指数型II的全二元函数

Q3 Mathematics
Andriy Ivanovych Bandura, F. Nuray
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引用次数: 2

摘要

设$f(z_{1},z_{2})$为二元整函数,$C$为正常数。如果$f(z_{1},z_{2})$对于非负整数$M$,对于所有非负整数$$,$l$满足以下不等式,使得$k+l\in\{0,1,2,\ldots,M\}$,对于某些整数$p\ge 1$和对于所有$(z_{1},z _{2})=(r_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\ttheta_{2}})$,$r_1$和$r_2$足够大:\ begin{collecte*}\sum_{i+j=0}^{M}\frac{\left(\ int_{0}^{2\pi}\ int_{0}^2 \pi}|f ^{(i+k,j+l)}(r_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\theta_{2}})|^{p}d\θ{1}\θ{2}\right_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\theta_{2}})|^{p}d\theta_{1}\thetau{2}\right)^{\frac{1}{p}}{i!j!},\end{collecte*}则$f(z_{1{,z_{2})$为指数型,不超过\[2+2\log\Big(1+\ frac{1}{C}\Big)+\log[(2M)!/M!].\]如果这个条件被相关的条件代替,那么$f$也是指数型的。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Entire Bivariate Functions of Exponential Type II
Let $f(z_{1},z_{2})$ be a bivariate entire function and $C$ be a positive constant. If $f(z_{1},z_{2})$ satisfies the following inequality for non-negative integer $M$, for all non-negative integers $k,$ $l$ such that $k+l\in\{0, 1, 2, \ldots, M\}$, for some integer $p\ge 1$ and for all $(z_{1},z_{2})=(r_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\theta_{2}})$ with $r_1$ and $r_2$ sufficiently large:\begin{gather*}\sum_{i+j=0}^{M}\frac{\left(\int_{0}^{2\pi}\int_{0}^{2\pi}|f^{(i+k,j+l)}(r_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\theta_{2}})|^{p}d\theta_{1}\theta_{2}\right)^{\frac{1}{p}}}{i!j!}\ge \\\ge \sum_{i+j=M+1}^{\infty}\frac{\left(\int_{0}^{2\pi}\int_{0}^{2\pi}|f^{(i+k,j+l)}(r_{1}e^{\mathbf{i}\theta_{1}},r_{2}e^{\mathbf{i}\theta_{2}})|^{p}d\theta_{1}\theta_{2}\right)^{\frac{1}{p}}}{i!j!},\end{gather*}then $f(z_{1},z_{2})$ is of exponential type not exceeding\[2+2\log\Big(1+\frac{1}{C}\Big)+\log[(2M)!/M!].\]If this condition is replaced by related conditions, then also $f$ is of exponential type.
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来源期刊
Matematychni Studii
Matematychni Studii Mathematics-Mathematics (all)
CiteScore
1.00
自引率
0.00%
发文量
38
期刊介绍: Journal is devoted to research in all fields of mathematics.
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