Convergence weak* in $L^{\infty}([0,T];L^2(\mathbb{T}^2))$ implies convergence weak* a.e in $L^2(\mathbb{T}^2)$? Suppose $x_n\rightharpoonup x$ in $L^{\infty}([0,T];L^2(\mathbb{T}^2))$ with the weak* topology, in other words, $\forall f\in L^{1}([0,T];L^2(\mathbb{T}^2))$ we have
$$\lim_{n \to\infty} \int _0 ^T <x_n(t)-x(t),y(t)>_{L^2}dt=0,$$
where $\mathbb{T}^2$ is the torus in $\mathbb{R}^2$.
Particularly,
$$\lim_{n \to\infty} \int _0 ^T <x_n(t)-x(t),z>_{L^2}dt=0  \;\;
\;
\;
 \forall z\in L^2(\mathbb{T}^2).$$
 I want to prove that $x_n(t)\rightharpoonup x(t)$ in $L^2(\mathbb{T}^2)$ for almost evere $t\in[0,T]$   with the weak* topology. Thus, i need help to show that
$$\int _0 ^T | \lim_{n \to\infty}  <x_n(t)-x(t),z>_{L^2}|dt=0. $$
 A: This cannot be true. For a counterexapmle, take 
$x_n(t)=\varphi\cdot {\rm sgn}\bigl(\sin{(nt)}\bigr)$ with some
$\varphi\in L^2(\mathbb{T}^2)$ such that $\|\varphi\|_{L^2}=1$. It is clear that
$x_n\rightharpoonup 0$ weakly$^{\ast}$, i.e.,
$$
\lim_{n\to\infty}\int_0^T\langle x_n(t),f(t)\rangle_{L^2}dt=0
\quad\forall\,f\in  L^{1}\bigl(0,T;L^2(\mathbb{T}^2)\bigr),
$$
while for $z=\varphi$ you get
$$
\bigl|\langle x_n(t),z \rangle_{L^2}\bigr|=\bigl|{\rm sgn}\bigl(\sin{(nt)}\bigr)\bigr|=1
\quad\forall\,n\in\mathbb{N},\;\forall t\in [0,T],\tag{1}
$$
with signum function defined as
$$
{\rm sgn}(\xi)=
\begin{cases}
\;\;\,1,\quad \xi\geqslant 0,\\
-1, \quad \xi<0.
\end{cases}
$$
Identity $(1)$ implies that the expected convergence $\langle x_n(t),\varphi \rangle_{L^2}={\rm sgn}\bigl(\sin{(nt)}\bigr)\to 0$ cannot hold for almost all $t\in [0,T]$, since if it converged to zero at some $t\in [0,T]$, its modulus would also converge to zero which would contradict (1). 
REMARK. Assume additionally that sequence $\{x_n\}$ is equicontinuous in 
$C\bigl([0,T];H^{-s}\bigr)$ with some $s>0$. Then
$$
\lim_{n\to\infty}\langle x_n(t)-x(t),z \rangle_{L^2}=0
\quad\forall\,z\in L^2(\mathbb{T}^2),\;\forall t\in [0,T].\tag{2}
$$
To prove $(2)$, notice that by the Banach-Steinhaus theorem a weakly${^\ast}$ convergent sequence in $L^{\infty}([0,T];L^2)$ is strongly uniformly bounded in $L^{\infty}([0,T];
L^2)$, i.e. in $C([0,T];L^2)$, and hence it is strongly uniformly bounded in 
$C([0,T];H^{-s})$, which implies that
$$
\lim_{n\to\infty}\langle x_n(t)-x(t),z \rangle=0
\quad\forall\,z\in H^s(\mathbb{T}^2),\;\forall t\in [0,T],\tag{3}
$$ 
since by assumption sequence $\{x_n\}$ is equicontinuous in $C\bigl([0,T];H^{-s}\bigr)$, while being weakly${^\ast}$ convergent in $L^{\infty}([0,T];L^2)$. Subspaqce $H^s(\mathbb{T}^2)$ is dense in $L^2(\mathbb{T}^2)$, which is why given any $z\in L^2
(\mathbb{T}^2)$, $\forall\,\varepsilon>0\;\exists\, z_{\varepsilon}\in H^s(\mathbb{T}^2)$ such that 
$$
|\langle x_n(t)-x(t),z-z_{\varepsilon} \rangle|<\varepsilon/2\quad
\forall\,n\in\mathbb{N},\;\forall t\in [0,T],\tag{4}
$$
due to the fact that $\{x_n\}$ is strongly uniformly bounded in $C([0,T];L^2)$. Due to $(3)$, $\forall\,\varepsilon>0\;\exists\,N_{\varepsilon}\in\mathbb{N}$ such that
$$
|\langle x_n(t)-x(t),z_{\varepsilon} \rangle|<\varepsilon/2\quad
\forall\,n>N_{\varepsilon}\,,\;\forall t\in [0,T],
$$
whence by $(4)$ follows 
$$
|\langle x_n(t)-x(t),z\rangle|<\varepsilon/2+\varepsilon/2=\varepsilon\quad
\forall\,n>N_{\varepsilon}\,,\;\forall t\in [0,T],
$$
which completes the proof of $(2)$.
