Show that the operator $(x_n)_n\mapsto (\frac{x_n}{n}) $ is compact

Question

I want to show that the following operator is compact:

$$T:\mathbb \ell^p\rightarrow \mathbb \ell^p, \text{ }(x_n)_n\mapsto(\frac{x_n}{n})_n \text{ } 1\leq p<\infty$$

Its the first time that I am trying to show that an operator is compact.

I know the following three definitions of a compact operator:

Let $T:X\rightarrow Y$ be a bounded operator, then $T$ is compact if

1) The image of the unit ball is relatively compact or

2) The image of any bounded set in X is relatively compact or

3) Any bounded sequence $(x_n)$ in $X$ has a subsequence such that $Kx_{n_k}$ converges.

But I feel like none of this definitions can help me to prove the compactness of the operator directly.

Is there something like a "general way" to show this? At least for Operators from $\mathbb \ell^p\rightarrow \ell^p$ ?

Thanks in advance

Answer 1

Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that $\|T-T_k\| \to 0$ for a suitable chosen sequence $\{ T_k\}$ of finite rank operators.

Try with $$ T_k: (x_1, x_2, \ldots)\mapsto (x_1, \frac{x_2}{2}, \ldots, \frac{x_k}{k}, 0, 0, \ldots ). $$ It is obvious that $rk(T_k)=k$, i.e., it is finite rank operator. Since $$ \| T-T_k\|^p=\sup\{ \sum_{n=k+1}^{\infty}\left|\frac{x_n}{n}\right|^p;\quad \sum_{n=1}^{\infty}|x_n|^p\leq 1\}\leq $$ $$\leq \frac{1}{(k+1)^p}\sup\{ \sum_{n=k+1}^{\infty}|x_n|^p;\quad \sum_{n=1}^{\infty}|x_n|^p\leq 1\} \leq \frac{1}{(k+1)^p}$$ we have $$ \| T-T_k\|\leq\frac{1}{k+1} \to 0\quad (k\to \infty). $$

Answer 2

The operator is of particular (diagonal ) form $$(Tx)_n=a_nx_n$$ Such operator is bounded iff $\sup_n|a_n|<\infty$. Moreover $\|T\|_{p\to p}=\sup|a_n|$, $1\le p\le\infty.$ It is compact iff $\lim_na_n=0.$