# Application Banach-Alaoglu Theorem

'' Let $f_n$ be a bounded sequence of functions in $L^p$. Then there exists a subsequence $f_{n_k}$ and an $f\in L^p$ such that $$\int f_{n_k}g\rightarrow\int fg$$ for all $g\in L^q$ (where $1/p+1/q=1$). The corresponeding result for $p=1$ is not true, as $L^1$ is not reflexive.''

I'm working in $\Omega\subseteq\mathbb{R}^n$. If $f\in L^1(\Omega)$, then $f:L^{\infty}(\Omega)\rightarrow \mathbb{C}$ defined by $f(g)=\int_{\Omega} fg$ is a bounded linear operator. Also, the norm of $f$ as a function coincides with the norm of $f$ as an operator by taking $$g(x)=\begin{cases} |f(x)|/f(x),\,f(x)\neq0 \\ 0,\,f(x)=0 \end{cases}\in L^{\infty}(\Omega)$$ (note that $fg=|f|$). Then, by Banach-Alaoglu Theorem, if $(f_n)\subseteq L^1(\Omega)$ is bounded, there exists a subsequence $f_{n_k}$ and $f\in L^1(\Omega)$ with $\int f_{n_k}g\rightarrow\int fg$ for all $g\in L^{\infty}(\Omega)$.

Then I don't understand the assertion from Wikipedia. I mean, it's true that $L^1(\Omega)^{**}\neq L^1(\Omega)$, but I'm not using that. Could you clarify all this for me?

EDIT: I think I understand the counterexample, but I'd like to know what's wrong in my reasoning.

When you say "then $f:L^{\infty}(\Omega)\rightarrow \mathbb{C}$ is bounded functional", it means $f\in (L^\infty)^*$, right? Then the convergence of $f_{n_k}\in L^1$ is going to be also as elements of $(L^\infty)^*$, and the limit may be not in $L^1$. (Note that $L^1$ is not closed in $(L^\infty)^*$ in the norm/weak topology.)
You cannot apply Banach-Alaoglu for $L^1(\Omega)$, since $L^1(\Omega)$ is not the dual space of a normed space. Rather you have to embed $L^1(\Omega)$ into larger spaces $L^\infty(\Omega)^*$ or $C(\bar\Omega)^*$ to obtain a weak-star convergent subsequence.
To see that for $p=1$ the assertion is not true, consider the sequence $f_n(x)=n \chi_{0,1/n}(x)$ on $(-1,1)$. Then $$\int_\Omega f_n g \to g(0)$$ for all continuous functions $g$. The functional $g\mapsto g(0)$ cannot be represented by an integral with $L^1$-functions.