WIMAN’S TYPE INEQUALITY FOR SOME DOUBLE POWER SERIES

A. Kuryliak, L. O. Shapovalovska, O. Skaskiv
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引用次数: 2

Abstract

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 $\mathbb{T}=\{z=(z_1,z_2)\in\mathbb C^2\colon|z_1|<1,\ |z_2|<+\infty\}=\mathbb{D}\times\mathbb{C}$ and$\frac{\partial}{\partial z_2}f(z_1,z_2)\not\equiv0$ in $\mathbb{T}.$ In this paper we prove some analogue of Wiman's inequalityfor analytic functions $f\in\mathcal{A}^2$. Let a function $h\colon \mathbb R^2_+\to \mathbb R_+$ be such that$h$ is nondecreasing with respect to each variables and $h(r)\geq 10$ for all $r\in T:=(0,1)\times (0,+\infty)$and $\iint_{\Delta_\varepsilon}\frac{h(r)dr_1dr_2}{(1-r_1)r_2}=+\infty$ for some $\varepsilon\in(0,1)$, where $\Delta_{\varepsilon}=\{(t_1, t_2)\in T\colon t_1>\varepsilon,\ t_2> \varepsilon\}$.We say that $E\subset T$ is a set of asymptotically  finite $h$-measure on\ ${T}$if $\nu_{h}(E){:=}\iint\limits_{E\cap\Delta_{\varepsilon}}\frac{h(r)dr_1dr_2}{(1-r_1)r_2}<+\infty$ for some $\varepsilon>0$. 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*}}
某些二重幂级数的Wiman型不等式
By $\mathcal{A}^2$ 表示形式为by的解析函数的类 $\mathcal{A}^2$ 表示如下形式的解析函数的类$f(z)=\sum_{n+m=0}^{+\infty}a_{nm}z_1^nz_2^m,$有 {the} 收敛域 $\mathbb{T}=\{z=(z_1,z_2)\in\mathbb C^2\colon|z_1|\varepsilon,\ t_2> \varepsilon\}$我们这么说 $E\subset T$ 一个集合是渐近有限的吗 $h$-测量 ${T}$如果 $\nu_{h}(E){:=}\iint\limits_{E\cap\Delta_{\varepsilon}}\frac{h(r)dr_1dr_2}{(1-r_1)r_2}0$. 因为 $r=(r_1,r_2)\in T$ 一个函数 $f\in\mathcal{A}^2$ 表示\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*}我们证明了以下定理:{\sl 让 $f\in\mathcal{A}^2$. 对于每一个 $\delta>0$ 存在一个集合 $E=E(\delta,f)$ 渐近有限的 $h$-测量 ${T}$ 对于所有人来说 $r\in (T\cap\Delta_{\varepsilon})\backslash E$ 我们有 \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*}}
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