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

and Applied Analysis 3 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. For 0 ≤ 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 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 . 2.1 Hence, Tf is a lattice homomorphism. Next, we show that Tf 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 TF |A x Tf ( x′′ ) x′′ · f f · x. 2.2 Thus, for each F ∈ A∼ n and x ∈ A, we see that (( Tf |A )′ F ) x F (( Tf |A ) x ) F ( f · x F · f x Tf F ) x , 2.3 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. 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 VF 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 ∣F · f∣ ∣F · f ∣ ∣F · f−∣ ∣Tf F ∣∣ ∣Tf− F ∣∣ Tf |F| Tf− |F| |F| · f |F| · f− |F| · ∣f∣. 2.4 4 Abstract and Applied Analysis Let f ⊥ g in A∼. Then we have ∣ ∣F · f∣ ∧ ∣G · g∣ |F| · ∣f∣ ∧ |G| · ∣g∣ ≤ ( |F| |G| · ∣f∣ ∧ ( |F| |G| · ∣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 Tf . 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∼ → 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 π ∈ Orth A∼ , πf ∣f∣, π∣f∣ 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 · ∣f∣ a F · πf a