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- Aleman, Alexandru, et al.
(författare)
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Beurling's theorem for the Bergman space
- 1996
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Ingår i: Acta Mathematica. - 0001-5962. ; 177:2, s. 275-310
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Tidskriftsartikel (refereegranskat)abstract
- A celebrated theorem in operator theory is A. Beurling's description of the invariant subspaces in $H^2$ in terms of inner functions [Acta Math. {\bf81} (1949), 239--255; MR0027954 (10,381e)]. To do the same thing for the Bergman space $L^2_a$ has been deemed virtually impossible by many analysts, in view of the fact that the lattice of invariant subspaces is so large, and that the invariant subspaces may have weird properties as viewed from the $H^2$ perspective. The size of the lattice can be appreciated from the known fact that essentially every operator on separable Hilbert space can be realized as the compression of the Bergman shift on $M\ominus N$, where $M$ and $N$ are invariant subspaces, $N\subset M$. But a Beurling-type theorem is precisely what the present paper delivers. Given an invariant subspace $M$ in $L^2_a$, consider the subspace $M\ominus TM$, where $T$ stands for multiplication by $z$. This makes sense because $TM$ is a closed subspace of $M$. In Beurling's $H^2$ case, $M\ominus TM$ is one-dimensional and spanned by an inner function. In the $L^2_a$ setting, the dimension of $M\ominus TM$ may be arbitrarily large, even infinite. However, with the correct analogous definition of inner functions in $L^2_a$, all vectors of unit norm in\break $M\ominus TM$ are $L^2_a$-inner. Following Halmos, the subspace $M\ominus TM$ is called the wandering subspace of $M$. Given an invariant subspace, a natural question is: which collections of elements generate it? In particular, one can ask for the least number of elements in a set of generators. It is known that the dimension of the wandering subspace represents a lower bound for the least number of generators. In the paper, it is shown that any orthonormal basis in the wandering subspace (which then consists of $L^2_a$-inner functions) generates $M$ as an invariant subspace. This settles the issue of the minimal number of generators. Let $P$ be the orthogonal projection $M\to M\ominus TM$, and let $L\colon M\to M$ be the operator such that $TL$ is the orthogonal projection $M\to TM$. Then, for $f\in M$, $f=Pf+TLf$. If we do the same for $Lf\in M$, we get $Lf=PLf+TL^2f$ and, inserting it into the original relation for $f$, we get $f=Pf+TPLf+T^2L^2f$. As we go on repeating this process, we get $f=Pf+TPLf+T^2PL^2f+\cdots+T^{n-1}PL^{n-1}f+T^nL^nf$. The point with this decomposition is that, apart from the last term, each term is of the form $T$ to some power times an element of $M\ominus TM$, so that $Pf+TPLf+T^2PL^2f+\cdots+T^{n-1}PL^{n-1}f$ is in\break $[M\ominus TM]$, the invariant subspace generated by $M\ominus TM$. If the operators $T^nL^n$ happened to be uniformly bounded, as they are in the case of $H^2$, $T^nL^nf$ would tend to $0$ in the weak topology, and $f$ would be in the weak closure of $[M\ominus TM]$, which by standard functional analysis coincides with $[M\ominus TM]$. However, for the Bergman space, it seems unlikely that the $T^nL^n$ are uniformly bounded for all possible invariant subspaces $M$, although no immediate counterexample comes to mind. For this reason, the authors try Abel summation instead, and consider for $0
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