# Convergence of the sign of a function in a Sobolev space

Consider a sequence of function $$\{f_n\}$$, where $$f_n:\mathbb{\Omega}\to \mathbb{R}$$, $$\Omega$$ is a bounded subset of $$\mathbb{R}^m$$ with a smooth boundary. Let $$D$$ be a countable dense subset of $$\Omega$$.

It is known that $$\text{sign}(f_n(x))\to 1\text{ }\forall x\in D,$$

Its also given that as $$n\to\infty$$, the Sobolev norm, $$\|f_n\|_{H^k(\Omega)} = O(1)$$, for some $$k> \frac{m}{2}$$.

I want to know If I can say as $$n\to\infty$$, $$\text{sign}(f_n(x)) \to 1\text{ }\forall x\in\Omega$$

What I know

If there was no $$\text{sign}$$ function over $$f$$, then through a direct application of Morrey's inequality, the result holds. But with the appearance of the sign function, the result doesn't seem to hold, but I am not sure.

Let $$\psi \colon \mathbb R^m \to \mathbb R$$ be smooth, compactly supported with $$\psi(x) \in [0,1]$$ and $$\psi(0) = 1$$. Let us assume $$0 \in \Omega$$.
Then, consider $$f_n(x) := 1 - \psi(x).$$ Then, $$\operatorname{sign}(f_n(x)) = 1$$ for all $$x \in \Omega \setminus\{0\}$$.