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An operator $T:X\rightarrow Y$ is said to be completely continuous if $T$ maps weakly convergent sequences to norm convergent sequences.

Let $Q: l_{1}\rightarrow l_{2}$ be any surjection and $J:l_{1}\rightarrow Y$ be an isomorphic embedding.

Question. Is there a completely continuous operator $S:Y\rightarrow l_{2}$ such that $Q=SJ$?

Thank you!

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    $\begingroup$ Maybe clarify: Are $X, Y$ Banach spaces? Are $T,Q,J$ linear bounded operators? $\endgroup$ Apr 9, 2018 at 9:26
  • $\begingroup$ @PietroMajer Yes. $X,Y$ are Banach spaces and all operators mean linear bounded operators. $\endgroup$ Apr 9, 2018 at 12:12
  • $\begingroup$ Giving an explicit example of $Q$ (bounded linear surjection $\ell_1\to \ell_2$) would help, I think. $\endgroup$ Apr 9, 2018 at 15:23
  • $\begingroup$ @JeanDuchon, this is standard. The map that sends the unit vector basis of $\ell_1$ to the unit vector basis of $\ell_2$ extends to a bounded linear operator, simply because the $\ell_1$ norm is a certain sense maximal. $\endgroup$ Apr 9, 2018 at 16:36
  • $\begingroup$ @TomekKania Isn't $\ell_1$ the space of summable sequences (and $\ell_2$ that of square summable ones)? The inclusion $\ell_1\subset\ell_2$ is not onto, so what did I miss? Is it obvious that a bounded linear surjection exists? I can't see any example... $\endgroup$ Apr 13, 2018 at 16:03

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By Grothendieck's theorem, $Q$ is absolutely summing, in particular $2$-summing. As such, it has a $2$-summing extension to $Y$, by Pietsch's factorisation theorem. But $2$-summing operators are completely continuous (again by Pietsch's theorem).

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