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

Authors: Ying Feng; Jin Xi Chen; Zi Li Chen
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider thef-orthomorphisms and f-linear operators on the order dual of an f-algebra. In particular, when the f-algebra has the factorization property (not necessarily unital), we prove that the orthomorphisms, f-orthomorphisms, and f-linear operators on the order dual are precisely the same class of operators.


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## Body

## 1. Introduction

LetA be an f-algebra with °(A~)={0}. Recall that we can define a multiplication on (A~)n~, the order continuous part of the order bidual of A, with respect to which (A~)n~ can also be made an f-algebra. This is done in three steps:  (1)
A×A~→A~(a,f)↦f·a:(f·a)(b)=f(ab) for b∈A,(2)
(A~)n~×A~→A~(F,f)↦F·f:(F·f)(a)=F(f·a) for a∈A,(3)
(A~)n~×(A~)n~→(A~)n~(F,G)↦F·G:(F·G)(f)=F(G·f) for f∈A~. With the so-called Arens multiplication defined in step (3), (A~)n~ is an Archimedean (and hence commutative) f-algebra. Moreover, if A has a multiplicative unit, then (A~)n~=(A~)~, the whole order bidual of A. The mapping V:(A~)n~→Orth(A~) defined by V(F)=VF for all F∈(A~)n~, where VF(f)=F·f for every f∈A~, is an algebra and Riesz isomorphism. See [1, 2] for details.LetA be an f-algebra. A Riesz space L with ∘(L~)={0} is said to be an (left) f-module over A (cf. [2, 3]) if L is a left module over A and satisfies the following two conditions:(i)
for eacha∈A+ and x∈L+, we have ax∈L+,(ii)
ifx⊥y, then for each a∈A, we have a·x⊥y.WhenA is an f-algebra with unit e, saying L is a unital f-module over A implies that the left multiplication satisfies e·x=x for all x∈L. From Corollary  2.3 in [2], we know that if L is an f-module over A, then L~ is an f-module over A (and (A~)n~). The f-module L over A with unit e is said to be topologically full with respect to A if for two arbitrary vectors x, y satisfying 0≤y≤x in L, there exists a net 0≤aα≤e in A such that aα·x→y in σ(L,L~). If L is topologically full with respect to A, then  L~ is topologically full with respect to (A~)n~ [2, Proposition 3.12].LetA be a unital f-algebra, and, L, M be f-modules over A. T∈Lb(L,M) is called an f-linear operator  if T(a·x)=a·Tx for each a∈A and x∈L. The collection of all f-linear operators will be denoted by Lb(L,M;A). For each x∈L and f∈L~, we can define ψx,f∈A~ by ψx,f(a)=f(a·x) for all a∈A. Let S(x):={ψx,f:f∈L~}. Then S(x) is an order ideal in A~ [2]. T∈Lb(L,M) is said to be an f-orthomorphism if S(Tx)⊆S(x) for each x∈L. The collection of all f-orthomorphisms will be denoted by Orth(L,M;A). Turan [2] showed that Orth(L,M;A)=Lb(L,M;A) whenever M is topologically full with respect to A.Clearly,A~  is an f-module over the f-algebras A and (A~)n~, respectively. If A is unital, then A is topologically full with respect to itself ([2, Proposition 2.6]). From the above remarks we know that A~ is topologically full with respect to (A~)n~, and hence, the f-orthomorphisms and f-linear operators are precisely the same class of operators, that is,
(*)Orth(A~,A~;(A~)n~)=Lb(A~,A~;(A~)n~).Anf-algebra A is said to be square-root closed whenever for any a∈A there exists b∈A such that |a|=b2. An immediate example is that a uniformly complete f-algebra with unit element is square-root closed [4]. However, a square-root closed f-algebra is not necessarily unital. For instance, c0, with the familiar coordinatewise operations and ordering, is a square-root closed f-algebra without unit. We recall that an Archimedean f-algebra A is said to have the factorization property if, given a∈A, there exist b,c∈A such that a=bc. It should be noted that if A is unital or square-root closed, then A has the factorization property.In this paper, we do not have to assume that thef-algebras are unital. We modify the definition of the f-orthomorphism introduced by Turan [2, Definition 3.7] and consider the f-orthomorphisms and f-linear operators on the order dual of an f-algebra. In particular, when the f-algebra with separating order dual has the factorization property, we prove that the orthomorphisms, f-orthomorphisms, and f-linear operators on the order dual are precisely the same class of operators, that is, the above equality (*) still holds.Our notions are standard. For the theory of Riesz spaces, positive operators, andf-algebras, we refer the reader to the monographs [5–7].


## 2.f-Orthomorphisms on the Order Dual

LetA be an f-algebra with separating order dual (and hence A Archimedean!) and f∈A~. We consider the mapping Tf:(A~)n~→A~ defined by Tf(F)=F·f for all F∈(A~)n~. It should be noted that the mapping V:(A~)n~→Orth(A~) defined by V(F)=VF for all F∈(A~)n~, where VF(f)=F·f for every f∈A~, is an algebra and Riesz isomorphism (cf. [2, Proposition 2.2]).Theorem 2.1.
For0≤f∈A~,  Tf is an interval preserving lattice homomorphism.Proof.
Clearly,Tf is linear and positive. Since the mapping V is a lattice homomorphism and VF,VG∈Orth(A~) for F,G∈(A~)n~, we have
(2.1)Tf(F∨G)=(F∨G)·f=VF∨G(f)=(V(F∨G))(f)=(V(F)∨V(G))(f)=(V(F)(f))∨(V(G)(f))=F·f∨G·f=Tf(F)∨Tf(G).
Hence, Tf is a lattice homomorphism.
Next, we show thatTf is an interval preserving operator. We identify x with its canonical image x′′ in (A~)n~ and denote the restriction of  Tf to A by Tf∣A. Then
(2.2)TF∣A(x)=Tf(x′′)=x′′·f=f·x.
Thus, for each F∈(A~)n~ and x∈A, we see that
(2.3)((Tf|A)′(F))(x)=F((Tf|A)(x))=F(f·x)=(F·f)(x)=(Tf(F))(x),
which implies that (Tf|A)' is the same as Tf on (A~)n~. Since (Tf|A)' is interval preserving (cf. [5, Theorem 7.8]), Tf is likewise an interval preserving operator.Corollary 2.2.
Forf∈A~,F∈(A~)n~, one has |F·f|=|F|·|f|. Furthermore, if f⊥g in A~, F·f⊥G·g holds for any F,G∈(A~)n~.Proof.
SinceVF is an orthomorphism on A~, we have VF(f+)⊥VF(f-) for each f∈A~, that is, F·(f+)⊥F·(f-). From Theorem 2.1, we know that
(2.4)|F·f|=|F·f+|+|F·f-|=|Tf+(F)|+|Tf-(F)|=Tf+(|F|)+Tf-(|F|)=|F|·f++|F|·f-=|F|·|f|.
Let f⊥g in A~. Then we have
(2.5)|F·f|∧|G·g|=|F|·|f|∧|G|·|g|≤((|F|+|G|)·|f|)∧((|F|+|G|)·|g|)=0,
which implies that  F·f⊥G·g for all F,G∈(A~)n~.Following the above discussion, we now considerR(f)={F·f:F∈(A~)n~}, the image of (A~)n~ under Tf.Corollary 2.3.
IfA is an f-algebra and f∈(A~), then R(f)=R(|f|), and R(f) is an order ideal in A~.Proof.
First, sinceT|f| is an interval preserving lattice homomorphism, we can easily see that R(|f|) is an order ideal in A~. By Corollary 2.2 we conclude that R(f)⊆R(|f|).
Now, to complete the proof we only need to prove thatR(|f|)⊆R(f). To this end, let P1:A~→Bf+, P2:A~→Bf- be band projections, where Bf+ and Bf- are the bands generated by f+ and f- in A~, respectively. If π=P1-P2, we have
(2.6)π∈Orth(A~),π(f)=|f|,π(|f|)=f.
In addition, π(f)·a=π(f·a) for all a∈A (cf. Theorem 3.1). Since π is an orthomorphism on A~ and hence order continuous (cf. [5, Theorem 8.10]), we have π′((A~)n~)⊆(A~)n~. For all a∈A and all F∈(A~)n~, from
(2.7)(F·|f|)(a)=(F·π(f))(a)=F(π(f)·a)=F(π(f·a))=(π′(F)·f)(a),
it follows that F·|f|=π′(F)·f for all F∈(A~)n~, which implies that R(|f|)⊆R(f), as desired.Next, we give a necessary and sufficient condition forR(f)⊥R(g) when A has the factorization property. First, we need the following lemma.Lemma 2.4.
LetA be an f-algebra with the factorization property, and  f∈A~. If f·x=0 for each x∈A, then f=0.Proof.
SinceA has the factorization property, for each a∈A+, there exist x,y∈A such that a=xy. Hence, from
(2.8)f(a)=f(xy)=(f·x)(y)=0,
it follows easily that f=0 holds.Theorem 2.5.
LetA be an f-algebra with the factorization property. If f,g∈A~, then f⊥g if and only if R(f)⊥R(g).Proof.
Iff⊥g in A~,  then it follows from Corollary 2.2 that F·f⊥G·g for all F,G∈(A~)n~. This implies that R(f)⊥R(g).
Conversely, ifR(f) and R(g) are disjoint, then for each F∈((A~)n~)+ we have
(2.9)F·(|f|∧|g|)=VF(|f|∧|g|)=VF(|f|)∧VF(|g|)=F·|f|∧F·|g|=|F·f|∧|F·g|=0.
In particular, for any x∈A, its canonical image x′′∈(A~)n~  also satisfies  x′′·(|f|∧|g|)=(|f|∧|g|)·x=0.  By the preceding lemma, we have |f|∧|g|=0, that is, f⊥g, as desired.Now, we give the definition of the so-calledf-orthomorphism.Definition 2.6.
LetA be an f-algebra and T∈Lb(A~). T is called an f-orthomorphism on A~ if R(Tf)⊆R(f) for each f∈A~. The collection of all f-orthomorphisms on A~ will be denoted by Orth(A~,A~;(A~)n~).
The next result deals with the relationship between thef-orthomorphisms and the orthomorphisms on the order dual of an f-algebra with the factorization property. Note that Orth(A~) is a band in Lb(A~).Theorem 2.7.
LetA be an f-algebra. Then Orth(A~,A~;(A~)n~) is a linear subspace of Lb(A~) and Orth(A~)⊆Orth(A~,A~;(A~)n~).
IfA, in addition, has the factorization property, then Orth(A~,A~;(A~)n~)=Orth(A~).Proof.
First, we can easily see thatOrth(A~,A~;(A~)n~) is a linear subspace of Lb(A~). To prove Orth(A~)⊆Orth(A~,A~;(A~)n~), let π∈Orth(A~). We claim that F·π(f)=π′(F)·f for all F∈(A~)n~ and all f∈A~. To this end, let F∈(A~)n~,  f∈A~, and x∈A be arbitrary. Since (A~)n~ is a commutative f-algebra, by Theorem 3.1, we have
(2.10)(π′(F)·f)(x)=π′(F)(f·x)=F(π(f·x))=F(π(x′′·f))=F(x′′·(π(f)))=(F·x′′)(π(f))=(x′′·F)(π(f))=x′′(F·π(f))=(F·π(f))(x).
Thus, F·π(f)=π′(F)·f. This implies that R(π(f))⊆R(f) for each f∈A~, that is, Orth(A~)⊆Orth(A~,A~;(A~)n~).
IfA has the factorization property, we prove that Orth(A~,A~;(A~)n~)⊆Orth(A~) holds. To this end, take T∈Orth(A~,A~;(A~)n~) and f,g∈A~ satisfying f⊥g in A~. Then, it follows from Theorem 2.5 that R(f)⊥R(g). Since T∈Orth(A~,A~;(A~)n~), we have R(T(f))⊂R(f). Therefore, R(T(f))⊥R(g), which implies that  T(f)⊥g, and hence T is an orthomorphism on A~, as desired.


## 3.f-Linear Operators on the Order Dual

LetA be an f-algebra with separating order dual and T∈Lb(A~). Recall that T is called to be f-linear with respect to (A~)n~ if  T(G·f)=G·T(f) for all f∈A~ and G∈(A~)n~. The set of all f-linear operators on A~ will be denoted by Lb(A~,A~;(A~)n~). It follows from [3, Lemma 4.4] that Lb(A~,A~;(A~)n~) is a band in Lb(A~).Theorem 3.1.
LetA be an f-algebra with separating order dual. Then Orth(A~)⊆Lb(A~,A~;(A~)n~).Proof.
ClearlyOrth(A~) is commutative since Orth(A~) is an Archimedean f-algebra. To complete the proof, let π∈Orth(A~). We have
(3.1)π(G·f)=π(VG(f))=VG(π(f))=G·(π(f)),
for all f∈A~ and G∈(A~)n~. Hence,  π∈Lb(A~,A~;(A~)n~).The following result deals with the order adjoint of anf-linear operator on the order dual of an f-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.
LetT∈Lb(A~,A~;(A~)n~). Then the order adjoint  T′ of  T satisfies T′(F)·f=F·T(f) for all F∈(A~)n~ and f∈A~. In particular, G·T′(F)=T′(G·F) for all F,G∈(A~)n~.Proof.
SinceT∈Lb(A~,A~;(A~)n~), and (A~)n~ is a commutative f-algebra, we have
(3.2)(T′(F)·f)(x)=T′(F)(f·x)=F(T(f·x))=F(T(x′′·f))=F(x′′·(T(f)))=(F·x′′)(T(f))=(x′′·F)(T(f))=x′′(F·T(F))=(F·T(f))(x),
for all F∈(A~)n~, f∈A~, and x∈A, which implies that  T′(F)·f=F·T(f).
LetF,G∈(A~)n~ be given. Then for f∈A~, from
(3.3)(G·T′(F))(f)=G(T′(F)·f)=G(F·T(f))=(G·F)(T(f))=(T′(G·F))(f),
it follows that G·T′(F)=T′(G·F). This completes the proof.Theorem 3.3.
(3.4)
L
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n
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⊆
Orth
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.Proof.
ForT∈Lb(A~,A~;(A~)n~), we know that |T| is also f-linear with respect to (A~)n~. Assume that 0≤G∈(A~)n~ and f∈A~. So by Lemma 3.2, we have
(3.5)0≤G·(|T(f)|)≤G·(|T||f|)=(|T|′(G))·|f|=T|f|(|T|′(G)).
Since T|f| is interval preserving, there exists F∈(A~)n~ such that 0≤F≤|T|′(G) and G·(|T(f)|)=F·|f|. It is now immediate that R(|T(f)|)⊆R(|f|), and hence, Lb(A~,A~;(A~)n~)⊆Orth(A~,A~;(A~)n~), as desired.Combining Theorems3.1, 3.3, and 2.7, we have the following result.Theorem 3.4.
IfA is an f-algebra with separating order dual, then
(3.6)Orth(A~)⊆Lb(A~,A~;(A~)n~)⊆Orth(A~,A~;(A~)n~).
In particular, if, in addition, A has the factorization property, then
(3.7)Orth(A~)=Lb(A~,A~;(A~)n~)=Orth(A~,A~;(A~)n~).



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*Source: 971560-2012-11-05.xml*