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Thread: Convergence in space of all square integrable functions on [-1,1]

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

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

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

    Pathetic Attempt:

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

    2. The limit $\displaystyle x(s)$ 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:

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

    I split the integral up in two for simplicity.

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

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

    Let n go to infinity and we have proved convergence.

    Still working on the other part though.
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