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

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!

• What have you tried? Do you know any relations between $W^{2,p}(\Omega)$ and $C(\overline\Omega)$ (say if $p$ is large enough) which might be of help here?
– ktoi
Jan 4, 2021 at 13:50
• @ktoi. I am not sure how to start. but I think $C(\overline{\Omega})$ is dense in $W^{2,p}(\Omega)$ ! and I think $p$ should be greater than $3$, I guess. Jan 4, 2021 at 14:45
• The result I was alluding to was the Sobolev inequality - I've posted an answer explaining this in detail. On MSE we ask users to post their thoughts and attempts on the questions first, hence my initial prompt.
– ktoi
Jan 5, 2021 at 8:35

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 $$H = \{u_n\} \subset C(\overline\Omega).$$

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.)