Weak star convergent sequence in $L^\infty(0,T; L^2(\Omega))$ Given a sequence $(u_n)_{n\in \mathbb{N}} \subseteq L^\infty (0,T; L^2(\Omega)) \cap H^1(0,T;L^2(\Omega))$ with
\begin{align*}
u_n \overset{\ast}{\rightharpoonup} u \,\, \text{ in } \,\, L^\infty (0,T; L^2(\Omega))
\end{align*}
where $T>0$ and $\Omega \subseteq \mathbb{R}^{42}$ is an open set, does the inequality
\begin{align*}
\liminf_{n \to \infty} \|u_n(T)\|_{L^2(\Omega)} \geq \|u(T)\|_{L^2(\Omega)}
\end{align*}
hold? I am thinking about the weak lower semicontinuity of $\|\cdot \|_{L^2(\Omega)}$, but for this I would need weak convergence of $(u_n(T))_{n\in \mathbb{N}}$ in $L^2(\Omega)$ which feels awkward.
Note that evaluating $u_n$ at the point $T$ makes sense because one has the embedding
\begin{align*}
H^1(0,T;L^2(\Omega)) \hookrightarrow \mathcal{C}([0,T],L^2(\Omega)).
\end{align*}
I am happy about any kind of help. Thanks in advance.
 A: I will use the  additional assumption (mentioned in your answer):
The sequence $(u_n')_{n \in \mathbb{N}} $ is bounded in $L^\infty(0,T;L^2(\Omega))$
Under this assumption, we can check that $u_n' \stackrel*\rightharpoonup u'$ in $L^\infty(0,T;L^2(\Omega))$.
Now, we have the identity
$$
u_n(T) = u_n(t) + \int_t^T u_n'(s) \, \mathrm{d}s.
$$
Integration over $t$ implies
$$
T \, u_n(T) = \int_0^T u_n(t) + \int_t^T u_n'(s) \, \mathrm{d}s \, \mathrm{d}t
= \int_0^T u_n(t)  \, \mathrm{d}t + \int_0^T s \, u_n'(s) \, \mathrm{d} s.
$$
From the weak-* convergence of $u_n$ and $u_n$, we can infer
$$
u_n(T) \rightharpoonup u(T)
$$
in $L^2(\Omega)$. This implies the desired inequality.
A: Apparently I have an additional assumption: 
The sequence $(u_n')_{n \in \mathbb{N}} $ is bounded in $L^\infty(0,T;L^2(\Omega))$
Having this assumption I will make a try of proving my desired inequality.
Given that the above sequence is bounded we have $u_n' \overset{\ast}{\rightharpoonup} v$ in $L^\infty(0,T;L^2(\Omega))$ for some $v \in L^\infty(0,T;L^2(\Omega))$. It then follows that $u \in H^1(0,T;L^2(\Omega))$ with $u'=v$. So now the evaluation of $u$ in the point $T$ makes sense because we can again use the embedding into the space of continuous functions.
The weak star convergence of $(u_n)_{n \in \mathbb{N}}$ and $(u_n')_{n \in \mathbb{N}}$ in $L^\infty(0,T;L^2(\Omega))$ implies weak convergence of $(u_n)_{n \in \mathbb{N}}$ and $(u_n')_{n \in \mathbb{N}}$ in $L^2(0,T;L^2(\Omega))$. I would now assume that 
\begin{align*}
u_n \rightharpoonup u \,\,\text{ in }\,\, H^1(0,T;L^2(\Omega))
\end{align*}
holds. I know that it holds in the real valued Sobolev space case and I guess that the proof can easily be adapted to the Bochner setting.
Now the compact embedding $H^1(0,T;L^2(\Omega)) \overset{c}{\hookrightarrow} L^2(0,T;L^2(\Omega))$ (here we should maybe upgrade $\Omega$ to be a bounded Lipschitz domain) implies 
\begin{align*}
u_n \rightarrow u \,\,\text{ in }\,\, L^2(0,T;L^2(\Omega))
\end{align*}
up to a subsequence. Finally we get
\begin{align*}
u_n(t) \rightarrow u(t) \,\,\text{ in }\,\, L^2(\Omega) \,\,\text{ for almost any t $\in$ [$0$,$T$]} 
\end{align*}
up to another subsequence and because of the continuity of $u$ and $u_n$ for every $n \in \mathbb{N}$ we get the convergence for all $t \in [0,T]$. Hence especially for $T$ so that the desired inequality even holds with equality.
