Does weak convergence of $u_m(t)$ in $L^2(0,T;X)$ imply weak convergence of a subsequence of $u_m(t_0)$ in $X$ for a.e. $t_0$ in $[0,T]$? In a book I'm reading (Navier Stokes Equations, by Constantin and Foias), the authors construct a sequence $u_m$ of functions in $L^2(0,T;V)$ which converge weakly to $u$ in $L^2(0,T;V)$.  They then claim that $u_m(t_0)$ converges weakly to $u(t_0)$ for a.e. $t_0\in [0,T]$.  Why is this true?
Here $V$ is the subset of $H^1_0(\Omega)$ ($\Omega\subset\mathbb{R}^n$ bounded, with nice boundary) whose elements are divergence-free; however, it seems unlikely to me that the definition of $V$ will matter for the question I am asking.  
 A: This is not true, even in the simplest case $V = \mathbb{R}$. To see this, fix $v \in V \setminus \{0\}$ and define $u_m(t) = v \, \sin(m \, t)$. Then, $u_m \rightharpoonup 0$ in $L^2(0,T;V)$, but it does not converge pointwise a.e.
One thing that ensures weak pointwise a.e. convergence, is Helly's selection theorem, see here and here.
A: I'd say that the definition of $V$ is important because we can prove a version of the desired result using the following theorem.

Theorem 1. Let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property (for example, $V$ is Hilbert or it's separable reflexive) and $1\leq p<\infty$. If
  $$u_n\rightharpoonup u\quad\text{em}\quad L^p(0,T;V),$$
  then
  $$\int_0^T\varphi u_n\;dt\rightharpoonup\int_0^T\varphi u\;dt\quad\text{em}\quad X,\qquad\forall\ \varphi\in C[0,T].$$

Theorem above implies the desired result provided that the sequence $\{u'_n\}$ of weak derivatives has sufficient regularity. More precisely:

Let $V$ be as in the hypothesis of Theorem 1. Assume that
  $$v_n\rightharpoonup v\quad\text{in}\quad L^2(0,T;V)$$
  and that $\{v_n'\}$ is bounded in $L^b(0,T; V)$ for some $1<b<\infty$. Then
  $$v_n(s)\rightharpoonup v(s)\quad\text{in}\quad V,\qquad\forall\ s\in(0,T).$$

Proof:
Let $s\in(0,T)$ and $f\in V'$. We want to prove that
$$f(v_n(s))\to f(v(s))\quad\text{in}\quad\mathbb{R}.$$
Define $u_n=v_n-v$. Then
$$u_n\rightharpoonup 0\quad\text{in}\quad L^2(0,T;V)\tag{1}$$
and we need to prove that
$$f(u_n(s))\to 0\quad\text{in}\quad\mathbb{R}.$$
Set $\phi\in C^1([0,T];\mathbb{R})$ such that $\phi(0)=-1$ and $\phi(T)=0$.
Take $\lambda\in(0,1-s/T)$ (to be fixed later) and define $w_n(t)=u_n(s+\lambda t)$. Then 
$$u_n(s)=w_n(0)=\phi(t)w_n(t)\Big|_0^T=\int_0^T(\phi w_n)'\;dt=\underset{\alpha_n}{\underbrace{\int_0^T\phi' w_n\;dt}}+\underset{\beta_n}{\underbrace{\int_0^T\phi w_n'\;dt}}$$
Notice that
$$\beta_n=\int_0^T\lambda\phi(t) u_n'(s+\lambda t)\;dt=\int_s^{s+\lambda T}\phi(\lambda^{-1}(\tau-s)) u_n'(\tau)\;d\tau.$$
Thus, by Hölder's inequality and by the boundedness of $\{v'_n\}$,
$$\begin{aligned}
\|\beta_n\|_{V}&\leq C\int_s^{s+\lambda T}\|u_n'(\tau)\|_V\;d\tau\\
&\leq C(\lambda T)^{1-\frac{1}{b}}\left(\int_s^{s+\lambda T}\|u_n'(\tau)\|_V^b\;d\tau\right)^{\frac{1}{b}}\\
&\leq C(\lambda T)^{1-\frac{1}{b}}\left(\int_0^T\|u_n'(\tau)\|_V^b\;d\tau\right)^{\frac{1}{b}}\\
%&= C(\lambda T)^{1-\frac{1}{b}}\|u_n'\|_{L^b(0,T;V)}\\
&\leq CT^{1-\frac{1}{b}}M\lambda^{1-\frac{1}{b}}
\end{aligned}$$
for some constant $M$. Therefore, given $\varepsilon>0$, we can choose $\lambda$ small enough  such that
$$\|\beta_n\|_{V}\leq\frac{\varepsilon}{2\|f\|},\quad \forall \ n\in\mathbb{N}.\tag{2}$$
Now, from $(1)$ we get
$$u_n|_{(s,s+\lambda T)}\rightharpoonup 0\quad\text{em}\quad L^2(s,s+\lambda T;V)$$
and thus, by Theorem 1,
\begin{equation}
\alpha_n=\int_0^T\phi' w_n\;dt=\lambda^{-1}\int_s^{s+\lambda T}\phi'(\lambda^{-1}(\tau-s)) u_n(\tau)\;d\tau\rightharpoonup 0\quad\text{in}\quad V
\end{equation}
which implies
$$f(\alpha_n)\to 0\quad\text{in}\quad\mathbb{R}.\tag{3}$$
From $(2)$ and $(3)$ we conclude that $f(u_n(s))=f(\alpha_n+\beta_n)\to 0$ as needed. $\blacksquare$
