Showing $l^p\to l^p,(x_n)_n\mapsto(a_nx_n)_n$ is compact iff $a_n\to0$

Question

Im studying functional analysis and I have to solve the following problem:

Let $T: l^p \rightarrow l^p$ the operator given by $T(x_j)=(x_j a_j)$. Show that $T$ is compact iff $\lim_{j \to \infty} a_j=0$. I have the following proof that if $a_j$ does not converge then $T$ is not compact but I don't know if it's correct:

Let $B$ be the unit ball in $l^p$. $T$ compact implies that $T(B)$ is totally bounded. Let $\varepsilon>0$. Suppose that $a_j$ does not tend to $0$. Suppose that $a_j$ does not tend to any limit, then $(a_j)$ is not a Cauchy sequence in $\mathbb{R}$, so $||a_n-a_m|| \geq \varepsilon$ for some $n,m$. Let $e_i=(0,...,1,0,...)$, then $T(e_i/2)=a_i/2$ and $||T(e_n/2)-T(e_m/2)||\geq \varepsilon/2$, so $T(B)$ is not totally bounded so $T$ is not compact, which is a contradiction.

I don't know how to procede from this to show that if $a_j$ tends to a limit $L\neq 0$ then $T$ is not compact or to show that if the limit is 0 then the operator is compact.

Answer 1

Your proof is almost okay.

What you need to do is go to subsequences assume that if $a_{n_{k}}$ is a subsequence such that $|a_{n_{k}}|>\epsilon$ for some $\epsilon>0$ , then consider the sequence $T(e_{n_{k}})$ . Due to compactness of $T$ , this sequence will have a convergent subsequence let's say $T(e_{n_{k_{l}}})$ .

But the problem is that $||T(e_{n_{k_{l}}})-T(e_{n_{k_{m}}})||=|a_{n_{k_{l}}}|e_{n_{k_{l}}}+|a_{n_{k_{m}}}|e_{n_{k_{m}}}>2\epsilon$ . Hence this sequence $T(e_{n_{k_{l}}})$ cannot be Cauchy and hence cannot be convergent.

Thus for all subsequences $a_{n_{k}}\to 0$ which means $a_{n}\to 0$ .

For the reverse, see that $\{T_{N}\}$ defined by $T_{N}((x_{k}))=\sum_{k=1}^{N}e_{k}a_{k}x_{k}$ are a sequence of bounded finite ranked(hence compact) operators such that for all $(x_{k})$ with unit norm $||T_{n}(x_{k})-T(x_{k})||\leq (\sup_{k\geq n}|a_{k}|)||(x_{k})||=\sup_{k\geq n}|a_{k}|$ .

Hence , as , for any $\epsilon>0$, there exists an $N$ such that $|a_{n}|<\epsilon$ for all $n\geq N$ , you have $||T_{n}-T||<\epsilon$ for all $n\geq N$.

Which shows that $T$ is a limit of compact operators and hence compact.

(I hope you know this result. Otherwise, try proving it by showing total boundedness of $T(B_{X})$.)

Answer 2

If $\{a_n\}$ does not tend to zero, then there exists an $\varepsilon>0$ and a subsequence $\{a_{k_n}\}$, with $$ |a_{k_n}|\ge\varepsilon. $$ If now $e_n=(0,\ldots,0,1,0,\ldots)$, with all elements equal to 0 and the $n$th one equal to 1, then $\{e_{k_n}\}\subset\ell^p$ is a bounded sequence, while $$ Te_{k_n}=a_{k_n}e_{k_n}, $$ does not have a converging subsequence.