Comparative growth of an entire function and the integrated counting function of its zeros

I. Andrusyak, P. Filevych
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引用次数: 0

Abstract

Let $(\zeta_n)$ be a sequence of complex numbers such that $\zeta_n\to\infty$ as $n\to\infty$, $N(r)$ be the integrated counting function of this sequence, and let $\alpha$ be a positive continuous and increasing to $+\infty$ function on $\mathbb{R}$ for which $\alpha(r)=o(\log (N(r)/\log r))$ as $r\to+\infty$. It is proved that for any set $E\subset(1,+\infty)$ satisfying $\int_{E}r^{\alpha(r)}dr=+\infty$, there exists an entire function $f$ whose zeros are precisely the $\zeta_n$, with multiplicities taken into account, such that the relation $$ \liminf_{r\in E,\ r\to+\infty}\frac{\log\log M(r)}{\log r\log (N(r)/\log r)}=0 $$ holds, where $M(r)$ is the maximum modulus of the function $f$. It is also shown that this relation is best possible in a certain sense.
整个函数的比较增长及其零点的综合计数函数
让$(\zeta_n)$是复数序列,使得$\zeta_n\to\infty$为$n\to\infty$,$N(r)$是这个序列的积分计数函数,让$\alpha$是$mathbb{R}$上的一个正连续且递增到$+\infty$的函数,对于这个函数,$\alpha(r)=o(\log (N(r)/\log r))$为$r\to\infty$。证明了对于满足$int_{E}r^{alpha(r)}dr=+\infty$的任意集合$E\subset(1,+\infty)$,存在一个整函数$f$,其零点正是$\zeta_n$、这样关系 $$ \liminf_{r\in E,\r\to+\infty}\frac{log\log M(r)}{log r\log (N(r)/\log r)}=0 $$ 成立,其中 $M(r)$ 是函数 $f$ 的最大模。研究还表明,这种关系在一定意义上是最可能的。
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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