Nassif Ghoussoub
email: nassif@math.ubc.ca

AND Saikat Mazumdar
email: saikat@math.iitb.ac.in

F R Éd

Éric Robert

Frédéric Robert
email: frederic.robert@univ-lorraine.fr

THE HARDY-SCHR ÖDINGER OPERATOR ON THE POINCAR É BALL: COMPACTNESS AND MULTIPLICITY

Keywords: 2010 Mathematics Subject Classification. Primary 35J35, Secondary 35J60, 35B44

Let Ω be a compact smooth domain containing zero in the Poincaré ball model of the Hyperbolic space B n (n ≥ 3) and let -∆ B n be the Laplace-Beltrami operator on B n , associated with the metric

. We consider issues of non-existence, existence, and multiplicity of variational solutions for the borderline Dirichlet problem,

where 0

), as we show that they exist whenever n ≥ 4, 0

-1 and λ > 0. The latter result also holds true for n ≥ 3 and γ >

(n-2) 2 4

-1 provided the domain has a positive "hyperbolic mass". On the other hand, the same analysis yields that if γ >

(n-2) 2 4

-1 and the mass is non vanishing, then there is a surprising stability of regimes where no variational positive solution exists. As for higher energy solutions to (1), we show that there are infinitely many of them provided n ≥ 5, 0 ≤ γ < (n-2) 2 4 -4 and λ > n-2 n-4 n(n-4) 4

-γ .

Introduction

Consider the Poincaré ball model of the Hyperbolic space B n , n ≥ 3, which is the Euclidean unit ball

where r = n i=1

x 2 i denotes the Euclidean distance from a point x to the origin.

The function 1 nωn-1 G(r) is then a fundamental solution of the Hyperbolic Laplacian ∆ B n u = div B n (∇ B n u). Moreover, if we consider the hyperbolic scaling of a given function u : B n → R, defined for λ > 0 by

u λ (r) = λ -1 2 u G -1 (λG(r)) ,
then for any radially symmetric u ∈ H 1 r (B n ) and p ≥ 1, one has the following invariance property:

|∇ B n u| 2 dv g B n .
The hyperbolic mass m γ,λ (Ω B n ) of a domain Ω B n , associated to the operator -∆ B n -γV 2 -λ was defined in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF] and is recalled below. The existence of a positive ground state solution for [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] has already been addressed in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF]. While our main objective in this paper is to study higher energy solutions, our methods improve in a couple of ways their results about ground state solutions, in particular in the critical dimensions 3 and 4. We get the following.

Theorem 1.1. Let Ω B n B n be a smooth compact domain containing 0 and let 0 < s < 2. Assume that 0 ≤ γ < (n-2) 2 4

Let λ ∈ R be such that the operator -∆ B n -γV 2 -λ is coercive. Then [START_REF] Berestycki | The principal eigenvalue and maximum principle for second-order elliptic operators in general domains[END_REF] For n ≥ 4 and γ ≤ (n-2) 2

4

-1, equation (9) has a positive ground state solution whenever λ > 0.

(2) For n ≥ 3 and γ > (n-2) 2

4

-1, equation (9) has a positive ground state solution whenever the mass m γ,λ (Ω B n ) > 0.

We note that statement (1) of the above theorem was established in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF] in the case where n ≥ 5,

0 ≤ γ ≤ (n-2) 2 4 -1 and λ > n -2 n -4 n(n -4) 4 -γ .
As for statement [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF], it was proved in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF] under additional assumptions on λ, at least in dimensions 3 and 4.

The above theorem is actually a consequence of the following more general compactness result for positive approximate solutions.

Theorem 1.2. Let Ω B n B n be a smooth compact domain containing 0 and let 0 < s < 2 and λ ∈ R such that the operator -∆ B n -γV 2 -λ is coercive. Let (p ) >0 be such that 0 ≤ p < 2 (s) -2 and lim →0 p = 0 and consider a sequence of functions (u ) >0 that is uniformly bounded in H 1 0 (Ω B n ) such that for each > 0, u is a solution to the equation:

   -∆ B n u -γV 2 u -λu = V 2 (s) u 2 (s)-1-p in Ω B n , u > 0 in Ω B n , u = 0 on ∂Ω B n .
Assume that 0 ≤ γ ≤ (n-2) 2

4

. Assuming one of the following conditions

• n ≥ 4, 0 ≤ γ ≤ (n-2) 2 4
-1 and λ > 0;

• γ > (n-2) 2

4

-1 and m γ,λ (Ω) B n > 0. Then the sequence (u ) >0 is pre-compact in the space H 1 0 (Ω B n ). We now recall the notion of hyperbolic mass of a domain Ω associated with a coercive operator -∆ B n -γV 2 -λ. This notion was introduced in the Euclidean case in [START_REF] Ghoussoub | The effect of curvature on the best constant in the Hardy-Sobolev inequalities[END_REF] and was extended to the hyperbolic setting in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF].

Theorem 1.3 (The hyperbolic interior mass). Let Ω B n B n (n ≥ 3) be a smooth compact domain containing 0. Let γ < (n -2) 2 4 and let λ ∈ R be such that the operator -∆ B n -γV 2 -λ is coercive on

Ω B n . Then, there exists H ∈ C ∞ (Ω B n \ {0}) such that    -∆ B n H -γV 2 H -λH = 0 in Ω B n \ {0} H > 0 in Ω B n \ {0} H = 0 on ∂Ω B n . ( 10
)
These solutions are unique up to a positive multiplicative constant, and there exists c + > 0 such that

H(x) x→0 c + |x| α+(γ) .
Moreover, when γ > max n(n -4) 4 , 0 , then there exists c -∈ R such that

H(x) = c + |x| α+(γ) + c - |x| α-(γ) + o 1 |x| α-(γ) as x → 0
where α + (γ), α -(γ) stand for

α ± (γ) := n -2 2 ± (n -2) 2 4 -γ. ( 11
)
We define the interior mass as m γ,λ (Ω B n ) := c + c - , which is independent of the choice of H.

This existence theorem will follow directly from Theorem 1.7 below.

In this paper we are concerned with showing the multiplicity of higher energy solutions. Here is our main result.

Theorem 1.4. Let Ω B n B n , n ≥ 5, be a smooth compact domain containing 0 and let 0 < s < 2. Let λ ∈ R be such that the operator -∆ B n -γV 2 -λ is coercive.

If 0 ≤ γ < (n -2) 2 4
-4, then equation (9) has an infinite number of solutions corresponding to higher energy critical levels for J γ,s whenever λ ≥ n -2 n -4 n(n -4) 4 -γ .

Note that -unlike the case of ground state solutions-we do not have a multiplicity result neither in low dimensions, nor when (n-2) 2

4

-4 ≤ γ ≤ (n-2) 2

4

-1, even under the assumption of positivity of the mass of the compact subdomain. Note also the condition λ ≥ n -2 n -4 n(n -4) 4 -γ , which is more restrictive than λ > 0.

In order to prove Theorem 1.4, we shall proceed as in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF] and use a conformal transformation

g B n = ϕ 4 n-2 Eucl where ϕ = 2 1 -r 2 n-2 2
, to reduce equation [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] to a Dirichlet boundary value problem on Euclidean space. We shall denote by Ω the subdomain Ω B n considered as a subset of R n . The following sharpens a few estimates obtained in [START_REF] Hardy Chan | Mass and extremals associated with the Hardy-Schrödinger operator on hyperbolic space[END_REF]. These will be needed when dealing with the critical (low dimensional) cases.

Lemma 1. u ∈ H 1 0 (Ω B n ) satisfies (9) if and only if v := ϕu ∈ H 1 0 (Ω Ω ) satisfies    -∆v - γ |x| 2 + h γ,λ (x) v = b(x) v 2 (s)-1 |x| s in Ω v = 0 on ∂Ω, (12)
where x → b(x) is a positive function in C 0,1 (Ω) with b(0) = (n -2)

2-s n-2 2 2-s satisfying the following estimates:

• For n = 3, we have that

We also have that h γ,λ ∈ C 1 Ω \ {0} in such a way that there exist c 3 , c 4 ∈ R such that (

h γ,λ (x) = h γ,λ (r) =                    4γ r + c 3 + O(
) 16
In addition, when n = 4 and γ = 0, we have that c 4 = 4(λ -2). Moreover, the hyperbolic operator -∆ B n -γV 2 -λ is coercive if and only if the corresponding Euclidean operator -∆ -γ |x| 2 + h γ,λ (x) is coercive.

We note that equation [START_REF] Druet | Stability of the Pohožaev obstruction in dimension 3[END_REF] have been studied extensively in the case where h γ,λ ≡ λ and b ≡ 1, that is

   -∆u -γ u |x| 2 -λu = |u| 2 (s)-2 u |x| s in Ω \ {0}, u = 0 on ∂Ω, (17)
and more so in the non-singular case (i.e., when γ = s = 0), with major contributions to the existence of ground state solutions by Brezis-Nirenberg [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF] when n ≥ 4 and λ > 0, and by Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF][START_REF]Optimal Sobolev inequalities and extremal functions. The three-dimensional case[END_REF] and Druet-Laurain [START_REF] Druet | Stability of the Pohožaev obstruction in dimension 3[END_REF] when n = 3. The multiplicity of solutions was established in that case by Devillanova-Solimini [START_REF] Devillanova | Concentration estimates and muliple solutions to ellipic problems at critical growth[END_REF] in dimension n ≥ 7 and 0 < λ < λ 1 (-∆).

The existence of ground state solutions when s = 0, 0 < γ ≤ (n-2) 2

4

-1, 0 < λ < λ 1 (-∆ -γ |x| 2 ), and 0 ∈ Ω was established by Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF], while the multiplicity of solutions for when 0 ≤ γ < (n-2) 2 4 -4 and n ≥ 7 was proved by Cao-Yan [START_REF] Cao | Infinitely many solutions for an elliptic problem involving critical Sobolev growth and Hardy potential[END_REF]. We also note that if Ω is the unit ball, then Catrina-Lavine [START_REF] Catrina | Radial solutions for weighted semilinear equations[END_REF] showed that [START_REF] Ghoussoub | Concentration estimates for Emden-Fowler equations with boundary singularities and critical growth[END_REF] admits no radial positive solutions [START_REF] Catrina | Radial solutions for weighted semilinear equations[END_REF] for γ > (N -2) 2 4 -1, while Esposito et al. show in [START_REF] Esposito | Sign-changing solutions for critical equations with Hardy potential[END_REF] that for λ small, problem [START_REF] Ghoussoub | Concentration estimates for Emden-Fowler equations with boundary singularities and critical growth[END_REF] has no radial sign-changing solutions in B whenever γ ≥ (N -2) 2 4 -4. There are also many results in the case where λ > λ 1 (-∆). They are summarized in [START_REF] Esposito | Sign-changing solutions for critical equations with Hardy potential[END_REF].

The cases where both γ and s are not zero, including the limiting cases were also studied by Ghoussoub-Robert [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] and many other authors [START_REF] Cao | A note on the sign-changing solutions to elliptic problems with critical Sobolev and Hardy terms[END_REF][START_REF] Peng | Infinitely many solutions for Hardy-Sobolev equation involving critical growth[END_REF][START_REF] Wang | Infinitely many solutions for Hardy-Sobolev-Maz'ya equation involving critical growth[END_REF][START_REF] Yan | Infinitely many solutions for an elliptic problem involving critical Sobolev and Hardy-Sobolev exponents[END_REF].

The case when 0 ∈ ∂Ω is more involved and was studied extensively in Ghoussoub-Robert [START_REF] Ghoussoub | Hardy-Singular Boundary Mass and Sobolev-Critical Variational Problems[END_REF] and by the above authors [START_REF] Ghoussoub | Multiplicity and stability of the Pohozaev obstruction for Hardy-Schrödinger equations with boundary singularity[END_REF]. This paper is mostly concerned with the influence of the potential h γ,λ on existence and multiplicity issues related to the equation [START_REF] Druet | Stability of the Pohožaev obstruction in dimension 3[END_REF].

Existence and Multiplicity of solutions. Our multiplicity result will therefore follow from the following more general Euclidean statement. We set

C 1 (Ω, |x| -θ ) :=    h ∈ C 1 (Ω \ {0}); ∃c ∈ R such that lim x→0 |x| θ h(x) = c & lim x→0 |x| θ (x, ∇h(x)) = -cθ. ( 18
)
The main multiplicity result in this paper is the following theorem Theorem 1.5. Consider a bounded, smooth domain Ω ⊂ R n (n ≥ 3) such that 0 ∈ Ω and assume that 0 < s < 2 and 0 ≤ θ < 2. Let h ∈ C 1 (Ω, |x| -θ ) be such that -∆ -γ |x| 2 -h(x) is coercive in Ω. Let b ∈ C 2 (Ω) be such that b ≥ 0, b(0) > 0 and ∇b(0) = 0, while if θ = 0, we shall assume in addition that ∇ 2 b(0) ≥ 0 in the sense of bilinear forms. 2 and lim x→0 |x| θ h(x) = K h > 0, then the boundary value problem

If 0 ≤ γ < (n -2) 2 4 -(2 -θ)
   -∆u -γ u |x| 2 -h(x)u = b(x) |u| 2 (s)-2 u |x| s in Ω \ {0}, u = 0 on ∂Ω, (19)
has an infinite number of possibly sign-changing solutions in H 2 1,0 (Ω). Moreover, these solutions belong to C 2 (Ω \ {0}) while around 0 they behave like

lim x→0 |x| n-2 2 - (n-2) 2 4 -γ u(x) = K ∈ R. (20)
Note that when h γ,λ ≡ λ and b ≡ 1, the above yields an infinite number of solutions for equation [START_REF] Ghoussoub | Concentration estimates for Emden-Fowler equations with boundary singularities and critical growth[END_REF], provided 0 < s < 2, n ≥ 5, 0 ≤ γ < (n -2) 2 4 -4 and 0 < λ < λ 1 (-∆ -γ |x| 2 ). Note that the fact that s > 0 allows for an improvement on the dimension n ≥ 7 established for s = 0 by Devillanova-Solimini [START_REF] Devillanova | Concentration estimates and muliple solutions to ellipic problems at critical growth[END_REF] when γ = 0 and Cao-Yan [START_REF] Cao | Infinitely many solutions for an elliptic problem involving critical Sobolev growth and Hardy potential[END_REF] when γ > 0.

The multiplicity result will follow from standard min-max arguments once we prove the required compactness, which relies on blow-up analysis techniques. The proof consists of analyzing the asymptotic behaviour of a family of solutions to the related subcritical equations -potentially developing a singularity at zero-as we approach the critical exponent.

Compactness of approximate solutions. For 0 ≤ θ < 2, let (h ) >0 be functions in C 1 (Ω \ {0}), and

h 0 ∈ C 1 (Ω, |x| -θ ) such that lim →0 sup x∈Ω |x| θ |h (x) -h 0 (x)| + |x| θ+1 |∇(h -h 0 )(x)| = 0. ( 21
)
Theorem 1.6. Consider a bounded, smooth domain Ω ⊂ R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2 and 0 ≤ θ < 2.

Let h 0 ∈ C 1 (Ω, |x| -θ ) be such that -∆ -γ |x| 2 -h 0 (x) is coercive in Ω and consider (h ) >0 such that (21) hold. Let b ∈ C 2 (Ω) be such that b ≥ 0, b (
0) > 0 and ∇b(0) = 0, while if θ = 0, we shall assume in addition that ∇ 2 b(0) ≥ 0 in the sense of bilinear forms.

Let (p ) >0 such that 0 ≤ p < 2 (s) -2 and lim →0 p = 0 and consider a sequence of functions (u ) >0 that is uniformly bounded in H 2 1,0 (Ω) and such that for each > 0, u is a solution to the equation:

   -∆u -γ u |x| 2 -h (x)u = b(x) |u | 2 (s)-2-p u |x| s in Ω \ {0}, u = 0 on ∂Ω. ( 22
)
If γ < (n-2) 2 4 -(2 -θ) 2 and lim |x|→0 |x| θ h 0 (x) = K h0 > 0, then the sequence (u ) >0 is pre-compact in the space H 2 1,0 (Ω).
The interior mass and compactness for approximate positive solutions at any energy level. Note that the function x → |x| -α is a solution of

-∆ - γ |x| 2 u = 0 on R n \ {0}, (23)
if and only if α ∈ {α -(γ), α + (γ)}, where α -(γ), α + (γ) are as in [START_REF] Druet | Stability of elliptic PDEs with respect to perturbations of the domain[END_REF]. Actually, one can show that any non-negative solution u ∈ C 2 (R n \ {0}) of ( 23) is of the form

u(x) = C -|x| -α-(γ) + C + |x| -α+(γ) for all x ∈ R n \ {0}, (24)
where C -, C + ≥ 0, see [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF].

For showing compactness in low-dimensions, the arguments are not any more local, but global. We are naturally led to introducing a notion of mass:

Theorem 1.7 (The Euclidean interior mass, see [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF]). Let Ω be a smooth bounded domain of R n such that 0 ∈ Ω is an interior point. Let γ < (n -2) 2 4 and let h ∈ C 1 (Ω, |x| -θ ) be such that the operator

-∆ - γ |x| 2 -h(x) is coercive in Ω. Then there exists H ∈ C ∞ (Ω \ {0}) such that      -∆H - γ |x| 2 H -h(x)H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω. (25)
These solutions are unique up to a positive multiplicative constant, and there exists c + > 0 such that

H(x) x→0 c + |x| α+(γ) . Moreover, when γ > (n -2) 2 4 -1 - θ 2 2
, there exists c -∈ R such that

H(x) = c + |x| α+(γ) + c - |x| α-(γ) + o 1 |x| α-(γ) as x → 0. ( 26
)
We define the interior mass as m γ,h (Ω) := c + c - , which is independent of the choice of H.

We then establish the following compactness result for positive solutions.

Theorem 1.8. Consider a bounded, smooth domain Ω ⊂ R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2 and

0 ≤ θ < 2. Let h 0 ∈ C 1 (Ω, |x| -θ ) be such that -∆ -γ |x| 2 -h 0 (x) is coercive in Ω and consider (h ) >0 such that (21) hold. Let b ∈ C 2 (Ω) be such that b ≥ 0, b (
0) > 0 and ∇b(0) = 0, while if θ = 0, we shall assume in addition that ∆b(0) ≥ 0.

Let (p ) >0 be such that 0 ≤ p < 2 (s) -2 and lim →0 p = 0 and consider a sequence of functions (u ) >0 that is uniformly bounded in H 2 1,0 (Ω) such that for each > 0, u is a solution to the equation:

       -∆u -γ u |x| 2 -h (x)u = b(x) u 2 (s)-1-p |x| s in Ω \ {0}, u > 0 in Ω, u = 0 on ∂Ω. ( 27
)
Assuming one of the following conditions

• 0 ≤ γ ≤ (n-2) 2 4 -(1 -θ 2 ) 2 and lim |x|→0 |x| θ h 0 (x) = K h0 > 0; • γ > (n-2) 2 4 -(1 -θ 2 ) 2 and m γ,h0 (Ω) > 0. Then the sequence (u ) >0 is pre-compact in the space H 2 1,0 (Ω).
Non-existence: Stability of the Pohozaev obstruction. To address issues of non-existence of solutions, we shall prove the following surprising stability of regimes where variational positive solutions do not exist.

Theorem 1.9. Let Ω be a smooth bounded domain in

R n (n ≥ 3) such that 0 ∈ Ω is an interior point. Assume that 0 < s < 2, 0 ≤ θ < 2 and γ < (n-2) 2 /4. Let h 0 ∈ C 1 (Ω, |x| -θ ) be such that -∆-γ|x| -2 -h 0 is coercive and let b ∈ C 2 (Ω) be such that b ≥ 0, b(0) > 0 and ∇b(0) = 0. Assume that γ > (n -2) 2 4 -1 - θ 2 2
, the mass m γ,h0 (Ω) is non-zero, and that there is no positive variational solution to the boundary value problem:

       -∆u -γ u |x| 2 -h 0 (x)u = b(x) u 2 (s)-1-p |x| s in Ω \ {0}, u > 0 in Ω, u = 0 on ∂Ω. (28)
Then, for all Λ > 0, there exists := (Λ, h 0 ) > 0 such that for any

h ∈ C 1 (Ω, |x| -θ ) satisfying sup x∈Ω (|x| θ |h(x) -h 0 (x)| + |x| θ+1 |∇(h -h 0 )(x)|) < , (29)
there is no positive solution to [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] such that ∇u 2 ≤ Λ.

The above result is surprising for the following reason: Assuming Ω is starshaped with respect to 0, then the classical Pohozaev obstruction (see Section 13) yields that [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] has no positive variational solution whenever

h 0 ∈ C 1 (Ω, |x| -θ ) satisfies h 0 (x) + 1 2 (x, ∇h 0 (x)) ≤ 0 for all x ∈ Ω. ( 30
)
We then get the following corollaries.

Corollary 1.

Let Ω be a smooth bounded domain in R n (n ≥ 3) that is starshaped around 0. Assume 0 < s < 2, and

(n -2) 2 4 -1 - θ 2 2 < γ < (n -2) 2 4 . If h 0 ∈ C 1 (Ω, |x| -θ
) is a potential satisfying (30), then for all Λ > 0, there exists (Λ, h 0 ) > 0 such that for all h ∈ C 1 (Ω, |x| -θ ) satisfying ( 29), there is no positive solution to (28) with b ≡ 1, such that ∇u 2 ≤ Λ.

Corollary 2.

Let Ω be a smooth bounded domain in R n (n ≥ 3) that is starshaped around 0. Assume 0 < s < 2, and

(n -2) 2 4 -1 < γ < (n -2) 2 4 .
Then, for all Λ > 0, there exists (Λ) > 0 such that for all λ ∈ [0, (Λ)), there is no positive solution to equation ( 17) with ∇u 2 ≤ Λ.

Remark 1. It is worth comparing these results to what happens in the nonsingular case, i.e., when

γ = s = θ = 0. Note that when γ = θ = 0, the condition γ > (n -2) 2 4 -1 - θ 2
2
reads n = 3. In contrast to the singular case, a celebrated result of Brezis-Nirenberg [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF] shows that, for γ = s = θ = 0, a variational solution to (17) always exists whenever n ≥ 4 and 0 < λ < λ 1 (Ω), with the geometry of the domain playing no role whatsoever. On the other hand, Druet-Laurain [START_REF] Druet | Stability of the Pohožaev obstruction in dimension 3[END_REF] showed that the geometry plays a role in dimension n = 3, still for γ = s = θ = 0, by proving that when Ω is star-shaped, then there is no solution to [START_REF] Ghoussoub | Concentration estimates for Emden-Fowler equations with boundary singularities and critical growth[END_REF] for all small values of λ > 0 (with no apriori bound on ∇u 2 ). Another point of view is that for n = 3, the nonexistence of solutions persists under small perturbations, but it does not for n ≥ 4, i.e., only for n = 3 that the Pohozaev obstruction is stable in the nonsingular case.

When 0 ∈ ∂Ω, the situation is different. Indeed, the authors showed in [START_REF] Ghoussoub | Multiplicity and stability of the Pohozaev obstruction for Hardy-Schrödinger equations with boundary singularity[END_REF] that there is no lowdimensional phenomenon and that in the singular case, the Pohozaev obstruction is stable in all dimensions.

Here are some other extensions related to this phenomenon.

• Our stability result still holds under an additional smooth perturbation of the domain Ω, just as was done by Druet-Hebey-Laurain [START_REF] Druet | Stability of elliptic PDEs with respect to perturbations of the domain[END_REF] when n = 3, γ = s = 0.

• On the other hand, the stability result of Druet-Laurain [START_REF] Druet | Stability of the Pohožaev obstruction in dimension 3[END_REF] is not conditional on an apriori bound ∇u 2 , while ours is. We expect however to be able to get rid of the apriori bound in the singular case, as long as 0 ∈ Ω and s > 0. This is a project in progress.

Setting the Blow-up

Throughout this paper, Ω will denote a bounded, smooth domain in R n , n ≥ 3, such that 0 ∈ Ω. We will always assume that γ < (n -2) 2 4 , s ∈ (0, 2). For > 0, we let p ∈ [0, 2 (s) -2) be such that

lim →0 p = 0. ( 31
)
We also consider

h 0 ∈ C 1 (Ω, |x| -θ ), 0 ≤ θ < 2 such that there exists K h0 ∈ R with lim x→0 |x| θ h 0 (x) = K h0 and lim x→0 |x| θ (x, ∇h 0 (x)) = -θK h0 , (32)
and -∆ -γ |x| 2 -h 0 (x) is coercive in Ω. We also let (h ) >0 be such that h ∈ C 1 (Ω \ {0}) for all > 0 and lim

→0 sup x∈Ω |x| θ |h (x) -h 0 (x)| + |x| θ+1 |∇(h -h 0 )(x)| = 0. ( 33
)
Concerning b, we assume that

b(x) ≥ 0 with b ∈ C 2 (Ω) and b(0) > 0, ∇b(0) = 0. ( 34
)
The exponents α ± (γ) will denote

α ± (γ) := n -2 2 ± (n -2) 2 4 -γ. ( 35
)
We consider a sequence of functions (u ) >0 in H 2 1,0 (Ω) such that for all > 0 the function u is a solution to the Dirichlet boundary value problem:

   -∆u -γ u |x| 2 -h (x)u = b(x) |u | 2 (s)-2-p u |x| s in H 2 1,0 (Ω), u = 0 on ∂Ω. (E )
where (p ), h (x) and b(x) is such that (31), ( 33) and (34) holds.

By the regularity Theorem 13.1, u ∈ C 2 (Ω \ {0}) and there exists K ∈ R such that lim

x→0 |x| α-(γ) u (x) =
K . In addition, we assume that the sequence (u ) >0 is bounded in H 2 1,0 (Ω) and we let Λ > 0 be such that

Ω |u | 2 (s)-p |x| s dx ≤ Λ for all > 0. ( 36
)
It then follows from the weak compactness of the unit ball of H 2 1,0 (Ω) that there exists

u 0 ∈ H 2 1,0 (Ω) such that u u 0 weakly in H 2 1,0 (Ω) as → 0. ( 37
)
Then u 0 is a weak solution to the Dirichlet boundary value problem

   -∆u 0 -γ u 0 |x| 2 -h 0 (x)u 0 = b(x) |u 0 | 2 (s)-2 u 0 |x| s in Ω \ {0}, u 0 = 0 on ∂Ω. (E 0 )
Again from the regularity Theorem 13.1 it follows that u 0 ∈ C 2,θ (Ω\{0}) and lim

x→0 |x| α-(γ) u 0 (x) = K 0 ∈ R. Fix τ ∈ R such that α -(γ) < τ < n -2 2 . ( 38
)
The following proposition shows that the sequence (u

) is pre-compact in H 2 1,0 (Ω) if x → |x| τ u is uniformly bounded in L ∞ (Ω). Proposition 1.
Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, γ < (n-2) 2

4

. We let (u ), (p ), h(x) and b(x) be such that (E ), (31), (33), (34) and (36) holds. Suppose there exists τ as in (38) and C > 0 such that |x| τ |u (x)| ≤ C for all x ∈ Ω \ {0} and for all > 0. Then up to a subsequence, lim

→0 u = u 0 in H 2 1,0 (Ω)
where u 0 is as in (37).

Proof: We have assumed that |x| τ |u (x)| ≤ C for all x ∈ Ω and for all > 0. So the sequence (u ) is uniformly bounded in L ∞ (Ω ) for any Ω ⊂⊂ Ω \ {0}. Then by standard elliptic estimates and from (37) it follows that u → u 0 in C 2 loc (Ω \ {0}). Now since |x| τ |u (x)| ≤ C for all x ∈ Ω and for all > 0 and since τ < n-2 2 , we have

lim δ→0 lim →0 B δ (0) b(x) |u | 2 (s)-p |x| s dx = 0 and lim δ→0 lim →0 B δ (0) |u | 2 |x| 2 dx = 0. ( 39
) Therefore lim →0 Ω b(x) |u | 2 (s)-p |x| s dx = Ω b(x) |u 0 | 2 (s)
|x| s dx and lim

→0 Ω |u | 2 |x| 2 dx = Ω |u 0 | 2 |x| 2 dx.
From (E ) and (37) we then obtain lim

→0 Ω |∇u | 2 -γ u 2 |x| 2 -h(x)u 2 dx = Ω |∇u 0 | 2 -γ u 2 0 |x| 2 -h(x)u 2 0 dx
and so then lim

→0 Ω |∇u | 2 = lim →0 Ω |∇u 0 | 2 .
And hence lim

→0 u = u 0 in H 2 1,0 (Ω).
From now on we shall assume that lim

→0 |x| τ u L ∞ (Ω) = +∞ where α -(γ) < τ < n -2 2 , (40)
and work towards obtaining a contradiction. We shall say that blow-up occurs whenever (40) holds.

Some Scaling Lemmas

We start with two scaling lemmas which we shall use many times in our analysis.

Lemma 2.

Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2 and γ < (n-2) 2

4

. Let (u ), (p ), h(x) and b(x) be such that (E ), (31), ( 33 From equation (E ) we obtain that v satisfies the equation

-∆v - κ 2 |y | 2 γ y |y | + κ |y | x 2 v -κ 2 h (y + κ x) v = b(y + κ x) |v | 2 (s)-2-p v y |y | + κ |y | x s
weakly in B 2ρ (0) for all > 0. With the help of (44), (33) and standard elliptic theory it then follows that there exists v ∈ C 1 (B 2ρ (0)) such that lim

→0 v = v in C 1 (B ρ (0)).
In particular,

v(0) = lim →0 v (0) = 1 ( 46
)
and therefore v ≡ 0.

On the other hand, change of variables and the definition of κ yields

Bρκ (y ) |u | 2 (s)-p |x| s dx = |u (y )| 2 (s)-p κ n |y | s Bρ(0) |v | 2 (s)-p y |y | + κ |y | x s dx = -1+ 2(2-s) 2 (s)-2-p |y | s( n-2 2 ) Bρ(0) |v | 2 (s)-p y |y | + κ |y | x s dx ≥ |y | s( n- 2
2 )

Bρ(0) |v | 2 (s)-p y |y | + κ |y | x s dx.
Using the equation (E ), (36), (43), (44) and passing to the limit → 0 we get that Bρ(0) |v| 2 (s) dx = 0, and so then v ≡ 0 in B ρ (0), a contradiction with (46). Thus (43) cannot hold. This proves that y = O( ) when → 0, which proves the lemma.

Lemma 3.

Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2 and γ < (n-2) 2

4

. Let (u ), (p ), h(x) and b(x) be such that (E ), (31), ( 33 For > 0 we rescale and define

w (x) := ν n-2 2 u ( x) for x ∈ -1 Ω \ {0}.
Assume that for any R > δ > 0 there exists C(R, δ) > 0 such that for all > 0

|w (x)| ≤C(R, δ) for all x ∈ B R (0) \ B δ (0). ( 47
)
Then there exists w

∈ D 1,2 (R n ) ∩ C 1 (R n \ {0}) such that w w weakly in D 1,2 (R n ) as → 0 w → w in C 1 loc (R n \ {0}) as → 0
And w satisfies weakly the equation

-∆w - γ |x| 2 w = b(0) |w| 2 (s)-2 w |x| s in R n \ {0}. Moreover if w ≡ 0, then R n |w| 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0) 2 (s) 2 (s)-2
and there exists t ∈ (0, 1] such that lim →0 ν p = t.

Proof of Lemma 3: The proof proceeds in four steps.

Step 3.1:

Let η ∈ C ∞ c (R n ).
One has that ηw ∈ H 1 0 (R n ) for > 0 sufficiently small. We claim that there exists

w η ∈ D 1,2 (R n ) such that upto a subsequence ηw w η weakly in D 1,2 (R n ) as → 0, ηw → w η (x)
a.e in R n as → 0.

We prove the claim. Let x ∈ R n , then

∇ (ηw ) (x) = w (x)∇η(x) + ν n-2 2 η(x)∇u ( x).
Now for any θ > 0, there exists C(θ) > 0 such that for any a, b > 0

(a + b) 2 ≤ C(θ)a 2 + (1 + θ)b 2
With this inequality we then obtain

R n |∇ (ηw )| 2 dx ≤ C(θ) R n |∇η| 2 w 2 dx + (1 + θ)ν n-2 2 R n η 2 |∇u ( x)| 2 dx
With Hölder inequality and a change of variables this becomes

R n |∇ (ηw )| 2 dx ≤ C(θ) ∇η 2 L n ν n-2   Ω |u | 2 dx   n-2 n + (1 + θ) ν n-2 Ω η x 2 |∇u | 2 dx. ( 48
) Since u H 2 1,0 (Ω) = O(1), so for > 0 small enough ηw D 1,2 (R n ) ≤ C η
Where C η is a constant depending on the function η. The claim then follows from the reflexivity of D 1,2 (R n ).

Step 3.2:

Let η 1 ∈ C ∞ c (R n ), 0 ≤ η 1 ≤ 1 be a smooth cut-off function, such that η 1 = 1 for x ∈ B 0 (1) 0 for x ∈ R n \B 0 (2) ( 49
)
For any R > 0 we let η R = η 1 (x/R). Then with a diagonal argument we can assume that upto a subsequence for any R > 0 there exists w

R ∈ D 1,2 (R n ) such that η R w w R weakly in D 1,2 (R n ) as → 0 η R w (x) → w R (x) a.e x in R n as → 0 Since ∇η R 2 n = ∇η 1 2
n for all R > 0, letting → 0 in (48) we obtain that

R n - |∇w R | 2 dx ≤ C for all R > 0 where C is a constant independent of R. So there exists w ∈ D 1,2 (R n ) such that w R w weakly in D 1,2 (R n ) as R → +∞ w R (x) → w(x)
a.e x in R n as R → +∞

Step 3.3: We claim that w ∈ C 1 (R n \ {0}) and it satisfies weakly the equation

-∆w - γ |x| 2 w = b(0) |w| 2 (s)-2 w |x| s in R n \ {0}.
We prove the claim. From (E ) it follows that for any > 0 and R > 0, η R w satisfies weakly the equation

-∆ (η R w ) - γ |x| 2 (η R w ) -2 h ( x) (η R w ) = b( x) | (η R w ) | 2 (s)-2-p (η R w ) |x| s . ( 50
)
From ( 47) and (33), using the standard elliptic estimates it follows that w R ∈ C 1 (B R (0) \ {0}) and that up to a subsequence lim

→0 η R w = w R in C 1 loc (B R (0) \ {0}) .
Letting → 0 in eqn (50) gives that w R satisfies weakly the equation

-∆w R - γ |x| 2 w R = b(0) |w R | 2 (s)-2-p w R |x| s . Again we have that |w R (x)| ≤ C(R, δ) for all x ∈ B R/2 (0) \ B 2δ (
0) and then again from standard elliptic estimates it follows that w ∈ C 1 (R n \ {0}) and lim R→+∞ wR = w in C 1 loc (R n \ {0}), up to a subsequence. Letting R → +∞ we obtain that w satisfies weakly the equation

-∆w - γ |x| 2 w = b(0) |w| 2 (s)-2 w |x| s .
This proves our claim.

Step 3.4: Coming back to equation (48) we have for R > 0

R n |∇(η R w )| 2 dx ≤ C(θ)    B0(2R)\B0(R) (η 2R w ) 2 * dx    n-2 n + (1 + θ) ν n-2 Ω |∇u | 2 dx. (51)
Since the sequence (u ) is bounded in H 2 1,0 (Ω), letting → 0 and then R → +∞ we obtain for some constant C

R n |∇w| 2 dx ≤ C lim →0 ν n-2 . ( 52
)
Now if w ≡ 0 weakly satisfies the equation

-∆w - γ |x| 2 w = b(0) |w| 2 (s)-2 w |x| s
using the definition of µ γ,s,0 (R n ) it then follows that

R n |w| 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0) 2 (s) 2 (s)-2 . Hence lim →0 ν > 0 which implies that t := lim →0 ν p > 0. ( 53
)
Since lim →0 ν = 0, therefore we have that 0 < t ≤ 1. This completes the lemma.

Construction and Exhaustion of the Blow-up scales

In this section we prove the following proposition:

Proposition 2.
Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that

0 < s < 2 and γ < (n -2) 2 4
. Let (u ), (p ), (h ) and b(x) be such that (E ), (31), (33), ( 34) and (36) holds. Assume that blow-up occurs, that is

lim →0 |x| τ u L ∞ (Ω) = +∞ where α -(γ) < τ < n -2 2 .
Then, there exists N ∈ N families of scales (µ i, ) >0 such that we have:

(A1) lim →0 u = u 0 in C 2 loc (Ω \ {0}
) where u 0 is as in(37).

(A2) 0 < µ 1, < ... < µ N, , for all > 0.

(A3)

lim →0 µ N, = 0 and lim →0 µ i+1, µ i, = +∞ for all 1 ≤ i ≤ N -1.
(A4) For any 1 ≤ i ≤ N and for > 0 we rescale and define ũi, (x

) := µ n-2 2 i, u (k i, x) for x ∈ k -1 i, Ω \ {0}
where

k i, = µ 1-p 2 (s)-2 i,
. Then there exists ũi ∈ D 1,2 (R n ) ∩ C 1 (R n \ {0}), ũi ≡ 0 such that ũi weakly solves the equation

-∆ũ i - γ |x| 2 ũi = b(0) |ũ i | 2 (s)-2 ũi |x| s (54) and ũi, -→ ũi in C 1 loc (R n \ {0}) as → 0, ũi, ũi weakly in D 1,2 (R n ) as → 0.
(A5) There exists C > 0 such that

|x| n-2 2 |u (x)| 1-p 2 (s)-2 ≤ C
for all > 0 and all x ∈ Ω \ {0}.

(A6) lim R→+∞ lim →0 sup Ω\B Rk N, (0) |x| n-2 2 |u (x) -u 0 (x)| 1-p 2 (s)-2 = 0. (A7) lim R→+∞ lim →0 sup B δk 1, (0)\{0} |x| n-2 2 u (x) -µ -n-2 2 1, ũ1 x k1, 1-p 2 (s)-2 = 0.
(A8) For any δ > 0 and any 1 ≤ i ≤ N -1, we have

lim R→+∞ lim →0 sup δki+1, ≥|x|≥Rki, |x| n-2 2 u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 1-p 2 (s)-2 = 0.
(A9) For any i ∈ {1, ..., N } there exists t i ∈ (0, 1] such that lim →0 µ p i, = t i . The proof of this proposition proceeds in five steps. Since s > 0, the subcriticality 2 (s) < 2 := 2 (0) of equations (E ) in Ω \ {0} along with (37) yields that u → u 0 in C 2 loc (Ω \ {0}). So the only blow-up point is the origin.

Step 4.1: The construction of the µ i, 's proceeds by induction. This step is the initiation.

By the regularity Theorem 13.1 and the definition of τ it follows that for any > 0 there exists

x 1, ∈ Ω \ {0} such that sup x∈Ω\{0} |x| τ |u (x)| = |x 1, | τ |u (x 1, )|. ( 55
)
We define µ 1, and k 1, > 0 as follows

µ -n-2 2 1, := |u (x 1, )| and k 1, := µ 1-p 2 -2 1, . (56)
Since blow-up occurs, that is (40) holds, we have lim

→0 µ 1, = 0
It follows that u satisfies the hypothesis (41) of Lemma 2 with y = x 1, , ν = µ 1, . Therefore

|x 1, | = O (k 1, ) as → 0.
Infact, we claim that there exists c 1 > 0 such that lim

→0 |x 1, | k 1, = c 1 . (57)
We argue by contradiction and we assume that |x 1, | = o(k 1, ) as → 0. We define for > 0 ṽ (x

) := µ n-2 2 1, u (|x 1, |x) for x ∈ |x 1, | -1 Ω \ {0}
Using (E ) we obtain that ṽ weakly satisfies the equation in

|x 1, | -1 Ω \ {0} -∆ṽ - γ |x| 2 ṽ -|x 1, | 2 h (|x 1, |x) ṽ = b(|x 1, |x) |x 1, | k 1, 2-s |ṽ | 2 (s)-2 ṽ |x| s .
The definition (55) yields

|x| τ |ṽ (x)| ≤ 1 for all x ∈ |x 1, | -1 Ω \ {0}. Standard elliptic theory then yield the existence of ṽ ∈ C 2 (R n \ {0}) such that ṽ → ṽ in C 2 loc (R n \ {0}) where -∆ṽ - γ |x| 2 ṽ = 0 in R n \ {0}.
In addition, we have that ṽ |x 1, | -1 x 1, = 1 and so ṽ ≡ 0. Also since

|x| τ |ṽ(x)| ≤ 1 in R n \ {0}, we have the bound |ṽ(x)| < 2|x| -α+(γ) + 2|x| -α-(γ) in R n \ {0}.
The classification of positive solutions of -∆v -γ

|x| 2 v = 0 in R n \ {0} (see (196)) yields the existence of A, B ∈ R such that ṽ(x) = A|x| -α+(γ) + B|x| -α-(γ) in R n \ {0}. Then the pointwise control |x| τ |ṽ(x)| ≤ 1 in R n \ {0} yields A = B =
0, contradicting ṽ ≡ 0. This proves the claim (57).

We rescale and define ũ1, (x

) := µ n-2 2 1, u (k 1, x) for x ∈ k -1 1, Ω \ {0}
It follows from ( 55) and (57) that ũ1, satisfies the hypothesis (47) of Lemma 3 with y = x 1, , ν = µ 1, . Then using lemma (3) we get that there exists ũ1

∈ D 1.2 (R n ) ∩ C 1 (R n \ {0}
) weakly satisfying the equation:

-∆ũ 1 - γ |x| 2 ũ1 = b(0) |ũ 1 | 2 (s)-2 ũ1 |x| s in R n \ {0}. and ũ1, ũ1 weakly in D 1,2 (R n ) as → 0 ũ1, → ũ1 in C 1 loc (R n \ {0}) as → 0
It follows from the definition that ũ1, x1, k1, = 1. From (57) we therefore have that ũ1 ≡ 0. And hence again from Lemma 3 we get that

R n |ũ 1 | 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0) 2 (s) 2 (s)-2
and there exists t 1 ∈ (0, 1] such that lim

→0 µ p 1, = t 1 . Next, since |x| α-(γ) ũ1 ∈ C 0 (R n ), we have lim δ→0 lim →0 sup B δk 1, (0)\{0} |x| n-2 2 µ -n-2 2 1, ũ1 x k 1, 1-p 2 (s)-2 = 0
and then using the definitions (55), (56) it follows that

lim δ→0 lim →0 sup B δk 1, (0)\{0} |x| n-2 2 u (x) -µ -n-2 2 1, ũ1 x k 1, 1-p 2 (s)-2 = 0.
Step 4.2: There exists C > 0 such that for all > 0 and all x ∈ Ω \ {0},

|x| n-2 2 |u (x)| 1-p 2 (s)-2 ≤ C. (58)
Proof of Step 4.2: We argue by contradiction and let (y

) >0 ∈ Ω \ {0} be such that sup x∈Ω\{0} |x| n-2 2 |u (x)| 1-p 2 (s)-2 = |y | n-2 2 |u (y )| 1-p 2 (s)-2 → +∞ as → 0. ( 59
)
By the regularity Theorem 13.1 it follows that the sequence (y ) >0 is well-defined and moreover lim

→0 y = 0, since u → u 0 in C 2 loc (Ω \ {0}). For > 0 we let ν := |u (y )| -2 n-2 , := ν 1-p 2 (s)-2 and κ := |y | s/2 2-s 2 .
Then it follows from (59) that lim

→0 ν = 0, lim →0 |y | = +∞ and lim →0 κ |y | = 0. ( 60
)
Let R > 0 and let x ∈ B R (0) be such that y + κ x ∈ Ω \ {0}. It follows from the definition (59) of y that for all > 0

|y + κ x| n-2 2 |u (y + κ x)| 1-p 2 (s)-2 ≤ |y | n-2 2 |u (y )| 1-p 2 (s)-2
and then, for all > 0

|u (y + κ x)| |u (y )| 1-p 2 (s)-2 ≤ 1 1 -κ |y | R n-2 2
for all x ∈ B R (0) such that y + κ x ∈ Ω \ {0}. Using (60), we get that there exists C(R) > 0 such that the hypothesis (41) of Lemma 2 is satisfied and therefore one has |y | = O( ) when → 0, contradiction to (60). This proves (58).

Let I ∈ N . We consider the following assertions:

(B1) 0 < µ 1, < ... < µ I, . (B2) lim →0 µ ,I = 0 and lim →0 µ ,i+1 µi, = +∞ for all 1 ≤ i ≤ I -1 (B3) For all 1 ≤ i ≤ I there exists ũi ∈ D 1,2 (R n ) ∩ C 2 (R n \ {0}
) such that ũi weakly solves the equation

-∆ũ i - γ |x| 2 ũi = b(0) |ũ i | 2 (s)-2 ũi |x| s in R n \ {0} with R n |ũ i | 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0) 2 (s) 2 (s)-2 , and ũi, -→ ũi in C 1 loc (R n \ {0}) as → 0, ũi, ũi weakly in D 1,2 (R n ) as → 0.
where for > 0 ũi, (x

) := µ n-2 2 i, u (k i, x) for x ∈ k -1 i, Ω \ {0} with k i, = µ 1-p 2 (s)-2 i,
.

(B4) For all 1 ≤ i ≤ I, there exists t i ∈ (0, 1] such that lim →0 µ p i, = t i . We say that H I holds if there exists I sequences (µ i, ) >0 , i = 1, ..., I such that points (B1), (B2) (B3) and (B4) holds. Note that it follows from Step 4.1 that H 1 holds. Next we show the following holds:

Step 4.3 Let I ≥ 1. We assume that H I holds. Then either H I+1 holds or lim

R→+∞ lim →0 sup Ω\B0(Rk I, ) |x| n-2 2 |u (x) -u 0 (x)| 1-p 2 (s)-2 = 0. Proof of Step 4.3: Suppose lim R→+∞ lim →0 sup Ω\B0(Rk I, ) |x| n-2 2 |u (x) -u 0 (x)| 1-p 2 (s)-2 = 0.
Then there exists a sequence of points (y

) >0 ∈ Ω \ {0} such that lim →0 |y | k I, = +∞ and lim →0 |y | n-2 2 |u (y ) -u 0 (y )| 1-p 2 (s)-2 = a > 0. ( 61
) Since u → u 0 in C 2 loc (Ω \ {0}) it follows that lim →0 y = 0.
Then by the regularity Theorem 13.1 and since

α -(γ) < n-2 2 , we get lim →0 |y | n-2 2 |u (y )| 1-p 2 (s)-2 = a > 0 (62)
for some positive constant a. In particular, lim

→0 |u (y )| = +∞. Let µ I+1, := |u (y )| -2 n-2 and k I+1, := µ 1-p 2 (s)-2 I+1,
.

As a consequence we have lim →0 µ I+1, = 0 and lim

→0 |y | k I+1, = a > 0. ( 63
)
We rescale and define ũI+1, (x

) := µ n-2 2 I+1, u (k I+1, x) for x ∈ k -1 I+1, Ω \ {0} It follows from (58) that for all > 0 |x| n-2 2 |ũ I+1, (x)| 1-p 2 (s)-2 ≤ C for x ∈ k -1 I+1
, Ω \ {0}. so hypothesis (47) of Lemma 3 is satisfied. Then using Lemma 3 we get that there exists ũI+1

∈ D 1,2 (R n ) ∩ C 1 (R n \ {0}
) weakly satisfying the equation:

-∆ũ I+1 - γ |x| 2 ũI+1 = b(0) |ũ I+1 | 2 (s)-2 ũI+1 |x| s in R n \ {0}. and ũI+1, ũI+1 weakly in D 1,2 (R n ) as → 0 ũI+1, → ũI+1 in C 1 loc (R n \ {0}) as → 0 We denote ỹ := y k I+1,
. From (63) it follows that that lim

→0 |ỹ | := |ỹ 0 | = a = 0. Therefore |ũ I+1 (ỹ 0 )| = lim →0 |ũ I+1, (ỹ )| = 1
, and hence ũI+1 ≡ 0. And hence again from Lemma 3 we get

R n |ũ I+1 | 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0)
and there exists t I+1 ∈ (0, 1] such that lim Hence the families (µ i, ) >0 , 1

≤ i ≤ I + 1 satisfy H I+1 .
The next step is equivalent to step 4.3 at intermediate scales.

Step 4.4 Let I ≥ 1. We assume that H I holds. Then for any 1 ≤ i ≤ I -1 and for any δ > 0, either

H I+1 holds or lim R→+∞ lim →0 sup B δk i+1, (0)\B Rk i, (0) |x| n-2 2 u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 1-p 2 (s)-2 = 0.
Proof of Step 4.4: We assume that there exists an i ≤ I -1 and δ > 0 such that lim

R→+∞ lim →0 sup B δk i+1, (0)\B Rk i, (0) |x| n-2 2 u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 1-p 2 (s)-2 > 0.
It then follows that there exists a sequence (y

) >0 ∈ Ω such that lim →0 |y | k i, = +∞, |y | ≤ δk i+1, for all > 0 (64) |y | n-2 2 u (y ) -µ -n-2 2 i+1, ũi+1 y k i+1, 1-p 2 (s)-2 = a > 0. ( 65
)
for some positive constant a. Note that a < +∞ since

|x| n-2 2 u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 1-p 2 (s)-2
is uniformly bounded for all x ∈ B δki+1, (0) \ B Rki, (0).

We let ỹ * ∈ R n be such that y = k i+1, ỹ * . It follows from (64) that |ỹ * | ≤ δ for all > 0. . We rewrite (65) as lim

→0 |ỹ * | n-2 2 |ũ i+1, (ỹ * ) -ũi+1 (ỹ * )| 1-p 2 (s)-2 = a > 0.
Then from point (B3) of H I it follows that ỹ * → 0 as → 0. And since

|x| α-(γ) ũi+1 ∈ C 0 (R n ), we get as → 0 |y | n-2 2 µ -n-2 2 i+1, ũi+1 y k i+1, 1-p 2 (s)-2 = O |y | k i+1, n-2 2 -α-(γ)
= o( 1)

Then (65) becomes lim →0 |y | n-2 2 |u (y )| 1-p 2 (s)-2 = a > 0. ( 66
)
In particular, lim →0 |u (y )| = +∞. We let

ν := |u (y )| -2 n-2 and := ν 1-p 2 (s)-2 .
Then we have lim

→0 ν = 0 and lim →0 |y | = a > 0. (67)
We rescale and define ũ (x) := ν n-2

2 u ( x) for x ∈ -1 Ω \ {0} It follows from (58) that for all > 0 |x| n-2 2 |ũ (x)| 1-p 2 (s)-2 ≤ C for x ∈ -1 Ω \ {0}.
so hypothesis (47) of Lemma 3 is satisfied. Then using lemma (3) we get that there exists ũ

∈ D 1.2 (R n ) ∩ C 1 (R n \ {0}
) weakly satisfying the equation:

-∆ũ - γ |x| 2 ũ = b(0) |ũ| 2 (s)-2 ũ |x| s in R n \ {0}. and ũ ũ weakly in D 1,2 (R n ) as → 0 ũ → ũ in C 1 loc (R n \ {0}) as → 0
We denote ỹ := y . From (66) it follows that that lim

→0 |ỹ | := |ỹ 0 | = a = 0. Therefore |ũ(ỹ 0 )| = lim →0 |ũ (ỹ )| = 1
, and hence ũ ≡ 0. And hence again from Lemma 3 we get

R n |ũ| 2 (s) |x| s ≥ µ γ,s,0 (R n ) b(0) 2 (s) 2 (s)-2
and there exists t ∈ (0, 1] such that lim →0 ν p = t. Moreover it follows from(66), (64) and since lim

→0 |y | ki+1, = 0, that lim →0 ν µ i, = +∞ and lim →0 µ i+1, ν = +∞.
Hence the families (µ 1, ),..., (µ i, ), (ν ), (µ i+1, ),..., (µ I, ) satisfy H I+1 .

The last step tells us that family {H I } is finite.

Step 4.5: Let N 0 = max{I : H I holds }. Then N 0 < +∞ and the conclusion of Proposition 2 holds with N = N 0 .

Proof of Step 4.5: Indeed, assume that H I holds. Since µ i, = o(µ i+1, ) for all 1 ≤ i ≤ N -1, we get with a change of variable and the definition of ũi, that for any R > δ > 0

Ω |u | 2 (s)-p |x| s dx ≥ I i=1 B Rk i, (0)\B δk i, (0) |u | 2 (s)-p |x| s dx ≥ I i=1 B R (0)\B δ (0) |ũ i, | 2 (s)-p |x| s dx.
Then from (36) we have Λ ≥

I i=1 B R (0)\B δ (0) |ũ i, | 2 (s)-p |x| s dx. (68)
Passing to the limit → 0 and then δ → 0, R → +∞ we obtain using point (B3) of

H I , that Λ ≥ µ γ,s,0 (R n ) b(0) I.
It then follows that N 0 < +∞.

We let families (µ 1, ) >0 ,..., (µ N0, ) >0 such that H N0 holds. We argue by contradiction and assume that the conclusion of Proposition 2 does not hold with N = N 0 . Assertions (A1), (A2), (A3), (A4), (A5), (A7) and (A9) holds. Assume that (A6) or (A8) does not hold. It then follows from Steps (4.3), (4.4) and (4.5) that H N +1 holds. A contradiction with the choice of N = N 0 and the proposition is proved.

Strong Pointwise Estimates

The objective of this section is the proof of the following strong pointwise control.

Proposition 3.

Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2 and γ < (n-2) 2

4

. Let (u ), (p ), (h ) and b(x) be such that (E ), ( 31), ( 33), ( 34) and (36) holds. Assume that blow-up occurs, that is

lim →0 |x| τ u L ∞ (Ω) = +∞ where α -(γ) < τ < n -2 2 .
Consider the µ 1, , ..., µ N, from Proposition 2. Then there exists C > 0 such that for all > 0

|u (x)| ≤ C   N i=1 µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |x| α-(γ) + |x| α+(γ) + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ)   (69) for all x ∈ Ω \ {0}.
The proof of this estimate proceeds in seven steps.

Step 5.1: We claim that for any τ > 0 small and any R > 0, there exists C(τ, R) > 0 such that for all > 0 sufficiently small

|u (x)| ≤ C(τ, R)    µ α + (γ)-α -(γ) 2 -τ N, |x| α+(γ)-τ + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ)+τ    for all x ∈ Ω \ B Rk N, (0). ( 70
) Proof of Step 5.1: We fix γ such that γ < γ < (n-2) 2 4 . Since the operator -∆ -γ |x| 2 -h 0 (x) is coercive, taking γ close to γ it follows that the operator -∆ -γ |x| 2 -h 0 (x) is also coercive in Ω. From Theorem 25 it get that there exists H ∈ C ∞ (Ω \ {0}) such that    -∆H -γ |x| 2 H -h 0 (x)H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω. (71)
And we have the following bound on H: there exists

δ 1 , C 1 > 0 such that 1 C 1 1 |x| β+(γ ) ≤ H(x) ≤ C 1 1 |x| β+(γ ) for all x ∈ B 2δ1 (0). ( 72
)
We let λ γ 1 > 0 be the first eigenvalue of the coercive operator -∆ -γ |x| 2 -h on Ω and we let ϕ ∈ H 2 1,0 (Ω) be the unique eigenfunction such that

   -∆ϕ -γ |x| 2 ϕ -h 0 (x)ϕ = λ γ 1 ϕ in Ω ϕ > 0 in Ω \ {0} ϕ = 0 on ∂Ω. (73)
It follows from the regularity result, Theorem 13.1 that there exists

C 2 , δ 2 > 0 such that 1 C 2 1 |x| β-(γ ) ≤ ϕ(x) ≤ C 2 1 |x| β-(γ ) for all x ∈ Ω ∩ B 2δ2 (0). ( 74
)
We define the operator

L := -∆ - γ |x| 2 + h -b(x) |u | 2 (s)-2-p |x| s .
Step 5.1.1: We claim that given any γ < γ < (n-2) 2 4 there exist δ 0 > 0 and R 0 > 0 such that for any 0 < δ < δ 0 and R > R 0 , we have for > 0 sufficiently small L H(x) > 0, and

L ϕ(x) > 0 for all x ∈ B δ (0) \ B Rk N, (0), if u 0 ≡ 0. L H(x) > 0 for all x ∈ Ω \ B Rk N, (0), if u 0 ≡ 0. ( 75
)
As one checks for all > 0 and x = 0

L H(x) H(x) = γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s and L ϕ(x) ϕ(x) = γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s + λ γ 1 . For > 0 sufficiently small h 0 -h ∞ ≤ γ -γ 4(1+sup Ω |x| 2 ) and we choose 0 < δ 0 < min{1, δ 1 , δ 2 } such that ||b|| L ∞ (Ω) δ (2 (s)-2)( n-2 2 -α-(γ)) 0 |x| α-(γ) u 0 || 2 (s)-2 L ∞ (Ω) ≤ γ -γ 2 2 (s)+3 (76)
It follows from point (A6) of Proposition 2 that, there exists R 0 > 0 such that for any R > R 0 , we have for all > 0 sufficiently small

|b(x)| 1 2 (s)-2 |x| n-2 2 |u (x) -u 0 (x)| 1-p 2 (s)-2 ≤ γ -γ 2 2 (s)+2 1 2 (s)-2 for all x ∈ Ω \ B Rk N, (0)
With this choice of δ 0 and R 0 we get that for any 0 < δ < δ 0 and R > R 0 , we have for > 0 small enough

|b(x)| |x| 2-s |u (x)| 2 (s)-2-p ≤ 2 2 (s)-1-p |x| 2-s |b(x)||u (x) -u 0 (x)| 2 (s)-2-p + 2 2 (s)-1-p |x| 2-s |b(x)||u 0 (x)| 2 (s)-2-p ≤ γ -γ 4 for all x ∈ B δ (0) \ B Rk N, (0), if u 0 ≡ 0 and |b(x)| |x| 2-s |u (x)| 2 (s)-2-p ≤ γ -γ 4 for all x ∈ Ω \ B Rk N, (0), if u 0 ≡ 0.
Hence we obtain that for > 0 small enough

L H(x) H(x) = γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s ≥ γ -γ |x| 2 + (h 0 -h ) - γ -γ 4|x| 2 > 0 for all x ∈ B δ (0) \ B Rk N, (0), if u 0 ≡ 0 and L H(x) H(x) > 0 for all x ∈ Ω \ B Rk N, (0), if u 0 ≡ 0. ( 77
)
And

L ϕ(x) ϕ(x) ≥ γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s . ≥ γ -γ |x| 2 + (h 0 -h ) - γ -γ 4|x| 2 > 0 for all x ∈ B δ (0) \ B Rk N, (0)). ( 78
)
Step 5.1.2: It follows from point (A4) of Proposition 2 that there exists C 1 (R) > 0 such that for all > 0 small

|u (x)| ≤ C 1 (R) µ β + (γ )-β -(γ ) 2 N, |x| β+(γ ) for all x ∈ ∂B Rk N, (0) (79)
By estimate (72) on H, we then have for some constant

C 1 (R) > 0 |u (x)| ≤ C 1 (R)µ β + (γ )-β -(γ ) 2 N, H(x) for all x ∈ ∂B Rk N, (0). ( 80
)
It follows from point (A1) of Proposition 2 and the regularity result (13.1), that there exists C 2 (δ) > 0 such that for all > 0 small

|u (x)| ≤ C 2 (δ) |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ) for all x ∈ ∂B δ (0), if u 0 ≡ 0. ( 81
)
And then by the estimate (74) on ϕ we then have for some constant C 2 (δ) > 0

|u (x)| ≤ C 2 (δ) |x| α-(γ) u 0 L ∞ (Ω) ϕ(x) for all x ∈ ∂B δ (0) if u 0 ≡ 0.. ( 82
)
We now let for > 0 ,

Ψ (x) := C 1 (R)µ β + (γ )-β -(γ ) 2 N, H(x) + C 2 (δ) |x| α-(γ) u 0 || L ∞ (Ω) ϕ(x) for x ∈ Ω \ {0}.
Then (82) and (80) implies that for all > 0 small

|u (x)| ≤ Ψ (x) for all x ∈ ∂(B δ (0) \ B Rk N, (0)), if u 0 ≡ 0. ( 83
)
and if u 0 ≡ 0 then

|u (x)| ≤ Ψ (x) for all x ∈ ∂(Ω \ B Rk N, (0)). ( 84
)
Therefore when u 0 ≡ 0 it follows from (75)) and ( 83) that for all > 0 sufficiently small

       L Ψ ≥ 0 = L u in B δ (0) \ B Rk N, (0) Ψ ≥ u on ∂(B δ (0) \ B Rk N, (0)) L Ψ ≥ 0 = -L u in B δ (0) \ B Rk N, (0) Ψ ≥ -u on ∂(B δ (0) \ B Rk N, (0)).
and from ( 75) and (84), in case u 0 ≡ 0, we have for > 0 sufficiently small

       L Ψ ≥ 0 = L u in Ω \ B Rk N, (0) Ψ ≥ u on ∂(Ω \ B Rk N, (0)) L Ψ ≥ 0 = -L u in Ω \ B Rk N, (0) Ψ ≥ -u on ∂(Ω \ B Rk N, (0)).
Since Ψ > 0 and L Ψ > 0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [START_REF] Berestycki | The principal eigenvalue and maximum principle for second-order elliptic operators in general domains[END_REF] that the operator L satisfies the comparison principle on B δ (0) \ B Rk N, (0). Therefore

|u (x)| ≤ Ψ (x) for all x ∈ B δ (0) \ B Rk N, (0), and |u (x)| ≤ Ψ (x) for all x ∈ Ω \ B Rk N, (0) if u 0 ≡ 0.
Therefore when u 0 ≡ 0, we have for all > 0 small

|u (x)| ≤ C 1 (R)µ β + (γ )-β -(γ ) 2 N, H(x) + C 2 (δ) |x| α-(γ) u 0 || L ∞ (Ω) ϕ(x)
for all x ∈ B δ (0) \ B Rk N, (0), for R large and δ small.

Then when u 0 ≡ 0, using the estimates (72) and (74) we have or all > 0 small

|u (x)| ≤ C 1 (R) µ β + (γ )-β -(γ ) 2 N, |x| β+(γ ) + C 2 (δ) |x| α-(γ) u 0 L ∞ (Ω) |x| β-(γ )
for all x ∈ B δ (0) \ B Rk N, (0), for R large and δ small.

And if u 0 ≡ 0, then all > 0 small and R > 0 large

|u (x)| ≤ C 1 (R) µ β + (γ )-β -(γ ) 2 N, |x| β+(γ ) for all x ∈ Ω \ B Rk N, (0).
Taking γ close to γ, along with points (A1) and (A4) of Proposition 2 it then follows that estimate (70) holds on Ω \ B Rk ,N (0) for all R > 0.

Step 5.2: Let 1 ≤ i ≤ N -1. We claim that for any τ > 0 small and any R, ρ > 0, there exists C(τ, R, ρ) > 0 such that all > 0.

|u (x)| ≤ C(τ, R, ρ)   µ α + (γ)-α -(γ) 2 -τ i, |x| α+(γ)-τ + 1 µ α + (γ)-α -(γ) 2 -τ i+1, |x| α-(γ)+τ   (85)
for all x ∈ B ρki+1, (0) \ B Rki, (0). Proof of Step 5.2: We let i ∈ {1, ..., N -1}. We emulate the proof of Step 5.1. Fix γ such that γ < γ < (n-2) 2

4

. Consider the functions H and ϕ defined in Step 5.1 satisfying (71) and ( 71) respectively. We define the operator

L := -∆ - γ |x| 2 + h -b(x) |u | 2 (s)-2-p |x| s .
Step 5.2.1: We claim that given any γ < γ < (n-2) 2 4

there exist ρ 0 > 0 and R 0 > 0 such that for any 0 < ρ < ρ 0 and R > R 0 , we have for > 0 sufficiently small L H(x) > 0, and L ϕ(x) > 0 for all x ∈ B ρki+1, (0) \ B Rki, (0) (86)

As one checks for all > 0 and x = 0

L H(x) H(x) = γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s , L ϕ(x) ϕ(x) ≥ γ -γ |x| 2 + (h 0 -h ) -b(x) |u | 2 (s)-2-p |x| s .
We choose 0 < ρ 0 < 1 such that

ρ 2 0 sup Ω |h 0 -h | ≤ γ -γ 4
for all > 0 small and

||b|| L ∞ ρ (2 (s)-2)( n-2 2 -α-(γ)) 0 |x| α-(γ) ũi+1 || 2 (s)-2 L ∞ (B2(0)∩R n ) ≤ γ -γ 2 2 (s)+3 (87)
It follows from point (A8) of Proposition 2 that there exists R 0 > 0 such that for any R > R 0 and any 0 < ρ < ρ 0 , we have for all > 0 sufficiently small

|b(x)| 1 2 (s)-2 |x| n-2 2 u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 1-p 2 (s)-2 ≤ γ -γ 2 2 (s)+2 1 2 (s)-2 for all x ∈ B ρki+1, (0) \ B Rki, (0).
With this choice of ρ 0 and R 0 we get that for any 0 < ρ < ρ 0 and R > R 0 , we have for > 0 small enough

|b(x)||x| 2-s |u (x)| 2 (s)-2-p ≤ 2 2 (s)-1-p |x| 2-s |b(x)| u (x) -µ -n-2 2 i+1, ũi+1 x k i+1, 2 (s)-2-p + 2 2 (s)-1-p |x| k i+1, 2-s |b(x)| ũi+1 x k i+1, 2 (s)-2-p ≤ γ -γ 4 for all x ∈ B ρki+1, (0) \ B Rki, (0). ( 88
)
Hence as in Step 5.1 we have that for > 0 small enough L H(x) H(x) > 0 and L ϕ(x) ϕ(x) > 0 for all x ∈ B ρki+1, (0) \ B Rki, (0).

Step 5.2.2: let i ∈ {1, ..., N -1}. It follows from point (A4) of Proposition 2 that there exists C 1 (R) > 0 such that for all > 0 small

|u (x)| ≤ C 1 (R) µ β + (γ )-β -(γ ) 2 i, |x| β+(γ ) for all x ∈ ∂B Rki, (0) (89)
And then by the estimate (72) on H we have for some constant

C 1 (R) > 0 |u (x)| ≤ C 1 (R)µ β + (γ )-β -(γ ) 2 i, H(x) for all x ∈ ∂B Rki, (0). ( 90
)
From point (A4) of Proposition 2 it follows that there exists C 2 (ρ) > 0 such that for all > 0 small

|u (x)| ≤ C 2 (ρ) 1 µ β + (γ )-β -(γ ) 2 i+1, |x| β-(γ ) for all x ∈ ∂B ρki+1, (0). ( 91
)
Then by the estimate (74) on ϕ we have for some constant

C 2 (δ) > 0 |u (x)| ≤ C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 i+1, for all x ∈ ∂B ρki+1, (0). ( 92
)
We let for all > 0

Ψ (x) := C 1 (R)µ β + (γ )-β -(γ ) 2 i, H(x) + C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 i+1, for x ∈ Ω \ {0}
Then (90) and (92) implies that for all > 0 small

|u (x)| ≤ Ψ (x) for all x ∈ ∂ B ρki+1, (0) \ B Rki, (0 . ( 93
)
Therefore it follows from (86) and (93) that > 0 sufficiently small

       L Ψ ≥ 0 = L u in B ρki+1, (0) \ B Rki, (0) Ψ ≥ u on ∂ B ρki+1, (0) \ B Rki, (0) L Ψ ≥ 0 = -L u in B ρki+1, (0) \ B Rki, (0) Ψ ≥ -u on ∂ B ρki+1, (0) \ B Rki, (0) .
Since Ψ > 0 and L Ψ > 0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [START_REF] Berestycki | The principal eigenvalue and maximum principle for second-order elliptic operators in general domains[END_REF] that the operator L satisfies the comparison principle on B ρki+1, (0) \ B Rki, (0). Therefore

|u (x)| ≤ Ψ (x) for all x ∈ B ρki+1, (0) \ B Rki, (0)).
So for all > 0 small

|u (x)| ≤ C 1 (R)µ β + (γ )-β -(γ ) 2 i, H(x) + C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 i+1,
for all x ∈ B ρki+1, (0) \ B Rki, (0), for R large and ρ small. Then using the estimates (72) and (74) we have or all > 0 small

|u (x)| ≤ C 1 (R) µ β + (γ )-β -(γ ) 2 i, |x| β+(γ ) + C 2 (ρ) µ β + (γ )-β -(γ ) 2 i+1, |x| β-(γ )
for all x ∈ B ρki+1, (0) \ B Rki, (0).

for R large and ρ small. Taking γ close to γ, along with point (A4) of Proposition 2 it then follows that estimate (85) holds on B ρki+1, (0) \ B Rki, (0) for all R, ρ > 0.

Step 5.3: We claim that for any τ > 0 small and any ρ > 0, there exists C(τ, ρ) > 0 such that all > 0.

|u (x)| ≤ C(τ, ρ) 1 µ α + (γ)-α -(γ) 2 -τ 1, |x| α-(γ)+τ for all x ∈ B ρk1, (0) \ {0}. ( 94
)
Proof of Step 5.3: Fix γ such that γ < γ < (n-2) 2

4

. Consider the function ϕ defined in Step 5.1 satisfying (71).

Step 5.3.1: We claim that given any γ < γ < (n-2) 2 4 there exist ρ 0 > 0 such that for any 0 < ρ < ρ 0 we have for > 0 sufficiently small

L ϕ(x) > 0 for all x ∈ B ρk1, (0) \ {0} (95)
We choose 0 < ρ 0 < 1 such that

ρ 2 0 sup Ω |h 0 -h | ≤ γ -γ 4
for all > 0 small and

||b|| L ∞ ρ (2 (s)-2)( n-2 2 -α-(γ)) 0 |x| α-(γ) ũ1 || 2 (s)-2 L ∞ (B2(0)∩R n ) ≤ γ -γ 2 2 (s)+3 (96)
It follows from point (A7) of Proposition 2 that for any 0 < ρ < ρ 0 , we have for all > 0 sufficiently small

|b(x)| 1 2 (s)-2 |x| n-2 2 u (x) -µ -n-2 2 1, ũ1 x k 1, 1-p 2 (s)-2 ≤ γ -γ 2 2 (s)+2 1 2 (s)-2 for all x ∈ B ρk1, (0) \ {0}.
With this choice of ρ 0 we get that for any 0 < ρ < ρ 0 we have for > 0 small enough

|b(x)||x| 2-s |u (x)| 2 (s)-2-p ≤ 2 2 (s)-1-p |x| 2-s |b(x)| u (x) -µ -n-2 2 1, ũ1 x k 1, 2 (s)-2-p + 2 2 (s)-1-p |x| k 1, 2-s |b(x)| ũ1 x k 1, 2 (s)-2-p ≤ γ -γ 4 for all x ∈ B ρk1, (0) \ {0}. (97)
Hence similarly as in Step 5.1, we obtain that for > 0 small enough

L ϕ(x) ϕ(x) > 0 > 0 for all x ∈ x ∈ B ρk1, (0) \ {0}. (98)
Step 5.3.2: It follows from point (A4) of Proposition 2 that there exists C 2 (ρ) > 0 such that for all > 0 small

|u (x)| ≤ C 2 (ρ) 1 µ β + (γ )-β -(γ ) 2 1, |x| β-(γ )
for all x ∈ ∂B ρk1, (0) (99) and then by the estimate (74) on ϕ we have for some constant

C 2 (δ) > 0 |u (x)| ≤ C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 1,
for all x ∈ ∂B ρk1, (0).

We let for all > 0

Ψ 0 (x) := C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 1, for x ∈ Ω \ {0}.
Then (100) implies that for all > 0 small

|u (x)| ≤ Ψ 0 (x) for all x ∈ ∂B ρk1, (0). ( 101
)
Therefore it follows from (95) and (101) that > 0 sufficiently small

       L Ψ 0 ≥ 0 = L u in B ρk1, (0) \ {0} Ψ 0 ≥ u on ∂B ρk1, (0) \ {0} L Ψ 0 ≥ 0 = -L u in B ρk1, (0) Ψ 0 ≥ -u on ∂B ρk1, (0).
Since the operator L satisfies the comparison principle on B ρk1, (0). Therefore

|u (x)| ≤ Ψ 0 (x) for all x ∈ B ρk1, (0).
And so for all > 0 small

|u (x)| ≤ C 2 (ρ) ϕ(x) µ β + (γ )-β -(γ ) 2 1,
for all x ∈ B ρk1, (0) \ {0}.

for ρ small. Using the estimate (74) we have or all > 0 small

|u (x)| ≤ C 2 (ρ) µ β + (γ )-β -(γ ) 2 1, |x| β-(γ )
for all x ∈ B ρk1, (0) \ {0}.

for ρ small. It then follows from point (A4) of Proposition 2 that estimate (94) holds on x ∈ B ρk1, (0) for all ρ > 0.

Step 5.4: Combining the previous three steps, it follows from (70), (85), (94) and Proposition 2 that for any τ > 0 small, there exists C(τ ) > 0 such that for all > 0 we have

|u (x)| ≤ C(τ )   N i=1 µ α + (γ)-α -(γ) 2 -τ i, µ (α+(γ)-α-(γ))-2τ i, |x| α-(γ)+τ + |x| α+(γ)-τ + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ)+τ   (102)
for all x ∈ Ω \ {0}.

Next we improve the above estimate and show that one can take τ = 0 in (102).

For small, we let G be the Green's function for the coercive operator -∆ -γ |x| 2 -h on Ω with Dirichlet boundary condition. Green's representation formula, the pointwise bounds on the Green's function (204) [see Ghoussoub-Robert [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF]] yields for any z ∈ Ω

u (z) = Ω G (z, x)b(x) |u (x)| 2 (s)-2-p u (x) |x| s dx |u (z)| ≤ Ω G (z, x)|b(x)| |u (x)| 2 (s)-1-p |x| s dx ≤ C Ω max{|z|, |x|} min{|z|, |x|} α-(γ) 1 |x -z| n-2 |u (x)| 2 (s)-1-p |x| s dx. (103)
Using (102) we then obtain with 0 < τ < 2 (s)-2 2 (s)-1

α+(γ)-α-(γ) 2 that |u (z)| ≤ C N i=1 Ω max{|z|, |x|} min{|z|, |x|} α-(γ) 1 |x -z| n-2 |x| s   µ α + (γ)-α -(γ) 2 -τ i, µ (α+(γ)-α-(γ))-2τ i, |x| α-(γ)+τ + |x| α+(γ)-τ   2 (s)-1-p dx + C |x| α-(γ) u 0 || 2 (s)-1-p L ∞ (Ω) Ω max{|z|, |x|} min{|z|, |x|} α-(γ) 1 |x -z| n-2 |x| s 1 |x| (α-(γ)+τ )(2 (s)-1-p ) dx. ( 104
)
The first term in the above integral was computed for each bubble in Ghoussoub-Robert in [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] when p = 0. The proof goes exactly the same with p > 0. The last last term is straightforward to estimate.

We then get that there exists a constant C > 0 such that for any sequence of points (z ) in Ω \ {0} we have

|u (z )| ≤ C   N i=1 µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |z | α-(γ) + |z | α+(γ) + |x| α-(γ) u 0 || L ∞ (Ω) |z | α-(γ)   . ( 105
)
This completes the proof of Proposition 3.

Sharp Blow-up rates and proof of Compactness

When the expression makes sense, we define

C n,s := 2-θ 2 1 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx 1 2 (s) n-s 2 (s) N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx (106)
The proof of compactness rely on the following two key propositions.

Proposition 4.

Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be such that (E ), (33), (31), (34) and (36) holds. Assume that blow-up occurs, that is

lim →0 |x| τ u L ∞ (Ω) = +∞ for some α -(γ) < τ < n -2 2 . ( 107
)
Consider the µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Suppose that

either { α + (γ) -α -(γ) > 4 -2θ } or {α + (γ) -α -(γ) > 2 -θ and u 0 ≡ 0}. ( 108
)
Then, we have the following blow-up rate:

lim →0 p µ 2-θ N, = - C n,s b(0)     K h0 + 1 θ=0 2 (s) n i,j=1 ∂ ij b(0) R n X i X j |ũ N | 2 (s) |X| s dX R n ũ2 N dx     . ( 109
)
Proposition 5 (The positive case). Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, 0 ≤ γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be as in Proposition 4. Assume that blow-up occurs as in (107). Consider µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Suppose in addition that u > 0 for all > 0.

Then, we have the following blow-up rates:

(1) When α + (γ) -α -(γ) > 4 -2θ, we have

lim →0 p µ 2-θ N, = - C n,s b(0) K h0 + 1 θ=0 (n -2)(n(n -4) -4γ) 4n(2n -2 -s) ∆b(0) b(0) .
(

) When α + (γ) -α -(γ) = 4 -2θ, we have lim →0 p µ 2-θ N, = - C n,s b(0) K h0 + 1 θ=0 (n -2)(n(n -4) -4γ) 4n(2n -2 -s) ∆b(0) b(0) if u 0 ≡ 0. lim →0 p µ 2-θ N, = - C n,s b(0) K h0 + K + 1 θ=0 (n -2)(n(n -4) -4γ) 4n(2n -2 -s) ∆b(0) b(0) if u 0 > 0, 2
for some K > 0.

(

) When α + (γ) -α -(γ) < 4 -2θ, we have u 0 ≡ 0 and lim →0 p µ 2-θ N, = - C n,s b(0) K h0 + 1 θ=0 (n -2)(n(n -4) -4γ) 4n(2n -2 -s) ∆b(0) b(0) if α + (γ) -α -(γ) > 2 -θ. lim →0 p µ 2-θ N, ln 1 µ N, = -C n,s K h0 b(0) if α + (γ) -α -(γ) = 2 -θ, 3
where

C n,s := 2-θ 2 K 2 ω n-1 1 2 (s) n-s 2 (s) N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s)
|x| s dx , with K as defined in (142).

lim →0 p µ α+(γ)-α-(γ) N, = -χ • m γ,h0 (Ω) if α + (γ) -α -(γ) < 2 -θ, (111)
where χ > 0 is a constant and m γ,h0 (Ω) is the mass of Ω associated with the operator -∆γ |x| 2 -h 0 (x), defined in Theorem 1.7.

Proof of Theorems 1.6 and 1.8: We argue by contradiction and assume that the family is not pre-compact. Then, up to a subsequence, it blows up. We then apply Propositions 4 and 5 to get the blow-up rate (that is nonegative). However, the hypothesis of Theorems 1.5, 1.6 and 1.8 yield exactly negative blow-up rates. This is a contradiction, and therefore the family is pre-compact. This proves the three compactness theorems.

We now establish Propositions 4 and 5. The proof is divided in 14 steps in Sections 7 to 8. These steps are numbered Steps P1, P2, etc.

Estimates on the localized Pohozaev identity

We let (u ), (h ), (p ) and b be such that (E ), (33), (31), (34) and (36) holds. Assume that blow-up occurs as in (107). Note that

γ < (n -2) 2 4 -1 - θ 2 ⇐⇒ α + (γ) -α -(γ) > 2 -θ,
In the following, we will use the following consequence of (A9) of Proposition 2: for all i = 1, ..., N , there exists c i > 1 such that

c -1 i µ i, ≤ k i, ≤ c i µ i, . (112)
Step P1 (Pohozaev identity). For δ 0 > 0 small, we define

ρ := δ 0 if u 0 ≡ 0, r := √ µ N, if u 0 ≡ 0 ( 113
)
and

F (x) := (x, ν) |∇u | 2 2 - γ 2
u 2 |x| 2 - h (x) 2 u 2 - b(x) 2 (s) -p |u | 2 (s)-p |x| s -x i ∂ i u + n -2 2 u ∂ ν u . ( 114
)
We then have

Bρ (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx + p 2 (s) n -s 2 (s) -p Bρ (0)\B k 2 1, (0) b(x) |u | 2 (s)-p |x| s dx + 1 2 (s) -p Bρ (0)\B k 2 1, (0) (x, ∇b(x)) |u | 2 (s)-p |x| s dx = - ∂Bρ (0) F (x) dσ + ∂B k 2 1, (0)
F (x) dσ. ( 115
)
Proof of Step P1: We apply the Pohozaev identity (193) with y 0 = 0 and U = B ρ (0) \ B k 2 1, (0). This ends the proof of Step P1.

We will estimate each of the terms in the above integral identities and calculate the limit as → 0.

Estimates of the L 2 (s) term in the localized Pohozaev identity.

Step P2. Assuming that b satisfies (34), we claim as → 0

Bρ (0)\B k 2 1, (0) b(x) |u | 2 (s)-p |x| s dx = N i=1 b(0) t n-2 2 (s)-2 i R n |ũ i | 2 (s) |x| s dx + o(1), (116)
where (ρ ) is as in (113) and the ũi 's are as defined in Proposition 2 (A4).

Proof of

Step P2: For any R, ρ > 0 we decompose the above integral as

Br (0)\B k 2 1, (0) b
(x) |u | 2 (s)-p |x| s dx = Br (0)\B Rk N, (0) b(x) |u | 2 (s)-p |x| s dx + N i=1 B Rk i, (0)\B ρk i, (0) b(x) |u | 2 (s)-p |x| s dx + N -1 i=1 B ρk i+1, (0)\B Rk i, (0) b(x) |u | 2 (s)-p |x| s dx + B Rk 1, (0)\B k 2 1, (0) b
(x) |u | 2 (s)-p |x| s dx.
We will evaluate each of the above terms and calculate the limit lim

R→+∞ lim ρ→0 lim →0 .
From the estimate (69), we get as → 0

Br (0)\B Rk N, (0) b(x) |u | 2 (s)-p |x| s dx ≤ C Br (0)\B Rk N, (0)    µ α + (γ)-α -(γ) 2 (2 (s)-p ) N, |x| α+(γ)(2 (s)-p )+s + 1 |x| α-(γ)(2 (s)-p )+s    dx ≤ C Br (0)\B Rk N, (0) µ α + (γ)-α -(γ) 2 (2 (s)-p ) N, |x| α+(γ)(2 (s)-p )+s dx + C Br (0)\B Rk N, (0) 1 |x| α-(γ)(2 (s)-p )+s dx ≤ C B r k N, (0)\B R (0) 1 |x| n+2 (s) α + (γ)-α -(γ) 2 -p α+(γ) dx + C B1(0)\B Rk N, r (0)
1 |x| n-2 (s) α + (γ)-α -(γ) 2 -p α-(γ) dx ≤C R -2 (s) α + (γ)-α -(γ) 2 -p α+(γ) + r 2 (s) α + (γ)-α -(γ) 2 +p α-(γ) . Therefore lim R→+∞ lim →0 Br (0)\B Rk N, (0) b(x) |u | 2 (s)-p |x| s dx = 0. ( 117
)
It follows from Proposition 2 with a change of variable x = k i, y, that for any 1

≤ i ≤ N lim R→+∞ lim ρ→0 lim →0 B Rk i, (0)\B ρk i, (0) b(x) |u | 2 (s)-p |x| s dx = b(0) t n-2 2 (s)-2 i R n |ũ i | 2 (s) |x| s dx. ( 118
) Let 1 ≤ i ≤ N -1.
In Proposition 3, we had obtained the following pointwise estimates: For any R, ρ > 0 and all > 0 we have

|u (x)| ≤ C µ α + (γ)-α -(γ) 2 i, |x| α+(γ) + C µ α + (γ)-α -(γ) 2 i+1, |x| α-(γ)
for all x ∈ B ρki+1, (0) \ B Rki, (0).

Then we have as → 0

B ρk i+1, (0)\B Rk i, (0) b(x) |u | 2 (s)-p |x| s dx ≤ C B ρk i+1, (0)\B Rk i, (0)   µ α + (γ)-α -(γ) 2 (2 (s)-p ) i, |x| α+(γ)(2 (s)-p )+s + µ - α + (γ)-α -(γ) 2 (2 (s)-p ) i+1, |x| α-(γ)(2 (s)-p )+s   dx ≤ C B ρk i+1, (0)\B Rk i, (0)   µ α + (γ)-α -(γ) 2 (2 (s)-p ) i, |x| α+(γ)(2 (s)-p )+s + µ - α + (γ)-α -(γ) 2 (2 (s)-p ) i+1, |x| α-(γ)(2 (s)-p )+s   dx ≤ C B ρk i+1, k i, (0)\B R (0) 1 |x| n+2 (s) α + (γ)-α -(γ) 2 -p α+(γ) dx + C B2ρ(0)\B Rk i, k i+1, (0)
1 |x| n-2 (s) α + (γ)-α -(γ) 2 -p α-(γ) dx ≤C R -2 (s) α + (γ)-α -(γ) 2 -p α+(γ) + ρ 2 (s) α + (γ)-α -(γ) 2 +p α-(γ) .
And so

lim R→+∞ lim ρ→0 lim →0 B ρk i+1, (0)\B Rk i, (0) b(x) |u | 2 (s)-p |x| s dx = 0. ( 119
)
Again, from the pointwise estimates of Proposition 3, we have as → 0

B ρk 1, (0)\B k 2 1, (0) b
(x) |u | 2 (s)-p |x| s dx ≤ C B ρk 1, (0)\B k 2 1, (0) µ
- α + (γ)-α -(γ) 2 (2 (s)-p ) 1, |x| α-(γ)(2 (s)-p )+s dx ≤ C Bρ(0)\B k 2 1, (0) 1 |x| n-2 (s) α + (γ)-α -(γ) 2 -p α-(γ) dx ≤C ρ 2 (s) α + (γ)-α -(γ) 2 +p α-(γ) . Therefore lim ρ→0 lim →0 B ρk 1, (0)\B k 2 1, (0) b
(x) |u | 2 (s)-p |x| s dx = 0. ( 120
)
Combining ( 117), ( 118), ( 119) and (120) we obtain (116).

We now prove (116) under the assumption that u 0 ≡ 0. We decompose the integral as

B δ 0 (0)\B k 2 1, (0) b(x) |u | 2 (s)-p |x| s dx = B δ 0 (0)\Br (0) b(x) |u | 2 (s)-p |x| s dx + Br (0)\B k 2 1, (0) b
(x) |u | 2 (s)-p |x| s dx,
with r := √ µ N, . From the estimate (69) and u 0 ≡ 0, we get as → 0

B δ 0 (0)\Br (0) b(x) |u | 2 (s)-p |x| s dx ≤ C B δ 0 (0)\Br (0)    µ α + (γ)-α -(γ) 2 (2 (s)-p ) N, |x| α+(γ)(2 (s)-p )+s    dx Since α + (γ)2 (s) + s > n, we then get that B δ 0 (0)\Br (0) b(x) |u | 2 (s)-p |x| s dx ≤ C µ N, r 2 (s) 2 (α+(γ)-α-(γ)) = o(1)
as → 0. Then with (116) we get (116) and this proves (116). This ends the proof of Step P2.

Step P3. Assuming that b satisfies (34) and taking (ρ ) as in (113), we obtain as → 0

• If 2 (s) 2 (α + (γ) -α -(γ)) > 2, then, we have that Bρ (0)\B k 2 1, (0)
(x, ∇b(x)) |u | 2 (s)-p |x| s dx = µ 2 N,   ∂ ij b(0) 2t n 2 (s)-2 N n i,j=1 R n X i X j |ũ N | 2 (s) |X| s dX   + o(µ 2 N, ). (121)
• If 2 (s) 2 (α + (γ) -α -(γ)) = 2 then, we have that

Bρ (0)\B k 2 1, (0)
(x, ∇b(x)) |u | 2 (s)-p |x| s dx = O µ 2 N, ln 1 µ N, . (122)
• If 2 (s) 2 (α + (γ) -α -(γ)) < 2, then Br (0)\B k 2 1, (0)
(x, ∇b(x)) |u | 2 (s)-p |x| s dx = o µ α+(γ)-α-(γ) N, (123)
Proof of Step P3: The proof is very similar to Step P4 and we only sketch it. For convenience, for any i = 1, ..., N , we define

B i, (x) := µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |x| α-(γ) + |x| α+(γ) . ( 124
)
We have the following useful estimates: Lemma 4. For any i ∈ {1, ..., N } and (r ) such that cµ i, ≤ r ≤ δ 0 , we have that for k ≥ 0

Br (0) |x| k B 2 (s)-p i, |x| s dx ≤ C        µ k i, if 2 (s) 2 (α + (γ) -α -(γ)) > k; µ k i, 1 + ln r µi, if 2 (s) 2 (α + (γ) -α -(γ)) = k; µ 2 (s) 2 (α+(γ)-α-(γ)) i, r k- 2 (s) 2 (α+(γ)-α-(γ)) if 2 (s) 2 (α + (γ) -α -(γ)) < k.
and Br (0)

|x| k 1 + ln 1 |x| B 2 (s)-p i, |x| s dx ≤ C          µ k i, ln 1 µi, if 2 (s) 2 (α + (γ) -α -(γ)) > k; µ k i, ln 1 µi, 1 + ln r µi, if 2 (s) 2 (α + (γ) -α -(γ)) = k; µ 2 (s) 2 (α+(γ)-α-(γ)) i, r k- 2 (s) 2 (α+(γ)-α-(γ)) 1 + ln 1 |r | if 2 (s) 2 (α + (γ) -α -(γ)) < k.
The proof follows from straightforward estimates and the change of variable x = µ i, z. Note that we have used that µ p i, → t i = 0 as → 0.

We start with the case 2 (s)

2 (α + (γ) -α -(γ)) > 2.
With the same change of variable, we get that (Br (0)\B Rk N, (0))∪B R -1 k N, (0)

|x| 2 B 2 -p N, |x| s dx ≤ (R)µ 2 N, , (125)
where lim R→∞ (R) = 0. With the change of variables x = k N, y, the definition (A4) of ũN and k i, and the asymptotic (A9) in Proposition 2, we get that

B Rk N, (0)\B R -1 k N, (0)
(x, ∇b(x)) |u | 2 (s)-p |x| s dx (126) = µ 2- n 2 (s)-2 p N, B R (0)\B R -1 (0) x, ∇b(k N, x) k ,N ) |ũ N, | 2 (s)-p |y| s dy. (127)
Since ∇b(0) = 0, we have that

(x, ∇b(x)) = ∂ ij b(0)x i x j + o(|x| 2 ) as x → 0. ( 128
)
Finally, with r := √ µ ,N , we have that

Br (0) |x| 2 |x| s |x| -α-(γ)(2 (s)-p ) dx = O µ 1+ 2 (s) 4 (α+(γ)-α-(γ)) N,
. Now, for R > 0, with the upper-bound (69), we get that

(Br (0)\B Rk N, (0))∪B R -1 k N, (0) (x, ∇b(x)) |u | 2 (s) |x| s dx ≤ C Br (0) |x| 2 |x| s |x| -α-(γ)(2 (s)-p ) dx + C i<N Br (0) |x| 2 B 2 -p i, |x| s dx +C (Br (0)\B Rk N, (0))∪B R -1 k N, (0) |x| 2 B 2 -p N, |x| s dx. (129)
Using (128), letting → 0 and then R → +∞ in (127) and (129) in the estimates above, we get (121). This is similar for (121). The estimates for the case 2 (s) 2 (α + (γ) -α -(γ)) ≤ 2 are obtained via a rough upper-bound. This ends Step P3.

Estimates of the L 2 term in the localized Pohozaev identity.

Step P4. We have, as → 0,

• When α + (γ) -α -(γ) > 2 -θ, Br (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = µ 2-θ N,   2 -θ 2
K h0 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx + o(1)   . ( 130
)
where K h0 is as in (32).

• When α + (γ) -α -(γ) = 2 -θ, Br (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = O µ 2-θ N, ln 1 µ ,N . ( 131
) • When α + (γ) -α -(γ) < 2 -θ, Br (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = O µ 2-θ+(α + (γ)-α -(γ)) 2 N, . (132)
Proof of Step P4: Due to the hypothesis (32) and (33), we have that

h (x) + (∇h , x) 2 ≤ C|x| -θ for all x ∈ Ω \ {0} and > 0. ( 133
) Case 1: α + (γ) -α -(γ) < 2 -θ.
We estimate roughly the integral with ( 69)

Br (0)\B k 2 1, (0)
|x| -θ u 2 dx ≤ C Br (0)\B k 2 1, (0)
|x| -θ N i=1 µ α+(γ)-α-(γ) i, |x| 2α+(γ) + 1 |x| 2α-(γ) dx ≤ Cµ α+(γ)-α-(γ) N, Br (0)\B k 2 1, (0) |x| -θ-2α+(γ) dx + C Br (0)\B k 2 1, (0) |x| -θ-2α-(γ) dx ≤ Cµ α+(γ)-α-(γ) N, r 0 r n-θ-2α+(γ)-1 dx + C r 0 r n-θ-2α-(γ)-1 dx ≤ Cµ α+(γ)-α-(γ) N, r 0 r 2-θ-(α+(γ)-α-(γ))-1 dr + C r 0 r α+(γ)-α+(γ)+2-θ-1 dr ≤ Cµ 2-θ+(α + (γ)-α -(γ)) 2 N, since α + (γ) -α -(γ) < 2 -θ and r = √ µ N, .
This proves (132).

Case 2: α + (γ) -α -(γ) = 2 -θ.
Here again since r = √ µ N, , we estimate roughly the integral with (69) to get

Br (0)\B k 2 1, (0)
|x| -θ u 2 dx ≤ C Br (0)\B k 2 1, (0) |x| -θ N i=1 µ α+(γ)-α-(γ) i, µ 2(α+(γ)-α-(γ)) i, |x| 2α-(γ) + |x| 2α+(γ) + 1 |x| 2α-(γ) dx ≤ C N i=1 µ 2-θ i, B µ -1 i, r (0) 1 |x| θ |x| 2α-(γ) + |x| 2α+(γ) dx + C Br (0) |x| -θ-2α-(γ) dx ≤ C N i=1 µ 2-θ i, ln r µ i, + Cµ 2-θ N, ≤ C N i=1 µ 2-θ i, ln 1 µ i, ≤ Cµ 2-θ N, ln 1 µ N, .
Since µ i, = o(µ N, ) for i < N , this proves (131).

Case 3: α + (γ) -α -(γ) > 2 -θ. It follows from point (2) of Theorem 13.1 that the for all 1 ≤ i ≤ N , the function ũi |x| θ/2 ∈ L 2 (R n ) in this case. For any R, ρ > 0 we decompose the integral as

Br (0)\B k 2 1, (0) h (x) + (∇h , x) 2 u 2 dx = Br (0)\B Rk N, (0) h (x) + (∇h , x) 2 u 2 dx + N i=1 B Rk i, (0)\B ρk i, (0) h (x) + (∇h , x) 2 u 2 dx + N -1 i=1 B ρk i+1, (0)\B Rk i, (0) h (x) + (∇h , x) 2 u 2 dx + B ρk 1, (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx.
From the estimate (69), we get as → 0

µ -(2-θ) N, Br (0)\B Rk N, (0) h (x) + (∇h , x) 2 u 2 dx ≤ C µ -(2-θ) N, Br (0)\B Rk N, (0) 1 |x| θ µ α+(γ)-α-(γ) N, |x| 2α+(γ) + 1 |x| 2α-(γ) dx ≤ C B r k N, (0)\B R (0) 1 |x| θ 1 |x| n+(α+(γ)-α-(γ)-2) dx + C B1(0)\B Rk N, r (0) 1 |x| θ µ α + (γ)-α -(γ)-(2-θ) 2 N, |x| n-(α+(γ)-α-(γ)+2) dx ≤ C R -(α+(γ)-α-(γ)-(2-θ)) + µ α + (γ)-α -(γ)-(2-θ) 2 N,
.

Therefore when α + (γ) -α -(γ) > 2 -θ lim R→+∞ lim →0 µ -(2-θ) N, Br (0)\B Rk N, (0) h (x) + (∇h , x) 2 u 2 dx = 0. ( 134
)
Since in this case ũi

|x| θ/2 ∈ L 2 (R n ) for any 1 ≤ i ≤ N , it follows from Proposition 2 with a change of variable x = k N, y, that lim R→+∞ lim ρ→0 lim →0     µ -(2-θ) i, B Rk i, (0)\B ρk i, (0) h (x) + (∇h , x) 2 u 2 dx     = 2 -θ 2
K h0 t n-θ 2 (s)-2 i R n ũ2 i |x| θ dx. ( 135
) Let 1 ≤ i ≤ N -1.
Using the pointwise estimates (69), for any R, ρ > 0 and all > 0 we have as → 0

µ -(2-θ) i+1, B ρk i+1, (0)\B Rk i, (0) h (x) + (∇h , x) 2 u 2 dx ≤ C µ -(2-θ) i+1, B ρk i+1, (0)\B Rk i, (0) 1 |x| θ µ α+(γ)-α-(γ) i, |x| 2α+(γ) + µ -(α+(γ)-α-(γ)) i+1, |x| 2α-(γ) dx ≤ C µ -(2-θ) i+1, B ρk i+1, k i, (0)\B R (0) 1 |x| θ µ 2-θ i, |x| n+(α+(γ)-α-(γ)-2) dx + C µ -(2-θ) i+1, Bρ(0)\B Rk i, k i+1, (0) 1 |x| θ µ 2-θ i+1, |x| n-(α+(γ)-α-(γ)+2) dx ≤ C µ -(2-θ) i+1, µ 2-θ i, R -(α+(γ)-α-(γ)-(2-θ)) + µ 2-θ i+1, ρ α+(γ)-α-(γ)+(2-θ) .
And so as α

+ (γ) -α -(γ) > 2 lim R→+∞ lim ρ→0 lim →0 µ -(2-θ) i+1, B ρk i+1, (0)\B Rk i, (0) h (x) + (∇h , x) 2 u 2 dx = 0. ( 136
)
Similarly from the pointwise estimates (69) of Theorem 3, we have as → 0

µ -(2-θ) 1, B ρk 1, (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx ≤ C µ -(2-θ) 1, B ρk 1, (0)\B k 2 1, (0)
1 |x| θ µ -(α+(γ)-α-(γ)) 1, |x| 2α-(γ) dx ≤ C Bρ(0) 1 |x| θ 1 |x| n-(α+(γ)-α-(γ)+2) dx ≤ C ρ α+(γ)-α-(γ)+(2-θ) . Therefore lim ρ→0 lim →0 µ -(2-θ) 1, B ρk 1, (0)\B k 2 1, (0) h (x) + (∇h , x) 2 u 2 dx = 0. ( 137
)
From ( 134), ( 135), ( 136), (137) and Proposition 2 we then obtain (130). This ends Step P4.

Step P5. Suppose that u 0 ≡ 0. We claim that, as → 0

• When α + (γ) -α -(γ) > 2 -θ, B δ 0 (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = µ 2-θ N,   2 -θ 2
K h0 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx + o(1)   . ( 138
)
where K h0 is as in (32);

• When α + (γ) -α -(γ) = 2 -θ, B δ 0 (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = O µ 2-θ N, ln 1 µ ,N . ( 139
)
• When α + (γ) -α -(γ) < 2 -θ, B δ 0 (0)\B k 2 1, (0)
h (x) + (∇h , x) 2 u 2 dx = O µ α+(γ)-α-(γ) N, . (140)
Proof of Step P5: It follows from (69) that

B δ 0 (0)\Br (0) u 2 |x| θ dx ≤ C N i=1 B δ 0 (0)\Br (0) 1 |x| θ µ α+(γ)-α-(γ) i, µ 2(α+(γ)-α-(γ)) i, |x| 2α-(γ) + |x| 2α+(γ) dx ≤ C N i=1 B δ 0 (0)\Br (0) µ α+(γ)-α-(γ) i, |x| 2α+(γ)+θ dx ≤ Cµ α+(γ)-α-(γ) N, δ0 r r 2-θ-(α+(γ)-α-(γ))-1 dr.
Combining this estimate with (130) and ( 131) yield ( 138) and (139): this proves Step P5 for α + (γ)α -(γ) ≥ 2 -θ. For the case α + (γ) -α -(γ) < 2 -θ, the same estimate as above yields

B δ 0 (0) u 2 |x| θ dx ≤ Cµ α+(γ)-α-(γ) ,N δ0 0 r 2-θ-(α+(γ)-α-(γ))-1 dr ≤ Cµ α+(γ)-α-(γ) N,
, which gives us (140). This ends Step P5.

Proof of the sharp blow-up rates when u 0 ≡ 0 and α

+ (γ) -α -(γ) = 2 -θ.
Step P6. We let (u ), (h ), (p ) and b be such that (E ), ( 33), ( 31), ( 34) and (36) holds. Assume that blow-up occurs as in (107). Suppose u > 0 for all > 0 and u 0 ≡ 0. We fix a family of parameters (λ ) >0 ∈ (0, +∞) such that lim

→0 λ = 0 and lim →0 µ N, λ = 0. ( 141
)
Then, for all x ∈ R n \ {0}, x = 0, we have that

lim →0 λ α+(γ) µ α + (γ)-α -(γ) 2 N, u (λ x) = K |x| α+(γ) ,
where

K := L γ,Ω t α + (γ) 2 (s)-2 N R n ũN (z) 2 (s)-1 |z| s+α-(γ) dz > 0 ( 142
)
and L γ,Ω is defined in (206). Moreover, this limit holds in C 2 loc (R n \ {0}).

Proof of

Step P6: We define

w (x) := λ α+(γ) µ α + (γ)-α -(γ) 2 N, u (λ x)
for all x ∈ R n \ {0} ∩ B δλ -1 (0). From (E ) it follows that for all > 0, we have that

-∆w -γ |x| 2 w -λ 2 h • (λ x)w = Ξ b(λ x) w 2 (s)-1-p |x| s in R n \ {0} ∩ B δλ -1 (0) w > 0 in R n \ {0} ∩ B δλ -1 (0). With Ξ :=    µ α + (γ)-α -(γ) 2 N, λ α+(γ)    2 -2-p λ 2-s .
Since µ p N, → t N > 0 (see (A9) of Proposition 2) and

α + (γ)(2 (s) -2) -(2 -s) = (2 (s) -2) α + (γ) -α -(γ) 2 ,
then using the hypothesis (141), we get that

   µ α + (γ)-α -(γ) 2 N, λ α+(γ)    2 -2-p λ 2-s ≤ C µ N, λ α+(γ)(2 (s)-2-p )-(2-s) = o(1)
as → 0. Since u 0 ≡ 0, it follows from the pointwise control (69) that there exists C > 0 such that 0 < w (x) ≤ C|x| -α+(γ) for all x ∈ R n \ {0} ∩ B δλ -1 (0). It then follows from standard elliptic theory that there exists

w ∈ C 2 (R n \ {0}) such that lim →0 w = w in C 2 loc (R n \ {0}) (143) with -∆w -γ |x| 2 w = 0 in R n \ {0} 0 ≤ w(x) ≤ C|x| -α+(γ) in R n \ {0}.
From Proposition 11 in Ghoussoub-Robert [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] it follows that there exists Λ ≥ 0 such that w(x) = Λ|x| -α+(γ) for all x ∈ R n \{0}. We are left with proving that Λ = K defined in (142). We fix x ∈ R n \{0}. Green's representation formula yields

w (x) = Ω λ α+(γ) µ α + (γ)-α -(γ) 2 N, G (λ x, y) u (y) 2 (s)-1-p |y| s dy = B Rk N, (0)\B δk N, (0) + Ω\(B Rk N, (0)\B δk N, (0)) ( 144
)
Here G is the Green's function for the coercive operator -∆ -γ |x| 2 -h , for small, on Ω with Dirichlet boundary condition.

Step P6.1: We estimate the first term of the right-hand-side. A change of variable yields

B Rk N, (0)\B δk N, (0) λ α+(γ) µ α + (γ)-α -(γ) 2 N, G (λ x, y) u (y) 2 (s)-1-p |y| s dy = Ξ (1) B R (0)\B δ (0) G (λ x, k N, z) ũN, (z) 2 (s)-1-p |z| s (1 + o(1)) dz with Ξ (1) := λ α+(γ) µ α + (γ)-α -(γ) 2 N, k n-s N, µ -n-2 2 (2 (s)-1-p ) N,
It follows from ( 206) that for any z ∈ R n \ {0}, we have that

G (λ x, k N, z) = (L γ,Ω + o(1)) 1 λ α+(γ) |x| α+(γ) • 1 k α-(γ) N, |z| α-(γ)
, and that the convergence is uniform with repect to z ∈ B R (0) \ B δ (0). Plugging this estimate in the above equality, using that k N, = µ

1-p /(2 (s)-2) N,
, µ p N, → t N > 0 and the convergence of ũN, to ũN (see Proposition 2), we get that

B Rk N, (0)\B δk N, (0) λ α+(γ) µ α + (γ)-α -(γ) 2 N, G (λ x, y) u (y) 2 (s)-1-p |y| s dy = L γ,Ω 1 |x| α+(γ) t - α + (γ) 2 (s)-2 N B R (0)\B δ (0) 1 |z| α-(γ) ũN (z) 2 (s)-1 |z| s dz + o(1)
as → 0. Therefore, lim

R→+∞,δ→0 lim →0 B Rk N, (0)\B δk N, (0) λ α+(γ) µ α + (γ)-α -(γ) 2 N, G (λ x, y) u (y) 2 (s)-1 |y| s dy = K |x| α+(γ) ( 145
)
where K is as in (142).

Step P6.2: With the control (204) on the Green's function and the pointwise control (69) on u , we get that

Ω\(B Rk N, (0)\B δk N, (0) λ α+(γ) µ α + (γ)-α -(γ) 2 N, G (λ x, y) u (y) 2 (s)-1-p |y| s dy ≤ N -1 i=1 A i, + B (R) + C (δ) ( 146
)
where

A i, := C λ α+(γ) µ α + (γ)-α -(γ) 2 N, B R 0 (0) (x, y) α-(γ) |λ x -y| n-2 |y| s   µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |y| α-(γ) + |y| α+(γ)   2 (s)-1-p dy B (R) := C λ α+(γ) µ α + (γ)-α -(γ) 2 N, B R 0 (0)\B Rk N, (0) (x, y) α-(γ) |λ x -y| n-2 µ α + (γ)-α -(γ) 2 (2 (s)-1-p ) N, |y| α+(γ)(2 (s)-1-p )+s dy C (δ) := C(x) λ α+(γ)+2-n+α-(γ) µ α + (γ)-α -(γ) 2 •(2 (s)-p ) N, B δk N, (0) dy |y| α-(γ)(2 (s)-1-p )+s+α-(γ)
where (x, y) := max{λ |x|,|y|} min{λ |x|,|y|} ,

Step P6.3. We first estimate C (δ). Since n > s+2 (s)α -(γ) (this is a consequence of α -(γ) < (n-2)/2), straightforward computations yield

C (δ) ≤ C(x)δ 2 (s) 2 (α+(γ)-α-(γ)) ,
and therefore

lim δ→0 lim →0 C (δ) = 0. ( 147
)
Step P6.4. We estimate B (R). We split the integral as

B (R) = Rk ,N <|y|<
n -(s + α + (γ)(2 (s) -1) + α -(γ)) = - 2 (s) -2 2 (α + (γ) -α -(γ)) < 0, straightforward computations yield Rk N, <|y|< λ |x| 2 I (y) dy ≤ C(x) λ α+(γ)+α-(γ)+2-n µ α + (γ)-α -(γ) 2 N, Rk N, <|y|< λ |x| 2 µ α + (γ)-α -(γ) 2 (2 (s)-1-p ) N, |y| α+(γ)(2 (s)-1-p )+s+α-(γ)-1 dy ≤ C(x)R -2 (s)-2 2 (α+(γ)-α-(γ)) ,
For the next term, a change of variable yields

λ |x| 2 <|y|<2λ |x| I (y) dy ≤ C(x) λ α+(γ) µ α + (γ)-α -(γ) 2 N, λ |x| 2 <|y|<2λ |x| |λ x -y| 2-n µ α + (γ)-α -(γ) 2 (2 (s)-1) N, |y| α+(γ)(2 (s)-1)+s dy ≤ C(x) µ N, λ 2 (s)-2 2 (α+(γ)-α-(γ)) |x| 2 <|z|<2|x| |x -z| 2-n dz = o(1) as → 0. Finally, since α + (γ) + α -(γ) = n -2 and n -s -α + (γ)2 (s) = 2 (s) 2 (α + (γ) -α -(γ))
, we estimate the last term

|y|>2λ |x| I (y) dy ≤ C(x)µ 2 (s)-2 2 (α+(γ)-α-(γ)) N, λ α+(γ)-α-(γ) |y|>2λ |x| |y| α-(γ)+1-n-s dy |y| α+(γ)(2 (s)-1) ≤ C(x) µ N, λ 2 (s)-2 2 (α+(γ)-α-(γ)) = o(1)
as → 0. All these inequalities yield

lim R→+∞ lim →0 B (R) = 0. ( 148
)
Step P6.5. We fix i ∈ {1, ..., N -1} and estimate A i, . As above, we split the integral as

A i, = |y|< λ |x| 2 J i, (y) dy + λ |x| 2 <|y|<2λ |x| J i, (y) dy + |y|>2λ |x| J i, (y) dy,
where J i, is the integrand. Since µ i, ≤ µ N, , as one checks, the second and the third integral of the right-hand-side are controled from above respectively by J i, (y) dy

≤ C(x) λ α+(γ)+α-(γ)+2-n µ α + (γ)-α -(γ) 2 N, × |y|< λ |x| 2   µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |y| α-(γ) + |y| α+(γ)   2 (s)-1-p dy |y| s+α-(γ) ≤ C(x) µ n-s-α-(γ)-n-2 2 (2 (s)-1-p ) i, µ α + (γ)-α -(γ) 2 N, × |z|< λ |x| 2µ i, 1 |z| α-(γ)+s 1 |z| α-(γ) + |z| α+(γ) 2 (s)-1-p dz ≤ C(x) µ i, µ N, α + (γ)-α -(γ) 2 since n > s + 2 (s)α -(γ) and n < α -(γ) + s + (2 (s) -1)α + (γ). Since µ i, = o(µ N,
) as → 0, we get that lim

→0 A i, = 0. ( 149
)
Step P6.6: Plugging (145), ( 147), ( 148) and ( 149) into ( 144) and ( 146) yields lim →0 w (x) = K |x| α+(γ) for all x ∈ R n \ {0}. With (143), we then get that Λ = K. This proves Step P6. Now we get the optimal asymptotic when α + (γ) -α -(γ) = 2 -θ:

Step P7. We let (u ), (h ), (p ) and b be such that (E ), ( 33), ( 31), ( 34) and (36) holds. Assume that blow-up occurs as in (107) and α + (γ) -α -(γ) = 2 -θ. Suppose u > 0 for all > 0 and u 0 ≡ 0. Then

B δ 0 (0)\B k 2 1, (0) h (x) + (∇h , x) 2 u 2 dx = 2 -θ 2 K 2 ω n-1 K h0 + o(1) µ 2-θ N, ln 1 µ N, . (150)
Proof of Step P7: Note that it follows from (33) that

h (x) + (∇h , x) 2 = (2 -θ)K h0 + o(1) 2|x| θ as x → 0 and → 0. (151)
We define θ :

= 1 | ln µ N, |
, α := µ θ N, and β := µ 1-θ N, . As one checks, we have that

µ N, = o(β ) β = o(α ) α = o(1) ln α β ln 1 µ N, ln β µ N, = o ln 1 µ N, ln α = o(ln µ N, )    (152)
as → 0. It then follows from (69) and the properties (152) that

B δ 0 (0)\Bα (0) u 2 |x| θ dx = O µ 2-θ N, ln 1 α = o µ 2-θ N, ln 1 µ N, ; B β (0)\B k 2 1, (0)
u 2 |x| θ dx = o µ 2-θ N, ln 1 µ N, .                (153)
Since µ N, = o(β ) and α = o(1) as → 0, it follows from Proposition P6 that lim

→0 sup x∈Bα (0)\B β (0) |x| 2α+(γ) u 2 (x) µ α+(γ)-α-(γ) N, -K 2 = 0 (154)
and therefore

Bα (0)\B β (0) u 2 |x| θ dx = (K 2 + o(1))µ α+(γ)-α-(γ) N, Bα (0)\B β (0) dx |x| θ+2α+(γ) = (K 2 ω n-1 + o(1))µ 2-θ N, ln 1 µ N, .
Plugging this last estimate into (151) yields (150). This ends the proof of Step P7.

Estimates of the boundary terms.

Step P8. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), (34) and (36) holds. We assume that blow-up occurs as in (107). For any > 0, we define

ṽ (x) := r α-(γ) u (r x) for x ∈ B 2δ0r -1 (0) \ {0},
with r := √ µ N, . We claim that there exists ṽ

∈ C 1 (R n ) such that lim →0 ṽ (x) = ṽ in C 1 loc (R n \ {0})
where ṽ is a solution of -∆ṽ -

γ |x| 2 ṽ = 0 in R n \ {0} (155)
Proof of Step P8: From (E ) it follows that for all > 0, the rescaled functions ṽ weakly satisfies the equation

-∆ṽ - γ |x| 2 ṽ -r 2 h • (r x) ṽ = r ϑ+p α-(γ) b(r x) |ṽ | 2 (s)-2-p ṽ |x| s . ( 156
)
with

ϑ := (2 (s) -2) α + (γ) -α -(γ) 2 > 0 .
Using the pointwise estimates (69) we obtain the bound, that as → 0 we have for

x ∈ R n |ṽ (x)| ≤ C r α-(γ) N i=1 µ α + (γ)-α -(γ) 2 i, µ α+(γ)-α-(γ) i, |x| α-(γ) + |x| α+(γ) + C r α-(γ) |x| α-(γ) u 0 | L ∞ (Ω) |x| α-(γ) ≤ C N i=1 µi, µ N, α + (γ)-α -(γ) 2 µi, √ µ N, α+ (γ)-α-(γ) x r α-(γ) + x r α+(γ)
+ C |x| α-(γ) u 0 | L ∞ (Ω) x r α-(γ) ≤ C     N i=1 µi, µ N, α + (γ)-α -(γ) 2 µi, √ µ N, α+(γ)-α-(γ) |x| α-(γ) + |x| α+(γ) + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ)     ≤ C 1 |x| α+(γ) + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ) .
Then passing to limits in the equation ( 156), standard elliptic theory yields the existence of ṽ ∈ C 2 (R n \ {0}) such that ṽ → ṽ in C 2 loc (R n \ {0}) and ṽ satisfies the equation:

-∆ṽ -

γ |x| 2 ṽ = 0 in R n \ {0}
and we have the following bound on ṽ

|ṽ(x)| ≤ C 1 |x| α+(γ) + |x| α-(γ) u 0 | L ∞ (Ω) |x| α-(γ) for all x ∈ R n \ {0}.
This ends the proof of Step P8.

Step P9. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), ( 34) and (36) holds. We assume that blow-up occurs as in (107). We claim that, as → 0,

∂Br (0) F (x) dσ = µ α + (γ)-α -(γ) 2 N, (F 0 + o(1))
and

∂B k 2 1, (0)
F (x) dσ = o µ α + (γ)-α -(γ) 2 N, , (157)
with

F 0 := ∂B1(0) (x, ν) |∇ṽ| 2 2 - γ 2 ṽ2 |x| 2 -x i ∂ i ṽ + n -2 2 ṽ ∂ ν ṽ dσ (158)
Proof of Step P9: We keep the notations of Step P8. With a change of variable and the definition of ṽ , and

ϑ := (2 (s) -2) α + (γ) -α -(γ) 2 > 0, we get ∂Br (0) F (x) dσ = r α+(γ)-α-(γ) ∂B1(0) (x, ν) |∇ṽ | 2 2 - γ 2 ṽ2 |x| 2 -x i ∂ i ṽ + n -2 2 ṽ ∂ ν ṽ dσ -r α+(γ)-α-(γ) ∂B1(0) (x, ν) r 2 h (r x) 2 ṽ2 -b(r x) r ϑ+α-(γ)p 2 (s) -p |ṽ | 2 (s)-p |x| s dσ.
Then from the convergence result of Step P8, we then get (157). This ends Step P9.

Step P10. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), (34) and (36) holds. Assume that blow-up occurs as in (107). Suppose u 0 ≡ 0. We define

ū := u µ α + (γ)-α -(γ) 2 N, . (159)
We claim that there exists ū ∈ C 2 (Ω \ {0}) such that

lim →0 ū = ū in C 2 loc (Ω \ {0}) with    -∆ū - γ |x| 2 + h 0 ū = 0 in Ω \ {0} ū = 0 in ∂Ω (160)
Proof of Step P10: Since u 0 ≡ 0, it follows from (69) that there exists C > 0 such that

|ū (x)| ≤ C|x| -α+(γ) for all x ∈ Ω \ {0} and > 0. (161)
Moreover, equation (E ) rewrites

-∆ū - γ |x| 2 + h ū = µ α + (γ)-α -(γ) 2 (2 (s)-2-p ) N, b(x) |ū | 2 (s)-2-p ū |x| s in Ω,
and ū = 0 on ∂Ω. It then follows from standard elliptic theory that the claim holds. This ends Step P10.

Step P11. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), ( 34) and (36) holds. Assume that blow-up occurs as in (107). Suppose that u 0 ≡ 0. We claim that

∂B δ 0 (0) F (x) dσ = (F δ0 + o(1)) µ α+(γ)-α-(γ) N,
and

∂B k 2 1, (0)
F (x) dσ = o µ α+(γ)-α-(γ) N, , (162)
where

F δ0 := ∂B δ 0 (0) (x, ν) |∇ū| 2 2 - γ |x| 2 + h 0 ū2 2 -x i ∂ i ū + n -2 2 ū ∂ ν ū dσ. (163)
Proof of Step P11: With a change of variable, the definition of ū and the convergence (160), we get

∂B δ 0 (0) F (x) dσ (164) = µ α+(γ)-α-(γ) N, ∂B δ 0 (0) (x, ν) |∇ū | 2 2 - γ |x| 2 + h ū2 2 dσ - µ α + (γ)-α -(γ) 2 (2 -p ) N, 2 -p ∂B δ 0 (0) (x, ν)b(x) |ū | 2 -ū |x| 2 dσ - ∂B δ 0 (0) x i ∂ i ū + n -2 2 ū ∂ ν ū dσ = µ α+(γ)-α-(γ) N, (F δ0 + o(1)) . (165)
where F δ0 is as above. And directly from the bound (69) we obtain

∂B k 2 1, (0)
F (x) dσ = o µ α+(γ)-α-(γ) N,
This ends Step P11.

Step P12. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), ( 34) and (36) holds. Assume that blow-up occurs as in (107). Suppose that u > 0 for all > 0. We have F 0 ≥ 0 and

F 0 > 0 ⇐⇒ u 0 > 0.
where F 0 is as in (158).

Proof of

Step P12: We let ṽ be defined as in Step P8. It then follows from Step P8 that ṽ ≥ 0 and ṽ satisfies (155) and we have the following bound on ṽ

|ṽ(x)| ≤ C 1 |x| α+(γ) + |x| α-(γ) u 0 || L ∞ (Ω) |x| α-(γ) for all x ∈ R n \ {0}. (166)
Given α ∈ R, we define v α (x) := |x| -α for all x ∈ R n \ {0}. Since ṽ ≥ 0, it follows from Proposition 6.4 in Ghoussoub-Robert [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] that there exists A, B ≥ 0 such that

ṽ := Av α+(γ) + Bv α-(γ) . (167)
Step P12.1: We claim that B = 0 when u 0 ≡ 0. This is a direct consequence of controling (167) with (166) when u 0 ≡ 0 and letting |x| → ∞.

Step P12.2: We claim that B > 0 when u 0 > 0.

We prove the claim. We fix x ∈ R n \ {0}. Green's representation formula yields ṽ

(x) = Ω r α-(γ) G (r x, y) u 2 (s)-1-p (y)
|y| s dy.

We fix D ⊂⊂ Ω \ {0}. Then there exists c(D) > 0 such that |y| ≥ c(D) for all y ∈ D. Moreover, the control (205) of the Green's function yields

ṽ (x) ≥ C D r α-(γ) 1 r α-(γ) |x| α-(γ) |c(D) -r |x|| -(n-2) u 2 (s)-1-p (y) |y| s dy,
and then, passing to the limit → 0, we get that ṽ

(x) ≥ C |x| α-(γ) D u 2 (s)-1 0 (y) |y| s dy, for all x ∈ R n \ {0}. As one checks, this yields B ≥ C D u 2 (s)-1 0 (y) |y| s
dy > 0 when u 0 > 0. This ends Step P12.2.

Step P12.3: We claim that A > 0.

The proof is similar to Step P12.2. We fix x ∈ R n \ {0} and D ⊂⊂ R n \ {0}. Green's representation formula and the pointwise control (205) yields ṽ (x)

≥ k N, D r α-(γ) G (r x, y) u 2 (s)-1 (y) |y| s dy ≥ c( D) D r α-(γ) G (r x, k N, y)k n N, u (k N, y) 2 (s)-1 |k N, y| s dy ≥ c( D) D r α-(γ) r |x| k N, |y| α-(γ) |r x -k N, y| 2-n k n-2 2 N, ũ ,N (y) 2 (s)-1 |y| s dy Since r := √ µ N, , letting → 0, we get with the convergence (A4) of Proposition 2 that ṽ (x) ≥ c( D) D r 2α-(γ)-(n-2) |x| α-(γ) x - k N, r y 2-n k n-2 2 -α-(γ) N, ũ ,N (y) 2 (s)-1 |y| s dy ≥ c( D) |x| α+(γ) D ũN (y) 2 (s)-1
|y| s dy for all x ∈ R n \ {0}. Therefore, as one checks, A ≥ c(ω) D ũN (y) 2 (s)-1 |y| s dy > 0. This ends Step P12.3.

Step P12.4: We claim that

F 0 = 2ω n-1 (n -2) 2 4 -γ • AB. (168)
We prove the claim. Definition (158) reads

F 0 := ∂B1(0) (x, ν) |∇ṽ| 2 2 - γ 2 ṽ2 |x| 2 -x i ∂ i ṽ + n -2 2 ṽ ∂ ν ṽ dσ (169)
For simplicity, we define the bilinear form:

H δ (u, v) = ∂B δ (0) (x, ν) (∇u, ∇v) -γ uv |x| 2 -x i ∂ i u + n -2 2 u ∂ ν v -x i ∂ i v + n -2 2 v ∂ ν u dσ
As one checks, using (167)

F 0 = 1 2 H 1 (Av α+(γ) + Bv α-(γ) , Av α+(γ) + Bv α-(γ) ) = A 2 2 H 1 (v α+(γ) , v α+(γ) ) + ABH 1 (v α+(γ) , v α-(γ) ) + B 2 2 H 1 (v α-(γ) , v α-(γ) )
In full generality, we compute H δ (v α , v β ) for all α, β ∈ R and all δ > 0. Consequently, straightforward computations yield

x i ∂ i v α + n -2 2 v α ∂ ν v β = -β n -2 2 -α v α v β |x| and (x, ν) (∇v α , ∇v β ) - γ |x| 2 v α v β = (αβ -γ) v α v β |x|
and then

H δ (v α , v β ) = ∂B δ (0) αβ -γ + β n -2 2 -α + α n -2 2 -β v α v β |x| dσ
Plugging all these identities together yields

H δ (v α , v β ) = -αβ -γ + (α + β) n -2 2 δ -α-β-1+n-1 ω n-1 Since α + (γ), α -(γ) are solutions to X 2 -(n -2)X + γ = 0, we get that H δ (v α-(γ) , v α-(γ) ) = H δ (v α+(γ) , v α+(γ) ) = 0. Since α + (γ) + α -(γ) = n -2 and α + (γ)α -(γ) = γ, we get that H δ (v α-(γ) , v α+(γ) ) = 2ω n-1 (n -2) 2 4 -γ .
Plugging all these results together yields (168). This ends Step P12.4.

These substeps end the proof of Step P12.

Step P13. We let (u ), (h ), (p ) and b(x) be such that (E ), (33), (31), ( 34) and (36) holds. Assume that blow-up occurs as in (107). If u > 0 for all > 0 then u 0 ≡ 0, when α 130), ( 131), ( 132) and ( 157) and ( 122) and (122) into the Pohozaev identity (115), we get as → 0

+ (γ) -α -(γ) < 4 -2θ. Proof of Step P13: Since α + (γ) -α -(γ) 2 < 2 -θ, plugging (116), (
p 2 (s) n -s 2 (s)   N i=1 b(0) t n-2 2 (s)-2 i R n |ũ i | 2 (s) |x| s dx + o(1)   = -(F 0 + o(1)) µ α + (γ)-α -(γ) 2 N, , (170)
where F 0 is as in (169). Therefore F 0 ≤ 0. Since u > 0, it then follows from (168) of Step P12 that u 0 ≡ 0. This proves Step P13.

Proof of the sharp blow-up rates

We now prove the sharp blow-up rates claimed in Propositions 4 and 5.

The case α + (γ) -α -(γ) > 2 -θ. It follows from the Pohozaev identity (115), ( 116), ( 121), (130) and

Step P9 when u 0 ≡ 0, and the Pohozaev identity (115), ( 118), ( 121), (138) and Step P11 when u 0 ≡ 0 that

µ 2-θ N,   2 -θ 2
K h0 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx + o(1)   . + p 2 (s) n -s 2 (s)   N i=1 b(0) t n-2 2 (s)-2 i R n |ũ i | 2 (s) |x| s dx + o(1)   + 1 θ=0 2 (s) µ 2 ,N   ∂ ij b(0) t n 2 (s)-2 N R n X i X j |ũ N | 2 (s) |X| s dX   + o(µ 2 ,N ). ( 171
) =    -µ α + (γ)-α -(γ) 2 N, (F 0 + o(1)) -µ α+(γ)-α-(γ) N, (F δ0 + o(1)) if u 0 ≡ 0.
We will use the following lemma: Lemma 5. Suppose we have α + (γ) -α -(γ) > 2, γ ≥ 0 and ũN > 0. Then

∂ ij b(0) 2 (s) R n X i X j |ũ N | 2 (s) |X| s dX R n ũ2 N |x| θ dx = ∆b(0) b(0) (n -2)[(α + (γ) -α -(γ)) 2 -4] 4n(2n -2 -s) . (172)
This lemma will be proved at the end of the present section. The case u > 0 and α + (γ) -α -(γ) < 2 -θ.

N, = - 2-θ 2 K h 0 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx + 1 θ=0 2 (s) ∂ij b(0) t n 2 (s)-2 N R n X i X j |ũ N | 2 (s) |X| s dX + F 0 1 2 (s) n-s 2 (s) N i=1 b(0) t n-2 2 (s)-2 i
Step P14. We let (u ), (h ), (p ) and b(x) be such that (E ), ( 33), ( 31), ( 34) and (36) holds. We assume that blow-up occurs as in (107). Suppose u > 0 for all > 0 and α + (γ) -α -(γ) < 2 -θ. Then (111) holds, that is

lim →0 p µ α+(γ)-α-(γ) N, = - 2ω n-1 2 (s) 2 (n-2) 2 4 -γ A 2 (n -s) N i=1 b(0) t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx • m γ,h (Ω) (173)
for some A > 0, where m γ,h (Ω) is the mass.

Proof of

Step P14: It follows from Step P13 that u 0 ≡ 0.

Step P14.1: We now claim that

p 2 (s) n -s 2 (s)   N i=1 b(0) t n-2 2 (s)-2 i R n |ũ i | 2 (s) |x| s dx + o(1)   = µ α+(γ)-α-(γ) N, (M δ0 + o(1))
where

M δ0 := - B δ 0 (0) h 0 (x) + (∇h 0 , x) 2 ū2 dx -F δ0 (174)
with F δ0 is as in (163) and ū is as in (160).

Indeed the convergence (159), (160), (161) and α + (γ) -α -(γ) < 2 -θ gives

B δ 0 (0) h (x) + (∇h , x) 2 u 2 dx (175) = µ α+(γ)-α-(γ) N,    B δ 0 (0) h 0 (x) + (∇h 0 , x) 2 ū2 dx + o(1)   
Then from the Pohozaev identity (115), (118), (121), (138) and Step P11 we obtain (174), which proves the claim and ends Step P14.1.

Fix δ < δ . Taking U := B δ (0) \ B δ (0), b ≡ 0 and u = ū in (193), and using (160), we get that M δ is independent of the choice of δ > 0 small enough.

Step P14.2: We claim that ū > 0.

We prove this claim. Since ū ≥ 0 is a solution to (160), it is enough to prove that ū ≡ 0. We argue as in the proof of Step P12. We fix x ∈ Ω \ {0}. Green's identity and u > 0 gives

u (x) = µ -(α+(γ)-α-(γ))/2 N, Ω G (x, y) u (y) 2 (s)-1-p |y| s dy ≥ µ -(α+(γ)-α-(γ))/2 N, D G (x, y) u (y) 2 (s)-1-p |y| s dy ≥ Cµ n-s-(α+(γ)-α-(γ))/2 N, D G (x, µ N, y) u (µ N, y) 2 (s)-1-p |y| s dy,
where

D := B 2µ N, (0) \ B µ N, (0), D := B 2 (0) \ B 1 (0)
. With the pointwise control (204), we get

u (x) ≥ C(x) D |x| |y| α-(γ) |x -µ N, y| 2-n u ,N (y) 2 (s)-1-p |y| s dy
where u ,N is as in Proposition 2. Letting → 0 and using the convergence (A4) of Proposition 2, we get that ū(x) ≥ C(x) 1 |x| α+(γ) for all x ∈ Ω.

And then ū > 0 in Ω. This proves the claim and Step P14.2.

We fix r 0 > 0 and η ∈ C ∞ (R n ) such that η(x) = 1 in B r0 (0) and η(x) = 0 in R n \ B 2r0 (0). It then follows from [START_REF] Ghoussoub | Hardy-Singular Boundary Mass and Sobolev-Critical Variational Problems[END_REF][START_REF] Ghoussoub | Sobolev inequalities for the Hardy-Schrödinger operator: extremals and critical dimensions[END_REF] that, for r 0 > 0 small enough, there exists A > 0 and β ∈ H 1 0 (Ω) such that ū(x) = A η(x) |x| α+(γ) + β(x) for all x ∈ Ω with

β(x) = m γ,h (Ω) η(x) |x| α-(γ) + o η(x) |x| α-(γ)
as → 0. Here, m γ,h0 (Ω) is the mass of Ω associated with the operator -∆ -γ |x| 2 -h 0 (x), defined in Theorem 1.7..

Step P14.3: We claim that lim δ→0 M δ = -2ω n-1 (n -2) 2 4 -γ A 2 • m γ,h (Ω) (176)
We prove the claim. Since ū is a solution to (160), it follows from standard elliptic theory that there exists C > 0 such that ū(x)+|x||∇ū(x)| ≤ C|x| -α+(γ) for all x ∈ B 2δ0 (0). Therefore, since α + (γ)-α -(γ) < 2-θ, we get that

lim δ→0    B δ (0) ū2 dx + ∂B δ (0) (x, ν)ū 2 dσ    = 0.
And therefore,

M δ = - A 2 2 Hδ (v α+(γ) + vα-(γ) , vα+(γ) + vα-(γ) ) + o(1)
as δ → 0, where Hδ (u, v) :=

∂B δ 0 (0) (x, ν) (∇u, ∇v) - γ |x| 2 uv -x i ∂ i u + n -2 2 u ∂ ν v -x i ∂ i v + n -2 2 v ∂ ν u dσ
with vα+(γ) (x) := η(x) |x| α+(γ) and vα-(γ) (x) = β(x) for all x ∈ Ω. We then get that

M δ = - A 2 2 Hδ (v α+(γ) , vα+(γ) ) -A 2 Hδ (v α+(γ) , vα-(γ) ) - A 2 2 Hδ (v α-(γ) , vα-(γ) ) + o(1)
as δ → 0. Since α + (γ) -α -(γ) < 2 -θ and α + (γ) + α -(γ) = n -2, we get with a change of variable that as δ → 0,

Hδ (v α+(γ) , vα+(γ) ) = H δ (v α+(γ) , v α+(γ) ) + O(δ 2-(α+(γ)-α-(γ)) ) Hδ (v α+(γ) , vα-(γ) ) = m γ,h (Ω) • H δ (v α+(γ) , v α-(γ) ) + o(δ 2 ) Hδ (v α-(γ) , vα-(γ) ) = O(δ 2+(α+(γ)-α-(γ)) ).
Using the computations performed in the proof of Step P12, we then get (176). This proves the claim and ends Step P14.3.

End of the proof of

Step P14: Since M δ is independent of δ small, we then get that M δ0 = -2ω n-1 (n -2) 2 4 -γ A 2 m γ,h (Ω Putting this estimate in (174), we then get (173). This end Step P14.

Proof of Lemma 5. We assume that ũN > 0. We define

U (x) := 1 |x| 2-s n-2 α-(γ) + |x| 2-s n-2 α+(γ) n-2 2-s
for all x ∈ R n \ {0}. As one checks, we have that

-∆U - γ |x| 2 U = 4(n -s) n -2 (n -2) 2 4 -γ U 2 (s)-1 |x| s = (n -s)(α + (γ) -α -(γ)) 2 n -2 U 2 (s)-1 |x| s . We define µ 2 (s)-2 = (n -s)(α + (γ) -α -(γ)) 2 (n -2)b(0) .
so that µU is a solution to (54).

We claim that there exists ν > 0 such that ũN = µν -n-2 2 U (ν -1 •). We prove the claim. It follows from the convergence (A4) and the pointwise control (69) that

ũN (x) ≤ C |x| α-(γ) + |x| α+(γ) for x ∈ R n \ {0}.
Since ũN solves (54), one has that

-div(|x| -2α-(γ) ∇(|x| α-(γ) ũN )) = |x| -α-(γ)2 (s)-s (|x| α-(γ) ũN ) 2 (s)-1 in R n \ {0} with |x| α-(γ) ũN ∈ C 2 (R n \ {0}) and |x| α-(γ) ũN ≤ C(1 + |x| α+(γ)-α-(γ) ) -1 in R n \ {0}.
Since s > 0, it then follows from Chou-Chu [START_REF] Chou | On the best constant for a weighted Sobolev-Hardy inequality[END_REF] that ũN is radially symmetrical when γ ≥ 0.

We define ϕ(t) := e -n-2 2 t ũN (e -t ) for t ∈ R. We get that -ϕ" + (n-2) 2

4

-γ ϕ = ϕ 2 (s)-1 in R, ϕ > 0 and lim t→±∞ ϕ(t) = 0. The ODE yields the existence of S ∈ R such that

(ϕ ) 2 2 + (n -2) 2 4 -γ ϕ 2 2 - ϕ 2 (s) 2 (s) = S. (177)
Letting t → +∞ yields S = 0. Since lim t→±∞ ϕ(t) = 0, there exists t 0 ∈ R such that ϕ (t 0 ) = 0, and (177) yields a unique possible value for ϕ(t 0 ). Therefore, the theory of ODEs yields uniqueness of ϕ up to translation. Since λU is also such a positive solution, we then get that there exists ν > 0 such that ũN = µν -n-2 2 U (ν -1 •). This proves the claim. With a change of variables and symmetry, we get that

∂ ij b(0) 2 (s) R n X i X j |ũ N | 2 (s) |X| s dX R n ũ2 N dx = ∆b(0) n 2 (s) R n |X| 2 |ũ N | 2 (s) |X| s dX R n ũ2 N dx = ∆b(0)µ 2 (s)-2 (n -2) 2n(n -s) R n |X| 2 U 2 (s) |X| s dX R n U 2 dx = ∆b(0) b(0) (α + (γ) -α -(γ)) 2 2n R n |X| 2 U 2 (s) |X| s dX R n U 2 dx
We estimate this ratio as in Jaber [START_REF] Jaber | Hardy-Sobolev equations on compact Riemannian manifolds[END_REF]. Passing to radial coordinates and taking the change of variable

t = r 2-s n-2 (α+(γ)-α-(γ)) , we get that R n |X| 2 U 2 (s) |X| s dX R n U 2 dx = ∞ 0 r 2-s+n-1 r 2-s n-2 α -(γ) +r 2-s n-2 α + (γ) 2(n-s) 2-s dr ∞ 0 r n-1 r 2-s n-2 α -(γ) +r 2-s n-2 α + (γ) 2(n-2) 2-s dr = I P+1 Q+2 I P Q where I P Q = ∞ 0 t P (1 + t) Q dt ; Q = 2(n -2) 2 -s and P = (n -2)(n -2α -(γ)) (2 -s)(α + (γ) -α -(γ)) - 1
Integrarting by parts, see Jaber [START_REF] Jaber | Hardy-Sobolev equations on compact Riemannian manifolds[END_REF], we have that

I P+1 Q+1 = P + 1 Q I P Q and I P Q+1 = Q -P -1 Q I P Q .
We then get that

R n |X| 2 U 2 (s) |X| s dX R n U 2 dx = (P + 1)(Q -P -1) Q(Q + 1) = (n -2)(α + (γ) -α -(γ) + 2)(α + (γ) -α -(γ) -2) 2(α + (γ) -α -(γ)) 2 (2n -2 -s) , so that ∂ ij b(0) 2 (s) R n X i X j |ũ N | 2 (s) |X| s dX R n ũ2 N dx = ∆b(0) b(0) (n -2)((α + (γ) -α -(γ)) 2 -4) 4n(2n -2 -s) when ũN > 0.
This completes Lemma 5.

9. Blow-up rates when b ∈ C 2 (Ω)

The function b that arises from the hyperbolic model (see Lemma 1) is not in C 2 when n = 3, 4, and therefore, the asymptotic rates of Propositions 4 and 5 do not apply. However, due to the behavior of (x, ∇b(x)) in ( 13) and ( 14), we are in position to get optimal rates.

The case of behavior like [START_REF] Ghoussoub | Duality and Perturbation Methods in Critical Point Theory[END_REF]. When the expressions make sense, we define

C n,s := 2-θ 2 1 t n-θ 2 (s)-2 N R n ũ2 N |x| θ dx n-s 2 (s) 2 N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s)
|x| s dx and D n,s :=

1 2 (s) 1 t n 2 (s)-2 N R n |x| 2 |ũ N | 2 (s) |x| s dx n-s 2 (s) 2 N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx (178)
The proof of compactness rely on the following two key propositions.

Proposition 6 (b ∈ C 1 \ C 2 ).
Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be such that (E ), ( 33), (31), and (36) holds. We assume that b ∈ C 1 (Ω) and that there exists

C 1 ∈ R such that (x, ∇b(x)) = C 1 |x| 2 ln 1 |x| + O(|x| 2 ) as x → 0.
Assume that blow-up occurs, that is

lim →0 |x| τ u L ∞ (Ω) = +∞ for some α -(γ) < τ < n -2 2 .
Consider the µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Then, we have the following blow-up rates:

lim →0 p µ 2-θ N, = - C n,s b(0) • K h0 if θ > 0 and either α + (γ) -α -(γ) > 4 -2θ or {α + (γ) -α -(γ) > 2 -θ and u 0 ≡ 0} lim →0 p µ 2 N, ln 1 µ N, = - D n,s b(0) • C 1 if θ = 0 and either α + (γ) -α -(γ) ≥ 4 or {α + (γ) -α -(γ) > 2 and u 0 ≡ 0} . Proposition 7 (b ∈ C 1 \ C 2 ,
The positive case). Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, 0 ≤ γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be as in Proposition 6. In particular

(x, ∇b(x)) = C 1 |x| 2 ln 1 |x| + O(|x| 2 ) as x → 0.
Assume that blow-up occurs as in (107). Consider µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Suppose in addition that u > 0 for all > 0.

Then, we have the following blow-up rates:

(1) When α + (γ) -α -(γ) = 4 -2θ and θ > 0, we have

lim →0 p µ 2-θ N, = - C n,s b(0) K h0 + K ,
for some K ≥ 0 such that K > 0 iff u 0 > 0. For θ = 0, see Proposition 6.

(2) When 2 -θ < α + (γ) -α -(γ) < 4 -2θ, we have u 0 ≡ 0 and Proposition 6 applies.

(

) When α + (γ) -α -(γ) = 2 -θ, we have u 0 ≡ 0 and lim →0 p µ 2-θ N, ln 1 µ N, = - C n,s b(0) • K h0 -1 θ=0 D n,s b(0) • C 1 3
where

C n,s := 2-θ 2 K 2 ω n-1 n-s 2 (s) 2 N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx , with K as defined in (142). (4) When α + (γ) -α -(γ) < 2 -θ, then u 0 ≡ 0 and lim →0 p µ α+(γ)-α-(γ) N, = -χ • m γ,h0 (Ω) if α + (γ) -α -(γ) < 2 -θ, (180)
where χ > 0 is a constant and m γ,h0 (Ω) is the mass of Ω associated with the operator -∆γ |x| 2 -h 0 (x), defined in Theorem 1.7.

The proof of the propositions goes as the proof of Propositions 4 and 5, with the use of the pointwise control of Lemma 4.

The case of behavior like [START_REF] Esposito | Sign-changing solutions for critical equations with Hardy potential[END_REF]. In this case, when writing the Pohozaev identity there are several terms to compare. For the sake of simplicity, we only deal here with the case θ = 1 that corresponds to the hyperbolic case when n = 3. When the expression makes sense, we define

E n,s := 1 2 1 t n-1 2 (s)-2 N R n ũ2 N |x| dx n-s 2 (s) 2 N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx (181)
The proof of compactness rely on the following two key propositions.

Proposition 8 (b ∈ C 0 \ C 1 ).
Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be such that (E ), (33), (31), and (36) holds. We assume that θ = 1, that b ∈ C 0,1 (Ω) and that there exists

C 2 ∈ R such that (x, ∇b(x)) = C 2 |x| + O(|x| 2 ) as x → 0. Assume that blow-up occurs, that is lim →0 |x| τ u L ∞ (Ω) = +∞ for some α -(γ) < τ < n -2 2 .
Consider the µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Assume that

{α + (γ) -α -(γ) > 2} or {α + (γ) -α -(γ) > 1 and u 0 ≡ 0}.
Then, we have the following blow-up rates:

lim →0 p µ N, = - E n,s b(0) •     K h0 + 2 2 (s) R n |x| |ũ N | 2 (s) |x| s dx R n ũ2 N |x| dx • C 2     Proposition 9 (b ∈ C 0 \ C 1 ,
The positive case for θ = 1). Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω and assume that 0 < s < 2, 0 ≤ γ < (n-2) 2

4

. Let (u ), (h ), (p ) and b be as in Proposition 8. In particular θ = 1 and

(x, ∇b(x)) = C 2 |x| + O(|x| 2 ) as x → 0.
Assume that blow-up occurs as in (107). Consider µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Suppose in addition that u > 0 for all > 0 and C 2 > 0 Then, we have the following blow-up rates:

(1) When α + (γ) -α -(γ) = 2, we have that

lim →0 p µ N, = - E n,s b(0) • K h0 + (n -2) 2 2(2n + 2 -s) (n -2) 2 -4γ -1 • C 2 b(0) + K for some K ≥ 0 such that K > 0 iff u 0 > 0. (2) When 1 < α + (γ) -α -(γ) < 2, we have u 0 ≡ 0 and Proposition 8 applies. (3) When α + (γ) -α -(γ) = 1, we have u 0 ≡ 0 and lim →0 p µ N, ln 1 µ N, = - E n,s b(0) • K h0 where E n,s := 1 2 K 2 ω n-1 n-s 2 (s) 2 N i=1 1 t n-2 2 (s)-2 i R n |ũi| 2 (s) |x| s dx , with K as defined in (142). (4) When α + (γ) -α -(γ) < 1, then u 0 ≡ 0 and lim →0 p µ α+(γ)-α-(γ) N, = -χ • m γ,h0 (Ω) (182)
where χ > 0 is a constant and m γ,h0 (Ω) is the mass of Ω associated with the operator -∆γ |x| 2 -h 0 (x), defined in Theorem 1.7.

The proof of the propositions goes as the proof of Propositions 4 and 5, with the use of the pointwise control of Lemma 4. We have also used that when ũN > 0, then

R n |x| |ũ N | 2 (s) |x| s dx R n ũ2 N |x| dx = (n -2)(n -s) 2b(0)(2n + 2 -s) (α + (γ) -α -(γ)) 2 -1 ,
which is proved like Lemma 5.

For the proof of compactness in dimension 3, the following proposition will be useful: its proof is similar to the other ones:

Proposition 10 (b ∈ C 0 \ C 1 ,
The positive case with θ = γ = 0). Let Ω be a smooth bounded domain of R n , n = 3, such that 0 ∈ Ω and assume that 0 < s < 2, γ = θ = 0. Let (u ), (h ), (p ) and b be as in Proposition 8. Assume that

(x, ∇b(x)) = C 2 |x| + O(|x| 2 ) as x → 0.
Assume that blow-up occurs as in (107). Consider µ 1, , ..., µ N, and t 1 , ..., t N from Proposition 2. Suppose in addition that u > 0 for all > 0 and that C 2 > 0. Then, α + (γ) -α -(γ) = 1, u 0 ≡ 0 and

lim →0 p µ α+(γ)-α-(γ) N, = -χ 1 • m 0,h0 (Ω) -χ 2 • C 2 (183)
where χ 1 , χ 2 > 0 are constants and m 0,h0 (Ω) is the mass of Ω associated with the operator -∆ -h 0 (x), defined in Theorem 1.7.

Proof of Multiplicity

Proof of Theorem 1.6: We fix γ < (n -2) 2 /4. Let h 0 ∈ C 1 (Ω, |x| -θ ) be such that -∆ -γ |x| 2 -h 0 (x) is coercive and let b satisfy (34). For each 2 < p ≤ 2 (s), we consider the C 2 -functional

I p,γ (u) = 1 2 Ω |∇u| 2 dx -γ |u| 2 |x| 2 -h 0 u 2 dx - 1 p Ω b(x) |u| p |x| s dx on H 2 1,0 (Ω), whose critical points are the weak solutions of    -∆u - γ |x| 2 u -h 0 u = b(x) |u| p-2 u |x| s on Ω u = 0 on ∂Ω. (184)
For a fixed u ∈ H 2 1,0 (Ω), u ≡ 0, we have that

I p,γ (λu) = λ 2 2 Ω |∇u| 2 dx -γ λ 2 2 Ω |u| 2 |x| 2 dx - λ 2 2 Ω h 0 u 2 dx - λ p p Ω b(x) |u| p |x| s dx.
Then, by coercivity, we have lim λ→∞ I p,γ (λu) = -∞, which means that for each finite dimensional subspace

E k ⊂ E := H 2 1,0 (Ω), there exists R k > 0 such that sup{I p,γ (u); u ∈ E k , u H 2 1 > R k } < 0 (185) as p → 2 (s). Let (E k ) ∞ k=1 be an increasing sequence of subspaces of H 2 1,0 (Ω) such that dim E k = k and ∪ ∞ k=1 E k = E := H 2 1,0 (
Ω) and define the min-max values:

c p,k = inf g∈H k sup x∈E k I p,γ (g(x)),
where

H k = {g ∈ C(E, E); g is odd and g(v) = v for v > R k for some R k > 0}.
Proposition 11. With the above notation and assuming n ≥ 3, we have:

(1) For each k ∈ N, c p,k > 0 and lim

p→2 (s) c p,k = c 2 (s),k := c k .
(

> 0 such that Ω |∇u| 2 - γ |x| 2 u 2 -h 0 u 2 dx ≥ Λ 0 Ω |∇u| 2 dx for all u ∈ H 2 1,0 (Ω). ( 186
)
With coercivity, the Hardy and the Hardy-Sobolev inequality, there exists C > 0 and α > 0 such that

I p,γ (u) ≥ Λ 0 2 ∇u 2 2 -C ∇u p 2 = ∇u 2 2 Λ 0 2 -C ∇u p-2 2 ≥ α > 0 for all u ∈ H 2 1,0 (
Ω) such that ∇u 2 = ρ > 0 is small enough. Then the sphere S ρ = {u ∈ E; u H 2 1,0 (Ω) = ρ} intersects every image g(E k ) by an odd continuous function g. It follows that

c p,k ≥ inf{I p,γ (u); u ∈ S ρ } ≥ α > 0.
In view of (185), it follows that for each g ∈ H k , we have that sup

x∈E k I pi,γ (g(x)) = sup x∈D k I p,γ (g(x))
where D k denotes the ball in E k of radius R k . Consider now a sequence p i → 2 (s) and note first that for each u ∈ E, we have that I pi,γ (u) → I 2 (s),γ (u). Since g(D k ) is compact and the family of functionals (I p,γ ) p is equicontinuous, it follows that sup

x∈E k I p,γ (g(x)) → sup x∈E k I 2 (s),γ (g(x)), from which follows that lim sup i∈N c pi,k ≤ sup x∈E k I 2 (s),γ (g(x)). Since this holds for any g ∈ H k , it follows that lim sup i∈N c pi,k ≤ c 2 (s),k = c k .
On the other hand, the function f (r) = 1 p r p -1 2 (s) r 2 (s) attains its maximum on [0, +∞) at r = 1 and therefore f (r) ≤ 1 p -1 2 (s) for all r > 0. It follows If now p < 2 (s), we are in the subcritical case, that is we have compactness in the Sobolev embedding H 2 1,0 (Ω) → L p (Ω; |x| -s dx) and therefore I p,γ satisfies the Palais-Smale condition. It is then standard to find critical points u p,k for I p,γ at each level c p,k (see for example the book [START_REF] Ghoussoub | Duality and Perturbation Methods in Critical Point Theory[END_REF]). Consider now the functional

I 2 (s),γ (u) = I p,γ (u) + Ω b(x) |x| s 1 p |u(x)| p - 1 2 (s) |u(x)| 2 (s) dx ≤ I p,γ (u) + Ω b(x) |x| s 1 p - 1
I p,0 (u) = 1 2 Ω |∇u| 2 dx - 1 p Ω b(x) |u| p |x| s dx and its critical values c 0 p,k = inf g∈H k sup x∈E k I p,0 (g(x)).
In [START_REF] Ghoussoub | Concentration estimates for Emden-Fowler equations with boundary singularities and critical growth[END_REF] it has been shown that (1), ( 2 it follows that the sequence (u pi,k ) i is uniformly bounded in H 2 1,0 (Ω). Moreover, since I pi (u pi,k ) = 0, letting p i → 2 (s) and using the compactness Theorem 1.6 we get a solution u k of (184) in such a way that I 2 (s),γ (u k ) = lim • Assume that γ < 0, so that α + (γ) -α -(γ) > 2. We fix θ ∈ (0, 2) small, so that α + (γ) -α -(γ) > 2 -θ and K h γ,λ = 0. Then Propositions 6 and 7 do not permit to conclude.

• Assume that γ = 0, so that α + (γ) -α -(γ) = 2 and we have that ũN (z) 2 (s)-1 |z| s dz since α + (γ) = n -2 = 2 and α -(γ) = 0. We first compute L 0,Ω . Since for x, y → 0, we have that the Green's function for -∆ -h 0,λ behaves like ((n -2)ω n-1 ) -1 |x -y| 2-n (see Robert [START_REF] Robert | Existence et asymptotiques optimales des fonctions de Green des opérateurs elliptiques d'ordre deux (Existence and optimal asymptotics of the Green's functions of second-order elliptic operators[END_REF]), we get that

L 0,Ω = 1 (n -2)ω n-1 = 1 2ω 3 .
Arguing as in the proof of Lemma 5 and using the same notations, we get that Integrating by parts as in Jaber [START_REF] Jaber | Hardy-Sobolev equations on compact Riemannian manifolds[END_REF], we get that for any q ≥ 0 1 -q p + 1 Therefore we get compactness when λ > 0.

• Assume that γ > 0, so that α + (γ) -α -(γ) < 2. We choose θ ∈ (0, 2) such that α + (γ) -α -(γ) < 2 -θ.

Then point (4) of Proposition 7 applies. Compactness for the sequence positive subcritical solutions then follows from the positivity of the mass as in the proof of Theorem 1.8. So in this case we get existence of minimizers when γ > 0 and the mass m γ,λ (Ω B n ) is positive. This proves point (2) of Theorem 1.4 for n = 4.

for some constant C > 0 and > 0 small, and then, since u > 0, there exists c 0 > 0 such that

Ω u 2 (s)
|x| s dx ≥ c 0 for all > 0 small. Passing to the limit yields u 0 ≡ 0. Therefore, u 0 > 0 is a solution to (189) with = 0. This is not possible by the hypothesis. The family (u ) is therefore not pre-compact and hence it blows-up with bounded energy. Let u 0 ∈ H 2 1,0 (Ω) be its weak limit, which is necessarily a solution to (190), and hence must be the trivial solution u 0 ≡ 0. Proposition 5 then yields that either here, this contradicts our assumption on the mass. Therefore no such a family of positive solutions (u ) >0 exists, which proves the theorem.

Proof of Corollary 1: First note that if h 0 satisfies h 0 (x) + 1 2 (∇h 0 (x), x) ≤ 0 for all x ∈ Ω, (

then by differentiating for any x ∈ Ω, the function t → t 2 h 0 (tx) (which is well defined for t ∈ [0, 1] since Ω is starshaped), we get that h 0 ≤ 0. Therefore -∆ -γ|x| -2 -h 0 is coercive. Assume now there is a positive variational solution u 0 of (189)corresponding to = 0. The Pohozaev identity (193) then gives ∂Ω (x, ν) (∂ ν u 0 ) 2 2 dσ -Ω h 0 + 1 2 (∇h 0 , x) u 2 0 dx = 0.

Hopf's strong comparison principle yields ∂ ν u 0 < 0. Since Ω is starshaped with respect to 0, we get that (x, ν) ≥ 0 on ∂Ω. Therefore, with (192), we get that (x, ν) = 0 for all x ∈ Ω, which is a contradiction since Ω is smooth and bounded. Since α + (γ) -α -(γ) < 2 -θ, we use Theorem 7.1 in Ghoussoub-Robert [START_REF] Ghoussoub | Hardy-Singular Boundary Mass and Sobolev-Critical Variational Problems[END_REF] to find K ∈ C 2 (Ω \ {0}) and A > 0 such that (199)

     -∆K - γ |x| 2 K -h 0 K = 0 in Ω K > 0 in Ω K = 0 on ∂Ω \ {0}.
Via the conformal transformation, the above regularity result will yield that the corresponding solutions for equation [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] in the Hyperbolic Sobolev Space H 1 0 (Ω B n ) will satisfy

lim |x|→0 u(x) G(|x|) α-= K ∈ R, (200)
where

α -(γ) = 1 2 - 1 4 - γ (n -2) 2 , ( 201
)
which amounts to the regularity claimed in Theorem 1.4.

Appendix C: Green's function. The next theorem describes the properties of the Green's function of the Hardy-Schrödinger operator in a bounded smooth domain. To prove this theorem one argue as in the case θ = 0 in [START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF].

Theorem 13.2 (Green's function). Let Ω be a smooth bounded domain of R n such that 0 ∈ Ω is an interior point. Let γ < (n-2) 2

4

, 0 ≤ θ < 2 and let h 0 ∈ C 1 (Ω, |x| -θ ) be such that the operator -∆ -γ (206)

B 1

 1 (0) := {x ∈ R n : |x| < 1} endowed with the metric g B n =

  (x, ∇b(x)) = (2 (s) + 2)b(0)|x| + O(|x| 2 ) as x → 0 (13) • For n = 4, we have b ∈ C 1 (Ω), ∇b(0) = 0 and (x, ∇b(x)) = 4(2 (s) + 2)b(0)|x| 2 ln 1 |x| + O(|x| 2 ) as x → 0 (14) • For n ≥ 5, we have b ∈ C 2 (Ω), ∇b(0) = 0 and ∆b(0) b(0) = 4n(2n -2 -s) n -4 .

  r) for n ≥ 5.

-ν = 0 .Fix a ρ > 0 . 2 2 1 y

 0021 ), (34) and (36) holds. Let (y ) ∈ Ω \ {0} and let ν Assume that for any R > 0 there exists C(R) > 0 such that for all > 0 |u (x)| ≤C(R) |y | τ |x| τ |u (y )| for all x ∈ B Rκ (y ) \ {0}. (41) Then |y | = O( ) as → 0. (42) Proof of Lemma 2: We proceed by contradiction and assume that lim →0 |y | = +∞. (43) Then it follows from the definition of κ that lim We define for all > 0 v (x) := ν nu (y + κ x) for x ∈ B 2ρ (0) Note that this is well defined since lim →0 |y | = 0 ∈ Ω and lim →0 κ |y | = 0. It follows from (41) that there exists C(ρ) > 0 such that all > 0 |v (x)| ≤ C(ρ) |y | + κ |y | x τ ∀x ∈ B 2ρ (0) (45) using (44) we then get as → 0 |v (x)| ≤ C(ρ) (1 + o(1)) ∀x ∈ B 2ρ (0).

-

  ), (34) and (36) holds. Let (y ) ∈ Ω \ {0} and let ν Suppose ν → 0 and |y | = O( ) as → 0.

  t I+1 . Moreover, it follows from (61) and (63) that lim →0 µ I+1, µ I, = +∞ and lim →0 µ I+1, = 0.

λ |x| 2 I

 2 (y) dy + λ |x| 2 <|y|<2λ |x| I (y) dy + |y|>2λ |x| I (y) dy where I (y) is the integrand. Since

I

  (y) dy that have been computed just above and go to 0 as → 0. We are then left with the first term. With a change of variables, we have that |y|< λ |x| 2

Case 1 :

 1 α + (γ) -α -(γ) > 2(2 -θ) or {u 0 ≡ 0 and α + (γ) -α -(γ) > 2 -θ}. Then (171) yields (109) and Proposition 4 is proved. When ũ > 0, then ũN ≥ 0 and ũN ≡ 0: it then follows from the strong comparison principle that ũN > 0 and then Lemma 5 holds. This yields Proposition 5 forα + (γ) -α -(γ) > 2(2 -θ) or {u 0 ≡ 0 and α + (γ) -α -(γ) > 2 -θ}.Case 2: u > 0 and α + (γ) -α -(γ) = 2(2 -θ). It then follows from (171) that lim →0 p µ 2-θ

  R n |ũi| 2 (s) |x| s dx . Then Step P12 and (172) yield Proposition 5 when α + (γ) -α -(γ) = 2(2 -θ).

Case 3 :

 3 u > 0 and 2 -θ < α + (γ) -α -(γ) < 2(2 -θ). It follows from Step P13 that u 0 ≡ 0. Then (171) and (172) yields Proposition 4 for 2 -θ < α + (γ) -α -(γ) < 2(2 -θ). The case u > 0 and α + (γ) -α -(γ) = 2 -θ. The Pohozaev identity (115), (118), (121), StepP7 and Step P11 yields Proposition 4 for α + (γ) -α -(γ) = 2 -θ.

) If 2 1 2n

 21 < p < 2 (s), there exists for each k, functions u p,k ∈ H 2 1,0 (Ω) such that I p,γ (u p,k ) = 0, andI p,γ (u p,k ) = c p,k . (3) For each 2 < p < 2 (s), we have c p,k ≥ D n,p k p+1 p-where D n,p > 0 is such that lim p→2 (s) D n,p = 0.(4) lim k→∞ c k = lim k→∞ c 2 (s),k = +∞. Proof: (1) Coercivity yields the existence of Λ 0

  2 (s) dx from which follows that c k ≤ lim inf i∈N c pi,k , and claim (1) is proved.

2 0 1 p- 2 0u.

 212 ) and (3) of Proposition 11 holds, with c 0 p,k and c 0 k replacing c p,k and c k respectively and b(x) ≡ 1. By similar arguments, lim k→∞ ),k = +∞. On the other hand, with the coercivity (186), we obtain I p,γ (u) ≥ Λ p p-I p,0 (v) for every u ∈ H 2 1,0 (Ω), where v = Λ -It then follows that lim k→∞ c k = lim k→∞ c 2 (s),k = +∞. To complete the proof of Theorem 1.6, notice that since for each k, we have lim pi→2 (s) I pi,γ (u pi,k ) = lim pi→2 (s) c pi,k = c k ,

11 . 4 . 4 - 4 .

 11444 p→2 (s) I p,γ (u p,k ) = lim p→2 (s) c p,k = c k . Since the latter sequence goes to infinity, it follows that (184) has an infinite number of critical levels with 2 < p ≤ 2 (s). Proof of Theorems 1.1, 1.2 and 1.4We reduce the problem to a smooth bounded domain of the Euclidean space R n .Case 1: n ≥ 5. It follows from Lemma 1 that finding solutions to[START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] amounts to finding solutions to   -∆v -γ |x| 2 + h γ,λ (x) v = b(x) v 2 (s)-1 |x| s in Ω v = 0 on ∂Ω, where h γ,λ ∈ C 1 (Ω) with h γ,λ (0) = 4λ + 4(n -2) n -4 γ -n(n -4) 4 . Moreover in this case, b ∈ C 2 (Ω), ∇b(0) = 0 and ∆b(0) b(0) = 4n(2n -2 -s) n -Since b is radial, we have that ∂ ij b(0) = ∆b(0) n δ ij .Therefore with θ = 0, the limit (109Since ∆b(0) > 0, then arguing as in the proof of Theorem 1.6, we get multiplicity of solutions as soon as we have: 4λ + 4(n -2) n -4 γ -n(n -4) 4 ≥ 0.We now consider the case of positive solutions. We have here θ = 0. Using the value of ∆b(0), it then follows from Lemma 5 and Proposition 5 that for γ ≥ 0lim when α + (γ) -α -(γ) > 2. (188)Therefore, for λ > 0 we get a contradiction and hence compactness for positive subcritical solutions. This yields compactness of minimizers and the existence of minimisers when λ > 0 for α + (γ) -α -(γ) > 2 and γ ≥ 0. When α + (γ) -α -(γ) = 2 and γ ≥ 0, we get the same result by using (110). When α + (γ) -α -(γ) < 2 and γ ≥ 0, we apply (111) to get existence of minimizers when the mass m γ,λ (Ω B n ) is positive.Case 2: n = 4. We prove point (2) of Theorem 1.4 for n = 4. By Lemma 1 here h γ,λ ∈ C 1 Ω \ {0} and h γ,λ (r) = 8γ ln 1 r + c 4 + O(r), as r → 0. In particular h γ,λ (x) = o(|x| -θ ) as x → 0 for all θ > 0.

  h γ,λ (r) = c 4 + O(r). Therefore we take θ = 0 and K h γ,λ = c 4 = 2(λ -2). Then point (3) of Proposition 7 withC 1 = 4(2 (s) + 2) yields lim →0 p µ 2 N, ln 1 µ N, = -C n,s b(0) • c 4 + D n,s b(0) • 4(2 (s) + 2) when u > 0.With the definitions of C n,s and D n,s , we have that C

R 4 |x| 2 |ũ N | 2

 422

4 ũN

 4 t) q+1 = R p+1 (p + 1)(1 + R) p+1 for all R > 0.Taking q = p + 1 and letting R → +∞ yields I p p+2 = 1 p+1 . With the same notations and the value of -∆U given in the proof of Lemma 5, we have that R

α

  + (γ) -α -(γ) ≥ 2 -θ and therefore K h0 = 0, or α + (γ) -α -(γ) < 2 -θ and therefore m γ,h0 (Ω)

  |x| α+(γ) + β(x) for all x ∈ Ω, where η ∈ C ∞ c (R n ) and β ∈ H 2 1,0 (Ω) are as in Step P14. We now apply the Pohozaev identity (193) to K on the domainU := B δ (0) . Using that K 2 ∈ L 1 (Ω) and (•, ν)(∂ ν K) 2 ∈ L 1 (∂Ω) when α + (γ) -α -(γ) < 2 -θ, 0 , x) K 2 dx = M δwhere M δ is defined in (174). With (176), we then get∂Ω (x, ν) (∂ ν K) 2 2 dσ -Ω h 0 + 1 2 (∇h 0 , x) K 2 dx = -2ω n-1 (n -2) 2 4 -γ A 2 • m γ,h0 (Ω).

( 1 )

 1 Let u ∈ H 2 1,0 (Ω) be a weak solution of∆u -γ + O(|x| τ ) |x| 2 u = f (x, u),(197)for some τ > 0. Then, there exists K ∈ R such thatlim x→0 u(x) |x| -α-(γ) = K. (198)Moreover, if u ≥ 0 and u ≡ 0, we have thatK > 0. (2) As a consequence, one gets that if u ∈ D 1,2 (R n ) is a weak solution for -∆u -γ |x| 2 u = |u| 2 (s)-2 u |x| s in R n \ {0}, then there exists K 1 , K 2 ∈ R such that u(x) ∼ |x|→0 K 1 |x| α-(γ)and u(x) ∼ |x|→+∞ K 1 |x| α+(γ) , and therefore there exists a constant C > 0 such that for all x in R n \ {0}, |u(x)| ≤ C |x| α-(γ) + |x| α+(γ) .

|x| 2 -

 2 h(x) is coercive in Ω. Then there existsG : (Ω \ {0}) 2 \ {(x, x)/ x ∈ Ω \ {0}} → R such that (i) For any p ∈ Ω \ {0}, G p := G(p, •) ∈ H 2 1 (Ω \ B δ (p)) for all δ > 0, G p ∈ C 2,θ (Ω \ {0, p}) (ii) For all f ∈ L p (Ω), p > n/2, and all ϕ ∈ H 2 1,0 (Ω) such that -∆ϕ -γ |x| 2 + h(x) ϕ = f in Ω ; ϕ |∂Ω = 0, we have ϕ(p) = Ω G(p, x)f (x) dx(202)In addition, G > 0 is unique and(iii) For all p ∈ Ω \ {0}, there exists C 0 (p) > 0 such thatG p (x) ∼ x→0 C 0 (p) |x| α-(γ) and G p (x) ∼ x→p 1 (n -2)ω n-1 |x -p| n-2(203)(iv) There exists C > 0 such that0 < G p (x) ≤ C max{|p|, |x|} min{|p|, |x|} α-(γ) |x -p| 2-n(204)(v) For all ω Ω, there exists C(ω) > 0 such thatC(ω) max{|p|, |x|} min{|p|, |x|} α-(γ) |x -p| 2-n ≤ G p (x) for all p, x ∈ ω \ {0}.(205)(v) There exists L γ,Ω > 0 such that for any(h i ) ∈ C 1 (Ω, |x| -θ ) satisfying lim i→+∞ sup x∈Ω |x| θ |h(x) -h 0 (x)| + |x| θ+1 |∇(h -h 0 )(x)| < ,and for any sequences of points (x i ) i , (y i ) i ∈ Ω withy i = o(|x i |) and x i = o(1) as i → +∞,we have as i → +∞G hi (x i , y i ) = L γ,Ω + o(1) |x i | α+(γ) |y i | α-(γ) .

(s) 2 (s)-2

This work was initiated when the second-named author held a postdoctoral position at the University of British Columbia under the supervision of the first-named author, that was partially supported by the Natural Sciences and Engineering Research Council of Canada (NSERC).

Case 3: n = 3. We prove point (2) of Theorem 1.4 for n = 3. In this case, by Lemma 1 we have h γ,λ ∈ C 1 Ω \ {0} and h γ,λ (r) = 4γ r + c 3 + O(r) as r → 0. Then θ = 1 and K h γ,λ = 4γ here. Moreover, we have (x, ∇b(x)) = C 2 |x| + O(|x| 2 ) with C 2 > 0.

• Assume that γ < 0, so that α + (γ) -α -(γ) > 1. Since θ = 1, it follows from Propositions 8 and 9 and

for some K ≥ 0. Since γ < 0, we cannot conclude.

• Assume that γ = 0, so that α + (γ) -α -(γ) = 1. It follows from Lemma 1 that we can take θ = 0. It then follows from Proposition 10 that lim

Compactness for the sequence positive subcritical solutions then follows from the positivity of the mass.

• Assume that γ > 0, so that α + (γ) -α -(γ) < 1. Then point (4) of Proposition 9 applies. Compactness for the sequence positive subcritical solutions then follows from the positivity of the mass as in the proof of Theorem 1.8. So in this case we get existence of minimizers when γ > 0 and the mass m γ,λ (Ω B n ) is positive. This proves point (2) of Theorem 1.4 for n = 3.

A quick analysis as above yields no compactness result for sign-changing solutions for n = 3, 4. To be more precise, when estimating the different terms in the Pohozaev identity, we get compactness when a combination of several terms is positive: however, it is not possible to give an apriori criterion to get the positivity.

12. Proof of the Non-existence result Proof of Theorem 1.9: We argue by contradiction. Let h 0 ∈ C 1 (Ω, |x| -θ ) be such that -∆-γ|x| -2 -h 0 is coercive and let b(x) be such that (34) holds. We fix γ < (n -2) 2 4 and Λ > 0. We assume that there is a family (u

with ∇u 2 ≤ Λ, and (32), (33) holds.

We claim that (u ) >0 is not pre-compact in H 2 1,0 (Ω). Otherwise, up to extraction, there would be

Ω) as → 0. Passing to the limit in the equation, we get that u 0 ≥ 0 and

The coercivity of -∆u 0 -γ|x| -2 -h 0 and the convergence of (h ) yields

Since Ω is star-shaped and h 0 satisfies (192), it follows that m γ,h0 (Ω) < 0 and Theorem 1.9 then applies to complete our corollary.

Appendices

Appendix A: Pohozaev identity. We let U ⊂ R n be a smooth bounded domain in R n . Let h ∈ C 1 (U ) and let b ∈ C 1 (U ). We suppose u ∈ C 2 (U ) and p ∈ [0, 2 (s) -2). The classical Pohozaev identity and integration by parts yields for any

where ν is the outer normal to the boundary ∂U .

Appendix B: Regularity. As to the regularity of the solutions, this will follow from the following result established by Ghoussoub-Robert in [START_REF] Ghoussoub | Hardy-Singular Boundary Mass and Sobolev-Critical Variational Problems[END_REF][START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF]. Assuming that γ < (n-2) 2

4

, note that the function

if and only if β ∈ {β -(γ), β + (γ)}, where

Actually, one can show that any non-negative solution u

where C -, C + ≥ 0.

We collect the following important results from the papers [START_REF] Ghoussoub | Hardy-Singular Boundary Mass and Sobolev-Critical Variational Problems[END_REF][START_REF] Ghoussoub | The Hardy-Schrödinger operator with interior singularity: The remaining cases[END_REF] which we shall use repeatedly in our work.