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 wellbehaved $f$)? 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?

$\begingroup$ $(\Delta)^{s}$ is positive if and only if $0<s\les 1$, and they generates positive heat semigroup $e^{t(\Delta)^{s}}$. $\endgroup$– user23078May 6, 2013 at 15:49

$\begingroup$ and I guess the intermediate formula involving "lower derivatives" looks like $$ \int\limits_{\mathbb{R}^d}f[(\Delta)^s g]=\int\limits_{\mathbb{R}^d}(\Delta)^{s/2} f(\Delta)^{s/2} g $$ ??? $\endgroup$– leo monsaingeonMay 6, 2013 at 18:44

$\begingroup$ These formulae are all correct, and the easiest way to realize this is to use the Fourier space representation, the usual function space is the $H^s$ space. $\endgroup$– Ray YangMay 7, 2013 at 17:29

$\begingroup$ For the usual fractional Laplacian $(\Delta)^{\frac{\alpha}{2}}$ $(0<\alpha<2)$. Is $(\Delta)^{\alpha} = (\Delta)^{\frac{\alpha}{2}}(\Delta)^{\frac{\alpha}{2}}$ true or not? $\endgroup$– CooLeeMay 19, 2015 at 16:58
1 Answer
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 $$ \int_{\mathbb{R}} (\Delta)^{1/2} f(x) g(x)dx=\int_\mathbb{R} g(x)P.V.\int_\mathbb{R} \frac{f(x)f(y)}{xy^2}dydx$$ $$ =\int_\mathbb{R} P.V.\int_\mathbb{R} \frac{g(y)(f(y)f(x))}{xy^2}dydx=\int_\mathbb{R} P.V.\int_\mathbb{R} \frac{g(y)(f(x)f(y))}{xy^2}dydx. $$ From here $$ \int_{\mathbb{R}} (\Delta)^{1/2} f(x) g(x)dx=\frac{1}{2}\int_\mathbb{R} P.V.\int_\mathbb{R} (g(x)g(y))\frac{f(x)f(y)}{xy^2}dydx $$ $$ =\frac{1}{2}\int_\mathbb{R} (\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int_{\mathbb{R}}P.V.\int_{\mathbb{R}} f(y)\frac{g(x)g(y)}{xy^2}dydx $$ $$ =\frac{1}{2}\int_\mathbb{R} (\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int_{\mathbb{R}}P.V.\int_{\mathbb{R}} f(y)\frac{g(x)g(y)}{xy^2}dydx. $$ Now you can change variables again in the last integral and conclude the result.

$\begingroup$ Thank you, very instructive. This is precisely the computation I wanted to see! $\endgroup$ May 16, 2013 at 20:26