Problem of Functional Analysis (weak convergence) I want to ask a problem of functional analysis, it says:
You have a sequence of $\{f_{n}\}\in L^{1}(\Omega)\cap L^{\infty}(\Omega)$ and you know that $\parallel f_{n}\parallel_{L^{1}(\Omega)} \le C$ (C is a real constant) and $\{f_{n}\}\rightharpoonup f \in L^{\infty} weak^{*}$. Prove that $f\in L^{1}(\Omega)$ and $\parallel f\parallel_{L^{1}(\Omega)}\le C$.
I don´t know how to prove that.
 A: First of all, I imagine it's not true if $\Omega$ is not $\sigma$-finite.  Indeed, it's not clear to me that in general $L^{\infty}(\Omega )$ has a predual, in which case it wouldn't make sense to speak of weak-$^*$ convergence.  (That said, I didn't try too hard to find an example of a measure space for which it was easy to prove that $L^{\infty}(\Omega )$ did not have a predual.)  Certainly though, if $\Omega$ is $\sigma$-finite, then $L^1(\Omega )^*\cong L^{\infty}(\Omega )$ canonically.
So, take $\Omega$ to be $\sigma$-finite, and write $\Omega =\bigcup _{m\in \mathbb{N}}\Omega _m$ for $\Omega _m\subseteq \Omega _{m+1}$ and $\mu (\Omega _m)<\infty$.
Now, try using the fact that for $g\in L^1(\Omega )$,
$$
\| g\| _1=\sup \left\{ \left| \int _{\Omega}gh\right| :\| h\| _{\infty}\leq 1\right\} .
$$
(In general, if $V$ is a normed vector space and $v\in V$ then $\| v\| =\sup \left\{ |\phi (v)|:\phi \in V^*,\| \phi \| \leq 1\right\}$.)
First, note that in fact
$$
\| g\| _1=\sup \left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\} ,
$$
for if $g\in L^1$ and $h\in L^{\infty}$, then there is a sequence $m\mapsto h_m\in L^1\cap L^{\infty}$ such that $\lim _m\int fh_m=\int fg$ (namely $h_m:=\chi _{\Omega _m}g$).  Indeed, this equality still holds even in the case $\| g\| _1=\infty$ (note that in this case if we only take $h\in L^{\infty}$ and not $h\in L^1\cap L^{\infty}$, $\int gh$ need not be defined), for in this case, without loss of generality $\int g_+=\infty$ (for simplicity, I'm taking $g$ to be real-valued), in which case we can the supremum is at least as large as $|\int g\chi _{\Omega _m}\chi _{\{ x\in \Omega :g(x)\geq 0\}}|=\int _{\Omega _m}g_+$, which diverges.
Now, for $h\in L^1\cap L^{\infty}$ with $\| h\| _{\infty}\leq 1$,
$$
\left| \int fh\right| =\left| \lim _m\int f_mh\right| =\lim _m\left| \int f_mh\right| \leq \liminf _m\int |f_mh|\leq \| h\| _{\infty}\liminf _m\int |f_m|\leq C.
$$
Taking the supremum over all such $h$, we obtain $\| f\| _1\leq C$.
UPDATE:  Below I provide more detail for the proof that $\| g\| _1=\sup \left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\}$.
First, suppose that $\| g\| _1<\infty$.  From $\| g\| _1=\sup \left\{ \left| \int _{\Omega}gh\right| :\| h\| _{\infty}\leq 1\right\}$, we have right away that $\| g\| _1$ is an upper bound of $\left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\}$.  To show that it is the least upper-bound, let $\varepsilon >0$.  Pick $h\in L^{\infty}$ with $\| h\| _{\infty}\leq 1$ and
$$
\| g\| _1-\varepsilon <\left| \int _{\Omega}gh\right| \leq \| g\| _1.
$$
The sequence $m\mapsto \chi _{\Omega _m}gh$ is then (essentially) bounded by the integrable function $g$ and converges pointwise to $gh$, and so by the Dominated Convergence Theorem, $\int _{\Omega}gh=\lim _m\int _{\Omega}gh\chi _{\Omega _m}$.  In particular, there is some $M\in \mathbb{N}$ such that
$$
\left| \int _{\Omega}gh\right| -\left| \int _{\Omega}gh\chi _{\Omega _m}\right| \leq \left| \left| \int _{\Omega}gh\chi _{\Omega _m}\right| -\left| \int _{\Omega}gh\right| \right| \leq \left| \int _{\Omega}gh\chi _{\Omega _m}-\int _{\Omega}gh\right| <\varepsilon
$$
for all $m\geq M$.  But then,
$$
\| g\| _1-2\varepsilon <\left| \int _{\Omega}gh\right|+\left( \left| \int _{\Omega}gh\chi _{\Omega _M}\right| -\left| \int _{\Omega}gh\right| \right) =\int _{\Omega}gh\chi _{\Omega _M}\leq \| g\| _1.
$$
But $h\chi _{\Omega _M}$ is in $L^1\cap L^{\infty}$ (and $\| h\chi _{\Omega_M}\| _{\infty}\leq 1$), giving us $\| g\| _1=\sup \left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\}$, as desired.
Now suppose that $\| g\| _1=\infty$.  Without loss of generality, suppose that $\| \Re (g)_+\| _1=\infty$ (here, for any real-valued function $h$, I am writing $h_+$ for the function that is equal to $h$ when $h\geq 0$ and equal to $0$ otherwise).  To condense notation, let us write $P:=\{ x\in \Omega :[\Re (g)](x)\geq 0\}$.  Then, $\chi _{\Omega _m}\chi _P\in L^1\cap L^{\infty}$ (and of course $\| \chi _{\Omega _m}\chi _P\| _{\infty}\leq 1$), and so
$$
\sup \left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\} \geq \left| \int _{\Omega}g\chi _{\Omega _m}\chi _P\right| \geq \int _{\Omega}\chi _{\Omega _m}\Re (g)_+.
$$
Taking the limit with respect to $m$, we see that $\sup \left\{ \left| \int _{\Omega}gh\right| :h\in L^1\cap L^{\infty},\| h\| _{\infty}\leq 1\right\} =\infty =\| g\| _1$, as desired.
