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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 896302, 4 pages
http://dx.doi.org/10.1155/2013/896302
Research Article

-Orthomorphisms and -Linear Operators on the Order Dual of an -Algebra Revisited

Research Laboratory of Algebra, Topology, Arithmetic and Order, Department of Mathematics, Tunis El Manar University, Tunisia

Received 19 March 2013; Accepted 14 May 2013

Academic Editor: Ferenc Hartung

Copyright © 2013 Jamel Jaber. 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 give a necessary and sufficient condition on an -algebra for which orthomorphisms, -linear operators, and -orthomorphisms on the order dual are the same class of operators.

1. Introduction and Preliminaries

First of all, we point out that the standard book [1] is adopted in this paper as a unique source of unexplained terminology and notation.

Let and be Riesz spaces. The operator is called order bounded if the image under of an order bounded set in is again an order bounded set in . An order bounded operator is said to be an orthomorphism if whenever for all . The set of all orthomorphisms on is denoted by .

If is a Riesz space , the first order dual of will be the Riesz space of all order bounded linear functionals on . The second order dual (or order bidual) is denoted by . It is well known that is a Dedekind complete Riesz space. Define for each the element by for all and then define the mapping by for all . If separates the points of then is injective and we can identify with the Riesz subspace . Throughout the paper, we only consider Archimedean Riesz spaces with point separating order dual .

From now on, denotes an -algebra, that is, a lattice ordered algebra in which implies that for all . Following constructions in [2], a multiplication can be introduced in the order bidual of . This is accomplished in three steps as explained next. For every and , we define by the following: Then, for and , we introduce by putting the following: Finally, let and define by the following

The latter equality defines the Arens multiplication in . Bernau and Huijsmans in [3] that is an -algebra with respect to the Arens multiplication. The band of all order continuous linear functionals on is denoted by and its disjoint complement in by . Observe that as is Dedekind complete.

For each , we define the mapping by the following: It is shown in [4, Theorem 5.2] that and the mapping are an algebra and lattice homomorphism between and .

The final paragraph of this introduction is devoted to the definition of the so-called -orthomorphism and -linear operator on the order dual.

Let be an -algebra. For all define the set by the following:

An order bounded operator is said to be as follows:(1)an -orthomorphism if for each , the collection of all -orthomorphisms on will be denoted by ;(2)an -linear with respect to if for all and , the set of all -linear operators on will be denoted by .

Turan showed [5, Proposition 2.6] that whenever is unital -algebra (since is topologically full with respect to itself). Recently, Feng et al. [6, Theorem 3.4] proved that for any -algebra we have the following:

This leads to a very natural question, namely, when does any -orthomorphisms is an orthomorphism. A partial answer was already obtained by Feng et al. in [6, Theorem 3.4]. In this regards they proved that if has the factorization property, then . In this paper, we give a complete answer to this question. In fact, we give a necessary and sufficient condition on an -algebra for which orthomorphisms, -orthomorphisms, and -linear operators on the order dual are precisely the same class of operators.

2. When -Orthomorphisms Are Orthomorphisms

Let be an -algebra with separating order dual. Let be the order ideal generated by all products of ; that is,

It is easy to see that is an -ideal (ring and order ideal) in . The annihilator is always a band in . Since is Dedekind complete Riesz space, then . We denote by the band projection.

As we will see later, the set plays a key role in the proof of the main result of this paper. We start our discussion by a study of the connection between -orthomorphism and orthomorphism in a special case.

Example 1. Suppose that the set ; hence . This implies that the product in is given by for all . It is easy to see that
Thus, if and only if every order bounded operator is an orthomorphism. Hence, .

From now on, we may assume that the ideal .

It is obvious that if has the factorization property, then . In the next example we show that the condition “” is strictly weaker than the condition “has the factorization property.” Recall that the condition “” is equivalent to the condition “ is a semi-prime -algebra” (see [4, Corollary 6.3]).

Example 2. Let be the Archimedean unital -algebra of all real-valued continuous functions on . A function is an eventual-polynomial if there is and a (unique) polynomial such that for all . Let be the set of all eventual-polynomials in . Now, put the following: and observe that is a semiprime -subalgebra of  . Clearly, does not satisfy the factorization property. It is shown in [7, Example 7] that is a unital -algebra, hence semiprime, -algebra. So .

We list some simple properties of the band . The proof of the next lemma is straightforward and therefore omitted.

Lemma 3. Let be an f-algebra and . Then(1) if and only if for all and .(2) for all .(3) for all .

The next result is important in the context of this work; it is already proved in [6] but for the sake of completeness we partially reproduce the proof.

Proposition 4. Let be an -algebra. Then.

Proof. Let ; we have to show that for all and . Since which is a commutative algebra we get the following: Thus, .
Let and . Pick and observe that Then for all we get the following:
Hence, where is the order adjoint mapping of defined by . Consequently, .

Another lemma turns out to be useful for later purposes.

Lemma 5. Let be an -algebra and . Then whenever in .

Proof. It suffices to show that for all . Let . Since we get the following: There exists such that . Consequently,

We have now gathered all of the ingredients for the proof of the principal theorem of this paper.

Theorem 6. Let be an -algebra. Then the following are equivalents:(i);(ii).

Proof. (i)(ii) Arguing by contradiction, suppose that . There exists a nonzero positive element such that for all . Let and such that . Define the mapping by . It is not hard to see that is positive and for all , we have the following:
Thus, . But is not an orthomorphism since and .
(ii)(i) In view of Proposition 4, it suffices to show that . To do this, pick and such that . According to Lemma 5,we obtain ; that is, , as required.

It follows from Theorem 6 that if then ; hence , is a semiprime -algebra.

We end this section with a consequence of Theorem 6. The notion of weak approximate unit plays a key role in the context of this study. Recall that an upward directed net of positive elements in an -algebra is said to be an approximate unit if for all . An approximate unit is called weak approximate unit if for all and (see [4, Definitions 2.2 and 7.1]). It is well known that if is a semi-prime -algebra then can be embedded as a ring and -subalgebra in the unital -algebra .

In Theorem 6, we gave two necessary and sufficient conditions for which -orthomorphisms are orthomorphisms. In the following proposition, we shall present a second one in terms of approximate unit. We have to impose, however, on an extra condition, namely, the Stone condition (i.e., whenever is positive in , where is the identity of ). It should be pointed out here that every uniformly complete semi-prime -algebra satisfies the Stone condition.

Proposition 7. Consider in a semi-prime -algebra the following conditions.

Theorem 8. (i)   has a weak approximate unit.
(ii)  .
(i)(ii) Moreover, if satisfies the Stone condition then (i)(ii).

Proof. (i)(ii) Assume that is a weak approximate unit. Let , it follows from for all that .
(ii)(i) It follows from Theorem 6, that is semi-prime. The result follows from [4, Theorem 7.2].

3. When -Linear Operators Are Orthomorphisms

In this section, we study the connection between -linear operators and orthomorphisms. From Proposition 4 and Theorem 6 if -linear operators on the dual of an -algebra are orthomorphisms and if then the two classes coincide. Next, we give an example of an -algebra such that in which -linear operators are orthomorphisms.

Example 9. Let equipped with the coordinatewise operations and ordering. Consider the multiplication defined by . It is easy to see that for all and we have and . Note that . Let an -linear operator. An easy computation shows that . Thus, is an orthomorphism.

It seems natural therefore to ask under what condition we have . The answer is given in the next theorem. First, let us discuss the ideal of nilpotents elements in the bidual of an -algebra . Let the absolute kernel or null ideal of is defined by the following: It is evident that is an order ideal. The disjoint complement of is called the carrier of and is always a band in . Given such that for all ( is nilpotent in ), then . Indeed, let , since is a band and is an orthomorphism in , then for all . But, by hypothesis , , so for all . This implies that . Consequently, and . For more information about the nilpotent elements in the bidual of -algebra, the reader is referred to [4].

We are now in position to prove the main theorem in this section.

Theorem 10. Let be an -algebra. The followings are equivalents.(i).(ii).

Proof. (i)(ii) Suppose that . We can find therefore positive elements such that . Let such that . Define the mapping by the following: Since for all and , we get . Thus is an -linear operator. Clearly, is not an orthomorphism since and .
(ii)(i) Let be an -linear mapping on and . First, we show that and . There exists such that . Since is an -linear operator, we derive that for all and . That is for all . By the remark above, we get for all . This implies that whenever . Now, If then This shows that , in particular where . Now, let in . We have to show that . Decompose and as and with and . The decomposition of is given by the following: Since , then and . On the other hand, according to Lemma 5, we have . So as . This completes the proof of the theorem.

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