Let $(a_n)\subset\mathbb{C}$ be bounded. Then $T_a\colon\ell^p\to\ell^p, \ (T_a x)_n:=a_n x_n$ compact iff $a_n\to0$. [duplicate]

(See title.) I think that I know how to prove $$\implies$$: If $$(x_n)$$ is a bounded sequence in $$\ell^p$$, then there is a $$c\in\mathbb{R}$$ such that $$\|x_n\|\leq c$$ for all $$n$$. So $$\|T_a x_n\|=\|a_n x_n\|\leq c\|a_n\|\to0,$$ from which it follows that any subsequence of $$(T_a x_n)$$ in $$\ell^p$$ converges (to $$0$$). (Is this correct?) However, for the converse, I have no idea. Any suggestions are greatly appreciated!

• See this. – David Mitra Nov 9 '19 at 12:03
• I don't see why the identity $\lVert (a_kx_{n,k}\,:\, k\in\Bbb N)\rVert=\lVert a_nx_n\rVert$ should hold. – Gae. S. Nov 9 '19 at 12:04
• Perhaps I have overlooked something, but it seems the proposed target duplicate has only one answer, consisting of a "hint" for the direction of implication that OP here says "that I know how to prove". So while both Questions pose an "iff" claim, the direction $\impliedby$ remains open. – hardmath Nov 9 '19 at 23:04