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About C^k and C^oo spaces

  1. May 7, 2007 #1
    I have a cauchy sequence (f_n) in the space of functions on [0,1] which are infinitely differentiable, C^oo ([0,1]) = INTERSECTION C^k ([0,1]), k from 0 to oo, with the metric d(f,g) = SUM 2^-i * ||f-g||_C^i / ( 1 + ||f-g||_C^i ), i from 0 to oo

    If we assume that we do not know whether ( C^oo, d ) is complete or not from the beginning, but that we know ( C^k, ||.|| ) are Banach spaces for all k,

    is it valid to claim that (f_n) is cauchy in C^k ([0,1]) for all k?

    reason:
    (f_n) cauchy in C^oo
    for all epsilon > 0 there exists N(depending on epsilon) in the set of natural numbers such that
    m,n >= N => d(f_m, f_n) < epsilon
    so there exists N s.t.
    epsilon > d(f_m, f_n) = SUM 2^-i * ||f_m - f_n||_C^i / ( 1 + ||f_m - f_n||_C^i ), i from 0 to oo
    i.e.
    epsilon > SUM 2^-i * [ 1 / ( 1 + ||f_m - f_n||_C^i ) - 1 ] , i from 0 to oo

    and so we can make the above sum as arbitrarily small as we desire,
    therefore we can make ||f_m - f_n||_C^i as arbitrarily small as we desire since this is the only "variable" in the above sum.
    i.e. ||f_m - f_n||_C^i < epsilon_i for all i

    and hence (f_n) is cauchy in C^k for all k








    Also with the same situation as before but this time I have a convergent sequence in C^oo, i.e. (f_n) -> f in C^oo

    Is it valid to claim that (f_n) is convergent in C^k for all k?

    reason:
    with exactly the same method as before except replace "f_m, f_n" by "f_n and f"

    or

    (f_n) -> f in C^oo means
    f_n - f -> 0, i.e.
    d(f_n, f) -> 0, i.e.
    SUM 2^-i * ||f_n - f||_C^i / ( 1 + ||f_n - f||_C^i ), i from 0 to oo, goes to zero, i.e.
    ||f_n - f||_C^i -> 0 for all i since 2^-i > 0 so the other part of the numerator needs to be zero
    i.e. (f_n) -> f in C^k for all k
     
    Last edited: May 7, 2007
  2. jcsd
  3. May 7, 2007 #2
    I'm not sure. I think you might have a problem with the N_k s, i.e. how far you have to go in the sequence before all terms are within epsilon of each other. If the N_k 's kept getting larger and larger, I'm not sure if you could say that f_n was cauchy in C^oo, because you might not be able to make that sum you mentioned as arbitraily small as possible. In other words, perhaps N_k needs an upper bound, or perhaps I'm simply being overly pedantic.

    If for every k the sequence is cauchy, that means that the functions are converging to a limit in C^k. Perhaps if you found this limit and proved that all the C^k limits were the same?.

    Actually, what if you could make all those terms in your sequence less than say epsilon*2^-k. Then the sum of the infinite geometric series would be less than epsilon. But woundn't you need an upper bound on N_k to do this.

    I don't have any analysis texts in front of me. Sorry.
     
  4. May 8, 2007 #3
    Okay, now I'm getting a bit confused...
    I want to show that any sequence that is cauchy in C^oo implies it is cauchy in C^k for all k, and that any sequence that is convergent in C^oo implies it is convergent in C^k for all k.
    So I can take any cauchy sequence in C^oo and can make that sum as small as desired, same with the second case where I can take any convergent sequence in C^oo...
     
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