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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 $\ell_\infty$?

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Yemon's answer probably does it. But I'm not quite sure what "extendible" means. For your second question, do you mean to map $\ell^1\rightarrow\ell^\infty$ by the formal identity map-- this is of course not an embedding... –  Matthew Daws Mar 13 '12 at 17:46
    
Do you mean "extendible to any larger space that contains $\ell^1$ as a closed subspace"? –  Yemon Choi Mar 13 '12 at 18:11

2 Answers 2

up vote 4 down vote accepted

@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.

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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.

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Why would the extension give a projection, Yemon? –  Bill Johnson Mar 13 '12 at 19:56
    
Anyway, the identity $i_{1,2}$ from $\ell_1$ to $\ell_2$ has an extension to any space containing $\ell_1$ isomorphically, because $i_{1,2}$ is $2$-summing (and thus factors through $\ell_\infty$), hence also the identity from $\ell_1$ to $c_0$ has an extension. –  Bill Johnson Mar 13 '12 at 20:06
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In his Memoirs "Extension of compact operators", on page 19 Lindenstrauss remarks that the formal identity from $\ell_2$ into $c_0$ does not have an extension to any copy of $\ell_\infty$ containing $\ell_2$ isomorphically. More generally, since $\ell_\infty$ has the Dunford-Pettis property and every bounded linear operator from $\ell_\infty$ to $c_0$ is weakly compact, there are severe restrictions on operators into $c_0$ which can be extended to a containing copy of $\ell_\infty$. –  Bill Johnson Mar 13 '12 at 20:49
    
Good points, Bill. I (mis)understood the question as asking for any extension to $E$ which contains a continuous linear image of $\ell^1$. See Matt's comment above. –  Yemon Choi Mar 13 '12 at 22:08
    
Thank you very much, Professor Johnson, for your beautiful answer. Joaquin –  Joaquin M. Gutierrez Mar 14 '12 at 16:15

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