Strict singularity of a Volterra-type integral operator on $H^p$

TL;DR

This paper demonstrates that for Volterra-type integral operators on Hardy spaces, strict singularity is equivalent to compactness, providing new insights into their structural properties and a novel proof of their weak compactness on H^1.

Contribution

It establishes the equivalence between strict singularity and compactness of Volterra-type operators on Hardy spaces, offering new proofs and deeper understanding.

Findings

Strict singularity coincides with compactness on H^p.

Provides a new proof for the equivalence of compactness and weak compactness on H^1.

Shows that non-compact operators fix an isomorphic copy of ll^p.

Abstract

We prove that a Volterra-type integral operator $T_gf(z) = \int_0^z f(\zeta)g'(\zeta)d\zeta, \, z \in \mathbb D,$ defined on Hardy spaces $H^p, \, 1 \le p < \infty,$ fixes an isomorphic copy of $\ell^p,$ if the operator $T_g$ is not compact. In particular, this shows that the strict singularity of the operator $T_g$ coincides with the compactness of the operator $T_g$ on spaces $H^p.$ As a consequence, we obtain a new proof for the equivalence of the compactness and the weak compactness of the operator $T_g$ on $H^1$.