# A question on characterizing a Banach space containing no copy of $l_{1}$

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 $X=c_{0},l_{p}(1<p<\infty)$.

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 $\ell_1$.
• 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 $\ell_1$. – Bill Johnson Sep 28 '15 at 22:31
• 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. – Dongyang Chen Sep 28 '15 at 23:33
• @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 $c_{0}$. – Dongyang Chen Sep 29 '15 at 14:12