Boundary values of Sobolev functions on the torus

Question

I am well aware of the trace theorem: Let $\Omega \subset \mathbb R^d$ be a bounded open region with $C^1$ boundary, then there exists a continuous map $T: H^{1,p}(\Omega ) \to L^p (\partial \Omega )$, called the trace, such that it returns the boundary values for $C^1$ functions.

I am about to solve a problem sheet where I need the torus Laplacian with the domain $\mathcal D (-\Delta ) = \lbrace f \in H^2 ([0,1]^d) : f\;\text{and its 1}^{\mathrm{st}}\text{-order derivatives are periodic} \rbrace $ (the problem is to prove that its resolvents are in certain Schatten classes; the focus is really not on properties of Sobolev spaces, but nevertheless I want to know hot to justify things).

Now $[0,1]^d$ does not have a $C^1$ boundary, hence the trace theorem in the above form is not applicable. I know that it is possible to define Sobolev spaces on differentiable manifolds, and perhaps this would solve the problem. However, I would like to know if it is possible to define the torus Laplacian elementarily.

Perhaps more generally: Is there a version of the trace theorem for singular boundaries? If not, then is there an ad-hoc approach for the torus?

Answer 1

I will give some more details on my comment, Jakob. As you correctly noticed later Lipschitz will do the job. Usually in textbooks it is the standard assumption made, though it is much stronger than what needed.

I give you here the sharp statements which you can find in Maz'ya's ``Sobolev spaces with applications to PDEs'' (Chapter 5 and 9, if I recall correctly).

Let $\Omega$ be an open set. It admits the classical continuous Sobolev (and $BV$) embeddings if and only if it is a domain of isoperimetry, i.e. there exists $k(\Omega)>0$ such that $$\min\{|E|; |\Omega\setminus E|\}^{n-1} \le kP(E)^n\,, \qquad \forall E\subset \Omega\,.$$ Additionally, if $|\Omega|<+\infty$, the classical compact embeddings hold as well.

On the other hand, for the existence of a linear, continuous operator which agrees with the trace of functions continuous up to the boundary, one has the following result.

Let $\Omega$ be an open set. It admits the classical trace operators if and only if $$P(\Omega) = \mathcal{H}^{n-1}(\partial \Omega)<+\infty$$ and there exists $K(\Omega)>0$ such that the Poincaré trace inequality holds, i.e. $$\min\{P(E,\partial \Omega); P(\Omega \setminus E; \partial \Omega)\} \le KP(E)\,, \qquad \forall E\subset \Omega\,.$$

In the above, by perimeter it is meant the one used in geometric measure theory: $$P(E;A) = |D\chi_E|(A)\,,$$ with the shorthand $P(E)$ when $A=\mathbb{R}^n$.

This one amounts to the total variation on $A$ of the characteristic function of the set $\chi_E$.

Answer 2

I just realized that there is a simple answer to my question without more sophisticated Sobolev theory: The trace theorem is not just true for $C^1$ boundaries, but also for Lipschitz domains. While $\partial [0,1]^d$ is certainly not $C^1$, it is clearly Lipschitz.