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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.

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