Totally boundedness of a compact operator Let $T:\ell_2(\mathbb N)\longrightarrow \ell_2(\mathbb N)$ bounded linear operator such that $$T(\{x_n\})=\{x_n/n\}.$$
I need to prove that $TB(\ell_2(\mathbb N))$, that is closed unit ball in $\ell_2(\mathbb N)$, is totally bounded.
 A: Since you requested a hint, here is a start:
Our operator is $T(\{x_n\}) = \{x_n/n\}$, and we are applying it to $B=B(l_2(\mathbb{N}))$.
We wish to show that for $\delta > 0$ the image $T(B)$ is coverable by a finite number of balls of radius $\delta$.  Since every element in B is of norm less than or equal to 1, we have that if $\{x_n\} \in B$ then $|x_n| < 1$ for every $n$.
Since we know that the square of the norm of every sequence $\{x_n/n\}$ is bounded entrywise by $\sum_{n=1}^\infty 1/n^2 = \pi^2/6$, there is an $M$ for which $$\left| \|Tx\|^2 - \sum_{n=1}^M |x_n|^2/n^2\right| < \delta$$ uniformly for all $x \in B$.
We know that the subset of $T(B)$ given by $$K = \{ Tx : \{x_n\} \in B \text{ with } x_n = 0 \text{ for } n>M\}$$ is a compact set, since it is a closed bounded subset of a finite dimensional space over $\mathbb{C}$.  We cover it with a finite number of balls of radius $\delta$ and centers $\{y_{in}\}$: $$N_{\delta}(\{y_{1n}\}),N_{\delta}(\{y_{2n}\}) ..., N_{\delta}(\{y_{kn}\}).$$
Try to complete the proof from there.
A: Let $x=(x_n)_{n\in\mathbb N^*}\in \mathrm{B}(\ell_2)=\{x=(x_n):|x|_2\leq 1\}$ so   $$|T(x)|_2^2\leq \sum_{n=1}^{+\infty} \left|\frac{x_n}{n}\right|^2\leq \sum_{n=1}^{+\infty} \frac{1}{n^2}=\frac{\pi^2}{6}.$$this implies that : $$|T(x)|_2\leq \frac{\pi}{\sqrt6}\hspace{2mm}\forall x\in\mathrm B(\ell_2).$$
(I think that is true I am not specialise in fonctionnel analysis !)
