A fully convex Banach space which does not have the Banach-Saks property
Let \(k\geq 2\) be an integer. A Banach space X is said to be fully k-convex if for every sequence \(\{x_ n\}\) in X, \(\lim_{n_ 1,...,n_ k\to \infty}(1/k)\| x_{n_ 1}+...+x_{n_ k}\| =1\) implies that \(\{x_ n\}\) is a convergent in X. Every uniformly convex space is fully 2-convex. A Banach space X is said to have the Banach-Saks property if for every bounded sequence \(\{x_ n\}\) in X, there is a subsequence \(\{y_ n\}\) of \(\{x_ n\}\) such that the Cesara mean, \(\{(1/n)(y_ 1+...+y_ n)\}\) of \(\{x_ n\}\) is convergent in X. It is known that every superreflexive space has the Banach-Saks property but there exists a reflexive space without the Banach -Saks property. In this paper, we show that there exists a fully 2-convex space which does not possess the Banach-Saks property. So the Milman's question whether every reflexive space is isomorphic to a fully 2-convex space is still open.
scientific article; zbMATH DE number 3973581