can somebody help me to prove that C2pi is complete

Discussion in 'General Math' started by Kal, Nov 21, 2004.

  1. Kal

    Kal Guest

    Hi
    Can somebody help me to prove that the space of 2pi-periodic continuous
    functions is complete and separable?
     
    Kal, Nov 21, 2004
    #1
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  2. Kal

    Lynn Kurtz Guest

    Hint 1: You need to specify a metric.

    Hint2: Do you have the fact that C[0, 2pi] is complete and separable?

    Hint3: If a sequence {f_n} is cauchy in the periodics, what about it
    in C[0, 2pi]? Would it have a limit? Would it be periodic? Etc.

    Hope that helps.

    --Lynn
     
    Lynn Kurtz, Nov 21, 2004
    #2
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  3. Kal

    Kal Guest

    can I use ||f(R)||=sup{abs(f(x)) where x is in R}?
    it's not given but I can assume it.
    it is not necesarily a cauchy but I can find a subsequence in C[0,2pi] that
    is cauchy
     
    Kal, Nov 21, 2004
    #3
  4. can I use ||f(R)||=sup{abs(f(x)) where x is in R}?[/QUOTE]

    What do you mean "can I"? It's your question.
     
    The World Wide Wade, Nov 22, 2004
    #4
  5. Kal

    Lynn Kurtz Guest

    Hey, it's your problem, use whatever metric you want to state the
    problem. It's nice, though, to use a metric for which your statement
    is true. Let's call P the set of periodic functions on R. You might
    note that for f in P, ||f|| also can be computed by taking the sup
    over [0, 2pi] as well as over R. In fact you might wish to think about
    the restriction of the functions in P to [0, 2pi]. After all, if a
    sequence of periodic functions converges there......

    The functions {f_n} are in S. Are you sure about that?

    --Lynn
     
    Lynn Kurtz, Nov 22, 2004
    #5
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