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Let $\Omega$ be a bounded domain in $\mathbb{R}^d$ with a smooth boundary. Consider the sequence $\{u_n(\cdot,s)\} \subset L^2(0,T;L^2(\Omega))$ such that $\lVert u_n(\cdot,s) \rVert_{ L^2(\Omega)} $ is Lipschitz continuous and $\lVert u_n(\cdot,s) \rVert_{ L^2(\Omega)} \leq 1$.

By Arzelà–Ascoli theorem, there is a subsequence such that $\lVert u_n(\cdot,s) \rVert_{ L^2(\Omega)} $ converges to $\lVert u_\infty(\cdot,s) \rVert_{ L^2(\Omega)} $ uniformly for $s \in [0,T]$, which implies $\lVert u_n \rVert_{ L^2(0,T;L^2(\Omega))} \to \lVert u_\infty \rVert_{ L^2(0,T;L^2(\Omega))}$. Since we also have $\lVert u_n \rVert_{ L^2(0,T;L^2(\Omega))} \leq T $, there is also a subsequence such that $u_n \to u_\infty$ weakly in $L^2(0,T;L^2(\Omega))$. Then the weak convergence together with the norm convergence imply $u_n(x,s) \to u_\infty(x,s)$ strongly. But I think that the arguments are problematic. Since we suppose that $u_n(x,s)$ is constant in $s$, then the assumptions reduce to only $\lVert u_n \rVert_{ L^2(\Omega)} \leq 1$, then it is impossible to induce a strong convergence subsequence. what is wrong?

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  • $\begingroup$ I apply Arzela-Ascoli theorem to the function $\lVert u_n(\cdot,s) \rVert_{ L^2(\Omega)} $ which maps $[0,T] \to \mathbb{R}$. $\endgroup$
    – mnmn1993
    Commented Jan 13, 2023 at 14:12

1 Answer 1

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The problem is that one does not know that the limit of $\|u_n\|_{L^2(0,T;L^2(\Omega)}$ is equal to the norm of the weak limit.

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  • $\begingroup$ What additional conditions allow us to ensure this? $\endgroup$
    – mnmn1993
    Commented Jan 15, 2023 at 6:53

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