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...

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Bibliographic Details
Main Authors: A. Bandura, F. Nuray
Format: Article
Language:deu
Published: Ivan Franko National University of Lviv 2023-06-01
Series:Математичні Студії
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Online Access:http://matstud.org.ua/ojs/index.php/matstud/article/view/288
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Summary: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}d\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}d\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.
ISSN:1027-4634
2411-0620