Bohr–Rogosinski radius for a certain class of close-to-convex harmonic mappings

Pub Date : 2023-01-31 DOI:10.4153/s0008439523000115
MOLLA BASIR AHAMED, VASUDEVARAO ALLU
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引用次数: 6

Abstract

Abstract Let $ \mathcal {B} $ be the class of analytic functions $ f $ in the unit disk $ \mathbb {D}=\{z\in \mathbb {C} : |z|<1\} $ such that $ |f(z)|<1 $ for all $ z\in \mathbb {D} $ . If $ f\in \mathcal {B} $ of the form $ f(z)=\sum _{n=0}^{\infty }a_nz^n $ , then $ \sum _{n=0}^{\infty }|a_nz^n|\leq 1 $ for $ |z|=r\leq 1/3 $ and $ 1/3 $ cannot be improved. This inequality is called Bohr inequality and the quantity $ 1/3 $ is called Bohr radius. If $ f\in \mathcal {B} $ of the form $ f(z)=\sum _{n=0}^{\infty }a_nz^n $ , then $ |\sum _{n=0}^{N}a_nz^n|<1\;\; \mbox {for}\;\; |z|<{1}/{2} $ and the radius $ 1/2 $ is the best possible for the class $ \mathcal {B} $ . This inequality is called Bohr–Rogosinski inequality and the corresponding radius is called Bohr–Rogosinski radius. Let $ \mathcal {H} $ be the class of all complex-valued harmonic functions $ f=h+\bar {g} $ defined on the unit disk $ \mathbb {D} $ , where $ h $ and $ g $ are analytic in $ \mathbb {D} $ with the normalization $ h(0)=h^{\prime }(0)-1=0 $ and $ g(0)=0 $ . Let $ \mathcal {H}_0=\{f=h+\bar {g}\in \mathcal {H} : g^{\prime }(0)=0\}. $ For $ \alpha \geq 0 $ and $ 0\leq \beta <1 $ , let $$ \begin{align*} \mathcal{W}^{0}_{\mathcal{H}}(\alpha, \beta)=\{f=h+\overline{g}\in\mathcal{H}_{0} : \mathrm{Re}\left(h^{\prime}(z)+\alpha zh^{\prime\prime}(z)-\beta\right)>|g^{\prime}(z)+\alpha zg^{\prime\prime}(z)|,\;\; z\in\mathbb{D}\} \end{align*} $$ be a class of close-to-convex harmonic mappings in $ \mathbb {D} $ . In this paper, we prove the sharp Bohr–Rogosinski radius for the class $ \mathcal {W}^{0}_{\mathcal {H}}(\alpha , \beta ) $ .
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一类近凸调和映射的Bohr-Rogosinski半径
摘要设$ \mathcal {B} $为单位圆盘$ \mathbb {D}=\{z\in \mathbb {C} : |z|<1\} $上的解析函数$ f $的类,使得$ |f(z)|<1 $对所有$ z\in \mathbb {D} $。如果$ f(z)=\sum _{n=0}^{\infty }a_nz^n $的形式为$ f\in \mathcal {B} $,那么$ |z|=r\leq 1/3 $和$ 1/3 $的形式为$ \sum _{n=0}^{\infty }|a_nz^n|\leq 1 $就不能改进。这个不等式叫做玻尔不等式这个量$ 1/3 $叫做玻尔半径。如果$ f\in \mathcal {B} $的形式是$ f(z)=\sum _{n=0}^{\infty }a_nz^n $,那么$ |\sum _{n=0}^{N}a_nz^n|<1\;\; \mbox {for}\;\; |z|<{1}/{2} $和半径$ 1/2 $是类$ \mathcal {B} $的最佳选择。这个不等式叫做玻尔-罗戈辛斯基不等式对应的半径叫做玻尔-罗戈辛斯基半径。设$ \mathcal {H} $为定义在单位盘$ \mathbb {D} $上的所有复值调和函数$ f=h+\bar {g} $的类,其中$ h $和$ g $在$ \mathbb {D} $中解析,归一化为$ h(0)=h^{\prime }(0)-1=0 $和$ g(0)=0 $。设$ \mathcal {H}_0=\{f=h+\bar {g}\in \mathcal {H} : g^{\prime }(0)=0\}. $对于$ \alpha \geq 0 $和$ 0\leq \beta <1 $,设$$ \begin{align*} \mathcal{W}^{0}_{\mathcal{H}}(\alpha, \beta)=\{f=h+\overline{g}\in\mathcal{H}_{0} : \mathrm{Re}\left(h^{\prime}(z)+\alpha zh^{\prime\prime}(z)-\beta\right)>|g^{\prime}(z)+\alpha zg^{\prime\prime}(z)|,\;\; z\in\mathbb{D}\} \end{align*} $$是$ \mathbb {D} $中的一类近凸调和映射。本文证明了该类$ \mathcal {W}^{0}_{\mathcal {H}}(\alpha , \beta ) $的尖锐Bohr-Rogosinski半径。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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