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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.

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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 $x \in \Omega \setminus\{0\}$.

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