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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.

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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 $t\in[0,T]$ .

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