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A friend of mine found a book in which the author said that the dual space of $L^\infty$ is $L^1$, of course not with the norm topology but with the weak-* topology. Does anyone know where I can find this result? Thanks.

Tomasz Kania
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    Perhaps the assertion is: the dual of $L^\infty$ with the weak* topology $\sigma(L^\infty,L^1)$ is $L^1$. More generally, for any Banach space $X$, the dual of $X^$ with its weak topology is $X$. – GEdgar Dec 16 '15 at 17:24
  • Hello, do you have a good reference for this fact? I have tried Rudin & Brezis but have not found anything about $\left(X^\right)^ \equiv X$ with the weak* topology. – siou0107 Aug 20 '20 at 21:55
  • I included a reference in my answer below. I know for a fact that Rudin's book does not cover the Mackey-Arens theorem, and I have seen many functional-analysis textbooks that do not either. – cruiser0223 Feb 17 '21 at 14:05
  • @siou0107 a bit late to the game, but that result is Theorem 3.10 in Rudin's Functional Analysis. He gives a nice short proof for it. – jet457 Feb 02 '25 at 16:23

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For any $C(K)$-space we have $C(K)^*\cong L_1(\mu)$ for some usually humongous measure $\mu$. See the proof of Proposition 4.3.8(iii) in

F. Albiac, N.J. Kalton, Topics in Banach Space Theory, Grad. Texts in Math. 233, Springer, 2006.

Of course, $L_\infty(\nu)\cong C(K)$ for some compact, Hausdorff space $K$. However, there is no clear relation between the measures $\mu$ and $\nu$. In fact, if $L_\infty(\nu)$ is infinite-dimensional, then $\mu$ is not even $\sigma$-finite.

Tomasz Kania
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  • Another way to see this is as follows: the dual of $C(K)$ is the space of all finite regular measures on $K$ with the total variation norm. This can indeed be represented as $L^1(\mu)$ - if the latter is defined appropriately for non-sigma-finite $\mu$. This $\mu$ can be, e.g., a "disjoint sum" of a maximal family of mutually singular measures on $K$, representing the "largest measure class". – Alexander Shamov Dec 16 '15 at 22:36
  • @AlexanderShamov, this is precisely the proof I am referring to. Another proof would be to notice that $C(K)^*$ is an abstract $L$-space, hence by Kakutani's representation theorem it is of the form $L_1(\mu)$. – Tomasz Kania Dec 17 '15 at 10:34
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There is a general fact from duality of linear spaces (see Proposition 4.28 in Fabian-Habala-Hajek-Montesinos-Pelant-Zizler, Functional Analysis and Infinite-Dimensional Geometry): If we consider a linear subspace $F$ in the space of linear functionals on $E$, then the space of linear functionals on $E$ continuous in the corresponding weak topology on $E$ coincides with $F$.

Mikhail Ostrovskii
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This is a consequence of the Mackey-Arens Theorem, see for instance Reed, Simon: Methods of modern mathematical physics, Theorem V.22. It says:

For a dual pair $(X,Y)$, a loc. conv. topology $\mathcal{T}$ is a dual-$(X,Y)$ topology if and only if it is stronger than the $\sigma(X,Y)$-topology (i.e. the weak topology) and weaker than the Mackey-topology (called the $\tau(X,Y)$-topology).

Now choose $X = L^\infty$ and $Y = L^1$, hence the $\sigma(L^\infty,L^1)$-topology is the dual-$(L^\infty,L^1)$ topology, i.e. the topological (or continuous) dual of $(L^\infty, \sigma(L^\infty,L^1)$) is $L^1$.

To notice that $(L^\infty,L^1)$ is a dual pair, see for instance this wikipedia article or the above textbook.

Since you asked for references, I am certain this result is present in most textbooks on topological vector spaces.

cruiser0223
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