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Boundedness and compactness of an integral operator in a mixed norm space on the polydisk

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Abstract

We study the following integral type operator

$$T_g (f)(z) = \int\limits_0^{z_{} } { \cdots \int\limits_0^{z_n } {f(\zeta _1 , \ldots ,\zeta _n )} g(\zeta _1 , \ldots ,\zeta _n )d\zeta _1 , \ldots ,\zeta _n } $$

in the space of analytic functions on the unit polydisk U n in the complex vector space ℂn. We show that the operator is bounded in the mixed norm space

, with p, q ∈ [1, ∞) and α = (α1, …, αn), such that αj > −1, for every j = 1, …, n, if and only if \(\sup _{z \in U^n } \prod\nolimits_{j = 1}^n {\left( {1 - \left| {z_j } \right|} \right)} \left| {g(z)} \right| < \infty \). Also, we prove that the operator is compact if and only if \(\lim _{z \to \partial U^n } \prod\nolimits_{j = 1}^n {\left( {1 - \left| {z_j } \right|} \right)} \left| {g(z)} \right| = 0\).

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Authors and Affiliations

  1. Mathematical Institute of the Serbian Academy of Sciences, Beograd, Serbia

    S. Stević

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  1. S. Stević
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Original Russian Text Copyright © 2007 Stevi’c S.

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Translated from Sibirskiĭ Matematicheskiĭ Zhurnal, Vol. 48, No. 3, pp. 694–706, May–June, 2007.

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Stević, S. Boundedness and compactness of an integral operator in a mixed norm space on the polydisk. Sib Math J 48, 559–569 (2007). https://doi.org/10.1007/s11202-007-0058-5

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  • DOI: https://doi.org/10.1007/s11202-007-0058-5

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