# Compactness of multiplication operator in $\ell^{p}$.

i have a bounded sequence $(\lambda_n)_{n=1}^{\infty}$ in $\mathbb{R}$ and $T \in \mathrm{L}(\ell^{p})$ with $1\leq p \leq \infty$ such that $Tx=(\lambda_{1}x_{1},\lambda_{2}x_{2},\dots)$ , where $x=(x_{1},x_{2},\dots)$ my question is if $\lambda_{n} \to 0$ then $T$ is a compact operator. A classmate gave me a tip to use Cantor diagonal argument. But i don't know how to do that. Any help?

$\mathrm{L}(\ell^{p})$ is the set of continuous linear operators in $\ell^{p}$.

Thanks

• Let $T_n x = \sum_{K=1}^n \lambda_k x_k$ which is finite rank and hence compact. Then show that $T_n \to T$. – copper.hat Feb 17 '17 at 4:00
• Sorry, i made a typo. The post has been edited. – C. Junior Feb 17 '17 at 4:05
• My comment is still valid. – copper.hat Feb 17 '17 at 4:07
• Nice. Thanks @cooper. – C. Junior Feb 17 '17 at 4:12
• You could give an idea to show the reciprocal of this result.? – C. Junior Feb 17 '17 at 4:12