Show that $T$ is compact if and only if $\lim_{n \rightarrow \infty} |\alpha_n|=0$

Question

Define $T: \ell^2(\mathbb{N}) \rightarrow \ell^2(\mathbb{N})$ by $$T(x_0,x_1,...)=(\alpha_0x_0,\alpha_1x_1,...)$$ where $(\alpha_n)_{n\geq0}$ is a bounded sequence of nonzero complex numbers. How do we show that $T$ is compact if and only if $\lim_{n \rightarrow \infty} |\alpha_n|=0$?

Answer 1

If $\alpha_n$ does not tend to $0$ then there exists $t>0$ and integers $n_1<n_2<..$ such that $|\alpha_{n_k}| >r$ for all $x$. Now $T(e_{n_k})=\alpha_{n_k} e_{n_k}$ and $(e_{n_k})$ is a bounded sequence. If $T$ is compact this would imply that some subsequence of $T(e_{n_k})$ is convergent. But $e_n \to 0$ weakly which implies $Te_n \to 0$ weakly. Hence some subsequence of $T(e_{n_k})$ converges to $0$ in the norm. We have arrived at a contradiction since $\|T(e_{n_k})=|\alpha_{n_k}| \geq r$.

Now suppose $\alpha_n \to 0$. Define $T_N(x_0,x_1,...)=(\alpha_0x_0,...,\alpha_N x_N,0,0,,,)$. For each $N$ this is a bounded operator of finite rank. Also $\|T-T_N\|\leq \sup_{n>N}|\alpha_n| \to 0$. [I will leave this simple verification to you]. Since finite rank operators and limits of compact operator are compact we are done.