# Sobolev Embedding into $L^{\infty}$

I tried to find the question here, but I couldn't. I'm a bit puzzled by Sobolev embeddings at the moment.

In my lecture notes, I found the statement

"If $$\Omega \subseteq \mathbb{R}^d$$ is bounded and open with $$\partial \Omega \in C^1$$, $$1 \leq p < \infty$$, $$k,m \in \mathbb{N}_0$$ with $$m > k$$, then the embedding $$W^{m,p}(\Omega) \hookrightarrow C^k(\overline{\Omega})$$ is compact, if $$m - \tfrac{d}{p} > k$$."

In my case, I have $$k = 0, \, p = 4, \, m = 1, d \leq 3$$. So the first question is:

$$1.)$$ From this theorem, I could conclude that functions in $$W^{1,4}(\Omega)$$ can be identified by functions in $$C^0(\overline{\Omega})$$, right? Because if I understand it correctly, this would imply, that I can always find a bounded representant in $$W^{1,4}(\Omega)$$.

But now I am puzzled, because the book where I wanted to refresh my knowledge about Sobolev Spaces says, that it is not possible to have a continous embedding of the form $$W^{m,p}(\Omega) \hookrightarrow C^0(\Omega).$$ (The book gives with $$d = 2$$ the function $$u(x) = \log(\log|x|)$$ as example, but I didn't understand that completely yet.)

$$2.)$$ If this statement is true, why does the embedding work if I take the closure of $$\Omega$$?

Thank you!

## 2 Answers

Since you have not received an answer yet, I thought I would share my thoughts on this.

The theorem you cite is correct. It´s statement and proof can be found in Adams, Sobolev Spaces, Theorem 6.2 Part III.

Furthermore, for $$\Omega \subset R^{n}$$ open, bounded by a sufficiently smooth boundary and $$mp > n, \; j\ge0$$ we have the (compact) imbedding:

$$W^{j+m,p}(\Omega) \subset \subset C_{B}^{j}(\Omega)$$ - this is Part II in the above theorem.

I am not familiar with the counterexample you provided, but I can assure you that for $$p > n$$ and sufficiently regular $$\Omega$$, we have $$W^{1,p}(\Omega) \rightarrow C^{0,\gamma}(\Omega)$$ - the latter being a Holder space.

A bounded set of Holder-continous functions can always be extended (holder-continousely) to the boundary of $$\Omega$$ - this of course is not true in general for continous functions! We can then use the Arzela-Ascoli-theorem to show compactness in $$C^{0}(\bar \Omega)$$, because in any such chain of (continous) imbeddings, it is always sufficient that only one of them is compact for the hole chain to be compact.

I hope this helps a bit, take a look at the section in Adams and you will probably find more answers.

Thank you so much for your answer!

I actually managed to resolve my confusion. When calculating the given example in the book, one comes to a point where the decisive question is: Is $$\int_0^a \frac{r}{(r \log(r))^p} \, \mathrm{d} r < \infty$$ for $$a < 1$$.

But this integral is only finite if $$p \leq 2$$, so actually the counterexample in the book holds only if $$p \leq d$$ without stating that at any point.

• those are known to be Bertran integral – Guy Fsone May 27 at 12:03