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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 ?

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.

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 ?

Thanks in advance.

Rafael
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  • Every separable Banach space is isometric to a subspace of $\ell_\infty$. See here. So your approach is wrong ("does not contain a copy of" does not mean a $T$ such as in your post exists). Also, where did you find the characterization of DPP mentioned in your second paragraph? As far as I know, only the reverse implication holds. – David Mitra Oct 23 '12 at 15:43
  • What is true is that the dual $X^*$ of a Banach space $X$ has the Schur property (weakly convergent sequences are norm convergent) if and only if $X$ has the DPP and does not contain a copy of $\ell_1$. – David Mitra Oct 23 '12 at 15:53
  • ohh.... thanks Mitra ! I found that result here. page 79:

    http://www.pg.im.ufrj.br/teses/Matematica/Mestrado/297.pdf

    it's in portuguese. So, since $\ell_\infty$ has the DPP and contains $\ell_1$ ... $\ell_\infty^*$ is not Schur ...

    I guess that's the wrong approach then ... ! any ideas ?

     – Rafael Oct 23 '12 at 16:09

2 Answers2

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Every $\mathcal{M}(K)=C(K)^*$ space has the Dunford Pettis Property (see, for example, Corollary $5.4.6$ of Albiac and Kalton's Topics in Banach Space Theory). Since $\ell_\infty=C(\beta\mathbb{N})$, the result is immediate.

Michael Causey
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Schaefer's Banach lattices and positive operators, II.9.9 shows that every AL-space has the Dunford-Pettis property. $\ell_\infty^*$ is an AL-space.

Denny Leung
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