about weak convergence in $L^2(0,T;H)$. Exercise
Suppose $H$ is Hilbert space and $u_k$ converges weakly to $u$ in $L^2(0,T;H)$.
Suppose further we have the uniform bounds
$\mathrm{esssup}_{0≤t≤T} ||u_k(t)||≤C$.
Then $\mathrm{esssup}_{0≤t≤T} ||u(t)||≤C$.
I cannot prove this question.
I think that $u_k(t)$ converges weakly to $u(t)$ for every $t$, but I cannot.
Please tell me this question.
 A: Hint: by Mazur's lemma, we can pick a convex  combination $v_n$ of $u_k$ such that $v_n$ converges strongly to $u$. Then there is a subsequence of $v_n$ which converges to $u$ almost everywhere and it's easy to see $v_n$ satisfies the uniform bounds, so is its almost everywhere limit
A: Evan's hint:
For all $v\in H$
$$
\int_0^T (v,u_k)\leq C||v|| T.
$$
Since $L^2(0,T;H)$ is Hilbert, the assumption  $u_k \rightarrow u$ weakly in $L^2(0,T;H)$
reads 
\begin{equation}
\int_0^T (v,u(t)) dt=\lim_{k\to\infty}\int_0^T(v,u_k(t)),  \,\,\forall v\in L^2(0,T;H)
\end{equation}
Consider the particular case, $v\in L^2(0,T;H)$ as $v= w$ with $w\in H$ non-depending on time.
Using the hint 
$$
\int_0^T( v,u_k(t))dt\leq ||v|| C T
$$
Take the limit for $k\to\infty $ in the l.h.s.
$$
\int_0^T(v,u(t))dt\leq||v||C T, \,\,\forall v\in H
$$
Take now on both members of the last inequality the supremum on $||v||\leq 1$, we get
$$
\int_0^T ||u(t)||dt=\int_0^T\sup_{||v||\leq 1}(v,u(t))\leq CT
$$
From which $||u(t)||\leq C$ for all $t\in [0,T]$, thus $\text{ess}\sup_{[0,T]}||u||\leq C.$
