Integration operators on Bergman spaces
(1997) In Indiana University Mathematics Journal 46(2). p.337-356- Abstract
- Let ${\bold D}$ denote the unit disk in the complex plane and let $m$ be the area Lebesgue measure on ${\bold D}$. Given a positive integrable function $w$ (a weight) on ${\bold D}$, let $L^p_{\rm a}(w)$ denote the collection of analytic functions $f$ on ${\bold D}$ such that $|f|^pw$ is integrable.
Given an analytic function $g$ on ${\bold D}$, the operator $T_g$ is defined by $T_g f(z) = \int_0^{z} f(\zeta)g'(\zeta)\,d\zeta$. The authors consider conditions on $g$ such that $T_g$ is bounded on $L^p_{\rm a}(w)$.
In many cases the derivative $D$ is an isomorphism between the subspace of $L^p_{\rm a}(w)$ consisting of functions vanishing at $0$ and some space $L^p_{\rm a}(v)$ with another weight $v$.... (More) - Let ${\bold D}$ denote the unit disk in the complex plane and let $m$ be the area Lebesgue measure on ${\bold D}$. Given a positive integrable function $w$ (a weight) on ${\bold D}$, let $L^p_{\rm a}(w)$ denote the collection of analytic functions $f$ on ${\bold D}$ such that $|f|^pw$ is integrable.
Given an analytic function $g$ on ${\bold D}$, the operator $T_g$ is defined by $T_g f(z) = \int_0^{z} f(\zeta)g'(\zeta)\,d\zeta$. The authors consider conditions on $g$ such that $T_g$ is bounded on $L^p_{\rm a}(w)$.
In many cases the derivative $D$ is an isomorphism between the subspace of $L^p_{\rm a}(w)$ consisting of functions vanishing at $0$ and some space $L^p_{\rm a}(v)$ with another weight $v$. Thus, the question of boundedness or compactness of $T_g$ becomes the corresponding question for the operator of multiplication by $g'$ acting from $L^p_{\rm a}(w)$ to $L^p_{\rm a}(v)$.
The authors consider only $w$ which are radial: $w(re^{i\theta}) = w(r)$. In the first part of the paper it is shown that $\int |f|^pw\,dm \le C\int |f'(z)|^p v(|z|) \,dm(z)$, $p \ge 1$, where $v(r) = \int_r^1 w(u)\,du$. Under the assumption that $v(r) \le C(1 - r)w(r)$, which is valid in particular for $w(r) \equiv (1 - r)^\alpha$, $\alpha > -1$, it follows that $T_g$ is bounded when $g'(z)(1 - |z|)$ is bounded.
The converse can be proved in a more general setting. It is obtained by estimating the norm of the linear functional $D_\lambda\colon f \mapsto f'(\lambda)$ in terms of that of the evaluation functional $L_\lambda\colon f \mapsto f(\lambda)$. Rather general hypotheses on a Banach space of analytic functions are obtained in order that $\| D_\lambda \|(1 - |\lambda|) \le C\| L_\lambda \|$. This leads immediately to the converse: if the operator $T_g$ is bounded, then $g'(z)(1 - |z|)$ is bounded. Several classes of weights are shown to satisfy the hypotheses required for both the necessity and the sufficiency of the condition.
In all cases, the problem of compactness of $T_g$ is also considered, and the solution involves the little-oh versions of the same conditions for boundedness. In the special case where $p = 2$ and $w(r) = (1 - r)^\alpha$ the Schatten class of $T_g$ is determined.
The techniques are briefly applied to the weight $w(r) = \exp[ -\beta(1 - r)^{-\alpha}]$, $\alpha > 0$. In this case $v(r) \le C(1 - r)^{\alpha + 1}w(r)$, and that leads to the conclusion that $T_g$ is bounded when $g'(z)(1 - |z|)^{\alpha + 1}$ is bounded. The necessity of this condition is left open. A theorem of V. L. Oleinik [Zap. Naučn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 47 (1974), 120--137, 187, 192--193; MR0369705 (51 #5937) (Theorem 3.3)] shows that it is necessary at least when $\alpha > 1$. (Less)
Please use this url to cite or link to this publication:
https://lup.lub.lu.se/record/1467195
- author
- Aleman, Alexandru LU and Siskakis, Aristomenis G
- organization
- publishing date
- 1997
- type
- Contribution to journal
- publication status
- published
- subject
- in
- Indiana University Mathematics Journal
- volume
- 46
- issue
- 2
- pages
- 337 - 356
- publisher
- Indiana University
- external identifiers
-
- scopus:0000157297
- ISSN
- 0022-2518
- DOI
- 10.1512/iumj.1997.46.1373
- language
- English
- LU publication?
- yes
- id
- 1c6f8fa0-a62a-4548-b371-9a7fdabb568a (old id 1467195)
- date added to LUP
- 2016-04-01 11:45:25
- date last changed
- 2022-04-05 04:33:45
@article{1c6f8fa0-a62a-4548-b371-9a7fdabb568a,
abstract = {{Let ${\bold D}$ denote the unit disk in the complex plane and let $m$ be the area Lebesgue measure on ${\bold D}$. Given a positive integrable function $w$ (a weight) on ${\bold D}$, let $L^p_{\rm a}(w)$ denote the collection of analytic functions $f$ on ${\bold D}$ such that $|f|^pw$ is integrable. <br/><br>
<br/><br>
Given an analytic function $g$ on ${\bold D}$, the operator $T_g$ is defined by $T_g f(z) = \int_0^{z} f(\zeta)g'(\zeta)\,d\zeta$. The authors consider conditions on $g$ such that $T_g$ is bounded on $L^p_{\rm a}(w)$. <br/><br>
<br/><br>
In many cases the derivative $D$ is an isomorphism between the subspace of $L^p_{\rm a}(w)$ consisting of functions vanishing at $0$ and some space $L^p_{\rm a}(v)$ with another weight $v$. Thus, the question of boundedness or compactness of $T_g$ becomes the corresponding question for the operator of multiplication by $g'$ acting from $L^p_{\rm a}(w)$ to $L^p_{\rm a}(v)$. <br/><br>
<br/><br>
The authors consider only $w$ which are radial: $w(re^{i\theta}) = w(r)$. In the first part of the paper it is shown that $\int |f|^pw\,dm \le C\int |f'(z)|^p v(|z|) \,dm(z)$, $p \ge 1$, where $v(r) = \int_r^1 w(u)\,du$. Under the assumption that $v(r) \le C(1 - r)w(r)$, which is valid in particular for $w(r) \equiv (1 - r)^\alpha$, $\alpha > -1$, it follows that $T_g$ is bounded when $g'(z)(1 - |z|)$ is bounded. <br/><br>
<br/><br>
The converse can be proved in a more general setting. It is obtained by estimating the norm of the linear functional $D_\lambda\colon f \mapsto f'(\lambda)$ in terms of that of the evaluation functional $L_\lambda\colon f \mapsto f(\lambda)$. Rather general hypotheses on a Banach space of analytic functions are obtained in order that $\| D_\lambda \|(1 - |\lambda|) \le C\| L_\lambda \|$. This leads immediately to the converse: if the operator $T_g$ is bounded, then $g'(z)(1 - |z|)$ is bounded. Several classes of weights are shown to satisfy the hypotheses required for both the necessity and the sufficiency of the condition. <br/><br>
<br/><br>
In all cases, the problem of compactness of $T_g$ is also considered, and the solution involves the little-oh versions of the same conditions for boundedness. In the special case where $p = 2$ and $w(r) = (1 - r)^\alpha$ the Schatten class of $T_g$ is determined. <br/><br>
<br/><br>
The techniques are briefly applied to the weight $w(r) = \exp[ -\beta(1 - r)^{-\alpha}]$, $\alpha > 0$. In this case $v(r) \le C(1 - r)^{\alpha + 1}w(r)$, and that leads to the conclusion that $T_g$ is bounded when $g'(z)(1 - |z|)^{\alpha + 1}$ is bounded. The necessity of this condition is left open. A theorem of V. L. Oleinik [Zap. Naučn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 47 (1974), 120--137, 187, 192--193; MR0369705 (51 #5937) (Theorem 3.3)] shows that it is necessary at least when $\alpha > 1$.}},
author = {{Aleman, Alexandru and Siskakis, Aristomenis G}},
issn = {{0022-2518}},
language = {{eng}},
number = {{2}},
pages = {{337--356}},
publisher = {{Indiana University}},
series = {{Indiana University Mathematics Journal}},
title = {{Integration operators on Bergman spaces}},
url = {{http://dx.doi.org/10.1512/iumj.1997.46.1373}},
doi = {{10.1512/iumj.1997.46.1373}},
volume = {{46}},
year = {{1997}},
}