Sobolev inequalities for the Hardy-Schr\"odinger operator: Extremals and critical dimensions

In this expository paper, we consider the Hardy-Schr\"odinger operator $-\Delta -\gamma/|x|^2$ on a smooth domain \Omega of R^n with 0\in\bar{\Omega}, and describe how the location of the singularity 0, be it in the interior of \Omega or on its boundary, affects its analytical properties. We compare the two settings by considering the optimal Hardy, Sobolev, and the Caffarelli-Kohn-Nirenberg inequalities. The latter rewrites: $C(\int_{\Omega}\frac{u^{p}}{|x|^s}dx)^{\frac{2}{p}}\leq \int_{\Omega} |\nabla u|^2dx-\gamma \int_{\Omega}\frac{u^2}{|x|^2}dx$ for all $u\in H^1_0(\Omega)$, where \gamma<n^2/4, s\in [0,2) and p:=2(n-s)/(n-2). We address questions regarding the explicit values of the optimal constant C, as well as the existence of non-trivial extremals attached to these inequalities. Scale invariance properties lead to situations where the best constants do not depend on the domain and are not attainable. We consider two different approaches to"break the homogeneity"of the problem: One approach was initiated by Brezis-Nirenberg and by Janelli. It is suitable for the case where 0 is in the interior of \Omega, and consists of considering lower order perturbations of the critical nonlinearity. The other approach was initiated by Ghoussoub-Kang , C.S. Lin et al. and Ghoussoub-Robert. It consists of considering domains where the singularity is on the boundary. Both of these approaches are rich in structure and in challenging problems. If 0\in \Omega, a negative linear perturbation suffices for higher dimensions, while a positive"Hardy-singular interior mass"is required in lower dimensions. If the singularity is on the boundary, then the local geometry around 0 plays a crucial role in high dimensions, while a positive"Hardy-singular boundary mass"is needed for the lower dimensions.

where γ < n 2 4 , s ∈ [0, 2) and 2 (s) := 2(n−s) n−2 . We address questions regarding the explicit values of the optimal constant C := µγ,s(Ω), as well as the existence of non-trivial extremals attached to these inequalities. Scale invariance properties often lead to situations where the best constants µγ,s(Ω) do not depend on the domain, and hence they are not attainable. We consider two different approaches for "breaking the homogeneity" of the problem, and restoring compactness.
One approach was initiated by Brezis-Nirenberg, when γ = 0 and s = 0, and by Janelli, when γ > 0 and s = 0. It is suitable for the case where the singularity 0 is in the interior of Ω, and consists of considering lower order perturbations of the critical nonlinearity. The other approach was initiated by Ghoussoub-Kang for γ = 0, s > 0, and by C.S. Lin et al. and Ghoussoub-Robert, when γ = 0, s ≥ 0. It consists of considering domains, where the singularity 0 is on the boundary.
Both of these approaches are rich in structure and in challenging problems. If 0 ∈ Ω, then a negative linear perturbation suffices for higher dimensions, while a positive "Hardy-singular interior mass" theorem for the operator Lγ is required in lower dimensions. If the singularity 0 belongs to the boundary ∂Ω, then the local geometry around 0 (convexity and mean curvature) plays a crucial role in high dimensions, while a positive "Hardy-singular boundary mass" theorem is needed for the lower dimensions. Each case leads to a distinct notion of critical dimension for the operator Lγ .
Elliptic problems with singular potential arise in quantum mechanics, astrophysics, as well as in Riemannian geometry, in particular in the study of the scalar curvature problem on the sphere S n . Indeed, if the latter is equipped with its standard metric whose scalar curvature is singular at the north and south poles, then by considering its stereographic projection of R n , the problem of finding a conformal metric with prescribed scalar curvature K(x) leads to finding solutions of the form −∆u − γ u |x| 2 = K(x)u 2 * −1 on R n . The latter is a simplified version of the nonlinear Wheeler-DeWitt equation, which appears in quantum cosmology (see [5,12,70,81] and the references cited therein).
Assuming now that 0 ∈ Ω, the first difficulty in these problems is due to the fact that 2 (s) is critical from the viewpoint of the Sobolev embeddings, in such a way that if Ω is bounded, then H 1 0 (Ω) is embedded in the weighted space L p (Ω, |x| −s ) for 1 ≤ p ≤ 2 (s), and the embedding is compact if and only if p < 2 (s). This lack of compactness defeats the classical minimization strategy to get extremals for µ γ,s (Ω). In fact, when s = 0 and γ = 0, this is the setting of the critical case in the classical Sobolev inequalities, which started this whole line of inquiry, due to its connection with the Yamabe problem on compact Riemannian manifolds [3], [63].
Another complicating feature of the problem is that the terms u |x| 2 and u 2 * (s)−1 |x| s are critical, in the sense that they have the same homogeneity as the Laplacian.
These results are due to Rodemich [76], Aubin [3] and Talenti [83]. We also refer to Lieb [64] and Lions [68,69] for other nice points of view. However, for general open subsets of R n , one can show by translating, scaling and cutting off u λ,x0 that µ 0,0 (Ω) = µ 0,0 (R n ) for all Ω open subset of R n , which means that if there is an extremal for µ 0,0 (Ω), then it is also an extremal for µ 0,0 (R n ) and has to be in the form of (9), which is impossible if Ω is bounded.
The above case has no singularities, which only appear when either γ = 0 or s > 0. But even in this case, we get the same phenomenon as soon as the singularity belongs to the interior of the domain, that is µ γ,s (Ω) = µ γ,s (R n ), which again means that µ γ,s (Ω) is not attained unless Ω is essentially equal to R n .
It is well known that if 0 is in the interior of Ω, then the best constant in the Hardy inequality, does not depend on the domain Ω ⊂ R n , is never achieved, and is always equal to Also, if 0 < s < 2, the constant µ 0,s (R n ) is again explicit, and the extremals are also known (see , Lieb [64], Catrina-Wang [17]). More precisely, Here, it is important to note the following asymptotics for u when → 0: lim →0 u (0) = +∞ and lim →0 u (x) = 0 for all x = 0.
In other words, the function u concentrates at 0 when → 0.
, the corresponding best constant is then See for example Beckner [10] or Dolbeault et al. [26]. The extremals for µ γ,s (R n ) are then given for ε > 0, by the functions Keep in mind that the radial function x → |x| −β is a solution of (−∆− γ |x| 2 )|x| −β = 0 on R n \ {0} if and only if β ∈ {β − (γ), β + (γ)}. Again, if 0 ∈ Ω, we have µ γ,s (Ω) = µ γ,s (R n ), and as above, there is no extremal for µ γ,s (Ω) if, for example, Ω is bounded. Now, in order to remedy the lack of compactness in this Euclidean setting, one can consider the subcritial case, by replacing 2 * (s) by a power p with 2 < p < 2 * (s). This direction, however, does not present any new idea or difficulty. In this paper, we shall describe two -more subtle-approaches for "breaking the homogeneity" of the problem, and restoring compactness: • One was initiated by Brezis-Nirenberg [7] when γ = 0 and considered by Ghoussoub-Yuan [45], Janelli [56], Kang-Peng [59][60][61] and many others [14][15][16] when γ > 0. It consists of considering lower order perturbations of the critical case. • The other approach was initiated by Ghoussoub-Kang [37] and developed by Ghoussoub-Robert [40][41][42] when s > 0 and γ = 0, and by C.S. Lin et al. [54,[65][66][67] and Ghoussoub-Robert [43] when γ = 0. It consists of considering domains, where the singularity 0 is on the boundary. Both of these approaches are rich in structure and in challenging problems. They both invoke the geometry of the domain (locally and globally), and introduce new critical dimensions to the problem. They also differ in many ways.

Linearly perturbed borderline variational problems with an interior singularity
The linear perturbation approach consists of considering equations of the form where 1 ≤ q < 2 * (s) and λ > 0 is small enough. For simplicity, we only discuss that case for q = 1. One then considers the quantity and use the fact that compactness is restored as long as (18) µ γ,s,λ (Ω) < µ γ,s (R n ).
This extremely important observation is due to Trudinger [84], when s = γ = λ = 0, in the case of Riemannian manifolds, where the geometry plays the crucial role. He was actually trying to salvage Yamabe's proof of his own conjecture. This kind of condition is now standard while dealing with borderline variational problems. See also Aubin [3], Brézis-Nirenberg [7]. The condition limits the energy level of minimizing sequences, prevents the creation of "bubbles" and hence insures compactness. We give below an idea of the proof based on Struwe's decomposition of non-convergent minimizing sequences. The idea of restoring compactness on Euclidean domains by considering linear perturbations was pioneered by Brezis-Nirenberg [7]. They studied the case where γ = 0, s = 0 and 0 < λ < λ 1 (Ω), the latter being the first eigenvalue of the Laplacian on H 1 0 (Ω), that is the equation They showed existence of extremals for n ≥ 4. The case n = 3 is special and involves a "positive mass" condition introduced by Druet [27], and inspired by the work of Shoen [78] 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 heavily on Ω, since the mass condition does. We shall elaborate further on this theme. The paper of Brezis-Nirenberg [7] generated lots of activities. Combined with the contribution of Druet [27], it contains most of the ingredients relevant to the case when 0 ∈ Ω, including the case when the Laplacian is replaced by the Hardy-Schrödinger operator L γ that we discuss below.
The proof again consists of testing the functional on minimizing sequences of the form ηU , where U is an extremal for µ γ,s (R n ) and η ∈ C ∞ c (Ω) is a cut-off function equal to 1 in a neigbourhood of 0, and showing that µ γ,s,λ (Ω) < µ γ,s (R n ).
Janelli and others had partial results for the remaining interval that is when , a gap that we proceeded to fill recently in [44]. In order to complete the picture, it was first important to know for which parameters γ and s, the best constant µ γ,s (R n ) is attained.
. Then, the best constant µ γ,s (R n ) is attained if either s > 0 or if {s = 0 and γ ≥ 0}. On the other hand, if s = 0 and γ < 0, then µ γ,s (R n ) is not attained.
A proof for general cones is given in section 5. Note that (14) gives explicit extremals for µ γ,s (R n ) under the conditions n ≥ 3, 0 ≤ s < 2 and 0 ≤ γ < (n−2) 2 4 . The next step was to define a notion of Hardy interior mass associated to the operator −∆ − γ |x| 2 − λ on a bounded domain Ω in R n containing 0.
These solutions are unique up to a positive multiplicative constant, and there exists c > 0 such that , then for any solution H ∈ C ∞ (Ω \ {0}) of (E), there exist c 1 > 0 and c 2 ∈ R such that The uniqueness implies that the ratio c 2 /c 1 is independent of the choice of H, hence the " Hardy-singular internal mass" of Ω associated to the operator L γ − a can be defined unambigously as One can then complete the picture as follows. Table 1. 0 ∈ Ω (Linearly perturbed problems), 0 ≤ λ < λ 1 (L γ ) and either s > 0 or {s = 0 and γ ≥ 0} Hardy term Dim. Sing. Analytic. cond. Ext.
As to the case when s = 0 and γ < 0, 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 Shoen-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 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 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 The following table summarizes the remaining situations. Table 2. 0 ∈ Ω (Linearly perturbed problems): 0 ≤ λ < λ 1 (L γ ) and s = 0, γ < 0 The following theorem summarizes the various situations.
• The function a → m γ,a (Ω) is continuous for the C 0 (Ω) norm.
Problem 2: Assuming such a λ 0 (L γ ) exists, can one show that there are no extremals if 0 < λ < λ 0 (L γ ). Note that this was verified for general domains by Druet [27] in the case γ = s = 0 and n = 3. If Ω is a unit ball B, one can then show -just as Janelli [56] did in the case when γ > 0, s = 0 -that this is indeed the case by showing that if (n−2) 2 In other words, λ 0 (L γ ) = λ * (L γ ). 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 µ γ,s,λ (Ω) is not attained.
One can then deduce that the critical dimension for L γ is 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.

Borderline variational problems with a boundary singularity
The situation changes dramatically and becomes much more interesting if the singularity 0 belongs to the boundary of the domain Ω. For one, the test functions ηU don't belong to H 1 0 (Ω) anymore, and one cannot mimic the arguments given above. Actually, the differences already start with the most basic properties of the Hardy-Schrödinger operator L γ = −∆ − γ |x| 2 . To begin with, recall that if 0 ∈ Ω, then L γ is positive if and only if γ < (n−2) 2 4 , while if 0 ∈ ∂Ω the operator L γ could be positive for larger value of γ, potentially reaching the maximal constant n 2 4 on convex domains. Moreover, if 0 ∈ Ω, we have already noted that the best constant in the Hardy inequality µ 0,2 (Ω) is then always equal to (n−2) 2 4 and is never achieved, while if 0 ∈ ∂Ω, the best constant µ 0,2 (Ω) can be anywhere in the interval ( (n−2) 2 4 , n 2 4 ], and it is achieved if µ 0,2 (Ω) < n 2 4 (See Ghoussoub-Robert [43]).
The situation changes further when 0 ≤ s < 2. Indeed, we had seen that whenever 0 ∈ Ω, µ γ,s (Ω) = µ γ,s (R n ), and is never achieved unless Ω is essentially equal to R n . The first indication that a new phenomenon may occur, when 0 ∈ ∂Ω, was given by the following surprising result of Egnell [30] even when γ = 0. He showed that if D is a nonempty connected domain of S n−1 , the unit sphere in R n , and C := {rθ; r > 0, θ ∈ D} is the cone based at 0 induced by D, then there are extremals for µ 0,s (C) whenever s > 0.
An important point to note here is that the cone C is not smooth at 0, unless it is R n + or R n . Actually, if a general domain Ω with 0 on its boundary is smooth, then it looks more like the half-space R n + around 0, and not like R n as in the case 0 ∈ Ω. One therefore has to compare µ γ,s (Ω) with µ γ,s (R n + ), which is strictly larger than µ γ,s (R n ). One can also easily show that if Ω is smooth bounded and 0 ∈ ∂Ω, then µ γ,s (R n ) < µ γ,s (Ω) ≤ µ γ,s (R n + ), and if Ω is convex (or if Ω ⊂ R n + ), then µ γ,s (Ω) = µ γ,s (R n + ) and again µ γ,s (Ω) has no extremals.
Another discrepancy with the case where 0 is in the interior, is the fact that the extremals for µ γ,s (R n ), which are the building blocks for the extremals on bounded domains, can often be written explicitly as seen above, while the ones for µ γ,s (R n + ) are not. So one then tries to understand as much as possible the profile of such extremals, which happen to solve the equation This was done in a recent analysis by Ghoussoub-Robert [43], where the needed information on the profile is given. The non-explicit solution has the following properties: • Symmetry: u • σ = u for all isometry of R n such that σ(R n + ) = R n + . In particular, there exists v ∈ C 2 (R + × R) such that for all x 1 > 0 and all • Asymptotic profile: If u ≡ 0, then there exist K 1 , K 2 > 0 such that Keep in mind that x → x 1 |x| −α is a solution of (−∆ − γ |x| 2 )x 1 |x| −α = 0 on R n \ {0} if and only if α ∈ {α − (γ), α + (γ)}. Note that α − (γ) < n 2 < α + (γ), which points to the difference between the "small" solution, namely x → x 1 |x| −α−(γ) , which is "variational", i.e. is locally in D 1,2 (R n + ), and the "large one" x → x 1 |x| −α+(γ) , which is not.
It also turned out that, unlike the case where 0 ∈ Ω, there are examples of domains with 0 ∈ ∂Ω such that µ γ,s (Ω) < µ γ,s (R n + ), which means that µ γ,s (Ω) has a good chance to be attained. This was first observed by Ghoussoub-Kang [37] in the most basic case, where 0 < s < 2 and γ = 0. Again, this condition limits the energy level of minimizing sequences, and therefore prevents the creation of bubbles (in this case around 0) and hence ensures compactness. There are many ways to see this, and we use the opportunity to introduce Struwe's approach via his famed decomposition [82].
Since ∂Ω is smooth at 0, there exists U, V open subsets of R n such that 0 ∈ U , 0 ∈ V and a C ∞ −diffeomorphism ϕ : U → V such that ϕ(0) = 0, Up to an affine transformation, we can assume that the differential of ϕ at 0 is the identity map. Letting η ∈ C ∞ c (U ) be such that η(x) ≡ 1 in a neighborhood of 0, and given ∈ (0, 2 (s) − 2), we consider the subcritical minimization problems: Since the exponent p := 2 (s)−ε is subcritical, the embedding H 1 0 (Ω) → L p (Ω, |x| −s ) is compact, and we therefore have a minimizer u ε ∈ H 1 0 (Ω) \ {0} where µ 0,s (Ω) is attained. Regularity theory then yields that u ε ∈ C ∞ (Ω \ {0}) ∩ C 1 (Ω) and we can assume that u ε solves the equation The "free energy" of the solutions then satisfy Ω The standard strategy is then to analyze what happens when we let → 0. This is not straightforward since the embedding H 1 0 (Ω) L 2 (s) (Ω; |x| −s ) is not compact. In the case s = 0, Struwe [82] gave a useful decomposition describing precisely this lack of compactness for minimzing sequences such as (u ε ) ε , which was extended to this situation by Ghoussoub-Kang [37]. It says that there exists Λ > 0 with Note that for any bubble, we have Ω , which means that if there is any bubble in the decomposition, then necessarily Since lim →0 µ 0,s (Ω) = µ 0,s (Ω), one then get that µ 0,s (Ω) ≥ µ 0,s (R n + ), which contradicts the initial energy hypothesis. It follows that there is no bubble and therefore lim →0 u ε = u 0 in H 1 0 (Ω), yielding that u 0 is an extremal for µ 0,s (Ω). The question now is what geometric condition on Ω insures that we have the analytic condition µ γ,s (Ω) < µ γ,s (R n + ). In view of the above, for any hope to find extremals, one has to avoid situations where Ω is convex or if it lies on one side of a hyperplane that is tangent at 0. This was first confirmed by Ghoussoub-Kang [37], who proved that this is indeed the case -and that extremals exist-provided n ≥ 4 and the principal curvatures of ∂Ω at 0 are all negative.
Concerning terminology, recall that the principal curvatures are the eigenvalues of the second fundamental form of the hypersurface ∂Ω oriented by the outward normal vector. The second fundamental form being where dn 0 is the differential of the outward normal vector at 0 and (·, ·) is the Euclidean scalar product.
The result of Kang-Ghoussoub was eventually improved later by Ghoussoub-Robert [40,41], who also proved it for n = 3 and by only requiring that the mean curvature, i.e., the trace of the second fundamental form, at 0, to be negative (see also Chern-Lin [22]). Qualitatively, this says that there are extremals for µ 0,s (Ω), whenever the domain at 0 has more concave directions than convex ones, in the sense that the negative principal directions dominate quantitatively the positive principal directions. This allows for new examples, which are neither convex nor concave at 0, and for which the extremals exist. Note that this result does not give any information about the value of the best constant.
We now illustrate how the mean curvature enters in the picture in the simplest case, namely when s > 0 and γ = 0. It consists of performing a more refined blow-up analysis on the minimizing sequences considered above. The proof -due to Ghoussoub-Robert [40]-uses the machinery developed in Druet-Hebey-Robert [29] for equations of Yamabe-type on manifolds. It also allows to tackle problems with arbitrary high energy and not just minima [41].
We consider again the solutions (u ) of the subcritical problems corresponding to p = 2 * (s) − with ∈ (0, 2 (s) − 2), in such a way that One then proves (see Ghoussoub-Robert [40]) that either u converges to an extremal of µ 0,s (Ω), or blow-up occurs in the following sense: u converges weakly to zero and there exists a solution v for while -modulo passing to a subsequence-we have where H Ω (0) is the mean curvature of the oriented boundary ∂Ω at 0. Note that if H Ω (0) < 0, such a blow-up cannot occur and we therefore end up with an extremal.
To sketch a proof of such a dichotomy, we start as before with the Struwe decomposition to write that either there exists , hence it is an extremal for µ 0,s (Ω), or there exists a bubble (B ) >0 such that and is a solution for (29). The idea is to prove that the family (u ε ) >0 behaves more or less like the bubble (B ) >0 . In fact (31) already indicates that these two families are equal up to the addition of a term vanishing in H 1 0 (Ω). But we actually need something more precise, like a pointwise description, as opposed to a weak description in Sobolev space. This requires a good knowledge of the bubbles: a difficult question since bubbles are not explicit here as in the case of R n . The proof has two main steps: First, one shows that there exists C 1 > 0 such that for all > 0, where (µ ) are involved in the definition (27) of the bubble (B ). The next step is to use the following Pohozaev identity, The left-hand-side is easy to estimate with (28). For the right-hand-side, one uses the optimal estimate (32) to obtain where II 0 is the second fondamental form at 0 defined on the tangent space of ∂Ω at 0 that we identify with ∂R n + . Finally, in view of the symmetry result mentioned above for the solution u, that is u(x 1 ,x) =ũ(x 1 , |x|) whereũ : R + × R → R, which means that the limit above rewrites as (30).
Optimal pointwise estimates like (32) have their origin in the work of Atkinson-Peletier [1] and Brézis-Peletier [8]. Pioneering work also include Han [51] and Hebey-Vaugon [53] in the case of a Riemannian manifold. For s = γ = 0, the general pointwise estimates are performed in the monograph [29] of Druet-Hebey-Robert. We also refer to Ghoussoub-Robert [41] for the optimal control with arbitrary high energy when s > 0 and γ = 0. Other methods developed to get pointwise estimates are due to Schoen-Zhang [80] and Kuhri-Marques-Schoen [62].
The negativity of the mean curvature at 0 turned out to be sufficient for the existence of extremals not only in the case where γ = 0, but also for a large range of γ > 0. Theorem 2.1. (Chern and Lin [22]) Let Ω be a smooth bounded domain such that 0 ∈ ∂Ω. Assume n ≥ 4, s ≥ 0, and 0 ≤ γ < (n−2) 2
The proof consists of testing the functional on minimizing sequences arising from suitably truncated extremals of µ γ,s (R n + ), whenever they are attained, and showing that µ γ,s,λ (Ω) < µ γ,s (R n + ). In [43] Ghoussoub-Robert consider the rest of the range left by Chern and Lin. In order to complete the picture, it was again important to know for which parameters γ and s, the best constant µ γ,s (R n + ) is attained. This is summarized in the following proposition, whose proof is given in section 5.
Ghoussoub-Robert first noted that the proof of Chern-Lin extends directly to the case when γ < n 2 −1 4 . The limiting case when γ = n 2 −1 4 is already quite more involved and requires precise information on the profile of the extremal for µ γ,s (R n + ). However, the case when γ > n 2 −1 4 turned out to be more intricate. The "local condition" of negative mean curvature at 0 is not sufficient anymore to ensure extremals for µ γ,s (Ω). One requires a positivity condition on the Hardy-singular boundary mass of Ω defined below. This new "global notion" associated with the operator L γ could be assigned to any smooth bounded domain Ω of R n with 0 ∈ ∂Ω, as long as n 2 −1 4 < γ < n 2 4 . Theorem 2.3. (Ghoussoub-Robert [43]) Assume Ω is a smooth bounded domain in R n with 0 ∈ ∂Ω in such a way that n 2 −1 4 < γ < γ H (Ω), the latter being the best Hardy constant for the domain Ω. Then, up to multiplication by a positive constant, there exists a unique function H ∈ C 2 (Ω \ {0}) such that Moreover, there exists c 1 > 0 and c 2 ∈ R such that The quantity b γ (Ω) := c2 c1 ∈ R, which is independent of the choice of H satisfying (33), will be referred to as the "Hardy-singular boundary mass" of Ω.
One can then complete the picture as follows. The following theorem summarizes the various situations Theorem 2.4. Let Ω be a smooth bounded domain in R n (n ≥ 3) such that 0 ∈ ∂Ω and let 0 ≤ s < 2 and γ < n  Table 3. Case where either s > 0 or {s = 0, γ > 0, and n ≥ 4}.
Hardy term Singularity Dim. Geometric condition Extremal Table 4. s = 0 and the remaining cases.
Hardy term Singularity Dim. Geometric condition Extremal Here are some of the remarkable properties of the Hardy-singular boundary mass.
• The map Ω → b γ (Ω) is a monotone increasing function on the class of domains having zero on their boundary, once ordered by inclusion. • One can also define the mass of unbounded sets as long as they can be "inverted" via a Kelvin transform into a smooth bounded domain. For example, b γ (R n + ) = 0 for any n 2 −1 4 < γ < n 2 4 , and therefore the mass of any one of its subsets having zero on its boundary is non-positive. In particular, b γ (Ω) < 0 whenever Ω is convex and 0 ∈ ∂Ω. In other words, the sign of the Hardy-singular boundary mass can be totally independent of the local properties of ∂Ω around 0, as illustrated by the following result.
Proposition 2.5. (Ghoussoub-Robert [43]) Let ω be a smooth open set of R n such that 0 ∈ ∂ω. Then, there exist two smooth bounded domains Ω + , Ω − of R n with Hardy constants > n 2 −1 4 , and there exists r 0 > 0 such that . The above analysis also leads to the following definition of another critical dimension for the operator L γ , which concerns domains having 0 on their boundary. It is the largest scalarn γ such that for every n <n γ , there exists a bounded smooth domain Ω ⊂ R n with 0 ∈ ∂Ω and with negative mean curvature at 0 such that µ γ,s (Ω) is not attained. Problem 3: An interesting question is to verify that if 0 ∈ ∂Ω, then the critical dimension for L γ is given by the formula Note that the above results yield thatn γ ≤ √ 4γ + 1 and that n < √ 4γ + 1 corresponds to when α Part 2. Caffarelli-Kohn-Nirenberg inequalities on R n and R n +

Inequalities of Hardy, Sobolev, and Caffarelli-Kohn-Nirenberg
We start by deriving these inequalities and show how they are interrelated.
The Hardy inequality: It states that which also yields that µ 0,2 (Ω) ≥ (n−2) 2 4 for all Ω ⊂ R n , and that µ γ,s (Ω) ≥ 0 for . An elementary proof of this inequality goes as follows: Associate to any smooth radial positive functions u ∈ C 2 c (B R ), where B R is the ball of radius R in R n the function v(r) = u(r)r (n−2)/2 where r = |x|. Denoting ω n−1 the volume of the unit sphere, one can estimate the quantity as follows: which is obviously non-negative.
If now u is a non-radial function on general domain Ω, we consider its symmetric decreasing rearrangement u * , defined by where for a general set A ⊂ R n , we denote by χ * A the characteristic function of a ball of volume |A| centered at the origin. the function u * is then symmetric-decreasing, and satisfies u * |x| p ≥ u |x| p for any p, since the rearrangement does not change the values of u, while only changing the places where these values occur. What is less obvious is that a proof of which can be found in [4].
Let now B R be a ball having the same volume as Ω with R = (|Ω|/ω n ) 1/n . If u ∈ H 1 0 (Ω), then u * ∈ H 1 0 (B R ), has the same L p -norm as u, while decreasing the Dirichlet energy. Hence, (35) holds for every u ∈ H 1 0 (Ω). To see that is not achieved, if the singularity 0 belongs to the interior of Ω, assume that u ≥ 0 is a weak solution of the corresponding Euler-Lagrange equation.
More recently, it was observed by Brezis-Vasquez [9] and others [35] that the inequality can be improved. The story here is the link -discovered by Ghoussoub-Moradifam [38,39]-between various improvements of this inequality confined to bounded domains and Sturm's theory regarding the oscillatory behavior of certain linear ordinary equations.
Following Ghoussoub-Moradifam [39], we say that a non-negative C 1 -function P defined on an interval (0, R) is a Hardy Improving Potential (abbreviated as HI-potential) if the following improved Hardy inequality holds on every domain Ω contained in a ball of radius R: It turned out that a necessary and sufficient condition for P to be an HI-potential on a ball B R , is for the following ordinary differential equation associated to P • P k,ρ (r) = 1 . e(k−times) ). This connection to the oscillatory theory of ODEs leads to a large supply of explicit Hardy improving potentials. One can show for instance that there is no c > 0 for which P (r) = cr −2 is an HI-potential, which means that (n−2) 2 4 is the best constant for γ H (Ω).
The Hardy-Sobolev inequalities: The basic Sobolev inequality states that there exists a constant C(n) > 0, such that in such a way that µ 0,0 (Ω) > 0 for every Ω ⊂ R n . Actually, the Sobolev inequality can be derived from Hardy's except for the value of the best constant, which we will discuss later. We first derive the inequality for radial decreasing functions. The general case follows from the properties of symmetric rearrangements noted above. The argument goes as follows: If u is radial and decreasing and p > 2, then for any y ∈ R n we have where ω n is the volume of the unit ball in R n . Now take this to the power 1 − 2 p , multiply by |u(y)| 2 |y| n(2−p) p and integrate over y to obtain It now suffices to take p := 2n n−2 and use Hardy's inequality to conclude. A Hölder-type interpolation between the Hardy and Sobolev inequalities yields the Hardy-Sobolev inequality, which states that for any s ∈ [0, 2], there exists C(s, n) > 0 such that where 2 (s) := 2(n−s) n−2 . In other words, µ 0,s (Ω) > 0 for every s ∈ [0, 2]. Indeed, by applying Hölder's inequality, then Hardy's and Sobolev's, we get It is remarkable that when s ∈ (0, 2), the Hardy-Sobolev inequality inherits the singularity at 0 from the Hardy inequality and the superquadratic exponent from the Sobolev inequality. Now what about the dependence on γ. Combining the above three inequalities, one obtain that for each γ < (n−2) 2 4 ≤ γ H (Ω) = µ 0,2 (Ω), the latter being the best constant in the Hardy inequality on Ω, we have that inequality (1) holds with C > 0, in other words, (42) µ γ,s (Ω) > 0 for all s ∈ [0, 2] and γ < (n−2) 2
We shall see later that this may hold true for values of γ beyond (n−2) 2
The Caffarelli-Kohn-Nirenberg inequalities: We now show that (42) also contains the celebrated Caffarelli-Kohn-Nirenberg inequalities [13], which state that there is a constant C := C(a, b, n) > 0 such that the following inequality holds: Indeed, by setting w(x) = |x| −a u(x), we see that for any u ∈ C ∞ 0 (Ω), with γ = a(n−2−a), and where the last equality is obtained by integration by parts. Now note that if a < n−2 2 , then by Hardy's inequality, Ω |x| −2a |∇u| 2 dx < +∞ if and only if both Ω |∇w| 2 dx < +∞ and Ω |w| 2 |x| 2 dx < +∞. Furthermore, where s = (b − a)q. This readily implies that (1) and (43) are equivalent under the above conditions on a, b, q, s, and γ.
There are many interesting examples of weights of the form −∆ρ ρ besides (n−2) 2 4|x| 2 , which could reflect the nature of the domain. Here is one that will concern us throughout this paper.
By Hölder-interpolating between the above general Hardy inequality and the Sobolev inequality, one gets the following generalized Caffarelli-Kohn-Nirenberg inequality.

5.
Attainability of the extremals on R n and R n + Let C be an open connected cone of R n , n ≥ 3, centered at 0, that is (53) C is a domain (that is open and connected) ∀x ∈ C, ∀r > 0, rx ∈ C.
Fix γ < γ H (C), and consider the question of whether there is an extremal u 0 ∈ D 1,2 (C) \ {0}, where µ γ,s (C) is attained. The question of the extremals on general cones has been tackled by Egnell [30] in the case {γ = 0 and s > 0}. Theorem 5.1 below has been noted in several contexts by Bartsche-Peng-Zhang [6] and Lin-Wang [22]. We shall sketch below an independent proof.
We isolate two corollaries. The first one is essentially what we need when C = R n + . The second deals with the case C = R n . There is no issue for n = 3 in the second corollary.
We let (ũ k ) k ∈ D 1,2 (R n + ) be a minimizing sequence for µ γ,s (C) such that C |ũ k | 2 (s) |x| s dx = 1 and lim For any k, there exists r k > 0 such that Br k (0)∩C Since C is a cone, we have that u k ∈ D 1,2 (C). We then have that We first claim that, up to a subsequence, Indeed, for k ∈ N and r ≥ 0, we define Q k (r) := Br(0)∩C |u k | 2 (s) |x| s dx.
Since 0 ≤ Q k ≤ 1 and r → Q k (r) is nondecreasing for all k ∈ N, then, up to a subsequence, there exists Q : [0, +∞) → R nondecreasing such that Q k (r) → Q(r) as k → +∞ for a.e. r > 0. Set It follows from (55) and (56) In particular, We claim that Indeed, for all x ∈ B R k (0) \ B R k (0), we have that R k ≤ |x| ≤ 3R k . Therefore, Hölder's inequality yields for all k ∈ N. Conclusion (59) then follows from (58).
Arguing as above, we get that for all x ∈ R n , we have that It then follows from the second identity of (56) that α 0 ≤ 1/2, and therefore α 0 = 0. Moreover, it follows from the first identity of (56) that there exist as most one point x 0 ∈ R n such that α x0 = 1. In particular x 0 = 0 since α 0 = 0. It then follows from Lions's second concentration compactness lemma [68,69] (see also Struwe [82] for an exposition in book form) that, up to a subsequence, there exists u ∞ ∈ D 1,2 (C), x 0 ∈ R n \ {0} and C ∈ {0, 1} such that u k u ∞ weakly in D 1,2 (C) and In particular, due to (56) and (57), we have that Since C ∈ {0, 1}, the claims in (60) and (61) follow. We now assume that u ∞ ≡ 0, and we claim that lim k→+∞ u k = u ∞ strongly in D 1,2 (C) and that u ∞ is an extremal for µ γ,s (C).
Indeed, it follows from (60) that C |u∞| 2 (s) |x| s dx = 1, hence Moreover, since u k u ∞ weakly as k → +∞, we have that Therefore, equality holds in this latest inequality, u ∞ is an extremal for µ γ,s (C) and boundedness yields the weak convergence of (u k ) to u ∞ in D 1,2 (C). This proves the claim.
We now assume that u ∞ ≡ 0 and show that as k → +∞, Indeed, since u k u ∞ ≡ 0 weakly in D 1,2 (C) as k → +∞, then for any 1 ≤ q < 2 := 2n n−2 , u k → 0 strongly in L q loc (C) when k → +∞. Assume by contradiction that s > 0, then 2 (s) < 2 and therefore, since x 0 = 0, we have that for δ > 0 small enough, contradicting (61). Therefore s = 0 and the first part of the claim is proved.
As noted in Remark 5.5, it is easy to see that if s = 0 and γ ≤ 0, then Moreover, if there are extremals then γ = 0.
We now show that in this case, there are extremals iff there exists z ∈ R n such that Indeed, the potential extremals for µ 0,0 (C) are extremals for µ 0,0 (R n ), and therefore of the form x → a(b + |x − z 0 | 2 ) 1−n/2 for some a = 0 and b > 0 (see Aubin [3] or Talenti [83]). Using the homothetic invariance of the cone, we get that there is an extremal of the form x → (1 + |x − z| 2 ) 1−n/2 for some z ∈ R n . Since an extremal has support in C, we then get that C = R n . This proves the claim.
Going further in the expansion, one can show the following: Note that C < +∞ if and only if γ < (n−2) 2 4 − 1, which happens if and only if β + (γ)−β − (γ) > 2. This explains the obstruction on the dimension in this situation, since the L 2 −concentration allows to overlook the role of the cut-off function.
Pushing the expansion to the limit, we have the following where C is a positive consatnt.
When γ > (n−2) 2 4 − 1, the above test functions do not suffice, and one needs more global test functions . We therefore let H ∈ C ∞ (Ω \ {0}) as in Proposition 1.3. Up to multiplying by a constant, we assume that C 1 = 1. We let ∈ H 1 0 (Ω) ∩ C 0 (Ω) be such that , where m γ,−λ (Ω) is the Hardy-interior mass. The test-functions can be taken in this case to be One can then show the following.
8. Existence of extremals when s = 0 and γ < 0 Recall from the introduction that R γ,λ (x 0 ) is the Robin function at x 0 , that is the value at x 0 of the regular part of the Green's function of −∆ − γ|x| −2 − λ at x 0 . We sketch the proof of the remaining cases.
On the other hand, one can argue as in Aubin [3] and prove the following Claim 1: If x 0 ∈ Ω \ {0} is such λ + γ |x0| 2 > 0 and n ≥ 5, then Note that C < +∞ iff n ≥ 4, in which case the L 2 −concentration again allows to overlook the cut-off function.
For n = 4 one needs to push the expansion further.
In order to deal with the case n = 3, global test-functions are again required. We let G x0 ∈ C ∞ (Ω \ {0}) be the Green's function of −∆ − λ − γ|x| −2 at x 0 . Up to multiplying by a constant, we may assume that Define now the test-function One can then show the following Claim 3: If x 0 ∈ Ω \ {0} is such λ + γ |x0| 2 > 0 and n = 3, then Part 4. When 0 is a boundary singularity for the operator L γ 9. Analytic conditions for the existence of extremals when 0 ∈ ∂Ω As mentioned in the introduction, the case when the singularity 0 ∈ ∂Ω is more intricate as far as the operator −∆ − γ |x| 2 is concerned. This is already apparent in the following linear situation. (4) For every > 0, there exists a smooth domain R n + ⊂ Ω ⊂ R n such that 0 ∈ ∂Ω and n 2 4 − ≤ γ H (Ω ) < n 2 4 .
The above mentioned properties of γ H were noted in [39] and [43]. We sketch the proofs. We have already noted in section 1, that γ H (R n ) = (n−2) 2 4 , while equation (48) yields that γ H (R n + ) = n 2 4 . It is also easy to see that if B r is a ball of radius r such that 0 ∈ ∂B r , then we also have γ H (B r ) = γ H (R n + ) = n 2 4 . If now ∂Ω is smooth at 0 ∈ ∂Ω, we can always find such a ball with B r ⊂ Ω, from which follows that γ H (Ω) ≥ γ H (B r ) = n 2 4 . To prove 3), one first shows that γ H (R n ) can be approached by the following nonsmooth conical domains. Let Ω 0 be a bounded domain of R n such that 0 ∈ Ω 0 (i.e., it is not on the boundary). Given δ > 0, define For δ > 0 small enough, 0 ∈ ∂Ω, and one can show that lim δ→0 γ H (Ω δ ) = (n−2) 2 4 . Note that this works for n ≥ 4. A different construction is needed for n = 3. Now to check the infimum for smooth domains, note that for each δ > 0 small, there exists Ω δ a smooth bounded domain of R n such that Ω δ ⊂ Ω δ and 0 ∈ Ω δ . Since Ω → γ H (Ω) is nonincreasing, we have that γ H (R n ) ≤ γ H (Ω δ ) ≤ γ H (Ω δ ) and therefore lim sup . Since ϕ ≥ 0, we have that R n + ⊂Ω t for all t > 0. It now suffices to take Ω ε :=Ω t for t small enough.
Case 3: s = 2. This is the easiest case, since then This completes the proof of (87) for all s ∈ [0, 2]. The following corollary is an easy consequence of the above.
Note that the case γ = n 2 4 is unclear as it seems that anything can happen at that value of γ. For example, if γ H (Ω) < n 2 4 then µ n 2 4 ,s (Ω) < 0, while if γ H (Ω) = n 2 4 then µ n 2 4 ,s (Ω) ≥ 0. It is our guess that many examples reflecting different regimes can be constructed. 10. Analysis of the operator L γ = −∆ − γ |x| 2 when 0 ∈ ∂Ω In the sequel, we shall be looking for geometric conditions on Ω that insures that µ γ,s (Ω) < µ γ,s (R n + ). As before, we need to compute the functional J Ω γ,s at bubbles modeled on extremals for µ γ,s (R n + ) and to make a Taylor expansion, hoping that one succeeds in getting below the energy threshold. But at this stage, a difficulty occurs: the extremals for µ γ,s (R n + ) are not explicit, and therefore the coefficients that appear in the estimate of J Ω γ,s at the bubbles are not explicit enough. One needs to know more about the profile of the solutions for the linear and nonlinear equations involving the operator L γ on R n + .
Assume γ < n 2 4 and let u ∈ D 1,2 (Ω) loc,0 be such that for some τ > 0, Then, there exists K ∈ R such that Moreover, if u ≥ 0 and u ≡ 0, we then have that K > 0.
The proof of this theorem is quite interesting since, unlike the regular case (i.e., when L γ = L 0 = −∆) or in the situation when the singularity 0 is in the interior of the domain Ω, the application of the standard Nash-Moser iterative scheme is not sufficient to obtain the required regularity. Indeed, the scheme only yields the existence of p 0 , with 1 < p 0 < n α−(γ)−1 such that u ∈ L p for all p < p 0 . Unfortunately, p 0 does not reach n α−(γ)−1 , which is the optimal rate of integration needed to obtain the profile (99) for u. However, the improved order p 0 is enough to allow for the inclusion of the nonlinearity f (x, u) in the linear term of (98). We are then reduced to the analysis of the linear equation, that is (98) with f (x, u) ≡ 0.
When u ≥ 0, u ≡ 0, we get the conclusion by constructing super-and sub-solutions to the linear equation behaving like (99).
As a corollary, one obtains a relatively detailed description of the profile of variational solutions of (6) on R n + , which improves greatly on a result of Chern-Lin [22], hence allowing us to construct sharper test functions and to prove existence of solutions for (6) when γ = n 2 −1 4 . In order to deal with the remaining cases for γ, that is when γ ∈ ( n 2 −1 4 , n 2 4 ), Ghoussoub-Robert [43] prove the following result which describes the general profile of any positive solution of L γ u = a(x)u, albeit variational or not.
Theorem 10.2 (Classification of singular solutions). Assume γ < n 2 4 and let u ∈ C 2 (B δ (0) ∩ (Ω \ {0})) be such that Then, there exists K > 0 such that In the first case, the solution u is variational; in the second case, it is not.
This result then allows us to completely classify all positive solutions to L γ u = 0 on R n + . One can therefore deduce the following.
Proposition 10.3. Assume γ < n 2 4 and let u ∈ C 2 (R n + \ {0}) be such that 11. The profile of the extremals for µ γ,s (R n + ) The following is a useful description of the solution profile for the extremals on R n + . We shall give below a proof of the symmetry.

4
-is key for the construction of test functions for µ γ,s (Ω) based on the solution U of (103), in the case when γ ≤ n 2 −1 4 . The proof of symmetry goes as follows. It was established by Chern-Lin [22]) for γ < 0 and by Ghoussoub-Robert [40] in the case when γ = 0, a proof which extends immediately to the case γ ≥ 0. Here is a sketch.
Denoting It then follows from standard regularity theory and Theorem 10.1 that v ∈ C 2 (D \ {0, e 1 }) and that there exists We now use the moving plane method to prove the symmetry property of v, which is defined on a ball. For µ ≥ 0 and x = (x , x n ) ∈ R n , where x ∈ R n−1 and x n ∈ R, we let x µ = (x , 2µ − x n ) and D µ = {x ∈ D/ x µ ∈ D}. It follows from Hopf's Lemma that there exists 0 > 0 such that for any µ ∈ (1 − 0 , 1), we have that D µ = ∅ and v(x) ≥ v(x µ ) for all x ∈ D µ such that x n ≤ µ. We let µ ≥ 0. We say that (P µ ) holds if: We let (106) λ := min {µ ≥ 0; (P ν ) holds for all ν ∈ (µ, 1)} .
We claim that λ = 0. Indeed, otherwise we have λ > 0, D λ = ∅ and that (P λ ) holds. We let for all x ∈ D λ ∩ {x n < λ}. Since (P λ ) holds, we have that w(x) ≥ 0 for all x ∈ D λ ∩ {x n < λ}. With the equation (105) of v and (P λ ), we get that 1 for all x ∈ D λ ∩ {x n < λ}. With straightforward computations, we have that for all x ∈ R n . It follows that −∆w(x) > 0 for all x ∈ D λ ∩ {x n < λ}. Note that we have used that λ > 0. It then follows from Hopf's Lemma and the strong comparison principle that By definition, there exists a sequence (λ i ) i∈N ∈ R and a sequence ( for all i ∈ N. Up to extraction a subsequence, we assume that there exists x ∈ D λ ∩ {x n ≤ λ} such that lim i→+∞ x i = x with x n ≤ λ. Passing to the limit i → +∞ in (108), we get that v(x) ≤ v(x λ ). It follows from this last inequality and (107) that v(x) − v(x λ ) = w(x) = 0, and then x ∈ ∂(D λ ∩ {x n < λ}). Case 1: If x ∈ ∂D. Then v(x λ ) = 0 and x λ ∈ ∂D. Since D is a ball and λ > 0, we get that x = x λ ∈ ∂D.
Since v is C 1 , we get that there exists Letting i → +∞, using that (x i ) n < λ i and (108), we get that ∂ n v(x) ≥ 0. On the other hand, we have that Therefore ∂v ∂ν (x) ≤ 0: this is a contradiction with Hopf's Lemma.
Since v(x λ ) = v(x), we then get that x λ ∈ D. Since x ∈ ∂(D λ ∩ {x n < λ}), we then get that x ∈ D ∩ {x n = λ}. With the same argument as in the preceding step, we get that ∂ n v(x) ≥ 0. On the other hand, with (107), we get that 2∂ n v(x) = ∂ n w(x) < 0. A contradiction.
This proves that λ = 0 in either one of the two cases considered above. It now follows from the definition (106) of λ that v(x , x n ) ≥ v(x , −x n ) for all x ∈ D such that x n ≤ 0. With the same technique, we get the reverse inequality, and then, we get that v(x , x n ) = v(x , −x n ) for all x = (x , x n ) ∈ D. In other words, v is symmetric with respect to the hyperplane {x n = 0}. The same analysis holds for any hyperplane containing e 1 . Coming back to the initial function u, this complete the proof of the symmetry of u.
We now outline the proof of the following existence result.
Proof. According to Theorem 9.2, in order to establish existence of extremals, it suffices to show that µ γ,s (Ω) < µ γ,s (R n + ). The rest of the section consists of showing that the above mentioned geometric conditions lead to such gap.
Note that for any k ≥ 0, we have The ultimate goal is to establish the following expansion as ε → 0.

The remaining 3-dimensional cases
It is easy to see that if s = 0 and γ ≤ 0, then µ γ,s (Ω) = µ 0,0 (R n ) and there is no extremal for µ γ,s (Ω). So the remaining case is when n = 3, s = 0 and γ > 0. But, we have seen that in this case, there may or may not be extremals for µ γ,0 (R n + ). If they do exist, we can then argue as before -using the same test functions-to conclude that there are extremals under the same conditions, that is if either γ ≤ n 2 −1 4 and the mean curvature of ∂Ω at 0 is negative, or γ > n 2 −1 4 and the Hardy-singular boundary mass b γ (Ω) is positive.
However, if no extremal exist for µ γ,0 (R n + ), then we have seen in section 5, that (123) µ γ,0 (R n + ) = inf u∈D 1,2 (R n )\{0} R n |∇u| 2 dx R n |u| 2 dx 2 2 , and therefore we are back to the case where the boundary singularity does not contribute anything. This means that one needs to resort to the standard notion of mass R γ,0 (Ω, x 0 ) for a domain Ω associated to an interior point x 0 ∈ Ω and construct test-functions in the spirit of Schoen.

Examples of domains with positive mass
We now assume that γ ∈ ( n 2 −1 4 , n 2 4 ) and would like to construct domains with either negative or positive mass. Since R n + is the main reference set in this theory, one needs to define a notion of mass for certain unbounded sets that include R n + . For that, define the following Kelvin transformation. For any x 0 ∈ R n , let The inversion i x0 is clearly the identity map on ∂B |x0| (x 0 ) (the ball of center x 0 and of radius |x 0 |), and in particular i x0 (0) = 0.
Definition 1. Say that a domain Ω ⊂ R n (0 ∈ ∂Ω) is conformally bounded if there exists x 0 ∈ Ω such that i x0 (Ω) is a smooth bounded domain of R n having both 0 and x 0 on its boundary ∂(i x0 (Ω)).
The following proposition shows that the notion of mass extends to unbounded domains that are conformally bounded. We define the mass b γ (Ω) := c2 c1 , which is independent of the choice of H in (125). One can easily check that R n + is a conformally bounded domain (take x 0 := (−1, 0, . . . , 0)), and the results of section 10 indicate that b γ (R n + ) = 0. Since the Hardy b-mass is strictly increasing and continuous, it follows that the mass is negative whenever Ω ⊂ R n + = T 0 ∂Ω. In particular, b γ (Ω) < 0 if Ω is convex and n 2 −1 4 < γ < n 2 4 . This also suggests that a conformally bounded set strictly containing R n + must have positive mass, which was proved by Ghoussoub-Robert [43]. and n 2 4 . The construction of such domains is technical but straightforward. Theorem 2.5 illustrates that one can construct smooth bounded domains with either positive or negative mass and having any type of behavior at 0.