when a bounded set in $W^{2,p}(\Omega)$ has a strong limit point in $C(\overline{\Omega})$?

Question

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!

Answer 1

I will assume $\Omega$ has sufficiently regular boundary (Lipschitz is sufficient).

There is a lot of unnecessary information in the question which makes it seems complicated, but this appears to boil down to whether the embedding $$ W^{2,p}(\Omega) \hookrightarrow C(\overline\Omega)$$ holds. By the general Sobolev inequality this is the case for $p > \frac32;$ you can see this follows from the Gagliardo-Nirenberg and Morrey inequalities noting that if $p \in (\frac32,3)$ we have $$ W^{2,p}(\Omega) \hookrightarrow W^{1,\frac{3p}{3-p}}(\Omega) \hookrightarrow C^{0,2-\frac3p}(\overline\Omega) \hookrightarrow C(\overline\Omega),$$ and if $p > 3$ we have $$ W^{2,p}(\Omega) \hookrightarrow C^{1,1-\frac3p}(\overline\Omega) \hookrightarrow C(\overline\Omega).$$ (If $p=3$ you can note that $W^{2,3}(\Omega) \hookrightarrow W^{2,2}(\Omega) \hookrightarrow C(\overline\Omega)$ for instance.)

Given this, for $p>\frac32$ if $H \subset W^{2,p}(\Omega)$ we have any limit point $u \in \overline{H} \subset W^{2,p}(\Omega)$ evidently lies in $C(\overline\Omega)$ by the above embedding. Therefore $p>\frac32$ is a sufficient condition.

If $p \leq \frac32,$ it well known that this embedding fails, and you can find $u \in W^{2,p}_0(\Omega) \setminus C(\overline\Omega).$ Then take a sequence $(u_n)$ of $C^{\infty}_c(\Omega)$ functions such that $u_n \to u \in W^{2,p}_0(\Omega),$ and consider $H = \{u_n\} \subset C(\overline\Omega).$

Answer 2

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.)