Weak star convergence in $L^\infty(0,T;L^2(\Omega))$ and pointwise weak limits

Let $\Omega$ be a bounded domain of $\mathbb{R}^n$ and $u_n \colon [0,T] \to L^2(\Omega)$ a sequence of functions such that $||u_n(t)||_{L^2(\Omega)} \leq C$ for every $n$ and for every $t \in [0,T]$.

We know that $u_n \rightharpoonup^* u$ in $L^\infty(0,T;L^2(\Omega))$ to some function $u \in L^\infty(0,T;L^2(\Omega))$, i.e. $$\int_0^T \int_\Omega u_n(t,x) v(t,x) \,dxdt \to \int_0^T \int_\Omega u(t,x) v(t,x) \,dxdt \quad \text{for all } v \in L^1(0,T;L^2(\Omega)).$$ At the same time, for each fixed $t \in [0,T]$, we have $u_n(t) \rightharpoonup w^t$ in $L^2(\Omega)$ to some function $w^t \in L^2(\Omega)$, i.e. $$\int_\Omega u_n(t,x)v(x)\,dx \to \int_\Omega w^t(x)v(x)\,dx \quad \text{for all } v \in L^2(\Omega).$$

Is it true that $u(t) = w^t$ for almost every $t \in [0,T]$?

Define the sequence $f_n(t)=\sin(n \pi t)$. Then it holds $f_n \rightharpoonup^*0$ in $L^\infty(0,1)^*$. The sequence converges pointwise to zero only on a set of zero measure. Now multiply $f_n$ with a not-null $L^2(\Omega)$ function $g$. Then $f_ng$ converges weak star to $0$. It converges to zero only on a set of zero measure.
• Thanks @daw So loosely speaking, taking $v \in L^2(\Omega)$, what prevents me from doing something like $\lim_n \int_0^T \int_\Omega u_n v \,dx\,dt= \int_0^T \lim_n \int_\Omega u_n(t) v\,dx\,dt = \int_0^T \int_\Omega w^t v \,dx\,dt$ is the fact that the sequence $F_n(t) = \int_\Omega u_n(t) v \,dx$ has no pointwise limit right? So in general I need some stronger control on $\{F_n\}_n$... – Mauro Feb 6 '18 at 8:54