Suppose $f_n \to f$ in $L^\infty(U)$ where $f_n, f \in C^{1,1}(\bar{U})$, (Holder space with $k=1, \alpha =1$, i.e, Lipschitz functions) where $U$ is an open and bounded in $\mathbb{R}^n$. Q: Can we say that $f_n \to f$ in $H^{1}(U)$?
I think it is indeed the case. Here is my attempt. Since $U$ is bounded, we can upper bound $L^2$-norm by $L^\infty$ so that \begin{align} \|f_n - f\|_{H^{1}(U)}^2 &= \|f_n - f\|_{L^2(U)}^2 +\|Df_n - Df\|_{L^2(U)}^2 \\ &\le \mu(U)\|f_n -f\|_{L^\infty(U)}^2 + \|Df_n - Df\|_{L^2(U)}^2. \end{align} Thus, it suffices to show the convergence of $\|Df_n - Df\|_{L^2(U)}^2$.
Since $e_n = f_n-f \in C^{1,1}(\bar{U})$, (if needed let assume $U$ is convex), the derivative of $e_n$ is bounded by its Lipschitz constant: $$ \|De_n\|_{L^\infty} \le \text{Lip}(e_n) = \sup_{x,y \in U, x\ne y} \frac{\|e_n(x)-e_n(y)\|}{\|x-y\|}. $$ Then for any $\epsilon > 0$, there exists $x', y' \in U, x'\ne y'$ such that $$ \text{Lip}(e_n) \le \epsilon + \frac{\|e_n(x')-e_n(y')\|}{\|x'-y'\|} \le \epsilon + \frac{2\|e_n\|_{L^\infty(U)}}{\|x'-y'\|} \to \epsilon \quad \text{as } n \to \infty. $$ Since $\epsilon$ was chosen arbitrarily, we conclude that $$ \lim_{n\to \infty} \|De_n\|_{L^\infty(U)} = 0, $$ which shows that $f_n \to f$ in $H^1(U)$. [Edited: This proof is false, as $x'$ and $y'$ depend on $n$].
It seems that the above derivation is correct. On one hand, the result seems quite strong. Normally, one cannot have the uniform convergence of $\{f_j'\}$ from the uniform convergence of $\{f_j\}$. At first glance, at least for me, this seems quite a surprising result. On the other hand, however, it sort of makes sense, as $\{f_j\}$ is Lipschitz and differentiable - $\{f_j'\}$ is somewhat well controlled. I would like to make sure of the rigourousness of the above argument.
Any comments/answers/suggestions will be very appreciated. Thanks!
[Here I added my second attempt.] Since $f_n \in C^{1,1}$, $f_n, f_n'$ are Lipschitz functions. By applying Arzela-Ascoli theorem in $\{f_n'\}$, we have a uniformly convergent subsequence $\{f_{n_k}'\}$. By invoking the standard theorem, that $\lim_{k\to \infty} f_{n_k}'(x) = f'(x)$. Since the limit is unique, we conclude that $\lim_{n\to \infty} f_n'(x)=f(x)$.
[2nd Edited: An additional Assumption is needed]; In order to ensure the uniform boundedness of Lipschitz constants of $\{f_n'\}$, it seems that I need to assume that $\{f_n\}$ is a bounded sequence in $C^{2,1}(\bar{U})$.
The claim is false:
Define $f=0$, $f_n(x) = n^{-1}\sin(n^2\pi x)$ on $(0,1)$. Then these functions are Lipschitz, $f_n \to f$ in $L^\infty$, but $(f_n)$ is even not bounded in $H^1$.