Sobolev spaces and the domain of fractional Laplacian

Question

I'm reading this paper on arxiv link.

So far OK.

Now this I don't understand. Take $s=\frac 12$. They say that by density the operator $(-\Delta)^s$ is defined on $\mathbb{H}^s(\Omega)$. Then they say that the theory of Hilbert scales shows that the domain of $(-\Delta)^s$ is $H^1_0(\Omega)$. But in the very next line they claim that $\mathbb{H}^s(\Omega)= H^{\frac 12}_{00}(\Omega)$, which is not $H^1_0(\Omega)$. What do I miss?

Answer 1

I think the authors are somewhat inconsistent in what they mean by the domain of an operator. Let's ignore the boundary conditions for now and focus on the order of smoothness.

• The domain of $(-\Delta)^s$ as an operator into $L^2$ is a Sobolev space of order $2s$. Indeed, $s$ fraction of the Laplacian is like $2s$ derivatives.
• The domain of $(-\Delta)^s$ as an operator from a space into its dual space is a Sobolev space of order $s$. This is because $\langle (-\Delta)^s f,g\rangle = \langle (-\Delta)^{s/2} f,(-\Delta)^{s/2} g\rangle$ so having $s/2$ fraction of Laplacian (hence $s$ portion of the gradient) is enough.

When they say that by density, $(-\Delta)^s$ can be extended to $\mathbb{H}^s$, this means as an operator from $\mathbb{H}^s$ to $(\mathbb{H}^s)'$. Indeed, you can see this from the definition of $\mathbb{H}^s$: if $u=\sum u_k \varphi_k$ then $(-\Delta)^su = \sum u_k \lambda_k^s\varphi_k$ which is not generally in $L^2$ since $\sum |u_k|^2\lambda_k^{2s}$ need not be finite.

An operator that maps $\mathbb{H}^s$ into $L^2$ is $(-\Delta)^{s/2}$. When they write $\operatorname{Dom}(-\Delta)^{s/2}$, the mean as an operator into $L^2$.

Summary:

• when they say "domain is", read it as an operator into $L^2$
• when they say "can be extended to", read it as an operator into the dual space.