# Global Schauder theory for
minimizers of the $H^{s}(\Omega)$ energy

Mouhamed Moustapha Fall  M.M.F.: African Institute for Mathematical Sciences
in Senegal, KM 2, Route de Joal, B.P. 14 18. Mbour, Sénégal
<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Xavier Ros-
Oton  X.R.: ICREA, Pg. Lluis Companys 23, 08010 Barcelona, Spain; and
Universitat de Barcelona, Departament de Matemàtiques i Informàtica, Gran Via
de les Corts Catalanes 585, 08007 Barcelona, Spain<EMAIL_ADDRESS>
###### Abstract.

We study the regularity of minimizers of the functional
$\mathcal{E}(u):=[u]_{H^{s}(\Omega)}^{2}+\int_{\Omega}fu$. This corresponds to
understanding solutions for the regional fractional Laplacian in
$\Omega\subset\mathbb{R}^{N}$. More precisely, we are interested on the global
(up to the boundary) regularity of solutions, both in the case of free
minimizers in $H^{s}(\Omega)$ (i.e., Neumann problem), or in the case of
Dirichlet condition $u\in H^{s}_{0}(\Omega)$ when $s>\frac{1}{2}$.

Our main result establishes the sharp regularity of solutions in both cases:
$u\in C^{2s+\alpha}(\overline{\Omega})$ in the Neumann case, and
$u/\delta^{2s-1}\in C^{1+\alpha}(\overline{\Omega})$ in the Dirichlet case.
Here, $\delta$ is the distance to $\partial\Omega$, and $\alpha<\alpha_{s}$,
with $\alpha_{s}\in(0,1-s)$ and $2s+\alpha_{s}>1$. We also show the optimality
of our result: these estimates fail for $\alpha>\alpha_{s}$, even when $f$ and
$\partial\Omega$ are $C^{\infty}$.

###### Key words and phrases:

Regional fractional Laplacian, regularity, Schauder estimates, Censored
processes.

###### 2010 Mathematics Subject Classification:

35B65, 35R11.

##  1\. Introduction and main results

Given $s\in(0,1)$ and a bounded domain $\Omega\subset\mathbb{R}^{N}$, we study
the regularity of minimizers of energy functionals of the type

$\mathcal{E}(u):=[u]_{H^{s}(\Omega)}^{2}+\int_{\Omega}fu\,dx,$ (1.1)

where $f\in L^{2}(\Omega)$,

$[u]_{H^{s}(\Omega)}^{2}=\frac{c_{N,s}}{4}\int_{\Omega}\int_{\Omega}\frac{\big{|}u(x)-u(y)\big{|}^{2}}{|x-y|^{N+2s}}\,dxdy,$

and
$c_{N,s}=\frac{s4^{s}\Gamma\left(\frac{N}{2}+s\right)}{\pi^{\frac{N}{2}}\Gamma(1-s)}$.

Notice that, as in the classical case $s=1$, one can study two types of
minimizers:

$\bullet$ Free minimizers $u\in H^{s}(\Omega)$ of (1.1) —i.e., minimizing
among all functions in $H^{s}(\Omega)$—, which corresponds to a Neumann
problem.

$\bullet$ Minimizers with prescribed zero boundary data —i.e., minimizing
among functions $u\in H^{s}_{0}(\Omega)$—, which corresponds to a Dirichlet
problem. Notice however that one needs $s>\frac{1}{2}$ in order to have a
trace operator111When $s\leq\frac{1}{2}$ we have
$H^{s}_{0}(\Omega)=H^{s}(\Omega)$ for any Lipschitz domain
$\Omega\subset\mathbb{R}^{N}$, see e.g. [14, Theorem 1.4.2.4]., thus _the
Dirichlet problem makes sense only for_ $s>\frac{1}{2}$.

The aim of this paper is to understand the global (up to the boundary)
regularity of solutions in both cases:

If $f$ and $\Omega$ are regular enough, what is the regularity of minimizers
$u$?

This is the question that motivates our work.

Minimizers of (1.1) solve the equation $(-\Delta)^{s}_{\Omega}u=f$ in $\Omega$
in the weak sense, where

$(-\Delta)^{s}_{\Omega}u(x)=c_{N,s}{\rm
p.v.}\int_{\Omega}\frac{u(x)-u(y)}{|x-y|^{N+2s}}\,dy$ (1.2)

is the so-called regional fractional Laplacian. Recall that this operator,
subjetcs to Dirichlet or Neumann boundary conditions are generated by Lèvy-
type processes which are censored to leave $\Omega$, see e.g. [3].

Problems involving the regional fractional Laplacian have been studied both
from the point of view of Probability [3, 7, 15, 16, 8] and of Analysis [17,
6, 2, 12]. The best known regularity results in this direction are those in
[12], where the first author established that for $\Omega$ sufficiently
regular and $f\in L^{\infty}(\Omega)$, solutions $u$ are
$C^{2s-\varepsilon}(\overline{\Omega})$ in the Neumann case and that
$u/\delta^{2s-1}\in C^{1-\varepsilon}(\overline{\Omega})$ for the Dirichlet
problem, for every $\varepsilon>0$.

Our main results here establish for the first time Schauder-type estimates for
such problem, which we show to be optimal. They can be summarized as follows:

######  Theorem 1.1 (Neumann problem).

Let $s\in(0,1)$, and let $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6, which
satisfies $2s+\alpha_{s}>1$. Let $\Omega\subset\mathbb{R}^{N}$ be any bounded
$C^{\max(1,2s+\alpha)}$ domain, and $f\in C^{\alpha}(\overline{\Omega})$ with
$\int_{\Omega}f\,dx=0$. Let $\alpha<\alpha_{s}$ with $\alpha+2s\neq 1$.

Let $u$ be the minimizer of (1.1) in $H^{s}(\Omega)$. Then, $u\in
C^{2s+\alpha}(\overline{\Omega})$ and

$\|u\|_{C^{2s+\alpha}(\overline{\Omega})}\leq
C\|f\|_{C^{\alpha}(\overline{\Omega})},$

with $C$ depending only on $N$, $s$, $\alpha$, and $\Omega$. In addition, if
$\alpha$ is such that $2s+\alpha>1$ then

$\partial_{\nu}u=0\quad\textrm{on}\quad\partial\Omega.$

Moreover, for $\alpha>\alpha_{s}$, solutions $u$ are in general not
$C^{2s+\alpha}(\overline{\Omega})$.

The critical exponent $\alpha_{s}>0$ is given by an explicit equation in terms
of Gamma functions; see Remark 1.4 below. For example, in the simplest case
$s=\frac{1}{2}$, it is the first positive solution of
$1+\alpha=\frac{1}{\pi}\tan(\pi\alpha)$.

Next, we notice that Theorem 1.1 implies that the minimizer $u$ satisfies
$(-\Delta)^{s}_{\Omega}u\in C^{\alpha}(\overline{\Omega})$ and thus it is in
fact a classical solution: $(-\Delta)^{s}_{\Omega}u(x)=f(x)$ for all
$x\in\overline{\Omega}$. This will follow from the results in Section 4 below.

######  Remark 1.2.

It is natural to wonder whether, for $s\in(0,\frac{1}{2}]$, solutions to the
Neumann problem are $C^{1}$ up to the boundary or not and have zero normal
derivative when $f$ and $\Omega$ are regular enough.

Our result answers this question for the first time: solutions are always
$C^{1}(\overline{\Omega})$, for every $s\in(0,1)$. Furthermore, since
$\alpha_{s}<1-s$, then solutions are in general not
$C^{2}(\overline{\Omega})$.

We think it is quite surprising that the answer is the same for all
$s\in(0,1)$.

In the Dirichlet case our result reads as follows:

######  Theorem 1.3 (Dirichlet problem).

Given $s\in(\frac{1}{2},1)$, let $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6.
Let $\Omega\subset\mathbb{R}^{N}$ be a $C^{2,\beta}$ domain and $f\in
C^{\alpha}(\overline{\Omega})$, with $\alpha<\min\\{\alpha_{s},2s-1,\beta\\}$
and $\alpha+2s\neq 1$.

Let $u$ be the minimizer of (1.1) in $H^{s}_{0}(\Omega)$. Then,
$u/\delta^{2s-1}\in C^{1+\alpha}(\overline{\Omega})$ and

$\|u/\delta^{2s-1}\|_{C^{1+\alpha}(\overline{\Omega})}\leq
C\|f\|_{C^{\alpha}(\overline{\Omega})},$

with $C$ depending only on $N$, $s$, $\alpha$, and $\Omega$. Here, $\delta\in
C^{2,\beta}(\overline{\Omega})$ is a positive function in $\Omega$ that
coincides with dist$(\cdot,\partial\Omega)$ near $\partial\Omega$.

In addition, if we denote $\psi:={u}/{\delta^{2s-1}}$, we have

$\partial_{\nu}\psi=-(N-1)H_{\partial\Omega}\,\psi\quad\textrm{on}\quad\partial\Omega,$
(1.3)

where $H_{\partial\Omega}$ is the mean curvature of $\partial\Omega$ and $\nu$
is the exterior normal of $\partial\Omega$.

Moreover, for $\alpha>\alpha_{s}$ the quotient $u/\delta^{2s-1}$ does not
belong in general to $C^{1+\alpha}(\overline{\Omega})$.

We point out that (1.3) does not hold in the local case $s=1$. Indeed, in that
case we have $\psi=-\partial_{\nu}u$ and hence
$\partial_{\nu}\psi=-(N-1)H_{\partial\Omega}\,\psi+f$ on $\partial\Omega$,
where we used $\Delta
u=\partial^{2}_{\nu}u+(N-1)H_{\partial\Omega}\partial_{\nu}u$ on
$\partial\Omega$.

A difficulty that arises in the proof of Theorem 1.3 in dimension $N\geq 2$ is
that $(-\Delta)^{s}_{\Omega}\delta^{2s-1}$ does not in general belong to
$L^{\infty}(\Omega)$ even if $\Omega$ is $C^{\infty}$. In fact near
$\partial\Omega$, we have that $(-\Delta)^{s}_{\Omega}\delta^{2s-1}(x)\asymp
g(x)\log(\delta)$, with $g|_{\partial\Omega}=(N-1)H_{\partial\Omega}$. This
leads to the assumption $\Omega$ to be of class $C^{2,\beta}$ with
$\beta>\alpha$. To overcome this difficulty we consider the correction
$\delta^{2s-1}+b\delta^{2s}$, for some appropriately chosen function $b$
defined on $\partial\Omega$. This yields also the identity (1.3).

######  Remark 1.4 (The 1D case).

Let us briefly explain what happens in dimension $1$.

Let us assume for example that $\Omega=\mathbb{R}_{+}$. Then, we will show
that any solution to $(-\Delta)^{s}_{\Omega}u=0$ in $(0,1)\subset\mathbb{R}$
has an expansion of the form

$\qquad\qquad
u(x)=c_{0}+a_{0}x^{2s-1}+a_{1}x^{\beta_{1}}+...+a_{k}x^{\beta_{k}}+...\quad\textrm{for}\quad
x\approx 0,$ (1.4)

where all $\beta_{k}$’s can be characterized as the positive solutions of

$\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)\Gamma(2s)}=\frac{1}{\pi}\frac{\sin\big{(}\pi(\beta-2s)\big{)}\sin(\pi
s)}{\sin\big{(}\pi(\beta-s)\big{)}}.$

By an appropriate analysis of such equation222Notice that when $s=\frac{1}{2}$
this equation becomes $\pi\beta=\tan(\pi\beta)$. In this case, it is easy to
visualize and prove all properties of the exponents $\beta_{k}$., we will show
that

$\beta_{k+1}>1+\beta_{k}\quad\textrm{for all}\quad k\geq 1,$

and $\beta_{1}=2s+\alpha_{s}$. In addition, we will see that
$\alpha_{s}\in(0,1-s)$ and $2s+\alpha_{s}>1$.

In case of zero Neumann boundary conditions on $\partial\Omega$, we must have
$a_{0}=0$ in (1.4), while in case of zero Dirichlet conditions we must clearly
have $c_{0}=0$. This already shows that the $C^{2s+\alpha}$ regularity result
from Theorem 1.1 cannot hold for $\alpha>\alpha_{s}$, as stated above.

In higher dimensions several difficulties arise, and we will have an expansion
of this type with some new terms, both coming from the tangential extra
variables and from the curvature of the domain.

We expect the assumptions on $\Omega$ in both Theorems 1.1 and 1.3 to be
optimal.

###  1.1. Preliminaries and definitions

Let us next give some important definitions that will be used throughout the
paper.

First, recall that given any open set $\Omega\subset\mathbb{R}^{N}$, the
regional fractional Laplacian in $\Omega$ is defined by (1.2), and its
associated bilinear form is given by

$D_{\Omega}(u,v)=\frac{c_{N,s}}{2}\int_{\Omega}\int_{\Omega}\frac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}\,dxdy$

The natural notion of weak solution is the following.

######  Definition 1.5.

We say that $u$ is a (weak) solution of

$(-\Delta)^{s}_{{\Omega}}u=f\quad\textrm{ in\quad$\Omega\cap B_{1}$}$ (1.5)

with zero _Neumann boundary condition_ on $\partial\Omega\cap B_{1}$ if $u\in
H^{s}(\Omega)$ and

$D_{\Omega}(u,\varphi)=\int_{\Omega}f\varphi\,dx\qquad\textrm{$\forall\varphi\in
C^{\infty}_{c}(B_{1})$.}$

On the other hand, when $s>\frac{1}{2}$, we say that $u$ is a (weak) solution
of (1.5) with zero _Dirichlet boundary condition_ on $\partial\Omega\cap
B_{1}$ if $u\in H^{s}_{0}(\Omega)$ and

$D_{\Omega}(u,\varphi)=\int_{\Omega}f\varphi\,dx\qquad\textrm{$\forall\varphi\in
C^{\infty}_{c}(B_{1}\cap\Omega)$.}$

By approximation, we can of course take any $\varphi\in H^{s}_{0}(\Omega\cap
B_{1})$.

We notice that being a weak solution is equivalent to being a minimizer of the
functional (1.1), for both the Neumann and the Dirichlet case.

###  1.2. Acknowledgements

The first author’s work is supported by the Alexander von Humboldt foundation.
The second author was supported by the European Research Council (ERC) under
Grant Agreement No 801867.

###  1.3. Organization of the paper

The paper is organized as follows. In Section 2 we completely characterize all
1D homogeneous solutions, and prove (1.4). In Section 3 we use this to prove
Liouville theorems in a half-space. Then, in Section 4 we provide some new
estimates for the regional fractional Laplacian in domains. Finally, in
Sections 5 and 6 we prove Theorems 1.1 and 1.3, respectively.

##  2\. 1D homogeneities

The aim of this section is to classify all possible homogeneities in dimension
1 and, as a consequence, to prove (1.4).

Roughly speaking, we want to classify all solutions $u\in
H^{s}_{\text{loc}}([0,\infty))$ of

$(-\Delta)^{s}_{\mathbb{R}_{+}}u=0\quad\textrm{in}\quad\mathbb{R}_{+}.$ (2.1)

To do so, we will define

$v(x):=\left\\{\begin{array}[]{ll}u(x)-u(0)&\quad\textrm{for}\quad x\geq 0\\\
0&\quad\textrm{for}\quad x\leq 0,\end{array}\right.$

and notice that we have333Here, $(-\Delta)^{s}$ denotes the fractional
Laplacian in $\mathbb{R}$, i.e., (1.2) with $\Omega=\mathbb{R}$.
$(-\Delta)^{s}v(x)=(-\Delta)^{s}_{\mathbb{R}_{+}}u(x)+u(x)c_{1,s}\int_{\mathbb{R}_{-}}|x-y|^{-1-2s}\,dy$.
Then by a direct computation, we find

$\begin{cases}(-\Delta)^{s}v-\displaystyle\frac{a_{s}}{x^{2s}}\,v=0&\qquad\textrm{
in $\mathbb{R}_{+}$}\\\ v=0&\qquad\textrm{ in $\mathbb{R}_{-}$,}\end{cases}$
(2.2)

where $a_{s}:=\frac{c_{1,s}}{2s}$. This allows us to use the Caffarelli-
Silvestre extension (see e.g. [4]).

Now note that, since our aim is to obtain $C^{2s+\alpha}$ regularity, we will
need to consider solutions to (2.2) that grow like $|x|^{2s+\alpha}$ at
infinity. Thus, the equations in (2.2) needs to be understood in a generalized
sense.
In the following of this paper we let $\chi_{R}\in C^{\infty}_{c}(B_{2R})$ be
such that $\chi_{R}\equiv 1$ in $B_{R}$.

######  Definition 2.1.

As in444This kind of definition was originally introduced in [10]. However,
the definition in [10] requires only pointwise convergence of $f_{R}$, while
here (as in [1]) we take uniform convergence. [1, Section 3], given a function
with polynomial growth

$\int_{\mathbb{R}^{N}}\frac{|u(x)|}{1+|x|^{N+2s+k}}\,dx<\infty,$

we say that

$(-\Delta)^{s}u\ {\stackrel{{\scriptstyle\mathclap{k}}}{{=}}}\
f\quad\textrm{in}\quad\Omega\subset\mathbb{R}^{N}$

if there exists a family of polynomials $p_{R}$ of degree at most $k-1$ and a
family of functions $f_{R}$ such that $(-\Delta)^{s}(u\chi_{R})=f_{R}+p_{R}$
in $\Omega\cap B_{R/2}$, with $f_{R}\to f$ uniformly in $\Omega$ as
$R\to\infty$.

Thanks to [1, Lemma 3.3], this is equivalent to saying that there exists an
extension $\tilde{u}(x,y)$ with polynomial growth in $\mathbb{R}^{N+1}_{+}$,
satisfying $\tilde{u}(x,0)=u(x)$ and ${\rm div}(t^{1-2s}\nabla\tilde{u})=0$ in
$\\{y>0\\}$, such that

$-\lim_{t\to
0}t^{1-2s}\partial_{t}\tilde{u}=\overline{\kappa}_{s}f\quad\textrm{on}\quad\Omega\cap\\{y=0\\}.$

It is easy to check that, if
$\int_{\mathbb{R}^{N}}|D^{k}u(x)|/(1+|x|^{N+2s})dx<\infty$, and
$(-\Delta)^{s}D^{k}u=D^{k}f$ in $\Omega$ for some function $f$, then
$(-\Delta)^{s}u\ {\stackrel{{\scriptstyle\mathclap{k}}}{{=}}}\ f$ in $\Omega$.

In the extended variables $(x,t)\in\mathbb{R}\times\mathbb{R}_{+}$ —and
denoting $v(x,t)$ the extension of $v(x)$—, (2.2) reads as

$\begin{cases}{\rm div}(t^{1-2s}\nabla v)=0&\qquad\textrm{ in $\\{t>0\\}$}\\\
\displaystyle-\lim_{t\to
0}t^{1-2s}\partial_{t}v=\overline{\kappa}_{s}\frac{a_{s}}{x^{2s}}v&\qquad\textrm{
on $\\{t=0,\ x>0\\}$}\\\ v=0&\qquad\textrm{ on $\\{t=0,\ x\leq
0\\}$,}\end{cases}$ (2.3)

where $\overline{\kappa}_{s}=\frac{\Gamma(1-s)}{2^{2s-1}\Gamma(s)}$.

If we look for homogeneous solutions

$\qquad\qquad v(r\cos\theta,r\sin\theta)=r^{\beta}\psi(\theta),$ (2.4)

with $r>0$ and $\theta\in[0,\pi]$, we are led to the the following eigenvalue
problem for the function $\psi$ in the space
$H^{1}\big{(}(0,\pi);\sin(\theta)^{1-2s}d\theta\big{)}$:

$\displaystyle\begin{cases}-\big{(}\sin(\theta)^{1-2s}\psi^{\prime}\big{)}^{\prime}=\lambda\sin(\theta)^{1-2s}\psi&\qquad\textrm{
for $\theta\in(0,\pi)$}\\\ -\lim_{\theta\to
0}\sin(\theta)^{1-2s}\psi^{\prime}(\theta)=\overline{\kappa}_{s}a_{s}\psi(0)&\\\
\psi(\pi)=0,&\end{cases}$ (2.5)

where

$\lambda=\beta(\beta-2s+1).$

Problem (2.5) possesses a sequence of increasing eigenvalues

$\lambda_{0}(s)<\lambda_{1}(s)\leq,\dots,$

with corresponding eigenfunctions $\psi_{k}\in
L^{2}\big{(}(0,\pi);\sin(\theta)^{1-2s}\big{)}$, normalized as

$\int_{0}^{\pi}\sin(\theta)^{1-2s}\psi^{2}_{k}(\theta)\,d\theta=1.$

Moreover, see e.g. [12], 555This follows from the Poincaré type inequality:
$\lambda_{0}(s)=\inf_{\stackrel{{\scriptstyle\psi\in
C^{1}([0,\pi])}}{{\psi(\pi)=0}}}\frac{\int_{0}^{\pi}\sin(\theta)^{1-2s}(\psi^{\prime}(\theta))^{2}\,d\theta-\overline{\kappa}_{s}a_{s}\psi(0)^{2}}{\int_{0}^{\pi}\sin(\theta)^{1-2s}\psi^{2}_{k}(\theta)\,d\theta}=0$.
we have $\lambda_{0}(s)=0$. Using this, we can prove the following.

######  Lemma 2.2.

Let $\lambda_{k}$ and $\psi_{k}$ as above, and let
$\beta_{k}=\frac{2s-1}{2}+\sqrt{\lambda_{k}(s)+(\frac{2s-1}{2})^{2}}$. Then,
the functions

$v_{k}(r,\theta)=r^{\beta_{k}}\psi_{k}(\theta)$

are homogeneous solutions of (2.3).

Moreover, if $w\in H^{1}\big{(}B_{\rho}^{+};t^{1-2s}dxdt\big{)}\cap
C(\overline{B_{\rho}^{+}})$ is any weak solution of

$\displaystyle\begin{cases}{\rm div}(t^{1-2s}\nabla w)=0&\qquad\textrm{ in
$B_{\rho}^{+}$}\\\ -\lim_{t\to
0}t^{1-2s}\partial_{t}w=\overline{\kappa}_{s}a_{s}x^{-2s}w&\qquad\textrm{ on
$\\{t=0,\ 0<x<\rho\\}$}\\\ w=0&\qquad\textrm{ on $\\{t=0,\ -\rho<x\leq
0\\}$},\end{cases}$ (2.6)

then

$w(r\cos\theta,r\sin\theta)=\sum_{k=0}^{\infty}\omega_{k}r^{\beta_{k}}\psi_{k}(\theta),$
(2.7)

with $\omega_{k}\in\mathbb{R}$.

###### Proof.

The first part of the lemma holds by construction of $\lambda_{k}$ and
$\psi_{k}$.

For the second part, notice that since $\psi_{k}$ are an orthonormal basis of
$L^{2}\big{(}(0,\pi);\sin(\theta)^{1-2s}d\theta\big{)}$, then we can write

$w(\rho\cos\theta,\rho\sin\theta)=\sum_{k\geq
0}\omega_{k}\rho^{{\beta_{k}}}\psi_{k}(\theta)$

for some constants $a_{k}\in\mathbb{R}$ (recall that here $\rho$ is a fixed
positive number).

But then, the function

$\tilde{w}(r\cos\theta,r\sin\theta):=\sum_{k\geq
0}\omega_{k}r^{{\beta_{k}}}\psi_{k}(\theta)$

is a solution of (2.6) that coincides with $w$ on $\partial B_{\rho}$.
Moreover, since $\lambda_{0}(s)=0$, then $\beta_{0}=\max({2s-1},0)$ so that by
continuity, $\omega_{0}=0$ for $s\leq 1/2$. This implies that $\tilde{w}\in
H^{1}(B_{\rho};t^{1-2s}dxdt)$. Now by uniqueness of solutions, the two
functions $\tilde{w}$ and $w$ must coincide. ∎

As a consequence, we find:

######  Corollary 2.3.

Let $u\in H^{s}([0,1])\cap C([0,1))$ be a function satisfying

$(-\Delta)^{s}_{\mathbb{R}_{+}}u=0\quad\textrm{in}\quad(0,1)\subset\mathbb{R},$

in the sense that $(-\Delta)^{s}v\stackrel{{\scriptstyle
k}}{{=}}a_{s}x^{-2s}v$ in $(0,1)$ for some $k\in\mathbb{N}$, with
$v:=(u-u(0))1_{\mathbb{R}_{+}}$. Then, near $x=0$ we have

$u(x)=\omega_{0}x^{\beta_{0}}+\omega_{1}x^{\beta_{1}}+\omega_{2}x^{\beta_{2}}+...,$

where $\omega_{i}\in\mathbb{R}$, and the exponents $\beta_{i}$ are given by
Lemma 2.2.

###### Proof.

It follows from Lemma 2.2. ∎

Our next goal is to find an explicit equation for these exponents $\beta_{k}$.
We start with the following

######  Lemma 2.4.

Let $s\in(0,1)$ and $\beta>-1$. Then, there exists a unique
$\beta$-homogeneous function $v_{\beta}$ satisfying

$\begin{cases}{\rm div}(t^{1-2s}\nabla v_{\beta})=0&\qquad\textrm{ in
$\mathbb{R}^{2}_{+}$}\\\ v_{\beta}(x,0)=(x_{+})^{\beta}&\qquad\textrm{ on
$\\{t=0\\}$,}\end{cases}$

Moreover, such function satisfies

$-\lim_{t\to
0}t^{1-2s}\partial_{t}v_{\beta}=\overline{\kappa}_{s}\,\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)}\frac{\sin(\pi(\beta-s))}{\sin(\pi(\beta-2s))}\,(x_{+})^{\beta-2s}.$

Equivalently, in $\mathbb{R}$ we have

$(-\Delta)^{s}(x_{+})^{\beta}\ {\stackrel{{\scriptstyle\mathclap{k}}}{{=}}}\
\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)}\frac{\sin(\pi(\beta-s))}{\sin(\pi(\beta-2s))}\,(x_{+})^{\beta-2s}\quad\textrm{for}\quad
x>0,$ (2.8)

where the equality must be understood in the sense of Definition 2.1, and with
$k>\beta-2s$.

###### Proof.

We will show666Equivalently, one could do these computations in the extended
variables, see [19, Appendix A] for a related result. (2.8). Assume first
$\beta\in(-1,2s)$. Then, at $x=1$ we have:

$(-\Delta)^{s}(x_{+})^{\beta}=c_{1,s}\int_{\mathbb{R}}\big{(}1-(1+y)_{+}^{\beta}\big{)}\frac{dy}{|y|^{1+2s}}=c_{1,s}\,\frac{2\pi\Gamma(-2s)}{\Gamma(-\beta)\Gamma(1-2s+\beta)}\frac{\cos(\pi
s)\sin(\pi(\beta-s))}{\sin(\pi\beta)\sin(\pi(\beta-2s))},$

where we used [9, Proposition 4.3].

Moreover, by definition of $(-\Delta)^{s}$, we have

$c_{1,s}=\frac{4^{s}\Gamma(s+{\textstyle\frac{1}{2}})}{\sqrt{\pi}|\Gamma(-s)|}=\frac{2\Gamma(2s)}{|\Gamma(-s)|\Gamma(s)}=\frac{2s}{\pi}\Gamma(2s)\sin(\pi
s),$

where we used

$\Gamma(z+{\textstyle\frac{1}{2}})\Gamma(z)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z),\qquad\Gamma(z+1)=z\Gamma(z),\qquad\textrm{and}\qquad\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi
z)}.$

Combining the previous expressions, we find that at $x=1$

$\begin{split}(-\Delta)^{s}(x_{+})^{\beta}&=\frac{2s}{\pi}\Gamma(2s)\sin(\pi
s)\,\frac{2\pi\Gamma(-2s)}{\Gamma(-\beta)\Gamma(1-2s+\beta)}\frac{\cos(\pi
s)\sin(\pi(\beta-s))}{\sin(\pi\beta)\sin(\pi(\beta-2s))}\\\
&=\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)}\frac{\sin(\pi(\beta-s))}{\sin(\pi(\beta-2s))},\end{split}$

where we used again the properties of the $\Gamma$ function and that
$\sin(2\pi s)=2\sin(\pi s)\cos(\pi s)$. Thus, by homogeneity, (2.8) follows in
case $\beta<2s$.

Once we have the result for $\beta\in(-1,2s)$, the general result follows by
induction on $\lfloor\beta\rfloor$, by noticing that we can take derivatives
in (2.8). ∎

As a consequence of the previous Lemma, we find:

######  Lemma 2.5.

Let $\beta>-1$, and $v_{\beta}$ be as in Lemma 2.4. Then, $v_{\beta}$ solves
(2.3) if and only if

$\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)\Gamma(2s)}=\frac{1}{\pi}\frac{\sin(\pi(\beta-2s))\sin(\pi
s)}{\sin(\pi(\beta-s))}.$ (2.9)

In particular, the exponents $\beta_{k}$ are characterized as solutions to
this equation.

###### Proof.

The statement is equivalent to (2.2) in the generalized sense. Thus, we need

$(-\Delta)^{s}(x_{+})^{\beta}=\frac{c_{1,s}}{2s}\,(x_{+})^{\beta-2s}.$

Thanks to the previous Lemma, this is equivalent to

$\frac{\Gamma(\beta+1)}{\Gamma(\beta-2s+1)}\frac{\sin(\pi(\beta-s))}{\sin(\pi(\beta-2s))}=\frac{1}{\pi}\Gamma(2s)\sin(\pi
s),$

and the result follows. ∎

Let us next analyze the equation (2.9).

######  Lemma 2.6.

There exists a sequence $0\leq\beta_{0}<\beta_{1}<\beta_{2}<...\to\infty$ of
solutions of (2.9). Moreover, they satisfy

$\beta_{0}=\max\\{2s-1,0\\},$
$\beta_{k}\in(k+\beta_{0},k+s)\qquad\textrm{and}\qquad\beta_{k+1}>1+\beta_{k}\quad\textrm{for}\quad
k\geq 1.$

In particular, we have $\beta_{1}=2s+\alpha_{s}$, with

$\alpha_{s}\in(0,1-s)\qquad\textrm{and}\qquad 2s+\alpha_{s}>1$

for all $s\in(0,1)$.

###### Proof.

Let us first notice that both $\beta=2s-1$ and $\beta=0$ are solutions of
(2.9). We let

$\beta_{0}:=\max\\{2s-1,0\\},$

and look for solutions $\beta>\beta_{0}$.

We consider the two functions

$h_{1}(\beta)=\frac{\Gamma(\beta-2s+1)\Gamma(2s)}{\Gamma(\beta+1)}$

and

$h_{2}(\beta)=\frac{\pi\sin(\pi(\beta-s))}{\sin(\pi(\beta-2s))\sin(\pi s)},$

and we need to study the intersection of their graphs for
$\beta\geq\beta_{0}$.

Let us first analyse the two functions separately, and then combine the
informations in order to see their intersections.

First notice that, by definition of the Beta function and its relation to the
Gamma function, for $\beta>2s-1$ we have

$h_{1}(\beta)=B(\beta-2s+1,2s)=\int_{0}^{1}t^{\beta-2s}(1-t)^{2s-1}dt.$

On the other hand, using $\sin(a+b)=\sin a\cos b+\sin b\cos a$ we find

$h_{2}(\beta)=\frac{\pi}{\tan(\pi(\beta-2s))}+\frac{\pi}{\tan(\pi s)}.$

The function $h_{1}$ satisfies:

$\lim_{\beta\downarrow 2s-1}h_{1}(\beta)=+\infty,\qquad
h_{1}(0)=\frac{\pi}{\sin(2\pi s)},\qquad\lim_{\beta\to+\infty}h_{1}(\beta)=0,$

and

$h_{1}(\beta)\ \textrm{is positive, decreasing, and convex in}\
(2s-1,\infty).$

The function $h_{2}$ satisfies:

$h_{2}(\beta)\ \textrm{is 1-periodic},\qquad\lim_{\beta\downarrow
2s-1}h_{2}(\beta)=+\infty,\qquad h_{2}(0)=\frac{\pi}{\sin(2\pi s)},$
$h_{2}(\beta)\ \textrm{is decreasing in the intervals }(2s-1,2s)+\mathbb{N},\
\,\textrm{convex in }(2s-1,2s-{\textstyle\frac{1}{2}})+\mathbb{N},$
$\textrm{it is positive in the intervals }(2s-1,s)+\mathbb{N},\ \,\textrm{and
negative in }(s,2s)+\mathbb{N}.$

We then split the analysis into three cases:

_Case 1_. Assume $s=\frac{1}{2}$. This is the easiest case, and we have
$h_{1}(\beta)=1/\beta$ and $h_{2}(\beta)=\pi/\tan(\pi(\beta-1))$. It is easy
to see that these two functions intersect exactly once at each interval
$(k,k+1)$, with $k\in\mathbb{N}$, at a point $\beta_{k}\in(k,k+\frac{1}{2})$
for $k\geq 1$ satisfying $\beta_{k+1}>1+\beta_{k}$.

_Case 2_. Assume $s>\frac{1}{2}$. In this case, $\beta_{0}=2s-1$. Then, we
claim that the functions $h_{1}$ and $h_{2}$ intersect exactly once in each of
the intervals $(2s-1+k,2s+k)$, $k\in\mathbb{N}$, $k\geq 1$.

Indeed, notice first that $h_{1}$ is continuous, positive, and decreasing in
$(2s-1,\infty)$, while $h_{2}$ is continuous, positive, and decreasing in
$(2s-1+k,s+k)$, $k\in\mathbb{N}$, with $h_{2}\to+\infty$ as $\beta\downarrow
2s-1+k$ and $h_{2}(s+k)=0$. Thus, these two functions intersect at least once
in each of the intervals $(2s-1+k,s+k)$, $k\geq 1$, and they never intersect
in $[s+k,2s+k]$ (since $h_{2}\leq 0$ therein).

It remains to see that they cannot intersect twice in one of these intervals.
For this, we notice that both $h_{1}$ and $h_{2}$ are differentiable in the
intervals $(2s-1+k,s+k)$. Moreover, for any $\beta\in(2s-1+k,s+k)$ and $k\geq
1$ we have

$h_{1}^{\prime}(\beta)\geq h_{1}^{\prime}(2s)=-\int_{0}^{1}|\log
t|(1-t)^{2s-1}dt\geq-1,$
$h_{2}^{\prime}(\beta)=-\frac{\pi^{2}}{\sin^{2}(-\pi(\beta-2s))}\leq-\pi^{2}.$

This means that $h_{2}^{\prime}<h_{1}^{\prime}$ in each of these intervals,
and therefore they cannot intersect twice in $(2s-1+k,s+k)$, for $k\geq 1$,
thanks to Rolle’s theorem.

It only remains to see that the two functions do not intersect in the first
interval $(2s-1,s)$. In that case, however, we are looking for solutions of
$(-\Delta)^{s}_{\mathbb{R}_{+}}x^{\beta}=0$ in $\mathbb{R}_{+}$, with
$\beta\in(2s-1,s)$, which we know they do not exist.

_Case 3_. Assume $s<\frac{1}{2}$. In this case, $\beta_{0}=0$. Then, we claim
that the functions $h_{1}$ and $h_{2}$ intersect exactly once in each of the
intervals $(2s-1+k,2s+k)$, $k\in\mathbb{N}$, $k\geq 1$.

Notice first that $h_{1}$ is continuous, positive, and decreasing in
$[0,\infty)$, while $h_{2}$ is continuous, positive, and decreasing in
$(2s-1+k,s+k)$, $k\in\mathbb{N}$, with $h_{2}\to+\infty$ as $\beta\downarrow
2s-1+k$ and $h_{2}(s+k)=0$. Thus, these two functions intersect at least once
in each of the intervals $(2s-1+k,s+k)$, $k\geq 1$, and they never intersect
in $[s+k,2s+k]$ (since $h_{2}\leq 0$ therein).

To see that they cannot intersect twice in one of these intervals, notice that
for $\beta\in(k+s,k+2s)$ the two functions cannot intersect —since $h_{2}$ is
negative therein—, while for $\beta\in(2s-1+k,k)$, $k\geq 1$, they cannot
intersect either —since $h_{1}<h_{1}(0)=h_{2}(0)=h_{2}(k)<h_{2}$ therein.

Instead, for $\beta\in(k,k+s)$, $k\geq 1$, we have777This can be seen by using
that $t^{1-2s}<1$ and that, since $|\log t|$ is decreasing and $(1-t)^{2s-1}$
is increasing, then $\int_{0}^{1}|\log t|(1-t)^{2s-1}dt\leq\int_{0}^{1}|\log
t|dt\int_{0}^{1}(1-t)^{2s-1}dt=\frac{1}{2s}$.

$h_{1}^{\prime}(\beta)\geq h_{1}^{\prime}(1)=-\int_{0}^{1}t^{1-2s}|\log
t|(1-t)^{2s-1}dt\geq-\frac{1}{2s},$ $h_{2}^{\prime}(\beta)\leq
h_{2}^{\prime}(1)=-\frac{\pi^{2}}{\sin^{2}(\pi(1-2s))}\leq-\frac{1}{s^{2}}.$

This means that in such intervals we have $h_{1}^{\prime}>h_{2}^{\prime}$, and
thus the functions cannot intersect twice in $(k,k+s)$ for all $k\geq 1$ by
Rolle’s theorem.

Notice also that we have showed that $\beta_{k}\in(k,k+s)$ for all $k\geq 1$.
Moreover, since $h_{1}$ is decreasing and $h_{2}$ is periodic, then
$\beta_{k+1}>1+\beta_{k}$.

Finally, notice that in the interval $(0,2s)$ we know that there are no
solutions to $(-\Delta)^{s}_{\mathbb{R}_{+}}x^{\beta}=0$ in $\mathbb{R}_{+}$
for $\beta\in(0,2s)$, and we are done. ∎

##  3\. Liouville theorems on the Half-space

###  3.1. Solutions with growth larger than $2s$

We denote
$\mathbb{R}^{N}_{+}=\\{x=(x^{\prime},x_{N})\in\mathbb{R}^{N-1}\times\mathbb{R}\,:\
x_{N}>0\\}$ and $B_{r}^{\prime}(z)$ the ball in $\mathbb{R}^{N-1}$ with radius
$r$ centred at $z$, while $B_{r}^{+}:=B_{r}\cap\mathbb{R}^{N}_{+}$.

######  Definition 3.1.

888This definition, is slightly different from Defnition 2.1. Here, we ask
more regularity on the reminder ${\mathcal{R}}_{R}$ to simplify the proofs.

Let $U$ be an open subset of $\mathbb{R}^{N}$ with $0\in\overline{U}$. Let
$u\in H^{s}(U)$, $f\in L^{1}_{loc}(U)$ and $\alpha<1$. We say that $u$ is a
solution to
$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}u\stackrel{{\scriptstyle\alpha}}{{=}}f$
in $U$ if for all $R\geq 1,$ there exist a constant $c_{R}\in\mathbb{R}$ and a
function ${\mathcal{R}}_{R}\in C^{0,1}(B_{R/2})$, such that

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}(\chi_{R}u)=f+c_{R}+{\mathcal{R}}_{R}\qquad\textrm{
in $U\cap B_{R/2}^{+}$},$ (3.1)

with ${\mathcal{R}}_{R}(0)=0$ and
$\|\nabla{\mathcal{R}}_{R}\|_{L^{\infty}(B_{R/2})}\leq C_{0}R^{\alpha-1}.$

######  Remark 3.2.

  1. (i)

We observe that if
$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}u\stackrel{{\scriptstyle\alpha}}{{=}}0$
in $U$ and $|u(x)|\leq C(1+|x|^{2s-\varepsilon})$, with $\varepsilon>0$, then
$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}u{=}0$ in $U$.

  2. (ii)

If $u$ satisfies (3.1) and $|u(x)|\leq C_{0}(1+|x|^{2s+\alpha})$. Then for any
open set $U^{\prime}\subset U$ with $U^{\prime}\cap B_{1}^{+}\not=\emptyset$,
there exists $C=C(N,s,\alpha,U^{\prime})$ such that

$|c_{R}|\leq C\left(C_{0}R^{\alpha}+\|f\|_{L^{1}(U^{\prime})}\right).$ (3.2)

Indeed, consider $\phi\in C^{\infty}_{c}(U^{\prime}\cap B_{1}^{+})$ with
$\int_{\mathbb{R}^{N}}\phi\,dx=1$. Then multiply (3.1) by $\phi$ and integrate
over $\mathbb{R}^{N}_{+}$ to get

$\displaystyle
c_{R}=\int_{\mathbb{R}^{N}}\chi_{R}u(-\Delta)^{s}_{\mathbb{R}^{N}_{+}}\phi\,dx-\int_{U^{\prime}}\phi
f\,dx-\int_{U^{\prime}}{\mathcal{R}}_{R}\phi\,dx.$

Since $|(-\Delta)^{s}_{\mathbb{R}^{N}_{+}}\phi(x)|\leq C(1+|x|)^{-N-2s}$, we
get (3.2).

The following result is contained in [12].

######  Lemma 3.3.

Let $v\in H^{s}(B_{2}^{+})\cap L^{\infty}(\mathbb{R}^{N})$ be a solution to

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}v=g\quad\text{in}\quad B_{2}^{+},$ (3.3)

with zero Dirichlet or Neumann boundary conditions on
$\partial\mathbb{R}^{N}_{+}\cap B_{2}$. Then

$\|v\|_{C^{\beta}(\overline{B_{1}^{+}})}\leq
C(N,s,\beta)(\|v\|_{L^{\infty}(\mathbb{R}^{N})}+\|g\|_{L^{\infty}(\mathbb{R}^{N})}),$

with $\beta=2s-1$ if $2s>1$ and $\beta\in(0,2s)$ if $2s\leq 1$. If moreover
$g\in C^{k}(B_{2})$, $k=1,2$, then

$\|\nabla_{x^{\prime}}^{k}v\|_{L^{\infty}(B_{1}^{+})}\leq
C(N,s)\left(\|v\|_{L^{\infty}(\mathbb{R}^{N})}+\|g\|_{C^{k}(B_{2})}\right).$

###  3.2. Liouville theorem

######  Lemma 3.4.

Let $u\in H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$ satisfy, for $\alpha<1$,
with $2s+\alpha\not=1$ and $2s+\alpha<2$,

$|u(x)|\leq C(1+|x|)^{2s+\alpha}$

and

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}u\stackrel{{\scriptstyle\alpha}}{{=}}0\qquad\textrm{
in }\mathbb{R}^{N}_{+},$

with zero Dirichlet or Neumann boundary conditions on
$\partial\mathbb{R}^{N}_{+}$. Then $u(x^{\prime},x_{N})=a\cdot
x^{\prime}+b(x_{N})$ in the Neumann case, while in the Dirichlet case
$u(x^{\prime},x_{N})=a\cdot x^{\prime}x_{N}^{2s-1}+\omega(x_{N})$. Here
$a\in\mathbb{R}^{N-1}$ is a constant and $\omega$ solves

$(-\Delta)^{s}_{{\mathbb{R}_{+}}}\omega\stackrel{{\scriptstyle\alpha}}{{=}}0\qquad\textrm{
in }\mathbb{R}_{+}.$

###### Proof.

By definition, we have

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}(\chi_{R}u)=c_{R}+{\mathcal{R}}_{R}\qquad\textrm{
in $B_{R/2}^{+}$}.$ (3.4)

Next, let $v(x)=(\chi_{R}u)(Mx)$, for $M=R/2$. Then

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}v=M^{2s}c_{R}+M^{2s}{\mathcal{R}}_{R}(M\cdot)\qquad\textrm{
in $B_{1}^{+}$}.$

Now by Lemma 3.3,

$\|\nabla_{x^{\prime}}v\|_{L^{\infty}(B_{1/2})}\leq
C(N,s)\left(\|v\|_{L^{\infty}(\mathbb{R}^{N})}+M^{2s}|c_{R}|+\|M^{2s}\nabla_{x^{\prime}}{\mathcal{R}}_{R}(Mx)\|_{L^{\infty}(B_{1})}\right).$

Hence, recalling the definition of ${\mathcal{R}}_{R}$, and Remark 3.2$(ii)$,
we get

$R\|\nabla_{x^{\prime}}(\chi_{R}u)\|_{L^{\infty}(B_{R/4})}\leq
C(N,s)R^{2s+\alpha}.$

Since $|u(x)\nabla_{x^{\prime}}\chi_{R}(x)|\leq CR^{2s+\alpha-1}$, we deduce
that, for all $R>1$,

$\|\nabla_{x^{\prime}}u\|_{L^{\infty}(B_{R/4})}\leq C(N,s)R^{2s+\alpha-1}.$

As a consequence

$|\nabla_{x^{\prime}}u(x)|\leq C(N,s)(1+|x|^{2s+\alpha-1})\qquad\textrm{ for
all $x\in\mathbb{R}^{N}_{+}$.}$ (3.5)

Differentiating (3.4) in $x^{\prime}$ and using Lemma 3.3, we obtain
$\nabla_{x^{\prime}}u\in H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$ and

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}(\chi_{R}\nabla_{x^{\prime}}u)=-(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}(u\nabla_{x^{\prime}}\chi_{R})+\nabla_{x^{\prime}}{\mathcal{R}}_{R}\qquad\textrm{
in $B_{R/2}^{+}$.}$ (3.6)

Thanks to (3.5) and the fact that
$|\nabla_{x^{\prime}}{\mathcal{R}}_{R}|+|(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}(u\nabla_{x^{\prime}}\chi_{R})|\leq
CR^{\alpha-1}$ in $B_{R/2}^{+}$, we can send $R\to\infty$ to deduce that

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}\nabla_{x^{\prime}}u=0\qquad\textrm{ in
$\mathbb{R}^{N}_{+}$.}$

with $\nabla_{x^{\prime}}u\in H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$
satisfies the same boundary condition as $u$. In view of (3.5) and since
$2s+\alpha-1<2s$, by applying the Liouville theorems in [12], we deduce that
$\nabla_{x^{\prime}}u(x^{\prime},x_{N})$ is a constant vector on
$\mathbb{R}^{N}_{+}$ in the Neumann case and is proportional to
$x_{N}^{\max(2s-1,0)}$ in the Dirichlet case. ∎

######  Lemma 3.5 (1D Liouville theorem).

Let $s\in(0,1)$, and let $\alpha_{s}$ be given by Lemma 2.6.

Let $\omega\in H^{s}_{loc}(\overline{\mathbb{R}_{+}})$ satisfy

$\qquad\qquad|\omega(x)|\leq
C(1+|x|)^{2s+\alpha},\qquad\textrm{with}\quad\alpha<\alpha_{s},$ (3.7)

and

$(-\Delta)^{s}_{{\mathbb{R}_{+}}}\omega\stackrel{{\scriptstyle\alpha}}{{=}}0\qquad\textrm{
in }\mathbb{R}_{+},$

with zero Dirichlet or zero Neumann at $\partial\mathbb{R}_{+}$. Then,
$\omega(x)=c_{1}+c_{2}x^{2s-1}$, for some $c_{1},c_{2}\in\mathbb{R}$.

###### Proof.

By Lemma 3.3, $(\omega-\omega(0))1_{\mathbb{R}_{+}}\in
H^{s}_{loc}([0,\infty))\cap C([0,\infty))$. In addition, we have that

$L_{s}\left[(\omega-\omega(0))1_{\mathbb{R}_{+}}\right]\stackrel{{\scriptstyle\alpha}}{{=}}0\qquad\textrm{
in $\mathbb{R}_{+}$},$

where

$L_{s}:=(-\Delta)^{s}-a_{s}x^{-2s}.$

By [1, Lemma 3.3], there exists $W\in
H^{1}_{loc}(\overline{\mathbb{R}^{2}_{+}};t^{1-2s}dxdt)\cap
C(\overline{\mathbb{R}_{2}^{+}})$ such that

$|W(x,t)|\leq C(1+(t+|x|)^{K}),$ (3.8)

for some constants $K\in\mathbb{N}$, $C>0$, and

$\displaystyle\begin{cases}\text{div}(t^{1-2s}\nabla W)=0&\qquad\textrm{in
}\mathbb{R}\times(0,\infty)\\\
-t^{1-2s}\partial_{t}W=\overline{\kappa}_{s}a_{s}x^{-2s}W&\qquad\textrm{on
$\mathbb{R}_{+}$}\times\\{0\\}\\\
W=(\omega-\omega(0))1_{\mathbb{R}_{+}}&\qquad\textrm{on
}\mathbb{R}\times\\{0\\}.\end{cases}$

Thanks to Lemma 2.2, for all $r>0$,

$W(r\cos\theta,r\sin\theta)=\sum_{k=0}^{\infty}b_{k}r^{\beta_{k}}\psi_{k}(\theta).$

By the Parseval identity and (3.8), we get

$\displaystyle\sum_{k=0}^{\infty}b_{k}^{2}r^{2\beta_{k}}=\int_{0}^{\pi}W^{2}(r\cos\theta,r\sin\theta)\sin(\theta)^{1-2s}\,d\theta\leq
C(1+r)^{2K}.$

Hence, letting $r$ to $\infty$ we find that
$W(r\cos\theta,r\sin\theta)=\sum_{\beta_{k}\leq
K}b_{k}r^{\beta_{k}}\psi_{k}(\theta)$. As a consequence

$W(r,0)=\sum_{\beta_{k}\leq K}b_{k}r^{\beta_{k}}\psi_{k}(0).$

Recall that $2s+\alpha<\beta_{1}<\beta_{k}$ for all $k>1$. Therefore since
$W(r,0)=\omega(r)-\omega(0)$ we then get from (3.7) and the above identity
that $b_{k}=0$ for all $k>1$, so that
$\omega(r)=\omega(0)+b_{0}r^{\beta_{0}}\psi_{0}(0)$.

∎

Combining Lemma 3.5, Lemma 3.4, and the regularity theory in [12, 2], where
have that $u\in C^{2s-1+\varepsilon}(\overline{B_{1}^{+}})$ in the zero Neuman
case, we get the

######  Theorem 3.6 (Liouville theorem: Neumann and Dirichlet).

Let $s\in(0,1)$ and let $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6. Let
$u\in H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$ satisfy,

$\qquad\qquad|u(x)|\leq
C_{0}(1+|x|)^{2s+\alpha},\qquad\textrm{with}\quad\alpha<\alpha_{s},$

and

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}u\stackrel{{\scriptstyle\alpha}}{{=}}0.$

  1. (i)

If $u$ satisfies zero Neumann boundary condition on
$\partial\mathbb{R}^{N}_{+}$, then $u(x^{\prime},x_{N})=B+A\cdot x^{\prime}$
for all $x\in\mathbb{R}^{N}_{+}$, for some constants $B\in\mathbb{R}$ and
$A\in\mathbb{R}^{N-1}$.

  2. (ii)

If $s>\frac{1}{2}$ and $u=0$ on $\partial\mathbb{R}^{N}_{+}$ then
$u(x)=(B+A\cdot x^{\prime})x_{N}^{2s-1}$ for all
$x=(x^{\prime},x_{N})\in\mathbb{R}^{N}_{+}$, for some constants
$B\in\mathbb{R}$ and $A\in\mathbb{R}^{N-1}$.

##  4\. Some estimates and formulas related to the regional fractional
laplacian

In this Section we establish some new estimates related to the regional
fractional Laplacian. To alleviate the notations, we assume in this section
that $c_{N,s}$, the constant in (1.2), is equal to 1. We start with the
following:

######  Lemma 4.1.

Let $s\in(0,1)$, $\alpha\in(0,1)$ be such that $1<2s+\alpha<2$,
$\Omega\subset\mathbb{R}^{N}$ be a $C^{2s+\alpha}$ domain, and $u\in
C^{2s+\alpha}(\overline{\Omega})$ satisfying $\partial_{\nu}u=0$ on
$\partial\Omega$. Then,

$\|(-\Delta)^{s}_{\Omega}u\|_{C^{\alpha}(\overline{\Omega})}\leq
C\|u\|_{C^{2s+\alpha}(\overline{\Omega})},$

with $C$ depending only on $N$, $s$, $\alpha$, and $\Omega$.

###### Proof.

It suffices to see that, for any $x^{\prime}\in B_{r}(x)$, where
$B_{2r}(x)\subset\Omega$, we have

$\big{|}(-\Delta)^{s}_{\Omega}u(x)-(-\Delta)^{s}_{\Omega}u(x^{\prime})\big{|}\leq
Cr^{\alpha};$

see e.g. [13, Appendix A].

For this, we write

$\begin{split}(-\Delta)^{s}_{\Omega}u(x^{\prime})&=\int_{(-x^{\prime}+\Omega)\cap
B_{r}}\frac{u(x^{\prime})-u(x^{\prime}+z)}{|z|^{N+2s}}\,dz+\int_{(-x^{\prime}+\Omega)\setminus
B_{r}}\frac{u(x^{\prime})-u(x^{\prime}+z)}{|z|^{N+2s}}\,dz\\\
&=:I_{1}(x^{\prime})+I_{2}(x^{\prime}),\end{split}$

for any $x^{\prime}\in B_{r}(x)$. Notice that $r>0$ is fixed, and that
$B_{r}(x^{\prime})\subset\Omega$ for all such $x^{\prime}$. Let us now treat
the two terms $I_{1}$ and $I_{2}$ separately.

For $I_{1}$ we have

$I_{1}(x^{\prime})=\int_{B_{r}}\frac{2u(x^{\prime})-u(x^{\prime}+z)-u(x^{\prime}-z)}{|z|^{N+2s}}\,dz=\int_{B_{r}}\int_{0}^{1}[\nabla
u(x^{\prime}+tz)-\nabla u(x^{\prime}-tz)]\cdot z\,\frac{dt\,dz}{|z|^{N+2s}}.$

Thus, since

$\big{|}\nabla u(x+tz)-\nabla u(x-tz)-\nabla u(x^{\prime}+tz)+\nabla
u(x^{\prime}-tz)\big{|}\leq C|z|^{2s+\alpha-1},$

we deduce that

$\big{|}I_{1}(x)-I_{1}(x^{\prime})\big{|}\leq
C\int_{B_{r}}|z|^{2s+\alpha-1}\frac{|z|dz}{|z|^{N+2s}}\leq Cr^{\alpha}.$

For $I_{2}$ we have

$\nabla I_{2}(x^{\prime})=\int_{\Omega\setminus B_{r}(x^{\prime})}\frac{\nabla
u(x^{\prime})-\nabla
u(y)}{|x^{\prime}-y|^{N+2s}}dy+\int_{\partial\Omega}\frac{u(x^{\prime})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y).$

The first term can be bounded as follows.

$\left|\int_{\Omega\setminus B_{r}(x^{\prime})}\frac{\nabla
u(x^{\prime})-\nabla u(y)}{|x^{\prime}-y|^{N+2s}}dy\right|\leq
C\int_{\Omega\setminus
B_{r}(x^{\prime})}\frac{|x^{\prime}-y|^{2s+\alpha-1}}{|x^{\prime}-y|^{N+2s}}dy\leq
Cr^{\alpha-1}.$

For the second term in $I_{2}$ we have notice that, since $\partial_{\nu}u=0$
on $\partial\Omega$, then

$\big{|}u(x^{\prime})-u(x^{*})\big{|}\leq Cd^{2s+\alpha},$

where $x^{*}\in\partial\Omega$ is the projection of $x^{\prime}$ onto
$\partial\Omega$, and $d={\rm dist}(x^{\prime},\partial\Omega)\geq r$. Then

$|u(x^{\prime})-u(y)|\leq Cd^{2s+\alpha}+|u(x^{*})-u(y)|,$

and therefore

$\left|\int_{\partial\Omega}\frac{u(x^{\prime})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y)\right|\leq
Cd^{\alpha-1}+\left|\int_{\partial\Omega}\frac{u(x^{*})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y)\right|.$

Using that $\partial\Omega$ is locally a $C^{2s+\alpha}$ graph, we will show
that

$\left|\int_{\partial\Omega}\frac{u(x^{*})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y)\right|\leq
Cd^{\alpha-1},$ (4.1)

and hence, combining the previous inequalities and using that $d>r$, we find

$|\nabla I_{2}(x^{\prime})|\leq Cr^{\alpha-1}.$

Thus, since this holds for any $x^{\prime}\in B_{r}(x)$, we deduce

$\big{|}I_{2}(x)-I_{2}(x^{\prime})\big{|}\leq C|x-x^{\prime}|r^{\alpha-1}\leq
Cr^{\alpha},$

and the result follows.

It only remains to prove (4.1). Clearly, it is enough to show

$\left|\int_{\partial\Omega\cap
B_{\rho_{\circ}}(x^{*})}\frac{u(x^{*})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y)\right|\leq
Cd^{\alpha-1},$ (4.2)

for some small $\rho_{\circ}>0$. For this, notice that

$u(y)=u(x^{*})+\nabla
u(x^{*})\cdot(y-x^{*})+O\big{(}|y-x^{*}|^{2s+\alpha}\big{)}$

and

$\nu(y)=\nu(x^{*})+O\big{(}|y-x^{*}|^{2s+\alpha-1}\big{)}.$

Therefore,

$\big{(}u(y)-u(x^{*})\big{)}\nu(y)=\nabla
u(x^{*})\cdot(y-x^{*})\nu(x^{*})+O\big{(}|y-x^{*}|^{2s+\alpha}\big{)}.$

Since $\nu(x^{*})$ and $\nabla u(x^{*})$ are fixed (and bounded), then

$\left|\int_{\partial\Omega\cap
B_{\rho_{\circ}}(x^{*})}\frac{u(x^{*})-u(y)}{|x^{\prime}-y|^{N+2s}}\,\nu(y)\,d\sigma(y)\right|\leq
C\left|\int_{\partial\Omega\cap
B_{\rho_{\circ}}(x^{*})}(y-x^{*})\,\frac{d\sigma(y)}{|x^{\prime}-y|^{N+2s}}\right|+Cd^{\alpha-1}.$

To bound the last integral, let us assume that $x^{*}=0$ and that
$\partial\Omega\cap B_{\rho_{\circ}}$ can be written as a graph in the
$e_{N}=-\nu(x^{*})$ direction, i.e., $\partial\Omega\cap
B_{\rho_{\circ}}(x^{*})=\\{x_{N}=g(x_{1},...,x_{N-1})\\}$. We denote
$z=(x_{1},...,x_{N-1})$, and then

$\int_{\partial\Omega\cap
B_{\rho_{\circ}}}(y-x^{*})\,\frac{d\sigma(y)}{|x^{\prime}-y|^{N+2s}}=\int_{B_{\rho_{\circ}}^{\prime}}\big{(}z,g(z)\big{)}\,\frac{J(z)dz}{\big{(}|d-g(z)|^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}},$

where $J(z)$ is the Jacobian of the change of variables, and
$B_{\rho_{\circ}}^{\prime}\subset\mathbb{R}^{N-1}$.

Since $g\in C^{2s+\alpha}$ and $g(0)=|\nabla g(0)|=0$, then $|g(z)|\leq
C|z|^{2s+\alpha}$, and the last component of such integral can be bounded by

$\int_{B_{\rho_{\circ}}^{\prime}}|g(z)|\,\frac{J(z)dz}{\big{(}|d-g(z)|^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\leq
C\int_{B_{\rho_{\circ}}^{\prime}}|z|^{2s+\alpha}\,\frac{dz}{\big{(}d^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\leq
Cd^{\alpha-1},$

where we used999It is easy to see that $|a-b|^{2}\geq\frac{1}{2}a^{2}-b^{2}$
for all $a,b\in\mathbb{R}$. that
$|d-g(z)|^{2}+|z|^{2}\geq\frac{1}{2}d^{2}-|g(z)|^{2}+|z|^{2}\geq\frac{1}{2}(d^{2}+|z|^{2})$
in $B_{\rho_{\circ}}$.

To control the first $n-1$ components, we symmetrize the integral, and then we
need to bound

$\frac{1}{2}\int_{B_{\rho_{\circ}}^{\prime}}z\left\\{\frac{J(z)}{\big{(}|d-g(z)|^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}-\frac{J(-z)}{\big{(}|d-g(-z)|^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\right\\}dz.$

As before, we have $|d-g(\pm z)|^{2}+|z|^{2}\asymp d^{2}+|z|^{2}$, and hence

$\left|\big{(}|d-g(z)|^{2}+|z|^{2}\big{)}^{-\frac{N+2s}{2}}-\big{(}|d-g(-z)|^{2}+|z|^{2}\big{)}^{-\frac{N+2s}{2}}\right|\leq
C\big{|}g(z)-g(-z)\big{|}\big{(}d^{2}+|z|^{2}\big{)}^{-\frac{N+2s+1}{2}}.$

Moreover,

$\big{|}J(z)-J(-z)\big{|}\leq
C|z|^{2s+\alpha-1}\qquad\textrm{and}\qquad\big{|}g(z)-g(-z)\big{|}\leq
C|z|^{2s+\alpha}.$

Therefore,

$\begin{split}\left|\int_{B_{\rho_{\circ}}^{\prime}}z\,\frac{J(z)dz}{\big{(}|d-g(z)|^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\right|&\leq
C\int_{B_{\rho_{\circ}}^{\prime}}|z|\left\\{\frac{|z|^{2s+\alpha}}{\big{(}d^{2}+|z|^{2}\big{)}^{\frac{N+2s+1}{2}}}+\frac{|z|^{2s+\alpha-1}}{\big{(}d^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\right\\}dz\\\
&\leq
C\int_{B_{\rho_{\circ}}^{\prime}}\frac{|z|^{2s+\alpha}}{\big{(}d^{2}+|z|^{2}\big{)}^{\frac{N+2s}{2}}}\,dz\leq
Cd^{\alpha-1}.\end{split}$

Hence, (4.2) follows, and we are done. ∎

In case $2s+\alpha<1$ we have the following.

######  Lemma 4.2.

Let $s,\alpha>0$ be such that $2s+\alpha<1$, let $\Omega\subset\mathbb{R}^{N}$
be any open set, and let $u\in C^{2s+\alpha}(\overline{\Omega})$. Then,

$\|(-\Delta)^{s}_{\Omega}u\|_{C^{\alpha}(\overline{\Omega})}\leq
C\|u\|_{C^{2s+\alpha}(\overline{\Omega})},$

with $C$ depending only on $N$, $s$, $\Omega$ and $\alpha$.

###### Proof.

It is easy to see that

$\|(-\Delta)^{s}_{\Omega}u\|_{L^{\infty}(\Omega)}\leq\sup_{x\in\Omega}\int_{\Omega}\frac{|u(x)-u(y)|}{|x-y|^{N+2s}}\,dy\leq
C\|u\|_{C^{2s+\alpha}(\overline{\Omega})}.$

Now, for $x,x^{\prime}\in\Omega$ let $r=2{|x-x^{\prime}|}$. We have

$\displaystyle
J(x,x^{\prime}):=(-\Delta)^{s}_{\Omega}u(x)-(-\Delta)^{s}_{\Omega}u(x^{\prime})=\int_{|x-y|<r}\frac{u(x)-u(y)}{|x-y|^{N+2s}}\,dy-\int_{|x-y|<r}\frac{u(x^{\prime})-u(y)}{|x^{\prime}-y|^{N+2s}}\,dy$
$\displaystyle+(u(x)-u(x^{\prime}))\int_{|x-y|>r}\frac{1}{|x^{\prime}-y|^{N+2s}}\,dy$
$\displaystyle-\int_{|x-y|>r}(u(x)-u(y))\left(\frac{1}{|x-y|^{N+2s}}-\frac{1}{|x^{\prime}-y|^{N+2s}}\right)\,dy.$

We observe that if $|y-x|<r$, then $|y-x^{\prime}|\leq 3r/2$ and if $|y-x|>r$,
then $|y-x^{\prime}|\geq r/2$. Moreover for all $t\in(0,1)$, if $|x-y|\geq r$,
then $|tx+(1-t)x^{\prime}-y|\geq|y-x|/2$. We then have

$\displaystyle|J(x,x^{\prime})|\leq Cr^{\alpha}$
$\displaystyle+Cr^{2s+\alpha}\int_{|x^{\prime}-y|>r/2}|x^{\prime}-y|^{-N-2s}\,dy+C\int_{0}^{1}\int_{|x-y|\geq
r}\frac{|x-x^{\prime}||x-y|^{2s+\alpha}dy}{|tx+(1-t)x^{\prime}-y|^{N+2s+1}}dt$
$\displaystyle\leq Cr^{\alpha}+C|x-x^{\prime}|\int_{|x-y|\geq
r}|x-y|^{2s+\alpha}|x-y|^{-N-2s-1}\leq Cr^{\alpha}.$

The proof is complete. ∎

Finally, we provide a new formula for the regional fractional Laplacian on
domains. This can be used to give a different proof of Lemma 4.1 in case
$s>\frac{1}{2}$.

######  Lemma 4.3.

Let $\Omega$ be a bounded $C^{\max(2s,1)}$ domain. Let $u\in
C^{2s+\alpha}(\overline{\Omega})$, with $2s+\alpha>1$ and $\partial_{\nu}u=0$
on $\partial\Omega$. Then, for $x\in\overline{\Omega}$ we have

$\displaystyle(-\Delta)^{s}_{\Omega}u(x)$
$\displaystyle=\int_{{\Omega}}\frac{u(x)-u(y)-\nabla
u(x)\cdot(x-y)}{|x-y|^{N+2s}}\,dy\,+$ $\displaystyle\hskip
128.0374pt+\frac{1}{N+2s-2}\int_{\partial\Omega}\frac{\big{(}\nabla
u(x)-\nabla u(y)\big{)}\cdot\nu(y)}{|x-y|^{N+2s-2}}\,d\sigma(y).$

where $\nu$ is the unit exterior normal of $\Omega$. Moreover, all the above
integrals converge absolutely.

Notice that, for simplicity, we establish the result in case of bounded
domains. Still, an analogous result could be proved for unbounded domains.

###### Proof of Lemma 4.3.

We have

$\displaystyle(-\Delta)^{s}_{\Omega}u(x)$
$\displaystyle=\int_{{\Omega}}\frac{u(x)-u(y)-\nabla
u(x)\cdot(x-y)}{|x-y|^{N+2s}}\,dy+p.v.\int_{{\Omega}}\frac{\nabla
u(x)\cdot(x-y)}{|x-y|^{N+2s}}\,dy.$

Define

$I_{\varepsilon}(x):=\int_{\Omega\cap\partial B_{\varepsilon}(x)}\frac{\nabla
u(x)\cdot\eta_{\varepsilon}(y)}{|x-y|^{N+2s-2}}d\sigma(y),$

where $\eta_{\varepsilon}$ is the unit exterior normal of
$B_{\varepsilon}(x)$. Using that
$\nabla_{y}|x-y|^{-N-2s+2}=(N+2s-2)(x-y)|x-y|^{-N-2s}$, the divergence theorem
and the fact that $\nabla u(y)\cdot\nu(y)=0$ for $y\in\partial\Omega$, we then
get

$\displaystyle(N+2s-2)p.v.\int_{{\Omega}}\frac{\nabla
u(x)\cdot(x-y)}{|x-y|^{N+2s}}\,dy$
$\displaystyle=p.v.\int_{\partial\Omega}\frac{\nabla
u(x)\cdot\nu(y)}{|x-y|^{N+2s-2}}d\sigma(y)$
$\displaystyle=\int_{\partial\Omega}\frac{[\nabla u(x)-\nabla
u(y)]\cdot\nu(y)}{|x-y|^{N+2s-2}}d\sigma(y)+\lim_{\varepsilon\to
0}I_{\varepsilon}(x).$

If $x\in\Omega$ then we immediately get $\lim_{\varepsilon\to
0}I_{\varepsilon}(x)=0$ and the result follows. We now consider the case
$x\in\partial\Omega$, then denote by $H_{x}:=T_{x}\partial\Omega$ the tangent
plane of $\partial\Omega$ at $x$ and $H^{+}_{x}$ the upper half space
containing $-\nu(x)$ with boundary $H_{x}$. The proof is now complete once we
show that

$I_{\varepsilon}(x)=-C(N)\varepsilon^{1-2s}\nabla
u(x)\cdot\nu(x)+o_{\varepsilon}(1)=o_{\varepsilon}(1),$ (4.3)

where $C(N)=\int_{S^{N-1}_{+}}\theta_{N}\,d\sigma(\theta)$. We have

$\displaystyle I_{\varepsilon}(x)=\int_{\Omega\cap\partial
B_{\varepsilon}(x)}\frac{\nabla
u(x)\cdot\eta_{\varepsilon}(y)}{|y-x|^{N+2s-2}}d\sigma(y)=\varepsilon^{-N-2s+1}\int_{\Omega\cap\partial
B_{\varepsilon}(x)}{\nabla u(x)\cdot(y-x)}d\sigma(y)$
$\displaystyle=\varepsilon^{-N-2s+1}\int_{H_{x}^{+}\cap\partial
B_{\varepsilon}(x)}{\nabla
u(x)\cdot(y-x)}d\sigma(y)-\varepsilon^{-N-2s+1}\int_{\Omega\bigtriangleup
H_{x}^{+}\cap\partial B_{\varepsilon}(x)}{\nabla u(x)\cdot(y-x)}d\sigma(y).$
(4.4)

By oddness, we have $\int_{H_{x}^{+}\cap\partial B_{\varepsilon}(x)}{\nabla
u(x)\cdot(y-x)}d\sigma(y)=-\varepsilon^{N}\nabla
u(x)\cdot\nu(x)\int_{S^{N-1}_{+}}\theta_{N}\,d\sigma(\theta)$. On the other
hand, since $\Omega$ is of class $C^{2s}$, we can write a neighbourhood of $x$
in $\partial\Omega$ as a graph of a function
$y_{N}=\gamma(y^{\prime})=o(|y^{\prime}|^{2s})$ over the tangent plane
$H_{x}$. We then observe that if $y\in\Omega\bigtriangleup
H_{x}^{+}\cap\partial B_{\varepsilon}(x)$ then $|y^{\prime}-x|\leq\varepsilon$
and $|y_{N}|\leq\gamma(y^{\prime})$, so that

$|\Omega\bigtriangleup H_{x}^{+}\cap\partial
B_{\varepsilon}(x)|=\varepsilon^{N-2}o(\varepsilon^{2s}).$

We thus conclude from this and (4.4) that (4.3) holds. ∎

######  Remark 4.4.

It follows from the above proof that the map

$x\mapsto\int_{\Omega\cap|x-y|>\varepsilon}\frac{u(x)-u(y)}{|x-y|^{N+2s}}\,dy$

converges in $C_{loc}(\overline{\Omega})$ to $(-\Delta)^{s}_{\Omega}u$, as
$\varepsilon\to 0$. The proof also shows that for $(-\Delta)^{s}_{\Omega}u(x)$
to be defined at $x\in\partial\Omega$, then necessarily,
$\partial_{\nu}u(x)=0$. See also [16], where these type of results were
obtained.

##  5\. Regional Neumann problem

In this section, we consider $\Omega$ a domain of class
$C^{\max(1,2s+\alpha)}$ with $0\in\partial\Omega$ and $\nu(0)=-e_{N}$. For,
$2>2s+\alpha>1$, we suppose that there exists a global diffeomorphism $\psi\in
C^{2s+\alpha}(\mathbb{R}^{N};\mathbb{R}^{N})$, satisfying

$\partial\Omega\cap B_{2}=\psi(B_{2}^{\prime}\times\\{0\\}),\qquad\Omega\cap
B_{2}=\psi(B_{2}^{+}),\qquad\psi(0)=0$ (5.1)

and

$-\frac{\partial_{y_{N}}\psi(y^{\prime},0)}{|\partial_{y_{N}}\psi(y^{\prime},0)|}=\nu(\psi(y^{\prime},0))\quad\textrm{
for $y^{\prime}\in B_{2}^{\prime}$},$ (5.2)

where $\nu$ is the exterior normal of $\partial\Omega$, see [1] for the
construction of such map.
In the case $2s+\alpha<1$, we simply assume that $\partial\Omega\cap B_{2}$ is
parameterized by a global diffeomorphism $\psi\in
C^{1}(\mathbb{R}^{N};\mathbb{R}^{N})$ satisfying (5.1).
Next, we define the operator ${\mathcal{L}}_{\psi}$ by

${\mathcal{L}}_{\psi}w(x):=c_{N,s}\int_{B_{2}^{+}}\frac{w(x)-w(y)}{|\psi(x)-\psi(y)|^{N+2s}}Jac_{\psi}(y)\,dy,$
(5.3)

so that for $u(x)=w(\psi(x))$, we get

$(-\Delta)^{s}_{\psi(B_{2}^{+})}u(\psi(x))={\mathcal{L}}_{\psi}w(x).$ (5.4)

We also define the bilinear form
$\mathcal{D}_{\psi}:H^{s}(\mathbb{R}^{N}_{+})\times
H^{s}(\mathbb{R}^{N}_{+})\to\mathbb{R}$ corresponding to this operator as

$\mathcal{D}_{\psi}(w,\varphi)=\frac{c_{N,s}}{2}\int_{B_{2}^{+}\times
B_{2}^{+}}\frac{(w(x)-w(y))(\varphi(x)-\varphi(y))}{|\psi(x)-\psi(y)|^{N+2s}}Jac_{\psi}(x)Jac_{\psi}(y)\,dydx.$

Let $A\in\mathbb{R}^{N-1}=\\{y_{N}=0\\}$ and define

$q_{A}(y)=A\cdot y=A\cdot y^{\prime}$ (5.5)

We now define

$Q_{A}:\mathbb{R}^{N}\to\mathbb{R}^{N},\qquad Q_{A}(x)=q_{A}(\psi^{-1}(x)).$

We have

$Q_{A}(\psi(y))=q_{A}(y)=A\cdot y^{\prime},$

and thus

$0=\partial_{y_{N}}q_{A}(y)=\nabla
Q_{A}(\psi(y))\cdot\partial_{y_{N}}\psi(y).$

This, together with (5.2), imply that

$\partial_{\nu}Q_{A}=0\qquad\textrm{ on $\partial\Omega\cap B_{2}$.}$

Hence, since $Q_{A}\in C^{2s+\alpha}(B_{2}\cap\overline{\Omega})$ for
$2s+\alpha>1$, by Lemma 4.1 and (5.4) we get:

######  Lemma 5.1.

Let ${\mathcal{L}}_{\psi}$ and $q_{A}$ be given by (5.4) and (5.5),
respectively. Then,

$\displaystyle\|{\mathcal{L}}_{\psi}q_{A}\|_{C^{\alpha}(\overline{B_{1}^{+}})}\leq
C|A|.$

with $C$ depending only on $\alpha,\|\psi\|_{C^{2s+\alpha}(B_{2})},N,s$.

###### Proof.

It follows from Lemma 4.1 and (5.4). ∎

Let

$\ell=\ell_{\alpha}=\begin{cases}1\quad\textrm{ for $2s+\alpha>1$}\\\
0\quad\textrm{ for $2s+\alpha<1$.}\end{cases}$ (5.6)

For $r>0$ and $w\in L^{2}_{loc}(\mathbb{R}^{N})$, let
$A_{r,w}\in\mathbb{R}^{N-1}$ such that

$\inf_{A\in\mathbb{R}^{N}}\|w-\ell A\cdot
y^{\prime}\|_{L^{2}(B_{r}^{+})}^{2}=\|w-\ell A_{r,w}\cdot
y^{\prime}\|_{L^{2}(B_{r}^{+})}^{2}.$ (5.7)

Then

$\ell A_{r,w}\cdot
e_{i}=\frac{1}{\int_{B_{r}^{+}}z_{i}^{2}\,dz}\int_{B_{r}^{+}}w(y)y_{i}\,dy.$
(5.8)

In particular

$\int_{B_{r}^{+}}(w(y)-\ell A_{r,w}\cdot y^{\prime})y_{i}\,dy=0\qquad\textrm{
for all $i=1,\dots,N-1$.}$ (5.9)

Before stating the main result of the present section, we record first the
following result from [12].

######  Lemma 5.2 ([12]).

Let $u\in H^{s}(\mathbb{R}^{N}_{+})$ and $f\in L^{\infty}(\mathbb{R}^{N}_{+})$
be a solution to
$\mathcal{D}_{\psi}(u,\varphi)=\int_{\mathbb{R}^{N}_{+}}u\varphi\,dx$ for all
$\varphi\in C^{\infty}_{c}(B_{2})$. Then, for all $\varrho\in(0,2s)$, there
exists $C=C(N,s,\varrho)>0$ such that

$\|u\|_{C^{2s-\varrho}(B_{1})}\leq
C\left(\|u\|_{L^{2}(\mathbb{R}^{N}_{+})}+\|f\|_{L^{\infty}(\mathbb{R}^{N}_{+})}\right).$

We next show the following.

######  Proposition 5.3.

Let $s\in(0,1)$ and $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6. Let
$\alpha\in(0,\alpha_{s})$, with $2s+\alpha\not=1$, $\psi$ be as above, with
$\|\psi\|_{C^{\max(1,2s+\alpha)}(\mathbb{R}^{N})}\leq c_{0}$. Let $f\in
C^{\alpha}(\mathbb{R}^{N}_{+})$ and $w\in H^{s}(\mathbb{R}^{N}_{+})\cap
C(\mathbb{R}^{N}_{+})$ such that $w(0)=0$ and

$\mathcal{D}_{\psi}(w,\varphi)=\int_{\mathbb{R}^{N}_{+}}f\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2})$}.$

Let $\varrho<2s+\alpha-1$ if $2s+\alpha>1$. Then, provided

$\ell|A_{r,w}|\leq c_{0}r^{-\varrho},$ (5.10)

there exists $C=C(N,s,\alpha,c_{0},\varrho,\beta)$ such that

$\sup_{r>0}r^{-2s-\beta}\|w-\ell A_{r,w}\cdot
y^{\prime}\|_{L^{\infty}(B_{r}^{+})}\leq
C(\|w\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}),$
(5.11)

where $\beta=\alpha-\varrho$, $\ell=\ell_{\alpha}$ and $A_{r,u}$ are given by
(5.6) and (5.7) respectively.

###### Proof.

Suppose that (5.11) does not hold. Then for all integer $n\geq 2$, there exist
$\psi_{n}\in C^{\max(1,2s+\alpha)}(\mathbb{R}^{N})$, $f_{n}\in
C^{\alpha}_{c}(\mathbb{R}^{N}_{+})$ and $w_{n}\in
H^{s}(\mathbb{R}^{N}_{+})\cap C(\mathbb{R}^{N}_{+})$, with $w_{n}(0)=0$, such
that

$\|\psi_{n}\|_{C^{\max(1,2s+\alpha)}(\mathbb{R}^{N})}\leq c_{0},$ (5.12)
$\|w_{n}\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f_{n}\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq
1$ (5.13)

such that

$\ell|A_{r,w_{n}}|\leq c_{0}r^{-\varrho},$ (5.14)
$\mathcal{D}_{\psi_{n}}(w_{n},\varphi)=\int_{\mathbb{R}^{N}_{+}}f_{n}\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2})$}.$ (5.15)

and

$\sup_{r>0}r^{-2s-\alpha}\|w_{n}-\ell A_{r,w_{n}}\cdot
y^{\prime}\|_{L^{\infty}(B_{r}^{+})}>n.$

Then, since $2s+\beta>1$ for $2s+\alpha>1$, by a well known (see e.g. [18]),
we can find sequences $r_{n}\to 0$ and $\theta_{n}\to\infty$ such that the
function

$v_{n}(y):=\frac{1}{r_{n}^{2s+\alpha}\theta_{n}}\left(w_{n}(r_{n}y)-r_{n}\ell
A_{r_{n},w_{n}}\cdot y^{\prime}\right)$

satisfies

$\|v_{n}\|_{L^{\infty}(B_{1}^{+})}\geq\frac{1}{4}$ (5.16)

and, in view of (5.14), for all $M\geq 1$ with $Mr_{n}\leq 1$,

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq CM^{2s+\beta},$ (5.17)

while for $2s+\alpha<1$

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq CM^{2s+\beta}\qquad\textrm{ for all
$M\geq 1$},$ (5.18)

Moreover, by construction (recalling (5.9)),

$v_{n}(0)=0,\qquad\int_{B_{1}^{+}}v_{n}(y)y_{i}\,dy=0\qquad\textrm{ for all
$i=1,\dots,N-1$.}$ (5.19)

Now for $Mr_{n}>1$ and $2s+\alpha>1$, we have

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq\frac{1}{r_{n}^{2s+\alpha}\theta_{n}}(1+r_{n}^{-\varrho})\leq
M^{2s+\beta}+M^{2s+\beta+\varrho}\leq 2M^{2s+\alpha}.$

In view of this, (5.17) and (5.18), we see that whenever $2s+\alpha\not=1$,

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq CM^{2s+\alpha}\qquad\textrm{ for all
$M\geq 1$}.$ (5.20)

Define $\widetilde{\psi}_{n}(x)=\frac{1}{r_{n}}\psi(r_{n}x)$ and let
$1<R<\frac{1}{r_{n}}$. Then by a change of variable in (5.15), we thus get

$\mathcal{D}_{\widetilde{\psi}_{n}}(\chi_{R}v_{n},\varphi)=\int_{\mathbb{R}^{N}_{+}}G^{n}_{R}\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{\frac{R}{2}})$},$ (5.21)

where

$G^{n}_{R}(x):=\frac{1}{r_{n}^{\beta}\theta_{n}}f_{n}(r_{n}x)+c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{(\chi_{R}(x)-\chi_{R}(y))v_{n}(y)}{|\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|^{N+2s}}J_{n}(y)\,dy-\frac{\ell}{r_{n}^{\alpha}\theta_{n}}{\mathcal{L}}_{\psi_{n}}[A_{r_{n},w_{n}}\cdot
y^{\prime}](r_{n}x),$

where $J_{n}(y):=Jac_{\widetilde{\psi}_{n}}(y)1_{B^{+}_{\frac{1}{r_{n}}}}(y).$
Since $\psi_{n}$ is a global diffeomorphism, by (6.15) and increasing $c_{0}$
if necessary, we obtain

$c_{0}|x-y|\geq|\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|\geq
c_{0}^{-1}|x-y|,\qquad|D\widetilde{\psi}_{n}(x)|\leq c_{0}\qquad\textrm{ for
all $x,y\in\mathbb{R}^{N}$. }$ (5.22)

Define, for $x\in B_{R/2}$,

$\displaystyle F_{R}^{n}(x)$
$\displaystyle:=c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{(\chi_{R}(x)-\chi_{R}(y))v_{n}(y)}{|\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|^{N+2s}}J_{n}(y)\,dy$
$\displaystyle=c_{N,s}\int_{\mathbb{R}^{N}_{+}\cap\\{|y|\geq
R\\}}\frac{(1-\chi_{R}(y))v_{n}(y)}{|\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|^{N+2s}}J_{n}(y)\,dy.$

Since $\widetilde{\psi}_{n}(0)=0$, by the mean value theorem, we have
$F_{R}^{n}(x)=F_{R}^{n}(0)+{\mathcal{R}}_{R}^{n}(x)$, where

${\mathcal{R}}_{R}^{n}(x):=-(N+2s)c_{N,s}\int_{0}^{1}\int_{\mathbb{R}^{N}_{+}}\frac{(\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y))\cdot\widetilde{\psi}_{n}(x)}{|\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|}\frac{(1-\chi_{R}(y))v_{n}(y)}{|t\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|^{N+2s+1}}J_{n}(y)\,dy\,dt.$

Note that by (5.22),
$|t\widetilde{\psi}_{n}(x)-\widetilde{\psi}_{n}(y)|\geq\frac{c_{0}}{2}|y|$
whenever $|y|\geq R$ and $|x|\leq\frac{R}{2c_{0}^{2}}$. Therefore, by (5.20)
and (5.22), there exists $C,n_{0}>0$ such that

$|\nabla{\mathcal{R}}_{R}^{n}(x)|\leq{C}R^{\alpha-1}\qquad\textrm{ for all
$R>1$, $|x|\leq\frac{R}{2c_{0}^{2}}$ and $n\geq n_{0}$}$ (5.23)

and by construction we have ${\mathcal{R}}_{R}^{n}(0)=0$. We can thus write

$G^{n}_{R}(x)=H^{n}_{R}(x)+c^{n}_{R}+{\mathcal{R}}^{n}_{R}(x),$ (5.24)

with

$H^{n}_{R}(x)=\frac{1}{r_{n}^{\beta}\theta_{n}}\left\\{f_{n}(r_{n}x)-f_{n}(0)+{\mathcal{L}}_{\psi_{n}}[A_{r_{n},w_{n}}\cdot
y^{\prime}](r_{n}x)-{\mathcal{L}}_{\psi_{n}}[A_{r_{n},w_{n}}\cdot
y^{\prime}](0)\right\\}$

and

$c^{n}_{R}=\frac{1}{r_{n}^{\beta}\theta_{n}}\left\\{f_{n}(0)+{\mathcal{L}}_{\psi_{n}}[A_{r_{n},w_{n}}\cdot
y^{\prime}](0)\right\\}+F_{R}^{n}(0).$

Recalling that $\|f_{n}\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq 1$, we then get
from (5.14) and Lemma 5.1 that

$|H^{n}_{R}(x)|\leq\frac{r_{n}^{\alpha}C}{r_{n}^{\beta}\theta_{n}}+\ell\frac{r_{n}^{\alpha-\varrho}}{r_{n}^{\beta}\theta_{n}}\leq\frac{C(R)}{\theta_{n}}\qquad\textrm{
for all $x\in B_{R/2}$.}$ (5.25)

Let us now show that $c_{R}^{n}$ is bounded. For this, we pick $\phi\in
C^{\infty}_{c}(B_{\frac{1}{2c_{0}^{2}}}^{+})$ with
$\int_{\mathbb{R}^{N}}\phi\,dx=1$. Then multiply (5.21) by $\phi$ and
integrate over $\mathbb{R}^{N}_{+}$ to get

$\displaystyle|c^{n}_{R}|\leq\int_{\mathbb{R}^{N}_{+}}|\chi_{R}(x)v_{n}(x)||{\mathcal{L}}_{\widetilde{\psi}_{n}}\phi(x)|\,dx+\int_{\mathbb{R}^{N}_{+}}|H_{R}^{n}(x)\phi(x)|dx+\int_{\mathbb{R}^{N}_{+}}|{\mathcal{R}}^{n}_{R}(x)\phi(x)|dx\leq
C(R).$ (5.26)

In view of this, (5.21), (5.24), (5.25) and (5.23), we can apply Lemma 5.2, to
deduce that $(v_{n})_{n}$ is bounded in
$C^{s+\delta}_{loc}(\overline{\mathbb{R}^{N}_{+}})$, for some $\delta>0$, and
converges to some $v\in
C^{s+\delta}_{loc}(\overline{\mathbb{R}^{N}_{+}})\subset
H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$. In addition passing to the limit
in (5.16), (5.20) and (5.19), we get

$|v(x)|\leq C(1+|x|^{2s+\alpha}),$ (5.27)
$\|v\|_{L^{\infty}(B_{1}^{+})}\geq\frac{1}{4},\qquad
v(0)=0,\qquad\int_{B_{1}^{+}}v(y)y_{i}\,dx=0,\textrm{ for $i=1,\dots,N-1$. }$
(5.28)

Passing to the limit in (5.21), (5.23) and using (5.26), we get, for all
$\phi\in C^{\infty}_{c}(B_{\frac{Rc_{0}^{2}}{2}})$,

$c_{N,s}\int_{\mathbb{R}^{2N}_{+}}\frac{((\eta_{R}v)(x)-(\eta_{R}v)(y))(\phi(x)-\phi(y))}{|Bx-
By|^{N+2s}}\textrm{det}(B)^{2}dydx=\int_{\mathbb{R}^{N}_{+}}(c_{R}^{\infty}+{\mathcal{R}}^{\infty}_{R}(x))\phi(x)\,dx,$

where
$B=\lim_{n\to\infty}D\widetilde{\psi}_{n}(0)=\lim_{n\to\infty}D\psi_{n}(0)$
and

$c^{\infty}_{R}\in\mathbb{R},\qquad{\mathcal{R}}_{R}^{\infty}(0)=0,\qquad|\nabla{\mathcal{R}}_{R}^{\infty}(x)|\leq{C}R^{\alpha-1}\qquad\textrm{
for all $R\geq 2$ and $x\in B_{\frac{R}{2c_{0}^{2}}}$.}$ (5.29)

Recall from the construction of $\psi_{n}$ that
$B=\textrm{diag}(1,\dots,1,\lambda)$ is a diagonal matrix, for some
$\lambda\not=0$. Hence by scaling and (6.39), we can assume that

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}v\stackrel{{\scriptstyle\alpha}}{{=}}0\qquad\textrm{
in $\mathbb{R}^{N}_{+}$}.$

Applying Theorem 3.4 and using (6.37), we deduce that $v(x)=b+a\cdot
x^{\prime}$. This is in contradiction with (5.28). ∎

As a consequence, we have :

######  Corollary 5.4.

Let $s\in(0,1)$ and $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6. Let
$\alpha\in(0,\alpha_{s})$, $\psi$ be as above, with
$\|\psi\|_{C^{\max(1,2s+\alpha)}(\mathbb{R}^{N})}\leq c_{0}$. Let $f\in
C^{\alpha}(\mathbb{R}^{N}_{+})$ and $w\in H^{s}(B_{2}^{+})\cap
L^{\infty}(\mathbb{R}^{N}_{+})$ such that

$\mathcal{D}_{\psi}(w,\varphi)=\int_{\mathbb{R}^{N}_{+}}f\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2})$}.$

Then, there exists $C=C(N,s,c_{0},\alpha)$ such that for all $z\in
B_{1/2}^{\prime}$ there exists a vector
$e=e_{z}\in\mathbb{R}^{N-1}=:\partial\mathbb{R}^{N}_{+}$, with $|e|\leq C$ and

$\sup_{r>0}r^{-2s-\alpha}\|w-w(z)-e\cdot(y^{\prime}-z)\|_{L^{\infty}(B_{r}^{+}(z))}\leq
C\left(\|w\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\right).$
(5.30)

In particular, if $2s+\alpha>1$ then $\partial_{y_{N}}w=0$ on
$B_{1/2}^{\prime}$.

###### Proof.

Thanks to Lemma 5.2, the function $v_{z}(x)=\chi_{2}(x)(w(x+z)-w(z))$
satisfies the hypothesis of Proposition 5.3 . We can thus prove (5.30) with
$z=0$. We assume for simplicity that
$\|w\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq
1.$
If $2s+\alpha_{s}<1$, then the result clearly holds with $e=0$. We start with
(5.30) and we distinguish the two cases $2s>1$ and $2s<1$. We start with the
case $2s>1$, where Lemma 5.2, yields $|w(y)|\leq C|y|$, so that (6.25) holds
with $\varrho=0$. By (5.11) and an (algebraic) iteration argument (see e.g.
[18, 5]), we can find $e\in\mathbb{R}^{N-1}$ such that

$\|w-e\cdot y^{\prime}\|_{L^{\infty}(B_{r}^{+})}\leq
Cr^{2s+\alpha},\qquad|e|\leq C.$ (5.31)

We finally consider the case $2s<1$. Here, (5.11) implies that $|w(y)|\leq
C|y|^{2s+\alpha^{\prime}}$ as soon as $2s+\alpha^{\prime}<1$ (i.e.
$\alpha^{\prime}<1-2s<\alpha$). As a consequence $w$ satisfies (6.25) with
$\varrho=1-2s-\alpha^{\prime}$. Therefore choosing $\alpha^{\prime}$ close to
$1-2s$ so that $2s+\alpha-\varrho>1$, we get from (5.11) and an iteration
argument, as above, that

$\|w-e\cdot y^{\prime}\|_{L^{\infty}(B_{r}^{+})}\leq
Cr^{2s+\alpha-\varrho},\qquad|e|\leq C,$ (5.32)

which again implies that $|w(y)|\leq C|y|$. Hence $w$ satisfies (5.33) with
$\varrho=0$ which, once again by an iteration argument, yields (5.32) with
$\varrho=0$.

If $2s=1$ then Lemma 5.2 implies that $|w(y)|\leq C|y|^{1-\varrho}$, for all
$\varrho\in(0,1)$ so that (6.25) holds with
$\varrho\in(0,\alpha_{\frac{1}{2}})$. We thus get, from (5.11) and an
iteration argument, as above, a vector $e\in\mathbb{R}^{N-1}$ such that

$\|w-e\cdot y^{\prime}\|_{L^{\infty}(B_{r}^{+})}\leq
Cr^{1+\alpha_{\frac{1}{2}}-\varrho},\qquad|e|\leq C.$ (5.33)

This gives $|w(y)|\leq C|y|$, so that $w$ satisfies (6.25) with $\varrho=0$.
Hence as above, we get (5.33) with $\varrho=0$. ∎

We will also need the following interior Schauder estimate.

######  Lemma 5.5.

Let $\Omega$ be an open set containing $B_{2}$ and $\alpha\in(0,1)$. Let $f\in
C^{\alpha}(B_{2})$ and $v\in H^{s}(\Omega)$ be a solution to

$(-\Delta)^{s}_{\Omega}v=f\qquad\text{in $B_{2}$}$

satisfying $C_{0}:=\int_{\Omega}\frac{|v(y)|dy}{1+|y|^{N+2s+1}}<\infty$. Then,
provided $2s+\alpha\not\in\mathbb{N}$,

$\|v\|_{C^{2s+\alpha}(B_{1})}\leq
C(N,s,\alpha)(C_{0}+\|v\|_{L^{2}(B_{2})}+\|f-f(0)\|_{C^{\alpha}(B_{2})}).$

###### Proof.

We have

$(-\Delta)^{s}_{\Omega}(\chi_{{2}}v)=f+g_{v}\qquad\text{in $B_{1}$, }$

where for $x\in B_{1}$

$g_{v}(x)=c_{N,s}\int_{\Omega}(1-\chi_{{2}}(y))|x-y|^{-N-2s}v(y)\,dy.$

Letting $\widetilde{v}=\chi_{{2}}v\in H^{s}(\mathbb{R}^{N})$, we get

$(-\Delta)^{s}_{\mathbb{R}^{N}}\widetilde{v}+\widetilde{v}V=f+g_{v}=c_{0}+(f-f(0))+(g_{v}-g_{v}(0))\qquad\text{in
$B_{1/2}$, }$

where $c_{0}=f(0)+g_{v}(0)$,
$V(x)=c_{N,s}\int_{\mathbb{R}^{N}\setminus\Omega}|x-y|^{-N-2s}\,dy$. Clearly
$\|V\|_{C^{3}(B_{1})}\leq C(N,s)$. Moreover
$\|g_{v}-g_{v}(0)\|_{C^{1}(B_{1/2})}\leq C(N,s)C_{0}$. Now as in Remark
3.2$(ii)$, we can estimate

$\displaystyle|c_{0}|$ $\displaystyle\leq
C(N,s)\left(C_{0}+\|\widetilde{v}V\|_{L^{1}(B_{1/4})}+\|g_{v}-g_{v}(0)\|_{L^{1}(B_{1/4})}+\|f-f(0)\|_{L^{1}(B_{1/4})}\right)$
$\displaystyle\leq
C(N,s)\left(C_{0}+\|v\|_{L^{2}(B_{2})}+\|f-f(0)\|_{L^{\infty}(B_{2})}\right).$

By the interior regularity estimates in [11], we have that

$\|\widetilde{v}\|_{C^{2s-\varepsilon}(B_{1/4})}\leq
C(N,s,\varepsilon)(C_{0}+\|v\|_{L^{2}(B_{2})}+\|f-f(0)\|_{L^{\infty}(B_{2})}).$

We can now apply the interior Schauder estimate in [18] and a standard
bootstrap (only necessary for $2s<1$ because $\alpha<1$) argument to get the
desired estimate. ∎

We can now prove the following.

######  Theorem 5.6.

Let $s\in(0,1)$ and $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6, which
satisfies $2s+\alpha_{s}>1$. Let $\Omega\subset\mathbb{R}^{N}$ be a
$C^{\max(1,2s+\alpha)}$ domain, with $0\in\partial\Omega$, and let $f\in
C^{\alpha}(\overline{\Omega}\cap B_{1})$, with $\alpha<\alpha_{s}$ and with
$\alpha+2s\neq 1$. Let $u\in H^{s}(\Omega)$ be such that

$D_{\Omega}(u,\varphi)=\int_{\Omega}f\varphi\,dx\qquad\textrm{$\forall\varphi\in
C^{\infty}_{c}(B_{1})$.}$

Then,

$\|u\|_{C^{2s+\alpha}(\overline{\Omega}\cap B_{1/2})}\leq
C\left(\|u\|_{L^{2}(\Omega)}+\|f\|_{C^{\alpha}(\overline{\Omega}\cap
B_{1})}\right).$ (5.34)

with $C$ depending only on $N$, $s$, $\alpha$, and $\Omega$.

Moreover, if $2s+\alpha>1$ then $\partial_{\nu}u=0$ on $\partial\Omega\cap
B_{1}$, where $\nu$ is the exterior normal of $\partial\Omega$.

###### Proof.

We assume for simplicity that
$\|u\|_{L^{2}(\Omega)}+\|f\|_{C^{\alpha}(\overline{\Omega})}\leq 1$ and that
$\Omega$ satisfies (5.1) and (5.2).

By Corollary 5.4 and the change of variable $w(y)=u(\psi(y))$ we have that,
for all $\zeta\in\partial\Omega\cap B_{1/2}$, there exists
$e=e_{\zeta}\in\mathbb{R}^{N-1}=:\partial\mathbb{R}^{N}_{+}$ with
$|e_{\zeta}|\leq C$ and such that

$\|u-u(\zeta)-\ell_{\alpha}e\cdot(\psi^{-1}(\cdot)-\zeta)\|_{L^{\infty}(B_{r}(\zeta)\cap\Omega)}\leq
Cr^{2s+\alpha}.$ (5.35)

In addition, using also (5.2), we have $\partial_{\nu}u=0$ on
$\partial\Omega\cap B_{1/2}$ as soon as $2s+\alpha>1$.

Let $x_{0}\in\Omega\cap B_{1}$ and $\zeta\in\partial\Omega$ be such that
$\textrm{dist}(x_{0},\partial\Omega)=|x_{0}-\zeta|$. Put
$\rho:=\frac{|x_{0}-\zeta|}{4}$ and
$\overline{x}=\frac{x_{0}-\zeta}{2}\in\Omega$ then
$B_{2\rho}(\overline{x})\subset\Omega$. We next, define

$v_{\rho}(y)=(u(\overline{x}+\rho
y)-u(\zeta))-\ell_{\alpha}e\cdot(\psi^{-1}(\rho y+\overline{x})-\zeta),\qquad
f_{\rho}(y)=\rho^{2s}f(\overline{x}+\rho y)$

and $\Omega_{\rho}=\frac{1}{\rho}(\Omega-\overline{x})$. Then

$(-\Delta)^{s}_{\Omega_{\rho}}v_{\rho}=f_{\rho}+F_{\rho}=:G_{\rho}\qquad\textrm{
in $B_{2}\subset\Omega_{\rho}$},$ (5.36)

where
$F_{\rho}(y):=\ell_{\alpha}\rho^{2s}[(-\Delta)^{s}_{\Omega}(e\cdot\psi^{-1})](\overline{x}+\rho
y))$. Since $f\in C^{\alpha}(\Omega)$, using Lemma 4.1, we then get
$[G_{\rho}]_{C^{\alpha}(B_{1})}\leq C\rho^{2s+\alpha}$. By (5.35), we have
that $\|v_{\rho}\|_{L^{\infty}(B_{1})}\leq C\rho^{2s+\alpha}$ and

$\|v_{\rho}(\cdot/\rho)\|_{L^{\infty}(B_{R\rho}\cap(\Omega-\overline{x}))}\leq\|v_{\rho}(\cdot/\rho)\|_{L^{\infty}(B_{R\rho+2\rho}\cap(\Omega-\zeta))}\leq
C(R\rho)^{2s+\alpha}\qquad\textrm{ for all $R\geq 1$. }$ (5.37)

Now applying Lemma 5.5 to (5.36), we obtain

$\|v_{\rho}\|_{C^{2s+\alpha}(B_{1/4})}\leq
C(N,s,\alpha)\left(\int_{\Omega_{\rho}\setminus
B_{1}}\frac{|v_{\rho}(y)|dy}{|y|^{N+2s+1}}+\|v_{\rho}\|_{L^{\infty}(B_{1})}+\|G_{\rho}-G_{\rho}(0)\|_{C^{\alpha}(B_{1})}\right).$

Thanks to (5.37) and a change of variable we get

$\displaystyle\int_{\Omega_{\rho}\setminus
B_{1}}\frac{|v_{\rho}(y)|dy}{|y|^{N+2s+1}}=\rho^{2s+1}\int_{(\Omega-\overline{x})\setminus
B_{\rho}}\frac{|v_{\rho}(x/\rho)|dx}{|x|^{N+2s+1}}$
$\displaystyle\leq\sum_{i=0}^{\infty}(2^{i}\rho)^{-N-2s-1}\rho^{2s+1}\int_{(\Omega-\overline{x})\\{|y|<2^{i+1}\rho\\}}{|v_{\rho}(x/\rho)|,dx}$
$\displaystyle\leq
C\sum_{i=0}^{\infty}(2^{i}\rho)^{-N-2s-1}\rho^{2s+1}(2^{i+1}\rho)^{N+2s+\alpha}\leq
C\rho^{2s+\alpha}.$

We conclude that $\|v_{\rho}\|_{C^{2s+\alpha}(B_{1/4})}\leq
C\rho^{2s+\alpha}.$ Therefore scaling and translating back, we get for
$2s+\alpha<1$ that

$[u]_{C^{2s+\alpha}(B_{\rho}(\overline{x}))}\leq C$ (5.38)

and thus $[u]_{C^{2s+\alpha}(B_{1/2}\cap\Omega)}\leq C$ because $\overline{x}$
is arbitrary.

If now $2s+\alpha>1$, we obtain

$[\nabla
u-\nabla{e}\cdot(\psi^{-1}(\cdot)-\zeta)]_{C^{2s+\alpha-1}(B_{\rho}(\overline{x}))}\leq
C.$

Hence, since
$[\nabla{e}\cdot(\psi^{-1}(\cdot)-\zeta)]_{C^{2s+\alpha-1}(B_{\rho}(\overline{x}))}\leq
C$, we get

$[\nabla u]_{C^{2s+\alpha-1}(B_{\rho}(\overline{x}))}\leq C.$

This implies

$[\nabla u]_{C^{2s+\alpha-1}(B_{1/2}\cap\overline{\Omega})}\leq C.$

∎

Finally, we give the:

###### Proof of Theorem 1.1.

The result follows from Theorem 5.6. ∎

##  6\. Regional Dirichlet problem

In this section, we consider $\phi\in C^{2,\beta}(\mathbb{R}^{N-1})$
satisfying $\|\phi\|_{C^{2,\beta}(\mathbb{R}^{N-1})}\leq\frac{1}{4}$ and
$\phi(0)=|\nabla\phi(0)|=0$. We define
$\Phi(x^{\prime},x_{N})=(x^{\prime},x_{N}+\phi(x^{\prime}))$. We start with
the following result.

######  Lemma 6.1.

Let $\eta\in C^{2s-1}_{c}(\mathbb{R})$ with $\eta=0$ on $\mathbb{R}_{-}$ and
$\ell(x^{\prime}):=a\cdot x^{\prime}+b$. Let
$f(x)=\eta(x_{N})\ell(x^{\prime})$. Then, for all $x\in\mathbb{R}^{N}_{+}$,

$\displaystyle
c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{f(x)-f(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=\ell(x^{\prime})T_{1}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}\eta(x_{N})+\ell(x^{\prime})T_{3}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}\eta)$
$\displaystyle-T_{2}(x^{\prime})\cdot
a\int_{\mathbb{R}}\frac{(t-x_{N})(\eta(t)-\eta(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt-
T_{3}(x^{\prime})\ell(x^{\prime})\left(\eta(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}+x_{N}(-\Delta)^{s}_{\mathbb{R}_{+}}\eta\right)+h(x),$

where $\|h\|_{C^{\min(\alpha,\beta)}(\overline{B_{1}^{+}})}\leq
C\|\eta\|_{C^{2s-1}(\mathbb{R})}$ for all $\alpha<2s-1$, $T_{1}(0)=1$,
$T_{2}(0)=0$ and $\|T_{j}\|_{C^{1,\beta}(B_{1}^{\prime})}\leq C(N,s,\beta)$,
for $j=1,2$. Moreover, $\|T_{3}\|_{C^{\beta}(B_{1}^{\prime})}\leq
C(N,s,\beta)$ and

$T_{3}(0)=-D^{2}\phi(0)[e_{i},e_{i}]=-\Delta\phi(0).$ (6.1)

###### Proof.

We have

$\displaystyle\int_{\mathbb{R}^{N}_{+}}\frac{f(x)-f(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z)}{|\Phi(x)-\Phi(x+z)|^{N+2s}}\,dz$
$\displaystyle=\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z)}{|D\Phi(x^{\prime})z|^{N+2s}}\,dz+\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z))}{|z|^{N+2s}}B(x^{\prime},|z^{\prime}|,z^{\prime}/|z^{\prime}|,z/|z|)\,dz,$

where

$B(x^{\prime},r,\theta^{\prime},\theta)=\frac{1}{|\int_{0}^{1}D\Phi(x^{\prime}+tr\theta^{\prime})\theta
dt|^{N+2s}}-\frac{1}{|D\Phi(x^{\prime})\theta|^{N+2s}}.$

Recall that for all $y\in\mathbb{R}^{N}$

$D\Phi(x+r\theta^{\prime})y=y+\left(\nabla\phi(x^{\prime}+r\theta^{\prime})\cdot
y^{\prime}\right)e_{N}.$

We can thus write

$B(x^{\prime},r,\theta^{\prime},\theta)=r\mu_{1}(x^{\prime},\theta^{\prime},\theta)+r\mathcal{O}(x^{\prime},r,\theta^{\prime},\theta),$
(6.2)

where

$\displaystyle\mu_{1}(x^{\prime},\theta^{\prime},\theta)=\partial_{r}B(x^{\prime},0,\theta^{\prime},\theta)$
$\displaystyle=-(N+2s)\frac{D^{2}\Phi(x^{\prime})[\theta^{\prime},\theta-(\theta\cdot
e_{N})e_{N}]\cdot
D\Phi(x^{\prime})\theta}{|D\Phi(x^{\prime})\theta|^{N+2s+2}},$

and

$\mathcal{O}(x^{\prime},r,\theta^{\prime},\theta)=\partial_{r}B(x^{\prime},r,\theta^{\prime},\theta)-\partial_{r}B(x^{\prime},0,\theta^{\prime},\theta).$

We then have $\mu_{1},\mathcal{O}\in C^{\beta}_{x}(B_{2})$,
$\mu_{1}(x^{\prime},-\theta^{\prime},-\theta)=-\mu_{1}(x^{\prime},\theta^{\prime},\theta)$,
$|\mathcal{O}(x^{\prime},r,\theta^{\prime},\theta)|\leq C\min(1,r^{\beta})$
and

$|\mathcal{O}(x^{\prime},r,\theta^{\prime},\theta)|\leq
C\min(1,r^{\beta}),\qquad|\mathcal{O}(x^{\prime},r,\theta^{\prime},\theta)-\mathcal{O}(\overline{x}^{\prime},r,\theta^{\prime},\theta)|\leq
C\min(|x^{\prime}-\overline{x}^{\prime}|,r^{\beta}).$ (6.3)

We thus get from (6.2),

$\displaystyle
c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{f(x)-f(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=c_{N,s}\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z)}{|D\Phi(x^{\prime})z|^{N+2s}}\,dz$
$\displaystyle+c_{N,s}\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z))}{|z|^{N+2s}}|z^{\prime}|\mu_{1}(x^{\prime},z^{\prime}/|z^{\prime}|,z/|z|)\,dz$
$\displaystyle+c_{N,s}\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{f(x)-f(x+z))}{|z|^{N+2s}}|z^{\prime}|\mathcal{O}(x^{\prime},|z^{\prime}|,z^{\prime}/|z^{\prime}|,z/|z|)\,dz$
$\displaystyle=I_{1}(x)+I_{2}(x)+I_{3}(x).$ (6.4)

Letting
$T_{1}(x^{\prime}):=\frac{1}{c_{1,s}}\int_{\mathbb{R}^{N-1}}\frac{c_{N,s}}{|D\Phi(x^{\prime})(z^{\prime},1)|^{N+2s}}\,dz^{\prime}$
and $T_{2}(x^{\prime})\cdot
e_{i}=c_{N,s}\int_{\mathbb{R}^{N-1}}\frac{z_{i}}{|D\Phi(x^{\prime})(z^{\prime},1)|^{N+2s}}\,dz^{\prime}$,
we then get

$\displaystyle I_{1}(x)$
$\displaystyle=\ell(x^{\prime})\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{\eta(x_{N})-\eta(x_{N}+z_{N})}{|D\Phi(x^{\prime})z|^{N+2s}}\,dz-\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{a\cdot
z^{\prime}\eta(x_{N}+z_{N})}{|D\Phi(x^{\prime})z|^{N+2s}}\,dz$
$\displaystyle=T_{1}(x^{\prime})\ell(x^{\prime})c_{1,s}\int_{-x_{N}}^{\infty}\frac{\eta(x_{N})-\eta(x_{N}+z_{N})}{|z_{N}|^{1+2s}}\,dz_{N}-T_{2}(x^{\prime})\cdot
a\int_{-x_{N}}^{\infty}z_{N}\frac{\eta(x_{N}+z_{N})}{|z_{N}|^{1+2s}}\,dz_{N}$
$\displaystyle={T_{1}(x^{\prime})}{}\ell(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}\eta(x_{N})-T_{2}(x^{\prime})\cdot
a\int_{\mathbb{R}^{+}}\frac{(t-x_{N})\eta(t)}{|x_{N}-t|^{1+2s}}\,dt.$

Since $\eta=0$ on $\mathbb{R}_{-}$, then

$\displaystyle\int_{\mathbb{R}^{+}}\frac{(t-x_{N})\eta(t)}{|x_{N}-t|^{1+2s}}\,dt$
$\displaystyle=\int_{\mathbb{R}}\frac{(t-x_{N})\eta(t)}{|x_{N}-t|^{1+2s}}\,dt=\int_{\mathbb{R}}\frac{(t-x_{N})(\eta(t)-\eta(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt.$

We thus conclude that

$\displaystyle I_{1}(x)$
$\displaystyle={T_{1}(x^{\prime})}\ell(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}\eta(x_{N})-T_{2}(x^{\prime})\cdot
a\int_{\mathbb{R}}\frac{(t-x_{N})(\eta(t)-\eta(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt.$
(6.5)

Since $D\Phi(0)=id$, we see that
${T_{1}(0)}=\frac{1}{c_{1,s}}\int_{\mathbb{R}^{N-1}}\frac{c_{N,s}}{(1+|z^{\prime}|^{2})^{\frac{N+2s}{2}}}dz^{\prime}=1$
and by oddness we have $T_{2}(0)=0$.

We consider next $I_{2}$. Similarly, letting
$T_{3}(x^{\prime})=\frac{c_{N,s}}{c_{1,s}}\int_{\mathbb{R}^{N-1}}\frac{|z^{\prime}|}{|(z^{\prime},1)|^{N+2s}}\mu_{1}(x^{\prime},\frac{z^{\prime}}{|z^{\prime}|},\frac{(z^{\prime},1)}{|(z^{\prime},1)|})\,dz^{\prime}$,
we obtain

$\displaystyle I_{2}(x)=$ $\displaystyle
T_{3}(x^{\prime})\ell(x^{\prime})c_{1,s}\int_{-x_{N}}^{\infty}z_{N}\frac{\eta(x_{N})-\eta(x_{N}+z_{N})}{|z_{N}|^{1+2s}}\,dz_{N}+h_{1}(x)$
$\displaystyle=T_{3}(x^{\prime})\ell(x^{\prime})c_{1,s}\int_{\mathbb{R}^{+}}\frac{(t-x_{N})(\eta(x_{N})-\eta(t))}{|x_{N}-t|^{1+2s}}\,dt+h_{1}(x)$
$\displaystyle={T_{3}(x^{\prime})}\ell(x^{\prime})\left((-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}\eta)-\eta(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}-x_{N}(-\Delta)^{s}_{\mathbb{R}_{+}}\eta\right)+h_{1}(x),$
(6.6)

where

$\displaystyle h_{1}(x)$
$\displaystyle:=c_{N,s}\int_{\mathbb{R}^{N-1}}\int_{-x_{N}}^{\infty}\frac{|z^{\prime}|a\cdot
z^{\prime}\eta(x_{N}+z_{N})}{|z|^{N+2s}}\mu_{1}(x^{\prime},z^{\prime}/|z^{\prime}|,z/|z|)\,dz$
$\displaystyle=c_{N,s}\int_{\mathbb{R}^{N-1}}\frac{|z^{\prime}|a\cdot
z^{\prime}}{|(z^{\prime},1)|^{N+2s}}\mu_{1}(x^{\prime},\frac{z^{\prime}}{|z^{\prime}|},\frac{(z^{\prime},1)}{|(z^{\prime},1)|})\,dz^{\prime}\int_{-x_{N}}^{\infty}\frac{z_{N}\eta(x_{N}+z_{N})}{|z_{N}|^{2s}}\,dz_{N}.$

Since $2s>1$, we can easily see that

$\|h_{1}\|_{C^{\min(\beta,2s-1)}(\overline{B_{1}^{+}})}\leq
C|a|\|\eta\|_{C^{2s-1}(\mathbb{R})}.$ (6.7)

Using that $D\Phi(0)=Id$, a change of variable and integration by part, we get

$\displaystyle
c_{1,s}T_{3}(0)=-(N+2s){c_{N,s}}{}\int_{\mathbb{R}^{N-1}}\frac{|z^{\prime}|^{2}D^{2}\phi(0)[z^{\prime}/|z^{\prime}|,z^{\prime}/|z^{\prime}|]}{|(z^{\prime},1)|^{N+2s+2}}\,dz^{\prime}$
$\displaystyle=-(N+2s){c_{N,s}}{}\int_{0}^{\infty}\frac{r^{N}}{(r^{2}+1)^{\frac{N+2s+2}{2}}}\,dr\sum_{i=1}^{N-1}D^{2}\phi(0)[e_{i},e_{i}]\int_{S^{N-2}}\theta_{1}^{2}d\sigma(\theta)$
$\displaystyle=-(N+2s)\frac{c_{N,s}|S^{N-2}|}{N-1}\int_{0}^{\infty}\frac{r^{N}}{(r^{2}+1)^{\frac{N+2s+2}{2}}}\,dr\sum_{i=1}^{N-1}D^{2}\phi(0)[e_{i},e_{i}]$
$\displaystyle=-{c_{N,s}|S^{N-2}|}\int_{0}^{\infty}\frac{r^{N-2}}{(r^{2}+1)^{\frac{N+2s}{2}}}\,dr\sum_{i=1}^{N-1}D^{2}\phi(0)[e_{i},e_{i}]=-c_{1,s}\sum_{i=1}^{N-1}D^{2}\phi(0)[e_{i},e_{i}].$
(6.8)

Here, we used that
${c_{N,s}}{}\int_{\mathbb{R}^{N-1}}\frac{1}{(1+|z^{\prime}|^{2})^{\frac{N+2s}{2}}}dz^{\prime}=c_{1,s}$.
Finally, using (6.3), we see that

$\|I_{3}\|_{C^{\min(\alpha,\beta)}(\overline{B_{1}^{+}})}\leq
C\|f\|_{C^{2s-1}(\mathbb{R})}\qquad\textrm{ for all $\alpha\in(0,2s-1)$.}$

From this together with (6.8), (6.7), (6), (6.5) and (6.4), we get the result.
∎

######  Lemma 6.2.

Let $\overline{\eta}\in C^{\infty}_{c}(-3,3)$ with $\overline{\eta}=1$ on
$(-2,2)$ and $\ell(x^{\prime}):=a\cdot x^{\prime}+b$. Define

$P(x)=\overline{\eta}(x_{N})(x_{N})^{2s-1}_{+}\ell(x^{\prime})-bT_{3}(0)\overline{\eta}(x_{N})(x_{N})^{2s}_{+},$

where $T_{3}(0)$ is given by Lemma 6.1. Then, for all $x\in B_{1}^{+}$,

$\displaystyle
c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{P(x)-P(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=g_{1}(x)\log(x_{N})+g_{2}(x),$

where $\|g_{i}\|_{C^{\min(\beta,\alpha)}(\overline{B_{1}^{+}})}\leq
C(N,s,\beta,\alpha,\overline{\eta})(|a|+|b|)$ for all $\alpha<2s-1$ and
$g_{1}(0)=0$.

###### Proof.

Recall from Lemma 6.1 that

$\displaystyle
c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{P(x)-P(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=L_{1}(x)+L_{2}(x)+h(x),$
(6.9)

where $h\in C^{\beta}(\overline{B_{1}^{+}})$,

$\displaystyle L_{1}(x)$
$\displaystyle=\ell(x^{\prime})T_{1}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s-1}\overline{\eta})+\ell(x^{\prime})T_{3}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})$
$\displaystyle-
bT_{3}(0)T_{1}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})-bT_{3}(0)T_{3}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s+1}\overline{\eta})$

and

$\displaystyle L_{2}(x)$ $\displaystyle=-T_{2}(x^{\prime})\cdot
a\int_{\mathbb{R}}\frac{(t-x_{N})(t^{2s-1}_{+}\overline{\eta}(t)-(x_{N})^{2s-1}_{+}\overline{\eta}(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt$
$\displaystyle-
T_{3}(x^{\prime})\ell(x^{\prime})\left(\overline{\eta}x_{N}^{2s-1}(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}+x_{N}(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s-1}\overline{\eta})\right)$
$\displaystyle-bT_{3}(0)T_{2}(x^{\prime})\cdot
a\int_{\mathbb{R}}\frac{(t-x_{N})(t^{2s}_{+}\overline{\eta}(t)-(x_{N})^{2s}_{+}\overline{\eta}(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt$
$\displaystyle-
bT_{3}(0)T_{3}(x^{\prime})\left(\overline{\eta}x_{N}^{2s}(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}+x_{N}(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})\right).$
(6.10)

We now write

$\displaystyle
L_{1}(x)=-bT_{3}(0)(T_{1}(x^{\prime})-1)(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})+b(T_{3}(x^{\prime})-T_{3}(0))(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})$
$\displaystyle+a\cdot
x^{\prime}T_{3}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})-bT_{3}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x^{2s+1}_{N}\overline{\eta})+\ell(x^{\prime})T_{1}(x^{\prime})(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s-1}\overline{\eta}).$

Moreover since $x_{N}\mapsto x^{2s+1}_{N}\overline{\eta}(x_{N})$ has zero
derivative at $x_{N}=0$, we get
$(-\Delta)^{s}_{\mathbb{R}_{+}}(x^{2s+1}_{N}\overline{\eta})\in
C^{\beta}([0,1])$ by Lemma 4.1. In addition, since
$(-\Delta)^{s}_{\mathbb{R}_{+}}x^{2s-1}=0$, we see that
$(-\Delta)^{s}_{\mathbb{R}_{+}}(x^{2s-1})\overline{\eta}\in C^{\beta}([0,1])$.
Therefore, recalling that $T_{1}(0)=1$, we thus obtain

$L_{1}(x)=g_{1}(x)H(x_{N})+g_{2}(x),$ (6.11)

for some functions $g_{i}$ as in the statement of the lemma and
$H(x_{N})=(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})(x_{N})$.
To estimate $L_{2}$, we first observe that

$\displaystyle
c_{1,s}\int_{\mathbb{R}}\frac{(t-x_{N})(t^{2s-1}_{+}\overline{\eta}(t)-(x_{N})^{2s-1}_{+}\overline{\eta}(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt$
$\displaystyle=-(-\Delta)^{s}(x_{N}(x_{N})^{2s-1}_{+}\overline{\eta})+x_{N}(-\Delta)^{s}((x_{N})^{2s-1}_{+}\overline{\eta})$
$\displaystyle=-(-\Delta)^{s}_{\mathbb{R}_{+}}((x_{N})^{2s}\overline{\eta})-a_{s}+x_{N}c_{N,s}\int_{\mathbb{R}_{+}}\frac{\overline{\eta}(x_{N})-\overline{\eta}(t))t^{2s-1}}{|x_{N}-t|^{1+2s}}\,dt,$

where we used that $(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}^{2s-1}=0$ and
$(-\Delta)^{s}x_{N}=0$ and the formula
$(-\Delta)^{s}(1_{\mathbb{R}_{+}}u)(x)=(-\Delta)^{s}_{\mathbb{R}_{+}}u+a_{s}x^{-2s}u$
in $\mathbb{R}_{+}$. We have that

$\displaystyle c_{1,s}$
$\displaystyle\int_{\mathbb{R}}\frac{(t-x_{N})(t^{2s}_{+}\overline{\eta}(t)-(x_{N})^{2s}_{+}\overline{\eta}(x_{N}))}{|x_{N}-t|^{1+2s}}\,dt=-(-\Delta)^{s}((x_{N})_{+}^{2s+1}\overline{\eta})+x_{N}(-\Delta)^{s}(x_{N})^{2s}_{+}\overline{\eta}$
$\displaystyle=-(-\Delta)^{s}((x_{N})_{+}^{2s+1}\overline{\eta})+x_{N}(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})+a_{s}x_{N}.$
(6.12)

Moreover $(-\Delta)^{s}_{\mathbb{R}_{+}}x_{N}=\kappa_{s}x_{N}^{1-2s}$, for
some $\kappa_{s}>0$. Next, using that
$(-\Delta)^{s}_{\mathbb{R}_{+}}x^{2s-1}=0$, we see that
$(-\Delta)^{s}_{\mathbb{R}_{+}}(x^{2s-1}\overline{\eta})\in C^{\beta}([0,1])$.
Since $T_{2}(0)=0$, we thus get from (6) and (6.13),

$L_{2}(x)=g_{1}(x)H(x_{N})+g_{2}(x),$ (6.13)

for some functions $g_{i}$ as in the statement of the lemma and
$H(x_{N})=(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s}\overline{\eta})(x_{N})$.
To conclude, we observe that

$\displaystyle
H^{\prime}(x_{N})=2s(-\Delta)^{s}(x_{N}^{2s-1}\overline{\eta})(x_{N})+(-\Delta)^{s}(x_{N}^{2s}\overline{\eta}^{\prime})(x_{N})-a_{s}\overline{\eta}^{\prime}(x_{N})$
$\displaystyle=2s(-\Delta)^{s}_{\mathbb{R}_{+}}(x_{N}^{2s-1}\overline{\eta})(x_{N})+a_{s}x_{N}^{-1}\overline{\eta}(x_{N})+(-\Delta)^{s}(x_{N}^{2s}\overline{\eta}^{\prime})(x_{N})-a_{s}\overline{\eta}^{\prime}(x_{N}).$

As a consequence, $x_{N}\mapsto H^{\prime}(x_{N})-a_{s}x_{N}^{-1}\in
C^{\infty}([0,1])$. It follows that for $x_{N}\in(0,1)$, we have
$H(x_{N})=-\int_{x_{N}}^{1/2}H^{\prime}(t)\,dt+H(1/2)=a_{s}\log
x_{N}+\widetilde{h}(x_{N})$, with $\widetilde{h}\in C^{\infty}([0,1])$. In
view of this, (6.13), (6.11) and (6.9), we get the result. ∎

###  6.1. Expansion near the boundary

In the remaining of this section, we define
$\mathcal{B}_{\Phi}:H^{s}(\mathbb{R}^{N}_{+})\times
H^{s}(\mathbb{R}^{N}_{+})\to\mathbb{R}$ by

$\mathcal{B}_{\Phi}(u,\varphi):=\frac{c_{N,s}}{2}\int_{\mathbb{R}^{N}_{+}\times\mathbb{R}^{N}_{+}}\frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dxdy.$

We start recalling the following result from [12].

######  Lemma 6.3 ([12]).

Let $2s>1$. Let $f\in L^{p}(B_{2}^{+})$ and $u\in
H^{s}_{0}(\mathbb{R}^{N}_{+})$ by a solution to

$\mathcal{B}_{\Phi}(u,\varphi)=\int_{\mathbb{R}^{N}_{+}}f\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2}^{+})$}.$

Then for all $p>N$, there exists $C=C(N,s,p)>0$ such that

$\|u/x_{N}^{2s-1}\|_{C^{1-N/p}(B_{1}^{+})}\leq
C\left(\|u\|_{L^{2}(\mathbb{R}^{N}_{+})}+\|f\|_{L^{p}(B_{2}^{+})}\right).$

We state the next result which is crucial for the proof of Theorem 6.7.

######  Proposition 6.4.

Let $s\in(\frac{1}{2},1)$, $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6 and
$\beta\in(0,\alpha_{s})$. Let $\alpha\in(0,\min(\beta,2s-1))$, with
$2s+\alpha\not=1$, $f\in C^{\alpha}(\mathbb{R}^{N}_{+})$ and $w\in
H^{s}_{0}(\mathbb{R}^{N}_{+})$ such that

$\mathcal{B}_{\Phi}(w,\varphi)=\int_{\mathbb{R}^{N}_{+}}f\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2}^{+})$}$

and

$\|w/x_{N}^{2s-1}\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq
1.$

Let $\overline{\eta}\in C^{\infty}_{c}(-3,3)$ with $\overline{\eta}=1$ on
$(-2,2)$. Then, there exists $C=C(N,s,\alpha,\beta,\overline{\eta})$ such that

$\sup_{r>0}r^{-1-\alpha}\left\|\frac{w}{x_{N}^{2s-1}}-\left(a_{r,w}\cdot
x^{\prime}+b_{r,{w}}(1-T_{3}(0)x_{N})\right)\overline{\eta}(x_{N})\right\|_{L^{\infty}(B_{r}^{+})}\leq
C,$ (6.14)

where $a_{r,w}:=A_{r,\,w/{x_{N}^{2s-1}}}$ is given by (5.7),

$b_{r,{w}}=\frac{1}{\int_{B_{r}^{+}}(1-T_{3}(0)y_{N})^{2}dy}\int_{B_{r}^{+}}\frac{w}{x_{N}^{2s-1}}(y)(1-T_{3}(0)y_{N})dy,$

and $T_{3}(0)=-\Delta\phi(0)$ is given by (6.1).

###### Proof.

Suppose on the contrary that (6.14) does not hold. Then for all integer $n\geq
2$, there exist $\phi_{n}\in C^{2,\beta}(\mathbb{R}^{N-1})$, $f_{n}\in
C^{\alpha}(\mathbb{R}^{N}_{+})$ and $w_{n}\in H^{s}_{0}(\mathbb{R}^{N}_{+})$,
such that

$\|\phi_{n}\|_{C^{2,\beta}(\mathbb{R}^{N-1})}\leq\frac{1}{4},\qquad\phi_{n}(0)=|\nabla\phi_{n}(0)|=0,$
(6.15)
$\|w_{n}/x_{N}^{2s-1}\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|f_{n}\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq
1$ (6.16)

with

$\mathcal{B}_{\Phi}(w_{n},\varphi)=\int_{\mathbb{R}^{N}_{+}}f_{n}\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{2}^{+})$}.$ (6.17)

and

$\sup_{r>0}r^{-1-\alpha}\left\|\frac{w_{n}}{x_{N}^{2s-1}}-\left(a_{r,w_{n}}\cdot
x^{\prime}+b_{r,{w_{n}}}(1-T_{3}^{n}(0)x_{N})\right)\overline{\eta}(x_{N})\right\|_{L^{\infty}(B_{r}^{+})}>n,$
(6.18)

where (recalling (6.1))

$T_{3}^{n}(0):=-\Delta\phi_{n}(0).$ (6.19)

Define

$Q_{r}(x)=a_{r,w_{n}}\cdot x^{\prime}+b_{r,{w_{n}}}(1-T_{3}^{n}(0)x_{N}).$

By (6.16), we can define the monotone nonincrease sequence of function
$\theta_{n}$ by

$\theta_{n}(\rho):=\sup_{r>\rho}r^{-1-\alpha}\left\|\frac{w_{n}}{x_{N}^{2s-1}}-Q_{r}(x)\overline{\eta}(x_{N})\right\|_{L^{\infty}(B_{r}^{+})}.$
(6.20)

Obviously, for $n\geq 2$, by (6.18), there exists $\rho_{n}>0$ such that

$\theta_{n}(\rho_{n})>n/2\geq 1.$ (6.21)

Hence, provided $n\geq 2$, by definition of $\theta_{n}(\rho_{n})$ as a
supremum and the monotonicity of $\theta_{n}$, there exists
$r_{n}\geq\rho_{n}$ such that

$\displaystyle\theta_{n}(r_{n})$ $\displaystyle\geq
r^{-1-\alpha}_{n}\left\|\frac{w_{n}}{x_{N}^{2s-1}}-Q_{r_{n}}(x)\overline{\eta}(x_{N})\right\|_{L^{\infty}(B_{r_{n}}^{+})}$
$\displaystyle\geq\theta_{n}(\rho_{n})-1/2\geq(1-1/2)\theta_{n}(\rho_{n})\geq\frac{1}{2}\theta_{n}(r_{n}),$
(6.22)

Also by (6.21) and the above estimate, we have that
$\theta_{n}(r_{n})\geq\frac{n}{2}-\frac{1}{2}\to\infty$ as $n\to\infty.$ We
now define $v_{n}\in H^{s}_{0}(\mathbb{R}^{N}_{+})$ by

$v_{n}(x):=\frac{1}{r_{n}^{2s+\alpha}\theta_{n}(r_{n})}\big{(}w_{n}(r_{n}x)-Q_{r_{n}}(r_{n}x)\overline{\eta}(r_{n}x_{N})(r_{n}x_{N})^{2s-1}\big{)}.$

In view of (6.22), it clearly satisfies

$\|v_{n}/x_{N}^{2s-1}\|_{L^{\infty}(B_{1}^{+})}\geq\frac{1}{2}.$ (6.23)

Moreover, by construction (recalling (5.9)),

$\int_{B_{1}^{+}}\frac{v_{n}}{x_{N}^{2s-1}}(y)(1-T_{3}^{n}(0)y_{N})dy=0,\qquad\int_{B_{1}^{+}}\frac{v_{n}}{x_{N}^{2s-1}}(y)y_{i}\,dy=0\qquad\textrm{
for all $i=1,\dots,N-1$.}$ (6.24)

Note that by Lemma 6.3, (6.17) and (6.16),

$|a_{r,w_{n}}|\leq c_{0}r^{-\varrho},\qquad\textrm{for all
$\varrho\in(0,\beta-\alpha)$.}$ (6.25)

By (6.15) and (6.19), for $M\leq 1$, we have

$\displaystyle\|Q_{2r}-Q_{r}\|_{L^{2}(B_{M})}^{2}\asymp|B_{M}|\left(M^{2}|a_{2r,w_{n}}-a_{r,w_{n}}|^{2}+|b_{2r,w_{n}}-b_{r,w_{n}}|^{2}\right).$
(6.26)

Therefore by (6.20), for $2r\leq 1$,

$\displaystyle 2r|a_{2r,w_{n}}-a_{r,w_{n}}|$
$\displaystyle+|b_{2r,w_{n}}-b_{r,w_{n}}|$ $\displaystyle\leq
C\left\|{w_{n}}/{x_{N}^{2s-1}}-Q_{2r}\right\|_{L^{\infty}(B_{2r}^{+})}+C\left\|{w_{n}}/{x_{N}^{2s-1}}-Q_{r}\right\|_{L^{\infty}(B_{r}^{+})}$
$\displaystyle\leq
C\left(\theta_{n}(2r)(2r)^{1+\alpha}+\theta_{n}(r)r^{1+\alpha}\right)\leq
C\theta_{n}(r)r^{1+\alpha}.$

Using the monotonicity of $\theta_{n}$, we see that for all $m\in\mathbb{N}$
such that $2^{m}r\leq 1$, we get

$\displaystyle\|Q_{2^{m}r}-Q_{r}\|_{L^{\infty}(B_{2^{m}r})}\leq
C\|Q_{2^{m}r}-Q_{r}\|_{L^{2}(B_{2^{m}r})}\leq
C\sum_{i=1}^{m}\|Q_{2^{i}r}-Q_{2^{i-1}}\|_{L^{2}(B_{2^{i}r})}$
$\displaystyle\leq
C\sum_{i=1}^{m}2^{i}r|a_{2^{i}r,w_{n}}-a_{2^{i-1}r,w_{n}}|+|b_{2^{i}r,w_{n}}-b_{2^{i-1}r,w_{n}}|\leq
C\theta_{n}(r)\sum_{i=1}^{m}2^{i(1+\beta)}r^{1+\alpha}$ $\displaystyle\leq
C\theta_{n}(r)(2^{m}r)^{1+\alpha}.$

Hence for all $M\geq 1$ such that $Mr\leq 1$, we get

$\|Q_{Mr}-Q_{r}\|_{L^{\infty}(B_{Mr})}\leq\theta_{n}(r)(Mr)^{1+\alpha}.$

From this, (6.20) and the monotonicity of $\theta_{n}$, we then have, for
$Mr\leq 1$,

$\displaystyle\left\|{v_{n}}/{x_{N}^{2s-1}}\right\|_{L^{\infty}(B_{M}^{+})}=\frac{1}{r_{n}^{1+\alpha}\theta_{n}(r_{n})}\|{w_{n}}/{x_{N}^{2s-1}}-Q_{r_{n}}\|_{L^{\infty}(B_{Mr_{n}}^{+})}$
$\displaystyle+\frac{1}{r_{n}^{1+\alpha}\theta_{n}(r_{n})}\|{w_{n}}/{x_{N}^{2s-1}}-Q_{Mr_{n}}\|_{L^{\infty}(B_{Mr_{n}}^{+})}\
+\frac{1}{r_{n}^{1+\alpha}\theta_{n}(r_{n})}\|Q_{Mr_{n}}-Q_{r_{n}}\|_{L^{\infty}(B_{Mr_{n}})}$
$\displaystyle\leq CM^{1+\alpha}.$

As a consequence,

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq CM^{2s+\alpha}\qquad\textrm{ whenever
$Mr_{n}\leq 1$.}$ (6.27)

Now for $Mr_{n}>1$ and using (6.15), we have (recall that $2s+\alpha>1$)

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq\frac{C}{r_{n}^{2s+\alpha}\theta_{n}(r_{n})}(1+r_{n}^{-\varrho}+Mr_{n}+|T_{3}^{n}(0)|Mr_{n})\leq
CM^{2s+\alpha+\varrho}.$

In view of this and (6.27) we deduce that

$\|v_{n}\|_{L^{\infty}(B_{M}^{+})}\leq CM^{2s+\beta}\qquad\textrm{ for all
$M\geq 1$}.$ (6.28)

Letting $\widetilde{\Phi}_{n}(x)=\frac{1}{r_{n}}\Phi(r_{n}x)$, we thus get

$\mathcal{B}_{\widetilde{\Phi}_{n}}(\chi_{R}v_{n},\varphi)=\int_{\mathbb{R}^{N}_{+}}G^{n}_{R}\varphi\,dx\qquad\textrm{
for all $\varphi\in C^{\infty}_{c}(B_{\frac{1}{r_{n}}}^{+})$},$ (6.29)

where

$G^{n}_{R}(x):=\frac{1}{r_{n}^{\beta}\theta_{n}(r_{n})}f_{n}(r_{n}x)+c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{(\chi_{R}(x)-\chi_{R}(y))v_{n}(y)}{|\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|^{N+2s}}\,dy-\frac{1}{r_{n}^{\beta}\theta_{n}(r_{n})}({\mathcal{L}}_{\Phi_{n}}P_{n})(r_{n}x),$
$P_{n}(x)=\overline{\eta}(x_{N})\left(a_{r_{n},w_{n}}\cdot
x^{\prime}+b_{r_{n},{w_{n}}}(1-T_{3}^{n}(0)x_{N})\right)x_{N}^{2s-1}$

and
$({\mathcal{L}}_{\Phi_{n}}P_{n})(x)=c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{P_{n}(x)-P_{n}(y)}{|\Phi_{n}(x)-\Phi_{n}(y)|^{N+2s}}\,dy$.
By Lemma 6.2,

$({\mathcal{L}}_{\Phi_{n}}P_{n})(r_{n}x)=g_{1,n}(r_{n}x)\log(r_{n}x_{N})+g_{2,n}(r_{n}x)$
(6.30)

and thus using (6.25) we see that

$\displaystyle\frac{1}{r_{n}^{\beta}\theta_{n}(r_{n})}|({\mathcal{L}}_{\Phi_{n}}P_{n})(r_{n}x)-(g_{2,n}(r_{n}x)-g_{2,n}(0))|$
$\displaystyle\leq\frac{Cr_{n}^{\beta}|x|^{\beta}}{r_{n}^{\alpha}\theta_{n}(r_{n})}\left(1+|\log
x_{N}|+|\log r_{n}|\right)(r_{n}^{-\varrho}+1)$
$\displaystyle\leq\frac{C|x|^{\beta}}{\theta_{n}(r_{n})}\left(1+|\log
x_{N}|\right).$ (6.31)

Since $|\nabla\phi_{n}|\leq\frac{1}{4}$ and
$\widetilde{\Phi}_{n}(x)=x+(0,r_{n}^{-1}\phi_{n}(r_{n}x))$, we obtain

$\displaystyle
2|x-y|\geq|\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|\geq\frac{1}{2}|x-y|,\qquad|D\widetilde{\Phi}_{n}(x)|\leq
2\qquad\textrm{ for all $x,y\in\mathbb{R}^{N}$. }$ (6.32)

Define, for $x\in B_{R/2}$,

$\displaystyle F_{R}^{n}(x)$
$\displaystyle:=c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{(\chi_{R}(x)-\chi_{R}(y))v_{n}(y)}{|\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|^{N+2s}}\,dy$
$\displaystyle=c_{N,s}\int_{\mathbb{R}^{N}_{+}\cap\\{|y|\geq
R\\}}\frac{(1-\chi_{R}(y))v_{n}(y)}{|\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|^{N+2s}}\,dy.$

By the mean value theorem and the fact that $\widetilde{\Phi}_{n}(0)=0$, we
have $F_{R}^{n}(x)=F_{R}^{n}(0)+{\mathcal{R}}_{R}^{n}(x)$, where

${\mathcal{R}}_{R}^{n}(x):=-(N+2s)c_{N,s}\int_{0}^{1}\int_{\mathbb{R}^{N}_{+}}\frac{(\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y))\cdot\widetilde{\Phi}_{n}(x)}{|\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|}\frac{(1-\chi_{R}(y))v_{n}(y)}{|t\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|^{N+2s+1}}\,dy\,dt.$

Note that by (6.32),
$|t\widetilde{\Phi}_{n}(x)-\widetilde{\Phi}_{n}(y)|\geq\frac{1}{4}|y|$
whenever $|y|\geq R$ and $|x|\leq\frac{R}{4}$. Therefore, by (5.20) and
(6.32), there exists $C>0$ such that

$|\nabla{\mathcal{R}}_{R}^{n}(x)|\leq{C}R^{\beta-1}\qquad\textrm{ for all
$R>1$, $|x|\leq\frac{R}{4}$ and $n\geq 2$}$ (6.33)

and by construction we have ${\mathcal{R}}_{R}^{n}(0)=0$. We can thus write

$G^{n}_{R}(x)=H^{n}_{R}(x)+c^{n}_{R}+{\mathcal{R}}^{n}_{R}(x),$ (6.34)

with (recalling (6.30))

$H^{n}_{R}(x)=\frac{1}{r_{n}^{\alpha}\theta_{n}(r_{n})}\left\\{f_{n}(r_{n}x)-f_{n}(0)-({\mathcal{L}}_{\Phi_{n}}P_{n})(r_{n}x)+(g_{2,n}(r_{n}x)-g_{2,n}(0))\right\\}$

and

$c^{n}_{R}=\frac{1}{r_{n}^{\alpha}\theta_{n}(r_{n})}\left(f_{n}(0)-g_{2,n}(0)\right)+F_{R}^{n}(0).$

Recalling that $\|f_{n}\|_{C^{\alpha}(\mathbb{R}^{N}_{+})}\leq 1$, we then get
from (6.1) that

$|H^{n}_{R}(x)|\leq\frac{C(R)}{\theta_{n}(r_{n})}(1+|\log
x_{N}|)\qquad\textrm{ for all $x\in B_{R/2}^{+}$,}$

so that

$\|H^{n}_{R}\|_{L^{p}(B_{R/2})}\leq\frac{C(R)}{\theta_{n}(r_{n})}\qquad\textrm{
for all $p>1$.}$ (6.35)

Let us now show that $(c_{R}^{n})_{n\in\mathbb{N}}$ is bounded. For this, we
pick $\phi\in C^{\infty}_{c}(B_{1/4}^{+})$ with
$\int_{\mathbb{R}^{N}}\phi\,dx=1$ and let $R\geq 2$. Then multiply (6.29) by
$\phi$ and integrate over $\mathbb{R}^{N}_{+}$ to get

$\displaystyle|c^{n}_{R}|\leq\int_{\mathbb{R}^{N}}|\chi_{R}(x)v_{n}(x)||{\mathcal{L}}_{\widetilde{\Phi}_{n}}\phi(x)|\,dx+\int_{B_{R/2}}|H^{n}_{R}(x)\phi(x)|dx+\int_{B_{R/2}}|{\mathcal{R}}^{n}_{R}(x)\phi(x)|dx\leq
C(R).$ (6.36)

In view of this, (6.29), (6.34), (6.35) and (6.33), we can apply Lemma 6.3, to
deduce that $v_{n}$ is bounded in
$C^{2s-1}_{loc}(\overline{\mathbb{R}^{N}_{+}})\cap
H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$ and converges to some $v\in
C^{2s-1}_{loc}(\overline{\mathbb{R}^{N}_{+}})\cap
H^{s}_{loc}(\overline{\mathbb{R}^{N}_{+}})$. Moreover $v_{n}/x_{N}^{2s-1}$
converges to $v/x_{N}^{2s-1}$ in
$C^{1-\varepsilon}_{loc}(\overline{\mathbb{R}^{N}_{+}})$, for all
$\varepsilon\in(0,1)$. In addition passing to the limit in (6.23), (6.28) and
(6.24), we get

$|v(x)|\leq C|x|^{2s+\beta}\qquad\textrm{ for all $x\in\mathbb{R}^{N}$,}$
(6.37)
$\|v/x^{2s-1}_{N}\|_{L^{\infty}(B_{1}^{+})}\geq\frac{1}{2},\qquad\int_{B_{1}^{+}}\frac{v}{x_{N}^{2s-1}}(1-T_{3}^{\infty}(0)x_{N})\,dx=\int_{B_{1}^{+}}\frac{v}{x_{N}^{2s-1}}x_{i}\,dx=0,$
(6.38)

for $i=1,\dots,N-1$, where $T_{3}^{\infty}(0)$ is the limit $T_{3}^{n}(0)$ as
$n\to\infty$ and satisfies $|T_{3}^{\infty}(0)|\leq 1/4$. Passing to the limit
in (6.29), (6.33) and using (6.36), we get, for all $\phi\in
C^{\infty}_{c}(B_{\frac{R}{4}})$,

$\int_{\mathbb{R}^{2N}_{+}}\frac{((\eta_{R}v)(x)-(\eta_{R}v)(y))(\varphi(x)-\varphi(y))}{|x-y|^{N+2s}}dydx=\int_{\mathbb{R}^{N}_{+}}(c_{R}^{\infty}+{\mathcal{R}}^{\infty}_{R}(x))\varphi(x)\,dx$

and

$c^{\infty}_{R}\in\mathbb{R},\qquad{\mathcal{R}}_{R}^{\infty}(0)=0,\qquad|\nabla{\mathcal{R}}_{R}^{\infty}(x)|\leq{C}R^{\beta-1}\qquad\textrm{
for all $R\geq 2$ and $x\in B_{\frac{R}{2}}$.}$ (6.39)

Consequently, since $\beta<\alpha_{s}<1$,

$(-\Delta)^{s}_{{\mathbb{R}^{N}_{+}}}v\stackrel{{\scriptstyle\beta}}{{=}}0\qquad\textrm{
in $\mathbb{R}^{N}_{+}$}.$

Applying Theorem 3.4 and using (6.37), we deduce that
$\frac{v}{x_{N}^{2s-1}}(x)=b+a\cdot x^{\prime}$. This is in contradiction with
(6.38). ∎

######  Corollary 6.5.

Under the assumptions of Proposition 6.4, there exist $a\in\mathbb{R}^{N-1}$
and $b\in\mathbb{R}$ with $|a|+|b|\leq C$ such that letting
$\psi=\frac{w}{x_{N}^{2s-1}},$ we get

$\left\|\psi-\left((a\cdot
x^{\prime}+b)-bT_{3}(0)x_{N}\right)\right\|_{L^{\infty}(B_{r}^{+})}\leq
Cr^{1+\alpha}\qquad\textrm{ for all $r>0$.}$ (6.40)

Moreover $a=\nabla_{x^{\prime}}\psi(0)$, $b=\psi(0)$ and
$\partial_{x_{N}}\psi(0)=-bT_{3}(0)$.

###### Proof.

Let $Q_{r}(x)=a_{r,w}\cdot x^{\prime}+b_{r,{w}}(1-T_{3}(0)x_{N}).$ Let
$0<\rho_{2}\leq\rho_{1}/4\leq 1/4$. Pick $k\in\mathbb{N}$ and
$\sigma\in[1/4,1/2]$ such that $\rho_{2}=\sigma^{k}\rho_{1}$. Then provided
$r\in(0,1)$, by (6.26) and (6.14) we get

$\displaystyle\rho_{1}|a_{\rho_{2},w}-a_{\rho_{1},w}|+|b_{\rho_{2},w}-b_{\rho_{1},w}|$
$\displaystyle\leq\sum_{i=0}^{k-1}\sigma^{i+1}\rho_{1}|a_{\sigma^{i+1}\rho_{1},w}-a_{\sigma^{i}\rho_{1},w}|+\sum_{i=0}^{k-1}|b_{\sigma^{i+1}\rho_{1},w}-b_{\sigma^{i}\rho_{1},w}|$
$\displaystyle\leq
C\frac{1}{|B_{\sigma^{i+1}\rho_{1}}|^{\frac{1}{2}}}\|Q_{\sigma^{i+1}\rho_{1}}-Q_{\sigma^{i}\rho_{1}}\|_{L^{2}(B_{\sigma^{i+1}\rho_{1}})}$
$\displaystyle\leq C\sum_{i=0}^{k-1}\left(\|\psi-
Q_{\sigma^{i}\rho_{1}}\|_{L^{\infty}\left(B_{\sigma^{i}\rho_{1}}\right)}+\|\psi-
Q_{\sigma^{i+1}\rho_{1}}\|_{L^{\infty}\left(B_{\sigma^{i+1}\rho_{1}}\right)}\right)$
$\displaystyle\leq
C\sum_{i=0}^{k-1}\left((\sigma^{i}\rho_{1})^{1+\alpha}+(\sigma^{i+1}\rho_{1})^{1+\alpha}\right)\leq
C\rho_{1}^{1+\alpha}.$

This implies that the limits $a:=\lim_{r\to 0}a_{r,w}$ and $b:=\lim_{r\to
0}b_{r,w}$ exist. Moreover, letting $\rho_{2}\to 0$ and $\rho_{1}=1/2$ in the
above estimate, we get $|a|+|b|\leq C$. Hence (6.40) follows.
It is by now easy to see from (6.40), that $b=\psi(0)$ and
$\nabla\psi(0)=(a,-bT_{3}(0))$. ∎

######  Corollary 6.6.

Under the hypothesis of Proposition 6.4, there exists a constant $C$ depending
only on $N,s,\alpha$ and $\beta$ such that for all ${\xi\in
B_{1}^{+}\cap\mathbb{R}e_{N}}$,

$[w/x_{N}^{2s-1}]_{C^{1+\alpha}({B_{\xi_{N}/2}(\xi)})}\leq C.$

###### Proof.

Let $\xi:=(0,2\rho)\in B_{1}^{+}$ with $\rho<1/2$. We define

$P(x)=\overline{\eta}(x_{N})(x_{N})^{2s-1}_{+}(a\cdot
x^{\prime}+b)-bT_{3}(0)\overline{\eta}(x_{N})(x_{N})^{2s}_{+},$

where $b$ and $a$ are given by Corollary 6.5 and $T_{3}(0)$ is given by Lemma
6.2. Put $v_{\xi}(y)=w(\xi+\rho y)-P(\xi+\rho y)$,
$f_{\xi}(y)=\rho^{2s}f(\xi+\rho y)$. We have

$\mathcal{L}_{\Phi_{\rho}}v_{\xi}=f_{\xi}(x)+\rho^{2s}(\mathcal{L}_{\Phi}P)(\xi+\rho
x)\qquad\textrm{ for all $\varphi\in C^{\infty}_{c}(B_{1}^{+})$,}$ (6.41)

where $\Phi_{\rho}(x)=\rho^{-1}\Phi(\rho
x+\xi)=(x^{\prime},x_{N}+\rho^{-1}\phi(\rho x^{\prime}))$ and

$\mathcal{L}_{\Phi_{\rho}}v(x)=c_{N,s}\int_{\\{y_{N}>-2\\}}\frac{v(x)-v(y)}{|\Phi_{\rho}(x)-\Phi_{\rho}(y)|^{N+2s}}\,dy.$

By Lemma 6.2, we have

$\displaystyle F(x):=(\mathcal{L}_{\Phi}P)(\xi+\rho x)$
$\displaystyle=g_{1}(\xi+\rho x)\log(2\rho+\rho x_{N})+g_{2}(\xi+\rho x)$

from which it follows that $\|F\|_{C^{\beta}([0,1))}\leq
C\rho^{\beta}|\log\rho|\leq C\rho^{\alpha}$. Note that $|u_{\xi}(x)|\leq
C(1+|x|^{2s+\alpha})$ and thus since $\Phi_{\rho}\in
C^{2}(\mathbb{R}^{N};\mathbb{R}^{N})$ is a global (volume preserving)
diffeomorphism, by a change of variable and (6.41), we obtain

$(-\Delta)^{s}_{\Omega_{\rho}}v_{\xi}\circ\Phi_{\rho}^{-1}{=}G_{\xi}\circ\Phi_{\rho}^{-1}\qquad\textrm{
in $\Omega_{\rho}$,}$

where $G_{\xi}(x):=f_{\xi}(x)+\rho^{2s}F(\xi+\rho x)$ and
$\Omega_{\rho}=\Phi_{\rho}^{-1}(\\{y_{N}>-2\\})$. We note that
$[G_{\xi}]_{C^{\alpha}(B_{1/2})}\leq\rho^{2s+\alpha}$,
$\|v_{\xi}\circ\Phi_{\rho}^{-1}\|_{L^{\infty}(B_{1/2})}\leq\rho^{2s+\alpha}$
and $B_{1}\subset\Omega_{\rho}$. We then apply Lemma 5.5 to deduce that

$\displaystyle\|v_{\xi}\circ\Phi_{\rho}^{-1}\|_{C^{2s+\alpha}(B_{1/4})}$
$\displaystyle\leq C\left(\int_{\Omega_{\rho}\setminus
B_{1/8}}|y|^{-N-2s-1}|v_{\xi}\circ\Phi_{\rho}^{-1}(y)|dy+\rho^{2s+\alpha}\right)$
$\displaystyle\leq C\left(\int_{\mathbb{R}^{N}_{+}\setminus
B_{1/16}}|x|^{-N-2s-1}|v_{\xi}(x)|dx+\rho^{2s+\alpha}\right)$
$\displaystyle\leq
C\left(\sum_{i=0}^{\infty}\int_{2^{i-3}>|x|>2^{i-4}}|x|^{-N-2s-1}|v_{\xi}(x)|dx+\rho^{2s+\alpha}\right)$
$\displaystyle\leq
C\left(\sum_{i=0}^{\infty}(2^{i}\rho)^{-N-2s-1}\rho^{2s+1}\int_{2^{i-3}\rho>|z|>2^{i-4}\rho}v_{\xi}(z/\rho)|dz+\rho^{2s+\alpha}\right)$
$\displaystyle\leq
C\left(\sum_{i=0}^{\infty}(2^{i}\rho)^{-N-2s-1}\rho^{2s+1}\int_{2^{i}\rho>|x|}|w(x)-P(x)|dx+\rho^{2s+\alpha}\right)$
$\displaystyle\leq
C\left(\sum_{i=0}^{\infty}2^{i{(\alpha-1)}}\rho^{2s+\alpha}+\rho^{2s+\alpha}\right)\leq
C\rho^{2s+\alpha},$

where we used (6.40) which yields $|w(x)-P(x)|\leq C|x|^{2s+\alpha}$. We
conclude that

$\|v_{\xi}\|_{C^{2s+\alpha}(B_{1/4})}\leq C\rho^{2s+\alpha}.$

Scaling and translating back, we then get

$\|w-P\|_{L^{\infty}(B_{\rho/4}(\xi))}\leq C\rho^{2s+\alpha},\qquad[\nabla
w-\nabla P]_{C^{\alpha}(B_{\rho/4}(\xi))}\leq C\rho^{2s-1}.$

Combining the above estimates with

$\|1/x_{N}^{2s-1}\|_{L^{\infty}(B_{\rho/4}(\xi))}\leq
C\rho^{1-2s},\qquad[\nabla(1/x_{N}^{2s-1})]_{C^{\alpha}(B_{\rho/4}(\xi))}\leq
C\rho^{-2s-\alpha}$

we finally get

$[\nabla(w/x^{2s-1})-\nabla(P/x^{2s-1})]_{C^{\alpha}(B_{\rho/4}(\xi))}\leq C.$

Since $[\nabla(P/x^{2s-1})]_{C^{\alpha}(B_{\rho/4}(\xi))}=0$, this yields

$[\nabla(w/x^{2s-1})]_{C^{\alpha}(B_{\rho/4}(\xi))}\leq C.$

and the proof is complete. ∎

We now prove the following.

######  Theorem 6.7.

Given $s\in(\frac{1}{2},1)$, let $\alpha_{s}\in(0,1-s)$ be given by Lemma 2.6.
Let $\Omega\subset\mathbb{R}^{N}$ be a $C^{2,\beta}$ domain, with
$0\in\partial\Omega$, and $f\in C^{\alpha}(\overline{\Omega}\cap B_{1})$ with
$\alpha<\min\\{\alpha_{s},2s-1,\beta\\}$ and $\alpha+2s\neq 1$. Let $u\in
H^{s}_{0}(\Omega)$ be such that

$D_{\Omega}(u,\varphi)=\int_{\Omega}f\varphi\,dx\qquad\textrm{$\forall\varphi\in
C^{\infty}_{c}(B_{1}\cap\Omega)$,}$

and let $\delta=\textrm{dist}(x,\partial\Omega)$. Then,

$\|u/\delta^{2s-1}\|_{C^{1+\alpha}(B_{\rho/2}\cap\overline{\Omega})}\leq
C\left(\|u\|_{L^{2}(\Omega)}+\|f\|_{C^{\alpha}(\overline{\Omega})}\right),$
(6.42)

with $C,\rho>0$ depending only on $N$, $s$, $\alpha$, and $\Omega$.

Moreover, letting $\psi=u/\delta^{2s-1}$, we have

$\partial_{\nu}\psi(\sigma)=-(N-1)H_{\partial\Omega}(\sigma)\psi(\sigma)\qquad\textrm{
for all $\sigma\in\partial\Omega\cap B_{1}$,}$ (6.43)

where $H_{\partial\Omega}$ is the mean curvature of $\partial\Omega$.

###### Proof.

We assume for simplicity that
$\|u\|_{L^{2}(\Omega)}+\|f\|_{C^{\alpha}(\overline{\Omega})}\leq 1.$ Hence by
Lemma 6.3, we get $\|u/\delta^{2s-1}\|_{L^{\infty}(\Omega\cap
B_{1/2})}+\|u\|_{C^{2s-1}(\overline{\Omega}\cap B_{1/2})}\leq C$.

Up to a scaling and a rotation, we may assume that $B_{1/4}\cap\partial\Omega$
is contained in the graph $\\{x_{N}=\gamma(x^{\prime})\\}$ with
$\gamma\in{C^{2,\beta}(B_{1/4}^{\prime})}$. For $q\in\partial\Omega$, let
$\nu(q)$ denote the unit exterior normal vector of $\Omega$ at $q$ and denote
$T_{q}\partial\Omega=\nu(q)^{\perp}$ the tangent plane of $\partial\Omega$ at
$q$.
By the local inversion theorem, there exist $r_{0}\in(0,1/4]$ depending only
on $N$, $\beta$ and $\|\gamma\|_{{C^{2,\beta}(B_{1}^{\prime})}}$ such that for
any $q\in B_{r_{0}}\cap\partial\Omega$, there exist an orthonormal basis
$(E_{1}(q)\dots,E_{N-1}(q))$ of $T_{q}\partial\Omega$ and $\widetilde{\phi}\in
C^{2,\beta}(B_{r_{0}}^{\prime})$ such that

$x^{\prime}\mapsto
q+\sum_{i=1}^{N-1}x_{i}E_{i}(q)-\widetilde{\phi}(x^{\prime})\nu(q):B_{r_{0}}^{\prime}\to\partial\Omega$

with $\|\widetilde{\phi}\|_{C^{2,\beta}(B_{r_{0}}^{\prime})}\leq
C(N,\beta,\gamma,r_{0})$. It is clear that
$\widetilde{\phi}(0)=|\nabla\widetilde{\phi}(0)|=0$. We let $\eta\in
C^{\infty}_{c}(B_{r_{0}}^{\prime})$ with $\eta\equiv 1$ on
$B_{r_{0}/2}^{\prime}$ and put $\phi=\eta\widetilde{\phi}$. For all $q\in
B_{r_{0}}\cap\partial\Omega$, we define

$\Phi:\mathbb{R}^{N}\to\mathbb{R}^{N},\qquad\Phi(x^{\prime},x_{N})=q+\sum_{i=1}^{N-1}x_{i}E_{i}(q)-\left(x_{N}+\phi(x^{\prime})\right)\nu(q).$
(6.44)

Next, decreasing $r_{0}$ if necessary, $\Phi$ is a global $C^{2,\beta}$
diffeomorphism and clearly $\textrm{Det}D\Phi(x)=1$ for all
$x\in\mathbb{R}^{N}$. We may assume that the (signed) distance function $d\in
C^{2,\beta}(\\{d<r_{0}/2\\})$. We also recall that the mean curvature of
$\partial\Omega$ at $q$ is given by

$H_{\partial\Omega}(q):=\frac{1}{N-1}\Delta\phi(0).$ (6.45)

We define $Q_{\delta}:=B_{\delta}^{\prime}\times(0,\delta)$ and
$\mathcal{Q}_{\delta}:=\Phi(Q_{\delta})\subset\Omega$. For any
$x_{0}\in\mathcal{Q}_{r_{0}/9}$, let $q\in\partial\Omega\cap B_{r_{0}}$ such
that $|x_{0}-q|=d(x_{0})$. Letting $v={\chi_{r_{0}/4}}u$, we have

$\displaystyle(-\Delta)^{s}_{\Omega}v=f+c_{N,s}\int_{\Omega}\frac{(1-\chi_{{r_{0}/4}}(y))u(y)}{|x-y|^{N+2s}}\,dy=:f_{1}.$

Clearly $\|f_{1}\|_{C^{\alpha}(\mathcal{Q}_{r_{0}/8})}\leq C$ for all
$\alpha\in(0,\min(2s-1,\beta))$. Now we get

$\displaystyle(-\Delta)^{s}_{\mathcal{Q}_{r_{0}/2}}v=f_{1}-(-\Delta)^{s}_{\Omega\setminus\mathcal{Q}_{r_{0}/2}}v=:f_{2},$

and obviously $\|f_{2}\|_{C^{\alpha}(\mathcal{Q}_{r_{0}/8})}\leq C$ because
$\|v\|_{C^{2s-1}(\mathcal{Q}_{r_{0}/8})}\leq C$. We define $w(x)=v(\Phi(x))$
and $g_{2}(x)=f_{2}(\Phi(x))$. Then by a change of variable, we get

$\displaystyle
c_{N,s}\int_{Q_{r_{0}/2}}\frac{w(x)-w(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=g_{2}(x).$

Hence $w\in H^{s}(\mathbb{R}^{N}_{+})$ satisfies
$\|w/x^{2s-1}_{N}\|_{L^{\infty}(\mathbb{R}^{N}_{+})}+\|w\|_{C^{2s-1}(\mathbb{R}^{N}_{+})}\leq
C$ and solves

$\displaystyle
c_{N,s}\int_{\mathbb{R}^{N}_{+}}\frac{w(x)-w(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy=g_{3}(x),$
(6.46)

with $g_{3}(x)=g_{2}(x)-c_{N,s}\int_{\mathbb{R}^{N}_{+}\setminus
Q_{r_{0}/2}}\frac{w(x)-w(y)}{|\Phi(x)-\Phi(y)|^{N+2s}}\,dy.$ We have that
$\|g_{3}\|_{C^{\alpha}(Q_{r_{0}/8})}\leq C.$

Recall that $x_{0}=q-|x_{0}-q|\nu(q)=\Phi(0,|x_{0}-q|).$ We can thus apply
Corollary 6.6 to the equation (6.46), to get

$[w/x_{N}^{2s-1}]_{C^{1+\alpha}(B_{r/2}(\xi))}=[u\circ\Phi/x_{N}^{2s-1}]_{C^{1+\alpha}({B_{r/2}(\xi)})}\leq
C,\quad\textrm{ with $\xi=\Phi^{-1}(x_{0})\in
Q_{\frac{r_{0}}{9}}\cap\mathbb{R}e_{N}$.}$ (6.47)

where $r=\xi_{N}/2=|x_{0}-q|/2$. In addition, Corollary 6.40, (6.1) and (6.45)
yield

$\partial_{x_{N}}[w/x_{N}^{2s-1}](0)=-T_{3}(0)[w/x_{N}^{2s-1}](0)=(N-1)H_{\partial\Omega}(q)[w/x_{N}^{2s-1}](0).$
(6.48)

In view of (6.44) and (6.47), by changing variables, we thus obtained

$[u/\delta^{2s-1}]_{C^{1+\alpha}({B_{r/8}(x_{0})})}\leq C,$

where we used that $x\mapsto\frac{d(\Phi(x))}{x_{N}}\in
C^{1+\alpha}(B_{r_{0}/2})$ and that $d=\delta$ in $\Omega$. This implies that
$[u/\delta^{2s-1}]_{C^{1+\alpha}(\mathcal{Q}_{r_{0}/9})}\leq C$ and (6.42)
follows.

Finally using (6.48) and the fact that
$d(\Phi(0,x_{N}))=d(0,x_{N})=x_{N}+O(x_{N}^{2})$, we conclude that

$\partial_{\nu}[u/\delta^{2s-1}](q)=-(N-1)H_{\partial\Omega}(q)[u/\delta^{2s-1}](q)$

and the proof is complete. ∎

We finally give the:

###### Proof of Theorem 1.3.

The result follows from Theorem 6.7 and a covering argument. ∎

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