# About an equality of fractional Laplacian on a bounded domain

Let $$0. Let $$\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!

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.

• 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? – Right Apr 20 '19 at 19:34
• It seems not trivial because it is probably wrong, sorry about that. But the rest of the answer stands. Let me correct – Giuseppe Negro Apr 20 '19 at 19:41