Nassif Ghoussoub
email: nassif@math.ubc.ca

F R Éd

Éric Robert

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

THE HARDY-SCHR ÖDINGER OPERATOR WITH INTERIOR SINGULARITY: THE REMAINING CASES

Keywords: Mathematics Subject Classification: 35J35, 35J60, 35B44

We consider the remaining unsettled cases in the problem of existence of energy minimizing solutions for the Dirichlet value problem Lγ u -λu = u 2 * (s)-1 |x| s on a smooth bounded domain Ω in R n (n ≥ 3) having the singularity 0 in its interior. Here γ <

n-2 and 0 ≤ λ < λ 1 (Lγ ), the latter being the first eigenvalue of the Hardy-Schrödinger operator Lγ := -∆ -γ |x| 2 . There is a threshold λ * (γ, Ω) ≥ 0 beyond which the minimal energy is achieved, but below which, it is not. It is well known that λ * (Ω) = 0 in higher dimensions, for

-1. Our main objective in this paper is to show that this threshold is strictly positive in "lower dimensions" such as when

, to identify the critical dimensions (i.e., when the situation changes), and to characterize it in terms of Ω and γ. If either s > 0 or if γ > 0, i.e., in the truly singular case, we show that in low dimensions, a solution is guaranteed by the positivity of the "Hardy-singular internal mass" of Ω, a notion that we introduce herein. On the other hand, and just like the case wnen γ = s = 0 studied by Brezis-Nirenberg [4] and completed by Druet [12], n = 3 is the critical dimension, and the classical positive mass theorem is sufficient for the merely singular case, that is when s = 0, γ ≤ 0.

Introduction

Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω and consider the following Dirichlet boundary value problem:

     -∆u -γ u |x| 2 -λu = u 2 * (s)-1 |x| s on Ω, u > 0 on Ω, u = 0 on ∂Ω, (1)
where γ < (n-2) 2

4

, 0 ≤ s < 2, 2 (s) := 2(n-s)

n-2

and 0 ≤ λ < λ 1 (L γ ), the latter being the first eigenvalue of the Hardy-Schrödinger operator L γ := -∆ -γ |x| 2 , that is

λ 1 (L γ , Ω) := inf    Ω |∇u| 2 dx -γ Ω u 2 |x| 2 dx Ω u 2 dx ; u ∈ H 1 0 (Ω) \ {0}    .
Equation ( 1) is essentially the Euler-Lagrange equation corresponding to the following energy functional on H 1 0 (Ω),

J Ω γ,s,λ (u) = Ω |∇u| 2 dx -γ Ω u 2 |x| 2 dx -λ Ω |u| 2 dx Ω u 2 * (s) |x| s dx 2 2 * (s)
It is also standard that µ γ,s,0 (Ω) = µ γ,s,0 (R n ) whenever Ω is a domain containing 0 in its interior, and hence µ γ,s,0 is not attained if Ω is bounded. The idea of restoring compactness by considering non-trivial negative linear perturbations was pioneered by Brezis-Nirenberg [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF] in the case when γ = 0, s = 0 and 0 < λ < λ 1 (Ω), the latter being the first eigenvalue of the Laplacian on H 1 0 (Ω). They showed that in this case (1) has a solution for n ≥ 4. The case n = 3 is special and involves a "positive mass" condition introduced 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 inspired by the work of Schoen [START_REF] Schoen | Conformal deformation of a Riemannian metric to constant scalar curvature[END_REF] on the Yamabe problem. The bottom line is that -at least for γ = 0-the geometry of Ω need not be taken into account in dimension n ≥ 4, while in dimension n = 3, the existence depends on the domain Ω via "a positive mass condition". We shall elaborate further on this theme. In this paper, we consider the case when the Laplacian is replaced by the Hardy-Schrödinger operator L γ . Here, the position of the singularity 0 within Ω matters a great deal. In [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF], we considered the case where 0 belongs to the boundary ∂Ω of the domain Ω. In this sequel, we deal with the case when 0 ∈ Ω, which was first considered by Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] in the case s = 0. It is already well known that there is a threshold λ * beyond which the infimum µ γ,s,λ (Ω) is achieved, and below which, it is not. It can be characterized as [START_REF] Bartsch | Existence and non-existence of solutions to elliptic equations related to the Caffarelli-Kohn-Nirenberg inequalities[END_REF] λ * (Ω) := sup{λ; µ γ,s,λ (Ω) = µ γ,s,0 (R n )}.

It is easy to see that 0 ≤ λ * (Ω) < λ 1 (L γ , Ω). It is also part of the folklore -that we sketch belowthat λ * (Ω) = 0 in higher dimensions. Our main objective in this paper is to show that this threshold is strictly positive in "lower dimensions," to identify the critical dimensions (i.e., when the situation changes), and to try to characterize it in terms of Ω and γ.

As opposed to Brezis-Nirenberg [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF] and Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF], we are dealing here with the case where 0 is an interior singularity, and our analysis below shows that the identification of λ * differ according to two distinct singularity regimes:

• The truly singular case, which corresponds to when either s > 0 or γ > 0. We note that in this case µ γ,s,0 (R n ) is achieved. • The merely singular case, which corresponds to the case when s = 0 and γ ≤ 0, a case where µ γ,s,0 (R n ) is not achieved, unless s = γ = 0.

The following three theorems are the main results of this paper. The first is rather standard. It deals with high dimensions and is included for completeness and comparison purposes. The second deals with the low dimensional cases, i.e., the remaining cases which are yet to be addressed in the literature.

Theorem 1. (The higher dimensional case) Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω. Assume that we are in the following situation:

• either in the truly singular case and γ < (n-2) 2

4

-1, • or in the merely singular case and n ≥ 4.

Then µ γ,s,λ (Ω) is achieved if and only if λ > λ * (Ω). Moreover, i) In the truly singular case (i.e, when either s > 0 or γ > 0), and if γ ≤ (n-2) 2 4 -1, then

λ * (Ω) = 0.

ii) In the merely singular case (i.e, when s = 0 and γ ≤ 0), and if n ≥ 4, then

* (Ω) = inf |γ| |x| 2 ; x ∈ Ω > 0 if γ < 0. (4) λ
Part (i) of Theorem 1 was proved by Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] in the case when s = 0. The case when s > 0 is not much different and was noted in several works such as [8-10, 21-23, 25]. Part (ii) of Theorem 1, that is the case when s = 0 and γ < 0, in dimension n ≥ 4 was also tackled by Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] and Ruiz-Willem [START_REF] Ruiz | Elliptic problems with critical exponents and Hardy potentials[END_REF]. Their proof, though not complete, essentially gives the above result. Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] also considered the lower dimensional case, that is

(n -2) 2 4 -1 < γ < (n -2) 2 4 ,
when Ω is the ball B centered at 0. He gave the following explicit value for λ * :

(5)

λ * (B) = inf    B |∇u| 2 |x| 2β + dx B u 2 |x| 2β + dx ; u ∈ H 1 0 (B) \ {0}    > 0,
where

β ± (γ) := n -2 2 ± (n -2) 2 4 -γ.
Note that the radial function x → |x| -β is a solution of (-∆ -γ |x| 2 )u = 0 on R n \ {0} if and only if β ∈ {β -(γ), β + (γ)}. In order to characterize the threshold λ * (Ω) for a general domain Ω, we need to define the notion of Hardy-singular interior mass associated to the operator -∆ -γ |x| 2 -λ on a bounded domain Ω in R n containing 0.

Theorem 2. (The Hardy singular internal mass) Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω. Suppose h is a C 2 -potential on Ω so that the operator -∆ -

( γ |x| 2 + h(x)) is coercive. i) There exists then H ∈ C ∞ (Ω \ {0}) such that (E)      -∆H -γ |x| 2 + h(x) H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω.
These solutions are unique up to a positive multiplicative constant, and there exists c > 0 such that H(x) x→0 c |x| β + (γ) . ii) If either h is sufficiently small around 0 or if (n-2) 2 4 -1 < γ < (n-2) 2

4

, then for any solution H ∈ C ∞ (Ω \ {0}) of (E), there exist c 1 > 0 and c 2 ∈ R such that

H(x) = c 1 |x| β+(γ) + c 2 |x| β-(γ) + o 1 |x| β-(γ)
as x → 0.

The uniqueness implies that the ratio c 2 /c 1 is independent of the choice of H, hence the "Hardysingular internal mass" of Ω associated to the operator L γ -h(x)I can be defined unambigously as m γ,h (Ω) := c 2 c 1 ∈ R.

For the merely singular case (s = 0 and γ ≤ 0) and the critical dimension n = 3, we need a more standard notion of mass associated to the operator L γ at an internal point x 0 ∈ Ω, which is reminiscent of Schoen-Yau's approach to complete the solution of the Yamabe conjecture in low dimensions. For that, one considers for a given γ ≤ 0, the corresponding Robin function or the regular part of the Green function with pole at x 0 ∈ Ω \ {0}. One shows that for n = 3, any solution

G of    -∆G -γ |x| 2 G -λG = 0 in Ω \ {x 0 } G > 0 in Ω \ {x 0 } G = 0 on ∂Ω,
is unique up to multiplication by a constant, and that there exists R γ,λ (Ω, x 0 ) ∈ R and c γ,λ (x 0 ) > 0 such that (6) G(x) = c γ,λ (x 0 ) 1 |x -x 0 | n-2 + R γ,λ (Ω, x 0 ) + o(1) as x → x 0 .

The quantity R γ,λ (Ω, x 0 ) is then well defined and will be called the internal mass of Ω at x 0 . We then define R γ,λ (Ω) = sup x∈Ω\{0} R γ,λ (Ω, x).

These will allow us to give an explicit value for λ * (γ, Ω) as follows.

Theorem 3. (The lower dimensional case) Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω. i) Assume we are • either in the truly singular case and (n-2) 2

4

-1 < γ < (n-2) 2 4 , • or in the merely singular case and n = 3. Then, there exists λ * (Ω) > 0 such that µ γ,s,λ (Ω) is not achieved for λ < λ * and µ γ,s,λ (Ω) is achieved for λ > λ * . ii) Moreover, in the truly singular case, with (n-2) 2 4 -1 < γ < (n-2) 2

4

, and under the assumption that µ γ,s,λ * (Ω) is not achieved, we have that m γ,λ * (Ω) = 0, and [START_REF] Caffarelli | First order interpolation inequalities with weights[END_REF] λ * (Ω) = sup{λ; m γ,λ (Ω) ≤ 0}.

iii) In the merely singular case, and with n = 3, then µ γ,s,λ (Ω) is not achieved and (8) λ * (Ω) = sup{λ; R γ,λ (Ω) ≤ 0}.

We conjecture that in all cases, µ γ,s,λ * (Ω) is never achieved, which means that (7) must hold unconditionally. Note that µ γ,s,λ * (Ω) = µ γ,s,0 (R n ), but we don't know whether this suffices to conclude that µ γ,s,λ * (Ω) is not achieved. When s = γ = 0 and n = 3, Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] proved that this is indeed the case by using a very elegant geometric argument. This extends to the merely singular case. In the truly singular case, the conjecture holds in the radially symmetric case, i.e., when Ω is a ball. This was verified by Janelli [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF]. Finally, we note that the above analysis lead to the following definition of a critical dimension for the operator L γ . It is the largest scalar n γ such that for n < n γ , there exists a bounded smooth domain Ω ⊂ R n and a λ ∈ (0, λ 1 (L γ , Ω)) such that there is a non-trivial minimiser satisfying [START_REF] Aubin | Problèmes isopérimétriques et espaces de Sobolev[END_REF]. µ γ,s,λ (Ω) is not attained.

n γ = 2 √ γ + 1 + 2 if γ ≥ -1 2 if γ < -1.
Note that n < n γ is exactly when

β + (γ) -β -(γ) < 2, which is the threshold where the radial function x → |x| -β+(γ) is locally L 2 -summable.
The proofs of Theorems 1 and 3 rely on a refined blow-up analysis for certain families of solutions of equation ( 1). We give -in Theorems 4 and 5 below-a complete description of how such blowups may occur. In particular, we show that in the truly singular case, the solutions necessarily concentrate at the singularity 0, while in the merely singular case, they do so at a point x 0 = 0 of the domain Ω. In the appendices, we establish several important properties of the Green function associated to the operator -∆ -γ|x| -2 , that are crucial for the proofs of Theorems 4 and 5.

The higher dimensional case

We recall the following facts, which by now are standard.

i) µ γ,s,λ (Ω) > 0 if and only λ < λ 1 (L γ , Ω). ii) µ γ,s,λ (Ω) = µ γ,s,0 (R n ) for all λ ≤ 0. iii) µ γ,s,λ (Ω) is attained if µ γ,s,λ (Ω) < µ γ,s,0 (R n ).
iv) The function λ → µ γ,s,λ (Ω) is continuous and nonincreasing. v) If µ γ,s,λ (Ω) is attained, then µ γ,s,λ (Ω) < µ γ,s,λ (Ω) for any λ > λ.

Writing λ * = λ * (Ω) for short, where λ * (Ω) is defined in (2), it follows from the above that vi) µ γ,s,λ (Ω) = µ γ,s,0 (R n ) for all λ ≤ λ * and µ γ,s,λ (Ω) < µ γ,s,0 (R n ) for all λ > λ * . vii) µ γ,s,λ (Ω) is not achieved for all λ < λ * . viii) µ γ,s,λ (Ω) is achieved for all λ > λ * .

It is clear that λ * ≥ 0. This section is devoted to show that λ * = 0 in "high dimensions," which in our case will depend on γ. The calculations are standard, and we include them for the convenience on the reader and for comparison to the other cases. As mentioned above in (iii), in order to show that extremals exist for µ γ,s,λ (Ω), it suffices to prove that µ γ,s,λ (Ω) < µ γ,s,0 (R n ), where µ γ,s,0 (R n ) := µ γ,s,0 (R n ). This kind of condition is now standard when dealing with borderline variational problems. See also Aubin [START_REF] Aubin | Problèmes isopérimétriques et espaces de Sobolev[END_REF], Brézis-Nirenberg [START_REF] Brezis | Positive solutions of nonlinear elliptic equations involving critical exponents[END_REF]. The condition limits the energy level of minimizing sequences, prevents the creation of "bubbles" and hence insures compactness.

To show the strict inequality, one needs to test the functional J Ω γ,s,λ on minimizing sequences of the form ηU , where U is an extremal for µ γ,s,0 (R n ) and η ∈ C ∞ c (Ω) is a cut-off function equal to 1 in a neigbourhood of 0. It is therefore important to know for which parameters γ and s, the best constant µ γ,s,0 (R n ) is attained. A proof of the following can be found in [START_REF]Sobolev inequalities for the Hardy-Schrödinger operator: extremals and critical dimensions[END_REF]. For explicit extremals, we refer to Beckner [START_REF] Beckner | On Hardy-Sobolev embedding[END_REF] or Dolbeault et al. [START_REF] Dolbeault | On the symmetry of extremals for the Caffarelli-Kohn-Nirenberg inequalities[END_REF].

Proposition 1. Assume γ < (n-2) 2 4 , n ≥ 3 and 0 ≤ s < 2. Then, i) µ γ,s,0 (R n ) is attained if either s > 0 or if {s = 0 and γ ≥ 0}. ii) If 0 ≤ γ < (n-2) 2 4
, then the extremals for µ γ,s,0 (R n ) are explicit and take the form u

(x) = c • -(n-2) 2 U ( x )
, where c = 0, > 0 and

(9) U (x) := 1 |x| (2-s)β -(γ) n-2 + |x| (2-s)β + (γ) n-2 n-2 2-s for x ∈ R n \ {0},
iii) On the other hand, if s = 0 and γ < 0, then µ γ,0 (R n ) is not attained and is equal to µ 0,0 (R n ), which is the best constant in the Sobolev inequality.

Subsection 2.1: The truly singular case

We now give a proof of Theorem 1. Assuming γ ≤ (n-2) 2

4

-1, we construct a minimizing sequence u in H 1 0 (Ω) \ {0} for the functional J γ,s,λ in such a way that µ γ,s,λ (Ω) < µ γ,s,0 (R n ).

Since either s > 0 or γ ≥ 0, then the infimum µ γ,s,0 (R n ) is achieved by the function

U (x) := 1 |x| (2-s)β -(γ) n-2 + |x| (2-s)β + (γ) n-2 n-2 2-s for x ∈ R n \ {0}.
In other words,

U ∈ D 1,2 (R n ) and J R n γ,s,0 (U ) = inf u∈D 1,2 (R n )\{0} J R n γ,s,0 (u), where J R n γ,s,0 (u) := R n |∇u| 2 -γ |x| 2 u 2 dx R n |u| 2 (s) |x| s dx 2 2 (s) for u ∈ D 1,2 (R n ) \ {0}.
In particular, there exists χ > 0 such that [START_REF] Cao | A note on the sign-changing solutions to elliptic problems with critical Sobolev and Hardy terms[END_REF] -∆U -

γ |x| 2 U = χ U 2 (s)-1 |x| s in R n \ {0}.
For convenience, we will write in the sequel, β + := β + (γ) and β -:= β -(γ). Note that the assumption that γ

≤ (n-2) 2 4 -1 is equivalent to β + (γ) -β -(γ) ≥ 2. Define a scaled version of U by U (x) := -n-2 2 U x =   2-s n-2 • β + -β - 2 2-s n-2 •(β+-β-) |x| (2-s)β - n-2 + |x| (2-s)β + n-2   n-2 2-s for x ∈ R n \ {0}. Fix now a function h ∈ C 0,θ (Ω), θ ∈ (0, 1), consider a cut-off function η ∈ C ∞ c (Ω) such that η(x) = 1
for x in a neighborhood of 0 contained in Ω, and define for > 0 the test-function u ∈ H 1 0 (Ω) by

u (x) := η(x)U (x) for x ∈ Ω \ {0}.
We now estimate J Ω γ,s,h (u ), where

J Ω γ,s,h (u) := Ω |∇u| 2 -γ |x| 2 + h(x) u 2 dx Ω |u| 2 (s) |x| s dx 2 2 (s) . Note first that 0 ≤ u (x) ≤ C β + -β - 2 |x| -β+ for all x ∈ Ω \ {0}. Therefore, since β + (γ) -β -(γ) ≥ 2, we have as → 0, Ω u 2 (s) |x| s dx = B δ (0) U 2 (s) |x| s dx + o 2 = B -1 δ (0) U 2 (s) |x| s dx + o 2 = R n U 2 (s) |x| s dx + o 2 . We also have Ω |∇u | 2 - γ |x| 2 u 2 dx = B δ (0) |∇U | 2 - γ |x| 2 U 2 dx + O β+-β- = B -1 δ (0) |∇U | 2 - γ |x| 2 U 2 dx + O β+-β- = χ B -1 δ (0) U 2 (s) |x| s dx + O β+-β- = χ R n U 2 (s) |x| s dx + O β+-β-. Finally, we estimate the last term as → 0, Ω h(x)u 2 dx = B δ (0) h(x)U 2 dx + O β+-β- = 2 B -1 δ (0) h( x)U 2 dx + O β+-β-. If γ < (n-2) 2 4 -1 and β + -β -> 2, the extremal U ∈ L 2 (R n ) and therefore Ω h(x)u 2 dx = h(0) R n U 2 dx 2 + o( 2 ) as → 0. If now γ = (n-2) 2 4 -1, then U (x) x→∞ |x| -n 2 and β + -β -= 2. Therefore Ω h(x)u 2 dx = h(0)ω n-1 2 ln 1 + o( 2 ln ) as → 0,
where ω n-1 is the volume of the canonical (n -1)-sphere. Combining the above estimates as → 0 yields

J Ω γ,s,h (u ) = J R n γ,s,0 (U ) -    h(0) U 2 2 U 2 2 (s),|x| -s 2 + o( 2 ) if γ < (n-2) 2 4 -1 h(0)ωn-1 U 2 2 (s),|x| -s 2 ln 1 + o( 2 ln ) if γ = (n-2) 2 4 -1.
In either case, if h(0) = λ > 0, we get that

µ γ,s,λ (Ω) ≤ J Ω γ,s,λ (u ) < J R n γ,s,0 (U ) = µ γ,s,0 (R n
), and we are done.

Subsection 2.2: The merely singular case

We now prove the second part of Theorem 1. Assuming that s = 0, γ < 0 and n ≥ 4, we shall prove that µ γ,s,λ (Ω) is attained if and only if λ (γ, Ω) < λ, where

λ (γ, Ω) = inf |γ| |x| 2 ; x ∈ Ω < λ 1 (L γ ).
Note that in this case, we have µ 0,0,0 (R n ) = µ γ,0,0 (R n ) as noted in [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF], that is inf

U ∈D 1,2 (R n )\{0} J R n γ,0,0 (U ) = 1 K(n, 2) 2 := inf U ∈D 1,2 (R n )\{0} R n |∇U | 2 dx R n |U | 2 dx 2 2
, and the infimum of µ γ,0 (R n \ {0}) is not achieved. Consider now the following known extremal for µ 0,0 (R n ),

U (x) := 1 (1 + |x| 2 ) n-2 2 for x ∈ R n .
Fix x 0 ∈ Ω, x 0 = 0, and define the test-function

(11) u (x) := η(x) -n-2 2 U ( -1 (x -x 0 )) for all x ∈ Ω, where η ∈ C ∞ c (Ω) is such that η(x) = 1 around x 0 ∈ Ω. A straightforward computation yields J Ω γ,0,λ (u ) = µ 0,0 (R n ) + o(1)
as → 0, which yields that µ γ,0,λ (Ω) ≤ µ 0,0 (R n ). Classical computations in the spirit of Aubin [START_REF] Aubin | Problèmes isopérimétriques et espaces de Sobolev[END_REF], which can be done by replacing -λ with a more general function h, yield that for n ≥ 4, there exists c n > 0, such that as → 0,

J Ω γ,h (u ) = 1 K(n, 2) 2 +    c n |γ| |x0| 2 -h(x 0 ) 2 + o( 2 ) if n ≥ 5 c 4 |γ| |x0| 2 -h(x 0 ) 2 ln( -1 ) + O( 2 ) if n = 4.
Therefore, if n ≥ 4 and assuming there exists

x 0 ∈ Ω \ {0} such that h(x 0 ) > |γ| |x0| 2 , we obtain that inf u∈H 1 0 (Ω)\{0} J Ω γ,h (u) < inf U ∈D 1,2 (R n )\{0} J R n γ,0 (
U ), and µ γ,h (Ω) is attained.

Conversely, if h(x) ≤ |γ| |x| 2 for all x ∈ Ω \ {0}, then -γ |x| 2 + h(x) ≥ 0 for all x ∈ Ω \ {0}
, hence µ γ,0,h (Ω) ≥ µ 0,0,0 (Ω) = µ 0,0,0 (R n ). We therefore have equality, and there is no extremal for µ γ,0,h (Ω) since the extremals on R n are rescaled and translated versions of U .

The Hardy-singular interior mass of a domain

This section is devoted to the construction of the singular interior mass, as stated in Theorem 2. We start with the following key result. Proposition 2. Assume Ω is a smooth bounded domain in R n and let h ∈ C 0,θ (Ω) with θ ∈ (0, 1).

If the operator -∆ -γ |x| 2 -h(x) is coercive, then there exists a solution H ∈ C ∞ (Ω \ {0}) for the linear problem (12)      -∆H -γ |x| 2 + h(x) H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω.
Moreover, there exists c > 0 such that

(13) H(x) x→0 c |x| β+(γ) . If H ∈ C ∞ (Ω \ {0}
) is another solution for [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF], then there exists λ > 0 such that H = λH.

Proof: First, we prove existence of a solution. For that, let η 1 ∈ C ∞ (R) be such that η 1 (t) = 0 for t < 1 and η 1 (t) = 1 for t > 2. For > 0, set η (x) := η 1 (|x|/ ) for all x ∈ R n . Then let H ∈ C ∞ (Ω \ {0}) be the Green's function for the operator -∆ -η (x) γ |x| 2 + h(x) that is singular at 0. In particular, we have that

     -∆ H -η (x) γ |x| 2 + h(x) H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω. Fix x 0 ∈ Ω \ {0} and define H (x) := H (x)
H (x0) for all x ∈ Ω \ {0}. For δ > 0 such that B 4δ (0) ⊂ Ω and δ < |x 0 |/4, we take ∈ (0, δ/2). We then have

     -∆H -γ |x| 2 + h(x) H = 0 in Ω \ B δ (0) H > 0 in Ω \ B δ (0) H = 0 on ∂Ω.
It follows from the boundary Harnack inequality (see for instance Ghoussoub-Robert [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF], Proposition 7.2) that there exists C δ > 0 such that

H (x) d(x, ∂Ω) ≤ C δ H (x 0 ) d(x 0 , ∂Ω) = C δ d(x 0 , ∂Ω) for all x ∈ Ω \ B 2δ (0).
Since this is valid for any δ > 0 small enough, it then follows from standard ellitpic theory that there exists

H ∈ C ∞ (Ω \ {0}) such that H → H in C k loc (Ω \ {0}) as → 0 for all k ∈ N. In particular, we have      -∆H -γ |x| 2 + h(x) H = 0 in Ω \ {0} H ≥ 0 in Ω \ {0} H = 0 on ∂Ω.
Since H(x 0 ) = 1, it follows from the strong maximum principle that H > 0, hence it satisfies [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF]. It then follows from Theorem 9 that there exists c > 0 such that

either H(x) x→0 c |x| β+(γ) or H(x) x→0 c |x| β-(γ) .
If [START_REF]Optimal Sobolev inequalities and extremal functions. The three-dimensional case[END_REF] does not hold, we the second case holds and H ∈ H 1 0 (Ω): since -∆ -γ|x| -2 -h is coercive, equation [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] then yield H ≡ 0, contradicting H > 0. This proves [START_REF]Optimal Sobolev inequalities and extremal functions. The three-dimensional case[END_REF].

To prove uniqueness, let H ∈ C ∞ (Ω\{0}) be such that (12) holds. Set λ 0 := max{λ ≥ 0; H ≥ λH}.

This is well defined and we set H

:= H -λ 0 H. Then H ≥ 0 satisfies -∆ H -γ |x| 2 + h(x) H = 0 in Ω \ {0}.
Therefore, if H ≡ 0, it follows from the maximum principle that H > 0. Then the asymptotic control (13) and Hopf's boundary comparison principle yield the existence of 0 > 0 such that H ≥ 0 H in Ω \ {0}, contradicting the definition of λ 0 . Therefore H ≡ 0 and H = λ 0 H, which proves the uniqueness statement. We now proceed with the proof of Theorem 2.

Proposition 3.

Let Ω be a smooth bounded domain in R n and fix h ∈ C 0,θ (Ω), θ ∈ (0, 1). Assume that the operator -∆ -

γ |x| 2 + h(x) is coercive and that γ > (n-2) 2 4 -1. If H ∈ C ∞ (Ω \ {0}) is a solution to (12), then there exist c 1 , c 2 ∈ R with c 1 > 0 such that (14) H(x) = c 1 |x| β+(γ) + c 2 |x| β-(γ) + o 1 |x| β-(γ) as x → 0.
The ratio c2 c1 ∈ R is independent of the choice of H. We can therefore define the mass as m γ,h (Ω) := c2 c1 .

Proof: Let η ∈ C ∞ c (Ω) be such that η(x) ≡ 1 around 0. Our first objective is to write H(x) := η(x)

|x| β + (γ) + f (x) for some f ∈ H 1 0 (Ω).
For that, we consider the function

g(x) := --∆ - γ |x| 2 + h(x) (η|x| -β+(γ) ) in Ω \ {0}.
Since η(x) = 1 around 0, we have that

(15) |g(x)| = | h(x) |x| β+(γ) | ≤ C|x| -β+(γ) around 0. Therefore, g ∈ L 2n n+2 (Ω) if and only if β + (γ) < n+2
2 , which holds if and only if γ > (n-2) 2

4

-4. The latter is guaranteed by our assumption on γ.

Since L 2n n+2 (Ω) = L 2n n-2 (Ω) ⊂ H 1 0 (Ω) , there exists f ∈ H 1 0 (Ω) such that -∆f - γ |x| 2 + h(x) f = g in H 1 0 (Ω).
By regularity theory, we have that f ∈ C 2 (Ω \ {0}). We now show that (16) |x| β-(γ) f (x) has a finite limit as x → 0.

Define

H(x) := η(x) |x| β+(γ) + f (x) for all x ∈ Ω \ {0}.
and note that H ∈ C 2 (Ω \ {0}) and is a solution to -∆ -γ+h(x)

|x| 2 H = 0.
Write g + (x) := max{g(x), 0} and g -(x) := max{-g(x), 0} so that g = g + -g -, and let f 1 , f 2 ∈ H 1 0 (Ω) be weak solutions to

(17) -∆f 1 - γ |x| 2 + h(x) f 1 = g + and -∆f 2 - γ |x| 2 + h(x) f 2 = g -in H 1 0 (Ω).
In particular, uniqueness, coercivity and the maximum principle yields f = f 1 -f 2 and f 1 , f 2 ≥ 0. Assume that f 1 ≡ 0, so that

f 1 > 0 in Ω \ {0}, fix α ∈ (β -(γ), β + (γ)), choose µ ∈ R such that µ(β -(γ)(n -2 -β -(γ)
) -γ) < 0, and define u -(x) := |x| -β-(γ) + µ|x| -α for all x = 0. As in the proof of the previous proposition, we get that for some δ > 0 small,

-∆ - γ + h(x) |x| 2 u -< 0 for x ∈ B δ (0) \ {0}, that is u -is a subsolution on B δ (0) \ {0}.
Fix now C > 0 such that f 1 ≥ Cu -on ∂B δ (0). Since f 1 and Cu -∈ H 1 0 (Ω) are respectively superand sub-solutions to -∆ -γ+h(x) |x| 2 u = 0, it follows from the comparison principle (via coercivity)

that f 1 ≥ Cu -in B δ (0) \ {0}, and therefore f 1 ≥ C |x| -β-(γ) in B δ (0) \ {0}. It then follows from (15) that g + (x) ≤ |g(x)| ≤ C 1 |x| -2+(2-(β+(γ)-β-(γ))) f 1 , and therefore (17) yields (18) -∆ - γ + O(|x| 2-(β+(γ)-β-(γ)) ) |x| 2 f 1 = 0 weakly in H 1 0 (Ω). Since γ > (n-2) 2 4 -1 if and only if τ := 2 -(β + (γ) -β -(γ)
) > 0, we can argue as in the proof of Proposition 2 (see also the regularity Theorem 8) and get that |x| β-(γ) f 1 (x) has a finite limit as x → 0. Similarly, |x| β-(γ) f 2 (x) has also a finite limit as x → 0, and therefore ( 16) is verified. It follows that there exists c 2 ∈ R such that

H(x) = 1 |x| β+(γ) + c 2 |x| β-(γ) + o 1 |x| β-(γ) as x → 0,
which proves the existence of a solution H to the problem with the relevant asymptotic behavior. The uniqueness result of Proposition 2 then yields the conclusion. The following proposition summarizes the properties of the mass. Proposition 4. Let Ω be a smooth bounded domain in R n and fix h ∈ C 0,θ (Ω) with θ ∈ (0, 1).

Assume that the operator -∆ -γ |x| 2 + h(x) is coercive and that γ > (n-2) 2

4

-1. The mass m γ,h (Ω) then satisfies the following properties:

i) m γ,0 (Ω) < 0, ii) If h ≤ h and h ≡ h , then m γ,h (Ω) < m γ,h (Ω), iii) If Ω Ω, then m γ,h (Ω ) < m γ,h (Ω). iv) The function h → m γ,h (Ω) is continuous for the C 0 (Ω) norm.
Proof: For any such h ∈ C 0,θ (Ω), we let H h be the unique solution to [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] such that [START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF] holds with c 1 = 1. In other words,

H h (x) = 1 |x| β+(γ) + m γ,h (Ω) |x| β-(γ) + o 1 |x| β-(γ) as x → 0. Since (-∆ -γ |x| 2 + h(x) )(H 0 (x) -|x| -β+(γ)
) = 0 and is negative on ∂Ω, it follows from the maximum principle that H 0 (x)-|x| -β+(γ) < 0 on Ω. It then follows from Theorem 8 that m γ,0 (Ω) < 0. This prove property (i) of the proposition. Property (iii) goes similarly.

For (ii), we define g := H h -H h . We have that g ∈ H 1 0 (Ω) and -∆g-γ |x| 2 + h(x) g = (h-h )H h ≤ 0, but ≡ 0. Therefore g < 0, and it follows from Theorem 8 in Appendix C that there exists K > 0 such that g(x)|x| β-(γ) → -K as x → 0, and therefore m γ,h (Ω) -m γ,h (Ω) = -K < 0, which proves the second part of the proposition.

Positive mass and the existence of extremals in lower dimensions

In this section, we show how the positivity of the Hardy-singular mass m γ,λ (Ω) in the truly singular case (resp., the mass in the merely singular case) yields that µ γ,s,λ (Ω) is attained in the corresponding low dimensions, i.e., (n-2)

2 4 -1 < γ < (n-2) 2 4
in the truly singular case, and n = 3 in the merely singular case. Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω. Assume either 0 < s < 2 or that {s = 0, γ > 0}.

If (n-2) 2 4 -1 < γ < (n-2) 2 4 and 0 < λ < λ 1 (L γ ) is such that the mass m γ,λ (Ω) is positive, then µ γ,s,λ (Ω) is attained. Proof: Assuming that (n-2) 2 4 -1 < γ < (n-2) 2 4
, we know that the mass m γ,λ (Ω) is defined. We need to show that if m γ,λ (Ω) > 0, then µ γ,s,λ (Ω) < µ γ,s,0 (R n ). Consider again for each > 0 the extremals

(19) U (x) := -n-2 2 U x =   2-s n-2 • β + -β - 2 2-s n-2 •(β+-β-) |x| (2-s)β - n-2 + |x| (2-s)β + n-2   n-2 2-s .
We shall first replace λ with any function h ∈ C 0,θ (Ω), where θ ∈ (0, 1) and the operator -∆ -

γ |x| 2 + h(x) is coercive . Consider again a test function η ∈ C ∞ c (Ω) such that η(x) = 1 for x ∈ Ω in a neighborhood of 0. Since γ > (n-2) 2 4
-1, it follows from Proposition 3 that there exists β ∈ H 1 0 (Ω) such that ( 20)

β(x) x→0 m γ,h (Ω) |x| β-,
and the function H(x) := η(x)

|x| β + + β(x) satisfies (21)      -∆H -γ |x| 2 + h(x) H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω. Define now u (x) := η(x)U (x) + β + -β - 2 β(x) for x ∈ Ω \ {0}.
It is clear that u ∈ H 1 0 (Ω) for all > 0. We now estimate J Ω γ,s,h (u ), where again J Ω γ,s,h is the functional on H 1 0 (Ω) defined by

J Ω γ,s,h (u) := Ω |∇u| 2 -γ |x| 2 + h(x) u 2 dx Ω |u| 2 (s) |x| s dx 2 2 (s) .
Thereafter, the notation "o δ (1)" will mean lim δ→0 lim →0 o δ (1) = 0.

Step 1:

Estimating Ω |∇u | 2 -γ |x| 2 + h(x) u 2 dx
Letting δ ∈ (0, dist(0, ∂Ω)), we start by estimating

Ω\B δ (0) |∇u | 2 -γ |x| 2 + h(x) u 2 dx. First note that (22) lim →0 u β + -β - 2 = H in C 2 loc (Ω \ {0}).
Therefore,

→0 Ω\B δ (0) |∇u | 2 -γ |x| 2 + h(x) u 2 dx β+-β- = Ω\B δ (0) |∇H| 2 - γ |x| 2 + h(x) H 2 dx. (23) lim
Integrating by parts and using equation [START_REF] Kang | Positive solutions for singular critical elliptic problems[END_REF] yields

Ω\B δ (0) |∇H| 2 - γ |x| 2 + h(x) H 2 dx = Ω\B δ (0) H -∆H - γ |x| 2 + h(x) H dx + ∂(Ω\B δ (0)) H∂ ν H dσ = - ∂B δ (0) H∂ ν H dσ. ( 24
)
Since β + + β -= n -2, using elliptic estimates, and the definition of H yields

H∂ ν H = -β + |x| -2β+-1 -(n -2)m γ,h (Ω)|x| -(n-1) + o(|x| -(n-1) ) as x → 0.
Therefore, plugging this expansion into ( 23) and ( 24) yields ( 25)

Ω\B δ (0) |∇u | 2 - γ + h(x) |x| 2 u 2 dx = β+-β-ω n-1 β + δ β+-β-+ (n -2)m γ,h (Ω) + o δ (1)
We now deal with the expression

B δ (0) |∇u | 2 -γ |x| 2 + h(x) u 2 dx. Take δ > 0 small enough such that η(x) = 1 for x ∈ B δ (0) ⊂ Ω. Therefore, u (x) = U (x) + β + -β - 2 β(x) for x ∈ B δ (0) and then B δ (0) |∇u | 2 - γ |x| 2 + h(x) u 2 dx = B δ (0) |∇U | 2 - γ |x| 2 + h(x) U 2 dx +2 β + -β - 2 B δ (0) ∇U ∇β - γ + h(x) |x| 2 U β dx + β+-β- B δ (0) |∇β| 2 - γ + h(x) |x| 2 β 2 dx.
Since U , β ∈ H 1 0 (Ω) and U is explicit, we integrate by parts the first and second term of the right-hand-side and we neglect the third term to get

B δ (0) |∇u | 2 - γ |x| 2 + h(x) u 2 dx = B δ (0) -∆U - γ |x| 2 U U dx + ∂B δ (0) U ∂ ν U dσ - B δ (0) h(x)U 2 dx +2 β + -β - 2 B δ (0) -∆U - γ |x| 2 + h(x) U β dx +2 β + -β - 2 ∂B δ (0) β∂ ν U dσ + o δ ( β+-β-). (26)
We estimate the terms of the right-hand-side separately. Note first that [START_REF] Cao | A note on the sign-changing solutions to elliptic problems with critical Sobolev and Hardy terms[END_REF] and [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] yield that as → 0,

B δ (0) -∆U - γ |x| 2 U U dx = λ B δ (0) U 2 (s) |x| s dx = χ B δ (0) U 2 (s) |x| s dx = χ R n U 2 (s) |x| s dx + O R n \B -1 δ (0) |x| -β+2 (s)-s dx = χ R n U 2 (s) |x| s dx + o β+-β-. (27)
The explicit expression of U in ( 19) yields ( 28)

∂B δ (0) U ∂ ν U dσ = -β + ω n-1 δ -(β+-β-) β+-β-+ o δ ( β+-β-) as → 0. Since γ > (n-2) 2 4
-1, we have that 2β + < n and therefore (29)

B δ (0) h(x)U 2 dx = O β+-β- B δ (0) |x| -2β+ dx = o δ β+-β-. Since β + + β -= n -2 < n, we also have that β + -β - 2 B δ (0) h(x)U β dx = O β+-β- B δ (0) |x| -β+ |x| -β-dx = o δ β+-β-. (30)
It follows from ( 10) and ( 19) that

B δ (0) -∆U - γ |x| 2 U β dx = χ B δ (0) U 2 (s)-1 |x| s β dx = χ β + -β - 2 B -1 δ (0) U 2 (s)-1 |x| s β-β( x) dx = O β + -β - 2 . (31)
Finally, using the expression [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] of U and the asymptotics (20) of β, we get that

∂B δ (0) β∂ ν U dσ = β + -β - 2 ∂B δ (0) β∂ ν |x| -β+ dσ + o β + -β - 2 = β + -β - 2 m γ,h (Ω) ∂B δ (0) |x| -β-∂ ν |x| -β+ dσ + o β + -β - 2 = - β + -β - 2 m γ,h (Ω)β + ω n-1 + o β + -β - 2 . ( 32
)
Plugging together ( 25), ( 26), ( 27), ( 28), ( 29), ( 30), ( 31) and (32) yields

Ω |∇u | 2 - γ |x| 2 + h(x) u 2 dx = χ R n U 2 (s) |x| s dx +(n -2 -2β + )m γ,h (Ω)ω n-1 β+-β-+ χ β + -β - 2 B δ (0) U 2 (s)-1 |x| s β dx + o( β+-β-). ( 33
)
Step 2: Estimating Ω u 2 (s) |x| s dx From ( 22) and the definition [START_REF] Jannelli | The Role played by Space Dimensions in Elliptic Critical Problems[END_REF] of U , we have as → 0 that

Ω u 2 (s) |x| s dx = B δ (0) u 2 (s) |x| s dx + o( β+-β-) = B δ (0) |U + β + -β - 2 β| 2 (s) |x| s dx + o( β+-β-) = B δ (0) U 2 (s) |x| s + 2 (s) U 2 (s)-1 |x| s β dx + B δ (0) O β+-β-U 2 (s)-2 |x| s β 2 + 2 (s) 2 (β+-β-) |β| 2 (s) dx + o( β+-β-) = B -1 δ (0) U 2 (s) |x| s dx + 2 (s) β + -β - 2 B δ (0) U 2 (s)-1 |x| s β dx + o( β+-β-).
Using the expression of U , we get that

Ω u 2 (s) |x| s dx = R n U 2 (s) |x| s dx + 2 (s) β + -β - 2 B δ (0) U 2 (s)-1 |x| s β dx + o( β+-β-). (34)
Therefore, plugging (33) and (34) into J Ω γ,h (u ) and using the equation [START_REF] Cao | A note on the sign-changing solutions to elliptic problems with critical Sobolev and Hardy terms[END_REF] satisfied by U yields

J Ω γ,s,h (u ) = J R n γ,s,0 (U ) 1 - 2ω n-1 β + (γ) -n-2 2 χ R n U 2 (s) |x| s dx m γ,h (Ω) β+-β-+ o( β+-β-) .
This readily shows that if h(x) = λ, where 0 < λ < λ 1 (L γ ), and if m γ,λ (Ω) > 0, then J Ω γ,s,λ (u ) < J R n γ,s,0 (U ) = µ γ,s,0 (R n ), and therefore µ γ,s,λ (Ω) is attained. This completes the proof of Proposition 5.

Subsection 4.2:

The merely singular case Proposition 6. Let Ω be a smooth bounded domain in R n , n = 3, such that 0 ∈ Ω. Assume that s = 0 and γ < 0.

If 0 < λ < λ 1 (L γ ) is such that the mass R γ,λ (Ω) is positive, then µ γ,0,λ (Ω) is attained.
Proof: This is by now classical, so we shall sketch a proof. For any x 0 ∈ Ω \ {0}, we let G x0 ∈ C ∞ (Ω \ {0}) be the Green's function for the operator -∆ -γ |x| 2 + h(x) at x 0 with Dirichlet boundary condition. Since n = 3, then up to multiplying by a constant, we have

G x0 (x) = 1 4π η(x) |x -x 0 | + β(x) for all x ∈ Ω \ {x 0 }, where β ∈ H 1 0 (Ω) ∩ C 0 (Ω), and there exists R γ,h (Ω, x 0 ) ∈ R such that G x0 (x) = 1 4π 1 |x -x 0 | + R γ,h (Ω, x 0 ) + o(1) as x → x 0 . Note that β(x 0 ) = R γ,h (x 0 ) is the Robin function at x 0 . Set now ũ (x) := u (x) + √ β(x) for all x ∈ Ω \ {0}
, where u are the functions defined in [START_REF] Dolbeault | On the symmetry of extremals for the Caffarelli-Kohn-Nirenberg inequalities[END_REF].

Then, classical computations in the spirit of Schoen [START_REF] Schoen | Conformal deformation of a Riemannian metric to constant scalar curvature[END_REF] yield

J Ω γ,h (ũ ) = 1 K(n, 2) 2 -c 3 R γ,h (Ω, x 0 ) n-2 + o n-2 as → 0. If now x 0 ∈ Ω \ {0} and 0 < λ < λ 1 (L γ ) are such R γ,λ (x 0 ) > 0, then J Ω γ,h (ũ ) = µ γ,0 (R n ) -c 3 R γ,λ (x 0 ) + o ( ) as → 0.
This implies that µ γ,0,λ (Ω) is attained.

Blow-Up analysis in the truly singular case

Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω is an interior point. Fix γ < (n -2) 2 /4 and recall that

µ γ,s,0 (R n ) := inf      R n |∇u| 2 -γ |x| 2 u 2 dx R n |u| 2 (s) |x| s dx 2 2 (s) ; u ∈ D 1,2 (R n ) \ {0}      , where 0 ≤ s < 2 and 2 (s) := 2(n-s) n-2 . Let (a α ) α∈N ∈ C 1 (Ω) be such that there exists a ∞ ∈ C 1 (Ω) with (35) lim α→+∞ a α = a ∞ in C 1 (Ω). Consider (λ α ) α ∈ (0, +∞) such that (36) lim α→+∞ λ α = µ γ,s,0 (R n ). Suppose (u α ) α ∈ H 1 0 (Ω) is a sequence of weak solutions to (37)      -∆u α -γ |x| 2 + a α u α = λ α u 2 (s)-1 α |x| s in Ω u α ≥ 0 a.e. in Ω u α = 0 on ∂Ω with (38) Ω u 2 (s) α |x| s dx = 1.
and such that (39) u α 0 as α → +∞ weakly in H 1 0 (Ω). We shall assume uniform coercivity, that is there exists c > 0 such that

Ω |∇ϕ| 2 - γ |x| 2 + a α ϕ 2 dx ≥ c Ω ϕ 2 dx for all ϕ ∈ H 1 0 (Ω).
Note that this is equivalent to the coercivity of -∆ -(γ|x| -2 + a ∞ ). The two following sections are devoted to the analysis of the Blow-up behavior of (u α ) as α → +∞. The present section deals mostly with the case {s > 0 or γ > 0}, for which there are extremals for µ γ,s,0 (R n ). The case {s = 0 and γ < 0} will be dealt with in the next section. The case s = γ = 0 has been extensively studied in the litterature, see for instance [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF][START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF] and the references therein.

Theorem 4. Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω is an interior point. Fix γ < (n -2) 2 /4
, and assume that either s > 0 or γ > 0. 36), (37), ( 38) and (39) hold. Then:

Let (a α ) α ∈ C 1 (Ω), (λ α ) α ∈ (0, +∞) and (u α ) α ∈ H 1 0 (Ω) such that (35), (
i) If γ ≤ (n-2) 2 4 -1, then a ∞ (0) = 0; ii) If (n-2) 2 4 -1 < γ < (n-2) 2 4
, then m γ,a∞ (Ω) = 0. In addition, there exists C > 0 such that

(40) u α (x) ≤ C µ β + (γ)-β -(γ) 2 α µ β+(γ)-β-(γ) α |x| β-(γ) + |x| β+(γ) for all x ∈ Ω \ {0},
where µ α → 0 as α → 0 is defined in (44) below.

The rest of this section is devoted to the proof of this theorem. We shall make frequent use of the following Pohozaev identity.

Proposition 7.

Let Ω ⊂ R n be a smooth bounded domain and let u ∈ C 2 (Ω), u ≥ 0. For any p ∈ R n , we have

Ω (x -p) i ∂ i u + n -2 2 u -∆u - γ |x| 2 u -c u 2 (s) |x| s dx (41) = ∂Ω (x -p, ν) |∇u| 2 2 - γu 2 2|x| 2 - cu 2 (s) 2 (s)|x| s -(x -p) i ∂ i u + n -2 2 u ∂ ν u dσ + Ω (p, x) |x| 2 γ u 2 |x| 2 + c su 2 (s) 2 (s)|x| s dx.
Proof: The classical Pohozaev identity yields

- Ω (x -p) i ∂ i u + n -2 2 u ∆u dx = ∂Ω (x -p, ν) |∇u| 2 2 -(x -p) i ∂ i u + n -2 2 u ∂ ν u dσ.
For any t ∈ [0, 2], integration by parts yields

Ω (x -p) i ∂ i u + n -2 2 u u 2 (t)-1 |x| t dx = - t 2 (t) Ω (p, x) |x| 2+t u 2 (t) dx + ∂Ω (x -p, ν)u 2 (t) 2 (t)|x| t dσ.
Putting together these two equalities yields the general identity claimed in the proposition.

To prove Theorem (4), we start by noting that regularity theory and Theorem 8 yield that for any α, there exists C α > 0 such that u α ∈ C2,θ (Ω \ {0}), and

(42)

u α (x) ∼ x→0 C α |x| β-(γ) and |∇u α (x)| ≤ C α |x| -β-(γ)-1 for x ∈ Ω \ {0}. Fix τ ∈ R such that β -(γ) < τ < n -2 2 .
It follows from (42) that for any α ∈ N, there exists

x α ∈ Ω \ {0} such that (43) sup x∈Ω\{0} |x| τ u α (x) = |x α | τ u α (x α ).
We now prove the following proposition, which is valid for any γ < (n -2) 2 /4. 36), ( 37) and (38) hold. Let (x α ) α ∈ Ω \ {0} be as in (43) and set

Proposition 8. Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω is an interior point. Fix γ < (n -2) 2 /4, and let (a α ) α ∈ C 1 (Ω), (λ α ) α ∈ (0, +∞) and (u α ) α ∈ H 1 0 (Ω) be such that (35), (
(44) µ α := u α (x α ) -2 n-2 .
Then,

(45) lim α→+∞ sup x∈Ω\{0} |x| τ u α (x) = +∞,
and therefore lim α→+∞ µ α = 0. In addition,

(46) lim α→+∞ d(x α , ∂Ω) µ α = +∞.
Proof of Proposition 8: If (45) does not hold, then there exists C > 0 such that, up to a subsequence, we have that |x| τ u α (x) ≤ C for all x ∈ Ω \ {0}. Since τ < n-2 2 , we then have that (47) lim

δ→0 lim α→+∞ B δ (0) u 2 (s) α |x| s dx = 0.
Since (u α ) is bounded uniformly in L ∞ outside 0, it then follows from (37) and (39) that u α → 0 in C 0 loc (Ω \ {0}). This limit and (47) yield Ω |x| -s u dx → 0 as α → +∞, contradicting (38). This proves (45). As a remark, note that when s > 0, the subcriticality 2 (s) < 2 and (39) yield u α → 0 in C 0 loc (Ω \ {0}). We now prove (46). Assume that d(x α , ∂Ω) = O(µ α ) as α → +∞, the above remark then yields s = 0. We let

x ∞ := lim α→+∞ x α such that x ∞ ∈ ∂Ω. Since Ω is smooth, we let δ > 0 and ϕ ∈ C ∞ (B δ (0), R n ) be a smooth diffeomorphism onto its image such that ϕ(0) = x ∞ , ϕ(B δ (0) ∩ {x 1 < 0}) = ϕ(B δ (0)) ∩ Ω and ϕ(B δ (0) ∩ {x 1 = 0}) = ϕ(B δ (0)) ∩ ∂Ω.
Up to a rotation and a rescaling, we can assume that dϕ

0 = Id. Let ((x α ) 1 , x α ) ∈ (-∞, 0) × R n-1 ∩ B δ (0) be such that ϕ((x α ) 1 , x α ) = x α . In particular, lim α→0 |(x α ) 1 | + |x α | = 0. Define ũα (x) := µ n-2 2 α u α • ϕ((0, x α ) + µ α x) for x ∈ {x 1 < 0} ∩ B µ -1 α δ/2 (0). We then have that -∆ gα ũα -µ 2 α γ |ϕ((0, x α ) + µ α x)| 2 + a α (ϕ((0, x α ) + µ α x)) = λ α ũ2 -1 α ,
where g α := (ϕ Eucl)((0, x α ) + µ α x). We have that ũα > 0 on

{x 1 < 0} ∩ B µ -1 α δ/2 (0), ũα = 0 on {x 1 = 0} ∩ B µ -1 α δ/2 (0), and ũα (µ -1 α (x α ) 1 , 0) = 1. Therefore, standard elliptic theory yields the existence of ũ ∈ C ∞ ({x 1 ≤ 0}) ∩ D 1,2 (R n ) such that -∆ũ = µ γ,s,0 (R n )ũ 2 -1 and ũ > 0 in {x 1 < 0}
and ũ = 0 on {x 1 = 0}. It follows from Theorem 1.3, chapter III in [START_REF] Struwe | Variational methods[END_REF] that this is a contradiction. This proves (46) and ends the proof of Proposition 8. In addiction to the hypothesis of Proposition 8, we now assume that either s > 0 or γ > 0. We claim that (48) lim

α→+∞ |x α | µ α = c > 0.
For that, we first show that (49)

|x α | = O(µ α ) as α → +∞.
Indeed, if not we can assume that µ -1 α |x α | → +∞ as α → +∞. By defining ũα := µ n-2 2 α u α (x α +µ α x), it follows from our assumption and Proposition 8 that for any R > 0, and for α large enough, ũα is defined on B R (0) and

-∆ũ α -    γ xα µα + x 2 + µ 2 α a α (x α + µ α x)    ũα = λ α ũ2 -s α xα µα + x s in B R (0).
It follows from ( 43) and the assumption that µ -1 α |x α | → +∞ as α → +∞, that there exists C(R) > 0 such that ũα ≤ C(R) on B R (0) and that ũα (0) = 1. It then follows from standard elliptic theory that ũα → ũ in C 2 loc (R n ) where 0 < ũ ≤ 1 and ( 50)

-∆ũ = µ γ,s,0 (R n )u 2 -1 if s = 0 and ∆ũ = 0 if s > 0.
By the Sobolev embedding, we have that

(51) B R (0) ũ2 α dx = B Rµα (xα) u 2 α dx ≤ Ω u 2 α dx ≤ C u α H 1 0 (Ω) ≤ C,
where we used the fact that B Rµα (x α ) ⊂ Ω since (46) holds. Therefore, by first passing to the limit as α → +∞ and then as R → +∞, we get that ũ ∈ L 2 (R n ).

Assuming that s > 0, and since 0 < ũ ≤ 1, it follows from (50) and Liouville's theorem that ũ ≡ 1, contradicting that ũ ∈ L 2 (R n ). In other words, (49) is proved when s > 0.

Assuming now that s = 0 but γ > 0, then by letting α → +∞ and R → +∞ in (51) and using(38), we get that

R n ũ2 dx ≤ 1. Equation (50) then yields µ 0,0 (R n ) ≤ R n |∇ũ| 2 dx R n ũ2 dx 2 2 = µ γ,0 (R n ) R n ũ2 dx 2 n ≤ µ γ,0 (R n ).
Since γ > 0, it follows from classical estimates (see [START_REF]Sobolev inequalities for the Hardy-Schrödinger operator: extremals and critical dimensions[END_REF]) that µ γ,0 (R n ) < µ 0,0 (R n ), yielding again a contradiction. In other words, (49) is proved when s = 0.

We now prove (48). We argue again by contradiction and assume that x α = o(µ α ) as α → +∞.

We define ũα (x

) := µ n-2 2 α u α (|x α |x) for x ∈ B |xα| -1 δ (0) and δ > 0 small enough. The definition (43) yields (|x α | • |x|) τ u α (|x α |x) ≤ |x α | τ u α (x α ), and therefore |x| τ ũα (x) ≤ 1 for all x ∈ B |xα| -1 δ (0) \ {0}. Equation (37) rewrites -∆ũ α - γ |x| 2 + |x α | 2 a α (|x α |x) ũα = λ α |x α | µ α 2-s ũ2 (s)-1 α |x| s in B |xα| -1 δ (0) \ {0}.
In addition, we have that ũα > 0 and ũα (|x α | -1 x α ) = 1. These estimates and standard elliptic theory then yield the existence of ũ

∈ C ∞ (R n \ {0}) such that ũα → ũ in C 2 loc (R n \ {0}) where -∆ũ - γ |x| 2 ũ = 0 in R n \ {0} ; ũ > 0 ; |x| τ ũ(x) ≤ 1 in R n \ {0}.

The classification of Proposition 11 yields the existence of

A, B ≥ 0 such that ũ(x) = A|x| -β+(γ) + B|x| -β-(γ) in R n \ {0}. The pointwise control |x| τ ũ(x) ≤ 1 in R n \ {0} yields A = B = 0, contra- dicting ũ > 0.
This completes the proof of (48).

We now define

v α (x) := µ n-2 2 α u α (µ α x) for x ∈ µ -1
α Ω \ {0}, and claim that there exists

U ∈ H 1 0 (Ω) ∩ C 2 (R n \ {0}) such that (52) lim α→+∞ v α = U in H 2 1,loc (R n ) ∩ C 2 loc (R n \ {0}).
For that, we first note that

-∆v α - γ |x| 2 + µ 2 α a α (µ α x) v α = λ α v 2 (s)-1 α |x| s in µ -1 α Ω \ {0}. Moreover, v α > 0 and |x| τ v α (x) ≤ C for all x ∈ µ -1 α Ω \ {0}. It then follows from standard elliptic theory that there exists U ∈ C ∞ (R n \ {0}), U ≥ 0, such that lim α→+∞ v α = U in C 2,θ loc (R n \ {0}) and (53) -∆U - γ |x| 2 U = µ γ,s,0 (R n ) U 2 (s)-1 |x| s in R n \ {0}.
Since v α (µ -1 α x α ) = 1, it then follows that U ≡ 0, and therefore U > 0. Moreover, we have that

B R (0)\B δ (0) U 2 (s) |x| s dx = lim α→+∞ B Rµα (0)\B δµα (0) u 2 (s) α |x| s dx ≤ 1.
Therefore, letting R → +∞ and δ → 0 yields

R n U 2 (s) |x| s dx ≤ 1. Similarly, R n U 2 |x| 2 dx < +∞ and R n |∇U | 2 dx < +∞. Therefore U ∈ D 1,2 (R n ),
and by integrating by parts, we obtain that

µ γ,s,0 (R n ) ≤ R n |∇U | 2 -γ |x| 2 U 2 dx R n U 2 (s) |x| s dx 2 2 (s) = µ γ,s,0 (R n ) R n U 2 (s) |x| s dx 2-s n-s ≤ µ γ,s,0 (R n ). Therefore R n U 2 (s)
|x| s dx = 1 and U ∈ D 1,2 (R n ) is an extremal for µ γ,s,0 (R n ). We now show that (54) lim

α→+∞ u α = 0 in C 0 loc (Ω \ {0}).
Indeed, when s > 0, we have already noted that the result follows from subcriticality. If however s = 0, it then follows from the convergence to U that lim

R→+∞ lim α→+∞ B Rµα (0)\B R -1 µα (0) u 2 (s) α |x| s dx (55) = lim R→+∞ lim α→+∞ B R (0)\B R -1 (0) v 2 (s) α |x| s dx = lim R→+∞ B R (0)\B R -1 (0) U 2 (s) |x| s dx = 1.
Therefore, for any δ > 0, we have that lim α→+∞ Ω\B δ (0)

u 2 (s)
α |x| s dx = 0. We then rewrite (37) as -∆u α = f α u α in Ω \ B δ (0) where lim α→0 f α n/2 = 0. It then follows from the classical deGiorgi-Nash-Moser iterative scheme that (u α ) is uniformly bounded in C 0 loc (Ω \ {0}). Elliptic theory and (39) then yield the convergence to 0. This proves (54).

We now claim that there exists C > 0 such that

(56) |x| n-2 2 u α (x) ≤ C for all x ∈ Ω \ {0} and α ∈ N.
We argue by contradiction and we let (y α ) α ∈ Ω \ {0} be such that (57) sup

x∈Ω\{0} |x| n-2 2 u α (x) = |y α | n-2 2 u α (y α ) → +∞ as α → +∞.
Note that it follows from (42) that (y α ) α is well-defined, and from (43), ( 44), ( 48) and ( 54) that where

ν α := u α (y α ) -2 n-2 → 0 as α → +∞. We define ũα (x) := ν n-2 2 α u α (y α +ν α x) for x ∈ ν -1 α Ω\{0}. Equation (37) rewrites (59) -∆ũ α - γ | yα να + x| 2 + ν 2 α a α (y α + ν α x) ũα = λ α ũ2 (s)-1 α | yα να + x| s in ν -1 α Ω \ {0}.
It follows from the definition (57) that for any R > 0, ũα ≤ 2 in B R (0) for α > 0 large enough. Since ũα (0) = 1, elliptic theory yields the existence of ũ ∈ C 2 (R n ) such that ũα → ũ > 0 in C 2 loc (R n ) as α → +∞. In addition, for all R > 0, we have with Sobolev's inequality that

B R (0) ũ2 α dx = B Rνα (yα) u 2 α dx ≤ Ω u 2 α dx ≤ C
and therefore, letting α → +∞ and R → +∞, we get that ũ ∈ L 2 (R n ). We now distinguish two cases:

If s > 0, then passing to the limit in (59), we get that ∆ũ = 0 in R n and ũ > 0 is bounded. Liouville's theorem then yields ũ ≡ ũ(0

) = 1, contradicting ũ ∈ L 2 (R n ).
If s = 0, then it follows from (38) and (55) that lim R→+∞ lim α→+∞ Ω\B Rµα (0)

u 2 α dx = 0.
It follows from (58) that for α > 0 large enough, then B Rνα (y α ) ∩ B Rµα (0) = ∅, and therefore, we have that lim R→+∞ lim α→+∞ B Rνα (yα) u 2 α dx = 0, which yields ũ ≡ 0, contradicting ũ(0) = 1. This proves (56).

We now claim that (60) lim

R→+∞ lim α→+∞ sup x∈Ω\B Rµα (0) |x| n-2 2 u α (x) = 0.
We just sketch the proof which is very similar to the proof of (56). Arguing by contradiction and letting (y

α ) α ∈ Ω be such that µ -1 α |y α | → +∞ as α → +∞ and |y α | n-2 2 u α (y α ) → c > 0.
We rescale at y α and we get that our hypothesis yields the persistence of some energy outside B Rµα (0) for R and α large, which is a contradiction.

We now prove that for any > 0 small, there exists C > 0 such that (61)

u α (x) ≤ C µ β + (γ)-β -(γ) 2 - α |x| β+(γ)- for all x ∈ Ω \ B µα (0).
Note first that in view of (56), it is enough to prove (61) in Ω \ B Rµα (0) for R > 0 large. For that, fix γ such that γ < γ < (n-2) 2

4

, and let Ω be a smooth bounded domain of R n such that Ω ⊂⊂ Ω is relatively compact in Ω . We extend (a α ) α and a ∞ on Ω such that (35) holds on Ω .

Let G α be the Green's function on Ω at x α of -∆ -γ |x| 2 + a ∞ + ν , where ν > 0 and Dirichlet boundary condition. Up to taking γ close to γ, ν small enough and Ω close to Ω, the operator is coercive and the Green's function is well defined on Ω \ {0, x α }. Theorem 6 in Appendix A then yields a C > 0 such that for any α ∈ N

(62) 0 < G α (x) ≤ C max{|x α |, |x|} min{|x α |, |x|} β-(γ ) |x -x α | 2-n for all x ∈ Ω \ {0, x α }.
Define the operator

L α := -∆ - γ |x| 2 + a α -λ α u 2 (s)-2 α |x| s . It follows from (48) that there exists R 0 > 1 such that |x α | ≤ (R 0 -1)µ α for all α ∈ N. It is easy to check that for x ∈ Ω \ B R0µα (0), L α G α G α (x) = γ -γ |x| 2 + (a ∞ (x) -a α (x)) + ν -λ α u 2 (s)-2 α (x) |x| s .
It follows from (35) that there exists α 0 > 0 such that a ∞ (x) -a α (x) ≥ -ν/2 for all α > α 0 and all x ∈ Ω. For a fixed δ > 0, (60) yields R > R 0 such that for α > 0 large enough, we have that u α (x) ≤ δ|x| -(n-2)/2 for x ∈ Ω \ B Rµα (0). Therefore, with (36), we get that for x ∈ Ω \ B Rµα (0),

L α G α G α (x) > 1 |x| 2 γ -γ -µ γ,s,0 (R n )δ 2 (s)-2s + o(1) .
Up to taking δ > 0 small enough, we then get that L α G α > 0 in Ω \ B Rµα (0). It follows from ( 52) and ( 130) that there exists c(R) > 0 such that

u α (x) ≤ c(R)µ n-2 2 α G α (x) for all x ∈ ∂B Rµα (0) and α ∈ N.
Therefore, defining

h α := c(R)µ n-2 2 α G α -u α , we get that L α h α > 0 in Ω \ B Rµα (0) and h α ≥ 0 in ∂(Ω\B Rµα (0)). Since G α > 0 in Ω \ B Rµα (0) and L α G α > 0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [3] that L α satisfies the comparison principle on Ω \ B Rµα (x α ). Therefore, u α ≤ c(R)µ n-2 2 α G α in Ω \ B Rµα (0)
. With the pointwise control (62), we then get that

u α (x) ≤ C(R) µ β + (γ )-β -(γ ) 2 α |x| β+(γ ) for all x ∈ Ω \ B Rµα (0)
Since this is valid for any γ > γ close to γ, with the remark made at the beginning of the proof, we get (61).

We now claim that there exists C > 0 such that (63)

u α (x) ≤ C µ β + (γ)-β -(γ) 2 α |x| β+(γ) for all x ∈ Ω \ B µα (0).
Indeed, as argued above, the result holds on B Rµα (0) \ B µα (0) for any R > 1. In order to establish (63), we will prove it for any sequence (z α ) α ∈ Ω such that (64) lim

α→+∞ |z α | µ α = +∞.
Let G α be the Green's function of -∆ -(γ|x| -2 + a α ) on Ω with Dirichlet boundary condition. Green's representation formula in Appendix A, and the pointwise control (61) yield

u α (z α ) = Ω G α (z α , x)λ α u 2 (s)-1 α (x) |x| s dy ≤ C Ω max{|z α |, |x|} min{|z α |, |x|} β-(γ) |x -z α | 2-n u 2 (s)-1 α (x) |x| s dx.
We split Ω into four subdomains. On D 1,α (R) := B Rµα (0), we have from (64), ( 61) and (52) that

D1,α max{|z α |, |x|} min{|z α |, |x|} β-(γ) |x -z α | 2-n u 2 (s)-1 α (x) |x| s dx ≤ C|z α | -β+(γ) Bµ α (0) u 2 (s)-1 α |x| s+β-dx ≤ C|z α | -β+(γ) µ β + (γ)-β -(γ) 2 α . Let D 2,α (R) := {Rµ α < |x| < 1 2 |z α |}, and note that |x -z α | > 1 2 |z α | for all x ∈ D 2,α (
R). Therefore, taking > 0 sufficienty small in (61), we have that

D2,α max{|z α |, |x|} min{|z α |, |x|} β-(γ) |x -z α | 2-n u 2 (s)-1 α (x) |x| s dx ≤ C|z α | -β+(γ) µ ( β + (γ)-β -(γ) 2 -)(2 (s)-1) α D2,α |x| -s-β-(γ)-(2 (s)-1)(β+(γ)-) dx ≤ θ(R)|z α | -β+(γ) µ β + (γ)-β -(γ) 2 α , as α → +∞, where lim R→+∞ θ(R) = 0. Set D 3,α := { 1 2 |z α | < |x| < 2|z α |},
and by using again (61) with > 0 sufficiently small, we get that

D3,α max{|z α |, |x|} min{|z α |, |x|} β-(γ) |x -z α | 2-n u 2 (s)-1 α (x) |x| s dx ≤ Cµ ( β + (γ)-β -(γ) 2 -)(2 (s)-1) α |z α | -s-(β+(γ)-)(2 (s)-1) D3,α |x -z α | 2-n dx ≤ Cµ β + (γ)-β -(γ) 2 α |z α | -β+(γ) µ α |z α | β + (γ)-β -(γ) 2 (2 (s)-2)-(2 (s)-1)
.

Finally, let D 4,α := {|x| ≥ 2|z α |} ∩ Ω. Since |x -z α | ≥ |x|/2, then using (61) with > 0 sufficiently small, we get that

D4,α max{|z α |, |x|} min{|z α |, |x|} β-(γ) |x -z α | 2-n u 2 (s)-1 α (x) |x| s dx ≤ C|z α | -β-(γ) µ ( β + (γ)-β -(γ) 2 -)(2 (s)-1) α D4,α |x| -s-β+(γ)2 (s)+ (2 (s)-1) dx ≤ Cµ β + (γ)-β -(γ) 2 α |z α | -β+(γ) µ α |z α | β + (γ)-β -(γ) 2 (2 (s)-2)-(2 (s)-1)
.

Plugging together these estimates yields (63).

Since U is a positive solution to (53) and U ∈ D 1,2 (R n ), it follows from the regularity Theorem 8 that there exists C 1 > 0 such that U (x) C 1 |x| -β-as s → 0. Taking the Kelvin transform Ũ (x) := |x| 2-n U (x|x| -2 ), we get that Ũ ∈ D 1,2 (R n ) is also a positive solution to (53), and enjoys a similar behavior at 0. Transforming back yields the existence of

C 1 , C 2 > 0 such that (65) U (x) x→0 C 1 |x| β-(γ) and U (x) |x|→∞ C 1 |x| β+(γ) .
We now show that there exists H ∈ C 2 (Ω \ {0}) such that (66) lim

α→+∞ u α µ β + (γ)-β -(γ) 2 α = H in C 2 loc (Ω \ {0}),
and H is a solution to (67)

     -∆H -γ |x| 2 + a ∞ H = 0 in Ω \ {0} H > 0 in Ω \ {0} H = 0 on ∂Ω. Define w α := µ - β + (γ)-β -(γ) 2
α u α . Equation (37) then rewrites as (68)

     -∆w α -γ |x| 2 + a α w α = λ α µ (2 (s)-2) β + (γ)-β -(γ) 2 α w 2 (s)-1 α |x| s
in Ω w α ≥ 0 a.e. in Ω w α = 0 on ∂Ω, and (63) yields that w α (x) ≤ C|x| -β+(γ) for all x ∈ Ω \ {0} and α ∈ N. It then follows from elliptic theory that there exists H ∈ C2 (Ω \ {0}) such that lim α→+∞ w α = H in C 2 loc (Ω \ {0}). Passing to the limit in (68) yields H ≥ 0 and

-∆H - γ |x| 2 + a ∞ H = 0 in Ω \ {0} and H = 0 on ∂Ω.
Fix x ∈ Ω \ {0}. Green's representation formula, the positivity of G α and a change of variable yields

u α (x) = Ω G α (x, y)λ α u 2 (s)-1 α (y) |y| s dy ≥ B2µ α (0)\Bµ α (0) G α (x, y)λ α u (y) |y| s dy ≥ µ n-2 2 α B2(0)\B1(0) G α (x, µ α y)λ α v α (y) 2 (s)-1 (y) |y| s dy.
The asymptotics (128) in Appendix A yields G α (x, z) ≥ c x |z| -β-(γ) for all α ∈ N and all z ∈ B |x|/2 (0). Therefore, we get that for all α ∈ N,

u α (x) ≥ c x µ β + (γ)-β -(γ) 2 α B2(0)\B1(0)
|y| -β-(γ) v α (y) 2 (s)-1 (y) |y| s dy

Passing to the limit as α → +∞ and using (52) yields H(x) > 0, which proves our claim in (67).

Let now δ > 0 be such that B δ (0) ⊂ Ω. For any 0 < < δ, the Pohozaev identity (41) with p = 0, and equation (37) yield

- B δ (0)\B (0) a α + x i ∂ i a α 2 u 2 α dx = ∂(B δ (0)\B (0)) B α (x) dx, (69)
where

B α (x) := (x, ν) |∇u α | 2 2 - γ |x| 2 + a α u 2 α 2 - λ α u 2 (s) α 2 (s)|x| s -x i ∂ i u α + n -2 2 u α ∂ ν u α .
Using the asymptotics (42), we pass to the limit as → 0 and get

- B δ (0) a α + x i ∂ i a α 2 u 2 α dx = ∂B δ (0)
B α (x) dx.

The limit (66) yields

lim α→+∞ µ -(β+(γ)-β-(γ)) α ∂B δ (0) B α (x) dσ (70) = ∂B δ (0) (x, ν) |∇H| 2 2 - γ |x| 2 + a ∞ H 2 2 -x i ∂ i H + n -2 2 H ∂ ν H dσ. Assuming now that β + (γ) -β -(γ) > 2, we show that (71) a ∞ (0) = 0.
Indeed, note first that in this case,

β + (γ) > n 2 . It follows from (63) that lim R→+∞ lim α→0 µ -2 α B δ (0)\B Rµα (0) a α + x i ∂ i a α 2 u 2 α dx = 0.
With a change of variable, we get that

B Rµα (0) a α + x i ∂ i a α 2 u 2 α dx = µ 2 α B R (0) a α + x i ∂ i a α 2 (µ α x)v 2 α dx.
The limit (52) and the compactness of the embedding

H 1 → L 2 yields the convergence of v α to U in L 2 loc (R n ). It follows from (65) that U ∈ L 2 (R n ), the two preceding identities therefore yield lim α→0 µ -2 α B δ (0) a α + x i ∂ i a α 2 u 2 α dx = a ∞ (0) R n U 2 dx.
Plugging this limit in the Pohozaev identity (69) and using the limit above yields that a ∞ (0) = O(µ

β+(γ)-β-(γ)-2 α
) + o(1) as α → +∞, and therefore a ∞ (0) = 0.

We now assume that β + (γ) -β -(γ) = 2, and we show again that (72) a ∞ (0) = 0.

Indeed, assume that a ∞ (0) = 0. Without loss of generality, we can suppose that a ∞ (0) > 0. Up to taking δ > 0 smaller and α large, we have that a α (x) + x i ∂iaα(x)

2

≥ a∞(0) 2 for x ∈ B δ (0). It then follows from (69) and (70) that there exists C > 0 such that

B Rµα (0) u 2 α dx ≤ B δ (0) u 2 α dx ≤ Cµ 2 α for all α ∈ N.
With a change of variable, the limit (52), letting α → +∞ and then R → +∞, we get that U ∈ L 2 (R n ), which is impossible due to (65) and 2β + (γ) = n. Therefore a ∞ (0) = 0.

Finally, we show that if

β + (γ) -β -(γ) < 2, then (73) m γ,a∞ (Ω) = 0,
where m γ,a∞ (Ω) is the Hardy-singular mass as defined in Proposition 3. Indeed, since 2β + (γ) < n, we have that

B δ (0) a α + x i ∂ i a α 2 u 2 α dx = O B δ (0) µ β+(γ)-β-(γ) α |x| -2β+(γ) dx = O µ β+(γ)-β-(γ) α δ n-2β+(γ) ,
uniformly with respect to α and δ > 0. Combining with (70), we get that (74) lim

δ→0 ∂B δ (0) (x, ν) |∇H| 2 2 - γ |x| 2 + a ∞ H 2 2 -x i ∂ i H + n -2 2 H ∂ ν H dσ = 0. Since β + (γ) -β -(γ) < 2
, it follows from the definition of the mass that there exists c > 0 such that

H(x) = c 1 |x| β+(γ) + m γ,a∞ (Ω) |x| β-(γ) + o 1 |x| β-(γ)
as x → 0.

Since H solve the equation ( 12), standard elliptic theory yields that this estimate can be differentiated. Therefore, putting it into (74) yields m γ,a∞ (Ω) = 0.

Theorem 4 is a consequence of (71), (72), and (73).

Blow-Up analysis in the merely singular case

In this section, we perform the blow-up analysis in the merely singular case, that is when s = 0 and γ < 0.

We let again (a α ) α∈N ∈ C 1 (Ω), a ∞ ∈ C 1 (Ω), (λ α ) α ∈ (0, +∞) such that (35) and (36) hold. We let (u α ) α ∈ H 1 0 (Ω) be a sequence of weak solutions to (37) such that (38) holds. In this case, (37) and (38) rewrite as:

(75)      -∆u α + |γ| |x| 2 -a α u α = λ α u 2 -1 α in Ω u α ≥ 0 a.e. in Ω u α = 0 on ∂Ω and (76) Ω u 2 α dx = 1.
We suppose that (77) u α 0 as α → +∞ weakly in H 1 0 (Ω). We let Ω be a smooth bounded domain of R n such that Ω ⊂⊂ Ω is relatively compact in Ω . We extend (a α ) α and a ∞ on Ω such that (35) holds on Ω and that the operator -∆ -(γ|x| -2 + a ∞ ) is coercive on Ω . This assumption is equivalent to saying that there exists c > 0 such that for α ∈ N large enough, we have

(78) λ 1 (-∆ -(γ|x| -2 + a α )) = inf ϕ∈H 1 0 (Ω)\{0} Ω |∇ϕ| 2 -γ |x| 2 + a α ϕ 2 dx Ω ϕ 2 dx ≥ c > 0.
This section is devoted to the proof of the following result:

Theorem 5.
Let Ω be a smooth bounded domain of R n , n ≥ 3, such that 0 ∈ Ω is an interior point. Fix γ < 0 and let (a α ) α ∈ C 1 (Ω), (λ α ) α ∈ (0, +∞) and (u α ) α ∈ H 1 0 (Ω) be such that (35), (36), (75) and (76) hold. We let (x α ) α ∈ Ω and (µ α ) α ∈ (0, +∞) be such that u α (x α ) := sup Ω u α = µ

-n-2 2 α .
Then lim α→+∞ x α = x 0 ∈ Ω, lim α→+∞ µ α = 0 and i) If n ≥ 4, then x 0 = 0 and a ∞ (x 0 ) + γ |x0| 2 = 0; ii) If n = 3, then x 0 ∈ Ω \ {0} and R γ,a∞ (Ω, x 0 ) = 0 (see [START_REF] Caffarelli | Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth[END_REF] for the definition). In addition, there exists C > 0 such that

u α (x) ≤ C µ α µ 2 α + |x -x α | 2 n-2 2
for all x ∈ Ω and α ∈ N.

Before delving into the proof, it is important to note a few observations that are relevant for the case s = 0 and γ < 0. First note that in this case β -(γ) < 0, and therefore, it follows from (42) that for any α ∈ N, u α can be extended continuously at 0 by 0, which means that we can and will consider u α ∈ C 0 (Ω). In the definition (43), we shall take τ := 0 and therefore, the sequence (x α ) α ∈ Ω will be such that Another remark is that (75) implies

-∆u α -a α u α ≤ λ α u 2 -1 α in Ω,
which means that u α is a subsolution of a nonlinear elliptic inequation with no Hardy potential term. We shall then be able to perform a blow-up analysis in the spirit of Druet-Hebey-Robert [START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF] to obtain a pointwise control of u α by a standard bubble. The conclusion of Theorem 5 will then follow from classical arguments via the Pohozaev identity and the analysis on the boundary in the spirit of Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF]. 75) and (76) then rewrites (81)

Set v α (x) := µ n-2 2 α u α (x α + µ α x) for all x ∈ µ -1 α (Ω -x α ). Equation (
-∆v α + |γ| | xα µα + x| 2 -µ 2 α a α (x α + µ α x) v α = λ α v 2 -1 α in Ω -x α µ α (82) Ω-xα µα v 2 α dx = 1.
We first claim that (83) lim

α→+∞ |x α | µ α = +∞, and (84)
lim α→+∞ v α = v :=   1 1 + |•| 2 n(n-2)K(n,2) 2   n-2 2 in C 2 loc (R n ) with R n v 2 dx = 1.
Indeed, it follows from (80) that for any R > 0, there exists

α 0 > 0 such that B R (0) ⊂ Ω-xα µα for all α > α 0 . Since (u α ) α is uniformly bounded in H 1 0 (Ω), then (v α ) α is bounded in H 1 loc (R n ). Up to extracting a subsequence, there exists v ∈ H 1 loc (R n ) such that v α v as α → +∞ weakly in H 1 loc (R n ) and strongly in L 2 loc (R n ). Since -∆v α -µ 2 α a α (x α + µ α x)v α ≤ λ α v 2 -1 α in B R (0),
and 0 ≤ v α ≤ 1, it follows from DeGiorgi-Nash-Moser iterative scheme (see for instance Theorem 4.1 in Han-Lin [START_REF] Han | Elliptic partial differential equations[END_REF]), that there exists C > 0 such that for all α > α 0 ,

1 = |v α (0)| ≤ C v α L 2 (B1(0))
and therefore, passing to the strong limit in L 2 , we get that 1 ≤ C v L 2 (B1(0)) , and hence v ≡ 0. Since 0

< v α ≤ v α (0) = 1, equation (81) and elliptic theory yields v ∈ C 2 (R n \ {θ ∞ }) and v α → v in C 2 loc (R n \ {θ ∞ }) with (85) -∆v + |γ| |x -θ ∞ | 2 v = µ γ,s,0 (R n )v 2 -1 in R n \ {θ ∞ } where θ ∞ := -lim α→+∞ µ -1 α x α if this limit is finite. Otherwise θ ∞ := ∞, in which case R n \{θ ∞ } := R n .
In addition, passing to the weak limit in (82) yields

R n v 2 dx = lim R→+∞ B R (0) v 2 dx ≤ lim R→+∞ lim α→+∞ B R (0) v 2 α dx ≤ 1. Since B R (0) |∇v α | 2 dx = B Rµα (xα) |∇u α | 2 dx ≤ C
uniformy for all R > 0 and α > 0 large enough, passing to the weak limit 85) by v and integrating, we obtain

yields |∇v| ∈ L 2 (R n ). Since v ∈ L 2 (R n ), classical arguments yield that v ∈ D 1,2 (R n ). Multiplying (
R n |∇v| 2 dx ≤ R n |∇v| 2 + |γ| |x -θ ∞ | 2 v 2 dx = µ γ,s,0 (R n ) R n v 2 dx.
Since v ≡ 0, the Sobolev inequality yields

R n |∇v| 2 dx R n v 2 dx 2 2 ≥ µ 0,0 (R n ) = µ γ,0 (R n ).

Since

R n v 2 dx ≤ 1 and |γ| > 0, putting these latest inequalities together yields

θ ∞ = ∞ and R n v 2 dx = 1.
We then get (86) lim

α→+∞ |x α | µ α = +∞ and lim α→+∞ v α = v in C 2 loc (R n ) where v ∈ D 1,2 (R n ) ∩ C 2 (R n ) is such that -∆v = µ 0,0 (R n )v 2 -1 in R n ; R n v 2 dx = 1 ; 0 ≤ v ≤ v(0) = 1.
Then ( 83) and (84) follow from (86), this latest assertion and the classification of Caffarelli-Gidas-Spruck [START_REF] Caffarelli | Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth[END_REF].

We now claim that there exists C > 0 such that (87)

u α (x) ≤ µ α µ 2 α + |x -x α | 2 n-2 2
for all α ∈ N and x ∈ Ω.

This estimate is by now standard and is in the spirit of similar results obtained by several authors. See for instance Druet-Hebey-Robert [START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF] and the several references therein. When possible, we shall only sketch an outline to the proof and we refer to these references for details. Note first that (88) lim

R→+∞ lim α→+∞ Ω\B Rµα (xα) u 2 α dx = 0.
Indeed, the convergence of (v α ) α to v in (84) yields that asymptotically, B Rµα (x α ) exhausts almost all the energy in (76).

Next, we claim that there exists C > 0 such that

|x -x α | n-2
2 u α (x) ≤ C for all α ∈ N and x ∈ Ω. Indeed, if not we find (y α ) α ∈ Ω that achieve the supremum of the left-hand-side and which go to +∞ as α → +∞. The same blow-up procedure as above at y α yields that asymptotically, B uα(yα) -2/(n-2) (y α ) carries a nonzero mass of u 2 α dx, contradicting (88), since this ball is disjoint from B Rµα (x α ) for R and α large. A similar argument -that we omit-yields that (89) lim

R→+∞ lim α→+∞ sup x∈Ω\B Rµα (xα) |x -x α | n-2 2 u α (x) = 0. Let now η 0 ∈ C ∞ (R) be such that 0 ≤ η 0 ≤ 1, η 0 (t) = 0 if t ≤ 1 and η 0 (t) = 1 if t ≥ 2.
We define η (x) := η 0 (|x|/ ) for x ∈ R n . We claim that there exists > 0 such that

(90) -∆ -η (x)γ|x| -2 -a ∞ -c/2 is coercive.
To prove this claim, we shall need the following continuity lemma for the first eigenvalue. Recall that for any V : Ω → R measurable such that for some C > 0, we have |x| 2 |V (x)| ≤ C for a.e.

x ∈ Ω, the following ratio

λ 1 (-∆ + V ) := inf ϕ∈H 1 0 (Ω)\{0} Ω (|∇ϕ| 2 + V ϕ 2 ) dx Ω ϕ 2 dx is well defined and is finite. Lemma 1. Let Ω ⊂ R n , n ≥ 3, be a smooth bounded domain. Let (V k ) k : Ω → R and V ∞ : Ω → R
be measurable functions and let (x k ) k ∈ Ω be a sequence of points. We assume that i)

lim k→+∞ V k (x) = V ∞ (x) for a.e. x ∈ Ω, ii) There exists C > 0 such that |V k (x)| ≤ C|x -x k | -2 for all k ∈ N and x ∈ Ω. iii) lim k→+∞ x k = 0 ∈ Ω. iv) For some γ 0 < (n -2) 2 /4, there exists δ > 0 such that |V k (x)| ≤ γ 0 |x -x k | -2 for all k ∈ N and x ∈ B δ (0) ⊂ Ω. v) The first eigenvalue λ 1 (-∆ + V k ) is achieved for all k ∈ N. Then lim k→+∞ λ 1 (-∆ + V k ) = λ 1 (-∆ + V ∞ ).
Proof of Lemma 1: We first claim that (λ 1 (-∆ + V k )) k is bounded. Indeed, fix ϕ ∈ H 1 0 (Ω) \ {0} and use the Hardy inequality to write for all k ∈ N,

λ 1 (-∆ + V k ) ≤ Ω (|∇ϕ| 2 + V k ϕ 2 ) dx Ω ϕ 2 dx ≤ Ω (|∇ϕ| 2 + C|x -x k | -2 ϕ 2 ) dx Ω ϕ 2 dx := M < +∞
For the lower bound, we have for any ϕ ∈ H 1 0 (Ω),

Ω (|∇ϕ| 2 + V k ϕ 2 ) dx = Ω |∇ϕ| 2 dx + B δ (0) V k ϕ 2 dx + Ω\B δ (0) V k ϕ 2 dx ≥ Ω |∇ϕ| 2 dx -γ 0 B δ (0) |x -x k | -2 ϕ 2 dx -4Cδ -2 Ω\B δ (0) ϕ 2 dx ≥ 1 -4γ 0 /(n -2) 2 Ω |∇ϕ| 2 dx -4Cδ -2 Ω ϕ 2 dx. ( 91
)
Since γ 0 < (n -2) 2 /4, we then get that λ 1 (-∆ + V k ) ≥ -4Cδ -2 for large k, which proves the lower bound. Up to a subsequence, we can now assume that (λ 1 (-∆ + V k )) k converges as k → +∞. We now show that

(92) lim inf k→+∞ λ 1 (-∆ + V k ) ≥ λ 1 (-∆ + V ∞ ). For k ∈ N, we let ϕ k ∈ H 1 0 (Ω) be a minimizer of λ 1 (-∆ + V k ) such that Ω ϕ 2 k dx = 1. In particular, (93) -∆ϕ k + V k ϕ k = λ 1 (-∆ + V k )ϕ k weakly in H 1 0 (Ω). Inequality (91) above yields the boundedness of (ϕ k ) k in H 1 0 (Ω). Up to a subsequence, we let ϕ ∈ H 1 0 (Ω) such that, as k → +∞, ϕ k ϕ weakly in H 1 0 (Ω), ϕ k → ϕ strongly in L 2 (Ω) (then Ω ϕ 2 dx = 1) and ϕ k (x) → ϕ(x) for a.e. x ∈ Ω. Letting k → +∞ in (93), the hypothesis on (V k ) allow us to conclude that -∆ϕ + V ∞ ϕ k = lim k→+∞ λ 1 (-∆ + V k )ϕ weakly in H 1 0 (Ω).
Since Ω ϕ 2 dx = 1 and we have extracted subsequences, we then get (92).

Finally, we prove the reverse inequality. For > 0, let ϕ ∈ H 1 0 (Ω) be such that

Ω (|∇ϕ| 2 + V ∞ ϕ 2 ) dx Ω ϕ 2 dx ≤ λ 1 (-∆ + a ∞ ) + .
We have

λ 1 (-∆ + V k ) ≤ λ 1 (-∆ + V ∞ ) + + Ω |V k -V ∞ |ϕ 2 dx Ω ϕ 2 dx
.

The hypothesis of Lemma 1 allow us to conclude that

Ω |V k -V ∞ |ϕ 2 dx → 0 as k → +∞. Therefore lim sup k→+∞ λ 1 (-∆ + V k ) ≤ λ 1 (-∆ + V ∞ ) + for all > 0.
Letting → 0, we get the reverse inequality and the conclusion of Lemma 1.

We now prove (90). First note that the coercivity property (78) yield

λ 1 (-∆ -γ|x| -2 -a ∞ -c/2) ≥ c/2. Define V (x) := -η (x)γ|x| -2 -a ∞ -c/2 for all x ∈ Ω and > 0. Since V ∈ C 0 (Ω), the eigenvalue λ 1 (-∆ + V ) is achieved. It then follows from Lemma 1 that λ 1 (-∆ + V ) → λ 1 (-∆ -γ|x| -2 -a ∞ - c/2) ≥ c/2 as → 0. Therefore, there exists > 0 such that λ 1 (-∆ -η (x)γ|x| -2 -a ∞ -c/2) > 0.
This proves (90).

Fix now ν ∈ (0, 1). We claim that there exists C ν , R ν > 0 such that (94)

u α (x) ≤ C ν µ n-2 2 -ν(n-2) α |x -x α | -(n-2)(1-ν) for all x ∈ Ω \ B Rν µα (x α ).
Since the proof is similar to Step 6.3 (p1228) in Ghoussoub-Robert [START_REF] Ghoussoub | The effect of curvature on the best constant in the Hardy-Sobolev inequalities[END_REF], we just give the main points and leave the details to the reader. We let G be the Green's function of the operator -∆ -η (x)γ|x| -2 -a ∞ -c/2 with Dirichlet boundary condition on Ω . Since x α ∈ Ω ⊂⊂ Ω for all α ∈ N, it follows from classical properties of the Green's function (see for instance [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]) that there exists c 1 > 0 and δ > 0 such that

(95) |∇G (x, x α )| G (x, x α ) ≥ c 1 |x -x α | for all x ∈ B δ (x α ) ⊂ Ω . and (96) c -1 1 ≥ |x -x α | n-2 G (x, x α ) ≥ c 1 for all x ∈ B δ (x α ) ⊂ Ω .
Consider the operator

L α := -∆ -(γ|x| -2 + a α ) -λ α u 2 -2 α . Straightforward computations yield L α G 1-ν (x α , •) G 1-ν (x α , •) = ν(1 -ν) |∇G (x, x α )| 2 G (x, x α ) 2 + |γ|(1 -η ) + ν|γ|η |x| 2 + c(1 -ν) 2 + a ∞ -a α -νa ∞ -λ α u 2 -2 α ≥ ν(1 -ν) |∇G (x, x α )| 2 G (x, x α ) 2 + c(1 -ν) 2 -a α -a ∞ ∞ -ν a ∞ ∞ -λ α u 2 -2 α .
Writing Ω\B Rµα (x α ) as a subset of the union of B δ (x α )\B Rµα (x α ) and Ω\B δ (x α ), and using ( 89) and (95), we get that there exists

R ν > 0 such that L α G 1-ν (x α , •) > 0 in Ω \ B Rν µα (x α ). It follows from
the convergence (84) and ( 96) that there exists

C ν > 0 such that u α ≤ C ν G 1-ν (x α , •)µ n-2 2 -ν(n-2) α on ∂(Ω \ B Rν µα (x α ))
. Since L α u α = 0, it then follows from the comparison principle of Beresticky-Nirenberg-Varadhan [START_REF] Berestycki | The principal eigenvalue and maximum principle for secondorder elliptic operators in general domains[END_REF] (with an extra care for the singular point 0) that

u α ≤ C ν G 1-ν (x α , •)µ n-2 2 -ν(n-2) α
on Ω \ B Rν µα (x α ). This combined with (96), yield (94).

We now prove the pointwise control claimed in (87).

As a preliminary remark, we note that (94) and the convergence (84) yield that for any ν ∈ (0, 1), there exists C ν > 0 such that (97)

u α (x) ≤ C ν µ n-2 2 -ν(n-2) α (µ α + |x -x α |) n-2-ν(n-2) for all x ∈ Ω.
Letting G α be the Green's function for -∆ -(γ|x| -2 + a α ) with Dirichlet boundary condition on Ω, we get from (129) that there exists C > 0 such that

0 < G α (x, y) ≤ C min{|x|, |y|} max{|x|, |y|} |β-(γ)| |x -y| 2-n ≤ C|x -y| 2-n
for all x, y ∈ Ω, x = y. Here, we have used that β -(γ) < 0 since γ < 0. Green's representation formula in Appendix A and (97) then yield

u α (x) = Ω G α (x, y)λ α u 2 -1 α (y) dy ≤ C Ω |x -y| 2-n µ n-2 2 -ν(n-2) α (µ α + |x -x α |) n-2-ν(n-2) 2 -1 dy.
By estimating this integral as in [START_REF] Ghoussoub | The effect of curvature on the best constant in the Hardy-Sobolev inequalities[END_REF], one gets the pointwise control (87). Now assume that lim α→+∞ x α = 0, and let r α := |x α |. We claim that there exists χ > 0 such that

(98) lim α→0 r n-2 α µ n-2 2 α u α (r α x) = χG θ∞ (x) for all x ∈ R n \ {0, θ ∞ }, where θ ∞ := lim α→+∞ xα |xα| . Moreover, this convergence holds in C 2 loc (R n \ {0, θ ∞ }).
Here, G θ∞ satisfies properties (i) to (iv) of Theorem 7. Indeed, our assumptions and ( 83 It follows from (75) and the pointwise control (87) that

-∆ũ α + |γ| |x| 2 -r 2 α a α (r α x) ũα = λ α µ α r α 2 ũ2 -1 α in Ω µ α , and
(100) 0 < ũα (x) ≤ C|x -θ α | 2-n with θ α := x α |x α | . Elliptic theory yields the existence of ũ ∈ C 2 (R n \ {0, θ ∞ }) such that ũα → ũ in C 2 loc (R n \ {0, θ ∞ }) (101) -∆ũ + |γ| |x| 2 ũ = 0 in R n \ {0, θ ∞ } (102) 0 ≤ ũ(x) ≤ C|x -θ ∞ | 2-n . ( 103
)
We are now aiming for a more precise control of ũ. For that, we consider G α , the Green's function for -∆ -(γ|x| -2 + a α ) in Ω with Dirichlet boundary condition. For x ∈ R n \ {0}, we have for all α ∈ N,

(104) u α (r α x) = Ω G α (r α x, y)λ α u 2 -1 α (y) dy. Since β -(γ) < 0, the estimate (129) yields ũα (x) ≤ C r n-2 α µ n-2 2 α B R (0) min{|r α x|, |y|} max{|r α x|, |y|} |β-(γ)| |r α x -y| 2-n µ α µ 2 α + |y -x α | 2 n+2 2 dy ≤ C r n-2 α µ n-2 2 α D 1 α F α (x, y) dy + C r n-2 α µ n-2 2 α D 2 α F α (x, y) dy,
where

F α (x, y) := min{|r α x|, |y|} max{|r α x|, |y|} |β-(γ)| |r α x -y| 2-n µ α µ 2 α + |y -x α | 2 n+2 2 , D 1 α := B R (0) ∩ |r α x -y| > |r α x -x α | 2 and D 2 α := B R (0) ∩ |r α x -y| ≤ |r α x -x α | 2 .
We estimate these two integrals separately. With a change of variable y = x α + µ α z, we get that

r n-2 α µ n-2 2 α D 1 α F α (x, y) dy ≤ |x -θ α | 2-n B 2Rµ -1 α (0) min{|x|, |θ α + µα rα z|} max{|x|, |θ α + µα rα z|} |β-(γ)| 1 1 + |y| 2 n+2 2
dy.

It follows from Lebesgue's convergence theorem that

(105) lim sup α→+∞ r n-2 α µ n-2 2 α D 2 α F α (x, y) dy ≤ C|x -θ ∞ | 2-n min{|x|, 1} max{|x|, 1} |β-(γ)|
For the second integral, we use that |y -

x α | ≥ |r α x -x α |/2 for all y ∈ D 2 α to write r n-2 α µ n-2 2 α D 1 α F α (x, y) dy (106) ≤ µ 2 α r 2 α |x -θ α | -n-2 B Cx (0) min{|x|, |z|} max{|x|, |z|} |β-(γ)| |x -z| 2-n dz,
for some C x > 0. Putting together (105) and (106) and letting α → +∞ yields

(107) ũ(x) ≤ C|x -θ ∞ | 2-n min{|x|, 1} max{|x|, 1} |β-(γ)| for all x ∈ R n \ {0, θ ∞ }.
We now prove a local reverse inequality. Since G α ≥ 0, Green's representation (104), the lower bound (130), the limit lim α→+∞ x α = 0 and a change of variable yield

ũα (x) ≥ r n-2 α µ n-2 2 α Bµ α (xα) G α (r α x, y)λ α u 2 -1 α (y) dy ≥ c r n-2 α µ n-2 2 α Bµ α (xα) min{|r α x|, |y|} max{|r α x|, |y|} |β-(γ)| |r α x -y| 2-n u 2 -1 α (y) dy ≥ c B1(0) min{|r α x|, |x α + µ α z|} max{|r α x|, |x α + µ α z|} |β-(γ)| |x -θ α -µ α r -1 α z| 2-n v 2 -1 α (z) dz Since µ α = o(r α ) as α → +∞, then, for x ∈ R n \ {0, θ ∞ }, as α → +∞, we get that (108) ũ(x) ≥ c min{|x|, 1} max{|x|, 1} |β-(γ)| |x -θ ∞ | 2-n for all x ∈ R n \ {0, θ ∞ }.
In particular, around θ ∞ = 0, ũ is controled from above and below by | • -θ ∞ | 2-n . It then follows from equation (102) and the classical classification of singular solutions of elliptic equations that there exists χ > 0 such that ũ(x) ∼ x→θ∞ χ (n-2)ωn-1|x-θ∞| n-2 . Integrating by parts, it follows from the pointwise control (107) and the equation (102) that

χϕ(θ ∞ ) = R n ũ(x) -∆ϕ - γ |x| 2 ϕ dx for all ϕ ∈ C ∞ c (R n ).
The uniqueness result of Theorem 7 then yields that ũ = χ • G θ∞ . This complete the proof of (98).

Next, we show that (109) lim

α→+∞ x α = x 0 = 0.
Indeed, otherwise we can assume that r α := |x α | → 0 as α → +∞, so that (98) applies. We define ũα as in (99). For δ ∈ (0, 1), the Pohozaev identity (41) applied on B δrα (x α ) ⊂⊂ Ω \ {0} with p := x α and combined with (75) yield

- B δrα (xα) γ (x, x α ) |x| 4 + a α + (x -x α ) i ∂ i a α 2 u 2 α dx = ∂B δrα (xα) (x -x α , ν) |∇u α | 2 2 - γ |x| 2 + a α u 2 α 2 - λ α u 2 α 2 dσ - ∂B δrα (xα) (x -x α ) i ∂ i u α + n -2 2 u α ∂ ν u α dσ = µ α r α n-2 ∂B δ (θα) (x -θ α , ν) |∇ũ α | 2 2 - γ |x| 2 + r 2 α a α (r α x) ũ2 α 2 dσ - µ α r α n ∂B δ (θα) (x -θ α , ν) λ α ũ2 α 2 dσ - µ α r α n-2 ∂B δ (θα) (x -θ α ) i ∂ i ũα + n -2 2 ũα ∂ ν u α dσ. ( 110
)
where θ α is defined in (100). In particular |θ α | = 1.

We first assume that n ≥ 4. The convergence (101) of ũα and δ < 1 yield (111)

B δrα (xα) γ (x, x α ) |x| 4 + a α + (x -x α ) i ∂ i a α 2 u 2 α dx = O µ α r α n-2
The change of variable x = x α + µ α y yield

B δrα (xα) γ (x, x α ) |x| 4 + a α + (x -x α ) i ∂ i a α 2 u 2 α dx (112) = µ 2 α r 2 α B δrα /µα (0) γ (θ α + µ α r -1 α y, θ α ) |θ α + µ α r -1 α y| 4 + r 2 α a α (x α + µ α y) v 2 α dx. + µ 2 α r 2 α B δrα/µα (0) r 2 α µ α y i ∂ i a α (x α + µ α y) 2 v 2 α dx. Since v α (x) ≤ C(1 + |x| 2 ) -1-n/2
from (87), and when n ≥ 5, Lebesgue's convergence theorem yields (113)

B δrα (xα) γ (x, x α ) |x| 4 + a α + (x -x α ) i ∂ i a α 2 u 2 α dx = γ R n v 2 dx + o(1) µ 2 α r 2 α
as α → +∞. Hence γ = 0, (111) and ( 113) yield

1 = O µ α r α n-4 as α → +∞,
which is a contradiction. Now if n = 4, we use that |γ| > 0, (111) and (112) to get that there exists C > 0 such that for any R > 0, we have that

B R (0) v 2 α dx ≤ C for all α.
Letting α → +∞ and then R → +∞ yields R n v 2 dx < +∞, a contradiction with (84) that settles the case n = 4.

We now deal with the case n = 3. With the pointwise control (98), we have that

B δrα (xα) γ (x, x α ) |x| 4 + a α + (x -x α ) i ∂ i a α 2 u 2 α dx ≤ Cr -2 α B δrα (xα) µ α |x -x α | -2 dx ≤ Cδ µ α r α
Plugging this inequality in (110), using the convergence (101) and letting α → +∞ yield that the expression (114)

∂B δ (θ∞) (x -θ ∞ , ν) |∇G θ∞ | 2 2 - γG 2 θ∞ 2|x| 2 -(x -θ ∞ ) i ∂ i G θ∞ + n -2 2 G θ∞ ∂ ν G θ∞ dσ is O(δ) as δ → 0. Since n = 3, there exists β θ∞ ∈ C 2 (R n \ {0}) such that G θ∞ (x) = 1 4π|x -θ ∞ | + β θ∞ (x) for all x ∈ R n \ {0, θ ∞ }.
Letting δ → 0 in (114), classical computations then yield (115)

β θ∞ (θ ∞ ) = 0.
We shall give an integral expression for β θ∞ . Since n = 3, it follows from the pointwise control (150) and from the definition that

β θ∞ ∈ D 1,2 (R n ) and is controled at ∞ by x → |x| -1 . Since -∆β θ∞ -γ|x| -2 β θ∞ = -γ|x| -2 |x -θ ∞ | -1 /4π, integrating by parts yields β ∞ (x) = -γ 4π R 3 G x (y) |y| 2 |y -θ ∞ | dy
for all x ∈ R n . Since γ < 0, we then get that β ∞ > 0, contradicting (115). This proves (109) also when n = 3.

We now show that (116) If x 0 ∈ ∂Ω, then n ≥ 4 and â∞ (x 0 ) = 0, where â∞ :

= a ∞ + γ| • | -2 . Indeed, let U, V ⊂ R n be open sets such that 0 ∈ U , x 0 ∈ V and ϕ : U → V a smooth diffeomorphism such that ϕ(0) = x 0 ; ϕ(U ∩ R n -) = ϕ(U ) ∩ Ω and ϕ(U ∩ ∂R n -) = ϕ(U ) ∩ ∂Ω.
Up to a rotation, we can assume that the differential of ϕ at 0 is

dϕ 0 = Id R n . Let (x α,1 , xα ) ∈ U ∩ R n -= U ∩ ((-∞, 0) × R n-1
) be such that x α = ϕ(x α,1 , xα ). In particular, we have that d(x α , ∂Ω) = ( 1 Moreover, the pointwise control (87) reads

ũα (x) ≤ C x - (x α,1 , 0) d α 2-n for x ∈ (U -(0, xα )) ∩ R n - d α .
It then follows from classical elliptic theory that there exists ũ

∈ C 2 (R n -\ {(-1, 0)}) such that lim α→+∞ ũα = ũ in C 2 loc (R n -\ {(-1, 0)}) ∆ũ = 0 in R n -\ {(-1, 0)} ; ũ|∂R n -≡ 0 0 ≤ ũ(x) ≤ C|x -(-1, 0)| 2-n for all x ∈ R n -\ {(-1, 0)}
By reflecting ũ along the hyperplane {x 1 = 0}, we get a harmonic function on R n \ {(±1, 0)}, which is nonnegative for x 1 < 0, nonpositive for x 1 > 0, and vanishing for x 1 = 0. Therefore, there exists c ≥ 0 such that

ũ(x) = c |x -(-1, 0)| 2-n -|x -(1, 0)| 2-n for all x ∈ R n -\ {(-1, 0)}.
A proof that is similar to the one for (108) and using the pointwise control of the Green's function in 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] -and that we omit it here-gives that c > 0. Fix now 0 < δ < 1 and define ûα (x) :=

d n-2 α µ n-2 2 α u α (x α + d α x) for x ∈ B δ (0).
It follows from the convergence result above that

(118) lim α→+∞ ûα = û := c| • | 2-n + φ in C 2 loc (B δ (0) \ {0}),
where φ(x) = -c|x -(2, 0)| 2-n for all x ∈ B δ (0). We now use the Pohozaev identity (41) on B δdα (x α ), equation ( 75) and an integration by parts, to obtain

- B δdα (xα) âα + (x -p) i ∂ i âα 2 u 2 α dx (119) = ∂B δdα (xα) (x -p, ν) |∇u α | 2 2 - âα u 2 α 2 - λ α u 2 α 2 dσ.
-

∂B δdα (xα) (x -p) i ∂ i u α + n -2 2 u α ∂ ν u α dσ.
Taking p := x α in this identity yields

- B δdα (xα) âα + (x -x α ) i ∂ i âα 2 u 2 α dx = ∂B δdα (xα) (x -x α , ν) |∇u α | 2 2 - âα u 2 α 2 - λ α u 2 α 2 -(x -x α ) i ∂ i u α + n -2 2 u α ∂ ν u α dσ.
With the change of variable x = x α + µ α y in the first integral, and x = x α + d α z in the second integral, we get that

-µ 2 α B δdα µ -1 α (0) âα (x α + µ α y) + µ α y i ∂ i âα (x α + µ α y) 2 v 2 α dx = µ α d α n-2 ∂B δ (0) (z, ν) |∇û α | 2 2 - d 2 α âα (x α + d α z)û 2 α 2 - λ α µ 2 α û2 α 2 d 2 α dσ - µ α d α n-2 ∂B δ (0) z i ∂ i ûα + n -2 2 ûα ∂ ν ûα dσ.
Fix i ∈ {1, ..., n} and differentiate (119) with respect to the i th variable p i to obtain

- B δdα (xα) ∂ i âα 2 u 2 α dx = ∂B δdα (xα) ν i |∇u α | 2 2 - âα u 2 α 2 - λ α u 2 α 2 dσ - ∂B δdα (xα) ∂ i u α ∂ ν u α dσ.
Performing the same changes of variables as above yields

-µ 2 α B δdα µ -1 α (0) ∂ i âα (x α + µ α y)v 2 α dx (120) = d -1 α µ α d α n-2 ∂B δ (0) ν i |∇û α | 2 2 - d 2 α âα (x α + d α z)û 2 α 2 - λ α µ 2 α û2 α 2 d 2 α dσ -d -1 α µ α d α n-2 ∂B δ (0) ∂ i ûα ∂ ν ûα dσ.
With the convergence (118) of ûα and an explicit computation, we get that

lim α→+∞ ∂B δ (0) (z, ν) |∇û α | 2 2 - d 2 α âα (x α + d α z)û 2 α 2 - λ α µ 2 α û2 α 2 d 2 α -z i ∂ i ûα + n -2 2 ûα ∂ ν ûα dσ = ∂B δ (0) (z, ν) |∇û| 2 2 -z i ∂ i û + n -2 2 û ∂ ν û dσ = (n -2) 2 c 2 φ(0) = -(n -2) 2 2 1-n c 2 .
Indeed, the limit is independent of δ since φ is harmonic. Similarly,

lim α→+∞ ∂B δ (0) ν i |∇û α | 2 2 - d 2 α âα (x α + d α z)û 2 α 2 - λ α µ 2 α û2 α 2 d 2 α -∂ i ûα ∂ ν ûα dσ = ∂B δ (0) ν i |∇û| 2 2 -∂ i û∂ ν û dσ (121) = (n -2)cω n-1 ∂ i φ(0) = -2 1-n (n -2) 2 c 2 ω n-1 δ i,1 .
We now divide the analysis in three cases.

Case 1: n ≥ 5. Since v α ≤ C(1 + |x| 2 ) 1-n/2 from (87), and v α → v in C 2 loc (R n ), then Lebesgue's theorem applied to the identities above yields

(122) µ 2 α â∞ (x 0 ) R n v 2 dx + o(1) = µ α d α n-2 • (n -2) 2 2 1-n c 2 + o(1)
and, with i = 1, (123)

µ 2 α ∂ 1 â∞ (x 0 ) R n v 2 dx + o(1) = d -1 α µ α d α n-2 • 2 1-n (n -2) 2 c 2 ω n-1 + o(1)
In particular, we get that â∞ (x 0 ) = 0. Case 2: n = 4. Arguing as in the case n ≥ 5, we get that

µ 2 α (â ∞ (x 0 ) + o(1)) B δdα µ -1 α (0) v 2 α dx + O(1) = µ α d α 2 • (n -2) 2 2 1-n c 2 + o(1)
and for i = 1,

µ 2 α (∂ 1 â∞ (x 0 ) + o(1)) B δdα µ -1 α (0) v 2 α dx + O(1) = d -1 α µ α d α 2 • 2 1-n (n -2) 2 c 2 ω n-1 + o(1) .
Since R n v 2 dx = +∞ when n = 4, here again, we get that â∞ (x 0 ) = 0. Case 3: n = 3. Here we need to show that x 0 / ∈ ∂Ω. Indeed, the uniform control v α ≤ C(1 + |x| 2 ) 1-n/2 , the estimates (120) and (121

) yield O(µ α d α ) = µα dα • -(n -2) 2 2 1-n c 2 and therefore 1 = O(d 2 α
), contradicting (117). The proof of (116) is complete.

Assume now that n ≥ 4 and x 0 ∈ Ω, set for convenience âα (x) := a α (x) + γ|x| -2 . Performing the Pohozaev identity (41) on B δ (x α ) assuming that B 2δ (x α ) ⊂ Ω, we get that

- B δ (xα) âα + (x -x α ) i ∂ i âα 2 u 2 α dx (124) = ∂B δ (xα) (x -x α , ν) |∇u α | 2 2 -âα u 2 α 2 - λ α u 2 α 2 dσ - ∂B δ (xα) (x -x α ) i ∂ i u α + n -2 2 u α ∂ ν u α dσ.
The pointwise control (87) and elliptic theory yield

u α (x) + |∇u α (x)| ≤ Cµ n-2 2 α
for x ∈ ∂B δ (x α ), and therefore as α → +∞,

B δ (xα) âα + (x -x α ) i ∂ i âα 2 u 2 α dx = O(µ n-2 α ).
Arguing as in the cases n ≥ 5 and n = 4 in the proof of ( 122) and (123) above, we then get that

â∞ (x 0 ) = a ∞ (x 0 ) + γ|x 0 | -2 = 0.
Finally, assume that n = 3. It follows from the above that x 0 = 0 and x 0 ∈ ∂Ω. Therefore (â α ) converges to â∞ in C 1 (B 2δ (x 0 )) for some small δ > 0. Passing to the limit as α → +∞ and δ → 0 in (124) above, and performing standard computations (see for instance Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF]), we get that the mass of the operator -∆ -(a ∞ + γ|x| -2 ) vanishes at x 0 . In other words, R γ,a∞ (x 0 ) = 0. This completes the proof of Theorem 5.

Proof of Theorem 3

Again, we start with the truly singular case and prove the following.

Proposition 9.

Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ Ω. Assume that either s > 0 or γ > 0.

If (n-2) 2 4 -1 < γ < (n-2) 2 4
, then (1) λ * (Ω) > 0.

(2) Moreover, if µ γ,s,λ * (Ω) is not achieved, then m γ,λ * (Ω) = 0, and λ * (Ω) = sup{λ; m γ,λ (Ω) ≤ 0}.

Proof: For λ > λ * , the infimum is achieved, and therefore, there exists u λ ∈ H 1 0 (Ω) such that (125)

     -∆u λ -γ |x| 2 + λ u λ = µ γ,s,λ (Ω) u 2 (s)-1 λ |x| s
in Ω u λ ≥ 0 a.e. in Ω u λ = 0 on ∂Ω and ( 126)

Ω u 2 (s) λ |x| s dx = 1.
As one checks, (u λ ) λ>λ * is bounded in H 1 0 (Ω), and therefore, up to extracting a sub-family, it has a weak limit u λ * as λ → λ * . If u λ * ≡ 0, then classical arguments yield it is a minimizer for µ γ,s,λ * (Ω). Suppose now λ * = 0, this means that u λ * is a minimizer for µ γ,s,0 (Ω) = µ γ,s,0 (R n ), which is impossible since u λ * has compact support, hence u λ * ≡ 0. It then follows from Theorem 4 that m γ,λ * (Ω) = m γ,0 (Ω) = 0. To get to a contradiction, we shall now prove that m γ,0 (Ω) < 0. Indeed, let H ∈ C 2 (Ω\{0}) be as in Proposition 3 for h ≡ 0. It follows from the definition of H and the expansion [START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF]

that x → H (x) := H(x) -|x| -β+(γ) ∈ H 2 1 (Ω) ∩ C 2 (Ω \ {0}), it satisfies -∆H -γ|x| -2 H = 0 in Ω \ {0}
and H (x) < 0 for x ∈ ∂Ω. It then follows from the comparison principle that H < 0 in Ω\{0}. Therefore, the expression ( 14) yields that c 2 < 0, and therefore m γ,0 (Ω) < 0. A contradiction that yields that λ * > 0. We now show (2) under the hypothesis that µ γ,s,λ * (Ω) is not achieved. Indeed, under such an assumption, the weak limit u λ * as λ → λ * is necessarily identically zero. It then follows from Theorem 4 that m γ,λ * (Ω) = 0. Finally, let λ := sup{λ; m γ,λ (Ω) ≤ 0}, and note that if λ > λ, then m γ,λ (Ω) > 0 and µ γ,s,λ * (Ω) is achieved in view of Theorem 6, which means that λ ≥ λ * . In other words, λ ≥ λ * . On the other hand, from the strict monotonicity of the mass, if λ > λ * , then m γ, λ(Ω) > m γ,λ * (Ω) = 0, which is a contradiction, hence λ = λ * . . An identical proof that uses Theorem 5 as opposed to Theorem 4, and the mass R γ,λ (Ω) as opposed to m γ,λ (Ω) gives the analogous result in the merely singular case. Note that in this case, the argument of Druet [START_REF] Druet | Elliptic equations with critical Sobolev exponents in dimension 3[END_REF] yields that µ γ,0,λ * (Ω) is not achieved for n = 3, and therefore this hypothesis is readily satisfied. In summary, we have shown the following result. Proposition 10. Assume n = 3, s = 0 and γ ≤ 0. Then

λ * (Ω) = sup{λ; R γ,λ (Ω) ≤ 0} > 0.
Appendix A: Green's function for -∆ -γ|x| -2 -h(x) on a bounded domain Theorem 6. Let Ω be a smooth bounded domain of R n such that 0 ∈ Ω is an interior point. We fix γ < (n-2) 2

4

. We let h ∈ C 0,θ (Ω) be such that -∆ -γ|x| -2 -h is coercive. Then there exists G

: (Ω \ {0}) 2 \ {(x, x)/ x ∈ Ω \ {0}} → R such that for all p ∈ Ω \ {0}, (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 2n n+2 (Ω) ∩ L p loc (Ω -{0}), p > n/2, and all ϕ ∈ H 1 0 (Ω) such that -∆ϕ - γ |x| 2 + h(x) ϕ = f in Ω ; ϕ |∂Ω = 0, then we have that (127) ϕ(p) = Ω G(p, x)f (x) dx
In addition, G > 0 is unique and

(iii) For all p ∈ Ω \ {0}, there exists c 0 (p) > 0 such that

(128) G p (x) ∼ x→0 c 0 (p) |x| β-(γ) and G p (x) ∼ x→p 1 (n -2)ω n-1 |x -p| n-2
(iv) There exists c > 0 such that

(129) 0 < G p (x) ≤ c max{|p|, |x|} min{|p|, |x|} β-(γ)
|x -p| 2-n for x ∈ Ω -{0, p}.

(v) For all ω Ω, there exists c(ω) > 0 such that

(130) c(ω) max{|p|, |x|} min{|p|, |x|} β-(γ) |x -p| 2-n ≤ G p (x) for all p, x ∈ ω \ {0}.
Proof: Fix δ 0 > 0 such that B δ0 (0) ⊂ Ω. We let η (x) := η( -1 |x|) for all x ∈ R n and > 0, where η ∈ C ∞ (R) is nondecreasing and such that η(t) = 0 for t < 1 and η(t) = 1 for t > 1. Set

L := -∆ - γη |x| 2 + h(x) .
It follows from Lemma 1 and the coercivity of -∆ -γ|x| -2 + h that there exists 0 > 0 and c > 0 such that such that for all ϕ ∈ H 1 0 (Ω) and ∈ (0, 0 ),

Ω |∇ϕ| 2 - γη |x| 2 + h(x) ϕ 2 dx ≥ c Ω ϕ 2 dx.
As a consequence, there exists c > 0 such that for all ϕ ∈ H 1 0 (Ω) and ∈ (0, 0 ), ( 131)

Ω |∇ϕ| 2 - γη |x| 2 + h(x) ϕ 2 dx ≥ c ϕ 2 D 1,2 .
Let G > 0 be the Green's function of -∆ -γη |x| -2 + h on Ω with Dirichlet boundary condition. The existence follows from the coercivity and the C 0,θ regularity of the potential for any > 0.

Step 1: Integral bounds for G . We claim that for all δ > 0 and 1 < q < n n-2 and δ ∈ (0, δ), there exists C(δ, q) > 0 and C(δ, δ ) > 0 such that (132) G (x, •) L q (Ω) ≤ C(δ, q) and G (x, •)

L 2n n-2 (Ω\B δ (x)) ≤ C(δ, δ ) for all x ∈ Ω, |x| > δ. Indeed, fix f ∈ C ∞ c (Ω) and let ϕ ∈ C 2,θ (
Ω) be the solution to the boundary value problem

(133) L ϕ = -∆ϕ -γη |x| 2 + h(x) ϕ = f in Ω ϕ = 0 on ∂Ω
Multiplying the equation by ϕ , integrating by parts on Ω, using (131) and Hölder's inequality, we get that

Ω |∇ϕ| 2 dx ≤ C f 2n n+2 ϕ 2n n-2
where C > 0 is independent of , f and ϕ . The Sobolev inequality

ϕ 2n n-2 ≤ C ∇ϕ 2 for ϕ ∈ H 1 0 (Ω) then yields ϕ 2n n-2 ≤ C f 2n n+2
where C > 0 is independent of , f and ϕ .

Fix p > n/2 and δ ∈ (0, δ 0 ) and δ 1 , δ 2 > 0 such that δ 1 + δ 2 < δ, and x ∈ Ω such that |x| > δ. It follows from standard elliptic theory that

|ϕ (x)| ≤ ϕ C 0 (B δ 1 (x)) ≤ C ϕ L 2 (B δ 1 +δ 2 (x)) + f L p (B δ 1 +δ 2 (x)) ≤ C f L 2n n+2 (Ω) + f L p (B δ 1 +δ 2 (x))
where C > 0 depends on p, δ, δ 1 , δ 2 , γ and h ∞ . Therefore, Green's representation formula yields

(134) Ω G (x, •)f dy ≤ C f L 2n n+2 (Ω) + f L p (B δ 1 +δ 2 (x)) for all f ∈ C ∞ c (Ω). It follows from (134) that Ω G (x, •)f dy ≤ C • f L p (Ω)
for all f ∈ C ∞ c (Ω) where p > n/2. It then follows from duality arguments that for any q ∈ (1, n/(n -2)) and any δ > 0, there exists C(δ, q) > 0 such that G (x, •) L q (Ω) ≤ C(δ, q) for all < 0 and x ∈ Ω \ B δ (0). Let δ ∈ (0, δ) and δ 1 , δ 2 > 0 such that δ 1 + δ 2 < δ . We get from (134) that ( 135)

Ω G (x, •)f dy ≤ C f L 2n n+2 (Ω\B δ (x)) for all f ∈ C ∞ c (Ω \ B δ (x)).
Here again, a duality argument yields (132), which proves the claim in Step 1.

Step

2: Convergence of G . Fix x ∈ Ω \ {0}. For 0 < < , since G (x, •), G (x, •) are C 2 outside x, we have -∆(G (x, •) -G (x, •)) - γη | • | 2 + h (G (x, •) -G (x, •)) = γ(η -η ) | • | 2 G (x, •)
in the strong sense. The coercivity (131) then yields

G (x, •) ≥ G (x, •) for 0 < < if γ ≥ 0,
and the reverse inequality if γ < 0. It then follows from the integral bound (132) and elliptic regularity that there exists G(x,

•) ∈ C 2,θ (Ω \ {0, x}) such that lim →0 G (x, •) = G(x, •) in C 2 loc (Ω -{0, x}).
In particular, G is symmetric and

(136) -∆G(x, •) - γ | • | 2 + h G(x, •) = 0 in Ω \ {0, x}.
Moreover, passing to the limit → 0 in (132) and using elliptic regularity, we get that for all δ > 0, 1 < q < n n-2 and δ ∈ (0, δ), there exist C(δ, q) > 0 and C(δ, δ ) > 0 such that for all x ∈ Ω, |x| > δ, (137) G(x, •) L q (Ω) ≤ C(δ, q) and G(x, •)

L 2n n-2 (Ω\B δ (x)) ≤ C(δ, δ ).
Moreover, for any f ∈ L p (Ω), p > n/2, let ϕ ∈ C 2 (Ω) be such that (133) holds, and fix x ∈ Ω \ {0}.

Passing to the limit → 0 in the Green identity ϕ

(x) = Ω G (x, •)f dy yields (138) ϕ(x) = Ω G(x, •)f dy for all x ∈ Ω \ {0} where ϕ ∈ H 1 0 (Ω) ∩ C 0 (Ω \ {0}) is the only weak solution to -∆ϕ -γ |x| 2 + h(x) ϕ = f in Ω ϕ = 0 on ∂Ω
In particular, the strong comparision principle yields G(x, •) > 0 for x ∈ Ω \ {0}.

Step 3: Upper bound for G(x, y) when one variable is far from 0. It follows from (136), elliptic theory and (137) that for any δ > 0, there exists C(δ) > 0 such that

(139) 0 < G(x, y) ≤ C(δ) for x, y ∈ Ω such that |x| > δ, |y| > δ, |x -y| > δ.
We claim that for any δ > 0, there exists C(δ) > 0 such that (140) 0 < |x -y| n-2 G(x, y) ≤ C(δ) for x, y ∈ Ω such that |x| > δ and |y| > δ.

Indeed, with no loss of generality, we can assume that δ ∈ (0, δ 0 ). Define now Ω δ := Ω \ B δ/2 (0), and fix x ∈ Ω such that |x| > δ. Let H x be the Green's function for -∆ -γ |x| 2 + h(x) in Ω δ with Dirichlet boundary condition. Classical estimates (see [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]) yield the existence of C(δ) > 0 such that |x -y| n-2 H x (y) ≤ C(δ) for all x, y ∈ Ω δ . It is easy to check that

     -∆(G x -H x ) -γ |x| 2 + h (G x -H x ) = 0 weakly in Ω δ G x -H x = 0 on ∂Ω G x -H x = G x on ∂B δ/2 (0).
Regularity theory then yields that G Indeed, fix δ 1 < δ and use (139) to deduce that G x (y) ≤ C(δ, δ 1 ) for all x ∈ Ω\B δ (0) and y ∈ ∂B δ1 (0). Since δ 1 < |x|, we have that

x -H x ∈ C 2,θ (Ω δ ). It follows from (139) that G x is
-∆G x -γ |x| 2 + h G x = 0 in H 2 1 (B δ1 (0)) 0 < G x ≤ C(δ, δ ) on ∂B δ1 (0).
It follows from (162) below that for δ 1 > 0 small enough, there exists u

β-∈ H 2 1 (B δ1 (0)) such that c 1 ≤ |z| β-(γ) u β-(z) < c 2 for all z ∈ B δ1 (0), and -∆u β-- γ |x| 2 + h u β-≥ 0 in H 2 1 (B δ1 (0)).
Therefore, there exists C(δ, δ ) > 0 such that G x (z) ≤ C(δ, δ )u β-(z) for all z ∈ ∂B δ1 (0). It then follows from the comparison principle that G x (y) ≤ C(δ, δ )u β-(y) for all y ∈ B δ1 (0) \ {0}.

Combining this with (139), we obtain (141). Note that by symmetry, we also get that for any 0 < δ < δ, there exists C(δ, δ ) > 0 such that (142) |x| β-(γ) G(x, y) ≤ C(δ, δ ) for x, y ∈ Ω such that |y| > δ > δ > |x| > 0.

Step 4: Upper bound for G(x, y) when both variables approach 0.

We claim first that for all c 1 , c 2 , c 3 > 0, there exists C(c 1 , c 2 , c 3 ) > 0 such that for x, y ∈ Ω such that Scaling back and using (143), we get (145) for x ∈ B δ0/2 (0) \ {0}. The general case is a consequence of (140). This ends the proof of (145). We now show that there exists C > 0 such that (146) |y| β-(γ) |x| β+(γ) G(x, y) ≤ C for x, y ∈ Ω such that |y| < 1 2 |x|.

Step 1: Construction of a positive kernel at a given point: For a fixed p 0 ∈ R n \ {0}, we show that there exists G ∈ C 2 (R n \ {0, p 0 }) such that (151) Indeed, let η ∈ C ∞ (R) be a nondecreasing function such that 0 ≤ η ≤ 1, η(t) = 0 for all t ≤ 1 and η(t) = 1 for all t ≥ 2. For > 0, set η (x) := η |x| for all x ∈ R n . For R > 0, we argue as in the proof of (131) to deduce that the operator -∆ -γη |x| 2 is coercive on B R (0) and that there exists c > 0 independent of R, > 0 such that

       -∆G -γ |x| 2 G = 0 in R n \ {0, p 0 } G > 0 G ∈ L
B R (0) |∇ϕ| 2 - γη |x| 2 ϕ 2 dx ≥ c B R (0)
|∇ϕ| 2 dx for all ϕ ∈ C ∞ c (B R (0)).

Consider R, > 0 such that R > 2|p 0 | and < |p0| 6 , and let G R, be the Green's function of -∆ -γη |x| 2 in B R (0) at the point p 0 with Dirichlet boundary condition. We have that G R, > 0 since the operator is coercive.

Fix R 0 > 0 and q ∈ (1, n n-2 ), then by arguing as in the proof of (132), we get that there exists C = C(γ, p 0 , q , R 0 ) such that

   G R, → G ≥ 0 in C 2 loc (R n \ {0, p 0 }) as R → +∞, → 0 -∆G -γ |x| 2 G = 0 in R n \ {0, p 0 } G ∈ L 2n n-2 (B δ (0)) Fix ϕ ∈ C ∞ c (R n ).
For R > 0 large enough, we have that ϕ(p 0 ) = R n G R, (-∆ϕ -γη |x| -2 ϕ) dx. With the integral bounds above, we then get that x → G(x)|x| -2 ∈ L 1 loc (R n ). Therefore, we get

(155) ϕ(p 0 ) = R n G(x) -∆ϕ - γ |x| 2 ϕ dx for all ϕ ∈ C ∞ c (R n ).
As a consequence, G > 0.

Step 2: Asymptotic behavior at 0 and p for solutions to (151). It follows from Theorem 9 below that either G behaves like |x| -β-(γ) or |x| -β+(γ) at 0. Since G ∈ L 2n n-2 (B δ (0)) for some small δ > 0 and β -(γ) < n-2 2 < β + (γ), we get that there exists c > 0 such that (156) lim We then have that H ∈ H 2 1,loc (R n \ {p 0 }) is such that

R n H (x) -∆ϕ - γ |x| 2 ϕ dx = 0 for all ϕ ∈ C ∞ c (R n ).
The ellipticity of the Laplacian then yields that H ∈ C ∞ (R n \ {0}). The pointwise bounds (158) yield that H ∈ D 1,2 (R n ). Multiplying -∆H -γ |x| 2 H = 0 by H , integrating by parts and using the coercivity yields that H ≡ 0, and therefore, G 1 -G 2 = c| • | -β-(γ) . This proves uniqueness.

Step 5: Existence. It follows from Step 3 that, up to substracting a multiple of | • | -β-(γ) , there exists G p0 > 0 satisfying (i), (ii) and the pointwise controls (iii) at p 0 . It is a consequence of (iii) that there exists c > 0 such that

Subsection 4 . 1 :

 41 The truly singular case Proposition 5.

  (79) u α (x α ) := sup x∈Ω u α (x) and µ α := u α (x α ) -2/(n-2) . It then follows from Proposition 8 that (80) lim α→+∞ µ α = 0 and lim α→+∞ d(x α , ∂Ω) µ α = +∞.

2 α

 2 u α (r α x) for x ∈ Ω rα .

2 α

 2 + o(1))|x α,1 | as α → +∞. Set d α := |x α,1 | = -x α,1 , and then x 0 ∈ ∂Ω and (46u α • ϕ((0, xα ) + d α x) for x ∈ (U -(0, xα )) ∩ R n d α ,and for convenience, define for each α ∈ N, the function âα (x) := a α (x) + γ|x| -2 for x ∈ Ω \ {0}. ũα -d 2 α âα • ϕ((0, xα ) + d α x)ũ α = µα dα 2

  bounded by a constant depending only on δ on ∂B δ/2 (0) for |x| > δ. The comparison principle then yields |G x (y) -H x (y)| ≤ C(δ) for y ∈ Ω δ and |x| > δ. The above bound for H x and (139) then yields (140). We now claim that for any 0 < δ < δ, there exists C(δ, δ ) > 0 such that (141) 0 < |y| β-(γ) G(x, y) ≤ C(δ, δ ) for x, y ∈ Ω such that |x| > δ > δ > |y| > 0.

γ |z| 2 ++ z 2 + 2 +

 222 c 1 |x| < |y| < c 2 |x| and |x -y| > c 3 |x|, we have (143) |x -y| n-2 G(x, y) ≤ C(c 1 , c 2 , c 3 ).Indeed, fix x ∈ B δ0/2 (0) \ {0} ⊂ Ω \ {0}, and defineH(z) := G x (|x|z) for z ∈ B δ0/|x| (0) \ 0, |x| 2 h(|x|z) H = 0 in B δ0/|x| (0) \ 0, x |x| .Since H > 0, it follows from the Harnack inequality that for all R > 0 large enough and α > 0 small enough, there exist δ 1 > 0 and C > 0 independent of |x| < δ 1 such thatH(z) ≤ CH(z ) for all z, z ∈ B R (0) \ B α (0) ∪ B α x |x| ,which rewrites as:(144) G x (y) ≤ CG x (y ) for all y, y ∈ B R|x| (0) \ B α|x| (0) ∪ B α|x| (x) .Let u β+ be a sub-solution to (163). In particular, for |x| < δ 2 small, there exists C > 0 such thatG x (z) ≥ c|x| β+(γ) inf ∂B R|x| (0) G x u β+ (z) for all z ∈ ∂B R|x| (0). Since -∆G x -(γ| • | -2 + h)G x =0 outside 0, it follows from coercivity and the comparison principle thatG x (z) ≥ c|x| β+(γ) inf ∂B R|x| (0) G x u β+ (z) for all z ∈ Ω \ B R|x| (0). Fix z 0 ∈ Ω \ {0}.Then for δ 3 small enough, it follows from (142) and the Harnack inequality (144) that there exists C > 0 independent of x such thatG x (y) ≤ C|x| -β+(γ)-β-(γ) for all y ∈ B R|x| (0) \ B α|x| (0) ∪ B α|x| (x)Taking α > 0 small enough and R > 0 large enough, we then get (143) for |x| < δ 3 . The general case for arbitrary x ∈ Ω \ {0} then follows from (140). This prove (143).Next we claim that for all c 1 , c 2 > 0, there existsC(c 1 , c 2 ) > 0 such that (145) |x -y| n-2 G(x, y) ≤ C(c 1 , c 2 ) for x, y ∈ Ω such that c 1 |x| < |y| < c 2 |x|.For that, we fix x ∈ B δ0/2 (0) \ {0} and setH(z) := |x| n-2 G x (x + |x|z) for all z ∈ B 1/2 (0) \ {0}.We have that H ∈ C 2 (B 1/2 (0) \ {0}) and satisfies |x| 2 h(x + |x|z)   H = δ 0 weakly in B 1/2 (0).We now argue as in the proof of (140). From (143), we have that |H(z)| ≤ C for all z ∈ ∂B 1/2 (0) where C is independent of x ∈ B δ0/2 (0) \ {0}. Let Γ 0 be the Green's function of |x| 2 h(x + |x|z) at 0 on B 1/2 (0) with Dirichlet boundary condition. Therefore, H-Γ 0 ∈ C 2 (B 1/2 (0)) and, via the comparison principle, it is bounded by its supremum on the boundary. Therefore |z| n-2 H(z) ≤ C for all B 1/2 (0) \ {0} where C is independent of x ∈ B δ0/2 (0) \ {0}.

2n n- 2 (

 2 B δ (0)) with δ := |p 0 |/4 G satisfies (ii).

2 (| 6 ,

 26 (152) G R, L q (B R 0 (0)) ≤ C for all R > R 0 and 0 < < |p 0 B δ 0 (0))≤ C for all R > R 0 and 0 < < |p 0where δ := |p 0 |/4. Arguing again as in Step 2 of the proof of Theorem 6, there exists G ∈ C 2 (R n \ {0, p 0 }) such that (154)

x→0 2 .

 2 |x| β-(γ) G(x) = c.In addition, Theorem 9 yields G ∈ H2 1,loc (R n \ {p 0 }). Since G is positive and smooth in a neighborhood of p, it follows from (155) and the classification of solutions to harmonic equations that(157) G(x) ∼ x→p0 1 (n -2)ω n-1 |x -p 0 | n-2 .Step 3: Asymptotic behavior at ∞ for solutions to (151): We letG(x) := 1 |x| n-2 G x |x| 2 for all x ∈ R n \ 0, p 0 |p 0 | 2 , be the Kelvin's transform of G. We have that -∆ Gγ |x| 2 G = 0 in R n \ 0, p 0 |p 0 | 2 .Since G > 0, it follows from Theorem 9 that there existsc 1 > 0 such that either G(x) ∼ x→0 c 1 |x| β-(γ) or G(x) ∼ x→0 c 1 |x| β+(γ) . Coming back to G, we get that either G(x) ∼ x→∞ c 1 |x| β+(γ) or G(x) ∼ |x|→∞ c 1 |x| β-(γ) .Assuming we are in the second case, for any c ≤ c 1 , we defineḠc (x) := G(x) -c |x| β-(γ) in R n \ {0, p 0 }, which satisfy -∆ Ḡγ |x| 2 Ḡ = 0 in R n \{0, p 0 }. It follows from (156) and (157) that for c < c 1 , Ḡc > 0 around p 0 and ∞. It then follows from the coercivity of -∆ -γ|x| -2 that Ḡc > 0 in R n \ {0, p} for c < c 1 . Letting c → c 1 yields Ḡc1 ≥ 0, and then Ḡc1 > 0 since it is positive around p 0 . Since Ḡc1 (x) = o(|x| -β-(γ) ) as |x| → ∞, performing again a Kelvin transform and using Theorem 9, we get that |x| β+(γ) Ḡc1 (x) → c 2 > 0 as |x| → ∞. Then there exists c 3 > 0 such thatlim x→0 |x| β-(γ) Ḡc1 (x) = c 3 > 0 and lim x→∞ |x| β+(γ) Ḡc1 (x) = c Since x → |x| -β-(γ) ∈ H 2 1,loc (R n ), we get that ϕ(p) = R n Ḡc1 (x) -∆ϕ -γ |x| 2 ϕ dx for all ϕ ∈ C ∞ c (R n ).Step 4: Uniqueness: Let G 1 , G 2 > 0 be 2 functions such that (i), (ii) hold for p := p 0 , and set H := G 1 -G 2 . It follows from Steps 2 and 3 that there exists c ∈ R such that H (x) := H(x) -c|x| -β-(γ) satisfies (158) H (x) = x→0 O |x| -β-(γ) and H (x) = |x|→∞ O |x| -β+(γ) .

c - 1 2 G p0 ρ - 1 p

 121 max{1, |x|} min{1, |x|} β-(γ) |x -p 0 | 2-n ≤ G p0 (x) ≤ c max{1, |x|} min{1, |x|} β-(γ) |x -p 0 | 2-nfor all x ∈ R n \ {0, p 0 }, c depending on p 0 . For p ∈ R n \ {0}, consider ρ p : R n → R n a linear isometry such that ρ p ( p0 |p0| ) = p |p| , and defineG p (x) := |p 0 | |p| n-|p 0 | |p| x for all x ∈ R n \ {0, p}.

(s) α

(s)-1 α

This work was carried out while N. Ghoussoub was visiting l'Institut Élie Cartan, Université de Lorraine in May 2015. He was partially supported by a research grant from the Natural Science and Engineering Research Council of Canada (NSERC). The paper was completed while Frédéric Robert was visiting the University of British Columbia in

Indeed, the proof goes essentially as in (141). Fix x ∈ B δ0/2 (0), x = 0, and set H(z) := |x| n-2 G x (|x|z) for z ∈ B 1/2 (0) \ {0}. We have that

Moreover, it follows from (143) that there exists C > 0 such that |H(z)| ≤ C for all z ∈ ∂B 1/2 (0). Then, as above, using a super-solution, we get that there exists C > 0 such that 0 < H(z) ≤ C|z| -β-(γ) for all z ∈ B 1/2 (0) \ {0}. Scaling back yields (146) when x ∈ B δ0/2 (0). The general case follows from (141). This proves (146). Again, by symmetry, we conclude that there exists C > 0 such that (147)

Finally, one easily checks that (129) is a direct consequence of ( 146), ( 147) and (145). When f ∈ C ∞ c (Ω), identity (127) is a consequence of (138). The general case follows from density and the integral controls on G. The behavior (128) is a consequence of the classification of solutions to harmonic equations and Theorem 9.

To conclude, we shall briefly sketch the proof of the lower bound (130). Indeed, in Steps 3 and 4, we repeatedly used the comparison principle to get the upper bound for G by considering domains on the boundary of which G was bounded from above. As one checks, in the case when x, y are in ω ⊂⊂ Ω, G is also bounded from below by some positive constant on the boundary of these domains. This yields the lower bound (130), and completes the proof of Theorem 6.

Appendix B: Green's function for -∆ -γ|x| -2 on R n

In this section, we prove the following:

. For all p ∈ R n \ {0}, there exists G :

Moreover, if G, G satisfy (i) and (ii) and are positive, then there exists

In addition, there exists one and only one function G := G p > 0 such that (i) and (ii) hold and

(iv) There exists c > 0 independent of p such that

Remark: Note that when γ = 0, we have

Proof: We shall again proceed with several steps.

It is easy to check that G p > 0 and that it satisfies (i), (ii), (iii) and (iv).

Appendix C: Singular solutions to -∆u -c(x)|x| -2 u = 0

We collect here a few results that should be classical, but quite difficult to find in the literature. These results and their proofs are closely related to the work of the authors in [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF], to which we shall frequently refer for details.

Theorem 8 (Optimal regularity and Generalized Hopf's Lemma). Fix γ < (n-2) 2 4 and let f :

for some θ > 0. Then, there exists K ∈ R such that

Moreover, if u ≥ 0 and u ≡ 0, we have that K > 0.

where c(x) = γ + O(|x| θ ) as x → 0 with γ < (n -2) 2 /4 and θ ∈ (0, 1). Then there exists α > 0 such that

In particular, u ∈ H 2 1 (B 1/2 (0)) if and only if the first case holds.

Then there exist λ -, λ + ≥ 0 such that

Proofs: The proofs of these results follow closely the proofs of Theorems 6.1 and 7.1 and Proposition 7.4 of [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF]. Here are the ingredients to adapt: Sub-and super-solutions: The first step is the following result: Proposition 12. Fix γ < (n -2) 2 /4 and θ ∈ (0, 1) and let c :

-∆u

The proof is as follows. For β ∈ {β -(γ), β + (γ)}, we define u β : x → |x| -β + λ|x| -β . A straightforward computation yields

as x → 0. Then, choosing β ∈ R such that 0 < β -β < θ and β (n -2 -β ) -γ = 0, we get either a sub-or a supersolution taking λ positive or negative. This proves the proposition.

Sub-solution with Dirichlet boundary condition: We let u β+(γ) as above be a super-solution on

Note that f vanishes around 0 and that it is in

Note that for δ > 0 small enough, -∆ -(γ + O(|x| θ ))|x| -2 is coercive on B δ (0), and therefore, the existence of v is ensured for small δ. Define

The definition of η and v yields (163) -∆u

Moreover, since -∆v -c(x)|x| -2 v = 0 around 0 and v ∈ D 1,2 (B δ (0)), it follows from Theorem 8 that there exists C > 0 such that |v(x)| ≤ C|x| -β-(γ) for all x ∈ B δ (0). Then it follows from the expression of u β+(γ) that

We then get a supersolution satisfying (163) with the above behavior at 0. This is similar for a subsolution.

In the proof above, it is important that the operator -∆ -c(x)|x| -2 is coercive on B δ (0) for δ > 0 small enough. Now let Ω be a smooth bounded domain of R n and let γ < (n -2) 2 /4 and h ∈ C 0,θ(Ω) be such that -∆ -(γ|x| -2 + h) is coercive on Ω. Arguing as above, we get that there exists u γ) . Similarly, we get a subsolution. These points are enough to adapt the proofs of the above-mentioned results of [START_REF] Ghoussoub | On the Hardy-Schrödinger operator with a boundary singularity[END_REF] to our context.