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: I. V. Andrusyak, P.V. Filevych
Format: Article
Language:deu
Published: Ivan Franko National University of Lviv 2020-12-01
Series:Математичні Студії
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Online Access:http://matstud.org.ua/ojs/index.php/matstud/article/view/160
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author I. V. Andrusyak
P.V. Filevych
author_facet I. V. Andrusyak
P.V. Filevych
author_sort I. V. Andrusyak
collection DOAJ
description 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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publisher Ivan Franko National University of Lviv
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spelling doaj-art-738e8677d735448094fe04e71dd267d02025-08-20T03:17:40ZdeuIvan Franko National University of LvivМатематичні Студії1027-46342411-06202020-12-0154214615310.30970/ms.54.2.146-153160The minimal growth of entire functions with given zeros along unbounded setsI. V. Andrusyak0P.V. Filevych1Lviv Polytechnic National University, Lviv, UkraineLviv Politechnic National University, LvivLet $l$ 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 $f$.http://matstud.org.ua/ojs/index.php/matstud/article/view/160entire function; maximum modulus; zeros; counting function
spellingShingle I. V. Andrusyak
P.V. Filevych
The minimal growth of entire functions with given zeros along unbounded sets
Математичні Студії
entire function; maximum modulus; zeros; counting function
title The minimal growth of entire functions with given zeros along unbounded sets
title_full The minimal growth of entire functions with given zeros along unbounded sets
title_fullStr The minimal growth of entire functions with given zeros along unbounded sets
title_full_unstemmed The minimal growth of entire functions with given zeros along unbounded sets
title_short The minimal growth of entire functions with given zeros along unbounded sets
title_sort minimal growth of entire functions with given zeros along unbounded sets
topic entire function; maximum modulus; zeros; counting function
url http://matstud.org.ua/ojs/index.php/matstud/article/view/160
work_keys_str_mv AT ivandrusyak theminimalgrowthofentirefunctionswithgivenzerosalongunboundedsets
AT pvfilevych theminimalgrowthofentirefunctionswithgivenzerosalongunboundedsets
AT ivandrusyak minimalgrowthofentirefunctionswithgivenzerosalongunboundedsets
AT pvfilevych minimalgrowthofentirefunctionswithgivenzerosalongunboundedsets