Every separable subspace, in particular, $\ell_p$ for $p\in (1,\infty)$ is a quotient of $\ell_1$. However, every map from $\ell_1$ to $\ell_p$ is strictly singular (as $\ell_1$ is self-saturated). What if we replace $\ell_1$ by $L_1$. Can there be a non-strictly singular surjective map $L_1\to \ell_p$ which is bounded below on some copy of $\ell_p$? ($p\in (1,2]$).
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1$\begingroup$ The answer is no because $L_1$ has the Dunford-Pettis property. More generally, if $X$ has the Dunford-Pettis property and $Y$ is reflexive, then every operator from $X$ to $Y$ is strictly singular. $\endgroup$– Philip BrookerCommented Mar 15, 2014 at 0:18
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The answer is 'no' because this would imply using the standard argument (like in Lindenstrauss-Tzafriri, volume 1), that $\ell_p$ is isomorphic to a complemented subspace in $L_1$. This is not the case for many reasons, e.g. this would imply that $\ell_q$ with $\frac1q+\frac1p=1$ is isomorphic to a complemented subspace in $L_\infty$, which is known not to be the case (see Lindenstrauss-Tzafriri, volume 1, again): all such subspaces are nonseparable.
answered Mar 15, 2014 at 1:39