A relationship between two almost f -algebra products

Abstract

Let A be an uniformly complete almost f-algebra. Then $ \Pi(A) = \{ab: a,b \in A\} $ is a positively generated ordered vector subspace of A with $ \Sigma (A) = \{a^2 : a \in A\} $ as a positive cone. If $ T : \Pi(A) \rightarrow A $ is a positive linear operator, we put $ \rho : A \rightarrow {\cal L}_b(A) $ the linear operator defined by $ \rho(a) = T_a $ with $ T_a(b) = T(ab) $ for all $ b \in A ({\cal L}_b(A) $ is the algebra of all order bounded linear operators of A). Let $ {\cal L}^T_b(A) $ denote the range of $ \rho $ and let's define a new product $ * $ by putting $ a * b = T(ab) $ for all $ a, b \in A $ . It is easily checked that if $ a \wedge b = 0 $ then $ a * b = T(ab) = 0 $ , this shows that if it happens that the product $ * $ is associative then A is an almost f- algebra with respect to this new product. It turns out that a necessarily and sufficient condition in order that $ * $ be an associative product is that $ {\cal L}^T_b(A) $ is a commutative subalgebra of $ {\cal L}_b(A) $ . We find necessarily and sufficient conditions on T in order that $ * $ is an almost f-algebra (respect.; d-algebra, f-algebra) product. Such conditions are described in terms of the algebraic and order structure of the algebra $ {\cal L}^T_b(A) $ .¶The converse problem is also studied. More precisely, let A be an uniformly complete almost f-algebra and assume that $ * $ is another almost f-algebra product on A. The aim is to find sufficient conditions in order that there exist $ T:\Pi(A)\rightarrow A $ such that $ a*b=T(ab) $ for all $ a,b\in A $ . It will be showed that a sufficient condition is that A is a d-algebra with respect to the initial product. An example is produced which shows that the condition "A is a d-algebra with respect to the initial product" can not be weakened.