TITLE: when a bounded set in $W^{2,p}(\Omega)$ has a strong limit point in $C(\overline{\Omega})$?
QUESTION [3 upvotes]: I just have a question in functional analysis. The question says:
If $\Omega \subset \mathbb{R}^3$ is bounded, and suppose that $H$ a bounded set in the Sobolev's space $W^{2,p}(\Omega)$. Then what values of $p \geq 1$ makes $H$ have a strong  strong limit point in $C(\overline{\Omega})$?
Thank you!

REPLY [0 votes]: On the negative side, for sets with bad boundary, it can fail: Let $\Omega = \mathbb B(0,1)\setminus([-1,1]\times \{0 \})\subset \mathbb R^2$ be the open unit disk with a horizontal cut. Then $\overline\Omega=\overline{\mathbb B(0,1)}$ but $f:\Omega\to\mathbb R$, defined by $$f(x,y)=\begin{cases} 1 &y>0\\ 0& y<0\end{cases}$$
belongs to every $W^{k,p}(\Omega)$, and not to $C(\overline \Omega)$. Its also easy to make an example with a connected set e.g. $B(0,1)\setminus([0,1]\times \{0 \})$, but this was easier to type.
In the above cases the boundary is not even locally a graph, but IIRC you can see a different behavior with $p$ if the boundary is only locally $\alpha$-Hölder: I believe one should check when $1/\sqrt{x^2+y^2}$ belongs to $W^{k,p}(\Omega)$ where now $\Omega = \mathbb B(0,1) \setminus \{ (x,y) : x\ge0,\ |y|^\alpha \le x\}$.
Also worth mentioning is that, even if $\Omega$ is as smooth as you like, $C(\overline\Omega)$ is not even a subset of $W^{k,p}(\Omega)$, $k\ge 1$, even if $k$ or $p$ is very large. So in particular, it is not dense in $W^{2,p}$ as claimed in the comments. The reason already appears in dimension 1 in the form of e.g. the devil's staircase. It's continuous, and  it's a.e. derivative (and therefore the only candidate for the weak derivative) is zero. But its not constant; so it has no weak derivative. (In fact, the derivative is just a measure.)
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