Let $s\geq 0$ be a real number, $\Omega$ bounded and smooth domain in $\mathbb R^d$. We can define the spectral power of the Dirichlet Laplacian $(-\Delta_\Omega)$ on $\Omega$. Then, on the whole space, the fractional laplacian $(-\Delta)^s$ is an operator defined via Fourier transform: $$\mathcal F((-\Delta)^s u)(\xi):= |\xi|^{2s} \mathcal F (u)(\xi)\, .$$ Then, do we have the following continuous embedding $$\text{dom}(-\Delta_\Omega)^s\subset \text{dom}(-\Delta)^s\, ?$$
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1$\begingroup$ No, this fails already in simple situations. If you take $s=d=1$, $\Omega=(a,b)$, then $D(-\Delta)=H^2(\mathbb R)$, $D(-\Delta_{\Omega})=H^2_0(a,b)$, but these latter functions need not have a second derivative at $a,b$ when you extend them. (I'm assuming you identify $L^2(\Omega)$ with a subspace of $L^2(\mathbb R^d)=L^2(\Omega)\oplus L^2(\Omega^c)$ in the obvious way.) $\endgroup$– Christian RemlingCommented Sep 7, 2017 at 15:47
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$\begingroup$ Agreed, except that $D(-\Delta_\Omega)=H^2\cap H^1_0\neq H^2_0$ (in usual notations)... $\endgroup$– Jean DuchonCommented Sep 7, 2017 at 19:37
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$\begingroup$ @JeanDuchon: Yes, my notation is sloppy, $D(-\Delta_{\Omega})$ is those $H^2$ functions with $u(a)=u(b)=0$, and I guess you wouldn't normally call this $H^2_0$. $\endgroup$– Christian RemlingCommented Sep 7, 2017 at 21:26
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$\begingroup$ Thank you for your comments. Can we say more generally that $\text{dom}(-\Delta_\Omega)^s=H^{2s}(\Omega)\cap H_0^1(\Omega)$? $\endgroup$– ThomasCommented Sep 8, 2017 at 7:34
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$\begingroup$ Obviously not for $s<1/2$, but otherwise yes, I think. $\endgroup$– Jean DuchonCommented Sep 8, 2017 at 10:50
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