Let $\Omega$ be a convex, open subset of $\mathbb{R}^d$, with a smooth boundary. Consider the set $$M = \{f/f:\Omega\to\mathbb{R} \bigwedge f\in C^0(\Omega)\bigwedge \lvert| f|\rvert_{W^{1,d+1}}\in\mathbb{R}\}$$
From Morrey's inequality, $f\in M \implies f\in C^{0,1/(d+1)}(\Omega)$. Lets take any function $g\in M$, and perturb its value at a single point $x_0 \in \Omega^o$ to get a new function $h$. Meaning
$h(x) = \begin{cases} g(x) + 10 & x = x_0 \\ g(x) & x \ne x_0 \end{cases} $
Given any sequence $\{f_n\},f_n \in M$, such that $f_n\to h$ in the norm $\lvert|.\rvert|_{W^{1,d+1}}$ then I'd like to prove that $f_n(x_0)$ cannot converge to $h(x_0)$.
I hope its a direct consequence of a well known theorem, but I am not aware it.
Since $h = g$ a.e., you also have $\|f_n - g\|_{W^{1,n+1}}\to0$. Now, Morrey even implies $\| f_n - g\|_{L^\infty} \to 0$. Now, $f_n$ and $g$ are continuous, hence, $\| f_n - g\|_{C^0} \to 0$, which implies uniform convergence. In particular, $f_n(x_0) \to g(x_0) \ne h(x_0)$.