#  Overcrowding for zeros of Hyperbolic Gaussian analytic functions

Keren Mor Waknin  School of Mathematical Sciences, Tel Aviv University, Tel
Aviv 69978, Israel<EMAIL_ADDRESS>
# Overcrowding for zeros of Hyperbolic Gaussian analytic functions

Keren Mor Waknin  School of Mathematical Sciences, Tel Aviv University, Tel
Aviv 69978, Israel<EMAIL_ADDRESS>
###### Abstract.

We consider the family $\\{f_{L}\\}_{L>0}$ of Gaussian analytic functions in
the unit disk, distinguished by the invariance of their zero set with respect
to hyperbolic isometries. Let $n_{L}\left(r\right)$ be the number of zeros of
$f_{L}$ in a disk of radius $r$. We study the asymptotic probability of the
rare event where there is an overcrowding of the zeros as $r\uparrow 1$, i.e.
for every $L>0$ and $\alpha>1$, we are looking for the asymptotics of the
probability
$\mathbb{P}\left[\left|n_{L}\left(r\right)-\mathbb{E}\left[n_{L}\left(r\right)\right]\right|\geq
V\left(r\right)\right]$ with
$V\left(r\right)\simeq\mathbb{E}\left[n_{L}\left(r\right)\right]^{\alpha}$.
Peres and Virág showed that for $L=1$ (and only then) the zero set forms a
determinantal point process, making many explicit computations possible.
Curiously, contrary to the much better understood planar model, it appears
that for $L<1$ the exponential order of decay of the probability of
overcrowding when $\alpha\downarrow 1$ is much less than the probability of a
deficit of zeros.

##  1\. Introduction

We study hyperbolic Gaussian analytic functions (GAFs) in the unit disk
$\mathbb{D}$

$f_{L}\left(z\right)=\sum_{n\geq
0}\xi_{n}\sqrt{\frac{L\left(L+1\right)\cdots\left(L+n-1\right)}{n!}}z^{n},\quad\text{for}\
L>0,$

where $\\{\xi_{n}\\}_{n\geq 0}$ are independent standard complex Gaussian
random variables. The zero set of this family has a distinguished property,
which is the invariance of the distribution of their zero set
$\mathcal{Z}_{f}=f^{-1}\\{0\\}$ with respect to Möbius transformations of the
unit disk. This property is unique in the sense that these are the only GAFs,
up to a multiplication by a deterministic nowhere vanishing analytic function,
whose zeros are isometry-invariant under the respective group of isometries
[9, Chap. 2]. Let $n_{L}\left(r\right)=\\#\\{z\ :\ f\left(z\right)=0\
\text{and}\ \left|z\right|\leq r\\}$ be the number of zeros of $f_{L}$ in the
disk of radius $r\in\left(0,1\right)$. One way to compute the expectation of
$n_{L}\left(r\right)$ (also called _first intensity_) is to use the Edelman-
Kostlan formula, and get that it has constant intensity which equal to
$\frac{L}{\pi}$, with respect to the hyperbolic measure
$\frac{dm\left(z\right)}{\left(1-|z|^{2}\right)^{2}}$ on the unit disk where
$dm\left(z\right)$ is the Lebesgue measure on the plane, see [18]. Buckley [1]
studied the asymptotic of the variance $v_{L}\left(r\right)$ and proved that
there is a transition in the asymptotics of the variance at $L=\frac{1}{2}$.
The case $L=1$ is special in the hyperbolic models. Peres and Virág [17]
discovered that the zero set of the Gaussian Taylor series
$f_{1}(z)=\sum_{n=0}^{\infty}\xi_{n}z^{n}$ is a _determinantal_ point process,
which allowed them to directly compute a lot of its characteristics (see [9,
Chapter 4] and also [9, Section 5.1]). Peres and Virág also showed that $L=1$
is the only instance when the zero set of $f_{L}$ is a determinantal point
process. In [3] Buckley, Nishry, Peled, and Sodin found the rate of decay of
the hole probability $p_{H}\left(r\right):=\mathbb{P}\left[\\{f_{L}\neq 0\
\text{on}\ r\bar{\mathbb{D}}\\}\right]$ as $r\uparrow 1$. Recently, Buckley
and Nishry [2] proved that the limiting (in law) behavior of the normalized
version of $n_{L}\left(r\right)$:

$\widehat{n}_{L}\left(r\right):=\frac{n_{L}\left(r\right)-\mathbb{E}\left[n_{L}\left(r\right)\right]}{\sqrt{v_{L}\left(r\right)}},\quad
r\uparrow 1,$

has a transition when $L=\frac{1}{2}$. They showed that $n_{L}\left(r\right)$
satisfies a CLT for $L\geq\frac{1}{2}$ and exhibits non-Gaussian limiting
behavior for $L<\frac{1}{2}$. The case $L=1$ was treated previously by Peres
and Virág [17]. For the hyperbolic Gaussian analytic functions Krishnnapur
[11] found some bounds for the probability of the overcrowding event
$\\{n_{L}\left(r\right)\geq m\\}$ with $r$ fixed as $m\rightarrow\infty$. In
this work we give matching lower and upper bounds for the overcrowding
probability when $r\uparrow 1$ (see Theorem 1.2).

The first result is about large deviations in the case $L=1$.

######  Theorem 1.1.

Let $\mu_{r},v_{1}\left(r\right)$ be the expectation and the variance of
$n_{1}\left(r\right)$ respectively. Then, for $r\uparrow 1$,

$-\log\mathbb{P}\left[\left|n_{1}(r)-\mu_{r}\right|\geq t\cdot
v_{1}\left(r\right)^{\alpha}\right]\sim c(t)v_{1}(r)^{2\alpha-1}$

where

$c(t)=\begin{cases}{\frac{t^{2}}{2}}&\frac{1}{2}<\alpha<1,\\\
\tfrac{1}{2}\left(t+2\right)^{2}+2\operatorname{Li}_{2}\left(1-\exp\left(\frac{t+2}{2}+W_{0}\left(-\frac{t+2}{2e^{\frac{t+2}{2}}}\right)\right)\right)+\left(t+2\right)W_{0}\left(-\frac{t+2}{2e^{\frac{t+2}{2}}}\right)&\alpha=1,\\\
{\frac{t^{2}}{4}}&\alpha>1,\end{cases}$

$\operatorname{Li}_{2}(z)=-\int_{0}^{z}\frac{\log(1-u)}{u}du$ is the
dilogarithm function [12], and $W_{0}$ is the principle branch of the Lambert
$W$ function [4].

We prove Theorem 1.1 in Appendix A. Our second result is the asymptotics of
the overcrowding probability for general $L>0$.

######  Theorem 1.2.

Fix $L>0$. Then there exists $C_{L}>0$ such that for all
$r\in\left[\tfrac{1}{2},1\right)$ and all $V>1$ satisfying

$V\geq\frac{C_{L}}{1-r}\log\frac{1}{1-r},$

we have

$-\log\mathbb{P}\left[n_{L}(r)\geq V\right]\simeq_{L}\left(1-r\right)V^{2}.$

######  Corollary 1.3.

For all $L>0$ and $\alpha>1$ we get as $r\uparrow 1$,

$-\log\mathbb{P}\left[n_{L}(r)\geq
v_{L}\left(r\right)^{\alpha}\right]\simeq_{L,\alpha}\begin{cases}\left(\frac{1}{1-r}\right)^{2\alpha-1}&L>\frac{1}{2},\\\
\left(\frac{1}{1-r}\right)^{2\alpha-1}\log\left(\frac{1}{1-r}\right)&L=\frac{1}{2},\\\
\left(\frac{1}{1-r}\right)^{2\alpha(2-2L)-1}&0<L<\frac{1}{2}.\end{cases}$

###  1.1. Related works

####  1.1.1. Comparison with the Gaussian entire function (GEF)

An analogue process to $f_{L}$ whose zero set distribution is invariant under
isometries of the plane is given by

$g_{L}\left(z\right)=\sum_{n\geq 0}\xi_{n}\sqrt{\frac{L^{n}}{n!}}z^{n}.$

with $L>0$. Note that it is enough to consider the case $L=1$ since other
values of $L$ are just scaled versions of this case (unlike the hyperbolic
model). The mean number of zeros is equal to $\frac{L}{\pi}$, with respect to
the Lebesgue measure $dm\left(z\right)$ on the plane. The asymptotics of the
variance in the planar $v\left(R\right)$ was computed by Forrester and Honner
in [7]:

$v\left(R\right)=cR+o\left(R\right),\quad R\rightarrow\infty.$

We use upper case $R$ for the radius to emphasize that the function’s domain
is $\mathbb{C}$. In [1], Buckley proved that for the hyperbolic model the
asymptotic of the variance is given by:

$v_{L}\left(r\right)\sim\begin{cases}\frac{c_{L}}{1-r}&\text{for }\
L>\frac{1}{2},\\\ \frac{c}{1-r}\log\frac{1}{1-r}&\text{for}\ L=\frac{1}{2},\\\
\frac{c_{L}}{(1-r)^{2-2L}}&\text{for }0<L<\frac{1}{2},\end{cases}r\rightarrow
1^{-},$

where the constants are explicitly computed and positive in all three cases.
Note that in the planar case, the expectation
$\mathbb{E}\left[n\left(R\right)\right]$ is in fact the Lebesgue measure of
the disc with radius $R$, which is much larger (by factor of $R$) than the
variance $v\left(R\right)$, which is proportional to the length of the
boundary of the disc. However, in the hyperbolic case, the hyperbolic area and
the length circumference of a disk of radius $r$ are approximately the same
and contrary to the plane, the variance is greater or equal (up to constants)
than the expectation (depending on the values of $L$). It is also quite
natural to study the probability of (large) fluctuations. We will restrict
ourselves to the case of a disk with large radius $r$ with respect to the
function’s domain (i.e in the unit disk $r$ is close to 1 and in the complex
plane $R$ tends to infinity) and study the probability:

$\mathbb{P}\left[\left|n_{L}\left(r\right)-\mathbb{E}\left[n_{L}\left(r\right)\right]\right|\geq
v_{L}\left(r\right)^{\alpha}\right],$

beyond what is known by asymptotic normality (i.e. $\alpha>\frac{1}{2}$). This
might be called moderate$\backslash$large$\backslash$huge deviations in
probability language, depending on $\alpha$ and the model we consider. For the
GEF there is a complete answer in terms of $\alpha$ given by:

######  Theorem 1.4 (_Jancovici-Lebowitz-Manificat Law_ [14, 11, 19]).

As $R\rightarrow\infty$

$\frac{\log\left(-\log\left(\mathbb{P}\left[\left|n\left(R\right)-R^{2}\right|>R^{\alpha}\right]\right)\right)}{\log
R}\rightarrow\begin{cases}2\alpha-1&\frac{1}{2}\leq\alpha\leq 1,\\\
3\alpha-2&1\leq\alpha\leq 2,\\\ 2\alpha&\alpha\geq 2.\end{cases}$

Here we can see that the asymptotic behavior changes for different values of
the exponent $\alpha$, unlike the case of the hyperbolic GAF Theorem 1.1. i.e.
considering the $\log\log$ asymptotic of this probability as $r\uparrow 1$, it
holds that the power remains $2\alpha-1$.

####  1.1.2. Determinantal point processes in the unit disk

Our result for the determinantal process in the hyperbolic model is related to
recent work of Fenzl and Lambert (see [6]), where they compute the precise
deviations for disk counting statistic of rotation invariant determinantal
point processes in the disk and the plane. In particular they proved that the
large deviations principle holds for the determinantal point process ($L=1$)
and provided a general formula for the rate of decay in terms of the rate
function. Here we provide an explicit expression to the rate function. Since
the only case when the hyperbolic GAF form a determinantal point process is
when $L=1$, we need to use other techniques for different values of $L$.

####  1.1.3. Overcrowding estimates.

In [11] Krishnapur studied the overcrowding probability for zeros of the
hyperbolic GAF. Our lower bound for
$\mathbb{P}\left[n_{L}\left(r\right)>V\right]$ is close to his. Our upper
bound is more precise, mainly because we found a more accurate estimate of the
asymptotic probability

$\mathbb{P}\left[\int_{r\mathbb{T}}\log|f(re^{i\theta})|d\mu\leq-x\right],\qquad\text{as}\
\,x\to\infty.$

Krishnapur conjectured that his lower bound gives the correct order of decay
of the overcrowding probability, and our Theorem 1.2 confirms his conjecture.

####  1.1.4. Hole probability

Consider the event when $f\left(z\right)\neq 0$ in $r\mathbb{D}$, sometimes
called the _hole event_. We denote
$p_{H}\left(r\right):=\mathbb{P}\left[\\{f_{L}\neq 0\ \text{on}\
r\bar{\mathbb{D}}\\}\right]$ and are interested in its decay as
$R\rightarrow\infty$ or $r\rightarrow 1^{-}$. There are results regarding the
rate of decay of the hole probability in both models - the planar and the
hyperbolic:

######  Theorem 1.5 ([15],[8],[19]).

Suppose $R\rightarrow\infty$. Then,

$-\log p_{H}\left(R\right)=\frac{e^{2}}{4}R^{4}+O\left(R^{2}\log^{2}R\right).$

######  Theorem 1.6 ([3, Theorem 1]).

Suppose $r\uparrow 1$. Then,

$-\log
p_{H}(r)\simeq_{L}\begin{cases}\frac{1}{\left(1-r\right)^{L}}\log\frac{1}{1-r}&\text{for
}\ 0<L<1,\\\ \frac{1}{1-r}\log\frac{1}{1-r}&\text{for}\ L=1,\\\
\frac{1}{1-r}\log^{2}\frac{1}{1-r}&\text{for }L>1,\end{cases}r\rightarrow
1^{-},$

where the constants are explicitly computed and positive.

####  1.1.5. Loss of the balance between excess of zeros and their deficit.

Consider the event of excess of zeros where
$\\{n_{L}\left(r\right)>\mu_{r}+v_{L}\left(r\right)\\}$ and the event of
deficit of zeroes $\\{n_{L}\left(r\right)<\mu_{r}-v_{L}\left(r\right)\\}$
where $\mu_{r}=\mathbb{E}\left[n_{L}\left(r\right)\right]$. Heuristically,
letting $\alpha\downarrow 1$ in Theorem 1.2 would imply

$\displaystyle-\log\mathbb{P}\left[n_{L}\left(r\right)>\mu_{r}+v\left(r\right)^{\alpha}\right]\simeq-\log\mathbb{P}\left[n_{L}\left(r\right)>v\left(r\right)^{\alpha}\right]$
$\displaystyle\simeq-v_{L}\left(r\right)^{2\alpha-1}$
$\displaystyle\overset{\alpha\downarrow
1}{\sim}\begin{cases}-v_{L}\left(r\right)&\frac{1}{2}\leq L<1,\\\
-v_{L}\left(r\right)^{\frac{3-4L}{2-2L}}&0<L<\frac{1}{2},\end{cases}$

while the hole probability results implies that

$\displaystyle-\log\mathbb{P}\left[n_{L}\left(r\right)<\mu_{r}-v\left(r\right)^{\alpha}\right]$
$\displaystyle\geq-\frac{1}{\left(1-r\right)^{L}}\log\frac{1}{1-r}\simeq\begin{cases}-v_{L}\left(r\right)^{L},&\frac{1}{2}\leq
L<1,\\\ -v_{L}\left(r\right)^{\frac{L}{2-2L}},&0<L<\frac{1}{2}\end{cases}$

up to some logarithmic factor which we neglect. I.e. the exponent is different
and we do not get the symmetry between these events (contrary to the planar
model), the probability for deficit is bigger.

###  1.2. Notation & Some special functions

  * –

$c$ and $C$ denote positive numerical constants that do not depend on
$r,\alpha,L$. By $c_{L}$ we denote positive constant that depends only on $L$.
In this work, the values of these constants are irrelevant and may vary from
line to line.

  * –

The notation $X\simeq Y$ means that there exists positive constants such that
$cX\leq Y\leq CX$. The notation $X\simeq_{L}Y$ is used when the constants
depend on $L$.

  * –

$e\left(t\right)=e^{2\pi it}$.

  * –

$\left[x\right]$ denotes the integer part of $x$.

  * –

$n\equiv k\left(N\right)$ means that $n\equiv k\ \text{modulo}\ N$.

  * –

$\mathbb{D}$ denotes the open unit disk, $\mathbb{T}$ denotes the unit circle.

  * –

We set $\delta:=1-r$. We always assume that $r$ is sufficiently close to $1$
(and hence $\delta$ close to $0$).

  * –

Let $f\left(r\right),g\left(r\right)$ be positive functions. The notation
$f\left(r\right)\sim g\left(r\right)$ as $r\uparrow 1$ means that
$\lim_{r\uparrow 1}\frac{f\left(r\right)}{g\left(r\right)}=1$. The notation
$f\left(r\right)\ll g\left(r\right)$ as $r\uparrow 1$ means that
$\lim_{r\uparrow 1}\frac{f\left(r\right)}{g\left(r\right)}=0$.

  * –

Consider the probability
$\mathbb{P}\left[\left|n_{L}\left(r\right)-\mathbb{E}\left[n_{L}\left(r\right)\right]\right|\geq
V\left(r\right)\right]$. In the hyperbolic case we define the types of
deviations as follows:

    * –

Moderate: $v_{L}^{\beta}<V<\mathbb{E}^{\alpha}$ with
$\frac{1}{2}<\beta\leq\alpha<1$ where $L<1-\frac{\alpha}{2\beta}$

    * –

Large: $V\simeq\mathbb{E}$

    * –

Huge: $V\simeq\mathbb{E}^{\alpha}$ with $\alpha>1$.

where $\mathbb{E}=\mathbb{E}\left[n_{L}\left(r\right)\right]$ is the expected
value.

### Acknowledgments

I am deeply grateful to my MSc advisors, Alon Nishry and Mikhail Sodin, for
their constant encouragement, endless patience and enlightening discussions
throughout this project. I would also like to thank Naomi Feldheim, Manjunath
Krishnapur, Ron Peled, and Oren Yakir for their valuable comments and fruitful
discussions. I thank Lakshmi Priya for reading this paper with great care and
making very insightful comments, which improved both the mathematics and the
presentation.

##  2\. Proof of Theorem 1.2

Fix $L>0$. Recall that we want to prove that there exists $C_{L}>0$ such that
for all $r\in\left[\tfrac{1}{2},1\right)$ and all $V>1$ satisfying

(1) $V\geq\frac{C_{L}}{1-r}\log\frac{1}{1-r},$

we have

$-\log\mathbb{P}\left[n_{L}(r)\geq V\right]\simeq_{L}\left(1-r\right)V^{2}.$

#### Several remarks on Theorem 1.2

  1. (1)

Assumption (1) is meant to ensure that the number of zeros $V$ is large enough
(compared with the expectation).

  2. (2)

Recall that

$v_{L}\left(r\right)=\textrm{Var}\left[n_{L}\left(r\right)\right]\sim\begin{cases}\frac{c_{L}}{1-r}&\text{for
}\ L>\frac{1}{2},\\\ \frac{c}{1-r}\log\frac{1}{1-r}&\text{for}\
L=\frac{1}{2},\\\ \frac{c_{L}}{(1-r)^{2-2L}}&\text{for
}0<L<\frac{1}{2}\end{cases}\quad,\quad r\uparrow 1.$

By the above, $v_{L}^{\alpha}\left(r\right)$ will satisfy assumption (1),
provided that either $L>0$, $\alpha>1$, or $L<\frac{1}{2}$,
$\alpha\in(\frac{1}{2-2L},1]$. In both cases
$v_{L}(r)^{\alpha}\gg\mu_{r}=\frac{Lr^{2}}{1-r^{2}}$ as $r\uparrow 1$, and
hence

$\log\mathbb{P}\left[\left|n_{L}(r)-\mu_{r}\right|\geq
v_{L}\left(r\right)^{\alpha}\right]\simeq\log\mathbb{P}\left[n_{L}\left(r\right)\geq
v_{L}\left(r\right)^{\alpha}\right].$

In order to prove Theorem 1.2 we find matching lower and upper bounds for the
$(\log)$ probability of the overcrowding event. We employ strategies somewhat
similar to the ones introduced in [11, 19, 3, 16]. Throughout the proof we
will assume that $r$ is sufficiently close to $1$, as we can do.

###  2.1. Lower Bound

######  Claim 2.1.

For $r\in[\tfrac{1}{2},1)$ we have

$\log\mathbb{P}\left[n_{L}\left(r\right)\geq V\right]\geq-
c_{L}\left(1-r\right)V^{2}.$

###### Proof.

Fix $r$ close to 1 and take $m=m\left(r\right)=\left[V\right]+1$. Consider the
following independent events:

  * •

$E_{1}\coloneqq\left\\{\left|\xi_{n}\right|\leq\sqrt{n}\ ,\ n\geq
2m+1\right\\},$

  * •

$E_{2}\coloneqq\left\\{\left|\xi_{n}\right|\binom{-L}{n}^{\frac{1}{2}}\leq\binom{-L}{m}^{\frac{1}{2}}r^{m}\
,\ m+1\leq n\leq 2m\right\\},$

  * •

$E_{3}\coloneqq\left\\{\left|\xi_{m}\right|\geq 4m\sqrt{1-r}\right\\},$

  * •

$E_{4}\coloneqq\left\\{\left|\xi_{n}\right|\binom{-L}{n}^{\frac{1}{2}}r^{n}\leq\sqrt{1-r}\binom{-L}{m}^{\frac{1}{2}}r^{m}\
,\ 0\leq n\leq m-1\right\\}$.

######  Claim 2.2.

Under the events $E_{1},E_{2},E_{3},E_{4}$, on $|z|=r$ we have,

$\left|f_{L}(z)-\xi_{m}\binom{-L}{m}^{\frac{1}{2}}z^{m}\right|<\left|\xi_{m}\binom{-L}{m}^{\frac{1}{2}}z^{m}\right|$

whence , $n_{L}\left(r\right)=m$.

######  Claim 2.3.

For all $1\leq j\leq 4$,

$\mathbb{P}\left[E_{j}\right]\geq\exp\left(-c_{L}\left(1-r\right)V^{2}\right).$

Overall, using the independence of these events we get the desired lower
bound, proving 2.1. It remains to prove these two claims.

###### Proof of 2.2.

It holds that

$\displaystyle\left|f_{L}\left(z\right)-\xi_{m}\binom{-L}{m}^{\frac{1}{2}}z^{m}\right|$
$\displaystyle\leq\left(\sum_{n=0}^{m-1}+\sum_{n=m+1}^{2m}+\sum_{n=2m+1}^{\infty}\right)\left|\xi_{n}\right|\binom{-L}{n}^{\frac{1}{2}}r^{n}$
$\displaystyle\leq
m\sqrt{1-r}\binom{-L}{m}^{\frac{1}{2}}r^{m}+\binom{-L}{m}^{\frac{1}{2}}r^{m}\cdot\frac{r^{m+1}(1-r^{m})}{1-r}$
$\displaystyle+\sum_{n=2m+1}^{\infty}\sqrt{n}\binom{-L}{n}^{\frac{1}{2}}r^{n}.$

So we need to show that the second and the third terms on the right hand side
are also bounded by $m\sqrt{1-r}\binom{-L}{m}^{\frac{1}{2}}r^{m}$.

  * –

2nd term:

$r^{m}=e^{-m\left|\log r\right|}\ll\sqrt{1-r}\quad\text{(assumption
\eqref{assumption})}$

whence,

$r^{m+1}\frac{1-r^{m}}{1-r}<mr^{m}\ll m\sqrt{1-r}.$

  * –

3rd term: Let $c_{n}=\sqrt{n}\binom{-L}{n}^{\frac{1}{2}}r^{n}$. Then
$\frac{c_{n+1}}{c_{n}}=\sqrt{\frac{L+n}{n}}\cdot r\rightarrow r$ as
$n\rightarrow\infty$. Choose $r$ sufficiently close to 1 such that
$\frac{c_{n+1}}{c_{n}}\leq\frac{1}{2}\left(1+r\right)$. Then,

$\sum_{n=2m+1}^{\infty}c_{n}\leq
c_{2m+1}\left(\frac{1+r}{2}\right)^{2m+1}\frac{1}{1-\frac{1+r}{2}}\ll
c_{2m+1}\ \text{(assumption \eqref{assumption})}$ $\lesssim c_{m}r^{m}\
\left(\text{since}\binom{-L}{2m+1}\simeq\binom{-L}{m}\right)$
$=\sqrt{m}r^{m}\binom{-L}{m}^{\frac{1}{2}}r^{m}\ll
m\sqrt{1-r}\binom{-L}{m}^{\frac{1}{2}}r^{m}.$

∎

###### Proof of 2.3.

We start with the event $E_{1}$. As $r\uparrow 1:$

$\displaystyle\mathbb{P}\left[\left|\xi_{n}\right|\leq\sqrt{n},\forall n\geq
2m+1\right]$ $\displaystyle\geq 1-\sum_{n=2m+1}^{\infty}e^{-n}$
$\displaystyle=1-e^{-\left(2m+1\right)}\frac{1}{1-e^{-1}}\geq\frac{1}{2}.$

For $E_{2}$, since
$\binom{-L}{n}\underset{n\rightarrow\infty}{\sim}\frac{n^{L-1}}{\Gamma(L)}$,
we get that, for $m+1\leq n\leq 2m$, it holds that
$\frac{\binom{-L}{m}^{\frac{1}{2}}}{\binom{-L}{n}^{\frac{1}{2}}}\simeq 1$
Therefore,

$\frac{\binom{-L}{m}^{\frac{1}{2}}}{\binom{-L}{n}^{\frac{1}{2}}}\cdot r^{m}\ll
1.$

Next, $|\xi_{n}|^{2}\sim\exp(1)$ implies that:

$\displaystyle\mathbb{P}\left[\left|\xi_{n}\right|\leq\frac{\binom{-L}{m}^{\frac{1}{2}}}{\binom{-L}{n}^{\frac{1}{2}}}\cdot
r^{m}\ ,\forall\ m+1\leq n\leq 2m\right]$
$\displaystyle\geq\prod_{n=m+1}^{2m}c_{L}r^{2m}$
$\displaystyle=c_{L}^{m}r^{2m^{2}}\geq e^{-c_{L}m^{2}\left(1-r\right)}.$

For $E_{3}$, $\mathbb{P}\left[|\xi_{m}|\geq
4m\sqrt{1-r}\right]=e^{-16m^{2}\left(1-r\right)}$. Finally, for $E_{4}$ we
have,

$\displaystyle\mathbb{P}\left[\left|\xi_{0}\right|\leq\sqrt{1-r}\frac{\binom{-L}{m}^{\frac{1}{2}}}{\binom{-L}{n}^{\frac{1}{2}}}\cdot
r^{m-n}\right]$
$\displaystyle\geq\mathbb{P}\left[\left|\xi_{0}\right|\leq\sqrt{1-r}\cdot
m^{\frac{L-1}{2}}r^{m}\right]$
$\displaystyle\geq\tfrac{1}{2}\left(1-r\right)m^{L-1}r^{2m}\geq
e^{-c_{L}m\left(1-r\right)},$

and for $1\leq n\leq m-1$ we have $\binom{-L}{n}\leq B_{L}n^{L-1}$, while
$\binom{-L}{m}\geq b_{L}m^{L-1}$ hence

$\displaystyle\mathbb{P}\left[\left|\xi_{n}\right|\leq\sqrt{1-r}\frac{\binom{-L}{m}^{\frac{1}{2}}}{\binom{-L}{n}^{\frac{1}{2}}}r^{m-n}\right]\geq\mathbb{P}\left[\left|\xi_{n}\right|\leq
A_{L}\sqrt{1-r}\left(\frac{m}{n}\right)^{\frac{L-1}{2}}r^{m-n}\right]$

Note that for $r$ sufficiently close to 1,

$\max_{1\leq n\leq
m-1}\left(\frac{m}{n}\right)^{\frac{L-1}{2}}r^{m-n}=\left(\frac{m}{m-1}\right)^{\frac{L-1}{2}}r\leq
D_{L}$

which implies that
$A_{L}\sqrt{1-r}\left(\frac{m}{n}\right)^{\frac{L-1}{2}}r^{m-n}\ll 1$, and
hence,

$\mathbb{P}\left[\left|\xi_{n}\right|\leq
A_{L}\sqrt{1-r}\left(\frac{m}{n}\right)^{\frac{L-1}{2}}\cdot
r^{m-n}\right]\geq
c_{L}\left(1-r\right)\left(\frac{m}{n}\right)^{L-1}r^{2\left(m-n\right)}$

So finally we get,

$\mathbb{P}\left[\left|\xi_{n}\right|\binom{-L}{n}^{\frac{1}{2}}r^{n}\leq\sqrt{1-r}\binom{-L}{m}^{\frac{1}{2}}r^{m}\
,\forall\ 0\leq n\leq m-1\right]$ $\displaystyle\geq
e^{-c_{L}m\left(1-r\right)}\prod_{n=1}^{m-1}\left(d_{L}\left(1-r\right)\left(\frac{m}{n}\right)^{L-1}r^{2\left(m-n\right)}\right)$
$\displaystyle=e^{-c_{L}m\left(1-r\right)}d_{L}^{m}\left(1-r\right)^{m}r^{b_{L}m^{2}}\prod_{n=1}^{m-1}\left(\frac{m}{n}\right)^{L-1}\geq
e^{-c_{L}m^{2}\left(1-r\right)},$

proving 2.3. ∎

Combining altogether, we get the desired lower bound. ∎

###  2.2. Upper Bound

######  Claim 2.4.

For $r\in[\tfrac{1}{2},1)$ we have

$\log\mathbb{P}\left[n_{L}\left(r\right)\geq
V\right]\leq-\tilde{c}_{L}\left(1-r\right)V^{2}.$

In this part we’ll use the following parameters:

$\delta\coloneqq 1-r\
,R\coloneqq\tfrac{1}{2}\left(1+r\right)=1-\tfrac{1}{2}\delta\ ,\
r_{0}\coloneqq 1-2\delta\ ,\ N\coloneqq\eta\left[V\right],$

where $\eta$ is a sufficiently small constant that will be chosen later on.
Recall the Poisson-Jensen formula, if
$f\in\text{Hol}\left(r\bar{\mathbb{D}}\right)$,
$\mathcal{Z}_{f}=f^{-1}\\{0\\}$ then,

$\log\left|f\left(z\right)\right|=\int_{0}^{2\pi}\log\left|f\left(re^{i\theta}\right)\right|\frac{r^{2}-\left|z\right|^{2}}{\left|re^{i\theta}-z\right|^{2}}d\mu\left(\theta\right)+\sum_{\lambda\in\mathcal{Z}_{f}\cap
r\bar{\mathbb{D}}}\log\left|\frac{r\left(z-\lambda\right)}{r^{2}-\bar{\lambda}z}\right|$

and also Jensen’s formula, for $z=0,f(0)\neq 0$

$\log\left|f\left(0\right)\right|+\int_{0}^{r}\frac{n\left(t\right)}{t}dt=\int_{0}^{2\pi}\log\left|f\left(re^{i\theta}\right)\right|d\mu\left(\theta\right).$

Hence we have,

(2)
$n_{L}\left(r\right)\log\frac{R}{r}\leq\left(\int_{R\mathbb{T}}-\int_{r\mathbb{T}}\right)\log\left|f\right|d\mu\leq\log
M_{f}\left(R\right)-\int_{r\mathbb{T}}\log|f|d\mu$

where $M_{f}\left(r\right)=\max_{r\mathbb{D}}\left|f\right|$.

######  Lemma 2.5 ([3, Lemma 2]).

Let $f$ be a GAF on $\mathbb{D}$, and $s\in\left(0,\delta\right)$. Put

$\sigma_{f}^{2}(r)=\max_{r\bar{\mathbb{D}}}\mathbb{E}[|f|^{2}].$

Then, for every $\lambda>0$,

$\mathbb{P}\big{[}M_{f}(r)>\lambda\sigma_{f}(r+s)\big{]}\leq\frac{C}{s}e^{-c\lambda^{2}}.$

For the hyperbolic GAF $f$, we have
$\sigma_{f}\left(t\right)=\left(1-t^{2}\right)^{-\tfrac{L}{2}}$. Hence

$2^{-\tfrac{L}{2}}\left(1-t\right)^{-\tfrac{L}{2}}\leq\sigma_{f}\left(t\right)\leq\left(1-t\right)^{-\tfrac{L}{2}},$

whence,

(3)
$\mathbb{P}\left[M_{f}\left(t\right)>\lambda\left(1-\left(t+s\right)\right)^{{-\tfrac{L}{2}}}\right]\leq\mathbb{P}\left[M_{f}\left(t\right)>\lambda\sigma_{f}\left(t+s\right)\right].$

Choose $t=R,s=\tfrac{1}{4}\delta,$ and $\lambda=\tfrac{1}{2}\sqrt{1-r}V$ in 3
and use Lemma 2.5 to get

(4)
$\mathbb{P}\left[M_{f}\left(R\right)>\left(\tfrac{\delta}{4}\right)^{\frac{1}{2}-\frac{L}{2}}V\right]\leq
C\delta^{-1}e^{-c\left(1-r\right)V^{2}}.$

Using that $\delta=1-r$ in (4), we get

(5)
$\mathbb{P}\left[M_{f}\left(R\right)>2^{L-1}\left(1-r\right)^{\tfrac{1-L}{2}}V\right]\leq
C\delta^{-1}e^{-c\left(1-r\right)V^{2}}.$

Denote by
$E_{r}^{(1)}:=\\{M_{f}\left(R\right)>2^{L-1}\left(1-r\right)^{\tfrac{1-L}{2}}V\\}$.
Then, outside $E_{r}^{(1)}$, the following holds:

(6) $\begin{split}\log M_{f}\left(R\right)&\leq\left(L-1\right)\log 2+\log
V+\tfrac{L-1}{2}\log\left(1-r\right)^{-1}\leq a_{L}\log V\end{split}$

where the last inequality follows from our assumption (1). Next, set
$Z_{r}=\left\\{n_{L}\left(r\right)\geq V\right\\}$; using (2) and (1), we see
that on $Z_{r}\backslash E_{r}^{(1)}$ we have:

$\int_{r\mathbb{T}}\log\left|f\right|\leq-d_{L}\left(1-r\right)V,$

where $d_{L}=-\frac{1}{2}+\frac{a_{L}}{C_{L}}$.

Choosing $N$ points $\\{r_{0}e^{i\theta_{j}}\\}_{j=1}^{N}$ with $r_{0}<r$,

$\displaystyle\int_{0}^{2\pi}\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|d\mu\left(\beta\right)$
$\displaystyle=\frac{1}{N}\sum_{j=1}^{N}\int_{0}^{2\pi}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|d\mu\left(\beta\right)$
$\displaystyle=\log\left|f\left(0\right)\right|+\int_{0}^{r_{0}}\frac{n_{L}\left(t\right)}{t}dt.$

By continuity away from zeros of $\log\left|f\right|$ we can find $\beta^{*}$
such that

(7) $\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta^{*}}\right)\right|=\int_{r_{0}\mathbb{T}}\log\left|f\right|d\mu\leq\int_{r\mathbb{T}}\log\left|f\right|d\mu\leq-
d_{L}\left(1-r\right)V.$

In the next step, we wish to avoid the dependence on $\beta^{*}$ by expanding
this estimate to a small arc on $r_{0}\mathbb{T}$ centered at
$r_{0}e^{i\beta^{*}}$. To this end, we will bound the angular derivative of
$\log\left|f\right|$ assuming that $f$ doesn’t vanishes in a thin ring around
the circle $r_{0}\mathbb{T}$.

Let $\left|z\right|=r_{0}$ and assume that $f\neq 0$ on
$r_{0}-\epsilon\leq\left|z\right|\leq r_{0}+\epsilon$ with
$\epsilon=\epsilon\left(r\right)\ll(1-r)^{3}$. Differentiating the Poisson-
Jensen, we get

$\displaystyle\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}$
$\displaystyle=\int_{0}^{2\pi}\frac{{2}r_{0}\cdot
r\left(r^{2}-r_{0}^{2}\right){\sin\left(\varphi-\theta\right)}}{\left|re^{i\theta}-r_{0}e^{i\varphi}\right|^{4}}\cdot\log\left|f\left(re^{i\theta}\right)\right|d\mu\left(\theta\right)$
$\displaystyle{+\sum_{\lambda\in\mathcal{Z}_{f}\cap
r\bar{\mathbb{D}}}r_{0}|\lambda|\sin\left(\theta-\varphi_{\lambda}\right)\left[\frac{1}{\left|r_{0}e^{i\varphi}-\lambda\right|^{2}}-\frac{r^{2}}{\left|r^{2}-\bar{\lambda}r_{0}e^{i\varphi}\right|^{2}}\right].}$

where $\lambda=|\lambda|e^{i\varphi_{\lambda}}$. Then,

$\displaystyle\left|\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}\right|$
$\displaystyle\leq\frac{{4}}{\left(r-r_{0}\right)^{3}}\int_{0}^{2\pi}\left|\log\left|f\left(re^{i\theta}\right)\right|\right|d\mu\left(\theta\right)$
$\displaystyle+\sum_{\lambda\in\mathcal{Z}_{f}\cap
r\bar{\mathbb{D}}}\left[\frac{1}{\underset{\lambda\in\mathcal{Z}_{f}\cap
r\bar{\mathbb{D}}}{\min}\left|r_{0}e^{i\varphi}-\lambda\right|^{2}}+\frac{1}{\left(r-r_{0}\right)^{2}}\right]$
$\displaystyle\leq\frac{{4}}{\left(r-r_{0}\right)^{3}}\int_{0}^{2\pi}\left|\log\left|f\left(re^{i\theta}\right)\right|\right|d\mu\left(\theta\right)+\frac{2n_{L}\left(r\right)}{\epsilon^{2}}$

Using (2) and (6), we bound each term on the right hand side.

The second summand is bounded by

$\displaystyle\frac{2n_{L}\left(r\right)}{\epsilon^{2}}$
$\displaystyle\leq\frac{2}{\epsilon^{2}}\left[\log
M_{f}\left(R\right)-\int_{r\mathbb{T}}\log\left|f\right|d\mu\right]\lesssim_{L}\frac{1}{\epsilon^{2}}\left[\log
V+\int_{r\mathbb{T}}\log\\_\left|f\right|d\mu\right]$
$\displaystyle\overset{\eqref{assumption}}{\lesssim}\frac{1}{\epsilon^{2}}\left[\left(1-r\right)V+\int_{r\mathbb{T}}\log\\_\left|f\right|d\mu\right].$

The first summand is bounded by

$\displaystyle\frac{4}{\left(r-r_{0}\right)^{3}}\int_{0}^{2\pi}\left|\log\left|f\left(re^{i\theta}\right)\right|\right|d\mu$
$\displaystyle\lesssim\frac{1}{\left(r-r_{0}\right)^{3}}\left[\log
M_{f}\left(r\right)+\int_{r\mathbb{T}}\log\\_\left|f\right|d\mu\right]$
$\displaystyle\lesssim\frac{1}{\left(r-r_{0}\right)^{3}}\left[\left(1-r\right)V+\int_{r\mathbb{T}}\log\\_\left|f\right|d\mu\right].$

Hence,

(8)
$\max_{\varphi}\left|\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}\right|\lesssim\frac{1}{\epsilon^{2}}\left(\left(1-r\right)V+\int_{r\mathbb{T}}\log\\_\left|f\right|d\mu\right).$

######  Lemma 2.6.

Set $\epsilon:=e^{-s\left(1-r\right)V^{2}}$ with some fixed constant $s>0$.
Then, outside a negligible event $E_{r}^{(2)}$, $f\neq 0$ in the annulus
$\left\\{r_{0}-\epsilon\leq\left|z\right|\leq r_{0}+\epsilon\right\\}$ .

###### Proof.

Applying Chebyshev’s inequality, we get

$\displaystyle\mathbb{P}\left[n_{L}\left(r_{0}+\epsilon\right)-n_{L}\left(r_{0}-\epsilon\right)\geq
1\right]$
$\displaystyle\leq\mathbb{E}\left[n_{L}\left(r_{0}+\epsilon\right)-n_{L}\left(r_{0}-\epsilon\right)\right]$
$\displaystyle=L\left(\frac{\left(r_{0}+\epsilon\right)^{2}}{1-\left(r_{0}+\epsilon\right)}-\frac{\left(r_{0}-\epsilon\right)^{2}}{1-\left(r_{0}-\epsilon\right)^{2}}\right)$
$\displaystyle\lesssim_{L}\epsilon\left(1-r\right)^{-2}.$

Therefore, for every $s^{\prime}<s$, we get (assuming $r$ is sufficiently
close to 1)

$\mathbb{P}\left[E_{r}^{(2)}\right]\leq e^{-s^{\prime}\left(1-r\right)V^{2}},$

proving the lemma. ∎

Next, to give an estimate for the integral
$\int_{r\mathbb{T}}\log_{-}\left|f\right|d\mu$, we want to make sure that
$M_{f}\left(r_{0}\right)$ is not too small.

######  Lemma 2.7.

Let $E_{r}^{(3)}=\left\\{\log
M_{f}\left(r_{0}\right)\leq-\left(1-r\right)V\right\\}$. Then

$\mathbb{P}\left[E_{r}^{\left(3\right)}\right]\leq
e^{-a_{L}\left(1-r\right)V^{2}}.$

###### Proof.

Assume that $M_{f}\left(r_{0}\right)\leq e^{-\left(1-r\right)V}$ , then by
Cauchy’s inequalities

$\left|\xi_{n}\right|\binom{-L}{n}^{\frac{1}{2}}\leq\frac{M_{f}\left(r_{0}\right)}{{r_{0}}^{n}}\
,\ n\geq 0.$

Since for all $n>0$, $\binom{-L}{n}^{\frac{1}{2}}\geq
c_{L}n^{\frac{1}{2}\left(L-1\right)}$ we get that,

$\left|\xi_{n}\right|\leq
B_{L}n^{\frac{1}{2}\left(1-L\right)}{r_{0}}^{-n}e^{-\left(1-r\right)V}.$

Recall that $r_{0}=1-2\delta$ so, when $r\geq\frac{1}{2}$,
$\delta\leq\frac{1}{2}$, hence $r_{0}^{-1}\leq e^{2\left(1-r\right)}$ and
then,

$\left|\xi_{n}\right|\leq
B_{L}n^{\frac{1}{2}\left(1-L\right)}e^{\left(1-r\right)\left(2n-V\right)}.$

In the range $1\leq n\leq\frac{1}{4}V$, the right-hand side doesn’t exceed
(assuming $C_{L}$ is large)

$e^{-b_{L}\left(1-r\right)V}$

and the probability that $\left|\xi_{n}\right|$ is smaller than this term
doesn’t exceed

$e^{-2b_{L}\left(1-r\right)V}.$

Overall,

$\mathbb{P}\left[E_{r}^{\left(3\right)}\right]\leq\left(e^{-2b_{L}\left(1-r\right)V}\right)^{\frac{1}{4}V}=e^{-a_{L}\left(1-r\right)V^{2}}.$

∎

We estimate $\int_{r\mathbb{T}}\log\\_\left|f\right|$ assuming that we are
outside of the event $E_{r}^{(1)}\cup E_{r}^{(2)}\cup E_{r}^{(3)}$. By Lemma
2.7, there exists $w\in r_{0}\mathbb{T}$ such that
$\log\left|f\left(w\right)\right|\geq-\left(1-r\right)V$. Let
$P\left(z,w\right)$ be the Poisson kernel for the disk $r{\mathbb{D}}$,
evaluated at $w$. Then, using subharmonicity of $\log\left|f\right|$ in
$r{\mathbb{D}}$ we have,

$-\left(1-r\right)V\leq\log\left|f\left(w\right)\right|\leq\int_{r\mathbb{T}}\log\left|f\right|P\left(\cdot,w\right)=\int_{r\mathbb{T}}\left(\log_{+}\left|f\right|-\log\\_\left|f\right|\right)P\left(\cdot,w\right),$

whence,

$\displaystyle\frac{r-r_{0}}{r+r_{0}}\int_{r\mathbb{T}}\log\\_\left|f\right|$
$\displaystyle\leq\int_{r\mathbb{T}}\log\\_\left|f\right|P\left(\cdot,w\right)$
$\displaystyle\leq\int_{r\mathbb{T}}\log_{+}\left|f\right|P\left(\cdot,w\right)+\left(1-r\right)V$
$\displaystyle\leq\frac{r+r_{0}}{r-r_{0}}\int_{r\mathbb{T}}\log_{+}\left|f\right|+\left(1-r\right)V\leq\frac{r+r_{0}}{r-r_{0}}\log
M_{f}\left(r\right)+\left(1-r\right)V$
$\displaystyle\lesssim_{L}\frac{r+r_{0}}{r-r_{0}}\left(1-r\right)V+\left(1-r\right)V.$

Therefore, assuming $r$ is sufficiently close to 1,

${\int_{r\mathbb{T}}\log\\_\left|f\right|\lesssim\frac{1}{\left(r-r_{0}\right)^{2}}\left(1-r\right)V.}$

Finally, we can plug this into (8) and obtain,

(9)
$\begin{split}\max_{\varphi}\left|\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}\right|\lesssim\frac{1}{\epsilon^{2}}\left(1-r\right)V\leq
e^{s^{\prime\prime}\left(1-r\right)V^{2}},&\quad s^{\prime\prime}>s,\\\
\text{where $r$ is sufficiently close to $1$.}\end{split}$

Combining this with (7) we get that for
${\left|\beta-\beta^{*}\right|\leq\Delta:=e^{-2s^{\prime\prime}V^{2}\left(1-r\right)}}$,

$\displaystyle\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|$
$\displaystyle-\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta^{*}}\right)\right|$
$\displaystyle\leq\left|\beta-\beta^{*}\right|\cdot\sup_{\varphi}\left|\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}\right|,$

which implies

(10) ${\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|\leq-\tfrac{d_{L}}{2}\left(1-r\right)V.}$

Now we are ready to complete the proof. Consider the product space
$\left(\mathbb{T},m\right)\times\left(\Omega,\mathbb{P}\right)$ with the
probability measure $\mathbb{Q}=m\times\mathbb{P}$. Here $\beta$ is chosen
uniformly in $[0,2\pi]$ and $\Omega$ is the probability space for our GAF $f$.
Consider the following events:

  * –

$H=\left\\{\left(\beta,\omega\right)\mid n_{L}\left(r\right)\geq V\right\\}$
$=\mathbb{T}\times Z_{r}$

  * –

$K=\left\\{\left(\beta,\omega\right)\mid\max_{\varphi}\left|\frac{\partial\log\left|f\left(r_{0}e^{i\varphi}\right)\right|}{\partial\varphi}\right|\leq
e^{s^{\prime\prime}\left(1-r\right)V^{2}}\right\\}\supset\mathbb{T}\times\left(Z_{r}\backslash
E_{r}^{(1)}\cup E_{r}^{(2)}\cup E_{r}^{(3)}\right)$

  * –

$J=\left\\{\left(\beta,\omega\right)\mid f\neq 0\ \text{in}\
\text{ann}\left(0,r_{0}-e^{-s\left(1-r\right)V^{2}},r_{0}+e^{-s\left(1-r\right)V^{2}}\right)\right\\}=\mathbb{T}\times\left(\Omega\backslash
E_{r}^{(3)}\right)$

  * –

$C=\left\\{\left(\beta,\omega\right)\mid\frac{1}{N}\sum_{j=1}^{N}\log|f(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta})|\leq-d_{L}\left(1-r\right)V\right\\}$

  * –

$D=\\{(\beta,\omega)\mid|\beta-\beta^{*}(\omega)|\leq\Delta=e^{-2s^{\prime\prime}\left(1-r\right)V^{2}}\\}$

######  Remark 2.8.

Recall that $\beta^{*}:\Omega\mapsto[0,2\pi)$, where $\beta^{*}$ was defined
in (7). We mention that $\beta^{*}$ may indeed be chosen to be measurable with
respect to $\Omega$. See Appendix B for a more detailed explanation.

Next, note that $H,K,J$ do not depend on $\beta$, and that
$\mathbbm{1}_{C}\left(\beta,\omega\right)\geq\mathbbm{1}_{H\cap J\cap K\cap
D}\left(\beta,\omega\right)$.
Also, the events $H\cap J\cap K$ and $D$ are independent. Indeed,

$\mathbbm{1}_{H\cap J\cap K\cap D}(\beta,\omega)=\mathbbm{1}_{H\cap J\cap
K}(\beta,\omega)\cdot\mathbbm{1}_{D}(\beta,\omega)=\mathbbm{1}_{H\cap J\cap
K}(0,\omega)\cdot\mathbbm{1}_{D}(\beta,\omega),$

and by Fubini’s theorem,

$\displaystyle\mathbb{Q}\left(H\cap J\cap K\cap D\right)$
$\displaystyle=\int_{\mathbb{T}\times\Omega}\mathbbm{1}_{D}\left(\beta,\omega\right)\mathbbm{1}_{H\cap
J\cap K}(0,\omega)d\mathbb{Q}$
$\displaystyle=\int_{\Omega}\left[\int_{\left\\{\left|\beta-\beta^{*}\left(\omega\right)\right|\right\\}}dm\left(\beta\right)\right]\mathbbm{1}_{H\cap
J\cap K}\left(0,\omega\right)d\mathbb{P}\left(\omega\right)$
$\displaystyle=2\Delta\int_{\Omega}\mathbbm{1}_{H\cap J\cap
K}\left(0,\omega\right)d\mathbb{P}\left(\omega\right)=\mathbb{Q}\left(D\right)\cdot\mathbb{Q}\left(H\cap
J\cap K\right)$

as desired. Then,
$\mathbbm{1}_{C}\left(\beta,\omega\right)\geq\mathbbm{1}_{H\cap J\cap
K}\left(0,\omega\right)\cdot\mathbbm{1}_{D}\left(\beta,\omega\right)$ and

$\displaystyle\mathbb{Q}\left(C\right)$ $\displaystyle\geq 2\Delta
P\left(H\cap J\cap K\right)\geq
2\Delta\left[\mathbb{Q}\left(H\right)-\mathbb{Q}\left(K^{\complement}\right)-\mathbb{Q}\left(J^{\complement}\right)\right]$
$\displaystyle\geq\Delta\left[\mathbb{P}\left(Z_{r}\right)-\mathbb{P}\left(E_{r}^{(1)}\right)-\mathbb{P}\left(E_{r}^{(2)}\right)-\mathbb{P}\left(E_{r}^{(3)}\right)\right]$

whence

$\mathbb{P}\left(Z_{r}\right)\leq\frac{1}{\Delta}\mathbb{Q}\left(C\right)+\mathbb{P}\left(E_{r}^{(1)}\right)+\mathbb{P}\left(E_{r}^{(2)}\right)+\mathbb{P}\left(E_{r}^{(3)}\right)$
(11) ${\leq
e^{2s^{\prime\prime}\left(1-r\right)V^{2}}\mathbb{Q}\left(C\right)+e^{-a_{L}\left(1-r\right)V^{2}}.}$

The following lemma is required to finish the proof of the upper bound.

######  Lemma 2.9.

For every fixed $\beta$,

(12)
$\mathbb{P}\left[\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|\leq-\tfrac{d_{L}}{2}\left(1-r\right)V\right]\leq
e^{-c_{L}\left(1-r\right)V^{2}}.$

Using this lemma and Fubini’s theorem we get,

$\mathbb{Q}(C)=\int\left[\int\mathbbm{1}_{C}(\beta,\omega)d\mathbb{P}(\omega)\right]dm\left(\beta\right)\leq
e^{-c_{L}\left(1-r\right)V^{2}}.$

Plugging in $\Delta$ together with this bound into Equation 11 we get,

$\mathbb{P}\left(Z_{r}\right)\leq
e^{\left(2s^{\prime\prime}-c_{L}\right)\left(1-r\right)V^{2}}+e^{-a_{L}V^{2}\left(1-r\right)}.$

Finally we choose $s^{\prime\prime}<\frac{1}{2}c_{L}$ and get the desired
upper-bound.

###### Proof of Lemma 2.9.

Fix $\beta$,$\zeta>0$ by Chebyshev’s inequality,

$\displaystyle\mathbb{P}\left[\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|\leq-\tfrac{d_{L}}{2}N\left(1-r\right)V\right]$
$\displaystyle=\mathbb{P}\left[\prod_{j=1}^{N}\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|^{-\zeta}\geq e^{\tfrac{d_{L}}{2}\zeta
N\left(1-r\right)V}\right]$ $\displaystyle\leq e^{-\tfrac{d_{L}}{2}\zeta
N\left(1-r\right)V}\cdot\mathbb{E}\big{[}\prod_{j=1}^{N}|f(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta})|^{-\zeta}\big{]}.$

To estimate the expectation on the right-hand side, we use several lemmas from
[3].

######  Lemma 2.10 ([3, Lemma 15]).

Suppose that $\left(\eta_{j}\right)_{1\leq j\leq N}$ are complex Gaussian
random variables with covariance matrix $\Sigma$. Then, for $0\leq\zeta<2$,

$\mathbb{E}\left[\prod_{j=1}^{N}\left|\eta_{j}\right|^{-\zeta}\right]\leq\frac{1}{\det\Sigma}\left(\Lambda^{\left(1-\frac{1}{2}\zeta\right)}\cdot\Gamma\left(1-\frac{1}{2}\zeta\right)\right)^{N},$

where $\Lambda$ is the largest eigenvalue of $\Sigma$.

######  Lemma 2.11 ([3, Lemma 10]).

Let $F$ be any Gaussian Taylor series with radius of convergence $R$ and let
$z_{j}=re\left(j/N\right)$ for $r<R$ and $j=0,\dots,N-1$. Consider the
covariance matrix $\Sigma=\Sigma\left(r,N\right)$ of the random variables
$F\left(z_{0}\right),\dots,F\left(z_{N-1}\right)$, that is,

$\Sigma_{jk}=\mathbb{E}\left[F\left(z_{j}\right)\overline{F\left(z_{k}\right)}\right]=\sum_{n\geq
0}a_{n}^{2}r^{2n}e\left(\left(j-k\right)n/N\right).$

Then, the eigenvalues of $\Sigma$ are

$\lambda_{m}=N\cdot\sum_{n\equiv m(N)}a_{n}^{2}r^{2n},\ \ m=0,\dots,N-1,$

where $n\equiv m(N)$ denotes that $n$ is equivalent to $m$ modulo $N$.

######  Lemma 2.12.

[3, Sec. 6.2] Let $\Sigma=\Sigma\left(r,N\right)$ be the covariance matrix of
the Gaussian random variables
$F\left(z_{0}\right),\dots,F\left(z_{N-1}\right)$, $z_{j}=re\left(j/N\right)$,
and let $\Lambda=\Lambda\left(r,N\right)$ be the largest eigen value of
$\Sigma$. Then,

$\Lambda\leq\begin{cases}\frac{C\cdot N}{\delta^{L-1}},&L>1\\\ C\cdot N,&L\leq
1\\\ \end{cases}\quad\text{and}\quad\det\Sigma\geq
e^{\left(1+o\left(1\right)\right)\left(LN\log N-2\delta N^{2}\right)}.$

###### Proof of Lemma 2.12.

We begin with a lower bound for $\text{det}\Sigma$. In our case
$a_{m}^{2}=\binom{-L}{m}\geq c\left(m+1\right)^{L-1}$, hence

$\displaystyle\text{det}\Sigma$
$\displaystyle=\prod_{m=0}^{N-1}\lambda_{m}\left(\Sigma\right)\geq\prod_{m=0}^{N-1}Na_{m}^{2}\left(1-2\delta\right)^{2m}\geq
N^{N}\left(1-2\delta\right)^{N(N-1)}\prod_{m=0}^{N-1}c\left(m+1\right)^{L-1}$
$\displaystyle=\left(cN\right)^{N}\left(N!\right)^{L-1}\left(1-2\delta\right)^{N\left(N-1\right)}=\exp\left(-4\delta
N^{2}+LN\log N+O\left(N\right)\right)$

For the upper bound for $\Lambda$, assume $L\leq 1$ (the case $L>1$ appears in
[3]). Since

$a_{m+jN}^{2}=\binom{-L}{m+jN}\sim_{L}\left(m+jN\right)^{L-1}\ \text{as}\
n\rightarrow\infty,$

we have for small $\delta$ and $j\geq 0$,

$\displaystyle\frac{a_{m+\left(j+1\right)N}^{2}r_{0}^{2\left(m+\left(j+1\right)N\right)}}{a_{m+jN}^{2}r_{0}^{2\left(m+jN\right)}}$
$\displaystyle\leq D_{L}\left(\frac{m+(j+1)N}{m+jN}\right)^{L-1}r_{0}^{2N}$
$\displaystyle=D_{L}\left(1+\frac{N}{m+jN}\right)^{L-1}r_{0}^{2N}$
$\displaystyle\leq e^{-4\delta N}.$

Take $r$ close enough to $1$, such that

$\frac{a_{m+(j+1)N}^{2}r_{0}^{2(m+(j+1)N)}}{a_{m+jN}^{2}r_{0}^{2(m+jN)}}\leq\frac{1}{2}.$

So,

$\lambda_{m}\left(\Sigma\right)\leq C\cdot N\cdot
a_{m}^{2}r_{0}^{2m}\sum_{j\geq 0}2^{-j}=C\cdot N\cdot a_{m}^{2}r_{0}^{2m}\leq
C\cdot N.$

(where the last inequality holds because $a_{m}^{2}\leq 1$). ∎

Now we readily complete the proof of Lemma 2.9,and therefore, of Theorem 1.2
Set $\zeta:=2-\delta$. Then, using $\Gamma\left(z\right)\sim(\frac{1}{z})$ as
$z\rightarrow 0$,
$\Gamma\left(1-\frac{1}{2}\zeta\right)=\Gamma\left(\frac{\delta}{2}\right)\leq\frac{2}{\delta}=\frac{2}{1-r}$
we get,

$\displaystyle\mathbb{E}\big{[}\prod_{j=1}^{N}|f(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta})|^{-\zeta}\big{]}$ $\displaystyle\leq e^{4\delta N^{2}-LN\log
N+O\left(N\right)}\left(\Lambda^{\frac{\delta}{2}}\cdot\frac{1}{\delta}\right)^{N}.$

Hence,

(13)
$\begin{split}\log\mathbb{E}\left[\prod_{j=1}^{N}|f(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta})|^{-\zeta}\right]&\leq 4\delta N^{2}-LN\log N+\delta
N\log\Lambda+N\log\frac{1}{1-r}+O\left(N\right)\\\ &\leq 4\delta
N^{2}+B_{L}\delta N\log(N)+N\log\frac{1}{1-r}+O\left(N\right).\end{split}$

Recall that $N=\eta\left[V\right]$ with a small constant $\eta$ and that $V$
satisfies assumption (1). Then the expression on the RHS of (13) is bounded by

$\left(4\eta^{2}+\frac{\max\\{L+1,2\\}}{C_{L}}\eta\right)\left(1-r\right)V^{2}$

assuming $r$ is sufficiently close to $1$. Also, recall that

$d_{L}=-\frac{1}{2}+\frac{a_{L}}{C_{L}}$

with fixed $a_{L}$. Therefore, since $\zeta\geq 1$ and $C_{L}$ is large, we
may choose $\eta$ sufficiently small and get

$\displaystyle\log\mathbb{P}\left[\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|\leq-\tfrac{d_{L}}{2}N\left(1-r\right)V\right]\leq-\tilde{c}_{L}\left(1-r\right)V^{2}.$

∎

##  Appendix A The Case $L=1$ \- proof of Theorem 1.1

Recall that for $L=1$, the random variable $n_{1}\left(r\right)$ is
distributed as a sum of independent Bernoulli random variables (see[17]),

(14)
$\begin{split}\mu_{r}&=\mathbb{E}\left[\sum_{k=1}^{\infty}X_{k}\right]=\sum_{k=1}^{\infty}r^{2k}=\frac{r^{2}}{1-r^{2}}\
\underset{r\uparrow 1}{\sim}\ \frac{1}{2}\cdot\frac{1}{1-r},\\\
v_{1}\left(r\right)&=v_{1}\left(r\right)=\textrm{Var}\left[\sum_{k=1}^{\infty}X_{k}\right]=\sum_{k=1}^{\infty}r^{2k}\left(1-r^{2k}\right)=\frac{r^{2}}{1-r^{4}}\
\underset{r\uparrow 1}{\sim}\ \frac{1}{4}\cdot\frac{1}{1-r}.\end{split}$

Here, it is not hard to calculate the moment generating function, hence we can
use large-deviations techniques to find the _rate function_ to be defined in
(17) and get the rate of decay of the large deviations probability (see [5,
Section 2.3] )

Now, in order to prove Theorem 1.1, we set:

$Z_{r}:=\frac{1}{v_{1}\left(r\right)^{\alpha}}\sum_{k=1}^{\infty}(X_{k}-r^{2k})\
,$ (15) $\epsilon_{r}:=\frac{1}{v_{1}\left(r\right)^{2\alpha-1}}\ ,$

where $r\in(0,1)$. Let $\Lambda_{r}\left(\cdot\right)$ be the logarithmic
moment generating function of $Z_{r}$ and

(16) $\Lambda\left(\lambda\right):=\underset{r\uparrow
1}{\lim}\epsilon_{r}\Lambda_{r}\left(\epsilon_{r}^{-1}\lambda\right).$

The rate function is defined as follows

(17)
$\Lambda^{*}\left(x\right)=\underset{\lambda\in\mathbb{R}}{\sup}\left\\{\lambda
x-\Lambda\left(\lambda\right)\right\\}$

###  A.1. Calculation of $\Lambda\left(\lambda\right)$

######  Lemma A.1.

For $\alpha>\frac{1}{2}$, the limit (16) exists, and differentiable for every
$\lambda\in\mathbb{R}$.

  1. (1)

For $\frac{1}{2}<\alpha<1$,

$\Lambda\left(\lambda\right)=\frac{\lambda^{2}}{2}$

for every $\lambda\in\mathbb{R}.$

  2. (2)

For $\alpha=1$

$\Lambda\left(\lambda\right)=-2\lambda-2\operatorname{Li}_{2}\left(1-e^{\lambda}\right)$

for every $\lambda\in\mathbb{R}$.

  3. (3)

For $\alpha>1$

$\Lambda(\lambda)=\begin{cases}\lambda^{2}&\lambda>0,\\\ 0&\lambda\leq 0.\\\
\end{cases}$

In order to apply the _Gärtner-Ellis_ theorem [5, Section 2.3], let us find
the rate function for each case.

###  A.2. The _Fenchel-Legendre transform_ $\Lambda^{*}\left(x\right)$

Using Lemma A.1, it is easy to compute $\Lambda^{*}\left(x\right)$ in case
$\frac{1}{2}<\alpha<1$ or $\alpha>1$, while the case $\alpha=1$ (which is more
complicated) follows from Lemma A.2 below.

  * •

For $\frac{1}{2}<\alpha<1$,

(18) $\Lambda^{*}\left(x\right)=\frac{x^{2}}{2}.$

  * •

For $\alpha=1$ (Lemma A.2),

(19) $\Lambda^{*}\left(x\right)=\begin{cases}+\infty&x<-2,\\\
\frac{\pi^{2}}{3}&x=-2,\\\
h\left(\lambda_{-1}\left(x\right),x\right)&-2<x<0,\\\
h\left(\lambda_{0}\left(x\right),x\right)&x\geq 0,\\\ \end{cases}$

where

$h\left(y,x\right)=y(x+2)+2\operatorname{Li}\left(1-e^{y}\right),$
$\lambda_{j}\left(x\right)=\frac{x+2}{2}+W_{j}\left(-\frac{x+2}{2}e^{-\frac{x+2}{2}}\right),\quad
j\in\\{-1,0\\}.$

$W_{j}$, $j\in\\{-1,0\\}$ is the principle$/$lower branches of Lambert $W$
function respectively.

Figure 1. The rate function $\Lambda^{*}$ for $\alpha=1$

Note that because $\Lambda^{*}$ is not symmetric the probability of the event
$\left|n_{1}\left(r\right)-\mu_{r}\right|\geq v_{1}\left(r\right)$ is
dominated by the event $n_{1}\left(r\right)\geq\mu_{r}+tv_{1}\left(r\right)$.
Furthermore, the lower bound for the special case of the hole event when
$t=-2,\alpha=1$ does not follow from the theory of large deviations, but can
be computed directly (see [9, Section 5.1]).

  * •

For $\alpha>1$,

(20) $\Lambda^{*}\left(x\right)=\begin{cases}\frac{x^{2}}{4}&x\geq 0,\\\
+\infty&x<0.\\\ \end{cases}$

###  A.3. Proof of Theorem 1.1

Next, assuming Lemma A.1, Theorem 1.1 follows from the Gartner-Ellis theorem:

###### Proof.

Notice that

$\displaystyle\mathbb{P}\left[\left|\sum_{k=1}^{\infty}X_{k}-\mu_{r}\right|\geq
t\cdot v_{1}\left(r\right)^{\alpha}\right]$
$\displaystyle=\mathbb{P}\left[\left|\sum_{k=1}^{\infty}\left(X_{k}-r^{2k}\right)\right|\geq
t\cdot v_{1}\left(r\right)^{\alpha}\right]$
$\displaystyle=\mathbb{P}\left[\left|Z_{r}\right|\geq
t\right]=\mathbb{P}\left[Z_{r}\in\left(-\infty,-t\right]\cup\left[t,\infty\right)\right]$

  * •

For $\frac{1}{2}<\alpha<1$, it holds that

$\underset{x\in\left(-\infty,-t\right)\cup\left(t,\infty\right)}{\inf}\Lambda^{*}\left(x\right)=\underset{x\in\left(-\infty,-t\right]\cup\left[t,\infty\right)}{\inf}\tfrac{1}{2}x^{2}=\frac{t^{2}}{2}$

thus,

$\underset{r\uparrow
1}{\lim}\epsilon_{r}\log\mu_{r}((-\infty,-t]\cup[t,\infty))=-\frac{t^{2}}{2},$

whence,

$-\log\mathbb{P}\left[\left|\sum_{k=1}^{\infty}X_{k}-\mu_{r}\right|\geq
v_{1}\left(r\right)^{\alpha}\right]\sim{\frac{t^{2}}{2}v_{1}\left(r\right)^{2\alpha-1}},\quad
r\uparrow 1.$

  * •

For $\alpha=1$, it holds that

$\underset{x\in\left(-\infty,-t\right)\cup\left(t,\infty\right)}{\inf}\Lambda^{*}\left(x\right)=\underset{x\in\left(-\infty,-t\right]\cup\left[t,\infty\right)}{\inf}\Lambda^{*}\left(x\right)=\min\left\\{\Lambda^{*}\left(-t\right),\Lambda^{*}\left(t\right)\right\\}=\Lambda^{*}\left(t\right)$

where the last equality also follows from Lemma A.2. Thus,

$\underset{r\uparrow
1}{\lim}\epsilon_{r}\log\mu_{r}\left(\left(-\infty,-t\right]\cup\left[t,\infty\right)\right)=-\Lambda^{*}\left(t\right)=-c(t),$

whence,

$-\log\mathbb{P}\left[\left|\sum_{k=1}^{\infty}X_{k}-\mu_{r}\right|\geq t\cdot
v_{1}\left(r\right)^{\alpha}\right]\sim{\frac{c(t)}{4}\cdot\left(\frac{1}{1-r}\right)^{2\alpha-1}},\quad
r\uparrow 1.$

  * •

For $\alpha>1$, it holds that
$\underset{x\in\left(-\infty,-t\right)\cup\left(t,\infty\right)}{\inf}\tfrac{1}{4}x^{2}=\frac{t^{2}}{4}$
thus,

$\underset{r\uparrow
1}{\lim}\epsilon_{r}\log\mu_{r}([t,\infty))=-\frac{t^{2}}{4},$

whence,

$-\log\mathbb{P}\left[\left|\sum_{k=1}^{\infty}X_{k}-\mu_{r}\right|\geq t\cdot
v_{1}\left(r\right)^{\alpha}\right]\sim{\frac{t^{2}}{4}v_{1}\left(r\right)^{2\alpha-1}},\quad
r\uparrow 1.$

∎

Finally, we turn back to the calculation of the logarithmic moment generating
function:

###### Proof of Lemma A.1.

Let $\lambda\in\mathbb{R}$,

$\displaystyle\Lambda_{r}\left(\epsilon_{r}^{-1}\lambda\right)$
$\displaystyle=\Lambda_{r}\left(v_{1}\left(r\right)^{2\alpha-1}\lambda\right)=\log\mathbb{E}\left[e^{v_{1}\left(r\right)^{2\alpha-1}\lambda
Z_{r}}\right]$
$\displaystyle=\log\mathbb{E}\left[e^{v_{1}\left(r\right)^{2\alpha-1}\lambda\frac{1}{v_{1}\left(r\right)^{\alpha}}\sum_{k=1}^{\infty}\left(X_{k}-r^{2k}\right)}\right]$
$\displaystyle=\log\mathbb{E}\left[e^{v_{1}\left(r\right)^{\alpha-1}\lambda\sum_{k=1}^{\infty}\left(X_{k}-r^{2k}\right)}\right]=\sum_{k=1}^{\infty}\log\mathbb{E}\left[e^{v_{1}\left(r\right)^{\alpha-1}\lambda\left(X_{k}-r^{2k}\right)}\right]$
$\displaystyle=\sum_{k=1}^{\infty}\log\left(r^{2k}e^{v_{1}\left(r\right)^{\alpha-1}\lambda\left(1-r^{2k}\right)}+\left(1-r^{2k}\right)e^{v_{1}\left(r\right)^{\alpha-1}\lambda\left(-r^{2k}\right)}\right)$
$\displaystyle=\sum_{k=1}^{\infty}\log\left(e^{v_{1}\left(r\right)^{\alpha-1}\lambda\left(-r^{2k}\right)}\left(r^{2k}e^{v_{1}\left(r\right)^{\alpha-1}\lambda}+1-r^{2k}\right)\right)$
$\displaystyle=\sum_{k=1}^{\infty}v_{1}\left(r\right)^{\alpha-1}\lambda\left(-r^{2k}\right)+\sum_{k=1}^{\infty}\log\left(1+r^{2k}\left(e^{v_{1}\left(r\right)^{\alpha-1}\lambda}-1\right)\right)$
$\displaystyle=-\mu_{r}v_{1}\left(r\right)^{\alpha-1}\lambda+\sum_{k=1}^{\infty}\log\left(1+r^{2k}\left(e^{v_{1}\left(r\right)^{\alpha-1}\lambda}-1\right)\right)=(\star).$

Put $T:=e^{v_{1}\left(r\right)^{\alpha-1}\lambda}-1$ and note that
$-1<T<+\infty$ ,
$\text{sgn}\left(T\right)=\text{sgn}\left(\lambda\right)=\text{sgn}\left(1+Tr^{2k}\right)$.
We look at the map $k\rightarrow\log\left(1+Tr^{2k}\right)$ on
$\left[0,\infty\right)$ and see that:

  * –

For $\lambda>0$ the mapping decreases and its range is between
$\log\left(1+T\right)=v_{1}\left(r\right)^{\alpha-1}\lambda$ to $0$.

  * –

For $\lambda<0$ the mapping increases and its range is between
$v_{1}\left(r\right)^{\alpha-1}\lambda$ to $0$.

Hence, we can write

$\sum_{k=1}^{\infty}\log\left(1+Tr^{2k}\right)=\int_{0}^{\infty}\log\left(1+Tr^{2x}\right)dx+O\left(v_{1}\left(r\right)^{\alpha-1}\right),$

and using the change of variables $u=-Tr^{2x},\ dx=\frac{1}{2\log
r}\frac{du}{u}$ we get

$\displaystyle\sum_{k=1}^{\infty}\log\left(1+Tr^{2k}\right)$
$\displaystyle=-\frac{1}{2\log
r}\int_{0}^{-T}\frac{\log\left(1-u\right)}{u}du+O\left(v_{1}\left(r\right)^{\alpha-1}\right)$
$\displaystyle=\frac{1}{2\log
r}\operatorname{Li}_{2}\left(-T\right)+O\left(v_{1}\left(r\right)^{\alpha-1}\right).$

Hence we get,

$(\star)=-\mu_{r}v_{1}\left(r\right)^{\alpha-1}\lambda+\frac{1}{2\log
r}\operatorname{Li}_{2}\left(-T\right)+O\left(v_{1}\left(r\right)^{\alpha-1}\right),$

therefore,

$\epsilon_{r}\Lambda_{r}\left(\epsilon_{r}^{-1}\lambda\right)=-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda+\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\operatorname{Li}_{2}\left(-T\right)+o(1).$

Finally, all that’s left is to calculate

$\lim_{r\uparrow
1}\left(-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda+\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\operatorname{Li}_{2}\left(-T\right)\right).$

We split the calculation into three cases based on the values of $\alpha$ and
use the asymptotic properties of the $\text{Li}_{2}$ function that we noted in
Section 1.2:

  1. (1)

For $\frac{1}{2}<\alpha<1$, $T\rightarrow 0$ when $r\uparrow 1$, hence $T\sim
v_{1}\left(r\right)^{\alpha-1}\lambda$. Thus in this case

$\operatorname{Li}_{2}\left(-T\right)\sim
v_{1}\left(r\right)^{\alpha-1}\lambda+\frac{1}{4}v_{1}\left(r\right)^{2\alpha-2}\lambda^{2},$

therefore,

$\displaystyle-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda+\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\operatorname{Li}_{2}\left(-T\right)$
$\displaystyle\sim-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda-\frac{v_{1}\left(r\right)^{-\alpha}}{2\log
r}\lambda-\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\cdot\frac{v_{1}\left(r\right)^{2\alpha-2}\lambda^{2}}{4}$
$\displaystyle\sim-\lambda
v_{1}\left(r\right)^{-\alpha}\left(\mu_{r}+\frac{1}{2\log
r}\right)+\frac{\lambda^{2}}{2}\sim\frac{\lambda^{2}}{2},\quad\text{as}\
r\uparrow 1.$

  2. (2)

For $\alpha=1$ it follows immediately that
$\Lambda\left(\lambda\right)=-2\lambda-2\operatorname{Li}_{2}\left(1-e^{\lambda}\right)$.

  3. (3)

For $\alpha>1$,

     * •

If $\lambda<0$, $T\rightarrow-1$ when $r\uparrow 1$, hence
$\operatorname{Li}_{2}\left(-T\right)\rightarrow\operatorname{Li}_{2}\left(1\right)=\frac{\pi^{2}}{6}$.
Thus

$\lim_{r\uparrow
1}-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda+\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\operatorname{Li}_{2}\left(-T\right)=\lim_{r\uparrow
1}-2v_{1}\left(r\right)^{1-\alpha}-2v_{1}\left(r\right)^{2-2\alpha}\frac{\pi^{2}}{6}=0.$

     * •

If $\lambda>0$, $T\rightarrow+\infty$ when $r\uparrow 1$ hence $T\sim
e^{v_{1}\left(r\right)^{\alpha-1}}$. So in this case

$\operatorname{Li}_{2}\left(z\right)\sim-\frac{1}{2}\log^{2}\left(T\right)\sim-\frac{1}{2}v_{1}\left(r\right)^{2\alpha-2}\lambda^{2},$

thus

$\displaystyle\lim_{r\uparrow
1}-\mu_{r}v_{1}\left(r\right)^{-\alpha}\lambda+\frac{v_{1}\left(r\right)^{-2\alpha+1}}{2\log
r}\operatorname{Li}_{2}\left(-T\right)$ $\displaystyle=\lim_{r\uparrow
1}-2v_{1}\left(r\right)^{1-\alpha}\lambda-\left(2v_{1}\left(r\right)^{2-2\alpha}\right)\left(-\frac{1}{2}v_{1}\left(r\right)^{2\alpha-2}\lambda^{2}\right)=\lambda^{2}.$

     * •

For $\lambda=0$,
$\epsilon_{r}\Lambda_{r}\left(\epsilon_{r}^{-1}\lambda\right)=0$,

proving the claim. ∎

During the proof we used the formula for $\Lambda^{*}$ in the case $\alpha=1$,
we calculate it as follows:

######  Lemma A.2.

$\Lambda^{*}\left(x\right)=\begin{cases}+\infty&x<-2,\\\
\frac{\pi^{2}}{3}&x=-2,\\\
h\left(\lambda_{-1}\left(x\right),x\right)&-2<x<0,\\\
h\left(\lambda_{0}\left(x\right),x\right)&x\geq 0,\\\ \end{cases}$

where

$h\left(y,x\right)=y(x+2)+2\operatorname{Li}\left(1-e^{y}\right),$
$\lambda_{j}\left(x\right)=\frac{x+2}{2}+W_{j}\left(-\frac{x+2}{2}e^{-\frac{x+2}{2}}\right),\quad
j\in\\{-1,0\\}.$

$W_{j}$, $j\in\\{-1,0\\}$ is the principle$/$lower branches of Lambert $W$
function respectively.

###### Proof.

Define
$h\left(\lambda\right)=\lambda\left(x+2\right)+2\operatorname{Li}_{2}\left(1-e^{\lambda}\right).$
It holds that

$\lim_{\lambda\rightarrow+\infty}h\left(\lambda\right)=\lim_{\lambda\rightarrow+\infty}\lambda\left(x+2\right)-\frac{\pi^{2}}{3}-\log^{2}\left(e^{\lambda}-1\right)=-\infty,$
$\lim_{\lambda\rightarrow-\infty}h\left(\lambda\right)=\lim_{\lambda\rightarrow-\infty}\lambda\left(x+2\right)+2\operatorname{Li}_{2}\left(1\right)=\begin{cases}2\operatorname{Li}_{2}\left(1\right)=\frac{\pi^{2}}{3}&x=-2,\\\
+\infty&x<-2,\\\ -\infty&x>-2.\\\ \end{cases}$

Consider the equation,

(21)
$ye^{y}=-\frac{1}{2}{\left(x+2\right)}e^{-\tfrac{1}{2}{\left(x+2\right)}},y=\lambda-\frac{1}{2}{\left(x+2\right)}.$

Hence,

$y=W\left(-\frac{1}{2}\left(x+2\right)e^{-\tfrac{1}{2}{\left(x+2\right)}}\right)\implies\lambda=\frac{1}{2}\left(x+2\right)+W\left(-\frac{1}{2}{\left(x+2\right)}e^{-\tfrac{1}{2}{\left(x+2\right)}}\right).$

$W_{0}\left(0\right)=0,W_{0}\left(-\frac{1}{e}\right)=-1$ and $W_{-1}$
decreases from $W_{-1}(-\frac{1}{e})=-1$ to
$W_{-1}\left(0^{-}\right)=-\infty$. Furthermore, it holds that:

$\lim_{x\rightarrow\infty}\frac{W_{0}\left(x\right)}{\log\left(x\right)}=1\ ,\
\lim_{x\rightarrow 0^{-}}\frac{W_{-1}\left(x\right)}{\log\left(-x\right)}=1.$

Observe that since $h^{\prime}$ is monotone decreasing, when $-2<x<0$,
$\lambda$ has to be negative hence by (A.3) and the fact that
$\frac{1}{2}{\left(x+2\right)}\leq 1$, $W$ has to be less than $-1$ so in this
case we’ll choose the branch $W_{-1}$. By the same argument when $x\geq 0$
we’ll choose $W_{0}$. Let us denote the solution according to the branches
$\lambda_{0/-1}(x)$ and overall we get-

$\Lambda^{*}\left(x\right)=\begin{cases}+\infty&x<-2,\\\
\frac{\pi^{2}}{3}&x=-2,\\\
h\left(\lambda_{-1}\left(x\right),x\right)&-2<x<0,\\\
h\left(\lambda_{0}\left(x\right),x\right)&x\geq 0.\\\ \end{cases}$

It holds that

$\displaystyle\lim_{x\downarrow-2}\Lambda^{*}\left(x\right)=\lim_{x\downarrow-2}\
\left(x+2\right)\left(\frac{1}{2}\left(x+2\right)+W_{-1}\left(-\frac{1}{2}\left(x+2\right)e^{-\frac{1}{2}{\left(x+2\right)}}\right)\right)$
$\displaystyle+2\operatorname{Li}_{2}\left(1-\exp\left(\frac{1}{2}\left(x+2\right)+W_{-1}\left(-\frac{1}{2}\left(x+2\right)e^{-\frac{1}{2}\left(x+2\right)}\right)\right)\right)$
$\displaystyle=\frac{\pi^{2}}{3},$

and

$\displaystyle\lim_{x\rightarrow+\infty}\Lambda^{*}\left(x\right)=\lim_{x\rightarrow+\infty}\
\left(x+2\right)\left(\frac{1}{2}{\left(x+2\right)}+W_{0}\left(-\frac{1}{2}{\left(x+2\right)}e^{-\tfrac{1}{2}{\left(x+2\right)}}\right)\right)$
$\displaystyle+2\operatorname{Li}_{2}\left(1-\exp\left({\frac{1}{2}{\left(x+2\right)}+W_{0}\left(-\frac{1}{2}{\left(x+2\right)}e^{-\frac{1}{2}{\left(x+2\right)}}\right)}\right)\right)$
$\displaystyle=+\infty,$

where the asymptotic of the functions $W_{0},W_{-1},\operatorname{Li}_{2}$ is
used for both limits. Note that $\Lambda^{*}\left(0\right)=0$ and that the
function $\Lambda^{*}$ is non-negative in a neighborhood of $0$ hence by
convexity, $0$ is the minimum of this function (see Figure 1), completing the
proof of Lemma A.2 and of Theorem 1.1.

∎

##  Appendix B Measurability of $\beta^{*}$

Recall that $\beta^{*}:\Omega\rightarrow[0,2\pi)$ is such that

$\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta^{*}}\right)\right|=\int_{r_{0}\mathbb{T}}\log\left|f\right|d\mu$

when $f\neq 0$ otherwise we set $\beta^{*}=0$. We understand our probability
space $\Omega$ as follows. Consider the complex plane $\mathbb{C}$ with
$\mathbb{C}_{\nu}$ the Borel sigma-algebra of $\mathbb{C}$. Then, we take the
countable product measure space $\mathbb{C}^{\infty}_{\nu}$, with the sigma
algebra generated by cylinder sets, i.e. $\mathbb{C}^{j-1}\times
U_{j}\times\mathbb{C}^{n-j}$ where
$U_{j}\in\mathcal{B}\left(\mathbb{C}\right)$, $n\geq 1$. Denote by
$\text{A}\left(\mathbb{D}\right)$ \- the space of analytic functions defined
on the unit disk. Endow this space with the topology of uniform convergence on
compact sets. This makes it a complete separable metric space. Now,
$\beta^{*}:\Omega\mapsto[0,2\pi)$ can be viewed as a composition of two
measurable functions. The map $\omega=\\{\omega_{n}\\}_{n=1}^{\infty}\mapsto
f\left(z,\omega\right)=\sum_{n=1}^{\infty}\xi_{n}\left(\omega\right)z^{n}$
where $\xi_{n}\left(\omega\right)=\omega_{n}$ is measurable as a limit of
measurable functions (projections are continuous). Then, we need to prove that
the map $\beta^{*}:A\left(\mathbb{D}\right)\mapsto[0,2\pi]$ defined
momentarily is measurable.

Consider all $\beta$’s such that

$\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|=\int_{r_{0}\mathbb{T}}\log\left|f\right|d\mu=:c_{f}$

(such $\beta$ exists as we’ve already mentioned by continuity, Equation 7). We
define
$\beta^{*}\left(f\right):\text{A}\left(\mathbb{D}\right)\rightarrow\left[0,2\pi\right]$
as follows:

$\beta^{*}\left(f\right)=\inf\\{\beta\in[0,2\pi)\ :\
\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|=c_{f}\\}.$

To see that $\beta^{*}$ is lower semi-continuous (and hence measurable), we
want to show that for all $a\in[0,2\pi)$,
$\\{f\in\text{A}\left(\mathbb{D}\right)\ :\
\beta^{*}\left(f\right)>a\\}\in\mathcal{B}_{\text{A}\left(r\bar{\mathbb{D}}\right)}$
where $\mathcal{B}_{\text{A}\left(r\bar{\mathbb{D}}\right)}$ is the Borel
sigma-algebra with respect to the sup metric. (since $r<1$ is fixed, we can
restrict our domain to $r\mathbb{D}$.) Indeed,

$\displaystyle\\{f\in\text{A}$ $\displaystyle\left(\mathbb{D}\right)\ :\
\beta^{*}\left(f\right)>a\\}$
$\displaystyle=\bigg{\\{}f\in\text{A}\left(\mathbb{D}\right)\ :\
\forall\beta\in\left[0,a\right]\ ,\
\frac{1}{N}\sum_{j=1}^{N}\log\left|f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\right|\neq c_{f}\bigg{\\}}$
$\displaystyle=\bigcup_{n}\,\bigg{\\{}f\in\text{A}\left(\mathbb{D}\right)\ :\
\forall\beta\in\left[0,a\right]\ ,\
\Big{|}\frac{1}{N}\sum_{j=1}^{N}\log\big{|}f\left(r_{0}e^{i\theta_{j}}\cdot
e^{i\beta}\right)\big{|}-c_{f}\Big{|}>\frac{1}{n}\bigg{\\}}$

So if $\sup_{r\bar{\mathbb{D}}}\left|f-g\right|<\delta$ then (by continuity)
the integrals are sufficiently close to each other
$\left|c_{f}-c_{g}\right|<\epsilon$ hence this set is open.

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