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Ascoli's theorem: A subspace F of C(X,R^n)

  1. Nov 15, 2009 #1
    I just read Ascoli's theorem: A subspace F of C(X,R^n) has compact closure if and only if F is equicontinuous and pointwise bounded.

    Then it says, As a corollary: If the collection {fn} of functions in C(X,R^k) is pointwise bounded and equicontinuous, then the sequence (fn) has a uniformly convergent subsequence.

    Can anybody tell me why the corollary follows from the theorem?
     
  2. jcsd
  3. Nov 16, 2009 #2
    Re: corollary

    in a metric space, when you have a compact subset A, then any sequence in A has a convergent subsequence. Now apply this to the appropriate metric space of functions, so that convergence in the metric is uniform convergence of functions.
     
  4. Nov 16, 2009 #3
    Re: corollary

    Thanks Edgar,
    The problem I see with your reply is that here the closure of the sequence is compact (by Ascoli's theorem), so as you say, this closure has a convergent subsequence. But we don't know that the sequence itself is compact, so how do we know that it has a convergent subsequence?
     
  5. Nov 16, 2009 #4

    quasar987

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    Re: corollary

    I think that here, we are taking F as the subspace generated by the f_n's. Then F has compact closure. Now, the f_n are in F and so in particular, they are in the closure of F. Therefor the sequence {f_n} has a convergent subsequence (in C(X,R^k)).

    AFAIK, it does not make sense to say that a sequence is compact.
     
  6. Nov 16, 2009 #5
    Re: corollary

    Of course!!! how could I miss that? ohh, I know.
    For some reason, instead of thinking that every sequence of a compact metric space has a convergent subsequence, I was thinking: every compact metric space has a convergent subsequence. I guess I was very tired.

    Thank you for clarifying.
     
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