Weak convergence in Bochner space and corresponding $L^p$ space

Question

I'm considering a sequence of functions $\{u_k\}$ which is defined on $\Omega\times(0,T)$ where $\Omega$ is a bounded domain with Lipschitz boundary.
Assume we now have a uniform bound that is
$$\sup_{t\in(0,T)}\{\lVert u_k(t)\rVert_{L^2(\Omega)}^2\}\leq M$$
where $M$ is a positive constant which does not depend on $k$ and $T$.
Then I think this means that for any time $t\in(0,T)$
$$\lVert u_k(t)\rVert_{L^2(\Omega)}^2\leq M,$$
so by reflexivity of $L^2$ space, for each time, we can find a weak limit (up to a subsequence but not relabel) say $u(t)$ i.e.
$$u_k(t)\rightharpoonup u(t)\mbox{ in }L^2(\Omega)\mbox{ for every }t\in(0,T).$$
Then I also think our uniform bound on $u_k$ implies that
$$\int_{0}^{T}\lVert u_k(t)\rVert_{L^2(\Omega)}^2dt\leq MT$$
which gives me a uniform bound in $L^2((0,T);L^2(\Omega))$. Then again by the reflexivity, we can have weak convergent subsequence, but still not relabel and name this weak limit $u^\prime$.
Hence we have that $$u_k\rightharpoonup u\mbox{ in }L^2(0,T;L^2(\Omega)).$$
Now I want to claim $u=u^\prime$ up to a set of zero measure in $\Omega\times(0,T)$.
Does my thought make sense or there is something wrong? Thank you for any reply!

Answer 1

The answer is no. For example let $\Omega=(0,1)$ and $u_{2k}(t,\omega)\equiv 1,\ u_{2k+1}(t,\omega)\equiv 0.$ Then for $0< t\le T/2 $ we choose the subsequence $u_{2k}$ which tends to $1.$ For $T/2<t<T$ we choose the subsequence $u_{2k+1}$ which tends to $0.$ Hence $$u(t,\omega)=\begin{cases} 1 & 0< t\le T/2\\ 0 & T/2<t< T \end{cases} $$ In the space $L^2((0,T)\times \Omega)$ we choose the subsequence $u_{2k},$ hence $u'(t,\omega)\equiv 1.$ Thus $u\neq u'.$

Remark The main obstacle is that in the first approach the choice of the subsequence depends on every single $t\in (0,T).$