Let $u\_n \to u$ in $C^0([0,T];H^{-1}(\Omega))$ and suppose $\lVert u\_n \rVert\_{L^\infty(0,T;L^\infty(\Omega))} \leq C$ for all $n$. 


It follows that for almost all $t$, $u\_n(t)$ is bounded in $L^\infty(\Omega)$, so we can extract a weak-\* convergent subsequence and after a bit of work we can show that
$$\text{for almost all $t$} \qquad u\_n(t) \rightharpoonup u(t) \text{ in $L^q(\Omega)$} \qquad\text{for all $q < \infty$.}$$


I want to show that this weak convergence holds for **all** $t$.


From personal correspondence, I have been told that



> 
> The key point is that one already has convergence in the larger space. So one already knows that pointwise, i.e. for **all** $t$, $u(t)$ converges in this larger space. But 
>  since it is also bounded in $L^\infty$, $u(t)$ **[for almost all $t$ -- riem's note]** will also converge weak-\* up to a subsequence, and the limit is the same are identified as 
>  distributions. The same is true for any $L^q$, $q<\infty$.
> 
> 
> 


Simply put, I don't get why it must hold for all $t$. I think we need to exploit the continuity of $u\_n$ and $u$ wrt. $t$ in the larger space $H^{-1}$ but once again $L^q$ is a stronger space than this..


Originally posted [on MSE](https://math.stackexchange.com/questions/867020/pointwise-in-time-convergence-in-h-1-implies-pointwise-weak-convergence-i).



Pointwise (in time) convergence in $H^{-1}$ implies pointwise weak convergence in $L^q$, why?
The explanation is that by the hypotheses you in fact have $\lVert u\_n \rVert\_{\ell^{+\infty}([0,T],L^{+\infty}(\Omega))} \leq C$ for all $n$ as for arbitrarily fixed $t\in[0,T]$ you can take a sequence of $t\_i\to t$ with $\lVert u\_n(t\_i) \rVert\_{L^{+\infty}(\Omega)} \leq C$ and $u\_n(t\_i)\to u\_n(t)$ in $H^{-1}(\Omega)$ . Then a subsequence converges also weakly$^\*$ in $L^{+\infty}(\Omega)$ and hence $u\_n(t)$ is also there with norm at most $C$ . So everything you have written "for almost all $t$ " in fact holds for all $t\in[0,T]$ .


