3
$\begingroup$

In this paper, the authors claim that for $s\in [0,1]$, $\left [ H_0^1(\Omega), L^2(\Omega)\right ]_\theta=\text{dom}(-\Delta)^{\frac s 2}$, where $\Omega$ is a smooth and bounded domain in $\mathbb R^d$, $(-\Delta)$ is the Dirichlet Laplacian in $\Omega$, and $\theta =1-s$. Here, the interpolation theory is taken from the book of Lions and Magenes "Problèmes aux limites non homogènes et applications", Vol 1. Is the following more general statement true?

$$\left [ H_0^m(\Omega), L^2(\Omega)\right ]_\theta=\text{dom}(-\Delta)^{\frac s 2}$$ for $m=\lfloor s \rfloor +1$, and $\theta=1-\frac s m$?

More precisely, my question is: How can I characterize $\text{dom}(-\Delta)^{\frac s 2}$ in terms of Sobolev spaces?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Please see the answer of Mateusz for some fundamental problems with your question. With a bit of good faith however, your question HAS a positive answer for $m=2$ in the sense that $$\bigl[H^2(\Omega) \cap H^1_0(\Omega),L^2(\Omega)\bigr]_{1-\frac{s}2} = \mathrm{dom}_{L^2(\Omega)}(-\Delta_D)^{\frac{s}{2}}.$$

This follows from the general formula $$\bigl[ \mathrm{dom}_{L^2(\Omega)}(-\Delta_D)^{\alpha},L^2(\Omega)\bigr]_\theta = \mathrm{dom}_{L^2(\Omega)}(-\Delta_D)^{(1-\theta)\alpha}$$ and the elliptic regularity result $$\mathrm{dom}_{L^2(\Omega)}(-\Delta_D) = H^2(\Omega) \cap H^1_0(\Omega),$$ which are valid for the Dirichlet Laplacian on sufficiently smooth domains (for the first identity you need to know that the operator admits bounded imaginary powers). In fact, your first cited interpolation equality also follows from the first formula for $\alpha = \frac12$ and $$\mathrm{dom}_{L^2(\Omega)}(-\Delta_D)^{\frac12} = H^1_0(\Omega),$$ which is the rather famous Kato square root property for the Dirichlet Laplacian.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you very much. Do we also have that $\left [H^m(\Omega) \cap H_0^1(\Omega) , L^2(\Omega)\right ]_{1-\frac s m}=\text{dom}(-\Delta_D)^{\frac s 2}$? $\endgroup$
    – Thomas
    Commented Sep 7, 2017 at 14:45
  • $\begingroup$ Just wondering: what can one tell about $\operatorname{dom}(-\Delta_D)^{s/2}$ for $s>1$? Is it always $H^s(\Omega) \cap H^1_0(\Omega)$? This seems to be true when $1<s<2$, but what about $s>2$? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 7, 2017 at 14:45
  • $\begingroup$ Even for mixed boundary conditions and suitable non-smooth bounded domains one gets $\operatorname{dom}(-\Delta)^{s/2}=H^s_D(\Omega)$ for some $s$ slightly above $1$, where the Laplacian is subject to mixed boundary conditions and the Bessel potential space embodies those boundary conditions as well. $\endgroup$
    – Sebastian Bechtel
    Commented Sep 7, 2017 at 20:35
  • $\begingroup$ @Hannes do you have a reference for your "general formula"? In particular one that holds for a large class of operators, not just the laplacian? $\endgroup$
    – Kore-N
    Commented Nov 25, 2020 at 14:44
  • $\begingroup$ @Kore-N sure, see for instance Chapter 4 in Interpolation Theory by Alessandra Lunardi. (Or Chapter 1.15.3 in the Triebel book.) $\endgroup$
    – Hannes
    Commented Dec 11, 2020 at 9:54
2
$\begingroup$

No. A short argument is that the domain of $-\Delta$ on $\Omega$ with Dirichlet boundary condition is not $H^2_0(\Omega)$.

To be specific, $\sin x$ is the eigenfunction of $\Delta$ on $\Omega = (0, \pi)$, but it is not in $H^2_0(\Omega)$. This means that $\sin x$ is in the domain of $(-\Delta)^s$ for any $s > 0$ (I understand that $(-\Delta)^{s/2}$ is a power of the Dirichlet Laplacian), but it does not belong to $H^s_0(\Omega)$ for any $s \ge 2$.

Two remarks:

  1. For a Lipschitz $\Omega$, $H^s_0(\Omega)$ is typically defined as an appropriate subset of $H^s(\mathbb{R}^d)$ and it is known to be the (comlex) interpolation space of $L^2(\Omega)$ and $H^n_0(\Omega)$ for any $n > s$.

  2. If $(-\Delta)^s$ is to be the fractional power of $-\Delta$ in full space, and then restricted to $\Omega$ by setting a "boundary" condition $0$ outside of $\Omega$, then already the domain of $(-\Delta)^{1/2}$ is different from $H^1_0(\Omega)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for your quick answer. If I understand correctly, in your argument $\sin x$ is not in $H_0^2(\Omega)$ since the derivative of this function is not zero on the boundary of the domain. In that case, if$\lambda_j$ are the eigenvalues of the Dirichlet laplacian, can we say $\left [ H_0^m(\Omega), L^2(\Omega)\right ]_\theta$ is the set of functions $f \in L^2(\Omega)$ such that $\sum_{i\geq 1} \lambda_i ^s a_i(f)^2<+\infty$, where $a_i(f)=\langle f, \phi_i \rangle$ and $\phi_i$ is the associated eigenfunction. $\endgroup$
    – Thomas
    Commented Sep 7, 2017 at 14:38
  • $\begingroup$ No, the condition $\sum \lambda_i^s |a_i(f)|^2 < \infty$ describes the domain of $(-\Delta)^{s/2}$. By the way, according to Encyclopedia of mathematics, the domain of $-\Delta$ is $H^2(\Omega) \cap H^1_0(\Omega)$ when $\Omega$ is a $C^2$ domain. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 7, 2017 at 14:41

You must log in to answer this question.

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