Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
$(-\Delta)^{s}$ is positive if and only if $0<s\les 1$, and they generates positive heat semigroup $e^{-t(-\Delta)^{s}}$. –  user23078 May 6 '13 at 15:49
    
and I guess the intermediate formula involving "lower derivatives" looks like $$ \int\limits_{\mathbb{R}^d}f[(-\Delta)^s g]=-\int\limits_{\mathbb{R}^d}(-\Delta)^{s/2} f(-\Delta)^{s/2} g $$ ??? –  leo monsaingeon May 6 '13 at 18:44
    
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. –  Ray Yang May 7 '13 at 17:29
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
Thank you, very instructive. This is precisely the computation I wanted to see! –  leo monsaingeon May 16 '13 at 20:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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