Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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: 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 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.