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.