Suppose $u_n$ is bounded in $L^\infty(\Omega)$. $\|u_n\|_{L^\infty(\Omega)}<M$.
Then $u_{n_k}\to u$ weak star in $L^\infty(\Omega)$ for some $n_k\uparrow \infty$ and $u\in L^\infty$.
I want to know why this is true? Some reference is welcomed.
Beside, can we get $\|u\|_{L^\infty(\Omega)}<M$? Like weak convergence norm has some lower semicontinuity like $\|u\|_{L^\infty(\Omega)}\le\mathop{\liminf}\limits_{{k\rightarrow\infty}}\|u_{n_k}\|_{L^\infty(\Omega)}$.
Thanks for any help.
Best Answer
Assuming that $L^1(\Omega)$ is separable, this is a special case of the sequential form of the Banach-Alaoglu theorem: the closed ball of radius $M$ in $L^\infty(\Omega)$ is then weak*-compact and metrizable, hence every bounded sequence of elements $u_n \in L^\infty(\Omega)$ such that $\lVert u_n \rVert_\infty \leq M$ has a weak*-convergent subsequence $u_{n_k} \to u$. This also ensures that $\lVert u \rVert \leq M$.
However, from $\lVert u_{n} \rVert_\infty \lt M$ for all $n$ we cannot conclude that $\lVert u \rVert_\infty \lt M$, only that $\lVert u \rVert_\infty \leq M$. Consider for example $u_{n} = (1-\frac{1}{n}) \chi_{\Omega}$.
Separability of $L^1(\Omega)$ holds whenever the $\sigma$-algebra is countably generated, so e.g. for subsets of $\mathbb{R}^n$ with the Borel $\sigma$-algebra.