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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 β∈L0\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, z∈Cpz\in \mathbb{C}^p, is analytic function in the bounded Reinhard domain G⊂CpG\subset \mathbb{C}^p, then the condition ∫R01α(ln⁡+MG(R,f))(1−R)2β(1/(1−R))d R<+∞,\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)∣ ⁣:z∈G},M_{G}(R,f)=\sup\{|F(Rz)|\colon z\in G\}, yields that ∑k=0+∞(α(k)−α(k−1))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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The convergence classes for analytic functions in the Reinhardt domains — Mathematical Frontier Network