| Popis: |
By $\mathcal{A}^2$ denote the class of analytic functions of the formBy $\mathcal{A}^2$ denote the class of analytic functions of the form$f(z)=\sum_{n+m=0}^{+\infty}a_{nm}z_1^nz_2^m,$with {the} domain of convergence $\mathbb{T}=\{z=(z_1,z_2)\in\mathbb C^2\colon|z_1|\varepsilon,\ t_2> \varepsilon\}$.We say that $E\subset T$ is a set of asymptotically finite $h$-measure on\ ${T}$if $\nu_{h}(E){:=}\iint\limits_{E\cap\Delta_{\varepsilon}}\frac{h(r)dr_1dr_2}{(1-r_1)r_2}0$. For $r=(r_1,r_2)\in T$ and a function $f\in\mathcal{A}^2$ denote\begin{gather*}M_f(r)=\max \{|f(z)|\colon |z_1|\leq r_1,|z_2|\leq r_2\},\\mu_f(r)=\max\{|a_{nm}|r_1^{n} r_2^{m}\colon(n,m)\in{\mathbb{Z}}_+^2\}.\end{gather*}We prove the following theorem:{\sl Let $f\in\mathcal{A}^2$. For every $\delta>0$ there exists a set $E=E(\delta,f)$ of asymptotically finite $h$-measure on\ ${T}$ such that for all $r\in (T\cap\Delta_{\varepsilon})\backslash E$ we have \begin{equation*} M_f(r)\leq\frac{h^{3/2}(r)\mu_f(r)}{(1-r_1)^{1+\delta}}\ln^{1+\delta} \Bigl(\frac{h(r)\mu_f(r)}{1-r_1}\Bigl)\cdot\ln^{1/2+\delta}\frac{er_2}{\varepsilon}. \end{equation*}} |
