$\ell^1$ vs. continuous dual of $\ell^{\infty}$ in ZF+AD 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?
 A: 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)
