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

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.

Answer

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}).

Attribution
Source : Link , Question Author : omar , Answer Author : Did

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