Prove that convex subspace of $l_2$ is compact Prove that the convex subspace of $l_2$ consisting of all sequences $\xi$ such that
$$
 \sum_{n=1}^{\infty} \xi_n^2 n^2 \le 1
$$
is compact. Have no idea how to proceed, any hints or suggestions?
 A: Being closed: $A=\{\xi|\sum_{n=1}^{\infty} \xi_n^2 n^2 \le 1\}=\cap A_n$, where $A_n=\{\xi|\sum_{k=1}^{n} \xi_k^2 k^2 \le 1\}$. Now define $T:l_2\to\mathbb{R}$, $T\xi=\sum_{k=1}^{n} \xi_k^2 k^2$. It's continuous (it is not linear, but that doesn't matter), so...
Theorem: A set $A$ is relatively compact in $l_p$ if and only if:


*

*There exists $(p_n)_{n\in\mathbb{N}}$ so that for every $x=(x_1,x_2,...)\in A$ we have $|x_n|\le p_n$

*For all $\epsilon$ there exists $n_0$ so that for all $x\in A$ series $\sum\limits_{k=n_0+1}^{\infty}|x_k|^p<\epsilon$.
A: To create notation which is less confusing for the problem at hand, define $x \in l^{2}$ to be a function $x : \mathbb{N}\rightarrow\mathbb{C}$ such that $\sum_{j=1}^{\infty}|x(j)|^{2} < \infty$. Let
$$M=\{ x \in l^{2} : \sum_{j=1}^{\infty}|jx(j)|^{2} \le 1\}.$$
If $x \in M$, then $|x(n)|\le 1/n$ for all $n \ge 1$. Suppose $\{ x_{k}\}_{k=1}^{\infty}\subset M$. Then $|x_{k}(n)| \le 1/n$ for all $k \ge 1$. So there exists a subsequence $\{ x_{k_{1,j}}\}_{j=1}^{\infty}$ such that $y_{1}=\lim_{j} x_{k_{1,j}}(1)$ exists. Next choose $\{ x_{k_{2,j}}\}_{j=1}^{\infty}$ which is a subsequence of $\{ x_{k_{1,j}}\}_{j=1}^{\infty}$ such that $\lim_{j} x_{k_{2,j}}(2)=y_{2}$ exists. Automatically $\lim_{j}x_{k_{2,j}}(1)=y_{1}$ exists. Continuing, one finds $\{ x_{k_{N,j}}\}_{j=1}^{\infty}$ which is a subsequence of $\{ x_{k_{N-1,j}}\}_{j=1}^{\infty}$ such that $\lim_{j}x_{k_{N,j}}(n)$ exists for all $1 \le n \le N$. This Cantor diagonalization process leads to the Cantor diagonal sequence $\{ x_{k_{n,n}}\}_{n=1}^{\infty}$ which has the property that $\lim_{n} x_{k_{n,n}}(j)=y_{j}$ for all $j=1,2,3,\cdots$.
Now we show that $x_{0} : \mathbb{N}\rightarrow\mathbb{C}$ defined by $x_{0}(j)=y_{j}$ is in $M$ and that $\{ x_{k_{n,n}}\}_{n=1}^{\infty}$ converges in $l^{2}$ to $x_{0}$, from which the compactness of $M$ follows. To do this, notice that, for $J \ge 1$,
$$
                       \sum_{j=1}^{J}|jx_{k_{n,n}}(j)|^{2} \le 1.
$$
Taking the limit with respect to $n$ then gives
$$
                  \sum_{j=1}^{J}|jx_{0}(j)|^{2} \le 1.
$$
Because the above holds for all $J$, then $x_{0} \in M$. Because $|x_{k_{n,n}}(j)|\le 1/j$ and $|x_{0}(j)|\le 1/j$ for all $j \ge 1$, and because $\lim_{n}x_{k_{n,n}}(j)=x_{0}(j)$ for all $j \ge 1$, then the Lebesgue dominated convergence theorem gives
$$
         \lim_{n} \sum_{j=1}^{\infty}|x_{k_{n,n}}(j)-x_{0}(j)|^{2} = 0.
$$
Therefore $M$ is compact in $l^{2}$.
