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Using the Fourier transform we can easily define the fractional Laplacian by $$(-\Delta)^{s/2}f(x)=(|\xi|^s\hat f(\xi))^\vee(x), \ \ f\in C_0^\infty. $$

However, I learned that there is another definition using the principal value of singular integral $$(-\Delta)^{s/2}f(x)=C_{n,s}P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(y)}{|x-y|^{n+s}}dy}, \ \ 0<s<2, $$ where $C_{n,s}$ is some constant. I am curious about the proof that these two definitions are equivalent, but I can not find the proof on the internet or textbooks. By the way, I found that one may symmetrize the integral above to regularize the integral $$ P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(y)}{|x-y|^{n+s}}dy}= P.V.\int_{\mathbb{R}^n}{\frac{f(x)-f(x-y)}{|y|^{n+s}}dy}\\=\frac12\int_{|y|\le h}{\frac{2f(x)-f(x+y)-f(x-y)}{|y|^{n+s}}dy}+\int_{|y|\ge h}{\frac{f(x)-f(x-y)}{|y|^{n+s}}dy}.$$

Ethan Bolker
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Mr.right
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1 Answers1

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As indicated by the now deleted answer, a full proof is available in this wonderful paper by Nezza-Palatucci-Valdinoci (Arxiv link here, published version here). I'll paraphrase it here to answer your question.

The equality is proven for Schwartz functions $\mathscr S$. Call the Fourier definition $Fu = \mathscr F^{-1}|\xi|^s \mathscr F u $ (I used $u$ instead of $f$ to avoid too many "f"s and to follow the above paper) and call the principal value definition $Pu$. The goal is to show $F=P$ on $\mathscr S$. Notice also that your second integral for $Pu$ can also be symmetrized to obtain $$ Pu = \frac12C_{n,s}\int_{\mathbb R^n} \frac{2u(x) - u(x+y)-u(x-y)}{|y|^{n+s}} \,\mathrm dy.$$ With $Pu$ rewritten in this way, it is enough to prove that their Fourier transforms are equal, namely $\mathscr F Pu = \mathscr F Fu = |\xi|^s\mathscr F u$. By the decay of $u\in \mathscr S$, and the cancellation at $y=0$ from the numerator, Fubini on $\mathbb R^{2n}$ applies to give us $ \mathscr F P u = P\mathscr F u$, which is in full, \begin{align} (\mathscr FPu)(\xi) & = \frac{1}{2}C_{n,s}\int_{\mathbb R^n} \frac{2\mathscr F u(\xi) - \mathscr F_x [u(x+y)](\xi)-\mathscr F_x [u(x-y)](\xi)}{|y|^{n+s}} \,\mathrm dy \\ & = \frac{1}{2}C_{n,s}\int_{\mathbb R^n } \frac{2 -e^{i\xi\cdot y}-e^{-i\xi\cdot y}}{|y|^{n+s}} \,\mathrm dy\, \mathscr F u (\xi)\\ & = \left[ C_{n,s} \int_{\mathbb R^n} \frac{1-\cos(\xi\cdot y)}{|y|^{n+s}} \,\mathrm dy \right] \mathscr F u (\xi). \end{align} Here, we used that translations correspond to modulations in the Fourier transform, i.e. $$ \mathscr F [u(\cdot +y) ](\xi) = e^{-i\xi\cdot y} \mathscr Fu(\xi).$$ This is already in the form of a multiplier. Moreover you don't seem to care what the constant is, so we might be satisfied by computing the dependence of our symbol $S(\xi):=\int_{\mathbb R^n} \frac{1-\cos(\xi\cdot y)}{|y|^{n+s}} dy$ on $|\xi|$. This is easy to see by performing the scaling $$ z = |\xi|y ⟹ dz = |\xi|^n \,\mathrm dy,$$ which gives $$ S(\xi) = |\xi|^s S\left(\frac{\xi}{|\xi|}\right). $$

Actually, it should be clear that $S$ is a radial function. The proof is identical in its steps to the proof that a radial function has radial Fourier transform. Hence, $S\left(\frac{\xi}{|\xi|}\right) = S\big((1,0,\dots ,0)\big)$ and depending on who you ask, we can even take $C_{n,s} := S\big((1,0,\dots ,0)\big)^{-1}$ which gives precisely the result.

Calvin Khor
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