Abstract and Applied Analysis

Volume 2013 (2013), Article ID 183786, 6 pages

http://dx.doi.org/10.1155/2013/183786

## The Order Continuity of the Regular Norm on Regular Operator Spaces

School of Mathematics, Southwest Jiaotong University, Chengdu 610031, China

Received 28 March 2013; Accepted 24 May 2013

Academic Editor: Aref Jeribi

Copyright © 2013 Zi Li Chen 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.

#### Abstract

We present here some sufficient conditions for the regular norm on to be order continuous, and for () to be a KB-space. In particular we deduce a characterization of the order continuity of the regular norm using L- and M-weak compactness of regular operators. Also we characterize when the space is an -space and is lattice isomorphic to an -space for . Some related results are also obtained.

#### 1. Introduction

For Banach lattices and , we use to denote the space of all continuous linear operators from into , and to denote the space of all regular operators from into , which is the linear span of the set of all positive operators from into . With respect to the operator norm the space is not complete in general (see, e.g., [1]), but there exists a natural norm on , the *regular norm *, which turns into a Banach space (see [2] for details). Namely,
In particular, . If is a vector lattice; then is a Banach lattice and for all . For instance, if is Dedekind complete, then is a Dedekind complete Banach lattice under the regular norm.

The natural and important questions are: if is a vector lattice (i.e., a Banach lattice), when is the regular norm on order continuous? When is a KB-space with respect to the regular norm?

Wickstead showed in [3] some characterizations of the space being (lattice isomorphic to) an AL- or AM-space. It is natural to ask that when is an -space or lattice isomorphic to an -space for .

The purpose of this work is to present some results involving the order continuity of the regular norm on and being a KB-space. Furthermore we will also present a complete description for being (lattice isomorphic to) an -space with . Some related results are included as well.

Recall that an operator is called L-*weakly compact* if is an L-weakly compact set in ; that is, for each disjoint sequence contained in the solid hull of . Also is called M-*weakly compact* if for each disjoint sequence , where denotes the unit ball of . See, for example, [2].

We refer to [2, 4] for any unexplained terms from the theory of Banach lattices and operators.

#### 2. Some General Results

We start with a necessary condition for the order continuity of the regular norm on spaces of regular operators.

Proposition 1. *Let and be Banach lattices. If the regular norm on is order continuous, then the norms both on and are order continuous. *

*Proof. *If the regular norm on is order continuous, for each increasing sequence , taking with and defining by
then and .

The order continuity of the regular norm implies that there is such that ; thus . Choosing with we have
It follows from Theorem 2.4.2 of [2] that the norm on is order continuous.

Similarly, for each increasing sequence , taking with and defining by
then and .

Again there is such that ; thus . Choosing with , it is easy to verify that
Theorem 2.4.2 of [2] yields that the norm on is order continuous.

Next result is a characterization of the order continuity of the regular norm on spaces of regular operators.

Theorem 2. *For Banach lattices and , the following statements are equivalent. *(1)* is a vector lattice and the regular norm on is order continuous. *(2)*Every positive operator is L- and M-weakly compact. *

*Proof. *. If the regular norm on is order continuous, then by the proposition above, the norms both on and are order continuous.

For , it suffices to show that is M-weakly compact (see Theorem 3.6.17 of [2]). Otherwise, there is a disjoint sequence such that for all . Note that weakly as weakly (see Theorem 2.4.14 of [2]).

By Corollary 2.3.5 of [2] there exists a sequence of naturals and a disjoint sequence such that and , where is any fixed number from . Let be the band projection; hereby denotes the band generated by in . It is easy to verify that and (); it follows that , and , where is the identity operator on . Now the order continuity of the regular norm implies that is a -Cauchy sequence; in particular, as . Therefore
This is impossible, so holds.

. For any and , let by . Clearly and the L- and M-weak compactness of yield the relatively weak compactness of both and . It follows from Theorem 2.4.2 of [2] that the norms both on and are order continuous; is certainly a (Dedekind complete) vector lattice.

For any decreasing sequence with , Proposition 3.6.19 of [2] yields that the operator norm, and hence the regular norm, on order interval is order continuous, which implies that . Now the order continuity of the regular norm is following from Theorem 2.4.2 of [2].

It is clear that the identity operator on a Banach lattice is M-weakly compact if and only if is finite dimensional. The next result should be no surprise.

Corollary 3. *Let be a Banach lattice. Then is a vector lattice and the regular norm on is order continuous if and only if . *

Theorem 4. *For Banach lattices and , the following statements are equivalent. *(1)* is a KB-space. *(2)* is a KB-space and on is order continuous. *(3)* is a KB-space and every positive operator is M-weakly compact. *

*Proof. *. If is a KB-space, then on certainly is order continuous. For a norm bounded increasing sequence , taking with and defining by for , then also is increasing and -bounded, so there is such that ; thus . Choosing with we have
It follows that is a KB-space.

is a consequence of Theorem 2. Now we show that . Clearly is a Banach lattice under the regular norm as is a KB-space. If is a -bounded increasing sequence, then for each , is norm convergent as it is a norm bounded increasing sequence in . It is easy to see that there is a such that with respect to the strong operator topology; it follows that and by hypothesis is M-weakly compact. Proposition 3.6.19 of [2] yields that which implies that is a KB-space.

It is obvious that if is regular then is also regular, and the converse is false in general. For example, let defined by , where is the th Rademacher function on . Then , , is regular (as it is order bounded) but is not regular. The following results will show some relationships between the order continuity of the regular norms in , and .

Theorem 5. *For Banach lattices and , the following assertions are equivalent. *(1)*The regular norm on is order continuous. *(2)* is a KB-space. *(3)*The regular norm on is order continuous. *(4)* is a KB-space. *

*Proof. *Let by for , where is the natural embedding. According to Theorem 5.6 of [5] the operator is regular if and only if is regular and .

Moreover is an order continuous isometric lattice isomorphism from onto . Thus is a simple consequence of these facts. Also the equivalences of and , and easily follow from Theorem 4 and the proof of Theorem 2 (remembering that the norm on is order continuous if and only if is a KB-space; compare Theorem 2.4.14 of [2]).

Corollary 6. *Let and be Banach lattices such that is reflexive. Then the following statements are equivalent. *(1)*The regular norm on is order continuous. *(2)* is a KB-space. *(3)*The regular norm on is order continuous. *(4)* is a KB-space. *

Theorem 7. *Let and be Banach lattices, and closed sublattices. Supposing that there is a positive projection from onto then the following statements hold.*(1)*If is a vector lattice and the regular norm on is order continuous, then also is a vector lattice and on is order continuous. *(2)*If is a KB-space then also is a KB-space. *

*Proof. *Suppose that is a vector lattice and the regular norm on is order continuous. For , then , Theorem 2 yields that is L- and M-weakly compact. For any disjoint sequence contained in the solid hull of in , then is a disjoint sequence in as is a sublattice of , which is contained in the solid hull of as , so that ; that is, is L-weakly compact. Also for each disjoint sequence , is disjoint as is a sublattice of ; it follows that , which implies that is M-weakly compact. Again by Theorem 2 is a vector lattice and the regular norm on is order continuous; that is, (1) holds.

If is a KB-space then it follows from Theorem 4 and (1) that is a KB-space, and hence , as a closed sublattice of a KB-space, also is a KB-space, and that is a Banach lattice with an order continuous norm. Again Theorem 4 yields that is a KB-space, so (2) holds.

Note that each Banach lattice can be identified with a closed sublattice of , and so, as a consequence of Theorems 4 and 7 we have the following result.

Corollary 8. *Let and be Banach lattices. If is a KB-space (equivalently the regular norm on is order continuous), then so is . *

*Remark 9. *The regular norm on may fail to be order continuous even if is a KB-space.

For example, let and . Clearly is a KB-space and . Define by
it is easy to see that is an isometric lattice homomorphism; that is, contains a closed sublattice isometrically lattice isomorphic to . Thus Theorem 2.4.14 of [2] implies that fails to be a KB-space (i.e., the norm on is not order continuous).

Now it follows from Theorem 12 (see next) that is a KB-space, but the regular norm on , and hence on , is not order continuous as the norm on is not order continuous (see the proof of Theorem 2).

#### 3. Some Concrete Sufficient Conditions

In this section we will present some sufficient conditions on Banach lattices and such that the regular norm on is order continuous, or is a KB-space.

Proposition 10. *Let be an AM-space with a strong order unit and a Banach lattice with an order continuous norm. Then the regular norm on is order continuous. *

*Proof. *We may assume that is equipped with the strong order unit norm and also the norm on is order continuous; clearly is a Banach lattice. For in then for each . It follows from the order continuity of the norm on that is norm convergent. So there is such that with respect to the strong operator topology and obviously . In particular
where is a strong order unit of . Therefore Theorem 2.4.2 of [2] yields that the regular norm on is order continuous.

*Remark 11. *If fails to possess a strong order unit the above result is false even if is an AM-space; , and are atomic with an order continuous norm. For example, let and then the regular norm on is not order continuous, compare also Corollary 3.

Recall that Banach lattice possesses the *positive Schur property* if every weakly null sequence in is norm convergent to 0.

Theorem 12. *Let be a Banach lattice such that possesses the positive Schur property, a Banach lattice. Then is a KB-space if and only if is a KB-space. *

* Proof. *The part of “only if” is obvious. If is a KB-space, by Theorem 4 it suffices to show that each positive operator is M-weakly compact. Indeed, if is not M-weakly compact then there is a disjoint sequence with and for . Note that weakly as weakly (see Theorem 2.4.14 of [2]); by Proposition 2.3.4 of [2] there exists a disjoint sequence , , satisfying
Also by Theorem 2.5.6 and 3.4.18 of [2], is weakly compact and so is by Gantmacher’s theorem (see Theorem 17.2 of [4]), so we may assume that is weakly convergent (replacing by a subsequence if necessary); say weakly, then for each
as in (see Corollary 2.4.3 of [2]); that is, weakly, so the positive Schur property of implies that and it follows that
This is impossible, thus is M-weakly compact.

The following result is a dual version of Theorem 12.

Theorem 13. *Let be a Banach lattice with the positive Schur property, a Banach lattice. Then is a KB-space if and only if the norm on is order continuous. *

* Proof. *The part of “only if” easily follows from the proof of Theorem 2. If the norm on is order continuous, for and each disjoint sequence , then weakly, and weakly. It follows from the positive Schur property of that as ; that is, is M-weakly compact; Theorem 4 yields that is a KB-space.

For a Banach lattice and , recall that has the *strong **-decomposition property* if there exists a constant such that for all disjoint elements in we have for and in case . The number has the strong -decomposition property} is call the *upper index* of .

Similarly has the *strong **-composition property* if there exists a constant such that for all disjoint elements in we have for and in case . The number has the strong -composition property} is called *lower index* of .

It is known that for any Banach lattice . If then has an order continuous norm. If then the norm on is order continuous. See [6] for details

Also recall that if the norm on a Banach lattice is -superadditive then ; and if has a -subadditive norm then ; see Proposition 2.8.2 of [2].

Theorem 14. *Let and be Banach lattices. If then is a KB-space. *

* Proof. *The norm on clearly is order continuous. Note that if then is a KB-space. Indeed, if is not a KB-space, then contains a sublattice lattice isomorphic to , which implies that as . Now the rest is a simple consequence of Theorem 4, Theorem 6.7 of [6], and Theorem 3.6.17 of [2].

Corollary 15. *Let and be Banach lattices. If the norm of is -subadditive, the norm of is -superadditive and , then is a KB-space. *

*Remark 16. *It is worth to point out that fails to be true in general even if is a KB-space, see [7, Example 3.6].

For and being - and -spaces, respectively, we have the following characterization.

Theorem 17. *Let and be infinite dimensional -space, and -space respectively, then is a KB-space if and only if . *

*Proof. *The part of “if” is a simple consequence of Corollary 15. To see the part of “only if”, we may first assume that
are sublattices of and , respectively. Suppose that then .

If there is a positive projection from onto (the existence of is following from Theorem 2.7.11 of [2]), then is not M-weakly compact, which, by Theorem 4, implies that is not a KB-space.

Also if , then and is not a KB-space, Theorem 4 again yields that is not a KB-space.

The next result shows that under the regular norm the spaces are rather rare to be -spaces.

Theorem 18. *Let and be non-zero Banach lattices and , with . Then the following assertions are equivalent. *(1)*The regular norm is -additive on . *(2)*One of following two conditions holds.(a) and the norm on is -additive (i.e., is an -space). (b) and the norm on is -additive (i.e., is an -space).*

*Proof. * is obvious. To see that , we assume that is -additive on . For any , pick with ; thus the -additivity of the regular norm yields that
which means that is an -space. A similar argument involving a fixed element of and two elements of shows that is an -space; hence is an -space (compare with Theorem 2.7.1 of [2]), where .

Now if both and hold we will obtain a contradiction. In fact, we may assume that and are 2-dimensional sublattices of and , respectively; define by
then . Let be a positive contractive projection from onto (see Theorem 2.7.11 of [2]); it follows that
Also it is easy to calculate that ; this is impossible. holds.

Theorem 19. *Let and be non-zero Banach lattices and , with . Then the following assertions are equivalent.*(1)*The regular norm on is equivalent to a -additive norm. *(2)*One of following two conditions holds.(a) and the norm on is equivalent to a -additive norm. (b) and the norm on is equivalent to a -additive norm. *

*Proof. *. Suppose that the regular norm on is equivalent to a -additive norm. We first show that the norms both on and are equivalent to -additive and -additive norms, respectively, where .

For each disjoint sequence with , fix with . It is easy to verify that is a disjoint sequence with for all . Corollary 2.8.12 of [2] yields that is equivalent to the natural basis of . Note that
for all and . It follows that is equivalent to the natural basis of , which by Corollary 2.8.12 of [2] implies that the norm on is equivalent to a -additive norm.

A similar argument involving a fixed element of and a disjoint sequence of elements of shows that the norm on is equivalent to a -additive norm; hence the norm on is equivalent to a -additive norm.

Now we show that either or . Otherwise, both and are infinite dimensional. Renorming and with equivalent -additive and -additive norms, respectively, the regular norm on is still equivalent to a -additive norm. Thus we may assume that the norms on and are - and -additive, and that and are sublattices, respectively. By Theorem 2.7.11 of [2] there is a positive contractive projection from onto . Consider the operators by , where is the element in and with th entry equals to 1 and all others are 0. Then it is easy to verify that is a disjoint sequence in with . Corollary 2.8.12 of [2] yields that is equivalent to the natural basis of . In particular, we have
for all , where and are constants. But
which easily shows that if and for . This is impossible for either or . So holds.

. Let with pairwise disjoint and . Then each corresponds to unique ; moreover, , satisfying the following conditions.(i) for .(ii) for all and .(iii) for all , and .(iv).

Since the norm on is equivalent to a -additive norm, for each disjoint sequence , by Corollary 2.8.12 of [2] there exist constants , such that
for all and . Now for any disjoint sequence with , note that the disjointness of for each ; it follows from (iii), (iv), and that
as . Also
as .

Therefore
for all and ; that is, is equivalent to the natural basis of . Corollary 2.8.12 of [2] again shows that the regular norm on is equivalent to a -additive norm, so holds.

The proof of is similar with . This completes the proof.

#### Acknowledgments

One of the authors would like to thank Professor A. W. Wickstead for useful discussions about the subjects of this paper. The authors were supported in part by the Fundamental Research Funds for the Central Universities (SWJTU11CX154, SWJTU12ZT13).

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