# On the Hang–Yang conjecture for GJMS equations on $\mathbb{S}^{n}$

Ali Hyder  TIFR Centre for Applicable Mathematics, Sharadanagar, Bangalore
560065, India<EMAIL_ADDRESS>and  Quốc Anh Ngô  University of Science,
Vietnam National University, Hanoi, Vietnam
ORCID iD: 0000-0002-3550-9689<EMAIL_ADDRESS>
(Date:  at )

###### Abstract.

This work concerns a Liouville type result for positive, smooth solution $v$
to the following higher-order equation

$\,{\mathbf{P}}^{2m}_{n}(v)=\frac{n-2m}{2}Q_{n}^{2m}(\varepsilon
v+v^{-\alpha})$

on $\mathbb{S}^{n}$ with $m\geq 2$, $3\leq n<2m$, $0<\alpha\leq(2m+n)/(2m-n)$,
and $\varepsilon>0$. Here $\,{\mathbf{P}}^{2m}_{n}$ is the GJMS operator of
order $2m$ on $\mathbb{S}^{n}$ and
$Q_{n}^{2m}=(2/(n-2m))\,{\mathbf{P}}^{2m}_{n}(1)$ is constant. We show that if
$\varepsilon>0$ is small and $0<\alpha\leq(2m+n)/(2m-n)$, then any positive,
smooth solution $v$ to the above equation must be constant. The same result
remains valid if $\varepsilon=0$ and $0<\alpha<(2m+n)/(2m-n)$. In the special
case $n=3$, $m=2$, and $\alpha=7$, such Liouville type result was recently
conjectured by F. Hang and P. Yang (Int. Math. Res. Not. IMRN, 2020). As a by-
product, we obtain the sharp (subcritical and critical) Sobolev inequalities

$\Big{(}\int_{\mathbb{S}^{n}}v^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}v\,{\mathbf{P}}^{2m}_{n}(v)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}$

for the GJMS operator $\,{\mathbf{P}}^{2m}_{n}$ on $\mathbb{S}^{n}$ under the
conditions $n\geq 3$, $n=2m-1$, and $\alpha\in(0,1)\cup(1,2n+1]$. A log-
Sobolev type inequality, as the limiting case $\alpha=1$, is also presented.

###### Key words and phrases:

GJMS operator; Sobolev inequality; moving plane method; compactness result,
Liouville result

###### 2000 Mathematics Subject Classification:

53C18, 58J05, 35A23, 26D15

##  1\. Introduction

Let $n\geq 3$ be an odd integer, $2m>n$, and $0<\alpha\leq(n+2m)/(2m-n)$. In
this work, we consider the following equation

$\displaystyle\,{\mathbf{P}}^{2m}_{n}(v)=\frac{n-2m}{2}Q_{n}^{2m}(\varepsilon
v+v^{-\alpha})\quad\text{in }\mathbb{S}^{n}.$ (1.1)ε

Here $\,{\mathbf{P}}^{2m}_{n}$ is the well-known GJMS operator on
$\mathbb{S}^{n}$ equipped with the standard metric $g_{\mathbb{S}^{n}}$, which
is given as follows

$\,{\mathbf{P}}^{2m}_{n}:=\prod_{i=0}^{m-1}\left(-\Delta_{g_{\mathbb{S}^{n}}}-(i+\frac{n}{2})(i-\frac{n}{2}+1)\right),$

see [GJMS92], and

$Q_{n}^{2m}:=\frac{2}{n-2m}\,{\mathbf{P}}^{2m}_{n}(1)=\frac{2}{n-2m}\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}$

is a non-zero constant representing the so-called $Q$-curvature of
$(\mathbb{S}^{n},g_{\mathbb{S}^{n}})$, namely

$\,{\mathbf{P}}^{2m}_{n}=(-\Delta_{g_{\mathbb{S}^{n}}})^{m}+\sum_{1\leq k\leq
m-1}c_{k}(-\Delta_{g_{\mathbb{S}^{n}}})^{k}+\frac{n-2m}{2}Q_{n}^{2m}$

for suitable constants $c_{k}$ with $1\leq k\leq m-1$. A special case of the
operator $\,{\mathbf{P}}^{2m}_{n}$, which has often been studied over the last
two decades, is the well-known Paneitz operator, which is of fourth order.
This example of a higher-order conformal operator gains interest because of
its role in conformal geometry; see [CGY02, HY16]. On
$(\mathbb{S}^{3},g_{\mathbb{S}^{3}})$, the Paneitz operator is given by

$\,{\mathbf{P}}^{4}_{3}=\Delta_{g_{\mathbb{S}^{3}}}^{2}+\frac{1}{2}\Delta_{g_{\mathbb{S}^{3}}}-\frac{15}{16},$

and therefore $Q_{3}^{4}=-2\Gamma(7/2)/\Gamma(-1/2)=15/8$. Using the above
recursive formula for $\,{\mathbf{P}}^{2m}_{n}$ we can compute higher
dimensional cases, for example

$\mathbf{P}_{3}^{6}=-\Delta_{g_{\mathbb{S}^{n}}}^{3}-\frac{23}{4}\Delta_{g_{\mathbb{S}^{n}}}^{2}-\frac{27}{16}\Delta_{g_{\mathbb{S}^{n}}}+\frac{315}{64}\quad\text{on
}(\mathbb{S}^{3},g_{\mathbb{S}^{3}})$

with $Q_{3}^{6}=-105/32$ and

$\mathbf{P}_{5}^{6}=-\Delta_{g_{\mathbb{S}^{n}}}^{3}+\frac{13}{4}\Delta_{g_{\mathbb{S}^{n}}}^{2}+\frac{93}{16}\Delta_{g_{\mathbb{S}^{n}}}-\frac{945}{64}\quad\text{on
}(\mathbb{S}^{5},g_{\mathbb{S}^{5}})$

with $Q_{5}^{6}=945/32$. One should pay attention on the sign difference of
$Q_{3}^{6}$ and $Q_{5}^{6}$.

Our motivation of working on the equation (1.1)ε traces back to a recent
conjecture by F. Hang and P. Yang in [HY20] that we are going to describe now.
This conjecture concerns the following sharp critical Sobolev inequality on
$\mathbb{S}^{3}$

$\|\phi^{-1}\|_{L^{6}(\mathbb{S}^{3})}^{2}\int_{\mathbb{S}^{3}}\Big{[}(\Delta_{g_{\mathbb{S}^{3}}}\phi)^{2}-\frac{1}{2}|\nabla_{g_{\mathbb{S}^{3}}}\phi|^{2}-\frac{15}{16}\phi^{2}\Big{]}d\mu_{\mathbb{S}^{3}}\geq-\frac{15}{16}|\mathbb{S}^{3}|^{4/3}$
(1.2)

for any $\phi\in H^{2}(\mathbb{S}^{3})$ with $\phi>0$, which was already
proved in [YZ04] by symmetrization argument and in [HY04] by variational
argument. Apparently, the inequality (1.2) can be rewritten as follows

$\|\phi^{-1}\|_{L^{6}(\mathbb{S}^{3})}^{2}\int_{\mathbb{S}^{3}}\phi\,{\mathbf{P}}^{4}_{3}(\phi)d\mu_{\mathbb{S}^{3}}\geq-\frac{15}{16}|\mathbb{S}^{3}|^{4/3}$
(1.3)

for any $0<\phi\in H^{2}(\mathbb{S}^{3})$, because the integral in (1.2) is
nothing but
$\int_{\mathbb{S}^{3}}\phi\,{\mathbf{P}}^{4}_{3}(\phi)d\mu_{\mathbb{S}^{3}}$.
In (1.3) and what follows, $|\mathbb{S}^{n}|$ denotes the surface area of
$\mathbb{S}^{n}$. Besides, by Morrey’s theorem, functions in
$H^{2}(\mathbb{S}^{3})$ are continuous and therefore the condition $\phi>0$ is
understood in pointwise sense. By direct calculation, one can easily verify
that equality in (1.3) occurs if $\phi$ is any positive constant. This tells
us that the Paneitz operator $\,{\mathbf{P}}^{4}_{3}$ on the standard sphere
$\mathbb{S}^{3}$ is no longer positive; see [XY02] for the assumption on the
positivity of the Paneitz operator on closed $3$-manifolds.

In an effort to provide a new proof for (1.3) with the sharp constant, the
authors in [HY20] propose a new way to prove the above Sobolev inequality by
considering the following minimizing problem

$\inf_{0<\phi\in
H^{2}(\mathbb{S}^{3})}\|\phi^{-1}\|_{L^{6}(\mathbb{S}^{3})}^{2}\Big{[}\int_{\mathbb{S}^{3}}\phi\,{\mathbf{P}}^{4}_{3}(\phi)d\mu_{\mathbb{S}^{3}}+\varepsilon\int_{\mathbb{S}^{3}}\phi^{2}d\mu_{\mathbb{S}^{3}}\Big{]}$
(1.4)

for small $\varepsilon>0$. Thanks to the small perturbation
$\varepsilon\|\phi\|_{L^{2}(\mathbb{S}^{3})}^{2}$, it is standard and
straightforward to verify that the extremal problem (1.4) has a minimizer.
Such a minimizer, denoted by $v_{\varepsilon}$, eventually solves

$\,{\mathbf{P}}^{4}_{3}(v_{\varepsilon})+\varepsilon
v_{\varepsilon}=-v_{\varepsilon}^{-7}$

on $\mathbb{S}^{3}$, up to a constant. Here is the key observation: if the
above equation only admits constant solution for small $\varepsilon>0$, namely
$v_{\varepsilon}\equiv\text{const.}$, then one immediately has

$\|\phi^{-1}\|_{L^{6}(\mathbb{S}^{3})}^{2}\Big{[}\int_{\mathbb{S}^{3}}\phi\,{\mathbf{P}}^{4}_{3}(\phi)d\mu_{\mathbb{S}^{3}}+\varepsilon\int_{\mathbb{S}^{3}}\phi^{2}d\mu_{\mathbb{S}^{3}}\Big{]}\geq|\mathbb{S}^{3}|^{1/3}\Big{[}\int_{\mathbb{S}^{3}}\,{\mathbf{P}}^{4}_{3}(1)d\mu_{\mathbb{S}^{3}}+\varepsilon|\mathbb{S}^{3}|\Big{]}$

for any $0<\phi\in H^{2}(\mathbb{S}^{3})$. Having this and as
$\,{\mathbf{P}}^{4}_{3}(1)=-(1/2)Q_{3}^{4}=-15/16$, letting
$\varepsilon\searrow 0$ yields (1.3). The novelty of this new approach is that
it automatically implies the sharp form of (1.3) with the precise sharp
constant.

The above observation leads Hang and Yang to propose the following conjecture.

The Hang–Yang conjecture ([HY20, page 3299]). Let $\varepsilon>0$ be a small
number. If $v$ is a positive smooth solution to

$\,{\mathbf{P}}^{4}_{3}(v)+\varepsilon v=-v^{-7}$

on $\mathbb{S}^{3}$, then $v$ must be a constant function.

In a recent work Zhang [Zha21] provides an affirmative answer to the above
conjecture. The idea behind Zhang’s proof is first to transfer the
differential equation on $\mathbb{S}^{3}$ to some differential equation on
$\mathbf{R}^{3}$ and then to classify solutions to that equation on
$\mathbf{R}^{3}$. More precisely, let
$\pi_{N}:\mathbb{S}^{3}\to\mathbf{R}^{3}$ be the stereographic projection from
the north pole $N$; see subsection 2.1 below. The pullback
$(\pi_{N}^{-1})^{*}$ enjoys

$(\pi_{N}^{-1})^{*}(g_{\mathbb{S}^{3}})=\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2}dx^{2}$

and for any smooth solution $v$ on $\mathbb{S}^{3}$ there holds

$\,{\mathbf{P}}^{4}_{3}(v)\circ\pi_{N}^{-1}=\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-7/2}\Delta^{2}\Big{(}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-1/2}v\circ\pi_{N}^{-1}\Big{)}.$

(Here and in the sequel, $\Delta$ is the usual Laplacian on Euclidean spaces.)
Setting

$u(x):=\big{(}\frac{1+|x|^{2}}{2}\big{)}^{1/2}\big{(}v\circ\pi_{N}^{-1}\big{)}(x),$
(1.5)

we see that if $v$ solves $\,{\mathbf{P}}^{4}_{3}(v)+\varepsilon v=-v^{-7}$ in
$\mathbb{S}^{3}$, then $u$ solves

$\Delta^{2}u(x)+\varepsilon\big{(}\frac{2}{1+|x|^{2}}\big{)}^{4}u(x)=-u^{-7}(x)\quad\text{in
}\mathbf{R}^{3}.$

Via a dedicated argument based on the method of moving planes and techniques
from potential theory, which are rather involved, it is proved that $u$ is
radially symmetric. Finally, with the help of a Kazdan–Warner type identity,
the function $v$ must be constant.

Inspired by the work of Zhang described above, we are interested in
Hang–Yang’s conjecture in higher dimensional cases, namely we want to seek for
a suitable Liouville type result for positive, smooth solution to equations
involving GJMS operators. This leads us to investigate solutions to (1.1)ε.
Very similar to situation studied by Hang and Yang, our motivation to study
the equation (1.1)ε comes from the higher-order sharp critical Sobolev
inequality; see Theorem 1.2 below. Using the perturbation approach introduced
in [HY20], we are able to establish a Liouville type result for solutions to
(1.1)ε.

Toward a suitable Liouville type result, let us first describe some
preliminary results on (1.1)ε. Our first observation concerns the admissible
range for $\varepsilon$. As the perturbation approach is being used, we
require the condition $\varepsilon\geq 0$; see the proof of Lemma 5.1. Now, by
integrating both sides of (1.1)ε over $\mathbb{S}^{n}$ and as $2m-n>0$ and
$Q_{n}^{2m}\neq 0$ we conclude that

$(1-\varepsilon)\int_{\mathbb{S}^{n}}vd\mu_{\mathbb{S}^{n}}=\int_{\mathbb{S}^{n}}v^{-\alpha}d\mu_{\mathbb{S}^{n}}.$

This immediately tells us that $\varepsilon<1$. Thus, the admissible range for
$\varepsilon$ is $0\leq\varepsilon<1$. Having this, let us now state the main
result of this paper.

######  Theorem 1.1.

Let $n\geq 3$ be odd and $m>n/2$. Then there exists $\varepsilon_{*}\in(0,1)$
such that under one of the following conditions

  1. (1)

either $\varepsilon\in(0,\varepsilon_{*})$ and $0<\alpha\leq(n+2m)/(2m-n)$

  2. (2)

or $\varepsilon=0$ and $0<\alpha<(n+2m)/(2m-n)$

any positive, smooth solution to (1.1)ε must be constant.

We have the following remarks:

  * •

The above result again confirms the Hang–Yang conjecture for the Paneitz
operator on $\mathbb{S}^{3}$, and generalizes the result of Zhang in the
critical setting in higher dimensional cases.

  * •

Theorem 1.1 can be compared with the Liouville type results obtained by
Bidaut-Véron and Véron in [BVV91, Theorem 6.1] for the Emden equation, see
also the work of Gidas and Spruck in [GS81]. Note that the condition
$\alpha<(n+2m)/(2m-n)$ is sharp for $\varepsilon=0$ as the result does not
hold if $\alpha=(n+2m)/(2m-n)$. This is because in this limiting case the
equation (1.1)0 is conformally invariant; see section 3.

  * •

The threshold $\varepsilon_{*}$ is given in Lemma 4.3.

  * •

Although for any $0\leq\varepsilon<1$, equation (1.1)ε always admits the
trivial solution $v_{\varepsilon}\equiv(1-\varepsilon)^{-1/(\alpha+1)}$, but
it is not clear whether or not the above Liouville type result still holds for
$\varepsilon\in[\varepsilon_{*},1)$. This seems to be an interesting open
question.

To prove Theorem 1.1, we adopt the strategy used by Zhang. Such strategy can
be formulated as the following two main steps: first to transfer (1.1)ε in
$\mathbb{S}^{n}$ to the equation (1.8)ε and the corresponding integral
equation in $\mathbf{R}^{n}$, then to study symmetry properties of solutions
to these equations for small $\varepsilon>0$. However, to be able to handle
higher-order cases, our approach is significantly different from Zhang. One
major reason is that less results is known for the higher-order cases compared
to the case $m=2$. For example, we do not know if the preliminary results of
Hang and Yang mentioned in [Zha21, section 2] are available for $m\geq 3$.
Because of this difficulty, instead of the differential equation (1.8)ε, we
mainly work on the corresponding integral equation on $\mathbf{R}^{n}$, and
directly prove compactness results and symmetry properties of solutions. As
pays off, our analysis is much simpler, and could handle higher-order cases
efficiently.

As the operator $\,{\mathbf{P}}^{2m}_{n}$ is conformally covariant, for any
smooth function $\varphi$ on $\mathbb{S}^{n}$ we have the following identity
($\pi$ denotes the stereographic projection from $\mathbb{S}^{n}$ to
$\mathbf{R}^{n}$ with respect to either the north or the south pole)

$\,{\mathbf{P}}^{2m}_{n}(\varphi)\circ\pi^{-1}=\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-\frac{n+2m}{2}}(-\Delta)^{m}\Big{(}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\frac{n-2m}{2}}\varphi\circ\pi^{-1}\Big{)};$

see e.g. [Han07, Section 2]. Then, similar to (1.5), by setting

$\displaystyle
u(x):=\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\frac{n-2m}{2}}\big{(}v\circ\pi^{-1}\big{)}$
(1.6)

and

$\displaystyle
F_{\varepsilon,u}(x):=\varepsilon\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}u(x)+\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\frac{n+2m}{2}+\alpha\frac{n-2m}{2}}u(x)^{-\alpha}$
(1.7)

we see that $u$ satisfies

$\displaystyle(-\Delta)^{m}u=\frac{n-2m}{2}Q_{n}^{2m}F_{\varepsilon,u}\quad\text{in
}\mathbf{R}^{n}.$ (1.8)ε

In view of (1.6), we know that the function $u$ on $\mathbf{R}^{n}$ has exact
growth $|x|^{2m-n}$ at infinity. This additional information allows us to
transfer the differential equation (1.8)ε into the following integral equation

$u(x)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{\varepsilon,u}(y)dy\quad\text{on
}\mathbf{R}^{n}$

for some constant $\gamma_{2m,n}>0$; see Theorem 2.2 below. Notice that in
general there might be more solutions to (1.8)ε than the above integral
equation, see e.g. [HW19] and [DN22].

Let us emphasize that transferring to an equivalent integral equation on
$\mathbf{R}^{n}$ also appears in the work of Zhang, but the proof provided in
[Zha21] does not seem to work in our case. Similar integral representation in
the fractional setting also appears in [FKT22]. In our work, by exploiting
some nice structures on $\mathbb{S}^{n}$ as well as some intriguing properties
of the stereographic projection, we offer a completely new argument, which is
surprisingly simpler; see section 2.

Having the above integral equation in hand, we use a variant of the method of
moving planes in the integral form to show that any positive smooth solution
$u$ to the above integral equation with exact growth $|x|^{2m-n}$ at infinity
must be radially symmetric. The symmetry of solutions to the integral equation
helps us to conclude that the corresponding function $v$, appeared as in
(1.6), must be constant. The strategy we just describe seems to be very simple
and straightforward at the first glance, but there are two major difficulties
that we want to highlight. First, it is worth emphasizing that the method of
moving planes and its variants work well in the case of equations with
positive exponents; unfortunately, our equations, both differential and
integral forms, have a negative exponent. Second, by analyzing the form of
$F_{\varepsilon,u}$ in (1.7), one immediately notices that because of our
special choice of perturbation, there are two powers of $u$, whose exponents
have opposite sign. Unless $\varepsilon=0$, otherwise to run the method of
moving planes, one needs to establish certain compactness result for solutions
to (1.1)ε for suitable small $\varepsilon$, which costs us some energy.

Concerning classification of solutions to (1.8)ε with $\varepsilon=0$ and with
the RHS depending only on $u$, that is equation of the form
$(-\Delta)^{m}u=cu^{-\alpha}$ we refer to [HW19, Ngo18, Li04] and the
references therein.

Finally, to illustrate our finding on a Liouville type result for solutions to
(1.1)ε, we revisit the sharp critical Sobolev inequality for
$\,{\mathbf{P}}^{2m}_{n}$ on $\mathbb{S}^{n}$ proved in [Han07]. In fact, we
offer both critical and subcritical inequalities at once.

######  Theorem 1.2.

Let $n\geq 3$ be an odd integer and $m=(n+1)/2$. Then, for any $\phi\in
H^{m}(\mathbb{S}^{n})$ with $\phi>0$ and any $\alpha\in(0,1)\cup(1,2n+1]$, we
have the following sharp Sobolev inequality

$\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}.$
(1.9)

Moreover, the equality occurs if $\phi$ is any positive constant.

Let us have some comments on Theorem 1.1 above.

######  Remark 1.3.

  * •

Although the condition $n=2m-1$ is not required in Theorem 1.1, but in our
proof of (1.9) we heavily use it as in this case we have the advantage of
$Q$-curvature $Q^{2m}_{n}$ being positive. In general, the inequality (1.9) is
not true for $n<2m-3$, see e.g. [FKT22].

  * •

Apparently, by chosing $\alpha=(n+2m)/(2m-n)=2n+1$, our inequality (1.9)
includes the following critical Sobolev inequality

$\Big{(}\int_{\mathbb{S}^{n}}\phi^{-\frac{2n}{2m-n}}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2m-n}{n}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}|\mathbb{S}^{n}|^{\frac{2m}{n}},$
(1.10)

which was already proved in [Han07], see also [HY04] and [FKT22].

  * •

The case $\alpha=1$ is excluded in Theorem 1.2 due to the presence of the term
$1/(\alpha-1)$. For $\alpha=1$, by a limiting argument one obtains the
inequality (1.11) below.

  * •

Our last comment concerns the order of the inequality (1.9) as $\alpha$
varies. It turns out that the subcritical case $0<\alpha<2n+1$ can be obtained
from the critical case $\alpha=2n+1$, see Section 5 for more details.

Note that our inequality (1.9) can be rewritten as

$\Big{(}\fint_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\fint_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)},$

where $\fint_{\mathbb{S}^{n}}:=|\mathbb{S}^{n}|^{-1}\int_{\mathbb{S}^{n}}$
denotes the average. Using this new form one can easily compute the limit as
$\alpha\searrow 1$ to obtain an inequality in the limiting case as shown in
the following corollary.

######  Corollary 1.4.

Let $n\geq 3$ be an odd integer and $m=(n+1)/2$. Then, for any $\phi\in
H^{m}(\mathbb{S}^{n})$ with $\phi>0$, we have the following sharp Sobolev
inequality

$\exp\Big{(}-2\fint_{\mathbb{S}^{n}}\log\phi
d\mu_{\mathbb{S}^{n}}\Big{)}\fint_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}.$
(1.11)

Moreover, the equality occurs if $\phi$ is any positive constant.

It turns out that without using any limit process, one can still obtain (1.11)
directly from (1.9); see Proposition 5.2. As such, we omit the proof of
(1.11). Without using averages, (1.11) can be rewritten as follows

$\exp\Big{(}-\frac{2}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\phi
d\mu_{\mathbb{S}^{n}}\Big{)}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}|\mathbb{S}^{n}|.$

To the best of our knowledge, the above inequality (or the inequality (1.11))
seems to be new.

Our final comment concerns a possible generalization to the fractional
setting. Indeed, it seems that part of our argument can be quickly extended to
the case of fractional operators of order $2s>n$ instead of GJMS operators of
integer order $2m>n$. However, to maintain our work in a reasonable length, we
leave this future research.

Before closing this section, let us mention the organization of the paper.

###### Contents

  1. 1 Introduction
  2. 2 Some auxiliary results
    1. 2.1 Basics of the stereographic projection
    2. 2.2 From differential equations to integral equations
    3. 2.3 Pohozaev-type identity
  3. 3 Compactness results
  4. 4 Moving plane arguments and proof of the main result
  5. 5 Application to the sharp Sobolev inequality

##  2\. Some auxiliary results

###  2.1. Basics of the stereographic projection

As routine, we denote by $\pi_{N}$ and $\pi_{S}$ the stereographic projections
from the north pole $N$ and from the south pole $S$ of the sphere
$\mathbb{S}^{n}$ respectively. If we denote by $(x,x_{n+1})$ a general point
in $\mathbf{R}^{n+1}=\mathbf{R}^{n}\times\mathbf{R}$, then we have the
following expressions for $\pi_{N}$

$\pi_{N}(x,x_{n+1})=\frac{x}{1-x_{n+1}},\quad\pi_{N}^{-1}(x)=\Big{(}\frac{2x}{|x|^{2}+1},\frac{|x|^{2}-1}{|x|^{2}+1}\Big{)}.$

Likewise, we also have similar expressions for $\pi_{S}$. But these
expressions for $\pi_{S}$ can be derived quickly from those for $\pi_{N}$ by
changing the sign of the last coordinate. In this sense, we arrive at

$\pi_{S}(x,x_{n+1})=\frac{x}{1+x_{n+1}},\quad\pi_{S}^{-1}(x)=\Big{(}\frac{2x}{|x|^{2}+1},-\frac{|x|^{2}-1}{|x|^{2}+1}\Big{)}.$

The following observation plays some role in our analysis.

######  Lemma 2.1.

There holds

$\pi_{N}^{-1}(x)=\pi_{S}^{-1}\big{(}\frac{x}{|x|^{2}}\big{)},\quad\pi_{S}^{-1}(x)=\pi_{N}^{-1}\big{(}\frac{x}{|x|^{2}}\big{)}$

in $\mathbf{R}^{n}\setminus\\{0\\}$.

###### Proof.

These identities follows from the above expressions for $\pi_{N}$ and
$\pi_{S}$.

$\mathbf{R}^{n}$$x_{n+1}$$N$$x$$S$$\pi_{N}^{-1}(x)\equiv\pi_{S}(\frac{x}{|x|^{2}})$$\frac{x}{|x|^{2}}$
Figure 1. Relation between $\pi_{N}^{-1}$ and $\pi_{S}$.

We leave the details for interested readers; also see Figure 1 above. ∎

###  2.2. From differential equations to integral equations

Let $v$ be a positive, smooth solution to (1.1)ε. Recall from (1.8)ε that the
projected function $u$, defined by (1.6), solves

$(-\Delta)^{m}u=\frac{n-2m}{2}Q_{n}^{2m}F_{\varepsilon,u}\quad\text{in
}\mathbf{R}^{n}.$

The main result of this subsection is to show that $u$ actually solves the
corresponding integral equation (2.1). To achieve this goal, we need certain
preparation including the introduction of a uniform constant that we are going
to describe now.

Since $n$ is an odd integer, for some dimensional constant $c_{2m,n}\neq 0$ we
have

$(-\Delta)^{m}\big{(}c_{2m,n}|x|^{2m-n}\big{)}=\delta_{0},$

where $\delta_{0}$ is the Dirac measure at the origin. For convenience, we
also set

$\gamma_{2m,n}:=c_{2m,n}\frac{n-2m}{2}Q_{n}^{2m}.$

For simplicity, throughout the paper, we often denote by $C$ a generic
constant whose value could vary from estimate to estimate. We now state our
main result in this subsection.

######  Theorem 2.2.

We have

$\gamma_{2m,n}>0$

and

$\displaystyle
u(x)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{\varepsilon,u}(y)dy$
(2.1)

where $F_{\varepsilon,u}$ is given by (1.7).

Notice that the integral in (2.1) is well-defined everywhere in
$\mathbf{R}^{n}$. Indeed, as $v$ is positive everywhere on $\mathbb{S}^{n}$,
we have from (1.6) that $u(x)\approx|x|^{2m-n}$ for $|x|\gg 1$, and hence

$\displaystyle(1+|x|^{2m-n})F_{\varepsilon,u}(x)\leq\frac{C}{1+|x|^{2n}}.$
(2.2)

In order to prove the above theorem we define the following functions
associated with the projections $\pi_{N}$ and $\pi_{S}$:

$u_{N}(x):=\big{(}\frac{1+|x|^{2}}{2}\big{)}^{\frac{2m-n}{2}}(v\circ\pi_{N}^{-1})(x)$

and

$u_{S}(x):=\big{(}\frac{1+|x|^{2}}{2}\big{)}^{\frac{2m-n}{2}}(v\circ\pi_{S}^{-1})(x)$

in $\mathbf{R}^{n}$. In view of the integral equation (2.1), we denote

$\widetilde{u}_{N}(x):=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{\varepsilon,u_{N}}(y)dy$

and

$\widetilde{u}_{S}(x):=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{\varepsilon,u_{S}}(y)dy$

in $\mathbf{R}^{n}$. Our aim is to show that $u_{N}\equiv\widetilde{u}_{N}$
and that $\gamma_{2m,n}>0$. This will be done through several steps. Our first
observation is as follows.

######  Lemma 2.3.

We have

$u_{S}(x)=|x|^{2m-n}u_{N}\big{(}\frac{x}{|x|^{2}}\big{)},\quad
u_{N}(x)=|x|^{2m-n}u_{S}\big{(}\frac{x}{|x|^{2}}\big{)}$

in $\mathbf{R}^{n}\setminus\\{0\\}$.

###### Proof.

This is elementary. Indeed, let us compute $u_{S}$. Clearly, with the help of
Lemma 2.1, we have

$\displaystyle u_{S}(x)$
$\displaystyle=\big{(}\frac{1+|x|^{2}}{2}\big{)}^{\frac{2m-n}{2}}v\big{(}\pi_{S}^{-1}(x)\big{)}$
$\displaystyle=\big{(}\frac{1+|x|^{2}}{2}\big{)}^{\frac{2m-n}{2}}v\big{(}\pi_{N}^{-1}\big{(}\frac{x}{|x|^{2}}\big{)}\big{)}$
$\displaystyle=|x|^{2m-n}\big{(}\frac{1+|x/|x|^{2}|^{2}}{2}\big{)}^{\frac{2m-n}{2}}v\big{(}\pi_{N}^{-1}\big{(}\frac{x}{|x|^{2}}\big{)}\big{)}$
$\displaystyle=|x|^{2m-n}u_{N}\big{(}\frac{x}{|x|^{2}}\big{)},$

which gives the desired formula for $u_{S}$. The identity for $u_{N}$ can be
verified similarly. ∎

Our next observation is similar to that in Lemma 2.3.

######  Lemma 2.4.

We have

$\widetilde{u}_{S}(x)=|x|^{2m-n}\widetilde{u}_{N}\big{(}\frac{x}{|x|^{2}}\big{)},\quad\widetilde{u}_{N}(x)=|x|^{2m-n}\widetilde{u}_{S}\big{(}\frac{x}{|x|^{2}}\big{)}$

in $\mathbf{R}^{n}\setminus\\{0\\}$.

###### Proof.

This is also elementary but rather involved. Indeed, let us verify the first
identity. With a change of variable $y=z/|z|^{2}$ and by Lemma 2.3 we easily
get

$\displaystyle|x|^{2m-n}\widetilde{u}_{N}\big{(}\frac{x}{|x|^{2}}\big{)}$
$\displaystyle=\gamma_{2m,n}|x|^{2m-n}\int_{\mathbf{R}^{n}}\Big{|}\frac{x}{|x|^{2}}-y\Big{|}^{2m-n}F_{\varepsilon,u_{N}}(y)dy$
$\displaystyle=\gamma_{2m,n}|x|^{2m-n}\int_{\mathbf{R}^{n}}\Big{|}\frac{x}{|x|^{2}}-\frac{z}{|z|^{2}}\Big{|}^{2m-n}F_{\varepsilon,u_{N}}(\frac{z}{|z|^{2}})\frac{dz}{|z|^{2n}}$
$\displaystyle=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-z|^{2m-n}F_{\varepsilon,u_{S}}(z)dz$
$\displaystyle=\widetilde{u}_{S}(x),$

where in the second last equality we have used the following facts:

$\left|\frac{x}{|x|^{2}}-\frac{z}{|z|^{2}}\right|=\frac{|x-z|}{|x||z|},\quad
F_{\varepsilon,u_{N}}(\frac{z}{|z|^{2}})=|z|^{2m+n}F_{\varepsilon,u_{S}}(z).$

The second identity can be verified similarly. ∎

Now we are able to examine $u_{N}-\widetilde{u}_{N}$ and
$u_{S}-\widetilde{u}_{S}$.

######  Lemma 2.5.

The following functions

$P_{N}:=u_{N}-\widetilde{u}_{N},\quad P_{S}:=u_{S}-\widetilde{u}_{S}$

are polynomials in $\mathbf{R}^{n}$ of degree at most $2m-n$.

###### Proof.

Before proving, we see that both $P_{N}$ and $P_{S}$ are well-defined
everywhere in $\mathbf{R}^{n}$. Now it follows from (2.2) that the function
$\widetilde{u}_{N}$ satisfies

$\widetilde{u}_{N}(x)\leq C(1+|x|^{2m-n})\quad\text{for }x\in\mathbf{R}^{n}.$

This together with the growth of $u_{N}$ implies that $|P_{N}(x)|\leq
C(1+|x|^{2m-n})$. Since

$\Delta^{m}P_{N}=\Delta^{m}u_{N}-\Delta^{m}\widetilde{u}_{N}=0,$

we conclude that $P_{N}$ is a polynomial in $\mathbf{R}^{n}$ of degree at most
$2m-n$; see [Mar09, Theorem 5]. A similar argument applies to $P_{S}$ yielding
the same conclusion for $P_{S}$. ∎

Finally, we are in a position to prove Theorem 2.2, which simply follows from
the next two lemmas.

######  Lemma 2.6.

There hold $u_{N}\equiv\widetilde{u}_{N}$ and $u_{S}\equiv\widetilde{u}_{S}$
everywhere.

###### Proof.

As

$u_{S}(x)=|x|^{2m-n}u_{N}\big{(}\frac{x}{|x|^{2}}\big{)},\quad\widetilde{u}_{S}(x)=|x|^{2m-n}\widetilde{u}_{N}\big{(}\frac{x}{|x|^{2}}\big{)}$

we obtain

$\displaystyle P_{S}(x)=|x|^{2m-n}P_{N}\big{(}\frac{x}{|x|^{2}}\big{)},$

which is a polynomial (of degree at most $2m-n$). Surely, as $n$ is odd, this
is impossible because $|x|^{2m-n}$ cannot be a polynomial unless $P_{N}\equiv
P_{S}\equiv 0$, which implies that $u_{N}\equiv\widetilde{u}_{N}$ and
$u_{S}\equiv\widetilde{u}_{S}$. This completes the proof. ∎

######  Lemma 2.7.

There hold $\gamma_{2m,n}>0$.

###### Proof.

The claim $\gamma_{2m,n}>0$ follows trivially by seeing its definition

$\gamma_{2m,n}=c_{2m,n}Q_{n}^{2m}.$

Note that $Q_{n}^{2m}>0$ and that $c_{2m,n}>0$ because $n<2m$ and $n$ is odd.
However, the claim can also be seen from the fact that $v\equiv 1$ is a
solution to (1.1)0. More precisely, making use of $v\equiv 1$ and (1.6) one
has the following identity

$\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\frac{n-2m}{2}}=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}\big{(}\frac{2}{1+|y|^{2}}\big{)}^{\frac{n+2m}{2}}dy$

everywhere in $\mathbf{R}^{n}$. ∎

We conclude this subsection by noting that our approach to prove Theorem 2.2
can be used for the case of equations with positive exponent. For example,
without using any super polyharmonic property, as in [CLS22], our new approach
offers a very simple and straightforward proof to convert differential
equations on $\mathbb{S}^{n}$ to the corresponding integral equations on
$\mathbf{R}^{n}$, detail will appear elsewhere.

###  2.3. Pohozaev-type identity

Our last auxiliary result is a Pohozaev-type identity, which shall be used in
the proof of a compactness type result; see section 3 below. For simplicity,
we let

$c_{\alpha}:=\alpha\frac{2m-n}{2}-\frac{2m+n}{2}\leq 0.$ (2.3)

For future usage, let us state our Pohozaev-type identity in a more general
framework.

######  Lemma 2.8.

Let $Q\in C^{1}(\mathbf{R}^{n})$ be such that

$|Q(x)|\lesssim(1+|x|)^{-n+(\alpha-1)(2m-n)-\delta},$

for some $\delta>0$. Let $u$ be a positive, regular solution to

$\displaystyle u(x)=\int_{\mathbf{R}^{n}}|x-y|^{2m-n}Q(y)u^{-\alpha}(y)dy,$
(2.4)

where $u$ satisfies

$u\gtrsim(1+|x|)^{2m-n}\quad\text{if }\alpha>1$

and that

$u\lesssim(1+|x|)^{2m-n}\quad\text{if }0<\alpha<1.$

Then, for $\alpha\neq 1$, there holds

$\int_{\mathbf{R}^{n}}(x\cdot\nabla
Q)u^{1-\alpha}dx=c_{\alpha}\int_{\mathbf{R}^{n}}Qu^{1-\alpha}dx,$

provided $(x\cdot\nabla Q)u^{1-\alpha}\in L^{1}(\mathbf{R}^{n})$.

###### Proof.

The proof given below is more or less standard. As $x=(1/2)(x+y+x-y)$ and

$\nabla_{x}(|x-y|^{2m-n})=(2m-n)|x-y|^{2m-n-2}(x-y),$

by differentiating under the integral sign in (2.4), we obtain

$x\cdot\nabla
u(x)=\frac{2m-n}{2}u(x)+\frac{2m-n}{2}\int_{\mathbf{R}^{n}}\frac{|x|^{2}-|y|^{2}}{|x-y|^{n+2-2m}}Q(y)u^{-\alpha}(y)dy.$

Multiplying the above identity by $Q(x)u^{-\alpha}(x)$, and then integrating
the resultant on $B_{R}$ we arrive at

$\displaystyle\frac{1}{1-\alpha}\int_{B_{R}}Q$ $\displaystyle(x\cdot\nabla
u^{1-\alpha})dx=\frac{2m-n}{2}\int_{B_{R}}Qu^{1-\alpha}$
$\displaystyle+\frac{2m-n}{2}\int_{B_{R}}Q(x)u^{-\alpha}(x)\Big{(}\int_{\mathbf{R}^{n}}\frac{|x|^{2}-|y|^{2}}{|x-y|^{n+2-2m}}Q(y)u^{-\alpha}(y)dy\Big{)}dx.$

Integration by parts leads to

$\displaystyle\int_{B_{R}}Q(x\cdot\nabla u^{1-\alpha})dx=$
$\displaystyle-\int_{B_{R}}(x\cdot\nabla
Q)u^{1-\alpha}dx-n\int_{B_{R}}Qu^{1-\alpha}dx$ $\displaystyle+R\int_{\partial
B_{R}}Qu^{1-\alpha}dx.$

Hence,

$\displaystyle\frac{R}{1-\alpha}\int_{\partial B_{R}}$ $\displaystyle
Qu^{1-\alpha}d\sigma-\frac{2m-n}{2}\int_{B_{R}}\int_{\mathbf{R}^{n}}\frac{|x|^{2}-|y|^{2}}{|x-y|^{n+2-2m}}Q(y)u^{-\alpha}(y)Q(x)u^{-\alpha}(x)dydx$
(2.5)
$\displaystyle=\frac{1}{1-\alpha}\Big{[}\frac{(2m+n)-\alpha(2m-n)}{2}\int_{B_{R}}Qu^{1-\alpha}dx+\int_{B_{R}}(x\cdot\nabla
Q)u^{1-\alpha}dx\Big{]}.$

Thanks to the decay assumption on $Q$ and the growth of $u$, we easily get

$\lim_{R\to\infty}\Big{(}R\int_{\partial B_{R}}Qu^{1-\alpha}d\sigma\Big{)}=0,$

and clearly

$\int_{\mathbf{R}^{n}}\int_{\mathbf{R}^{n}}\frac{|x|^{2}-|y|^{2}}{|x-y|^{n+2-2m}}Q(y)u^{-\alpha}(y)Q(x)u^{-\alpha}(x)dydx=0$

due to the antisymmetry of the integrand. Furthermore, under the assumptions
on $Q$ and on $u$, there holds $Qu^{1-\alpha}\in L^{1}(\mathbf{R}^{n})$.
Hence, by sending $R\nearrow+\infty$, we conclude that the LHS of (2.5)
vanishes, giving the desired identity. This completes the proof. ∎

Let us now discuss how to use our Pohozaev-type identity in the current
setting. Recall that the solution $v$ to (1.1)ε is positive and smooth on
$\mathbb{S}^{n}$. Thanks to (1.6) we deduce that $u$ enjoys the upper and
lower growths as in Lemma 2.8. Hence, we have a Pohozaev-type identity for $u$
whenever $\alpha\neq 1$. We shall use this identity in the proof of Lemma 3.2
below.

##  3\. Compactness results

This section is devoted to a compactness type result for solutions to (1.1)ε,
which is of interest itself; see Theorem 3.1 below. Heuristically, one should
study the compactness result for fixed $\varepsilon$ and $\alpha$. However, to
derive useful estimates for our analysis, one needs certain compactness result
which is independent of $\varepsilon$; see the proof of Lemmas 4.2 and 4.3
below.

######  Theorem 3.1.

Let $\varepsilon^{*}\in(0,1)$ and $\alpha\in(0,(2m+n)/(2m-n)]$ be arbitrary
but fixed. Assume that $v_{k}=v_{\varepsilon_{k}}$ is a sequence of positive
regular solutions to (1.1)${}_{\varepsilon_{k}}$ for some
$\varepsilon_{k}\in(0,\varepsilon^{*})$. Then there exists
$C=C(\varepsilon^{*})>0$ such that

$\frac{1}{C}\leq v_{k}\leq C\quad\text{in }\mathbb{S}^{n}$

for all $k$. The same conclusion holds true for
$\varepsilon_{k}\in[0,\varepsilon^{*})$ if $\alpha\in(0,(2m+n)/(2m-n))$.

It is worth noting that the above compactness fails for solutions to (1.1)0 in
the case $\alpha=(n+2m)/(2m-n)$ due to the conformally invariant property of
the underlying equation. More specifically, fixing any solution $v$ to

$\,{\mathbf{P}}^{2m}_{n}(v)=\frac{n-2m}{2}Q_{n}^{2m}v^{\frac{n+2m}{2m-n}}\quad\text{in
}\mathbb{S}^{n}$

and let

$v_{\phi}=(v\circ\phi)|\det(d\phi)|^{-\frac{1}{2n}},$

where $\phi$ is any conformal transformation on $\mathbb{S}^{n}$. Then, it is
well-known that $v_{\phi}$ solves the same equation in $\mathbb{S}^{n}$.
Hence, if one choose a sequence of $\phi$ in such a way that
$|\det(d\phi)|\searrow 0$, then the sequence $v_{\phi}$ is unbounded in
$\mathbb{S}^{n}$.

In order to prove the above theorem we first need to rule out the possibility
that the sequence $v_{k}$ will eventually touch zero. This in particular
implies the lower estimate in the theorem.

######  Lemma 3.2.

Under the hypothesis of Theorem 3.1 we have

$\inf_{k\geq 1}\min_{\mathbb{S}^{n}}v_{k}>0.$

###### Proof.

We assume by contradiction that the lemma is false. Then, up to a subsequence,
we assume that

$\min_{\mathbb{S}^{3}}v_{k}\to 0\quad\text{as }k\to\infty.$

Without loss of generality we can further assume that the minimum of $v_{k}$
is attained at the south pole. Let $u_{k}$ be defined by (1.6) using
$\pi_{N}$, and let $F_{k}:=F_{\varepsilon_{k},u_{k}}$ as in (1.7). In view of
(1.6) and $2m>n$, the function $u_{k}$ achieves its minimum at $0$. By Theorem
2.2, the function $u_{k}$ satisfies

$\displaystyle
u_{k}(x)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{k}(y)dy.$ (3.1)

To show that this is also not the case, we use the Pohozaev-type identity of
Lemma 2.8 and the role played by $\varepsilon_{k}$ and $\alpha$. Indeed, as
$F_{k}>0$ we first obtain

$u_{k}(0)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|y|^{2m-n}F_{k}(y)dy=o(1)_{k\to\infty}.$
(3.2)

Using this one can show that

$\lim_{k\to\infty}u_{k}(x)=\infty\quad\text{for each
}x\in\mathbf{R}^{n}\setminus\\{0\\}.$ (3.3)

Indeed, by way of contradiction suppose that there is some
$x_{0}\in\mathbf{R}^{n}\setminus\\{0\\}$ such that
$u_{k}(x_{0})=O(1)_{k\to\infty}$. As

$\displaystyle\frac{u_{k}(x_{0})}{\gamma_{2m,n}}$
$\displaystyle=\int_{\mathbf{R}^{n}}|x_{0}-y|^{2m-n}F_{k}(y)dy$
$\displaystyle\geq
2^{-2m+n+1}\int_{\mathbf{R}^{n}}|x_{0}|^{2m-n}F_{k}(y)dy-\int_{\mathbf{R}^{n}}|y|^{2m-n}F_{k}(y)dy$

we obtain

$\int_{\mathbf{R}^{n}}F_{k}(y)dy=O(1)_{k\to\infty},$

thanks to $u_{k}(0)=O(1)_{k\to\infty}$. Hence

$\int_{\mathbf{R}^{n}}(1+|y|^{2m-n})F_{k}(y)dy=O(1)_{k\to\infty}.$ (3.4)

Consequently, for any $x\in\mathbf{R}^{n}$, one can estimate

$\displaystyle\frac{u_{k}(x)}{\gamma_{2m,n}}$
$\displaystyle=\int_{\mathbf{R}^{n}}|x-y|^{2m-n}F_{k}(y)dy\leq
2^{2m-n-1}\int_{\mathbf{R}^{n}}(|x|^{2m-n}+|y|^{2m-n})F_{k}(y)dy,$

which leads to

$u_{k}(x)\leq C(1+|x|^{2m-n})\quad\text{in }\mathbf{R}^{n}$

for some constant $C>0$. Having this, one can bound $F_{k}$ from below near
the origin. For example, for any $x\in B_{2}$, we easily get

$\displaystyle F_{k}(x)$
$\displaystyle\geq\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}u_{k}(x)^{-\alpha}\geq\frac{1}{C^{\alpha}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\frac{n+2m}{2}}\geq\frac{1}{C^{\alpha}}\Big{(}\frac{2}{5}\Big{)}^{\frac{n+2m}{2}},$

thanks to $u_{k}(x)\leq C(1+|x|^{2})^{(2m-n)/2}$ in $\mathbf{R}^{n}$. However,
this violates the fact that $u_{k}(0)=o(1)_{k\to\infty}$. Indeed,

$\displaystyle\frac{u_{k}(0)}{\gamma_{2m,n}}$
$\displaystyle\geq\int_{B_{2}\setminus
B_{1}}|y|^{2m-n}F_{k}(y)dy\geq\frac{1}{C^{\alpha}}\Big{(}\frac{2}{5}\Big{)}^{\frac{n+2m}{2}}\int_{B_{2}\setminus
B_{1}}|y|^{2m-n}dy>0$

for all $k$. Thus, no such a point $x_{0}$ could exist, and hence (3.3) must
hold. Notice that the above proof also reveals the fact that

$\lim_{k\to\infty}\int_{\mathbf{R}^{n}}F_{k}(y)dy=\infty,$ (3.5)

otherwise by (3.2) one would again have (3.4) and again this leads to a
contradiction. Now we normalize $u_{k}$ and $F_{k}$ as follows

$\widetilde{u}_{k}:=\frac{u_{k}}{\gamma_{2m,n}\int_{\mathbf{R}^{n}}F_{k}dy},\quad\widetilde{F}_{k}:=\frac{F_{k}}{\int_{\mathbf{R}^{n}}F_{k}dy}.$

Then

$\widetilde{u}_{k}(x)=\int_{\mathbf{R}^{n}}|x-y|^{2m-n}\widetilde{F}_{k}(y)dy,\quad\int_{\mathbf{R}^{n}}\widetilde{F}_{k}dy=1.$

Having (3.5), it is clear that $\widetilde{u}_{k}(0)\to 0$ and

$|\nabla\widetilde{u}_{k}(x)|\leq(2m-n)\int_{\mathbf{R}^{n}}|x-y|^{2m-n-1}\widetilde{F}_{k}(y)dy\leq
C(1+|x|^{2m-n-1})\quad\text{in }\mathbf{R}^{n}.$

Notice that because of (3.5) for large $k$ there holds
$\widetilde{F}_{k}(x)\leq F_{k}(x)$ everywhere. This and (3.2) now implies the
following

$\lim_{k\to\infty}\int_{\mathbf{R}^{n}\setminus
B_{\delta}}\widetilde{F}_{k}(y)dy\to 0\quad\text{for any fixed }\delta>0.$

Once we have the above limit in hand and seeing $\widetilde{u}$ as a
convolution, by standard argument, we get that

$\displaystyle\widetilde{u}_{k}\to\widetilde{u}:=|x|^{2m-n}\quad\text{in
}C^{0}_{\rm loc}(\mathbf{R}^{n})$ (3.6)

and at the same time

$\displaystyle\frac{1}{C}|x|^{2m-n}\leq\widetilde{u}_{k}\leq
C|x|^{2m-n}\quad\text{in }\mathbf{R}^{n}\setminus B_{1}$ (3.7)

for some $C>0$. Notice that we can write $F_{k}$ as

$F_{k}=\left(\varepsilon_{k}f^{2m}u_{k}^{1+\alpha}+f^{-c_{\alpha}}\right)u_{k}^{-\alpha}=:Q_{k}u_{k}^{-\alpha},$

where we denote

$f(x):=\frac{2}{1+|x|^{2}}.$

By the Pohozaev-type identity in Lemma 2.8, we get

$\displaystyle\int_{\mathbf{R}^{n}}(x\cdot\nabla
Q_{k})u_{k}^{1-\alpha}dx=c_{\alpha}\int_{\mathbf{R}^{n}}Q_{k}u_{k}^{1-\alpha}dx.$
(3.8)

(Here, the multiplicative constant $\gamma_{2m,n}\neq 0$ cancels out from the
both sides, thanks to Theorem 2.2.) Let us first compute

$\nabla\big{(}\varepsilon_{k}f^{2m}u_{k}^{1+\alpha}\big{)}=2m\varepsilon_{k}f^{2m-1}u_{k}^{1+\alpha}\nabla
f+\frac{1+\alpha}{2}\varepsilon_{k}f^{2m}u_{k}^{\alpha-1}\nabla u_{k}^{2}$

and

$\nabla(f^{-c_{\alpha}})=-c_{\alpha}f^{-c_{\alpha}-1}\nabla f,$

leading us to

$x\cdot\nabla
Q_{k}=\Big{[}\left(2m\varepsilon_{k}f^{2m-1}u_{k}^{2}-c_{\alpha}f^{-c_{\alpha}-1}u_{k}^{1-\alpha}\right)(x\cdot\nabla
f)+\frac{1+\alpha}{2}\varepsilon_{k}f^{2m}(x\cdot\nabla
u_{k}^{2})\Big{]}u_{k}^{\alpha-1}.$

Therefore, from (3.8) we get

$\displaystyle
c_{\alpha}\int_{\mathbf{R}^{n}}\big{[}\varepsilon_{k}f^{2m}u_{k}^{2}+f^{-c_{\alpha}}u_{k}^{1-\alpha}\big{]}dx$
$\displaystyle=\int_{\mathbf{R}^{n}}\big{[}2m\varepsilon_{k}f^{2m-1}u_{k}^{2}-c_{\alpha}f^{-c_{\alpha}-1}u_{k}^{1-\alpha}\big{]}(x\cdot\nabla
f)dx$
$\displaystyle\quad+\frac{1+\alpha}{2}\varepsilon_{k}\int_{\mathbf{R}^{n}}f^{2m}(x\cdot\nabla
u_{k}^{2})dx$
$\displaystyle=\int_{\mathbf{R}^{n}}m\varepsilon_{k}(1-\alpha)f^{2m-1}u_{k}^{2}(x\cdot\nabla
f)dx$
$\displaystyle\quad+\int_{\mathbf{R}^{n}}\varepsilon_{k}\frac{1+\alpha}{2}u_{k}^{2}(x\cdot\nabla
f^{2m})dx$ $\displaystyle\quad-
c_{\alpha}\int_{\mathbf{R}^{n}}f^{-c_{\alpha}-1}u_{k}^{1-\alpha}(x\cdot\nabla
f)dx$
$\displaystyle\quad+\frac{1+\alpha}{2}\varepsilon_{k}\int_{\mathbf{R}^{n}}f^{2m}(x\cdot\nabla
u_{k}^{2})dx.$

By integration by parts, we note that

$\displaystyle\int_{\mathbf{R}^{n}}\big{[}u_{k}^{2}(x\cdot\nabla f^{2m})$
$\displaystyle+f^{2m}(x\cdot\nabla u_{k}^{2})\big{]}dx$
$\displaystyle=\lim_{R\to\infty}\sum_{i=1}^{n}\Big{[}\int_{B_{R}}\big{[}-u_{k}^{2}f^{2m}\big{]}dx+\frac{1}{R}\int_{\partial
B_{R}}x_{i}^{2}f^{2m}u_{k}^{2}d\sigma\Big{]}$
$\displaystyle=\lim_{R\to\infty}\Big{[}-n\int_{B_{R}}u_{k}^{2}f^{2m}dx+R\int_{\partial
B_{R}}f^{2m}u_{k}^{2}d\sigma\Big{]}$
$\displaystyle=-n\int_{\mathbf{R}^{n}}f^{2m}u_{k}^{2}dx.$

Putting the above estimates together we arrive at

$\displaystyle\varepsilon_{k}\int_{\mathbf{R}^{n}}f^{2m-1}u_{k}^{2}\Big{[}m(1-\alpha)(x\cdot\nabla
f)$ $\displaystyle-\Big{(}\frac{n(1+\alpha)}{2}+c_{\alpha}\Big{)}f\Big{]}dx$
(3.9)
$\displaystyle=c_{\alpha}\int_{\mathbf{R}^{n}}f^{-c_{\alpha}-1}u_{k}^{1-\alpha}\left(x\cdot\nabla
f+f\right)dx.$

Since

$x\cdot\nabla f+f=f\frac{1-|x|^{2}}{1+|x|^{2}},$

and

$m(1-\alpha)+n\frac{1+\alpha}{2}+c_{\alpha}=0,$

the identity (3.9) can be rewritten as

$\displaystyle\varepsilon_{k}m(1-\alpha)\int_{\mathbf{R}^{n}}f^{2m}u_{k}^{2}\frac{1-|x|^{2}}{1+|x|^{2}}dx=c_{\alpha}\int_{\mathbf{R}^{n}}f^{-c_{\alpha}}u_{k}^{1-\alpha}\frac{1-|x|^{2}}{1+|x|^{2}}dx.$
(3.10)

Our next step is to show that for large $k$, the two integrals in (3.10) are
non-zero with different sign.

Estimate of the LHS of (3.10). Concerning the integral on the LHS of (3.10), a
simple calculation shows that

$\displaystyle\frac{1}{M_{k}^{2}}\int_{\mathbf{R}^{n}}f^{2m}(x)$
$\displaystyle u_{k}^{2}(x)\frac{1-|x|^{2}}{1+|x|^{2}}dx$
$\displaystyle=\int_{\mathbf{R}^{n}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}\widetilde{u}_{k}^{2}(x)\frac{1-|x|^{2}}{1+|x|^{2}}dx$
$\displaystyle=\int_{B_{1}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}\frac{1-|x|^{2}}{1+|x|^{2}}\Big{(}\widetilde{u}_{k}^{2}(x)-|x|^{4m-2n}\widetilde{u}_{k}^{2}\big{(}\frac{x}{|x|^{2}}\big{)}\Big{)}dx,$

here we have converted the integral on $\mathbf{R}^{n}\setminus B_{1}$ into
$B_{1}$ using Kelvin’s transformation. In $B_{1}\setminus\\{0\\}$, it follows
from (3.6) and (3.7) that

$\widetilde{u}_{k}^{2}(x)-|x|^{4m-2n}\widetilde{u}_{k}^{2}\big{(}\frac{x}{|x|^{2}}\big{)}\to|x|^{4m-2n}-1\leq
0\quad\text{as }k\to\infty.$

Notice that

$\displaystyle\lim_{k\to\infty}\int_{B_{1}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}$
$\displaystyle\frac{1-|x|^{2}}{1+|x|^{2}}\Big{(}\widetilde{u}_{k}^{2}(x)-|x|^{4m-2n}\widetilde{u}_{k}^{2}\big{(}\frac{x}{|x|^{2}}\big{)}\Big{)}dx$
$\displaystyle=\int_{B_{1}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}\frac{1-|x|^{2}}{1+|x|^{2}}\left(|x|^{4m-2n}-1\right)dx<0.$

This and $\varepsilon_{k}>0$ imply that the LHS of (3.10) is strictly negative
for large $k$.

Estimate of the RHS of (3.10). Reasoning as in the previous step we should
have

$\displaystyle\frac{1}{M_{k}^{1-\alpha}}$
$\displaystyle\int_{\mathbf{R}^{n}}f^{-c_{\alpha}}(x)u_{k}^{1-\alpha}(x)\frac{1-|x|^{2}}{1+|x|^{2}}dx$
$\displaystyle=\int_{B_{1}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}\frac{1-|x|^{2}}{1+|x|^{2}}\Big{(}\widetilde{u}_{k}^{1-\alpha}(x)-|x|^{-2c_{\alpha}-2n}\widetilde{u}_{k}^{1-\alpha}\big{(}\frac{x}{|x|^{2}}\big{)}\Big{)}dx.$

In $B_{1}$, it follows from (3.6) that

$\widetilde{u}_{k}^{1-\alpha}(x)-|x|^{-2c_{\alpha}-2n}\widetilde{u}_{k}^{1-\alpha}\big{(}\frac{x}{|x|^{2}}\big{)}\to|x|^{(2m-n)(1-\alpha)}-1\geq
0\quad\text{as }k\to\infty.$

Now observe that for $\alpha>1$

$\displaystyle\int_{B_{1}}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}\frac{1-|x|^{2}}{1+|x|^{2}}\big{(}|x|^{(2m-n)(1-\alpha)}-1\big{)}dx>0,$

which imply that the the RHS of (3.10) is strictly positive for large $k$ (for
certain $\alpha>1$, the preceding integral could be infinity). Now going back
to (3.10), we easily obtain a contradiction for $\alpha>1$. Indeed, we have
two possible cases. First, if $\varepsilon_{k}>0$ for large $k$, then as
$\varepsilon_{k}m(1-\alpha)<0$, the LHS of (3.10) becomes strictly positive.
However, as $c_{\alpha}\leq 0$, the RHS of (3.10) becomes non-positive. This
is a contradiction. In contrary, we have $\varepsilon_{k}=0$ for a sequence of
$k$. However, under $\varepsilon_{k}=0$ the LHS of (3.10) vanishes but as
$c_{\alpha}<0$ the RHS of (3.10) becomes strictly negative. This is again a
contradiction. And this completes our proof of the compactness for $\alpha>1$.

Finally we consider the case $0<\alpha\leq 1$. We set

$\eta_{k}(x):=\frac{u_{k}(r_{k}x)}{u_{k}(0)},\quad
r_{k}:=u_{k}(0)^{\frac{1+\alpha}{2m}}\to 0.$

Then $\eta_{k}$ satisfies $\eta_{k}\geq\eta_{k}(0)=1$, and

$\displaystyle\eta_{k}(x)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}|x-y|^{2m-n}\left(\varepsilon_{k}r_{k}^{2m}f^{2m}(r_{k}y)\eta_{k}(y)+\frac{f^{-c_{\alpha}}(r_{k}y)}{\eta_{k}^{\alpha}(y)}\right)dy.$
(3.11)

Then it follows that

$\displaystyle\int_{\mathbf{R}^{n}}|y|^{2m-n}\left(\varepsilon_{k}r_{k}^{2m}f^{2m}(r_{k}y)\eta_{k}(y)+\frac{f^{-c_{\alpha}}(r_{k}y)}{\eta_{k}^{\alpha}(y)}\right)dy=\frac{\eta_{k}(0)}{\gamma_{2m,n}}\leq
C,$ (3.12)

and together with $\eta_{k}\geq 1$,

$\displaystyle\int_{\mathbf{R}^{n}}\left(1+|y|^{2m-n}\right)\frac{f^{-c_{\alpha}}(r_{k}y)}{\eta_{k}^{\alpha}(y)}dy\leq
C.$ (3.13)

Therefore,

$\eta_{k}(x)=\gamma_{2m,n}\varepsilon_{k}r_{k}^{2m}\int_{B_{1}}|x-y|^{2m-n}f^{2m}(r_{k}y)\eta_{k}(y)dy+O(1)\quad\text{for
}x\in B_{1}.$

Integrating the above identity with respect to $x$ in $B_{1}$, and using that
$f(r_{k}y)=2+o(1)$ on $B_{1}$, we obtain

$\int_{B_{1}}\eta_{k}(x)dx=o(1)\int_{B_{1}}\eta_{k}(y)dy+O(1),$

and hence

$\int_{B_{1}}\eta_{k}dx\leq C.$

Combining the above estimates

$\displaystyle\int_{\mathbf{R}^{n}}\left(1+|y|^{2m-n}\right)\left(\varepsilon_{k}r_{k}^{2m}f^{2m}(r_{k}y)\eta_{k}(y)+\frac{f^{-c_{\alpha}}(r_{k}y)}{\eta_{k}^{\alpha}(y)}\right)dy\leq
C.$ (3.14)

This yields

$\displaystyle|\nabla\eta_{k}(x)|\leq
C(1+|x|^{2m-n-1}),\quad\frac{1}{C}\left(1+|x|^{2m-n}\right)\leq\eta_{k}(x)\leq
C\left(1+|x|^{2m-n}\right).$ (3.15)

Hence, up to a subsequence,

$\eta_{k}\to\eta\quad\text{in }C^{0}_{loc}(\mathbf{R}^{n}).$

From Fatou’s lemma, we get that

$\int_{\mathbf{R}^{n}}\frac{|y|^{2m-n}}{\eta^{\alpha}(y)}dy<\infty,$

thanks to (3.15). Since $\eta$ satisfies the second estimate in (3.15), we
necessarily have that

$\displaystyle(\alpha-1)(2m-n)>n,$

a contradiction to $0<\alpha\leq 1$. ∎

We are now in a position to prove Theorem 3.1.

###### Proof of Theorem 3.1.

Since $\varepsilon_{k}\in[0,\varepsilon^{*})$ and $0<\varepsilon^{*}<1$,
integrating (1.1) on $\mathbb{S}^{n}$ we get that

$0\leq\int_{\mathbb{S}^{n}}v_{k}d\mu_{\mathbb{S}^{n}}\leq\frac{1}{1-\varepsilon^{*}}\int_{\mathbb{S}^{n}}v_{k}^{-\alpha}d\mu_{\mathbb{S}^{n}}=O(1)_{k\to\infty},$

thanks to Lemma 3.2. Therefore, we arrive at

$\,{\mathbf{P}}^{2m}_{n}(v_{k})-\varepsilon_{k}\frac{n-2m}{2}Q_{n}^{2m}v_{k}=O(1)_{k\to\infty}\quad\text{in
}\mathbb{S}^{n}$

with $\|v_{k}\|_{L^{1}(\mathbb{S}^{n})}=O(1)_{k\to\infty}$. The theorem
follows from standard elliptic estimates. ∎

##  4\. Moving plane arguments and proof of the main result

This section is devoted to the proof of Theorem 1.1. To obtain the symmetry of
solutions, our approach is based on the method of moving planes with some new
ingredients. The major difficulty is how to handle the negative exponent. As
far as we know, although the method of moving planes can be effectively
applied to nonlinear equations with positive exponents, see [CL91, WX99,
CLO06, CLS22] and the references therein, its applications to equations with
negative exponents are very rare.

Let us recall some notation and convention often used in the method of moving
planes; see Figure 2 below. For $\lambda\in\mathbf{R}$ we set

$\Sigma_{\lambda}:=\\{x\in\mathbf{R}^{n}:x_{1}>\lambda\\},\quad
T_{\lambda}:=\partial\Sigma_{\lambda}.$

Also for any $\lambda\in\mathbf{R}$ we let $x^{\lambda}$ be the reflection of
$x\in\mathbf{R}^{n}$ about the plane $T_{\lambda}$, namely

$x^{\lambda}:=(2\lambda-x_{1},x_{2},x_{3},\dots,x_{n}).$

Also for any function $f$ we let $f_{\lambda}$ be the reflection of $f$ about
the plane $T_{\lambda}$, namely

$f_{\lambda}(x):=f(x^{\lambda})=f(2\lambda-x_{1},x_{2},x_{3},\dots,x_{n}).$

$x_{1}$$T_{\lambda}$$\mathbf{R}^{n-1}$$x_{1}^{\lambda}$$x^{\lambda}$$x$$x_{1}$$y$$y^{\lambda}$$\Sigma_{\lambda}$
Figure 2. Reflection in the method of moving planes

Throughout this section we let $u=u_{\varepsilon}>0$ be a (smooth) solution to
(2.1) with $F_{\varepsilon}:=F_{\varepsilon,u}$ as in (1.7) for fixed
$0<\alpha\leq(n+2m)/(2m-n)$ and fixed $0\leq\varepsilon<\varepsilon^{*}$ with
an additional assumption that $\alpha<(n+2m)/(2m-n)$ if $\varepsilon=0$. For
simplicity, we set

$w_{\varepsilon,\lambda}(x):=u_{\varepsilon}(x)-u_{\varepsilon}(x^{\lambda})\quad\text{for
all }x\in\mathbf{R}^{n}.$

To start moving planes, the following lemma is often required.

######  Lemma 4.1.

There hold

$w_{\varepsilon,\lambda}(x)=\gamma_{2m,n}\int_{\mathbf{R}^{n}}\big{[}|x-y|^{2m-n}-|x^{\lambda}-y|^{2m-n}\big{]}F_{\varepsilon}(y)dy$
(4.1)

and

$w_{\varepsilon,\lambda}(x)=\gamma_{2m,n}\int_{\Sigma_{\lambda}}\big{[}|x^{\lambda}-y|^{2m-n}-|x-y|^{2m-n}\big{]}[F_{\varepsilon}(y^{\lambda})-F_{\varepsilon}(y)]dy$
(4.2)

for any $\lambda\in\mathbf{R}$.

###### Proof.

The first identity is obvious from the definition of
$w_{\varepsilon,\lambda}$. The second identity follows from variable changes.
Indeed, one can write

$\displaystyle u_{\varepsilon}(x)$
$\displaystyle=\Big{(}\int_{\Sigma_{\lambda}}+\int_{\mathbf{R}^{n}\setminus\Sigma_{\lambda}}\Big{)}|x-y|^{2m-n}F_{\varepsilon}(y)dy$
$\displaystyle=\int_{\Sigma_{\lambda}}|x-y|^{2m-n}F_{\varepsilon}(y)dy+\int_{\Sigma_{\lambda}}|x-y^{\lambda}|^{2m-n}F_{\varepsilon}(y^{\lambda})dy$
$\displaystyle=\int_{\Sigma_{\lambda}}|x-y|^{2m-n}F_{\varepsilon}(y)dy+\int_{\Sigma_{\lambda}}|x^{\lambda}-y|^{2m-n}F_{\varepsilon}(y^{\lambda})dy.$

Similarly, one has

$\displaystyle u_{\varepsilon}(x^{\lambda})$
$\displaystyle=\int_{\Sigma_{\lambda}}|x^{\lambda}-y|^{2m-n}F_{\varepsilon}(y)dy+\int_{\Sigma_{\lambda}}|x-y|^{2m-n}F_{\varepsilon}(y^{\lambda})dy.$

By putting the above identities together we arrive at the second identity. ∎

Our next step is to show that the method of moving planes can start from a
very large $\lambda_{0}>0$, where $\lambda_{0}$ is independent of
$\varepsilon$.

######  Lemma 4.2.

Let $\varepsilon^{*}\in(0,1)$ be fixed. Then there exists $\lambda_{0}\gg 1$
such that for every $\varepsilon\in[0,\varepsilon^{*}]$ we have

$w_{\varepsilon,\lambda}(x)\geq 0\quad\text{in }\Sigma_{\lambda}$

for $\lambda\geq\lambda_{0}$.

###### Proof.

We start the proof by observing the existence of some constant $C>0$ such that
for each $\varepsilon\in[0,\varepsilon^{*}]$ we have

$\displaystyle\frac{1}{C}\frac{1}{1+|y|^{2m+n}}\leq F_{\varepsilon}(y)\leq
C\frac{1}{1+|y|^{2m+n}}\quad\text{in }\mathbf{R}^{n};$ (4.3)

see (2.2) for a similar estimate. In the case $\varepsilon>0$, this simply
follows from the uniform bound for $v_{\varepsilon}$ with respect to
$\varepsilon\in(0,\varepsilon^{*}]$ as given by Theorem 3.1. In the case
$\varepsilon=0$, the above estimate is trivial because $u(x)\approx|x|^{2m-n}$
for $|x|\gg 1$. By a simple algebraic computations we have

$\displaystyle|x-y|^{2m-n}-|x^{\lambda}-y|^{2m-n}=\frac{|x-y|^{2}-|x^{\lambda}-y|^{2}}{|x-y|^{2m-n}+|x^{\lambda}-y|^{2m-n}}\widetilde{P}_{\lambda}(x,y),$

where the function $\widetilde{P}_{\lambda}$ is given by

$\widetilde{P}_{\lambda}(x,y):=\sum_{k=0}^{2m-n-1}|x-y|^{2(2m-n-1-k)}|x^{\lambda}-y|^{2k}.$

(It is clear that $\widetilde{P}_{\lambda}\equiv 1$ if $2m-n=1$.) Using (4.1)
and

$|x-y|^{2}-|x^{\lambda}-y|^{2}=4(x_{1}-\lambda)(\lambda-y_{1})$

we can write

$|x|^{2+n-2m}\frac{w_{\varepsilon,\lambda}(x)}{x_{1}-\lambda}=\int_{\mathbf{R}^{n}}(\lambda-
y_{1})P_{\lambda}(x,y)F_{\varepsilon}(y)dy=:U_{\varepsilon}(x),$

where

$\displaystyle
P_{\lambda}(x,y):=4\gamma_{2m,n}\frac{|x|^{2+n-2m}}{|x-y|^{2m-n}+|x^{\lambda}-y|^{2m-n}}\widetilde{P}_{\lambda}(x,y).$
(4.4)

For later use, we note that for $x,y\in\Sigma_{\lambda}$ there holds

$\displaystyle P_{\lambda}(x,y)$ $\displaystyle\leq
C|x|^{2+n-2m}\frac{|x-y|^{2(2m-n-1)}+|x^{\lambda}-y|^{2(2m-n-1)}}{|x-y|^{2m-n}+|x^{\lambda}-y|^{2m-n}}$
$\displaystyle\leq C\left\\{\begin{aligned} &\frac{|x|}{|x-y|}&&\text{for
}2m-n=1\\\ &1+|x|^{2+n-2m}|y|^{2m-n-2}&&\text{for }2m-n\geq
3\end{aligned}\right.$ (4.5) $\displaystyle\leq C\left\\{\begin{aligned}
&\frac{|x|}{|x-y|}&&\text{for }2m-n=1\\\ &|y|^{2m-n-2}&&\text{for }2m-n\geq
3.\end{aligned}\right.$

To conclude the lemma, it suffices to show the existence of $\lambda_{0}\gg 1$
such that

$U_{\varepsilon}(x)>0\quad\text{for any }x\in\Sigma_{\lambda}\cup T_{\lambda}$

and for every $\lambda\geq\lambda_{0}$. With the help of (4.3) we can roughly
estimate

$\displaystyle U_{\varepsilon}(x)$ $\displaystyle=\int_{B_{1}}(\lambda-
y_{1})P_{\lambda}(x,y)F_{\varepsilon}(y)dy+\int_{\mathbf{R}^{n}\setminus
B_{1}}(\lambda-y_{1})P_{\lambda}(x,y)F_{\varepsilon}(y)dy$
$\displaystyle\geq\frac{1}{C}\int_{B_{1}}(\lambda-
y_{1})P_{\lambda}(x,y)dy+\int_{y_{1}>\lambda}(\lambda-
y_{1})P_{\lambda}(x,y)F_{\varepsilon}(y)dy$
$\displaystyle\geq\frac{1}{C}\int_{B_{1}}(\lambda-
y_{1})P_{\lambda}(x,y)dy-C\int_{y_{1}>\lambda}\frac{P_{\lambda}(x,y)}{1+|y|^{2m+n-1}}dy$
$\displaystyle=:I_{1}(x)-I_{2}(x).$

Here to get the term $I_{2}$ we have used the estimates $0\leq
y_{1}-\lambda\leq y_{1}\leq|y|$ in the region
$\\{y\in\mathbf{R}^{n}:y_{1}>\lambda\\}$ and

$\frac{|y|}{1+|y|^{2m+n}}\leq\frac{2}{1+|y|^{2m+n-1}}\quad\text{for all }y.$

Next, we estimate $I_{1}$ from below and $I_{2}$ from above. For $I_{1}$, we
note that

$P_{\lambda}(x,y)\geq\frac{1}{C}\quad\text{for }y\in
B_{1},\,x\in\Sigma_{\lambda},\,\lambda\geq\lambda_{0}\gg 1.$

From this we deduce

$I_{1}(x)\geq\frac{\lambda}{C}.$

We now estimate $I_{2}$. For $2m-n\geq 3$ and as

$\frac{|y|^{2m-n-2}}{1+|y|^{2m+n-1}}\leq\frac{2}{1+|y|^{2n+1}}\quad\text{for
all }y$

and $|y|\geq y_{1}>\lambda$ we can estimate

$\displaystyle I_{2}(x)\leq
C\int_{y_{1}>\lambda}\frac{|y|^{2m-n-2}dy}{1+|y|^{2m+n-1}}\leq
C\int_{y_{1}>\lambda}\frac{dy}{1+|y|^{2n+1}}\leq\frac{C}{\lambda^{n+1}}\leq
C.$

For $2m-n=1$, we split $\\{y_{1}>\lambda\\}$ as follows

$\\{y_{1}>\lambda\\}\subset A_{1}\cup A_{2}\cup A_{3}$

where

$A_{1}:=\big{\\{}y:\lambda<|y|\leq|x|/2\big{\\}},\quad
A_{2}:=B_{2|x|}\setminus B_{|x|/2},\quad A_{3}:=\mathbf{R}^{n}\setminus
B_{2|x|}.$

(Although $|x|>\lambda$ as $x\in\Sigma_{\lambda}$, the set $A_{1}$ could be
empty if $|x|<2\lambda$, but it is not important.) Since $|x-y|\geq|x|/2$ on
$A_{1}\cup A_{3}$ and again $|y|\geq y_{1}>\lambda$, we can estimate

$\int_{A_{1}\cup
A_{3}}\frac{|x|}{|x-y|}\frac{dy}{1+|y|^{2m+n-1}}\leq\frac{C}{\lambda^{2m-1}}.$

On the remaining set $A_{2}$ as $|x|/2\leq|y|\leq 2|x|$ we easily get

$\int_{A_{2}}\frac{|x|}{|x-y|}\frac{dy}{1+|y|^{2m+n-1}}\leq\frac{C}{|x|^{2m+n-2}}\int_{A_{2}}\frac{dy}{|x-y|}\leq\frac{C}{|x|^{2m-1}}\leq\frac{C}{\lambda^{2m-1}}\leq
C.$

Putting the above estimate together, we arrive at

$U_{\varepsilon}(x)\geq I_{1}(x)-I_{2}(x)\geq\frac{\lambda}{C}-C$

for some constant $C>0$. Thus, the lemma follows by letting $\lambda_{0}$
large enough. ∎

In Lemma 4.2, we have compared $u_{\varepsilon}(x)$ and
$u_{\varepsilon}(x^{\lambda})$, via $w_{\varepsilon,\lambda}(x)$, in
$\Sigma_{\lambda}$. As there was no restriction on $\varepsilon^{*}\in(0,1)$,
our comparison requires large $\lambda>0$ to hold. In the next lemma, we
compare $F_{\varepsilon}(x)$ and $F_{\varepsilon}(x^{\lambda})$ in
$\Sigma_{\lambda}$. As there will be no restriction on $\lambda>0$, our
comparison now requires small $\varepsilon>0$, and this is the place where the
constant $\varepsilon_{*}$ appears. Due to the form of $F_{\varepsilon}$ to
achieve the goal we need the compactness result established earlier; see
section 3.

######  Lemma 4.3.

There exists $\varepsilon_{*}\in(0,\varepsilon^{*})$ small enough such that
for arbitrary $\lambda\in(0,\lambda_{0}]$ but fixed, the conclusion if

$w_{\varepsilon,\lambda}\geq 0\quad\text{in }\Sigma_{\lambda},$ (4.6)

then

$F_{\varepsilon}(x)-F_{\varepsilon}(x^{\lambda})\leq 0\quad\text{in
}\Sigma_{\lambda}$ (4.7)

holds for each $\varepsilon\in[0,\varepsilon_{*})$. In addition, if the
inequality (4.6) is strict, then so is the inequality (4.7).

###### Proof.

Let us first be interested in the existence of $\varepsilon_{*}$ and
$\varepsilon\in(0,\varepsilon_{*})$. As $|x^{\lambda}|<|x|$ for $\lambda>0$
and $x\in\Sigma_{\lambda}$, we obtain

$\displaystyle F_{\varepsilon}(x)-F_{\varepsilon}(x^{\lambda})$
$\displaystyle=\varepsilon\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}u_{\varepsilon}(x)-\varepsilon\big{(}\frac{2}{1+|x^{\lambda}|^{2}}\big{)}^{2m}u_{\varepsilon}(x^{\lambda})$
$\displaystyle\quad+\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}\frac{1}{u_{\varepsilon}^{\alpha}(x)}-\big{(}\frac{2}{1+|x^{\lambda}|^{2}}\big{)}^{-c_{\alpha}}\frac{1}{u_{\varepsilon}^{\alpha}(x^{\lambda})}$
$\displaystyle\leq\varepsilon\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}(u_{\varepsilon}(x)-u_{\varepsilon}(x^{\lambda}))$
$\displaystyle\quad+\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}\Big{(}\frac{1}{u_{\varepsilon}^{\alpha}(x)}-\frac{1}{u_{\varepsilon}^{\alpha}(x^{\lambda})}\Big{)},$

where the constant $c_{\alpha}\leq 0$ is already given in (2.3). Hence, to
prove (4.7) in $\Sigma_{\lambda}$, it suffices to prove that

$\displaystyle\frac{u_{\varepsilon}^{\alpha}(x)-u_{\varepsilon}^{\alpha}(x^{\lambda})}{u_{\varepsilon}(x)-u_{\varepsilon}(x^{\lambda})}\frac{1}{u_{\varepsilon}^{\alpha}(x)u_{\varepsilon}^{\alpha}(x^{\lambda})}\geq\varepsilon\big{(}\frac{2}{1+|x|^{2}}\big{)}^{(2m-n)\frac{1+\alpha}{2}}\quad\text{in
}\Sigma_{\lambda},$ (4.8)

where we have used that

$2m+c_{\alpha}=(2m-n)\frac{1+\alpha}{2}.$

To this end, for some $R\gg 1$ to be specified later, we first split
$\Sigma_{\lambda}$ into two parts as follows:

$\Sigma_{\lambda}=\big{[}\Sigma_{\lambda}\cap
B_{R}\big{]}\cup\big{[}\Sigma_{\lambda}\setminus B_{R}\big{]}.$

In the region $\Sigma_{\lambda}\setminus B_{R}$, there exists some
$\varepsilon_{1}>0$ such that (4.8) holds. To see this we need to use uniform
bounds with respect to $\varepsilon>0$, see Theorem 3.1, to obtain

$\frac{u_{\varepsilon}^{\alpha}(x)-u_{\varepsilon}^{\alpha}(x^{\lambda})}{u_{\varepsilon}(x)-u_{\varepsilon}(x^{\lambda})}\geq\varepsilon_{1}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{(2m-n)\frac{1-\alpha}{2}}$

and

$\frac{1}{u_{\varepsilon}^{\alpha}(x)u_{\varepsilon}^{\alpha}(x^{\lambda})}\geq\varepsilon_{1}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{\alpha}$

for some small $\varepsilon_{1}\in(0,1)$. This is mainly because when $R$ is
large enough, we have $|x|\approx|x^{\lambda}|$ for $|x|>R$ and
$\lambda\in(0,\lambda_{0}]$. In the region $\Sigma_{\lambda}\cap B_{R}$, by
the smoothness of $u_{\varepsilon}$, there exists some small
$\varepsilon_{2}\in(0,1)$ such that

$\displaystyle\frac{u_{\varepsilon}^{\alpha}(x)-u_{\varepsilon}^{\alpha}(x^{\lambda})}{u_{\varepsilon}(x)-u_{\varepsilon}(x^{\lambda})}\frac{1}{u_{\varepsilon}^{\alpha}(x)u_{\varepsilon}^{\alpha}(x^{\lambda})}\geq\varepsilon_{2}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{(2m-n)\frac{1+\alpha}{2}}$
(4.9)

for any $x\in B_{R}$. Hence, combining (4.8) and (4.9) yields the desired
estimate (4.7) with

$\varepsilon_{*}=\frac{1}{2}\min\\{\varepsilon_{1},\varepsilon_{2}\\}.$

Now we consider the remaining case $\varepsilon=0$. However, this case is
trivial because

$\displaystyle F_{0}(x)-F_{0}(x^{\lambda})$
$\displaystyle=\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}\Big{(}\frac{1}{u_{0}^{\alpha}(x)}-\frac{1}{u_{0}^{\alpha}(x^{\lambda})}\Big{)}\leq
0$

whenever $w_{0,\lambda}(x)=u_{0}(x)-u_{0}(x^{\lambda})\geq 0$. Finally, from
the above calculation, it is clear that if the inequality (4.6) is strict,
then the inequality (4.7) is also strict. Hence, the lemma is proved. ∎

Thanks to Lemma 4.2, for each $\varepsilon>0$ we can set

$\overline{\lambda}_{\varepsilon}:=\inf\big{\\{}\lambda>0:w_{\varepsilon,\mu}\geq
0\;\text{ in }\;\Sigma_{\mu}\;\text{ for every }\;\mu\geq\lambda\big{\\}}.$

Then, still by Lemma 4.2, we necessarily have

$0\leq\overline{\lambda}_{\varepsilon}\leq\lambda_{0}.$

Our goal is to show that $\overline{\lambda}_{\varepsilon}=0$. This can be
done through two steps. First we show that if
$\overline{\lambda}_{\varepsilon}>0$, then we must have
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\equiv 0$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$; see Lemma 4.5. Finally, we show
that $\overline{\lambda}_{\varepsilon}=0$; see Lemma 4.6.

Our next lemma is of importance to achieve the first step as it allows us to
move $\lambda$ to the left.

######  Lemma 4.4.

Let $\varepsilon\in[0,\varepsilon_{*})$ and $\bar{\lambda}\in(0,\lambda_{0}]$
be such that

$0\not\equiv w_{\varepsilon,\bar{\lambda}}\geq 0\quad\text{in
}\Sigma_{\bar{\lambda}}.$

Then, there exist $R\gg 1$ and $\delta>0$ small, both may depend on
$w_{\varepsilon,\bar{\lambda}}$, such that for every
$\lambda\in(\bar{\lambda}-\delta,\bar{\lambda})$ we have

$w_{\varepsilon,\lambda}>0\quad\text{in }\Sigma_{\lambda}\setminus B_{R}.$

###### Proof.

Using the representation (4.2), and as in the first part of the proof of Lemma
4.2, we have

$\displaystyle w_{\varepsilon,\lambda}(x)\frac{|x|^{2+n-2m}}{x_{1}-\lambda}$
$\displaystyle=\int_{\Sigma_{\lambda}}(y_{1}-\lambda)P_{\lambda}(x,y)[F_{\varepsilon}(y^{\lambda})-F_{\varepsilon}(y)]dy,$
(4.10)

where $P_{\lambda}$ is given by (4.4). In view of (4.10), it suffices to show
that its RHS is positive in $\Sigma_{\lambda}\setminus B_{R}$ for suitable
$R>0$. For convenience, we recall the following formula for $P_{\lambda}$

$\displaystyle
P_{\lambda}(x,y)=4\gamma_{2m,n}\frac{|x|^{2+n-2m}}{|x-y|^{2m-n}+|x^{\lambda}-y|^{2m-n}}\sum_{k=0}^{2m-n-1}|x-y|^{2(2m-n-1-k)}|x^{\lambda}-y|^{2k}.$

Hence, there exists some $\theta>0$ such that for every $R_{1}>0$ fixed

$\displaystyle P_{\lambda}(x,y)\rightrightarrows\theta\quad\text{ uniformly in
}y\in B_{R_{1}}$ (4.11)

as $|x|\to\infty$. This is because $|x|\approx|x-y|\approx|x^{\lambda}-y|$ for
large $|x|$. From (4.7) we know that

$0\not\equiv F_{\varepsilon}(y^{\bar{\lambda}})-F_{\varepsilon}(y)\geq
0\quad\text{for }y\in\Sigma_{\bar{\lambda}},$

which implies

$\int_{\Sigma_{\bar{\lambda}}}(y_{1}-\bar{\lambda})[F_{\varepsilon}(y^{\bar{\lambda}})-F_{\varepsilon}(y)]dy\geq
2c_{0}>0,$

for some small constant $c_{0}>0$. Thus, by the dominated convergence theorem,
we can find some $\delta>0$ such that

$\displaystyle\int_{\Sigma_{\lambda}}(y_{1}-\lambda)[F_{\varepsilon}(y^{\lambda})-F_{\varepsilon}(y)]dy\geq
c_{0}>0,$ (4.12)

for every $|\lambda-\bar{\lambda}|<\delta$. To obtain the positivity of the
right hand side of (4.10), we split the integral $\int_{\Sigma_{\lambda}}$
into two parts as follows

$\int_{\Sigma_{\lambda}}=\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}+\int_{\Sigma_{\lambda}\cap B_{R_{2}}}$

for some $R_{2}>0$ to be determined later and estimate these integrals term by
term; see the two estimates (4.14) and (4.15) below. Our aim is to show that
the integral $\int_{\Sigma_{\lambda}\setminus B_{R_{2}}}$ is negligible.

We assume for a moment that such a constant $R_{2}$ exists. We now estimate
the integral $\int_{\Sigma_{\lambda}\setminus B_{R_{2}}}$. First we choose a
$R_{0}\gg 1$ in such a way that $|y_{1}-\lambda|<2|y|$ for all $|y|\geq
R_{0}$. Then we find some $R_{1}\gg R_{0}$ such that in
$\Sigma_{\lambda}\setminus B_{R_{1}}$ we have

$\int_{\Sigma_{\lambda}\setminus
B_{R_{1}}}\frac{dy}{1+|y|^{2m+n-1}}\leq\frac{\theta c_{0}}{16C}$ (4.13)

and

$F_{\varepsilon}(y)+F_{\varepsilon}(y^{\lambda})\leq\frac{C}{1+|y|^{2m+n}}$

for some $C>0$ because $|y|\approx|y^{\lambda}|$. By the estimate (4) for
$P_{\lambda}$, we now claim that there are some $R_{3}\gg 1$ and $R_{2}\gg
R_{1}$ such that

$\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}(y_{1}-\lambda)P_{\lambda}(x,y)[F_{\varepsilon}(y^{\lambda})+F_{\varepsilon}(y)]dy\leq\frac{\theta
c_{0}}{4}$ (4.14)

for $|x|\geq R_{3}$. To see this, for clarity, we consider the two cases
$2m-n=1$ and $2m-n\geq 3$ separately.

Case 1. Suppose $2m-n=1$. In this case our estimate for $P_{\lambda}$ becomes
$P_{\lambda}(x,y)\leq C|x|/|x-y|$. Consequently, there holds

$\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}(y_{1}-\lambda)P_{\lambda}(x,y)[F_{\varepsilon}(y^{\lambda})+F_{\varepsilon}(y)]dy\leq
C\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}\frac{|x|}{|x-y|}\frac{|y|}{1+|y|^{2m+n}}dy.$

For $|x|\geq R_{3}\gg 2R_{2}$ to be determined later, we now split
$\int_{\Sigma_{\lambda}\setminus B_{R_{2}}}$ as follows

$\int_{\Sigma_{\lambda}\setminus B_{R_{2}}}=\int_{[\Sigma_{\lambda}\setminus
B_{R_{2}}]\cap[B_{|x|/2}\cup(\mathbf{R}^{n}\setminus
B_{2|x|})]}+\int_{[\Sigma_{\lambda}\setminus
B_{R_{2}}]\setminus[B_{|x|/2}\cup(\mathbf{R}^{n}\setminus B_{2|x|})]}.$

Thanks to (4.13), we get

$C\int_{[\Sigma_{\lambda}\setminus
B_{R_{2}}]\cap[B_{|x|/2}\cup(\mathbf{R}^{n}\setminus
B_{2|x|})]}\frac{|x|}{|x-y|}\frac{|y|}{1+|y|^{2m+n}}dy<\frac{\theta
c_{0}}{8}.$

For the remaining integral on $[\Sigma_{\lambda}\setminus
B_{R_{2}}]\setminus[B_{|x|/2}\cup(\mathbf{R}^{n}\setminus B_{2|x|})]$ which is
a subset of $B_{2|x|}\setminus B_{|x|/2}$ because $|x|\geq 2R_{2}$, we
estimate as follows

$C\int_{[\Sigma_{\lambda}\setminus
B_{R_{2}}]\setminus[B_{|x|/2}\cup(\mathbf{R}^{n}\setminus
B_{2|x|})]}\leq\frac{C|x|^{2}}{1+|x|^{2m+n}}\int_{B_{2|x|}\setminus
B_{|x|/2}}\frac{dy}{|x-y|}.$

Since the last integral is of order $|x|^{n}$ and $m\geq 2$ we can find some
$R_{3}\gg 1$ such that

$\frac{C|x|^{2}}{1+|x|^{2m+n}}\int_{B_{2|x|}\setminus
B_{|x|/2}}\frac{dy}{|x-y|}\leq\frac{\theta c_{0}}{8}$

for all $x\in\Sigma_{\lambda}\cap B_{R_{3}}$. Combining the two estimates
above gives (4.14). This completes the first case.

Case 2. Suppose $2m-n\geq 3$. This case is easy to handle. Recall that our
estimate for $P_{\lambda}$ becomes $P_{\lambda}(x,y)\leq C|y|^{2m-n-2}$.
Consequently, there holds

$\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}(y_{1}-\lambda)P_{\lambda}(x,y)[F_{\varepsilon}(y^{\lambda})+F_{\varepsilon}(y)]dy\leq
C\int_{\Sigma_{\lambda}\setminus
B_{R_{2}}}\frac{|y|^{2m-n-1}}{1+|y|^{2m+n}}dy.$

Seeing (4.13) or as in the proof of Lemma 4.2, we easily obtain the desired
estimate.

Hence, up to this point, we have already shown that there are some $R_{2}\gg
1$ and $R_{3}\gg 1$ such that the estimate (4.14) holds for $|x|\geq R_{3}$.
Now we estimate the integral $\int_{\Sigma_{\lambda}\cap B_{R_{2}}}$. Keep
using the constant $R_{2}$. By the uniform convergence in (4.11), we can
choose $R_{4}\gg R_{2}$ such that

$P_{\lambda}(x,y)\geq\frac{1}{2}\theta\quad\text{for }|x|\geq R_{4}\text{ and
}|y|\leq R_{2}.$

This and (4.12) imply that

$\int_{\Sigma_{\lambda}\cap
B_{R_{2}}}(y_{1}-\lambda)P_{\lambda}(x,y)[F_{\varepsilon}(y^{\lambda})-F_{\varepsilon}(y)]dy\geq\frac{\theta
c_{0}}{2}$ (4.15)

for $|x|\geq R_{4}$. We conclude the lemma by combing the two estimates (4.14)
and (4.15) and choosing $R=\max\\{R_{3},R_{4}\\}$. ∎

We are now in a position to complete the first step, namely, to show that
$\overline{\lambda}_{\varepsilon}=0$. To this purpose, we must rule out the
case $\overline{\lambda}_{\varepsilon}>0$ and this is the content of the next
two lemmas. First, we characterize the function
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}$ in case
$\overline{\lambda}_{\varepsilon}>0$.

######  Lemma 4.5.

If $\overline{\lambda}_{\varepsilon}>0$ for some
$\varepsilon\in[0,\varepsilon_{*})$, then
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\equiv 0$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$. In other words, the function
$u_{\varepsilon}$ is symmetric with respect to the hyperplane
$\\{x\in\mathbf{R}^{n}:x_{1}=\overline{\lambda}_{\varepsilon}\\}$.

###### Proof.

Let $\overline{\lambda}_{\varepsilon}>0$ for some
$\varepsilon\in[0,\varepsilon_{*})$ and assume by contradiction that
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\not\equiv 0$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$. This and the definition of
$\overline{\lambda}_{\varepsilon}$ imply that

$0\not\equiv w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\geq
0\quad\text{in }\Sigma_{\overline{\lambda}_{\varepsilon}}.$

By Lemma 4.4, there exist $R\gg 1$ and $\delta>0$ small enough such that

$w_{\varepsilon,\lambda}>0\quad\text{in }\Sigma_{\lambda}\setminus
B_{R}\quad\text{for every
}\lambda\in(\overline{\lambda}_{\varepsilon}-\delta,\overline{\lambda}_{\varepsilon}).$

Then there exists a sequence $\mu_{k}\nearrow\overline{\lambda}_{\varepsilon}$
such that $w_{\varepsilon,\mu_{k}}$ is negative somewhere in
$\Sigma_{\mu_{k}}$. Since outside $B_{R}$, the function
$w_{\varepsilon,\mu_{k}}$ is strictly positive, for each $k$ there is some
$x_{k}\in\Sigma_{\mu_{k}}\cap\overline{B_{R}}$ such that

$w_{\varepsilon,\mu_{k}}(x_{k})=\min_{\Sigma_{\mu_{k}}}w_{\varepsilon,\mu_{k}}<0.$

In particular, there holds

$\frac{w_{\varepsilon,\mu_{k}}(x_{k})}{(x_{k})_{1}-\mu_{k}}<0.$

Obviously, the sequence $(x_{k})$ is bounded as $x_{k}\in\overline{B_{R}}$.
Also note that
$\Sigma_{\overline{\lambda}_{\varepsilon}}\subset\Sigma_{\mu_{k}}$ and
$\Sigma_{\mu_{k}}\searrow\Sigma_{\overline{\lambda}_{\varepsilon}}$ as
$k\nearrow+\infty$. Therefore, up to a subsequence, we have

$\Sigma_{\overline{\lambda}_{\varepsilon}}\cup
T_{\overline{\lambda}_{\varepsilon}}\ni x_{\infty}:=\lim_{k\to\infty}x_{k}.$

In particular, by passing to the limit as $k\to\infty$, there holds
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}(x_{\infty})\leq 0$. This and
(4.10) implies that

$\displaystyle 0\geq
w_{\varepsilon,\overline{\lambda}_{\varepsilon}}(x_{\infty})\frac{|x_{\infty}|^{2+n-2m}}{(x_{\infty})_{1}-\overline{\lambda}_{\varepsilon}}$
$\displaystyle=\int_{\Sigma_{\overline{\lambda}_{\varepsilon}}}(y_{1}-\overline{\lambda}_{\varepsilon})P_{\overline{\lambda}_{\varepsilon}}(x_{\infty},y)[F_{\varepsilon}(y^{\overline{\lambda}_{\varepsilon}})-F_{\varepsilon}(y)]dy\geq
0,$

thanks to $|x_{\infty}|>0$ and
$F_{\varepsilon}(y^{\overline{\lambda}_{\varepsilon}})\geq F_{\varepsilon}(y)$
in $\Sigma_{\overline{\lambda}_{\varepsilon}}$ by Lemma 4.3. Thus, we must
have

$F_{\varepsilon}(y^{\bar{\lambda}\varepsilon})-F_{\varepsilon}(y)=0\quad\text{for
any }y\in\Sigma_{\overline{\lambda}_{\varepsilon}},$

which, by (4.10), now yields
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\equiv 0$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$. However, this is a contradiction.
Once we have $w_{\varepsilon,\overline{\lambda}_{\varepsilon}}\equiv 0$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$, the symmetry of $u_{\varepsilon}$
follows from the definition of
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}$. The proof is complete. ∎

From the characterization of
$w_{\varepsilon,\overline{\lambda}_{\varepsilon}}$ in the case
$\overline{\lambda}_{\varepsilon}>0$ and the role of the size of $\varepsilon$
and $\alpha$, we are able to show that in fact the case
$\overline{\lambda}_{\varepsilon}>0$ cannot happen.

######  Lemma 4.6.

Let $\varepsilon\in[0,\varepsilon_{*})$. There holds
$\overline{\lambda}_{\varepsilon}=0$. In particular, the function
$u_{\varepsilon}$ is symmetric with respect to the hyperplane
$\\{x\in\mathbf{R}^{n}:x_{1}=0\\}$.

###### Proof.

By way of contradiction, assume that $\overline{\lambda}_{\varepsilon}>0$. In
view of Lemma 4.5, we must have

$0=w_{\varepsilon,\overline{\lambda}_{\varepsilon}}(x)=u_{\varepsilon}(x)-u_{\varepsilon}(x^{\overline{\lambda}_{\varepsilon}})$

in $\Sigma_{\overline{\lambda}_{\varepsilon}}$. This and (4.2) tell us that

$\displaystyle\int_{\Sigma_{\overline{\lambda}_{\varepsilon}}}\big{[}|x^{\overline{\lambda}_{\varepsilon}}-y|^{2m-n}-|x-y|^{2m-n}\big{]}[F_{\varepsilon}(y^{\overline{\lambda}_{\varepsilon}})-F_{\varepsilon}(y)]dy=0$

for any $x\in\Sigma_{\overline{\lambda}_{\varepsilon}}$, thanks to
$\gamma_{2m,n}\neq 0$, see Theorem 2.2. But this cannot happen because
$|x-y|\leq|x^{\overline{\lambda}_{\varepsilon}}-y|$ for any
$x,y\in\Sigma_{\overline{\lambda}_{\varepsilon}}$ and

$\displaystyle
F_{\varepsilon}(x)-F_{\varepsilon}(x^{\overline{\lambda}_{\varepsilon}})$
$\displaystyle=\varepsilon\Big{[}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{2m}-\big{(}\frac{2}{1+|x^{\overline{\lambda}_{\varepsilon}}|^{2}}\big{)}^{2m}\Big{]}u_{\varepsilon}(x)$
$\displaystyle\quad+\Big{[}\big{(}\frac{2}{1+|x|^{2}}\big{)}^{-c_{\alpha}}-\big{(}\frac{2}{1+|x^{\overline{\lambda}_{\varepsilon}}|^{2}}\big{)}^{-c_{\alpha}}\Big{]}\frac{1}{u_{\varepsilon}^{\alpha}(x)}$
$\displaystyle<0,$

everywhere in $\Sigma_{\overline{\lambda}_{\varepsilon}}$, thanks to the
estimates $u_{\varepsilon}>0$, $-c_{\alpha}\geq 0$, and
$|x|\leq|x^{\overline{\lambda}_{\varepsilon}}|$ in
$\Sigma_{\overline{\lambda}_{\varepsilon}}$. (Here we also use the fact that
if $\varepsilon=0$, then $\alpha<(n+2m)/(2m-n)$ in order to guarantee
$-c_{\alpha}>0$.) Thus, we must have $\overline{\lambda}_{\varepsilon}=0$. In
particular, we have from the definition of $\overline{\lambda}_{\varepsilon}$
the following

$u_{\varepsilon}(x_{1},x_{2},...,x_{n})\geq
u_{\varepsilon}(-x_{1},x_{2},...,x_{n}).$

We now apply the method of moving planes in the opposite direction, namely
$\lambda<0$, to get

$u_{\varepsilon}(x_{1},x_{2},...,x_{n})\leq
u_{\varepsilon}(-x_{1},x_{2},...,x_{n}).$

Hence

$u_{\varepsilon}(x_{1},x_{2},...,x_{n})=u_{\varepsilon}(-x_{1},x_{2},...,x_{n}).$

This establishes the symmetry of $u_{\varepsilon}$ with respect to the
hyperplane $\\{x\in\mathbf{R}^{n}:x_{1}=0\\}$. The proof is now complete. ∎

We now have a quick note. In the proof of Lemma 4.6 above, we crucially use
the hypothesis that either $\varepsilon>0$ and $\alpha\leq(n+2m)/(n-2m)$ or
$\varepsilon=0$ and $\alpha<(n+2m)/(n-2m)$. For the latter case, if
$\varepsilon=0$ and $\alpha=(n+2m)/(n-2m)$, then we cannot claim that
$\overline{\lambda}_{0}=0$. Therefore, we could only claim that $u_{0}$ is
radially symmetric with respect to some point not necessarily the origin. This
leads to explicit form of non-trivial solutions to (1.1)0 in the conformaly
invariant case.

As a consequence of Lemma 4.6 above, we obtain a Liouville type result for
positive, smooth solution to (1.1)ε for small $\varepsilon>0$, hence proving
Theorem 1.1.

######  Lemma 4.7.

Any positive, smooth solution $v_{\varepsilon}$ to (1.1)ε for small
$\varepsilon$ must be constant.

###### Proof.

Let $\varepsilon\in[0,\varepsilon_{*})$ be arbitrary. From Lemma 4.6 we know
that the corresponding solution $u_{\varepsilon}$ is symmetric with respect to
the hyperplane $\\{x\in\mathbf{R}^{n}:x_{1}=0\\}$. This together with the
relation

$u_{\varepsilon}(x)=\Big{(}\frac{1+|x|^{2}}{2}\Big{)}^{\frac{2m-n}{2}}\big{(}v_{\varepsilon}\circ\pi_{N}^{-1}\big{)}(x)$

tells us that $v_{\varepsilon}$ depends only on the last coordinate $x_{n+1}$.
However, as the $x_{n+1}$-axis is freely chosen, we conclude that
$v_{\varepsilon}$ must be constant. This completes the proof. ∎

Before closing this section, we have a remark. To obtain the symmetry of
solutions to (1.1)ε for small $\varepsilon$, our approach is based on the
method of moving planes in the integral form. A natural question is weather or
not one can use the method of moving spheres; see [LZ95, Li04]. Due to the
presence of the weight $2/(1+|x|^{2})$ in (1.7), it is natural to ask whether
or not the method of moving spheres can still be used. Toward a possible
answer to this question, we refer the reader to the work [JLX08].

##  5\. Application to the sharp Sobolev inequality

This section is devoted to a proof of Theorem 1.2 which concerns a sharp
(critical or subcritical) Sobolev inequality. Let $\varepsilon\in(0,1)$ and
inspired by (1.4) consider the following variational problem

$\mathcal{S}_{\varepsilon}=\inf_{0<\phi\in
H^{m}(\mathbb{S}^{n})}\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\big{[}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}\phi^{2}\big{]}d\mu_{\mathbb{S}^{n}}$
(5.1)

with $m=(n+1)/2$ and $\alpha\in(0,1)\cup(1,2n+1]$. We note that although the
constant $(n-2m)/2$ becomes $-1/2$ in the present case, we intent to keep it
in various calculation below for convenience. Similar convention also applies
for $\,{\mathbf{P}}^{2m}_{n}$ instead of $\mathbf{P}_{n}^{n+1}$, etc. Now as

$\,{\mathbf{P}}^{2m}_{n}(1)-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}=(1-\varepsilon)\frac{n-2m}{2}Q_{n}^{2m}\neq
0$

by testing (5.1) with constant functions we conclude from (5.1) that

$\mathcal{S}_{\varepsilon}\leq(1-\varepsilon)\frac{n-2m}{2}Q_{n}^{2m}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}<0,$

however, $\mathcal{S}_{\varepsilon}$ could be $-\infty$. Next we show that
$\mathcal{S}_{\varepsilon}$ is finite and is achieved by some smooth positive
function.

######  Lemma 5.1.

Assume that $m=(n+1)/2$ and $\alpha\in(0,1)\cup(1,2n+1]$. Then, the constant
$\mathcal{S}_{\varepsilon}$ in (5.1) is finite and there exists some
$v_{\varepsilon}\in C^{\infty}(\mathbb{S}^{n})$ such that $v_{\varepsilon}>0$
and

$\Big{(}\int_{\mathbb{S}^{n}}v_{\varepsilon}^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\big{[}v_{\varepsilon}\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{\varepsilon}^{2}\big{]}d\mu_{\mathbb{S}^{n}}=\mathcal{S}_{\varepsilon}.$

In particular, $v_{\varepsilon}$ solves

$\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{\varepsilon}=S_{\varepsilon}v_{\varepsilon}^{-\alpha}$

in $\mathbb{S}^{n}$ with

$S_{\varepsilon}=\frac{\mathcal{S}_{\varepsilon}}{\|v_{\varepsilon}^{-1}\|_{L^{\alpha-1}(\mathbb{S}^{n})}^{\alpha+1}}.$

###### Proof.

Let $(v_{k})_{k}$ be a positive, smooth minimizing sequence in
$H^{2m}(\mathbb{S}^{n})$, that is

$\Big{(}\int_{\mathbb{S}^{n}}v_{k}^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\big{[}v_{k}\,{\mathbf{P}}^{2m}_{n}(v_{k})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{k}^{2}\big{]}d\mu_{\mathbb{S}^{n}}\searrow\mathcal{S}_{\varepsilon}$

as $k\to\infty$. By the scaling invariant we can assume
$\max_{\mathbb{S}^{n}}v_{k}=1$ which then yields

$\|v_{k}\|_{L^{2}(\mathbb{S}^{n})}^{2}\leq|\mathbb{S}^{n}|.$

As $\,{\mathbf{P}}^{2m}_{n}$ is a monic polynomial of
$-\Delta_{g_{\mathbb{S}^{n}}}$, the coefficient of the highest degree is equal
to $1$, it is easy to get that

$\int_{\mathbb{S}^{n}}v_{k}\,{\mathbf{P}}^{2m}_{n}(v_{k})d\mu_{\mathbb{S}^{n}}\geq
c_{1}\|v_{k}\|_{H^{m}(\mathbb{S}^{n})}^{2}-c_{2}\|v_{k}\|_{L^{2}(\mathbb{S}^{n})}^{2}\geq
c_{1}\|v_{k}\|_{H^{m}(\mathbb{S}^{n})}^{2}-c_{2}|\mathbb{S}^{n}|$

for some $c_{1}>0$ and $c_{2}>0$. Note that $\mathcal{S}_{\varepsilon}<0$ and
$Q^{2m}_{n}>0$ would imply

$\int_{\mathbb{S}^{n}}v_{k}\,{\mathbf{P}}^{2m}_{n}v_{k}d\mu_{\mathbb{S}^{n}}<0.$

Therefore, the previous estimate leads to

$c_{1}\|v_{k}\|_{H^{m}(\mathbb{S}^{n})}^{2}\leq c_{2}|\mathbb{S}^{n}|,$

giving the boundedness of the sequence $(v_{k})$ in $H^{m}(\mathbb{S}^{n})$.
Hence, after passing to a subsequence if necessary, there exists some
$v_{\varepsilon}\in H^{m}(\mathbb{S}^{n})$ such that

$v_{k}\to v_{\varepsilon}\geq 0\text{ uniformly in }C(\mathbb{S}^{n})$

by Morrey’s inequality and the Arzelà–Ascoli lemma, and

$v_{k}\rightharpoonup v_{\varepsilon}\text{ weakly in }H^{m}(\mathbb{S}^{n}).$

In particular, there holds $\max_{\mathbb{S}^{n}}v_{\varepsilon}=1$. As
$v_{\varepsilon}\geq 0$, there are two possibilities. First, let us assume
that $v_{\varepsilon}$ vanishes somewhere on $\mathbb{S}^{n}$. By assuming
this we shall obtain a contradiction, therefore we must have
$v_{\varepsilon}>0$. Indeed, as $n=2m-1$, we can make use of [Han07, Corollary
3.1] to conclude that

$\int_{\mathbb{S}^{n}}v_{\varepsilon}\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})d\mu_{\mathbb{S}^{n}}\geq
0.$

This together with $\varepsilon\frac{n-2m}{2}Q_{n}^{2m}<0$ and
$\int_{\mathbb{S}^{n}}v_{\varepsilon}^{2}d\mu_{\mathbb{S}^{n}}>0$ help us to
get

$\displaystyle 0$
$\displaystyle<\int_{\mathbb{S}^{n}}\big{[}v_{\varepsilon}\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{\varepsilon}^{2}\big{]}d\mu_{\mathbb{S}^{n}}$
$\displaystyle\leq\liminf_{k\nearrow+\infty}\int_{\mathbb{S}^{n}}\big{[}v_{k}\,{\mathbf{P}}^{2m}_{n}(v_{k})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{k}^{2}\big{]}d\mu_{\mathbb{S}^{n}}.$

This is a contradiction to $\mathcal{S}_{\varepsilon}<0$. Thus,
$v_{\varepsilon}>0$ everywhere. Then, this allows us to gain

$v_{k}^{-1}\to v_{\varepsilon}^{-1}\quad\text{uniformly in }C(\mathbb{S}^{n})$

and consequently

$\int_{\mathbb{S}^{n}}v_{k}^{1-\alpha}d\mu_{\mathbb{S}^{n}}\to\int_{\mathbb{S}^{n}}v_{\varepsilon}^{1-\alpha}d\mu_{\mathbb{S}^{n}}.$

Putting these facts together, we obtain

$\displaystyle\mathcal{S}_{\varepsilon}$
$\displaystyle\leq\Big{(}\int_{\mathbb{S}^{n}}v_{\varepsilon}^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\big{[}v_{\varepsilon}\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{\varepsilon}^{2}\big{]}d\mu_{\mathbb{S}^{n}}$
(5.2)
$\displaystyle\leq\liminf_{k\nearrow+\infty}\Big{[}\Big{(}\int_{\mathbb{S}^{n}}v_{k}^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\big{[}v_{k}\,{\mathbf{P}}^{2m}_{n}(v_{k})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{k}^{2}\big{]}d\mu_{\mathbb{S}^{n}}\Big{]}$
$\displaystyle=\mathcal{S}_{\varepsilon}.$

Hence, on one hand implies that $\mathcal{S}_{\varepsilon}$ must be finite, on
the other hand, yields that $v_{\varepsilon}$ is a minimizer for (5.1). Rest
of the proof follows immediately. ∎

Now we are in a position to give a proof of Theorem 1.2.

###### Proof of Theorem 1.2.

Let $\varepsilon>0$ and $\alpha\in(0,1)\cup(1,2n+1]$. By Lemma 5.1, there is
some positive, smooth function $v_{\varepsilon}$ satisfying

$\int_{\mathbb{S}^{n}}v_{\varepsilon}^{1-\alpha}d\mu_{\mathbb{S}^{n}}=1$

and

$\int_{\mathbb{S}^{n}}\big{[}v_{\varepsilon}\,{\mathbf{P}}^{2m}_{n}(v_{\varepsilon})-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}v_{\varepsilon}^{2}\big{]}d\mu_{\mathbb{S}^{n}}=\mathcal{S}_{\varepsilon}.$

Then, up to a constant multiple, $v_{\varepsilon}$ solves (1.1)ε in
$\mathbb{S}^{n}$. Therefore, for small $\varepsilon>0$, it follows from
Theorem 1.1 that $v_{\varepsilon}$ is constant. Keep in mind that $\alpha\neq
1$. Hence, on one hand, as $((n-2m)/2)Q_{n}^{2m}=\,{\mathbf{P}}^{2m}_{n}(1)$,
we can compute to get

$\mathcal{S}_{\varepsilon}=(1-\varepsilon)\frac{n-2m}{2}Q_{n}^{2m}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}},$

on the other hand, by the definition of $\mathcal{S}_{\varepsilon}$ we get

$\displaystyle\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}$
$\displaystyle\int_{\mathbb{S}^{n}}\big{[}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)-\varepsilon\frac{n-2m}{2}Q_{n}^{2m}\phi^{2}\big{]}d\mu_{\mathbb{S}^{n}}$
$\displaystyle\geq(1-\varepsilon)\frac{n-2m}{2}Q_{n}^{2m}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}$

for any $\phi\in H^{m}(\mathbb{S}^{n})$ with $\phi>0$. Now letting
$\varepsilon\searrow 0$ we obtain

$\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{n-2m}{2}Q_{n}^{2m}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}.$

Recall that

$\frac{n-2m}{2}Q_{n}^{2m}=\,{\mathbf{P}}^{2m}_{n}(1)=\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}.$

This completes the proof of Theorem 1.2. ∎

Before closing this section, let us revisit the last comment in Remark 1.3.
For convenience, let us relabel (1.9) as follows

$\displaystyle\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}|\mathbb{S}^{n}|^{\frac{\alpha+1}{\alpha-1}}.$
(5.3)α

We shall establish the following, which has its own interest.

######  Proposition 5.2.

There holds

(5.3)2n+1 $\longrightarrow$ (5.3)β with $\beta\in(1,2n+1)$ $\longrightarrow$
(1.11) $\longrightarrow$ (5.3)α with $\alpha\in(0,1)$,

where the notation $A\longrightarrow B$ means we can obtain $B$ from $A$.

Before proving Proposition 5.2 we observe that, as
$\Gamma(n/2+m)/\Gamma(n/2-m)<0$, if $n=2m-1$, our related inequalities are
only meaningful if

$\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}<0.$
(5.4)

Therefore, from now on we always assume the above inequality. Besides, one can
simplify the computation below by normalizing the measure on $\mathbb{S}^{n}$
in such a way that $|\mathbb{S}^{n}|=1$. However, we intend to keep it for
clarity.

###### Proof of Proposition 5.2.

Let us establish all $\longrightarrow$ each by each.

Proof of (5.3)2n+1 $\longrightarrow$ (5.3)β with $\beta\in(1,2n+1)$. Let
$\beta\in(1,2n+1)$ be arbitrary but fixed. We wish to derive (5.3)β from
(5.3)2n+1. Thanks to $0<\beta-1<2n$, we can apply Hölder’s inequality in the
following way

$\int_{\mathbb{S}^{n}}\big{(}\phi^{-1}\big{)}^{\beta-1}d\mu_{\mathbb{S}^{n}}\leq|\mathbb{S}|^{\frac{2n+1-\beta}{2n}}\Big{(}\int_{\mathbb{S}^{n}}\big{(}\phi^{-1}\big{)}^{2n}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{\beta-1}{2n}}$

to get

$\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\beta}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\beta-1}}\leq|\mathbb{S}|^{\frac{2n+1-\beta}{n(\beta-1)}}\Big{(}\int_{\mathbb{S}^{n}}\phi^{-2n}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{1}{n}}.$

From this and (5.4) one immediately obtains

$\displaystyle\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\beta}$ $\displaystyle
d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\beta-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}$
$\displaystyle\geq|\mathbb{S}^{n}|^{\frac{2n+1-\beta}{n(\beta-1)}}\Big{(}\int_{\mathbb{S}^{n}}\phi^{-2n}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{1}{n}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}.$

With help of (5.3)2n+1 and the identity

$\frac{2n+1-\beta}{n(\beta-1)}+\frac{2m}{n}=\frac{\beta+1}{\beta-1}$

we obtain (5.3)β as claimed. (Keep in mind that $2m=n+1$.) This shows the
first $\longrightarrow$ from the left.

Proof of (5.3)β with $\beta\in(1,2n+1)$ $\longrightarrow$ (1.11). We now
consider arbitrary but fixed $\beta\in(1,2n+1)$ and we wish to derive (1.11)
from (5.3)β. By Jensen’s integral inequality of the form

$\frac{1}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\psi\,d\mu_{\mathbb{S}^{n}}\leq\log\Big{(}\frac{1}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\psi\,d\mu_{\mathbb{S}^{n}}\Big{)}$
(5.5)

we know by choosing $\psi=\phi^{-\gamma}$ that

$\exp\Big{(}-\frac{2}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\phi\,d\mu_{\mathbb{S}^{n}}\Big{)}\leq\Big{(}\frac{1}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\phi^{-\gamma}\,d\mu_{\mathbb{S}^{n}}\Big{)}^{2/\gamma}$
(5.6)

for any $\gamma\in\mathbf{R}$. In (5.6) we choose $\gamma=\beta-1$ and
together with (5.4) we eventually get

$\displaystyle\exp\Big{(}-\frac{2}{|\mathbb{S}^{n}|}$
$\displaystyle\int_{\mathbb{S}^{n}}\log\phi\,d\mu_{\mathbb{S}^{n}}\Big{)}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}$
$\displaystyle\geq|\mathbb{S}^{n}|^{\frac{2}{1-\beta}}\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\beta}\,d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\beta-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}.$

With help of (5.3)β we quickly obtain the inequality (1.11). Hence we have the
second $\longrightarrow$ from the left. (We should point out that the above
argument works for any $\beta>1$ as long as (5.3)β is available. In
particular, it works for $\beta=2n+1$. However, we intend to keep $\beta<2n+1$
since we want to show that the limiting case can be derived from the
subcritical case.)

Proof of (1.11) $\longrightarrow$ (5.3)α with $\alpha\in(0,1)$. Let us now
consider arbitrary but fixed $\alpha\in(0,1)$ and we wish to derive (5.3)α
from (1.11). Indeed, still by Jensen’s integral inequality (5.5) applied for
$\psi=\phi^{\gamma}$, we obtain

$\exp\Big{(}\frac{2}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\phi\,d\mu_{\mathbb{S}^{n}}\Big{)}\leq\Big{(}\frac{1}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\phi^{\gamma}\,d\mu_{\mathbb{S}^{n}}\Big{)}^{2/\gamma}$
(5.7)

for any $\gamma\in\mathbf{R}$. In (5.7) we choose $\gamma=1-\alpha$ and
reverse the resulting inequality to get

$\exp\Big{(}-\frac{2}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\phi\,d\mu_{\mathbb{S}^{n}}\Big{)}\geq\Big{(}\frac{1}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}\,d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}.$

Combining this with (5.4) gives

$\displaystyle\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}\,$ $\displaystyle
d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{2}{\alpha-1}}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}$
$\displaystyle\geq|\mathbb{S}^{n}|^{\frac{2}{\alpha-1}}\exp\Big{(}-\frac{2}{|\mathbb{S}^{n}|}\int_{\mathbb{S}^{n}}\log\phi\,d\mu_{\mathbb{S}^{n}}\Big{)}\int_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}.$

With the help of (1.11) we are now able to obtain the inequality (5.3)α. This
establishes the last $\longrightarrow$ from the left, hence completes our
proof. ∎

Again we should point out that the above argument for the last
$\longrightarrow$ from the left works for any $\alpha<1$, namely we have the
following sharp inequality

$\fint_{\mathbb{S}^{n}}\phi\,{\mathbf{P}}^{2m}_{n}(\phi)d\mu_{\mathbb{S}^{n}}\geq\frac{\Gamma(n/2+m)}{\Gamma(n/2-m)}\Big{(}\fint_{\mathbb{S}^{n}}\phi^{\gamma}d\mu_{\mathbb{S}^{n}}\Big{)}^{2/\gamma}$

for any $\gamma:=1-\alpha>0$. This makes sense because
$\Gamma(n/2+m)/\Gamma(n/2-m)<0$.

In the final discussion, we show that we can actually compare the two
inequalities (5.3)β with $\beta\in(1,2n+1)$ and (5.3)α with $\alpha\in(0,1)$
without using the limiting inequality (1.11). Indeed, by decomposing the
constant $1$ as

$1=\phi^{\frac{(1-\alpha)(\beta-1)}{\beta-\alpha}}\phi^{-\frac{(1-\alpha)(\beta-1)}{\beta-\alpha}}$

and applying Hölder’s inequality in the following way

$|\mathbb{S}^{n}|\leq\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{\beta-1}{\beta-\alpha}}\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\beta}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{1-\alpha}{\beta-\alpha}}$

we arrive at

$\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\alpha}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{1}{\alpha-1}}\leq|\mathbb{S}|^{\frac{\alpha-\beta}{(\beta-1)(1-\alpha)}}\Big{(}\int_{\mathbb{S}^{n}}\phi^{1-\beta}d\mu_{\mathbb{S}^{n}}\Big{)}^{\frac{1}{\beta-1}}.$

Combining (5.3)β with (5.4) and the identity

$\frac{\beta+1}{\beta-1}+\frac{2(\alpha-\beta)}{(\beta-1)(1-\alpha)}=\frac{\alpha+1}{\alpha-1}$

gives (5.3)α.

## Data availability

Data sharing is not applicable to this article as no dataset was generated or
analysed during the current study.

## Acknowledgments

This work was initiated and finalized when QAN was visiting the Vietnam
Institute for Advanced Study in Mathematics (VIASM) in 2021 and in 2022. QAN
would like to thank VIASM for hospitality and financial support. The research
of QAN is funded by Vietnam National Foundation for Science and Technology
Development (NAFOSTED) under grant number 101.02-2021.24 “On several real and
complex differential operators”. AH was partially supported by the SNSF grant
no. P4P4P2-194460. Last but not least, the authors would like to thank
Professor Dong Ye and an anonymous referee for their careful reading and
useful comments leading to some improvement of the paper. Proposition 5.2 and
the discussion near the end of section 5 are actually due to them.

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