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Sibirskii Matematicheskii Zhurnal, 2007, Volume 48, Number 3, Pages 694–706
(Mi smj58)
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This article is cited in 76 scientific papers (total in 76 papers)
Boundedness and compactness of an integral operator in a mixed norm space on the polydisk
S. Stević Mathematical Institute, Serbian Academy of Sciences and Arts
Abstract:
We study the following integral type operator
$$
T_g(f)(z)=\int\limits_0^{z_1}\dots\int\limits_0^{z_n}f(\zeta_1,\dots,\zeta_n)g(\zeta_1,\dots,\zeta_n)\,d\zeta_1\dots\zeta_n
$$
in the space of analytic functions on the unit polydisk $U^n$ in the complex vector space $\mathbb C^n$. We show that the operator is bounded in the mixed norm space
$${\mathscr A}^{p,q}_\alpha(U^n)=\biggl\{f\in H(U^n)\mid\int\limits_{[0,1)^n}M_p^q(f,r)\prod_{j=1}^n(1-r_j)^{\alpha_j}\,dr_j<\infty\biggr\},
$$
with $p,q\in[1,\infty)$ and $\alpha=(\alpha_1,\dots,\alpha_n)$, such that $\alpha_j>-1$, for every $j=1,\dots,n$, if and only if $\sup\limits_{z\in U^n}\prod\limits_{j=1}^n(1-|z_j|)|g(z)|<\infty$. Also, we prove that the operator is compact if and only if $\lim\limits_{z\to\partial U^n}\prod\limits_{j=1}^n(1-|z_j|)|g(z)|=0$.
Keywords:
analytic function, mixed norm space, integral operator, polydisk, boundedness, compactness.
Received: 23.11.2005
Citation:
S. Stević, “Boundedness and compactness of an integral operator in a mixed norm space on the polydisk”, Sibirsk. Mat. Zh., 48:3 (2007), 694–706; Siberian Math. J., 48:3 (2007), 559–569
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