# Some problems in the application of Arzelà–Ascoli theorem

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?

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

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.