Bergman spaces on disconnected domains
(1996) In Canadian Journal of Mathematics 48(2). p.225243 Abstract
 For a bounded open set $U$ in the complex plane, we consider the Bergman space $L^p_a(U)$ of $p$thpower areaintegrable analytic functions on $U$, with $p$ assumed to be in the range $1\le p<+\infty$, where the lower bound assures that $L^p_a(U)$ is a Banach space. Fix a bounded domain $G$ and a compact subset $K$ of $G$ with zero area measure. Let $M$ be a closed subspace of $L^p_a(G\sbs K)$. We say that $M$ is invariant provided that $zf\in M$ whenever $f\in M$. Aleman, Richter, and Ross study those invariant subspaces $M$ of $L^p_a(G\sbs K)$ that contain $L^p_a(G)$. They also add the requirement that $M$ have index one, which is taken to mean that $(z\lambda)M$ has codimension $1$ in $M$, for all $\lambda\in G\sbs K$. Natural... (More)
 For a bounded open set $U$ in the complex plane, we consider the Bergman space $L^p_a(U)$ of $p$thpower areaintegrable analytic functions on $U$, with $p$ assumed to be in the range $1\le p<+\infty$, where the lower bound assures that $L^p_a(U)$ is a Banach space. Fix a bounded domain $G$ and a compact subset $K$ of $G$ with zero area measure. Let $M$ be a closed subspace of $L^p_a(G\sbs K)$. We say that $M$ is invariant provided that $zf\in M$ whenever $f\in M$. Aleman, Richter, and Ross study those invariant subspaces $M$ of $L^p_a(G\sbs K)$ that contain $L^p_a(G)$. They also add the requirement that $M$ have index one, which is taken to mean that $(z\lambda)M$ has codimension $1$ in $M$, for all $\lambda\in G\sbs K$. Natural examples of such invariant subspaces are those of the form $M=L^p_a(G\sbs E)$, where $E$ is a closed subset of $K$. The authors show that for $p<2$, these are indeed all such invariant subspaces which can be found. For $p\ge2$, this is not so, but nevertheless a complete classification can be found in terms of quasiclosed subsets $E$ of $K$.
If the condition that the index of $M$ is one is dropped, then the structure of such invariant subspaces can be extremely complicated. If, however, $G\sbs K$ is connected, the authors show that $M$ automatically has index one (due to the assumption that $M$ should contain $L_a^p(G)$). (Less)
Please use this url to cite or link to this publication:
http://lup.lub.lu.se/record/1467233
 author
 Aleman, Alexandru ^{LU} ; Richter, Stefan and Ross, William T
 publishing date
 1996
 type
 Contribution to journal
 publication status
 published
 subject
 in
 Canadian Journal of Mathematics
 volume
 48
 issue
 2
 pages
 225  243
 publisher
 Canadian Mathematical Society
 external identifiers

 scopus:0030504629
 ISSN
 0008414X
 language
 English
 LU publication?
 no
 id
 8b2528f146da4cb3b0d99eae2db94884 (old id 1467233)
 date added to LUP
 20090916 13:17:53
 date last changed
 20181121 20:41:36
@article{8b2528f146da4cb3b0d99eae2db94884, abstract = {For a bounded open set $U$ in the complex plane, we consider the Bergman space $L^p_a(U)$ of $p$thpower areaintegrable analytic functions on $U$, with $p$ assumed to be in the range $1\le p<+\infty$, where the lower bound assures that $L^p_a(U)$ is a Banach space. Fix a bounded domain $G$ and a compact subset $K$ of $G$ with zero area measure. Let $M$ be a closed subspace of $L^p_a(G\sbs K)$. We say that $M$ is invariant provided that $zf\in M$ whenever $f\in M$. Aleman, Richter, and Ross study those invariant subspaces $M$ of $L^p_a(G\sbs K)$ that contain $L^p_a(G)$. They also add the requirement that $M$ have index one, which is taken to mean that $(z\lambda)M$ has codimension $1$ in $M$, for all $\lambda\in G\sbs K$. Natural examples of such invariant subspaces are those of the form $M=L^p_a(G\sbs E)$, where $E$ is a closed subset of $K$. The authors show that for $p<2$, these are indeed all such invariant subspaces which can be found. For $p\ge2$, this is not so, but nevertheless a complete classification can be found in terms of quasiclosed subsets $E$ of $K$. <br/><br> <br/><br> If the condition that the index of $M$ is one is dropped, then the structure of such invariant subspaces can be extremely complicated. If, however, $G\sbs K$ is connected, the authors show that $M$ automatically has index one (due to the assumption that $M$ should contain $L_a^p(G)$).}, author = {Aleman, Alexandru and Richter, Stefan and Ross, William T}, issn = {0008414X}, language = {eng}, number = {2}, pages = {225243}, publisher = {Canadian Mathematical Society}, series = {Canadian Journal of Mathematics}, title = {Bergman spaces on disconnected domains}, volume = {48}, year = {1996}, }