# weak* convergence for sequence in $L^\infty$

Let $\Omega \subset \mathbb{R}^d$ be a bounded and open set. Suppose $\{f_n\} \subset L^{\infty} (\Omega) , f \in L^{\infty} (\Omega)$. Prove that

$f_n \rightharpoonup^* f \ \ \text{in} \ \ L_{\infty} (\Omega) \Longleftrightarrow \left( \sup\limits_{n \in \mathbb{N} } \| f_n \|_{L^{\infty}} < + \infty \ \ \text{and} \ \ { \displaystyle \forall_{ V \subset \Omega} \ \lim\limits_{n \to \infty} \ \int\limits_V \ (f_n(x) - f(x)) =0} \right) \\ \text{where } V= \Pi^d_{i=1}(a_i,b_i).$

Can anyone give me any clue? I would appreciate any help.

• First prove this for hypercubes $\Omega = \prod_{i=1}^d (a_i, b_i)$, then try to generalise using that every bounded open set in $\mathbb R^d$ can be exhausted by open hypercubes. Dec 18 '14 at 21:18

## 1 Answer

$\Rightarrow$: a weak* convergent sequence is bounded. For the remaining part, use the definition of weak* convergence with the characteristic function of $V$.

$\Leftarrow$: if $g$ is integrable on $\Omega$, we can approximate in $\mathbb L^1(\Omega)$ this function by linear combinations of sums of charactistic functions of cubes.

• Worth mentioning in $\impliedby$ that boundedness is used to carry through the approximation. Jul 22 '17 at 0:07