Talk:Finite rank operator

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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)