Operator norm limits of order continuous operators
Let \(X\) and \(Y\) be Banach lattices, and denote by \({\mathcal L}^b(X, Y)\) the space of order bounded linear operators from \(X\) into \(Y\) equipped with the order bound norm, a norm introduced by the second author [in: Functional analysis and economic theory, Samos, Greece, July 1996, 109--118 (1998; Zbl 0912.47018)]. Moreover, let \({\mathcal L}^r(X, Y)\) stand for the subspace of \({\mathcal L}^b(X, Y)\) consisting of regular operators equipped with the regular norm. Let \((T_n)\) be a sequence convergent to \(T\) in \({\mathcal L}^b(X, Y)\) [resp., \({\mathcal L}^r(X,Y)\)] or in the operator norm \(\|\cdot\|\). Assuming that \(T_n\), \(n= 1,2,\dots\), are continuous with respect to what the authors call weak order convergence (for order bounded nets), which amounts to order convergence if the vector lattice in question is Dedekind complete, they prove that \(T\) has the same continuity property. In the case of \(\|\cdot\|\), this is done under the extra condition that the Lorenz seminorm of \(Y\) be a norm. A counterexample is constructed to show that this condition may not be dispensed with entirely. In a note added in proof, the authors refer to an independent paper by \textit{Yu.\,Abramovich} and \textit{G.\,Sirotkin} [Positivity 9, No.\,3, 287--292 (2005; Zbl 1110.46002)], where some variants of the notion of order convergence are studied. Reviewer's remark: The presentation of the counterexample needs some corrections. In particular, in the definition of \(\psi\), one should replace ``\(2^t\)'' by ``\(2^n t\)''. Moreover, the description of \(\psi^{-1}(x)\) is wrong and the definition of \(S\) on p.\,353 is unclear.
scientific article; zbMATH DE number 5070778
scientific article