classical spaces of holomorphic functions · spaces of holomorphic functions 1 in these notes we...

CLASSICAL SPACES OF HOLOMORPHIC FUNCTIONS

MARCO M. PELOSO

Contents

1. Hardy Spaces on the Unit Disc 11.1. Review from complex analysis 11.2. Hardy spaces 51.3. Harmonic Hardy classes 71.4. Fatou’s theorem 91.5. The zero sets of functions in Hp 131.6. Boundary behaviour of functions in Hardy spaces 161.7. The Cauchy–Szego projection 192. Bergman spaces on the unit disc 222.1. Function spaces with reproducing kernel 222.2. The Bergman spaces 252.3. Biholomorphic invariance 282.4. Lp-boundedness of a family of integral operators 292.5. The Bergman kernel and projection on the unit disc 333. The Paley–Weiner and Bernstein spaces 353.1. The Fourier transform 353.2. The Paley–Wiener theorems 403.3. The Paley–Wiener spaces 453.4. The Bernstein spaces 494. Function theory on the upper half plane 514.1. Hardy spaces on the upper-half plane 514.2. Factorization and boundary behaviour of functions in Hp(U). 544.3. H2 and the Paley–Wiener theorem revisited 554.4. H1, the atomic decomposition and the space of bounded mean oscillations 554.5. Weighted Bergman spaces on the upper-half plane 554.6. The Paley–Wiener theorem for Bergman spaces 555. Further topics 565.1. Hardy spaces on tube domains in Cn 565.2. Weighted Bergman spaces on the unit ball in Cn 565.3. The Cauchy integral along Lipschitz curves 565.4. Real variable Hardy spaces 565.5. Hankel and Toeplitz operators 56References 56

Appunti per il corso Argomenti Avanzati di Analisi Complessa per i Corsi di Laurea in Matematicadell’Universita di Milano, a.a. 2011/12.

– November 12, 2011.

SPACES OF HOLOMORPHIC FUNCTIONS 1

In these notes we consider spaces of holomorphic functions defined on domains of the complexplane or, toward the end of the notes, on domains of the n-dimensional complex space Cn.

We begin with the Hardy spaces on the unit disc.

1. Hardy Spaces on the Unit Disc

1.1. Review from complex analysis. Let C denote the field of complex numbers z = x+ iy,where x, y ∈ R are the real and imaginary part, respectively, of z. We will denote by D the unitdisc, that is,

D =z ∈ C : |z| < 1

.

Here and in what follows, we will denote by D(z0, r) the disc centered at z0 ∈ C and radiusr > 0; we will sometimes abbreviate Dr to denote the disc D(0, r).

If f is holomorphic on D then it can be written as sum of a converging power series

f(z) =+∞∑k=0

akzk .

From Cauchy’s formula we know that, if γ is a simple closed curve contained in D and z liesin the interior of γ (which is a well-defined domain by the Jordan curve theorem), then

f(z) =1

2πi

∫γ

f(ζ)ζ − z

dζ .

If the function f is actually holomorphic in a slightly larger disc DR with R > 1, then wemay take the unit circonference ∂D as curve of integration and obtain

f(z) =1

2πi

∫∂D

f(ζ)ζ − z

dζ ,

for all z ∈ D.One of the main themes of this course is trying to recover the holomorphic function f (in a

given class) from its boundary values, once these values are suitably defined.Setting z = reiη, we rewrite the above identity as

f(reiη) =1

2π

∫ 2π

0f(eiθ)

eiθ

eiθ − reiηdθ

=1

2π

∫ 2π

0f(eiθ)

11− rei(η−θ)

dθ

=1

2π

∫ 2π

0f(ei(η−θ))

11− reiθ

dθ , (1.1)

using the periodicity of the exponential.Notice that

11− reiθ

=+∞∑n=0

rneinθ

=: Cr(eiθ) , (1.2)

where Cr(eiθ) is the Cauchy kernel (thought of as a convolution kernel, see (1.4) below).

2 M. M. PELOSO

Therefore, using the uniform convergence on compact subsets of D of the power series, wehave that (1.1) gives the identity

f(reiη) =+∞∑n=0

rn1

2π

∫ 2π

0f(eiθ)ein(η−θ) dθ

=+∞∑n=0

rneinη1

2π

∫ 2π

0f(eiθ)e−inθ dθ . (1.3)

We recall that the unit circle ∂D = ζ ∈ C : ζ = eiθ inherits the multiplicative structurefrom C and it is homeomorphic (as a topological group) to the interval [−π, π] (or any interval oflength 2π) once we identify the two end points of the interval. This group T = R/2πZ is calledthe torus. Of course, the identification between T and ∂D is given by the mapping t 7→ eit.

Given f, g ∈ L1(T) (where, we recall, T is identified with the unit circle) the convolution onT is defined as

f ∗ g(eiη) =1

2π

∫ 2π

0f(eiθ)g(ei(η−θ)) dθ . (1.4)

Therefore, the identity (1.1) can be expressed as convolution on the group T as

f(reiη) = (f ∗ Cr)(eiη) . (1.5)

Moreover, in (1.3) we recognize the n-th Fourier coefficient of the restriction of f on the unitcircle ∂D (function that, at this stage, is assumed to be continuous)

f(n) =1

2π

∫ 2π

0f(eiθ)e−inθ dθ .

It is clear that in order to define the Fourier coefficients is sufficient to assume that f is anL1-function on the unit circle.

Another integral formula, with its relative integral kernel, that will play a key role in thesenotes is the Poisson reproducing formula for the unit disc D

u(reiη) =1

2π

∫ 2π

0u(eiθ)

1− r2

|1− rei(η−θ)|2dθ

= (Pu)(reiη) , (1.6)

where u is a function harmonic in a domain containing D and 0 < r < 1. The operator P iscalled the Poisson operator and Pu the Poisson integral of the function u defined on the unitcircle.

More generally, the solution of the Dirichlet problem on the unit D (see [CA-notes]) showsthat if we start with a continuous function g on the unit circle ∂D, the function u := Pg isharmonic in D, continuous on the closure of the unit disc and coincides with g on the unit circle.

It is worth to observe that Pg is well defined for all g ∈ L1(T) and that Pg is a harmonicfunction in D.

One of the main questions that we are going to address is if, and in which sense, the functionPg admits boundary values and, in this case, if such values coincide with the function g. In


other words, a classical question is whether the Dirichlet problem (for the Laplacian on the unitdisc)

∆f = 0 on Df = g on ∂D ,

(1.7)

where g ∈ Lp(∂D) is an assigned function, 1 ≤ p <∞, admits a solution and in what sense thesolution f admits g as values on the boundary.

As in the case of the reproducing formula (1.5) we may think of the Poisson integral as aconvolution integral operator on T by writing

Pr(eiθ) =1− r2

|1− reiθ|2, (1.8)

so that the Poisson integral of a function g ∈ L1(T) is defined by the formula

(Pg)(reiη) =1

2π

∫ 2π

0g(eiθ)

1− r2

|1− rei(η−θ)|2dθ = (Pr ∗ g)(eiη) . (1.9)

In complex analysis we were used to think of the Poisson kernel, that we denote temporarelyas Pz(eiθ) as a function of the variables z = reiη ∈ D and eiθ ∈ ∂D, where

Pz(eiθ) =1− r2

|eiθ − reiη|2=

1− r2

|1− rei(η−θ)|2(1.10)

=1− r2

1− 2r cos(θ − η) + r2. (1.11)

In the current setting it is more convenient to think of the Poisson kernel as a family offunctions Pr defined on the torus T, hence depending on the variable eiη, and of the Poissonintegral of a function g as the convolution of g with Pr, as in (1.9). According to this point ofview, the collection of functions Pr0<r<1 on T is a summability kernel (or a approximationof the identity), A family of functions Φtt>0 is called a summability kernel if they satisfy thefollowing properties:

(1)1

2π

∫ 2π

0Φt(eiθ) dθ = 1 for t > 0;

(2) ‖Φt‖L1 ≤ C with C independent of t;

(3) for every δ > 0, limt→+∞

∫δ≤|θ|≤π

|Φt(eiθ)| dθ = 0.

We recall that if Φtt>0 is a summability kernel on T and g ∈ Lp(T), 1 ≤ p < ∞, thenΦt ∗ g → g in the Lp-norm, and that if g ∈ C(T), then Φt ∗ g → g in the sup-norm.

We collect here the basic properties of the Poisson kernel and integral.

Proposition 1.1. The function P (reiη) = Pr(eiη) is harmonic in D as function of z = reiη

and1

2π

∫ 2π

0Pr(eiη) dη = 1

for all 0 < r < 1.Moreover, as collection Pr of functions on T we have

(i) for 1 ≤ p <∞, ‖f ∗ Pr‖Lp ≤ ‖f‖Lp;(ii) for 1 ≤ p <∞, limr→1− ‖f ∗ Pr − f‖Lp = 0;

4 M. M. PELOSO

(iii) if g ∈ C(T), then limr→1− ‖g ∗ Pr − g‖L∞ = 0.Finally, for f ∈ Lp(T), 1 ≤ p ≤ ∞, P(f) is harmonic in D.

Proof. It is clear that it suffices to show Pr is a summability kernel and that P (reiη) isharmonic in the variable z = reiη ∈ D.

The latter fact is well known, as well as that 12π

∫ 2π0 Pr(eiη) dη = 1.

In order to prove that Pr is a summability kernel, it remains to show that condition (3) issatisfied. This is easily verified and we leave the details to the reader. 2

Lemma 1.2. For 0 < r < 1 we have that

Pr(eiη) =+∞∑

k=−∞r|k|eikη ,

and the convergence is uniform on T.

Proof. Recall that we always identify the unit circle ∂D with T.We can write

1− r2

|1− reiη|2=

11− reiη

+1

1− re−iη− 1 (1.12)

=+∞∑k=0

rkeikη ++∞∑k=0

rke−ikη − 1

=+∞∑

k=−∞r|k|eikη .

The statement about the uniform convergence is clear, by the Weierstrass M -test. 2

Notice that by (1.12) we obtain a relation between the Poisson and Cauchy kernels, that is,

Pr = 2ReCr − 1 . (1.13)

Lemma 1.3. Let g ∈ L1(∂D). Then for every 0 < r < 1 we have that(Pr ∗ g

)(eiη) =

+∞∑k=−∞

g(k)r|k|eikη .

Proof. Using the uniform convergence of the series development of Pr we see that(Pr ∗ g

)(eiη) =

12π

∫ 2π

0g(eiθ)

+∞∑k=−∞

r|k|eik(η−θ) dθ

=+∞∑

k=−∞

( 12π

∫ 2π

0g(eiθ)e−ikθ dθ

)r|k|eikη

=+∞∑

k=−∞g(k)r|k|eikη ,

and the convergence is uniform on T. 2


1.2. Hardy spaces. We are finally ready to define the classical Hardy spaces. We will denoteby H(Ω) the space of holomorphic functions on a given domain Ω.

Definition 1.4. For 0 < p <∞ and 0 < r < 1, for a function f defined on D we set

Mp(f, r) =( 1

2π

∫ 2π

0|f(reiθ)|p dθ

)1/p.

We define the Hardy space Hp = Hp(D) as

Hp =f ∈ H(D) : sup

0<r<1Mp(f, r) <∞

,

and for f ∈ Hp we set‖f‖Hp = sup

0<r<1Mp(f, r) .

Furthermore, we define H∞ as the space of holomorphic functions that are bounded on the unitdisc, endowed with the sup-norm.

We mention in passing that, also when 0 < p < 1 we set

‖f‖Lp =( 1

2π

∫ 2π

0|f(eiθ)|p dθ

)1/p

and, with an abuse of language, we call it the Lp-norm. However, setting d(f, g) = ‖f − g‖pLpLp becomes a complete metric space. Notice that ‖f + g‖Lp ≤ 2(1−p)/p(‖f‖Lp + ‖g‖Lp).

Remark 1.5. Aside from the case p = ∞, also the space H2 can be described at once. For,if f is holomorphic on D, then it admits power series expansion f(z) =

∑+∞n=0 anz

n. Using theuniform convergence on compact subsets of D we have

M2(f, r)2 =1

2π

∫ 2π

0|f(reit)|2 dt

=1

2π

∫ 2π

0

+∞∑n,m=0

anamrn+meit(n−m) dt

=+∞∑n=0

r2n|an|2 .

Therefore,

sup0<r<1

M2(f, r) =( +∞∑n=0

|an|2)1/2

,

that is, f ∈ H2 if and only if∑+∞

n=0 |an|2 is finite. In particular, it follows that H2 is a Hilbertspace.1

Observe that, in particular, a function f in H2 can be extended to the boundary ∂D, havingas “boundary values” the function f ∈ L2(∂D) given by

f(eit) =+∞∑n=0

aneint .

1Exercise I.2.

6 M. M. PELOSO

Moreover, by Lemma 1.3, P(f)(r·) = Pr∗f =∑+∞

n=0 anrnein(·) = f(r·). We can call f “boundary

values” since f(r·) = Pr ∗ f → f in L2(T) as r → 1−.This fundamental relation is a first indication of the deep connection between the theories of

Fourier series and of Hardy spaces.

Remark 1.6. For 0 < p < q ≤ ∞ we have 0 ( Hq ( Hp.For, notice that for 0 < p < q ≤ ∞ we clearly have Hq ⊆ Hp. Moreover, all holomorphic

polynomials are bounded, hence in all the Hp spaces and these are non-trivial. Moreover, usingLemma 2.15 (that we will see later on) one can show that

f(z) =1

(1− z)a∈ Hp if and only if p <

1a.

In these notes we will essentially restrict ourselves to the case 1 ≤ p ≤ ∞, although the Hardyspaces turn out to be of great interest also and, from a certain point of view, especially, in thecase 0 < p ≤ 1. We point out that the Hardy space H1 can be considered as a limit case forboth theories 1 < p ≤ ∞ and 0 < p < 1 and we will say more about it in Section 4.4 when wedeal with the Hardy spaces on the upper half-plane.

The first observation we make is that the “sup” in the definition of the Hp-norm is actuallya “lim”.

Remark 1.7. Given f holomorphic in D, the function Mp(f, r) is increasing in r, 0 < r < 1.This statement holds true for the full range 0 < p ≤ ∞, but its proof is elementary only in

the case p ≥ 1 and we will restrict to this case.For 0 < % < 1 and a function f defined in D we write f% := f(%·) to denote a function on the

unit circle. For 0 < r, % < 1, given a function f holomorphic in D, notice that fr is holomorphicin a ngbh of D so that

f(r%eiη) = (fr ∗ P%)(eiη) .Hence, we have

Mp(f, r%) = ‖fr ∗ P%‖Lp(T)

≤ ‖fr‖Lp(T) ‖P%‖L1(T) = Mp(f, r) ,

that is, Mp(f, r) is increasing in r and

sup0<r<1

Mp(f, r) = limr→1−

Mp(f, r) .

Proposition 1.8. For 1 ≤ p ≤ ∞ Hp is a Banach space.

Proof. It is clear that ‖ · ‖Hp is a norm, so we only need to prove that it is complete.Let 0 < r, % < 1 and notice that for f holomorphic on D we have

|f(r%eit)| =∣∣fr ∗ P%(eit)∣∣ ≤ ‖fr‖Lp(T)‖P%‖Lp′ (T)

≤ C%‖f‖Hp .

This shows thatsup|z|≤%|f(z)| ≤ C%‖f‖Hp ,


and therefore the convergence in the Hp-norm implies the uniform convergence on compactsubsets.

Thus, let fn be a Cauchy sequence in the Hp-norm, and let f be the function uniform limiton compact subsets of D. Then f is holomorphic and for 0 < r < 1 fixed, using the uniformconvergence, for p <∞ we have

Mp(fn − f, r)p =1

2π

∫ 2π

0|fn(reit)− f(reit)|p dt

= limm→+∞

12π

∫ 2π

0|fn(reit)− fm(reit)|p dt

= limm→+∞

Mp(fn − fm, r)p

≤ limm→+∞

‖fn − fm‖pHp .

Therefore,‖fn − f‖Hp ≤ lim

m→+∞‖fn − fm‖Hp < ε ,

for n sufficiently large. The case p =∞ is similar and we leave the details to the reader.2 2

1.3. Harmonic Hardy classes. We now analyze the action of the Poisson integral on thespaces Lp(T). This analysis has interest on its own right, but also will serve for studying theboundary behaviour of functions in (the holomorphic) Hardy spaces.

We consider now harmonic functions on D that satisfy the Hp-growth condition, 0 < p <∞,that is,

sup0<r<1

Mp(f, r) <∞ , (1.14)

and in an obvious way when p =∞.

Proposition 1.9. Let f be harmonic in D and satisfy (1.14), 1 < p ≤ ∞. Then there existsf ∈ Lp(∂D) such that f = P(f), that is,

f(reit) = (f ∗ Pr)(eit) =1

2π

∫ 2π

0f(eiθ)

1− r2

|1− rei(t−θ)|2dθ .

Moreover, if 1 < p <∞, fr → f in the Lp-norm.

We observe that this statement is false when p = 1. For instance, f(reiη) = Pr(eiη) satisfiesthe hypotheses with p = 1, but there exists no L1-function g on ∂D such that Pr → g andPr ∗ g = Pr. We leave the details as an exercise.3

Proof. Let fr = f(r·), 0 < r < 1. Let rn be a strictly increasing sequence in (0, 1), rn → 1−.Then frn is a bounded set in Lp(T). By the Banach-Alaoglu theorem (see [Ru]), there exists

a subsequence frnj converging in the weak-∗ topology to a funtion f ∈ Lp(T), 1 < p ≤ ∞,

that is, for every g ∈ Lp′(T) we have that∫ 2π

0frnj (e

iθ)g(eiθ) dθ →∫ 2π

0f(eiθ)g(eiθ) dθ as j → +∞ .

2Exercise I.4.3Exercise I.5.

8 M. M. PELOSO

(Here we need the restriction p > 1, since the Banach-Alaoglu theorem requires that the spacecontaining fr to be the dual space of another Banach space– and L1(T) is not.)

Let 0 < r < 1 and let j0 be such rnj > r for j ≥ j0. Then, frnj is continuous on D andharmonic in D, hence it satisfies the Poisson reproducing formula

f(reit) = frnj((r/rnj )e

it)

= (frnj ∗ Pr/rnj )(eit)

=1

2π

∫ 2π

0frnj (e

iθ)Pr/rnj (ei(t−θ)) dθ

=1

2π

∫ 2π

0frnj (e

iθ)Pr(ei(t−θ)) dθ +1

2π

∫ 2π

0frnj (e

iθ)[Pr/rnj (e

i(t−θ))− Pr(ei(t−θ))]dθ .

(1.15)

Since Pr ∈ C(∂D) ⊂ Lp′(∂D) for all 0 < r < 1, the first term on the right hand side above

converges to1

2π

∫ 2π

0f(eiθ)Pr(ei(t−θ)) dθ .

Now we show that the second term on the right hand side of (1.15) tends to 0 as j → +∞. For,gj := Pr/rnj − Pr → 0 in the Lp

′-norm, since gj ∈ C(T) and tend to 0 uniformly. Therefore,∣∣∣ ∫ 2π

0frnj (e

iθ)gj(ei(t−θ)) dθ∣∣∣ ≤ ‖frnj ‖Lp‖gj‖Lp′ ≤ C‖gj‖Lp′ → 0

as j → +∞. Notice that we have used the assumption that f satisfies (1.14).From (1.15) it follows that

f(reit) =1

2π

∫ 2π

0f(eiθ)Pr(ei(t−θ)) dθ ,

as we wished to show.The last part of the statement is clear: fr = f ∗ Pr as functions on ∂D and since Pr is a

summability kernel, fr = f ∗ Pr → f in the Lp-norm, for 1 < p <∞. 2

Remark 1.10. Notice that, in particular Prop. 1.9 applies to Hp-functions.Given f ∈ Hp, we have constructed the boundary value function f as weak-∗ limit of a

sequence fnj, where nj → 1−. It is easy to see that this construction is independent of thechoice of the sequence fnj. (This is obvious in particular when 1 < p <∞ since f is the limitin the Lp-norm of the whole family fr. )

Moreveor, if fr → g in Lp(∂D), it also easy to see that g = f . In particular, the boundaryvalue function of an H2-function identified in Remark 1.5 coincides with the one constructed inProp. 1.9.

We now prove a partial converse of Prop. 1.9.

Proposition 1.11. Let g ∈ Lp(∂D), 1 ≤ p < ∞. Then the Poisson integral of g, P(g) isharmonic in D, satisfies the Hp-growth condition (1.14), and (in the notation of Prop. 1.9)

P(g) = g .


Proof. Notice that, as functions on ∂D, fr = f(r·) := g ∗Pr satisfy the norm estimate ‖fr‖Lp ≤‖g‖Lp‖Pr‖L1 = ‖g‖Lp . Therefore,

sup0<r<1

Mp(f, r) ≤ ‖g‖Lp .

This shows that the harmonic function P(g) satisfies the Hp-growth condition (1.14).Next, if 1 < p <∞, by the previous proposition, f := P(g) admits a boundary value function

f such that P(f) = f , that is, P(f − g) = 0. Therefore, Pr ∗ (f − g) = 0 in Lp(T) for all0 < r < 1, since Pr is a summability kernel, 0 = Pr ∗ (f − g)→ f − g in Lp(T). (Notice thathere we need to restrict to the case p <∞.)

Finally, in the case p = 1, we simply notice that fr := Pr ∗ g → g in L1(∂D) as r → 1−, sothat we may define again the “boundary values” of f as the limit of the family fr, that is,f = g. This concludes the proof. 2

Remark 1.12. Recall the Dirichlet problem (1.7). If the boundary datum g ∈ C(∂D) we knowthat the solution f = P(g) is harmonic, continuous in the closure of the unit disc and f restrictedto ∂D coincides with g (e.g. see [CA-notes]).

Prop. 1.11 can be restated by saying that if g ∈ Lp(∂D), 1 ≤ p < ∞, then f = P(g) isharmonic in D and can be extdended to a boundary value function f that coincides with g.Then, we say that f = P(g) is (the) solution of the Dirichlet problem (1.7) with boundarydatum in Lp.

We conclude this part with the following consequence of the last two propositions.

Corollary 1.13. Define

hp(D) =f harmonic on D : ‖f‖hp := sup

0<r<1Mp(f, r) <∞

.

Then, for 1 < p <∞,P : Lp(∂D)→ hp(D)

is an isometric isomorphism.

1.4. Fatou’s theorem. Perhaps, the main result about the Hardy spaces is that if f ∈ Hp with0 < p ≤ ∞, then f , as in the case p = 2, admits “boundary values”, that is,

limr→1−

f(reit) = f(eit) exists for a.e. t ∈ [0, 2π) ,

and f ∈ Lp(∂D). Actually, more is true: the function f converges to f when z approaches apoint eit ∈ ∂D, for a.e. t ∈ [0, 2π), within the non-tangential approach regions.

Definition 1.14. We define the subset of the unit disc

Γα(eit) =z ∈ D : |z − eit| < α(1− |z|)

. (1.16)

The set Γα(eit) is also called the Stoltz region, with vertex in eit and aperture α > 1.

Remark 1.15. We need to make a few comments on the definition of Γα(eit).(i) It is easy to see that the regions Γα(eit) are increasing with α and that they would be

empty for α ≤ 1.

10 M. M. PELOSO

(ii) The region Γα(eit) is also called a non-tangential approach region, since outside a fixedcompact subset of D, it is contained in a cone with vertex in eit and aperture Cαα, where Cα > 1is a suitable constant.

For, assume that 1 − εα ≤ r < 1, where cα is a (small) positive constant, that we are goingto fix momentarily, and let z = reiη ∈ Γα(eit). Then

|reiη − eit| =∣∣eiη − eit − (1− r)eiη

∣∣ ≥ |eiη − eit| − (1− r) ,

so that|eiη − eit| ≤ (α+ 1)(1− r) .

If cα is chosen sufficientily small, then we must have |η − t| < π/4 (that is, z ∈ Γα(eit) andr ≥ 1− cα implies |η − t| < π/4).

Next, |eiη − eit| = 2 sin(|η − t|/2

)≥ 2

√2

π |η − t|, as long as |η − t| ≤ π/4, which is the case.Therefore, z = reiη ∈ Γα(eit) and r ≥ 1− cα implies

|η − t| ≤ π

2√

2(α+ 1)(1− r) = Cαα(1− r) , (1.17)

as claimed. In particular, we may take Cα = 3.

Theorem 1.16. (Fatou) Let f ∈ Hp, 1 < p ≤ ∞. Then

limΓα(eit)3z→eit

f(z) = f(eit) ,

where f is the function as in Prop. 1.9, and Γα(eit) is as in (1.16).

In order to prove Fatou’s theorem, we need to prove a result concerning the Hardy–Littlewoodmaximal function, defined (in the case of the torus T) as

Mf(eit) = sup0<δ≤π

12δ

∫|θ|<δ

|f(ei(t−θ))| dθ .

We recall that M is weak-type (1, 1), that is, it satisfies the following estimate: there exists aconstant C > 0 such that for every λ > 0,∣∣t : Mf(eit) > λ

∣∣ ≤ C

λ‖f‖L1(T) .

The next result shows the reason for our interest in the Hardy–Littlewood maximal function.

Proposition 1.17. Let g ∈ L1(∂D), α > 1 a fixed parameter. Then there exists Cα > 0 suchthat for every eit ∈ ∂D

supz∈Γα(eit)

|P(g)(z)| ≤ CαMg(eit) .

Proof. The proof is divided in a few steps.

(Step 1.) Assume first that z = reiη varies in a fixed compact subset of D, namely assumethat 1− r ≥ cα, with cα as in Remark 1.15 (ii).

Notice that for all θ

Pr(eiθ) =1− r2

|1− reiθ|2≤ 1− r2

(1− r)2≤ 2

1− r. (1.18)

12 M. M. PELOSO

Using the inequalities in Step 2 we have that∫E0

|g(ei(η−θ))|Pr(eiθ) dθ ≤2

1− r

∫|θ|≤3α(1−r)

|g(ei(η−θ))| dθ

≤ Cα1

6α(1− r)

∫|θ|≤3α(1−r)

|g(ei(t−θ))| dθ

≤ CαMg(eit) ,

using the change of variables θ 7→ θ − t+ η.Next, using again the change of variables θ 7→ θ − t+ η,∫

EN+1

|g(ei(η−θ))|Pr(eiθ) dθ ≤ 21

2π

∫|θ|≤π


= 21

2π

∫|θ|≤π

|g(ei(t−θ))| dθ

≤ cMg(eit) .

Finally, we estimate the central part of the integral:

N∑j=1

∫Ej

|g(ei(η−θ))|Pr(eiθ) dθ ≤ 2N∑j=1

122jα2(1− r)

∫Ej


≤ 2N∑j=1

122jα2(1− r)

∫|θ|≤2j3α(1−r)


≤ 2N∑j=1

122jα2(1− r)

∫|θ|≤2j+13α(1−r)

|g(ei(t−θ))| dθ

≤ CαN∑j=1

2−j1

2j+13α(1− r)

∫|θ|≤2j+13α(1−r)

|g(ei(t−θ))| dθ

≤ CαN∑j=1

2−jMg(eit)

≤ CαMg(eit) .

This concludes the proof. 2

We are now ready to prove Thm. 1.16.

Proof of Thm. 1.16. Since H∞ ⊂ Hp for all p <∞, it suffices to deal with the case p <∞.Let f ∈ Lp(∂D) be as in Prop. 1.9 (so that P(f) = f) and for a given γ > 0 let g ∈ C(∂D) be

such that ‖f−g‖Lp ≤ γ; notice that here we need p <∞. By the solution of the Dirichlet problemwe know that P(g)(z)→ g(eit) as D 3 z → eit (even without the restriction z ∈ Γα(eit)).


We use Prop. 1.17 and argue as in the proof of the Lebesgue differentiation theorem,∣∣∣eit ∈ ∂D : lim supΓα(eit)3z→eit

|f(z)− f(eit)| ≥ ε∣∣∣

≤∣∣∣eit ∈ ∂D : lim sup

Γα(eit)3z→eit|f(z)− P(g)(z)| ≥ ε/3

∣∣∣+∣∣∣eit ∈ ∂D : lim sup

Γα(eit)3z→eit|P(g)(z)− g(eit)| ≥ ε/3

∣∣∣+∣∣∣eit ∈ ∂D : |g(eit)− f(eit)| ≥ ε/3

∣∣∣≤∣∣∣eit ∈ ∂D : CαM(f − g)(eit)| ≥ ε/3

∣∣∣+∫eit∈∂D:|g(eit)−f(eit)|≥ε/3

(3|g(eit)− f(eit)|ε

)pdt

≤ C

ε‖g − f‖L1 +

C ′

εp‖g − f‖pLp ≤

C ′′

εpγp

≤ Cε ,

if we choose γ sufficiently small. This shows that∣∣∣eit ∈ ∂D : lim supΓα(eit)3z→eit

|f(z)− f(eit)| ≥ ε∣∣∣ ≤ Cε ,

which implies thatlim

Γα(eit)3z→eitf(z) = f(eit) ,

a.e., and the statement follows. 2

1.5. The zero sets of functions in Hp. In this section we describe the zero sets of functionsin the Hardy spaces, for the full range 0 < p ≤ ∞. Surprisingly, it turns out that these arethe same for all values of p. The description of such sets will allow us to prove a factorizationtheorem for Hp-functions that will have important consequences.

We begin by recalling a result from complex analysis.

Proposition 1.18. (Jensen’s formula) Let r > 0 and let f be holomorphic in a ngbh ofD(0, r). Let ζ1, . . . , ζN be the zeros of f in D(0, r), counting multiplicity. Assume that f doesnot vanish at the origin and on the circle of radius r. Then

log |f(0)|+ logN∏j=1

r

|ζj |=

12π

∫ 2π

0log |f(reiθ)| dθ .

For a proof we refer the reader to any classical text in complex analysis, as well as for thetheory of infinite products.

A consequence of this result is the following corollary.

Corollary 1.19. Let f ∈ H(D), f(0) 6= 0 and ζ1, ζ2, . . . its zeros counting multiplicity. Then,

log |f(0)|+ log+∞∏j=1

1|ζj |≤ sup

0<r<1

12π

∫ 2π

0log+ |f(reiθ)| dθ ,

where, for t > 0, log+ t = max(log t, 0).

14 M. M. PELOSO

Proof. Clearly, for every 0 < r < 1,

log |f(0)|+ logN∏j=1

r

|ζj |≤ 1

2π

∫ 2π

0log+ |f(reiθ)| dθ

≤ sup0<r<1

12π

∫ 2π

0log+ |f(reiθ)| dθ .

Letting r → 1− we obtain the result. 2

Corollary 1.20. Let f ∈ Hp, 0 < p ≤ ∞, f 6≡ 0, and let ζ1, ζ2, . . . its zeros countingmultiplicity. Then

+∞∑j=1

(1− |ζj |) <∞ . (1.19)

Proof. If f vanishes of order k at the origin, we consider f(z)/zk, which is still in Hp and it iszero-free at the origin. Notice that

0 ≤ log+ |f(reiθ)| ≤ |f(reiθ)|p .Then we applying Cor. 1.19 to obtain that

0 < log+∞∏j=1

1|ζj |

<∞ ,

which implies that∏+∞j=1

1|ζj | converges; hence

∏+∞j=1 |ζj | converges. But this is equivalent to∑+∞

j=1(1− |ζj |) <∞. 2

We recall that for ζ ∈ D the function

ϕζ(z) =z − ζ1− ζz

is a biholomorphic mapping of D onto itself. In this setting ϕζ is also called a Blaschke factor.

Proposition 1.21. Let ζ1, ζ2, . . . be points in D satisfying condition (1.19). Then the infiniteproduct

+∞∏j=1

−ζj|ζj |

ϕζj

converges uniformly on compact subsets of D.

Proof. It suffices to show that, for r < 1 be fixed and |z| ≤ r, the series+∞∑j=1

∣∣∣1 +ζj|ζj |

ϕζj (z)∣∣∣ <∞

converges uniformly. A simple calculation now shows that∣∣∣1 +ζj|ζj |

ϕζj (z)∣∣∣ =

∣∣|ζj |(1− |ζj |) + ζjz(1− |ζj |)∣∣

|ζj ||1− ζjz|

≤ 1 + r

1− r(1− |ζj |) .


By the Weierstrass M -test the convergence is uniform in |z| ≤ r, and the conclusion follows. 2

Definition 1.22. Let 0 < p ≤ ∞, f ∈ Hp and let ζ1, ζ2, . . . its zeros different from the origin,counting multiplicity. We set

B(z) = zk+∞∏j=1

−ζj|ζj |

ϕζj (z) ,

where k ≥ 0 is the order of zero of f at the origin. Then, B is a well-defined holomorphicfunction on D, and there exists a holomorphic, zero-free function F on D such that f/B = F .

The function B is called a Blaschke product and f = FB is called the canonical factorizationof f ∈ Hp.

Proposition 1.23. Let 0 < p ≤ ∞, f ∈ Hp and let f = FB be its canonical factorization.Then F ∈ Hp and ‖F‖Hp = ‖f‖Hp.

Proof. Notice that B = zk∏+∞j=1 −

ζj|ζj |ϕζj and every factor satisfies

∣∣ ζj|ζj |ϕζj

∣∣ ≤ 1, so that |B| ≤ 1.Therefore, |f(z)| ≤ |F (z)| for all z ∈ D, so that ‖f‖Hp ≤ ‖F‖Hp .

We set BN =∏Nj=1−

ζj|ζj |ϕζj and FN = f/BN . We recall that ϕζj is defined and holomorphic

in the disc D(0, 1/|ζj |) which is strictly larger than the unit disc D and that |ϕζj (eiη)| = 1 forall η. Therefore, BN converge uniformly on D to a function BN as r → 1−, with |BN | = 1 on∂D.

Hence,

‖FN‖pHp = limr→1−

Mp(FN , r)p

= limr→1−

12π

∫ 2π

0|(f/BN )(reiθ)|p dθ

≤ limr→1−

1infθ |BN (reiθ)|

12π

∫ 2π

0|(f)(reiθ)|p dθ

= ‖f‖pHp .

Moreover,

Mp(F, r)p = limN→+∞

Mp(FN , r)p

≤ limN→+∞

‖FN‖pHp = ‖f‖pHp .

Therefore, ‖F‖Hp ≤ ‖f‖Hp , and the equality follows. 2

Corollary 1.24. Let ζ1, ζ2, . . . be points in D satisfying the condition∑+∞

j=1(1 − |ζj |) < ∞

and let B = zk∏+∞j=1 −

ζj|ζj |ϕζj be the corresponding Blaschke product. Then B ∈ H∞ and |B| = 1

a.e. on ∂D.

Proof. In the proof of Prop. 1.23 we have already noticed that B ∈ H∞ and that its boundaryvalue function B is such that |B| ≤ 1.

Since B ∈ H∞, its canonical factorization is B = 1 ·B. By Prop. 1.23 again it follows that

1 = ‖1‖2H2 = ‖B‖2H2 =1

2π

∫ 2π

0|B(eiθ)|2 dθ .

16 M. M. PELOSO

Together with the estimate |B| ≤ 1, it implies that |B| = 1 a.e. on ∂D. 2

We conclude this part with the extension of Thm. 1.16 to the range 0 < p ≤ 1.

Theorem 1.25. (Fatou) Let f ∈ Hp, 0 < p ≤ 1. Then

limΓα(eit)3z→eit

f(z) =: f(eit) ,

where Γα(eit) is as in Def. 1.16, exists a.e. The function f ∈ Lp(∂D) and

‖f‖Lp = ‖f‖Hp .

Proof. Notice that, for this range of p’s, the existence of the boundary values f of f is part ofthe statement, as well as the equality of the norms of f and f .

By Prop. 1.23, f admits the canonical factorization f = FB, and the function F ∈ Hp and iszero-free. Therefore, F p/2 is well defined, holomorphic, and in H2. By Thm. 1.16, F p/2 admitsboundary values F p/2 a.e. Therefore, also F admits non-tangential boundary values a.e.

By Cor. 1.24 B admits boundary values B and |B| = 1 a.e. Hence f admits non-tangentialboundary values f =

(F p/2

)2/pB. Finally,

‖f‖pLp =1

2π

∫ 2π

0

∣∣(F p/2)2/pB(eiθ)|p dθ

=1

2π

∫ 2π

0

∣∣F p/2(eiθ)∣∣2 dθ = ‖F p/2‖2H2

= sup0<r<1

12π

∫ 2π

0|F (reiθ)|p dθ

= ‖F‖pHp = ‖f‖Hp ,

by Prop. 1.23 again. This proves the theorem. 2

1.6. Boundary behaviour of functions in Hardy spaces. The goal of this section is todescribe the boundary behaviour of functions in Hp and to characterize them in terms of theFourier coefficients of their boundary values.

In the previous Section 1.3 we studied the classes of functions that are Poisson integrals ofLp-functions on ∂D. In this case, we had to restrict to the case p > 1.

In the section we study the boundary behaviour of functions in Hp, and we will be ableto remove the restriction p > 1. In this analysis, we will consider the action of the Cauchykernel, that is more closely related to Hp-functions, since, in particular, it produces holomorphicfunctions, unlike the Poisson integral.

In the previous section we learned that for 0 < p ≤ ∞, if f ∈ Hp, then f converges pointwisenon-tangentially a.e. to a boundary function that, with a momentary abuse of notation, wedenote by f . We also know that fr → f in Lp(∂D) when 1 < p < ∞. We now show that thisholds true also in the case 0 < p ≤ 1.

Proposition 1.26. Let 0 < p ≤ 1, f ∈ Hp, and f be its boundary value function, as in Thm.1.25. Then fr → f in Lp as r → 1−.


Proof. We first prove the case p = 1.Let f ∈ H1 and let f = BF its canonical factorization. Set

g = F 1/2 , h = BF 1/2 . (1.20)

Then both g and h are in H2 and gr → g, hr → h in L2(∂D) as r → 1−.Then,

‖fr − f‖L1 = ‖grhr − gh‖L1 ≤ ‖(gr − g)hr‖L1 + ‖g(hr − h)‖L1

≤ ‖gr − g‖L2‖hr‖L2 + ‖g‖L2‖hr − h‖L2 → 0

as r → 1−. This proves the case p = 1.When 0 < p < 1 we proceed similarly, but we have to use an iteration process. Let f ∈ Hp

and define g and h as before. Then, g, h ∈ H2p. If the results holds true in H2p, that is, gr → g,hr → h in L2p(∂D) then the previous arguments applies (with a constant appearing in thetriangular inequality) and therefore the results holds true in Hp. Thus, after k ≥ log2 1/p stepswe can prove the results for Hp. 2

Recall that we denote by Cr(eiθ) the Cauchy kernel. We will denote by C(g) the Cauchyintegral of a given function g on ∂D. Recall that, for g ∈ C(∂D), C(g) is a holomorphic in Dand arguing as in (1.3) (using the uniform convergence of the power series expansion of Cr) wehave that

C(g)(reiη) = (Cr ∗ g)(eiη) =+∞∑n=0

g(n)rneinη .

Hence, we see that if g ∈ C(∂D) and g(n) = 0 when n < 0, then C(g) = P(g).More generally we have

Lemma 1.27. Let g ∈ Lp(∂D), 1 ≤ p ≤ ∞. Then P(g) = C(g) if and only if g(n) for all n < 0.

Proof. It suffices to assume that g ∈ L1(∂D).We use the uniform convergence of the power series expansion of Pr and Cr. We have

P(g)(reiη) = (Pr ∗ g)(eiη) =1

2π

∫ 2π

0g(eiθ)

+∞∑n=−∞

rnei(η−θ) dθ

=+∞∑

n=−∞g(n)rneinη .

Analogously,

C(g)(reiη) = (Cr ∗ g)(eiη) =1

2π

∫ 2π

0g(eiθ)

+∞∑n=0

rnei(η−θ) dθ

=+∞∑n=0

g(n)rneinη .

It follows that P(g) = C(g) if and only if g(n) for all n < 0. 2

18 M. M. PELOSO

Theorem 1.28. Let f ∈ Hp, 1 ≤ p ≤ ∞ and let f be its boundary value function. Thenˆf(n) = 0 for n < 0 and

P(f) = C(f) = f .

On the other hand, if 1 ≤ p ≤ ∞ and g ∈ Lp(∂D) is such that g(n) = 0, then f := P(g) = C(g)is in Hp and f = g.

Proof. Let f(z) =∑+∞

n=0 anzn be the power series expansion of f in D, f ∈ Hp, 1 ≤ p ≤ ∞.

Notice that it suffices to consider the case p = 1. Then

fr(eiη) =+∞∑n=0

anrneinη

and for all 0 < r < 1, fr(n) = anrn, by the uniform convergence. Since by Prop. 1.26 fr → f

in L1(∂D), fr(n)→ f(n) as r → 1−, so that

f(n) =

an for n ≥ 00 for n < 0 .

This clearly implies that

f(reiη) =+∞∑n=0

( 12π

∫ 2π

0f(eiθ)e−inθ dθ

)rneinη

= (Pr ∗ f)(reiη) ,

that is, f = P(f). The fact that C(f) = P(f) follows from the previous Lemma 1.27.When 1 ≤ p < ∞, he second part of the statement follows immediately from Lemma 1.27,

Prop. 1.11 and the holomorphicity of the Cauchy integrals.If p = ∞, we obtain C(g) = P(g) as before, and clearly f = P(g) = C(g) ∈ H∞ and the

equality f = g follows from the case p <∞. 2

The next result is now obvious, and we state it for sake of completeness.

Corollary 1.29. For 1 ≤ p ≤ ∞ we denote by Hp(∂D) the subspace of Lp(∂D) of the functionsthat are boundary values of functions in Hp.

Then,Hp(∂D) =

g ∈ Lp(∂D) : g(n) = 0 for n < 0

.

We denote by P the set of the (complex) polynomials in z, and by A(D) the space ofholomorphic functions on D that are continuous up to the boundary, that is,

A(D) = H(D) ∩ C(D) ,

endowed with the sup-norm.

Corollary 1.30. For 1 ≤ p < ∞ the polynomials are dense in Hp, while the closure of P inthe sup-norm is A(D) and A(D) ( H∞. Furthermore, if f ∈ A(D), then fr → f uniformly on∂D.


Proof. We denote by σN (g) the Fejer trigonometric polynomial of an integrable function g onT = ∂D, that is,

σN (g)(eiη) =N∑

k=−N

(1− |k|

N + 1

)g(k)eikη .

It is well known (see [Ka] or your Real Analysis lecture notes) that σN (g) → g in Lp(∂D), asN → +∞, when 1 ≤ p <∞.

Now, let 1 ≤ p <∞, f ∈ Hp, f its boundary function. Then,

σN (f)(eiη) =N∑k=0

(1− |k|

N + 1

) ˆf(k)eikη ,

by Prop. 1.28. Clearly σN (f) is the restriction to the boundary of a polynomial pN and it holdsthat pN = C(σN (f)). Then, we obtain a sequence of polynomials in D such that

‖f − pN‖Hp = ‖f − pN‖Lp(∂D) = ‖f − σN (f)‖Lp(∂D) → 0

as N → +∞. Thus, P is dense in Hp, 1 ≤ p <∞.

Next, it is clear that A(D) ⊂ H∞ and that the closure of P in the sup-norm is contained inA(D) = H(D) ∩ C(D). The same proof as before shows that P is dense in A(D). For,

supz∈D|f(z)− pN (z)| = sup

eiη∈∂D|f(eiη)− pN (eiη)|

= supeiη∈∂D

|f(eiη)− σN (f)(eiη)|

< ε

choosing N large enough.Moreover, if f ∈ A(D), then f ∈ C(∂D), f = P(f) and therefore, fr = Pr ∗ f → f uniformly

on ∂D.It only remains to show that A(D) is strictly contained in H∞. It suffices to consider an

infinite Blaschke product B, as defined in Def. 1.22. By Cor. 1.24 we know that |B| = 1 a.e.on ∂D. Let ζj be the (infinite) sequence of the zeros of B in D, then they must accumulateat boundary points. Clearly B cannot be continuous at those points. 2

1.7. The Cauchy–Szego projection.

Definition 1.31. We define the Cauchy–Szego projection on L2(∂D) as the Hilbert spaceorthogonal projection as

S(g)(eiη) = S( +∞∑n=−∞

g(n)ein(·))

(eiη) =+∞∑n=0

g(n)einη .

It is clear that S is a projection (that is, S2 = S) and that it is self-adjoint (that is, S∗ = S);hence an orthogonal projection.

The next result follows from the previous lemma.

Proposition 1.32. For g ∈ L2(∂D) we have that

S(g) = C(g) .

20 M. M. PELOSO

In particular we have obtained a characterization of S as (boundary limit of an) integraloperator. Namely,

S(g)(eiη) = limr→1−

12π

∫ 2π

0g(ei(η−θ))

11− reiθ

dθ . (1.21)

With an abuse of notation, we will simply write

S(g)(eiη) =1

2π

∫ 2π

0g(ei(η−θ))

11− eiθ

dθ . (1.22)

Notice however that the integral is not absolutely convergent since g ∈ L2 and the function1/(1−eiθ) has a non-integrable singularity in θ = 0. The integral in (1.22) has to be interpretedas in (1.21).

We wish to extend the action of S to all the spaces Lp(∂D) and prove its Lp-boundedness,for the appropriate range of p’s.

The following result, of which we will not provide a complete proof, answers the question.

Theorem 1.33. The operator S, initially defined on the dense subset L2 ∩ Lp, extends to anoperator

S : Lp(∂D)→ Lp(∂D)that is of weak-type (1, 1) and is bounded for 1 < p <∞.

Sketch of the proof. Recall the definition S(g) = limr→1− Cr ∗ g and the relation between theCauchy and Poisson kernels (1.13). From the properties of the Poisson kernel we see that

S1(g) = limr→1−

ReCr ∗ g

extends to a bounded linear operator for 1 ≤ p <∞.Thus, in order to prove the theorem, it suffices to prove the statement for (twice) the imaginary

part of Cr, where

2ImCr(eiη) =1i

[ 11− reiη

− 11− re−iη

]=

2r sin η|1− reiη|2

=2r sin η

1− 2r cos η + r2(1.23)

= −i+∞∑

n=−∞sgn(n)r|n|einη := Qr(eiη) , (1.24)

(where the last equality follows from the power series expansions of the right hand side of thefirst line in display).

The kernel Qr is called the conjugate Poisson kernel, the reason being that for every 0 < r < 1,Qr is the harmonic function conjugate to the real harmonic function Pr on the unit disc.

The family of functions on T Qr does not form a summability kernel, and the propertiesof the mapping g 7→ limr→1− Qr ∗ g are substantially different from the ones we obtain when Qris replaced by the Poisson kernel.

We give a name to the operator g 7→ limr→1− Qr ∗ g, we write

Q(g) = limr→1−

Qr ∗ g , (1.25)


and call Q the Hilbert transform on T. (In fact, this operator is the closely related to theclassical Hilbert transform on the real line. We will return to the latter operator in Section 4.3).

The operator Q belongs to a fundamental class of operators, called singular integrals. Us-ing the fact that the (convolution) kernels Qr are odd and therefore they have built in somecancellation, it is indeed possible to show that Q is Lp-bounded for 1 < p <∞.

Notice that the pointwise limit of the kernels

limr→1−

Qr(eiη) =1 sin η

1− cos η= cot(η/2) := Q(eiη)

and that Q 6∈ L1(∂D). It turns out (and it is not hard to see) that,

Q(g)(eiη) = p.v.(Q ∗ g)(eiη) =: limε→0+

∫ε<|θ|≤π

g(ei(η−θ))Q(θ) dθ .

We will not prove that Q is weak-type (1, 1) and bounded on Lp, 1 < p < ∞, but refer to[Ka], Ch. 3, e.g., or also [Gr]. 2

It can be shown that the Cauchy–Szego projection is not bounded from L1(∂D) to itself.4

Thm. 1.33 also gives an answer to the problem of the harmonic conjugate. Suppose u is a realharmonic function on the unit disc. Then we know that u admits a harmonic conjugate v, thatis, a real harmonic function such that u + iv is holomorphic in D. The function v is uniquelydetermined modulo constants, so we require v to vanish at the origin z = 0, i.e. we set v(0) = 0.

In fact, it is not difficult to see that the boundedness of the Cauchy– Szego projection onLp(∂D) is equivalent to the boundedness of the harmonic conjugate operator T .

Corollary 1.34. Let u be a real harmonic function satisfying the Hp-growth condition (1.14),1 < p <∞, and let v be its harmonic conjugate. Then, also v satisfies the Hp-growth condition(1.14) and the mapping T : u 7→ v is bounded on Lp(∂D), 1 < p <∞.

Proof. By Prop. 1.9 u = Pr ∗ u, with u ∈ Lp(∂D). Now, vr = Qr ∗ u and v = Q(u). By theprevious theorem ‖v‖Lp = ‖Q(u)‖Lp ≤ C‖u‖Lp , and we are done. 2

There exists an anologous theory for the Hardy on the upper half-plane. The two theory havemany common aspects, but they also present significant differencies, since the upper half-plane isunbounded, while D is not. Moreover, the symmetry properties of the two domains are different(although analogous) and certain formulas appear simpler in one setting or another.

We will present some further aspects of this theory in the setting of the upper half-plane inSections 4.1 - 4.4.

4Exercise.

22 M. M. PELOSO

2. Bergman spaces on the unit disc

2.1. Function spaces with reproducing kernel. In this section we briefly describe the theoryof reproducing kernels.

LetH be a Hilbert space of functions defined on a set Ω and suppose that the point evaluationsare bounded (linear) functionals on H, that is, for each z ∈ Ω, there exists a constant Cz > 0such that for all f ∈ H we have

|f(z)| ≤ Cz‖f‖H . (2.1)This implies that the linear functional

Lz(f) = f(z) , f ∈ H ,is bounded. By the Riesz–Fisher theorem, there exists kz ∈ H such that, for all f ∈ H we have

〈f, kz〉H = f(z) .

We define a kernel function K : Ω× Ω→ C by setting

K(w, z) = kz(w) . (2.2)

Such kernel K is called the reproducing kernel for H.

Proposition 2.1. The kernel K satisfies the following properties:(i) for all f ∈ H and z ∈ Ω we have f(z) = 〈f,K(·, z)〉H;(ii) for all z, w ∈ Ω, K(w, z) = K(z, w).

Proof. Property (i) is obvious by construction. For (ii), using the reproducing property of K wehave

K(w, z) = kz(w) = 〈kz,K(·, w)〉H = 〈K(·, w), kz〉H= K(z, w)

and we are done. 2

The following result establishes conditions that characterize the reproducing kernel of H.

Lemma 2.2. Let H(z, w) be a function on Ω× Ω such that(i) H(·, w) ∈ H for all w ∈ Ω fixed;(ii) 〈f,H(·, z)〉H = f(z) for all f ∈ H and z ∈ Ω.

Then, H(z, w) coincides with the reproducing kernel K(z, w) of H.

Proof. This is simple: using the reproducing property of H(·, z) and K(·, z), for every w ∈ Ωfixed we have

H(z, w) = 〈H(·, w),K(·, z)〉H = 〈K(·, z), H(·, w)〉H = K(w, z)

= K(z, w) ,

as we wished to show. 2

Notice that (i) and (ii) together imply the hermitian-symmetric property H(w, z) = H(z, w)for all z, w ∈ Ω, as in the proof of (ii) in Prop. 2.1.

In this notes we will be concerned only with spaces H for which (2.1) holds and that consistof holomorphic functions.


It will also always be the case that, for each compact subset E of Ω there exists C = CE > 0such that for all f ∈ H we have

supz∈E|f(z)| ≤ C‖f‖H . (2.3)

If (2.3) holds, then clearly the convergence in H implies the uniform convergence on compactsubsets of Ω.

Proposition 2.3. Let H be a Hilbert space of holomorphic functions on Ω for which condition(2.3) holds. Let ϕj be an orthonormal basis for H. Then the series

+∞∑j=1

ϕj(z)ϕj(w)

converges uniformly on compact subsets of Ω× Ω to the reproducing kernel K(z, w) of H.

Proof. It suffices to consider the case of subsets of the form E×E, where E is a compact subsetof Ω. By the Riesz-Fisher theorem for `2, for z ∈ E we have( +∞∑

j=1

|ϕj(z)|2)1/2

=∥∥∥ϕj(z)∥∥∥

`2= sup‖aj‖`2=1

∣∣∣ +∞∑j=1

ajϕj(z)∣∣∣

= sup‖f‖H=1

|f(z)|

≤ sup‖f‖H=1

supz∈E|f(z)|

≤ CE .Therefore,

supz∈E

( +∞∑j=1

|ϕj(z)|2)1/2

≤ CE ,

so that the Cauchy–Schwarz inequality gives that+∞∑j=1

∣∣ϕj(z)ϕj(w)∣∣ ≤ C2

E ,

and the convergence is uniform on E × E.Next, for z ∈ Ω fixed, the previous argument also shows that ϕj(z) ∈ `2, so that

∑+∞j=1 ϕj(z)ϕj

converges in H to a function hz. By the theory of Fourier series in H we have that, for all f ∈ H

〈f, hz〉H =+∞∑j=1

〈f, ϕj〉Hϕj(z) = f(z) .

Therefore, the function H(w, z) = hz(w) =∑+∞

j=1 ϕj(z)ϕj(w) satisfies the conditions ofLemma 2.2 and hence coincides with the reproducing kernel K(w, z).

Notice that the argument above also shows that the series∑+∞

j=1 ϕj(z)ϕj converges in theH-norm, for all z ∈ Ω fixed. 2

An interesting consequence of this proposition is the following result that shows that thereproducing kernel K satisfies an extremal property.

24 M. M. PELOSO

Corollary 2.4. Let H be a space of holomorphic functions on Ω for which condition (2.3) holdsand let K(z, w) be its reproducing kernel. Then

K(z, z) = supf∈H,‖f‖H=1

|f(z)|2 .

Proof. Let ϕk be an orthonormal basis for H. We have

K(z, z) =∑k

|ϕk(z)|2

=(

supak∈`2, ‖ak‖`2=1

∣∣∣∑k

akϕk(z)∣∣∣)2

= supf∈H,‖f‖H=1

|f(z)|2 ,

as claimed. 2

Notice that the reproducing kernel K satisfies the condition K(z, z) ≥ 0 and in fact K(z, z) >0 unless f(z) = 0 for all f ∈ H.

We conclude this introduction on Hilbert spaces with reproducing kernels with a concreteexample, that we have in fact already encountered.

Proposition 2.5. The Hardy space H2(D) is a Hilbert space with reproducing kernel, with innerproduct 〈f, g〉H2 = 〈f , g〉L2(∂D). Its reproducing kernel is the Cauchy–Szego kernel

K(z, w) =1

1− wz.

Proof. We showed that ‖f‖H2 = ‖f‖L2(∂D). Therefore, H2 is a Hilbert space with inner product〈f, g〉H2 = 〈f , g〉L2(∂D), for f, g ∈ H2. (We also showed that we can describe the inner productin H2 by setting

〈f, g〉H2 = 〈an, bn〉`2 ,

where f(z) =∑+∞

n=0 anzn and g =

∑+∞n=0 bnz

n. However, we will not use this characterization atthis stage.)

We now show that the point evalutions are bounded functionals on H2. Let z ∈ D and let|z| < r < 1. By the Cauchy formula

f(z) =1

2πi

∫|ζ|=r

f(ζ)ζ − z

dζ =1

2π

∫ 2π

0

f(reiθ)reiθ − z

reiθ dθ , (2.4)

and by the Cauchy–Schwartz inequality we have

|f(z)| ≤( 1

2π

∫ 2π

0|f(reiθ)|2 dθ

)1/2( 12π

∫ 2π

0

r2

|reiθ − z|2dθ)1/2

≤ r

r − |z|‖f‖H2 .

This shows that H2 is a Hilbert space with reproducing kernel.


Consider now the kernel K(z, w) = 11−wz . It is immediate to see that K(·, w) ∈ H2 for every

w ∈ D fixed. Moreover,

〈f,K(·, z)〉H2 = 〈f , K(·, z)〉L2(∂D) =1

2π

∫ 2π

0

f(eiθ)1− e−iθz

dθ

= limr→r1−

12π

∫ 2π

0

f(reiθ)reiθ − z

reiθ dθ

= f(z) ,

by (2.4). The conclusion now follows from Lemma 2.2. 2

2.2. The Bergman spaces. Let Ω be a domain in C. We denote by dA the Lebesgue measureon C, which of course is identified with real euclidean space R2, and we write |E| to denotethe Lebesgue measure of a measurable set E. For 0 < p <∞ we denote by Lp(Ω) the space ofthe Lebesgue measurable functions that are p-integrable with respect to dA, and by L∞(Ω) thespace of measurable functions that are essentially bounded.

Definition 2.6. For 0 < p ≤ ∞ we define the Bergman space Ap as Ap(Ω) = Lp(Ω) ∩H(Ω).

Proposition 2.7. Let Ω ⊂ C be a domain and let 0 < p ≤ ∞. Then the space Ap(Ω) is a closedsubspace of Lp(Ω). When p = 2, A2 is a Hilbert space with reproducing kernel.

Proof. We show that the functions in the Bergman space Ap satisfy (2.3). Let E be a compactsubset of Ω and let δ be half the distance of E from ∂Ω, i.e. δ = 1

2dist(E, ∂Ω).For every z ∈ E, D(z, δ) ⊂ Ω. From the mean value property

f(z) =1

2π

∫ 2π

0f(z + reiθ) dθ ,

(valid for every r > 0 such that D(z, r) ⊂ Ω) multiplying by r both sides and integrating inr ∈ [0, δ] we obtain the identity5

f(z) =1

|D(z, δ)|

∫∫D(z,δ)

f(w) dA(w) .

Therefore, for every z ∈ E, using Holder’s inequality, for 1 ≤ p ≤ ∞,

|f(z)|p ≤ 1|D(z, δ)|

∫∫D(z,δ)

|f(w)|p dA(w) (2.5)

≤ 1|D(z, δ)|

‖f‖pLp =1πδ2‖f‖pLp .

We remark that the above inequality (2.5) holds also for 0 < p < 1, but this is a consequenceof the fact that |f |p is subharmonic when f is holomorphic, and to avoid introducing anothernotion, we will not prove this fact here; see [Kr], e.g.

Since E can be covered by a finite number of such discs

supz∈E|f(z)| ≤ CE‖f‖Lp . (2.6)

5 This identity is called the mean value property and it is valid for all holomorphic functions f and discs D(z, δ)contained in its region of holomorphicity.

26 M. M. PELOSO

This inequality show that if fn is a sequence of holomorphic functions converging in theLp-norm to a function f , then fn converges to f also uniformly on compact subsets. Thisproves that Ap is closed 1 < p <∞. The case p =∞ is obvious.

When p = 2 (2.6) is exactly (2.3) and this concludes the proof.Notice that the constant CE in inequality (2.6), depends on the distance of the set E from

the boundary of Ω, and, more precisely CE ∼ δ−2/p, as δ = 12dist(E, ∂Ω)→ 0. 2

Definition 2.8. Given a domain Ω ⊂ C, let K = KΩ be its Bergman kernel, and for z ∈ Ωwrite kz = K(·, z). We define the Bergman projection on Ω the mapping B defined on L2(Ω) as

(Bf)(z) = 〈f, kz〉 =∫

Ωf(w)kz(w) dA(w)

=∫

Ωf(w)K(z, w) dA(w) .

Proposition 2.9. The Bergman projection is the Hilbert space orthogonal projection of L2(Ω)onto its closed subspace A2(Ω).

Proof. First of all, notice that Bf is well defined for all f ∈ L2, since for all z ∈ Ω, kz ∈ A2 ⊂ L2,and the L2-pairing 〈f, kz〉 makes sense (and converges).

Next, we use the characterization of the Bergman kernel given by Prop. 2.3, that is, kz =∑j ϕj(z)ϕj , where ϕj is an orthonormal basis for A2, where the convergence is in the L2-norm,

for all z ∈ Ω fixed.Therefore,

(Bf)(z) = 〈f, kz〉 =⟨f,∑j

ϕj(z)ϕj⟩

=∑j

ϕj(z)〈f, ϕj〉 =∑j

ajϕj(z) ,

where aj = 〈f, ϕj〉 is the j-th Fourier coefficient of f , w.r.t. the orthonormal system ϕj,(which is not complete in L2). Clearly, aj ∈ `2 and

∑j |aj |2 ≤ ‖f‖2L2 , by Bessel’s inequality.

Hence, the series∑

j〈f, ϕj〉ϕj converges in L2 and therefore

‖Bf‖L2 ≤ ‖f‖L2 .

It is also clear that Bf ∈ A2 since A2 is closed, so that B : L2 → A2 and it is bounded. Itremains to show that B is the orthogonal projection, that is, that B2 = B and B∗ = B.

The former identity follows by the reproducing property of the Bergman kernel when actingon functions in A2, while the latter one follows from the hermitian symmetry property: forf, g ∈ L2

〈Bf, g〉 =⟨∑

j

〈f, ϕj〉ϕj , g⟩

=∑j

〈f, ϕj〉〈ϕj , g〉

=⟨f,∑j

〈g, ϕj〉ϕj⟩

= 〈f,Bg〉 .

This shows that B∗ = B and we are done. 2


Remark 2.10. We remark that, up to this point, we have not discussed whether, on a givendomain Ω ⊂ C, the Bergman space A2(Ω) is trivial, i.e. reduces to 0, is infinite dimensional,or even if it is not trivial, but finite dimensional. We mention that all these cases can occur. Itis easy to see that A2(C) = 0, and the finite dimensional examples are also fairly elementary;we refer the reader to [Wi].

However, it is important to point out that, an application of the Weierstrass approxima-tion theorem6 shows that, if Ω is a bounded domain, the space of holomorphic polynomials is(contained and) dense in A2(Ω).

We conclude this remark by observing that, as a consequence of the extremal property in Cor.2.4 we have that, it Ω ⊆ Ω′ are domains in C and K,K ′ denote their Bergman kernels resp.,then

K ′(z, z) ≤ K(z, z)

for all z ∈ Ω.

Before proceeding any further, we present an example of Bergman space and kernel.

Proposition 2.11. Let D be the unit disc. Then, the set of functions ϕk, where ϕk =√(k + 1)/πzk, k = 0, 1, 2, . . . , is a complete orthonormal system in A2(D) and the Bergman

kernel has the expression

K(z, w) =1π

1(1− zw)2

.

Proof. It is clear that ϕk is an orthonormal system:

〈ϕj , ϕk〉 =

√(j + 1)(k + 1)

π

∫ 1

0

∫ 2π

0ei(j−k)θ dθ rj+k+1 dr

= δjk2(k + 1)∫ 1

0r2k+1 dr

= δjk ,

where δjk denotes the Kronecker’s delta.By the uniqueness of the Taylor expansion, it is also clear that the holomorphic polinomials

(i.e. the polynomials in z) are dense in A2(D). For, let f ∈ A2(D) be orthogonal to all the zk’sand let f(z) =

∑n anz

n be its power series expansion in D. By integrating over the closed disc|z| ≤ r for some fixed r < 1 and interchanging the summation and integration orders, we obtainthat an = 0 for all n; hence f = 0.

Thus, ϕk is a complete orthonormal system for A2(D) and, by Prop. 2.3 the Bergmankernel is given by the sum

∑k ϕk(z)ϕk(w), that we easily compute:∑k

k + 1π

zkwk =1π

∑k

(k + 1)(zw)k

=1π

1(1− zw)2

,

6True? Recall? Reference?

28 M. M. PELOSO

as we wished to show.7 2

Notice the Bergman kernel for the unit disc K(z, w) satisfies the condition K(z, w) 6= 0 andK(z, z) > 0 for all z, w ∈ D.

2.3. Biholomorphic invariance. One of the main features of the Bergman kernel is its invari-ance under conformal mappings. This invariance is the content of the next proposition.

Proposition 2.12. Let Ω1,Ω2 ⊂ C be domains, Φ : Ω1 → Ω2 be a biholomorphic mapping, andlet KΩ1 ,B1 and KΩ2 ,B2 be the Bergman kernels and projections of Ω1 and Ω2, resp. Then wehave

KΩ1(z, w) = Φ′(z)KΩ2

(Φ(z),Φ(w)

)Φ′(w) , (2.7)

and, for f ∈ A2(Ω2) we

B1

(Φ′(f Φ)

)= Φ′

((B2f) Φ

). (2.8)

Proof. We begin by observing that if Φ : Ω1 → Ω2 is a biholomorphic mapping, then thedeterminant jacobian of Φ thought as a mapping between domains in R2 equals |Φ′(z)|2. For,if we write Φ = u+ iv, u, v real functions, then Φ is identified with the mapping

Ω 3 z = x+ iy ≡ (x, y) 7→(u(x, y), v(x, y)

).

Then

Jac Φ =(∂xu ∂yu∂xv ∂yv

),

so that, using the Cauchy-Riemann equations,

det(Jac Φ) = (∂xu)2 + (∂xv)2 = |∂xΦ|2 = |Φ′|2 .

This identity implies that the mapping f 7→ Φ′(f Φ) is an isometric isomorphism of A2(Ω2)onto A2(Ω1).

Notice also the analogous statements hold true with Φ−1 in place of Φ and with the roles ofΩ1 and Ω2 switched.

For z, w ∈ Ω1 we set

H(z, w) = Φ′(z)KΩ2

(Φ(z),Φ(w)

)Φ′(w) ,

and easily see that H satisfies the conditions in Lemma 2.2, thus implying that H coincides withthe Bergman kernel on Ω1. 8

7Exercise. Using the residue theorem, show that for every f ∈ A2(D) and z ∈ D we have the identity

f(z) =1

π

ZZD

f(w)

(1− zw)2dA(w) .

Once you have completed your argument, explain where you use the assumption f ∈ A2(D). Can this assumptionbe relaxed?

8Exercise.


Finally, (2.8) follows easily from (2.7). Let z ∈ Ω1

B1

(Φ′(f Φ)

)(z) =

∫Ω1

Φ′(ζ)(f Φ)(ζ)KΩ1(z, ζ) dA(ζ)

= Φ′(z)∫

Ω1

f(Φ(ζ)

)KΩ2

(Φ(z),Φ(ζ)

)|Φ′(ζ)|2 dA(ζ)

= Φ′(z)∫

Ω2

f(ω)KΩ2

(Φ(z), ω

)dA(ω)

= Φ′(z)(

(B2f) Φ)

(z) ,

as we wanted to prove. 2

2.4. Lp-boundedness of a family of integral operators. In the next section we will provethe boundedness of the Bergman projection B when acting on Lp, 1 < p <∞ as a consequenceof a more general result concerning the boundedness of a family of operators on weighted Lp-spaces. This greater generality does not introduce extra arguments in the proof and at the sametime provides a variety of possible applications.

We begin by recalling a general result concerning the boundedness of integral operators withpositive kernels.

Proposition 2.13. (Schur’s test) Let (X , dµX ), (Y, dµY) be measure spaces. Let T be theintegral operator given by

Tf(x) =∫YK(x, y)f(y) dµY(y) ,

where K is a measurable positive kernel on X × Y. Let 1 < p, p′ < ∞ be conjugate exponents.Suppose there exist positive functions ψ : Y → (0,+∞), ϕ : X → (0,+∞), such that

(i)∫YK(x, y)ψp

′(y) dµY(y) ≤ Cϕ(x)p

′;

(ii)∫XK(x, y)ϕp(x) dµX (x) ≤ Cψ(y)p.

Then T : Lp(Y)→ Lp(X ) is bounded.

The proof is a fairly simple application of Holder’s inequality. We leave the details as anexercise.

For ν > −1 we consider the weights (1− |w|2)ν and the kernels

Kν(z, w) =1π

1|1− zw|2+ν

.

We are interested in the boundedness of the integral operators

T : Lp((1− |w|2)νdA

)→ Lp

((1− |w|2)νdA

),

where

Tf(z) =∫Df(w)Kν(z, w) (1− |w|2)ν dA(w) .

30 M. M. PELOSO

We begin with the following integral estimate. Given two positive quantities A and B depend-ing on a variable x, we write A ≈ B as x→ x0 if there exists c > 0 such that 1

c ≤ A(x)/B(x) ≤ cas x→ x0.

Lemma 2.14. Let γ > −1, t ∈ R and for z ∈ D set

It(z) =∫D

(1− |w|2)γ

|1− zw|2+γ+tdA(w) .

Then,

It(z) ≈

1 if t < 0log[1/(1− |z|)

]if t = 0

(1− |z|2)−t if t > 0 ,

for |z| → 1−.

Notice that the function It is bounded on every compact subsets of D. Therefore, the be-haviour of It(z) is non-trivial only for z approaching the boundary.

The proof of the lemma is based on another integral estimate.

Lemma 2.15. Let τ ∈ R and for % ∈ (0, 1) set

Jτ (%) =∫ 2π

0

1|1− %eiθ|1+τ

dθ .

Then, as %→ 1−,

Jτ (%) ≈

1 if τ < 0log[1/(1− %)

]if τ = 0

(1− %)−τ if τ > 0 .

Proof. Notice that, for all θ ∈ [0, 2π),

(1− %) ≤ |1− %eiθ| ≤ 1 + % ,

so that the region of integration [0, 2π) is contained in the union of the disjoint sets Ej , j =0, . . . , N , where

Cj =θ : 2j(1− %) ≤ |1− %eiθ| < 2j+1(1− %)

and N = blog2

[(1 + %)/(1− %)

]c.9 We also observe that, if we set

Ej =θ : |1− %eiθ| < 2j(1− %)

,

then Cj = Ej+1 \ Ej , j = 0, . . . , N . Moreover,

|Ej | ≤ c2j(1− %)

for some constant c > 0 (and independent of j and %).10

9We denote by btc the integral part of the real number t.10For, |1− %eiθ|2 < 22(j+1)(1− %)2 if and only if 1− A ≤ cos θ, where A = (22(j+1) − 1)(1− %)2/2%. But this

is equivalent to |θ| ≤ cos−1(1−A) ≈ (1− %), as %→ 1−.


Then, if τ > 0

Jτ (%) ≤N∑j=0

∫Cj

1|1− %eiθ|1+τ

dθ

≤N∑j=0

1(2j(1− %)

)1+τ |Ej |

≤ Cτ (1− %)−τ .

When τ = 0 we need to refine the above estimate. We obtain

Jτ (%) ≤N∑j=0

1(2j(1− %)

) |Cj |≤

N∑j=0

1(2j(1− %)

) |Ej | ≤ cN≤ c log(1/(1− %)) .

When τ < 0 is clear that Jτ is bounded as %→ 1− since |1− eiθ|−1−τ is integrable on [0, 2π).Finally, the reverse inequalities are easy to prove and we leave the details to the reader.11 2

Proof of Lemma 2.14. Integrating in polar coordinates w = reiθ and using the invariance in theintegral over the circle we see that

It(z) =∫ 1

0

∫ 2π

0

1∣∣1− |z|reiθ∣∣2+γ+t dθ(1− r)γr dr .

If t > 0, then τ := 1 + γ + t is also > 0 and Lemma 2.15 gives that

It(z) ≈∫ 1

0

(1− r)γ(1− |z|r

)1+γ+t r dr =∫ 1

0

(1− r)γ((1− r) + (1− |z|)r

)1+γ+t r dr .

We split the latter integral into the regions G1 = r : (1 − r) ≤ (1 − |z|)r and G2 = r :(1− r) > (1− |z|)r, that is,

[0, 1/(2− |z|)

]and

(1/(2− |z|), 1

). Then,

It(z) ≤∫ 1/(2−|z|)

0

1(1− r)1+t

r dr +∫ 1

1/(2−|z|)

(1− r)γ((1− |z|)r

)1+γ+t r dr

≤∫ 1/(2−|z|)

0

1(1− r)1+t

dr +C

(1− |z|)1+γ+t

∫ 1

1/(2−|z|)(1− r)γ dr

≤ Cγ,t(1− |z|)t

.

This proves the estimate from above in the case t > 0. It is not hard to see that the estimatefrom below follows from the same argument.

The case t < 0 is easy, so is the case t = 0, whose details are left to the reader. 2

11Exercise.

32 M. M. PELOSO

We finally come to the main result of this section. For ν > −1 define the measure on the unitdisc dAν(w) = (1− |w|2)νdA(w).

Theorem 2.16. For α, ν > −1 consider the integral operator Tν defined as

Tνf(z) =∫∫

Df(w)

(1− |w|2)ν

|1− wz|2+νdA(w) .

Then, if α+ 1 < p(ν + 1), Tν : Lp(D, dAα)→ Lp(D, dAα) is bounded.

Proof. We are going to use Schur’s test Prop. 2.13 and the integral estimates in Lemma 2.14.Notice that in this case the pairs (Y, dµY) and (X , dµX ) are both equal to (D, dAα) and that

we write the operator Tν as

Tν(f)(z) =∫∫

Df(w)

(1− |w|2)ν−α

|1− wz|2+νdAα(w) ,

that is, as having integral kernel H(z, w) = (1−|w|2)ν−α

|1−wz|2+ν w.r.t. the measure dAα.Let a > 0 to be chosen later and define ψ(w) = (1 − |w|2)−a. When ν − ap′ > −1, we may

apply Lemma 2.14 with γ = ν − ap′ and t = ap′ and obtain

Tν(ψp′)(z) =

∫∫Dψp′(w)H(z, w) dAα(w)

=∫∫

D

(1− |w|2)ν−ap′

|1− wz|2+νdA(w)

≤ C(1− |z|2)−ap′

= Cψp′(z) .

Notice that this argument is possible if the parameter a satisfies

0 < a <ν + 1p′

. (2.9)

On the other hand, we apply Lemma 2.14 again with γ = α−ap > −1 and t = ν−(α−ap) > 0,and obtain

Tν(ψp)(w) =∫∫

Dψp(z)H(z, w) dAα(z)

= (1− |w|2)ν−α∫∫

D

(1− |z|2)α−ap

|1− wz|2+νdA(z)

≤ C(1− |w|2)−ap = Cψp(w) .

Again, this argument is possible if the parameter a satisfiesα− νp

< a <α− νp

. (2.10)

Therefore, we can apply Prop. 2.13 if we can find a satisfying both (2.9) and (2.10), that is,if the intervals (A,B) = (0, ν+1

p′ ) and (C,D) = (α−νp , α−νp ) have non-empty intersection. SinceC > 0, this happens if and only if C < B, that is,

α− νp

<ν + 1p′

.

Recalling that 1/p′ = 1− 1/p, the statement follows. 2


2.5. The Bergman kernel and projection on the unit disc. In this section we consider theBergman projection B on the unit disc D. Since C(D) ⊂ L∞(D) ⊂ Lp(D) for all 0 < p < ∞,we have that B is well defined on a dense subset of Lp if 0 < p ≤ 2, and on all Lp if 2 < p ≤ ∞.

Main goal of this section is proving the follwing result.

Theorem 2.17. Let 1 ≤ p ≤ ∞. Then, the Bergman projection B is bounded B : Lp(D) →Ap(D) if and only if 1 < p <∞.

Proof. The sufficiency follows easily from Thm. 2.16. Observe that the bounedness of theoperator T with kernel |K(z, w)| implies the boundedness of P . Thus, it suffices to apply Thm.2.16 with ν = α = 0.

To prove the necessity, observe that if B : L1(D, dA) → A1(D, dA) were bounded, B :L∞(D) → A∞(D) would also be bounded. Thus, it suffices to show that B is not boundedon L∞(D). Consider the family of functions fz(w) = (1 − |w|2)2+ν(1 − zw)−(2+ν). Then,‖fz‖L∞ = 1 for all z ∈ D. But, for all ν > −1,

‖Tνfz‖L∞ ≥ |Tνfz(z)| =∣∣ ∫∫

D

(1− |w|2)2+2ν

|1− wz|2(2+ν)dA(w)

∣∣≥ C log(1− |z|)−1 ,

that clearly is not bounded. Therefore, B cannot be bounded on L∞ and on L1 and we aredone. 2

We now discuss some consequences of the boundedness of the Bergman projection. The firstone is the question of the description of the dual space of Ap(D). It is clear that A2(D) is aHilbert space and can be identified with its dual space, that is,(

A2(D))′ = A2(D) ,

where the identification is w.r.t. to its own inner product.By Holder’s inequality we also immediately have taht any g ∈ Ap′(D) defines a linear func-

tional on Ap(D), again via the L2(D)-inner product

Lg(f) =∫∫

Df(z)g(z) dA(z) .

However, it is not apriori clear that every g ∈ Ap′(D), g 6= 0, gives rise to a non-trivial

functional on Ap(D).

Theorem 2.18. Let 1 < p < ∞, then the dual space of Ap(D) can be identified with Ap′(D),

where p′ is the conjugate exponent.

Proof. Let L be a continuous linear functional on Ap(D), that is, L ∈(Ap(D)

)′. By the Hahn–Banach theorem, L extends to a linear functional on Lp(D), so that there exists g ∈ Lp

′(D)

such that L(f) =∫∫

D f(z)g(z) dA(z) for all f ∈ Ap(D). Using the fact that B is bounded on Lp

and the density of L2 ∩ Lq in Lq for q = p, p′, we see that

L(f) = 〈f, g〉 = 〈B(f), g〉 = 〈f,B(g)〉 =: 〈f, h〉 ,

with h ∈ Ap′(D).Hence,(Ap(D)

)′ ⊆ Ap′(D).It only remains to see that a non-zero element of Ap

′(D) cannot be identically zero on Ap(D).

By symmetry we may assume that p < 2 < p′. Then Ap′ ⊂ A2 ⊂ Ap. If h ∈ Ap′, h 6= 0, and

34 M. M. PELOSO

〈f, h〉 = 0 for all f ∈ Ap, then h ∈ A2 and in particular 〈f, h〉 = 0 for all f ∈ A2. Hence h = 0,a contradiction. 2

We show that the problem of the conjugate function has a simpler solution w.r.t. the case ofthe Hardy spaces.

Given a real harmonic function u in D we denote by v = Tu the harmonic conjugate in Dsuch that v(0) = 0. Hence, f = u+ iv is holomorphic in D and Im f(0) = 0.

Theorem 2.19. For 1 < p <∞, the mapping u 7→ v is bounded in the Lp-norm, that is, thereexists a constant C > 0 such that

‖v‖Lp(D) ≤ C‖u‖Lp(D)

for all real harmonic functions u on D.

We remark that, using the weighted Bergman projection Bν with ν ≥ 1, it is possible to provethat the the harmonic conjugation mapping u 7→ v is also bounded in the L1(dA)-norm.12

Proof. By Thm. 2.17 we know that there exist c > 0 such that ‖B(g)‖Lp ≤ c‖g‖Lp , for allg ∈ Lp(D). Next, if f ∈ Ap, by the mean value property applied to the function f(w)(1− zw)−2

which is holomorphic for w ∈ D, we have

B(f)(z) =1π

∫∫D

f(w)(1− zw)2

dA(w)

=1π

(∫∫D

f(w)(1− zw)2

dA(w))

= f(0) .

Therefore, if f ∈ Ap, f = u+ iv and Im f(0) = 0 we have |v| ≤ |u+ iv| = |f + f(0)| so that

‖Im f‖Lp ≤ ‖f + f(0)‖Lp = ‖B(f + f)‖Lp = 2‖B(Re f)‖Lp≤ 2c‖Re f‖Lp .

This proves the statement under the initial a priori assumption that f ∈ Ap, that is, both u andv are in Lp.

Suppose now that u is real harmonic in Lp(D). For r < 1 consider ur. Then ur is harmonic inthe disc |z| < 1/r, so it is continuous in the closure of D; hence ur ∈ Lp(D) and ‖ur‖Lp → ‖u‖Lpas r → 1− monotonically from below. It is easy to check that for each r < 1, the functionharmonic conjugate of ur is exactly vr, where v is the harmonic conjugate of u. Therefore, vrare also harmonic in the disc |z| < 1/r and in Lp and they converge pointwise to v. Hence,‖vr‖Lp ≤ c‖ur‖Lp ≤ c‖u‖Lp . The conclusion now follows from the dominated convergencetheorem. 2

12Exercise.


3. The Paley–Weiner and Bernstein spaces

The growth of the modulus of an entire function can be described in terms of the decay of theFourier transform of its restriction to the real line. With this approach in mind, we introduce(or recall) the main properties of the Fourier transform. The reader who is already familiar withthis subject, can skip the next two sections.

3.1. The Fourier transform. In this section we briefly review a few basic facts about theFourier transform and the space of rapidly decreasing functions, the Schwartz space. For sake ofsimplicity, we will limit ourselves to the real line R, although the extension to several variables,that is, to Rn, is straightforward. There are many excellent references that serve as introductionto the Fourier transform– for simplicity we mention only [Ka] or [Gr] and the references therein.

For f ∈ L1(R) we define the Fourier transform of f as

f(ξ) =∫f(x)e−ixξ dx .

We will also denote it by Ff to emphasyse the mapping F : L1(R) → C(R) ∩ L∞(R). It isclear that f is well defined for f ∈ L1 and that ‖f‖L∞ ≤ ‖f‖L1 . Moreover, f is continuous, asan immediate application of the dominated convergence theorem.

We now see the first elementary properties of the Fourier transform.Given a function f on R we write

τyf(x) = f(x− y) .

Moreover, we write Dx to denote the differentiation operator d/dx.

Proposition 3.1. Let f, g ∈ L1. Then, the following hold.

(i) For any y, η ∈ R, F(τyf)(ξ) = e−iyξ f(ξ) and τηf(ξ) = F(eixηf)(ξ).(ii) F(f ∗ g) = f g.

(iii) If xjf ∈ L1 for 0 ≤ j ≤ k, then f ∈ Ck and Djξ f(ξ) = F

((−ix)jf

)(ξ).

(iv) Suppose f ∈ Ck, Djxf ∈ L1 for 0 ≤ j ≤ k, then F(Dj

xf)(ξ) = (iξ)j f(ξ).

Proof. (i) is elementary:

(τyf ) (ξ) =∫f(x− y)e−ixξ dx =

∫f(x)e−i(x+y)ξ dx

= e−iyξ f(ξ) ,

and analogously for the other part of the statement.(ii) is also elementary. By Fubini’s theorem,

(f ∗ g) (ξ) =∫∫

f(x− y)g(y) dye−ixξ dx

=∫∫

f(x− y)e−i(x−y)ξ dxg(y)e−iyξ dy

= f(ξ)g(ξ) .

(iii) is also elementary. It suffices to pass the differentiation under the integral sign.

36 M. M. PELOSO

In order to prove (iv), we first assume that f is such that Djxf ∈ C(0), where we denote by

C(0) the space of continuous functions that vanish at ∞ (that is lim|x|→+∞ g(x) = 0).13 Underthis assumption, assume that j = 1 first. Then, since f ′ vanishes at ∞, by integration by partswe have ∫

f ′(x)e−ixξ dx = −∫f(x)(−iξ)e−ixξ dx

= iξf(ξ) .

The case when j > 1 follows by induction. The general case now follows by approximatingf ∈ L1 with f0 ∈ C0 and using Lemma 3.2. 2

Lemma 3.2. The Fourier transform maps L1 continuously into C(0)(R):

F : L1(R)→ C(0)(R) .

Proof. We only need to show that for f ∈ L1, f vanishes at ∞ (that is lim|x|→+∞ f(x) = 0).For f ∈ L1 ∩ C1

0 we have that

|ξ||f(ξ)| ≤∣∣ξ ∫ f(x)e−ixξ dx

∣∣≤ C0 ,

i.e. |ξ|f is bounded, that is f ∈ C(0). Finally, the result follows by the density of L1 ∩C10 in L1.

2

We now compute a fundamental Fourier transform.

Lemma 3.3. Let a > 0 and f(x) = e−ax2. Then

f(ξ) =√π

ae−ξ

2/4a .

Proof. By the previous proposition, we can differentiate under the integral sign and get

(f)′(ξ) = F(−ixe−aπx2)

(ξ) =( i

2πa

)F((e−aπx

2)′)(ξ)

=( i

2πa

)(iξ)f(ξ)

= − ξ

2πaf(ξ) .

Therefore, the function g = eξ2/4πaf(ξ) satisfies the differential equation

d

dξg = 0 ,

i.e. it is a constant. In order to compute the constant, we select ξ = 0 and recall that∫e−t

2dt =√

π, so that

f(0) =∫e−ax

2dx =

√π

a.

13The one we are adopting is not the standard notation for the space of continuous functions that vanish at∞. Typically, this latter space is denoted by C0. We have decided to keep the subscript “0” to denote functionswith compact support.


Remark 3.4. We will consider two types of dilations, that in fact are dual to each other in theintegral pairing.

Given a function f on R and t > 0 we define

ft(x) = t−1f(x/t) and f t(x) = f(tx) . (3.1)

Notice that, if f ∈ L1 then∫ft(x) dx =

∫t−1f(x/t) dx =

∫f(y) dy. Moreover, we have that

ft(ξ) = f(tξ) = (f)t(ξ) and f t(ξ) = t−1f(ξ/t) = (f)t(ξ) , (3.2)

(We leave the details as an Exercise.)Finally, notice that, if the integral is well defined (for instance if f, g ∈ L2)∫

ft(x)g(x) dx =∫f(x)gt(x) dx ,

as a simple change of variables shows.

We now prove

Theorem 3.5. (Inversion Theorem) Let f ∈ L1 be such that f ∈ L1. Then

f(x) =1

2π

∫eiξxf(ξ) dξ .

Proof. Let t > 0 and setψ(ξ) = eixξ−tξ

2.

Then,

(Fψ)(y) = τx(Fe−tξ2)(y) =√π

tτx(e−y

2/4t)

= ϕ√t(x− y) ,

where ϕ(x) =√πe−x

2/4.Next notice that, if f, g ∈ L1, then∫

f(x)g(x) dx =∫∫

f(x)g(ξ)e−ixξ dxdξ =∫f(ξ)g(ξ) dξ . (3.3)

Therefore, ∫eixξ−tξ

2f(ξ) dξ =

∫ψ(ξ)f(ξ) dξ

=∫ϕ√t(x− y)f(y) dy = f ∗ ϕ√t(x) .

Since∫ϕ(x) dx =

√π∫e−x

2/4 dx = 2π, we have that f ∗ ϕ√t → 2πf in the L1 norm as t → 0.On the left hand side of the equalities above we can use the dominated convergence theorem toshow that

∫eixξ−tξ

2f(ξ) dξ →

∫eixξ f(ξ) dξ as t→ 0. 2

As an immediate consequence of the Inversion Theorem we have the following.

Corollary 3.6. If f ∈ L1 and f = 0, then f = 0.

We now extend the definition of the Fourier transform to functions in L2.

38 M. M. PELOSO

Theorem 3.7. (Plancherel Theorem) Let f ∈ L1 ∩ L2. Then f ∈ L2 and F|L1∩L2 extendsuniquely to a unitary isomorphism of L2 onto L2(dξ/2π); in other words, for all f ∈ L2(R),

‖f‖L2 =1

2π‖f‖L2 .

Proof. Consider the space X = f ∈ L1 : f ∈ L1. We use the facts that f ∈ L1 implies f ∈ L∞and that L1∩L∞ ⊂ L2 to see that C∞0 ⊂ X ⊂ L2. (We denote by C∞0 the space of C∞ functionshaving compact support.) Since C∞0 is dense in L2, also X is. Notice that, for g ∈ X , by theinversion theorem, setting h = g/2π

h(ξ) =1

2π

∫e−ixξ g(x) dx =

12π

∫eixξ g(x) dx = g(ξ) .

Therefore, ∫fg dx =

∫fh, dx =

∫fh dξ =

12π

∫f g dξ .

Thus, F restricted to X preserves the L2-scalar products, so that, by taking f = g we see that‖f‖2 = 1

2π‖f‖2. Since F(X ) = X and X is dense in L2, F extends by continuity to a uniqueunitary isomorphism of L2 onto L2(dξ/2π).

Finally, one has to show that such an extension coincides with F on L1 ∩ L2. Supposef ∈ L1 ∩ L2 and let ϕ(x) = e−π|x|

2. Consider f ∗ ϕt. Then f ∗ ϕt ∈ L1 ∩ L2, F(f ∗ ϕt) =

fF(e−πt|·|2) ∈ L1 (since f is bounded). Therefore, f ∗ϕt ∈ X and f ∗ϕt → f both in L1 and L2

norms, and F(f ∗ ϕt)→ f both uniformly and in L2 norm. 2

We concluded this section with a brief introduction to a class of functions that are smoothand rapidly decreasing. This class turns out to be of fundamental importance, it is called theSchwartz space and it is denoted by S(R) (or simply as S).

We define the seminorm %(j,k) as

%(j,k)(f) = supx∈R|xjDk

xf(x)|

and defineS = f ∈ C∞(R) : %(j,k)(f) <∞, for all integers j, k .

It is cleat that S is contained in all the Lp-spaces, and that S contains C∞0 .It is also clear that if ϕ ∈ S, p is a polinomial and j a positive integer, then Dj

x

(pϕ) ∈ S. We

will soon see that this correspondence is continuous on S in the topology given by the seminormsabove, see Prop. 3.9.

The family of seminorms %(j,k) defines a topology on S as follows. A ngbh basis at a pointf ∈ S is given by the sets U(j1,k1),...,(jN ,kN );ε

U(j1,k1),...,(jN ,kN );ε =g ∈ S : %(jh,kh)(g − f) < ε

,

for N a positive integer and ε > 0.A consequence of Prop. 3.1 and the Inversion Theorem is the following.

Proposition 3.8. The Fourier transform F : S → S is a bijection.


We remark that in fact F is in fact a continuous bijection.14 We will prove the continuity inProp. 3.9

We remark that, as a consequence of these facts, we obtain a simple proof that the convolutionof two Schwartz functions is still a Schwartz function. For, ϕ ∗ ψ = F−1(ϕψ), where ϕψ ∈ S.

Proposition 3.9. The following are continuous mappings from S into itself:(i) for any polinomial p and j ∈ N, T : S 3 ϕ 7→ Dj

x

(pϕ)∈ S;

(ii) the Fourier transform F : S → S.

Proof. Exercise. (Notice that, in order to prove the continuity of F , and therefore the one ofF−1, we need to show that for every semi-norm %(j,k), there exist integers j1, k1, . . . , jN , kN andC > 0 such that for every F ∈ S,

%(j,k)

(F(f)

)≤ C

∑j=1

%(jj ,kj)(f) .

But,

%(j,k)

(F(f)

)= sup

ξ∈R|ξjDk

ξ (f)|

and then we use Prop. 3.1.) 2

Proposition 3.10. The space S is a complete metric space (i.e. a Frechet space) in the topologydefined by the seminorms %(j,k). Moreover, C∞0 is dense in S.

Proof. We only need to prove that S is complete in the given topology.Let fn be a Cauchy sequence in S, then %(j,k)(fm − fn) → 0 as m,n → +∞ for all (j, k) .

In particular, Dkxfn converges uniformly to a function gk for all k. Notice that

g0(x+ t)− g0(x) = limk→+∞

fn(x+ t)− fn(x)

= limk→+∞

∫ t

0f ′n(x+ s) ds

=∫ t

0g1(x+ s) ds .

Therefore, Dxg0 = g1, and by induction on k we obtain that gk = Dkxg0. Finally,

|xj ||Dkxfn(x)−Dk

xg0(x)| = limm→+∞

|xj ||Dkxfn(x)−Dk

xfm(x)|

≤ limm→+∞

%(j,k)(fn − fm)

≤ ε ,

for j, k ≥ k0. Then %(j,k)(fn − g0)→ 0 as n→ +∞.

14We remark that, in ordet to prove the continuity of a linear mapping T : S → S we need to show that forevery semi-norm %(j,k), there exist pair of integers (j1, k1), . . . , (jN , kN ) and C > 0 such that for every ϕ ∈ S,

%(j,k)

`T (ϕ)

´≤ C

Xj=1

%(jj ,kj)(ϕ) .

40 M. M. PELOSO

In order to show that C∞0 is dense in S, observe that, if η ∈ C∞0 with η(0) = 1 and ϕ ∈ S,then, for every ε > 0, ηεϕ is in C∞0 , where ηε(x) = η(εx). Now it is easy to check that ηεϕ→ ϕin the S-topology, as ε→ 0 that is, %(j,k)(ηεϕ− ϕ)→ 0, as ε→ 0. 2

3.2. The Paley–Wiener theorems. We now study the connection between the Fourier trans-form and the growth of (the maximum modulus of) certain holomorphic functions. We identifythe real line R with the subset of the complex plane z : Im z = 0.

Our presentation of the Paley–Wiener theorem begins with the simple observation that if fis in L1(R) and has compact support, then its Fourier transform f can be extended to an entirefunction. For, by definition

f(ξ) =∫ b

af(x)e−iξx dx

and, if we set

G(ζ) =∫ b

af(x)e−iζx dx

we immediately see that: G is holomorphic in C, i.e. G is entire, since we can differentiate under the integral sign

and in particular obtain that

G′(ζ) =∫ b

a−ixf(x)e−iζx dx ;

G extends f , that is, G∣∣R = f ;

finally, we can also estimate |G| by

|G(ζ)| ≤ supx∈[a,b]

|e−iζx|∫ b

a|f(x)| dx = eM |η|‖f‖L1 ,

where M = max(|a|, |b|) and η = Im ζ; hence, G is bounded in every horizontal stripS = z = x+ iy : a ≤ y ≤ b.

In few words, we will see that a function on the real line whose Fourier transform has somesupport property is the restriction of a function holomorphic on a certain domain and satisfyingsome growth condition.

Since we are more interested in the Hilbert space theory, we will concentrate on the case offunctions that are L2 on the real line.

Theorem 3.11. (Paley–Wiener Thm. for the strip) Let f ∈ L2(R) and a > 0. Then thefollowing are equivalent:

(i) f is the restriction to the real line of a function F holomorphic in the strip Sa = z =x+ iy : |y| < a and

sup|y|<a

∫R|F (x+ iy)|2 dx <∞ ; (3.4)

(ii) ea|ξ|f ∈ L2(R).


Proof. (ii) ⇒ (i). Assume that (ii) holds and for z ∈ Sa define

F (z) =1

2π

∫Rf(ξ)eizξ dξ . (3.5)

Since |e−yξ f(ξ)| ≤ ea|ξ|f(ξ) ∈ L2(R), F is well defined and F|R = f by the inversion theorem.Moreover,

|Dz(f(ξ)eizξ)| ≤ C|f(ξ)|ea|ξ|

for z ∈ Sa and some constant C > 0. Then, it is easy to see that we pass the differentiationunder the integral sign to show that F ′(z) exists in Sa; hence, F is holomorphic in Sa.

Notice that, writing Fy = F (·+ iy), Fy is the inverse Fourier transform of e−y(·)f , so that byPlancherel’s theorem we see that∫

R|F (x+ iy)|2 dx =

12π

∫R|f(ξ)|2e−2ξy dξ

≤ 12π

∥∥f ea|·|∥∥2

L2(R).

(i)⇒ (ii). Let Fy = F (·+ iy). We wish to show that Fy = e−y(·)f , where f = F|R ≡ F0, sinceby Plancherel’s theorem we would then have

‖ea|·|f‖2L2(R) = sup|y|<a

‖e−y(·)f‖2L2(R)

= sup|y|<a

2π∫R|F (x+ iy)|2 dx <∞ .

Suppose first that f = F|R has compact support. Then ea|·|f ∈ L2(R) and if we set

G(z) =1

2π

∫Rf(ξ)eizξ dξ

by the first part we obtain a function that is holomorphic in the strip Sa and that coincides withF on the real line. Hence, G = F on Sa and

Fy = Gy = e−y(·)f .

In the general case, let ϕ ∈ S be such that ϕ has compact support and it is identically 1 onthe interval [−1, 1]. (Notice that

∫ϕ = ϕ(0) = 1 so that ϕε is a summability kernel, and that

ϕ(εξ) = 1 for ξ ∈ [−1/ε, 1/ε].) Then, for ε > 0 consider

G(x+ iy) =(ϕε ∗ F (·+ iy)

)(x) .

Then G it is easy to see that holomorphic in the strip Sa and for |y| < a

sup|y|<a

‖G(·+ iy)‖L2(R) ≤ sup|y|<a

‖ϕε‖L1‖F (·+ iy)‖L2(R) <∞ ,

by the assumption on F .Now, G|R = ϕ(ε·)f has compact support and therefore the previous argument applies to give

thatGy = G|Re

−y(·) .

Therefore,ϕ(ε·)Fy = Gy = G|Re

−y(·) = ϕ(ε·)f e−y(·) ,

42 M. M. PELOSO

which implies thatFy = f e−y(·) ,

for |ξ| ≤ 1/ε. We get the desired conclusion by letting ε→ 0.15 2

We callU =

z = x+ iy : y > 0

(3.6)

the upper half-plane.The domain U is conformally equivalent to the unit disc D, unbounded and its boundary

is the real line. We will concentrate most of our upcoming efforts on the analysis of functionspaces on such domain. We recall also that, if F is a function defined on U, we denote by Fythe function R 3 x 7→ F (x+ iy).

Theorem 3.12. (Paley–Wiener Thm. for the upper half-plane) The following propertieshold true.

(I) Let F be a function holomorphic on the upper half-plane U such that

supy>0

∫R|F (x+ iy)|2 dx <∞ . (3.7)

Then, there exists f ∈ L2(R) such that Fy → f in L2, as y → 0+, i.e.

limy→0+

∫R|F (x+ iy)− f(x)|2 dx = 0 ; (3.8)

andf(ξ) = 0 for ξ < 0 . (3.9)

(II) Let f ∈ L2(R) be such that (3.9) holds. For z ∈ U we define F (z) by

F (z) =1

2π

∫Rf(ξ)eizξ dξ . (3.10)

Then F in holomorphic in U and satisfies (3.7) and (3.8).

The carefull reader has certainly realized the similarity of (3.7) with the H2-growth condition.In fact, we anticipate the definition of H2(U): we define

H2(U) =F ∈ H(U) : sup

y>0‖Fy‖L2(R) <∞

,

and set‖F‖H2(U) = sup

y>0‖Fy‖L2(R) .

We will come back to this theorem in Section 4.3 where we study the Hardy space H2(U).

Proof of Thm. 3.15. (II) Define F as in (3.10) for z ∈ U, i.e. for y > 0. If is clear that Fis holomorphic in U and that, writing Fy = F (· + iy), Fy is the inverse Fourier transform ofe−y(·)f . By Plancherel’s theorem we have

‖Fy‖2L2(R) =1

2π

∫R|f(ξ)|2e−2yξ dξ ≤ ‖f‖2L2 .

15Exercise: Find and prove the Paley–Wiener theorem in the hypothesis that f ∈ L2 has support in [a, b].


This shows that (3.7) holds. Moreover,∫R|Fy(x)− f(x)|2 dx =

12π

∫R

∣∣f(ξ)(e−yξ − 1

)∣∣2 dξ → 0

as y → 0+, by the dominated convergence theorem; that is, (3.8) holds too.

(I) Consider F1 = F (·+ i) and set f1 = F1|R. Then F1 satisfies the hypotheses in Thm. 3.11(i) with a = 1, since

sup|y|<1

∫R|F1(x+ iy)|2 dx = sup

|y|<1

∫R|F (x+ i(1 + y))|2 dx ≤ sup

y′>0

∫R|F (x+ iy′)|2 dx <∞ . (3.11)

Therefore, arguing as in the proof of the implication (i) ⇒ (ii) of Thm. 3.11, F(F1(· + iy)

)=

e−y(·)f1 for |y| < 1, that is,F(F (·+ i(1 + y))

)= e−y(·)f1 ,

and morevoer, eξ f1 ∈ L2. We define f ∈ L2 by setting

f = eξ f1 = limy→−1−

F(F (·+ i(1 + y))

)= lim

y→0+F(F (·+ iy)

), (3.12)

where the limits above are in the L2-norm. Hence, (3.8) holds true.On the other hand, from (3.11) it follows that

12π

∫R|f1(ξ)e−yξ|2 dξ = ‖F1(·+ iy)‖2L2(R) =

∫R|F (x+ i(1 + y))|2 dx <∞

uniformly for y > −1. Letting y → +∞, since the left hand side above remain bounded, weobtain that f1(ξ) = 0 for ξ < 0.16 By (3.12) it follows that e−ξ f(ξ) = 0 for ξ < 0, that is,f(ξ) = 0 for ξ < 0. 2

The following corollary is essentially a restatement of the previous theorem. We state for sakeof completeness and leave the details to the reader.

Corollary 3.13. Let f ∈ L2(R). Then the following are equivalent.(i) There exists a function F holomorphic on the upper half-plane U such that

supy>0

∫R|F (x+ iy)|2 dx <∞

andlimy→0+

∫R|F (x+ iy)− f(x)|2 dx = 0 ;

(ii)f(ξ) = 0 for ξ < 0 .

Before presenting the last version of the Paley–Wiener theorem (for the whole complex plane),we need to recall a result from classical complex analysis, a consequence of the maximum modulusprinciple.

For its proof see e.g. [CA-notes].17

16Make sure you agree with this assertion.17Not yet though.

44 M. M. PELOSO

Theorem 3.14. (Phragmen–Lindelof) Let F be holomorphic in U and continuous on itsclosure. Suppose that |F (x)| ≤M for x ∈ R and that for every ε > 0

|F (z)| ≤ e(a+ε)|z| ,

for z ∈ U and |z| sufficiently large, where a ≥ 0 is given. Then

|F (z)| ≤Meay

for all z ∈ U.

Theorem 3.15. (Paley–Wiener, third version) Let F be an entire function and a > 0.Then the following are equivalent.

(i) F|R ∈ L2(R) and

|F (z)| = o(ea|z|) as |z| → +∞ ; (3.13)

(ii) there exists f ∈ L2(R) with supp f ⊆ [−a, a] such that (3.10) holds, i.e.,

F (z) =1

2π

∫Rf(ξ)eizξ dξ .

Proof. (ii) ⇒ (i). Assume that (ii) holds, then

|F (z)| =∣∣∣ 12π

∫ a

−af(ξ)eizξ dξ

∣∣∣ ≤ 12π

∫ a

−a|f(ξ)|e−yξ dξ

≤( 1

2π

∫ a

−a|f(ξ)|2 dξ

)1/2( 12π

∫ a

−ae−2yξ dξ

)1/2

= ‖f‖L2(R)

( 12πy

sinh(2ya))1/2

.

Notice that, for y ∈ R, the function sinh(2ya)/y is C∞, even and positive, and that∣∣ sinh(2ya)/y

∣∣ ≤e2a|y|/|y|, as |y| → +∞. Therefore, as |y| → +∞,

|F (z)| ≤ ‖f‖L2(R)ea|y|√2π|y|

,

which is stronger than (3.13). Finally, by the Fourier inversion theorem, F|R = f ∈ L2(R). (i)then follows.

(i) ⇒ (ii). Assume that (i) holds, and suppose first that F|R is bounded. Condition (3.13)implies that the hypoheses of the Phragmen–Lindelof theorem are satisfied so that |F (z)| ≤Meay

in U. Set G(z) = eiazF (z). Then G is entire and |G(z)| ≤M in U.We now claim that G satisfies (3.7), the H2-growth condition on U.Assuming the claim for now we finish the proof. By Thm. 3.15 we obtain that supp g ⊂

[0,+∞), where g = G|R. Since f(ξ) = g(ξ + a), supp f ⊂ [−a,+∞), where f = F|R.We may repeat the previous argument for G(z) = e−iazF (z) in the lower half-space and obtain

that supp f ⊂ (−∞, a], so thatsupp f ⊂ [−a, a] . (3.14)

Now set

H(z) =1

2π

∫Rf(ξ)eizξ dξ .


Then H is an entire function (by the first part of the proof) and, by the inversion theorem,coincides with F on the real line, hence on all of C.

This prove the assertion under the assumption F|R bounded (and modulo the claim). If F isnot bounded, consider

Fϕ(z) =∫RF (z − u)ϕ(u) du ,

where ϕ be an arbitrary continuous function with compact support in R. The function Fϕsatisfies condition (3.13) and Fϕ|R is bounded, as it is easy to check. It follows then that

suppF(Fϕ|R

)⊂ [−a, a] .

But F(Fϕ|R

)= f ϕ, and by the arbitrariness of ϕ we obtain (3.14).

It only remains to prove the claim. Let ϕ be a continuous function with compact support inR, ‖ϕ‖L2 ≤ 1. We define

Gϕ(z) =∫RG(z − u)ϕ(u) du .

Using the fact that the ϕ has compact support, by differentiating under the integral sign it iseasy to see that Gϕ is holomorphic on the closure of U. Moreover, Gϕ is bounded, since

|Gϕ(z)| ≤M‖ϕ‖L1 ,

and it holds that|Gϕ|R(x)| ≤ ‖G|R‖L2‖ϕ‖L2 ≤ ‖G|R‖L2 .

Then we can apply the Phragmen–Lindelof theorem (with a = 0) to obtain that

|Gϕ(z)| ≤ ‖G|R‖L2

for all z ∈ U.This in particular implies that∣∣∣ ∫

RG(x+ iy)ϕ(−x) dx

∣∣∣ ≤ ‖G|R‖L2

for y > 0. Therefore,supy>0‖Gy‖L2 ≤ ‖G|R‖L2

This proves the claim and therefore the theorem. 2

3.3. The Paley–Wiener spaces. Our starting point is Thm. 3.15 where we dealt with entirefunctions with a prescribed order of growth.

Definition 3.16. Let F be an entire function. We say that F is an entire function of exponentialtype if there exists constants A,B > 0 such that, for all z ∈ C

|F (z)| ≤ AeB|z| .If F is entire function of exponential type, we call a the type of F where

a = lim supr→+∞

logM(r)r

,

and where M(r) = sup|z|=r |F (z)|.We denote by Ea the space of functions of exponential type at most a.

46 M. M. PELOSO

Remark 3.17. Consider the following conditions on an entire function F :

(i) for every ε > 0 there exists Cε > 0 such that

|F (z)| ≤ Cεe(a+ε)|z| ;

(ii) there exists C > 0 such that

|F (z)| ≤ Cea|z| ;

(iii) as |z| → +∞|F (z)| = o(ea|z|) .

Then clearly, (iii) ⇒ (ii) ⇒ (i) ⇒ “F is of exponential type at most a”.

We recall that an entire function F is said to be of finite order if there exists % > 0 such that

|F (z)| = O(e|z|%) as |z| → +∞ .

The infimum of the values % for which such relation holds is called the order of F . Thus, anentire function of exponential type is of finite order 1, although the converse is not true.

Definition 3.18. Let a > 0, 1 ≤ p ≤ ∞. We define the Paley–Wiener space PW pa as

PW pa =

f : f(x) =

∫ a

−ag(y)e−ixy dy , where g ∈ Lp(−a, a)

,

and we set‖f‖PW p

a= 2π‖g‖Lp .

In other words, PW pa is the image via the Fourier transform of the Lp-functions that are

supported in [−a, a].

Remark 3.19. In these notes we will essentially first concentrate on the case p = 2, in whichcase we simply write PWa to denote the Paley–Wiener space PW 2

a . Notice that, it follows fromthe Plancherel formula that

‖f‖PW 2a

= ‖g‖PW 2a

= 2π‖g‖L2 = ‖g‖L2 = ‖f‖L2 .

Hence, by polarization, for f, ϕ ∈ PWa,

〈f, ϕ〉PWa = 〈f, ϕ〉L2 . (3.15)

Forthermore, we remark that, from Thm. 3.15 we obtain the following characterization ofPWa:

“A function f is in PWa if and only if f ∈ L2(R) and f = F|R (that is, f is the restrictionto the real line of a function F ), where F is an entire function of exponential type such that|F (z)| = o(ea|z|) for |z| → +∞.”

It follows that PWa coincides with the subspace of Ea of the functions whose restriction tothe real line are in L2. For, if F ∈ Ea and F|R ∈ L2 then F satisfies (3.13) in (i) of Thm. 3.15 forany a′ > a. By Thm. 3.15 itself it then follows that suppF(F|R) ⊆ [−a′, a′], for every a′ > a.Hence suppF(F|R) ⊆ [−a, a] and by Thm. 3.15 again |F (z)| = o(ea|z|) for |z| → +∞. Theremaining implication is trivial.


Theorem 3.20. The Paley–Wiener space PWa is a Hilbert space with reproducing kernel w.r.t.the inner product given by (3.15). Its reproducing kernel is the function

K(x, y) =a

πsinc(a(x− y)) ,

where sinc t = sin t/t.18 Hence, for every f ∈ PWa and x ∈ R

f(x) =a

π

∫Rf(y) sinc(a(x− y)) dy . (3.16)

Proof. Let f ∈ PWa. It is clear that f ∈ L2(R), f = F|R for an entire function F as in Remark3.19, and that f = g, for some g ∈ L2(R) with supp g ⊆ [−a, a], so that

f(x) =1

2π

∫ a

−af(ξ)eixξ dξ . (3.17)

Notice that f ∈ L1 so that the integral converges absolutely. Then,

|f(x)| ≤ 12π‖f‖L1(−a,a) ≤

12π

(2a)1/2‖f‖L2 = (2a)1/2‖f‖L2 ,

and it is immediate to check that PWa is closed. This shows that PWa, as a subspace of L2(R),is a Hilbert space with reproducing kernel.

Next, notice that from (3.17), if we could switch the integration order (e.g. if f ∈ L1(R)), wewould obtain

f(x) =1

2π

∫ a

−a

∫Rf(y)e−iyξ dyeixξ dξ

=∫Rf(y)

12π

∫ a

−aei(x−y)ξ dξ dy . (3.18)

Therefore, the candidate for the reproducing kernel of PWa is the function

K(x, y) =1

2π

∫ a

−aei(x−y)ξ dξ .

However, we may switch the integration order if we approximate f with fϕε where ϕ ∈ C∞0 ,ϕ identically 1 in a ngbh of the origin (so that ϕε(x) = ϕ(εx)→ 1 as ε→ 0).

Since K(·, y) = F−1(χ(−a,a))(· − y), K(·, y) ∈ PWa for every y ∈ R. Since K satisfies theconditions of Lemma 2.2, K(x, y) is the reproducing kernel for PWa.

Notice also that the resulting identity (3.18) gives an absolutely convergent integral sinceK(x, ·) ∈ L2(R), for every x ∈ R.

Finally, since1

2π

∫ a

−aeitξ dξ =

1π

sin att

=a

πsinc at ,

we haveK(x, y) =

a

πsinc(a(x− y)) .

This proves the theorem. 2

18This function is called the (unnormalized) “cardinal sine”, or “sinc” function for short. It is clear from theformulas above that the case a = π is particularly simple. The function sinc(π·) is called the “normalized” sincfunction.

48 M. M. PELOSO

Theorem 3.21. The following properties hold.(1) The set ( aπ )1/2 sinc

(a(x− nπa )

)n∈Z is an orthonormal basis for PWa.

(2) For every f ∈ PWa we have the orthogonal expansion

f(x) =∑n∈Z

f(nπ

a

)sinc

(a(x− nπ

a)). (3.19)

The series converges absolutely, it holds also when x ∈ C and it converges uniformly on compactsubsets of C.

(3) For every f ∈ PWa we have the Parseval relation

‖f‖2L2 =π

a

∑n∈Z|f(nπ

a

)|2 . (3.20)

Proof. Recall that 1√2πF is a unitary isomorphism of L2 onto itself. Therefore, 1√

2πF maps an

orthonormal basis of L2(−a, a) onto an orthonormal basis of PWa.Let

ϕn(t) =1√2aein

πat n ∈ Z .

Then setting ψn = 1√2πϕn, ψnn∈Z becomes an orthonormal basis for PWa, where

ψn(ξ) =1

2√aπ

∫ a

−aeit(n

πa−ξ) dt

=1√aπ

1nπa − ξ

sin(a(n

π

a− ξ)

)=√a

πsinc

(a(ξ − nπ

a)).

This proves (1).

(2) By the theory of abstract Fourier series, from (1) it follows that, for every f ∈ PWa,

f =∑n∈Z〈f, ψn〉ψn ,

where the series converges in the Hilbert space norm.Then, applying (3.16) we have

f(x) =a

π

∑n∈Z

(∫Rf(y) sinc

(a(n

π

a− y)

)dy)

sinc(a(x− nπ

a))

=∑n∈Z

f(nπ

a

)sinc

(a(x− nπ

a)),


where the convergence is the L2-norm. In order to prove the absolute convergence we first prove(3). The convergence is the L2-norm allows to switch the summation and integration orders

‖f‖2L2 =∫R

∑m,n∈Z

f(mπ

a

)f(nπ

a

)sinc

(a(x−mπ

a))

sinc(a(x− nπ

a))dx

=∑m,n∈Z

f(mπ

a

)f(nπ

a

)πaδm,n

=π

a

∑n∈Z|f(nπ

a

)|2 .

This proves (3). Now, using the Cauchy–Schwarz inequality, it easy to see that the series in (2)converges absolutely since∑

n∈Z

∣∣f(nπa

)sinc

(a(x− nπ

a))∣∣ ≤√π

a‖f‖L2

(∑n∈Z

∣∣ sinc(a(x− nπ

a))∣∣2)1/2

= ‖f‖L2

∥∥ sinc(a(x− ·)

)∥∥L2

=√π

a‖f‖L2 .

Finally, the functions on the two sides of (3.19) are restrictions of functions in Ea and theycoincides on R, hence they equal when x is let vary on the whole complex plane. 2

3.4. The Bernstein spaces. We now define a scale of spaces closely related to the Paley–Wiener spaces.

Definition 3.22. Let a > 0, 1 ≤ p ≤ ∞. We define the Bernstein space Bpa as

Bpa =

f : f ∈ Lp(−a, a) , f = F|R for some F ∈ Ea

,

and we set‖f‖Bpa = ‖f‖Lp .

In other words, Bpa consists of the subspace of Lp(R) given of the restriction of entire function

of exponential type at most a.

By the Paley–Wiener theorem, we clearly have that B2a = PWa.

In order to prove they are complete we need the following important result proved by Planchereland Polya. We remark that it holds true also when 0 < p < 1, but we prove only in the case1 ≤ p <∞ to keep our presentation within reasonable space.

Theorem 3.23. (Plancherel–Polya) Let 1 < p <∞, a > 0. Then, for every f ∈ Bpa we have∫

R|f(x+ iy)|p dx ≤ eap|y|

∫R|f(x)|p dx . (3.21)

The proof is based on an application of the Phragmen–Lindelof theorem. However, it issomewhat technical and we refer to [Yo], p. 80.

Corollary 3.24. For 1 < p <∞ and a > 0, Bpa is a Banach space.

50 M. M. PELOSO

Proof. We only need to prove that Bpa is closed in Lp.

We first observe that, for every compact set E ⊂ C there exists a constant C = CE > 0 suchthat for every f ∈ Bp

a we havesupz∈E|f(z)| ≤ CE‖f‖Lp . (3.22)

For, we may assume that E = Q(z0, δ) is a closed square,

Q(z0, δ) =z = x+ iy : |x− x0| , |y − y0| ≤ δ

,

for some δ > 0. Notice that, for every z ∈ Q(z0, δ), D(z, δ) ⊂ Q(z, δ) ⊂ Q(z0, 2δ). By themean-value inequality (2.5), for z ∈ we obtain

|f(z)|p ≤ 1πδ2

∫∫Q(z,δ)

|f(w)|p dA(w)

≤ 1πδ2

∫∫Q(z0,2δ)

|f(w)|p dA(w)

≤ Cδ∫ x0+2δ

x0−2δ

∫R|f(u+ iv)|p du dv

≤ Cδ∫ x0+2δ

x0−2δeap|v|

∫R|f(u)|p du dv

≤ CE‖f‖pLp .

This clearly implies that the convergence in Bpa forces the uniform convergence on compact

subsets in C.Therefore, if fn ∈ Bp

a, n = 1, 2, . . . , and fn → f in Lp, then fn → f uniformly on compactsubsets of C. Hence, f is entire, of exponential type at most a, that is, f ∈ Bp

a. 2

We now consider the orthogonal projection Pa : L2(R)→ B2a which is given by the formula

Pa(f) =a

πf ∗ sinc(a·) , f ∈ L2 .

We do not have the necessary tools to prove the next result, and we refer to any classical textin Fourier Analysis, or the notes [HA-notes]. The argument essentially relies on the boundednessof the Hilbert transform on R.

Theorem 3.25. Let a > 0, 1 < p < ∞. Then, the operator P (f) = f ∗ sinc, initially definedon the dense subset L2 ∩ Lp, extends to bounded operator Pa : Lp → Bp

a.

Proof. It is obvious that Pa(Lp) ⊂ Ea. We notice that Pa = cTχ(−a,a) , where Tm denotes theFourier multiplier operator, with multiplier m = χ(−a,a). The fact that Tχ(−a,a) is a boundedoperator follows as a simple consequence of the Lp-boundedness of Hilbert transform. 2

Dipartimento di Matematica, Universita degli Studi di Milano, Via C. Saldini 50, 20133 Milano,Italy

E-mail address: [email protected]

URL: http://wwww.mat.unimi.it/~peloso/

classical spaces of holomorphic functions · spaces of holomorphic functions 1 in these notes we...

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