The wandering subspace property and Shimorin’s condition of shift operator on the weighted Bergman spaces

In the present paper, we first study the wandering subspace property of the shift operator on the Ia\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I_{a}$$\end{document} type zero based invariant subspaces of the weighted Bergman spaces La2(dAn)(n=0,2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_{a}^{2}(dA_{n})(n=0,2)$$\end{document} via the spectrum of some Toeplitz operators on the Hardy space H2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H^{2}$$\end{document}. Second, we give examples to show that Shimorin’s condition for the shift operator fails on the Ia\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I_{a}$$\end{document} type zero based invariant subspaces of the weighted Bergman spaces La2(dAα)(α>0)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L_{a}^{2}(dA_{\alpha })(\alpha >0)$$\end{document}.

where f (z) = ∑ ∞ n=0 a n z n is the power series representation of f. Let be the boundary of the unit disk . A function (z) in H 2 is called inner if | (z)| = 1 a.e. on . The famous Beurling theorem [2] says that every invariant subspace of the multiplication operator T z on the Hardy space H 2 other than {0} has the form H 2  Let dA be the normalized Lebesgue area measure on , and let: The space L 2 ( , dA ) consists of complex valued functions f on such that: It is well-known that L 2 ( , dA ) is a Hilbert space with the above norm. For any > −1 , we define: then L 2 a (dA ) is a closed subspace of L 2 ( , dA ) . These spaces will be called the weighted Bergman spaces. Let B denote the shift operator on L 2 a (dA ) which maps every f ∈ L 2 a (dA ) to zf. If = 0 , for convenience, the Bergman space L 2 a (dA 0 ) and the Bergman shift B 0 will be denoted by L 2 a and B , respectively. In 1996, Aleman, Richter and Sundberg [1] proved that the Beurling type theorem holds for B on L 2 a . Later, different proofs of the Beurling theorem for B on L 2 a were given in [6,7,9]. In [9], Shimorin proved the following theorem.  (i) ‖Tx + y‖ 2 ≤ 2 � ‖x‖ 2 + ‖Ty‖ 2 � for allx, y ∈ H , and (ii) ∩ ∞ n=1 T n H = {0} , then T possesses the wandering subspace property on H.

Remark 1.2 The above condition (i) is equivalent to say that T is bounded and bounded below on H and satisfies the following equation:
If T satisfies the condition (ii) on H, then we say that T is analytic on H. Definition 1. 3 We say that Shimorin's condition for T holds on H if T satisfies the above conditions (i) and (ii) (see [12]).
Shimorin's theorem implies that T possesses the wandering subspace property on H if Shimorin's condition for T holds on H. As an application of Shimorin's theorem, it is proved that for any −1 < ≤ 0 the Beurling type theorem holds for B on L 2 a (dA ) , and then as a corollary, they gave a simpler proof of the Beurling type theorem on the Bergman space. One year later, Shimorin proved that for any −1 < ≤ 1 the Beurling-type theorem holds for B on L 2 a (dA ) (see [10]). The first step to solve Question 1 is to verify whether Shimorin's condition for the operator T holds on the invariant subspaces. It is always difficult to verify directly according to Definition 1.3, for example, the reproducing kernel Hilbert spaces with the complicate kernel functions. If Shimorin's condition for the operator T fails on some invariant subspaces, then we must try other ways to solve Question 1.
The famous example is the zero-based invariant subspaces of the weighted Bergman spaces L 2 a (dA )( > −1).

Definition 1.4
For any n ≥ 1, > −1 , now suppose the sequence A = {a 1 , a 2 , … , a n , …} ⊂ (may be same). Let: then I A is an invariant subspace of B . The subspace I A is called a zero based invariant subspace of L 2 a (dA ) (see [3]). When A = {a 1 , a 2 , … , a n } is a finite set of points inside , for distinguishing the difference, we denote by: In particular, let A = {a} , the kernel function for the I a type zero based invariant subspaces of L 2 a (dA ) (see [3]) is: In the present paper, we give the following theorem to show that Shimorin's condition for the shift operator fails on the I a type zero-based invariant subspaces of the weighted Bergman spaces L 2 a (dA )( > 0).
I a 1 ,a 2 ,…,a n = I A . We study the zero based invariant subspaces of the weighted Bergman spaces is also due to the following facts: Hedenmalm and Zhu [4] have showed that for any > 4 there exists some I a type zero based invariant subspaces of L 2 a (dA ) don't possess the wandering subspace property. So in this case, the Beurling type theorem for B on L 2 a (dA ) fails. In [9], Shimorin conjectured that the critical value for the Beurling type theorem on the weighted Bergman spaces L 2 a (dA ) is = 1 . In 2004, Hedenmalm and Perdomo [3] essentially proved that the Beurling type theorem fails in L 2 a (dA ) for > c , where c ≈ 1.04 . It is natural to discuss the following question: Question 2 Does Shimorin's condition for B hold on all I a type zero based invariant subspaces of L 2 a (dA ) with ∈ (1, +∞)?
gives a negative answer to Question 2. Then we must try other ways to solve the following natural question: Question 3 Does B possess the wandering subspace property on the I a type zero based invariant subspaces of L 2 a (dA ) with ∈ (1, +∞)?
In [12], the authors gave a positive answer for the case of = 2 of Question 3 and proved the case of = 2 of Theorem 1.5. To solve Question 3, let 2 , 2 be the bidisk and torus which are the Cartesian product of 2 copies of and , respectively. The Hardy space H 2 ( 2 ) over the bidisk 2 consists of holomorphic functions f on 2 satisfying: where is the normalized Haar measure on 2 . It is also well-known that H 2 ( 2 ) is a Hilbert space. Let M be a closed subspace of H 2 ( 2 ) , we say that M is a submodule if M is invariant under multiplication operators T z and T w . In particular, for any subset X ⊂ H 2 ( 2 ) , let: where A( 2 ) is the bidisk algebra, that is the closure of polynomials in z and w under the norm of H ∞ ( 2 ) . Then [X] is a submodule, it is called the submodule generated [X] = clos span A( 2 )X , by X. The orthonormal complement of a submodule is called a quotient module. Let (S z , S w ) be the two variable Jordan block which are the compression operators defined on a quotient module N, more precisely, where P N is the projection of H 2 ( 2 ) onto N. In particular, let K 1 = [z − w] be the submodule generated by z − w , and let N 1 = H 2 ( 2 ) ⊖ K 1 be the related quotient module. As we know, the quotient module N 1 plays a great role in many situations since the compression operator S z on N 1 is unitarily equivalent to the Bergman shift on L 2 a . In [11], by lifting the Bergman shift as S z on N 1 , Sun and Zheng gave a proof of the Beurling type theorem for B on L 2 a . In [12], the authors considered the submodule K 0 = [(z − w) 2 ] and the related quotient , and the following lemma [12] holds: , then S 1 z and B 2 are unitarily equivalent.
The space L ∞ ( ) is the collection of all essentially bounded measurable functions on . For each (e i ) ∈ L ∞ ( ) , the Toeplitz operator with symbol (e i ) is the operator T defined by: where P is the orthogonal projection of L 2 onto H 2 . By some statements in [12], we get the following theorem: The last part of the proof of Theorem 8 in [12] has in fact proved that a ∉ (T ) . We believe that Theorem 1.7 is a general phenomenon in the case of all zero based invariant subspaces of L 2 a (dA )( > 1) . We also believe that B does not possess the wandering subspace property on some zero-based invariant subspaces of L 2 a (dA )( > 1) and it will substantiate Shimorin's conjecture. Our basic problem is to find and prove the general phenomenon, and then, we prove the wandering subspace property of B on all zero based invariant subspaces of L 2 a (dA )( > 1) via the spectrum of some Toeplitz operators on the Hardy space H 2 . It is obvious that Question 3 is a special case of our basic problem. In the present paper, we prove the similar phenomenon in the case of Bergman space and reprove the wandering subspace property of the Bergman shift on the I a type zero-based invariant subspaces. The following is our main theorem: Through Theorems 1.7 and 1.8, we conjecture that the following proposition holds for the general case of the weighted Bergman spaces L 2 a (dA n )(n = 0, 1, 2, …) . We believe that the proof of Conjecture 1 will make progress in solving Question 3, but we cannot prove it in this paper. This paper is arranged as follows. In Sect. 2, we prove Theorem 1.7. In Sect. 3, we give an equivalent condition that S z does not possess the wandering subspace property on any fixed invariant subspace of S z on N 1 . In this part of the research process, we simply get the same characterization of M 00 as in [11]. In Sect. 4, we give the proof of Theorem 1.8, and then as a corollary, we reprove that the Bergman shift possesses the wandering subspace property on all I a type zero based invariant subspaces. In Sect. 5, we prove Theorem 1.5.
In this paper, for a Hilbert space H and a bounded linear operator T on it, we denote by Lat(T) the lattice of invariant subspaces for T on H.

The proof of theorem 1.7
Let then {̃1,̃2, …} is an orthonormal basis of H 1 0 (see [12]). In [12], the authors have proved the following proposition: if and only if there exists a nonzero solution for the following where for any i ≥ 0 , and ̃1,̃2, … is an orthonormal basis of H 1 0 as mentioned above.
In this section, we first point out that Proposition 2.1 can be written in the following simple form:

and only if there exists a nonzero solution for the following equations of
for any i ≥ 0 , and ̃1,̃2, … is an orthonormal basis of H 1 0 as mentioned above.
Proof The condition (i) in Proposition 2.1 is equivalent to C i 2 = 0 for any i ≥ 0 (see [12]).
Then by (iii), we can calculate that In general, we can prove that (ii) and (iii) are equivalent to: that is (ii). If (ii) and (iii) hold, then (2.2) holds for k = 2 . We assume that (2.2) holds for any fixed k ≥ 2 , then by (iii),

and only if there exists a nonzero solution for the following equations of
Proof of Theorem 1.7 By the proof of Theorem 8 in [12], we get that (2.5) has a nonzero solution {C i 3 } i≥0 satisfying: which is equivalent to being an eigenvalue of Toeplitz operator By Lemma 2.3, we get Theorem 1.7. ◻

Beurling type theorem for S z on N 1
In this section, we use the techniques in [12].

Proposition 3.1 S z is left invertible and analytic on N 1 [5].
Proof Note that an orthonormal basis of N 1 is: where Then, we get the matrix A of S z under the above basis: And one calculates A * and A * A , respectively: , and therefore, S z is left invertible. Since, for any f ∈ N 1 , there exists a sequence {a n } ∞ n=0 such that: Then the coordinate vector of S z f is Now if g ∈ ∩ ∞ n=1 S n z N 1 , there exists f n ∈ N 1 such that: For any fixed n ≥ 1 , comparing the coordinate vectors of g and S n z f n , we get that the fist n numbers of coordinate vector of g are zero, thus It is easy to verify that M is an invariant subspace for T z in H 2 ( 2 ) . In Theorem 6.8 of [8], Richter showed that the mapping ∶ M →M is one-to-one correspondence between invariant subspaces of S z and invariant subspaces of T z containing [z − w] . Let LM be the wandering space of T z on M , we easily get the following theorem: , we have the following decomposition: A(a 0 , a 1 , … , a n , …) Thus: The above implies that: Conversely, for any g in Since, then f ∈ K 1 ⊖ zK 1 . By above calculations, we have f = zP N 1 g − wg(w) and this completes the proof. where m k ∈ M 0 and u n = −S z B * M S z −1 P M T * z f n + f n ∈ M 00 . Since, q ∈ N and q ⟂ ∨ n≥0 S n z M 0 , taking inner product of q with S k zmk gives This implies that m k = 0 for all k ≥ 0 . Thus each function q in N has the following form: Let: Ñ = � ∑ ∞ n=0 z n u n ∶ u n ∈ M 00 � , by Corollary 3.6, we have N ⊂ � N . Let

Proof
Step 1. For any f in Ñ , let f = ∑ ∞ n=0 z n u n , where u n ∈ M 00 . On the one hand, by Theorem 3.2, we have f ∈M . Thus f ∈ M ⇔ f ⟂ K 1 . On the other hand, assume f ∈ M . Since for any k ≥ 0,m k ∈ M 0 , we have: ] . This implies that: Thus, we have the following equivalent relation: Step 2. It is obvious that f ⟂ K 1 is equivalent to And it is easy to verify that: Since f 0 n ⟂ K 1 , so we get: Note that f n ⟂ zK 1 and (z − w)z i−n w j ∈ zK 1 (n < i) , we have: Hence, f ∈ N is equivalent to: Step 3. When j = 0 , since 1 ∈ N 1 , so 1 ⟂ f n (n ≥ 0) , thus, we get: On the other hand, it is easy to verify that: Then, in this case, (3.13) is equivalent to C i 0 = 0(i ≥ 0). Step 4. When j ≥ 1 , it is easy to verify that: (3.14) and Then, in this case, (3.13) is equivalent to (ii).
is a nonzero solution for the above equations.
Conversely, assume {C i k } k,i≥0 is a nonzero solution for the above equations. Let: As we know, the Beurling type theorem holds for S z on N 1 , thus we have the following corollary through Theorem 3.8 which reflect the common property of invariant subspaces of S z . Corollary 3.9 Let M ⊂ N 1 and M ∈ Lat(S z ) , then the following equations of C i k (k ≥ 0, i ≥ 0) has only zero solution: where for any i ≥ 0 , and ẽ 0 ,ẽ 1 , … is an orthonormal basis of N 1 as mentioned above.
The proof of Theorem 1.8 Let: then {E k } k≥0 is an orthonormal basis of L 2 a . For any fixed a ∈ , the reproducing kernel of L 2 a is Define the operator U ∶ L 2 a → N 1 such that where (4.10) (4.14) And one calculates that: since, that is equivalent to: Then, Hence, and, By the above calculates, we can choose A a being any positive constant, such that: Note that: where C a,k is a positive constant depending only on a, k. Then, Proof For any fixed a ∈ , let M = M a in Theorem 3.8, we easily calculate that: and Then, Then (ii) in Theorem 3.8 is: If C i 0 = 0(i ≥ 0) , let k = 1 in (4.27), we get: Hence, (4.27) is equivalent to holding the following equations: By the induction, we easily get that Eq. (4.30) is equivalent to the following equations: (4.28) (4.31) where And 1 a n , when 0 ≤ m ≤ n − 1, n ≥ 1; x nm = 0, when 0 ≤ n − 1 < m.
where C is a positive constant, so we have: Then, (4.32) is equivalent to the following form: That is also equivalent to the following matrix form: where Since, the matrix: is the matrix of the Toeplitz operator with symbol (e i ) ∈ L ∞ ( ) with respect to the basis {e in } ∞ n=0 of H 2 , and let k be the kth Fourier coefficient of (e i ) , then Thus, we get: Note that (4.36) has a nonzero solution {C i 1 } i≥0 such that: which is equivalent to a being an eigenvalue of Toeplitz operator T ( f = ∑ ∞ i=0 C i 1 z i is an eigenvector). This completes the proof. ◻ . By Theorem 1.8, in the following process, we need to prove that a is not an eigenvalue of Toeplitz operator T for any 0 ≠ a ∈ , where, and Note that: then (T * ) = (T̄) =̄( ) . If a ∈ p (T ) , since a is real, then a ∈ (T * ) . Note that    where b = |a| 2 , x = 1 1−āz . That is: Since 0 < b < 1 and |āz| < 1 for all z ∈ , then (4.42) fails for any z ∈ , i.e., a −̄(z) ≠ 0 on . But a −̄(z) is continuous on , hence This implies a ∉̄( ) , i.e., a ∉ (T * ) , and this completes the proof.
that is: Repeating the same process of the proof of Proposition 4.3, we get the following proposition: