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.


2 Answers 2


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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    $\begingroup$ Quite a while later but what is "Ba" here? $\endgroup$ Commented Dec 10, 2015 at 10:47
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    $\begingroup$ 'Ba' stands for bounded additive -- look here for more info. $\endgroup$
    – ec92
    Commented Mar 18, 2016 at 18:19
  • $\begingroup$ 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 Radon-Nikodym? 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? $\endgroup$ Commented Jul 25, 2021 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 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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    $\begingroup$ Of 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 $(L^{\infty})^{\ast}$). $\endgroup$
    – t.b.
    Commented Jun 24, 2011 at 13:19
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    $\begingroup$ @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. $\endgroup$ Commented Jun 24, 2011 at 14:51
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    $\begingroup$ @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. $\endgroup$
    – t.b.
    Commented Jun 24, 2011 at 14:58
  • $\begingroup$ @Theo: Thanks for clearing that up. I'm upvoting your comment anyway because it makes some good points. $\endgroup$ Commented Jun 24, 2011 at 15:01

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