about weak convergence in $L^2(0,T;H)$.

Question

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.

Answer 1

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

Answer 2

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 $\text{ess}\sup_{[0,T]}||u||\leq C.$