bergman spaces

7/31/2019 Bergman Spaces

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arXiv:0801

.1512v1

[math.CV

]9Jan2008

Bergman Spaces Seminar

Lecture Notes

Department of Mathematics

University of Crete

Heraklion 2005
http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1http://arxiv.org/abs/0801.1512v1


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Contents

1 Introduction to Bergman Spaces 5

1.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Growth ofAp Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 The Bergman Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Some Density Matters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 The Bergman Projection 21

2.1 The Bergman Projection on Ap() . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 A Bounded Projection ofL1() onto A1() . . . . . . . . . . . . . . . . . . . . . 252.3 A characterization ofAp in terms of derivatives . . . . . . . . . . . . . . . . . . . 32

3


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Chapter 1

Introduction to Bergman Spaces

1.1 Basic Definitions

We begin our discussion with the Bergman Spaces on the unit disc of the complex plane

= {z C

: |z| < 1}. (1.1)These are defined as

Ap() = {f H() : fpAp() =

|f(z)|pdm(z) < } (1.2)

where dm(z) = 1

dA(z) is the two dimensional Lebesgue measure normalized in and H()

is the space of analytic functions on the unit disc.

However, the story begins with Stephan Bergman and around 1970 (The Kernel function

and Conformal Mapping) theres already been some progress in the study of the spaces

Ap() = {f H() : fpAp(()) =

|f(z)|pdA(z) < }, 0 < p < (1.3)

where C is an open connected set, for the case p = 2. The interest was then focused onthe case = .

The study of the Bergman Spaces is inspired from that of the Hardy Spaces

Hp() =

f H() : fpHp() = sup

r[0,1)

1

2

20

|f(rei))|pd <

, 0 < p < (1.4)

and

H() = {f H() : fH() = supz

|f(z)| < } (1.5)

since Hp() Ap(). Observe however that H() = A() so hencefotrth we will restrictour interest to the case p 0.

(iv) Fix a z . Then, for every function f A2() satisfying f(z) = 1 we have the estimate

fA2() K(z, )1A2(). (1.12)


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To see this, use formula (1.9) to write

|f(z)| =

f()K(z, )dA()

|f()||K(z, )|dA()

f2A2()

|K(z, )|2dA() = f2A2()K(z, )2A2().

It is easy to see that estimate (1.12) is sharp. Indeed, fix a z

and define

f() =K(, z)

K(z, z)=

K(, z)

K(z, )2A2()

.

Then, f(z) = 1 and

f2A2() =1

K(z, )2A2()

|K(, z)|2dA()

=K(z, )2A2()K(z, )4

A2()

= K(z, )2A2().

(v) The Bergman Kernel K(z, ) is the only function in A2() which reproduces the value ofevery function f

A2() in the sense of (1.9). This is just a consequence of the definition of

the Bergman Kernel by means of the Riesz representation theorem.

We next turn to the question of calculating the Bergman Kernel in A2(). We consider an

orthonormal base {n}n=0 of A2(). To simplify notation we write , for the inner productin A2(). Then

n, m = nm =

1, if n = m

0, if n = m .The usual Hilbert space identities hold:

(i) Fourier series: Every f A2() has a series representation

f =

n=0

cnn, (1.13)

where cn = f, n and the series in (1.13) converges with respect to the A2() norm. Now,Lemma (1.1) implies that the partial sums

Nn=0 cnn converge to f uniformly on the compact

subsets of .

(ii) Parsevals Identity: For every in f A2() we have thatn=0

|cn|2 = f2A2(). (1.14)

(iii) A formula for the Bergman Kernel: For every z, , the Bergman Kernel can becalculated as

K(z, ) =

n=0

n(z)n() (1.15)

where the series in (1.15) converges with respect to the A2() norm and hence uniformly on

the compact sets of .


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To see this, write the series representation of the function K(, ) as in (1.13). The Fouriercoefficients of the function K(, ) are

cn = K(, ), n =

K(w, )n(w)dA(w) =

K(w, )n(w)dA(w)

=

K(, w)n(w)dA(w) = n().

Writing the series representation of the function K(, )

K(z, ) =

n=0

cnn(z) =

n=0

n(z)n()

we get (1.15).

The next step is to try to calculate the Bergman Kernel in the case of the unit disc = .

In order to do this we will employ formula (1.15). First of all we have to define an orthonormal

base in A2().

Proposition 1.4. The set{n(z) =

n + 1 zn}n=0 is an orthonormal base of A2().

Proof. The proof will be done in two steps:step 1 The set {n}n=0 is orthonormal.

Indeed, for every n, m N we have that

n, m =

n + 1

m + 1

znzmdm(z)

=

n + 1

m + 1

10

20

(rei)n(rei)m1

drdr

=1

n + 1

m + 1

10

rn+m+1dr

20

ei(nm)d

= 1 , if n = m0 , if n = m .

step 2 The set {n}n=0 is a base ofA2().It is equivalent to show that Parsevals formula

fA2() =n=0

|f, n|2 (1.16)

holds true for every f A2(). If we write the expansion of f in its Taylor series

f(z) =n=0

anzn

it is clear that we have to show that

fA2() =n=0

|an|2n + 1

. (1.17)


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Consider the partial sums of the Taylor series of f, SN(f)(z) =N

n=0 anzn and the disc of

radius , centered at the origin, where 0 < < 1,

= {z C : |z| < }.

Then,

|SN(f)(z)|

2

dm(z) =

Nn=0

anz

n2 Nn=0

an z

n2dm(z)

=N

n=0

an

Nm=0

an

0

rn+m+11

20

ei(nm)ddr

= 2

Nn=0

|an|20

r2n+1dr =

Nn=0

|an|2 2(n+1)

n + 1.

Since SN(f) converges to f uniformly in as n , it is clear that

limN

|SN(f)(z)|2dm(z) =

|f(z)|2dm(z).

However,

limN

Nn=0

|an|2 2(n+1)

n + 1=

n=0

|an|2 2(n+1)

n + 1

and so, combining the last two formulas we get

|f(z)|2dm(z) =n=0

|an|2 2(n+1)

n + 1.

Letting 1 yields (1.17).

Having defined an orthonormal base of A2(),it is now easy to calculate an exact formula

for the Bergman Kernel. Indeed, employing (1.15), we have that

K(z, ) =n=0

n(z)n() =n=0

(n + 1)(z)n =1

(1 z)2 .

We have actually established the following.

Theorem 1.3. (i) The Bergman Kernel in A2() is given by the formula

K(z, ) =1

(1 z )2 . (1.18)

(ii) For every function f A2() we have the representation

f(z) =

f()

(1 z)2dm() . (1.19)

We close this section with a theorem that relates the Bergman Kernels of two open connected

sets through a univalent mapping.


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Theorem 1.4. Let , D be open and connected subsets ofC and : D, (z) = w, bea univalent mapping of onto D. Suppose further that J(w, ) is the Bergman Kernel in D.Then, the Bergman Kernel in is given by

K(z, ) = J((z), ())

(z)(). (1.20)

Proof. Define the operator T : A2(D) A2() as

T(f)(z) = (f )(z)(z).

It easy to check that T is an isometry. Indeed,

f2A2(D) =D

|f(w)|2dA(w) =()

|f(w)|2dA(w) =

|f((z))|2|(z)|2dA(w)

=

|T(f)(z)|2dA(z) = T f2A2().

If g A2() then define the function f on D by the formula

f(w) = g(1(w))(1)

(w).

Then f A2(D) by a simple change of variable and T(f) = g which shows that T is onto.Now, consider any g A2() and f A2(D) such thath g = T(f). Then

f(w) =

D

J(w, )f()dA().

and replacing w by (z) in the above we get

f((z)) =

D

J((z), )f()dA()

for z

. Making the change of variable = () results to

f((z)) =

J((z), ())f(())| (z)|2dA()

and multiplying by

(z),

f((z))

(z) =

J((z), ())

(z)() f(())

()dA().

Remembering that g(z) = T(f)(z) we get

g(z) =

J((z), ())

(z)

() g()dA()

which is the desired result.

An application of Theorem 1.4 is contained in the Corollary below.


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Corollary 1.1. Let be an open, connected, proper subset ofC and K(z, ) be the Bergman

Kernel for . From Riemanns mapping theorem we know that for every there exists aconformal mapping of onto with () = 0 and

() > 0. Then, for every z ,

(z) =

K(, )K(z, ). (1.21)

Proof. Using theorem 1.4, the Bergman Kernel of is written as

K(z, ) = J((z), ())

(z)

()

where

J(w, ) =1

1

(1 w)2is the Bergman Kernel for the unit disc (the constant 1

is there because we have considered

the normalised Lebesgue measure on the unit disc). Combining tha last two relations we get

K(z, ) =1

1

(1 (z)())2

(z)() =1

(z)

()

since

() > 0 and () = 0. Therefore,

K(, ) =1

(

())2

and hence

K(z, ) =1

(z)(K(, ))12

which is just (1.21).

1.4 Some Density Matters

It is obvious from Proposition 1.4 that the polynomials are dense in A2(). In fact, some-

thing stronger is true.

Theorem 1.5. The set of polynomials is dense inA2(). Whatsmore, every function in A2()

can be approached in the A2() norm by the partial sums of its Taylor series, that is

limN

SN(f) fA2() = 0 (1.22)

where SN(f) =N

n=0 anzn is the partial sum of the Taylor series of f.

In order to prove this we shall need a simple lemma that relates the Taylor coefficients of

f with its Fourier coefficients with respect to the orthonormal base {n}.Lemma 1.2. Suppose that f

A2() has the Taylor expansion f(z) = n=0 anzn and that

n = n + 1 zn. Then, for every n N,f, n = an

n + 1(1.23)


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Proof. Consider the disc = {z C : |z| < } and fix some n N. Then, for N > n we havethat

SN(f)(z)n(z)dm(z) =

Nk=0

akzk(z)

n(z)dm(z)

=

n + 1

N

k=0

ak zkzndm(z)

=

n + 1 2 an

0

r2n+1dr =

n + 1an2n+1

n + 1.

Since SN(f) converges to f, uniformly on as N , we get

SN(f)(z)n(z)dm(z) =

n + 1an2n+1

n + 1.

Letting 1 we get (1.23).

Proof of Theorem 1.5. We write the Taylor series of f as

f(z) =

n=0 a

nz

n

=

n=0

an

n + 1

n + 1z

n

=

n=0

an

n + 1 n(z) .Then, for N N,

|f(z) SN(f)(z)|2dm(z) = f SN(f), f SN(f)

= f2A2() f, SN(f) f, SN(f) + SN(f)2A2()

= f2A2() N

n=0

anf, zn N

n=0

anf, zn +N

n=0

|an|2n + 1

= f2A2() N

n=0an

n + 1f, n

N

n=0an

n + 1f, n +

N

n=0|an|2n + 1

.

Employing Lemma 1.2, the right hand side is equal to

f2A2() N

n=0

|an|2n + 1

N

n=0

|an|2n + 1

+N

n=0

|an|2n + 1

= f2A2() N

n=0

|an|2n + 1

.

However,

f2A2() =n=0

|an|2n + 1

from Parsevals identity and so

f SN(f)A2() 0 as N

and this finishes the proof.

Remark. It is not true that the polynomials are dense in A2() for every simply connected

set C. Consider for example the set = [0, 1]. This is obviously simply connected


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and gives rise to the space A2(). On one can consider the function f(z) = z12 which is

clearly analytic square integrable. However, it is not possible to approach f by polynomials in

the A2() sense.

To see this, suppose that one could find a sequence of polynomials {Pn}nN, such that

limn

f PnA2() = 0 .

Then the sequence {Pn}nN is Cauchy in

A2

() and hence inA2

() since Pn

PmA

2

() =Pn PmA2(). This means that the sequence {Pn}nN is uniformly Cauchy on the compactsubsets of which in turn means that it converges uniformly on the compact subsets of

to some A2() function, say g. Since {Pn}nN also converges to f uniformly on the compactsubsets of , it turns out that f g in . But this means that f has an analytic extension tothe whole of , a contradiction.

We will be able to show next that the polynomials are dense in Ap() for general p,

0 < p < . However, we wont be able to approach a function in Ap() by the partial sums ofits Taylor series.

Theorem 1.6. The polynomials are dense in Ap(), 0 < p < .Proof. Let f

Ap(). We consider the function f = f(z), 0 < < 1. The function f is

analytic inside the disc = {z C : |z| < 1} . We deduce that the partial sums ofthe function f, SN(f), converge to f uniformly on the compact subsets of and hence

uniformly in . That is

SN(f) f uniformly in as N . (1.24)

It is immidiate from (1.24) that

limN

SN(f) fpAp() = 0. (1.25)

It suffices to show that

lim1

f fpAp() = 0. (1.26)

Indeed, assuming for a moment (1.25), we have that

f SN(f)pAp() 2pf fAp() + 2pf SN(f))Ap().

Let > 0. We chose close to 1 so that 2pf fAp() < 2 (this is possible because of(1.26)). Then, for N large enough, 2pf SN(f))Ap() < 2 because of (1.25) and so

f SN(f)pAp() <

and so SN(f) is the seeked for polynomial.

It remains to prove equation (1.26). We have that

f fpAp() =

|f(z) f(z)|pdm(z)

=

10

1

20

|f(rei) f(rei)|pdrdr


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For the inner integral we have the estimate

20

|f(rei) f(rei)|pd 20

(|f(rei)| + |f(rei)|)d

2p20

|f(rei)|p + |f(rei)|p

d

= 2p2

0

|f(rei)|pd +20

|f(rei)|pd

2p+120

|f(rei)|pd < .

The last inequality is due to the fact the function

F(r) =

20

|f(rei)|pd, 0 r < 1,

is an increasing function of r. Thus, since 0 < r < 1 (0 < < 1), we have that

F(r) < F(r).

This is a simple lemma which well prove after the proof of this theorem. Thus we get20

|f(rei) f(rei)|pd 2p+120

|f(rei)|pd (1.27)

We show next that f converges to f uniformly on the compact subsets of as 1. IfK is a compact subset of and z K then

|f(z) f (z)| = |n=0

an(1 n)zn| n=0

|an||1 n||z|n n=0

|an||1 n|Mn

for some M < 1. Now, Lebesgues dominated convergence theorem for series yields that f

converges to f uniformly on the compact subsets of as 1. Thus, for r < 1 fixed, we getlim1

|f(rei) f(rei)| = 0

and so

lim1

20

|f(rei) f(rei)|d = 0. (1.28)

Combining (1.27) with (1.28) and Lebesgues dominated convergence theorem we get (1.26).

We now give the proof of the Lemma we already used in the proof of Theorem 1.6.

Lemma 1.3. Suppose g is a nonnegative subharmonic function in and 0 r < 1. Then thefunction

F(r) = 1

2

0

g(rei)d.

is a decreasing function of r.


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Proof. Since g is a subharmonic function, for every open connected set B with B , thereexists a function U, harmonic in B, such that g(z) = U(z) in B and g(z) U(z) in B.

Suppose that 0 r1 < r2 < 1. Set B = {z C : |z| < r2}. Then

F(r1) =1

20

g(r1ei)d 1

20

U(r1ei)d = 2U(0)

by the mean value theorem. Again by the mean value theorem we have that

2U(0) =1

20

U(r2ei)d

and so

F(r1) 1

20

U(r2ei)d =

1

20

g(r2ei)d = F(r2)

which shows that F is increasing.

Remark. In theorem 1.6 we used the fact that the function

F(r) =

20

|f(rei)|pd, 0 r < 1,

is an increasing function of r. This is an immidiate consequence of lemma 1.3 since the function

|f(z)|p is a subharmonic function.


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Chapter 2

The Bergman Projection

2.1 The Bergman Projection on Ap()

Let us recall the formula of the main theorem of the previous chapter, that is Theorem 1.3.

The latter states that for every function f

A2() we have the representation

f(z) =

f()

(1 z)2 dm(). (2.1)

Although the discussion that led to Theorem 1.3 strongly depends on the Hilbert space structre

of A2(), one can try to see if the above formula has a meaning in a more general context.

This is the content of the following proposition.

Proposition 2.1. For f Lp(), 1 p < , define the function P(f) on as

P(f)(z) =

f()

(1 z)2 dm(), z . (2.2)

Then,

(i) The function P(f) is a well defined analytic function on .

(ii) If in addition f Ap(), 1 p < and z , we have that

P(f)(z) = f(z). (2.3)

Proof. Let us first note that the integral in (2.2) is well defined for every f Lp(), 1 p < .Indeed, fix a z and let 1 p < and q be the conjugate exponent of p, that is, 1

p+ 1

q= 1.

Then,

|f()||1 z|2 dm()

1

|1 z|2q dm() 1

q fLp().

However, for z fixed and , the function 1(1z)2

is a bounded function of and hence in

every Lq(), 1 < q

. Since the integral in (2.2) defines an analytic function on whenever

it exists, this proves (i).

For (ii) observe that the spaces Ap() are nested so Theorem 1.3 implies that formula (2.3)

holds true for every function f Ap(), 2 p < . Its easy to extend this formula to A1()

21


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and hence to every Ap(), 1 p < . Indeed, consider an f A1(). For (0, 1) definethe function f(z) = f(z). Now, f H() A2() and therefore we can write

f(z) =

f()

(1 z)2 dm() =1

f()

(1

z)2() d

where is the characteristic function of the disc of radius centered at the origin. For every

and (0, 1) we have that|f()|

|1

z|2()

1

|1 z|2 |f()| L1().

Empolying Lebesgues dominated convergence theorem yields formula (2.3) for f A1().

Definition 2.1. The linear operator P is called the Bergman Projection.

In the stronger L2() case, one can easily see that the Bergman Projection is the orthogonal

projection ofL2() onto the closed subspace A2().

Proposition 2.2. The Bergman Projection is the orthogonal projection of L2() onto A2().

Proof. Since A2() is a closed subspace of L2(), there exists an orthogonal projection, say

Po, of L2() onto A2(). Then, for every f L2(), Po(f) A2() and so

Po(f)(z) = Po(f), Kz = f, Po(Kz) = f, Kz = P(f).

Consequently, Po coincides with P which shows the proposition.

When p = 2 there is no orthogonal projection. However, since P(Lp()) Ap(), it isnatural to ask whether the Bergman Projection is a bounded operator from Lp() to Ap()

which would also show that P(Lp()) = Ap().

Let us first show that this is not the case when p = 1.

Proposition 2.3. The Bergman projection is not bounded from L1() to L1().

Proof. We will actually show that the adjoint operator of P, P, is not a bounded operator on

L(). We therefore need to find a formula for the adjoint operator. To that end, consider

f L1() and h L(). Writing down the definition of P we have that

f, P(h) = P(f), h =

P(f)(z)h(z)dm(z) =

f()

(1 z)2 dm()

h(z) dm(z)

=

f()

h(z)

(1 z)2 dm(z)

dm(),

where the last equality follows by applying Fubinis theorem. Thus, we have established theformula

P(h) =

h()

(1 z )2 dm(), h L().


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Suppose now, for the sake of contradiction, that P is a bounded operator on L1(), that

is, that P B(L1())1. This is equivalent to saying that P B(L()). For z anda (0, 1) define the functions ga as

ga() =(1 a)2|1 a|2 .

Clearly ga

L() and

ga

L() = 1. However,

P(ga)(a) =

ga()

(1 z )2 dm() =

1

|1 a|2 dm()

= 2

10

1

2

20

1

|1 arei|2 d rdr = 210

n=0

a2nr2nrdr

= 2

10

n=0

a2nr2n+1dr =n=0

1

n + 1a2n = log

1

1 a2 .

Now the hypothesis that P B(L) implies that

log1

1 |a|2 PgaL() gaL() = 1.

Since this must hold for every a (0, 1) we ger a contradiction as a 1.Having got rided of the bad case p = 1 we will now show that the Bergman projection is

indeed a bounded operator from Lp() onto Ap() for all 1 < p < .Theorem 2.1. The Bergman projection is a bounded linear operator from Lp() onto Ap()

for every 1 < p < .Proof. We have already showed that P is onto since P(f) = f for every f Ap() Lp().Now, fix 1 < p < and an f Lp(). It is clear that P(f) defines an anlytic function in theunit disc so it remains to show that P(f) is in Lp(). For z and q the conjugate exponentof p we have the estimate

|P(f)(z)|

|f()||1 z |2 dm() =

(1 ||2

) 1

pq

|1 z| 2q |f()|(1 ||

2

)

1pq

|1 z | 2pdm()

(1 ||2) 1p|1 z|2 dm()

1q

(1 ||2) 1q|1 z |2 |f()|

p dm()

1p

= [I1(z)]1q [I2(z)]

1p .

Taking pth powers and integrating we get

|P(f)(z)|p dm(z)

[I1(z)]pq I2(z) dm(z). (2.4)

At this point we need to estimate integrals of the form (1||2)

|1z|dm() for suitable

choices of, R. Instead of doing so in the special case were interested in, we will state andprove a general lemma which will come in handy in many cases through-out the text.

1We denote by B(L) the set of bounded linear operators from L to L.


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Lemma 2.1. Lets, t R with 1 < t < s. Then there exists a constantC = C(s, t), dependingonly on s, t, such that

(1 ||2)t2|1 z|s dm() C(1 |z|

2)ts, (2.5)

for every z .Postponing the proof of this lemma for a while, lets see how we can apply it to complete

the proof of the theorem. For I1 take t = 2

1

p

and s = 2 in the lemma above to get

I1(z) C(1 |z|2)1p . Plugging this estimate into formula (2.4) we get

|P(f)(z)|p dm(z) C

(1 |z|2) 1q

(1 ||2) 1q|1 z |2 |f()|

p dm()

dm(z)

= C

|f()|p(1 ||2) 1q

(1 |z|2) 1q|1 z |2 dm(z)

dm(),

where the last equation comes from an application of Fubinis theorem. Now, using Lemma 2.1

on more time with t = 2 1q

and s = 2 we get

|P(f)(z)|p dm(z) C

|f()|p(1 ||2) 1q (1 ||2) 1q dm() = CfpL()

which finishes the proof.

We now prove Lemma 2.1.

Proof of Lemma 2.1. Let s, t R with 1 < t < s and set I = (1||2)t2|1z|s dm(). Using thefact that 1

(1z)s2

=

n=0(n+ s2 )

n!( s2 )znn and writing I in polar coordinates we have

I =

10

1

20

1

|(1 zrei) s2 |2 d(1 r2)t2r dr

= 2n=0

(n + s2)

n!( s2)

2 10

r2n+1(1 r2)t2 dr|z|2n

=n=0

(n +

s

2)n!( s2)

2 10

rn(1 r)t2 dr|z|2n

=n=0

(n + s2)

n!( s2)

2(n + 1)(t 1)

(n + t)|z|2n

=(t 1)

( s2 )2

n=0

(n + s2)

n!

2n!

(n + t)|z|2n.

Using standard estimates for the Gamma function we see that(n+ s2 )

n!

2n!

(n+t) is of the order(n+st)

n! . But this means that

I C(s, t)n=0

(n + s

t)

n! |z|2n = C(1 |z|2)ts

which is just the statement of the lemma.


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2.2 A Bounded Projection ofL1() onto A1()

The Bergman projection seems to be the natural operator that maps Lp() onto Ap(),

that is, it defines for every Lp() function, its analytic counterpart on the unit disc and it

is the identity when restricted to Ap(). The image ofLp() under the Bergman projection

is exactly Ap() which reflects the fact that, well, its a projection! This nice theory however,

fails to provide with a bounded projection of L1() onto A1() since all these nice properties

hold for 1 < p < . We therefore set ourselves the task to find out if such a projection exists onL1() and, if it does, to describe it. As a sideresult, we will also define the weighted Bergman

spaces which arise naturally in the seek of such a projection.

First, recall the representation formula

f(z) =

f()

(1 z)2 dm(), z ,

for a nice function f, say f H(). Although this identiy also holds for A1 functions,it does not define a bounded operator. As a first step, we will try to construct a family of

representation formulas, at least for nice functions. This family will be more general, in the

sense that it will include the above as a special case. Then we will see under what hypothesis

this new family may define a bounded projection of L1() onto A1().A first remark is that it suffices to represent a specific value of a function f, say f(0). It is

then easy to use a disc automorphism that carries 0 to any z and automatically obtain arepresentation formula for the values of f at any z . In this spirit, we have the followinglemmas.

Lemma 2.2. Letf H() and > 1. Then

f(0) = ( + 1)

f()(1 ||2) dm(). (2.6)

Proof. First of all notice that the integral is well defined since f is bounded and > 1. Now,if f(z) = n=0 anzn is the Taylor series of f, the right hand side of (2.6) equals

n=0

ann(1 ||2) dm() =

10

n=0

1

20

eind rn

(1 r2)rdr

= 2a0

10

r(1 r2)dr = f(0)10

r(1 r)dr

= f(0)1

+ 1.

Multiplying by + 1 we get the lemma.

We now use a disc automorphism to get a representation formula for f at any z .

Lemma 2.3. Letf H

() and > 1. Thenf(z) = ( + 1)

f()

(1 z )+2 (1 ||2) dm(). (2.7)


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Proof. For z define the disc automorphism z : as

z(w) =z w

1 zw .

Clearly z(0) = z. Whatsmore, its easy to establish the properties

1z = z,

z(w) = 1 |z|2

(1 zw)2 ,

1 |z(w)|2 = |z(w)|(1 |w|2).

Using these properties and Lemma 2.2, we can write, for any z ,

f(z) = (f z)(0) = ( + 1)

(f z)()(1 ||2) dm()

= ( + 1)

z()

f()|z()|2(1 |z()|2) dm()

= ( + 1)

f()(1 |z|2)2|1 z|4

(1 |z|2)(1 ||2)|1 z|2 dm()

= ( + 1)(1 |z|2)+2

f() (1 ||2)|1 z |2+4 dm() .

Now, write the above identity for the function g, defined as g() = (1 z)+2f(), in the placeof f to get

(1 |z|2)+2f(z) = ( + 1)(1 |z|2)+2

f()(1 ||2)(1 z)+2 dm() .

This proves the lemma.

So formula (2.7) defines the family of representations we were seeking for, at least for nice

functions in H(). The next step is to try to extend this formula to some bigger space

that will hopefully incude the spaces Ap

(), for 1 p < . This gives rise to the weightedBergman spaces.

Definition 2.2. For > 1 we define the family of measures dm(z) as

dm(z) = ( + 1)(1 |z|2)dm(z). (2.8)

Let0 < p < . The weighted Bergman spaces Ap() are defined as

Ap() =

f H() : fp

Ap()

=

|f(z)|pdm(z) <

. (2.9)

The spaces Ap() consist of the functions that are in Lp() = L

p(, dm) and are analytic

on the unit disc . Thus, the spaces Ap

() are closed subspaces of Lp

() and one can write

down a series of results analogous to the ones we have seen for the usual Bergman spaces. Let us

first extend formula (2.7) to the case of Ap() functions. This is the analogous of Proposition

2.1 for the weighted case.


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Proposition 2.4. Let > 1 and f Lp(), 1 p < . Define the function P(f) on as

P(f)(z) =

f()

(1 z)+2 dm(), z . (2.10)

Then,

(i) The function P(f) is a well defined analytic function on .

(ii) If in addition f

Ap(), 1

p 0 and f L

1

(). This ismore promising! What we really need is a thorough description of when the operator P is a

bounded projection of Lp() onto Ap(). We already know this description when = 0 (this

is Theorem 2.1). The following Theorem gives an answer in the general case > 1.Theorem 2.2. Let > 1 and 1 p < . For f Lp() we define the function P(f) as

P(f) = ( + 1)

f()

(1 z)+2 (1 ||2) dm().

Then P is a bounded operator from Lp() onto Ap() if and only if p( + 1) > 1.

Remarks. (i) When = 0, this is just a repetition of the statement of Theorem 2.1, that is

that the Bergman Projection is bounded on Lp

() if and only if 1 < p < .(ii) The theorem tells us that if > 0 then P carries boundedly L

p() onto Ap() for any

1 p < . More specifically this means that there exists a bounded projection of L1() ontoA1() which is what we were seeking for. Remember that this is not the case for the Hardy

space H1().

(iii) There exists an even more general version of this theorem that says that if 1 < , < then the operator P : L

p() Ap() is a bounded projection of Lp() onto Ap() if and

only if + 1 < ( + 1)p. Since we wont need the result in this generality, we will only give the

proof for the case = 0.

Proof. Case p = 1. Let us first show that if P is bounded on L1() if and only if > 0. To

that end we will use the fact that the adjoint operator of P, P, is bounded on L

() if and

only if P is bounded on L1(). The operator P is defined by means of the inner product

f, g =

f(z)g(z)dm(z)


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28

where f L1() and g L(). That is, we define the operator P on L() so that forevery pair of functions f L1() and g L() we have that

P(f), g = f, P(g).

The left hand-side inner product can be calculated as follows.

P

(f), g

=

P

(f)()g()dm() =

f(z)

(1 z)2+dm

(z)g() dm()

=

f(z)

g()

(1 z)2+ dm() dm(z)

=

f(z)

( + 1)g()(1 |z|2)(1 z)2+ dm() dm(z).

We have thus found an explicit formula for the operator P,

P(g)(z) =

( + 1)g()(1 |z|2)(1 z)2+ dm(). (2.12)

It is now easy to see that P is bounded on L

() if and only if

supz

(1 |z|2)

dm()

|1 z|2+ < (2.13)

(just test P against the function g() =(1z)2+

|1z|2+).

Let us next show that equation (2.13) holds if and only if > 0. Indeed, suppose that

> 0. Then,

(1 |z|2)

dm()

|1 z |2+ = (1 |z|2)

10

1

20

1

|1 zrei|2+ d rdr

= (1 |

z

|2)

n=0

(n +2 + 1)

n!(2 + 1)

2 1

0

r2n+1dr

|z

|2n

=(1 |z|2)

2

n=0

(n + 2 + 1)n!(2 + 1)

2 1n + 1

|z|2n

C (1 |z|2)

2

(2 + 1)2

n=0

(n + 1)1|z|2n C,

where C, C are absolute constants. On the other hand, if = 0, we have that

1

|1 z |2 dm() =n=0

2

10

r2n+1dr|z|2n =n=0

1

n + 1|z|2n c log 1

1 |z| ,

for some absolute constant c. Finally, if1 < < 0 then

(1 |z|2)

dm()

|1 z|2+ (1 |z|2)

dm()

(1 + |z|)2+ 1

22+(1 |z|2).


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29

This shows that if P is bounded on L1() then we must have that > 0.

Now let > 0. We will show that P is bounded on L1(). We have that

|P(f)(z)|dm(z) =

( + 1)

f()(1 ||2)(1 z )2+ dm()

dm(z) ( + 1)

|f()||1 z |2+ (1 ||

2)dm()dm(z)

= ( + 1)

|f()|

1|1 z|2+ dm(z)(1 ||2)dm().

However,

1

|1 z|2+ dm(z) =

1

|(1 z ) 2+2 |2dm(z)

=1

(2 + 1)2

n=0

(n + 2 + 1)

n!

21

n + 1||2n

c ()(2 + 1)

2

n=0

(n + )

n!()||2n = ()

(2 + 1)

1

(1 ||2) .

Hence,

|P(f)(z)|dm(z) C fL1(),for some constant C depending only on . This finishes the case p = 1.

Case 1 < p < . Suppose now that P B(Lp(, dm)). This implies that P B(Lq(, dm))where q = p

p1 is the conjugate exponent of p. We will show that p( + 1) > 1. Suppose, for

the sake of contradiction, that p( + 1) 1 which is equivalent to 1q

.

Remember that the operator P is described by equation (2.12). Taking g() = 1 Lq(, dm) we have

P(g)(z) =

( + 1)(1 |z|2)(1 z)2+ dm()

= ( + 1)(1 |z|2

)

10

1

20

1

(1 zrei)2+ d rdr= ( + 1)(1 |z|2)

10

1

20

n=0

(n + + 2)

n!( + 2)rnzneindrdr

= ( + 1)(1 |z|2)10

n=0

1

20

eindrnznrdr

= ( + 1)(1 |z|2)10

rdr = ( + 1)(1 |z|2).

However, the function ( +1)(1 |z|2) is not in Lq(, dm) and hence we have a contradiction.Indeed, if < 1

qthen

( + 1)q

(1 |z|2)qdm(z) = ( + 1)q 10

xqdx

= ( + 1)q

1

q+ 1

1 lim

x0

1

x1q

= .


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On the other hand, if = 1q

then

( + 1)q

(1 |z|2)1dm(z) = ( + 1)q10

1

xdx = .

In order to complete the proof we shall need a boundedness criterion known as Schurs Test.

Theorem 2.3. (Schurs Test) Let(X, ) be a measure space and K : XX R+ a nonnegativemeasurable function. For 1 < p < and f Lp(X, d) we define

T(f)(x) =

X

K(x, y)f(y)d(y), x X.

Suppose that there exists a positive constant C > 0 and a positive measurable function h onX

such that

(a)For almost every x X,

X K(x, y)h(y)qd(y) Ch(x)q .

(b)For almost every y X,X

K(x, y)h(x)pd(x) Ch(y)p .

Then, T B(Lp(X, d)) and T C.

Remark. Suppose that (X, ) is a measure space and K : X X R+ a nonnegativemeasurable function. Suppose further that there exists some positive constant C and some

positive measurable function h, defined on X, such that

(i)X K(x, y)h(y)d(y) Ch(x) for -almost every x

X.

(ii)X

K(x, y)h(x)d(x) Ch(y) for -almost every y X.Then, the operator T(f)(x) =

X

K(x, y)f(y)d(y) B(L2(X, d)) with T C.

We postpone the proof of Schurs test until after the end of the proof of Theorem 2.2

Lets fix some p (1, ) and set q = pp1 . Suppose that p( + 1) > 1. We will now use

Schurs test in order to show that P is a bounded operator from Lp() to Ap().

We set d() = (1 ||2)dm(), h(z) = (1 |z|2) 1pq and K(z, ) = 1|1z|2+

. We have

that

K(z, )h()

q

d() =

1

|1 z|2+ h()q

(1 ||2

)

dm()

=

(1 ||2) 1p|1 z |2+ d().


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31

However, since 1(1z)

2+2

=

n=0(n+2+1)

n!(2+1)znn, we get that,

(1 ||2) 1p|1 z|+2 dm()

n=0

(n + 2 + 1)

n!(2 + 1)

2 10

rn(1 r) 1p dr|z|2n

=

n=0

(n + 2 + 1)

n!(2 + 1)

2 (n + 1)( + 1 1p

)

(n + + 2 1p

)|z|2n

=( + 1

q)

(2 + 1)2

n=0

(n + 2 + 1)

n!

2 1(n++1+ 1

q)

n!

|z|2n

C ( + 1

q)

(2 + 1)2

n=0

(n + 1 1q

)

n!|z|2n = C

( + 1q

)

(2 + 1)2

1

(1 |z|) 1p= Ch(z)

q,

for some constant C, depending only on .

Using a similar calculation, we show that

K(z, )h(z)pd(z) Ch()p.

Now, Schurs test tells us that P B(Lp(), Ap()) and that P C.

Proof of theorem 2.3. Let f Lp(X, d). Then,

|T(f)(x)| X

K(x, y)h(y)1

h(y)|f(y)|d(y).

Holders inequality now yields

|T(f)(x)|

X

K(x, y)h(y)qd(y)

1q

X

K(x, y)|f(y)|ph(y)p

d(y)

1p

C1q

h(x)

X K(x, y)h(y)

p

|f(y)|p

d(y)

1p

(2.14)

for -almost every x X. Integrating with respect to x and using Fubini we getX

|T(f)(x)|pd(x) CpqX

h(x)pX

K(x, y)h(y)p|f(y)|pd(y) d(x)

= Cpq

X

X

K(x, y)h(x)pd(x)h(y)p|f(y)|pd(y)

Cpq + 1X

|f(y)|pd(y),

where the last inequality follows by (2.14). Thus we have that

T(f)pLp(Xd) Cp

q+1fpLp(Xd).Taking p-th roots copletes the proof.


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2.3 A characterization ofAp in terms of derivatives

Suppose that f is an analytic function and fix some p [0, ). Let n be a positve integer.We want to find a condition for f(n) that assures that f Ap(). We actually get somethingbetter, that is, a characterization of the space Ap() in terms of derivatives. Our main result

for this section is the following.

Theorem 2.4. Let n be a positive integer greater than 1 and 1 p < . Suppose thatf H(). Then f Ap() if and only if (1 |z|2)f(n)(z) Lp().

The proof will be done in several steps. Lets begin with the necessity of the condition of

Theorem 2.4.

Lemma 2.4. Letn be a positive integer greater than 1 and1 p < . Suppose thatf Ap().Then (1 |z|2)nf(n) Lp().Proof. Case p = 1. Suppose that f A1(). For = 1, equation (2.10) of Proposition 2.4emplies

f(z) = P(f)(z) =

f()

(1

z)+2

dm() = 2

(1 ||2)(1

z)3

f()dm().

Differentiating n times we get

f(n)(z) = (n + 2)!

()n(1 ||2)(1 z)n+3 f()dm().

Thus,

(1 |z|2)n|f(n)(z)|dm(z) (n + 2)!

(1 |z|2)n

(1 ||2)|1 z |n+3 |f()|dm()dm(z)

(n + 2)!

(1 ||2)|f()|

(1 |z|2)n|1 z |n+3 dm(z)dm()

C

(1

|

|2)

|f()

|

1

(1 ||)dm()

= CfA1().Case 1 < p < . Suppose that f Ap(). By Proposition 2.1 we have that

f(z) =

f()

(1 z )2 dm().

Differentiating n times we get

(1 |z|2)nf(n)(z) = (n + 1)!(1 |z|2)n

()nf()

(1 z)n+2 dm() = n!(Pn Sn)(f)(z),

where Sn(f)(z) = znf(z) and Pn is described by equation (2.12). Consequently

|1 |z|2|pn|f(n)(z)|pdm(z) n!Pn Sn fAp() CnfAp().

Remember that Pn B(Lp()) by Theorem 2.2.


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To prove the sufficiency in Theorem 2.4 we need the following lemma.

Lemma 2.5. Letf H() and n be a positive integer greater than 1, such that(i) (1 |z|2)nf(n)(z) L1(), and(ii)f(0) = f(0) = f(0) = = f(2n1)(0) = 0.Then, for every z

,

f(z) =1

n!

(1 ||2)nf(n)()()n(1 z)2 dm().

Proof. First notice that condition (ii) implies

f(n)(z) =

m=2n

m(m 1) . . . (m n + 1)amzmn

= zn

m=2n

m(m 1) . . . (m n + 1)amzm2n.

As a result,f(n)(z)

zn H().For z , we write

(1 ||2)|f(n)()||1 z|2|n| dm()

1

(1 |z|)2

(1 ||2)nf(n)()

n

dm() 1

(1 |z|)2

12

(1 ||2)nf(n)()n

dm()+

1

2

(1 ||2)nf(n)()n

dm()

.

For the first integral notice that sup 12

f(n)()n < . Hence

12

(1 ||2)n|f(n)()|dm() < .

On the other hand 1

2

(1 ||2)nf(n)()n

dm() 2n 1

2

(1 ||2)n|f(n)()|dm() <

due to hypothesis (i). This shows that the integral

(1||2)nf(n)()

()n(1z)2dm() exists. What we

actually showed is that the functionF

(z

) =

(1|z|2)nf(n)(z)

zn

is inL1

().

But this means thatthe integral

g(z) =

F()

(1 z)2 dm() =1

n!

(1 ||2)nf(n)()(1 z)2n dm()


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34

defines an analytic function on . Hence

g(n)(z) = (n + 1)

(1 ||2)nf(n)()(1 z)n+2n

ndm()

= (n + 1)

(1 ||2)nf(n)()(1 z )n+2 dm() = Pn(f

(n))(z).

Since f(n)

A1n(), part (ii) of Proposition 2.4 gives that Pn(f

(n)

)(z) = f(n)

(z). This meansthat f(n)(z) = g(n)(z). However, for 0 k n 1, we have that

g(k)(z) =(k + 1)!

n!

f(n)()(1 ||2)k

(1 z )n+2n dm().

But this implies

g(k)(0) =(k + 1)!

n!

f(n)()(1 ||2)nk

dm()

=(k + 1)!

n!

10

m=2nm(m 1) . . . (m n + 1)rm+k2n 1

2

20

ei(mk)d rdr

= 0 = f(k)(0).

Hence f = g and the proof is complete.

We are now ready to prove Theorem 2.4

Proof of Theorem 2.4. Let 1 p < and n be a postive integer, greater than 1. Suppose thatf H() is such that (1 |z|2)nf(n)(z) Lp(). Suppose further that

f(0) = f(0) = f(0) = = f(2n)(0) = 0.

As weve seen in Lemma 2.5, if we set F(z) =(1|z|2)nf(n)(z)

znis in L1(), then

f(z) = 1n!

(1 ||2)nf(n)()(1 z)2n dm() = P(f)(z).

If 1 < p < , then P B(Lp(, dm)) and hence f Ap().Let p = 1. Then, for > 0, the operator

P(f)(z) = ( + 1)

f()

(1 z )+2 (1 ||2)dm()

is bounded on L1().


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