A note on summability in Banach spaces

José Rodríguez¹

¹Universidad de Castilla-La Mancha, Dpto. de Matemáticas, E.T.S. de Ingenieros Industriales de Albacete

Annales Fennici Mathematici, Tome 50 (2025) no. 1, p. 49–58

Résumé

Let $Z$ and $X$ be Banach spaces. Suppose that $X$ is Asplund. Let $\mathcal{M}$ be a bounded set of operators from $Z$ to $X$ with the following property: a bounded sequence $(z_n)_{n\in \mathbb{N}}$ in $Z$ is weakly null if, for each $M\in\mathcal{M}$, the sequence $(M(z_n))_{n\in\mathbb{N}}$ is weakly null. Let $(z_n)_{n\in\mathbb{N}}$ be a sequence in $Z$ such that: (a) for each $n\in\mathbb{N}$, the set $\{M(z_n)\colon M\in \mathcal{M}\}$ is relatively norm compact; (b) for each sequence $(M_n)_{n\in\mathbb{N}}$ in $\mathcal{M}$, the series $\sum_{n=1}^\infty M_n(z_n)$ is weakly unconditionally Cauchy. We prove that if $T\in \mathcal{M}$ is Dunford–Pettis and $\inf_{n\in\mathbb{N}}\|T(z_n)\|\|z_n\|^{-1}>0$, then the series $\sum_{n=1}^\infty T(z_n)$ is absolutely convergent. As an application, we provide another proof of the fact that a countably additive vector measure taking values in an Asplund Banach space has finite variation whenever its integration operator is Dunford–Pettis.

Détail

DOI : 10.54330/afm.156613

Keywords: Absolutely convergent series, Dunford–Pettis operator, vector measure, Schauder basis