# The Fractional Malmheden Theorem

Serena Dipierro  Giovanni Giacomin and Enrico Valdinoci Department of
Mathematics and Statistics, University of Western Australia, 35 Stirling
Highway, WA6009 Crawley, Australia.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
SD and EV are members of INdAM and AustMS. Supported by the Australian
Laureate Fellowship FL190100081 ‘‘Minimal surfaces, free boundaries and
partial differential equations’’ and by the Australian Research Council DECRA
DE180100957 ‘‘PDEs, free boundaries and applications’’.

###### Abstract

We provide a fractional counterpart of the classical results by Schwarz and
Malmheden on harmonic functions. From that we obtain a representation formula
for $s$-harmonic functions as a linear superposition of weighted classical
harmonic functions which also entails a new proof of the fractional Harnack
inequality. This proof also leads to optimal constants for the fractional
Harnack inequality in the ball.

To Neil, the Master of us all.

##  1 Introduction

In 1934, Harry William Malmheden [6] proved a simple algorithm to compute the
value of a harmonic function at a point of $B_{1}$, knowing its value on the
boundary.

The Malmheden Theorem makes use of two fundamental geometric ingredients:

  1. 1.

the notion of affine interpolation between the values of a given function at
two different points of the space,

  2. 2.

the projections of a point inside a ball to the boundary in a given direction.

Hence, to state the Malmheden Theorem explicitly, we now formalize these two
notions into a precise mathematical setting. We start by introducing a
notation for the affine interpolation between the values of some given
function. That is, given a set $K\subseteq\mathbb{R}^{n}$, a function
$f:K\to\mathbb{R}$, two distinct points $a$, $b\in K$, and a point $x$ on the
segment $L$ joining $a$ and $b$, we define $\mathcal{L}_{f}^{a,b}(x)$ as the
affine function on $L$ such that $\mathcal{L}_{f}^{a,b}(a)=f(a)$ and
$\mathcal{L}_{f}^{a,b}(b)=f(b)$.

Of course, one can write this affine function explicitly by using the analytic
expression

$\mathcal{L}_{f}^{a,b}(x)=\frac{(x-a)\cdot e}{|b-a|}\,f(b)+\frac{(b-x)\cdot
e}{|b-a|}\,f(a),\qquad{\mbox{where }}e:=\frac{b-a}{|b-a|}.$ (1.1)

One can call $\mathcal{L}_{f}^{a,b}(x)$ the “affine function of $f$ with
extrema $a$ and $b$ evaluated at the point $x$”.

Figure 1: The projections $Q_{\pm}^{x}(e)$ used in the Malmheden Theorem.

Now we discuss the notation regarding the projections of a point inside a ball
to the boundary of the ball in a given direction. For this, given a point
$x\in B_{1}$ and a direction $e\in\partial B_{1}$ we consider the
intersections $Q_{+}^{x}(e)$ and $Q_{-}^{x}(e)$ of $\partial B_{1}$ with the
straight line passing through $x\in B_{1}$ with direction $e$, with the
convention that $Q_{+}^{x}(e)-Q_{-}^{x}(e)$ has the same orientation of $e$,
see Figure 1.

Clearly, from the analytic point of view, one can write explicitly these
projections in the form

$\begin{split}&Q_{+}^{x}(e):=x+r_{+}^{x}(e)e\\\ {\mbox{and
}}\qquad&Q_{-}^{x}(e):=x+r_{-}^{x}(e)e,\end{split}$ (1.2)

where

$\begin{split}&r_{+}^{x}(e)=-x\cdot e+\sqrt{(x\cdot e)^{2}-|x|^{2}+1}\\\
{\mbox{and }}\qquad&r_{-}^{x}(e)=-x\cdot e-\sqrt{(x\cdot
e)^{2}-|x|^{2}+1}.\end{split}$ (1.3)

We note from equations (1.2) and (1.3) that $Q_{\pm}^{x}(e)$ are continuous
functions in $(x,e)\in B_{1}\times\partial B_{1}$. Moreover,

$\lim_{x\to 0}Q_{\pm}^{x}(e)=\pm e$ (1.4)

for each $e\in\partial B_{1}$. This tells us that the maps $Q_{\pm}^{x}$
simply reduce to $\pm id_{\partial B_{1}}$ when $x=0$.

Given a boundary datum $f:\partial B_{1}\to\mathbb{R}$, the core of the
Malmheden Theorem is thus to consider, for every point $x\in B_{1}$ and every
direction $e\in\partial B_{1}$, the affine function of $f$ with extrema
$Q_{-}^{x}(e)$ and $Q_{+}^{x}(e)$, namely the function

$\mathcal{L}_{f}^{Q_{-}^{x}(e),Q_{+}^{x}(e)}(x)$ (1.5)

and then to average in all directions $e$.

The remarkable result by Malmheden is that this averaging procedure of linear
interpolations produces precisely the solution of the classical Dirichlet
problem in $B_{1}$ with boundary datum $f$, according to the following
classical statement (see [6]):

######  Theorem 1.1 (Malmheden Theorem).

Let $n\geqslant 2$ and $f:\partial B_{1}\rightarrow\mathbb{R}$ be continuous.
Then

$u_{f}(x):=\fint_{\partial
B_{1}}\mathcal{L}_{f}^{Q_{-}^{x}(e),Q_{+}^{x}(e)}(x)\,dH_{e}^{n-1}$ (1.6)

is the harmonic function in $B_{1}$ with boundary datum $f$.

As usual, here above and in the rest of this paper, we denoted by $H^{n-1}$
the $(n-1)$-Hausdorff measure (hence, the integral on the right hand side of
(1.6) is simply the spherical integral along $\partial B_{1}$; we kept the
explicit notation with the Hausdorff measure to have a typographical evidence
of the surface integrals, to be distinguished by the classical volume ones).

We remark that Theorem 1.1 contains the Mean Value Theorem for harmonic
functions as a particular case: indeed, in light of (1.1) and (1.4), if we
take $x:=0$ then (1.6) reduces to

$u_{f}(0)=\fint_{\partial
B_{1}}\mathcal{L}_{f}^{-e,e}(0)\,dH_{e}^{n-1}=\fint_{\partial
B_{1}}\left(\frac{f(e)}{2}+\frac{f(-e)}{2}\right)\,dH_{e}^{n-1}=\fint_{\partial
B_{1}}f(e)\,dH_{e}^{n-1},$ (1.7)

which is the content of the Mean Value Theorem.

We also stress that an elegant result such as Theorem 1.1 is specific for
balls and cannot be extended in general to other domains, as pointed out in
[5].

Interestingly, Theorem 1.1 contains as a particular case a classical result
due to Hermann Amandus Schwarz [MR0392470] about the Dirichlet problem in the
plane and related to conformal mappings in the complex framework.

Figure 2: The reflection $Q^{x}(\omega)$ used in the Schwarz Theorem.

To state Schwarz result it is convenient to introduce the reflection of a
point $\omega\in\partial B_{1}$ through a point $x\in B_{1}$, see Figure 2.
More precisely, given $x\in B_{1}$ and $\omega\in\partial B_{1}$ we define

$Q^{x}(\omega):=\omega-2\frac{(x-\omega)\cdot\omega}{|x-\omega|^{2}}(x-\omega).$
(1.8)

Comparing with (1.2), one sees that if $e:=\frac{\omega-x}{|\omega-x|}$ then
$Q_{+}^{x}(e)=\omega$ and $Q_{-}^{x}(e)=Q^{x}(\omega)$.

In this setting, the result by Schwarz is that the average of the boundary
datum composed with the above reflection returns the solution of the Dirichlet
problem in the ball. More explicitly:

######  Theorem 1.2 (Schwarz Theorem).

Let $n=2$ and $f:\partial B_{1}\rightarrow\mathbb{R}$ be continuous. Then

$u_{f}(x):=\fint_{\partial B_{1}}f(Q^{x}(e))\,dH_{e}^{n-1}$ (1.9)

is the harmonic function in $B_{1}$ with external datum $f$.

Theorem 1.2 can be proved in several ways using either complex or real
analysis (see e.g. [zbMATH02701995, MR86885, MR1272823]), but it is also a
direct consequence of Theorem 1.1, see e.g. [2] for a detailed presentation of
this classical argument.

######  Example 1.3.

A very neat application of Theorem 1.2 (see e.g. [MR1446490]) consists in the
determination of the stationary temperature $u$ at a point $x$ in a plate (say
$B_{1}$) when the temperature along the boundary of the plate is $1$ along
some arc $\Sigma$ and $0$ outside. In this case, the reflection in (1.8) sends
$\Sigma$ into an arc $\Sigma^{\prime}$ (the symmetric of $\Sigma$ through $x$,
see Figure 3) and it therefore follows from Theorem 1.2 that

$u(x)=\frac{|\Sigma^{\prime}|}{2\pi},$

where $|\Sigma^{\prime}|$ is the length of the arc $\Sigma^{\prime}$, thus
providing an elementary geometric construction to solve a problem of physical
relevance.

Figure 3: The geometric construction to detect the temperature of a plate at
the point $x$.

The objective of this paper is to obtain a fractional counterpart for the
Malmheden and Schwarz Theorems.

We will thus replace the notion of harmonic functions in $B_{1}$ with that of
$s$-harmonic functions, namely functions whose fractional Laplacian vanishes
in $B_{1}$, that is, for all $x\in B_{1}$,

$\int_{\mathbb{R}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+2s}}\,dy=0,$ (1.10)

where the integral above is intended in the principal value sense. Here above
and throughout the paper the fractional parameter $s\in(0,1)$.

Rather than a boundary value along $\partial B_{1}$, as usual in the nonlocal
setting, we complement (1.10) with an external condition of the type $u=f$ in
$\mathbb{R}^{n}\setminus B_{1}$.

We recall that in general $s$-harmonic functions behave way more wildly that
their classical counterparts, see e.g. [MR3626547]. Therefore, in principle
one cannot easily expect that a “simple formulation” such as the one in
Theorems 1.1 and 1.2 accounts for all the complex situations arising in the
fractional setting.

However, we will prove that a counterpart of Theorems 1.1 and 1.2 carries over
to the case of the fractional Laplacian, considering the following structural
modifications:

  1. 1.

the classical spherical averages are replaced by suitable weighted averages on
spheres of radius larger than $1$,

  2. 2.

the geometric transformations in (1.2) and (1.8) are scaled in dependence of
the radius of each of these spheres.

To clarify these points, and thus reconsider (1.5) in a nonlocal setting,
given $\rho>1$ and $f:\mathbb{R}^{n}\setminus B_{1}\to\mathbb{R}$, for all
$x\in\partial B_{1}$ we define

$f_{\rho}(x):=f(\rho x).$ (1.11)

Hence, in the notation of (1.1), we define

$\mathcal{L}_{f,e,\rho}(x):=\mathcal{L}_{f_{\rho}}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right).$
(1.12)

Notice that when $\rho=1$ the above setting reduces to (1.5), otherwise one is
considering here a similar framework but for a rescaled version of the
function $f$ and rescaled points.

To detect the long-range effect of the fractional Laplacian, it is also useful
to consider the kernel

$B_{1}\times(1,+\infty)\ni(x,\rho)\longmapsto{\mathcal{E}}(x,\rho):=c(n,s)\,\frac{\rho\,(1-|x|^{2})^{s}}{(\rho^{2}-1)^{s}\,(\rho^{2}-|x|^{2})},$
(1.13)

where

$c(n,s):=\frac{\displaystyle\Gamma\left(\frac{n}{2}\right)\,\sin(\pi
s)}{\pi^{\frac{n}{2}+1}}.$ (1.14)

For our purposes, the kernel ${\mathcal{E}}$ will play the role of a suitable
spherical average111More precisely, the intuition behind ${\mathcal{E}}$ is
that it satisfies, for each $x\in B_{1}$ and $\rho\in(1,+\infty)$,
${\mathcal{E}}(x,\rho)=\rho^{n-1}\fint_{\partial
B_{\rho}}P(x,y)\,dH^{n-1}_{y},$ where $P$ is the Fractional Poisson Kernel.
This relation can be obtained as a consequence of our Fractional Malmheden
Theorem. Though our approach does not pass explicitly through this identity,
for the sake of completeness we provide an independent proof in Appendix A.1.
of a Fractional Poisson Kernel and the constant $c(n,s)$ is merely needed for
normalization purposes.

We also define the space

$L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{1}):=\left\\{f:\mathbb{R}^{n}\to\mathbb{R}{\mbox{ measurable
}}\,:\,\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|f(x)|}{|x|^{n+2s}}\,dx<\infty\right\\}.$ (1.15)

With this, we can state the main result of this paper as follows:

######  Theorem 1.4 (Fractional Malmheden Theorem).

Let $n\geqslant 2$, $s\in(0,1)$, $R>1$ and $f\in L^{\infty}(B_{R}\setminus
B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$.

Then, the unique solution (up to a zero measure subset of
$\mathbb{R}^{n}\setminus B_{1}$) to the problem

$\begin{cases}\begin{split}(-\Delta)^{s}u&=0\;\;\textrm{in}\;\;B_{1},\\\
u&=f\;\;\textrm{in}\;\;\mathbb{R}^{n}\setminus B_{1}\end{split}\end{cases}$
(1.16)

can be written, for each $x\in B_{1}$, as

$u_{f}^{(s)}(x):=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}{\mathcal{E}}(x,\rho)\,\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho.$
(1.17)

As a fractional counterpart of the observation in (1.7), we point out that
Theorem 1.4 entails as a straightforward consequence the Mean Value Formula
for $s$-harmonis functions. Indeed, by the changes of variable
$e:=\omega/|\omega|$ and $y:=\rho\omega/|\omega|$,

$\displaystyle u_{f}^{(s)}(0)$ $\displaystyle=$
$\displaystyle\int_{1}^{\infty}\left(\int_{\partial
B_{1}}{\mathcal{E}}(0,\rho)\,\mathcal{L}_{f,e,\rho}(0)\,dH_{e}^{n-1}\right)\,d\rho$
$\displaystyle=$ $\displaystyle c(n,s)\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\frac{1}{\rho(\rho^{2}-1)^{s}}\,\mathcal{L}_{f_{\rho}}^{-e,e}(0)\,dH_{e}^{n-1}\right)\,d\rho$
$\displaystyle=$ $\displaystyle c(n,s)\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\frac{1}{\rho(\rho^{2}-1)^{s}}\,\left(\frac{f(\rho e)}{2}+\frac{f(-\rho
e)}{2}\right)\,dH_{e}^{n-1}\right)\,d\rho$ $\displaystyle=$ $\displaystyle
c(n,s)\int_{1}^{\infty}\left(\int_{\partial
B_{\rho}}\frac{1}{\rho^{n}(\rho^{2}-1)^{s}}\,\left(\frac{f(\omega)}{2}+\frac{f(-\omega)}{2}\right)\,dH_{\omega}^{n-1}\right)\,d\rho$
$\displaystyle=$ $\displaystyle c(n,s)\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{1}{|y|^{n}(|y|^{2}-1)^{s}}\,\left(\frac{f(y)}{2}+\frac{f(-y)}{2}\right)\,dy$
$\displaystyle=$ $\displaystyle c(n,s)\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{f(y)}{|y|^{n}(|y|^{2}-1)^{s}}\,dy,$

which is the Mean Value Formula for $s$-harmonis functions, see e.g. formula
(1.3) in [MR4149297].

We consider Theorem 1.4 as the natural fractional counterpart of Theorem 1.1
and we mention that indeed one can “recover” Theorem 1.1 in the limit as
$s\nearrow 1$, according to the following result:

######  Proposition 1.5.

Let $n\geqslant 2$, $s_{0}\in(0,1)$, $R>1$ and $f\in C(B_{R}\setminus
B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ for each $s\in(s_{0},1]$.
Then, for each $x\in B_{1}$, it holds that

$\lim_{s\nearrow 1}u_{f}^{(s)}(x)=u_{f}(x),$

where $u_{f}^{(s)}$ and $u_{f}$ are defined in (1.17) and (1.6), respectively.

As a straightforward consequence of the classical Malmheden Theorem (Theorem
1.1) and its fractional formulation Theorem (Theorem 1.4), we deduce the
following result.

######  Theorem 1.6 (An $s$-harmonic function is the superposition of
classical harmonic functions).

Let $n\geqslant 2$, $s\in(0,1)$ and $f\in C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. For each $\rho>1$ we define
${u}_{f_{\rho}}$ as the unique solution to the Dirichlet problem

$\begin{cases}\begin{split}\Delta u&=0\;\;\textrm{in}\;\;B_{1},\\\
u|_{\partial B_{1}}&=f_{\rho}|_{\partial B_{1}},\end{split}\end{cases}$ (1.18)

where $f_{\rho}$ is defined in (1.11).

Then the unique solution $u_{f}^{(s)}$ to (1.16) can be written as

$u_{f}^{(s)}(x)=|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho.$
(1.19)

The interest of Theorem 1.6 is that it allows us to write an $s$-harmonic
function in $B_{1}$ as a weighted integral of classical harmonic functions,
where the weight coincide with $\mathcal{E}(x,\rho)$. Besides being
interesting in itself, this result is very useful to deduce properties of
$s$-harmonic functions, as the Harnack inequality (see Section 4), starting
from their local counterpart.

As a matter of fact, as a consequence of Theorem 1.6 one obtains a new proof
of the Harnack inequality for $s$-harmonic functions in $B_{1}$. The result
goes as follows:

######  Theorem 1.7 (Harnack inequality).

Let $n\geqslant 2$, $s\in(0,1)$, $R>1$ and $u$ be non negative, $s$-harmonic
in $B_{1}$ and such that $u\in L^{\infty}(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$.

Then, for each $r\in(0,1)$ and $x\in B_{r}$,

$\frac{(1-r^{2})^{s}}{(1+r)^{n}}u(0)\leqslant
u(x)\leqslant\frac{(1-r^{2})^{s}}{(1-r)^{n}}u(0).$ (1.20)

The constants in (1.20) are optimal, and for $s\nearrow 1$ they converge to
the optimal constants of the classical Harnack inequality in $B_{r}$ for
harmonic functions in $B_{1}$.

For different proofs of the fractional Harnack inequality see [MR1941020, 8,
MR2817382] and the references therein.

Another consequence of Theorem 1.4 is the fractional version of Schwarz
result:

######  Theorem 1.8 (Fractional Schwarz Theorem).

Let $n=2$, $s\in(0,1)$, $R>1$ and $f\in L^{\infty}(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$. Then, the unique solution (up to a
zero measure subset of $\mathbb{R}^{2}\setminus B_{1}$) to the problem

$\begin{cases}\begin{split}(-\Delta)^{s}u&=0\;\;\textrm{in}\;\;B_{1},\\\
u&=f\;\;\textrm{in}\;\;\mathbb{R}^{2}\setminus B_{1}\end{split}\end{cases}$

can be written, for each $x\in B_{1}$, as

$u_{f}^{(s)}(x):=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}{\mathcal{E}}(x,\rho)\,f_{\rho}(Q^{x/\rho}(e))\,dH_{e}^{1}\right)\,d\rho.$
(1.21)

This is a fractional counterpart of Theorem 1.2, in the sense of Proposition
1.9 below. Proposition 1.9 is a straightforward consequence of Theorems 1.2
and 1.8 and Proposition 1.5.

######  Proposition 1.9.

Let $n=2$, $s_{0}\in(0,1)$, $R>1$ and $f\in C(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$ for each $s\in(s_{0},1]$. Then, for
each $x\in B_{1}$, it holds that

$\lim_{s\nearrow 1}u_{f}^{(s)}(x)=u_{f}(x)$

where $u_{f}^{(s)}$, $u_{f}$ are defined in (1.21) and (1.9), respectively.

######  Remark 1.10.

It is worth pointing out that from Theorem 1.4 we can evince the identity

$\int_{1}^{\infty}\mathcal{E}(x,\rho)\,d\rho=\frac{1}{|\partial B_{1}|},$
(1.22)

for each $x\in B_{1}$. Indeed, if we consider as external data $f=1$ in
$\mathbb{R}^{n}\setminus B_{1}$, then the unique solution to the problem
(1.16) is $u=1$ in $\mathbb{R}^{n}$. Therefore, according to (1.17) and the
fact that in this case the linear interpolation $\mathcal{L}_{1,e,\rho}(x)=1$
for each $x\in B_{1}$, we obtain that

$1=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{1,e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho=\int_{1}^{\infty}|\partial
B_{1}|\,\mathcal{E}(x,\rho)\,d\rho,$

which gives (1.22).

As an application of Theorem 1.8, we have:

######  Example 1.11.

Let $n=2$ and take an arc $\Sigma\subset\partial B_{1}$. Consider the function
defined on $\mathbb{R}^{2}\setminus B_{1}$ as

$\tilde{\chi}_{\Sigma}(y):=\begin{cases}\begin{split}1\qquad\textrm{if}\;\;&\displaystyle\frac{y}{|y|}\in\Sigma,\\\
0\qquad\textrm{if}\;\;&\displaystyle\frac{y}{|y|}\in\partial
B_{1}\setminus\Sigma.\end{split}\end{cases}$ (1.23)

It is clear that $\tilde{\chi}_{\Sigma}$ is positively homogeneous of degree
zero, and furthermore $\tilde{\chi}_{\Sigma}\in
L^{\infty}(\mathbb{R}^{2}\setminus B_{1})\subset
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$. Then by Theorem 1.8 we get that for
each $x\in B_{1}$

$u_{\tilde{\chi}_{\Sigma}}^{(s)}(x)=\int_{1}^{\infty}\mathcal{E}(x,\rho)|\Sigma_{x/\rho}^{\prime}|\,d\rho$
(1.24)

where $\Sigma_{x/\rho}^{\prime}$ is the projected arc of $\Sigma$ on $\partial
B_{1}$ through the focal point $x/\rho$, as constructed in Example 1.3. We
denoted with $|\Sigma_{x/\rho}^{\prime}|$ its length.

This gives a simple geometrical procedure to compute the solution of

$\begin{cases}\begin{split}(-\Delta)^{s}u&=0\;\;\textrm{in}\;\;B_{1},\\\
u&=\tilde{\chi}_{\Sigma}\;\;\textrm{in}\;\;\mathbb{R}^{2}\setminus
B_{1}\end{split}\end{cases}$

at a point $x$ of the two dimensional disc when the non local boundary
condition is given by (1.23). Note that as $\rho$ is getting larger, the
measure of $\Sigma_{x/\rho}^{\prime}$ reaches the one of $\Sigma$, or more
precisely

$\lim_{\rho\to\infty}|\Sigma_{x/\rho}^{\prime}|=|\Sigma|.$

If $x=0$, formula (1.24) boils down to

$u_{\tilde{\chi}_{\Sigma}}^{(s)}(0)=c(n,s)|\Sigma|\int_{1}^{\infty}\frac{1}{\rho(\rho^{2}-1)^{s}}\,d\rho=\frac{|\Sigma|}{2\pi},$
(1.25)

where we have applied identity (1.22). This example can be seen as the
fractional counterpart of Example 1.3.

This paper is organized as follows. In Section 2 we give some preliminary
results on the $s$-harmonic function written as a convolution with the
Fractional Poisson Kernel.

Section 3 is devoted to the proofs of the fractional Malmheden and Schwarz
results, that is Theorems 1.4 and 1.8, and of the convergence result in
Proposition 1.5.

In Section 4 we use these results to provide a proof of the well-known Harnack
inequality for $s$-harmonic functions under some regularity assumptions on the
external datum $f:\mathbb{R}^{n}\setminus B_{1}\rightarrow\mathbb{R}$, that is
we prove Theorem 1.7.

##  2 Preliminary results on the Fractional Poisson Kernel

In this section, we revisit the well-established result according to which
fractional harmonic functions can be represented as an integral of the datum
outside the domain against a suitable Poisson Kernel. For completeness, we
extend this result to the case in which the datum is not necessarily
continuous, so to be able to present the results of this paper in a suitable
generality. Notice that the extension to functions that are not necessarily
continuous is also useful for us to comprise situations as in Example 1.11.

The framework that we consider is the following. For $n\geqslant 2$ and
$s\in(0,1)$, we consider the space $L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$
as defined in (1.15). Given $f\in L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$,
we denote the norm on $L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ by

$\|f\|_{L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{1})}:=\int_{\mathbb{R}^{n}\setminus B_{1}}\frac{|f(x)|}{|x|^{n+2s}}\,dx.$

Furthermore, we define the following Fractional Poisson Kernel in the unit
ball

$P(x,y):=c(n,s)\left(\frac{1-|x|^{2}}{|y|^{2}-1}\right)^{s}\frac{1}{|x-y|^{n}}$
(2.1)

for $x\in B_{1}$ and $y\in\mathbb{R}^{n}\setminus B_{1}$, and $c(n,s)$ is the
normalizing constant in (1.14).

As customary, the role of the constant $c(n,s)$ is to normalize the Poisson
Kernel, namely we have that

$\int_{\mathbb{R}^{n}\setminus B_{1}}P(x,y)\,dy=1,$ (2.2)

see e.g. formula (1.14) and Lemma A.5 in [1].

We also remark that

$P(\cdot,y)\in C^{\infty}(B_{1})$

and, for every $\rho\in(0,1)$, $\alpha\in\mathbb{N}^{n}$ and
$y\in\mathbb{R}^{n}\setminus B_{1}$,

$\sup_{x\in
B_{\rho}}|D_{x}^{\alpha}P(x,y)|\leqslant\frac{C_{\rho}}{(|y|-1)^{s}\,|y|^{n+s+|\alpha|}},$
(2.3)

where $C_{\rho}>0$ depends only on $\rho$, $n$ and $s$ and, as usual, we have
denoted the length of the multi-index $\alpha$ as
$|\alpha|:=\alpha_{1}+\dots+\alpha_{n}$.

Then, we define

$u_{f}^{(s)}(x):=\begin{cases}\begin{split}\displaystyle\int_{\mathbb{R}^{n}\setminus
B_{1}}P(x,y)f(y)\,dy\;\;&\textrm{if}\;x\in B_{1},\\\
f(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&\textrm{if}\;x\in\mathbb{R}^{n}\setminus
B_{1},\end{split}\end{cases}$ (2.4)

and we have the following result on the representation of $s$-harmonic
functions:

######  Theorem 2.1.

Let $n\geqslant 2$, $s\in(0,1)$ and $f\in C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. Then the function in (2.4) is the
unique pointwise continuous solution to the problem

$\begin{cases}\begin{split}(-\Delta)^{s}u_{f}^{(s)}&=0\;\;\textrm{in}\;\;B_{1},\\\
u_{f}^{(s)}&=f\;\;\textrm{in}\;\;\mathbb{R}^{n}\setminus
B_{1}.\end{split}\end{cases}$ (2.5)

For a proof of Theorem 2.1 see e.g. Theorem 2.10 in [1].

We now generalize Theorem 2.1 by allowing external data that are not
necessarily continuous:

######  Proposition 2.2.

Let $n\geqslant 2$, $s\in(0,1)$, $R>1$ and $f\in L^{\infty}(B_{R}\setminus
B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. Then the function
defined in (2.4) is the unique solution (up to a zero measure subset of
$\mathbb{R}^{n}\setminus B_{1}$) to the problem in (2.5).

###### Proof.

We argue by approximation, owing to Theorem 2.1. The gist is indeed to take a
sequence of continuous functions $f_{k}$ approaching $f$ as $k\to+\infty$, use
Theorem 2.1 and then pass to the limit. To implement this idea, one needs to
take care of some regularity issues.

The details of this technical argument go as follows. By (2.3), for each $x\in
B_{1}$ and multi-index $\alpha$ we have that

$D_{x}^{\alpha}P(x,\cdot)f(\cdot)\in
L^{1}(\mathbb{R}^{n}\setminus\overline{B}_{1}).$

As a consequence, we obtain that $u_{f}^{(s)}(x)$ in (2.4) is well defined and
smooth inside $B_{1}$.

To complete the proof of Proposition 2.2, we need to show that $u_{f}^{(s)}$,
as defined in (2.4), is the unique solution of (2.5). To do so, we start by
checking that $u_{f}^{(s)}$ is $s$-harmonic in $B_{1}$. We consider a sequence
$\\{f_{k}\\}_{k}\subset C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$, such that

$f_{k}\rightarrow f$ in $L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ as
$k\to+\infty$. (2.6)

More specifically, we take $f_{k}:=(\chi_{B_{k}}\tilde{f})*\eta_{\frac{1}{k}}$
with $k\geqslant 2$, where $\tilde{f}$ is defined as

$\tilde{f}(x):=\begin{cases}\begin{split}f(x)\;\;\textrm{if}\;\;&x\in\mathbb{R}^{n}\setminus
B_{1},\\\ 0\;\;\textrm{if}\;\;&x\in B_{1},\end{split}\end{cases}$

and $\eta_{\frac{1}{k}}$ is a mollifier of radius $\frac{1}{k}$, while
$\chi_{B_{k}}$ is the characteristic function of $B_{k}$. We also let
$u_{f_{k}}^{(s)}$ be the unique pointwise continuous solution to the problem
(2.5), according to Theorem 2.1.

Then we have that for each multi-index $\alpha$

$\|D^{\alpha}u_{f_{k}}^{(s)}-D^{\alpha}u_{f}^{(s)}\|_{L_{\textrm{loc}}^{\infty}(B_{1})}\to
0\quad{\mbox{ as~{}$k\to+\infty$}}.$ (2.7)

Indeed for each multi-index $\alpha$ and $g\in L^{\infty}(B_{R}\setminus
B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ one finds that

$D^{\alpha}u_{g}(x)=\int_{\mathbb{R}^{n}\setminus
B_{1}}D_{x}^{\alpha}P(x,y)g(y)\,dy$

for each $x\in B_{1}$, and therefore, choosing $R_{0}\in(1,R)$, we see that,
for every $x\in B^{\prime}$ with $B^{\prime}\Subset B_{1}$,

$\begin{split}&|D^{\alpha}u_{f_{k}}^{(s)}(x)-D^{\alpha}u_{f}^{(s)}(x)|\\\
&\qquad\leqslant\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}|D_{x}^{\alpha}P(x,y)||f_{k}(y)-f(y)|\,dy+\int_{B_{R_{0}}\setminus
B_{1}}|D_{x}^{\alpha}P(x,y)||f_{k}(y)-f(y)|\,dy\\\ &\qquad\leqslant
c\,\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|f_{k}(y)-f(y)|}{|y|^{n+2s}}\,dy+\int_{B_{R_{0}}\setminus
B_{1}}|D_{x}^{\alpha}P(x,y)||f_{k}(y)-f(y)|\,dy,\end{split}$ (2.8)

where $c$ is a positive constant depending on $\alpha$, $R_{0}$, $n$, $s$ and
$B^{\prime}$. The first term in the third line in (2.8) converges to zero as
$k\to+\infty$, thanks to (2.6). We also observe that, if $y\in
B_{R_{0}}\setminus B_{1}$, then $|f_{k}(y)-f(y)|\leqslant
2\|f\|_{L^{\infty}(B_{R}\setminus B_{1})}$, and therefore, by the Dominated
Convergence Theorem, we have that also the second term in the third line in
(2.8) converges to zero as $k\to+\infty$. These considerations prove (2.7).

Furthermore note that if $\alpha=0$, taking $R_{0}\in(1,R)$ and using also
(2.2), we have that, for all $x\in B_{1}$,

$\begin{split}|u_{f_{k}}^{(s)}(x)|\leqslant\;&\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}|P(x,y)||f_{k}(y)|\,dy+\int_{B_{R_{0}}\setminus
B_{1}}|P(x,y)||f_{k}(y)|\,dy\\\ \leqslant\;&C\,\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|f_{k}(y)|}{|y|^{n+2s}}\,dy+\int_{B_{R_{0}}\setminus
B_{1}}|P(x,y)||f_{k}(y)|\,dy\\\
\leqslant\;&C\|f_{k}\|_{L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{1})}+\|f\|_{L^{\infty}(B_{R}\setminus B_{1})},\end{split}$ (2.9)

where $C$ is a positive constant depending on $R_{0}$, $n$ and $s$. Now, we
observe that the sequence $\|f_{k}\|_{L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{1})}$ is uniformly bounded, thanks to (2.6). Accordingly, from (2.9) we see
that

$u_{f_{k}}^{(s)}$ is uniformly bounded in $B_{1}$. (2.10)

Now, if $x\in B_{1}$, taking $\delta\in\left(0,1-|x|\right)$, we have that

$\begin{split}(-\Delta)^{s}u_{f_{k}}^{(s)}(x)-(-\Delta)^{s}u_{f}^{(s)}(x)=\;&\int_{\mathbb{R}^{n}}\frac{u_{f_{k}}^{(s)}(x)-u_{f_{k}}^{(s)}(y)-u_{f}^{(s)}(x)+u_{f}^{(s)}(y)}{|x-y|^{n+2s}}\,dy\\\
=\;&A+B+C+D+E+F,\end{split}$ (2.11)

where

$\displaystyle A:=\displaystyle\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{u_{f_{k}}^{(s)}(x)-u_{f}^{(s)}(x)}{|x-y|^{n+2s}}\,dy,\qquad\qquad$
$\displaystyle B:=\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{u_{f_{k}}^{(s)}(y)-u_{f}^{(s)}(y)}{|x-y|^{n+2s}}\,dy,$
$\displaystyle
C:=\displaystyle\int_{B_{\delta}(x)}\frac{u_{f_{k}}^{(s)}(x)-u_{f_{k}}^{(s)}(y)}{|x-y|^{n+2s}}\,dy,\qquad\qquad$
$\displaystyle
D:=\int_{B_{\delta}(x)}\frac{u_{f}^{(s)}(y)-u_{f}^{(s)}(x)}{|x-y|^{n+2s}}\,dy,$
$\displaystyle E:=\displaystyle\int_{B_{1}\setminus
B_{\delta}(x)}\frac{u_{f_{k}}^{(s)}(x)-u_{f}^{(s)}(x)}{|x-y|^{n+2s}}\,dy\qquad{\mbox{
and }}\qquad$ $\displaystyle F:=\int_{B_{1}\setminus
B_{\delta}(x)}\frac{u_{f}^{(s)}(y)-u_{f_{k}}^{(s)}(y)}{|x-y|^{n+2s}}\,dy.$

Notice that

$|A+E|\leqslant\int_{\mathbb{R}^{n}\setminus
B_{\delta}(x)}\frac{|u_{f_{k}}^{(s)}(x)-u_{f}^{(s)}(x)|}{|x-y|^{n+2s}}\,dy\leqslant|u_{f_{k}}^{(s)}(x)-u_{f}^{(s)}(x)|\int_{\mathbb{R}^{n}\setminus
B_{\delta}}\frac{dz}{|z|^{n+2s}}\leqslant\frac{C}{\delta^{2s}}|u_{f_{k}}^{(s)}(x)-u_{f}^{(s)}(x)|,$

which converges to zero as $k\to+\infty$, thanks to (2.7).

Furthermore, we observe that if $y\in\mathbb{R}^{n}\setminus B_{1}$ then

$|x-y|\geqslant|y|-|x|=\delta|y|+(1-\delta)|y|-|x|\geqslant\delta|y|+1-\delta-|x|\geqslant\delta|y|,$

and thus

$|B|\leqslant\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|f_{k}(y)-f(y)|}{|x-y|^{n+2s}}\,dy\leqslant\frac{1}{\delta^{n+2s}}\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|f_{k}(y)-f(y)|}{|y|^{n+2s}}\,dy$

which, in light of (2.6), converges to zero as $k\to+\infty$.

Moreover, from (2.7) and the Dominated Convergence Theorem, we see that the
quantity $C+D$ converges to zero as $k\to+\infty$.

Finally, recalling (2.10) and making again use of the Dominated Convergence
Theorem, we have that $F$ converges to zero as $k\to+\infty$.

These considerations and (2.11) give that $(-\Delta)^{s}u_{f_{k}}^{(s)}(x)$
converges to $(-\Delta)^{s}u_{f}^{(s)}(x)$ as $k\to+\infty$ for every $x\in
B_{1}$. Since, by Theorem 2.1, we know that
$(-\Delta)^{s}u_{f_{k}}^{(s)}(x)=0$ for each $x\in B_{1}$, we conclude that
$(-\Delta)^{s}u_{f}^{(s)}(x)=0$. This proves that $u_{f}^{(s)}$ solves (2.5).

It is only left to show the uniqueness statement. Suppose that there exists
$u_{1}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ satisfying

$\begin{cases}\begin{split}(-\Delta)^{s}u_{1}&=0\;\;\textrm{in}\;\;B_{1},\\\
u_{1}&=f\;\;\textrm{in}\;\;\mathbb{R}^{n}\setminus
B_{1}.\end{split}\end{cases}$

Then both $v:=u_{f}^{(s)}-u_{1}$ and $-v=u_{1}-u_{f}^{(s)}$ are solutions to

$\begin{cases}\begin{split}(-\Delta)^{s}u&=0\;\;\textrm{in}\;\;B_{1},\\\
u&=0\;\;\textrm{in}\;\;\mathbb{R}^{n}\setminus B_{1},\end{split}\end{cases}$

and therefore by the maximum principle for the fractional Laplacian (see e.g.
Theorem 2.3.2. in [7]) we have that $v=0$ in $B_{1}$, leading to uniqueness. ∎

##  3 Proof of the Fractional Malmheden and Schwarz Theorems

In this section we provide the proofs of the Fractional Malmheden and Schwarz
results, as stated in Theorems 1.4 and 1.8, and of the convergence result in
Proposition 1.5.

We start with the main argument to prove Theorem 1.4. For this, we employ the
following change of variable result (see Lemma 2.13.3 in [2] for the proof of
it):

######  Lemma 3.1.

Let $n\geqslant 2$, $x\in B_{1}$ and $Q_{\pm}^{x}$ and $r_{\pm}^{x}$ be
defined as in (1.2) and (1.3), respectively. Then for each $e\in\partial
B_{1}$ it holds that

$|\det DQ_{\pm}^{x}(e)|=\frac{(\pm
r_{\pm}^{x}(e))^{n}}{1-|x|^{2}-r_{\pm}^{x}(e)x\cdot e},$

and for each continuous $f:\partial B_{1}\rightarrow\mathbb{R}$ we have that

$\int_{\partial B_{1}}f(e)\,dH_{e}^{n-1}=\int_{\partial
B_{1}}f(Q_{\pm}^{x}(e))\frac{(\pm
r_{\pm}^{x}(e))^{n}}{1-|x|^{2}-r_{\pm}^{x}(e)x\cdot e}\,dH_{e}^{n-1}.$

With this notation, Theorem 1.4 will be a consequence of the following
statement.

######  Theorem 3.2.

Let $n\geqslant 2$, $s\in(0,1)$, $R>1$ and $f\in L^{\infty}(B_{R}\setminus
B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. Let $u_{f}^{(s)}$ be as
in (2.4).

Then, for each $x\in B_{1}$,

$u_{f}^{(s)}(x)=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}{\mathcal{E}}(x,\rho)\,\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho,$
(3.1)

where the notation in (1.12) and (1.13) has been used.

Furthermore, if $f$ is positively homogeneous of degree $\gamma$ for some
$\gamma\geqslant 0$, then we have that

$u_{f}^{(s)}(x)=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\rho^{\gamma}{\mathcal{E}}(x,\rho)\,\mathcal{L}_{f,e,1}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}\right)\,d\rho.$
(3.2)

###### Proof.

We first suppose that $f\in C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. Let $x\in B_{1}$, then, using polar
coordinates, from (2.1) and (2.4) we get that

$\begin{split}u_{f}^{(s)}(x)&=\int_{\mathbb{R}^{n}\setminus
B_{1}}P(x,y)f(y)\,dy\\\ &=c(n,s)(1-|x|^{2})^{s}\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{1}{(|y|^{2}-1)^{s}}\frac{f(y)}{|x-y|^{n}}\,dy\\\
&=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{1}{(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{\rho}}\frac{f(\omega)}{|x-\omega|^{n}}\,dH_{\omega}^{n-1}\right)\,d\rho\\\
&=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{\rho^{n-1}}{(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\frac{f(\rho e)}{|x-\rho e|^{n}}\,dH_{e}^{n-1}\right)\,d\rho\\\
&=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{1}{\rho(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\frac{f(\rho
e)}{\big{|}\frac{x}{\rho}-e\big{|}^{n}}\,dH_{e}^{n-1}\right)\,d\rho\\\
&=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\frac{f(\rho
e)}{\big{|}\frac{x}{\rho}-e\big{|}^{n}}\left(1-\frac{|x|^{2}}{\rho^{2}}\right)\,dH_{e}^{n-1}\right)\,d\rho\\\
&=:{\mathcal{I}}.\end{split}$ (3.3)

Hence, defining

$g(e):=\frac{f(\rho
e)}{\big{|}\frac{x}{\rho}-e\big{|}^{n}}\left(1-\frac{|x|^{2}}{\rho^{2}}\right)$

and applying Lemma 3.1 we obtain that

$\begin{split}&\frac{{\mathcal{I}}}{c(n,s)(1-|x|^{2})^{s}}\\\
&\quad=\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}g(e)\,dH_{e}^{n-1}\right)\,d\rho\\\
&\quad=\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}g(Q_{-}^{x/\rho}(e))\frac{(-r_{-}^{x/\rho}(e))^{n}}{1-|x/\rho|^{2}-(x/\rho\cdot
e)r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}\right)\,d\rho\\\
&\quad=\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\frac{f(\rho
Q_{-}^{x/\rho}(e))}{\big{|}\frac{x}{\rho}-Q_{-}^{x/\rho}(e)\big{|}^{n}}\frac{\left(1-|x/\rho|^{2}\right)(-r_{-}^{x/\rho}(e))^{n}}{1-|x/\rho|^{2}-(x/\rho\cdot
e)r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}\right)\,d\rho.\end{split}$ (3.4)

From equations (1.2) and (1.3) we deduce that

$\left|\frac{x}{\rho}-Q_{-}^{x/\rho}(e)\right|=|r_{-}^{x/\rho}(e)|=-r_{-}^{x/\rho}(e),$

and also (see formula (2.13.25) in [2])

$\frac{1-|x/\rho|^{2}}{1-|x/\rho|^{2}-(x/\rho\cdot
e)r_{-}^{x/\rho}(e)}=\frac{2r_{+}^{x/\rho}(e)}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}$

which, together with (3.4), gives that

${\mathcal{I}}=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\frac{2r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}\right)\,d\rho.$
(3.5)

We now observe that

$\begin{split}&\int_{\partial B_{1}}\frac{2r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}\\\
=&\int_{\partial B_{1}}\frac{r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}-\int_{\partial
B_{1}}\frac{r_{-}^{x/\rho}(e)f(\rho
Q_{+}^{x/\rho}(e)))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}\\\
=&\int_{\partial B_{1}}\frac{r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))-r_{-}^{x/\rho}(e)f(\rho
Q_{+}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}.\end{split}$
(3.6)

From (1.2), (1.3), (1.11) and (1.12) we also deduce that

$\displaystyle\frac{r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))-r_{-}^{x/\rho}(e)f(\rho
Q_{+}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}$ $\displaystyle=$
$\displaystyle\frac{r_{+}^{x/\rho}(e)f_{\rho}(Q_{-}^{x/\rho}(e))-r_{-}^{x/\rho}(e)f_{\rho}(Q_{+}^{x/\rho}(e))}{|Q_{+}^{x/\rho}(e)-Q_{-}^{x/\rho}(e)|}$
$\displaystyle=$
$\displaystyle\frac{\left(Q_{+}^{x/\rho}(e)-\frac{x}{\rho}\right)\cdot
e\,f_{\rho}(Q_{-}^{x/\rho}(e))+\left(\frac{x}{\rho}-Q_{-}^{x/\rho}(e)\right)\cdot
e\,f_{\rho}(Q_{+}^{x/\rho}(e))}{|Q_{+}^{x/\rho}(e)-Q_{-}^{x/\rho}(e)|}$
$\displaystyle=$
$\displaystyle\mathcal{L}_{f_{\rho}}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)$
$\displaystyle=$ $\displaystyle\mathcal{L}_{f,e,\rho}(x).$

Plugging this information into (3.6) we find that

$\int_{\partial B_{1}}\frac{2r_{+}^{x/\rho}(e)f(\rho
Q_{-}^{x/\rho}(e))}{r_{+}^{x/\rho}(e)-r_{-}^{x/\rho}(e)}\,dH_{e}^{n-1}=\int_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1},$

which in turn, together with (3.5), gives that

${\mathcal{I}}=c(n,s)(1-|x|^{2})^{s}\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho.$

Thus, recalling (1.13), this and (3.3) establish the desired result in (3.1)
under the additional assumption that $f$ is continuous in
$\mathbb{R}^{n}\setminus B_{1}$.

Now we remove the continuity assumption on $f$ by an approximation argument.
Given $f\in L^{\infty}(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$, we consider a sequence of functions
$\\{f_{k}\\}_{k}\subset C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ as in (2.6) and we let
$u_{f_{k}}^{(s)}$ be the unique pointwise continuous solution to the problem
(2.5), according to Theorem 2.1.

By (2.7),

$\|u_{f_{k}}^{(s)}-u_{f}^{(s)}\|_{L_{\textrm{loc}}^{\infty}(B_{1})}\to 0.$
(3.7)

Therefore we have that, for each $x\in B_{1}$,

$u_{f}^{(s)}(x)=\lim_{k\to\infty}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f_{k},e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho.$

From this, we claim that there exists a subsequence $\\{f_{k_{j}}\\}_{j}$,
such that

$\begin{split}u_{f}^{(s)}(x)&=\lim_{j\to\infty}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f_{k_{j}},e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho\\\
&=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho.\end{split}$
(3.8)

For the convenience of the reader, the technical proof of (3.8) can be found
in Appendix A.2.

The claim in (3.8) establishes the desired result in (3.1) also for non
continuous functions.

To prove the claim in (3.2), we use (1.1) and (1.12) to see that

$\displaystyle\mathcal{L}_{f,e,\rho}(x)$ $\displaystyle=$
$\displaystyle\mathcal{L}_{f_{\rho}}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)$
$\displaystyle=$
$\displaystyle\frac{\left(\frac{x}{\rho}-Q^{x/\rho}_{-}(e)\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f_{\rho}\big{(}Q^{x/\rho}_{+}(e)\big{)}+\frac{\left(Q^{x/\rho}_{+}(e)-\frac{x}{\rho}\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f_{\rho}\big{(}Q^{x/\rho}_{-}(e)\big{)}$
$\displaystyle=$
$\displaystyle\frac{\left(\frac{x}{\rho}-Q^{x/\rho}_{-}(e)\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f\big{(}\rho
Q^{x/\rho}_{+}(e)\big{)}+\frac{\left(Q^{x/\rho}_{+}(e)-\frac{x}{\rho}\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f\big{(}\rho
Q^{x/\rho}_{-}(e)\big{)}$ $\displaystyle=$
$\displaystyle\rho^{\gamma}\left[\frac{\left(\frac{x}{\rho}-Q^{x/\rho}_{-}(e)\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f\big{(}Q^{x/\rho}_{+}(e)\big{)}+\frac{\left(Q^{x/\rho}_{+}(e)-\frac{x}{\rho}\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,f\big{(}Q^{x/\rho}_{-}(e)\big{)}\right]$
$\displaystyle=$
$\displaystyle\rho^{\gamma}{\mathcal{L}}_{f}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)$
$\displaystyle=$
$\displaystyle\rho^{\gamma}{\mathcal{L}}_{f,e,1}\left(\frac{x}{\rho}\right).$

The claim in (3.2) then follows from this and (3.1). This concludes the proof
of Theorem 3.2. ∎

###### Proof of Theorem 1.4.

From Proposition 2.2, we know that, under the hypotheses of Theorem 1.4, the
function defined in (2.4) is the unique solution (up to a zero measure subset
of $\mathbb{R}^{n}\setminus B_{1}$) to the problem (1.16). Then, the desired
result in Theorem 1.4 follows from this and Theorem 3.2. ∎

We now give the proof of the Schwarz result in Theorem 1.8.

###### Proof of Theorem 1.8.

We first suppose that $f\in C(\mathbb{R}^{2}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$. Under these assumptions we can
apply Theorem 1.6 in dimension $2$ and get

$u_{f}^{(s)}(x)=2\pi\int_{1}^{\infty}\mathcal{E}(x,\rho)\tilde{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho,$

where $\tilde{u}_{f_{\rho}}$ has been defined in the statement of Theorem 1.6.
Therefore, when $n=2$ we can apply Theorem 1.2 to $\tilde{u}_{f_{\rho}}$ and
get

$\tilde{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)=\frac{1}{2\pi}\int_{\partial
B_{1}}f_{\rho}(Q^{x/\rho}(e))\,dH_{e}^{n-1},$

which leads to (1.21) in the case in which $f\in C(\mathbb{R}^{2}\setminus
B_{1})$.

Suppose now that $f\in L^{\infty}(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$, and consider a sequence of
functions $\\{f_{k}\\}_{k}\subset C(\mathbb{R}^{2}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{2}\setminus B_{1})$ as in (2.6). We let
$u_{f_{k}}^{(s)}$ be the unique pointwise continuous solution to the problem
(2.5), according to Theorem 2.1. From the previous step, we have that, for
each $x\in B_{1}$,

$u_{f_{k}}^{(s)}(x)=\int_{1}^{\infty}\mathcal{E}(x,\rho)\left(\int_{\partial
B_{1}}f_{k}(\rho Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho.$

By this and (3.7), we have that, for each $x\in B_{1}$,

$u_{f}^{(s)}(x)=\lim_{k\to\infty}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)f_{k}(\rho
Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho.$

From this, one sees that there exists a subsequence $\\{f_{k_{j}}\\}_{j}$ such
that

$\begin{split}u_{f}^{(s)}(x)&=\lim_{j\to\infty}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)f_{k_{j}}(\rho
Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho\\\
&=\int_{1}^{\infty}\left(\int_{\partial B_{1}}\mathcal{E}(x,\rho)f(\rho
Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho.\end{split}$ (3.9)

For the facility of the reader, a detailed proof of (3.9) is given in Appendix
A.3.

We observe that the proof of Theorem 1.8 is completed, thanks to (3.9). ∎

We now deal with the convergence result in Proposition 1.5.

###### Proof of Proposition 1.5.

Let $f\in C(B_{R}\setminus B_{1})\cap L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{1})$, for each $s\in(s_{0},1]$. Furthermore let $u_{f}^{(s)}$ and ${u}_{f}$
as in the statement of Proposition 1.5. Then, Theorem 1.1 implies that the
following identity holds for each $\rho\in(1,R)$

${u}_{f_{\rho}}\left(\frac{x}{\rho}\right)=\fint_{\partial
B_{1}}\mathcal{L}_{f_{\rho}}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}=\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1},$ (3.10)

thanks to (1.12), where ${u}_{f_{\rho}}$ is the unique solution to the
classical Dirichlet problem (1.18).

In particular, we have that

${u}_{f}\left(\frac{x}{\rho}\right)=\fint_{\partial
B_{1}}\mathcal{L}_{f}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}=\fint_{\partial
B_{1}}\mathcal{L}_{f,e,1}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}.$ (3.11)

Now, using (1.17), (1.22) and (3.10) we obtain for each $x\in B_{1}$ and
$R_{0}\in(1,R)$ the following identity

$\begin{split}&u_{f}^{(s)}(x)-{u}_{f}(x)\\\ =\;&\int_{1}^{\infty}|\partial
B_{1}|\mathcal{E}(x,\rho)\left(\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}-{u}_{f}(x)\right)\,d\rho\\\
=\;&\int_{1}^{R_{0}}|\partial
B_{1}|\mathcal{E}(x,\rho)\left({u}_{f_{\rho}}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right)\,d\rho\\\
&\qquad+\int_{R_{0}}^{\infty}|\partial
B_{1}|\mathcal{E}(x,\rho)\left(\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}-{u}_{f}(x)\right)\,d\rho.\end{split}$
(3.12)

By the continuity of ${u}_{f}$ and $f$, we have that for each $\delta>0$ there
exists some $R_{0}\in(1,R)$ such that for each $\rho\in(1,R_{0})$

$\begin{split}&\left|{u}_{f}\left(\frac{x}{\rho}\right)-u_{f}(x)\right|\leqslant\delta\quad{\mbox{
for all }}x\in B_{1}\\\ \qquad{\mbox{ and
}}&\qquad\|f_{\rho}-f\|_{L^{\infty}(\partial
B_{1})}\leqslant\delta.\end{split}$ (3.13)

Also, we point out that

$\displaystyle\mathcal{L}_{f_{\rho},e,1}\left(\frac{x}{\rho}\right)$
$\displaystyle=$
$\displaystyle\mathcal{L}_{f_{\rho}}^{Q^{x/\rho}_{-}(e),Q^{x/\rho}_{+}(e)}\left(\frac{x}{\rho}\right)={\mathcal{L}}_{f,e,\rho}\left(x\right),$

thanks to (1.12).

Therefore, from this, (3.10), (3.11), and (3.13), for all $x\in B_{1}$, we
deduce that, if $\rho\in(1,R_{0})$,

$\begin{split}\left|{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|&\leqslant\left|{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)-{u}_{f}\left(\frac{x}{\rho}\right)\right|+\left|{u}_{f}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|\\\
&=\left|\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)-\mathcal{L}_{f,e,1}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}\right|+\left|{u}_{f}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|\\\
&=\left|\fint_{\partial
B_{1}}\mathcal{L}_{f_{\rho}-f,e,1}\left(\frac{x}{\rho}\right)\,dH_{e}^{n-1}\right|+\left|{u}_{f}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|\\\
&\leqslant\fint_{\partial
B_{1}}\left|\mathcal{L}_{f_{\rho}-f,e,1}\left(\frac{x}{\rho}\right)\right|\,dH_{e}^{n-1}+\delta.\end{split}$
(3.14)

Now, if $x\in B_{1}$ and $\rho\in(1,R_{0})$, we see that

$\begin{split}&|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|=2\sqrt{\left(\frac{x}{\rho}\cdot
e\right)^{2}-\frac{|x|^{2}}{\rho^{2}}+1}\geqslant
2\sqrt{1-\frac{|x|^{2}}{\rho^{2}}}\\\
&\qquad\qquad=\frac{2}{\rho}\sqrt{\rho^{2}-|x|^{2}}\geqslant\frac{2}{R_{0}}\sqrt{1-|x|^{2}}\end{split}$
(3.15)

and thus, according to (1.1),

$\begin{split}&\left|\mathcal{L}_{f_{\rho}-f,e,1}\left(\frac{x}{\rho}\right)\right|\\\
=\;&\left|\frac{\left(\frac{x}{\rho}-Q^{x/\rho}_{-}(e)\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,(f_{\rho}-f)\big{(}Q^{x/\rho}_{+}(e)\big{)}+\frac{\left(Q^{x/\rho}_{+}(e)-\frac{x}{\rho}\right)\cdot
e}{|Q^{x/\rho}_{+}(e)-Q^{x/\rho}_{-}(e)|}\,(f_{\rho}-f)\big{(}Q^{x/\rho}_{-}(e)\big{)}\right|\\\
\leqslant\;&4\,\frac{R_{0}}{2\sqrt{1-|x|^{2}}}\|f_{\rho}-f\|_{L^{\infty}(\partial
B_{1})}\\\ \leqslant\;&\frac{2R_{0}\,\delta}{\sqrt{1-|x|^{2}}},\end{split}$
(3.16)

and therefore, plugging this information into (3.14), we obtain that, if $x\in
B_{1}$ and $\rho\in(1,R_{0})$,

$\left|{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|\leqslant\left(\frac{2R_{0}}{\sqrt{1-|x|^{2}}}+1\right)\delta.$
(3.17)

Furthermore, employing the change of variable $e:=\omega/|\omega|$ and
recalling (1.13),

$\begin{split}&\int_{R_{0}}^{\infty}\mathcal{E}(x,\rho)\left(\int_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}-|\partial
B_{1}|{u}_{f}(x)\right)\,d\rho\\\
=\;&\int_{R_{0}}^{\infty}\frac{\mathcal{E}(x,\rho)}{\rho^{n-1}}\left(\int_{\partial
B_{\rho}}\mathcal{L}_{f,\omega/|\omega|,|\omega|}(x)\,dH_{\omega}^{n-1}-|\partial
B_{1}|{u}_{f}(x)\right)\,d\rho\\\
=\;&c(n,s)(1-|x|^{2})^{s}\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|y|^{2}}{(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})|y|^{n}}\left(\mathcal{L}_{f,y/|y|,|y|}(x)-|\partial
B_{1}|{u}_{f}(x)\right)dy.\end{split}$ (3.18)

We also deduce from (1.1) the following pointwise estimate

$\begin{split}&\left|\mathcal{L}_{f,y/|y|,|y|}(x)\right|\\\
=\;&\left|\frac{\left(\frac{x}{|y|}-Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\cdot\frac{y}{|y|}}{|Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)-Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)|}\,f_{|y|}\left(Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)+\frac{\left(Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)-\frac{x}{|y|}\right)\cdot\frac{y}{|y|}}{|Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)-Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)|}\,f_{|y|}\left(Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|\\\
&\qquad\leqslant\frac{1}{\sqrt{1-|x|^{2}}}\left(\left|f_{|y|}\left(Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|+\left|f_{|y|}\left(Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|\right).\end{split}$
(3.19)

We claim that

the right hand side in (3.19) is $L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$
for each $s\in(s_{0},1]$. (3.20)

Indeed, if we define the following function

$\begin{split}F_{\pm}:\mathbb{R}^{n}\setminus&B_{1}\rightarrow\mathbb{R}^{n}\setminus
B_{1}\\\ &y\mapsto|y|Q_{\pm}^{x/|y|}(y/|y|),\end{split}$ (3.21)

we see that it is $C^{1}$ and invertible. Note that, recalling also the limits
in (1.4), one finds that

$\|\det(DF_{\pm}^{-1}(z)\|_{L^{\infty}(\mathbb{R}^{n}\setminus
B_{1})}\leqslant C,$ (3.22)

for some positive constant $C$, depending on $x$. Therefore by applying
Theorem 2 in Section 3.3.3 of [3], we obtain that

$\begin{split}&\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{\left|f_{|y|}\left(Q_{\pm}^{x/|y|}(y/|y|)\right)\right|}{\left|y\right|^{n+2s}}\,dy=\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{\left|f_{|y|}\left(Q_{\pm}^{x/|y|}(y/|y|)\right)\right|}{\left||y|Q_{\pm}^{x/|y|}(y/|y|)\right|^{n+2s}}\,dy\\\
&\qquad\qquad=\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|f(z)|}{|z|^{n+2s}}|\det(DF_{\pm}^{-1}(z))|\,dz\leqslant
C\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|f(z)|}{|z|^{n+2s}}\,dz.\end{split}$ (3.23)

This and the fact that $f\in L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ for
each $s\in(s_{0},1]$ entail that (3.20) holds true.

As a consequence of (3.19) and (3.20) we have that the integrals in (3.18) are
finite and bounded in $s$.

Using this information and (3.17), we deduce from (3.12) that for each
$\delta>0$ there exists some $R_{0}\in(1,R)$ such that for each
$\rho\in(1,R_{0})$ we have

$\begin{split}&|u_{f}^{(s)}(x)-{u}_{f}(x)|\\\
\leqslant\;&\int_{1}^{R_{0}}|\partial
B_{1}|\mathcal{E}(x,\rho)\left|{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)-{u}_{f}(x)\right|\,d\rho\\\
&\qquad\qquad+\left|\int_{R_{0}}^{\infty}|\partial
B_{1}|\mathcal{E}(x,\rho)\left(\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}-{u}_{f}(x)\right)\,d\rho\right|\\\
\leqslant\;&\int_{1}^{R_{0}}|\partial
B_{1}|\mathcal{E}(x,\rho)\left(\frac{2R_{0}}{\sqrt{1-|x|^{2}}}+1\right)\delta\,d\rho\\\
&\qquad\qquad+\left|\int_{R_{0}}^{\infty}|\partial
B_{1}|\mathcal{E}(x,\rho)\left(\fint_{\partial
B_{1}}\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}-{u}_{f}(x)\right)\,d\rho\right|\\\
\leqslant\;&C(x,R)\delta+c(n,s)(1-|x|^{2})^{s}\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|y|^{2}}{(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})|y|^{n}}\left(\mathcal{L}_{f,y/|y|,|y|}(x)-|\partial
B_{1}|{u}_{f}(x)\right)dy\\\
\leqslant\;&C(x,R)\delta+c(n,s)(1-|x|^{2})^{s}\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|y|^{2}}{(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})|y|^{n}}\\\
&\qquad\times\left(\frac{1}{\sqrt{1-|x|^{2}}}\left(\left|f_{|y|}\left(Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|+\left|f_{|y|}\left(Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|\right)-|\partial
B_{1}|{u}_{f}(x)\right)dy\end{split}$

where ${C}(x,R_{0})$ depends only on $x$ and $R_{0}$.

By taking the limit as $s\nearrow 1$, we see that

$\displaystyle\lim_{s\nearrow
1}c(n,s)(1-|x|^{2})^{s}\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|y|^{2}}{(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})|y|^{n}}$
$\displaystyle\qquad\times\left(\frac{1}{\sqrt{1-|x|^{2}}}\left(\left|f_{|y|}\left(Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|+\left|f_{|y|}\left(Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|\right)-|\partial
B_{1}|{u}_{f}(x)\right)dy=0$

since $c(n,s)\to 0$ for $s\nearrow 1$ by (1.14). As a consequence

$\lim_{s\nearrow 1}|u_{f}^{(s)}(x)-{u}_{f}(x)|\leqslant C(x,R)\delta.$

This and the arbitrariness of $\delta$ give the desired claim in Proposition
1.5. ∎

##  4 Harnack inequality

In this section we provide a new proof of the Harnack inequality for
$s$-harmonic functions as stated in Theorem 1.7. Our strategy is to use the
Fractional Malmheden Theorem to show that this result can be directly inferred
from the classical Harnack inequality for harmonic functions.

###### Proof of Theorem 1.7.

For convenience we call $u|_{\mathbb{R}^{n}\setminus B_{1}}=f$. Let us first
assume that $f\in C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$. Under this assumption, we can apply
Theorem 1.6 and obtain that

$u(x)=|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho){u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho,$

for each $x\in B_{1}$, where ${u}_{f_{\rho}}$ has been defined in the
statement of Theorem 1.6. Therefore, we have that

$u(0)=c(n,s)|\partial
B_{1}|\int_{1}^{\infty}\frac{{u}_{f_{\rho}}(0)}{\rho(\rho^{2}-1)^{s}}\,d\rho.$
(4.1)

Now we fix $r\in(0,1)$ and we consider $x\in B_{r}$. Applying the Harnack
inequality for classical harmonic functions to ${u}_{f_{\rho}}$, we have that

${u}_{f_{\rho}}(0)\leqslant\frac{(1+|x|/\rho)^{n-1}}{1-|x|/\rho}{u}_{f\rho}\left(\frac{x}{\rho}\right).$

From this, (1.13) and (4.1) we obtain that

$\begin{split}u(0)\leqslant\;&c(n,s)|\partial
B_{1}|\int_{1}^{\infty}\frac{1}{\rho(\rho^{2}-1)^{s}}\frac{(1+|x|/\rho)^{n-1}}{1-|x|/\rho}\,{u}_{f\rho}\left(\frac{x}{\rho}\right)\,d\rho\\\
=\;&|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)\frac{(\rho^{2}-|x|^{2})}{\rho^{2}\,(1-|x|^{2})^{s}}\frac{(\rho+|x|)^{n-1}}{\rho^{n-2}(\rho-|x|)}\,{u}_{f\rho}\left(\frac{x}{\rho}\right)\,d\rho\\\
=\;&|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)\frac{(\rho+|x|)^{n}}{\rho^{n}(1-|x|^{2})^{s}}\,{u}_{f\rho}\left(\frac{x}{\rho}\right)\,d\rho\\\
=\;&|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)\,g(\rho,t)\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho,\end{split}$
(4.2)

where for convenience we have called $t:=|x|$ in the last line and defined

$g(\rho,t):=\frac{(\rho+t)^{n}}{\rho^{n}(1-t^{2})^{s}},$

with $(\rho,t)\in[1,\infty)\times[0,r]$.

Since $g(\rho,t)$ is decreasing in $\rho$ and increasing in $t$, we have that

$\frac{(1+r)^{n}}{(1-r^{2})^{s}}=\sup_{(\rho,t)\in[1,\infty)\times[0,r]}g(\rho,t).$

Therefore, from this, (1.19) and (4.2) we obtain that

$\displaystyle u(0)\leqslant|\partial
B_{1}|\frac{(1+r)^{n}}{(1-r^{2})^{s}}\int_{1}^{\infty}\mathcal{E}(x,\rho)\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho=\frac{(1+r)^{n}}{(1-r^{2})^{s}}u(x),$

which establishes the first inequality in (1.20).

To prove the second inequality in (1.20), we make use of the Harnack
inequality for harmonic functions, thus obtaining that

${u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\leqslant\frac{1+|x|/\rho}{(1-|x|/\rho)^{n-1}}{u}_{f_{\rho}}(0).$

Using this and (1.13) into (4.1), we find that

$\displaystyle u(0)$ $\displaystyle=$ $\displaystyle c(n,s)|\partial
B_{1}|\int_{1}^{\infty}\frac{{u}_{f_{\rho}}(0)}{\rho(\rho^{2}-1)^{s}}\,d\rho$
$\displaystyle\geqslant$ $\displaystyle c(n,s)|\partial
B_{1}|\int_{1}^{\infty}\frac{1}{\rho(\rho^{2}-1)^{s}}\frac{(1-|x|/\rho)^{n-1}}{1+|x|/\rho}\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho$
$\displaystyle=$ $\displaystyle c(n,s)|\partial
B_{1}|\int_{1}^{\infty}\frac{(\rho-|x|)^{n-1}}{\rho^{n-1}(\rho^{2}-1)^{s}(\rho+|x|)}\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho$
$\displaystyle=$ $\displaystyle|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)\frac{(\rho-|x|)^{n}}{\rho^{n}(1-|x|^{2})^{s}}\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho.$

Using again the notation $t:=|x|$, we define the following function

$g_{1}(\rho,t):=\frac{(\rho-t)^{n}}{\rho^{n}(1-t^{2})^{s}},$ (4.3)

with $(\rho,t)\in[1,\infty)\times[0,r]$, and we see that

$u(0)\geqslant|\partial
B_{1}|\int_{1}^{\infty}\mathcal{E}(x,\rho)g_{1}(\rho,t)\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho.$
(4.4)

Since $g_{1}$ is increasing in $\rho$, we have that, for all
$(\rho,t)\in[1,\infty)\times[0,r]$,

$g_{1}(\rho,t)\geqslant
g_{1}(1,t)=\frac{(1-t)^{n}}{(1-t^{2})^{s}}=\frac{(1-t)^{n-s}}{(1+t)^{s}}=:g_{2}(t).$

Notice also that $g_{2}$ is decreasing, and therefore, for all
$(\rho,t)\in[1,\infty)\times[0,r]$,

$g_{1}(\rho,t)\geqslant
g_{2}(r)=\frac{(1-r)^{n-s}}{(1+r)^{s}}=\frac{(1-r)^{n}}{(1-r^{2})^{s}}.$

Plugging this information into (4.4) and recalling (1.19), we get

$u(0)\geqslant|\partial
B_{1}|\frac{(1-r)^{n}}{(1-r^{2})^{s}}\int_{1}^{\infty}\mathcal{E}(x,\rho)\,{u}_{f_{\rho}}\left(\frac{x}{\rho}\right)\,d\rho\geqslant\frac{(1-r)^{n}}{(1-r^{2})^{s}}u(x),$

which completes the proof of (1.20) under the additional continuity assumption
on $f$.

To deal with the general case, we perform an approximation argument, whose
details go as follows. If $f\in L^{\infty}(B_{R}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$, we take a sequence
$\\{f_{k}\\}_{k}\subset C(\mathbb{R}^{n}\setminus B_{1})\cap
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ as in (2.6). Then for
$u_{f_{k}}^{(s)}$ the two-sided inequality in (1.20) holds true, thanks to the
previous step. Also, by (3.7), we have the local uniform convergence

$\|u_{f_{k}}^{(s)}-u\|_{L_{\textrm{loc}}^{\infty}(B_{1})}\to 0\quad{\mbox{ as
}}k\to+\infty,$

and therefore we deduce the two sided inequality (1.20) also in this case.

It is only left to prove that the constants provided in equation (1.20) are
optimal. To show this let us fix some direction $e\in\partial B_{1}$, a
constant $\epsilon>0$ and the function

$f_{\epsilon}(y):=\begin{cases}\begin{split}0\;\;\;\;\;\;\;\;\;\;\;\;\;\;&\textrm{if}\;\;y\in\mathbb{R}^{n}\setminus
B_{\epsilon}((1+\epsilon)e),\\\ (|y|^{2}-1)^{s}\;\;&\textrm{if}\;\;y\in
B_{\epsilon}((1+\epsilon)e).\end{split}\end{cases}$

Then the function

$u_{f_{\epsilon}}^{(s)}(x):=c(n,s)\int_{B_{\epsilon}((\epsilon+1)e)}\frac{(1-|x|^{2})^{s}}{|y-x|^{n}}\,dy$

is $s$-harmonic in $B_{1}$, as a consequence of Proposition 2.2. Therefore, if
we fix $x=-re$ for $r\in(0,1)$, we have that

$\frac{u_{f_{\epsilon}}^{(s)}(0)}{u_{f_{\epsilon}}^{(s)}(-re)}=\frac{\displaystyle\int_{B_{\epsilon}((\epsilon+1)e)}\frac{dy}{|y|^{n}}}{\displaystyle\int_{B_{\epsilon}((\epsilon+1)e)}\frac{(1-r^{2})^{s}}{|y+re|^{n}}\,dy},$

and thus, by Lebesgue Differentiation Theorem, we conclude that

$\lim_{\epsilon\to
0}\frac{u_{f_{\epsilon}}^{(s)}(0)}{u_{f_{\epsilon}}^{(s)}(-re)}=\frac{(1+r)^{n}}{(1-r^{2})^{s}}.$

This proves that the constant on the left hand side inequality in (1.20) is
optimal.

Similarly, taking $x=re$, one sees that

$\lim_{\epsilon\to
0}\frac{u_{f_{\epsilon}}^{(s)}(re)}{u_{f_{\epsilon}}^{(s)}(0)}=\frac{(1-r^{2})^{s}}{(1-r)^{n}},$

which shows that the constant on the right hand side inequality in (1.20) is
also optimal. This concludes the proof of Theorem 1.7. ∎

##  Appendix A Appendices

###  A.1 Averaging the Fractional Poisson Kernel

Here we give a direct proof of the identity pointed out in footnote 1 on page
1. Namely, we establish that, if $x\in B_{1}$ and $\rho>1$,

${\mathcal{E}}(x,\rho)=\rho^{n-1}\fint_{\partial
B_{\rho}}P(x,y)\,dH^{n-1}_{y},$ (A.1)

being $P$ the Fractional Poisson Kernel in (2.1).

The identity in (A.1) has its own interest and it can be deduced from our
Fractional Malmheden Theorem, by taking a datum $f$ concentrating along a
given sphere $\partial B_{\rho}$. For the sake of completeness however, we
provide here an independent proof, only based on elementary computations and
standard integral formulas.

More specifically, we aim at showing that

$\fint_{\partial
B_{1}}\frac{dH^{n-1}_{\omega}}{|x-\rho\omega|^{n}}=\frac{\rho^{2-n}}{\rho^{2}-|x|^{2}}.$
(A.2)

Indeed, once (A.2) is established, we deduce from it, (1.13) and (2.1) that

$\displaystyle\fint_{\partial
B_{\rho}}P(x,y)\,dH^{n-1}_{y}=c(n,s)\;(1-|x|^{2})^{s}\fint_{\partial
B_{\rho}}\frac{dH^{n-1}_{y}}{(|y|^{2}-1)^{s}\;|x-y|^{n}}$
$\displaystyle\qquad=\frac{c(n,s)\;(1-|x|^{2})^{s}}{|\partial
B_{1}|\,\rho^{n-1}}\int_{\partial
B_{1}}\frac{\rho^{n-1}\,dH^{n-1}_{\omega}}{(\rho^{2}-1)^{s}\;|x-\rho\omega|^{n}}$
$\displaystyle\qquad=\frac{c(n,s)\;(1-|x|^{2})^{s}}{(\rho^{2}-1)^{s}}\fint_{\partial
B_{1}}\frac{dH^{n-1}_{\omega}}{|x-\rho\omega|^{n}}=c(n,s)\,\frac{\rho^{2-n}\,(1-|x|^{2})^{s}}{(\rho^{2}-1)^{s}\,(\rho^{2}-|x|^{2})}$
$\displaystyle\qquad=\rho^{1-n}{\mathcal{E}}(x,\rho)$

and this would complete the proof of (A.1).

Hence, we focus now on proving (A.2). To this end, we use spherical
coordinates on $\partial B_{1}$, with $\theta$, $\theta_{1}$, $\dots$,
$\theta_{n-3}\in[0,\pi]$ and $\theta_{n-2}\in[0,2\pi)$, see e.g. equation
(A.23) in [1] or pages 60-61 in [MR0185399], which correspond to
$\omega=\omega(\theta)$ of the form

$\begin{cases}\omega_{1}=\sin\theta\,\sin\theta_{1}\dots\sin\theta_{n-3}\sin\theta_{n-2},\\\
\omega_{2}=\sin\theta\,\sin\theta_{1}\dots\sin\theta_{n-3}\cos\theta_{n-2},\\\
\omega_{3}=\sin\theta\,\sin\theta_{1}\dots\cos\theta_{n-3},\\\ \vdots\\\
\omega_{n}=\cos\theta\end{cases}$

and produce a surface element of the form

$\sin^{n-2}\theta\;\sin^{n-3}\theta_{1}\;\dots\;\sin^{2}\theta_{n-4}\;\sin\theta_{n-3}\,d\theta\,d\theta_{1}\,\dots\,d\theta_{n-2}.$

Also, up to a rotation, to prove (A.2) we can assume that $x=(0,\dots,0,|x|)$.
In this way, we find that

$|x-\rho\omega|^{2}=|x|^{2}+\rho^{2}-2\rho
x\cdot\omega=|x|^{2}+\rho^{2}-2\rho|x|\,\omega_{n}=|x|^{2}+\rho^{2}-2\rho|x|\,\cos\theta,$

whence it follows that

$\displaystyle\int_{\partial
B_{1}}\frac{dH^{n-1}_{\omega}}{|x-\rho\omega|^{n}}=\int_{{\theta,\theta_{1},\dots,\theta_{n-3}\in[0,\pi]}\atop{\theta_{n-2}\in[0,2\pi)}}\frac{\sin^{n-2}\theta\;\sin^{n-3}\theta_{1}\;\dots\;\sin^{2}\theta_{n-4}\;\sin\theta_{n-3}\,d\theta\,d\theta_{1}\,\dots\,d\theta_{n-2}}{\big{(}|x|^{2}+\rho^{2}-2\rho|x|\,\cos\theta\big{)}^{\frac{n}{2}}}$
$\displaystyle\qquad=2\pi\left(\prod_{j=1}^{n-3}\int_{0}^{\pi}\sin^{n-j-2}\theta_{j}\;d\theta_{j}\right)\int_{\theta\in[0,\pi]}\frac{\sin^{n-2}\theta\,d\theta}{\big{(}|x|^{2}+\rho^{2}-2\rho|x|\,\cos\theta\big{)}^{\frac{n}{2}}}.$

Thus, we use the notation $\tau:=\frac{\rho}{|x|}\in(1,+\infty)$ and, by
Proposition A.9 in [1], we deduce that

$\displaystyle\int_{\partial
B_{1}}\frac{dH^{n-1}_{\omega}}{|x-\rho\omega|^{n}}$ $\displaystyle=$
$\displaystyle\frac{2\pi}{|x|^{n}}\left(\prod_{k=1}^{n-3}\int_{0}^{\pi}\sin^{k}\vartheta\,d\vartheta\right)\int_{0}^{\pi}\frac{\sin^{n-2}\theta\,d\theta}{\big{(}\tau^{2}+1-2\tau\,\cos\theta\big{)}^{\frac{n}{2}}}$
$\displaystyle=$
$\displaystyle\frac{2\pi}{|x|^{n}\,\tau^{n-2}(\tau^{2}-1)}\left(\prod_{k=1}^{n-3}\int_{0}^{\pi}\sin^{k}\vartheta\,d\vartheta\right)\int_{0}^{\pi}\sin^{n-2}\alpha\,d\alpha$
$\displaystyle=$
$\displaystyle\frac{2\pi}{|x|^{n}\,\tau^{n-2}(\tau^{2}-1)}\left(\prod_{k=1}^{n-2}\int_{0}^{\pi}\sin^{k}\vartheta\,d\vartheta\right).$

This and Proposition A.10 in [1] yield that

$\displaystyle\int_{\partial
B_{1}}\frac{dH^{n-1}_{\omega}}{|x-\rho\omega|^{n}}=\frac{2\pi^{\frac{n}{2}}}{|x|^{n}\,\tau^{n-2}(\tau^{2}-1)\;\Gamma\displaystyle\left(\frac{n}{2}\right)}=\frac{|\partial
B_{1}|}{|x|^{n}\,\tau^{n-2}(\tau^{2}-1)}=\frac{|\partial
B_{1}|\;\rho^{2-n}}{\rho^{2}-|x|^{2}}.$

The proof of (A.2) is thereby complete.

###  A.2 Proof of (3.8)

We recall (1.13) and we employ the change of variable $e:=\omega/|\omega|$ to
see that

$\begin{split}&\frac{1}{c(n,s)(1-|x|^{2})^{s}}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f_{k},e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho\\\
=\;&\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-1)^{s}(\rho^{2}-|x|^{2})}\left(\int_{\partial
B_{1}}\mathcal{L}_{f_{k},e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho\\\
=\;&\int_{1}^{\infty}\frac{\rho^{2}}{\rho^{n}(\rho^{2}-1)^{s}(\rho^{2}-|x|^{2})}\left(\int_{\partial
B_{\rho}}\mathcal{L}_{f_{k},\omega/|\omega|,|\omega|}(x)\,dH_{\omega}^{n-1}\right)\,d\rho\\\
=\;&\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{|y|^{2}}{|y|^{n}(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})}\mathcal{L}_{f_{k},y/|y|,|y|}(x)\,dy.\end{split}$
(A.3)

It also follows from (2.6) that, for a.e. $y\in\mathbb{R}^{n}\setminus B_{1}$,

$\mathcal{L}_{f_{k},y/|y|,|y|}(x)\to\mathcal{L}_{f,y/|y|,|y|}(x)\quad{\mbox{
as }}k\to+\infty.$ (A.4)

Now we take $R_{0}\in(1,R)$ and we deduce from (A.3) that

$\begin{split}&\frac{1}{c(n,s)(1-|x|^{2})^{s}}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f_{k},e,\rho}(x)\,dH_{e}^{n-1}\right)\,d\rho\\\
=\;&\int_{B_{R_{0}}\setminus
B_{1}}\frac{|y|^{2}}{|y|^{n}(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})}\mathcal{L}_{f_{k},y/|y|,|y|}(x)\,dy\\\
&\qquad+\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{|y|^{2}}{|y|^{n}(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})}\mathcal{L}_{f_{k},y/|y|,|y|}(x)\,dy.\end{split}$
(A.5)

Recalling the computation in (3.16), for $k$ large enough we have that

$\|\mathcal{L}_{f_{k},y/|y|,|y|}(x)\|_{L^{\infty}(B_{R_{0}}\setminus
B_{1})}\leqslant\frac{2R_{0}}{\sqrt{1-|x|^{2}}}\|f\|_{L^{\infty}(B_{R}\setminus
B_{1})}.$

Consequently, using this, (A.4) and the Dominated Convergence Theorem,

$\begin{split}&\lim_{k\to+\infty}\int_{B_{R_{0}}\setminus
B_{1}}\frac{|y|^{2}}{|y|^{n}(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})}\mathcal{L}_{f_{k},y/|y|,|y|}(x)\,dy\\\
&\qquad\qquad=\int_{B_{R_{0}}\setminus
B_{1}}\frac{|y|^{2}}{|y|^{n}(|y|^{2}-1)^{s}(|y|^{2}-|x|^{2})}\mathcal{L}_{f,y/|y|,|y|}(x)\,dy.\end{split}$
(A.6)

Also, we claim that there exists a subsequence $\\{f_{k_{j}}\\}_{j}$ such that

$\|\mathcal{L}_{f_{k_{j}},y/|y|,|y|}(x)-\mathcal{L}_{f,y/|y|,|y|}(x)\|_{L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{R_{0}})}\to 0\quad{\mbox{ as }}j\to+\infty.$ (A.7)

To show (A.7), we recall (3.15) and we observe that, for every $x\in B_{1}$
and $y\in\mathbb{R}^{n}\setminus B_{R_{0}}$,

$\begin{split}&\left|\mathcal{L}_{f_{k},y/|y|,|y|}(x)\right|\\\
=\;&\left|\frac{\left(\frac{x}{|y|}-Q^{x/|y|}_{-}\left(\frac{y}{|y|}\right)\right)\cdot\frac{y}{|y|}}{|Q^{x/|y|}_{+}\left(\frac{y}{|y|}\right)-Q^{x/|y|}_{-}\left(\frac{y}{|y|}\right)|}\,f_{k}\left(|y|Q^{x/|y|}_{+}\left(\frac{y}{|y|}\right)\right)\right.\\\
&\qquad\qquad\left.+\frac{\left(Q^{x/|y|}_{+}\left(\frac{y}{|y|}\right)-\frac{x}{|y|}\right)\cdot\frac{y}{|y|}}{|Q^{x/|y|}_{+}\left(\frac{y}{|y|}\right)-Q^{x/|y|}_{-}\left(\frac{y}{|y|}\right)|}\,f_{k}\left(Q^{x/|y|}_{-}\left(\frac{y}{|y|}\right)\right)\right|\\\
\\\
\leqslant\;&\frac{R_{0}}{\sqrt{1-|x|^{2}}}\bigg{[}\left|f_{k}\left(|y|Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|+\left|f_{k}\left(|y|Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\right|\bigg{]}.\end{split}$
(A.8)

Moreover, by (2.6) there exists a subsequence $\\{f_{k_{j}}\\}_{j}$ and a
function $h\in L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ such that
$|f_{k_{j}}(y)|\leqslant h(y)$ for a.e. $y\in\mathbb{R}^{n}\setminus B_{1}$
(see for instance Theorem 4.9 in [4]). Therefore, using this information into
(A.8), we have

$\left|\mathcal{L}_{f_{k_{j}},y/|y|,|y|}(x)\right|\leqslant\frac{R_{0}}{\sqrt{1-|x|^{2}}}\bigg{[}h\left(|y|Q_{-}^{x/|y|}\left(\frac{y}{|y|}\right)\right)+h\left(|y|Q_{+}^{x/|y|}\left(\frac{y}{|y|}\right)\right)\bigg{]}$
(A.9)

for a.e. $y\in\mathbb{R}^{n}\setminus B_{R_{0}}$.

Now we recall the map $F_{\pm}$ defined in (3.21), which is $C^{1}$ and
invertible, and therefore, by Theorem 2 in Section 3.3.3 of [3] and (3.22), we
get that

$\displaystyle\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h\left(|y|Q_{\pm}^{x/|y|}(y/|y|)\right)}{\left|y\right|^{n+2s}}\,dy=\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h\left(|y|Q_{\pm}^{x/|y|}(y/|y|)\right)}{\left||y|Q_{\pm}^{x/|y|}(y/|y|)\right|^{n+2s}}\,dy$
$\displaystyle\qquad\qquad=\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h(z)}{|z|^{n+2s}}|\det(DF_{\pm}^{-1}(z))|\,dz\leqslant
C\int_{\mathbb{R}^{n}\setminus B_{R_{0}}}\frac{h(z)}{|z|^{n+2s}}\,dz.$

Accordingly, we deduce that

$h\left(|y|Q_{\pm}^{x/|y|}(y/|y|)\right)\in L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{R_{0}}).$

This, the bound in (A.9), the pointwise convergence in (A.4) and the Dominated
Convergence Theorem lead to (A.7), as desired.

Hence, putting together (LABEL:doweu34imsnsdlif), (A.6) and (A.7), we obtain
that

$u_{f}(x)=\int_{1}^{\infty}\int_{\partial
B_{1}}\mathcal{E}(x,\rho)\mathcal{L}_{f,e,\rho}(x)\,dH_{e}^{n-1}\,d\rho$

for each $x\in B_{1}$, which completes the proof of (3.8). $\Box$

###  A.3 Proof of (3.9)

The proof of (3.9) is similar to the one of (3.8). We provide here the details
for the convenience of the reader.

From (1.13) with $n=2$ and the change of variable $e:=\omega/|\omega|$,

$\begin{split}&\frac{1}{c(2,s)(1-|x|^{2})^{s}}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)f_{k}(\rho
Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho\\\
=\;&\int_{1}^{\infty}\frac{\rho}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{1}}f_{k}(\rho Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho\\\
=\;&\int_{1}^{\infty}\frac{1}{(\rho^{2}-|x|^{2})(\rho^{2}-1)^{s}}\left(\int_{\partial
B_{\rho}}f_{k}(|\omega|Q^{x/|\omega|}(\omega/|\omega|))\,dH_{\omega}^{n-1}\right)\,d\rho\\\
=\;&\int_{\mathbb{R}^{n}\setminus
B_{1}}\frac{1}{(|y|^{2}-|x|^{2})(|y|^{2}-1)^{s}}f_{k}(|y|Q^{x/|y|}(y/|y|))\,dy\end{split}$
(A.10)

By (2.6), we have that, for a.e. $y\in\mathbb{R}^{n}\setminus B_{1}$,

$f_{k}(|y|Q^{x/|y|}(y/|y|))\to f(|y|Q^{x/|y|}(y/|y|))\quad{\mbox{ as
}}k\to+\infty.$ (A.11)

Now we take $R_{0}\in(1,R)$ and we get from (A.10) that

$\begin{split}&\frac{1}{c(2,s)(1-|x|^{2})^{s}}\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)f_{k}(\rho
Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho\\\ =\;&\int_{B_{R_{0}}\setminus
B_{1}}\frac{1}{(|y|^{2}-|x|^{2})(|y|^{2}-1)^{s}}f_{k}(|y|Q^{x/|y|}(y/|y|))\,dy\\\
&\qquad\qquad+\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{1}{(|y|^{2}-|x|^{2})(|y|^{2}-1)^{s}}f_{k}(|y|Q^{x/|y|}(y/|y|))\,dy\end{split}$
(A.12)

Notice that, for $k$ large enough,

$\|f_{k}\|_{L^{\infty}(B_{R_{0}}\setminus
B_{1})}\leqslant\|f\|_{L^{\infty}(B_{R}\setminus B_{1})}.$

As a consequence,

$\begin{split}&\lim_{k\to+\infty}\int_{B_{R_{0}}\setminus
B_{1}}\frac{1}{(|y|^{2}-|x|^{2})(|y|^{2}-1)^{s}}f_{k}(|y|Q^{x/|y|}(y/|y|))\,dy\\\
&\qquad\qquad=\int_{B_{R_{0}}\setminus
B_{1}}\frac{1}{(|y|^{2}-|x|^{2})(|y|^{2}-1)^{s}}f(|y|Q^{x/|y|}(y/|y|))\,dy\end{split}$
(A.13)

Furthemore, recalling (2.6) (see also Theorem 4.9 in [4]) we deduce the
existence of a subsequence $\\{f_{k_{j}}\\}_{j}$ and of a function $h\in
L_{s}^{1}(\mathbb{R}^{n}\setminus B_{1})$ such that

$\left|f_{k_{j}}(|y|Q^{x/|y|}(y|y|))\right|\leqslant h(|y|Q^{x/|y|}(y/|y|))$
(A.14)

for a.e. $y\in\mathbb{R}^{n}\setminus B_{R_{0}}$.

Furthermore, we claim that

${\mbox{$h(|y|Q^{x/|y|}(y/|y|))$ belongs to $L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{R_{0}})$}}.$ (A.15)

Indeed, the function

$\begin{split}F:\mathbb{R}^{n}\setminus&B_{R_{0}}\to\mathbb{R}^{n}\setminus
B_{R_{0}}\\\ &y\mapsto|y|Q^{x/|y|}(y/|y|)\end{split}$

is $C^{1}$ and invertible. Moreover, since

$\lim_{|y|\to\infty}Q^{x/|y|}(y/|y|)=id_{\partial B_{1}},$

we find that

$\|\det(DF^{-1}(z))\|_{L^{\infty}(\mathbb{R}^{n}\setminus
B_{R_{0}})}\leqslant\tilde{C},$

for some positive constant $\tilde{C}>0$.

From this and Theorem 2 in Section 3.3.3 of [3] we have that

$\begin{split}&\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h(|y|Q^{x/|y|}(y/|y|))}{|y|^{2+2s}}\,dy=\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h(|y|Q^{x/|y|}(y/|y|))}{||y|Q^{x/|y|}(y/|y|)|^{2+2s}}\,dy\\\
&\qquad\qquad=\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h(z)}{|z|^{{2+2s}}}|\det(DF^{-1}(z))|\,dz\leqslant\tilde{C}\int_{\mathbb{R}^{n}\setminus
B_{R_{0}}}\frac{h(z)}{|z|^{{2+2s}}}\,dz,\end{split}$

which establishes (A.15).

From (A.11), (A.14) and (A.15) and the Dominated Convergence Theorem, we
deduce that

$\|f_{k_{j}}(|y|Q^{x/|y|}(y|y|))-f(|y|Q^{x/|y|}(y|y|))\|_{L_{s}^{1}(\mathbb{R}^{n}\setminus
B_{R_{0}})}\to 0\quad{\mbox{ as }}j\to+\infty.$

Gathering together this, (A.12) and
(LABEL:swie23uri3poiuytrlkjhgfvbnjkvfgdszfxdghfkeg2233), we conclude that

$u_{f}^{(s)}(x)=\int_{1}^{\infty}\left(\int_{\partial
B_{1}}\mathcal{E}(x,\rho)f(\rho Q^{x/\rho}(e))\,dH_{e}^{n-1}\right)\,d\rho.$

This finishes the proof of (3.9). $\Box$

## References