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Let $0<s<1$. Let $\Omega\subset\mathbb{R}^n$ be a bounded domain.

We know that $$\|(-\Delta)^{s/2}u\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}dxdy$$ See e.g. Hitchhiker's guide to the fractional Sobolev spaces, page 16.

I am wondering if the equality still holds when $\mathbb{R}^n$ is replaced by $\Omega$. Namely $$\|(-\Delta)^{s/2}u\|_{L^2(\Omega)}^2=\int_{\Omega}\int_{\Omega}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}dxdy\ ?$$

We may assume that $u\in L^2(\mathbb{R}^n)$ and supp $u$ $\subset\Omega$.

Thanks!

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Nice question, with negative answer (unless $s=0$ or $2$, of course).

(I am unsure about this formula. The non-shaded part of this post is fine). The correct answer is $$\|(-\Delta)^{s/2} f\|_{L^2(\Omega)}^2=\int_\Omega \int_{\mathbb R^n} \frac{ |f(x)-f(y)|^2}{|x-y|^{n+2s}}\, dxdy; $$ note that only one of the integrals is on $\Omega$.

The point is that the fractional Laplacian is non-local; the value of $(-\Delta )^{s/2}f $ at a point depends on $f$ at all points. In particular, a norm of $(-\Delta)^{s/2} f$ must take into account the values of $f$ everywhere; you cannot arbitrarily choose to consider only the values in $\Omega$.


There are different versions of the fractional Laplacian adapted to domains. If you take a complete orthonormal system of eigenfunctions $\phi_0, \phi_1, \phi_2\ldots$ for the Dirichlet problem on $\Omega$, you can define the so-called Dirichlet Laplacian by $$ (-\Delta)^{s}_{\mathrm{Dir}} f:=\sum_{\ell=0}^\infty \lambda_\ell^s \hat{f}(\ell) \phi_\ell, $$ where $-\Delta \phi_\ell=\lambda_\ell \phi_\ell$, and $$ \hat{f}(\ell):=\int_{\Omega} f \phi_\ell\, dx.$$ This is a version of the fractional Laplacian that is localized on $\Omega$.

I don't know what is the precise relationship between the Dirichlet Laplacian and the free-space fractional Laplacian.

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  • $\begingroup$ Thanks! But the equality you pointed out seems not trivial to me. I cannot prove it directly from the definition or modifying the proof of the $\mathbb{R}^n$ case. Could you add more details? $\endgroup$ – Right Apr 20 '19 at 19:34
  • $\begingroup$ It seems not trivial because it is probably wrong, sorry about that. But the rest of the answer stands. Let me correct $\endgroup$ – Giuseppe Negro Apr 20 '19 at 19:41

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