Interpretation of the singular integral for the definition of fractional Laplacian in classical case $s=1$

Question

Fractional Laplacian is often defined via next principal value integral (assume dimension one for the simplicity) $$(-\Delta)^su(x):=c_sP.V.\int_\mathbb{R} \frac{u(x)-u(y)}{|x-y|^{1+2s}}\, dy.$$ This improper integral makes sense when $s<1$, but how do we interpret this singular integral when $s=1$? It should correspond to classical local Laplacian, but the integral itself is not well-defined even as a principal value sense. Of course classical Laplacian is a local operator so definition via integral should not make sense as long as the kernel is not some singular distribution, but what can be said more about the interpretation of this improper integral? For me I do not see any immediate relation between $1/|x|^3$ and singular distributions (second order Dirac deltas). Also residue theorem pops up in my head, but I am not sure if there is any clear relation to it.

Thank you im advance.

Answer 1

One way to interpret the singular integral for $s=1-\epsilon$ is by Fourier transformation, to check that it tends to $k^2 \hat{u}(k)$ when $\epsilon\downarrow 0$.

I will make use of the fact that the coefficient $c_s$, given in Wikipedia, has the expansion $$c_{1-\epsilon}=2\epsilon+{\cal O}(\epsilon^2).$$

First rewrite the right-hand-side in a way that I can remove the principal value (see arXiv:1104.4345, lemma 3.2) $$F_\epsilon(x)\equiv c_{1-\epsilon}\operatorname{P.V.} \int_{-\infty}^\infty \frac{u(x)-u(y)}{|x-y|^{3-2\epsilon}}\,dy$$ $$\qquad=\tfrac{1}{2}c_{1-\epsilon}\int_{-\infty}^\infty \frac{2u(x)-u(x+y)-u(x-y)}{|y|^{3-2\epsilon}}\,dy, \tag{*} \label{PV}$$ and Fourier transform to obtain $$\hat{F}_\epsilon(k)\equiv\int_{-\infty}^\infty e^{ikx}F_\epsilon(x)\,dx=c_{1-\epsilon}\,\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos ky}{|y|^{3-2\epsilon}}\,dy=$$ $$\qquad=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k)\int_{-\infty}^\infty \frac{1-\cos z}{|z|^{3-2\epsilon}}\,dz=c_{1-\epsilon}\,k^{2-2\epsilon}\hat{u}(k) \; 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)$$ $$\qquad\rightarrow k^2\hat{u}(k) \;\;\text{for}\;\;\epsilon\downarrow 0,$$ since $ 2 \cos( \pi \epsilon) \Gamma (2\epsilon-2)=\frac{1}{2\epsilon}+{\cal O}(1)$.

Answer 2

Just for sake of sharing, I add my elementary idea: For any $f\in C^\infty_0(\mathbb{R})$ (for more general function it is a common practice to define via distribution $\langle (-\Delta)^sf,g\rangle:=\langle f,(-\Delta)^s g\rangle$ for all $g\in C_0^\infty(\mathbb{R})$ with some given conditions on $f$), we have

$$ \begin{align*} &(-\Delta)^sf(0)\\ &=c(s)P.V.\int_{-\infty}^\infty\frac{f(0)-f(x)}{|x|^{1+2s}}\,dx\\ &=c(s)P.V.\int_{B_\epsilon(0)}\frac{f(0)-f(x)}{|x|^{1+2s}}\,dx\\ &+c(s)\int_{B_\epsilon^c(0)}\frac{f(0)-f(x)}{|x|^{1+2s}}\,dx\\ &=c(s)P.V.\int_{B_\epsilon(0)}-f'(0)\textrm{sign}(x)|x|^{-2s}-\frac{f''(0)}{2}|x|^{1-2s}+O(|x|^{2-2s})\,dx+c(s)C_{\epsilon,f}O(1)\\ &=-f''(0)\frac{c(s)\epsilon^{2-2s}}{2-2s}+c(s)O(\epsilon^{4-2s})+c(s)C_{\epsilon,f}O(1)\\ &\xrightarrow{s\rightarrow 1} -f''(0), \end{align*} $$ because $c(s)$ has order one zero at $s=1$, i.e. $c(s)\sim 2(1-s)$, and the second and third term just vanishs in the limit of $s\rightarrow 1-$. I did not verify that $\lim_{s\rightarrow 1-}\frac{c(s)\epsilon^{2-2s}}{2-2s}=1$ with the precise definition of $c(s)$, but I guess it should be true if $c_s$ is choosen properly. So we just have the classical Laplacian as we hoped.

EDIT: For the completeness I add the interpretation for $(-\Delta)^{s}\xrightarrow{s\rightarrow 0+}Id$ in the limit $s\rightarrow 0+$ as well. In this case all terms vanish except the one multiplied with $f(0)$.

$$ \begin{align*} &(-\Delta)^sf(0)=c(s)P.V.\int_{B_\epsilon(0)}\frac{f(0)-f(x)}{|x|^{1+2s}}\,dx\\ &+c(s)\int_{B_\epsilon^c(0)}\frac{f(0)-f(x)}{|x|^{1+2s}}\,dx\\ &=c(s)C_{\epsilon,f}O(1)+f(0)\frac{c(s)\epsilon^{-2s}}{2s}\\ &\xrightarrow{s\rightarrow 0+}f(0), \end{align*} $$ because $c(s)\sim 2s$ near $s\sim 0$. Note that we have used $\int_{B_\epsilon^c(0)}\frac{f(x)}{|x|^{1+2s}}=C_{\epsilon,f}O(1)$.