Can we identify the dual space of $l^\infty$ with another "natural space"? If the answer is yes, what can we say about $L^\infty$? By the dual space I mean the space of all continuous linear functionals.

1This paper could be helpful, I hope https://docs.google.com/file/d/0ByszALzBG7kUlRDUlpwUERyeUE/edit?usp=sharing – Oct 22 '13 at 21:58
2 Answers
Yes, if $(\Omega, \Sigma, \mu)$ is a (complete) $\sigma$finite measure space then $(L^{\infty}(\Omega,\Sigma,\mu))^{\ast}$ is the space $\operatorname{ba}(\Omega, \Sigma,\mu)$ of all finitely additive finite signed measures defined on $\Sigma$, which are absolutely continuous with respect to $\mu$, equipped with the total variation norm. The proof is relatively easy and can be found e.g. in DunfordSchwartz, Linear Operators I, Theorem IV.8.16, page 296.
I should add that in that theorem DunfordSchwartz treat the general case as well, which is a bit messier to state. The duality is the one you would expect, namely "integration". As far as I know, the bidual space of $L^{\infty}$ does not have an explicit analytic description, however (whatever that should mean precisely).
Moreover, the canonical embedding $L^{1}(\Omega,\Sigma,\mu) \to \operatorname{ba}{(\Omega,\Sigma,\mu)}$ is of course the map sending $f$ to the signed measure $d\nu = f\,d\mu$. In the $\sigma$finite case, the image of this map can be recovered by looking at the $\sigma$additive measures via the RadonNikodym theorem (and $\sigma$additivity corresponds of course precisely to weak$^{\ast}$continuity of the functionals on $L^{\infty} = (L^1)^{\ast}$).

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4'Ba' stands for bounded additive  look [here](https://en.wikipedia.org/wiki/Ba_space) for more info. – ec92 Mar 18 '16 at 18:19

Can this description of the dual of $L^\infty$ be leveraged to prove the existence of conditional expectation of an integrable random variable without invoking RadonNikodym? I asked a question about a "functional analytic" proof of existence of conditional expectations yesterday and this thread was linked in an answer. I guess yes, since you say that weak${}^*$ continuity $\iff$ $\sigma$additivity and $\sigma$additivity is clear for $Z\mapsto\mathbf{E}[ZX]$. Can you recommend a source for this material? – Olivier Bégassat Jul 25 '21 at 12:18
This is perhaps not such a nice characterization as finitely additive measures on $\mathbb N$, but it might be worth mentioning. (You can look it as follows: We obtain nicer measures  $\sigma$additive instead of finitely additive  on a more complicated space  $\beta\mathbb N$ instead of $\mathbb N$.)
The space $\ell_\infty$ is isometrically isomorphic to $C(\beta\mathbb N)$, hence the dual is isomorphic to $C^*(\beta\mathbb N)$.
More details about the correspondence between $\ell_\infty$ and the StoneCech compactification of integers can be found at wikipedia or in chapter 15 of Carothers' book A short course on Banach space theory.
Now, $C^*(\beta\mathbb N)$ is the space of regular Borel measures on $\beta\mathbb N$ by Riesz representation theorem.
In fact, Carothers goes the other way round. First, the dual of $C^*(\beta\mathbb N)$ is described via finitely additive measures. And then he uses this to prove Riesz representation theorem for compact spaces. (This proof of Riesz' representation theorem is due to Garling.)
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4Of course, I should have mentioned that. In fact, every $L^{\infty}$space is a space of the form $C(K)$ where $K$ is compact and extremally disconnected (= hyperstonian), thus every dual of $L^\infty$ can be implemented as space of measures on *some* compact space. I prefer the proof via $C^{\ast}$theory: look at the space $K$ of states on the commutative unital $C^{\ast}$algebra $L^{\infty}$ (this is of course somewhat circular, as $K$ is a subset of $(L^{\infty})^{\ast}$). – t.b. Jun 24 '11 at 13:19

4@Theo: I might misunderstand what you're saying, but if $L^\infty\cong C(K)$, then $K$ is the maximal ideal space (or character space), not the state space. – Jonas Meyer Jun 24 '11 at 14:51

1@Jonas: No, that was a plain mistake on my part (I simply confused characters and states  I'm no operator theorist). Sorry about that and thanks for the correction. – t.b. Jun 24 '11 at 14:58

@Theo: Thanks for clearing that up. I'm upvoting your comment anyway because it makes some good points. – Jonas Meyer Jun 24 '11 at 15:01