Getting Acquainted with the Fractional Laplacian

Abstract

These are the handouts of an undergraduate minicourse at the Università di Bari (see Fig. 1), in the context of the 2017 INdAM Intensive Period “Contemporary Research in elliptic PDEs and related topics”. Without any intention to serve as a throughout epitome to the subject, we hope that these notes can be of some help for a very initial introduction to a fascinating field of classical and modern research.

Appendices

Appendix A: Confirmation of (2.7)

We write Δ_x to denote the Laplacian in the coordinates $x\in \mathbb {R}^n$. In this way, the total Laplacian in the variables $(x,y)\in \mathbb {R}^n\times (0,+\infty )$ can be written as

$$\displaystyle \begin{aligned}\Delta=\Delta_x+\partial^2_y.\end{aligned} $$

(A.1)

Given a (smooth and bounded, in the light of footnote 3 on page 5) $u:\mathbb {R}^n\to \mathbb {R}$, we take U := E _u be (smooth and bounded) as in (2.6).

We also consider the operator

$$\displaystyle \begin{aligned} Lu(x):=-\partial_y E_u(x,0)\end{aligned} $$

(A.2)

and we take V (x, y) := −∂ _y U(x, y). Notice that ΔV = −∂ _y ΔU = 0 in $\mathbb {R}^n\times (0,+\infty )$ and V (x, 0) = Lu(x) for any $x\in \mathbb {R}^n$. In this sense, V  is the harmonic extension of Lu and so we can write V = E _Lu and so, in the notation of (A.2), and recalling (2.6) and (A.1), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle L (Lu)(x) = -\partial_y E_{Lu}(x,0) =-\partial_y V(x,0)= \partial_y^2 U(x,0)\\ &\displaystyle &\displaystyle \qquad= \Delta U(x,0)-\Delta_x U(x,0)=-\Delta_x U(x,0)=-\Delta u(x). \end{array} \end{aligned} $$

This gives that L ² = − Δ, which is consistent with L = (− Δ)^1∕2, thanks to (2.5).

Appendix B: Proof of (2.10)

Let $u\in {\mathcal {S}}$. By (2.9), we can write

$$\displaystyle \begin{aligned} \sup _{x\in \mathbb{R}^{n}}(1+|x|{}^{n})\left|u(x)\right| + \sup _{x\in \mathbb{R}^{n}}(1+|x|{}^{n+2 })\left|D^{2}u(x)\right|\leqslant\,{\mathrm{const}}\,.\end{aligned} $$

(B.1)

Fixed $x\in \mathbb {R}^n$ (with |x| to be taken large), recalling the notation in (2.3), we consider the map y↦δ _u(x, y) and we observe that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \delta_u(x,0)=0,\\ &\displaystyle &\displaystyle \nabla_y \delta_u(x,y)=\nabla u(x+y)-\nabla u(x-y),\\{\mbox{and }} &\displaystyle &\displaystyle D^2_y \delta_u(x,y)=D^2 u(x+y)+D^2 u(x-y).\end{array} \end{aligned} $$

Hence, if $|Y|\leqslant |x|/2$ we have that $|x\pm Y|\geqslant |x|-|Y|\geqslant |x|/2$, and thus

$$\displaystyle \begin{aligned}\big| D^2_y \delta_u(x,Y)\big|\leqslant 2\sup_{|\zeta|\geqslant|x|/2} \big|D^2 u(\zeta)\big| \leqslant 2\sup_{|\zeta|\geqslant|x|/2}\frac{(2|\zeta|)^{n+2}\big|D^2 u(\zeta)\big|}{|x|{}^{n+2}} \leqslant \frac{\,{\mathrm{const}}\,}{|x|{}^{n+2}},\end{aligned}$$

thanks to (B.1).

Therefore, a second order Taylor expansion of δ _u in the variable y gives that, if $|y|\leqslant |x|/2$,

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle \big| \delta_u(x,y)\big| \leqslant\sup_{|Y|\leqslant|x|/2} \left| \delta_u(x,0)+\nabla \delta_u(x,0)\cdot y+\frac{D^2 \delta_u(x,Y)\,y\cdot y}{2}\right|\\ &\displaystyle &\displaystyle \qquad=\sup_{|Y|\leqslant|x|/2} \left| \frac{D^2 \delta_u(x,Y)\,y\cdot y}{2}\right|\leqslant\frac{\,{\mathrm{const}}\,\,|y|{}^2}{|x|{}^{n+2}}. \end{array} \end{aligned} $$

Consequently,

$$\displaystyle \begin{aligned} \left| \int_{B_{|x|/2}} \frac{\delta_u(x,y)}{|y|{}^{n+2s}}\,dy\right|\leqslant \frac{\,{\mathrm{const}}\,}{|x|{}^{n+2}}\int_{B_{|x|/2}}\frac{|y|{}^2}{|y|{}^{n+2s}}\,dy \leqslant\frac{\,{\mathrm{const}}\,\,|x|{}^{2-2s}}{|x|{}^{n+2}}=\frac{\,{\mathrm{const}}\,}{|x|{}^{n+2s}}.\end{aligned} $$

(B.2)

Moreover, by (B.1),

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left| \int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{\delta_u(x,y)}{|y|{}^{n+2s}}\,dy\right|\\ &\displaystyle \leqslant&\displaystyle \int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{|u(x+y)|}{|y|{}^{n+2s}}\,dy +\int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{|u(x-y)|}{|y|{}^{n+2s}}\,dy +2\int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{|u(x)|}{|y|{}^{n+2s}}\,dy\\ &\displaystyle \leqslant&\displaystyle \int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{|u(x+y)|}{(|x|/2)^{n+2s}}\,dy +\int_{\mathbb{R}^n\setminus B_{|x|/2}} \frac{|u(x-y)|}{(|x|/2)^{n+2s}}\,dy +\frac{\,{\mathrm{const}}\,\,|u(x)|}{|x|{}^{2s}}\\ &\displaystyle \leqslant&\displaystyle \frac{\,{\mathrm{const}}\,}{|x|{}^{n+2s}} \int_{\mathbb{R}^n} |u(\zeta)|\,d\zeta+ \frac{\,{\mathrm{const}}\,\,|u(x)|}{|x|{}^{2s}}\\ &\displaystyle \leqslant&\displaystyle \frac{\,{\mathrm{const}}\,}{|x|{}^{n+2s}}. \end{array} \end{aligned} $$

This and (B.2), recalling (2.3), establish (2.10).

Appendix C: Proof of (2.14)

Let $M:=\frac 1{2n}\,\left (1+\sup _{B_1}|f|\right )$ and v(x) := M(1 −|x|²) − u(x). Notice that v = 0 along ∂B ₁ and

$$\displaystyle \begin{aligned}\Delta v = -2n M -\Delta u \leqslant -M-f\leqslant -M+\sup_{B_1}|f|\leqslant0\end{aligned}$$

in B ₁. Consequently, v⩾0 in B ₁, which gives that $u(x)\leqslant M(1-|x|{ }^2)$.

Arguing similarly, by looking at $\tilde v(x):= M(1-|x|{ }^2)+u(x)$, one sees that $-u(x) \leqslant M(1-|x|{ }^2)$. Accordingly, we have that

$$\displaystyle \begin{aligned}|u(x)|\leqslant M(1-|x|{}^2)\leqslant M(1+|x|)(1-|x|)\leqslant 2M(1-|x|).\end{aligned}$$

This proves (2.14).

Appendix D: Proof of (2.17)

The idea of the proof is described in Fig. 7. The trace of the function in Fig. 7 is exactly the function u _1∕2 in (2.16). The function plotted in Fig. 7 is the harmonic extension of u _1∕2 in the halfplane (like an elastic membrane pinned at the halfcircumference along the trace). Our objective is to show that the normal derivative of such extended function along the trace is constant, and so we can make use of the extension method in (2.6) and (2.7) to obtain (2.17).

In further detail, we use complex coordinates, identifying $(x,y)\in \mathbb {R}\times (0,+\infty )$ with $z:=x+iy\in \mathbb {C}$ with ℑ(z) > 0. Also, as customary, we define the principal square root in the cut complex plane

$$\displaystyle \begin{aligned}\mathbb{C}_\star:=\{ z=re^{{i\varphi }}{\text{ with }}r>0 {\mbox{ and }} -\pi <\varphi < \pi \}\end{aligned}$$

by defining, for any $z=re^{{i\varphi }}\in \mathbb {C}_\star$,

$$\displaystyle \begin{aligned} \displaystyle{\surd}({z}) := \sqrt{r} \, e^{i \varphi / 2},\end{aligned} $$

(D.1)

see Fig. 8 (for typographical convenience, we distinguish between the complex and the real square root, by using the symbols √(⋅) and $\sqrt {\cdot }$ respectively).

Fig. 8

Real and imaginary part of the complex principal square root

The principal square root function is defined using the nonpositive real axis as a “branch cut” and

$$\displaystyle \begin{aligned} \big( \displaystyle{\surd}({z}) \big)^2= r \, e^{i \varphi }=z. \end{aligned} $$

(D.2)

Moreover,

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} &\displaystyle {\mbox{the function}~\displaystyle{\surd}\mbox{ is holomorphic in}~\mathbb{C}_\star} \end{array} \end{aligned} $$

(D.3)

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} &\displaystyle {\mbox{and }}\;\partial_z\displaystyle{\surd}({z})=\frac 1{2\displaystyle{\surd}(z)}. \end{array} \end{aligned} $$

(D.4)

To check these facts, we take $z\in \mathbb {C}_\star$: since $\mathbb {C}_\star$ is open, we have that $z+w\in \mathbb {C}_\star$ for any $w\in \mathbb {C}\setminus \{0\}$ with small module. Consequently, by (D.2), we obtain that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle w=(z+w)-z=\big( \displaystyle{\surd}({z+w}) \big)^2-\big( \displaystyle{\surd}({z}) \big)^2\\ &\displaystyle &\displaystyle \qquad= \Big( \displaystyle{\surd}({z+w})+\displaystyle{\surd}({z})\Big) \Big( \displaystyle{\surd}({z+w})-\displaystyle{\surd}({z})\Big). \end{array} \end{aligned} $$

Dividing by w and taking the limit, we thus find that

$$\displaystyle \begin{aligned} \begin{aligned} 1\;&=\lim_{w\to0}\Big( \displaystyle{\surd}({z+w})+\displaystyle{\surd}({z})\Big)\,\frac{ \displaystyle{\surd}({z+w})-\displaystyle{\surd}({z})}{w}\\ &= 2\displaystyle{\surd}({z})\;\lim_{w\to0}\frac{ \displaystyle{\surd}({z+w})-\displaystyle{\surd}({z})}{w} \end{aligned}\end{aligned} $$

(D.5)

Since $\mathbb {C}_\star \subseteq \mathbb {C}\setminus \{0\}$, we have that z≠0, and thus √(z)≠0. As a result, we can divide (D.5) by 2√(z) and conclude that

$$\displaystyle \begin{aligned}\lim_{w\to0}\frac{ \displaystyle{\surd}({z+w})-\displaystyle{\surd}({z})}{w}=\frac 1{ 2\displaystyle{\surd}({z}) },\end{aligned}$$

which establishes, at the same time, both (D.3) and (D.4), as desired.

We also remark that

$$\displaystyle \begin{aligned} {\mbox{if}~z\in\mathbb{C}\mbox{ with}~\Im (z)>0,\mbox{ then}~1-z^2\in\mathbb{C}_\star.} \end{aligned} $$

(D.6)

To check this, if z = x + iy with y > 0, we observe that

$$\displaystyle \begin{aligned}1-z^2 =1-(x+iy)^2 = 1-x^2+y^2-2ixy.\end{aligned} $$

(D.7)

Hence, if 1 − z ² lies on the real axis, we have that xy = 0, and so x = 0. Then, the real part of 1 − z ² in this case is equal to 1 + y ² which is strictly positive. This proves (D.6).

Thanks to (D.6), for any $z\in \mathbb {C}$ with ℑ(z) > 0 we can define the function √(1 − z ²). From (D.7), we can write

$$\displaystyle \begin{aligned} \begin{aligned} & 1-z^2 = r(x,y)\, e^{{i\varphi(x,y) }},\\{\mbox{where }}\;& r(x,y)=\big( (1-x^2+y^2)^2+4x^2y^2 \big)^{1/2},\\& r(x,y)\,\cos\varphi(x,y)=1-x^2+y^2\\ {\mbox{and }}&r(x,y)\,\sin\varphi(x,y)=2xy. \end{aligned}\end{aligned}$$

Notice that

As a consequence,

This says that, if x ² > 1 then

while if x ² < 1 then

On this account, we deduce that

(D.8)

and therefore, recalling (D.1),

(D.9)

This implies that

(D.10)

Now we define

$$\displaystyle \begin{aligned}z=x+iy\mapsto \Re\Big( \displaystyle{\surd}(1-z^2)+i z\Big)=:U_{1/2}(x,y).\end{aligned}$$

The function U _1∕2 is the harmonic extension of u _1∕2 in the halfplane, as plotted in Fig. 7. Indeed, from (D.10),

Furthermore, from (D.3), we have that U _1∕2 is the real part of a holomorphic function in the halfplane and so it is harmonic.

These considerations give that U _1∕2 solves the harmonic extension problem in (2.6), hence, in the light of (2.7),

(D.11)

Now, recalling (D.4), we see that, for any x ∈ (−1, 1) and small y > 0,

$$\displaystyle \begin{aligned}\partial_y\displaystyle{\surd}(1-z^2)=\partial_z\displaystyle{\surd}(1-z^2)\;\partial_y z= \frac 1{2\displaystyle{\surd}(1-z^2)}\;\partial_z(1-z^2)\;\partial_y (x+iy)= -\frac{iz}{\displaystyle{\surd}(1-z^2)}. \end{aligned} $$

(D.12)

We stress that the latter denominator does not vanish when x ∈ (−1, 1) and y > 0 is small. So, using that $\Re (ZW)=\Re Z\Re W-\Im Z\Im W$ for any Z, $W\in \mathbb {C}$, we obtain that

$$\displaystyle \begin{aligned} \begin{aligned} & y= \Re\big(-i(x+iy)\big)= \Re(-iz)= \Re\Big(\displaystyle{\surd}(1-z^2)\;\partial_y\displaystyle{\surd}(1-z^2)\Big)\\ &\qquad= \Re\Big(\displaystyle{\surd}(1-z^2)\Big)\;\Re\Big(\partial_y\displaystyle{\surd}(1-z^2)\Big) - \Im\Big(\displaystyle{\surd}(1-z^2)\Big)\;\Im\Big(\partial_y\displaystyle{\surd}(1-z^2)\Big) .\end{aligned}\end{aligned} $$

(D.13)

From (D.9), for any x ∈ (−1, 1) we have that

This and the fact that ∂ _y √(1 − z ²) is bounded (in view of (D.12)) give that, for any x ∈ (−1, 1),

This, (D.9) and (D.13) imply that, for any x ∈ (−1, 1),

and therefore

(D.14)

Plugging this information into (D.11), we conclude the proof of (2.17), as desired.

Appendix E: Deducing (2.19) from (2.15) Using a Space Inversion

From (2.15), up to a translation, we know that

$$\displaystyle \begin{aligned} {\mbox{the function}~\mathbb{R}\ni x\mapsto v_s(x):=(x-1)_+^s\mbox{ is }s\mbox{-harmonic in}~(1,+\infty).} \end{aligned} $$

(E.1)

We let w _s be the space inversion of v _s induced by the Kelvin transform in the fractional setting, namely

$$\displaystyle \begin{aligned}w_s(x):=|x|{}^{2s-1}\, v_s\left(\frac{x}{|x|{}^2}\right) =|x|{}^{2s-1}\,\left( \frac{x}{|x|{}^2}-1\right)_+^s= \left\{\begin{matrix} x^{s-1}(1-x)^s & {\mbox{ if }}x\in(0,1),\\ 0 & {\mbox{ otherwise}}. \end{matrix} \right. .\end{aligned}$$

By (E.1), see Corollary 2.3 in [63], it follows that w _s(x) is s-harmonic in (0, 1). Consequently, the function

$$\displaystyle \begin{aligned}w^\star_s(x):=w_s(1-x) =\left\{\begin{matrix} x^{s}(1-x)^{s-1} & {\mbox{ if }}x\in(0,1),\\ 0 & {\mbox{ otherwise}}. \end{matrix} \right. \end{aligned}$$

is also s-harmonic in (0, 1). We thereby conclude that the function

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle W^\star_s(x):= w_s(x)-w^\star_s(x) =\left\{\begin{matrix} x^{s-1}(1-x)^s-x^{s}(1-x)^{s-1} &\displaystyle {\mbox{ if }}x\in(0,1), \\0 &\displaystyle {\mbox{ otherwise}}. \end{matrix} \right. \end{array} \end{aligned} $$

is also s-harmonic in (0, 1). See Fig. 9 for a picture of w _s and $W^\star _s$ when s = 1∕2. Let now

$$\displaystyle \begin{aligned}U_s(x):=x_+^s(1-x)_+^s =\left\{\begin{matrix} x^{s}(1-x)^{s} & {\mbox{ if }}x\in(0,1), \\0 & {\mbox{ otherwise}}. \end{matrix} \right.\end{aligned}$$

and notice that U _s is the primitive of $s W^\star _s$. Since the latter function is s-harmonic in (0, 1), after an integration we thereby deduce that (− Δ)^s U _s =  const in (0, 1). This and the fact that

$$\displaystyle \begin{aligned}U_s\left(\frac{x+1}2\right)=2^{-s}\,u_s(x)\end{aligned}$$

imply (2.19).

Fig. 9

The functions w _1∕2 and $W^\star _{1/2}$

Fig. 10

Harmonic extension in the halfplane of the function $\mathbb {R}\ni x\mapsto (1-x^2)^{-1/2}_+$

Appendix F: Proof of (2.21)

Fixed $y\in \mathbb {R}^n\setminus \{0\}$ we let ${\mathcal {R}}^y$ be a rotation which sends $\frac {y}{|y|}$ into the vector e ₁ = (1, 0, …, 0), that is

$$\displaystyle \begin{aligned} \sum_{k=1}^n {\mathcal{R}}_{ik}^y\, y_k =|y|\delta_{i1}, \end{aligned} $$

(F.1)

for any i ∈{1, …, n}. We also denote by

$$\displaystyle \begin{aligned}K(y):=\frac{y}{|y|{}^2}\end{aligned}$$

the so-called Kelvin Transform. We recall that for any i, j ∈{1, …, n},

$$\displaystyle \begin{aligned}\partial_{y_i} K_j(y)=\frac{\delta_{ij}}{|y|{}^2}-\frac{2y_iy_j}{|y|{}^4}\end{aligned}$$

and so, by (F.1),

$$\displaystyle \begin{aligned}\Big( {\mathcal{R}}^y\;(DK(y))\;({\mathcal{R}}^y)^{-1}\Big)_{ij}= \sum_{k,h=1}^n {\mathcal{R}}^y_{ik} \,\partial_{y_k} K_h(y)\, {\mathcal{R}}^y_{jh}=\frac{\delta_{ij}}{|y|{}^2} -\frac{2\delta_{i1}\delta_{j1}}{|y|{}^2}.\end{aligned}$$

This says that ${\mathcal {R}}^y\;(DK(y))\;({\mathcal {R}}^y)^{-1}$ is a diagonal⁶ matrix, with first entry equal to $-\frac {1}{|y|{ }^2}$ and the others equal to $\frac {1}{|y|{ }^2}$.

As a result,

$$\displaystyle \begin{aligned} \big|\det(DK(y))\big|=\Big| \det\big( {\mathcal{R}}^y\;(DK(y))\;({\mathcal{R}}^y)^{-1}\big)\Big|=\frac{1}{|y|{}^{2n}}. \end{aligned} $$

(F.2)

The Kelvin Transform is also useful to write the Green function of the ball B ₁, see e.g. formula (41) on p. 40 and Theorem 13 on p. 35 of [62]. Namely, we take n⩾3 for simplicity, and we write

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle G(x,y):= \,{\mathrm{const}}\,\left( \frac{1}{|y-x|{}^{n-2}}-\frac{1}{\big|\,|x|(y-K(x))\big|{}^{n-2}}\right)\\ &\displaystyle &\displaystyle \qquad\qquad= \,{\mathrm{const}}\,\left( \frac{1}{|x-y|{}^{n-2}}-\frac{1}{\big|\,|y|(x-K(y))\big|{}^{n-2}}\right)=G(y,x) \end{array} \end{aligned} $$

and, for a suitable choice of the constant, for any x ∈ B ₁ we can write the solution of (2.20) in the form

$$\displaystyle \begin{aligned}u(x)=\int_{B_1}f(y)\,G(x,y)\,dy.\end{aligned}$$

see e.g. page 35 in [62].

On this account, we have that, for any x ∈ B ₁,

$$\displaystyle \begin{aligned} \begin{array}{rcl} |\nabla u(x)|&\displaystyle \leqslant&\displaystyle \int_{B_1}|f(y)|\,|\partial_x G(x,y)|\,dy \\ &\displaystyle \leqslant&\displaystyle \,{\mathrm{const}}\,\sup_{B_1}|f| \int_{B_1}\left( \frac{1}{|x-y|{}^{n-1}}+\frac{1}{|y|{}^{n-2}\big|x-K(y)\big|{}^{n-1}} \right)\,dy\\ &\displaystyle \leqslant&\displaystyle \,{\mathrm{const}}\,\sup_{B_1}|f| \left( \int_{B_2} \frac{d\zeta}{|\zeta|{}^{n-1}} + \int_{\mathbb{R}^n\setminus B_1} \frac{d\eta}{|\eta|{}^{n+2}\big|x-\eta\big|{}^{n-1}} \right) \\&\displaystyle \leqslant&\displaystyle \,{\mathrm{const}}\,\sup_{B_1}|f| \left( 1+ \int_{B_2\setminus B_1} \frac{d\eta}{|x-\eta|{}^{n-1}} + \int_{\mathbb{R}^n\setminus B_2} \frac{d\eta}{|\eta|{}^{n+2}} \right) \\&\displaystyle \leqslant&\displaystyle \,{\mathrm{const}}\,\sup_{B_1}|f|. \end{array} \end{aligned} $$

Notice that here we have used the transformations ζ := x − y and η := K(y), exploiting also (F.2). The claim in (2.21) is thus established.

Appendix G: Proof of (2.24) and Probabilistic Insights

We give a proof of (2.24) by taking a derivative of (2.17). To this aim, we claim⁷ that

$$\displaystyle \begin{aligned} \begin{aligned} & \frac{d}{dx} \int_{\mathbb{R}} \frac{u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x)}{|y|{}^{2}}\,dy\\ =\;& -\int_{\mathbb{R}} \frac{(x+y) u_{-1/2}(x+y)+(x-y) u_{-1/2}(x-y)-2x u_{-1/2}(x)}{|y|{}^{2}} \,dy.\end{aligned} \end{aligned} $$

(G.1)

To this end, we fix x ∈ (−1, 1) and $h\in \mathbb {R}$. We define

$$\displaystyle \begin{aligned}\ell_x:=\min\{ |x-1|,\,|x+1|\}>0.\end{aligned}$$

In the sequel, we will take |h| as small as we wish in order to compute incremental quotients, hence we can assume that

$$\displaystyle \begin{aligned} |h|<\frac{\ell_x}4.\end{aligned} $$

(G.2)

We also define

$$\displaystyle \begin{aligned} I_x(h) := \Big\{ y\in\mathbb{R} {\mbox{ s.t. }} \min\{ |(x+y)-1|,\,|(x-y)-1|,\, |(x+y)+1|,\,|(x-y)+1| \}\leqslant 2|h| \Big\}. \end{aligned} $$

(G.3)

Since I _x(h) ⊆ (x−1−2|h|, x−1+2|h|)∪(x+1−2|h|, x+1+2|h|)∪(1−x−2|h|, 1−x+2|h|)∪(−1−x−2|h|, −1−x+2|h|), we have that

$$\displaystyle \begin{aligned} {\mbox{the measure of }I_x\mbox{ is less than }\,{\mathrm{const}}\,|h|.} \end{aligned} $$

(G.4)

Furthermore,

$$\displaystyle \begin{aligned} I_x(h) \subseteq \left\{ y\in\mathbb{R} {\mbox{ s.t. }} |y|\geqslant \frac{\ell_x}2 \right\}. \end{aligned} $$

(G.5)

To check this, let y ∈ I _x(h). Then, by (G.3), there exist σ _1,x,y, σ _2,x,y ∈{−1, 1} such that

$$\displaystyle \begin{aligned}|x+\sigma_{1,x,y} y +\sigma_{2,x,y}|\leqslant2|h|\end{aligned}$$

and therefore

$$\displaystyle \begin{aligned}|y|=|\sigma_{1,x,y} y| \geqslant |x+\sigma_{2,x,y}| - |x+\sigma_{1,x,y} y+\sigma_{2,x,y}|\geqslant\ell_x-2|h|\geqslant\frac{\ell_x}2,\end{aligned}$$

where the last inequality is a consequence of (G.2), and this establishes (G.5).

Now, we introduce the following notation for the incremental quotient

$$\displaystyle \begin{aligned}Q_h(x,y):=\frac{\substack{\Big(\big( u_{1/2}(x+y+h)+u_{1/2}(x-y+h)-2u_{1/2}(x+h)\big)\\ - \big(u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x)\big)\Big)} }{h}\end{aligned}$$

and we observe that, since u _1∕2 is globally Hölder continuous with exponent 1∕2, it holds that

$$\displaystyle \begin{aligned} \begin{aligned} |Q_h(x,y)|\;&\leqslant \frac{\substack{\Big(\big| u_{1/2}(x+y+h)-u_{1/2}(x+y)\big|+\big| u_{1/2}(x-y+h)-u_{1/2}(x-y)\big|\\ +2\,\big| u_{1/2}(x+h)-u_{1/2}(x)\big|\Big)}}{|h|}\\&\leqslant\frac{\,{\mathrm{const}}\, |h|{}^{1/2}}{|h|}\\&=\frac{\,{\mathrm{const}}\,}{|h|{}^{1/2}}, \end{aligned}\end{aligned}$$

for any x, $y\in \mathbb {R}$. Consequently, recalling (G.4) and (G.5), we conclude that

$$\displaystyle \begin{aligned} \lim_{h\to0} \left| \int_{I_x(h)} \frac{Q_h(x,y)}{|y|{}^{2}}\,dy\right| \leqslant \lim_{h\to0} \int_{I_x(h)} \frac{\,{\mathrm{const}}\,}{|h|{}^{1/2}\,\ell_x^{2}}\,dy\leqslant \lim_{h\to0} \frac{\,{\mathrm{const}}\, |h|}{|h|{}^{1/2}\,\ell_x^{2}}=0. \end{aligned} $$

(G.6)

Now we take derivatives of u _1∕2. For this, we observe that, for any ξ ∈ (−1, 1),

$$\displaystyle \begin{aligned}u_{1/2}^{\prime}(\xi)=-\xi(1-\xi^2)^{-1/2}=-\xi u_{-1/2}(\xi).\end{aligned}$$

Since the values outside (−1, 1) are trivial, this implies that

$$\displaystyle \begin{aligned} u_{1/2}^{\prime}(\xi)=-\xi u_{-1/2}(\xi)\quad{\mbox{ for any }} \xi\in\mathbb{R}\setminus\{-1,1\}. \end{aligned} $$

(G.7)

Now, by (G.3), we know that if $y\in \mathbb {R}\setminus I_x(h)$ we have that $x+y+t\in \mathbb {R}\setminus \{-1,1\}$ for all $t\in \mathbb {R}$ with |t| < |h| and therefore we can exploit (G.7) and find that

$$\displaystyle \begin{aligned} \lim_{h\to0} \frac{u_{1/2}(x+y+h)-u_{1/2}(x+y)}{h}= -(x+y) u_{-1/2}(x+y).\end{aligned}$$

Similar arguments show that, for any $y\in \mathbb {R}\setminus I_x(h)$,

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \lim_{h\to0} \frac{u_{1/2}(x-y+h)-u_{1/2}(x-y)}{h}= -(x-y) u_{-1/2}(x-y)\\ {\mbox{and }}&\displaystyle &\displaystyle \lim_{h\to0} \frac{u_{1/2}(x+h)-u_{1/2}(x)}{h}= -x u_{-1/2}(x). \end{array} \end{aligned} $$

Consequently, for any $y\in \mathbb {R}\setminus I_x(h)$,

$$\displaystyle \begin{aligned} \lim_{h\to0} \frac{Q_h(x,y)}{|y|{}^{2}}= -\frac{(x+y) u_{-1/2}(x+y)+(x-y) u_{-1/2}(x-y)-2x u_{-1/2}(x)}{|y|{}^{2}}. \end{aligned} $$

(G.8)

Now we set

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Xi_h(x,y)&\displaystyle :=&\displaystyle \frac{Q_h(x,y)\,\chi_{\mathbb{R}\setminus I_x(h)}(y)}{|y|{}^{2}}\\&\displaystyle =&\displaystyle \frac{1}{h\,|y|{}^{2}}\,\Big( \big( u_{1/2}(x+y+h)+u_{1/2}(x-y+h)-2u_{1/2}(x+h)\big)\\&\displaystyle &\displaystyle \qquad- \big(u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x)\big)\Big)\,\,\chi_{\mathbb{R}\setminus I_x(h)}(y) \end{array} \end{aligned} $$

and we claim that

$$\displaystyle \begin{aligned} |\Xi_h(x,y)|\leqslant C_x\,\left[\chi_{(-3,3)}(y)\,\left( \frac{1}{|1-(x+y)^2|{}^{1/2}}+\frac{1}{|1-(x-y)^2|{}^{1/2}}\right)+\frac{ \chi_{\mathbb{R}\setminus (-3,3)}(y)}{|y|{}^2} \right], \end{aligned} $$

(G.9)

for a suitable C _x > 0, possibly depending on x. For this, we first observe that if |y|⩾3 then |x ± y|⩾1 and also |x ± y + h|⩾1. This implies that if |y|⩾3, then u _1∕2(x ± y) = u _1∕2(x ± y + h) = 0 and therefore

$$\displaystyle \begin{aligned}\Xi_h(x,y)= \frac{1}{h\,|y|{}^{2}}\,\big( 2u_{1/2}(x)-2u_{1/2}(x+h)\big).\end{aligned}$$

This and the fact that u _1∕2 is smooth in the vicinity of the fixed x ∈ (−1, 1) imply that (G.9) holds true when |y|⩾3. Therefore, from now on, to prove (G.9) we can suppose that

$$\displaystyle \begin{aligned} |y|<3. \end{aligned} $$

(G.10)

We will also distinguish two regimes, the one in which $|y|\leqslant \frac {\ell _x}4$ and the one in which $|y|>\frac {\ell _x}4$.

If $|y|\leqslant \frac {\ell _x}4$ and $|t|\leqslant h$, we have that

$$\displaystyle \begin{aligned}|(x+y+t)+1|\geqslant |x+1|-|y|-|t|\geqslant\ell_x-|y|-|h|\geqslant \frac{\ell_x}2,\end{aligned}$$

due to (G.2), and similarly $|(x-y+t)-1|\geqslant \frac {\ell _x}2$. This implies that

$$\displaystyle \begin{aligned}|u_{1/2}(x+y+t)+u_{1/2}(x-y+t)-2u_{1/2}(x+t)|\leqslant C_x\,|y|{}^2,\end{aligned}$$

for some C _x > 0 that depends on ℓ _x. Consequently, we find that if $|y|\leqslant \frac {\ell _x}4$ then

$$\displaystyle \begin{aligned} |\Xi_h(x,y)|\leqslant\frac{\,{\mathrm{const}}\,\,C_x\,|y|{}^2}{|y|{}^{2}}=\,{\mathrm{const}}\, C_x.\end{aligned} $$

(G.11)

Conversely, if $y\in \mathbb {R}\setminus I_x(h)$, with $|y|>\frac {\ell _x}4$, then we make use of (G.7) and (G.10) to see that

$$\displaystyle \begin{aligned} \begin{aligned} & |u_{1/2}(x+y+h)-u_{1/2}(x+y)|\leqslant \int_0^{|h|} |u_{1/2}^{\prime}(x+y+\tau)|\,d\tau\\ &\qquad= \int_0^{|h|} |x+y+\tau|\, |u_{-1/2}(x+y+\tau)|\,d\tau \leqslant 5\,\int_0^{|h|} |u_{-1/2}(x+y+\tau)|\,d\tau\\ &\qquad\leqslant5\,\int_0^{|h|} \frac{d\tau}{|1-(x+y+\tau)^2|{}^{1/2}}. \end{aligned}\end{aligned} $$

(G.12)

Also, if $y\in \mathbb {R}\setminus I_x(h)$ we deduce from (G.3) that $|1\pm (x+y)|\geqslant 2 |h|$ and therefore, if $|\tau |\leqslant |h|$, then

$$\displaystyle \begin{aligned}|1\pm(x+y+\tau)|\geqslant |1\pm (x+y)|-|\tau|\geqslant |1\pm (x+y)|-|h|\geqslant\frac{|1\pm (x+y)|}{2}.\end{aligned}$$

Therefore

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle |1-(x+y+\tau)^2|= |1+(x+y+\tau)|\,|1-(x+y+\tau)|\\&\displaystyle &\displaystyle \qquad\geqslant \frac{1}{4}\, |1+(x+y)|\,|1-(x+y)|=\frac{1}{4}\,|1-(x+y)^2|.\end{array} \end{aligned} $$

Hence, we insert this information into (G.12) and we conclude that

$$\displaystyle \begin{aligned} |u_{1/2}(x+y+h)-u_{1/2}(x+y)|\leqslant \,{\mathrm{const}}\, \int_0^{|h|} \frac{d\tau}{|1-(x+y)^2|{}^{1/2}}=\frac{\,{\mathrm{const}}\, |h|}{|1-(x+y)^2|{}^{1/2}} . \end{aligned} $$

(G.13)

Similarly, one sees that

$$\displaystyle \begin{aligned} |u_{1/2}(x-y+h)-u_{1/2}(x-y)|\leqslant \frac{\,{\mathrm{const}}\, |h|}{|1-(x-y)^2|{}^{1/2}} . \end{aligned} $$

(G.14)

In view of (G.13) and (G.14), we get that, for any $y\in \mathbb {R}\setminus I_x(h)$ with $|y|>\frac {\ell _x}4$,

$$\displaystyle \begin{aligned} \begin{array}{rcl} |\Xi_h(x,y)|&\displaystyle \leqslant&\displaystyle \frac{1}{h\,|y|{}^2}\, \left( \,{\mathrm{const}}\,|h|+\frac{\,{\mathrm{const}}\, |h|}{|1-(x+y)^2|{}^{1/2}}+ \frac{\,{\mathrm{const}}\, |h|}{|1-(x-y)^2|{}^{1/2}}\right)\\&\displaystyle \leqslant&\displaystyle \frac{\,{\mathrm{const}}\,}{\ell_x^2}\, \left( 1+\frac{1}{|1-(x+y)^2|{}^{1/2}}+ \frac{1}{|1-(x-y)^2|{}^{1/2}}\right).\end{array} \end{aligned} $$

Combining this with (G.11), we obtain (G.9), up to renaming constants.

Now, we point out that the right hand side of (G.9) belongs to $L^1(\mathbb {R})$. Accordingly, using (G.9) and the Dominated Convergence Theorem, and recalling also (G.7), it follows that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \lim_{h\to0} \int_{\mathbb{R}\setminus I_x(h)} \frac{1}{h|y|{}^{2}}\,\Big( \big( u_{1/2}(x+y+h)+u_{1/2}(x-y+h)-2u_{1/2}(x+h) \big)\\ &\displaystyle &\displaystyle \qquad- \big( u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x) \big)\Big) \,dy \\ &\displaystyle &\displaystyle \qquad\qquad= \lim_{h\to0} \int_{\mathbb{R}} \Xi_h(x,y)\,dy\\ &\displaystyle &\displaystyle \qquad\qquad=\int_{\mathbb{R}} \lim_{h\to0} \Xi_h(x,y)\,dy=\int_{\mathbb{R}} \frac{ u_{1/2}^{\prime}(x+y)+u_{1/2}^{\prime}(x-y)-2u_{1/2}^{\prime}(x) }{|y|{}^{2}}\,dy\\ &\displaystyle &\displaystyle \qquad\qquad=-\int_{\mathbb{R}} \frac{(x+y) u_{-1/2}(x+y)+(x-y) u_{-1/2}(x-y)-2x u_{-1/2}(x)}{|y|{}^{2}} \,dy, \end{array} \end{aligned} $$

where the last identity is a consequence of (G.8).

From this and (G.6), the claim in (G.1) follows, as desired.

Now, we rewrite (G.1) as

$$\displaystyle \begin{aligned} \begin{aligned} & \frac{d}{dx} \int_{\mathbb{R}} \frac{u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x)}{|y|{}^{2}}\,dy\\ =\;& -{\mathcal{J}}(x)-x\int_{\mathbb{R}} \frac{ u_{-1/2}(x+y)+ u_{-1/2}(x-y)-2 u_{-1/2}(x)}{|y|{}^{2}} \,dy\\ &\qquad{\mbox{ where }} {\mathcal{J}}(x):= \int_{\mathbb{R}} \frac{y \,\big(u_{-1/2}(x+y)- u_{-1/2}(x-y)\big)}{|y|{}^{2}} \,dy\\ &\qquad \qquad \qquad \quad \ = \int_{\mathbb{R}} \frac{u_{-1/2}(x+y)- u_{-1/2}(x-y)}{y} \,dy. \end{aligned} \end{aligned} $$

(G.15)

We claim that

$$\displaystyle \begin{aligned} {\mathcal{J}}(x)=0. \end{aligned} $$

(G.16)

This follows plainly for x = 0, since u _−1∕2 is even. Hence, from here on, to prove (G.16) we assume without loss of generality that x ∈ (0, 1). Moreover, by changing variable y↦ − y, we see that

$$\displaystyle \begin{aligned}-\,{\mathrm{P.V.}}\,\int_{\mathbb{R}} \frac{ u_{-1/2}(x-y)}{y} \,dy=\,{\mathrm{P.V.}}\,\int_{\mathbb{R}} \frac{ u_{-1/2}(x+y)}{y} \,dy\end{aligned}$$

and therefore

$$\displaystyle \begin{aligned} \begin{aligned} & {\mathcal{J}}(x)= 2\;\,{\mathrm{P.V.}}\,\int_{\mathbb{R}}\frac{u_{-1/2}(x+y)}{y}\;dy\ =\ 2\;\,{\mathrm{P.V.}}\,\int_{-1-x}^{1-x}\frac{dy}{y\sqrt{1-(x+y)^2}}\\ &\qquad\qquad=\ 2\;\,{\mathrm{P.V.}}\,\int_{-1}^1\frac{dz}{(z-x)\sqrt{1-z^2}}. \end{aligned}\end{aligned} $$

(G.17)

Now, we apply the change of variable

$$\displaystyle \begin{aligned} \xi:=\frac{1-\sqrt{1-z^2}}{z},\qquad{\mbox{hence }} z=\frac{2\xi}{1+\xi^2}. \end{aligned}$$

We observe that when z ranges in (−1, 1), then ξ ranges therein as well. Moreover,

$$\displaystyle \begin{aligned} \sqrt{1-z^2}=1-\xi z=\frac{1-\xi^2}{1+\xi^2} ,\end{aligned}$$

thus, by (G.17),

$$\displaystyle \begin{aligned} & {\mathcal{J}}(x) =2\;\,{\mathrm{P.V.}}\,\int_{-1}^1\frac 1{(\frac{2\xi}{1+\xi^2}-x)\frac{1-\xi^2}{1+\xi^2}} \cdot\frac{2-2\xi^2}{(1+\xi^2)^2}\;d\xi \\ &\qquad=4\;\,{\mathrm{P.V.}}\,\int_{-1}^1\frac{d\xi}{2\xi-x(1+\xi^2)} =4x\;\,{\mathrm{P.V.}}\,\int_{-1}^1\frac{d\xi}{1-x^2-(1-x\xi)^2}. \end{aligned} $$

We now apply another change of variable

$$\displaystyle \begin{aligned} \eta:=\frac{1-x\xi}{\sqrt{1-x^2}} \end{aligned}$$

which gives

$$\displaystyle \begin{aligned} {\mathcal{J}}(x)= \frac{4}{\sqrt{1-x^2}}\;\,{\mathrm{P.V.}}\,\int_{a_-}^{a^+}\frac{d\eta}{1-\eta^2}, \end{aligned} $$

(G.18)

where

$$\displaystyle \begin{aligned} a_+:=\sqrt{\frac{1+x}{1-x}}\; {\mbox{ and }} \; a_-:=\sqrt{\frac{1-x}{1+x}}=\frac 1{a_+}. \end{aligned}$$

Now we notice that

$$\displaystyle \begin{aligned} \,{\mathrm{P.V.}}\,\int_{a_-}^{a_+}\frac{d\eta}{1-\eta^2}= \frac 12\ln\left|\frac{(1+a_+)(1-a_-)}{(1-a_+)(1+a_-)}\right|=0. \end{aligned}$$

Inserting this identity into (G.18), we obtain (G.16), as desired.

Then, from (G.15) and (G.16) we get that

$$\displaystyle \begin{aligned} &\frac{d}{dx} \int_{\mathbb{R}} \frac{u_{1/2}(x+y)+u_{1/2}(x-y)-2u_{1/2}(x)} {|y|{}^{2}}\,dy \\ &= -x\int_{\mathbb{R}} \frac{ u_{-1/2}(x+y)+ u_{-1/2}(x-y)-2 u_{-1/2}(x)}{|y|{}^{2}} \,dy\end{aligned} $$

that is

$$\displaystyle \begin{aligned}\frac{d}{dx}(-\Delta)^{1/2} u_{1/2}=-x\,(-\Delta)^{1/2} u_{-1/2}\qquad{\mbox{in }}(-1,1).\end{aligned}$$

From this and (2.17) we infer that x (− Δ)^1∕2 u _−1∕2 = 0 and so (− Δ)^1∕2 u _−1∕2 = 0 in (−1, 1).

These consideration establish (2.24), as desired. Now, we give a brief probabilistic insight on it. In probability—and in stochastic calculus—a measurable function $f:\mathbb {R}^n\to \mathbb {R}$ is said to be harmonic in an open set $D\subset \mathbb {R}^n$ if, for any D ₁ ⊂ D and any x ∈ D ₁,

$$\displaystyle \begin{aligned} \begin{aligned} & f(x)\ =\ \mathbb E_x\left[f(W_{\tau_{D_1}})\right], \\ & \qquad\text{where }W_t\mbox{ is a Brownian motion and }\tau_{D_1} \mbox{is the first exit time from }D_1\mbox{, namely} \\ & \qquad\tau=\inf\{t>0:W_t\not\in D_1\}. \end{aligned} \end{aligned} $$

(G.19)

Notice that, since W _t has (a.s.) continuous trajectories, then (a.s.) $W_{\tau _{D_1}}\in \partial D_1$. This notion of harmonicity coincides with the analytic one.

If one considers a Lévy-type process X _t in place of the Brownian motion, the definition of harmonicity (with respect to this other process) can be given in the very same way. When X _t is an isotropic (2s)-stable process, the definition amounts to having zero fractional Laplacian (− Δ)^s at every point of D and replace (G.19) by

$$\displaystyle \begin{aligned} f(x)\ =\ \mathbb E_x[f(X_{\tau_{D_1}})], \qquad {\mbox{for any }}D_1\subseteq D. \end{aligned}$$

In this identity, we can consider a sequence $\{D_j:D_j\subset D,j\in \mathbb {N}\}$, with , and equality

$$\displaystyle \begin{aligned} f(x)\ =\ \mathbb E_x[f(X_{\tau_{D_j}})],\qquad \text{for any }j\in\mathbb{N}. \end{aligned} $$

(G.20)

When f = 0 in $\mathbb {R}^n\setminus D$, the right-hand side of (G.20) can be not 0 (since $X_{\tau _{D_j}}$ may also end up in D ∖ D _j), and this leaves the possibility of finding f which satisfies (G.20) without vanish identically (an example of this phenomenon is exactly given by the function u _−1∕2 in (2.24)).

It is interesting to observe that if f vanishes outside D and does not vanish identically, then, the only possibility to satisfy (G.20) is that f diverges along ∂D. Indeed, if $|f|\leqslant \kappa$, since $f(X_{\tau _{D_j}})\neq 0$ only when x ∈ D ∖ D _j and as j →∞, we would have that

$$\displaystyle \begin{aligned}\lim_{j\to+\infty}\mathbb E_x[f(X_{\tau_{D_j}})]\leqslant\lim_{j\to+\infty}\,{\mathrm{const}}\, \kappa \,|D\setminus D_j|=0,\end{aligned}$$

and (G.20) would imply that f must vanish identically.

Of course, the function u _−1∕2 in (2.23) embodies exactly this singular boundary behavior.

Appendix H: Another Proof of (2.24)

Here we give a different proof of (2.24) by using complex analysis and extension methods. We use the principal complex square root introduced in (D.2) and, for any $x\in \mathbb {R}$ and y > 0 we define

$$\displaystyle \begin{aligned}U_{-1/2}(x,y):=\Re \left( \frac 1{\displaystyle{\surd}{1-z^2}}\right) ,\end{aligned}$$

where z := x + iy.

The function U _−1∕2 is plotted in Fig. 10. We recall that the function U _−1∕2 is well-defined, thanks to (D.6). Also, the denominator never vanishes when y > 0 and so U _−1∕2 is harmonic in the halfplane, being the real part of a holomorphic function in such domain.

Furthermore, in light of (D.9), we have that

and therefore

This gives that U _−1∕2 is the harmonic extension of u _−1∕2 to the halfplane. Therefore, by (2.6), (2.7), and (D.14), for any x ∈ (−1, 1) we have that

that is (2.24).

Appendix I: Proof of (2.29) (Based on Fourier Methods)

When n = 1, we use (2.28) to find that⁸

$$\displaystyle \begin{aligned} \begin{aligned} {\mathcal{G}}_{1/2}(x)\;&= \int_{\mathbb{R}} e^{-|\xi|} \,e^{ix\xi}\,d\xi =\lim_{R\to+\infty}\int_0^{R} e^{-\xi} \,e^{ix \xi}\,d\xi +\int_{-R}^0 e^{\xi} \,e^{ix \xi}\,d\xi\\ &= \lim_{R\to+\infty}\frac{e^{R(ix-1)} -1 }{ix-1}+\frac{1-e^{-R(ix+1)}}{ix+1} = -\frac{1 }{ix-1}+\frac{1}{ix+1}=\frac{2}{x^2+1}. \end{aligned}\end{aligned} $$

(I.1)

This proves (2.28) when n = 1.

Let us now deal with the case n⩾2. By changing variable Y := 1∕y, we see that

$$\displaystyle \begin{aligned}\int_0^{+\infty} e^{-\frac{|\xi|\,\left(y-\frac 1y\right)^2}{2}}\,dy =\int_0^{+\infty} e^{-\frac{|\xi|\,\left(Y-\frac 1Y\right)^2}{2}}\,\frac{dY}{Y^2}.\end{aligned} $$

Therefore, summing up the left hand side to both sides of this identity and using the transformation $\eta :=y-\frac 1y$,

$$\displaystyle \begin{aligned} \begin{array}{rcl}2\int_0^{+\infty} e^{-\frac{|\xi|\,\left(y-\frac 1y\right)^2}{2}}\,dy &\displaystyle =&\displaystyle \int_0^{+\infty} \left(1+\frac 1{y^2}\right)\,e^{-\frac{|\xi|\,\left(y-\frac 1y\right)^2}{2}}\,dy\\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\,\int_0^{+\infty} e^{-\frac{|\xi|\,\eta^2}{2}}\,d\eta\\&\displaystyle =&\displaystyle \frac{\,{\mathrm{const}}\,}{\sqrt{|\xi|}}.\end{array} \end{aligned} $$

As a result,

$$\displaystyle \begin{aligned} \begin{array}{rcl}e^{-|\xi|} = \frac{\,{\mathrm{const}}\, e^{-|\xi|}\,\sqrt{|\xi|}}{\sqrt{|\xi|}} &\displaystyle =&\displaystyle \,{\mathrm{const}}\, e^{-|\xi|}\,\sqrt{|\xi|}\int_0^{+\infty} e^{-\frac{|\xi|\,\left(y-\frac 1y\right)^2}{2}}\,dy \\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\, e^{-|\xi|}\,\sqrt{|\xi|}\int_0^{+\infty} e^{-\frac{|\xi|\,\left(y^2+\frac 1{y^2}-2\right)}{2}}\,dy \\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\, \sqrt{|\xi|}\int_0^{+\infty} e^{-\frac{|\xi|\,\left(y^2+\frac 1{y^2}\right)}{2}}\,dy \\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\, \int_0^{+\infty} \frac{1}{\sqrt{t}}\;e^{-\frac{t}{2}}\, e^{-\frac{|\xi|{}^2}{2t}}\,dt, \end{array} \end{aligned} $$

where the substitution t := |ξ| y ² has been used.

Accordingly, by (2.28), the Gaussian Fourier transform and the change of variable τ := t(1 + |x|²),

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathcal{G}}_{1/2}(x)&\displaystyle =&\displaystyle \int_{\mathbb{R}^n} e^{-|\xi|} \,e^{ix\cdot\xi}\,d\xi\\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\, \iint_{\mathbb{R}^n\times(0,+\infty)} \frac{1}{\sqrt{t}}\; e^{-\frac{t}{2}}\, e^{-\frac{|\xi|{}^2}{2t}}\,e^{ix\cdot\xi} \,d\xi\,dt\\ &\displaystyle =&\displaystyle \,{\mathrm{const}}\, \int_{(0,+\infty)} t^{\frac{n-1}{2}} e^{-\frac{t}{2}}\, e^{-\frac{ t|x|{}^2}2} \,dt \\&\displaystyle =&\displaystyle \,{\mathrm{const}}\, \int_{(0,+\infty)} \left(\frac\tau{1+|x|{}^2}\right)^{\frac{n-1}{2}} e^{-\frac\tau2} \,\frac{d\tau}{1+|x|{}^2}\\&\displaystyle =&\displaystyle \frac{\,{\mathrm{const}}\,}{ \big(1+|x|{}^2\big)^{\frac{n+1}{2}} }. \end{array} \end{aligned} $$

This establishes (2.29).

Appendix J: Another Proof of (2.29) (Based on Extension Methods)

The idea is to consider the fundamental solution in the extended space and take a derivative (the time variable acting as a translation and, to favor the intuition, one may keep in mind that the Poisson kernel is the normal derivative of the Green function). Interestingly, this proof is, in a sense, “conceptually simpler”, and “less technical” than that in Appendix I, thus demonstrating that, at least in some cases, when appropriately used, fractional methods may lead to cultural advantages⁹ with respect to more classical approaches.

For this proof, we consider variables $X:=(x,y)\in \mathbb {R}^n\times (0,+\infty )\subset \mathbb {R}^{n+1}$ and fix t > 0. We let Γ be the fundamental solution in $\mathbb {R}^{n+1}$, namely

$$\displaystyle \begin{aligned}\Gamma(X):=\left\{ \begin{matrix} -\,{\mathrm{const}}\,\log|X| &{\mbox{ if }}n=1,\\ \displaystyle\frac{\,{\mathrm{const}}\,}{|X|{}^{n-1}}&{\mbox{ if }}n\geqslant2. \end{matrix} \right.\end{aligned}$$

By construction Δ Γ is the Delta Function at the origin and so, for any t > 0, we have that $\tilde \Gamma (X;t)=\tilde \Gamma (x,y;t):=\Gamma (x,y+t)$ is harmonic for $(x,y)\in \mathbb {R}^n\times (0,+\infty )$. Accordingly, the function $U(x,y;t):=\partial _y\tilde \Gamma (x,y;t)$ is also harmonic for $(x,y)\in \mathbb {R}^n\times (0,+\infty )$. We remark that

$$\displaystyle \begin{aligned} U(x,y;t)&=\partial_y \Gamma(x,y+t)= \displaystyle\frac{\,{\mathrm{const}}\,}{|(x,y+t)|{}^{n}} \partial_y \sqrt{|x|{}^2+(y+t)^2} = \displaystyle\frac{\,{\mathrm{const}}\,(y+t)}{|(x,y+t)|{}^{n+1}}\\ &= \displaystyle\frac{\,{\mathrm{const}}\,(y+t)}{\big(|x|{}^2+(y+t)^2\big)^{\frac{n+1}{2}}}. \end{aligned} $$

This function is plotted in Fig. 11 (for the model case in the plane). We observe that

As a consequence, by (2.6) and (2.7) (and noticing that the role played by the variables y and t in the function U is the same),

This shows that u solves the fractional heat equation, with u approaching a Delta function when . Hence

$$\displaystyle \begin{aligned}{\mathcal{G}}_{1/2}(x)=u(x,1)=\displaystyle\frac{\,{\mathrm{const}}\,}{\big(1+|x|{}^2\big)^{\frac{n+1}{2}}} ,\end{aligned}$$

that is (2.29).

Fig. 11

Harmonic extension in the halfplane of the function $\mathbb {R}\ni x\mapsto\frac {1}{1+|x|{ }^2}$

Fig. 12

Harmonic extension in the halfplane of the function $\mathbb {R}\ni x\mapsto\frac 2\pi \,\arctan x$

Appendix K: Proof of (2.36)

First, we construct a useful barrier. Given A > 1, we define

$$\displaystyle \begin{aligned}w(t):=\left\{ \begin{matrix} A & {\mbox{ if }} |t|\leqslant 1,\\ t^{-1-2s} & {\mbox{ if }} |t|>1. \end{matrix} \right.\end{aligned}$$

We claim that if A is sufficiently large, then

$$\displaystyle \begin{aligned} (-\Delta)^s w(t)<-3w(t) \quad{\mbox{ for all }}t\in\mathbb{R}\setminus (-3,3). \end{aligned} $$

(K.1)

To prove this, fix t⩾3 (the case $t\leqslant -3$ being similar). Then, if |ξ − t| < 1, we have that

$$\displaystyle \begin{aligned}\xi\geqslant t-1=\frac{2t}{3}+\frac{t}3-1\geqslant\frac{2t}{3}.\end{aligned}$$

As a consequence, if |τ − t| < 1,

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \big| w(t)-w(\tau)+\dot w(t)(\tau-t)\big|\leqslant \sup_{|\xi-t|<1} |\ddot w(\xi)|\,|t-\tau|{}^2\\ &\displaystyle &\displaystyle \qquad \leqslant\,{\mathrm{const}}\, \sup_{\xi\geqslant 2t/3} \xi^{-3-2s}\,|t-\tau|{}^2 \leqslant\,{\mathrm{const}}\, t^{-3-2s}\,|t-\tau|{}^2.\end{array} \end{aligned} $$

This implies that

$$\displaystyle \begin{aligned} \begin{aligned} & \int_{\{|\tau-t|<1\}} \frac{w(t)-w(\tau)}{|t-\tau|{}^{1+2s}}\,d\tau= \int_{\{|\tau-t|<1\}} \frac{w(t)-w(\tau)+\dot w(t)(\tau-t)}{|t-\tau|{}^{1+2s}}\,d\tau\\ &\qquad\leqslant \,{\mathrm{const}}\, t^{-3-2s}\int_{\{|\tau-t|<1\}} \frac{|t-\tau |{}^2}{|t-\tau|{}^{1+2s}}\,d\tau= \,{\mathrm{const}}\, t^{-3-2s}\\&\qquad \leqslant \,{\mathrm{const}}\, t^{-1-2s}=\,{\mathrm{const}}\, w(t). \end{aligned} \end{aligned} $$

(K.2)

On the other hand,

$$\displaystyle \begin{aligned} \int_{\{|\tau-t|\geqslant1\}\cap\{|\tau|>1\}} \frac{w(t)-w(\tau)}{|t-\tau|{}^{1+2s}}\,d\tau\leqslant \int_{\{|\tau-t|\geqslant1\}} \frac{w(t)}{|t-\tau|{}^{1+2s}}\,d\tau\leqslant\,{\mathrm{const}}\, w(t) .\end{aligned} $$

(K.3)

In addition, if |τ|⩽1 then $|\tau -t|\geqslant t-\tau \geqslant 3-1>1$, hence

$$\displaystyle \begin{aligned}\{|\tau-t|\geqslant1\}\cap\{|\tau|\leqslant1\}=\{|\tau|\leqslant1\}.\end{aligned}$$

Accordingly,

$$\displaystyle \begin{aligned} \int_{\{|\tau-t|\geqslant1\}\cap\{|\tau|\leqslant1\}} \frac{w(t)-w(\tau)}{|t-\tau|{}^{1+2s}}\,d\tau = \int_{\{|\tau|\leqslant1\}} \frac{t^{-1-2s}-A}{|t-\tau|{}^{1+2s}}\,d\tau\leqslant \int_{\{|\tau|\leqslant1\}} \frac{1-A}{|t-\tau|{}^{1+2s}}\,d\tau. \end{aligned} $$

(K.4)

We also observe that if |τ|⩽1 then $|t-\tau |\leqslant t+1\leqslant 2t$ and therefore

$$\displaystyle \begin{aligned}\int_{\{|\tau|\leqslant1\}} \frac{d\tau}{|t-\tau|{}^{1+2s}}\geqslant\frac{\,{\mathrm{const}}\,}{t^{1+2s}}= {\,{\mathrm{const}}\,}w(t).\end{aligned}$$

So, we plug this information into (K.4), assuming A > 1 and we obtain that

$$\displaystyle \begin{aligned} \int_{\{|\tau-t|\geqslant1\}\cap\{|\tau|\leqslant1\}} \frac{w(t)-w(\tau)}{|t-\tau|{}^{1+2s}}\,d\tau \leqslant -{(A-1)\,\,{\mathrm{const}}\,}w(t). \end{aligned} $$

(K.5)

Thus, gathering together the estimates in (K.2), (K.3) and (K.5), we conclude that

$$\displaystyle \begin{aligned}\int_{\mathbb{R}}\frac{w(t)-w(\tau)}{|t-\tau|{}^{1+2s}}\,d\tau\leqslant\,{\mathrm{const}}\, w(t) -{(A-1)\,\,{\mathrm{const}}\,}w(t)\leqslant -4 w(t)<-3w(t),\end{aligned}$$

as long as A is sufficiently large. This proves (K.1).

Now, to prove (2.36), we define $v:=\dot u>0$. From (2.40), we know that

$$\displaystyle \begin{aligned} (-\Delta)^s v=(1-3u^2) v\geqslant -3u^2v\geqslant -3v. \end{aligned} $$

(K.6)

Given ε > 0, we define

$$\displaystyle \begin{aligned}w_\varepsilon(t):= \frac{\iota}{A}\,w(t)-\varepsilon,\qquad{\mbox{ where }} \iota:=\min_{t\in[-3,3]} v(t).\end{aligned}$$

We claim that

$$\displaystyle \begin{aligned} w_\varepsilon\leqslant v. \end{aligned} $$

(K.7)

Indeed, for large ε, it holds that w _ε < 0 < v and so (K.7) is satisfied. In addition, for any ε > 0,

$$\displaystyle \begin{aligned} \lim_{t\to+\infty} w_\varepsilon(t)=-\varepsilon<0\leqslant\inf_{t\in\mathbb{R}} v(t).\end{aligned} $$

(K.8)

Suppose now that ε _⋆ > 0 produces a touching point between $w_{\varepsilon _\star }$ and v, namely $w_{\varepsilon _\star }\leqslant v$ and $w_{\varepsilon _\star }(t_\star )=v(t_\star )$ for some $t_\star \in \mathbb {R}$. Notice that, if |τ|⩽3,

$$\displaystyle \begin{aligned}w_{\varepsilon_\star}(\tau)\leqslant \frac{\iota}{A}\,\sup_{t\in\mathbb{R}}w(t)-\varepsilon\leqslant \iota-\varepsilon =\min_{t\in[-3,3]} v(t)-\varepsilon\leqslant v(\tau)-\varepsilon<v(\tau),\end{aligned}$$

and therefore |t _⋆| > 3. Accordingly, if we set $v_\star :=v-w_{\varepsilon _\star }$, using (K.1) and (K.6), we see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle 0=-3 v_\star(t_\star)= -3v(t_\star)+3w_{\varepsilon_\star}(t_\star) \leqslant (-\Delta)^s v(t_\star)-(-\Delta)^s w_{\varepsilon_\star}(t_\star) \\ &\displaystyle &\displaystyle \qquad\quad\qquad =(-\Delta)^s v_\star(t_\star) =\int_{\mathbb{R}} \frac{ v_\star(t_\star)-v_\star(\tau) }{|t_\star-\tau|{}^{1+2s}}\,d\tau =-\int_{\mathbb{R}} \frac{v_\star(\tau) }{|t_\star-\tau|{}^{1+2s}}\,d\tau. \end{array} \end{aligned} $$

Since the latter integrand is nonnegative, we conclude that v _⋆ must vanish identically, and thus $w_{\varepsilon _\star }$ must coincide with v. But this is in contradiction with (K.8) and so the proof of (K.7) is complete.

Then, by sending in (K.7) we find that $v\geqslant \frac \iota {A}\,w$, and therefore, for t⩾1 we have that $\dot u(t)= v(t)\geqslant \kappa t^{-1-2s}$, for all t > 1, for some κ > 0.

Consequently, for any t > 1,

$$\displaystyle \begin{aligned} 1-u(t) &=\lim_{T\to+\infty} u(T)-u(t) =\lim_{T\to+\infty} \int^T_t \dot u(\tau)\,d\tau\\ &=\int^{+\infty}_t \dot u(\tau)\,d\tau\geqslant \kappa \int^{+\infty}_t \tau^{-1-2s}\,d\tau =\frac{\kappa}{2s}\,t^{-2s},\end{aligned} $$

and a similar estimates holds for 1 + u(t) when t < −1.

These considerations establish (2.36), as desired.

Appendix L: Proof of (2.38)

Here we prove that (2.38) is a solution of (2.37). The idea of the proof, as showed in Fig. 12, is to consider the harmonic extension of the function $\mathbb {R}\ni x\mapsto \frac 2\pi \,\arctan x$ in the halfplane $\mathbb {R}\times (0,+\infty )$ and use the method described in (2.6) and (2.7).

We let

$$\displaystyle \begin{aligned} U(x,y):=\frac{2}{\pi}\arctan\frac{x}{y+1}.\end{aligned}$$

The function U is depicted¹⁰ in Fig. 12. Of course, it coincides with u when y = 0 and, for any $x\in \mathbb {R}$ and y > 0,

$$\displaystyle \begin{aligned} \frac\pi2\,\Delta U(x,y)= -\frac{2 x (1 + y)}{(x^2 + (1 + y)^2)^2}+ \frac{2 x (1 + y)}{(x^2 + (1 + y)^2)^2}=0. \end{aligned} $$

(L.1)

Hence, the setting in (2.6) is satisfied and so, in light of (2.7). we have

(L.2)

Also, by the trigonometric Double-angle Formula, for any $\theta \in \left (-\frac \pi 2,\frac \pi 2\right )$,

$$\displaystyle \begin{aligned}\sin{}(2\theta )=2\sin \theta \cos \theta ={\frac {2\tan \theta }{\tan ^{2}\theta +1}}.\end{aligned}$$

Hence, taking $\theta :=\arctan x$,

$$\displaystyle \begin{aligned}\sin{}(\pi u(x))=\sin{}(2\arctan x )={\frac {2x }{x^2+1}}.\end{aligned}$$

This and (L.2) show that (2.38) is a solution of (2.37).

Appendix M: Another Proof of (2.38) (Based on (2.29))

This proof of (2.38) is based on the fractional heat kernel in (2.29). This approach has the advantage of being quite general (see e.g. Theorem 3.1 in [27]) and also to relate the two “miraculous” explicit formulas (2.29) and (2.38), which are available only in the special case s = 1∕2.

For this, we let P = P(x, t) the fundamental solution of the heat flow in (2.25) with n = 1 and s = 1∕2. Notice that, by (2.29), we know that

$$\displaystyle \begin{aligned} P(x,1)={\mathcal{G}}_{1/2}(x)=\frac{c}{1+x^2},\end{aligned} $$

(M.1)

with

$$\displaystyle \begin{aligned}c:=\left( \int_{\mathbb{R}} \frac{dx}{1+x^2}\right)^{-1}=\frac 1\pi.\end{aligned}$$

Also, by scaling,

$$\displaystyle \begin{aligned} P(x,t)=t^{-1} P(t^{-1}x,1)=t^{-1}{\mathcal{G}}_{1/2}(t^{-1}x) .\end{aligned} $$

(M.2)

For any $x\in \mathbb {R}$ and any t > 0, we define

$$\displaystyle \begin{aligned} U(x,t):= 2\int_0^x P(\eta, t+1)\,d\eta.\end{aligned} $$

(M.3)

In light of (M.2), we see that

$$\displaystyle \begin{aligned}|U(x,t)|\leqslant 2(t+1)^{-1} \int_0^x {\mathcal{G}}_{1/2}\big((t+1)^{-1}\eta\big)\,d\eta =2 \int_0^{(t+1)^{-1}x} {\mathcal{G}}_{1/2}(\zeta)\,d\zeta,\end{aligned}$$

which is bounded in $\mathbb {R}\times [0,+\infty )$, and infinitesimal as t → +∞ for any fixed $x\in \mathbb {R}$.

Notice also that

$$\displaystyle \begin{aligned}\partial^2_t P = \partial_t(\partial_t P)=\partial_t (-\Delta)^{1/2}P= (-\Delta)^{1/2}\partial_t P=(-\Delta)^{1/2}(-\Delta)^{1/2} P=-\partial^2_x P,\end{aligned}$$

by (2.5). As a consequence,

$$\displaystyle \begin{aligned} \begin{aligned} \frac 12(\partial^2_x+\partial^2_t)U(x,t)\,&= \partial_x P(x,t+1)+ \int_0^x \partial^2_t P(\eta, t+1)\,d\eta \\ &= \partial_x P(x,t+1)- \int_0^x \partial^2_x P(\eta, t+1)\,d\eta\\ &= \partial_x P(0,t+1)\\&=0, \end{aligned}\end{aligned} $$

(M.4)

where the last identity follows from (M.2).

Besides, from (M.2) we have that

$$\displaystyle \begin{aligned}\partial_t P(x,t)= \partial_t\Big( t^{-1}{\mathcal{G}}_{1/2}(t^{-1}x) \Big) = -t^{-2}{\mathcal{G}}_{1/2}(t^{-1}x) -t^{-3} x{\mathcal{G}}_{1/2}^{\prime}(t^{-1}x)\end{aligned}$$

and so

$$\displaystyle \begin{aligned}-\partial_t P(x,1)={\mathcal{G}}_{1/2}(x) +x{\mathcal{G}}_{1/2}^{\prime}(x) = \partial_x\big( x\,{\mathcal{G}}_{1/2}(x)\big). \end{aligned}$$

In view of this, we have that

$$\displaystyle \begin{aligned} \partial_t U(x,0)= 2\int_0^x \partial_t P(\eta, 1)\,d\eta= 2\int_0^x \partial_\eta\big( \eta\,{\mathcal{G}}_{1/2}(\eta)\big)\,d\eta= 2x\,{\mathcal{G}}_{1/2}(x). \end{aligned} $$

(M.5)

Accordingly, from (M.4) and (M.5), using the extension method in (2.6) and (2.7) (with the variable y called t here), we conclude that, if

$$\displaystyle \begin{aligned} u(x):=U(x,0),\end{aligned}$$

then

$$\displaystyle \begin{aligned} (-\Delta)^{1/2}u(x)=2x\,{\mathcal{G}}_{1/2}(x).\end{aligned} $$

(M.6)

We remark that, by (M.1) and (M.3),

$$\displaystyle \begin{aligned} u(x)= 2c \int_0^x \frac{d\eta}{1+ x^2} = \frac{2}\pi\,\arctan x. \end{aligned} $$

(M.7)

This, (M.1) and (M.6) give that

$$\displaystyle \begin{aligned}(-\Delta)^{1/2}u(x)=\frac{1}{\pi}\,\frac{2x}{1+x^2}=\frac{1}{\pi}\,\sin{}(2\arctan x)= \frac{1}{\pi}\,\sin\big(\pi u( x)\big), \end{aligned}$$

that is (2.38), as desired.

Appendix N: Proof of (2.46)

Due to translation invariance, we can reduce ourselves to proving (2.46) at the origin. We consider a measurable $u:\mathbb {R}^n\to \mathbb {R}$ such that

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^n}\frac{|u(y)|}{1+|y|{}^{n+2}}\ <\ +\infty.\end{aligned}$$

Assume first that

$$\displaystyle \begin{aligned} {u(x)=0\mbox{ for any }x\in B_r,} \end{aligned} $$

(N.1)

for some r > 0. As a matter of fact, under these assumptions on u, the right-hand side of (2.46) vanishes at 0 regardless the size of r. Indeed,

This proves (2.46) under the additional assumption in (N.1), that we are now going to remove. To this end, for r ∈ (0, 1), denote by χ _r the characteristic function of B _r, i.e. χ _r(x) = 1 if x ∈ B _r and χ _r(x) = 0 otherwise. Consider now u ∈ C ^{2, α}(B _r), for some α ∈ (0, 1), with

$$\displaystyle \begin{aligned}u(0)=|\nabla u(0)|=0\end{aligned} $$

(N.2)

for simplicity (note that one can always modify u by considering $\tilde u(x)=u(x)-u(0)-\nabla u(0)\cdot x$ and without affecting the operators in (2.46)). Then, the right hand side of (2.46) becomes in this case

The second addend is trivial for any r ∈ (0, 1), in view of the above remark, since u(1 − χ _r) is constantly equal to 0 in B _r. For the first one, we have

$$\displaystyle \begin{aligned} & \int_{\mathbb{R}^n}\frac{u(2y)\chi_r(2y)-4u(y)\chi_r(y)}{{|y|}^{n+2}}\;dy\ =\ \int_{B_{r/2}}\frac{u(2y)-4u(y)}{{|y|}^{n+2}}\;dy -4\int_{B_r\setminus B_{r/2}}\frac{u(y)}{{|y|}^{n+2}}\;dy. \end{aligned} $$

(N.3)

Now, we recall (N.2) and we notice that

$$\displaystyle \begin{aligned} |u(2y)-4u(y)|\leqslant\|u\|{}_{C^{2,\alpha}(B)}|y|{}^{2+\alpha}, \end{aligned} $$

which in turn implies that

$$\displaystyle \begin{aligned} \left|\int_{B_{r/2}}\frac{u(2y)-4u(y)}{{|y|}^{n+2}}\;dy\right|\ \leqslant\ \,{\mathrm{const}}\,\|u\|{}_{C^{2,\alpha}(B)}r^\alpha. \end{aligned} $$

(N.4)

On the other hand, a Taylor expansion and (N.2) yield

(N.5)

in view of (1.1), for some $\eta :B_r\to \mathbb {R}$ such that $|\eta (x)|\leqslant c|x|{ }^\alpha$. From this, (N.3) and (N.4) we deduce that

which finally justifies (2.46).

It is interesting to remark that the main contribution to prove (2.46) comes in this case from the “intermediate ring” in (N.5).

Appendix O: Proof of (2.48)

Take for instance Ω to be the unit ball and $\bar u=1-|x|{ }^2$. Suppose that $\|\bar u-v_\varepsilon \|{ }_{C^2(\Omega )}\leqslant \varepsilon$. Then, for small ε, if $x\in \mathbb {R}^n\setminus B_{1/2}$ it holds that

$$\displaystyle \begin{aligned}v_\varepsilon(x)\leqslant \bar u(x)+\varepsilon= 1-|x|{}^2+\varepsilon\leqslant \frac 34+\varepsilon\leqslant\frac 45,\end{aligned}$$

while

$$\displaystyle \begin{aligned}v_\varepsilon(0)\geqslant\bar u(0)-\varepsilon=1-\varepsilon\geqslant\frac 56.\end{aligned}$$

This implies that there exists $x_\varepsilon \in \overline {B_{1/2}}$ such that

$$\displaystyle \begin{aligned}v_\varepsilon(x_\varepsilon)=\sup_{B_1} v_\varepsilon \geqslant\frac 56>\frac 45\geqslant\sup_{B_1\setminus B_{1/2}}v_\varepsilon.\end{aligned}$$

As a result,

$$\displaystyle \begin{aligned} \,{\mathrm{P.V.}}\, \int_{\Omega} \frac{v_\varepsilon(x_\varepsilon)-v_\varepsilon(y)}{|x_\varepsilon-y|{}^{n+2s}}\,dy&\geqslant \int_{B_1\setminus B_{3/4}} \frac{v_\varepsilon(x_\varepsilon)-v_\varepsilon(y)}{|x_\varepsilon-y|{}^{n+2s}}\,dy\\ &\geqslant \int_{B_1\setminus B_{3/4}} \left(\frac 56-\frac 45\right)\,dy \geqslant\,{\mathrm{const}}\,. \end{aligned} $$

This says that (− Δ)^s v _ε cannot vanish at x _ε and so (2.48) is proved.

Appendix P: Proof of (2.52)

Let us first notice that the identity

$$\displaystyle \begin{aligned} \lambda^s\ =\ \frac{s}{\Gamma(1-s)} \int_0^\infty\frac{1-e^{-t\lambda}}{t^{1+s}}\;dt \end{aligned} $$

(P.1)

holds for any λ > 0 and s ∈ (0, 1), because

$$\displaystyle \begin{aligned} \int_0^\infty\frac{1-e^{-t}}{t^{1+s}}\;dt =\left.\frac{1-e^{-t}}{-s\,t^s}\right|{}_0^\infty+\frac 1s\int_0^\infty\frac{e^{-t}}{t^s}\;dt=\frac{\Gamma(1-s)}s. \end{aligned} $$

We also observe that when $u\in C^\infty _0(\Omega )$, the coefficients $\hat u_j$ decay fast as j →∞: indeed

$$\displaystyle \begin{aligned} \hat u_j=-\frac 1{\mu_j}\int_\Omega u\,\Delta\psi_j=-\frac 1{\mu_j}\int_\Omega \psi_j\,\Delta u =\ldots=(-1)^k\frac 1{\mu_j^k}\int_\Omega \psi_j\,\Delta^k u. \end{aligned}$$

Therefore, applying equality (P.1) to the μ _j’s in (2.51) we obtain¹¹

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {(-\Delta)^s_{N,\Omega}} u &\displaystyle =&\displaystyle \ \frac{s}{\Gamma(1-s)}\sum_{j=0}^{+\infty}\int_0^\infty\frac{\hat u_j\psi_j-e^{-t\mu_j}\hat u_j\psi_j}{t^{1+s}}\;dt\\ &\displaystyle =&\displaystyle \frac{s}{\Gamma(1-s)}\int_0^\infty \frac{u-e^{t{\Delta_{N,\Omega}}u}}{t^{1+s}}\;dt,\quad u\in C^\infty_0(\Omega) \end{array} \end{aligned} $$

(P.2)

where $\{ e^{t \Delta _{N,\Omega } }\}_{t>0}$ stands for the heat semigroup associated to Δ_N,Ω. i.e. $e^{t{\Delta _{N,\Omega }}}u$ solves

$$\displaystyle \begin{aligned}\left\{ \begin{matrix} \partial_t v(x,t)=\Delta v(x,t) {\mbox{ in }}\Omega\times(0,\infty) \\ \partial_\nu v(x,t)=0 \text{ on }\partial\Omega\times(0,\infty) \\ v(x,0)=u(x)\text{ on }\Omega\times\{0\}. \end{matrix} \right.\end{aligned}$$

To check (P.2), it is sufficient to notice that

$$\displaystyle \begin{aligned} \partial_t\left(\sum_{j=0}^{+\infty}e^{-t\mu_j}\hat u_j\psi_j\right)= -\sum_{j=0}^{+\infty}\mu_je^{-t\mu_j}\hat u_j\psi_j= \sum_{j=0}^{+\infty}e^{-t\mu_j}\hat u_j\Delta\psi_j= \Delta\left(\sum_{j=0}^{+\infty}e^{-t\mu_j}\hat u_j\psi_j\right) \end{aligned}$$

and that

$$\displaystyle \begin{aligned} \left.\left(\sum_{j=0}^{+\infty}e^{-t\mu_j}\hat u_j\psi_j\right)\right|{}_{t=0} =\sum_{j=0}^{+\infty}\hat u_j\psi_j=u. \end{aligned}$$

Under suitable regularity assumptions on Ω, write now the heat semigroup in terms of the heat kernel $p_{N}^\Omega$ as

$$\displaystyle \begin{aligned} e^{t{\Delta_{N,\Omega}}}u(x)\ =\ \int_\Omega p_{{N}}^\Omega(t,x,y)\,u(y)\;dy, \qquad x\in\Omega,\ t>0 \end{aligned} $$

(P.3)

where the following two-sided estimate on $p_{{N}}^\Omega$ holds (see, for example, [102, Theorem 3.1])

$$\displaystyle \begin{aligned} \frac{c_1\:e^{-c_2|x-y|{}^2/t}}{t^{n/2}} \ \leqslant\ p_{{N}}^\Omega(t,x,y)\ \leqslant\ \frac{c_3\:e^{-c_4|x-y|{}^2/t}}{t^{n/2}}, \qquad x,y\in\Omega,\ t,c_1,c_2,c_3,c_4>0. \end{aligned} $$

(P.4)

Recall also that $p_{N}^\Omega (t,x,y)=p_{N}^\Omega (t,y,x)$ for any t > 0 and x, y ∈ Ω, and that

$$\displaystyle \begin{aligned} \int_\Omega p_{{N}}^\Omega(t,x,y)\;dy\ =\ 1, \qquad x\in\Omega,\ t>0, \end{aligned} $$

(P.5)

which follows from noticing that, for any $u\in C^\infty _0(\Omega )$,

$$\displaystyle \begin{aligned} \partial_t\int_\Omega e^{t {\Delta_{N,\Omega}} }u=\int_\Omega\partial_t e^{t {\Delta_{N,\Omega}} }u= \int_\Omega\Delta e^{t {\Delta_{N,\Omega}} }u=-\int_{\partial\Omega}\partial_\nu e^{t {\Delta_{N,\Omega}} }u=0 \end{aligned}$$

and therefore

$$\displaystyle \begin{aligned} \int_\Omega u(x)\;dx&=\int_\Omega e^{t {\Delta_{N,\Omega}} }u(x)\;dx =\int_\Omega\int_\Omega p_{{N}}^\Omega(t,x,y)\,u(y)\;dy\;dx\\ &=\int_\Omega u(y) \int_\Omega p_{N}^\Omega(t,x,y)\;dx\;dy. \end{aligned} $$

By (P.5), for any x ∈ Ω and t > 0,

$$\displaystyle \begin{aligned} & u(x)-e^{t {\Delta_{N,\Omega}} }u(x)=\int_\Omega p_{{N}}^\Omega(t,x,y) \left(u(x)-u(y)\right)dy \end{aligned} $$

and, exchanging the order of integration in (P.2) (see below for a justification of this passage), one gets

$$\displaystyle \begin{aligned} {(-\Delta)^s_{N,\Omega}} u(x) &= \frac{s}{\Gamma(1-s)}\int_0^\infty\frac{u(x)-e^{t {\Delta_{N,\Omega}} } u(x)}{t^{1+s}}\;dy\\ &= \frac{s}{\Gamma(1-s)}\int_0^\infty\frac{\int_\Omega p_{{N}}^\Omega(t,x,y) \left(u(x)-u(y)\right)dy}{t^{1+s}}\;dt \\ &= \frac{s}{\Gamma(1-s)}\;\,{\mathrm{P.V.}}\,\int_\Omega\left(u(x)-u(y)\right) \int_0^\infty\frac{p_{{N}}^\Omega(t,x,y)}{t^{1+s}}\;dt\;dy\\ &=\,{\mathrm{P.V.}}\,\int_\Omega\frac{\left(u(x)-u(y)\right)k(x,y)}{{|x-y|}^{n+2s}}\;dy ,\end{aligned} $$

where, in view of (P.4), we have

$$\displaystyle \begin{aligned} k(x,y) &:=\frac{s}{\Gamma(1-s)}\;|x-y|{}^{n+2s} \int_0^\infty\frac{p_{{N}}^\Omega(t,x,y)}{t^{1+s}}\;dt\\ &\simeq |x-y|{}^{n+2s} \int_0^\infty\frac{e^{-|x-y|{}^2/t}}{t^{n/2+1+s}}\,dt \simeq\int_0^\infty\frac{e^{-1/t}}{t^{n/2+1+s}}\,dt \simeq 1. \end{aligned} $$

These considerations establish (2.52). Note however that in the above computations there is a limit exiting the integral in the t variable, namely:

(P.6)

To properly justify this we are going to build an integrable majorant in t and independent of ε of the integrand

$$\displaystyle \begin{aligned} \frac{\int_{\Omega\setminus B_\varepsilon(x)} p_{{N}}^\Omega(t,x,y) \left(u(x)-u(y)\right)dy}{t^{1+s}}. \end{aligned} $$

(P.7)

To this end, first of all we observe that, by the boundedness of u and (P.5),

$$\displaystyle \begin{aligned} \left|\frac{\int_{\Omega\setminus B_\varepsilon(x)} p_{{N}}^\Omega(t,x,y) \left(u(x)-u(y)\right)dy}{t^{1+s}}\right|\leqslant \frac{2\|u\|{}_{L^\infty(\Omega)}}{t^{1+s}} \int_{\Omega\setminus B_\varepsilon(x)} p_{{N}}^\Omega(t,x,y)\;dy\leqslant \frac{2\|u\|{}_{L^\infty(\Omega)}}{t^{1+s}} \end{aligned}$$

which is integrable at infinity. So, to obtain an integrable bound for (P.7), we can now focus on small values of t, say t ∈ (0, 1). For this, we denote by p the heat kernel in $\mathbb {R}^N$ and we write

$$\displaystyle \begin{aligned} &\int_{\Omega\setminus B_\varepsilon(x)} p_{{N}}^\Omega(t,x,y) \left(u(x)-u(y)\right)dy\ =\\ &=\ \int_{\Omega\setminus B_\varepsilon(x)} p(t,x,y) \left(u(x)-u(y)\right)dy\\ &- \int_{\Omega\setminus B_\varepsilon(x)}(p_{{N}}^\Omega(t,x,y)-p(t,x,y))(u(x)-u(y))\;dy=:A+B. \end{aligned} $$

We first manipulate A. We reformulate u as

$$\displaystyle \begin{aligned} u(y)=u(x)+\nabla u(x)\cdot(y-x)+\eta(y)|x-y|{}^2, \qquad y\in \mathbb{R}^n,\ \|\eta\|{}_{L^\infty(\mathbb{R}^n)}\leqslant \|u\|{}_{C^2(\Omega)}, \end{aligned}$$

so that

$$\displaystyle \begin{aligned} \begin{aligned} & \int_{\Omega\setminus B_\varepsilon(x)} p(t,x,y)\left(u(x)-u(y)\right)dy =\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} p(t,x,y)\left(u(x)-u(y)\right)dy\\ &\qquad-u(x)\int_{\mathbb{R}^n\setminus \Omega} p(t,x,y)\;dy \\ & =\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} p(t,x,y)\nabla u(x)\cdot(x-y)\;dy\\ &\qquad-\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} p(t,x,y)\,\eta(y)|x-y|{}^2\;dy-u(x)\int_{\mathbb{R}^n\setminus \Omega} p(t,x,y)\;dy. \end{aligned}\end{aligned} $$

(P.8)

In the last expression, the first integral on the right-hand side is 0 by odd symmetry, while for the second one

$$\displaystyle \begin{aligned} \begin{aligned} & \left|\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} p(t,x,y) \,\eta(y)|x-y|{}^2\;dy\right| \leqslant \|u\|{}_{C^2(\Omega)}\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} p(t,x,y)|x-y|{}^2\;dy \\ & \leqslant \,{\mathrm{const}}\,\|u\|{}_{C^2(\Omega)}t^{-/2}\int_{\mathbb{R}^n\setminus B_\varepsilon(x)} e^{-|x-y|{}^2/(4t)}|x-y|{}^2\;dy\\ &\leqslant \,{\mathrm{const}}\,\|u\|{}_{C^2(\Omega)}t\int_{\mathbb{R}^n\setminus B_{\varepsilon/\sqrt{4t}}} e^{-|z|{}^2}|z|{}^2\;dz \\ & \leqslant \,{\mathrm{const}}\,\|u\|{}_{C^2(\Omega)}t. \end{aligned} \end{aligned} $$

(P.9)

As for the last integral in (P.8), we have that

$$\displaystyle \begin{aligned} \begin{aligned} & |u(x)|\int_{\mathbb{R}^n\setminus\Omega}p(t,x,y)\;dy\leqslant\,{\mathrm{const}}\,|u(x)|t^{-n/2} \int_{\mathbb{R}^n\setminus\Omega} e^{-|x-y|{}^2/(4t)}\;dy\leqslant \\ & \leqslant\,{\mathrm{const}}\,|u(x)|t^{-n/2} \int_{\mathbb{R}^n\setminus B_{\text{dist}(x,\partial\Omega)}} e^{-|y|{}^2/(4t)}\;dy \leqslant\,{\mathrm{const}}\,|u(x)| \int_{\mathbb{R}^n\setminus B_{\text{dist}(x,\partial\Omega)/\sqrt{4t}}} e^{-|z|{}^2}\;dz \\ & \leqslant \,{\mathrm{const}}\,|u(x)|e^{-\text{dist}(x,\partial\Omega)/\sqrt{4t}}. \end{aligned} \end{aligned} $$

(P.10)

Equations (P.9) and (P.10) imply that

$$\displaystyle \begin{aligned} \frac{|A|}{t^{1+s}}\leqslant\,{\mathrm{const}}\, t^{-s},\qquad t\in(0,1), \end{aligned}$$

which is integrable for t ∈ (0, 1).

We turn now to the estimation of B which we rewrite as

$$\displaystyle \begin{aligned} B=\int_{\Omega}\big(p_{{N}}^\Omega(t,x,y)-p(t,x,y)\big)\big(u(x)-u(y)\big)\chi_{\Omega\setminus B_\varepsilon(x)}(y)\;dy \end{aligned}$$

where χ _U stands for the characteristic function of a set $U\subset \mathbb {R}^n$. By definition, B solves the heat equation in Ω with zero initial condition. Moreover, since u is supported in a compact subset K of Ω, B is satisfying the (lateral) boundary condition

$$\displaystyle \begin{aligned} \Big|B\big|{}_{\partial B}\Big|&\leqslant \int_{\Omega}\big|p_{{N}}^\Omega(t,x,y)-p(t,x,y)\big||u(y)|\chi_{\Omega\setminus B_\varepsilon(x)}(y)\;dy\\ &\leqslant \,{\mathrm{const}}\, t^{-n/2}\int_K e^{-c_1|x-y|{}^2/t}|u(y)|\;dy \\ & \leqslant \,{\mathrm{const}}\,\|u\|{}_{L^1(\Omega)}t^{-n/2} e^{-c_2/t} \end{aligned} $$

for some c ₁, c ₂ > 0, in view of (P.4) and that, for x ∈ ∂ Ω and y ∈ K, $|x-y|\geqslant \text{dist}(K,\partial \Omega )>0$. Then, by the parabolic maximum principle (see, for example, Section 7.1.4 in [62]),

$$\displaystyle \begin{aligned} \frac{|B|}{t^{1+s}}\leqslant \,{\mathrm{const}}\, t^{-n/2-1-s} e^{-c_2/t}, \end{aligned}$$

which again is integrable for t ∈ (0, 1). These observations provide an integrable bound for the integrand in (P.8), thus completing the justification of the claim in (P.6), as desired.

Appendix Q: Proof of (2.53)

If u is periodic, we can write it in Fourier series as

$$\displaystyle \begin{aligned}u(x)=\sum_{k\in\mathbb{Z}^n} u_k\,e^{2\pi ik\cdot x},\end{aligned}$$

and the Fourier basis is also a basis of eigenfunctions. We have that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_{\mathbb{R}^n} \frac{u(x+y)+u(x-y)-2u(x)}{|y|{}^{n+2s}}\,dy\\ &\displaystyle &\displaystyle \qquad=\sum_{k\in\mathbb{Z}^n} \int_{\mathbb{R}^n} \frac{u_k e^{ 2\pi ik\cdot(x+y)}+ u_ke^{2\pi ik\cdot(x-y)}-2u_ke^{2\pi ik\cdot x}}{|y|{}^{n+2s}}\,dy \\&\displaystyle &\displaystyle \qquad =\sum_{k\in\mathbb{Z}^n} u_k e^{ 2\pi ik\cdot x} \int_{\mathbb{R}^n} \frac{e^{ 2\pi ik\cdot y}+ e^{-2\pi ik\cdot y}-2}{|y|{}^{n+2s}}\,dy\\ &\displaystyle &\displaystyle \qquad=\sum_{k\in\mathbb{Z}^n} u_k e^{ 2\pi ik\cdot x}\,|k|{}^{2s} \int_{\mathbb{R}^n} \frac{e^{ 2\pi i\frac{k}{|k|}\cdot Y}+ e^{-2\pi i\frac{k}{|k|}\cdot Y}-2}{|Y|{}^{n+2s}}\,dY\\ \\&\displaystyle &\displaystyle \qquad =\sum_{k\in\mathbb{Z}^n} u_k e^{ 2\pi ik\cdot x}\,|k|{}^{2s} \int_{\mathbb{R}^n} \frac{e^{ 2\pi iY_1}+ e^{-2\pi iY_1}-2}{|Y|{}^{n+2s}}\,dY\\ &\displaystyle &\displaystyle \qquad=\,{\mathrm{const}}\,\sum_{k\in\mathbb{Z}^n} u_k e^{ 2\pi ik\cdot x}\,|k|{}^{2s} \end{array} \end{aligned} $$

and this shows (2.53).

Appendix R: Proof of (2.54)

We fix $\bar k\in \mathbb {N}$. We consider the $\bar k$th eigenvalue $\lambda _{\bar k}>0$ and the corresponding normalized eigenfunction $\phi _{\bar k}=:\bar u$. We argue by contradiction and suppose that for any ε > 0 we can find v _ε such that $\|\bar u-v_\varepsilon \|{ }_{C^2(B_1)}\leqslant \varepsilon$, with $(-\Delta )^s_{D,\Omega } v_\varepsilon =0$ in B ₁.

Using the notation in (2.49), we have that $\bar u_k=\delta _{k\bar k}$ and therefore

$$\displaystyle \begin{aligned} \begin{aligned} & \left\| (-\Delta)^s_{D,\Omega} \bar u-(-\Delta)^s_{D,\Omega} v_\varepsilon\right\|{}^2_{L^2(\Omega)}= \left\| (-\Delta)^s_{D,\Omega} \bar u\right\|{}^2_{L^2(\Omega)} =\left\| (-\Delta)^s_{D,\Omega}\phi_{\bar k} \right\|{}^2_{L^2(\Omega)}\\ &\qquad=\left\| \lambda_{\bar k}^s\,\phi_{\bar k} \right\|{}^2_{L^2(\Omega)}=\lambda_{\bar k}^{2s}. \end{aligned}\end{aligned} $$

(R.1)

Furthermore

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left\| (-\Delta)^s_{D,\Omega} \bar u-(-\Delta)^s_{D,\Omega} v_\varepsilon\right\|{}^2_{L^2(\Omega)} =\left\| (-\Delta)^s_{D,\Omega} (\bar u- v_\varepsilon)\right\|{}^2_{L^2(\Omega)}\\ &\displaystyle &\displaystyle \qquad\qquad=\left\| \sum_{k=0}^{+\infty}\lambda_k^s (\bar u-v_\varepsilon)_k\, \phi_k \right\|{}^2_{L^2(\Omega)}\\ &\displaystyle &\displaystyle \qquad\qquad= \sum_{k=0}^{+\infty}\lambda_k^{2s} (\bar u-v_\varepsilon)_k^2\leqslant \,{\mathrm{const}}\,\sum_{k=0}^{+\infty}\lambda_k^{2} (\bar u-v_\varepsilon)_k^2\\ &\displaystyle &\displaystyle \qquad\qquad=\,{\mathrm{const}}\,\left\| \Delta(\bar u- v_\varepsilon)\right\|{}^2_{L^2(\Omega)}\\ &\displaystyle &\displaystyle \qquad\qquad\leqslant\,{\mathrm{const}}\, \| \bar u- v_\varepsilon\|{}^2_{C^2(\Omega)}\leqslant\,{\mathrm{const}}\,\varepsilon. \end{array} \end{aligned} $$

Comparing this with (R.1), we obtain that $\lambda _{\bar k}^{2s}\leqslant \,{\mathrm {const}}\,\varepsilon$, which is a contradiction for small ε. Hence, the proof of (2.54) is complete.

Appendix S: Proof of (2.60)

Let

$$\displaystyle \begin{aligned}v(t):=\int_0^t \frac{\dot u (\tau) }{(t-\tau)^{s}} \, d\tau.\end{aligned}$$

Then, by (2.59) and writing 𝜗 := ω (t − τ), we see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathcal{L}} v(\omega)&\displaystyle =&\displaystyle \int_{0}^{+\infty } \left[ \int_0^t \frac{\dot u (\tau) }{(t-\tau)^{s}} \, d\tau\right] e^{-\omega t}\,dt =\int_0^{+\infty} \left[ \int_\tau^{+\infty} \frac{\dot u (\tau) e^{-\omega t}}{(t-\tau)^{s}} \,dt\right]\,d\tau\\ &\displaystyle =&\displaystyle \omega^{s-1} \int_0^{+\infty} \left[ \int_0^{+\infty} \frac{\dot u (\tau) e^{-\omega \tau}\,e^{-\vartheta}}{\vartheta^{s}} \,d\vartheta\right]\,d\tau\\ &\displaystyle =&\displaystyle \Gamma(1-s)\, \omega^{s-1} \int_0^{+\infty} \dot u (\tau) e^{-\omega \tau}\,d\tau\\ &\displaystyle =&\displaystyle \Gamma(1-s)\, \omega^{s-1} \int_0^{+\infty} \left(\frac{d}{d\tau}\big( u (\tau) e^{-\omega \tau}\big) +\omega u (\tau) e^{-\omega \tau}\right) \,d\tau\\ &\displaystyle =&\displaystyle \Gamma(1-s)\, \omega^{s-1} \left( -u (0)+ \omega \int_0^{+\infty} u (\tau) e^{-\omega \tau} \,d\tau \right) \\ &\displaystyle =&\displaystyle \Gamma(1-s)\, \omega^{s-1} \left( -u (0)+ \omega{\mathcal{L}}u(\omega)\right), \end{array} \end{aligned} $$

where Γ denotes here the Euler’s Gamma Function. This and (2.56) give (2.60), up to neglecting normalizing constants, as desired.

It is also worth pointing out that, as , formula (2.60) recovers the classical derivative, since, by (2.59),

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathcal{L}}\dot u(\omega)&\displaystyle =&\displaystyle \int_{0}^{+\infty } \dot u(t)e^{-\omega t}\,dt\\ &\displaystyle =&\displaystyle \int_{0}^{+\infty } \left(\frac{d}{dt} \big( u(t)e^{-\omega t}\big) +\omega u(t)e^{-\omega t} \right)\,dt\\ &\displaystyle =&\displaystyle -u(0)+ \omega\int_{0}^{+\infty } u(t)e^{-\omega t}\,dt\\ &\displaystyle =&\displaystyle -u(0)+\omega\,{\mathcal{L}}u(\omega), \end{array} \end{aligned} $$

which is (2.60) when s = 1.

Appendix T: Proof of (2.61)

First, we compute the Laplace Transform of the constant function. Namely, by (2.59), for any $b\in \mathbb {R}$,

$$\displaystyle \begin{aligned} {\mathcal{L}} b(\omega)= b\,\int_{0}^{+\infty } e^{-\omega t}\,dt=\frac{b}{\omega}. \end{aligned} $$

(T.1)

We also set

$$\displaystyle \begin{aligned}\Psi(t):= \int_0^t \frac{f(\tau)}{(t-\tau)^{1-s}}\,d\tau\end{aligned}$$

and we use (2.59) and the substitution 𝜗 := ω (t − τ) to calculate that

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathcal{L}} \Psi(\omega)&\displaystyle =&\displaystyle \int_{0}^{+\infty } \left[ \int_0^t \frac{f(\tau)}{(t-\tau)^{1-s}}\,d\tau \right]\,e^{-\omega t}\,dt\\ &\displaystyle =&\displaystyle \int_{0}^{+\infty } \left[ \int_\tau^{+\infty} \frac{f(\tau) \,e^{-\omega t} }{(t-\tau)^{1-s}}\,dt \right]\,d\tau\\ &\displaystyle =&\displaystyle \omega^{-s}\int_{0}^{+\infty } \left[ \int_0^{+\infty} \frac{f(\tau) \,e^{-\omega \tau}\,e^{-\vartheta} }{\vartheta^{1-s}}\,d\vartheta \right]\,d\tau\\ &\displaystyle =&\displaystyle \Gamma(s)\, \omega^{-s}\,\int_{0}^{+\infty } f(\tau) \,e^{-\omega \tau}\,d\tau = \Gamma(s)\, \omega^{-s}\,{\mathcal{L}} f(\omega), \end{array} \end{aligned} $$

where Γ denotes here the Euler’s Gamma Function.

Exploiting this and (T.1), and making use also of (2.60), we can write the expression $\partial ^s_{C,t} u=f$ in terms of the Laplace Transform as

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \omega^s \Big( {\mathcal{L}} u(\omega)- {\mathcal{L}} b(\omega)\Big)= \omega^s {\mathcal{L}} u(\omega)-\omega^{s-1} u(0)= {\mathcal{L}} (\partial^s_{C,t} u)(\omega)\\ &\displaystyle &\displaystyle \qquad\qquad= {\mathcal{L}} f(\omega) = \frac{\omega^s}{\Gamma(s)} {\mathcal{L}} \Psi(\omega), \end{array} \end{aligned} $$

with b := u(0). Hence, dividing by ω ^s and inverting the Laplace Transform, we obtain that

$$\displaystyle \begin{aligned}u(t)-b= \frac{1}{\Gamma(s)} \Psi(t), \end{aligned}$$

which is (2.61).

Appendix U: Proof of (2.62)

We take G to be the fundamental solution of the operator “identity minus Laplacian”, namely

$$\displaystyle \begin{aligned} G-\Delta G =\delta_0 \quad{\mbox{ in }}\mathbb{R}^n,\end{aligned} $$

(U.1)

being δ ₀ the Dirac’s Delta. The study of this fundamental solution can be done by Fourier Transform in the sense of distributions, and this leads to an explicit representation in dimension 1 recalling (I.1); we give here a general argument, valid in any dimension, based on the heat kernel

$$\displaystyle \begin{aligned}g(x,t):= {\frac {1}{(4\pi t)^{n/2}}}e^{-\frac{|x|{}^{2}}{4t}}.\end{aligned}$$

Notice that ∂ _t g =  Δg and g(⋅, 0) = δ ₀(⋅). Let also

$$\displaystyle \begin{aligned} G(x):=\int_0^{+\infty} e^{-t} g(x,t)\,dt.\end{aligned} $$

(U.2)

Notice that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta G(x) &\displaystyle &\displaystyle = \int_0^{+\infty} e^{-t} \Delta g(x,t)\,dt =\int_0^{+\infty} e^{-t} \partial_t g(x,t)\,dt\\ &\displaystyle &\displaystyle =\int_0^{+\infty} \Big(\partial_t(e^{-t} g(x,t))+e^{-t} g(x,t)\Big)\,dt \\ &\displaystyle &\displaystyle = -\delta_0(x)+ \int_0^{+\infty} e^{-t} g(x,t)\,dt = -\delta_0(x)+G(x),\end{array} \end{aligned} $$

hence G, as defined in (U.2) solves (U.1).

Notice also that G is positive and bounded, due to (U.2). We also claim that

$$\displaystyle \begin{aligned} {\mbox{for any}~x\in\mathbb{R}^n\setminus B_1,\mbox{ it holds that }G(x)\leqslant C e^{-c|x|},} \end{aligned} $$

(U.3)

for some c, C > 0. To this end, let us fix $x\in \mathbb {R}^n\setminus B_1$ and distinguish two regimes. If t ∈ [0, |x|], we have that $\frac {|x|{ }^{2}}{t}\geqslant |x|$ and thus

$$\displaystyle \begin{aligned}g(x,t)\leqslant {\frac {1}{(4\pi t)^{n/2}}}e^{-\frac{|x|{}^{2}}{8t}}e^{-\frac{|x|}8}.\end{aligned}$$

Consequently, using the substitution $\rho :=\frac {|x|{ }^{2}}{8t}$,

$$\displaystyle \begin{aligned} \int_0^{|x|} e^{-t} g(x,t)\,dt &\leqslant \int_0^{|x|} {\frac {1}{(4\pi t)^{n/2}}}e^{-\frac{|x|{}^{2}}{8t}}e^{-\frac{|x|}8}\,dt\\ &= \int_{|x|/8}^{+\infty}\frac{C\rho^{n/2}}{|x|{}^n}\,e^{-\rho} \,e^{-\frac{|x|}8}\,\frac{|x|{}^2\,d\rho}{\rho^2}\leqslant C|x|\,e^{-\frac{|x|}8}, \end{aligned} $$

(U.4)

for some C > 0 possibly varying from line to line. Furthermore

$$\displaystyle \begin{aligned}\int_{|x|}^{+\infty} e^{-t} g(x,t)\,dt\leqslant \int_{|x|}^{+\infty} e^{-\frac{|x|}2}\,e^{-\frac{t}2} g(x,t)\,dt\leqslant e^{-\frac{|x|}2}\int_1^{+\infty} {\frac {e^{-\frac{t}2}}{(4\pi t)^{n/2}}}\,dt\leqslant C\,e^{-\frac{|x|}2},\end{aligned}$$

for some C > 0. This and (U.4) give that

$$\displaystyle \begin{aligned}\int_0^{+\infty} e^{-t} g(x,t)\,dt\leqslant C|x|\,e^{-\frac{|x|}8},\end{aligned}$$

up to renaming C, which implies (U.3) in view of (U.2).

Now we compute the Laplace Transform of t ^s: namely, by (2.59),

$$\displaystyle \begin{aligned} {\mathcal{L}}(t^s)(\omega)= \int_0^{+\infty} t^s e^{-\omega t}\,dt=\omega^{-1-s} \int_0^{+\infty} \tau^s e^{-\tau}\,d\tau=C\omega^{-1-s}. \end{aligned} $$

(U.5)

We compare this result with the Laplace Transform of the mean squared displacement related to the diffusion operator in (2.62). For this, we take u to be as in (2.62) and, in the light of (2.42), we consider the function

$$\displaystyle \begin{aligned} v(\omega):={\mathcal{L}}\left( \int_{\mathbb{R}^n} |x|{}^2\, u(x,t)\,dx \right)(\omega)=\int_{\mathbb{R}^n} |x|{}^2\, {\mathcal{L}} u(x,\omega)\,dx .\end{aligned} $$

(U.6)

In addition, by taking the Laplace Transform (in the variable t, for a fixed $x\in \mathbb {R}^n$) of the equation in (2.62), making use of (2.60) we find that

$$\displaystyle \begin{aligned} \omega^s {\mathcal{L}} u(x,\omega)-\omega^{s-1} \delta_0(x)=\Delta {\mathcal{L}} u(x,\omega).\end{aligned} $$

(U.7)

Now, we let

$$\displaystyle \begin{aligned} W(x,\omega):= \omega^{1-\frac{sn}2} \,{\mathcal{L}} u(\omega^{-s/2}x,\omega).\end{aligned} $$

(U.8)

From (U.7), we have that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \Delta W(x,\omega)&\displaystyle =&\displaystyle \omega^{1-\frac{sn}2} \,\omega^{-s}\,\Delta{\mathcal{L}} u(\omega^{-s/2}x,\omega) \\&\displaystyle =&\displaystyle \omega^{1-\frac{sn}2} \,\omega^{-s}\,\Big( \omega^s {\mathcal{L}} u(\omega^{-s/2} x,\omega)-\omega^{s-1} \delta_0(\omega^{-s/2} x) \Big)\\ &\displaystyle =&\displaystyle W(x,\omega)-\omega^{-\frac{sn}2}\delta_0(\omega^{-s/2} x)\\ &\displaystyle =&\displaystyle W(x,\omega)-\delta_0(x), \end{array} \end{aligned} $$

and so, comparing with (U.1), we have that W(x, ω) = G(x).

Accordingly, by (U.8),

$$\displaystyle \begin{aligned}{\mathcal{L}} u(x,\omega)= \omega^{\frac{sn}2-1} \,W(\omega^{s/2}x,\,\omega)= \omega^{\frac{sn}2-1} \,G(\omega^{s/2}x). \end{aligned}$$

We insert this information into (U.6) and we conclude that

$$\displaystyle \begin{aligned}v(\omega)=\omega^{\frac{sn}2-1} \, \int_{\mathbb{R}^n} |x|{}^2\, G(\omega^{s/2}x)\,dx= \omega^{-1-s} \, \int_{\mathbb{R}^n} |y|{}^2\, G(y)\,dy. \end{aligned}$$

We remark that the latter integral is finite, thanks to (U.3), hence we can write that

$$\displaystyle \begin{aligned}v(\omega)=C\omega^{-1-s},\end{aligned}$$

for some C > 0.

Therefore, we can compare this result with (U.5) and use the inverse Laplace Transform to obtain that the mean squared displacement in this case is proportional to t ^s, as desired.

Appendix V: Memory Effects of Caputo Type

It is interesting to observe that the Caputo derivative models a simple memory effect that the classical derivative cannot comprise. For instance, integrating a classical derivative of a function u with u(0) = 0, one obtains the original function “independently on the past”, namely if we set

$$\displaystyle \begin{aligned} M_u(t):=\int_0^t \dot u(\vartheta)\,d\vartheta , \end{aligned} $$

(V.1)

we just obtain in this case that M _u(t) = u(t) − u(0) = u(t). On the other hand, an expression as in (V.1) which takes into account the Caputo derivative does “remember the past” and takes into account the preceding events in such a way that recent events “weight” more than far away ones. To see this phenomenon, we can modify (V.1) by defining, for every s ∈ (0, 1),

$$\displaystyle \begin{aligned} M_u^s(t):= \int_0^t \partial^s_{C,t} u(\vartheta)\,d\vartheta . \end{aligned} $$

(V.2)

To detect the memory effect, for the sake of concreteness, we take a large time $t:=N\in \mathbb {N}$ and we suppose that the function u is constant on unit intervals, that is u = u _k in [k − 1, k), for each k ∈{1, …, N}, with $u_k\in \mathbb {R}$, and u(0) = u ₁ = 0. We see that $M_u^s$ in this case does not produce just the final outcome u _N, but a weighted average of the form

$$\displaystyle \begin{aligned} M_u^s(N)=\sum_{k=0}^{N-1} c_k\, u_{N-k}, \qquad{\mbox{with }c_j>0\mbox{ decreasing and }}c_j\simeq\frac 1{j^s} {\mbox{ for large }j.} \end{aligned} $$

(V.3)

To check this, we notice that, for all τ ∈ (0, N),

$$\displaystyle \begin{aligned}\dot u(\tau)=\sum_{k=2}^N (u_{k}-u_{k-1})\delta_{k-1}(\tau),\end{aligned}$$

and hence we exploit (2.56) and (V.2) to see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} M_u^s(N)&\displaystyle =&\displaystyle \int_0^N \left[ \int_0^\vartheta \frac{\dot u (\tau) }{(\vartheta-\tau)^{s}} \, d\tau\right] \,d\vartheta \\ &\displaystyle =&\displaystyle \sum_{k=2}^N \int_0^N\left[ \int_0^\vartheta (u_{k}-u_{k-1})\delta_{k-1}(\tau) \frac{d\tau}{(\vartheta-\tau)^{s}}\right] \,d\vartheta\\ &\displaystyle =&\displaystyle \sum_{k=2}^N \int_{k-1}^N \frac{(u_{k}-u_{k-1})}{(\vartheta-k+1)^{s}} \,d\vartheta \\ &\displaystyle =&\displaystyle \sum_{k=2}^N u_{k} \int_{k-1}^N \frac{d\vartheta}{(\vartheta-k+1)^{s}} - \sum_{k=2}^N u_{k-1} \int_{k-1}^N \frac{d\vartheta}{(\vartheta-k+1)^{s}} \\&\displaystyle =&\displaystyle \sum_{k=2}^N u_{k} \int_{k-1}^N \frac{d\vartheta}{(\vartheta-k+1)^{s}} - \sum_{k=1}^{N-1} u_{k} \int_k^N \frac{d\vartheta}{(\vartheta-k)^{s}}\\ &\displaystyle =&\displaystyle \sum_{k=1}^N u_{k} \left[ \int_{k-1}^N \frac{d\vartheta}{(\vartheta-k+1)^{s}} - \int_k^N \frac{d\vartheta}{(\vartheta-k)^{s}} \right]\\ &\displaystyle =&\displaystyle \sum_{k=1}^N u_{k} \,\frac{(N-k+1)^{1-s}-(N-k)^{1-s}}{1-s}\\ &\displaystyle =&\displaystyle \sum_{k=2}^N c_{N-k} \,u_{k}\\ &\displaystyle =&\displaystyle \sum_{k=0}^{N-2} c_{k} \,u_{N-k} ,\end{array} \end{aligned} $$

with

$$\displaystyle \begin{aligned} \begin{array}{rcl} c_{j}&\displaystyle :=&\displaystyle \frac{(j+1)^{1-s}-j^{1-s}}{1-s} \,.\end{array} \end{aligned} $$

This completes the proof of the memory effect claimed in (V.3).

Appendix W: Proof of (3.7)

Since M is bounded and positive and u is bounded, it holds that

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^n\setminus B_1} \frac{| u(x)-u(x-y)| }{|M(x-y,y)\, y|{}^{n+2s}}\,dy\leqslant\,{\mathrm{const}}\, \int_{\mathbb{R}^n\setminus B_1} \frac{dy }{|y|{}^{n+2s}}\,dy\leqslant \frac{\,{\mathrm{const}}\,}{s}. \end{aligned} $$

(W.1)

Moreover, for y ∈ B ₁,

$$\displaystyle \begin{aligned} u(x-y)=u(x)-\nabla u(x)\cdot y +\frac 12 D^2 u(x)\,y\cdot y+O(|y|{}^3).\end{aligned} $$

(W.2)

To simplify the notation, we now fix $x\in \mathbb {R}^n$ and we define ${\mathcal {M}}(y):= M(x-y,y)$. Then, for y ∈ B ₁, we have that

$$\displaystyle \begin{aligned} M(x-y,y)\, y = {\mathcal{M}}(y)\,y= {\mathcal{M}}(0)\,y + \sum_{i=1}^n \partial_i {\mathcal{M}}(0) \, y\,y_i+O(|y|{}^3)\end{aligned}$$

and so

$$\displaystyle \begin{aligned} \begin{array}{rcl} |M(x-y,y)\, y|{}^2 &\displaystyle =&\displaystyle |{\mathcal{M}}(0)\,y|{}^2 + 2 \sum_{i=1}^n ({\mathcal{M}}(0)\,y)\cdot(\partial_i {\mathcal{M}}(0) \, y)\,y_i +O(|y|{}^4).\end{array} \end{aligned} $$

Consequently, since ${\mathcal {M}}(0)=M(x,0)$ is non-degenerate, we can write

$$\displaystyle \begin{aligned}{\mathcal{E}}(y):= 2 \sum_{i=1}^n ({\mathcal{M}}(0)\,y)\cdot(\partial_i {\mathcal{M}}(0) \, y)\,y_i = O(|y|{}^3)\end{aligned}$$

and

$$\displaystyle \begin{aligned} \begin{aligned} & |M(x-y,y)\, y|{}^{-n-2s} \\ =\;& \left( |{\mathcal{M}}(0)\,y|{}^2 + {\mathcal{E}}(y) +O(|y|{}^4) \right)^{-\frac{n+2s}2} \\ =\;& |{\mathcal{M}}(0)\,y|{}^{-n-2s} \left( 1 + |{\mathcal{M}}(0)\,y|{}^{-2}\,{\mathcal{E}}(y) +O(|y|{}^2) \right)^{-\frac{n+2s}2} \\ =\;& |{\mathcal{M}}(0)\,y|{}^{-n-2s} \left( 1 -\frac{n+2s}2\, |{\mathcal{M}}(0)\,y|{}^{-2}\,{\mathcal{E}}(y) +O(|y|{}^2) \right) \\ =\;& |{\mathcal{M}}(0)\,y|{}^{-n-2s} -\frac{n+2s}2\, |{\mathcal{M}}(0)\,y|{}^{-n-2s-2}\,{\mathcal{E}}(y) +O(|y|{}^{2-n-2s}) . \end{aligned}\end{aligned} $$

(W.3)

Hence (for smooth and bounded functions u, and y ∈ B ₁) we obtain that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{ u(x)-u(x-y) }{|M(x-y,y)\,y|{}^{n+2s}}\\ &\displaystyle =&\displaystyle \frac{ u(x)-u(x-y) }{|{\mathcal{M}}(0)\,y|{}^{n+2s} } -\frac{n+2s}2\, \frac{ \big( u(x)-u(x-y)\big)\,{\mathcal{E}}(y) }{ |{\mathcal{M}}(0)\,y|{}^{n+2s+2}} +O(|y|{}^{3-n-2s}). \end{array} \end{aligned} $$

Thus, since the map $y\mapsto \frac {\nabla u(x)\cdot y}{|{\mathcal {M}}(0)\,y|{ }^{n+2s} }$ is odd, recalling (W.2) we conclude that

(W.4)

Now we observe that, for any α⩾0,

$$\displaystyle \begin{aligned} &{\mbox{if }\varphi\mbox{ is positively homogeneous of degree }2+\alpha\mbox{ and}~T\in{\text{Mat}}(n\times n),\mbox{ then}} \\ & (1-s)\,\int_{B_1} \frac{\varphi(y)}{|Ty|{}^{n+2s+\alpha}}\,dy= \frac 12\,\int_{S^{n-1}} \frac{\varphi(\omega)}{|T\omega|{}^{n+2s+\alpha}}\,d{\mathcal{H}}^{n-1}_{\omega}. \end{aligned} $$

(W.5)

Indeed, using polar coordinates and the fact that φ(ρω) = ρ ^2+α φ(ω), for any ρ⩾0 and ω ∈ S ⁿ⁻¹, thanks to the homogeneity, we see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_{B_1} \frac{\varphi(y)}{|Ty|{}^{n+2s+\alpha}}\,dy= \iint_{(0,1)\times S^{n-1}} \frac{\rho^{n-1}\,\varphi(\rho\omega)}{\rho^{n+2s+\alpha}\, |T\omega|{}^{n+2s+\alpha}}\,d\rho\,d{\mathcal{H}}^{n-1}_{\omega}\\ &\displaystyle &\displaystyle \qquad= \iint_{(0,1)\times S^{n-1}} \frac{\rho^{1-2s}\varphi(\omega)}{ |T\omega|{}^{n+2s+\alpha}}\,d\rho\,d{\mathcal{H}}^{n-1}_{\omega}= \frac{1}{2(1-s)} \int_{S^{n-1}} \frac{\varphi(\omega)}{ |T\omega|{}^{n+2s+\alpha}}\,d{\mathcal{H}}^{n-1}_{\omega}, \end{array} \end{aligned} $$

which implies (W.5).

Using (W.5) (with α := 0 and α := 2), we obtain that

and

Thanks to this, (W.1) and (W.4), we find that

(W.6)

with

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle a_{ij}(x):= \frac 14\,\int_{S^{n-1}} \frac{\omega_i\, \omega_j}{ |{\mathcal{M}}(0)\,\omega|{}^{n+2} }\,d{\mathcal{H}}^{n-1}_{\omega} =\frac 14\,\int_{S^{n-1}} \frac{\omega_i\, \omega_j}{ |M(x,0)\,\omega|{}^{n+2} }\,d{\mathcal{H}}^{n-1}_{\omega} \\{\mbox{and }}&\displaystyle &\displaystyle b_j(x):=\frac{n+2}{2}\,\sum_{i=1}^n \int_{S^{n-1}} \frac{ \omega_i\,\omega_j \,\big( ({\mathcal{M}}(0)\,\omega)\cdot(\partial_i {\mathcal{M}}(0) \, \omega)\big) }{ |{\mathcal{M}}(0)\,\omega|{}^{n+4}} \,d{\mathcal{H}}^{n-1}_{\omega} .\end{array} \end{aligned} $$

We observe that

$$\displaystyle \begin{aligned} b_j=\sum_{i=1}^n \partial_i a_{ij}(x). \end{aligned} $$

(W.7)

To check this, we first compute that

$$\displaystyle \begin{aligned} \begin{aligned} \sum_{i=1}^n \partial_i a_{ij}(x) \;&=\frac 14\,\sum_{i=1}^n \partial_{x_i} \left( \int_{S^{n-1}} \frac{\omega_i\, \omega_j}{ |M(x,0)\,\omega|{}^{n+2} }\,d{\mathcal{H}}^{n-1}_{\omega} \right)\\ &=-\frac{n+2}{4}\,\sum_{i=1}^n \int_{S^{n-1}} \frac{\omega_i\, \omega_j\,\big( (M(x,0)\,\omega)\cdot(\partial_{x_i} M(x,0)\,\omega)\big)}{ |M(x,0)\,\omega|{}^{n+4} }\,d{\mathcal{H}}^{n-1}_{\omega} .\end{aligned} \end{aligned} $$

(W.8)

Now, we write a Taylor expansion of M(x, y) in the variable y of the form

$$\displaystyle \begin{aligned}M_{\ell m}(x,y)=A_{\ell m}(x) + B_{\ell m}(x)\cdot y+O(y^2),\end{aligned}$$

for some $A_{\ell m}:\mathbb {R}^n\to \mathbb {R}$ and $B_{\ell m}:\mathbb {R}^n\to \mathbb {R}^n$. We notice that

$$\displaystyle \begin{aligned} \partial_{x_i} M_{\ell m}(x,0)=\partial_{x_i} A_{\ell m}(x). \end{aligned} $$

(W.9)

Also,

$$\displaystyle \begin{aligned} \begin{aligned} \partial_{i} {\mathcal{M}}_{\ell m}(0)\,&=\lim_{y\to0} \partial_{y_i}\big( M_{\ell m}(x-y,y) \big) \\ &=\lim_{y\to0} \partial_{y_i}\big( A_{\ell m}(x-y) + B_{\ell m}(x-y)\cdot y+O(y^2) \big)\\&= -\partial_{x_i}A_{\ell m}(x) + B_{\ell m}(x)\cdot e_i.\end{aligned} \end{aligned} $$

(W.10)

Furthermore, we use the structural assumption (3.6), and we see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle A_{\ell m}(x) - B_{\ell m}(x)\cdot y+O(y^2) =M(x,-y)\\ &\displaystyle &\displaystyle \qquad=M(x-y,y)= A_{\ell m}(x-y) + B_{\ell m}(x-y)\cdot y+O(y^2) \\ &\displaystyle &\displaystyle \qquad=A_{\ell m}(x) -\nabla A_{\ell m}(x)\cdot y + B_{\ell m}(x)\cdot y+O(y^2). \end{array} \end{aligned} $$

Comparing the linear terms, this gives that

$$\displaystyle \begin{aligned}2B_{\ell m}(x)=\nabla A_{\ell m}(x).\end{aligned}$$

This and (W.10) imply that

$$\displaystyle \begin{aligned} \partial_{i} {\mathcal{M}}_{\ell m}(0)= -\partial_{x_i}A_{\ell m}(x) + \frac 12\nabla A_{\ell m}(x)\cdot e_i =-\frac 12 \partial_{x_i}A_{\ell m}(x) .\end{aligned}$$

Comparing this with (W.9), we see that

$$\displaystyle \begin{aligned} \partial_{x_i} M_{\ell m}(x,0)=-2\partial_{i} {\mathcal{M}}_{\ell m}(0). \end{aligned}$$

So, we insert this information into (W.8) and we conclude that

$$\displaystyle \begin{aligned} \sum_{i=1}^n \partial_i a_{ij}(x) =\frac{n+2}{2}\,\sum_{i=1}^n \int_{S^{n-1}} \frac{\omega_i\, \omega_j\,\big( (M(x,0)\,\omega)\cdot(\partial_{i} {\mathcal{M}}(0)\,\omega)\big)}{ |M(x,0)\,\omega|{}^{n+4} }\,d{\mathcal{H}}^{n-1}_{\omega}.\end{aligned}$$

This establishes (W.7), as desired.

Then, plugging (W.7) into (W.6), we obtain the equation in divergence form¹² which was claimed in (3.7).

Fig. 13

A nice representation of nonlocal effects

Appendix X: Proof of (3.12)

First we observe that

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^n\setminus B_1} \frac{ |u(x)-u(x-y)| }{|M(x,y)\,y|{}^{n+2s}}\,dy\leqslant \,{\mathrm{const}}\, \int_{\mathbb{R}^n\setminus B_1} \frac{ dy }{|y|{}^{n+2s}}\leqslant \frac{\,{\mathrm{const}}\,}{s}. \end{aligned} $$

(X.1)

Furthermore, for y ∈ B ₁,

$$\displaystyle \begin{aligned} M(x,y)\,y = M(x,0)\,y+O(|y|{}^2).\end{aligned} $$

Consequently,

$$\displaystyle \begin{aligned} |M(x,y)\,y|{}^2=|M(x,0)\,y|{}^2+O(|y|{}^3)\end{aligned} $$

and so, from the non-degeneracy of M(⋅, ⋅),

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle |M(x,y)\,y|{}^{-n-2s} =\big( |M(x,0)\,y|{}^2+O(|y|{}^3)\big)^{-\frac{n+2s}2} \\&\displaystyle &\displaystyle \qquad=|M(x,0)\,y|{}^{-n-2s}\big( 1+O(|y|)\big)^{-\frac{n+2s}2}= |M(x,0)\,y|{}^{-n-2s}\big( 1-O(|y|)\big). \vspace{-2pt}\end{array} \end{aligned} $$

Using this and the expansion in (W.2), we see that, for y ∈ B ₁,

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{ u(x)-u(x-y)-\nabla u(x)\cdot y }{|M(x,y)\,y|{}^{n+2s}}\\ &\displaystyle =&\displaystyle |M(x,0)\,y|{}^{-n-2s}\big( 1-O(|y|)\big)\left( -\frac 12 D^2 u(x)\,y\cdot y+O(|y|{}^3)\right)\\ &\displaystyle =&\displaystyle |M(x,0)\,y|{}^{-n-2s}\left( -\frac 12 D^2 u(x)\,y\cdot y+O(|y|{}^3)\right). \end{array} \end{aligned} $$

Thus, since, in the light of (3.11), we know that the map $y\mapsto \frac { \nabla u(x)\cdot y }{|M(x,y)\,y|{ }^{n+2s}}$ is odd, we can write that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{B_1} \frac{ u(x)-u(x-y) }{|M(x,y)\,y|{}^{n+2s}}\,dy&\displaystyle =&\displaystyle \int_{B_1} \frac{ u(x)-u(x-y)-\nabla u(x)\cdot y }{|M(x,y)\,y|{}^{n+2s}}\,dy\\ &\displaystyle =&\displaystyle -\frac 12 \int_{B_1}\frac{D^2 u(x)\,y\cdot y}{ |M(x,0)\,y|{}^{n+2s} }\,dy +\frac{O(1)}{3-2s}\\ &\displaystyle =&\displaystyle -\frac{\,{\mathrm{const}}\,}{1-s} \int_{ S^{n-1}}\frac{ D^2 u(x)\,\omega\cdot\omega}{ |M(x,0)\,\omega|{}^{n+2s} }\,d{\mathcal{H}}^{n-1}_{\omega} +\frac{O(1)}{3-2s} ,\end{array} \end{aligned} $$

where the last identity follows by using (W.5) (with α := 0). From this and (X.1) we obtain that

which gives (3.12).