Disjointly homogeneous Banach lattices and compact products of operators
A Banach lattice \(E\) is called disjointly homogeneous if, for two arbitrary disjoint sequences in \(E\), there exist subsequences which are equivalent. Examples of disjointly homogeneous spaces include the spaces \(L_p(\mu)\) for \( 1\leq p \leq \infty \) and Lorentz spaces \(L_{p,q }(\mu)\). In general, a Banach lattice is called \(p\)-disjointly homogeneous if every normalized disjoint sequence has a subsequence equivalent to the unit vector basis of \(l^p\) \((c_0{\text{ if}}\,\, p = \infty \)). An interesting example is given to show that not every disjointly homogeneous Banach lattice is \(p\)-disjointly homogeneous. If \(E\) is a disjointly homogeneous Banach lattice, then either \(E\) or \(E'\) or both is a KB-space. If \(E\) is an \(\infty\)-disjointly homogeneous Banach lattice, then \(E'\) is a 1-disjointly homogeneous Banach lattice. Examples are provided to show that this duality is not true in general. In the second part of the paper, the authors study regular operators on disjointly homogeneous Banach lattices. The main result of this part is the following Theorem. Suppose that \(E\) is a disjointly homogeneous Banach lattice with order continuous norm and a weak unit. Let \(S,T\) be regular operators on \(E\) such that \(S\) is disjointly strictly singular and \(T\) is AM-compact. Then if \(E'\) has order continuous norm then \(ST\) is a compact operator. If \(E'\) does not have order continuous norm, then \(TS\) is a compact operator.
scientific article; zbMATH DE number 5555223
scientific article