On positive-off-diagonal operators on ordered normed spaces
Let \((X,K, \| . \|)\) be an ordered normed space, where \(X\) is a real vector space, \(\| . \|\) is a norm on \(X\) and \(K\) is a cone in \(X\) with \(K \cap (-K)= \{0\}\). Let \(X'\) denote the vector space of all continuous linear functionals on \(X\) and \(L(X)\) the vector space of all continuous linear operators on \(X\) and \(I \in L(X)\) the identity operator. \(K'\) denotes the wedge of all positive functionals in \(X'\). An operator \(A \in L(X)\) is called positive-off-diagonal if \(x \in K\) and \(f \in K'\) with \(f(x)=0\) imply \(f(Ax) \geq 0.\) In the paper under review, the main result is that if \((X,K, \| .\|)\) is an ordered normed space that satisfies the Riesz decomposition property and \(K\) is a closed normal non-flat \(b\)-generating cone and there exists a total set of extremals of \(K'\), then for any operator \(A \in L(X)\) the following statements are equivalent: (i) \(A\) is a positive-off-diagonal operator; (ii) \(\| A \| I + A \geq 0 .\) Among other things, the author gives two examples. The first example shows that a positive-off-diagonal operator \(A\) in general cannot be represented as \(A=sI + B\) with a positive operator \(B\) and a real number \(s\), even if \(A\) operates on a finite-dimensional space. In the second example, an ordered normed space that is not a vector lattice satisfying all assumptions of the main result is given.
scientific article; zbMATH DE number 1895124