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?
Best Answer
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.