Abstract and Applied AnalysisVolume 2011 (2011), Article ID 813723, 13 pagesdoi:10.1155/2011/813723

Research Article

Spatiality of Derivations of Operator Algebras in Banach Spaces

Quanyuan Chen1,2 and Xiaochun Fang¹

¹Department of Mathematics, Tongji University, Shanghai 200092, China²Department of Information Engineering, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi 333403, China

Received 5 June 2011; Revised 17 August 2011; Accepted 9 September 2011

Academic Editor: Wolfgang Ruess

Copyright © 2011 Quanyuan Chen and Xiaochun Fang. 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.

Abstract.

Suppose that is a transitive subalgebra of and its norm closure contains a nonzero minimal left ideal . It is shown that if is a bounded reflexive transitive derivation from into , then is spatial and implemented uniquely; that is, there exists such that for each , and the implementation of is unique only up to an additive constant. This extends a result of E. Kissin that “if contains the ideal of all compact operators in , then a bounded reflexive transitive derivation from into is spatial and implemented uniquely.” in an algebraic direction and provides an alternative proof of it. It is also shown that a bounded reflexive transitive derivation from into is spatial and implemented uniquely, if is a reflexive Banach space and contains a nonzero minimal right ideal .

1. Introduction

Throughout this paper, is a Banach space (and will be replaced by if it is a Hilbert space) and is a subalgebra of , the Banach algebra of all bounded operators on . Suppose that and is an -bimodule. A linear map from into is called a derivation if Then is called the domain of and denoted by . The derivation is called inner (resp., spatial) if there exists an operator (resp., ) such that If the operator is not bounded, then is said to be quasispatial. More precisely, if there exists a densely defined, closed operator such that then the derivation is called quasispatial, and the operator is an implementation of . Compared to the spatiality, the quasispatiality is a slightly weaker notion.

Given a bounded derivation on an operator algebra, the natural question is whether the derivation is inner (or spatial). The spatiality of derivations is a classical problem when formulated for self-adjoint algebras and non-self-adjoint reflexive operator algebras. And it has been extensively studied in the literature in a large variety of situations, and some interesting results have been obtained [1–13]. For example, every derivation of a -algebra is spatial [12], every derivation of a von Neumann algebra is inner [13], and so is the derivation of a nest algebra [14]. Every derivation from an atom Boolean subspace lattice algebra into its ideal is quasispatial [7]. A necessary and sufficient condition is given for a derivation on CDC algebras to be quasispatial [6]. In [10], the quasispatiality of derivations on CSL algebras is studied.

As to general operator algebras, it is well known that every derivation of is inner [9] and that every derivation of a standard operator subalgebra on a normed space is spatial [4]. Since these operator algebras are transitive, the above question for a reflexive transitive derivation in a Banach space is raised naturally as follows.

Problem 1. Suppose that is a transitive subalgebra of . Let be a bounded reflexive transitive derivation from into . Does there always exist such that for each Is the implementation unique only up to an additive constant if there is any

In the case when is a Hilbert space, it is Problem 2.12 of [5]. Although the problem is still open, some strong conditions have been found by Kissin to imply that such derivations are spatial and implemented uniquely (Proposition 2.11, [5]). In particular, Kissin has proved that the answer to Problem 1 is affirmative under the conditions that is a transitive subalgebra of , and contains the ideal of all compact operators in . As far as we know, there are no other solution to Problem 1.

The purpose of this paper is to investigate the quasispatiality of derivations and to address the above question in a Banach space . The paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we investigate the quasispatiality and spatiality of derivations. The main result (Theorem 3.1) shows that if is a transitive operator algebra on a Banach space and contains a nonzero minimal left ideal , then a bounded reflexive transitive derivation from into is spatial and implemented uniquely. As an application, the quasispatiality of the adjoint of a derivation is discussed in Section 4. The main result (Theorem 4.2) in Section 4 shows that under the conditions that is a reflexive Banach space and contains a nonzero minimal right ideal , is also spatial and implemented uniquely if is a bounded reflexive transitive derivation from into . As another application, Proposition 2.11 (ii) of [5] can be proved by using Theorem 3.1 of this paper, which is Corollary 4.3. Since Theorem 2.5 and Proposition 2.11 of [5] hold in a Hilbert space and they are not valid in a Banach space without the approximation property, Theorems 3.1 and 4.2 of this paper extend the result of Kissin to Banach spaces in an algebraic direction. The role of “compact operators” is replaced by that of “minimal one-sided ideals”. The proof of our results relies on the quasispatiality of the derivation and Banach algebra techniques.

This paper is a continuation of [5]. Some definitions and notations can be found in [5].

2. Preliminaries

Throughout this paper, is a complex Banach space, and is the topological dual space of , the Banach space of all continuous linear functionals on . We denote by the algebra of all finite-rank operators on . If a subalgebra contains , then is called a standard operator algebra. For a bounded operator on , denote by the lattice of all closed invariant subspaces of and the adjoint operator of . For a subalgebra of , denote by the lattice of all closed subspaces invariant under every operator in . For a set of subspaces of , denote by the algebra of all operators in which leave all subspaces in invariant. An operator algebra is transitive if ; is reflexive if

For and , the rank-one operator acts on by for . Let be an operator on with . If and , then . Let be a nonempty subset of and a nonempty subset of . The annihilator of and the preannihilator of are defined as follows [15]: , . It is obvious that is a weak*-closed subspace of and is a norm-closed subspace of . For a subalgebra and a closed, densely defined operator with domain , we say that commutes with , if and for any and any . For a subalgebra of , let in notation.

A subset of an algebra is a left ideal of if , a right ideal if , and a two-sided ideal if it is both a left and a right ideal. A left ideal of is minimal if every left ideal of included in is either or , similarly for minimal right ideals.

A derivation is bounded (resp., closed) if the map is bounded (resp., closed) in the operator norm topology. The derivation is transitive if its domain is a transitive operator algebra; is reflexive if is a reflexive operator algebra on . Denote by the set of all closed, densely defined operators implementing the derivation as in (1.3). For a densely defined, closed operator with domain , we can define the derivation with domain by

If a densely defined, closed operator implements a derivation , then the derivation is an extension of the derivation . In fact, implies and for any by (1.3) and (2.4). For any set of derivations (), define the derivation by In particular,

Remark 2.1. is a reflexive transitive derivation for any densely defined, closed operator .Indeed, is a subalgebra of , and is a derivation.If and , for any . So . Since , so that is bounded on . It follows that for any and .Let with . Then for any and . Since is dense, there exists such that by Hahn-Banach Theorem. It follows that for any so that . Thus . Therefore is transitive.For the derivation , . It is easy to see that and for any . Let and . Then . Since , for any . It follows that . Thus and . Since , for any and any . Then and for any and any . It follows that and for all . Since is dense in and , . So and on . Therefore, is bounded on . It follows that and . Thus . It follows that is a reflexive algebra. Therefore is a reflexive transitive derivation.

3. Main Results

In this section, we discuss the quasispatiality of bounded transitive derivations on operator algebras in a Banach space. The main result is as follows.

Theorem 3.1. Suppose that is a transitive subalgebra of and , the norm closure of , contains a nonzero minimal left ideal . If is a bounded reflexive transitive derivation from into , then is spatial and implemented uniquely, or, more precisely, there exists such that for each and the implementation is unique only up to an additive constant.

The proof of Theorem 3.1 will proceed through several lemmas, in each of which we maintain the same notation. The results similar to the following two lemmas (Lemmas 3.2 and 3.3) can be found in [5]. For the sake of completeness, we outline the proof.

Lemma 3.2. Let be a reflexive transitive derivation from into . Then is quasispatial.

Proof. If is nonempty, then the lemma is trivially true. Therefore, for the rest of the argument, we assume that is empty.First, we have that if is a transitive derivation from into , then where is the graph of .Indeed, it is easy to see that , , , and all are invariant subspaces of . For the converse, suppose that such that , , and . If there is a vector of form with , then for any . It follows that by the transitivity of . If contains no vector of form with , then ; otherwise, there is a vector of form with , then we have that . Therefore . It follows that there is a closed operator such that As , by (3.1), It is a contradiction to the assumption that is reflexive, which shows that Therefore is quasispatial.

Lemma 3.3. Let be a reflexive transitive derivation from into . Then .

Proof. By (3.4), . And by of [5], As both and are transitive derivations, by (3.1). Since both of these derivations are reflexive, Hence

Lemma 3.4. Let be a transitive subalgebra of . If contains a nonzero minimal left ideal , then is also a transitive subalgebra of .

Proof. We complete the proof step by step.Step 1. We have that Indeed, suppose that . Let . Then . It follows that and . Set . It is obvious that is an -invariant closed subspace of . Hence since is a transitive operator algebra. Since , . It follows that so that . We can choose () such that . It is obvious that is an -invariant linear manifold of and . It follows that is dense in . However, is contained in the null space of , , which is a contradiction.Step 2. Since , it follows from Lemma and Corollary of [16] that there exists an idempotent in such that and is a division algebra consisting of scalar multiples of (with identity ), that is, Then is a rank-one operator; that is, there exist and such that Indeed, let such that and let . It is obvious that and is an -invariant linear manifold of . Hence is dense in . Since for any , the restriction of on has one-dimensional range. As is bounded, it also has one-dimensional range. Hence , where is a continuous functional.Step 3. We have . As is an algebra, the set of rank-one operators lies in . For each , the linear manifold is dense in . Indeed, if is not dense in , by Hahn-Banach theorem, there exists such that for all . Hence If for all , then vanishes on the linear manifold . However, as is transitive, the manifold is dense in , so that . This contradiction shows that there is such that . Then (3.13) implies that for all . Repeating the above argument, we obtain that . This contradiction shows that any manifold is dense in . Therefore the algebra is transitive.

Lemma 3.5. Suppose that is a transitive subalgebra of and contains a nonzero minimal left ideal . Then the operators commuting with are scalars. More precisely, if is a closed, densely defined operator such that and for any and any , then for some complex scalar .

Proof. For any , there exists a net of operators such that by the transitivity of . Then , since as in (3.10). It follows that Suppose that is a closed, densely defined operator commuting with . As for any , then . Since is dense in and , there is such that . So It follows that . Set . Then , as required.

Lemma 3.6. Suppose that is a subalgebra of and is its norm closure. Then .

Proof. Clearly, . Conversely, let . For any , there exists a net of operators such that . Then for any . Since and is a closed subspace of , . It follows that so that . Therefore . The Lemma follows.

Let be a subalgebra of and be a bounded derivation from () into . For any , there exists a net of operators such that . If is a uniformly convergent net in ; then is also a uniformly convergent net in since for any indexes . Also, if two nets of operators in converge uniformly to the same limit, then , converge uniformly to the same limit, since . Therefore a linear map can be unambiguously defined by for any and any net in such that , where is the limit of in the operator norm topology. It is obvious that if is bounded.

Proposition 3.7. Let be a bounded transitive derivation from into . Then is a bounded transitive derivation from into and

Proof. Since is transitive, is also transitive. It is obvious that is a linear map. For any , there are two nets of operators in such that . Then and It follows that the linear map is a transitive derivation from into .By (2.2), and . So that by (3.16) and by Lemma 3.6. By (3.1), It follows that .

Proof of Theorem 3.1. By (3.4), . Suppose that . Then is a closed, densely defined operator on and Let be the derivation (with the same domain of ) from into , that is, for any . If with domain , then by (3.17), and that is, for any . Then for some scalar by Lemma 3.5. It follows that Let be any closed, densely defined operators implementing . Then by (3.17). Then is a nonzero -invariant linear manifold of by (1.3); that is, is a nonzero -invariant linear manifold of . Therefore for any . If , by (3.14) and for any . Since is dense in and , there is such that . So . Since is arbitrary chosen, it follows that , the whole space. Then also holds since is an arbitrary implementation of . It is obvious that . Therefore the operator is everywhere defined.Set . Then is closable.Indeed, suppose that , and . Since , and for any and any . So for any . As by (3.10) and is an algebra, . Then for any . Since , for any . Thus vanishes on the linear manifold . The transitivity of assures that or is dense in . Since , and .It follows that is a closable operator with domain , the whole space. So that is a closed operator. Clearly, . Then for some scalar by (3.22), that is, . It follows that Since , by (3.7), Since is closed and , is bounded by the closed graph theorem. It follows from (2.4), (3.24), and (3.25) that for each and the implementation is unique only up to an additive constant.

4. Applications

Suppose that is a subalgebra of and is a derivation from () into . We can define the adjoint of by

Lemma 4.1. Let be a derivation from into . (1)The adjoint of is a derivation from into .(2)Furthermore, suppose that is a reflexive Banach space and is a subalgebra of . Then is a reflexive resp., transitive derivation from into if and only if is a reflexive resp., transitive derivation from into . And .

Proof. (1) It is obvious that is a linear map from into . If , then It follows that is a derivation.(2) Suppose that is transitive. Then is transitive and . Since , We have that is transitive, that is, is transitive.Conversely, suppose that is transitive. We obtain, as above, that is transitive. Since is a reflexive Banach space, for any bounded operator . It follows from (4.1) that is transitive by the definition of .Set , where is the identity operator on and 0 is the zero operator on . Since the two operator algebras and are similar. We have that It is obvious that If , for any , since Conversely, if such that , then So that the functional is a bounded functional on . Therefore and holds on by (4.8). As and are bounded functionals on and is dense in , on . It follows that Hence It follows from (3.4) that Suppose that is reflexive. Then is a reflexive operator algebra on . Since , the two operator algebras and are similar. It follows that is reflexive if and only if is reflexive. Since is reflexive, is also reflexive by the reflexivity of the Banach space . Thus is reflexive, that is, is reflexive.Conversely, suppose that is reflexive. We obtain, as above, that is reflexive. It follows that is reflexive.

Theorem 4.2. Suppose that is a reflexive Banach space. Let be a transitive subalgebra of and let contain a nonzero minimal right ideal . If is a bounded reflexive transitive derivation from into , then is spatial and implemented uniquely, or more precisely, there exists such that for each , and the implementation is unique only up to an additive constant.

Proof. Consider the derivation . is a bounded reflexive transitive derivation from into by Lemma 4.1.Set . Since is a right ideal of , for any and . For any , , that is, , , we have that . Then . It follows that is a left ideal of . If there exists a nonzero left ideal of included in , then is a nonzero right ideal of included in . Thus and . It follows that is a minimal left ideal of .By (3.24), there exists a bounded operator such that where is the identity operator on . By (4.11), where is the identity operator on . Set . Then is bounded with . By (3.7) and (4.13), . It follows from (2.4) that for each , and the implementation is unique only up to an additive constant.

Corollary 4.3 (Proposition 2.11, [5]). Suppose that is a transitive subalgebra of . Let be a bounded reflexive transitive derivation from into . If , then there exists such that for each , and the implementation is unique only up to an additive constant.

Proof. Since , for a fixed vector , is a minimal left ideal of , where is the rank-one operator on with for . By Theorem 3.1, there exists such that for each , and the implementation is unique only up to an additive constant.

Remark 4.4. In a Banach space without the approximation property (there exists such space as this, e.g., [17]), not all compact operators can be approximated by finite-rank operators in the norm operator topology. Therefore, Theorems 3.1 and 4.2 of this paper improve [5, Proposition 2.11].

Acknowledgments

This paper was supported by the National Nature Science Foundation of China (Grant no. 11071188). The first author would like to express deep gratitude to Professor Kissin for helpful answer to the questions run across during the work. The authors are greatly indebted to the referees for carefully reading the paper and providing many helpful comments which lead to great improvement of the paper, and pointing out a major error in the original version of this paper.

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