Dual space of $L^\infty$ is $L^1$ with the weak-* topology?

Question

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.

Answer 1

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.

Answer 2

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 $F$.

Answer 3

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.