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Math Help - Convergence in space of all square integrable functions on [-1,1]

  1. #1
    Member Mollier's Avatar
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    Convergence in space of all square integrable functions on [-1,1]

    Problem:
    1. Given the sequence x_n(s) = (|s|+\frac{1}{n})^{-1/3} and function x(s)=|s|^{-1/3}. show that x_n\rightarrow x
    in L_2(-1,1)

    2. How can we use this to deduce that C[-1,1] (with the max norm) is not complete.

    Pathetic Attempt:

    1. lim_{n\rightarrow \infty}||x_n(s)-x(s)||_{L_2} = lim_{n\rightarrow \infty} \int^1_{-1} |x_n(s)-x(s)|^2 = 0
    because lim_{n\rightarrow \infty} 1/n = 0.

    2. The limit x(s) isn't in C[-1,1] so its not complete...

    Very bad I know.
    Any suggestions are welcome.
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  2. #2
    Member Mollier's Avatar
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    Re: Convergence in space of all square integrable functions on [-1,1]

    A somewhat better attempt:

     \int^1_{-1} |x_n(s) - x(s)|^2 ds

    I split the integral up in two for simplicity.

    \int^1_{-1} (|s|+1/n)^{-1/3}ds = 6(1/n+1)^{1/3} - 6(1/n)^{1/3}

    \int^1_{-1} |s|^{-1/3}ds = 6

    Let n go to infinity and we have proved convergence.

    Still working on the other part though.
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