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

Austin Anderson, Mirjana Jovovic, Wayne Smith

Abstract

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.

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

Abstract

Let and be analytic on the unit disc . The integral operator is defined by , . The problem considered is characterizing those symbols for which acting on , the space of bounded analytic functions on , is bounded or compact. When the symbol is univalent, these become questions in univalent function theory. The corresponding problems for the companion operator, , acting on are also studied.
Paper Structure (4 sections, 23 theorems, 73 equations, 2 figures)

This paper contains 4 sections, 23 theorems, 73 equations, 2 figures.

Key Result

Lemma 2.1

Let $X$ and $Y$ be Banach spaces of analytic functions, $z \in \mathbb{D}$, and let $\lambda_z$ and $\lambda_z'$ be linear functionals defined by $\lambda_zf = f(z)$ and $\lambda'_zf = f'(z)$ for $f \in X \cup Y$. Suppose $\lambda_z$ and $\lambda'_z$ are bounded on $X$ and $Y$. (i) If $S_g$ maps $X$ (ii) If $T_g$ maps $X$ boundedly into $Y$, then

Figures (2)

  • Figure 1: $g \notin A$, $g \in T[H^{\infty}]$
  • Figure 2: $g \in A$, $g \notin T[H^{\infty}]$

Theorems & Definitions (44)

  • Lemma 2.1
  • proof
  • Proposition 2.2
  • proof
  • Proposition 2.3
  • proof
  • Proposition 2.4
  • Proposition 2.5
  • Lemma 2.6
  • proof
  • ...and 34 more