# $H_0^1(\Omega, D) \hookrightarrow L^2(D)$ is compact, for $\Omega$ quasi-open in $D$ - Proof verification

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.

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.

• 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 $H_0^1(\Omega,D)$? Jul 2, 2021 at 19:47
• I believe quasi-everywhere and quasi-open rely on the notion of capacity, see e.g. en.wikipedia.org/wiki/Capacity_of_a_set Jul 2, 2021 at 22:18
• @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) Jul 3, 2021 at 0:43
• 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. Jul 3, 2021 at 13:05
• Thank you very much for your effort :D Jul 3, 2021 at 13:09

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 $$L^2(D)$$.
• 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 Jul 6, 2021 at 12:50
• 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.
• @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. Jul 7, 2021 at 14:05
• (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. Jul 7, 2021 at 14:06