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Math Help - Showing a space is not Banach

  1. #1
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    Showing a space is not Banach

    Let C[0,1] be the space of continuous real valued functions on [0.1] which has a continuous derviative i.e. its C1.

    Define the norm ||f|| = |f(0)| + sup |f '(t)| for t in [0,1]

    How do you show that this is not a Banach space?




    I think I might I have a counterexample though I cannot prove by defn of Cauchy sequences that it does not converge.

    So we would want to find a Cauchy sequence that does not converge.
    i.e. ||f_n - f_m|| = |f_n(0) - f_m(0)| + sup |f '_n(0) - f '_m(0)|

    I was thinking f_n(t) = t^n would work but then I could not figure it out.
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  2. #2
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    Quote Originally Posted by vinnie100 View Post
    Let C[0,1] be the space of continuous real valued functions on [0.1] which has a continuous derviative i.e. its C1.

    Define the norm ||f|| = |f(0)| + sup |f '(t)| for t in [0,1]

    How do you show that this is not a Banach space?




    I think I might I have a counterexample though I cannot prove by defn of Cauchy sequences that it does not converge.

    So we would want to find a Cauchy sequence that does not converge.
    i.e. ||f_n - f_m|| = |f_n(0) - f_m(0)| + sup |f '_n(0) - f '_m(0)|

    I was thinking f_n(t) = t^n would work but then I could not figure it out.
    Maybe I'm missing something but I think this is a Banach space:

    Let (u_n) be a Cauchy sequence then (u_n(0)) is a Cauchy sequence and thus converges and (u'_n) is a unif. Cauchy sequence so it also converges, and by a standard theorem we have that u is a C^1 function and u'_n \rightarrow u' if u_n\rightarrow u
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