# Liouville results for semilinear integral equations with conical diffusion

Isabeau Birindelli, Lele Du and Giulio Galise  Dipartimento di Matematica
Guido Castelnuovo, Sapienza Università di Roma, Piazzale Aldo Moro 5, 00185,
Roma, Italy<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Nonexistence results for positive supersolutions of the equation

$-Lu=u^{p}\quad\text{in $\mathbb{R}^{N}_{+}$}$

are obtained, $-L$ being any symmetric and stable linear operator, positively
homogeneous of degree $2s$, $s\in(0,1)$, whose spectral measure is absolutely
continuous and positive only in a relative open set of the unit sphere of
$\mathbb{R}^{N}$. The results are sharp: $u\equiv 0$ is the only nonnegative
supersolution in the subcritical regime $1\leq p\leq\frac{N+s}{N-s}\,$, while
nontrivial supersolutions exist, at least for some specific $-L$, as soon as
$p>\frac{N+s}{N-s}$.
The arguments used rely on a rescaled test function’s method, suitably adapted
to such nonlocal setting with weak diffusion; they are quite general and also
employed to obtain Liouville type results in the whole space.

##  1\. Introduction

In this paper we obtain Liouville theorems for supersolutions of semilinear
integral equations in the half-space
$\mathbb{R}^{N}_{+}:=\left\\{x\in\mathbb{R}^{N}:\,x_{N}>0\right\\}$ with the
following class of nonlocal operators

(1.1)
$Lu\left(x\right)=\left(1-s\right)\int_{\mathbb{R}^{N}}\frac{u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy,$

where $0<s<1$ and $a\left(\theta\right)\in
L^{\infty}\left(\mathbb{S}^{N-1}\right)$ is a nonnegative and even function on
the unit sphere $\mathbb{S}^{N-1}$ of $\mathbb{R}^{N}$. We will be more
precise later, but let us emphasize immediately that the function
$a\left(\theta\right)$ can be chosen so that the operator diffuses only along
a cone.

The main result will be about nonexistence of classical solutions, besides the
trivial one, $u\in C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$ of
the problem

(1.2) $\displaystyle\left\\{\begin{aligned} -Lu&\geq
u^{p},&&x\in\mathbb{R}_{+}^{N},\\\ u&\geq
0,&&x\in\mathbb{R}^{N},\end{aligned}\right.$

where $\mathcal{L}_{s}$ is the natural functional space required that will be
defined precisely later.

Nonexistence of entire bounded harmonic functions, goes back to Cauchy (1844)
and Liouville (1850), but since the acclaimed works of Gidas and Spruck [24]
much attention has been given to so called semilinear Liouville theorems in
the whole space and in half-spaces. Indeed these nonexistence results in
unbounded domains correspond to existence results in bounded domains thanks to
the priori bounds they imply and the use of Leray-Schauder degree theory.

A particular role has been given in these decades to determine the range of
existence, or better lack of it, when instead of considering solutions, one
concentrates on supersolutions. This has been done for very different cases:
the case of cone-like domains for linear and quasilinear operators appeared
for instance in [1, 6, 5, 23, 26], existence and nonexistence for positive
supersolutions in the whole space, half-spaces and exterior domains for fully
nonlinear inequalities can be found e.g. in [2, 17, 27], systems and sub-
Laplacians have been respectively addressed in [10, 29] and [4, 7], while as
far as nonlinear degenerate elliptic operators are concerned we refer to [3,
8].

We cannot recall all the results in this area but the interested readers can
refer to the book [30] and the references therein. We just remind that, for
the Laplacian, if

$1\leq p\leq\frac{N+1}{N-1}$

then the inequality

$\Delta u+u^{p}\leq 0\quad\text{in $\mathbb{R}^{N}_{+}$}$

admits only the trivial solution (in fact such result is still true for
$p\geq-1$, but in this paper we only consider the superlinear case $p\geq 1$).
Furthermore, the bound $\frac{N+1}{N-1}$ is sharp in the sense that for
$p>\frac{N+1}{N-1}$ it is possible to construct explicit positive
supersolutions.

In the realm of nonlocal operators, the results are more recent but
nonetheless really numerous, see e.g. [11, 13, 19, 35] concerning entire
$s$-harmonic functions and [14, 18, 21, 34] for positive solutions and
supersolutions of Lane-Emden type equations/systems in $\mathbb{R}^{N}$ or in
exterior domains. Since we mainly concern here with supersolutions of
semilinear nonlocal operators in half-spaces, we will recall results directly
related with this cases. In [15], using the method of moving planes in
integral forms, Chen, Fang and Yang have proved that for
$1<p\leq\frac{N+2s}{N-2s}$ any nonnegative and locally bounded distributional
solution of

$\displaystyle\left\\{\begin{aligned}
(-\Delta)^{s}u&=u^{p},&&x\in\mathbb{R}_{+}^{N},\\\
u&=0,&&x\in\overline{\mathbb{R}_{-}^{N}},\end{aligned}\right.$

where $\mathbb{R}^{N}_{-}:=\left\\{x\in\mathbb{R}^{N}:\,x_{N}<0\right\\}$, is
identically zero. The result has been extended in [16] for a larger class of
operators.

Instead, for supersolutions, the proofs and the range for which the validity
of the Liouville theorem holds are different and very few results are
available in the half-spaces. We wish to mention the result for nonlocal fully
nonlinear operators by Nornberg, dos Prazeres and Quaas [28] where they study
the existence of fundamental solutions in conical domains for fully nonlinear
integral operators of Isaacs type and, as application, they obtain Liouville
type results in subcritical regimes. In the special case where the cone is the
half-space $\mathbb{R}^{N}_{+}$ and the diffusion operator is $(-\Delta)^{s}$
they find in particular that for

$1\leq p\leq\frac{N+s}{N-s}$

the only nonnegative (in $\mathbb{R}^{N}$) viscosity supersolutions of the
problem

$(-\Delta)^{s}u\geq u^{p}\quad\text{in $\mathbb{R}^{N}_{+}$}$

is the trivial one (even in this case the results is still valid for
$p\geq-1$).

In the works mentioned above, even in the nonlinear case, the kernels of
operators considered are, up to some constant, bounded above and below by that
of the fractional Laplacian. Here instead, the operators $L$ we consider are
part of the larger class of infinitesimal generators of stable Lévy processes

(1.3)
$Lu\left(x\right)=(1-s)\int_{\mathbb{S}^{N-1}}\int_{0}^{+\infty}\frac{u\left(x+t\theta\right)+u\left(x-t\theta\right)-2u\left(x\right)}{t^{1+2s}}dtd\mu,$

where the spectral measure $\mu=\mu\left(\theta\right)$ on $\mathbb{S}^{N-1}$
satisfies the ellipticity conditions

$\displaystyle
0<\lambda\leq\inf_{\nu\in\mathbb{S}^{N-1}}\int_{\mathbb{S}^{N-1}}\left|\nu\cdot\theta\right|^{2s}d\mu\quad\text{and}\quad\int_{\mathbb{S}^{N-1}}d\mu\leq\Lambda<\infty.$

When $\mu$ is absolutely continuous and $d\mu=a\left(\theta\right)d\theta$,
then (1.1) and (1.3) coincide.
At this stage it is worth to point out that the operator (1.1) converges, as
$s\to 1^{-}$, to a second order linear uniformly elliptic operators with
constant coefficients, in the sense that for any $u\in
C^{2}_{0}(\mathbb{R}^{N})$, then

$\lim_{s\to
1^{-}}(1-s)Lu(x)=\sum_{i,j=1}^{N}a_{ij}\partial^{2}_{ij}u(x)\qquad\forall
x\in\mathbb{R}^{N},$

where the constant coefficients $a_{ij}$ depend on the function $a(\theta)$.

Moreover, let us point out that, since the normalizing constant $1-s$ in (1.1)
is irrelevant for the issues that we address in the present paper, henceforth
it will be omitted.

We refer to the previous works of Ros-Oton and Serra [32] and in particular to
the very interesting paper [33] where, in order to give boundary estimates,
some linear Liouville theorems in half-spaces are proved under the further
assumption that the function $a(\theta)$ is positive bounded away from zero
everywhere.

Contrarily to the cases where $a(\theta)$ is positive bounded away from zero
everywhere, here we only suppose that $a(\theta)$ is positive in some relative
open set on $\mathbb{S}^{N-1}$. Clearly if $a(\theta)$ is bounded and bounded
away from zero then the kernel of the operator (1.1) can be compared from
above and below by that of the fractional Laplacian, which corresponds to the
case $a(\theta)$ is a constant function. Other results concerning
classifications of entire solutions and Hölder regularity for nonlocal
operators with anisotropic diffusion can be found respectively in [20, 25].

We can now give the precise conditions on the function $a(\theta)$. Let us fix
the following notation: for any vector $\nu\in\mathbb{S}^{N-1}$ and
$0<\tau\leq 1$, we define the closed two fold cone
$\Sigma_{\nu,\tau}\left(x\right)$ of aperture
$\arccos(1-\tau)\in\left(0,\frac{\pi}{2}\right]$, with vertex
$x\in\mathbb{R}^{N}$ and axis $\nu$, by

(1.4)
$\displaystyle\Sigma_{\nu,\tau}\left(x\right):=\left\\{y\in\mathbb{R}^{N}:\;\left|\left(y-x\right)\cdot\nu\right|\geq\left(1-\tau\right)\left|y-x\right|\right\\}.$

In all the paper, we assume that for some constants $0<d<D$,

(1.5) $\displaystyle 0\leq a(\theta)\leq D\quad\text{in $\mathbb{S}^{N-1}$}$

and that there exist $\nu_{0}\in\mathbb{S}^{N-1}$ and $0<\tau_{0}\leq 1$ such
that

(1.6) $\displaystyle a(\theta)\geq d>0\quad\text{in
$\Sigma_{\nu_{0},\tau_{0}}\left(0\right)\cap\mathbb{S}^{N-1}$}.$

Hence in all the proofs we can only use that the operator (1.1) diffuses along
a fixed cone. This feature will induce some delicate geometric constructions.

Let

$\displaystyle\mathcal{L}_{s}=\left\\{u\in
L^{1}_{loc}\left(\mathbb{R}^{N}\right):\,\limsup_{\left|x\right|\rightarrow+\infty}\frac{\left|u\right|}{\left|x\right|^{2s-\delta}}<+\infty\text{\;
for some\,}\delta\in(0,2s]\right\\}.$

Our main result, is the following

######  Theorem 1.1.

Let $s\in(0,1)$ and let $L$ be any operator of the form (1.1) satisfying
(1.5)-(1.6). If $1\leq p\leq\frac{N+s}{N-s}$ and $u\in
C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$ is a solution of
(1.2), then $u\equiv 0$.

We were inspired by the pioneering proof used by Berestycki, Capuzzo Dolcetta
and Nirenberg in [5] for supersolutions of semilinear Liouville Theorem in
cones. But it is important to notice that the key ingredients they use need to
be completely reconsidered due both to the nonlocal character of the operators
we consider and their weak diffusion. Even the simple integrations by parts
formula, since the test functions we shall use in the proof of Theorem 1.1 are
not smooth on $\partial\mathbb{R}^{N}_{+}$, needs to be proved (see
Proposition 2.2). The other novelty is in the ad hoc construction of the test
functions. Indeed we construct a sequence of nonnegative compactly supported
test functions that converge to 1 in the whole half space, but their supports
depend on the cone $\Sigma_{\nu_{0},\tau_{0}}\left(0\right)$ where the
function $a(\theta)$ is supported.

Interestingly the bound on $p$, does not depend on the size of the cone.
Furthermore, at least in the case of the fractional Laplacian, the bound is
optimal: for any $p>\frac{N+s}{N-s}$ the problem

$\displaystyle\left\\{\begin{aligned} \left(-\Delta\right)^{s}u&\geq
u^{p},&&x\in\mathbb{R}_{+}^{N},\\\
u&=0,&&x\in\mathbb{R}_{-}^{N}\end{aligned}\right.$

admits positive solutions $u\in
C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$. This is proved in
Theorem 4.1.

The arguments used in the proof of Theorem 1.1 are quite flexible and they
also apply to get the equivalent Liouville result in the whole space
$\mathbb{R}^{N}$, in a simplified form due to the absence of the boundary of
the domain which, instead, poses additional difficulties in the case of half-
spaces. We obtain nonexistence of positive solutions provided $1\leq
p\leq\frac{N}{N-2s}$, see Theorem 5.1. This range is known to be optimal for
the fractional Laplacian and even more, as showed by Felmer and Quaas in [22],
for Pucci’s nonlinear operators ${\mathcal{M}}^{\pm}$, which are extremal
operators among the class of linear integral operators whose kernels,
differently to the cases treated in the present paper, are comparable to that
of the fractional Laplacian. The critical exponents associated to Pucci’s
operators are $p=\frac{N^{\pm}}{N^{\pm}-2s}$, where $N^{\pm}$ are dimensional-
like numbers which reduce to the dimension $N$ when
$-{\mathcal{M}}^{\pm}=(-\Delta)^{s}$.

The paper is organized as follows. Section 2 contains several technical
results that will be used throughout the paper. Section 3 is devoted to the
proof of Theorem 1.1 and Section 4 is concerned with the optimality of the
exponent $p=\frac{N+s}{N-s}$. In the last Section 5, we prove a Liouville
theorem in the whole space. Let us mention that the case $N=1$ is much simpler
and some remarks concerning the results in that case are scattered along the
paper (See e.g. Remarks 4.2 and 5.2).

##  2\. preliminary

In this section we prove several technical results that will be used in the
proof of Theorem 1.1. We start by the following simple lemma, where it is
proved that the nonnegativity assumption of $u$ in
$\overline{\mathbb{R}^{N}_{-}}$ can be, in fact, replaced by the simplest one
$u=0$ in $\overline{\mathbb{R}_{-}^{N}}$. This reduction is useful in the
integration by parts formula of Proposition 2.2.

Henceforth we denote by $e_{N}=(0,\ldots,0,1)$ and we use the notation
$C=C(\cdot)$ to denote positive constants depending on given quantities.

######  Lemma 2.1.

For any nonnegative classical solution $u$ of (1.2), the translation-
truncation function (see Figure 1)

(2.1) $\displaystyle\widetilde{u}\left(x\right)=\left\\{\begin{aligned}
&u\left(x+e_{N}\right),&&x\in\mathbb{R}_{+}^{N},\\\
&0,&&x\in\overline{\mathbb{R}_{-}^{N}}\end{aligned}\right.$

is a nonnegative classical solution of the problem

(2.2) $\displaystyle\left\\{\begin{aligned} -Lu&\geq
u^{p},&&x\in\mathbb{R}_{+}^{N},\\\
u&=0,&&x\in\overline{\mathbb{R}_{-}^{N}}.\end{aligned}\right.$

Figure 1. The graph of $\widetilde{u}$.

###### Proof.

For $x\in\mathbb{R}_{+}^{N}$ and any $y\in\mathbb{R}^{N}$ it holds that

$\displaystyle\widetilde{u}\left(x\pm y\right)\leq u\left(x+e_{N}+y\right).$

Hence

$\displaystyle-L\widetilde{u}\left(x\right)$
$\displaystyle=-\int_{\mathbb{R}^{N}}\frac{\widetilde{u}\left(x+y\right)+\widetilde{u}\left(x-y\right)-2\widetilde{u}\left(x\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\geq-\int_{\mathbb{R}^{N}}\frac{u\left(x+e_{N}+y\right)+u\left(x+e_{N}-y\right)-2u\left(x+e_{N}\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=-Lu\left(x+e_{N}\right).$

Since $u$ is a solution of (1.2), we have

$\displaystyle-L\widetilde{u}\left(x\right)\geq-Lu\left(x+e_{N}\right)\geq
u^{p}\left(x+e_{N}\right)=\widetilde{u}^{p}\left(x\right).$

∎

For any $\widetilde{u}$ as in Lemma 2.1, we find that $\widetilde{u}$
satisfies the following regularity condition up to the boundary of
$\mathbb{R}^{N}_{+}$:

(2.3)
$\displaystyle\left\|u\right\|_{C^{2}\left(K\cap\overline{\mathbb{R}^{N}_{+}}\right)}<+\infty,\quad\forall\text{
compact }K\subseteq\overline{\mathbb{R}_{+}^{N}}.$

It is well known that $L$ is a self adjoint operator but we need to take care
of the “integration by part” since the test function is not smooth. This is
the object of the next Proposition.

######  Proposition 2.2.

Let $(2s-1)_{+}<\alpha<2s$, $\psi\in C^{2}_{0}\left(\mathbb{R}^{N}\right)$ and
$v_{\alpha}\left(x\right)=\left(x_{N}\right)_{+}^{\alpha}\psi\left(x\right)$.
Then for any $u\in C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$
satisfying (2.3) and $u=0$ in $\overline{\mathbb{R}_{-}^{N}}$, we have

$\displaystyle\int_{\mathbb{R}^{N}}uLv_{\alpha}dx=\int_{\mathbb{R}^{N}}v_{\alpha}Ludx.$

###### Proof.

Assume that supp $\psi\subseteq B_{R}$, $R>1$. Let us define two maps

$\displaystyle
F_{1},F_{2}:\mathbb{R}^{N}\times\left(\mathbb{R}^{N}\backslash\left\\{0\right\\}\right)\rightarrow\mathbb{R}$

by

$\displaystyle
F_{1}\left(x,y\right):=\frac{v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)u\left(x\right)$

and

$\displaystyle
F_{2}\left(x,y\right):=\frac{u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)v_{\alpha}\left(x\right).$

What needs to be proved in order to get (2.3) is

(2.4)
$\displaystyle\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left|F_{1}\right|dy\right)dx<+\infty$

and

(2.5)
$\displaystyle\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left|F_{2}\right|dy\right)dx<+\infty.$

Once (2.4) and (2.5) are proved i.e. that $F_{1},F_{2}\in
L^{1}\left(\mathbb{R}^{N}\times\mathbb{R}^{N}\right)$, we may apply Fubini-
Tonelli’s theorem and it is classical that

$\displaystyle\int_{\mathbb{R}^{N}}uLv_{\alpha}dx=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}F_{1}dy\right)dx=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}F_{1}dx\right)dy$
$\displaystyle=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left[v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right]u\left(x\right)dx\right)\frac{1}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left[u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)\right]v_{\alpha}\left(x\right)dx\right)\frac{1}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}F_{2}dx\right)dy=\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}F_{2}dy\right)dx=\int_{\mathbb{R}^{N}}v_{\alpha}Ludx.$

We first prove (2.4). Note that

(2.6) $\displaystyle
F_{1}\left(x,y\right)=0\quad\forall\left(x,y\right)\in\overline{\mathbb{R}_{-}^{N}}\times\left(\mathbb{R}^{N}\backslash\left\\{0\right\\}\right),$

we only consider
$\left(x,y\right)\in\mathbb{R}_{+}^{N}\times\left(\mathbb{R}^{N}\backslash\left\\{0\right\\}\right)$.
For any $x\in\mathbb{R}^{N}_{+}\cap\left\\{\left|x\right|\geq 2R\right\\}$,
there is $v_{\alpha}\left(x\right)=0$. Moreover, if $\left|y\pm x\right|\geq
R$, then $v_{\alpha}\left(x\pm y\right)=0$, while if $\left|y\pm x\right|\leq
R$, we have

$\displaystyle\left|y\right|\geq\left|x\right|-\left|y-x\right|\geq\frac{\left|x\right|}{2}.$

Therefore, for any $x\in\mathbb{R}^{N}_{+}\cap\left\\{\left|x\right|\geq
2R\right\\}$,

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=\int_{B_{R}\left(x\right)}\frac{\left|v_{\alpha}\left(x-y\right)\right|}{\left|y\right|^{N+2s}}dy+\int_{B_{R}\left(-x\right)}\frac{\left|v_{\alpha}\left(x+y\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=\int_{B_{R}\left(x\right)}\frac{\left|\left(x_{N}-y_{N}\right)_{+}^{\alpha}\psi\left(x-y\right)\right|}{\left|y\right|^{N+2s}}dy+\int_{B_{R}\left(-x\right)}\frac{\left|\left(x_{N}+y_{N}\right)_{+}^{\alpha}\psi\left(x+y\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle\leq
2^{N+2s}R^{\alpha}\left\|\psi\right\|_{L^{\infty}\left(\mathbb{R}^{N}\right)}\left(\left|B_{R}\left(x\right)\right|+\left|B_{R}\left(-x\right)\right|\right)\frac{1}{\left|x\right|^{N+2s}}$
$\displaystyle=C\left(N,s,\alpha,R,\left\|\psi\right\|_{L^{\infty}\left(\mathbb{R}^{N}\right)}\right)\frac{1}{\left|x\right|^{N+2s}}.$

For any $x\in\mathbb{R}^{N}_{+}\cap B_{2R}$, letting $z=\frac{y}{x_{N}}$, we
have

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{\left|\left(x_{N}+y_{N}\right)_{+}^{\alpha}\psi\left(x+y\right)+\left(x_{N}-y_{N}\right)_{+}^{\alpha}\psi\left(x-y\right)-2x_{N}^{\alpha}\psi\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=x_{N}^{\alpha}\int_{\mathbb{R}^{N}}\frac{\left|\left(1+\frac{y_{N}}{x_{N}}\right)_{+}^{\alpha}\psi\left(x+y\right)+\left(1-\frac{y_{N}}{x_{N}}\right)_{+}^{\alpha}\psi\left(x-y\right)-2\psi\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=\left(\int_{\mathbb{R}^{N}}\frac{\left|A\left(x,z\right)\right|}{\left|z\right|^{N+2s}}dz\right)x_{N}^{\alpha-2s},$

where

$\displaystyle
A\left(x,z\right)=\left(1+z_{N}\right)_{+}^{\alpha}\psi\left(x+x_{N}z\right)+\left(1-z_{N}\right)_{+}^{\alpha}\psi\left(x-x_{N}z\right)-2\psi\left(x\right).$

Since $A$ satisfies

$\displaystyle\left\\{\begin{aligned} &A\left(x,0\right)=0&&\forall
x\in\mathbb{R},\\\
&A\left(x,z\right)=A\left(x,-z\right)&&\forall\left(x,z\right)\in\mathbb{R}^{N}\times\mathbb{R}^{N},\\\
&A\in
C^{2}\left(\mathbb{R}^{N}\times\overline{B_{\frac{1}{2}}}\right),\end{aligned}\right.$

and $\alpha>0$, then for any $x\in\mathbb{R}^{N}_{+}\cap B_{2R}$ a second
order Taylor expansion yields

(2.7) $\displaystyle\left|A\right|\leq
C\left(\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&\left|z\right|^{2},&&\left|z\right|<\frac{1}{2},\\\
&\left|z\right|^{\alpha},&&\left|z\right|\geq\frac{1}{2}.\end{aligned}\right.$

Therefore, for $x\in\mathbb{R}^{N}_{+}\cap B_{2R}$, by the condition
$\alpha<2s$,

(2.8)
$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|A\right|}{\left|z\right|^{N+2s}}dz\leq
C\left(N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right).$

We conclude that

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
(2.9) $\displaystyle\leq
C\left(N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&\frac{1}{x_{N}^{2s-\alpha}},&&x\in\mathbb{R}^{N}_{+}\cap B_{2R},\\\
&\frac{1}{\left|x\right|^{N+2s}},&&x\in\mathbb{R}^{N}_{+}\cap\left\\{\left|x\right|\geq
2R\right\\}.\end{aligned}\right.$

Let us notice that by the assumption $u\in\mathcal{L}_{s}$, $u$ satisfies
(2.3) and $u=0$ in $\overline{\mathbb{R}_{-}^{N}}$, we deduce that there
exists $\beta=\beta\left(u\right)>0$ such that

(2.10)
$\displaystyle\left|u\right|\leq\beta\left(1+\left|x\right|^{2s-\delta}\right)\quad\forall
x\in\mathbb{R}^{N}.$

Since $\alpha>2s-1$, by (1.5), (2.6), (2) and (2.10),

$\displaystyle\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left|F_{1}\right|dy\right)dx$
$\displaystyle=\int_{\mathbb{R}_{+}^{N}}\left|u\right|\left(\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}\left|a\left(\frac{y}{\left|y\right|}\right)\right|dy\right)dx$
$\displaystyle\leq
D\int_{\mathbb{R}_{+}^{N}}\left|u\right|\left(\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy\right)dx$
$\displaystyle\leq
C\left(D,N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\left[\int_{\mathbb{R}^{N}_{+}\cap
B_{2R}}\frac{\left|u\right|}{x_{N}^{2s-\alpha}}dx+\int_{\mathbb{R}^{N}_{+}\cap\left\\{\left|x\right|\geq
2R\right\\}}\frac{\left|u\right|}{\left|x\right|^{N+2s}}dx\right]$
$\displaystyle\leq
C\left(D,N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\left[\left\|u\right\|_{L^{\infty}\left(B_{2R}\right)}\int_{\mathbb{R}^{N}_{+}\cap
B_{2R}}\frac{1}{x_{N}^{2s-\alpha}}dx\right.$
$\displaystyle\quad\left.+\beta\int_{\mathbb{R}^{N}_{+}\cap\left\\{\left|x\right|\geq
2R\right\\}}\frac{1+\left|x\right|^{2s-\delta}}{\left|x\right|^{N+2s}}dx\right]<+\infty.$

Next, we prove (2.5). Since

(2.11) $\displaystyle
F_{2}\left(x,y\right)=0\quad\forall\left(x,y\right)\in\left(\overline{\mathbb{R}^{N}_{-}}\cup\left\\{\left|x\right|\geq
R\right\\}\right)\times\left(\mathbb{R}^{N}\backslash\left\\{0\right\\}\right),$

we need to focus on $\left(x,y\right)\in\left(\mathbb{R}_{+}^{N}\cap
B_{R}\right)\times\left(\mathbb{R}^{N}\backslash\left\\{0\right\\}\right)$.
For any $x\in\mathbb{R}^{N}_{+}\cap B_{R}$, by (2.10),

(2.12) $\displaystyle\left|u\left(x\pm
y\right)\right|\leq\beta\left(1+\left|x\pm y\right|^{2s-\delta}\right)\leq
C\left(s,\delta,\beta,R\right)\left(1+\left|y\right|^{2s-\delta}\right)\quad\forall
y\in\mathbb{R}^{N}.$

Moreover, by (2.3), we obtain $u\in
C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap B_{2R}\right)$ and together with
(2.12), we infer that

$\displaystyle\frac{\left|u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)\right|}{\left|y\right|^{N+2s}}\leq\left\\{\begin{aligned}
&\left\|u\right\|_{C^{2}\left(B_{x_{N}}\left(x\right)\right)}\frac{1}{\left|y\right|^{N+2s-2}},&&\left|y\right|<x_{N},\\\
&C\left(s,\delta,\beta,R\right)\frac{1+\left|y\right|^{2s-\delta}}{\left|y\right|^{N+2s}},&&\left|y\right|\geq
x_{N}.\end{aligned}\right.$

Consequently, for any $x\in\mathbb{R}^{N}_{+}\cap B_{R}$, we get
$\left\|u\right\|_{C^{2}\left(B_{x_{N}}\left(x\right)\right)}\leq\left\|u\right\|_{C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap
B_{2R}\right)}$ and then

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle\leq\left\|u\right\|_{C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap
B_{2R}\right)}\int_{\left|y\right|<x_{N}}\frac{1}{\left|y\right|^{N+2s-2}}dy+C\left(s,\delta,\beta,R\right)\int_{\left|y\right|\geq
x_{N}}\frac{1+\left|y\right|^{2s-\delta}}{\left|y\right|^{N+2s}}dy$
$\displaystyle\leq
C\left(s,\delta,\beta,R,\left\|u\right\|_{C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap
B_{2R}\right)}\right)\left(x_{N}^{2-2s}+\frac{1}{x_{N}^{2s}}+\frac{1}{x_{N}^{\delta}}\right)$
(2.13) $\displaystyle\leq
C\left(s,\delta,\beta,R,\left\|u\right\|_{C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap
B_{2R}\right)}\right)\frac{1}{x_{N}^{2s}}.$

Reasoning that $\alpha>2s-1$, by (1.5), (2.11) and (2),

$\displaystyle\int_{\mathbb{R}^{N}}\left(\int_{\mathbb{R}^{N}}\left|F_{2}\right|dy\right)dx$
$\displaystyle=\int_{\mathbb{R}^{N}_{+}\cap
B_{R}}x_{N}^{\alpha}\left|\psi\right|\left(\int_{\mathbb{R}^{N}}\frac{\left|u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)\right|}{\left|y\right|^{N+2s}}\left|a\left(\frac{y}{\left|y\right|}\right)\right|dy\right)dx$
$\displaystyle\leq
D\left\|\psi\right\|_{L^{\infty}\left(\mathbb{R}^{N}\right)}\int_{\mathbb{R}^{N}_{+}\cap
B_{R}}x_{N}^{\alpha}\left(\int_{\mathbb{R}^{N}}\frac{\left|u\left(x+y\right)+u\left(x-y\right)-2u\left(x\right)\right|}{\left|y\right|^{N+2s}}dy\right)dx$
$\displaystyle\leq
C\left(D,s,\delta,\beta,R,\left\|\psi\right\|_{L^{\infty}\left(\mathbb{R}^{N}\right)},\left\|u\right\|_{C^{2}\left(\overline{\mathbb{R}^{N}_{+}}\cap
B_{2R}\right)}\right)\int_{\mathbb{R}^{N}_{+}\cap
B_{R}}\frac{1}{x_{N}^{2s-\alpha}}dx<+\infty.$

∎

######  Remark 2.3.

For the sake of completeness we point out that the assumption $\alpha<2s$ in
Proposition 2.2 can be in fact removed. Such condition has been used in
(2.7)-(2.8). On the other hand, since the function $\psi$ is compactly
supported in the ball $B_{R}$, a more precise estimate of (2.7) can be
obtained. Indeed, since for any $x\in\mathbb{R}^{N}_{+}\cap B_{2R}$ and
$|z|\geq\frac{3R}{x_{N}}$ we have that $\psi(x\pm x_{N}z)=0$, then

$\displaystyle\left|A\right|\leq
C\left(\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&\left|z\right|^{2},&&\left|z\right|<\frac{1}{2},\\\
&\left|z\right|^{\alpha},&&\frac{1}{2}\leq\left|z\right|<\frac{3R}{x_{N}},\\\
&1,&&\left|z\right|\geq\frac{3R}{x_{N}},\end{aligned}\right.$

and

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|A\right|}{\left|z\right|^{N+2s}}dz\leq
C\left(N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&1+\frac{1}{x_{N}^{\alpha-2s}},&&\alpha>2s,\\\
&1+\left|\log{x_{N}}\right|,&&\alpha=2s,\\\
&1,&&\alpha<2s.\end{aligned}\right.$

Therefore for any $x\in\mathbb{R}^{N}_{+}\cap B_{2R}$,

$\displaystyle\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy$
$\displaystyle=\left(\int_{\mathbb{R}^{N}}\frac{\left|A\right|}{\left|z\right|^{N+2s}}dz\right)x_{N}^{\alpha-2s}$
$\displaystyle\leq
C\left(N,s,\alpha,R,\left\|\psi\right\|_{C^{2}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&1,&&\alpha>2s,\\\ &1+\left|\log{x_{N}}\right|,&&\alpha=2s,\\\
&\frac{1}{x_{N}^{2s-\alpha}},&&\alpha<2s\end{aligned}\right.$

and as a result

$\displaystyle\int_{\mathbb{R}^{N}_{+}\cap
B_{2R}}\left(\int_{\mathbb{R}^{N}}\left|F_{1}\right|dy\right)dx$
$\displaystyle\leq
D\left\|u\right\|_{L^{\infty}\left(B_{2R}\right)}\int_{\mathbb{R}^{N}_{+}\cap
B_{2R}}\left(\int_{\mathbb{R}^{N}}\frac{\left|v_{\alpha}\left(x+y\right)+v_{\alpha}\left(x-y\right)-2v_{\alpha}\left(x\right)\right|}{\left|y\right|^{N+2s}}dy\right)dx<+\infty.$

We are now concerned with the computation of the operator (1.1) acting on
barrier type functions. The result of the next Lemma is partially known, in
particular the fact that for $\alpha=s$, $\left(x_{N}\right)_{+}^{\alpha}$ is
harmonic see e.g. [12, 33]. But a complete result for $\alpha\in(0,2s)$
doesn’t seem to be available, so we decided to put here the statement and a
short proof.

######  Lemma 2.4.

Let $0<\alpha<2s$ and
$w_{\alpha}\left(x\right)=\left(x_{N}\right)_{+}^{\alpha}$. For any
$x\in\mathbb{R}^{N}_{+}$,

$\displaystyle Lw_{\alpha}\left(x\right)=C_{\alpha}x_{N}^{\alpha-2s},$

where

$\displaystyle C_{\alpha}\left\\{\begin{aligned} &<0,&&0<\alpha<s,\\\
&=0,&&\alpha=s,\\\ &>0,&&s<\alpha<2s.\end{aligned}\right.$

###### Proof.

For any $x\in\mathbb{R}^{N}_{+}$,

$\displaystyle Lw_{\alpha}\left(x\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{\left(x_{N}+y_{N}\right)_{+}^{\alpha}+\left(x_{N}-y_{N}\right)_{+}^{\alpha}-2x_{N}^{\alpha}}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=\int_{\mathbb{S}^{N-1}}\left(\int_{0}^{+\infty}\frac{\left(x_{N}+r\theta_{N}\right)_{+}^{\alpha}+\left(x_{N}-r\theta_{N}\right)_{+}^{\alpha}-2x_{N}^{\alpha}}{r^{1+2s}}dr\right)a\left(\theta\right)d\theta$
$\displaystyle=x_{N}^{\alpha}\int_{\mathbb{S}^{N-1}}\left(\int_{0}^{+\infty}\frac{\left(1+\frac{\theta_{N}}{x_{N}}r\right)_{+}^{\alpha}+\left(1-\frac{\theta_{N}}{x_{N}}r\right)_{+}^{\alpha}-2}{r^{1+2s}}dr\right)a\left(\theta\right)d\theta.$

By the change of variable $t=\frac{\left|\theta_{N}\right|}{x_{N}}r$, with
$\theta_{N}\neq 0$, we obtain

$\displaystyle\int_{0}^{+\infty}\frac{\left(1+\frac{\theta_{N}}{x_{N}}r\right)_{+}^{\alpha}+\left(1-\frac{\theta_{N}}{x_{N}}r\right)_{+}^{\alpha}-2}{r^{1+2s}}dr$
$\displaystyle=\left(\frac{\left|\theta_{N}\right|}{x_{N}}\right)^{2s}\int_{0}^{+\infty}\frac{\left(1+t\right)_{+}^{\alpha}+\left(1-t\right)_{+}^{\alpha}-2}{t^{1+2s}}dt$
$\displaystyle=:c_{\alpha}\left|\theta_{N}\right|^{2s}x_{N}^{-2s}.$

Moreover, the above equalities are still true for $\theta_{N}=0$. Thus we have

$\displaystyle Lw_{\alpha}\left(x\right)=C_{\alpha}x_{N}^{\alpha-2s}$

with

$C_{\alpha}:=c_{\alpha}\left(\int_{\mathbb{S}^{N-1}}\left|\theta_{N}\right|^{2s}a\left(\theta\right)d\theta\right).$

By the assumption (1.6) it turns out that

$\int_{\mathbb{S}^{N-1}}\left|\theta_{N}\right|^{2s}a\left(\theta\right)d\theta>0.$

Hence the sign of $C_{\alpha}$ is given by $c_{\alpha}$. For this, by the same
arguments of [12, Lemma 2.4] concerning the case $\alpha=s$ (see also [9,
Lemma 2.3]), we get that

$\displaystyle c_{\alpha}\left\\{\begin{aligned} &<0,&&0<\alpha<s,\\\
&=0,&&\alpha=s,\\\ &>0,&&s<\alpha<2s.\end{aligned}\right.$

Hence the result follows. ∎

######  Lemma 2.5.

Let $u,g,h\in C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$.

  * (1)

For any $R>0$ and $u_{R}=u\left(\frac{x}{R}\right)$,

$\displaystyle
Lu_{R}\left(x\right)=R^{-2s}Lu\left(\frac{x}{R}\right)\quad\forall
x\in\mathbb{R}_{+}^{N}.$

  * (2)

For $u=gh$,

$\displaystyle
Lu\left(x\right)=g\left(x\right)Lh\left(x\right)+h\left(x\right)Lg\left(x\right)+l\left[g,h\right]\left(x\right)\quad\forall
x\in\mathbb{R}_{+}^{N},$

where

$\displaystyle l\left[g,h\right]\left(x\right)=$
$\displaystyle\int_{\mathbb{R}^{N}}\left\\{\frac{\left[g\left(x+y\right)-g\left(x\right)\right]\left[h\left(x+y\right)-h\left(x\right)\right]}{\left|y\right|^{N+2s}}\right.$
$\displaystyle\left.+\,\frac{\left[g\left(x-y\right)-g\left(x\right)\right]\left[h\left(x-y\right)-h\left(x\right)\right]}{\left|y\right|^{N+2s}}\right\\}a\left(\frac{y}{\left|y\right|}\right)dy.$

###### Proof.

For any $R>0$,

$\displaystyle Lu_{R}\left(x\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{u\left(\frac{x+y}{R}\right)+u\left(\frac{x-y}{R}\right)-2u\left(\frac{x}{R}\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=R^{-2s}\int_{\mathbb{R}^{N}}\frac{u\left(\frac{x}{R}+y\right)+u\left(\frac{x}{R}-y\right)-2u\left(\frac{x}{R}\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle=R^{-2s}Lu\left(\frac{x}{R}\right),$

hence we get Lemma 2.5-(1). Next, for $x\in\mathbb{R}_{+}^{N}$ and any
$y\in\mathbb{R}^{N}$,

$\displaystyle g$
$\displaystyle\left(x+y\right)h\left(x+y\right)+g\left(x-y\right)h\left(x-y\right)-2g\left(x\right)h\left(x\right)$
$\displaystyle=\left[g\left(x+y\right)+g\left(x-y\right)-2g\left(x\right)\right]h\left(x\right)+\left[h\left(x+y\right)+h\left(x-y\right)-2h\left(x\right)\right]g\left(x\right)$
$\displaystyle\quad+\left[g\left(x+y\right)-g\left(x\right)\right]\left[h\left(x+y\right)-h\left(x\right)\right]+\left[g\left(x-y\right)-g\left(x\right)\right]\left[h\left(x-y\right)-h\left(x\right)\right].$

Lemma 2.5-(2) is a direct result of the above identity. ∎

We conclude this section with a continuity property of the function $l$
introduced in Lemma 2.5-(2) that will be used in Theorem 1.1.

######  Lemma 2.6.

Let $(2s-1)_{+}<\alpha<\min\left\\{1,2s\right\\}$, $\varphi\in
C^{1}_{0}\left(\mathbb{R}^{N}\right)$ and
$w_{\alpha}\left(x\right)=\left(x_{N}\right)_{+}^{\alpha}$. For any
$\left\\{x_{n}\right\\}\subset\mathbb{R}_{+}^{N}$ and $x_{n}\rightarrow
x_{0}\in\overline{\mathbb{R}_{+}^{N}}$ as $n\rightarrow+\infty$, we have

$\displaystyle\lim_{n\rightarrow+\infty}l\left[w_{\alpha},\varphi\right]\left(x_{n}\right)=l\left[w_{\alpha},\varphi\right]\left(x_{0}\right).$

###### Proof.

By Lemma 2.5-(2),

$\displaystyle l\left[w_{\alpha},\varphi\right]\left(x_{n}\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\left\\{\frac{\left[\left(\left(x_{n}\right)_{N}+y_{N}\right)_{+}^{\alpha}-\left(x_{n}\right)_{N}^{\alpha}\right]\left[\varphi\left(x_{n}+y\right)-\varphi\left(x_{n}\right)\right]}{\left|y\right|^{N+2s}}\right.$
$\displaystyle\quad\left.+\,\frac{\left[\left(\left(x_{n}\right)_{N}-y_{N}\right)_{+}^{\alpha}-\left(x_{n}\right)_{N}^{\alpha}\right]\left[\varphi\left(x_{n}-y\right)-\varphi\left(x_{n}\right)\right]}{\left|y\right|^{N+2s}}\right\\}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle:=\int_{\mathbb{R}^{N}}H_{n}\left(y\right)dy.$

Since $0<\alpha<1$, for any $n\in\mathbb{N}$ and $y\in\mathbb{R}^{N}$ we have

$\displaystyle\left|\left(\left(x_{n}\right)_{N}\pm
y_{N}\right)_{+}^{\alpha}-\left(x_{n}\right)_{N}^{\alpha}\right|\leq\left|y_{N}\right|^{\alpha}.$

Moreover,

$\displaystyle\left|\varphi\left(x_{n}\pm
y\right)-\varphi\left(x_{n}\right)\right|\leq
C\left(\left\|\varphi\right\|_{C^{1}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&|y|,&&\left|y\right|<1,\\\ &1,&&\left|y\right|\geq 1.\end{aligned}\right.$

Hence we get

$\displaystyle\left|H_{n}\left(y\right)\right|\leq
H\left(y\right)=C\left(D,\left\|\varphi\right\|_{C^{1}\left(\mathbb{R}^{N}\right)}\right)\times\left\\{\begin{aligned}
&\frac{1}{\left|y\right|^{N+2s-\alpha-1}},&&\left|y\right|<1,\\\
&\frac{1}{\left|y\right|^{N+2s-\alpha}},&&\left|y\right|\geq
1.\end{aligned}\right.$

By the assumption $2s-1<\alpha<2s$, we have $H\in
L^{1}\left(\mathbb{R}^{N}\right)$. By Lebesgue’s dominated convergence
theorem, we reach the conclusion. ∎

##  3\. Nonexistence in the half space $\mathbb{R}_{+}^{N}$

This section is devoted to the proof of Theorem 1.1. Observe that Theorem 1.1
is true if $N=1$, in this case the operators (1.1) satisfying (1.5)-(1.6)
reduce, up to a positive multiplicative constant, to the $1$-dimensional
fractional Laplacian. The proof can be done in a similar way to the case
$N\geq 2$ but every step is much simpler, details are left to the reader. So
from now on we suppose that $N\geq 2$ unless otherwise specified.

We start with some geometric considerations concerning the cones
$\Sigma_{\nu,\tau}\left(x\right)$, see (1.4) for their definition. Next lemma
is basic, we report a proof for the reader’s convenience.

######  Lemma 3.1.

Given $\nu\in{\mathbb{S}}^{N-1}$ and $\tau\in(0,1]$, then for any
$x\in\partial B_{1}$ it holds

$\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap B_{1}\right|>0.$

###### Proof.

If $x\cdot\nu\neq 0$, there exists $t_{0}\in\mathbb{R}$ and
$\varepsilon_{0}>0$ such that
$y_{0}=x+t_{0}\nu\in\Sigma_{\nu,\tau}\left(x\right)\cap B_{1}$ and
$B_{\varepsilon_{0}}\left(y_{0}\right)\subset\left(\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\right)$, hence

$\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\right|>\left|B_{\varepsilon_{0}}\left(y_{0}\right)\right|>0.$

If $x\cdot\nu=0$, then we can pick $\widetilde{\nu}\in\mathbb{S}^{N-1}$ such
that $x\cdot\widetilde{\nu}\neq 0$ and
$0<\left|\widetilde{\nu}-\nu\right|\leq\frac{\tau}{2}$. For any
$y\in\Sigma_{\widetilde{\nu},\frac{\tau}{2}}\left(x\right)$ we have

$\displaystyle\left|\left(y-x\right)\cdot\nu\right|$
$\displaystyle\geq\left|\left(y-x\right)\cdot\widetilde{\nu}\right|-\left|\left(y-x\right)\cdot\left(\widetilde{\nu}-\nu\right)\right|$
$\displaystyle\geq\left(1-\frac{\tau}{2}\right)\left|y-x\right|-\frac{\tau}{2}\left|y-x\right|=\left(1-\tau\right)\left|y-x\right|.$

Then
$\Sigma_{\widetilde{\nu},\frac{\tau}{2}}\left(x\right)\subseteq\Sigma_{\nu,\tau}\left(x\right)$
and by the first part of the proof we conclude that

$\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\right|\geq\left|\Sigma_{\widetilde{\nu},\frac{\tau}{2}}\left(x\right)\cap
B_{1}\right|>0.$

∎

######  Remark 3.2.

In the whole space $\mathbb{R}^{N}$, for any $x\in\partial B_{1}$, it is
obvious that

$\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\cap\mathbb{R}^{N}\right|=\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\right|>0$

as showed in Lemma 3.1, see also Figure 2.

Figure 2. The blue area $\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\right|>0$.

However, in the half space $\mathbb{R}_{+}^{N}$, there might be that for some
$x\in\partial\mathbb{R}_{+}^{N}\cap\partial B_{1}$

$\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\cap\mathbb{R}^{N}_{+}\right|=0,$

see Figure 3.
Our strategy is therefore the following: since, as a consequence of Lemma 3.1,

$\left|\Sigma_{\nu,\tau}\left(0\right)\cap B_{1}\left(e_{N}\right)\right|>0,$

we can then slightly move the center of the ball $B_{1}\left(e_{N}\right)$
from $e_{N}=(0,\ldots,0,1)$ to
$\left(1-\gamma\right)e_{N}=\left(0,...,0,1-\gamma\right)$, with $\gamma>0$
sufficiently small, in order to guarantee that

$\displaystyle\left|B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap{\mathbb{R}^{N}_{-}}\right|>0$

and that for any $x\in\partial\mathbb{R}_{+}^{N}\cap\partial
B_{1}\left(\left(1-\gamma\right)e_{N}\right)$

(3.1) $\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap\mathbb{R}^{N}_{+}\right|>0,$

see Figure 4. In fact, condition (3.1) is still true for any $x\in\partial
B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap\overline{\mathbb{R}^{N}_{+}}$.

Figure 3. $\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\cap\mathbb{R}^{N}_{+}\right|=0$. Figure 4. The blue area
$\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap\mathbb{R}^{N}_{+}\right|>0$.

The above considerations are summarized in the following

######  Corollary 3.3.

Given $\nu\in{\mathbb{S}}^{N-1}$ and $\tau\in(0,1]$, then there exist
$\gamma=\gamma\left(\nu,\tau\right)\in\left(0,1\right)$ such that for any
$x\in\partial
B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap\overline{\mathbb{R}^{N}_{+}}$,

$\displaystyle\left|\Sigma_{\nu,\tau}\left(x\right)\cap
B_{1}\left(\left(1-\gamma\right)e_{N}\right)\cap\mathbb{R}^{N}_{+}\right|>0.$

Next theorem is the main result from which Theorem 1.1 easily follows, as
proved at the end of this section.

######  Theorem 3.4.

Assume $1\leq p\leq\frac{N+s}{N-s}$. If $u\in
C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}$ is a nonnegative
solution of (2.2) satisfying (2.3), then $u\equiv 0$.

###### Proof.

For the $\nu_{0},\tau_{0}$ given in (1.6), by Corollary 3.3, there exists
$0<\gamma_{0}<1$ such that for any $x\in\partial
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)\cap\overline{\mathbb{R}^{N}_{+}}$,

(3.2) $\displaystyle\left|\Sigma_{\nu_{0},\tau_{0}}\left(x\right)\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)\cap\mathbb{R}^{N}_{+}\right|>0.$

We choose $\varphi\in C^{\infty}_{0}\left(\mathbb{R}^{N}\right)$ such that
$0<\varphi\leq 1$ in $B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$ and

$\displaystyle\varphi\left(x\right)=\left\\{\begin{aligned} &1,&&x\in
B_{1-\frac{\gamma_{0}}{2}}\left(\left(1-\gamma_{0}\right)e_{N}\right),\\\
&0,&&x\notin
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right),\end{aligned}\right.$

see Figure 5.

Figure 5. The graph of $\varphi$.

For any $s\leq\alpha<\min\left\\{1,2s\right\\}$, we define

$\displaystyle\phi_{\alpha}\left(x\right):=w_{\alpha}(x)\varphi\left(x\right):=\left(x_{N}\right)_{+}^{\alpha}\varphi\left(x\right).$

Step 1. For any given $\alpha_{0}\in\left(s,\min\left\\{1,2s\right\\}\right)$,
we show that there exists $M>0$ such that

(3.3)
$\displaystyle-L\phi_{\alpha_{0}}\left(x\right)-L\phi_{s}\left(x\right)\leq
M\phi_{s}\left(x\right)\quad\forall x\in\mathbb{R}_{+}^{N}.$

For this we first note that for any
$x\in\mathbb{R}_{+}^{N}\cap\left\\{\left|x-\left(1-\gamma_{0}\right)e_{N}\right|\geq
1\right\\}$, by the assumption (1.5) we infer that

$\displaystyle-L\phi_{\alpha_{0}}\left(x\right)-L\phi_{s}\left(x\right)=$
$\displaystyle-\int_{\mathbb{R}^{N}}\frac{\phi_{\alpha_{0}}\left(x+y\right)+\phi_{\alpha_{0}}\left(x-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle-\int_{\mathbb{R}^{N}}\frac{\phi_{s}\left(x+y\right)+\phi_{s}\left(x-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy\leq
0=M\phi_{s}\left(x\right).$

Hence we only need to show that

(3.4) $\displaystyle\inf_{x\in\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)}\frac{L\phi_{\alpha_{0}}\left(x\right)+L\phi_{s}\left(x\right)}{\phi_{s}\left(x\right)}>-\infty.$

Assuming that (3.4) is not true, then there exists a convergent sequence
$\left\\{x_{n}\right\\}_{n}\subset\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$ such that

(3.5)
$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)}{\phi_{s}\left(x_{n}\right)}=-\infty.$

Let $x_{n}\rightarrow x_{\infty}\in\overline{\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)}$ as $n\rightarrow+\infty$.
We distinguish different cases, each of them producing a contradiction to
(3.5).

Case 1: $x_{\infty}\in\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$. We obtain

$\displaystyle\lim_{n\rightarrow+\infty}L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)=L\phi_{\alpha_{0}}\left(x_{\infty}\right)+L\phi_{s}\left(x_{\infty}\right)$

and

$\displaystyle
0<\lim_{n\rightarrow+\infty}\phi_{s}\left(x_{n}\right)=\phi_{s}\left(x_{\infty}\right)<+\infty,$

then

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)}{\phi_{s}\left(x_{n}\right)}=\frac{L\phi_{\alpha_{0}}\left(x_{\infty}\right)+L\phi_{s}\left(x_{\infty}\right)}{\phi_{s}\left(x_{\infty}\right)}.$

Case 2: $x_{\infty}\in\mathbb{R}_{+}^{N}\cap\partial
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$. Recalling that $a\geq 0$,
by (1.6) and (3.2) we have

$\displaystyle\lim_{n\rightarrow+\infty}L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{\phi_{\alpha_{0}}\left(x_{\infty}+y\right)+\phi_{\alpha_{0}}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\quad+\int_{\mathbb{R}^{N}}\frac{\phi_{s}\left(x_{\infty}+y\right)+\phi_{s}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\geq
d\int_{\Sigma_{\nu_{0},\tau_{0}}\left(0\right)}\frac{\phi_{\alpha_{0}}\left(x_{\infty}+y\right)+\phi_{\alpha_{0}}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}dy$
$\displaystyle\quad+d\int_{\Sigma_{\nu_{0},\tau_{0}}\left(0\right)}\frac{\phi_{s}\left(x_{\infty}+y\right)+\phi_{s}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}dy>0.$

Moreover, $\phi_{s}\left(x_{n}\right)>0$ for any $n$ and

$\displaystyle\lim_{n\rightarrow+\infty}\phi_{s}\left(x_{n}\right)=0,$

thus

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)}{\phi_{s}\left(x_{n}\right)}=+\infty.$

Case 3: $x_{\infty}\in\partial\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$. By Lemmas 2.4 and 2.5-(2),
for any $x\in\mathbb{R}^{N}_{+}$

(3.6)
$\begin{split}L\phi_{\alpha_{0}}\left(x\right)+L\phi_{s}\left(x\right)&=C_{\alpha_{0}}\varphi\left(x\right)x^{\alpha_{0}-2s}_{N}+x_{N}^{\alpha_{0}}L\varphi\left(x\right)+x_{N}^{s}L\varphi\left(x\right)\\\
&\quad+l\left[w_{\alpha_{0}},\varphi\right]\left(x\right)+l\left[w_{s},\varphi\right]\left(x\right),\end{split}$

where $C_{\alpha_{0}}>0$ and

$\displaystyle
l\left[w_{\alpha_{0}},\varphi\right]\left(x\right)+l\left[w_{s},\varphi\right]\left(x\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\left\\{\frac{\left[\varphi\left(x+y\right)-\varphi\left(x\right)\right]\left[\left(x_{N}+y_{N}\right)_{+}^{\alpha_{0}}+\left(x_{N}+y_{N}\right)_{+}^{s}-x_{N}^{\alpha_{0}}-x_{N}^{s}\right]}{\left|y\right|^{N+2s}}\right.dy$
$\displaystyle\quad\left.+\frac{\left[\varphi\left(x-y\right)-\varphi\left(x\right)\right]\left[\left(x_{N}-y_{N}\right)_{+}^{\alpha_{0}}+\left(x_{N}-y_{N}\right)_{+}^{s}-x_{N}^{\alpha_{0}}-x_{N}^{s}\right]}{\left|y\right|^{N+2s}}\right\\}a\left(\frac{y}{\left|y\right|}\right)dy.$

We see that

$\displaystyle\lim_{n\rightarrow+\infty}C_{\alpha_{0}}\varphi\left(x_{n}\right)\cdot\left(x_{n}\right)_{N}^{\alpha_{0}-2s}=+\infty$

and

$\displaystyle\lim_{n\rightarrow+\infty}\left(x_{n}\right)_{N}^{\alpha_{0}}L\varphi\left(x_{n}\right)+\left(x_{n}\right)_{N}^{s}L\varphi\left(x_{n}\right)=0.$

Moreover Lemma 2.6 yields

$\displaystyle\lim_{n\rightarrow+\infty}l\left[w_{\alpha_{0}},\varphi\right]\left(x_{n}\right)+l\left[w_{s},\varphi\right]\left(x_{n}\right)=l\left[w_{\alpha_{0}},\varphi\right]\left(x_{\infty}\right)+l\left[w_{s},\varphi\right]\left(x_{\infty}\right),$

hence by (3.6),

$\displaystyle\lim_{n\rightarrow+\infty}L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)=+\infty.$

By the fact

$\displaystyle\lim_{n\rightarrow+\infty}\phi_{s}\left(x_{n}\right)=0,$

and $\phi_{s}\left(x_{n}\right)>0$ for any $n$, we get

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)}{\phi_{s}\left(x_{n}\right)}=+\infty.$

Case 4: $x_{\infty}\in\partial\mathbb{R}_{+}^{N}\cap\partial
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)$. Notice that by (3.6), since
$\varphi(x_{n})>0$ for any $n$,

(3.7)
$\begin{split}L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)&\geq\left(x_{n}\right)_{N}^{\alpha_{0}}L\varphi\left(x_{n}\right)+\left(x_{n}\right)_{N}^{s}L\varphi\left(x_{n}\right)\\\
&\quad+l\left[w_{\alpha_{0}},\varphi\right]\left(x_{n}\right)+l\left[w_{s},\varphi\right]\left(x_{n}\right).\end{split}$

It is obvious that

(3.8)
$\displaystyle\lim_{n\rightarrow+\infty}\left(x_{n}\right)_{N}^{\alpha_{0}}L\varphi\left(x_{n}\right)+\left(x_{n}\right)_{N}^{s}L\varphi\left(x_{n}\right)=0.$

Since $a\geq 0$, by (1.6), Lemma 2.6 and (3.2) we have

$\displaystyle\lim_{n\rightarrow+\infty}l\left[w_{\alpha_{0}},\varphi\right]\left(x_{n}\right)+l\left[w_{s},\varphi\right]\left(x_{n}\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{\phi_{\alpha_{0}}\left(x_{\infty}+y\right)+\phi_{\alpha_{0}}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\quad+\int_{\mathbb{R}^{N}}\frac{\phi_{s}\left(x_{\infty}+y\right)+\phi_{s}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\geq
d\int_{\Sigma_{\nu_{0},\tau_{0}}\left(0\right)}\frac{\phi_{\alpha_{0}}\left(x_{\infty}+y\right)+\phi_{\alpha_{0}}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}dy$
(3.9)
$\displaystyle\quad+d\int_{\Sigma_{\nu_{0},\tau_{0}}\left(0\right)}\frac{\phi_{s}\left(x_{\infty}+y\right)+\phi_{s}\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}dy>0.$

Combining (3.7), (3.8) and (3) with the fact

$\displaystyle\lim_{n\rightarrow+\infty}\phi_{s}\left(x_{n}\right)=0$

and $\phi_{s}\left(x_{n}\right)>0$ for any $n$, even in this case we get :

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\phi_{\alpha_{0}}\left(x_{n}\right)+L\phi_{s}\left(x_{n}\right)}{\phi_{s}\left(x_{n}\right)}=+\infty.$

In conclusion, we have reached a contradiction to (3.5) for any
$x_{\infty}\in\overline{\mathbb{R}_{+}^{N}\cap
B_{1}\left(\left(1-\gamma_{0}\right)e_{N}\right)}$, then inequality (3.3)
holds.

Step 2. For any $R>0$, we make the following rescaling

$\displaystyle\varphi_{R}\left(x\right)=\varphi\left(\frac{x}{R}\right)=\left\\{\begin{aligned}
&1,&&x\in
B_{R\left(1-\frac{\gamma_{0}}{2}\right)}\left(R\left(1-\gamma_{0}\right)e_{N}\right),\\\
&0,&&x\notin
B_{R}\left(R\left(1-\gamma_{0}\right)e_{N}\right).\end{aligned}\right.$

For any $s\leq\alpha<\min\left\\{1,2s\right\\}$, we define

$\displaystyle\phi_{\alpha,R}\left(x\right)=\left(x_{N}\right)_{+}^{\alpha}\varphi_{R}\left(x\right).$

Since $u$ is solution of (2.2), by Proposition 2.2 and Lemma 2.5-(1) we obtain

$\displaystyle\int_{\mathbb{R}_{+}^{N}}u^{p}\phi_{s,R}dx$
$\displaystyle\leq-R^{s}\int_{\mathbb{R}^{N}}\left(\frac{x_{N}}{R}\right)_{+}^{s}\varphi_{R}Ludx$
$\displaystyle\leq-R^{s}\int_{\mathbb{R}^{N}}\left(\frac{x_{N}}{R}\right)_{+}^{s}\varphi_{R}Ludx-R^{s}\int_{\mathbb{R}^{N}}\left(\frac{x_{N}}{R}\right)_{+}^{\alpha_{0}}\varphi_{R}Ludx$
$\displaystyle=-R^{-s}\int_{\mathbb{R}^{N}}uL\phi_{s}\left(\frac{x}{R}\right)dx-R^{-s}\int_{\mathbb{R}^{N}}uL\phi_{\alpha_{0}}\left(\frac{x}{R}\right)dx$
$\displaystyle=R^{-s}\int_{\mathbb{R}_{+}^{N}}u\left[-L\phi_{\alpha_{0}}\left(\frac{x}{R}\right)-L\phi_{s}\left(\frac{x}{R}\right)\right]dx.$

Now we use (3.3) to infer that

$\displaystyle\int_{\mathbb{R}_{+}^{N}}u\left[-L\phi_{\alpha_{0}}\left(\frac{x}{R}\right)-L\phi_{s}\left(\frac{x}{R}\right)\right]dx\leq
M\int_{\mathbb{R}_{+}^{N}}u\phi_{s}\left(\frac{x}{R}\right)dx=MR^{-s}\int_{\mathbb{R}_{+}^{N}}u\phi_{s,R}dx$

and that

(3.10) $\displaystyle\int_{\mathbb{R}_{+}^{N}}u^{p}\phi_{s,R}dx\leq
MR^{-2s}\int_{\mathbb{R}_{+}^{N}}u\phi_{s,R}dx.$

Since $0\leq\phi_{s,R}\leq(x_{N})_{+}^{s}$ and
$B_{R}\left(R\left(1-\gamma_{0}\right)e_{N}\right)\subset B_{2R}$, we may
apply the Hölder inequality to get, for any $p>1$,

$\displaystyle\int_{\mathbb{R}_{+}^{N}}u\phi_{s,R}dx$
$\displaystyle\leq\left(\int_{\mathbb{R}_{+}^{N}\cap
B_{R}\left(R\left(1-\gamma_{0}\right)e_{N}\right)}u^{p}\phi_{s,R}dx\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}_{+}^{N}\cap
B_{R}\left(R\left(1-\gamma_{0}\right)e_{N}\right)}\phi_{s,R}dx\right)^{\frac{p-1}{p}}$
$\displaystyle\leq
C\left(N,p\right)R^{\frac{\left(N+s\right)\left(p-1\right)}{p}}\left(\int_{\mathbb{R}_{+}^{N}}u^{p}\phi_{s,R}dx\right)^{\frac{1}{p}}.$

Therefore, by (3.10),

(3.11) $\displaystyle\int_{\mathbb{R}_{+}^{N}}u^{p}\phi_{s,R}dx\leq
C\left(N,M,p\right)R^{N+s-\frac{2sp}{p-1}}.$

Step 3. In the case $p=1$, we immediately obtain $u\equiv 0$ by letting
$R\rightarrow+\infty$ in (3.10).
We now consider $p>1$. Since $\varphi_{R}\rightarrow\varphi_{\infty}\equiv 1$
in $\mathbb{R}^{N}_{+}$ and $\phi_{s,R}\rightarrow\phi_{s,\infty}=x_{N}^{s}$
as $R\rightarrow+\infty$, if we let $R\rightarrow+\infty$ in (3.11), we have
$u\equiv 0$ provided $p<\frac{N+s}{N-s}$.
If instead $p=\frac{N+s}{N-s}$, then we infer that

(3.12) $\displaystyle\int_{\mathbb{R}_{+}^{N}}x^{s}_{N}u^{p}dx<+\infty.$

We can rewrite

$\displaystyle\int_{\mathbb{R}_{+}^{N}}u\phi_{s,R}dx=\int_{\mathbb{R}_{+}^{N}\cap
B_{\sqrt{R}}}u\phi_{s,R}dx+\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}u\phi_{s,R}dx.$

Using once again the fact $0\leq\phi_{s,R}\leq(x_{N})_{+}^{s}$, by the Hölder
inequality we obtain

$\displaystyle\int_{\mathbb{R}_{+}^{N}\cap B_{\sqrt{R}}}u\phi_{s,R}dx$
$\displaystyle\leq\left(\int_{\mathbb{R}_{+}^{N}\cap
B_{\sqrt{R}}}x_{N}^{s}u^{p}dx\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}_{+}^{N}\cap
B_{\sqrt{R}}}x_{N}^{s}dx\right)^{\frac{p-1}{p}}$ $\displaystyle\leq
C\left(N,p\right)R^{\frac{\left(N+s\right)\left(p-1\right)}{2p}}\left(\int_{\mathbb{R}_{+}^{N}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}$
$\displaystyle=C\left(N,p\right)R^{s}\left(\int_{\mathbb{R}_{+}^{N}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}$

and

$\displaystyle\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}u\phi_{s,R}dx$
$\displaystyle\leq\left(\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}x_{N}^{s}u^{p}dx\right)^{\frac{1}{p}}\left(\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}x_{N}^{s}dx\right)^{\frac{p-1}{p}}$ $\displaystyle\leq
C\left(N,p\right)R^{\frac{\left(N+s\right)\left(p-1\right)}{p}}\left(\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}$
$\displaystyle=C\left(N,p\right)R^{2s}\left(\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}.$

By (3.10), we get

$\displaystyle\int_{\mathbb{R}_{+}^{N}}u^{p}\phi_{s,R}dx\leq
C\left(N,M,p\right)\left[R^{-s}\left(\int_{\mathbb{R}_{+}^{N}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}+\left(\int_{\mathbb{R}_{+}^{N}\cap\left\\{\sqrt{R}\leq\left|x\right|\leq
2R\right\\}}x^{s}_{N}u^{p}dx\right)^{\frac{1}{p}}\right].$

Therefore, by (3.12) the right-hand side of the above inequality goes to $0$
as $R\rightarrow+\infty$ and we again conclude that $u\equiv 0$. ∎

Now we are ready to prove Theorem 1.1.

###### Proof of Theorem 1.1.

Let $u$ be any nonnegative classical solution of (1.2). By Lemma 2.1, the
function $\widetilde{u}$, defined in (2.1), is a nonnegative classical
solution of (2.2). Theorem 3.4 yields $\widetilde{u}\equiv 0$ in
$\mathbb{R}^{N}$, hence

$\displaystyle u\left\\{\begin{aligned} &\geq 0,&&x_{N}<1,\\\ &=0,&&x_{N}\geq
1.\end{aligned}\right.$

For any $\overline{x}$ with $\overline{x}_{N}=1$, we have

$\displaystyle 0=u^{p}\left(\overline{x}\right)\leq-
Lu\left(\overline{x}\right)=-\int_{\mathbb{R}^{N}}\frac{u\left(\overline{x}+y\right)+u\left(\overline{x}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy\leq
0$

and $Lu\left(\overline{x}\right)=0$. Since $u\in
C^{2}\left(\mathbb{R}_{+}^{N}\right)$, then $u\equiv 0$ in
$\Sigma_{\nu_{0},\tau_{0}}\left(\overline{x}\right)\cap\mathbb{R}^{N}_{+}$.
The arbitrariness of $\overline{x}$ implies $u\equiv 0$ in
$\mathbb{R}^{N}_{+}$ and moreover $u=0$ a.e. in
$\overline{\mathbb{R}_{-}^{N}}$, see Figure 6.

Figure 6.

∎

##  4\. Optimality of the critical exponent $\frac{N+s}{N-s}$

We prove that the exponent $p=\frac{N+s}{N-s}$ in Theorem 1.1 is sharp, in the
sense that there is at least some operator in the class (1.1) for which
problem (1.2) has nontrivial solutions as soon as $p>\frac{N+s}{N-s}$. In
fact, this occurs whenever $a(\theta)$ is any positive constant function, in
which case $-L$ coincides, up to a multiplicative constant, with
$(-\Delta)^{s}$.

######  Theorem 4.1.

For any $p>\frac{N+s}{N-s}$ and $N\geq 2$, the problem

(4.1) $\displaystyle\left\\{\begin{aligned} \left(-\Delta\right)^{s}u&\geq
u^{p},&&x\in\mathbb{R}_{+}^{N},\\\
u&=0,&&x\in\mathbb{R}_{-}^{N}\end{aligned}\right.$

admits positive classical solutions.

###### Proof.

Let

$\displaystyle w_{\alpha}\left(x\right)=\left(x_{N}\right)^{\alpha}_{+},\quad
0<\alpha<s.$

Consider the Kelvin transform of $w_{\alpha}$, given by

$\displaystyle\overline{w}_{\alpha}\left(x\right):=\frac{1}{\left|x\right|^{N-2s}}w_{\alpha}\left(\frac{x}{\left|x\right|^{2}}\right)=\frac{\left(x_{N}\right)^{\alpha}_{+}}{\left|x\right|^{N-2s+2\alpha}}\in
C^{2}\left(\mathbb{R}_{+}^{N}\right)\cap\mathcal{L}_{s}.$

By the general property of the Kelvin transformation

$\left(-\Delta\right)^{s}\overline{w}_{\alpha}(x)=\frac{1}{\left|x\right|^{N+2s}}\left(-\Delta\right)^{s}w_{\alpha}\left(\frac{x}{\left|x\right|^{2}}\right),$

see e.g. [31, Proposition A.1], from Lemma 2.4 we infer that for any
$x\in\mathbb{R}^{N}_{+}$,

$-\left(-\Delta\right)^{s}\overline{w}_{\alpha}\left(x\right)=C_{\alpha}\frac{1}{\left|x\right|^{N+2s}}\left(\frac{x_{N}}{\left|x\right|^{2}}\right)^{\alpha-2s}=C_{\alpha}\frac{x_{N}^{\alpha-2s}}{\left|x\right|^{N-2s+2\alpha}},$

where $C_{\alpha}<0$.

For $0<\varepsilon\leq\left(-C_{\alpha}\right)^{\frac{1}{p-1}}$, we define the
function

$\displaystyle
u\left(x\right)=\varepsilon\overline{w}_{\alpha}\left(x\right).$

For any $x\in\mathbb{R}_{+}^{N}$, then we have

$\displaystyle-\left(-\Delta\right)^{s}u\left(x\right)+u^{p}\left(x\right)$
$\displaystyle=\varepsilon
C_{\alpha}\frac{x_{N}^{\alpha-2s}}{\left|x\right|^{N-2s+2\alpha}}+\varepsilon^{p}\frac{x_{N}^{\alpha
p}}{\left|x\right|^{\left(N-2s+2\alpha\right)p}}$ (4.2)
$\displaystyle=\varepsilon\frac{x_{N}^{\alpha-2s}}{\left|x\right|^{N-2s+2\alpha}}\left(C_{\alpha}+\varepsilon^{p-1}\frac{x_{N}^{\alpha
p-\alpha+2s}}{\left|x\right|^{\left(N-2s+2\alpha\right)\left(p-1\right)}}\right).$

When $\frac{N+s}{N-s}<p<\frac{N}{N-2s}$, we can choose
$\alpha\in\left(0,s\right)$ in such a way

$\displaystyle\alpha p-\alpha+2s=\left(N-2s+2\alpha\right)\left(p-1\right),$

i.e. $\alpha=\frac{N-\left(N-2s\right)p}{p-1}$. Then, from (4), for any
$x\in\mathbb{R}_{+}^{N}$ we have

(4.3)
$\displaystyle-\left(-\Delta\right)^{s}u\left(x\right)+u^{p}\left(x\right)\leq\varepsilon\frac{x_{N}^{\alpha-2s}}{\left|x\right|^{N-2s+2\alpha}}\left(C_{\alpha}+\varepsilon^{p-1}\right)\leq
0.$

Thus $u$ is a classical solution of (4.1) when
$\frac{N+s}{N-s}<p<\frac{N}{N-2s}$.

If $p\geq\frac{N}{N-2s}$, then for any $\alpha\in\left(0,s\right)$ one has

$\displaystyle\alpha p-\alpha+2s<\left(N-2s+2\alpha\right)\left(p-1\right).$

In view (4), $u$ satisfies (4.3) for any $x_{N}\geq 1$. As a consequence, the
function

$\displaystyle\widetilde{u}\left(x\right)=\left\\{\begin{aligned}
&u\left(x+e_{N}\right),&&x\in\mathbb{R}_{+}^{N},\\\
&0,&&x\in\overline{\mathbb{R}_{-}^{N}}\end{aligned}\right.$

is in turn a solution of (4.1). ∎

######  Remark 4.2.

The existence result of Theorem 4.1 remains valid even if $N=1$. We briefly
discuss this aspect by distinguishing the cases $0<s<\frac{1}{2}$ and
$\frac{1}{2}\leq s<1$.
If $0<s<\frac{1}{2}$, using the fact that the function $|x|^{-(1-2s)}$ is
$s$-harmonic in $\mathbb{R}\backslash\left\\{0\right\\}$, one can argue as in
the proof of Theorem 4.1 with $N$ replaced by $1$.
If instead $\frac{1}{2}\leq s<1$, the function $|x|^{2s-1}$ is still
$s$-harmonic in $\mathbb{R}\backslash\left\\{0\right\\}$ (if $s=\frac{1}{2}$
we simply mean the constant $1$) but unbounded at infinity if $s>\frac{1}{2}$.
The equivalent formulation of (4) for the function

$u(x)=\varepsilon\frac{(x)_{+}^{\alpha}}{|x|^{2\alpha-2s+1}}$

is

$\displaystyle-\left(-\Delta\right)^{s}u\left(x\right)+u^{p}\left(x\right)$
$\displaystyle=\frac{\varepsilon
C_{\alpha}}{x^{1+\alpha}}+\frac{\varepsilon^{p}}{x^{\left(1+\alpha-2s\right)p}}$
$\displaystyle=\frac{\varepsilon}{x^{1+\alpha}}\left(C_{\alpha}+\frac{\varepsilon^{p-1}}{x^{\left(1+\alpha-2s\right)p-1-\alpha}}\right)\qquad\forall
x\in\mathbb{R}_{+},$

where $C_{\alpha}<0$. In this case we can choose
$\alpha=\frac{1+(2s-1)p}{p-1}$ is such a way

$(1+\alpha-2s)p=\alpha+1$

and, differently from the previous cases $N=1$ and $0<s<\frac{1}{2}$ or $N\geq
2$, the exponent $\alpha$ is now always positive since, for any
$p>\frac{1+s}{1-s}$, it turns out that
$\frac{1+\left(2s-1\right)p}{p-1}\in\left(2s-1,s\right)$. Hence

$-\left(-\Delta\right)^{s}u\left(x\right)+u^{p}\left(x\right)\leq
0\quad\forall x\in\mathbb{R}_{+}.$

##  5\. Nonexistence in the whole space $\mathbb{R}^{N}$

We prove in this section that below the nonlocal Serrin exponent, the only
entire non negative supersolution in $\mathbb{R}^{N}$ is the trivial one. The
proof is similar to the case of the half space but simpler and it is given for
completeness and clarity sake.

######  Theorem 5.1.

Let $N\geq 2$ and $1\leq p\leq\frac{N}{N-2s}$. If $u\in
C^{2}\left(\mathbb{R}^{N}\right)\cap\mathcal{L}_{s}$ is a nonnegative solution
of

(5.1) $\displaystyle-Lu\geq u^{p}\quad\text{in $\mathbb{R}^{N}$},$

then $u\equiv 0$.

###### Proof.

We choose $\varphi\in C^{\infty}_{0}\left(\mathbb{R}^{N}\right)$ such that
$0<\varphi\leq 1$ in $B_{2}$ and

$\displaystyle\varphi\left(x\right)=\left\\{\begin{aligned} &1,&&x\in
B_{1},\\\ &0,&&x\notin B_{2}.\end{aligned}\right.$

We claim that there exists $M>0$ such that

(5.2) $\displaystyle-L\varphi\left(x\right)\leq
M\varphi\left(x\right)\quad\forall x\in\mathbb{R}^{N}.$

By the assumption (1.5), for any $\left|x\right|\geq 2$

$\displaystyle-L\varphi\left(x\right)=-\int_{\mathbb{R}^{N}}\frac{\varphi\left(x+y\right)+\varphi\left(x-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy\leq
0=M\varphi\left(x\right).$

Hence (5.2) is equivalent to

(5.3) $\displaystyle\inf_{x\in
B_{2}}\frac{L\varphi\left(x\right)}{\varphi\left(x\right)}>-\infty.$

Assume (5.3) does not hold, then there exists a convergent sequence
$\left\\{x_{n}\right\\}_{n}\subset B_{2}$ such that

(5.4)
$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\varphi\left(x_{n}\right)}{\varphi\left(x_{n}\right)}=-\infty.$

Let $x_{n}\rightarrow x_{\infty}\in\overline{B_{2}}$ as $n\rightarrow+\infty$.
We distinguish two cases.

Case 1: $\left|x_{\infty}\right|<2$. In this case

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\varphi\left(x_{n}\right)}{\varphi\left(x_{n}\right)}=\frac{L\varphi\left(x_{\infty}\right)}{\varphi\left(x_{\infty}\right)}$

which is a finite quantity since $\varphi\in
C^{\infty}_{0}\left(\mathbb{R}^{N}\right)$ and
$\varphi\left(x_{\infty}\right)>0$.

Case 2: $\left|x_{\infty}\right|=2$. By the assumptions (1.5)-(1.6) and using
Lemma 3.1, we have

$\displaystyle\lim_{n\rightarrow+\infty}L\varphi\left(x_{n}\right)$
$\displaystyle=\int_{\mathbb{R}^{N}}\frac{\varphi\left(x_{\infty}+y\right)+\varphi\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}a\left(\frac{y}{\left|y\right|}\right)dy$
$\displaystyle\geq
d\int_{\Sigma_{\nu_{0},\tau_{0}}(0)}\frac{\varphi\left(x_{\infty}+y\right)+\varphi\left(x_{\infty}-y\right)}{\left|y\right|^{N+2s}}dy>0.$

Thus

$\displaystyle\lim_{n\rightarrow+\infty}\frac{L\varphi\left(x_{n}\right)}{\varphi\left(x_{n}\right)}=+\infty.$

Therefore, the assumption (5.4) can not occur, so that (5.2) holds.

For any $R>0$, we consider the rescaled test-function

$\displaystyle\varphi_{R}\left(x\right)=\varphi\left(\frac{x}{R}\right).$

Multiplying (5.1) by $\varphi_{R}$, integrating by parts, then using Lemma
2.5-(1) and (5.2) we have

(5.5)
$\displaystyle\int_{\mathbb{R}^{N}}u^{p}\varphi_{R}dx\leq-\int_{\mathbb{R}^{N}}uL\varphi_{R}dx\leq
MR^{-2s}\int_{\mathbb{R}^{N}}u\varphi_{R}dx.$

By the Hölder inequality,

$\displaystyle\int_{\mathbb{R}^{N}}u\varphi_{R}dx$
$\displaystyle\leq\left(\int_{B_{2R}}u^{p}\varphi_{R}dx\right)^{\frac{1}{p}}\left(\int_{B_{2R}}\varphi_{R}dx\right)^{\frac{p-1}{p}}$
$\displaystyle\leq
C\left(N,p\right)R^{\frac{N\left(p-1\right)}{p}}\left(\int_{\mathbb{R}^{N}}u^{p}\varphi_{R}dx\right)^{\frac{1}{p}}.$

Thus, by (5.5), we obtain that for any $p>1$

(5.6) $\displaystyle\int_{\mathbb{R}^{N}}u^{p}\varphi_{R}dx\leq
C\left(N,M,p\right)R^{N-\frac{2sp}{p-1}}.$

In the case $p=1$, we let $R\rightarrow+\infty$ in (5.5) to get $u\equiv 0$.
When $1<p<\frac{N}{N-2s}$, letting $R\rightarrow+\infty$ in (5.6), we infer
that $u\equiv 0$. If $p=\frac{N}{N-2s}$, then (5.6) yields, in the limit as
$R\rightarrow+\infty$, that

(5.7) $\displaystyle\int_{\mathbb{R}^{N}}u^{p}dx<+\infty.$

Moreover, in this case, we can rewrite

$\displaystyle\int_{\mathbb{R}^{N}}u\varphi_{R}dx=\int_{\left|x\right|\leq\sqrt{R}}u\varphi_{R}dx+\int_{\sqrt{R}\leq\left|x\right|\leq
2R}u\varphi_{R}dx.$

By the Hölder inequality

$\displaystyle\int_{\left|x\right|\leq\sqrt{R}}u\varphi_{R}dx$
$\displaystyle\leq
C\left(N,p\right)R^{s}\left(\int_{\mathbb{R}^{N}}u^{p}dx\right)^{\frac{1}{p}}$

and

$\displaystyle\int_{\sqrt{R}\leq\left|x\right|\leq 2R}u\varphi_{R}dx$
$\displaystyle\leq
C\left(N,p\right)R^{2s}\left(\int_{\sqrt{R}\leq\left|x\right|\leq
2R}u^{p}dx\right)^{\frac{1}{p}}.$

Therefore, by (5.5), it follows that

$\displaystyle\int_{\mathbb{R}^{N}}u^{p}\varphi_{R}dx\leq
C\left(N,M,p\right)\left[R^{-s}\left(\int_{\mathbb{R}^{N}}u^{p}dx\right)^{\frac{1}{p}}+\left(\int_{\sqrt{R}\leq\left|x\right|\leq
2R}u^{p}dx\right)^{\frac{1}{p}}\right].$

In view of (5.7), we again conclude that $u\equiv 0$ by letting
$R\rightarrow+\infty$. ∎

######  Remark 5.2.

If $N=1$, Theorem 5.1 reads as follows: if $s\geq\frac{1}{2}$, then $u\equiv
0$ is the only nonnegative solution of (5.1) for any $p\geq 1$; if instead
$s<\frac{1}{2}$, then the same conclusion holds provided $1\leq
p\leq\frac{1}{1-2s}$.

## Acknowledgements

I.B and G.G. are partially supported by INdAM-GNAMPA. The authors wish to
thank Sapienza University of Rome for the financial support from the research
project Ateneo 2022 “At the edge of reaction-diffusion equations: from
population dynamics to geometric analysis, a qualitative approach”,

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