Analytic Gaussian functions in the unit disc: probability of zeros absence

Abstract

In the paper we consider a random analytic function of the form $$f(z,\omega )=\sum\limits_{n=0}^{+\infty}\varepsilon_n(\omega_1)\xi_n(\omega_2)a_nz^n.$$ Here $(\varepsilon_n)$ is a sequence of inde\-pendent Steinhaus random variables, $(\xi_n)$ is a sequence of indepen\-dent standard complex Gaussian random variables, and a sequence of numbers $a_n\in\mathbb{C}$ such that $a_0\neq0,\ \varlimsup\limits_{n\to+\infty}\sqrt[n]{|a_n|}=1,\ \sup\{|a_n|\colon n\in\mathbb{N}\}=+\infty.$ We investigate asymptotic estimates of the probability $p_0(r)=\ln^-P\{\omega\colon f(z,\omega )$ has no zeros inside $r\mathbb{D}\}$ as $r\uparrow1$ outside some set $E$ of finite logarithmic measure. Denote $ N(r):=\#\{n\colon |a_n|r^n>1\},$ $ s(r):=2\sum_{n=0}^{+\infty}\ln^+(|a_n|r^{n}), $ $ \alpha:=\varliminf\limits_{r\uparrow1}\frac{\ln N(r)}{\ln\frac{1}{1-r}}. $ The article, in particular, proves the following statements: 1) if $\alpha>4$ then $\displaystyle\lim\limits_{\stackrel{r\uparrow1}{r\notin E}}\frac{\ln(p_0(r)- s(r))}{\ln N(r)}=1$; 2) if $\alpha=+\infty$ then $\displaystyle 0\leq\varliminf\limits_{\stackrel{r\uparrow1}{r\notin E}}\frac{\ln(p_0(r)- s(r))}{\ln s(r)},\quad \varlimsup\limits_{\stackrel {r\uparrow1}{r\notin E}}\frac{\ln(p_0(r)- s(r))}{\ln s(r)}\leq\frac1{2}.$ Here $E$ is a set of finite logarithmic measure. The obtained asymptotic estimates are in a certain sense best possible. Also we give an answer to an open question from \!\cite[p. 119]{Nishry2013} for such random functions.

Keywords

• gaussian analytic functions; steinhaus analytic functions, zeros distribution of random analytic functions.