Bibasic sequences and norming basic sequences

Abstract:

It is shown that every infinite dimensional Banach space X contains a basic sequence $({x_n})$ having biorthogonal functionals $({f_n}) \subset {X^\ast }$ such that $({f_n})$ is also basic. If $[{f_n}]$ norms $[{x_n}]$ then $({f_n})$ is necessarily basic. If $[{f_n}]$ norms $[{x_n}]$ then $[{x_n}]$ norms $[{f_n}]$. In order that $[{f_n}]$ norms $[{x_n}]$ it is necessary and sufficient that the operators ${S_n}x = \Sigma _1^n{f_i}(x){x_i}$ be uniformly bounded. If $[{f_n}]$ norms $[{x_n}]$ then ${X^\ast }$ has a complemented subspace isomorphic to ${[{x_n}]^\ast }$. Examples are given to show that $({f_n})$ need not be basic and, if $({f_n})$ is basic, still $[{f_n}]$ need not norm $[{x_n}]$.