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hi, I'm trying to prove that $\ell_\infty^*$ has the Dunford–Pettis property. It's enough to show that $\ell_\infty$ does not contain a copy of $\ell_1$ … but I'm having some trouble doing that. Can anyone help me ?

Thanks in advance.

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  • $\begingroup$ Homework? Maybe you should take this question over to Math.SE instead. It is more likely to be positively received there. Also, when you do ask, say a bit about what approaches you have tried. $\endgroup$
    – Harald Hanche-Olsen
    Commented Oct 23, 2012 at 7:39
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    $\begingroup$ Here is an addition to your homework assignment: Prove that $\ell_\infty$ has a subspace isometrically isomorphic to $\ell_1(2^{\aleph_0})$. $\endgroup$
    – Bill Johnson
    Commented Oct 23, 2012 at 13:57
  • $\begingroup$ right. Here's my approach. I found this result that says that $X^*$ has DPP iff $X$ has $DPP$ and doesn't contais a copy of $\ell_1$. So I'm trying to show that $\ell_\infty$ does not contain a copy of $\ell_1$ . I found a result that says that if $T\colon X \to \ell_1$ is bounded linear and onto then $X$ contains a complemented copy of $\ell_1$. It looked like I could use this to prove the result by contradiction, showing that $\ell_\infty$ has a complemented copy of $\ell_1$, which is absurd, since $\ell_\infty$ is prime. But I got stuck, I'm afraid I got the wrong road. $\endgroup$
    – user27480
    Commented Oct 23, 2012 at 14:53
  • $\begingroup$ Suppose $T\colon \ell_\infty \to \ell_1$ is an isomorphism onto it's image. Then $T^{-1}\colon \mathrm{Im} \; T \to \ell_1$ is onto, bounded and linear. Therefore $\mathrm{Im} \; T$ contains a complemented copy of $\ell_1$. So I can see that $\ell_1$ is a complemented subspace of a (not complemented) subspace of $\ell_\infty$. I can't seem to close the gap. Is this the right approach ? $\endgroup$
    – user27480
    Commented Oct 23, 2012 at 15:07
  • $\begingroup$ The question was posted at MSE: math.stackexchange.com/q/219454/8297 $\endgroup$
    – Martin Sleziak
    Commented Oct 23, 2012 at 15:15

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