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Math Help - show compactness

  1. #1
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    show compactness

    Let \mathbf{X}= L^2(\mathbf{N})= \{\ x=\{\ x_n \}\ | \sum_{n=1} ^\infty = x_n ^2 < \infty \}, \| x \| = ( \sum_{n=1} ^\infty x _n ^2 )^{1/2} and \mathbf{K} \subset \math{X} be closed and bounded with the property \forall \epsilon > 0 there exists N such that \forall n >N, \sum_{n=N} ^\infty x_n ^2 < \epsilon \forall x \in \mathbf{K}. Show that \mathbf{K} is compact. (show every sequence \mathbf{K} has a convergent subsequence.)


    Let x \in \mathbf{K}. then \sum_{n=1} ^\infty = x_n ^2 < \infty. So x_n \rightarrow 0 as n \rightarrow \infty. So x converges to 0. and x is a subsequence of itself.

    Is it not enough to show that K is compact?
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  2. #2
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    Re: show compactness

    A hint: X is reflexive so K has weakly compact (weak) closure, in particular any sequence (x^{(n)}) in K has a (weak) limit point say x , use this to show that for k\in \mathbb{N} you can have \sum_{i=1}^{k} |x^{(n)}_i-x_i| <\varepsilon for n big enough. Now use your last condition to show that we can also control the tails uniformly in n.
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  3. #3
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    Re: show compactness

    The solution using the weak compactness works, but we can only use the following result: a subset of a metric space is compact if and only if it's precompact and complete. Since K is closed and \ell^2 is complete, K is complete. The precompactness can be shown using the fact that K is bounded and the property.
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  4. #4
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    Re: show compactness

    Quote Originally Posted by Jose27 View Post
    A hint: X is reflexive so K has weakly compact (weak) closure, in particular any sequence (x^{(n)}) in K has a (weak) limit point say x , use this to show that for k\in \mathbb{N} you can have \sum_{i=1}^{k} |x^{(n)}_i-x_i| <\varepsilon for n big enough. Now use your last condition to show that we can also control the tails uniformly in n.
    any sequence {x_n} in K converges to 0, right?
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