0
$\begingroup$

Let $X$ and $Y$ two banach space. We say that $L:X\longrightarrow Y$ is a compact operator if $L(U)$ is relatively compact whenever $U$ is bounded.

I recall that a set $E$ is relatively compact if its closure $\bar E$ is compact.

I suppose that if we give such definition for compact operator, it's because there are set that are bounded but s.t. the closure is not compact (otherwise the definition would be that $L:X\longrightarrow Y$ is compact if $L(U)$ is bounded whenever $U$ is bounded), but I can't find such set. So do you have an example of bounded set s.t. the closure is not compact ? If not, why such a definition of compact operator ?

$\endgroup$
2
$\begingroup$

It is well known that the closed unit ball in a normed vector space is compact if and only if the space is finite dimensional. In particular, if $X$ is an infinite dimensional Banach space then $B_X = \{ x \in X : \|x\| \leq 1\}$ is bounded, closed and is not compact.

$\endgroup$
  • $\begingroup$ I dindn't know that theorem. Thanks a lot. $\endgroup$ – user330587 Apr 14 '18 at 9:29
  • $\begingroup$ @user330587 In that case, attempting to prove it is a good exercise in functional analysis! The key is to try to apply Riesz' Lemma to construct a sequence in $B_X$ with no convergent subsequence (Hence $B_X$ is not sequentially compact and therefore isn't compact). $\endgroup$ – Rhys Steele Apr 14 '18 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.