# The Duals of l^\inftyl^\infty and L^{\infty}L^{\infty}

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

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 $L^{\infty} = (L^1)^{\ast}$).