The question is the following:
Let $\left(\alpha_{n}\right)$ be a sequence in $\mathbb{R}_{+}^{*}$ such that $\alpha_{n} \rightarrow+\infty$ as $n \rightarrow+\infty$. We set $$ V=\left\{\left(u_{n}\right)_{n \in \mathbb{N}}: \sum_{n \in \mathbb{N}} \alpha_{n}\left|u_{n}\right|^{2}<+\infty\right\} \subset l^{2}(\mathbb{N}) $$ $V$ is a Hilbert space for the inner product defined by $$ \langle u, v\rangle_{V}=\sum_{n \in \mathbb{N}} \alpha_{n} u_{n} \overline{v_{n}}, \quad u=\left(u_{n}\right), v=\left(v_{n}\right) $$ Prove that $V$ is compactly embedded in $l^{2}(\mathbb{N})$
My attempt is as follows:
Consider the operator $P_k$ maps $(u_n)_{n \in \mathbb{N}}$ to its first $k$ terms of the sequence. Note that $P_k$ converges to a limit $P$ which is the inclusion map, and it suffices to prove that $P$ is a compact operator. Since if a sequence of compact operators converges to a bounded operator in the operator norm, then the bounded operator is compact. Now, $P_k$ is compact since it is bounded with finite rank. $P$ is bounded because the norm of both spaces are bounded. Therefore, the proof is complete.
I feel like there are plenty of holes in my attempt. I am not sure of the arguments that $P$ and $P_k$ are bounded are trivial(or maybe they are?). And I am not sure of the statement that $P_k$ converges to $P$ in operator norm. Also, the condition $\alpha_n \to \infty$ as $n \to \infty$ seems to be negligible, so I might do something wrong.
Consider the operator $$(Tu)_n={1\over \sqrt{\alpha_n}}u_n$$ Since $\alpha_n^{-1}\to 0^+$ the operator $T$ is compact, as the norm limit of the finite rank operators $$(T_Nu)_n={1\over \sqrt{\alpha_n}}u_n,\ n\le N,\quad (T_Nu)_n=0,\quad n>N$$ Observe that $T$ maps $\ell^2$ onto $V\subset \ell^2.$ Hence the set $$V_1:=\{Tu\,:\, \|u\|_2\le 1\} $$ is relatively compact in $\ell^2.$ Observe that $$V_1=\{ v\in V\,:\, \sum_{n=1}^\infty \alpha_n|v_n|^2\le 1\}$$