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)