How do I prove that this operator is compact. Functional analysis

Question

We have $$K:l_2\to l_2$$ $$Kx=\left(x_1, \frac{1}{2}x_2,\frac{1}{3}x_3,...\right)$$ where $x=(x_1,x_2,x_3,...)\in l_2$.

I want to use the definition of a compact operator:

$K$ is compact if $\overline{K(B(0,1))}$ is a compact set in $l_2$

We have $$B(0,1)=\{x \in l_2 : ||x||_2<1\}$$

so we need to check $\overline{K(B(0,1))}$ is a compact set in $l_2$ ?

How to continue ? some hint please?

Using the definition of a compact set, some characterization of a compact set in $l_2$?

Answer 1

Let $T_n :\ell_2 \to \ell_2 , $ $T_n (x) =\left(x_1 ,\frac{x_2}{2} , \frac{x_3 }{3} ,...,\frac{x_n }{n} ,0,0,0,...\right) .$ Then $T_n $ is finite rank operator because $\mbox{dim} (T(\ell_2 )) =n<\infty .$ Moreover $$||T_n -T ||_{\ell_2 \to \ell_2} =\sup_{||x||_2 =1 } ||(T_n -T)(x) ||_2 \leq \sqrt{\sum_{j=n+1}^{\infty} \frac{1}{j^2}}\to 0.$$ Hence $T$ is limit of finite rank operators and therefore compact.