diposit.ub.edudiposit.ub.edu/dspace/bitstream/2445/125123/2/memoria.pdfContents Abstract iii...

Treball final de grau

GRAU DE MATEMÀTIQUES

Facultat de Matemàtiques i InformàticaUniversitat de Barcelona

ZEROS OF RANDOM ANALYTICFUNCTIONS

Autor: Alexis Arraz Almirall

Director: Dr. Joaquim Ortega Cerdà

Realitzat a: Departament de Matemàtiques i Informàtica

Barcelona, 27 de juny de 2018

Contents

Abstract iii

Acknowledgements v

Introduction vii

1 Gaussian analytic functions 11.1 Complex Gaussian distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Gaussian analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Isometry-invariant zero sets 112.1 The complex plane C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The sphere S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 The Hyperbolic Plane D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.4 The Paley - Wiener space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Distribution and intensity of zeros of a GAF 253.1 The Edelman - Kostlan formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.1.1 The Fock space in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.1.2 The space of polynomials in S2 . . . . . . . . . . . . . . . . . . . . . . . 343.1.3 The weighted Bergman space in D . . . . . . . . . . . . . . . . . . . . . 353.1.4 The Paley - Wiener space with NC(0, 1) random variables . . . . . . . 353.1.5 The Paley - Wiener space with NR(0, 1) random variables . . . . . . . 36

3.2 Calabi’s rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 GAF computation 414.1 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2

FL. . . 41

4.2 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2PL

. . . 444.3 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2

BL. . . 45

4.4 GAF in the finite Paley - Wiener space . . . . . . . . . . . . . . . . . . . . . . . 48

5 Annex 515.1 Preliminary explanations and concepts . . . . . . . . . . . . . . . . . . . . . . 51

i

5.2 Zeros of a GAF in the finite space of polynomial endowed with the norm‖ · ‖2

FL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.3 First intensity of a GAF in the finite space of polynomials endowed with thenorm ‖ · ‖2

FL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.4 Zeros of a GAF in the finite space of polynomials endowed with the norm‖ · ‖2

PL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.5 Zeros of a GAF in the finite space of polynomials endowed with the norm‖ · ‖2

BL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.6 First intensity of a GAF in the finite space of polynomial endowed with thenorm ‖ · ‖2

BL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.7 Zeros of a GAF in the finite Paley - Wiener space with NR(0, 1) randomvariables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.8 First intensity of a GAF in the finite Paley - Wiener space with NR(0, 1) ran-dom variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.9 First intensity of a GAF in the finite Paley - Wiener space with NC(0, 1) ran-dom variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Bibliography 69

Abstract

In this project we deal with random analytic functions. Here we specifically use Gaus-sian analytic functions. Without technicalities, a GAF f (for short) is a random holomorphicfunction on a region of C such that ( f (z1), ..., f (zn)) is a random vector with normal distri-bution. One way to generate them is using linear combinations of holomorphic functionswhose coefficients are Gaussian random variables in C (or in R in special cases). For find-ing the zero set of a GAF we work on four isometric - invariant Hilbert spaces of analyticfunctions: the Fock space in C, the finite space of polynomials in S2, the weighted Bergmanspace in D and the Paley - Wiener space. The first intensity determines the average of thedistribution of the zero set of a GAF, and the Edelman - Kostlan formula gives an explicitexpression of it. A result of uniqueness, called Calabi’s Rigidity, concludes that the firstintensity determines the distribution of the zero set of a GAF. At the end, some examplesmade in C++ and gnuplot clarify the theory in these Hilbert spaces.

2010 Mathematics Subject Classification. 30B20, 30C15, 60G55

iii

iv Abstract

Acknowledgements

To my advisor, Dr. Joaquim Ortega Cerdà, for his weekly dedication and endless pa-tience.

v

vi Acknowledgements

Introduction

Most of the physical phenomena that we witness every day can be modeled by randomprocesses. One of them could be the arrival of people in a queue of a supermarket or thedistribution of trees in a forest. This effort to model phenomenons using probability theoryalso includes the physic of quantum particles. In the last century, one of the most rele-vant and challenging problems was creating point processes to simulate the distribution offermions, particles characterized by its repulsion. The Poisson process was one of the can-didates, but it was discarded because it does not have any repulsion behavior. Then, it wasmandatory to generate a random point process that had an anti-clumping behavior, that itwas indifferent where to study this process and, obviously, that it simulates the distribu-tion of the particles described before. For this, random analytic functions were consideredin determined spaces of functions. The advantage after assuming all the properties aboveand some concepts explained in this project was that the random process is formed by thepoints where a random analytic function f is zero. This last type of functions can be, forexample, of the form:

f (z) = anzn + an−1zn−1 + ... + a1z + a0,

where (an)n are random variables, for all n ∈N∪ ∞.In Chapter 1 we see that the random variables or vectors to consider are Gaussian ones inC, and these are useful to define the functions that are the keystone of this project, calledGaussian Analytic Functions (or GAF, for short). At the end of this chapter, we give thegeneral expression of the covariance kernel in a Hilbert space.In Chapter 2 we introduce the main Hilbert spaces to study the zero sets of a GAF, andthose are the Fock space in C, the space of polynomials of finite degree in S2, the weightedBergman space in D and the space of Paley - Wiener. We prove that these spaces haveinteresting isometric properties with regard to Möbius transformations.In Chapter 3 we deal with the first intensity for a point process, and the Edelman - Kost-lan formula is one of the most important results of this project, because it allows to us tounderstand how the zero set of a GAF is distributed. Finally, a uniqueness theorem calledCalabi’s Rigidity establishes that the first intensity determines the distribution of the zeroset of a GAF.In Chapter 4 we put theory into practice. C++ coding has been used here, and graphicsmade with gnuplot clarify the explanations. Here, we consider finite Hilbert spaces and wecompute the zero set and the first intensity of a GAF in these spaces.In the Annex, a brief theory section precedes the C++ programs used in the last chapter.

vii

viii Introduction

In the execution of this project, Zeros of Gaussian Analytic Functions and Determinantal PointProcesses by John Ben Hough, Manjunath Krishnapur, Yuval Peres and Bálint Virág [2] hasbeen the step-by-step book that has guided me through these pages, and the majority ofthe results and definitions has been extracted from there. Other few results and proce-dures in Chapter 3 are from the magnificent paper Zeroes of Gaussian Analytic Functions withtranslation - invariant distribution, by Naomi Feldheim [8]. Also the book Anàlisi complexa,by Joaquim Bruna and Julià Cufí [4] has been an excellent and helpful source. For thecomputing section I used C++ and the graphic program gnuplot.

Chapter 1

Gaussian analytic functions

1.1 Complex Gaussian distribution

Before introducing this section, we should establish some terminology. Let be X arandom variable. We say that X follows a normal distribution if its density function is

f (x) =1√2πσ

e−(x−µ)2

2σ2 ,

where µ ∈ R and σ ∈ (0,+∞). We denote X ∼ NR

(µ, σ2) when the random variable X

follows a normal distribution. In addition, µ represents the mean of the random variableand σ2 its variance.

Definition 1.1. A standard complex Gaussian is a complex-valued random variable with densityfunction

f (z) =1π

e−|z|2

with regard to the Lebesgue measure on the complex plane and z is a complex value.

To define the latter density function we may write the random variable Z as X + iY,where X and Y are independent identically distributed (i.i.d. for short) random variableswith NR

(0, 1

2

)distribution. Indeed, considering the random vector (X, Y) we have that

fX(x) =1√π

e−x2

andfY(y) =

1√π

e−y2.

Using the fact that X and Y are independent, we obtain:

f(X,Y)(x, y) = fX(x) fY(y) =1π

e−(x2+y2).

Thus, if z = x + iy, we have that |z|2 = x2 + y2, and the density function is

fZ(z) =1π

e−|z|2.

1

2 Gaussian analytic functions

Definition 1.2. Let Zk be an i.i.d. standard complex Gaussian random variables, where 1 ≤ k ≤ n.Then Z := (Z1, ..., Zn)

t is a standard complex Gaussian random vector. If M is a complexmatrix of dimension m× n, then W = MZ + µ is an m-dimensional complex Gaussian vector withmean µ, which is an m× 1 vector, and covariance Σ = MM∗, which is an m× n matrix and M∗ isthe conjugate transpose of M. We denote the distribution of W as Nm

C (µ, Σ).

By the last definition, the density function of a standard complex Gaussian vector Z is:

fZ(z) = f(Z1,...,Zn)(z1, ..., zn) =n

∏k=1

fZk(zk) =1

πn

n

∏k=1

e−zkzk =1

πn e−‖z‖2,

where z = (z1, ..., zn)t is a complex vector of n components. Recall that the second equalityof the formula above is a consequence of the independence of the random components ofthe vector Z.Following the book [11] (page 77, chapter 3), if Z follows a Nn

C (µ, Σ) distribution, where

µ := E [Z]

andΣ := E

[(Z− µ)(Z− µ)t] ,

then Z has density function with regard to the Lebesgue measure on C:

fZ(z) =1

πn1

det(Σ)e−(Z−µ)tΣ−1(Z−µ),

remarking that z is an n-dimensional complex vector, Σ is regular, (Z − µ)tΣ−1(Z − µ) isreal and fZ is a real-valued function.

The following results can be applied to complex Gaussian random variables or vectors,but we will prove them in the real case, because, as we saw before, a complex Gaussianrandom variable is a two-component random vector, and those components are the realand imaginary part of the complex random variable. Moreover, each of them has a realGaussian distribution and they are independent. To get the main results and proofs inmultidimensional normal distribution, see pages from 126 to 130 in [14].

Proposition 1.3. Let Zn be a random variable that has NR(µn, Σn) distribution. Then, Zn con-verges in distribution to a random variable Z with NR(µ, Σ) distribution if and only if the sequences(µn)n and (Σn)n converge respectively to µ and Σ. In other words, weak limits of real Gaussiansare real Gaussians.

Proof. Each element of (Zn)n has NR(µn, Σn) distribution. Then, the characteristic func-tion of Zn is, for all s of R:

ϕZn(s) = eisµn− 12 s2Σn .

If we suppose that (µn)n and (Σn)n converge respectively to µ and Σ, we get:

ϕZn(s) = eisµn− 12 s2Σn n→+∞−→ ϕZ(s) = eisµ− 1

2 s2Σ.

1.1 Complex Gaussian distribution 3

The characteristic function is continuous at the point 0. By Paul-Lévy’s theorem, the se-quence (Zn)n converges in distribution to Z and this last random variable has NR(µ, Σ)distribution.For the converse, if the sequence of random variables (Zn)n converges in distribution tothe random variable Z, and this last one has NR(µ, Σ) distribution, then their characteristicfunctions converge; that is:

ϕZn(s) = eisµn− 12 s2Σn n→+∞−→ ϕZ(s) = eisµ− 1

2 s2Σ,

for all real s. Therefore the sequences (µn)n and (Σn)n respectively converge to µ and Σ.

Thus we have the following result with complex Gaussian.

Proposition 1.4. Let Zn be a random variable that has NC(µn, Σn) distribution. Then, Zn convergesin distribution to a random variable Z with NC(µ, Σ) distribution if and only if the sequences (µn)n

and (Σn)n converge respectively to µ and Σ. In other words, weak limits of complex Gaussians arecomplex Gaussians.

Proposition 1.5. Let Z be an n-dimensional random vector that has NR(µ, Σ) distribution. If A isa matrix of order m× n, then the random vector AZ has NR(Aµ, AΣAt) distribution.

Proof. For all m-dimensional real vector s we have:

ϕAZ(s) = ϕZ(Ats) = eist Aµ− 12 (Ats)tΣ(Ats) = eist(Aµ)− 1

2 st(AΣAt)s.

Then AZ has NR(Aµ, AΣAt) distribution, because it is the characteristic function of a ran-dom vector that has NR(Aµ, AΣAt) distribution.

The last proposition assures us that if Z is an n-dimensional random vector such thatall linear combination of its components has normal distribution, then Z has a multidimen-sional normal distribution.With this, we have the analogous result for complex random vectors with the same remark.

Proposition 1.6. Let Z be an n-dimensional random vector that has NC(µ, Σ) distribution. If A isa matrix of order m× n, then the random vector AZ has NC(Aµ, AΣA∗) distribution.

Proposition 1.7. The mean and the covariance of a real Gaussian random vector determines itsdistribution.


Proof. Let Y be n - dimensional random vector such that Y = AX +µ, where A is a matrix oforder n, µ an n - dimensional vector and X a random vector that has NR(0, Id) distribution.We want to see that Y has a normal distribution with mean µ and covariance matrix Σ =

AAt. Using characteristic functions, we have, for any n - dimensional real vector s:

ϕY(s) = ϕAX+µ(s) = E[eist(AX+µ)

]= eistµ ϕAX(s)

(∗)= eistµe−

12 st AIdAts = eistµ− 1

2 stΣs,

where in (∗) we use Proposition 1.5. Then Y has a NR(µ, Σ) distribution.

Thus, by this proposition we have:

Proposition 1.8. The mean and the covariance of a complex Gaussian random vector determines itsdistribution.

1.2 Gaussian analytic functions

To introduce and prove all the statements of this section we must consider the spaceof analytic functions on a region Ω ⊂ C with the topology of uniform convergence oncompact sets. Let be A(Ω) the space before described. Now, given f , g of C(Ω) (the spaceof continuous functions in Ω) and following [7], we can define the norm:

dK( f , g) :=‖ f − g ‖L∞(K):= supz∈K| f (z)− g(z)|,

and the metric:

ρ( f , g) :=+∞

∑n=1

12n

dKn( f , g)1 + dKn( f , g)

,

where K and Kn are compact sets of Ω. In addition, it exists a sequence of compact sets

(Kn)n of Ω such that Ω =⋃+∞

n=1 Kn and Kn ⊂

Kn+1 ⊂ Kn+1.We have the next:

Theorem 1.9. The space (A(Ω), ρ) is a separable, complete and metric space.

For the proof, Corollary 2.3, page 152, in [7] gives us that (A(Ω), ρ) is a completemetric space since (C(Ω), ρ) also is (for the last statement see Proposition 1.12. in [7], page145). Separability is accomplished in disks or annulus due to the uniform convergence oncompact sets of Ω of the Taylor series and Laurent ones, respectively. For other spaces,Runge’s theorem is required (see page 198 in [7]).

Definition 1.10. Let f be a random variable on a probability space taking values in the space ofanalytic functions on a region Ω of C. We say that f is a Gaussian analytic function (or GAF, forshort) on Ω if the random vector ( f (z1), ..., f (zn)) has a mean zero complex Gaussian distributionfor every n ≥ 1 and every z1, ..., zn of Ω. In addition, the components of the last random vector arenot necessarily independent.

1.2 Gaussian analytic functions 5

As a remark, for any z1, ..., zn of Ω for all n ≥ 1, the random vector ( f (z1), ..., f (zn)) hasNn

C(0, Σ) distribution, where Σ is the covariance matrix K(zi, zj) for all i, j ≤ n. Then, byProposition 1.8, the covariance kernel K determines the distribution of f .An interesting question to formulate is how can we generate Gaussian analytic functions.The next result addresses this problem.

Proposition 1.11. Let ( fn)n be a sequence of analytic functions in Ω and let (ζn)n be i.i.d. standardcomplex Gaussian random variables, that is, each ζn follows a NC(0, 1) distribution. If ∑+∞

n=1 | fn(z)|2converges uniformly on compact sets on Ω, then f (z) = ∑+∞

n=1 ζn fn(z) converges uniformly almostsurely on compact sets on Ω and f defines a GAF. Furthermore, f has covariance kernel K f (z, w) =

∑+∞n=1 fn(z) fn(w).

Lemma 1.12. (Kolmogorov’s inequality) Under the same hypothesis than the last proposition,and assuming that K is a compact subset of Ω and the random variable Xn = ∑n

k=1 ζk fk is definedin the Lebesgue space L2(K), it holds the inequality:

P

(sup

1≤j≤n‖ Xj ‖L2(K)≥ ε

)≤ 1

ε2

n

∑j=1‖ f j ‖2

L2(K),

for a given ε > 0.

Proof. (see proof of Lemma 2.2.3. in [2], pages 16 and 17) We define, for all k < n, the setZ = ζ j : j ≤ k. It holds that:

E[‖ Xn ‖2

L2(K) | Z]=‖ Xk ‖2

L2(K) +n

∑j=k+1

‖ f j ‖2L2(K) .

Indeed,

E[‖ Xn ‖2

L2(K) | Z]

(∗)= E

[n

∑j=1‖ ζ j f j ‖2

L2(K) | Z

](1.1)

= E

[k

∑j=1‖ ζ j f j ‖2

L2(K) +n

∑j=k+1

‖ ζ j f j ‖2L2(K) | Z

](1.2)

= E

[k

∑j=1‖ ζ j f j ‖2

L2(K) | Z

]+ E

[n

∑j=k+1

‖ ζ j f j ‖2L2(K) | Z

](1.3)

=k

∑j=1‖ ζ j f j ‖2

L2(K) +n

∑j=k+1

‖ f j ‖2L2(K) E

[|ζ j|2

]︸︷︷︸1

(1.4)

=k

∑j=1‖ ζ j f j ‖2

L2(K) +n

∑j=k+1

‖ f j ‖2L2(K) (1.5)

=‖ Xk ‖2L2(K) +

n

∑j=k+1

‖ f j ‖2L2(K) . (1.6)


In (∗) we use the independence of (ζn)n. The first sum of (1.4) is due to the next property:if Y is an integrable random variable, then E [Y|Y] = Y. The second sum of the same lineis given because the sequence (ζ j)j>k is independent of the sequence (ζ j)j≤k. Now, definingthe stopping time τ = infn ∈N : ‖ Xn ‖L2(K)> ε, we have that:

E[‖ Xn ‖2

L2(K)

]≥

n

∑k=1

E[‖ Xn ‖2

L2(K) 1τ=k

]=

n

∑k=1

E[E[‖ Xn ‖2

L2(K) 1τ=k | Z]]

(1.7)

=n

∑k=1

E

[(‖ Xk ‖2

L2(K) +n

∑j=k+1

‖ f j ‖2L2(K)

)1τ=k

](1.8)

=n

∑k=1

E[‖ Xk ‖2

L2(K) 1τ=k

]+ E

[n

∑j=k+1

‖ f j ‖2L2(K) 1τ=k

]︸︷︷︸

≥0

(1.9)

≥n

∑k=1

E[‖ Xk ‖2

L2(K) 1τ=k

]≥

n

∑k=1

ε2E[1τ=k

]= ε2

n

∑k=1

P (τ = k) (1.10)

= ε2P

(n⊎

k=1

τ = k)

= ε2P (τ ≤ n) . (1.11)

The equality of line (1.7) maintains because, given integrable random variables X, Y, itsatisfies the equation E [E [X|Y]] = E [X]. Thus,

P

(sup

1≤j≤n‖ Xj ‖L2(K)≥ ε

)≤ 1

ε2

n

∑j=1‖ f j ‖2

L2(K) .

Proof of Proposition 1.11 (see proof of Lemma 2.2.3. in [2], pages from 16 to 18). First ofall, we must prove that Xn is a Cauchy sequence in L2(K) almost surely. For this, we cansee that, for a given natural n0 such that n, m ≥ n0:

P

[sup

n,m≥n0

‖ Xm − Xn ‖L2(K)≥ 2ε

]n0→+∞−→ 0.

Indeed, take the sequence (Xn0+n − Xn0)n. We have

P

[sup

n,m≥n0

‖ Xm − Xn ‖L2(K)≥ 2ε

]≤ P

[supn≥1‖ Xn0+n − Xn0 ‖L2(K)≥ ε

]

= limk→∞

P

[sup

1≤n≤k‖ Xn0+n − Xn0 ‖L2(K)≥ ε

](∗)≤ lim

k→∞

1ε2

n0+k

∑j=n0+1

‖ f j ‖2L2(K)=

1ε2

+∞

∑j=n0+1

‖ f j ‖2L2(K),


which the last expression tends to zero when n0 tends to infinite and the last lemma isused in (∗). Therefore it exists a natural number n0 such that for all n ≥ n0, the sequence(Xn0+n − Xn0)n satisfies

P[‖ Xn0+n − Xn0 ‖L2(K)≤ ε

]= 1,

and Xn is a Cauchy sequence in L2(K). An important remark is that K is not fixed, it can beany compact set of Ω. Indeed, by an exhaustive sequence of compact sets of Ω, the compactKn contains the sequence of compact sets (Ki)i<n for all natural n. If we apply the result wejust proved to Kn, the property remains true with any compact of the sequence (Ki)i<n forall natural n.For the uniform convergence on compact sets of Ω, we have to consider the disk D(z0, 4R)on Ω. Since the sequence ( fn)n is analytic, Xn is also analytic on Ω for all natural n.Therefore, by Cauchy’s integral formula:

Xn(z) =1

2πi

∫|w−z0|=r

Xn(w)

w− zdw,

where w = z0 + reiθ , |z− z0| < r, 0 ≤ θ ≤ 2π and 2R < r < 3R. For a given z of the diskD(z0, R), consider the annulus A = A(z0; 2R, 3R). Then

Xn(z) =1

2πi

∫|w−z0|=r

Xn(w)

w− zdw =

12πiR

∫ 3R

2R

∫|w−z0|=r

Xn(w)

w− zdwdr

=1

2πiR

∫ 3R

2R

∫ 2π

0

Xn(z0 + reiθ)

z0 − z + reiθ ireiθdθdr =1

2π

∫A

Xn(w)φz(w)dm(w),

where dm(w) = rdθdr and

φz(w) =eiθ

R(w− z).

Observe that the collection φzz is uniformly bounded in the Lebesgue space L2(A). In-deed:

‖ φz(w) ‖2L2(A) =

∫A

1R2

1|w− z|2 dm(w) =

1R2

∫ 3R

2R

∫ 2π

0

r|w− z|2 dθdr

≤ 1R2

∫ 3R

2R

∫ 2π

0

rR2 dθdr =

2π

R4

∫ 3R

2Rrdr =

2π

R45R2

2=

5π

R2 ,

where the inequality above is given by |z0− z + reiθ |2 ≥ R2, recalling the fact that z belongsto D(z0, R) and 2R < r < 3R. Thus

‖ φz(w) ‖L2(A)≤√

5π

R.

Now, let K be the disk D(z0, 4R). As we saw before, Xn is a Cauchy sequence in L2(K).Then it exists a random variable X of L2(K) such that ‖ X− Xn ‖L2(K) tends to zero when ntends to infinite and

12π

∫A

Xn(w)φz(w)dm(w)n→+∞−→ 1

2π

∫A

X(w)φz(w)dm(w)


uniformly for all z of D(z0, R). Indeed, for all ε > 0 there exists a natural n0 such that forall n, m ≥ n0 it holds ‖ Xn − Xm ‖L2(A)< ε. Then, we can use the last n0 to state that:

supz∈D(z0,R)

|Xn(z)− Xm(z)| ≤‖ Xn − Xm ‖L2(A)‖ φz ‖L2(A)≤ ε

√5π

R.

Since ε is arbitrary, we can tend ε to zero and the uniform convergence on D(z0, R) issatisfied.Therefore Xn converges uniformly to X on compact sets of Ω almost surely, and X is ananalytic function on Ω by Weierstrass’ theorem.By hypothesis, (ζn)n are complex Gaussian with mean zero, therefore Xn is also a complexGaussian with mean zero, by Proposition 1.6, and defines a GAF by definition. Since limitsof complex Gaussians are complex Gaussians, by Proposition 1.4, X is also a GAF.For the covariance formula, we have:

K f (z, w) = E[

f (z) f (w)]= E

[(+∞

∑n=1

ζn fn(z)

)(+∞

∑n=1

ζn fn(w)

)]

= E

[+∞

∑n=1

n

∑k=1

ζk fk(z)ζn−k fn−k(w)

]=

+∞

∑n=1

n

∑k=1

E[ζk fk(z)ζn−k fn−k(w)

]=

+∞

∑n=1

n

∑k=1

fk(z) fn−k(w)E[ζkζn−k

] (∗)=

+∞

∑m=1

fm(z) fm(w),

where in (∗) it holds the following. If k = n− k, then E[ζkζk

]= E

[|ζk|2

]= 1. Otherwise,

if k 6= n− k, then, by the independence of (ζn)n and remembering that ζn ∼ NC(0, 1), wehave that E

[ζkζn−k

]= E [ζk]E

[ζn−k

]= 0. Thus

K f (z, w) =+∞

∑m=1

fm(z) fm(w).

We notice that in Proposition 1.11 there is no restriction about which space we must deter-mine a GAF. Although, in this project we will calculate them in Hilbert spaces, concretelyin Hilbert spaces of analytic functions on a region Ω. We denote this space as H.We have the next result:

Proposition 1.13. Consider the space H as described before. Let ( fn)n be an orthonormal basis ofH and assume that for all z of Ω, there exists a positive constant Cz such that | f (z)| ≤ Cz ‖ f ‖Hfor all function f of H. Moreover, assume that the norm is continuous at z. Then ∑+∞

n=1 | fn(z)|2converges uniformly on compact sets of Ω.

Proof. For all z of Ω we consider the punctual evaluation operator at z, defined as:

Ez : A(Ω) −→ Ωf 7−→ Ez( f ) = f (z)


This operator is linear and continuous. Indeed, for any f and g of H we have

Ez( f + g) = ( f + g)(z) = f (z) + g(z) = Ez( f ) + Ez(g),

and for a complex scalar λ we get

Ez(λ f ) = (λ f )(z) = λ f (z) = λEz( f ).

The continuity is straightforward because | f (z)| ≤ Cz ‖ f ‖H. Indeed:

|Ez( f )| = | f (z)| ≤ Cz ‖ f ‖H .

Thus ‖ Ez ‖H≤ Cz and the operator is continuous.For the convergence result, we have to see that:

‖ Ez ‖2H=

+∞

∑n=1| fn(z)|2.

Since ( fn)n is an orthonormal basis, by Fischer - Riesz theorem:

f =+∞

∑n=1

( f , fn)H fn,

and Parseval’s equality assures that

‖ f ‖2H=

+∞

∑n=1|( f , fn)H|2.

We have that:

|Ez( f )|2 = | f (z)|2 =

∣∣∣∣∣+∞

∑n=1

( f , fn)H fn(z)

∣∣∣∣∣2

≤+∞

∑n=1|( f , fn)H|2

+∞

∑n=1| fn(z)|2

=‖ f ‖2H

+∞

∑n=1| fn(z)|2.

Then

|Ez( f )| ≤‖ f ‖H

(+∞

∑n=1| fn(z)|2

) 12

.

Therefore:

‖ Ez ‖2H≤

+∞

∑n=1| fn(z)|2.

For the other inequality, we consider the function g(·) = ∑+∞n=1 fn(z) fn(·). We have that

‖ g ‖2H=

+∞

∑n=1| fn(z)|2 and g(z) =

+∞

∑n=1| fn(z)|2.


Knowing the fact that ‖ Ez ‖H is the minimum constant value that holds:

|g(z)| ≤‖ Ez ‖H‖ g ‖H,

thus:+∞

∑n=1| fn(z)|2 ≤‖ Ez ‖H

(+∞

∑n=1| fn(z)|2

) 12

,

and we arrive to the other inequality:+∞

∑n=1| fn(z)|2 ≤‖ Ez ‖2

H .

Since Ez is a bounded linear functional then, by Riesz’s representation theorem, it existsa unique function Kz of H for all z of Ω such that

f (z) = ( f , Kz)H,

for all f of H. The function Kz is the reproducing kernel of H at z.Following [13] (page 375), if we start with a Hilbert space H of analytic functions on aregion Ω with reproducing kernel Kz for all z of Ω, we can compute a GAF as

f (z) =+∞

∑n=1

ζn fn(z),

where the sequence ( fn)n is an orthonormal basis of H and the elements of the sequence(ζn)n are random variables with NC(0, 1) distribution. Furthermore, the GAF has covari-ance kernel

K f (z, w) =+∞

∑n=1

fn(z) fn(w).

This construction does not depend of the chosen basis of H. For this we will prove thatKz = K f or, similarly, that ( f , Kz)H = ( f ,K f )H, for any function f of H. We know that( f , Kz)H = f (z) for z of Ω. On the other hand, for any w of Ω, we have that:

( f ,K f (z, w))H =

(f ,

+∞

∑n=1

fn(z) fn(w)

)H=

+∞

∑n=1

fn(z)( f , fn)H = f (z),

where in the last equation we use the Fischer - Riesz theorem by the fact that ( fn)n is anorthonormal basis. Therefore we got the equality we wanted.Now, as a curiosity, there is an interesting theorem in the theory of reproducing kernelson Hilbert spaces. We proved before that for a given Hilbert space we can determinethe reproducing kernel, but the converse is also true, as is explained in the paper [1] byAronszajn:

Theorem 1.14. (Moore - Aronszajn’s theorem) Let K be a hermitian, positive definite covariancekernel on a region Ω. If for a given complex values z and w in Ω, K(z, w) is holomorphic in z andanti-holomorphic in w and K(z, w) is bounded on compact sets of Ω, then there is a unique Hilbertspace H of holomorphic functions in Ω such that K is its reproducing kernel.

Chapter 2

Isometry-invariant zero sets

In this chapter we will use the concepts explained before in four Hilbert spaces offunctions on different domains, and those domains are the complex plane, the Riemannsphere and the hyperbolic plane. For each one of them we will concrete the collectionof Hilbert spaces given by a parameter, prove that the chosen basis is an orthonormalone, compute the covariance kernel and see that each GAF is invariant under suitabletransformations.For results about isometry-invariant zero sets consult Section 2.3. in Chapter 2 of [2].

2.1 The complex plane C

Fixing a real parameter L > 0, we consider the Fock space, defined as

FL :=

f ∈ A(C) : ‖ f ‖2FL=

Lπ

∫C| f (z)|2 e−L|z|2 dm(z) < +∞

.

Let us denote:

en =

√Ln√

n!zn.

Then we have that (en)+∞n=0 is an orthonormal basis of FL. We must see that (en, en)FL = 1

and (en, em)FL = 0, for all n 6= m.On one hand we have:

(en, en)FL =Lπ

∫C

Ln|z|2n

n!e−L|z|2 dm(z) = 2

Ln+1

n!

∫ +∞

0r2n+1e−Lr2

dr

=Ln+1

n!

∫ +∞

0r2ne−Lr2

2rdr(∗)=

Ln+1

n!

∫ +∞

0

tn

Ln+1 e−tdt(∗∗)=

Ln+1

n!Γ(n + 1)

Ln+1 = 1,

where in (∗) we use the change of variables t = Lr2 and dt = 2Lrdr; in (∗∗) we observethat

∫ ∞0 tne−tdt = n! = Γ(n + 1). On the other hand, for n 6= m:

(en, em)FL =Lπ

∫C

√Ln√

n!

√Lm√

m!znzme−L|z|2 dm(z)

=Lπ

√Ln√

n!

√Lm√

m!

∫ 2π

0

∫ +∞

0rn+m+1eiθ(n−m)e−Lr2

drdθ.

11

12 Isometry-invariant zero sets

However: ∫ 2π

0eiθ(n−m)dθ =

[eiθ(n−m)

i(n−m)

]θ=2π

θ=0

= 0.

Then (en, em)FL = 0. For completeness, if for a given function f of FL it holds ( f , en)FL = 0,then f ≡ 0. Following a procedure in [22], we can expand f as a Taylor power series:

f (z) =+∞

∑m=0

cmzm,

and we know that this series converges uniformly over compact subsets of C. We have:

( f , en)FL =L√

Ln

π√

n!

∫C

f (z)zne−L|z|2 dm(z)

=L√

Ln

π√

n!lim

R→+∞

∫ R

0

∫ 2π

0f (reiθ)rn+1e−inθe−Lr2

dθdr

=L√

Ln

π√

n!lim

R→+∞

∫ R

0

∫ 2π

0

+∞

∑m=0

cmrn+m+1eiθ(m−n)e−Lr2dθdr

=L√

Ln

π√

n!lim

R→+∞

+∞

∑m=0

∫ R

0

∫ 2π

0cmrn+m+1eiθ(m−n)e−Lr2

dθdr

= 2L√

Ln√

n!cn lim

R→+∞

∫ R

0r2n+1e−Lr2

dr = 0,

where the commutativity of the integrals with the sum is due to the uniform convergenceof the power series of f over the compact subsets of C. Therefore, if ( f , en)FL = 0, the valuecn is the unique one that vanishes. Hence f ≡ 0, as we wanted to see. Thus (en)

+∞n=0 is an

orthonormal basis of FL.Now by Propositions 1.13 and 1.11 we define the GAF:

f (z) =+∞

∑n=0

ζn

√Ln√

n!zn,

for all real L > 0. Then f has covariance kernel:

K f (z, w) =+∞

∑n=0

Ln

n!znwn =

+∞

∑n=0

(Lzw)n

n!= eLzw.

Once we have calculated this, we are going to see that the functions of the Fock space areinvariant under translations, and the points of the function and its translated are almost

equal in distribution, noted as d=. More precisely:

Proposition 2.1. Let f be a GAF in FL over the complex plane C. The point sets of f are invariantunder the transformation

ϕa(z) = z− a,

where z and a are complex values.

2.1 The complex plane C 13

Proof. By hypothesis:

f (z) =+∞

∑n=0

ζn

√Ln√

n!zn.

We know that the covariance kernel of f is

K f (z, w) = eLzw.

If fa(z) = f (ϕa(z)), fa has covariance kernel:

K fa(z, w) = K f (ϕa(z), ϕa(w)) = eL(z−a)(w−a) = eLzw−Lza−Law+L|a|2 .

Now, we have that:f (z) d

= fa(z)eLza− L2 |a|2 .

To show this, if we denoteTa f (z) = fa(z)eLza− L

2 |a|2 ,

we must prove thatK f (z, w) = KTa f (z, w).

Indeed:

KTa f (z, w) = K fa(z, w)eLza− L2 |a|2 eLaw− L

2 |a|2 = eLzw−Lza−Law−L|a|2 eLza+Law+L|a|2 = eLzw

= K f (z, w).

Therefore:f (z) d

= Ta f (z).

Proposition 2.2. Using the same notation as in the last proof, f and Ta f are isometric, that is:

‖ f ‖2FL=‖ Ta f ‖2

FL.

Proof. Indeed, we have that:

‖ Ta f ‖2FL=

Lπ

∫C| f (z− a)|2

∣∣∣eLaz− L2 |a|2

∣∣∣2 e−L|z|2 dm(z).

The exponential factor can be written as:∣∣∣eLaz− L2 |a|2

∣∣∣2 =(

eLaz− L2 |a|2

) (eLaz− L

2 |a|2)=(

eLaz− L2 |a|2

) (eLaz− L

2 |a|2)= eLaz+Laz−L|a|2 .

Returning to the integral and applying the change w = z− a and dm(w) = dm(z):

‖ Ta f ‖2FL

=Lπ

∫C| f (z− a)|2

∣∣∣eLaz− L2 |a|2

∣∣∣2 e−L|z|2 dm(z)

=Lπ

∫C| f (w)|2eLa(w+a)+La(w+a)−L|a|2−L|w+a|2 dm(w)

=Lπ

∫C| f (w)|2eLa(w+a)+La(w+a)−L|a|2−L|w|2−Lwa−Law−L|a|2 dm(w)

=Lπ

∫C| f (w)|2e−L|w|2 dm(w) =‖ f ‖2

FL.


As a remark, let f be a function of the Fock space. If f is a GAF, Ta f is also a GAF,since it is a composition of f with the translation ϕa and multiplied by the corrector factorηa(z) = eLza− L

2 |a|2 . The factor ηa is a deterministic one. It is a term that we impose to get theequality of covariance kernels of f and Ta f and to obtain the isometric property. Moreover,the last results assures us the equality in distribution of f and Ta f for every point z of theregion Ω, and that includes the zero sets of both functions. In this case it is simple, becausesince the exponential function does not vanish at any point, f and fa have the same zerosets in distribution. In addition, the theorems to find zeros with meromorphic functionssuch that Rouché’s theorem or the Argument Principle can be also applied to Ta f . As aconclusion, the last proved results are powerful tools to find zero sets of functions. Indeed,if f is a GAF whose zeros are tough to determine, then, with a translation to the origin, forexample, we can erase this difficulty.

2.2 The sphere S2

For every natural number L, we endow the Riemann’s sphere S2 with the Hilbert spaceof polynomials:

PL :=

p ∈ PL[C] : ‖ p ‖2PL=

L + 1π

∫C

|p(z)|2(1 + |z|2)L+2 dm(z) < +∞

,

where PL[C] is the vectorial space of polynomials of degree at most L with complex coeffi-cients.Let us denote:

en =

√L(L− 1)...(L− n + 1)√

n!zn.

Then we have that (en)Ln=0 is an orthonormal basis on PL. We must prove that (en, en)PL = 1

and (en, em)PL = 0 for all n 6= m.Before facing this, we should remark some results to clarify future calculations. First of all,it holds the next equality:

L(L− 1)...(L− n + 1)n!

=Γ(L + 1)

Γ(n + 1)Γ(L− n + 1).

Second of all, the beta function is defined, for all Re(x), Re(y) > 0, as:

B(x, y) =Γ(x)Γ(y)Γ(x + y)

=∫ 1

0sx−1(1− s)y−1ds.

We have the next property:

Proposition 2.3. For all Re(x), Re(y) > 0, it satisfies:

B(x, y) =∫ 1

0sx−1(1− s)y−1ds =

∫ +∞

0

sx−1

(1 + s)x+y ds.

2.2 The sphere S2 15

Proof. Using the change of variables s = uu+1 and ds = du

(1+u)2 we get:

∫ 1

0sx−1(1− s)y−1ds =

∫ +∞

0

ux−1

(1 + u)x+y−2du

(1 + u)2 =∫ +∞

0

ux−1

(1 + u)x+y du.

Now we can return to the original problem. On one hand we have:

(en, en)PL =L + 1

π

∫C

L(L− 1)...(L− n + 1)n!

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

=L + 1

π

Γ(L + 1)Γ(n + 1)Γ(L− n + 1)

∫ 2π

0

∫ +∞

0

r2n+1

(1 + r2)L+2 drdθ

=(L + 1)Γ(L + 1)

Γ(n + 1)Γ(L− n + 1)

∫ +∞

0

r2n

(1 + r2)L+2 2rdr

(∗)=

(L + 1)Γ(L + 1)Γ(n + 1)Γ(L− n + 1)

∫ +∞

0

tn

(1 + t)L+2 dt

(∗∗)=

(L + 1)Γ(L + 1)Γ(n + 1)Γ(L− n + 1)

Γ(n + 1)Γ(L− n + 1)Γ(L + 2)

=(L + 1)Γ(L + 1)

Γ(L + 2)=

(L + 1)!(L + 1)!

= 1,

where in (∗) we use the change of variables t = r2 and dt = 2rdr. The (∗∗) equality is givenby the last proposition.On the other hand, for all n 6= m:

(en, em)PL =L + 1

π

∫C

√L(L− 1)...(L− n + 1)√

n!

√L(L− 1)...(L−m + 1)√

m!znzm

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

=L + 1

π

√L(L− 1)...(L− n + 1)√

n!

√L(L− 1)...(L−m + 1)√

m!

∫ 2π

0

∫ +∞

0

rn+m+1eiθ(n−m)

(1 + r2)L+2 drdθ,

but we already know that: ∫ 2π

0eiθ(n−m)dθ = 0.

Hence, (en, em)PL = 0. Completeness is direct, because (zn)Ln=0 generates the polynomials of

PL and forms an orthonormal basis over that space. Thus (en)Ln=0 is an orthonormal basis

of PL.To conclude this section, by Propositions 1.13 and 1.11 we define the GAF:

f (z) =L

∑n=0

ζn

√L(L− 1)...(L− n + 1)√

n!zn,

for all natural L. Then f has covariance kernel:

K f (z, w) =L

∑n=0

L(L− 1)...(L− n + 1)n!

znwn =L

∑n=0

(Ln

)znwn = (1 + zw)L.


Similarly to the last section, the point sets of functions in the Hilbert space of polynomialsof degree at most L on S2 are invariant under Möbius transformations. The next propositionguarantees this:

Proposition 2.4. Let f be a GAF in PL over the Riemann’s sphere S2. The point sets of f areinvariant under the Möbius transformation

ϕa(z) =z− a

1 + az,

where z and a are values of C.

Proof. By hypothesis, the GAF f is of the form:

f (z) =L

∑n=0

ζn

√L(L− 1)...(L− n + 1)√

n!zn,

and has covariance kernel:K f (z, w) = (1 + zw)L.

Let fa be the function:fa(z) = f (ϕa(z)).

Then fa has covariance kernel:

K fa(z, w) = K f (ϕa(z), ϕa(w)) =

(1 +

z− a1 + az

w− a1 + aw

)L

=

((1 + |a|2)(1 + zw)

(1 + az)(1 + aw)

)L

.

Now, we have that:

f (z) d= fa(z)

(1 + |a|2(1 + az)2

)− L2

.

To show this, if we denote

Ta f (z) = fa(z)(

1 + |a|2(1 + az)2

)− L2

,


Indeed:

KTa f (z, w) = K fa(z, w)

(1 + |a|2(1 + az)2

)− L2(

1 + |a|2(1 + aw)2

)− L2

=

((1 + |a|2)(1 + zw)

(1 + az)(1 + aw)

)L ( 1 + |a|2(1 + az)(1 + aw)

)−L

= (1 + zw)L = K f (z, w).

Therefore:f (z) d

= Ta f (z).

2.2 The sphere S2 17


‖ f ‖2PL=‖ Ta f ‖2

PL.


‖ Ta f ‖2PL

=L + 1

π

∫C

|Ta f (z)|2(1 + |z|2)L+2 dm(z) =

L + 1π

∫C

| fa(z)|2(1 + |z|2)L+2

∣∣∣∣ 1 + |a|2(1 + az)2

∣∣∣∣−L

dm(z).

By the change of variables

w =z− a

1 + azthis implies:

z =w + a

1− aw.

Also we have that:

dm(z) =∣∣∣∣ ∂

∂w

(w + a

1− aw

)∣∣∣∣2 dm(w) =(1 + |a|2)2

|1− aw|4 dm(w) =(1 + |a|2)2

(1− aw)2(1− aw)2 dm(w).

Therefore:

1 + |z|2 = 1 +∣∣∣∣ w + a1− aw

∣∣∣∣2 = 1 +|w|2 + aw + aw + |a|2(1− aw)(1− aw)

=(1 + |a|2)(1 + |w|2)(1− aw)(1− aw)

.

This implies that:1

(1 + |z|2)L+2 =(1− aw)L+2(1− aw)L+2

(1 + |a|2)L+2(1 + |w|2)L+2 .

Also we have:

1 + az = 1 + aw + a

1− aw=

1 + |a|21− aw

,

and then:1 + |a|2(1 + az)2 =

(1− aw)2

1 + |a|2 .

In addition: ∣∣∣∣ (1− aw)2

1 + |a|2

∣∣∣∣−L

=|1− aw|−2L

(1 + |a|2)−L =(1− aw)−L(1− aw)−L

(1 + |a|2)−L .

Hence, by the last equalities:

‖ Ta f ‖2PL

=L + 1

π

∫C

| fa(z)|2(1 + |z|2)L+2

∣∣∣∣ 1 + |a|2(1 + az)2

∣∣∣∣−L

dm(z)

=L + 1

π

∫C| f (w)|2 (1− aw)L+2(1− aw)L+2

(1 + |a|2)L+2(1 + |w|2)L+2

∣∣∣∣ (1− aw)2

1 + |a|2

∣∣∣∣−L(1 + |a|2)2

(1− aw)2(1− aw)2 dm(w)

=L + 1

π

∫C| f (w)|2 (1− aw)L+2(1− aw)L+2(1− aw)−L(1− aw)−L(1 + |a|2)2

(1 + |a|2)L+2(1 + |w|2)L+2(1 + |a|2)−L(1− aw)2(1− aw)2 dm(w)

=L + 1

π

∫C

| f (w)|2(1 + |w|2)L+2 dm(w) =‖ f ‖2

PL.


For every function f of PL, we remark that if f is a GAF, then Ta f is also a GAF becauseit is a composition with the transformation ϕa and the deterministic corrector factor

ηa(z) =(

1 + |a|2(1 + az)2

)− L2

.

Given the fact that the point sets of f and Ta f are equal in distribution, the zero sets ofthe last functions also satisfy this. Moreover, the zeros sets of fa are equal in distributionwith f . Indeed, if ηa(z) = 0, we would have |a|2 = −1, which is a contradiction since themodulus of a complex value is non-negative. As a conclusion, the zero sets of fa are thesame in distribution than f .

2.3 The Hyperbolic Plane D

Giving a real parameter L > 0 we define the weighted Bergman space over D as:

BL :=

f ∈ A(D) : ‖ f ‖2BL=

Lπ

∫D| f (z)|2(1− |z|2)L−2dm(z) < +∞

.

Let us denote:

en =

√L(L + 1)...(L + n− 1)√

n!zn.

In this case, (en)+∞n=0 is an orthonormal basis on BL. For this we must prove that (en, en)BL = 1

and (en, em)BL = 0. Before going forward, we remark that:

L(L + 1)...(L + n− 1)n!

=Γ(L + n)

Γ(n + 1)Γ(L).

Now, we have that:

(en, en)BL =Lπ

∫D

L(L + 1)...(L + n− 1)n!

|z|2n(1− |z|2)L−2dm(z)

=Lπ

Γ(L + n)Γ(n + 1)Γ(L)

∫ 2π

0

∫ 1

0r2n+1(1− r2)L−2drdθ

=LΓ(L + n)

Γ(n + 1)Γ(L)

∫ 1

0r2n(1− r2)L−22rdr

(∗)=

LΓ(L + n)Γ(n + 1)Γ(L)

∫ 1

0tn(1− t)L−2dt

(∗∗)=

LΓ(L + n)Γ(n + 1)Γ(L)

Γ(n + 1)Γ(L− 1)Γ(n + L)

= LΓ(L− 1)

Γ(L)=

Γ(L)Γ(L)

= 1,

where in (∗) we use the change of variables t = r2 and dt = 2rdr. The (∗∗) equality holdsby the beta function described in the section before.Now, for all n 6= m:

(en, em)BL =Lπ

∫D

√L(L + 1)...(L + n− 1)√

n!

√L(L + 1)...(L + m− 1)√

m!znzm(1− |z|2)L−2dm(z)

=Lπ

√L(L + 1)...(L + n− 1)√

n!

√L(L + 1)...(L + m− 1)√

m!

∫ 2π

0

∫ 1

0

rn+m+1eiθ(n−m)

(1− r2)2−L drdθ.

2.3 The Hyperbolic Plane D 19

We know that: ∫ 2π

0eiθ(n−m)dθ = 0.

Hence, (en, em)BL = 0. For completeness, if for a given function f of BL it holds ( f , en)BL = 0,then f ≡ 0. Following a procedure in [22], we can expand f as a Taylor power series:

f (z) =+∞

∑m=0

cmzm,

and this power series converges uniformly over compact subsets of D. We have:

( f , en)BL =L√

L(L + 1)...(L + n− 1)π√

n!

∫D

f (z)zn(1− |z|2)L−2dm(z)

=L√

L(L + 1)...(L + n− 1)π√

n!limR→1

∫ R

0

∫ 2π

0f (reiθ)rn+1e−inθ(1− r2)L−2dθdr

=L√

L(L + 1)...(L + n− 1)π√

n!limR→1

∫ R

0

∫ 2π

0

+∞

∑m=0

cmrn+m+1(1− r2)L−2eiθ(m−n)dθdr

=L√

L(L + 1)...(L + n− 1)π√

n!limR→1

+∞

∑m=0

∫ R

0

∫ 2π

0cmrn+m+1(1− r2)L−2eiθ(m−n)dθdr

= 2L√

L(L + 1)...(L + n− 1)√n!

cn limR→1

∫ R

0r2n+1(1− r2)L−2dr = 0,

where the commutativity of the integrals with the sum is due to the uniform convergenceof the power series of f over D. Therefore, if ( f , en)BL = 0, the value cm is the unique onethat vanishes. Hence f ≡ 0, as we wanted to see. Thus (en)

+∞n=0 is an orthonormal basis on

BL.By Propositions 1.13 and 1.11 we define the GAF:

f (z) =+∞

∑n=0

ζn

√L(L + 1)...(L + n− 1)√

n!zn,

for all real L > 0. Then f has covariance kernel:

K f (z, w) =+∞

∑n=0

L(L + 1)...(L + n− 1)n!

znwn =+∞

∑n=0

(L + n− 1

n

)znwn = (1− zw)−L,

where the last equality is given by:

+∞

∑n=0

(n + a

n

)xn =

1(1− x)a+1 ,

for real a and |x| < 1. The equality remains true since the sum term is the Taylor series ofthe function of the right side of the equality.

Likewise the other Hilbert spaces, the point sets of functions in BL on D are also invari-ant under transformations.


Proposition 2.6. Let f be a GAF in BL over the hyperbolic plane D. The point sets of f areinvariant under the transformation

ϕa(z) =z− a

1− az,

where z and a are values of D.


f (z) =+∞

∑n=0

ζn

√L(L + 1)...(L + n− 1)√

n!zn,

and has covariance kernel:K f (z, w) = (1− zw)−L.

Let fa be the function:fa(z) = f (ϕa(z)).

Then fa has covariance kernel:

K fa(z, w) = K f (ϕa(z), ϕa(w)) =

(1− z− a

1− azw− a

1− aw

)−L

=

((1− |a|2)(1− zw)

(1− az)(1− aw)

)−L

.

Now, we have that:

f (z) d= fa(z)

(1− |a|2(1− az)2

) L2

.

To show this, if we denote

Ta f (z) = fa(z)(

1− |a|2(1− az)2

) L2

,


Indeed:

KTa f (z, w) = K fa(z, w)

(1− |a|2(1− az)2

) L2(

1− |a|2(1− aw)2

) L2

=

((1− |a|2)(1− zw)

(1− az)(1− aw)

)−L ( 1− |a|2(1− az)(1− aw)

)L

= (1− zw)−L = K f (z, w).

Therefore:f (z) d

= Ta f (z).


‖ f ‖2BL=‖ Ta f ‖2

BL.

2.3 The Hyperbolic Plane D 21


‖ Ta f ‖2BL

=Lπ

∫D|Ta f (z)|2(1− |z|2)L−2dm(z)

=Lπ

∫D| fa(z)|2(1− |z|2)L−2

∣∣∣∣ 1− |a|2(1− az)2

∣∣∣∣L dm(z).

By the change of variables

w =z− a

1− azthis implies:

z =w + a

1 + aw.

Also we have that:

dm(z) =∣∣∣∣ ∂

∂w

(w + a

1 + aw

)∣∣∣∣2 dm(w) =(1− |a|2)2

|1 + aw|4 dm(w) =(1− |a|2)2

(1 + aw)2(1 + aw)2 dm(w).

Therefore:

1− |z|2 = 1−∣∣∣∣ w + a1 + aw

∣∣∣∣2 = 1− |w|2 + aw + aw + |a|2(1 + aw)(1 + aw)

=(1− |a|2)(1− |w|2)(1 + aw)(1 + aw)

.

This implies that:

(1− |z|2)L−2 =(1− |a|2)L−2(1− |w|2)L−2

(1 + aw)L−2(1 + aw)L−2 .

Also we have:

1− az = 1− aw + a

1 + aw=

1− |a|21 + aw

,

and then:1− |a|2(1− az)2 =

(1 + aw)2

1− |a|2 .

In addition: ∣∣∣∣ (1 + aw)2

1− |a|2

∣∣∣∣L =|1 + aw|2L

(1− |a|2)L =(1 + aw)L(1 + aw)L

(1− |a|2)L .

Hence, by the last equalities:

‖ Ta f ‖2BL

=Lπ

∫D| fa(z)|2(1− |z|2)L−2

∣∣∣∣ 1− |a|2(1− az)2

∣∣∣∣L dm(z)

=Lπ

∫D| f (w)|2 (1− |a|

2)L−2(1− |w|2)L−2

(1 + aw)L−2(1 + aw)L−2

∣∣∣∣ (1 + aw)2

1− |a|2

∣∣∣∣L (1− |a|2)2

(1 + aw)2(1 + aw)2 dm(w)

=Lπ

∫D| f (w)|2 (1− |a|

2)L−2(1− |w|2)L−2(1 + aw)L(1 + aw)L(1− |a|2)2

(1 + aw)L−2(1 + aw)L−2(1− |a|2)L(1 + aw)2(1 + aw)2 dm(w)

=Lπ

∫D| f (w)|2(1− |w|2)L−2dm(w) =‖ f ‖2

BL.


For every GAF f of BL, the function Ta f is also a GAF because it is a composition withthe disk automorphism ϕa and multiplied by the deterministic corrector factor:

ηa(z) =(

1− |a|2(1− az)2

) L2

.

As we just saw, the point sets of f are equal in distribution than those of Ta f , therefore thezero sets are also equal in distribution. Moreover, the zero sets of f are equal in distributionthan those of fa. Indeed, if ηa(z) = 0, we would have |a|2 = 1, which is a contradictionsince a is a value from D. As a conclusion, the zero sets of fa are the same in distributionthan those of f .

2.4 The Paley - Wiener space

For a real parameter L > 0, the Paley - Wiener space is given by:

PWL :=

f ∈ A(C) : | f (z)| ≤ CeL|Im(z)|, ‖ f ‖2PWL

=∫

R| f (x)|2dx < +∞

.

Let us denote:

en =sin π(n− Lz)

π(n− Lz).

We have that (en)n∈Z is an orthonormal basis on PWL. For this we will use Fourier trans-formations, defined as

f (z) =∫ +∞

−∞f (x)e−2πixzdx,

where f is an integrable function in R; and we will use also the Parseval’s identity to extendthe definition to L2(R). First of all we will see that the function f (x) = 1

L χ[− L2 , L

2 ](x)e

2πinxL is

the Fourier transformation of en for all integer n, real x and complex z. Indeed:

f (z) =1L

∫ L2

− L2

e2πinx

L e−2πixzdx =1L

∫ L2

− L2

e2πinx

L e−2πiLxz

L dx =1L

∫ L2

− L2

e2πix(n−Lz)

L dx

=L

2πiL(n− Lz)

[e

2πix(n−Lz)L

]x= L2

x=− L2

=sin π(n− Lz)

π(n− Lz).

Returning to the problem, we must see that (en, en)PWL = 1 and (en, em)PWL = 0 for allinteger numbers n 6= m. Indeed:

(en, en)PWL =∫ +∞

−∞

∣∣∣∣sin π(n− Lz)π(n− Lz)

∣∣∣∣2 dx(∗)=

1L

∫ +∞

−∞χ[− L

2 , L2 ](x)|e 2πixn

L |2dx =1L

∫ L2

− L2

dx = 1,

where in (∗) we use the Parseval’s identity.Now, for all integer n 6= m:

(en, em)PWL =∫ +∞

−∞

sin π(n− Lz)π(n− Lz)

sin π(n− Lz)π(n− Lz)

dx(∗∗)=

1L2

∫ +∞

−∞χ[− L

2 , L2 ](x)e

2πi(n−m)xL dx

=1L2

∫ L2

− L2

e2πi(n−m)x

L dx =eπi(n−m) − e−πi(n−m)

2πiL(n−m)=

sin π(n−m)

πL(n−m)= 0,

2.4 The Paley - Wiener space 23

where in (∗∗) we use the Parseval’s identity again. For completeness, if f is a function ofPWL and ( f , en)PWL = 0, then f ≡ 0. Indeed, the space of Paley - Wiener can be written as:

CSL :=

f ∈ L2(R) : supp f ⊂[−L

2,

L2

],

and this is true by the next theorem:

Theorem 2.8. The complex Fourier transformation establishes an isometry between PWL and CSL.

The proof of this result is long, but in pages 566 to 570 of reference [4] it is well -explained and the isometry is a consequence of Theorem 12.11, page 570.Therefore, by Parseval’s identity:

( f , en)L2(R) = ( f , en)L2(R) =

(f ,

1L

χ[− L2 , L

2 ]e−

2πinxL

)L2(R)

.

However, we know that the collection (e2πinx

L )n∈Z is an orthonormal basis on L2([− L

2 , L2

]).

Hence, ( f , 1L χ[− L

2 , L2 ]

e−2πinx

L )L2(R) = 0, and that implies that f ≡ 0. So, the function f also

vanishes. As a conclusion, (en)n∈Z is an orthonormal basis on PWL.By Propositions 1.13 and 1.11 we define the GAF:

f (z) =+∞

∑n=−∞

ζnsin π(n− Lz)

π(n− Lz),

where (ζn)n are complex i.i.d. Gaussian random variables with zero mean and unit varianceand L > 0 is a real parameter. Denoting g ≡ f , the covariance kernel of f can be calculatedas:

(g,K f )PWL = g(z) =(

f ,1L

e2πizx

L

)L2([− L

2 , L2 ])

.

Therefore:

K f (z, w) =1L

∫ L2

− L2

e2πi(z−w)x

L dx =1

2πi(z− w)

(eπi(z−w) − e−πi(z−w)

)=

sin π(z− w)

π(z− w).

As we did before, we will prove the invariance under transformations of the point sets ofa GAF in PWL. But there is a difference from the rest of the cases: the transformation isunder the real line. Then, we have the next proposition:

Proposition 2.9. Let f be a GAF in PWL. The point sets of f are invariant under the transformation

ϕa(z) = z− a,

where z is a complex value and a is a real value. Also we have that

‖ f ‖2PWL

=‖ fa ‖2PWL

.



f (z) =+∞

∑n=−∞

ζnsin π(n− Lz)

π(n− Lz),

and has covariance kernel:

K f (z, w) =sin π(z− w)

π(z− w).

If we denote fa(z) = f (ϕa(z)), then fa has covariance kernel:

K fa(z, w) = K f (ϕa(z), ϕa(w)) =sin π(z− a− w + a)

π(z− a− w + a)(∗)=

sin π(z− w)

π(z− w),

where in (∗) we use that a = a since a is real. Therefore it holds that:

K f (z, w) = K fa(z, w),

and the point sets of f are equal in distribution than fa. Also it is obvious that:

‖ f ‖2PWL

=‖ fa ‖2PWL

.

As a direct consequence of this proposition, the zero sets of f are equal in distributionthan fa thanks to ϕa.

Chapter 3

Distribution and intensity of zeros ofa GAF

Once we saw the main properties of Gaussian analytic functions and the invarianceunder Möbius transformation of the GAFs considered before, we would like to study thezero sets of those GAFs. We will see that the average of the distribution of the zero pointsis directly determined by the covariance kernel. And what is more impressive, if two GAFshave the same first intensity, then these two functions are equal in distribution.

3.1 The Edelman - Kostlan formula

Let Ω be a region of C and let H be a Hilbert space of analytic functions on Ω. If f is aGAF of H, we define Z f as the zero set of f . In other words:

Z f = f−1(0).

Intuitively, we can understand the counting measure ν f as a way to know how many zerosf has. In this context, ν f is called the empirical measure. Thus, if A is a set of Ω:

ν f (A) = #(Z f ∩ A

).

Definition 3.1. Let X be a point process in Ω. If for all ϕ of C∞c (Ω) satisfies:

E

(∫Ω

ϕdν f

)=∫

Ωϕdρ1,

then the deterministic measure ρ1 is called the first intensity.

Our main goal in this section is to determine a formula to express the first intensity ofa GAF. We are going to compute this using Green’s second identity. Although, we shouldintroduce a definition from [4] (page 315) before going deeper. There, Rn is considered, butsince C is isomorphic to R2, we can state:

25

26 Distribution and intensity of zeros of a GAF

Definition 3.2. Let Ω be a region of C and let µ be a measure such that is finite over compactsubsets of Ω. We say that a function u of L1

loc(Ω) is a solution of

∆u = µ

on Ω in a distributional sense if for all function ψ of C∞c (Ω) it holds∫

Ωu(z)∆ψ(z)dm(z) =

∫Ω

ψ(z)dµ(z).

The Laplacian ∆ should be understood in the distributional sense.

We are going to show a proposition for analytic functions, not only for GAFs.

Proposition 3.3. Using the same notation as before, it satisfies:

ν f =1

2π∆ log | f |,

where f is a function of A(Ω) and ∆ is taken in the distributional sense.

Proof. (See Subsection 2.4.1. in [2], page 24). By the last definition, we must see that:∫Ω

12π

log | f (z)|∆ψ(z)dm(z) =∫

Ωψ(z)dν f (z),

where ψ is a function of C∞c (Ω). Since ψ is compactly supported, the zeros of f that are

laying in the compact support of ψ are finite. Thus, in a neighbourhood of the support ofψ:

f (z) = g(z)n

∏k=1

(z− zk)mk ,

where zk are the zeros of f with multiplicity mk and g is an analytic function, since f is,with no zeros in the support of ψ. Therefore it holds:

log | f (z)| = log |g(z)|+n

∑k=1

mk log |z− zk|,

and also we have:

∆ log | f (z)| = ∆ log |g(z)|+n

∑k=1

mk∆ log |z− zk|.

Since g is an element of A(Ω) and g(z) 6= 0 for all z of Ω in a neighbourhood in thecompact support of ψ, we have that log |g(z)| is a harmonic function, that is, ∆ log |g(z)| =0. Indeed, let u ≡ u(x, y) and v ≡ v(x, y) be real functions of C2(Ω). Then we can writeg(z) = g(x, y) = u(x, y) + iv(x, y). We want to see that:

∆ log |g(x, y)| = ∂2

∂x2 log |g(x, y)|+ ∂2

∂y2 log |g(x, y)| = 0.

3.1 The Edelman - Kostlan formula 27

We have:

∂2

∂x2 log |g(x, y)| = (u2x + uuxx + v2

x + vvxx)(u2 + v2)− (uux + vvx)(2uux + 2vvx)

(u2 + v2)2 ,

and:

∂2

∂y2 log |g(x, y)| =(u2

y + uuyy + v2y + vvyy)(u2 + v2)− (uuy + vvy)(2uuy + 2vvy)

(u2 + v2)2 .

By the analyticity of g, we can use the Cauchy - Riemann equations and from here itholds that uxx + uyy = 0 and vxx + vyy = 0. Making suitable changes, we conclude that∆ log |g(z)| = 0.Now, let us declare G(z) = 1

2π log |z− zk|, which is integrable in a neighbourhood of thecompact support of ψ. Indeed, if we consider a disk D centered in zk and radii ρ, we have:∫

DG(z)dm(z) =

12π

∫D

log |z− zk|dm(z) =∫ ρ

0r log rdr =

[r2

4(2 log(r)− 1)

]r=ρ

r=0

=ρ2

4(2 log(ρ)− 1) < +∞.

Now we are going to see that:∫Ω

12π

log |z|∆ψ(z)dm(z) = ψ(0).

Since we are operating with a logarithm, we should extract the disk D(0, ε), with ε > 0,from Ω. Let us denote Ωε = Ω \D(0, ε). Using Green’s second identity:∫

Ωε

12π

log |z|∆ψ(z)dm(z) =∫

Ωε

12π

∆ log |z|ψ(z)dm(z)

−∫

∂D(0,ε)−

(1

2πlog ε

∂ψ

∂n(z)− 1

2π

∂ log |z|∂n ||z|=ε

ψ(z))

ds,

where the integral over ∂Ωε is 0 because ψ is compactly supported. The first integral of theright side of the equality is zero because ∆ log |z| = 0. For the second integral we have:∫

∂D(0,ε)−

12π

log ε∂ψ

∂n(z)ds =

12π

log ε∫

∂D(0,ε)−

∂ψ

∂n(z)ds ≤ ε log ε ‖ ∇ψ ‖∞

ε→0−→ 0,

and:∫∂D(0,ε)−

12π

∂ log |z|∂n ||z|=ε

ψ(z)ds =1

2π

∫ 2π

0ψ(εeiθ)dθ =

12π

∫ 2π

0

(ψ(εeiθ)− ψ(0) + ψ(0)

)dθ

= ψ(0) +1

2π

∫ 2π

0

(ψ(εeiθ)− ψ(0)

)dθ.

Notice that limε→0 ψ(εeiθ) = ψ(0) and that ψ(εeiθ) is bounded because ψ is from C∞c (Ω).

Then, by the Dominated Convergence Theorem, we have:

limε→0

(ψ(0) +

12π

∫ 2π

0

(ψ(εeiθ)− ψ(0)

)dθ

)= ψ(0).


Since Ωεε→0−→ Ω, it holds: ∫

Ω

12π

log |z|∆ψ(z)dm(z) = ψ(0).

Evaluating the last integral in ψ(z + zk), we get:∫Ω

12π

log |z− zk|∆ψ(z)dm(z) = ψ(zk).

Now, we know that f (z) = g(z)∏nk=1(z− zk)

mk . Since g has no zeros in a neighbourhoodof the compact support of ψ, g is a harmonic function in there. Taking logarithms andthe Laplacian operator and using again Green’s second identity,

∫Ω log |g(z)|∆ψ(z)dm(z)

vanishes. Then, we get∫Ω

12π

log | f (z)|∆ψ(z)dm(z) =∫

Ωψ(z)dν f (z),

and then:ν f =

12π

∆ log | f |,

as we wanted to see.

The next theorem is one of the keystones of this chapter. It will explain that the expec-tation of the distributed zeros values of a GAF is directly related to the covariance kernelof the GAF that we are considering. The formula is called the Edelman - Kostlan formula.

Theorem 3.4. (The Edelman - Kostlan formula) Let H be a Hilbert space of analytic functionson a region Ω of C. Let f be a Gaussian analytic function with mean zero and covariance kernelK f (z, w), for all z and w of Ω. The first intensity of the zeros of f is given by:

ρ1(z) =1

4π∆ logK f (z, z),

where the Laplacian ∆ must be interpreted in the distributional sense.

Proof. (See Subsection 2.4.1. in [2], pages 24 and 25). By Proposition 3.3 and using the samenotation than its proof we have:∫

Ωψ(z)dν f (z) =

∫Ω

12π

log | f (z)|∆ψ(z)dm(z).

Taking expectations at both sides we have:

E

[∫Ω

ψ(z)dν f (z)]= E

[∫Ω

12π

log | f (z)|∆ψ(z)dm(z)]

. (3.1)

To finish the procedure we must be able to use Fubini’s theorem at the right side of theequation. Therefore, we must see that:

E

[∫Ω

∣∣∣∣ 12π

log | f (z)|∆ψ(z)dm(z)∣∣∣∣] < +∞.


By the linearity of the expectation we obtain:

E

[∫Ω

∣∣∣∣ 12π

log | f (z)|∆ψ(z)dm(z)∣∣∣∣] = 1

2π

∫Ω|∆ψ(z)|E [|log | f (z)||] dm(z).

For a fixed z of Ω, f (z) is a complex Gaussian random variable with zero mean and varianceK f (z, z). Hence:

E [| log | f (z)||] = E

∣∣∣∣∣∣log

∣∣∣∣∣∣ f (z)√K f (z, z)

√K f (z, z)

∣∣∣∣∣∣∣∣∣∣∣∣

= E

∣∣∣∣∣∣log

∣∣∣∣∣∣ f (z)√K f (z, z)

∣∣∣∣∣∣∣∣∣∣∣∣+ log

∣∣∣√K f (z, z)∣∣∣ .

Denoting ζ = f (z)√K f (z,z)

, the random variable ζ is a standard complex Gaussian random

variable for all z of Ω. Therefore:

E [| log | f (z)||] = E [| log |ζ||] + log∣∣∣√K f (z, z)

∣∣∣=∫

C| log |z|| e

−|z|2

πdm(z) +

12

log |K f (z, z)|

(∗)=∫ +∞

02r| log(r)|e−r2

dr +12

log |K f (z, z)|

(∗∗)=∫ +∞

0| log(ρ)|e−ρdρ +

12

log |K f (z, z)| = C +12

log |K f (z, z)|,

where C is a constant value, in (∗) we use a polar coordinate change and in (∗∗) we applythe change of variable ρ = r2 and dρ = 2rdr. The factor log |K f (z, z)| is locally integrablefor all z of Ω, but there is a problem when K f (z0, z0) = 0 for z0 of Ω. In such a case, K f canbe expressed as:

K f (z, z) = |z− z0|2mH(z, z),

where H is a function that is different of zero at z0 and m a natural number. Indeed, weknow that the covariance kernel of a GAF f is:

K f (z, z) =+∞

∑n=1| fn(z)|2,

and each fn can be expressed as:

fn(z) = (z− z0)mn hn(z),

where mn is the multiplicity of the value z0 and hn is a function that does not vanish at z0.Hence:

K f (z, z) =+∞

∑n=1|z− z0|2mn |hn(z)|2.


Now, if we denote m by the minimum value that takes the sequence (mn)n∈N, we have:

K f (z, z) = |z− z0|2m+∞

∑n=1|z− z0|2mn−2m|hn(z)|2 = |z− z0|2mH(z, z),

where H is a function that is not zero at z0 because hn is not either and there is a value mn

that is equal to m. Therefore it holds that:

K f (z, z) = |z− z0|2mH(z, z).

With this, logK f (z, z) is integrable in a neighbourhood of z0. Thus:

E

[∫Ω

∣∣∣∣ 12π

log | f (z)|∆ψ(z)dm(z)∣∣∣∣] < +∞,

and we can apply Fubini’s theorem at (3.1) to obtain:

E

[∫Ω

ψ(z)dν f (z)]=∫

Ω

12π

E [log | f (z)|]∆ψ(z)dm(z) =∫

Ω

12π

∆E [log | f (z)|]ψ(z)dm(z).

Now, since ζ = f (z)√K f (z,z)

is a standard complex Gaussian random variable:

E [log | f (z)|] = E

log

∣∣∣∣∣∣ f (z)√K f (z, z)

√K f (z, z)

∣∣∣∣∣∣ = E

log

∣∣∣∣∣∣ f (z)√K f (z, z)

∣∣∣∣∣∣+ log

√K f (z, z)

= E [log |ζ|] + 12

logK f (z, z) = C +12

logK f (z, z).

where C is a constant value. Therefore:

E

[∫Ω

ψ(z)dν f (z)]=∫

Ω

14π

∆ logK f (z, z)ψ(z)dm(z),

and by the definition of the first intensity we have, with respect to the Lebesgue measure:

ρ1(z) =1

4π∆ logK f (z, z).

The coefficients of the linear combination of holomorphic functions of a GAF in thePaley - Wiener can be real or complex Gaussian random variables. The Edelman - Kostlanformula is valid in the case of using complex random variables. The real case is slightlydifferent. The paper [8], by Naomi D. Feldheim, will give us all the tools to prove the otherversion of the Edelman - Kostlan formula. Before introducing the theorem, we must definewhat is a symmetric GAF.

Definition 3.5. Let f be a random analytic function on a symmetric domain Ω, that is Ω = Ω.We say that f is a symmetric GAF if f = ∑+∞

n=1 ζn fn, where ζn are independent random variableswith NR(0, 1) distribution and ( fn)n is a sequence of analytic functions such that ∑+∞

n=1 | fn(z)|2converges uniformly on compact sets of the domain Ω and ( fn)n is symmetric with respect to thereal axis, that is fn(z) = fn(z) for all z of Ω.


Theorem 3.6. Let f be a symmetric Gaussian analytic function with mean zero and covariancekernel K f (z, w), for all z and w of Ω. The first intensity of the zeros of f is given by:

ρ1(z) =1

4π∆ log

(K f (z, z) +

√K f (z, z)2 − |K f (z, z)|2

),

where the Laplacian ∆ must be interpreted in the distributional sense.

As a remark, the covariance kernel in the real field is equal to the complex one if weconsider (ζn)n ∼ NR(0, 1) or (ζn)n ∼ NC(0, 1).

Proof. (see Section 4 in [8]) We must see that:∫Ω

ψ(z)ρ1(z) = E

[∫Ω

ψ(z)dν f (z)]=∫

Ω

12π

E [∆ log | f (z)|ψ(z)] dm(z)

=∫

Ω

12π

∆E [log | f (z)|]ψ(z)dm(z),

where we applied Fubini’s theorem. The integrability of the terms of the integrals is due tothe same reason than in Theorem 3.4. We have to compute E [log | f (z)|]. First of all, if f isa GAF as described and ( fn)n is a sequence of analytic functions in Ω , we can write:

f (z) =+∞

∑n=1

ζn fn(z) =+∞

∑n=1

ζnun(z) + i+∞

∑n=1

ζnvn(z) = u(z) + iv(z),

where (ζn)n are random variables that follow a real standard Gaussian distribution and un

and vn are real functions that denote the real and imaginary part of fn respectively. There-fore the random vector (u, v) follows a real centered Gaussian distribution with covariancematrix:

Λ =

(E[(

∑+∞n=1 ζnun

) (∑+∞

n=1 ζnun)]

E[(

∑+∞n=1 ζnun

) (∑+∞

n=1 ζnvn)]

E[(

∑+∞n=1 ζnun

) (∑+∞

n=1 ζnvn)]

E[(

∑+∞n=1 ζnvn

) (∑+∞

n=1 ζnvn)])

(∗)=

(∑+∞

n=1 u2n ∑+∞

n=1 unvn

∑+∞n=1 unvn ∑+∞

n=1 v2n

),

where in (∗) we applied the linearity of the mean and that (ζn)n ∼ NR(0, 1).Now, we claim that the eigenvalues of Λ are

λ1 =K f (z, z)∓ |K f (z, z)|

2, λ2 =

K f (z, z)± |K f (z, z)|2

.

Indeed, for every fn = un + ivn we have the equations, which are simple to check:

u2n =

12(| fn|2 + Re( f 2

n))

, v2n =

12(| fn|2 − Re( f 2

n))

, unvn =12

Im( f 2n).

Therefore the covariance matrix can be expressed as:

Λ =

(12

(∑+∞

n=1 | fn|2 + Re(∑+∞

n=1 f 2n)) 1

2 Im(∑+∞

n=1 f 2n)

12 Im

(∑+∞

n=1 f 2n) 1

2

(∑+∞

n=1 | fn|2 − Re(∑+∞

n=1 f 2n))) .


Thus:

det Λ =14

(+∞

∑n=1| fn|2

)2

−(

Re

(+∞

∑n=1

f 2n

))2

−(

Im

(+∞

∑n=1

f 2n

))2 (3.2)

=14(K f (z, z)2 − |K f (z, z)|2

), (3.3)

trΛ =+∞

∑n=1| fn|2 = K f (z, z). (3.4)

And knowing the fact that det Λ = λ1λ2 and trΛ = λ1 + λ2 we have that det Λ =

(trΛ− λ2) λ2. Then:

14(K f (z, z)2 − |K f (z, z)|2

)= K f (z, z)λ2 − λ2

2

=⇒ λ22 −K f (z, z)λ2 +

14(K f (z, z)− |K f (z, z)|2

)= 0

=⇒ λ2 =K f (z, z)±

√K f (z, z)−K f (z, z) + |K f (z, z)|2

2

=⇒ λ2 =K f (z, z)± |K f (z, z)|

2,

and this implies that:

λ1 =K f (z, z)∓ |K f (z, z)|

2.

Now:

E [log | f (z)|] = 12π√

det Λ

∫R2

log(√

x2 + y2

)e−

12 (x y)Λ−1(x y)t

dxdy. (3.5)

To compute the factor (x y)Λ−1(x y)t we will use the change of variable:

(x y)t = M(x y)t =⇒ M−1(x y)t = (x y)t,

where M is a two - dimensional orthogonal matrix, that is, M satisfies Mt = M−1. Also wehave that x2 + y2 = x2 + y2, dxdy = dxdy and the determinant of the Jacobian matrix is 1.Knowing the fact that the covariance matrix Λ can be written as Λ = MtDM, where D is adiagonal matrix whose elements are the eigenvalues of Λ, then:

(x y)Λ−1(x y)t = (x y)M−tΛ−1M−1(x y)t = (x y)MΛ−1Mt(x y)t

=1

λ1λ2(x y)

(λ2 00 λ1

)(x y)t =

λ2 x2 + λ1y2

λ1λ2= λ−1

1 x2 + λ−12 y2.


From (3.5) we obtain:

E [log | f (z)|] = 12π√

det Λ

∫R2

log(√

x2 + y2

)e−

12 λ−1

1 x2+λ−12 y2

dxdy

(∗)=

12π

∫R2

log(√

λ1u2 + λ2v2)

e−12 (u

2+v2)dudv

(∗∗)=

12π

∫ +∞

0

∫ 2π

0log(√

λ1r2 cos2 θ + λ2r2 sin2 θ

)re−

r22 dθdr

=∫ +∞

0r log(r)e−

r22 dr +

12π

∫ +∞

0re−

r22 dr︸︷︷︸

1

∫ 2π

0log(√

λ1 cos2 θ + λ2 sin2 θ

)dθ

=1

2π

∫ 2π

0log(√


)dθ + C1,

where in (∗) we use the change of variable x = u√

λ1, y = v√

λ2 with Jacobian√

λ1λ2 =

det Λ. In (∗∗) we apply polar coordinates u = r cos θ and v = r sin θ. The factor C1 is aconstant value that will vanish with the Laplacian operator. We have that:√

λ1 cos2 θ + λ2 sin2 θ =∣∣∣√λ1 cos θ + i

√λ2 sin θ

∣∣∣ ,

and: √λ1 cos θ + i

√λ2 sin θ =

√λ1

eiθ + e−iθ

2+ i√

λ2eiθ − e−iθ

2i

=12

[eiθ(√

λ1 +√

λ2) + e−iθ(√

λ1 −√

λ2)]

=

√λ1 +

√λ2

2e−iθ

(e2iθ +

√λ−√

λ2√λ1 +

√λ2

).

Therefore: ∣∣∣√λ1 cos θ + i√

λ2 sin θ∣∣∣ = ∣∣∣∣√λ1 +

√λ2

2

∣∣∣∣∣∣∣∣∣e2iθ +

√λ−√

λ2√λ1 +

√λ2

∣∣∣∣∣ .

This leads to:

E [log | f (z)|] = 12π

∫ 2π

0log(√


)dθ + C1

=1

2π

∫ 2π

0log

(√λ1 +

√λ2

2

∣∣∣∣∣e2iθ +

√λ−√

λ2√λ1 +

√λ2

∣∣∣∣∣)

dθ + C1

= log(√

λ1 +√

λ2

)+

12π

∫ 2π

0log

∣∣∣∣∣e2iθ +

√λ−√

λ2√λ1 +

√λ2

∣∣∣∣∣ dθ + C2,

where C2 is another constant that will also vanish with the Laplacian operator. To computethe last integral we use Jensen’s formula. Let us denote h(z) = z2 + k, where k =

√λ−√

λ2√λ1+√

λ2<

1. Hence, since h has two simple roots in D, let them be z1 and z2, we have:

12π

∫ 2π

0log |h(eiθ)|dθ = log |h(0)| − log |z1| − log |z2| = log |k| − 2 log

(√|k|)= 0.


Then, by (3.3) and (3.4), it is straightforward that:

E [log | f (z)|] = log(√

λ1 +√

λ2

)=

12

log(

λ1 + λ2 +√

4λ1λ2

)=

12

log(K f (z, z) +

√K f (z, z)2 − |K f (z, z)|2

).

Therefore:

E

[∫Ω

ψ(z)dν f (z)]=∫

Ω

14π

∆ log(K f (z, z) +

√K f (z, z)2 − |K f (z, z)|2

)ψ(z)dm(z),

and by the definition of the first intensity we have, with respect to the Lebesgue measure:

ρ1(z) =1

4π∆ log

(K f (z, z) +

√K f (z, z)2 − |K f (z, z)|2

).

Now we can compute the first intensity for all the Hilbert spaces of analytic functionsdescribed in the last chapter.

3.1.1 The Fock space in C

The covariance kernel of a GAF f in the Fock space is

K f (z, w) = eLzw,

for all real L > 0. Thus:

∆ log(K f (z, z)

)= 4

∂

∂z∂

∂z(Lzz) = 4L,

and the first intensity is

ρ1(z) =Lπ

.

As a remark, in Chapter 2 we saw that the zero set of a GAF in the Fock space was invariantunder translations. Thus, it is not a surprise that the first intensity does not depend of thepoint we are computing this value. In addition, the first intensity is also invariant undertranslations in C and it is proportional to the Lebesgue measure.

3.1.2 The space of polynomials in S2

The covariance kernel of a GAF f in the space of polynomial of the degree at most L,for a natural L, is

K f (z, w) = (1 + zw)L.

Thus:

∆ log(K f (z, z)

)= 4L

∂

∂z∂

∂z[log(1 + |z|2)

]= 4L

∂

∂z

(z

1 + |z|2

)=

4L(1 + |z|2)2 ,



ρ1(z) =Lπ

1(1 + |z|2)2 .

In Chapter 2 we proved that the zero set of a GAF in the space of polynomials with finitedegree was invariant under the Möbius transformation

ϕa(z) =z− a

1 + az,

for z and a values of C. In this case, if we multiply the first intensity by the Lebesguemeasure over the sphere and we apply the inverse mapping of the stereographic projection,we get that the first intensity is invariant under rotations. Indeed, following the notation ofthe proof of Proposition 2.5, we have that:

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

(1 + |a|2)2(1− aw)2(1− aw)2

(1− aw)2(1− aw)2(1 + |a|2)2(1 + |w|2)2 dm(w) =dm(w)

(1 + |w|2)2 .

3.1.3 The weighted Bergman space in D

The covariance kernel of a GAF f in the weighted Bergman space is

K f (z, w) = (1− zw)−L,

for all real L > 0. Thus:

∆ log(K f (z, z)

)= −4L

∂

∂z∂

∂z[log(1− |z|2)

]= 4L

∂

∂z

(z

1− |z|2

)=

4L(1− |z|2)2 ,


ρ1(z) =Lπ

1(1− |z|2)2 .

In this case, the first intensity in the weighted Bergman space is proportional to the hyper-bolic measure

dm(z)(1− |z|2)2 .

It is straightforward that ρ1 is invariant under automorphisms in D. The proof of this isanalogous than the one in the space of polynomials in S2.

3.1.4 The Paley - Wiener space with NC(0, 1) random variables

The covariance kernel of a GAF f in the Paley - Wiener space is


π(z− w).

First of all we should rewrite the covariance kernel K f (z, z) into another expression. If wedenote z as x + iy for all real values x = Re(z) and y = Im(z), we have:

sin π(z− z)π(z− z)

=sin(2πiy)

2πiy=

e2πy − e−2πy

4πy=

sinh(2πy)2πy

.


For the following we are going to consider only that y > 0, because the last function is aneven one. Taking logarithms we have:

logK f (y, y) = log sinh(2πy)− log(2πy),

and applying the Laplacian ∆(x,y) ≡ ∂2xx + ∂2

yy at both sides:

∆ logK f (y, y) = ∆ log sinh(2πy)− ∆ log(2πy) =∂2

∂y2 log sinh(2πy)− ∂2

∂y2 log(2πy)

=∂

∂y2π

cosh(2πy)sinh(2πy)

− ∂

∂y1y=

∂

∂y2π coth(2πy)− ∂

∂y1y

(∗)= −4π2csch2(2πy) +

1y2 ,

where in (∗) we use that d/dx(coth x) = −csch2x. Therefore, the first intensity is:

ρ1(y) =1

4πy2 − πcsch2(2πy),

and we can see that is invariant under translations on the real line because it only dependsof Im(z). At first glance, it seems we have a potential problem at y = Im(z) = 0. Howeverwe can specify the value of the first intensity at that point. The series of csch2(2πy) at aneighbourhood of 0 is:1

csch2(2πy) =1

4π2y2 −13+

4π2y2

15+O(y4).

Then:

limy→0

ρ1(y) = limy→0

(1

4πy2 −1

4πy2 +π

3− 4π3y2

15+O(y4)

)=

π

3.

3.1.5 The Paley - Wiener space with NR(0, 1) random variables

The covariance kernel of a GAF f in the Paley - Wiener space is


π(z− w).

By Theorem 3.6, the first intensity is:

ρ1(z) =1

4π∆ log

(K f (z, z) +

√K f (z, z)2 − |K f (z, z)|2

).

We are going to calculate the first intensity in terms of the imaginary part of z. DenotingIm(z) = y for short, κ(y) = K f (z, z), K f (z, z) = 1 (this value is due to the definition of thecardinal sine) and following a procedure of [8], we have:

ρ1(y) =1

4π∆ log

(κ(y) +

√κ(y)2 − 1

)=

14π

∂

∂yκ′(y)√

κ(y)2 − 1.

1Expression extracted from [29].


Indeed:

∂2

∂y2 log(

κ(y) +√

κ(y)2 − 1)=

∂

∂y

κ′(y) + 2κ(y)κ′(y)2√

κ(y)2−1

κ(y) +√

κ(y)2 − 1

=∂

∂yκ′(y)

√κ(y)2 − 1 + κ(y)κ′(y)

(κ(y) +√

κ(y)2 − 1)(√

κ(y)2 − 1)

=∂

∂yκ′(y)√

κ(y)2 − 1.

In our case, if

κ(y) =sin(2πiy)

2πiy,

we have, using the identities sinh y = −i sin iy and cosh y = cos iy, that

κ′(y) =−4π2y cos(2πiy)− 2πi sin(2πiy)

−4π2y2 =2πy cosh(2πy)− sinh(2πy)

2πy2 ,

and

ρ1(y) =1

4π

∂

∂y2πy cosh(2πy)− sinh(2πy)

y√

sinh2(2πy)− 4π2y2.

Thus:2

ρ1(y) =(3 + 48π2y2 + 64π4y4) sinh(2πy)− sinh(6πy)− 64π3y3 cosh(2πy)

16y3(4π3y2 − π sinh2(2πy))√

sinh2(2πy)y2 − 4π2

,

and it is invariant under translations on the real line because it only depends of Im(z). Tosee what is the value of the first intensity at neighbourhood of 0, we can write κ as a Taylorseries near of 0. Since κ is an even function, we can consider only Im(z) = y > 0. We havethat:

κ(y) = 1 +23

π2y2 +O(y4), κ2(y) = 1 +43

π2y2 +O(y4).

Then:

ρ1(y) =1

4π∆ log

(1 +

23

π2y2 +O(y4) +

√1 +

43

π2y2 − 1 +O(y4)

)

=1

4π∆ log

(1 +

23

π2y2 +O(y4) +2√3

πy√

1 +O(y2)

)=

14π

∆ log(

1 +23

π2y2 +O(y4) +2√3

πy[

1 +12O(y2)

])=

14π

∆ log(

1 +2√3

πy +23

π2y2 +O(y3)

)=

14π

∆(

2√3

πy +O(y3)

),

2Expression extracted from [30].


where in the last equality we applied the Taylor series of log(1+ y) for y in a neighbourhoodof 0, and we omitted long computations. At the end, the first intensity in a neighborhoodof the origin takes low values. Considering any value of y at a neighbourhood of 0, it isclear that the expression 2√

3π|y|+O(y3) is not differentiable. However, using the concept

of derivation in the distributional sense we will conclude that d2/dx2(|x|) = 2δ0. The deltaof Dirac δ with a test function τ is

(δ, τ) = τ(0),

and the derivative in a distributional sense for a locally integrable function f is:

( f ′, τ) =∫

Rf ′(x)τ(x)dx = −

∫R

f (x)τ′(x)dx = −( f , τ′).

The derivative of the absolute value can be described as:

H(x) = sgn(x) =d|x|dx

=

1, if x > 0,

−1, if x < 0.

The function H is the so-called Heaviside function. Now we have:

(H′, τ) = −(H, τ′) = −∫

RH(x)τ′(x)dx = −2

∫ +∞

0τ′(x)dx = 2τ(0) = 2(δ, τ).

Then the derivative of the absolute value behaves like 2δ0, and the first intensity we wantedto compute at y = 0 is 1√

3δ0.

3.2 Calabi’s rigidity

In this section we will enunciate and prove the Calabi’s rigidity. It guarantees that twoGAFs with the same first intensity are equal in distribution. Therefore, this theorem ofuniqueness concludes that the zero set of a GAF is determined by its first intensity.Before going to the main result, we have to see the next lemma:

Lemma 3.7. Let F(z, w) be a function in a region Ω that is holomorphic in z and anti-holomorphicin w, for z and w of Ω. If F(z, z) = 0 for all z of Ω, then F(z, w) = 0 for z and w of Ω.

Proof. (See proof of Lemma 2.5.1 in [2], page 30) It is only necessary to see that F is zero ina neighbourhood of (0, 0) in Ω. Since F is a real analytic function, we can express F as:

F(z, w) =+∞

∑n=1

+∞

∑m=1

cn,mznwm, (3.6)

for complex values cn,m. Therefore it also holds that:

F(z, z) =+∞

∑n=1

+∞

∑m=1

cn,mznzm.

3.2 Calabi’s rigidity 39

We state that:∂n+m

∂zn∂zm zjzk |z=0= δ(n,m)(j,k)n!m!,

where

δ(n,m)(j,k) =

1, if j = n and k = m,

0, otherwise.

Rewriting (3.6) as:

F(z, z) =+∞

∑j=1

+∞

∑k=1

cj,kzjzk,

and using the F(z, z) = 0 hypothesis, we arrive to:

0 =∂n+m

∂zn∂zm F(z, z) |z=0= cn,mn!m!

Thus, cn,m = 0 and F(z, w) = 0 for all z and w of Ω.

Now we are ready for Calabi’s rigidity theorem.

Theorem 3.8. (Calabi’s rigidity) Let Ω be a simply connected domain of C. Let f and g be GAFsin Ω. If the first intensity of f and g are equal, there exists a deterministic analytic function ψ on

Ω such that it does not vanish at any point and it holds that f d= ψg. In addition, Z f

d= Zg. As a

conclusion, the first intensity of a GAF determines its zero set.

Proof. (See proof of Theorem 2.5.2. in [2], pages 30 and 31) For all z of Ω, we have that z isalmost surely a zero of a GAF f if and only if it is a zero of K f (z, z), and the multiplicityof z is the same for f and the covariance kernel. Indeed; let ( fn)n be a sequence of analyticfunctions on Ω. Since:

K f (z, z) =+∞

∑n=1

fn(z) fn(z) =+∞

∑n=1| fn(z)|2 and f (z) =

∞

∑n=1

ζn fn(z),

for ζn ∼ NC(0, 1), we have, almost surely, that:

+∞

∑n=1| fn(z)|2 = 0⇔ fn(z) = 0⇔ f (z) =

∞

∑n=1

ζn fn(z) = 0⇔ f (z) = 0.

Since we assumed that f and g have the same first intensity, the discrete set of deterministiczeros, let us call it D, is the same for both GAFs, whatever are the multiplicities of thesezeros. However, we are interested in the random zeros of f and g. Then, we subtract theset D of Ω and we conclude that f and g do not vanish anywhere in Ω \ D, or which is thesame, K f (z, z) and Kg(z, z) are not zero for all z of Ω \ D.Since K f (z, z) and Kg(z, z) are not zero for all z of Ω \ D, by hypothesis we have that:

14π

∆ logK f (z, z) =1

4π∆ logKg(z, z) =⇒ ∆

[log(K f (z, z))− log(Kg(z, z))

]= 0

=⇒ ∆ log(K f (z, z)Kg(z, z)

)= 0,


and this implies that log(K f (z,z)Kg(z,z)

)is a harmonic function in Ω \ D. Let us denote the last

expression as u, which is obvious that u is a harmonic function in Ω \ D. Then:

log(K f (z, z)Kg(z, z)

)= u(z) =⇒

K f (z, z)Kg(z, z)

= eu(z) =⇒ K f (z, z) = eu(z)Kg(z, z).

Since Ω is simply connected there exists an analytic function h in Ω \D such that 2Re(h) =u. Thus, by

eu(z) = e2Re(h(z)) = |eh(z)|2 = eh(z)eh(z) = ψ(z)ψ(z),

we have thatK f (z, z) = ψ(z)ψ(z)Kg(z, z), (3.7)

Then, the functions K f (z, w) and ψ(z)ψ(w)Kg(z, w) are equal on the diagonal by (3.7). Sinceboth last functions satisfy the hypothesis of Lemma 3.7, it holds true that:

K f (z, w) = ψ(z)ψ(w)Kg(z, w),

and this implies that f d= ψg. Since ψ is the exponential function, it never vanishes, and we

get that Z fd= Zg.

Chapter 4

GAF computation

In this last chapter of the project we will obtain the first experimental results using C++coding and the plotting program gnuplot. The codes are specified in Annex.

4.1 GAF in the finite space of polynomials endowed with the norm‖ · ‖2

FL

In this section we will consider the finite space of polynomials endowed with the normof the Fock space, ‖ · ‖2

FL, so let m be a natural number such that the orthonormal basis is

(en)mn=0, where:

en =

√Ln√

n!zn.

Before calculating the first intensity of a GAF f , we should introduce the incomplete gammafunction, that is, for a given positive real a and non-negative x:

Γ(a, x) =∫ +∞

xta−1e−tdt.

If n is a natural number, then:

Γ(n, x) = (n− 1)!e−xn−1

∑k=0

xk

k!.

Now, let us calculate the first intensity step by step. We have:

K f (z, z) =m

∑n=0

Ln

n!|z|2n =

m

∑n=0

(L|z|2)n

n!=

Γ(m + 1, L|z|2)Γ(m + 1)

eL|z|2 .

Taking logarithms at both sides:

logK f (z, z) = log(

Γ(m + 1, L|z|2)Γ(m + 1)

eL|z|2)= log

(Γ(m + 1, L|z|2)

Γ(m + 1)

)+ L|z|2,

41

42 GAF computation

and taking the Laplacian operator at each side of the equality:

∆ logK f (z, z) = ∆ log(

Γ(m + 1, L|z|2)Γ(m + 1)

)+ 4L.

Thus, the first intensity is:

ρ1(z) =1

4π∆ log

(Γ(m + 1, L|z|2)

Γ(m + 1)

)+

Lπ

.

Since m is a natural number, Γ(m + 1) will vanish with the Laplacian operator. Hence:

ρ1(z) =1

4π∆ log Γ(m + 1, L|z|2) + L

π. (4.1)

We have that:

∂

∂zΓ(m + 1, L|z|2) = −Lm!ze−L|z|2

m

∑n=0

Lnznzn

n!+ m!e−L|z|2

m

∑n=1

Lnznzn−1

(n− 1)!

= −Lzm!e−L|z|2m

∑n=0

Ln|z|2n

n!+ Lzm!e−L|z|2

m−1

∑n=0

Ln|z|2n

n!

= LzmΓ(m, L|z|2)− LzΓ(m + 1, L|z|2).

Then:

∆ log Γ(m + 1, L|z|2) = 4∂

∂z∂

∂zlog Γ(m + 1, L|z|2) (4.2)

= 4∂

∂z

(LzmΓ(m, L|z|2)− LzΓ(m + 1, L|z|2)

Γ(m + 1, L|z|2)

)(4.3)

= 4Lm∂

∂z

(z

Γ(m, L|z|2)Γ(m + 1, L|z|2)

)− 4L. (4.4)

From now on, we are going to use the following notation to shorten further expressions:

Γk(z) ≡ Γ(k, L|z|2),

for whatever natural number k is. Now:

∂

∂z

(z

Γm(z)Γm+1(z)

)=

Γm(z)Γm+1(z)

+ z∂

∂zΓm(z)

Γm+1(z).

The derivative in the right side is:

∂

∂zΓm(z)

Γm+1(z)=

∂zΓm(z)Γm+1(z)− Γm(z)∂zΓm+1(z)Γ2

m+1(z)

=(−LzΓm(z) + Lz(m− 1)Γm−1(z))Γm+1(z)− Γm(z)(LzmΓm(z)− LzΓm+1(z))

Γ2m+1(z)

=Lz(m− 1)Γm−1(z)Γm+1(z)− LzmΓ2

m(z)Γ2

m+1(z).

4.1 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2FL

43

Thus:

∂

∂z

(z

Γm(k)Γm+1(z)

)=

Γm(z)Γm+1(z)

+L|z|2(m− 1)Γm−1(z)Γm+1(z)− L|z|2mΓ2

m(z)Γ2

m+1(z)(4.5)

=Γm(z)Γm+1(z) + L|z|2(m− 1)Γm−1(z)Γm+1(z)− L|z|2mΓ2

m(z)Γ2

m+1(z). (4.6)

Finally, using (4.6) to (4.4) and plugging the result in (4.1) we obtain:

ρ1(z) =Lmπ

Γm(z)Γm+1(z) + L|z|2(m− 1)Γm−1(z)Γm+1(z)− L|z|2mΓ2m(z)

Γ2m+1(z)

.

Using the program in 5.2 for the zero set (left) and 5.3 for the first intensity (right) andsetting the values n = 100 and L = 1, we obtain:

For n = 150 and L = 0.5:

44 GAF computation

For n = 30 and L = 5:

As we can see in every example, the first intensity is constant at the region where we canfind zeros almost surely. At the boundary of the circle that generates the zero set, the curvedecays to the origin line, indicating that there are no zeros beyond the circle almost surely.


PL

Since the space of polynomials PL, for L a natural number, is already finite, the firstintensity for a GAF in the finite space of polynomials endowed with the norm ‖ · ‖2

PLis

the one we calculated in the last chapter. Now, using the program 5.4 for the zero set andgnuplot for the first intensity and setting n=100, we get:

The first intensity, which is the picture of the right, matches with the zero set distribution(left picture), because all the zeros of the GAF are gathered nearby the origin and thefunction takes its maximum values at a neighbourhood of zero.In the left picture it seems that there is no pattern in the distribution of the zeros, that thereis only a cluster of points. However, using the inverse of the stereographic projection map,we obtain a sphere whose points are distributed uniformly over the surface of S2.

4.3 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2BL

45

Now it is visually comprehensible that the distribution of zeros of a GAF in this space isinvariant under rotations.


BL

For any z in D, we define the hypergeometric function:

2F1(a, b; c; z) =+∞

∑n=0

(a)n(b)n

(c)n

zn

n!,

where (a)n is called the Pochhammer symbol and it is defined as:

(a)n :=

a(a + 1)...(a + n− 1), n > 0,

1, n = 0.

We want to compute the covariance kernel of a GAF in the finite space of polynomialsendowed with the norm ‖ · ‖2

BL. Let m be a natural number. We have:

(1− |z|2)−L =+∞

∑n=0

(L + n− 1

n

)|z|2n =

m

∑n=0

(L + n− 1

n

)|z|2n +

+∞

∑n=m+1

(L + n− 1

n

)|z|2n,

therefore:

m

∑n=0

(L + n− 1

n

)|z|2n = (1− |z|2)−L −

+∞

∑n=m+1

(L + n− 1

n

)|z|2n.

46 GAF computation

We must rewrite the second sum to obtain the hypergeometric function. Making the changeof variables k = n−m− 1 and knowing that (1)k = k! = Γ(k + 1):

+∞

∑n=m+1

(L + n− 1

n

)|z|2n =

+∞

∑k=0

(L + k + mk + m + 1

)|z|2k+2m+2 = |z|2m+2

+∞

∑k=0

(L + k + mk + m + 1

)|z|2k

= |z|2m+2+∞

∑k=0

(L + mm + 1

)(L + m + 1)k

(m + 2)k|z|2k

= |z|2m+2

(L + mm + 1

)+∞

∑k=0

(1)k(L + m + 1)k

(m + 2)k

|z|2k

k!

= |z|2m+2

(L + mm + 1

)2F1(1, L + m + 1; m + 2; |z|2).

Thus, the covariance kernel of a GAF f in this space is:

K f (z, z) =m

∑n=0

(L + n− 1

n

)|z|2n = (1−|z|2)−L−|z|2m+2

(L + mm + 1

)2F1(1, L+m+ 1; m+ 2; |z|2).

Let us calculate the first intensity. First of all:

logK f (z, z) = log

((1− |z|2)−L − |z|2m+2

(L + mm + 1

)2F1(1, L + m + 1; m + 2; |z|2)

).

The following equations are long to compute, so we are going to use another notation tosimplify terms, and also we are going to use polar coordinates. We have |z|2 = r2, for0 < r < 1. Let us denote:

2F1(r) ≡ 2F1(1, L + m + 1; m + 2; r2), ∂r ≡∂

∂r, ∂2

rr ≡∂2

∂r2 .

Now,

∂r logK f (r) =

2Lr(1− r2)−L−1 −(

L + mm + 1

) [(2m + 2)r2m+1

2F1(r) + r2m+2∂r2F1(r)]

K f (r).

In the disk of convergence, we have:

∂r2F1(r) =+∞

∑k=1

(L + m + 1)k

(m + 2)k2kr2k−1 =

2r

+∞

∑k=1

(L + m + 1)k

(m + 2)kkr2k

(∗)=

2r

+∞

∑l=0

(L + m + 1)l+1

(m + 2)l+1(l + 1)r2l+2 = 2r

+∞

∑l=0

(L + m + 1)l+1

(m + 2)l+1(l + 1)r2l

= 2rL + m + 1

m + 2

+∞

∑l=0

(L + m + 2)l

(m + 3)l

(2)l

l!r2l = 2r

L + m + 1m + 2 2F1(2, L + m + 2; m + 3; r2),

4.3 GAF in the finite space of polynomials endowed with the norm ‖ · ‖2BL

47

where in (∗) we use the change l = k − 1 and remarking that (2)l = (l + 1)! = Γ(l + 2).Following an analogous reasoning we have that:

∂2rr2F1(r) = 2

L + m + 1m + 2 2F1(2, L + m + 2; m + 3; r2)

+ 8r2 (L + m + 1)(L + m + 2)(m + 2)(m + 3) 2F1(3, L + m + 3; m + 4; r2).

And also we get:

∂rK f (r) = 2Lr(1− r2)−L−1 −(

L + mm + 1

) [(2m + 2)r2m+1

2F1(r) + r2m+2∂r2F1(r)]

.

Then, since the function that we want to compute does not depend of the angle, the Lapla-cian is ∆(r,θ) ≡ ∂2

rr + r−1∂r, and:

∆ logK f (r) =2L[(1− r2)−L−1 + 2(L + 1)r2(1− r2)−L−2]

K f (r)

−

(L + mm + 1

) [(2m + 2)(2m + 1)r2m

2F1(r) + (2m + 2)r2m+1∂r2F1(r)]

K f (r)

−

(L + mm + 1

) [(2m + 2)r2m+1∂r2F1(r) + r2m+2∂2

rr2F1(r)]

K f (r)−(∂rK f (r)

)2(K f (z, z)

)2

+

2L(1− r2)−L−1 −(

L + mm + 1

) [(2m + 2)r2m

2F1(r) + r2m+1∂r2F1(r)]

K f (r).

and the first intensity is:

ρ1(r) =2L[(1− r2)−L−1 + 2(L + 1)r2(1− r2)−L−2]

4πK f (r)

−

(L + mm + 1

) [(2m + 2)(2m + 1)r2m

2F1(r) + (2m + 2)r2m+1∂r2F1(r)]

4πK f (r)

−

(L + mm + 1

) [(2m + 2)r2m+1∂r2F1(r) + r2m+2∂2

rr2F1(r)]

4πK f (r)−

(∂rK f (r)

)2

4π(K f (z, z)

)2

+

2L(1− r2)−L−1 −(

L + mm + 1

) [(2m + 2)r2m

2F1(r) + r2m+1∂r2F1(r)]

4πK f (r).

Using the programs 5.5 and 5.6 for n = 100 and L = 1, we obtain:

48 GAF computation

As we can observe, most of the zero points of a GAF in the finite space of polynomialsendowed with the norm ‖ · ‖2

BLare distributed at the boundary of D. The first intensity,

represented at the picture of the right, takes elevated values when the curve is near of -1and 1, which is obvious with what we explained before.

4.4 GAF in the finite Paley - Wiener space

For the left picture in Figure 4.1 we use the program 5.7 for n = 15 and L = 1. Theright one was made in gnuplot. For Figure 4.2, the left picture was made with the program5.9 and the other with the program 5.8. In the left picture of Figure 4.1, the red dots are

Figure 4.1: [Left picture] Plot of a GAF with NR(0, 1) random variables.[Right picture] Plot of log

(κ(y) +

√κ(y)2 − 1

), where κ(y) = sin(2πiy)/(2πiy).

the real zeros of a GAF with NR(0, 1) random variables, and such a function is representedin green. It is highly remarkable that considering NR(0, 1) or NC(0, 1) random variablescould change the first intensity, as we also saw in the last chapter. In Figure 4.2 there is thefirst intensity for both cases. We see that in the complex one there are no real zeros, butin the real one there are, as we can spot in the picture of the left in Figure 4.1. But, whatgenerates this phenomenon? In Figure 4.2, right picture, we see that the curve intersectswith the imaginary axis at the origin point, and this generates the real zeros since the

4.4 GAF in the finite Paley - Wiener space 49

Figure 4.2: [Left picture] First intensity for NC(0, 1) random variables.[Right picture] First intensity for NR(0, 1) random variables.

curve log(

κ(y) +√

κ(y)2 − 1)

, where κ(y) = sin(2πiy)/(2πiy), behaves as an absolutevalue (see right picture in Figure 4.1), and this implies that the first intensity at y = 0 isproportional to a delta of Dirac at 0. From here it emerges two observations. The first oneis the form of the curve of the first intensity for NR(0, 1) random variables. Comparingboth plots of Figure 4.2, in the real case there are two horns near the origin point. Sincethe intersection generates the real zeros of this GAF, the complex ones move away from aneighbourhood of the origin. Heuristically, the complex zeros are making room for the realones. The second remark is the expected number of zeros in a real interval. If we observeagain Figure 4.1, the left picture, it seems that for almost every interval of length two,there are two real zeros. With this, one could think that the expected number of zeros in aninterval follows this rule: if a and b are real values such that a < b, then ρ1([a, b]) = C|b− a|,where C is a positive real constant and |b− a| is the length of the interval [a, b].

50 GAF computation

Chapter 5

Annex

In this annex we will discuss and expose the C++ programs used to illustrate the chapterGAF computation.

5.1 Preliminary explanations and concepts

Let f be a GAF in the finite space of polynomials endowed with one of the followingnorms: ‖ · ‖2

FL, ‖ · ‖2

PL, or ‖ · ‖2

BL. We know that f is expressed as:

f (z) = ζnenzn + ζn−1en−1zn−1 + ... + ζ1e1z + ζ0e0,

where (ζn)n ∼ NC(0, 1), z is a value that depends of the space before mentioned and (en)n

is the orthonormal basis of the last spaces. What we want to compute is the zero set of f .For this we must introduce the next:

Definition 5.1. Let p be a monic polynomial of degree n:

p(x) = xn + an−1xn−1 + ... + a1x + a0,

where (an)n can be complex or real values. The companion matrix of p is an n× n square matrix

M =

0 0 · · · 0 −a0

1 0 · · · 0 −a1

0 1 · · · 0 −a2...

.... . . . . . · · ·

0 0 · · · 1 −an−1

such that the eigenvalues of M are the zeros points of p.

With this definition, one of the strategies to determine the zero set of a GAF couldbe generate the coefficients of every factor (zk)k, for 0 ≤ k ≤ n, obtain the companionmatrix M and calculate the eigenvalues of this matrix. For this, the C++ library Armadillo1

1Web page: http://arma.sourceforge.net/

51

52 Annex

is a highly useful linear algebra tool that is based on LAPACK (focused on linear algebraoperations). For the finite Paley - Wiener space is rough to explicitly compute the zeroset due to the sine function, independently of using complex or real standard Gaussianrandom variables. Then, we will calculate the position of the zeros of a GAF only with realGaussian random variables by the bisection method.In the program to obtain the first intensity in the finite space of polynomials endowed bythe ‖ · ‖2

BLnorm I used the numerical library gsl 2 (GNU Scientific Library) to use the

hypergeometric function.In the program to obtain the first intensity in the finite space of polynomials endowed bythe ‖ · ‖2

FLnorm I used the library boost 3 to use the incomplete gamma function.

Comments of some non-standard functions that I used in the programs:

• eig_gen(eigval,M)Let eigval be a vector and let M be a general square matrix. This function determinesthe eigenvalues of M and it stores them in eigval. This function is from the libraryArmadillo.

• boost::math::tgamma (double a, double x)Returns the incomplete gamma function described at the beginning of Section 4.1.This function is from the library boost.

• gsl_sf_hyperg_2F1(double a, double b, double c, double x)Returns the hypergeometric function described at the beginning of Section 4.3. Thisfunction is from the library gsl.

5.2 Zeros of a GAF in the finite space of polynomial endowed withthe norm ‖ · ‖2

FL

// Name: Alexis Arraz Almirall// Project: ZEROS OF RANDOM ANALYTIC FUNCTIONS

#include <iostream>#include <armadillo>#include <random>#include <complex>#include <cmath>#include <chrono>

using namespace std;using namespace arma;

2Web page: https://www.gnu.org/software/gsl/3Web page: https://www.boost.org/

5.2 Zeros of a GAF in the finite space of polynomial endowed with the norm ‖ · ‖2FL

53

int main()

int i, n;double L, prod;default_random_engine gen;gen.seed(std::chrono::system_clock::now().time_since_epoch().count());normal_distribution <double> re_rv(0,1./sqrt(2.)), im_rv(0,1./sqrt(2.));ofstream d;d.open("fock_zeros.dad");

cout << "Degree: ";cin >> n;

cout << "Parameter: ";cin >> L;

cx_vec z(n);

z.zeros();

complex<double> val(re_rv(gen), im_rv(gen));

z(0) = val;

prod = 1;

for(i=1; i<n; i++)


z(i) = val * sqrt(prod);

prod *= L/i;

for(i=0; i<(n-1); i++)z(i) = z(i)/z(n-1);

cx_mat M(n,n);

54 Annex

M.zeros();M.diag(-1).ones();for(i=0; i<(n-1); i++)

M.row(i).col(n-1)= -z(i);

cx_vec eigval;

eig_gen(eigval,M);

for(i=0; i<n; i++)d << real(eigval(i)) << "\t" << imag(eigval(i)) << endl;

d.close();

return 0;

5.3 First intensity of a GAF in the finite space of polynomials en-dowed with the norm ‖ · ‖2

FL


#include <iostream>#include <armadillo>#include <random>#include <boost/math/special_functions/gamma.hpp>#include <cmath>

#define _USE_MATH_DEFINES


int main()

int n;double L, gn, g, gN;

ofstream d;

5.4 Zeros of a GAF in the finite space of polynomials endowed with the norm ‖ · ‖2PL

55

d.open("fock_first_int.dad");



double x = 0, r;

dogn = boost::math::tgamma(n-1,L*x*x);g = boost::math::tgamma(n,L*x*x);gN = boost::math::tgamma(n+1,L*x*x);

r = ((L*n)/M_PI)*((g*gN + L*x*x*(n-1)*gn*gN-L*x*x*n*g*g)/(gN*gN));

d << x << "\t" << r << endl;

x += 0.001;

while(x<=6);

d.close();

return 0;

As a remark, the condition at the do - while varies at every set zero we want to compute.In this case, the number 6 corresponds to n = 30 and L = 5. As you can see, the libraryboost has been used here to apply the incomplete gamma function.

5.4 Zeros of a GAF in the finite space of polynomials endowedwith the norm ‖ · ‖2

PL


#include <iostream>#include <armadillo>#include <random>

56 Annex

#include <complex>#include <cmath>#include <chrono>


int main()

int i;double n;default_random_engine gen;gen.seed(std::chrono::system_clock::now().time_since_epoch().count());normal_distribution <double> re_rv(0, 1./sqrt(2.)), im_rv(0, 1./sqrt(2.));ofstream d, g;d.open("poly_zeros.dad");g.open("poly_zeros_3D.dad");


cx_vec z(n);

z.zeros();

double p = 1;

for(i=1; i<=n; i++)


p *= (n-i+1)/i;z(i-1) = val*sqrt(p);

for(i=0; i<(n-1); i++)z(i) = z(i)/z(n-1);

cx_mat M(n,n);

M.zeros();

5.5 Zeros of a GAF in the finite space of polynomials endowed with the norm ‖ · ‖2BL

57

M.diag(-1).ones();for(i=0; i<(n-1); i++)

M.row(i).col(n-1)= -z(i);

cx_vec eigval;

eig_gen(eigval,M);

double comp;

for(i=0; i<n; i++)d << real(eigval(i)) << "\t" << imag(eigval(i)) << endl;comp = real(eigval(i))*real(eigval(i)) + imag(eigval(i))*imag(eigval(i));g << 2*real(eigval(i))/(1+comp) << "\t" << 2*imag(eigval(i))/(1+comp)<< "\t" << (comp-1)/(1+comp) << endl;

d.close();g.close();

return 0;

As a remark, I divided above the line g « 2*real(eigval(i))/(1+comp) « "\t"« 2*imag(eigval(i))/(1+comp) « "\t" « (comp-1)/(1+comp) « endl; into two due toits length.

5.5 Zeros of a GAF in the finite space of polynomials endowedwith the norm ‖ · ‖2

BL


#include <iostream>#include <armadillo>#include <random>#include <complex>#include <cmath>#include <chrono>

58 Annex


int main()

int i, n;double L;default_random_engine gen;gen.seed(std::chrono::system_clock::now().time_since_epoch().count());normal_distribution <double> re_rv(0, 1./sqrt(2.)), im_rv(0, 1./sqrt(2.));ofstream d;d.open("bergman_zeros.dad");



cx_vec z(n);

z.zeros();

double p = 1;

for(i=1; i<=n; i++)


p *= (L+i-1)/i;z(i-1) = val*sqrt(p);

for(i=0; i<(n-1); i++)z(i) = z(i)/z(n-1);

cx_mat M(n,n);

M.zeros();M.diag(-1).ones();

5.6 First intensity of a GAF in the finite space of polynomial endowed with the norm‖ · ‖2

BL59

for(i=0; i<(n-1); i++)M.row(i).col(n-1)= -z(i);

cx_vec eigval;

eig_gen(eigval,M);

for(i=0; i<n; i++)d << real(eigval(i)) << "\t" << imag(eigval(i)) << endl;

d.close();

return 0;

5.6 First intensity of a GAF in the finite space of polynomial en-dowed with the norm ‖ · ‖2

BL


#include <iostream>#include <armadillo>#include <random>#include <cmath>#include <gsl/gsl_sf_hyperg.h>



double comb (double, int);double dhyge (double, double, int);double ddhyge (double, double, int);double k (double, double, int);double dk (double, double, int);double n1 (double, double, int);double n2 (double, double, int);

60 Annex

double n3 (double, double, int);double n4 (double, double, int);double frac1 (double, double, int);double frac2 (double, double, int);

int main()

int n;double L;

ofstream d;d.open("bergman_first_intensity.dad");



double x = -0.99;

do

d << x << "\t" << frac1(x,L,n) + frac2(x,L,n) << endl;

x += 0.01;

while(x<1);

d.close();

return 0;

double comb (double l, int m)

return tgamma(l+m+1)/(tgamma(m+2)*tgamma(l));

double dhyge (double z, double l, int m)

5.6 First intensity of a GAF in the finite space of polynomial endowed with the norm‖ · ‖2

BL61

return 2*z*((l+m+1)/(m+2))*gsl_sf_hyperg_2F1(2,l+m+2,m+3,z*z);

double ddhyge (double z, double l, int m)

return 2*((l+m+1)/(m+2))*gsl_sf_hyperg_2F1(2,l+m+2,m+3,z*z)+ 8*z*z*(((l+m+1)*(l+m+2))/((m+2)*(m+3)))*gsl_sf_hyperg_2F1(3,l+m+3,m+4,z*z);

double k (double z, double l, int m)

return pow(1-z*z,-l) - pow(z,2*m+2) * comb(l,m)* gsl_sf_hyperg_2F1(1,l+m+1,m+2,z*z);

double dk (double z, double l, int m)

return 2*l*z*pow(1-z*z,-l-1)-comb(l,m)*(2*m+2)*pow(z,2*m+1)*gsl_sf_hyperg_2F1(1,l+m+1,m+2,z*z)-comb(l,m)*pow(z,2*m+2)*dhyge(z,l,m);

double n1 (double z, double l)

return 2*l*pow(1-z*z,-l-1)+4*z*z*l*(l+1)*pow(1-z*z,-l-2);

double n2 (double z, double l, int m)

return (2*m+2)*(2*m+1)*pow(z,2*m)*gsl_sf_hyperg_2F1(1,l+m+1,m+2,z*z)+ (2*m+2)*pow(z,2*m+1)*dhyge(z,l,m);


return (2*m+2)*pow(z,2*m+1)*dhyge(z,l,m)+pow(z,2*m+2)*ddhyge(z,l,m);


return dk(z,l,m)*dk(z,l,m);

62 Annex

double frac1 (double z, double l, int m)

return ((n1(z,l)-comb(l,m)*n2(z,l,m)-comb(l,m)*n3(z,l,m))*k(z,l,m)- n4(z,l,m))/(4*M_PI*k(z,l,m)*k(z,l,m));

double frac2 (double z, double l, int m)

return dk(z,l,m)/(4*M_PI*z*k(z,l,m));

Remark: I separated into two the return line in the functions ddhyge, k, dk, n2 and frac1here due to their length. In this program I used the gsl library to apply the hypergeometricfunction. If one implements the definition of this function, the error in the operations isremarkable due to the multiple number of sums and products. Using the library gsl forthis function, the error decreases significantly because it uses a recursive algorithm, insteadof power series.

5.7 Zeros of a GAF in the finite Paley - Wiener space with NR(0, 1)random variables


#include <iostream>#include <random>#include <armadillo>#include <cmath>#include <chrono>



double bisection (double, double, int, double, vec);double f(double, int, double, vec);

int main()

int i, n;

5.7 Zeros of a GAF in the finite Paley - Wiener space with NR(0, 1) random variables 63

double L, h = 0.0001;default_random_engine gen;gen.seed(std::chrono::system_clock::now().time_since_epoch().count());normal_distribution <double> rv(0,1);ofstream d, g;d.open("pw_real_zeros.dad");g.open("pw_real_graphic.dad");



vec rvec(n);

for(i=0; i<n; i++)rvec(i) = rv(gen);

double z0 = -10, z1 = z0 + h, p;

do

g << z0 << "\t" << f(z0,n,L,rvec) << endl;

if(f(z0,n,L,rvec)*f(z1,n,L,rvec) > 0) z0 = z1;z1 += h;

elsep = bisection(z0,z1,n,L,rvec);d << p << "\t" << f(p,n,L,rvec) << endl;z0 = z1;z1 += h;

while(z1 <= 10);

d.close();g.close();

return 0;

64 Annex

double f(double z, int n, double L, vec v)

int i;double sum = 0.0;

for(i=0; i<n; i++)

if(fabs(M_PI*(i-L*z)) > 1.e-10)sum += (v(i)*(sin(M_PI*(i-L*z)))/(M_PI*(i-L*z)));

elsesum += v(i);

return sum;

double bisection (double a, double b, int n, double L, vec v)

int bn = 0;double c;

do

c = (a+b)*0.5;

if(fabs(f(c,n,L,v)) < 1.e-10) return c;

else

if(f(a,n,L,v)*f(c,n,L,v) < 0) b = c;if(f(c,n,L,v)*f(b,n,L,v) < 0) a = c;

bn += 1;

while(fabs(a-b) > 1.e-10 || bn <= 200);

5.8 First intensity of a GAF in the finite Paley - Wiener space with NR(0, 1) randomvariables 65

5.8 First intensity of a GAF in the finite Paley - Wiener space withNR(0, 1) random variables


#include <iostream>#include <armadillo>#include <cmath>#include <math.h>



double num (double);double den (double);

int main()

ofstream d;d.open("pw_real_first_int.dad");

double x = -1.5;

do

d << x << "\t" << num(x)/den(x) << endl;

x += 0.01;

while(x<=1.5);

d.close();

return 0;

66 Annex

double num (double x)

return -64*M_PI*M_PI*M_PI*x*x*x*cosh(2*M_PI*x)+(64*M_PI*M_PI*M_PI*M_PI*x*x*x*x+48*M_PI*M_PI*x*x+3)*sinh(2*M_PI*x)-sinh(6*M_PI*x);

double den (double x)

return 16*x*x*x*(4*M_PI*M_PI*M_PI*x*x-M_PI*sinh(2*M_PI*x)*sinh(2*M_PI*x)) * sqrt(((sinh(2*M_PI*x)*sinh(2*M_PI*x))/(x*x))-4*M_PI*M_PI);

Remark: The return in both functions has been separated here in multiple lines due totheir length.

5.9 First intensity of a GAF in the finite Paley - Wiener space withNC(0, 1) random variables


#include <iostream>#include <armadillo>#include <cmath>#include <math.h>



int main()

ofstream d;d.open("pw_cx_first_int.dad");

5.9 First intensity of a GAF in the finite Paley - Wiener space with NC(0, 1) randomvariables 67

double x = -6;

doif(fabs(x)>0.2)

d << x << "\t" << (1./(4*M_PI*M_PI*x*x))-M_PI*(1./(sinh(2*M_PI*M_PI*x)*sinh(2*M_PI*M_PI*x))) << endl;

elsed << x << "\t" << (M_PI/3) - ((4*M_PI*M_PI*M_PI*x*x)/15) << endl;

x += 0.01;

while(x<=6);

d.close();

return 0;

As a remark, I separated into two lines here the exit of the if due to its length.

68 Annex

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http://math.iisc.ernet.in/~manju/GAF_book.pdf

[3] Patrick Billingsley, Probability and measure, New York: Wiley, cop. 1995, Third edition,Wiley series in probability and mathematical statistics collection.

[4] Joaquim Bruna, Julià Cufí, Anàlisi Complexa, Universitat Autònoma de Barcelona,Servei de Publicacions, Manuals de Matemàtiques UAB. Primera edició, 2008.

[5] Jeremiah Buckley, Random zero sets of analytic functions and traces of functions in Fockspaces, PhD Thesis, Facultat de Matemàtiques, Universitat de Barcelona. Barcelona,Publication date: June, 6th 2013. Link: http://hdl.handle.net/2445/45245

[6] Joan Cerdà, Análisis Real, Segunda edición, 2000. Col.lecció UB 23, Edicions Univer-sitat de Barcelona.

[7] John B. Conway, Functions of one complex variable I, Second Edition, Graduate Texts inMathematics, Springer, 1978.

[8] Naomi D. Feldheim, Zeroes of Gaussian Analytic Functions with translation - invariantdistribution, arXiv:1105.3929v3 [math.PR]

[9] Harumi Hattori, Partial Differential Equations: Methods, Applications and Theories, WestVirginia University, USA. Published by World Scientific, 2013.

[10] Jean-Pierre Kahane, Some random series of functions, Cambridge University Press, 1985.Second Edition. Cambridge studies in advanced mathematics.

[11] Kenneth S. Miller, Complex Stochastic Processes. An Introduction to Theory and Applica-tion, Reading, Mass.: Addison-Wesley, Advanced Book Program, 1974.

69

70 BIBLIOGRAPHY

[12] Kenneth S. Miller, Multidimensional Gaussian distributions, New York: Wiley, 1964.Collection: SIAM series in Applied Mathematics.

[13] Fedor Nazarov, Mikhail Sodin, What is... a Gaussian Entire Function?, Notices of theAMS (March, 2010), Volume 57, Number 3, pp. 375 - 377.

[14] D. Nualart, M. Sanz, Curs de Probabilitats, PPU, Barcelona, 1990, Primera Edició.Col.lecció Estadística y Análisis de Datos.

[15] Joaquim Ortega - Cerdà, Sèries de potències (aleatòries), Butlletí de la Societat Catalanade Matemàtiques, Vol. 30, núm. 2, 2015. pp. 193 - 205.

[16] Vern I. Paulsen and Mrinal Raghupathi, An Introduction to the theory of reproducing ker-nel Hilbert spaces, Cambridge University Press, 2016. Collection: Cambridge studiesin advanced mathematics.

[17] Walter Rudin, Análisis real y complejo, Madrid: Alhambra, 1979. Primera edición es-pañola, 1979.

[18] Conrad Sanderson, Ryan Curtin, Armadillo: a template-based C++ library for linear alge-bra, Journal of Open Source Software, Vol. 1, pp. 26, 2016.

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BIBLIOGRAPHY 71

[29] Weisstein, Eric W. i Wolfram Research Inc. Wolfram Mathworld [recurs en línia delliure accés]. Champaign, Illinois: Wolfram Research Inc, 1998. Web page:

http://www.wolframalpha.com/input/?i=csch%5E2(2*pi*y)

[30] Weisstein, Eric W. i Wolfram Research Inc. Wolfram Mathworld [recurs en línia delliure accés]. Champaign, Illinois: Wolfram Research Inc, 1998. Web page:

https://www.wolframalpha.com/input/?i=laplacian+(1%2F(4*pi))*log((sin(2*pi*i*y))%2F(2*pi*i*y)+%2B+sqrt(((sin(2*pi*i*y))%2F(2*pi*i*y))%5E2-1))

[The last line has been separated into two due to its length]

[All the electronic sources of the project have been visited for the last time on June,26th 2018.]

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