Some integral operators acting on $H^{\infty}$

Fuente: arXiv
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Auteurs principaux: Anderson, Austin, Jovovic, Mirjana, Smith, Wayne
Format: Preprint
Publié: 2024
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_version_ 1866916121282609152
author Anderson, Austin
Jovovic, Mirjana
Smith, Wayne
author_facet Anderson, Austin
Jovovic, Mirjana
Smith, Wayne
contents Let $f$ and $g$ be analytic on the unit disc $\mathbb{D}$. The integral operator $T_g$ is defined by $ T_g f(z) = \int_0^z f(t)g'(t)\,dt$, $z \in \mathbb{D}$. The problem considered is characterizing those symbols $g$ for which $T_g$ acting on $H^\infty$, the space of bounded analytic functions on $\mathbb{D}$, is bounded or compact. When the symbol is univalent, these become questions in univalent function theory. The corresponding problems for the companion operator, $ S_g f(z)= \int_0^z f'(t)g(t)\, dt$, acting on $H^\infty$ are also studied.
format Preprint
id arxiv_https___arxiv_org_abs_2402_06774
institution arXiv
publishDate 2024
record_format arxiv
spellingShingle Some integral operators acting on $H^{\infty}$
Anderson, Austin
Jovovic, Mirjana
Smith, Wayne
Complex Variables
Functional Analysis
47B38 (Primary) 47G10, 30H05 (Secondary)
Let $f$ and $g$ be analytic on the unit disc $\mathbb{D}$. The integral operator $T_g$ is defined by $ T_g f(z) = \int_0^z f(t)g'(t)\,dt$, $z \in \mathbb{D}$. The problem considered is characterizing those symbols $g$ for which $T_g$ acting on $H^\infty$, the space of bounded analytic functions on $\mathbb{D}$, is bounded or compact. When the symbol is univalent, these become questions in univalent function theory. The corresponding problems for the companion operator, $ S_g f(z)= \int_0^z f'(t)g(t)\, dt$, acting on $H^\infty$ are also studied.
title Some integral operators acting on $H^{\infty}$
topic Complex Variables
Functional Analysis
47B38 (Primary) 47G10, 30H05 (Secondary)
url https://arxiv.org/abs/2402.06774