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

Abstract

We consider the 𝑓-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.

1. Introduction

Let 𝐴 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)=𝑉_𝐹 for all 𝐹∈(𝐴^∼)^∼_𝑛, where 𝑉_𝐹(𝑓)=𝐹⋅𝑓 for every 𝑓∈𝐴^∼, is an algebra and Riesz isomorphism. See [1, 2] for details.

Let 𝐴 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 each 𝑎∈𝐴⁺ and 𝑥∈𝐿⁺, we have 𝑎𝑥∈𝐿⁺,(ii)if 𝑥⟂𝑦, then for each 𝑎∈𝐴, we have 𝑎⋅𝑥⟂𝑦.

When 𝐴 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≤𝑎_𝛼≤𝑒 in 𝐴 such that 𝑎_𝛼⋅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].

Let 𝐴 be a unital 𝑓-algebra, and, 𝐿, 𝑀 be 𝑓-modules over 𝐴. 𝑇∈𝐿_𝑏(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 𝐿_𝑏(𝐿,𝑀;𝐴). For each 𝑥∈𝐿 and 𝑓∈𝐿^∼, we can define 𝜓_𝑥,𝑓∈𝐴^∼ by 𝜓_𝑥,𝑓(𝑎)=𝑓(𝑎⋅𝑥) for all 𝑎∈𝐴. Let 𝑆(𝑥)∶={𝜓_𝑥,𝑓∶𝑓∈𝐿^∼}. Then 𝑆(𝑥) is an order ideal in 𝐴^∼ [2]. 𝑇∈𝐿_𝑏(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)=𝐿_𝑏(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, Orth𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛=𝐿_𝑏𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛. (∗)

An 𝑓-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 the 𝑓-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, and 𝑓-algebras, we refer the reader to the monographs [5–7].

2. 𝑓-Orthomorphisms on the Order Dual

Let 𝐴 be an 𝑓-algebra with separating order dual (and hence 𝐴 Archimedean!) and 𝑓∈𝐴^∼. We consider the mapping 𝑇_𝑓∶(𝐴^∼)^∼_𝑛→𝐴^∼ defined by 𝑇_𝑓(𝐹)=𝐹⋅𝑓 for all 𝐹∈(𝐴^∼)^∼_𝑛. It should be noted that the mapping 𝑉∶(𝐴^∼)^∼_𝑛→Orth(𝐴^∼) defined by 𝑉(𝐹)=𝑉_𝐹 for all 𝐹∈(𝐴^∼)^∼_𝑛, where 𝑉_𝐹(𝑓)=𝐹⋅𝑓 for every 𝑓∈𝐴^∼, is an algebra and Riesz isomorphism (cf. [2, Proposition 2.2]).

Theorem 2.1. For 0≤𝑓∈𝐴^∼,  𝑇_𝑓 is an interval preserving lattice homomorphism.

Proof. Clearly, 𝑇_𝑓 is linear and positive. Since the mapping 𝑉 is a lattice homomorphism and 𝑉_𝐹,𝑉_𝐺∈Orth(𝐴^∼) for 𝐹,𝐺∈(𝐴^∼)^∼_𝑛, we have 𝑇_𝑓(𝐹∨𝐺)=(𝐹∨𝐺)⋅𝑓=𝑉_𝐹∨𝐺𝑓=(𝑉(𝑓=(𝑉(𝐹)(𝐺𝑓=𝑉(𝐹)𝑓∨𝑉(𝐺)𝑓𝐹∨𝐺))∨𝑉))=𝐹⋅𝑓∨𝐺⋅𝑓=𝑇_𝑓(𝐹)∨𝑇_𝑓(𝐺). (2.1) Hence, 𝑇_𝑓 is a lattice homomorphism.
Next, we show that 𝑇_𝑓 is an interval preserving operator. We identify 𝑥 with its canonical image 𝑥^′′ in (𝐴^∼)^∼_𝑛 and denote the restriction of 𝑇_𝑓 to 𝐴 by 𝑇_𝑓∣_𝐴. Then 𝑇_𝐹∣_𝐴(𝑥)=𝑇_𝑓𝑥^′′=𝑥^′′⋅𝑓=𝑓⋅𝑥. (2.2) Thus, for each 𝐹∈(𝐴^∼)^∼_𝑛 and 𝑥∈𝐴, we see that 𝑇_𝑓|_𝐴^′(𝐹)(𝑥)=𝐹𝑇_𝑓|_𝐴(𝑥)=𝐹𝑓⋅𝑥=𝐹⋅𝑓(𝑥)=𝑇_𝑓(𝐹)(𝑥), (2.3) which implies that (𝑇_𝑓|_𝐴)′ is the same as 𝑇_𝑓 on (𝐴^∼)^∼_𝑛. Since (𝑇_𝑓|_𝐴)′ is interval preserving (cf. [5, Theorem 7.8]), 𝑇_𝑓 is likewise an interval preserving operator.

Corollary 2.2. For 𝑓∈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. Since 𝑉_𝐹 is an orthomorphism on 𝐴^∼, we have 𝑉_𝐹(𝑓⁺)⟂𝑉_𝐹(𝑓⁻) for each 𝑓∈𝐴^∼, that is, 𝐹⋅(𝑓⁺)⟂𝐹⋅(𝑓⁻). From Theorem 2.1, we know that 𝐹⋅𝑓=𝐹⋅𝑓⁺+𝐹⋅𝑓⁻=𝑇_𝑓⁺(𝐹)+𝑇_𝑓⁻(𝐹)=𝑇_𝑓⁺(|𝐹|)+𝑇_𝑓⁻𝐹=|𝐹|(||)⋅𝑓⁺+|𝐹|⋅𝑓⁻=|𝐹|⋅𝑓. (2.4) Let 𝑓⟂𝑔 in 𝐴^∼. Then we have ∧=|𝐹|⋅𝑓∧|𝐺|⋅𝑔≤𝐹|+|𝐺⋅𝑓∧𝐹|+|𝐺⋅𝑔𝐹⋅𝑓𝐺⋅𝑔(||)(||)=0, (2.5) which implies that 𝐹⋅𝑓⟂𝐺⋅𝑔 for all 𝐹,𝐺∈(𝐴^∼)^∼_𝑛.

Following the above discussion, we now consider 𝑅(𝑓)={𝐹⋅𝑓∶𝐹∈(𝐴^∼)^∼_𝑛}, the image of (𝐴^∼)^∼_𝑛 under 𝑇_𝑓.

Corollary 2.3. If 𝐴 is an 𝑓-algebra and 𝑓∈(𝐴^∼), then 𝑅(𝑓)=𝑅(|𝑓|), and 𝑅(𝑓) is an order ideal in 𝐴^∼.

Proof. First, since 𝑇_|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 that 𝑅(|𝑓|)⊆𝑅(𝑓). To this end, let 𝑃₁∶𝐴^∼→𝐵_𝑓⁺, 𝑃₂∶𝐴^∼→𝐵_𝑓⁻ be band projections, where 𝐵_𝑓⁺ and 𝐵_𝑓⁻ are the bands generated by 𝑓⁺ and 𝑓⁻ in 𝐴^∼, respectively. If 𝜋=𝑃₁−𝑃₂, we have 𝜋∈Orth(𝐴^∼),𝜋𝑓=𝑓,𝜋𝑓=𝑓. (2.6) 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 𝑓(𝑎)=𝑓(𝑎)𝜋𝑓𝜋=𝜋𝐹⋅𝐹⋅𝜋=𝐹⋅𝑎=𝐹𝑓⋅𝑎_′(𝐹)⋅𝑓(𝑎), (2.7) 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 for 𝑅(𝑓)⟂𝑅(𝑔) when 𝐴 has the factorization property. First, we need the following lemma.

Lemma 2.4. Let 𝐴 be an 𝑓-algebra with the factorization property, and 𝑓∈𝐴^∼. If 𝑓⋅𝑥=0 for each 𝑥∈𝐴, then 𝑓=0.

Proof. Since 𝐴 has the factorization property, for each 𝑎∈𝐴⁺, there exist 𝑥,𝑦∈𝐴 such that 𝑎=𝑥𝑦. Hence, from 𝑓(𝑎)=𝑓𝑥𝑦=𝑓⋅𝑥𝑦=0, (2.8) it follows easily that 𝑓=0 holds.

Theorem 2.5. Let 𝐴 be an 𝑓-algebra with the factorization property. If 𝑓,𝑔∈𝐴^∼, then 𝑓⟂𝑔 if and only if 𝑅(𝑓)⟂𝑅(𝑔).

Proof. If 𝑓⟂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, if 𝑅(𝑓) and 𝑅(𝑔) are disjoint, then for each 𝐹∈((𝐴^∼)^∼_𝑛)⁺ we have 𝐹⋅𝑓∧𝑔=𝑉_𝐹𝑓∧𝑔=𝑉_𝐹𝑓∧𝑉_𝐹𝑔𝑓𝑔=∧=𝐹⋅∧𝐹⋅𝐹⋅𝑓𝐹⋅𝑔=0. (2.9) 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-called 𝑓-orthomorphism.

Definition 2.6. Let 𝐴 be an 𝑓-algebra and 𝑇∈𝐿_𝑏(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 the 𝑓-orthomorphisms and the orthomorphisms on the order dual of an 𝑓-algebra with the factorization property. Note that Orth(𝐴^∼) is a band in 𝐿_𝑏(𝐴^∼).

Theorem 2.7. Let 𝐴 be an 𝑓-algebra. Then Orth(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛) is a linear subspace of 𝐿_𝑏(𝐴^∼) and Orth(𝐴^∼)⊆Orth(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛).
If 𝐴, in addition, has the factorization property, then Orth(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛)=Orth(𝐴^∼).

Proof. First, we can easily see that Orth(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛) is a linear subspace of 𝐿_𝑏(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 𝜋^′(𝐹)⋅𝑓(𝑥)=𝜋^′(𝐹)𝑓⋅𝑥=𝐹𝜋𝑓⋅𝑥=𝐹𝜋𝑥^′′𝑥⋅𝑓=𝐹^′′⋅𝜋𝑓=𝐹⋅𝑥^′′𝜋𝑓=𝑥^′′⋅𝐹𝜋𝑓=𝑥^′′𝐹⋅𝜋𝑓=𝐹⋅𝜋𝑓(𝑥). (2.10) 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).
If 𝐴 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

Let 𝐴 be an 𝑓-algebra with separating order dual and 𝑇∈𝐿_𝑏(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 𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛). It follows from [3, Lemma 4.4] that 𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛) is a band in 𝐿_𝑏(𝐴^∼).

Theorem 3.1. Let 𝐴 be an 𝑓-algebra with separating order dual. Then Orth(𝐴^∼)⊆𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛).

Proof. Clearly Orth(𝐴^∼) is commutative since Orth(𝐴^∼) is an Archimedean 𝑓-algebra. To complete the proof, let 𝜋∈Orth(𝐴^∼). We have 𝜋𝐺⋅𝑓=𝜋𝑉_𝐺𝑓=𝑉_𝐺𝜋𝑓=𝐺⋅𝜋𝑓, (3.1) for all 𝑓∈𝐴^∼ and 𝐺∈(𝐴^∼)^∼_𝑛. Hence, 𝜋∈𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛).

The following result deals with the order adjoint of an 𝑓-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. Let 𝑇∈𝐿_𝑏(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. Since 𝑇∈𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛), and (𝐴^∼)^∼_𝑛 is a commutative 𝑓-algebra, we have 𝑇^′(𝐹)⋅𝑓(𝑥)=𝑇^′(𝐹)𝑇𝑇𝑥𝑓⋅𝑥=𝐹𝑓⋅𝑥=𝐹^′′𝑥⋅𝑓=𝐹^′′⋅𝑇𝑓=𝐹⋅𝑥^′′𝑇𝑓=𝑥^′′⋅𝐹𝑇𝑓=𝑥^′′(𝐹⋅𝑇(𝐹))=𝐹⋅𝑇𝑓(𝑥), (3.2) for all 𝐹∈(𝐴^∼)^∼_𝑛, 𝑓∈𝐴^∼, and 𝑥∈𝐴, which implies that 𝑇^′(𝐹)⋅𝑓=𝐹⋅𝑇(𝑓).
Let 𝐹,𝐺∈(𝐴^∼)^∼_𝑛 be given. Then for 𝑓∈𝐴^∼, from 𝐺⋅𝑇^′(𝐹)𝑓=𝐺𝑇^′(𝐹)𝑓=()𝑇𝑓=𝑇⋅𝑓=𝐺𝐹⋅𝑇𝐺⋅𝐹_′(𝐺⋅𝐹)𝑓, (3.3) it follows that 𝐺⋅𝑇^′(𝐹)=𝑇^′(𝐺⋅𝐹). This completes the proof.

Theorem 3.3. 𝐿_𝑏𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛⊆Orth𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛. (3.4)

Proof. For 𝑇∈𝐿_𝑏(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 0≤𝐺⋅𝑇𝑓≤𝐺⋅|𝑇|𝑓=|𝑇|^′(𝐺)⋅𝑓=𝑇_|𝑓||𝑇|^′(𝐺). (3.5) 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, 𝐿_𝑏(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛)⊆Orth(𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛), as desired.

Combining Theorems 3.1, 3.3, and 2.7, we have the following result.

Theorem 3.4. If 𝐴 is an 𝑓-algebra with separating order dual, then Orth(𝐴^∼)⊆𝐿_𝑏𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛⊆Orth𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛. (3.6) In particular, if, in addition, 𝐴 has the factorization property, then Orth(𝐴^∼)=𝐿_𝑏𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛=Orth𝐴^∼,𝐴^∼;(𝐴^∼)^∼_𝑛. (3.7)

Funding

The authors were supported in part by the Fundamental Research funds for the Central Universities (SWJTU11CX154, SWJTU12ZT13).

Acknowledgment

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 “𝐴 has the factorization property,” which is weaker than the hypothesis “𝐴 is square-root closed” used originally in this paper.