1. ## 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?

2. ## 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 $n$.

3. ## 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.

4. ## Re: show compactness

Originally Posted by Jose27
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?