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The convergence classes for analytic functions in the Reinhardt domains

T.M. Salo, O.Yu. Tarnovecka

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Source: Crossref

Published: Dec 31, 2018

DOI: 10.15330/cmp.10.2.408-411

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Source abstract

Let L0L^0 be the class of positive increasing on [1,+)[1,+\infty) functions ll such that l((1+o(1))x)=(1+o(1))l(x)l((1+o(1))x)=(1+o(1))l(x) (x+)(x\to +\infty). We assume that α\alpha is a concave function such that α(ex)L0\alpha(e^x)\in L^0 and function βL0\beta\in L^0 such that 1+α(x)β(x)dx<+\displaystyle\int_1^{+\infty}\frac{\alpha(x)}{\beta(x)}dx<+\infty. In the article it is proved the following theorem: if f(z)=n=0+anzn\displaystyle f(z)=\sum_{\|n\|=0}^{+\infty}a_nz_n, zCpz\in \mathbb{C}^p, is analytic function in the bounded Reinhard domain GCpG\subset \mathbb{C}^p, then the condition R01α(ln+MG(R,f))(1R)2β(1/(1R))dR<+,\displaystyle \int\limits_{R_0}^{1} \frac{\alpha(\ln^{+} M_{G}(R,f))} {(1-R)^2\beta(1/(1-R))}d\,R<+\infty, MG(R,f)=sup{F(Rz) ⁣:zG},M_{G}(R,f)=\sup\{|F(Rz)|\colon z\in G\}, yields that k=0+(α(k)α(k1))β1(k/ln+Ak)<+,\sum_{k=0}^{+\infty}(\alpha(k)-\alpha(k-1)) \beta_1\left({k}/{\ln^{+}|A_k|}\right)<+\infty, β1(x)=x+dtβ(t),Ak=max{an ⁣:n=k}.\beta_1(x)= \int\limits_{x}^{+\infty} \frac{dt}{\beta(t)},\quad A_k=\max\{|a_n|\colon\|n\|=k\}.

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