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

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 ?ⁿ. We show that the operator is bounded in the mixed norm space

Open image in new windowp, q ∈ 1, ∞) and α = (α₁, …, α_n), such that α_j > ?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 \(\lim _{z \to \partial U^n } \prod\nolimits_{j = 1}^n {\left( {1 - \left| {z_j } \right|} \right)} \left| {g(z)} \right| = 0\).