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?

  • $\begingroup$ $(-\Delta)^{s}$ is positive if and only if $0<s\les 1$, and they generates positive heat semigroup $e^{-t(-\Delta)^{s}}$. $\endgroup$
    – user23078
    Commented May 6, 2013 at 15:49
  • $\begingroup$ and I guess the intermediate formula involving "lower derivatives" looks like $$ \int\limits_{\mathbb{R}^d}f[(-\Delta)^s g]=-\int\limits_{\mathbb{R}^d}(-\Delta)^{s/2} f(-\Delta)^{s/2} g $$ ??? $\endgroup$ Commented May 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 Yang
    Commented May 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? $\endgroup$
    – CooLee
    Commented May 19, 2015 at 16:58

1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.