Talk:Finite-rank operator

Finite rank do not imply boundeness. It must be required by definition. For a counter-example, take $X\,$ an infinite dimensional Banach space. By the axiom of choice, there is an unbounded linear funtional $l:X\to \mathbb {R} \,$. Define $T:X\to X\,$ as $Tx=l(x)y\,$, where $0\neq y\in X\,$.Lechatjaune 23:24, 23 May 2007 (UTC)