Subnormality of Powers of Multivariable Weighted Shifts

Lee, Sang Hoon; Lee, Woo Young; Yoon, Jasang

doi:https://doi.org/10.1155/2020/5678795

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Research Article | Open Access

Volume 2020 | Article ID 5678795 | https://doi.org/10.1155/2020/5678795

Subnormality of Powers of Multivariable Weighted Shifts

Sang Hoon Lee,¹Woo Young Lee,²and Jasang Yoon³

Academic Editor: Nikolai L. Vasilevski

Received23 Jul 2020

Accepted04 Nov 2020

Published28 Nov 2020

Abstract

Given a pair of commuting subnormal Hilbert space operators, the Lifting Problem for Commuting Subnormals (LPCS) asks for necessary and sufficient conditions for the existence of a commuting pair of normal extensions of and ; in other words, is a subnormal pair. The LPCS is a longstanding open problem in the operator theory. In this paper, we consider the LPCS of a class of powers of -variable weighted shifts. Our main theorem states that if a “corner” of a 2-variable weighted shift is subnormal, then is subnormal if and only if a power is subnormal for some . As a corollary, we have that if is a 2-variable weighted shift having a tensor core or a diagonal core, then is subnormal if and only if a power of is subnormal.

1. Introduction

For a Hilbert space operator, a subnormal operator means an operator admitting a normal extension, i.e., an extension which is a normal operator. As a lifting problem of operators, many researchers of operator theory have considered necessary and sufficient conditions for a pair of subnormal operators on a Hilbert space to admit commuting normal extensions: more concretely, given a pair of commuting subnormal operators on a Hilbert space, find a necessary and sufficient condition for the existence of commuting normal extensions and of and , respectively. This problem is referred to as the Lifting Problem for Commuting Subnormals (LPCS). A pair of subnormal operators admitting commuting normal extensions is called a subnormal pair.

For a bounded linear operator on a complex Hilbert space , it is well known that the subnormality of implies the subnormality of powers . However, its converse is not true in general; in fact, Stampfli [1, p. 378] showed that the subnormality of all powers does not necessarily imply the subnormality of , even if is a unilateral weighted shift. It is also well known that the hyponormality (i.e., is positive semidefinite) of does not imply the hyponormality of [2]. However, for a unilateral weighted shift , the hyponormality of (detected by the condition for all when ) clearly implies the hyponormality of all powers .

On the other hand, Franks [3] showed that given a pair of commuting subnormal operators, if is subnormal for all 2-variable polynomials with , then is a subnormal pair. Clearly, if is a subnormal pair and if , then is also a subnormal pair. Motivated by Stampfli’s work [1], it is natural to ask whether the subnormality of for each implies the subnormality of . For the 2-variable weighted shifts, we may consider these analogous results. The standard assumption on a pair is that each component is subnormal . With this in mind, these analogous results are highly nontrivial. In the works [4–8], it was considered whether there is a -variable weighted shift such that is subnormal for some , but is not subnormal.

For , a bounded sequence of positive real numbers (called weights), a weighted shift is defined by (all ), where is the canonical orthonormal basis in . In this case, we write

Now, for , consider the weighted shift

Then, is hyponormal (detected by the condition for all ) but not subnormal. However, all powers are subnormal. If in , then the following statements are equivalent: (a) is subnormal(b) is subnormal for all (c) is subnormal for some

In [5, 6], we have examined the above results for the class of 2-variable weighted shifts . More concretely, for the class of 2-variable weighted shifts with a core of tensor form, denoted [5], or with a core of diagonal form, denoted [6], we have shown that if , then the following statements are equivalent: (a) is subnormal(b) is subnormal for all (c) is subnormal for some

In spite of the above facts for 1 or 2-variable weighted shifts and consideration of the recent works ([4–8]), we have guessed that there exists a class of 2-variable weighted shifts such that is not subnormal but is subnormal for all , under a more general condition that a “corner” of is subnormal. In this paper, we show that this guess is not right and that the above three statements are equivalent whenever a corner of is subnormal. In the below, we will notice that is a very special corner of .

On the other hand, the reason why we take 2-variable weighted shifts for examining the subnormality of powers for pairs of operators is that 2-variable weighted shifts play an important role in detecting properties such as subnormality, via the Lambert-Lubin Criterion ([9, 10]): a commuting pair of injective operators acting on a Hilbert space admits a commuting normal extension if and only if for every nonzero vector , the 2-variable weighted shift with weights has a normal extension.

The organization of this paper is as follows. In Section 2, we give preliminary notions and state the main theorem. In Section 3, we provide a proof of the main theorem.

2. Preliminaries and the Main Theorem

Let be a complex Hilbert space and let denote the algebra of bounded linear operators on . For let . We say that an -tuple of operators on is (jointly) hyponormal if the operator matrix is positive semidefinite on the direct sum of copies of (cf. [11, 12]). The -tuple is said to be normal if is commuting and each is normal, and is subnormal if is the restriction of a normal -tuple to a common invariant subspace. For , a commuting pair is said to be -hyponormal ([13]) if is hyponormal, or equivalently

Clearly, normal ⇒ subnormal ⇒ -hyponormal for -tuples of operators. The Bram-Halmos criterion states that an operator is subnormal if and only if the -tuple is hyponormal for all .

Let be a weighted shift with weights . The moments of are given as

It is easy to see that is never normal and that it is hyponormal if and only if . Similarly, consider double-indexed positive bounded sequences and and let be the Hilbert space of square-summable complex sequences indexed by . Recall that is canonically isometrically isomorphic to . We define the 2-variable weighted shift by a pair of operators acting on the Hilbert space given by for each . For all , we clearly have

For a commuting 2-variable weighted shift , the moment of of order is

We remark that, due to the commutativity condition (9), can be computed using any nondecreasing path from to . For a detailed discussion of the 2-variable weighted shifts, the reader may refer to ([4, 13, 14] [8]).

We now recall a well-known characterization of subnormality for multivariable weighted shifts [15], due to C. Berger (cf. [2, III.8.16]) and independently established by Gellar and Wallen [16]) in the single variable case: admits a commuting normal extension if and only if there is a probability measure , called the Berger measure of , defined on the 2-dimensional rectangle (where such that

Observe that and are subnormal, with Berger measures and , respectively, where denotes the point-mass probability measure with support from the singleton set .

Throughout this paper, we write and for . For a 2-variable weighted shift , a corner of is defined by which is a restriction of to the invariant subspace . The core of , denoted by , is defined by a corner with , i.e.,

Thus, for 2-variable weighted shifts, the core is a special form of a corner. A 2-variable weighted shift is said to be of tensor form if it is of the form . If a tensor form is subnormal, then the corresponding Berger measure is given by a Cartesian product where and are the Berger measure of and , respectively (cf. [5]). Also, for strictly increasing weight sequences , consider a 2-variable weighted shift on given by the double-indexed weight sequences

This -variable weighted shift induced by a -variable weighted shift is said to be of the diagonal form. If a diagonal form is subnormal, then the corresponding Berger measure is given by a measure supported in the diagonal (cf. [6]). The class of all -variable weighted shifts whose core is of the tensor form will be denoted by ; in symbols,

(see Figure 1(i)).

Also, the class of all 2-variable weighted shifts whose core is of the diagonal form will be denoted by ; in symbols,

(see Figure 1(ii)).

In [5, 6], it was shown that if , then is subnormal for some if and only if is subnormal. Now, it is natural to consider that given a 2-variable weighted shift , whether or not is subnormal if and only if is subnormal. In other words, we ask the following:

Problem 1 ([6, 8]). Given a 2-variable weighted shift , assume that is subnormal for all . Does it follow that is subnormal?

For the class of 2-variable weighted shifts , it is often the case that the powers are less complex than the initial pair; thus, it becomes especially significant to unravel the invariance of subnormality under the action . The aim of this paper is to shed new light on some of the intricacies associated with LPCS and powers of commuting subnormals in .

Our main theorem now states:

Theorem 2. Let . If a corner of is subnormal, i.e., is subnormal for some , then the following are equivalent: (a) is subnormal(b) is subnormal for all (c) is subnormal for some

As we observed before, is a special corner of . Indeed, if , then is subnormal, so that satisfies the condition of Theorem 2. Therefore, if , then three conditions of Theorem 2 are equivalent. Thus, as immediate corollaries of Theorem 2, we can recapture the both main results of [5, Theorem 7.1] and [6, Theorem 3.2].

Briefly stated, our key idea to prove the main results is as follows: (i) we split the ambient space as an orthogonal direct sum ; (ii) when is subnormal, we show that for some is subnormal if and only if is subnormal by using the backward extension of subnormality; (iii) when is subnormal, we show that for some is subnormal if and only if is subnormal by using (i) and the backward extension of subnormality; and (iv) by combining (ii) and (iii), we have that if is subnormal for some , then is subnormal for some if and only if is subnormal.

3. The Proof of the Main Theorem

We will first establish several auxiliary lemmas and then prove the main theorem (Theorem 2).

To study subnormality for powers of multivariable weighted shifts, we recall that, in one variable, the -th power of a weighted shift is unitarily equivalent to the direct sum of weighted shifts. First, we need some terminology. Let . Given integers and , define ; clearly, . Following the notation in [17], for a weight sequence , we let that is, denotes the shift with the weight sequence given by the products of weights in adjacent packets of size , beginning with . For example, given a weight sequence , we have , , , etc. For and , we note that is unitarily equivalent to Therefore, is unitarily equivalent to Thus, we have (cf. [17]) that if is subnormal with the Berger measure , then is subnormal with the Berger measure , where and furthermore,

For , we let denote the invariant subspace obtained by removing the first vectors in the canonical orthonormal basis of . Thus, if is subnormal with Berger measure , then is subnormal for each , and the Berger measure of is given by

Something similar happens in two variables, as we will see it below. For a 2-variable weighted shift on , we observe a new direct sum decomposition for powers of 2-variable weighted shifts which parallels the decomposition used in [17] to analyze the subnormality for powers of (one-variable) weighted shifts. Specially, we split the ambient space as an orthogonal direct sum , where for and where is a 2-variable weighted shift on . Then, each of reduces and . Also, is subnormal if and only if each is subnormal. Similarly, for , consider on and let

In a similar fashion to (13) and (14), we can define

We thus have:

Lemma 3. Let be a 2-variable weighted shift. Then, for a fixed and , we have and for , , we have where means a unitary equivalence of two operators.

Lemma 4. Let be a 2-variable weighted shift in . If is subnormal with Berger measure , then is subnormal for all . Furthermore, for each , the Berger measure of is given by

Proof. Observe first that if is subnormal then , where each direct summand is a subnormal 2-variable weighted shift.
For the second assertion, observe Thus, for , so that which gives the result.

To detect the subnormality of -variable weighted shifts, we introduce some definitions. (i)For a regular Borel measure on we say that is positive if for all Borel subset , or equivalently, if and only if for all such that on . Similarly, we say that is positive (denoted by ) if for all such that on . For positive two measures and on we say that on if is positive(ii)Let be a probability measure on , and assume that The extremal measure (which is also a probability measure) on is given by (iii)Given a measure on , the marginal measure is given by , where is the canonical projection onto . Thus, for every , or equivalently, .

Lemma 5 [14, Proposition 3.10] (subnormal backward extension). Let be a 2-variable weighted shift, and assume that is subnormal with associated Berger measure and that is subnormal associated with Berger measure . Then, is subnormal if and only if Moreover, if then . In the case when is subnormal, the Berger measure of is given by

We also recall:

Lemma 6 [18]. For a positive measure on , let . Then, and where and is the canonical projection onto .

Given a 2-variable weighted shift , and given , we let be the -th horizontal slice of with associated Berger measure ; similarly, we let be the -th vertical slice of with associated Berger measure . Clearly, and are the unilateral weighted shifts associated with the 0th row and 0-column in the weight diagram for , respectively.

Then, we have:

Lemma 7. Let . If is subnormal for some , then is also subnormal.

Proof. It is enough to show that if or is subnormal, then is subnormal. Without loss of generality, assume that is subnormal. Since , we have that is subnormal for all . By Lemma 5, we have . Let be the Berger measure of . By Lemma 6, since and , it follows from Lemma 5 that , that is, Moreover, and . Thus, by Lemma 5 again, we have that is subnormal, as desired.

Next, we have:

Lemma 8. Assume and is subnormal with Berger measure . Assume Let be the Berger measure of subnormal shift which is in the zero level of . We also let If is subnormal for some , then . Moreover, if is subnormal, then its Berger measure is

Proof. We first claim that To see (40), we note that the positivity of depends on the densities of and . There are three cases to consider. (i)If and are finite atomic measures, then it is clear(ii)If and are continuous measures, then by a change of variables, letting , which goes in both directions because is an invertible function on the positive axis. That is, one can change the name of the variable from to and then relabel as (iii)If and are any probability measures, then by the above arguments, we have the claim (40).Now, suppose that is subnormal with the Berger measure . Since and are subnormal, we thus reconstruct the subnormality of as a backward extension of by applying Lemma 5. We let be the Berger measure of Since is the Berger measure of , by Lemma 4, we have Thus, we obtain Also, by Lemma 4 and (44), we have Since is the Berger measure of and is the Berger measure of , by Lemma 4, (44) and (45), we have We let be the Berger measure of the following subnormal shift: Since is the Berger measure of , by Lemma 6, we have If in the zero level of is subnormal with the Berger measure , then is subnormal with the Berger measure by (18).
Since and is subnormal, it follows from Lemma 5, (22), (44), (47), and (49) that This proves the first assertion. The second assertion is clear from Lemma 4 and analyzing the moments of

The following corollary is an immediate result of Lemma 8.

Corollary 9. Assume and is subnormal with Berger measure . Let Then (a) is subnormal (b)For any , is subnormal is subnormal. Hence, in particular

Proof. (a) This follows from Lemma 5. (b) If is subnormal for some , then is subnormal. By Lemma 8, we have and hence is subnormal by (a). Clearly, if is subnormal then is subnormal for all .

Remark 10. We remark that if and is subnormal with Berger measure , then because is subnormal.

We next have:

Corollary 11. Let . If there exists such that is subnormal, then is subnormal for some if and only if is subnormal.

Proof. It suffices to consider the case of . In the case, if is subnormal for some , then is subnormal. Thus, by Corollary 9, is subnormal, and therefore, is subnormal. The converse is clear.

We now have:

Lemma 12. Let and let be subnormal. Then,

Proof. Let be the Berger measure of and suppose that is subnormal for fixed . Then, is subnormal. Let be its Berger measure. By Lemma 3, we have Hence, by Lemma 4 and (54), we have Observe We now characterize the subnormality of as a backward extension of after applying Lemma 5. To do so, let in the zero level of be subnormal with the Berger measure . Then, by (57), we have that By again Lemma 5 and (59), therefore, is subnormal.

By Corollary 9 and Lemma 12, we have:

Corollary 13. Let . If or is subnormal, then the following are equivalent: (a) is subnormal for some (b) is subnormal for all (c) is subnormal for some (d) is subnormal for all (e)T is subnormal

The following theorem is a core of our main result.

Theorem 14. If is subnormal, then for some

Proof. . For fixed , suppose that is subnormal. Then, is also subnormal. Let be the Berger measure of and let be the Berger measure of . For each , using Lemma 5, we characterize the subnormality of as a backward extension of (in the direction).
First observe that from (26) in Lemma 3, for , By Lemma 4 and (62), we have Observe Let be the -th horizontal slice of with Berger measure for . By (65) and a similar way to (59), for , we have that where is the Berger measure of the subnormal shift: By Lemma 5, (66), and a similar way to the proof of Lemma 12, for , we have Therefore, by (68), we get the result.
Since , it follows that is subnormal whenever is subnormal.

Corollary 15. Let . If is subnormal, then the following are equivalent: (a) is subnormal for some (b) is subnormal(c) is subnormal

We are ready to give a proof of our main theorem (Theorem 2). For convenience, we restate Theorem 2:

Theorem 16. Let . If a corner of is subnormal, i.e., is subnormal for some , then the following are equivalent: (a) is subnormal(b) is subnormal for all (c) is subnormal for some

Proof. (a) (b): This is clear from the functional calculus. (b) (c): Clear. (c) (a): Suppose is subnormal for some and is subnormal for some . By Lemma 7, we have is subnormal. Also, is subnormal. We thus have Therefore, by Corollary 15, we can see that is subnormal.

We conclude by revealing examples to illustrate Theorem 2. Their proofs are given from a straightforward calculation. We will omit their proofs.

Example 1. Let and let be subnormal with Berger measure Also, let be the Berger measure of . We then have: (a) is subnormal(b) is subnormal(c) is subnormal

Example 2. Let and let be subnormal with Berger measure

Also, let be the Berger measures on of and be the Berger measures on of . We then have: (a) is subnormal(b) is subnormal

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no competing interests.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgments

The first author was partially supported by NRF (Korea) grant No. 2020R1A2C1A0100584611. The second author was partially supported by NRF (Korea) grant No. 2018R1A2B6004116. The third author was partially supported by a grant from the University of Texas System and the Consejo Nacional de Ciencia y Tecnologa de México (CONACYT).

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Copyright © 2020 Sang Hoon Lee 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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