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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 971560, 7 pages
-Orthomorphisms and -Linear Operators on the Order Dual of an -Algebra
Department of Mathematics, Southwest Jiaotong University, Chengdu 610031, China
Received 1 August 2012; Revised 24 September 2012; Accepted 27 September 2012
Academic Editor: Alberto Fiorenza
Copyright © 2012 Ying Feng 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.
We consider the -orthomorphisms and -linear operators on the order dual of an -algebra. In particular, when the -algebra has the factorization property (not necessarily unital), we prove that the orthomorphisms, -orthomorphisms, and -linear operators on the order dual are precisely the same class of operators.
Let be an -algebra with . Recall that we can define a multiplication on , the order continuous part of the order bidual of , with respect to which can also be made an -algebra. This is done in three steps: (1) for , (2) for , (3) for . With the so-called Arens multiplication defined in step (3), is an Archimedean (and hence commutative) -algebra. Moreover, if has a multiplicative unit, then , the whole order bidual of . The mapping defined by for all , where for every , is an algebra and Riesz isomorphism. See [1, 2] for details.
Let be an -algebra. A Riesz space with is said to be an (left) -module over (cf. [2, 3]) if is a left module over and satisfies the following two conditions:(i)for each and , we have ,(ii)if , then for each , we have .
When is an -algebra with unit , saying is a unital -module over implies that the left multiplication satisfies for all . From Corollary 2.3 in , we know that if is an -module over , then is an -module over (and ). The -module over with unit is said to be topologically full with respect to if for two arbitrary vectors , satisfying in , there exists a net in such that in . If is topologically full with respect to , then is topologically full with respect to [2, Proposition 3.12].
Let be a unital -algebra, and, , be -modules over . is called an -linear operator if for each and . The collection of all -linear operators will be denoted by . For each and , we can define by for all . Let . Then is an order ideal in . is said to be an -orthomorphism if for each . The collection of all -orthomorphisms will be denoted by Orth. Turan  showed that whenever is topologically full with respect to .
Clearly, is an -module over the -algebras and , respectively. If is unital, then is topologically full with respect to itself ([2, Proposition 2.6]). From the above remarks we know that is topologically full with respect to , and hence, the -orthomorphisms and -linear operators are precisely the same class of operators, that is,
An -algebra is said to be square-root closed whenever for any there exists such that . An immediate example is that a uniformly complete -algebra with unit element is square-root closed . However, a square-root closed -algebra is not necessarily unital. For instance, , with the familiar coordinatewise operations and ordering, is a square-root closed -algebra without unit. We recall that an Archimedean -algebra is said to have the factorization property if, given , there exist such that . It should be noted that if is unital or square-root closed, then has the factorization property.
In this paper, we do not have to assume that the -algebras are unital. We modify the definition of the -orthomorphism introduced by Turan [2, Definition 3.7] and consider the -orthomorphisms and -linear operators on the order dual of an -algebra. In particular, when the -algebra with separating order dual has the factorization property, we prove that the orthomorphisms, -orthomorphisms, and -linear operators on the order dual are precisely the same class of operators, that is, the above equality still holds.
2. -Orthomorphisms on the Order Dual
Let be an -algebra with separating order dual (and hence Archimedean!) and . We consider the mapping defined by for all . It should be noted that the mapping defined by for all , where for every , is an algebra and Riesz isomorphism (cf. [2, Proposition 2.2]).
Theorem 2.1. For , is an interval preserving lattice homomorphism.
Proof. Clearly, is linear and positive. Since the mapping is a lattice homomorphism and for , we have
Hence, is a lattice homomorphism.
Next, we show that is an interval preserving operator. We identify with its canonical image in and denote the restriction of to by . Then Thus, for each and , we see that which implies that is the same as on . Since is interval preserving (cf. [5, Theorem 7.8]), is likewise an interval preserving operator.
Corollary 2.2. For , one has . Furthermore, if in , holds for any .
Proof. Since is an orthomorphism on , we have for each , that is, . From Theorem 2.1, we know that Let in . Then we have which implies that for all .
Following the above discussion, we now consider , the image of under .
Corollary 2.3. If is an -algebra and , then , and is an order ideal in .
Proof. First, since is an interval preserving lattice homomorphism, we can easily see that is an order ideal in . By Corollary 2.2 we conclude that .
Now, to complete the proof we only need to prove that . To this end, let , be band projections, where and are the bands generated by and in , respectively. If , we have In addition, for all (cf. Theorem 3.1). Since is an orthomorphism on and hence order continuous (cf. [5, Theorem 8.10]), we have . For all and all , from it follows that for all , which implies that , as desired.
Next, we give a necessary and sufficient condition for when has the factorization property. First, we need the following lemma.
Lemma 2.4. Let be an -algebra with the factorization property, and . If for each , then .
Proof. Since has the factorization property, for each , there exist such that . Hence, from it follows easily that holds.
Theorem 2.5. Let be an -algebra with the factorization property. If , then if and only if .
Proof. If in , then it follows from Corollary 2.2 that for all . This implies that .
Conversely, if and are disjoint, then for each we have In particular, for any , its canonical image also satisfies . By the preceding lemma, we have , that is, , as desired.
Now, we give the definition of the so-called -orthomorphism.
Definition 2.6. Let be an -algebra and . is called an -orthomorphism on if for each . The collection of all -orthomorphisms on will be denoted by .
The next result deals with the relationship between the -orthomorphisms and the orthomorphisms on the order dual of an -algebra with the factorization property. Note that is a band in .
Theorem 2.7. Let be an -algebra. Then is a linear subspace of and .
If , in addition, has the factorization property, then .
Proof. First, we can easily see that is a linear subspace of . To prove , let . We claim that for all and all . To this end, let , , and be arbitrary. Since is a commutative -algebra, by Theorem 3.1, we have
Thus, . This implies that for each , that is, .
If has the factorization property, we prove that holds. To this end, take and satisfying in . Then, it follows from Theorem 2.5 that . Since , we have . Therefore, , which implies that , and hence is an orthomorphism on , as desired.
3. -Linear Operators on the Order Dual
Let be an -algebra with separating order dual and . Recall that is called to be -linear with respect to if for all and . The set of all -linear operators on will be denoted by . It follows from [3, Lemma 4.4] that is a band in .
Theorem 3.1. Let be an -algebra with separating order dual. Then .
Proof. Clearly is commutative since is an Archimedean -algebra. To complete the proof, let . We have for all and . Hence, .
The following result deals with the order adjoint of an -linear operator on the order dual of an -algebra. It should be noted that the order adjoint of an order-bounded operator is order continuous (cf. [5, Theorem 5.8]).
Lemma 3.2. Let . Then the order adjoint of satisfies for all and . In particular, for all .
Proof. Since , and is a commutative -algebra, we have
for all , , and , which implies that .
Let be given. Then for , from it follows that . This completes the proof.
Proof. For , we know that is also -linear with respect to . Assume that and . So by Lemma 3.2, we have Since is interval preserving, there exists such that and . It is now immediate that , and hence, , as desired.
Theorem 3.4. If is an -algebra with separating order dual, then In particular, if, in addition, has the factorization property, then
The authors were supported in part by the Fundamental Research funds for the Central Universities (SWJTU11CX154, SWJTU12ZT13).
The authors would like to thank the reviewer for his/her kind comments and valuable suggestions which have improved this paper. Corollary 2.3 in its present formulation and the proof are essentially due to the reviewer. In particular, he/she suggested that the authors should consider their questions under the condition of “ has the factorization property,” which is weaker than the hypothesis “ is square-root closed” used originally in this paper.
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