uniformly bounded sequence in $L^2(0,T;H^1(\Omega))$ and $L^{\infty}(0,T;L^2(\Omega))$

Question

I'm considering a sequence $(v_\epsilon)$ which is $\epsilon-$uniformly bounded in both $L^2(0,T;H^1(\Omega))$ and $L^{\infty}(0,T;L^2(\Omega))$ where $\Omega$ is a bounded domain with sufficient regular boundary.
Then by the reflexivity and diagonalization argument, I can find a subsequence (but not relabel) which is convergent in the following sense: $$v_k\rightharpoonup v\mbox{ in }L^2(0,T;H^1(\Omega))$$ $$v_k\overset{*}{\rightharpoonup} v\mbox{ in }L^{\infty}(0,T;L^2(\Omega))$$ Then for any time $t\in(0,T)$, I want to compare the $L^2-$norm of $(v_\epsilon)$ and $v$ i.e. some relation like

$$\lVert v(t)\rVert_2\leq\liminf_{\epsilon\to0}\lVert v_\epsilon(t)\rVert_2$$

Is it possible to compare so?

Answer 1

No, this is not possible. You can even consider the easy situation in which $v_k$ does not depend on the spatial variable. Then, you have $v_k \stackrel*\rightharpoonup v$ in $L^\infty(0,T)$ and this is not enough to get $$ |v(t)| \le \liminf_{k\to\infty}|v_k(t)|.$$