Let Y be a closed linear subspace of X and suppose that Y does not have copy of l1 .Does each weakly p-summable sequence in X/Y has a subsequence that's the image of a weakly p-summable sequence in X under the natural quotient map?
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$\begingroup$ What copy are you excluding? Could you make that more precise? What exactly do you mean by weakly $p$-summable? $\endgroup$– user1688Commented Feb 3, 2016 at 12:09
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$\begingroup$ @anton Here is my interpretation. (1) Y does not have any closed linear subspace that is TVS isomorphic to $\ell^1$ (2) a sequence $(v_n)_{n\geq 1}$ in a Banach space $E$ should be weakly p-summable if for all $\psi\in E^*$ the sequence $(\psi(v_n))_{n\geq 1}$ belongs to $\ell^p$. $\endgroup$– Yemon ChoiCommented Feb 3, 2016 at 12:25
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A sequence $(x_n)$ is weakly $p$ summable if the mapping $e_n\mapsto x_n$ extends to a bounded linear operator from $\ell_{p^*}$ into $X$. IMO, this is the right way to look at weakly $p$ summable sequences. So your question asks whether you can lift a certain linear operator into $X/Y$ back into $X$ through the quotient map. Take $X=L_r$ with $2<r<\infty$. Then $\ell_{p^*}$ is a quotient of $L_r$ if $r^*<p<2$, so you are asking for the identity on $\ell_{p^*}$ to lift to $L_r$ through the quotient map. This would force $\ell_{p^*}$ to embed isomorphically into $L_r$, which it does not.
answered Feb 4, 2016 at 0:04