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Compact operator in reflexive space compact

  1. Sep 10, 2014 #1
    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. jcsd
  3. Sep 10, 2014 #2
    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.
     
  4. Sep 10, 2014 #3
    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
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