Convergence weak* in $L^{\infty}([0,T];L^2(\mathbb{T}^2))$ implies convergence weak* a.e in $L^2(\mathbb{T}^2)$?

Question

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 $$\int _0 ^T | \lim_{n \to\infty} <x_n(t)-x(t),z>_{L^2}|dt=0. $$

Answer 1

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 $(2)$.