On the ``function'' and ``lattice'' definitions of a narrow operator
Let \(E\) be a Köthe Banach space on a finite atomless measure space \((\Omega , \Sigma , \mu )\), \(X\) be a Banach space. A bounded linear operator \(T : E \to X\) is called {\parindent=6mm \begin{itemize}\item[--] function narrow if, for every \(\varepsilon > 0\) and every \(A \in \Sigma\), there is a decomposition of \(A\) into a disjoint union of two subsets \(B, C \in \Sigma\) in such a way that \(\mu(B) = \mu(C)\) and \(\|T(\mathbf 1_B - \mathbf 1_C)\| \leq \varepsilon\); \item[--] function weakly narrow if, for every \(\varepsilon > 0\) and every \(A \in \Sigma\), there is a decomposition of \(A\) into a disjoint union of \(B, C \in \Sigma\) in such a way that \(\|T(\mathbf 1_B - \mathbf 1_C)\| \leq \varepsilon\); \item[--] lattice narrow if, for every \(\varepsilon > 0\) and every \(e \in E^+\), there is \(x \in E\) with \(|x|= e\) such that \(\|Tx\| \leq \varepsilon\). \end{itemize}} Theorem. Let \(E\) be as above, then the following assertions are equivalent: {\parindent=6mm \begin{itemize}\item[(1)] For every Banach space \(X\), the classes of function weakly narrow and lattice narrow operators \(T : E \to X\) are the same; \item[(2)] the subspace of simple functions on \(\Omega\) is dense in \(E\). \end{itemize}} Moreover, if (2) does not hold, then there is a Banach space \(X\) and a function narrow operator \(T : E \to X\) which is not lattice narrow. This theorem together with an example of \(E\) for which (2) does not hold answers Problem 10.3 from \textit{M. Popov} and \textit{B. Randrianantoanina} [Narrow operators on function spaces and vector lattices. Berlin: de Gruyter (2013; Zbl 1258.47002)].
scientific article; zbMATH DE number 6861650