Weak limit in Bochner-Sobolev space

Question

Let "$V\subset H\subset V^*$" be an evolution triple. Let $(u_n)$ be a sequence of elements from $W^{1,2}(0,T;V,H)$ such that

$$u_n(t)\to u_1(t)\qquad\text{weakly in} \qquad V, $$ $$u_n'(t)\to u_2(t)\qquad\text{weakly in} \qquad V^*, $$ for a.e. $t\in (0,T)$. Does it imply that $$u_1(t)=\int_{0}^tu_2(s)\,ds+u_1(0)$$ and the weak limit of $(u_n)$ belongs to $W^{1,2}(0,T;V,H)?$

Answer 1

You need the following result. The assumptions in the comment is enough. In the question itself it is not enough with the almost everywhere convergence. You would need $L^1$ convergence in time (as stated below). You can also get that using the Lebesgue dominated convergence theorem in your case, if the sequence is assumed to be bounded in $W^{1,1}(0,T;V,H)$.

Theorem: If there is a sequence $\{u_n\}_{n\in \mathbb{N}} \subset W^{1,1}(0,T;V,H)$ such that $$\begin{aligned} &u_n \rightharpoonup u \quad\text{ weakly in } L^1(0,T;V), \\ &u_n' \rightharpoonup v \quad \text{ weakly in } L^1(0,T;V'),\end{aligned}$$ then $v=u'$.

We have

$$ \underbrace{\int_0^T \eta'(t)\, u_n(t) \text{ d}t}_{(1)} = -\underbrace{\int_0^T \eta(t) \, u_n'(t) \text{ d}t}_{(2)} \tag{$\star$}$$

in $V'$ for all $\eta \in C_c^\infty(0,T)$. Now, let $v \in V$ be arbitrary. Then it follows $v \eta, v \eta' \in L^\infty(0,T;V)$ and due to the assumption of weak convergence on $u_n, u_n'$ we have after applying $v$ to the integral in $(\star)$

$$\begin{aligned}\langle (1),v \rangle_{V',V}=\int_0^T (u_n, \eta' v )_H \text{ d}t &\to \int_0^T (u,\eta' v)_H\text{ d}t, \\ \langle (2),v \rangle_{V',V}=\int_0^T \langle u_n', \eta v \rangle_{V',V} \text{ d}t &\to \int_0^T \langle v,\eta v\rangle_{V',V} \text{ d}t \end{aligned}.$$

Hence applying the limit on both sides in $(\star)$ gives $u'=v$.