# Show that the operator is compact. [duplicate]

Show that if $$\lambda_n \to 0$$ so the operator

$$T: l^p \to l^p$$ , $$T (x_1, \cdots , x_n, \cdots)= (\lambda_1 x_1, \cdots, \lambda_n x_n, \cdots)$$

is compact to $$1\leq p < \infty$$.

I am tryng show that if $$x_n$$ is bounded so $$T (x_n)$$ is precompact.

• Maybe you could see if $T$ is the limit of finite rank operators? – copper.hat Nov 8 '19 at 19:07
• Neither of the above links have an answer to this. – David Mitra Nov 8 '19 at 19:56