# Classification of solutions to equations involving Higher-order fractional
laplacian

Zhuoran Du  School of Mathematics, Hunan University, Changsha 410082, PRC
<EMAIL_ADDRESS>,  Zhenping Feng  School of Mathematics, Hunan University,
Changsha 410082, PRC<EMAIL_ADDRESS>,  Jiaqi Hu  School of Mathematics,
Hunan University, Changsha 410082, PRC<EMAIL_ADDRESS>and  Yuan Li
School of Mathematics, Hunan University, Changsha 410082, PRC
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we are concerned with the following equation involving higher-
order fractional Lapalacian

$\left\\{\begin{aligned}
&(-\Delta)^{p+{\frac{\alpha}{2}}}u(x)=u_{+}^{\gamma}~{}~{}\mbox{ in
}\mathbb{R}^{n},\\\
&\int_{\mathbb{R}^{n}}u_{+}^{\gamma}dx<+\infty,\end{aligned}\right.$

where $p\geq 1$ is an integer, $0<\alpha<2$, $n>2p+\alpha$ and
$\gamma\in(1,\frac{n}{n-2p-\alpha})$. We establish an integral representation
formula for any nonconstant classical solution satisfying certain growth at
infinity. From this we prove that these solutions are radially symmetric about
some point in ${\mathbb{R}}^{n}$ and monotone decreasing in the radial
direction via method of moving planes in integral forms.

2020 Mathematics Subject Classification: 35B06; 35B08; 35J91.
Keywords: Higher-order fractional Lapalcian, Super poly-harmonic properties,
Moving planes in integral forms, Radial symmetry.

##  1\. Introduction

In this paper, we study the following higher-order fractional equation

(1) $\left\\{\begin{aligned}
&(-\Delta)^{p+\frac{\alpha}{2}}u(x)=u_{+}^{\gamma},\ \ \ \
x\in\mathbb{R}^{n},\\\
&\int_{\mathbb{R}^{n}}u_{+}^{\gamma}dx<+\infty,\end{aligned}\right.$

where $\gamma\in(1,\frac{n}{n-2p-\alpha})$, $n>2p+\alpha$, $0<\alpha<2$,
$p\geq 1$ is an integer, $u_{+}=\max\\{u,0\\}$ and the higher-order fractional
Laplacian is defined by
$(-\Delta)^{p+\frac{\alpha}{2}}:=(-\Delta)^{p}(-\Delta)^{\frac{\alpha}{2}}$.
It is well-known that the integral form of fractional Laplacian
$(-\Delta)^{\frac{\alpha}{2}}$ $(0<\alpha<2)$ is

$(-\Delta)^{\frac{\alpha}{2}}u(x)=C_{n,\alpha}P.V.\int_{{\mathbb{R}}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+\alpha}}dy,$

where the constant
$C_{n,\alpha}=\frac{2^{\alpha}\Gamma(\frac{n+\alpha}{2})}{-\pi^{n/2}\Gamma(-\frac{\alpha}{2})}$.

Denote

$\mathcal{L}_{\alpha}({\mathbb{R}}^{n}):=\Big{\\{}u:{\mathbb{R}}^{n}\to{\mathbb{R}}\big{|}\int_{{\mathbb{R}}^{N}}\frac{|u(x)|}{1+|x|^{n+\alpha}}dx<+\infty\Big{\\}}.$

It is known that $(-\Delta)^{\frac{\alpha}{2}}u$ is well defined for $u\in
C_{loc}^{[\alpha],\\{\alpha\\}+\varepsilon}\cap\mathcal{L}_{\alpha}({\mathbb{R}}^{n})$,
where $[\alpha]$ denotes the integer part of $\alpha$,
$\\{\alpha\\}:=\alpha-[\alpha]$ and any $\varepsilon>0$. In order to guarantee
that $(-\Delta)^{\frac{\alpha}{2}}u\in C^{2p}({\mathbb{R}}^{n})$, we have to
assume $u\in
C_{loc}^{2p+[\alpha],\\{\alpha\\}+\varepsilon}\cap\mathcal{L}_{\alpha}({\mathbb{R}}^{n})$
(see [14], [19]), and hence $u$ is a classical solution of the equation in (1)
in the sense that $(-\Delta)^{p+\frac{\alpha}{2}}u$ is point-wise well-defined
in the whole $\mathbb{R}^{n}$.

For decades, many researchers are interested in the classification of
solutions to semi-linear elliptic equations. We now recall some relevant
results. The following conformally invariant equations

(2)
$(-\Delta)^{\frac{\beta}{2}}u=u^{\frac{n+\beta}{n-\beta}}~{}\mbox{in}~{}\mathbb{R}^{n},~{}0<\beta<n$

have been extensively studied (see [2, 5, 7, 8, 9, 10, 12, 13, 18] and the
reference therein). In case $\beta=2$, (2) becomes the well-known Yamabe
equation. Cafferelli, Gidas and Spruck [2] (see also [6]) classified all
positive solutions of Yamabe equation. For general $0<\beta<n$, Chen, Li and
Ou in [8] (see also [18]) classified all positive
$L_{loc}^{\frac{2n}{n-\beta}}$ solutions to the equivalent integral equation
of (2) by the method of moving planes in integral forms. For $0<\beta<2$,
Chen, Li and Li in [7] developed a direct method of moving planes for
fractional Laplace equation and complete the classification of all nonnegative
solutions. When $\beta=3$ is an odd integer, Dai and Qin in [10] derived the
classification of nonnegative classical solutions to (2) with the assumption
$\int_{{\mathbb{R}}^{n}}\frac{u^{\frac{n+3}{n-3}}}{|x|^{n-3}}dx<+\infty$.
Recently, Cao, Dai and Qin in [3] extended the result of [10] to general case
$0<\beta<n$. Their classification results completely improved the results in
[10] without assumption on integrability. Precisely they proved the super
poly-harmonic properties for nonnegative solutions by making full use of the
the Poisson representation formula for
$(-\Delta)^{\frac{\alpha}{2}}(0<\alpha<2)$ and developing some new integral
estimates on the outer-spherical average
$\int_{R}^{+\infty}\frac{R^{\alpha}}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr$ and
iteration techniques. Based on super poly-harmonic properties, they
established integral representation formula, Liouville’s Theorem and
classification of classical solutions to higher-order equations involving
fractional Laplacian.

For (1), classification of nonconstant solutions have been established for
some particular $p,\alpha$ and $\gamma$. For $p=1,\alpha=0$ and
$\gamma=\frac{n}{n-2}$, Wang and Ye [17] classified all nonconstant solutions
of (1). Suzuki and Takahashi [15] extended the results of [17] from the
exponent $\frac{n}{n-2}$ to more general exponent. Precisely, they considered
the problem

(3) $-\Delta v=v^{\gamma}_{+}~{}~{}\mbox{ in
}{\mathbb{R}^{n}},~{}~{}n>2,~{}~{}\int_{\mathbb{R}^{n}}v_{+}^{\frac{n(\gamma-1)}{2}}dx<+\infty,$

where $\gamma\in(1,\frac{n+2}{n-2})$. For the case $p=2,\alpha=0$ and
$\gamma\in(1,\frac{n}{n-4}]$, Chammakhi, Harrabi and Selmi [4] completed the
classification of all sign-changing solutions of (1). The first, second and
fourth authors [11] extended the result of [4] to the corresponding
polyharmonic equation, namely for the case of any integer $p$ larger than $1$.
Particularly for the case $p=2$, they obtain the same results for more general
exponent $\gamma\in(1,\frac{n+4}{n-4})$. They also considered nonconstant
solutions of the following fractional equation (namely $p=0$ corresponds to
the equation in (1))

(4) $(-\Delta)^{\frac{\alpha}{2}}v=v^{\gamma}_{+}~{}~{}\mbox{ in
}{\mathbb{R}^{n}},~{}~{}\alpha\in(0,2),~{}~{}n\geq
2,~{}~{}\int_{\mathbb{R}^{n}}v_{+}^{\frac{n(\gamma-1)}{\alpha}}dx<+\infty,~{}~{}~{}~{}~{}$

where $\gamma\in(1,\frac{n+\alpha}{n-\alpha})$, and obtained the corresponding
classification results.

To classify nonconstant solutions of (1) for general $0<\alpha<2$, $p\geq 1$
and $\gamma\in(1,\frac{n}{n-2p-\alpha})$, we need to establish the following
super poly-harmonic properties.

######  Theorem 1.

Assume $\gamma\in(1,\frac{n}{n-2p-\alpha})$, $n>2p+\alpha$, $0<\alpha<2$ and
$p\geq 1$ is an integer. Suppose $u$ is a nonconstant classical solution to
(1) satisfying $u(x)=o(|x|^{\alpha+\frac{4}{\alpha}})$. Then for every
$i=0,1,\ldots,p-1$,

$(-\Delta)^{i+\frac{\alpha}{2}}u(x)\geq 0,\ \ \ \
\forall~{}x\in{\mathbb{R}}^{n}.$

As well-known, establishing equivalent integral equation of (1.1) is
beneficial to classifying solutions to higher order equations. As a
consequence of super poly-harmonic properties for equation (1), we can derive
the equivalence between the differential equation (1) and the following
integral equation

(5)
$u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2p+\alpha,n}}{|x-y|^{n-2p-\alpha}}u_{+}^{\gamma}(y)dy+C_{0},$

where $C_{0}<0$ and the Riesz potential’s constant
$R_{\mu,n}=\frac{\Gamma(\frac{n-\mu}{2})}{\pi^{\frac{n}{2}}2^{\mu}\Gamma(\frac{\mu}{2})}$
for $0<\mu<n$ (see [16]).

######  Theorem 2.

Under the assumptions of Theorem 1, suppose u is a nonconstant classical
solution to (1) satisfying $u(x)=o(|x|^{\alpha+\frac{4}{\alpha}})$. Then u is
a solution to integral equation (5), and vice verse. Moreover, the support of
$u_{+}$ is compact.

Apparently, for (1), any non-positive constant is its solution and any
positive constant is not its solution. Hence we only consider nonconstant
solutions. Actually, there is no positive solutions to (1) from the
nonexistence of positive solutions in the subcritical exponent case of Theorem
1.9 in [3]. All negative nonconstant solutions of (1) can be ruled out by the
Liouville theorem for fractional poly-harmonic functions (see Theorem 1.3 in
[3]). Consequently any nonconstant solutions of (1) are sign-changing
solutions. From this observation and Theorem 2, the following classification
results can be obtained by applying the method of moving planes in integral
forms.

######  Theorem 3.

Under the assumptions of Theorem 1, suppose $u$ is a sign-changing classical
solution to equation (1) satisfying $u(x)=o(|x|^{\alpha+\frac{4}{\alpha}})$ at
infinity, then $u$ is symmetric about some point $x_{0}\in{\mathbb{R}}^{n}$
and $\frac{\partial u}{\partial r}<0$, where $r=|x-x_{0}|$.

Throughout this paper, we will use $C$ to denote a positive constant which may
change from line to line and even within the same line.

##  2\. Preliminaries

The purpose of this section is to introduce several useful properties which
will be crucial to the forthcoming sections.

######  Proposition 1.

[1] Let $R>0$, $x_{0}\in{\mathbb{R}}^{n}$, $h\in
C^{\alpha+\varepsilon}(B_{R}(x_{0}))\cap C(\overline{B}_{R}(x_{0}))$,
$0<\alpha<2$ and let

$u(x):=\left\\{\begin{aligned}
&\int_{B_{R}(x_{0})}G_{R}^{\alpha}(x,y)h(y)dy,&\ \ \ \ &x\in B_{R}(x_{0}),\\\
&0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus B_{R}(x_{0}).\end{aligned}\right.$

Then $u$ is the unique point-wise continuous solution of the following problem

$\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}u(x)=h(x),&\ \ \ \ &x\in
B_{R}(x_{0}),\\\ &u(x)=0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(x_{0}).\end{aligned}\right.$

The Green function is defined by

$G_{R}^{\alpha}(x,y):=\frac{C_{n,\alpha}}{|x-y|^{n-\alpha}}\int_{0}^{\frac{t_{R}}{s_{R}}}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db,\
\ \ \ \ x,y\in B_{R}(x_{0})$

with $s_{R}=\frac{|x-y|^{2}}{R^{2}}$,
$t_{R}=\Big{(}1-\frac{|x-x_{0}|^{2}}{R^{2}}\Big{)}\Big{(}1-\frac{|y-x_{0}|^{2}}{R^{2}}\Big{)}$,
and $G_{R}^{\alpha}(x,y)=0$ if $x$ or $y\in{\mathbb{R}}^{n}\setminus
B_{R}(x_{0})$, where

$C_{n,\alpha}=\frac{\Gamma(\frac{n}{2})}{2^{\alpha}\pi^{\frac{n}{2}}\Gamma^{2}(\frac{\alpha}{2})}\
\mbox{if}\ \ n\neq\alpha,\ \ \ \ C_{1,1}=\frac{1}{\pi}\ \mbox{if}\
n=\alpha=1.$

######  Proposition 2.

[1] Let $R>0$, $x_{0}\in{\mathbb{R}}^{n}$,
$g\in\mathcal{L}_{\alpha}({\mathbb{R}}^{n})\cap C({\mathbb{R}}^{n})$,
$0<\alpha<2$ and let

$u_{g}(x):=\left\\{\begin{aligned} &\int_{{\mathbb{R}}^{n}\setminus
B_{R}(x_{0})}P_{R}^{\alpha}(x,y)g(y)dy,&\ \ \ \ &x\in B_{R}(x_{0}),\\\
&g(x),&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(x_{0}).\end{aligned}\right.$

Then $u_{g}$ is the unique point-wise continuous solution of the following
problem

$\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}u(x)=0,&\ \ \ \ &x\in
B_{R}(x_{0}),\\\ &u(x)=g(x),&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(x_{0}).\end{aligned}\right.$

The Poisson kernel $P_{R}^{\alpha}$ is defined by

$P_{R}^{\alpha}(x,y):=\frac{\Gamma(\frac{n}{2})}{\pi^{\frac{n}{2}+1}}\sin\frac{\pi\alpha}{2}\Big{(}\frac{R^{2}-|x-x_{0}|^{2}}{|y-x_{0}|^{2}-R^{2}}\Big{)}^{\frac{\alpha}{2}}\frac{1}{|x-y|^{n}}.$

######  Proposition 3.

[7] Let $\Omega$ be a bounded domain in ${\mathbb{R}}^{n}$. Assume that
$u\in\mathcal{L}_{\alpha}\cap C_{loc}^{1,1}(\Omega)$ and is lower semi-
continuous on $\bar{\Omega}$. If

$\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}u(x)\geq 0,&\ \ \ \
&x\in\Omega,\\\ &u(x)\geq 0,&\ \ \ \
&x\in{\mathbb{R}}^{n}\setminus\Omega,\end{aligned}\right.$

then $u(x)\geq 0$ in $\Omega$. Moreover, if $u=0$ at some point in $\Omega$,
then $u=0$ a.e. in ${\mathbb{R}}^{n}$. These conclusions also holds for
unbounded domain $\Omega$, if we further assume that
$\mathop{\liminf}\limits_{|x|\to+\infty}u(x)\geq 0$.

##  3\. Proof of Theorem 1

In this section we will complete the proof of Theorem 1.

Proof We borrow the idea in [3] to prove this theorem. Denote
$v_{i}:=(-\Delta)^{i+\frac{\alpha}{2}}u(x)$ for $i=0,1,\ldots,p-1$. It follows
from (1) that

(6) $\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}u(x)=v_{0},&\ \ \ \
&x\in\mathbb{R}^{n},\\\ &-\Delta v_{0}=v_{1},&\ \ \ \ &x\in\mathbb{R}^{n},\\\
&\cdots\cdots\\\ &-\Delta v_{p-1}=u_{+}^{\gamma},&\ \ \ \
&x\in\mathbb{R}^{n}.\end{aligned}\right.$

Assume that Theorem 1 is not true, then there must exists a largest integer
$0\leq k\leq p-1$ and a point $x_{0}\in{\mathbb{R}}^{n}$ such that

$v_{k}(x_{0})=(-\Delta)^{k+\frac{\alpha}{2}}u(x_{0})<0.$

For any $r>0$, define

$\bar{g}(r,x_{0}):=\frac{1}{|\partial B_{r}(x_{0})|}\int_{\partial
B_{r}(x_{0})}g(x)d\sigma,$

where $|\partial B_{r}(x_{0})|$ denotes the area of the sphere $\partial
B_{r}(x_{0})$. For simplicity, we denote $\bar{g}(r,x_{0})$ as $\bar{g}(r)$.

First, we will illustrate that $0\leq k\leq p-1$ is even by contradiction.
Assume $k$ is odd. From the well-known property $\overline{\Delta
u}(r)=\frac{1}{r^{n-1}}(r^{n-1}\bar{u}^{\prime}(r))^{\prime}$ and (6), we have

$\bar{v}_{k}(r)\leq\bar{v}_{k}(0):=-c_{0}<0,\ \ \ \ \forall r>0.$

Simple calculation shows that

$\bar{v}_{k-1}(r)\geq\bar{v}_{k-1}(0)+\frac{c_{0}}{2n}r^{2},\ \ \ \ \forall
r>0,$

and

$\bar{v}_{k-2}(r)\leq\bar{v}_{k-2}(0)-\frac{\bar{v}_{k-1}(0)}{2n}r^{2}-\frac{c_{0}}{8n(n+2)}r^{4},\
\ \ \ \forall r>0.$

Repeating the above argument, we derive

(7)
$\bar{v}_{0}(r)\geq\bar{v}_{0}(0)+c_{1}r^{2}+c_{2}r^{4}+\cdots+c_{k}r^{2k},\ \
\ \ \forall r>0,$

where $c_{k}>0$. From (7), we deduce that that there exists a $r_{0}$ large
enough such that

(8) $\bar{v}_{0}(r)\geq\frac{1}{2}c_{k}r^{2k}\ \ \ \ \forall r>r_{0}.$

The first equation in (6), Proposition 1 and Proposition 2 imply that for
arbitrary $R>0$,

$u(x)=\int_{B_{R}(x_{0})}G_{R}^{\alpha}(x,y)v_{0}(y)dy+\int_{|y-x_{0}|>R}P_{R}^{\alpha}(x,y)u(y)dy,\
\ \ \ \forall x\in B_{R}(x_{0}).$

Therefore, we have

$\displaystyle+\infty>u(x_{0})$
$\displaystyle=\int_{B_{R}(x_{0})}\frac{C_{n,\alpha}}{|x_{0}-y|^{n-\alpha}}\Big{(}\int_{0}^{\frac{R^{2}}{|y-x_{0}|^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\Big{)}v_{0}(y)dy$
(9)
$\displaystyle+C^{\prime}_{n,\alpha}\int_{|y-x_{0}|>R}\frac{R^{\alpha}}{(|y-x_{0}|^{2}-R^{2})^{\frac{\alpha}{2}}}\frac{u(y)}{|y-x_{0}|^{n}}dy$
$\displaystyle=:\mathrm{I}+\mathrm{II}.$

Observe that $0<r\leq\frac{R}{2}$ implies
$3\leq\frac{R^{2}}{r^{2}}-1<+\infty$. Thus

(10)
$\int_{0}^{3}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\leq\int_{0}^{\frac{R^{2}}{r^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\leq\int_{0}^{+\infty}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db.$

From (8)-(10), we conclude that for any $R>2r_{0}$ there exist $c_{i}>0$
$(i=1,...,6)$ independent of $R$ such that

$\displaystyle\mathrm{I}$ $\displaystyle=C_{n,\alpha}|\partial
B_{1}|\int_{0}^{R}r^{\alpha-1}\Big{(}\int_{0}^{\frac{R^{2}}{r^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\Big{)}\bar{v}_{0}(y)dy$
$\displaystyle\geq
c_{1}\int_{r_{0}}^{\frac{R}{2}}r^{\alpha-1}\bar{v}_{0}(r)dr-
c_{2}\int_{0}^{r_{0}}r^{\alpha-1}|\bar{v}_{0}(r)|dr$ $\displaystyle\geq
c_{3}\int_{r_{0}}^{\frac{R}{2}}r^{2k+\alpha-1}dr-c_{4}$ (11)
$\displaystyle\geq c_{5}R^{2k+\alpha}-c_{6}.$

Due to $u(x)=\mbox{o}(|x|^{\alpha+\frac{4}{\alpha}})$ at infinity, we derive
that for sufficiently large $R$

$\displaystyle C_{n,\alpha}^{\prime}$
$\displaystyle\int_{R<|y-x_{0}|<R+R^{1-\frac{4}{\alpha}}}\frac{R^{\alpha}}{(|y-x_{0}|^{2}-R^{2})^{\frac{\alpha}{2}}}\frac{u(y)}{|y-x_{0}|^{n}}dy$
$\displaystyle=CR^{\alpha}\int_{R}^{R+R^{1-\frac{4}{\alpha}}}\frac{\bar{u}(r)}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr$
$\displaystyle\leq\frac{2-\alpha}{2^{\frac{\alpha}{2}+\frac{4}{\alpha}+2}}c_{5}R^{\alpha}\int_{R}^{R+R^{1-\frac{4}{\alpha}}}\frac{r^{\alpha+\frac{4}{\alpha}}}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr$
$\displaystyle\leq\frac{2-\alpha}{8}c_{5}R^{\frac{3}{2}\alpha+\frac{4}{\alpha}-1}\int_{R}^{R+R^{1-\frac{4}{\alpha}}}(r-R)^{-\frac{\alpha}{2}}dr$
(12) $\displaystyle\leq\frac{c_{5}}{4}R^{2+\alpha}.$

Note that if $|y-x_{0}|\geq R+R^{1-\frac{4}{\alpha}}$, one has

$\frac{R^{2}}{|y-x_{0}|^{2}}\leq\frac{R^{2}}{(R+R^{1-\frac{4}{\alpha}})^{2}}\leq\frac{R^{2}}{R^{2}+2R^{2-\frac{4}{\alpha}}},$

which gives that

$|y-x_{0}|^{2}-R^{2}\geq\frac{2R^{2-\frac{4}{\alpha}}}{R^{2}+2R^{2-\frac{4}{\alpha}}}|y-x_{0}|^{2}.$

Obviously for large $R$, $|y-x_{0}|\geq\frac{|y|}{2}$ holds if $|y-x_{0}|\geq
R+R^{1-\frac{4}{\alpha}}$. Therefore for $R$ large enough, we have

$\displaystyle C_{n,\alpha}^{\prime}$
$\displaystyle\int_{{\mathbb{R}}^{n}\setminus
B_{R+R^{1-\frac{4}{\alpha}}}(x_{0})}\frac{R^{\alpha}}{(|y-x_{0}|^{2}-R^{2})^{\frac{\alpha}{2}}}\frac{u(y)}{|y-x_{0}|^{n}}dy$
$\displaystyle\leq CR^{\alpha}\int_{{\mathbb{R}}^{n}\setminus
B_{R+R^{1-\frac{4}{\alpha}}}(x_{0})}\frac{(R^{2}+2R^{2-\frac{4}{\alpha}})^{\frac{\alpha}{2}}|u(y)|}{(R^{2-\frac{4}{\alpha}})^{\frac{\alpha}{2}}|y|^{n+\alpha}}dy$
$\displaystyle\leq CR^{2\alpha-(\alpha-2)}\int_{{\mathbb{R}}^{n}\setminus
B_{R+R^{1-\frac{4}{\alpha}}}(x_{0})}\frac{|u(y)|}{|y|^{n+\alpha}}dy$ (13)
$\displaystyle\leq\frac{c_{5}}{4}R^{\alpha+2},$

where the last inequality holds due to
$u\in\mathcal{L}_{\alpha}({\mathbb{R}}^{n})$. Combining (3) with (3), we
conclude

(14) $\mathrm{II}\leq\frac{c_{5}}{2}R^{\alpha+2}.$

From (3), (3) and (14), we derive that for $R$ large enough,

(15) $+\infty>u(x_{0})\geq
c_{5}R^{2k+\alpha}-c_{6}-\frac{c_{5}}{2}R^{2+\alpha}.$

Letting $R\to+\infty$ in (15), we get immediately a contradiction. Thus, $k$
must be even.

Next, we will show that $k=0$. Otherwise, suppose that $2\leq k\leq p-1$ is
even, using the same arguments as in deriving (7), we deduce

(16) $\bar{v}_{0}(r)\leq\bar{v}_{0}(0)-c_{1}r^{2}-c_{2}r^{4}-\cdots-
c_{k}r^{2k},\ \ \ \ \forall r>0,$

where $c_{k}>0$. From (16), we infer that there exists $r_{1}$ large enough
such that

(17) $\bar{v}_{0}(r)\leq-\frac{1}{2}c_{k}r^{2k}\ \ \ \ \forall r>r_{1}.$

Note that if $\frac{R}{2}<r<R$, then $0<\frac{R^{2}}{r^{2}}-1<3$, and hence

(18)
$\int_{0}^{\frac{R^{2}}{r^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\geq\int_{0}^{\frac{R^{2}}{r^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{2^{n}}db\geq
C\Big{(}\frac{R^{2}}{r^{2}}-1\Big{)}^{\frac{\alpha}{2}}.$

From (3), (10), (17) and (18), we get for any $R>2r_{1}$,

$\begin{split}\int_{R}^{+\infty}&\frac{|\partial
B_{1}|R^{\alpha}\bar{u}(r)}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr=-C_{n,\alpha}\int_{0}^{R}r^{\alpha-1}\Big{(}\int_{0}^{\frac{R^{2}}{r^{2}}-1}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db\Big{)}\bar{v}_{0}(r)dr+u(x_{0})\\\
&\geq
C\int_{r_{1}}^{\frac{R}{2}}r^{2k+\alpha-1}dr-\widetilde{C}\int_{0}^{r_{1}}r^{\alpha-1}|\bar{v}_{0}(r)|dr+C\int_{\frac{R}{2}}^{R}r^{2k+\alpha-1}\Big{(}\frac{R^{2}}{r^{2}}-1\Big{)}^{\frac{\alpha}{2}}dr+u(x_{0})\\\
&\geq CR^{2k+\alpha}-\widetilde{C},\end{split}$

where $C$ and $\widetilde{C}$ are independent of $R$. Thus there exists a
$r_{2}>2r_{1}$ large enough such that

(19)
$\int_{R}^{+\infty}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}\geq
CR^{2k+\alpha},\ \ \ \ \forall R>r_{2}.$

Due to $u\in\mathcal{L}_{\alpha}({\mathbb{R}}^{n})$, we obtain

(20)
$\int_{1}^{+\infty}\frac{\overline{|u|}(r)}{r^{1+\alpha}}dr=C\int_{|x-x_{0}|>1}\frac{|u(x)|}{|x-x_{0}|^{n+\alpha}}dx<+\infty.$

Therefore for any $\delta>0$,

$\begin{split}\int_{1}^{+\infty}&\frac{1}{R^{1+\alpha+\delta}}\int_{R}^{+\infty}\frac{R^{\alpha}|\overline{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}drdR=\int_{1}^{+\infty}\frac{|\overline{u}(r)|}{r}\int_{1}^{r}\frac{1}{R^{1+\delta}(r^{2}-R^{2})^{\frac{\alpha}{2}}}dRdr\\\
&\leq
C\int_{1}^{+\infty}\frac{|\overline{u}(r)|}{r^{1+\alpha}}\int_{1}^{\frac{r}{2}}\frac{1}{R^{1+\delta}}dRdr+C\int_{1}^{+\infty}\frac{|\overline{u}(r)|}{r^{2+\delta}}\int_{\frac{r}{2}}^{r}\frac{1}{r^{\frac{\alpha}{2}}(r-R)^{\frac{\alpha}{2}}}dRdr\\\
&\leq
C_{\delta}\int_{1}^{+\infty}\frac{|\overline{u}(r)|}{r^{1+\alpha}}dr+C\int_{1}^{+\infty}\frac{|\overline{u}(r)|}{r^{1+\alpha+\delta}}dr\\\
&\leq C\int_{1}^{+\infty}\frac{{|\bar{u}}(r)|}{r^{1+\alpha}}dr\\\ &\leq
C\int_{1}^{+\infty}\frac{\overline{|u|}(r)}{r^{1+\alpha}}dr<+\infty,\end{split}$

which leads to a contradiction with (19). Thus $k=0$.

From $k=0$, we conclude that

(21) $\bar{v}_{0}(r)\leq\bar{v}_{0}(0):=-c_{0}<0,\ \ \ \ \forall r>0.$

Thus for any $R>0$, (3), (10), (18) and (21) yield that

$|\partial
B_{1}|\int_{R}^{+\infty}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr\geq
C\int_{0}^{\frac{R}{2}}r^{\alpha-1}dr+C\int_{\frac{R}{2}}^{R}r^{\alpha-1}\Big{(}\frac{R^{2}}{r^{2}}-1\Big{)}^{\frac{\alpha}{2}}dr+u(x_{0})\geq
CR^{\alpha}-\tilde{C},$

where positive constants $C$ and $\tilde{C}$ are independent of $R$. Thus
there exists $R_{0}$ large enough such that

(22)
$\int_{R}^{+\infty}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr\geq
CR^{\alpha},\ \ \ \ \forall R>R_{0}.$

Obviously

$\int_{N}^{+\infty}\frac{\overline{|u|}(r)}{r^{1+\alpha}}=C\int_{|x-x_{0}|>N}\frac{|u(x)|}{|x-x_{0}|^{n+\alpha}}dx=\mbox{o}_{N}(1)$

as $N\to+\infty$ due to $u\in\mathcal{L}_{\alpha}({\mathbb{R}}^{n})$.
Therefore

(23)
$\int_{2R}^{+\infty}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr\leq
CR^{\alpha}\int_{2R}^{+\infty}\frac{{|\bar{u}}(r)|}{r^{1+\alpha}}\leq\mbox{o}_{R}(1)R^{\alpha},$

as $R\to+\infty$. It follows from (22) and (23) that there exists $R_{1}\geq
R_{0}$ large enough such that

(24)
$\int_{R}^{2R}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}dr\geq\frac{C}{2}R^{\alpha},\
\ \ \ \forall R>R_{1}.$

On the other hand, from (20), we derive that

$\begin{split}\int_{1}^{+\infty}\frac{1}{R^{1+\alpha}}\int_{R}^{2R}\frac{R^{\alpha}|\bar{u}(r)|}{r(r^{2}-R^{2})^{\frac{\alpha}{2}}}drdR&\leq\int_{1}^{+\infty}\frac{|\bar{u}(r)|}{r}\int_{\frac{r}{2}}^{r}\frac{1}{R(r^{2}-R^{2})^{\frac{\alpha}{2}}}dRdr\\\
&\leq
C\int_{1}^{+\infty}\frac{|\bar{u}(r)|}{r^{2+\frac{\alpha}{2}}}\int_{\frac{r}{2}}^{r}\frac{1}{(r-R)^{\frac{\alpha}{2}}}dRdr\\\
&\leq
C\int_{1}^{+\infty}\frac{{|\bar{u}}(r)|}{r^{1+\alpha}}<+\infty,\end{split}$

which contradicts with (24). Hence Theorem 1 is proved. $\square$

##  4\. Proof of Theorem 2

To complete the proof of Theorem 2, we need to establish the following lemmas.

######  Lemma 1.

Under the assumptions of Theorem 1, suppose u is a nonconstant classical
solution to (1.1) satisfying $u(x)=o(|x|^{\alpha+\frac{4}{\alpha}})$ at
infinity. Then $(-\Delta)^{\frac{\alpha}{2}}u$ satisfies the following
integral equation

(25)
$\displaystyle(-\Delta)^{\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2p,n}}{|x-y|^{n-2p}}u_{+}^{\gamma}(y)dy+C_{p},$

where $C_{p}$ is a nonnegative constant.

###### Proof.

Firstly, we prove that

(26)
$(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}u_{+}^{\gamma}(y)dy,\
\ \ \ \forall x\in{\mathbb{R}}^{n}.$

To this end, for arbitrary $R>0$, denote $f_{1}(x)=u_{+}^{\gamma}(x)$ and

$v_{1}^{R}(x):=\int_{B_{R}(x_{0})}G_{R}^{2}(x,y)f_{1}(y)dy,$

where the Green’s function for $-\Delta$ on $B_{R}(0)$ is given by

$G_{R}^{2}(x,y)=R_{2,n}\Big{[}\frac{1}{|x-y|^{n-2}}-\frac{1}{(|x|\cdot|\frac{Rx}{|x|^{2}}-\frac{y}{R}|)^{n-2}}\Big{]},\
\ \ \ x,y\in B_{R}(0),$

and $G_{R}^{2}(x,y)=0$ if $x$ or $y\in{\mathbb{R}}^{n}\setminus B_{R}(0)$.
Then we derive that $(-\Delta)^{p-1+\frac{\alpha}{2}}u-v_{1}^{R}$ is harmonic
in $B_{R}(0)$ and continuous up to $\partial B_{R}(0)$. Hence $v_{1}^{R}(x)\in
C^{2}(B_{R}(0))\cap C({\mathbb{R}}^{n})$ and satisfies

(27) $\left\\{\begin{aligned} &-\Delta v_{1}^{R}(x)=u_{+}^{\gamma},&\ \ \ \
&x\in B_{R}(0),\\\ &v_{1}^{R}(x)=0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(0).\end{aligned}\right.$

Set $w_{1}^{R}(x):=(-\Delta)^{p-1+\frac{\alpha}{2}}u-v_{1}^{R}$. Combining
Theorem 1 with (27), we obtain $w_{1}^{R}(x)\in C^{2}(B_{R}(0))\cap
C({\mathbb{R}}^{n})$ and satisfies

$\left\\{\begin{aligned} &-\Delta w_{1}^{R}(x)=0,&\ \ \ \ &x\in B_{R}(0),\\\
&w_{1}^{R}(x)\geq 0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(0).\end{aligned}\right.$

The maximum principle implies that

(28) $w_{1}^{R}(x)=(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)-v_{1}^{R}(x)\geq 0,\ \
\ \ \forall x\in{\mathbb{R}}^{n}.$

Now for each fixed $x\in{\mathbb{R}}^{n}$, letting $R\to+\infty$ in (28), we
have

(29)
$(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)\geq\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{1}(y)dy=:v_{1}(x)\geq
0.$

Since $v_{1}-v_{1}^{R}$ is smooth in $B_{R}(0)$ for any $R>0$, one can obtain
that $v_{1}\in C^{2}({\mathbb{R}}^{n})$ satisfies

(30) $-\Delta v_{1}(x)=u_{+}^{\gamma}(x),\ \ \ \ \forall
x\in{\mathbb{R}}^{n}.$

Set $w_{1}(x):=(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)-v_{1}(x)$. From (1), (29)
and (30), we have that $w_{1}(x)\in C^{2}({\mathbb{R}}^{n})$ and satisfies

$\left\\{\begin{aligned} &-\Delta w_{1}(x)=0,&\ \ \ \
&x\in{\mathbb{R}}^{n},\\\ &w_{1}(x)\geq 0,&\ \ \ \
&x\in{\mathbb{R}}^{n}.\end{aligned}\right.$

Then by the Liouville Theorem for harmonic function, we can infer that

$w_{1}(x)=(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)-v_{1}(x)\equiv C_{1}\geq 0.$

Therefore

(31)
$(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}u_{+}^{\gamma}(y)dy+C_{1}=:f_{2}(x)\geq
C_{1}\geq 0.$

Next, for arbitrary $R>0$, denote

$v_{2}^{R}(x):=\int_{B_{R}(x_{0})}G_{R}^{2}(x,y)f_{2}(y)dy.$

Then we have

(32) $\left\\{\begin{aligned} &-\Delta v_{2}^{R}(x)=f_{2}(x),&\ \ \ \ &x\in
B_{R}(0),\\\ &v_{2}^{R}(x)=0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(0).\end{aligned}\right.$

Set $w_{2}^{R}(x):=(-\Delta)^{p-2+\frac{\alpha}{2}}u-v_{2}^{R}$. By Theorem 1,
(31) and (32), we obtain

$\left\\{\begin{aligned} &-\Delta w_{2}^{R}(x)=0,&\ \ \ \ &x\in B_{R}(0),\\\
&w_{2}^{R}(x)\geq 0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(0).\end{aligned}\right.$

The maximum principle implies that for any $R>0$

(33) $w_{2}^{R}(x)=(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)-v_{2}^{R}(x)\geq 0,\ \
\ \ \forall x\in{\mathbb{R}}^{n}.$

Now for each fixed $x\in{\mathbb{R}}^{n}$, letting $R\to+\infty$ in (33), we
have

(34)
$(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)\geq\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{2}(y)dy=:v_{2}(x)\geq
0.$

Taking $x=0$ in (34) and using (31), we get

$\int_{{\mathbb{R}}^{n}}\frac{C_{1}}{|y|^{n-2}}dy\leq\int_{{\mathbb{R}}^{n}}\frac{f_{2}(y)}{|y|^{n-2}}dy<+\infty.$

It follows immediately that $C_{1}=0$, and hence we shows that (26) holds,
that is

(35)
$(-\Delta)^{p-1+\frac{\alpha}{2}}u(x)=f_{2}(y)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}u_{+}^{\gamma}(y)dy,\
\ \ \ \forall x\in{\mathbb{R}}^{n}.$

One can easily observe that $v_{2}$ is a solution of

(36) $-\Delta v_{2}(x)=f_{2}(x),\ \ \ \ \forall x\in{\mathbb{R}}^{n}.$

Set $w_{2}(x):=(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)-v_{2}(x)$. Then it solves

$\left\\{\begin{aligned} &-\Delta w_{2}(x)=0,&\ \ \ \
&x\in{\mathbb{R}}^{n},\\\ &w_{2}(x)\geq 0,&\ \ \ \
&x\in{\mathbb{R}}^{n}.\end{aligned}\right.$

Applying the Liouville Theorem for harmonic functions again, we can infer that

$w_{2}(x)=(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)-v_{2}(x)\equiv C_{2}\geq 0.$

Therefore, we deduce that

(37)
$(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{2}(y)dy+C_{2}=:f_{3}(x)\geq
C_{2}\geq 0.$

By the same methods as above, we can prove that $C_{2}=0$, and hence

$(-\Delta)^{p-2+\frac{\alpha}{2}}u(x)=f_{3}(y)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{2}(y)dy,\
\ \ \ \forall x\in{\mathbb{R}}^{n}.$

Define

(38)
$f_{k+1}(x):=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{k}(y)dy$

for $k=1,2,\ldots,p$. Repeating the above argument we can derive that

(39)
$(-\Delta)^{p-k+\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{k}(y)dy,\
\ \ \ \forall x\in{\mathbb{R}}^{n}$

for $k=1,2,\ldots,p-1$ and

(40)
$(-\Delta)^{\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2,n}}{|x-y|^{n-2}}f_{p}(y)dy+C_{p}=f_{p+1}(x)+C_{p}\geq
C_{p}\geq 0,\ \ \ \ \forall x\in{\mathbb{R}}^{n}.$

Owing to the properties of the Riesz potential, for any
$\alpha_{1},\alpha_{2}\in(0,n)$ such that $\alpha_{1}+\alpha_{2}\in(0,n)$, it
is easy to know that

(41)
$\int_{{\mathbb{R}}^{n}}\frac{R_{\alpha_{1},n}}{|x-y|^{n-\alpha_{1}}}\cdot\frac{R_{\alpha_{2},n}}{|y-z|^{n-\alpha_{2}}}dy=\frac{R_{\alpha_{1}+\alpha_{2},n}}{|x-z|^{n-(\alpha_{1}+\alpha_{2})}}.$

In particular, it follows from (38)-(41) and Fubini’s theorem that

$(-\Delta)^{p-k+\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2k,n}}{|x-y|^{n-2k}}u_{+}^{\gamma}(y)dy,\
\ \ \ \forall x\in{\mathbb{R}}^{n}$

for $k=1,2,\ldots,p-1$ and

$(-\Delta)^{\frac{\alpha}{2}}u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{2p,n}}{|x-y|^{n-2p}}u_{+}^{\gamma}(y)dy+C_{p},\
\ \ \ \forall x\in{\mathbb{R}}^{n}.$

Thus the lemma holds. $\square$

######  Lemma 2.

Under the assumptions of Theorem 1, if $u\in
C_{loc}^{2p+[\alpha],\\{\alpha\\}+\varepsilon}({\mathbb{R}}^{n})$ satisfies
$u_{+}^{\gamma}\in L^{1}({\mathbb{R}}^{n})$ and (25), then $\Delta u\in
L^{\infty}({\mathbb{R}}^{n})$.

###### Proof.

From (25), (41) and the formula of fundamental solution for
$(-\Delta)^{\frac{\alpha}{2}}$, we obtain

$\displaystyle-\Delta u(x)$
$\displaystyle=(-\Delta)^{1-\frac{\alpha}{2}}\Big{(}\int_{{\mathbb{R}}^{n}}\frac{R_{2p,n}}{|x-y|^{n-2p}}u_{+}^{\gamma}(y)dy+C_{p}\Big{)}$
$\displaystyle=(-\Delta)^{1-\frac{\alpha}{2}}\Big{(}\int_{{\mathbb{R}}^{n}}u_{+}^{\gamma}(y)\int_{{\mathbb{R}}^{n}}\frac{R_{2-\alpha,n}}{|x-z|^{n-2+\alpha}}\frac{R_{2p-2+\alpha,n}}{|z-y|^{n-2p+2-\alpha}}dzdy\Big{)}$
$\displaystyle=(-\Delta)^{1-\frac{\alpha}{2}}\Big{(}\int_{{\mathbb{R}}^{n}}\frac{R_{2-\alpha,n}}{|x-z|^{n-2+\alpha}}\int_{{\mathbb{R}}^{n}}\frac{R_{2p-2+\alpha,n}}{|z-y|^{n-2p+2-\alpha}}u_{+}^{\gamma}(y)dydz\Big{)}$
(42)
$\displaystyle={R_{2p-2+\alpha,n}}\int_{{\mathbb{R}}^{n}}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy.$

Then analogous to Lemma 2.3 in [11], we can immediately infer that $\Delta
u\in L^{\infty}({\mathbb{R}}^{n})$. Indeed, from (4), we derive that

(43) $\displaystyle~{}\Delta
u(x)+R_{2p-2+\alpha,n}\int_{\mathbb{R}^{n}}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy=0.$

Denote $\beta:=\int_{\mathbb{R}^{n}}u_{+}^{\gamma}(y)dy$. For any
$x_{0}\in{\mathbb{R}^{n}}$, consider the solution $h_{1}$ of the boundary
value problem

(46) $\displaystyle\left\\{\begin{array}[]{ll}-\Delta
h_{1}(x)=R_{2p-2+\alpha,n}\int_{B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy{}{}&\mbox{
in }{B_{{R}}(x_{0})},\\\
h_{1}=0{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}&\mbox{
on }{\partial B_{{R}}(x_{0})},\end{array}\right.$

where ${R}$ will be determined later. Set

$v_{1}(x)=C_{n,p,\alpha}\int_{B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy~{}~{}\mbox{
for any }x\in B_{{R}}(x_{0}),$

where $C_{n,p,\alpha}=\frac{R_{2p-2+\alpha,n}}{(n-2p-\alpha)(2p-2+\alpha)}$.
It’s obvious that $v_{1}(x)\geq 0$ in $B_{{R}}(x_{0})$ and

(47) $\displaystyle-\Delta
v_{1}(x)=R_{2p-2+\alpha,n}\int_{B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy.$

The comparison principle allows us to conclude that

$|h_{1}(x)|\leq v_{1}(x),~{}~{}x\in B_{{R}}(x_{0}).$

Therefore there exists $t>1$ and a constant $C>0$ independent of $x_{0}$ such
that

(48) $\displaystyle\int_{B_{{R}}(x_{0})}|h_{1}(x)|^{t\gamma}dx\leq C.$

Indeed

(49)
$\displaystyle\begin{split}\int_{B_{{R}}(x_{0})}|h_{1}(x)|^{t\gamma}dx\leq\int_{B_{{R}}(x_{0})}|v_{1}(x)|^{t\gamma}dx=\int_{B_{{R}}(x_{0})}\left|{C}_{n,p,\alpha}\int_{B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy\right|^{t\gamma}dx.\end{split}$

Owing to $\gamma\in(1,\frac{n}{n-2p-\alpha})$, we know that there exists $t>1$
such that $t\gamma<\frac{n}{n-2p-\alpha}$. Denote
$d\mu=u_{+}^{\gamma}dy/\int_{B_{{R}}(x_{0})}u_{+}^{\gamma}(y)dy$, then the
assumption $u_{+}^{\gamma}(y)\in L^{1}(\mathbb{R}^{n})$ and Jensen’s
inequality imply that

(50)
$\displaystyle\begin{split}\int_{B_{{R}}(x_{0})}\left|{\int_{B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy}\right|^{t\gamma}dx\leq&C\int_{B_{{R}}(x_{0})}\left(\int_{B_{{R}}(x_{0})}\frac{1}{|x-y|^{n-2p-\alpha}}d\mu\right)^{t\gamma}dx\\\
\leq&C\int_{B_{{R}}(x_{0})}\int_{B_{{R}}(x_{0})}\frac{1}{|x-y|^{(n-2p-\alpha)t\gamma}}dxd\mu\\\
\leq&C,~{}~{}~{}~{}\end{split}$

where the final inequality used the fact that $t\gamma(n-2p-\alpha)<n$. From
this and (49), we obtain (48).

Now, we consider the function $q(x):=u(x)-h_{1}(x)$ in the smaller ball
$B_{{R}-1}(x_{0})$. First we observe that

$\displaystyle\begin{split}\Delta q(x)&=\Delta u(x)-\Delta
h_{1}(x)=-R_{2p-2+\alpha,n}\int_{\mathbb{R}^{n}\setminus
B_{{R}}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy.\end{split}$

If $x\in B_{{R}-1}(x_{0})$ and $y\in\mathbb{R}^{n}\setminus B_{{R}}(x_{0})$,
then $|x-y|\geq 1$. Therefore we have

(51) $\displaystyle 0\leq-\Delta q(x)\leq R_{2p-2+\alpha,n}\beta.$

Hence, it follows from weak Harnack principle that

(52) $\displaystyle\sup\limits_{B_{{R}-2}(x_{0})}q(x)\leq
C[\|q_{+}\|_{L^{\gamma}(B_{{R}-1}(x_{0}))}+\|\Delta
q\|_{L^{\infty}(B_{{R}-1}(x_{0}))}].$

(51) shows that the second term of the right hand side of the above inequality
is bounded independent of $x_{0}$. By using (48) and $u_{+}^{\gamma}\in
L^{1}(\mathbb{R}^{n})$, we obtain that

$\int_{B_{{R}-1}(x_{0})}(q_{+}(x))^{\gamma}dx\leq C,$

where $C$ independent of $x_{0}$. Therefore, it follows that
$u(y)=q(y)+h_{1}(y)\leq C+|h_{1}(y)|$ in the smaller ball $B_{{R}-2}(x_{0})$.
This and (48) yield the estimate

(53) $\displaystyle\int_{B_{{R}-2}(x_{0})}u_{+}^{t\gamma}(y)dy\leq C.$

Now consider the solution $h_{2}$ of the boundary value problem

(56) $\displaystyle\left\\{\begin{array}[]{ll}-\Delta
h_{2}(x)=R_{2p-2+\alpha,n}\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy{}{}&\mbox{
in }{B_{{R}-2}(x_{0})},\\\
h_{2}=0{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}&\mbox{
on }{\partial B_{{R}-2}(x_{0})}.\end{array}\right.$

Set

$v_{2}(x)=C_{n,p,\alpha}\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy~{}~{}\mbox{
for any }x\in B_{{R}-2}(x_{0}),$

and we have that $|h_{2}(x)|\leq v_{2}(x)$ in $B_{{R}-2}(x_{0})$. Simple
computation shows that

$\displaystyle\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy$
$\displaystyle\leq
C\left(\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{t\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy\right)^{\frac{1}{t}}\left(\int_{B_{{R}-2}(x_{0})}\frac{1}{|x-y|^{n-2p-\alpha}}dy\right)^{\frac{t-1}{t}}$
$\displaystyle\leq
C\left(\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{t\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy\right)^{\frac{1}{t}}.$

Hence

(57)
$\displaystyle\begin{split}\int_{B_{{R}-2}(x_{0})}|h_{2}(x)|^{t^{2}\gamma}dx&\leq\int_{B_{{R}-2}(x_{0})}|v_{2}(x)|^{t^{2}\gamma}dx\\\
&=C\int_{B_{{R}-2}(x_{0})}\left(\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy\right)^{t^{2}\gamma}dx\\\
&\leq
C\int_{B_{{R}-2}(x_{0})}\left(\int_{B_{{R}-2}(x_{0})}\frac{u_{+}^{t\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy\right)^{t\gamma}dx.\end{split}$

Let $d\mu=u_{+}^{t\gamma}dy/\int_{B_{{R}-2}(x_{0})}u_{+}^{t\gamma}(y)dy$.
Therefore (53), (57) and Jensen’s inequality imply that there exists a
constant $C>0$ independent of $x_{0}$ such that

$\displaystyle\begin{split}\int_{B_{{R}-2}(x_{0})}|h_{2}(x)|^{t^{2}\gamma}dx\leq
C.\end{split}$

Similar argument as the above, we can obtain that there exists a constant
$C>0$ independent of $x_{0}$ such that

(58) $\displaystyle\int_{B_{{R}-4}(x_{0})}u_{+}^{t^{2}\gamma}(y)dy\leq C.$

Let $h_{3}$ be the solution of the equation

(61) $\displaystyle\left\\{\begin{array}[]{ll}-\Delta
h_{3}(x)=R_{2p-2+\alpha,n}\int_{B_{{R}-4}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p+2-\alpha}}dy{}{}&\mbox{
in }{B_{{R}-4}(x_{0})},\\\
h_{3}=0,{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}&\mbox{
on }{\partial B_{{R}-4}(x_{0})},\end{array}\right.$

and

$v_{3}(x)=C_{n,p,\alpha}\int_{B_{{R}-4}(x_{0})}\frac{u_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy~{}~{}\mbox{
for all }x\in B_{{R}-4}(x_{0}).$

Similar argument as before yields that

$\int_{B_{{R}-6}(x_{0})}u_{+}^{t^{3}\gamma}(y)dy\leq C.$

Repeating the process, we derive that for any $k\in\mathbb{N}$ satisfying
$R-2k>0$,

(62) $\displaystyle\int_{B_{{R}-2k}(x_{0})}u_{+}^{t^{k}\gamma}(y)dy\leq C.$

Choose $k>0$ large enough such that $\frac{1}{t^{k}}+\frac{1}{s}=1$ where $s$
satisfies $s(n-2p+2-\alpha)<n$. For such $k$, we choose ${R}$ large enough
such that ${{R}-2k}>2$. From (43), we have

$\displaystyle\begin{split}|\Delta
u(x_{0})|&=R_{2p-2+\alpha,n}\int_{\mathbb{R}^{n}\setminus{B_{{R}-2k}(x_{0})}}\frac{u_{+}^{\gamma}(y)}{|x_{0}-y|^{n-2p+2-\alpha}}dy+R_{2p-2+\alpha,n}\int_{{B_{{R}-2k}(x_{0})}}\frac{u_{+}^{\gamma}(y)}{|x_{0}-y|^{n-2p+2-\alpha}}dy\\\
&\leq\frac{R_{2p-2+\alpha,n}}{4}\int_{\mathbb{R}^{n}}u_{+}^{\gamma}(y)dy+C\left(\int_{{B_{{R}-2k}(x_{0})}}u_{+}^{t^{k}\gamma}(y)dy\right)^{\frac{1}{t^{k}}}\left(\int_{{B_{{R}-2k}(x_{0})}}\frac{1}{|x_{0}-y|^{(n-2p+2-\alpha)s}}dy\right)^{\frac{1}{s}}\\\
&\leq\frac{R_{2p-2+\alpha,n}}{4}\int_{\mathbb{R}^{n}}u_{+}^{\gamma}(y)dy+C\left(\int_{{B_{{R}-2k}(x_{0})}}u_{+}^{t^{k}\gamma}(y)dy\right)^{\frac{1}{t^{k}}},\end{split}$

where the final inequality used the fact that $(n-2p+2-\alpha)s<n$. (62) and
the integral constraint in (1) show that $|\Delta u(x_{0})|\leq C$, where
$C>0$ is independent of $x_{0}$. Therefore, we finish the proof of this lemma.
$\hfill\square$

######  Lemma 3.

If $\gamma\in(1,\frac{n}{n-2p-\alpha})$, $n>2p+\alpha$, $p\geq 1$ is an
integer and $u$ is a solution of (1) satisfying
$u(x)=o(|x|^{\alpha+\frac{4}{\alpha}})$ at infinity. then there exists a
constant $M>0$ such that $\sup\limits_{\mathbb{R}^{n}}u\leq M$.

Proof From Lemma 2, we have that there exists $A>0$ such that $|\Delta u|\leq
A$ in ${\mathbb{R}^{n}}.$ Denote $h(x)=-\Delta u(x)$. Given
$x_{0}\in{\mathbb{R}^{n}}$, let $u_{1}$ be the solution of

(65) $\displaystyle\left\\{\begin{array}[]{rl}-\Delta v=h,{}{}{}&\mbox{ in
}B_{1}(x_{0}),\\\ v=0,{}{}{}{}{}{}{}{}{}{}&\mbox{ on }\partial
B_{1}(x_{0}).\end{array}\right.$

It follows from the elliptic theory that $|u_{1}|\leq C$, where $C>0$
independent of $x_{0}$. Denote $u_{2}=u-u_{1}$, then $(u_{2})_{+}\leq
u_{+}+|u_{1}|$. Since $|u_{1}|\leq C$ in $B_{1}(x_{0})$ and
$\int_{\mathbb{R}^{n}}u_{+}^{\gamma}(x)dx<+\infty$, we derive

$\int_{B_{1}(x_{0})}(u_{2})_{+}^{\gamma}(x)\leq C.$

Note that $\Delta u_{2}=0$ in $B_{1}(x_{0})$. For the subharmonic function
$(u_{2})_{+}$, we have

$\|(u_{2})_{+}\|_{L^{\infty}(B_{1/2}(x_{0}))}\leq
C\int_{B_{1}(x_{0})}(u_{2})_{+}(x)dx\leq
C\left(\int_{B_{1}(x_{0})}(u_{2})_{+}^{\gamma}(x)dx\right)^{\frac{1}{\gamma}}\leq
C.$

where $C$ is independent of $x_{0}$. Recalling that $u=u_{1}+u_{2}$ and the
arbitrariness of $x_{0}$, we derive that there exists $M>0$ independent of
$x_{0}$ such that $u_{+}(x)\leq M$. Hence, $\sup\limits_{\mathbb{R}^{n}}u\leq
M$. $\hfill\square$

Set

$\zeta(x)=-\int_{{\mathbb{R}}^{n}}\frac{R_{2p+\alpha,n}}{|x-y|^{n-2p-\alpha}}u_{+}^{\gamma}(y)dy.$

Then it is easy to obtain some asymptotic behaviors about $\zeta(x)$ at
infinity as follows.

######  Lemma 4.

$\zeta(x)$ satisfies

$\lim_{|x|\to+\infty}(-\Delta)^{i}\zeta(x)|x|^{n-2p-\alpha+2i}=a_{i},\ \ \ \
i=0,1,\ldots,p-1,$

where $a_{0}=-R_{2p+\alpha,n}\int_{{\mathbb{R}}^{n}}u_{+}^{\gamma}(y)dy$ and
$a_{i+1}=a_{i}(n-2p-\alpha+2i)(2p+\alpha-2i-2)$, $i=0,1,\ldots,p-2$.

Moreover,

$\lim_{|x|\to+\infty}(-\Delta)^{i+\frac{\alpha}{2}}\zeta(x)|x|^{n-2p+2i}=b_{i},\
\ \ \ i=0,1,\ldots,p-1,$

where $b_{0}=-R_{2p,n}\int_{{\mathbb{R}}^{n}}u_{+}^{\gamma}(y)dy$ and
$b_{i+1}=b_{i}(n-2p+2i)(2p-2i-2)$, $i=0,1,\ldots,p-2$.

We are now ready to complete the proof of the equivalence between (1) and (5),
that is proof of Theorem 2.

Proof of Theorem 2 From Lemma 1, we derive (25) holds. Next we will show that
$C_{p}=0$.

For arbitrary $R>0$, let

$v_{p+1}^{R}(x):=\int_{B_{R}(x_{0})}G_{R}^{\alpha}(x,y)(f_{p+1}(y)+C_{p})dy,$

where $G_{R}^{\alpha}(x,y)$ is the Green’s function for
$(-\Delta)^{\frac{\alpha}{2}}$ with $0<\alpha<2$ on $B_{R}(0).$ Then, we can
get

(66) $\left\\{\begin{aligned}
&(-\Delta)^{\frac{\alpha}{2}}v_{p+1}^{R}(x)=f_{p+1}(x)+C_{p},&\ \ \ \ &x\in
B_{R}(0),\\\ &v_{p+1}^{R}(x)=0,&\ \ \ \ &x\in{\mathbb{R}}^{n}\setminus
B_{R}(0).\end{aligned}\right.$

Denote $w_{p+1}^{R}(x):=M-u(x)+v_{p+1}^{R}(x)$. From (25), (66) and Lemma 3,
we have

$\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}w_{p+1}^{R}(x)=0,&\ \ \
\ &x\in B_{R}(0),\\\ &w_{p+1}^{R}(x)\geq 0,&\ \ \ \
&x\in{\mathbb{R}}^{n}\setminus B_{R}(0).\end{aligned}\right.$

By maximum principle, we can deduce that for any $R>0$

(67) $w_{p+1}^{R}(x)=M-u(x)+v_{p+1}^{R}(x)\geq 0,\ \ \ \ \forall
x\in{\mathbb{R}}^{n}.$

Now for each fixed $x\in{\mathbb{R}}^{n}$, letting $R\to+\infty$ in (67), we
have

(68)
$M-u(x)\geq-\int_{{\mathbb{R}}^{n}}\frac{R_{\alpha,n}}{|x-y|^{n-\alpha}}(f_{p+1}(y)+C_{p})dy=:-v_{p+1}(x).$

Thus as $R\to+\infty$, we have

$\left\\{\begin{aligned} &(-\Delta)^{\frac{\alpha}{2}}(M-u(x)+v_{p+1}(x))=0,&\
\ \ \ &x\in B_{R}(0),\\\ &M-u(x)+v_{p+1}(x)\geq 0,&\ \ \ \
&x\in{\mathbb{R}}^{n}\setminus B_{R}(0).\end{aligned}\right.$

By Liouville Theorem, we obtain

$M-u(x)+v_{p+1}(x)\equiv\widetilde{C}\geq 0.$

That is,

(69)
$u(x)=\int_{{\mathbb{R}}^{n}}\frac{R_{\alpha,n}}{|x-y|^{n-\alpha}}(f_{p+1}(y)+C_{p})dy+C.$

Taking $x=0$ in (69), we get

$\int_{{\mathbb{R}}^{n}}\frac{C_{p}}{|y|^{n-\alpha}}dy\leq\int_{{\mathbb{R}}^{n}}\frac{f_{p+1}(y)+C_{p}}{|y|^{n-\alpha}}dy<+\infty.$

It follows immediately that $C_{p}=0$. Thus, from (38), (41) and (69), we
derive that (5) holds. Moreover, we assert that $C_{0}<0$. Indeed, if
$C_{0}\geq 0$, we have $u(x)\geq 0$ in ${\mathbb{R}}^{n}$, which is impossible
from $\gamma\in(1,\frac{n+2p+\alpha}{n-2p-\alpha})$ and the results of Theorem
1.9 in [3]. Hence (5) holds.

From Lemma 4 and the fact $C_{0}<0$, we obtain that the support of $u_{+}$ is
compact. Meanwhile, it is obvious that if $u$ is a solution of (5), then it
satisfies equation (1). The proof of Theorem 2 is completed. $\square$

##  5\. Proof of Theorem 3

In this section, we verify Theorem 3 by taking advantage of the method of
moving planes in integral forms.

Proof  To complete the proof of Theorem 3, it’s enough to show that $\zeta$ is
symmetric about some point $x_{0}\in{\mathbb{R}^{n}}$ and
$\frac{\partial\zeta}{\partial r}>0$ where $r=|x-x_{0}|$. From Lemmas 2-4, we
obtain that $\lim\limits_{|x|\rightarrow\infty}\zeta(x)=0$ and

(70)
$\zeta(x)=-R_{2p+\alpha,n}\int_{\mathbb{R}^{n}}\frac{(C_{0}-\zeta)_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy,~{}~{}x\in{\mathbb{R}^{n}}.$

For $x=(x_{1},x_{2},...,x_{n})\in{\mathbb{R}}^{n}$ and
$\lambda\in{\mathbb{R}}$, we define
${\mathrm{T}}_{\lambda}=\\{x\in{\mathbb{R}}^{n}|x_{1}=\lambda\\}$,
$\Sigma_{\lambda}=\\{x\in{\mathbb{R}}^{n}|x_{1}<\lambda\\}$,
$x^{\lambda}=(2\lambda-x_{1},x_{2},...,x_{n})$ and
$\zeta_{\lambda}(x)=\zeta(2\lambda-x_{1},x_{2},...,x_{n})=\zeta(x^{\lambda})$.
Set $w_{\lambda}(x)=\zeta(x)-\zeta_{\lambda}(x)$. It’s obvious that

$\zeta_{\lambda}(x)=-R_{2p+\alpha,n}\int_{\mathbb{R}^{n}}\frac{(C_{0}-\zeta_{\lambda})_{+}^{\gamma}(y)}{|x-y|^{n-2p-\alpha}}dy,~{}~{}x\in{\mathbb{R}^{n}}.$

From this and (70), we have

(71)
$\displaystyle\zeta_{\lambda}(x)-\zeta(x)=R_{2p+\alpha,n}\int_{\Sigma_{\lambda}}\left(\frac{1}{|x-y|^{n-2p-\alpha}}-\frac{1}{|x-y^{\lambda}|^{n-2p-\alpha}}\right)\left((C_{0}-\zeta)_{+}^{\gamma}(y)-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}(y)\right)dy.$

$\mathbf{Step~{}1}$: We claim that for $\lambda$ sufficiently negative,

(72) $\displaystyle w_{\lambda}(x)>0,~{}~{}~{}~{}~{}x\in\Sigma_{\lambda}.$

Due to $\lim\limits_{|x|\rightarrow\infty}\zeta(x)=0$ and $C_{0}<0$, we have
for $\lambda$ sufficiently negative

$(C_{0}-\zeta)_{+}^{\gamma}-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}=-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}\leq
0,~{}~{}~{}~{}x\in\Sigma_{\lambda}.$

From this, (71) and the fact that $u$ is a sign-changing classical solution of
(1), we have $\zeta_{\lambda}(x)-\zeta(x)<0$ for any $x\in\Sigma_{\lambda}$.
Thus (72) holds.

$\mathbf{Step~{}2}$: Step 1 provides a starting point, from which we can now
move the plane $T_{\lambda}$ to the right as long as (72) holds to its
limiting position. Define

$\lambda_{0}=\sup\\{\lambda|w_{\mu}(x)>0,\forall
x\in\Sigma_{\mu},\mu\leq\lambda\\}.$

It’s obviously that $\lambda_{0}<+\infty$ and

$w_{\lambda_{0}}(x)\geq 0,~{}~{}~{}x\in\Sigma_{\lambda_{0}}.$

We will show that $w_{\lambda_{0}}(x)\equiv 0$ for $x\in\Sigma_{\lambda_{0}}$.

Otherwise if $w_{\lambda_{0}}\geq 0$ and $w_{\lambda_{0}}\not\equiv 0$, we
must have

(73) $\displaystyle w_{\lambda_{0}}(x)>0,~{}~{}~{}x\in\Sigma_{\lambda_{0}},$

where (73) follows from (71). We can derive that there exists $R$ large enough
such that

(74) $(C_{0}-\zeta)_{+}\equiv 0,~{}~{}x\in\mathbb{R}^{n}\setminus B_{R}(0),$

due to $\lim\limits_{|x|\rightarrow\infty}\zeta(x)=0$ and $C_{0}<0$. Fixing
this $R$, we have there exists constant $\delta>0$ and $c>0$ such that

(75) $\displaystyle w_{\lambda_{0}}(x)\geq
c,~{}~{}~{}x\in\overline{\Sigma_{\lambda_{0}-\delta}\cap B_{R}(0)}.$

Therefore by the continuity of $w_{\lambda}$ in $\lambda$ there exists
$\varepsilon>0$ and $\varepsilon<\delta$ such that for all
$\lambda\in[\lambda_{0},\lambda_{0}+\varepsilon)$, we have

$w_{\lambda}(x)\geq 0,~{}~{}~{}x\in\overline{\Sigma_{\lambda_{0}-\delta}\cap
B_{R}(0)}.$

We will show that for sufficiently small $0<\varepsilon<\delta$ and any
$\lambda\in[\lambda_{0},\lambda_{0}+\varepsilon)$

(76) $\displaystyle w_{\lambda}(x)\geq 0,~{}~{}~{}x\in\Sigma_{\lambda},$

which contradicts with the definition of $\lambda_{0}$. Therefore we must have
$w_{\lambda_{0}}\equiv 0$. Define

$\Sigma_{\lambda}^{-}=\\{x\in\Sigma_{\lambda}|w_{\lambda}(x)<0\\}.$

Next we claim that $\Sigma_{\lambda}^{-}$ must be measure zero.

For $y\in\Sigma_{\lambda}^{-}$, we can obtain that

(77)
$(C_{0}-\zeta)_{+}^{\gamma}(y)-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}(y)\leq\gamma(C_{0}-\zeta)_{+}^{\gamma-1}(y)|w_{\lambda}(y)|.$

Thus for $x\in\Sigma_{\lambda}$,

(78) $\displaystyle\begin{split}\zeta_{\lambda}(x)-\zeta(x)&\leq
R_{2p+\alpha,n}\int_{\Sigma_{\lambda}^{-}}\left(\frac{1}{|x-y|^{n-2p-\alpha}}-\frac{1}{|x-y^{\lambda}|^{n-2p-\alpha}}\right)\left((C_{0}-\zeta)_{+}^{\gamma}(y)-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}(y)\right)dy\\\
&\leq
R_{2p+\alpha,n}\int_{\Sigma_{\lambda}^{-}}\left(\frac{1}{|x-y|^{n-2p-\alpha}}-\frac{1}{|x-y^{\lambda}|^{n-2p-\alpha}}\right)\gamma(C_{0}-\zeta(y))_{+}^{\gamma-1}|w_{\lambda}(y)|dy.\end{split}$

Applying Hardy-Littlewood-Sobolev inequality [10] and Hölder inequality to
(78) we obtain that

(79)
$\displaystyle\begin{split}\|\zeta_{\lambda}(x)-\zeta(x)\|_{L^{\frac{2n}{n-2p-\alpha}}(\Sigma_{\lambda}^{-})}&\leq
C\left(\int_{\Sigma_{\lambda}^{-}}\left((C_{0}-\zeta)_{+}^{\gamma-1}(y)|w_{\lambda}(y)|\right)^{\frac{2n}{n+2p+\alpha}}dy\right)^{\frac{n+2p+\alpha}{2n}}\\\
&\leq
C\left(\int_{\Sigma_{\lambda}^{-}}\left((C_{0}-\zeta)_{+}^{\gamma-1}(y)\right)^{\frac{n}{2p+\alpha}}dy\right)^{\frac{2p+\alpha}{n}}\left(\int_{\Sigma_{\lambda}^{-}}|w_{\lambda}(y)|^{\frac{2n}{n-2p-\alpha}}dy\right)^{\frac{n-2p-\alpha}{2n}}.\end{split}$

Recall that
$\Sigma_{\lambda}^{-}\subset((\Sigma_{\lambda}\setminus\Sigma_{\lambda_{0}-\delta})\cap
B_{R})\cup(\Sigma_{\lambda}\setminus B_{R})$ and $-\zeta$ is bounded above, we
can choose $\delta$ sufficiently small such that

$C\left(\int_{{\Sigma_{\lambda}^{-}}\cap
B_{R}}\left((C_{0}-\zeta)_{+}^{\gamma-1}(y)\right)^{\frac{n}{2p}}dy\right)^{\frac{2p}{n}}\leq\frac{1}{2}.$

From this and (74), we have

$C\left(\int_{\Sigma_{\lambda}^{-}}\left((C_{0}-\zeta)_{+}^{\gamma-1}(y)\right)^{\frac{n}{2p}}dy\right)^{\frac{2p}{n}}\leq\frac{1}{2}.$

Now (79) implies that
$\|w_{\lambda}\|_{L^{\frac{2n}{n-2p-\alpha}}(\Sigma_{\lambda}^{-})}=0$ and
therefore $\Sigma_{\lambda}^{-}$ must be measure zero.

This verifies (76). Thus we must have $w_{\lambda_{0}}\equiv 0$.

$\mathbf{Step~{}3}$: We show that $\frac{\partial\zeta}{\partial x_{1}}<0$ for
$x\in\Sigma_{\lambda_{0}}$.

In fact, from the definition of $\lambda_{0}$ we have for any
$\lambda<\lambda_{0}$,

(80) $w_{\lambda}(x)>0,~{}~{}~{}~{}~{}x\in\Sigma_{\lambda}.$

Simple calculation gives that for any $x\in T_{\lambda}$ with
$\lambda<\lambda_{0}$,

$\begin{split}\zeta_{x_{1}}(x)=&R_{2p+\alpha,n}(n-2p-\alpha)\int_{\mathbb{R}^{n}}\frac{(C_{0}-\zeta)_{+}^{\gamma}(y)(x_{1}-y_{1})}{|x-y|^{n-2p+1-\alpha}}dy\\\
=&R_{2p+\alpha,n}(n-2p-\alpha)\int_{\Sigma_{\lambda}}\frac{\left((C_{0}-\zeta)_{+}^{\gamma}(y)-(C_{0}-\zeta_{\lambda})_{+}^{\gamma}(y)\right)(x_{1}-y_{1})}{|x-y|^{n-2p+1}}dy\\\
<&0,\end{split}$

where the last inequality follows from (80). Thus the claim holds.

Since the problem is invariant with respect to rotation, we can take any
direction as the $x_{1}$ direction. Hence we have that $\zeta$ is radially
symmetric about some $x_{0}\in\mathbb{R}^{n}$ and
$\frac{\partial\zeta}{\partial r}>0$ where $r=|x-x_{0}|$. $\hfill\square$

Actually, we may also prove Theorem 3 by applying moving plane method to the
function $(-\Delta)^{p-1+\frac{\alpha}{2}}u$, after asymptotic behaviors at
infinity of this function and its first-order derivatives are established.

## References

  * [1] C. Bucur. Some observations on the Green function for the ball in the frational Laplace framework. Commun. Pure Appl. Anal. 15(2016), no.2, 657-699.
  * [2] L. Caffarelli, B. Gidas and J. Spruck. Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth. Comm. Pure Appl. Math. 42(1989), no.3, 271-297.
  * [3] D. Cao, W. Dai and G. Qin. Super poly-harmonic properties, Liouville Theorems and classification of nonnegative solutions to equations involving higer-order fractional Laplacians. T Am Math Soc. 374(2021), no.7: 4781-4813.
  * [4] R. Chammakhi, A. Harrabi and A. Selmi. A classification of solutions of a fourth order semi-linear elliptic equation in ${\mathbb{R}}^{n}$. Differ. Integral Equ. 30(2017), no.7-8, 569-586.
  * [5] S. Chang and P. Yang. On uniqueness of solutions of nth order differential equations in conformal geometry. Math. Res. Lett. 4 (1997), no. 1, 91-102.
  * [6] W. Chen and C. Li. A priori estimates for prescribing scalar curvature equations. Ann. of Math. 145(1997), no.3, 547-564.
  * [7] W. Chen, C. Li and Y. Li. A direct method of moving planes for the fractional Laplacian. Adv. Math. 308(2016), 404-437.
  * [8] W. Chen, C. Li and B. Ou. Classification of solutions for an integral equation. Comm. Pure Appl. Math. 59(2006), no.3, 330-343.
  * [9] W. Dai and G. Qin, Classification of positive smooth solutions to third-order PDEs involving fractional Laplacians, Pacific J. Math. 295 (2018), no. 2, 367-383.
  * [10] W. Dai and G. Qin. Classification of nonnegative classical solutions to third-order equations. Adv. Math. 328(2018), 822-857.
  * [11] Z. Du, Z. Feng and Y. Li. Classification of solutions to several semi-linear polyharmonic equations and fractional equations. Arxiv:2201.00917.
  * [12] B. Gidas, W. Ni and L. Nirenberg. Symmetry of positive solutions of nonlinear elliptic equations in ${\mathbb{R}}^{n}$. Adv. in Math. Suppl. Stud. 1981, 369-402.
  * [13] C. Lin. A classification of solutions of a conformally invariant fourth order equation in ${\mathbb{R}}^{n}$. Comment. Math. Helv. 73(1998), no.2, 206-231.
  * [14] L. Silvestre. Regularity of the obstacle problem for a fractional power of Laplace operator. Comm. Pure Appl. Math. 60(2007), no. 1, 67-112.
  * [15] T. Suzuki and R. Takahashi. Critical blow up exponent to a class of semilinear elliptic equations with constraints in higher dimension-local properties. Ann. Mat. Pura Appl. 195(2016), no.4, 1123-1151.
  * [16] M. Stein. Singular integrals and differentiability properties of functions. Princeton Mathematical Series. No. 30, Princeton University Press, Princeton, N.J., 1970, MR0290095.
  * [17] G. Wang and D. Ye. On a nonlinear elliptic equation arising in a free boundary problem. Math. Z. 244(2003), no.3, 531-548.
  * [18] J. Wei and X. Xu. Classification of solutions of higher order conformally invariant equations. Math. Ann. 313(1999), no.2, 207-228.
  * [19] C. Yang, M. Ge and Y. He (eds.). Topology and physics, World Scientific Publishing Co. Pte. Ltd. Hackenensack, NJ, (2019), MR2196360.