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Thread: show compactness

  1. #1
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    show compactness

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


    Let $\displaystyle x \in \mathbf{K}$. then $\displaystyle \sum_{n=1} ^\infty = x_n ^2 < \infty$. So $\displaystyle x_n \rightarrow 0$ as $\displaystyle n \rightarrow \infty$. So $\displaystyle x$ converges to 0. and $\displaystyle x$ 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: $\displaystyle X$ is reflexive so $\displaystyle K$ has weakly compact (weak) closure, in particular any sequence $\displaystyle (x^{(n)})$ in $\displaystyle K$ has a (weak) limit point say $\displaystyle x$ , use this to show that for $\displaystyle k\in \mathbb{N}$ you can have $\displaystyle \sum_{i=1}^{k} |x^{(n)}_i-x_i| <\varepsilon$ for $\displaystyle n$ big enough. Now use your last condition to show that we can also control the tails uniformly in $\displaystyle n$.
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  3. #3
    Super Member girdav's Avatar
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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 $\displaystyle K$ is closed and $\displaystyle \ell^2$ is complete, $\displaystyle K$ is complete. The precompactness can be shown using the fact that $\displaystyle K$ 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: $\displaystyle X$ is reflexive so $\displaystyle K$ has weakly compact (weak) closure, in particular any sequence $\displaystyle (x^{(n)})$ in $\displaystyle K$ has a (weak) limit point say $\displaystyle x$ , use this to show that for $\displaystyle k\in \mathbb{N}$ you can have $\displaystyle \sum_{i=1}^{k} |x^{(n)}_i-x_i| <\varepsilon$ for $\displaystyle n$ big enough. Now use your last condition to show that we can also control the tails uniformly in $\displaystyle n$.
    any sequence {x_n} in K converges to 0, right?
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