THE HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY: THE REMAINING CASES

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1 THE HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY: THE REMAINING CASES NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT Abstract. We consider the remaining unsettled cases in the problem of existence of energy minimizing solutions for the Dirichlet value problem L u λu = u s 1 s on a smooth bounded domain in R n n 3 having the singularity 0 in its interior. Here < n, 0 s <, s := n s and 0 λ < λ n 1 L, the latter being the first eigenvalue of the Hardy-Schrödinger operator L :=. There is a threshold λ, 0 beyond which the minimal energy is achieved, but below which, it is not. It is well known that λ = 0 in higher dimensions, for example if 0 n 1. Our main objective in this paper is to show that this threshold is strictly positive in lower dimensions such as when n 1 < < n, to identify the critical dimensions i.e., when the situation changes, and to characterize it in terms of and. If either s > 0 or if > 0, i.e., in the truly singular case, we show that in low dimensions, a solution is guaranteed by the positivity of the Hardy-singular internal mass of, a notion that we introduce herein. On the other hand, and just like the case when = s = 0 studied by Brezis-Nirenberg [] and completed by Druet [1], n = 3 is the critical dimension, and the classical positive mass theorem is sufficient for the merely singular case, that is when s = 0, 0. Contents 1. Introduction. The higher dimensional case 5 3. The Hardy-singular interior mass of a domain 8. Positive mass and the existence of extremals in lower dimensions Blow-Up analysis in the truly singular case Blow-Up analysis in the merely singular case 5 7. Proof of Theorem 3 36 Appendix A: Green s function for hx on a bounded domain 37 Appendix B: Green s function for on R n Appendix C: Singular solutions to u cx u = 0 5 References 7 Date: August, 017. This work was carried out while N. Ghoussoub was visiting l Institut Élie Cartan, Université de Lorraine in May 015. He was partially supported by a research grant from the Natural Science and Engineering Research Council of Canada NSERC. The paper was completed while Frédéric Robert was visiting the University of British Columbia in July Mathematics Subject Classification: 35J35, 35J60, 35B. 1

2 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT 1. Introduction Let be a smooth bounded domain in R n n 3 such that 0 and consider the following Dirichlet boundary value problem: 1 u u s 1 λu = u s on, u > 0 on, u = 0 on, where < n, 0 s <, s := n s n and 0 λ < λ 1 L, the latter being the first eigenvalue of the Hardy-Schrödinger operator L :=, that is λ 1 L, := inf u dx u dx ; u H0 1 \ {0}. u dx Equation 1 is essentially the Euler-Lagrange equation corresponding to the following energy functional on H0 1, J,s,λu = u dx u dx λ u dx, u s dx s s where H0 1 is the completion of Cc for the norm u u. We shall therefore study whether the following minimization problem µ,s,λ := inf { J,s,λu; u H 1 0 \ {0} }, is attained, that is if µ,s,λ = J,s,λ u 0 for some u 0 H 1 0. For convenience, we define D 1, R n := H 1 0 R n, that is the completion of C c R n for the norm u u. Note that the fact that µ,s,0 R n > 0 is equivalent to the critical case of the Caffarelli-Kohn- Nirenberg inequalities [7]. In particular, see for instance Ghoussoub-Robert [17], µ,s,0 R n is achieved iff {s > 0} or {s = 0 and 0}. It is also standard that µ,s,0 = µ,s,0 R n whenever is a domain containing 0 in its interior, and hence µ,s,0 is not attained if is bounded. The idea of restoring compactness by considering non-trivial negative linear perturbations was pioneered by Brezis-Nirenberg [] in the case when = 0, s = 0 and 0 < λ < λ 1, the latter being the first eigenvalue of the Laplacian on H 1 0. They showed that in this case 1 has a solution for n. The case n = 3 is special and involves a positive mass condition introduced by Druet [1, 13], and inspired by the work of Schoen [6] on the Yamabe problem. The bottom line is that at least for = 0 the geometry of need not be taken into account in dimension n, while in dimension n = 3, the existence depends on the domain via a positive mass condition. We shall elaborate further on this theme. In this paper, we consider the case when the Laplacian is replaced by the Hardy-Schrödinger operator L. Here, the position of the singularity 0 within matters a great deal. In [16], we considered the case where 0 belongs to the boundary of the domain. In this sequel, we deal with the case when 0, which was first considered by Janelli [19] in the case s = 0. It is already well known that there is a threshold λ beyond which the infimum µ,s,λ is achieved, and below which, it is not. It can be characterized as λ := sup{λ; µ,s,λ = µ,s,0 R n }. It is easy to see that 0 λ < λ 1 L,. It is also part of the folklore that we sketch below that λ = 0 in higher dimensions. Our main objective in this paper is to show that this threshold is strictly positive in lower dimensions, to identify the critical dimensions i.e., when the situation changes, and to try to characterize it in terms of and.

3 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 3 As opposed to Brezis-Nirenberg [] and Druet [1], we are dealing here with the case where 0 is an interior singularity, and our analysis below shows that the identification of λ differ according to two distinct singularity regimes: The truly singular case, which corresponds to when either s > 0 or > 0. We note that in this case µ,s,0 R n is achieved. The merely singular case, which corresponds to the case when s = 0 and 0, a case where µ,s,0 R n is not achieved, unless s = = 0. The following three theorems are the main results of this paper. The first is rather standard. It deals with high dimensions and is included for completeness and comparison purposes. The second deals with the low dimensional cases, i.e., the remaining cases which are yet to be addressed in the literature. Theorem 1. The higher dimensional case Let be a smooth bounded domain in R n n 3 such that 0. Assume that we are in the following situation: either in the truly singular case and < n 1, or in the merely singular case and n. Then µ,s,λ is achieved if and only if λ > λ. Moreover, i In the truly singular case i.e, when either s > 0 or > 0, and if n 1, then 3 λ = 0. ii In the merely singular case i.e, when s = 0 and 0, and if n, then { } λ = inf ; x > 0 if < 0. Part i of Theorem 1 was proved by Janelli [19] in the case when s = 0. The case when s > 0 is not much different and was noted in several works such as [8 10, 1 3, 5]. Part ii of Theorem 1, that is the case when s = 0 and < 0, in dimension n was also tackled by Janelli [19] and Ruiz-Willem [5]. Their proof, though not complete, essentially gives the above result. Janelli [19] also considered the lower dimensional case, that is n 1 < < n, when is the ball B centered at 0. He gave the following explicit value for λ : u 5 λ B dx B = inf β + u B dx; u H1 0 B \ {0} > 0, β + where β ± := n n ±. Note that the radial function x β is a solution of u = 0 on R n \ {0} if and only if β {β, β + }. In order to characterize the threshold λ for a general domain, we need to define the notion of Hardy-singular interior mass associated to the operator λ on a bounded domain in R n containing 0. Theorem. The Hardy singular internal mass Let be a smooth bounded domain in R n n 3 such that 0. Suppose h is a C -potential on so that the operator + hx is coercive.

4 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT i There exists then H C \ {0} such that H + hx H = 0 in \ {0} E H > 0 in \ {0} H = 0 on. These solutions are unique up to a positive multiplicative constant, and there exists c > 0 such that Hx x 0 c β +. ii If either h is sufficiently small around 0 or if n 1 < < n, then for any solution H C \ {0} of E, there exist c 1 > 0 and c R such that Hx = c 1 β+ + c β + o 1 β as x 0. The uniqueness implies that the ratio c /c 1 is independent of the choice of H, hence the Hardysingular internal mass of associated to the operator L hxi can be defined unambigously as m,h := c c 1 R. For the merely singular case s = 0 and 0 and the critical dimension n = 3, we need a more standard notion of mass associated to the operator L at an internal point x 0, which is reminiscent of Schoen-Yau s approach to complete the solution of the Yamabe conjecture in low dimensions. For that, one considers for a given 0, the corresponding Robin function or the regular part of the Green function with pole at x 0 \ {0}. One shows that for n = 3, any solution G of G G λg = 0 in \ {x 0 } G > 0 in \ {x 0 } G = 0 on, is unique up to multiplication by a constant, and that there exists R,λ, x 0 R and c,λ x 0 > 0 such that 1 6 Gx = c,λ x 0 x x 0 n + R,λ, x 0 + o1 as x x 0. The quantity R,λ, x 0 is then well defined and will be called the internal mass of at x 0. We then define R,λ = sup x \{0} R,λ, x. These will allow us to give an explicit value for λ, as follows. Theorem 3. The lower dimensional case Let be a smooth bounded domain in R n n 3 such that 0. i Assume we are either in the truly singular case and n 1 < < n, or in the merely singular case and n = 3. Then, there exists λ > 0 such that µ,s,λ is not achieved for λ < λ and µ,s,λ is achieved for λ > λ. ii Moreover, in the truly singular case, with n 1 < < n, and under the assumption that µ,s,λ is not achieved, we have that m,λ = 0, and 7 λ = sup{λ; m,λ 0}. iii In the merely singular case, and with n = 3, then µ,s,λ is not achieved and 8 λ = sup{λ; R,λ 0}.

5 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 5 We conjecture that in all cases, µ,s,λ is never achieved, which means that 7 must hold unconditionally. Note that µ,s,λ = µ,s,0 R n, but we don t know whether this suffices to conclude that µ,s,λ is not achieved. When s = = 0 and n = 3, Druet [1] proved that this is indeed the case by using a very elegant geometric argument. This extends to the merely singular case. In the truly singular case, the conjecture holds in the radially symmetric case, i.e., when is a ball. This was verified by Janelli [19]. Finally, we note that the above analysis lead to the following definition of a critical dimension for the operator L. It is the largest scalar n such that for n < n, there exists a bounded smooth domain R n and a λ 0, λ 1 L, such that there is a non-trivial minimiser satisfying 1. µ,s,λ is not attained. { if 1 n = if < 1. Note that n < n is exactly when β + β <, which is the threshold where the radial function x β+ is locally L -summable. The proofs of Theorems 1 and 3 rely on a refined blow-up analysis for certain families of solutions of equation 1. We give in Theorems and 5 below a complete description of how such blowups may occur. In particular, we show that in the truly singular case, the solutions necessarily concentrate at the singularity 0, while in the merely singular case, they do so at a point x 0 0 of the domain. In the appendices, we establish several important properties of the Green function associated to the operator, that are crucial for the proofs of Theorems and 5.. The higher dimensional case We recall the following facts, which by now are standard. i µ,s,λ > 0 if and only λ < λ 1 L,. ii µ,s,λ = µ,s,0 R n for all λ 0. iii µ,s,λ is attained if µ,s,λ < µ,s,0 R n. iv The function λ µ,s,λ is continuous and nonincreasing. v If µ,s,λ is attained, then µ,s,λ < µ,s,λ for any λ > λ. Writing λ = λ for short, where λ is defined in, it follows from the above that vi µ,s,λ = µ,s,0 R n for all λ λ and µ,s,λ < µ,s,0 R n for all λ > λ. vii µ,s,λ is not achieved for all λ < λ. viii µ,s,λ is achieved for all λ > λ. It is clear that λ 0. This section is devoted to show that λ = 0 in high dimensions, which in our case will depend on. The calculations are standard, and we include them for the convenience on the reader and for comparison to the other cases. As mentioned above in iii, in order to show that extremals exist for µ,s,λ, it suffices to prove that µ,s,λ < µ,s,0 R n, where µ,s,0 R n := µ,s,0 R n. This kind of condition is now standard when dealing with borderline variational problems. See also Aubin [1], Brézis-Nirenberg []. The condition limits the energy level of minimizing sequences, prevents the creation of bubbles and hence insures compactness. To show the strict inequality, one needs to test the functional J,s,λ on minimizing sequences of the form ηu ɛ, where U ɛ is an extremal for µ,s,0 R n and η Cc is a cut-off function equal to 1 in a neigbourhood of 0. It is therefore important to know for which parameters and s, the best constant µ,s,0 R n is attained. A proof of the following can be found in [17]. For explicit extremals, we refer to Beckner [5] or Dolbeault et al. [11]. Proposition 1. Assume < n, n 3 and 0 s <. Then, i µ,s,0 R n is attained if either s > 0 or if {s = 0 and 0}.

6 6 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT ii If 0 < n, then the extremals for µ,s,0 R n are explicit and take the form u ɛ x = c ɛ n U x ɛ, where c 0, ɛ > 0 and 9 Ux := 1 sβ n + sβ + n n s for x R n \ {0}, iii On the other hand, if s = 0 and < 0, then µ,0 R n is not attained and is equal to µ 0,0 R n, which is the best constant in the Sobolev inequality. Subsection.1: The truly singular case We now give a proof of Theorem 1. Assuming n 1, we construct a minimizing sequence u ɛ in H 1 0 \ {0} for the functional J,s,λ in such a way that µ,s,λ < µ,s,0 R n. Since either s > 0 or 0, then the infimum µ,s,0 R n is achieved by the function Ux := 1 sβ n + sβ + n n s for x R n \ {0}. In other words, U D 1, R n and J Rn,s,0U = inf u D 1, R n \{0} J Rn,s,0u, where J Rn,s,0u := u R n u dx u s dx s R n s In particular, there exists χ > 0 such that s 1 for u D 1, R n \ {0}. 10 U U = χu s in R n \ {0}. For convenience, we will write in the sequel, β + := β + and β := β. Note that the assumption that n 1 is equivalent to β + β. Define a scaled version of U by U ɛ x := ɛ n x U = ɛ ɛ s n β+ β ɛ s n β+ β sβ n + sβ + n n s for x R n \ {0}. Fix now a function h C 0,θ, θ 0, 1, consider a cut-off function η C c such that ηx = 1 for x in a neighborhood of 0 contained in, and define for ɛ > 0 the test-function u ɛ H 1 0 by We now estimate J,s,h u ɛ, where u ɛ x := ηxu ɛ x for x \ {0}. J,s,hu := u u s s + hx u dx s dx.

7 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 7 Note first that 0 u ɛ x Cɛ β + β β+ for all x \ {0}. Therefore, since β + β, we have as ɛ 0, We also have u s ɛ s dx = u ɛ u ɛ dx = = = = = χ = χ Finally, we estimate the last term as ɛ 0, hxu ɛ dx = B δ 0 B ɛ 1 δ 0 Rn U s B δ 0 B ɛ 1 δ 0 B δ 0 = ɛ U s ɛ s dx + o ɛ U s s dx + o ɛ s dx + o ɛ. U ɛ B ɛ 1 δ 0 Rn U s U U s s U ɛ dx + O ɛ β+ β U dx + O ɛ β+ β s dx + O ɛ β+ β. hxu ɛ dx + O ɛ β+ β B ɛ 1 δ 0 hɛxu dx + O ɛ β+ β. If < n 1 and β + β >, the extremal U L R n and therefore hxu ɛ dx = h0 U dxɛ + oɛ as ɛ 0. R n If now = n 1, then Ux x n and β + β =. Therefore 1 hxu ɛ dx = h0ω n 1 ɛ ln + oɛ ln ɛ as ɛ 0, ɛ dx + O ɛ β+ β where ω n 1 is the volume of the canonical n 1 sphere. Combining the above estimates as ɛ 0 yields h0 U ɛ + oɛ if < n J,s,hu U ɛ = J,s,0U Rn s, s 1 h0ω n 1 ɛ ln 1 U s, s ɛ + oɛ ln ɛ if = n 1. In either case, if h0 = λ > 0, we get that and we are done. µ,s,λ J,s,λu ɛ < J Rn,s,0U = µ,s,0 R n,

8 8 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT Subsection.: The merely singular case We now prove the second part of Theorem 1. Assuming that s = 0, < 0 and n, we shall prove that µ,s,λ is attained if and only if λ, < λ, where { } λ, = inf ; x < λ 1 L. Note that in this case, we have µ 0,0,0 R n = µ,0,0 R n as noted in [16], that is inf J R n 1,0,0U = U D 1, R n \{0} Kn, := inf U dx R n U D 1, R n \{0} U R n dx, and the infimum of µ,0 R n \ {0} is not achieved. Consider now the following known extremal for µ 0,0 R n, 1 Ux := for x R n. 1 + n Fix x 0, x 0 0, and define the test-function 11 u ɛ x := ηxɛ n Uɛ 1 x x 0 for all x, where η C c is such that ηx = 1 around x 0. A straightforward computation yields J,0,λu ɛ = µ 0,0 R n + o1 as ɛ 0, which yields that µ,0,λ µ 0,0 R n. Classical computations in the spirit of Aubin [1], which can be done by replacing λ with a more general function h, yield that for n, there exists c n > 0, such that as ɛ 0, J,hu 1 ɛ = Kn, + c n x 0 hx 0 ɛ + oɛ if n 5 c x 0 hx 0 ɛ lnɛ 1 + Oɛ if n =. Therefore, if n and assuming there exists x 0 \ {0} such that hx 0 > x 0, we obtain that inf u H 1 0 \{0} J,h u < inf U D 1, R n \{0} J,0U, Rn and µ,h is attained. Conversely, if hx for all x \ {0}, then + hx 0 for all x \ {0}, hence µ,0,h µ 0,0,0 = µ 0,0,0 R n. We therefore have equality, and there is no extremal for µ,0,h since the extremals on R n are rescaled and translated versions of U. 3. The Hardy-singular interior mass of a domain This section is devoted to the construction of the singular interior mass, as stated in Theorem. We start with the following key result. Proposition. Assume is a smooth bounded domain in R n and let h C 0,θ with θ 0, 1. If the operator hx is coercive, then there exists a solution H C \ {0} for the linear problem H + hx H = 0 in \ {0} 1 H > 0 in \ {0} H = 0 on. Moreover, there exists c > 0 such that 13 Hx x 0 c β+. If H C \ {0} is another solution for 1, then there exists λ > 0 such that H = λh.

9 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 9 Proof: First, we prove existence of a solution. For that, let η 1 C R be such that η 1 t = 0 for t < 1 and η 1 t = 1 for t >. For ɛ > 0, set η ɛ x := η 1 /ɛ for all x R n. Then let H ɛ C \{0} be the Green s function for the operator η ɛ x + hx that is singular at 0. In particular, we have that H ɛ η ɛ x + hx Hɛ = 0 in \ {0} H ɛ > 0 in \ {0} H ɛ = 0 on. Fix x 0 \ {0} and define H ɛ x := H ɛx H for all x \ {0}. For δ > 0 such that B ɛx 0 δ0 and δ < x 0 /, we take ɛ 0, δ/. We then have H ɛ + hx H ɛ = 0 in \ B δ 0 H ɛ > 0 in \ B δ 0 H ɛ = 0 on. It follows from the boundary Harnack inequality see for instance Ghoussoub-Robert [16], Proposition 7. that there exists C δ > 0 such that H ɛ x dx, C H ɛ x 0 δ dx 0, = C δ dx 0, for all x \ B δ0. Since this is valid for any δ > 0 small enough, it then follows from standard ellitpic theory that there exists H C \ {0} such that H ɛ H in Cloc k \ {0} as ɛ 0 for all k N. In particular, we have H + hx H = 0 in \ {0} H 0 in \ {0} H = 0 on. Since Hx 0 = 1, it follows from the strong maximum principle that H > 0, hence it satisfies 1. It then follows from Theorem 9 that there exists c > 0 such that c either Hx x 0 or Hx c β+ x 0. β If 13 does not hold, we the second case holds and H H 1 0 : since h is coercive, equation 1 then yield H 0, contradicting H > 0. This proves 13. To prove uniqueness, let H C \{0} be such that 1 holds. Set λ 0 := max{λ 0; H λh}. This is well defined and we set H := H λ 0 H. Then H 0 satisfies H + hx H = 0 in \ {0}. Therefore, if H 0, it follows from the maximum principle that H > 0. Then the asymptotic control 13 and Hopf s boundary comparison principle yield the existence of ɛ 0 > 0 such that H ɛ 0 H in \ {0}, contradicting the definition of λ 0. Therefore H 0 and H = λ 0 H, which proves the uniqueness statement. We now proceed with the proof of Theorem. Proposition 3. Let be a smooth bounded domain in R n and fix h C 0,θ, θ 0, 1. Assume that the operator + hx is coercive and that > n 1. If H C \ {0} is a solution to 1, then there exist c 1, c R with c 1 > 0 such that c 1 1 Hx = + c β+ + o 1 as x 0. β β The ratio c c 1 R is independent of the choice of H. We can therefore define the mass as m,h := c c 1.

10 10 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT Proof: Let η Cc be such that ηx 1 around 0. Our first objective is to write Hx := ηx + fx for some f H 1 β + 0. For that, we consider the function gx := + hx η β+ in \ {0}. Since ηx = 1 around 0, we have that hx 15 gx = β+ C β+ around 0. Therefore, g L n n+ if and only if β+ < n+ n, which holds if and only if >. The latter is guaranteed by our assumption on. Since L n n n+ = L n H0 1, there exists f H0 1 such that f + hx f = g in H0 1. By regularity theory, we have that f C \ {0}. We now show that 16 β fx has a finite limit as x 0. Define Hx := ηx + fx for all x \ {0}. β+ and note that H C \ {0} and is a solution to +hx H = 0. Write g + x := max{gx, 0} and g x := max{ gx, 0} so that g = g + g, and let f 1, f H0 1 be weak solutions to 17 f 1 + hx f 1 = g + and f + hx f = g in H0 1. In particular, uniqueness, coercivity and the maximum principle yields f = f 1 f and f 1, f 0. Assume that f 1 0, so that f 1 > 0 in \ {0}, fix β, β +, choose µ R such that µβ n β < 0, and define u x := β + µ for all x 0. As in the proof of the previous proposition, we get that for some δ > 0 small, + hx u < 0 for x B δ 0 \ {0}, that is u is a subsolution on B δ 0 \ {0}. Fix now C > 0 such that f 1 Cu on B δ 0. Since f 1 and Cu H0 1 are respectively superand sub-solutions to +hx u = 0, it follows from the comparison principle via coercivity that f 1 Cu in B δ 0 \ {0}, and therefore f 1 C β in B δ 0 \ {0}. It then follows from 15 that g + x gx C 1 + β+ β f 1, and therefore 17 yields 18 + O β+ β f 1 = 0 weakly in H0 1. Since > n 1 if and only if τ := β + β > 0, we can argue as in the proof of Proposition see also the regularity Theorem 8 and get that β f 1 x has a finite limit as x 0. Similarly, β f x has also a finite limit as x 0, and therefore 16 is verified. It follows that there exists c R such that Hx = 1 β+ + c β + o 1 β as x 0,

11 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 11 which proves the existence of a solution H to the problem with the relevant asymptotic behavior. The uniqueness result of Proposition then yields the conclusion. The following proposition summarizes the properties of the mass. Proposition. Let be a smooth bounded domain in R n and fix h C 0,θ with θ 0, 1. Assume that the operator + hx is coercive and that > n 1. The mass m,h then satisfies the following properties: i m,0 < 0, ii If h h and h h, then m,h < m,h, iii If, then m,h < m,h. iv The function h m,h is continuous for the C 0 norm. Proof: For any such h C 0,θ, we let H h be the unique solution to 1 such that 1 holds with c 1 = 1. In other words, 1 H h x = + m,h β+ + o 1 as x 0. β β Since + hx H 0 x β+ = 0 and is negative on, it follows from the maximum principle that H 0 x β+ < 0 on. It then follows from Theorem 8 that m,0 < 0. This prove property i of the proposition. Property iii goes similarly. For ii, we define g := H h H h. We have that g H0 1 and g + hx g = h h H h 0, but 0. Therefore g < 0, and it follows from Theorem 8 in Appendix C that there exists K > 0 such that gx β K as x 0, and therefore m,h m,h = K < 0, which proves the second part of the proposition.. Positive mass and the existence of extremals in lower dimensions In this section, we show how the positivity of the Hardy-singular mass m,λ in the truly singular case resp., the mass in the merely singular case yields that µ,s,λ is attained in the corresponding low dimensions, i.e., n 1 < < n in the truly singular case, and n = 3 in the merely singular case. Subsection.1: The truly singular case Proposition 5. Let be a smooth bounded domain in R n n 3 such that 0. Assume either 0 < s < or that {s = 0, > 0}. If n 1 < < n and 0 < λ < λ 1 L is such that the mass m,λ is positive, then µ,s,λ is attained. Proof: Assuming that n 1 < < n, we know that the mass m,λ is defined. We need to show that if m,λ > 0, then µ,s,λ < µ,s,0 R n. Consider again for each ɛ > 0 the extremals 19 U ɛ x := ɛ n x U = ɛ ɛ s n β+ β ɛ s n β+ β sβ n + sβ + n We shall first replace λ with any function h C 0,θ, where θ 0, 1 and the operator + hx is coercive. Consider again a test function η C c such that ηx = 1 for x in n s.

12 1 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT a neighborhood of 0. Since > n 1, it follows from Proposition 3 that there exists β H 1 0 such that 0 βx x 0 m,h β, and the function Hx := ηx + βx satisfies β + 1 Define now H + hx H = 0 in \ {0} H > 0 in \ {0} H = 0 on. u ɛ x := ηxu ɛ x + ɛ β + β βx for x \ {0}. It is clear that u ɛ H0 1 for all ɛ > 0. We now estimate J,s,h u ɛ, where again J,s,h functional on H0 1 defined by u J,s,hu + hx u dx := u. s dx s s is the Thereafter, the notation o δ 1 will mean lim δ 0 lim ɛ 0 o δ 1 = 0. Step 1: Estimating u ɛ + hx u ɛ dx Letting δ 0, dist0,, we start by estimating \B δ u 0 ɛ + hx u ɛ dx. First note that lim ɛ 0 Therefore, 3 lim ɛ 0 \B δ 0 \B δ 0 u ɛ ɛ β + β = H in C loc \ {0}. u ɛ + hx u ɛ dx = H ɛ β+ β \B δ 0 + hx H dx. Integrating by parts and using equation 1 yields H + hx H dx = \B δ 0 + = \B δ 0 B δ 0 H H + hx H dx H ν H dσ H ν H dσ. Since β + + β = n, using elliptic estimates, and the definition of H yields H ν H = β + β+ 1 n m,h n 1 + o n 1 as x 0. Therefore, plugging this expansion into 3 and yields 5 u ɛ + hx u β+ ɛ dx = ɛ β+ β ω n 1 δ + n m,h + o δ 1 β+ β \B δ 0

13 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 13 We now deal with the expression B δ u 0 ɛ + hx u ɛ dx. Take δ > 0 small enough such that ηx = 1 for x B δ 0. Therefore, u ɛ x = U ɛ x + ɛ β + β βx for x B δ 0 and then u ɛ + hx u ɛ dx = B δ 0 B δ 0 +ɛ β + β +ɛ β+ β U ɛ + hx B δ 0 B δ 0 U ɛ dx U ɛ β dx U ɛ β + hx β + hx β dx. Since U ɛ, β H0 1 and U ɛ is explicit, we integrate by parts the first and second term of the right-hand-side and we neglect the third term to get u ɛ B δ 0 + hx u ɛ dx = U ɛ B δ 0 U ɛ U ɛ dx + U ɛ ν U ɛ dσ hxuɛ dx 6 B δ 0 +ɛ β + β +ɛ β + β We estimate the terms of the right-hand-side separately. Note first that 10 and 19 yield that as ɛ 0, 7 B δ 0 U ɛ U ɛ U ɛ dx = λ B δ 0 = χ = χ U s ɛ s Rn U s s Rn U s B δ 0 B δ 0 B δ 0 U ɛ + hx β ν U ɛ dσ + o δ ɛ β+ β. U ɛ β dx U s dx = χ B δɛ 0 s dx β+ s s dx R n \B ɛ 1 δ 0 dx + O s dx + o ɛ β+ β. The explicit expression of U ɛ in 19 yields 8 U ɛ ν U ɛ dσ = β + ω n 1 δ β+ β ɛ β+ β + o δ ɛ β+ β as ɛ 0. B δ 0 Since > n 1, we have that β + < n and therefore 9 hxu ɛ dx = O ɛ β+ β β+ dx = o δ ɛ β + β. B δ 0 B δ 0 Since β + + β = n < n, we also have that ɛ β + β hxu ɛ β dx = O ɛ β+ β β+ β dx B δ 0 B δ 0 30 = o δ ɛ β + β.

14 1 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT It follows from 10 and 19 that U ɛ U ɛ β dx = χ 31 B δ 0 B δ 0 = χɛ β + β U s 1 ɛ s B ɛ 1 δ 0 β dx U s 1 s ɛ β βɛx dx = O Finally, using the expression 19 of U ɛ and the asymptotics 0 of β, we get that β ν U ɛ dσ = ɛ β + β β ν β+ dσ + o ɛ β + β 3 B δ 0 B δ 0 = ɛ β + β m,h B δ 0 β ν β+ dσ + o = ɛ β + β m,h β + ω n 1 + o ɛ β + β. ɛ β + β ɛ β + β. Plugging together 5, 6, 7, 8, 9, 30, 31 and 3 yields Rn u ɛ + hx u U s ɛ dx = χ s dx +n β + m,h ω n 1 ɛ β+ β + χɛ β + β U s 1 ɛ 33 s β dx + oɛ β+ β. Step : Estimating u s ɛ s From and the definition 19 of U ɛ, we have as ɛ 0 that u s ɛ u s ɛ s dx = s dx + oɛ β+ β = = = B δ 0 B δ 0 B δ 0 + B δ 0 B ɛ 1 δ 0 dx B δ 0 U ɛ + ɛ β + β β s s dx + oɛ β+ β U s ɛ s + s U s 1 ɛ s β dx O ɛ U s ɛ β+ β s β + ɛ s U s s dx + sɛ β+ β Using the expression of U, we get that u s Rn ɛ U s 3 s dx = s dx + sɛ β+ β β + β β s B δ 0 B δ 0 dx + oɛ β+ β U s 1 ɛ s β dx + oɛ β+ β. U s 1 ɛ s β dx + oɛ β+ β. Therefore, plugging 33 and 3 into J,h u ɛ and using the equation 10 satisfied by U yields J,s,hu ɛ = J,s,0U Rn 1 ω n 1 β+ n χ m,h ɛ β+ β + oɛ β+ β. dx R n U s s This readily shows that if hx = λ, where 0 < λ < λ 1 L, and if m,λ > 0, then J,s,λ u ɛ < J,s,0U Rn = µ,s,0 R n, and therefore µ,s,λ is attained. This completes the proof of Proposition 5.

15 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 15 Subsection.: The merely singular case Proposition 6. Let be a smooth bounded domain in R n, n = 3, such that 0. Assume that s = 0 and < 0. If 0 < λ < λ 1 L is such that the mass R,λ is positive, then µ,0,λ is attained. Proof: This is by now classical, so we shall sketch a proof. For any x 0 \ {0}, we let G x0 C \ {0} be the Green s function for the operator + hx at x 0 with Dirichlet boundary condition. Since n = 3, then up to multiplying by a constant, we have G x0 x = 1 ηx π x x 0 + βx for all x \ {x 0 }, where β H0 1 C 0, and there exists R,h, x 0 R such that G x0 x = 1 1 π x x 0 + R,h, x 0 + o1 as x x 0. Note that βx 0 = R,h x 0 is the Robin function at x 0. Set now ũ ɛ x := u ɛ x + ɛ βx for all x \ {0}, where u ɛ are the functions defined in 11. Then, classical computations in the spirit of Schoen [6] yield J,hũ ɛ = 1 Kn, c 3R,h, x 0 ɛ n + o ɛ n as ɛ 0. If now x 0 \ {0} and 0 < λ < λ 1 L are such R,λ x 0 > 0, then This implies that µ,0,λ is attained. J,hũ ɛ = µ,0 R n c 3 R,λ x 0 ɛ + o ɛ as ɛ Blow-Up analysis in the truly singular case Let be a smooth bounded domain of R n, n 3, such that 0 is an interior point. < n / and recall that u R µ,s,0 R n := inf n u dx u ; u D 1, R n \ {0} s dx s, R n where 0 s < and s := n s n. Let a N C 1 be such that there exists a C 1 with 35 lim + a = a in C 1. Consider λ 0, + such that 36 lim + λ = µ,s,0 R n. Suppose u H0 1 is a sequence of weak solutions to u u + a u = λ s 1 s 37 u 0 u = 0 with 38 s u s s dx = 1. in a.e. in on Fix

16 16 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT and such that 39 u 0 as + weakly in H 1 0. We shall assume uniform coercivity, that is there exists c > 0 such that ϕ + a ϕ dx c ϕ dx for all ϕ H0 1. Note that this is equivalent to the coercivity of + a. The two following sections are devoted to the analysis of the Blow-up behavior of u as +. The present section deals mostly with the case {s > 0 or > 0}, for which there are extremals for µ,s,0 R n. The case {s = 0 and < 0} will be dealt with in the next section. The case s = = 0 has been extensively studied in the litterature, see for instance [1, 1] and the references therein. Theorem. Let be a smooth bounded domain of R n, n 3, such that 0 is an interior point. Fix < n /, and assume that either s > 0 or > 0. Let a C 1, λ 0, + and u H 1 0 such that 35, 36, 37, 38 and 39 hold. Then: i If n 1, then a 0 = 0; ii If n 1 < < n, then m,a = 0. In addition, there exists C > 0 such that 0 u x C β + β µ µ β+ β β + where µ 0 as 0 is defined in below. β+ for all x \ {0}, The rest of this section is devoted to the proof of this theorem. We shall make frequent use of the following Pohozaev identity. Proposition 7. Let R n be a smooth bounded domain and let u C, u 0. For any p R n, we have x p i i u + n u u s 1 u cu s dx [ u = x p, ν u cu s s s x p i i u + n ] u ν u dσ p, x + u + c su s s s dx. Proof: The classical Pohozaev identity yields x p i i u + n = [x p, ν u For any t [0, ], integration by parts yields x p i i u + n = t t u u dx x p i i u + n u u t 1 u t dx p, x u t dx + +t ] ν u dσ. x p, νu t t t dσ. Putting together these two equalities yields the general identity claimed in the proposition.

17 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 17 To prove Theorem, we start by noting that regularity theory and Theorem 8 yield that for any, there exists C > 0 such that u C,θ \ {0}, and u x x 0 C β and u x C β 1 for x \ {0}. Fix τ R such that β < τ < n. It follows from that for any N, there exists x \ {0} such that 3 sup τ u x = x τ u x. x \{0} We now prove the following proposition, which is valid for any < n /. Proposition 8. Let be a smooth bounded domain of R n, n 3, such that 0 is an interior point. Fix < n /, and let a C 1, λ 0, + and u H 1 0 be such that 35, 36, 37 and 38 hold. Let x \ {0} be as in 3 and set µ := u x n. Then, 5 lim sup + x \{0} and therefore lim + µ = 0. In addition, τ u x = +, dx, 6 lim = +. + µ Proof of Proposition 8: If 5 does not hold, then there exists C > 0 such that, up to a subsequence, we have that τ u x C for all x \ {0}. Since τ < n, we then have that u s 7 lim lim δ 0 + s dx = 0. B δ 0 Since u is bounded uniformly in L outside 0, it then follows from 37 and 39 that u 0 in Cloc 0 \ {0}. This limit and 7 yield s u s dx 0 as +, contradicting 38. This proves 5. As a remark, note that when s > 0, the subcriticality s < and 39 yield u 0 in Cloc 0 \ {0}. We now prove 6. Assume that dx, = Oµ as +, the above remark then yields s = 0. We let x := lim + x such that x. Since is smooth, we let δ > 0 and ϕ C B δ 0, R n be a smooth diffeomorphism onto its image such that ϕ0 = x, ϕb δ 0 {x 1 < 0} = ϕb δ 0 and ϕb δ 0 {x 1 = 0} = ϕb δ 0. Up to a rotation and a rescaling, we can assume that dϕ 0 = Id. Let x 1, x, 0 R n 1 B δ 0 be such that ϕx 1, x = x. In particular, lim 0 x 1 + x = 0. Define ũ x := µ n u ϕ0, x + µ x for x {x 1 < 0} B µ 1 δ/ 0. We then have that g ũ µ ϕ0, x + µ x + a ϕ0, x + µ x = λ ũ 1, where g := ϕ Eucl0, x + µ x. We have that ũ > 0 on {x 1 < 0} B µ 1 δ/ 0, ũ = 0 on {x 1 = 0} B µ 1 δ/ 0, and ũ µ 1 x 1, 0 = 1. Therefore, standard elliptic theory yields the existence of ũ C {x 1 0} D 1, R n such that ũ = µ,s,0 R n ũ 1 and ũ > 0 in {x 1 < 0}

18 18 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT and ũ = 0 on {x 1 = 0}. It follows from Theorem 1.3, chapter III in [9] that this is a contradiction. This proves 6 and ends the proof of Proposition 8. In addiction to the hypothesis of Proposition 8, we now assume that either s > 0 or > 0. We claim that x 8 lim = c > 0. + µ For that, we first show that 9 x = Oµ as +. Indeed, if not we can assume that µ 1 x + as +. By defining ũ := µ n u x +µ x, it follows from our assumption and Proposition 8 that for any R > 0, and for large enough, ũ is defined on B R 0 and ũ x µ + x + ũ µ a x + µ x ũ = λ x µ + x s in B R 0. It follows from 3 and the assumption that µ 1 x + as +, that there exists CR > 0 such that ũ CR on B R 0 and that ũ 0 = 1. It then follows from standard elliptic theory that ũ ũ in C loc Rn where 0 < ũ 1 and 50 ũ = µ,s,0 R n u 1 if s = 0 and ũ = 0 if s > 0. By the Sobolev embedding, we have that 51 ũ dx = B R 0 B Rµ x s u dx u dx C u H 1 0 C, where we used the fact that B Rµ x since 6 holds. Therefore, by first passing to the limit as + and then as R +, we get that ũ L R n. Assuming that s > 0, and since 0 < ũ 1, it follows from 50 and Liouville s theorem that ũ 1, contradicting that ũ L R n. In other words, 9 is proved when s > 0. Assuming now that s = 0 but > 0, then by letting + and R + in 51 and using38, we get that ũ dx 1. Equation 50 then yields R n µ 0,0 R n ũ dx R n ũ R n dx = µ,0 R n n dx µ,0 R n. R n ũ Since > 0, it follows from classical estimates see [17] that µ,0 R n < µ 0,0 R n, yielding again a contradiction. In other words, 9 is proved when s = 0. We now prove 8. We argue again by contradiction and assume that x = oµ as +. We define ũ x := µ n u x x for x B 1 δ0 and δ > 0 small enough. The definition 3 yields x τ u x x x τ u x, and therefore τ ũ x 1 for all x B 1 δ0 \ {0}. Equation 37 rewrites s ũ + x x ũ s 1 a x x ũ = λ µ s in B 1 δ0 \ {0}. In addition, we have that ũ > 0 and ũ x 1 x = 1. These estimates and standard elliptic theory then yield the existence of ũ C R n \ {0} such that ũ ũ in Cloc Rn \ {0} where ũ ũ = 0 in Rn \ {0} ; ũ > 0 ; τ ũx 1 in R n \ {0}.

19 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 19 The classification of Proposition 11 yields the existence of A, B 0 such that ũx = A β+ + B β in R n \ {0}. The pointwise control τ ũx 1 in R n \ {0} yields A = B = 0, contradicting ũ > 0. This completes the proof of 8. We now define v x := µ n u µ x for x µ 1 \ {0}, and claim that there exists U H0 1 C R n \ {0} such that 5 lim + v = U in H 1,locR n C locr n \ {0}. For that, we first note that v + v s 1 µ a µ x v = λ s in µ 1 \ {0}. Moreover, v > 0 and τ v x C for all x µ 1 \ {0}. It then follows from standard elliptic theory that there exists U C R n \ {0}, U 0, such that lim + v = U in C,θ loc Rn \ {0} and 53 U U = µ,s,0r n U s 1 s in R n \ {0}. Since v µ 1 x = 1, it then follows that U 0, and therefore U > 0. Moreover, we have that U s u s s dx = lim + s dx 1. B R 0\B δ 0 B Rµ 0\B δµ 0 Therefore, letting R + and δ 0 yields U s R n dx 1. Similarly, U s R n dx < + and U dx < +. Therefore U D 1, R n, and by integrating by parts, we obtain that R n U R µ,s,0 R n n U dx Rn = µ,s,0 R n U s s n s U s dx s s dx µ,s,0 R n. R n Therefore R n We now show that s U s s dx = 1 and U D 1, R n is an extremal for µ,s,0 R n. 5 lim + u = 0 in C 0 loc \ {0}. Indeed, when s > 0, we have already noted that the result follows from subcriticality. If however s = 0, it then follows from the convergence to U that lim lim u s 55 R + + B Rµ 0\B R 1 µ 0 s dx = lim lim v s R + + B R 0\B R 1 0 s dx U s = lim R + s dx = 1. B R 0\B R 1 0 Therefore, for any δ > 0, we have that lim + \B δ dx = 0. We then rewrite 37 as 0 u = f u in \ B δ 0 where lim 0 f n/ = 0. It then follows from the classical degiorgi- Nash-Moser iterative scheme that u is uniformly bounded in Cloc 0 \ {0}. Elliptic theory and 39 then yield the convergence to 0. This proves 5. We now claim that there exists C > 0 such that u s s 56 n u x C for all x \ {0} and N.

20 0 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT We argue by contradiction and we let y \ {0} be such that 57 sup n u x = y n u y + as +. x \{0} Note that it follows from that y is well-defined, and from 3,, 8 and 5 that 58 lim + y = 0, where ν := u y n Equation 37 rewrites y y lim = + and lim = +, + µ + ν 0 as +. We define ũ x := ν n 59 ũ y ν + x + ν a y + ν x ũ s 1 ũ = λ y ν u y +ν x for x ν 1 \{0}. in ν 1 + x s \ {0}. It follows from the definition 57 that for any R > 0, ũ in B R 0 for > 0 large enough. Since ũ 0 = 1, elliptic theory yields the existence of ũ C R n such that ũ ũ > 0 in Cloc Rn as +. In addition, for all R > 0, we have with Sobolev s inequality that B R 0 ũ dx = B Rν y u dx u dx C and therefore, letting + and R +, we get that ũ L R n. We now distinguish two cases: If s > 0, then passing to the limit in 59, we get that ũ = 0 in R n and ũ > 0 is bounded. Liouville s theorem then yields ũ ũ0 = 1, contradicting ũ L R n. If s = 0, then it follows from 38 and 55 that lim lim R + + \B Rµ 0 u dx = 0. It follows from 58 that for > 0 large enough, then B Rν y B Rµ 0, and therefore, we have that lim R + lim + dx = 0, which yields ũ 0, contradicting ũ0 = 1. This proves 56. We now claim that 60 lim B Rν y u lim sup R + + x \B Rµ 0 n u x = 0. We just sketch the proof which is very similar to the proof of 56. Arguing by contradiction and letting y be such that µ 1 y + as + and y n u y c > 0. We rescale at y and we get that our hypothesis yields the persistence of some energy outside B Rµ 0 for R and large, which is a contradiction. We now prove that for any ɛ > 0 small, there exists C ɛ > 0 such that β + β ɛ µ 61 u x C ɛ for all x \ B β+ ɛ µ 0. Note first that in view of 56, it is enough to prove 61 in \ B Rµ 0 for R > 0 large. For that, fix such that < < n, and let be a smooth bounded domain of R n such that is relatively compact in. We extend a and a on such that 35 holds on. Let G be the Green s function on at x of + a + ν, where ν > 0 and Dirichlet boundary condition. Up to taking close to, ν small enough and close to, the operator is

21 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 1 coercive and the Green s function is well defined on \ {0, x }. Theorem 6 in Appendix A then yields a C > 0 such that for any N β max{ x, } 6 0 < G x C x x n for all x \ {0, x }. min{ x, } Define the operator L := + a u s λ s. It follows from 8 that there exists R 0 > 1 such that x R 0 1µ for all N. It is easy to check that for x \ B R0µ 0, L G u s x x = G + a x a x + ν λ s. It follows from 35 that there exists 0 > 0 such that a x a x ν/ for all > 0 and all x. For a fixed δ > 0, 60 yields R > R 0 such that for > 0 large enough, we have that u x δ n / for x \ B Rµ 0. Therefore, with 36, we get that for x \ B Rµ 0, L G x > 1 G µ,s,0 R n δ s s + o1. Up to taking δ > 0 small enough, we then get that L G > 0 in \ B Rµ 0. It follows from 5 and 130 that there exists cr > 0 such that u x crµ n G x for all x B Rµ 0 and N. Therefore, defining h := crµ n G u, we get that L h > 0 in \ B Rµ 0 and h 0 in \B Rµ 0. Since G > 0 in \ B Rµ 0 and L G > 0, it follows from the comparison principle of Berestycki-Nirenberg-Varadhan [3] that L satisfies the comparison principle on \ B Rµ x. Therefore, u crµ n G in \ B Rµ 0. With the pointwise control 6, we then get that β+ u x CR µ β β+ for all x \ B Rµ 0 Since this is valid for any > close to, with the remark made at the beginning of the proof, we get 61. We now claim that there exists C > 0 such that β+ β 63 u x C µ for all x \ B β+ µ 0. Indeed, as argued above, the result holds on B Rµ 0 \ B µ 0 for any R > 1. In order to establish 63, we will prove it for any sequence z such that z 6 lim = +. + µ Let G be the Green s function of + a on with Dirichlet boundary condition. Green s representation formula in Appendix A, and the pointwise control 61 yield u s 1 x u z = G z, xλ s dy β max{ z, } s 1 n u x C x z min{ z, } s dx.

22 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT We split into four subdomains. On D 1, R := B Rµ 0, we have from 6, 61 and 5 that β max{ z, } s 1 n u x x z D 1, min{ z, } s dx C z β+ u s 1 dx C z β+ β + β µ s+β. B µ 0 Let D, R := {Rµ < < 1 z }, and note that x z > 1 z for all x D, R. Therefore, taking ɛ > 0 sufficienty small in 61, we have that β max{ z, } s 1 n u x x z D, min{ z, } s dx C z β+ β + β ɛ µ s 1 θr z β+ β + β µ, as +, where lim R + θr = 0. s β s 1β + ɛ dx D, Set D 3, := { 1 z < < z }, and by using again 61 with ɛ > 0 sufficiently small, we get that β max{ z, } s 1 n u x x z D 3, min{ z, } s dx β + β ɛ Cµ s 1 z s β+ ɛ s 1 x z n dx D 3, β + β β + β Cµ z β+ µ s ɛ s 1. z Finally, let D, := { z }. Since x z /, then using 61 with ɛ > 0 sufficiently small, we get that β max{ z, } s 1 n u x x z D, min{ z, } s dx C z β β + β ɛ µ s 1 s β+ s+ɛ s 1 dx D, β + β β + β Cµ z β+ µ s ɛ s 1. z Plugging together these estimates yields 63. Since U is a positive solution to 53 and U D 1, R n, it follows from the regularity Theorem 8 that there exists C 1 > 0 such that Ux C 1 β as s 0. Taking the Kelvin transform Ũx := n Ux, we get that Ũ D1, R n is also a positive solution to 53, and enjoys a similar behavior at 0. Transforming back yields the existence of C 1, C > 0 such that 65 Ux x 0 C 1 β and Ux We now show that there exists H C \ {0} such that 66 lim + u β + β µ C 1 β+. = H in C loc \ {0},

23 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 3 and H is a solution to 67 H + a H = 0 in \ {0} H > 0 in \ {0} H = 0 on. Define w := µ 68 β + β w w 0 w = 0 u. Equation 37 then rewrites as + a w = λ µ β + β s w s 1 s in a.e. in on, and 63 yields that w x C β+ for all x \ {0} and N. It then follows from elliptic theory that there exists H C \ {0} such that lim + w = H in Cloc \ {0}. Passing to the limit in 68 yields H 0 and H + a H = 0 in \ {0} and H = 0 on. Fix x \ {0}. Green s representation formula, the positivity of G and a change of variable yields u s 1 y u x = G x, yλ y s dy u s 1 y G x, yλ B µ 0\B µ 0 y s dy µ n v y s 1 y G x, µ yλ y s dy. B 0\B 10 The asymptotics 18 in Appendix A yields G x, z c x z β for all N and all z B / 0. Therefore, we get that for all N, β + β u x c x µ y β v y s 1 y y s dy B 0\B 10 Passing to the limit as + and using 5 yields Hx > 0, which proves our claim in 67. Let now δ > 0 be such that B δ 0. For any 0 < ɛ < δ, the Pohozaev identity 1 with p = 0, and equation 37 yield a + xi i a 69 u dx = B x dx, B δ 0\B ɛ0 B δ 0\B ɛ0 where B x := x, ν u u + a λ u s s s Using the asymptotics, we pass to the limit as ɛ 0 and get a + xi i a u dx = The limit 66 yields 70 = B δ 0 lim + µ β+ β B δ 0 [ x, ν H B δ 0 B x dσ + a H B δ 0 x i i u + n u ν u. B x dx. x i i H + n H ] ν H dσ.

24 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT Assuming now that β + β >, we show that 71 a 0 = 0. Indeed, note first that in this case, β + > n. It follows from 63 that lim lim R + 0 µ a + xi i a u dx = 0. B Rµ 0 B δ 0\B Rµ 0 With a change of variable, we get that a + xi i a u dx = µ B R 0 a + xi i a µ xv dx. The limit 5 and the compactness of the embedding H 1 L yields the convergence of v to U in L loc Rn. It follows from 65 that U L R n, the two preceding identities therefore yield lim 0 µ a + xi i a u dx = a 0 U dx. R n B δ 0 Plugging this limit in the Pohozaev identity 69 and using the limit above yields that a 0 = Oµ β+ β + o1 as +, and therefore a 0 = 0. We now assume that β + β =, and we show again that 7 a 0 = 0. Indeed, assume that a 0 0. Without loss of generality, we can suppose that a 0 > 0. Up to taking δ > 0 smaller and large, we have that a x + xi ia x a 0 for x B δ 0. It then follows from 69 and 70 that there exists C > 0 such that u dx u dx Cµ for all N. B Rµ 0 B δ 0 With a change of variable, the limit 5, letting + and then R +, we get that U L R n, which is impossible due to 65 and β + = n. Therefore a 0 = 0. Finally, we show that if β + β <, then 73 m,a = 0, where m,a is the Hardy-singular mass as defined in Proposition 3. Indeed, since β + < n, we have that a + xi i a u dx = O B δ 0 B δ 0 µ β+ β β+ dx = O µ β+ β δ n β+, uniformly with respect to and δ > 0. Combining with 70, we get that [ H H 7 lim x, ν δ 0 + a x i i H + n ] H ν H dσ = 0. B δ 0 Since β + β <, it follows from the definition of the mass that there exists c > 0 such that 1 Hx = c + m,a 1 + o as x 0. β+ β β Since H solve the equation 1, standard elliptic theory yields that this estimate can be differentiated. Therefore, putting it into 7 yields m,a = 0. Theorem is a consequence of 71, 7, and 73.

25 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 5 6. Blow-Up analysis in the merely singular case In this section, we perform the blow-up analysis in the merely singular case, that is when s = 0 and < 0. We let again a N C 1, a C 1, λ 0, + such that 35 and 36 hold. We let u H0 1 be a sequence of weak solutions to 37 such that 38 holds. In this case, 37 and 38 rewrite as: 75 and 76 We suppose that u + u 0 u = 0 a u = λ u 1 u dx = 1. in 77 u 0 as + weakly in H 1 0. a.e. in on We let be a smooth bounded domain of R n such that is relatively compact in. We extend a and a on such that 35 holds on and that the operator + a is coercive on. This assumption is equivalent to saying that there exists c > 0 such that for N large enough, we have 78 λ 1 + a = inf ϕ H0 1\{0} ϕ This section is devoted to the proof of the following result: + a ϕ dx c > 0. ϕ dx Theorem 5. Let be a smooth bounded domain of R n, n 3, such that 0 is an interior point. Fix < 0 and let a C 1, λ 0, + and u H0 1 be such that 35, 36, 75 and 76 hold. We let x and µ 0, + be such that u x := sup u = µ n. Then lim + x = x 0, lim + µ = 0 and i If n, then x 0 0 and a x 0 + x 0 = 0; ii If n = 3, then x 0 \ {0} and R,a, x 0 = 0 see 6 for the definition. In addition, there exists C > 0 such that u x C µ µ + x x n for all x and N. Before delving into the proof, it is important to note a few observations that are relevant for the case s = 0 and < 0. First note that in this case β < 0, and therefore, it follows from that for any N, u can be extended continuously at 0 by 0, which means that we can and will consider u C 0. In the definition 3, we shall take τ := 0 and therefore, the sequence x will be such that 79 u x := sup u x and µ := u x /n. x It then follows from Proposition 8 that 80 lim + µ = 0 and Another remark is that 75 implies dx, lim = +. + µ u a u λ u 1 in,

26 6 NASSIF GHOUSSOUB AND FRÉDÉRIC ROBERT which means that u is a subsolution of a nonlinear elliptic inequation with no Hardy potential term. We shall then be able to perform a blow-up analysis in the spirit of Druet-Hebey-Robert [1] to obtain a pointwise control of u by a standard bubble. The conclusion of Theorem 5 will then follow from classical arguments via the Pohozaev identity and the analysis on the boundary in the spirit of Druet [1]. Set v x := µ n u x + µ x for all x µ 1 x. Equation 75 and 76 then rewrites 81 v + 8 We first claim that x µ + x µ a x + µ x x µ v dx = 1. x 83 lim = +, + µ and 8 lim + v = v := nn Kn, n v = λ v 1 in C locr n with in x µ v R n dx = 1. Indeed, it follows from 80 that for any R > 0, there exists 0 > 0 such that B R 0 x µ for all > 0. Since u is uniformly bounded in H 1 0, then v is bounded in H 1 loc Rn. Up to extracting a subsequence, there exists v H 1 loc Rn such that v v as + weakly in H 1 loc Rn and strongly in L loc Rn. Since v µ a x + µ xv λ v 1 in B R 0, and 0 v 1, it follows from DeGiorgi-Nash-Moser iterative scheme see for instance Theorem.1 in Han-Lin [18], that there exists C > 0 such that for all > 0, 1 = v 0 C v L B 10 and therefore, passing to the strong limit in L, we get that 1 C v L B 10, and hence v 0. Since 0 < v v 0 = 1, equation 81 and elliptic theory yields v C R n \ {θ } and v v in C loc Rn \ {θ } with 85 v + x θ v = µ,s,0r n v 1 in R n \ {θ } where θ := lim + µ 1 x if this limit is finite. Otherwise θ :=, in which case R n \{θ } := R n. In addition, passing to the weak limit in 8 yields v R n dx = lim R + B R 0 v dx lim lim R + + B R 0 v dx 1. Since B R 0 v dx = B Rµ x u dx C uniformy for all R > 0 and > 0 large enough, passing to the weak limit yields v L R n. Since v L R n, classical arguments yield that v D 1, R n. Multiplying 85 by v and integrating, we obtain v dx v + R n R x θ n v dx = µ,s,0 R n v dx. R n

27 HARDY-SCHRÖDINGER OPERATOR WITH INTERIOR SINGULARITY 7 Since v 0, the Sobolev inequality yields R n v dx R n v dx µ 0,0 R n = µ,0 R n. Since v dx 1 and > 0, putting these latest inequalities together yields R n θ = and dx = 1. We then get R n v x 86 lim = + and lim + µ v = v in ClocR n + where v D 1, R n C R n is such that v = µ 0,0 R n v 1 in R n ; v R n dx = 1 ; 0 v v0 = 1. Then 83 and 8 follow from 86, this latest assertion and the classification of Caffarelli-Gidas- Spruck [6]. We now claim that there exists C > 0 such that µ 87 u x µ + x x n for all N and x. This estimate is by now standard and is in the spirit of similar results obtained by several authors. See for instance Druet-Hebey-Robert [1] and the several references therein. When possible, we shall only sketch an outline to the proof and we refer to these references for details. Note first that 88 lim lim u dx = 0. R + + \B Rµ x Indeed, the convergence of v to v in 8 yields that asymptotically, B Rµ x exhausts almost all the energy in 76. Next, we claim that there exists C > 0 such that x x n u x C for all N and x. Indeed, if not we find y that achieve the supremum of the left-hand-side and which go to + as +. The same blow-up procedure as above at y yields that asymptotically, B /n y uy carries a nonzero mass of u dx, contradicting 88, since this ball is disjoint from B Rµ x for R and large. A similar argument that we omit yields that 89 lim lim sup R + + x \B Rµ x x x n u x = 0. Let now η 0 C R be such that 0 η 0 1, η 0 t = 0 if t 1 and η 0 t = 1 if t. We define η ɛ x := η 0 /ɛ for x R n. We claim that there exists ɛ > 0 such that 90 η ɛ x a c/ is coercive. To prove this claim, we shall need the following continuity lemma for the first eigenvalue. Recall that for any V : R measurable such that for some C > 0, we have V x C for a.e. x, the following ratio is well defined and is finite. λ 1 + V := inf ϕ H 1 0 \{0} ϕ + V ϕ dx ϕ dx