If $u_n \in L^\infty(0,T;L^\infty)$ and $u_n \rightharpoonup^* u$ in $L^\infty((0,T)\times \Omega)$, is $u \in L^\infty(0,T;L^\infty)$?

Question

Let $\Omega$ be a smooth bounded domain. If $u_n \in L^\infty(0,T;L^\infty(\Omega))$ and $u_n \rightharpoonup^* u$ in $L^\infty((0,T)\times \Omega)$, is $u \in L^\infty(0,T;L^\infty(\Omega))$?

Note that $L^\infty(0,T;L^\infty(\Omega)) \subset L^\infty((0,T)\times\Omega)$ so it is not obvious.

The problem I think is whether $t \mapsto \lVert u(t) \rVert_{L^\infty(\Omega)}$ is measurable. We know $\lVert u(t)\rVert_{L^\infty(\Omega)}=\lim_{p \to \infty} \lVert u(t) \rVert_{L^p(\Omega)}$, so I believe it is measurable as the pointwise limit of measurable functions. Hence $u \in L^\infty(0,T;L^\infty(\Omega))$.

Answer 1

The problem is that $t \mapsto u(t) \in L^\infty(\Omega)$ might fail to be separably valued, i.e., it cannot be measurable in the Bochner sense.

The main counterexample is $$ u(t) = \chi_{[0,t]} $$ with $T = 1$ and $\Omega = (0,1)$.

Now consider the sequence of simple functions $$ u_n(t) = \chi_{[0, \lfloor n \, t\rfloor/n]}. $$ These functions belong to $L^\infty(0,1;L^\infty(0,1))$ and converge towards $u$ in the weak-* sense. Hence, the answer to your question is "no".