# Weak-* convergence in $L^\infty((0,T)\times\Omega)$ implies weak-* convergence in $L^\infty(\Omega)$ for a.e. $t \in (0,T)$?

Let $$\Omega$$ be a bounded and smooth domain. Suppose I have a sequence of non-negative functions $$u_n \in L^\infty((0,1)\times \Omega) \cap L^\infty((0,1);L^\infty(\Omega))$$ with $$0 \leq u_n \leq 1 \quad \text{a.e}$$ and $$u_n \rightharpoonup^* u \quad\text{in L^\infty((0,1)\times \Omega)}$$ to some $$u$$.

Is it possible to conclude that for a subsequence, $$u_{n_j}(t) \rightharpoonup^* u(t)$$ in $$L^\infty(\Omega)$$ for a.e. $$t$$? The subsequence $$\{n_j\}$$ should not depend on the point $$t$$.

• For each $$n$$, $$u_n$$ is a piecewise constant function, i.e., $$u_n = \sum_{k=1}^n a_{kn}\chi_{I_{kn}}(t)$$ holds for a partition $$\{I_{kn}\}$$ and $$a_{kn} \in L^\infty(\Omega)$$.
• The sequence $$u_n(t)$$ is bounded uniformly in $$L^\infty(\Omega)$$ for a.e. $$t$$.
• If I understood correctly the question, this is not true. If $u_n(t,x)=\sin (2\pi n\, t)f(x)$ for a fixed $f \neq 0$, then $u_n \to 0$, $w^*$ in $(t,x)$ by Riemann Lebesgue, but for fixed $t\neq 0,1/2,1$ does not converge $w^*$ in $x$. Similarly, there is no way to find a sequence $n_j$ such that $\sin (n_j t)$ converges a.e. Positivity does not help: tafe $f$ positive and add 1 to the sinus. – Giorgio Metafune Feb 7 '20 at 15:02
If I understood correctly the question, this is not true. If $$u_n(t,x)=sin(2πnt)f(x)$$ for a fixed $$f\neq 0$$, then $$u_n \to 0$$, $$w^*$$ in $$(t,x)$$ by Riemann Lebesgue, but for fixed $$t\neq 0,1/2,1$$ does not converge $$w^*$$ in $$x$$. Similarly, there is no way to find a sequence $$n_j$$ such that $$sin(n_jt)$$ converges a.e. Positivity does not help: take f positive and add 1 to the sinus.