Theory of Hp Spaces

Then U E 11', and there is a constant AP depending only on p such that II Ullp

:0::::

Apllcpllp·

PROOF. The proof depends upon a theorem on "maximal rearrangements " of functions, which is discussed in Appendix B. Extend cp periodically with period 2n, and let

.41(0) = ..H(O; cp) =

sup O
I.!_ ( T

0

cp(O

+ t) dt

I·

12

1 HARMONIC AND SUBHARMONIC FUNCTIONS

By Theorem B.3, .A e I! and l
II.AIIp::;; Cpii«PIIP'

U(O) is a measurable function, since it is the upper envelope of the functions lu(r, 0)1 as r runs through the sequence of rational numbers in the interval (0, 1). Hence it will suffice to show that O::;;r< 1.

lu(r, 0)1 ::;; 2.A(O),

(7)

This can be seen by writing the Poisson formula in the form 1

J"

u(r, 0) = P(r, t)cp(O + t) dt 2n -" and integrating by parts. For fixed 0, let

Jcp(O + u) du. t

(t) =

0

Then

1 u(r, 0) = - [P(r, 2n

t)(t)]~"

J"

1 a (t) - t-;-- {P(r, t)} dt. 2n -" ut t

From this (7) follows, since l(t)/tl::;; .A(O) and

:_ r It ot~

I

P(r, t) dt =

2n -"

~ < 1. 1+ r

It should be observed that (7) remains true in the case p = 1, although the theorem is then false. (See Exercise 6.) However, the corresponding theorem for analytic functions is true for all pin the range 0 < p ::;; oo. THEOREM 1.9 (Hardy-Littlewood).

LetfeHP, 0

F(O) = suplf(rei 9)1. r< 1

Then F e I! and

II FliP::;; Bpllfllp' where BP depends only on p and 11/IIP =lim Mp(r,f). r -1

PROOF. For each fixed R, 0

< R < 1, the function g(z) = 1/(Rz)IP/ 2 is

subharmonic in lzl < 1, so it follows from Theorem 1.8 that

G(O)

= suplg(rei8)1 e L2 r< 1

NOTES

13

and

In other words, where FR(8) = suplf(re;6)\. r
Since FR(8) increases to F(8) as R ~ 1, the proof is completed by appeal to the Lebesgue monotone convergence theorem. EXERCISES

1. Show that (1 -z)- 1 is in HP for every p < 1, but is not in H 1 .

2. Show that if f(z) is analytic in lzl < 1 and its range is contained in a sector of angle ex (0 < ex ::o:;;2n), then/ E HP for every p < n/ex. 3. Show that the function 1 - {1 -l o g -1-}-c 1-zz 1-z belongs to H 1 if c > 1. 4. Show directly (i.e., by refining the proof of Theorem 1.2) that a PoissonStieltjes integral has a nontangential limit almost everywhere. 5. Use Littlewood's subordination theorem to show that MP(r,f) is a nondecreasing function of r. 6. Use the modified Poisson kernel u(r, 8)

=

R2- r2 2

R - 2Rr cos 8

+ r2

,

R > 1,

to show that Theorem 1.8 is false for p = 1. 7. Show that if f(z) and g(z) are subharmonic, then so is max{f(z), g(z)}. NOTES

In the language of functional analysis, the Belly selection theorem simply asserts the weak-star sequential compactness of the unit sphere of the dual space of the space C of continuous functions over the interval [a, b]. That is, ifcpnEC* and supllcfJnll
14

1 HARMONIC AND SUBHARMONIC FUNCTIONS

The paper of Fatou [1] is one of the earliest studies of the boundary behavior of analytic and harmonic functions in the disk. Fatou showed, among other things, that a Poisson-Stieltjes integral has a nontangential limit almost everywhere. The existence of such a limit for a bounded analytic function is known as " Fatou's theorem. " Theorem 1.1 goes back at least to Plessner [1 ], although Herglotz [1] had earlier found the representation for positive harmonic functions. Hardy's convexity theorem is in his paper [2], which is now regarded as the historical starting point of the theory of HP spaces. (Thus the "H" in "HP" is for Hardy.) The elegant proof of Hardy's theorem via subharmonic functions is due to F. Riesz [2]. A. E. Taylor [1] found another proof using Banach space methods. Littlewood's subordination theorem is in his paper [1]; again F. Riesz [5] vastly simplified the proof using subharmonic functions. For an extension, see Gabriel [1 ,2]. See also Ryff [1]. For a thorough discussion of subharmonic functions, see Rado [1]. The Hardy-Littlewood maximal theorems are in their paper [4]. They also proved more general theorems in which the supremum is taken not over a radial segment, but over a "kite-shaped region" S~~.(8) with vertex at a boundary point.

BASIC STRUCTURE OF HP FUNCTIONS

CHAPTER 2

In this chapter we consider, among other things, the boundary behavior and the zeros of HP functions. As may be expected, the growth restriction on an HP function imposes a restriction on the density of its zeros. This leads to the formation of" Blaschke products" and the canonical factorization theorems, which play a vital role in later work. Another important result is the mean convergence of an HP function to its boundary function. Two new classes of functions, N and N+, arise in the course of the discussion, and their properties are explored. Finally, an equivalent definition of HP is given in terms of harmonic majorants. 2.1. BOUNDARY VALUES

We have seen that a harmonic function of class h 1 must have a radial limit in almost every direction. The same is therefore true for analytic functions of class H 1• The theorem remains valid for HP functions with p < 1, despite the fact (to be noted in Chapter 4) that the corresponding statement for harmonic functions is false. Surprisingly, the generalization to p < 1 is obtained most simply by dealing with an even wider class of functions. This is the Nevanlinna

16

2 BASIC STRUCTURE OF HP FUNCTIONS

class N, also known as the class of functions of bounded characteristic. A function J(z) analytic in lzl < 1 is said to be of class N if the integrals 27t

Jo

log+ lf(rew)l de

are bounded for r < 1. Since log+ 1/1 is subharmonic whenever f is analytic, these integrals always increase with r. It is clear that N contains HP for every p >0. The following theorem is basic to the further development of the theory. THEOREM 2.1 (F. and R. Nevanlinna). A function analytic in the unit disk belongs to the class N if and only if it is the quotient of two bounded analytic functions.

PROOF. Suppose first that f(z) = cp(z)/IIJ(z), where cp and 1/J are analytic and bounded in lzl < 1. There is no loss of generality in assuming Icp(z)l ::;; 1, 1!/J(z)l::;; 1, and 1/J(O) ¥- 0. Then

But by Jensen's formula (see Ahlfors [2], p. 206),

-2n1 J

2"

logi!/J(re;6 )1 de = logi!/J(O)I

+ L

lznl
0

r log -1- 1, Zn

where zn are the zeros of 1/J. This shows that Jlogll/11 increases with r, so.f EN. Conversely, let .f(z) =i- 0 be of class N. Let .f have a zero of multiplicity m ;;::: 0 at the origin, so that z-mf(z) ~ex¥- 0 as z ~ 0. Let zn be the other zeros of J, repeated according to multiplicity and arranged so that 0 < lz1 1 ::;; Iz 2 1 ::;; · · · < 1. If f(z) ¥- 0 on the circle lzl = p < 1, the function

pm F(z) = log{J(z) lri z

TI (pp( z_ _ zz)} n) lznl

is analytic in lzl ::;; p, and Re{F(z)} = loglf(z)l on lzl = p. Hence, by analytic completion of the Poisson formula,

J2"

.

1 peit + z F(z) = loglf(pe'1)1 - ;1 dt 2n o pe - z

+ iC.

This is sometimes called the Poisson-Jensen formula. After exponentiation, it takes the formf(z) = ((Jp(z)/111 p(z), where

2.1 BOUNDARY VALUES

zm p(z- z) { 1 cp,lz) =--;;; ll 2 n • exp - 2n p l•nl

= exp { -

f

J

2"

•

log-lf(pe'')l

0

peit + z it dt pe - z

17

+ iC },

piit + z dt }. " • -21 log+ IJ(pe'')l no pe -z 2

1

Now choose a sequence {pk} increasing to I, such thatf(z) ¥- 0 on the circles lzl = Pk· Let

Thenf(Pk z) = k(z)/'1\(z) for lzl < I. But the functions k and '1\ are analytic in the unit disk, and lk(z)l::::;; I, l'l'k(z)l::::;; 1 there. Hence {k} and {'l'k} are normal families, and there is a sequence {k;} such that k,(z) ~ cp(z) and 'l'k.(z)--+ 1/f(z) uniformly in each disk lzl ::::;; R < 1. The functions cp and 1/1 are analytic in the unit disk, and lcp(z)l::::;; 1, 11/f(z)l::::;; I. According to the definition of 1/1P, the fact that Jlog+ 1/1 is bounded gives a uniform estimate l'l'k(O)I ~ ~ > 0, so 1/J(z) ¢. 0. Thus .f = cp/1/1, and the proof is complete. The importance of this representation theorem is that it allows properties of functions inN to be deduced from the corresponding properties of bounded analytic functions. The boundary behavior, for example, can now be discussed.

fEN, the nontangential limit f(ei 9 ) exists almost everywhere, and logl.f(eil)l is integrable unless f(z) = 0. If jEW for some p > 0, then.f(ei 9) E I!. THEOREM 2.2. For each function

PROOF. Assuming f(z) ¢. 0, represent it in the fmm cp(z)fl/l(z), where lcp(z)l ::::;; 1 and 11/!(z)l ::::;; I. Since cp and 1/1 are bounded analytic functions, they have nontangentiallimits cp(ei9 ) and l/f(ei 9 ) almost everywhere, by Theorem 1.3. Appealing to Fatou's lemma, we find

But Jloglcp(re 19)1 d() increases with r, by Jensen's theorem. Hence loglcp(ei 9 )1

E

Ll, and similarly for 1/f. In particular, ljl(ei 9) cannot vanish on a set of positive measure. The radial limit f(ei 9) therefore exists almost everywhere, and loglf(eil)l E I!. Another application of Fatou's lemma shows that f(ei 9) E I! ifjEHP. The theorem says that iff EN and ifj(ei 9 ) = 0 on a set of positive measure, thenf(z) = 0. In other words, a function of class N is uniquely determined by its boundary values on any set of positive measure.

18

2 BASIC STRUCTURE OF HP FUNCTIONS

f = cp/111 that

It is evident from the representation

Jlog-lf(rei )1 de

is bounded if/EN. (Here log- x =max{ -log x, 0}.) Hence/EN if and only if Jlloglf(rei6)11 de is bounded. 6

2.2. ZEROS

The zeros of an analytic function cannot cluster inside the domain of analyticity unless the function vanishes identically. More is true if the function satisfies a growth restriction: the zeros must tend "rapidly" to the boundary. The following theorem illustrates this principle. THEOREM 2.3. Let f(z) =i- 0 be analytic in lzl < 1, and let z 1 , z 2 , ••• be its zeros, repeated according to multiplicity, in lzl < 1. Then g" loglf(rei 6)1 de is bounded if and only if 1 (1 - lznl) < oo.

L:'=

PROOF.

rxzm

Letfhave a zero of multiplicity m ~ 0 at the origin, so thatj(z) = Call the other zeros an, and let them be ordered so that la 2 1 :::;:; · • • . By Jensen's formula,

+ · · · , ex ¥- 0.

0 < la 1 1 :::;:;

-1 2n

J loglf(rem)l de= 2

"

0

L

\an\
log 1- r I + log(lcxlr'n). an

This increases monotonically with r; suppose it is bounded from above by some constant C. Then, fixing an integer N, we have for all r > laNI

Letting r tend to 1, we find

nlanl. N

0 < lcxle-c :::;:;

n=J

so that L (1 - lanl) < oo. To prove the converse, we have only to note that Jensen's theorem implies

COROLLARY.

The zeros {zn} of a function of class N must satisfy

One might well expect a stronger growth condition such asfE HP to imply a correspondingly stronger statement about the rate at which the zeros

2.2 ZEROS

19

approach the boundary. Even if fE H 00 , however, nothing more can be said about its zeros. In fact, given an arbitrary sequence {a,} of numbers in the unit disk such that L (1 - la,i) < oo, one can produce a bounded analytic function whose zeros are precisely {a,}. It is natural to construct such a function as an infinite product. The factors will be essentially the functions (z- a,Jf(l - a,z), which map the unit disk onto itself and vanish at a,. THEOREM 2.4. Let a 1 , a 2 ,

0<

l"tl :<;;;

la 2 1:<;;; ••• < 1 and

•••

be a sequence of complex numbers such that 1 (1 -Ia,!) < oo. Then the infinite product

L:-'=

converges uniformly in each disk lzl :<;;; R < 1. Each a, is a zero of B(z), with multiplicity equal to the number of times it occurs in the sequence; and B(z) has no other zeros in lzl < 1. Finally, IB(z)l < 1 in lzl < 1, and IB(e;9 )1 = 1 a. e. PROOF. In the disk

lzl

:<;;;

R,

a, -_z I =I (a,+ la,lz)~- Ia,!) I:-;;;; 2(1 - Ia,!). ll _ ~ a, 1-a,z a,(l-a,z) 1-R

Since L (I - la,l) < oo, it follows that the infinite product converges uniformly in each disk lzl :<;;; R
Now apply this to the function/= B/B,, where

20

2 BASIC STRUCTURE OF HP FUNCTIONS

But Bn(z) -.B(z) uniformly on lzl =r, so 2n

2tr ~I IB(ei')l dt. 0

Because IB(e; )1 ~ 1 a.e., this shows IB(e; 9)1 = 1 a.e. 9

A function of the form B(z) = zm

TI _lanl _a_n-_z

(1)

n an 1 -On Z

is called a Blaschke product. Herem is a nonnegative integer and L (1 - Iani) < oo. The set {an} may be finite, or even empty. If {an} is empty, it is understood thatB(z) =zm. 2.3. MEAN CONVERGENCE TO BOUNDARY VALUES

We have seen that every HP function f(rei 9 ) converges almost everywhere to an I! boundary functionf(e; 9). For the further development of the theory, it is important to know that f(rei 9) always tends to f(e;~ in the sense of the I! mean. The following factorization theorem will enter into the proof and will become a standard tool in later chapters. THEOREM 2.5 (F. Riesz). Every functionf(z) ¢. 0 of class HP (p > 0) can be factored in the formf(z) =B(z)g(z), where B(z) is a Blaschke product and g(z) is an HP function which does not vanish in lzl < 1. Similarly, eachfe N has a factorization f = Bg, where g is a nonvanishing function of class N. PROOF. We may suppose thatf(z) has infinitely many zeros, since otherwise the theorem is trivial. Let

Bn(z) = zm

fi lakl 1-

ak -

k= 1 ak

z akz

denote the partial Blaschke product, and let gn(z) = f(z)f Bn(z). For fixed n and e > 0, IBn(z)l > 1 - e for lzl sufficiently close to 1. Therefore, 2n

I

0

ign(ri9 )1P d() ~ (1-

e)-pI

2n

0

lf(rei9 )1P d() ~ (1-

erP M

for r sufficiently large; hence, by monotonicity, for all r. Letting e---. 0, we find

2.3 MEAN CONVERGENCE TO BOUNDARY VALUES

21

for all r < I and all n. By Theorem 2.4, however, gn(z) tends to g(z) = f(z)/B(z) uniformly on each circle lzl = R < I. Thus g is in HP, and it has no zeros. (The proof for feN is similar.) We are now ready for the mean convergence theorem. THEOREM 2.6. If fe HP (0

lim r-+1

J

2"

!f(rei8 )!P d(} =

0

J

IJ(e 18 )!P d(}

(2)

0

and 21<

lim r-+1

J !f(re'

8) -

f(e' 8 )!P d(} = 0.

PROOF. First let us prove (3) for p = 2. If f(z) =

L la,l

2

(3)

0

L an~

is in H 2 , then

< oo. But by Fatou's lemma, 2n

J lf(re'

8 )-

f(e; 8Wd(} ~lim inf

0

p-+ 1

2n

J

IJ(re'8 ) - f(pe' 8Wd(}

0

00

= 2n

L 1anl 2(1 -

rn) 2 ,

n=1

which tends to zero as r -+ I. This proves (3), and hence (2), in the case p = 2. If/e HP(O

J

IJ{re;e)IP d(} ~

0

21<

J 0

21<

!g(rere)!P d(}-+

J

21<

!g(ere)!P d(} =

0

J

IJ{e;e)IP d(}.

0

This together with Fatou's lemma proves (2). The following lemma can now be applied to deduce (3) from (2).

n be a measurable subset of the real line, and let E'(!l), 0 < p < oo; n = I, 2, .... As n-+ oo, suppose qJn(x)-+ qJ(x) a. e. onO.and LEMMA 1. Let

(/In E

Jn !({Jn(x)IP dx-+ Jn !q>(x)IP dx < oo. Then

22

2 BASIC STRUCTURE OF HP FUNCTIONS

PROOF.

h lrpiP.

Let

For a measurable set E c Then

E = n- E.

n,

let Jn(E) = h IIPniP and J(E) =

J(E) ::;; lim inf Jn(E)::;; lim sup Jn(E) n __,co

n->oo

This shows that Jn(E) -+ J(E) for each E c n. Given e > 0, choose a set F c n of finite measure such that J(F') < e. Choose {J > 0 so that J(Q) < e for every set Q c F of measure m(Q) < {J, By Egorov's theorem, there exists a set Q c F with m(Q) < {J, such that IPn(x)-+ rp(x) uniformly on E = F- Q. Thus

< (2P · 6 + l)e for n sufficiently large, since Jn(f')-+ J(F') and Jn(Q)-+ J(Q). This proves the lemma, and Theorem 2.6 follows. COROLLARY. Iff E H' for some p > 0, then 2n

lim ,__, 1

J Ilog+ lf(re; )1 -log+ lf(ew)li de= 0. 6

0

The corollary is an immediate consequence of Theorem 2.6 and the following lemma. LEMMA 2. Fora~O,b;;=::O,andforO
llog+ a- log+ bl ::;; (lfp) Ia- hiP. PROOF. It

is enough to assume 1 ::;; b < a. The result then follows from the

inequality log x::;; (1/p) (x- l)P,

X;;::: 1,

by setting x = afb. The latter inequality can be proved by noting that the difference of the two expressions has a positive derivative and vanishes at X= 1. It is tempting to think that the corollary holds generally for all f E N. This isfa/se; the functionf(z) = exp{(l + z)/(1- z)} is a counterexample.

2.4 CANONICAL FACTORIZATION

23

There is also a " short" proof of the mean convergence theorem (Theorem 2.6) which does not use the Riesz factorization theorem but appeals instead to the Hardy-Littlewood maximal theorem (Theorem 1.9). If fe HP, then f(re 1~-+f(i 6) almost everywhere, andf(e 19) e JJ'. But by the maximal theorem, IJ(re 1~1 ::;;; F(O), where Fe JJ'. Hence Theorem 2.6 follows at once from the Lebesgue dominated convergence theorem! It should be observed at this point that HP is a normed linear space if I ~ p ~ oo. The norm is defined as the lJ' norm of the boundary function. Thus if I ::;;; p < oo, IIJIIP

1

= {-2

J lf(e )IP d() n 2n

} lfp

16

0

=lim Mp(r,J); r->1

while 11/llco = sup IJ(z)l = ess sup lf(e19)1. 1~1<1

Os9<2n

2.4. CANONICAL FACTORIZATION

The Riesz factorization (Theorem 2.5) can be refined to produce a canonical factorization which is of supreme importance both for the theory of HP spaces and for its applications. This refinement rests upon the following inequality. THEOREM 2.7. lffe HP, p >0, then

1

2n

J

loglf(re19)1 ::<;;;P(r, ()- t) loglf(elt)l dt. 2n o PROOF. After factoring out the Blaschke product, whose presence would only strengthen the inequality, we may assume f(z) =F 0 in lzl < 1. Then loglf(z)l is harmonic in lzl < 1, and

J2"

loglf(pre19)1 = 1P(r, () - t) loglf(pe1')1 dt, 2n o

r < p < 1.

By the corollary to Theorem 2.6, 211:

lim p-+1

J

27t

P(r, ()- t) log+IJ(pt')l dt

=

0

J P(r, ()- t) log+IJ(e ')1 dt. 1

0

On the other hand, by Fatou's lemma,

lim p-+1

J27< P(r, () -

t) log-IJ(pt')l dt ~

0

Subtraction gives the desired result.

J2"P(r, () 0

t) log-lf(~')l dt.

24

2 BASIC STRUCTURE OF HP FUNCTIONS

The theorem is false for the class N, as the example exp{(l + z)/(1 - z)} again shows. The reciprocal of this function reveals, incidentally, that strict inequality may occur even iff(z) is bounded and has no zeros. Returning to the problem of factorization, Jetf(z) ¢. 0 be of class HP for some p > 0. According to Theorem 2.2, f(e 16) E IJ' and loglf(e 16)1 E I!. Consider the analytic function 1

J -e + z 1'

2x

.

}

F(z) = exp {-2 1,-loglf(e'')l dt . n o e - z

(4)

Let.f(z) = B(z)g(z) as in Theorem 2.5; thus g(z) =F 0 and lg(ei6)1 = lf(ei6)1 a.e. By Theorem 2.7, lg(z)l ::;;IF(z)l in lzl < 1. Also, lg(e 16)1 = IF(ei 6)1 a.e., by Corollary 2 to Theorem 1.2. Hence if e1Y = g(O)flg(O)I, the function S(z) = e- 1Yg(z)/F(z) is analytic in lzl < 1 and has the properties 0 < IS(z) I ::;; 1;

IS(ei6 )1 = I a.e.;

S(O) > 0.

This shows that -logiS(z)l is a positive harmonic function which vanishes almost everywhere on the boundary. Thus by the Herglotz representation and Theorem 1.2, -logiS(z)l can be represented as a Poisson-Stieltjes integral with respect to a bounded nondecreasing function Jl(t), and ll'(t) = 0 a.e. Since S(O) > 0, analytic completion gives

{ J -ee'',-+, -zz dJl(t)}.

S(z) = exp -

2x

o

Putting everything together, we have the factorizationf(z)

(5) = e'YB(z)S(z)F(z).

We now introduce some terminology. An outer function for the class HP is a function of the form } 1 2" ei' + z F(z) = e1Y exp {-ilog t{!(t) dt , (6) 1 2n o e - z

J

where ')'is a real number, t{!(t) ;;:.: 0, log t{!(t) E I!, and t{!(t) E IJ'. Thus (4) is an outer function. An inner function is anyfunctionf(z) analytic in lzl < 1, having the properties lf(z)l ::;; l and lf(e 16)1 = l a.e. We have shown that every inner function has a factorization eiy B(z)S(z), where B(z) is a Blaschke product and S(z) is a function of the the form (5), Jl(t) being a boundea nondecreasing singular function (!l'(t) = 0 a.e.). Such a function S(z) is called a singular inner function. THEOREM 2.8 (Canonical factorization theorem). Every functionf(z) ¢. 0 of class HP (p > 0) has a unique factorization of the form.f(z) = B(z)S(z)F(z), where B(z) is a Blaschke product, S(z) is a singular inner function, and F(z)

2.5 THE ClASS N+

25

is an outer function for the class HP (with 1/J(t) = !.f(ei 1)!). Conversely, every such product B(z)S(z)F(z) belongs to HP. PROOF. We have already shown that everyf E HP can be factored as claimed, and the uniqueness is obvious. To prove the converse, it suffices to show that an outer function (6) must belong to HP. Applying the arithmetic-geometric mean inequality (see Exercise 2), we find

jF(z)jP

1 :$;-

J2"P(r, 8 -

2n o

t)[l/J(t)JP dt.

Thus 27t

J

27t

!F(rei6 )!P de ::-::;

0

J

[1/J(t)]P dt.

0

There is a similar factorization for/EN. A function F(z) of the form (6), where 1/J(t) ;;::: 0 and log 1/J(t) E L1 , will be called an outer function for the class N. (Note that the condition 1/1 E I! has been dropped.) THEOREM 2.9.

Every functionf(z) oj. 0 of class N can be expressed in the

form

f(z) = B(z)[S1 (z)/S2 (z)]F(z),

(7)

where B(z) is a Blaschke product, S 1 (z) and S 2 (z) are singular inner functions, and F(z) is an outer function for the class N (with 1/J(t) = !f(ei1)1). Conversely, every function of the form (7) belongs to N. PROOF. Letf(z) =eiYB(z)g(z), whereg E N,g(z) of. Oin lzl < 1, andg(O) > 0. Since loglg(z)l E h 1 , it has a representation as a Poisson-Stieltjes integral

J2"

1 loglg(z)l = P(r, 8- t) dv(t) 2n v

with respect to a function v(t) of bounded variation. Analytic completion and separation of v(t) into its absolutely continuous and singular components gives the desired representation. The converse follows directly from the fact that every Poisson-Stieltjes integral is of class h1 • 2.5. THE CLASS N+

The two preceding theorems point out the sharp structural difference between functions in the classes HP and N. In factoring functions of class N, it is recessary not only to enlarge the class of admissible outer functions, but also to replace the singular factor by a quotient of two singular inner functions.

26

2 BASIC STRUCTURE OF H" FUNCTIONS

This allows, for instance, our "pathological" example exp{(l + z)/(1 - z)}, which we now recognize as the reciprocal of a singular inner function. It is useful to distinguish the class N+ of all functions fEN for which S 2 (z) l. That is,.f EN+ if it has the form/= BSF, where B is a Blaschke product, Sis a singular inner function, and F is an outer function for the class N. In a sense, N+ is the natural limit of HP asp-.. 0. The proper inclusions HP c N+ c N are obvious.

=

THEOREM 2.10. A functionfE Nbelongs to the class N+ if and only if 2x

lim

J

2x

log+lf(rei 6)1

d()

=

0

r~1

J

log+lf(ei 6)1

d(),

(8)

0

PROOF. Suppose first that/EN+, so thatf=BSF. Then, in view of (4),

log+lf(rei 6)1

~ log+IF(rei 6)1 ~ 2~ S:xP(r, ()- t) log+lf(e;')l dt.

Hence 2x

lim r~l

J

log+l/(re16)1 d() ~

0

2x

J

log+lf(e;')l dt.

0

Fatou's lemma gives the reverse inequality. It is more difficult to prove the sufficiency of the condition (8). We first observe that for an arbitrary Blaschke product B(z), 2x

lim r--+

1

J

logiB(re; 6)1 d() = 0.

(9)

0

This is obvious if B(z) has only a finite number of factors, so we may assume

By Jensen's theorem, l 2n

f2x logiB(rei6)1 d() = L

r

log-+ logiB(O)I. \an\ <• Iani

0

Holding N fixed, we then have for all r > laNI 1 2-11:

2x

J 0

logiB(rei6)1

d()

~

N

L

r

oo

log - 1+ L loglanl n=l lan n=l 00

= N log r +

L loglanl· n=N+l

2.5

THE CLASS N+

27

Consequently, co

2n

L

2n

J

loglanl :5: lim

n=N+1

r-+1

loglB(rei 6 )l d8 ::;;; 0,

0

and (9) follows by letting N ~ oo. Continuing the proof of the theorem, let the given function/ E N be expressed in the formf(z) = B(z)g(z), where g EN and g(z) -:1 0 in izl < 1. Since logJB(z)l

+ log+/g(z)l::;;; log+lf(z)l::;;; log+lg(z)J,

combination of (8) and (9) gives

J

2n

lim r-+1

Since logJg(z)l

E

J

2"

log+Jg(rei 6)l d8 =

0

1og+Jg(e16)l d8.

(10)

0

h 1, it has a representation 1 27t logJg(z)l = P(r, 8 -t) dv(t) 2n o

J

(11)

with respect to a function v(t) of bounded variation. Recalling the proof of Theorem 1.1, we see that v can be chosen to have the form 6

v(8) = lim n-+oo

J logJg(rn ei')l dt,

0::;;; 8::;;; 2n,

0

where {r"} is an appropriate sequence increasing to 1. On the other hand, by Fatou's lemma, 6

6

f log+Jg(~')l dt :s; lim inf J log+Jg(rn~')l dt = v+(8), 0

n-+co

(12)

0

say. If there were strict inequality for some 8, then a similar application of Fatou's lemma in [8, 2n] and addition of the two results would give a contradiction to (10). Equality therefore holds in (12) for all 8, which shows that v+(8) is absolutely continuous. On the other hand, 6

v _(8) = v +(8) - v(8) =lim inf n-+co

J log-Jg(rn ~')I dt 0

is nondecreasing. In view of(ll), this shows thatg = SG, where Sis a singular inner function and G is an outer function. Hence fEN+, which was to be shown. The following useful result is an easy consequence of the factorization theorems.

28

2 BASIC STRUCTURE OF HP FUNCTIONS

THEOREM 2.11.

If/EN+ andf(ei 6)EI!forsomep>0,thenjEHP.

The a priori assumption that fEN+ cannot be relaxed. The reciprocal of any (nontrivial) singular inner function is bounded on lzl = 1, but is not of classN+. 2.6. HARMONIC MAJORANTS

We noted in Section 1.3 that if f(z) is analytic in a domain D, then lf(z)IP is subharmonic in D. This means that in each disk contained in D, lf(z)IP is dominated by a harmonic function, the Poisson integral of its boundary function. However, there may not be a single harmonic function which dominates lf(z)IP throughout D. In general, a function g(z) is said to have a harmonic majorant in D if there is a function U(z) harmonic in D such that g(z) :$; U(z) for all z in D. If g is continuous and has a harmonic majorant, it is obviously subharmonic; but the converse is false. THEOR.EM 2.12. lf/(z) is analytic in lzl <1, thenfEHP (O
If U(z) is a harmonic majorant of 1/(z)IP, then by the mean value

theorem Mp(r,f)

:$;

[U(0)] 11 P.

Conversely, if fE HP, it follows from Theorem 2.7 and the arithmeticgeometric mean inequality (Exercise 2) that

1 :$;

2,.

t

27t

P(r, 8- t)lf(e;')IP dt.

In other words, 1/(z)IP is dominated by the Poisson integral U(z) of its boundary function. It is easy to see, in fact, that this is the least harmonic majorant of lf(z)IP. That is, if V(z) is any other harmonic majorant, then U(z) :$; V(z) for all z, lzl < 1. Indeed, for any p < 1.

-2n f 1

o

f 2n 1

27t

P(r, 8 - t)lf(pe;')IP dt

:$; -

27t

o

P(r,

= V( pz). As p ~ 1, this shows U(z)

:$;

V(z).

e-

t) V(pe '') dt

EXERCISES

29

If f E HP (0 < p < oo) and if cp is analytic and satisfies < 1 in lzl < 1, then g(z) = f(cp(z)) E HP and

COROLLARY. l~p(z)l

'p

1 + lcp(O)I} 1 Mp(r, g):::;:; {1 -lcp(O)I Mp(l,f). PROOF. Let U(z) be the Poisson integral of lf(ei')IP. Then 1/(z)IP:::;:; U(z), so that U(cp(z)) is a harmonic majorant of Ig(z)IP. Hence the mean value theorem gives p 1 + lcp(O)I p {Mp(r, g)} :::;:; U(cp(O)):::;:; 1 _ lcp(O)I {Mp(l,f)} ,

sinceP(r, 8):::;:; (1

+ r)/(1

-r).

EXERCISES

J

1. Prove in detail that fEN if and only if jloglf(re'6)lj de is bounded. 2. (Arithmetic-geometric mean inequality.) Letf(x) ~ 0 be integrable with respect to a nonnegative measure d11- of unit total mass. Let 21(/) = Jf(x) d11-(x) be the "arithmetic mean" off, and let CfJ(f) = exp{21(logf)} be the "geometric mean." Prove that CfJ(f) :::;:; 2I(f), with equality if and only if f(x) is constant a.e. (Hint: Note that log t:::;:; t- 1, t ~ 0, and set t = f(x)f21(f). This proof is due to F. Riesz [7].) 3. Show that the classical inequality

is a special case of the general arithmetic-geometric mean inequality. 4. (Jensen's inequality) Let cp(u) be a convex function in an interval a< u < b, and suppose a
~

),.(u - ex),

a< u
This proof is due to Zygmund [4].) Interpret Jensen's inequality for the case in which the unit measure d11- is concentrated at a finite number of points. 5. Show that if t/J(u) is a nondecreasing convex function and g(z) is subharmonic, then t/J(g(z)) is subharmonic.

30

2 BASIC STRUCTURE OF HP FUNCTIONS

6. Show that a function f(z) analytic in lzl < 1 is a Blaschke product (aside from a constant factor of modulus one) if and only if 27t

lim r--+

1

J

jloglf(re;6)lj de = 0.

0

7. Show that if f(z) is an inner function and rp(z) is a conformal mapping of the unit disk onto itself, thenf(rp(z)) and rp(f(z)) are inner functions.

8. Letf(z) be an inner function, and suppose lexl < 1. Prove that for all ex outside a set of planar measure zero, the function f(z)- ex 1- rif(z) is a Blaschke product! (Frostman [1].) 9. Prove the corollary to Theorem 2.6 under the weaker hypothesis that fEN+. 10. Show that for functions/analytic in the unit disk,/ENif and only if log+lf(z)l has a harmonic majorant.

NOTES

Fatou [1] showed that each f E H 00 has a nontangentiallimit f(ei 6) almost everywhere, and that f(ei 6 ) cannot vanish on an arc eo : :;:; e : :;:; eo + (j unless f(z) 0. F. and M. Riesz [1] improved this to H 1 and showed thatf(z) 0 if its boundary function vanishes on a set of positive measure. Later, Szego [1] showed that iff E H 2 and f(z) 1- 0, then loglf(e; 6)1 E I!. After F. Riesz [3] proved his factorization theorem (Theorem 2.5), he was able to show that each f E HP (p > 0) has a radial limit almost everywhere, and that loglf(e; 6)1 E L1 unless f(z) 0. He also proved the mean convergence of f(rei') to f(ei') (Theorem 2.6). Lemma 1, which simplifies the proof of this theorem, is also due to F. Riesz [6], at least for p ~ 1. The proof given in the text is in Littlewood's book [5]. A somewhat different approach to the structure of HP functions is in the paper of M. and G. Weiss [1]. For further information on the boundary behavior of HP functions, see Tanaka [1]. Theorem 2.1 is due to F. and R. Nevanlinna [1]; it unifies the presentation of the HP theory. A paper of Ostrowski [1] is also relevant here. Blaschke [1] introduced "Blaschke products" and proved Theorem 2.4. Regarding matters of priority, however, see Landau [3]. The canonical factorization theorems (Theorems 2.8 and 2.9) are due to Smirnov [2], who also noted a weaker form of Theorem 2.11 : iff E HP and f(e; 6) E Lq for some q > p, then

=

=

=

NOTES

31

fe Hq. Beurling [1] coined the terms "inner function" and" outer function." Smimov [3] introduced the class N+ (called "D" in his paper and in subsequent Soviet literature) and cited Theorem 2.10, which he attributed to "Madame P. Kotchine". More about the class N+ may be found in Privalov's book [4]. The condition (9) actually characterizes Blaschke products among all analytic functions with 1/(z)l :::;;; l (see Exercise 6). This theorem is due to M. Riesz, but was first published in Frostman [1]. (See also Zygmund [4], Vol. I, p. 281.) Theorem 2.12 was mentioned in passing by Smirnov ([3], p. 341). There is a large literature on the boundary behavior of Blaschke products; see, for example, Frostman [2], Cargo [1, 2], and Somadasa [1]. Paley and Zygmund [2] have shown that if the hypothesis f e N is "slightly relaxed" (which they interpret in two different ways), thenf(z) need not have a radial limit on any set of positive measure.

This Page Intentionally Left Blank

APPLICATIONS

CHAPTER 3

We now turn to some applications of the HP theory to such diverse topics as measure theory (the F. and M. Riesz theorem), Cauchy and Poisson integrals, and conformal mapping. Some of the results obtained will be useful in the later development of the HP theory. Further applications will appear in subsequent chapters.

3.1. POISSON INTEGRALS AND H 1

In Chapter 1, the harmonic functions u E h1 were characterized as PoissonStieltjes integrals. We shall now show that if the harmonic conjugate of u also belongs to h 1 , then the representing function 11-(t) is absolutely continuous. This result will have a number of interesting consequences. NOTE.

u(z)

The harmonic conjugate of u is any function v such that f(z) =

+ it(z) is analytic in the disk. We shall speak of" the" harmonic conjugate

even though it is determined only up to an additive constant. Later it will be convenient to normalize v by the requirement v(O) = 0.

34

3 APPLICATIONS

THEOREM 3.1.

A functionf(z) analytic in lzl < 1 is representable in the

form 1

27t

f(z) = 2,.

Jo

P(r, 8 - t)rp(t) dt

(1)

as the Poisson integral of a function rp E I! if and only iff E H 1 • In this case, rp(t) = f(ei') a.e. PROOF. If an analytic function f(z) has the form (1 ), then 27t

J

27t

lf(rei 6)1 d8 :::;:;

0

J

Irp(t)l dt,

0

so thatjE H 1 • Conversely, supposefEH 1 , and write

f 2n

1 (z) = -

2n

o

P(r, 8- t)f(e;') dt.

For any fixed p, 0 < p < 1, f(pz)

=1 -

J2"P(r, 8 -

2n o

t)f(pe 1') dt.

But, by Theorem 2.6, Jlf(pe 11 ) - f(e 11)1 dt ~ 0 as p ~ 1, so f(pz) ~ (z). Hence (z) = f(z), and the theorem is proved. Let u E h 1 , so that it is the Poisson-Stieltjes integral of a function 11-(t) of bounded variation. If the conjugate harmonic function v E h 1 , then 11-(t) is absolutely continuous. COROLLARY 1.

COROLLARY 2. A functionf(z) analytic in lzl < 1 is the Poisson integral of a function rp E IJ' (1 :::;:; p :::;; oo) if and only iff E HP. COROLLARY

then S 1 (z)/S2 (z)

3. Letf(z) =B(z)[S1 (z)/S2 (z)]F(z) EN. Iflog[f(z)/B(z)] EH\

=1.

Corollary 2 follows from Theorem 2.11. We shall prove in Chapter 4 that the harmonic conjugate of any function u E hP (1 < p < oo) is also of class hP. This cannot be true for p = 1, as Corollary 1 shows, even if u is a positive harmonic function. In fact, the Poisson kernel P(r, 8) itself is a counterexample. The following weaker theorem, however, remains true. THEOREM 3.2. Every analytic functionf(z) with positive real part in lzl < 1 is of class HP for all p < 1.

3.2 DESCRIPTION OF THE BOUNDARY FUNCTIONS

35

PROOF. Without loss of generality, suppose f(O) =I. The range off is contained in the right half-plane, sofis subordinate to

l+z 1 _ z = P(r, lJ) + iQ(r, lJ), where P(r, lJ) is the Poisson kernel and

2r sin lJ Q (} (r, ) = 1 - 2r cos(}+ r2 is the conjugate Poisson kernel. It follows from Littlewood's subordination theorem (Theorem I. 7) that

f2"1f(reie)IP d(J ::::; J2" 111 +- rere:: iP dlJ ~ J2"1Q(1, lJ)IP d(J < oo o

o

o

for any p < I, since (I + z)/(1 - z) is in HP and P(l, lJ) = 0 for(} # 0. COROLLARY. If u

e h 1 , then its harmonic conjugate belongs to h11 for all

p
We have seen that each function fe HP (0 < p::::; oo) has a nontangential limit f(ei 9 ) at almost every boundary point. It is of interest to characterize these boundary functions in a simple way. Let .:ifP denote the set of boundary functions f(e 19 ) of functions f e HP. As usual, two functions are identified if they coincide almost everywhere, so the elements of .:lfP are actually equivalence classes. We know that .:ifP c IJ', and in fact that .:ifP is a vector subspace of IJ': it is closed under addition and scalar multiplication. We shall see presently that .:ifP is also topologically closed. For qJ e IJ', let 1

11(/JII = 11(/JIIp = {2 7t

t

27t

i({J(t)IP dt

}1/p

•

It is convenient to retain the norm notation even for p < 1, when II liP is not a genuine norm. The notation llfiiP for fe HP will refer to the boundary function. Thus 11!11 = II fliP= Mp(1,f) =lim Mp(r,f). r-+1

36

3 APPLICATIONS

LEMMA. If fE HP (0 < p < oo), then

lf(z)l::;; 2 11PIIf11P(l- r)- 11 P,

r

=

lzl.

PROOF. If p = 1, the estimate follows easily from the Poisson integral representation of H 1 functions (Theorem 3.1). For p "# 1, we may use the Riesz factorization (Theorem 2.5) of a function fE HP in the form f = Bg, where B is a Blaschke product and g is a nonvanishing HP function. Since lf(z)l ::;; lg(z)l and llfiiP = llgiiP' it suffices to prove the lemma for nonvanishing HP functions. But if g E HP and g(z) "# 0, some branch of [g(z)JP is single-valued and belongs to H 1 • Thus the desired result follows from the H 1 case.

We are now ready to characterize :TfP. It is clear that :TfP contains all functions of the form

where the ak are complex constants. These functions will be called polynomials in e;8. They are dense in :TfP, as we shall now show. THEOREM 3.3. For 0

nomials in e; 8 . PROOF. For 0 < p < 1, let fp(z) = f(pz). Given fE HP and e > 0, choose p < 1 such that

llfP -fliP < ef2. (This is possible, by Theorem 2.6.) Now let Sn(z) denote the nth partial sum of the Taylor series off at the origin. Since Sn(z) --. f(z) uniformly on the circle lzl = p, IISnp- fpllp < ef2 for n sufficiently large. Thus, using Minkowski's inequality in the case p we find

~

1,

IISnp- fliP< e. This shows that the boundary functionf(e;'1 belongs to the IJ' closure of the polynomials in ei 8 . For p < 1, the inequality (a+ b)P ::;; 2P(aP

gives the same result.

+ bP),

a> 0, b > 0,

3.2 DESCRIPTION OF THE BOUNDARY FUNCTIONS

37

To complete the proof, we have to show that :TfP is closed. But suppose that a sequence Un(e; 8)} of :TfP functions converges in I! mean to qJ(8) E I!. Then by the lemma, Un(z)} is uniformly bounded in each disk lzl ::;; R < 1. Thus {j~} is a normal family, so a diagonalization argument produces a subsequence {J,,(z)} which converges uniformly in each disk lzl ::;; R < 1 to an analytic function f(z). It is clear that fE HP. We wish to show that qJ(8) = f(ei 8 ) a.e. But given e > 0, choose N such that 11/n-fmiiP < e for n, m;;::: N. Then for m;;::: Nand r < 1, Mp(r,f- fm) =lim Mp(r,fn,- fm)::;; lim supll/n,- fmllp <E. k--+oo

k--+oo

Letting r--. 1, we find II/- fmiiP < e for all m ;;::: N; thus II/- /,liP--. 0, and rp(B) = f(eiB) a.e. The same argument, in simpler form, shows that :Tt 00 is closed. COROLLARY 1.

If 1 ::;; p::;; oo, HP is a Banach space.

If p < 1, then II I!P is not a true norm and in fact the space HP is not normable. However, the expression d(f, g)= 11/- gil/

defines a metric on HP if p < 1. This can be verified using the inequality

a> 0, b > 0, which is valid for 0 < p < 1. (See Section 4.2.) Furthermore, the theorem shows that HP is complete under the topology induced by this metric. COROLLARY 2.

If 0 < p < 1, HP is a complete metric space.

Theorem 3.3 is false for p = oo, since :Tt 00 contains functions which do not coincide almost everywhere with continuous functions. One example is the singular inner function exp{(z + 1)/(z -1)}, whose boundary function is exp{ - i cot(0/2)}, e =F 0. There is another approach, however, which leads to a description of :TfP for 1 ::;; p ::;; oo. The Fourier coefficients of a function qJ E I! are the numbers n

= 0, ± 1, ± 2, ....

It is important to note that the Taylor coefficients of a function /E HP (1 ::;; p::;; oo) coincide with the Fourier coefficients of its boundary function. The following theorem expresses this more precisely.

38

3 APPLICATIONS

THEOREM 3.4. Let f(z) = L,""=o anzn belong to H 1 , and let {en} be the Fourier coefficients of its boundary function f(e; 1). Then en= an for n ~ 0, and en = 0 for n < 0. Furthermore, JfP (I ::5; p ::5; oo) is exactly the class of I! functions whose Fourier coefficients vanish for all n < 0.

PROOF. The Taylor coefficients off can be expressed in the form

O
::$;

llf.. -Jilt - 0

as r- l, by Theorem 2.6. [Here, as above, f..(z) = f(rz).] This shows that en = an for n ~ 0. Similarly, en = 0 for n < 0. Suppose now that cp e 11' (I ::5; p ::5; oo) and that its Fourier coefficients en vanish for all n < 0. Let

1

2n

J 2n o

f(z) = -

P(r, ()- t)cp(t) dt,

be the Poisson integral of cp. Since the Poisson kernel has the expansion CXl

P(r, t) = l +

L rn(int + e-int), n=l

it follows that CXl

f(z) =

L en zn,

lzl < l.

n=O

This shows that f(z) is analytic in lzl
Jff(z) = L an zn E H 1 , then a,.- 0 as n- oo.

From this point on we will usually suppress the notation JfP for the space of boundary functions. It is convenient, and customary in the literature, to identify JfP with HP, and thus to regard HP as a subspace of 11'.

3.3 CAUCHY AND CAUCHY-STIELTJES INTEGRALS

39

3.3. CAUCHY AND CAUCHY-STIELTJES INTEGRALS

For each complex-valued function qJ(O integrable over the unit circle 1'1 the Cauchy integral qJ(() d( 1 2 " e; 1qJ(e 11 ) 1 F(z) = --- =- . - - dt 2ni ICI=l (-z 2n o e''-z

= 1,

J

J

represents a pair of functions: one analytic in lzl < l, the other in lzl > l. More generally, the same is true for a Cauchy-Stieltjes integral

F(z) =

_!_ J2n ei: d~J.(t) 2n o

e"- z

with respect to a complex-valued function 11-(t) of bounded variation. THEOREM 3.5. If a function F(z) can be represented in lzl < l as a CauchyStieltjes integral, then it belongs to HP for all p < l. PROOF. It suffices to assume IJ.(t) is real and nondecreasing. Then Re{F(z)} > 0, since

e11

1 - r cos( t - 0) 0; 2 > 1 - 2r cos( t - 0) + r

}

Re {.....,.,.-- =

e'1 - z

and the result follows from Theorem 3.2. The converse is false: a function can be in HP for all p < l, yet not be representable as a Cauchy-Stieltjes integral. (See Exercise 2.) We shall also see that a Cauchy-Stieltjes integral, or even a Cauchy integral, can easily fail to be of class H 1 . A Cauchy-Stieltjes integral is related in a simple way to the PoissonStieltjes integral formed with respect to the same function 11-(t):

F(z)- F

(i1) = 21n t P(r, 0- t) d11-(t), 21t

lzl < 1.

(2)

In view of Theorem 3.5 and Theorem 2.2, the radial limit lim F(re18 ) r-+1-

exists almost everywhere (taken from inside the unit circle). But as r--+ l, the right-hand side of (2) tends to 11-'(0) a.e., by Theorem 1.2. Thus the exterior radial limit also exists a.e., and lim F(re 18 ) r-+1-

-

lim F(re18) = 11-'(0) r-+1+

a.e.

40

3 APPLICATIONS

THEOREM 3.6. Every function fE H 1 can be expressed as the Cauchy integral of its boundary function. In fact,

r f (k)< z ) = ~ 2 •),

J
lzl <

m !CI=l (Y~- z )k+l'

Each of these integrals vanishes identically in

1;

k = 0, 1, 2, ....

lzl > 1.

PROOF. It suffices to establish the result fork= 0; the rest will then follow by the familiar procedure of "differentiation under the integral sign." (See, for example, Ahlfors [2], p. 121.) Let F(z) be the Cauchy integral of/(e; 1), and let {Pn(z)} be a sequence of polynomials such that II/- Pnll 1 --. 0. (See Theorem 3.3.) Writing

~ f

F(z) =

f(O-

2mJIC/=1

Pn(O d(,

(-z

lzl >

1,

=

we find that F(z) 0 in lzl > 1. Hence it follows from the relation (2) with d11-(t) = f(e; 1) dt, and from Theorem 3.1, that

F(z)

J2"P(r, 8- t)f(ei') dt

=1 -

2n o

=

f(z),

lzl <

1.

(3)

This completes the proof. The relation (2) can also be used to prove the following theorem. THEOREM 3.7. For a complex-valued function 11-(t) of bounded variation, the following three statements are equivalent: 27t

J

(i)

einr d~-L(t) = 0,

n = 1, 2, ....

0

(ii) The Cauchy-Stieltjes integral

J2lt

1 eir d11-(t) F(z) = 2n o e' 1 - z

vanishes identically in lzl >I. (iii) The Poisson-Stieltjes integral 1 27t f(z) = P(r, 8- t) d11-(t) 2n o

f

is analytic in

lzl < I.

PROOF. (i) -e:. (ii). This follows by expanding the kernel e; 1(ei 1

powers of Ifz and integrating term by term.

-

z)- 1 in

3.3. CAUCHY AND CAUCHY-STIELTJES INTEGRALS

41

(ii) =;.(iii). Since the Cauchy-Stieltjes integral F(z) is always analytic in lzl < 1, this follows at once from (2). (iii)= (ii). The Cauchy-Stieltjes integral F(z) is analytic in lzl > 1. If the Poisson-Stieltjes integral f(z) is analytic in lzl < 1, then (2) shows that F(z) is also analytic in lzl > 1. Thus by the Cauchy-Riemann equations, F(z) is identically constant in lzl > 1. Since F(z)--. 0 as z--. oo, the constant must be zero. Hence F(z) 0 in lzl > 1.

=

A complex-valued function 11-(t) of bounded variation over [0, 2n] will be called normalized if 11-(t) = 11-(t+) for 0 :o::; t < 2n; that is, if 11- is continuous from the right. We can now use the preceding results to establish an important measure-theoretic theorem. THEOREM 3.8 (F. and M. Riesz). Let~-L(t)beanormalizedcomplex-valued

function of bounded variation on [0, 2n], with the property 27t

Jo

einr d11-(t) = 0,

n = 1, 2, ....

Then 11-(t) is absolutely continuous. PROOF. By Theorem 3.7, the hypothesis implies the corresponding PoissonStieltjes integral f(z) is analytic in lzl < 1 ; hence f E H 1 • But according to Theorem 3.1, every H 1 function can be represented in the form

f(z)

=1 -

J2"P(r, 8- t)f(ei

1)

2n o

dt.

Since the Poisson representation is unique (Section 1.2), d11-(t) = f(ei 1) dt. The essential step in the proof was the use of Theorem 3.1, the representation of H 1 functions by Poisson integrals. In fact, this result is in a sense equivalent to the theorem of F. and 1\1. Riesz. Below is another variant which is sometimes useful. The proof is left as an exercise. THEOREM 3.9. If a functionf(z) analytic in lzl < 1 can be represented in any one of the following four ways: (i) as a Cauchy-Stieltjes integral with F(z) 0 in lzl > 1; (ii) as a Cauchy integral with F(z) 0 in lzl > 1 ; (iii) as a Poisson-Stieltjes integral; (iv) as a Poisson integral, then it can be represented in each of the other three ways. The class of functions so representable is H 1 .

=

=

42

3 APPLICATIONS

3.4. ANALYTIC FUNCTIONS CONTINUOUS IN

izi:::::; 1

In general, a continuous function of bounded variation need not be absolutely continuous. For example, the familiar Lebesgue function over the Cantor set is monotonic, continuous, and purely singular. For the boundary values of functions analytic in the disk, however, continuity is equivalent to absolute continuity. In fact, a somewhat stronger statement can be made, as follows. THEOREM 3.10. If /E H 1 and its boundary functionf(ei 8) is equal almost

everywhere to a function of bounded variation, then f(z) is continuous in lzl :::::; 1 and f(ei 8 ) is absolutely continuous. Suppose f(ei 8) = p.(8) a.e., where p. is a normalized function of bounded variation and p.(O) = p.(2n). By Theorem 3.4, PROOF.

2~

f

2~

ein8p.(8) d8 =

J

0

ein8f(ei8) d8 = 0,

n = 1, 2, ....

0

Integrating by parts, we find 2~

J

ein8 dp.(8) = 0,

n = 1, 2, ....

0

By the F. and M. Riesz theorem, then, p.(8) is absolutely continuous. On the other hand,.f(z) can be represented as a Poisson integral of p.(8), by Theorem 3.1 and the fact that f(ei 8) = p.(8) a.e. Thus the radial limit .f(ei8) exists everywhere, coincides with p.(8), and is absolutely continuous. We shall now show that the functions just considered-analytic functions with absolutely continuous boundary values-are characterized by the simple condition/' E H 1 . The expressionf'(ei'1 then can have two possible meanings. It may indicate the radial limit of f'(z); or it may indicate the derivative with respect toe of the boundary function, apart from a factor iei 8 . It is a remarkable fact that the two are essentially the same. THEOREM 3.11. A functionf(z) analytic in lzl < 1 is continuous in lzl:::::; 1 and absolutely continuous on lzl = 1 if and only iff' E H 1. Iff' E H 1, then

'8 = ie''8 limf'(re') '8 -d f(e') d8 r->1

a.e.

(4)

PROOF. If f(z) is analytic in lzl < 1 and continuous in lzl::;;; 1, it is the Poisson integral of its boundary function:

1

f(z) = 2n

2~

Ia

P(r, 8 - t)f(ei') dt.

3.5 APPLICATIONS TO CONFORMAL MAPPING

43

Differentiate with respect to 8:

J

" aP izf'(z) = 1 - (r, 2n o 2

ae

e- t)f(ei') dt;

and integrate by parts [using the absolute continuity of f(ei')] to obtain

J2"

1 izf'(z) = P(r, 8- t)iei'f'(ei') dt. 2n o

(5)

Thus izf'(z) E H 1, which implies f' E H 1 • Conversely, iff' E H 1, then izf'(z) can be represented in the form (5) as the Poisson integral of its boundary function. Here it is understood that f'(ei') denotes lim, ..., 1 f'(rei'). Since the function 8

g(8) =

J iei'f'(ei') dt 0

is absolutely continuous in [0, 2n], and g(O) = g(2n) = 0, integration by parts in (5) gives

aea {f(reiB)} = aea{12n fo27tP(r, e- t)g(t) dt } . Thus

J2"

f(rei 8 ) = 1 P(r, 2n o

e- t)g(t) dt + C(r).

Since C(r) is the difference of two (complex-valued) harmonic functions, it is itself harmonic: C"(r)

+ (1/r)C'(r) = 0.

Thus C(r) =a log r + b, where a and b are constants. To ensure continuity at the origin, a must be zero. Hencef(z) is the Poisson integral of the continuous function [g(t) + b], and is therefore continuous in lzl ::;:; 1 and has boundary valuesf(e;8) = g(8) +b. The relation (4) now follows from the definition of g(8). 3.5. APPLICATIONS TO CONFORMAL MAPPING

The theorems just given, together with a few general facts about HP functions, can be applied to obtain some rather deep results in the theory of conformal mapping. A Jordan curve (or a simple closed curve) C is the image of a continuous complex-valued function w = w(t) (0::;:; t::;:; 2n) such that w(O) = w(2n) and

44

3 APPLICATIONS

w(td "# w(t2 ) for 0::;:; t 1 < t2 < 2n. The curve C is said to be rectifiable if w(t) is of bounded variation. Its length Lis then defined as the total variation ofw(t): n

L =sup

L lw(tk)- w(tk-1)1, k=1

where the supremum is taken over all finite partitions 0 = t0 < t 1 < · · · < tn = 2n of [0, 2n]. It is easily seen that Ldepends only on the curve C, and is invariant under a change of parameter. If w(t) is absolutely continuous, the length of the arc of C corresponding to an arbitrary interval a ::;:; t ::;:; b is given by b

J lw'(t)l dt. a

(A proof may be found in Natanson [1 ], Vol. II, p. 227.) It follows that any Lebesgue measurable set E c [0, 2n] has an image on C of measure

tlw'(t)l dt. Indeed, the formula is obviously valid for open sets E, hence for G6 sets (countable intersections of open sets). Therefore, since an arbitrary measurable set E is contained in a G6 set with the same measure, the formula holds generally. A famous theorem of Caratheodory asserts that every conformal mapping w = f(z) of lzl < 1 onto the interior of a Jordan curve C has a one-one continuous extension to lzl ::;:; 1. In particular, w = f(ei 8 ) is a parametrization of C. Applying Theorems 3.10 and 3.11, we therefore obtain the following important result. THEOREM 3.12. Let f(z) map lzl < 1 conformally onto the interior of a Jordan curve C. Then C is rectifiable if and only iff' E H 1.

This theorem is highly plausible, perhaps even "obvious," when viewed geometrically. It says that the lengths

of the images of the circles lzl = r are bounded if and only if the boundary has finite length. In the presence of a rectifiable boundary, this result puts the general HP theory at our disposal. The derivative of the mapping function has an angular

3.5 APPLICATIONS TO CONFORMAL MAPPING

45

Iimitf'(ei 8 ) almost everywhere on the boundary, and loglf'(e; 8)1 is integrable. The function f'(ei 8 ) is related as in (4) to the derivative of the absolutely continuous boundary function f(e; 8 ). A measurable set Eon the unit circle is carried onto a subset of C with measure

Consequently, a subset of lzl = I has measure zero if and only if its image on C has measure zero. In other words, the boundary sets of measure zero are

preserved under the conformal mapping. This remains true (in view of the Riemann mapping theorem) for a conformal mapping between any two Jordan domains with rectifiable boundaries. More can be said. If f(z) is a conformal mapping of lzl < 1 onto the interior of a rectifiable Jordan curve C, then its continuous extension to lzl ::::;; I is conformal at almost every boundary point. To be precise, let y be a continuous curve in lzl < 1 which terminates at a point z 0 = eiBa in a well-defined direction not tangent to the unit circle; i.e., the limit of arg{z- z 0 } is to exist as z-+ z 0 along y, and is not to equal e0 ± nf2. The image of y is then a curve r inside C, terminating atf(z 0 ). Since (djde){f(e; 8 )} exists a.e., C has a tangent direction at almost every point. We assert that for almost every e0 , the angle between rand the tangent to C atf(z 0 ) exists and is equal to the angle between y and the tangent to the unit circle at z 0 . In other words, the mapping preserves angles at almost every boundary point. We have to show that } arg{ [.!!.._ f(ei 8 )] e~ea de

-

lim arg{f(z)- f(z 0 )}

z->zo Z E y

=eo+ nj2- Jim arg{z -

Zo}•

z--+zo zEy

In view of relation (4), it is eno'.lgh to prove lim f(z)- J(zo) = f'(zo) z-za

(6)

Z - Zo

zEy

wherever the angular limitf'(z 0 ) off'(z) exists; hence almost everywhere. But

f(z)- J(zo) z- z 0

=

_1_ z- z 0

Jz f'(O d(,

(7)

za

the integration being performed along the segment joining z 0 and z. If f'(z 0 ) exists, the right-hand side of(7) approachesf'(z 0 ) as z-+ z 0 along y, which was assumed to be a nontangential path. This proves (6).

46

3 APPLICATIONS

3.6. INEQUALITIES OF FEJ~R-RIESZ, HILBERT, AND HARDY

We shall now discuss some interesting inequalities which will have applications in later chapters. THEOREM 3.13 (Fejer-Riesz inequality). If /E HP (0 < p < oo), then the integral of lf(x)IP along the segment -1 ::;; x::;; 1 converges, and

(8)

t

The constant

is best possible.

PROOF. Consider first the case p = 2, and suppose for the moment that f(z) is real on the real axis. By Cauchy's theorem with a semi-circular path,

r

+i

[f(x)] 2 dx

r

[f(rei 8 )] 2 ei 8 de

=0

0

-r

for each r < 1. It follows that

Adding a similar inequality for the lower semicircle, we find 2n

r

2

J

[f(x)] 2 dx::;;

J

2n

lf(re;8)1 2 de::;;

0

-r

J

IJ(e;8 )1 2 de.

0

The desired inequality is now obtained by letting r tend to 1. More generally, we may expreso; an arbitrary H 2 functionf(z) in the form co

f(z)

=

co

n~o

n~o

where g and h are in proved,

1

co

L (cxn + i{Jn)zn = L CXnZn + in=O L f1nzn = g(z) + ih(z),

2~

: ; -2 J 0

H2

and are real on the real axis. By what we have just

1

lg(e;8)1 2 de+2

2~

J

lh(e;8)1 2 de

0

1 J2~ lf(e;a)l2 de--i J2~[h(e;a)g(e-ia)- g(e;a)h(e-;a)] de. 2 0 2 0

=-

But the last integral must vanish, since the integrand is an odd function of e. Hence inequality (8) is established in the case p = 2. To deduce the result for general p, we use a familiar device. lf/(z) belongs

3.6 FEJtR-RIESZ. HILBERT. AND HARDY INEQUALITIES

47

to HP and vanishes nowhere in lzl < 1, then [f(z)]Pf 2 E H 2 , and (8) follows from the special case p = 2 already proved. If f(z) has zeros, we factor out the Blaschke product B(z) to obtain a nonvanishing function g(z) = f(z)/B(z) which is again of class HP. Thus 1

f_

1

1/(x)IP dx 1

:$;

f_

27t

lg(x)IP dx 1

:$;

t

Jo

27t

lg(e; 8)IP de =

t

Jo

lf(ei8)IP de.

In view of Theorem 3.12, there is an interesting geometric application. COROLLARY. If the unit disk lzl < 1 is mapped conformally onto the interior of a rectifiable Jordan curve C, the image of any diameter has length at most half the length of C.

We may now finish the proof of the theorem by showing that tis the best possible constant. Let w = qJ(z) map lzl < 1 conformally onto the interior of the rectangle with vertices ± 1 ± ie, the diameter - 1 :$; z :$; 1 corresponding to the real segment - 1 :$; w :$; 1. The ratio of the length of this segment to the perimeter of the rectangle is 2[4(1 + e)r\ which tends to } as e __. 0. The functionf(z) = [qJ'(z)] 1 fP E HP therefore shows that the constant cannot be improved. Two further inequalities, named after Hilbert and Hardy, lie in the same circle of ideas. We shall prove them in generalized form, with a view to later applications. For a complex vector x = (x 0 , x 1 , ..• , xN), let N

llxll 2 THEOREM 3.14.

Let 1/J

E

L lxnl

=

2•

n~o

L"' and n = 0, 1, 2, ....

Let N

AN(x, y) =

L

A.n+m

xn Ym.

n,m=O

Then

IAN(x, x)l

= 1 2171: Jo27t[P(t)] 2 1/J(t) dt :$;

1 111/JIIoo 2

I

27t

nfo

IP(t)l 2 dt = 11!/JIIoollxll 2 •

(9)

48

3 APPLICATIONS

To deduce the more general result, observe that AN(x, y)

= -!-AN(x + y, x + y) - -!-AN(x - y, x - y).

Thus

Y)l::;; Hl/llloo{llx + Yll 2+ llx- Yll 2} = tlll/111 {11xll 2+ llyll 2 }. This shows that IAN(x,y)l::;; 111/ll oo if llxll = IIYII = 1, which is enough to IAN(x,

00

prove the theorem. COROLLARY (Hilbert's inequality).

In.t~o n :n~m+ 1 1::;; nllxiiiiYII. PROOF. Choose 1/l(t) = ie- ir(n

Let

THEOREM 3.15.

f(z) =

L anzn

E

- t), so that .A.n = (n + 1) - 1and 111/111 = n. 00

.A.n ~ 0 be given by (9) for some

1/1

E

L00 • Then if

H\ CXl

L .A.nlanl::;; 111/JIIooll/111· n=O PROOF. Every /E H 1 has a factorization f = gh, where g and h are H 2 functions with llgll/ = llhll/ = 11/11 1. Indeed,/= BqJ, where B is a Blaschke product and qJ is a nonvanishing H 1 function (Theorem 2.5). Let g = BqJ 1f 2 and h = qJ 1'2. Now let g(z) = bn~ and h(z) = cnzn; then

L

L

n

an=Lbkcn-k· k~O

Hence N

N

n

L An Iani ::;; n=O L An L lbkllcn-kl n=O k=O

N

::;; L

k,m=O

.A.k+mlbkllcml::;; lll/JIIoollgll2llhll2 = 111/JIIooll/111•

by Theorem 3.14. COROLLARY (Hardy's inequality).

Iff(z) =

L anzn

Ia I L ~1 ::;; nllfll1· n=O n + CXl

E

H\ then

3.7 SCHLICHT FUNCTIONS

49

Hardy's inequality is proved with the same choice of 1/J that gave Hilbert's inequality. One interesting consequence should be mentioned. Suppose f(z) = a"z" is analytic in lzl < 1. If lanl < oo, then f has a continuous extension to lzl::;:; 1, but the converse is false (see Exercise 7). Hardy's inequality shows, however, that iff' E H 1 (or equivalently, in light of Theorem 3.11, ifjis continuous in lzl::;:; 1 and absolutely continuous on lzl = 1), then la.l < oo. In particular, lanl < oo iff is a conformal mapping of the unit disk onto a Jordan domain with rectifiable boundary.

L

L

L

L

3.7. SCHLICHT FUNCTIONS

A function analytic in a domain is said to be sch/icht (or univalent) if it does not take any value twice; that is, if/(z 1 ) "# f(z 2 ) whenever z1 "# z2 . Our main aim in this section is to prove that every schlicht function in the unit disk is of class HP for all p <-!-.This will show, for instance, that a conformal mapping of the unit disk onto an arbitrary simply connected domain, however pathological the boundary, automatically has a finite radial limit in almost every direction. The Koebe function

z k(z) =

(1 - z)

co

2

=

L nz", n~ 1

which maps lzl < 1 onto the full plane slit along the negative real axis from -t to oo, shows that a schlicht function need not belong to H 1 ' 2 • We say thatjE Y if /is analytic and schlicht in lzl < 1 and is normalized so that f(O) = 0 and f'(O) = 1. The first of the following lemmas is very well known, and the proof is omitted here. (See, for example, Hayman [1], p. 4.) LEMMA 1. Iff E Y, then

r ~ r ( 1 + r)2 ::;:; lf(re' )I ::;:; ( 1 _ r)2 · LEMMA 2. Let C 1 and C 2 be twice continuously differentiable Jordan curves in the plane, surrounding the origin, with C 1 in the interior of C 2 • Then

f

c,

rP de::;:;

f

c2

rP de,

p> 0,

where (r, e) are polar coordinates and both curves are traversed in the positive direction.

50

3 APPLICATIONS

PROOF. Let D be the ring domain between C 1 and C 2 , let C be its boundary, and let (x, y) denote rectangular coordinates. Applying Green's theorem and the Cauchy-Riemann equations

a(log r) ae ay =-ax'

a(log r) ae. ax = ay' we find

fc rP de ff = p

rr 1

[axar ayae

ar ae] dx dy ay ax

-- ---

D

This proves the lemma. We are now ready to state the main theorem. THEOREM 3.16. If f(z)

is analytic and schlicht in lzl < 1, then f E H P for

allp< 1/2. PROOF. Without loss of generality, we may assume/E Y, so thatf(z) "# 0 unless z = 0. Now set

f(reiB) =

Rei~~>

(r > 0)

and apply the Cauchy-Riemann equation

r

a( log R) a ar = ae

to obtain

-d J

2"

dr o

lf(re; 8)IP de =

Jo -;ora;-- (RP) de 2"

PJ2"RPa de = Pf -

=-

r

0

ae

r r,

RP d,

where r, is the image underfofthe circle lzl = r. Thus if M(r) = the maximum of lf(z)l on lzl = r, then for each e > 0 the circle lwl surrounds r,' and it follows from Lemma 2 that d p -d {M/(r,f)}::;:;- [M(r) r

r

+ e]P.

M~(r,f)

is

= M(r) + e

EXERCISES

51

Now let e ~ 0, integrate from 0 to r, and apply Lemma 1 :

M/(r,f) ~ p [ [M(r)]P dr o r

~p

JrP- (1- r)1

1

2P

dr < oo

0

if 0 < p < ! . This concludes the proof. As a function of class HP(p < -t), each schlicht function factorization f(z) = B(z)S(z)F(z)

f has a canonical

(see Theorem 2.8). The Blaschke product B(z) obviously has at most one factor. Less obvious is the fact thatfcan have no singular part. THEOREM 3.17. lf/(z) is analytic and schlicht in lzl

factor S(z)

=1.

< 1, then its singular

PROOF. Iff does not vanish in the open disk, then 1/fis analytic and schlicht, so 1/f E HP for all p < -t. Hence by the uniqueness of the canonical factorization of a function of class N (Theorem 2.9), S(z) = 1. If /(0 = 0 for some (, lei < 1, then the function

g(z) = (1 -

t(t: c~)

1'1 2 )- 1 [/'<0r 1

belongs toY. By Lemma 1, then, lu(z)/zl ~ (1 + r)- 2 >

!·

From this it follows that Blf E H 00 , where

z-C

B(z) = 1 _ (z. In particular, f cannot have a (nontrivial) singular factor. EXERCISES

1. Prove: If f(z) is analytic and Re{f(z)} > 0 in lzl < 1, thenf is an outer function. (Hint: Show thatf.(z) = e + f(z) is outer and let e ~ 0.) 2. Showthat

f(z) = - 1-log(-1- )

1-z

1-z

belongs to HP for all p < 1, but has unbounded Taylor coefficients. In particular, show thatf(z) is not a Cauchy-Stieltjes integral, so that the converse to Theorem 3.4 is false.

52

3 APPLICATIONS

3. Prove Theorem 3.9. 4. Prove the integral analogue of Hilbert's inequality:

IJooJoo 0

0

f(x)g(y) dx dy

X+ y

I~ nii!II2IIYII2

iff and g are in L2 (0, oo ). 5. It is a natural conjecture that iff EN andf(ei 8)

27t

J

ein 8f(e; 8 ) de = 0,

E

L1 , then

n = 1, 2, ....

0

Show this is false. 6. Construct an analytic functionf(z) =

f' ¢ Hl.

L anzn such that L lanl < oo, but L

7. Give an example of a function f(z) = anzn analytic in lzl < 1 and continuous in lzl ~ 1, such that lanl = oo. (Suggestion: Let f map the unit disk conformally onto a Jordan domain constructed in such a way that the radius 0 ~ z < 1 corresponds to a curve of infinite length.)

L

NOTES

Theorems 3.1 and 3.8 are essentially due to F. and M. Riesz [1]. They also showed that the coefficients of an H 1 function tend to zero. Theorem 3.4 and the falsity of its converse (Exercise 2) may be found in Smirnov [2]. Theorem 3.9 is due to Fichtenholz [1 ]. Privalov's book [4] discusses these results. Helson [1] has given a" soft" proof of the F. and M. Riesz theorem, making no use of complex function theory. Rudin [6] obtained a generalization of this theorem in which Jeinr dp. = 0 only outside a certain "thin" set of positive integers. Theorem 3.11 is due to Privalov [1, 2], as are most of the results in Section 3.5. Theorems 3.10 and 3.12 appear in a paper of Smirnov [4]. The theorem of Caratheodory mentioned before Theorem 3.12 may be found in Goluzin [3], Chap II, Sec. 3, or in Zygmund [4], Chap. VII, Sec. 10. Golubev [1] gave the first example of a conformal mapping of the unit disk onto a Jordan domain which carries a boundary set of measure zero onto a set of positive measure on the (nonrectifiable) boundary of the image domain. Theorem 3.13 is in a paper of Fejer and Riesz [1]. Hardy [1] had essentially proved it for p = 2. Many proofs of Hilbert's inequality have been given; see Hardy, Littlewood, and P6lya [1]. "Hardy's inequality" seems to have appeared first in a paper of Hardy and Littlewood [1]. Theorem 3.16 is essentially in a paper of Prawitz [1]; see also Goluzin [3], Chap. IV, Sec. 6. Another proof is in Littlewood [5]. Theorem 3.17 is due to Lohwater and Ryan [1].

CONJUGATE FUNCTIONS

CHAPTER 4

If a harmonic function has a certain property, must the same be true of its conjugate? Questions of this kind have been widely investigated, both for their own interest and for their importance in applications. In this chapter and the next, we consider growth and smoothness properties of functions harmonic in the unit disk. Generally speaking, a harmonic function and its conjugate behave alike, but there are some rather surprising exceptions.

4.1. THEOREM OF M. RIESZ

Given a real-valued function u(z) harmonic in lzl < 1, let v(z) be its harmonic conjugate, normalized so that v(O) = 0. Thusf(z) = u(z) + iv(z) is analytic in lzl < 1, andf(O) is real. If f(z) = Len zn and en= an- ibn, then co

u(z) = ao

+ L rn(an cos ne + bn sin ne), n=l

co

v(z) =

L rn<- bn cos ne + an sin ne).

n= 1

(I)

54

4 CONJUGATE FUNCTIONS

By the Parseval relation,

Hence (2)

with equality if and only if a0 = 0. It was the discovery of M. Riesz that something similar is true for every pin the range 1 < p < oo. THEOREM 4.1 (M. Riesz). If u E hP for some p, 1 < p < oo, then its harmonic conjugate v is also of class hP. Furthermore, there is a constant AP, depending only on p, such that

0::;; r < 1,

(3)

for all u E hP. PROOF. First let us observe that only the case 1 < p::;; 2 needs to be considered. Indeed, if the theorem holds for some index p ( 1 < p < oo ), then it is also true for the "conjugate index" q defined by lfp + lfq = 1, and in fact with Aq = AP. To show this, let N

g(z) = g(rei 8) = ao

+ L r"(an cos ne + bn sin ne) n=l

be an arbitrary harmonic polynomial, and let N

h(z) =

L rn<- bn cos ne + an sin ne) n=l

be the conjugate function, normalized by h(O) = 0. Thus g(z) + ih(z) is an ordinary algebraic polynomial; and for fixed p (0 < p < 1), the function

F(z) is analytic in 27t

J

lzl ::;;

=

[u(pz)

+ iv(pz)][g(z) + ih(z)]

1. The mean value theorem applied to Im{F(z)} gives

[u(pe; 8)h(ei8) + v(pei'1g(e;'1J de= 2n[u(O)h(O) + v(O)g(O)]

=

0.

0

Assuming the theorem has been proved for the index p, it follows that 12171:

s:

I

"v(peiB)g( ei8) de = 12171:

s:

"u(pei8)h( ei8) de

::;; llhllpMq(p, u)::;; AP IIYIIpMq(p, u).

I

4.1 THEOREM OF M. RIESZ

55

Now take the supremum of l(l/2n)J vgl over all g with llgiiP:::;; l, and use the fact that trigonometric polynomials are dense in LP. The result is

Mq((J, v) :::;; AP Mq(p, u). Thus we may assume l < p in the form r

f

2" 0

s 2. The next step is to apply Green's theorem 0(/) or d(}

ff V cp dx dy, 2

=

lzls r

where V2 denotes the Laplacian. We suppose for the moment that u(z) > 0, and compute V2{uP} = p(p- l) l/'12uP-2; V2{1fiP} = P2 1f'l 2 1/lp- 2· Hence

By Green's theorem, then,

f2"

-d lf(re;8 )1P dO dr o

d f2" lu(re;8 )1P dO. s -p- p- 1 dr o

Integrating from 0 to r, and recalling that v(O) = 0, we find

Thus

ifu(z) > 0. For general u E hP, fix r < l and set

Then

u(re 18)

= U1 (0)- U2 (0);

lu(re; )1P = IU1(0)IP 8

+ IUiO)IP.

(4)

56

4 CONJUGATE FUNCTIONS

For 0 :s; p < 1, define

f 2n 1

ui(pei 8 )

=-

27t

P(p,

o

e- t)Uit) dt,

(5)

j = 1, 2.

Let vi be the harmonic conjugate of ui. Then u(pre;o)

= ut(pe;o)-

uipe;o);

v(preio) = vt (pe;o) - vipe;o). Since u1 and u2 are positive harmonic functions, it follows from what has been proved that ("1v(pre;8 )IP de :s; 2P{("Iv 1 (pe; 8)1P de+ ("1v2(pei 8)IP de} 0

0

0

2Pp { :s; - -

J

p- 1

27t

lut(pei8 )IP de+

0

J

27t

lu2(pe;oW de

}

.

0

Letting p--. 1 and using relations (4) and (5), one now obtains (3) with

- (-p)lfp.

AP- 2

p-1

The factor 2 can be removed by a more refined argument (see Exercise 5). As pointed out in Section 3.1, Riesz's theorem breaks down for p = 1. The Poisson kernel P(r, e) E h 1 , but its analytic completion (1 + z)/(1 - z) ¢ H 1 . The next two sections will be concerned with weaker theorems which may be said to replace Riesz's theorem in the case p = 1. Let us note that Riesz's theorem also fails in the case p = oo. Consider, for example, the function f(z) = u(z)

1+z + iv(z) = i log 1 _ z,

which maps lzl < 1 conformally onto the vertical strip -n/2 < u < n/2. Here u is bounded, but v is not. On the other hand, Riesz's theorem guarantees that f E H P for all p < oo, a fact not easy to verify by direct calculation. 4.2. KOLMOGOROV'S THEOREM

Although the harmonic conjugate of an h 1 function u(z) need not be in h 1 , it does belong to hP for all p < 1. We have already proved this (Corollary to Theorem 3.2) by appeal to Littlewood's subordination theorem. We now give an independent proof of a slightly stronger result. The following lemma will be needed.

4.2 KOLMOGOROV'S THEOREM

57

LEMMA. For arbitrary positive numbers a and b,

O I. PROOF. It is enough to find the supremum of

(1 + x)P g(x) = 1 + xP

for x

~

I. Differentiation gives

Thus g(x) is decreasing in the interval I ~ x < oo if p > 1, increasing if 0 < p < I. Since g(l) = 2p-l and g(x)-> I as x-> oo, the lemma is proved. For p > I, the result also follows easily from the fact that xP is a convex function. THEOREM 4.2 (Kolmogorov). If u E h 1, then its conjugate v E hP for all

p < I. Furthermore, there is a constant BP, depending only on p, such that O~r<

I,

for all u E h1 . PROOF. Assume first u(z) > 0, and set

f(z) = u(z)

+ iv(z) =

Re; 41 ,

I\ <

n/2.

Then F(z) = U(z)]P = RPeip<J> is analytic in lzl < I. By the mean value theorem,

-2n1 Jo F(rei Z7t

8)

dO= F(O)

= [u(O)]P =

[M 1(r, u)]P.

Hence

In view of the inequalities lv(z)l < R and 0 < cos pn/2 < cos p, it follows that Mp(r, v) < (secpn/2) 11PM1 (r, u)

for positive u. The constant is in fact best possible for u(z) > 0 (see Exercise 3).

58

4 CONJUGATE FUNCTIONS

The extension to general u E h 1 is similar to that in the proof of Riesz's theorem. Using the same notation, we find by the lemma that

::;; sec pnf2{[M 1(p, u 1 )]P

+ [M 1(p, u2)]P}.

The other part of the lemma now gives (since 1/p > 1)

Mp(pr, v)::;; BP[M1 (p, u 1)

+ M 1(p, u2 )],

where

BP = 2°fpl- 1 (sec pn/2) 1 1P. The proof is completed by letting p --. 1. 4.3. ZYGMUND'S THEOREM

Since u E h1 is not enough to ensure v E h1, but the stronger hypothesis u E hP (for some p > 1) is sufficient, it is natural to ask for the "minimal" growth restriction on u which will imply v E h1• Such a condition is the boundedness of 27t

J

lu(re; 6)llog+ lu(rei 6)1

de,

r < 1.

0

We shall denote by h log+ h the class of harmonic functions u(z) for which these integrals are bounded. Clearly, hP c h log+ h for all p > 1. THEOREM 4.3 (Zygmund).

If u E h log+ h, then its conjugate vis of class

h 1 , and 27t

J

27t

lv(rei6)1

0

de::;;

J

lu(re; 6)llog+ lu(rei6)1

de+ 6ne,

O::;;r
0

PROOF. The proof is similar to that of Riesz's theorem. Assuming first u(z) ;;::: e, one finds

Thus V2 {1/l}::;; V2 {u log u}, and an application of Green's theorem gives, as before, 27t

J 0

27t

lf(rei6)1

de- 2nu(O)::;;

J 0

u(rei6) log u(rei6) de- 2nu(O) log u(O).

4.3 ZYGMUND'S THEOREM

~

From this we conclude [since u(O)

e]

provided u(z);;:: e. For general u(z), fix r < 1 and let

U 1((J) = max{u(rei8 ), e},

Ui9) =max{ -u(rei8 ), e},

U3(9) = u(re;8 ) - [U 1(9)- Ui9)]. Thus IU3 (9)1 ::;; e. For 0 ::;; p < 1, let

ui(pe;8 ) = 217t

J2" P(p, 9- t)Ui(t) dt, 0

j

= 1, 2, 3,

and let vi be the respective harmonic conjugates. Then

+ u3(pe;e), vt(pe;e)- v2(pe;e) + v3(pe;e).

u(preie) = ut(pe;e)- u2(peiB) v(pre;e) =

By what we have already proved,

M 1(pr, v)::;; M 1(p, v1) 1 2 ::;; 2 7t ~1

+ M 1(p, v2) + M 1(p, v3)

J211 lui(pei8)llog+ lui(pe18)1 d9 + M 0

1 (p,

v3).

On the other hand, applying the Schwarz inequality and (2), we find M 1(p, v3 )::;; Mip, v3 )::;; Mip, u3 )::;; e. Now let p tend to I, with the result 2

21tM 1 (r, v)::;; i~t

27t

Jo

1Ui(9)1log+ 1Ui(9)1 d9

+ 27te.

To complete the proof, consider the subsets of [0, 27t] E 1 = {9: u(re18 );;:: e}

E 2 = {9: u(rei8 )::;; -e}.

and

Then 2

_L J=l

f

27t

0

2

IUi9)1log+IUj(9)1 d9::;;

.L

J=l

f.

IU/9)1log+IU/9)1 d9

+ 41te

EJ

: ; J iu(rei~llog+lu(rei8)1 d9 + 4ne. 27t

0

59

60

4 CONJUGATE FUNCTIONS

Zygmund's theorem is, in a sense, best possible: the growth restriction it imposes on u(z) cannot be weakened. The following is a partial converse. THEOREM 4.4. If both u and its conjugate v belong to h 1 , and if u(z) > C for some constant C, then u e h log+ h. PROOF. We may assume u(z) > I, since addition of a constant to u does not affect v. Set

f(z) = u(z)

+ iv(z) =

Rei41 ,

1<1>1

< 7t/2.

The function f(z) logf(z) is analytic in lzl < 1. Applying the mean value theorem to its real part, we find

f

2n

R cos log R dO =

0

f

2n

R sin dO

+ 21ru(O) log u(O).

0

Thus

f

J u log R dO 7t J2" lv(rei )1 dO+ 21tu(O) log u(O). <2

2n

2n

u(rei0 ) log u(rei6 ) dO::;;

0

0

6

0

COROLLARY. Let cp(x) ;;;:: 0 be a convex nondecreasing function of x;;;:: 0, such that cp(x) = o(x log x) as x- oo. Then there exists a harmonic function u(z), with conjugate z:(z), such that

J cp(lu(rei )1) dO 2"

6

0

is bounded for r < 1, while v ¢ h1 • PROOF. Choose an integrable function U(t);;;:: 0 on [0, 27t] such that cp( U(t)) is integrable but U(t) log+ U(t) is not. (We will show in a moment that this is possible.) Let u(z) be the Poisson integral of U(t). Then u(z) > 0, and an application of Jensen's inequality (Chapter 2, Exercise 4) shows that f cp(u(re'~) dO is bounded. By the preceding theorem, v e h 1 would imply u e h log+ h, which is not the case (by Fatou's lemma). Thus v ¢ h1 • It remains to show that a function U(t) can be constructed with the required properties. Suppose, more generally, that cp(x) and t/J(x) are any continuous, nonnegative, nondecreasing functions (convex or not), defined on 0 ::;; x < oo, such that t/J(x)- oo and cp(x)Jt/l(x)- 0 as x- oo. Choose increasing numbers

4.4 TRIGONOMETRIC SERIES

61

x.(n = 1, 2, ... ) such that 1/J(x.) = 2". Then 2-"tp(xJ--. 0; select a subsequence {nk} for which co

L 2-"•rp(x•.) < oo. k=1 Now define t0 = 0 and tk = tk_ 1 + 2-"k, k ~ l. Notice that 0 < t 1 < t 2 < · · · ::;; 1. Let T = lim tk. Finally, let U(t) be the step function with the value x•• in the interval tk_ 1 ::;; t < tk and with U(t) = 0 in T :$; t :$; 2n. Observe that 27t

Jo

co

tp(U(t)) dt = k"'i;yk- tk_ 1 )tp(x•• ) + (2n- T)tp(O) co

=

L 2-"•rp(x•.) + (2n- T)tp(O) < oo. k=1

On the other hand, 1/J( U(t)) is not integrable because 2 -••1/J(x• .)

=1.

4.4. TRIGONOMETRIC SERIES

The preceding results can also be expressed in the language of trigonometric series. We shall now briefly indicate the connection between the two theories. A formal trigonometric series

ao -

+

2

~

L..

(a. cos nO

. + b n Sill nO)

(6)

n= 1

is called a Fourier series if there exists an integrable (real-valued) function tp(t) such that 1 27t n = 0, 1, 2, ... ; a. = cos nt rp(t) dt,

J

n o

1

(7)

27t

J sin nt rp(t) dt, n o

b. =-

n = 1, 2, ....

It is called a Fourier-Stieltjes series if there is a function 11-(t) of bounded

variation such that (7) holds with rp(t) dt replaced by d11-(t). The formal series co

L (-b. cos nO + a. sin nO)

(8)

n=1

is called the conjugate trigonometric series of (6). The use of the word conjugate is easily justified. Assuming that (6) is the Fourier series of rp(O), consider the Poisson integral

J2"

u(z) = 1 P(r, 0- t)tp(t) dt. 2n o

62

4 CONJUGATE FUNCTIONS

Let

f(z) = u(z)

+ iv(z) =

J2"

1 eir + z - i1rp(t) dt 2n o e - z

(9)

be its analytic completion, normalized as usual by v(O) = 0. The expansion

eir

+z

oo

e - z

•

L e-•n'z"

-;,-- = 1 + 2

n=1

and the relations (7) now give

Taking real and imaginary parts, we find

a u(z) = ~ 2

oo

+ L r"Can cos ne + bn sin ne), n= 1

CXl

v(z) =

L r"C- bn cos ne + an sin ne). n=1

These series are convergent for every r < 1. On the other hand

u(rei6 )

--+

rp(8)

a.e.,

by Theorem 1.2. In the language of summability theory, we may therefore say that the series (6) is Abel summable to rp(8) for almost every e. It is known, however, that if rp is merely integrable, the series (6) need not converge anywhere in the usual sense. A similar statement may be made about the conjugate series (8). If (6) is a Fourier series, or even a Fourier-Stieltjes series, then u E h 1• Hence, by Kolmogorov's theorem, v E hP for all p < 1, and (by Theorem 2.2) v(z) has a radial limit a.e. The function

cp(8) =lim v(rei 6) r __, 1

is called the conjugate function of rp(8). Thus the conjugate series of the Fourier series of rp is Abel summable almost everywhere to cp(8), and cp E LP for all p < 1. Must cp be integrable? If so, then v E h 1 (Theorem 2.11), and v(z) is the Poisson integral of cp; thus (8) is actually the Fourier series of cp. Conversely, if (8) is the Fourier series of some integrable function, that function must be cp. Thus cp E L 1 if and only if the conjugate series (8) is itself a Fourier series, or if and only if v(z) E h 1• If cp is not integrable, then the conjugate series is not even

4.5 THE CONJUGATE OF AN h' FUNCTION

63

a Fourier-Stieltjes series (by Theorem 1.1). An example constructed in Section 4.5 will show that this can actually happen. If

2_ J2 " 2n o

cot( 8 - t)
Unfortunately, this last integral need not converge, because of the singularity at t The formula is correct, however, if the integral is interpreted in the Cauchy principal-value sense. That is, after changing variables and extending

=e.

J"

cp(8) =lim -21 cot(tf2)[
+ t)] dt

a.e.

We omit the proof, since we shall make no use of this formula. Proofs may be found in most books on trigonometric series; for example, in Bary [I], Vol. 2, pp. 60-62. 4.5. THE CONJUGATE OF AN h' FUNCTION

If u(z) and its conjugate v(z) both belong to h1 , then u(z) is a Poisson integral (Theorem 3.1); that is, the function 11-(t) in its Poisson-Stieltjes representation must be absolutely continuous. The converse is false. Even if u(z) is a Poisson integral, v(z) need not be of class h1 • We are about to present a counterexample. In view of Hardy's inequality (Corollary to Theorem 3.15), the function oo

'f(z)

z"

=I -¢ H n=21og n

1•

However, its real part belongs to h 1 , and is in fact a Poisson integral. To see this, it is enough to observe that

f

cos nt n=2 log n

64

4 CONJUGATE FUNCTIONS

is the Fourier series of an integrable function. But this is a particular case of the following more general proposition. THEOREM 4.5. If the real sequence {a0 , a 1 ,

.•. }

is convex and a.-+ 0 as

n -+ oo, then the series (10)

converges for all t # 0 in [ -n, n] to a nonnegative integrable function q>(t), and (I 0) is the Fourier series of q>(t). PROOF. Let A a.= an+l- a. and A2 a"= A(A a"). The convexity of {a"} means that A 2 a.~ 0; thus {A a"} is an increasing sequence. Since {a.} converges, A a.-+ 0; hence A a. :s;; 0. Since a.-+ 0, it follows that a.~ 0. Let SN(t) denote the Nth partial sum of (10). After two summations by parts, we find N

SN(t) =

L (A2 a.)(n + 1)K.(t)

n=O

(11)

where 1 N sin(N + -!)t DN(t) = -2 + L cos nt = 2 . 12 n= 1 sm t

and 1

KN(t) = N

N

+ 1 n~O

D.(t)

sin 2 (N + l)t/2 1) sin 2 (t/2)

= 2(N +

are the Dirichlet and the Fejer kernel, respectively. For each t # 0, the sequences {DN(t)} and {NKN_ 1(t)} are bounded; hence the last two terms in (11) tend to zero. Consequently, 00

SN(t)-+ q>(t) =

L (A 2 a.)(n + 1)K.(t).

(12)

n=O

Because (n + l)K.(t) is uniformly bounded in each set 0 < {J :S iti ~ n, the series (12) converges uniformly there, and so represents a function q>(t) continuous for t # 0. Since the terms of the series are all nonnegative, so is q>(t). By the Lebesgue monotone convergence theorem,

f,

q>(t) cos mt dt =

Jo

(A 2 a.)(n

+ 1) f,K"(t) cos mt dt

00

= n L (A 2 a.)(n- m + 1), n=m

m

= 0, 1, 2, ....

4.6 THE CASE p < 1 : A COUNTEREXAMPLE

65

This last series may be evaluated through summation by parts: k

L (A

k

LA an+ (k- m + 1) A ak+

an)(n- m + 1) = -

2

n=m

1

+ (k - m + 1) A ak+ 1 • the right-hand side approaches a,., and so =am- ak+ 1

Ask.-. oo,

J" rp(t) cos mt dt = am'

-1 1T.

m = 0, 1, 2, ....

-n

[In particular, rp(t) is integrable.] Thus (l 0) is a Fourier series, which was to be proved. The fact that k A ak-+ 0 follows from a classical theorem of Abel: If {bd is a monotonically decreasing sequence and L bk converges, then kbk-+ 0. For a proof, observe that 2"

2nb2" =::; 2 k=

4.6. THE CASE p

< 1:

L2"-

bk · 1

A COUNTEREXAMPLE

For 1 < p < oo, the class hP is preserved under conjugation. This is false for h1, but it is "almost true" in the sense that the conjugates of h 1 functions belong to hP for all p < 1. If the hypothesis is further weakened by requiring only that u E hP for some p < 1, hardly a trace of the theorem remains. The conjugate function v may not belong to hq for any positive q. In fact, we are about to construct an analytic functionf(z) = u(z) + iv(z) such that u E hP for allp < 1, yet v rt hP for any p > 0. The example is co

J(z) = u(z)

+ iv(z) =:Len n=1

z2"

1-z

(13)

zn+l,

We are going to show that for every choice of the signs en, u E hP for all p < 1; while for" almost every" sequence of signs,f(z) has a radial limit on no set of positive measure. In particular, some choice of the e. gives a function f(z) which is not even of class N, but whose real part belongs to hP for all p < 1. LEMMA. For each p

f" (1 -n

> t,

dO

2r cos

e + r2 )P =

( 1 ) 0 (1 - r) 2 P 1

as

r-+ I.

66

4 CONJUGATE FUNCTIONS

PROOF.

~

Since sin x

(2/n)x for 0 :-:::; x :-:::; n/2,

I - 2r cos()+ r 2 =(I - rf + 4r sin 2 ()j2 ~(I - r) 2 + (4rfn 2)()2. Hence, for r

~

t, the integral does

f-x" [(1- r)2 d()+ 2n

not exceed 1

The last integral is convergent because p >

dt

JCX)

2()2]p < (1- r)2P

1

-oo [1

+ 2n-2t2]p"

f.

In showing u E hP for all p < I, we may suppose p > begin by computing

t.

With z = rei 6, we

e{ z 2 " } _ Rn(l - Rn 2 ) cos 2n() R 1 - z2 "+' - 1 - 2Rn 2 cos 2n+ 1 () + Rn 4 ' From this we find, using the lemma, 2>r

f

CX)

iu(re;6 )1P d() :-:::;

0

d()

2><

L= 1R/(1- R/)P J0

n

(

1 - 2Rn

2

CX)

:-:::;A

COS (} CX)

L R/(1- R/)

1

-p :-:::; 2 1 -pA

n=1

+ Rn

4

)P

L R/(1- Rn) 1 -p, n=1

where A is a constant. But R/(1- Rn) 1 -p < r 2 "P2n< 1 -P>(1- r) 1 -P,

since I- rm =(I - r)(I + r + · · · + rm- 1 ) < m(I - r). Furthermore, 1- rP

~ =

p + 0(1- r)

(r-+ 1),

by the binomial expansion of rP =[I -(I - r)]P. We will have shown u E hP, then, if we prove CX)

L 2n'"r2 " =

IX> 0.

0((1 - r)-'"),

n-1

It is equivalent to estimate the integral

fo 2xar 2"' dx OCI

1 log 2

= --

fOCI

eulogru«-l

du

(u = 2"').

1

Since log r < r - I, the second integral is majorized by

which proves (14). Thus u E hP for all p < I.

(14)

EXERCISES

67

To prove the nonexistence of radial limits we appeal to the theory of Rademacher functions, as developed in Appendix A, and to Theorem A.4 in particular. For the present application we have

z 2" 9n(z)= 1 2n+t· -z Thus co

L lgn(z)l n~N

co

2

L R/(1- 2R/ cos 2n+le + Rn

=

4 )- 1

n~N

1

>-

co

L

- 4 n=N+ 1

r2"-+ oo

as r-+ 1. Since the other hypotheses of Theorem A.4 are obviously satisfied, we conclude that for almost every choice of signs, the analytic function (13) has a radial limit almost nowhere, which is what we wanted to prove. EXERCISES

1. Show that if cp E U (1 < p < oo) and co

cp( t) "'

L

an eint

n==- oo

L:'=

then its "analytic projection" f(z) = 0 an zn belongs to H P, and IIJIIP::;:; APIIcpiiP for some constant AP independent of cp. 2. Show that the Poisson kernel P(r, 0) does not belong to hP for any p > 1.

3. Show that the constant BP =(sec pnf2) 1fP is best possible for the special case of Kolmogorov's theorem in which u(z) > 0. (Hint: Try the Poisson kernel.) 4. Use Green's theorem to prove Kolmogorov's theorem with 2lfp-1(1- p)-lfp.

5. Letf(z) = u(z) + iv(z) be analytic in lzl < l, and suppose u p, 1 < p ::;:; 2. For fixed e > 0, set G(z) = {[u(z)] 2

Show that

+ e}Pf2 ;

H(z) = {lf(z)l 2

E

+ e}P 12 •

BP =

hP for some

68

4 CONJUGATE FUNCTIONS

Apply Green's theorem, integrate, and let

Mp(r,

f)~

E-+

0 to obtain

p ) 1/p (p _ Mp(r, u), 1

thus proving the theorem of M. Riesz with a smaller constant. (This idea is due toW. K. Hayman. It is an open problem to find the best possible constant, for p = 2, while the best even for u(z) > 0. The above constant reduces to possible constant in this case is I.)

J2

6. For the example of Section 4.6, show that there is a choice of signs such that

NOTES

Theorem 4.1, or an equivalent form of it, is in the paper of M. Riesz [1 ]. The proof based on Green's theorem, as presented in the text, is due to P. Stein [1]. This approach has the advantage of leading to a relatively good value of the constant AP. Calderon [1] has given still another proof; see also Zygmund [4], Chap. VII. Shortly after Kolmogorov [1] proved Theorem 4.2, Littlewood [2] suggested the proof using his subordination theorem. Hardy [3} then discovered the elementary argument given in the text as a second proof. Theorem 4.3 and the converse results (Theorem 4.4 and its corollary) are due to Zygmund [1, 4]. Another proof is in a paper of Littlewood [3]. Theorem 4.5 and similar results may be found in Zygmund [4], Chap. V. Phenomena of the type illustrated in Sec. 4.6 are discussed in the paper of Hardy and Littlewood [6]. Previously, Littlewood [1] had constructed a harmonic function which belongs to hP for all p < 1, yet has a radial limit almost nowhere. The example (13) given in Sec. 4.6 is essentially due to Hardy and Littlewood [6], who showed by a highly nonelementary argument that a function similar to this (but with En== I) fails to have a radial limit on some set of positive measure. Paley and Zygmund [2] introduced Rademacher functions and "constructed" an example of the type (13) for which/¢ HP for all p < 1. The argument given in the text is considerably simpler than theirs. Hardy and Littlewood [6] also proved that if u E hP for some p ~ 1, then its conjugate v satisfies r-+ 1.

For p = 1, }, !, ... , they showed by an elementary example that this estimate is best possible. Whether or not it can be improved for other values of p

NOTES

69

remains an open question, although Swinnerton-Dyer [I] has shown that it cannot be improved to 1-). Mp(r, v)= o(log1- r (Unfortunately, Hardy and Littlewood stated the positive result incorrectly in the introduction to their paper, and Swinnerton-Dyer reproduced this error.) Gwilliam [1] simplified some of the work of Hardy and Littlewood in this area.

This Page Intentionally Left Blank

MEAN GROWTH AND SMOOTHNESS

CHAPTER 5

If f(z) is analytic in the unit disk, there is a very close relation between the mean growth of the derivative f'(z) and the smoothness of the boundary functionf(e; 6). This principle takes several precise forms, as we shall see in the present chapter. Other topics to be discussed are the relations between the growth of Mp(r,f) and Mp(r,f'), between the growth of Mir,f) and Mq(r,f), and between the growth and smoothness of a harmonic function and its conjugate. Some of these results will be applied repeatedly in later chapters. We shall begin by discussing several ways to measure smoothness and exploring the connections between them. 5.1. SMOOTHNESS CLASSES

Let rp(x) be a complex-valued function defined on - oo < x < oo and periodic with period 2n. The modulus of continuity of rp is the function w(t) = w(t; rp) =

sup !rp(x)- rp(y)!. !x- Yl :51

Thus rp is continuous if and only if w(t)-+ 0 as t-+ 0. We say that rp is of class

72

5 MEAN GROWTH AND SMOOTHNESS

A,. (0 < oe::;; I) if w(t) = O(tj as t-+ 0. Alternatively, A,. is the class of functions which satisfy a Lipschitz condition of order oe: iq>(x)- q>(y)l ::;; A l(x- YW·

The definition is of no interest for oe > I, since A,. would then contain only the constant functions. It is clear that Ap c A,. if oe < p. A continuous function q>(x) is said to be of class A* if there is a constant A such that (I) Iq>(x + h) - 2q>(x) + q>(x - h)l ::;; Ah for all x and for all h > 0. The condition (I) alone does not imply continuity;

indeed, it is well known that there exist nonmeasurable functions such that q>(x

+ y) = q>(x) + q>(y).

For any oe < I, the proper inclusions AI cA. c A,.

hold. In fact, every q> E A* has modulus of continuity ro(t) = O(t log 1/t), as we shall see later. (See also Exercise 14.) It often turns out that A., not A1, is the "natural limit" of A,. as oe-+ I. Another class, apparently larger than A,., could be defined by requiring that the left-hand side of (I) be O(h"). However, this class is actually the same as A,. if oe < I. For q> E £P = LP(O, 2n), I ::;; p < oo, the function wp(t) = wp(t; q>) = sup {t"lq>(x +h)- q>(x)l Pdx}l/p O
o

is called the integral modulus of continuity of order p. For every £P function q>, wp(t)-+ 0 as t-+ 0, by a theorem of F. Riesz (see, for example, Titchmarsh [I], p. 397). If wp(t) = O(t"), 0 < oe::;; I, we say that q> belongs to the class A/. Because the LP means increase with p, it is evident that A,.q c A/ if p < q. If q> is continuous, wP(t)-+ w(t) asp-+ oo. The classes A/ may therefore be viewed as generalizations of A,.. Obviously, A,. c A/. In the proof of Theorem 5.4 we shall need the following result, which is of some interest in itself. LEMMA 1 (Hardy-Littlewood). If q> is of bounded variation over [0, 2n], then q> E A1 1 • Conversely, every function q> E A 11 is equal almost everywhere to a function of bounded variation. PROOF. Suppose first that q>(x) is of bounded variation, and let V(x) be the total variation of q> on [0, x]. Then for small h > 0,

5.1 SMOOTHNESS CLASSES

2x

73

2x

J i
0

2x+h

h

2x

0

The converse is more difficult. Assuming

=

r

E A1 1 ,

let


0

and
+ lfn)- F(x)].

Then 0 we have

IPn(x +h) -
E

[
+~+

t)

-
+ t)]

dt.

A 1 1, it follows that

S:xiiPn(x

+ 11) -
n

s: S:x i
-
+ t)i

dx

::::; Ch,

where the constant C is independent of n. Thus, by Fatou's lemma

Now let 0 = x 0 < x 1 < · · · < xm = 2n be an arbitrary finite partition of the interval [0, 2n]. Bearing in mind that
k~ 1 1
t

2x

IPn(xk-t)l::::;

IIP.'(x)l dx::::; C.

But
L i

(2)

k:l

It can now be shown from (2), by imitating the classical argument (see, for example, Rudin [7]) that
74

5 MEAN GROWTH AND SMOOTHNESS

5.2. SMOOTHNESS OF THE BOUNDARY FUNCTION

Generally speaking, it is reasonable to expect an analytic function to be smooth on the boundary if its derivative grows slowly, and conversely. For the unit disk, this principle can be expressed in surprisingly precise form, as follows. THEOREM 5.1 (Hardy-Littlewood). Let f(z) be a function analytic in lzl < 1. Then f(z) is continuous in lzl ::;::; 1 and f(ei 6 ) E A~ (0 < oc::;::; 1), if and only if

PROOF. First let us dispose of the case oc = 1. By Theorem 3.11, the Lipschitz condition on the boundary implies that f' E H 1 and that f'(ei 6 ) E L<XJ. Hence/' E H<XJ, by Theorem 2.11. Conversely, iff' E H<XJ, another application of Theorem 3.11 shows thatf(z) is continuous in lzl::;::; l,f(ei 9) is absolutely continuous, and (dfd0)f(ei 6 ) E L <XJ. Integration of the derivative therefore gives f(eie) E A,. Now letf(z) be continuous in lzl::;::; 1, and supposef(ei9) E A~, 0 < oc < 1. By the Cauchy formula,

,

-

1

f (z) -2n

J2" [f(ei')- f(eie)]eit dt o

Thus

( eit - z )2

z

'

= rei 6•

J"

1 lf(ei
(3)

Using the Lipschitz condition and the relation 1- 2r cost+ r 2

= (1- r) 2 +

. t 4rt 2 4r sm 2 2' ~ (1- r) 2 + 7 ,

0 :::; t:::; n, we find A

"

If (z)l:::; 2n f_, (1 I

IW dt - r) 2

+ 4r(tfn)2 = 0

( .(1 -

r) 1 -~

1

The assumption oc < 1 assures the convergence of the integral

f<XJ 1u"'+duu o

which arises after the substitution u Conversely, suppose

2'

= tf(l - r).

) •

5.2 SMOOTHNESS OF THE BOUNDARY FUNCTION

lf'(z)l

$; (

l

_~)I

75

O
-a •

Then the radial limit

J f'(rei R

f(ei 6 )

= f(O) +lim R-+1

6)

dr

0

exists for every (). Furthermore, /E H 00 , so f(z) is the Poisson integral of f(ei 6). Hence the continuity of f(e; 9) would imply the continuity of f(z) in lzl $; I. We shall prove the continuity by showing f(ei 0) E 1\. For this purpose, choose () and cp with 0 < cp - () < I. Fix p, 0 < p < I, and let r be the contour consisting of the radial segment from ei 6 to pei 6 , the arc of the circle lzl = p from pei 6 to pei"', and the radial segment from pei"' to ei"'· (See Fig. 2.) Then

pe ''P

Figure 2

f(ei"')- f(e;e) =

Jf'
Breaking up the integral into its three components, we find 1

lf(ei"')- J(ei9)1

$;

1

rp

JIJ'(rew)l dr + JIJ'(rei"')l dr + J IJ'(pei')l dt p

p

5

dr $; 2C P (l _ r)l-a 1

C(cp - ()) + (l _ p)1 a·

0

76

5 MEAN GROWTH AND SMOOTHNESS

With the choice p

= 1-

(qJ - (}), this gives

if(e 1
j(e'~l ~

c( 1 + ~) i((J- Oia,

and the proof is complete. The method may be used to prove other theorems of the same type. We mention one example, leaving the proof as an exercise. THEOREM 5.2.

Letf(z) be analytic in lzl < 1, and suppose f'(z)

= a(log-1-). 1- r

Thenf(z) is continuous in izl ~ 1, andf(e 16) has modulus of continuity

w(t) = o(t log~)· According to Theorem 5.1 ,f' E H 00 if and only if f(e 16) E A1 . If the growth condition is slightly weakened to f"(z) = 0(( I - r) - t ), Theorem 5.2 says that f(e 16) must still be "almost" of class A1 : w(t) = O(t log lft). This degree of smoothness, however, does not ensure thatf"(z) = 0((1 - r)- 1 ). The necessary and sufficient condition is that the boundary function belong to the class A* introduced in Section 5.1. The precise result is as follows. THEOREM 5.3 (Zygmund). Let f(z) be analytic in lzl < l. Then f(z) is continuous in izl ~ 1 and f(e 16) E A* if and only if

f"(z)

=

a(-

1 )· 1- r

(4)

If f(z) is analytic in lzl < 1, continuous in izl ~ 1, and A* thenf(e 16) has modulus of continuity ro(t) = O(t log 1/t).

COROLLARY.

f(e 16)

E

PROOF OF THEOREM.

If f(z) is continuous in izl

~

1, it can be represented

as a Poisson integral: 1 " f(z) = 2 1l f_/(r, (}- t)J(e 1') dt,

Since the second derivative P 06(r, (})is an even function of(} and J"P 66 (r, t) dt 0

= P 6(r, n)- P 6(r, 0) = 0,

(5)

5.2 SMOOTHNESS OF THE BOUNDARY FUNCTION

77

it follows that fee(z) =

2~ S:P

The hypothesis f(e' 6 )

E

66

+ f(ei(e-t))}

(r, t){J(ei(e+r>)- 2J(ei6 )

dt.

A* therefore implies

lfee(z)l::.:;; A (tPeo(r, t) dt

= A[P(r, 0)- P(r, n)] =

oC ~ r)·

-.J3,

since P6 e(r, 8) 2:: 0 for 0::.:;; e::.:;; n and r 2:: 2 as a calculation shows. On the other hand, (5) and the boundedness ofj(e;') easily show fe(z) =

o(~)-

Thus f"(z) = r- 2 e- 2 ' 6 {ife(z)- fee(z)} =

oC ~ r)·

The proof of the converse is more difficult. In view of Theorem 5.1, the condition (4) implies f(z) is continuous in lzl ::.:;; I; it must be shown that f(ei 6 ) EA •. For 0 < h < I, let us use the notation llh G(O) = G(O +h)- G(O), llh 2 G(O) = G(O + 2h)- 2G(8 +h)+ G(O).

We are required to show that ll/ f(ei 6 ) = O(h), uniformly in 0, as h-+ 0. Our strategy is to write ll/ f(e;e)

= llh2{J(e;o)-

f(pe;e)}

+ ll/f(pe;e)

(6)

(0 < p < 1), to set p = I - h, and to show that as h-+ 0, each of the two terms in (6) is uniformly O(h). The identity

+ e2i6 J (1 1

J(eie)- f(pei9) = (1 - p)ewf'(peie)

- r)J"(rew) dr

(7)

p

is easily verified through integration by parts. Now set p = I -h. Under the hypothesis (4), the integral in (7) is then uniformly O(h). Thus llh 2{f(e' 6 ) - J(pei 0 )}

= h ll/{e 16J'(pe;6)} + O(h).

(8)

But llh{e;ef'(pie)} = llh{e'e}f'(pe'(6+1rl)

= uniformly in

+ e;e llh{f'(pe;e)}

o(h log~) + s: ipei6

ei(e+t)J"(pei(e+t)) dt = 0(1),

e. Hence the expression (8) is uniformly O(h).

(9)

78

5 MEAN GROWTH AND SMOOTHNESS

Finally, I1/J(pei(6-hl) = ip

J{ei(6+r>f'(pei(6+r>) _ ei(e-r>f'(pei(6-t))} dt. h

0

Analyzing the integrand as in (9), we see that it is dominated by et{log

k+ ~}.

where Cis independent of e. Hence 11h2 f(pe 16 ) is uniformly O(h), and the proof is complete. The next theorem may be viewed as the LP analogue of Theorem 5.1. It is included here for the sake of completeness; we shall make no use of it in this book. THEOREM 5.4 (Hardy-Littlewood). Let f(z) be analytic in /zl < I. Then /E HP andf(e 16) E A/ (l ::;p < oo; 0
Mp(r, f') =

o((l- ~-)~- . )·

PROOF. We first show that the condition (10) for any

It is convenient to assume f(O)

(10)

a> 0 implies/E HP.

= 0, so that .r

J(re;e) =

J f'(pe;e)e;e dp. 0

Applying the continuous form of Minkowski's inequality, we find

t r

Mp(r, f) 5

t

~

r

Mp(p, f') dp 5 C

(l _ p)l-cz <

C

~;

hence/E HP. Now let us deduce from (l 0) that f(e 16 ) E A/. Let 0 < p < r < I and 0 < rp - 8 < 2n. Then J(re 1"')- J(re 16 ) =

Jr f'(O dC,

where the contour r goes radially from re 16 to pe 16 , along /zl = p to pe 1"', then radially to re 1"'. It follows that /J(ri"')- f(re 16 )/5 J'lf'(se16)/ ds p

+

(1f'(se 1"')/ ds

•p

+ J"'if'(pe1')1 dt. 6

5.3 GROWTH OF FUNCTION AND DERIVATIVE

79

Now choose h, 0 < h <-},and let cp = e + h. Applying Minkowski's inequality in both discrete and continuous form, we obtain (11)

The parameters r and p are still free. Assuming r > 1 - h, set p = r - h. Using the hypothesis (1 0), we conclude that for all r > 1 - h, the left-hand side of (11) is no larger than Ch~, where the constant Cis independent of r. Letting r-+ 1, it follows thatf(e;~ E A/. To prove the converse, we begin with the estimate (3) for lf'(z)l. If f(ei~ E A/, Minkowski's inequality gives

Mp(r, f')::;:; C

" J-" 1 -

2

IWdt r cost+ r

2"

Hence, as in the proof of Theorem 5.1, we obtain (10) for ex< 1. If ex= 1, it must be shown that fE HP and f(ei 6) E A 1 P imply f' E HP, 1::;:; p < oo. Consider first the case p = 1. Then fE H 1 and by Lemma 1, f(e;~ is equivalent to a function of bounded variation. Invoking Theorems 3.10 and 3.11, we conclude thatf(e; 6) is absolutely continuous,.(' E H 1 , and the radiallimitf'( ei 6) is obtained by differentiation of f(ei 6). If it is also known thatf(ei 6 ) E A1 P for some p > 1, then f(ew) IP f2" If(ei
It then follows by use of Fatou's lemma that the H 1 functionf'(z) has an U boundary function. Therefore f' E H P, by Theorem 2.11. 5.3. GROWTH OF A FUNCTION AND ITS DERIVATIVE

Our next object is to explore the relation between the mean growth of an analytic function and that of its derivative. We begin with a lemma which is especially useful in dealing with the case p < 1. LEMMA 2. Any function/E HP (0 < p::;:; oo) can be expressed in the form f(z) = h (z) + / 2 (z), where h and / 2 are nonvanishing H P functions such that 11/,IIP::;:; 211/IIP, n = 1, 2. PROOF. If f(z) "# 0, we may take h = / 2 = !f Iff has zeros, we apply Theorem 2.5 to write f(z) = B(z)g(z), where B is a Blaschke product and g is an HP function with no zeros. Thus

f(z) = [B(z) - l]g(z)

+ g(z).

80

5 MEAN GROWTH AND SMOOTHNESS

< p :s; oo and {3 > 0, and let/(z) be analytic in

THEOREM 5.5. Suppose 0

lzl
if and only if (13) PROOF.

First let 1 :s; p :s; oo, and assume that (12) holds. By the Cauchy

formula,

f

I

_1_ (z)- 2 . -

J

JW r

d')2 -£ J2" - 2

nz 1'/=p(.,-z

f(pei(r+Bl)ei(r-6)

no

(

,, )2 pe-r

dt,

where p =!(I + r). Minkowski's inequality (in continuous form) then gives 1 Mp(r, f) :s; 2n I

J

2"

0

p2

-

Mp(p,f)dt 2pr .cost+ r 2

_ M p(p, f) _ O ( -

p2- r2 -

1 ) (1- r)ll+l .

Conversely, if (13) holds, we apply Minkowski's inequality to the relation

and obtain

We remark that the proof could also have been based on Jensen's inequality. Suppose now that 0 < p < 1 and that

Mp(r, f)

s

c

(1 - r)p·

Under the preliminary assumption that f(z) =I= 0 m lzl < 1, the function F(z) = [f(z)JP is analytic and

5.3 GROWTH OF FUNCTION AND DERIVATIVE

With p =~(I

+ r),

81

this implies , M1(p, F) 4CP M1(r,F)~ pz - r 2 ~( 1 - r)IIP+1'

On the other hand, f'(z)

= ! [F(z)] 11P- 1F'(z); p

thus HOlder's inequality gives

M p(r,

f') ~ ~ {M 1(r, F)} 11 p- 1M 1( r, F') ~ (/~~~+ 1 •

If f(z) has zeros, we fix p (0 < p < 1) and appeal to the lemma to write f(pz) = ft(z)

+ fz(z),

where ft and / 2 are nonvanishing H P functions such that

n = 1, 2, and the constant C is independent of p. Hence, by the result just obtained,

, < (8/p)C Mp(r, fn)- (1 - p)/1(1 - r)'

n = 1, 2.

Since it follows that

z

M p(p ,

'

K

f ) ~ ( 1 - p)II+ 1'

where K is independent of p. This easily gives (13). The proof that (13) implies (12) when 0 < p < 1 is much more difficult, and is omitted here (see Notes). The theorem says, roughly, thatf' grows faster thanfby a factor (1 - r)- 1 • The same phenomenon is reflected in the following more delicate result, which will be needed later in this chapter. THEOREM 5.6. Let f(z) be analytic 1 < a < oo, and -1 < b < oo,

In

lzl < 1. Then for 1 < p < oo,

J:(l- r)b{Mp(r, JW dr ~ c{f:o- rt+b{Mp(r, f'W dr + lf(OW}. where C is a constant independent off

82

5 MEAN GROWTH AND SMODTHNESS

First suppose fis analytic in the closed unit disk, and integrate by

PROOF.

parts: 1

J (1 -

r)b{Mp(r,

JW dr

0

= b

+1 1 1J(O)Ia + b +1 1

J(1- r)b+ 1

1

0

a

ar {Mp(r, f)Y dr.

(14)

But

a a a -a {Mp(r, !W = - {Mp(r, f)y-p a- {Mp(r, f)}P, r p r and a calculation based on the Cauchy-Riemann equations gives

I:r If(re; IP I: ; pI f(rei 6)

6)

lp- 1 If'(rei 6) 1.

Thus it follows from Holder's inequality that

I:r {Mp(r, nYI::;; a{Mp(r, f)}a-

1 Mp(r,

f').

Consequently, the integral on the right-hand side of (14) is dominated by 1

a

J (1- r)b+

1

{Mp(r, f)}a- 1 Mp(r, f') dr

0

where Holder's inequality has been used again. This easily gives the desired inequality. If f(z) is not analytic in lzl ::;; 1, replace it by f(pz), where 0 < p < 1, and let p tend to 1 to obtain the result by means of the Lebesgue monotone convergence theorem. 5.4. MORE ON CONJUGATE FUNCTIONS

We saw in Chapter 4 that the class hP is preserved under harmonic conjugation if 1 < p < oo, but not if 0 < p::;; 1 or if p = oo. Nevertheless, the symmetry is restored in these latter cases if instead of the boundedness of the means one considers their order of growth. We shall confine attention to 1 ::;; p::;; oo. The theorem remains true for 0 < p < 1, but the proof is much more difficult.

5.4 MORE ON CONJUGATE FUNCTIONS

THEOREM 5.7.

Let f(z)

Mp(r, u) =

= u(z) + iv(z)

be analytic in

a((l ~ r)fl ).

1~ p

~

oo,

83

lzl < 1, and suppose {3 > 0.

Then

PROOF. For 1 < p < oo, we could apply the theorem of M. Riesz, but we shall give a proof which makes no appeal to this deeper result. Let p = 1{1 +r), and express f(z) by the Poisson formula:

1 2" peir + z . f(z) = ir u(pe'1) dt 2n o pe - z

J

+ iy.

Then 1 2" iu(peHB+rl)l dt lf'(re; 6)1 ~2 n o p - 2pr cos t + r 2

J

.

By Minkowski's inequality, M ( P

r'

f') < ~ - n

f21t 0

p2

-

Mp(p, u) dt 2pr cos t + r 2

_ 2Mp(p, u) _ - p2 _ r2 -

a(

1 ) ( 1 _ r)/1+1 ·

Theorem 5.5 now gives the desired conclusion. Generally speaking, then, a harmonic function and its conjugate have the same rate of growth. We are now in a position to show that they also have the same degree of smoothness on the boundary. THEOREM 5.8. Let f(z) = u(z) + iv(z) be analytic in lzl < 1, and suppose u(z) is continuous in lzl ~ 1. If u(ei6) E A~ (ex < 1), then v(z) is continuous in lzl ~ 1 and v(ei 6) E A~. If u(e;6) E A*, then v(ei'1 E A*.

PROOF. If we represent f(z) as a Poisson integral of u(e; 1) and follow the proof of Theorem 5.1, we see that the weaker condition u(e 16 ) E A~ still implies f'(z) = 0((1 - rY- 1J- Thus f(ew) E A~, by Theorem 5.1. To show that A* is

also preserved under conjugation, we may express u(z) as the Poisson integral ofu(ei 1) and follow the proof of Theorem 5.3 to see that u 6 o(z) = 0((1 - r)- 1).

84

5 MEAN GROWTH AND SMOOTHNESS

Theorem 5.7 then implies v99(z) = 0((1 - r)- 1), so thatf"(z) = 0((1 - r)- 1). Hence f(e; 9) E A*, by Theorem 5.3. COROLLARY. Every function q>(x) of class A* has modulus of continuity w{t) = O(t log lft). PROOF. Let u(z) be the Poisson integral of the given function q> E A*, and let f(z) = u(z) + iv(z) be its analytic completion. Then f(z) is continuous in lzl :::;; I, and f(ei 9) E A*. The result therefore follows from the special case already noted as a corollary to Theorem 5.3.

5.5. COMPARATIVE GROWTH OF MEANS

If f(z) E H P (0 < p < oo ), it is possible to give a sharp estimate on the growth of Mq(r,f) for any q > p. This theorem has a number of interesting applications. The proof will make use of the following lemma, which can be established by essentially the same argument used to prove the lemma in Section

4.6. LEMMA 3. If a> I and p

t

= !(! + r), then

2"

lpeit- ri-a dt

= 0((1

- r) 1 -a),

r-+ 1.

THEOREM 5.9 (Hardy-Littlewood). Let f(z) be analytic in lzl

< I, and

suppose 0
00,

p ; : : 0.

= K(p, p) independent off such that . KC Mq(r, f):::;; (1 - r)P+ 1fp 1/q, p < q:::;; oo.

(15)

Then there is a constant K

If (p

p = 0 (i.e., if /E HP), then Mq(r,f) = + 1/p- lfq) is best possible in all cases.

(16)

o((l - r) 1fq- 1fP). The exponent

PROOF. Let us first observe that it suffices to consid.er the case q = oo. Indeed, suppose that (16) has been proved for q = oo, and suppose for convenience that K;;::: 1. Then for p < q < oo,

Mq(r, f)= {21n

s:"lf(re;8)1Pif(rei~lq-p dOtq

:::;; {M 00 ( r, !)} 1-pfq{Mp( r, J)}Pfq:::;; ( 1 ~~).1.,

5.5 COMPARATIVE GROWTH OF MEANS

85

where )c

= {lpjq + ({3 + 1/p)(l - pfq) = {3 + 1/p - 1/q.

Similar remarks apply to the "o" part of the theorem. Hence we may confine attention to the case q = oo. Suppose now that 1 < p < oo, andletp'betheconjugateindex: 1/p+ 1/p'= 1. By the Cauchy formula,

J2"

- p f(peit)eir f(z) - it dt, 2n o pe - z

r
Applying Holder's inequality and the hypothesis (15), we find

J

. c {1 2 " dt }l{p. lf(re'o)l ::;:; (1 - p)fl 2n o ipeir- reiolp" We now set p = -}(1 + r) and use Lemma 3 to obtain (16) for q = oo. The proof for p = 1 is similar but easier. Turning now to the "o" assertion, let f E HP, 1 ::;:; p < oo. Given e > 0, there exists (j > 0 such that

f

,j

lf(ei(O+r)w dt <

for all 8.

Bp

-ol

On the other hand, the Cauchy formula gives .

1

J"

f(re' 6 ) = 2n -"

f(eHo+rl)eir 1 {Jo~ ir dt = e - r 2n -o~

+ J" + Jo~ }. o~

-"

For each fixed f>, the last two integrals are obviously bounded as r--. 1. If p > 1, the integral over ( -f>, f>) is dominated by

{S:)f(8 + t)IP dt} lfp{f" lei'~ riP·} lfp" < (1 ::)lfp' where the lemma of Section 4.6 has been used. Similarly for p = 1. Hence f(z) = o((l - r) -liP), which is what we wanted to show. It remains to deal with the case p < 1. If f(z) #- 0 in lzl < 1, the function F(z) = [f(z)]P is analytic, and (15) gives

CP M 1(r, F)= {Mp(r, f)Y::;:; ( 1 _ r)liP. Thus, by what we have already proved,

KCP

lf(z)IP

= IF(z)l::;:; ( 1 _ r)flp+l,

86

5 MEAN GROWTH AND SMOOTHNESS

which is the desired result. The same argument showsf(z) = o((l - r)- 1fP) for nonvanishing HP functions. lf/(z) has zeros, we fix p < 1 and use Lemma 2 to write

where / 1 and /

2

do not vanish and

n = 1, 2. Since J.(z) "# 0, it follows that •a

•a

lf(p 2 e'")l::;; lf1(pe'")l

•a

4KC

+ lf2(pe'")l::;; ( 1 _

p)fl+1fp'

The constant K is independent of p. Finally, if f E H P, the general result f(z) = o((l - r)_ 1,P) follows directly from Lemma 2. The functionf(z) = (1 - z)-Y, for suitable y > 0, shows that the exponent ({3 + lfp- lfq) cannot be improved. We leave the details to the reader. The "o" assertion can also be reduced to the special case p = 2, which is easily proved as follows. Let f(z) = a. z" be in H 2., and fix N so that

L

Then

IJ(z)l::;; J~oa.z·J + L=~+11a.12f'l=~+/2"f'2 ::;; 0(1)

+ e(l- r)- 1 ' 2 ,

which shows thatf(z) = o((l - r)- 112 ). It is natural to ask whether Theorem 5.9 has a converse. In particular, if p < q and Mq(r,f)-> oo sufficiently slowly, must/be in HP? Unfortunately, no such thing can be true, as the following theorem shows. THEOREM 5.10. Let 1/J(r) be continuous and increasing on 0::;; r < 1, with 1/1(0) = 1 and 1/J(r)-> oo as r-> 1. Then there is a function f(z) analytic in izl < 1, satisfying lf(z)i ::;; 1/J(Izl), yet having a radial limit on no set of positive measure. In particular, f(z) is not of class HP for any p > 0.

PROOF. The proof rests upon the following elementary observation.

5.5 COMPARATIVE GROWTH OF MEANS

87

LEMMA 4. Given a function t/J(r) as in Theorem 5.1 0, there is a sequence of integers 0 < n1 < n2 < · · · such that 00

L rn• ~ t/J(r),

O~r<

1.

k=1

Momentarily taking the lemma for granted, let 00

f(z)

=

L ek znk, k=1

Then 1/(z)l ~ t/J(r), and it follows from Theorem A.5 (in Appendix A) that for almost every choice of signs {en},f(z) has a radial limit almost nowhere. This proves the theorem. PROOF OF LEMMA. Let r0 = 0, and let r 1 , r 2 , ... be determined by t/J(rk) = k + l. Then 0 < r 1 < r2 < · · · < l. Choose n1 such that ri 1 ~ -!. Generally, select nk > nk_ 1 such that r;:• ~ rk. Given r in [0, 1), lets be the index for which r._ 1 ~ r < r •. Then co

oo

s-1

oo

L rn• ~ k=1 L r:• = k=1 L r:• + L r:• k=1 k=s

00

~(s-1)+ L';:·~(s-1)+r•+ 1 k=s

Although Theorem 5.9 is in a sense best possible, it can be slightly sharpened in one direction to produce a very useful result. THEOREM 5.11

(Hardy-Littlewood). If O
feHP, A.";;::.p,

and ex = 1/p - ljq, then

J(1- r);.~- 1 {Mq(r, f)};. dr < oo. 1

0

PROOF. It suffices to take A. = p, since by Theorem 5.9, 1

f (1- r);.~- 1 {Mq(r, f)};. dr ~ 0

f (1- ,y~- 1 {Mq(r, fW dr. 1

c;.-p

0

Next, suppose the theorem were proved for A. (0

= [f(z)]P 12 e H 2,

= p = 2.

Then if

f

eHP

88

5 MEAN GROWTH AND SMOOTHNESS

and the desired result (with A.= p) would follow. The general theorem could then be deduced by writing jE HP as the sum of two non-vanishing HP functions, as in Lemma 2. Hence the theorem reduces to the special case A.= p = 2. To prove this, suppose first that 2 < q < oo, and let

f(z)

"' anzn E =L

H 2•

n=O

Then by Theorem 5.6,

((1- r)- 21q{Mq(r, f)V dr

:5:

c{la 0 12 +

((1- r) 2- 21q{Mq(r, f')} 2dr}. (17)

But it follows from Theorem 5.9 that

M q( r, f') :5: K(1 - r)1fq-1f2 M z(r1f2' f'), so the second integral in (17) is dominated by a constant multiple of

J(1 1

r){M 2(r 112 , f')V dr =

0

L n 2 lanl 2 J(I "'

n=1

=

1

r)r"- 1 dr

0

n L"' - lanl n+1

2

<

00.

n=1

Finally, the result for q = oo is easily deduced from the case q < oo by Theorem 5.9. 5.6. FUNCTIONS WITH HP DERIVATIVE

According to Theorem 3.11, every function which has an H 1 derivative is continuous in the closed disk and absolutely continuous on the boundary. In particular, f' E H 1 implies /E H"'. This latter result has an interesting generalization.

f' E HP for some p < l, then - p). For each value of p, the index q is best possible.

THEOREM 5.12 (Hardy-Littlewood). If

jE Hq, where q

= pf(l

For convenience, assume f(O) = 0. Case 1: t :5: p :5: l. Since q;::::: l, Minkowski's inequality may be applied (as in the proof of Theorem 5.5) to give PROOF.

Mq(r, f) :5: rMq(s, f') ds. 0

5.6 FUNCTIONS WITH HP DERIVATIVE

89

But iff' E H P and q = p/(1 - p), it follows from Theorem 5.11 that 1

J Mq(s, f') ds < oo. 0

Case II: p <

t.

Fix ex in the range lf2p <ex< lfp;

then ex > 1. In view of Lemma 2, we may assume that f'(z) has no zeros in izl < 1. Letting g(z) = [f'(z)] 11a, we have lf(rew)l

:$;

{ig(sei6Wds

:$;

[G(8)]a- 1h(r, 8),

0

where

G(8)

=

sup ig(rei 6)1 O
and

Hence

Now set {3 = ex(1 - p) > 1, ex-1

and let {3'

= {3/({3 -

1) be the conjugate index. By Holder's inequality,

{Mq(r,

!W :$; IIGII::'11 { 21,.

J:" [h(r, 8)]q

11 '

de} 1111 ·,

since (ex- l)qfl = exp. By the Hardy-Littlewood maximal theorem (Theorem 1.9), G EL«P. On the other hand, exp qfl' = - - > 1; 1- exp thus Minkowski's inequality gives { 21,.

J2" [h(r,8)]q 'de} 0

11

1fqfl'

:$;

J' Mq r(s,g)ds, 0

1

which remains bounded as r-> 1, by Theorem 5.11. ThusfE Hq.

90

5 MEAN GROWTH AND SMOOTHNESS

To show that q = p/(1 - p) cannot be replaced by any larger index, one need only consider f'(z) = (1 - z)'-lfp for small e > 0. EXERCISES

1. Letf(z) = L~o anz" be analytic in izl < 1. Show that if N

L n lanl = O(N 1-«), n:l

0 <ex::;:; 1,

then f(z) is continuous in lzl ::;:; 1 and f(ew) E A«. Show further that the converse is true if ex < 1 and an ;;::: 0, n = 1, 2, . . . . 2. Letf(z)

= L~o anz" be

analytic in izl < 1. Show that if

N

L

n lanl = O(log N), n=l thenf(z) is continuous in lzl ::;:; 1 andf(ei 6) has modulus of continuity w(t) = O(t log 1/t). 3. Show that if N

L n 2 1anl = n=l

O(N),

then f(z) is continuous in lzl ::;:; 1 and f(ei 6 ) true if an ;;::: 0, n = 1' 2, ....

E

A*. Show that the converse is

4. Show that ifj(z) is continuous in lzl ::;:; 1 andf(ei 6)

E

A«, 0 < ex::;:; 1, then

N

L n21anl2 = O(N2(1-«l).

n=l If f(ei 6 )

E

A*, show that N

Ln n=l

3

1anl 2 = O(N);

hence N

L n 2 1anl = O(N n=l

312 ).

(Duren, Shapiro, and Shields [1]). 5. Show that if m is an integer greater than 1, and if 0 < ex < 1, the function

f(z)

=

is continuous in izl ::;:; 1 andf(ei 6)

E

co

L m-k«zmk

k=l A«.

NOTES

91

6. Show that the first two estimates in Exercise 4 are best possible in the sense that "0" cannot be replaced by "o." (The third estimate is also best possible; see Duren, Shapiro, and Shields [1].)

7. Show that if f(z) is analytic in Jzl < 1, continuous in Jzl :$; 1, and f(ei'1 E A« for some ex> 1, then L:'=o Janl < oo. Show further that the index -1 is best possible: the series L:': 1 Iani may diverge if only f(e 1'1 E A112 (Theorem of S. Bernstein). 8. Show that iff(z) is continuous in lzl :$; 1 andf(e;'1 E A«, 0 <ex:$; 1, then 1 < oo for all y > 2/(2ex + 1). (Hint: Show that I, n11 Janl 2 < oo for all {3 < 2ex.)

h = 1 Jan I 00

9. Show that iff' E H P, 1 < p :$; oo, then f(z) is continuous in Jzl :$; 1 and f(e 16 ) E A«, ex = 1 - lfp. 10. Show that if rp(t) is continuous and Jrp(t +h)- 2rp(t)

+ rp(t- h)l :$; Ah«,

0 < ex< 1,

then rp

E A«. (Suggestion: Imitate the proof of Theorem 5.1.) 11. For 1 :$; p < oo, show that the class A/ contains only the constant functions if ex > 1. 12. Letf(z) = u(z) + iv(z) be analytic in Jzl < 1. Suppose u E hP (1 < p < oo) and u(e;'1 E A/ (0 < ex< 1). Show that v(ei 6 ) E A/. In other words, show that the class A/ is self-conjugate. 13. Prove that iff E H P (1 < p < oo) and f(ei'1 E A/ (1/p < ex :$; 1), then f(z) is continuous in lzl :$; 1 and f(ei 6 ) E A11 , {3 = ex - lfp. 14. Show that the function

rp(x)

= x log 1/x,

X

>0,

belongs to A* but not to A1 .

15. Prove Theorem 5.2. 16. Using the example f(z) = (1 - z)- 1 , show that the exponent ({3 + 1/p - lfq) in Theorem 5.9 is best possible. NOTES

Almost everything in this chapter is due to Hardy and Littlewood. Lemma 1 is in their paper [2). Theorems 5.1, 5.4-5.6, 5.9, and 5.11 occur in their paper [5]. The proofs of Theorems 5.5 and 5.7 for the case p < 1 may be found in their paper [6]. The simple but enormously useful remark that every HP function is the sum of two nonvanishing ones (Lemma 2) can be traced to their

92

5 MEAN GROWTH AND SMOOTHNESS

paper [1]. Further results and generalizations are in their papers [5] and [8]. The "o" growth condition given in Theorem 5.9 is best possible; see G. D. Taylor [1] and Duren and Taylor [1]. Flett [2] has recently based a proof of Theorem 5.11 on the Marcinkiewicz interpolation theorem. Zygmund [2, 4] introduced the class A* and proved Theorem 5.3 in somewhat different form. The self-conjugacy of A« (Theorem 5.8) goes back to Privalov [1]. Zygmund [2] showed that A* is self-conjugate. For a direct proof that every A* function has modulus of continuity which is O(t log 1/t), see Zygmund [4], Vol. I, p. 44. The converse to Theorem 5.12 is totally false. The "Bloch-Nevanlinna conjecture" asserted that if/EN, thenf' EN, but this has been disproved in many ways. In fact, there exist functions /E H 00 continuous in the closed disk, such that f'(z) has a radial limit almost nowhere. See, for example, Lohwater, Piranian, and Rudin [1] and Duren [4]. Hayman [3] has recently shown that f' E N does not imply f E N, an unexpected result in view of Theorem 5.12. Caughran [2] has obtained results comparing the canonical factorizations off and f' in case f' E H 1 •

TAYLOR COEFFICIENTS

CHAPTER 6

L

If a functionf(z) = anzn belongs to a certain HP space, what can be said about its Taylor coefficients an? Clearly, one can hope to describe only the "eventual behavior" of {an} as n-> oo, since any finite number of coefficients can be changed arbitrarily without upsetting the fact thatfis in HP. It is also interesting to ask how an HP function can be recognized by the behavior of its Taylor coefficients. Ideally, one would like to find a condition on the an which is both necessary and sufficient for fto be in HP. For p = 2, of course, the problem is completely solved: /E H 2 if and only if lanl 2 < oo. But the general situation is much more complicated, and no complete answer is available. If 1 < p < oo, the problem is equivalent to that of describing the Fourier coefficients of LP functions, as the M. Riesz theorem shows. This chapter contains some scattered information about the coefficients of HP functions. Curiously, the results are most complete in the case 0 < p < 1.

L

6.1. HAUSDORFF-YOUNG INEQUALITIES

Any available information about the Fourier coefficients of LP functions can be applied, in particular, to the Taylor coefficients of HP functions. In fact, the

94

6 TAYLOR COEFFICIENTS

two sets of coefficient sequences are essentially the same if 1 < p < oo. The Hausdorff-Young theorem states that if rp(x) e U = U[O, 2n], 1 < p ~ 2, then its sequence {c,} of Fourier coefficients is in tq (1/p + lfq = 1) and

Conversely, every tP sequence {c,} of complex numbers (1 < p ~ 2) is the sequence of Fourier coefficients of some rp e Lq (1/p + lfq = 1), and

This result may also be expressed in HP language, as follows. THEOREM 6.1. If 00

f(z) = then {a,} e

tq (lfp + 1/q =

L a,z" e HP n=O

(1

~

p ~ 2),

1) and

I {a,}llq ~ 11/llp. Conversely, if {a,} is any tP sequence of complex numbers (1 = a,z" is in Hq (lfp + 1/q = 1) and

f(z)

L

(1) ~

p ~ 2), then

11/llq ~ II {a,} liP.

(2)

PROOF. The first statement follows easily from the Hausdorff-Young theorem. Indeed, if fe HP, then the radiallimitf(ei9) is in U and {a,} is its sequence of Fourier coefficients. On the other hand, if {a,} e tP (1 ~ p ~ 2), then fe H 2 , and the numbers a, are the Fourier coefficients of f(ei 9 ). The Hausdorff-Young theorem then tells us thatf(e 19 ) e Lq, which implies f e Hq, and (2) follows.

Neither part of the theorem remains valid if p > 2. In fact, the hypothesis thatfe H P for some p > 2 implies nothing more about the Ia, I than {a,} e t 2 • And if the condition {a,} e t 2 is weakened to {a,} e tP for some p > 2, nothing reasonable can be said aboutf(z). These claims are justified by Theorem A.5 of Appendix A. According to this theorem, the mere assumption that {a,} et 2 implies that, for almost every choice of signs {e,}, the function e, a,z" is in HP for allp < oo. On the other hand, if {a,}¢ t 2 , almost every choice of signs produces a function e, a, z" having a radial limit almost nowhere.

L

L

6.2 THEOREM OF HARDY AND LITTLEWOOD

95

6.2. THEOREM OF HARDY AND LITTLEWOOD

The following theorem provides further information about the coefficients of HP functions. THEOREM 6.2 (Hardy-Littlewood).

If

co

f(z) = I, anzn

E

HP,

0< p::;:; 2,

n=O

{J (n + I)p- 1an1pf'p::;:; cpllfllp. 2

0

where

cp

(3)

depends only on p.

PROOF. We suppose 1 ::;:; p ::;:; 2, postponing the case 0 < p < 1 to Section 6.4. If p = 1, the theorem reduces to Hardy's inequality (Corollary to Theorem 3.15). If p = 2, it follows from the Parseval relation. A proof for intermediate values of p can be based on the Marcinkiewicz interpolation theorem, but the argument given here will be self-contained. Let p. be the measure defined on the set of integers by

p.(n) = (lnl + 1)- 2 ,

n

= 0, ± 1, ± 2, ....

If gEL and 2

co

g(t)""'

I.

bn elnr,

n=- oo

let g(n) =(In I + l)bn,

n

= 0, ± 1, ±2, ....

For s > 0, let £~

= {n: l§(n)l > s}.

Then ... co

s 2p.(E,)::;:;

co

I, lii(nWp.(n) = n=- oo

I, lbnl 2

= IIYII/.

(4)

n= -oo

Furthermore, since lhnl ::;:; 11911 1,

E, c F, = {n: (lnl + l)llgll1 > s}; thus p.(E,)::;:;p.(F,)= I,(lnl+l)- 2 ::;;cs- 1 ilgil 1. neF5

(5)

96

6 TAYLOR COEFFICIENTS

Now write

g(t) = q>,(t)

+ t/l,(t),

where

ig(t)i??:s lg(t)i < s. Then 00

00

I
cx(s) = .u({n: llP.(n)l > s}), P(s) = .u({n: ltfr.(n)l > s}).

Then in view of (4) and (5),

f

n=-oo

llP.(n)IP.u(n) = - Joo sPdcx(s) = p Joo sr 1cx(s) ds

~

0

pC

Jooo

pC =-

oo J sp- J lcp.(t)l dt ds 2n o o pC

sP- 2 IIcp.ll 1

f2"

2n o

0

ds

2n

2

=-

2n

lg(t)l

lg(t)l

Jo

Jo

sr 2 ds dt = AP

lg(t)IP dt,

where AP is a constant depending only onp. Similarly, by (4),

f sP- p(s) ds ~PJ sp- llt/1.1i/ds=-J sp- J lt/l.(tWdtds o 2n o o

n~ooltfr.(n)IP.u(n) = - IaoosP dp(s) = p

00

p

2n

1

27t

00

3

p = -2

00

0

3

oo

J lg(tW J n o

sp- 3 ds dt ~ BP

lg(tJI

2n

Jo lg(t)IP dt.

Combining these two estimates with (6), we have (7)

6.2 THEOREM OF HARDY AND LITTLEWOOD

97

lf/E H P (1 < p < 2), the desired estimate (3) now follows after approximating the boundary functionf(ei') by 9M(t)

1/(e'')l ::;; M, 1/(ei')l > M,

= {f(ei'), 0,

applying (7) to gM, and Jetting M--. oo. As with the Hausdorff-Young theorem, the converse to Theorem 6.2 is false if p < 2. However, the converse is true for indices larger than 2, and can be deduced from Theorem 6.2 by a duality argument. The exact statement is as follows. THEOREM 6.3. Let {an} be a sequence of complex numbers such that

L nq-

1anlq < oo for some q, 2::;; q < oo. Then the functionf(z) is in Hq, and 2

llfllq::;; where

cq{~o(n + l)q-

2

= L:'=o anzn

lanlq} lfq'

cq depends only on q.

PROOF. Let p = qf(q- 1), and let

G(eio)

n

L ck e'ko

=

k=-n

be an arbitrary trigonometric polynomial with II GIIP::;; 1. By the M. Riesz theorem, its " analytic projection " n

g(eio) =

L ck eiko k=O

n

sn(z)= :Lak~k=O

Then

=

n

L lckl(k + l)(pk=O

2

J!Piakl(k

+ l)(q- 2 Jfq

98

6 TAYLOR COEFFICIENTS

Taking the supremum over all G with IIGIIP:::;; 1, we have

Mq(r, s,.):::;; CPAP{X0(k

+ 1)q- 2 lakiqf'q,

and the result follows by letting n--+ oo. 6.3. THE CASE p,:::;; 1

As pointed out in the corollary to Theorem 3.4, the Taylor coefficients of an H 1 function must tend to zero, by the Riemann-Lebesgue lemma. The simple example (1 - z)- 1 shows this is false for HP functions withp < l, but a sharp asymptotic estimate for the coefficients can be given as follows. THEOREM 6.4. If co

f(z) =

L a,.z" E HP,

0< p:::;; 1,

n=O

then (8)

and (9)

Furthermore, the estimate (8) is best possible for each p: given any positive sequence {~,.}tending to zero, there existsfe HP such that

a,. =1-

O(~,.ntfrt).

We have already observed that a,.= o(l) iffe H 1 . This statement cannot be improved, even iffe H"". Indeed, given a sequence{~,.} tending to zero, choose integers 0 < n1 < n2 < · · · such that ~"k < 2-k. Define PROOF.

- {k

a,.- 0 ~11k

if n = nk, otherwise.

Then L la,.i < oo, sof(z) = L a,.z" is continuous in lzl:::;; 1, yet a,. =1- 0(~,.). Now suppose f E HP, p < 1. The coefficient a,. has the representation

a,.= -21 . f 1tl

Jl•! =r

~~~~ dz,

O
Z

Hence

O
6.4 MULTIPLIERS

99

But by Theorem 5.9, M 1 (r,f) = o((l- r) 1 -

1 fP).

With the choice r = 1 - 1/n, (8) follows. Theorem 5.9 also gives (9) in the same way. Functions of the form (1 - z)- 1 show the exponent in (8) cannot be reduced. The proof that the estimate is best possible will be postponed to Section 6.4. 6.4. MULTIPLIERS

A complex sequence {A.n} is said to be a multiplier of HP into the sequence space tq if {A.n an} E tq whenever I, an zn E HP. Similarly, {A.n} is a multiplier of HP into Hq if I, anzn E HP implies I. A.nanzn E Hq. One way to give information about the coefficients of HP functions is to identify the multipliers of HP into various spaces. Results of this type have a number of interesting consequences. We shall content ourselves with describing the multipliers of HP (0 < p < 1) into tq (p::; q::; oo) and of H 1 into t 1 and into t 2 (alias H 2 ). Some preliminary remarks are in order. First of all, even though HP and {Pare not Banach spaces if p < 1, they are complete topological vector spaces whose topologies are induced by the translation invariant metrics II/- ull/ and

llx- Yll/ = L lxn- YniP, respectively (see Section 3.2). In other words, HP and tP are £-spaces, in the terminology introduced by Banach [1]. The closed graph theorem therefore applies to mappings between these spaces (see Banach [1] or Dunford and Schwartz [1], Chap. II). A sequence {A.n} may be regarded as inducing a linear operator

A : I, an zn--+ {A.n an} from HP to tq. The domain of A is the linear manifold

If p < oo, PA(A) is dense in HP, since it contains all polynomials. Furthermore, it is clear that PA(A) = HP if and only if {A.n} is a multiplier of HP into tq. The closed graph theorem therefore tells us that A is a bounded operator from HP to tq if {A.n} is a multiplier. We have only to check that A is closed. To show A is a closed operator, suppose

h. E PA(A), and

h. --+/E HP,

100

6 TAYLOR COEFFICIENTS

Let

fk(z) =

L a~klzn,

f(z) =

L anzn,

and

g = {bn}.

Then by Theorem 6.4, a~k)-+ an, n = 0, 1, .... Since also A.n a~k)-+ bn, this shows that A.n an= bn, n = 0, 1, .... In other words, fE PA(A) and A(/)= g, which proves that A is closed. THEOREM 6.5.

Suppose 0 < p::;:; 1. Then {A.n} is a multiplier of HP into t"'

if and only if (10)

L

PROOF. If /(z) = anzn is in HP and (10) holds, then {A.n an} E f'Z', by Theorem 6.4. (In fact, A.n an-+ 0.) Conversely, if {A.n} is a multiplier, then by the closed graph theorem (as discussed above), A is a bounded operator from HP to f'X'. In other words,

(11)

sup JA.nanl :S CJIJIIP' n

Now let

g(z) = (1- z)-1-1/p =

L bnzn,

where bn"' Bn 1 1P; and choosef(z) =g(rz) for fixed r < 1. Then by (11) and the lemma in Section 4.6,

l.A.nln 11Pyft :S C(1 - r)- 1 • The choicer= 1 - 1/n now gives (10). As a corollary, we can now show that the estimate (8) in Theorem 6.4 is best possible. COROLLARY. If {dn} is a sequence of positive numbers such that an = O(dn) for every function anzn in HP(p::;:; 1), then there is an e > 0 such that

L

n

= 1, 2, ....

PROOF. If an= O(dn) for every /E HP, then {1/dn} multiplies HP into f'Z'. Thus 1/dn = O(n 1_ 1,P), as claimed. THEOREM 6.6.

of HP into

SupposeD< p < 1 andp::;:; q < oo. Then {A.n} is a multiplier

tq if and only if N

L nqfpJA.nlq = n= 1

O(Nq).

(12)

6.4 MULTIPLIERS

101

Before passing to the proof, let us apply the result to establish the HardyLittlewood theorem (Theorem 6.2) for p < 1. COROLLARY. If

L

anzn E

HP (0 < p < 1), then co

L np-

2

lanJP < oo.

n=1

PROOF. Let q = p, and observe that A., = n1 - 21P then satisfies (12). Hence {n 1 - 2 1P} is a multiplier of HP into tP. The closed graph theorem now gives the inequality (3) stated in Theorem 6.2.

The following lemma will be needed in the proof of Theorem 6.6. LEMMA. Suppose bn;;::: 0 and 0

< {3 <ex. Then

N

L n«bn =

(13)

O(Nfl)

n=1

if and only if co

L bn =

(14)

O(NfJ-}.

n=N

M

M-1

n=N

n=N

L bn = L Sn[n-«- (n + 1)-«J + SMM-«- SN_

1 N-«

M-1

:$;

C

L nP-«- 1 + CMP-«.

n=N

Letting M--+ oo, we obtain (14). Conversely, assume (14) and let Rn Then N

N

n= 1

n= 1

L n«bn = L [n«- (n- 1tJRn- N«RN+

PROOF OF THEOREM 6.6.

1 :$;

=

Lk'=n

bk.

CNfl.

By the lemma, the condition (12) implies

co

L JA.nJq = O(Nq(1-1fpl).

(15)

n=N

Assume without loss of generality that A.n;;::: 0 and and

Ln~ 1

n = 2, 3, ... ,

A.nq = 1. Let s 1 = 0

102

6 TAYLOR COEFFICIENTS

where y = q(lfp- 1). Note that fE HP (0 < p < l) implies

Sn

increases to l as n-+ oo. By Theorem 5.11,

J(1- r)Y- M q(r,f) dr < oo, 1

1

1

p 5:, q

<

00.

0

Since M 1(r,f) ~ rnlanl, wheref(z) = oo >

L anzn, it follows that

f f"+ (1- r)Y- M q(r,f) dr 1

n == 1

1

1

Sn

by the definition of Sn. But by (15), 00

{

"A.q ~ n

}1/7 <-c

k=n

-

n'

which shows that (snrq ~ (l- Cfn)nq-+e-cq > 0.

Since these factors (snr are eventually bounded away from zero, we have shown that {A.n} is a multiplier of HP into tq if it satisfies (12). Conversely, if {~} is a multiplier, then by the closed graph theorem,

L~0 l.A.n anlq rq 5:, Cllfllp. Choosingf(z) = g(rz) as in the proof of Theorem 6.5, we now find 00

L nqfPJA.nlqrnq 5:, C(l -

r)-q.

n=1

Hence N

r"q L

nqfPJ~Iq 5:,

C(l - r)-q,

n-1

and (12) follows upon setting r = l - lfN. Note that the argument shows (12) is necessary even if p ~ l or q < p. The condition (12) also characterizes the multipliers of H 1 into tq, if 2 5:, q < oo. We give the proof first in the case q = 2.

6.4 MULTIPLIERS

103

THEOREM 6.7. The sequence {A.n} is a multiplier of H 1 into H 2 if and only

if N

Ln n=1

2

(16)

l.A.nl 2 = O(N 2 ).

PROOF. The necessity of (16) was observed in the proof of Theorem 6.6. In proving the sufficiency, it is enough to consider functions without zeros, since by the lemma in Section 5.3, every H 1 function is the sum of two nonvanishing H 1 functions. Thus supposef(z) = anzn is an H 1 function which 2 does not vanish in Jzl < 1, and write/= cp , where cp(z) = bnzn is in H 2 • Then

L

L

n

an=

L bk bn-k. k=O

L

It is to be shown that (16) and lhnl 2 < oo imply of generality, suppose A.n;;::: 0 and bn;;::: 0. Then

L IA.nanl

2

< oo. Without loss

n

an::;:; 2

L k=

bk bn-k,

[n{2]

where [x] denotes the greatest integer ::;:;x. Now apply the Cauchy-Schwarz inequality: co

a/ : ;:; 4 L b/

n

L b/ = k=[nf2]

k=O

n

L

C b/ = Cfln, k=[nf2]

say. It follows that N

N

L A./a/::;:; Cn=L A./fln · n= 1

1

But a summation by parts gives N

N-1

L A./fln = N- 2sN flN- n=L sn A(nn= 1

2

/1n),

1

where

and A xn = Xn+ 1 - Xn. The term N - 2SN flN is bounded, by (16) and the assumption b/ < oo, so it remains only to show that

L

co

Ln n=1 But

2

IA(n- 2 /1n)l < oo.

(17)

104 6 TAYLOR COEFFICIENTS

Since

L b/ <

oo, it is clear that

L - +n 1)2 IAPnl < oo; oo

(

n=1 n and

I n2PniA(n- 2)1 ~ 2n=f 1 n~ k=[n/21 I bk

2

n-1

2k+ 1 1

00

~2

00

'L b/ n=k 'L -n ~ 6 I

k~ 1

bk2 < oo.

k= 1

This establishes (17) and completes the proof of the theorem. As a first illustration, let A.n

= n- 112 • Then ( 16) is satisfied, so 2

lan1 L -
for every f E H 1 • However, this follows from Hardy's inequality and the fact that an__. 0. There is another application which is more interesting. Let nt> n 2 , ••• be a lacunary sequence of integers in the sense that nk+1/nk ~

Q > l.

Let

A. n

If nk

~

=

{l0

if n=nk otherwise.

N < nk+ 1, N

k

L n21A.nl2 = L n/ ~ nk2{l + Q-2 + ... + Q20-kl}:::;; CN2.

n= 1

j~

1

We thus obtain PALEY'S THEOREM.

If 00

f(z)= LanznEH 1, n=O then for every lacunary sequence {nd,

6.4 MULTIPLIERS

105

As a final corollary of Theorem 6.7, we now show that, more generally, {A.,} is a multiplier of H 1 into tq (2::;:; q < oo) if(and only if) it satisfies (12). Let li-n = IA.nlqf 2 , and observe that (12) gives a condition equivalent to N

L n2p./ = O(N2).

n=1

Thus {p.n} multiplies into t 2 , which implies (since the coefficients of an H 1 function are bounded) that {A.n} multiplies H 1 into tq. The multipliers from H 1 to t 1 are more difficult to describe. Hardy's inequality shows that the sequence {(n + 1)- 1 } is one example. It is possible to characterize all the multipliers, but only by a condition difficult to verify in most situations. Thus the following theorem, interesting though it may be, is really more a translation of the problem than a solution. H1

THEOREM 6.8. The sequence {A.n} is a multiplier of H 1 into t 1 if and only if there is a function 1/J E L oo such that

n

= 0,

1, 2, ....

(18)

PROOF. The sufficiency of the condition (18) was established in Theorem 3.15. Conversely, suppose {A.n} is a multiplier from H 1 to t 1. Then by the closed graph theorem, the mapping

L anzn-+ {A.n an} is a bounded operator from H 1 to t 1 : CXl

L IA.nanl::;:; Cll/1!1, n=O In particular, CXl

r/J(f)

= n=O L IA.nlan

is a bounded linear functional on H 1 • By the Hahn-Banach theorem,¢ can be extended to a bounded linear functional on L 1 • Thus by the Riesz representation theorem, there exists 1/J E L oo with 1 (f) = -

27t

-

J f(e'')!/J(t) dt, 2n o

106

6 TAYLOR COEFFICIENTS

Choosing the H 1 function f(z)

=

zn (n

=

0, 1, 2, ... ), we have

IA.nl = ¢(!) = (f)(f) = 21n

J2" einrt/l(t) - dt, 0

which becomes (18) after conjugation. EXERCISES

1. Show that Theorem 6.2 is best possible if 0 < p < I: For each positive sequence {kn} increasing to infinity, there exists L anzn E HP with

L knnp-21aniP = 00.

•

(This is also true if I :$ p :$ 2; see Duren and Taylor [I].) 2. Show that Theorem 6.2 is false for p > 2. 3. Show that the converse to Theorem 6.2 is false (0 < p :$ 2).

4. Show that Paley's theorem does not generalize to Fourier series: There existsfe L 1 with Fourier series L cne"' 9 , for which L lc 2 ,.1 2 = oo. (Suggestion: Try L (log n)-a cos ne, a> 0. See Theorem 4.5.)

5. Show thatf(z) may be analytic in lzl < I and continuous in izl :$ I, yet f'(z) have a radial limit almost nowhere. (This disproves the" Bloch-Nevanlinna conjecture" thatfe N implies/' eN. See Duren [4].) 6. A function/(z) analytic in izl < I is said to have finite Dirichlet integral (feD) if

~ lzlJJ

<1

lf'(zW dx dy =

I nlanl

n= 1

2

< oo.

Show that/' e H 1 impliesfe D, but thatfneed not be in D iff' e HP for all p
For the Hausdorff-Young theorem, see Zygmund [4]. Hardy and Littlewood [1] gave a long and difficult proof of Theorem 6.2 in their original paper. The idea to prove it for 0 < p < 1 via Theorem 5.11 is due to Flett [2]. The fact that an= o(n 11 P- 1 ) for fe HP (p < I) is due to Hardy and Littlewood [5], although in the Soviet literature it is often ascribed to G. A. Fridman, who rediscovered it in 1949. Evgrafov [1] gave a direct construction to show that

NOTES

107

this estimate is best possible; a simpler argument was given by Duren and Taylor [1 ]. Theorem 6.5 (and its corollary) and Theorem 6.6 are due to Duren and Shields [1, 2]. Hardy and Littlewood [7, 8] gave a sufficient condition for multipliers from HP to Hq (1 ::-::; p ::-::; 2 ::-::; q < oo) which includes the "if" part of Theorem 6. 7; the proofin the text is theirs. They also stated without proof a sufficient condition for multipliers from HP to Hq (0 < p < 1 ::-::; q < oo ). A proof appears in Duren and Shields [2]. The closed graph theorem has often been applied to multiplier problems in various spaces . .T. H. Wells pointed out to the author that the necessity of (16) could be proved by this method. Paley [1] proved "Paley's theorem" and inspired the original Hardy-Littlewoad work on multipliers. A converse to Paley's theorem, which Rudin [4] proved by explicit construction, is a simple consequence of the necessity of the multiplier condition (16); see Stein and Zygmund [1]. Hedlund [1] recently gave a sufficient condition for multipliers from HP (1 < p < 2) to H 2 which extends that of Hardy and Littlewood. The condition (12) is necessary but not sufficient if p = 1 and q < 2; see Duren and Shields [2]. E. M. Stein [1] has given a sufficient condition for multipliers from HP to HP (0 < p < 1). For further information on multipliers, see Kaczmarz [1], Caveny [1, 2], Wells [1], Duren and Shields [1, 2], and Duren [3]. The coefficients of inner functions have been studied by Newman and Shapiro [1]. They show that no inner function except a finite Blaschke product can have coefficients o(l ln), although the coefficients of an infinite Blaschke product can be 0(1/n). The coefficients of a singular inner function cannot be o(n- 3 14 ), but they can be O(n- 3 14 ). Information is also available on the coefficients of functions of bounded characteristic. Mergelyan has shown that if f(z) = I an zn E N, then lim sup n -l/ 2 loglanl ::-::; C, where C can be given explicitly in terms off (See Privalov [4].) Cantor [1] recently proved that if/EN and An denotes the matrix (a;+ J, 0 ::-::; i,j ::-::; n, then ldet Anllfn--+ 0 as n--+ oo.

This Page Intentionally Left Blank

HP AS A LINEAR SPACE

CHAPTER 7

In order to solve problems concerning HP functions, it is often advantageous to view HP as a linear space and to use the methods of functional analysis. We have already seen in Chapter 6 that the closed graph theorem is an effective tool for describing coefficient multipliers. The same tool will be used in Chapter 9 to discuss interpolation problems. In the present chapter, we shall study the linear space structt,re of HP in some detail. One major objective is to represent the continuous linear functionals on HP both for p ;;::: 1 and in the more interesting case p < 1. The results for p ;;::: 1 help to solve an approximation problem (Section 7.3) and prepare the ground for a full discussion of extremal and interpolation problems in Chapters 8 and 9. The dual space structure of HP with p < 1 is applied to demonstrate the failure of the HahnBanach theorem in a non-locally convex space with" reasonable" properties. This chapter concludes with a description of the extreme points of the unit sphere in Hi, a topic also of interest in the general context of linear space theory.

110

7 HP AS A LINEAR SPACE

7.1. QUOTIENT SPACES AND ANNIHILATORS

Let us begin by recalling a few general Banach space concepts. Let X be a Banach space, and let S be a (closed) subspace. A coset of X modulo S is a subset ~ = x + S consisting of all elements of the form x + y, where xis some fixed member of X and y ranges over S. Two cosets are either identical or disjoint. The quotient space X/S has as its elements all distinct cosets of X modulo S. With the natural definitions of addition and scalar multiplication, X IS is a linear space. To be specific, (x 1 + S) + (x 2 + S) = (x 1 + x 2 ) + S and cx(x + S) = cxx + S. The zero element of X/S is the coset S. Finally, the norm of a coset~ = x + S is defined by WI= inf llx + Yll. yeS

It is easy to check that this is a genuine norm. Since S is closed, 11~11 = 0 implies ~ = S. The relations WI 2::0, llcx~ll = lcxl WI. and 11~ 1 + ~ 2 11::;:; ll~1ll + 11~ 2 11 are obvious. Under the given norm, XfS is complete, and therefore is itself a Banach space. To see this, let {~n} be a Cauchy sequence of cosets. Choose a sequence of integers 0 < n1 < n 2 < · · · such that

~~~nk- ~nk+,ll::;:; T\

Then choose xk E ~nk (k

=

k = 1, 2, ....

1, 2, ... ) such that

llxk-

Xk+1ll < 2ll~nk- ~nk+,11.

With this construction, {xk} is a Cauchy sequence, so xk tends to a limit x E X. Let ~ = x + S. Then ~nk-> ~. because (by definition of the norm) ll~nk- ~II

::;:; llxk- xll.

Finally, since {~n} is a Cauchy sequence, this implies ~n-> ~. The annihilator of the subspace S is the set S.l of all linear functionals ¢EX* such that ¢(x) = 0 for all xES. It can be easily verified that S.l is a subspace of X*. The following results play an essential role in the theory of extremal problems (Chapter 8). THEOREM 7.1. The quotient space X*/S.l is isometrically isomorphic to S*. Furthermore, for each fixed ¢EX*,

sup

1¢(x)l =min II¢+ 1/111,

xeS, l[x[[S1

1/leS~

where "min" indicates that the infimum is attained.

7.1 QUOTIENT SPACES AND ANNIHILATORS

111

THEOREM 7.2. The space (X/S)* is isometrically isomorphic to S1.. Furthermore, for each fixed x EX,

max

ill/Ill Sl

l/leS~,

11/l(x)l = inf llx + yll, yeS

where "max'' indicates that the supremum is attained. For each fixed 1/1 E S*, the class of all extensions

PROOF OF THEOREM 7.1.

eX* is a coset in X*/S1.. It is clear that this correspondence between S* and X*/SJ. is an isomorphism. It is also an isometry. In fact, 111/111 ~ l c/>11 for cJ>

every extension cJ>; and, by the Hahn-Banach theorem, there is at least one extension for which 111/111 = l c/>11. In other words, for the coset of extensions of 1/1, the infimum defining the norm is attained and is equal to 111/111. PROOF OF THEOREM 7.2.

a linear functional c/> in

sl.

In terms of a given C1> E(X/S)*, one can define unambiguously by cj>(x)

= (J)(x + S).

Conversely, given c/> E sl., this relation defines a linear functional cp on X/S. The correspondence is clearly an isomorphism. We assert that the boundedness of either functional implies that of the other, and in fact l c/>11 = IIC1>11This follows from the relations IC1>(x

lc/>(x)l + S)l

= IC1>(x

+ S)l

~

= lc/>(x)l = lc/>(x

I C1>11 llx + Sli ~ I C1>11 llxll; + Y)l ~ l c/>11 llx + yll, YES.

Hence 51. and (X/S)* are isometrically isomorphic. To prove the rest of the theorem, observe first that for any c/> e 51. with l c/>11 ~ 1, and for any x EX andy e S, lc/>(x)l = lc/>(x

+ Y)l

~ llx

+ Yll·

Thus sup

,Pes~.II
lc/>(x)l ~ inf llx

+ Yll·

(1)

yeS

On the other hand, given x eX, a corollary of the Hahn-Banach theorem (see Dunford and Schwartz [1], p. 65) shows the existence of C1> e (X/S)* such that

IC1>(x + S)l

= llx

+ Sil

and

I C1>11

= 1.

Now let c/> e 51. correspond to C1> as above, so that cj>(x) = C1>(x + S) and l c/>11 = I C1>11 = 1. Equality therefore holds in (1), and the supremum is attained.

112

7 HP AS A LINEAR SPACE

7.2. REPRESENTATION OF LINEAR FUNCTIONALS

As we saw in Chapter 3, HP is a Banach space if 1 :::;;, p:::;;, oo, with norm 11/11 = Mp(l,f). The polynomials are dense inHP ifO

In particular, if eachfEHP is identified with its boundary function, HP can be regarded as a subspace of IJ', 0 < p :::;;, oo. According to the Riesz representation theorem, every bounded linear functional ¢ on IJ' (1 :::;;, p < oo) has a unique representation gEH,

(2)

where 1/p + lfq = 1. In fact, 11¢11 = 119llq, and (IJ')* is isometrically isomorphic to H. Since HP is a subspace of IJ', then, Theorem 7.1 can be used to describe (HP)* if the annihilator of HP in (IJ')* can be determined. But if 9 E H annihilates every HP function, then surely n

= 0, 1, 2, ....

Therefore 9(ei 6 ) is the boundary function of some 9(z) E Hq, and 9(0) = 0. Call this class of functions H 0 q. Conversely, if 9 E H 0 q, it is clear that 27t

J f(eio)g(eio) de= 0 0

for every fE HP. Hence H 0q is the annihilator of HP, and it follows from Theorem 7.1 that (HP)* is isometrically isomorphic to HfH 0 q. Actually, we may as well replace HfH 0 q by HfHq, since the correspondence ~-ei 6 ~ between cosets of the two spaces is an ;sometric isomorphism. It is even possible to give a canonical representation of the bounded linear functionals on HP. Any ¢ E (HP)* can be extended (by the Hahn-Banach theorem) to a functional on IJ', and hence may be represented in the form (2) for some 9 E H. This representation is certainly not unique. Two functions 9 1 and 9 2 belonging to the same coset of HfH 0 q, so that co

g l(eio) - g2(eio)"'

L en eino, n=l

obviously represent the same functional ¢ on HP. Functions in different cosets, however, generate different functionals. Thus for p > 1, the representation becomes unique if we distinguish in each coset that function 9 for which

7.3 BEURLING'S APPROXIMATION THEOREM

t

21t

e-!n9g(ei9) d() = 0,

113

n = 1, 2, ....

Equivalently, there is a unique function g E Hq for which

¢(/) =

2_ J2"f(e;e)g(e;e) d() 2n o

(3)

for all f E HP, I < p < oo. Since I < q < oo, the M. Riesz theorem (Theorem 4.1) guarantees that the" analytic projection" g of the original Lq function is in Hq (see Chapter 4, Exercise 1). In summary: THEOREM 7.3. For 1 ~ p < oo, the space (HP)* is isometrically isomorphic to IJfHq, where 1/p + lfq = 1. Furthermore, if 1 < p < oo, each ¢ E (HP)* is representable in the form (3) by a unique function g E Hq, while each¢ E (H 1)* can be represented in the form (3) by some g E L00 •

We might have chosen to put g(e- ;9) instead of g(ei9) in (3). This would have the advantage of setting up an isomorphism between (HP)* and Hq. But in either case the correspondence need not be an isometry. In fact, 11¢11 is equal to the norm of the coset determined by g(ei 9) (or by g(e-; 9)), so that only the inequality

I

The representation of linear functionals can be applied to prove an interesting theorem on polynomial approximation in HP, 1 ~ p < oo. Recall that a functionfE HP has a canonical factorization

f(z) = B(z)S(z)F(z), where B(z) is a Blaschke product, S(z) is a singular inner function generated by a nondecreasing singular function Jl(t), and F(z) is an outer function. The inner function f 0 (z) = B(z)S(z) is called the inner factor of f(z). Let g 0 (z) be another inner function, with an associated singular function v(t). fo is said to be a divisor of g 0 if g 0 (z)/f0 (z) is an inner function. This is clearly the case if and only if every zero of f 0 (z) in lzl < 1 is also a zero of g 0 (z) (with the same or higher multiplicity) and [v(t)- Jl(t)] is nondecreasing.

114

7 Ht> AS A LINEAR SPACE

For fixed f e HP, let &'[/] denote the subspace generated by the functions z"f(z), n = 0, I, 2, .... Thus .9'[/] consists of all HP functions which can be approximated by polynomial multiples off The problem is to characterize &'[/]. It is already known that .9'[1] = HP, since the polynomials are dense in HP. More generally, &'[/] can be described as follows. THEOREM 7.4 (Beurling). Let f and g be HP functions (1 .::;; p < oo), not identically zero, with inner factors/0 and g 0 , respectively. Then g e .9' [f] if and only if / 0 is a divisor of g 0 • COROLLARY 1 .

.9'[/] = .9'[/0 ]. If .9'[/0 ]

=

.9'[g0 ], then/0 = g 0 •

PROOF OF THEOREM. Under the assumption thatf0 divides g 0 , it is to be shown that g e &'[!]. Since

liP/- ull = IIPF- (uo//o)GII. it suffices to prove that &'[F] = HP for every outer function F(z). If &'[F] is not the whole space, there is a nontrivial bounded linear functional which annihilates it. That is, there exists a function h(e 16 ) e IJ (lfp + 1/q = 1) such that h( ei 6 ) ¢ H 0 q and

f

2"

-

ein6F(el~h(ei6) d() =

0,

n = 0, 1, 2, ....

0

Thus F(e 16)h(ei~ = k(e;~. where k(z) e H 0 1 • Since k(e 1 ~/F(e 16) e IJ, it follows from Theorem 2.11 that

a contradiction. Conversely, if g e .9'[/] it is obvious that g must vanish in lzl < 1 wherever f does, since norm convergence implies pointwise convergence. Thus h(z) = g 0 (z)jf0 (z) is analytic. It has to be proved that his an inner function. But since g e &'[!], there is a sequence of polynomials {Pn} for which IIPnf- ull -4 0. Division by f 0 (ei 6) then shows that h(e 16 )G(e 16 ) is an I! function approximable in norm by HP functions. Thus hG e HP, which implies that he HP. This shows (by Theorem 2.11) that his an inner function, completing the proof. COROLLARY 2.

Let 0

=

fo ·HP.

This is true even for 0

7.4 LINEAR FUNCTIONALS ON HP. 0 < p
7.4. LINEAR FUNCTIONALS ON W, 0
115

<1

Although HP is not normable in the case p < 1, its bounded linear functionals can still be defined in the usual manner. Thus a linear functional ¢ on HP is said to be bounded if 11¢11 = sup 1¢(/)1 < oo. II fliP=

1

It is easy to verify that a linear functional is bounded if and only if it is continuous. It can also be checked that the bounded linear functionals on

HP form a Banach space under the given norm. We observed in Section 6.4 that HP is an F-space even if p < 1. Hence the principle of uniform boundedness (Banach-Steinhaus theorem) still applies: every pointwise bounded sequence of bounded linear functionals on HP is uniformly bounded. (See Dunford and Schwartz [1], Chapter II.) We are about to derive a complete representation of the bounded linear functionals on HP,p < 1. First, however, we need to introduce some notation. Let A denote the class of functions analytic in lzl < 1 and continuous in lzl:::;; 1. It is convenient to writefEA~ to indicate thatfE A and its boundary functionf(ei 6) belongs to the Lipschitz class A~, 0 <ex:::;; 1. Similarly,JE A* will mean thatjE A andf(ei 6) EA*. (See Section 5.1.) THEOREM 7.5. To each bounded linear functional ¢ on HP, 0 < p < 1, there corresponds a unique function g E A such that

(4) lf(n+l)- 1
Given r/J E (HP)*, let bk = r/J(z~, k = 0, 1, .... Then

lhkl

:5:

11¢11 llzkll = 11¢11.

so the function co

g(z)= :Lbkzk k=O

is well defined and analytic in

lzl < 1. Suppose now that co

f(z) =

L ak~ E HP.

k=O

116 7HPASAUNEARSPACE

For fixed p, 0 < p < 1, let fp(z) = f(pz). Since /p is the uniform limit on lzl = 1 of the partial sums of its power series, and since tjJ is continuous, it follows that

t/J(fp)=

!~~t/JCtoakpkz") = Joakbkpk.

ButfP __.fin HP norm asp__. 1, so CXl

t/J(f) =lim p-+ 1

L akbkpk.

(5)

k~o

To deduce (4) from (5), it would suffice to show g e H 1• But for fixed (, !CI<1,1et

f(z)

= (1

CXl

- (z)-1

= L (kzk. k=O

Then by (5), CXl

t/JU) =

:L Ckbk = g(C).

k~O

Hence luWI::.:;; 111/JIIII/IIp::.:;; llt/JIIIIC1- z)- 1 llp; that is, g e H"". This proves (4), with g e H"". Now suppose (n + 1) - 1 < p < n- 1 , and let

n! zn F(z) = (1- Czr+ 1 '

1'1 <

1.

In view of (4) and the Cauchy formula, we have

t/J(F) = gml::.:;; 111/JII IIFIIP = 0((1 -IW 11p-n- 1), so that geAa with a= 1/p-n, by Theorem 5.1. If p=(n+ 1)- 1 , a similar argument shows g(0 = 0((1 - 1(1)- 1), which implies ge A*, by Theorem 5.3. To prove the converse, we first suppose (n + 1)- 1 < p < n - 1 and that g(z) = bkz" is given with ge Aa, a= 1fp- n. It is to he shown that for every f(z) = akzk in HP, the function

L

L

CXl

t/J(r)

=

L ak bk rk k=O

has a limit as r--+ 1, and that lim 11/J(r)l ::.:;; r-+1

CllfiiP ·

7.4 LINEAR FUNCTIONAL$ ON H:r>. 0

<

p

< 1 117

We shall prove the existence of the limit by showing that 1

Jo lt/l'(r)l dr < oo.

(6)

Setting h(z)

= z"- 1g(z)

and

v =0, 1, ... ' where/0 (z)

=

f(z), we have the relation

rnt/J'(r2) = 217t (" e-inBfn-1(reiB)h(nl(re-i8) dO,

which can be checked by comparing power series. By Theorem 5.12,

v = 1, 2, ... , n, where p 0

= p and Pv+1

Pv - Pv

= -1 - - .

Simple induction shows that

p Pv = 1 -- - , -vp

v = 1, 2, ... , n.

Hence

P < P1 < · · · < Pn-1 < I < Pn < oo · On the other hand, the fact that g
< C lh(nl(reiB)I -(1r)1

a'

by Theorem 5.1. It therefore follows from (7) that rnlt/l'(r 2 )1 ~ C(1 - r)"- 1 M 1(r.Jn-1),

!X= 1/p- n.

But according to Theorem 5.11,

t

1

(1- r) 11M 1(r.Jn_ 1) dr < oo,

where 1 f3=--2=!X-l. Pn-1

This proves (6) for (n

+ l)- 1

(7)

118

7 HP AS A

LIN EAR

SPACE

If p = (n + 1) - 1 and g
¢(!) =lim

L ak bk r"

r-->1 k=O

exists for every f(z) =

L akzk in HP, then¢ E (HP)*. For fixed r < 1, let co

¢,(!) =

L ak bk r".

k=O

By Theorem 6.5, ¢, E (HP)* for each r. But for each fixedfE HP, supl¢,(!)1 < oo. Thus by the principle of uniform boundedness, supll¢,11 < oo, which implies ¢ E (HP)*. This concludes the proof. We remark that the principle of uniform boundedness need not have been used. The proofs of the various auxiliary theorems actually give norm estimates which could have been carried through the proof to establish the boundedness of¢ directly. The function g(z) = (1- (z)- 1 , 1(1 < 1, provides an interesting example. Here the integral in (4) reduces simply to f(r(), so that ¢(/) = f((). Point evaluation is therefore a bounded linear functional on HP, p < 1. (This can also be proved directly; see the lemma in Section 3.2.) In particular, there are enough linear functionals on HP to distinguish elements of the space. This contrasts sharply with the situation for IJ' (p < 1), where there are no bounded linear functionals at all except the zero functional. 7.5. FAILURE OF THE HAHN-BANACH THEOREM

We have just seen that HP has enough continuous linear functionals to distinguish elements of the space, even if p < 1. On the other hand, we shall now show that there are not always enough functionals to separate points from subspaces. That is, given a proper subspace M of HP (p < 1) and an HP function/¢ M, there may not be any functional¢ E (HP)* such that r/J(M) = 0

7.5. FAILURE OF THE HAHN-BANACH THEOREM

119

and ¢(f) "# 0. To show this, we shall construct a proper subspace of HP which is annihilated by no functional ¢ E (HP)* except the zero functional. Before turning to the actual construction, we have to develop some background material which is of interest itself. Let

{ J -e ;+-, -z d11-(t)

S(z) = exp -

2"

eir

z

}

o

be a singular inner function; and let w(t; 11-) denote the modulus of continuity of11-. LEMMA 1. If w(t; 11-) a and C such that

= O(t log 1ft), then there exist positive constants

IS(z)l

~

C(l -

rt,

lzl < 1.

PROOF. We have 8+1t

-logiS(z)l =

J

P(r, t- 8) d11-(t)

8-lt

= P(r, n){11-(8

+ n) -

11-(8 - n)}

- Jo"aP -(r, t){11-(8 + t) -11-(8- t)} dt at ::;; k1 +

aP) l11-(8 + t) Jo".(-at

::::;_ k 2 + k 3

11-(8 - t)l dt

aP) t log-1 dt. Jo -at t 1fe (

But another integration by parts shows this last integral is less than k4 +

f

1fe

0

1 P(r, t) log- dt t

1 1-r 1 ::;; k4 +log -1 + P(r, t) log- dt - r o t

J

1 2 ::;; k 4 + log -1 + --r 1-r 1

= (k 4 + 2) + 3 log--.

1-r

1 log- dt

1-r

J 0

t

120

7 HP AS A LINEAR SPACE

Hence 1 -logiS(z)l :$; k +a log--, 1- r

which proves the lemma. We now introduce the space B~ (ex> 0) of functionsf(z) = L a.z" analytic in lzl < 1, such that co 11!11~2 = L (n + 1)-~la.l 2 < oo. n=O

B~

is a Hilbert space with inner product co

-

L (n + 1)-~a. b.,

(f, g)=

n=O

where g(z) = L b.z". Obviously, the polynomials are dense in B~, for each ex > 0. A straightforward calculation shows that

C1llfll~2 :$;

27t

1

J J (1- rY0

1 If(rej 6

)1 2 dr de

0

(8)

:$; C2llfll/.

for some positive constants C1 and C2 depending only on ex. A function fEB~ will be called an ex-outer function if the set of all polynomial multiples Pfis dense in B~. Clearly,fis an ex-outer function if and only if 1 belongs to the B~ closure of the polynomial multiples off LEMMA 2. If w(t; p.) = O(t log 1/t), then S is an ex-outer function, for

every ex> 0. PROOF. By Lemma 1 and (8), s-lfm

Given

E

E

B~

for some positive integer m.

> 0, it follows from (8) that IIPSlfm- Ill~:$;

CliP- s-l{mll~ < E

for a suitable polynomial P. This shows that S 1 fm is an ex-outer function. To conclude that Sis an ex-outer function, we need only observe that if fE Hco and g E B~ are ex-outer functions, so is their productfg. Indeed, by (8), IIPQ[g- Ill~:$; liPf(Qg- 1)11~ + liP/- Iii~

:$; CIIP/11 coil Qg- Ill~+ liP/- Iii~. Choose P to make the second term small; then choose Q to make the first term small. This shows thatfg is an ex-outer function, and the proof is complete.

7.5 FAILURE OF THE HAHN-BANACH THEOREM

121

We now consider the Hilbert space H 2 and denote by Ml. the orthogonal complement of a subspace M of H 2 . For fixed /E H 00 , fHP will denote the subspace of HP (0 < p ::;; oo) consisting of all multiples fg, g E HP. THEOREM 7.6. Let S be a singular inner function such that w(t; p.) = O(t log lft). Then (SH 2 )l. contains no nonnull function g(z) = L b.z" such that 00

L nY!b.l 2 < n=l

PROOF. Let S(z) = La.~- Suppose g is orthogonal to zkS(z), k = 0, 1, ... ; so 00

I

-

a.-kb.

n=k

= o,

y > 0.

oo,

(9)

H 2 is orthogonal to SH 2 • Then g

E

k

= 0, 1, ....

(10)

Now define If (9) is satisfied, then 00

h(z) =

L c.z" EBy; n=O

while the conditions (10) are equivalent to k

= 0, 1, ....

This says that his orthogonal to zkS(z) (k = 0, 1, ... ) in the space BY. But by Lemma 2, the functions ~S(z) span BY. Hence h = 0, which implies g = 0. One final lemma is needed. LEMMA 3. If g(z) =

L b.z" EA~ for some ex> 0, then for all y < 2cx.

PROOF. By Theorem 5.1,

lg'(z)l::;; C(l-

r)~- 1 .

Thus

It follows that N

r2N

L n21h.l2::;; C2(1- r)2~-2. n=l

122

7 HP AS A LINEAR SPACE

=l-

Setting r

= 2, 3, ... , we find

lfN, N

N

SN

=

L n21bnl2:::::; KN2-2«, n~1

A summation by parts now gives N

N

L nYibnl 2 = L1 n2lbnl 2 · ny- 2 n= 1 n=

N

L s.[nr

=

2 -

(n

+ 1)Y- 2 ] + sN(N + 1f- 2 =

0(1)

n=1

ify < 2a.

We are now ready to give the construction promised at the beginning of this section. THEOREM 7.7.

LetS be a singular inner function with w(t; Jl)

= O(t log

lft).

Then for each p (0 < p < l ), the only bounded linear functional on HP which annihilates the subspace SHP is the zero functional. PROOF. Suppose

¢ e (HP)* annihilates SHP. Then, in particular, n = 0, 1, ... ,

where g(z) = L bnz" eA" for some a> 0 (see Theorem 7.5). This implies that g, regarded as an element of H 2 , is orthogonal to the subspace SH 2 • Therefore, by Theorem 7.6, 00

L nYibnl 2 = n=1 unless g

00

for each y > 0,

= 0. But this contradicts Lemma 3, since g e A".

We remark that SHP is a proper subspace unless S

= l.

COROLLARY 1. If Sis a singular inner function as described in the theorem, the quotient space HPf(SHP) has no continuous linear functionals except the zero functional, for each p < l.

7.6 EXTREME POINTS

123

COROLLARY 2. If p < 1, there is a subspace M of HP and a bounded linear functional on M which cannot be extended to a bounded linear functional onHP.

The proofs are left as exercises.

7.6. EXTREME POINTS

A setS in a linear space X is said to be convex if whenever x 1 and x 2 are inS, every proper convex combination 0
llxll

~

1}.

Every extreme point obviously lies on the boundary of S; that is, it has unit norm. It is useful to observe that if a point x with llxll = 1 is not an extreme point, it can be represented as the midpoint of the segment joining two distinct points (x + y) and (x- y) inS. Furthermore,

llx+yll = llx-yll =

1,

(11)

since

llxll = Hllx + Yll + llx- Yll} ~ 1. Hence a point x with llxll = 1 is an extreme point if and only if (11) implies 1=

y=O. If X is a Hilbert space, it is not hard to show that every boundary point of S is an extreme point. More generally, the same is true if X is a uniformly convex space; that is, if for each E > 0 there exists a fJ > 0 such that llxll = IIYII = 1 and llx- Yll > E imply ll(x + Y)/211 < 1- fJ. If 1

124

7 HP AS A LINEAR SPACE

THEOREM 7.8 (deleeuw-Rudin). A function /E H 1 with 11/11 = 1 is an extreme point of the unit sphere in H 1 if and only iff is an outer function.

PROOF. Suppose first that f is an outer function and 11/11 = 1. To show f is an extreme point, suppose g E H 1 and

(12)

llf+gll = 11/-gll = 1. Let h =gff Then hE H 1 and 27t

Jo

{11 + h(ei 6)1 + 11 - h(ei6 )1 - 2}1f(ei6 )1 de= 0.

The integrand is nonnegative, since the sum of the distances from h(ei 6) to 1 and ( -1) is not less than 2. Sincef(ei 6) does not vanish on any set of positive measure, it follows that a.e.

(13)

Thus, in view of the Poisson representation for H functions (Theorem 3.1), h(z) is real everywhere in lzl < 1, hence is identically constant. But (12) and the fact that 11/11 = 1 then give 1

11 +hi= 11-hl,

=

soh= 0. This shows g(z) 0, proving thatfis an extreme point. Conversely, let f =IF be an H 1 function with nonconstant inner factor I and II f II = 1. To show that f is not an extreme point, we shall construct a function g ¥- 0 in H 1 such that II/+ gil = II/- gil = 1. This will be achieved if h = gffhas the properties (13) and 27t

J

h(ei 6 )lf(ei 6 )1 de= 0.

(14)

0

Let J(z) = ei~I(z) and define g

= tf(J + lfJ).

(15)

Then g E H 1, and (13) is obviously satisfied. For (14) also to hold, we have only to choose ex such that 27t

J

Re{e;~I(ei 6 )}lf(ei 6 )1 de= 0.

(16)

0

This is possible, since the integral is a real-valued continuous function of ex which either vanishes identically or changes sign in the interval 0 ::;:; ex ::;:; 11:. Hence the proof is complete.

7.6 EXTREME POINTS

125

COROLLARY. (i) If /E Hi, 11/11 = 1, and f is not an extreme point of S, then there are two extreme points.t;_ andf2 such thatf = t(/1 + / 2). (ii) If 11/11 < 1, thenfis a convex combination of two extreme points.

PROOF. (i) If 11/11 = 1 and/= IF is not an outer function, define g by (15) and (16). Then II/± gil = 1, so it will be enough to show thatf ± g are outer functions. But we claim that f ± A.g are outer functions whenever ). ;;::: 1. Indeed,

f ± A.g = !J [2J ± A.(J 2 + 1)] 2 +~J+ =~e-i~F[+J 2 ). -

1]

where). cos {3 = 1. The last two factors are outer functions, because they have positive real part (see Exercise 1, Chapter 3). Thus f ± ).g is outer, since it is the product of outer functions. (ii) If 11/11 < 1 andfis not an outer function, let g be as above and choose A.1 > 1 and A. 2 > 1 such that

II/+ A.1gll = II/- A.2gll =

1.

This is possible, by continuity, since 11/11 < 1. Then (f + A.1 g) and(/- A. 2 g) are extreme points, as shown in (i). Iff is outer, the functions ±.1711/11 are extreme points. The next theorem identifies the extreme points of the unit sphere in H 00 • By way of motivation, let us first note that f E H 00 is an extreme point if 11/11 = 1 and lf(ei 6)1 = 1 on some set E of positive measure. Indeed, if g E Hoo and II/+ gil oo = II/- gil oo = I, theng(ei 6) = 0 a.e. onE, which implies g=O. A function f E H 00 with unit sphere in H"" if and only if THEOREM 7.9.

I f I = 1 is an extreme point of the

27t

Jo PROOF. Letg

E

log(1 - lf(ei 6)1) de= - oo.

H 00 such that

II/+ gil= II/ -gil= 1.

(17)

126

7 HP AS A LINEAR SPACE

Then /g(z)l 2

:::;;

1 - lf(zW :::;; 2(1 - lf(z)l),

and it follows from (17) that

t

21t

loglg(e;~l

d() = -

oo,

which implies g = 0. Hence fis an extreme point. Conversely, if the integral (17) converges, let

J

. } 1 2 " eir + z g(z) = exp {-2 -it-log(l - lf(e'')l) dt . 1t o e -z

Then lg(z)l :::;; 1 and lg(ei 6)1

:::;; 1 -

lf(e; 6)1

a.e.

Thus II!+ ull:::;; 1 and II/- ull:::;; 1, sofis not an extreme point. EXERCISES

1'1 < 1, and for each positive integer n, = Jco is a bounded linear functional on HP, p < 1.

1. Show that for each fixed (,

¢(!)

2. Prove Corollary 1 to Theorem 7.7. 3. Prove Corollary 2 to Theorem 7.7. 4. Show that every Hilbert space is uniformly convex. 5. Show that in a uniformly convex space, every boundary point of the unit sphere is an extreme point. 6. Show thatf E Leo with 11/11 = 1 is an extreme point of the unit sphere in Leo if and only if lf(ei 6)1 = 1 a.e. 7. Show that the unit sphere in I! has no extreme points. NOTES

A. E. Taylor [2, 3] was among the first to study HP (1 :::;; p:::;; oo) as a Banach space. He represented the linear functionals (Theorem 7.3) in his paper [3]. Beurling proved Theorem 7.4 for H 2 in his fundamental paper [1]. The proof given in the text is essentially his, suitably extended to 1 :::;; p < oo. Actually, Beurling showed that every subspace of H 2 invariant under multiplication by z has the form gl'[/0 ] for some (unique) inner function fo ; and he described

NOTES

127

the lattice structure of these invariant subspaces. Equivalently, he described the invariant subspaces of the shift operator on t 2 • Subsequently, a large literature has evolved on invariant subspaces. References up to 1964 may be found in Belson [3]. Gamelin [1) has extended Beurling's theory to HP with O
where ex is some complex number of modulus one and tp is some conformal mapping of the unit disk onto itself. The Hilbert space H 2 of course has many other isometries. Forelli also describes the isometries of HP into itself for p "# 2. These are more complicated.

This Page Intentionally Left Blank

EXTREMAL PROBLEMS

CHAPTER 8

We are now prepared to discuss the theory of extremal problems in HP spaces. It turns out that extremal problems come in pairs, and that even a partial solution to one problem may help solve the other. We first discuss the existence and uniqueness of solutions to the two types of problems, and establish an important "duality relation." Then we find the qualitative form of the solutions in the case of a rational kernel, which is general enough to include most of the interesting examples. These results offer a systematic and practical method for solving various extremal problems. At the end of the chapter, a few examples are discussed to illustrate the technique.

8.1. THE EXTREMAL PROBLEM AND ITS DUAL

The most general bounded linear functional on HP (1 sed in the form

¢(!) = -21 . 1t1

JJzl

:$;

f(z)k(z) dz, =1

p < oo) can be expres-

(1)

130

8 EXTREMAL PROBLEMS

where k( e16 ) e H, l /p + l Jq = I. The functionals of greatest interest are generated by kernels k(e;6 ) which are boundary values of rational functions. Some examples are given below. (i)

n

k(z) =

C·

:L i=1z-Pi

IPjl < 1;

1- ,

(ii)

k(z)=n!(z-p)-"- 1, IPI<1;

(iii)

k(z)= Lciz-i- 1 ;

n

¢(!) = L" ciai,

j=O

j=O

wheref(z) = LJ=O aizi. For fixed k e H, the typical extremal problem is to find

11¢11

sup

=

feHP,

II/ lipS 1

!¢(f)!.

(2)

In addition to finding the value of the supremum, several questions are to be considered. Is the supremum attained, and if so, is the extremal function unique? What are the extremal functions? A function he Lq is said to be equivalent to the given kernel k (written h"' k) if h and k belong to the same coset of Lq/Hq; that is, if h- k e Hq. Thus h and k determine the same functional on HP if and only if h "' k. An application of Theorem 7.1 gives at once the important relation sup

/eHP,II/IIPS1

1 2 7t

IJ

lzl~1

f(z)k(z) dzl =min

Ilk- gllq

(3)

geH•

for l ::5; p < oo. This is called the duality relation. It connects the original extremal problem with what is called the dual extremal problem: to find the function g e Hq which is closest to the given kernel k e Lq (or, equivalently, to find the function h "'k of minimal norm). According to Theorem 7.1, the minimum is actually attained; that is, the dual extremal problem always has a solution if l < q ::5; oo. The argument breaks down for p = oo (q = 1), since the conjugate space of L"' is larger than I!. However, the dual extremal problem still has a solution in this case, and the duality relation remains true. To prove this, let C be the subspace of L"' consisting of all continuous periodic functions, and let P be the subspace of C generated by 1, e 16 , e2 i 6, •••• The idea is to view ¢ as a functional on C. By Theorem 7.1, sup feP,

ll/lloos1

1¢(!)1 = min II¢+ t/111, o/lePJ.

(4)

8.1 THE EXTREMAL PROBLEM AND ITS DUAL

131

where Pl_ indicates the subspace of C* which annihilates P. It is easy to describe Pl_. According to the Riesz representation theorem, each 1/J E C* has the form

for some function p.(8) of bounded variation. But 1/J E pl_ implies 1/J(eino) = 0, n = 0, 1, 2, .... Hence it follows from the theorem of F. and M. Riesz that so that 1/J(f)

= -21 . J m

lzl~l

f(z)g(z) dz,

Conversely, every 1/J of this form annihilates P. Thus the relation (4) takes the form sup

1¢(!)1 =min Ilk+ gtl 1 •

feP, 11/lloo.,;;l

geH 1

(5)

In particular, the dual extremal problem has a solution in the case q = 1. On the other hand, for every f E H"' with 11/11 :$; 1 and for every g E H 1, it is trivially true that 1¢(!)1 :$; Ilk+ gll 1 . Hence sup

1¢(!)1 :$;

feP,IIJII.,;;l

1¢(!)1

sup feH"",IIJII.,;;l

min

:$;

Ilk+ gtl.

geH 1

In view of (5), then, equality must hold throughout, and the duality relation is true even for p = oo. Let us now return to the original extremal problem (2) and show that a solution always exists for 1 < p :$; oo. In other words, for each Lq function k(ei6 ), the supremum in (3) is attained. The idea of the proof is to regard 1/J(k)

=~ J 2m

f(z)k(z) dz

(6)

lzl=l

as a bounded linear functional on Lq generated by a fixed function f(ei 6) which belongs (more generally) to I!. Then Ill/Ill= llftl. For 1 :$; q < oo, each 1/J E (m* has the form (6) for some f E I!; and those 1/J which annihilate Hq are precisely the functionals generated by functions f E HP. This implies that the extremal problem in question has a solution, since

11/J(k)l

sup o/1 e(H•)~.

is attained for each fixed k

E

llo/111

:S 1

Lq, by Theorem 7.2.

132

8 EXTREMAL PROBLEMS

The proof fails in the case p = 1, and for a very good reason: the extremal problem (2) need not have a solution in H 1 • A counterexample will be given in Section 8.3. A solution does exist in H 1 if the given kernel k(e 19 ) is continuous. To prove this, consider again the space C of continuous periodic functions and the subspace P which is the uniform closure of the polynomials in ei6 • Each function f e L1 defines, by (6), a functional t/1 e C* with II t/1 II = llfll. The argument given above, using the F. and M. Riesz theorem, shows that the annihilator of P consists of all t/1 e C* of the form (6) withfe H 1 • Theorem 7.2 may therefore be invoked to conclude that the supremum in (3) is attained, which was to be shown. Observe that periodicity of the kernel is not really necessary; continuity on the interval [0, 2n] is sufficient. The conclusion obviously may be extended to any kernel which is the sum of a continuous function and an H 00 function. 8.2. UNIQUENESS OF SOLUTIONS

It will be convenient to reserve the term extrema/function to indicate a solution to the original extremal problem (2). A function K(ei 6 ) " ' k(ei 6 ) for which

IIKIIq = inf llhllq = inf Ilk- gliq h-k

geH•

will be called an extremal kernel. Having established the existence of extremal functions and extremal kernels, it is natural to ask about uniqueness. Since l¢(e 1"l')l = 1¢(/)1 for every real ex, some normalization clearly must be imposed before an extremal function can be unique. It is convenient to single out the extremal functions for which ¢(!) = II¢ II; these will be called normalized extremal functions. To avoid trivial complications, it is always assumed that ¢ is not the zero functional; i.e., k ¢ Hq. The main results on existence and uniqueness can be stated as follows. THEOREM 8.1 (Main existence and uniqueness theorem). For each p (I ~ p ~ oo) and for each function k(e 10 ) e IJ (1/p + 1/q = 1) with k ¢ Hq, the duality relation sup 1¢(!)1 = inf Ilk- gliq JeHP,I\J\1,.,;1

geH•

holds, where ¢(/) is defined by (l ). If p > 1, there is a unique extremal function f for which ¢(!) > 0. If p = 1 and k(ei 6 ) is continuous, at least one extremal function exists. If p > 1 (q < oo ), the dual extremal problem has a unique solution. If p = 1 (q = oo), the dual extremal problem has at least one solution; it is unique if an extremal function exists.

8.2 UNIQUENESS OF SOLUTIONS

133

The existence of solutions was established in Section 8.1; only the uniqueness assertions remain to be proved. The argument is based upon the following simple observation. LEMMA. In order that a function FE HP, with IIFII = 1 and cp(F) > 0, be an extremal function and that K (K,...., k) be an extremal kernel, it is necessary and sufficient that

(7) and that 1K(ei 6 )1 = IIKII

if p

a.e.

IF(e 16 )1P = IIKII-q1K(ei 6 )lq IF(e; 0)1

=1

a.e.

a.e.

on {8: K(ei 6 )

if ::1=

0}

= 1;

l
if p

=

(8)

00.

PROOF OF LEMMA. It is clear from the duality relation that F is a normalized extremal function and K an extremal kernel if and only if.¢(F) = IIKII. The lemma then merely expresses the conditions for equality in Holder's inequality. In the condition for p = 1, the fact is used that F(e 16 ) cannot vanish on a set of positive measure without vanishing identically. PROOF OF THEOREM. Let F and K be a normalized extremal function and an extremal kernel, respectively. Since k E Hq, K(ei 6 ) must be different from zero on a set E of positive measure. Through relations (7) and (8), K determines both sgn F(ei 6 ) and IF(e 16)1 almost everywhere onE, provided 1 < p ~ oo. This means that if 1 < p ::::;; oo, all normalized extremal functions coincide on some set of positive measure, hence coincide almost everywhere. In other words, F is unique. For p = 1, the argument shows at least that sgn F(ew) is determined almost everywhere, if any extremal functions F exist. To prove the uniqueness of K, let F be a fixed normalized extremal function (uniqueness is not used), K any extremal kernel. According to (7),

Re{ie 16F(e 16 )K(e 16 )} = 0

a.e.

Let G = k - K, so that G E Hq; and let h(z) = izF(z)G(z). Then Re{h(e 16)} = Re{ie 16F(e 1')k(e 16 )}

a.e.

But since hE H 1 and h(O) = 0, the knowledge of Re{h(e 16)} completely determines h(z), through the Poisson formula. This shows that G, and hence K, is uniquely determined. The proof is valid in the case p = 1 (q = oo) if an extremal function F exists.

134 8 EXTREMAL PROBLEMS

8.3. COUNTEREXAMPLES IN THE CASE p

=

1

The existence and uniqueness of extremal functions and extremal kernels has now been established for 1 < p :s; oo. The situation is entirely different, however, in the case p = 1. An extremal function need not exist in H 1, and if it does, it need not be unique up to normalization. At least one extremal kernel must exist, as shown above, but it is not in general unique. Counterexamples will now be given to justify the three negative statements. (i) To show that an extremal function may fail to exist, consider the example 1 1 ¢(!) = 2 - . f(x) dx, 1tl -1

I

The existence of the integral (in the principal value sense) can be proved by applying Cauchy's theorem with a contour consisting of the segment -r $ x :s; rand the semicircle z = rei 9 , 0 :s; () :s; rr. Sincefe H 1 , it follows that lim r-+1

r

f(x) dx = - i {f(ej 9)ei9 d(),

-r

0

so that¢(!) has the usual form (1) with a kernel

An equivalent kernel is K(z) = k(z) + !-; hence 11¢11 $ \IK\1 =-!-- This is also a consequence of the Fejer-Riesz inequality (Theorem 3.13). In fact, the example used to prove the sharpness of the Fejer-Riesz inequality, a conformal mapping of izl < l onto an elongated rectangle, shows that 11¢11 =!. Thus K is an extremal kernel. If there exists an extremal function F, normalized so that ¢(F)= t, then by (7) e19 F(e 19 )K(ei9) ;;::: 0

a.e.

Therefore, -rr<()
Since F e H 1 , it follows that 0=

r

ei<•+1J9F(/e) d()

-n

=

I

0

-n

ei"91F(ei9)1 d() -

Je;"eiF(e;e)l d(), "

0

n = 0, 1, 2, ....

8.3 COUNTEREXAMPLES IN THE CASE p = 1

135

Consequently, 0

J

de= JeinBIF(e'B)I de, ~

einBIF(e'B)I

-~

n = 0, ± 1, ±2, ....

0

In other words, the two functions

-n::>;e::>;O

o < e < n,

and

have identical Fourier coefficients. This implies I/J 1 (e) which is obviously impossible. (ii)

= I/J 2 (e)

a.e., so F(z) = 0,

The elementary example 1

2~

J 2n

¢(!) = -

f(e's)e-is

de,

0

shows that a normalized extremal function need not be unique. Here it is obvious that ll¢ll = l, and thatf(z) = z is an extremal function. But a simple calculation shows that for every complex constant ex, ( ) _ (z

f z -

+ ex)(1 + ciz) 1 + lexl 2

is also extremal. (iii) A more elaborate example is needed to demonstrate the nonuniqueness of an extremal kernel. In view of the main theorem, this will automatically be another case in which no extremal function exists. Let

I, k(e 18 ) =

-1,

0,

1t

O::>;e::>;2 1t

2 <e::>;n

-n < e <

o,

and let ¢([) be the functional with kernel k, defined by (1). Obviously, II¢ II ::>; I. To show that II¢ I = l, consider the function g(z) which maps lzl < I conformally onto the rectangle with vertices ± I ± iE, in such a way that g(l) = -1 - iE, g(i) = l, and g( -I)= -I + iE. Let eia. (0 <ex< n/2) be the

136

8 EXTREMAL PROBLEMS

point carried into (1 - iE). Then g(ei
io){ ~o.0

::;:;,

e <ex, e < 71:,

0< 1t -

ex <

while ei 6g'(e 16 ) ~ 0 for ex < 8 < n - ex. Thus

2nicp(g') =

f

rt/2

"

ie 16 g'(e 16 ) de-

J

iei 6g'(e 16 ) d8

~2

0

= (2 + iE) - (iE - 2) = 4. On the other hand,

Therefore, lc/J(g')l llg'll1

= _1_ -4 1 1+E

as

E ____.

0,

which implies II¢ II = 1. It now follows from the duality theorem that k is itself an extremal kernel, since Ilk II oo = 1. The object is to produce another extremal kernel; that is, another function K"' k with IlK II oo = 1. Let K = k + h, where h(z) maps lzl < 1 conformally onto the half-disk lwl < 1, lm{w} > 0, with h(l) = -1, h(i) = 0, h( -1) = 1. Obviously, K"' k. But lh(ei 6)1 = 1, IK(e 16)1 = ( 1 + h(ew) ::;:; 1, 1-1 + h(ei6 )1::;:; 1, Hence IlK II oo

-n<8<0; o::;:;e::;:;n/2; n/2 < e::;:; 71:.

= 1.

8.4. RATIONAL KERNELS

If the kernel k(ei 6 ) is the restriction to the unit circle of a rational function k(z), it is possible to describe the extremal functions and extremal kernels more or less explicitly. This is of some interest, because most of the functionals which arise in practice have rational kernels. Some examples were given at the beginning of the chapter. Let k(z) be analytic in lzl < 1 except for poles at fl~o {3 2 , ••• , {3. (lflkl < 1), each pole being repeated according to multiplicity. Suppose k(z) is continuous elsewhere in lzl ::;:; 1. Then by Theorem 8.1, a normalized extremal function F

8.4 RATIONAL KERNELS

137

exists for each p, 1 :o:; p :o:; oo. F is unique for 1 < p :o:; oo, but not necessarily for p = 1. An extremal kernel K exists for each p, 1 :o:; p :o:; oo, and is unique (even for p = 1, since F exists). Consider the function R(z) = zF(z)K(z), which is analytic in tzl < 1 except perhaps for poles at the points fh. According to the lemma of Section 8.2, R(ei 6) ;:.::: 0 a.e., because F and K are extremal. Furthermore, since b

lim r-+ 1

J IR(re'

6) -

R(e' 6)1 d8 = 0

a

for any arc (a, b) on lzl = 1, the classical technique of Schwarz reflection may be applied. (See, for instance, Nehari [5], pp. 183-185.) The conclusion is that R(z) has a continuation onto the extended z-plane as a function analytic except perhaps for poles at {3 1, ... , fln; 1/{3 1, ... , 1/fln. Hence R is a rational function. If R(a.) = 0 for Ia. I < 1, then 1/ii. is a zero of the same multiplicity. Furthermore, any zero on the boundary tzl = 1 must be of even multiplicity, since R(z) ~ 0 there. Thus there are points a. 1, ... , a.m, la.;l :o:; 1, in terms of which R has the structure R(z)

Tir= 1 (z - a.;)(l -ex; z) = Cz TI~= 1 (z- {3;)(1- fl,z) .

By the reflection property, the order of the zero (or pole) of R at the origin must be the same as the order of the zero (or pole) at infinity. If 1-L of the a.; and v of the {3, are equal to zero, this statement is expressed by the equation 1-L -

v

+ 1 = (2n

- v) - (2m - /-L) - 1.

Hence m = n- 1. It follows, incidentally, that C > 0, smce R(z) ~ 0 on lzl = 1. Some of the a.; which are zeros ofF may possibly coincide with numbers {3;. However, K has a pole at each {3,, and in fact its principal part at each pole must coincide with that of k. Let the a.; be renumbered, if necessary, so that a. 1, ... , a., are the zeros of K and a.,+ 1, ... , a.a those ofF in lzl < 1. Then Ia.; I= 1 fori= u + 1, ... , n- 1. The functions -

K(z)

• 1-~z" z-{3, - - TI - = i=1z- a.; i=11-{3;z

= K(z) TI

and -

F(z)

= F(z)

a 1-ii;z TI i=s+1 Z - CY.;

are analytic and nonvanishing in tzl < 1. FE HP, while K(z) is continuous in tzl :o:; 1. Furthermore, F and K are outer functions. To prove this, it suffices

138

8 EXTREMAL PROBLEMS

to show that the product R(z) = F(z)K(z) has no singular factor. In view of the structure of R(z), however, R(z) is a rational function without zeros or poles in Jzl ~ I, except perhaps for zeros on Jzl = I. Thus the problem reduces to showing that (z-oe) is an outer function if Joel = I. But this follows from the fact that (z- oe)- 1 E HP for p < I. Consequently,

F(z)

= e'Y. exp {-1

J

2x

eir

+z

.

}

(9)

-i-r-logJF(e'r)l dt ,

2n o e - z

and similarly for K(z). The relations (8) between F and K are yet to be used. Suppose first that 1

a.e.

on

izl

= I.

= R(z) on izl = I, so IF(z)l

= IIKII- 11 P[R(z)]l 1P = (-

C )1/pn-1

n

fl 11- a;zi 21 Pfl 11- /i;zi- 21 P

IIKII

i=1

i=1

a.e. on Jzl = I. Combination of this expression with (9) gives

z

t1

F(z) =A

_(X·

n-1

n

fl ------=:!- fl (1- iX;z) 21 Pfl (1- /f;z)- 21P,

i=s+11-oe;Zi=1

i=1

(10)

where A is a complex constant. Since zF(z)K(z) = R(z), it follows that

K(z)

s

= B fl

z

(X

n-1

n

- ; fl (1- a;z)2tq fl

i=11-oe;Zi=1

(1

i=1

-

{J-z)1-2jq i

.

(11)

z-{J;

Here the fact has been used that z-oe 1- iiz

if Joel = 1.

--==-(X

(12)

The structural formulas (10) and (I I) remain valid in the cases p = I and p = oo, with the understanding that 1/oo = 0. This may be verified by similar calculations based upon the appropriate relations (8). For p = oo, the extremal function takes the simple form •

F(z) = e'Y

fl t1

z-

i=s+ 1 1 -

!X; !X;

z

The practical value of the formulas (IO) and (I I) will in general depend upon the actual calculation of the parameters oe;. This can be a very difficult problem.

8.5 EXAMPLES

139

Suppose, however, that an integer u::::; n- 1 and numbers Band cx 1 , cx 2 , . . . , cx,_ 1 can be found with lcx;l < 1 forl::::; i::::; uand lcx;l = lforu + 1::::; i::::; n- 1, such that for somes::::; u the function K given by the formula (11) is equivalent to the original kernel k. Then K is the extremal kernel and, with an appropriate choice of the constant A, F defined by (10) is a normalized extremal function (unique if p > 1). This follows from the lemma of Section 8.2. The problem therefore reduces to an interpolation problem: to find numbers Band ex;, subject to the above restrictions, such that K has the same principal part ask at each of the poles {3;. This problem is guaranteed to have a unique solution if 1 < p::::; oo, since there is a unique kernel K of the form (11) and a unique normalized extremal function F with the structure (10). The interpolation problem has at least one solution in the case p = 1. Here it is a question of finding numbers B and cx 1 , . . . , ex, with lcx;l < 1 and s:::; n- 1, such that

has the given principal parts at the poles {3;. This modified extremal problem will have a unique solution, because there is only one extremal kernel. If it turns out that s = n - 1, the extremal function F is uniquely determined by (10), with the first product missing. Otherwise (if 0 ::::; s < n - 1), the remaining ex; can be chosen arbitrarily in the disk lcxl ::::; 1. If lcx;l < 1 for s + 1 ::::; i::::; u and lex; I = 1 for u + 1 ::::; i::::; n - 1, the function (10) (with p = 1) is necessarily an extremal function, and every extremal function is obtained in this manner. This characterizes the family of extremal functions in the indeterminate case. 8.5. EXAMPLES

(i) A coefficient problem and its dual. Given complex constants Co, C1, ... , Cn (en # 0), let it be required to find

Jcoao + c1a1 + ... +en ani•

sup

(13)

feHP, II/1Jp,;1

where f(z) = Li=o aizi. This coefficient problem was posed and elegantly solved by E. Landau in the special case p = oo and c0 = c 1 = · · · =en = 1. From the present point of view, it is an extremal problem with the rational kernel

k(z)

n

L ciz-i-

=

j=O

The dual extremal problem is inf geH•

Ilk- gllq

1•

140

8 EXTREMAL PROBLEMS

which is equivalent to the problem

In other words, the dual extremal problem is to find inf llhll among all Hq functions of the form CXl

h(z)=c,.+cn_ 1z+···+c 0 zn+

L

j=n+1

aizi.

The latter is a "minimal interpolation" problem at the origin. According to the general theory, there is always a unique solution (I ::::; q ::::; oo ), and the value of the minimum is equal to the supremum in (13). The problem (13) also has a unique normalized solution if 1 < p::::; oo, and at least one solution ifp = 1. In fact, these solutions take the forms (II) and (10), respectively. The parameters B and cxi must be determined so that Jcx;l ::::; 1, Jcxil < 1 for 1 ::::; i::::; s, and

B

n 1z-- n(1- CX;Z) CXj

s

i= 1

n

-

2

fq = Cn + Cn-1Z +···+Co~+···.

(14)

CX; Z i = 1

Assuming p > 1 and recalling that en "# 0, let CXl

( c + cn-1 z + ... + c0 zn)qfl = "i...J ).1. zi 11

j:O

for small

lzl. The numbers ). 0 , A1 ,

c0

Cn . If

, C1 , ... ,

. . . , ).n

can be expressed m terms of

P,.(z) = ).0 + ). 1z + · · · + ).,zn

does not vanish in

izl < 1, then n

Pn(z)

= ).0

TI (1- ii;z),

i=1

and n

[Pn(z)]2fq =en

TI (1- ii;z)2fq

i= 1 is the required solution to (14). Thus n

K(z) =en

TI (1- ii;z)21qz-n-1

i= 1

is the extremal kernel, and the extremal function has the form a

F(z) =A

TI

z-

CX·

~

n

TI (1- ii;z) 21P,

i=1 1- CX;Z i=1

8.5 EXAMPLES

141

where a 1 , . . . , a,. are the reflections of the zeros of Pn(z), if any, which lie outside the unit circle. The functions K and F are not so easily found if Pn(z) vanishes in izl < I. Various criteria are available, however, for concluding that P.(z) has no zeros inside the unit circle. (See M. Marden [I].) For example, the classical Enestrom-Kakeya theorem asserts that this is the case if A-0 ~ A- 1 ~

•· ·

~

A.. > 0.

If A- 0 > A- 1 > · · · > A.. > 0,

all the zeros of P.(z) lie strictly outside the unit circle. (ii) Minimal interpolation. Let distinct points z 1 , •.. , z. be given in 1. It is required to find an HP function (I :::;; p :::;; oo) of minimal norm which takes prescribed values

izl <

j

=

l, ... , n,

at the given points. The problem could be generalized by prescribing the values of the function and its first few derivatives at the points zi, but the more special problem will serve to illustrate the main ideas. It is not immediately clear that a solution exists at all. However, the problem can be cast in the form of a dual extremal problem with rational kernel, which will ensure the existence and uniqueness of a solution. Let

n~ i=ll-ziz n

B(z)=

Z - Z·

Define h(z)=

"

Bj(z)

L wi-(-)' Bizi

i=l

Then h(z) is analytic in izl :::;; I and h(z) HP function taking the values wi at zi is f(z)

= wi, J =

l, 2, ... , n. The general

= h(z)- B(z)g(z),

where g is an arbitrary HP function. The problem of minimizing fore equivalent to the problem of finding min ilk-

geHP

where

g\IP'

1\fll

is there-

142

8 EXTREMAL PROBLEMS

This is a dual extremal problem with the rational kernel k. The associated functional on Hq (lfp + 1/q = 1) is

¢(!) = -21 . 1l!

J

f

=

k(z)f(z) dz

Jzl='l

wi(~~ l;il 2 ) f(zj). j Zj

j='1

The duality relation (3) therefore gives the curious result

II fliP=

min /eHP,f(zj)=wJ

max /eH•,IJ/IJ 0 Sl

If

wj{ 1 - lzil 2 ) f(zi),. Bj(zj)

j=1

(iii) Extremal problems in H 2 • If the kernel is rational and p the extremal kernel (11) takes the form s

K(z)

= B IT (z- o:;) 1=1

n-1

IT

(15)

= q = 2,

11

(1

-a; z) IT (z -

/3;)- 1 •

(16)

i=l

l=s+1

The parameters must be chosen so that this function K is equivalent to the given kernel k. But the" natural kernel" k*, defined as the sum of the pdncipal parts of k at each of the poles f3 i> has precisely the form (16): n

k*(z) = Q(z)

IT (z -

{3;)- 1 ,

!=1

where Q is a polynomial of degree (n - 1) or less. Hence the natural kernel is always the extremal kernel in the H 2 case. Knowing the extremal kernel, it is easy to find the extremal function F as given by (10), without actually determining the roots o:;. Indeed, in view of (12), __

s

a

zn- 1 Q(1/z) =cIT (1- ii[z) i='

1

n-1

IT 1(z- 0:;) IT 1(1- il;z).

i=s+

i=a+

Hence F(z) =

A~- 1 Q(1/z)

IT (1 - ~ z)-

1•

i= 1

(iv) The coefficient problem in H 1 • Let us return now to the coefficient problem (13) and its dual, the minimal interpolation problem at the origin. Let mP be the value of the maximum in (13). Obviously, m2 =

{lcol 2+ lc1l 2+ · · · + lcnl 2}112 .

It is also possible to find m 1 , at least up to an algebraic computation. Any H 1 functionf(z) = L an zn can be expressed as the product of two H 2 functions gandhwith ·

11/111 = llgll2 llhll2.

143

EXERCISES

If CXl

=

g(z)

CXl

L xkz\ k=O

h(z) =

L ykz\ k=O

then Thus

L" ckak= j=O L" k=O L" ci+kxiyk,

k=O

j+ksn

and the problem reduces to finding m1

where

=

max

[[x[[ S1,Jiy[[ S1

I f k=Of A j=O

ik xi Yk \'

llxll 2 = lx0 12 + · · · + lxnl 2 and (A ik) is the Hankel matrix (Aik) =

[~: ~~ c"

c2::: ~"]·

0

··· 0

The maximum of this symmetric bi-quadratic form can be characterized as the greatest absolute value of the eigenvalues of the matrix (A ik). By the duality relation, this is also the norm of the minimal interpolating function. EXERCISES

1. Show that if f(z)=l.,akzkeHP (1
2. Show that if f(z) = only if /has the form

L akzk E Hi, then

f(z) =A

la"l

TI" (z- cx;)(l -

5,

lanl::::; 11/IIP, with

11/11 1, with equality if and

CX;z),

i= 1

where the

ex, are arbitrary complex constants. distinct points z1 = 0, z 2 , ••• , z" in lzl <

3. For find the HP function of minimum norm such that f(O)

= 1;

What is the minimum norm?

1, and for 0 < p

f(z2) = · · · = f(z") = 0.

5,

oo,

144

8 EXTREMAL PROBLEMS

4. Show that iff e HP (0 < p ::::; oo ), then lf(z)l ::::; (1 - Jzi 2)- 11PJJ/IIP,

izl < 1.

Show that the inequality is sharp for each fixed z. 5. For fixed z, Jzl < 1, and for each positive integer n, find the maximum of JJC"l(z)J over all H 2 functions f with 11/11 2 ::::; 1. Determine the extremal functions. 6. For fe H 2 and for lz1l < 1, lz2l < 1, and z1 "# z 2 , prove

I {

} 112 llfll f(zd- j(z2) 1- Jz1z2l 2 2' 2 2 2 Z1- Z2 :5: 11- zlz21 (1 -lzll )(1 -lz2l ) ' and show that the constant is best possible. 7. For each fixed z, Jzl < 1, and for each odd integer n =2m+ 1, prove the sharp inequality

and identify the extremal functions. (Hint: Show that the natural kernel is extremal.) (Szasz [2]; Macintyre and Rogosinski [2].) 8. Let M. denote the maximum of Ja 0 + a 1 + · · · + a.l among all H"' functions f(z) = L ak-!' with 11/11 00 ::::; 1. Show that

k(-t) (2k)! k =4k(k!)2"

n

M. =

L (J.k)2,

k=O

.J.k=(-1)

Hence M."' 1/n log n as n-> oo. (Hint: Observe that

{f ).kzk}2

=

1+ Z

+ ... + z" + bn+lzn+l + .... )

k=O

(Landau [1, 2].) 9. Find the distance from H 1 to the function

k(z)=(l-z-n-1)(1-z)-I; that is, find the minimum of II k -/11 1 for all f 10. Show that for f(z) = . max Ja 0

llflltsl

E

H 1•

L akzk E H 1, + a 1 + · · · + a.l

1 (n + 1)n =-sec . 2 2n + 3

NOTES

i45

(Hint: Observe that the eigenvalue problem Xn-k

+ Xn-k+l ·+- · • · +X,. = }.xk •

k=O,l, ... ,n,

has the solution

1 (n + l)n A.=- sec ; 2 2n + 3

xk

. (k + l)n =sin 2 , n+ 3

k= 0, I, ... , n.)

(Egervary [1 ].) NOTES

The theory of extremal problems in 1-fP spaces has evolved from a long history of scattered work on special examples. Landau [1, 2] solved the coefficient problem (i) in Section 8.5 in the case p = oo and c0 = c 1 = · · · = c,. = 1. Szasz [3, 4] discussed the more general problem in 1-f"'. Caratheodory and Fejer [2] proposed the "minimal interpolation" problem to minimize II! II oo among all f E H 00 whose Taylor coefficients a0 , a 1 , ••• , an at the origin are prescribed. They proved the existence and uniqueness of a solution, and they gave an algebraic method for computing the minimum. Gronwall [1] simplified this work. F. Riesz [1] discussed the same problem in I-1 1 , proved the existence and uniqueness of a solution, and characterized it using a variati anal method. Fejer [1] solved the coefficient problem (i) of Section 8.5 in H 1 , as well as its dual, the Caratheodory-Fejer problem, essentially by the method described in (iv) of Section 8.5. Egerv;iry [1] obtained a completely explicit solution in the case c0 = c 1 =···=en= 1. Kakeya [1, 2] considered the more general interpolation problem (ii) of Section 8.5 and found the form of the solution. In the H 00 case, this problem was also discussed by Pick [I, 2), Schur [I], and R. Nevanlinna [1]. Geronimus [1, 2, 3) treated some problems similar to those of Caratheodory-Fejer and Riesz. Doob [1] considered the problem of approximating an arbitrary rational kernel by an H 1 function, proved the existence and uniqueness of a solution, and found its gualita1ivc form. Goluzin [1, 2] discussed a wide variety of extremal problems. The first systematic attempt to unify the theory and to extend it to 1-fP (I ::;:; p ::;:; oo) was the paper of Macintyre and Rogosi nski [1 ]. They formulated the duality relation in full generality, proved the existence and uniqueness of extremal functions and extremal kernels, described their form in the case of a rational kernel, and applied the theory to obtain explicit solutions to a great many extremal problems. Nevertheless, the theory achieved complete unity and elegance only wlten Havinson [1, 2] and (independently) Rogosinski and Shapiro [1] introduced

146

8 EXTREMAL PROBLEMS

the methods of functional analysis. A. L. Shields discusses this development in a historical note following his translation of Havinson's paper [2]. Subsequent refinements by H. S. Shapiro [1 ], Havinson [3), Bonsall [1], and others have led to the main existence and uniqueness theorem as stated and proved in Sections 8.1 and 8.2. The three counterexamples given in Section 8.3 are essentially due to Rogosinski and Shapiro [1]; the third is given in a simplified form suggested by Caughran [1). Extremal problems for HP spaces over multiply connected domains (see Chapter 10) have been studied by Ahlfors [1], Garabedian [1], Nehari [2, 3, 4], Rudin [1, 2], Havinson [3], Penez [1], Lax [1], Tumarkin and Havinson [8], and others. Akutowicz and Carleson [1] have studied the analytic continuation properties of extremal interpolatory functions.

INTERPOLATION THEORY

CHAPTER 9

Let z1 , z 2 , ••• be distinct points in the open unit disk, and let 1r1 , w2 , ••• be arbitrary complex numbers. The general interpolation problem is to characterize the pairs of sequences {zk} and {wk} for which there exists a function f E HP withf(zk) = wk, k = 1, 2, ... , and to find all of the interpolating functions. But in such general form the problem is difficult, and has not yet been fully solved (see Notes). In this chapter we confine attention to the more modest problem of describing the "universal interpolation sequences." These are the sequences {zk} which admit an H"' interpolation for every bounded sequence {wk}. This restricted problem has a natural generalization to HP (0 < p < oo) which we also discuss. The chapter concludes with a theorem of Carleson which generalizes a special result used in proving the main interpolation theorem. Carleson's theorem plays an important part in the proof of the corona theorem (Chapter 12), and it has other applications. 9.1. UNIVERSAL INTERPOLATION SEQUENCES

Consider the problem of finding the most general function jE HP such that f(zJ = wk, k = 1, 2, .... If a particular solution f can be found, it is an easy

148 9 INTERPOL4TION THEORY

matter to describe the general solution. Indeed, the interpolation f is unique if and only if:L (I -lzkl) = oo, since the difference of two interpolations would vanish at the points zk. If L (I - lzkl) < oo, then the general solution is f + Bg, where B is the Blaschke product with zeros {zk} and g is an arbitrary HP function. In discussing the existence of an HP interpolation, it is helpful to adopt a more abstract point of view. Fixing p (0