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

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 Un in the complex vector space ℂn. We show that the operator is bounded in the mixed norm space Open image in new window , 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 \(\lim _{z \to \partial U^n } \prod\nolimits_{j = 1}^n {\left( {1 - \left| {z_j } \right|} \right)} \left| {g(z)} \right| = 0\).