So here is my problem,
I try to show that following operator is compact,
\begin{align} J: h_1 & \rightarrow\ell^1 \\ (x_n) & \mapsto(x_n) \end{align} where $$h_1:=\left\{x_n\in \ell^2: \sum_{n\geq 1} n^2|x_n|^2<\infty,\; \lvert\lvert x_n \rvert \rvert:=\left(\sum_{n\geq 1} n^2|x_n|^2\right)^{\frac{1}{2}}\right\}.$$ I first tried to approximate $J$ with a sequence of finite ranked operator but I failed finding a sequence which converges in the operator norm.
Could someone give me a hint how to proceed?
Thanks in advance!
First, let's check that $J : h_1 \to \ell^1$ is indeed bounded. Let $(x_n) \in h_1$; by definition of $h_1$, observe that $\|(x_n)\|_{h_1} = \|(n|x_n|)\|_2$. Then, by Cauchy–Schwarz, $$ \|Jx\|_1 = \sum_{n=1}^\infty |x_n| = \sum_{n=1}^\infty \frac{1}{n} n|x_n| = \langle (n^{-1}),(n|x_n|) \rangle_2 \leq \|(n^{-1})\|_2\| \|(n|x_n|)\|_2 = \frac{\pi^2}{6}\|(x_n)\|_{h_1} $$ so that $J(x_n)$ converges absolutely for each $(x_n)$, and $J$ is bounded with operator norm $\|J\| \leq \tfrac{\pi^2}{6}$.
Now, for each $N \in \mathbb{M}$, define the finite rank operator $J_N : h_1 \to \ell^1$ by $$ J_N : (x_N) \mapsto (x_1,x_2,\ldots,x_N,0,\ldots). $$ The claim is that $\lim_{N \to \infty} J_N = J$ in the operator norm topology on $B(h_1,\ell^1)$. However, $$ \|(J-J_N)(x)\|_1 = \|(0,\ldots,0,x_{N+1},x_{N+2},\ldots)\| = \sum_{n = N+1}^\infty |x_n| = \sum_{n=N+1}^\infty \frac{1}{n}n|x_n|; $$ since $(n|x_n|) \in \ell^2$ and $\xi_N := (0,\ldots,0,(N+1)^{-1},(N+2)^{-2},\ldots) \in \ell^2$, it follows by Cauchy–Schwarz that $$ \|(J-J_N)(x)\|_1 = \sum_{n=N+1}^\infty \frac{1}{n}n|x_n| = \langle \xi_N, (n|x_n|)\rangle_2 \leq \|\xi_N\|_2\|(n|x_n|)\|_2 = \|\xi_N\|_2\|(x_n)\|_{h_1}, $$ and hence that $\|J-J_N\| \leq \|\xi_N\|_2$. So, what happens to $\xi_N$ as $N \to +\infty$?