let $H$ a hilbert space, $\{e_{n}\}_{n=1}^{\infty}$ an orthonormal basis for $H$, and $\operatorname{A}\in B(H)$
$\forall n\geqslant 0$ , $\operatorname{E_{n}}$ is the projection onto $\operatorname{span}\{e_{1},....,e_{n}\}$
$\operatorname{A}$ is compact $\Leftrightarrow$ $\lim_{n\rightarrow \infty} \left \| (\operatorname{Id}-\operatorname{E}_{n})\operatorname{A}(\operatorname{Id}-\operatorname{E_{n}}) \right \|=0$
any idea, or book where i can read this
i think forward direction $\Rightarrow$ is here
thanks
Edit:
i was thinking and
$\| (\operatorname{Id}-\operatorname{E}_{n})\operatorname{A}(\operatorname{Id}-\operatorname{E_{n}}) \| = \| \operatorname{A}-\operatorname{AE_{n}} -\operatorname{E_{n} A}+\operatorname{E_{n} A E_{n}} \| $
and
$-\operatorname{AE_{n}} +\operatorname{E_{n} A}-\operatorname{E_{n} A E_{n}}$
are sum of three finite's rank operator, so are compact $\forall n\geqslant 0$
so $A$ is compact for be limit of compact's operator .
other way for forward direction $\Rightarrow$
using that $\lim_{n\rightarrow \infty} ||A(e_n)||= 0$ , i can go to
$||A(Id-E_{k})||=\delta \cdot sup_{||x||\leqslant 1} \sum_{j=k+1}^{\infty}|<x,e_{j}> |$
where k is the natural for $||A(e_{j})||<\delta$, $\forall j>k$
but how delimit sumatory?