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 $n = 1$.

**Comments**

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.

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.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.