We already know the following facts： $$\displaystyle (-\Delta)^su(x):=c_{n,s}\text{P.V.}\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+2s}}\,dy, $$ where $s\in (0,1)$. $$\int_{\mathbb{R}^N} (-\Delta)^{s} u(x) \cdot v(x)dx=\frac{c_{n,s}}{2}\int_{\mathbb{R}^N} P.V.\int_{\mathbb{R}^N} \frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}dydx.$$ $$\int_{\mathbb{R}^N} (-\Delta)^{s} u(x) \cdot v(x)dx=\int_{\mathbb{R}^N} (-\Delta)^{s/2} u(x) (-\Delta)^{s/2} v(x)dx.$$
But I have a puzzle. According the definition of $(-\Delta)^s$, I think $(-\Delta)^{s/2}$ should be defined as follows: $$(-\Delta)^{s/2}u(x)=c_{n,s}\text{P.V.}\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+s}}\,dy,$$ so $$\int_{\mathbb{R}^N} (-\Delta)^{s/2} u(x) (-\Delta)^{s/2} v(x)dx=c_{n,s}\int_{\mathbb{R}^N}\text{P.V.}\!\!\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+s}}\,dy \text{P.V.}\int_{\mathbb{R}^N}\frac{v(x)-v(y)}{|x-y|^{N+s}}\,dy\,dx.$$ How did the third identity hold? How to prove it by this definition, not using the Fourier transform?