the theory of univalent functions we refer the reader to Duren [19, Ch. 6]. We will consider two examples which cannot be found in [19].

4.4.2 Theorem (Kolmogorov/Smirnov) If f ∈ H(D) and Re f ∈ h1, then f ∈ hp, i.e., 1 Z π sup |f(reiθ)|p dθ < ∞ for every p ∈ (0, 1). r<1 2π −π

Proof. We can assume that f(0) ∈ R. Assume ﬁrst that Re f > 0. Then f is subordinate to the univalent function F (z) = c(1 + z)/(1 − z), c = Re f(0). Applying the subordination principle we get the following:

Z π Z π iθ p Z π iθ p 1 iθ p p 1 1 + re p 1 1 + e |f(re )| dθ 6 c iθ dθ 6 c iθ dθ 2π −π 2π −π 1 − re 2π −π 1 − e 2 Z π/2 u(0)p = cp (cot θ)p dθ = (0 < p < 1). π 0 cos(pπ/2) If f is arbitrary, then we can use Theorem 3.2.5(b) to reduce the proof to the preceding case. 2

4.4.3 Remark The Kolmogorov/Smirnov theorem can be proved in the following p p p way. Let Re f > 0. Then Re(f ) = |f| cos(arg f) > |f| cos(πp/2), and hence Z π Z π 1 iθ p 1 1 iθ p 1 p f(re ) dθ 6 Re f(re ) dθ = Re{f(0) }. 2π −π cos(πp/2) 2π −π cos(πp/2)

Our next example is the case p 6 1 of the following theorem of Ahern [1].

4.4.4 Theorem If f ∈ H(D) and 0 < |f(z)| < 1 for all z ∈ D, then for every p > 0 we have Z π 1 iθ p 1/2 (1 − |f(re )|) dθ > cp(1 − r) , 2π −π where cp is a positive constant.

Ahern’s proof is based on a highly nontrivial analysis of singular measures, which enabled him to treat the case p > 1. Here we use the subordination principle to discuss the case p = 1/2. It turns out that then Ahern’s theorem can be improved. On the other hand, it seems that application of the subordination principle is limited to the case p 6 1. 66 4 Subharmonic functions

4.4.5 Theorem With the hypotheses of the previous theorem, we have Z π 1 iθ 1/2 1/2 2 1 − |f(re )| dθ > c(1 − r) log , 2π −π 1 − r where c is a positive constant.

Proof. The analytic function a(z) = − log f(z) maps D into the right half-plane and therefore is subordinate to the function λ(1 + z)/(1 − z), where λ = a(0) > 0. It follows that f(z) is subordinate to 1 + z S (z) = exp − λ . λ 1 − z The function −(1 − |z|)1/2 is subharmonic for |z| < 1, and therefore, by the subordination principle, Z π Z π 1 iθ 1/2 1 iθ 1/2 1 − |f(re )| dθ > 1 − |Sλ(re )| dθ. 2π −π 2π −π In order to estimate this integral we use the inequality x 2x 1 − e−x , x > 0. 1 + x 6 6 1 + x It follows that 1 Z π Z π λP (r, θ) 1/2 Z π (1 − r)1/2 (1 − |S (reiθ)|)1/2 dθ dθ dθ λ 2 1/2 2π −π 0 1 + λP (r, θ) 0 (1 − r + θ ) √ Introducing the change θ = t 1 − r we conclude the proof. 2

Miscellaneous

4.4.6 (Harmonic Schwarz lemma) If u ∈ h(D) is real valued, |u| 6 1, and u(0) = 0, then u is subordinate to the function 2 1 + z U(z) = arg . π 1 − z 4 Hence |u(z)| arctan |z|, and, by the subordination principle, M (r, u) r, 6 π 2 6 0 < r < 1.

4.4.7 (Rogosinski’s theorem [19, §6.2]) Let f(z) = F (ω(z)), where F is ana- Pn k Pn k lytic in D, and ω is as above. Let Fn(z) = k=0 Fb(k)z and fn(z) = k=0 fb(k)z . n+1 Then Fn(ω(z)) = fn(z) + O(z ). Therefore, by the subordination principle and Parseval’s formula, If f is subordinate to F ∈ H(D), then n n X 2 2k X 2 2k |fb(k)| r 6 |Fb(k)| r , 0 < r < 1. k=0 k=0 4.5 The Riesz measure 67

4.4.8 With the hypotheses of Theorem 4.4.4, we have Z π 1 iθ p p (1 − |f(re )|) dθ > cp(1 − r) for 0 < p < 1/2. 2π −π

4.4.9 [19] If U = |f|p, 0 < p < ∞, then strict equality holds for 0 < r < 1 in (4.4) unless f is constant or ω(z) = αz, |α| = 1.

4.5 The Riesz measure

2 2 Let C0 (D) be the set of all C -functions with compact support in D. If u is subharmonic in D, then the functional Z 2 ϕ 7→ u∆ϕ dm, ϕ ∈ C0 (D) D is positive and therefore there exists a unique positive Borel measure µ on D such that Z Z 2 ϕ dµ = u∆ϕ dm, ϕ ∈ C0 (D). D D This measure is called the Riesz measure of u. If u is of class C2, then

dµ = (∆u) dm.

The Riesz measure of the function log |z − a| is equal to 2πδa, where δa denotes the Dirac measure at the point a. The existence and uniqueness of the Riesz measure can be proved by using Riesz’ theorem on representation of positive linear (†) functionals on C0(D).

Green’s formula Here we will consider in some detail a generalization of Green’s formula.

4.5.1 Theorem Let u be subharmonic in DR = {z : |z| < R} and let u(0) > −∞. Then 1 Z π 1 Z r u(reiθ) dθ − u(0) = log dµ(z), (4.5) 2π −π 2π |z|

Observe that log(r/|z|) = 0 for |z| = r, so it does not matter what stands in (4.5): |z| < r or |z| 6 r. Besides, the formula holds in the case u(0) = −∞ as well, which shows that the integral on the right-hand side is ﬁnite iﬀ u(0) > −∞.

(†)For more information on the Riesz measure we refer to H¨ormander [30] and Hayman/Kennedy [27]. For a discussion on Riesz’ representation of positive linear functionals on C0(D), see Rudin [86, Ch. 2]. 68 4 Subharmonic functions

Jensen’s formula The well known Jensen’s formula is one of special cases of (4.5). Namely, if a function f is analytic in DR and f(0) 6= 0, then the Riesz measure P of the function log |f(z)| is equal to 2π k δak , where ak are the zeros of f. This and (4.5) give Jensen’s formula,

1 Z π X r log |f(reiθ)| dθ = log |f(0)| + log , 2π −π |ak| |ak|

which, of course, can easily be proved without appealing to Theorem 4.5.1.

Proof of Theorem 4.5.1

Fix R0 < R and let 0 < ρ < r < R0. Choose a decreasing sequence {un} of 2 subharmonic functions of class C in the disk |z| < R0 such that un tends to u (Theorem 4.1.15). Then Green’s formula for smooth functions gives

1 Z I(r, un) − I(ρ, un) = Gρ(z)∆un(z) dm(z), 2π DR

where r r G (z) = min log , log , |z| < r, ρ ρ |z|

2 Gρ(z) = 0 for |z| > r. Let g be a C -function in DR such that g = 0 in a neighborhood of the circle |z| = r and g 6 Gρ in DR. Then 1 Z 1 Z I(r, un) − I(ρ, un) > g(z)∆un(z) dm(z) = un(z)∆g(z) dm(z). 2π DR 2π DR

Here we apply the dominated convergence theorem, which is possible because |un| 6 |u| + |u1| and the functions u and u1 are locally integrable (Theorem 4.1.12), to get

1 Z 1 Z I(r, u) − I(ρ, u) > u(z)∆g(z) dm(z) = g(z) dµ(z). 2π DR 2π DR

Now we take an increasing sequence of the functions g which tends to Gρ and get

1 Z I(r, u) − I(ρ, u) > Gρ(z) dµ(z). 2π DR

The reverse inequality is proved in a similar way. Finally, let ρ tend to 0. 2

Riesz’ representation formula A slight modiﬁcation of the above proof yields the following variant of Theo- rem 4.5.1. 4.5 The Riesz measure 69

4.5.2 Theorem If u is subharmonic in a neighborhood of the closed unit disk, then 1 Z u(z) = h(z) + G(z, w) dµ(w), (4.6) 2π D for every z ∈ D, where 1 Z π h(reiθ) = P (r, θ − t)u(eit) dt, 2π −π w − z µ is the Riesz measure of u, and G(z, w) = log . 1 − wz¯

Miscellaneous

4.5.3 Let u be subharmonic in D. Then for every ε ∈ (0, 1/4) there exists a, |a| < 1/4, such that Z π Z π 1 iθ 2 1 iθ u(a + εe ) dθ − u(a) 6 Cε u(e ) dθ − u(0) , 2π −π 2π −π where C is an absolute constant.

4.5.4 Let ϕ : D 7→ C be a conformal mapping and let u be subharmonic in D. Then Z Z −1 f(z) dµu◦ϕ(z) = f(ϕ (w)) dµu(w), D ϕ(D) where f is an arbitrary positive (Borel) function on D, and µ is the Riesz measure of the corresponding function. For instance, if v(z) = u(a + rz)(a ∈ C, r > 0), then Z Z f(z) dµv(z) = f (w − a)/r dµu(w). D a+rD 4.5.5 If u subharmonic in the unit disk, then there holds the formula Z Z f(z) dµu◦ϕa (z) = f ϕa(w) dµu(w)(|a| < 1), D D a − z where ϕ (z) = and f is positive and Borel. a 1 − az¯ 4.5.6 If f is analytic and p > 0, then the Riesz measure of |f|p is absolutely continuous and equal to p2 |f|p−2|f 0|2 dm. 2 p−2 0 2 If p > 2, then the function |f| |f | is subharmonic; from this and Green’s formula we can deduce that the function 1 Z π √ r 7→ |f( reiθ)|p dθ 2π −π is convex. 70 4 Subharmonic functions

4.6 A Littlewood/Paley theorem

For a function u deﬁned in D we write I(u) = sup I(r, u), where, as above, 0 1) via the Riesz measure of u (not of uq as in Green’s formula).

4.6.1 Theorem [79] Let u > 0 be subharmonic in D and let µ be the Riesz q measure of u. If q > 1 and I(u ) < ∞, then there holds the inequality Z −1 q q q (1 − |z|) µ(Eε(z)) dm(z) 6 Cq I(u ) − u(0) , (4.7) D

where ε = 1/6 and Eε(z) = {w : |w − z| < ε(1 − |z|)}. If in addition ∆u (u ∈ C2) is a subharmonic function, then Z 2 2 µ(Eε(z)) = ∆u dm > πε (1 − |z|) ∆u(z), Eε(z) which leads to the following:

2 4.6.2 Theorem Let u > 0 be a subharmonic function of class C (D) such that q its Laplacian is a subharmonic function. If q > 1 and I(u ) < ∞, then Z 2q−1 q q q (1 − |z|) ∆u(z) dm(z) 6 Cq I(u ) − u(0) . (4.8) D The classical inequality of Littlewood and Paley (Theorem 3.5.1) is a special p case of (4.8). Namely, if p > 2 and I(|h| ) < ∞, where f is a real-valued harmonic function in D, then we take u = h2 and q = p/2, and get Z p−1 p p p (1 − |z|) |∇h| dm 6 Cp I(|h| − |h(0)| ) . D In the case q < 1 we have the following theorem, the proof of which is omitted here (cf. [79]).

2 q 4.6.3 Theorem Let 0 < q < 1 and let u > 0 be a C -function such that both u and ∆u are subharmonic. If Z (1 − |z|)2q−1(∆u)q dm < ∞, D then I(uq) < ∞ and there holds the inequality Z q q 2q−1 q I(u ) − u(0) 6 Cq (1 − |z|) (∆u) dm(z). D 4.6 A Littlewood/Paley theorem 71

Jevti´c[32] extended the above theorems to the case of M-harmonic functions on the unit ball in Cn.

Local estimates for the Riesz measure In what follows we suppose that u is a nonnegative subharmonic function deﬁned in D and denote by µ the Riesz measure of u. As we have seen, there holds the formula 1 Z r I(r, u) − u(0) = log dµ(z) (0 < r < 1) (4.9) 2π |z| rD (see Theorem 4.5.1).

4.6.4 Lemma There holds the equality 1 Z 1 I(u) − u(0) = log dµ(z). 2π |z| D Proof. Write (4.9) as 1 Z r I(r, u) − u(0) = K (z) log dµ(z), 2π r |z| D where Kr(z) is the characteristic function of the disk rD. Since Kr(z) log(r/|z|) increases with r, we have 1 Z r lim I(r, u) − u(0) = lim Kr(z) log dµ(z). r→1− 2π r→1− |z| D

Since I(r, u) increases, we see that I(u) = limr→1− I(r, u). 2

q 4.6.5 Lemma Let q > 1 and let µq be the Riesz measure of u . Then q {µ(E)} 6 Cq µq(5E) (4.10) for every disk E such that 6E ⊂ D. The constant Cq depends only of q. If E is a disk of radius R, then rE denotes the concentric disk of radius Rr. Proof. By translation, the proof reduces to the case where E is centered at zero. Then, since µ(E) = ν((1/r)E), where ν is the Riesz measure of the function u(rz), we can assume that the radius of E is ﬁxed, e.g., E = εD, ε = 1/6. Using the simple inequalities

q q q I(r, u) − u(0) 6 I(r, u) − u(0)

q q and I(r, u) 6 I(r, u ), which hold because q > 1, we see from (4.9) (applied to u and uq) that

1 Z r q 1 Z r log dµ(z) log dµ (z). 2π |z| 6 2π |z| q rD rD 72 4 Subharmonic functions

Letting r = 4ε, we get Z q −1 {µ(2εD)} 6 C |z| dµq(z), (4.11) 4εD where we have applied the estimate log(4ε/|z|) > log 2 for |z| < 2ε and log(4ε/|z|) 6 1/|z|. Therefore, in order to prove (4.10) we have to remove |z|−1. To do this, we translate the “center” of (4.11) to get Z q −1 {µ(2εDa)} 6 C |z − a| dµq(z) 4εDa

for a ∈ εD, where Da = {z : |z − a| < 1}. Since εD ⊂ 2εDa and 4εDa ⊂ 5εD, we see that Z q −1 {µ(εD)} 6 C |z − a| dµq(z). 4εDa Now we integrate this inequality over the disk εD, with respect to dm(a), and apply Fubini’s theorem, which ﬁnishes the proof because Z sup |z − a|−1dm(a) < ∞. 2 z∈D εD Proof of Theorem 4.6.1 From (4.10) it follows that Z Z −1 q −1 (1 − |z|) µ(Eε(z)) dm(z) 6 C (1 − |z|) µq(E5ε(z)) dm(z). (4.12) D D Further, from Z µq(E5ε(z)) = dµq(w) E5ε(z) and Fubini’s theorem it follows that the right-hand side of (4.12) equals Z Z −1 dµq(w) (1 − |z|) dm(z), D G(w) where G(w) = {z : |z − w| < 5ε(1 − |z|)}. Since z ∈ G(w) implies |z| − |w| < 5ε(1 − |z|), whence 1 − |w| < (1 + 5ε)(1 − |z|), we see that Z −1 −1 (1 − |z|) dm(z) 6 (1 + 5ε) m(G(w)) (1 − |w|) . G(w)

0 2 And since (1−5ε)(1−|z|) < 1−|w| for z ∈ G(w), we have m(G(w)) 6 C (1−|w|) , where C0 = π(5ε/(1 − 5ε))2. Combining all these results we see that Z Z −1 q (1 − |z|) µ(Eε(z)) dm 6 Cq (1 − |w|) dµq(w). D D This completes the proof of (4.7) because of Lemma 4.6.4 and the inequality 1−|w| 6 log(1/|w|). 5 Classical Hardy spaces

p There are various equivalent deﬁnitions of H -spaces. If p > 1, the shortest way to introduce Hp is by identifying it with a subspace of Lp(T),

p p H (T) = {φ ∈ L (T): φb(−n) = 0 for n > 1}. Thus Hp(T) coincides with the closure in Lp(T) of the set of all analytic polynomials; this can be used to define Hp(T) for 0 < p < 1. In this text, we define Hp as a subclass of H(D), see (5.1). In view of Riesz’ projection theorem 6.2.1, Hp is isomorphic with hp for 1 0, which is used to extend Hp-theory to several real variables (cf. [93]). This chapter contains the standard facts on radial limits and factorization; an exception is Section 5.5, where we consider the composition of an Hp-function with an inner function. Our approach slightly differs from that in other texts [18, 22, 46, 83, 86, 99] in that we first prove the Hardy/Littlewood decomposition lemma (Lemma 5.1.7), and then deduce the radial limits theorem and F. and M. Riesz’ theorems, without appealing to the Blaschke products.

5.1 Basic properties

p p The Hardy space H (0 < p 6 ∞) is deﬁned as the subspace of h consisting of analytic functions,

p H = {f ∈ H(D): kfkp = supr<1 Mp(r, f) < ∞}. (5.1)

Here we can replace “sup” by “lim” because Mp(r, f) increases with r for every p p > 0. If p > 1, then H is a Banach space, and if 0 < p < 1, it is a p-Banach space. The completeness is proved in the standard way. The ﬁrst step is the continuity of the inclusion Hp ⊂ H(D), which follows from the following lemma.

p 5.1.1 Lemma If f ∈ H , 0 < p 6 ∞, then 2 −1/p |f(z)| 6 (1 − |z| ) kfkp. (5.2) The space Hp is not normable for p < 1. On the other hand, it follows from the lemma that the dual of Hp separates points in Hp.

5.1.2 Corollary If f ∈ Hp, then

2 1/q−1/p Mq(r, f) 6 (1 − r ) kfkp (q > p). (5.3)

73 74 5 Classical Hardy spaces

Proof. This follows from (5.2) and the following:

Z π q−p q 1 iθ q−p iθ p iθ p Mq (r, f) = |f(re )| |f(re )| dθ 6 sup |f(re )| Mp (r, f). 2 2π −π θ Taking q = 1 > p and r = 1 − (1/n) in (5.3), we get, via the inequality n M1(r, f) > |fb(n)|r , one of many results of Hardy and Littlewood.

p 1/p−1 5.1.3 Corollary If f ∈ H (0 < p < 1), then |fb(n)| 6 Cp kfkp (n + 1) , where Cp depends only on p.

Proof of Lemma 5.1.1. Assume that f is analytic in a neighborhood of the closed disk. Then, for p < ∞, 1 Z kfkp = |f(ζ)|p |dζ|. p 2π T z − w For ﬁxed z ∈ let ϕ(w) = . By the substitution ζ = ϕ(ξ) we get D 1 − zw¯ 1 Z 1 Z kfkp = |f(ϕ(ξ))|p |ϕ0(ξ)| |dξ| = |f(ϕ(ξ))ϕ0(ξ)1/p|p |dξ|. p 2π 2π T T The function inside the last integral is subharmonic and therefore

p 0 1/p p p 2 kfkp > |f(ϕ(0))ϕ (0) | = |f(z)| (1 − |z| ),

which was to be proved. 2

5.1.4 Theorem For every p > 0 the space Hp is complete.

p Proof. Let {fn} be a Cauchy sequence in H . This means that for every ε > 0 there exists N such that Mp(r, fn −fm) < ε for every r and m, n > N. From (5.1.1) it follows {fn} is a Cauchy sequence in H(D) and hence {fn} converges uniformly on compact subsets to some function f ∈ H(D). Letting m tend to ∞, we get Mp(r, fn − f) 6 ε for n > N and all r, which implies kfn − fkp 6 ε for n > N. 2

5.1.5 Exercise If 0 0, then f =const.

5.1.6 Exercise For a ﬁxed z ∈ D, equality occurs in (5.2) iﬀ

1 − |z|2 1/p f(w) = c , (1 − zw¯ )2

where c is a constant. 5.1 Basic properties 75

A decomposition lemma Various properties of a zero-free Hp-function can be deduced from the corresponding properties of the function f 2/p ∈ H2. The following lemma of Hardy and Littlewood is often used to reduce the general case to the zero-free case.

5.1.7 Lemma If f ∈ Hp, p > 0, then there exist functions g and h without zeros in D such that f = g + h, kgkp 6 kfkp and khkp 6 kfkp.

p For example, in proving an inequality of the form kT fkX 6 Ckfkp, f ∈ H , where X is a quasinormed space and T : Hp 7→ X a linear operator, we can suppose that f has no zeros in D. Proof. Assume, at ﬁrst, that f 6≡ 0 is analytic in a neighborhood of the closed disk and that f has at least one zero in D. Then the number of zeros of f is ﬁnite; denote the zeros by a1, . . . , am (counting multiplicity). Let

m Y z − ak A(z) = 1 − a¯kz k=0 and deﬁne g and h as follows: g = (A − 1)f/2A, h = (A + 1)f/2A; we have f = g + h. Neither g nor h have zeros in D because |A| < 1 in D and the function f/A has no zeros in D. Both h and g are analytic in a neighborhood of the closed disk because so are A−1, A+1 and f/A. And since |A| = 1 on T, we have |g| 6 |f| and |h| 6 |f| on T, which proves the lemma in that special case. If f is arbitrary, let fn(z) = f(rnz), where rn = 1 − 1/n (or any sequence tending to 1). By the preceding, we have a decomposition fn = gn + hn with the the desired properties. Since

kgnkp 6 kfnkp 6 kfkp (5.4)

(and similarly for h), then, according to Lemma 5.1.1, the sequences {gn} and {hn} are bounded on compact subsets of D. Therefore, passing to subsequences, we can assume that gn and hn tend uniformly on compact subsets to analytic functions g and h, respectively. By Hurwitz’ theorem, the function g is either without zeros or g ≡ 0 because gn have no zeros. The same holds for h. But it is not true that g ≡ 0 because this and f = g + h imply f ≡ h, which is impossible because f 6≡ 0 and f has zeros. Finally, from (5.4) it follows that Mp(r, gn) 6 kfkp for every r < 1, and hence Mp(r, g) 6 kfkp. 2

Radial limits Since H1 ⊂ h1, we see from the Riesz/Herglotz theorem and Fatou’s theorem that every function f ∈ H1 has radial limits almost everywhere. However, the hypothesis that f is analytic improves the properties of the boundary function substantially (see, e.g., Theorems 5.1.8 and 5.2.1). 76 5 Classical Hardy spaces

p 5.1.8 Theorem Let f ∈ H , 0 < p 6 ∞. Then almost everywhere on T there iθ iθ exist radial limits f∗(e ) = lim f(re ) , and there hold the relations r→1−

kfkp = kf∗kp (p 6 ∞) (5.5) Z π 1 iθ iθ p lim |f(re ) − f∗(e )| dθ = 0 (p < ∞). (5.6) r→1− 2π −π Proof. When p > 1, we may appeal to Theorem 3.4.2. Let f ∈ Hp and 1/2 1, the function g, so the function f = g2, iθ iθ has radial limits. Let f∗ denote the boundary function and fr(e ) = f(re ). Then

kfr − f∗kp = k(gr − g∗)(gr + g∗)kp 6 kgr − g∗k2pkgr + g∗k2p,

where we have applied the Cauchy/Schwarz inequality. Since kgr − g∗k2p tends to 0 and kgr + g∗k2p is bounded, we can conclude that there holds (5.6) for p > 1/2. In the same way we reduce the case p > 1/4 to the case p > 1/2, etc. 2 p The set of all harmonic polynomials is not dense in h for p 6 1. However, we have:

5.1.9 Theorem If 0 < p < ∞, then the set of all (analytic) polynomials is dense in Hp.

Proof. This follows from (5.6) and the fact that kfr − snfrkp → 0 (n → ∞), for every ﬁxed r ∈ (0, 1), where sng denotes the partial sum of the Taylor series of g. 2

The Poisson integral of log |f∗|

If f is analytic in a neighborhood of the closed disk, then we have log |f| 6 P [log |f∗|] because of the subharmonicity of log |f|.

p 1 5.1.10 Theorem Let f 6≡ 0 belong to H (p > 0). Then log |f∗| ∈ L (T) and there holds Z π iθ 1 it log |f(re )| 6 P (r, θ − t) log |f∗(e )| dt, 0 6 r < 1. (5.7) 2π −π Before proving this theorem we note two consequences.

Smirnov’s maximum principle There are unbounded functions f ∈ H(D) for 1 + z which the boundary function belongs to L∞( ); one of them is f(z) = exp . T 1 − z However:

p ∞ ∞ 5.1.11 Theorem (Smirnov) If f ∈ H , p > 0, and if f∗ ∈ L , then f ∈ H and kfk∞ = kf∗k∞. 5.1 Basic properties 77

In the case p > 1 this theorem is contained in Corollary 3.4.4, while in the general case it is a consequence of (5.7), or of the weaker inequality Z π iθ p 1 it p |f(re )| 6 P (r, θ − t)|f∗(e )| dt (0 6 r < 1) (5.8) 2π −π and (5.5). It is worthwhile to note that (5.8) can be deduced immediately from (5.6) p p and the inequality |f(ρz)| 6 P [ |fρ| ](z). In fact, inequality (5.7), together with Jensen’s inequality for the function x 7→ ex, implies a more general fact, namely:

p q q 5.1.12 Theorem (Smirnov) If f ∈ H and f∗ ∈ L for some q > p, then f ∈ H .

Uniqueness theorem An immediate consequence of Theorem 5.1.10:

p iθ 5.1.13 Theorem If f ∈ H and f∗(e ) = 0 on a set of positive measure, then f(z) = 0 for every z ∈ D.

Proof of Theorem 5.1.10

p + p Let f ∈ H and, say, f(0) 6= 0. Since log x 6 x /p, x > 0, it follows that Z π 1 + iθ log |f(re )| dθ 6 C (0 < r < 1), 2π −π where C is a constant. We also have Z π Z π Z π 1 iθ 1 + iθ 1 iθ log |f(re )| dθ = 2 log |f(re )| dθ − log |f(re )| dθ. 2π −π 2π −π 2π −π

The last summand is 6 − log |f(0)| because of the subharmonicity of log |f|. Hence, Z π 1 iθ log |f(re )| dθ 6 2C − log |f(0)|, (5.9) 2π −π 1 and Fatou’s lemma concludes the proof that log |f∗| ∈ L (T). To prove (5.7) we start from the inequality

iθ iθ log |f(ρz)| 6 P [log |fρ|](z), where fρ(e ) = f(ρe ), (5.10)

ρ < 1, z ∈ D, which holds for an arbitrary f ∈ H(D) because log |f| is subharmonic. Since log x = log+ x − log− x, we have, from (5.10),

+ − log |f(z)| 6 lim sup P [log |fρ|](z) − lim inf P [log |fρ|](z). (5.11) ρ→1 ρ→1

+ + p And since | log x − log y| 6 |x − y| /p, x, y > 0, it follows from (5.6) that Z π 1 + iθ + iθ lim log |f(ρe )| − log |f∗(e )| dθ = 0; ρ→1− 2π −π 78 5 Classical Hardy spaces

+ + therefore lim sup P [log |fρ|](z) = P [log |f∗|](z). Now we deduce (5.7) from (5.11) ρ→1 by means of Fatou’s lemma. 2 Remark. For another proof, see 7.1.5.

5.2 The space H1

The results of this section, due to F. and M. Riesz, Privalov, and Smirnov(∗), show how much H1 diﬀers from h1.

The Poisson integral of the boundary function A function belonging to h1 need not be equal to the Poisson integral of the boundary function. However:

1 1 5.2.1 Theorem If f ∈ H , then f∗ ∈ L and f = P [f∗].

This is easily deduced from the relation f(rz) = P [fr](z)(r < 1), by means of (5.6).

5.2.2 Exercise (Cauchy’s integral formula) If f ∈ H1, then

1 Z f (ζ) f(z) = ∗ dζ (z ∈ ). 2πi ζ − z D T Now we are in position to prove the famous theorem of F. and M. Riesz:

5.2.3 Theorem Let µ be a complex Borel measure on such that R ζn dµ(ζ) = 0 T T for every n = 1, 2,... . Then µ is absolutely continuous.

1 Proof. Let f = P [µ]. Then f ∈ h and fb(k) = µb(k) for every k ∈ Z (Proposi- tion 3.2.1). Hence, the condition of the theorem implies that f is analytic. Hence 1 f ∈ H so, according to Theorem 5.2.1, we have f = P [f∗]. In view of the injec- iθ iθ tivity of the Poisson integral (Theorem 3.2.2), it follows that dµ(e ) = f∗(e ) dt. 2

Bounded variation =⇒ absolute continuity

5.2.4 Theorem If f ∈ H1 and if the boundary function is almost everywhere equal to a function of bounded variation, then f has absolutely continuous extension to D. (†)

(∗)Further information, as well as references and historical comments, can be found in Zygmund [100, Ch. VII§§8–10] and Duren [18, Ch. III] (†) i.e., a continuous extension that is absolutely continuous on T. 5.2 The space H1 79

Proof. Let f∗ = γ a.e., γ ∈ BV [−π, π]. Then f = P [γ], by Theorem 5.2.1. We have ∂f g(reiθ) := = PS[γ](reiθ) − k · P (r, θ + π), ∂θ where PS[γ] is the Poisson/Stiltjes integral of γ (see (3.10) and (3.13) ). Then, 1 by the Riesz/Herglotz theorem, g ∈ H and, by Theorem 5.2.1, g = P [g∗]; thus g = PS[G], where Z θ it G(θ) = g∗(e )(θ ∈ R). 0 Applying (3.13) again, with the obvious change of notation, and taking into account that the function G is 2π-periodic because g(0) = 0, we get

∂f ∂ 1 Z π = PS[G](reiθ) = P (r, θ − t)G(t) dt. ∂θ ∂θ 2π −π It follows that 1 Z π f(reiθ) = const + P (r, θ − t)G(t) dt, 2π −π which concludes the proof. 2 In a similar way one proves the following:

5.2.5 Theorem The derivative of a function f ∈ H(D) belongs to H1 iﬀ f has absolutely continuous extension to D. The boundary function of the function iθ iθ 0 iθ 1 iθ (∂/∂θ)f(re ) = ire f (re ), if f ∈ H , is equal to (d/dθ)f∗(e ).

Conformal mappings

5.2.6 Theorem Let f be a conformal mapping of D onto a domain G whose boundary, ∂G, is a Jordan curve. Then f 0 ∈ H1 iff ∂G is rectifiable. If ∂G is rectifiable, then Z π |∂G| = |f 0(eiθ)| dθ (5.12) −π P∞ and |∂G| > 2 n=0 |fb(n)|, where |∂G| is the length of ∂G. The last inequality follows from (5.12) and the inequality

∞ X |fb(n)| πkfk , (5.13) n + 1 6 1 n=0 due to Hardy (see 5.3.7). Proof. Let f 0 ∈ H1. The function f can be extended as a continuous function to D, and the extended function is a homeomorphism between D and G (theorem of Carathe´odory, [100, Ch. VII, §10]). By Theorem 5.2.5, this extension is absolutely 80 5 Classical Hardy spaces

continuous on T and therefore ∂G is parameterized by the absolutely continuous iθ function Γ(θ) = f∗(e )(|θ| 6 π). Hence the length of ∂G is equal to Z π |Γ0(θ)| dθ. −π

Now formula (5.12) follows from theorem 5.2.5. Conversely, let ∂G be rectiﬁable, |∂G| = 2π, and let γ(θ) (0 6 θ 6 2π) be the arclength parameterization of ∂G. The function f has the radial limits ϕ(θ) = iθ limr→1− f(re ) on a set S ⊂ [0, 2π], |S| = 2π. Since the extended mapping is homeomorphic, the function ϕ “increases”, i.e., there exists an increasing function t : S 7→ [0, 2π] such that ϕ(θ) = γ(t(θ)), θ ∈ S. We extend the function t(θ) as an increasing function on [0, 2π], and the corresponding extension of ϕ is of bounded variation on [0, 2π]. Now Theorem 5.2.4 shows that f has absolutely continuous extension to D. Finally, f 0 ∈ H1, by Theorem 5.2.5.(‡) 2

5.3 Blaschke product

If a function f ∈ H(D) has an inﬁnite number of zeros, a1, a2,... , then 1−|an| → 0. If f ∈ Hp, there holds more:

p 5.3.1 Theorem If {an} (n > 1) is the sequence of zeros of a function f ∈ H P∞ (p > 0), then the Blaschke condition is satisﬁed: n=1(1 − |an|) < ∞. Conversely, if a sequence {an} ⊂ D satisﬁes the Blaschke condition, then the product ∞ Y an − z |an| B(z) = 1 − a¯ z a n=1 n n

converges in D and the function B is analytic and has the properties: (a) The sequence of zeros of B, including repetitions for multiplicities, coincides iθ with {an}; (b) |B(z)| 6 1 for |z| < 1; (c) |B(e )| = 1 almost everywhere.

The function B(z) is called a Blaschke product; in the case an = 0 the Q∞ ratio |an|/an is interpreted as −1. Note that B(0) = n=1 |an|, and this product converges iﬀ the Blaschke condition is satisﬁed. By the term a Blaschke product we also mean a function of the form

k Y an − z |an| B(z) = 1 − a¯ z a n=1 n n

with k > 1 as well as the function B(z) ≡ 1.

(‡)so we have proved the implication r“... =⇒ f 0 ∈ H1” without appealing to the theorem of Carathe´odory. 5.3 Blaschke product 81

Proof. Let f ∈ Hp (p > 0) and, say, f(0) 6= 0. From the inequality Z π Z π 1 iθ p 1 iθ p log |f(re )| dθ 6 log |f(re )| dθ , 2π −π 2π −π which is obtained by an application of Jensen’s inequality for the concave function log x, it follows that Z π 1 iθ log |f(re )| dθ 6 log kfkp. 2π −π On the other hand, there holds (Jensen’s) formula

1 Z π X r log |f(reiθ)| dθ = log |f(0)| + log . 2π −π |ak| |ak|

an − z 1 + |z| 1 − 6 (1 − |an|) . 1 − a¯nz 1 − |z| (Details are omitted.) It is clear that B(z) has property (b). In order to prove (c), observe that B(z)/Bk(z), where

k Y an − z |an| B (z) = , k 1 − a¯ z a n=1 n n is a Blaschke product as well. It follows that

|B(0)| 1 Z π |B(eiθ)| 6 iθ dθ. |Bk(0)| 2π −π |Bk(e )|

iθ And since |Bk(e )| = 1 and B(0)/Bk(0) → 1 (k → ∞), we have Z π 1 iθ |B(e )| dθ > 1. 2π −π

iθ Finally, since |B(e )| 6 1 almost everywhere, we have (c). 2

Riesz’ factorization theorem

p 5.3.2 Theorem Let {an} be the sequence of zeros of f ∈ H (p > 0) and B(z) p the corresponding Blaschke product. Then f/B belongs to H and kf/Bkp = p kfkp. Consequently, every H -function can be represented as f = Bg, where B is a Blaschke product (ﬁnite or inﬁnite), the function g has no zeros in D and kfkp = kgkp. 82 5 Classical Hardy spaces

Proof. With the above hypotheses, let

k Y an − z |an| B (z) = . k 1 − a¯ z a n=1 n n

p Then the function f/Bk belongs to H because it is analytic in D and |Bk(z)| > 1/2 near T. Since |f/Bk| = |f| on T, we have kf/Bkkp = kfkp. Hence

1 Z π |f(reiθ)| iθ dθ 6 kfkp 2π −π |Bk(re )| for every r < 1. By Fatou’s lemma we get

1 Z π |f(reiθ)| iθ dθ 6 kfkp, 2π −π |B(re )|

which implies kf/Bkp 6 kfkp. The reverse holds because |f/B| > |f|. 2 5.3.3 Exercise If f ∈ Hp, then there exist functions g, h ∈ H2p such that f = gh.

Some inequalities Isoperimetric inequality Let G be a domain with rectiﬁable boundary. Then there holds the (isoperimetric) 2 inequality |G| 6 |∂G| /4π, and equality occurs iﬀ G is a disk. This inequality can be rewritten as Z 1 Z π 2 |f 0|2 dA |f 0(eiθ)| dθ , 6 2π D −π where f is a conformal mapping of D onto G. Here dA denotes the Lebesgue measure on D normalized so that the measure of D is 1. Thus the isoperimetric inequality is a consequence of the following theorem of Carleman.

5.3.4 Theorem If f ∈ Hp, p > 0, then Z 2p 2p |f| dA 6 kfkp . (5.14) D Equality occurs iﬀ f(z) = c(1 − az)−2/p, where c and a are complex constants, |a| < 1.

Proof. Let f ∈ Hp. Then we can write f = Bg2/p, where B is a Blaschke product and g is in H2 and has no zeros in D. Then Z Z 2p 2 2 2p 2 |f| dA 6 |g | dA and kfkp = kgk2. D D 5.3 Blaschke product 83

We have

Z ∞ n 2 2 2 X An X g dA = , where An = g(k)g(n − k) . n + 1 b b D n=0 k=0

Since n X 2 2 An 6 (n + 1) |gb(k)| |gb(n − k)| , k=0 we have Z ∞ n 2 2 X X 2 2 2 g dA 6 |gb(k)| |gb(n − k)| = kgk2. D n=0 k=0 This proves the inequality. If in (5.14) equality holds, then we see from the preceding that B = 1 and that there exists a sequence λn such that gb(k)gb(n − k) = λn −1 for 0 6 k 6 n. This implies that g(z) = c(1 − az) ; etc. 2

5.3.5 Exercise [62] If f ∈ Hp, p > 0, and n = 2, 3,... , then

Z 1/np np 2 n−2 (n − 1) |f(z)| (1 − |z| ) dA(z) 6 kfkp. D It is not known whether this holds for other values of n.

5.3.6 Exercise Let q > p > 0 and r = pp/q. If q/p is an integer and f ∈ Hp, then Z π 1/q 1 iθ q |f(re )| dθ 6 kfkp. 2π −π

Inequalities of Riesz/Fej´erand Hilbert

If g is a function analytic in D, then a special case of the Riesz/Zygmund theorem (see (6.5)) states that Z 1 0 0 |g (r)| dr 6 πkg k1. −1 Replacing here g0 by f and using Riesz’ factorization we get the Riesz/Fejer inequality: Z 1 p p p |f(r)| dr 6 πkfkp (f ∈ H , p > 0). (5.15) −1 P∞ 2n In particular, if p = 2 and f(z) = n=0 anz (an > 0), then (5.15) yields

∞ X aman X π |a |2. (5.16) m + n + (1/2) 6 n m,n>0 n=0 84 5 Classical Hardy spaces

This inequality, known as Hilbert’s inequality, can be deduced immediately from the equality Z 1 Z π f(r)2 dr = i f(eiθ)2eiθ dθ, −1 0 a consequence of Cauchy’s integral theorem. From (5.16) it follows that ∞ 1/2 ∞ 1/2 X ambn X X π |a |2 |b |2 . (5.17) m + n + (1/2) 6 n n m,n>0 n=0 n=0

Hardy’s inequality From Hilbert’s inequality we can obtain a slightly improved version of Hardy’s inequality (5.13). Namely:

5.3.7 Theorem If f ∈ H1, then ∞ X |fb(n)| πkfk . (5.18) n + (1/2) 6 1 n=0 Proof. Let f ∈ H1 and let f = Bg be the Riesz’ factorization of f. Then the 1/2 1/2 2 2 2 functions F = Bg and G = g belong to H and kfk1 = kF k2 = kGk2. Let ak = |Fb(k)| and bk = |Gb(k)|. Then we have ∞ ∞ n X |fb(n)| X 1 X X ambn a b = . n + (1/2) 6 n + (1/2) k n−k m + n + (1/2) n=0 n=0 k=0 m,n>0 Now we use (5.17) to get

∞ X |fb(n)| πkF k kGk = πkfk . 2 n + (1/2) 6 2 2 1 n=0 5.3.8 Remark In the case p = 2 the isoperimetric inequality (5.3.4) can be written as ∞ X |fb(n)|2 kfk2. (5.19) n + 1 6 1 n=0 It is interesting to compare this inequality with (5.18). In general, convergence P of the series b|f(n)|/(n + 1), with f ∈ H(D), does not imply convergence of P 2 1 |fb(n)| /(n + 1). However, if f ∈ H , then |fb(n)| 6 kfk1, and therefore (5.18) implies a weak form of (5.19), namely

∞ X |fb(n)|2 πkfk2. n + 1 6 1 n=0

2 2 0 2 On the other hand, (5.19) implies kfk1 − |f(0)| > (1/2)|f (0)| , which cannot be deduced from (5.18). 5.4 Inner and outer functions 85

5.4 Inner and outer functions

Inner-outer factorization Riesz’ factorization theorem can be reﬁned by introducing inner and outer functions. Suppose that a function f ∈ Hp has no zeros in D. Then log |f| is harmonic and belongs to h1, which follows from (5.9). Nevertheless inequality (5.7) may be 1 + z strict; this is the case if, e.g., f is the so called atomic function exp − . The 1 − z atomic function satisﬁes the following: iθ (a) |S(z)| 6 1 for z ∈ D; (b) |S∗(e )| = 1 almost everywhere. An analytic function satisfying (a) and (b) is called an inner function.

Singular inner functions If an inner function has no zeros, then it is called a singular inner function. The following theorem describes a connection between singular inner functions and singular measures.

5.4.1 Theorem A function S ∈ H(D) is a singular inner function iﬀ there exists a nonnegative singular measure σ on T such that 1 Z ζ + z S(z) = eic exp − dσ(ζ) , (5.20) 2π ζ − z T where c is a real constant.

Proof. If S is inner, then the function u = log |S| is negative and, by the Riesz/Herglotz theorem, there exists a nontrivial positive measure σ ∈ M(T) such that u = −P [σ]. The measure is singular because u(reiθ) → 0, r → 1 (Corol- lary 3.3.5). Then, by passing to “analytic completion”, we get

1 Z ζ + z log S(z) = − dσ(ζ) + ic, 2π ζ − z T where c is a real constant, and this implies (5.20). The rest of the proof is simpler and we omit it. 2

Outer functions A function F is called an outer function if 1 Z ζ + z F (z) = eic exp log ψ(ζ) |dζ| , (5.21) 2π ζ − z T

1 where ψ > 0 is a measurable function such that log ψ ∈ L (T). By the theorems of Fatou and Privalov/Plessner, F has radial limits and we have |F∗(ζ)| = ψ(ζ), ζ ∈ T a.e. p If f ∈ H , then, in view of (5.7), we can deﬁne F by (5.21) with ψ = |f∗|. The function ω = f/F is then inner and we have the factorization f = ωF . This leads to Smirnov’s factorization theorem. 86 5 Classical Hardy spaces

5.4.2 Theorem Every function f ∈ Hp, p > 0, admits a representation f = BSF , where B is a Blaschke product, S is a singular inner function and F is an outer function. We have kfkp = kF kp, |f(z)| 6 |F (z)|, and |f∗| = |F∗|. We call F and I = BS the outer factor and the inner factor of f, respectively. The factorization f = IF is called the inner-outer factorization of f. This factorization is unique if we require, for instance, that I(0) is a positive real number.

Exercises 5.4.3 A positive, nonconstant harmonic function u is equal to the Poisson integral of a singular measure iﬀ there exists an inner function ω such that

1 + ω(z) u(z) = Re . 1 − ω(z)

5.4.4 If f is a nonconstant inner function, then there holds strict inequality in (5.7). On the other hand, if a function f ∈ Hp is outer, then

Z π iθ 1 it log |f(re )| = P (r, θ − t) log |f∗(e )| dt (0 6 r < 1). 2π −π

Conversely, if this equality holds for a ﬁxed reiθ ∈ D, then f is outer. In particular, a function f ∈ Hp is outer iﬀ Z π 1 it log |f(0)| = log |f∗(e )| dt. 2π −π 5.4.5 Every outer function can be represented as the ratio of two bounded outer functions. Consequently, every Hp-function is the ratio of two bounded analytic functions.

5.4.6 If f ∈ Hp and 1/f ∈ Hp for some p > 0, then f is outer. By Theorem 4.4.2, f is in Hp if Re f > 0. Since Re(1/f) = f/¯ |f|2, we see that f is outer if Re f > 0.

Addendum: Riesz’ representation theorem The factorization theorems of Riesz and Smirnov are closely related to the Riesz’ representation theorem for subharmonic functions. We formulate this theorem following H¨ormander [30, §3.3].

5.4.7 Theorem (a) A function u 6≡ −∞, subharmonic in D, can be represented in the form Z 1 w − z u(z) = h(z) + log dµ(w), (5.22) 2π 1 − wz¯ D where h is a function harmonic in D and µ is a positive measure in D, iﬀ the function I(r, u) (0 < r < 1) is bounded from above, which is equivalent to the requirement 5.4 Inner and outer functions 87

that u possesses a harmonic majorant in D. The function h and the measure µ are uniquely determined by u; h is the smallest harmonic majorant of u and µ is equal to the Riesz measure of u. (b) Let µ be a positive measure on D. If the integral in (5.22) converges for some z ∈ , then D Z (1 − |z|) dµ(z) < ∞. (5.23) D Conversely, this condition implies that the integral in (5.22) deﬁnes a subharmonic function 6≡ −∞, the smallest harmonic majorant of which is ≡ 0.

When specialized to the case u = log |f|, f ∈ H(D), this theorem yields the following extension of Theorem 5.3.2.

5.4.8 Theorem Let f ∈ H(D), f 6≡ 0, and Z π 1 iθ log |f(re )| dt 6 C (0 < r < 1). (5.24) 2π −π

If {an} is the sequence of zeros of f, then the Blaschke condition is satisﬁed and f = Bg, where B is the corresponding Blaschke product and g has no zeros in D. The smallest harmonic majorant of log |B| is ≡ 0. Further, if f admits a factorization f = f1f2 with |f1(z)| 6 1 and f2(z) 6= 0 for z ∈ D, then (5.24) holds and |f1(z)| 6 |B(z)|, |f2(z)| > |g(z)|.

Here we only note that if u = log |f|, f ∈ H(D), then condition (5.23) reduces P to the Blaschke condition because the Riesz measure of log |f| is equal to 2π δan .

Beurling’s approximation theorem This theorem can be viewed as a generalization of the fact that the set, Q, of all polynomials is dense in Hp for 0 < p < ∞.

5.4.9 Theorem Let f ∈ Hp, 0 < p < ∞. Then the closed linear span in Hp of the set Qf = {qf : q ∈ Q} is equal to BSHp, where BS is the inner factor of f. In particular, if f is outer, then the set Qf is dense in Hp.

Proof. [30] Since |BS| 6 1, we see that it suﬃces to prove that QF is dense in Hp. Proving this reduces to proving that Q ⊂ QF because Q is dense in Hp. Now we see that the proof of the theorem reduces to the proof that 1 ∈ QF . Let

1 Z ζ + z F (z) = eic exp λ(ζ) |dζ| , 2π ζ − z T Z ic 1 ζ + z Fs(z) = e exp λ(ζ) |dζ| . 2π λ(ζ)

~~The functions Fs have the properties: (a) |Fs(ζ)| 6 1 for s < 0 and Fs(ζ) → 1 as ∞ s → −∞; (b) Fs = FGs, where Gs ∈ H . 88 5 Classical Hardy spaces~~

It follows from (a) that kFs − 1kp → 0 as s → −∞ so, by (b), it remains to prove that Fs ∈ QF . To prove this let Hρ(z) = Gs(ρz), ρ < 1. Then Z π p 1 iθ p iθ iθ p kFHρ − Fskp = |F (e )| |Gs(ρe ) − Gs(e )| dθ → 0, ρ → 1, 2π −π

~~by the dominated convergence theorem. Finally, if ρ < 1, then the Taylor series of Hρ converges uniformly on the unit circle, which implies that FHρ ∈ QF . The proof is complete. 2~~

~~5.5 Composition with inner functions~~

~~Throughout this section we consider nonconstant inner functions. If ω is such a function, then we put~~

~~ω∗(ζ) = lim ω(z), ^ z→ζ~~

~~for those ζ ∈ T for which this limit exists and belongs to T; then we extend ω∗ to a function from T to T in an arbitrary way. Our main purpose is to prove the validity of the relations~~

~~f ∈ Hp ⇐⇒ f ◦ ω ∈ Hp, and~~

~~kf ◦ ωkp = kfkp if ω(0) = 0,~~

~~due to Stephenson [94] (see Theorems 5.5.5 and 5.5.6). These relations as well as all other assertions in this section become obvious when specialized to the case n ω(z) = z , n > 1.~~

5.5.1 Proposition Let ω be an inner function. If φ and g are Borel measurable functions on T such that φ = g a.e., then φ ◦ ω∗ = g ◦ ω∗ a.e. Consequently, if φn are Borel functions on T such that φn → f a.e., then φn ◦ ω∗ → f ◦ ω∗ a.e. Proof. If ω(0) = a, then ω = ϕ ◦ (ϕ ◦ ω), where ϕ(z) = (a − z)/(1 − az¯ ) and ϕ ◦ ω(0) = 0. Therefore we can assume that ω(0) = 0. Let

~~iθ iθ iθ E = {e ∈ T: φ(e ) 6= g(e )}.~~

~~This is a set of measure zero, and we have to prove that the set~~

iθ iθ F = {e ∈ T: ω∗(e ) ∈ E} S is of measure zero. To prove this let ε > 0, let Eε = n In, where In ⊂ T are closed P iθ iθ arcs such that E ⊂ Eε, n |In| < ε, and let Fε = {e ∈ T: ω∗(e ) ∈ Eε} We shall prove that Z 2π Z 2π iθ iθ Kn(ω∗(e )) dθ = Kn(e ) dθ = |In|, (5.25) 0 0 5.5 Composition with inner functions 89

~~where Kn is the characteristic function of In. This implies that~~

Z 2π X iθ |Fε| 6 Kn(ω∗(e )) dθ < ε, n 0 and this implies that F is of measure zero, because F ⊂ Fε for all ε > 0. To prove (5.25), when n is ﬁxed, we choose a sequence φj ∈ C(T) such that it it it it φj(e ) tends to Kn(e ) for every t and |φj(e )| 6 1 for every t. Then φj(ω∗(e )) → it Kn(ω∗(e )), as j → ∞, so we can apply the dominated convergence theorem to reduce the proof to the formula

~~Z 2π Z 2π iθ iθ φ(ω∗(e )) dθ = φ(e ) dθ, φ ∈ C(T). (5.26) 0 0~~

~~Finally, this is reduced to the case where φ is a trigonometric polynomial. The details are left to the reader. 2 The formula (5.26) extends to arbitrary φ ∈ L1(T).~~

~~5.5.2 Theorem [85, 94] If φ is a Borel function of class L1(T) and ω is an inner 1 function with ω(0) = 0, then φ ◦ ω∗ ∈ L (T) and~~

~~Z 2π Z 2π iθ iθ φ(ω∗(e )) dθ = φ(e ) dθ. 0 0~~

Proof. In order to reduce the proof to the case φ ∈ C(T), we can suppose that φ is a positive real function. The sequence min{φ(ζ), n} increases to φ(ζ) everywhere, so the proof reduces to the case where φ is bounded. If φ is bounded, then we choose a bounded sequence φn ∈ C(T) such that φn → φ a.e.; by Proposition 5.5.1, we have φn ◦ ω∗ → φ ◦ ω∗ a.e. The result follows. 2

~~5.5.3 Theorem If φ is a Borel function of class L1(T) and ω is an inner function, then P [φ ◦ ω∗] = P [φ] ◦ ω.~~

Proof. It suﬃces to consider the case where φ ∈ C(T). Then the functions P [φ ◦ ω∗] and P [φ] ◦ ω are harmonic and bounded so it suﬃces to prove that their boundary functions coincide almost everywhere. Since P [φ] is continuous on the iθ iθ closed disk, we have limr→1 P [φ](ω(re )) = φ(ω∗(e )) a.e. On the other hand, iθ iθ limr→1 P [φ ◦ ω](re ) = (φ ◦ ω∗)(e ) a.e., and this completes the proof. 2

~~5.5.4 Corollary If ω and I are inner functions, then so is the composition I ◦ ω, and (I ◦ ω)∗ = I∗ ◦ ω∗ a.e. If in addition I is singular, then so is I ◦ ω.~~

~~Proof. This follows from the relations: P [I∗ ◦ ω∗] = P [I∗] ◦ ω = I ◦ ω and P [(I ◦ ω)∗] = I ◦ ω. 2 90 5 Classical Hardy spaces~~

~~Stephenson’s theorems Combining the above results one easily proves the following.~~

5.5.5 Theorem If f ∈ Hp, p > 0, and ω is inner with ω(0) = 0, then f ◦ ω ∈ Hp, (f ◦ ω)∗ = f∗ ◦ ω∗, and kf ◦ ωkp = kfkp. If f = IF is the inner-outer factorization of f, then f ◦ ω = (I ◦ ω)(F ◦ ω) is the inner-outer factorization of f ◦ ω.

~~We conclude this section by proving the implication f ◦ ω ∈ Hp =⇒ f ∈ Hp.~~

~~5.5.6 Theorem If f ∈ H(D) and ω is an inner function, then f ◦ ω ∈ Hp implies f ∈ Hp.~~

Proof. Assume that ω(0) = 0. Let f ◦ ω ∈ Hp, p > 0, let u = |f|p and p v = |f ◦ ω| , and let h = P [v∗]. We know that v 6 h; see (5.8). Let D = rD, where r, 0 < r < 1, is ﬁxed. Then u 6 M on D for some constant M < ∞. Put −1 Ω = ω (D). For 0 < ρ < 1, let Eρ = {ζ ∈ T: ω(ρζ) ∈ D}. Since 1 − |ω(ρζ)| → 0 a.e. as ρ → 1, we see, using Egorov’s theorem, that limρ→1 |Eρ| = 0. Hence we can choose ρ so that the following is true: If ϕ is the bounded harmonic function in the disk ρD whose values are M on ρEρ and 0 on the rest of ρEρ, then

~~ϕ(0) < 1. (5.27)~~

Since u is subharmonic and continuous in D, there exists a function u1 ∈ C(D) harmonic in D such that u 6 u1 6 M and u1 = u at every point of ∂D. By the mean value property, Z π Z π 1 iθ 1 iθ p u1(0) = u(re ) dθ = |f(re )| dθ. (5.28) 2π −π 2π −π

Since u ◦ ω 6 h in D, we have that u1 ◦ ω 6 h on ∂Ω ∩ ρD. Consider the function u1 ◦ ω − h on the closure of the set Ωρ = Ω ∩ ρD. At boundary points of Ωρ that lie in ρD we have u1 ◦ω −h 6 0. The other boundary points of Ωρ lie in ρEρ, and there u1 ◦ ω − h 6 u1 ◦ ω 6 M. Thus u1 ◦ ω − h 6 ϕ on ∂Ωρ. Since these functions are harmonic in Ωρ and Ωρ 3 0, it follows from (5.27) that u1(0) = u1(ω(0)) 6 h(0)+1. Now the desired result follows from (5.28). 2

~~Approximation by inner functions Let ω be an inner function with ω(0) = 0. If f ∈ H2, then, by Rogosinski’s theorem (see 4.4.7) and Stephenson’s theorem, we have~~

~~∞ ∞ X 2 X 2 |fb(k)| > |Fb(k)| , k=n k=n where f = F ◦ ω. This fact can be expressed in terms of best approximation. 5.5 Composition with inner functions 91~~

Let Qn, n > 0, denote the set of all holomorphic polynomials of degree at most p n. For a function f ∈ H , let En(f)p = infg∈Qn kf −gkp. Then the above inequality 2 can be stated as En(f ◦ ω)2 > En(f)2, f ∈ H . It is interesting that this extends to the case 1 6 p 6 ∞.

p 5.5.7 Theorem [64] Let 1 6 p 6 ∞, f ∈ H , and let ω be an inner function with ω(0) = 0. Then En(f ◦ ω)p > En(f)p. Proof. Let 1 Z π (f, h) = f(eiθ)h(eiθ) dθ. 2π −π We identify Hp with Hp(T). From the general theory of best approximation in Banach spaces we know that

q En(f)p = sup{ |(f, h)|: h ∈ L (T), (Qn, h) = 0, khkq 6 1} (1/p + 1/q = 1), where (Qn, h) = 0 means that (g, h) = 0 for every g ∈ Qn. Now we apply this formula to f ◦ ω and use the following facts: (a) If h ∈ Lq(T), then h ◦ ω ∈ Lq(T) and kh ◦ ωkq = 1, and (b) if (h, Qn) = 0, then h ◦ ω = 0. Fact (a) follows from Theorem 5.5.2, while the proof of (b) is straightforward—it is enough to observe that (h, Qn) = 0 iﬀ bh(j) = 0 for 0 6 j 6 n. We get

~~q En(f ◦ ω)p > sup{ |(f ◦ ω, h ◦ ω)|: h ∈ L (T), (Qn, h) = 0, khkq 6 1}. This concludes the proof because |(f ◦ω, h◦ω)| = |(f, h)|, by Theorem 5.5.2. 2~~

p 5.5.8 Remark If we change the notation and denote by En(f)p the best L p approximation of f ∈ L (T) by trigonometric polynomials of degree 6 n, then Theorem 5.5.7 remains valid. However, the above method heavily depends on the Hahn/Banach theorem and cannot be applied to the case p < 1. It would be interesting to study this case. 6 Conjugate functions

The main theorems of this chapter are the Privalov/Plessner theorem on the existence of radial limits of conjugate functions and the existence of the Hilbert operator (Theorem 6.1.1), and the Riesz theorem that if u is in hp, 1 < p < ∞, then so is its harmonic conjugate (Section 6.2). The rest of the chapter is devoted to some related results. The equality Lp(T) = Hp(T) + Hp(T), 0 < p < 1, due to Aleksandrov, is in Section 6.4. Section 6.5 contains a theorem on strong convergence in H1. In Section 6.6 we characterize harmonic quasiconformal homeomorphisms of the unit disk via the Hilbert transformation of the derivative of the boundary function.

~~6.1 Harmonic conjugates~~

~~To each f ∈ h(D) there corresponds the harmonic conjugate fe ∈ h(D), ∞ X fe(reiθ) = −i (sign n) fb(n)r|n|einθ. n=−∞~~

~~If f is real-valued, then fe is uniquely determined by the conditions: (a) fe is real- valued, (b) f + ife is analytic, and (c) fe(0) = 0. For an arbitrary f ∈ h(D) there holds~~

~~fee = −f + f(0). (6.1)~~

And if f is analytic, then fe = −i(f − f(0)), and Regf = Im(f − f(0)). The function conjugate to P [φ], φ ∈ L1(T), equals Z π Z π iθ 1 it 1 Pe[φ](re ) = Pe(r, θ − t)φ(e ) dt = Pe(r, t) φ(θ − t) − φ(θ + t) dt, 2π −π 2π 0 where we write φ(x) instead of φ(eix). Here Pe denotes the conjugate Poisson kernel, 2z 1 + z 2r sin θ Pe(z) = Im = Im , i.e., Pe(r, θ) = Pe(reiθ) = . 1 − z 1 − z 1 + r2 − 2r cos θ This kernel does not belong to h1. The kernels P and Pe are connected by the formula 1 1 − r P (r, θ) Pe(r, θ) − = − . (6.2) tan(θ/2) tan(θ/2) 1 + r

~~92 6.1 Harmonic conjugates 93~~

Note that 1/tan(θ/2) = Pe(1, θ). It is known that Pe[φ] need not be in h1 (see6.1.4) but the Kolmogorov/Smirnov theorem 4.4.2 says that there holds the implication f ∈ h1 =⇒ fe ∈ hp, 0 < p < 1.(∗)

~~The Privalov/Plessner theorem~~

~~6.1.1 Theorem If φ ∈ L1(T), then there exist and are equal the following limits (a.e.): 1 Z π φ(θ − t) − φ(θ + t) lim Pe[φ](reiθ) and lim dt. r→1− ε→0+ π ε 2 tan(t/2)~~

iθ The existence of limr→1− Pe[φ](re ) is contained in Corollary 3.3.9. This theorem guarantees the existence of the improper integral 1 Z π φ(θ − t) − φ(θ + t) φe(eiθ) = dt. π 0+ 2 tan(t/2) There exists such a function φ ∈ C(T) that this integral converges absolutely for no θ. It is even more interesting that there exists a function φ ∈ C(T) such that the improper integral Z π φ(θ + t) − φ(θ) dt 0+ 2 tan(t/2) diverges for every θ (see [100, p. 133–4]).

The Hilbert operator The function φe is said to be conjugate with φ and the operator H taking φ to φe is called the Hilbert operator.(†) The Hilbert operator maps L1 into Lp, for every p < 1, but not into L1, so in the general case the Poisson integral of φe has no sense. However, as we will prove later on (see Theorem 6.1.3), if φe ∈ L1, then P [φe] = Pe[φ]. Proof of Theorem 6.1.1. It suﬃces to prove the relation 1 Z π φ(θ − t) − φ(θ + t) lim Pe[φ](reiθ) − dt = 0, (6.3) r→1− 2π 1−r tan(t/2) under the hypothesis that θ is a Lebesgue point of φ. We write the diﬀerence under limr→1− in (6.3) as I1(r) + I2(r), where Z 1−r 1 I1(r) = Pe(r, t) φ(θ − t) − φ(θ + t) dt. 2π 0

Since |Pe(r, t)| 6 2/(1 − r), |t| 6 1 − r, we have Z 1−r 1 |I1(r)| 6 φ(θ − t) − φ(θ + t) dt → 0 (r → 1), π(1 − r) 0 (∗)See also Kolmogorov’s theorem 7.1.10. (†)Usually 2 tan(t/2) is replaced by t. 94 6 Conjugate functions

~~because θ is a Lebesgue point. In the case of the integral Z π 1 1 I2(r) = Pe(r, t) − φ(θ − t) − φ(θ + t) dt 2π 1−r tan(t/2) we use the formula (6.2); it follows that~~

1 Pe(r, t) − const.P (r, t) (1 − r < |t| < π). tan(t/2) 6 Thus Z π 1 |I2(r)| 6 C P (r, t) φ(θ − t) − φ(θ + t) dt. 2π −π Now the hypothesis that θ is a Lebesgue point and the following lemma imply that I2(r) → 0 (r → 1), and this completes the proof. 2

6.1.2 Lemma If θ is a Lebesgue point of a function ψ ∈ L1, then 1 Z π lim sup P (r, t)|ψ(θ + t) − ψ(θ)| dt = 0. r→1− 2π −π Proof. This can be deduced from Proposition 3.3.1 by taking Z t γ(t) = |ψ(θ + x) − ψ(θ)| dx. 2 0

The Poisson integral of the conjugate function 6.1.3 Theorem If φ ∈ L1 and φe ∈ L1, then P [φe] = Pe[φ]. Proof. Assume that φ is real-valued. By the Kolmogorov/Smirnov theorem, the function f = P [φ] + iPe[φ] belongs to Hp for p < 1. Now Smirnov’s theo- 1 rem 5.1.12 tells us that f ∈ H , and hence f = P [f∗], by Theorem 5.2.1. Finally, since f∗ = φ + iφe, by the theorems of Fatou and Privalov/Plessner, we see that

~~P [φ] + iPe[φ] = P [f∗] = P [φ] + iP [φe], and the result follows. 2~~

~~Miscellaneous~~

6.1.4 If {ak} is a convex sequence tending to 0, then the sum of the series P∞ 1 n=1 an cos nθ is positive for every θ ∈ (−π, π), and its sum belongs to L (T) (see [42, Theorem 4.1]). In particular, the function

~~∞ X φ(eiθ) = (log n)−1 cos nθ n=2 1 P∞ −1 is in L , while the function conjugate to φ is equal to n=2(log n) sin nθ and is not in L1. 6.1 Harmonic conjugates 95~~

~~6.1.5 Using (6.1) one can prove the following: If φe ∈ L1, then 1 Z π φe = −φ + φ(eiθ) dθ. 2π −π The Privalov/Plessner theorem can be stated in the following form:~~

~~6.1.6 Theorem Let φ ∈ L1(T) and let Φ(θ) be the indeﬁnite integral of the function θ 7→ φ(eiθ). Then the improper integral~~

1 Z π Φ(θ + t) + Φ(θ − t) − 2Φ(θ) − 2 dt π 0+ 4 sin (t/2) exists for all θ and is equal to φe(θ) almost everywhere. A more general variant states: Let γ ∈ BV [−π, π] and let γ(t + 2π) − γ(t) = const. Then the integral

1 Z π γ(θ + t) + γ(θ − t) − 2γ(θ) − 2 dt π 0+ 4 sin (t/2) and the limit lim PSg[γ](reiθ) r→1− exist and are equal almost everywhere. See Zygmund [100, Ch. III §§7-8, IV §3 and VII §1].

~~The Riesz/Zygmund inequality~~

~~1 1 As we have seen, if g ∈ h , then the conjugate function ge need not belong to h .A result of Riesz and Zygmund [100, Ch. IV, (6.28)] states that if g ∈ h1, then~~

~~Z 1 g(reit) e dr 6 πkgk1. (6.4) −1 r In other words:~~

~~6.1.7 Theorem If g ∈ h(D) and (∂g/∂θ) ∈ h1, then Z 1 ∂g ∂g it (re ) dr 6 π . (6.5) −1 ∂r ∂θ 1~~

~~Proof. We can assume that g is harmonic in a neighborhood of D. The function v = r∂g/∂r is conjugate to the function u = ∂g/∂θ and therefore~~

~~1 Z π v(reiθ) = Im F (reit)u(ei(θ−t)) dt, (6.6) 2π −π 96 6 Conjugate functions~~

~~where F (z) = 2z/(1 − z). Now (6.5) can be deduced from (6.6) and the equation~~

Z 1 |r−1 Im F (reit)| dr = π (0 < |t| < π), (6.7) −1 by using Fubini’s theorem. In order to prove (6.7) let 0 < t < π, and apply Cauchy’s integral theorem to iθ the function F (z)/z on the semidisk {re : 0 6 r 6 1, t 6 θ 6 t + π} (which does not contain the point z = 1). We get

~~Z 1 F (reit) Z t+π dr = −i F (eiθ) dθ. −1 r t Hence Z 1 Im F (reit) Z t+π dr = − Re F (eiθ) dθ. −1 r t Now (6.7) follows from Re F (eiθ) = −1 and Im F (reit)/r > 0. 2~~

Geometric interpretation Let γ be a function continuous and of bounded variation on [−π, π], let γ(π) = γ(−π), and f(reiθ) = P [γ]. The function f can be treated as a continuous mapping of the closed disk to the complex plane.

~~6.1.8 Corollary If f is a homeomorphism, then~~

~~length of f([−1, 1]) 6 (1/2) × length of ∂f(D).~~

~~The hypothesis that the curve z = γ(t), π 6 t 6 π, is a Jordan curve is not suﬃcient for f to be a homeomorphism. A suﬃcient condition is described by Choquet’s theorem [13]:~~

~~6.1.9 Theorem A suﬃcient condition for f to be a homeomorphism is that the curve z = γ(t)(|t| 6 π) is a convex Jordan curve.~~

~~6.2 Riesz projection theorem~~

~~The operator R+ acting from h(D) to H(D) according to the rule (R+u)(z) = P∞ n n=0 ub(n)z is called the Riesz projector.~~

~~The projection theorem~~

~~It is a direct consequence of Parseval’s theorem that R+ acts as an orthogonal projection from h2 onto H2. This fact was generalized by M. Riesz in the following way.~~

~~p 6.2.1 Theorem If 1 < p < ∞, then R+ acts as a bounded projection from h onto Hp. 6.2 Riesz projection theorem 97~~

By the Kolmogorov/Smirnov theorem, we can treat R+ as an operator from 1 p p L (T) to H (T) = {f∗ : f ∈ H } ⊂ L0(T), for p < 1. From the projection theorem and Theorem 3.4.2 it follows that for every φ ∈ Lp(T) (1 0 and ψb(n) = 0 p p for n < 0. This enables us to treat R+ as an operator from L (T) to H (T), 1 < p < ∞. However, the Riesz projection does not map L1 into H1(T) (see 6.1.4); furthermore, H1 is not complemented in L1, i.e., there is no bounded projection from L1 to H1 (see [87]).(‡)

~~The conjugate functions theorem~~

~~Since the Riesz projector is connected with conjugate function in a simple way, namely R+u = u(0) + (u + iue)/2, the Riesz theorem can be stated as follows:~~

~~p p 6.2.2 Theorem If u ∈ h , p > 1, then ue ∈ h and there exists a constant Cp such that kuekp 6 Cpkukp.~~

~~In view of the connection between conjugate functions and the Hilbert operator (Privalov/Plessner theorem, 6.1.1), we have:~~

~~6.2.3 Theorem The Hilbert operator maps Lp(T) to Lp(T) for 1 < p < ∞.~~

In the case p = 2, Theorems 6.2.2 and 6.2.1 follow from Parseval’s formula; we 2 2 2 have kuek2 = kuk2 − |u(0)| . If a proof is known either for 1 2, then the general case can be treated by duality. Let us mention four “short proofs.”

~~1 (i) The operator R+ : L (T) 7→ L0(T) is of strong type (2, 2) and, by Kol- mogorov’s theorem (Theorem 7.1.10), of weak type (1, 1). Therefore we can apply Marcinkiewicz’s theorem.~~

~~(ii) If 1 < p < 2, we can use Theorem 2.6.2 and the existence of the radial limits of ue.~~

~~(iii) If 1 < p < 2, then, as noted after Theorem 7.2.1, we can use Theorems 7.2.1 and 7.1.2.~~

(iv) If p = 2n, for some positive integer n, then we can easily deduce the validity of Theorem 6.2.2 from the case p = 2; see the proof of Theorem 6.2.6. Then, for arbitrary p > 2, we can apply the Riesz/Thorin theorem.

~~Here we present an elementary proof, due to P. Stein, based on the Hardy/Stein identities.~~

~~(‡)If every subspace of a Banach space X is complemented in X, then X is isomorphic to a Hilbert space [51]. 98 6 Conjugate functions~~

Hardy/Stein identities There is a way to express the Hp-norm as a double integral. For example, from Green’s formula (3.3) and the formula ∆(|f|2) = 4|f 0|2 it follows that Z 1 kfk2 = |f(0)|2 + 2 |f 0(z)|2 log dA(z). (6.8) 2 |z| D Concerning other values of p, we consider only the cases that are suﬃcient to prove Riesz’ projection theorem.

~~6.2.4 Lemma Let 0 < p < ∞. A function f ∈ H(D) belongs to Hp if and only if p2 Z 1 HS (f) := |f(0)|p + |f|p−2|f 0|2 log dA < ∞. p 2 |z| D p Moreover we have kfkp = HSp(f).~~

Proof. If f has no zeroes in D, then it suﬃces to apply (6.8) to the function p/2 p 2 f . If p > 2, then the function |f| is of class C so we can apply Green’s formula. In the general case one applies Green’s formula to the functions (|f|2 +ε)p/2, ε > 0, and then let ε tend to 0 (see also 4.5.6). 2 p The formula kfkp = HSp(f) is known as the Hardy/Stein identity. There holds an analogous formula for real-valued harmonic functions. We only need the case of positive functions.

6.2.5 Lemma Let f ∈ H(D) and let u = Re f belong to hp, 1 0 this reduces to Lemma 3.5.4. In the general case one considers the functions (u2 + ε)p/2. 2

~~Proof of Riesz’ theorems We shall prove Theorem 6.2.2. We may suppose that u is real-valued, and then the theorem can be stated as follows.~~

~~6.2.6 Theorem Let f be analytic in D, and let 1 < p < ∞. If Re f ∈ hp, then f ∈ Hp and there holds the inequality~~

p p p kfkp 6 Cp(k Re fkp + |f(0)| ). If f is a conformal mapping of the disk onto the domain G = {z : 0 < Re z < 1}, then Re f ∈ h∞ but f is not in H∞; therefore, the theorem does not hold for p = ∞. For the case p = 1 see 6.1.4. Proof of Theorem 6.2.6. Consider ﬁrst the case 1 0 and, as in the proof of 6.3 Applications of the projection theorem 99

Lemma 5.1.7, that f is analytic in a neighborhood of D. Then from Lemmas 6.2.4 p−2 p−2 and 6.2.5, together with the inequality u > |f| , it follows that p kfkp − |f(0)|p (kukp − |u(0)|p), p 6 p − 1 p which gives the desired result for 1 < p 6 2. Let 2 < p 6 4 and let f be analytic in a neighborhood D. Then the function g = −if 2 belongs to Hq, q = p/2, and we have Re g = 2uv, where v = Im f, p q q and therefore, by the preceding case, kfkp = kgkq 6 Cpk2uvkq. On the other q p/2 p hand, Cauchy/Schwarz inequality gives kuvkq 6 (kukpkvkp) . Hence kfkp 6 q p/2 p/2 p/2 Cp2 kukp kfkp . Since kfkp is ﬁnite, we can divide both sides by kfkp , and this yields the result for 2 6 p 6 4; etc. 2 6.2.7 Remark If 0 < p < 1, then (6.9) does not hold, but we still have p(1 − p) Z 1 up−2|f 0|2 log dA = u(0)p − M p(1, u) u(0)p, 2 |z| p 6 D provided u is positive and continuous on the closed disk. When combined with Lemma 6.2.4, this yields another proof of Theorem 4.4.2.

~~6.3 Applications of the projection theorem~~

The projection theorem has many important applications. For example, the trigonometric system is a Shauder basis in Lp(T) for 1 < p < ∞; in other words, the system of the functions r|n|einθ is a Shauder basis in hp.(§)

~~n p iθ P ikθ 6.3.1 Theorem Let φ ∈ L (T), 1 < p < ∞, and φm,n(e ) = φb(k)e , where m and n are integers, m < n. Then k=m~~

~~kφm,nkp 6 Cpkφkp (6.10) kφ − φm,nkp → 0 as n → ∞, m → −∞. (6.11)~~

iθ ikθ Proof. Let ek(e ) = e . Then φm,n = em R+(e−m φ) − en R+(e−n φ). From this and Theorem 6.2.1 we obtain (6.10), and from (6.10) and the Weierstrass approximation theorem we obtain (6.11). 2 Now we can determine the dual of Hp for 1 < p < ∞.

~~6.3.2 Theorem If 1 < p < ∞, then the dual of Hp is isomorphic to Hq (1/p + 1/q = 1) with respect to the bilinear form~~

~~∞ 1 Z π X (f, g) = lim f(re−iθ)g(reiθ) dθ = lim fb(n)g(n)r2n. r→1− 2π r→1− b −π n=0~~

~~(§)However, there are spaces, e.g., Bergman, in which this system is not a basis although there holds the analogue of the projection theorem. 100 6 Conjugate functions~~

p Proof. Let Λ be a bounded linear functional on H and Λ1 be the Hahn/Banach extension to hp. By the projection theorem and Theorem 3.4.2, there exists a q p function g ∈ h such that kΛ1k = kg1kq and Λ1f = (f, g1) for f ∈ h . Hence, p q Λf = (f, g1) for f ∈ H . The function g = R+g1 belongs to h and we have p (f, g) = (f, g1) for f ∈ H (because f is analytic), and this proves the inclusion (Hp)∗ ⊂ Hq. The reverse inclusion is proved by using H¨older’s inequality. 2

~~6.3.3 Exercise (isomorphism Lp to Hp) If 1 < p < ∞, then the formula~~

~~∞ ∞ X 2n X 2n−1 (T u)(z) = ub(n)z + ub(−n)z n=0 n=1 deﬁnes an isomorphism of hp onto Hp.~~

~~6.3.4 Exercise (Parseval’s formula) If f ∈ Lp(T) and g ∈ Lq(T), where 1/p + 1/q = 1 and 1 < p < ∞, then the series~~

~~∞ X |n| inθ fb(n)gb(n) r e n=−∞~~

~~converges uniformly in D, and there holds Parseval’s formula: ∞ 1 Z π X f(eiθ)g(eiθ) dθ = fb(n)g(n). 2π b −π n=−∞~~

~~6.4 Aleksandrov’s theorem~~

Relation (5.5) shows that Lp(T) contains an isometric copy of Hp (p > 0); denote p p p p this subspace by H (T). Thus H (T) = {f∗ : f ∈ H }. If p > 1, then H (T) can be described in the following way:

~~p p H (T) = {φ ∈ L (T): φb(−n) = 0 for n > 1}. In the case p < 1, this fact does not hold, simply because the Fourier coeﬃcients are not deﬁned; then Hp(T) is equal to the Lp-closure of~~

T+ = {φ ∈ T : φb(−n) = 0 for n > 1}, where T is the set of all trigonometric polynomials. Let Hp(T) = {φ : φ ∈ Hp(T)}. One of consequences of the projection theorem is that Lp(T) = Hp(T)+Hp(T), 1 < p < ∞. This fact was extended by Aleksandrov [4, 3] to the case p < 1. However, in that case, the decomposition is not unique (up to an additive constant) because the intersection Hp ∩ Hp is equal to the linear span of the set of the functions ga(ζ) = 1/(1 − aζ)(a ∈ T, ζ ∈ T) (see [4, 3]).

~~6.4.1 Theorem (Aleksandrov) If f ∈ Lp(T), p < 1, then there are functions p p f1 ∈ H (T), f2 ∈ H (T), such that f = f1 + f2 and kf1kp + kf2kp 6 Cpkfkp. 6.5 Strong convergence in H1 101~~

Proof. Let X denote the direct sum of the spaces Hp and Hp. Consider the p operator T : X 7→ L , T (f1, f2) = f1 + f2. For every trigonometric polynomial f, kfkp 6 1, we will ﬁnd (f1, f2) so that f = T (f1, f2), where k(f1, f2)k 6 Cp (and Cp depends only of p) and then the result will follow from Theorem 1.3.1. P ikθ n ¯ p Let f = ake and γ(ζ) = ζ . Then γf and γf belong to H . Put |k|6n p p it ϕ = γ/(γ − 1). Then ϕ ∈ H ∩ H and ϕ +ϕ ¯ = 1. Now let ϕt(ζ) = ϕ(ζe ), ¯ p ¯ t ∈ R, and gt = fϕt, ht = fϕt. Then gt and ht are H and f = gt + ht. Routine calculation shows that Z π 1 p p p kgtkp dt = kϕkp kfkp, 2π −π which means that there exists t such that kgtkp 6 kϕkp. For this value of t we p p p p ¯ have khtkp 6 kfkp + kgtkp 6 1 + kϕkp. Finally, we take f1 = gt, f2 = ht, and this completes the proof. 2

~~Kalton’s theorem Let Bp denote the space of functions f ∈ H(D) such that~~

Z 1/p 0 p p−1 kfkBp := |f(0)| + |f (z)| (1 − |z|) dA(z) < ∞. D By the Littlewood/Paley theorem, we have Bp ⊂ Hp for 1 < p < 2. This inclusion remains valid for p 6 1, which follows from the inequality p p −np p 0 −n Mp (rn+1, f) − Mp (rn, f) 6 Cp2 Mp (rn+1, f ), rn = 1 − 2 , see Proposition 7.1.6. Therefore the following result of Kalton [37] improves The- orem 6.4.1.

~~6.4.2 Theorem If f ∈ Lp(T), p < 1, then there are functions g ∈ Bp, h ∈ Bp, such that f = g∗ + h∗ and kgkBp + khkBp 6 Cpkfkp.~~

~~6.5 Strong convergence in H1~~

For a function f analytic in D let n n 1 X 1 X 1 P f = s f, where A = (n = 0, 1, 2,...) n A j + 1 j n j + 1 n j=0 j=0 and sjf are the partial sums of the Taylor series of f. It is well known that ksnfk 6 CAnkfk and that An is “best possible”. A direct consequence is that n 1 X 1 ks fk Ckfk log n (n 2). (6.12) log n j + 1 j 6 > j=0 102 6 Conjugate functions

~~where C is an absolute constant. It turns out, however, that there holds the stronger inequality n 1 X 1 ks fk Ckfk (f ∈ H1, n 2). (6.13) log n j + 1 j 6 > j=0~~

~~Moreover, we have the following characterization of the space H1.~~

~~6.5.1 Theorem [92, 76] For a function f analytic in D the following assertions are equivalent:~~

~~f ∈ H1 ; n 1 X 1 sup ksjfk < ∞ ; (6.14) n A j + 1 n j=0~~

~~sup kPnfk < ∞. (6.15) n~~

Remark. It follows from the proof that the quantities occurring in (6.14) and (6.15) are “proportional” to the original norm in H1; in particular there holds (6.13). Since the polynomials are dense in H1, we have the following consequence:

~~6.5.2 Theorem If f ∈ H1, then~~

~~n 1 X 1 lim kf − sjfk = 0 n A j + 1 n j=0~~

~~and, consequently, n 1 X 1 lim ksjfk = kfk. n A j + 1 n j=0~~

~~1 6.5.3 Corollary [76] If f ∈ H , then lim inf kf − snfk = 0. n→∞~~

1 There are functions φ ∈ L such that limn kφ − snφk = ∞; such an example is iθ P∞ −1/2 −1/2 given by φ(e ) = j=2(log j) cos jθ. Since the sequence (log j) is convex, 1 the function belongs to L (see 6.1.4). Furthermore, one can show that kf −snfk > c(log n)1/2, c = const. > 0. We omit the details. By means of Fatou’s lemma, from Corollary 6.5.3 we obtain:

~~1 iθ iθ 6.5.4 Corollary [76] If φ ∈ H ( ), then lim inf |φ(e ) − snφ(e )| = 0 a.e. T n→∞~~

~~On the other hand, there exists a function φ ∈ H1(T) whose Fourier series diverges almost everywhere (see [100, Ch. VIII, theorem (3.5)]). 6.5 Strong convergence in H1 103~~

Konyagin’s theorem The above corollary does not extend to L1; Konyagin [45] proved the following improvement of Kolmogorov’s theorem: If {ψ(m)} is a sequence of positive numbers such that √ √ ψ(m) = o( ln m/ ln ln m) as m → ∞, then there exists a function φ ∈ L1(T) such that iθ lim sup smφ(e )/ψ(m) = ∞ for all θ ∈ T. m→∞

Proof of Theorem 6.5.1 It is obvious that (6.14) implies (6.15). To prove that the condition f ∈ H1 1 implies (6.14) let f ∈ H and for ﬁxed n > 2 and w ∈ D deﬁne the function g ∈ H1 by −1 g(z) = (1 − rz) f(rwz)(|z| 6 1), P∞ j j where r = 1−1/n. We have g(z) = j=0 sjf(w)r z . Applying Hardy’s inequality (Theorem 5.3.7) we get

~~∞ ∞ X 1 X 1 |s f(w)|rj = |g(j)| πkgk. j + 1 j j + 1 b 6 j=0 j=0~~

n X 1 Z 2π ks fk (1/2c)kfk |1 − reit|−1dt, j + 1 j 6 j=0 0 where we have used Fubini’s theorem. Finally, using the estimate Z 2π it −1 1 |1 − re | dt 6 C log = C log n, 0 1 − r we see that (6.13) holds and therefore that (6.14) is implied by f ∈ H1. Let f be analytic in D. From the uniform convergence of snf on compact sets it follows that Pnf 7→ f uniformly on compact subsets of D. Assuming that kPnfk 6 1 for each n, we have M1(r, Pnf) 6 1 for all n and r < 1. This implies, via the uniform convergence of Pnf on the circles |z| = r, that M1(r, f) 6 1 for every r < 1, which means that kfk 6 1. Thus we have proved that (6.15) implies f ∈ H1, and this completes the proof. 2 104 6 Conjugate functions

~~Remarks~~

6.5.5 Inequality (6.12) is optimal in L1 in the sense that log n cannot be replaced by any ψ(n) (independent of f) such that ψ(n) = o(log n). To see this one takes f to be the Poisson kernel, then let r tend to 1 and use the norm estimate for the Dirichlet kernel.

6.5.6 Using Fejer’s theorem one shows, by summation by parts, that if f ∈ h1, then supn kPnfk < ∞, where Pn is extended to harmonic functions in the obvious 1 way. Conversely, if f is harmonic in D and supn kPnfk < ∞, then f ∈ h .

~~6.6 Quasiconformal harmonic homeomorphisms(§)~~

Throughout this section we denote by ϕ a continuous increasing function on R such that ϕ(t + 2π) − ϕ(t) ≡ 2π, so that the function γ(t) = eiϕ(t) is 2π-periodic and continuous, and of bounded variation on [0, 2π]. We consider the harmonic mapping f deﬁned on D = {z : |z| < 1} by

~~1 Z π f(z) = P (r, θ − t)γ(t) dt (z = reiθ). (6.16) 2π −π~~

By Choquet’s theorem (Theorem 6.1.9), f is a homeomorphism of D onto D. Con- versely, every orientation-preserving homeomorhism f : D 7→ D, harmonic in D, can be represented in the form (6.16).(¶) A consequence of Choquet’s theorem and a result of Lewy [48] is that the Jacobian of f is strictly positive in D, i.e.,

~~2 ¯ 2 Jf (z) = |∂f(z)| − |∂f(z)| > 0 (z ∈ D). (6.17)~~

Being harmonic, the mapping f can be represented as f(z) = h(z) + g(z), g(0) = 0, where h and g are analytic in D and uniquely determined by f. We can rewrite (6.17) as g0(z) < 1 (z ∈ ). (6.18) h0(z) D

~~Characterization theorem~~

~~We characterize those ϕ for which f is quasiconformal, i.e., for which (6.18) can be improved to g0(z) k = sup < 1. (6.19) 0 z∈D h (z)~~

~~(§)This section is almost identical to the paper [80]. (¶)Various properties of harmonic homeomorphisms are described in [14]. 6.6 Quasiconformal harmonic homeomorphisms 105~~

6.6.1 Theorem The mapping f is quasiconformal iff the function ϕ is bi-Lipschitz and the Hilbert transformation of ϕ0 is essentially bounded on R. In other words, f is quasiconformal iff ϕ is absolutely continuous and satisfies the conditions:

~~ess inf ϕ0 > 0, (6.20) ess sup ϕ0 < ∞, (6.21) Z π 0 0 ϕ (θ + t) − ϕ (θ − t) ess sup dt < ∞. (6.22) θ∈R +0 t~~

The proof that the three conditions are suﬃcient is short; we simply compute the radial limits of the modulus of the bounded analytic function g0/h0 and apply the maximum modulus principle. The necessity proof is more complicated and depends on Mori’s theorem in theory of quasiconformal mappings (cf. Ahlfors [2]), which states that if Φ is a quasiconformal homeomorphism of D, then

~~α |Φ(z1) − Φ(z2)| 6 C|z1 − z2| (z1, z2 ∈ D), (6.23) where~~

1 − k ∂¯Φ(z) α = , k = sup 1 + k z∈D ∂Φ(z) and C depends only on f(0)(k). The mapping |z|α(z/|z|) shows that the exponent α cannot be improved in the class of arbitrary k-quasiconformal homeomorphisms. However, it follows from our proof (see (6.32) ) that if Φ is harmonic, then it satisﬁes the ordinary Lipschitz condition (with Lipschitz constant depending on k). 0 2 0 2 2 Combining this with Heinz’ inequality [29] |h (z)| + |g (z)| > 1/π , z ∈ D, which holds if f(0) = 0, we get the following.

~~6.6.2 Theorem If the mapping f is quasiconformal, then it is bi-Lipschitz, i.e., there is a constant L < ∞ such that~~

~~1 f(z1) − f(z2) 6 6 L (z1, z2 ∈ D). L z1 − z2 Note that an arbitrary bi-Lipschitz homeomorphism is quasiconformal.~~

Boundary values of the derivatives In calculating the boundary values of the analytic functions h0 and g0 it is useful to use the formulas 1 f (z) h0(z) = ∂f(z) = e−iθ f (z) − i θ (6.24) 2 r r 1 f (z) g0(z) = ∂f¯ (z) = eiθ f (z) + i θ , (6.25) 2 r r (k)C = 16 if Φ(0) = 0. 106 6 Conjugate functions

where fθ = ∂f/∂θ, fr = ∂f/∂r. The derivatives fr and fθ are connected by the simple but fundamental fact that the function rfr is equal to the harmonic conjugate of fθ. It follows from (6.16) that fθ equals the Poisson–Stieltjes integral of γ = eiϕ : Z π iθ 1 fθ(re ) = P (r, θ − t) dγ(t). 2π −π

Hence, by Fatou’s theorem, the radial limits of fθ exist almost everywhere and iθ 0 limr→1− fθ(re ) = γ0(θ) a.e., where γ0 is the absolutely continuous part of γ. It iθ 0 turns out that if γ is absolutely continuous, then limr→1− fr(re ) = H(γ )(θ), a.e.

Absolute continuity The function γ, of course, need not be absolutely continuous. However: If Z π 1 iθ sup fr(ρe ) dθ < ∞, ρ<1 2π −π then γ is absolutely continuous and, moreover, the functions h(eiθ) and g(eiθ) are absolutely continuous. This is one of possible formulations of Theorem 5.2.5. Using (6.24) and (6.25) we can easily show that (6.19) implies

1 − k rfr(z) 1 + k 6 6 (z ∈ D). 1 + k fθ(z) 1 − k Thus: If f is quasiconformal, then ϕ is absolutely continuous. From now on we suppose that ϕ is absolutely continuous. Then there hold the iθ 0 0 iϕ(θ) formulas fθ(e ) = γ (θ) = iϕ (θ)e and, by Theorem 6.1.6, Z π iθ 0 1 γ(θ + t) + γ(θ − t) − 2γ(θ) fr(e ) = H(γ )(θ) = − 2 dt. π +0 4 sin (t/2)

~~−iϕ(θ) iθ By straightforward computation we ﬁnd that e fr(e ) = A(θ) + iB(θ), where~~

1 Z π 2 − cos ϕ(θ + t) − ϕ(θ) − cos ϕ(θ − t) − ϕ(θ) A(θ) = 2 dt π +0 4 sin (t/2) 1 Z π sin ϕ(θ + t)/2 − ϕ(θ)/22 = dt, 2π −π sin(t/2) 1 Z π sin ϕ(θ + t) − ϕ(θ) + sin ϕ(θ − t) − ϕ(θ) B(θ) = − 2 dt. π +0 4 sin (t/2)

~~Then using (6.24) and (6.25) we get 1 |h0(eiθ)|2 = (A(θ) + ϕ0(θ))2 + B(θ)2, (6.26) 4 1 |g0(eiθ)|2 = (A(θ) − ϕ0(θ))2 + B(θ)2. 4 6.6 Quasiconformal harmonic homeomorphisms 107~~

~~Since the function g0/h0 is analytic and bounded, by (6.18), we see that~~

0 2 0 2 2 2 0 2 g (z) ϕ (θ) + A(θ) + B(θ) − 2ϕ (θ)A(θ) k = sup 0 = ess sup 0 2 2 2 0 . z∈D h (z) θ ϕ (θ) + A(θ) + B(θ) + 2ϕ (θ)A(θ) Hence: The mapping f is quasiconformal iﬀ ϕ0(θ)2 + A(θ)2 + B(θ)2 K := ess sup 0 < ∞. (6.27) θ∈R 2ϕ (θ)A(θ) K − 11/2 There holds the formula k = . K + 1

~~Proof of the characterization theorem Now it is easy to show that conditions (6.20), (6.21) and (6.22) imply that f is quasiconformal. We have only to note that condition (6.21) implies~~

~~0 0 2 kB − Hϕ k∞ 6 Ckϕ k∞, (6.28) where C is an absolute constant; this inequality is deduced from Theorem 6.1.6 by using the relation x − sin x = O(x3).~~

~~The necessity proof. Let f be quasiconformal. Then K < ∞ (see (6.27) ), i.e.,~~

0 2 2 2 0 ϕ (θ) + A(θ) + B(θ) 6 2Kϕ (θ)A(θ). (6.29) 2 0 It follows that A(θ) 6 2Kϕ (θ)A(θ) and therefore 1 ϕ0(θ) A(θ). (6.30) > 2K Since Z π 1 A(θ) > 1 − cos(ϕ(θ + t) − ϕ(θ)) dt 4π −π 1 1 = 1 − Re e−iϕ(θ)f(0) 1 − |f(0)|, 2 > 2 we get ess inf ϕ0(θ) > 0. Thus condition (6.20) is satisﬁed. In order to verify (6.21) we use the inequality Z π 2 0 ϕ(θ + t) − ϕ(θ) ϕ (θ) 6 C dt (6.31) −π t (C is an absolute constant) which is obtained from (6.29). Assume ﬁrst that ϕ is of class C2 and choose θ so that ϕ0(θ) = max ϕ0 =: M. Let 0 < β < 1. It follows from (6.31) that Z π 2−β ϕ(θ + t) − ϕ(θ) β M 6 C M dt, −π t 108 6 Conjugate functions

~~whence Z π 2−β 1−β ϕ(θ + t) − ϕ(θ) M 6 C dt. −π t Now we apply Mori’s inequality (6.23) to deduce that~~

Z π 1−β α−12−β 1 − k M 6 C1 t dt, α = . 0 1 + k Choose β so that (α − 1)(2 − β) > −1, which is possible because (α − 1)(2 − β) → − 0 α − 1 > −1 as β → 1 , to get max ϕ 6 C2, where C2 depends only on K. From 0 iθ this and (6.30) we get A(θ) 6 2KC2 and hence, by (6.29) and (6.26), |h (e )| 6 C3. The function h0(z) is continuous on the closed disk because the function γ = eiϕ is C2, so we have 0 |h (z)| 6 C3 (z ∈ D), (6.32)

~~and the constant C3 depends only on K. In the general case consider the mappings fn, of D onto D, deﬁned by~~

~~fn(z) = f(wn(z))/rn = hn(z) + gn(z)(rn = 1 − 1/n, n > 2),~~

−1 0 where wn is the conformal mapping of D onto Gn = f (rnD), wn(0) = 0, wn(0) > 0. Since the boundary of Gn is an analytic Jordan curve, the mapping wn can be continued analytically across ∂D, which implies that fn has a harmonic extension across ∂ . Since also D 0 0 0 gn (g ◦ wn)wn 0 = 0 0 6 k, hn (h ◦ wn)wn 0 0 we can appeal to the preceding special case to conclude that |h (wn(z))| |wn(z)|/rn 6 C3, where C3 is independent of n and z. And since Gn ⊂ Gn+1 and ∪Gn = D, we can apply Carath´eodory’s convergence theorem (Theorem 6.6.3 below): wn(z) 0 tends to z, uniformly on compacts, whence wn(z) → 1 (n → ∞). Thus inequal- 0 ity (6.32) holds in the general case. Using this and (6.26) we get ϕ (θ)+|B(θ)| 6 C4. Finally, it remains to apply (6.28).

~~The Carath´eodory convergence theorem Here we prove the simplest variant of Carath´eodory’s theorem; for the general form see [23, Ch. II§5] as well as [19].~~

~~6.6.3 Theorem Let fn : D 7→ D be a sequence of univalent functions such that 0 S fn(0) = 0, fn(0) > 0, fn(D) ⊂ fn+1(D) for every n, and fn(D) = D. Then fn(z) → z uniformly on compact subsets.~~

Proof. The set {fn} is relatively compact in H(D) and hence it is enough to prove that every H(D)-convergent subsequence of {fn} converges to the function ϕ(z) = z. Therefore we can assume that fn tends, uniformly on compact subsets, 0 0 to some function f ∈ H(D). Clearly f(D) ⊂ D and fn(0) → f (0). Let Dρ = 6.6 Quasiconformal harmonic homeomorphisms 109

{z : |z| 6 ρ}, ρ < 1. Since ∪fn(D) = D and Dρ is compact, we see that fn(D) −1 contains Dρ for n > n0, where n0 is large enough. The function g(w) = fn (ρw) 0 0 maps D into D and g(0) = 0 and hence g (0) = ρ/fn(0) 6 1, for n > n0. It follows 0 0 that fn(0) → 1 and hence f (0) = 1. Hence f(z) = z, by Schwarz’ lemma, and the proof is complete. 2

~~A problem~~

Let QCH = {f∗ : f is a q.c. harmonic homeomorhism of D}. Is QCH a group with respect to composition? The set of all quasiconformal harmonic homeomorphisms of D is not a group because the composition of two harmonic functions need not be harmonic. On the other hand, the set of all quasiconformal transformations of D is a group (cf. [2]).

~~Miscellaneous~~

6.6.4 [56] The mapping f = P [eiϕ] is quasiconformal if ϕ ∈ C1(R), min ϕ0 > 0 and Z π ω(t) dt < ∞, (6.33) 0 t where ω(t) = sup{ |ϕ0(x) − ϕ0(y)| : |x − y| < t} is the modulus of continuity of ϕ0. Condition (6.33), known as Dini’s condition (applied to ϕ0), is suﬃcient but not necessary for the Hilbert transformation of ϕ0 to belong to L∞.

6.6.5 Let G = ϕ(D), where ϕ: D 7→ C is a univalent function such that ϕ0(0) > 0. Let fn : D 7→ G be a sequence of univalent functions such that fn(0) = ϕ(0), 0 fn(0) > 0, fn(D) ⊂ fn+1(D) for every n, and ∪fn(D) = G. Then fn(z) → ϕ(z) uniformly on compact subsets.

6.6.6 Martio [56] proved that condition (6.21) implies that Jf (z) is bounded above iθ on D. The converse is true because Jf (z) = Im( fr(z)fθ(z)), whence Jf (e ) = 0 ϕ (θ)A(θ). Therefore f satisfies a Lipschitz condition on the boundary iff Jf (z) 6 C (z ∈ D), where C is a constant, or, equivalently, |f(E)| 6 C|E| for any measurable set E ⊂ D. Clearly, this does not imply a Lipschitz condition on D. The mapping f satisfies a Lipschitz condition on D iff both ϕ0 and Hϕ0 are bounded. 7 Maximal functions, interpolation, coefficients

~~This chapter concerns two fundamental facts:~~

~~(1) (maximal theorem) The operator Mrad deﬁned on h(D) by~~

~~iθ iθ (Mradu)(e ) = sup |u(re )| 0~~

maps hp to Lp(T) for p > 1, and Hp to Lp(T) for p > 0. (2) (maximal characterization) Let p ∈ (0, ∞). An analytic function f belongs to p p H iff Mrad(Re f) belongs to L (T). In Section 7.3 we state a theorem of Hardy and Littlewood on (C, α)-means in Hp and prove an analogous result concerning “smooth” Cesáromeans. A Marcinki- ewicz theorem for Hp, 0 < p < ∞, with applications to Taylor coefficients, is in Section 7.4. In Section 7.5 we consider some extensions of Hardy and Littlewood’s theorem on Taylor coefficients. Section 7.6 contains some remarks on the dual of H1.

~~7.1 Maximal theorems~~

~~Radial maximal function~~

~~Let u be a complex-valued function deﬁned on D. The radial maximal function of u is the function Mradu deﬁned on T by~~

~~(Mradu)(ζ) = sup |u(rζ)| (ζ ∈ T). 0~~

~~In other words, Mradu is the smallest dominant of the family ur(ζ) = u(rζ).~~

~~1 7.1.1 Proposition If φ ∈ L (T) and f = P [φ], then Mradf(ζ) 6 Mφ(ζ), ζ ∈ T, where M is the maximal operator of Hardy and Littlewood deﬁned by (2.12) and (2.13).~~

~~R t ix Proof. Let ζ = 1, γ(t) = 0 φ(e ) dx and Z t Z t F (t) = φ(eix) + φ(e−ix) dx = φ(eix) dx = γ(t) − γ(−t). 0 −t~~

~~110 7.1 Maximal theorems 111~~

~~From (3.11)(θ = 0), substituting −t for t, we get~~

~~1 1 Z π f(r) = P (r, π)F (π) − P 0(r, t)F (t) dt, 2π 2π 0 where P 0(r, t) = ∂P/∂t. It follows that Z π 1 − r 1 0 |f(r)| 6 Mφ(0) + Mφ(0) |t P (t)| dt. 1 + r π 0~~

~~R π 0 Now the result follows from 0 |t P (t)| dt = 2r/(1 + r). 2~~

Radial maximal theorems The following theorem, due to Hardy and Littlewood, is an immediate consequence of Theorem 2.3.1 and Proposition 7.1.1. This theorem provides additional information on the convergence in (3.17): if p > 1, the convergence is dominated.

1 p 7.1.2 Theorem (radial maximal) The operator Mrad maps h (resp. h , p > 1) into L1,∞(T) (resp. Lp(T)), and is continuous. It is worthwhile to note that this theorem can be proved without appealing to the main maximal theorem. Namely, we can use Nikishin’s theorem, Banach’s principle, and Fatou’s theorem, but Proposition 7.1.1 is of independent interest. Using the properties of subharmonic functions, we can easily deduce the “subharmonic” version of Theorem 7.1.2.

~~7.1.3 Theorem (subharmonic maximal) Let u ba a nonnegative, continuous, subharmonic function on D, and let Z π iθ p sup u(re ) dθ =: Ap < ∞, where p > 1. 0~~

~~p 1/p Then Mradu ∈ L (T) and we have kMradukp 6 Cp(Ap) , where Cp is a constant depending only on p.~~

~~In the case of analytic functions, Theorem 7.1.2 extends to all p > 0; it is enough to take u = |f|p/2 and apply Theorem 7.1.3.~~

7.1.4 Theorem (complex maximal) For f ∈ Hp, p > 0, we have Z π 1 iθ p p Mradf(e ) dθ 6 Cpkfkp. 2π −π 7.1.5 Remark We can use the complex maximal theorem to deduce (5.7) iθ iθ p from (5.10). Namely, from the inequality log |f(ρe )| 6 Mradf(e ) /p, and p the complex maximal theorem it follows that Mradf is an integrable majorant of the family log |fρ| and therefore we may apply Fatou’s lemma to inequality (5.10). 112 7 Maximal functions, interpolation, coefficients

~~A further example:~~

~~7.1.6 Proposition If f ∈ H(D) and 0 < p < 1, then~~

~~p p p p 0 Mp (ρ, f) − Mp (r, f) 6 Cp(ρ − r) Mp (ρ, f ) (0 < r < ρ < 1). Proof. From the inequality~~

~~iθ iθ 0 iθ |f(ρe )| − |f(re )| 6 (ρ − r) sup{|f (se )| : r < s < ρ}, it follows that~~

~~iθ p iθ p p 0 iθ p |f(ρe )| − |f(re )| 6 (ρ − r) sup{|f (se )| : r < s < ρ}. Then the desired inequality is obtained by integration in θ and using the maximal theorem. 2~~

~~Nontangential maximal function Theorems 7.1.2, 7.1.3, and 7.1.4 remain true if we replace the radial maximal function by the nontangential maximal function M∗u,~~

~~M∗u(ζ) = sup |u(z)| (ζ ∈ T), z∈Uζ~~

~~where r0 < 1 is ﬁxed and Uζ = the convex hull of {|z| 6 r0} ∪ {ζ}. These variants are proved by means of the following.~~

~~1 7.1.7 Proposition If φ ∈ L (T) and f = P [φ], then M∗f(ζ) 6 C Mφ(ζ), ζ ∈ T, where C depends only on r0.~~

~~The proof is very similar to that of Proposition 7.1.1 and we omit it; see the proof of Theorem 3.3.7. iθ iθ Although the pointwise estimate M∗u(e ) 6 const Mradu(e ) is not valid, we still have:~~

7.1.8 Theorem (Feﬀerman/Stein [21]) If 0 0 is subharmonic in D, then Z 2π Z 2π iθ p iθ p {M∗u(e )} dθ 6 Cp {Mradu(e )} dθ. 0 0 Proof. The proof is based on the inequality Z q Cq q u(z) 6 2 u(w) dm(w), δ |w−z|<δ

~~due to Hardy and Littlewood; see Theorem 9.1.1. For z = reiθ ∈ D let~~

~~it ∆z = {ρe : |ρ − r| < δ, |t − θ| < δ}, δ = 1 − r. 7.1 Maximal theorems 113~~

~~iη Let z ∈ Uζ , ζ = e ∈ T, and 0 < q < p. It is easy to deduce from the preceding inequality that~~

~~Z Z θ+δ Z r+δ q Cq it q it Cq it q u(z) 6 2 u(ρe ) dm(ρe ) = 2 u(ρe ) ρdρ dt. δ ∆z δ θ−δ r−δ~~

~~Hence Z θ+δ q 2Cq it q u(z) 6 g(t) dt, where g(t) = {Mradu(e )} . δ θ−δ~~

~~iθ Since z = re ∈ Uζ , we have |θ − η| 6 κ(1 − r), κ = const., which implies that~~

~~(θ − δ, θ + δ) ⊂ (η − ε, η + ε), ε = (1 + κ)δ.~~

It turns out that iη q q 0 {M∗u(e )} = sup u(z) 6 CqMg(η), z∈Uζ whence iη p 00 p/q {M∗u(e )} 6 Cq {Mg(η)} . Integrating this inequality from η = 0 to 2π, and using Theorem 2.3.1 we get the desired result. 2

~~Kolmogorov’s theorem It follows from the Theorem 4.4.2 and the complex maximal theorem that if f ∈ H(D) and Re f ∈ h1, then~~

~~kMradfkp 6 Cpk Re fk1, for every p ∈ (0, 1).~~

~~This is improved by the following result of Kolmogorov.~~

~~1 1,∞ 7.1.9 Theorem If f ∈ H(D) and Re f ∈ h , then Mradf ∈ L , and there is a constant C such that kMradfk1,∞ 6 Ck Re fk1.~~

~~This can also be stated as follows.~~

~~7.1.10 Theorem The operator u 7→ Mradue acts as a continuous operator from h1 into L1,∞(T).~~

This theorem is easily deduced from Corollary 3.3.9, Banach’s principle 2.7.1, and the Nikishin/Stein theorem; details are left to the reader. Another form of Kolmogorov’s theorem concerns the maximal Hilbert operator. 114 7 Maximal functions, interpolation, coefficients

~~7.1.11 Theorem Let~~

~~iθ iθ Hmaxφ(e ) = sup |Hεφ(e )|, 0<ε<π where Z π iθ 1 φ(θ − t) − φ(θ + t) Hεφ(e ) = dt. π ε 2 tan(t/2)~~

~~Then Hmax is of weak type (1, 1).~~

~~7.1.12 Remark The maximal Hilbert operator maps Lp to Lp, 1 < p < ∞. See Garnett [22, Ch. 3, Ex. 11].~~

~~7.2 Maximal characterization of Hp~~

~~The following theorem, due to Burkholder, Gundy and Silverstein [11] enables us to treat Hp as a space of harmonic functions and can be used to extend Hp-theory to several real variables (cf. [21]).~~

~~7.2.1 Theorem Let 0 < p < ∞. A function f = u + iv ∈ H(D) belongs to Hp if p and only if the function Mradu belongs to L (T); there holds the inequality~~

p (1/Cp)kfkHp 6 kMradukLp + |v(0)| 6 CpkfkHp . (7.1) Before proving the theorem some remarks are in order. The right-hand side inequality is a consequence of the complex maximal theorem. In the case p > 1, the left-hand side inequality is a formal consequence of the radial maximal theorem p and Riesz’ inequality (1/Cp)kfkHp 6 kukLp +|v(0)| (Theorem 6.2.6). On the other hand, if 1 2 can then be discussed in various ways (for example, as in the proof of Theorem 6.2.6). Proof. By the preceding remarks, it suﬃces to prove the inequality

~~Z 2π Z 2π iθ p + p |v(e )| dθ 6 Cp (u (θ)) dθ, 0 < p < 2, 0 0~~

+ iθ where u (θ) = Mradu(e ), supposing that f is a polynomial and f(0) = 0. By + ∗ ∗ iθ Theorem 7.1.8, we can replace u by u , u (θ) = supζ∈A |u(ζe )|, where √ A = convex hull of {1} ∪ { ζ : |ζ| 6 1/ 2 }. So it suﬃces to prove

~~Z 2π Z 2π iθ p ∗ p |v(e )| dθ 6 Cp (u (θ)) dθ, 0 < p < 2, (7.2) 0 0 where u + iv is a polynomial. 7.2 Maximal characterization of Hp 115~~

∗ ∗ Let Eλ = |{θ ∈ [0, 2π]: u (θ) 6 λ}|, Gλ = {θ ∈ [0, 2π]: u (θ) > λ}, and m(λ) = |Gλ|. Assume we have proved that Z Z 2 2 2 v dθ 6 u dθ + 2λ m(λ), λ > 0. (7.3) Eλ Eλ Multiplying this by qλ−q−1, q = 2 − p > 0, and then integrating from λ = 0 to ∞, and using Fubini’s theorem, we get Z 2π Z 2π Z 2π 2 ∗ −q 2 ∗ −q 2q ∗ 2−q v (u ) dθ 6 u (u ) dθ + (u ) dθ 0 0 2 − q 0 Hence Z 2π Z 2π 2 ∗ p−2 ∗ p v (u ) dθ 6 Cp (u ) dθ, p < 2, (7.4) 0 0 R 2π ∗ p where Cp = 1 + 2(2 − p)/p. To obtain (7.2) assume that 0 (u ) dθ = 1. Then, by Jensen’s inequality,

~~Z 2π 2/p Z 2π p ∗ −p ∗ p p ∗ −p 2/p ∗ p |v| (u ) (u ) dθ 6 {|v| (u ) } (u ) dθ 0 0 Z 2π = |v|2(u∗)p−2 dθ; 0 now (7.2) follows from (7.4).~~

~~Proof of (7.3). We can suppose that 0 < |Eλ| < 2π. Let~~

~~F (z) = sup |u(ζz)|, z ∈ C. ζ∈A~~

The set Fλ = {z ∈ C : F (z) 6 λ} ∩ D is nonempty and simply connected because the function sup{F (z), |z|/λ} is subharmonic in C. Also, Fλ contains the set Hλ = S iθ {e A: θ ∈ Eλ}. Let Γλ = (∂Hλ) r Eλ. By Cauchy’s integral formula, we have 1 Z f(z)2 dz = f(0)2 = 0, 2πi ∂Hλ z whence Z Z − f(eiθ)2 dθ = f(eiθ)2 (dθ + idr/r). Eλ Γλ Taking the real parts we get Z Z Z (v2 − u2) dθ = (u2 − v2) dθ − 2uv dr/r. Eλ Γλ Γλ

~~2 2 Since |dr/rdθ| 6 1 on Γλ and 2|uv| 6 u + v , we conclude that Z Z Z 2 2 2 v dθ 6 u dθ + 2u dθ Eλ Eλ Γλ 116 7 Maximal functions, interpolation, coefficients~~

Now the desired result follows from the inequality Z Z Z 2 2 2 2 u dθ 6 λ dθ = λ dθ = λ m(λ). Γλ Γλ Gλ This completes the proof of the theorem. 2 Remark. The above proof is a slight modification of Koosis’ proof (see Garnett [22]). The only difference is in that Koosis uses (7.3) to show that 2 Z λ |{θ : |v(θ)| > λ}| 6 m(λ) + 2 s m(s) ds. λ 0 7.3 “Smooth” Cesàromeans

In contrast to the case 1 1/p − 1 (p 6 1) (for a proof, see [67]). Instead, we shall construct the “smooth” Cesàrobasis, which has an advantage in that it is “universal”, i.e., independent of p (Theorem 7.3.4). Before, we state the theorem of Hardy and Littlewood. The Cesàro means of order α > −1 of an analytic function f are defined by n Γ(n + 1) X Γ(α + n + 1 − k) σαf(z) = fb(k)zk, n Γ(α + n + 1) Γ(n + 1 − k) k=0 where Γ is the Euler gamma function. In particular n X k σ1 f(z) = 1 − fb(k)zk. n n + 1 k=0

~~Deﬁne the maximal operator σmax by α α (σmaxf)(ζ) = sup |σn f(ζ)| (ζ ∈ T). n~~

It should be noted that the nontangential maximal function M∗f is dominated α by a constant multiple of σmaxf; in the case α = 1, this follows from the inequality |1 − z|2 |f(ζz)| (σ1 f)(ζ), 6 (1 − |z|)2 max and this follows from the formula ∞ 2 X 1 n f(ζz) = (1 − z) (n + 1)(σnf)(ζ) z . n=0 p α 7.3.1 Theorem If 0 1/p−1, and f ∈ H , then kσmaxfkp 6 Cpkfkp, α α iθ iθ lim kf − σn fkp = 0, and lim σn f(e ) = f(e ) a.e. n→∞ n→∞ In the case 1 < p < ∞, there is a much deeper result, due to Carleson[12] and Hunt[31], which states that the above relations hold for the partial sums of f ∈ hp (see p. 47). 7.3 “Smooth” Ces`aromeans 117

~~The polynomials Wn Let ψ be a complex-valued C∞-function with compact support in R. Deﬁne the trigonometric polynomials Wn, n > 1, by X k W (eit) = W ψ(eit) = ψ eikt. n n n |k|<∞~~

~~7.3.2 Lemma For every positive integer N there holds~~

~~iθ −N 1−N |Wn(e )| 6 CN min{n, |θ| n } (|θ| < π), where CN depends only of N and ψ.~~

~~Proof. Suppose that supp ψ is contained in, say, the interval [−2, 2]. Then~~

∞ X k (1 − eiθ)N W (eiθ) = ψ (1 − eiθ)N eikθ n n k=−∞ ∞ N X k X N = ψ (−1)mei(m+k)θ n m k=−∞ m=0 N ∞ X N X k = (−1)m ψ ei(m+k)θ m n m=0 k=−∞ N ∞ X N X k − m = (−1)m ψ eikθ m n m=0 k=−∞ ∞ N ! X X N k − m = (−1)m ψ eikθ m n k=−∞ m=0 N+2n N ! X X N k − m = (−1)m ψ eikθ. m n k=−2n m=0

The inner sum in the last expression, denote it by Sk,N , is equal to the symmetric diﬀerence of order N of the function ψ(x/n). By Lagrange’s theorem, for every k −N (N) (N) there exists ξm,N such that Sk,n = n ψ (ξm,N ). Since the derivative ψ is iθ −N 1−N bounded, we see that |Wn(e )| 6 C|θ| n , for −π < θ < π, with C depending only on N and ψ. On the other hand, from the deﬁnition of Wn it follows that iθ |Wn(e )| 6 Kn, where K = 2 max |ψ|. This concludes the proof. 2

~~The operator Wmax~~

~~For a function f ∈ H(D), we deﬁne the maximal function Wmaxf by~~

~~(Wmaxf)(ζ) = sup |Wn ∗ f(ζ)| (ζ ∈ T). n 118 7 Maximal functions, interpolation, coefficients~~

~~7.3.3 Lemma If 0 < q 6 1, then~~

~~q q q (Wmaxf) 6 Cq M(|f| ), f ∈ H . (7.5) Proof. Let supp(ψ) ⊂ [−2, 2]. Then~~

~~Z π 2n 1 it X −k −ikt |Wn ∗ f(ζ)| f(rζe ) r ψ(k/n)e dt 6 2π −π k=−2n −2n Z π 2n r it X 2n−k (2n−k)it f(rζe ) r ψ(k/n)e dt. 6 2π −π −2n~~

For ﬁxed ζ , n put 2n X g(z) = f(zζ) ψ(k/n) z2n−k k=−2n −2n and rewrite the preceding inequality as |Wn ∗ f(ζ)| 6 r M1(r, g). The function 2 1−1/q g is analytic and hence M1(r, g) 6 (1 − r ) Mq(1, g) (see Corollary 5.1.2). Putting r = 1 − 1/(n + 1), we get Z π q 1−q iθ q |Wn ∗ f(ζ)| 6 C n |g(e )| dθ −π Z π 1−q iθ q iθ q = C n |f(ζe )| |Wn(e )| dθ. −π Hence, by Lemma 7.3.2, we get Z it q 1−q q q |Wn ∗ f(e )| 6 Cn |f(θ + t)| n dθ |θ|<1/n Z + Cn1−q |f(θ + t)|q n(1−N)q|θ|−Nq dθ 1/n<|θ|<π Z = Cn |f(θ + t)|q dθ |θ|<1/n Z π + Cn1−Nq {|f(t + θ)|q + |f(t − θ)|q} θ−Nq dθ. 1/n

~~The ﬁrst summand in the latter sum is dominated by CM(|f|q)(eit). Write the integral in the second summand in the form~~

~~Z π π Z π −Nq −Nq−1 J = θ dF (θ) = F (θ) + Nq F (θ)θ dθ, 1/n 1/n 1/n~~

~~where Z θ F (θ) = |f(t + x)|q + |f(t − x)|q dx. 0 7.4 Interpolation of operators on Hardy spaces 119~~

~~From this we obtain Z π F (θ) −Nq J 6 F (π) + Nq sup θ dθ 0<θ<π θ 1/n~~

~~Nq F (θ) Nq−1 6 F (π) + sup n Nq − 1 θ θ~~

~~Nq−1 q it 6 Cn M(|f| )(e ), where we choose N so that Nq > 1. This completes the proof. 2~~

~~The “W -maximal” theorem From the preceding lemma, by the maximal theorem and the theorem on convergence almost everywhere (Theorem 2.7.2), we get the following.~~

~~7.3.4 Theorem For every p ∈ (0, ∞] we have~~

~~p kWmaxfkp 6 Cpkfkp (f ∈ H ). (7.6) If ψ(0) = 1 and p < ∞, then~~

~~lim kf − Wn ∗ fkp = 0, (7.7) n→∞ iθ iθ lim Wn ∗ f(e ) = f(e ) almost everywhere (7.8) n→∞ for every f ∈ Hp.~~

p q s Proof. Let 0 < q < p and s = p/q. Then (Wmaxf) 6 C M(|f| ) , because of (7.5), so we get (7.6) from the maximal theorem 2.3.1, applied to Ls. The relation (7.7) holds if f is a polynomial; that (7.7) holds for all f ∈ Hp, follows from kWn ∗ fkp 6 Ckfkp and Theorem 1.2.2. Finally, (7.8) is proved by using Theorem 2.7.2. 2

~~7.3.5 Exercise Let σ = s + it be a complex number and p > 0. Then~~

~~4n 4n X σ k s X k k akz n akz .~~

~~k=n p k=n p~~

~~This can be proved by taking ψ(x) = xσϕ(x), where ϕ is a C∞-function such that supp(ϕ) ⊂ (0, ∞) and ϕ(x) = 1 for x ∈ [1, 4].~~

~~7.4 Interpolation of operators on Hardy spaces~~

7.4.1 Theorem Let µ be a sigma-ﬁnite measure over a set Ω, let 0 < p < q < ∞, and let T be a quasilinear operator from Hp to the set of all µ-measurable functions. 120 7 Maximal functions, interpolation, coefficients

Assume there exist constants C1 and C2, independent of f, such that p kT fkp,∞ 6 C1kfkp , f ∈ H , (7.9) q kT fkq,∞ 6 C2kfkq , f ∈ H . Then for every s ∈ (p, q) there exists a constant C independent of f such that s kT fks 6 Ckfks , f ∈ H . Observe that the case q = ∞ is now excluded. In that case the things lie much deeper, as one can see in [7, Ch. 5]. Proof. The idea is the same as in the case of the classical Marcinkiewicz’s theorem. The unique obstacle is in that we cannot use the old decomposition p of f because gλ need not be in H (T). Fortunately, we have the decomposition f = gλ + hλ, where Z p p kgλkp 6 A |f| dσ, |f|>λ Z Z q q 2q −q khλkq 6 A |f| dσ + Aλ |f| dσ, |f|6λ |f|>λ where A=const, and dσ is the normalized measure on T (Lemma 7.4.2 below). Assuming that T is subadditive and C1 = C2 = 1, we have

µ(T f, λ) 6 µ(T gλ, λ/2) + µ(T hλ, λ/2) Z Z p p q q 6 A(2/λ) |f| dσ + A(2/λ) |f| dσ |f|>λ |f|6λ Z q −q + A(2λ) |f| dσ = I1(λ) + I2(λ) + I3(λ). |f|>λ Now we multiply this by sλs−1 and integrate these three summands from λ = 0 to ∞. For instance, we have Z ∞ Z Z |f| A2q Z s I (λ)λs−1 dλ = A2q |f|−q dσ λqλs−1 dλ = |f|−q|f|q+s dσ. 3 q + s 0 T 0 T In the case of I1 and I2 we proceed similarly. 2 The following lemma is essentially due to Bourgain [9]. The following presenta- tion is taken from Kislyakov/Xu [43].

~~7.4.2 Lemma If f ∈ Hp (0 0, then there are functions h ∈ H∞ and g ∈ Hp such that~~

~~|f∗| λ |h∗| 6 Cλ min , and λ |f∗| Z p p kgkp 6 C |f∗(ζ)| |dζ|, ζ∈T, |f∗(ζ)|>λ where C depends only on p. 7.4 Interpolation of operators on Hardy spaces 121~~

Proof. Let λ > 0 and deﬁne the functions α on T, and F on D by |f |p/2 1 α = max 1, ∗ and F = . λ P [α] + iPe[α] Since P [α] > 1 in D, we have 0 < |F | 6 1 in D. Therefore the function G = 1 − (1 − F 4/p)2/p is well deﬁned, analytic and bounded in D. We claim that the functions h = Gf and g = (1 − G)f satisfy the desired conditions. 4/p Since |F∗| 6 1/α and |G| 6 C|F | (by Schwarz’ lemma), we have |f |−2 |h | C|f | min 1, ∗ , ∗ 6 ∗ λ and this gives the desired estimate for h. On the other hand, 2/p |g∗| 6 C|1 − F∗| |f∗| α − 1 |H(α − 1)|2/p C + |f | 6 α α ∗ 2/p 2/p 6 C(1 − 1/α) |f∗| + Cλ |H(α − 1)| , where H is the Hilbert operator. Since α = 1 on the set {|f∗| < λ}, and H is bounded on L2, we see that Z Z p 2 p p 2 kgkp 6 C (1 − 1/α) |f∗| + Cλ |H(α − 1)| T T Z Z p p 2 6 C |f∗| + Cλ (α − 1) |f∗|>λ T Z p 6 C |f∗| . |f∗|>λ This completes the proof. 2

~~Application to Taylor coeﬃcients As an example, consider the operator p (T f)(n) = (n + 1) sup |fb(k)|, f ∈ H , n > 0. 06k6n By Corollary 5.1.3, for every p ∈ (0, 1) we have~~

1/p p (T f)(n) 6 Cp (n + 1) kfkp, f ∈ H . Define the measure µ on Ω = {0, 1, 2,...} by µ({n}) = (n + 1)−2. Then, arguing as in the proof of Theorem 2.2.5 we find that T satisfies (7.9) for every p ∈ (0, 1). Hence, by Theorem 7.4.1, T maps Hp into Lp(Ω, µ), for every p ∈ (0, 1). Thus we have proved: 122 7 Maximal functions, interpolation, coefficients

~~∞ 7.4.3 Theorem [57] If f ∈ Hp, 0 < p < 1, then P (n + 1)p−2 sup |fb(k)|p < ∞. n=0 k6n~~

~~As a corollary we have part (a) of the following theorem of Hardy and Little- wood.~~

7.4.4 Theorem (a) If f ∈ Hp, 0 0, then there is a constant c > 0 such that ψn > cn .

∞ Proof. To prove (b) assume that |fb(n)| = O(ψn). Consider the space `ψ of all −1 scalar sequences {an} for which k{an}kψ := supn ψn |an| < ∞. By the hypothesis p ∞ we have H ⊂ `ψ . To prove that the inclusion operator is continuous assume p ∞ that (1) fj → f in H , and (2) {fbj(n)} → {an} in `ψ , as j → ∞. It follows from assertion (a) and (1) that fbj(n) → fb(n), j → ∞, for every n. On the other hand, (2) implies that fbj(n) → an for every n. Hence an = fb(n) for every n, and therefore the closed graph theorem tells us that the inclusion is continuous. In particular, −1 −2/p we have supn ψn |fbr(n)| 6 Cpkfrkp, 0 < r < 1, where fr(z) = (1 − rz) and Cp depends only on p. Since −2/p n n 2/p−1 2 −1/p fbr(n) = (−1) r cn and kfrkp = (1 − r ) , n >

~~−1 2/p−1 n −1/p we conclude that ψn n r 6 C(1 − r) , where C is independent of r and n. Now the desired result is obtained by taking r = 1 − 1/n. 2~~

~~q L -integrability of Mp(r, f) We have proved that if f ∈ Hp, p < ∞, then~~

~~ 1/q−1/p Mq(r, f) = O (1 − r) (r → 1) for q > p, (7.10)~~

see (5.3). Then, using the fact that the polynomials are dense in Hp, we can prove that (7.10) remains valid if we replace “O” by “o”. This is further improved by inequality (7.11) below because Mp(r, f) increases with r.

~~p 7.4.5 Theorem (Hardy/Littlewood) If f ∈ H , 0 q > p, then~~

~~Z 1 −p/q p p (1 − r) Mq (r, f) dr 6 Ckfkp, (7.11) 0 where C depends only on p, q.~~

~~From (5.3) and (7.11) we get the following: 7.5 On the Hardy/Littlewood inequality 123~~

~~p 7.4.6 Corollary If f ∈ H , 0 q > p and s > p, then Z 1 sα−1 s s (1 − r) Mq (r, f) dr 6 Ckfkp, 0 where α = 1/p − 1/q (> 0) and C is independent of f.~~

~~In the case s = q, this can be written in the form:~~

7.4.7 Corollary If f ∈ Hp, 0 p, then Z q 2 q/p−2 q |f(z)| (1 − |z| ) dA(z) 6 Cp,q kfkp. D In connection with this corollary see 5.3.5. Proof of Theorem 7.4.5. We can suppose that q < ∞ because

~~−1/q √ M∞(r, f) 6 (1 − r) Mq( r, f).~~

~~Deﬁne the (quasilinear) operator T : Hp 7→ C(0, 1) by~~

~~−1/q (T f)(r) = (1 − r) Mq(r, f), 0 < r < 1.~~

We will prove that T maps Hs into Ls,∞(0, 1) for s = p and s = q. By Theo- rem 7.4.1, this will imply that T maps Hs into Ls(0, 1) for p < s < q; this will conclude the proof. In the case s = p we use inequality (5.3); we get

~~−1/p T f(r) 6 A(1 − r) kfkp, (7.12) where A is a positive constant. If kfkp = 1, then it follows from (7.12) that~~

−1/p p {r ∈ (0, 1): T f(r) > λ} 6 {r ∈ (0, 1): A(1 − r) > λ} = min(1, (A/λ) ), which proves that T is of weak type (p, p). In the case s = q the desired conclusion follows from the inequality Mq(r, f) 6 kfkq. The proof is complete. 2 Remark. For a more elementary proof, see [18].

~~7.4.8 Exercise If f is harmonic in D, then (7.11) holds for q > p > 1.~~

~~7.5 On the Hardy/Littlewood inequality~~

~~As a consequence of Theorems 7.4.3 and 2.2.4 we have another result of Hardy and Littlewood. 124 7 Maximal functions, interpolation, coefficients~~

7.5.1 Theorem If f ∈ Hp, 0 2, then f ∈ Hp p and kfkp 6 CpK. In the case 1 p, inequality (7.11) can be written as

Z 1 ∞ p/2 −p/2 X 2 2n p (1 − r) |fb(n)| r dr 6 Cpkfkp. (7.14) 0 n=0 Let f ∈ Hp, 0 < p < 2. We use (7.14) and Lemma 7.5.4 below to get ∞ p/2 X −n(1−p) −n X 2 p 2 2 |fb(k)| 6 Cpkfkp. (7.15) n=0 k∈In

n p/2 Since card(In) = 2 for n > 1, and the function t 7→ t , t > 0, is concave, we have p/2 −n X 2 −n X p 2 |fb(k)| > 2 |fb(k)| , k∈In k∈In and this together with the previous relation implies (7.13). To discuss the case p > 2 we consider the space Xp ⊂ H(D) deﬁned by condi- ∗ tion (7.13). It is easily seen that (Xp) = Xq, 1/p + 1/q = 1, with respect to the bilinear form ∞ X (f, g) = fb(n)gb(n) n=0 Since (Hp)∗ = Hq, 1 2. The proof is complete. 2 Inequality (7.15) can be improved by combining Theorem 7.4.5 with the Haus- dorﬀ/Young theorem. For example, in the case p = 1 we get:

~~7.5.2 Theorem [57] If f ∈ H1, then~~

~~∞ 1/q X X 2−n |fb(k)|q < ∞, 0 < q < ∞. (7.16)~~

~~n=0 k∈In Note that the quantity in (7.16) increases with q. In the case q = 1, (7.16) reduces to ∞ X |fb(n)| < ∞. n + 1 n=0 7.5 On the Hardy/Littlewood inequality 125~~

Addendum: Littlewood’s conjecture In connection with the so called Littlewood’s conjecture, Z 2π n X iλ θ e k dθ c log n, > 0 k=1 the following extension of Theorem 7.5.1 (p = 1) was proved in [65].

∞ 7.5.3 Theorem If {λn}1 is a strictly increasing sequence of positive integers, 1 P∞ −1 and f ∈ H is such that supp fb ⊂ {λn : n > 1}, then n=1 n |fb(λn)| 6 Ckfk1, where C is independent of {λn} and f.

~~Lp-integrability of power series with positive coeﬃcients~~

n n+1 7.5.4 Lemma Let α > −1, 0 < q < ∞, and In = {k : 2 6 k < 2 } for n > 1, I0 = {0, 1}. If {an} is a sequence of nonnegative numbers such that the P∞ n series G(r) = n=0 anr converges for every r ∈ (0, 1), then the following three conditions are equivalent and the corresponding quantities are “proportional”: Z 1 (1 − r)αG(r)q dr < ∞, 0 ∞ q X −(α+1)n X 2 ak < ∞,

n=0 k∈In ∞ n q X −α−2 X (n + 1) ak < ∞. n=0 k=0 n P Pn In the case of the function G(r) = sup anr the expressions ak, ak n>0 k∈In k=0 should be replaced by sup ak, sup ak, respectively. k∈In 06k6n This lemma is an immediate consequence of its special case:

7.5.5 Lemma Let {ak} be a sequence of nonnegative real numbers such that P∞ 2k the series F (r) = k=0 akr converges for every r ∈ (0, 1). Let α > −1 and 0 < q < ∞. Then the conditions Z 1 ∞ α q X −(α+1) q A := (1 − r) F (r) dr < ∞ and B := 2 ak < ∞ 0 k=0 are equivalent. There holds the inequality B/C 6 A 6 CB, where C depends only 2k on α, q. The same holds for the function F (r) = supk akr .

~~Proof. The proof of the inequality A > B/C is very simple (see, e.g., the proof of Theorem 11.1.1, p. 171)(∗). To prove the reverse inequality, observe ﬁrst that~~

~~(∗)It is, however, diﬃcult to prove that this inequality remains hold if we drop the hypothesis an > 0; see Theorems 11.4.1 and 11.4.2. 126 7 Maximal functions, interpolation, coefficients~~

~~the case q 6 1 is trivial; we integrate the inequality ∞ α q α X q q2k (1 − r) F (r) 6 (1 − r) akr . k=0~~

~~In the case q > 1 we can use Jensen’s inequality (as in [58]), or proceed as follows. Write the inequality A 6 CB as kT ({an})(r)kLq (µ) 6 Cqk{an}k`q , where~~

~~∞ β X βn 2n T ({an})(r) = (1 − r) 2 anr , β > 0, n=0~~

~~and dµ(r) = dr/(1 − r). It is easily veriﬁed that T maps `q to Lq(µ) for q = 1, ∞, so we can apply the Riesz/Thorin theorem. 2~~

~~Miscellaneous~~

~~7.5.6 If α > 0 and let {an} be a nonnegative sequence, then the following conditions are equivalent:~~

~~∞ X n −α − X αn anr = O (1 − r) (r → 1 ), and ak = O(2 )(n → ∞).~~

~~n=0 k∈In~~

~~The equivalence remains true if we replace “O” by “o.”~~

p ∞ 7.5.7 If f ∈ H , 1 < p < 2, and {λn}1 is a strictly increasing sequence of positive P∞ p−2 p integers, then n=1 n |fb(λn)| < ∞. 7.5.8 [59] Lemma 7.5.4 can be generalized in the following way: P k For an analytic function f let ∆nf(z) = fb(k)z . Let 1 0. Then k∈In

~~Z 1 ∞ qα−1 q X −αn q (1 − r) Mp (r, f) dr < ∞ if and only if 2 k∆nfkp < ∞. 0 n=0 In the case q = ∞ there holds the equivalence~~

~~−α αn Mp(r, f) = O(1 − r) ⇐⇒ k∆nfkp = O(2 ).~~

~~This can be deduced from Lemma 7.5.5 in the following way. We have~~

~~∞ X Mp(r, f) 6 Mp(r, ∆nf), 1 6 p 6 ∞, n=0~~

~~and, by Riesz’ projection theorem, Mp(r, f) cp sup Mp(r, ∆nf), 1 n>0 It remains to apply 4.1.4. 7.6 On the dual of H1 127~~

7.5.9 Let 1 < p < ∞, f ∈ H(D), and let the sequence fb(n) be decreasing. Then, f ∈ Hp if and only if ∞ X (n + 1)p−2|fb(n)|p < ∞. n=0 This can be deduced from 7.5.8 and the Littlewood/Paley theorem.

7.5.10 Assertion 7.5.8 does not hold for p ∈ (0, 1) ∪ {∞}. Then we can use Theorem 7.3.4 to replace ∆nf by Γn ∗ f, where Γn is a sequence of polynomials Γn, n > 0, satisfying, for all p > 0,

~~n−1 n+2 supp(Γbn) ⊂ [2 , 2 ], n > 1, ∞ X f = Γn ∗ f, f ∈ H(D), n=0 p kΓn ∗ fkp 6 Cpkfkp, f ∈ H , n(1−1/p) kΓnkp 2 .~~

A more complicated polynomials can be constructed by means of Theorem 7.3.1 (cf. [69]). Such polynomials play a central role in calculating multipliers for various spaces of analytic functions (cf. [33, 70]).

~~7.6 On the dual of H1~~

A function φ ∈ L1(T) is called a function of bounded mean oscillation if Z Z 1 it 1 it sup φ(e ) − φI dt = kφk∗ < ∞, where φI = φ(e ) dt, I⊂T |I| I |I| I and the supremum is taken over all subarcs of T. The class BMO = {φ: kφk∗ < ∞} is normed by kφkBMO = kφkL1 + kφk∗. The intersection of BMO with H1(T) is denoted by BMOA. The famous theorem of Fefferman states that BMOA is isomorphic to the dual of H1(T). This means the following: A function φ ∈ H1(T) belongs to BMOA if and only if 1 Z π sup φ(eit)Q(e−it) dt = kφk0 < ∞, (7.17) 2π ∗ kQk161 −π where the supremum is taken over the set of all polynomials in H1. Moreover, the 0 norm k · k∗ is equivalent to the original norm in BMOA. The proof of Fefferman’s theorem as well as of various other properties of BMOA can be found in Garnett [22, Ch. VI]. Here we can accept (7.17) as a definition of BMOA. Also, we can treat BMOA as a space of analytic functions on D, via the Poisson integral. Then Theorem 7.5.2 leads to the following. 128 7 Maximal functions, interpolation, coefficients

∞ 7.6.1 Theorem [57] Let {an}0 be a sequence of complex numbers satisfying 1/q n −n X q sup 2 2 |ak| < ∞, for some q > 1. (7.18) n 0 > k∈In P∞ n Then the function f(z) = 0 anz belongs to BMOA.

~~Note that (7.18) is satisﬁed if an = O(1/n).~~

~~7.6.2 Remark It is interesting that condition (7.18) is suﬃcient for the validity of the implication~~

~~∞ ∞ X n X lim anr exists =⇒ an converges; r→1− n=0 n=0 see Theorem 11.2.2 and Corollary 11.2.3.~~

Miscellaneous 7.6.3 The space BMOA is closely related to two important spaces of functions— the space H∞ and the Bloch space B = {f ∈ H( ): sup (1 − |z|2)|f 0(z)| < ∞}. D z∈D 0 ∞ It is clear from (7.17) that kfk∗ 6 kfk∞ so H ⊂ BMOA. This inclusion is proper P∞ n ∞ because the function f(z) = n=1 z /(n + 1) is in BMOA but not in H . The inclusion BMOA ⊂ B can be deduced from the easily proved relations 1 Z π Z 1 φ(eit)Q(e−it) dt = f(0)Q(0) + 2 f 0(z)Q0(¯z) log dA(z), 2π |z| −π D and Z 0 kQk1 6 |Q(0)| + C |Q (z)| dA(z), D where C is an absolute constant. The inclusion is proper. Namely, the function n f(z) = P z2 is in B. On the other hand, since H1 ⊂ Hp for 1 < p < ∞, and 0 (Hp)∗ = Hp (Theorem 6.3.2), we see that BMOA ⊂ Hp for every p < ∞ and therefore f is not in BMOA because f is not in H2.

~~7.6.4 Feﬀerman’s duality theorem can be expressed in terms of multipliers, namely:~~

~~1 ∞ ∞ 1 BMOA = (H ,H ) := {g ∈ H(D): g ∗ f ∈ H for all f ∈ H }. It was proved in [61] that (H1, BMOA) = B.~~

7.6.5 If f ∈ H1 and g ∈ BMOA, then f(eit)g(e−it) need not be integrable on [−π, π]. Example: 1 − z 1 − z −2 1 − z f(z) = z2 log , g(z) = log . 1 + z 1 + z 1 + z 8 Bergman spaces: Atomic decomposition

In this chapter we consider the Coifman/Rochberg theorem on the atomic decomposition of Bergman spaces (Theorems 8.3.1, 8.3.4) as well as some applications, due to Kalton [37], to the theory of vector-valued analytic functions (Theorems 8.4.2, 8.4.3). Some technical simpliﬁcations in the proof of the Coif- man/Rochberg theorem are made. Kalton’s results are formulated in a more precise form (Theorems 8.4.3, 8.4.5). Further topics in the theory of Bergman spaces can be found in the book of Hedenmalm, Korenblum and Zhu [28].

~~8.1 Bergman spaces~~

~~Let 0 < p < ∞. The Bergman space Ap is the subspace of Lp(D, dA) consisting of analytic functions. The quasinorm can be written as~~

~~Z 1 Z π p p 1 iθ p kfk = kfkAp = 2 Ip(r, f) rdr, where Ip(r, f) = |f(re )| dθ. 0 2π −π~~

8.1.1 Proposition For p ∈ (0, ∞) there hold the following: (a) For every z ∈ D the functional f 7→ f(z) is continuous on Ap; (b) The space Ap is complete; p p (c) If f ∈ A , and f%(z) = f(%z), then f% ∈ A and kf − f%k → 0 (% → 1); (d) The set of all polynomials is a dense subset Ap.

~~Proof. The function |f|p is subharmonic, which means that Z p 1 p |f(z)| 6 2 |f(w)| dA(w) R |w−z|~~

~~129 130 8 Bergman spaces: Atomic decomposition~~

p And since fn → f in L (K), where K is any compact subset of D, we have f = g almost everywhere. (c) Let f ∈ Ap, 0 < p < 1. Then Z 1 p kf − f%k = 2 Ip(r, f − f%) rdr. 0 Since Ip(r, f − f%) 6 Ip(r, f) + Ip(r, f%) and Ip(r, f%) = Ip(r%, f) 6 Ip(r, f), the function r 7→ 2Ip(r, f) is an integrable dominant so we can apply the dominated convergence theorem. The case p > 1 is discussed in a similar way. p (d) Let f ∈ A , ε > 0 and 0 < % < 1. The Taylor series of f% converges uniformly on D, which implies kf% − snk 6 ε for n large enough, where sn are the partial sums of the Taylor series of fρ. Now (d) can easily be deduced from (c). 2

Exercises 8.1.2 The assertion (c) of Proposition 8.1.1 holds and for an arbitrary function f ∈ Lp(D). 8.1.3 The inclusion Aq ⊂ Ap is compact for q > p. Remark. It is known that Ap is isomorphic to `p (see Theorem 1.5.1). And since every operator from `q to `p, q > p, is compact (see Theorem 1.5.6), every operator from Aq u Ap, q > p, is compact.

2 −2/p 8.1.4 There holds the inequality |f(z)| 6 (1 − |z| ) ||f||Ap . See Lemma 5.1.1. Pn k 8.1.5 For every p > 0 and every polynomial f(z) = k=m akz there holds the inequality 1 1 kfk p kfk p kfk p . n + 1 H 6 A 6 m + 1 H 8.2 Reproductive kernels

Let 1 K (w, z) = (z, w ∈ ). p (1 − wz¯ )2/p+1 D 2 2/p−1 Let dµp denote the measure deﬁned over D by dµp(w) = cp(1 − |w| ) dA(w), where 1 Z p = (1 − |w|2)2/p−1dA(w) = . c 2 p D The following theorem shows that da Kp has the reproducing property.

p 8.2.1 Theorem Let 0 < p 6 1 and f ∈ A . Then for every z ∈ D the function 1 w 7→ f(w)Kp(w, z) belongs to L (D, dµp) and there holds the formula Z f(z) = Kp(w, z)f(w) dµp(w). (8.2) D 8.2 Reproductive kernels 131

Proof. If kfkAp = 1 and z ∈ D, then, according to (8.1), p 1−p p −2/p+2 |f(w)| = |f(w)| |f(w)| 6 |f(w)| (1 − |w|) , 1 which implies that the function w 7→ f(w)Kp(w, z) is in L (D, dµp). It follows that the functional Z Λzf = Kp(w, z)f(w) dµp(w) D is well deﬁned and bounded on Ap. Since the functional f 7→ f(z) is bounded (Proposition 8.1.1(a)) and the polynomials are dense in Ap, we see that the proof n of (8.2) reduces to the proof that Λzf = f(z) for f(z) = z , n > 0. However this follows from the formulas ∞ 1 X Γ(α + k + 1) = zk, (1 − z)α+1 Γ(α + 1)Γ(k + 1) k=0 Z Γ(α + 1)Γ(k + 1) Z (1 − |w|2)α−1|w|2kdA(w) = (1 − |w|2)α−1dA(w), Γ(α + k + 1) D D which are true for α > 0, and the formula R wmw¯n dA(w) = 0 (m 6= n). 2 D Remark. A more interesting proof of (8.2) can be found in [63].

p 8.2.2 Lemma Let f ∈ A , 0 < p 6 1 and let Z g(z) = |Kp(w, z)| |f(w)| dµp(w)(z ∈ D). D p Then g ∈ L (D, dA) and there holds the inequality kgkLp 6 Cp kfkAp . Proof. Let E be a partition of the unit disk into disjoint sets E with the following properties: 1 diam(E) C (w ∈ E), C 6 1 − |w| 6 1 area(E) C (w ∈ E), C 6 (1 − |w|)2 6 1 1 − |ζ| C (ζ, w ∈ E), C 6 1 − |w| 6 where C is an absolute constant. Such a family consists of the sets n 1 1 2πj 2π(j + 1) o z : < 1 − |z| , arg z < , 2k+1 6 2k 2k+2 6 2k+2 k+2 where k = 0, 1, 2,... , 0 6 j < 2 . Let E = {En : n > 1}. Then ∞ Z X 2/p−1 g(z) 6 C (1 − |wn|) |f(w)Kp(w, z)| dA(w) n=1 En ∞ X 2/p+1 6 C (1 − |wn|) sup |f(w)Kp(w, z)|, n=1 w∈En 132 8 Bergman spaces: Atomic decomposition

~~where {wn} is an arbitrary sequence such that wn ∈ En and C is a constant depending only on p. It follows that~~

~~∞ p X 2+p p g(z) 6 C (1 − |wn|) sup |f(w)Kp(w, z)| . n=1 w∈En For a ﬁxed z the function p p f(w) F (w) = |f(w)Kp(w, z)| = (1 − wz¯)2/p+1~~

~~is subharmonic in D because the function f(w)/(1 − wz¯)2/p+1 is analytic with respect to w. Therefore 1 Z F (w) 6 F (ζ) dA(ζ), A(Dw) Dw~~

~~where Dw ⊂ D is an arbitrary disk centered at w. From this and the properties of the family E we get 1 Z |f(w)|p sup F 6 C 2 2+p dA(w), En (1 − |wn|) Bn |1 − wz¯ |~~

where Bn is the union of those E ∈ E for which the set E ∩ En is nonempty. Combining these with the inequality Z 1 C dA(z) |1 − wz¯ |2+p 6 (1 − |w|)p D (see Lemma 8.2.3 below), we get Z ∞ Z p X p g(z) dA(z) 6 C |f(w)| dA(w). D n=1 Bn

~~Now our result follows from the easily checked fact that each Bn contains at most N members of the family E, with N being independent of n. 2~~

~~8.2.3 Lemma Let 1 Z π Z I (a) = |1 − ae¯ iθ|−s dθ, J (a) = |1 − aw¯ |−s dA(w). s 2π s −π D There hold the relations(∗):~~

~~ −s+1 (1 − |a|) for s > 1  2 Is(a) log 1−|a| for s = 1 (a ∈ D), 1 for s < 1~~

~~(∗)“A B ” means that the ratio A/B lies between two positive constants. 8.3 The Coifman/Rochberg theorem 133~~

~~ −s+2 (1 − |a|) for s > 2  2 Js(a) log 1−|a| for s = 2 (a ∈ D). 1 for s < 2~~

~~R 1 Proof. Since Js(a) = 2 0 Is(ra) dr, it suﬃces to discuss the case of Is. Let ρ = |a| < 1. We have~~

~~Z π 1 2 2 −s/2 Is(a) = (1 − r) + 4r sin (θ/2) dθ π 0 1 Z π (1 − r + θ)−sdθ. 2 π 0~~

~~8.3 The Coifman/Rochberg theorem~~

It is the idea of Coifman and Rochberg [15] to represent a member of Ap as a sum of “atoms” by replacing the integral in (8.2) by a Riemannian sum over a suﬃciently ﬁne partition of the disk. They proved the “decomposition theorem” for every p > 0 and for a class of domains in Cn, in particular on balls. Here we consider the case p < 1 because this case, as was shown by Kalton [37], is of fundamental importance in the theory of vector-valued analytic functions.

8.3.1 Theorem (on atomic decomposition) Let 0 < p 6 1. Then (a) There exists a sequence {wn} in D and a constant C such that for every p p f ∈ A there exists a sequence {an} ⊂ ` with the properties

~~∞ 2 X 1 − |wn| f(z) = a (8.3) n (1 − w¯ z)2/p+1 n=1 n ∞ Z X p p |an| 6 C |f| dA. n=1 D~~

~~p p (b) The function f of the form (8.3), where {an} ∈ ` , belongs to A and there holds kfkAp 6 Ck{an}kp.~~

~~For the proof we need a lemma.~~

~~8.3.2 Lemma There exists a constant C such that~~

~~|w − ζ| |K (w, z) − K (ζ, z)| C |K (w, z)| p p 6 diam E p for all E ∈ E, w, ζ ∈ E and z ∈ D. 134 8 Bergman spaces: Atomic decomposition~~

~~Proof. By the mean value theorem, we have~~

~~−2/p−2 |Kp(w, z) − Kp(ζ, z)| 6 (2/p + 1)|w − ζ| sup |1 − az¯ | a∈E |w − ζ| 6 (2/p + 1) sup |Kp(a, z)| 1 − |a| a∈E |w − ζ| 6 Cp sup |Kp(a, z)|. diamE a∈E On the other hand,~~

~~1 − wz¯ (w − a)z |w − a| 1 − = C. 1 − az¯ 1 − az¯ 6 1 − |a| 6~~

Hence |1 − wz¯ | 6 C|1 − az¯ | and hence |Kp(a, z)| 6 C|Kp(w, z)| for a, w ∈ E. The result follows. 2 Proof of Theorem 8.3.1. Assertion (b) follows from the boundedness of the sequence kψkkAp , where 1 − |w |2 ψ (z) = k . k 1+2/p (1 − w¯kz) Let us prove (a). Let ε > 0. Dividing each E ∈ E into N subsets, where N is a suﬃciently large integer independent of E, we can represent D as a disjoint union D1 ∪ D2 ∪ ... , where D1,D2,... are subsets of D with the properties: 2 ε diam(Dn) ε area(Dn) 2 6 6 C1ε and 6 2 6 C1ε (w ∈ Dn). C1 1 − |w| C1 (1 − |w|)

~~Let {wn} be a sequence such that wn ∈ Dn. Deﬁne the operator T by~~

∞ X 2 T f(z) = bn(1 − |wn| )Kp(wn, z)(z ∈ D), n=1 where 1 Z bn = 2 f(w) dµp(w). 1 − |wn| Dn Proceeding as in the proof of Lemma 8.2.2, we can prove that T maps Ap into Ap. In order to conclude the proof it suﬃces to prove that T is an isomorphism for ε small enough and that ∞ Z X p p |bn| 6 C |f| dA. n=1 D The proof of the latter is similar to that of Lemma 8.2.2. To prove the rest we start from the relation ∞ X Z f(z) − T f(z) = (Kp(w, z) − Kp(wn, z))f(w) dµp(w). n=1 D 8.3 The Coifman/Rochberg theorem 135

~~From this, by Lemma 8.3.2, we get~~

~~∞ X Z |TF (z) − f(z)| 6 Cε |f(w)| |Kp(z, w)| dµp(w) n=1 Dn Z = Cε |f(w)| |Kp(z, w)| dµp(w). D~~

~~Now Lemma 8.2.2 shows that ||T f − f|| 6 Cpε||f||. Finally we take ε = 1/2Cp and apply Proposition 1.2.1. 2~~

~~It is of importance that the sequence wn in (8.3) can be chosen from the annulus {w : r < |w| < 1}, where r < 1 is ﬁxed.~~

~~8.3.3 Theorem (Kalton [38]) Let 0 < r < 1 and ψ ∈ Ap. Then ψ can be represented as~~

~~∞ 2 ∞ 1/p X 1 − |ζk| X p p ψ(z) = αk 2 , where r 6 |ζk| < 1, |αk| 6 C||ψ||A , p +1 k=0 (1 − ζkz) k=0 and C depends only on r and p.~~

In contrast to Theorem 8.3.1, here we do not assert {ζk} is independent of ψ (though this is probably true). Proof. If Γ is an arbitrary set, then the space `p(Γ) consists of those functions ω 7→ aω for which 1/p X p kak := |aω| < ∞. ω∈Γ p p Let Γ = {w : r 6 |w| < 1}. Deﬁne the operator S : ` (Γ) 7→ A by

~~X 1 − |w|2 S({a })(z) = a . w w (1 − wz¯ )2/p+1 w∈Γ~~

Let E = S(`p) and Λ an arbitrary (not necessarily continuous) linear functional on Ap such that Λ(E) = 0. We want to prove Λ = 0 on Ap, which implies that S is onto; this proves the desired result, according to the open mapping theorem. 2/p+1 Let F be the linear hull of the vectors ϕk(z) = 1/(1 − w¯kz) , |wk| < r, p where {wk} is the sequence from Theorem 8.3.1. If f ∈ A , then f = g + h, where g ∈ E, h ∈ F and ||g|| + ||h|| 6 C||f||. Since dim(F ) < ∞, we see that Λ is bounded on F , i.e., there exists a constant C1 such that |Λh| 6 C1||h||, which p implies that |Λf| = |Λh| 6 C2||f||, and this means that Λ is bounded on A . Take 2/p+1 φ(w) = Λ(ψw), where ψw(z) = 1/(1 − wz¯ ) . The function φ is antianalytic and φ(w) = 0 for r 6 |w| < 1 because Λ(E) = 0. It follows that φ(w) = 0 for every w ∈ D and in particular Λϕk = 0 for every k. By Theorem 8.3.1, this implies Λ = 0 on Ap, which was to be proved. 2 136 8 Bergman spaces: Atomic decomposition

~~The case p > 1~~

~~If 1 < p < ∞, then the formulation of the Coifman/Rochberg theorem is similar to that of Theorem 8.3.1; we only have to replace (8.3) by~~

~~∞ 2 2/q X (1 − |wn| ) f(z) = a , n (1 − w¯ z)2 n=1 n~~

~~where 1/p + 1/q = 1 (see [99]).~~

~~The case of weighted spaces~~

~~p The weighted Bergman space Aβ, 0 −1, consists of those f ∈ H(D) for which~~

~~Z 1/p p 2 β kfkp,β := |f(z)| (1 − |z| ) dA(z) < ∞. D~~

~~8.3.4 Theorem Let 0 −1 and γ > 0. p (a) There exists a sequence {wn} in D and a constant C such that every f ∈ Aβ can be represented as~~

~~∞ 2 γ X (1 − |wn| ) f(z) = a (8.4) n (1 − w¯ z)γ+(β+2)/p n=1 n~~

~~with k{an}k`p 6 Ckfkp,β. p p (b) Every function f of the form (8.4) with {an} ∈ ` belongs to Aβ and kfkp,β 6 Ck{an}k`p .~~

~~This theorem is proved in the same way as its particular case, Theorem 8.3.1 (β = 0, γ = 1); the key is in the formula~~

~~Z f(w)(1 − |w|2)s−2 f(z) = c dA(w)(s = γ + (β + 2)/p), s (1 − wz¯ )s D p which holds for f ∈ Aβ.~~

~~Envelops of Hardy spaces~~

~~We have deﬁned the q-Banach envelope of a quasi-Banach space (see p. 10). Now we can give a nontrivial example.~~

~~8.3.5 Theorem If 0 < p < q 6 1, then the q-Banach envelope of the Hardy space p q H is equal to Aβ, β = q/p − 2. 8.4 Coeﬃcients of vector-valued functions 137~~

p q Proof. The space X = H is embedded into the q-Banach space Y = Aβ P∞ (see 7.4.7). On the other hand, every f ∈ Y can be represented as f = n=1 fn, where (1 − |w |2)γ f = a n n n γ+1/p (1 − w¯nz)

P q 1/q and ( |an| ) 6 CkfkY . Lemma 8.2.3 shows that Z π 1 C iθ γp+1 dθ 6 γp , −π |1 − w¯ne | (1 − |wn|) which implies ∞ ∞ X q X q kfnkX 6 C |an| 6 CkfkY . n=1 n=1 Now the result follows from Proposition 1.2.5. 2

~~8.4 Coeﬃcients of vector-valued analytic functions~~

A function F :Ω 7→ X, where Ω is a domain in C, is said to be analytic if every point in Ω admits a neighborhood in which f can be expanded into a power series with X-valued coeﬃcients. In the case where Ω is the unit disk, it turns out that the P∞ n analyticity implies the existence of vectors Fb(n) such that F (z) = n=0 Fb(n) z , |z| < 1, with uniform convergence on compact subsets. The vectors Fb(n), uniquely determined by F , are called the Taylor coeﬃcients and, as one expects, satisfy the 1/n condition lim supn→∞ kFb(n)k 6 1. On the other hand, if {fn} is a sequence of vectors in X, then the condition

~~1/n lim sup kfnk 6 1 (8.5) n→∞~~

P∞ n is necessary and suﬃcient for the series n=0 fn z to converge for every z ∈ D. In the case of convergence, the sum of that series is analytic in D. Therefore the set of the functions F : D 7→ X analytic in D can be identiﬁed with the set of the formal power series satisfying (8.5), i.e., with the set of the power series converging in D. We will denote this set by H(D,X).

~~Derivatives The derivative of a function f ∈ H(D,X) is deﬁned by means of 0 P∞ n−1 power series, F (z) = n=1 nFb(n) z , or by the formula~~

~~F (w) − F (z) F 0(z) = lim . (8.6) w→z w − z~~

~~It should be noted, however, that the existence of the limit (8.6) for an arbitrary function F does not guarantee that F is analytic. 138 8 Bergman spaces: Atomic decomposition~~

~~Hadamard product~~

~~The Hadamard product of a (scalar) function ψ ∈ H(D) and a function F ∈ H(D,X) is deﬁned by ∞ X (ψ ∗ F )(z) = ψb(n)Fb(n) zn. n=0~~

~~The proof that ψ ∗ F belongs to H(D,X) is straightforward. There holds z z F 0(z) = (ψ ∗ F )(z), where ψ(z) = . (1 − z)2~~

8.4.1 Proposition Let F ∈ H(D,X), where X is a p-Banach space, and let P∞ the series k=1 ψk(z) = ψ(z), where ψj ∈ H(D), converge uniformly on compact subsets. Then ∞ X (ψ ∗ F )(z) = (ψk ∗ F )(z)(|z| < 1). k=1 PN P∞ Proof. Let |z|=r <1, FN (z) = j=1(ψj ∗ F )(z), and RN (z) = j=N+1 ψj(z). Then ∞ p p X kFN (z) − ψ ∗ F (z)k = kRN ∗ F (z)k 6 AN (j), (8.7) j=0

p p jp where AN (j) = |RbN (j)| kFb(j)k |z| . The sequence RN (z) is uniformly bounded on compact subsets and therefore for every ρ < 1 there exists a constant C = C(ρ) j such that |RbN (j)| 6 C/ρ for all N and j. From this and the inequality kFb(j)k 6 j 2 jp p p K(ρ)/ρ , we get AN (j) 6 M(ρ)(r/ρ ) where M(ρ) = C(ρ) K(ρ) . Thus by 2 √ taking ρ = r, we see that the sequence AN has a summable majorant. On the other hand, from the hypothesis RN (z) → 0, uniformly on compact subsets, it follows AN (j) → 0 (N → ∞), for every j. Therefore we can apply the dominated convergence theorem:

~~∞ ∞ X X lim AN (j) = lim AN (j) = 0. N→∞ N→∞ j=0 j=0 Now the desired assertion follows from (8.7). 2~~

~~Inequalities for the coeﬃcients~~

Let H∞(X) denote the set of all bounded functions F ∈ H(D,X); the quasinorm is ∞ given by kF k∞,X = sup|z|<1 kF (z)kX . If F ∈ H (X), where X is a Banach space, then one can use the Hahn/Banach theorem to deduce the inequality kFb(n)k 6 kF k∞,X from the corresponding inequality for scalar-valued functions. However if X is p-Banach, then this inequality does not hold, and proving that

1/p−1 kFb(n)k 6 Cp(n + 1) kF k∞,X 8.4 Coeﬃcients of vector-valued functions 139 is quite nontrivial. Following Kalton [37], we will prove the latter by means of the Coifman/Rochberg theorem. We ﬁrst consider a weak variant of Schwarz lemma.

~~8.4.2 Theorem Let F ∈ H∞(X), where X p-Banach space. Then there holds the inequality 0 2 −1 kF (z)kX 6 Cp (1 − |z| ) kF k∞,X (z ∈ D), where Cp is a constant depending only on p.~~

~~Proof. Let ψ be a scalar-valued analytic function belonging to Lp(D). Accord- ing to Theorem 8.3.1, there holds the formula~~

∞ 2 X 1 − |zk| ψ(w) = α , k (1 − z w)2/p+1 k=1 k where {zk} is a sequence in D and {αk} is a sequence of complex numbers such p that k{αk}kp 6 CpkψkL (D), where Cp is independent of ψ (zk are independent of ψ as well). The series converges uniformly on compact subsets so we can apply Proposition 8.4.1; we get ∞ X 2 2/p ψ ∗ G(w) = αk(1 − |zk| )D G(wzk)(w ∈ D), (8.8) k=1 where G ∈ H(D,X) and ∞ X (s + 1)(s + 2)...(s + j) DsG(z) = G(0) + Gb(j) zj. j! j=1 Now choose G so that D2/pG = F and put ψ(w) = w in (8.8). We get

Fb(1) X w = α (1 − |z |2)F (z w). 2/p + 1 k k k Hence 0 p p X p 2 p p kF (0)kX |w| 6 (1 + 2/p) |αk| (1 − |zk| ) kF (zkw)k p X p p p 6 (1 + 2/p) |αk| kF k∞,X 6 CpkF k∞,X . This proves the result for z = 0. If z ∈ D is arbitrary, then we apply this special case to the function F1(w) = F (z + (1 − |z|)w); the derivative can be found by the 0 0 formula (8.6): F1(0) = (1 − |z|)F (z). We need however to show that the function F1(w) is analytic in D; this is true because ∞ n X X n n−j j j F1(w) = Fb(n) z (1 − |z|) w j n=0 j=0

~~∞  ∞  X X n = Fb(n) zn−j(1 − |z|)j wj,  j  j=0 n=j 140 8 Bergman spaces: Atomic decomposition~~

~~where we have used 1.1.5. 2~~

~~∞ 8.4.3 Theorem Let F ∈ H (X), where X is a p-Banach space, 0 < p 6 1. Then there exists a constant Cp depending only on p such that~~

1/p−1 kFb(n)kX 6 Cp n kF k∞,X (n > 1). (8.9) Before proving the theorem consider a simple example. Let f ∈ Hp, 0 < p < 1. Deﬁne the function F : D 7→ Hp by F (w)(z) = f(zw)(z, w ∈ D). The function p P∞ n n F belongs to H(D,H ) because F (w) = n=0 fb(n)en w , en(z) = z , and the series converges for |w| < 1. The function F is bounded; moreover kF k∞ = kfkHp . 1/p−1 An application of Theorem 8.4.3 gives |fb(n)| 6 Cp (n + 1) kfkHp because kenkHp = 1. Thus, in this special case, Theorem 8.4.3 reduces to the well known result of Hardy and Littlewood (Corollary 5.1.3). Proof of Theorem. In the case n = 1 the assertion is contained in Theo- 0 rem 8.4.2 because F (0) = Fb(1). Let n > 2. We use the atomic decomposition again. In this situation we choose G in (8.8) so that D2/pG = F 0. Then we take ψ(w) = wn in (8.8) and get

~~n X 2 0 AnFb(n + 1)w = αk(1 − |zk| )F (zkw),~~

~~where (n + 1)! A = . n (2/p + 1)...(2/p + n) 1−2/p Since An behaves as n , it follows that~~

~~p−2 p X p 2 p 0 p n kFb(n + 1)k 6 C |αk| (1 − |zk| ) kF (zk)k . From this and Lemma 8.4.2 it follows~~

~~p−2 p X p p n kFb(n + 1)k 6 |αk| kF k∞,X . Finally since X Z 2c |α |p C |z|np dA(z) = k 6 np + 2 D we get (8.9). 2~~

Vector-valued Hardy spaces Theorem 8.4.3 can be used for its own improvement. For a quasi-Banach space X let Hp(X) (0 < p < ∞) denote the space of analytic functions F : D 7→ X such that Z π 1/p 1 iθ p kF kp,X = sup kF (re )k dθ < ∞. 0

p 8.4.4 Theorem Let F ∈ H (X), where X is a p-Banach space (0 1. Proof. Let Y be the completion of the space of those continuous functions g : T 7→ X for which the function kgk belongs to Lp(T). The space Y is p-Banach. For a ﬁxed r, 0 < r < 1, deﬁne the function G : D 7→ Y by G(z)(eit) = F (rzeit). Then ∞ it n int G ∈ H (Y ) and kGk∞,Y 6 kF kp,X . Since Gb(n)(e ) = r Fb(n)e , we can apply n 1/p−1 Theorem 8.4.3 to get r kFb(n)kX 6 Cpn kF kp,X , where Cp is independent of r. This concludes the proof. 2

~~Inequalities for a Hadamard product~~

~~The Hadamard product of a function Q ∈ L1(T) and a function F ∈ H(D,X) is deﬁned by ∞ X Q ∗ F (z) = Qb(k)Fb(k) zk. k=−∞~~

~~Since |Qb(n)| 6 kQk1, the function Q ∗ F (z) belongs to H(D,X).~~

8.4.5 Theorem Let F ∈ H∞(X), where X is p-Banach, and let Q ∈ L1(T) be such that supp(Qb) is contained in (−∞, n] for some positive integer n. Then there 1/p−1 exists a constant Cp such that kQ ∗ F k∞,X 6 Cp n kQkpkF k∞,X .

n Note that Q ∗ F (z) = P Qb(k)Fb(k) zk. Note that in the special case Q(eiθ) = k=0 eint we have Theorem 8.4.3; on the other hand, we will deduce Theorem 8.4.5 from Theorem 8.4.3 by means of Theorem 8.4.4. P∞ k Proof. Let ak = Qb(n − k) for k > 0 and A(z) = k=0 akz . Then Q ∗ F (w) = Pn k k=0 an−kFb(k)w . Thus Q ∗ F (w) is the n-th coeﬃcient in the Taylor expansion of the function A(z)F (zw), which implies, by Theorem 8.4.4,

~~Z 2π p 1−p it p it p kQ ∗ F (w)kX 6 Cn sup |A(re )| kF (re w)k dt. 0~~

~~Z 2π Z 2π |A(eit)|p dt = |Q(eit)|p dt. 0 0 The result follows. 2 As an immediate consequence of Theorem 8.4.5 and Lemma 7.3.2 we have: 142 8 Bergman spaces: Atomic decomposition~~

~~8.4.6 Theorem Let F ∈ H∞(X), where X is a quasi-Banach space. Let~~

~~X k W (eit) = W ψ(eit) = ψ eikt, n n n |k|<∞~~

~~where ψ is a complex-valued C∞-function with compact support in R. Then there holds the inequality kWn ∗ F k∞,X 6 CkF k∞,X , n > 1, where C is independent of n and F .~~

~~Miscellaneous~~

~~8.4.7 If a sequence Fn ∈ H(D,X) converges uniformly on compact subsets of D, then its limit belongs to H(D,X).~~

8.4.8 If the function φ : D 7→ D is analytic and F ∈ H(D,X), then the composition F ◦ φ belongs to H(D,X) and there holds the standard formula for the derivative. 8.4.9 If X is quasi-Banach, then Hp(X) is complete for every p ∈ (0, ∞].

~~8.4.10 (Kalton [38]) If X is quasi-Banach, then for every ρ ∈ (0, 1) there exists a constant C = C(ρ) such that kF (0)k 6 C sup kF (z)k. ρ<|z|<1~~

8.4.11 A function P : X 7→ [−∞, ∞) is said to be plurisubharmonic if it is upper semicontinuous and Z π 1 iθ P (x) 6 P (x + ye ) dθ 2π −π for all x, y ∈ X. The space X is said to be PL-convex if its quasinorm is plurisubharmonic. Since convex functions are plurisubharmonic, Banach spaces are PL- convex. Some of quasi-Banach spaces are PL-convex, e.g., Lp (p > 0), and some are not. A quasi-Banach space X is PL-convex iﬀ kf(0)k 6 maxkf(ζ)k for every con- |ζ|=1 tinuous function f : D 7→ X analytic in D. Further information can be found in [68, 16, 38, 3, 74].

~~8.4.12 Let X be quasi-Banach space and σ = s+it an arbitrary complex number. Then 4n 4n X σ k s X k k akz n akz ({ak} ⊂ X).~~

~~k=n ∞,X k=n ∞,X This generalizes 7.3.5. 9 Subharmonic behavior~~

The basic topological properties of the spaces hAp = h(D) ∩ Lp and hp, p < 1, can be deduced from the inequality Z p p |g(0)| 6 Cp |g| dA, g harmonic, (†) D essentially due to Hardy and Littlewood. In this chapter we give two proofs of this inequality (Section 9.1) as well as some generalizations (Section 9.3). Using (†) we prove the Hardy and Littlewood’s theorem that hAp is “self-conjugate” for every p > 0 (Theorem 9.1.2). In Section 9.2 we apply (†) to prove four estimates, due to Hardy and Littlewood, concerning hp (Theorems 9.2.1 and 9.2.2).

~~9.1 Subharmonic behavior of |u|p, p < 1, and Bergman spaces~~

The space hAp = h(D) ∩ Lp(D, dA), 0 1 can be proved by appealing to the subharmonicity of |g|p, g ∈ h(D) (see the proof of Proposition 8.1.1). However, as we already noted (see 4.1.5), if 0 0, and this can be used to prove that hAp is complete for every p.

9.1.1 Theorem If u = |g|, where g is harmonic in D, and 0 0 is an arbitrary subharmonic function in D. Recall that we have used this theorem in proving the important Theorem 7.1.8. Of course, if p > 1, then (9.1) holds with Cp = 1; if u = |f|, f analytic, then Cp = 1 for every p > 0. In the case where u = |g|, g harmonic, Theorem 9.1.1 is a formal consequence of the following theorem of Hardy and Littlewood [26]:

~~9.1.2 Theorem (Hardy/Littlewood) If g = Re f, where f is analytic in D, then Z Z p p p |f| dA 6 Cp | Im f(0)| + |g| dA . (9.2) D D~~

~~143 144 9 Subharmonic behavior~~

Indeed, since the function |f|p is subharmonic for every p > 0, and since |g(0)|p = |f(0)|p, we see that (9.2) implies (9.1). However, it seems more natural ﬁrst to prove Theorem 9.1.1 and then deduce Theorem 9.1.2.

~~Proof of Theorem 9.1.1~~

First proof.(∗) If we apply (9.1) to the disk of radius 1 − |z| centered at z ∈ D, we get Z p −2 p u(z) 6 Cp(1 − |z|) u dA. (9.3) D Hence, it is natural to consider the maximum of the function F (z) = (1−|z|)2u(z)p, which exists under the condition that u is bounded on D. We shall assume that u is bounded(†) and that Z up dA = 1. (9.4) D Then the function F is above semicontinuous on the closed disk and equal to zero on the boundary. Consequently, F attains its maximum at a point a ∈ D. Now we apply the sub-mean-value property of u on the disk Da of radius (1 − |a|)/2 centered at a. It follows that Z −2 u(a) 6 4(1 − |a|) u dA. Da

p p−1 2 1−p Writing u = u u , we see that, in view of (9.4), (1−|a|) u(a) 6 4(Ma) , where Ma is the supremum of u on Da. Assume we have found a constant Kp such that 2 1−p 1−p Ma 6 Kpu(a); then (1 − |a|) u(a) 6 4(Kp) u(a) , i.e., after multiplying by u(a)p−1, p 1−p u(0) = F (0) 6 F (a) 6 4(Kp) , which implies (9.1). In order to prove Ma 6 Kpu(a), let z ∈ Da. Then |z|−|a| 6 (1−|a|)/2, whence 1 − |a| 6 2(1 − |z|). By the deﬁnition of a, there holds the inequality F (z) 6 F (a), i.e., 1 − |a|2 u(z)p u(a)p. 6 1 − |z|

~~1/p From these two inequalities it follows that Ma 6 4 u(a), which completes the proof. 2 Remark. Let u : D 7→ R be a Borel function that satisﬁes the condition Z p C p u(a) 6 2 u dA r Dr (a)~~

(∗)In the case of the modulus of a harmonic function, the proof is given by Kuran [47] and Feﬀerman and Stein [21]. Feﬀerman and Stein’s proof can be found in Koosis [46] and Garnett [22]. Here we present two proofs from [75]. (†)If u is not bounded, then we apply (9.1) to the functions u(ρz). 9.1 Subharmonic behavior and Bergman spaces 145

for some p > 0, provided that Dr(a) = {z : |z − a| < r} ⊂ D. Then this condition is satisﬁed for every p > 0. Second proof. This proof is applicable only to the case where u = |g|, g harmonic. This time we start from the inequality K |∇g(a)| sup{|g(z)| : |z − a| < r}, (9.5) 6 r which is valid whenever the disk Dr(a) = {z : |z − a| < r} is contained in D. (The constant K is independent of r and a.) From Lagrange’s theorem and inequality (9.5) it follows that

u(a) 6 u(z) + K(t/r) sup u (s = t + r) Ds(a) provided z ∈ Dt(a) and Ds(a) ⊂ D. The point a is chosen in the same way as in the ﬁrst proof (with other hypothesis for u). Then, we choose t and r so that p p s = (1 − |a|)/2. Then u(w) 6 4u(a) for w ∈ Ds(a) and therefore

~~u(a) 6 u(z) + Kp(t/r)u(a), z ∈ Dt(a),~~

1/p where Kp = 4 . Next we choose t and r so that Kp(t/r) = 1/2, and get the p p p inequality u(a) 6 2 u(z) , z ∈ Dt(a). Integrating this inequality in z ∈ Dt(a), we get Z 2 p p p p t u(a) 6 2 u dA 6 2 . Dt(a) p 2 p p 2 Since t = cp(1 − |a|), we see that u(0) 6 (1 − |a|) u(a) 6 2 /cp , which concludes the proof. 2

Proof of Theorem 9.1.2. Let p < 1. We start from the inequality Z 0 p p |f (0)| 6 C |g(w)| dA(w)(ε = 1/2), (9.6) εD which follows from (9.5) and (9.1); write this as Z 0 p p |f (0)| 6 C |g(w)| dτ(w)(ε = 1/2), εD where dτ(w) = (1 − |w|2)−2dA(w). The measure dτ invariant with respect to conformal automorphisms, which means that Z Z G ◦ ϕa dτ = G dτ D D 146 9 Subharmonic behavior

for every positive Borel function G, where a − z ϕ (z) = (|a| < 1). a 1 − az¯ The automorphism ϕa has the property ϕa ϕa(z) ≡ z. Hence, applying (9.6) to the function f ◦ ϕa, we get Z Z 0 p 2 p p p |f (a)| (1 − |a| ) 6 C |g(ϕa(w)| dτ(w) = C |f(z)| dτ(z), (9.7) εD Hε(a)

where Hε(a) = ϕa(εD) = {z : |ϕa(z)| < ε}. Integrating inequality (9.7) and changing the order of integration, we obtain Z Z Z 0 p 2 p p |f (a)| (1 − |a| ) dA(a) 6 C |f(z)| dτ(z) dA(a). (9.8) D D Hε(z)

The set Hε(z) is a hyperbolic disk centered at z. The euclidean area of Hε(z) is “proportional” to (1 − |z|2)2, so (9.8) implies Z Z 0 p 2 p p |f (a)| (1 − |a| ) dA(a) 6 Cp |g(a)| dA(a). (9.9) D D Now it is enough to prove the inequality Z Z p p 0 p 2 p |f| dA 6 Cp |f(0)| + |f (a)| (1 − |a| ) dA(a) ; (9.10) D D indeed, (9.2) is implied by (9.9), (9.10) and (9.1). In order to prove (9.10), we start from

~~Z Z 1 ∞ p p X −n p |f| dA = Mp (r, f)r dr 6 2 2 Mp (rn, f) D 0 n=0 ∞ p X −n p p = 2|f(0)| + 2 Mp (rn+1, f) − Mp (rn, f) , n=0~~

~~−n where rn = 1 − 2 . Hence, by Proposition 7.1.6, Z ∞ p p X −n −np p 0 |f| dA 6 2|f(0)| + Cp 2 2 Mp (rn+1, f ), D n=0~~

from which (9.10) is obtained immediately. In the case p > 1 we can proceed in a similar way; we only have to modify the proof of (9.10) (see [61]). Also, we can proceed as in the proof of Riesz’ 2 theorem 6.2.6, i.e., if 1 6 p 6 2 apply the preceding result to the function f ; etc. 2 9.2 The space hp, p < 1 147

9.1.3 Remark As Theorem 9.1.2 shows, the space hAp is “self-conjugate” for every p > 0, in contrast to the case of hp, which is self-conjugate only for p > 1. Using Riesz’ theorem on conjugate functions, we deduced that the sequence iθ |n| inθ p en(re ) = r e is a Shauder basis of h (Theorem 6.3.1), which implies that p p {en} is a basis of hA for p > 1. However, this sequence is not a basis of hA for p p 6 1. More precisely, for every p 6 1 there exists a function f ∈ A the Taylor series of which does not converge in Ap.

9.1.4 Remark The space hAp is equal to the direct sum of its subspaces Ap and ¯p ¯ p p A0 = {f : f ∈ A , f(0) = 0}, for every p > 0. This can be used to show that hA and Ap are isomorphic for every p > 0 (see 6.3.3).

~~9.2 The space hp, p < 1~~

~~Recall that the harmonic Hardy space hp, 0 < p < ∞, is deﬁned by the requirement~~

kukp := sup Mp(r, u) < ∞. r<1 In contrast to the case of Bergman spaces, the space hp for p < 1 diﬀers very much from its analytic analogue; for information we refer to [89]. However, there are similarities; for example:

~~9.2.1 Theorem For u ∈ hp (0 < p < 1) we have:~~

~~−1/p |u(z)| 6 Cp(1 − |z|) kukp (z ∈ D), (9.11) 1/p−1 |ub(n)| 6 Cp(|n| + 1) kukp (n ∈ Z). (9.12) Proof. By (9.3) we have~~

~~Z Z 1 iθ p −2 p p |u(re )| 6 Cp(1 − r) |u(w)| dA(w) 6 2Cp Mp (ρ, u)ρ dρ 2r−1<|w|<1 2r−1~~

~~(for 1/2 < r < 1), which implies (9.11). Hence~~

1 Z π 1 Z π |u(reiθ)| dθ = |u(reiθ)|p |u(reiθ)|1−p dθ 2π −π 2π −π 1−p p 1−p −1/p+1 p 6 {M∞(r, u)} kukp 6 (Cp) (1 − r) kukp. This gives (9.12) via the inequality Z π |n| 1 iθ r |ub(n)| 6 |u(re )| dθ, r = 1 − 1/(|n| + 1). 2π −π The proof is complete. 2 In a similar way one proves the following estimates. 148 9 Subharmonic behavior

~~9.2.2 Theorem (Hardy/Littlewood [26]) If u = Re f, f ∈ H(D) and u ∈ hp (0 < p 6 1), then 2 1/p M (r, f 0) C (1 − r)−1kuk ,M (r, f) C log kuk (0 < r < 1). p 6 p p p 6 p 1 − r p~~

~~Two open problems As far I know, the following two problems are still unsolved. Problem A. Whether there exists a function u ∈ hp (p < 1) such that~~

1/p−1 |ub(n)| > c(|n| + 1) (c = const > 0, n ∈ Z) ? (9.13) Problem B. Whether there exists a function f ∈ H(D) such that u = Re f ∈ hp and 2 M p(r, f) c log (0 < r < 1) ? (9.14) p > 1 − r Hardy and Littlewood [26] proved that the answer is positive provided that p = 1/N, N = 2, 3,... ; their example is ∂N P u(z) = DN P (z) = (reiθ), ∂θN where P is the Poisson kernel. It is clear that u satisfies condition (9.13) for p = 1/N. That condition (9.14) is satisfied follows from the estimate (e.g., [89]) 1 − |z|2 |DN P (z)| C . 6 |1 − z|N+1 It should be noted that solving Problem A leads to solution of B. Namely, condition (9.13) implies condition (9.14). Indeed, if (9.13) is satisfied, then |fb(n)| > c(n + 1)1/p−1, so the conclusion follows from the inequality

~~∞ p X p−2 p np Mp (r, f) > c (n + 1) |fb(n)| r n=0 (see Theorem 7.5.1).~~

~~9.2.3 Exercise (Hardy/Littlewood) Let u = Re f, f ∈ H(D), α > 0, and 0 < p < ∞. Then the following three conditions are equivalent:~~

~~−α −α 0 −α−1 Mp(r, f) = O(1 − r) ; Mp(r, u) = O(1 − r) ; Mp(r, f ) = O(1 − r) .~~

~~9.3 Subharmonic behavior of smooth functions~~

In this section we present some results that are closely related to Theorem 9.1.1. Unless otherwise stated, we consider real-valued functions deﬁned on a proper subdomain G of C. For the proofs of slightly more general results we refer to [75, 77]. For generalizations to the case of the so called M-harmonic functions, see [73, 34]. 9.3 Subharmonic behavior of smooth functions 149

~~The class SH(G). For a constant K > 1, let SHK (G) denote the class of nonnegative, continuous functions u on G such that Z −2 u(z) 6 Kr u dm Dr (z)~~

~~S (‡) whenever Dr(z) := {w : |w − z| < r} ⊂ G. We put SH(G) = SHK (G). K>1 As remarked on page 144, the ﬁrst proof of Theorem 9.1.1 gives the following result:~~

~~p 9.3.1 Proposition Let p > 0. If u ∈ SHK (G), then u ∈ SHC (G), where C is a constant depending only on K and p.~~

Classes of C1 functions The class HC1(G) The second proof of Theorem 9.1.1 (p. 145) was based on the property −1 |∇f(z)| 6 Kr sup |f|,Dr(z) ⊂ G, (9.15) Dr (z) 1 where K > 1 is a constant independent of Dr(z) ⊂ G. We denote by HCK (G) the class of all locally Lipschitz functions f on G satisfying (9.15). Note that (9.15) is implied by |∇f(z)| 6 K|f(z)|/δG(z), δG(z) = dist(z, ∂G), (9.16) which is a restriction on the growth of f and is therefore stronger than (9.15).

9.3.2 Example (a) It is a simple but important fact that condition (9.16) is satisﬁed if f is a positive function harmonic in G. (b) If f = |g0|, where g is a univalent function in G = D, then the proof of Theorem 4.3.5 shows that 3f(z) |∇f(z)| = |g00(z)| . 6 1 − |z|2

~~1 1 The class OCK (G) is the subclass of HC2K (G) consisting of those f such that~~

~~−1 |∇f(z)| 6 Kr Of(z, r),Dr(z) ⊂ G, where Of(z, r) is the oscillation of f on Dr(z),~~

~~Of(z, r) = sup{ |f(w) − f(z)| : w ∈ Dr(z) }.~~

1 9.3.3 Example If a function f : G 7→ R is convex or harmonic, then f ∈ OCK (G) for some K independent of f. In particular the function f(x+iy) = ex is in HC1(G), where G is the right half-plane, but f does not satisfy (9.16).

~~(‡)In deﬁning other classes we proceed in a similar way. 150 9 Subharmonic behavior~~

~~9.3.4 Example As a further example, we have that |f| ∈ OC1(G), if f is analytic in G. (See Lemma 10.2.4.)~~

~~1 9.3.5 Theorem (a) If f ∈ HCK (G), then |f| ∈ SHC (G), where the constant C depends only on K. (b) If f ∈ OC1(G), then both |f| and |∇f| belong to SH(G).~~

~~We note two consequences of assertion (a).~~

9.3.6 Corollary Let p > 0. A function f ∈ C1(G) belongs to HC1(G) if and only if there is a constant K such that Z p −2−p p |∇f(z)| 6 Kr |f| dm, 0 < r < δG(z). Dr (z) Let Z 1/p n 1 p o Opf(z, r) = |f(w) − f(z)| dm(w) , |Dr(z)| Dr (z) p the L -oscillation over Dr(z).

9.3.7 Corollary Let p > 0. A function f belongs to OC1(G) if and only if −1 |∇f(z)| 6 Kr Opf(z, r), 0 < r < δG(z), for some constant K. This is deduced from the preceding corollary by considering the functions f − const. Proof of Theorem 9.3.5. The proof of (a) is the same as the second proof of Theorem 9.1.1, p. 145. It remains to prove that f ∈ OC1(G) implies |∇f| ∈ SH(G). 1 Let f ∈ OCK (G). By Theorem 9.3.1, it suﬃces to prove that, for some q, the function |∇f|q belongs to SH(G). This can be reduced to proving that Z q q |∇f(0)| 6 C |∇f| dm D provided that D ⊂ G. By Corollary 9.3.7, we have Z |∇f(0)| 6 K |f(z) − f(0)| dm(z). D On the other hand, Z 1 |f(z) − f(0)| 6 |z| |∇f(rz)| dr, 0 whence Z Z 1 Z |f(z) − f(0)| dm(z) 6 dr |∇f(rz)| |z| dm(z). D 0 D Hence, by the change z = w/r and Fubini’s theorem,

~~Z Z 1 Z −3 −1 |∇f(0)| 6 K |∇f(w)| dm(w) r |w| dr 6 K |∇f(w)||w| dm(w). D |w| D 9.3 Subharmonic behavior of smooth functions 151~~

~~Now the required inequality is proved by H¨older’sinequality with the indices q = 3 and q0 = 3/2, using the fact that the function w 7→ |w|−1 belongs to the space L3/2(D, dm). 2~~

~~Classes of C2-functions The class HC2(G) consists of those f ∈ C2(G) for which~~

~~−1 −2 |∆f(z)| 6 Kr sup |∇f| + K0r sup |f| (9.17) Dr (z) Dr (z) for some constants K, K0. The condition~~

~~−1 −2 |∆f(z)| 6 K|∇f(z)|δG(z) + K0|f(z)|δG(z) implies (9.17) with the same values of K and K0.~~

~~The class OC2(G) consists of those f ∈ C2(G) such that~~

~~−1 |∆f(z)| 6 Kr sup |∇f| (9.18) Dr (z) for some constant K. In particular, OC2(G) contains every function f for which~~

~~|∆f(z)| 6 K|∇f(z)|/δG(z), z ∈ G. By Lagrange’s theorem, f belongs to OC2(G) iﬀ~~

~~−1 −2 |∆f(z)| 6 Kr sup |∇f| + K0r O(f, r) for some K and K0. Dr (z)~~

9.3.8 Example Condition (9.18) is satisﬁed if f is a harmonic function on an arbitrary domain G. Let f be an eigenfunction of ∆, i.e., ∆f ≡ λf for some constant λ. Assuming that the closure of rD = Dr(0) is contained in G, we have 1 Z π d Z 2 f(reiθ) dθ = r−1 ∆f dm. 2π dr −π rD Hence 1 Z π d Z r∆f(0) = 2 f(reiθ) dθ − r−1 (f − f(0)) dm, 2π dr −π rD and hence −1 |∆f(0)| 6 2r sup |∇f| + r|λ| sup |∇f|. rD rD Applying this to the function w 7→ f(w+z) we conclude that f ∈ OC1(G) provided G is bounded. On the other hand, if f(x+iy) = sin x, and G is the right half-plane, then ∆f = −f but f is not in OC2(G). 152 9 Subharmonic behavior

~~9.3.9 Theorem The following inclusions hold: (a) HC2(G) ⊂ HC1(G); and (b) OC2(G) ⊂ OC1(G).~~

9.3.10 Corollary A function f ∈ C2(G) belongs to HC2(G) if and only if there −2 is a constant K such that |∆f(z)| 6 Kr sup |f|. Dr (z) The proof of Theorem 9.3.9 is based on the following lemmas.

2 9.3.11 Lemma If f : Dr(z) → R is a C -function, then −1 |∇f(z)| 6 2r sup |f| + (2/3) r sup |∆f|. (9.19) Dr (z) Dr (z) This inequality is a consequence of the formula ∂f 1 Z π 1 Z (0) = f(eiθ)e−iθ dθ − ∆f(w)(1 − |w|2)w−1 dm(w), ∂z 2π 4π −π D where f is a C2-function deﬁned in a neighborhood of the closed unit disk. The veriﬁcation of this formula can be reduced to the case of f(z) = zmz¯n, where m, n are nonnegative integers.

~~9.3.12 Lemma Let F1, F2 and F3 be nonnegative, continuous functions on G such that, for some constant K,~~

~~−1 F1(z)/K 6 r sup F2 + r sup F3 and (9.20) Dr (z) Dr (z)~~

~~−1 −2 F3(z)/K 6 r sup F1 + r sup F2 (9.21) Dr (z) Dr (z)~~

~~whenever Dr(z) ⊂ G. Then there is a constant C = C(K) such that~~

−1 F1(z) 6 Cr sup F2. (9.22) Dr (z) Proof. By translations the proof of (9.22) reduces to the case z = 0. Let the closure of Dε(0) be contained in G and F2 6 1 on Dε(0). (In the general case we consider the functions Fi/A, where A is chosen so that F2(z) 6 A for all z ∈ Dε(0).) Choose z ∈ Dε(0) so that F1(w)(ε − |w|) 6 F1(z)(ε − |z|) for all w ∈ Dε(0). This implies that F1(w) 6 2F1(z) for w ∈ Dδ(z), where δ = (ε − |z|)/2. Now we use the −1 −2 hypotheses to ﬁnd w in Dδ(z) so that F1(z)/K 6 r +(Kr/t)F1(w)+Krt for all −1 −2 r, t > 0 such that r +t = δ, which implies F1(z)/K 6 r +(2Kr/t)F1(z)+Krt . Now choose r, t so that r + t = δ and 2Kr/t = 1/2K, which implies that r = c1(ε − |z|), t = c2(ε − |z|) for some ci = ci(K), to obtain F1(z)/K 6 F1(z)/2K + −1 −1 −2 K1(ε − |z|) , where K1 = c1 + c1c2 . Hence F1(0)ε 6 F1(z)(ε − |z|) 6 2KK1, and this concludes the proof. 2

~~Proof of Theorem 9.3.9. Let f satisfy (9.17). We may assume that K > 2 and K0 > 2. Deﬁne the functions F1(z) = |∇f(z)|,F2(z) = |f(z)|,F3(z) = 9.3 Subharmonic behavior of smooth functions 153~~

|∆f(z)|. Then (9.20) is satisfied because of (9.19), and (9.21) is satisfied because of (9.17). Hence f ∈ HC1(G), by Lemma 9.3.12. This proves assertion (a). To prove (b) let f ∈ OC2(G). Applying (a), together with its proof, to the functions f − c we find a constant K1 independent of z, r, c so that |∇f(z)| 6 −1 K1r supDr (z) |f − c|. Finally we take c = f(z) to finish the proof. 2

Vector-valued functions The preceding notions and results can be extended to 2 vector-valued functions. For example, if f = (f1, f2) is a function from G to R , then we replace |∇f(z)| by kDf(z)k, where Df(z): R2 7→ R2 is the derivative of f at z. The Laplacian of f is deﬁned as ∆f = (∆f1, ∆f2). Applying Lemma 9.3.11 2 2 to the functions af1 + bf2, a + b = 1, we ﬁnd that

~~−1 kDf(z)k 6 2r sup |f| + (2/3) r sup |∆f|. Dr (z) Dr (z)~~

~~It turns out that Theorem 9.3.9 remains valid. As an application we note a suﬃcient condition for a C3-function to be in OC2(G).~~

~~9.3.13 Theorem A real valued C3-function f belongs to OC2(G) if there are constants K1 and K2 such that~~

~~−1 −2 |∇(∆f)(z)| 6 K1r sup ∇2(f) + K2r sup |∇f|. (9.23) Dr (z) Dr (z)~~

Here ∂2f 2 ∂2f 2 ∂2f 21/2 ∇ (f) = + + . 2 ∂x2 ∂y2 ∂x∂y Proof. Suppose that a real-valued function f satisﬁes (9.23). Since ∇(∆f) = ∆(∇f) and kD(∆f)k ∇2(f), we see that condition (9.23) means that ∇f ∈ HC2(G). Therefore (9.23) implies ∇f ∈ HC1(G), by Theorem 9.3.9, which means kD(∇f)(z)k Kr−1 sup |∇f| for some constant K. Since obviously |∆f| 6 Dr (z) 6 const · kD(∇f)k, it follows that f ∈ OC2(G). 2 Remark. Condition (9.23) implies the formally stronger condition

~~−2 |∇(∆f)(z)| 6 Kr sup |∇f|. Dr (z)~~

9.3.14 Corollary A C4-function f : G → R belongs to OC2(G) if so does ∆f. Consequently a C∞-function f belongs to OC2(G) if so does ∆kf for some integer k. In particular every polyharmonic function of ﬁnite order belongs to OC2.

A function f is polyharmonic of order k, where k is a positive integer, if ∆kf ≡ 0. For the theory of polyharmonic functions we refer to [5]. Proof. Let ∆f ∈ OC2. Then ∆f ∈ HC1, by Theorem 9.3.9, which means that |∇(∆f)(z)| Kr−1 sup |∆f|. Now the desired conclusion follows from 6 Dr (z) Theorem 9.3.13. 2 10 Lipschitz spaces

In this chapter we are concerned with some results which relate the modulus of smoothness of n-th order of a function g ∈ C(T) with the n-th tangential derivative of the Poisson integral of g. In Section 10.1 we consider the case n = 1 (Theorem 10.1.1); as an application we prove Privalov’s theorem on conjugate functions (Theorem 10.1.3 and 10.1.4). In Section 10.2 we use Section 10.1 to prove some results on the Lipschitz condition for the modulus of an analytic function. The case n > 2 is considered in the next two sections.

~~10.1 Lipschitz spaces of ﬁrst order~~

The space Λα(K) If K is a bounded subset of C, then, by deﬁnition, Λα(K) (0 < α 6 1) is the set of all complex-valued functions g on K such that α |g(z) − g(w)| 6 C |z − w| (z, w ∈ K), where C is a constant independent of z, w.

The space Lip(ω, K) More generally, let ω be a majorant, i.e., a continuous increasing function on [0, t0], where t0 is large enough, such that ω(0) = 0 and that ω(t)/t decreases (t > 0). Then the space Lip(ω, K) is deﬁned by the requirement

~~|g(z) − g(w)| 6 C ω(|z − w|). (10.1)~~

The norm is given by Cg + |g(a)|, where Cg = C (> 0) is the smallest constant satisfying (10.1) and a is any ﬁxed point from K; with this norm the space Lip(ω, K) is Banach. Since maxK |g| 6 |g(a)| + Cg ω(diam K), we see that the inclusion Lip(ω, K) ⊂ L∞(K) is continuous.

~~Lipschitz condition and the tangential derivative~~

it iθ In the case K = T condition (10.1) is equivalent to |g(e ) − g(e )| 6 C ω(|t − θ|), t, θ ∈ R (the value of C need not be the same). In particular, this means that the “classical” Lipschitz space Λ1(T) consists of absolutely continuous functions g ∈ C(T) with the bounded derivative (d/dθ)g(eiθ), and that therefore is isomorphic to L∞(T). Since Z π d ∂ Z π P (r, θ − t) g(eit) dt = P (r, θ − t)g(eit) dt, −π dt ∂θ −π ∞ we can conclude that g belongs to Λ1(T) iﬀ D(P [g]) belongs to h , where D is the operator of “tangential” diﬀerentiation, (Du)(reiθ) := ∂u(reiθ)/∂θ. This can be generalized in the following way.

~~154 10.1 Lipschitz spaces of ﬁrst order 155~~

10.1.1 Theorem Let ω be a majorant such that ω(t) is almost increasing for some α > 0. (10.2) tα Then a function u, harmonic in D, is equal to the Poisson integral of some function g ∈ Lip(ω, T) iff there exists a constant C such that ω(1 − |z|) |Du(z)| C for all z ∈ . (10.3) 6 1 − |z| D A real function ϕ, defined on some interval, is called almost increasing if there exists a constant C such that x < y implies ϕ(x) < Cϕ(y). (An almost decreasing function is defined similarly.) Let A(D) denote the disk-algebra, i.e., the set of functions that are analytic in D and have a continuous extension to the boundary. Since Df(reiθ) = ireiθ f 0(reiθ), f ∈ A(D), we have the following consequence.

~~10.1.2 Corollary Let ω be as in the theorem and let f ∈ A(D). The boundary function f∗ belongs to Lip(ω, T) iﬀ ω(1 − |z|) |f 0(z)| C (z ∈ ). 6 1 − |z| D~~

This last condition implies f ∈ A(D). Theorem 10.1.1 is an immediate consequence of the following assertions which will be proved later on in a more general situation (see Propositions 10.4.4 and 10.4.5); the direct proof is rather simple. −1 If u = P [g], g ∈ C(T), then M∞(r, Du) 6 C(1 − r) ω(g, 1 − r), 0 < r < 1, where ω(g, t) is the modulus of continuity of g. R 1 On the other hand, if u ∈ h(D) and 0 M∞(r, Du) dr < ∞, then u = P [g] for some g ∈ C(T) satisfying Z 1 ω(g, t) 6 C M∞(r, Du) dr, 0 < t < 1. 1−t

~~Privalov’s theorem on conjugate functions~~

The condition g ∈ Λ1(T) does not imply that the radial derivative ∂P [g]/∂r is bounded (in contrast to the tangential derivative) because the functions ∂u/∂θ and r∂u/∂r represent an arbitrary pair of harmonic conjugates. However, under an additional hypothesis on the majorant we have the following theorem of Privalov:

~~10.1.3 Theorem Let (10.2) be satisﬁed and suppose that ω(t)/tβ is almost decreasing for some β < 1. (10.4)~~

~~If u = P [g] and g ∈ Lip(ω, T), then the conjugate function ue has a continuous extension to D and its boundary function belongs to Lip(ω, T). 156 10 Lipschitz spaces~~

In other words, If conditions (10.2) and (10.4) are satisﬁed, then the Hilbert operator maps Lip(ω, T) to Lip(ω, T). Proof of Theorem 10.1.3. Let u be real-valued. The function v = Du is −1 harmonic and therefore |∇v(z)| 6 2R sup{ |v(w)| : |w − z| < R}, where R = (1 − |z|)/2. Using the hypotheses and Theorem 10.1.1, we get ω(1 − |z|) |∇v(z)| C . 6 (1 − |z|)2

2 2 2 2 2 2 2 Since |∇v| > |∂v/∂θ|/r = |∂ u/∂θ |/r and ∂ u/∂θ = −r ∂ u/∂r , it follows that ∂2u ω(1 − r) (reiθ) C , 0 < r < 1, ∂r2 6 (1 − r)2 where C is independent of reiθ. Integrating this from r = 0 and using (10.4) we get ∂u ω(1 − r) (reiθ) C , ∂r 6 1 − r which together with (10.3) gives ω(1 − |z|) |∇u(z)| = |∇u(z)| C . (10.5) e 6 1 − |z|

Finally, (10.2) and (10.5) imply ue ∈ Lip(ω, D) (see Lemma 10.1.6 below). 2 According to the above proof, we have the following theorem on extension of a Lipschitz condition from T to D. 10.1.4 Theorem (a) Let ω satisfy (10.2) and (10.4). Then the function g belongs to Lip(ω, T) iﬀ the Poisson integral of g belongs to Lip(ω, D). (b) If f ∈ A(D), then (10.2) is suﬃcient for the validity of the implication

f∗ ∈ Lip(ω, T) ⇒ f ∈ Lip(ω, D). 10.1.5 Remark There are cases where a Lipschitz condition on the circle extends to the disk with saving the corresponding Lipschitz constant. For example, if f ∈ α α A(D), 0 < α 6 1 and |f(ζ)−f(η)| 6 |ζ −η| , ζ, η ∈ T, then |f(z)−f(w)| 6 |z−w| , z, w ∈ D. The latter is equivalent to |z − w| |f(z) − f(w)| (z, w ∈ ), 6 |1 − wz¯ |1−α D

~~0 2 α−1 and this implies |f (z)| 6 (1 − |z| ) , z ∈ D.~~

10.1.6 Lemma If a C1-function u: D 7→ C satisfies (10.5) and ω satisfies the (Dini) condition Z x ω(t) ω1(x) = dt < ∞, x > 0, (10.6) 0 t then u ∈ Lip(ω1, D). 10.1 Lipschitz spaces of first order 157

~~Proof (cf. [84, Lemma 6.4.8]). Let~~

~~ω(1 − |z|) |∇u(z)| , z ∈ . 6 1 − |z| D~~

Let |a| 6 |b| 6 1. By Lagrange’s theorem, ω(1 − |c|) |u(a) − u(b)| |a − b|, 6 1 − |c| where c = (1 − λ)a + λb for some λ ∈ (0, 1). Since |c| 6 |b| and ω(t)/t decreases, we see that ω(1 − |c|) ω(1 − |b|) , 1 − |c| 6 1 − |b| hence |u(a) − u(b)| 6 ω(|a − b|) 6 ω1(|a − b|), under the condition |a − b| 6 1 − |b|. 0 0 If 1 − |b| 6 |a − b| 6 1 − |a|, then |u(a) − u(b)| 6 |u(a) − u(b )| + |u(b ) − u(b)|, where b0 = (1 − δ)b/|b|, δ = |a − b|. Using Lagrange’s theorem as above we get

~~ω(1 − |b0|) ω(δ) |u(a) − u(b0)| |a − b0| = |a − b0| ω(δ) ω (δ). 6 1 − |b0| δ 6 6 1~~

~~In the case of |u(b0) − u(b)|, we have~~

~~Z |b| Z 1 0 ω(1 − t) ω(1 − t) |u(b ) − u(b)| 6 dt 6 dt = ω1(δ). |b0| 1 − t 1−δ 1 − t~~

~~10.1.7 Remark If ω satisﬁes (10.6), then ω1 is a concave majorant. Let~~

~~Z 1 ω(t) ω2(x) = x 2 dt, 0 < x < 1. x t~~

This function is not increasing but has the properties: ω2 is concave, ω2(t)/t is decreasing and ω2(0+) = 0. It follows that ω2 is a majorant near 0. We have Z x ω2(t) dt = ω1(x) + ω2(x), 0 < x < 1. 0 t Then Lemma 10.1.6 and the proof of Privalov’s theorem yield: Suppose that ω satisﬁes the Dini condition. (a) If u = P [g] and g ∈ Lip(ω, T), then the conjugate function ue has a continuous extension to D, and, moreover, u and ue belong to Lip(ω3, D), where ω3(x) = ω1(x) + ω2(x). (b) If f ∈ A(D) and f∗ ∈ Lip(ω, T), then f ∈ Lip(ω1, D). 158 10 Lipschitz spaces

10.1.8 Remark A majorant ω is said to be regular if there exists a constant C such that Z x ω(t) Z 1 ω(t) dt + x 2 dt 6 Cω(x) (0 < x < 1). 0 t x t It is easily verified that condition (10.2)&(10.4) implies that ω is regular. The converse is true as well. Moreover, ω satisfies (10.2) iff ω1(x) 6 Cω(x) for some constant C; and ω satisfies (10.4) iff ω2(x) 6 Cω(x) for some constant C.

~~10.2 Lipschitz condition for the modulus~~

0 A function f ∈ H(D) satisfies the condition |f(z) − f(w)| 6 |z − w| in D iff |f | 6 1 in D. On the other hand, the corresponding Lipschitz condition for |f| is satisfied 0 iff ∇|f| 6 1. Since ∇|f| = |f |, we conclude that there holds the relation

~~f ∈ Λ1(D) ⇐⇒ |f| ∈ Λ1(D), f ∈ H(D). This is the simplest case of the following theorem.~~

~~10.2.1 Theorem Let ω satisfy the Dini condition (10.6). If f ∈ H(D) and |f| ∈ Lip(ω, D), then f ∈ Lip(ω1, D).~~

The theorem states, in particular, that if |f| ∈ Lip(ω, D), and ω satisﬁes (10.6), then f ∈ A(D). On the other hand, there exists a function f ∈ H(D) r A(D) such that the function |f| has a continuous extension to the closed unit disk. To show this, we use the known fact that there exists a bounded analytic function u + iv such that u is continuous on D, while v has no continuous extension to D. Then there are a point ζ ∈ T, two sequences {zn} ⊂ D and {wn} ⊂ D tending to ζ, and two points a, b ∈ C (a 6= b) such that v(zn) → a and v(wn) → b. We can assume that eia 6= eib since otherwise we can consider the function (u + iv)/λ for a suitable λ > 0. Then the desired function is f = exp(u + iv). As a consequence of the above theorem we have the following result of Dyakonov.

10.2.2 Theorem [20] Let ω satisfy (10.2). A function f ∈ H(D) belongs to the space Lip(ω, D) iﬀ |f| belongs to the same space. Before stating another theorem of Dyakonov recall that the function |f| is subharmonic and consequently P [|f∗|](z) − |f(z)| > 0, for all z ∈ D. We also know that P [|f∗|](z) − |f(z)| → 0 as |z| → 1, (†) so it is a natural question how fast the convergence in (†) can be. It turns out that this is closely related to Lipschitz condition for f, i.e., to growth of the ﬁrst derivative.

~~10.2.3 Theorem (Dyakonov [20]) Let ω satisfy (10.2) and (10.4), f ∈ A(D). Then the following conditions are equivalent: 10.2 Lipschitz condition for the modulus 159~~

~~iθ iθ (i) |f∗| ∈ Lip(ω, T) and |f(e )| − |f(re )| 6 Cω(1 − r) for some constant C;~~

(ii) |f∗| ∈ Lip(ω, T) and P [|f∗|](z) − |f(z)| 6 Cω(1 − |z|) for some constant C; (iii) f ∈ Lip(ω, D). Dyakonov’s proofs are based on theorems on pseudoanalytic continuation and theorems on division by inner functions. Here we present the proof from [78]. The key is in connection between the modulus of the ﬁrst derivative and the oscillation of the modulus of the function.

~~10.2.4 Lemma Let Dz = {w : |w − z| 6 1 − |z|}, f ∈ A(D). Then~~

~~1 0 (1 − |z|)|f (z)| 6 sup |f(w)| − |f(z)| (z ∈ D). 2 w∈Dz~~

Proof. Let Mz = sup{|f(w)|: w ∈ Dz}. If z = 0 and M0 = 1, then Schwarz’ 0 2 lemma gives |f (0)| 6 1 − |f(0)| 6 2(1 − |f(0)|), which is the required inequality in a special case. In the general case we apply this special case to the function F (ζ) = f(z + ζ(1 − |z|))/Mz, ζ ∈ D. 2 Proof of Theorem 10.2.1 Assuming |f| ∈ Lip(ω, D), we have |f(w)| − |f(z)| 6 Cω(|w − z|) 6 Cω(1 − |z|) for every z ∈ D and w ∈ Dz. Taking the supremum over w ∈ Dz and using Lemma 0 10.2.4, we get |f (z)|(1 − |z|) 6 2Cω(1 − |z|). Now the result follows from Lemma 10.1.6. 2 Proof of Theorem 10.2.3.[78] The implication (iii) ⇒ (i) is trivial. Consider the implication (i) ⇒ (ii). Assuming (i), let h(z) = P [|f∗|](z), f ∈ Lip(ω, D) =: X and |f∗| ∈ Lip(ω, T). Then h ∈ X, by Theorem 10.1.4, and |f| ∈ X because f ∈ X. Hence h(z) − |f(z)| = h(z) − |f(z/|z|)| + |f(z/|z|)| − |f(z)| = h(z) − h(z/|z|) + |f(z/|z|)| − |f(z)| 6 Cω(1 − |z|) for z ∈ D r {0}, which implies (ii). In order to prove that (ii) implies (iii), we start from the inequality

~~|f(w)| − |f(z)| 6 h(w) − |f(z)| = h(w) − h(z) + h(z) − |f(z)|, valid for z ∈ D and w ∈ Dz. From the hypothesis |f∗| ∈ Lip(ω, T) and Theo- rem 10.1.4 it follows that~~

~~h(w) − h(z) 6 Cω(|w − z|) 6 Cω(1 − |z|), w ∈ Dz. By (ii), we have h(z) − |f(z)| 6 Cω(1 − |z|), so we get~~

~~|f(w)| − |f(z)| 6 Cω(1 − |z|), w ∈ Dz. The proof is now concluded as in the case of Theorem 10.2.2. 2 160 10 Lipschitz spaces~~

~~10.3 Lipschitz spaces of higher order~~

~~n Moduli of smoothness If h is a complex-valued function deﬁned on R, then ∆t h (n is a positive integer, t ∈ R) denotes the n-th symmetric diﬀerence with step t :~~

~~1 n 1 n ∆t h(θ) = h(θ + t) − h(θ)(θ ∈ R), and ∆t h = ∆t ∆t h (n > 2).~~

~~2 In particular, ∆t h(θ) = h(θ + 2t) − 2h(θ + t) + h(θ). In the general case there holds the formula n X n ∆nh(θ) = (−1)n−kh(θ + kt). t k k=0~~

n n iθ n If g is a function on the unit circle, then ∆t g is defined by ∆t g(e ) = ∆t h(θ), iθ n where h(θ) = g(e ). For fixed n and t, ∆t is a linear operator which preserves n n C(T); we have k∆t gk 6 2 kgk, where k · k = k · k∞ denotes the max-norm in C(T). n The modulus of smoothness of order n is defined by ωn(g, t) = sup{k∆s gk : |s| < t}, t > 0, g ∈ C(T).

~~Lipschitz spaces Let φ be a positive function on (0, π] and let n be an integer. The Lipschitz space Lipn(φ) consists of those functions g ∈ C(T) for which~~

~~ωn(g, t) = O(φ(t)), t → 0;~~

~~the norm is deﬁned by~~

~~ω (g, t) kgk = |g(0)| + sup n . φ,n b φ(t) 0~~

~~The analogous space deﬁned by the “little oh” condition will be denoted by lipn(φ).~~

Zygmund classes In the case where n = 2 and ω(t) = t the spaces Lipn(φ) and lipn(φ) are known as the Zygmund classes (cf. [100, II.§§3,4]) and are denoted by Λ∗ and λ∗, respectively. A function g ∈ λ∗ is called a smooth function because the existence of the left derivative at a point implies the existence of the right as well as that they are equal. The set of points where g is diﬀerentiable may be of measure zero although must be everywhere dense and of the power of the continuum. See [100, p. 43–48]. The Weierstrass function

~~∞ X g(θ) = b−nα cos(bnθ), n=1~~

where b = 2, 3,... and α > 0, belongs to Λ∗ for α = 1, and to Λα(T) for α < 1. It is known that for α 6 1 the function g is nowhere diﬀerentiable (Weierstrass, Hardy). 10.3 Lipschitz spaces of higher order 161

The spaces h∞,n(ψ) Let h∞(ψ) denote the class of the functions u ∈ h(D) for − which u(z) = O (ψ (1/(1 − r))), r = |z| → 1 , where ψ(x), x > 1, is a positive nonincreasing function that grows slower than some power of x. The latter means that, for some β > 0, the function ψ(x)/xβ is almost decreasing. Such functions α γ are, for example ψ(x) = x log(2x) , where α > 0 and β ∈ R, or α = 0 and (∗) β > 0. In the simplest case, when ψ(x) ≡ 1, we have h∞(ψ) = C(T), and then, as we have seen, the Poisson integral acts as an isometrical isomorphism from C(T) onto hC(D) = C(D) ∩ h(D) (see Theorem 3.1.7). Hence every subclass of C(T) can be regarded as a subclass of hC(D), and conversely. Following the paper [71], we shall show that any h∞(ψ) is isomorphic to some Lipschitz space via the tangential derivative of suﬃciently large order. Under the above hypothesis on ψ, we deﬁne

~~M(r, u) kukψ = |u(0)| + sup , 0~~

~~n h∞,n(ψ) = {u ∈ h(D): kD ukψ < ∞}, n = 0, 1, 2,..., where~~

~~∞ ∂nu X (Dnu)(reiθ) = (reiθ) = (ij)nuˆ(j)r|j|eijθ, reiθ ∈ . ∂θn D j=−∞~~

~~n − The subspace of h∞,n(ψ) deﬁned by M(r, D u) = o (ψ (1/(1 − r))), r → 1 , will be denoted by ho,n(ψ).~~

~~Conditions for majorants From now on we shall assume that ψ is a positive almost increasing function on [1, ∞) satisfying the condition:~~

~~there exists a constant C < ∞ such that ψ(2x) 6 Cψ(x), x > 1. (U)~~

~~This is equivalent to the existence of a positive constant α such that~~

~~ψ(x) is almost decreasing (x 1). (U ) xα > α~~

~~Concerning the function φ, we shall assume that it positive and almost increasing on (0, 1] and that there exists β > 0 such that~~

~~φ(t) is almost decreasing (0 < t < 1). (U 0) tβ β~~

~~(∗)with respect to ψ 162 10 Lipschitz spaces~~

~~10.4 Growth of derivatives~~

~~The following theorem is classical and is due Hardy and Littlewood, Zygmund, and Privalov (see [18, Ch. 2, §§1,2], [100], and [67]).~~

~~10.4.1 Theorem Let 0 < a 6 n (n = 1, 2,... ) and u ∈ h(D). Then the following two conditions are equivalent:~~

~~a u ∈ hC(D)& ωn(u∗, t) = O(t ) ; (10.7) Dnu(z) = O((1 − |z|)a−n). (10.8)~~

~~If in addition a < n and u = Re f, f ∈ H(D), then the condition f (n)(z) = O((1 − |z|)a−n) is equivalent to each of the preceding.~~

~~Recall that hC(D) = C(D) ∩ h(D). In this section we extend Theorem 10.4.1 in the following way.~~

10.4.2 Theorem If φ(t)/tn almost decreases and φ(t)/tb almost increases, for some b > 0, then the Poisson integral acts as an isomorphism of Lipn(φ) onto n h∞,n(ψ), where ψ(x) = x φ(1/x), x > 1.

~~Remark. The analogous assertion for the pair lipn(φ), ho,n(ψ) is also valid. This theorem is contained in the following.~~

~~10.4.3 Theorem Let n be a positive integer and let ψ and φ satisfy conditions 0 (U) and (Un). Then the following assertions are equivalent:~~

~~(a) Lipn(φ) = h∞,n(ψ);~~

~~(b) There are constants α < n and C such that ψ satisﬁes (Uα) and~~

~~n φ(t)/C 6 t ψ(1/t) 6 Cφ(t), 0 < t < 1.~~

0 Condition (Un) is natural because the modulus of smoothness has the property n that ωn(g, t)/t almost decreases (Lemma 10.4.7). However, the theorem cannot be applied to the case 1 φ(t) = (b > 0), (1 + log(1/t))b

~~and we do not know whether the space Lipn(φ) is isomorphic to some of the spaces h∞(ψ).~~

~~Proof of Theorem 10.4.3 The implication (b)⇒(a) is a consequence of the following two propositions, which will be proved later on. 10.4 Growth of derivatives 163~~

~~10.4.4 Proposition If u ∈ hC(D), then n −n M(r, D u) 6 C(1 − r) ωn(u∗, 1 − r), 0 < r < 1, (10.9) where C < ∞ depends only on n (n = 1, 2,... ).~~

~~For the proof see page 166.~~

10.4.5 Proposition If u ∈ h(D) and Z 1 (1 − r)n−1M(r, Dnu) dr < ∞, (10.10) 0 then u ∈ hC(D) and Z 1 n−1 n ωn(u∗, t) 6 C (1 − r) M(r, D u) dr, 0 < t < 1, (10.11) 1−t where C depends only on n.

~~For the proof see page 168.~~

Let n be a ﬁxed integer. Condition (Uα) from the preceding section can be written as α ψ(y) 6 C(y/x) ψ(x), y > x > 1. (Uα) Hence (Uα), α < n, implies Z ∞ −n−1 −n ψ(y)y dy 6 Cx ψ(x), x > 1, (An) x which is part of the following lemma.(†)

~~10.4.6 Lemma The function ψ satisfies (An) iff there exists α < n such that ψ satisfies (Uα).~~

Proof. We need to prove that (An) implies (Uα) for some α < n. Let ψ satisfy (An), and let Z ∞ −n−1 F (x) = ψ(y)y dy, x > 1. x −n It is easy to see that cF (x) 6 x ψ(x) 6 CF (x), x > 1, and hence it suﬃces b to ﬁnd b > 0 such that x F (x) is nonincreasing for x > 1. Choose b so that −n 0 F (x) 6 (1/b)ψ(x)x , x > 1, which can be written as F (x) 6 −(1/b)xF (x). This b implies that the derivative of x F (x) is 6 0, which concludes the proof. 2 Now the implication (b)⇒(a) of Theorem 10.4.3 is obtained immediately from Propositions 10.4.4, 10.4.5, Lemma 10.4.6 and the formula Z ∞ Z 1 ψ(y)y−n−1 dy = (1 − r)n−1ψ(1/(1 − r)) dr, 0 < t < 1. 1/t 1−r For the proof of the implication (a)⇒(b) we need some more lemmas.

~~(†)Such assertions are often encountered in the theory of regularly varying functions (see [88]). 164 10 Lipschitz spaces~~

~~n 10.4.7 Lemma If g ∈ C(T), then the function ωn(t)/t is almost decreasing for t > 0.~~

~~Proof. The lemma is easily deduced from the inequality~~

~~n ωn(g, 2t) 6 2 ωn(g, t), t > 0, (10.12) while (10.12) can be proved by means of the formula~~

~~n iθ X ijt n ijθ ∆t g(e ) = gˆ(j)(e − 1) e (10.13) |j|<∞ (g is a trigonometric polynomial), from which one gets~~

~~n X X n ∆n g(eiθ) = (eijt + 1)ngˆ(j)(eijt − 1)neijθ = ∆ng(ei(θ+kt)). 2 2t k t j k=0~~

~~iθ ikθ 10.4.8 Lemma If g ∈ C(T) and gk(e ) = g(e ), where k = 1, 2,... , then~~

|gˆ(0)| + ωn(gk, π/k) > kgk∞. Proof. From (10.13) it follows that Z π n iθ 1 n iθ (−1) g(e ) − gˆ(0) = (∆t g)(e ) dt. 2π −π Hence Z π 1 n kg − gˆ(0)k∞ 6 k∆t k∞ dt 6 ωn(g, π) = ωn(gk, π/k). 2 2π −π

10.4.9 Lemma Let X = Lipn(φ) or h∞,n(ψ). Then there hold the assertions: (i) If a sequence {um} ⊂ X converges in norm to u, then um(z) → u(z) uniformly on compact subsets of D. (ii) X is complete.

~~Proof. Let X = Lipn(φ). According to Lemmas 10.4.8 (k = 1) and 10.4.7,~~

kukX > |u(0)| + ωn(u∗, π)/φ(π) > cku∗k∞ = ckuk∞, u ∈ X. This shows that X is continuously embedded into C(T) = hC(D), which implies (i). The same fact can be used to prove the completeness of X. In the case when X = h∞,n(ψ) the proof is even simpler. 2

0 10.4.10 Lemma Let φ and ψ satisfy the conditions (U) and (Un). Let uk(z) = k z , where |z| 6 1 and k = 1, 2,... Then kn kDnu k (k 1), (10.14) k ψ ψ(k) > 1 kuk (k 1). (10.15) φ,n φ(1/k) > 10.4 Growth of derivatives 165

~~Proof. The proof of (10.14) is straightforward. In order to prove (10.15) we n k ikt n use the equality (∆t uk∗(w)) = w (e − 1) , w ∈ T. Hence~~

~~n n ωn(uk∗, t) = 2 sup{| sin(ks/2)| : 0 < s 6 t}, t > 0, and therefore~~

~~n n kukkφ,n = 2 sup{| sin(ks/2)| /φ(t): s 6 t 6 1, 0 < s 6 1},~~

~~Since φ(t) > φ(s)/C for 0 < s 6 t 6 1, we have~~

~~n kukkφ,n 6 C sup{| sin(ks/2)| /φ(s) : 0 < s 6 1},~~

~~n where C is independent of k. If 1/k 6 s 6 1, then | sin(ks/2)| /φ(s) 6 C/φ(1/k) because the function 1/φ is almost decreasing. If 0 < s 6 1/k, then~~

~~n −n n n n n | sin(ks/2)| /φ(s) 6 2 k s /φ(s) 6 C k (1/k) /φ(1/k)~~

n because s /φ(s) is almost increasing. Thus kukφ,n 6 C/φ(1/k). The proof of the reverse inequality is simpler. 2 Now we ready to prove the the implication (a)⇒(b) in Theorem 10.4.3. Let Lipn(φ) = h∞,n(ψ). It follows from Lemma 10.4.9 and the closed graph theorem n that kD ukψ kukkφ,n, k > 1, where uk is as in Lemma 10.4.10. Hence, by n Lemma 10.4.10, φ(1/k) (1/k) ψ(k), k > 1, which yields

n φ(t) t ψ(1/t), 0 < t 6 1, (10.16) and this is part of (b). In order to prove that (a) implies (Uα) for for some α < n, we consider the functions ∞ −n k X −n jk Uk(z) = k ψ(k)z + (jk) ψ(jk) − ψ((j − 1)k) z , z ∈ D. j=2

~~Assume, as we may, that ψ is nondecreasing. Then~~

∞ n k X jk M(r, D Uk) 6 ψ(k)r + ψ(jk) − ψ((j − 1)k) r j=2 ∞ jk−1 k X X p+1 6 ψ(k)r + ψ(p + 1) − ψ(p) r j=2 p=(j−1)k ∞ X = ψ(k)rk + ψ(p + 1) − ψ(p) rp+1 j=k ∞ X p = (1 − r) ψ(p) r 6 C ψ 1/(1 − r) . p=k 166 10 Lipschitz spaces

~~It follows that {Uk} is a norm bounded sequence in h∞,n(ψ). Now we use the inclusion h∞,n(ψ) ⊂ Lipn(φ) to conclude that the functions Uk are continuous on the closed disk and~~

~~ωn(Uk∗, t) 6 Cφ(t), 0 < t < 1, (10.17) where C is independent of t, k. On the other hand, by Lemmas 10.4.7 and 10.4.8,~~

~~Cωn(Uk∗, 1/k) > ωn(Uk∗, π/k) > kUk∗k∞ ∞ X = k−nψ(k) + k−n j−nψ(jk) − ψ((j − 1)k) j=2 ∞ X = k−n j−n − (j + 1)−nψ(jk). j=1~~

~~Hence, by (10.17), (10.16) and (U),~~

~~∞ −n X −n−1 −n k j ψ (j + 1)k 6 Cφ(1/k) 6 Ck ψ(k) j=1~~

~~and therefore Z ∞ Z ∞ −n−1 −n −n−1 −n y ψ(y) dy = k y ψ(yk) dy 6 Ck ψ(k), k = 1, 2,.... k 1~~

~~It is easily veriﬁed that this implies (An). Thus ψ satisﬁes (Uα) for some α < n (Lemma 10.4.6), and this concludes the proof of Theorem 10.4.3.~~

~~Proof of Proposition 10.4.4 In proving Proposition 10.4.4 and Proposition 10.4.5 we shall use the inequalities~~

~~n+1 −1 n M(r, D f) 6 C(1 − r) M((1 + r)/2,D u) (10.18) (n+1) −1 n M(r, f ) 6 C(1 − r) M((1 + r)/2,D u),~~

~~n where u = Re f, f ∈ H(D), and n > 0. Equivalently: if D u is bounded in D, then~~

n+1 −1 n (n+1) −1 n |D f(z)| 6 C(1 − |z|) kD uk∞, |f (z)| 6 C(1 − |z|) kD uk∞, where C is independent of f and z. The proof is left to the reader as an exercise. In proving Proposition 10.4.4 we may assume that u is real-valued and harmonic iθ in a neighborhood of the closed disk. For fixed r < 1 let h(θ) = ur(θ) = u(re ). Then Z n (n) (∆t h)(θ) = h (θ + x1 + ··· + xn) dx1 . . . dxn, (10.19) tE 10.4 Growth of derivatives 167 where tE is the n-dimensional cube [0, t]n. Hence h(n)(θ) = (Dnu)(reiθ)tn Z n (n) (n) = (∆t )(θ) − h (θ + x1 + ··· + xn) − h (θ) dx1 . . . dxn. tE Since Z x (n) (n) (n+1) n+1 |h (θ +x)−h (θ)| = h (θ +y) dy 6 M(r, D )x, x = x1 +. . . xn, 0 we get Z n+1 n n+1 M(r, D u) 6 k∆t urk∞ + M(r, D u)(x1 + ··· + xn) dx1 . . . dxn tE n n+1 n+1 = k∆t urk∞ + (n/2)M(r, D )t , 0 < r < 1, t > 0. n n iθ n The function ∆t u defined by (∆t u)(re ) = (∆t ur)(θ) is harmonic on the closed n n disk and therefore k∆t urk∞ 6 k∆t u∗k 6 ωn(u∗, t), t > 0. These inequalities together with (10.18) yield n −n −1 n M(r, D u) 6 t ωn(u∗, t) + Kt(1 − r) M((1 + r)/2,D u) (10.20) (t > 0, 0 < r < 1,) where K depends only on n. Let A(r) = (1 − r)−nM(r, Dnu), 0 < r < 1. It follows from (10.20) that −n n n −1 A(r) 6 t (1 − r) ω(t) + 2 Kt(1 − r) A((1 + r)/2), n m where ω(t) = ω(u∗, t). Choose an integer m so that 2 K 6 (1/4)2 and take −m −m t = a(1 − r), a = 2 . Then we have A(r) 6 a ω(1 − r) + (1/4)A((1 + r)/2), 0 < r < 1. Integrating this inequality from ρ (< 1) to 1, and introducing appropriate substitutions, we get Z 1 Z 1−ρ Z 1 −m A(r) dr 6 a ω(t) dt + (1/2) A(r) dr ρ 0 (1+ρ)/2 Z 1−ρ Z 1 −m 6 a ω(t) dt + (1/2) A(r) dr. 0 ρ R 1 Hence, since the integral ρ A(r) dr is finite, Z 1 Z 1−ρ −m (1/2) A(r) dr 6 a ω(t) dt. ρ 0 Now (10.9) follows from the inequalities Z 1−ρ Z 1 n+1 n+1 ω(t) dt 6 (1 − ρ) ω(1 − ρ),M(r, D u)(1 − ρ) 6 (n + 1) A(r) dr, 0 ρ which are valid because the functions ω and M are increasing. Thus the proof of Proposition 10.4.4 is finished. 168 10 Lipschitz spaces

~~Proof of Proposition 10.4.5~~

~~Let u = Re f, where f is analytic u D. Then n X f (k)(rz) 1 Z 1 f(z) = zk(1 − r)k + (1 − s)nzn+1f (n+1)(sz) ds (10.21) k! n! k=0 r~~

(z ∈ D, 0 < r < 1). Denoting the sum by fr,n we have Z 1 1 n (n+1) |f(z) − fr,n(z)| 6 (1 − s) M(s, f ) ds. n! r − From this and (10.18) it follows that (10.10) implies kf − fr,nk∞ → 0 (r → 1 ). Since the functions fr,n (r < 1) are continuous on the closed disk, we see that (10.10) implies the continuity of f, and consequently of u, on the closed disk. iθ In order to prove (10.11) let ur(θ) = u(re ), 0 < r 6 1. Then (10.11) is equivalent to Z 1 n n−1 n k∆t u1k∞ 6 C (1 − s) M(s, D u) ds, 0 < t < 1. (10.22) 1−t n n n Let r = 1 − 2t, 0 < t < 1/4. Then k∆t u1k 6 k∆t (u1 − ur)k + k∆t urk. It follows from (10.19) and the “increasing” property of M(r, Dnu) that Z 1 n n n n−1 n k∆t urk 6 t M(r, D u) 6 n (1 − s) M(s, D u) ds, 1−t n and therefore we have to prove that k∆t (u1 − ur)k is dominated by the right-hand n n side of (10.22). Since k∆t (u1 − ur)k 6 k∆t (f1 − fr)k, it is enough to prove that Z 1 n n−1 n k∆t (f1 − fr)k 6 C (1 − s) M(s, D u) ds. 1−t To prove this write (10.21) in the form

n X k f1(θ) − fr(θ) = H(θ) + hk(θ)(1 − r) /k!, where k=1 Z 1 1 n i(n+1)θ (n+1) iθ (k) iθ ikθ H(θ) = (1 − s) e f (se ) ds, and hk(θ) = f (re )e . n! r We have n Z 1 n n 2 n (n+1) k∆t Hk 6 2 kHk 6 (1 − s) M(s, f ) ds n! r Z 1 n−1 n 6 C (1 − s) M((1 + s)/2,D u) ds r Z 1 = 2nC (1 − s)n−1M(s, Dnu) ds, 1−t 10.4 Growth of derivatives 169

~~n where we have applied (10.18). In order to estimate k∆t hkk, let m = n − k + 1 (1 6 k 6 n) and observe that (10.19) implies~~

~~n k−1 m k−1 m k−1 m (m) k∆t hkk = k∆t ∆t hkk 6 2 k∆t hkk 6 2 t khk k.~~

(m) −1 n From this and the inequality khk k 6 C(1 − r) M((1 + r)/2,D u) (see (10.18)) it follows that Z 1 n n−k n −k n−1 n k∆t hkk 6 Ct M(1 − t, D u) 6 Ct (1 − s) M(s, D u) ds, 1−t where C is independent of t. Combining all the above results yields (10.11) for 0 < t < 1/4. If t > 1/4, we can apply Lemma 10.4.7 to reduce (10.11) to the case 0 < t < 1/4, and this completes the proof. 2

~~Misclellaneous~~

~~10.4.11 (conjugate functions) If ψ(x), x > 1, satisﬁes (U) (p. 161) and there exists a constant β > 0 such that~~

ψ(x)/xβ almost increases (10.23) then the space h∞(ψ) is “self-conjugate”, i.e., there holds u ∈ h∞(ψ) ⇒ ue ∈ P∞ n h∞(ψ), or, what is the same, the Riesz projector (R+u)(z) = n=0 ub(n)z acts from h∞(ψ) to h∞(ψ). From this and Theorem 10.4.3 it follows that the same α holds for Lipn(φ) provided there exist constants β < n and α > 0 such that φ(t)/t is almost increasing and φ(t)/tβ is almost decreasing. It was proved in [91] that

~~if (U) holds, then (10.23) is necessary for h∞(ψ) to be self-conjugate.~~

10.4.12 If a < n, then the derivative Dnu in Theorem 10.4.1 may be replaced by each of the derivatives ∂nu/∂jr∂n−jθ (j = 0, 1, . . . , n), see 10.4.11. If 0 < a < n, then Theorem 10.4.1 remains true when “O” is replaced by “o”.

~~10.4.13 A real function g ∈ C(T) belongs to Λ∗ iﬀ the Cauchy integral 1 Z π u(eiθ) f(z) := −it dθ, 2π −π 1 − ze satisﬁes C |f 00(z)| (|z| < 1). 6 1 − |z|~~

~~This condition implies that f ∈ A(D) and that the boundary function f∗ belongs to Λ∗. 11 Lacunary series~~

In the ﬁrst section we consider Paley’s theorems on lacunary series in Hp (The- orems 11.1.1 and 11.1.3). The rest of the chapter is devoted to the proof of a generalized version of the Gurarij/Matsaev theorem on Lp(0, 1)-integrability of lacunary power series (see Theorems 11.4.2 and 11.3.1). Since the proof is based on the ideas from Karamata’s proof of Littlewood’s tauberian theorem, we included a short discussion of Karamata’s ideas (Section 11.2).

~~11.1 Lacunary series in Hp~~

~~A sequence {n } of positive real numbers is called lacunary if it satisﬁes the k k>1 condition n inf k+1 > 1. k>1 nk~~

~~P nk The corresponding series akx is then called a lacunary series.~~

~~11.1.1 Theorem (Paley) Let nk be a lacunary sequence of positive integers. If f ∈ H1, then ∞ 1/2 X 2 kfk1 > c |fb(nk)| , k=1 where c > 0 is a constant independent of f.~~

~~This theorem does not extend to the case of L1(T), and this can be seen from 6.1.4. Theorem 11.1.1 will be deduced from the following result.~~

~~11.1.2 Theorem (Hardy/Littlewood) Let f be analytic in D. Then there hold the assertions: p (a) If f ∈ H , 0 < p 6 2, then Z 1 2 0 K := Mp (r, f )(1 − r) dr < ∞ (11.1) 0~~

~~2 and there exists a constant Cp such that K 6 Cpkfkp. p 2 2 (b) If 2 6 p < ∞, then (11.1) imply f ∈ H and kfkp 6 Cp(K + |f(0)| ).~~

~~Proof. (a) Let 0 < p 6 2. In view of Lemma 5.1.7, it is enough to discuss the case where f has no zeros in D. Further, we can assume that kfkp = 1. Let~~

~~170 11.1 Lacunary series in Hp 171~~

~~p/2 p 2 g = f . Then kfkp = kgk2 and~~

~~Z π p 0 p 1 iθ 2−p 0 iθ p Mp (r, f ) = (2/p) |g(re )| |g (re )| dθ. 2π −π~~

~~Hence, by H¨older’s inequality with the indices 2/(2 − p), 2/p, we get~~

~~p 0 p 2−p p 0 Mp (r, f ) 6 (2/p) M2 (r, g)M2 (r, g ).~~

~~2 0 2 2 0 Since M2(r, g) 6 kgk2 = 1, we see that Mp (r, f ) 6 (2/p) M2 (r, g ), and this reduces the proof to the easily proved inequality~~

~~Z 1 2 0 2 M2 (r, g )(1 − r)dr 6 Ckgk2. 0~~

~~(b) As remarked in the proof of Lemma 6.2.4, the function |f|p is of class C2 so we can apply Green’s formula to prove that the Hardy/Stein identity,~~

~~p2 Z 1 kfkp = |f(0)|p + |f|p−2|f 0|2 log dA, p 2 |z| D holds for every f ∈ Hp. Let g(z) = f(ρz), 0 < ρ < 1. By H¨older’s inequality we get~~

~~Z π 1 iθ p−2 0 iθ 2 p−2 2 0 p−2 2 0 |g(re ) |g (re )| dθ 6 Mp (r, g)Mp (r, g ) 6 kgkp Mp (r, g ). 2π −π~~

~~Hence Z 1 p 2 p−2 2 p−2 2 0 1 kgkp 6 |g(0)| kgkp + p kgkp Mp (r, g )r log dr. 0 r 2−p Multiplying this inequality by kgkp and then letting ρ tend to 1, we get the desired result. 2~~

nk+1 Proof of Theorem 11.1.1. Let λ = infk 1 . From Theorem 11.1.2 it > nk follows that ∞ Z rm+1 X 2 0 kfk1 > c1 M1 (r, f )(1 − r) dr, m=1 rm −m m m+1 where rm = 1 − λ , c1=const> 0. For each m, the block Im = [λ , λ ) contains at most one member of the sequence {nk}. Therefore we can suppose that 0 n−1 nm ∈ Im for all m. Since M1(r, f ) > n|fb(n)|r , we have

~~Z rm+1 Z rm+1 2 0 2 2 2nm−2 M1 (r, f )(1 − r) dr > nm|fb(nm)| r (1 − r) dr. rm rm Now the proof is easily completed. 2 172 11 Lacunary series~~

P∞ nk 11.1.3 Theorem (Paley) Let the series f(z) = k=1 akz , where {nk} is a p P∞ 2 lacunary sequence, converge in D. Then f ∈ H (0 0 such that

~~−1 C k{ak}k2 6 kfkp 6 Ck{ak}k2. (11.2)~~

Proof. In the case 1 6 p < 2, the result is an immediate consequence of Paley’s theorem. Let 0 < p < 1. Let f be analytic in a neighborhood of the closed disk. Then, by means of the Cauchy/Schwarz inequality, we get Z p/2 1−p/2 p/2 (2−p)/2 p/2 (2−p)/2 kfk1 = |f| |f| 6 kfkp kfk2−p 6 kfkp kfk2 .

p/2 p/2 Since kfk1 > ckfk2, we see that c kfk2 6 kfkp . If f is arbitrary, then we apply this inequality to the functions fρ (ρ → 1) and this completes the proof in the case 0 < p < 2. Let 2 < p < ∞ and q = p/(p − 1). It follows from Paley’s theorem that

P nk q 2 the operator P ,(P f)(z) = fb(nk)z , is bounded from H to H . The dual operator P ∗ is formally equal to P , and since (Hq)∗ = Hp (Theorem 6.3.2), we 2 have kP fkp 6 Cpkfk2 for f ∈ H . Hence kfkp 6 Cpkfk2 if P f = f. In the case p > 2, the theorem can also be deduced from Theorem 11.1.2(b) by using 7.5.5. 2

Miscellaneous 11.1.4 Inequality (11.2), known as Paley’s inequality, holds for every p ∈ (0, ∞) under the more general assumption that f is an arbitrary trigonometric polynomial with lacunary coeﬃcients.

~~On the other hand, in the case p = ∞, there holds Sidon’s theorem [100, Theorem VI.(6.1)]:~~

~~11.1.5 Theorem A trigonometric series with lacunary coefficients is the Fourier series of a bounded function iff the sequence of coefficients is absolutely summable.~~

~~For further properties of lacunary series see Zygmund [100, Ch. V, §§6,7].~~

~~11.2 Karamata’s theorem and Littlewood’s theorem~~

~~Recall Abel’s theorem: Let ∞ ∞ X k X n f(z) := Ak z = (1 − z) sn z (|z| < 1, z ∈ C), (11.3) k=0 n=0 11.2 Karamata’s theorem and Littlewood’s theorem 173~~

where {Ak} (k > 0) is a sequence of vectors in a Banach space (X, | · |), and sn = Pn k=0 Ak. If there exists the limit limn→∞ sn =: s ∈ X, then limr→1− f(r) = s. In other words, convergence of a series implies its summability by Abel’s method; the converse is not true. Tauber proved that if the “tauberian“ condition limn→∞ n An = 0 is satisﬁed, then the existence of limr→1 f(r) =: l implies limn→∞ sn = l. Littlewood [54] improved Tauber’s theorem by replacing Tauber’s condition by (∗) sup (k + 1)|Ak| < ∞. (11.4) k>0 The original proof of Littlewood is very complicated. Karamata [39] found a simple approach to Littlewood’s theorem, which enabled him to improve and generalize some more tauberian theorems, mainly due to Hardy and Littlewood [39, 41, 40].

~~Karamata’s theorem P∞ k Condition (11.4) and boundedness of the function f(r) = (1−r) k=0 skr on (0, 1) imply boundedness of the sequence sn (Lemma 11.2.4).~~

11.2.1 Theorem (Karamata) If the sequence sn is bounded and the condition ∞ P n lim (1 − r) snr = S is satisﬁed, then r→1− n=0 ∞ Z 1 X k k lim (1 − r) sk ψ(r )r = S ψ(t) dt, (11.5) r→1− k=0 0 where ψ is an arbitrary Riemann(†) integrable function on [0, 1].

~~Proof of Littlewood’s theorem. In order to deduce Littlewood’s theorem from Karamata’s theorem, i.e., to prove that the conditions (11.4) and~~

∞ X k lim Akr = S (11.6) r→1− k=0 P imply convergence of the series Ak, we choose ψ so that rψ(r) = 1/(λ − 1) for −λ −1 e 6 r < e , ψ(r) = 0 otherwise, where λ > 1. Then from (11.5) we get, by taking r = e−1/n, [λn] 1 X lim sk = S. (11.7) n→∞ (λ − 1)n k=n+1 On the other hand,

~~[λn] [λn] 1 X [λn] − n 1 X s − s = − 1 s + (s − s ). (λ − 1)n k n (λ − 1)n n (λ − 1)n k n k=n+1 k=n+1~~

~~(∗)See Theorem 11.2.2. (†)“Riemann” cannot be replaced by “Lebesgue”. 174 11 Lacunary series~~

~~From this and condition (11.4) it follows that~~

~~ [λn] 1 X lim sup sk − sn 6 lim sup max |sk − sn| 6 C log λ. n→∞ (λ − 1)n n→∞ n 1 is arbitrary. 2 Proof of Karamata’s theorem. Observe that the expression~~

~~∞ ∞ X X (1 − r) ψ(rk) rk = ψ(rk)(rk − rk+1) k=0 k=0 is a Riemannian sum of the function ψ(r). The diameter of the underlying partition is equal to 1 − r and therefore~~

~~∞ X Z 1 lim (1 − r) ψ(rk) rk = ψ(t) dt. r→1− k=0 0 It follows that condition (11.6) can be written as~~

~~∞ X k lim (1 − r) (sk − S)r = 0. r→1− k=0 Replacing r by r1+α, α > 0, we ﬁnd that~~

~~∞ X αk k lim (1 − r) (sk − S)r r = 0, r→1− k=0 and this implies ∞ X k k lim (1 − r) (sk − S)P (r ) r = 0, r→1− k=0 where P is an arbitrary polynomial. Hence~~

∞ X k k lim sup (1 − r) (sk − S)ψ(r ) r − r→1 k=0 ∞ X k k k 6 lim sup (1 − r) (sk − S) ψ(r ) − P (r ) r − r→1 k=0 ∞ X k k k 6 lim sup (1 − r) ψ(r ) − P (r ) r − r→1 k=0 Z 1

~~= M ψ(t) − P (t) dt, 0~~

~~where M = sup |sn|. This concludes the proof because the polynomials are n>0 dense in L1(0, 1). 2 11.2 Karamata’s theorem and Littlewood’s theorem 175~~

~~On Littlewood’s theorem~~

The proof of Littlewood’s theorem shows that condition An = O(1/n) can be weakened. Here we state a special case of a very general result of Karamata [40]. P∞ 11.2.2 Theorem Let the function (11.3) have property (11.6). The series 0 An converges if there exists a function δ(λ), λ > 1, such that limλ→1+ δ(λ) = 0, and k X lim sup max Aj 6 δ(λ) for all λ > 1. (11.8) n→∞ n

~~2n 1/q 1 X C |A |q for some q > 1, (11.9) n k 6 n k=n+1 where C is independent of n.~~

~~It should be observed that condition (11.8) is necessary for convergence of an arbitrary series and is weaker than the condition~~

~~[λn] X lim sup |Aj| 6 δ(λ), n→∞ j=n+1 while the latter is weaker than (11.9); namely, if (11.9) holds, then by H¨older’s inequality,~~

[λn] [λn] 1/q X X q 1−1/q 1−1/q |Aj| 6 |Aj| ([λn] − n) 6 C(λ − 1) , j=n+1 j=n+1 so we can take δ(λ) = C(λ − 1)1−1/q. Theorem 11.2.2 is an immediate consequence of the above proof of Littlewood’s theorem and the following lemma.

~~11.2.4 Lemma If the function f is bounded and~~

~~j X sup max Ak < ∞, (11.10) n 0 n j 2n > 6 6 k=n then the sequence sn bounded. We leave the proof of this lemma to the interested reader. We only note that one can start from the inequality~~

~~2n+1 ∞ X j X j n |f(rn) − s2 | 6 Aj(1 − rn) + Ajrn , (11.11) j=0 j=2n 176 11 Lacunary series~~

~~−n P2n where rn = 1 − 2 . If (11.10) is strengthened to sup |Aj| < ∞, then n>0 j=n the proof becomes shorter; namely, then it easy to show that the right-hand side of (11.11) is bounded.~~

Miscellaneous 11.2.5 The “uniform” version of Dirichlet/Jordan test says: The Fourier series of a 2π-periodic continuous function f(θ) of bounded variation converges uniformly on R. Let Ak(θ) = ak cos kθ + bk sin kθ (θ ∈ R, k > 1), where ak, bk are the Fourier coeﬃcients of f(θ). From Theorem 3.1.6 it follows that

~~∞ a0 X k lim + Ak r − f = 0 r→1− 2 k=1 ∞~~

~~If in addition f is of bounded variation on [0, 2π], then kAkk∞ = O(1/k), so the conclusion of the test can be obtained by Littlewood’s theorem.~~

11.2.6 Equality (11.5) holds if the sequence sn is real and bounded from above. In his proof Karamata appeals to the following: If ψ is a Riemann integrable function, then for every ε > 0 there are polynomials P (r) and Q(r) such that R 1 P (r) < ψ(r) < Q(r) and 0 Q(r) − P (r) dr < ε. Pn 11.2.7 Let the sequence sn = k=0 Ak be bounded and let condition (11.6) be satisﬁed. If ϕ is an absolutely continuous function on the segment [0, 1] and ϕ(0) = 0, then ∞ X k lim Ak ϕ(r ) = ϕ(1) lim f(r). r→1− r→1− k=0

~~11.3 Lacunary series in C[0, 1]~~

~~We continue to denote by {Ak} a sequence of vectors in a Banach space X. We consider the series ∞ X λk L(r) = Ak r , (11.12) k=0~~

~~where λk is a lacunary sequence, i.e., a sequence satisfying λ inf k+1 = q > 1. (11.13) k>0 λk The following theorem is taken from Gurarij/Matsaev [24].~~

~~11.3.1 Theorem If there exists the limit S := limr→1− L(r), and is ﬁnite, then P∞ the series k=0 Ak converges. 11.3 Lacunary series in C[0, 1] 177~~

~~Proof. First we prove that the hypotheses imply~~

~~sup |Ak| 6 CM, (11.14) k>0 where M = sup0~~

~~2 n P (r) = p1r + p2r + ··· + pnr .~~

~~Then ∞ X λk n Ak P (r ) = p1L(r) + ··· + pnL(r ), k=0 whence ∞ X λk Ak P (r ) (|p1| + ··· + |pn|) M. 6 k=0~~

−q Choose P so that P (1/2) = 1, |P (r)| 6 δr for 0 < r < 2 , and |P (r)| 6 δ(1 − r) for 2−1/q < r < 1, where δ is small enough; for example, we can take P (r) = N {4r(1−r)} , N large enough. Next, if Ak → 0, choose ν so that |Aν | = sup |Ak|. k>0 Then

~~λν X λk |Aν | P (r ) 6 (|p1| + ··· + |pn|)M + |Ak| P (r ) k6=ν~~

~~X λk 6 (|p1| + ··· + |pn|)M + |Aν | P (r ) . k6=ν~~

~~Let r = 2−1/λν . We have~~

~~X λ X −λ /λν X −λ /λν P (r k ) 6 δ 1 − 2 k + δ 2 k k6=ν k<ν k>ν ν−1 ∞ X X −qk X k−ν X −qk 6 δ λk/λν + δ 2 6 δ q + 2 . k<ν k>ν k=0 k=1~~

Now choose δ so that the latter is less than 1/2. We get (1/2)|Aν | 6 (|p1| + ... |pn|) M, which is just the desired assertion in the case that Ak → 0. In the general case, we consider the function L(ρr) for 0 < ρ < 1; since the constant C = |p1| + ... |pn| is independent of {Ak}, we see, by the preceding, that 2k (1/2)|Ak|ρ 6 C supr |L(ρr)| 6 CM for all ρ and k, which completes the proof of inequality (11.14). Pn Further, let us prove that |sn| 6 C1M, where sn = k=0 Ak, and C1 is an 178 11 Lacunary series

whence, by condition (11.13), |L(rn) − sn| 6 const. M. Finally, let CX [0, 1] be the space of continuous functions from [0, 1] to X and let Y be the subspace of CX [0, 1] consisting of the functions g of the form g = L with the property that there exists the limit limr→1 g(r). By what we have proved, the formula Λn(g) = sn deﬁnes a bounded sequence of linear operators from Y to X. On the other hand, because of uniform continuity of functions g ∈ Y , we have limρ→1− kg − gρk∞ = 0. Since gρ ∈ Y and limn→∞ Λn(gρ) exists, we see that limn→∞ Λn(g) exists. 2

~~11.4 Lp-integrability of lacunary series on (0, 1)~~

~~With the above notation, we denote by F (x, y) (0 6 x 6 1, y > 0) a nonnegative real function that satisﬁes the condition~~

~~a c b d λ µ F (x, y) 6 F (λx, µy) 6 λ µ F (x, y) (0 < λ, µ < 1), (11.15)~~

where a, b, c, d are positive constants (a > b, c > d). We write F ∈ ∆(a, b; c, d) or F ∈ ∆ according to whether the values of a, b, c, d are or are not important. The simplest example is F (x, y) = xαyp (α > 0, p > 0). Further examples can be found by considering the functions F (x, y) = xαΦ(xβy), where Φ : [0, ∞) 7→ [0, ∞) is a function for which there are positive constants γ and δ such that Φ(t)/tγ decreases and Φ(t)/tδ increases.

~~11.4.1 Theorem If F ∈ ∆, and L is given by (11.12), then the following conditions are equivalent:~~

~~Z 1 (1 − r)−1F (1 − r, |L(r)|) dr < ∞; (11.16) 0 ∞ X 1 F , |A | < ∞. (11.17) λ n n=0 n~~

~~In the case where F (x, y) = xyp, this reduces to the following theorem of Gurarij and Matsaev [24]: 11.4 Lp-integrability of lacunary series on (0, 1) 179~~

~~11.4.2 Theorem For every p ∈ (0, ∞), the following conditions are equivalent:~~

~~Z 1 ∞ p X p |L(r)| dr < ∞; (1/λn)|An| < ∞. 0 n=0~~

~~We shall deduce Theorem 11.4.1 from a weaker result, namely:~~

~~11.4.3 Proposition There holds the inequality~~

~~Z 1 −1 (1 − r) F (1 − r, |L(r)|) dr > c0 sup F (1/λn, |An|), (11.18) 0 n>0 where c0 is a positive constant.~~

Proof. Let F ∈ ∆(a, b; c, d). The integral in (11.18) is > Z 1 (1 − r)−1F (1 − r, |L(r)|r1/d) dr/r, 0 and since the series L(r)r1/d is lacunary, it follows that in (11.18) we can write −1 r dr instead of dr. Next, assuming that λk+1/λk > q > 1 for every k, let [r(1 − r)]N P (r) = , [ρ(1 − ρ)]N where N is an integer and ρ satisﬁes the condition 2−q < ρ < 1/2 < 1 − ρ < 2−1/q. Choose ε so small that

~~2−q(1−ε) < ρ < 2−(1+ε), 2−(1−ε) < 1 − ρ < 2−(1+ε)/q.~~

~~For every δ > 0 we can choose N so that~~

~~−(1+ε) −(1−ε) P (r) > 1, for 2 < r < 2 , −q(1−ε) 0 < P (r) 6 δr, for 0 < r < 2 , −(1+ε)/q 0 < P (r) 6 δ(1 − r), for 2 < r < 1. It is easily checked that~~

Z 1 ∞ dr −1 X λk Int : = (1 − r) F 1 − r, AkP (r ) r 0 0 Z 1 −1 dr 6 C (1 − r) F 1 − r, |L(r)| , 0 r where C depends only of P and F . Therefore to prove (11.18) it is enough to ﬁnd a polynomial P so that Int c0 sup F (1/λn, |An|). Suppose that An → 0 > n>0 180 11 Lacunary series

~~and take ν so that F (1/λν , |Aν |) > F (1/λn, |An|) for all n. We shall prove the (formally) stronger inequality~~

~~Intν (F ) > c0 sup F (1/λn, |An|), (11.19) n>0 where Z ∞ dr −1 X λk Intν (F ) = (1 − r) F 1 − r, AkP (r ) , r Jν 0~~

~~−(1+ε)/λν −(1−ε)/λν Jν = 2 , 2 . There hold the implication~~

λk −q(1−ε) k > ν, r ∈ Jν =⇒ r 6 2 , (11.20) λk −(1+ε)/q k < ν, r ∈ Jν =⇒ r > 2 . Now we pass to the proof of (11.19). Assume ﬁrst that F ∈ ∆(a, b; 1, d). This means that the function y 7→ F (x, y)/y (y > 0) decreases, which implies that F (x, y1 + y2) 6 F (x, y1) + F (x, y2), whence Z dr −1 λν Intν (F ) > (1 − r) F 1 − r, P (r )|Aν | Jν r X Z dr − (1 − r)−1F 1 − r, P (rλk )|A | . k r k6=ν Jν By the relations (11.20) and the properties of P , for δ small enough, one can prove that the subtrahend is small with respect to F (1/λν , |Aν |), while the minuend is > KF (1/λν , |Aν |), where K is independent of δ, and this concludes the proof in the special case. (Details are omitted.) On the other hand, applying Minkowski’s inequality to the preceding inequality, we get, for p > 1,

1/p 1/p Z dr p −1 p λν Intν (F ) > (1 − r) F 1 − r, P (r )|Aν | Jν r 1/p X Z dr − (1 − r)−1F p1 − r, P (rλk )|A | , k r k6=ν Jν which is enough to complete the proof because an arbitrary function of class ∆ can be represented as F p, where F ∈ ∆(a, b; 1, d). 2 In order to deduce Theorem 11.4.1 from Proposition 11.4.3 we need two technical lemmas.

~~11.4.4 Lemma If α > 0, β > 0, then Z 1 α−1 β−1 α+β−1 (1 − ρ) (ρ − r) dρ 6 Cα,β(1 − r) (0 < r < 1). (11.21) r 11.4 Lp-integrability of lacunary series on (0, 1) 181~~

Proof. Let α < 1 and β < 1.(‡) We split the interval (0, 1) by the point √ √ α−1 β−1 √ α−1 β−1 ρ = r. If r < ρ < r, then (1 − ρ) (ρ − r) 6 (1 − r) (ρ − r) . If √ α−1 β−1 α−1 √ β−1 r < ρ < 1, then (1 − ρ) (ρ − r) 6 (1 − ρ) ( r − r) . The result follows by integration of these inequalities over the corresponding intervals. 2

11.4.5 Lemma If G is a positive measurable function on (0, 1) and γ > 0, then Z 1 Z 1 Z ρ γ−1 γ/2−1 γ/2−1 (1 − r) G(r) dr > c0 (1 − ρ) dρ (ρ − r) G(r) dr, 0 0 0 where c0 is a positive constant depending only on γ. Proof. We can apply Fubini’s theorem and appeal to inequality (11.21). 2 Proof of Theorem 11.4.1. Let F satisfy condition (11.15) and assume there holds (11.16). Put G(r) = (1 − r)−bF (1 − r, |L(r)|). According to Lemma 11.4.5 we have Z 1 Z 1 Z ρ −1 b/2−1 b/2−1 (1 − r) F (1 − r, |L(r)|) dr > c0 (1 − ρ) dρ (ρ − r) G(r) dr. 0 0 0 On the other hand, the inner integral equals Z 1 ρb/2 (1 − r)b/2−1G(ρr) dr 0 Z 1 = ρb/2 (1 − r)b/2−1(1 − ρr)−bF (1 − ρr, |L(ρr)|) dr, which is 0 Z 1 b/2 −b/2−1 > ρ (1 − r) F (1 − r, |L(ρr)|) dr. 0 (Here we used the fact that x−bF (x, y) increases with x, which follows from (11.15).) Further, the function x−b/2F (x, y) belongs to ∆(a − b/2, b/2; c, d) so we can apply Proposition 11.4.3; we get Z 1 −b/2−1 b/2 λn (1 − r) F (1 − r, |L(ρr)|) dr > c λn F (1/λn, |An|ρ ). 0 From these inequalities and the estimates ∞ −γ X γ λn (1 − ρ) > c0 λnρ , where γ = 1 − b/2 > 0, n=0 we get Z 1 Z 1 ∞ −1 X 1−b/2 λn+b/2 b/2 λn (1−r) F (1−r, |L(r)|) dr > c0 dρ λn ρ λn F (1/λn, |An|ρ ), 0 0 n=0 whence, by condition (11.15), we see that (11.17) holds. The proof of the implication (11.17) =⇒ (11.16) is much simpler and we omit it. The reader could see the papers [60] and [58]. 2

~~(‡)Only this case will be used in the proof of the theorem. Bibliography~~

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(An), 163 Formula Ap, Bergman space, 16 Cauchy integral, 78 p Aβ , weighted Bergman space, 136 Green, 39, 67 absolute continuity of f∗, 78 Jensen, 68, 81 approximation by inner Parseval, 100 functions, 91 Riesz representation, 69 approximation by smooth Function functions, 58 almost increasing, 155 atomic decomposition, 133, 135, 136 atomic, 85 conjugate, 93, 97 BV [a, b], 44 harmonic conjugate, 92 B, Bloch space, 128 inner, 85, 89 BMO, 127 inner, singular, 85 Banach envelope, 10 logarithmically convex, 61 Blaschke condition, 80 main maximal, 25 Blaschke product, 80 nontangential maximal, 112 Bourgain’s lemma, 120 of bonded mean oscillation, 127 outer, 85, 86 2 C0 (D), 67 polyharmonic, 153 c0, 16 Rademacher, 28 complemented subspaces of `p, 17 radial maximal, 110 conformal mappings, 79 semicontinuous, 55 convexity of mean values, 60, 61 subharmonic, 55 superharmonic, 56 Dr(z), 149 Weierstrass, 160 Du = ∂u/∂θ, 154 ∆(a, b; c, d), 178 H, Hilbert operator, 93 n ∆t , symmetric diﬀerence, 160 H(D,X), all X-valued analytic functions, dµp, 130 137 dA = dm/π, 82 H(Ω), all analytic functions, 39 decomposition of Hp functions, 75 HC1, 149 density of polynomials in Hp, 76 HC2(G), 151 density of rational functions in Lp, 27 H∞(X), 138 Dini condition, 156 Hp, Hardy space, 73 p Dirichlet problem, 41 H (T), 100 Dirichlet/Jordan test, 176 Hp(T), 100 disk-algebra, 155 h(Ω), all harmonic functions, 39 p dm, Lebesgue measure in C, 39 hA , harmonic Bergman space, 143 hC(D) = C(D) ∩ h(D), 42 1 Eε(z), 70 h , 43 E, 131 hp, harmonic Hardy space, 49 h∞(ψ), 161 fe, conjugate function, 92, 93 h∞,n(ψ), 161 f∗, boundary function, 50 Hadamard product, 138, 141 F -norm, 14 Hardy/Stein identity, 98

187 harmonic Schwarz lemma, 53, 66 M, maximal function, 25 Heine/Borel property, 39 majorant, 154 maximum principle, 57 I(r, u), mean values, 60 Smirnov, 76 I(u), 70 mean value property, 39 Ip(r, f), 129 moduli of smoothness, 160 Inequality multiplier, 15 Chebyshev, 22 multipliers, 35 Hardy, 79, 84

Hardy/Littlewood, 124 Nq(f), 10 Harnack, 44 N (D), Nevanlinna class, 15 + Heinz, 105 N (D), Smirnov class, 15 Hilbert, 83, 84 N0(f), quasinorm in L0, 30 isoperimetric, 82, 83 Nevanlinna class, 15 Khintchine, 29 Littlewood/Paley, 51, 70 OC1, 149 Paley, 30, 172 OC2(G), 151 Riesz/Fejer, 83 Of(z, r), 149 Riesz/Zygmund, 95 Opf(z, r), 150 integral means, 49 ωn(g, t), modulus of smoothness, 160 of univalent functions, 62 Operator interpolation for Hardy spaces, 119 bilinear, 13 p p isometry L to h , 50 from `p to `q, 17 p p isomorphism A to ` , 16 Hilbert, 93, 97, 156 invertible, 10, 12 J (z), 104 f of strong type, 23 J (r, f), 62 p of weak type, 23 quasilinear, 22 Kp(w, z), 130 subadditive, 22 Lp,∞, weak Lebesgue space, 22 sublinear, 31 Λ∗, Zygmund class, 160 P (z) = P (r, θ), Poisson kernel, 40 Λα(K), Lipschitz space, 154 Lip(ω, K), Lipschitz space, 154 PS, Poisson/Stieltjes integral, 44 P [·], Poisson integral, 41 Lipn(φ), Lipschitz space, 160 λ∗, little Zygmund class, 160 Pnf, 101 L(r), 176 Pe, conjugate Poisson kernel, 92 L0, all measurable functions, 30 p-norm, 7 lacunary series, 125, 172, 176 Poisson integral, 41 Lp-integrability, 179 of f∗, 50, 78 Lebesgue point, 26, 27 of a Borel measure, 42 Lebesgue set, 26, 27 of log|f∗|, 76 Littlewood’s conjecture, 125 of the conjugate function, 94 Poisson kernel, 40, 51 M(T), Borel measures, 42 conjugate, 92 M∗u, nontangential maximal Poisson/Stieltjes integral, 44 function, 112 Principle Mp(r, f), integral mean, 49 Banach, 36 Mradu, radial maximal function, 110 Littlewood subordination, 64 µ(f, λ), distribution function, 22 maximum, 57, 76

188 maximum modulus, 40 on coefficients in BMO, 128 uniform boundedness, 13 on radial and nontangential limits, 48 q-Banach envelope, 10 Abel, 172 of the space Hp, 136 Ahern, 65 quasinorm, 7 Aleksandrov, 100 Aoki/Rolewicz, 8 R+, Riesz projector, 96 Banach/Steinhauss, 13 rj (t), Rademacher functions, 28 Beurling approximation, 87 Radial limits Bieberbach, 63 of hp functions, 50 Burkholder, Gundy of Hp-functions, 76 and Silverstein, 114 of conjugate function, 93 Carathéodory convergence, 108 of the Poisson integral, 45, 46 Choquet, 96 reproductive kernels, 130 closed graph, 14 Riesz measure, 67 Coifman/Rochberg, 133 Riesz projector, 96 complex maximal, 111 Dyakonov, 158 SH, 149 Fatou on nontangential limits, 47 α σmaxf, 116 Fatou on radial limits, 46 α σn f, 116 Fefferman, 127 Shauder basis, 13, 99 Fefferman/Stein, 112 Smirnov class, 15 Gurarij/Matsaev, 176, 179 Space Hadamard’s three circles, 61 M(T), 42 Hardy, 61 L0, 30 Hardy/Littlewood, 122, 170 p hC(D), 42 on h , p < 1, 147, 148 h1, 43 on Cesaro means, 116 h∞,n(ψ), 161 on coefficients, 24, 122 ho,n(ψ), 161 on conjugate functions, 148 F -space, 14 on conjugates, 143 h∞,n(ψ), 161 Harnack, 44 p-Banach, 9 Hausdorff/Young, 21 Bergman, 16, 129 Hurwitz, 44 Bergman harmonic, 143 inner-outer factorization, 86 Bergman, weighted, 136 Kalton, 101 BMO, 127 Karamata, 173, 176 Hardy, 73 Koebe, 63 Hardy harmonic, 49 Kolmogorov, 113 Lipschitz, 154 Kolmogorov/Smirnov, 65 Lipschitz, of higher order, 160 Konyagin, 103 of type p, 31 Lebesgue points, 26 quasi-Banach, 9 Lindelöf, 48 weak Lebesgue, 22 Liouville, 61 Zygmund, 160 Littlewood tauberian, 173, 175 subordination, 64 Littlewood/Paley, 51, 70 n symmetric difference, ∆t , 160 local integrability, 58 Marcinkiewicz, 23, 24, 119 Theorem maximal, 26, 116, 119

189 McGehee, Pigno and Smith, 125 Mori, 105 Nikishin, 31 Nikishin/Stein, 34 on a.e. convergence, 37 on coefficients in F ∈ H∞(X), 140 on coefficients in H1, 124 on quasiconformal harmonic homeomorphisms, 105 on subharmonic behavior, 143 on the dual of Hp, 99 open mapping, 12 Paley on Fourier coefficients, 25 on lacunary series, 170 Prawitz, 62 Privalov on conjugates, 155, 157 Privalov/Plessner, 93, 95 radial maximal, 111 Riesz factorization, 81 Riesz on conjugate functions, 97 Riesz projection, 96 Riesz representation, 86 Riesz the brothers, 78 Riesz/Herglotz, 43, 44 Riesz/Thorin, 18, 20 Rogosinski, 66 Sidon, 172 Stephenson on composition, 90 subharmonic maximal, 111 Tauber, 173 uniqueness, 38, 77 Weierstrass approximation, 41

~~(U), 161 (Uα), 161 0 (Uβ ), 161~~

~~Wn (polynomials), 117 Wmax, 117 weak type, 23~~

~~[X]q, 11~~

~~190~~