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

Authors: Jamel Jaber
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We give a necessary and sufficient condition on anf-algebra A for which orthomorphisms, f-linear operators, and f-orthomorphisms on the order dual A~ are the same class of operators.


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## Body

## 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.LetL and M be Riesz spaces. The operator T:L→M is called order bounded if the image under T of an order bounded set in L is again an order bounded set in M. An order bounded operator π:L→L is said to be an orthomorphism if |πx|∧|y|=0 whenever |x|∧|y|=0 for all x,y∈L. The set of all orthomorphisms on L is denoted by Orth(L).IfL is a Riesz space L~, the first order dual of L will be the Riesz space of all order bounded linear functionals on L. The second order dual (or order bidual) is denoted by L~~. It is well known that L~ is a Dedekind complete Riesz space. Define for each x∈L the element x~~∈L~~ by x~~(f)=f(x) for all f∈L~ and then define the mapping κ:L→L~~ by κ(x)=x~~ for all x∈L. If L~ separates the points of L then κ is injective and we can identify L with the Riesz subspace κ(L). Throughout the paper, we only consider Archimedean Riesz spaces L with point separating order dual L~.From now on,A denotes an f-algebra, that is, a lattice ordered algebra in which a∧b=0 implies that a·c∧b=0 for all 0≤c∈A. Following constructions in [2], a multiplication can be introduced in the order bidual A~~ of A. This is accomplished in three steps as explained next. For every f∈A~ and x∈A, we define f·x∈A~ by the following:
(1)(f·x)(y)=f(xy)(y∈A).
Then, for G∈A~~ and f∈A~, we introduce G·f∈A~ by putting the following:
(2)(G·f)(x)=G(f·x)(x∈A).
Finally, let F,G∈A~~ and define FG∈A~~ by the following
(3)(FG)(f)=F(G·f)(f∈A~).The latter equality defines theArens multiplication in A~~. Bernau and Huijsmans in [3] that A~~ is an f-algebra with respect to the Arens multiplication. The band of all order continuous linear functionals on A~ is denoted by (A~)n~ and its disjoint complement in A~~ by (A~)s~. Observe that
(4)A~~=(A~)n~⊕(A~)s~,
as A~~ is Dedekind complete.For eachF∈(A~)n~, we define the mapping VF:A~→A~ by the following:
(5)VF(f)=F·f∀f∈A~.
It is shown in [4, Theorem 5.2] that VF∈Orth(A~) and the mapping V:F→V(F)=VF are an algebra and lattice homomorphism between (A~)n~ and Orth(A~).The final paragraph of this introduction is devoted to the definition of the so-calledf-orthomorphism and f-linear operator on the order dual.LetA be an f-algebra. For all f∈A~ define the set R(f) by the following:
(6)R(f)={F·f:F∈(A~)n~}.An order bounded operatorT:A~→A~ is said to be as follows:(1)
anf-orthomorphism 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~);(2)
anf-linear with respect to (A~)n~ if T(F·f)=F·Tf  for all f∈A~ and F∈(A~)n~, the set of all f-linear operators on A~ will be denoted by ℒb(A~,A~,(A~)n~).Turan showed [5, Proposition 2.6] that Orth(A~,A~,(A~)n~)=ℒb(A~,A~,(A~)n~) whenever A is unital f-algebra (since A is topologically full with respect to itself). Recently, Feng et al. [6, Theorem 3.4] proved that for any f-algebra A we have the following:
(7)Orth(A~)⊆ℒb(A~,A~,(A~)n~)⊆Orth(A~,A~,(A~)n~).This leads to a very natural question, namely, when does anyf-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 A has the factorization property, then Orth(A~)=ℒb(A~,A~,(A~)n~)=Orth(A~,A~,(A~)n~). In this paper, we give a complete answer to this question. In fact, we give a necessary and sufficient condition on an f-algebra A for which orthomorphisms, f-orthomorphisms, and f-linear operators on the order dual are precisely the same class of operators.


## 2. Whenf-Orthomorphisms Are Orthomorphisms

LetA be an f-algebra with separating order dual. Let J be the order ideal generated by all products of A; that is,
(8)J={x∈A:|x|≤y·zforsomey,z∈A}.It is easy to see thatJ is an ℓ-ideal (ring and order ideal) in A. The annihilator
(9)J0={f∈A~:f(x)=0∀x∈J}
is always a band in A~. Since A~ is Dedekind complete Riesz space, then A~=J0⊕(J0)d. We denote by P0:A~→J0 the band projection.As we will see later, the setJ0 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 f-orthomorphism and orthomorphism in a special case.Example 1.
Suppose that the setJ={0}; hence J0=A~. This implies that the product in A is given by x·y=0 for all x,y∈A. It is easy to see that
(10)ℒb(A~,A~,(A~)n~)=Orth(A~,A~,(A~)n~)=ℒb(A~).
Thus,Orth(A~)=Orth(A~,A~,(A~)n~) if and only if every order bounded operator is an orthomorphism. Hence, dim⁡A~=dim⁡A=1.From now on, we may assume that the idealJ≠{0}.It is obvious that ifA has the factorization property, then J0={0}. In the next example we show that the condition “J0={0}” is strictly weaker than the condition “has the factorization property.” Recall that the condition “J0={0}” is equivalent to the condition “(A~)n~ is a semi-prime f-algebra” (see [4, Corollary 6.3]).Example 2.
LetC([0,1)) be the Archimedean unital f-algebra of all real-valued continuous functions on [0,1). A function f∈C([0,1)) is an eventual-polynomial if there is rf∈[0,1) and a (unique) polynomial Pf such that f(r)=Pf(r) for all r∈[rf,1). Let Y be the set of all eventual-polynomials in C([0,1)). Now, put the following:
(11)A={f∈Y:Pf(0)=0},
and observe that A is a semiprime f-subalgebra of  Y. Clearly, A does not satisfy the factorization property. It is shown in [7, Example 7] that (A~)n~ is a unital f-algebra, hence semiprime, f-algebra. So J0={0}.We list some simple properties of the bandJ0. The proof of the next lemma is straightforward and therefore omitted.Lemma 3.
LetA be an f-algebra and F∈A~~. Then(1)
f∈J0 if and only if f·x for all f∈A~and x∈A.(2)
F·f=0 for all f∈J0.(3)
F·f∈(J0)d for all f∈A~.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.
LetA be an f-algebra. ThenOrth (A~)⊆ℒb(A~,A~,(A~)n~)⊆Orth (A~,A~,(A~)n~).Proof.
Letπ∈Orth(A~); we have to show that π(F·f)=F·π(f) for all F∈(A~)n~ and f∈A~. Since VF∈Orth(A~) which is a commutative algebra we get the following:
(12)π(F·f)=π∘VF(f)=VF∘π(f)=F·π(f).
Thus, π∈ℒb(A~,A~,(A~)n~).
LetT∈ℒb(A~,A~,(A~)n~) and f∈A~. Pick x∈A and observe that
(13)T(f·x)=T(x~~·f)=x~~·Tf=Tf·x.
Then for all F∈(A~)n~ we get the following:
(14)F·Tf(x)=F(Tf·x)=F(T(f·x))=F∘T(f·x)=F∘T·f(x)∀x∈A.
Hence,F·Tf=T~(F)·f  where T~ is the order adjoint mapping of T defined by T~(F)=F∘T. Consequently, T∈Orth(A~,A~,(A~)n~).Another lemma turns out to be useful for later purposes.Lemma 5.
LetA be an f-algebra and T∈Orth(A~,A~,(A~)n~). Then |Tf|∧g∈J0 whenever f∧g=0 in A~.Proof.
It suffices to show that(|Tf|∧g)·x=0 for all 0≤x∈A. Let 0≤x∈A. Since Vx~~∈Orth(A~) we get the following:
(15)(|Tf|∧g)·x=x~~·(|Tf|∧g)=x~~·|Tf|∧x~~·g=|x~~·Tf|∧x~~·g.
There exists G∈(A~)n~ such that x~~·Tf=G·f. Consequently,
(16)(|Tf|∧g)·x=|G·f|∧x~~·g≤G·|f|∧x~~·g≤(G+x~~)·(|f|∧g)=0.We have now gathered all of the ingredients for the proof of the principal theorem of this paper.Theorem 6.
LetA be an f-algebra. Then the following are equivalents:(i)
Orth(A~)=ℒb(A~,A~,(A~)n~)=Orth(A~,A~,(A~)n~);(ii)
J0={0}.Proof.
(i)⇒(ii) Arguing by contradiction, suppose that J0≠{0}. There exists a nonzero positive element h∈A~ such that h(a)=0 for all a∈J. Let 0≤g∈(J0)d and 0≤x∈A such that g(x)≠0. Define the mapping T:A~→A~ by T(f)=f(x)h. It is not hard to see that T is positive and for all F∈(A~)n~, f∈A~ we have the following:
(17)F·Tf=f(x)F·h=0∈R(f).
Thus,T∈Orth(A~,A~,A~). But T is not an orthomorphism since g∧h=0 and Tg∧h=g(x)h∧h≠0.
(ii)⇒(i) In view of Proposition 4, it suffices to show that Orth(A~,A~,(A~)n~)⊆Orth(A~). To do this, pick T∈Orth(A~,A~,(A~)n~) and f,g∈A~ such that f∧g=0. According to Lemma 5,we obtain |Tf|∧g∈J0={0}; that is, |Tf|∧g=0, as required.It follows from Theorem6 that if Orth(A~)=Orth(A~,A~,(A~)n~) then (A~)n~; hence A, is a semiprime f-algebra.We end this section with a consequence of Theorem6. The notion of weak approximate unit plays a key role in the context of this study. Recall that an upward directed net {aτ:τ∈T} of positive elements in an f-algebra A is said to be an approximate unit if sup⁡τaτ·x=x for all 0≤x∈A. An approximate unit {aτ} is called weak approximate unit if sup⁡τf(aτx)=f(x) for all 0≤f∈A~ and 0≤x∈A (see [4, Definitions 2.2 and 7.1]). It is well known that if A is a semi-prime f-algebra then A can be embedded as a ring and f-subalgebra in the unital f-algebra Orth(A).In Theorem6, we gave two necessary and sufficient conditions for which f-orthomorphisms are orthomorphisms. In the following proposition, we shall present a second one in terms of approximate unit. We have to impose, however, on A an extra condition, namely, the Stone condition (i.e., x∧I∈A whenever x is positive in A, where I is the identity of Orth(A)). It should be pointed out here that every uniformly complete semi-prime f-algebra A satisfies the Stone condition.Proposition 7.
Consider in a semi-primef-algebra A the following conditions.Theorem 8.
(i)A has a weak approximate unit.
(ii)Orth(A~)=ℒb(A~,A~,(A~)n~)=Orth(A~,A~,(A~)n~).
(i)⇒(ii) Moreover, if A satisfies the Stone condition then (i)⇔(ii).Proof.
(i)⇒(ii) Assume that (aτ)τ is a weak approximate unit. Let 0≤f∈J0, it follows from f(aτx)=0 for all 0≤x∈A that f(x)=0.
(ii)⇒(i) It follows from Theorem 6, that (A~)n~ is semi-prime. The result follows from [4, Theorem 7.2].


## 3. Whenf-Linear Operators Are Orthomorphisms

In this section, we study the connection betweenf-linear operators and orthomorphisms. From Proposition 4 and Theorem 6 if f-linear operators on the dual A~ of an f-algebra A are orthomorphisms and if J0={0} then the two classes coincide. Next, we give an example of an f-algebra such that J0≠{0} in which f-linear operators are orthomorphisms.Example 9.
LetA=ℝ2 equipped with the coordinatewise operations and ordering. Consider the multiplication defined by (a,b)·(c,d)=(ac,0). It is easy to see that for all f=(f1,f2)∈A~ and F=(F1,F2) we have f·(a,b)=(f1a1,0) and F·f=(F1·f1,0). Note that J0={f=(0,f2):f2=0}. Let T:=(αβγσ) an f-linear operator. An easy computation shows that β=γ=0. Thus, T is an orthomorphism.It seems natural therefore to ask under what condition we haveOrth(A~)=ℒb(A~,A~,(A~)n~). The answer is given in the next theorem. First, let us discuss the ideal of nilpotents elements in the bidual of an f-algebra A. Let F∈A~~ the absolute kernel or null ideal NF of F is defined by the following:
(18)NF={f∈A~:|F|(|f|)=0}.
It is evident that NF is an order ideal. The disjoint complement of CF=NFd is called the carrier of F and is always a band in A~. Given 0≤F∈A~~ such that F·f=0 for all f∈A~ (F is nilpotent in A~~), then (J0)d⊆NF. Indeed, let f∈CF, since CF is a band and Vx~~ is an orthomorphism in A~, then f·x∈CF for all x∈A. But, by hypothesis f·x∈NF, f·x∈NF, so f·x=0 for all x∈A. This implies that f∈J0. Consequently, CF⊆J0 and (J0)d⊆NF. For more information about the nilpotent elements in the bidual of f-algebra, the reader is referred to [4].We are now in position to prove the main theorem in this section.Theorem 10.
LetA be an f-algebra. The followings are equivalents.(i)
Orth(A~)=ℒb(A~,A~,(A~)n~).(ii)
dim⁡J0=1.Proof.
(i)⇒(ii) Suppose that dim⁡J0≥2. We can find therefore positive elements h,k∈J0 such that h∧k=0. Let 0≤a∈A such that k(a)≠0. Define the mapping T:A~→A~ by the following:
(19)T(f)=(P0∘f)(a)·h.
Since F·f∈(J0)d for all F∈(A~)n~ and f∈A~, we get T(F·f)=F·Tf=0. Thus T is an f-linear operator. Clearly, T is not an orthomorphism since h∧k=0 and T(k)∧h=k(a)h∧h≠0.
(ii)⇒(i) Let T be an f-linear mapping on A~ and 0≤h∈J0. First, we show that T((J0)d)⊂(J0)d and T(J0)⊂J0. There exists φ∈A~~ such that P0∘T(f)=φ(f)·h. Since P0∘T is an f-linear operator, we derive that φ(F·f)=0 for all F∈(A~)n~ and f∈A~. That is φ·f=0 for all f∈A~. By the remark above, we get φ(f)=0 for all f∈(J0)d. This implies that T(f)∈(J0)d whenever f∈(J0)d. Now, If f∈J0 then
(20)Tf·x=x~~·Tf=T(x~~·f)=T(f·x)=0∀x∈A.
This shows that T(f)∈J0, in particular T(h)=t·h where t∈ℝ. Now, let f∧g=0 in A~. We have to show that |Tf|∧g=0. Decompose f and g as f=αh+f1 and g=βh+g1 with α,β∈ℝ and f1,g1∈(J0)d. The decomposition of Tf is given by the following:
(21)Tf=αt·h+T(f2).
Since f∧g=0, then α·β=0 and f1∧g1=0. On the other hand, according to Lemma 5, we have |Tf|∧g∈J0. So
(22)|Tf|∧g=(αt∧β)·h=0
as α·β=0. This completes the proof of the theorem.



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