Reducing Subspaces of the Dual Truncated Toeplitz Operator

Hu, Yinyin; Deng, Jia; Yu, Tao; Liu, Liu; Lu, Yufeng

doi:https://doi.org/10.1155/2018/7058401

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Research Article | Open Access

Volume 2018 | Article ID 7058401 | https://doi.org/10.1155/2018/7058401

Reducing Subspaces of the Dual Truncated Toeplitz Operator

Yinyin Hu,¹Jia Deng,²Tao Yu,²Liu Liu,²and Yufeng Lu²

Academic Editor: Raúl E. Curto

Received01 Mar 2018

Accepted23 Apr 2018

Published03 Jun 2018

Abstract

We define the dual truncated Toeplitz operators and give some basic properties of them. In particular, spectrum and reducing subspaces of some special dual truncated Toeplitz operator are characterized.

1. Introduction

Let denote the open unit disk in the complex plane and denote the unit circle. As usual, denotes the Hilbert space of Lebesgue square integral functions on with the inner product: where , is the normalized Lebesgue measure. As usual, will be identified with the subspace of consisting of the functions whose Fourier coefficients with negative indices vanish. Let denote the projection from to , which is given explicitly by the Cauchy integral: For , the Toeplitz operator on , with symbol , is defined by where Let denote the unilateral shift operator on Its adjoint, the backward shift, is given by

For the remainder of the paper, will denote a nonconstant inner function. The subspace is a proper nontrivial invariant subspace of , the most general one by the well-known theorem of A. Beurling [1]. Truncated Toeplitz operators (TTO) are compressions of the standard Toeplitz operators on the Hardy space to its coinvariant, the so-called model space Let denote the orthogonal projection from onto the subspace For , the truncated Toeplitz operator with symbol is defined by on the dense subset of the space The symbol is not unique [2].

Although the truncated Toeplitz operators share many fundamental properties of classical Toeplitz operators on the Hardy space, they differ in many crucial ways. For example, compact Toeplitz operators on Hardy space are zero, but there are many nonzero compact truncated Toeplitz operators. In part motivated by several of the problems posed in the aforementioned article [2], the area has undergone vigorous development during the past several years [3–22]. While several of the initial questions raised by Sarason have now been resolved, the study of truncated Toeplitz operators has nevertheless proven to be fertile ground, spawning both new questions and unexpected results.

Recently, Gu [23] defined truncated Hankel operator (THO) as the compression of Hankel operator to invariant subspaces for the backward shift and proved a number of algebraic properties of them. Some of the properties in his paper reveal the relation between the THOs and TTOs. Later, Kang and Kim [24] characterized the pairs of truncated Hankel operator on the model spaces whose products result in truncated Toeplitz operators when the inner function has a certain symmetric property.

In this paper, we will define dual truncated Toeplitz operators and introduce some algebraic properties of them. As is well known, the orthogonal complement of , , which is the subspace of with Let , and the dual truncated Toeplitz operator on with symbol is defined by where For , the kernel function in for the functional of evaluation at will be denoted by , which is

The paper is organized as follows. In Section 2, we give some basic properties of dual truncated Toeplitz operator. In Section 3, we characterize the spectrum and reducing subspaces of some special dual truncated Toeplitz operator.

2. Basic Properties

An operator on is called C-symmetric if , where is an antiunitary involution on defined by In fact, it is easily seen that maps to and to .

Lemma 1. Dual truncated Toeplitz operators are C-symmetric.

Proof. Let be in with bounded. For and in , we have This completes the proof.

In the following, we will discuss the boundedness and compactness of dual Truncated Toeplitz operators.

Theorem 2. For , we have

Proof. Let denote the approximate point spectrum of the bounded linear operator . It is well known that multiplication operators on the Hardy space , , and
Assuming that , then there exists a sequence of functions in with such that By removing a “tail” of small norm and renormalizing, for each there is of norm that has only a finite number of nonzero Fourier coefficients corresponding to positive indices and satisfy Then Let , and, for each , there exists a positive integer such that is in Since is unitary and commute with , it follows that and For each , define Then each is in , Since we have , which implies that Thus we get that Hence On the other hand, which completes the proof.

In this paper, let denote the orthonormal basis. For , we have For , we get

Theorem 3. The only compact dual truncated Toeplitz operator is zero.

Proof. Let denote a compact dual Truncated Toeplitz operator. For nonnegative integers and , since converges weakly to zero as , and it is obtained that Let , where is the Fourier coefficient of , and it follows that Hence for all nonnegative integers and So for all integer which implies that Thus

With the above theorem, we get the following corollary.

Corollary 4. A dual truncated Toeplitz operator is self-adjoint if and only if the symbol is real-valued almost everywhere.

3. Spectrum and Reducing Subspace

In this section, we discuss the spectrum and reducing subspace of some special dual truncated Toeplitz operator. Let denote the spectrum of the linear operator and denote the essential spectrum and denote the point spectrum.

Theorem 5. For dual Truncated Toeplitz operator , if , then

Proof. It is clear that the spectrum of is contained in First, we will show that how the dual Truncated Toeplitz operator act on the orthonormal basis A direct computation givesFrom the above, if , is the bilateral shift. By corollary 24.4 in [25], we have the spectrum of that is the unit circle. In the following, assume that Then is a special weight shift with the sequence of scalar , which for and for
By proposition 27.7(c) in [22], we have Then we will see the point spectrum of Suppose that , Let , and it is obtained that and From above, we get that implies almost everywhere. Suppose that , we have Since and , we have that almost everywhere. Hence we have that the point spectrum of is
It suffices for us to show that any point in is not the spectrum of the dual truncated Toeplitz operator Since it is known that , for any , it follows that is Fredholm operator. Let be an operator defined on as follows where is in It is clear that is an unitary operator and Hence is invertible. It is obtained that A direct computation gives that hence we have As , it follows that So we get that is invertible and This completes the proof.

In the rest of this section, we characterize the structure of lattice of reducing subspaces for and . To do so, we need to give some notations.

Let be a bounded linear operator on a Hilbert space . A closed subspace is said to be a reducing subspace for , if and . Or equivalently, is a reducing subspace for if and only if and , where is the orthogonal projection from onto .

In addition, is called minimal if there is no nonzero reducing subspace contained in properly. The operator is said to be completely reducible if its lattice of reducing subspaces has no nonzero minimal elements [26].

By some easy computations, we have the following equations:

Moreover, we have

Let . Let denote the reducing subspace generated by , that is, the minimal reducing subspace containing . Denote by , all the nonnegative integers and positive integers, respectively. Write if and .

Theorem 6. If and , then has no nontrivial reducing subspaces on .

Proof. Suppose there exists a reducing subspace for and denote the orthogonal projection from onto . Then . Moreover, from (32) and (34), we haveIn the following, we shall prove that or .
Let Taking the two equalities above into (35) and (36), along with (31), (32), (33) and (34), we deduceandComparing the coefficients of in (38) and (39), we getLikewise, comparing the coefficients of , we obtainIf , then immediately we have for and for ; or else, if , it follows from (40) and (41) that for and for since .
Considering the coefficients of , we haveandwhere (42) holds since . By some computations, from (43), we obtain and If , obviously we have ; if not, since if and only if , along with the assumption , we also have .
Associated with (42), we deduce If , then ; otherwise, we have which proves that .
Suppose . Then the following statements hold:(i) since .(ii) and show that .(iii)On the basis of (ii), we have since . These give that , which shows that . Similarly, we can demonstrate if we assume , and thus .
The proof is complete.

Theorem 7. If and , then (i)if is a reducing subspace for and , then ;(ii)if , then has no nontrivial reducing subspaces;(iii)if , then is a minimal reducing subspace for .

Proof. Suppose there exists a reducing subspace for and denote the orthogonal projection from onto . Then .
The assumption in (i) indicates that . implies . By the analysis in the proof of Theorem 6, we have .
Suppose is a reducing subspace for such that . Let where . By (40), (41), (43) and the assumption , we obtain Therefore, If , then combining (42) with the assumption , we obtain Since and , we have . Because if and only if , surely we have . shows that . By the analysis in the proof of Theorem 7, we get . Associated with (i), (ii) is established.
If , from (31), one can easily get that Suppose is a reducing subspace for contained in . Set indicates that shows for and for . Thus, . Therefore, if , then , forcing ; otherwise, if , similarly we have . Thus, (iii) holds.

Theorem 8. If is an inner function, then the following statements hold: (i)If , and , then is a minimal reducing subspace for if and only if has the following forms:(1);(2);(3), where satisfying , .(ii)If , and , then has 3 minimal reducing subspaces.(iii)If and if is a reducing subspace for , then is minimal if and only if or where satisfying and .

Proof. Suppose is a reducing subspace for and let denote the orthogonal projection from onto . Then .
To show (i), firstly we show that are two minimal reducing subspaces for . Obviously, and are reducing subspaces generated by , respectively. Moreover, they are both minimal for . In fact, suppose there exists a reducing subspace for . Then and . Write By (33), (34), along with , we get It follows that which forces that for , and for and since . Therefore, we have . If , then , showing , that is, ; if not, shows that , implying . In conclusion, is minimal for . Using a similar method, we can also prove is minimal for .
Next, supposeBy the assumption, we have Along with (63) and the fact that , we have for and for . Thus, we obtain thatSimilarly, we getCase 1 (). Then, from (66) and (67), we have . By the analysis above, we obtain that and are the only two minimal reducing subspaces for .
Case 2 (). Again from (66) and (67), we have that is, . Moreover, shows that . By the assumption, we have is real and .
If , then . That is, . Suppose , then . By (66) and (67), we have and . Otherwise, , imply that and . In this way, we still obtain that and are the only two minimal reducing subspaces for .
On the other hand, if , it is clear that . Immediately, we have , where , . Then, contains a minimal reducing subspace like Furthermore, is a minimal reducing subspace for if and only if where , .
The assumption in Condition (ii) and a similar method indicate that where is supposed to be a reducing subspace for .
Note that there is always a reducing subspace satisfying . Thus, shows that . Therefore, is definitely a minimal reducing subspace for . Hence, shows that if is a reducing subspace such that , then , . Similarly, is a minimal reducing subspace for . Denote by . We shall prove is also minimal. Assume is a reducing subspace for and From (34), we see that . Therefore, which forces that and thus . Hence, we have or .
By the similar way in (ii), the statement in (iii) shows that are four minimal reducing subspaces for . Using a similar method as in (i), along with (27)-(34), we obtain that where . A similar discussion as in (i) leads to the desired conclusion.

Theorem 9. If , and , then is completely reducible.

Proof. Under the assumptions, along with (31)-(34), is a normal operator on . By the Spectral Theorem of the normal operators, the range of is the eigenspace , where is the spectral measure for . Thus, for every , if and only if has no minimal reducing subspaces, that is, is completely reducible.
Given , suppose there exists such thatAssumeTaking this into (76), we get from which we can obtain and Then, (77) turns to be Thus, we must have , or is infinite. Therefore, we conclude . Then, the desired result follows.

Remark 10. Notice that the assumption in Theorem 9 implies that is a constant function with module , and hence the conclusion in this theorem is trivial.

Remark 11. If the condition in Theorem 8 (ii) is changed to and , then the result is different. In fact, , are still both minimal for . As given in the proof of Theorem 8, Since , , is normal on . By the same method in Theorem 9, we obtain that is completely reducible when restricted to .

The proofs in Theorems 6–9 apply to characterize the structure of the reducing subspaces for . By some computations, we have Moreover,It is easy to verify that if , then has no nontrivial reducing subspaces.

If , by (83), is a normal operator on . Theorem 5 gives . Associated with the proof in Theorem 9, is completely reducible on

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work was financially supported by NSFC no. 11271059.

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Copyright © 2018 Yinyin Hu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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