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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?

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  • $\begingroup$ $(-\Delta)^{s}$ 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

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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.

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

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