{"title":"Note to the behavior of the maximal term of Dirichlet series absolutely convergent in half-plane","authors":"M. Sheremeta","doi":"10.30970/ms.56.2.144-148","DOIUrl":null,"url":null,"abstract":"By $S_0(\\Lambda)$ denote a class of Dirichlet series $F(s)=\\sum_{n=0}^{\\infty}a_n\\exp\\{s\\lambda_n\\} (s=\\sigma+it)$ withan increasing to $+\\infty$ sequence $\\Lambda=(\\lambda_n)$ of exponents ($\\lambda_0=0$) and the abscissa of absolute convergence $\\sigma_a=0$.We say that $F\\in S_0^*(\\Lambda)$ if $F\\in S_0(\\Lambda)$ and $\\ln \\lambda_n=o(\\ln |a_n|)$ $(n\\to\\infty)$. Let$\\mu(\\sigma,F)=\\max\\{|a_n|\\exp{(\\sigma\\lambda_n)}\\colon n\\ge 0\\}$ be the maximal term of Dirichlet series. It is proved that in order that $\\ln (1/|\\sigma|)=o(\\ln \\mu(\\sigma))$ $(\\sigma\\uparrow 0)$ for every function $F\\in S_0^*(\\Lambda)$ it is necessary and sufficient that $\\displaystyle \\varlimsup\\limits_{n\\to\\infty}\\frac{\\ln \\lambda_{n+1}}{\\ln \\lambda_n}<+\\infty. $For an analytic in the disk $\\{z\\colon |z|<1\\}$ function $f(z)=\\sum_{n=0}^{\\infty}a_n z^n$ and $r\\in (0, 1)$ we put $M_f(r)=\\max\\{|f(z)|\\colon |z|=r<1\\}$ and $\\mu_f(r)=\\max\\{|a_n|r^n\\colon n\\ge 0\\}$. Then from hence we get the following statement: {\\sl if there exists a sequence $(n_j)$ such that $\\ln n_{j+1}=O(\\ln n_{j})$ and $\\ln n_{j}=o(\\ln |a_{n_{j}}|)$ as $j\\to\\infty$, then the functions $\\ln \\mu_f(r)$ and $\\ln M_f(r)$ are or not are slowly increasing simultaneously.","PeriodicalId":37555,"journal":{"name":"Matematychni Studii","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2021-12-27","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Matematychni Studii","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.30970/ms.56.2.144-148","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"Mathematics","Score":null,"Total":0}
引用次数: 0
Abstract
By $S_0(\Lambda)$ denote a class of Dirichlet series $F(s)=\sum_{n=0}^{\infty}a_n\exp\{s\lambda_n\} (s=\sigma+it)$ withan increasing to $+\infty$ sequence $\Lambda=(\lambda_n)$ of exponents ($\lambda_0=0$) and the abscissa of absolute convergence $\sigma_a=0$.We say that $F\in S_0^*(\Lambda)$ if $F\in S_0(\Lambda)$ and $\ln \lambda_n=o(\ln |a_n|)$ $(n\to\infty)$. Let$\mu(\sigma,F)=\max\{|a_n|\exp{(\sigma\lambda_n)}\colon n\ge 0\}$ be the maximal term of Dirichlet series. It is proved that in order that $\ln (1/|\sigma|)=o(\ln \mu(\sigma))$ $(\sigma\uparrow 0)$ for every function $F\in S_0^*(\Lambda)$ it is necessary and sufficient that $\displaystyle \varlimsup\limits_{n\to\infty}\frac{\ln \lambda_{n+1}}{\ln \lambda_n}<+\infty. $For an analytic in the disk $\{z\colon |z|<1\}$ function $f(z)=\sum_{n=0}^{\infty}a_n z^n$ and $r\in (0, 1)$ we put $M_f(r)=\max\{|f(z)|\colon |z|=r<1\}$ and $\mu_f(r)=\max\{|a_n|r^n\colon n\ge 0\}$. Then from hence we get the following statement: {\sl if there exists a sequence $(n_j)$ such that $\ln n_{j+1}=O(\ln n_{j})$ and $\ln n_{j}=o(\ln |a_{n_{j}}|)$ as $j\to\infty$, then the functions $\ln \mu_f(r)$ and $\ln M_f(r)$ are or not are slowly increasing simultaneously.