In their paper The spectral drop problem, Buttazzo and Velichkov state that the embedding $H_0^1(\Omega, D) \hookrightarrow L^2(D)$ is compact, where $\Omega \subset D$ is quasi-open, $D \subset \mathbb R^N$ is a Lipschitz open domain and $|\Omega| < |D|$ and $$ H_0^1(\Omega, D) = \{u \in H^1(D) \ : \ u = 0 \text{ quasi-everywhere in } D \setminus \Omega\}. $$

They claim that this is a well-known fact and follows from the Lipschitz condition on $D$, but I wasn't able to find a proof.

At first I thought about adapting the classical proof of the Rellich-Kondrachov theorem, but I failed.

Can you recommend any reference where I can find a detailed proof of this fact, or (preferably) any hints?

I am beginning to learn about capacities and quasi-things, and would like to see the proof to get more acquainted with the theory.

Thanks in advance.

EDIT - Related notions

The capacity of a set $A \subset \mathbb R^N$ is $$ \text{cap}(A) = \inf\left\{\int_{\mathbb R^N} (|\nabla u|^2 + u^2) \ dx \ : \ u \in H^1(\mathbb R^N), \ u \geq 1 \text{ in a neighborhood of } A\right\} $$

A set $\Omega \subset \mathbb R^N$ is quasi-open if, for every $\varepsilon > 0$, there exists an open set $\omega_\varepsilon$ such that $\text{cap}(\omega_\varepsilon) \leq \varepsilon$ and $\Omega \cup \omega_\varepsilon$ is open.

A property is said to hold quasi-everywhere if it holds everywhere except, at most, in a set of zero capacity.

EDIT 2 - Attempt of proof

I think I have solved the problem, but I have no confidece. Will appreciate comments on my argument.

I tried to adapt the proof of the Rellich-Kondrachov theorem presented by Evans (Partial Differential Equations, 2nd Ed, Theorem 1 in Section 5.7). Instead of writing all the proof again, I will highlight the differences.

We observe that if $A$ has $0$ capacity, then clearly the Lebesgue measure of $A$ is $0$, for there exists a sequence $v_n \xrightarrow{H^1(\mathbb R^N)} 0$ with $v_n \geq 1$ in a neighborhood of $A$, and so $$ |A| = \int_A 1 \ dx \leq \int_A v_n \ dx \leq \|v_n\|_{H^1(\mathbb R^N)} \to 0 $$

Step 1: $H^1_0(\Omega, D) \subset L^2(D)$ by definition.

Step 2: We can use the extension theorem 13.17 in Leoni's A First Course in Sobolev Spaces to consider $D = \mathbb R^N$. Let $u_m$ be the bounded (in $H_0^1(\Omega, D)$) sequence whose convergence we want to prove. By the observation above, the support of $u_m$ has finite measure (since $|\Omega| < |D| = \infty$). Suppose the supports are in some set $V$ of finite measure.

Step 3: We can use mollifiers to regularize the sequence $u_m$, obtaining $u_m^\varepsilon$. $V$ has finite measure.

Step 4: $V$ having finite measure is all we need for the estimates to hold. We obtain that $$ u_m^\varepsilon \to u_m \quad \text{ in } L^2(V), \text{ uniformly in } m $$

Step 5: Again using that $V$ has finite measure, we conclude that $u_m^\varepsilon$ is uniformly bounded and equicontinuous, for all $\varepsilon > 0$

Steps 6, 7: By the Arzelà-Ascoli Theorem for $u_m^\varepsilon$ and the previous results, we obtain the convergence. We note, in particular, that since the supports have finite measure, the uniform boundedness required for the A-A Theorem hold.

  • $\begingroup$ Could you add some more detail? What does it mean for a set to be quasi-open; does quasi-everywhere mean almost-everywhere? Also, is there a typo in the definition of $H_0^1(\Omega,D)$? $\endgroup$
    – Leo Moos
    Jul 2, 2021 at 19:47
  • 1
    $\begingroup$ I believe quasi-everywhere and quasi-open rely on the notion of capacity, see e.g. en.wikipedia.org/wiki/Capacity_of_a_set $\endgroup$ Jul 2, 2021 at 22:18
  • $\begingroup$ @LeoMoos there was a problem, indeed, thanks. I fixed it. I will add the related definitions tomorrow (I'm on my mobile device now), but one can see, for example the article I mentioned, or the book Shape Variation and Optimization, by Henrot and Pierre. Unfortunately, the link by leo monsaingeon is not really helpful (but thank you, leo) $\endgroup$ Jul 3, 2021 at 0:43
  • 1
    $\begingroup$ I appreciate your edits. For what it's worth, this may be well-known in their area, but I've never heard of these extensions of the Sobolev embeddings, and I couldn't find anything in Evans-Gariepy either. If you don't get an answer here, I don't think there'd be any shame in reaching out directly and asking the authors for a reference. $\endgroup$
    – Leo Moos
    Jul 3, 2021 at 13:05
  • $\begingroup$ Thank you very much for your effort :D $\endgroup$ Jul 3, 2021 at 13:09

1 Answer 1


You have that $H_0^1(\Omega, D)$ is a subspace of $H^1(D)$ (equipped with the same norm). Due to the Lipschitz condition on $D$, $H^1(D)$ is compactly embedded in $L^2(D)$ (standard-Rellich-Kondrachov). Thus, $H_0^1(\Omega, D)$ is also compactly embedded in $L^2(D)$.

  • 1
    $\begingroup$ Thanks for your answer, but the standard compact embedding requires the set to be bounded, no? Here I'm allowing for $D$ to be unbounded, possibly with infinite measure $\endgroup$ Jul 6, 2021 at 12:50
  • $\begingroup$ Oh, sorry, I missed that $D$ might fail to be bounded. But then, the keyword might be "uniformly Lipschitz". However, there no definition of this term appears in the paper. $\endgroup$
    – gerw
    Jul 6, 2021 at 14:07
  • 1
    $\begingroup$ @DaniloGregorinAfonso The answer by gerw is nearly correct I think; the uniform Lipschitz condition gives an extension operator $E \colon H^1(D) \to H^1(\mathbb{R}^N)$ such that the support of an extension is contained in a set $F \supset D$ which is only slightly larger than $D$. Then the support of the extension of an $H^1_0(\Omega,D)$ function is contained in a set $\Lambda$ slightly larger than $\Omega$, in particular, it is still of of finite volume. $\endgroup$
    – Hannes
    Jul 7, 2021 at 14:05
  • 1
    $\begingroup$ (cont'd) Then one can use Rellich-Kondrachov for $\Lambda$ with finite volume as in Lair: A Rellich Compactness Theorem for Sets of Finite Volume, I think. $\endgroup$
    – Hannes
    Jul 7, 2021 at 14:06
  • 1
    $\begingroup$ @DaniloGregorinAfonso The supports of an extension can usually be made arbitrarily close to the original set because the extensions are constructed locally with cut-off functions (partition of unity) and reflection. It is stated explicitly e.g. in Gilbarg/Trudinger Theorem 7.25. $\endgroup$
    – Hannes
    Jul 12, 2021 at 8:39

You must log in to answer this question.

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