Beurling's theorem for the Bergman space
(1996) In Acta Mathematica 177(2). p.275-310- 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... (More) - 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 $0<s<1$ the operators $R_s\colon M\to [M\ominus TM]$ given by $R_s=P+sTPL+s^2T^2PL^2+s^3T^3PL^3+\cdots$; it is easy\break to show that the series converges in norm. To prove that $M=\break [M \ominus TM]$, it suffices to check that (a) $\|R_s\|\le C$ for some constant independent of $s$, and (b) $R_sf\to f$ in the topology of uniform convergence on compacts in the unit disk. This is so because $R_sf\in [M\ominus TM]$ then converges weakly to $f\in M$ as $s\to1$. Part (b) is easy; the trick is to obtain (a). The authors show at an early stage that for $\lambda\in\bold D$, $M=(M\ominus TM)+\break (T-\lambda)M$, and that the sum is direct (in the Banach space sense). They write $Q_\lambda f$ for the skewed projection operator $M\to M\ominus TM$ associated with this decomposition. The operator $Q_\lambda$ has a convergent Taylor series expansion $Q_\lambda=P+\lambda PL+\lambda^2 PL^2+\lambda^3PL^3+\cdots$, which very much resembles the expression for $R_s$. In fact, one sees that $R_sf(w)=Q_{sw}f(w)$. Now one observes that $Q_\lambda f(\lambda)=f(\lambda)$, because the element of $(T-\lambda)M$ which one has to add to $Q_\lambda f$ to obtain $f$ vanishes at the point $\lambda$. It follows that $R_sf(w)\to f(w)$ normally in $\bold D$ as $s\to1$.
We come to the hard part: obtaining the uniform boundedness of the operators $R_s$. What is needed is a concrete representation formula for the norm in $M$ which makes it possible to compare the norms of $f$ and $R_sf$. Here, the factorization theory due to the reviewer [J. Reine Angew. Math. 422 (1991), 45--68; MR1133317 (93c:30053)], and subsequent improvements by Duren-Khavinson-Shapiro-Sundberg, come to assistance. Let $\varphi$ be an $L^2_a$-inner function, and let $f\in[\varphi]$, the invariant subspace generated by $\varphi$. The norm of $f$ can then be expressed in terms of the quotient $f/\varphi$. That formula suggests the norm identity (1) $\|f\|^2=\pi^{-1}\int_\bold D\|Q_w f\|^2dA(w)+\pi^{-2}\int_{\bold D\times\bold D}\Gamma(z,w) \Delta_z\Delta_w|Q_wf(z)|^2\,dA(z)\,dA(w)$, which turns out to be valid for general invariant subspaces $M$, where $\Gamma(z,w)$ is the biharmonic Green function for the unit disk (suitably normalized), which is known to be positive. One first checks that the formula (1) holds for all $f$ in $[M\ominus TM]$. Second, a related integral formula for the norm in $M$, in terms of integrals along concentric circles, is established for all $f\in M$. Then an intricate argument, involving Green's formula and rather subtle analysis of signs of functions, shows that for $f\in M$, we have at least a $\ge$ inequality in (1). To get the identity (1) for general $f\in M$, it is necessary to show first that $M=[M\ominus TM]$.
Formula (1), which we know to hold for $f\in [M\ominus TM]$ and with $\ge$ for general $f\in M$, applies to $R_s f\in[M\ominus TM]$ with equality, and if we notice that $Q_wR_sf(z)=Q_{sw}f(z)$, we get (2) $\|R_sf\|^2= \pi^{-1}\int_\bold D\|Q_{sw} f\|^2dA(w)+\pi^{-2}\int_{\bold D\times\bold D} \Gamma(z,w)\Delta_z\Delta_w|Q_{sw}f(z)|^2\,dA(z)\,dA(w)$. An almost radial monotonicity property of $\Gamma(z,w)$ in the $w$ variable then shows that the last term on the right-hand side of (2) has a $\limsup$ as $s\to1$ bounded by twice the value one gets when $s$ is set equal $1$. Moreover, the first term on the right-hand side of (2) has a $\limsup$ as $s\to 1$ which equals what one gets when $s=1$ is plugged in. Thus, $\limsup_s\|R_sf\|\le 2\|R_1f\|\le2\|f\|$. It follows that $R_sf\to f$ weakly, which completes the proof. In the paper, it is even shown that $R_sf\to f$ in norm as $s\to1$. Incidentally, the proof also answers affirmatively one of the conjectures raised by the reviewer concerning polynomial approximation in certain weighted Bergman spaces [in Linear and complex analysis. Problem book 3. Part II, 114, Lecture Notes in Math., 1574, Springer, Berlin, 1994].
The above-mentioned theorem is new also in the case when the wandering subspace $M\ominus TM$ is one-dimensional.
The theorem seems to represent a breakthrough in our understanding of the invariant subspaces of the Bergman space. What is desirable and remains to be developed is a better understanding of the wandering subspaces. (Less)
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- author
- Aleman, Alexandru ^{LU} ; Richter, Stefan and Sundberg, Carl
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- Acta Mathematica
- volume
- 177
- issue
- 2
- pages
- 275 - 310
- publisher
- Springer
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- 0001-5962
- DOI
- 10.1007/BF02392623
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- English
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@article{09f1fbd7-45e5-4e5d-bc9f-157df52c9576, 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, $N\subset M$. <br/><br> <br/><br> 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 $M$. <br/><br> <br/><br> 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. <br/><br> <br/><br> 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. <br/><br> <br/><br> For this reason, the authors try Abel summation instead, and consider for $0<s<1$ the operators $R_s\colon M\to [M\ominus TM]$ given by $R_s=P+sTPL+s^2T^2PL^2+s^3T^3PL^3+\cdots$; it is easy\break to show that the series converges in norm. To prove that $M=\break [M \ominus TM]$, it suffices to check that (a) $\|R_s\|\le C$ for some constant independent of $s$, and (b) $R_sf\to f$ in the topology of uniform convergence on compacts in the unit disk. This is so because $R_sf\in [M\ominus TM]$ then converges weakly to $f\in M$ as $s\to1$. Part (b) is easy; the trick is to obtain (a). The authors show at an early stage that for $\lambda\in\bold D$, $M=(M\ominus TM)+\break (T-\lambda)M$, and that the sum is direct (in the Banach space sense). They write $Q_\lambda f$ for the skewed projection operator $M\to M\ominus TM$ associated with this decomposition. The operator $Q_\lambda$ has a convergent Taylor series expansion $Q_\lambda=P+\lambda PL+\lambda^2 PL^2+\lambda^3PL^3+\cdots$, which very much resembles the expression for $R_s$. In fact, one sees that $R_sf(w)=Q_{sw}f(w)$. Now one observes that $Q_\lambda f(\lambda)=f(\lambda)$, because the element of $(T-\lambda)M$ which one has to add to $Q_\lambda f$ to obtain $f$ vanishes at the point $\lambda$. It follows that $R_sf(w)\to f(w)$ normally in $\bold D$ as $s\to1$. <br/><br> <br/><br> We come to the hard part: obtaining the uniform boundedness of the operators $R_s$. What is needed is a concrete representation formula for the norm in $M$ which makes it possible to compare the norms of $f$ and $R_sf$. Here, the factorization theory due to the reviewer [J. Reine Angew. Math. 422 (1991), 45--68; MR1133317 (93c:30053)], and subsequent improvements by Duren-Khavinson-Shapiro-Sundberg, come to assistance. Let $\varphi$ be an $L^2_a$-inner function, and let $f\in[\varphi]$, the invariant subspace generated by $\varphi$. The norm of $f$ can then be expressed in terms of the quotient $f/\varphi$. That formula suggests the norm identity (1) $\|f\|^2=\pi^{-1}\int_\bold D\|Q_w f\|^2dA(w)+\pi^{-2}\int_{\bold D\times\bold D}\Gamma(z,w) \Delta_z\Delta_w|Q_wf(z)|^2\,dA(z)\,dA(w)$, which turns out to be valid for general invariant subspaces $M$, where $\Gamma(z,w)$ is the biharmonic Green function for the unit disk (suitably normalized), which is known to be positive. One first checks that the formula (1) holds for all $f$ in $[M\ominus TM]$. Second, a related integral formula for the norm in $M$, in terms of integrals along concentric circles, is established for all $f\in M$. Then an intricate argument, involving Green's formula and rather subtle analysis of signs of functions, shows that for $f\in M$, we have at least a $\ge$ inequality in (1). To get the identity (1) for general $f\in M$, it is necessary to show first that $M=[M\ominus TM]$. <br/><br> <br/><br> Formula (1), which we know to hold for $f\in [M\ominus TM]$ and with $\ge$ for general $f\in M$, applies to $R_s f\in[M\ominus TM]$ with equality, and if we notice that $Q_wR_sf(z)=Q_{sw}f(z)$, we get (2) $\|R_sf\|^2= \pi^{-1}\int_\bold D\|Q_{sw} f\|^2dA(w)+\pi^{-2}\int_{\bold D\times\bold D} \Gamma(z,w)\Delta_z\Delta_w|Q_{sw}f(z)|^2\,dA(z)\,dA(w)$. An almost radial monotonicity property of $\Gamma(z,w)$ in the $w$ variable then shows that the last term on the right-hand side of (2) has a $\limsup$ as $s\to1$ bounded by twice the value one gets when $s$ is set equal $1$. Moreover, the first term on the right-hand side of (2) has a $\limsup$ as $s\to 1$ which equals what one gets when $s=1$ is plugged in. Thus, $\limsup_s\|R_sf\|\le 2\|R_1f\|\le2\|f\|$. It follows that $R_sf\to f$ weakly, which completes the proof. In the paper, it is even shown that $R_sf\to f$ in norm as $s\to1$. Incidentally, the proof also answers affirmatively one of the conjectures raised by the reviewer concerning polynomial approximation in certain weighted Bergman spaces [in Linear and complex analysis. Problem book 3. Part II, 114, Lecture Notes in Math., 1574, Springer, Berlin, 1994]. <br/><br> <br/><br> The above-mentioned theorem is new also in the case when the wandering subspace $M\ominus TM$ is one-dimensional. <br/><br> <br/><br> The theorem seems to represent a breakthrough in our understanding of the invariant subspaces of the Bergman space. What is desirable and remains to be developed is a better understanding of the wandering subspaces.}, author = {Aleman, Alexandru and Richter, Stefan and Sundberg, Carl}, issn = {0001-5962}, language = {eng}, number = {2}, pages = {275--310}, publisher = {Springer}, series = {Acta Mathematica}, title = {Beurling's theorem for the Bergman space}, url = {http://dx.doi.org/10.1007/BF02392623}, volume = {177}, year = {1996}, }