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Weakly l.s.c. function attains its min on weakly compact set?

  1. Mar 2, 2008 #1


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    1. The problem statement, all variables and given/known data
    I'm reading the proof of a theorem and the author claims w/o justification that a weakly lower semi-continuous function (w.l.s.c.) f:C-->R attains its min on the convex weakly compact subset C of a normed space E.

    At first I though I saw why: Let a be the inf of f on C and x_n be a sequence in C such that f(x_n) --> a. Since C is weakly compact, we can find a weakly convergent subsequence x_n_k-->x, and because f is w.l.s.c., we will have f(x)<=a, thus f(x)=a.

    But what reason do we have to believe that C is weakly sequentially compact, so that the bold part above is justified??

    (By "weakly" I mean "under the weak topology [tex]\sigma(C,C^*)[/tex]".)
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2


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    That a weakly compact set C is weakly sequentially compact is true, and follows from the Eberlein-Smulian theorem. But the Eberlein-Smulian is highly nontrivial, so there's probably an easier way to see why what you said is true.
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