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Let the base field be the real numbers or the complex numbers (I don't think it will matter).

Let (\ell^{\infty})' be the continuous dual of the Banach space $\ell^{\infty}$.
Let \: f : \ell^1 \to (\ell^{\infty})' \: be the obvious embedding.

Does ZF+AD prove that $f$ is surjective?
Does ZF+DC+AD prove that $f$ is surjective?

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    @Ricky: I was in Jerusalem today attending a course by Menachem Magidor about descriptive set theory. I asked him this question during the break, he couldn't give me an answer, and said he'll think about it. $A$s luck would ha$v$e it, I'm seeing him on Friday as well. If he won't have an answer by then, I'm going to bet that there isn't a written one (I did try and search for it in several places). If you are not in a rush, by the end of the academic year I should be capable to attempt this proof on my own.2012-01-30

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By a random chance, in this answer t.b. posted an answer which deals with models of $\mathrm{ZF+DC+PM_\omega}$. The latter in fact states that $\ell^1$ is reflexive.

Martin Väth, The dual space of $L^\infty$ is $L^1$, Indag. Mathem., N.S., 9 (4), 1998, 619–625.

It is mentioned that the axiom $\mathrm{PM_\omega}$ holds in Solovay's model. The paper itself cites both Solovay's original paper as well a paper by David Pincus which I was not able to find online (MR link).

Assuming $\mathrm{AD}$ holds in $L(\mathbb R)$ implies it is indeed a Solovay model, so the above should be applicable (since $\mathrm{AD}+V=L(\mathbb R)$ implies $\mathrm{DC}$).

Lastly, one of the remarks was that it is the fact that every set of real numbers has the Baire property which implies $\mathrm{PM}_\omega$, this was later shown consistent without an inaccessible cardinal, by Shelah.

Saharon Shelah, Can you take Solovay's inaccessible away?, Israel Journal of Mathematics 48 (1): 1–47.


While the above answers perfectly the second question, it can be done in a clearer way via Fremlin's Measure Theory, and in particular Vol. 5, Ch. 6 whose last section deals with $\mathrm{ZF+AD}$.

It is not clear to me whether or not the arguments brought in that chapter are sufficient to answer the first question positively, though.

(the results themselves are available in a .ps file which you can convert to .pdf here)

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    this should be a comment to Asaf's answer but I don't have the privilege to comment. @Ricky: It is always the definition of reflexivity that the *canonical map* $X \to X^{\ast\ast}$ given by evaluation is an isomorphism. Assuming Hahn-Banach and in many explicit examples, this reduces to checking surjectivity. But even with full choice having just *some isomorphism* is not good enogh. Look up the *James space*, which is isometrically isomorphic to its bidual but not reflexive.2012-02-24