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Let $s\in(0,1)$, let $u\in H^s(\mathbb{R}^n)$. For all $\psi\in\mathcal{S}(\mathbb{R}^n)$, let: $$ (-\Delta)^s\psi(x)=c(n,s)\lim_{\epsilon\to0^+} \int_{\mathbb{R}^n\setminus B_\epsilon(0)}\frac{\psi(x)-\psi(y)}{|x-y|^{n+2s}}\,dy \quad\forall x\in\mathbb{R}^n,$$ the fractional laplacian of $\psi$. Or equivalently: $$ (-\Delta)^s\psi(\xi)=\mathcal{F}^{-1}(|\xi|^{2s}\mathcal{F}\psi)(\xi),\quad\forall\xi\in\mathbb{R}^n. $$ I think that the following equality holds: $$ \int_{\mathbb{R}^n}u(x)(-\Delta)^s\psi(x)\,dx=\frac{c(n,s)}{2}\int_{\mathbb{R}^n\times\mathbb{R}^n} \frac{(u(x)-u(y))(\psi(x)-\psi(y))}{|x-y|^{n+2s}}\,dx\,dy\tag{1} $$ Can you help me to prove or improve (1)? I have no idea on how to go on. Any help would be appreciated.

I have proved that, using Plancherel: $$ \int_{\mathbb{R}^n}u(x)(-\Delta)^s\psi(x)\,dx=\int_{\mathbb{R}^n}|\xi|^{2s}\mathcal{F}u(\xi)\mathcal{F}\psi(\xi)\,d\xi, $$ where $\mathcal{F}$ is the Fourier transform.

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  • $\begingroup$ I am a bit confused. Can you enlighten me on the definition of $c(n,s)$? $\endgroup$
    – Bombyx mori
    Commented Nov 1, 2020 at 3:10
  • $\begingroup$ The constant $c(n,s)$ is the dimensional constant in the definition of fractional laplacian, that is: $$ c(n,s)=\biggl\{\int_{\mathbb{R}^n}\frac{1-\cos \zeta_1}{|\zeta|^{n+2s}}\,d\zeta\biggr\}^{-1} $$ $\endgroup$
    – inoc
    Commented Nov 1, 2020 at 6:43

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