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.
Best Answer
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.