arXiv subjects
Publications and source records attributed to G. Martínez-Cervantes.
Building on a recent construction of G. Plebanek and the third named author, it is shown that a complemented subspace of a Banach lattice need not be linearly isomorphic to a Banach lattice. This solves a long-standing open question in Banach lattice theory.
Let $X$ be a Banach space and $Y \subseteq X$ be a closed subspace. We prove that if the quotient $X/Y$ is weakly Lindelöf determined or weak Asplund, then for every $w^*$-convergent sequence $(y_n^*)_{n\in \mathbb N}$ in $Y^*$ there exist a subsequence $(y_{n_k}^*)_{k\in \mathbb N}$ and a $w^*$-convergent sequence $(x_k^*)_{k\in \mathbb N}$ in $X^*$ such that $x_k^*|_Y=y_{n_k}^*$ for all $k\in \mathbb N$. As an application we obtain that $Y$ is Grothendieck whenever $X$ is Grothendieck and $X/Y$ is reflexive, which answers a question raised by González and Kania.
We study the class of Banach spaces $X$ such that the locally convex space $(X,μ(X,Y))$ is complete for every norming and norm-closed subspace $Y \subset X^*$, where $μ(X,Y)$ denotes the Mackey topology on $X$ associated to the dual pair $\langle X,Y\rangle$. Such Banach spaces are called fully Mackey complete. We show that fully Mackey completeness is implied by Efremov's property ($\mathcal{E}$) and, on the other hand, it prevents the existence of subspaces isomorphic to $\ell_1(ω_1)$. This extends previous results by Guirao, Montesinos and Zizler [J. Math. Anal. Appl. 445 (2017), 944-952] and Bonet and Cascales [Bull. Aust. Math. Soc. 81 (2010), 409-413]. Further examples of Banach spaces which are not fully Mackey complete are exhibited, like $C[0,ω_1]$ and the long James space $J(ω_1)$. Finally, by assuming the Continuum Hypothesis, we construct a Banach space with $w^*$-sequential dual unit ball which is not fully Mackey complete. A key role in our discussion is played by the (at least formally) smaller class of Banach spaces $X$ such that $(Y,w^*)$ has the Mazur property for every norming and norm-closed subspace $Y \subset X^*$.