Is there an integration by parts formula for fractional laplacians in $L^p(\mathbb{R}^N)$, something like 
 $$
 s\in(0,1),\qquad\int\limits\_{\mathbb{R}^N}f[(-\Delta)^sg] =\int\limits\_{\mathbb{R}^N}[(-\Delta)^{s}f]g 
 $$
or an intermediate formula involving "lower derivatives"?
Typically, I would like to know if
 $$
 \int\limits\_{\mathbb{R}^N}f\cdot[(-\Delta)^sf] dx\geq 0
 $$
still holds true as for the usual Laplacian (say for well-behaved $f$)? Computing formally in the Fourier space with $\widehat{(-\Delta)^sf}(\xi)=|\xi|^{2s}\hat{f}(\xi)$ it seems obvious, but it is not clear to me from the Riesz potential representation of $(-\Delta)^{-s}f$. Also, what kind of regularity/decay at infinity do I need in order not to bother with boundary terms at infinity?



integration by parts for the fractional Laplacian
You can integrate by parts:


$$
\int\_{\mathbb{R}^d} (-\Delta)^s f(x) g(x)dx=\int\_{\mathbb{R}^d} (-\Delta)^s g(x) f(x)dx. 
$$
Using Fourier and $L^2$ the equality is obvious. Let's do "by hand" in $d=1$ and $s=1/2$ (the other cases follow the same idea:


You have
$$
\int\_{\mathbb{R}} (-\Delta)^{1/2} f(x) g(x)dx=\int\_\mathbb{R} g(x)P.V.\int\_\mathbb{R} \frac{f(x)-f(y)}{|x-y|^2}dydx$$
$$
=\int\_\mathbb{R} P.V.\int\_\mathbb{R} \frac{g(y)(f(y)-f(x))}{|x-y|^2}dydx=-\int\_\mathbb{R} P.V.\int\_\mathbb{R} \frac{g(y)(f(x)-f(y))}{|x-y|^2}dydx.
$$
From here
$$
\int\_{\mathbb{R}} (-\Delta)^{1/2} f(x) g(x)dx=\frac{1}{2}\int\_\mathbb{R} P.V.\int\_\mathbb{R} (g(x)-g(y))\frac{f(x)-f(y)}{|x-y|^2}dydx
$$
$$
=\frac{1}{2}\int\_\mathbb{R} (-\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int\_{\mathbb{R}}P.V.\int\_{\mathbb{R}} -f(y)\frac{g(x)-g(y)}{|x-y|^2}dydx
$$
$$
=\frac{1}{2}\int\_\mathbb{R} (-\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int\_{\mathbb{R}}P.V.\int\_{\mathbb{R}} -f(y)\frac{g(x)-g(y)}{|x-y|^2}dydx.
$$
Now you can change variables again in the last integral and conclude the result.


