# Compact operator in reflexive space compact

1. Sep 10, 2014

### DavideGenoa

Hi, friends! I find an interesting unproven statement in my functional analysis book saying the image of the closed unit sphere through a compact linear operator, defined on a linear variety of a Banach space $E$, is compact if $E$ is reflexive.
Do anybody know a proof of the statement?
$\infty$ thanks!!!

Last edited: Sep 10, 2014
2. Sep 10, 2014

### economicsnerd

Some facts which may be useful:
- Given a Banach space, the unit ball of its dual space is weak*-compact.
- Given a reflexive Banach space, the the weak and weak* topologies on its dual space coincide.
- Any closed convex subset of a Banach space is weakly closed.
- Any continuous map with compact domain has compact range.

3. Sep 10, 2014

### DavideGenoa

I have found a proof what I didn't know of the propositions you quote here. My text states a lot of things of a more advanced level than the elements of theory that it explains, but I'm curious and want to try to understand as much as I can...
While searching I have also found this page, which would resolve the problem if I understood why Because A is compact, $Ax_{n_j}\to Ax$ strongly. I only know that $\forall f\in X^{\ast}$ $f(x_{n_j})\to f(x)$ strongly...
I $\infty$-ly thank you and anybody wishing to join the thread!!!

Last edited: Sep 10, 2014