Is every weakly compact operator from $\ell_1$ into $c_0$ extendible to any larger space? Equivalently, is every weakly compact operator from $\ell_1$ into $c_0$ extendible to $\ell_\infty$?

@Joaquin: This one pushed me. It is, IMO, one of the nicest problems on Banach space theory asked on MO. The answer is no. For a counterexample, take any weakly compact operator $T:\ell_1 \to c_0$ that preserves $\ell_1^n$ uniformly for all $n$. In fact, since $\ell_1^n$ embeds isometrically into $\ell_\infty^{2^n}$, it is easy to construct one that has norm one so that for each $n$, there is a subspace of $\ell_1$ isometric to $\ell_1^n$ on which the operator acts isometrically. Embed $\ell_1$ isometrically into $\ell_\infty$ and assume that $T$ extends to an operator $S$ from $\ell_\infty$ into $c_0$. The operator $S$ necessarily preserves (copies of) $\ell_\infty^n$ uniformly for all $n$. A soft way of seeing this is to pass to an ultrapower, which will be an operator from some $C(K)$ space that is an isomorphism on a copy of $\ell_1$, hence cannot be weakly compact, whence preserves a copy of $c_0$ by Pelczynski's classical theorem. Taking adjoints, we see that $S^*$ is an operator from $\ell_1$ into an $L_1$ space that preserves $\ell_1^n$s uniformly, hence preserves a copy of $\ell_1$, whence is not weakly compact. But every operator from $\ell_\infty$ into a separable space is weakly compact. 


Edit: it seems that I probably misunderstood the question, see Bill Johnson's comments below. No. The identity map factors through $\ell^2$, so it is weakly compact (no doubt one can also see that directly). If it extended continuously to $\ell^\infty$ then this would give a projection of $\ell^\infty$ onto $c_0$, which is impossible. On the other hand, every weakly compact operator from $X$ to $Y$ extends to $X^{**}$ in a natural way. 

