This paper investigates the existence, multiplicity, and stability of solutions to Hardy-Schrödinger equations with boundary singularities, revealing new conditions under which solutions exist or do not exist, especially near boundary points with specific curvature properties.
Contribution
The authors perform sharp blow-up analysis to establish new multiplicity and stability results for solutions, extending classical Pohozaev obstruction insights to boundary singular cases.
Findings
01
Infinite solutions when certain curvature conditions are met.
02
Non-existence of solutions in star-shaped domains with small potentials.
03
Stability of solutions under perturbations when mass is non-zero.
Abstract
Let Ω be a smooth bounded domain in Rn (n≥3) such that 0∈∂Ω. In this memoir, we consider issues of non-existence, existence, and multiplicity of variational solutions in H1,02(Ω) for the borderline Dirichlet problem, −Δu−γ∣x∣2u−h(x)u=∣x∣s∣u∣2⋆(s)−2u in Ω, where 0<s<2, 2⋆(s):=n−22(n−s), γ∈R and h∈C0(Ω). We use sharp blow-up analysis on --possibly high energy-- solutions of corresponding subcritical problems to establish, for example, that if γ<4n2−1 and the principal curvatures of ∂Ω at 0 are non-positive but not all of them vanishing, then the above equation has an infinite number of (possibly sign-changing) solutions in H1,02(Ω). This complements results of the first and…
\displaystyle\left\{\begin{array}[]{llll}-\Delta U-\gamma\frac{U}{|x|^{2}}&=&\frac{|U|^{2^{\star}(s)-2}U}{|x|^{s}}&\text{ in }\mathbb{R}_{-}^{n},\\
\hfill U&=&0&\text{on }\partial\mathbb{R}_{-}^{n}.\end{array}\right.
\displaystyle\left\{\begin{array}[]{llll}-\Delta U-\gamma\frac{U}{|x|^{2}}&=&\frac{|U|^{2^{\star}(s)-2}U}{|x|^{s}}&\text{ in }\mathbb{R}_{-}^{n},\\
\hfill U&=&0&\text{on }\partial\mathbb{R}_{-}^{n}.\end{array}\right.
\displaystyle\left\{\begin{array}[]{llll}-\Delta U-\gamma\frac{U}{|x|^{2}}&=&0&\text{ in }\mathbb{R}_{-}^{n},\\
\hfill U&=&0&\text{on }\partial\mathbb{R}_{-}^{n}.\end{array}\right.
\displaystyle\left\{\begin{array}[]{llll}-\Delta U-\gamma\frac{U}{|x|^{2}}&=&0&\text{ in }\mathbb{R}_{-}^{n},\\
\hfill U&=&0&\text{on }\partial\mathbb{R}_{-}^{n}.\end{array}\right.
β±(γ):=2n±4n2−γ for γ<4n2.
β±(γ):=2n±4n2−γ for γ<4n2.
γ>4n2−1,
γ>4n2−1,
\qquad\qquad\left\{\begin{array}[]{ll}-\Delta\mathcal{H}-\frac{\gamma}{|x|^{2}}\mathcal{H}+h(x)\mathcal{H}=0&\hbox{ in }\Omega\\
\hfill\mathcal{H}>0&\hbox{ in }\Omega\\
\hfill\mathcal{H}=0&\hbox{ on }\partial\Omega\setminus\{0\}.\end{array}\right.
\qquad\qquad\left\{\begin{array}[]{ll}-\Delta\mathcal{H}-\frac{\gamma}{|x|^{2}}\mathcal{H}+h(x)\mathcal{H}=0&\hbox{ in }\Omega\\
\hfill\mathcal{H}>0&\hbox{ in }\Omega\\
\hfill\mathcal{H}=0&\hbox{ on }\partial\Omega\setminus\{0\}.\end{array}\right.
\left\{\begin{array}[]{cl}-\Delta u-\gamma\frac{u}{|x|^{2}}-\lambda u=\frac{u^{2^{\star}(s)-1}}{|x|^{s}}&\text{ in }\Omega,\\
u>0&\hbox{ in }\Omega\\
u=0&\hbox{ on }\partial\Omega\setminus\{0\}\end{array}\right.
\left\{\begin{array}[]{cl}-\Delta u-\gamma\frac{u}{|x|^{2}}-\lambda u=\frac{u^{2^{\star}(s)-1}}{|x|^{s}}&\text{ in }\Omega,\\
u>0&\hbox{ in }\Omega\\
u=0&\hbox{ on }\partial\Omega\setminus\{0\}\end{array}\right.
uϵ>0
uϵ>0
ϵ→0lim∥uϵ∥2=0.
ϵ→0lim∥uϵ∥2=0.
C(∫Ω∣x∣s∣u∣2⋆(s)dx)2/2⋆(s)≤∫Ω∣∇u∣2dx−γ∫Ω∣x∣2u2dx for all u∈H1,02(Ω).
C(∫Ω∣x∣s∣u∣2⋆(s)dx)2/2⋆(s)≤∫Ω∣∇u∣2dx−γ∫Ω∣x∣2u2dx for all u∈H1,02(Ω).
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Full text
Multiplicity and stability of the Pohozaev obstruction for Hardy-Schrödinger equations with boundary singularity
Nassif Ghoussoub
Nassif Ghoussoub: Department of Mathematics, University of British Columbia, Vancouver, V6T 1Z2 Canada
Let Ω be a smooth bounded domain in Rn (n≥3) such that 0∈∂Ω. We consider issues of non-existence, existence, and multiplicity of variational solutions in H1,02(Ω) for the borderline Dirichlet problem,
[TABLE]
where 0<s<2, 2⋆(s):=n−22(n−s), γ∈R and
h∈C0(Ω).
We use sharp blow-up analysis on –possibly high energy– solutions of corresponding subcritical problems to establish, for example, that if γ<4n2−1 and the principal curvatures of ∂Ω at [math] are non-positive but not all of them vanishing, then Equation (E) has an infinite number of high energy (possibly sign-changing) solutions in H1,02(Ω). This complements results of the first and third authors, who showed in [gr4] that if γ≤4n2−41 and the mean curvature of ∂Ω at [math] is negative, then (E) has a positive least energy solution.
On the other hand, the sharp blow-up analysis also allows us to show that if the mean curvature at [math] is nonzero and the mass, when defined, is also nonzero, then there is a surprising stability of regimes where there are no variational positive solutions under C1-perturbations of the potential h. In particular, and in sharp contrast with the non-singular case (i.e., when γ=s=0), we prove non-existence of such solutions for (E) in any dimension, whenever Ω is star-shaped and h is close to [math], which include situations not covered by the classical Pohozaev obstruction.
This work was initiated when the second-named author held a postdoctoral position at the University of British Columbia under the supervision of the first-named author, that was partially supported by the Natural Sciences and Engineering Research Council of Canada (NSERC)
This manuscript is the continuation of a long-time project initiated by the first and the third author in [gr1] about nonlinear critical equations involving the Hardy potential when the singularity is located on the boundary of the domain under study.
Let Ω be such a smooth bounded domain in Rn, n≥3, with 0∈∂Ω. We fix s∈(0,2) and define the critical Sobolev exponent 2⋆(s):=n−22(n−s). For γ∈R and h0∈C1(Ω), we consider in the sequel issues of non-existence, existence, and multiplicity of variational solutions in H1,02(Ω) for the borderline Dirichlet problem,
[TABLE]
By solutions, we mean here functions u∈H1,02(Ω), i.e., the completion of Cc∞(Ω) for the L2-norm of the gradient ∥∇u∥2. This problem has by now a long history starting with the fact that when γ=s=0 and h0 is a constant, it is the counterpart of the Yamabe problem [aubin, LeeParker, schoen1] in Euclidian space, as initiated by Brezis-Nirenberg [bn], with important contributions in the critical dimension n=3, by Druet [d2], and for multiplicity results for n≥7, by Devillanova-Solimini [11], among many others.
The case dealing with least energy solutions for s>0 but γ=0, when the singularity [math] is on the boundary of the domain was initiated by Ghoussoub-Kang [22] and developed by Ghoussoub-Robert [gr1]. The case involving the Hardy potential, i.e., when γ>0, was introduced by Lin-Wadade [LW3] with a follow-up contribution by Ghoussoub-Robert [gr4]. This paper addresses remaining issues about the multiplicity of solutions, but also about obstructions to the existence of solutions and their stability under small perturbations.
The existence of solutions is related to the coercivity of the operator−Δ−∣x∣2γ−h0(x). It is clear that the operator −Δ−∣x∣2γ is coercive on H1,02(Ω) whenever γ<γH(Ω), where γH(Ω) is the Hardy constant associated to the domain Ω, that is
[TABLE]
which has been extensively studied (see for example [GM.book] and [gr4]). We recall that if 0∈Ω, then
[TABLE]
When 0∈∂Ω, the situation is extremely different. For non-smooth domains modeled on cones, we refer to Egnell [17], and the more recent works of Cheikh-Ali [HCA, HCA2]. If Ω is smooth, then, around [math], the domain is modeled on the half-space R−n:={x∈Rn;x1<0}. We then get that (see [gr4])
[TABLE]
Note that when h0≡0, (3) is the Euler-Lagrange equation for the following Hardy-Sobolev variational problem: For γ<γH(Ω) and 0≤s≤2, there exists μγ,s(Ω)>0 such that
[TABLE]
Note that when s=2 and γ=0, this is the Hardy inequality mentioned above, while if s=0 and γ=0, it is the Sobolev inequality. If Ω=Rn, s∈[0,2] and γ∈(−∞,4(n−2)2), (7) contains – after a suitable change of variables –
the Caffarelli-Kohn-Nirenberg inequalities [ckn]. The latter state that there is a constant C:=C(a,b,n)>0 such that,
[TABLE]
where
[TABLE]
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 H1,02(Ω) is embedded in the weighted space Lp(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 (7). 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 [aubin], [schoen1], [LeeParker]. Another complicating feature of the problem is that the term ∣x∣2u is as critical as ∣x∣su2∗(s)−1 in the sense that they have the same homogeneity as the Laplacian. These difficulties are summarized by the invariance of the problem under conformal transformation. Indeed, for a function u:Ω→R and r>0, let
[TABLE]
and note that Equation (3) is then ”essentially” invariant under the transformation u↦ur in the sense that
[TABLE]
This ”invariance” is behind the lack of compactness in the embeddings associated to the variational formulation of (3), which prohibits the use of general abstract topological or variational methods. However, as one notices, the invariance is not complete, since the potential h has changed, and the domain itself was transformed. As we shall see, both the geometry of the domain and -to a lesser extent- the potential h break the invariance enough that one will be able to recover compactness for (3).
Another important aspect of this problem is the singularity at [math] and its location within the domain since the Hardy potential does not belong to the Kato class. 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 standard sphere. 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 Rn, the problem of finding a conformal metric with prescribed scalar curvature K(x) leads to finding solutions of the form −Δu−γ∣x∣2u=K(x)u2⋆(0)−1 on Rn. The latter is a simplified version of the nonlinear Wheeler-DeWitt equation, which appears in quantum cosmology (see [BB, BE, LZ, SmetsTAMS] and the references cited therein).
This paper deals specifically with the case where [math] belongs to the boundary of a smooth domain Ω. We shall see that the boundary at [math] plays an important role, and our starting point is the existence Theorem 1 below for least energy solutions. It was first established by Ghoussoub-Robert [gr1] when γ=0, then by Lin-Wadade [LW3] when 0<γ<4(n−2)2 under the assumption that the mean curvature at [math] is negative. The result was extended to the range γ≤4n2−1 in [gr4], but more importantly, it was shown there that in the remaining range (4n2−1,4n2), the curvature condition does not suffice anymore and a more global condition is needed: the boundary mass mγ,h(Ω) of a domain associated to γ and h, that we now recall.
1.1. The models and the definition of the mass
Letting formally r→0 in (13), we get that u should behave like solutions to
[TABLE]
To the best of our knowledge, no explicit positive solution of (16) is known. This was the reason why a specific blowup analysis was carried out in [gr1], which relied on the symmetry properties and a precise description of the asymptotic behavior of such solutions –also established in [gr1]. On the other hand, the asymptotic behavior of such nonlinear problems is governed by the solutions to the linear problem
[TABLE]
One can then easily see that a function of the form u(x)=x1∣x∣−β is a solution to (19) if and only if β∈{β−(γ),β+(γ)}, where
[TABLE]
Theorem-Definition 1** ([gr4]).**
Let Ω be a smooth bounded domain of Rn(n≥3) such that 0∈∂Ω. Suppose γ<4n2 and let h∈C1(Ω) be such that the operator −Δ−γ∣x∣−2−h is coercive. Assuming that
[TABLE]
then there exists H∈C2(Ω∖{0}) such that
[TABLE]
Then, there exist constants c1,c2∈R with c1>0 such that
[TABLE]
as x→0. In the spirit of Schoen-Yau [SY], we define the boundary mass as
[TABLE]
which is independent of the choice of H.
The problem of existence of least energy solutions can now be summarized in the following theorem, whose proof can also be deduced from the refined blow-up techniques developed in this paper.
Let Ω be a smooth bounded domain in Rn(n≥3) such that the singularity [math] belongs to the boundary ∂Ω. Suppose that 0<s<2
and fix h0∈C1(Ω) such that −Δ−γ∣x∣−2−h0 is coercive. Assume one of the following two conditions:
•
γ≤4n2−1* and the mean curvature of ∂Ω at [math] is negative.*
•
4n2−1<γ<4n2* and the boundary mass mγ,h0(Ω) is positive.*
Then, there is a positive solution to (3) that is a minimizer for the associated variational problem,
[TABLE]
Our focus in this project, is to investigate the extent to which the above local curvature condition at [math] and the global (mass) condition on the domain are necessary for the existence of positive solutions. Most importantly, we give results pertaining to the persistence of the lack of positive solutions for (3) under C1-perturbations of the potential h. We will also show that, under suitable curvature conditions, this equation has an infinite number of non-necessarily positive solutions.
Both existence and non-existence results will follow from a sharp blow-up analysis of solutions to perturbations of Equation (3). More precisely, we consider
[TABLE]
and a family (hϵ)ϵ>0∈C1(Ω) such that
[TABLE]
We then perform a blow-up analysis, as ϵ go to zero, on a sequence of functions (uϵ)ϵ>0 in H1,02(Ω) such that uϵ is a solution to the Dirichlet boundary value problems:
[TABLE]
The novelty and delicacy of our analysis stem from the fact that the sequence (uϵ)ϵ>0 might blow up along excited states, as opposed to a unique ground state in [gr1]. Moreover, the sequence (uϵ)ϵ>0 is not assumed to have a fixed sign.
1.2. Non-existence: stability of the Pohozaev obstruction.
Starting with issues of non-existence of solutions, we shall prove the following surprising stability of regimes where variational positive solutions do not exist.
Theorem 2**.**
Let Ω be a smooth bounded domain in Rn(n≥3) such that the singularity [math] belongs to the boundary ∂Ω. Assume that 0<s<2 and γ<n2/4. Fix h0∈C1(Ω) such that −Δ−γ∣x∣−2−h0 is coercive, and assume that one of the following conditions hold:
•
γ≤4n2−1* and the mean curvature at [math] is non-zero;*
•
γ>4n2−1* and the boundary mass mγ,h0(Ω) is non-zero.*
If there is no positive variational solution to (3) with h=h0, then for all Λ>0, there exists ϵ:=ϵ(Λ,h0)>0 such that for any h∈C1(Ω) with
∥h−h0∥C1(Ω)<ϵ,
there is no positive solution to (3) such that ∥∇u∥2≤Λ.
The above result is surprising for the following reason: Assuming Ω is starshaped with respect to [math], then the classical Pohozaev obstruction (see Section 11) yields that (3) has no positive variational solution whenever
[TABLE]
We then get the following corollaries.
Corollary 1**.**
Let Ω be a smooth bounded domain in Rn(n≥3) such that 0∈∂Ω. Assume Ω is starshaped with respect to [math], 0<s<2 and γ<γH(Ω). If γ≤4n2−1, we shall also assume that the mean curvature at [math] is non-vanishing. If h0 is a potential satisfying (24), then for all Λ>0, there exists ϵ(Λ,h0)>0 such that for all h∈C1(Ω) satisfying
∥h−h0∥C1(Ω)<ϵ(Λ,h0),
there is no positive solution to (3) such that ∥∇u∥2≤Λ.
Corollary 2**.**
Let Ω be a smooth bounded domain in Rn(n≥3), such that 0∈∂Ω. We fix 0<s<2 and γ<γH(Ω), the Hardy constant defined in (4). Assume that
[TABLE]
When γ≤4n2−1, we assume that the mean curvature at [math] is positive. Then for all Λ>0, there exists ϵ(Λ)>0 such that for all λ∈[0,ϵ(Λ)), there is no positive solution to
[TABLE]
with ∥∇u∥2≤Λ.
It is worth comparing these results to what happens in the nonsingular case. Indeed, in contrast to the singular case, a celebrated result of Brezis-Nirenberg [bn] shows that, for γ=s=0, a variational solution to (25) always exists whenever n≥4 and 0<λ<λ1(Ω), with the geometry of the domain playing no role whatsoever. On the other hand, Druet-Laurain [dl] showed that the geometry plays a role in dimension n=3, still for γ=s=0, by proving that when Ω is star-shaped, then there is no solution to (25) for all small values of λ>0 (with no apriori bound on ∥∇u∥2). Another point of view is that for n=3, the nonexistence of solutions persists under small perturbations, but it does not for n≥4: the Pohozaev obstruction is stable only for n=3 in the nonsingular case.
This is in stark contrast with the situation here, i.e. when 0∈∂Ω and s>0. In this case, for both the existence and non-existence results, the geometry plays a role in all dimensions: it is either the local geometry at [math] (i.e., depending on whether the mean curvature at [math] is
vanishing or not) in high dimensions, or the global geometry of the domain (i.e., depending on whether the mass is positive or the domain is star-shaped) in low dimensions. Corollaries 1 and 2 show that the Pohozaev obstruction is stable in all dimensions in the singular case.
Let us discuss some extensions related to this absence or not of low/large dimension phenomenon.
•
Our stability result still holds under an additional smooth perturbation of the domain Ω, as was done by Druet-Hebey-Laurain [dhl] when n=3, γ=s=0.
•
In the forthcoming paper [GMR2], we tackle the case of the interior singularity 0∈Ω, where the results are much more in the spirit of Brezis-Nirenberg and Druet-Laurain concerning the dichotomy between low and high dimensions.
•
On of the main features of the stability result of Druet-Laurain [dl] is the absence of any apriori control on ∥∇u∥2. In the interor case 0∈Ω, we expect to get rid also of the apriori bound in the singular case s>0. In the boundary case 0∈∂Ω, bypassing the apriori bound by Λ is more delicate and will require extra care. These issues are projects in progress.
The proof of Theorem 2 (and Corollaries 1 and 2) relies on the blow-up analysis. Namely, arguing by contradiction, we assume the existence of solutions (uϵ)ϵ to (25) with pϵ≡0 and (hϵ)ϵ→h0 in C1 with a control on the Dirichlet energy. Due to the ”invariance” under the conformal transformation (10), the uϵ’s might concentrate on some peaks at [math]. The formation of these peaks is described via blow-up analysis in Proposition 3. Then Proposition 6 applies which yields vanishing of the mean curvature or the mass, depending on the dimension, contradicting the hypothesis of Theorem 2. Concerning Corollaries 1 and 2, the hypothesis imply that the mass is negative when defined.
1.3. Multiplicity of sign-changing solutions
As to the question of multiplicity, we shall prove the following result, which uses that in the subcritical case, i.e., when pϵ>0, there is an infinite number of higher energy solutions for such ϵ. Again, the core of the proof is a sharp blow-up analysis of such solutions as pϵ→0.
Theorem 3** (The general case).**
Let Ω be a smooth bounded domain in Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2. Let h0∈C1(Ω) and (hϵ)ϵ>0∈C1(Ω) be such that (23) holds, and
let (pϵ)ϵ>0 be such that (22) holds.
Consider a sequence of functions (uϵ)ϵ>0 that is uniformly bounded in H1,02(Ω) such that for each ϵ>0, uϵ satisfies
Equation (Eϵ). Then,
(1)
If γ<4n2−1 and the principal curvatures of ∂Ω at [math] are non-positive but not all of them vanish, then the sequence (uϵ)ϵ>0 is pre-compact in H1,02(Ω).
2. (2)
In particular, Equation (3) has an infinite number of (possibly sign-changing) solutions in H1,02(Ω).
The above result was established by Ghoussoub-Robert [gr2] in the case when γ=0. The main challenge here is to prove the compactness of the subcritical solutions at high energy levels, as the nonlinearities approach the critical exponent. The multiplicity result then follows from standard min-max methods. The proof
relies heavily on pointwise blow-up analysis techniques in the spirit of Druet-Hebey-Robert [dhr] and Druet [druet.jdg], though our situation adds considerable difficulties to carrying out the program.
1.4. Compactness Theorems and blow-up analysis
As mentioned above, the central tool is an analysis of the formation of peaks on families (uϵ)ϵ of solutions to equations like (3) when blow-up occurs. This long analysis yields Propositions 5 and 6 that describe the blow-up rate. When blowup does not occur, there is compactness. The following theorems are immediate consequences of these propositions.
We note that the restrictions on both γ and on the curvature at [math] are more stringent than for the existence of a ground state solution in Theorem 1. The stronger assumptions turned out to be due to the potentially sign-changing approximate solutions -actually solutions of subcritical problems- and not because they are not necessarily minimizing. Indeed, the following theorem does not assume any smallness of the energy bound as long as the approximate solutions are positive. It therefore yields another proof for Theorem 1, which does not rely on the existence of minimizing sequence below the energy level of a single bubble.
Theorem 4** (The non-changing sign case).**
Assume in addition to the hypothesis of Theorem 3,
that the solutions (uϵ)ϵ>0 satisfy for all ϵ>0,
[TABLE]
Then, the sequence (uϵ)ϵ>0 is pre-compact in H1,02(Ω), provided one of the following conditions is satisfied:
•
γ≤4n2−1* and the mean curvature of ∂Ω at [math] is negative.*
•
4n2−1<γ<4n2* and the boundary mass mγ,h0(Ω) is positive.*
Our method also shows that if the –possibly sign-changing– sequence is weakly null, then the compactness result in Theorem 3 will still hold for γ up to 4n2−41:
Theorem 5** (The case of a weak null limit).**
Assume in addition to the hypothesis of Theorem 3,
that the solutions (uϵ)ϵ>0 satisfy,
[TABLE]
If γ<4n2−1 and the principal curvatures of ∂Ω at [math] are non-positive but not all of them vanishing, then the sequence (uϵ)ϵ>0 converges strongly to [math] in H1,02(Ω).
1.5. Structure of the manuscript
This paper is organized as follows. Section 2 consists in preliminary material in order to introduce the sequence of functions that will be thoroughly analyzed in Sections 3 to 8 in the case where they ”blow-up”. Section 9 contains the proof of the multiplicity result and Section 10 will have the applications to non-existence regimes and their stability under perturbations. We then have five relevant appendices. The first (Appendix A, Section 11) introduces the Pohozaev identity in our setting. The second (Appendix B, Section 12) contains a technical lemma on the continuity of the first eigenvalue λ1(Δ+V) with respect to variations of the potential V. Appendix C (Section 13) recalls regularity results established in [gr4] about the regularity and behavior at [math] of solutions of equations involving the Hardy-Schrödinger operator on bounded domains having [math] on their boundary. In Appendix D (Section 14), we construct the Green functions associated to the operators −Δ−∣x∣2γ−h on such domains, and exhibit some of their properties needed throughout the memoir. The last Appendix E (Section 15) does the same but for the Hardy-Schrödinger operator −Δ−∣x∣2γ on R−n.
2. Setting up the blow-up
Throughout this paper, Ω will always be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω. We will always assume that γ<4n2 and s∈(0,2). We set 2⋆(s):=n−22(n−s). When γ<γH(Ω), then the
following Hardy-Sobolev inequality holds on Ω: there exists C>0 such that
[TABLE]
For each ϵ>0, we consider pϵ∈[0,2⋆(s)−2) such that
[TABLE]
Let h0∈C1(Ω) and consider a family (hϵ)ϵ>0∈C1(Ω) such that (23) holds. Consider a sequence of functions (uϵ)ϵ>0 in H1,02(Ω) such that for all ϵ>0 the function uϵ is a solution to the Dirichlet boundary value problem:
[TABLE]
By the regularity result Theorem 6 in Appendix B, we have uϵ∈C2(Ω∖{0}) and there exists Kϵ∈R such that limx→0d(x,∂Ω)∣x∣β−(γ)uϵ(x)=Kϵ. In addition, we assume that the sequence (uϵ)ϵ>0 is bounded in H1,02(Ω) and we let Λ>0 be such that
[TABLE]
It then follows from the weak compactness of the unit ball of H1,02(Ω) that there exists u0∈H1,02(Ω) such that as ϵ→0
[TABLE]
Note that u0 is a solution to the Dirichlet boundary value problem:
[TABLE]
From the regularity Theorem 6 we have u0∈C2(Ω∖{0}) and x→0limd(x,∂Ω)∣x∣β−(γ)u0(x)=K0∈R. It then follows that
Ωsupd(x,∂Ω)∣x∣β−(γ)u0(x) and hence ∥∣x∣β−(γ)−1u0(x)∥L∞(Ω) is finite.
We fix τ∈R such that
[TABLE]
The following proposition shows that the sequence (uϵ)ϵ is pre-compact in H1,02(Ω) if (∣x∣τuϵ)ϵ>0 is uniformly bounded in L∞(Ω).
Proposition 1**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23) and (29) holds. Suppose that there exists C>0 such that ∣x∣τ∣uϵ(x)∣≤C for all x∈Ω and for all ϵ>0. Then up to a subsequence, ϵ→0limuϵ=u0 in H1,02(Ω), where u0 is as in (31).
Proof of Proposition 1: The sequence (uϵ) is clearly uniformly bounded in L∞(Ω′) for any Ω′⊂⊂Ω∖{0}. Then by standard elliptic estimates and from (31) it follows that uϵ→u0 in Cloc2(Ω∖{0}). Now since ∣x∣τ∣uϵ(x)∣≤C for all x∈Ω and for all ϵ>0, and since τ<2n−2, we have
And hence ϵ→0limuϵ=u0 in H1,02(Ω). This proves Proposition 1. ∎
From now on, we assume that
[TABLE]
We shall say that blow-up occurs whenever (34) holds.
3. Scaling Lemmas
In this section we state and prove two scaling lemmas which we shall use many times in our analysis. We start by describing a parametrization around a point of the boundary ∂Ω. Let p∈∂Ω. Then there exists U,V open in Rn, there exists I⊂R an open interval, there exists U′⊂Rn−1 an open subset, and there exist a smooth diffeomorphism T:U⟶V and T0∈C∞(U′), such that upto a rotation of coordinates if necessary
[TABLE]
This boundary parametrization will be throughout useful during our analysis. An important remark is that
[TABLE]
Lemma 1**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) holds.
Let (yϵ)ϵ∈Ω and let
[TABLE]
Suppose ϵ→0limyϵ=0 and ϵ→0limνϵ=0. Assume that for any R>0 there exists C(R)>0 such that for all ϵ>0
[TABLE]
Then
[TABLE]
Proof of Lemma 1: We proceed by contradiction and assume that
[TABLE]
Then it follows from the definition of κϵ that
[TABLE]
Case 1: We assume that there exists ρ>0 such for all ϵ>0
that
[TABLE]
We define for all ϵ>0
[TABLE]
Note that this is well defined for ϵ>0 small enough. It follows from (48) that there exists
C(ρ)>0 such that all ϵ>0
weakly in B2ρ(0) for all ϵ>0. With the help of (50) and standard elliptic theory it then follows that there exists v∈C1(B2ρ(0)) such that
[TABLE]
In particular,
[TABLE]
and therefore v≡0.
On the other hand, a change of variables and the definition of κϵ yields
[TABLE]
Using the equation (Eϵ), (30), (49), (50) and passing to the limit ϵ→0 we get that
[TABLE]
and so then v≡0 in Bρ(0), a contradiction with (52). Thus (49) cannot hold in that case.
Case 2: We assume that, up to a subsequence,
[TABLE]
Note that ϵ→0limyϵ=0. Consider the boundary map T:U→V as in (46), where U,V are both open neighbourhoods of [math]. We let u~ϵ=uϵ∘T, which is defined in U∩R−n. For any i,j=1,...,n, we let gij=(∂iT,∂jT), where (⋅,⋅) denotes the Euclidean scalar product on Rn, and we consider g as a metric on Rn. We let Δg=divg(∇) the Laplace-Beltrami
operator with respect to the metric g. As easily checked, using (Eϵ) we get that for all ϵ>0
[TABLE]
weakly in U∩R−n. We let zϵ∈∂Ω be such that
[TABLE]
We let y~ϵ,z~ϵ∈U such that
[TABLE]
It follows from the properties (46) of the boundary map T that
[TABLE]
We rescale and define for all ϵ>0
[TABLE]
With 56), we get that v~ϵ is defined on BR(0)∩{x1<0} for all R>0, for ϵ is small enough. Then for all ϵ>0 the functions v~ϵ satisfies the equation:
[TABLE]
weakly in BR(0)∩{x1<0}. In this expression, g~ϵ=g(z~ϵ+κϵx) and Δg~ϵ is the Laplace-Beltrami operator with respect to the metric g~ϵ. With (53), (54) and (55), we get for all ϵ>0
[TABLE]
where, there exists CR>0 such that ∣OR(1)∣≤CR for all x∈BR(0)∩{x1≤0}. With (50), we then get that
[TABLE]
It follows from (48) that there exists C′(R)>0 such that all ϵ>0
[TABLE]
Using (50) and the propoerties of the boundary map T we then get as ϵ→0
[TABLE]
With the help of (50) and standard elliptic theory it then follows that there exists v~∈C1(BR(0)∩{x1≤0}) such that
[TABLE]
Since v~ϵ vanishes on BR(0)∩{x1=0} and (57) holds, it follows that
In particular, v~(0)=1, contradiction with (58). Thus (49) cannot hold in Case 2 also.
In both cases, we have contradicted (49) . This proves that yϵ=O(ℓϵ) when ϵ→0,
which proves the Lemma. ∎
Lemma 2**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) such that (Eϵ), (23), (29) and (30) holds. Let (yϵ)ϵ∈Ω and let
[TABLE]
Suppose νϵ→0 and ∣yϵ∣=O(ℓϵ) as ϵ→0.
Since 0∈∂Ω, we let T:U→V as in (46) with y0=0, where U,V are open neighborhoods of [math].
For ϵ>0 we rescale and define
[TABLE]
Assume that for any R>δ>0 there exists C(R,δ)>0 such that for all ϵ>0
[TABLE]
Then there exists w~∈H1,02(R−n)∩C1(R−n∖{0}) such that
[TABLE]
And w~ satisfies weakly the equation
[TABLE]
Moreover if w~≡0, then
[TABLE]
and there exists t∈(0,1] such that ϵ→0limνϵpϵ=t,
where μγ,s(R−n) is as in (7).
Proof of Lemma 2:
The proof proceeds in four steps.
Step 2.1:
Let η∈Cc∞(Rn). One has that ηw~ϵ∈H0,12(R−n) for ϵ>0 sufficiently small. We claim that there exists w~η∈H1,02(R−n) such that upto a subsequence
[TABLE]
We prove the claim. Let x∈R−n, then
[TABLE]
In this expression, DxT is the differential of the function T at x.
Now for any θ>0, there exists C(θ)>0 such that for any a,b>0
[TABLE]
With this inequality we then obtain
[TABLE]
Since D0T=IRn we have as ϵ→0
[TABLE]
With Hölder inequality and a change of variables this becomes
[TABLE]
Since ∥uϵ∥H1,02(Ω)=O(1), so for ϵ>0 small enough
[TABLE]
Where Cη is a constant depending on the function η. The claim then follows from the reflexivity of H1,02(R−n).
Step 2.2:
Let η1∈Cc∞(Rn), 0≤η1≤1 be a smooth cut-off function, such that
[TABLE]
For any R>0 we let ηR=η1(x/R). Then with a diagonal argument we can assume that upto a subsequence for any R>0 there exists w~R∈H1,02(R−n) such that
[TABLE]
Since ∥∇ηR∥n2=∥∇η1∥n2 for all R>0, letting ϵ→0 in (\refb′ddontheintegral1) we obtain that
[TABLE]
where C is a constant independent of R. So there exists w~∈H1,02(R−n) such that
[TABLE]
Step 2.3:
We claim that w~∈C1(R−n∖{0}) and it satisfies weakly the equation
[TABLE]
We prove the claim. For any i,j=1,...,n, we let (g~ϵ)ij=(∂iT(ℓϵx),∂jT(ℓϵx)),
where (⋅,⋅) denotes the Euclidean scalar product on Rn. We
consider g~ϵ as a metric on Rn. We let Δg=divg(∇) the Laplace-Beltrami
operator with respect to the metric g. From (Eϵ) it follows that for any ϵ>0 and R>0, ηRw~ϵ satisfies weakly the equation
[TABLE]
and note that ηRw~ϵ≡0 on BR(0)∖{0}∩∂R−n.
From (46), (59) and using the standard elliptic estimates it follows that w~R∈C1(BR(0)∖{0}∩R−n) and that up to a subsequence
[TABLE]
Letting ϵ→0 in eqn \eqrefblowupeqn1 gives that wR satisfies weakly the equation
[TABLE]
Again we have that ∣w~R(x)∣≤C(R,δ) for all x∈BR/2(0)∖B2δ(0) and then again from standard elliptic estimates it follows that w~∈C1(R−n∖{0})
and R→+∞limw~R=w~ in Cloc1(R−n∖{0}), up to a subsequence. Letting R→+∞ we obtain that w~ satisfies weakly the equation
[TABLE]
This proves our claim.
Step 2.4:
Coming back to equation \eqrefb′ddontheintegral1 we have for R>0
[TABLE]
Since the sequence (uϵ)ϵ is bounded in H1,02(Ω), letting ϵ→0 and then R→+∞ we obtain for some constant C
[TABLE]
Now if w≡0 weakly satisfies the equation
[TABLE]
using the definition of μγ,s(R−n) it then follows that
[TABLE]
Hence ϵ→0lim(ℓϵνϵ)>0 which implies that
[TABLE]
Since ϵ→0limνϵ=0, therefore we have that 0<t≤1. This completes the lemma. ∎
4. Construction and exhaustion of the blow-up scales
In this section we prove the following proposition in the spirit of Druet-Hebey-Robert [dhr]:
Proposition 2**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) holds. Assume that blow-up occurs, that is
[TABLE]
Then there exists N∈N⋆ families of scales (μi,ϵ)ϵ>0 such that we have:
(A1)
ϵ→0limuϵ=u0* in Cloc2(Ω∖{0}) where u0 is as in(31). *
2. (A2)
0<μ1,ϵ<...<μN,ϵ*, for all ϵ>0. *
3. (A3)
ϵ→0limμN,ϵ=0 and ϵ→0limμi,ϵμi+1,ϵ=+∞ for all 1≤i≤N−1.* *
4. (A4)
For any 1≤i≤N and for ϵ>0 we rescale and define
[TABLE]
where ki,ϵ=μi,ϵ1−2⋆(s)−2pϵ.
Then there exists u~i∈H1,02(R−n)∩C1(R−n∖{0}), u~i≡0 such that u~i weakly solves the equation
For any i∈{1,...,N}, there exists ti∈(0,1] such that
limϵ→0μi,ϵpϵ=ti.
The proof of this proposition is inspired by [dhr] and proceeds in five steps.
Since s>0, the subcriticality 2⋆(s)<2⋆(s) of equations (Eϵ) along with (31) yields that uϵ→u0 in Cloc2(Ω∖{0}). So the only blow-up point is the origin.
Step 4.1:
The construction of the μi,ϵ’s proceeds by induction. This step is the initiation.
By the regularity Theorem 6 and the definition of τ in (32) it follows that for any ϵ>0 there exists x1,ϵ∈Ω∖{0} such that
[TABLE]
We define μ1,ϵ and k1,ϵ>0 as follows
[TABLE]
Since blow-up occurs, that is (34) holds and since uϵ→u0 in Cloc2(Ω∖{0}), we have that
[TABLE]
It follows that uϵ satisfies the hypothesis (48) of Lemma 1 with yϵ=x1,ϵ, νϵ=μ1,ϵ. Therefore
[TABLE]
In fact, we claim that there exists c1>0 such that
[TABLE]
We argue by contradiction and we assume that ∣x1,ϵ∣=o(k1,ϵ) as ϵ→0. Let x~1,ϵ:=T−1(x1,ϵ)∈R−n. Since ∣x1,ϵ∣=o(k1,ϵ) as ϵ→0, so also ∣x~1,ϵ∣=o(k1,ϵ) as ϵ→0.
We define for ϵ>0
[TABLE]
Using (Eϵ) we obtain that v~ϵ satisfies the equation
[TABLE]
The definition (69) yields as ϵ→0, ∣x∣τ∣v~ϵ(x)∣≤2 for all x∈R−n.
Standard elliptic theory then yields the existence of v~∈C2(R−n∖{0}) such that v~ϵ→v~ in Cloc2(R−n∖{0}) where
[TABLE]
In addition, we have that v~ϵ(∣x~1,ϵ∣−1x~1,ϵ)=1 and so v~≡0. Also since ∣x∣τ∣v~(x)∣≤2 in R−n∖{0}, we have the bound that
[TABLE]
which implies that
[TABLE]
Therefore x↦V~(x):=4∣x∣β+(γ)∣x1∣+4∣x∣β−(γ)∣x1∣−v~(x) is a positive solution to −ΔV~−∣x∣2γV~=0 in R−n. Proposition 9 yields the existence of A,B∈R such that
[TABLE]
But the pointwise control (72) then implies A=B=0 by letting ∣x∣→0 and →∞. This contradicts v~≡0. This proves Claim (71).
We rescale and define for all ϵ>0
[TABLE]
It follows from (69) and (71) that u~1,ϵ satisfies the hypothesis (59) of Lemma 2 with yϵ=x1,ϵ, νϵ=μ1,ϵ. Then using Lemma 2 we get that there exists u~1∈H1,02(R−n)∩C1(R−n∖{0}) weakly satisfying the equation:
[TABLE]
and
[TABLE]
It follows from the definition that u~io,ϵ(k1,ϵx~1,ϵ)=1. From (71) we therefore have that u~1≡0. And hence again from Lemma 2 we get that
[TABLE]
Moreover, there exists t1∈(0,1] such that ϵ→0limμ1,ϵpϵ=t1.
Since ∣x1∣∣x∣β−(γ)u~1∈C0(Rn), we get as ϵ→0
[TABLE]
and
[TABLE]
∎
Step 4.2: We claim that there exists C>0 such that
[TABLE]
We argue by contradiction and let (yϵ)ϵ>0∈Ω be such that
[TABLE]
By the regularity Theorem 6, it follows that the sequence (yϵ)ϵ>0 is well-defined and moreover ϵ→0limyϵ=0, since uϵ→u0 in Cloc2(Ω∖{0}).
For ϵ>0 we let
Let R>0 and let x∈BR(0) be such that yϵ+κϵx∈Ω. It follows from the definition (74) of yϵ that for all ϵ>0
[TABLE]
and then, for all ϵ>0
[TABLE]
for all x∈BR(0) such that yϵ+κϵx∈Ω. Using (75), we get that there exists C(R)>0 such that the hypothesis (48) of Lemma 1 is satisfied and therefore one has ∣yϵ∣=O(ℓϵ) when ϵ→0, contradiction to (75). This proves (73).
∎
Let I∈N⋆. We consider the following assertions:
(B1)
0<μ1,ϵ<...<μI,ϵ.
2. (B2)
limϵ→0μϵ,I=0 and limϵ→0μi,ϵμi+1,ϵ=+∞ for all 1≤i≤I−1
3. (B3)
For all 1≤i≤I, there exists u~i∈H1,02(R−n)∩C2(R−n∖{0}) such that u~i weakly solves the equation
[TABLE]
with
[TABLE]
and
[TABLE]
where for ϵ>0, we have set ki,ϵ=μi,ϵ1−2⋆(s)−2pϵ and
[TABLE]
4. (B4)
For all 1≤i≤I, there exists ti∈(0,1] such that
limϵ→0μi,ϵpϵ=ti.
We shall then say that (HI) holds if there exists I sequences (μi,ϵ)ϵ>0, i=1,...,I such that items (B1), (B2) (B3) and (B4) holds. Note that it follows from Step 4.1 that (H1) holds.
Next we show the following:
Step 4.3
Let I≥1. We assume that (HI) holds.
Then, either
[TABLE]
or HI+1 holds.
Proof of Step 4.3:
Suppose
R→+∞limϵ→0limsupΩ∖BRkI,ϵ(0)∣x∣2n−2∣uϵ(x)−u0(x)∣1−2⋆(s)−2pϵ=0.
Then, there exists a sequence of points (yϵ)ϵ>0∈Ω such that
[TABLE]
Since uϵ→u0 in Cloc2(Ω∖{0}) it follows that ϵ→0limyϵ=0. Then by the regularity Theorem 6 and since β−(γ)<2n−2, we get
[TABLE]
for some positive constant a.
In particular, ϵ→0lim∣uϵ(yϵ)∣=+∞. Let
and so hypothesis (59) of Lemma 2 is satisfied. Using Lemma 2, we then get that there exists u~I+1∈H1,02(R−n)∩C1(R−n∖{0}) that satisfies weakly the equation:
[TABLE]
while
[TABLE]
as ϵ→0.
We denote y~ϵ:=kI+1,ϵT−1(yϵ)∈R−n. From (78) it follows that that ϵ→0lim∣y~ϵ∣:=∣y~0∣>a/2=0.
Therefore
[TABLE]
Since u~I+1≡0 on ∂R−n∖{0} so y~ϵ∈/∂R−n and hence u~I+1≡0. Hence again from Lemma 2, we get
[TABLE]
and there exists tI+1∈(0,1] such that
ϵ→0limμI+1,ϵpϵ=tI+1. Moreover, it follows from (76) and (78) that
[TABLE]
Hence the families (μi,ϵ)ϵ>0, 1≤i≤I+1 satisfy HI+1.∎
The next step is equivalent to step 4.3 at intermediate scales.
Step 4.4
Let I≥1. We assume that (HI) holds. Then, for any 1≤i≤I−1 and for any δ>0, either
[TABLE]
or (HI+1) holds.
Proof of Step 4.4:
We assume that there exist an i≤I−1 and δ>0 such that
[TABLE]
It then follows that there exists a sequence (yϵ)ϵ>0∈Ω such that
[TABLE]
for some positive constant a. Note that a<+∞ since
[TABLE]
is uniformly bounded for all x∈Ω∩Bδki+1,ϵ(0)∖BRki,ϵ(0).
Let y~ϵ∗∈R−n be such that T−1(yϵ)=ki+1,ϵy~ϵ∗. It follows that ∣y~ϵ∗∣≤δ for all ϵ>0. We rewrite (80) as
[TABLE]
Then from point (B3) of HI it follows that y~ϵ∗→0 as ϵ→0. Since ∣x1∣∣x∣β−(γ)u~i+1∈C0(Rn), we get as ϵ→0
so that hypothesis (59) of Lemma 2 is satisfied. We can then use it to get that there exists u~∈D1.2(R−n)∩C1(R−n∖{0}) that satisfies weakly the equation:
[TABLE]
while
[TABLE]
We denote y~ϵ:=ℓϵT−1(yϵ)∈R−n. From (81) it follows that that ϵ→0lim∣y~ϵ∣:=∣y~0∣>a/2=0.
Therefore
[TABLE]
Since u~≡0 on ∂R−n∖{0} so y~ϵ∈/∂R−n and hence u~≡0. Hence again from Lemma 2 we get
[TABLE]
and there exists t∈(0,1] such that
ϵ→0limνϵpϵ=t. Moreover, from (81), (79), and since ϵ→0limki+1,ϵ∣yϵ∣=0, it follows that
[TABLE]
Hence the families (μ1,ϵ),…, (μi,ϵ), (νϵ), (μi+1,ϵ),…, (μI,ϵ) satisfy (HI+1).∎
The last step tells us that the process of constructing {HI} stops after a finite number of steps.
Step 4.5: Let N0=max{I:(HI) holds }. Then N0<+∞ and the conclusion of Proposition 2 holds with N=N0.
Proof of Step 4.5:
Indeed, assume that (HI) holds. Since μi,ϵ=o(μi+1,ϵ) for all 1≤i≤N−1, we get with a change of variable and the
definition of u~i,ϵ that for any R>δ>0
[TABLE]
Here gi,ϵ is the metric such that (gϵ,i)qr=(∂qT(ki,ϵx),∂rT(ki,ϵx)) for all q,r∈{1,...,n}. Then from (30) we have
[TABLE]
Passing to the limit ϵ→0 and then δ→0, R→+∞ we obtain using point (B3) of HI, that
[TABLE]
from which it follows that N0<+∞. □
To complete the proof, we let families (μ1,ϵ)ϵ>0,…, (μN0,ϵ)ϵ>0 be such that HN0 holds. We argue
by contradiction and assume that the conclusion of Proposition 2 does not hold with N=N0. Assertions (A1), (A2), (A3),(A4), (A5) , (A7) and (A9) hold. Assume that (A6) or (A8) does not hold. It then follows from Steps 4.3, 4.4 and 4.5 that HN+1 holds. A contradiction with the choice of N=N0. Hence the proposition is proved. ∎
5. Strong pointwise estimates
The objective of this section is to obtain pointwise controls on uϵ and ∇uϵ. The core is the proof of the following proposition in the spirit of Druet-Hebey-Robert [dhr]:
Proposition 3**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) holds. Assume that blow-up occurs, that is
[TABLE]
Consider μ1,ϵ,...,μN,ϵ from Proposition 2. Then, there exists C>0 such that for all ϵ>0
[TABLE]
for all x∈Ω.
The proof of this estimate, inspired by the methodology of [dhr], proceeds in seven steps.
Step 5.1: We claim that for any α>0 small and any R>0, there exists C(α,R)>0 such that for all ϵ>0 sufficiently small, we have for all x∈Ω∖BRkN,ϵ(0),
[TABLE]
Proof of Step 5.1: We fix γ′ such that γ<γ′<4n2. Since the operator −Δ−∣x∣2γ−h0(x) is coercive, taking γ′ close to γ it follows that the operator −Δ−∣x∣2γ′−h0 is also coercive in Ω. From Theorem 7, there exists H∈C2(Ω∖{0}) such that
[TABLE]
And we have the following bound on H, that there exists C1>0 such that
[TABLE]
We let λ1γ′>0 be the first eigenvalue of the coercive operator −Δ−∣x∣2γ′−h0 on Ω and we let
φ∈C2(Ω∖{0})∩H1,02(Ω) be the unique eigenfunction such that
[TABLE]
It follows from the regularity result, Theorem 6 that there exists C2>0 such that
[TABLE]
We define the operator
[TABLE]
Step 5.1.1: We claim that given any γ<γ′<4n2 there exist δ0>0 and R0>0 such that for any 0<δ<δ0 and R>R0, we have for ϵ>0 sufficiently small
[TABLE]
We prove the claim. As one checks for all ϵ>0 and x∈Ω
[TABLE]
and
[TABLE]
One has for ϵ>0 sufficiently small ∥h0−hϵ∥∞≤4(1+Ωsup∣x∣2)γ′−γ and we choose 0<δ0<1 such that
[TABLE]
This choice is possible thanks to (23) and the regularity Theorem 6 respectively. It follows from point (A6) of Proposition 2 that, there exists R0>0 such that for any R>R0, we have for all ϵ>0 sufficiently small
[TABLE]
With this choice of δ0 and R0 we get that for any 0<δ<δ0 and R>R0, we have for ϵ>0 small enough
[TABLE]
for all x∈Bδ(0)∖BRkN,ϵ(0)∩Ω, if u0≡0, and
[TABLE]
Hence we obtain that for ϵ>0 small enough
[TABLE]
Similarly we have
[TABLE]
∎
Step 5.1.2: It follows from point (A4) of Proposition 2 that there exists C1′(R)>0 such that for all ϵ>0 small
[TABLE]
By estimate (90) on H, we then have for some constant C1(R)>0
[TABLE]
It follows from point (A1) of Proposition 2 and the regularity Theorem 6, that there exists C2′(δ)>0 such that for all ϵ>0 small
[TABLE]
And then by the estimate (95) on φ we have for some constant C2(δ)>0
Therefore when u0≡0 it follows from (5) and (102) that for all ϵ>0 sufficiently small
[TABLE]
and from (5) and (103), in case u0≡0, we have for ϵ>0 sufficiently small
[TABLE]
Since Ψϵ>0 and LϵΨϵ>0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [bnv] that the operator Lϵ satisfies the comparison principle on Bδ(0)∖BRkN,ϵ(0)∩Ω. Therefore
[TABLE]
Therefore when u0≡0 we have for for all ϵ>0 small
[TABLE]
for all x∈Bδ(0)∖BRkN,ϵ(0)∩Ω, for R large and δ small.
Then, when u0≡0, using the estimates (90) and (95), we have or all ϵ>0 small
[TABLE]
for all x∈Bδ(0)∖BRkN,ϵ(0)∩Ω, for R large and δ small.
Similarly if u0≡0, then all ϵ>0 small and R>0 large
[TABLE]
Taking γ′ close to γ, along with points (A1) and (A4) of Proposition 2, it then follows that estimate (85) holds on Ω∖BRkϵ,N(0) for all R>0.∎
Step 5.2: Let 1≤i≤N−1. We claim that for any α>0 small and any R,ρ>0, there exists C(α,R,ρ)>0 such that all ϵ>0.
[TABLE]
for all x∈Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω.
Proof of Step 5.2: We let i∈{1,...,N−1}. We emulate the proof of Step 5.1. Fix γ′ such that γ<γ′<4n2. Consider the functions H and φ defined in Step 5.1 satisfying (89) and (89) respectively.
Step 5.2.1:
We claim that given any γ<γ′<4n2 there exist ρ0>0 and R0>0 such that for any 0<ρ<ρ0 and R>R0, we have for ϵ>0 sufficiently small
We prove the claim. As one checks for all ϵ>0 and x∈Ω
[TABLE]
We choose 0<ρ0<1 such that
[TABLE]
It follows from point (A8) of Proposition 2 that there exists R0>0 such that for any R>R0 and any 0<ρ<ρ0, we have for all ϵ>0 sufficiently small
[TABLE]
for all x∈Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω.
With this choice of ρ0 and R0 we get that for any 0<ρ<ρ0 and R>R0, we have for ϵ>0 small enough
[TABLE]
Hence as in Step 5.1 we have that for ϵ>0 small enough
[TABLE]
Step 5.2.2: Let i∈{1,...,N−1}. It follows from point (A4) of Proposition 2 that there exists C1′(R)>0 such that for all ϵ>0 small
[TABLE]
And then by the estimate (90) on H we have for some constant C1(R)>0
[TABLE]
Again from point (A4) of Proposition 2 it follows that there exists C2′(ρ)>0 such that for all ϵ>0 small
[TABLE]
and then by the estimate (95) on φ we have for some constant C2(δ)>0
[TABLE]
We let for all ϵ>0
[TABLE]
Then (107) and (108) implies that for all ϵ>0 small
[TABLE]
Therefore it follows from (105) and (109) that ϵ>0 sufficiently small
[TABLE]
Since Ψ~ϵ>0 and LϵΨ~ϵ>0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [bnv] that the operator Lϵ satisfies the comparison principle on Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω. Therefore
[TABLE]
So for all ϵ>0 small
[TABLE]
for all x∈Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω, for R large and ρ small. Then using the estimates (90) and (95) we have or all ϵ>0 small
[TABLE]
for all x∈Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω, for R large and ρ small.
Taking γ′ close to γ, along with point (A4) of Proposition 2 it then follows that estimate (104) holds on Bρki+1,ϵ(0)∖BRki,ϵ(0)∩Ω, for all Rρ>0.∎
Step 5.3: We claim that for any α>0 small and any ρ>0, there exists C(α,ρ)>0 such that all ϵ>0.
[TABLE]
Proof of Step 5.3: Fix γ′ such that γ<γ′<4n2. Consider the function φ defined in Step 5.1 satisfying (89).
Step 5.3.1:
We claim that given any γ<γ′<4n2 there exist ρ0>0 such that for any 0<ρ<ρ0 we have for ϵ>0 sufficiently small
Therefore it follows from (111) and (113) that ϵ>0 sufficiently small
[TABLE]
Since the operator Lϵ satisfies the comparison principle on Bρk1,ϵ(0). Therefore
[TABLE]
And so for all ϵ>0 small
[TABLE]
for ρ small. Using the estimate (95) we have or all ϵ>0 small
[TABLE]
for ρ small. It then follows from point (A4) of Proposition 2 that estimate (110) holds on x∈Bρk1,ϵ(0)∩Ω for all ρ>0.∎
Step 5.4: Combining the previous three steps, it follows from (85), (104), (110) and Proposition 2 that for any α>0 small, there exists C(α)>0 such that for all ϵ>0 we have for all x∈Ω,
[TABLE]
Next we improve the above estimate and show that one can take α=0 in (114).
We let Gϵ be the Green’s function for the coercive operator −Δ−∣x∣2γ−hϵ on Ω with Dirichlet boundary condition. Green’s representation formula, the pointwise bounds on the Green’s function (243) and the regularity Theorem 6, yields for any z∈Ω,
[TABLE]
and therefore,
[TABLE]
To estimate the above integral we break it into three parts.
Step 5.5: There exist C>0 such that for any sequence (zϵ) with zϵ∈Ω∖BkN,ϵ(0), we have
[TABLE]
Proof of Step 5.5: To estimate the right-hand-side of (5) in this case, we split Ω into four subdomains as: Ω=i=1⋃4Di,ϵN where
In our next result we obtain a pointwise control on the gradient.
Proposition 4**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) holds. Assume that blow-up occurs, that is
[TABLE]
Consider the μ1,ϵ,...,μN,ϵ from Proposition 2. Then there exists C>0 such that for all ϵ>0
[TABLE]
for all x∈Ω∖{0}.
Proof of Proposition 4: We let Gϵ be the Green’s function of the coercive operator −Δ−∣x∣2γ−hϵ on Ω with Dirichlet boundary condition. Differentiating the Green’s representation formula, and then using the pointwise bounds on the gradient Green’s function (245) and the regularity result Theorem 6 yields for any z∈Ω
[TABLE]
Then using the pointwise estimates (84) the proof goes exactly as in Proposition 3.
□
6. Sharp blow-up rates and the proof of Compactness
The proof of compactness rely on the following two key propositions.
Proposition 5**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2. Let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) holds. Assume that blow-up occurs, that is
[TABLE]
Consider the μ1,ϵ,...,μN,ϵ and t1,...,tN from Proposition 2. Suppose that
[TABLE]
Then, we have following blow-up rates:
[TABLE]
Here II0 denotes the second fundamental form of ∂Ω at 0∈∂Ω and
[TABLE]
Proposition 6** (The positive case).**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω and assume that 0<s<2, γ<4n2.
Let (uϵ), (hϵ) and (pϵ) be as in Proposition 5 and let H(0) denote the mean curvature of ∂Ω at [math]. Assume that blow-up occurs as in (133).
Consider the μ1,ϵ,...,μN,ϵ and t1,...,tN from Proposition 2. Suppose in addition that
[TABLE]
We define
[TABLE]
Then, we have the following blow-up rates:
1) When β+(γ)−β−(γ)≥2, then
[TABLE]
[TABLE]
for some K~>0.
2) When β+(γ)−β−(γ)<2, then u0≡0 and
[TABLE]
[TABLE]
[TABLE]
where
[TABLE]
the constant K is as in (205), χ>0 is a constant and mγ,h(Ω) is the boundary mass defined in Theorem 1.
Proof of Theorems 3, 5 and 4: We argue by contradiction and assume that the family is not pre-compact. Then, up to a subsequence, it blows up. We then apply Propositions 5 and 6 to get the blow-up rate (that is nonegative). However, the hypothesis of Theorems 3, 5 and 4 yield exactly negative blow-up rates. This is a contradiction, and therefore the family is pre-compact. This proves the Theorems.∎
We now establish Propositions 5 and 6. The proof is divided in 13 steps in Sections 7 to 8. These steps are numbered Steps P1, P2, etc.
7. Estimates on the localized Pohozaev identity
In the sequel, we let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. Note that
[TABLE]
and
[TABLE]
In the sequel, we will permanently use the following consequence of (A9) of Proposition 2: for all i=1,...,N, there exists ci>1 such that
[TABLE]
Step P1** (Pohozaev identity).**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We define
[TABLE]
We let T be a chart at [math] as in (46). We define rϵ:=μN,ϵ. Then
[TABLE]
and, for δ0>0 small enough,
[TABLE]
Proof of Step P1: We apply the Pohozaev identity (11) with y0=0 and
[TABLE]
This yields
[TABLE]
It follows from the properties of the boundary map that
[TABLE]
Since for all ϵ>0, uϵ≡0 on ∂Ω, identity (7) yields (P1). Concerning (P1), we apply the Pohozaev identity (11) with y0=0 and
[TABLE]
The argument is similar. This ends the proof of Step P1.∎
We will estimate each of the terms in the above integral identities and calculate the limit as ϵ→0.
7.1. Estimates of the L2⋆(s) and L2−terms in the localized Pohozaev identity
Step P2**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We claim that, as ϵ→0
[TABLE]
and
[TABLE]
Proof of Step P2: For any R,ρ>0 we decompose the above integral as
[TABLE]
We will evaluate each of the above terms and calculate the limit R→+∞limρ→0limϵ→0lim.
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We claim that
[TABLE]
And if u0≡0
[TABLE]
Proof of Step P3:
From estimate (84) and after a change of variables, we get as ϵ→0,
[TABLE]
Case 1: Assuming that β+(γ)−β−(γ)<2, we then have the following rough bound from (159),
[TABLE]
Case 2: Assuming β+(γ)−β−(γ)≥2, then via a change of variable in (159), we get
[TABLE]
Therefore, if β+(γ)−β−(γ)>2, then
[TABLE]
When β+(γ)−β−(γ)=2, we get that
[TABLE]
Since μN,ϵ→0 and limϵ→0μi,ϵ/μN,ϵ is finite for all i=1,...,N−1, we get that
[TABLE]
since β+(γ)−β−(γ)=2.
Inequality (159) put together with (160), (161) and (162) yield (154).
When u0≡0 we decompose the integral and proceed as in the proof of (146) to obtain (158). This ends Step P3.∎
7.2. Estimate of the curvature term in the Pohozaev identity when β+(γ)−β−(γ)>1
Step P4**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs and that β+(γ)−β−(γ)>1. We claim that, as ϵ→0
[TABLE]
Here, see Proposition 5, II0 denotes the second fundamental form. Moreover, when u0≡0, we claim that as ϵ→0,
We consider the second fundamental form associated to ∂Ω,
II0(x,y)=(dνpx,y)
for 0∈∂Ω and all x,y∈T0∂Ω (ν is the outward normal vector at the hypersurface ∂Ω). In the canonical basis of ∂R−n=T0∂Ω, the matrix of the bilinear form II0 is −D02T0, where D02T0 is the Hessian matrix of T0 at [math]. Using the expression of T (see (46)), we can write for all x∈U∩∂R−n
[TABLE]
With the expression of T, we then
get that
[TABLE]
And so for all x∈U∩∂R−n.
[TABLE]
Since T0(0)=0 and ∇T0(0)=0 (see (46)), we then get as ∣x∣→0
[TABLE]
and therefore for all ϵ>0 and all x∈BR(0)∩∂R−n
[TABLE]
where ϵ→0limBR(0)∩{x1=0}sup∣θϵ,R∣=0 for any R>0.
Step P4.1:
Let 1≤i≤N−1. In Proposition 4 we have obtained the pointwise estimates, that for any R,ρ>0 and all ϵ>0 we have for all x∈Bρki+1,ϵ(0)∖BRki,ϵ(0),
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We fix a chart T as in (46) and, for any ϵ>0, we define
[TABLE]
where rϵ:=μN,ϵ. We claim that there exists v~∈C1(R−n∖{0}) such that
[TABLE]
where v~ is a solution of
[TABLE]
Proof of Step P5: For any i,j=1,...,n, we let (g~ϵ)ij=(T∗Eucl)(rϵx)ij=(∂iT(rϵx),∂jT(rϵx)),
where (⋅,⋅) denotes the Euclidean scalar product on Rn. We consider g~ϵ as a metric on Rn. In the sequel, we let Δg=divg(∇) be the Laplace-Beltrami
operator with respect to a metric g. From (Eϵ) it follows that for all ϵ>0, the rescaled functions v~ϵ weakly satisfies the equation
[TABLE]
with θ:=(2⋆(s)−2)2β+(γ)−β−(γ)>0 and v~ϵ≡0 on ∂R−n∖{0}.
Using the pointwise estimates (84) we obtain the bound, that as ϵ→0 we have for x∈R−n
[TABLE]
Then passing to limits in the equation (176), standard elliptic theory yields the existence of v~∈C2(R−n∖{0}) such that v~ϵ→v~ in Cloc2(R−n∖{0}) and v~ satisfies the equation:
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that u0≡0. We define
[TABLE]
We claim that
there exists uˉ∈C2(Ω∖{0}) such that
[TABLE]
Proof of Step P7: Since u0≡0, it follows from (84) that there exists C>0 such that
[TABLE]
Moreover, equation (Eϵ) rewrites
[TABLE]
and uˉϵ=0 on ∂Ω. It then follows from standard elliptic theory that the claim holds. This ends Step P7.∎
Step P8**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that u0≡0. We claim that
[TABLE]
and
[TABLE]
where
[TABLE]
Proof of Step P8: The second term has already been estimated in (179). We are left with the first term.
With a change of variable, the definition of uˉϵ and the convergence (182), we get
[TABLE]
where Fδ0 is as above.
Arguing as in the proof of (180), we get that
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that uϵ>0 for all ϵ>0. Then F0≥0 and
Proof of Step P9: We let v~ be defined as in Step P5. It follows from Step P5 that v~ satisfies (175) and we have the following bound on v~
[TABLE]
Given α∈R, we define vα(x):=x1∣x∣−α for all x∈R−n. Since v~≥0, it follows from Proposition 6.4 in Ghoussoub-Robert [gr4] that there exists A,B≥0 such that
In full generality, we compute Hδ(vα,vβ) for all α,β∈R and all δ>0. As one checks, for any i=1,...,n, we have that ∂ivα=(δi,1−α∣x∣2x1xi)∣x∣−α for all x∈R−n. Moreover, for x∈∂Bδ(0), we have that ∂νvα=∣x∣xi∂ivα. Consequently, straightforward computations yield
[TABLE]
and
[TABLE]
and then
[TABLE]
We have that
[TABLE]
and
[TABLE]
Plugging all these identities together yields
[TABLE]
Since β+(γ),β−(γ) are solutions to X2−nX+γ=0, we get that
[TABLE]
Since β+(γ)+β−(γ)=n and β+(γ)β−(γ)=γ, we get that
[TABLE]
Plugging all these results together yields (192). This ends Step P9.4.
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that β+(γ)−β−(γ)<2 and uϵ>0 for all ϵ>0. Then u0≡0.
Indeed, if β+(γ)−β−(γ)>1, the claim follows from (163) and 1>2β+(γ)−β−(γ). If now β+(γ)−β−(γ)<1, then (167) and the control (132) yield that
[TABLE]
as ϵ→0. The limit case β+(γ)−β−(γ)=1 is similar. This proves the claim.
Plugging (145), (154), (177), (179) and (194) into the Pohozaev identity (P1), we get
[TABLE]
as ϵ→0, where F0 is as in (193). Therefore F0≤0. Since uϵ>0, it then follows from (192) of Step P9 that u0≡0. This proves Step P10.∎
8. Proof of the sharp blow-up rates
We now prove the sharp blow-up rates claimed in Propositions 5 and 6. We start with the case when β+(γ)−β−(γ)=1. As a preliminary estimate, we claim that
[TABLE]
as ϵ→0, where F0 is as in (178); and, when u0≡0, we claim that
We prove the claim. Collecting the first estimate of Step P2, (154), (177) and (179) of the terms of the Pohozaev identity (P1) gives (196). Similarly, the second estimate of Step P2, (158), (184) and (185) of the terms of the Pohozaev identity (P1) gives (8).
8.1. Proof of the sharp blow-up rates when β+(γ)−β−(γ)=1
We first assume uϵ>0 and β+(γ)−β−(γ)<1.
Step P11**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that uϵ>0 and β+(γ)−β−(γ)<1. Then (139) holds, that is
[TABLE]
for some A>0, where mγ,h(Ω) is the boundary mass.
Proof of Step P11: It follows from Step P10 that u0≡0.
Indeed, the Pohozaev identity (P1), the convergence (181), (183), (182) and β+(γ)−β−(γ)<1 yield
[TABLE]
With u0≡0 and the control (132), we get that ∣∇uϵ(x)∣≤CμN,ϵ2β+(γ)−β−(γ)∣x∣−β+(γ) for all ϵ>0 and x∈Ω. Therefore, with (181) and (182), we get that
[TABLE]
as ϵ→0. Plugging (179), (200) and (8.1) into (P1), we get (199). This proves the claim and ends Step P11.1.
We fix δ<δ′. Taking U:=T(R−n∩Bδ′(0)∖Bδ(0)), K=0 and u=uˉ in (11), and using (182), we get that Mδ is independent of the choice of δ>0 small enough.
We prove the claim. Since uˉ≥0 is a solution to (182), it is enough to prove that uˉ≡0. We argue as in the proof of Step P9. We fix x∈Ω. Green’s identity anduϵ>0 yield
[TABLE]
where Aϵ:=T(R−n∩B2μN,ϵ(0)∖BμN,ϵ(0)), A:=R−n∩B2(0)∖B1(0). With the pointwise control (243), we get
[TABLE]
where uϵ,i is as in Proposition 2. Letting ϵ→0 and using the convergence (A4) of Proposition 2, we get that
[TABLE]
And then uˉ>0 in Ω. This proves the claim and Step P11.2.
We fix r0>0 and η∈C∞(Rn) such that η(x)=1 in Br0(0) and η(x)=0 in Rn∖B2r0(0). It then follows from [gr4, gr5] that, for r0>0 small enough, there exists A>0 and β∈H01(Ω) such that
We prove the claim. Since uˉ is a solution to (182), it follows from standard elliptic theory that there exists C>0 such that uˉ(x)+∣x∣∣∇uˉ(x)∣≤C∣x∣1−β+(γ) for all x∈Ω. Therefore, since β+(γ)−β−(γ)<1, we get that
[TABLE]
Therefore,
[TABLE]
as δ→0, where
[TABLE]
and
[TABLE]
We then get that
[TABLE]
as δ→0. For any x∈R−n∩Bδ(0), with the chart T and the definition of β, we get
[TABLE]
[TABLE]
Moreover, elliptic theory yields
[TABLE]
[TABLE]
where vβ is defined in the proof of Step P9. Since β+(γ)−β−(γ)<1 and β+(γ)+β−(γ)=n, we get with a change of variable that as δ→0,
[TABLE]
Using the computations performed in the proof of Step P9, we then get (202). This proves the claim and ends Step P11.3.
End of the proof of Step P11: Since Mδ is independent of δ small, we then get that Mδ0=−nωn−1(4n2−γ)A2mγ,h(Ω). Putting this estimate in (199), we then get (198). This end Step P11.∎
Proof of Proposition 5 when β+(γ)−β−(γ)>2: Plugging (163) into (196) and using that β+(γ)−β−(γ)>2, we obtain
Proof of Proposition 5 when β+(γ)−β−(γ)>1 and u0≡0. Plugging (164) into (8) and using that β+(γ)−β−(γ)>1, we obtain also (135).
Proof of Proposition 6 when β+(γ)−β−(γ)>1. Since uϵ>0, we get that u~N>0. Therefore, it follows from Ghoussoub-Robert [gr4] that uˉN(x1,x′)=UˉN(x1,∣x′∣) for all (x1,x′)∈(0,+∞)×Rn−1. Due to this symmetry, when β+(γ)−β−(γ)>1, we get that
[TABLE]
When β+(γ)−β−(γ)>2 or {β+(γ)−β−(γ)=2 and u0≡0}, Proposition 6 follows from (135) and (203). When {β+(γ)−β−(γ)=2 and u0>0}, Proposition 6 follows from (196), (192) of Step P9, (163) and (203). When 1<β+(γ)−β−(γ)<2, Proposition 6 follows from Step P10, (8), (164) and (203).
Proof of Proposition 6 when β+(γ)−β−(γ)<1: This is a direct consequence of Steps P10 and P11.
8.2. Proof of the sharp blow-up rates when β+(γ)−β−(γ)=1
We start with the following refined asymptotics when uϵ>0, β+(γ)−β−(γ)=1 and u0≡0.
Step P12**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that uϵ>0 and u0≡0. We fix a family of parameters (λϵ)ϵ>0∈(0,+∞) such that
for all x∈R−n∩λϵ−1U. As in the proof of (177), for any i,j=1,...,n, we let (g~ϵ)ij=(∂iT(rϵx),∂jT(rϵx)),
where (⋅,⋅) denotes the Euclidean scalar product on Rn. We consider g~ϵ as a metric on Rn. We let Δg=divg(∇), the Laplace-Beltrami
operator with respect to the metric g. From (Eϵ) it follows that for all ϵ>0, we have that
[TABLE]
With
[TABLE]
Since μN,ϵpϵ→tN>0 (see (A9) of Proposition 2) and
as ϵ→0. Since u0≡0, it follows from the pointwise control (84) that there exists C>0 such that 0<wϵ(x)≤C∣x1∣⋅∣x∣−β+(γ) for all x∈R−n∩λϵ−1U. It then follows from standard elliptic theory that there exists w∈C2(R−n∖{0}) such that
[TABLE]
with
[TABLE]
It follows from Lemma 4.2 in Ghoussoub-Robert [gr4] (see also Pinchover-Tintarev [PT]) that there exists Λ≥0 such that w(x)=Λ∣x1∣⋅∣x∣−β+(γ) for all x∈R−n. We are left with proving that Λ=K defined in (205). We fix x∈R−n. Green’s representation formula yields
[TABLE]
Step P12.1: We estimate the first term of the right-hand-side. Since D0T=IRn, a change of variable yields
[TABLE]
with
[TABLE]
It follows from (246) that for any z∈R−n, we have that
[TABLE]
and that the convergence is uniform with repect to z∈R−n∩(BR(0)∖Bδ(0)). Plugging this estimate in the above equality, using that kN,ϵ=μN,ϵ1−pϵ/(2⋆(s)−2), μN,ϵpϵ→tN>0 and the convergence of u~N,ϵ to u~N (see Proposition 2), we get that
Step P12.2: With the control (243) on the Green’s function and the pointwise control (84) on uϵ, we get that
[TABLE]
where
[TABLE]
[TABLE]
[TABLE]
where ℓϵ(x,y):=min{λϵ∣x∣,∣y∣}max{λϵ∣x∣,∣y∣}, and rϵ(x,y)=min{1,∣T(λϵx)−y∣2λϵ∣x1∣⋅∣y∣}.
Step P12.3. We first estimate Cϵ(δ). Since n>s+2⋆(s)(β−(γ)−1) (this is a consequence of β−(γ)<n/2), straightforward computations yield
[TABLE]
and therefore
[TABLE]
Step P12.4. We estimate Bϵ(R). We split the integral as
[TABLE]
where Iϵ(y) is the integrand. Since
[TABLE]
straightforward computations yield
[TABLE]
For the next term, a change of variable yields
[TABLE]
as ϵ→0. Finally, since β+(γ)+β−(γ)=n and n−s−(β+(γ)−1)2⋆(s)=22⋆(s)(β+(γ)−β−(γ)), we estimate the last term
[TABLE]
as ϵ→0. All these inequalities yield
[TABLE]
Step P12.5. We fix i∈{1,...,N−1} and estimate Ai,ϵ. As above, we split the integral as
[TABLE]
where Ji,ϵ is the integrand. Since μi,ϵ≤μN,ϵ, as one checks, the second and the third integral of the right-hand-side are controled from above respectively by ∫2λϵ∣x∣<∣y∣<2λϵ∣x∣Iϵ(y)dy and ∫∣y∣>2λϵ∣x∣Iϵ(y)dy that have been computed just above and go to [math] as ϵ→0. We are then left with the first term. With a change of variables, we have that
[TABLE]
since n>s+(2⋆(s)(β−(γ)−1)) and n<(β−(γ)−1)+s+(2⋆(s)−1)(β+(γ)−1). Since μi,ϵ=o(μN,ϵ) as ϵ→0, we get that
[TABLE]
Step P12.6: Plugging (208), (210), (211) and (212) into (207) and (209) yields limϵ→0wϵ(x)=K∣x∣β+(γ)∣x1∣ for all x∈R−n. With (206), we then get that Λ=K. This proves Step P12.
Now we can prove Proposition 6 when β+(γ)−β−(γ)=1 in the case when uϵ>0.
Step P13**.**
We let (uϵ), (hϵ) and (pϵ) be such that (Eϵ), (23), (29) and (30) hold. We assume that blow-up occurs. We assume that uϵ>0 and β+(γ)−β−(γ)=1. Then u0≡0 and
[TABLE]
The case β+(γ)−β−(γ)=1 of Proposition 6 is a consequence of Step P13.
Proof of Step P13: First remark that since β+(γ)+β−(γ)=n, we then have that
[TABLE]
It follows from Step P10 that u0≡0. We use (8) that writes
[TABLE]
where Tϵ:=T(∂R−n∩Bδ0(0)∖Bk1,ϵ3(0)).
It follows from (168) that
[TABLE]
With the control (132) and β+(γ)−β−(γ)=1, we get that
[TABLE]
We need an intermediate result. We let (sϵ)ϵ,(tϵ)ϵ∈[0,+∞) such that 0≤sϵ≤tϵ, and μϵ,N=o(tϵ) as ϵ→0. We claim that
[TABLE]
Indeed, with the pointwise control (132), u0≡0 and 2β+(γ)=n+1, we get that
[TABLE]
Distinguishing the cases sϵ≤μi,ϵ and sϵ≥μi,ϵ, we get (217). This proves the claim.
We define θϵ:=∣lnμN,ϵ∣1, αϵ:=μN,ϵθϵ and βϵ:=μN,ϵ1−θϵ. As one checks, we have that
[TABLE]
as ϵ→0. It then follows from (217) and the properties (218) that
[TABLE]
Since μN,ϵ=o(βϵ) and αϵ=o(1) as ϵ→0, it follows from Proposition P12 that
[TABLE]
We fix i,j∈{2,...,n}. It follows from (220) and β+(γ)−β−(γ)=1 that
[TABLE]
Independently, with a change of variable and 2β+(γ)=n+1, we get that
[TABLE]
where ωn−2 is the volume of the round (n−2)−unit sphere. This equality, (221) and the properties (218) yield
[TABLE]
Therefore, plugging (216), (219) and (8.2) into (215) yields
[TABLE]
Plugging this latest estimate into (214) yields (213). This ends the proof of Step P13.∎
9. Proof of multiplicity
Proof of Theorem 3: We fix γ<n2/4 and h∈C1(Ω) such that −Δ−γ∣x∣−2−h is coercive. For each 2<p≤2⋆(s), we consider the C2-functional
[TABLE]
on H1,02(Ω), whose critical points are the weak solutions of
[TABLE]
For a fixed u∈H1,02(Ω), u≡0, we have that
[TABLE]
Then, since coercivity holds, we have that that limλ→∞Ip,γ(λu)=−∞, which means
that for each finite dimensional subspace Ek⊂E:=H1,02(Ω), there
exists Rk>0 such that
[TABLE]
when p→2⋆(s)(s). Let (Ek)k=1∞ be an increasing sequence of subspaces of
H1,02(Ω) such that
dimEk=k and ∪k=1∞Ek=E:=H1,02(Ω) and define
the min-max values:
[TABLE]
where
[TABLE]
Proposition 7**.**
With the above notation and assuming n≥3, we have:
(1)
For each k∈N, cp,k>0 and p→2⋆(s)limcp,k=c2⋆(s),k:=ck.
2. (2)
If 2<p<2⋆(s), there exists for each k, functions up,k∈H1,02(Ω) such that Ip,γ′(up,k)=0, and Ip,γ(up,k)=cp,k.
3. (3)
For each 2<p<2⋆(s), we have
cp,k≥Dn,pkp−1p+1n2 where Dn,p>0 is
such that
p→2⋆(s)limDn,p=0.
4. (4)
k→∞limck=k→∞limc2⋆(s),k=+∞.
Proof: (1) Coercivity yields the existence of a0>0 such that
[TABLE]
With (225), the Hardy and the Hardy-Sobolev inequality (28), there exists C>0 and α>0 such that
[TABLE]
for all u∈H1,02(Ω) such that provided ∥∇u∥2=ρ for some ρ>0 small enough.
Then the sphere Sρ={u∈E;∥u∥H1,02(Ω)=ρ} intersects every image g(Ek) by an odd continuous function g. It follows that
[TABLE]
In view of (224), it follows that for each g∈Hk, we have
that
[TABLE]
where Dk denotes the
ball in Ek of radius Rk.
Consider now a sequence pi→2⋆(s) and note
first that for each u∈E, we have that Ipi,γ(u)→I2⋆(s),γ(u). Since g(Dk) is compact and the family of
functionals (Ip,γ)p is equicontinuous, it follows that
x∈EksupIp,γ(g(x))→x∈EksupI2⋆(s),γ(g(x)), from which follows that
i∈Nlimsupcpi,k≤x∈EksupI2⋆(s),γ(g(x)). Since this holds for any g∈Hk, it follows that
[TABLE]
On the other hand, the function f(r)=p1rp−2⋆(s)1r2⋆(s)
attains its maximum on [0,+∞) at r=1 and therefore
f(r)≤p1−2⋆(s)1 for all r>0. It follows
[TABLE]
from which follows that ck≤i∈Nliminfcpi,k, and
claim (1) is proved.
If now p<2⋆(s), we are in the subcritical case, that is we have
compactness in the Sobolev embedding H1,02(Ω)→Lp(Ω;∣x∣−sdx)
and
therefore Ip,γ has the Palais-Smale condition. It is then standard to
find critical points up,k for Ip,γ at each level cp,k (see for
example the book [Gh]). Consider now the functional
[TABLE]
and its critical values
[TABLE]
It has been shown in [gr2] that (1), (2) and (3) of Proposition 7 hold, with
cp,k0 and ck0 replacing cp,k and ck respectively. In particular, k→∞limck0=k→∞limc2⋆(s),k0=+∞.
On the other hand, with the coercivity (225), we have that
[TABLE]
where v=a0−p−21u. It then follows that k→∞limck=k→∞limc2⋆(s),k=+∞.
To complete the proof of Theorem 3, notice that since for each k, we
have
[TABLE]
it follows that the sequence (upi,k)i is
uniformly bounded in H1,02(Ω). Moreover, since Ipi′(upi,k)=0, it
follows from the compactness result that by letting pi→2⋆(s), we
get a solution uk of (223) in such a way that
I2⋆(s)(s),γ(uk)=p→2⋆(s)limIp,γ(up,k)=p→2⋆(s)limcp,k=ck. Since the latter sequence goes to infinity, it follows
that (223) has an infinite number of critical levels.
10. Proof of the non-existence result
Proof of Theorem 2: We argue by contradiction. We fix γ<γH(Ω)≤4n2 and Λ>0. We assume that there is a family (uϵ)ϵ>0∈H1,02(Ω) of solutions to
[TABLE]
with ∥∇uϵ∥2≤Λ and limϵ→0hϵ=h0 in C1(Ω).
We claim that (uϵ)ϵ>0 is not pre-compact in H1,02(Ω). Otherwise, up to extraction, there would be u0∈H1,02(Ω), u0≥0, such that uϵ→u0 in H1,02(Ω) as ϵ→0. Passing to the limit in the equation, we get that u0≥0 and
[TABLE]
The coercivity of −Δu0−γ∣x∣−2−h0 and the convergence of (hϵ)ϵ yield
[TABLE]
for small ϵ>0, and then, since uϵ>0, there exists c0>0 such that
[TABLE]
for all ϵ>0. Passing to the limit yields u0≡0. Therefore, u0>0 is a solution to (226) with ϵ=0.
This is not possible simply by the hypothesis.
The family (uϵ)ϵ is not pre-compact and it therefore blows-up with bounded energy. Let u0∈H1,02(Ω) be its weak limit, which is necessarily a solution to (227), and hence must be the trivial solution u0≡0. Proposition 6 then yields that
either
[TABLE]
or
[TABLE]
It now suffices to note that when γ≤(n2−1)/4 then β+(γ)−β−(γ)≥1 and the above contradicts our assumption that H(0)=0. Similarly, if γ>(n2−1)/4, then β+(γ)−β−(γ)<1 and the above contradicts our assumption that the mass is non-zero. In either case, this means that no such a family of positive solutions (uϵ)ϵ>0 exist. ∎
Proof of Corollary 1: First note that if h0 satisfies
[TABLE]
then by differentiating for any x∈Ω, the function t↦t2h0(tx) (which is well defined for t∈[0,1] since Ω is starshaped), we get that h0≤0. Therefore −Δ−γ∣x∣−2−h0 is coercive.
Assume now there is a positive variational solution u0 corresponding to h0, the Pohozaev identity (11) then gives
[TABLE]
Hopf’s strong comparison principle yields ∂νu0<0. Since Ω is starshaped with respect to [math], we get that (x,ν)≥0 on ∂Ω. Therefore, with (230), we get that (x,ν)=0 for all x∈Ω, which is a contradiction since Ω is smooth and bounded.
If now γ≤(n2−1)/4, the result follows from Theorem 2 since we have assumed that H(0)=0.
If γ>(n2−1)/4, we use
Theorem 7.1 in Ghoussoub-Robert [gr4] to find K∈C2(Ω∖{0}) and A>0 such that
[TABLE]
and such that
[TABLE]
where η∈Cc∞(Rn) and β∈H1,02(Ω) are as in Step P11. We now apply the Pohozaev identity (11) to K on the domain U:=Ω∖T(Bδ(0)) for T as in (46): using that K2∈L1(Ω) and (⋅,ν)(∂νK)2∈L1(∂Ω) when β+(γ)−β−(γ)<1, we get that
[TABLE]
where Mδ is defined in (199). With (202), we then get
[TABLE]
Since Ω is star-shaped and h0 satisfies (230), it follows that mγ,h0(Ω)<0 and Theorem 2 then applies to complete our corollary.
11. Appendix A: The Pohozaev identity
Proposition 8**.**
Let U⊂Rn be a smooth bounded domain and let u∈C2(U) be a solution of
[TABLE]
Then, we have
[TABLE]
where
[TABLE]
Proof:
For any y0∈Rn, the classical Pohozaev identity yields
[TABLE]
where ν is the outer normal to the boundary ∂U.
One has for 1≤j≤n
[TABLE]
So
[TABLE]
Then integration by parts yields
[TABLE]
[TABLE]
Similarly,
[TABLE]
[TABLE]
and
[TABLE]
Combining the above, we obtain for any K and any y0∈Rn,
[TABLE]
We conclude by taking y0=0 and using that u satisfies (231) on U.
12. Appendix B: A continuity property of the first eigenvalue of Schrödinger operators
Lemma 3**.**
Let Ω⊂Rn, n≥3, be a smooth bounded domain. Let (Vk)k:Ω→R and V∞:Ω→R be measurable functions and let (xk)k∈Ω be a sequence of points. We assume that
i)
k→+∞limVk(x)=V∞(x)* for a.e. x∈Ω,*
2. ii)
There exists C>0 such that ∣Vk(x)∣≤C∣x−xk∣−2 for all k∈N and x∈Ω.
3. iii)
k→+∞limxk=0∈∂Ω.**
4. iv)
For some γ0<n2/4, there exists δ>0 such that ∣Vk(x)∣≤γ0∣x−xk∣−2 for all k∈N and x∈Bδ(0)∩Ω.
5. v)
The first eigenvalue λ1(−Δ+Vk) is achieved for all k∈N.
Then,
[TABLE]
Proof:
We first claim that (λ1(−Δ+Vk))k is bounded. Indeed, fix φ∈H1,02(Ω)∖{0} and use the Hardy inequality to write for all k∈N,
[TABLE]
For the lower bound, we have for any φ∈H1,02(Ω),
[TABLE]
Since γ0<n2/4, we then get that λ1(−Δ+Vk)≥−4Cδ−2 for large k, which proves the lower bound.
Up to a subsequence, we can now assume that (λ1(−Δ+Vk))k converges as k→+∞. We now show that
[TABLE]
For k∈N, we let φk∈H1,02(Ω) be a minimizer of λ1(−Δ+Vk) such that ∫Ωφk2dx=1. In particular,
[TABLE]
Inequality (234) above yields the boundedness of (φk)k in H1,02(Ω). Up to a subsequence, we let φ∈H1,02(Ω) such that, as k→+∞, φk⇀φ weakly in H1,02(Ω), φk→φ strongly in L2(Ω) (then ∫Ωφ2dx=1) and φk(x)→φ(x) for a.e. x∈Ω. Letting
k→+∞ in (236), the hypothesis on (Vk) allow us to conclude that
[TABLE]
Since ∫Ωφ2dx=1 and we have extracted subsequences, we then get (235).
Finally, we prove the reverse inequality. For ϵ>0, let φ∈H1,02(Ω) be such that
[TABLE]
We have
[TABLE]
The hypothesis of Lemma 3 allow us to conclude that ∫Ω∣Vk−V∞∣φ2dx→0 as k→+∞. Therefore limsupk→+∞λ1(−Δ+Vk)≤λ1(−Δ+V∞)+ϵ for all ϵ>0. Letting ϵ→0, we get the reverse inequality and
the conclusion of Lemma 3.∎
13. Appendix C: Regularity and the Hardy-Schrödinger operator on R−n
In this section, we collect some important results from the paper [gr4] used in the proof of the compactness theorems. First we state the following regularity result:
Theorem 6** ([gr4], see also [ff] ).**
Let Ω be a smooth bounded domain of Rn(n≥3) such that 0∈∂Ω. We fix γ<4n2 and f:Ω×R→R is a Caratheodory function such that
[TABLE]
Let u∈H1,02(Ω) be a weak solution of
[TABLE]
for some θ>0. Then there exists K∈R such that
[TABLE]
Moreover, if u≥0 and u≡0, we have that K>0.
The following result characterizes the positive solution to the singular global equation
Proposition 9** ([gr4]).**
Let γ<4n2 and let u∈C2(Rn∖{0}) be a nonnegative function such that
[TABLE]
Then there exist C−,C+≥0 such that
[TABLE]
Next, we recall the existence and behaviour of the singular solution to the homogeneous equation.
Theorem 7** ([gr4]).**
Let Ω be a smooth bounded domain of Rn(n≥3) such that 0∈∂Ω. Fix γ<4n2 and h∈C1(Ω) be such that the operator Δ−γ∣x∣−2−h is coercive. There exists then H∈C2(Ω∖{0}) such that
[TABLE]
These solutions are unique up to a positive multiplicative constant, and there exists c>0 such that
H(x)≃x→0c∣x∣β+(γ)d(x,∂Ω).
14. Appendix D: Green’s function for the Hardy-Schrödinger operator with boundary singularity on a bounded domain
Definition 1**.**
Let Ω be a smooth bounded domain of Rn, n≥3, such that 0∈∂Ω. We fix γ<n2/4 and h∈C0,θ(Ω), θ∈(0,1) such that −Δ−(γ∣x∣−2+h) is coercive. We say that G:Ω×Ω∖{(x,x)/x∈Ω} is a Green’s function for −Δ−γ∣x∣−2−h if
∙* For any p∈Ω, Gp:=G(p,⋅)∈L1(Ω).*
∙* For all f∈Cc∞(Ω) and all p∈Ω, then*
[TABLE]
where φ∈H1,02(Ω)∩C0(Ω) is the unique solution to
[TABLE]
This appendix is devoted to the proof of the following result.
Theorem 8**.**
Let Ω be a smooth bounded domain of Rn such that 0∈∂Ω. We fix γ<4n2. We let h∈C0,θ(Ω) be such that −Δ−γ∣x∣−2−h is coercive. Then there exists a unique Green’s function G for −Δ−γ∣x∣−2−h. Moreover:
I. Properties of G.* The Green’s function G is such that*
(a)* Gp∈C2,θ(Ω∖{0,p}) and Gp>0 for all p∈Ω.*
(b)* For all p∈Ω and all η∈Cc∞(Rn∖{p}), we have that ηGp∈H1,02(Ω).*
(c)* For all f∈Ln+22n(Ω)∩Lq(Ω∖Bδ(0)), for all δ>0 and some q>n/2, we have for any p∈Ω*
[TABLE]
where φ∈H1,02(Ω)∩C0(Ω) is the unique solution to
[TABLE]
In particular,
[TABLE]
II. Asymptotics.* G satisfies the following properties:*
(d)* For all p∈Ω∖{0}, there exists c0(p)>0 such that*
[TABLE]
where
[TABLE]
(e)* There exists c>0 depending only on γ, the coercivity constant and an upper-bound for ∥h∥C0,θ such that*
[TABLE]
where
[TABLE]
And
[TABLE]
(f)* There exists Lγ,Ω>0 such that for any (hi)i∈C0,θ(Ω) such that i→+∞limhi=h in C0,θ, then for any sequences (xi)i,(yi)i∈Ω such that*
[TABLE]
then, as i→+∞ we have that
[TABLE]
Notations: In order to simplify notations, we will often drop the dependence in the domain Ω and the dimension n≥3. If F:A×B→R is a function, then for any x∈A, we define Fx:B→R by Fx(y):=F(x,y) for all y∈B. Finally, we will write Diag(A):={(x,x)/x∈A} for any set A.
We split the proof into several parts.
14.1. Proof of existence and uniqueness of the Green function
We let ηϵ(x):=η~(ϵ−1∣x∣) for all x∈Rn and ϵ>0, where η~∈C∞(R) is nondecreasing and such that η~(t)=0 for t<1 and η~(t)=1 for t>1. It follows from Lemma 3 (see Appendix B) and the coercivity of −Δ−(γ∣x∣−2+h) that there exists ϵ0>0 and c>0 such that such that for all φ∈H1,02(Ω) and ϵ∈(0,ϵ0),
[TABLE]
As a consequence, there exists c>0 such that for all φ∈H1,02(Ω) and ϵ∈(0,ϵ0),
[TABLE]
Let Gϵ>0 be the Green’s function of −Δ−(γηϵ∣x∣−2+h) on Ω with Dirichlet boundary condition. The existence follows from the coercivity and the C0,θ regularity of the potential for any ϵ>0 (see Robert [rob.green]). In particular, we have that
[TABLE]
Step 14.1: Integral bounds for Gϵ. We claim that for all δ>0 and 1<q<n−2n and δ′∈(0,δ), there exists C(δ,q)>0 and C(δ,δ′)>0 such that
[TABLE]
for all x∈Ω, ∣x∣>δ. We prove the claim. We fix f∈Cc∞(Ω) and let φϵ∈C2,θ(Ω) be the solution to the boundary value problem
[TABLE]
Multiplying the equation by φϵ, integrating by parts on Ω, using (247) and Hölder’s inequality, we get that
[TABLE]
where C>0 is independent of ϵ, f and φϵ. The Sobolev inequality ∥φ∥n−22n≤C∥∇φ∥2 for φ∈H1,02(Ω) then yields
[TABLE]
where C>0 is independent of ϵ, f and φϵ. Fix p>n/2 and δ∈(0,δ0) and δ1,δ2>0 such that δ1+δ2<δ, and x∈Ω such that ∣x∣>δ. It follows from standard elliptic theory that
[TABLE]
where C>0 depends on p,δ,δ1,δ2, γ and ∥h∥∞. Therefore, Green’s representation formula yields
for all f∈Cc∞(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 ∥Gϵ(x,⋅)∥Lq(Ω)≤C(δ,q) for all ϵ<ϵ0 and x∈Ω∖Bδ(0).
Let δ′∈(0,δ) and δ1,δ2>0 such that δ1+δ2<δ′. We get from (251) that
[TABLE]
for all f∈Cc∞(Ω∖Bδ′(x)). Here again, a duality argument yields (249), which proves the claim in Step 14.1.
Using the same method, we can get an improvement of the control, the cost being the integrability exponent q. When q∈(1,n/(n−1)), we get that p>n. Then, ∥φϵ∥C1(Bδ1(x)∩Ω) is controled by the Lp and Ln+22n norms. Moreover, ∣φϵ(x)∣≤∥φϵ∥C0(Bδ1(x)∩Ω)d(x,∂Ω). The argument above then yields
[TABLE]
Step 14.2: Convergence of Gϵ. Fix x∈Ω∖{0}. For 0<ϵ<ϵ′, since Gϵ(x,⋅), Gϵ′(x,⋅) are C2 outside x, (248) yields
[TABLE]
in the strong sense. The coercivity (247) then yields Gϵ(x,⋅)≥Gϵ′(x,⋅) for 0<ϵ<ϵ′ if γ≥0, and the reverse inequality if γ<0. It then follows from the integral bound (249) and elliptic regularity that there exists G(x,⋅)∈C2,θ(Ω∖{0,x}) such that
[TABLE]
In particular, G is symmetric and
[TABLE]
Moreover, passing to the limit ϵ→0 in (249), (253) and using elliptic regularity, we get that for all δ>0, 1<q<n−2n and δ′∈(0,δ), there exist C(δ,q)>0 and C(δ,δ′)>0 such that for all x∈Ω, ∣x∣>δ,
[TABLE]
and
[TABLE]
In particular, for any x∈Ω∖{0}, G(x,⋅)∈Lk(Ω) for all 1<k<n/(n−2) and G(x,⋅)∈L2n/(n−2)(Ω∖Bδ(x)) for all δ>0. Moreover, for any f∈Ln+22n(Ω)∩Lq(Ω∖Bδ(0)) for all δ>0 with q>n/2, let φϵ∈H1,02(Ω) be such that (250) holds. It follows from elliptic theory that φϵ∈C0,τ(Ω∖{0}) for some τ∈(0,1) and that for all δ1>0, there exists C(δ1)>0 such that ∥φϵ∥C0,τ(Ω∖Bδ1(0))≤C(δ1). We fix x∈Ω∖{0}. Passing to the limit ϵ→0 in the Green identity φϵ(x)=∫ΩGϵ(x,⋅)fdy yields
[TABLE]
where φ∈H1,02(Ω)∩C0(Ω∖{0}) is the only weak solution to
[TABLE]
Since G(x,⋅)≥0, (255) and the strong comparison principle yield G(x,⋅)>0. These points prove that G is a Green’s function for the operator and that (c) holds.
We now prove point (b). We fix η∈Cc∞(Rn−{x}) such that η(y)=1 when y∈Bδ(0) for some δ>0. Then ηGϵ(x,⋅)∈C2,θ(Ω)∩H1,02(Ω). It follows from (248) and (254) that
[TABLE]
where ∥fϵ∥C0(Ω)≤C for some C>0 and all ϵ>0. Therefore, with the coercivity (247) and the convergence (254), we get that
[TABLE]
for all ϵ>0. Reflexivity yields convergence of (ηGϵ(x,⋅)) in H1,02(Ω)∩L2(Ω) as ϵ→0 up to extraction. The convergence in C2 and uniqueness then yields ηG(x,⋅)∈H1,02(Ω) and ηGϵ(x,⋅)→ηG(x,⋅) in H1,02(Ω) as ϵ→0. The case of a general η is a direct consequence. This proves point (b).
For the uniqueness, we suppose G′ be another Green’s function. We fix x∈Ω and we define Hx:=Gx−Gx′. Then Hx∈L1(Ω) and for any f∈Cc∞(Ω), we have that ∫ΩHxfdy=0. Approximating a compactly supported function by smooth fonctions with compact support, we get that this equality holds for all f∈Cc0(Ω). Integration theory then yields Hx≡0, and then Gx′≡Gx. This proves uniqueness. This finishes the proof of (a).
This proves existence and uniqueness of the Green’s function in Theorem 8(I).
14.2. Proof of the upper bound
The behavior (242) is a consequence of the classification of solutions to harmonic equations and Theorem 4.1 in Ghoussoub-Robert [gr4].
In the proof, we will often use sub- and super-solutions to the linear problem. The following existence result is contained in Proposition 4.3 of [gr4]:
Proposition 10**.**
Let Ω be a smooth domain and h∈C0(Ω) be a continuous fonction. We fix γ<4n2 and β∈{β−(γ),β+(γ)}. Then, there exist r>0, and uβ,uβ∈C∞(Ω∖{0}) such that
[TABLE]
Moreover, for some τ>0, we have that, as x→0, x∈Ω,
[TABLE]
Step 14.3: Upper bound for G(x,y) when one variable is far from [math].
Step 14.3.1: It follows from (255), elliptic theory, (257) and (256) that for any δ>0, there exists C(δ)>0 such that
[TABLE]
Step 14.3.2:
We claim that for any δ>0, there exists C(δ)>0 such that
[TABLE]
Indeed, with no loss of generality, we can assume that δ∈(0,δ0). Let Ωδ be a smooth domain of Rn be such that Ω∖B3δ/4(0)⊂Ωδ⊂Ω∖Bδ/2(0). We fix x∈Ω such that ∣x∣>δ. Let Hx be the Green’s function for −Δ−(∣x∣2γ+h(x)) in Ωδ with Dirichlet boundary condition. Classical estimates (see [rob.green]) yield the existence of C(δ)>0 such that
[TABLE]
It is easy to check that
[TABLE]
Regularity theory then yields that Gx−Hx∈C2,θ(Ωδ). It follows from (261) that Gx(y)≤C1(δ)d(y,∂Ω)d(x,∂Ω) on (∂Ωδ)∩B3δ/4(0) for ∣x∣>δ. The comparison principle then yields Gx(y)−Hx(y)≤C1(δ)d(y,∂Ω)d(x,∂Ω) for y∈Ωδ and ∣x∣>δ. The above bound for Hx and (261) then yields (262).
Step 14.3.3:
We now claim that for any 0<δ′<δ, there exists C(δ,δ′)>0 such that
[TABLE]
We let δ1∈(0,δ′) that will be fixed later. We use (261) to deduce that Gx(y)≤C(δ,δ1)d(x,∂Ω)d(y,∂Ω) for all x∈Ω∖Bδ(0) and y∈∂Bδ1(0)∩Ω. Since δ1<∣x∣, we have that
[TABLE]
We choose a supersolution uβ−(γ) as in (259) of Proposition 10. It follows from (260) and (261) that for δ1>0, there exists C(δ,δ1)>0 such that Gx(z)≤C(δ,δ1)d(x,∂Ω)uβ−(z) for all z∈∂(Ω∩Bδ1(0)). It then follows from the comparison principle that Gx(y)≤C(δ,δ1)d(x,∂Ω)uβ−(y) for all y∈(Ω∩Bδ1(0))∖{0}. Combining this with (261) and (259), we obtain (263).
Note that by symmetry, we also get that for any 0<δ′<δ, there exists C(δ,δ′)>0 such that
[TABLE]
**Step 14.4: Upper bound for G(x,y) when both variables approach [math].
**
We claim first that for all c1,c2,c3>0, there exists C(c1,c2,c3)>0 such that for x,y∈Ω such that c1∣x∣<∣y∣<c2∣x∣ and ∣x−y∣>c3∣x∣, we have
[TABLE]
When one of the variables stays far from [math], (265) is a consequence of (261). We now consider a chart T at [math] as in (46). In particular, there is δ0>0, 0∈V⊂Rn and T:B2δ0(0)→V a smooth diffeomorphism such that T(0)=0 and
[TABLE]
Moreover, D0T=IRn and
[TABLE]
We fix X∈R−n such that 0<∣X∣<3δ0/2. We define
[TABLE]
so that
[TABLE]
where gX:=(T⋆Eucl)X is the pulled-back metric of the Euclidean metric Eucl via the chart T at the point X. Since H>0, it follows from the Harnack inequality on the boundary (see Proposition 6.3 in Ghoussoub-Robert [gr4]) that for all R>0 large enough and r>0 small enough, there exist δ1>0 and C>0 independent of ∣X∣<3δ0/2 such that
[TABLE]
which, via the chart T, yields
[TABLE]
for all x∈Ω such that ∣x∣<δ0. We let W be a smooth domain of Rn such that for some λ>0 small enough, we have
[TABLE]
We choose a subsolution uβ+(γ) as in (259) of Proposition 10. It follows from (260) and (261) that for ∣x∣<δ2 small
[TABLE]
Since −ΔGx−(γ∣⋅∣−2+h)Gx=0 outside [math], it follows from coercivity and the comparison principle that
[TABLE]
We fix z0∈W∖{0}. Then for δ3 small enough, when ∣x∣<δ3, it follows from (264) and the Harnack inequality (268) that there exists C>0 independent of x such that
[TABLE]
Taking r>0 small enough and R>0 large enough, we then get (265) for ∣x∣<δ3. The general case for arbitrary x∈Ω∖{0} then follows from (262). This completes the proof of (265).
Step 14.4.2: We claim that for all c1,c2>0, there exists C(c1,c2)>0 such that
[TABLE]
for all x,y∈Ω s.t. c1∣x∣<∣y∣<c2∣x∣. To prove (270), we distinguish three cases:
Case 1: We assume that
[TABLE]
We define
[TABLE]
Note that this definition makes sense since for such z, x+∣x∣z∈Ω. We then have that H∈C2(B1/(2C1)(0)∖{0}) and
[TABLE]
We now argue as in the proof of (262). From (265), we have that ∣H(z)∣≤C for all z∈∂B1/(2C1)(0) where C is independent of x∈Ω∖{0} satisfying (271). Let Γ0 be the Green’s function of −Δ−(∣∣x∣x+z∣2γ+∣x∣2h(x+∣x∣z)) at [math] on B1/(2C1)(0) with Dirichlet boundary condition. Therefore, H−Γ0∈C2(B1/(2C1)(0)) and, via the comparison principle, it is bounded by its supremum on the boundary. Therefore ∣z∣n−2H(z)≤C for all B1/(2C1)(0)∖{0} where C is independent of x∈Ω∖{0} satisfying (271). Scaling back and using (265), we get ∣x−y∣n−2Gx(y)≤C for all x,y∈Ω∖{0} such that c1∣x∣<∣y∣<c2∣x∣ and (271) holds. This proves (270) if d(x,∂Ω)d(y,∂Ω)≥∣x−y∣2. If d(x,∂Ω)d(y,∂Ω)<∣x−y∣2, we get that d(x,∂Ω)<2∣x−y∣, and then (271) yields ∣x∣≤2C1∣x−y∣, and (270) is a consequence of (265).
Case 2: By symmetry, (270) also holds when ∣y∣≤C1d(y,∂Ω).
Case 3: We assume that d(x,∂Ω)≤C1−1∣x∣ and d(y,∂Ω)≤C1−1∣y∣. We consider a chart at [math], that is δ0>0, 0∈V⊂Rn and T:B2δ0(0)→V a smooth diffeomorphism such that T(0)=0 and that (266) and (267) hold. We fix x′∈Rn−1 such that 0<∣x′∣<3δ0/2.
We assume that r≤c0∣x′∣. We define
[TABLE]
We then have that Hy∈C2(BR0(0)∩R−n∖{0,y}) and
[TABLE]
where gr:=(T⋆Eucl)(0,x′)+rz is the pulled-back metric of the Euclidean metric Eucl via the chart T at the point (0,x′)+rz. We now argue as in the proof of (262). From (214), we have that ∣Hy(z)∣≤C for all z∈∂BR0(0)∩R−n where C is independent of y∈BR0/2(0) and r∈(0,δ0/4). Let Γy be the Green’s function of −Δgr−((r∣T((0,x′)+rz))2γ+r2h(T((0,x′)+rz))) at y on Bc0/2(0)∩R−n with Dirichlet boundary condition. Therefore, Hy−Γy∈C2(Bc0/2(0)∩R−n) and, via the comparison principle, it is bounded by its supremum on the boundary. It follows from (214) and elliptic estimates for Γy (see for instance [rob.green]) that ∣Hy−Γy∣(z)≤C∣y1∣⋅∣z1∣ for z∈∂(Bc0/2(0)∩R−n) and y∈Bc0/4(0)∩R−n. Applying elliptic estimates, we then get that ∣Hy−Γy∣(z)≤C∣y1∣⋅∣z1∣ for z∈Bc0/2(0)∩R−n and y∈Bc0/4(0)∩R−n, and since
[TABLE]
(see [rob.green]), we get that
[TABLE]
where C is independent of x′∈Bδ0/2(0)∖{0}. This yields
[TABLE]
for ∣x′∣<δ0/3, r≤c0∣x′∣ and ∣y∣,∣z∣≤c0/4.
We now prove (270) in the last case. We fix x∈Ω∖{0} such that ∣x∣<δ0/3. We assume that d(x,∂Ω)≤C1−1∣x∣,d(y,∂Ω)≤C1−1∣y∣ and ∣x−y∣≤ϵ0∣x∣.
We let (x1,x′),(y1,y′)∈Bδ0(0) be such that x=T(x1,x′) and y=T(y1,y′). Taking the norm ∣(x1,x′)∣=∣x1∣+∣x′∣, we define r:=max{d(x,∂Ω),∣x−y∣}. Using that ∣X∣/2≤∣T(X)∣≤2∣X∣ for X∈Bδ0(0), up to taking ϵ0>0 small and C1,c0>1 large enough, we get that
[TABLE]
Therefore, (272) applies and we get (270) in Case 3.
We are now in position to conclude. Inequality (270) is a consequence of Cases 1, 2, 3, (262) and (214). This ends the proof of (270).
Step 14.4.3: We now show that there exists C>0 such that
[TABLE]
The proof goes essentially as in (263). For ∣x∣<δ with δ>0 small, we have that
[TABLE]
It follows from (214) that Gx(y)≤C∣x∣−nd(x,∂Ω)d(y,∂Ω) in Ω∩∂B∣x∣/3(0). We choose a supersolution uβ−(γ) as in (259) of Proposition 10. It follows from (260) and (214) that there exists C>0 such that
[TABLE]
The comparison principle yields that this inequality holds on Ω∩B∣x∣/3(0).
Step 14.4.4: By symmetry, we conclude that there exists C>0 such that
[TABLE]
Step 14.5: Finally, it follows from (273), (274) and (270) that there exists c>0 such that
[TABLE]
for all x,y∈Ω, x=y. This proves the upper bound in (243) of Theorem 8. The lower-bound and the control of the gradient will be proved in Section 14.4.
14.3. Behavior at infinitesimal scale
We prove three convergence results to get a comprehensive behavior of the Green’s function. Throughout this subsection, we assume Ω is a smooth bounded domain of Rn such that 0∈∂Ω. We fix γ<4n2 and let h∈C0,θ(Ω) be such that −Δ−γ∣x∣−2−h is coercive. We consider G to be the Green’s function of −Δ−γ∣x∣−2−h with Dirichlet boundary condition on ∂Ω.
Lemma 4**.**
Let (xi)i∈Ω and (ri)i∈(0,+∞) be such that
[TABLE]
Then, for all X,Y∈Rn such that X=Y, we have that
[TABLE]
Moreover, the convergence holds in Cloc2((Rn)2∖Diag(Rn)).
To deal with the case when the points approach the boundary, we consider a chart T as in (46). In particular, D0T=IRn.
Lemma 5**.**
Let (xi)i∈∂Ω and (ri)i∈(0,+∞) and x0∈∂Ω be such that
[TABLE]
We let T be a chart at x0 as in (46). We define xi′∈Rn−1 such that xi=T(0,xi′). Then, for all X,Y∈R−n such that X=Y, we have that
[TABLE]
where (Y1,Y′)∗=(−Y1,Y′) for (Y1,Y′)∈R×Rn−1. Moreover, the convergence holds in Cloc2((R−n)2∖Diag(R−n)}).
Lemma 6**.**
Let (ri)i∈(0,+∞) be such that limi→+∞ri=0. We let T be a chart at [math] as in (46). Then, for all X,Y∈R−n∖{0} such that X=Y, we have that
[TABLE]
where G(X,Y)=GX(Y) is the Green’s function for −Δ−γ∣x∣−2 on R−n with Dirichlet boundary condition. Moreover, the convergence holds in Cloc2((R−n∖{0})2∖Diag(R−n∖{0})).
Proof of Lemma 4: We let (ri)i∈(0,+∞) and (xi)i∈Ω as in the statement of the lemma. For any X,Y∈Rn, X=Y, we define
[TABLE]
for all i∈N. Since ri=o(d(xi,∂Ω)) as i→+∞, for any R>0, there exists i0∈N such that this definition makes sense for any X,Y∈BR(0). Equation (241) yields
for all X,Y∈BR(0) such that X=Y. Since 0∈∂Ω, we have that d(xi,∂Ω)≤∣xi∣, and therefore ri=o(∣xi∣) as i→+∞. Equation (276) and inequality (277) yield
[TABLE]
where θi→0 uniformly in Cloc0((Rn)2) and 0<Gi(X,Y)≤c∣X−Y∣2−n for all X,Y∈BR(0) such that X=Y. It then follows from standard elliptic theory that, up to a subsequence, there exists G∞(X,⋅)∈C2(Rn∖{X}) such that Gi(X,⋅)→G∞(X,⋅)≥0 in Cloc2(Rn∖{X}) and
[TABLE]
It then follows from the classification of positive harmonic functions that there exists λ>0 such that G∞(X,Y)=λ∣X−Y∣2−n for all X,Y∈Rn, X=Y.
We fix φ∈Cc∞(Rn). We define φi(x):=φ(ri−1(x−xi)) for x∈Ω (this makes sense for i large enough). It follows from (240) that
[TABLE]
Via a change of variable, and passing to the limit, we get that
[TABLE]
Since G∞(X,Y)=λ∣X−Y∣2−n, we get that λ=1/((n−2)ωn−1). Since the limit is unique, the convergence holds without extracting a subsequence. The convergence in Cloc2((Rn)2∖Diag(Rn)) follows from the symmetry of G and elliptic theory.∎
Proof of Lemma 5: The proof goes as in the proof of lemma 4, except that we have to take a chart due to the closeness of the boundary. We let (ri)i∈(0,+∞), (xi)i∈∂Ω and x0∈∂Ω as in the statement of the lemma. We let T be a chart at x0 as in (46) (in particular D0T=IRn) and we set xi′∈Rn such that xi=T(0,xi′). In particular, limi→+∞xi′=0. For any X,Y∈R−n, X=Y, we define
[TABLE]
for all i∈N. Here again, provided X,Y remain in a given compact set, the definition of Gi makes sense for large i. Equation (241) then rewrites
[TABLE]
where
[TABLE]
and gi=T⋆Eucl((0,xi′)+ri⋅) is the pull-back of the Euclidean metric. In particular, since D0T=IRn, we get that gi→Eucl in Cloc2(Rn). Since ri=o(∣xi∣), we get that ri=o(∣xi′∣) as i→+∞, and, using again that D0T=IRn, we get that θ^i→0 uniformly in BR(0)∩R−n. The pointwise control (275) rewrite Gi(X,Y)≤c∣X−Y∣2−n for all X,Y∈R−n, X=Y. With the same arguments as above, we get that for any X∈R−n, there exists G∞(X,⋅)∈C2(R−n∖{X}) such that
[TABLE]
[TABLE]
and
[TABLE]
with 0≤G∞(X,Y)≤c∣X−Y∣2−n for all X,Y∈R−n, X=Y. Define
[TABLE]
As one checks (see for instance [rob.green]), ΓR−n satisfies the same properties as G∞. We set f:=G∞(X,⋅)−ΓR−n(X,⋅). As one checks, f∈C∞(R−n∖{X}), −Δf=0 in the distribution sense in R−n, ∣f∣≤C∣X−⋅∣2−n in R−n∖{X} and f∂R−n=0. Hypoellipticity yields f∈C∞(R−n). Multiplying −Δf by f and integrating by parts, we get that f≡0, and then G∞(X,⋅)=ΓR−n(X,⋅). As above, this proves the convergence without any extraction. The convergence in Cloc2((R−n)2∖Diag(R−n)) follows from the symmetry of G and elliptic theory.∎
Proof of Lemma 6: Here again, the proof is similar to the two preceding proofs. We let (ri)i∈(0,+∞) such that limi→+∞ri=0. We let T be a chart at [math] as in (46) (in particular D0T=IRn). For any X,Y∈R−n∖{0}, we define
with Gi(X,⋅)≡0 on BR(0)∩∂R−n, where gi=T⋆Eucl(ri⋅) is the pull-back of the Euclidean metric. In particular, since D0T=IRn, we get that gi→Eucl in Cloc2(Rn). The pointwise control (275) writes
[TABLE]
It then follows from elliptic theory that Gi(X,⋅)→G∞(X,⋅) in Cloc2(R−n∖{0,X}). In particular, G∞(X,⋅) vanishes on ∂R−n∖{0} and
[TABLE]
Moreover, passing to the limit in Green’s representation formula, we get that
[TABLE]
Since G(x,⋅) is locally in H1,02(Ω) (see (b) in Theorem 8), we get that (ηGi(X,⋅))i is uniformly bounded in H1,02(R−n) for all η∈Cc∞(Rn∖{X}). Up to another extraction, we get weak convergence in H1,02(R−n), and then ηG∞(X,⋅)∈H1,02(R−n) for all η∈Cc∞(Rn∖{X}). It then follows from Theorem 9 and (279) that G∞(X,⋅)=GX is the unique Green’s function of −Δ−γ∣x∣−2 on R−n with Dirichlet boundary condition. Here again, the convergence in C2 follows from elliptic theory.∎
14.4. A lower bound for the Green’s function
We let Ω, γ, h be as in Theorem 8. We let G be the Green’s function for −Δ−(γ∣x∣−2+h) on Ω with Dirichlet boundary condition. We let (xi),(yi)i∈N be such that xi,yi∈Ω and xi=yi for all i∈N. We also assume that there exists x∞,y∞∈Ω such that
for x,y∈Ω, x=y. Note that c1<+∞ by (275). We claim that
[TABLE]
The lower bound in (243) and the upper bound in (245) both follow from (280).
This section is devoted to proving (280). We distinguish several cases:
Case 1: x∞=y∞, x∞,y∞∈Ω. As one checks, we then have that
[TABLE]
Therefore, we get that c1∈(0,+∞). Concerning the gradient, limi→+∞∣∇Gxi(yi)∣=∣∇Gx∞(y∞)∣≥0 and this yields c2<+∞. This proves (280) in Case 1.
Case 2: x∞∈Ω and y∞∈∂Ω∖{0}. Since x∞,y∞ are distinct and far from [math], we have that G(xi,yi)=d(yi,∂Ω)(−∂νGx∞(y∞)+o(1)) as i→+∞, where ∂νGx∞(y∞) is the normal derivative of Gx∞>0 at the boundary point y∞. Hopf’s Lemma then yields ∂νGx∞(y∞)<0. As one checks, we have that H(xi,yi)=(c+o(1))d(yi,∂Ω) as i→+∞. This then yields 0<c1<+∞. Concerning the gradient, we get that limi→+∞∣∇Gxi(yi)∣=∣∇Gx∞(y∞)∣≥0 and limi→+∞Γ(xi,yi)∈(0,+∞), which yields c2<+∞. This proves (280) in Case 2.
Case 3: x∞∈Ω and y∞=0∈∂Ω. It follows from Case 2 above that there exists c>0 such that Gxi(y)≥cd(y,∂Ω)∣y∣−β−(γ) for all y∈∂(Ω∩Br0(0)). We take the subsolution uβ−(γ) defined in Proposition 10. With (260), there exists c′>0 such that Gxi(y)≥c1uβ−(γ)(y) for all y∈∂(Ω∩Br0(0)). Since Gxi is locally in H1,02 around [math], the comparison principle and (260) yields Gxi(y)≥c"d(y,∂Ω)∣y∣−β−(γ) for all y∈Ω∩Br0(0). This yields c1>0.
We deal with the gradient. We let T be a chart at [math] as in (46) and we define
[TABLE]
with ri→0. It follows from (275) that Gi(y)≤C∣y1∣⋅∣y∣−β−(γ) for all y∈R−n∩B2(0). It follows from (241) that −ΔgiGi−(γ∣⋅∣2+o(1))Gi=0 in R−n∩B2(0) where gi:=T⋆Eucl(ri⋅) and o(1)→0 in Lloc∞(Rn). Elliptic regularity then yields ∣∇Gi(y)∣≤C for y∈R−n∩B3/2(0). We now let ri:=∣y~i∣ where yi:=T(y~i), so that ri→0. We then have that ∣∇Gi(yi~/ri)∣≤C, which rewrites ∣∇Gxi(yi)∣≤C∣yi∣−β−(γ). By estimating Γ(xi,yi), we then get that c2<+∞. This proves (280) in Case 3.
Case 4: x∞=y∞, x∞,y∞∈∂Ω∖{0}. Since x∞,y∞ are distinct and far from [math], we have that G(xi,yi)=d(yi,∂Ω)d(xi,∂Ω)(∂νx∂νyGx∞(y∞)+o(1)) as i→+∞, where ∂νx is the normal derivative along the first coordinate, and ∂νy is the normal derivative along the second coordinate. Since y↦Gx(y) is positive for x,y∈Ω, x=y, and solves (241), Hopf’s maximum principle yields −∂νyG(x,y∞)>0 for x∈Ω. Moreover, it follows from the symmetry of G that −∂νyG(x,y∞)>0 solves also (241). Another application of Hopf’s principle yields ∂νx∂νyGx∞(y∞)>0. Estimating independently H(xi,yi), we get that 0<c1<+∞.
We deal with the gradient. We have that ∣∇yGxi(yi)∣=∣∇y(Gxi−Gxi~)(yi)∣ where x~i∈∂Ω is the projection of xi on ∂Ω. The C2−control then yields ∣∇yGxi(yi)∣≤Cd(xi,∂Ω). Estimating independently Γ(xi,yi), we get that c2<+∞. This proves (280) in Case 4.
Case 5: x∞=y∞, x∞∈∂Ω∖{0} and y∞=0. It follows from Cases 2 and 4 that Gxi(y)≥Cd(xi,∂Ω)d(yi,∂Ω) for all y∈∂(B∣x∞∣/2(0)∩Ω). Using a sub-solution as in Case 3, we get that
Gxi(y)≥cd(xi,∂Ω)d(y,∂Ω)∣y∣−β−(γ) for all y∈∂(B∣x∞∣/2(0)∩Ω). This yields 0<c1.
For the gradient estimate, we choose a chart T around y∞=0 as in (46), and we let ri:=∣y~i∣→0 where yi=T(y~i)we define Gi(y):=riβ−(γ)−1Gxi(T(riy))/d(xi,∂Ω) for y∈R−n∩B2(0) where ri→0 . The pointwise control (275) and equation (241) yields the convergence of (Gi) in Cloc1(R−n∩B2(0)∖{0}) as i→+∞. The boundedness of ∣∇Gi∣ yields c2<+∞. This proves (280) in Case 5.
Since G is symmetric, it follows from Cases 1 to 5 that (280) holds when x∞=y∞.
We now deal with the case x∞=y∞, which rewrites limi→+∞∣xi−yi∣=0. Via a rescaling, we are essentially back to the case x∞=y∞ via the convergence Theorems 4, 5 and 6.
Case 6: ∣xi−yi∣=o(d(xi,∂Ω)) as i→+∞. We set ri:=∣xi−yi∣→0 as i→+∞ and we define
[TABLE]
It follows from Theorem 4 that Gi→cn∣⋅∣2−n in Cloc2(Rn∖{0}) as i→+∞, with cn:=((n−2)ωn−1)−1. We define Yi:=∣yi−xi∣yi−xi, and we then get that ∣yi−xi∣n−2G(xi,yi)=Gi(Yi)→cn as i→+∞. Estimating H(xi,yi) (and noting that d(xi,∂Ω)≤∣xi−0∣=∣xi∣), we get that 0<c1<+∞.
The convergence of the gradient yields ∣∇Gi(Yi)∣≤C for all i. With the original function G and points xi, yi, this yields c2<+∞. This proves (280) in Case 6.
Case 7: d(xi,∂Ω)=O(∣xi−yi∣) and ∣xi−yi∣=o(∣xi∣) as i→+∞. Then limi→+∞xi=x∞∈∂Ω. We let T be a chart at x∞ as in (46), in particular D0T=IRn. We let xi=T(xi,1,xi′) and yi=T(yi,1,yi′) where (xi,1,xi′),(yi,1,yi′)∈(−∞,0)×Rn−1 are going to [math] as i→+∞. In particular d(xi,∂Ω)=(1+o(1))∣xi,1∣ and d(yi,∂Ω)=(1+o(1))∣yi,1∣ as i→+∞. We define ri:=∣(yi,1,yi′)−(xi,1,xi′)∣. In particular ri=(1+o(1))∣xi−yi∣ as i→+∞. The hypothesis of Case 7 rewrite xi,1=O(ri) and ri=o(∣(xi,1,xi′)∣). Consequently, we have that yi,1=O(ri) and ri=o(∣xi′∣) as i→+∞. We define
[TABLE]
for X,Y∈R−n such that X=Y. It follows from Theorem 5 that
[TABLE]
for all X,Y∈R−n, X=Y, and this convergence holds in Cloc2. We define Xi:=(ri−1xi,1,0) and Yi:=(ri−1yi,1,ri−1(yi′−xi′)): the definition of ri yields Xi→X∞∈R−n and Yi→Y∞∈R−n as i→+∞. Therefore, we get that
[TABLE]
as i→+∞, and
[TABLE]
Case 7.1: X∞,1=0 and Y∞,1=0. We then get that limi→+∞∣xi−yi∣n−2G(xi,yi)=Ψ(X∞,Y∞)>0. Moreover, it follows from (281) that d(xi,∂Ω)d(yi,∂Ω)=(c+o(1))∣xi−yi∣2 as i→+∞ for some c>0. Since ∣xi∣=(1+o(1))∣yi∣ as i→+∞ (this follows from the assumption of Case 7), we get that limi→+∞∣xi−yi∣n−2H(xi,yi)∈(0,+∞). Then 0<c1<+∞.
Case 7.2: X∞,1=0 and Y∞,1=0. Then Yi,1→0 as i→+∞, and then, there exists (τi)i∈(0,1) such that Gi(Xi,Yi)=Yi,1∂Y1Gi(Xi,(τiYi,1,Yi′)). Letting i→+∞ and using the convergence of Gi in C1, we get that
[TABLE]
as i→+∞. As one checks, ∂Y1Ψ(X∞,Y∞)<0. Arguing as in Case 7.1, we get that 0<c1<+∞.
Case 7.3: X∞,1=Y∞,1=0. As in Case 7.2, there exists (τi)i,(σi)i∈(0,1) such that
Gi(Xi,Yi)=Yi,1Xi,1∂Y1∂X1Gi((σiXi,1,Xi′)Xi,(τiYi,1,Yi′)). We conclude as above, noting that ∂Y1∂X1Ψ(X∞,Y∞)>0. Then 0<c1<+∞.
The gradient estimate is proved as in Cases 1 to 6. This proves (280) in Case 7.
Case 8: d(xi,∂Ω)=O(∣xi−yi∣), ∣xi∣=O(∣xi−yi∣) and ∣yi∣=O(∣xi−yi∣) as i→+∞. In particular, x∞=y∞=0. We take a chart at [math] as in Case 7, and we define (xi,1,xi′),(yi,1,yi′) similarly. We define ri:=∣(yi,1,yi′)−(xi,1,xi′)∣=(1+o(1))∣xi−yi∣ as i→+∞. We define
[TABLE]
for X,Y∈R−n. It follows from Theorem 6 that Gi→G in Cloc2((R−n∖{0})2∖Diag(R−n∖{0})), where G is the Green’s function for −Δ−γ∣⋅∣−2 in R−n. Then
[TABLE]
as i→+∞.
Case 8.1: We assume that X∞,1=0 and Y∞,1=0. Then we get 0<c1<+∞ as in Case 7.1.
Case 8.2: We assume that X∞∈R−n and Y∞∈∂R−n∖{0} or X∞,Y∞∈∂R−n∖{0}. Then we argue as in Cases 7.2 and 7.3 to get 0<c1<+∞ provided {∂Y1G(X∞,Y∞)<0 if X∞∈R−n and Y∞∈∂R−n} and {∂Y1∂X1G(X∞,Y∞)>0 if X∞,Y∞∈∂R−n}. So we are just left with proving these two inequalities.
We assume that X∞∈R−n. It follows from Theorem 9 below that G(X∞,⋅)>0 is a solution to (−Δ−γ∣⋅∣−2)G(X∞,⋅)=0 in R−n−{X∞}, vanishing on ∂R−n∖{0}. Hopf’s maximum principle then yields −∂Y1G(X∞,Y∞)>0 for Y∞∈∂R−n∖{0}.
We fix Y∞∈∂R−n∖{0}. For X∈R−n, we then define H(X):=−∂Y1G(X,Y∞)>0 by the above argument. Moreover, (−Δ−γ∣⋅∣−2)H=0 in R−n, vanishing on ∂R−n∖{0,Y∞}. Hopf’s maximum principle yields −∂X1H(X∞)=∂Y1∂X1G(X∞,Y∞)>0 for X∞,Y∞∈∂R−n∖{0}
Case 8.3: we assume that X∞=0 or Y∞=0. Since ∣X∞−Y∞∣=1, without loss of generality, we can assume that X∞=0. It follows from Cases 8.1 and 8.2 that there exists C>0 such that
[TABLE]
for all y∈∂(B∣xi∣/2(0)∩Ω). We let uβ−(γ) be the sub-solution given by Proposition 10. Arguing as in Case 3, it then follows from the comparison principle that (282) holds for y∈B∣xi∣/2(0)∩Ω. Since ∣yi∣=o(∣xi∣), we then get that (282) holds with y:=yi. Estimating H(xi,yi), we then get that 0<c1<+∞.
The gradient estimate is proved as in Cases 1 to 6. This proves (280) in Case 8.
Since G is symmetric, it follows from Cases 7 and 8 that (280) holds when x∞=y∞.
In conclusion, we get that (280) holds, which proves the initial claim. As noted previously, both the lower bound in (243) and the upper bound in (245) follow from these results.
We are now left with proving (246). We let (x~i)i,(y~i)i∈Ω be such that
[TABLE]
and (hi)i∈C0,θ(Ω) such that i→+∞limhi=h in C0,θ. It follows from (243) that, up to extraction, there exists l>0 such that
[TABLE]
From now on, to avoid unnecessary notations, the extraction is fixed. We define
It follows from (287), (288) and standard elliptic theory that there exists G∈C2(R−n∖{0}) such that, up to a subsequence,
[TABLE]
with
[TABLE]
[TABLE]
It the follows from Proposition 6.4 in [gr4] that there exists λ>0 such that
[TABLE]
We claim that λ=l. We prove the claim. It follows from (283) and the definition (284) of τi that
[TABLE]
Case 1: we assume that τ∞∈R−n∖{0}, that is τ∞,1=0. Passing to the limit in (291), using the convergence (289) and the explicit form (290), we get that
[TABLE]
and therefore, since τ∞,1=0, we get that λ=l.
Case 2: we assume that τ∞∈∂R−n∖{0}, that is τi,1→0 as i→+∞. With a Taylor expansion, we get that there exists a sequence (ti)i∈N∈(0,1) such that Gi(τi)=∂1Gi(tiτi,1,θi′)τi,1 for all i∈N. With the convergence (289) of Gi to G in C1, we get that
[TABLE]
Since τi,1=0 for all i∈N, it follows form (291) that λ=l.
Therefore, in both cases, we have proved that λ=l. It follows from this uniqueness that the convergence of Gi holds with no extraction.
We now prove (286). We let (zi)i∈R−n∖{0} be such that zi→z∞∈R−n∖{0}. Then Gi(zi)→G(z∞) as i→+∞. Therefore, if z∞,1=0, we get that Gi(zi)=(1+o(1))G(zi) as i→+∞. We now assume that z∞,1=0, that is zi,1→0 as i→+∞. We use the C1−convergence of (Gi) and argue as in Case 2 above to get that limi→+∞∣zi,1∣−1Gi(zi)=−∂1G(z∞)=0. As one checks, this yields also Gi(zi)=(1+o(1))G(zi) as i→+∞. As noticed above, this proves (285) and ends Step P14.∎
Step P15**.**
We fix R>0. We claim that
[TABLE]
uniformly for y∈Ω∩T(BRsi(0)).
Proof of Step P15: For r>0 small, we choose uˉβ−(γ)∈C2(Ω∩Br(0)) a supersolution to −Δuˉβ−(γ)−(γ∣x∣−2+hi)uˉβ−(γ)>0 as in (259) and (260). Note that, due to the convergence of (hi) to h in C0, the choice of uˉβ−(γ) can be made independently of i. We fix ϵ>0. It follows from the convergence (285) of Step P14 and (260) that there exists i0∈N
[TABLE]
Note that Ghi(x~i,⋅),uˉβ−(γ)∈H12(Ω∩T(BRsi(0))) (these are variational super- or sub-solutions) and that the operator −Δ−(γ∣x∣−2+hi) is coercive. Since Ghi(x~i,⋅) is a solution and uˉβ−(γ) is a supersolution to −Δu−(γ∣x∣−2+hi)u=0, it follows from the comparison principle that (293) holds for y∈Ω∩T(BRsi(0)). With (260), we get that there exists i1∈N such that
[TABLE]
Using a subsolution uβ−(γ) as in (259) and (260) and arguing as above, we get that
[TABLE]
The inequalities (294) and (295) put together yield (292). This ends Step P15.∎
We now vary the x−variable.
Step P16**.**
We fix R,R′>0. We claim that
[TABLE]
Proof of Step P16: We fix a sequence (yi)i∈Ω such that yi∈T(BRsi(0)) for all i∈N. For z∈B2R′∖B(2R′)−1, we define
It follows from (298), (299) and standard elliptic theory that there exists G~∈C2(R−n∖{0}) such that, up to a subsequence,
[TABLE]
with
[TABLE]
[TABLE]
It the follows from Proposition 6.4 in [gr4] that there exists μ>0 such that
[TABLE]
We claim that μ=l. We prove the claim. It follows from (292) and the definition (284) of θi that
[TABLE]
Case 1: we assume that θ∞∈R−n∖{0}, that is θ∞,1=0. Passing to the limit in (302), using the convergence (300) and the explicit form (301), as in Case 1 of Step P14, we get that l∣θ∞,1∣⋅∣θ∞∣−β−(γ)=μ∣θ∞,1∣⋅∣θ∞∣−β−(γ), and therefore, since θ∞,1=0, we get that μ=l.
Case 2: we assume that θ∞∈∂R−n∖{0}, that is θi,1→0 as i→+∞. With a Taylor expansion, we get that there exists a sequence (t~i)i∈N∈(0,1) such that G~i(θi)=∂1G~i(t~iθi,1,θi′)θi,1 for all i∈N. With the convergence (300) of G~i to G~ in C1, we get that
[TABLE]
Since θi,1=0 for all i∈N, it follows form (302) that μ=l.
Therefore, in both cases, we have proved that μ=l. It follows from this uniqueness that the convergence of G~i holds with no extraction. As for Step P14, we get (285). This ends Step P16.∎
Step P17**.**
We fix R,R′>0. We claim that
[TABLE]
uniformly for y∈Ω∩T(BRsi(0)) and x∈Ω∖T(B(R′)−1ri(0)).
Proof of Step P17: The differs from Step P15 since one works on domains exteriors to the ball of radius ri. Here again, we choose (yi)i such that yi∈T(BRsi(0)). For r>0 small, we choose uˉβ+(γ)∈C2(Ω∩Br(0)) a supersolution to −Δuˉβ+(γ)−(γ∣x∣−2+hi)uˉβ+(γ)>0 as in (259) and (260). Note that, due to the convergence of (hi) to h in C0, the choice of uˉβ−(γ) can be made independently of i. We fix ϵ>0. It follows from the convergence (296) of Step P16 and (260) that there exists i0∈N
[TABLE]
We fix δ>0 such that δ<r. We choose a supersolution uˉβ−(γ) as in (259) and (260). It follows from the upper bound (243) that for some i1∈N, there exists C>0 such that
[TABLE]
Therefore,
[TABLE]
where
[TABLE]
and, since uˉβ+(γ),uˉβ−(γ) are supersolution,
[TABLE]
Since −Δ−(γ∣x∣−2+hi) is coercive, the maximum principle holds and (306) holds on Ω∩T(Bδ(0)∖B(R′)−1ri(0)). With (260), we get that there exists i2∈N such that
[TABLE]
for all x∈Ω∩T(Bδ(0)∖B(R′)−1ri(0)) for all i≥i2. Using subsolutions and arguing as above, we get that for some i3∈N
[TABLE]
for all x∈Ω∩T(Bδ(0)∖B(R′)−1ri(0)) for all i≥i3. The inequalities (307) and (308) put together yield (303). This ends Step P17.∎
Step P18**.**
We let (Xi)i,(Yi)i∈Ω such that ∣Yi∣=o(∣Xi∣) and Xi=o(1) as i→+∞. We assume that there exists l′>0 such that
We let (zi)i,(ti)i∈Ω such that c1σi≤∣zi∣≤c2σi and c1ρi≤∣ti∣≤c2ρi for all i∈N. Since ∣zi∣=O(si), ri=O(∣ti∣) and ti→0 as i→+∞, it follows from (303) that
[TABLE]
In addition, since ∣zi∣=O(∣Y∣i), ∣Xi∣=O(∣ti∣) and ti→0 as i→+∞, it follows from (303) that
We let (Xi)i,(Yi)i∈Ω such that ∣Yi∣=o(∣Xi∣) and Xi=o(1) as i→+∞. Then
[TABLE]
Proof of Step P19: We argue by contradiction and we assume that there exists ϵ0>0 and a subsequences (φ(i))i such that ∣Uφ(i)−l∣≥ϵ0 for all i∈N where
[TABLE]
Since (Uφ(i)) is bounded, up to another extraction, there exists l′′>0 such that Uφ(i)→l′′ as i→+∞. Therefore, ∣l−l′′∣≥ϵ0 and l′′=l. Since (283) holds for the subfamily (φ(i)), it then follows from Step P18 that l′′=l, contradicting l′′=l. This ends Step P19.
We are now in position to prove (246), that is the convergence with no extraction of subsequence. It follows from (283) and Step P18 applied to (hi)i and to the null function that there exists a subsequence (hφ(i)) and l,Lγ,Ω>0 such that for any (xi)i,(yi)i∈Ω such that ∣yi∣=o(∣xi∣) and xi=o(1) as i→+∞, then
[TABLE]
and
[TABLE]
as i→+∞. We fix a sequence (xi)i∈Ω such that xi→0 and d(xi,∂Ω)≥∣xi∣/2 as i→+∞. In the distribution sense, we have that
[TABLE]
in the distribution sense and Ghφ(i)(xi,⋅)−G0(xi,⋅)=0 on ∂Ω. It follows from (243) that for any 1<p<n−2n, we have that ∥G0(xi,⋅)∥p≤C(p) for all i∈N. It then follows from elliptic theory that Ghφ(i)(xi,⋅)−Gh(xi,⋅)∈W2,p(Ω) and that
[TABLE]
For 1<p<min{n/2;n/(n−2)}, we define q:=n−2pnp. Sobolev embeddings then yield
[TABLE]
We let (ϵi)>0 such that ϵi→0 as i→+∞. We define αi:=ϵi∣xi∣ so that αi=o(∣xi∣) as i→+∞. We have that
[TABLE]
It then follows from (309), (310) and the boundedness of (hi) in C0 that
[TABLE]
We assume by contradiction that l=Lγ,Ω, so that
[TABLE]
If n≤q(1−β−(γ)), then the integral is infinite. This is a contradiction. Therefore n>q(1−β−(γ)). Estimating the integral and using that ∣xi∣≤2d(xi,∂Ω), we get that
[TABLE]
With αi=ϵi∣x∣i, β−(γ)+β+(γ)=n and the definition of q, we get that
[TABLE]
Since ∣xi∣→0, with a suitable choice of ϵi→0, we get a contradiction.
Therefore l=Lγ,Ω that is independent of the choice of the sequence (hi). This proves (246) and ends the proof of Theorem 8.
15. Appendix E: Green’s function for the Hardy-Schrödinger operator on R−n
In this section, we prove the following:
Theorem 9**.**
Fix γ<4n2. For all p∈R−n∖{0}, there exists Gp∈L1(R−n) such that
(i)* ηGp∈H1,02(R−n) for all η∈Cc∞(Rn−{p}),*
(ii)* For all φ∈Cc∞(R−n), we have that*
[TABLE]
Moreover, if Gp,Gp′ satisfy (i) and (ii) and are positive, then there exists C∈R such that Gp(x)−Gp′(x)=C∣x1∣⋅∣x∣−β−(γ) for all x∈R−n∖{0,p}.
In particular, there exists one and only one function Gp=G(p,⋅)>0 such that (i) and (ii) hold with Gp=Gp and
*(iii) *** Gp(x)=O(∣x∣β+(γ)∣x1∣) as ∣x∣→+∞.
We then say that G is the Green’s function for −Δ−γ∣x∣−2 on R−n with Dirichlet boundary condition.
In addition, G satisfies the following properties:
*(iv) *** For all p∈Rn∖{0}, there exists c0(p),c∞(p)>0 such that
[TABLE]
and
[TABLE]
*(v) *** There exists c>0 independent of p such that
[TABLE]
where
[TABLE]
Proof of Theorem 9:
We shall again proceed with several steps.
Step 15.1: Construction of a positive kernel at a given point: For a fixed p0∈Rn∖{0}, we show that there exists Gp0∈C2(R−n∖{0,p0}) such that
[TABLE]
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 ϵ>0, set ηϵ(x):=η~(ϵ∣x∣) for all x∈Rn.
We let Ω1 be a smooth bounded domain of Rn such that R−n∩B1(0)⊂Ω1⊂R−n∩B3(0). We define ΩR:=R⋅Ω1 so that
R−n∩BR(0)⊂ΩR⊂R−n∩B3R(0).
We argue as in the proof of (247) to deduce that the operator −Δ−∣x∣2γηϵ is coercive on ΩR and that there exists c>0 independent of R,ϵ>0 such that
[TABLE]
Consider R,ϵ>0 such that R>2∣p0∣ and ϵ<6∣p0∣, and let GR,ϵ be the Green’s function of −Δ−∣x∣2γηϵ in ΩR with Dirichlet boundary condition. We have that GR,ϵ>0 since the operator is coercive.
Fix R0>0 and q′∈(1,n−2n), then by arguing as in the proof of (249), we get that there exists C=C(γ,p0,q′,R0) such that
[TABLE]
and
[TABLE]
where δ:=∣p0∣/4. Arguing again as in Step 14.2 of the proof of Theorem 8, there exists Gp0∈C2(R−n∖{0,p0}) such that
[TABLE]
and ηGp0∈H1,02(R−n) for all η∈Cc∞(Rn∖{p0}). Fix φ∈Cc∞(R−n). For R>0 large enough, we have that φ(p0)=∫R−nGR,ϵ(p0,⋅)(−Δφ−γηϵ∣x∣−2φ)dx. The integral bounds above yield x↦Gp0(x)∣x∣−2∈Lloc1(R−n). Therefore, we get
[TABLE]
As a consequence, Gp0>0.
Step 15.2: Asymptotic behavior at [math] and p0 for solutions to (316). It follows from Theorem 6.1 in Ghoussoub-Robert [gr4] that either Gp0 behaves like ∣x1∣⋅∣x∣−β−(γ) or ∣x1∣⋅∣x∣−β+(γ) at [math]. Since Gp0∈Ln−22n(Bδ(0)∩R−n) for some small δ>0 and β−(γ)<2n<β+(γ), we get that there exists c0>0 such that
[TABLE]
Since Gp0 is positive and smooth in a neighborhood of p0, it follows from (320) and the classification of
solutions to harmonic equations that
[TABLE]
Step 15.3: Asymptotic behavior at ∞ for solutions to (316): We let
[TABLE]
be the Kelvin’s transform of G. We have that
[TABLE]
Since G~p0>0, it follows from Theorem 6.1 in [gr4] that there exists c1>0 such that
[TABLE]
Coming back to Gp0, we get that
[TABLE]
Assuming we are in the second case, for any c≤c1, we define
[TABLE]
which satisfy −ΔGˉc−∣x∣2γGˉc=0 in R−n∖{0,p0}. It follows from (323) and (322) that for c<c1, Gˉc>0 around p0 and ∞. Using that ηGˉc∈H1,02(R−n) for all η∈Cc∞(Rn∖{p0}), it follows from the coercivity of −Δ−γ∣x∣−2 that Gˉc>0 in R−n∖{0,p0} for c<c1. Letting c→c1 yields Gˉc1≥0, and then Gˉc1>0. Since Gˉc1(x)=o(∣x1∣⋅∣x∣−β−(γ)) as ∣x∣→∞, another Kelvin transform and Theorem 6.1 in [gr4] yield ∣x1∣−1∣x∣β+(γ)Gˉc1(x)→c2>0 as ∣x∣→∞ for some c2>0. Then there exists c3>0 such that
[TABLE]
Since x↦∣x1∣⋅∣x∣−β−(γ)∈H1,loc2(Rn), we get that
[TABLE]
Step 15.4: Uniqueness: Let G1,G2>0 be 2 functions such that (i),(ii) hold for p:=p0, and set H:=G1−G2. It follows from Steps 2 and 3 that there exists c∈R such that H′(x):=H(x)−c∣x1∣⋅∣x∣−β−(γ) satisfies
[TABLE]
We then have that ηH′∈H1,02(R−n) for all η∈Cc∞(Rn∖{p0}) and
[TABLE]
The ellipticity of the Laplacian then yields H′∈C∞(R−n∖{0}). The pointwise bounds (325) yield that H′∈H1,02(R−n). Multiplying −ΔH′−∣x∣2γH′=0 by H′, integrating by parts and the coercivity yield H′≡0, and therefore, (G1−G2)(x)=c∣x1∣⋅∣x∣−β−(γ) for all x∈R−n. This proves uniqueness.
Step 15.5: Existence. It follows from Steps 2 and 3 that, up to substracting a multiple of x↦∣x1∣⋅∣x∣−β−(γ), there exists a unique function Gp0>0 satisfying (i), (ii) and the pointwise control (iii). Moreover, (321), (322) and (324) yield (312) and (313). As a consequence, (314) holds with p=p0.
For p∈Rn∖{0}, consider ρp:R−n→R−n a linear isometry fixing R−n such that ρp(∣p0∣p0)=∣p∣p, and define
[TABLE]
As one checks, Gp>0 satisfies (i), (ii), (iii), (312), (313) and (314).
The definition of Gp is independent of the choice of ρp. Indeed, for any linear isometry ρp0:R−n→R−n fixing p0 and R−n, Gp0∘ρp0−1 satisfies (i), (ii), (iii), and therefore Gp0∘ρp0−1=Gp0. The argument goes similarly of any isometry fixing p.
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