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 $X=c\_{0},l\_{p}(1<p<\infty)$.



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