The minimal growth of entire functions with given zeros along unbounded sets
Let $l$ be a continuous function on $\mathbb{R}$ increasing to $+\infty$, and $\varphi$ be a positive function on $\mathbb{R}$. We proved that the condition $$ \varliminf_{x\to+\infty}\frac{\varphi(\ln[x])}{\ln x}>0 $$ is necessary and sufficient in order that for any complex sequence $(\zeta_n)$...
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| Main Authors: | , |
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| Format: | Article |
| Language: | German |
| Published: |
Ivan Franko National University of Lviv
2020-12-01
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| Series: | Математичні Студії |
| Subjects: | |
| Online Access: | http://matstud.org.ua/ojs/index.php/matstud/article/view/160 |
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| Summary: | Let $l$ be a continuous function on $\mathbb{R}$ increasing to $+\infty$, and $\varphi$ be a positive function on $\mathbb{R}$. We proved that the condition
$$
\varliminf_{x\to+\infty}\frac{\varphi(\ln[x])}{\ln x}>0
$$
is necessary and sufficient in order that for any complex sequence $(\zeta_n)$ with $n(r)\ge l(r)$, $r\ge r_0$, and every set $E\subset\mathbb{R}$ which is unbounded from above there exists an entire function $f$ having zeros only at the points $\zeta_n$ such that
$$
\varliminf_{r\in E,\ r\to+\infty}\frac{\ln\ln M_f(r)}{\varphi(\ln n_\zeta(r))\ln l^{-1}(n_\zeta(r))}=0.
$$
Here $n(r)$ is the counting function of $(\zeta_n)$, and $M_f(r)$ is the maximum modulus of $f$. |
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| ISSN: | 1027-4634 2411-0620 |