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Is it true that $$\int_{\mathbb{R}^N}(-\Delta)^su(x) dx = 0,$$ where $(-\Delta)^s$ is the fractional Laplacian?

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    $\begingroup$ Seems clear from the Fourier-transform definition of $(-\Delta)^s$: the integral over ${\bf R}^n$ of $(-\Delta)^s u$ is the value at zero of the Fourier transform of $(-\Delta)^s u$; but this Fourier transform is $\xi \mapsto |\xi|^{2s} \, \widehat u(\xi)$, so it vanishes at $\xi = 0$, QED. $\endgroup$ Mar 16, 2018 at 0:46

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Yes, under certain assumptions. For $0<\alpha<2$ and $\varphi\in\mathscr{S}_n$ we have \begin{eqnarray*} (-\Delta)^{\alpha/2}\varphi & = & \frac{1}{2\gamma(-\alpha)} \int_{\mathbb R^n} \frac{\varphi(x+y)+\varphi(x-y)-2\varphi(x)}{|y|^{n+\alpha}}\, dy, \end{eqnarray*} where $$ \gamma(-\alpha)= \frac{\pi^{\frac{n}{2}}2^{-\alpha}\Gamma\left(-\frac{\alpha}{2}\right)}{\Gamma\left(\frac{n+\alpha}{2}\right)}\,. $$ One can show that $$ (x,y)\mapsto\frac{\varphi(x+y)+\varphi(x-y)-2\varphi(x)}{|y|^{n+\alpha}}\in L^1(\mathbb R^n\times\mathbb R^n). $$ Then integrating $(-\Delta)^{\alpha/2}\varphi$ over $\mathbb R^n$ and applying Fubini theorem yields zero. The same applies to $s=\frac{\alpha}{2}+k$, where $k$ is a positive integer since \begin{eqnarray*} (-\Delta)^{\frac{\alpha}{2}+k}\varphi & = & \frac{(-1)^k}{2\gamma(-\alpha)} \int_{\mathbb R^n} \frac{\Delta^k\varphi(x+y)+\Delta^k\varphi(x-y)-2\Delta^k\varphi(x)}{|y|^{n+\alpha}}\, dy. \end{eqnarray*} You can find proofs for example here (Theorem 7.13 and Corollary 7.16). The argument does not apply to negative powers of the Laplacian. The negative powers of compactly supported smooth functions are rarely integrable (I think).

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  • $\begingroup$ OK. My answer is correct, but overly complicated (see comment by Noam D. Elkies). However, some people use the integral fromula from my solution as a definition of the fractional Laplacian. $\endgroup$ Mar 16, 2018 at 1:10
  • $\begingroup$ Does the result hold for $\varphi \in W^{2,\infty}$? $\endgroup$
    – user60665
    Mar 21, 2018 at 23:00
  • $\begingroup$ @Dal: The explicit example in my answer is a counter-example (unless $n = 1$ and $\alpha \geqslant 1$; then $f'$ is a counter-example). You do need some kind of integrability of $\varphi$ at infinity. $\endgroup$ Mar 22, 2018 at 13:20
  • $\begingroup$ @Dal: A simplest example $n=1$, $s=1$, $u(x)=\sin x$, $-\Delta u = -u''=u$ is not integrable. $\endgroup$ Mar 22, 2018 at 18:09
  • $\begingroup$ Thanks. Then what is the lightest assumption on $\varphi$ that makes the result true? $\endgroup$
    – user121481
    Mar 24, 2018 at 20:51
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Noam D. Elkies and Piotr Hajlasz give a positive answer to the original question, with two different proofs. While their arguments are completely fine, they are somewhat restrictive: the integral of $(-\Delta)^{\alpha/2} f$ is zero if $f$ is integrable (and regular enough). The general answer is no!


Suppose that $\alpha < n$, and let $g$ be any continuous, compactly supported (or merely integrable) function. Let $$ f(x) = c_{n,\alpha} \int_{\mathbb{R}^n} \frac{g(y)}{|x - y|^{n - \alpha}} \, dy $$ be the Riesz potential of $g$. Then $(-\Delta)^{\alpha/2} f(x) = g(x)$ for every $x \in \mathbb{R}^n$ (for the proof, see, for example, Samko's book Hypersingular Integrals and Their Applications), and so the integral of $(-\Delta)^{\alpha/2} f$ need not be zero.


A nice explicit example is $$f(x) = (1 + |x|^2)^{\alpha/2 - n/2},$$ which corresponds to $$g(x) = c_{n,\alpha} (1 + |x|^2)^{-\alpha/2 - n/2}.$$ This follows, for example, from Corollary 1 here.

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