# Weak-star lower semicontinuity in $L^\infty$

Let $u_n \rightharpoonup^* u$ in $L^\infty(\Omega)$. Do we get something like $$\lVert u \rVert_{L^\infty} \leq \liminf_{n \to \infty} \lVert u_n \rVert_{L^\infty}$$ i.e. a weak-star lower semicontinuity?

This is because I want to know if $\lVert u_n \rVert_{L^\infty} \leq C$ for all $n$ if the limit $u$ also satisfies this.

Is this property true for Banach spaces in general?

Suppose $u_n \stackrel*\rightharpoonup u$ in some $X^*$. Given $\epsilon > 0$ choose some $x\in X$ with $\|x\| = 1$ and $|u(x)| \ge \|u\|-\epsilon$. We have $$\lim |u_n(x)| = |u(x)| \ge \|u\| - \epsilon$$ and on the other hand $$\lim |u_n(x)| \le \liminf \|u_n\|\|x\| = \liminf \|u_n\|$$ So $$\|u\| - \epsilon \le \liminf\|u_n\|$$ for each $\epsilon$ and we are done.
• Thanks... Did you really mean $X^*$ in your ifirst sentence? – asda Jul 17 '14 at 12:31
• @asda Yes. Otherwise weakly$^*$-convergence does not make sense. In the $L^\infty$-case, $X= L^1$, of course. – martini Jul 17 '14 at 13:09
• So if $u_n$ are uniformly bounded in $L^\infty$, they are uniformly bounded in the dual of $L^1$, so by Banach-Alagalou $u_n$ weak-star convergence in the dual of $L^1$, which is $L^\infty$. Thanks for clarying I forgot about it all. – asda Jul 17 '14 at 13:17