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An operator $T:X\to X$, where $X$ is a Banach space, is strictly singular if no restriction to a closed, infinite dimensional subspace is an isomorphism. It is well known they form a closed ideal in $\mathcal{B}(X)$. When $X=l_2$, the only proper closed ideal is the ideal $\mathcal{K}(l_2)$ of compact operators, therefore every strictly singular operator on $l_2$ is compact (the converse is always true). My questions are:

  1. Is there a "simple" direct proof of the fact that every strictly singular operator on $l_2$ is compact?
  2. Is there a known characterization of Banach spaces for which the above is true, in other words for which the ideals of compact and strictly singular operators coincide?

Thank you!

Markus
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    You can prove this for all $\ell_p$, $1\le p<\infty$, using basic sequences (in these spaces, a normalized, weakly null sequence $(x_n)$ has a subsequence equivalent to the standard unit vector basis of the space). See this for a bit more detail. – David Mitra May 10 '20 at 10:56

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