Orthosymmetric Archimedean-Valued Vector Lattices

Abstract

We introduce and study the notion of orthosymmetric spaces over an Archimedean vector lattice as a generalization of finite-dimentional Euclidean inner spaces. A special attention has been paid to linear operators on these spaces. Example 1.1 As usual, R denotes the Archimedean vector lattice of all real numbers. Pick n ∈ N = {1, 2, ...} and suppose that the vector space R n is equipped with its usual structure of Euclidean space. In particular,f, e k g, e k for all f, g ∈ R n ,where (e 1 , ..., e n ) is the canonical (orthogonal) basis of R n . Simultaneously, R n is a vector lattice with respect to the coordinatewise ordering. That is,Therefore, f, g = 0 meaning that the inner product on R n is an R-valued orthosymmetric product. Thus, the Euclidean space R n is an orthosymmetric space over R.Beginning with the next lines, we shall impose the blanket assumption that all orthosymmetric spaces under consideration are taken over the fixed Archimedean vector lattice V (unless otherwise stated explicitly).The following property is useful for later purposes.Lemma 1.2 Let L be an orthosymmetric space. Then,Hence,This is the desired result.Example 1.6 Assume that the Euclidean plan R 2 is endowed with its lexicographic ordering. Hence, R 2 is a non-Archimedean vector lattice. Put f, g = x 1 y 1 for all f = (x 1 , x 2 ) , g = (y 1 , y 2 ) in R 2 .Since R 2 is totally ordered (i.e., linearly ordered), this formula defines an R-valued orthosymmetric product on R 2 . This means that R 2 is a non-Archimedean orthosymmetric space over R.The orthosymmetric space L is said to be definite if its neutral part is trivial. That is to say, L is definite if and only if f, f = 0 in V implies f = 0 in L.Definite orthosymmetric spaces have a better behavior as explained in the following.Proposition 1.7 Any definite orthosymmetric space is Archimedean.