It is well known that the dual space of $L^\infty([0,1])$ is pretty large and I do not really have a feeling for it to be honest. Currently, I am interested in the following question:
Suppose that you have a bounded sequence $(f_n)_n$ in $L^\infty([0,1])$ that converges almost everywhere to some function $f \in L^\infty([0,1])$. Does it follow that $f_n \to f$ weakly?
I know that $\langle f_n, g \rangle \to \langle f, g \rangle$ for each $L^1([0,1])$ by Lebesgue's dominated convergence theorem but that suffice to show that $f_n \to f$ weakly? In particular, is $L^1$ dense in the dual space of $L^\infty$? If so, does the result still hold in for the Bochner space $L^\infty([0,1]; X)$, where $X$ is some Banach space? It would be also nice to see some reference, if the answer to that question is well known.
Best Answer
No. (But of course this does not work in ZF, it requires Hahn-Banach.)
We may consider the subspace $C[0,1]$ of continuous functions, $C[0,1] \subset L^\infty[0,1]$. The usual sup norm on $C[0,1]$ agrees with the norm obtained by restricting the norm on $L^\infty[0,1]$.
Here is a linear functional $\phi$ on $C[0,1]$: $$ \phi(f) = f(0)\qquad\text{for all } f \in C[0,1]. $$ This functional has norm $1$ as a functional on $C[0,1]$.
By the Hahn-Banach theorem, there is an extension: $\Phi$ is a linear functional on $L^\infty[0,1]$, $\Phi(f) = \phi(f)$ if $f \in C[0,1]$, and $\Phi$ has norm $1$.
Now consider the sequence $f_n \in L^\infty[0,1]$ $$ f_n(x) = \begin{cases}1-nx,\quad & 0\le x \le \frac{1}{n},\\ 0,\qquad & \frac{1}{n} < x \le 1.\end{cases} $$ Also consider the constant function $f(x) = 0$ for all $x$.
Note that $f_n, f$ are in $L^\infty[0,1]$ and $f_n(x) \to f(x)$ for almost all $x \in [0,1]$.
But also $f_n, f \in C[0,1]$ so $\Phi(f_n) = \phi(f_n) = 1$ for all $n$ while $\Phi(f) = \phi(f) = 0$. So $f_n$ does not converge weakly to $f$.