Thesis Submitted in Partial Fulﬁllment of the Requirements for the Degree of Master of Arts

FOURIER REPRESENTATIONS IN BERGMAN SPACES

Pranav Upadrashta

A thesis submitted in partial fulﬁllment of the requirements for the degree of Master of Arts

Department of Mathematics

Central Michigan University Mount Pleasant, Michigan April 2018 ACKNOWLEDGEMENTS

I am glad to have Prof. Debraj Chakrabarti supervising my Master’s thesis, who, along with being a brilliant teacher, is also a wonderful professional mentor. I thoroughly enjoyed discussing math (and a host of other topics) with him, and still do. I would like to thank Prof. David Barrett and Prof. Sivaram Narayan for their careful ex- amination of this thesis and for providing insightful remarks on the subject matter. I am fortunate to have them on my thesis defense committee. I’d also like to thank Prof. Meera Mainkar for having me at dinner on occasions where I lost track of time working on this thesis - she was a godsend!

ii ABSTRACT

FOURIER REPRESENTATIONS IN BERGMAN SPACES

by Pranav Upadrashta

It is a well known classical theorem of Paley and Wiener that every function in the Hardy space of the upper half plane is the holomorphic Fourier transform of a function square integrable on the positive real line. This correspondence establishes an isometric isomorphism between the Hardy space of the upper half plane and the Hilbert space of square integrable functions on the positive real line. Under this isometric isomorphism, the action of the group of translation symmetries of the upper half plane on the Hardy space corresponds to multiplication by a function in the Hilbert space of square integrable functions. In this thesis we obtain similar results for Bergman spaces of certain domains which we call polynomial half spaces. The upper half plane and Siegel upper half space in higher dimensions are familiar examples of polynomial half spaces. Like the upper half plane, these domains have three one-parameter subgroups of symmetries, namely rotations, translations and scalings. Corresponding to each of the one-parameter subgroup of symmetries of a polynomial half space above, we obtain an isometric isomorphism between the Bergman space of the polynomial half space and a Hilbert space of square integrable functions. This isometric isomorphism respects the action of the corresponding one-parameter subgroup of symmetries on the Bergman space of polynomial half space analogous to the classical Paley-Wiener theorem. We give complete and detailed proofs of the three Fourier representations. Some similar results can be found in literature for the translation group, though perhaps not for the scaling group. Finally we use the Fourier representations of Bergman space of polynomial half spaces to obtain integral representations of the Bergman kernel of these domains. For the Siegel upper half space, the classical formula for its Bergman kernel is recaptured using our method.

iii TABLE OF CONTENTS

LIST OF FIGURES ...... vi

LIST OF SYMBOLS ...... vii

CHAPTER

I. INTRODUCTION ...... 1 I.1. Motivation ...... 1 I.1.1. Motivating Example ...... 1 I.1.2. Relation with the Translation Group ...... 1 I.2. A Class of Domains with Large Automorphism Group ...... 2 I.2.1. Bergman Space of a Domain ...... 2 I.2.2. Polynomial Half Space ...... 3 I.3. Automorphisms of Polynomial Half Spaces...... 4 I.3.1. Rotation Subgroup ...... 4 I.3.2. Translation Subgroup ...... 5 I.3.3. Scaling Subgroup ...... 6 I.4. Results ...... 6 I.4.1. Rotation Subgroup ...... 7 I.4.2. Translation Subgroup ...... 9 I.4.3. Scaling Subgroup ...... 10 I.4.4. Bergman Kernel of Polynomial Half Spaces ...... 12 I.5. Outline of this Thesis ...... 14 I.6. Historical Note ...... 15

II. PRELIMINARIES ...... 16 II.1. Bergman Spaces ...... 16 II.2. Geometry of Polynomial Half Spaces ...... 18 II.2.1. Siegel Upper Half Space ...... 21 II.2.2. Bedford-Pinchuk Domains...... 21 II.3. Direct Integral Representations ...... 22 II.3.1. Direct Integral Representation of Hp ...... 23 II.3.2. Direct Integral Representation of Xp ...... 27

III. FOURIER REPRESENTATIONS ...... 35 III.1. Maximal Compact Subgroup ...... 35 2 III.1.1. A Fourier Representation of A (Ep) ...... 36 III.2. Translation Subgroup ...... 39 III.2.1. The case of the strip S(a,b) ...... 39 2 III.2.2. A Fourier Representation of A (Up) ...... 47 III.3. Scaling Subgroup ...... 51

iv III.3.1. The case of the sector V(a,b) ...... 51 2 III.3.2. Another Fourier Representation of A (Up) ...... 55

IV. BERGMAN KERNEL OF POLYNOMIAL HALF SPACES ...... 61 IV.1. Reproducing Kernel Hilbert Spaces ...... 61 IV.1.1. Reproducing Kernel Hilbert Spaces ...... 61 IV.1.2. Computing Reproducing Kernels ...... 62 IV.2. Asymptotic Behavior of the Bergman Kernel ...... 66 IV.3. A Formula for the Bergman Kernel of Ep ...... 70 IV.4. A Formula for the Bergman Kernel of Up ...... 72 IV.5. Another Formula for the Bergman Kernel of Up ...... 76 IV.6. An Application: the Siegel Upper Half Space ...... 78

REFERENCES ...... 82

v LIST OF FIGURES

FIGURE PAGE

1. Action of the Rotation Subgroup on D ...... 5

2. Action of the Translation Subgroup on U ...... 6

3. Action of the Scaling Subgroup on U ...... 7

4. The Rectangle SN ...... 45

5. Conformal Equivalence of a Sector with a Strip ...... 51

vi LIST OF SYMBOLS

SYMBOL PAGE

m ...... 3

p ...... 3

Up ...... 3

Ep ...... 3

Bp ...... 7

Wp(k) ...... 7

Yp ...... 8

Hp ...... 9

Xp ...... 10

λ(p(ζ),t) ...... 10

ρbγ ...... 11 1/µ ...... 11

M ...... 23

Sp(t) ...... 24

δζ ...... 28

Qp(t) ...... 30

S(a,b) ...... 39

ωa,b ...... 39

LTrans(p) ...... 47

V(a,b) ...... 51

Cp ...... 53

LScal(p) ...... 55

vii RH ...... 61

viii CHAPTER I INTRODUCTION

I.1. Motivation

I.1.1. Motivating Example

Let U = {z ∈ C|Imz > 0} denote the upper half plane in C. Recall that the Hardy space H2(U) of the upper half plane consists of functions which are holomorphic in U and uniformly square integrable on horizontal lines in U. More precisely, H2(U) is the Banach space of functions F ∈ O(U) such that Z 2 2 kFkH2(U) := sup |F(x + iy)| dx < ∞. y>0 R A classical theorem of Paley and Wiener (see [21, Theorem 19.2]) states that every function in H2(U) arises as the holomorphic Fourier transform of a square integrable function on the positive real line (0,∞). More precisely, the map T : L2(0,∞) → H2(U) given by

Z ∞ T f (z) = f (t)ei2πzt dt (I.1) 0 is an isometric isomorphism of Banach spaces. Consequently, H2(U) is a Hilbert space.

I.1.2. Relation with the Translation Group

For θ ∈ R and F ∈ H2(U), let γ(θ)F ∈ H2(U) be given by (γ(θ)F)(z) = F(z+θ). Since the translation z 7→ z + θ is a biholomorphic automorphism of U, θ 7→ γ(θ) is a unitary action of

R on H2(U). i.e., 2 γ : R → U (H (U) is a unitary representation of R in the Hilbert Space H2(U). Thus, we have in (I.1) ∞ Z (γ(θ)F)(z) = F(z + θ) = ei2πθt f (t)ei2πzt dt = T(ei2πθ(·) f )(z), 0

1 for all t ∈ (0,∞). That is, for each θ ∈ R, the following diagram commutes

L2(0,∞) T H2(U)

χθ γ(θ) L2(0,∞) T H2(U)

2 i2πθ(·) where χθ denotes the operator which multiplies a function in L (0,∞) by e . We think of T as a linear change of co-ordinates in the vector space H2(U) which diagonalizes all the linear operators {γ(θ)|θ ∈ R}. When such simultaneous diagonalization holds, we say that the isometric isomorphism T : L2(0,∞) → H2(U) in (I.1) a Fourier representation of H2(U) corresponding to the translation group, the subgroup of Aut(U) consisting of maps of the form z 7→ z + θ for real θ.

I.2. A Class of Domains with Large Automorphism Group The goal of this thesis is to study some analogs of (I.1) for Bergman spaces of certain

n domains in C admitting large groups of automorphisms. We ﬁrst recall the notion of Bergman space of a domain, and then introduce the class of domains we want to study.

I.2.1. Bergman Space of a Domain

n Let Ω be a domain in C , and let λ be a positive continuous function on Ω. The space L2(Ω,λ) consists of all measurable complex valued functions f on Ω such that

Z 2 2 k f kL2(Ω,λ) := | f (z)| λ(z) dV(z) < ∞, Ω where we identify functions that are almost everywhere equal. This is a Hilbert space of L2 functions on Ω with the natural inner product. The closed subspace A2(Ω,λ) of L2(Ω,λ) (see Proposition II.2) consisting of functions holomorphic in Ω is called the weighted Bergman space of Ω with weight λ. When λ ≡ 1, we denote A2(Ω,1) = A2(Ω). The Hilbert space A2(Ω) is called the Bergman space of Ω.

2 I.2.2. Polynomial Half Space

n We consider a class of domains in C which generalize the upper half plane U ⊂ C. Let n m = (m1,··· ,mn) be a given tuple of positive integers. Given a tuple α ∈ N of nonnegative integers deﬁne the weight of α with respect to m to be

n αi wtm(α) := ∑ . i=1 2mi

n A real polynomial p : C → R is then called a weighted homogeneous balanced polynomial (with n respect to the tuple m ∈ N ) if p is of the form

α β p(w1,··· ,wn) = ∑ Cα,β w w . (I.2) wtm(α)=wtm(β)=1/2

n n Let p : C → R be a weighted homogeneous balanced polynomial such that p ≥ 0 on C . n+1 The polynomial half space Up deﬁned by p is the unbounded domain in C

n Up = {(z,w) ∈ C × C | Im z > p(w)}. (I.3)

We show in Lemma II.6 that a polynomial half space Up is biholomorphically equivalent to the

n+1 bounded domain Ep ⊂ C , where

n 2 Ep = {(z,w) ∈ C × C | |z| + p(w) < 1}. (I.4)

We call the domain Ep a polynomial ellipsoid. In the trivial case n = 0, the unit disk D = {z ∈ C| |z| < 1} is the only example of a polynomial ellipsoid corresponding to the only weighted homogeneous polynomial p ≡ 0 on the 0-dimensional space C0 = {0}.The domain D is biholomorphically equivalent to the upper half plane U ⊂ C, which is a polynomial half space corresponding n to the polynomial p = 0. The unit ball in C is another familiar example of polynomial ellipsoid (see Section II.2), which is biholomorphically equivalent to a polynomial half space called the Siegel upper half space.

3 Henceforth, m = (m1,...,mn) will denote a tuple of positive integers and p will be a nonnegative weighted homogeneous balanced polynomial with respect to m, as in (I.2). It is very natural to consider polynomial half spaces in the problem of ﬁnding Fourier representations of Bergman spaces, thanks to the following result.

Theorem 1 ([2]). Let Ω be a bounded convex domain with smooth boundary, and of ﬁnite type in sense of D’Angelo. Then Aut(Ω) is noncompact if and only if Ω is biholomorphic to a polynomial ellipsoid.

The type of a bounded domain Ω in sense of D’Angelo (see [6, Pg 118]) at a point p ∈ ∂Ω is the maximum order of contact of ambient one-dimensional complex varieties with ∂Ω at p. The automorphism group of polynomial half spaces (and therefore polynomial ellipsoids) are noncompact, and contain one-parameter subgroups (see Section I.3 below) with respect to which we can obtain Fourier representations of Bergman spaces of these domains. This explains our interest in these domains.

I.3. Automorphisms of Polynomial Half Spaces.

In this section, we describe the subgroups of Aut(Up), the group of biholomorphic automorphisms of Up that are of interest to us. These subgroups described below do not necessarily generate the group Aut(Up).

I.3.1. Rotation Subgroup

In case of the upper half plane U ⊂ C, the action of the rotation subgroup of Aut(U) is easier to understand in the bounded model D, the unit disk in C. Similarly, in the case of polynomial half space Up, the action of the compact one-parameter subgroup of automorphisms is easier to understand in the bounded model Ep, which we now describe.

It is apparent that for θ ∈ R, the map σθ : Ep → Ep given by

iθ n σθ (z,w) = (e z,w), for z ∈ C,w ∈ C such that (z,w) ∈ Ep (I.5)

4 n 2 is an automorphism of Ep = {(z,w) ∈ C × C ||z| + p(w) < 1}. We call such an automorphism a rotation of Ep. The rotations of Ep form a compact one-parameter subgroup of the group Aut(Ep) of biholomorphic automorphisms of Ep, isomorphic to the circle group. Since there is a biholomorphic equivalence Λ : Up → Ep (see Lemma II.6), therefore, for each θ ∈ R, the map −1 Λ ◦ σθ ◦ Λ : Up → Up is an automorphism of the polynomial half space Up.

Figure 1. Action of the Rotation Subgroup on D

I.3.2. Translation Subgroup

We next describe a noncompact one-parameter subgroup of automorphism group Aut(Up) of the polynomial half space Up.

For θ ∈ R, the map τθ : Up → Up given by

n τθ (z,w) = (z + θ,w), for z ∈ C,w ∈ C such that (z,w) ∈ Up (I.6)

n is an automorphism of Up = {(z,w) ∈ C × C | Imz > p(w)}, since Im(z + θ) = Imz > p(w).

We call τθ a translation of Up. The translations form a one-parameter subgroup of Aut(Up) isomorphic to R.

5 Figure 2. Action of the Translation Subgroup on U

I.3.3. Scaling Subgroup We now describe another non-compact one-parameter subgroup of automorphism group

Aut(Up) of a polynomial half space Up.

Recall that m = (m1,...,mn) is the tuple of positive integers with respect to which p is a n n weighted homogeneous balanced polynomial. For θ > 0, let ρbθ : C → C be given by 1/2m1 1/2mn n ρbθ (w) = θ w1,...,θ wn , for w ∈ C . (I.7)

Then the map ρθ : Up → Up given by

1/2m1 1/2mn ρθ (z,w) = (θz,ρbθ (w)) = θz,θ w1,··· ,θ wn for (z,w) ∈ Up (I.8) can be shown to be an automorphism of Up (see Lemma II.5). We call such an automorphism a scaling of Up. The scalings of Up form a one-parameter subgroup of Aut(Up) isomorphic to the multiplicative group of positive real numbers.

Note that, in case of the upper half plane U ⊂ C, these consist of dilations z 7→ θz, where θ > 0.

I.4. Results In this thesis, we obtain Fourier representations analogous to (I.1) corresponding to the action of the rotation, translation and scaling groups on Bergman spaces of polynomial half spaces.

6 Figure 3. Action of the Scaling Subgroup on U

n ∗ Recall that for a domain Ω ⊂ C , if φ : Ω → Ω is an automorphism then the map φ from A2(Ω) to itself given by

φ ∗ f (z) = f (φ(z)) · detφ 0(z), for all z ∈ Ω, f ∈ A2(Ω) (I.9) is an isometric isomorphism of A2(Ω) with itself (see Lemma II.3). In other words, the map

φ 7→ φ ∗ is a unitary representation of Aut(Ω) in A2(Ω). The rotation group being compact, in this case (Theorem 2) we obtain a Fourier repre-

2 sentation of A (Ep) with respect to this group as an inﬁnite series instead of as an integral as in Theorems 3 and 4. We include this theorem here for completeness. In Theorems 3 and 4 we

2 2 n obtain Fourier representations of A (Up) as a Hilbert space of L functions on a subset of R×C , such that the functions in the Hilbert space are holomorphic in the complex parameter.

I.4.1. Rotation Subgroup

n Let Bp ⊂ C be the domain given by

n Bp = {w ∈ C | p(w) < 1}. (I.10)

For k ∈ N, let Wp(k) be the weighted Bergman space on Bp with respect to the weight w 7→ (1 − p(w))k+1, i.e.,

2 k+1 Wp(k) = A Bp,(1 − p)

7 When n = 0, the constant polynomial p = 0 is the only weighted homogeneous balanced polyno-

0 k+1 mial, so Bp = C = {0}. Then the weight w 7→ (1 − p(w)) is identically 1 for all k ∈ N, and 0 we endow C with the counting measure. Then we interpret Wp(k) as the space of all maps from 0 0 C to C that are square integrable on C , i.e., Wp(k) = C for all k ∈ N. ∞ Let Yp be the Hilbert space of sequences a = (ak)k=0 where for each k ∈ N, ak ∈ Wp(k) and ∞ 1 kak2 := ka k2 < ∞. (I.11) Yp π ∑ k Wp(k) k=0 k + 1 A convenient language to express constructions like (I.11) is that of direct integrals (see [9, Sec

7.4]). Let µ be the measure on N which assigns to each nonnegative integer k, the mass µ(k) = π k+1 . Notice that the function a ∈ Yp is such that for each k ∈ N, ak ∈ Wp(k). Then the space Yp is the direct integral Z ⊕ Yp = Wp(k) dµ(k). (I.12)

Notice that in this case the direct integral reduces to a weighted direct sum of Hilbert spaces.

2 Theorem 2. The map T : Yp → A (Ep) given by

∞ k Ta(z,w) = ∑ ak(w)z , for all (z,w) ∈ Ep (I.13) k=0 is an isometric isomorphism of Hilbert spaces.

Note that when n = 0, this reduces to a familiar result about Taylor expansions of a function in Bergman space A2(D) of the unit disc D (see Proposition III.1). 2 Let σθ be a rotation of Ep as in (I.5), when θ ∈ R. Then for each F ∈ A (Ep), the map ∗ 2 ∗ F 7→ σθ F determines an action of the rotation subgroup of Aut(Ep) on A (Ep), where σθ F is deﬁned as in (I.9). By Theorem 2, there is a sequence a = (a0,a1,...) ∈ Yp such that F = Ta, where T is as in (I.13). Thus, for (z,w) ∈ Ep, we have

∞ ! ∗ 0 iθ k iθ σθ F(z,w) = F(σθ (z,w))detσθ (z,w) = ∑ ak(w)(e z) · e k=0

8 iθ i2θ i3θ = T(a0e ,e a1,e a2,...). (I.14)

2 Therefore T is a Fourier representation of A (Ep) corresponding to the rotation subgroup of Aut(Ep). It follows from equation (I.14) that the isometric isomorphism T in Theorem 2 diag-

2 onalizes the action of rotation subgroup of Aut(Ep) on A (Ep).

I.4.2. Translation Subgroup

2 2 We now introduce Hp, which is a space of L functions, such that functions in A (Up) are holomorphic Fourier transforms of functions in Hp (see Theorem 3). n Let Hp be the Hilbert space of measurable functions g on (0,∞) × C such that

Z Z ∞ e−4π p(w)t kgk2 := |g(t,w)|2 dV(w) dt < , (I.15) Hp ∞ Cn 0 4πt and ∂g = 0 in the sense of distributions, 1 ≤ j ≤ n, ∂w j n where w1,...,wn are co-ordinates of C . In other words, functions in Hp are square integrable on n − p(w)t (0,∞)×C with respect to the weight (t,w) 7→ e 4π /4πt, and are holomorphic in the variable n w ∈ C .

2 Theorem 3. The map TS : Hp → A (Up) given by

Z ∞ i2πzt TS f (z,w) = f (t,w)e dt, for all (z,w) ∈ Up (I.16) 0 is an isometric isomorphism of Hilbert spaces.

2 For θ ∈ R, let τθ be a translation of Up as in (I.6). Then for each F ∈ A (Up) the map ∗ 2 ∗ F 7→ τθ F determines an action of the translation subgroup of Aut(Up) on A (Up), where τθ F is deﬁned analogous to (I.9). By Theorem 3, there is an f ∈ Hp such that F = TS f , where TS is as in

(I.16). Thus, for (z,w) ∈ Up we have

Z ∞ ∗ 0 i2π(z+θ)t i2πθ(·) τθ F(z,w) = F(τθ (z,w)) · detτθ (z,w) = f (t,w)e dt · 1 = TS(e f ). 0 9 2 Therefore (I.16) is a Fourier representation of A (Up) corresponding to the translation subgroup of

Aut(Up) as the isometric isomorphism TS in Theorem 3 diagonalizes the action of the translation 2 subgroup of Aut(Up) on A (Up).

I.4.3. Scaling Subgroup

To study Fourier representations with respect to the scaling group we introduce Xp, another

2 Hilbert space isometrically isomorphic to A (Up) (see Theorem 4). For s ∈ (−1,1) and t ∈ R let λ be given by   1 −4π sin−1(s)t −4π(π−sin−1(s))t  e − e , if t 6= 0 λ(s,t) = 4πt (I.17)  −1 π − 2sin s, if t = 0.

−1 Since limt→0 λ(s,t) = π −2sin (s), we see that for all s ∈ (−1,1), the function λ(s,·) is continuous on R. n Recall from (I.10) that Bp = {w ∈ C | p(w) < 1}. Let Xp be the Hilbert space consisting of measurable functions f on R × Bp such that

Z Z k f k2 := | f (t, )|2 (p( ),t) dV( ) dt < , Xp ζ λ ζ ζ ∞ R Bp and such that ∂ f = 0 in the sense of distributions, 1 ≤ j ≤ n, ∂w j where w1,...,wn are co-ordinates of Bp. That is, Xp is the space of measurable functions on

R × Bp which are holomorphic in the variable ζ ∈ Bp and square integrable with respect to the weight (t,ζ) 7→ λ(p(ζ),t), where for ζ ∈ Bp,  !  1 −1 −1  exp −4πt sin p(ζ) − exp − 4πt π − sin p(ζ) , if t 6= 0 4πt λ (p(ζ),t) =  −1 π − 2sin (p(ζ)), if t = 0. (I.18)

10 n Let γ ∈ C, and suppose that γ does not lie on the negative real axis. Let the map ρbγ : C → n C be given by

1/2m1 1/2mn n ρbγ (w1,...,wn) = γ w1,...,γ wn for all w ∈ C . (I.19) where the powers of γ are deﬁned using a branch of the logarithm which coincides with the natural logarithm on the positive real line. Note that this extends the map (I.7) on page 8 to the complex plane.

2 Theorem 4. Let 1/µ = 1/m1 + ··· + 1/mn. For (z,w) ∈ Up, the map TV : Xp → A (Up) given by

Z zi2πt T g(z,w) = gt,ρ (w) dt (I.20) V b1/z 1+1/2µ R z Z w w zi2πt = g t, 1 ,··· , n dt 1/2m 1/2m 1+1/2µ R z 1 z n z is an isometric isomorphism of Hilbert spaces.

2 Let ρθ ∈ Aut(Up) be as in (I.8), for θ > 0. Then for each F ∈ A (Up), the map F 7→ ∗ 2 ∗ ρθ F determines the action of the scaling subgroup of Aut(Up) on A (Up), where ρθ F is deﬁned analogous to (I.9). By Theorem 4, there is an f ∈ Xp such that F = TV f , where TV is as in (I.20).

Thus, for (z,w) ∈ Up we have

∗ 0 ρθ F(z,w) = F(ρθ (z,w)) · detρθ (z,w) Z (θz)i2πt = f t,ρ (ρ (w)) dt · θ 1+1/2µ b1/θz bθ 1+1/2µ R (θz) i2π(·) = TV θ f .

2 Thus, TV is a Fourier representation of A (Up) corresponding to the scaling subgroup of Aut(Up) as the isometric isomorphism TV in Theorem 4 diagonalizes the action of the scaling subgroup of

2 Aut(Up) on A (Up).

11 I.4.4. Bergman Kernel of Polynomial Half Spaces

Recall that associated to a weighted Bergman space A2(Ω,λ) is a function K : Ω×Ω → C such that for all f ∈ A2(Ω,λ), we have

Z f (z) = f (w)K(z,w)λ(w) dV(w) for all z ∈ Ω. Ω

The reproducing kernel K for the weighted Bergman space A2(Ω,λ) (see Proposition IV.1) is the unique function with this property that is holomorphic in the variable z and anti-holomorphic in the variable w. Also, recall that the reproducing kernel for A2(Ω) is called the Bergman kernel of Ω (see [15, Chapter 1]). The Bergman kernel of a domain Ω can sometimes be determined explicitly when Ω has a large group of symmetries, as in the case of the ball. For the polynomial half space Up, the asymptotic behavior of the Bergman kernel on the diagonal is determined by the symmetries of

Up, as we show in Theorem 5 below. This result is a special case of a more general theorem in [4].

Theorem 5. Let Up be a polynomial half space and for c > 0, let Γc be the subset of Up,

Γc = {(z,w) ∈ Up | |z| > c|w|}. (I.21)

Let K be the Bergman kernel for Up and for (z,w) ∈ Up, let

Kdiag(z,w) = K(z,w;z,w) be the Bergman kernel of Up on the diagonal of Up × Up. Then we have for each c > 0,

2+1/µ lim Kdiag(z,w)d(z,w) = Kdiag(i,0), (I.22) (z,w)→(0,0) (z,w)∈Γc

n where 1/µ = ∑ j=1 1/m j and d(z,w) is the distance of a point (z,w) ∈ Up to ∂Up.

Remark. The limit in the left hand side of equation (I.22) corresponds to the nontangential convergence of (z,w) ∈ Up to the point (0,0) ∈ ∂Up.

12 Recall from (I.12) that Yp admits the following direct integral representation

Z ⊕ Yp = Wp(k) dµ(k),

k+1 where for k ∈ N, Wp(k) is the weighted Bergman space on Bp with weight w 7→ (1 − p(w)) . 2 Using this together with the fact that A (Ep) is isometrically isomorphic to Yp allows us to express the Bergman kernel of Ep in terms of the reproducing kernels for Wp(k) below.

Theorem 6. For k ∈ N, let Yp(k;·,·) be the reproducing kernel for the weighted Bergman space n n Wp(k) on the domain Bp ⊂ C . Then for z,Z ∈ C and w,W ∈ C such that (z,w),(Z,W) ∈ Ep, the

Bergman kernel B of Ep is given by

∞ k + 1 k k B(z,w;Z,W) = ∑ Yp(k;w,W)z Z . k=0 π

n −4π p(w)t For t > 0, let Sp(t) be the weighted Bergman space on C with weight w 7→ e , i.e.,

2 n −4π pt Sp(t) = A C ,e . (I.23)

Then we have the following direct integral representation (see Proposition II.7, Section II.3) of

Hp Z ⊕ dt Hp = Sp(t) , (I.24) 4πt where the integral is taken with respect to the Haar measure dt/4πt on (0,∞). Then we may

2 represent the Bergman kernel for A (Up) in terms of reproducing kernel for Sp(t) as below.

Theorem 7. Let the Bergman kernel for Sp(t) be denoted by Hp(t;·,·). Then for (z,w),(Z,W) ∈

Up, the Bergman kernel K of Up is given by

Z ∞ i2π(z−Z)t K(z,w;Z,W) = 4π tHp(t;w,W)e dt (I.25) 0

13 For t ∈ R let Qp(t) be the weighted Bergman space on Bp with weight w 7→ λ(p(w),t), where λ is as in equation (I.18) i.e.,

2 Qp(t) = A (Bp,λ(p,t)). (I.26)

Then the Hilbert space Xp admits the following direct integral representation (see Proposition II.9, Section II.3) Z ⊕ Xp(t) = Qp(t) dt, (I.27)

2 where the integral is taken over R. Then we may represent the Bergman kernel for A (Up) in terms of reproducing kernel for Qp(t) as below.

Theorem 8. Let the reproducing kernel for Qp(t) be denoted by Xp(t;·,·). Then for (z,w),(Z,W) ∈

Up, the Bergman kernel K of Up is given by

Z z2πitZ−2πit K(z,w;Z,W) = X (t;w,W) dt. (I.28) p 1+1/2µ R (zZ)

I.5. Outline of this Thesis In Chapter 2, we recall basic facts about Bergman spaces and present proofs of proposi- tions that we will need to prove our main results in Chapter 3 and Chapter 4. We will also discuss geometric properties of polynomial ellipsoids and give a few examples. Finally, we discuss the direct integral representation of Hilbert spaces Hp and Xp. In Chapter 3, we present proofs of Theorems 2, 3, and 4. We ﬁrst prove analogous results in one variable, and use these in the proofs of Theorems 2, 3, and 4. Finally in Chapter 4, we recall notions about Hilbert spaces that are necessary to obtain integral representations of Bergman kernel of Up as in Theorems 7 and 8, and complete the proofs of Theorems 6, 7 and 8.

14 I.6. Historical Note The classical Paley-Wiener theorem was obtained in [17] for the Hardy space of the upper

n half plane. A Fourier representation for Hardy space of tube domains in C was obtained by S. Bochner ([5]); tube domains are a different set of generalizations of the upper half plane in C to higher dimensions. A Fourier representation of Bergman space of tube domains over self dual cones was obtained by O. Rothaus ([20]) and this was generalized to Bergman space of tubes over more general cones by A. Koranyi ([13]). The Fourier type representation of Bergman space of the upper half plane with respect to the translation group has been rediscovered many times (see [10, 22]), most recently by P. Duren, et. al ([7]). A. Koranyi along with E. Stein also obtained a Fourier representation for Hardy spaces of generalized half spaces ([14]). R. Ogden and S. Vagi´ ([16]) obtained Fourier representations of functions in Hardy space of Siegel domains of type II, with respect to the translation group.

n For Bergman space of the Siegel upper half space Un ⊂ C , M. Peloso et al., ([1]) obtained Fourier representations of Bergman space of Un with respect to the Heisenberg group. N. Vasilevski et al., ([18]) obtained representations of Bergman space of Siegel upper half space

n 2 Un ⊂ C as L spaces, with respect to maximal abelian subgroups of Aut(Un). Their representation however is not holomorphic in the complex parameter. Integral representation of the Bergman kernel of polynomial half space as in (I.25) was already obtained by F. Haslinger ([12]), by differentiating a similar formula for the Szego¨ kernel. It seems that Fourier representations of Bergman spaces of polynomial half spaces (even for the upper half plane) with respect to the scaling group have not been noted before.

15 CHAPTER II PRELIMINARIES In this chapter we state and prove familiar properties of the weighted Bergman space

n A2(Ω,λ) of a domain Ω ⊂ C . We ﬁrst show that the weighted Bergman space A2(Ω,λ) of Ω 2 2 2 is a closed subspace of L (Ω,λ). Then we show that Bergman spaces A (Ω1) and A (Ω2) of n biholomorphically equivalent domains Ω1 and Ω2 in C are isometrically isomorphic. We also show that a polynomial half space Up is biholomorphically equivalent to a polynomial ellipsoid

Ep. Then we give a few examples of polynomial half spaces. Finally we obtain the direct integral representations (II.5) and (II.19) of the Hilbert spaces Hp and Xp introduced in Theorems 3 and 4 respectively.

II.1. Bergman Spaces

Proposition II.1 (Bergman Inequality). Let A2(Ω,λ) be the weighted Bergman space of Ω with weight λ. Then, for each compact set K ⊂ Ω, there is a constant CK > 0 such that

sup| f (z)| ≤ CK k f kA2(Ω,λ) . (II.1) z∈K

n n Proof. Deﬁne the number r > 0 in the following way. If Ω = C , let r = 1, and if Ω 6= C , 1 r = 2 (dist(K,∂Ω)). Let n Kr = {z ∈ C | dist(z,K) ≤ r}.

Then Kr ⊂ Ω and for all z ∈ K, B(z,r) ⊂ Kr. Here B(z,r) is the closed ball of radius r and center z. Then, by mean value property of harmonic functions applied to real and imaginary parts of f ∈ A2(Ω,λ), we have 1 Z f (z) = f (ζ) dV(ζ). Vol (B(z,r)) B(z,r) Consequently, we have

1 Z | f (z)| = f (ζ) dV(ζ) Vol (B(z,r)) B(z,r)

16 1 Z ≤ | f (z)| dV(ζ) Vol (B(z,r)) B(z,r) 1 Z dV(ζ)1/2 Z 1/2 ≤ | f (z)|2 λ(ζ) dV(ζ) Vol (B(z,r)) B(z,r) λ(ζ) B(z,r) 1/2 1 1 ≤ k f k 2 1/2 L (B(z,r),λ) (Vol B(z,r)) λ B(z,r) 1/2 1 1 ≤ k f k 2 1/2 A (Ω,λ) (Vol B(z,r)) λ Kr 1/2 1 1 = k f k 2 1/2 n A (Ω,λ) (Vol B(0,1)) r λ Kr

:= CK k f kA2(Ω,λ) .

Here C > 0 is ﬁnite, because K is compact and is strictly positive on so k1/ k , the K r λ Ω λ Kr minimum value of λ on Kr is a positive quantity.

n Corollary II.2. Let Ω be a domain in C . The weighted Bergman space A2(Ω,λ) of Ω with weight λ is a closed subspace of L2(Ω,λ).

∞ 2 2 Proof. Let { fn}n=1 be a sequence of functions in A (Ω,λ), which converges in L (Ω,λ). Then by ∞ Bergman inequality (II.1) it follows that the sequence of holomorphic functions { fn}n=1 converges uniformly on compact sets of Ω, and consequently the limiting function is holomorphic in Ω.

Thus, the limiting function is in A2(Ω,λ) which shows that A2(Ω,λ) is closed in L2(Ω,λ).

Lemma II.3 (Biholomorphic Invariance of Bergman Space). Let Φ : Ω1 → Ω2 be a biholomorphic n map between two domains Ω1 and Ω2 in C . The map Φ induces an isometric isomorphism ∗ 2 2 2 Φ : A (Ω2) → A (Ω1) between the Bergman spaces of Ω2 and Ω1, where for all F ∈ A (Ω2) and all z ∈ Ω1 we have (Φ∗F)(z) = F (Φ(z)) · detΦ0(z).

2 Proof. Let F ∈ A (Ω2). Then we have

Z 2 2 kFk 2 = |F(ζ)| dV(ζ). A (Ω2) Ω2 17 Applying the change of variables formula to the map Φ,

Z 2 2 0 2 kFk 2 = |F(Φ(z))| detΦ (z) dV(z) A (Ω2) Ω1 Z 0 2 = F(Φ(z))detΦ (z) dV(z) Ω1 0 2 = (F ◦ Φ) · detΦ 2 . A (Ω1)

2 0 2 ∗ This shows that F ∈ A (Ω2) if and only if (F ◦Φ)·detΦ ∈ A (Ω1). This also shows that Φ is an 2 isometry and injective. Let Ψ : Ω2 → Ω1 be the inverse of Φ. Given F ∈ A (Ω1), let G : Ω2 → C be given by

G(z) = F(Ψ(z)) · detΨ0(z).

Then we have

G ◦ Φ(z) · detΦ0(z) = F(Ψ(Φ(z))) · detΨ0(Φ(z)) · detΦ0(z) = F(z).

0 2 2 ∗ This shows that (G ◦ Φ) · detΦ ∈ A (Ω1). Thus, it follows that G ∈ A (Ω2) and Φ G = F. Con- sequently, Φ∗ is surjective and an isometric isomorphism as claimed.

II.2. Geometry of Polynomial Half Spaces

n Lemma II.4. Let γ ∈ C \{z ∈ C|Imz = 0,Rez ≤ 0}, and let p : C → R be a weighted homoge- n n neous balanced polynomial. Recall from (I.19) that ρbγ : C → C is given by 1/2m1 1/2mn n ρbγ (ζ1,...,ζn) = γ ζ1,...,γ ζn , for ζ ∈ C .

Then we have p(ρbγ (ζ)) = |γ| p(ζ).

Proof. The function L : C \{z ∈ C|Imz = 0,Rez ≤ 0} → C given by

L(γ) = log|γ| + iargγ, −π < argγ < π

18 is a well deﬁned branch of logarithm, using which we deﬁne the powers γ1/2m1 ,...,γ1/2mn . Con- sequently, for a multi-index α = (α1,...,αn) we have

α1 αn α 1/2m1 1/2mn ρbγ (ζ) = γ ζ1 ... γ ζn

n ∑ j=1 α j/2m j α1 αn = γ ζ1 ···ζn

= γwtm(α)ζ α .

Thus, for a weighted homogeneous balanced polynomial p we have

α β p ρbγ (ζ) = ∑ Cα,β ρbγ (ζ) ρbγ (ζ) wtm(α)=wtm(β)=1/2 β wtm(α) α wtm(β) = ∑ Cα,β γ ζ γ ζ wtm(α)=wtm(β)=1/2 β wtm(α) α wt(β) = ∑ Cα,β γ ζ γ ζ wtm(α)=wtm(β)=1/2

1/2 1/2 α β = ∑ Cα,β γ γ ζ ζ wtm(α)=wtm(β)=1/2

= |γ| p(ζ).

For a weighted homogeneous balanced polynomial p, recall that the polynomial half space

n+1 Up is a domain in C given by

n Up = {(z,w) ∈ C × C | Imz > p(w)}, and the polynomial ellipsoid deﬁned by p is given by

n 2 Ep = {(z,w) ∈ C × C | |z| + p(w) < 1}.

n Corollary II.5. Recall from (I.7) that for θ > 0, and w = (w1,...,wn) ∈ C , ρbθ is given by 1/2m1 1/2mn ρbθ (w1,··· ,wn) = θ w1,··· ,θ wn .

19 Then the map ρb : Up → Up given by (as in (I.8))

ρθ (z,w) = (θz,ρbθ (w)) for (z,w) ∈ Up is an automorphism of Up.

Proof. By Lemma II.4 we have p(ρθ (w)) = p(ρbθ (w)) = θ p(w), since θ is a positive number. Thus, we have

Imρθ (z) = Imθz = θ Imz > θ p(w) = p(ρθ (w)).

This shows that ρθ (z,w) is in Up. Since

ρθ ◦ ρ1/θ (z,w) = ρ1/θ ◦ ρθ (z,w) = (z,w), it follows that ρθ is a bijection. Since ρθ is a linear map, it is also holomorphic in Up. Thus, ρθ is an automorphism of Up, as it is a holomorphic bijection (see [19, Theorem 2.14]).

Lemma II.6. The map Λ : Up → Ep given by 1 + iz/4 w1 wn Λ(z,w1,...,wn) = , ,··· , (II.2) 1 − iz/4 (1 − iz/4)1/m1 (1 − iz/4)1/mn 1 + iz/4 = ,ρ −2 (w) , 1 − iz/4 b(1−iz/4) where ρb is as in (I.19), is a biholomorphic equivalence.

Proof. Since Imz > p(w) > 0, (1 − iz/4) ∈ C \{z ∈ C|Imz = 0,Rez ≤ 0}. Thus, by Lemma II.4 we have w1 wn p(w) p ,··· , = p ρb(1−iz/4)−2 (w) = . (1 − iz/4)1/m1 (1 − iz/4)1/mn |1 − iz/4|2

Consequently we have

2 2 1 + iz/4 w1 wn |1 + iz/4| + p(w) + p ,··· , = 1 − iz/4 (1 − iz/4)1/m1 (1 − iz/4)1/mn |1 − iz/4|2 1 + |z/4|2 − Imz/2 + p(w) = 1 + |z/4|2 + Imz/2 20 1 + |z/4|2 − Imz/2 + Imz ≤ 1 + |z/4|2 + Imz/2 = 1.

−1 This shows that Λ(z,w) ∈ Ep. For (z,w) ∈ Ep, the inverse of Λ is the map Λ : Ep → Up given by ! 1/m1 1/mn −1 1 − z 2 w1 2 wn Λ (z,w1,··· ,wn) = 4i , ,··· , . 1 + z (1 + z)1/m1 (1 + z)1/mn

Since Λ is a holomorphic map and a set theoretic bijection, it follows that Λ is a biholomorphism (see [19, Theorem 2.14] for instance).

We now give some examples of polynomial half spaces.

II.2.1. Siegel Upper Half Space

n 2 Let p(w) = ∑ j=1 w j . Then p is a weighted homogeneous balanced polynomial with n respect to the tuple (1,...,1) ∈ N . We see that the polynomial half space in this case is

n 2 Un+1 = {(z,w) ∈ C × C | Imz > |w| }.

n+1 n+1 2 2 Un+1 is biholomorphically equivalent to the unit ball B = {(z,w) ∈ C | |z| + |w| < 1}, n+1 and the bihilomorphism Λ : Un+1 → B in Lemma II.6 is the Cayley map given by

1 − iz/4 w w Λ(z,w ,...,w ) = , 1 ,··· , n . 1 n 1 + iz/4 1 − iz/4 1 − iz/4

Note that when n = 0, the domain U1 is the upper half plane U in C and the map above reduces to the familiar one variable Cayley map between U1 = U and the disc D.

II.2.2. Bedford-Pinchuk Domains.

Let m be a positive integer. Then the solution α ∈ N to the equation wtm(α) = 1/2 is α = m. Then a weighted homogeneous balanced polynomial p with respect to this integer is of

21 the form p(w) = C |w|2m. Taking C = 1, we get

2 2m Em = {(z,w) ∈ C | Imz > |w| }.

The polynomial half space Em is biholomorphically equivalent to the Bedford-Pinchuk domain

2 2 2m Em = {(z,w) ∈ C | |z| + |w| < 1} (II.3) where the bihilomorphism Λ : Em → Em in Lemma II.6 is the given by

1 − iz/4 w Λ(z,w) = , . 1 + iz/4 (1 − iz/4)1/m

2 2 Interestingly, the domains Em ⊂ C are the only ‘nice’ domains in C that have large automorphism groups as the following theorem due to Bedford and Pinchuk shows.

Theorem 9 ([3]). Let Ω ⊂ C2 be a bounded pseudoconvex domain with real analytic boundary. If

Aut(Ω) is noncompact, then it is biholomorphically equivalent to the domain Em given by (II.3).

II.3. Direct Integral Representations

In this section, we express the Hilbert spaces Hp and Xp as a direct integral of weighted n Bergman spaces on domains in C . In view of this, we may interpret Theorems 3 and 4 as n+1 representing the Bergman space of the polynomial half space Up in C as a direct integral n of weighted Bergman spaces on domains in C . We now introduce the notion of a direct integral of Hilbert spaces.

Let (X,M , µ) be a σ-ﬁnite measure space, i.e., µ is a σ-ﬁnite measure on the σ-algebra

M of sets in X. Let {Ht}t∈X be a collection of separable Hilbert spaces of equal dimension.

A section f of the collection {Ht}t∈X is a function on X such that f (t) ∈ Ht for each t ∈ X.A

∞ countable set of sections {e j} j=1 is called a measurability structure on the collection of Hilbert spaces {Ht}t∈X if the functions t 7→ he j(t),ek(t)iHt are measurable for all j,k ∈ N and the linear

22 ∞ span of {e j(t)} j=1 is dense in Ht for each t ∈ X. A section f is said to be measurable if the function t 7→ he j(t), f (t)iHt is measurable for each j ∈ N. Suppose the following data is given: (1) a σ-ﬁnite measure space (X,M , µ), (2) a collection {Ht}t∈X of separable Hilbert spaces of equal dimension, (3) a measurability structure on

{Ht}t∈X . Then the direct integral of Ht’s with respect to the measure µ denoted by

Z ⊕ Ht dµ(t) is the space of equivalence classes of almost everywhere equal measurable sections f for which

Z 2 2 k f k = k f (t)kH dµ(t) < ∞. X t

It can be shown that the direct integral of a collection of Hilbert spaces is a Hilbert space with the natural inner product. Recall that the p is a nonnegative weighted homogeneous balanced polynomial with respect to the tuple m = (m1,...,mn) of positive integers. With this notation let

M = l.c.m(2,m1,··· ,mn).

Then we call a polynomial of the form

α n g(w) = ∑ Cα w , w ∈ C (II.4) wtmα=k/M α∈Nn a weighted homogeneous polynomial of weighted degree k/M.

II.3.1. Direct Integral Representation of Hp n Recall that the Hilbert space Hp consists of measurable functions f : (0,∞) × C → C such that Z ∞ Z e−4π p(w)t k f k2 = | f (t,w)|2 dV(w) dt < Hp ∞ 0 Cn 4πt

23 and ∂ f = 0 in the sense of distributions for 1 ≤ j ≤ n, ∂w j n where w1,··· ,wn are co-ordinates of C .

n Proposition II.7. For t > 0, let Sp(t) be the weighted Bergman space on C with the weight w 7→ e−4π p(w)t, i.e.,

2 n −4π pt Sp(t) = A (C ,e ).

n Then {Sp(t)}t>0 is a collection of Hilbert spaces indexed by (0,∞). For a tuple α ∈ N , let qα be given by

α n qα (t)(w) = w , for all t > 0,w ∈ C .

Then qα is a section on the collection {Sp(t)}t>0 of Hilbert spaces, i.e., qα (t) ∈ Sp(t) for all t > 0, which is a measurability structure on {Sp(t)}t>0. Then the Hilbert space Hp (as above) admits the direct integral representation

Z ⊕ dt Hp = Sp(t) . (II.5) dπt

−4π p(w)t Proof. It is not difﬁcult to verify that qα (t) ∈ Sp(t) for all t > 0, because the weight w 7→ e decays faster than any monomial qα (t) grows. We now show that the collection of sections

{qα }α∈Nn is a measurability structure on Sp(t). n For multi-indices α,β ∈ N , we have

Z α β −4π p(w)t qα (t),qβ (t) S (t) = w w e dV(w). p Cn

By the dominated convergence theorem it follows that the map t 7→ hqα (t),qβ (t)iSp(t) is continuous and hence measurable on (0,∞). It remains to be seen that for all t ∈ (0,∞) the linear span of

{qα (t)}α∈Nn , i.e., the polynomials are dense in Sp(t). n First, we show that if α,β ∈ N are two multi-indices such that wtm(α) 6= wtm(β), then for all t ∈ (0,∞), the monomial qα (t) is orthogonal to the monomial qβ (t) in Sp(t). Fix a θ ∈ (−π,π).

24 Then by Lemma II.4 we have

iθ p(ρbeiθ (ζ)) = e p(ζ) = p(ζ), where ρbeiθ is as in Lemma II.4. Making a change of variables ζ = ρbeiθ (w) below,we obtain Z α β −4π p(ζ)t qα (t),qβ (t) S (t) = ζ ζ e dV(ζ) p Cn Z β α −4π p(ρb iθ (w))t = (ρbeiθ (w)) (ρbeiθ (w)) e e dV(w) Cn Z = eiθ(wtm(α)−wtm(β))wα wβ e−4π p(w)t dV(w). Cn

Thus we have iθ(wtm(α)−wtm(β)) 1 − e qα (t),q (t) = 0, β Sp(t) from which it follows that qα (t) is orthogonal to qβ (t) in Sp(t) since wtm(α) 6= wtm(β). It follows from this that a weighted homogeneous polynomial of weighted degree k/M is orthogonal to a weighted homogeneous polynomial of weighted degree j/M in Sp(t) for all t > 0.

Now, we are ready to show that polynomials are dense in Sp(t). Suppose f ∈ Sp(t) is a function which is orthogonal to every polynomial. Then, we wish to show that f = 0, almost

n everywhere in C . Since f is entire, it has a power series expansion that converges normally on n compact subsets of C . We may rearrange such a series to obtain a representation of the form

∞ f (ζ) = ∑ fk(ζ), k=0

n that converges to f uniformly on compact subsets of C , and where fk is a weighted homogeneous polynomial of weighted degree k/M of the form (II.4). Since f is orthogonal to all weighted homogeneous polynomials f j of weighted degree j/M, for all j ∈ N, we have

0 = f , f j (II.6) Sp(t) Z −4π p(ζ)t = f (ζ) f j(ζ)e dV(ζ) Cn

25 Z −4π p(ζ)t = lim f (ζ) f j(ζ)e dV(ζ) r→∞ {|ζ|

n which shows that f j = 0 almost everywhere in C . Since j ∈ N was arbitrary, it follows that f = 0 n almost everywhere in C . This shows that the polynomials are dense in Sp(t) for all t ∈ (0,∞), which shows that the collection of sections {qα }t>0 is a measurability structure on {Sp(t)}t>0. R ⊕ dt Suppose f ∈ Hp. We then show that f ∈ Sp(t) 4πt . Since f ∈ Hp, we have Z ∞ Z e−4π p(w)t k f k2 = | f (t,w)|2 dV(w) dt < . Hp ∞ 0 Cn 4πt

Thus, by Fubini’s theorem, for almost all t > 0 we have

Z | f (t,w)|2 e−4π p(w)t dV(w) < ∞, Cn

n − pt n i.e., f (t,·) ∈ L2(C ,e 4π ). We now show that f (t,·) is holomorphic in C for almost all t > 0.

To show this, it sufﬁces to show that f (t,·) is holomorphic in each of the variables w1,...,wn, by Hartog’s theorem on separate analyticity. By Weyl’s lemma, it sufﬁces to show that for all 1 ≤ j ≤ n, ∂ f = 0 in the sense of distributions. ∂w j ∞ n ∞ Let α ∈ Cc ((0,∞) × C ) be of the form α(t,w) = α1(t)α2(w), where α1 ∈ Cc ((0,∞)), and ∞ n ∂ f α ∈ C ( ). Since f ∈ Hp, for 1 ≤ j ≤ n, = 0 in the sense of distributions the action of the 2 c C ∂w j distributional derivative of f with respect to w j on the test function α above gives

Z ∞ Z ∂α 0 = − f (t,w) dV(w) dt 0 Cn ∂w j ∞ Z Z ∂α2 = α1(t) − f (t,w) dV(w) dt. 0 Cn ∂w j 26 ∞ Since this holds for all α1 ∈ Cc ((0,∞)), we have

Z ∂α − f (t,w) 2 dV(w) = 0, Cn ∂w j

∞ n for almost all t ∈ (0,∞) and all α2 ∈ Cc (C ). This shows that for almost all t ∈ (0,∞), we have ∂ f (t,·) = 0 in the sense of distributions and thus, f (t,·) ∈ Sp(t). Moreover, ∂w j

Z ∞ dt Z ∞ Z e−4π p(w)t k f (t,·)k2 = | f (t,w)|2 dV(w) dt Sp(t) 0 4πt 0 Cn 4πt = k f k2 < . (II.7) Hp ∞

R ⊕ dt This shows that f ∈ Sp(t) 4πt . Suppose f ∈ R ⊕ S (t) dt . In view of (II.7) above, it follows that k f k < . Moreover, it p 4πt Hp ∞ 1 n n follows from (II.7) that f ∈ Lloc((0,∞)×C ), and since f (t,·) ∈ O(C ) for almost every t ∈ (0,∞) it follows that ∂ f = 0, in the sense of distributions for j = 1,...,n, ∂w j and consequently, f ∈ Hp.

II.3.2. Direct Integral Representation of Xp Recall that p is a nonnegative weighted homogeneous balanced polynomial with respect to

n the tuple m = (m1,...,mn) and M = l.c.m(2,m1,...,mn). Also recall that Bp ⊂ C is the domain

n Bp = {w ∈ C | p(w) < 1}.

We now state and prove a result that we need to obtain the direct integral representation of

Xp.

Lemma II.8. A function f ∈ O(Bp) admits a series expansion in weighted homogeneous polynomials, that is we may write ∞ f (z) = ∑ fk(z), for all z ∈ Bp (II.8) k=0

27 where for each k ≥ 0, fk is a weighted homogeneous polynomial of weighted degree k/M as in

(II.4). The series (II.8) converges uniformly on compact subsets of Bp.

n n Proof. For ζ = (ζ1,...,ζn) ∈ C , let δζ : C → C be the map

M/2m1 M/2mn δζ (µ) = µ ζ1,··· , µ ζn , for µ ∈ C. (II.9)

Note that since M = l.c.m(2,m1,...,mn), the powers of µ above are all positive integers, and a calculation shows that M p(δζ (µ)) = |µ| p(ζ). (II.10)

Fix a z ∈ Bp \{0}, and let ! z1 zn ω = δz(1/|z|) = ,··· , . |z|M/2m1 |z|M/2mn

We now show that the set of all µ ∈ C such that δω (µ) ∈ Bp is given by

D(0,R(ω)) = {µ ∈ C | |µ| < R(ω)}, (II.11) where R(ω) is a positive number depending on ω. Indeed, if µ ∈ C is such that ζ = δω (µ) lies in

Bp, we must have

p(δω (µ)) < 1. (II.12)

On the other hand, by using (II.10) we get

µ M p(δ (µ)) = |µ|M p(ω) = |µ|M p(δ (1/|z|)) = p(z), ω z |z| so equation (II.12) becomes

|z| 1 |µ| < = := R(ω). (II.13) (p(z))1/M (p(ω))1/M

Since f is in O(Bp), in a neighborhood of 0, it admits a power series expansion

α f (ζ) = ∑ Aα ζ . α∈Nn 28 This series converges normally in an open set containing 0, so we rearrange the terms to get

∞ f (ζ) = ∑ fk(ζ), (II.14) k=0 where fk is a weighted homogeneous polynomial of weighted degree k/M. To show that the series n in (II.14) converges at the point z ∈ Bp we restrict f to the complex analytic disc {ζ ∈ C | ζ =

δω (µ), µ ∈ D(0,R(ω))} to obtain

∞ ∞ ∞ M k/M k ϕ(µ) := f (δω (µ)) = ∑ fk(δω (µ)) = ∑ fk(ω)(|µ| ) = ∑ fk(ω)|µ| , (II.15) k=0 k=0 k=0 where we used the fact that fk is a weighted homogeneous polynomial of weighted degree k/M to obtain the penultimate equality above. Since f ∈ O(Bp), ϕ is holomorphic in the disc D(0,R(ω)) given by (II.11). Now, it follows from equation (II.13) that |z| < R(ω), and consequently the series

(II.15) converges for µ = |z|. Since ϕ(|z|) = f (δω (|z|)) = f (z), it follows that the series (II.14) converges at the point z. Now we wish to show that the series (II.14) converges uniformly on compact subsets of

Bp. Suppose a compact subset K of Bp is given. Then, there are numbers 0 < s,q < 1 such that n M K ⊂ {z ∈ C | p(z) < q s}. Letting ω = δz(1/|z|) as before, we use (II.10) to get

M p(z) = p(δω (|z|)) = |z| p(ω).

Then for every z ∈ K, we have p(z) = |z|M p(ω) < qMr from which it follows that

s 1/M |z| < q := qr(ω). (II.16) p(ω)

Then we have

k/M M k k | fk(z)| = | fk(δω (|z|))| = | fk(ω)| |z| ≤ | fk(ω)|r (ω)q . (II.17)

29 It follows from equations (II.13) and (II.16) that r(ω) < R(ω). Thus, to estimate | fk(ω)|, we note that it is the coefﬁcient of µk in series (II.15) and apply Cauchy integral formula to get

1 Z f (δ (µ)) f k ω k(ω) = k+1 dµ. (II.18) 2πi |µ|=r(ω) µ

The set {ζ ∈ Bp | ζ = δω (µ), µ ∈ C and |µ| = r(ω)} is a compact set in Bp, so there exists an

N > 0 such that fk(δµ (ω)) ≤ N for all µ ∈ C such that |µ| = r(ω). Using this, we obtain Cauchy estimates in (II.18) as N | f (δ (µ))| ≤ , k ω rk(ω) and thus, equation (II.17) reduces to

k | fk(z)| ≤ Nq .

This shows that the series (II.14) converges uniformly on K, since 0 < q < 1 is independent of the choice of z.

Recall from equation (I.18) that for ζ ∈ Bp, λ(p(ζ),t) is given by   1 −4πt sin−1 p(ζ) −4πt(π−sin−1 p(ζ))  e − e if t 6= 0 λ(p(ζ),t) = 4πt  −1 π − 2sin (p(ζ)) if t = 0.

Also recall that the Hilbert space Xp consists of all measurable functions f such that

Z Z k f k2 := | f (t, )|2 (p( ),t) dV( ) dt < Xp ζ λ ζ ζ ∞ R Bp and ∂ f = 0, in the sense of distributions for 1 ≤ j ≤ n, ∂ζ j where ζ1,··· ,ζn are co-ordinates of Bp.

Proposition II.9. For t ∈ R, let Qp(t) be the weighted Bergman space on Bp with the weight ζ 7→ λ(p(ζ),t), i.e.,

2 Qp(t) = A (Bp,λ(p,t)). 30 n Then {Qp(t)}t∈R is a collection of Hilbert spaces indexed by R. For a tuple α ∈ N , let qα be given by

α qα (t)(w) = w ,for all t ∈ R,w ∈ Bp.

Then qα is a section of the collection {Qp(t)}t∈R of Hilbert spaces, i.e., qα (t) ∈ Qp(t) for each t ∈ R, which is a measurability structure on {Qp(t)}t∈R. Then the Hilbert space Xp (as above) has the direct integral representation

Z ⊕ Xp = Qp(t) dt. (II.19)

Proof. It is not difﬁcult to verify that qα (t) ∈ Qp(t) for all t ∈ R, because the weight λ and the function qα (t) are both continuous upto the boundary of Bp. We now show that the collection of sections {qα }α∈Nn is a measurability structure on {Qp(t)}t∈R. n For multi-indices α,β ∈ N , we have

Z α β hqα (t),qβ (t)iQp(t) = w w λ(p(w),t) dV(w). Bp

By the dominated convergence theorem it follows that the map t 7→ hqα (t),qβ (t)iQp(t) is continuous and hence measurable on R. It remains to be seen that for all t ∈ R, the linear span of

{qα (t)}α∈Nn , i.e., the polynomials are dense in Qp(t). n First we show that if α,β ∈ N are two multi-indices such that wtm(α) 6= wtm(β), then qα (t) is orthogonal to qβ (t) in Qp(t) for all t ∈ R. Fix a θ ∈ (−π,π). Then by Lemma II.4 we have

iθ p(ρbeiθ (ζ)) = e p(ζ) = p(ζ), where ρbeiθ is as in Lemma II.4. Making a change of variables ζ = ρbeiθ (w), we obtain Z α β hqα (t),qβ (t)iQp(t) = ζ ζ λ(p(ζ),t) dV(ζ) Bp Z α β = (ρbeiθ (w)) (ρbeiθ (w)) λ(p(ρbeiθ (w)),t) dV(w) Bp

31 Z = eiθ(wtm(α)−wtm(β))wα wβ λ(p(w),t) dV(w). Bp

Thus we have iθ(wtm(α)−wtm(β)) 1 − e hqα (t),qβ (t)iQp(t) = 0, from which it follows that qα (t) is orthogonal to qβ (t) in Qp(t), since wtm(α) 6= wtm(β). It follows from this that a weighted homogeneous polynomial of weighted degree k/M is orthogonal to a weighted homogeneous polynomial of weighted degree j/M in Qp(t) for all t ∈ R.

Now, we show that polynomials are dense in Qp(t). Suppose f ∈ Qp(t) is a function which is orthogonal to every polynomial. Then, we wish to show that f = 0, almost everywhere in Bp. By Lemma II.8 we may expand f in a series of weighted homogeneous polynomials, i.e., we may write ∞ f (ζ) = ∑ fk(ζ), k=0 where fk is a weighted homogeneous polynomial of weighted degreee k/M as in (II.4). The series above converges to f uniformly on compact subsets of Bp. Since f is orthogonal to all weighted homogeneous polynomials f j of weighted degree j/M, for all j ∈ N, we have

0 = f , f j (II.20) Qp(t) Z = f (ζ) f j(ζ)λ(p(ζ),t) dV(ζ) Bp Z = lim f (ζ) f j(ζ)λ(p(ζ),t) dV(ζ) r→1− {p(ζ)

32 R ⊕ Suppose f ∈ Xp. We then show that f ∈ Qp(t) dt. Since f ∈ Xp, we have

Z Z k f k2 = | f (t, )|2 (p( ),t) dV( ) dt < . Xp ζ λ ζ ζ ∞ R Bp

Thus, by Fubini’s theorem we see that for almost all t ∈ R,

Z | f (t,ζ)|2 λ(p(ζ),t) dV(ζ) < ∞ Bp

2 i.e., f (t,·) ∈ L (Bp,λ(p,t)). We now show that f (t,·) is holomorphic in Bp for almost all t ∈ R.

To show this it sufﬁces to show that f (t,·) is holomorphic in each variable ζ1,...,ζn by Hartogs’s theorem on separate analyticity. To show that f (t,·) is holomorphic in the variable ζ j, by Weyl’s lemma it sufﬁces to show that

∂ f (t,·)/∂ζ j = 0 in the sense of distributions.

∞ ∞ ∞ Let ϕ ∈ Cc (R×Bp) be of the form ϕ(t,ζ) = ϕ1(t)ϕ2(ζ), where ϕ1 ∈ Cc (R), and ϕ2 ∈ Cc (Bp). ∂ f Since f ∈ Xp, = 0 in sense of distributions, the pairing of the distributional derivative of f ∂ζ j with respect to ζ j with the test function ϕ above gives

Z Z ∂ϕ 0 = − f (t,ζ) dV(ζ) dt R Bp ∂ζ j ! Z Z ∂ϕ2 = ϕ1(t) − f (t,ζ) dV(ζ) dt. R Bp ∂ζ j

∞ Since this holds for all ϕ1 ∈ Cc (R), we have

Z ∂ϕ − f (t,ζ) 2 dV(ζ) = 0, Bp ∂ζ j

∞ ∂ f (t,·) for almost all t ∈ R and all ϕ2 ∈ Cc (Bp).This shows that for almost all t ∈ R, we have = 0 ∂ζ j in sense of distributions and hence f (t,·) ∈ Qp(t).

33 Moreover we have ,

Z Z Z | f (t,·)|2 dt = | f (t, )|2 (p( ),t) dV( ) dt Qp(t) ζ λ ζ ζ R R Bp = k f k2 < . (II.21) Xp ∞

R ⊕ This shows that f ∈ Qp(t) dt. Now suppose that f ∈ R ⊕ Q (t) dt. In view of (II.21) above, it follows that k f k < . p Xp ∞ 1 n Moreover, it follows from equation (II.21) that f ∈ Lloc(R × C ), and since f (t,·) ∈ O(Bp) for almost every t ∈ R, it follows that

∂ f = 0, in the sense of distributions for j = 1,...,n, ∂ζ j and consequently f ∈ Xp.

34 CHAPTER III FOURIER REPRESENTATIONS In this chapter, we provide proofs of Theorems 2, 3 and 4. In order to obtain Fourier

2 2 representations of A (Ep) with respect to the rotation group, and those of A (Up) with respect to the translation and scaling groups we obtain analogous results for domains in C having rotation, translation and scaling symmetries, namely discs, strips and sectors respectively.

III.1. Maximal Compact Subgroup

2 Let D(r) = {z ∈ C | |z| < r} be the disk of radius r having center at the origin in C. Let `r ∞ be the Hilbert space of complex sequences a = (ak)k=0 such that

∞ 2k+2 2 r 2 kak 2 := π |a | < ∞. `r ∑ k k=0 k + 1

The following proposition included for completeness is a well known result.

2 2 Proposition III.1. The map T : `r → A (D(r)) given by

∞ k Ta(z) = ∑ akz , for all z ∈ D(r) (III.1) k=0 is an isometric isomorphism of Hilbert spaces.

2 2 j Proof. First we show that T is an isometry of `r with A (D(r)). Note that the monomial z is k 2 2 orthogonal to the monomial z in A (D(r)), whenever j 6= k. For a ∈ `r , we have

∞ 2 2 k kTakA2( (r)) = akz (By Parseval’s formula) D ∑ A2( (r)) k=0 D ∞ 2+2 r 2 = π ∑ |ak| k=0 k + 1 2 = kak 2 . `r

Thus, T is an isometry into L2(D(r)) and consequently injective. Moreover, this shows that the N k 2 2 2 partial sums z 7→ ∑k=0 akz converge to Ta in L (D(r)). Since A (D(r)) is closed in L (D(r)) we see that Ta ∈ A2(D(r)). 35 k ∞ 2 To show that T is surjective, we show that the set {z }k=0 in A (D(r)) is a complete orthogonal system. Since we already know that z j is orthogonal to zk when j 6= k, it sufﬁces to

k ∞ 2 k show that {z }k=0 is complete. To see this, let f ∈ A (D(r)) be a function that is orthogonal to z for all k ∈ N. Since f is holomorphic in the disk D(r), it has a series expansion

∞ k f (z) = ∑ akz , (III.2) k=0 that converges uniformly on compact subsets of D(r). Then for all k ∈ N, we have

k = h f ,a z i 2 0 k A (D(r)) Z k = f (z)akz dV(z) D(r) Z k = lim f (z)akz dV(z) (By Dominated Convergence Theorem) − s→r D(s) ∞ Z j k = lim a jz akz dV(z) (Uniform Convergence) − ∑ s→r j=0 D(s) r2k+2 = π |a |2 . k + 1 k

k ∞ This shows that ak = 0 for all k ∈ N. Thus, f ≡ 0 on D(r). This shows that {z }k=0 is a complete orthogonal system in A2(D(r)), which completes the proof.

2 III.1.1. A Fourier Representation of A (Ep)

2 We recall the notation necessary for obtaining the Fourier representation of A (Ep) with

n+1 respect to the rotation group, where Ep is the domain in C given by

n 2 Ep = {(z,w) ∈ C × C | |z| + p(w) < 1}.

n For a weighted homogeneous balanced polynomial p, the domain Bp ⊂ C is given by

n Bp = {w ∈ C | p(w) < 1}.

36 k+1 For k ∈ N, Wp(k) is the Bergman space on Bp with respect to the weight w 7→ (1− p(w)) , i.e.,

2 k+1 Wp(k) = A Bp,(1 − p)

∞ The Hilbert space Yp is the of sequences a = (ak)k=0 such that ak ∈ Wp(k) for each k ∈ N, and

∞ 1 kak2 = ka k2 < ∞. Yp π ∑ k Wp(k) k=0 k + 1

We make use of the following fact when integrating over Ep: for each w ∈ Bp such that (z,w) ∈ Ep, p we have z ∈ D( 1 − p(w)).

2 Proof of Theorem 2. We ﬁrst show that T is an isometry into L (Ep), and hence injective. Since a ∈ Yp we have

∞ 1 Z ∞ 1 kak2 = ka k2 = |a (w)|2 (1 − p(w))k+1 dV(w) < ∞, Yp π ∑ k Wp(k) π ∑ k k=0 k + 1 Bp k=0 k + 1 where we were allowed to interchange summation and integration by Tonelli’s theorem. Thus, for almost all w ∈ Bp we have

∞ k+1 (1 − p(w)) 2 π ∑ |ak(w)| < ∞. (III.3) k=0 k + 1

Recall from (I.13) that for (z,w) ∈ Ep, the function Ta is given by

∞ k Ta(z,w) = ∑ ak(w)z . k=0

2 p If w ∈ Bp is such that (III.3) holds, by Proposition III.1 the function Ta(·,w) is in A (D( 1 − p(w))) and ∞ k+1 Z (1 − p(w)) 2 2 π ∑ |ak(w)| = √ |Ta(z,w)| dV(z). (III.4) k=0 k + 1 D( 1−p(w)) Integrating (III.4) over Bp we get

Z Z 2 2 kTak 2 = |Ta(z,w)| dV(z) dV(w) L (Ep) √ Bp D( 1−p(w))

37 ∞ Z 1 2 k+1 = π ∑ |ak(w)| (1 − p(w)) dV(w) k=0 k + 1 Bp = kak2 . (III.5) Yp

2 We now show that the image of T is contained in A (Ep). Let

N k TNa(z,w) = ∑ ak(w)z . (III.6) k=0

k 2 2 For each k ∈ N, the function (z,w) 7→ ak(w)z is in A (Ep). Consequently TNa is in A (Ep) for all N > 0. Since a = (a0,a1,...) ∈ Yp, we know that the partial sums

N 1 ka k2 π ∑ k Wp(k) k=0 k + 1

2 2 converge to kak . Since T is an isometry, it follows that kT a − Tak 2 → 0. Since A (E ) is Yp N L (Ep) p 2 2 2 closed in L (Ep), it follows that Ta ∈ A (Ep). Thus, T is an isometry of Yp into A (Ep).

2 Now we show that the map T is in fact surjective. Suppose F ∈ A (Ep). Then we have

Z Z 2 2 kFk 2 = |F(z,w)| dV(z) dV(w) < ∞. A (Ep) √ Bp D( 1−p(w))

Thus, for almost all w ∈ Bp, we have

Z √ |F(z,w)|2 dV(z) < ∞. D( 1−p(w))

2 For such a w ∈ p, by Proposition III.1, there is an a(w) ∈ `√ such that B 1−p(w)

∞ k p F(z,w) = ∑ ak(w)z , for all z ∈ D( 1 − p(w)) k=0 and Z ∞ k+1 2 (1 − p(w)) 2 √ |F(z,w)| dV(z) = π ∑ |ak(w)| . (III.7) D( 1−p(w)) k=0 k + 1

38 p By Cauchy’s integral formula applied to F(·,w) in the disk D( 1 − p(w)) we obtain a formula for the coefﬁcients ak(w), for each k ∈ N as

k! Z F(ζ,w) a w r p p w k( ) = k+1 dζ, for 0 < < 1 − ( ). (III.8) 2πi |ζ|=r ζ

Differentiating under the integral in (III.8) with respect to w j, where 1 ≤ j ≤ n, shows that ak is holomorphic in the variables w1,...,wn. Thus, by Hartogs’s theorem on separate analyticity it follows that ak ∈ O(Bp) for each k ∈ N.

Finally integrating equation (III.7) on Bp on both sides yields

∞ 1 Z kak2 = |a (w)|2 (1 − p(w))k+1 dV(w) Yp π ∑ k k=0 k + 1 Bp Z ∞ k+1 (1 − p(w)) 2 = π ∑ |ak(w)| dV(w) Bp k=0 k + 1 Z Z = √ |F(z,w)|2 dV(z) dV(w) < ∞. Bp D( 1−p(w))

This shows that a ∈ Yp and that T is surjective.

III.2. Translation Subgroup III.2.1. The case of the strip S(a,b)

For −∞ ≤ a < b ≤ ∞, let S(a,b) be the strip

S(a,b) = {z ∈ C | a < Imz < b}. (III.9)

For a,b ∈ R, let ωa,b : R → (0,∞) be given by

e−4πat − e−4πbt ωa,b(t) = , for all t ∈ R. (III.10) 4πt

2 Let L (R,ωa,b) be the Hilbert space of measurable functions f on R such that

Z 2 2 k f k 2 := | f (t)| ω (t) dt < ∞. L (R;ωa,b) a,b R

39 2 Theorem 10 (Paley-Wiener theorem for Bergman space of the strip). The mapping TS : L (R;ωa.b) → A2(S(a,b)) given by Z i2πzt TS f (z) = f (t)e dt, for all z ∈ S(a,b) (III.11) R −1 2 2 is an isometric isomorphism of Hilbert spaces. Let TS : A (S(a,b)) → L (R,ωa,b) be the inverse 2 of TS. Then, for each function F ∈ A (S(a,b)) and for t ∈ R we have

Z −1 2πct −i2πxt TS F(t) = F(x + ic)e e dx, R for any c ∈ (a,b).

2 Proof. We ﬁrst show that for all z ∈ S(a,b) and all f ∈ L (R,ωa,b), the integral in (III.11) converges. We begin by noting that whenever a < y < b, i.e., when z ∈ S(a,b),

2 Z ei2πzt Z 4πte−4πyt (t) t = t ωa,b d −4πat −4πbt d (III.12) R ωa,b(t) R e − e Z 0 Z ∞ ≤ −4πte4π(b−y)t dt + 4πte−4π(y−a)t dt (III.13) −∞ 0 < ∞,

2 since both integrals in (III.13) clearly converge. Thus for all f ∈ L (R,ωa,b) and all z ∈ S(a,b), by the Cauchy-Schwarz inequality we have

1/2 Z Z 1/2 Z i2πzt 2 ! i2πzt 2 e f (t)e dt ≤ | f (t)| ωa,b(t) ωa,b(t) dt (III.14) R R R ωa,b(t) which shows that the integral in (III.11) converges. Let K ⊂ S(a,b) be a compact set. It is easy to see that the integral in (III.12) is uniformly bounded for all z ∈ K. Now integrating (III.14) with

i zt respect to z on K shows that the function (t,z) 7→ f (t)e 2π ∈ L1(R × K). 2 Now we show that for all f ∈ L (R,ωa,b), the function TS f given by (III.11) is holomorphic in the strip S(a,b). To see this, consider a closed triangle Γ ⊂ S(a,b). Since ∂Γ is a compact set

i zt it follows that (t,z) 7→ f (t)e 2π belongs to L1(R × ∂Γ). Thus, we may use Fubini’s theorem to

40 interchange order of integration below

Z Z Z Z Z Z i2πzt i2πzt TS f (z) dz = f (t)e dt dz = f (t) e dz dt = f (t) · 0 dt = 0. ∂Γ ∂Γ R R ∂Γ R

2 This shows that TS f is holomorphic in S(a,b) by Morera’s theorem. We now compute the A norm of TS f :

Z b 2 2 kTS f kA2(S(a,b)) = kTS f (· + iy)kL2( ) dy a R Z b −2πy(·) 2 = ke f kL2( ) dy (By Plancherel’s Theorem) a R Z Z b = | f (t)|2 e−4πyt dy dt (By Tonelli’s theorem) R a Z e−4πat − e−4πbt = | f (t)|2 dt R 4πt 2 = k f k 2 < ∞ L (R;ωa,b)

2 2 This shows that TS f ∈ A (S(a,b)) and also that TS is an isometry of L (R,ωa,b) with a subspace of A2(S(a,b)). Let z ∈ S(a,b) be written as z = x+iy. Given a function F ∈ A2(S(a,b)) and c ∈ (a,b), let

Fc : R → C be given by Fc(x) = F(x + ic) for all x on the horizontal line y = c. To show that TS is surjective, we show that for each F in A2(S(a,b)), we may choose a c ∈ (a,b) and obtain

Z Z 2πc(·) 2πct i2πzt i2π(z−ic)t F(z) = TS(Fbce )(z) = Fbc(t)e e dt = Fbc(t)e dt, R R

2 2πc(·) 2 where Fbc is the L Fourier transform of the function Fc, and such that Fbce ∈ L (R;ωa,b). To see this, we note that

Z b Z Z b 2 2 2 kFk 2 = |F(x + iy)| dx dy = Fy dy < ∞, A (S(a,b)) L2(R) a R a

2 2 which shows that Fy ∈ L (R) for almost all y ∈ (a,b). We choose a c ∈ (a,b) such that Fc ∈ L (R) and let Z i2πc(·) i2π(z−ic)t G(z) = TS(Fbce )(z) = Fbc(t)e dt. (III.15) R 41 2 First suppose that Fbc is compactly supported. Since Fc ∈ L (R), by Plancherel’s theorem 2 i2π(z−ic)t it follows that Fbc ∈ L (R). Since the function (t,z) 7→ e is continuous, it follows that the i2π(z−ic)t 1 2 function (t,z) 7→ Fbc(t)e ∈ L (R×K) for any compact K contained in S(a,b), as Fbc ∈ L (R) and is compactly supported in R. We claim that the function G in (III.15) is holomorphic in S(a,b). To see this consider

i2π(z−ic)t a closed triangle Γ ⊂ S(a,b). Since ∂Γ is a compact set it follows that (t,z) 7→ Fbc(t)e belongs to L1(R × ∂Γ). Thus, we may use Fubini’s theorem to interchange order of integration below

Z Z Z Z Z i2π(z−ic)t i2π(z−ic)t G(z) dz = Fbc(t)e dt dz = Fbc(t) e dz dt = 0, ∂Γ ∂Γ R R ∂Γ and by Morera’s theorem, G is holomorphic in S(a,b).

By the Fourier inversion formula, Gc = Fc on the line y = c, and, by the identity theorem for holomorphic functions, we must have G = F on S(a,b). Thus we see that

2πc(·) 2πc(·) F = TS(Fce ) and Fce = kFk 2 , (III.16) b b 2 A (S(a,b)) L (R;ωa,b) where we used Plancherel’s theorem to arrive at the second equality in (III.16).

ε When Fbc is not compactly supported, for each ε > 0, we construct a function G ∈ A2(S(a,b)) with the following properties:

2πyt ε 1. The measurable function t 7→ e Gcy(t) is independent of the choice of y ∈ (a,b).

ε 2. For each y ∈ (a,b), the function Gcy → Fby uniformly on compact subsets of R as ε → 0.

For then, by choosing a c ∈ (a,b) such that Gε ∈ L2(R) it follows from property (1) of Gε that for almost all t ∈ R, we get 2πct ε 2πyt ε e Gcc (t) = e Gcy (t), for all y ∈ (a,b). Letting ε & 0, and using property (2) of Gε , we get

−2π(y−c)t Fby(t) = e Fbc(t), for almost allt ∈ R 42 −1 −1 −2π(y−c)(·) =⇒ F Fby = F e Fbc

2πc(·) =⇒ Fy = TS e Fbc , which is precisely what we wanted. Here F −1 denotes the inverse L2 Fourier transform.

We now construct such a family of functions Gε for each ε > 0. Let φ ∈ C ∞(R) be a ∞ −1 function such that φb ∈ Cc (R) and φb ≡ 1 on the interval [−1,1]. Let φε (x) = ε φ(x/ε), so that R ε (t) = ( t). Note that (x)dx = k k 1 . Finally for each > 0, and z ∈ S(a,b) let G be φbε φb ε R φε φ L (R) ε given by

ε ε G (z) = Gy (x) = (φε ∗ Fy)(x).

It is apparent that property (2) above is satisﬁed. Indeed consider any compact set K ⊂

ε ε R. Since Gy (x) = (φε ∗ Fy)(x), we have Gcy = φbε Fby. Now choose ε0 small enough so that K ⊂ ε [−1/ε0,1/ε0]. For ε < ε0, φbε ≡ 1 so Gcy → Fby uniformly K. It remains to be shown that for each ε > 0, property (1) holds and that Gε ∈ A2(S(a,b)).

ε ∞ First we show that G is holomorphic in S(a,b) for every ε > 0. For N > 0, let ψN ∈ Cc (R) be a function such that ψN ≡ 1 on [−N,N] and 0 ≤ ψN < 1 outside [−N,N]. For z ∈ S(a,b) let Gε,N be given by

Z Z ε,N ε,N G (z) = Gy (x) = (ψNφε ∗ Fy)(x) = Fy(x −t)ψN(t)φε (t) dt = F(z −t)ψN(t)φε (t) dt. R R

∂F Since the function (z,t) 7→ ∂z (z−t)ψN(t)φε (t) is continuous on S(a,b)×R, we may differentiate under the integral sign to see that Gε,N is holomorphic in S(a,b). Note that by Young’s inequality for convolution, we get the following estimate

ψNφ ∗ Fy ≤ kψNφ k 1 Fy ≤ kφ k 1 Fy ε L2(R) ε L (R) L2(R) ε L (R) L2(R)

= kφk 1 Fy . (III.17) L (R) L2(R)

43 Now we compute the A2 norm of Gε,N:

Z b ε,N 2 ε,N 2 G A2(S(a,b)) = Gy 2 dy a L (R) Z b 2 = ψNφε ∗ Fy L2( ) dy a R Z b 2 2 ≤ kφkL1( ) Fy L2( ) dy (By estimate (III.17)) R a R Z 2 2 = kφkL1( ) |F(z)| dV(z) R S(a,b)

= kφk2 kFk2 < ∞. L1(R) A2(S(a,b))

This shows that Gε,N ∈ A2(S(a,b)) for all ε,N > 0. We now show that Gε,N → Gε in L2(S(a,b)). Note, that by Young’s inequality for convolution, we have

(φ − ψNφ ) ∗ Fy ≤ k(1 − ψN)φ k 1 Fy . (III.18) ε ε L2(R) ε L (R) L2(R)

Thus, we have

Z b ε ε,N 2 ε ε,N G − G L2(S(a,b)) = Gy − Gy 2 dy a L (R) Z b 2 = (φε − ψNφε ) ∗ Fy L2( ) dy a R Z b 2 2 ≤ k(1 − ψN)φε kL1( ) Fy L2( ) dy R a R Z 2 2 = k(1 − ψN)φε kL1( ) |F(z)| dz R S(a,b) 2 = k(1 − ) k 1 kFk . ψN φε L (R) A2(S(A,b))

1 Note that (1−ψN)φε → 0 pointwise on R and |(1 − ψN)φε | ≤ |φε |. Since φε ∈ L (R), by the dom- ε,N ε inated convergence theorem we see that k(1 − ) k 1 → 0 as N → . Thus, G → G in ψN φε L (R) ∞ L2(S(a,b)) as claimed. Consequently, Gε ∈ A2(S(a,b)) since A2(S(a,b)) is closed in L2(S(a,b)). Since Z b ε 2 ε 2 kG kA2(S(a,b)) = Gy 2 dy < ∞, a L (R)

44 b

−N + ic N + ic

−N + iy N + iy

Figure 4. The Rectangle SN

ε ε ε we may choose a c ∈ (a,b) such that kG k 2 < ∞. Since G = F , the function G is com- c L (R) cc φbε bc cc pactly supported and (III.16) holds for such a Gε in A2(S(a,b)) we obtain by our choice of c,

Z ε ε 2πc(·) ε 2πct i2πzt G (z) = T Gcc e (z) = Gcc (t)e e dt, R and

ε 2πc(·) ε kG e k 2 = kG k 2 . cc L (R) A (S(a,b))

2πc(·) ε 2 ε ε −2π(y−c)(·) 1 Thus e Gcc ∈ L (R), and since Gcc is compactly supported Gcc e belongs to L (R), for all y ∈ (a,b). Thus, an application of the Riemann-Lebesgue lemma shows that

Z ε ε ε −2π(y−c)t i2πNt lim G (N + iy) = lim Gy (N) = lim Gcc (t)e e dt = 0, |N|→∞ |N|→∞ |N|→∞ R for all y ∈ (a,b).

We now show that property (1) holds. For N > 0, let SN be the rectangle with vertices −N + ic,N + ic,N + iy, and −N + iy where y ∈ (a,b).

Let ∂SN be the boundary of SN oriented clockwise. Then for all t ∈ R we have by Cauchy’s theorem,

Z 0 = Gε (z)e−i2πzt dz (III.19) ∂SN Z N Z y = Gε (x + ic)e2πcte−i2πxt dx + Gε (N + iη)e2πηte−i2πNt dη −N c

45 Z −N Z c + Gε (x + iy)e2πyte−i2πxt dx + Gε (−N + iη)e2πηtei2πNt dη N y

ε Since limN→∞ G (N + iη) = 0 there exists N0 > 0 such that

ε |G (N + iη)| < 1 for all N > N0 and for all η ∈ (a,b).

Thus, we have the following estimate

|Gε (N + iη)e2πηte−i2πNt| ≤ |Gε (N + iη)e2πηt| < e2πηt.

Note that e2πηt belongs to L1([c,y]), since it is a continuous function on a compact set. Then by the dominated convergence theorem, we have

Z y Z y 2 t −i2 Nt 2 t −i2 Nt lim Gε (N + iη)e πη e π dη ≤ lim Gε (N + iη)e πη e π dη N→∞ c N→∞ c Z y 2 t −i2 Nt = lim Gε (N + iη)e πη e π dη c N→∞ Z y = lim |Gε (N + iη)|e2πηt dη c N→∞ Z y = 0 dη = 0. c

R c ε 2πηt i2πNt Similarly limN→∞ y G (−N + iη)e e dη = 0. Letting N → ∞, we get larger and larger

SN. Using the fact that Z c 2 t i2 Nt lim Gε (N + iη)e πη e π dη = 0, |N|→∞ y in (III.19) gives

Z N Z −N 0 = lim Gε (x + ic)e2πcte−i2πxt dx + lim Gε (x + iy)e2πyte−i2πxt dx N→∞ −N N→∞ N Z N Z N = lim Gε (x + ic)e2πcte−i2πxt dx − lim Gε (x + iy)e2πyte−i2πxt dx N→∞ −N N→∞ −N Z N Z N ε 2πct −i2πxt ε 2πyt −i2πxt = lim Gc (x)e e dx − lim Gy (x)e e dx N→∞ −N N→∞ −N Z Z ε 2πct −i2πxt ε 2πyt −i2πxt = Gc (x)e e dx − Gy (x)e e dx R R 2πct ε 2πyt ε = e Gcc (t) − e Gcy (t). 46 Thus, property (1) holds as claimed and this completes the proof.

2 III.2.2. A Fourier Representation of A (Up) n Let LTrans(p) be the Hilbert space of measurable functions g : (0,∞) × C → C such that

Z Z ∞ e−4π p(w)t kgk2 := |g(t,w)|2 dV(w) dt < ∞. LTrans(p) Cn 0 4πt

Recall that Hp is the closed subspace of LTrans(p) consisting of functions g ∈ LTrans(p) such that

∂g = 0in the sense of distributions, 1 ≤ j ≤ n. ∂w j

n We make use of the following fact when integrating over Up: for each w ∈ C such that n (z,w) ∈ Up = {(z,w) ∈ C×C | Imz > p(w)}, we have z ∈ S(p(w),∞). where we set S(p(w),∞) = {z ∈ C | Imz > p(w)} as in (III.9).

2 Proof of Theorem 3. We ﬁrst show that the map TS is an isometry from Hp into L (Up). Since f ∈ Hp, we have

Z Z ∞ e−4π p(w)t k f k2 = | f (t,w)|2 dt dV(w) < . Hp ∞ Cn 0 4πt

n Thus, for almost all w ∈ C , we have

Z ∞ −4π p(w)t 2 2 e k f (·,w)kL2( ,ω ) = | f (t,w)| dt < ∞. R p(w),∞ 0 4πt

n For such a w ∈ C , by Theorem 10, the function TS f (·,w) given by

Z ∞ i2πzt TS f (z,w) = f (t,w)e dt, for all z ∈ S(p(w),∞) 0 is well deﬁned and

Z Z ∞ −4π p(w)t 2 2 2 e kTS f (·,w)kA2(S(p(w),∞)) = |TS f (z,w)| dV(z) = | f (t,w)| dt. (III.20) S(p(w),∞) 0 4πt

47 n Integrating equation (III.20) over C , we get

Z Z Z Z ∞ −4π p(w)t 2 2 e |TS f (z,w)| dV(z) dV(w) = | f (t,w)| dt dV(w) Cn S(p(w),∞) Cn 0 4πt 2 2 i.e., kT f k 2 = k f k . S L (Up) Hp

2 ∞ We now show that the image of TS lies in A (Up). Let ϕN ∈ Cc ([0,∞)) be such that ϕN ≡ 1 n on [0,N] and 0 ≤ ϕ ≤ 1 on [0,∞). Then ϕN f is a measurable function on (0,∞) × C and for

(z,w) ∈ Up, the function TS(ϕN f ) is given by

Z ∞ i2πzt TS(ϕN f )(z,w) = ϕN(t) f (t,w)e dt. (III.21) 0

Note that for N > 0, we have k f k ≤ k f k and consequently f ∈ H , since it is clear that ϕN Hp Hp ϕN p

∂(ϕ f ) N = 0 in the sense of distributions for j = 1,...,n. ∂w j

n Thus, we see that for almost all w ∈ C ,

kϕN f (·,w)k 2 ≤ k f (·,w)k 2 < ∞ L (R,ωp(w),∞) L (R,ωp(w),∞)

n and by Theorem 10 we see that for almost all w ∈ C , the function TS(ϕN f )(·,w) belongs to 2 A (S(p(w),∞)) which shows that TS(ϕN f ) is holomorphic in the variable z. Since f is square n − p(w)t integrable on L2((0,∞) × C ) with respect to the weight (t,w) 7→ e 4π /4πt, it follows that 1 n ∞ n f ∈ Lloc((0,∞)×C ). Let α ∈ Cc (C ), then the action of the distributional derivative of TS(ϕN f ) with respect to w j on the test function α above gives

Z Z Z ∞ ∂α i2πzt ∂α −TS(ϕN f )(z,w) dV(w) = −ϕN(t) f (t,w)e dt dV(w) Cn ∂w j Cn 0 ∂w j Z ∞ Z ∂α i2πzt = − f (t,w)ϕN(t) e dV(w) dt (III.22) 0 Cn ∂w j = 0,

48 since

∂α i2πzt ∞ n ϕN(t) e ∈ Cc ((0,∞) × C ), ∂w j and the integral in (III.22) is the pairing of the distributional derivative of f with respect to w j with the test function above. By Weyl’s lemma we see that TS(ϕN f ) is holomorphic in the variable w j for 1 ≤ j ≤ n. Thus, by Hartogs’s theorem on separate analyticity we see that TS(ϕN f ) ∈ O(Up) 2 2 and consequently TS(ϕN f ) ∈ A (Up). We claim that TS(ϕN f ) → TS f in L (Up). Since TS is an 2 isometry, to show that TS(ϕN f ) → TS f in L (Up) it sufﬁces to show that ϕN f → f in Hp. Indeed, n we have | f − ϕN f | < | f | on (0,∞) × C and | f − ϕN f | → 0, as N → ∞. Thus, by dominated convergence theorem we obtain k f − f k → 0, as N → , which proves the claim. Since ϕN Hp ∞ 2 2 2 A (Up) is closed in L (Up) it follows that TS f ∈ A (Up). 2 We now show that the map T is surjective. Suppose F ∈ A (Up). We have

Z Z 2 2 kFk 2 = |F(z,w)| dV(z) dV(w) < ∞. A (Up) Cn S(p(w),∞)

n By Fubini’s theorem, for almost all w ∈ C , we have

Z |F(z,w)|2 dV(z) < ∞, (III.23) S(p(w),∞)

n where S(p(w),∞) = {z ∈ C | Imz > p(w)}. Thus, by Theorem 10, for such a w ∈ C , there is a 2 measurable function f (·,w) ∈ L (R,ωp(w),∞)

Z ∞ F(z,w) = f (t,w)ei2πzt dt 0 and

Z 2 2 kF(·,w)kA2(S(p(w),∞)) = |F(z,w)| dV(z) S(p(w),∞) Z ∞ e−4π p(w)t = | f (t,w)|2 dt. (III.24) 0 4πt

49 Using notation of Theorem 10, f (·,w) is given by

Z −1 i2πct −i2πxt f (t,w) = TS F(·,w) = F(x + ic,w)e e dx, R

n for any c > p(w). Integrating (III.24) over C , we get

Z 2 2 kFk 2 = kF(·,w)k 2 A (Up) A (S(p(w),∞)) Cn Z Z ∞ e−4π p(w)t = | f (t,w)|2 dt Cn 0 4πt = k f k2 . (III.25) LTrans(p)

−1 2 ∞ This, shows that TS : A (Up) → LTrans(p) is an isometry. Let φN ∈ Cc (R) be such that φ ≡ 1 on n [−N,N] and 0 ≤ φ ≤ 1. For t ∈ (0,∞) and w ∈ C , let fN be given by

Z −1 2πct −i2πxt fN(t,w) = TS (φNF) = φN(x)F(x + ic,w)e e dx, (III.26) R

2πct −i2πxt Note that (x,t,w) 7→ φN(x)F(x + ic,w)e e is a smooth function in each of the variables.

2πct −i2πxt 1 Consequently, the function (x,t,w) 7→ φN(x)F(x + ic,w)e e belongs to Lloc(R × (0,∞) × n C ), and by Fubini’s theorem we may interchange the order of integration below. For (t,w) ∈ n ∞ n (0,∞) × C , let α ∈ Cc ((0,∞) × C ) be of the form

∞ ∞ n α(t,w) = α1(t)α2(w), whereα1 ∈ Cc (R),α2 ∈ Cc (C ).

Then, the action of the distributional derivative with respect to w j of fN as in (III.26) on the test function α gives

∞ Z Z ∂α2 fN(t,w)α1(t) − (w) dV(w) dt 0 Cn ∂w j Z ∞ Z Z −i2πxt 2πct ∂α2 = −φN(x)F(x + ic,w)e dx e α1(t) (w) dV(w) dt 0 Cn R ∂w j Z Z ∞ Z ∂α2 2πct −i2πxt = φN(x) −F(x + ic,w) (w) dV(w) e e α1(t) dt dx R 0 Cn ∂w j = 0,

50 ∂ fN since F is holomorphic in the variable w j. This shows that = 0, in the sense of distributions ∂w j for all 1 ≤ j ≤ n.

2 −1 vergence theorem it follows that φNF → F in A (Up). Since TS is an isometry it follows that,

∂ fN ∂ f fN → f in L (p) and consequently → as distributions for 1 ≤ j ≤ n. This shows that Trans ∂w j ∂w j f ∈ Hp and completes the proof.

III.3. Scaling Subgroup III.3.1. The case of the sector V(a,b)

For 0 ≤ a < b ≤ π, let V(a,b) be the sector

V(a,b) = {z ∈ C | a < arg z < b}. (III.27)

Figure 5. Conformal Equivalence of a Sector with a Strip

Recall from (III.10) that for a,b as above, ωa,b : R → (0,∞) is given by

e−4πat − e−4πbt ωa,b(t) = , for t ∈ R. 4πt

Theorem 11 (Paley-Wiener Theorem for the Bergman Space of the Sector). The mapping TV :

2 2 L (R,ωa,b) → A (V(a,b)) given by

Z zi2πt T f (z) = f (t) dt, for all z ∈ V(a,b) (III.28) V z R is an isometric isomorphism of Hilbert spaces.

51 Proof. The mapping Φ : V(a,b) → S(a,b) given by Φ(z) = log|z|+iargz where a < argz < b for all z ∈ V(a,b) is a conformal equivalence. By Lemma II.3, this conformal equivalence establishes an isometric isomorphism

Φ∗ : A2(S(a,b)) → A2(V(a,b)) given by Φ∗(F) = (F ◦ Φ) · (Φ0). (III.29)

2 2 By Theorem 10, we know that TS : L (R;ωa,b) → A (S(a,b)) given by

Z i2πzt TS( f ) = f (t)e dt R is an isometric isomorphism.

∗ 2 2 Thus, Φ ◦TS is the desired isometric isomorphism of L (R;ωa,b) with A (V(a,b)), which we compute below

Z ∗ ∗ i2πzt Φ ◦ TS( f )(z) = Φ f (t)e dt R Z = f (t)ei2πΦ(z)t dt · Φ0(z) R Z 1 = f (t)e(logz)i2πt dt · R z Z 1 = f (t)zi2πt dt · R z Z zi2πt = f (t) dt. (III.30) z R This completes the proof.

Corollary III.2. Let (Φ∗)−1 : A2(V(a,b)) → A2(S(a,b)) be the inverse of the isometric isomor-

∗ ∗ 2 −1 phism Φ , where Φ is as in (III.29). Then, for F ∈ A (V(a,b)), the function f = TV F ∈ 2 L (R;ωa,b) is given by

Z −1 ∗ −1 2πct −i2πxt f (t) = TV F(t) := (Φ ) F(x + ic)e e dx, for all t ∈ R R

52 Moreover, we have

∗ −1 k f kL2( , ) = (Φ ) F = kFkA2(V(a,b)) . R ωa,b A2(S(a,b))

2 Proof. It follows from equation (III.30) that given f ∈ L (R,ωa,b) such that TV f = F is given by

Z −1 ∗ −1 −1 ∗ −1 ∗ −1 2πct −i2πxt f = (TS ◦ (Φ ) )F = TS (Φ ) F = (Φ ) F(x + ic)e e dx, R

−1 ∗ −1 where the last equality follows from Theorem 10. Since TS and (Φ ) are isometries, we obtain

−1 ∗ −1 ∗ −1 k f k 2 = T (Φ ) F = (Φ ) F = kFk 2 . L (R,ωa,b) S 2 2 A (V(a,b)) L (R,ωa,b) A (S(a,b))

n+1 We now introduce a domain Cp ⊂ C which is biholomorphically equivalent to Up (see

Lemma III.3 below), whose properties we use in the proof of Theorem 3. Let Cp be the domain

n Cp = {(γ,ζ) ∈ C × C | Im γ > p(ζ)|γ|}, (III.31)

n and recall that Bp = {ζ ∈ C | p(ζ) < 1}.

Since (z,w) ∈ Up, we have Im z > p(w) ≥ 0, so there exists a branch of logarithm deﬁned n on {z ∈ C | Im z > 0} independently of w ∈ C . We take this branch of logarithm to be the one that coincides with the natural logarithm on the real line and deﬁne the powers z1/2m1 ,··· ,z1/2mn with respect to this branch.

Lemma III.3. The map Ψ : Up → Cp given by w1 wn Ψ(z,w) = z, ,··· , = z,ρ1/z(w) , for all (z,w) ∈ Up, z1/2m1 z1/2mn b where ρb1/z is as in Lemma II.4, is a biholomorphic equivalence.

Proof. First, we show that Ψ(z,w) ∈ Cp. Since (z,w) ∈ Up we have Im z > p(w). We then get

Im z > p(w) = p ρbz(ρb1/z(w)) = p ρb1/z(w) |z| 53 where we used Lemma II.4 to arrive at the last equality. This then shows that Ψ(z,w) ∈ Cp. It is easy to see that Ψ is one-to-one. If Ψ(z,w) = Ψ(z0,w0) we get

w w w0 w0 z, 1 ,··· , n = z0, 1 ,··· , n =⇒ (z,w) = (z0,w0). z1/2m1 z1/2mn z01/2m1 z01/2mn

We now show that the function Ψ is onto. Note that for (γ,ζ) ∈ Cp, we have p(ζ)|γ| < Im γ, which gives Im γ p(ζ) < < 1, |γ| n Thus, for all ζ ∈ C such that (γ,ζ) ∈ Cp, we have p(ζ) < 1, i.e., ζ ∈ Bp. For each ζ ∈ Bp, such that (γ,ζ) ∈ Cp, we have

−1 −1 Im γ > p(ζ)|γ| i.e. γ ∈ Vζ = γ ∈ C | sin p(ζ) < arg z < π − sin p(ζ) .

1/2m1 1/2mn Vζ is a subset of the upper half plane, and hence we deﬁne the powers γ ,··· ,γ using the branch of the logarithm that coincides with the natural logarithm on the positive real line,

1/2m independent of ζ ∈ Bp. Let (γ,ζ) ∈ Cp be given. Let z = γ and w j = γ j ζ j, for j = 1,...,n. Then ! 1/2m1 1/2mn 1/2m1 1/2mn γ ζ1 γ ζn Ψ(z,w) = Ψ(γ,γ ζ1,...,γ ζn) = γ, ,..., = (γ,ζ). γ1/2m1 γ1/2mn

This shows that Ψ is a biholomorphism between Up and Cp.

∗ 2 2 2 By Lemma II.3 Ψ induces an isomorphism Ψ : A (Cp) → A (Up). For f ∈ A (Cp) the

∗ 2 function Ψ f ∈ A (Up) is given by

∗ 0 Ψ f (z,w) = f (Ψ(z,w)) · det Ψ (z,w), for (z,w) ∈ Up. (III.32)

54 2 III.3.2. Another Fourier Representation of A (Up)

Let LScal(p) be the Hilbert space of measurable functions f : R × Bp → C such that

Z Z k f k2 := | f (t,ζ)|2 λ(p(ζ),t) dV(ζ) dt < ∞. LScal(p) R Bp

Then Xp is the closed subspace of LScal(p) consisting of functions f such that

∂ f = 0in the sense of distributions, 1 ≤ j ≤ n. ∂w j

We use the following fact about Cp: if ζ ∈ Bp is such that (γ,ζ) ∈ Cp then γ belongs to the domain

Vζ given by −1 −1 Vζ = {γ ∈ C | sin p(ζ) < argγ < π − sin p(ζ)}.

2 Proof of Theorem 4. We ﬁrst show that the map TeV : Xp → A (Cp) given by

Z γ2πit TeV f (γ,ζ) = f (t,ζ) dt, for all (γ,ζ) ∈ Cp, R γ

2 is an isometric isomorphism from Xp into A (Cp). We then show that composing this isomorphism

∗ 2 2 with Ψ : A (Cp) → A (Up) in (III.32) gives the isometric isomorphism (I.20).

2 We now show that the map TeV as given above is an isometry from Xp to L (Cp). Since f ∈ Xp we have

Z Z k f k2 = | f (t, )|2 (p( ),t) dt dV( ) < , Xp ζ λ ζ ζ ∞ Bp R where we were allowed to interchange the order of integration by Tonelli’s theorem. Thus, for almost all ζ ∈ Bp, we have Z | f (t,ζ)|2 λ(p(ζ),t) dt < ∞. R 55 −1 −1 Let a(ζ) = sin p(ζ) and b(ζ) = π − sin p(ζ). Recall that Vζ is the planar sector

−1 −1 Vζ := V (a(ζ),b(ζ)) = {γ ∈ C | sin p(ζ) < argγ < π − sin p(ζ)}.

For such a ζ ∈ Bp, by Theorem 11, the function TeV f (·,ζ) given by

Z γ2iπt TeV f (γ,ζ) = f (t,ζ) dt, for all γ ∈ Vζ R γ is well deﬁned and

2 Z 2 Z 2 TV f (·,w) = TV f (γ,ζ) dV(z) = | f (t,w)| ω (t) dt e 2 e a(ζ),a(ζ) A (Vζ ) Vζ R Z = | f (t,w)|2 λ(p(ζ),t) dt, (III.33) R where the last equality follows from deﬁnitions (III.10) and (I.18) of ωa(ζ),b(ζ)(t) and λ(p(ζ),t) respectively. Integrating equation (III.33) over Bp, we get

2 Z Z 2 TV f = TV f (γ,ζ) dV(γ) dV(ζ) e 2 e L (Cp) Bp Vζ Z Z = | f (t,w)|2 λ(p(ζ),t) dt dV(ζ) Bp R = k f k2 . Xp

2 ∞ We now show that the image of TeV lies in A (Cp). Let φN ∈ Cc (R) be such that φN ≡ 1 on [−N,N] n and 0 ≤ φ ≤ 1 on R. Then, φN f is a measurable function on R × C and for (γ,ζ) ∈ Cp we have

Z γi2πt TeV (φN f )(γ,ζ) = φN(t) f (t,ζ) dt. (III.34) R γ

Note that for N > 0, k f k ≤ k f k and thus, f ∈ X since it is clear that φN Xp Xp φN p

∂(φ f ) N = 0 in the sense of distributions for j = 1,...,n. ∂ζ j

Thus we see that for almost all ζ ∈ Bp,

kφN f (·,ζ)k 2 ≤ k f (·,ζ)k 2 < ∞, L (R,ωa(ζ),b(ζ)) L (R,ωa(ζ),b(ζ))

56 2 and by Theorem 11 we see that for almost all ζ ∈ Bp, the function TeV (φN f )(·,ζ) ∈ A (Vζ ) which shows that T˜V (φN f ) is holomorphic in the variable γ. Since f is square integrable on R ×Bp with 1 ∞ a weight λ, it follows that f ∈ Lloc(R × Bp).Let α ∈ Cc (Bp), then the action of the distributional derivative of TeV (φN f ) with respect to ζ j on the test function α gives ! ! Z ∂α Z Z γi2πt ∂α −TeV (φN f )(γ,ζ) dV(ζ) = −φN(t) f (t,ζ) dt dV(ζ) Bp ∂ζ j Bp R γ ∂ζ j Z Z ∂α γi2πt = − f (t,ζ)φN(t) dV(ζ) dt (III.35) R Bp ∂ζ j γ = 0, since i2πt ∂α γ ∞ φN(t) ∈ Cc (R × Bp), ∂ζ j γ and since the integral in (III.35) is the pairing of the distributional derivative of f with respect to ζ j with the test function above. By Weyl’s lemma we see that TeV (φN f ) is holomorphic in the variable

ζ j for 1 ≤ j ≤ n. Thus, by Hartogs’s theorem on separate analyticity we see that TeV (φN f ) ∈ O(Cp), 2 2 and consequently TeV (φN f ) ∈ A (Cp). We claim that TeV (φN f ) → TeV f in L (Cp). Since TeV is an 2 isometry, to show that TeV (φN f ) → Tge in L (Cp) it sufﬁces to show that (φN f ) → f in Xp. Indeed

| f − φN f | < | f | on R × Bp, and | f − φN f | → 0 as N → ∞. Thus, by dominated convergence theorem we see that k f − f k → 0, as N → . Since A2(C ) is closed in L2(C ), this shows φN Xp ∞ p p 2 that TeV f is in A (Cp). 2 Now we show that the map TeV is surjective. Suppose F ∈ A (Cp). Then we have

Z Z 2 2 kFk 2 = |F(γ,ζ)| dV(γ) dV(ζ) < ∞, A (Cp) Bp Vζ

−1 −1 where Vζ = {γ ∈ C | sin (p(ζ)) < argγ < π − sin (p(ζ))}. Thus, for almost all ζ ∈ Bp, we have Z |F(γ,ζ)|2 dV(γ) < ∞. (III.36) Vζ

57 2 Thus, by Theorem 11, for such a ζ ∈ Bp, there exists an f (·,ζ) ∈ L (R,ωα(ζ),β(ζ)) such that

Z γi2πt F(γ,ζ) = f (t,ζ) dt, for all γ ∈ Vζ R γ and Z Z |F(γ,ζ)|2 dV(γ) = | f (t,ζ)|2 λ(p(ζ),t) dt. (III.37) Vζ R Using the notation of Theorem 11, by Corollary III.2, the measurable function f (·,ζ) is given by

Z ∗ −1 2πct −i2πxt −1 f (t,ζ) = (Φ ) F(x + ic,ζ)e e dx := TeV F(t,ζ), for t ∈ R, (III.38) R where c ∈ (sin−1 p(ζ),π − sin−1 p(ζ)).

We now show that the distributional derivative with respect to ζ j of the function f above ∞ vanishes, for 1 ≤ j ≤ n. Let φN ∈ Cc (R) be such that φ ≡ 1 on [−N,N] and 0 ≤ φ ≤ 1. For every t ∈ R and every ζ ∈ Bp, let fN(t,ζ) be given by

Z ∗ −1 2πct −i2πxt fN(t,ζ) = φN(x)(Φ ) F(x + ic,ζ)e e dx. (III.39) R

∗ −1 2πct −i2πxt The function (x,t,ζ) 7→ φN(x)(Φ ) F(x + ic,ζ)e e is a smooth function in each of the variables, and consequently is locally integrable on R × R × Bp. For (t,ζ) ∈ R × Bp, let α ∈ ∞ Cc (R × Bp) be of the form

∞ ∞ α(t,ζ) = α1(t)α2(ζ), whereα1 ∈ Cc (R),α2 ∈ Cc (Bp).

Then the action of the distributional derivative with respect to ζ j of fN as in equation (III.39) on the test function α gives

Z Z −∂α2 fN(t,ζ)α1(t) (ζ) dV(ζ) dt R Bp ∂ζ j Z Z Z x+ic x+ic −i2πxt 2πct −∂α2 = φN(x)F(e ,ζ)e e dx e α1(t) (ζ) dV(ζ) dt R Bp R ∂ζ j Z Z ∞ Z −∂α2 x+ic −i2πxt 2πct = φN(x) F(x + ic,w) dV(w) e e e α1(t) dt dx R 0 Cn ∂w j

58 = 0,

∂ fN since F is holomorphic in the variable ζ j. This shows that = 0 in the sense of distributions ∂ζ j for all 1 ≤ j ≤ n and thus fN ∈ Xp for all N > 0.

∗ −1 ∗ −1 Since (Φ ) F(·,ζ) − φN(Φ ) F(·,ζ) A2(S(a,b)) → 0 as N → ∞, it follows from Corol- lary III.2 that k f (·,ζ) − fN(·,ζ)k 2 → 0 as N → ∞, for almost every ζ ∈ p. Conse- L (R,ωα(ζ),β(ζ)) B quently we obtain

Z Z 2 2 k f − fNk = | f (t,ζ) − fN(t,ζ)| λ(p(ζ),t) dt dV(ζ) LScal(p) Bp R Z Z 2 = | f (t,ζ) − fN(t,ζ)| ωα(ζ),β(ζ)(t) dt dV(ζ) Bp R Z 2 = k f (·,ζ) − fN(·,ζ)k 2 dV(ζ) → 0, L (R,ωα(ζ),β(ζ)) Bp

∂ fN ∂ f as N → ∞. This shows that fN → f in LScal(p) and consequently, → in sense of distribu- ∂ζ j ∂ζ j 2 tions for 1 ≤ j ≤ n. This shows that TeV f ∈ Xp and that TeV is surjective and thus TeV : Xp → A (Cp) −1 is an isometric isomorphism. The map TeV given by equation (III.38) is the inverse of TeV . ∗ 2 2 ∗ Recall that Ψ : A (Cp) → A (Up) as in (III.32) is an isometric isomorphism. Then Ψ ◦TeV 2 ∗ is an isometric isomorphism from Xp to A (Up). We now show that TV = Ψ ◦TeV . For f ∈ Xp and

(z,w) ∈ Up we have

∗ 0 Ψ (TeV f )(z,w) = TeV f (Ψ(z,w)) · det Ψ (z,w) Z w w z2πit 1 = f t, 1 ,..., n dt 1/2m 1/2m 1/2µ R z 1 z n z z

= TV f (z,w), where we used the fact that for (z,w) ∈ Up,

1 det Ψ0(z,w) = . z1/2µ

This completes the proof of the theorem.

59 −1 Remark. The biholomorphism Ψ : Up → Cp of Lemma III.3 has an inverse Ψ : Cp → Up given by −1 1/2m1 1/2mn Ψ (γ,ζ) = γ,γ ζ1,...,γ ζn , for (γ,ζ) ∈ Cp.

The biholomorphic map Ψ−1 induces an isometric isomorphism induces an isometric isomor-

−1 ∗ 2 2 −1 phism (Ψ ) : A (Up) → A (Cp). Since Ψ ◦ Ψ is the identity automorphism of Up, it follows

−1 ∗ ∗ 2 −1 that the unitary automorphism (Ψ ) ◦ Ψ of A (Up) induced by Ψ ◦ Ψ is the identity map.

∗ −1 −1 ∗ ∗ This shows that (Ψ ) = (Ψ ) . Since TV = Ψ ◦ TeV , it follows that

−1 −1 ∗ −1 −1 −1 ∗ TV = TeV ◦ (Ψ ) = TeV ◦ (Ψ ) .

2 Thus, for F ∈ A (Up) we have

−1 −1 −1 ∗ TV F(t,ζ) = TeV (Ψ ) F(γ,ζ) −1 1/2m1 1/2mn 1/2µ = TeV F(γ,γ ζ1,...,γ ζn)γ Z = e(x+ic)/2µ (Φ∗)−1F(x + ic,ζ)e2πcte−i2πxt dx R

∗ −1 where the map Φ is as in the proof of Theorem 11. Here we used the formula (III.38) for TeV .

60 CHAPTER IV BERGMAN KERNEL OF POLYNOMIAL HALF SPACES In this chapter we ﬁrst recall the notion of a Reproducing Kernel Hilbert space, and also

n that a weighted Bergman space on a domain in C is a reproducing kernel Hilbert space. We then describe a method in Proposition IV.2 to compute the reproducing kernel of a reproducing kernel Hilbert space. Finally we compute the well known formula for Bergman kernel of Siegel Upper

n+ Half space in C 1.

IV.1. Reproducing Kernel Hilbert Spaces

IV.1.1. Reproducing Kernel Hilbert Spaces Let H be a Hilbert space. By the Riesz representation theorem, for each bounded linear

∗ functional ϕ ∈ H , there is an element RH(ϕ) ∈ H such that

ϕ( f ) = h f ,RH(ϕ)iH, f ∈ H. (IV.1)

∗ We call the map RH : H → H the Riesz map. The map RH is a conjugate linear isometry of Hilbert spaces.

Given a set Ω, let H be a Hilbert space of functions on Ω. Recall that H is called a reproducing kernel Hilbert space on Ω if, for each z in Ω, the evaluation map ez : H → C given by ez f = f (z) is a bounded linear functional on H. The function K : Ω × Ω → C given by

K(z,Z) = RH(ez)(Z), (IV.2) is called the reproducing kernel for H, where RH is as in (IV.1). For each z ∈ Ω and f ∈ H we have, D E f ,K(z,·) = h f ,RH(ez)iH = f (z), H which is why K is called the reproducing kernel for H. The following proposition gives the most important example of a reproducing kernel Hilbert space for the purposes of this thesis. 61 n Proposition IV.1. Let Ω be a domain in C and A2(Ω,λ) be the weighted Bergman space on Ω with weight the continuous function λ > 0. Then A2(Ω,λ) is a reproducing kernel Hilbert space.

Proof. Note that A2(Ω,λ) is a Hilbert space of functions on Ω. To show that A2(Ω,λ) is a

2 reproducing kernel Hilbert space, it sufﬁces to show that the evaluation map ez : A (Ω) → C given by

2 ez( f ) = f (z), f ∈ A (Ω,λ)

2 is a bounded linear functional on A (Ω,λ). It is clear that ez is a linear functional. To see that it is bounded, note that for each f ∈ A2(Ω,λ) we have

|ez( f )| = | f (z)| ≤ Cz k f kA2(Ω,λ) , where the last inequality follows from the Bergman inequality (II.1) applied to the compact set {z}.

n In particular, A2(Ω) the Bergman space on Ω ⊂ C is a reproducing kernel Hilbert space. The reproducing kernel for A2(Ω) is called the Bergman kernel of Ω.

IV.1.2. Computing Reproducing Kernels Proposition IV.2. (cf. [8, Lemma IX.3.5]) Let H be a reproducing kernel Hilbert space on a set

Ω. Suppose L is another Hilbert space such that there is an isometric isomorphism T : L → H.

Then the function K : Ω × Ω → C given by

K(z,Z) = hRL(eZ ◦ T),RL(ez ◦ T)iL , (IV.3) is the reproducing kernel for H.

Proof. The map ez ◦ T : L → C is given by

(ez ◦ T) f = ez(T f ) = T f (z). (IV.4)

62 It is clear that ez ◦ T is bounded and linear since each of ez and T are bounded and linear. Conse- quently, ez ◦ T is a bounded linear funtional on L. Thus, we have

T f (z) = (ez ◦ T) f (from equation (IV.4))

= h f ,RL(ez ◦ T)iL (by (IV.1))

= hT f ,TRL(ez ◦ T)iH. (T is an isometry) where RL is the Riesz map for the Hilbert space L. That is for every g ∈ H and each z ∈ Ω, we have

g(z) = hg,TRL(ez ◦ T)iH, i.e., TRL(ez ◦ T) = RH(ez) (IV.5)

Thus, the reproducing kernel K for the Hilbert space H is given by

K(z,Z) = RH(ez)(Z) (By (IV.2))

= TRL(ez ◦ T)(Z) (By equation (IV.5))

= eZ(TRL(ez ◦ T))

= hTRL(ez ◦ T),TRL(eZ ◦ T)iH (By equation (IV.1))

= hTRL(eZ ◦ T),TRL(ez ◦ T)iH

= hRL(eZ ◦ T),RL(ez ◦ T)iL. (T is an isometry)

If we take T : H → H to be the identity map in Proposition IV.2, we have the alternative representation of the reproducing kernel

K(z,Z) = hRH(eZ),RH(ez)iH . (IV.6)

It immediately follows that the reproducing kernel has the Hermitian symmetry

K(z,Z) = K(Z,z). (IV.7)

63 n Let K denote the reproducing kernel for the weighted Bergman space A2(Ω,λ) on Ω ⊂ C with weight λ. Then the following identity is an immediate consequence of (IV.6)

Z K(z,Z) = K(z,ζ)K(Z,ζ)λ(ζ) dV(ζ). (IV.8) Ω

Indeed, for each z and each w in Ω, by (IV.6), we have

D E K(z,Z) = RA2(Ω,λ)(eZ),RA2(Ω,λ)(ez) A2(Ω,λ) Z = RA2(Ω,λ)(eZ)(ζ)RA2(Ω,λ)(ez)(ζ)λ(ζ) dV(ζ) Ω Z = K(Z,ζ)K(z,ζ)λ(ζ) dV(ζ). (By equation (IV.2)) Ω

We now show that Proposition IV.2 can be used to recapture the familiar formula for the reproducing kernel of a weighted Bergman space.

n ∞ 2 Corollary IV.3. Let Ω be a domain in C and let φ j j=1 be an orthonormal basis for A (Ω,λ). Then K, the reproducing kernel for A2(Ω,λ) is given by

∞ K(z,Z) = ∑ φ j(z)φ j(Z). (IV.9) j=1

Proof. Let `2 be the Hilbert space of square summable sequences, and consider the map T : `2 →

A2(Ω,λ) given by ∞ ∞ a := (a j) j=1 7→ ∑ a jφ j := Ta. j=1 The map T is an isometry, for we have

∞ 2 2 2 kak`2 = ∑ a j = kTakA2(Ω) , j=1 where the last equality follows from Parseval’s formula. Thus, T is injective. Moreover T is

∞ 2 surjective since {φ j} j=1 is a basis for A (Ω,λ) and consequently T is an isometric isomorphism.

64 2 2 Let z ∈ Ω be ﬁxed. The map ez ◦ T : ` → C is given for a ∈ ` by

∞ (ez ◦ T)a = ∑ a jφ j(z). j=1

∞ 2 2 Denoting the sequence of complex numbers φ j(z) in ` by φ(z), we see that for a ∈ ` , j=1 D E (ez ◦ T)a = a,φ(z) . `2

Thus, by deﬁnition (IV.1) of the Riesz map, we have

R`2 (ez ◦ T) = φ(z).

Applying Proposition IV.2 to T : `2 → A2(Ω,λ) gives

D E ∞ K(z,Z) = hR 2 (eZ ◦ T),R 2 (ez ◦ T)i 2 = φ(Z),φ(z) = φ j(z)φ j(Z). ` ` ` 2 ∑ ` j=1

The well known transformation formula for Bergman kernel under biholomorphic maps can also be obtained by using Proposition IV.2 as we show below.

Corollary IV.4 (Transformation of the Bergman Kernel). Let Φ : Ω1 → Ω2 be a biholomorphic n map between two domains Ω1 and Ω2 in C . Then, the Bergman kernels K1 of Ω1 and K2 of Ω2 are related as

0 0 K1(z,Z) = detΦ (z)K2(Φ(z),Φ(Z))detΦ (Z).

Proof. Recall from Lemma II.3, that the biholomorphic map Φ : Ω1 → Ω2 induces an isometric ∗ 2 2 isomorphism Φ : A (Ω2) → A (Ω1) given by

Φ∗F(z) = F(Φ(z)) · detΦ0(z).

∗ 2 Thus, the map ez ◦ Φ : A (Ω2) → C is given by

∗ 0 D 0 E (ez ◦ Φ )F = F(Φ(z)) · detΦ (z) = F,R 2 (e )detΦ (z) , (IV.10) A (Ω2) Φ(z) 2 A (Ω2) 65 2 for each F ∈ A (Ω2). Comparing equation (IV.10) with deﬁnition (IV.1) of Riesz map, we see that

∗ 0 R 2 (e ◦ Φ ) = R 2 (e ) · detΦ (z). A (Ω2) z A (Ω2) Φ(z)

∗ 2 2 Applying Proposition IV.2 to Φ : A (Ω2) → A (Ω1), we get

D 0 0 E K (z,Z) = R 2 (e )detΦ (Z),R 2 (e )detΦ (z) 1 A (Ω2) Φ(Z) A (Ω2) Φ(z) 2 A (Ω2)

0 D E 0 = detΦ (Z) · R 2 (e ),R 2 (e ) · detΦ (z) A (Ω2) Φ(Z) A (Ω2) Φ(z) 2 A (Ω2)

0 D E 0 = detΦ (Z) · R 2 (e ),R 2 (e ) · detΦ (z) A (Ω2) Φ(z) A (Ω2) Φ(Z) 2 A (Ω2)

0 0 = detΦ (Z) · R 2 (e )(Φ(Z)) · detΦ (z) A (Ω2) Φ(z)

0 0 = detΦ (Z) · K2(Φ(z),Φ(Z)) · detΦ (z).

IV.2. Asymptotic Behavior of the Bergman Kernel We ﬁrst prove a result that is needed in the proof of Theorem 5.

N− N Proposition IV.5. Let g ∈ C 1(R 1) and U be a domain in R given by

0 0 N xN − g(x ) > 0, for (x1,...,xN−1,xN) = (x ,xN) ∈ R .

N Suppose that g ≥ 0 on R , g(0) = 0 and 5g(0) = 0. If d(x,∂U) denotes the distance of a point N x ∈ R to the boundary ∂U of U, then we have

x lim N = 1. x→0 d(x,∂U) x∈U

Proof. For x ∈ U, let y(x) ∈ ∂U be the point on the boundary of U closest to x. First we show that the set of points in ∂U closest to x is non-empty. Let δ = inf{|z − x| | z ∈ ∂U}. Then, there is a

∞ sequence {zn}n=1 ∈ ∂U such that |zn − x| → δ as n → ∞. Then there is an N > 0 such that for all

66 n > N we have |zn − x| < δ + 1. Then for such n, we have

|zn| ≤ |zn − x| + |x| ≤ |x| + δ + 1,

∞ ∞ which shows that the sequence {zn}n=1 is bounded. Thus, there is a subsequence {znk }k=1 that converges to a point y(x) ∈ ∂U, since ∂U is closed. This shows that there is at least one point y(x) in ∂U that is closest to x ∈ U. Then we have

|y(x)| ≤ |y(x) − x| + |x|

0 0 = (y (x),yN(x)) − (x ,xN) + |x|

0 0 0 0 0 ≤ (x ,g(x )) − (x ,xN) + |x| (since (x ,g(x )) ∈ ∂U)

0 = xN − g(x ) + |x|

≤ |xN| + |x| (since g ≥ 0)

≤ 2|x|.

Since the vector x−y(x) is normal to the boundary ∂U at y(x), there is a λ ∈ R for which we have

∂g 0 x j − y j(x) = −λ (y ), 1 ≤ j ≤ N − 1 ∂x j

xN − yN(x) = λ.

Thus, we get

d(x,∂U) = |x − y(x)| q = λ 2 + λ 2 |5g(y0)|2 q 0 2 = |xN − yN(x)| 1 + |5g(y )| .

Since |y(x)| ≤ 2|x|, we get y(x) → 0 as x → 0. Thus, we obtain

xN xN lim = lim q x→0 d(x,∂U) x→0 0 2 x∈U x∈U |xN − y(x)| 1 + |5g(y )|

67 = 1, as y(0) = 0 and 5g(0) = 0.

Corollary IV.6. Let Up be a polynomial half space. For (z,w) ∈ Up, let d(z,w) be the distance of

(z,w) to the boundary of Up. Then we have

Imz lim = 1. (z,w)→0 d(z,w)

2n+2 Proof. The domain Up ⊂ R is given by y − p(u1,v1,...,un,vn) > 0, where z = x + iy and w j = u j + iv j for 1 ≤ j ≤ n. Since p is a positive polynomial with no constant terms, we have p ≥ 0, p(0) = 0 and

n+ p ∈ C 1(R2 1), and the result of previous proposition applies.

Recall that for c > 0, the domain Γc is the subset of Up given by

Γc{(z,w) ∈ Up | |z| > c|w|}.

Proof of Theorem 5. For a point (z,w) ∈ Up, write z = x + iy. Recall that for (z,w) ∈ Up, the translation τ−x : Up → Up given by

τ−x(z,w) = (z − x,w) is an automorphism of Up. Thus, an application of Corollary IV.4 now shows that

0 2 Kdiag(z,w) = detτ−x(z,w) Kdiag(τ−x(z,w))

= Kdiag(iy,w), (IV.11)

0 where we used the fact that detτ−x(z,w) = 1 to get the last equality. Also recall that ρy : Up → Up, for (z,w) ∈ Up given by

1/2m1 1/2mn ρy(z,w1,··· ,wn) = yz,y w1,··· ,y wn

68 is an automorphism of Up, where y = Imz. Now, applying Corollary IV.4 we get

2 Kdiag(iy,w) = detρy(iy,w) Kdiag(ρy(iy,w))

2+1/µ = y Kdiag(ρy(iy,w)) (IV.12)

1+1/2µ n because detρy(iy,w) = y where 1/µ = ∑ j=1 1/m j. From equation (IV.12), we get

−1 2+1/µ Kdiag(i,ρy (w)) = y Kdiag(iy,w). (IV.13)

−1 Note that ρy (w) = ρ1/y(w), so we have

2 2 2 |w1| |wn| lim ρ1/y(w1,...,wn) = lim + ··· (iy,w)→(0,0) (iy,w)→(0,0) y1/m1 y1/mn (iy,w)∈Γc (iy,w)∈Γc |w|2 ≤ lim (iy,w)→(0,0) y (iy,w)∈Γc |w|2 = lim y · (iy,w)→(0,0) y2 (iy,w)∈Γc 1 ≤ lim y · (Since (iy,w) ∈ Γc, we get |iy| > c|w|) (iy,w)→(0,0) c2 (iy,w)∈Γc

= 0.

Using this and equations (IV.11) and (IV.13) we may write

−1 Kdiag(i,0) = lim Kdiag(i,ρy (w)) (iy,w)→(0,0) (iy,w)∈Γc

2+1/µ = lim y Kdiag(iy,w) (iy,w)→(0,0) (iy,w)∈Γc

2+1/µ = lim y Kdiag(z,w). (z,w)→(0,0) (z,w)∈Γc

69 Combining this with Corollary IV.6, we obtain

2+1/µ lim (d(z,w)) Kdiag(z,w) = Kdiag(i,0), (z,w)→(0,0) (z,w)∈Γc which completes the proof.

IV.3. A Formula for the Bergman Kernel of Ep

Recall that for k ∈ N, Wp(k) is the weighted Bergman space on Bp with respect to the weight w 7→ (1 − p(w))k+1, i.e.,

2 k+1 Wp(k) = A Bp,(1 − p) .

∞ Also recall that Yp is the Hilbert space of sequences a = (ak)k=0 such that ak ∈ Wp(k) for each k ∈ N, and ∞ 1 kak2 = ka k2 < ∞. Yp π ∑ k Wp(k) k=0 k + 1 2 Proof of Theorem 6. Let T : Yp → A (Ep) be the map as in Theorem 2, which for a sequence a = (a0,a1,...) ∈ Yp is given by

k Ta(z,w) = ∑ ak(w)z , for (z,w) ∈ Ep. k=0

Let e(z,w) be the evaluation functional at a point (z,w) ∈ Ep. Then, for a ∈ Yp the map e(z,w) ◦ T :

Yp → C is given by

(e(z,w) ◦ T)a = Ta(z,w) ∞ k = ∑ ak(w)z k=0 ∞ D E = ak,Yp(k;w,·) ∑ W (k) k=0 p ∞ Z k π (k + 1)z k+1 = ∑ ak(ζ)Yp(k;w,ζ) (1 − p(ζ)) dV(ζ) , (IV.14) k=0 k + 1 Bp π

70 where the last equality follows from the reproducing property of the kernel Yp(k;·,·). On the other ∗ ∗ hand, the image RYp e(z,w) ◦ T of the functional e(z,w) ◦ T ∈ Yp under the Riesz map RYp : Yp →

Yp is given by

(e(z,w) ◦ T)a = a,RY (e(z,w) ◦ T) , a ∈ Yp p Yp ∞ Z π k+1 = ∑ ak(ζ)RYp (e(z,w) ◦ T)(k,ζ)(1 − p(ζ)) dV(ζ) (IV.15) k=0 k + 1 Bp

For every k ∈ N and every ζ ∈ Bp we claim that

(k + 1)zk Yp(k;w,ζ) = R (e ◦ T)(k,ζ). π Yp (z,w)

Let φ : N × Bp → C be the difference of the two quantities above, i.e.,

(k + 1)zk φ(k,ζ) = Yp(k;w,ζ) − R (e ◦ T)(k,ζ). π Yp (z,w)

It then follows from equations (IV.14) and (IV.15) that

∞ Z π k+1 ∑ ak(ζ)φ(k,ζ)(1 − p(ζ)) dV(ζ) = 0 (IV.16) k=0 k + 1 Bp for all a ∈ Yp. Let a = (a0,a1,...,) ∈ Yp be such that ak ∈ Wp(k) for a k ∈ N, and a j ≡ 0 on Bp, for j ∈ N and j 6= k. For this choice of the sequence a in Yp, equation (IV.16) reduces to

Z k+1 π ak(ζ)φ(k,ζ)(1 − p(ζ)) dV(ζ) = 0, Bp for any ak ∈ Wp(k). Thus, we see that φ(k,ζ) = 0 for almost all ζ ∈ Bp. Since k was arbitrary we see that φ(k,ζ) = 0 for all k ∈ N and almost all ζ ∈ Bp. This proves the claim, and we have

k (k + 1)z RYp e(z,w) ◦ T (k,ζ) = Yp(k;w,ζ) , (k,ζ) ∈ N × Bp. π

2 Applying Proposition IV.2 to isometric isomorphism T : Yp → A (Ep), we get

B(z,w;Z,W) = RY (e(Z,W) ◦ T),RY (e(z,w) ◦ T) p p Yp

71 ∞ Z k k ! π (k + 1)z (k + 1)Z k+1 = ∑ Yp(k;w,ζ) Yp(k;W,ζ) (1 − p(ζ)) dV(ζ) k=0 k + 1 Bp π π ∞ Z k + 1 k+1 k k = ∑ Yp(k;w,ζ)Yp(k;W,ζ)(1 − p(ζ)) dV(ζ) z Z k=0 π Bp ∞ k + 1 k k = ∑ Yp(k;w,W)z Z , k=0 π where we used (IV.8) to arrive at the last equality.

IV.4. A Formula for the Bergman Kernel of Up To prove Theorem 7, we will need the following lemma.

n Lemma IV.7. For t > 0, let Sp(t) be the weighted Bergman space on C with weight w 7→ −4π p(w)t n n e . Let the reproducing kernel for Sp(t) be denoted by Hp(t;·,·). Let ρbt : C → C be given by 1/2m1 1/2mn n ρbt(w1,...,wn) = t w1,··· ,t wn , for w ∈ C .

n Then for all w,W ∈ C , we have

1/µ Hp(t;w,W) = t Hp (1;ρbt(w),ρbt(W)), (IV.17)

n where for the multi-index m = (m1,...,mn) we let 1/µ = ∑ j=1 1/m j.

Proof. We ﬁrst show that the map Dt : Sp(1) → Sp(t) given by

1/2µ Dt f (ζ) = t f (ρbt(ζ)) (IV.18) is an isometric isomorphism. Let w = ρbt(ζ). Then by Lemma II.4 we have p(w) = p(ρbt(ζ)) = t p(ζ) and dV(w) = t1/µ dV(ζ).

Further we have

Z k f k2 = | f (w)|2 e−4π p(w) dV(w) Sp(1) Cn 72 Z 2 −4π p(ζ)t 1/µ = | f (ρbt(ζ))| e t dV(ζ) Cn Z 2 1/2µ −4π p(ζ)t = t f (ρbt(ζ)) e dV(ζ). Cn = kD f k2 (IV.19) t Sp(t)

It follows from equation (IV.19) that f ∈ Sp(1) if and only if Dt f ∈ Sp(t). Moreover, if f ∈ Sp(1), then it follows from equation (IV.19) that

k f k = kD f k , Sp(1) t Sp(t)

n which shows that Dt is an isometry and hence injective. For F ∈ Sp(t) let the function f : C → C −1/2µ be given by f (ζ) = t F ρb1/t(ζ) . We have

1/2µ 1/2µ −1/2µ Dt f (ζ) = t · f (ρbt(ζ)) = t ·t F ρb1/t(ρbt(ζ)) = F(ζ).

Since Dt f = F ∈ Sp(t) it follows that f ∈ Sp(1). This shows that Dt is surjective, and hence an isometric isomorphism.

We now apply Lemma IV.2 to the isometric isomorphism Dt in (IV.18) to obtain the equation (IV.17). Applying equation (IV.1) (which deﬁnes the Riesz map) to the functional eζ ◦ Dt : Sp(1) → C we obtain

D E f ,RSp(1)(eζ ◦ Dt) = (eζ ◦ Dt) f Sp(1)

= eζ (Dt f )

= Dt f (ζ)

1/2µ = f (ρbt(ζ))t D E 1/2µ = f ,RSp(1)(eρt (ζ)) t b Sp(1)

D 1/2µ E = f ,t RSp(1)(eρt (ζ)) . b Sp(1)

73 This shows that R (e ◦ D ) = t1/2µ R (e ). Sp(1) ζ t Sp(1) ρbt (ζ)

Applying Proposition IV.2 to the isometric isomorphism Dt in equation (IV.18), we see that the reproducing kernels Hp(1;·,·) and Hp(t;·,·) are related as

D E Hp(t;w,W) = RSp(1)(eW ◦ Dt),RSp(1)(ew ◦ Dt) Sp(1)

D 1/2µ 1/2µ E = t RSp(1)(eρt (W)),t RSp(1)(eρt (w)) b b Sp(1)

1/µ D E = t RSp(1)(eρt (W)),RSp(1)(eρt (w)) b b Sp(1)

1/µ = t Hp (1;ρbt(w),ρbt(W)), where the last equality follows from (IV.6).

2 Proof of Theorem 7. Let TS : Hp → A (Up) be the isometric isomorphism given by

Z ∞ i2πzt TS f (z,w) = f (t,w)e dt, for (z,w) ∈ Up 0 as in Theorem 7. We begin by showing that the image RHp e(z,w) ◦ TS of the functional e(z,w) ◦ ∗ ∗ TS ∈ Hp under the Riesz map RHp : Hp → Hp is given by

−i2πzt RHp e(z,w) ◦ TS (t,ζ) = 4πtHp(t;w,ζ)e .

n For each (z,w) ∈ Up, consider the map χz,w : (0,∞) × C given by

i2πzt n χz,w(t,ζ) = 4πtHp(t;w,ζ)e , for (t,w) ∈ (0,∞) × C .

Letting z = x + iy, we obtain

Z ∞ Z e−4π p(ζ)t kχ k = |χ (t,ζ)|2 dV(ζ) dt z,w Hp z,w 0 Cn 4πt Z ∞ Z −4πyt 2 −4π p(ζ)t = 4πte Hp(t;w,ζ) e dV(ζ) dt 0 Cn

74 Z ∞ Z −4yt 2 2/µ −4π p(ρt (ζ)) = 4π te Hp(1;ρbt(w),ρbt(ζ) t e b dV(ζ) dt (From lemma IV.7) 0 Cn Z ∞ Z 1+1/µ −4πyt 2 −4π p(ρt (ζ)) 1/µ = 4π t e Hp(1;ρbt(w),ρbt(ζ) e b t dV(ζ) dt 0 Cn Z ∞ Z 1+1/µ −4πyt 2 −4π p(ρt (ζ)) = 4π t e Hp(1;ρbt(w),ρbt(ζ) e b dV(ρbt(ζ)) dt 0 Cn Z ∞ 1+1/µ −4πyt 2 = 4π t e Hp(1;ρbt(w),·) S (1) dt 0 p Z ∞ 2 1+1/µ −4πyt = 4π Hp(1;ρbt(w),·) S (1) t e dt p 0 < ∞

R ∞ 1+1/µ −4πyt because Hp(1;ρbt(w),·) ∈ Sp(1) and the integral 0 t e dt converges (as y > 0). This shows that χz,w ∈ Hp and we have

Z ∞ Z −4π p(ζ)t 2πizt e h f , χz,wiHp = f (t,ζ) 4πtHp(t,w,ζ)e dV(ζ) dt (IV.20) 0 Cn 4πt Z ∞ Z −4π p(ζ)t i2πzt = f (t,ζ)Hp(t;w,ζ)e dV(ζ) e dt 0 Cn Z ∞ = f (t,w)ei2πzt dt. (IV.21) 0

Here the integral on the left hand side of (IV.20) is ﬁnite, as it is the inner product of two functions in Hp. Thus, by Fubini’s theorem we were justiﬁed in evaluating the double integral on the left hand side of (IV.20) iteratively. The equation (IV.21) follows by the reproducing property of the kernel Hp(t;w,ζ). This shows that the map RHp e(z,w) ◦ TS is given by

−2πizt RHp e(z,w) ◦ TS (t,ζ) = χz,w(t,ζ) = 4πtHp(t;w,ζ)e .

We now apply Proposition IV.2 to the isometric isomorphism TS above to obtain equation (I.25). By Lemma (IV.2), we get

K(z,w;Z,W) = RH e(Z,W) ◦ TS ,RH e(z,w) ◦ TS p p Hp Z ∞ Z −4π p(ζ)t i2πzt −2πiZt e = 4πtHp(t;w,ζ)e 4πtHp(t;W,ζ)e dV(ζ) dt 0 Cn 4πt 75 Z ∞ Z i2π(z−Z)t −4π p(ζ)t = 4π te Hp(t;w,ζ)Hp(t;W,ζ)e dV(ζ) dt 0 Cn Z ∞ i2π(z−Z)t = 4π tHp(t;w,W)e dt, 0 where we used (IV.8) to obtain the last equality.

IV.5. Another Formula for the Bergman Kernel of Up

2 Proof of Theorem 8. Let TV : Xp → A (Up) be the isometric isomorphism given by

Z w w z2πit T f (z,w) = f t, 1 ,··· , n dt, V 1/2m 1/2m 1+1/2µ R z 1 z n z

n as in Theorem 4, where z ∈ C and w ∈ C . For each (z,w) in Up, we ﬁrst show that the image of ∗ the functional e(z,w) ◦ TV ∈ Xp under the Riesz map RXp is given by

−2πit z RX e ◦ TV (t,ζ) = Xp(t;w,ζ) . p (z,w) z1+1/2µ

For TV as above, we represent the map e(z,w) ◦ TV : Xp → C in two different ways. One one hand,

e(z,w) ◦ TV f = TV f (z,w) Z w w z2πit = f t, 1 ,··· , n dt 1/2m 1/2m 1+1/2µ R z 1 z n z i2πt Z Z ζ1 ζn z = f t, ,··· , Xp(t;w,ζ)λ(p(ζ),t) dV(ζ) dt, 1/2m1 1/2mn 1+1/2µ R Bp z z z (IV.22) where the last equality follows from the reproducing property of the kernel Xp(t;·,·). On the ∗ other hand RXp e(z,w) ◦ TV , the image of the functional e(z,w) ◦ TV ∈ Xp under the Riesz map ∗ RXp : Xp → Xp is given by

e(z,w) ◦ TV f = f ,RX e(z,w) ◦ TV p Xp Z Z ζ1 ζn = f t, ,..., RXp e(z,w) ◦ TV (t,ζ)λ(p(ζ),t) dV(ζ) dt. 1/2m1 1/2mn R Bp z z (IV.23)

76 The integral in equation (IV.23) is ﬁnite as it represents an inner product of two functions in Xp and hence maybe evaluated iteratively. We claim that 2πit z Xp(t;w,ζ) = R e ◦ TV (t,ζ). z1+1/2µ Xp (z,w)

Let φ : R × Bp → C be the difference of the two quantities above, i.e.,

2πit z φ(t,ζ) = Xp(t;w,ζ) − R e ◦ TV (t,ζ). z1+1/2µ Xp (z,w)

It then follows from equations (IV.22) and (IV.23) that the iterated integral

Z Z ζ ζ f t, 1 ,··· , n φ(t,ζ)λ(p(ζ),t) dV(ζ) dt = 0, (IV.24) 1/2m1 1/2mn R Bp z z for all f ∈ Xp.

Let g(t,ζ) = g1(t)q(ζ), where g1 ∈ Cc(R) and q is a polynomial. Since g1 is compactly supported and q is a continuous function on Bp, g ∈ Xp and it follows from equation (IV.24) that

Z Z 0 = g(t,ζ)φ(t,ζ)λ(p(ζ),t) dV(ζ) dt R B Z Z = g1(t) q(ζ)φ(t,ζ)λ(p(ζ),t) dV(ζ) dt. (IV.25) R Bp

Since equation (IV.25) holds for all continuous compactly supported g1, we must have for almost every t ∈ R, Z q(ζ)φ(t,ζ)λ(p(ζ),t) dV(ζ) = 0, Bp

Since q is a polynomial and since polynomials are dense in Qp(t) for all t ∈ R, (by Proposition II.9) it follows that for almost all t ∈ R,

φ(t,ζ) ≡ 0 on Bp, i.e., for almost all t,ζ ∈ R × Bp we have

i2πt z Xp(t;w,ζ) = R e ◦ TV (t,ζ). z1+1/2µ Xp (z,w) 77 Applying Proposition (IV.2) to the isometric isomorphism TV above, we get

K(z,w;Z,W) = RX e(Z,W) ◦ TV ,RX e(z,w) ◦ TV p p Xp ! Z Z z2iπt Z−2πit = X (t;w,ζ) X (t;W,ζ) λ(p(ζ),t) dV(ζ) dt p 1+1/2µ p 1+1/2µ R Bp z Z Z Z z2πitZ−2πit = X (t;w,ζ)X (t;W,ζ)λ(p(ζ),t) dV(ζ) dt p p 1+1/2µ R Bp (zZ) Z z2πitZ−2πit = X (t;w,W) dt, p 1+1/2µ R (zZ) where we used (IV.8) to obtain the last equality.

IV.6. An Application: the Siegel Upper Half Space In this section, we apply Theorem 7 to compute the Bergman kernel for the Siegel Upper

n+1 2 Half space Un+1 in C . We restrict our attention to the polynomial p(w) = |w| , which is a n weighted homogeneous balanced polynomial with respect to the tuple m = (1,...,1) ∈ N .

n n Lemma IV.8. Let the weighted homogeneous balanced polynomial p : C → C be given by 2 n n p(w) = |w| for w ∈ C . Let Sn(1) be the weighted Bergman space on C with weight w 7→ −4π|w|2 n e . Let Hn(1;·,·) be the reproducing kernel for Sn(1). Then for w,W ∈ C , we have

n 4πw·W Hn(1;w,W) = 4 e , (IV.26)

n n where w ·W = ∑ j=1 w jW j is the Hermitian inner product of w,W in C .

Remark. The weighted Bergman space Sn(1) above is called the Segal-Bargmann space, which arises in connection to quantum mechanics (see [9] for instance).

n n α Proof. For a multi-index α ∈ N , and w ∈ C , let qα (w) = w . If α 6= β, then the monomials qα and qβ are orthogonal to each other in Sn(1). We now compute the Sn(1) norm of qα .

Z 2 kq k2 = wα wα e−4π|w| dV(w) α Sn(1) Cn 78 Z 2 Z 2 2α1 −4π|w1| 2αn −4π|wn| = |w1| e dV(w1) ··· |wn| e dV(wn) C C

Thus, to compute the norm of qα it sufﬁces to evaluate the following integral

∞ Z 2 Z 2α 2 2α j −4π|w j| j −4πr j w j e dV(w j) = 2π r j e r j dr j (Using Polar co-ordinates) C 0 Z ∞ α 1 u j −u 2 = e du (Letting u = −4πr j ) 4 0 4π 1 = Γ(α j + 1) 4 · (4π)α j α ! = j . 4 · (4π)α j

Thus, it follows that 2 α1! · α2!···αn! kqα kS (1) = . n 4n · (4π)α1 · (4π)α2 ···(4π)αn

Since polynomials are dense in Sn(1) by Theorem II.7, it follows that the set ( ) qα kqα kS (1) n α∈Nn n is then an orthonormal basis for Sn(1). Then by Lemma IV.2, we see that for w,W ∈ C , the reproducing Kernel Hn(1;·,·) for Sn(1) is given by

q (w) q (W) H (1;w,W) = α · α n ∑ kq k kq k α∈Nn α Sn(1) α Sn(1) √ α √ √ α √ αn 2n · (2 π)α1 w 1 ···(2 π)α1 wαn 2n · (2 π)α1W 1 ···(2 π)α1W = ∑ √ 1 n · √ 1 n n α !···α ! α !···α ! α∈N 1 n 1 n (4πw W )α1 ···(4πw W )αn = 4n ∑ 1 1 n n n α !···αn! α∈N 1 = 4ne4πw1W 1 ···e4πwnW n

= 4ne4πw·W .

79 Now using Lemma IV.7 along with Theorem 7 gives us the following well known formula for the Bergman kernel of the Siegel Upper half space Un+1.

n+1 Theorem 12. Let Un+1 ⊂ C be the Siegel Upper Half space given by

n 2 Un+1 = {(z,w) ∈ C × C | Imz > |w| }, and let K be its Bergman kernel. Then, for (z,w),(Z,W) ∈ Un+1, we have " #−n−2 (n + 1)! i n K(z,w;Z,W) = n+1 (Z − z) − ∑ w jW j . 4π 2 j=1

Proof. Recall that the balanced weighted homogeneous polynomial p(w) = |w|2 corresponds to

n n the tuple m = (1,...,1). Then for t > 0, using the notation of Lemma IV.7, the map ρbt : C → C becomes,

√ 1/2m1 1/2mn 1/2 1/2 n ρbt(w) = (t w1,...,t wn) = (t w1,...,t wn) = tw, for w ∈ C .

In addition to this, we see that 1/µ = 1/m1 + ··· + 1/mn = n. Thus, by Lemma IV.7 we see that the reproducing kernels Hn(t;·,·) and Hn(1;·,·) for weighted Bergman spaces Sn(t) and Sn(1) respectively are related as

1/µ n Hn(t;w,W) = t Hn(1;ρbt(w),ρbt(W)) (for w,W ∈ C ) n √ √ = t Hn(1; tw, tW)

= 4ntne−4πtw·W (By Lemma IV.8)

For (z,w),(Z,W) ∈ Un+1, Theorem 7 gives the representation for Bergman Kernel K of Un+1 as

Z ∞ i2π(z−Z)t K(z,w;Z,W) = 4π Hn(t;w,W)e dt 0 Z ∞ = 4n+1π tn+1eαt dt, (IV.27) 0

80 where α = 2πi(z − Z) + 4πw ·W. Note that we have

αt 2πi(z−Z)t 4πtw·W e = e e = e−2π(Imz+ImZ)te4πtRe(w·W)

≤ e−2π(Imz+ImZ)te4πt|w·W|

≤ e−2π(Imz+ImZ)te4πt|w||W| (By Cauchy Schwarz Inequality)

2 2 ≤ e−2π(Imz+ImZ)te2πt(|w| +|W| ) (By AM-GM inequality)

2 2 = e−2πt(Imz−|w| )e−2πt(ImZ−|W| ).

2 2 αt Since (z,w),(Z,W) ∈ Un+1, we have Imz−|w| > 0 and ImZ −|W| > 0. This shows that e → 0 as t → ∞ and also that the integral in (IV.27) converges for all (z,w),(Z,W) ∈ Un+1. Thus, we obtain ∞ Z ∞ tn+1eαt (n + 1) Z ∞ tn+1eαt dt = − tneαt dt 0 α α 0 0 (n + 1) Z ∞ = − tneαt dt, α 0 where the ﬁrst term vanishes because eαt → 0 as t → ∞. Integrating by parts successively in this manner, we obtain

Z ∞ K(z,w;Z,W) = 4n+1π tn+1eαt dt 0 (−1)n(n + 1)! = 4n+1π αn+2 n+2 n+1 (−1) (n + 1)! = 4 π n+2 2πi(z − Z) + 4πw ·W 4n+1π (−1)n+2(n + 1)! = 4n+2πn+2 i n+2 2 (z − Z) + w ·W " #−n−2 (n + 1)! i n = n+1 (Z − z) − ∑ w jW j . 4π 2 j=1

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