Showing that a subset of a Hilbert space is compact My question is about this previous question by somebody.
Here, we know that a separable Hilbert space is isometrically isomorphic to $l^2,$ so it can be identified with $l^2.$ Also, it suffices to show that $K$ is sequentially compact, that is, any sequence in $K$ has a convergent subsequence. But I cannot follow the answer's last line: "Using the fact that $(x^{k,k})_k \subset K$ show that $x^{k,k}$ also converges in norm." Any elaboration would be appreciated.
 A: Let $(x^{k,k})_{k \in \mathbb{N}} \subseteq K$ be the sequence (of sequences in $\ell^2(\mathbb{R})$) as described in the linked question, and let $x = (x_j)_{j \in \mathbb{N}}$ be the constructed pointwise limit such that $x^{k,k}_j \to x_j$ as $j \to \infty$.
First, we can check that $(x_j)_{j \in \mathbb{N}}$ is in $K$ (e.g. since for each $k$, $\sum^\infty_{j=M_s} |x^{k,k}_j|^2 \leq \ell_s$, we can take the limit as $k \to \infty$ to deduce that $\sum^\infty_{j=M_s} |x_j|^2 \leq \ell_s$ for all $s$).
Therefore, for each $s$, $\sum^\infty_{j=M_s} |x^{k,k}_j - x_j|^2 \leq \sum^\infty_{j=M_s} |x^{k,k}_j|^2 + \sum^\infty_{j=M_s} |x_j|^2 \leq 2 \ell_s$. Hence,
$$
\lVert x^{k,k} - x \rVert_2^2 = \sum^\infty_{j=1} |x^{k,k}_j - x_j|^2 = \sum^{M_s-1}_{j=1} |x^{k,k}_j - x_j|^2 + \sum^\infty_{j=M_s} |x^{k,k}_j - x_j|^2 \leq \sum^{M_s-1}_{j=1} |x^{k,k}_j - x_j|^2 + 2\ell_s . 
$$
Taking the limit as $k \to \infty$ shows that $\lim_{k \to \infty} \lVert x^{k,k} - x \rVert_2^2 \leq 2\ell_s$ by pointwise convergence of our constructed sequence. Since  this holds for all $s$, and $\ell_s$ tends to zero, this shows that $x^{k,k}$ also converges to $x$ in norm.
