Available online at www.sciencedirect.com ScienceDirect Indagationes Mathematicae ( ) – www.elsevier.com/locate/indag Multiplicative structure of biorthomorphisms and embedding of orthomorphisms✩ Karim Boulabiar, Wael Brahmi ∗ Research Laboratory of Algebra, Topology, Arithmetic, and Order, Department of Mathematics, Faculty of Mathematical, Physical and Natural Sciences of Tunis, Tunis-El Manar University, 2092-El Manar, Tunisia Received 25 February 2015; received in revised form 23 December 2015; accepted 25 January 2016 Communicated by B. de Pagter Abstract Let X be an Archimedean vector lattice. A biorthomorphism on X is a bilinear map from X × X into X which is an orthomorphism on X in each variable separately. The set of such biorthomorphisms is denoted by Orth (X , X ). We prove that if Orth (X , X ) is not trivial then Orth (X , X ) is equipped with a structure of f -algebra, giving thus a complete answer to a question asked quite recently by Buskes, Page, and Yilmaz. On the other hand, we assume that X is a semiprime f -algebra and we show that if X is either Dedekind-complete or uniformly-complete with a weak order unit, then the set of all orthomorphisms on X has an order ideal copy in Orth (X , X ). Notice that the Dedekind-complete case has been obtained again by Buskes, Page, and Yilmaz in a completely different way. c 2016 Published by Elsevier B.V. on behalf of Royal Dutch Mathematical Society (KWG). ⃝ Keywords: Biorthomorphism; Dedekind-complete; f -algebra; Order ideal; Orthomorphism; Uniformly complete; Vector lattice; Weak order unit 1. Introduction Let X be an Archimedean vector lattice over the field of all real numbers R. By a biorthomorphism on X we mean a bilinear map from X × X into X which is an orthomorphism ✩ This research is supported by Research Laboratory LATAO Grant LR11ES12. ∗ Corresponding author. E-mail addresses: karim.boulabiar@ipest.rnu.tn (K. Boulabiar), brahmiwael@yahoo.com (W. Brahmi). http://dx.doi.org/10.1016/j.indag.2016.01.010 c 2016 Published by Elsevier B.V. on behalf of Royal Dutch Mathematical Society (KWG). 0019-3577/⃝ 2 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – on X in each variable separately. The set of all biorthomorphisms on X is denoted by Orth (X , X ). Biorthomorphisms have been introduced in some form or other in [13] by Yilmaz and Rowlands. However, their first systematic investigation appeared more recently in [4] by Buskes, Page, and Yilmaz. In this regard, they proved that Orth (X , X ) is again a vector lattice with respect to the pointwise addition and ordering (see [3] for a lattice-ordered group version). This fact motivated them to ask the following question (see Question 3.13 [4]). Problem 1. When is Orth (X , X ) an f -algebra? Pointing out that the problem is quite unclear, they provided a partial solution, imposing some extra completeness and approximation conditions (for more details, see [4, Proposition 3.14]). The main objective of this paper is to give a complete answer to the above question. Surprisingly enough, we will prove that Orth (X , X ) can always be equipped with a non-trivial f -algebra structure, unless Orth (X , X ) itself is trivial. Again in the aforementioned paper [4, Proposition 3.14], it is shown that if X is a semiprime f -algebra, then the f -algebra Orth (X ) of all orthomorphisms on X can be embedded in Orth (X , X ) as a vector sublattice. This leads the authors to look for conditions on X for Orth (X ) to be an order ideal in Orth (X , X ). In this prospect, they proved that if X is in addition Dedekind-complete, then Orth (X ) has an order ideal copy in Orth (X , X ). As we shall see next, we will obtain this result in an alternative way. In spite of that, we shall give a completely different condition under which the result remains valid. A short account of the content of this paper seems now to be in order. Let X be a non-trivial f -algebra. For every e ∈ X with 0 < e, we introduce a multiplication ∗e on Orth (X , X ) by putting ( f ∗e g) (x, y) = f (x, g (e, y)) for all f, g ∈ Orth (X , X ) and x, y ∈ X . We prove that ∗e is well-defined and that Orth (X , X ) is an f -algebra with respect to ∗e . It turns out that there exists at least one element e > 0 such that the f -algebra Orth (X , X ) is non-trivial. This answers completely the above Problem 1. Moreover, we show that if e is a weak order unit in X , then Orth (X , X ) is semiprime. This allows us to prove that if X is a uniformly complete f -algebra with a weak order unit, then Orth (X ) can be embedded as an order ideal in Orth (X , X ). The proof of the result is based upon a representation theorem of biorthomorphisms on uniformly complete semiprime f -algebras. Indeed, we shall prove that if X is such an f -algebra, then for every biorthomorphism A on X there exists a unique orthomorphism f from some vector sublattice of X into X such that A (x, y) = f (x y) for all x, y ∈ X . Finally, we take it for granted that the reader is familiar with the notions of vector lattices, order ideals, and order bounded operators. For unexplained terminology and notations we refer the reader to the classical books [1] and [12]. 2. Preliminaries Throughout this section, X stands for an Archimedean vector lattice over R. The first paragraph of this preliminaries section is devoted to the notion of biorthomorphisms on X as introduced by Buskes, Page, and Yilmaz in [4]. This concept relies heavily on the theory of orthomorphisms. Our main reference in this regard is the Chapter 20 in the classical book [14] by Zaanen. However, the reader can alternatively consult the monograph [11] by Luxemburg. An order bounded operator f : X → X is called an orthomorphism on X if | f (x)| ∧ |y| = 0 for all x, y ∈ X with |x| ∧ |y| = 0. K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – 3 The set Orth (X ) of all orthomorphisms on X is an Archimedean vector lattice. The lattice operations in Orth (X ) are given pointwise. Namely, if f, g ∈ Orth (X ) and x ∈ X + then and ( f ∧ g) (x) = f (x) ∧ g (x) ( f ∨ g) (x) = f (x) ∨ g (x) . Moreover, orthomorphisms on X commute (see, for instance, Theorem 4.13 in [11]). Following Buskes and van Rooij in [5], we will call the restriction of any orthomorphism on a vector sublattice Y of X an orthomorphism from Y into X . For a bilinear map A : X × X → X and x ∈ X , two operators A (x, .) : X → X and A (., x) : X → X can be defined in a natural way by putting A (x, .) (y) = A (x, y) and A (., x) (y) = A (y, x) for all y ∈ X . The bilinear map A : X × X → X is called a biorthomorphism on X if A (x, .) ∈ Orth (X ) and A (., x) ∈ Orth (X ) for all x ∈ X . The set Orth (X , X ) of all biorthomorphisms on X is again an Archimedean vector lattice with respect to the pointwise addition and multiplication. Hence, A ∈ Orth (X , X ) is positive if and only if A (x, y) ∈ X + for all x, y ∈ X + . By the way, the bilinear map A : X × X → X is said to be positive if A (x, y) ∈ X + for all x ∈ X + . Moreover, the lattice operations in Orth (X , X ) are given again pointwise. More precisely, if A, B ∈ Orth (X , X ) and x, y ∈ X + then (A ∧ B) (x, y) = A (x, y) ∧ B (x, y) and (A ∨ B) (x, y) = A (x, y) ∨ B (x, y) . In particular, A+ (x, y) = A (x, y)+ and A− (x, y) = A (x, y)− for all x, y ∈ X + (see [4] for proofs). Next, we collect some useful properties of biorthomorphisms on Archimedean vector lattices. First, recall that if A : X × X → X is a bilinear map then A is said to be orthosymmetric if |x| ∧ |y| = 0 implies A (x, y) = 0. Buskes and Van Rooij proved in [6] that if A is positive and orthosymmetric then A is symmetric, that is, A (x, y) = A (y, x) for all x, y ∈ X . On the other hand, any biorthomorphism on X is clearly an orthosymmetric map. Hence, it is symmetric being in fact the difference of two positive biorthomorphisms. Moreover, A is said to be separately disjointness preserving if, for every x ∈ X , the operators A (x, .) and A (., x) preserve disjointness. This means that if y, z ∈ X with |y| ∧ |z| = 0, then |A (x, y)| ∧ |A (x, z)| = |A (y, x)| ∧ |A (z, x)| = 0. 4 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – Any positive separately disjointness preserving A : X × X → X is called a lattice (or, Riesz) bimorphism (see [8]). Now, it is not hard to see that if A ∈ Orth (X , X ) then A is simultaneously orthosymmetric and separately disjointness preserving (in particular, any positive biorthomorphism is a lattice bimorphism). Therefore, if A ∈ Orth (X , X ) and x ∈ X then   A x + , x − = 0. So, A (x, x) = A (|x| , |x|) for all x ∈ X . (1) This together with the following equalities will be needed for later calculations. Indeed, if A, B ∈ Orth (X , X ) and x, y, z ∈ X then A (B (x, y) , z) = A (x, B (y, z)) = B (x, A (y, z)) . (2) Actually, these equalities can be obtained quite easily by a suitable use of the commutativity of orthomorphisms. The next lines deal with the notion of f -algebras. We follow by and large the notation and terminology of [1] and [14]. Assume that X is an associative algebra over R. We call X a lattice-order algebra if X is simultaneously a vector lattice such that the ordering and the multiplication are compatible, that is, xy ∈ X+ for all x, y ∈ X + . The lattice-ordered algebra X is called an f -algebra if X satisfies the extra condition that x ∧ y = 0 and 0 ≤ z imply (x z) ∧ y = (zx) ∧ y = 0. For instance, if X is an Archimedean vector lattice, then Orth (X ) is an f -algebra with the composition as multiplication (see Theorem 2.59 in [1]). The f -algebra X is said to be trivial if xy = 0 for all x, y ∈ X. Otherwise, we speak about a non-trivial f -algebra. Any Archimedean f -algebra is commutative and has positive squares. Actually, |x|2 = x 2 and |x y| = |x| |y| for all x, y ∈ X. The set of all nilpotent elements of an Archimedean f -algebra X is denoted by N (X ). It is well-known that   N (X ) = x ∈ X : x 2 = 0 . If the equality N (X ) = {0} holds, then X is said to be semiprime. Hence, X is semiprime if and only if X has no nonzero nilpotent elements. Furthermore, if X is a semiprime f -algebra and x, y ∈ X then xy = 0 if and only if |x| ∧ |y| = 0. If X is a semiprime f -algebra and x, y ∈ X with zx = zy for all z ∈ X , then x = y. For, (x − y) x = (x − y) y and so (x − y)2 = 0. Since X is semiprime, we get x − y = 0 and thus x = y. K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – 5 We end by emphasizing that most of the content of this preliminary section will be used systematically throughout the paper, that is, without further mention. 3. Multiplicative structure of biorthomorphisms In this section, we give our complete answer to the main question around which this paper is developed (see Problem 1). First of all, recall that if X is an Archimedean vector lattice then Orth (X , X ) is again an Archimedean vector lattice, the lattice operations of which are given pointwise. Theorem 2. Let X be an Archimedean vector lattice. If e ∈ X + , then Orth (X , X ) is an Archimedean f -algebra with respect to the multiplication defined by (A ∗e B) (x, y) = A (x, B (e, y)) for all A, B ∈ Orth (X , X ) and x, y ∈ X . Proof. Choose e ∈ X + and observe that if A, B ∈ Orth (X , X ) then the map A ∗e B : X ×X → X defined by (A ∗e B) (x, y) = A (x, B (e, y)) for all x, y ∈ X is bilinear. Moreover, if x ∈ X then (A ∗e B) (x, .) = A (x, .) ◦ B (e, .) ∈ Orth (X ) and (A ∗e B) (., x) = A (., B (e, x)) ∈ Orth (X ) . It follows that A ∗e B ∈ Orth (X , X ) . Furthermore, it is clear that the map Te : Orth (X , X ) × Orth (X , X ) → Orth (X , X ) given by Te (A, B) = A ∗e B for all A, B ∈ Orth (X , X ) is a positive bilinear map. On the other hand, if A, B, C ∈ Orth (X , X ) and x, y ∈ X then ((A ∗e B) ∗e C) (x, y) = (A ∗e B) (x, C (e, y)) = A (x, B (e, C (e, y))) = A (x, (B ∗e C) (e, y)) = A ∗e (B ∗e C) (x, y) . Accordingly, Orth (X , X ) is a lattice-ordered algebra with respect to the multiplication ∗e . From the fact that biorthomorphisms are symmetric, it follows that the lattice-ordered algebra Orth (X , X ) is commutative. Pick A, B, C ∈ Orth (X , X ) with A ∧ B = 0 and 0 ≤ C. We claim that (A ∗e C) ∧ B = (C ∗e A) ∧ B = 0. To this end, let x, y ∈ X + and observe that from A (x, y) ∧ B (x, y) = 0 it follows that C (e, A (x, y)) ∧ B (x, y) = 0 6 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – (where we use that C (e, .) ∈ Orth (X )). However, using the symmetry of biorthomorphisms and the equalities (2), we get C (e, A (x, y)) = C (x, A (e, y)) = (C ∗e A) (x, y) . Thus, (C ∗e A) (x, y) ∧ B (x, y) = 0 and so (C ∗e A) ∧ B = 0. Since ∗e is commutative (as observed above), we derive that (A ∗e C) ∧ B = 0. This shows that ∗e makes Orth (X , X ) an f -algebra and completes the proof of the theorem.  For every A ∈ Orth (X , X ), put K (A) = {x ∈ X : A (x, x) = 0} . The following lemma will be useful for later purposes. Lemma 3. Let X be an Archimedean vector lattice and A ∈ Orth (X , X ). Then K (A) = {x ∈ X : A (x, y) = 0 for all y ∈ X } . In particular, A is an order ideal in X . Proof. Let x ∈ K (A) and y ∈ X . For the sake of convenience, we put u = |x| and v = |y| . From A (x, x) = 0 and (1) it follows that A+ (u, u) = A+ (x, x) = A− (x, x) = A− (u, u) . Thus,   A+ (u, u) = A+ (u, u) ∧ A− (u, u) = A+ ∧ A− (u, u) = 0. Therefore, A+ (u, u) = A− (u, u) = 0. We claim that A+ (u, v) = 0. To this end, pick n ∈ {1, 2, . . .} and observe that 0 ≤ A+ (nu − v, nu − v) = n 2 A+ (u, u) − 2n A+ (u, v) + A+ (v, v) = A+ (v, v) − 2n A+ (u, v) . Whence, 0 ≤ 2n A+ (u, v) ≤ A+ (v, v) for all n ∈ {1, 2, . . .} . Since X is Archimedean, we obtain A+ (u, v) = 0, as required. Using the lattice operations in Orth (X , X ), we get  +   A (x, y) = A+ (u, v) = 0. K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – 7 Hence, A+ (x, y) = 0 and, analogously, A− (x, y) = 0. It follows that A (x, y) = 0. Now, we prove that K (A) is an order ideal in X . By the first part, we derive quickly that K (A) is a vector subspace of X . Moreover, from (1) it follows that K (A) is a vector sublattice of X . It remains to show that K (A) is a solid in X . Let x, y ∈ X with 0≤x ≤y and y ∈ K (A) . Since A (x, .) ∈ Orth (X ), it is an order bounded disjointness preserving operator on X . By Proposition 1.2 in [9], we get 0 ≤ |A (x, x)| ≤ |A (x, y)| = 0. Consequently, x ∈ K (A) and the lemma follows.  The next lemma is also a key tool for the proof of our main result. Lemma 4. Let X be an Archimedean vector lattice and A ∈ Orth (X , X ). If x ∈ X then A (x, x) = 0 if and only if A (x, x) ∈ K (A). Proof. It is obvious that K (A) contains 0. Conversely, suppose that A (x, x) ∈ K (A). In view of (1) and Lemma 3, we may assume that x and A are positive. Therefore, if n ∈ {1, 2, . . .} then +  (n A (x, x) − x)+ ∧ nx − n 2 A (x, x) = 0. Since A (x, .) ∈ Orth (X ), we obtain   +  = 0. (n A (x, x) − x)+ ∧ A x, nx − n 2 A (x, x) Moreover, by Lemma 3, we have A (x, A (x, x)) = 0. Whence, (n A (x, x) − x)+ = (n A (x, x) − x)+ ∧ n A (x, x) = (n A (x, x) − x)+ ∧ n A (x, x)+ +   = (n A (x, x) − x)+ ∧ A x, nx − n 2 A (x, x)   +  + 2 = 0. = (n A (x, x) − x) ∧ A x, nx − n A (x, x) Therefore, 0 ≤ n A (x, x) ≤ x. But then A (x, x) = 0 as X is Archimedean. The proof is complete.  We are in position at this point to prove the central result of this paper. 8 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – Theorem 5. Let X be an Archimedean vector lattice. Then the following are equivalent. (i) Orth (X , X ) ̸= {0}. (ii) X is a non-trivial f -algebra. (iii) Orth (X , X ) is a non-trivial f -algebra. Proof. (i) ⇒ (iii) Assume by contradiction that the only f -algebra multiplication in Orth (X , X ) is trivial. Choose x ∈ X + and observe that ∗x is an f -algebra multiplication in Orth (X , X ) (where we use Theorem 2). Hence, A ∗x A = 0 for all A ∈ Orth (X , X ) . Pick A ∈ Orth (X , X ) and observe that A (A (x, x) , A (x, x)) = (A ∗x A) (A (x, x) , x) = 0. It follows that A (x, x) ∈ K (A) . Using Lemmas 3 and 4, we derive that A = 0. This yields that Orth (X , X ) = {0}, which contradicts (i) and completes the proof of (iii). (iii) ⇒ (ii) Since Orth (X , X ) is a nonzero vector lattice, there exists a positive biorthomorphism A on X such that A ̸= 0. But then the multiplication in X defined by x y = A (x, y) for all x, y ∈ X makes X a non-trivial f -algebra. Indeed, if x, y, z ∈ X then from (2) it follows that x (yz) = A (x, A (y, z)) = A (A (x, y) , z) = (x y) z. This means that the multiplication defined above is associative. The other algebraic properties of an f -algebra are readily verified. (ii) ⇒ (i) Suppose that X is a non-trivial f -algebra and define a map A : X × X → X by A (x, y) = x y for all x, y ∈ X . Obviously, A is a nonzero biorthomorphism on X . This finishes the proof of the theorem.  As a matter of fact, Theorem 5 eliminates any reason for considering biorthomorphisms on other than non-trivial f -algebras. Still, there exist Archimedean vector lattices with only zero biorthomorphisms. An example in this direction follows. Example 6. Let X be the Archimedean vector lattice of all real-valued piecewise linear continuous functions on the real interval [0, 1]. Using Theorem 6 in [15], we have Orth (X ) = {r I : r ∈ (−∞, ∞)} , where I is the identity operator on X . So, let A ∈ Orth (X , X ) and observe that if x ∈ X then there exists rr ∈ (−∞, ∞) such that A (x, .) = r x I. Pick x ∈ X with x ̸= 0 and choose y ∈ X such that x and y are linearly independent. We now have r x y = A (x, .) (y) = A (x, y) = A (y, x) = r y x. We derive that r x = 0. Thus A (x, .) = 0 and so 0 is the only biorthomorphism on X . K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – 9 According to Theorem 5, the vector lattice in Example 6 cannot be endowed with a non-trivial structure of f -algebra. We end this section with the following result, which will turn out to be crucial for a proof of one of the central results of this work. Corollary 7. Let X be an Archimedean vector lattice with a weak order unit 0 < e. Then the vector lattice Orth (X , X ) is a semiprime f -algebra with respect to the multiplication ∗e . Proof. We already know, by Theorem 2, that Orth (X , X ) is an f -algebra with respect to the multiplication ∗e . Let us see that it is semiprime. For, choose A ∈ Orth (X , X ) such that A ∗e A = 0. Thus, A (x, A (e, y)) = 0 for all x, y ∈ X . This is equivalent to saying that A (e, y) ∈ K (A) for all y ∈ X . By Lemma 4, we derive that A (e, y) = 0 for all y ∈ X . But then A (., y) = 0 for every y ∈ X because e is a weak order unit in X and A (., y) ∈ Orth (X ) (see, for instance, [14, Corollary 140.6]). This means that A = 0 and the corollary follows.  4. Embedding of orthomorphisms in biorthomorphisms Let X be a semiprime f -algebra. Hence, X enjoys a useful property, namely, if f ∈ Orth (X ) then x f (y) = f (x y) = y f (x) for all x, y ∈ X . (3) This follows directly from (2). Moreover, it is not hard to see that if f ∈ Orth (X ) then the map  f : X × X → X defined by  f (x, y) = f (x y) for all x, y ∈ X is a biorthomorphism on X . It follows that a map I : Orth (X ) → Orth (X , X ) can be defined by I(f) =  f for all f ∈ Orth (X ) . One may check easily that I is a one-to-one lattice homomorphism (see Proposition 3.11(i) in [4]). Accordingly, Orth (X ) can be embedded as a vector sublattice of Orth (X , X ) and so any f ∈ Orth (X ) could be identified with its image  f ∈ Orth (X , X ). The main purpose of this section is to look for conditions on X under which Orth (X ) becomes an order ideal in Orth (X , X ). This problem was first considered by Buskes, Page, and Yilmaz in [4]. They proved in this direction that Dedekind-completeness is indeed a sufficient condition on X for Orth (X ) to be an order ideal in Orth (X , X ). In the next theorem, we show how this result can be obtained alternatively as a consequence of a representation theorem for biorthomorphisms. Let us proceed to the details. Suppose that X is in addition uniformly complete and put X ⊙ = {x y : x, y ∈ X } . 10 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – ([12] contains an exhaustive study of uniformly complete vector lattices.) It follows quickly from   Lemma 8 in [6] that X ⊙ is a vector sublattice of X with the set x 2 : x ∈ X as positive cone. The representation theorem we were talking about is the following. Theorem 8. Let X be a uniformly complete semiprime f -algebra and A : X × X → X be a map. Then A is a (positive) biorthomorphism A on X if and only if there exists a unique (positive) orthomorphism A⊙ : X ⊙ → X such that A (x, y) = A⊙ (x y) for all x, y ∈ X . Proof. The ‘if’ part being obvious, let us prove the ‘only if’ part. First, assume that A is a positive biorthomorphism on X . In particular, A is an orthosymmetric lattice bimorphism. Define another orthosymmetric lattice bimorphism ⊙ : X × X → X by putting ⊙ (x, y) = x y for all x, y ∈ X . By Theorem 8 in [7], the vector lattice X ⊙ equipped with ⊙ is in fact the square of X as introduced by Buskes and Van Rooij in the same paper (see Definition 3). So, there exists a unique lattice homomorphism A⊙ : X ⊙ → X such that A (x, y) = A⊙ (x y) for all x, y ∈ X . We claim that A⊙ is an orthomorphism. To this end, let x ∈ X and v ∈ X ⊙ such that |x|∧|v| = 0. As A is orthosymmetric, we obtain A (x, v) = 0. On the other hand, there exists y, z ∈ X such that v = yz. By (3), we have x A (y, z) = A (x y, z) = y A (x, z) = A (x, yz) = A (x, v) = 0. But then     |x| ∧  A⊙ (v) = |x| ∧  A⊙ (yz) = |x| ∧ |A (y, z)| = 0 because X is semiprime. This shows that A⊙ is a positive orthomorphism, as desired. Now, suppose that A is an arbitrary biorthomorphism on X . Since Orth (X , X ) is a vector lattice, there exist two positive biorthomorphisms on X , say B and C, such that A = B −C. Using the positive case, there exist two positive orthomorphisms B ⊙ , C ⊙ : X ⊙ → X such that B (x, y) = B ⊙ (x y) and C (x, y) = C ⊙ (x y) for all x, y ∈ X. Hence, if x, y ∈ X then   A (x, y) = B (x, y) − C (x, y) = B ⊙ (x y) − C ⊙ (x y) = B ⊙ − C ⊙ (x y) . Clearly, B ⊙ − C ⊙ is an orthomorphism from X ⊙ into X , which completes the proof.  We are in position to give an alternative proof of [4, Proposition 3.11 (ii)] by Buskes, Page, and Yilmaz. Corollary 9 (Buskes–Page–Yilmaz). Let X be a Dedekind-complete semiprime f -algebra. Then Orth (X ) is an order ideal in Orth (X , X ). Proof. Since Orth (X ) is a vector sublattice of Orth (X , X ), it suffices to prove that Orth (X ) is a solid in Orth (X , X ). So, let A ∈ Orth (X , X ) and assume that the inequalities 0 ≤ A ≤ f hold K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – 11 for some f ∈ Orth (X ). We claim that A ∈ Orth (X ). To this end, notice that X is uniformlycomplete. So, by Theorem 8, there exists a unique positive orthomorphism A⊙ : X ⊙ → X such that A (x, y) = A⊙ (x y) for all x, y ∈ X . Thus, if x ∈ X then     0 ≤ A⊙ x 2 = A (x, x) ≤ f x 2 . In other words, 0 ≤ A⊙ (v) ≤ f (v) for all v ∈ X ⊙ positive. This together with Theorem 1.26 in [1] yields that A⊙ extends to a positive operator, denoted again by A⊙ , such that 0 ≤ A⊙ ≤ f.  ⊙ = A. Consequently, From these inequalities it follows that A⊙ ∈ Orth (X ) and, clearly, A A ∈ Orth (X ), which gives the conclusion of the theorem.  Now, we intend to prove that the result holds under a completely different condition on X . To do this, we need the following proposition, which seems interesting in its own right. Proposition 10. If the vector lattice X is uniformly-complete, then so is the Orth (X , X ). Proof. Some steps of the proof are routine and technical in nature and then will be omitted. Recall that the vector lattice Orth (X ) is again uniformly complete (see, for instance, Exercise 140.12 in [14]). Consider a uniformly Cauchy sequence (An ) in Orth (X , X ). There exists a positive biorthomorphism B on X such that, for every ϵ ∈ (0, ∞) there is a natural number Nϵ for which    A p − Aq  ≤ ϵ B for all p, q ∈ {Nϵ , Nϵ + 1, . . .} . Choose ϵ ∈ (0, ∞) and p, q ∈ {Nϵ , Nϵ + 1, . . .} such that the inequality    A p (x, .) − Aq (x, .) ≤ ϵ B (x, .) holds in Orth (X ) for any x ∈ X + . In other words, if x ∈ X + then the sequence (An (x, .)) is a uniformly Cauchy sequence in Orth (X ) and so it converges to an orthomorphism f x on X . It is not hard to see that f x (y) = f y (x) for all x, y ∈ X + . Define a map A : X + × X + → X by A (x, y) = f x (y) for all x, y ∈ X + . By standard arguments (as in the proof of the classical Kantorovich Theorem 1.10 in [1]), f extends uniquely to a bilinear map from X × X into X , denoted again by A. Now, it is quite straightforward to check that A ∈ Orth (X , X ) and that A is the uniform limit of the sequence (An ). Consequently, Orth (X , X ) is uniformly-complete.  12 K. Boulabiar, W. Brahmi / Indagationes Mathematicae ( ) – We arrive to the last result of this paper, in which we prove that Orth (X ) is an order ideal in Orth (X , X ) whenever the semiprime f -algebra X is uniformly-complete and has a weak order unit. By the way, it would be convenient for the reader to keep the paper [10] by Huijsmans and de Pagter handy. Theorem 11. Let X be a uniformly-complete semiprime f -algebra with a weak order unit. Then Orth (X ) is an order ideal in Orth (X , X ). Proof. First, keep in mind that Orth (X , X ) is a uniformly-complete semiprime f -algebra (where we use Proposition 10). Now, let e denote a positive weak order unit in X . According to Corollary 7, Orth (X , X ) is a semiprime f -algebra with respect to the multiplication ∗e . Let us see that Orth (X ) is a ring ideal in Orth (X , X ). Indeed, choose f ∈ Orth (X ) and A ∈ Orth (X , X ). From f ◦ A (e, .) ∈ Orth (X ) and (3) it follows that if x, y ∈ X then    f (x, A (e, y)) f ∗e A (x, y) =  = f (x A (e, y)) = x f (A (e, y)) = x ( f ◦ A (e, .)) (y) = f ◦ A (e, .) (x, y) . We derive that  f ∗e A = f ◦ A (e, .) ∈ Orth (X ) . Consequently, Orth (X ) is a ring ideal in Orth (X , X ), as desired. On the other hand, Orth (X ) is a uniformly-complete f -algebra with unit and so Orth (X ) is square-root closed (see Theorem 4.2 in [2]). 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