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Siberian Mathematical Journal

, Volume 48, Issue 3, pp 559–569 | Cite as

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

  • S. Stević
Article

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 \(\lim _{z \to \partial U^n } \prod\nolimits_{j = 1}^n {\left( {1 - \left| {z_j } \right|} \right)} \left| {g(z)} \right| = 0\).

Keywords

analytic function mixed norm space integral operator polydisk boundedness compactness 

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Copyright information

© Springer Science+Business Media, Inc. 2007

Authors and Affiliations

  • S. Stević
    • 1
  1. 1.Mathematical Institute of the Serbian Academy of SciencesBeogradSerbia

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