# Concentration analysis for elliptic critical equations with no boundary
control: ground-state blow-up

Hussein Mesmar  Hussein Mesmar, Institut Élie Cartan, Université de Lorraine,
BP 70239, F-54506 Vandœuvre-les-Nancy, France<EMAIL_ADDRESS>and  Frédéric Robert  Frédéric Robert, Institut Élie Cartan, Université de
Lorraine, BP 70239, F-54506 Vandœuvre-lès-Nancy, France  frederic.robert@univ-
lorraine.fr

(Date: October 2nd, 2023. MSC: 35J60, 58J10)

###### Abstract.

We perform the apriori analysis of solutions to critical nonlinear elliptic
equations on manifolds with boundary. The solutions are of minimizing type.
The originality is that we impose no condition on the boundary, which leads us
to assume $L^{2}-$concentration. We also analyze the effect of a non-
homogeneous nonlinearity that results in the fast convergence of the
concentration point.

Dedicated to Yihong Du on the occasion of his 60th birthday

##  1\. Introduction

###  1.1. Context and main results

Let $(M,g)$ be a Riemannian manifold of dimension $n\geq 3$, with or without
boundary $\partial M$. When $\partial M\neq\emptyset$, $M$ denotes the
interior of the manifold and $\overline{M}$ denotes its closure, so that
$\overline{M}=M\cup\partial M$: in particular, $M$ is open in $\overline{M}$.
We let $a,f\in C^{0}(\overline{M})$ be functions and we consider $u\in
C^{2}(M)$ solution to

(1) $\Delta_{g}u+au=fu^{2^{\star}-1}\,;\,u>0\hbox{ in }M.$

where $\Delta_{g}:=-\hbox{div}_{g}(\nabla)$ is the Laplacian with minus sign
convention and $2^{\star}:=\frac{2n}{n-2}$ is critical for the Sobolev
embeddings $H_{1}^{2}(M)\hookrightarrow L^{2^{\star}}(M)$. Here, the Sobolev
space $H_{1}^{2}(M)$ is the completion of $\\{u\in
C^{\infty}(M)/\,\|u\|_{H_{1}^{2}}<\infty\\}$ for the norm
$\|\cdot\|_{H_{1}^{2}}:=\|\nabla\cdot\|_{2}+\|\cdot\|_{2}$. In the case of a
Euclidean smooth domain $\Omega\subset\mathbb{R}^{n}$, then $a,f\in
C^{0}(\bar{\Omega})$ and we consider $u\in C^{2}(\Omega)$ solution to

(2) $\Delta u+au=fu^{2^{\star}-1}\,;\,u>0\hbox{ in }\Omega.$

where $\Delta:=-\hbox{div}(\nabla)$ is the Euclidean Laplacian. Due to the
critical exponent $2^{\star}$, there might be families of solutions to (1)
that are not relatively compact in $C^{2}_{loc}(M)$. For instance, given
$x_{0}\in\mathbb{R}^{n}$ and $\mu>0$, define the Bubble as

$x\mapsto
U_{\mu,x_{0}}(x):=\left(\frac{\mu}{\mu^{2}+\frac{c|x-x_{0}|^{2}}{n(n-2)}}\right)^{\frac{n-2}{2}}.$

Then for any domain $\Omega\subset\mathbb{R}^{n}$, $U_{\mu,x_{0}}$ is a
solution to (2) when $a\equiv 0$ and $f\equiv c$. Moreover, if
$x_{0}\in\bar{\Omega}$, then $\sup_{\Omega}U_{\mu,x_{0}}\to+\infty$ as
$\mu\to+\infty$. In the Riemannian context, for $x\in M$ and $\mu>0$, the
Bubble writes

$x\mapsto
U^{(M,g)}_{\mu,x_{0}}(x):=\left(\frac{\mu}{\mu^{2}+\frac{cd_{g}(x,x_{0})^{2}}{n(n-2)}}\right)^{\frac{n-2}{2}}.$

Concerning terminology, we say that a family $(u_{\epsilon})_{\epsilon}\in
C^{0}(M)$ blows-up if $\lim_{\epsilon\to 0}\|u_{\epsilon}\|_{\infty}=+\infty$.

When $\partial M=\emptyset$, the description of blowing-up families of (1)
with bounded $L^{2^{\star}}-$norm has been performed by Druet-Hebey-Robert
[dhr]. The main result in [dhr] is that blowing-up families are controled from
above by
$\left(\sum_{i}U_{\mu_{i,\epsilon},x_{i,\epsilon}}\right)_{\epsilon}$, for
given families $(x_{i,\epsilon})_{\epsilon}\in M$ and
$(\mu_{i,\epsilon})_{\epsilon}\to 0$. With this control, it is possible to
give informations on the localization of the limits
$x_{i,\infty}:=\lim_{\epsilon\to 0}x_{i,\epsilon}$: see Druet [druet:jdg].
This analysis extends to manifolds with boundary provided a boundary condition
like Dirichlet (see Ghoussoub-Mazumdar-Robert [gmr]) or Neumann (see Druet-
Robert-Wei [drw]). See Premoselli [premoselli] for a more recent point of
view.

The first objective of the present work is to perform an analysis similar to
[dhr] and [druet:jdg] without condition on $\partial M\neq\emptyset$. Tackling
such generality requires additional assumption: the relevant notion here is
$L^{2}-$concentration that already appeared in Djadli-Druet [dd] (see (7)
below).

Our second objective is to analyse the effect of a nonconstant function $f$ in
(1). In the case of a single peak, concentration occurs at a critical point.
We prove that when this critical point is nondegenerate, then the family of
concentration points converges very fast to its limit (see (6) below): this
does not generally happen for a constant function $f$. A similar control
appears in Malchiodi-Mayer [mm].

As was shown by Aubin [aubin0problemes], below a threshold, blow-up cannot
occur. In this manuscript, we are considering solutions $(u_{\epsilon})$ that
carry the minimal energy for blow-up, namely ground-state type solution. The
minimal energy is given by the best constant in Sobolev embeddings:

(3) $\displaystyle\frac{1}{K_{0}(n)}=\inf_{\varphi\in
D_{1}^{2}(\mathbb{R}^{n})\setminus\\{0\\}}\frac{\int_{\mathbb{R}^{n}}|{\nabla}\varphi|^{2}\mathrm{d}X}{\left(\int_{\mathbb{R}^{n}}|\varphi|^{2^{\star}}\mathrm{d}X\right)^{\frac{2}{2^{\star}}}},$

where $D_{1}^{2}(\mathbb{R}^{n})$ is the completion of
$C_{c}^{\infty}(\mathbb{R}^{n})$ for the norm
$\varphi\mapsto\|\nabla\varphi\|_{2}$. Aubin [aubin0problemes] and Talenti
[talenti0best] have computed this best constant and have showed that the
extremals are exactly $C\cdot U_{\mu,x_{0}}$ for $C\neq 0$, $\mu>0$ and
$x_{0}\in\mathbb{R}^{n}$. Our main theorem for ground-state solutions is the
following:

######  Theorem 1.

Let $(M,g)$ be a smooth compact Riemannian manifold of dimension $n\geq 4$
with nonempty boundary $\partial M\neq\emptyset$. We fix $f\in
C^{2}(\overline{M})$ such that $f>0$. We consider a family
$(h_{\epsilon})_{\epsilon}\in C^{1}(\overline{M})$ and $f\in
C^{2}(\overline{M})$, $f>0$, such that there exists $h\in C^{1}(\overline{M})$
and such that $\Delta_{g}+h$ is coercive and

(4) $\lim_{\epsilon\to 0}h_{\epsilon}=h\hbox{ in }C^{1}(\overline{M}).$

We let $(u_{\epsilon})_{\epsilon}\in C^{2}(\overline{M})$ be a family of
solutions to

(5)
$\Delta_{g}u_{\epsilon}+h_{\epsilon}u_{\epsilon}=fu_{\epsilon}^{2^{\star}-1}\hbox{
in }M.$

Let $x_{\epsilon}\in\overline{M}$ and $\mu_{\epsilon}>0$ be such that

$u_{\epsilon}(x_{\epsilon})=\sup_{\overline{M}}u_{\epsilon}=\mu_{\epsilon}^{1-\frac{n}{2}}$

We assume that

  * •

$u_{\epsilon}\to 0$ in $L^{2}(M)$,

  * •

$\lim_{\epsilon\to 0}x_{\epsilon}=x_{0}\in M$ is an interior point of $M$,

  * •

The solution has minimal-type energy, that is

$\lim_{\epsilon\to
0}\int_{M}fu_{\epsilon}^{2^{\star}}\,dv_{g}=\frac{1}{K_{0}(n)^{\frac{n}{2}}f(x_{0})^{\frac{n-2}{2}}}$

  * •

The Hessian $\nabla^{2}f(x_{0})$ is nondegenerate.

Then $x_{0}$ is a critical point of $f$ and

(6) $\displaystyle d_{g}(x_{\epsilon},x_{0})=o(\mu_{\epsilon})\hbox{ as
}\epsilon\to 0,$

and for all $\omega\subset M$ such that $\overline{\omega}\subset M$ and
$\delta_{0}>0$, there exists $C(\omega,\delta_{0})>0$ such that

$u_{\epsilon}(x)\leq
C(\omega,\delta_{0})\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+d_{g}(x,x_{0})^{2}}\right)^{\frac{n-2}{2}}+C(\omega,\delta_{0})\sup_{\partial
B_{\delta_{0}}(x_{0})}u_{\epsilon}$

for all $x\in\omega$. In addition, assuming that for all $\delta>0$, we have
that

(7) $\lim_{\epsilon\to 0}\frac{\int_{M\setminus
B(x_{0},\delta)}u_{\epsilon}^{2}\,dv_{g}}{\int_{M}u_{\epsilon}^{2}\,dv_{g}}=0\,\hbox{
for }\,n\in\\{4,5,6\\},$

then

(8) $\displaystyle
h(x_{0})=\frac{n-2}{4(n-1)}\left(\hbox{Scal}_{g}(x_{0})-\frac{n-4}{2}\cdot\frac{\Delta_{g}f(x_{0})}{f(x_{0})}\right),$

where $\hbox{Scal}_{g}$ is the scalar curvature of $(M,g)$.

Remark: Theorem 1 applies to the case of a bounded domain of $\mathbb{R}^{n}$
endowed with the Euclidean metric $g:=\hbox{Eucl}$. In this situation,
$M=\Omega\subset\mathbb{R}^{n}$ is a domain,
$\Delta_{g}=-\sum_{i}\partial_{ii}$, $d_{g}(x,y)=|x-y|$ is the usual Euclidean
norm for $x,y\in\mathbb{R}^{n}$ and $\hbox{Scal}_{g}=0$.

The control (6) is remarkable since it does not hold when $f$ is degenerate.
Indeed, when $f\equiv 1$ there is an abundance of blowup profiles with various
speeds of convergence of the $(x_{\epsilon})$’s to their limit, see for
instance Premoselli [premo:jgea].

The restriction of dimension $n\geq 4$ is not surprising: indeed, see
Corollary 6.4 in Druet-Hebey [AB], (7) does not hold in general for $n=3$. It
is known since Aubin and Schoen that for $n=3$, blowup cannot be characterized
by local arguments and involves global arguments, like the mass. In the
general local context of Theorems 1, no information is known regarding the
boundary, which forbids to get any global information.

###  1.2. Application to supercritical problems with symmetries

A natural set application of Theorem 1 is in the context of manifolds
invariant under a group of isometries. We consider a compact Riemannian
manifold $(X,g)$ of dimension $n\geq 3$, but without boundary ($\partial
X=\emptyset$). The critical exponent $2^{\star}$ can be improved by imposing
invariance under the action of an isometry group. Let $G$ be a compact
subgroup of isometries of $(X,g)$: we say that a function $u:X\to\mathbb{R}$
is $G-$invariant if $u\circ\sigma=u$ for all $\sigma\in G$. It follows from
Hebey-Vaugon [hebey0sobolev] that the critical exponent in this setting is
$2^{\star}(k):=\frac{2(n-k)}{n-k-2}$ where $k:=\min_{x\in X}\hbox{dim }Gx$ and
assuming that $1\leq k<n-2$. We refer to Hebey-Vaugon [hebey0sobolev],
Saintier [saintier0blow] and Faget [faget2002best] for extensive
considerations on problems invariant under isometries. In general, the
quotient $X/G$ is not a manifold of dimension $n-k$. Following Saintier
[saintier0blow], we make the following assumption on $G$:

Assumption (H): For any $x_{0}\in X$ such that the orbit $Gx_{0}$ is of
dimension $k=\min\limits_{x\in X}dimGx\geq 1$ and of volume $V_{m}=\min_{x\in
X}\\{\hbox{Vol}_{g}(Gx)/\,\hbox{dim }Gx=k\\}$, there exists $\delta>0$, and
$G^{\prime}$ a closed subgroup of $Isom_{g}(X)$ such that:

  1. (1)

$G^{\prime}x_{0}=Gx_{0}$;

  2. (2)

For all $x\in B_{\delta}(Gx_{0}):=\\{y\in X/d_{g}(y;Gx_{0})<\delta\\}$, then
$G^{\prime}x$ is principal and $G^{\prime}x\subset Gx$.

In particular $dim\,G^{\prime}x=dim\,Gx_{0}=k$, $\forall x\in
B_{\delta}(Gx_{0})$.

This assumption ensures that $B_{\delta}(Gx_{0})/G^{\prime}$ is a Riemannian
manifold of dimension $m:=n-k$ with a nontrivial boundary. In the sequel, for
any $p\in\mathbb{N}$, we define $C^{p}_{G}(X)$ as the space of $G-$invariant
functions of $C^{p}(X)$. We prove the following in the spirit of Faget
[faget:ade].

######  Theorem 2.

Let $(X,g)$ be a compact Riemannian manifold of dimension $n$ without
boundary, and let $G$ be a compact subgroup of isometries of $X$ which
satisfies Assumption $(H)$ and such that $1\leq k<n-2$. Let
$(h_{\epsilon})_{\epsilon}\in C^{1}_{G}(X)$ and $h\in C^{1}_{G}(X)$ be such
that $\Delta_{g}+h$ is coercive and

(9) $\lim_{\epsilon\to 0}h_{\epsilon}=h>0\hbox{ in }C^{1}_{G}(X).$

Let $(u_{\epsilon})_{\epsilon}\in C^{2}_{G}(X)$ be a family of solutions to

(10)
$\Delta_{g}u_{\epsilon}+h_{\epsilon}u_{\epsilon}=\lambda_{\epsilon}u_{\epsilon}^{2^{\star}(k)-1}\,;\,u_{\epsilon}>0\hbox{
in }X,\;\int_{X}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}=1$

We assume that

  * •

$u_{\epsilon}\to 0$ strongly in $L^{2}(X)$,

  * •

The energy is of minimal type, that is

(11) $\lim_{\epsilon\to
0}\lambda_{\epsilon}=\frac{V_{m}^{1-\frac{2}{2^{\star}(k)}}}{K_{0}(n-k)},\hbox{
where }V_{m}:=\min_{x\in X}\\{\hbox{Vol}_{g}(Gx)/\,\hbox{dim }Gx=k\\}.$

  * •

For all point $z_{0}\in X$ such that $\hbox{dim }Gz_{0}=k$ and
$\hbox{Vol}_{g}(Gz_{0})=V_{m}$, then the function

$\left\\{\begin{array}[]{cccc}\bar{v}:&B_{\delta}(Gz_{0})/G^{\prime}&\to&\mathbb{R}\\\
&G^{\prime}x&\to&\hbox{Vol}_{g}(G^{\prime}x)\end{array}\right\\}\hbox{ is
nondegenerate at }Gz_{0}.$

This latest assumption makes sense due to Assumption (H). Let
$(x_{\epsilon})_{\epsilon}\in X$ be such that
$u_{\epsilon}(x_{\epsilon})=\max_{X}u_{\epsilon}$ and define
$\mu_{\epsilon}^{-\frac{n-k-2}{2}}=u_{\epsilon}(x_{\epsilon})$. Then there
exists $x_{0}\in X$ such that $\hbox{dim }Gx_{0}=k$ and
$\hbox{Vol}_{g}(Gx_{0})=V_{m}$ such that $\lim_{\epsilon\to
0}x_{\epsilon}=x_{0}$ and

(12) $\displaystyle d_{g}(x_{\epsilon},Gx_{0})=o(\mu_{\epsilon}).$

Moreover, there exists $C>0$ such that

(13) $u_{\epsilon}(x)\leq
C\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+d_{g}(x,Gx_{0})^{2}}\right)^{\frac{n-2}{2}}+\left\\{\begin{array}[]{cc}o(\mu_{\epsilon})&\hbox{
if }n-k\geq 5\\\
o\left(\mu_{\epsilon}\sqrt{\ln\frac{1}{\mu_{\epsilon}}}\right)&\hbox{ if
}n-k=4\end{array}\right.$

and

(14)
$h(x_{0})=\frac{n-k-2}{4(n-k-1)}\left(\hbox{Scal}_{\bar{g}}(\bar{x}_{0})+3\frac{\Delta_{\bar{g}}\bar{v}(\bar{x}_{0})}{\bar{v}(\bar{x}_{0})}\right)\hbox{
when }\,n-k\geq 4,$

where $\bar{g}$ is the metric on $B_{\delta}(Gx_{0})/G^{\prime}$ such that the
canonical projection
$(B_{\delta}(Gx_{0}),g)\to(B_{\delta}(Gx_{0})/G^{\prime},\bar{g})$ is a
Riemannian submersion.

##  2\. Pointwise control

We consider $(u_{\epsilon})\in C^{2}(\overline{M})$, $(h_{\epsilon})\in
C^{1}(\overline{M})$, $h\in C^{1}(\overline{M})$, $f\in C^{2}(\overline{M})$,
$(x_{\epsilon})_{\epsilon}\in M$ and
$(\mu_{\epsilon})_{\epsilon}\in(0,+\infty)$ as in the statement of Theorem 1.
In the sequel, we let $i_{g}(M,x)>0$ be the injectivity radius of $(M,g)$ at
an interior point $x\in M$.

######  Claim 1.

Set $\delta\in(0,i_{g}(M,x_{0}))$ and define

$w_{\epsilon}(X):=\mu_{\epsilon}^{\frac{n-2}{2}}u_{\epsilon}(\exp_{x_{\epsilon}}(\mu_{\epsilon}X))\hbox{
for any }X\in{B_{\frac{\delta}{\mu_{\epsilon}}}(0)}\subset\mathbb{R}^{n}$

Then

(15) $\lim_{\epsilon\to
0}w_{\epsilon}(X)=w(X)=\left(\frac{1}{1+\frac{f(x_{0})|X|^{2}}{n(n-2)}}\right)^{\frac{n-2}{2}}\hbox{
for all }X\in\mathbb{R}^{n}.$

Moreover, the convergence holds in $C^{2}_{loc}(\mathbb{R}^{n})$. In addition,

$\lim\limits_{R\to+\infty}\lim\limits_{{\epsilon}\to
0}\int_{B_{x_{\epsilon}}(R\mu_{\epsilon})}fu_{\epsilon}^{2^{\star}}\,dv_{g}=\frac{1}{K_{0}(n)^{\frac{n}{2}}f(x_{0})^{\frac{n-2}{2}}}.$

In particular

(16) $\lim\limits_{R\to+\infty}\lim\limits_{{\epsilon}\to 0}\int_{M\setminus
B_{x_{\epsilon}}(R\mu_{\epsilon})}fu_{\epsilon}^{2^{\star}}\,dv_{g}=0.$

Proof of Claim 1: We define the metric
$g_{\epsilon}:=\exp_{x_{\epsilon}}^{\star}g(\mu_{\epsilon}\cdot)$ in
$B_{\frac{\delta}{\mu_{\epsilon}}}(0)\subset\mathbb{R}^{n}$. Since,
$\mu_{\epsilon}\rightarrow 0$ when ${\mathbb{\epsilon}}\rightarrow 0$, then
$g_{\epsilon}\to\xi$ in $C^{2}_{loc}(\mathbb{R}^{n})$ as $\epsilon\to 0$ where
$\xi$ is the Euclidean metric. The function $w_{\epsilon}$ satisfies the
equation

(17)
$\Delta_{g_{\epsilon}}w_{\epsilon}+\mu_{\epsilon}^{2}\tilde{h}_{\epsilon}w_{\epsilon}=\tilde{f}_{\epsilon}w_{\epsilon}^{2^{\star}-1}\hbox{
in }B_{\frac{\delta}{\mu_{\epsilon}}}(0)$

where
$\tilde{h}_{\epsilon}(X)=h_{\epsilon}\left(\exp_{x_{\epsilon}}(\mu_{\epsilon}X)\right)$
and $\tilde{f}_{\epsilon}(X)=f(\exp_{x_{\epsilon}}(\mu_{\epsilon}X))$ for all
$X\in B_{\frac{\delta}{\mu_{\epsilon}}}(0)$. Since $0<w_{\epsilon}\leq
w_{\epsilon}(0)=1$, there exists $w\in C^{2}\left(\mathbb{R}^{n}\right)$ such
that the sequence $w_{\epsilon}\to w$ in $C^{2}_{loc}(\mathbb{R}^{n})$ as
$\epsilon\to 0$ up to extraction. Passing to the limit in (17), we get that

(18) $\Delta_{\xi}{w}=f(x_{0}){w}^{2^{\star}-1}\hbox{ in
}\mathbb{R}^{n},\,0\leq w(0)=1.$

It follows from Cafarelli-Gidas-Spruck [cgs] that
$w(X)=\left(1+\frac{f(x_{0})|X|^{2}}{n(n-2)}\right)^{-\frac{n-2}{2}}$ for all
$X\in\mathbb{R}^{n}$. The change of variable
$x=\hbox{exp}_{x_{\epsilon}}(\mu_{\epsilon}X)$ yields

$\displaystyle\int_{B_{x_{\epsilon}}(R\mu_{\epsilon})}fu_{\epsilon}^{2^{\star}}\,dv_{g}=\int_{B_{R}(0)}f(\hbox{exp}_{x_{\epsilon}}(\mu_{\epsilon}X)w_{\epsilon}^{2^{\star}}\mathrm{d}v_{g_{\epsilon}}.$

Therefore,

$\displaystyle\lim_{R\to+\infty}\lim_{\epsilon\to
0}\int_{B_{x_{\epsilon}}(R\mu_{\epsilon})}fu_{\epsilon}^{2^{\star}}\,dv_{g}$
$\displaystyle=$ $\displaystyle\lim_{R\to+\infty}\lim_{\epsilon\to
0}\int_{B_{R}(0)}f(\hbox{exp}_{x_{\epsilon}}(\mu_{\epsilon}x)w_{\epsilon}^{2^{\star}}\mathrm{d}v_{g_{\epsilon}}$
$\displaystyle=$ $\displaystyle
f(x_{0})\int_{\mathbb{R}^{n}}w^{2^{\star}}\,dx=\frac{1}{K_{0}(n)^{\frac{n}{2}}f(x_{0})^{\frac{n-2}{2}}},$

where we have used that $w$ is a solution to (18) and is an extremal for the
Sobolev inequality (3). This proves Claim 1.∎

######  Claim 2.

$u_{\epsilon}\to 0$ in $C^{0}_{loc}(M\setminus\\{x_{0}\\})$.

Proof of the claim: It follows from (16) and $f>0$ that for all $\delta>0$, we
have that

$\lim_{\epsilon\to 0}\int_{M\setminus
B_{x_{0}}(\delta)}u_{\epsilon}^{2^{\star}}\,\,dv_{g}=0.$

Let us fix $\omega\subset M$ such that $\overline{\omega}\subset
M\setminus\\{x_{0}\\}$. We let $\omega^{\prime}$ open such that
$\bar{\omega}\subset\omega^{\prime}$ and $\bar{\omega^{\prime}}\subset
M-\\{x_{0}\\}$. Let $\eta\in C^{\infty}_{c}(\omega^{\prime})$ such that
$\eta(x)=1$ for all $x\in\omega$. Let us take $l>1$ to be fixed later.
Integrating by parts as in Druet-Hebey ([AB], Theorem 6.1), we get that

$\displaystyle\int_{M}\eta^{2}u_{\epsilon}^{l}\Delta_{g}u_{\epsilon}\,dv_{g}$
$\displaystyle=$ $\displaystyle\int_{M}\nabla(\eta^{2}u_{\epsilon}^{l})\nabla
u_{\epsilon}\,dv_{g}=\int_{M}l\eta^{2}u_{\epsilon}^{l-1}|\nabla
u_{\epsilon}|^{2}\,dv_{g}+\int_{M}\nabla\eta^{2}\nabla\frac{u_{\epsilon}^{l+1}}{l+1}\,dv_{g}$
$\displaystyle=$ $\displaystyle\frac{4l}{(l+1)^{2}}\int_{M}\eta^{2}|\nabla
u_{\epsilon}^{\frac{l+1}{2}}|_{g}^{2}\,dv_{g}+\int_{M}\frac{\Delta\eta^{2}}{l+1}u_{\epsilon}^{l+1}\,dv_{g}$

Independently, for any $v\in C^{1}(M)$, integrating also by parts, we get that

$\int_{M}(|\nabla(\eta v)|_{g}^{2}-\eta^{2}|\nabla
v|_{g}^{2})\,dv_{g}=-\int_{M}\eta v^{2}\Delta_{g}\eta\,dv_{g}.$

Plugging these integrals together yields

$\displaystyle\int_{M}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}=\frac{(l+1)^{2}}{4l}\int_{M}\eta^{2}u_{\epsilon}^{l}\Delta_{g}u_{\epsilon}\,dv_{g}+\frac{l+1}{2l}\int_{M}\left(|{\nabla}\eta|_{g}^{2}+\frac{l-1}{l+1}\eta\Delta_{g}\eta\right)u_{\epsilon}^{l+1}\,dv_{g}$

We then get that

$\displaystyle\int_{M}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}$
$\displaystyle=\frac{(l+1)^{2}}{4l}\int_{M}\eta^{2}u_{\epsilon}^{l}\Delta_{g}u_{\epsilon}\,dv_{g}+\frac{l+1}{2l}\int_{M}\left(|{\nabla}\eta|_{g}^{2}+\frac{l-1}{l+1}\eta\Delta_{g}\eta\right)u_{\epsilon}^{l+1}\,dv_{g}$
$\displaystyle=\frac{(l+1)^{2}}{4l}\lambda_{\epsilon}\int_{M}\eta^{2}u_{\epsilon}^{l+2^{\star}-1}\,dv_{g}-\frac{(l+1)^{2}}{4l}\int_{M}\eta^{2}{h_{\epsilon}}u_{\epsilon}^{l+1}\,dv_{g}$
$\displaystyle+\frac{l+1}{2l}\int_{M}\left(|{\nabla}\eta|_{g}^{2}+\frac{l-1}{l+1}\eta\Delta_{g}\eta\right)u_{\epsilon}^{l+1}\,dv_{g}$
$\displaystyle\leq\frac{(l+1)^{2}}{4l}\lambda_{\epsilon}\left(\int_{M}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}}\,dv_{g}\right)^{\frac{2}{2^{\star}}}\left(\int_{B_{x}(2\delta)}\eta
u_{\epsilon}^{2^{\star}}\,dv_{g}\right)^{1-\frac{2}{2^{\star}}}+C\int_{\Omega}u_{\epsilon}^{l+1}\,dv_{g}$

It follows from the Sobolev inequality that there exists
$C(\omega^{\prime})>0$ independent of $\epsilon$ such that

$\displaystyle\left(\int_{\omega^{\prime}}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}}\,dv_{g}\right)^{\frac{2}{2^{\star}}}\leq
C(\omega^{\prime})\left(\int_{\omega^{\prime}}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}+B\int_{\omega^{\prime}}\eta^{2}u_{\epsilon}^{l+1}\,dv_{g}\right)$

Combining these inequalities yields

$\left(\int_{\omega}u_{\epsilon}^{\frac{l+1}{2}2^{\star}}\,dv_{g}\right)^{\frac{2}{2^{\star}}}\leq\left(\int_{\omega^{\prime}}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}}\,dv_{g}\right)^{\frac{2}{2^{\star}}}\leq
C\int_{\omega^{\prime}}u_{\epsilon}^{l+1}\,dv_{g}$

for $\epsilon>0$ small enough and where $C$ is independent of $\epsilon$.
Taking $1<l<2^{\star}-1$, we then get that $u_{\epsilon}\to 0$ in
$L^{q}(\omega)$ for some $q>2^{\star}$. Since $u_{\epsilon}$ satisfies (5), it
is classical that $u_{\epsilon}\to 0$ in $C^{0}_{loc}(\omega)$. This proves
Claim 2.∎

######  Claim 3.

For all $\omega\subset M$ such that $\overline{\omega}\subset M$, there exists
$C(\omega)$ such that

(19) $d_{g}(x,x_{\epsilon})^{\frac{n-2}{2}}u_{\epsilon}(x)\leq C(\omega)\hbox{
for all }\epsilon>0\hbox{ and }x\in\omega.$

Moreover,

(20) $\lim\limits_{R\to 0}\lim\limits_{{\epsilon}\to
0}\sup\limits_{x\in\omega\setminus
B_{x_{\epsilon}}(R\mu_{\epsilon})}d_{g}(x,x_{\epsilon})^{\frac{n-2}{2}}|u_{\epsilon}(x)|=0$

Proof of the Claim: We argue by contradiction and we let
$(y_{\epsilon})_{\epsilon}\in\overline{\omega}$ be such that

$d_{g}(y_{\epsilon},x_{\epsilon})^{\frac{n-2}{2}}u_{\epsilon}(y_{\epsilon})=\sup_{x\in\overline{\omega}}d_{g}(x,x_{\epsilon})^{\frac{n-2}{2}}u_{\epsilon}(x)\to+\infty\hbox{
as }\epsilon\to 0.$

It follows from Claim 2 that $\lim_{\epsilon\to 0}y_{\epsilon}=x_{0}$. Arguing
as in Step 2 of Chapter 4 in Druet-Hebey-Robert [dhr] and using (16), we get
(19). The second estimate (20) follows also from [dhr].∎

We now state and prove the main result of this section:

######  Proposition 1.

Let $\delta>0$ be such that $B_{2\delta}(x_{0})\subset M$. Under the
assumptions of Theorem 1, we have that

(21)
$u_{\epsilon}(y_{\epsilon})=\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+\frac{f(x_{0})}{n(n-2)}d_{g}(x_{\epsilon},y_{\epsilon})^{2}}\right)^{\frac{n-2}{2}}(1+o(1))+O(\theta_{\epsilon})\hbox{
when }\lim_{\epsilon\to 0}y_{\epsilon}=x_{0}.$

Moreover, there exists $C(\delta)>0$ independent of ${\mathbb{\epsilon}}$ such
that

(22) $\displaystyle u_{\epsilon}(x)$ $\displaystyle\leq
C\frac{\mu_{\epsilon}^{\frac{n-2}{2}}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{n-2}}+C\theta_{\epsilon}$
(23) $\displaystyle|{\nabla}u_{\epsilon}|(x)$ $\displaystyle\leq
C\frac{\mu_{\epsilon}^{\frac{n-2}{2}}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{n-1}}+C\theta_{\epsilon}$

for all $x\in B_{\delta}(x_{0})$ where

(24) $\theta_{\epsilon}:=\sup_{x\in\partial
B_{\delta}(x_{0})}u_{\epsilon}(x)\to 0\hbox{ as }\epsilon\to 0.$

Proof of Proposition 1: We let $\nu\in(0,1)$ to be fixed later. We let
$\alpha_{0}>0$ such that ${\Delta_{g}}+\frac{h-\alpha_{0}}{1-{\nu}}$ is
coercive on $B_{2\delta}(x_{0})$ where $h$ is as in (4): up to taking
$\delta>0$ small, this is always possible. We let $\tilde{G}_{\nu}$ be the
Green’s function of ${\Delta_{g}}+\frac{h-\alpha_{0}}{1-{\nu_{1}}}$ on
$B_{2\delta}(x_{0})$ with Dirichlet boundary condition. It follows from Robert
[robert0existence] that there exists $c_{1},c_{2}>0$ such that

(25) $c_{1}d_{g}(x,y)^{2-n}\leq\tilde{G}_{\nu}(x,y)\leq
c_{2}d_{g}(x,y)^{2-n}\hbox{ for all }x,y\in B_{\delta}(x_{0}),\,x\neq y.$

We define the operator

$u\mapsto
L_{\epsilon}u:=\Delta_{g}u+h_{\epsilon}u-fu_{\epsilon}^{2^{\star}-2}u,$

so that (5) reads $L_{\epsilon}u_{\epsilon}=0$. A straightforward computation
yields

(26)
$\displaystyle\frac{L_{\epsilon}{\tilde{G}_{\nu}^{1-\nu}}}{{\tilde{G}_{\nu}^{1-\nu}}}({x},x_{\epsilon})$
$\displaystyle=$
$\displaystyle\alpha_{0}+h_{\epsilon}(x)-h({x})+{\nu}(1-{\nu})\left|\frac{{\nabla}{\tilde{G}_{\nu}}}{{\tilde{G}_{\nu}}}\right|_{g}^{2}-fu_{\epsilon}^{2^{\star}-2}$

By standard properties of Green’s function [robert0existence], there exists
$c_{1},\rho>0$, such that

(27)
$\frac{|{\nabla_{g,x}}{\tilde{G}_{\nu}}|_{g}}{{\tilde{G}_{\nu}}}(x,x_{\epsilon})\geq\frac{c_{1}}{{d_{g}(x,x_{{\epsilon}})}}\hbox{
for all }x\in B_{\rho}(x_{\epsilon})-\\{x_{\epsilon}\\}.$

Since $u_{\epsilon}\to 0$ in $C_{loc}^{0}(M\setminus\\{x_{0}\\})$ and
$h_{\epsilon}\to h$ in $C_{loc}^{0}(M\setminus\\{x_{0}\\})$, (26) yields

$\displaystyle L_{\mathbb{\epsilon}}{\tilde{G}_{\nu}^{1-\nu}}\geq 0\hbox{ in
}B_{2\delta}(x_{0})\setminus B_{\rho}(x_{0})$

Let $R>0$ to be fixed later. It follows from (20) that

$\displaystyle{d_{g}(x,x_{{\epsilon}})}^{2}u_{\epsilon}^{2^{\star}-2}({x})\leq\eta(R)\hbox{
for all }{x}\in B({x_{{\mathbb{\epsilon}}}},\rho)\setminus
B({x_{{\mathbb{\epsilon}}}},R\mu_{\epsilon}),$

where $\lim_{R\to+\infty}\eta(R)=0$. Now, using $h_{\epsilon}\to h$ in
$C^{0}(M)$, (26) and (27), for any $x\in B(x_{\epsilon},\rho)\setminus
B(x_{\epsilon},R\mu_{\epsilon})$, we get that

$\displaystyle\frac{L_{\epsilon}{\tilde{G}_{\nu}^{1-\nu}}}{{\tilde{G}_{\nu}^{1-\nu}}}({x},x_{\epsilon})$
$\displaystyle\geq\frac{\alpha_{0}}{2}+{\nu}(1-{\nu})\frac{c_{1}^{2}}{{d_{g}(x,x_{{\epsilon}})}^{2}}-fu_{\epsilon}^{2^{\star}-2}({x})$
$\displaystyle\geq\frac{\alpha_{0}}{2}+\frac{{\nu}(1-{\nu})c_{1}^{2}-f\eta(R)}{{d_{g}(x,x_{{\epsilon}})}^{2}}\geq
0$

for $R>0$ large enough. Therefore, we get that

(28) $\displaystyle
L_{\mathbb{\epsilon}}{\tilde{G}_{\nu}^{1-\nu}}({x},{x_{{\mathbb{\epsilon}}}})\geq
0\hbox{ for all }x\in B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon}).$

We fix $\nu_{1}\in(0,1)$. It follows from (25) and
${\mathbb{||}}{u_{\mathbb{\epsilon}}}{\mathbb{||}}_{\infty}=\mu_{\epsilon}^{1-n/2}$,
that there exists $c_{3}>0$ such that

(29) $u_{\epsilon}(x)\leq
c_{3}\mu_{\epsilon}^{\frac{n-2}{2}-{\nu_{1}}(n-2)}{\tilde{G}_{\nu_{1}}^{1-\nu_{1}}}({x},x_{\epsilon})\hbox{
for all }x\in\partial B_{R\mu_{\epsilon}}(x_{\epsilon}).$

We set $\theta_{\epsilon}:=\sup_{x\in\partial
B_{\delta}(x_{0})}u_{\epsilon}(x)$. It follows from Claim 2 that
$\lim_{\epsilon\to 0}\theta_{\epsilon}=0$. We fix $\nu_{2}\in(0,1)$ and we
consider the Green’s function $\tilde{G}_{\nu_{2}}$. It follows from (25) that
there exists $c_{4}>0$ such that

(30) $\displaystyle u_{\epsilon}({x})\leq
c_{4}{{\theta}_{\epsilon}}{\tilde{G}_{\nu_{2}}^{1-\nu_{2}}}(x,x_{\epsilon})\hbox{
for all }x\in\partial B_{\delta}(x_{0}).$

We define

$H_{\epsilon}(x):=c_{3}\mu_{\epsilon}^{\frac{n-2}{2}-{\nu_{1}}(n-2)}{\tilde{G}_{\nu_{1}}^{1-\nu_{1}}}(x,x_{\epsilon})+c_{4}{{\theta}_{\epsilon}}{\tilde{G}_{\nu_{2}}^{1-\nu_{2}}}(x,x_{\epsilon})\hbox{
for }x\in B_{2\delta}(x_{\epsilon})-\\{x_{\epsilon}\\}.$

It follows from (28), (29) and (30) that

$\left\\{\begin{array}[]{cc}L_{\epsilon}u_{\epsilon}=0\leq
L_{\epsilon}H_{\epsilon}&\hbox{ in }B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon}))\\\ 0<u_{\epsilon}\leq H_{\epsilon}&\hbox{
on }\partial\left(B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon})\right)\end{array}\right.$

Since $L_{\epsilon}u_{\epsilon}\geq 0$ in
$\overline{B_{\delta}(x_{0})\setminus B_{R\mu_{\epsilon}}(x_{\epsilon})}$, it
follows from [berestycki1994principal0principe0de0max] that

$u_{\epsilon}\leq H_{\epsilon}\hbox{ in }B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}({x_{{\mathbb{\epsilon}}}}).$

Using the pointwise control (25) and that
$\|u_{\epsilon}\|_{\infty}=\mu_{\epsilon}^{1-n/2}$, we get that for all
$\nu_{1},\nu_{2}\in(0,1)$, there exists $C_{\nu_{1},\nu_{2}}>0$ such that

(31) $u_{\epsilon}({x})\leq
C_{\nu_{1},\nu_{2}}\left(\frac{\mu_{\epsilon}^{\frac{n-2}{2}-{\nu_{1}}(n-2)}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{(n-2)(1-\nu_{1})}}+\theta_{\epsilon}{d_{g}(x,x_{{\epsilon}})}^{(2-n)(1-{\nu_{2}})}\right)$

for all $x\in B_{\delta}(x_{0})$. Our next step is to prove (21). We let
$(y_{\epsilon})_{\epsilon}\in M$ such that $\lim_{\epsilon\to
0}y_{\epsilon}=x_{0}$.

We first assume that $d_{g}(x_{\epsilon},y_{\epsilon})=O(\mu_{\epsilon})$ as
$\epsilon\to 0$. Then, (21) is a direct consequence of (15).

From now on, we assume that

$\lim_{\epsilon\to 0}d_{g}(x_{\epsilon},y_{\epsilon})=0\hbox{ and
}\lim_{\epsilon\to
0}\frac{d_{g}(x_{\epsilon},y_{\epsilon})}{\mu_{\epsilon}}=+\infty.$

We let $G_{\epsilon}$ be the Green’s function for $\Delta_{g}+h_{\epsilon}$ in
$B_{\delta}(x_{0})$ with Dirichlet boundary condition. We let
$(y_{\epsilon})_{\epsilon}\in B_{\delta/2}(x_{0})$. Green’s representation
formula yields

(32)
$u_{\epsilon}(y_{\epsilon})=\int_{B_{\delta}(x_{0})}G_{\epsilon}(y_{\epsilon},x)(\Delta_{g}u_{\epsilon}+h_{\epsilon}u_{\epsilon})(x)\,dv_{g}(x)-\int_{\partial
B_{\delta}(x_{0})}\partial_{\vec{n}}G_{\epsilon}(y_{\epsilon},z)u_{\epsilon}(z)\,d\sigma_{g}(z).$

It follows from Robert [robert0existence] that there exists $c_{5},c_{6}>0$
such that

(33)
$d_{g}(x,y)^{n-2}|G_{\epsilon}(x,y)|+d_{g}(x,y)^{n-1}|\nabla_{x}G_{\epsilon}(x,y)|\leq
c_{5}\hbox{ for all }x,y\in B_{\delta}(x_{0}),\,x\neq y$

for all $\epsilon>0$. Combining these estimates with equation (5) and (24), we
get that

(34) $\displaystyle u_{\epsilon}(y_{\epsilon})$ $\displaystyle=$
$\displaystyle\int_{B_{R\mu_{\epsilon}}(x_{\epsilon})}G_{\epsilon}(y_{\epsilon},x)f(x)u_{\epsilon}^{2^{\star}-1}(x)\,dv_{g}(x)+A_{\epsilon}(R)+B_{\epsilon}$

where

$|A_{\epsilon}(R)|\leq C\int_{B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon})}d_{g}(x,y_{\epsilon})^{2-n}u_{\epsilon}^{2^{\star}-1}(x)\,dv_{g}(x)$
$\hbox{ and }|B_{\epsilon}|\leq C\int_{\partial
B_{\delta}(x_{0})}d_{g}(z,y_{\epsilon})^{1-n}u_{\epsilon}(z)\,d\sigma_{g}(z)\leq
C\theta_{\epsilon}.$

We deal with the first term of (34). With a change of variable and (15), we
get that

$\displaystyle\int_{B_{R\mu_{\epsilon}}(x_{\epsilon})}G_{\epsilon}(y_{\epsilon},x)f(x)u_{\epsilon}^{2^{\star}-1}(x)\,dv_{g}(x)$
$\displaystyle=\mu_{\epsilon}^{n-2}\int_{B_{R}(0)}G_{\epsilon}(y_{\epsilon},\hbox{exp}_{x_{\epsilon}}(\mu_{\epsilon}X))f(\hbox{exp}_{x_{\epsilon}}(\mu_{\epsilon}X))w_{\epsilon}^{2^{\star}-1}(X)\,dv_{g_{\epsilon}}(X)$

It follow from [robert0existence] that for any $(z_{\epsilon})_{\epsilon}\in
M$ such that $\lim_{\epsilon\to 0}d_{g}(z_{\epsilon},x_{\epsilon})=0$ we have
that

$\lim_{\epsilon\to
0}d_{g}(x_{\epsilon},z_{\epsilon})^{n-2}G_{\epsilon}(x_{\epsilon},z_{\epsilon})=\frac{1}{(n-2)\omega_{n-1}}.$

Since $\mu_{\epsilon}=o(d_{g}(x_{\epsilon},y_{\epsilon}))$, we then get that

$\displaystyle\int_{B_{R\mu_{\epsilon}}(x_{\epsilon})}G_{\epsilon}(y_{\epsilon},x)f(x)u_{\epsilon}^{2^{\star}-1}(x)\,dv_{g}(x)$
$\displaystyle=\frac{f(x_{0})\mu_{\epsilon}^{n-2}}{(n-2)\omega_{n-1}d_{g}(x_{\epsilon},y_{\epsilon})^{n-2}}\left(\int_{B_{R}(0)}w^{2^{\star}-1}(X)\,dX+o(1)\right)$
(35)
$\displaystyle=\frac{f(x_{0})\mu_{\epsilon}^{n-2}}{(n-2)\omega_{n-1}d_{g}(x_{\epsilon},y_{\epsilon})^{n-2}}\left(\int_{\mathbb{R}^{n}}w^{2^{\star}-1}(X)\,dX+o(1)+\eta(R)\right)$

where $\lim_{R\to+\infty}\eta(R)=0$. With (18) and (15), we get that

(36) $\displaystyle f(x_{0})\int_{\mathbb{R}^{n}}w^{2^{\star}-1}(X)\,dX$
$\displaystyle=$ $\displaystyle\lim_{R\to+\infty}\int_{B(0,R)}\Delta
w\,dX=\lim_{R\to+\infty}\int_{\partial B(0,R)}(-\partial_{\nu}w)\,d\sigma$
$\displaystyle=$
$\displaystyle\left(\frac{n(n-2)}{f(x_{0})}\right)^{\frac{n-2}{2}}(n-2)\omega_{n-1}.$

We now deal with $A_{\epsilon}(R)$. Using the pointwise control (31), we get
that

(37) $\displaystyle|A_{\epsilon}(R)|$ $\displaystyle\leq$ $\displaystyle
C_{\nu_{1},\nu_{2}}\int_{B_{\delta}(x_{0})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon})}d_{g}(x,y_{\epsilon})^{2-n}\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{(n+2)(1-\nu_{1})}}\,dv_{g}(x)$
$\displaystyle+C_{\nu_{1},\nu_{2}}\theta_{\epsilon}^{2^{\star}-1}\int_{B_{\delta}(x_{0})}d_{g}(x,y_{\epsilon})^{2-n}{d_{g}(x,x_{{\epsilon}})}^{-(n+2)(1-{\nu_{2}})}\,dv_{g}(x)+C\theta_{\epsilon}.$

It follows from Giraud’s lemma (see Appendix A of [dhr] for instance) that for
$\alpha,\beta\in(0,n-2)$ such that $\alpha+\beta>n$, there exists $C>0$ such
that

$\int_{B_{\delta}(x_{0})}d_{g}(y,x)^{\alpha-n}d_{g}(x,y)^{\beta-n}\,dv_{g}(x)\leq
C\hbox{ for all }y,z\in B_{\delta}(x_{0}).$

Taking $1-\nu_{2}>0$ close to $0$, we then get that

(38)
$\theta_{\epsilon}^{2^{\star}-1}\int_{B_{\delta}(x_{0})}d_{g}(x,y_{\epsilon})^{2-n}{d_{g}(x,x_{{\epsilon}})}^{-(n+2)(1-{\nu_{2}})}\,dv_{g}(x)\leq
C\theta_{\epsilon}^{2^{\star}-1}\leq C\theta_{\epsilon}.$

We now deal with the remaining term of (37). We split the domain
$B_{\delta}(x_{0})=D_{\epsilon}^{1}\cup D^{2}_{\epsilon}$ where

$D_{\epsilon}^{1}:=\\{x\in B_{\delta}(x_{0})\hbox{ s.t.
}d_{g}(x,y_{\epsilon})\geq d_{g}(x_{\epsilon},y_{\epsilon})/2\\}$ $\hbox{ and
}D_{\epsilon}^{2}:=\\{x\in B_{\delta}(x_{0})\hbox{ s.t.
}d_{g}(x,y_{\epsilon})<d_{g}(x_{\epsilon},y_{\epsilon})/2\\}.$

We fix $R>0$. With the change of variable
$x:=\hbox{exp}_{g}(\mu_{\epsilon}X)$, we get that

$\displaystyle\int_{D_{\epsilon}^{1}\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon})}d_{g}(x,y_{\epsilon})^{2-n}\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{(n+2)(1-\nu_{1})}}\,dv_{g}(x)$
$\displaystyle\leq
Cd_{g}(x_{\epsilon},y_{\epsilon})^{2-n}\int_{B_{2\delta}(x_{\epsilon})\setminus
B_{R\mu_{\epsilon}}(x_{\epsilon})}\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{(n+2)(1-\nu_{1})}}\,dv_{g}(x)$
$\displaystyle\leq
Cd_{g}(x_{\epsilon},y_{\epsilon})^{2-n}\mu_{\epsilon}^{\frac{n-2}{2}}\int_{\mathbb{R}^{n}\setminus
B_{R}(0)}\frac{1}{(1+|X|)^{(n+2)(1-\nu_{1})}}\,dX$ (39)
$\displaystyle\leq\eta(R)d_{g}(x_{\epsilon},y_{\epsilon})^{2-n}\mu_{\epsilon}^{\frac{n-2}{2}}\hbox{
where }\lim_{R\to+\infty}\eta(R)=0$

when $\nu_{1}<2/(n+2)$. Concerning the other integral, note that for all $x\in
D_{\epsilon}^{2}$, we have that $d_{g}(x,x_{\epsilon})\geq
d_{g}(x_{\epsilon},y_{\epsilon})/2$. Therefore

$\displaystyle\int_{D_{\epsilon}^{2}}d_{g}(x,y_{\epsilon})^{2-n}\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{(n+2)(1-\nu_{1})}}\,dv_{g}(x)$
$\displaystyle\leq
C\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{d_{g}(y_{\epsilon},x_{\epsilon})^{(n+2)(1-\nu_{1})}}\int_{d_{g}(x,y_{\epsilon})<d_{g}(x_{\epsilon},y_{\epsilon})/2}d_{g}(x,y_{\epsilon})^{2-n}\,dv_{g}(x)$
$\displaystyle\leq
C\frac{\mu_{\epsilon}^{\frac{n+2}{2}-{\nu_{1}}(n+2)}}{d_{g}(y_{\epsilon},x_{\epsilon})^{(n+2)(1-\nu_{1})}}d_{g}(x_{\epsilon},y_{\epsilon})^{2}$
(40) $\displaystyle\leq
C\frac{\mu_{\epsilon}^{\frac{n-2}{2}}}{d_{g}(x_{\epsilon},y_{\epsilon})^{n-2}}\left(\frac{\mu_{\epsilon}}{d_{g}(y_{\epsilon},x_{\epsilon})}\right)^{2-\nu_{1}(n+2)}=o\left(\frac{\mu_{\epsilon}^{\frac{n-2}{2}}}{d_{g}(x_{\epsilon},y_{\epsilon})^{n-2}}\right)$

when $\nu_{1}<2/(n+2)$. Putting (35), (36), (37), (38), (39) and (40) together
yields

$u_{\epsilon}(y_{\epsilon})=\left(\frac{n(n-2)}{f(x_{0})}\right)^{\frac{n-2}{2}}\frac{\mu_{\epsilon}^{\frac{n-2}{2}}}{d_{g}(x_{\epsilon},y_{\epsilon})^{n-2}}(1+o(1))+O(\theta_{\epsilon})$

This yields (21) since
$d_{g}(x_{\epsilon},y_{\epsilon})/\mu_{\epsilon}\to+\infty$ as $\epsilon\to
0$.

When $d_{g}(x_{\epsilon},y_{\epsilon})=o(1)$, (22) is a direct consequence of
(21). Since $u_{\epsilon}\to 0$ in $C^{0}_{loc}(M-\\{x_{0}\\})$ and
$u_{\epsilon}>0$, it follows from Harnack’s inequality that there exists
$c(\tau)>0$ such that $u_{\epsilon}(x)\leq c(\tau)u_{\epsilon}(y)$ for all
$x,y\in B_{2\delta}(x_{0})\setminus B_{\tau}(x_{0})$. Therefore, if
$(y_{\epsilon})_{\epsilon}\in B_{2\delta}(x_{0})$ is such that
$y_{\epsilon}\not\to x_{0}$, we have that
$u_{\epsilon}(y_{\epsilon})=O(\theta_{\epsilon})$. This proves (22) when $x$
is far from $x_{0}$. This proves (21) holds in all cases, which yields (22).

Concerning the gradient estimate, differentiate Green’s representation formula
(32) to obtain

$\nabla
u_{\epsilon}(y_{\epsilon})=\int_{B_{\delta}(x_{0})}\nabla_{y}G_{\epsilon}(y_{\epsilon},x)(\Delta_{g}u_{\epsilon}+h_{\epsilon}u_{\epsilon})(x)\,dv_{g}(x)-\int_{\partial
B_{\delta}(x_{0})}\partial_{\vec{n}}\nabla_{y}G_{\epsilon}(y_{\epsilon},z)u_{\epsilon}(z)\,d\sigma_{g}(z).$

Using the pointwise control (33), we then get that

$|\nabla u_{\epsilon}(y_{\epsilon})|\leq
C\int_{B_{\delta}(x_{0})}d_{g}(y_{\epsilon},x)^{1-n}u_{\epsilon}^{2^{\star}-1}(x)\,dv_{g}(x)+C\int_{\partial
B_{\delta}(x_{0})}d_{g}(y_{\epsilon},z)^{-n}u_{\epsilon}(z)\,d\sigma_{g}(z).$

We get (23) arguing as in the proof of (22). This proves Proposition 1.∎

##  3\. Speed of convergence of $(x_{\epsilon})_{\epsilon}$

Let $\Omega\subset{\mathbb{R}}^{n}$ be a smooth bounded domain. Let $u\in
C^{2}(\bar{\Omega})$, $u>0$, and $f\in C^{1}(\bar{\Omega})$ be functions and
$c\in{\mathbb{R}}$. Then for all $z\in\mathbb{R}^{n}$, the Pohozaev identity
writes

(41)
$\displaystyle\int_{\Omega}\left((x-z)^{i}\partial_{i}u+\frac{n-2}{2}u\right)\left(\Delta_{\xi}u-cfu^{2^{\star}-1}\right)dx$
$\displaystyle=\int_{\partial\Omega}\left[(x-z,\nu)\left(\frac{|\nabla
u|_{\xi}^{2}}{2}-\frac{cfu^{2^{\star}}}{2^{\star}}\right)-\left((x-z)^{i}\partial_{i}u+\frac{n-2}{2}u\right)\partial_{\nu}u\right]\mathrm{d}\sigma$
$\displaystyle+\frac{1}{2^{\star}}\int_{\Omega}c\langle{\nabla}f(x),x-z\rangle_{\xi}u^{2^{\star}}dx$

Differentiating with respect to $z$, we get that for any $j\in\\{1,...,n\\}$

(42)
$\displaystyle-\int_{\Omega}\partial_{j}u\left(\Delta_{\xi}u-cfu^{2^{\star}-1}\right)\mathrm{dx}$
$\displaystyle=\int_{\partial\Omega}\left[-\nu_{j}\left(\frac{|\nabla
u|_{\xi}^{2}}{2}-\frac{cfu^{2^{\star}}}{2^{\star}}\right)+\partial_{j}u\,\partial_{\nu}u\right]\mathrm{d}\sigma-\frac{c}{2^{\star}}\int_{\Omega}{\partial_{j}}f(x)u^{2^{\star}}\mathrm{d}x$

We refer to Ghoussoub-Robert [gmr] for a proof. We fix
$\delta\in(0,i_{g}(M,x_{0}))$. We define

$\hat{u}_{\epsilon}(X):=u_{\epsilon}\left(\exp_{x_{\epsilon}}(X)\right)\hbox{
for all }X\in B_{\delta}(0)\subset{\mathbb{R}}^{n}.$

Therefore, equation (5) rewrites

$\Delta_{\hat{g}_{\epsilon}}\hat{u}_{\epsilon}+\hat{u}_{\epsilon}u_{\epsilon}=\hat{f}_{\epsilon}\hat{u}_{\epsilon}^{2^{\star}-1}\hbox{
in }B_{\delta}(0).$

where $\hat{h}_{\epsilon}(X):=h_{\epsilon}\left(\exp_{x_{\epsilon}}(X)\right)$
and $\hat{f}_{\epsilon}(X):=f\left(\exp_{x_{\epsilon}}(X)\right)$ for all
$X\in B_{\delta}(0)\subset{\mathbb{R}}^{n}$ and
$\hat{g}_{\epsilon}:=\hbox{exp}^{\star}_{x_{\epsilon}}g$ is the pull-back of
$g$ via the exponential map.

######  Lemma 1.

Let $(\phi_{\epsilon})_{\epsilon}\in C^{0}(B_{\delta}(0))$ such that

$\left\\{\begin{array}[]{c}\lim_{\epsilon\to
0}\phi_{\mathbb{\epsilon}}(0)=s\in\mathbb{R},\\\
|\phi_{\epsilon}(X)-\phi_{\epsilon}(0)|\leq C|X|\hbox{ for all }X\in
B_{\delta}(0)\hbox{ and }\epsilon>0.\end{array}\right.$

We fix $p\geq 0$ and $q\geq 1$. Then

$\displaystyle\int_{B_{\delta}(0)}\phi_{\epsilon}|X|^{p}\hat{u}_{\epsilon}^{q}\,dX$
$\displaystyle=\left\\{\begin{array}[]{cc}\mu_{\epsilon}^{n+p-\frac{q(n-2)}{2}}\left(s\int_{{\mathbb{R}}^{n}}|X|^{p}{w}^{q}\mathrm{d}X+o(1)\right)+O(\theta_{\epsilon}^{q})&\hbox{
if }(n-2)q>p+n,\\\
\left(s\left(\frac{n(n-2)}{f(x_{0})}\right)^{q\frac{n-2}{2}}\omega_{n-1}+o(1)\right)\mu_{\epsilon}^{\frac{q(n-2)}{2}}\ln\left(\frac{1}{\mu_{\epsilon}}\right)+O(\theta_{\epsilon}^{q})&\hbox{
if }(n-2)q=p+n,\\\
O(\mu_{\epsilon}^{q\frac{n-2}{2}})+O(\theta_{\epsilon}^{q})&\hbox{ if
}(n-2)q<p+n\end{array}\right.$

Moreover, for any family $(\delta_{\epsilon})_{\epsilon}\in(0,1)$ such that
$\lim_{\epsilon\to 0}\delta_{\epsilon}=\lim_{\epsilon\to
0}\frac{\mu_{\epsilon}}{\delta_{\epsilon}}=0$, we have that

(44)
$\int_{B_{\delta_{\epsilon}}(0)}\phi_{\epsilon}|X|^{p}\hat{u}_{\epsilon}^{q}\,dX=\mu_{\epsilon}^{n+p-\frac{q(n-2)}{2}}\left(s\int_{{\mathbb{R}}^{n}}|X|^{p}{w}^{q}\mathrm{d}X+o(1)\right)+O(\theta_{\epsilon}^{q})\hbox{
if }q>\frac{p+n}{n-2}.$

Proof: We fix $\nu>0$. It follows from (21) and (22) that there exists
$\alpha\in(0,\delta)$ such that

$\left|\hat{u}_{\epsilon}^{q}(X)-\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+\frac{f(x_{0})}{n(n-2)}|X|^{2}}\right)^{q\frac{n-2}{2}}\right|\leq\nu\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+\frac{f(x_{0})}{n(n-2)}|X|^{2}}\right)^{q\frac{n-2}{2}}+C\theta_{\epsilon}^{q}$

for all $X\in B_{\rho}(0)\setminus B_{\alpha}(0)$. Note that for all
$\alpha\in(0,\delta)$, it follows from the Harnack inequality that

$\int_{B_{\delta}(0)\setminus
B_{\alpha}(0)}\phi_{\epsilon}|X|^{p}\hat{u}_{\epsilon}^{q}\,dX=O(\theta_{\epsilon}^{q}).$

We then get that

$\displaystyle\left|\int_{B_{\delta}(0)}\phi_{\epsilon}|X|^{p}\hat{u}_{\epsilon}^{q}\,dX-\int_{B_{\alpha}(0)}\phi_{\epsilon}|X|^{p}\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+\frac{f(x_{0})}{n(n-2)}|X|^{2}}\right)^{q\frac{n-2}{2}}\,dX\right|$
$\displaystyle\leq
C\nu\int_{B_{\delta}(0)}|X|^{p}\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+|X|^{2}}\right)^{q\frac{n-2}{2}}\,dX+C\theta_{\epsilon}^{q}$

We the get Lemma 1 when $q(n-2)<n+p$. With the change of variable
$X=\mu_{\epsilon}Y$, we get that

$\displaystyle\int_{B_{\alpha}(0)}\phi_{\epsilon}|X|^{p}\left(\frac{\mu_{\epsilon}}{\mu_{\epsilon}^{2}+\frac{f(x_{0})}{n(n-2)}|X|^{2}}\right)^{q\frac{n-2}{2}}\,dX$
$\displaystyle=\mu_{\epsilon}^{n+p-q\frac{n-2}{2}}\int_{B_{\alpha/\mu_{\epsilon}}(0)}\phi_{\epsilon}(\mu_{\epsilon}Y)|Y|^{p}\left(\frac{1}{1+\frac{f(x_{0})}{n(n-2)}|Y|^{2}}\right)^{q\frac{n-2}{2}}\,dY$
$\displaystyle=\mu_{\epsilon}^{n+p-q\frac{n-2}{2}}\left\\{\begin{array}[]{cc}s\int_{\mathbb{R}^{n}}|Y|^{p}w^{q}(Y)\,dY+o(1)&\hbox{
if }q(n-2)>p+n\\\
s\left(\frac{n(n-2)}{f(x_{0})}\right)^{q\frac{n-2}{2}}\omega_{n-1}\ln\left(\frac{1}{\mu_{\epsilon}}\right)+o(\ln\mu_{\epsilon})&\hbox{
if }q(n-2)=p+n\end{array}\right.$

The proof of (44) is similar by taking $\alpha:=\delta_{\epsilon}$. Putting
these estimates together yields Lemma 1.∎

We now prove (6). We fix $l\in\\{1,...,n\\}$. We define

$\delta_{\epsilon}:=\mu_{\epsilon}^{\frac{1}{n-1}}.$

Pohozaev’s identity (42) applied to $\hat{u}_{\epsilon}$ reads

(46) $\displaystyle A_{\epsilon}=-B_{\epsilon}+C_{\epsilon}-D_{\epsilon}$

With

$\displaystyle A_{\epsilon}$
$\displaystyle:=-\frac{1}{2^{\star}}\int_{{B_{\delta_{\epsilon}}(0)}}{\partial_{l}}{{\hat{f}}_{\epsilon}}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X$
$\displaystyle B_{\epsilon}$
$\displaystyle:=\int_{\partial{B_{\delta_{\epsilon}}(0)}}-\left[\nu_{l}\left(\frac{|{\nabla}{{\hat{u}}_{{\epsilon}}}|^{2}}{2}-\frac{{\hat{f}}_{\epsilon}{\hat{u}}_{\epsilon}^{2^{\star}}}{2^{\star}}\right)+\partial_{l}{{\hat{u}}_{{\epsilon}}}\partial_{\nu}{{\hat{u}}_{{\epsilon}}}\right]\mathrm{d}\nu$
$\displaystyle C_{\epsilon}$
$\displaystyle:=\int_{{B_{\delta_{\epsilon}}(0)}}\partial_{l}{{\hat{u}}_{{\epsilon}}}{{\hat{h}}_{{\epsilon}}}{{\hat{u}}_{{\epsilon}}}\mathrm{d}X\hbox{
and
}D_{\epsilon}:=\int_{{B_{\delta_{\epsilon}}(0)}}\partial_{l}{{\hat{u}}_{{\epsilon}}}(\Delta_{\xi}{{\hat{u}}_{{\epsilon}}}-\Delta_{{{\hat{g}}_{{\epsilon}}}}{{\hat{u}}_{{\epsilon}}})\mathrm{d}X$

We estimate these terms separately. It follows from (22) and (23) that

$B_{\epsilon}=O\left(\mu_{\epsilon}\left(\left(\frac{\mu_{\epsilon}}{\delta_{\epsilon}}\right)^{n-3}+\frac{\mu_{\epsilon}^{n-1}}{\delta_{\epsilon}^{n+1}}+\frac{\delta_{\epsilon}^{n-1}}{\mu_{\epsilon}}\theta_{\epsilon}^{2}\right)\right)=o(\mu_{\epsilon})\hbox{
as }\epsilon\to 0.$

Concerning $C_{\epsilon}$, integrating by parts, we have that

$C_{\epsilon}=\int_{{B_{\delta_{\epsilon}}(0)}}{\partial_{l}}{{\hat{u}}_{{\epsilon}}}{{\hat{h}}_{{\epsilon}}}{{\hat{u}}_{{\epsilon}}}\mathrm{d}X=-\int_{{B_{\delta_{\epsilon}}(0)}}\partial_{l}\frac{{{\hat{h}}_{{\epsilon}}}}{2}{{\hat{u}}_{{\epsilon}}}^{2}\mathrm{d}X+\int_{\partial{B_{\delta_{\epsilon}}(0)}}{{\hat{h}}_{{\epsilon}}}\frac{{{\hat{u}}_{{\epsilon}}}^{2}}{2}\vec{\nu}_{l}\mathrm{d}\sigma$

With (22) and Lemma 1, we then get that

$\displaystyle
C_{\epsilon}=O\left(\mu_{\epsilon}\left(o(1)+\left(\frac{\mu_{\epsilon}}{\delta_{\epsilon}}\right)^{n-3}+\frac{\delta_{\epsilon}^{n-1}}{\mu_{\epsilon}}\theta_{\epsilon}^{2}\right)\right)=o(\mu_{\epsilon})\hbox{
as }\epsilon\to 0.$

We now estimate $D_{\mathbb{\epsilon}}$. We write

$-(\Delta_{{{\hat{g}}_{{\mathbb{\epsilon}}}}}-\Delta_{\xi})=({{\hat{g}}_{\mathbb{\epsilon}}^{ij}}-{\delta^{ij}}){\partial_{ij}}-{{\hat{g}}_{\mathbb{\epsilon}}^{ij}}{\hat{\Gamma}_{ij}^{k}}(\hat{g}_{\epsilon}){\partial_{k}}$

where the ${\hat{\Gamma}_{ij}^{k}}$’s are the Christoffel symbols of the
metric $\hat{g}_{\epsilon}$. The following lemma is reminiscent in such
problems:

######  Lemma 2.

Let $\Omega$ be a smooth domain of $\mathbb{R}^{n}$. For any
$i,j,k\in\\{1,...,n\\}$, let us consider $a^{ijk}\in C^{1}(\mathbb{R}^{n})$.
We assume that $a^{ijk}=a^{jik}$ for all $i,j,k\in\\{1,...,n\\}$. Then for all
$u\in C^{2}(\mathbb{R}^{n})$, we have that

$\displaystyle\int_{\Omega}a^{ijk}\partial_{ij}u\partial_{k}u\,dx$
$\displaystyle=$
$\displaystyle\int_{\Omega}\left(-\partial_{l}a^{lij}+\frac{1}{2}\partial_{l}a^{ijl}\right)\partial_{i}u\partial_{j}u\,dx$
$\displaystyle+\int_{\partial\Omega}\left(-\frac{1}{2}a^{ijl}\vec{\nu}_{l}+a^{lji}\vec{\nu}_{l}\right)\partial_{i}u\partial_{j}u\,d\sigma$

where $\vec{\nu}$ is the outer normal vector at $\partial\Omega$ and
Einstein’s summation convention has been used.

The proof is by integrations by parts and goes back to Hebey-Vaugon [hv:mz]
and can also be found in Cheikh-Ali [hca:pjm]. It follows from Lemma 2 that

$\displaystyle\int_{B_{\delta_{\epsilon}}(0)}(\hat{g}_{\epsilon}^{ij}-\delta^{ij})\partial_{ij}\hat{u}_{\epsilon}\partial_{l}\hat{u}_{\epsilon}\,dx$
$\displaystyle=$
$\displaystyle\int_{B_{\delta_{\epsilon}}(0)}\left(-\partial_{m}\hat{g}_{\epsilon}^{mj}\delta_{j,l}+\frac{1}{2}\partial_{l}\hat{g}_{\epsilon}^{ij}\delta_{m,l}\right)\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx$
$\displaystyle+\int_{\partial
B_{\delta_{\epsilon}}(0)}\left(-\frac{1}{2}(\hat{g}_{\epsilon}^{ij}-\delta^{ij})\delta_{m,l}\vec{\nu}_{m}+(\hat{g}_{\epsilon}^{mj}-\delta^{mj})\delta_{i,l}\vec{\nu}_{l}\right)\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,d\sigma$

Using (22) to control the boundary terms, we get that

$\displaystyle D_{\epsilon}$ $\displaystyle=$
$\displaystyle\int_{B_{\delta_{\epsilon}}(0)}\left(-\partial_{m}\hat{g}_{\epsilon}^{mj}\delta_{j,l}+\frac{1}{2}\partial_{m}\hat{g}_{\epsilon}^{ij}\delta_{m,l}\right)\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx$
$\displaystyle-\int_{B_{\delta_{\epsilon}}(0)}\hat{g}_{\epsilon}^{ij}\Gamma_{ij}^{k}(\hat{g}_{\epsilon})\partial_{k}\hat{u}_{\epsilon}\partial_{l}\hat{u}_{\epsilon}\,dX+\underbrace{O\left(\mu_{\epsilon}\left(\left(\frac{\mu_{\epsilon}}{\delta_{\epsilon}}\right)^{n-3}+\frac{\delta_{\epsilon}^{n-1}}{\mu_{\epsilon}}\theta_{\epsilon}^{2}\right)\right)}_{=o(\mu_{\epsilon})}$

as $\epsilon\to 0$. Since

$\Gamma_{ij}^{k}(\hat{g}_{\epsilon})=\frac{1}{2}\hat{g}_{\epsilon}^{km}\left(\partial_{i}(\hat{g}_{\epsilon})_{jm}+\partial_{j}(\hat{g}_{\epsilon})_{im}-\partial_{m}(\hat{g}_{\epsilon})_{ij}\right)$

and $\hat{g}_{\epsilon}$ is normal at $0$ (that is
$\partial_{m}(\hat{g}_{\epsilon})_{ij}(0)=0$ for all $i,j,m\in\\{1,...,n\\}$),
we then get that there exists $a_{ij\alpha}\in\mathbb{R}$,
$i,j,\alpha\in\\{1,...,n\\}$ such that

$\displaystyle D_{\epsilon}$ $\displaystyle=$
$\displaystyle\int_{B_{\delta_{\epsilon}}(0)}a_{ij\alpha}X^{\alpha}\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx+O\left(\int_{B_{\delta_{\epsilon}}(0)}|X|^{2}|\nabla\hat{u}_{\epsilon}|^{2}\,dX\right)+o(\mu_{\epsilon})$

With (23), we get that

$\displaystyle\int_{B_{\delta_{\epsilon}}(0)}|X|^{2}|\nabla\hat{u}_{\epsilon}|^{2}\,dX$
$\displaystyle\leq$ $\displaystyle
C\int_{B_{\delta_{\epsilon}}(0)}|X|^{2}\frac{\mu_{\epsilon}^{n-2}}{(\mu_{\epsilon}+|X|)^{2(n-1)}}\,dX+C\theta_{\epsilon}^{2}\delta_{\epsilon}^{n}$
$\displaystyle\leq$ $\displaystyle
C\mu_{\epsilon}^{2}\int_{B_{\delta_{\epsilon}/\mu_{\epsilon}}(0)}\frac{|X|^{2}\,dX}{(1+|X|)^{2(n-1)}}+C\theta_{\epsilon}^{2}\delta_{\epsilon}^{n}=o(\mu_{\epsilon})$

With (23), given $R>0$, using (51) and $n>3$, we have that

$\displaystyle\left|\int_{B_{\delta_{\epsilon}}(0)\setminus
B_{R\mu_{\epsilon}}(0)}X^{\alpha}\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx\right|\leq
C\int_{B_{\delta_{\epsilon}}(0)\setminus
B_{R\mu_{\epsilon}}(0)}\frac{\mu_{\epsilon}^{n-2}|X|\,dX}{(\mu_{\epsilon}+|X|)^{2(n-1)}}+C\theta_{\epsilon}^{2}\delta_{\epsilon}^{n}$
$\displaystyle\leq C\mu_{\epsilon}\int_{\mathbb{R}^{n}\setminus
B_{R}(0)}\frac{|Y|\,dX}{(1+|Y|)^{2(n-1)}}+C\theta_{\epsilon}^{2}\delta_{\epsilon}^{n}\leq\eta(R)\mu_{\epsilon}+o(\mu_{\epsilon})$

where $\lim_{R\to+\infty}\eta(R)=0$. Using the change of variable
$X=\mu_{\epsilon}Y$, the convergence (15) and the radial symmetry of $w$, we
have that

$\displaystyle\int_{B_{R\mu_{\epsilon}}(0)}X^{\alpha}\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx$
$\displaystyle=$
$\displaystyle\mu_{\epsilon}\int_{B_{R}(0)}Y^{\alpha}\partial_{i}w_{\epsilon}\partial_{j}w_{\epsilon}\,dY$
$\displaystyle=$
$\displaystyle\mu_{\epsilon}\left(\int_{B_{R}(0)}X^{\alpha}\partial_{i}w\partial_{j}w\,dY+o(1)\right)=o(\mu_{\epsilon})\hbox{
since }n>3.$

Therefore, we get that
$\int_{B_{\delta_{\epsilon}}(0)}X^{\alpha}\partial_{i}\hat{u}_{\epsilon}\partial_{j}\hat{u}_{\epsilon}\,dx=o(\mu_{\epsilon})$,
and then

$D_{\epsilon}=o(\mu_{\epsilon})\hbox{ for }n\geq 4.$

We now deal with $A_{\epsilon}$. With a Taylor expansion of $f$, we get

$\displaystyle A_{\epsilon}$ $\displaystyle=$
$\displaystyle-\frac{1}{2^{\star}}\int_{{B_{\delta_{\epsilon}}(0)}}{\partial_{l}}{{\hat{f}}_{\epsilon}}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X$
$\displaystyle=$
$\displaystyle-\frac{1}{2^{\star}}{\partial_{l}}{{\hat{f}}_{\epsilon}}(0)\int_{{B_{\delta_{\epsilon}}(0)}}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X-\frac{1}{2^{\star}}{\partial_{lj}}{{\hat{f}}_{\epsilon}}(0)\int_{{B_{\delta_{\epsilon}}(0)}}X^{j}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X$
$\displaystyle+O\left(\int_{{B_{\delta_{\epsilon}}(0)}}|X|^{2}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X\right)$

Arguing as above, we get that

$\int_{{B_{\delta_{\epsilon}}(0)}}|X|^{2}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X=o(\mu_{\epsilon})\hbox{
and
}\int_{{B_{\delta_{\epsilon}}(0)}}X^{j}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X=o(\mu_{\epsilon})\hbox{
as }\epsilon\to 0.$

With (21), we get that there exists $C_{0}>0$ such that

$\int_{{B_{\delta_{\epsilon}}(0)}}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X=C_{0}+o(1)\hbox{
as }\epsilon\to 0.$

Therefore, we get that

$A_{\epsilon}=\left(-\frac{C_{0}}{2^{\star}}+o(1)\right){\partial_{l}}{{\hat{f}}_{\epsilon}}(0)+o(\mu_{\epsilon})\hbox{
as }\epsilon\to 0.$

Putting the estimates of $A_{\epsilon}$, $B_{\epsilon}$, $C_{\epsilon}$ and
$D_{\epsilon}$ into (46) yields

(47) $\partial_{l}\hat{f}_{\epsilon}(0)=o(\mu_{\epsilon})\hbox{ as
}\epsilon\to 0\hbox{ for all }l\in\\{1,...,n\\}.$

Passing to the limit, we get that $\nabla f(x_{0})=0$. We now express
$\partial_{l}\hat{f}_{\epsilon}(0)$ more precisely. We write

$\hat{f}_{\epsilon}(X)=f\circ\exp_{x_{\epsilon}}(X)=\tilde{f}\circ\varphi(X_{\mathbb{\epsilon}},X)\hbox{for}X\in\mathbb{R}^{n}$

where $\tilde{f}:=f\circ\exp_{x_{0}}$ and
$\varphi(Z,X):=\exp_{x_{0}}^{-1}\circ\exp_{\exp_{x_{0}}(Z)}(X)$ for
$X,Z\in\mathbb{R}^{n}$. We set
$X_{\mathbb{\epsilon}}:=\exp_{{x_{{0}}}}^{-1}({x_{\epsilon}})$. Since
$\nabla\tilde{f}(0)=0$, we get that

${\partial_{l}}(f\circ\exp_{x_{\epsilon}})(0)=\frac{\partial^{2}(f\circ\exp_{x_{0}})}{\partial
x_{l}\partial x_{j}}(0)X_{\epsilon}^{j}+o(|X_{\epsilon}|)$

therefore, with (47), we get that

$\frac{\partial^{2}(f\circ\exp_{x_{0}})}{\partial x_{l}\partial
x_{j}}(0)X_{\epsilon}^{j}=o(|X_{\epsilon}|)+o(\mu_{\epsilon})\hbox{ as
}\epsilon\to 0\hbox{ for all }l=1,..,n.$

Since $\nabla^{2}f(x_{0})$ is nondegenerate, we then get that
$|X_{\epsilon}|=o(\mu_{\epsilon})$, in other words
$d_{g}(x_{\epsilon},x_{0})=o(\mu_{\epsilon})$ as $\epsilon\to 0$. This proves
(6).

######  Lemma 3.

Under the assumptions of Theorem 1, we have that

(51) $\displaystyle\theta_{\epsilon}=\left\\{\begin{array}[]{ll}o(1)&\hbox{ if
}n\geq 7,\\\ o(\mu_{\epsilon})&\hbox{ if }n\in\\{5,6\\}.\\\
o\left(\mu_{\epsilon}\sqrt{\ln(\frac{1}{\mu_{\epsilon}}}\right)&\hbox{ if
}n=4.\end{array}\right.$

Proof: The case $n\geq 7$ is simply (24). It follows from (22) that

$\displaystyle\int_{B_{\delta}(x_{0})}u_{\epsilon}^{2}\,dv_{g}$
$\displaystyle\leq$
$\displaystyle\int_{B_{2\delta}(x_{\epsilon})}\frac{\mu_{\epsilon}^{n-2}}{(\mu_{\epsilon}+d_{g}(x,x_{\epsilon}))^{2(n-2)}}\,dv_{g}+C\theta_{\epsilon}^{2}$
$\displaystyle\leq$
$\displaystyle\mu_{\epsilon}^{2}\int_{B_{2\delta/\mu_{\epsilon}}(0)}\frac{1}{(1+|X|))^{2(n-2)}}\,dv_{g}+C\theta_{\epsilon}^{2}$
$\displaystyle\leq$ $\displaystyle
C\left\\{\begin{array}[]{ll}\mu_{\epsilon}^{2}&\hbox{ if }n\geq 5\\\
\mu_{\epsilon}^{2}\ln(\frac{1}{\mu_{\epsilon}})&\hbox{ if }n=4\\\
\end{array}\right.$

Equation (5) rewrites
$\Delta_{g}u_{\epsilon}+(h_{\epsilon}-fu_{\epsilon}^{2^{\star}-2})u_{\epsilon}=0$
in $M$. Since $u_{\epsilon}\to 0$ in $C^{0}_{loc}(M-\\{x_{0}\\})$ and
$u_{\epsilon}>0$, it follows from Harnack’s inequality that there exists $c>0$
such that $u_{\epsilon}(x)\leq cu_{\epsilon}(y)$ for all $x,y\in
B_{2\delta}(x_{0})\setminus B_{\delta/3}(x_{0})$. Therefore, with the
definition (24) of $\theta_{\epsilon}$, we get that

$\int_{B_{2\delta}(x_{0})\setminus
B_{\delta}(x_{0})}u_{\epsilon}^{2}\,dv_{g}\geq c^{-2}\theta_{\epsilon}^{2}.$

When $4\leq n\leq 6$, it follows from $L^{2}-$concentration assumption (7)
that

$\int_{B_{2\delta}(x_{0})\setminus
B_{\delta}(x_{0})}u_{\epsilon}^{2}\,dv_{g}\leq\int_{M\setminus
B_{\delta}(x_{0})}u_{\epsilon}^{2}\,dv_{g}=o\left(\int_{B_{\delta}(x_{0})}u_{\epsilon}^{2}\,dv_{g}\right)\hbox{
as }\epsilon\to 0.$

Putting these inequalities together, we get (51). This proves Lemma 3.∎

##  4\. Interaction with the scalar curvature: proof of (8)

This part is strongly inspired by Cheikh-Ali [hca:pjm]. We define

$\delta_{\epsilon}:=\left\\{\begin{array}[]{cc}\mu_{\epsilon}^{\frac{2}{n-2}}&\hbox{
if }n\geq 7\\\ \delta&\hbox{ if }n\in\\{4,5,6\\}.\end{array}\right.$

Writing the Pohozaev identity (41) for $\hat{u}_{\epsilon}$ that satisfies
(5), we get that

(53) $\displaystyle A_{\epsilon}+B_{\epsilon}=C_{\epsilon}+D_{\epsilon}$

where

$\displaystyle B_{\epsilon}$
$\displaystyle:=\int_{\partial{B_{\delta_{\epsilon}}(0)}}\left[(X,\nu)\left(\frac{|{\nabla}{{\hat{u}}_{{\epsilon}}}|_{\xi}^{2}}{2}-\frac{{{\hat{f}}_{\epsilon}}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}}{2^{\star}}\right)-\left(X^{l}\partial_{l}{{\hat{u}}_{{\epsilon}}}+\frac{n-2}{2}{{\hat{u}}_{{\epsilon}}}\right)\partial_{\nu}{{\hat{u}}_{{\epsilon}}}\right]\mathrm{d}\nu$
$\displaystyle C_{\epsilon}$
$\displaystyle:=-\int_{{B_{\delta_{\epsilon}}(0)}}\left(X^{l}\partial_{l}{{\hat{u}}_{{\epsilon}}}+\frac{n-2}{2}{{\hat{u}}_{{\epsilon}}}\right){{\hat{h}}_{{\epsilon}}}{{\hat{u}}_{{\epsilon}}}\mathrm{d}X$
$\displaystyle D_{\epsilon}$
$\displaystyle:=-\int_{{B_{\delta_{\epsilon}}(0)}}\left(X^{l}\partial_{l}{{\hat{u}}_{{\epsilon}}}+\frac{n-2}{2}{{\hat{u}}_{{\epsilon}}}\right)(\Delta_{{{\hat{g}}_{{\epsilon}}}}{{\hat{u}}_{{\epsilon}}}-\Delta_{\xi}{{\hat{u}}_{{\epsilon}}})\mathrm{d}X$
$\displaystyle A_{\epsilon}$
$\displaystyle:=\frac{1}{2^{\star}}\int_{{B_{\delta_{\epsilon}}(0)}}({\nabla}{{\hat{f}}_{\epsilon}},X){{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X$

Following Cheikh-Ali [hca:pjm] and using the pointwise controls (21), (22),
(23) and the control (51) on $(\theta_{\epsilon})_{\epsilon}$ when
$n\in\\{4,5,6\\}$, we get that

$B_{\epsilon}=\left\\{\begin{array}[]{cc}o(\mu_{\epsilon}^{2})&\hbox{ if
}n\geq 5\\\ o\left(\mu_{\epsilon}^{2}\ln\frac{1}{\mu_{\epsilon}}\right)&\hbox{
if }n=4.\end{array}\right.\hbox{ as }\epsilon\to 0,$ $\displaystyle
C_{\epsilon}=\left\\{\begin{array}[]{ll}h(x_{0})\mu_{\epsilon}^{2}\ln\left(\frac{1}{\mu_{\epsilon}}\right)\left(\frac{8}{f(x_{0})}\right)^{2}\omega_{3}+o(\mu_{\epsilon}^{2}\ln\frac{1}{\mu_{\epsilon}})&\hbox{
if }n=4\\\
h(x_{0})\mu_{\epsilon}^{2}\int_{{\mathbb{R}}^{n}}{w}^{2}\mathrm{d}X+o(\mu_{\epsilon}^{2})&\hbox{
if }n\geq 5\end{array}\right.$ $\displaystyle
D_{\epsilon}=\left\\{\begin{array}[]{ll}-\mu_{\epsilon}^{2}\ln\left(\frac{1}{\mu_{\epsilon}}\right)\frac{1}{6}\hbox{Scal}_{g}(x_{\epsilon})\left(\frac{8}{f(x_{0})}\right)^{2}\omega_{3}+o(\mu_{\epsilon}^{2}\ln\frac{1}{\mu_{\epsilon}})&\hbox{
if }n=4\\\
-\mu_{\epsilon}^{2}\frac{n-2}{4(n-1)}\hbox{Scal}_{g}(x_{\epsilon})\int_{{\mathbb{R}}^{n}}{w}^{2}\mathrm{d}X+o(\mu_{\epsilon}^{2})&\hbox{
if }n\geq 5\end{array}\right.$

We are then left with estimating $A_{\epsilon}$. With a Taylor expansion of
$\hat{f}_{\epsilon}$, we get that

$\displaystyle A_{\epsilon}$ $\displaystyle=$
$\displaystyle\frac{1}{2^{\star}}\int_{{B_{\delta_{\epsilon}}(0)}}{\partial_{i}}{{\hat{f}}_{\epsilon}}(0)X^{i}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X+\frac{1}{2^{\star}}{\partial_{ij}}{{\hat{f}}_{\epsilon}}(0)\int_{{B_{\delta_{\epsilon}}(0)}}X^{i}X^{j}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X$
$\displaystyle+O\left(\int_{{B_{\delta_{\epsilon}}(0)}}|X|^{3}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X\right)$

Since $\hat{f}_{\epsilon}:=f\circ\hbox{exp}_{x_{\epsilon}}$ and $\nabla
f(x_{0})=0$, we get that
$\nabla\hat{f}_{\epsilon}(0)=O(d_{g}(x_{\epsilon},x_{0}))$. With (6), we then
get that $\nabla\hat{f}_{\epsilon}(0)=o(\mu_{\epsilon})$. It follows from
Lemma 1 that
$\int_{{B_{\delta_{\epsilon}}(0)}}|X|^{3}{{\hat{u}}_{{\mathbb{\epsilon}}}}^{2^{\star}}\mathrm{d}X=o(\mu_{\epsilon}^{2})$
and
$\int_{{B_{\delta_{\epsilon}}(0)}}|X|{{\hat{u}}_{{\mathbb{\epsilon}}}}^{2^{\star}}\mathrm{d}X=O({\mu_{\mathbb{\epsilon}}})$.
Therefore, we get that

$A_{\epsilon}=\frac{1}{2^{\star}}{\partial_{ij}}{{\hat{f}}_{\epsilon}}(0)\int_{{B_{\delta_{\epsilon}}(0)}}X^{i}X^{j}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X+o(\mu_{\epsilon}^{2})$

Arguing as in the proof of Lemma 1, we get

$\displaystyle\int_{{B_{\frac{\delta_{\epsilon}}{\mu_{\epsilon}}}(0)}}X^{i}X^{j}{{\hat{u}}_{{\epsilon}}}^{2^{\star}}\mathrm{d}X=\mu_{\epsilon}^{2}\int_{{\mathbb{R}}^{n}}X^{i}X^{j}w^{2^{\star}}\mathrm{d}X+o(\mu_{\epsilon}^{2})\hbox{
when }n\geq 4.$

Since $w$ is radially symmetric, we get that
$\int_{{\mathbb{R}}^{n}}X^{i}X^{j}w^{2^{\star}}\mathrm{d}X=\frac{\delta_{ij}}{n}\int_{{\mathbb{R}}^{n}}|X|^{2}w^{2^{\star}}\mathrm{d}X$.
Since $\hat{g}_{\epsilon}$ is normal at $0$, we have that
$\Delta_{g}f(x_{\epsilon})=-\sum_{i}\partial_{ii}\hat{f}_{\epsilon}(0)$, which
yields We claim that

$\displaystyle
A_{\epsilon}=-\frac{1}{2^{\star}n}\Delta_{g}f(x_{0})\mu_{\epsilon}^{2}\int_{{\mathbb{R}}^{n}}|X|^{2}w^{2^{\star}}\mathrm{d}X+o(\mu_{\epsilon}^{2})$
$\displaystyle\hbox{ if }n\geq 4$

By Jaber [jaber2015optimal] we have that

$\frac{\int_{{\mathbb{R}}^{n}}|X|^{2}w^{2^{\star}}\mathrm{d}X}{\int_{{\mathbb{R}}^{n}}w^{2}\mathrm{d}X}=\frac{n^{2}(n-4)}{4(n-1)f(x_{0})}\hbox{
for }n\geq 5.$

Putting the expressions of $A_{\epsilon}$, $B_{\epsilon}$, $C_{\epsilon}$ and
$D_{\epsilon}$ in (53) and letting $\epsilon\to 0$ yield (8). This ends the
proof of Theorem 1.

##  5\. Application to a super-critical problem: proof of Theorem 2

We follow the notations and assumptions of Theorem 2. We consider a family
$(u_{\epsilon})_{\epsilon>0}\in C^{2}_{G}(X)$ of $G-$invariant solutions to
the problem

(56)
$\Delta_{g}u_{\epsilon}+h_{\epsilon}u_{\epsilon}={{\mathbb{\lambda}}_{\epsilon}}u_{\epsilon}^{2^{\star}(k)-1}\,,\,\int_{X}u_{\epsilon}^{2^{\star}(k)}{\mathrm{d}v_{g}}=1\,,\,\|u_{\epsilon}\|_{2}\to
0\hbox{ as }\epsilon\to 0$

where $(h_{\epsilon})_{\epsilon>0}\in C^{1}_{G}(X)$ is such that there exists
$h\in C^{1}_{G}(X)$ such that (9) holds and $(\lambda_{\epsilon})_{\epsilon}$
is such that (11) holds.

######  Claim 4.

There exists $x_{0}\in X$ such that

(57) $\lim_{\epsilon\to
0}\int_{B_{\delta}(Gx_{0})}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}=1\hbox{ for all
}\delta>0.$

Proof: We fix a point $z_{0}\in X$. We choose $\eta_{0}\in
C^{\infty}({\mathbb{R}})$ such that $\eta_{0}(t)=1$ for $t\leq 1$ and
$\eta_{0}(t)=0$ for $t\geq 2$. Given $\delta>0$, we define
$\eta(x):=\eta_{0}(\frac{d_{{g}}(Gx_{0},x)}{\delta})$ for all $x\in X$. For
$\delta>0$ small enough, we have that $\eta\in C^{\infty}_{G}(X)$. Multiply
(10) by $\eta^{2}u_{\epsilon}^{l}$ for some $l>1$ and integrate over $X$, we
get that

(58)
$\int_{M}\eta^{2}u_{\epsilon}^{l}\Delta_{g}u_{\epsilon}{\mathrm{d}v_{g}}+\int_{M}\eta^{2}{h_{\mathbb{\epsilon}}}u_{\epsilon}^{l+1}{\mathrm{d}v_{g}}={{\mathbb{\lambda}}_{\mathbb{\epsilon}}}\int_{M}\eta^{2}u_{\epsilon}^{l+2^{\star}(k)-1}{\mathrm{d}v_{g}}$

As in the proof of Claim 2, we get that

$\displaystyle\int_{M}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}=\frac{(l+1)^{2}}{4l}\int_{M}\eta^{2}u_{\epsilon}^{l}\Delta_{g}u_{\epsilon}\,dv_{g}+\frac{l+1}{2l}\int_{M}\left(|{\nabla}\eta|_{g}^{2}+\frac{l-1}{l+1}\eta\Delta_{g}\eta\right)u_{\epsilon}^{l+1}\,dv_{g}$

Using (58) and Hölder’s inequality, we get

(59) $\displaystyle\int_{X}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}\leq
C\int_{X}u_{\epsilon}^{l+1}\,dv_{g}$
$\displaystyle+\frac{(l+1)^{2}}{4l}\lambda_{\epsilon}\left(\int_{X}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}\left(\int_{B_{2\delta}(Gz_{0})}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}\right)^{1-\frac{2}{2^{\star}(k)}}$

It follows from Faget [faget2002best] that for all $\alpha>0$, there exists
$B>0$ such that, for all ${\mathbb{\epsilon}}$,

$\displaystyle\left(\int_{X}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}\leq\frac{K_{0}(n-k)(1+\alpha)}{V_{m}^{1-\frac{2}{2^{\star}(k)}}}\int_{X}\left|{\nabla}\right(\eta
u_{\epsilon}^{\frac{l+1}{2}}\left)\right|_{g}^{2}\,dv_{g}+B\int_{X}\eta^{2}u_{\epsilon}^{l+1}\,dv_{g}$

where $K_{0}(n-k)$ is as in (3) and $V_{m}=\hbox{min}_{x\in
X}\\{Vol_{g}(Gx)/\,\hbox{dim }Gx=k\\}$. By combining this inequality with
(59), we obtain:

(60) $\left(\int_{X}\left(\eta
u_{\epsilon}^{\frac{l+1}{2}}\right)^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}\chi_{\epsilon}\leq
C\|u_{\epsilon}\|_{l+1}^{l+1}$

where

$\chi_{\epsilon}:=1-\frac{(l+1)^{2}}{4l}\lambda_{\epsilon}\frac{K_{0}(n-k)(1+\alpha)}{V_{m}^{1-\frac{2}{2^{\star}(k)}}}\left(\int_{B_{2\delta}(Gz_{0})}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}\right)^{1-\frac{2}{2^{\star}(k)}}$

Assume that, up to extraction,

$\lim_{\epsilon\to
0}\int_{B_{2\delta}(Gz_{0})}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}<1.$

Using (11), there exists $1<l<2^{\star}(k)-1$ such that
$\chi_{\epsilon}\geq\beta>0$ for all $\epsilon>0$ up to taking $\alpha$ small.
As $u_{\epsilon}\to 0$ in $L^{l+1}(X)$ since $l+1<2^{\star}(k)$, with (60), we
then get that $\lim_{\epsilon\to
0}\int_{B_{\delta}(Gz_{0})}u_{\epsilon}^{\frac{l+1}{2}2^{\star}(k)}\,dv_{g}=0$.
With similar arguments, we get that for all $\delta^{\prime}<\delta$,
$u_{\epsilon}\to 0$ in $L^{q}(B_{\delta^{\prime}}(Gz_{0})$ for all $q\geq 1$.
It then follows from (10) and elliptic theory that $u_{\epsilon}\to 0$ in
$C^{0}(B_{\delta^{\prime}}(Gz_{0}))$. Since
$\int_{X}u_{\epsilon}^{2^{\star}}\,dv_{g}=1$ and $X$ is compact, the existence
of $x_{0}\in X$ such that (57) holds follows. This proves the claim.∎

######  Claim 5.

We have that $\hbox{dim }Gx_{0}=k$ and $\hbox{Vol}_{g}(Gx_{0})=V_{m}$.

Proof: We follow Faget [faget2002best]. Assume that $\hbox{dim }Gx_{0}>k$.
Therefore, there exists $\delta>0$ such that $\hbox{dim }Gx\geq k_{1}>k$ for
all $x\in B_{2\delta}(Gx_{0})$. It then follows from Hebey-Vaugon
[hebey0sobolev] that $H_{1,G}^{2}(B_{\delta}(Gx_{0}))\hookrightarrow
L^{p}(B_{\delta}(Gx_{0}))$ is compact for $1\leq p<2^{\star}(k_{1})$. Since
$2^{\star}(k)<2^{\star}(k_{1})$ and $u_{\epsilon}\to 0$ in $L^{2}(X)$, we get
that $u_{\epsilon}\to 0$ strongly in $L^{2^{\star}(k)}(B_{\delta}(Gx_{0}))$,
contradicting (57). Therefore $\hbox{dim }Gx_{0}=k$. It follows from Faget
(formula (8) in [faget2002best]) that for all $\alpha>0$, there exists
$\delta_{\alpha}>0$ such that

$\left(\int_{B_{\delta_{\alpha}}(Gx_{0})}|v|^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}\leq(1+\alpha)\frac{K_{0}(n-k)}{\hbox{Vol}_{g}(Gx_{0})^{1-\frac{2}{2^{\star}(k)}}}\int_{B_{\delta_{\alpha}}(Gx_{0})}|\nabla
v|^{2}\,dv_{g}$

for all $v\in C^{1}_{G}(B_{\delta_{\alpha}}(Gx_{0}))$ with compact support in
$B_{\delta_{\alpha}}(Gx_{0})$. Let us fix $\eta_{\alpha}\in C^{\infty}_{G}(X)$
with compact support in $B_{\delta_{\alpha}}(Gx_{0})$ and such that
$0\leq\eta_{\alpha}\leq 1$ and $\eta_{\alpha}(x)=1$ for
$d_{g}(x,Gx_{0})<\delta_{\alpha}/2$. We then get that

(61)
$\displaystyle\left(\int_{B_{\delta_{\alpha}/2}(Gx_{0})}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}$
$\displaystyle\leq$
$\displaystyle\left(\int_{B_{\delta_{\alpha}}(Gx_{0})}(\eta_{\alpha}u_{\epsilon})^{2^{\star}(k)}\,dv_{g}\right)^{\frac{2}{2^{\star}(k)}}$
$\displaystyle\leq$
$\displaystyle(1+\alpha)\frac{K_{0}(n-k)}{\hbox{Vol}_{g}(Gx_{0})^{1-\frac{2}{2^{\star}(k)}}}\int_{X}|\nabla(\eta_{\alpha}u_{\epsilon})|^{2}\,dv_{g}$

Integrating by parts and using $\|u_{\epsilon}\|_{2}\to 0$, we get that

(62)
$\int_{X}|\nabla(\eta_{\alpha}u_{\epsilon})|_{g}^{2}\,dv_{g}=\int_{X}\eta_{\alpha}^{2}|\nabla
u_{\epsilon}|_{g}^{2}\,dv_{g}+\int_{X}\eta(\Delta_{g}\eta)u_{\epsilon}^{2}\,dv_{g}\leq\int_{X}|\nabla
u_{\epsilon}|_{g}^{2}\,dv_{g}+o(1)$

Multiplying (56) by $u_{\epsilon}$, integrating and using again
$\|u_{\epsilon}\|_{2}\to 0$, we get that

(63)
$\lambda_{\epsilon}=\int_{X}\lambda_{\epsilon}u_{\epsilon}^{2^{\star}(k)}\,dv_{g}=\int_{X}|\nabla
u_{\epsilon}|^{2}\,dv_{g}+\int_{X}h_{\epsilon}u_{\epsilon}^{2}\,dv_{g}=\int_{X}|\nabla
u_{\epsilon}|^{2}\,dv_{g}+o(1).$

Putting together (61), (62), (63) and (57), we get that

$1\leq(1+\alpha)\frac{K_{0}(n-k)}{\hbox{Vol}_{g}(Gx_{0})^{1-\frac{2}{2^{\star}(k)}}}\lambda_{\epsilon}+o(1).$

Using (11), letting $\epsilon\to 0$ and then $\alpha\to 0$ yields
$\hbox{Vol}_{g}(Gx_{0})\leq V_{m}$. Therefore $\hbox{Vol}_{g}(Gx_{0})=V_{m}$
and the claim is proved.∎

Claim: The following $L^{2}-$concentration holds

(64) $\lim_{\epsilon\to 0}\frac{\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}}{\int_{X}u_{\epsilon}^{2}\,dv_{g}}=0\hbox{
for }n-k\geq 4.$

We prove the claim by arguing as in Djadli-Druet [dd]. We have that

$\displaystyle\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\leq\left(\sup_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}\right)\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}\,dv_{g}$

Since $u_{\epsilon}\to 0$ in $C^{0}_{loc}(X\setminus Gx_{0})$, Harnack’s
inequality yields $c>0$ such that

$\displaystyle\int_{X\setminus B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}$
$\displaystyle\leq c\inf_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}\,dv_{g}$ $\displaystyle\leq
c\left(\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\right)^{\frac{1}{2}}\int_{X}u_{\epsilon}\,dv_{g}\leq
c\|u_{\epsilon}\|_{2}\int_{X}u_{\epsilon}\,dv_{g}$

Integrating (10) yields
$\int_{X}{h_{\epsilon}}u_{\epsilon}\,dv_{g}=\int_{X}u_{\epsilon}^{2^{\star}(k)-1}\,dv_{g}$.
It follows from (9) that there exists $\beta>0$ such that
$h_{\epsilon}\geq\beta$ for all $\epsilon>0$. Therefore we get that

$\displaystyle\int_{X\setminus B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\leq
c\beta^{-1}\|u_{\epsilon}\|_{2}\|u_{\epsilon}\|_{2^{\star}(k)-1}^{2^{\star}(k)-1}.$

if $n-k\geq 6$, we have $2^{\star}(k)-1\leq 2$. Using Hölder inequalities, we
have $\|u_{\epsilon}\|_{2^{\star}(k)-1}^{2^{\star}(k)-1}\leq
c\|u_{\epsilon}\|_{2}^{2^{\star}(k)-1}$. Since $u_{\epsilon}\to 0$ in
$L^{2}(X)$, we get that

$\displaystyle\int_{X\setminus B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\leq
c\|u_{\epsilon}\|_{2}^{2^{\star}(k)}=o(\|u_{\epsilon}\|_{2}^{2})$

if $4\leq n-k\leq 5$, we have $2^{\star}(k)-1=2\omega+(1-\omega)2^{\star}(k)$
with $\omega=\frac{n-k-2}{4}>0$. Hölder’s inequality yields
$\|u_{\epsilon}\|_{2^{\star}(k)-1}^{2^{\star}(k)-1}\leq\|u_{\epsilon}\|_{2}^{2\omega}\|u_{\epsilon}\|_{2^{\star}(k)}^{(1-\omega)2^{\star}(k)}$.
As a result

$\int_{X\setminus
B_{\delta}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\leq\|u_{\epsilon}\|_{2^{\star}(k)-1}^{2^{\star}(k)-1}\|u_{\epsilon}\|_{2}\leq\|u_{\epsilon}\|_{2}^{1+2\omega}\|u_{\epsilon}\|_{2^{\star}(k)}^{(1-\omega)2^{\star}(k)}=o(\|u_{\epsilon}\|_{2}^{2}).$

This proves the claim.∎

We are now in position to take the quotient. Since $\hbox{dim }Gx_{0}=k$, we
choose $\delta>0$ and $G^{\prime}\subset G$ as in Assumption $(H)$. Then
$M:=B_{\delta}(Gx_{0})/G^{\prime}$ is a manifold of dimension $n-k$ that is
endowed with the metric $\bar{g}$ on $B_{\delta}(Gx_{0})/G^{\prime}$ such that
the projection
$(B_{\delta}(Gx_{0}),g)\to(B_{\delta}(Gx_{0})/G^{\prime},\bar{g})$ is a
Riemannian submersion. We define $\bar{u}_{\epsilon}\in C^{2}(M)$,
$\bar{h}_{\epsilon}\in C^{1}(M)$ and $\bar{v}\in C^{2}(M)$ be such that

$\bar{u}_{\epsilon}(\bar{x})=u_{\epsilon}(x)\,,\,\bar{h}_{\epsilon}(\bar{x})=h_{\epsilon}(x)\hbox{
and }\bar{v}(\bar{x})=\hbox{Vol}_{g}(G^{\prime}x)\hbox{ for all }x\in
B_{\delta}(Gx_{0}).$

Let us first rewrite equation (56) as in Saintier [saintier0blow]. Let
$\bar{\varphi}\in C^{\infty}_{c}(M)$ be a function on
$M=B_{\delta}(Gx_{0})/G^{\prime}$. Define $\varphi(x):=\bar{\varphi}(\bar{x})$
for all $x\in B_{\delta}(Gx_{0})$: as one checks, $\varphi\in
C^{\infty}_{c}(B_{\delta}(Gx_{0}))$ and is $G-$invariant. It then follows from
(56) that

$\int_{X}(\nabla
u_{\epsilon},\nabla\varphi)_{g}\,dv_{g}+\int_{X}h_{\epsilon}u_{\epsilon}\varphi\,dv_{g}=\lambda_{\epsilon}\int_{X}u_{\epsilon}^{2^{\star}(k)-1}\varphi\,dv_{g}.$

We define $\tilde{g}:=\bar{v}^{\frac{2}{n-k-2}}\bar{g}$. Since
$u_{\epsilon},\varphi$ are $G-$invariant, we get that

$\int_{X}(\nabla
u_{\epsilon},\nabla\varphi)_{g}\,dv_{g}=\int_{B_{\delta}(Gx_{0})/G^{\prime}}\bar{v}(\bar{x})(\nabla\bar{u}_{\epsilon},\nabla\bar{\varphi})_{\bar{g}}\,dv_{\bar{g}}=\int_{B_{\delta}(Gx_{0})/G^{\prime}}(\nabla\bar{u}_{\epsilon},\nabla\bar{\varphi})_{\tilde{g}}\,dv_{\tilde{g}}.$

Performing the same computations for the remaining terms, setting
$\tilde{h}_{\epsilon}:=\bar{v}^{-\frac{2}{n-k-2}}\bar{h}_{\epsilon}$,
$\tilde{f}:=\bar{v}^{-\frac{2}{n-k-2}}$ and
$\tilde{u}_{\epsilon}:=\lambda_{\epsilon}^{\frac{1}{2^{\star}-2}}\bar{u}_{\epsilon}$,
we get that

$\Delta_{\tilde{g}}\tilde{u}_{\epsilon}+\tilde{h}_{\epsilon}\tilde{u}_{\epsilon}=\tilde{f}\tilde{u}_{\epsilon}^{2^{\star}(k)-1}\hbox{
in }M.$

We deal with the $L^{2^{\star}(k)}-$norm. The definitions of $\tilde{f}$ and
$\tilde{g}$ and (57) yield

$\lim_{\epsilon\to
0}\int_{M}\tilde{f}\tilde{u}_{\epsilon}^{2^{\star}(k)}\,dv_{\tilde{g}}=\lim_{\epsilon\to
0}\lambda_{\epsilon}^{\frac{2^{\star}(k)}{2^{\star}(k)-2}}=\frac{1}{K_{0}(n-k)^{\frac{n-k}{2}}\tilde{f}(\bar{x_{0}})^{\frac{n-k-2}{2}}}.$

Concerning the $L^{2}-$concentration, it follows from (64) that for any
$r<\delta$, for $n-k\geq 4$, we have that

$\int_{B_{\delta}(Gx_{0})\setminus
B_{r}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\leq\int_{X\setminus
B_{r}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}=o\left(\int_{B_{r}(Gx_{0})}u_{\epsilon}^{2}\,dv_{g}\right).$

Taking the quotient, we get that

$\int_{M\setminus
B_{r}(\bar{x}_{0})}\tilde{u}_{\epsilon}^{2}\,dv_{\tilde{g}}=o\left(\int_{B_{r}(\bar{x}_{0})}\tilde{u}_{\epsilon}^{2}\,dv_{\tilde{g}}\right)\hbox{
when }n-k\geq 4,$

which yields the $L^{2}-$concentration on $M$. We apply Theorem 1. Taking
$(x_{\epsilon})_{\epsilon}\in X$ such that
$\|u_{\epsilon}\|_{\infty}=u_{\epsilon}(x_{\epsilon})=\mu_{\epsilon}^{-\frac{n-k-2}{2}}$,
we get (12) and (13). Equation (8) rewrites

$\tilde{h}(\bar{x}_{0})=\frac{n-k-2}{4(n-k-1)}\left(\hbox{Scal}_{\tilde{g}}(\bar{x}_{0})-\frac{n-k-4}{2}\cdot\frac{\Delta_{\tilde{g}}\tilde{f}(\bar{x}_{0})}{\tilde{f}(\bar{x}_{0})}\right)$

where $\tilde{h}:=\lim_{\epsilon\to 0}\tilde{h}_{\epsilon}$. Using the
invariance of the conformal Laplacian, that is

$\Delta_{\tilde{g}}\varphi+\frac{m-2}{4(m-1)}\hbox{Scal}_{\tilde{g}}\varphi=\omega^{-\frac{m+2}{m-2}}\left(\Delta_{\bar{g}}(\omega\varphi)+\frac{m-2}{4(m-1)}\hbox{Scal}_{\bar{g}}\omega\varphi\right)$

for any $\varphi\in C^{2}(M)$ and where
$\tilde{g}=\omega^{\frac{4}{m-2}}\bar{g}$, $m=n-k$, we get (14). This proves
Theorem 2.

Acknowledgement:  The initial version of this article required the
$L^{2}-$concentration (7) for all dimensions $n\geq 4$. The authors are
grateful to the anonymous referee who noticed that this concentration could be
bypassed for $n\geq 7$.

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