Let $X$ be a Banach space. My question is: $X$ contains no copy of $l_{1}$ if and only if any operator from $X$ to $l_{1}$ is compact? I guess that the necessary part may be true. But is the sufficient part true? At least, the sufficient part is true for $X=c_{0},l_{p}(1<p<\infty)$.
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asked Sep 28, 2015 at 21:47
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Since weakly compact operators into $\ell_1$ are compact, and since by a result of Kadec and Pelczynski every non-weakly compact operator into $\ell_1$ fixes a copy of $\ell_1$, we have that if $X$ contains no copy of $\ell_1$ then every operator from $X$ to $\ell_1$ is compact.
However, the converse is not true in general. Since $ C [0,1] $ is universal for separable Banach spaces, it contains a copy of $\ell_1$, but every operator from $ C [0,1] $ is compact; indeed, Pelczynski showed that non-weakly compact operators from a $ C (K) $ space fix a copy of $ c_0$, but there is no copy of $ c_0$ in $\ell_1$.
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answered Sep 28, 2015 at 22:27
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4$\begingroup$ There is a non compact operator from $X$ into $\ell_1$ iff and only if $\ell_1$ is isomorphic to a complemented subspace of $X$. For that you need one additional comment; namely, that every subspace of $\ell_1$ contains a small complemented subspace that is isomorphic to $\ell_1$. $\endgroup$ Commented Sep 28, 2015 at 22:31
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$\begingroup$ Could you give a sketch of the proof of the necessary part? I am not sure. $\endgroup$ Commented Sep 28, 2015 at 23:06
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$\begingroup$ Thanks, Philip. Your counterexample shows that the condition "$X$ contains no copy of $l_{1}$" does not characterize that any operator from $X$ into $l_{1}$ is compact. $\endgroup$ Commented Sep 28, 2015 at 23:33
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$\begingroup$ @BillJohnson Could you give a sketch of the proof of your characterization? I am not sure. $\endgroup$ Commented Sep 29, 2015 at 12:23
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$\begingroup$ @BillJohnson I am able to prove your characterization. I show that if there is a non-compact operator from $X$ into $l_{1}$, then $X^{*}$ contains a copy of $c_{0}$. $\endgroup$ Commented Sep 29, 2015 at 14:12