Does weak convergance in $L^2(0,T,V)$ imply weak convergance in $V$ a.e. $t$? Suppose $u_n\to u$  in $L^2(0,T,V)$ (in here $V$ is a reflexive Banach space). It is well known, that the latter implies  (at least for a subsequance)
$$u_n(t)\to u(t)\quad\text{in}\quad V\, \text{a.e.}\,\, t\in (0,T).$$
Now, suppose that $u_n\to u$ weakly in $L^2(0,T,V)$. Can we deduce (at least for a subsequance) that 
$$u_n(t)\to u(t)\quad\text{weakly in}\quad V\, \text{a.e.}\,\, t\in (0,T)?$$

The statement is invalid. Let $V=\mathbb{R}$ and consider a sequence $u_{n}(t)=\sin(2\pi n t)$. Then $u_n$ is weakly convergent to $0$ in $L^2(0,T;V)$. Indeed; for any $v'\in L^2(0,T;V')$, we have
$$|\langle v', u_n\rangle|=| \int_{0}^{T}\langle v'(t), u_n(t)\rangle_{V'\times V}\, dt |\leq |\int_{0}^{T}\sup_{t\in[0,T]}|v'(t)|\cdot u_{n}(t)\, dt |\\\le c\cdot |\int_{0}^{T}\sin (2\pi nt)\,dt|=\frac{1}{2\pi n}|\int_{0}^{2\pi n}\sin(x)\,dx|\le \frac{1}{2\pi n}\cdot C\to 0.$$
Now, the sequence $(u_{n}(t))$ (with fixed $t$) has a convergent subseguence (for exapmle, $t=1$), but the set of such $t\in (0,T)$ has measure $0$ which is opposite to the statement of the theorem (in $\mathbb{R}$ weak and strong topologies coincide). This reveals a possible mistake in Jan Bohr answer - does the set $N$ has measure $0$?
