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.
The Duals of $l^\infty$ and $L^{\infty}$
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1This paper could be helpful, I hope https://docs.google.com/file/d/0ByszALzBG-7kUlRDUlpwUERyeUE/edit?usp=sharing – 2013-10-22
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 Dunford-Schwartz, Linear Operators I, Theorem IV.8.16, page 296.
I should add that in that theorem Dunford-Schwartz 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 Radon-Nikodym 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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2'Ba' stands for bounded additive -- look [here](https://en.wikipedia.org/wiki/Ba_space) for more info. – 2016-03-18
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 Stone-Cech 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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0@Theo: Thanks for clearing that up. I'm upvoting your comment anyway because it makes some good points. – 2011-06-24