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.

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    $\begingroup$ 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 $X$. $\endgroup$ – GEdgar Dec 16 '15 at 17:24

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.

  • $\begingroup$ 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". $\endgroup$ – Alexander Shamov Dec 16 '15 at 22:36
  • $\begingroup$ @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 $L_1(\mu)$. $\endgroup$ – Tomek Kania Dec 17 '15 at 10:34

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 $F$.


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