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Introduction to Hardy spaces

1

CHAPTER 1

Preliminaries

1. Sequences and families of holomorphic functions

Given any sequence (fn) of holomorphic functions in the open set U ⊂C which converges uniformly on every compact subset of U to a functionf : U 7→ C, it follows that f is also holomorphic on U . We shalllist some basic results about uniform convergence on compacts. Theirproofs and additional information is defered to Appendix A.

Theorem 1.1. Let (fn) be a sequence of holomorphic functions on theopen connected set U ⊂ C that converges uniformly on compacts to the(holomorphic) function f . Let V be an open subset of U such that thereexists n0 ≥ 1 with fn(z) 6= 0 for all z ∈ V and n ≥ n0. Then either fis the constant function 0, or f has no zeros in V .

Definition 1.1. A family F of holomorphic functions on the openset U ⊂ C is called normal if every sequence (fn) in F contains asubsequence (fnk) that converges uniformly on compact subsets to some(holomorphic) function f (not necessarily in F !)

Theorem 1.2. (Montel). A family F of holomorphic functions on theopen set U ⊂ C is normal if it is uniformly bounded on compacts, thatis, for every compact set K ⊂ U there exists a positive constant CKsuch that for all f ∈ F and all z ∈ K we have

|f(z)| ≤ CK .

For uniform convergence on compacts we have the following useful cri-terion.

Theorem 1.3. (Vitali) Let (fn) be a sequence of holomorphic functionson the open connected set U ⊂ C. Suppose that {fn, n = 1, 2, . . .} isuniformly bounded on compacts and also that there is a nondiscretesubset A of U such that (fn) converges pointwise on A. Then (fn)converges uniformly on compacts.

We want to apply this theorem to infinite products of bounded analyticfunctions.

3

4 1. PRELIMINARIES

Proposition 1.1. Let (fn) be a sequence of holomorphic functions onthe open set U ⊂ C with the property that |fn(z)| ≤ 1 for all z ∈ U andall n ∈ N. Let (Fn) be defined by

Fn(z) =n∏k=1

fk(z), z ∈ U

and assume that there is z0 ∈ U such that (Fn(z0)) has a nonzero limit.Then (Fn) converges uniformly on compacts to a nonzero holomorphicfunction F on U and each zero of F is the zero of some fn, n ∈ N.

Proof. Note that the family {Fn, n ≥ 1} is uniformly boundedin U by 1, hence, it is a normal family by Montel’s theorem. As inthe proof of Vitali’s theorem it suffices to show that if (Fnk) and (Fn′k)subsequences that converge uniformly on compacts to G and H respec-tively, then G = H. Since |Fn| ≥ |Fn+1|, we can choose a subsequence(n′′k) of (nk) such that n′′k ≥ n′k which implies

|G(z)| = limk→∞|Fn′′k (z)| ≤ lim

k→∞|Fn′k(z)| = |H(z)|.

Analogously, one shows that |H| ≤ |G| and we obtain that |G| = |H|.This implies that G = αH for some α ∈ C with |α| = 1. But, byhypothesis, we have also G(z0) = H(z0) 6= 0 which implies α = 1and G = H. To see the assertion about the zeros, note first that ifK ⊂ U is compact then K ∩ (∪nf−1

n ({0})) is finite, otherwise F wouldbe identically zero. Then if z is a zero of F that is not a zero of any fnthere is an open neighborhood of z that contains no zeros of fn, n ≥ 1and this leads to a contradiction by Theorem 1.1. �

Let us denote from now on by D the (open) unit disc (i.e. the disccentered at the origin and of radius one). Our next result will be fre-quently used in what follows and is a direct consequence of Proposition1.4 combined with the following simple but important identity:

If z, w ∈ C with wz 6= 1 then

(1.2) 1−∣∣∣∣ z − w1− wz

∣∣∣∣2 =(1− |z|2)(1− |w|2)

|1− wz|2.

Indeed, this can be verified by a direct calculation (exercise!).

Corollary 1.1. Let (an) be a sequence in D\{0}. If∑

n(1−|an|) =∞then the infinite product

B(z) =∞∏n=1

−an|an|

z − an1− anz

1. SEQUENCES AND FAMILIES OF HOLOMORPHIC FUNCTIONS 5

converges uniformly to zero on compact subsets of D. If∑

n(1−|an|) <∞ then the product converges uniformly on compacts to a nonzero holo-morphic function in D. Moreover, in this case, the zeros of B are pre-cisely the points an, n ≥ 1 and for fixed m, the multiplicity of the zeroam equals the number of occurences of am in the sequence.

Proof. Assume that∑

n(1− |an|) =∞ and let w ∈ D. Note that

N∏n=1

|w − an|2

|1− anw|2= exp

(N∑n=1

log|w − an|2

|1− anw|2

)≤ exp

(N∑n=1

(|w − an|2

|1− anw|2− 1

)).

By (1.2) we have

1− |w − an|2

|1− anw|2=

(1− |w|2)(1− |an|2)

|1− anw|2≥ (1− |w|2)(1− |an|2)

(1 + |w|)2

which implies that

N∑n=1

(|w − an|2

|1− anw|2− 1

)≤ − 1− |w|2

(1 + |w|)2

N∑n=1

(1− |an|2)→ −∞

when N →∞, and hence, the product converges to zero uniformly oncompacts.

Now assume that∑

n(1 − |an|) < ∞. From (1.2) we deduce that thefactors fn(z) = −an

|an|z−an1−anz satisfy

1− |fn(z)|2 =(1− |z|2)(1− |an|2)

|1− anz|2> 0

for all z ∈ D, that is |fn(z)| < 1, z ∈ D. If the above sum converges,it is a standard matter to conclude that limn→∞

∏nk=1 |ak| exists and

is nonzero. Thus, the two assumptions in Proposition 1.1 are satisfiedwith z0 = 0 and the convergence of the product follows. The statementabout the zeros of B is also an immediate consequence of Proposition1.1. �

The functions considered in Corollary 1.1 are called Blaschke productsand they play an important role in the theory of analytic functions inthe unit disc. If we want to allow Blaschke products to have zeros atthe origin as well (and we should) then the general definition reads asfollows: A Blaschke product is a holomorphic function in D of the form

(1.3) B(z) = zm∞∏n=1

−an|an|

z − an1− anz

,

6 1. PRELIMINARIES

where m is a nonnegative integer, (an) is a sequence in D with∑

n(1−|an|) <∞.

Exercise 1. Show that the following families of holomorphic functionsare normal.

(i) F1 = {fn, fn(z) = zn, |z| < 1, n ∈ N}.(ii) The family F2 of all power series f(z) =

∑n≥0 anz

n with |an| ≤4, n ≥ 0.

(iii) The family F3 of all holomorphic functions f in the open set U ⊂ Cwith the property that ∫

U

|f(x+ iy)|dxdy < 5.

Exercise 2. Show that if F is a normal family of analytic functions inthe open set U then the families F (n) = {f (n), f ∈ F} are also normalfor all n ∈ N.

Exercise 3. Prove the converse of Montel’s theorem.

Exercise 4. Let f be analytic in D and continuous in D such that|f(z)| ≤ 147 for all z ∈ D. Show that if a1, a2, . . . , an are zeros of fand B is the (finite) Blaschke product with zeros ak, 1 ≤ k ≤ n, then|f(z)| ≤ 147|B(z)| for all z ∈ D.

2. Boundary behavior of power series

This section is a primary on the subject. It is intended to provide someintuition and motivation for the results that follow. For this reason, itwill contain a number (four) unproved results that are important forour discussion.

Given a power series with a positive radius of convergence (in otherwords, an analytic function in a disc) we would like to investigate itsbehavior near the boundary points of the disc of convergence. As itis well known from calculus in 2 variables, our observations may de-pend heavily on the way we approach such points and this does indeedhappen as the following example shows.

Example 2.1. Let f(z) = exp(−1+z1−z ), z ∈ D. Then:

2. BOUNDARY BEHAVIOR OF POWER SERIES 7

(i) f is bounded in D and

limr→1−

f(r) = 0,

(ii) For every angle A with vertex at 1, symmetric w.r.t. the segment[0, 1) and with opening less than π we have

limz→1z∈A∩D

f(z) = 0,

(iii) For every w ∈ D there exists a sequence (zn) in D with zn → 1 and

limn→∞

f(zn) = w.

To prove these assertions, let g(z) = 1+z1−z , z ∈ D. Then clearly, |g(z)| →

∞ if and only if z → 1. Let us show first that g(D) equals the righthalf plane. Indeed, if g(z) = w and |z| < 1 then

Rew = Re1 + z

1− z=

Re(1 + z)(1− z)

|1− z|2=

1− |z|2

|1− z|2> 0.

Conversely, if Re w > 0, a similar computation shows that |z| < 1.

(i) Clearly, |f | = exp(−Re g) ≤ 1 if Re g > 0. The second assertion isobvious since Re g(r) = 1+r

1−r →∞ when r → 1.

(ii) Every z = reit ∈ A ∩ D can be written in the form z = 1 + ρeiθ,where ρ > 0 and |θ − π| ≤ α < π/2. Then

r = (1 + ρ2 + 2ρ cos θ)1/2, eit =1 + ρeiθ

(1 + ρ2 + 2ρ cos θ)1/2

andeit − 1

1− r2= − 1 + ρeiθ − (1 + ρ2 + 2ρ cos θ)1/2

(1 + ρ2 + 2ρ cos θ)1/2ρ(ρ+ 2 cos θ)

Now the inequality satisfied by the angle θ implies that this quantitystays bounded when ρ→ 0, or equivalently, z → 1, z ∈ A ∩ D. This isfurther equivalent to the fact that

(2.1) lim supreit→1reit∈A∩D

|t|1− r

<∞.

With this fact at hand, the proof of (ii) is immediate. Indeed, fromabove we have that

Reg(reit) =1− r2

|1− reit|2=

1− r2

1 + r2 − 2r cos t=

1− r2

(1− r)2 + 4r sin2( t2)

8 1. PRELIMINARIES

and from (2.1) it follows that

lim infreit→1reit∈A∩D

Reg(reit) =∞.

(iii) Clearly, it suffices to prove the statement for w ∈ D. Write w = reiθ

and let wn = − log r + i(θ + 2nπ). Then Rewn > 0 and there existszn ∈ D with g(zn) = wn and since wn → ∞ when n → ∞, it followsthat limn→∞ zn = 1. Obviously, f(zn) = w for all n.

Motivated by the above example we introduce the following terminol-ogy.

Definition 2.1. Let f be a complex-valued function defined on D andlet ζ ∈ ∂D.

(i) We say that f has the radial limit L at ζ if

limr→1

f(rζ) = L.

(ii) We say that a sequence (zn) in D converges nontangentially toa point ζ ∈ ∂D if there is an angle with vertex at ζ, symmetric w.r.t.the ray joining ζ to the origin and with opening less than π such thatzn belongs to this angle for all n sufficiently large.

(ii) We say that the function f has the nontangential limit L at thepoint ζ ∈ ∂D if for every sequence (zn) in D that converges nontangen-tially to ζ we have

limn→∞

f(zn) = L

To avoid confusions about the sets involved in the definition of non-tangential convergence one usually considers Stolz angles which are setsobtained in the following way: Take a disc of radius 0 < σ < 1 centeredat the origin and denote by Γσ(ζ) the union of all line segments thatjoin a point in that disc and ζ. In other words, Γσ(ζ) is the convex hullof the above disc and the point ζ. Then a sequence (zn) in D convergesnontangentially to a point ζ ∈ ∂D if there exists 0 < σ < 1 such thatzn ∈ Γσ(ζ) for all n sufficiently large.

Remark 2.1. Recall that in general, 1− |z| ≤ |z− ζ| if z ∈ D and ζ ∈∂D and of course, if z → ζ it may happen that 1−|z| goes much fasterto 0 than |z − ζ|. The point is that if z approaches ζ nontangentiallythis cannot happen, i.e. the quotient (1 − |z|)/|z − ζ| stays boundedbelow. This statement is equivalent to the one in the next exercise anda proof was already pointed out in the previous example.


Exercise 1. Prove in detail that a sequence (zn) in D with zn = rneiθn,

converges nontangentially to a point ζ = eiθ ∈ ∂D if and only if |θn−θ|1−rn

stays bounded when n→∞.

Exercise 2. Show that the power series

f(z) =∞∑n=0

zn!, z ∈ D

has no radial limit at any point eit with t/π ∈ Q.

Throughout in what follows m will denote the normalized arclength-measure on the unit circle T = ∂D, which can be identified with thenormalized Lebesgue measure on [0, 2π].

It turns out that the radial limits of a power series provide much lessinformation about the function inside the disc than the nontangentiallimits. Thus, for example, it is known that there exist analytic func-tions f in the unit disc that are not identically zero and yet satisfy

limr→1

f(reit) = 0 a.e. on [0, 2π]

(see [Col]). An even more dramatic example has been constructed byKahane and Katznelson [KK]. They show that there exist such ”bad”functions f that satisfy in addition a growth restriction of the form

|f(z)| ≤ C(1− |z|)−α

for α > 0 fixed but arbitrary!

On the other hand, this cannot happen with nontangential limits. Thisis essentially the content of a famous theorem of Privalov that is statedbelow without proof (for a proof see [Koo] ).

Theorem 2.1. (Privalov) If f is analytic in D and has the nontangen-tial limit zero on a set of positive measure on the unit circle T, then fvanishes identically.

Exercise 3. Given an analytic function f in D show that the set ofpoints t ∈ [0, 2π] with the property that f has a nontangential limit ateit is measurable.

Example 2.2. If µ > 1 is fixed then the infinite product

f(z) =∞∏n=0

(1 + µzn)

10 1. PRELIMINARIES

converges uniformly on compacts on D to a nonzero holomorphic func-tion with the property that the set of points t ∈ [0, 2π] such that f hasa nontangential limit at eit has measure zero.

To prove the convergence note first that every compact subset of D iscontained in some disc centered at the origin with radius r < 1. Nowfor such r and |z| < r we have |1 + µzn| < 1 + µrn ≤ exp(µrn), whichimplies that the functions

fN(z) =N∏n=0

(1 + µzn)

satisfy

|fN(z)| < exp

(∞∑n=0

µrn

)= exp(

µ

1− r).

Thus, the family {fN , N ≥ 1} is uniformly bounded on compactsand hence normal, by Montel’s theorem. Moreover, exactly the sameargument as above shows that (fN(r)) converges for every 0 < r < 1and by Vitali’s theorem the convergence assertion follows. Finally,using the inequality 1 + x ≥ e−x we see that for 0 < r < 1

f(r) ≥ exp

(−∞∑n=0

µrn

)= exp(− µ

1− r),

i.e. f is not identically zero.

On the other hand, every point

znk = µ−1/ne(2k+1)πi

n , n, k ∈ N

is a zero of the function f because znnk = −1/µ. We claim that forevery t ∈ [0, 2π] there exists a subsequence of (znk)n,k≥1 that convergesnontangentially to eit. Indeed, given n ≥ 1 we can find 0 ≤ kn < n

such that |t − 2kn+1n

π| < πn. Then clearly, znkn = µ−1/ne

(2kn+1)πin → eit

and since|t− 2kn+1

nπ|

1− µ−1/n<

π

n(1− µ−1/n)→ π

log µ

it follows by Exercise 1 that (znkn) converges nontangentially to eit.This shows that whenever f has a nontangential limit at some point eit

the limit must be zero. But by Privalov’s theorem the set of these pointsmust have measure zero (recall that the set in question is measurable!).

Example 2.3. There exist analytic functions f in D that have a radiallimit but no nontangential limit at almost every boundary point!


Indeed, if we consider a nonconstant function f that has radial limitszero almost everywhere on the boundary, the set of boundary pointswhere f has nontangential limits must have measure zero otherwisethis would contradict Privalov’s theorem.

It will turn out that such an erratic behavior near the boundary pointsas described in the previous examples, becomes impossible if functionsin question satisfy an appropriate growth restriction. This is an impor-tant point of view because, as we shall see in the sequel, these growthrestrictions are very easy to define and provide large classes (spaces) ofanalytic functions that behave nicely near the boundary.

Theorem 2.2. Let f be analytic and bounded in D. If f has a radiallimit at a boundary point eit then f has a nontangential limit at thatpoint and these limits coincide.

Proof. By a translation and a rotation it will be sufficient to provethe following statement. For a > 0, 0 < b < π/2 let Γa,b be the set ofpoints z = reis ∈ C with 0 < r < a and |s| < b. If f is analytic andbounded in Γa,b and limr→0 f(r) = L then for 0 < c < b

limz→0z∈Γa,c

f(z) = L.

To this end, consider the sequence (fn) with fn(z) = f( zn), z ∈ Γa,b and

apply Vitali’s theorem. The family {fn, n ≥ 1} is uniformly boundedin Γa,b and if 0 < r < a then

limn→∞

fn(r) = limn→∞

f(r

n) = L.

Since the set (0, a) is not discrete in Γa,b Vitali’s theorem implies that(fn) converges uniformly on compacts to the constant function L. Nowlet (zk) be a sequence in Γa,c, 0 < c < b that converges to 0. Wewant to show that f(zk)→ L. Choose a sequence (nk) of integers suchthat nk|zk| ∈ (a/2, a) for sufficiently large k (for example, the choicenk+1 = integer part of a

|zk|will do). Then the points wk = nkzk satisfy

a/2 < |wk| < a, | argwk| < c for sufficiently large k which shows thatthese points wk lie in a compact subset of Γa,b. Then by the abovereasoning we have

L = limk→∞

fnk(wk) = limk→∞

f(wknk

) = limk→∞

f(zk)

and the result follows. �

12 1. PRELIMINARIES

Exercise 3. Prove the following statement which provides an improve-ment of Theorem 2.2: Let γ be an arc (continuous image of an interval)with one endpoint eit and all other points contained in some Stolz angleΓσ(eit). If f is bounded and analytic in D and f(z) has the limit L asz approaches eit along γ, then f has the nontangential limit L at eit.

The natural and basic question that arises is when do nontangentiallimits of an analytic function exist? The following result is a classicalsufficient condition for the existence of such a limit due to Abel.

Theorem 2.3. (Abel) Let f(z) =∑

n≥0 anzn be a convergent power

series in D. Assume that ζ ∈ ∂D is such that the series∑

n≥0 anζn

converges and let L be the value of the sum. Then f has the nontan-gential limit L at ζ.

Proof. We may assume without loss of generality that ζ = 1. LetSn =

∑∞k=n an and note that by hypothesis S0 = L and Sn → 0, n →

∞. In particular, (Sn) is bounded, say |Sn| ≤M for all n. Now write

f(z) =∞∑n=0

anzn =

∞∑n=0

(Sn − Sn+1)zn.

Since |Sn| ≤ M it follows that∑∞

n=0 Snzn converges for all z ∈ D and

thus,

f(z) =∞∑n=0

(Sn−Sn+1)zn =∞∑n=0

Snzn−

∞∑n=0

Sn+1zn = S0+

∞∑n=1

Sn(zn−zn−1).

This trick is frequently called summation by parts. Let ε > 0 and n0

be such that |Sn| < ε for n ≥ n0. Recall also that S0 = L. Then fromthe last equality we get

|f(z)−L| ≤ |z−1|∞∑n=1

|Sn||zn−1| ≤M |z−1|n0∑n=1

|z|n−1+ε|z−1|∞∑

n=n0+1

|z|n−1

≤Mn0|z − 1|+ ε|z − 1|(1− |z|)−1.Now let (zk) be a sequence in D that converges nontangentially to 1.Recall that in this case there is a constant K > 0 such that |zk−1|(1−|zk|)−1 < K and use the last estimate to obtain

lim supk→∞

|f(zk)− L| ≤ εK,

for every ε > 0, which finishes the proof. �


Of course, at the first sight the condition in Abel’s theorem seems diffi-cult to test and has little to do with a growth restriction. Neverthelessthere is a deep result in analysis that yields a global existence the-orem for nontangential limits of power series with square summablecoefficients. The following famous theorem was proved by L. Carleson.

Theorem 2.4. (Carleson) Let f(z) =∑

n≥0 anzn be a power series

with square summable coefficients, that is,∑n≥0

|an|2 <∞.

Then the series∑

n≥0 anζn converges for almost every ζ ∈ ∂D.

It is a simple exercise to show that such power series converge in theunit disc. Thus, by Abel’s theorem every power series with square sum-mable coefficients has a nontangential limit almost everywhere on theunit circle. Let us now show that the condition on the coefficients reallyis a growth restriction. This is a consequence of Parseval’s formula.

Theorem 2.5. Let f(z) =∑

n≥0 anzn be a convergent power series in

D. Then∑n≥0

|an|2 = sup0<r<1

∫ 2π

0

|f(reit)|2 dt2π

= limr→1

∫ 2π

0

|f(reit)|2 dt2π.

Proof. Write

|f(reit)|2 =

(∞∑n=0

anrneint

)(∞∑n=0

anrne−int

)=

∞∑m,n=0

anamrm+nei(n−m)t

and note that the last sum converges uniformly on [0, 2π]. But thenwe can interchange limit and integration to obtain∫ 2π

0

|f(reit)|2 dt2π

=∞∑

m,n=0

anamrm+n

∫ 2π

0

ei(n−m)t dt

2π=∞∑n=0

|an|2r2n,

because∫ 2π

0eikt dt

2π= δk0, that is, it equals 0 if k 6= 0 and 1 if k = 0.

The rest of the proof is a routine exercise based on the fact that theintegrals involved here increase with r as the last identity shows. �

Thus, we can reformulate the result on nontangential limits in termsof the growth restriction we just found.

Corollary 2.1. Let f be analytic in D and suppose that

sup0<r<1

∫ 2π

0

|f(reit)|2 dt2π

<∞.

14 1. PRELIMINARIES

Then f has nontangential limits a.e. on the unit circle.

The above result is classical, and, as we shall see in the sequel, its proofdoes not need the ”heavy artillery” provided by Carleson’s theorem.We close this section with a result which shows that, in terms of Taylorcoefficients, the square summability condition in Corollary 2.6 cannotbe improved. The theorem below is a special case of a much strongerresult of Khinchin and Kolmogorov (see [Du],p.).

Theorem 2.6. Let f(z) =∑

n≥0 anzn be a convergent power series in

D such that ∑n≥0

|an|2 =∞.

Then there exists a sequence (εn) with εn = ±1, n = 0, 1, 2 . . . suchthat the power series

g(z) =∑n≥0

εnanzn

has no radial limit almost everywhere on the unit circle.

CHAPTER 2

Poisson integrals

1. Harmonic functions

Recall that a twice differentiable complex-valued function u definedon some open set G ⊂ C is called harmonic if it satisfies the Laplaceequation

(1.1) ∆u =∂2u

∂x2+∂2u

∂y2= 0 in G.

Of course, holomorphic functions are harmonic. Moreover, a very im-portant class of examples of real-valued harmonic functions are thereal (and imaginary) parts of holomorphic functions in G. This followsfrom a very simple computation with the Cauchy-Riemann equations.Conversely, if G is simply connected (roughly speaking, this means ithas no holes), it turns out that every real-valued harmonic function inG is the real part of a holomorphic function in G. This is a classical re-sult that needs to be proved. In order to avoid technicalities regardingthe definition of a simply connected domain we will use an equivalentproperty.

1.1. Theorem Suppose that G is an open connected set in C withthe property that every holomorphic function on G has a primitive inG. Then every real-valued harmonic function in G is the real part of aholomorphic function in this open set.

Proof. Observe first that if u is harmonic on G then the functionh = ∂u

∂x− i∂u

∂ysatisfies the Cauchy-Riemann equations in G. Indeed,

since u is harmonic we have

∂

∂x

∂u

∂x=∂2u

∂x2= −∂

2u

∂y2=

∂

∂y

(−∂u∂y

),

and∂

∂y

∂u

∂x=∂2u

∂xy= − ∂

∂x

(−∂u∂y

).

15

16 2. POISSON INTEGRALS

Let f be a primitive of the holomorphic function h, i.e. f ′ = h. Then

∂

∂xRef = Ref ′ =

∂u

∂x

and

∂

∂yRef = −Imf ′ =

∂u

∂y.

Thus, Ref − u is constant in G and the result follows.

The simplest example that comes to mind of a domain G as above, isa disc.

The theorem has the following reformulation: If G is as above and u isharmonic in G then there exists a harmonic function v in G such thatu+ iv is analytic in G.

Exercise 1. Show that v is harmonic and it is uniquely determined upto an additive constant (i.e. if v1, v2 have this property then v1 − v2 isconstant in G).

Such a function v is called a harmonic conjugate of u, while the asso-ciation u 7→ f = u+ iv is sometimes called analytic completion.

Exercise 2 Given function of the form

u(z) =N∑

k,n=0

aknzkzn, z ∈ C

show that u is harmonic if and only if akn = 0 whenever k and n areboth nonzero.

2. Representation by Poisson integrals

Since harmonic functions are so close to the analytic ones it is naturalto try to find an analogue of Cauchy’s formula for these functions. Hereis a first attempt for harmonic functions in the unit disc.If u is harmonic and real-valued in a disc containing D and f is ananalytic completion of u then by Cauchy’s formula we have for all

2. REPRESENTATION BY POISSON INTEGRALS 17

z ∈ D

u(z) = Ref(z) = Re

(1

2πi

∫|ζ|=1

f(ζ)dζ

ζ − z

)(2.1)

= Re

(1

2π

∫ 2π

0

f(eit)eitdt

eit − z

).

This is an integral representation of u which, unfortunately, involvesone of its harmonic conjugates and therefore needs to be improved.

Lemma 2.1. If u is harmonic in a disc containing D then for all z ∈ Dwe have

u(z) =1

2π

∫ 2π

0

1− |z|2

|eit − z|2u(eit)dt.

Proof. The trick is to insert in the last integral in (2.1) the harm-less term zf(eit)/(e−it−z), where f is again an analytic function whosereal part equals u. Note that for every integer n ≥ 0 we have∫ 2π

0

eintdt

e−it − z=

∫ 2π

0

ei(n+1)tdt

1− eitz= −i

∫|ζ|=1

ζndζ

1− ζz= 0,

by Cauchy’s Theorem. Thus, if we write

f(eit) =∞∑n=0

aneint

then from the fact that the power series converges uniformly on theunit circle we obtain∫ 2π

0

f(eit)zdt

e−it − z=∞∑n=0

an

∫ 2π

0

eintzdt

e−it − z= 0,

which shows that the term inserted in the last integral in (2.1) is indeedharmless. Therefore, from (2.1) we obtain

u(z) = Re

(1

2π

∫ 2π

0

f(eit)

(eit

eit − z+

z

e−it − z

)dt

).

Now note also that

(2.2)eit

eit − z+

z

e−it − z=

1− |z|2

|eit − z|2, z 6= eit,

and that the right hand side of this equality is always real (it becomesalso positive if |z| < 1). Then

u(z) =1

2π

∫ 2π

0

1− |z|2

|eit − z|2Ref(eit)dt



Observe that this lemma gives a much better integral representationof a harmonic function. The only problem that remains is to weakenthe very restrictive hypothesis that u is actually harmonic in a largerdisc. A first step in this direction is an immediate application of thetheorem. is given in the next theorem.

Corollary 2.1. Let u be harmonic in D and continuous in the closeddisc D. Then

u(z) =1

2π

∫ 2π

0

1− |z|2

|eit − z|2u(eit)dt.

Proof. For 0 < r < 1 fixed but arbitrary, consider the dilation urof u defined in D by ur(z) = u(rz). Clearly, ur is harmonic in the discof radius 1/r centered at the origin, so that Lemma 2.1 gives for everyz ∈ D

ur(z) =1

2π

∫ 2π

0

1− |z|2

|eit − z|2ur(e

it)dt.

Now let r → 1 and note that ur(z) → u(z), z ∈ D and also, that thefunctions t 7→ ur(e

it), t ∈ [0, 2π] converge uniformly to t 7→ u(eit), t ∈[0, 2π], by the continuity assumption. This implies that for fixed z ∈ D∫ 2π

0

1− |z|2

|eit − z|2ur(e

it)dt→∫ 2π

0

1− |z|2

|eit − z|2u(eit)dt, r → 1

and we are done. �

The continuity assumption on u can be relaxed using some results fromintegration theory and functional analysis.

Theorem 2.1. Let u be harmonic in D and suppose that there existconstants C > 0 and 1 ≤ p ≤ ∞ such that for all r ∈ (0, 1) we have∫ 2π

0

|u(reit)|p dt2π≤ C,

if p <∞, or

supt∈[0,2π]

|u(reit)| ≤ C,

if p =∞. Then:(i) If p > 1, there exists a unique u ∈ Lp(m) such that

u(z) =

∫ 2π

0

1− |z|2

|eit − z|2u(eit)

dt

2π.

2. REPRESENTATION BY POISSON INTEGRALS 19

(ii) If p = 1, there exists a unique finite Borel measure µ on T suchthat

u(z) =

∫T

1− |z|2

|ζ − z|2dµ(ζ).

Proof. We shall prove first the existence part. The argumentgoes exactly as the one in Corollary 2.1. More precisely, we considerthe dilations ur, 0 < r < 1 of u, apply Lemma 2.1 and let r → 1.Clearly, for fixed z ∈ D ur(z)→ u(z), when r → 1.(i) Assume first that p <∞, and use the assumption in the statementtogether with the fact that Lp(m) is reflexive, to find a sequence (rn)tending to 1 such that (urn) converges weakly in Lp(m) to u ∈ Lp(m).Since the functions

Pz(eit) =

1− |z|2

|eit − z|2are bounded in t, we obtain the representation in the statement. Whenp =∞ we only use weak-star sequential compactness instead of reflex-ivity to arrive at the same result. (ii) Here we use the fact that forfixed z ∈ D the function Pz(·) is continuous on T and since the mea-sures urdm have bounded total variation, we can find a sequence thatconverges weak-star in the dual of C(T) to some finite Borel measureµ on T.It remains to show that u, µ are unique. A direct computation showsthat

Pz(eit) = 1 + 2Re

ze−it

1− ze−it= 1 + 2Re

∞∑n=0

zne−int,

where the series converges uniformly on [0, 2π]. Since the equality inthe statement holds for all z ∈ D we see that the Fourier coefficients ofu, µ are uniquely determined, i.e. u, µ are unique. �

The kernel

Pz(eit) =

1− |z|2

|eit − z|2is called the Poisson kernel, and a harmonic function of the form de-scribed in the above theorem is called the Poisson integral of u, or µ.The simplest example that (i) cannot hold for p = 1 are the Poissonkernels themselves. For example, Pz(1) satisfies the theorem with p = 1and is the Poisson integral of the Dirac measure at 1 which is uniquelydetermined by this assertion.

Corollary 2.2. Let u be harmonic and nonnegative in D. Then thereexists a unique nonnegative Borel measure µ on T such that µ(T) =


u(0) and

u(z) =

∫T

1− |z|2

|ζ − z|2dµ(ζ).

Proof. A nonnegative harmonic function u in D satisfies the as-sumption in part (ii) of Theorem 2.1 since∫ 2π

0

|u(reit)| dt2π

=

∫ 2π

0

u(reit)dt

2π= u(0).

Since the measure µ goiven by the theorem is the weak-star limit ofnonnegative measures, it will be nonnegative as well. �

Corollary 2.3. Let f be analytic with nonnegative real part in D.Then there exists a unique nonnegative Borel measure µ on T suchthat µ(T) = Ref(0) and

f(z) =

∫T

ζ + z

ζ − zdµ(ζ).

The above result is called the Herglotz representation of functions withpositive real part.

Exercise 1 Prove Corollary 2.3.

Exercise 2 Prove that a harmonic function in an open connected sub-set G of C which has a local minimum, or a local maximum in G isconstant.

3. The nontangential maximal function

Given a function f : D→ C and σ ∈ (0, 1), the nontangential maximalfunction of f is defined on T by

(3.3) f ∗(ζ) = supz∈Γσ(ζ)

|f(z)|.

The definition depends on σ, but is is customary to drop this parameterin the notation. In all assertions σ ∈ (0, 1) is fixed but arbitrary, andthe constants apearing in estimates will depend on the choice of σ. Theaim of this section is to compare the nontangential maximal function ofPoisson integrals of functions h ∈ L1(m) with the well-known Hardy-Littlewood maximal function of h, defined by

(3.4) h†(eiθ) = supt∈[0,π]

1

2t

∫ θ+t

θ−t|h(eis)ds.

3. THE NONTANGENTIAL MAXIMAL FUNCTION 21

The Hardy-Littlewood maximal function plays a crucial role in Lp-estimates as it is shown by the following classical theorem essentiallydue to Hardy and Littlewood, which we state without proof.

Theorem 3.1. For p > 1, h† ∈ Lp(m), whenever h ∈ Lp(m), and thereexists Cp > 0 such that

‖h†‖p ≤ Cp‖h‖p.

If h ∈ L1(m) there exists C1 > 0 such that for all t > 0 we have

m({ζ ∈ T : h†(ζ) > t}) < C1‖h‖1

t.

We begin with some simple properties of the Poisson kernel which arelisted below.

Proposition 3.1. (i) Pz(eit) > 0, for all z ∈ D and t ∈ [0, 2π].

(ii) For all t, θ ∈ [0, 2π) with t 6= θ

limz→eiθ

Pz(eit) = 0.

In fact, if we regard Pz(·) as a family of functions on T then the aboveconvergence is uniform on any set of the form {ζ ∈ T : |ζ − eit| > δ},where δ > 0 is fixed.(iii) For all z ∈ D, s ∈ [0, 2π], , 0 < r < 1, we have∫ 2π

0

Pz(eit)dt

2π=

∫ 2π

0

Preiθ(eis)dθ

2π= 1.

(iv) If z = reiθ then Pz(ei(θ+t)) = Pz(e

i(θ−t)).(v) If ζ ∈ T and z = reit ∈ Γσ(ζ), there exists C > 0 depending onlyon σ such that for all λ ∈ T we have

Pz(λ) ≤ CPrζ(λ).

(vi) If r ∈ [0, 1), t ∈ [−π, 2π] then

t∂

∂tPr(e

it) ≤ 0,

and consequently, there exists C > 0 such that∫ π

−π

∣∣∣∣t ∂∂tPr(eit)∣∣∣∣ dt2π

≤ C,

for all r ∈ [0, 1).


Proof. (i), (ii), and (iv) are obvious. The first integral in (iii)equals 1 by an application of Lemma 2.1u ≡ 1, the second equalityfollows from Preiθ(t) = Preit(θ). To see (v), write

Pz(λ) =1− r2

|λ− z|2=

1− r2

|λ− rζ|2|λ− rζ|2

|λ− z|2= Prζ(λ)

|λ− rζ|2

|λ− z|2.

Moreover,

|λ− rζ||λ− z|

≤ |λ− z|+ |z − rz|+ r|z − ζ|λ− z|

≤ 1 + r + r|z − ζ|1− r

,

and the right hand side is bounded within Γσ(ζ) by a constant thatdepends only on σ. Finally, the first part of (vi) is immediate since

t∂

∂tPt(e

it) = −2r(1− r2)t sin t

|eit − r|2,

while the second part follows from this together with integration byparts: ∫ π

−π

∣∣∣∣t ∂∂tPr(eit)∣∣∣∣ dt2π

= −∫ π

−πt∂

∂tPr(e

it)dt

2π

= −Pr(−1) +

∫ π

−πPr(e

it)dt

2π

= −1− r1 + r

+ 1.

�

We can now turn to the main result of the section.

Theorem 3.2. There exists C > 0 such that, given h ∈ L1(T) withPoisson integral u, we have

u∗(ζ) ≤ Ch†(ζ), ζ ∈ T.

In particular, if h ∈ Lp(m) for some p ∈ (1,∞) then u∗ ∈ Lp(m), andthere exists Cp > 0, depending only on p, such that

‖u∗‖Lp(m) ≤ Cp‖h‖Lp(m)

Proof. Let z ∈ Γσ(eiθ), with |z| = r ∈ [0, 1), and use Proposition3.1 (v) to obtain the estimate

(3.5) |u(z)| ≤∫TPz(λ)|h(λ)|dm(λ) ≤ C

∫TPreiθ(λ)|h(λ)|dm(λ).

4. BOUNDARY BEHAVIOR OF POISSON INTEGRALS 23

Using again this result we can write∫TPreiθ(λ)|h(λ)|dm(λ) =

∫ π

−πPr(e

it)|h((ei(θ+t))| dt2π

=

∫ π

0

Pr(eit)[|h((ei(θ+t))| − |h((ei(θ−t))|] dt

2π.

Finally, integration by parts gives∫TPreiθ(λ)|h(λ)dm(λ) = Pr(−1)

∫ π

−π|h((ei(θ+t))| dt

2π

−∫ π

0

∂

∂tPr(e

it)

∫ t

−t|h((ei(θ+s))| ds

2π.

Clearly,∫ π

−π|h((ei(θ+t))| dt

2π≤ h†(eiθ),

∫ t


2π≤ t

πh†(eiθ).

By Proposition 3.1 (vi) we have that

−∫ π

0

∂

∂tPr(e

it)

∫ t


2π≤ C ′h†(eiθ),

hence, ∫TPreiθ(λ)|h(λ)|dm(λ) ≤ C ′′h†(eiθ),

and the result follows by the estimate (3.5). �

4. Boundary behavior of Poisson integrals

In this section we will use the maximal function estimate to derive aseries of results about Poisson integrals of integrable functions.

Theorem 4.1. If u is the Poisson integral of h ∈ L1(m), then u hasthe nontangential limit h(ζ) at almost every ζ ∈ T.

Proof. Let ε > 0, and let g ∈ C(T) with ‖h − g‖1 < ε. By theHardy-Littlewood theorem it follows that

m({ζ ∈ T : (h− g)†(ζ) >√ε}) < C1

√ε.

Let v denote the Poisson integral of g. By Theorem 3.2 we have that

m({ζ ∈ T : (u− v)∗(ζ) > C√ε}) < C1

√ε,

where C > 0 is the constant given in that theorem. For a fixed integern > 0 consider the set An ⊂ T consisting of those points ζ for whichthe nontangential ”limsup” of |u − h(ζ)| at ζ exceeds 1

n. Since v is


continuous in D and equals g on T, it follows that if ε is sufficientlysmall, An is contained in the set

{ζ ∈ T : (u− v)∗(ζ) > C√ε} ∪ {ζ ∈ T : |h(ζ)− g(ζ)| >

√ε}.

This gives that m(An) < C1

√ε, and since ε was arbitrary we conclude

that m(An) = 0, which completes the proof. �

The result can be extended to Poisson integrals of finite Borel measures,but this requires an additional step.

Lemma 4.1. If u is the Poisson integral of a finite Borel measure µon T which is singular w.r.t. m, then u has nontangential limits zerom-a.e.

Proof. We may assume without loss of generality that the mea-sure µ is positive. The the existence of the nontangential limits folowsfrom Theorem 4.1. Indeed, if f is analytic in D with Ref = u, then e−f

is analytic in and bounded D, in particular it is a bounded harmonicfunction. Then it is the Poisson integral of a function in L∞(m), henceby the above result, it has nontangential limits a.e. on T. This impliesthat u = − log |f | has nontangential limits u(ζ) ≥ 0 for almost everyζ ∈ T. Now recall that our measure µ is the weak-star limit of themeasures urdm, where ur(z) = u(rz), i.e. for every continous functionh on T we have that

limr→1−

∫Thurdm =

∫hdµ.

On the other hand, by Fatou’s lemma we obtain for every nonnegativecontinuous function h on T,∫

Tu(ζ)h(ζ)dm(ζ) ≤ lim inf

r→1−

∫Thurdm =

∫hdµ.

Since the characteristic function of any compact subset of T can beapproximated pointwise by a sequence of nonnegative continuous func-tions, another application of Fatou’s lemma leads to∫

A

udm ≤ µ(A),

for all compact subsets of T. From the fact that µ is singular we deducethat u = 0, m-a.e. �

If u is the Poisson integral of the finite Borel measure µ on T we canwrite

dµ = hdm+ dν,

5. hp-SPACES 25

where h = dµdm∈ L1(m) is the Radon-Nykodim derivative of µ w.r.t.

m, and ν is singular w.r.t. m. By Theorem 4.1 and the above lemmawe obtain:

Corollary 4.1. If u is the Poisson integral of the finite Borel measureµ on T then u has the nontangential limit dµ

dm(ζ) at almost every ζ ∈ T.

5. hp-spaces

Given 0 < p < ∞ we denote by hp the vector space of all harmonicfunctions u in D with

(5.6) ‖u‖pp = sup0<r<1

∫ 2π

0

|u(reit)|p dt2π

<∞.

h∞ will denote the space of bounded harmonic functions u in D with

‖u‖∞ = supz∈D|u(z)|.

We will be interested in the case p ≥ 1, when (5.6) defines a norm onhp.If p > 1, we know from the previous section that every u ∈ hp is thePoisson integral of its boundary function defined a.e. on T by u(ζ) =nontangential limit of u at ζ. We shall keep this notation for theboundary values throughout in what follows. If u ∈ h1 we only knowthat u is a Poisson integral of a (unique) finite Borel measure on T,which we denote by µu. Its total variation is ‖µu‖ = |µu|(T).

Theorem 5.1. (i) If 1 < p ≤ ∞ and u ∈ hp we have

‖u‖pp = ‖u‖pLp(m) =

∫T|u|pdm,

and if p = 1, then

‖u‖1 = ‖µu‖.For 1 ≤ p ≤ ∞, hp is a Banach space with the norm (5.6).(iii) If 1 < p <∞, and u ∈ hp, ur(z) = u(rz), 0 ≤ r < 1, then

limr→1−

‖ur − u‖p = 0.

Proof. (i) The case p =∞ is immediate

|u(z)| ≤∫TPz(ζ)|u(ζ)dm(ζ) ≤ ‖u‖∞

∫TPz(ζ)dm(ζ) = ‖u‖∞.


If 1 < p < ∞, note that u(ζ) = limr→1− u(rζ), m−a.e. on T, henceFatou’s lemma gives ∫

T|u|pdm ≤ ‖u‖pp.

On the other hand, since u is the Poisson integral of its boundaryvalues, we can apply by Holder’s inequality to obtain

|u(z)|p =

∣∣∣∣∫TPz(ζ)u(ζ)dm(ζ)

∣∣∣∣p≤(∫

TPz(ζ)|u(ζ)|pdm(ζ)

)(∫TPz(ζ)dm(ζ)

)p−1

=

∫TPz(ζ)|u(ζ)|pdm(ζ),

so that∫ 2π

0

|u(reit)|p dt2π≤∫ 2π

0

∫TPz(ζ)|u(ζ)|pdm(ζ)

dt

2π=

∫T|u|pdm.

The case when p = 1 is similar. Since µu is the weak-star limit of asequence urndm, it follows that ‖µu‖ ≤ ‖u‖1. On the other hand, byFubini’s theorem∫ 2π

0

|u(reit)| dt2π≤∫ 2π

0

∫TPz(ζ)d|µ|(ζ)dm(ζ)

dt

2π= ‖µu‖.

(ii) assume (un) is a Cauchy sequence in hp, and use (i) to concludethat the corresponding sequence of boundary functions, or measures isCauchy in Lp(m), respectively in the space of finite Borel measures onT. Since these are complete spaces, we have that (un) converges to u inLp(m), or (µun) converges in norm to µ and again by(i), (un) convergesin hp to the Poisson integral of u, or µ. (iii) Almost everywhere on Twe have

|ur(ζ)− u(ζ)| ≤ u∗(ζ), limr→1−

ur(ζ) = u(ζ),

and the result follows by Theorem 3.2 together with the dominatedconvergence theorem. �

Exercise 1 Show that part (iii) of Theorem 5.1 fails for p =∞ when-ever u has a discontinuous boundary function, or when p = 1, and u isthe Poisson integral of a measure with a nonzero singular part.

In the above proof we have used the important estimate

(5.7) |u(z)|p ≤∫TPz(ζ)|u(ζ)|pdm(ζ),

6. EMBEDDINGS OF hp INTO Lp(ν) 27

which has on obvious version in the case when p = 1

(5.8) |u(z)| ≤∫TPz(ζ)d|µu|(ζ).

I)n particular, these estimates show that evaluation at a fixed pointz ∈ D is a bounded linear functional on hp.

Corollary 5.1. If z ∈ D and u ∈ hp, 1 ≤ p <∞, then

|u(z)| ≤(

1 + |z|1− |z|

)1/p

‖u‖p.

In particular, for 1 ≤ p ≤ ∞, convergence in hp implies uniform con-vergence on compact subsets of D.

Proof. The result follows from the simple estimate

Pz(ζ) ≤ 1− |z|2

(1− |z|)2=

1 + |z|1− |z|

.

�

6. Embeddings of hp into Lp(ν)

This is an important problem with far-reaching consequences. We wantto characterize the positive Borel measures ν on D with the propertythat hp ⊂ Lp(ν). An application of the closed graph theorem showsthat if this inclusion holds, then the inclusion map must be bounded,that is

hp ⊂ Lp(ν) ⇐⇒ ‖u‖p ≤ C‖u‖Lp(ν),

for some C > 0 and all u ∈ hp. Indeed, if (un) converges to u in hp andto v in Lp(ν), then by Corollary 5.1, for each z ∈ D, (un(z)) convergesto u(z). Also, there exists a subsequence (unk) which converges ν-a.e.to v, hence u = v ν−a.e.. For fixed λ ∈ D, let

uλ(z) =(1− |λ|2)1/p

(1− λz)2/p, z ∈ D.

uλ is analytic, hence harmonic in D and for ζ ∈ T we have |uλ(ζ)| =Pλ(ζ). Then the above inequality yields the following necessary condi-tion for the inclusion hp ⊂ Lp(ν)

(6.9) supλ∈D

(1− |λ|2)

∫|1− λz|−2dν(z) <∞.


A similar but unfortunately stronger condition is sufficient and followsimmediately from the estimates (5.8) and (5.7), namely

(6.10) supζ∈T

∫1− |z|2

|1− ζz|2dν(z) <∞.

For example, (5.7) and Fubini’s theorem give∫|u|pdν ≤

∫ ∫TPz(ζ)|u(ζ)|pdm(ζ)dν(z)

=

∫T|u(ζ)|p

∫1− |z|2

|1− ζz|2dν(z)

≤ ‖u‖pp supζ∈T

∫1− |z|2

|1− ζz|2dν(z).

Exercise 1 Show that if ν is the one-dimensional Lebesgue measure onthe segment (−1, 1) then (6.9) holds, but (6.10) fails .

It turns out that the necessary condition above is also sufficient for ourembedding. Let us rewrite this condition in a more transparent way.

Proposition 6.1. Given 0 < h < 1, and θ ∈ [0, 2π], let

Sh(eiθ) = {reit : 1− h < r < 1, |t− θ| < h}.

A positive Borel measure ν on D satisfies (6.9) if and only if

sup0<h<1θ∈[0,2π]

ν(Sh(eiθ))

h<∞, or sup

ζ∈Tr>0

ν({|z − ζ| < r})r

<∞.

Proof. We leave it as an exercise to show that if one supremumabove is finite, then so is the other. Fix λ ∈ D \ {0} and note that if|z − λ

|λ| | < (1− |λ|, then

|1− λz| ≤ |λ||z − λ

|λ||+ 1− |λ| < (|λ|+ 1)(1− |λ|),

hence

(1− |λ|2)

∫|1− λz|−2dν(z) >

(1− |λ|2)ν({|z − λ|λ| | < (1− |λ|)})

4(1− |λ|)2

≥ν({|z − λ

|λ| | < (1− |λ|)})4(1− |λ|)

,


which shows that if (6.9) holds, then the second supremum above isfinite. Conversely, fix λ ∈ D with |λ| ≥ 1/2, and let

En = {z ∈ D : 2n(1− |λ|2) ≤ |1− λ| < 2n+1(1− |λ|2)}.Then

(1− |λ|2)

∫|1− λz|−2dν(z) ≤

∑n

ν(En)

22n(1− |λ|2).

Moreover, if z ∈ En, then

1

2|z − λ

|λ|| < |1− λz|+ 1− |λ| < 2n+2(1− |λ|2),

hence,

ν(En) ≤ 2n+2(1− |λ|2) supζ∈Tr>0

ν({|z − ζ| < r})r

,

which gives

(1− |λ|2)

∫|1− λz|−2dν(z) ≤ C sup

ζ∈Tr>0

ν({|z − ζ| < r})r

,

when λ ∈ D with |λ| ≥ 1/2. Since the integrals on the left are obviouslybounded or |λ| < 1/2, the result follows. �

The sets Sh have been used by Carleson who was the first to studyembeddings of this type. They will be called Carleson boxes, and themeasures satisfying one of the conditions in Proposition 6.1 are calledCarleson measures. The next theorem is essentially due to Carleson.

Theorem 6.1. If ν is a finite positive Borel measure on D, and 1 0 such that if t > 0 issufficiently large then

ν({z ∈ D : |u(z)| > t}) ≤ km({ζ ∈ T : u∗(ζ) > t}).If this is achieved, it follows that∫

|u|pdν = p

∫ ∞0

tp−1ν({z ∈ D : |u(z)| > t})dt

≤ kp

∫ ∞0

tp−1m({ζ ∈ T : u∗(ζ) > t})dt

= k

∫T(u∗)pdm,


hence, by Theorem 3.2, we have u ∈ Lp(ν).To see the claim, let ε > 0, and let G be an open subset of T such that

{ζ ∈ T : u∗(ζ) > t} ⊂ G, m(G) < m({ζ ∈ T : u∗(ζ) > t}) + ε.

If z ∈ D \ {0} is such that |u(z)| > t then z lies outside any Stolz angleΓσ(ζ) (here σ is fixed), with u∗(ζ) ≤ t. In particular, z

|z| ∈ G. Recall

that G is a countable union of disjoint open arcs In, n ≥ 1 so that, if

At = {z ∈ D : |u(z)| > t}, At,n = {z ∈ D : |u(z)| > t,z

|z|∈ In},

we have

At =⋃n≥1

At,n.

Moreover, if z ∈ At,n, and ζn is the endpoint of In closest to z then

|z − ζn| <m(In)

2+ 1− |z|.

Also, recall that z /∈ Γσ(ζn). By Corollary 5.1 it follows that if t issufficiently large, then |z| > σ, so there exists c > 1, depending onlyon σ, with |z− ζ| > c(1− |z|. By replacing this in the last estimate weobtain

1− |z| < m(In)

2(c− 1), z ∈ At,n.

Then if m(G) is sufficiently small , which happens when t is large andε is small, we can conclude that

At,n ⊂ Shn(ζ ′n),

where ζ ′n is the middlepoint of In, and hn = c′m(In), for some fixedc′ > 0. Then by the assumption on ν we have

ν(At,n ≤ Cm(In),

and

m(At) =∑n≥1

m(At,n) ≤ C(m(G)) < m({ζ ∈ T : u∗(ζ) > t}) + ε,

which implies the claim and completes the proof of the theorem. �

Example 6.1. As pointed out in Exercise 1, the one-dimensional Lebesguemeasure on the segment (−1, 1) is a Carleson measure, that is, thereexists C > 0 such that ∫ 1

−1

|u(x)|pdx ≤ C‖u‖pp,


for all u ∈ hp, 1 < p ≤ ∞. If we consider only analytic functionsthe best constant is C = 1

2and the estimate is called the Fejer-Riesz

inequality .

Remark 6.1. It is important to note that for a finite positive Borelmeausre ν the quantities considered in Proposition 6.1 are essentiallyrelated to the behavior of µ near the boundary. In fact, it is an im-mediate consequence of Corollary 5.1 that ν is a Carleson measure ifand only if its restriction to any annulus {z : 0 < r < |z| < 1} is aCarleson meausre.

An important class of examples of Carleson measures are those of theform

dν = |∇u|2(z) log1

|z|dA,

where u ∈ h∞, A is the normalized 2-dimensional Lebesgue measurerestricted to D, dA = dxdy

π, and ∇u stands for the gradient of u, so that

|∇u|2 =

∣∣∣∣∂u∂x∣∣∣∣2 +

∣∣∣∣∂u∂y∣∣∣∣2 .

This is a consequence of a more general result based on the followingidentity of Littlewood and Paley.

Proposition 6.2. Let u ∈ h2 alnd λ ∈ D. Then

(6.11)

∫T|u(ζ)− u(λ)|2Pλ(ζ)dm(ζ) =

∫D|∇u|2(z) log

∣∣∣∣1− λzz − λ

∣∣∣∣ dA(z).

Proof. The proof is essentially based on the following direct ap-plication of Green’s formula: If ϕ : D→ C is twice continuously differ-entiable then

(6.12)

∫Tϕdm = ϕ(0) +

∫D

∆ϕ(z) log1

|z|dA(z).

Moreover, another direct computation shows that if v is harmonic inD, and ϕ(z) = |v(z)|2, then

∆ϕ(z) = |∇v|2(z).

Assume first that u is harmonic in a larger disc, and set v = u −u(0), ϕ = |v|2. Then (6.12) immediately implies (6.11) when λ = 0.Thus, for any u ∈ h2 and 0 < r < 1, (6.11) holds for the dilationur(z) = u(rz) and λ = 0. Also, if 0 < r, ρ < 1 we can apply (6.11) toobtain

‖ur − uρ‖22 =

∫D|∇ur(z)−∇uρ(z)|2 log

1

|z|dA(z),


and by Theorem 5.1 (iii) it follows that ∇ur−∇uρ converges to zero inL2(D, log 1

|z|dA), when r, ρ → 1−. This easily shows that when λ = 0,

(6.11) holds for every u ∈ h2.If λ ∈ D is arbitrary, let φλ(z) = z−λ

1−λz . Clearly, φλ is an analytic

bijection from D onto itself. By what we have proved above, we havefor every u ∈ h2,∫

T|u ◦ φλ − u(λ)|2dm =

∫D|∇u ◦ φλ|2(z) log

1

|z|dA(z).

A direct computation similar to the above yields

|∇u ◦ φλ|2(z) = |∇u|2(φλ(z))|φ′λ(z)|2,i.e. ∫

T|u ◦ φλ − u(λ)|2dm =

∫D|∇u|2(φλ(z))|φ′λ(z)|2 log

1

|z|dA(z).

Then (6.11) follows directly from the change of variables ξ = φλ(ζ), w =φλ(z). �

Exercise 1 Prove (6.12).

Corollary 6.1. Let u ∈ h2. Then the measure |∇u|2(z) log 1|z|dA, is

a Carleson measure if and only if

supλ∈D

∫T|u(ζ)− u(λ)|2Pλ(ζ)dm(ζ) <∞.

Proof. If the condition in the statement holds, we apply Propo-sition 6.2 together with the inequality log 1

x≥ 1 − x, 0 < x ≤ 1, to

obtain

(6.13) supλ∈D

∫D|∇u|2(z)

(1− |λ|2)(1− |z|2)

|1− λz|2dA(z) <∞,

that is, (6.9) holds for the measure |∇u|2(z)(1 − |z|2)dA. Obviously,this measure is comparable to the measure |∇u|2(z) log 1

|z|dA outside

the disc centered at the origin and of radius 12, hence by Remark 6.1,

|∇u|2(z) log 1|z|dA, is a Carleson measure. To see the converse, use

again the inequality log 1x≥ 1 − x, 0 < x ≤ 1, to conclude that

|∇u|2(z)(1 − |z|2)dA is a Carleson measure, that is, (6.13) holds. Wewant to prove that

supλ∈D

∫D|∇u|2(z) log

∣∣∣∣1− λzz − λ

∣∣∣∣ dA(z) <∞.


After the change of variables w = φλ(z) in both integrals above, we seethat it suffices to verify the estimate∫

D|∇u|2(z) log

1

|z|dA(z) ≤ C

∫D|∇u|2(z)(1− |z|2)dA(z),

for some C > 0 and all u ∈ h2. This is a simple exercise based onParseval’s formula and is left to the reader. �

Obviously, bounded harmonic functions have the property stated inCorollary 6.1, but there are also unbounded functions with this prop-erty. An easy example whose verification is left to the reader is theanalytic function

f(z) = log(1− z), z ∈ D.There is an important intrinsic characterization of these functions interms of the boundary values alone. The Poisson integral u of a functionh ∈ L1(m) satisfies the conditions in Corollary ?? if and only if

supI

1

m(I)

∫I

∣∣∣∣h− 1

m(I)

∫I

hdm

∣∣∣∣ dm <∞,

where the supremum is taken over all arcs I ⊂ T. We say that such afunction has bounded mean oscillation and denote the space of all suchfunctions by BMO.The proof of the above assertions relies on a real-variable argumentwhich will not be covered here.

CHAPTER 3

Hardy spaces

The Hardy space Hp, 0 < p ≤ ∞, is the subspace of hp consisting ofanlytic functions in D. A direct application of Corollary 5.1 shows thatwhen 1 ≤ p ≤ ∞, Hp is a closed subspace of hp, in particular, it isa Banach space. These spaces are the object of our study, and someof the tools are provided by the material presented in the previouschapter.

1. Outer functions.

In this section we shall apply the results on Poisson integrals in order toconstruct analytic functions with prescribed boundary values of theirmodulus.

Proposition 1.1. Let h be a nonnegative function on the unit circlesuch that ∫

T| log h|dm <∞.

Then function F defined in D by

F (z) = exp

(∫T

ζ + z

ζ − zlog h(ζ)dm(ζ)

)has nontangential limits almost everywhere on the unit circle. More-over, if we denote by F (ζ) the nontangential limit of F at ζ ∈ T (when-ever it exists) then |F (ζ)| = h(ζ) m−a.e. on T.

Proof. We leave it as an exercise to prove that F is analytic. Alsonote that this function satisfies

|F (z)| = exp Re

(∫T

ζ + z


)= exp

(∫TPz(ζ) log h(ζ)dm(ζ)

).

Consequently, by the results about Poisson integrals, |F | has the non-tangential limit exp log h(ζ) = h(ζ), m−a.e.. It remains to show that Fhas nontangential limits almost everywhere as well. This follows from

35

36 3. HARDY SPACES

the simple observation that if h is bounded above then the function Fis bounded in D. Indeed, if h ≤M then

|F (z)| = exp


)≤ exp

(∫TPz(ζ) logMdm(ζ)

)= M.

This means that the functions F1, F2 defined by

F1(z) = exp

(∫T

ζ + z

ζ − zlog min{h(ζ), 1}dm(ζ)

),

and

F1(z) = exp

(∫T

ζ + z

ζ − zlog max{h(ζ), 1}dm(ζ)

),

are analytic and bounded in D, hence both functions have nontangentiallimits almost everywhere on the unit circle. By the above computationwe also know that these limits are nonzero m− a.e.. Thus F = F1/F2

F has nontangential limits almost everywhere on the unit circle aswell. �

Of course, the function F constructed above, is not the unique ana-lytic functions such that the modulus of its nontangential limits takethe prescribed value h a.e. If B is any finite Blaschke product then|B| extends continuously to the boundary and takes the value 1 there.Thus, BF shares the same property as F . It is also possible to con-struct another analytic function G in D such that |G| has the samenontangential limits as |F | a.e. and in addition, G has no zeros in D.Indeed, if g(z) = exp(−1+z

1−z ) then G = gF has the required propertybecause g extends continuously to ∂D \ {1} with |g(ζ)| = 1 for allζ ∈ ∂D \ {1}. Nevertheless, the function F constructed in the proof ofthe above theorem plays a central role in what follows.

Definition 1.1. An analytic function F in D is called an outer func-tion if there exists a nonnegative function h on the unit circle suchthat ∫

T| log h|dm <∞

and

F (z) = exp

(∫T

ζ + z


).

In this case, F is the outer function whose modulus equals h a.e. onthe boundary.

1. OUTER FUNCTIONS. 37

Exercise 1. Show that the product and quotient of two outer functionsis outer. Moreover, prove that if F is outer with |F (eit)| ≥ a > 0a.e. then |F (z)| ≥ a for all z ∈ D. Finally, show that any functionF that is analytic in a larger disc (than D) and has no zeros in D isan outer function, but the statement is no longer true if we replace theassumption ”F analytic in a larger disc” by ”F continuous in D.

Proposition 1.2. Let h be a nonnegative function on the unit circlesuch that ∫

T| log h|dm <∞

and let Fh be the outer function whose modulus equals h a.e. on theboundary. Then

|Fh(z)| ≤∫TPz(ζ)h(ζ)dm(ζ).

The proof is essentially based on the following simple lemma.

Lemma 1.1. If u, v : [a, b] 7→ R are integrable functions, v is nonnega-

tive with∫ bav(x)dx = 1 then

e∫ ba u(t)v(t)dt ≤

∫ b

a

eu(t)v(t)dt.

Proof. The inequality is equivalent to

1 ≤∫ b

a

v(t)(eu(t)−

∫ ba u(x)v(x)dx

)dt.

Since for all T ∈ R we have eT ≥ T + 1, we deduce that

eu(t)−∫ ba u(x)v(x)dx ≥ u(t)−

∫ b

a

u(x)v(x)dx+ 1 , t ∈ [a, b].

Integration on [a, b] (w.r.t t) now yields∫ b

a

v(t)(eu(t)−

∫ ba u(x)v(x)dx

)dt ≥

∫ b

a

v(t)

(u(t)−

∫ b

a

u(x)v(x)dx+ 1

)dt = 1


Exercise 2. Prove the following generalization of Lemma 1.1 which iscalled Jensen’s inequality.

Let φ : R 7→ R be a differentiable convex function, that is, φ satisfies

φ(x)− φ(y) ≥ φ′(y)(x− y), x, y ∈ R.

38 3. HARDY SPACES

If u, v : [a, b] 7→ R are integrable functions, with v ≥ 0 and∫ bav(x)dx =

1 then

φ

(∫ b

a

u(t)v(t)dt

)≤∫ b

a

φ(u(t))v(t)dt.

Proof of Proposition 1.2. Recall first that∫T Pzdm = 1 for all z ∈ D

and set u(t) = log h(eit), v(t) = 12πPz(e

it). Then by Lemma 1.1 weobtain

|Fh(z)| = exp


)≤∫TPz(ζ)h(ζ)dm(ζ).

Corollary 1.1. Let h be a nonnegative function on the unit circlesuch that ∫

T| log h|dm <∞

and let Fh be the outer function whose modulus equals h a.e. on theboundary. Then for 0 0.The inequality

‖Fh‖p ≥ ‖h‖Lp(m),

follows by an application of Fatou’s lemma and Proposition 1.1. Thereverse inequality follows by Proposition 1.2 applied to the outer func-tion F p

h , which is obviously well defined. �

2. Zeros of Hp-functions

It turns out that in terms of zeros, the condition defining membershipin Hp is quite restrictive, and in fact, it is equivalent to the Blaschkecondition. This has a number of important consequences for the devel-opment of the theory.We begin with a lemma on the integrals involved in the definition ofHp norms.

Lemma 2.1. If f is analytic in D and 0 1, the result can be obtained using properties ofPoisson integrals. The argument below is valid for all values p > 0. Itis easy to verify that for 0 < r < 1,∫

T| log |f(rζ)||dm(ζ) <∞,

so that we can consider the outer function Fr with |Fr(ζ)| = |f(rζ)|,m-a.e. on T . Moreover, if f has no zeros on rT, then |Fr| is boundedbelow, hence fr/Fr is bounded in D, where, as usual fr(z) = f(rz), z ∈D. Since the nontangential limits of fr/Fr have modulus one a.e. onthe circle, it follows by the Poisson representation that |fr/Fr| ≤ 1,that is,

|Fr(z)| ≥ |f(rz)|, z ∈ D.Then for 0 < ρ < 1 we have by Corollary 1.1

Mp(rρ, f) ≤ ‖fr‖pp ≤ ‖Fr‖pp = Mp(r, f).

Thus we have shown the inequality Mp(rρ, f) ≤ Mp(r, f) for all ρ ∈(0, 1) and all r in a dense subset of [0, 1]). The result follows. �

Theorem 2.1. Let 0 < p ≤ ∞, and assume that f ∈ Hp is notidentically zero. Let a1, a2, . . . be the zeros of f in D, repeated accordingto their multiplicity. Then∑

n≥1

(1− |an|) <∞.

Moreover, if B is the Blaschke product with zeros a1, a2, . . . then f/B ∈Hp, and

||f/B||p = ||f ||p.

Proof. Denote by BN the partial products

BN(z) = zmN∏

n=m+1

−an|an|

z − an1− anz

.

Then f/BN is analytic in D. We claim that f/BN ∈ Hp with

‖f/BN‖p = ‖f‖p.

If p =∞ this follows immediately from the maximum principle togetherwith the fact that BN is continuous in the closed unit disc and |BN | = 1on T. Indeed, if we pick a a sequence (zn), |zn| → 1 such that

||f/BN ||∞ = limn→∞

|f(zn)||BN(zn)|

,

40 3. HARDY SPACES

from |BN(zn)| → 1, n→∞, we obtain

||f/BN ||∞ = limn→∞

|f(zn)| ≤ ||f ||∞.

The reverse inequality is abvious, since |BN | ≤ 1 in D. The case when0 < p <∞ is similar, but the maximum principle is replaced by Lemma2.1. By this result we have

‖f/BN‖pp = limr→1−

Mp(r, f/BN) ≤ lim supr→1−

Mp(r, f) max|z|=r

(|BN(z)|−p = ‖f‖p,

and, again the reverse inequality is immediate from |BN | ≤ 1 in D.With the claim in hand, we see that (BN) cannot converge to zerouniformly on compact subsets of D, since f does not vanish identically,hence this would contradict the estimates

Mp(r, f/BN) ≤ ‖f‖pp, max|z|=r|f/BN(z)| ≤ ‖f‖∞,

valid for all r ∈ (0, 1). Thus the Blaschke condition in the statementholds true. Letting N → infty, we obtain from the last estimates

Mp(r, f/B) ≤ ‖f‖pp, max|z|=r|f/B(z)| ≤ ‖f‖∞,

i.e. f/B ∈ Hp and ‖f/B‖p ≤ ‖f‖p. The reverse inequality followsagain from |b(z)| < 1, z ∈ D, and the proof is complete. �

Exercise 1. In the case when p = ∞ Theorem 2.1 belongs to a veryclassical circle of ideas which includes the famous Schwarz Lemma.Prove the following assertions:

(i) (Schwarz Lemma ) Let f be analytic in D and satisfy |f(z)| ≤ 1there. If f(0) = 0 then

|f(z)| ≤ |z|, z ∈ D,

and

|f ′(0)| ≤ 1.

Moreover, if equality occurs in one of the above inequalities then f is arotation of the unit disc, i.e. f(z) = αz for some α ∈ C with |α| = 1.

(ii) (Schwarz-Pick Lemma ) Let f be analytic in D and satisfy|f(z)| ≤ 1 there. Then if z, w ∈ D, z 6= w we have

|f(z)− f(w)||1− f(w)f(z)|

≤ |z − w||1− wz|

3. APPLICATIONS 41

with equality if and only if f has the form f(z) = α z−a1−az for some fixed

a ∈ D and α ∈ ∂D. In particular, for all z ∈ D we have also

|f ′(z)| ≤ 1

1− |z|2.

Corollary 2.1. Let (an) be a sequence in D such that∑n≥1

(1− |an|) <∞

and let B be the corresponding Blaschke product. Then B has nontan-gential limits of modulus one almost everywhere on the unit circle.

Proof. Clearly B is in H∞ with ‖B‖∞ ≤ 1. Let f be any boundedanalytic function in D and apply Theorem 2.1 to the function Bf . Thenthe conclusion reads

||fB/B||∞ = ||f ||∞ = ||Bf ||∞.Now suppose there exist ε > 0 and a measurable set E ⊂ T, m(E) > 0such that the nontangential limits B(ζ) satisfy |B(ζ)| < 1−ε for ζ ∈ E.Let f be the outer function whose modulus equals 1 a.e. on E and 1

2a.e. on its complement. Then by Corollary 5.3, Chapter I we have that||f ||∞ = 1, but

||Bf ||∞ = essup|Bf | < max{1

2, 1− ε} < 1

which contradicts the previous equality. Thus, for all ε > 0 the set ofpoints ζ where |B(ζ)| < 1− ε has measure zero and hence, the set

{ζ, |B(ζ)| < 1} =∞⋃n=1

{ζ, |B(ζ)| < 1− 1/n}

has measure zero as well. �

3. Applications

Theorem 2.1 yields a very useful factorization of Hp-functions, whichhas a number of important consequences. According to this result,every nonzero f ∈ Hp, 0 < p ≤ ∞ cna be written in the form

f = Bg,

where B is a Blachke product and g ∈ Hp is zero-free in D with ‖g‖p =‖f‖p.

42 3. HARDY SPACES

3.1. Boundary behavior. With the above factorization in hand,we can easily apply the results obtained for Poisson integrals in Section5 of the previous chapter. Indeed, if f = Bg ∈ Hp, the zero-free factor gbelongs toHp so that, gp/2 belongs toH2 ⊂ h2. This simple observationleads to a quick proof of the following theorem.

Theorem 3.1. (i) If f ∈ Hp, 0 0 such that

‖f ∗‖Lp(m) ≤ Cp‖f‖p.

Proof. (i) As pointed out above, if f = Bg with B a Blachkeproduct and g ∈ Hp zero-free in D, then gp/2 ∈ H2, which impliesthat gp/2, and hence g has nontangential limits m-a.e.. Since B ∈ H∞,f = BG has nontangential limits m-a.e.. By Corollary 2.1 we have|f | = |g|, m−a.e. on T, and using also Theorem 5.1 we obtain

‖f‖p = ‖g‖p = ‖gp/2‖2/p2 = ‖gp/2‖2/p

L2(m) = ‖f‖Lp(m).

(ii) If f = Bg as above f ∗ ≤ g∗ = (gp/2)∗, and the result follows againfrom the fact that gp/2 ∈ H2. �

The norm convergence of dilations also extends for all values of 0 <p <∞ as the following result shows.

Theorem 3.2. If f ∈ Hp, 0 < p < ∞, and for 0 ≤ r < 1, fr(z) =f(rz), z ∈ D, then

limr→1−

‖f − fr‖p = 0.

Proof. By Theorem 3.1 (i) we have

‖f − fr‖pp =

∫T|f − fr|pdm,

and since |f(ζ) − f(rζ)| ≤ 2f ∗(ζ), ζ ∈ T, the result follows by anapplication of the dominated convergence theorem �

As pointed out in the previous chapter, the result fails when p =∞.

Corollary 3.1. (i) For 1 ≤ p <∞, Hp is a separable Banach space.(ii) For 0 < p < 1, Hp is a complete separable space w.r.t. the metric

dp(f, g) = ‖f − g‖pp.

3. APPLICATIONS 43

Proof. If f ∈ Hp, and 0 < r < 1, then fr can be approximateduniformly by polynomials, so that polynomials with rational coefficientsform a countable dense subset of Hp. The verification of the remainingassertions in (ii) is straightforward. �

3.2. Poisson representation of H1 and the F. & M. Riesztheorem. Since Hp ⊂ hp, it follows that for 1 < p ≤ ∞, every functionin Hp is the Poisson integral of its boundary function. Interestingenough, this result continues to hold for H1 as well, even if it fails forthe larger space h1.

Theorem 3.3. If f ∈ H1 then

f(z) =

∫TPz(ζ)f(ζ)dm(ζ), ζ ∈ D.

Proof. This is a direct application of Theorem 3.2. Since for 0 <r < 1 fr ∈ H∞, we have

f(rz) =

∫TPz(ζ)f(rζ)dm(ζ), ζ ∈ D.

Let r → 1−, and apply Theorem 3.2 together with the fact that forfixed z ∈ D, the Poisson kernel Pz(·) belongs to L∞(m). �

The next theorem is a famous result due to the brothers Riesz with wideapplications in function theory. The original proof is quite differentfrom the one below and is of interest in its own right.

Theorem 3.4. Let µ be a finite Borel measure on T with the propertythat ∫

Tζndµ(ζ) = 0,

for all nonnegative integers n. Then µ is absolutely continuous w.r.t.m and there exists f ∈ H1 with f(0) = 0 such that dµ

dm= f .

Proof. Let u be the Poisson integral of µ, and recall that u de-termines µ uniquely. Clearly, u ∈ h1, and u(0) = 0. We show that u isanalytic in D, hence u ∈ H1. For z ∈ D,

2u(z) = 2

∫TPz(ζ)dµ(ζ) =

∫T

ζ + z

ζ − zdµ(ζ) + intT

ζ + z

ζ − zdµ(ζ).

If ζ ∈ Tζ + z

ζ − z=

1 + zζ

1− zζ= 1 + 2

∞∑n=1

znζn,

44 3. HARDY SPACES

and the series converges uniformly on T, for fixed z ∈ D. Thus byassumption, ∫

T

ζ + z

ζ − zdµ(ζ) = 0,

and

2u(z) = +intTζ + z

ζ − zdµ(ζ),

so that u ∈ H1. Then the result follows by Theorem 3.3. �

3.3. Inner-outer factorization. The factorization used in theprevious section can be further refined. The crucial step is to constructfor nonzero f ∈ Hp the outer function F|f |. To this end we need toshow that log |f | ∈ L1(m), a quite remarkable property ofHp-functions.The classical approach to this fact is via the so-caled Jensen’s formula,but we shall follow a different path here and begin with the followinginequality, which is an analogue of (5.7), Chapter 2.

Proposition 3.1. If f ∈ Hp, 0 < p <∞, then for all z ∈ D,

|f(z)|p ≤∫TPz(ζ)|f(ζ)|pdm(ζ).

In particular,

|f(z)| ≤(

1 + |z|1− |z|

)1/p

‖f‖p, z ∈ D.

Proof. Write again f = Bg, as in the previous proofs, with Ba Blachke product and g ∈ Hp zero-free in D, and recall that |g| =|f |, m−a.e. on T. Since gp/2 ∈ H2 ⊂ h2 we can apply the estimate(5.7) to obtain

|f(z)|p ≤ |gp/2(z)|2 ≤∫TPz(ζ)|g(ζ)|pdm(ζ) =

∫TPz(ζ)|f(ζ)|pdm(ζ).

The second part follows by the standard estimate of the Poisson kernel.�

This inequality exhibits a harmonic majorant for Hp-functions, andit turns out that such harmonic majorants can be used to give analternative definition of Hardy spaces which extends to more generalplane domains as well. We shall not pursue this matter here.

Exercise 1. Show that if f ∈ Hp, 0 < p <∞, the harmonic function

uf (z) =

∫TPz(ζ)|f(ζ)|pdm(ζ)

3. APPLICATIONS 45

is the least harmonic majorant of |f |p in D, that is, if v is harmonicin D with v ≥ |f |p, then v ≥ uf .

With Proposition 3.1 in hand we can turn to our original goal.

Proposition 3.2. If f ∈ Hp, 0 < p ≤ ∞ is not identically zerothen log |f | ∈ L1(m). Moreover, if F is the outer function with F | =log |f |, m−a.e. on T then

|f(z)| ≤ |F (z)|, z ∈ D.

Proof. We start with the case when p = ∞ and assume withoutloss of generality that ‖f‖∞ ≤ 1. Write f = Bg with B a Blachkeproduct and g ∈ Hp zero-free in D, and note that

| log |f(ζ)|| = − log |f(ζ)| = − log |g(ζ)|,m−a.e. on T. By Fatou’s lemma

−∫T

log |g(ζ)|dm(ζ) ≤ lim infr→1−

−∫T

log |g(rζ)|dm(ζ) = − log |g(0)|,

because log |g| is harmonic in D. If p ∈ (0,∞), let G be the outerfunction with |G(ζ)| = exp(|f(ζ)|p), m−a.e. on T and note that:1) |G(z)| ≥ 1, z ∈ D,2) |f/G| is essentially bounded on T (ex

p> cpx),

From the first inequality we see that f/G ∈ Hp and if we apply Propo-sition 3.1 to f/G it follows easily that f/G ∈ H∞. Then by the firstpart of the proof we have that

log |f | − log |G| = log |f | − |f |p ∈ L1(m),

i.e. log |f | ∈ L1(m). To see the second part of the statement, we con-sider the sequence of outer functions (Fn) with |Fn(ζ)| = log(|f(ζ)| +1n), m−a.e. on T. Exactly the same reasoning as before shows thatf/Fn ∈ H∞ with ‖f/Fn‖∞ ≤ 1. Using the first part of the proofwe can easily show that Fn(z) → F (z) for all z ∈ D, which gives thedesired result. �

The outer function F in Proposition 3.2 will be called the outer factorof f .

Definition 3.1. A bounded analytic function I in D is called inner ifits nontangential limits satisfy |I(ζ)| = 1, m− a.e. on T.

Clearly, inner functions are bounded by 1 in D. Note that unimodularconstant are inner functions. Less trivial examples are provided byBlaschke products. According to Proposition 3.2 I = f/F is inner,

46 3. HARDY SPACES

whenever f ∈ Hp. This function will be called the inner factor off ∈ Hp. and the factorization

(3.14) f = IF

will be retfered to as the inner-outer factorization of f .

Proposition 3.3. The inner outer factorizarion (3.14) of f ∈ Hp, 0 <p ≤ ∞ is unique.

Proof. If f = IF = JG, with I, J inner and F,G outer then

I

J=G

F,

which implies that the nontangential limits of G/F are unimodular a.e.on T. Since G/F is outer it follows by definition that G/F = 1. �

At its turn, by Theorem 2.1 the inner factor can be further decomposedas I = BS, where B is a Blaschke product and a function S which iszero-free in D. Here we consider B = 1 if I has no zeros to begin with.

Definition 3.2. An inner function without zeros in D is called sin-gular inner.

Singular inner functions have a very special form. If S is such a functionthen log |S| is harmonic and negative in D, in particular it belongs toh1. Thus, by Corollary 2.2 in Chapter 2, there is a nonnegative finiteBorel measure µ on T such that

log |S(z)| = −∫TPz(ζ)dµ(ζ).

Since log |S| has nontangential limits zero a.e. on T, it follows byCorollary 4.1 in Chapter 2, that the measure µ is singular w.r.t. m.This argument together with analytic completion and exponentiationyields the following representation formula for singular inner functions.

Corollary 3.2. If S is a singular inner function then there existsα ∈ R and a nonnegative finite Borel measure µ on T which is singularw.r.t. m, such that

S(z) = exp

(iα−

∫T

ζ + z

ζ − zdµ(ζ)

).

The above argument refines the inner outer factorization f = IF , off ∈ Hp. The final result is stated below, but already proved above.

3. APPLICATIONS 47

Theorem 3.5. Let f ∈ Hp, 0 < p ≤ ∞. If f is not identically zero, itcan be written uniquely in the form

f = BSF,

where B is a Blaschke product (or B = 1), S is a singular inner func-tion, and F is the outer factor of f .

Exercise 2. Show that if f, 1/f ∈ Hp for some p > 0, then f is outer.

Exercise 3. Show that a singular inner function cannot extend contin-uously to the closed unit disc unless it is constant. Construct a functionthat extends continuosly to D and yet has a nontrivial singular innerfactor. Similarily one can construct such a function that has infinitelymany zeros. Hint: Take the Blaschke product with zeros 1− n−2, andmultiply by (1− z). Then show that the product is continuous.

Exercise 4. What is the cannonical factorization given by Theorem3.5 of the following functions?

a) f(z) = (2z − 1)ez, z ∈ D.

b) g(z) = exp( 1z2−1

), z ∈ D.

c) h(z) = sin z, z ∈ D.

3.4. Littlewood subordination. Given f, g analytic in D, we saythat f is subordinate to g, and write f ≺ g if there exists ϕ : D → Danalytic with ϕ(0) = 0 such that f = g ◦ ϕ. Note that in this casef(D) ⊆ g(D), and by the Schwarz lemma we have f(rD) ⊆ g(rD),for all r ∈ [0, 1]. J. E. Littlewood proved long ago that subordinationdecreases the Hp-norm, and his theorem is part of the following result.

Theorem 3.6. Let ϕ : D→ D be analytic. Then for every f ∈ Hp, 0 <p <∞, we have f ◦ ϕ ∈ Hp and

‖f ◦ ϕ‖p ≤(

1 + |ϕ(0)|1− |ϕ(0)|

)1/p

‖f‖p.

Equality holds for some 0 < p <∞ and all f ∈ Hp, if and only if ϕ isinner with ϕ(0) = 0.

Proof. The norm inequality in the statement is a direct conse-quence of Proposition 3.1, since by that inequality we have for everyf ∈ Hp,

|f(ϕ(z))|p ≤∫TPϕ(z)(ζ)|f(ζ)|pdm(ζ).

48 3. HARDY SPACES

Then ∫T|f(ϕ(rz))|pdm(z) ≤

∫T

∫TPϕ(rz)(ζ)|f(ζ)|pdm(ζ)dm(z)

=

∫T

∫T|f(ζ)|pPϕ(rz)(ζ)dm(z)dm(ζ),

and since z → Pϕ(rz)(ζ) is harmonic in r−1D, the inner integral is∫TPϕ(rz)(ζ)dm(z) = Pϕ(0)(ζ) ≤ 1 + |ϕ(0)|

1− |ϕ(0)|,

which proves the first part of the theorem.To see the second, assume first that equality holds for some 0 < p <∞and all f ∈ Hp. Let ζ ∈ T be fixed but arbitrary, and let f(z) =(ζ + z)2/p. Then

2 = ‖f‖pp = ‖f ◦ ϕ‖pp =1 + |ϕ(0)|1− |ϕ(0)|

∫T|ζ + ϕ(z)|2dm(z)

=1 + |ϕ(0)|1− |ϕ(0)|

(1 + 2Re ζϕ(0) + 1).

Clearly, the last expression on the right depends on ζ and cannot equalu2, unless ϕ(0) = 0. To see that ϕ is inner, let f(z) = z, to obtain

1 = ‖f‖pp = ‖f ◦ ϕ‖pp =

∫T|ϕ(ζ)|pdm(ζ),

which immediately |ϕ(ζ)| = 1, m−a.e., because, by the assumption ofthe theorem we have |ϕ(ζ)| ≤ 1, m−a.e..Conversely, assume that ϕ is inner with ϕ(0) = 0, and let Q(z) =∑

n anzn be any polynomial. Then

‖Q ◦ ϕ‖22 =

∑n,k

anak

∫Tϕn(ζ)ϕk(ζ)dm(ζ) =

∑n

|an|2 = ‖Q‖22,

since for n > k∫Tϕn(ζ)ϕk(ζ)dm(ζ) =

∫Tϕn−k(ζ)dm(ζ) = 0.

If f ∈ H2 is arbitrary, we approximate f in the norm of H2 by asequence of polynomials (Qn). By the first part of the theorem Qn◦ϕ→f ◦ ϕ in H2, which gives that ‖f‖2 = ‖f ◦ ϕ‖2. In particular, for everyinner function I we have

1 = ‖I‖2 = ‖I ◦ ϕ‖2,

3. APPLICATIONS 49

which implies that I ◦ϕ is inner as well. If 0 < p <∞ is arbitrary, andf ∈ Hp, we write f = Bg with B a Blaschke product and gp/2 ∈ H2.Then B ◦ ϕ is inner and

‖f ◦ ϕ‖p = ‖gp/2 ◦ ϕ‖2/p2 = ‖gp/2‖2/p

2 = ‖f‖p,and the proof is complete. �

This result, especially the first part of it has numerous applications.We shall only mention one of these below.

Corollary 3.3. Let f be analytic in D with Re f(z) ≥ 0, z ∈ D.Then f is outer and belongs to Hp for all p < 1.

Proof. If φ is an automorphism of the disc with f(0)1+φ(0)1−φ(0)

, then

f ≺ g = 1+φ1−φ . Indeed, g maps the disc onto the right half-plane, hence

g−1◦f maps D into itself with g−1◦f(0) = 0. We claim that g ∈ Hp, 0 <p < 1. In view of Theorem 3.6 it suffices to show that z → 1+z

1−z belongsto Hp, 0 < p < 1, which follows by a direct calculation. Thus, anotherapplication of Theorem 3.6 gives that f ∈ Hp, 0 < p < 1. Note that1/f has nonnegative real part as well, so 1/f ∈ Hp, 0 < p < 1, henceby Exercise 2 in the previous subsection, f is outer. �

3.5. The Phragmen-Lindelof Principle. The type of problemwe want to investigate in this section is illustrated by the followingexample.Let 0 < a < π, and suppose that f is analytic in the the angle Γ ={z = reit, r > 0, |t| < a}, and continuous in its closure, and assumethere exists M > 0 such that

|f(re±ia)| ≤M, r ≥ 0.

It is not difficult to show by means of examples that such a functionf does not need to be bounded in Γ, i.e. the ”naive” form of themaximum principle fails in this domain. Of course one would liketo add conditions that are as weak as possible, under which one canconclude that |f | ≤M in Γ.A more general situation of this type can be described in the unit disc.Consider an analytic function f in D which has the nontangential limitf(ζ), for almost every ζ ∈ T. It turns out that the properties of f aredifficult to deduce from the behavior of the boundary function. Forexample, if S is a nonconstant singular inner function, then f = 1/Shas unimodular nontangential limits a.e. on T, yet f is unbounded, infact, it doesn’t belong to any Hp, p > 0. This is easily seen from theinner-outer factorization, since 1/S = JF implies F = 1, hence SJ = 1,

50 3. HARDY SPACES

and by the maximum principle, both S and J would be constant.It turns out that this type of situation can be addressed with help ofthe Phragmen-Lindelof principle which we present here in the mostgeneral form-

Theorem 3.7. Let F be an outer function and f be an analytic func-tion in D such that

|f(z)| ≤ |F (z)|for all z ∈ D. Then f has the nontangential limit f(ζ), for almostevery ζ ∈ T, and if f ∈ Lp(m), for some p ∈ (0,∞], then f ∈ Hp.

Proof. By assumption we have that f/F ∈ H∞, so that f/F , andhence f has nontangential limits a.e. on T. By Proposition 1.1 F hasnonzero nontangential limits a.e. on T, hence, so has f . The inner-outer factorization of f/F gives f/F = IG with I inner and G outer,so that f = IGF , and

|f | = |GF |,m−a.e. on T. Then by Corollary 1.1 GF ∈ Hp, and since |f | ≤ |GF |in D, it follows that f ∈ Hp. �

Corollary 3.4. Let 1 ≤ p ≤ ∞. The map which associates to f ∈ Hp

its boundary function is a linear isometry from Hp onto the closedsubspace of Lp(m) consisting of functions g with∫

Tζng(ζ)dm(ζ) = 0, n ≥ 1.

Its inverse is given by the Cauchy formula

f(z) =

∫T

ζf(ζ)

ζ − zdm(ζ).

Proof. It is clear that the boundary functions of Hp-functionsbelong to the subspace described in the statement , and also that thissubspace is closed. If g ∈ Lp(m) with∫

Tζng(ζ)dm(ζ) = 0, n ≥ 1,

then it is the boundary function of an H1-function, and the resultfollows by Theorem 3.7. The Cauchy formula obviously holds for thedilated function fr, f ∈ Hp, and the formula in the statement followsby letting r → 1−. �

We close with an application of the theorem to the situation describedat the beginning of the paragraph.

3. APPLICATIONS 51

Theorem 3.8. Suppose that f is analytic in the the angle Γ = {z =reit, r > 0, |t| < a}, where 0 < a < π is fixed. Assume there existsM > 0 such that for every boundary point ζ = re±ia we have

lim supz→ζz∈Γ

|f(z)| ≤M.

If f satisfies in addition an inequality of the form

|f(z)| ≤ b exp(c|z|d)for some constants b, c > 0, 0 < d < π

2aand all z ∈ Γ then f is bounded

in the whole angle with |f(z)| ≤M, z ∈ Γ

Proof. Under our assumption on the constant d, there is c1 > 0such that |z|d ≤ Re zd, hence

|f(z)| ≤ b exp(c2Re zd), z ∈ Γ.

Let ψ : D → Γ, ψ(z) = (1+z1−z

2a/π. The ψ maps D conformally onto Γ,

so that it suffices to show that f ◦ ψ ∈ H∞ with ‖f ◦ ψ‖∞ ≤ M . Bythe first inequality in this proof we have that

|f ◦ ψ(z)| ≤ |eψd(z)| = |F (z)|.We claim that F is outer. If the claim holds, the result follows byTheorem 3.7, since the nontangential limits of ψ lie on ∂Γ a.e. andby assumption it follows that the nontangential limits of |f | are ≤ Ma.e.. To see the claim, recall from the previous subsection that ψπ/2a ∈Hp, 0 1, henceit is the Poisson integral of its boundary function, which is obviouslyintegrable. Thus F is outer and the proof is complete. �

CHAPTER 4

The dual of Hp

1. Basic reduction to the study of Cauchy integrals

We shall be concerned with the continuous linear functionals on theBanach spaces Hp, 1 ≤ p < ∞, that is the linear maps l : Hp → Cwith

|l(f)| ≤ C‖f‖p,for some C > 0 and all f ∈ Hp.The description of these functionals is particulary simple when p = 2,since H2 is a Hilbert space with respect to the scalr prduct in L2(m)

〈f, g〉 =

∫Tfgdm, f, g ∈ H2.

According to the Riesz representation theorem, every continuous linearfunctional l on H2 has the form

l(f) = 〈f, g〉 =

∫Tfgdm, f ∈ H2,

for some fixed g ∈ H2.When p 6= 2, general functional analysis provides less information.However we can use Corollary 3.4 to embed Hp ito Lp(m) and then usethe Hahn-Banach theorem to represent every continuous linear func-tional l on Hp, 1 ≤ p <∞ by

l(f) =

∫Tfgdm,

with g ∈ Lq(m), 1p

+ 1q

= 1. The problem that arises here, is that this

representation is not unique, it leads to a representation of the dual ofHp as a quotient space. This is a well known fact in functional analysisand follows by another application of the Hahn-Banach theorem. Theproof of the proposition below is left as an exercise. As usual we denotethe dual of the normed space X by X ′ and if S ⊂ X,

S⊥ = {l ∈ X ′ : l(S) = {0}}.

53

54 4. THE DUAL OF Hp

Proposition 1.1. Let X be a Banach space and let Y be a closedsubspace. Then the map J : X ′/Y ⊥ → Y ′,

J [l] = l|Y,

is a linear isometry.

According to this result, (Hp)′ is isometrically isomorphic to the quo-tient space Lq(m)/(Hp)⊥. Moreover, (Hp)⊥ consists of functions g ∈Lq(m) with ∫

Tζng(ζ)dm(ζ) = 0, n ≥ 0,

hence by the F.& M. Riesz theorem, it follows easily that (Hp)⊥ = Hq0 ,

the subspace of Hq consisting of functions that vanish at the origin.Can we ”single out” elements in the cosets [f ] ∈ Lq(m)/Hq

0 in a mean-ingful way?There is a simple but powerful idea that can be used here, namely toconsider Cauchy integrals. Note that if g ∈ (Hp)⊥ then∫

T

ζg(ζ)

ζ − zdm(ζ) = 0, z ∈ D

since

ζ

ζ − z=∞∑n=0

zn

ζn=∞∑n=0

znζn, ζ ∈ T, z ∈ D,

and the series converges uniformly on T when z ∈ D is fixed. Thisgives the following result which is the starting point for the material ofthis chapter.

Proposition 1.2. Let l be a continuous linear functional on Hp, 1 ≤p <∞. Then there exists a unique analytic function h in D of the form

h(z) =

∫T

ζg(ζ)

ζ − zdm(ζ), z ∈ D,

where g ∈ Lq(m), ‖g‖Lq(m) ≤ 2‖l‖, such that

l(f) = limr→1−

∫Tf(ζ)h(rz)dm(z) f ∈ Hp.

Proof. We can write

l(f) =

∫Tf(ζ)g(ζ)dm(ζ),

1. BASIC REDUCTION TO THE STUDY OF CAUCHY INTEGRALS 55

with [g] ∈ Lq(m)/(Hp)⊥ and ‖g‖Lq(m) ≤ 2‖[g]‖ = ‖l‖. Moreover, sincefr → f in Hp, when r → 1− we obtain easily that l(fr)→ l(f), i.e.

l(f) = limr→1−

∫Tf(rζ)g(ζ)dm(ζ).

Using the Cauchy formula as in Corollary 3.4 we have for ζ ∈ T

f(rζ) =

∫T

zf(z)

z − rζdm(z), ζ ∈ T, 0 < r < 1,

and by an application of Fubini’s theorem we obtain∫Tf(rζ)g(ζ)dm(ζ) =

∫Tzf(z)

∫T

g(ζ)

z − rζdm(ζ)dm(z)

=

∫Tf(z)

∫T

ζg(ζ)

ζ − rzdm(ζ)dm(z).

Then the result follows with

h(z) =

∫T

ζg(ζ)

ζ − zdm(ζ), z ∈ D,

using the considerations preceeding the proposition. �

According to this result, one way to describe the continuous linearfunctionals on Hp is to understand Cauchy integrals of Lq(m)-functions1p

+ 1q

= 1, and this is precisely what will be done in the sequel.

The Cauchy integral, or the Cauchy transform of h ∈ L1(m) is denoted

by h and is defined by

(1.15) h(z) =

∫T

h(ζ)

ζ − zdm(ζ), z ∈ C \ T.

This formula gives 2 analytic functions, one in D, and the other in C\D,but we will be mostly concerned with the first, since the properties ofthe second are very similar. In fact their boundary values are relatedby the so-called ”jump theorem”.

Theorem 1.1. If h ∈ L1(m)

limr→1−

h(rz)− h(1

rz) = zh(z)

m−a.e. on T.

Proof. It suffices to check the simple identity

1

ζ − rz− 1

ζ − r−1z= ζPz(ζ)

and apply the results about boundary behavior of Poisson integrals. �


2. The M. Riesz theorem

In view of the representation as Cauchy integrals of Hp-functions, 1 ≤p ≤ ∞, proved in Corollary 3.4, a natural question that comes tomind is whether the Cauchy transforms of Lp(m)-functions belong toHp, 1 ≤ p ≤ ∞?For real-valued functions, the above question turns into a questionabout harmonic conjugation. Indeed, if h ∈ L1(m) is real-valued withPoisson integral u, then the analytic function

f(z) =

∫T

ζ + z

ζ − zh(ζ)dm(ζ), z ∈ D,

satisfies

Ref(z) = u(z), z ∈ D, , f(0) ∈ R.Thus for such functions, we can reformulate the above in terms of realparts:If f is analytic in D and Ref ∈ hp, 1 ≤ p ≤ ∞, does f ∈ Hp (orImf ∈ hp)?Obviously the original question reduces to real-valued functions, so thatthe two are equivalent.It turns out that the answer is negative for the ”endpoints” p = 1 andp =∞. Indeed, if

f1(z) =1 + z

1− zf2(z) = i log(1− z), z ∈ D,

then Ref1(z) = Pz(1), so that Ref1 ∈ h1, and Ref2 ∈ h∞. However,f1 /∈ h1, because its boundary function is not integrable, and f2 is ob-viously unbounded.

Nevertheless, the questions discussed above have an affirmative answerfor all 1 0 such that

‖h‖p ≤ Cp‖h‖Lp(m).

Equivalently, if f is analytic in D, f(0) ∈ R and Ref ∈ hp, thenf ∈ Hp and there exists C ′p > 0 such that

‖f‖p ≤ C ′p‖Ref‖p.

2. THE M. RIESZ THEOREM 57

Proof. When 1 < p ≤ 2 we shall prove the second statement. Letu = Ref, v = Imf . We use the Green formula in the form (6.12), i.e.∫

Tϕdm = ϕ(0) +

∫D

∆ϕ(z) log1

|z|dA(z).

to compute for 0 < r < 1, ‖fr‖pp, ‖ur‖pp. A straightforward computationbased on the Cauchy-Riemann equations yields

∆|fr|p = p2|fr|p−2|f ′r|2 ,

∆|ur|p = p(p− 1)up−2r

[(∂ur∂x

)2

+

(∂ur∂y

)2]

= p(p− 1)up−2r |f ′|2.

Thus for 1 < p ≤ 2 we have

∆|fr|p ≤p

p− 1∆|ur|p,

hence by (6.12) and the fact that v(0) = 0 we obtain

‖fr‖pp ≤p

p− 1‖ur‖pp,

and the result follows letting r → 1−.In the remaining case when 2 < p <∞, we shall prove the first state-ment by duality. Let h ∈ Lp(m), and let g ∈ Lq(m), where 1

p+ 1

q= 1.

For 0 < r < 1 we have by Fubini’s theorem,∫Th(rζ)g(ζ)dm(ζ) =

∫Th(z)

∫T

g(ζ)

z − rζdm(ζ)dm(z).

As in the proof of Proposition 1.2 we have for z ∈ T,∫T

g(ζ)

z − rζdm(ζ) = z

∫T

ζg(ζ)

ζ − rzdm(ζ) = zg1(rz),

where g1(ζ) = ζg(ζ). Using the first part of the proof, the fact that1 < q < 2, and the equivalence of the two statements, it follows thatG(z) = zg1(z) belongs to Hq with

‖G‖qq ≤q

q − 1‖g‖qLq(m).

Thus by Holder’s inequality∣∣∣∣∫Th(rζ)g(ζ)dm(ζ)

∣∣∣∣ ≤ ‖h‖Lp(m)‖G‖q ≤(

q

q − 1

)1/q

‖h‖Lp(m)‖g‖Lq(m),


which gives

‖(h)r‖p ≤(

q

q − 1

)1/q

‖h‖Lp(m),

and the result follows letting r → 1−. �

Combining the M. Riesz theorem with Proposition 1.2 we obtain im-mediately a nice description of the dual of Hp for 1 < p <∞.

Corollary 2.1. For every continuous linear functional l on Hp, 1 <p <∞, there exists a unique g ∈ Hq, 1

p+ 1

q= 1, such that

l(f) =

∫Tf(ζ)g(ζ)dm(z), f ∈ Hp.

Moreover, there exists a positive constant ap such that

ap‖g‖q ≤ ‖l‖ ≤ ‖g‖q.

Exercise 2. Prove the M. Riesz theorem in the case p = 2 using theParseval formula.

3. The dual of H1

According to the discussion at the beginning of this chapter, in order torepresent the dual of H1 ia useful way, as a space of analytic functions,we need to understand the Cauchy integrals of L∞(m)-functions. Aswe have seen in the previous section, such functions may fail to bebounded, because the M. Riesz theorem fails for p = ∞. However,a description of Cauchy integrals of L∞(m)-functions can be obtainedusing the following computation based on conformal invariance.

Lemma 3.1. Let h ∈ L1(m) and let h be its Cauchy transform. Ifa ∈ D, and φa(z) = a−z

1−az , z ∈ D, then

h(φa(z))− h(a) = zh1(z), z ∈ D,

where h1(ζ) = φa(ζ)h(φa(ζ)), ζ ∈ T.

Proof. Start with

1

ζ − φa(z)− 1

ζ − a=

φa(z)− a(ζ − φa(z))(ζ − a)

,

and

φa(z)− a = −(1− |a|2)z

1− az.

3. THE DUAL OF H1 59

Moreover,

ζ − φa(z) =ζ − aζz − a+ z

1− az=

1− aζ1− az

(−φa(ζ) + z).

From these three computations we obtain

1

ζ − φa(z)− 1

ζ − a=

(1− |a|2)z

(φa(ζ)− z)(1− aζ)(ζ − a), z ∈ D , ζ ∈ T,

and since ζ ∈ T, we have ζ = 1ζ, i.e.

1

ζ − φa(z)− 1

ζ − a=

zζ

(φa(ζ)− z)

(1− |a|2)

|1− aζ|2=

zζ

(φa(ζ)− z)|φ′a(ζ)|.

Thus,

h(φa(z))− h(a) = z

∫T

ζ

(φa(ζ)− z)h(ζ)|φ′a(ζ)|dm(ζ).

Now use the fact that φ−1a = φa, together with the change of variable∫Tg ◦ φa|φ′a|dm =

∫Tgdm,

to obtain

h(φa(z))− h(a) = z

∫T

φa(ζ)h(φa(ζ)

ζ − zdm(ζ),

which is the formula in the statement. �

This useful result reveals a remarkable property of Cauchy transformsof L∞-functions. Note first that by the M. Riesz theorem, these Cauchytransforms belong to Hp, for all p > 0.

Proposition 3.1. There exists C > 0, such that if h ∈ L∞(m), andφa(z) = a−z

1−az , a, z ∈ D, then

supa∈D‖h− h(a)‖2 ≤ C‖h‖L∞(m).

In other words, the boundary function of h belongs to BMO.

Proof. Apply the M.Riesz theorem in the case p = 2, togetherwith Lemma 3.1 to obtain

‖h− h(a)‖2 ≤ C‖φah ◦ φa‖L2(m) ≤ C‖h‖L∞(m).

�


Let us denote by BMOA the space of analytic functions in BMO,more precisely, the space of functions in H2, whose boundary valuedbelong to BMO. The more striking result is that Cauchy transformsof L∞(m)-functions actually coincide with BMOA.

Theorem 3.1. A function belongs to BMOA if and only if it is theCauchy transform of an element of L∞(m).

Proof. The fact that the Cauchy transform of an element of L∞(m)belongs to BMOA has already been established in the previous propo-sition. To see the converse, we use Proposition 1.2 from the beginningof this chapter. According to this result, a function g ∈ BMOA is theCauchy transform of an L∞(m)-function if the limit

l(f) = limr→1−

∫Tfgrdm, f ∈ H1,

exists and defines a bounded linear functional on H1. Observe thatsince g ∈ H2 we can apply Parseval’s formula to obtain∫

Tfgrdm =

∫Tfrgdm, f ∈ H1, , 0 < r < 1.

Finally, let us note that it suffices to show an estimate of the form

(3.16)

∣∣∣∣∫Tfrgdm

∣∣∣∣ ≤ C‖f‖1,

for some C > 0 and all f ∈ H1, 0 < r < 1. Indeed, if (3.16) holds thenfor r, ρ→ 1−, r > ρ, we have fr − fρ = (f − fρ/r)r, hence∣∣∣∣∫

T(fr − fρ)gdm

∣∣∣∣ ≤ C‖f − fρ/r‖1 → 0,

i.e. the limit l(f) exists.

We proceed in three steps. First we use the Littlewood-Paley formulaProposition 6.2 in Chapter 2 in polarized form to obtain∫

Tfrgdm = f(0)g(0) +

∫T(fr − f(0))g − g(0)dm

= f(0)g(0) +

∫D∇fr · ∇g log

1

|z|dA

= f(0)g(0) + 2

∫Df ′rg′ log

1

|z|dA.

3. THE DUAL OF H1 61

Next we write f ∈ H1 as a product f = uv, with u, v ∈ H2, ‖u|22 =‖v‖2

2 = ‖f‖1. This follows easily if we write f = Bh, with B a Blaschkeproduct and h ∈ H1, zero-free in D, and set u = Bh1/2, v = h1/2. Then∫

Tfrgdm = f(0)g(0) + 2

∫Du′rvrg

′ log1

|z|dA+ 2

∫Durv

′rg′ log

1

|z|dA.

Obviously, it suffices to estimate the first integral on the right handside, since the second is completely analogous. By the Cauchy-Schwarzinequality∣∣∣∣∫

Du′rvrg

′ log1

|z|dA

∣∣∣∣2 ≤ ∫D|u′r|2 log

1

|z|dA

∫D|vr|2|g′|2 log

1

|z|dA.

Another application of the littlewood-Paley formula gives∫D|u′r|2 log

1

|z|dA ≤ 1

2‖ur‖2

2 ≤1

2‖u‖2

2 = ‖f‖1.

To see the last integral, apply Corollary 6.1 in Chapter 2 to concludethat if g ∈ BMOA, the measure

|∇g(z)|2 log1

|z|dA(z) = 2|g′(z)|2 log

1

|z|dA(z),

is a Carleson measure, so that∫D|vr|2|g′|2 log

1

|z|dA ≤ C‖vr‖2

2 ≤ C‖v‖22 = ‖f‖1.

The proof is complete. �

From the discussion at the beginning of the chapter we now obtain thefollowing description of the dual of H1. Until now, for the definition ofBMOA we have used the seminorm

supa∈D‖g ◦ φa − g(a)‖2.

A natural norm on this space is

‖g‖∗ = |g(0) + supa∈D‖g ◦ φa − g(a)‖2.

Corollary 3.1. A linear functional l : H1 → C is continuous if andonly if it has the form

l(f) = limr→1−

∫Tfgrdm,

for some function g ∈ BMOA. Moreover, g is uniquely determined byl and there exist absolute constants a, b > 0 such that

a‖l‖ ≤ ‖g‖∗ ≤ b‖l‖.


4. The John-Nirenberg inequality

BMOA is a very interesting space of functions with a number of sur-prising properties. Let us note from the begining that the study of theanalytic functions with bounded mean oscillation, is not too restrictive,since many of their properties continue to hold for all such functions.The reason is provided below.

Proposition 4.1. The function f belongs to BMOA if and only if theboundary function of its real part belongs to BMO.

Proof. Let u = Ref . Since |∇u|2 = 2|f ′|2, the measure |∇u|2 log 1|z|dA

is Carleson if and only if |f ′|2 log 1|z|dA is. �

Exercise 1. Show that every function in BMOA is a linear combina-tion of four functions in BMOA whose real part is bounded.

Recall that by the M. Riesz theorem it follows that BMOA ⊂⋂p>0H

p,but it is not contained in H∞. Since any real-valued function in BMOequals the boundary values of the real part of a BMOA function iffollows also that BMO ⊂

⋂p>0 L

p(m). The estimate below is muchstronger than that and has far-reaching consequences.

Theorem 4.1. Let g ∈ BMOA. Then there exists c > 0 such thatecg ∈ H2. If h ∈ BMO then there exists d > 0 such that ed|h| ∈ L2(m).

Proof. The slick argument below is due to Pommerenke. Giveng ∈ BMOA, and assume without loss of generality that g(0) = 0.Consider the linear map defined on H2 by

Tgf(z) =

∫ z

0

f(ζ)g′(ζ)dζ, z ∈ D, , f ∈ H2.

We claim that Tg is a bounded linear operator from H2 into itself.Indeed, by the Littlewood-Paley formula and the fact that g ∈ BMOAwe obtain

‖Tgf‖22 = 2

∫D|(Tgf)′|2 log

1

|z|dA = 2

∫D|fg′|2 log

1

|z|dA ≤ C‖f‖2

2,

and the claim follows. Then by elementary functional analysis it followsthat for |λ| > ‖Tg‖, λI − Tg is invertible on H2, i.e., for every F ∈ H2

there is a unique solution f ∈ H2 of

λf − Tgf = F.

4. THE JOHN-NIRENBERG INEQUALITY 63

A simple calculation shows that for F = 1, the solution f is given by

f = eg/λ,

and if we choose λ > 0, the first part follows. Also note that

e−g/λ , e±ig/λ ∈ H2,

as well. This gives that e±Reg/λ, e±Img/λ ∈ L2(m), and together withthe above discussion, this immediately implies the second part. �

APPENDIX A

Sequences and families of holomorphic functions

One of the basic applications of the Cauchy formula is the fact thatthe set of holomorphic functions in a given region in the complex planeis closed with respect to uniform convergence on compacts. That is,given any sequence (fn) of holomorphic functions in the open set U ⊂ Cwhich converges uniformly on every compact subset of U to a functionf : U 7→ C, it follows that f is also holomorphic on U . This is animportant result which, in particular, enables us to construct highlynontrivial examples of holomorphic functions. Some of these are listedbelow as exercises.

Exercise 1. Show that the following expressions define holomorphicfunctions in the specified domain:

(i) ζ(z) =∞∑n=1

1

nz, Re z > 1, (Zeta function),

(ii) Γ(z) =

∫ ∞0

e−ttzdt

t, Re z > 0, (Gamma function),

(iii) f(z) =∞∑n=1

1

(z − n)2, z ∈ C \ N, (no special name),

(iv) ℘(z) =1

z2+

∞∑m,n∈Z

m2+n2 6=0

[1

(z −m− in)2− 1

(m+ in)2

],

z ∈ C \ Z + iZ, (Weierstrass function).

Theorem 0.2. Let (fn) be a sequence of holomorphic functions on theopen connected set U ⊂ C that converges uniformly on compacts to the(holomorphic) function f . Let V be an open subset of U such that there

65

66 A. SEQUENCES AND FAMILIES OF HOLOMORPHIC FUNCTIONS

exists n0 ≥ 1 with fn(z) 6= 0 for all z ∈ V and n ≥ n0. Then either fis the constant function 0, or f has no zeros in V .

Proof. Suppose f is not identically zero, but has a zero z0 ∈ V .Choose a small disc ∆ centered at z0 such that ∆ ⊂ V and that f hasno zero on ∂∆. Since for n ≥ n0 fn has no zero in ∆ we have∫

∂∆

f ′n(z)dz

fn(z)= 0, n ≥ n0.

On the other hand, f ′nfn

converges uniformly on ∂∆ to f ′

f, so that

limn→∞

1

2πi

∫∂∆

f ′n(z)dz

fn(z)=

1

2πi

∫∂∆

f ′(z)dz

f(z)≥ 1

which gives a contradiction and the theorem is proved. �

Let us now turn our attention to an important property of families ofholomorphic functions that resembles to (relative) compactness of setsin Rn. We begin with the following general definition.

Definition 0.1. A family F of holomorphic functions on the openset U ⊂ C is called normal if every sequence (fn) in F contains asubsequence (fnk) that converges uniformly on compact subsets to some(holomorphic) function f (not necessarily in F !)

The following classical result due to Montel characterizes the normalfamilies of holomorphic functions.

Theorem 0.3. (Montel). A family F of holomorphic functions on theopen set U ⊂ C is normal if it is uniformly bounded on compacts, thatis, for every compact set K ⊂ U there exists a positive constant CKsuch that for all f ∈ F and all z ∈ K we have

|f(z)| ≤ CK .

Proof. The result will follow by a standard diagonalization pro-cess, once we show that the family F is equicontinuous at every pointz0 ∈ U , that is, for every ε > 0 there exists δ > 0 such that for everyz ∈ U with |z − z0| < δ and all f ∈ F we have

|f(z)− f(z0)| < ε.

Assume this claim for a moment and proceed as follows. Choose acountable dense set {zn, n = 1, 2, . . .} in U (i.e. for every z ∈ U andevery neighborhood V of z there exists n such that zn ∈ V ). Since thesequences (fn(zj))n≥1 are bounded in C we can extract a subsequence

A. SEQUENCES AND FAMILIES OF HOLOMORPHIC FUNCTIONS 67

(fn1k) such that (fn1

k(z1)) converges. This sequence has at its turn a

subsequence (fn2k) such that (fn2

k(z2)) converges. If we continue this

process indefinitely we obtain subsequences (fnjk) such that (fnjk

(zl))

converges for 1 ≤ l ≤ j. If we set nk = nkk then clearly, the subsequence(fnk) has the property that (fnk(zj)) converges for all j = 1, 2, . . .. Thisis called a diagonalization process.

Let us now show that (fnk) converges uniformly on compacts. Notefirst that if w ∈ U and ε > 0 then by the claim (equicontinuity) thereis δ > 0 such that

|fnk(z)− fnk(w)| < ε

for all k ≥ 1 and z ∈ U with |z − w| < δ. Then, if z, ζ ∈ U satisfy|z − w| < δ and |ζ − w| < δ, we have

|fnk(z)− fnk(ζ)| ≤ |fnk(z)− fnk(w)|+ |fnk(ζ)− fnk(w)| < 2ε.

Fix a point zj (j depends on w) in the above set, such that |zj−w| < δ.Then there exists k0 depending only on w such that for k, p ≥ k0

|fnk(zj)− fnp(zj)| < ε.

Putting this together we obtain for all z ∈ U with |z − w| < δ that

|fnk(z)−fnp(z)| ≤ |fnk(z)−fnk(zj)|+|fnk(zj)−fnp(zj)|+|fnp(zj)−fnp(z)| < 3ε

for all k, p ≥ k0.

Finally, let K ⊂ U be compact ε, δ > 0 as above and cover K by finitelymany sets of the form {z, |z − w| < δ} ∩ U, w ∈ K. Then the abovereasoning shows that there exists k1 ≥ 1 such that for all z ∈ K andk, p ≥ k1 we have

|fnk(z)− fnp(z)| < 3ε.

Since K was arbitrary we obtain that (fnk) converges pointwise to somefunction f and from the above inequality, letting p → ∞ we get thatfor every compact set K ⊂ U there exists k1 ≥ 1 such that for allz ∈ K and k ≥ k1 we have

|fnk(z)− f(z)| < 3ε,

i.e. (fnk) converges uniformly on compacts to f .

It remains to prove the claim from the beginning of the proof, i.e. theequicontinuity of the family F . For z0 ∈ U let ∆0 be a disc of radiusr0 > 0 with ∆0 ⊂ U . By hypothesis, we know that there exists aconstant C0 such that

supz∈∆0

|f(z)| ≤ C0

68 A. SEQUENCES AND FAMILIES OF HOLOMORPHIC FUNCTIONS

for all f ∈ F . By Cauchy’s formula we have

f(z)−f(z0) =1

2πi

∫∂∆0

f(ζ)

(1

ζ − z− 1

ζ − z0

)dζ =

z − z0

2πi

∫∂∆0

f(ζ)dζ

(ζ − z)(ζ − z0)

for all f ∈ F and all z ∈ ∆0. This leads to the estimate

|f(z)− f(z0)| ≤ |z− z0| sup|ζ−z0|=r0

|f(ζ)| 1

|ζ − z|≤ C0|z− z0|

1

r0 − |z − z0|.

Thus, if ε > 0 is given, choose 0 < δ < r0 such that C0δ/(r0 − δ) < ε.Then for |z − z0| < δ and f ∈ F we have

|f(z)− f(z0)| ≤ C0δ1

r0 − δ< ε.

The claim now follows and the proof is complete. �

This is a famous theorem with a large number of applications in com-plex analysis. One of these applications is a very useful criterion foruniform convergence on compacts.

Theorem 0.4. (Vitali) Let (fn) be a sequence of holomorphic functionson the open connected set U ⊂ C. Suppose that {fn, n = 1, 2, . . .} isuniformly bounded on compacts and also that there is a nondiscretesubset A of U such that (fn) converges pointwise on A. Then (fn)converges uniformly on compacts.

Proof. By Montel’s theorem there is at least a subsequence (fnk)that converges uniformly on compacts to a holomorphic function fon U . Suppose the sequence itself does not converge uniformly oncompacts to f . Then there exists a compact subset K of U , an ε > 0and a subsequence (fn′p) of (fn) such that

(0.17) supz∈K|fn′p(z)− f(z)| ≥ ε

But, again by Montel’s theorem, (fn′p) contains at its turn a subse-quence (fn′′p ) that converges uniformly on compacts to some holomor-phic function g on U . Then letting p→∞ in (1.1) we see that g 6= f .On the other hand, if A is the set in the statement we have, by hy-pothesis, for all z ∈ A

g(z) = limp→∞

fn′′p (z) = limn→∞

fn(z) = limk→∞

fnk(z) = f(z).

Since A is not discrete we obtain g = f by the identity theorem andwe arrive at a contradiction that proves our result. �

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