Prove $Tx=(r_1x_1, r_2x_2, r_3x_3,...)$ is compact, $T:l^2\to l^2$, $r\in l^2$

Question

Here is my question:

Fix $r=(r_1,r_2,...)\in l^2$. Define $T:l^2\to l^2$ by

$$Tx=(r_1x_1, r_2x_2, r_3x_3,...)$$

Prove that $T$ is compact.

Here is what I have, input would be appreciated:

Let $\{x_n\}\in l^2$ be a bounded sequence, and $r\in l^2$ be fixed. Then there exists some $M>0$ such that $\|x_n\|\leq M, \forall n$. Then

$\|Tx_n\|=\|(r_1x^1_n,r_2x^2_n,r_3x^3_n,...)\|\leq M\|(r_1,r_2,r_3,...)\|=\|(Mr_1,Mr_2,Mr_3,...)\|$,

and therefore $\{Tx_n\}$ converges to $(Mr_1,Mr_2,Mr_3,...)\in l^2$, therefore $T$ is compact.

Answer 1

Unfornately, we cannot deduce that $T(x_n)\to M(r_j)_j$ only from the inequality. The argument shows that $T$ is a bounded operator.

In order to show that $T$ is compact, we have to show that the range of the unit ball $B$ has a compact closure. To this aim, we can show that $T(B)$ is totally bounded. Fix $\varepsilon$: for some $N$, $\sum_{n\geqslant N+1}|r_n|^2\lt \varepsilon$. Then we use the fact that $[-1,1]^N$ is compact.