Integration operators on Bergman spaces
(1997) In Indiana University Mathematics Journal 46(2). p.337356 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 littleoh 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), 120137, 187, 192193; 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:
http://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
 19435258
 DOI
 10.1512/iumj.1997.46.1373
 language
 English
 LU publication?
 yes
 id
 1c6f8fa0a62a4548b3719a7fdabb568a (old id 1467195)
 date added to LUP
 20090916 12:55:49
 date last changed
 20170528 03:32:21
@article{1c6f8fa0a62a4548b3719a7fdabb568a, 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 littleoh 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), 120137, 187, 192193; MR0369705 (51 #5937) (Theorem 3.3)] shows that it is necessary at least when $\alpha > 1$.}, author = {Aleman, Alexandru and Siskakis, Aristomenis G}, issn = {19435258}, language = {eng}, number = {2}, pages = {337356}, 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}, volume = {46}, year = {1997}, }