Point-wise convergence but not weakly

Question

Let $X$ be a reflexive Banach space and $x_n\in C(0,\tau;X)$ be a bounded sequence. We know that a subsequence of $x_n$, denote it by the same symbol, converges weakly to $x$ in $L^2(0,\tau;X)$. Can we conclude that for almost every $t\in [0,\tau]$, $x_n(t)$ converges weakly to $x(t)$ in $X$?

Answer 1

No. Let us take $X = \mathbb{R}$, $\tau = 1$ and $$ x_n(t) = \operatorname{sign}(\sin(n \, t)). $$ It can be shown that $x_n \rightharpoonup 0$ in $L^2(0,1)$, but $x_n(t) = \pm 1$ a.e.

Edit: Ups, this is not quite continuous. But you can modify $x_n$ on a set of very small measure, lets say $\varepsilon \, 2^{-n}$ to make it continuous. Then, there is a set $A \subset (0,1)$ with measure $1-\varepsilon$, such that $x_n(t) = \pm 1$ on $A$.