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 2


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-continuous functions can always be extended (holder-continuousely) to the boundary of $\Omega$ - this of course is not true in general for continuous functions! We can then use the Arzela-Ascoli-theorem to show compactness in $C^{0}(\bar \Omega)$, because in any such chain of (continuous) 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.

  • 1
    $\begingroup$ those are known to be Bertran integral $\endgroup$
    – Guy Fsone
    May 27, 2019 at 12:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .