In several places in the literature (e.g. this paper of Caffarelli and Silvestre), I've seen an integral formula for fractional Laplacians. I'd like to understand it. In this question, I'll stick to the case of the square root.

The formula I've seen is this: $$((-\triangle)^{1/2}f)(x)= C_n \int_{\mathbb{R}^n}\frac{f(x) - f(y)}{\|x - y\|^{n + 1}}\ dy. $$ Here $x \in \mathbb{R}^n$ and $C_n$ is a constant. Also, $f$ is a function $\mathbb{R}^n \to \mathbb{R}$, but I'm not sure what regularity assumptions it's supposed to satisfy.

For this notation to be justified, it must surely be the case that $$ (-\triangle)^{1/2} \bigl((-\triangle)^{1/2} f\bigr) = -\triangle f $$ for all nice enough $f$. My question is: why? I haven't been able to prove this identity even in the case $n = 1$.


  1. It's clearly the case that the Laplace operator has a square root defined by $$ \widehat{((-\triangle)^{1/2}f)}(\xi) = \|\xi\| \hat{f}(\xi). $$ The paper linked to says that this operator $(-\triangle)^{1/2}$ is the same as the operator $E$ defined by the integral formula. If I'm understanding correctly, proving this is equivalent to proving (i) that $E$ really is a square root of the Laplacian, and (ii) that $E$ is a positive operator on functions of compact support.

  2. I've seen a couple of references to Landkof's 1972 book Foundations of Modern Potential Theory. Unfortunately, those citing Landkof's book don't say which part of the book they're referring to, and I've been unable to find the relevant part myself. I'd be happy for someone to simply tell me where in that book to look.

  3. I can see that the integral formula has something to do with Laplacians. Switching to spherical coordinates, the formula is $$ ((-\triangle)^{1/2}f)(x) = \text{const}\cdot\int_0^\infty \frac{\int_{S^{n-1}} f(x + ru)\ du - f(x)}{r^2}\ dr $$ where $du$ is surface area measure on $S^{n-1}$ normalized to a probability measure. The integrand converges to $(\triangle f)(x)$ as $r \to 0$ (up to a constant factor). Also, the integrand is identically zero if $f$ is harmonic, which is promising.

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    $\begingroup$ Hi Tom, A basic reference for this sort of things is the classic paper of Seeley "Complex powers of an elliptic operator". Since this is one of the early papers on pseudo-differential operators, I'm sure you can find something about this in a book (Hôrmander or Taylor are good bets). $\endgroup$ – alvarezpaiva May 9 '12 at 6:13

A very nice way to see all of this is to look start with the Poisson semi-group $f \mapsto e^{-t\sqrt{-\Delta}}f$ for $t>0$. These operators are defined by Fourier Transform as $$\widehat{e^{-t\sqrt{-\Delta}}f}(p)=e^{-t|p|}\widehat{f}(p),$$ for any function $f$ so that the right hand side makes sense. Then for $f$, say in $ L^1$, we have $$e^{-t\sqrt{-\Delta}}f(x)=\int_{\mathbb{R}^n} P_t(x-y)f(y) dy$$ where $P_t(x)$ is the Poisson kernel $$P_t(x)=\frac{1}{(2\pi)^n} \int_{\mathbb{R}^n} e^{ix\cdot p} e^{-t|p|} dp = C_n \frac{t}{(t^2 +|x|^2)^\frac{n+1}{2}}. $$ (The computation of $P_t(x)$ is carried out in the first chapter of Stein and Weiss, Fourier Analysis on Euclidean Spaces.) The formula for $\sqrt{-\Delta}$ follows by taking the limit $$\sqrt{-\Delta}f = \lim_{t\downarrow 0} \frac{f- e^{-t\sqrt{-\Delta}}f}{t}$$ whenever this limit exists in a suitable sense. Because $\int P_t(x)dx=1$ we have $$\frac{1}{t} \left (f(x)-e^{-t\sqrt{-\Delta}}f(x) \right ) = C_n \int_{\mathbb{R}^n} \frac{f(x)-f(y)}{(t^2 +|x-y|^2)^{\frac{n+1}{2}}}dy.$$ Your identity now follows whenever $f$ is smooth enough and decays fast enough for the integral to make sense.


The Fourier transform of a radially symmetric function is a radially symmetric function. Basic scaling shows that the Fourier transform of $\|\xi\|$ must be $\|x\|^{-n-1}$ in some sense. Some care must be taken in interpreting this, since this function is not integrable. A good reference for this is Gelfand and Shilov, Generalized Functions.

Edit by Tom Leinster Here I'll try to flesh out Michael's answer. Michael: please feel free to edit.

We begin by finding the Fourier transform of $\|\xi\|$. A priori this doesn't make sense as $\|\xi\|$ isn't integrable, but I'll proceed formally anyway, hoping that there's a world in which this is all kosher. Write $g(\xi) = \|\xi\|$ and $g_a(\xi) = g(a\xi)$ (for $a > 0$). For completely general reasons, $$ \widehat{g_a}(x) = a^{-n} \hat{g}(x/a). $$ Also, for this particular $g$ we have $g_a = a g$, so $\widehat{g_a} = a\hat{g}$. Hence $\hat{g}(x) = a^{-(n+1)} \hat{g}(x/a)$, giving $$ \hat{g}(x) = \|x\|^{-(n+1)} g(x/\|x\|). $$ On the other hand, $g$ is spherically symmetric, so $\hat{g}$ is too; hence $\hat{g}$ has constant value $C$ on the unit sphere. So the Fourier transform of $\|\xi\|$ is $C\|x\|^{-(n+1)}$.

Now we take the Fourier transform of each side of the equation $(\widehat{(-\Delta)^{1/2}}f)(\xi) = \|\xi\|\hat{f}(\xi)$. This gives $$ ((-\Delta)^{1/2}f)(x) = C\|x\|^{-(n+1)} * f = C\int_{\mathbb{R}^n} \frac{f(y)}{\|x-y\|^{n+1}} \ dy. $$ That seems good, but now I have three problems. First, this isn't the integral formula I was looking for. (Maybe I'm missing a simple trick.) Second, I don't know the value of $C$. Third, I don't know how to make this all rigorous: what are some sufficient conditions on $f$ guaranteeing that $(-\Delta)^{1/2}\bigl((-\Delta)^{1/2}f\bigr)$ is defined (in the sense of the integrals existing) and equal to $f$?

Edit by Michael Renardy:

The problem is that $\|x\|^{-n-1}$ is not integrable at zero. Therefore it needs to be replaced by a regularization. The theory of such regularizations is developed in detail in Section 3 of Chapter I in Gelfand and Shilov.

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    $\begingroup$ Thanks for the answer, Michael. I'd been slightly frustrated at people citing Landkof's book of 400+ pages without saying which part of it they were referring to. But you've gone one better, by citing a three-volume collection without saying which part of it you're referring to :-) $\endgroup$ – Tom Leinster May 7 '12 at 23:19
  • $\begingroup$ More seriously, could you elaborate? I think I see what you mean and can guess what strategy you have in mind, but when I carry it out, it leads me to an integral formula different from the one above. $\endgroup$ – Tom Leinster May 7 '12 at 23:46
  • $\begingroup$ I'll provide chapter numbers after I get to my office where I can look it up. But it will be the morning before that happens. If you can look at the table of contents before that, good for you! $\endgroup$ – Michael Renardy May 7 '12 at 23:53
  • $\begingroup$ Great, thanks. I have it on my shelf, but I don't really know what I'm looking for. $\endgroup$ – Tom Leinster May 7 '12 at 23:55
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    $\begingroup$ The relevant sections are Chapter I, Section 3.9 and Chapter II, Section 3.3 in Volume I. $\endgroup$ – Michael Renardy May 8 '12 at 13:03

Denote integral operator as follows:$$Lu=P.V.\int_{\mathbb{R}^n}\frac{u(x) - u(y)}{\|x - y\|^{n + 1}}\ dy$$when $u\in \varphi$. It can also be written as (after changing variable $y$ to $-y$ and then taking average) a more symmetric form $$Lu=\int_{\mathbb{R}^n}\frac{u(x+y) +u(x-y)- 2u(x)}{\|y\|^{n + 1}}\ dy.$$ Thus the singularity can be cancelled. We are looking for its symbol (or multipler), that is $$\mathcal {F}(Lu)(\xi)=m(\xi)\mathcal {F}u$$and we want to prove that $m(\xi)$ is exactly $c_{n}|\xi|$, where $c_{n}$ is a constant determinated later which is the answer of the second question of Tom Leinster. First for $u\in \varphi$, we have $$\frac{u(x+y) +u(x-y)- 2u(x)}{\|x - y\|^{n + 1}}\in L^{1}(\mathbb{R}^n).$$ By Fubini's theorem (we exchange the integral in $y$ with the Fourier transform in $x$), we obtain $$ \begin{align*} \mathcal {F}(Lu)&=\int_{\mathbb{R}^n}\frac{\mathcal {F}(u(x+y) +u(x-y)- 2u(x))}{\| y\|^{n + 1}}\ dy \\ &= \int_{\mathbb{R}^n}\frac{e^{i\xi\cdot y}+e^{-i\xi \cdot y}-2}{\|y\|^{n + 1}}\ dy \cdot \mathcal {F}(u).\end{align*}$$ Hence, in order to get the desired result, it suffices to show that $\int_{\mathbb{R}^n}\frac{e^{i\xi y}+e^{-i\xi y}-2}{\|y\|^{n + 1}}\ dy=c_{n}|\xi|$. Define $$I(\xi)=\int_{\mathbb{R}^n}\frac{1-\cos(y \xi)}{\|y\|^{n + 1}}\ dy.$$ Since $I(\xi)$ is rotationally invariant, $I(\xi)=I(|\xi|e_{1})$ where $e_{1}$ denotes the first direction vector. So $$I(\xi)=I(|\xi|e_{1})=|\xi|\int\frac{1-\cos(z_{1})}{\|z\|^{n + 1}}dz$$so we take $c_{n}=\int\frac{1-\cos(z_{1})}{\|z\|^{n + 1}}\ dz$. And we have $(-\triangle)^{1/2}=\frac{1}{c_{n}}L$ as desired.

  • $\begingroup$ it's really nice to start from the poisson semigroup,but the mathod pressented here is applied for any $0<\alpha<1$ with $(-\triangle)^{\alpha}$ $\endgroup$ – user23078 May 8 '12 at 23:17
  • $\begingroup$ Good job. Your answer also clarifies the meaning of the original integral, which of course does not converge as a Lebesgue integral but only if interpreted as a singular integral or a principal value. $\endgroup$ – Piero D'Ancona May 9 '12 at 8:17

So you believe expression (1)? Have you tried taking the inverse Fourier transform of the right side of this expression? This will give you the convolution of f with the inverse Fourier transform of $|\xi|$, at least morally speaking. The inverse Fourier transform of $|\xi|^{\alpha}$ can be made sense of.

When $n=1$, for example, and $-1<\alpha<0$, this is relatively easy to compute directly, and is something like $C_\alpha |x|^{-\alpha-1}$. This kind of computation would probably be done in a textbook in a section like 'tempered distributions'. You can see that that $\alpha=1$ doesn't fit into this range, which is clear since the formulas wouldn't match up in such a case. But if one naively input this value, we'd get a convolution with $|x|^{-2}$, which is nice. So you clearly should be guessing that the inverse Fourier transform of $|\xi|$ is convolution with $|x|^{-2}$ plus a delta function, modulo constants.

This can be made rigorous, but the computation involves knowing some stuff about distributions. Try looking up 'homogeneous distribution'. I think actually with the information on the wikipedia pages for Fourier transform and homogeneous distribution, you should be able to do the computation. This older mathoverflow question might also be helpful: The fourier transform of homogeneous distribution and related topics

  • $\begingroup$ Thanks, Peter. My edit to Michael's answer shows my attempt to do what you're suggesting in your first paragraph. I'll go and look up homogeneous distributions. I learned about tempered distributions and their Fourier transforms from Friedlander and Joshi's text, but the treatment there is quite concise. $\endgroup$ – Tom Leinster May 8 '12 at 14:42
  • $\begingroup$ Yes, much of this stuff is fairly disjointed, as I recall, which is unfortunate. Many of the basic computations are not terribly difficult (just tricky), so I think they are not written down in many places. The issue you are having in the computation in your comment comes from having to use some analytic continuation technique to extend the range of possible $\alpha$s in the inverse transform of $|\xi|^\alpha$. $|x|^{-n-1}$ is not locally integrable, as Tom pointed out, so its a bit more delicate. $\endgroup$ – Peter Luthy May 8 '12 at 15:05
  • $\begingroup$ Now that I think about it, the analytic continuation might need some additional work... there might be some poles at certain integers $\alpha$, so you would need to add an additional term to balance things out. This computation has to be written down someplace. Sorry that I don't know of a good place. $\endgroup$ – Peter Luthy May 8 '12 at 16:12

This answer is a modified version of Tom's edit on Michael's answer just to make it a bit more rigorous.

Take $b(\xi)$ to be a smooth non negative radially symmetric function supported in $1< |\xi| <2$. Then we see that $|\xi|^\alpha = c \int_0^\infty t^{\alpha-1} b(\xi/t) dt$.

Now, let us say $b$ is the fourier transform of some kernel $k$, which will also be smooth, radially symmetric, and its integral is zero because $b(0)=0$. Then $b \cdot \hat u = \widehat{k \ast u}$. Moreoever, since k has integral zero and is symmetric $$ b \cdot \hat u = \left(\int k(y) (u(x+y)-u(x)) dy\right)^\wedge. $$

Let us assume that $\alpha \in (0,2)$.

I want to use the identity above for $|\xi|^\alpha$ as an integral of scaled versions of $b$. For that recall that $b(\cdot/t)^{\wedge} = t k(t\cdot)$. So, we have $$ |\xi|^\alpha \cdot \hat u = \left( c \int_0^\infty t^{\alpha-1} \int t k(ty) (u(x+y)-u(x)) \ dy dt \right)^\wedge$$ Exchanging the order of integration we have that $$ |\xi|^\alpha \cdot \hat u = \left( \int (u(x+y)-u(x)) K(y) dy \right)^\wedge$$ for $$ K(y) = c \int_0^\infty t^{\alpha} k(ty) dt = c_0 |y|^{-n-\alpha}.$$ The last identity is obtained from the change of variables $s = t|y|$.

There are still a few things that should be clarified, like the application of Fubini, and that the integrals may be principal values when $\alpha \geq 1$. But everything should be fine if $u$ is smooth enough.


Perhaps you could see what the right-hand-side evaluates to for the eigen-functions of the Laplacian (I don't know what they are offhand but I would guess that the eigenbasis for the Laplacian on L^2(R) is well known). You should get that the eigen-functions are still eigen-functions of the right-hand-side but with each of the eigen-values raised to the 1/2 power.

Also, your question leads me to a question of my own: From the probabilistic perspective, the right-hand-side integral operator is associated to a Markov jump process. So this statement seems to imply that in some way "The square of a jump process gives a true diffusion" or maybe "if you jump and jump again you get a true diffusion". I am wondering if anyone knows anymore on this line of thought or a reference where it is made precise?

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    $\begingroup$ That is it: there are no eigenfunctions here unfortunately. And you should ask your question separately, it will not be noticed here. $\endgroup$ – András Bátkai May 8 '12 at 20:26

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