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Borderline variational problems for fractional Hardy-Schrödinger operators Shakerian, Shaya 2017

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Borderline variational problems forfractional Hardy-Schro¨dingeroperatorsbyShaya ShakerianB.Sc., University of Tehran, 2009M.Sc., Sharif University of Technology, 2012A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)July 2017c© Shaya Shakerian 2017AbstractIn this thesis, we study properties of the fractional Hardy-Schro¨dinger op-erator Lγ,α := (−∆)α2 − γ|x|α both on Rn and on its bounded domains. Thefollowing functional inequality is key to our variational approach.C(∫Rn|u|2∗α(s)|x|s dx) 22∗α(s)≤∫Rn|(−∆)α4 u|2dx− γ∫Rn|u|2|x|αdx, (0.1)where 0 ≤ s < α < 2, n > α, 2∗α(s) := 2(n−s)n−α and γ < γH(α), the latter beingthe best fractional Hardy constant on Rn. We address questions regardingthe attainability of the best constant C > 0 attached to this inequality.This allows us to establish the existence of non-trivial weak solutions for thefollowing doubly critical problem on Rn,Lγ,αu = |u|2∗α−2u+ |u|2∗α(s)−2u|x|s in Rn where 2∗α := 2∗α(0).We then look for least-energy solutions of the following linearly perturbednon-linear boundary value problem on bounded subdomains of Rn contain-ing the singularity 0:(Lγ,α − λI)u = u2∗α(s)−1|x|s on Ω. (0.2)We show that if γ is below a certain threshold γcrit(α), then such solutionsexist for all 0 < λ < λ1(Lγ,α), the latter being the first eigenvalue of Lγ,α.On the other hand, for γcrit(α) < γ < γH(α), we prove existence of suchsolutions only for those λ in (0, λ1(Lγ,α)) for which Ω has a positive fractionalHardy-Schro¨dinger mass mαγ,λ(Ω). This latter notion is introduced by wayof an invariant of the linear equation (Lγ,α − λI)u = 0 on Ω.We then study the effect of non-linear perturbation h(x)uq−1, where h ∈C0(Ω), h ≥ 0 and 2 < q < 2∗α. Our analysis shows that the existence ofsolutions is guaranteed by this perturbation whenever 0 ≤ γ ≤ γcrit(α),while for γcrit(α) < γ < γH(α), it depends on both the perturbation andthe geometry of the domain.iiLay SummaryMany physical and probabilistic phenomena can be described by either de-terministic or stochastic differential equations. In this thesis, we focus ona class of partial differential equation driven by non-local operators. Thesedescribe long range interactions between the objects under study, and notonly by those interacting with their closest neighbours. These equationshave deep connections to probability theory and geometry, and they ap-pear in many physical applications such as elasticity, thin obstacle, phasetransition, flames propagation, as well as mathematical finance.iiiPrefaceMuch of this dissertation is adapted from three of the author’s research pa-pers: [38], [39] and [66]. In particular, Chapter 3, which evolves around theproof of Theorem 3.1 and 3.2, form the main content of [39], Borderline vari-ational problems involving fractional Laplacians and critical singularities.All of Chapter 4 are adapted from [38], Mass and asymptotics associated tofractional Hardy-Schro¨dinger operators in critical regimes. Chapter 5, wherethe proof of Theorem 5.1 is presented, is in accordance with [66], Existenceresults for non-linearly perturbed Hardy-Schro¨dinger problems: Local andnon-local cases. The first manuscript [39] (joint work with Dr. Ghoussoub)was published in Advanced Non-linear Studies Journal. The second paper[38] (joint work with Dr. Ghoussoub, Dr. Robert and Dr. Zhao) has beensubmitted, and the third one [66] has been written and will be submittedfor publication soon.ivTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . vAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Preliminaries and a Description of The Functional Setting 82.1 The fractional Laplacian operator . . . . . . . . . . . . . . . 82.2 The fractional Sobolev spaces . . . . . . . . . . . . . . . . . 112.2.1 The classical fractional Sobolev space . . . . . . . . . 112.2.2 The variational fractional Sobolev space . . . . . . . 132.3 The α-harmonic extension and weighted Sobolev space . . . 142.4 Fractional Hardy-Sobolev type inequalities . . . . . . . . . . 153 Borderline Variational Problems Involving Fractional Lapla-cian and Critical Singularities . . . . . . . . . . . . . . . . . . 183.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Proof of Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . 223.3 Proof of Theorem 3.2 . . . . . . . . . . . . . . . . . . . . . . 323.3.1 Existence of a suitable Palais-Smale sequence . . . . . 333.3.2 Analysis of the Palais-Smale sequences . . . . . . . . 363.3.3 End of proof of Theorem 3.4 . . . . . . . . . . . . . . 454 Mass and Asymptotics Associated to Fractional Hardy-Schro¨dingerOperators in Critical Regimes . . . . . . . . . . . . . . . . . 484.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.2 The fractional Hardy-Schro¨dinger operator Lγ,α on Rn . . . 54vTable of Contents4.3 Profile of solutions . . . . . . . . . . . . . . . . . . . . . . . . 574.4 Analytic conditions for the existence of extremals . . . . . . 654.5 The fractional Hardy singular interior mass of a domain inthe critical case . . . . . . . . . . . . . . . . . . . . . . . . . 684.6 Existence of extremals . . . . . . . . . . . . . . . . . . . . . . 754.6.1 General estimates for ηu . . . . . . . . . . . . . . . . 774.6.2 The test functions for the non-critical case . . . . . . 794.6.3 The test function for the critical case . . . . . . . . . 804.6.4 Proof of Theorem 4.4 . . . . . . . . . . . . . . . . . . 885 Existence Results for Non-linearly Perturbed Fractional Hardy-Schro¨dinger Problems . . . . . . . . . . . . . . . . . . . . . . . 895.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.2 The Palais-Smale condition below a critical threshold . . . . 915.3 Mountain pass geometry and existence of a Palais-Smale se-quence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.4 Proof of Theorem 5.1 . . . . . . . . . . . . . . . . . . . . . . 96Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104viAcknowledgementsI would like to express my deep gratitude and appreciation to my super-visor, Professor Nassif Ghoussoub, for his invaluable guidance, direction,encouragement but also constructive criticism during the course of this dis-sertation. I thank him for not only caring about my current and futureacademic career, but also for his genuine interest, concern, and support formy well-being outside my academic life.I would also like to acknowledge the profound gratitude I have for Pro-fessor Fre´de´ric Robert, from whom i also learned so much. His role as amentoring collaborator, and his persistent call for rigour throughout theprogress of our work was absolutely critical for my quest to become a re-searcher.Finally, I would like to thank my committee members Professor Jun-cheng Wei and Professor Stephen Gustafson for their instructive comments,and especially Professor Jun-cheng Wei, with whom I have had several mo-tivating and inspiring discussions.viiChapter 1IntroductionThe main focus of research in this thesis is the study of borderline variationalproblems and corresponding functional inequalities involving the fractionalHardy-Schro¨dinger operator. This thesis is based on three research papers[38, 39, 66] that can be found in Chapters 3, 4 and 5. Each of the chaptersbegins with a detailed introduction to the results it contains. In this intro-duction, I will give a general description of the problems I address in thisthesis.Equations driven by non-local operators appear in many areas of mathe-matics such as probability theory, fluid mechanics, and geometry, as well asother sciences such as physics, and economics. In analysis and PDE, thesephenomena originate in potential theory, conformal geometry, and a wideclass of physical systems modeled by α-stable Levy processes and relatedstochastic interfaces. Primary examples of non-local operators are those rep-resented by the fractional Laplacians (−∆)α2 for 0 < α < 2, which are cleargeneralizations of the well-studied Laplace operator (α = 2). Over the lastdecade or so, many well-known properties of standard elliptic and parabolicequations have been extended to their non-local counterparts. Much of theprogress can be summarized as follows:• Representation and regularity results such as those in Silvestre [62],Caffarelli-Silvestre [13, 14], Ros-Oton and Serra [58, 59] and others.• Variatonal formulations and methods, such as in Servadei [65], Bisci-Radulescu-Servadei [6], Servadei-Valdinoci [64] and others.• Conformal Geometry and Yamabe type problems, such as in Chang-Gonzalez [16], Gonzalez-Qing [42] and others.In the first part of this thesis, we study problems of existence of equa-tions involving the fractional Hardy-Schro¨dinger operator on Rn. In [39],N. Ghoussoub and I consider issues of existence of the variational solutionsof the following borderline problem associated with the fractional Hardy-Schro¨dinger operator Lγ,α := (−∆)α2 − γ|x|α on Rn :{(−∆)α2 u− γ u|x|α = u2∗α(s)−1|x|s in Rnu ≥ 0 in Rn, (1.1)1Chapter 1. Introductionwhere 0 < α < 2, 0 ≤ s < α < n, 2∗α(s) = 2(n−s)n−α , and γ < γH(α) =2αΓ2(n+α4)Γ2(n−α4)is the best fractional Hardy constant on Rn. For any α ∈ (0, 2),the fractional space Hα20 (Rn) is defined as the completion of C∞0 (Rn) underthe norm‖u‖Hα20 (Rn)=∫Rn|2piξ|α|Fu(ξ)|2dξ =∫Rn|(−∆)α4 u|2dx,where F(u)(ξ) = ∫Rn e−2piix·ξu(x)dx denotes the Fourier transform of u.In order to study problem (1.1), we address the attainability of the bestconstant corresponding to the fractional Hardy-Sobolev inequality in Rn,that isµγ,s,α(Rn) := infu∈Hα20 (Rn)\{0}∫Rn |(−∆)α4 u|2dx− γ ∫Rn |u|2|x|αdx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s). (1.2)In the following, we check for which parameters γ and s, the best constantµγ,s,α(Rn) is attained. Using the Caffarelli-Silvestre representation [12] andEkeland’s variational principle [25], we establish the following.Theorem 1.1 (Ghoussoub-Shakerian [39]). Suppose 0 < α < 2, 0 ≤ s <α < n, and γ < γH(α) := 2α Γ2(n+α4)Γ2(n−α4).1. If either {s > 0} or {s = 0 and γ ≥ 0}, then µγ,s,α(Rn) is attained.2. If s = 0 and γ < 0, then there are no extremals for µγ,s,α(Rn).3. If either {0 < γ < γH(α)} or {0 < s < α and γ = 0}, then anynon-negative minimizer for µγ,s,α(Rn) is positive, radially symmetric,radially decreasing, and approaches zero as |x| → ∞.The results in Theorem 1.1 allow us to show the existence of non-trivialweak solutions for the following doubly critical problem on Rn,(−∆)α2 u− γ u|x|α = |u|2∗α−2u+|u|2∗α(s)−2u|x|s in Rn, (1.3)where 2∗α :=2nn−α is the critical α-fractional Sobolev exponent, and 0 ≤γ < γH(α). We used the Caffarelli-Silvestre representation [12] and theMountain Pass lemma [3] to prove the following result.2Chapter 1. IntroductionTheorem 1.2 (Ghoussoub-Shakerian [39]). Let 0 < α < 2, 0 < s < α < nand 0 ≤ γ < γH(α). Then, there exists a non-trivial weak solution of (1.3).In the next step, we use the ground state representation introduced byFrank-Lieb-Seiringer [34], and Moser iteration, to establish the followingasymptotic properties of the extremals of µγ,s,α(Rn) at zero and infinity.Theorem 1.3 (Ghoussoub-Robert-Shakerian-Zhao [38]). Assume 0 ≤ s <α < 2, n > α and 0 ≤ γ < γH(α). Then, any positive extremal u ∈ Hα20 (Rn)for µγ,s,α(Rn) satisfies u ∈ C1(Rn \ {0}) andlimx→0|x|β−(γ)u(x) = λ0 and lim|x|→∞ |x|β+(γ)u(x) = λ∞, (1.4)where λ0, λ∞ > 0 and β−(γ) (resp., β+(γ)) is the unique solution in(0, n−α2)(resp., in(n−α2 , n− α)) of the equationΨn,α(t) := 2α Γ(n−t2 )Γ(α+t2 )Γ(n−t−α2 )Γ(t2)= γ.In particular, there exist C1, C2 > 0 such thatC1|x|β−(γ) + |x|β+(γ) ≤ u(x) ≤C2|x|β−(γ) + |x|β+(γ) for all x ∈ Rn \ {0}.Remark 1.4. We point out that the functions u1(x) = |x|−β−(γ) and u2(x) =|x|−β+(γ) are the fundamental solutions for the fractional Hardy-Schro¨dingeroperator Lγ,α := (−∆)α2 − γ|x|α on Rn. Indeed, a straightforward computationyields (see Section 4.2)Lγ,α|x|−β = (Ψn,α(β)− γ)|x|−β = 0 on Rn \ {0} for β ∈ {β−(γ), β+(γ)},which implies that β+(γ) and β−(γ) satisfy Ψn,α(β) = γ.We then tackle the more challenging problems on bounded domains Ωof Rn with an interior singularity by considering the role of linear and non-linear perturbations on the existence of positive solutions. More precisely,if Ω is now a smooth bounded domain in Rn containing 0 in its interior, wethen consider the fractional Sobolev space Hα20 (Ω) as the closure of C∞0 (Ω)with respect to the norm‖u‖2Hα20 (Ω)=cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy where Cn,α =2αΓ(n+α2)pin2∣∣Γ (−α2 )∣∣ ,3Chapter 1. Introductionas well as the best constant in the fractional Hardy-Sobolev inequality ondomain Ω, namely,µγ,s,α(Ω) := infu∈Hα20 (Ω)\{0}cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ω|u|2|x|αdx(∫Ω|u|2∗α(s)|x|s dx)22∗α(s). (1.5)As in the local case, one can show by translating, scaling and cutting offthe extremals of µγ,s,α(Rn) that µγ,s,α(Ω) = µγ,s,α(Rn), which means thatµγ,s,α(Ω) has no extremals if Ω is bounded. We therefore resort to a set-ting popularized by Brezis-Nirenberg [11] in the local case, where one de-homogenizes the problem by considering the following equation:{(−∆)α2 u− γ u|x|α = u2∗α(s)−1|x|s + λu in Ωu = 0 in Rn \ Ω, (1.6)where 0 < λ < λ1(Lγ,α) and λ1(Lγ,α) is the first eigenvalue of the operatorLγ,α = (−∆)α2 − γ|x|α with the Dirichlet boundary condition. One thenconsiders the quantityµγ,s,α,λ(Ω) = infu∈Hα20 (Ω)\{0}KΩ(u),whereKΩ(u) :=Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ωu2|x|αdx− λ∫Ω u2dx(∫Ωu2∗α(s)|x|s dx)22∗α(s),and uses the fact that compactness is restored as long as µγ,s,α,λ(Ω) <µγ,s,α(Rn); see [11, 37]. Janelli [45] showed that the behaviour of problem(1.6) - in the case when α = 2 and s = 0 - is deeply influenced by thevalue of the parameter γ. Roughly speaking, when γ is sufficiently near toγH(α) then problem (1.6) becomes critical, in the sense of Pucci-Serrin [55].Following [45], we prove that there exists a constant 0 < γcrit(α) < γH(α)such that the operator Lγ,α becomes critical when γ ∈ (γcrit(α), γH(α)).We also show that the existence results in [64] can be extended to thecase when 0 < s < α and 0 < γ < γH(α) as long as the operator Lγ,α is notcritical.The critical case is more delicate and the existence of solutions requiresthe domain Ω to satisfy a positive mass condition, defined as follows. Indeed,following the work of [36, 37] in the local setting, we define the fractionalHardy singular interior mass of a domain is the following way.4Chapter 1. IntroductionTheorem 1.5 (Ghoussoub-Robert-Shakerian-Zhao [38]). Let Ω be a boundedsmooth domain in Rn (n > α) and consider, for 0 < α < 2, the boundaryvalue problem(−∆)α2H −(γ|x|α + a(x))H = 0 in Ω \ {0}H > 0 in Ω \ {0}H = 0 in Rn \ Ω,(1.7)where a(x) ∈ C0,τ (Ω) for some τ ∈ (0, 1). Assuming the operator (−∆)α2 −( γ|x|α +a(x)) is coercive, there exists then a threshold −∞ < γcrit(α) < γH(α)such that for any γ with γcrit(α) < γ < γH(α), there exists a unique solutionto (1.7) (in the sense of Definition 4.6) H : Ω→ R, H 6≡ 0, and a constantc ∈ R such thatH(x) =1|x|β+(γ) +c|x|β−(γ) + o(1|x|β−(γ))as x→ 0.We define the fractional Hardy-singular internal mass of Ω associated to theoperator Lγ,α to bemαγ,a(Ω) := c ∈ R.We then establish the following results.Theorem 1.6 (Ghoussoub-Robert-Shakerian-Zhao [38]). Let Ω be a smoothbounded domain in Rn(n > α) such that 0 ∈ Ω, and let 0 ≤ s < α, 0 ≤ γ <γH(α), and 0 < λ < λ1(Lγ,α). Then, there are extremals for µγ,s,α,λ(Ω)under one of the following two conditions:(1) 0 ≤ γ ≤ γcrit(α)(2) γcrit(α) < γ < γH(α) and mαγ,λ(Ω) is positive.One can then complete the picture as follows.Linearly perturbed problem (1.6) with 0 ∈ Ω and 0 < λ < λ(Lγ,α)Hardy term Dimension Singularity Analytic. cond. Ext.0 ≤ γ ≤ γcrit(α) n ≥ 2α s ≥ 0 λ > 0 Yesγcrit(α) < γ < γH(α) n ≥ 2α s ≥ 0 mαγ,λ(Ω) > 0 Yes0 ≤ γ < γH(α) α < n < 2α s ≥ 0 mαγ,λ(Ω) > 0 Yes5Chapter 1. IntroductionIn this direction, we address the following non-linearly perturbed prob-lem associated with the operator Lγ,α − λI on bounded domains Ω :(−∆)α2 u− γ u|x|α − λu =u2∗α(s)−1|x|s + h(x)uq−1 in Ωu ≥ 0 in Ω,u = 0 in Rn \ Ω,(1.8)where h ∈ C0(Ω), h ≥ 0 and q ∈ (2, 2∗α). We study the combined effect ofthe non-linear perturbation (i.e., h(x)uq−1) and the geometry of the domain(i.e., the mass introduced in Theorem 1.5) on the existence of a positivesolution for (1.8).Inspired by the work of Jaber [44], in a Riemannian context, we in-vestigate the existence of solutions using the Mountain Pass Lemma ofAmbrosetti- Rabinowitz [3] (see Lemma 3.6). It turns out that the exis-tence of a solution for (1.8) depends only on the non-linear perturbationwhenever the operator Lγ,α is non-critical (i.e., when 0 ≤ γ ≤ γcrit(α)).On the other hand, in the critical case (i.e., when γcrit(α) < γ < γH(α)),solvabiliy can either depend on the non-linear perturbation or the globalgeometry of the domain, or both. The difference between these regimes isdetermined by another threshold, but this time on the values of q in (2, 2∗α).We shall utilize ideas of [38] and [44] to prove the following.Theorem 1.7 (Shakerian[66]). Let Ω be a smooth bounded domain in Rn(n> α) such that 0 ∈ Ω, and let 2∗α(s) := 2(n−s)n−α , 0 ≤ s < α, −∞ < λ <λ1(Lγ,α), and 0 ≤ γ < γH(α). We also assume that 2 < q < 2∗α, h ∈ C0(Ω)and h ≥ 0. Then, there exists a non-negative solution u ∈ Hα20 (Ω) to (1.8)under one of the following conditions:(1) 0 ≤ γ ≤ γcrit(α) and h(0) > 0.(2) γcrit(α) < γ < γH(α) andh(0) > 0 if q > qcritc1h(0) + c2mαγ,λ(Ω) > 0 if q = qcritmαγ,λ(Ω) > 0 if q < qcrit.Here c1, c2 are two positive constant that can be computed explicitly (seeSection 5.4), while qcrit = 2∗α − 2β+−β−n−α ∈ (2, 2∗α).In order to summarize the results in Theorem 1.7, we set γcrit := γcrit(α),γH := γH(α) and mαγ,λ := mαγ,λ(Ω), and assume that 0 ≤ s < α and 2 < q <2∗α.6Chapter 1. IntroductionOne can then complete the picture as follows.Non-linearly perturbed problem (1.8): n > α and λ < λ(Lγ,α)Hardy term Dim. q λ Analytic. cond Ext.0 ≤ γ ≤ γcrit n ≥ 2α > 2 > −∞ h(0) > 0 Yesγcrit < γ < γH0 ≤ γ < γHn ≥ 2αn < 2α> qcrit > −∞ h(0) > 0 Yesγcrit < γ < γH0 ≤ γ < γHn ≥ 2αn < 2α= qcrit > −∞ c1h(0)+c2mαγ,λ > 0 Yesγcrit < γ < γH0 ≤ γ < γHn ≥ 2αn < 2α< qcrit > 0 mαγ,λ > 0 Yes7Chapter 2Preliminaries and aDescription of TheFunctional SettingIn this chapter, we introduce the concepts and the classical results relatedto the non-local framework that will be the focus of this dissertation. Ourmain sources for this chapter are [6] and [20].The basic operator is the so-called fractional Laplacian (−∆)α2 withα ∈ (0, 2). This chapter is devoted to introducing its various formulations,its associated fractional Sobolev spaces as well as the corresponding frac-tional Hardy-Sobolev inequalities. Since our main focus in this thesis is thestudy of fractional laplacians with Dirichlet boundary data via variationalmethods, we first introduce suitable function spaces required for the varia-tional principles to apply. In Section 2.3, we introduce a weighted Sobolevspace in higher dimension Rn × (0,∞), which will provide a suitable func-tion space for our approach. Its construction is based on the α-extensionformula that Caffarelli-Silvestre [12] associate to equations involving frac-tional laplacians. We later introduce functional inequalities involving thefractional Hardy-Schro¨dinger operator Lγ,α, which play a crucial role in thenon-local problems under study.2.1 The fractional Laplacian operatorIn this section, we start by presenting the different definitions for the frac-tional Laplacian operator (−∆)α2 . First, consider the Schwartz space S(Rn)which consists of all rapidly decreasing functions in C∞(Rn), equipped withthe following norm: IfZn+ = {a = (a1, a2, ..., an) : ai ≥ 0 and ai ∈ Z for i = 1, 2, 3, ..., n},and∂a := (∂∂x1)a1(∂∂x2)a2 ...(∂∂xn)an ,82.1. The fractional Laplacian operatorthen, the Schwartz space S(Rn) consists of all functions ψ ∈ C∞(Rn) suchthat‖ψ‖a,b = supx∈Rn|xa∂bψ(x)|is finite for every pair of multi-indices a, b ∈ Zn+.Remark 2.1. An element of S(Rn) is essentially a smooth function ψ(x)such that ψ(x) and all of its derivatives exist everywhere on R and decayfaster than any inverse power of x, as |x| → ±∞. In particular, S(Rn) is asubspace of the function space C∞(Rn) of infinitely differentiable functions.The Fourier transform of a function ψ ∈ S(Rn) is then defined asFψ(ξ) :=∫Rne−2piix·ξψ(x)dx. (2.1)Note that for every ψ ∈ S(Rn), we have Fψ ∈ S(Rn). It is easy to verifythat the Fourier transform and the inverse Fourier transform, given byF−1ψ(x) :=∫Rne2piiξ·xψ(ξ)dξare both continuous on S(Rn) to S(Rn). Moreover, we haveFF−1ψ = F−1Fψ = ψ,which implies that each of F and F−1 is an isomorphism and a homomor-phism of S(Rn) onto S(Rn).The dual space of S(Rn), denoted by S ′(Rn), is the space of continuouslinear functionals T : S(Rn) → R. The elements of S ′(Rn) are called tem-pered distributions. The space S ′(Rn) is a linear space under the pointwiseaddition and scalar multiplication of functionals. If T ∈ S ′(Rn), the Fouriertransform of T can be defined as the tempered distribution given by〈FT, ψ〉 = 〈T,Fψ〉 for every ψ ∈ S(Rn),where 〈., .〉 denotes the usual duality bracket between S(Rn) and S ′(Rn).One can use the definition (2.1) to prove the Parseval-Plancherel formula(2.2), which will be crucial in what follows for proving the equivalence be-tween the fractional spaces will be defined in the next section; see formula(2.6).‖u‖2L2(Rn) = ‖Fu‖2L2(Rn) for all u ∈ L2(Rn). (2.2)92.1. The fractional Laplacian operatorFor a detailed introduction to the classical theory of distribution and Fouriertransform, we refer to the monograph [60] and to the recent book [18] forseveral applications to elliptic problems of linear and nonlinear functionalanalysis.We now present various definitions for the fractional Laplacian operatoron Schwartz space S(Rn). In the first definition, we define the operator(−∆)α2 as a singular integral.Definition 2.2. For any α ∈ (0, 2), the fractional Laplacian operator (−∆)α2is defined on the Schwartz class S(Rn) as(−∆)α2 u = Cn,αP.V.∫Rnu(x)− u(y)|x− y|n+α dxdy= Cn,α lim→0+∫Rn\B(x)u(x)− u(y)|x− y|n+α dxdy,where P.V. stands for the principal-value andCn,α =(∫Rn1− cos(ζ1)ζdζ)−1.We refer the readers to [6, Section 1.3.1] and [20] for properties of the con-stant Cn,α. It has been shown there thatCn,α =(∫Rn1− cos(ζ1)ζdζ)−1=2αΓ(n+α2 )pin2 |Γ(−α2 )|.The operator (−∆)α2 can be also defined via the Fourier transform. Itwas shown in [20, Proposition 3.3] that the fractional Laplacian operator(−∆)α2 can be viewed as a pseudo-differential operator as follows.Definition 2.3. Let α ∈ (0, 2). For any u ∈ S(Rn), we have(−∆)α2 u = F−1(|2piξ|α(Fu)) ∀ξ ∈ Rn. (2.3)There is another way to define the fractional Laplace operator via theinverse Fourier transform (2.3). In fact, in the case α = 1, the half-Laplacianacting on a function u in the whole space Rn can be computed as the normalderivative on the boundary of its harmonic extension to the upper half-spaceRn+1 := Rn× (0,+∞). In other words, it is the Dirichlet to Neumann oper-ator (see [12]). The operator (−∆)α2 can be characterized in a similar way,by defining its α-harmonic extension to the upper half-space (See formula2.5). Indeed, Caffarelli-Silvestre [12] proved the following:102.2. The fractional Sobolev spacesDefinition 2.4. Let α ∈ (0, 2). The fractional Laplacian operator (−∆)α2on Rn can be expressed on the higher dimension Rn+1+ = Rn × (0,∞) in thefollowing way:(−∆)α2 u(x) = ∂w∂να:= −kα limy→0+y1−α∂w∂y(x, y), (2.4)where kα =Γ(α2)21−αΓ(1−α2), and w : Rn+1+ → R is the α-harmonic extension ofu, that solves {div (y1−α∇w) = 0 in Rn+1+w = u on Rn × {y = 0}. (2.5)See also [6, 20] and references therein for the basics on the fractionalLaplacian.2.2 The fractional Sobolev spacesIn this section, we introduce the classical fractional Sobolev space on Rnand its bounded subsets.2.2.1 The classical fractional Sobolev spaceFor α ∈ (0, 2), the classical fractional Sobolev space of order α2 is defined asHα2 (Rn) :={u ∈ L2(Rn) such that∫Rn(1 + |2piξ|α)|Fu(ξ)|2dξ <∞}equipped with the norm‖u‖Hα2 (Rn) = ‖Fu(ξ)‖L2(Rn) + ‖|2piξ|α2Fu(ξ)‖L2(Rn).One can use Parseval-Plancherel’s formula (2.2) to rewrite this norm as‖u‖2Hα2 (Rn)= ‖u‖2L2(Rn) +∫Rn|2piξ|α|F(u)(ξ)|2dξ.The space Hα2 (Rn) is well defined and is a Hilbert space for every α > 0.The following relation between the fractional Laplacian operator (−∆)α2 andthe fractional Sobolev space Hα2 (Rn) was proved in [34, Lemma 3.1]; see also[20, Proposition 3.4]:∫Rn|2piξ|α|Fu(ξ)|2dξ = Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy, (2.6)112.2. The fractional Sobolev spacesfor all u ∈ H α2 (Rn). Using this relation, one can obtain a new expression forthe norm of the space Hα2 (Rn) :‖u‖2Hα2 (Rn)= ‖u‖2L2(Rn) +Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy. (2.7)Consequently, the space Hα2 (Rn) can be defined as the linear space ofLebesgue measurable functions u from Rn to R such that the norm definedin (2.7) is finite.As in the classical case, any function in the fractional Sobolev space Hα2 (Rn)can be approximated by a sequence of smooth functions with compact sup-port. Indeed, for 0 < α < 2,Hα2 (Rn) := C∞0 (Rn)‖ . ‖Hα2 (Rn) .In addition, the operator (−∆)α2 can be extended by density from S(Rn) toHα2 (Rn) . In this way, the associated scalar product can be formulated asfollows〈u, v〉Hα2 (Rn) := 〈u, v〉+ (u, v)=Cn,α2∫Rn∫Rn(u(x)− u(y))(v(x)− v(y))|x− y|n+α dxdy +∫Rnuvdx.Clearly, the norm ‖ . ‖Hα2 (Rn) defined in (2.7) induced by this scalar product.The following proposition states a crucial formulation which will be usedfreely in this thesis and it is a direct consequence of (2.6) coupled withProposition 3.6 in [20].Proposition 2.5. Let 0 < α < 2 and u ∈ H α2 (Rn). Then,∫Rn|2piξ|α|Fu(ξ)|2dξ =∫Rn|(−∆)α4 u|2dx = Cn,α2∫(Rn)2|u(x)− u(y)|2|x− y|n+α dxdy.Let now α ∈ (0, 2) be fixed and Ω be an open-bounded subset of Rn with(n > α). We define Hα2 (Ω) as the space of measurable functions u such thatthe following norm is finite‖u‖2Hα2 (Ω)= ‖u‖2L2(Ω) +Cn,α2∫Ω∫Ω|u(x)− u(y)|2|x− y|n+α dxdy. (2.8)We also define Hα20 (Ω) to be the closure of C∞0 (Ω) with respect to the norm‖ . ‖Hα2 (Ω), that isHα20 (Ω) := C∞0 (Ω)‖ . ‖Hα2 (Ω) .One can find many properties of the above spaces in [20] and [6, Section 1].122.2. The fractional Sobolev spacesRemark 2.6. It is well known that if u ∈ H1(Ω), which is the classicalSobolev space corresponding to the case α = 2, then its null extension outsideΩ, u¯, is in H1(Rn) and ‖u‖H1(Ω) = ‖u¯‖H1(Rn). This is, however, not true ingeneral for functions in Hα2 (Ω), for 0 < α < 2.In order to deal with problems on bounded domain, we shall need func-tion spaces where null extensions are controlled. This will be done in thenext section.2.2.2 The variational fractional Sobolev spaceWe now consider the following Hilbert space on Rn:Definition 2.7. For 0 < α < 2, we define the fractional Sobolev spaceHα20 (Rn) as the completion of C∞0 (Rn) under the norm‖u‖2Hα20 (Rn)=∫Rn|(−∆)α4 u|2dx = Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy.Let O = Rn × Rn \ (CΩ × CΩ), where CΩ := Rn \ Ω, and define H˙ α2 (Ω) asthe linear space of all Lebesgue measurable functions u from Rn to R suchthat the restriction to Ω of any function u in H˙α2 (Ω) belongs to L2(Ω), andthe mapp (x, y) 7→ (u(x)− u(y))|x− y| 12 is in L2(O, dxdy).The norm in H˙α2 (Ω) is defined as follows:‖u‖2H˙α2 (Ω)= ‖u‖2L2(Ω) +Cn,α2∫O|u(x)− u(y)|2|x− y|n+α dxdy. (2.9)Remark 2.8. Note that the norms in (2.8) and (2.9) are not the samebecause Ω× Ω is strictly contained in O: this makes the classical fractionalSobolev space approach not sufficient for studying the nonlocal problem weconsider in this thesis.We are now ready to introduce a suitable function space on a boundeddomain Ω for the variational setting that will be needed later in this thesis.One can easily check that if u ∈ H˙ α2 (Ω), then the null extension of u outsideΩ is in Hα2 (Rn). This allows us to define the desired space asHα20 (Ω) ={u ∈ H˙ α2 (Ω) such that u = 0 a.e. in CΩ}.132.3. The α-harmonic extension and weighted Sobolev spaceObserve that since u = 0 in CΩ, the integral in formula (2.9) can be extendedto Rn; that is for any Hα20 (Ω), the norm can be written as:‖u‖2Hα20 (Ω)= ‖u‖2L2(Rn) +Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy. (2.10)It has been shown in [6] that the norm on Hα20 (Ω) defined by (2.10) isequivalent to the following norm – still denoted by ‖ . ‖Hα20 (Ω):‖u‖2Hα20 (Ω)=Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy.The density properties of smooth and compactly supported functions inspace Hα20 (Ω) is studied in [6], and it was proved there that the space C∞0 (Ω)is a dense subspace of Hα20 (Ω); see [6, Theorem 2.6]. So, it follows from thisdensity result that the space Hα20 (Ω) eventually can be expressed in thefollowing way:Definition 2.9. Let 0 < α < 2. We define the fractional Sobolev spaceHα20 (Ω) on the smooth bounded domain Ω as the completion of C∞0 (Ω) underthe norm‖u‖2Hα20 (Ω)=Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy. (2.11)2.3 The α-harmonic extension and weightedSobolev spaceIn this subsection, we present an useful representation given in [5] and [9]for the fractional Laplacian (−∆)α2 as a trace class operator, as well as acorresponding representation for the space Hα20 (Rn).For a function u ∈ Hα20 (Rn), let w = Eα(u) be its α-harmonic extensionto the upper half-space, Rn+1+ , that is the solution of:{div (y1−α∇w) = 0 in Rn+1+w = u on Rn × {y = 0}.Recall that the extension function w = Eα(u) satisfies (2.4) and belongsto the Hilbert space Xα(Rn+1+ ) defined as the closure of C∞0 (Rn+1+ ) for thenorm‖w‖Xα(Rn+1+ ) :=(kα∫Rn+1+y1−α|∇w|2dxdy) 12,142.4. Fractional Hardy-Sobolev type inequalitieswhere kα =Γ(α2)21−αΓ(1−α2)is a normalization constant chosen in such a waythat the extension operator Eα(u) : Hα20 (Rn)→ Xα(Rn+1+ ) is an isometry,that is, for any u ∈ Hα20 (Rn), we have‖Eα(u)‖Xα(Rn+1+ ) = ‖u‖H α20 (Rn) = ‖(−∆)α4 u‖L2(Rn). (2.12)Conversely, for a function w ∈ Xα(Rn+1+ ), we denote its trace on Rn×{y = 0}as Tr(w) := w(., 0). This trace operator is also well defined and satisfies‖w(., 0)‖Hα20 (Rn)≤ ‖w‖Xα(Rn+1+ ). (2.13)We shall frequently use the following useful fact: Since α ∈ (0, 2), theweight y1−α belongs to the Muckenhoupt class A2; [56], which consists of allnon-negative functions w on Rn satisfying for some constant C, the estimatesupB(1|B|∫Bwdx)(1|B|∫Bw−1dx) ≤ C, (2.14)where the supremum is taken over all balls B in Rn.If Ω ⊂ Rn+1 is an open domain, we denote by L2(Ω, |y|1−α) the space ofall measurable functions on Ω such that ‖w‖2L2(Ω,|y|1−α) =∫Ω |y|1−α|w|2dxdy <∞, and by H1(Ω, |y|1−α) the weighted Sobolev spaceH1(Ω, |y|1−α) = {w ∈ L2(Ω, |y|1−α) : ∇w ∈ L2(Ω, |y|1−α)} .It is remarkable that most of the properties of classical Sobolev spaces,including the embedding theorems have a weighted counterpart as long asthe weight is in the Muckenhoupt class A2; see [28] and [41]. Note thatH1(Rn+1+ , y1−α) -up to a normalization factor- is also isometric to Xα(Rn+1+ ).2.4 Fractional Hardy-Sobolev type inequalitiesThe starting point of the study of existence of weak solutions of borderlinevariational problems (1.1), (1.3) and (1.6) is based on the following fractionalHardy-Sobolev inequalities which guarantee that the associated functionalsare well defined and sometimes bounded below on the right function spaces.We start with the fractional Sobolev inequality [19], which asserts thatfor n > α and 0 < α < 2, there exists a constant C(n, α) > 0 such that(∫Rn |u|2∗αdx)22∗α ≤ C(n, α) ∫Rn |(−∆)α4 u|2dx for all u ∈ H α20 (Rn), (2.15)152.4. Fractional Hardy-Sobolev type inequalitieswhere 2∗α =2nn−α .Another important inequality is the fractional Hardy inequality (see [34]and [43]), which states that under the same conditions on n and α, we haveγH(α)∫Rn|u|2|x|αdx ≤∫Rn |(−∆)α4 u|2dx for all u ∈ Hα20 (Rn), (2.16)where γH(α) is the best constant in the above inequality on Rn, that isγH(α) := inf∫Rn |(−∆)α4 u|2dx∫Rn|u|2|x|αdx; u ∈ Hα20 (Rn) \ {0} . (2.17)It has also been shown there that γH(α) = 2α Γ2(n+α4)Γ2(n−α4). Note that γH(α)converges to the best classical Hardy constant γH(2) =(n−2)24 when α→ 2.By interpolating these inequalities via Ho¨lder’s inequality, one gets the fol-lowing fractional Hardy-Sobolev inequalities.Lemma 2.10 (Fractional Hardy-Sobolev Inequalities). Assume that 0 <α < 2 and 0 ≤ s ≤ α < n. Then, there exists positive constant c such that(∫Rn|u|2∗α(s)|x|s dx)22∗α(s) ≤ c∫Rn|(−∆)α4 u|2dx for all u ∈ Hα20 (Rn). (2.18)Moreover, if γ < γH(α) = 2α Γ2(n+α4)Γ2(n−α4), thenC(∫Rn|u|2∗α(s)|x|s dx)22∗α(s) ≤∫Rn|(−∆)α4 u|2dx− γ∫Rn|u|2|x|αdx, (2.19)for all u ∈ Hα20 (Rn) and some positive constant C.Proof of Lemma 2.10. Note that for s = 0 (resp., s = α) the first inequalityis just the fractional Sobolev (resp., the fractional Hardy) inequality. Wetherefore have to only consider the case where 0 < s < α in which case2∗α(s) > 2. By applying Ho¨lder’s inequality, then the fractional Hardy and162.4. Fractional Hardy-Sobolev type inequalitiesthe fractional Sobolev inequalities, we have∫Rn|u|2∗α(s)|x|s dx =∫Rn|u| 2sα|x|s |u|2∗α(s)− 2sα dx≤ (∫Rn|u|2|x|αdx)sα (∫Rn|u|(2∗α(s)− 2sα ) αα−sdx)α−sα= (∫Rn|u|2|x|αdx)sα (∫Rn|u|2∗αdx)α−sα≤ C1(∫Rn|(−∆)α4 u|2dx) sαC2(∫Rn|(−∆)α4 u|2dx) 2∗α2.α−sα≤ c(∫Rn|(−∆)α4 u|2dx) n−sn−α= c(∫Rn|(−∆)α4 u|2dx) 2∗α(s)2 .From the definition of γH(α), it follows that for all u ∈ Hα20 (Rn),∫Rn |(−∆)α4 u|2dx− γ ∫Rn |u|2|x|αdx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s)≥ (1− γγH(α))∫Rn |(−∆)α4 u|2dx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s).Hence (2.18) implies (2.19) whenever γ < γH(α).Remark 2.11. One can use (2.12) and (2.13) to rewrite inequalities (2.16),(2.18) and (2.19) as the following trace class inequalities:γH(α)∫Rn|w(x, 0)|2|x|α dx ≤ ‖w‖2Xα(Rn+1+ ),(∫Rn|w(x, 0)|2∗α(s)|x|s dx)22∗α(s) ≤ c ‖w‖2Xα(Rn+1+ ),andC(∫Rn|w(x, 0)|2∗α(s)|x|s dx)22∗α(s) ≤ ‖w‖2Xα(Rn+1+ )− γ∫Rn|w(x, 0)|2|x|α dx,for all w ∈ Xα(Rn+1+ ) and γ < γH(α).17Chapter 3Borderline VariationalProblems InvolvingFractional Laplacian andCritical Singularities3.1 IntroductionIn this chapter, we consider the problem of existence of nontrivial weak solu-tions to the following doubly critical problem on Rn involving the fractionalHardy-Schro¨dinger operator:(−∆)α2 u− γ u|x|α = |u|2∗α−2u+|u|2∗α(s)−2u|x|s in Rn, (3.1)where 0 ≤ s < α < 2, n > α, 2∗α := 2nn−α , 2∗α(s) := 2(n−s)n−α and γ ∈ R.Problems involving two non-linearities have been studied in the case oflocal operators such as the Laplacian −∆, the p-Laplacian −∆p and theBiharmonic operator ∆2 (See [8], [33], [47] and [68]). Problem (3.1) above isthe non-local counterpart of the one studied by Filippucci-Pucci-Robert in[33], who treated the case of the p-Laplacian in an equation involving boththe Sobolev and the Hardy-Sobolev critical exponents.Questions of existence and non-existence of solutions for fractional el-liptic equations with singular potentials were recently studied by severalauthors. All studies focus, however, on problems with only one critical ex-ponent –mostly the non-linearity u2∗α−1– and to a lesser extent the criticalHardy-Sobolev singular term u2∗α(s)−1|x|s (see [19], [31], [70] and the referencestherein). These cases were also studied on smooth bounded domains (seefor example [5], [7], [9], [29], [65] and the references therein). In general, thecase of two critical exponents involve more subtleties and difficulties, evenfor local differential operators.183.1. IntroductionThe variational approach that we adopt here, relies on the followingfractional Hardy-Sobolev type inequality:C(∫Rn|u|2∗α(s)|x|s dx)22∗α(s) ≤∫Rn|(−∆)α4 u|2dx− γ∫Rn|u|2|x|αdx ∀u ∈ Hα20 (Rn),(3.2)where γ < γH(α) := 2α Γ2(n+α4)Γ2(n−α4)is the best fractional Hardy constant onRn. Recall that the fractional space Hα20 (Rn) is defined as the completion ofC∞0 (Rn) under the norm‖u‖2Hα20 (Rn)=∫Rn|2piξ|α|Fu(ξ)|2dξ =∫Rn|(−∆)α4 u|2dx.The best constant in the above fractional Hardy-Sobolev inequality is definedas:µγ,s,α(Rn) := infu∈Hα20 (Rn)\{0}∫Rn |(−∆)α4 u|2dx− γ ∫Rn |u|2|x|αdx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s). (3.3)One step towards addressing Problem (3.1) consists of proving the ex-istence of extremals for µγ,s,α(Rn), when s ∈ [0, α) and γ ∈ (−∞, γH(α)).Note that the Euler-Lagrange equation corresponding to the minimizationproblem for µγ,s,α(Rn) is –up to a constant factor– the following:{(−∆)α2 u− γ u|x|α = u2∗α(s)−1|x|s in Rnu ≥ 0 in Rn. (3.4)When α = 2, i.e., in the case of the standard Laplacian, the aboveminimization problem (3.3) has been extensively studied. See for example[15], [17], [33], [35], [36] and [40]. The non-local case has also been thesubject of several studies, but in the absence of the Hardy term, i.e., whenγ = 0. In [31], Fall, Minlend and Thiam proved the existence of extremalsfor µ0,s,α(Rn) in the case α = 1. Recently, J. Yang in [70] proved thatthere exists a positive, radially symmetric and non-increasing extremal forµ0,s,α(Rn) when α ∈ (0, 2). Asymptotic properties of the positive solutionswere given by Y. Lei [48], Lu and Zhu [54], and Yang and Yu [71].In section 3.2, we consider the remaining cases in the problem of de-ciding whether the best constant in the fractional Hardy-Sobolev inequalityµγ,s,α(Rn) is attained. We use Ekeland’s variational principle to show thefollowing.193.1. IntroductionTheorem 3.1. Suppose 0 < α < 2, 0 ≤ s < α < n and γ < γH(α) :=2αΓ2(n+α4)Γ2(n−α4).1. If either {s > 0} or {s = 0 and γ ≥ 0}, then µγ,s,α(Rn) is attained.2. If s = 0 and γ < 0, then there are no extremals for µγ,s,α(Rn).3. If either {0 < γ < γH(α)} or {0 < s < α and γ = 0}, then anynon-negative minimizer for µγ,s,α(Rn) is positive, radially symmetric,radially decreasing, and approaches zero as |x| → ∞.In section 3.3, we consider problem (3.1) and use the Mountain Passlemma to establish the following result.Theorem 3.2. Let 0 < α < 2, 0 < s < α < n and 0 ≤ γ < γH(α). Then,there exists a non-trivial weak solution of (3.1).We say u ∈ Hα20 (Rn) is a weak solution of (3.1), if we have for all ϕ ∈Hα20 (Rn),∫Rn(−∆)α4 u(−∆)α4 ϕdx =∫Rnγu|x|αϕdx+∫Rn|u|2∗α−2uϕdx+∫Rn|u|2∗α(s)−2u|x|s ϕdx.The standard strategy to construct weak solutions of (3.1) is to findcritical points of the corresponding functional on Hα20 (Rn). However, (3.1)is invariant under the following conformal one parameter transformationgroup,Tr : Hα20 (Rn)→ Hα20 (Rn); u(x)→ Tr[u](x) = rn−α2 u(rx) r > 0, (3.5)which means that the convergence of Palais-Smale sequences is not a given.As it was argued in [33], there is an asymptotic competition between theenergy carried by the two critical nonlinearities. Hence, the crucial stephere is to balance the competition to avoid the domination of one termover another. Otherwise, there is vanishing of the weakest one, leading toa solution for the same equation but with only one critical nonlinearity. Inorder to deal with this issue, we choose a suitable minimax energy level, insuch a way that after a careful analysis of the concentration phenomena, wecould eliminate the possibility of a vanishing weak limit for these well chosenPalais-Smale sequences, while ensuring that none of the two nonlinearitiesdominate the other.203.1. IntroductionWith representation (2.4), the non-local problem (3.1) can then be writ-ten as the following local problem:{ −div (y1−α∇w) = 0 in Rn+1+∂w∂να = γw(.,0)|x|α + w(., 0)2∗α−1 + w(.,0)2∗α(s)−1|x|s on Rn,(3.6)for which w ∈ Xα(Rn+1+ ). Recall that the Hilbert space Xα(Rn+1+ ) definedas the closure of C∞0 (Rn+1+ ) for the norm‖w‖Xα(Rn+1+ ) :=(kα∫Rn+1+y1−α|∇w|2dxdy) 12,where kα =Γ(α2)21−αΓ(1−α2)is a normalization constant.A function w ∈ Xα(Rn+1+ ) is said to be a weak solution to (3.6), if forall ϕ ∈ Xα(Rn+1+ ),kα∫Rn+1+y1−α〈∇w,∇ϕ〉dxdy = γ∫Rnw(x, 0)|x|α ϕdx+∫Rn|w(x, 0)|2∗α−2w(x, 0)ϕdx+∫Rn|w(x, 0)|2∗α(s)−2w(x, 0)|x|s ϕdx.Note that for any weak solution w in Xα(Rn+1+ ) to (3.6), the function u =w(., 0) defined in the sense of traces (see Section 2.3), is in Hα20 (Rn) and isa weak solution to problem (3.1). The energy functional corresponding to(3.6) isΦ(w) =12‖w‖2Xα(Rn+1+ )− γ2∫Rn|w(x, 0)|2|x|α dx−12∗α∫Rn|w(x, 0)|2∗α dx− 12∗α(s)∫Rn|w(x, 0)|2∗α(s)|x|s dx.Hence, the associated trace of any critical point w of Φ in Xα(Rn+1+ ) is aweak solution for (3.1).It follows from Remark 2.11 that there exist positive constants c, C suchthat the following fractional trace inequalities hold for all w ∈ Xα(Rn+1+ )and γ < γH(α) :213.2. Proof of Theorem 3.1γH(α)∫Rn|w(x, 0)|2|x|α dx ≤ ‖w‖2Xα(Rn+1+ ), (3.7)(∫Rn|w(x, 0)|2∗α(s)|x|s dx)22∗α(s) ≤ c ‖w‖2Xα(Rn+1+ ), (3.8)andC(∫Rn|w(x, 0)|2∗α(s)|x|s dx)22∗α(s) ≤ ‖w‖2Xα(Rn+1+ )− γ∫Rn|w(x, 0)|2|x|α dx. (3.9)The best constant µγ,s,α(Rn) can then be written as:S(n, α, γ, s) = infw∈Xα(Rn+1+ )\{0}kα∫Rn+1+y1−α|∇w|2dxdy − γ ∫Rn |w(x,0)|2|x|α dx(∫Rn|w(x,0)|2∗α(s)|x|s dx)22∗α(s).We shall therefore investigate whether there exist extremal functionswhere this best constant is attained. Theorems 3.1 and 3.2 can therefore bestated in the following way:Theorem 3.3. Suppose 0 < α < 2, 0 ≤ s < α < n and γ < γH(α). Wethen have the following:1. If {s > 0} or {s = 0 and γ ≥ 0}, then S(n, α, γ, s) is attained inXα(Rn+1+ ).2. If s = 0 and γ < 0, then there are no extremals for S(n, α, γ, s) inXα(Rn+1+ ).Theorem 3.4. Let 0 < α < 2, 0 < s < α < n and 0 ≤ γ < γH(α). Then,there exists a non-trivial weak solution to (3.6) in Xα(Rn+1+ ).3.2 Proof of Theorem 3.1We shall minimize the functionalIγ,s(w) =kα∫Rn+1+y1−α|∇w|2dxdy − γ ∫Rn |w(x,0)|2|x|α dx(∫Rn|w(x,0)|2∗α(s)|x|s dx)22∗α(s)on the space Xα(Rn+1+ ). Whenever S(n, α, γ, s) is attained at some w ∈Xα(Rn+1+ ), then it is clear that u = Tr(w) := w(., 0) will be a function inHα20 (Rn), where µγ,s,α(Rn) is attained.223.2. Proof of Theorem 3.1Note first that inequality (3.7) asserts that Xα(Rn+1+ ) is embedded inthe weighted space L2(Rn, |x|−α) and that this embeding is continuous. Ifγ < γH(α), it follows from (3.7) that‖w‖ :=(kα∫Rn+1+y1−α|∇w|2dxdy − γ∫Rn|w(x, 0)|2|x|α dx) 12is well-defined on Xα(Rn+1+ ). Set γ+ = max{γ, 0} and γ− = −max{γ, 0}.The following inequalities then hold for any u ∈ Xα(Rn+1+ ),(1− γ+γH(α))‖w‖2Xα(Rn+1+ )≤ ‖w‖2 ≤ (1 + γ−γH(α))‖w‖2Xα(Rn+1+ ). (3.10)Thus, ‖ . ‖ is equivalent to the norm ‖ . ‖Xα(Rn+1+ ).We start by considering the case when s > 0. Ekeland’s variationalprinciple [25] applied to the functional I(w) := Iγ,s(w) yields the existenceof a minimizing sequence (wk)k∈N for S(n, α, γ, s) such that as k →∞,∫Rn|wk(x, 0)|2∗α(s)|x|s dx = 1, (3.11)I(wk) −→ S(n, α, γ, s), (3.12)andI ′(wk)→ 0 in (Xα(Rn+1+ ))′, (3.13)where (Xα(Rn+1+ ))′ denotes the dual of Xα(Rn+1+ ). Consider the functionalsJ,K : Xα(Rn+1+ ) −→ R byJ(w) :=12‖w‖2 = kα2∫Rn+1+y1−α|∇w|2dxdy − γ2∫Rn|w(x, 0)|2|x|α dx,andK(w) :=12∗α(s)∫Rn|w(x, 0)|2∗α(s)|x|s dx.Straightforward computations yield that as k →∞,J(wk) −→ 12S(n, α, γ, s),andJ ′(wk)− S(n, α, γ, s)K ′(wk) −→ 0 in (Xα(Rn+1+ ))′. (3.14)233.2. Proof of Theorem 3.1Consider now the Levy concentration functions Q of |wk(x,0)|2∗α(s)|x|s , defined asQ(r) =∫Br|wk(x, 0)|2∗α(s)|x|s dx for r > 0,where Br is the ball of radius r in Rn. Since∫Rn|wk(x,0)|2∗α(s)|x|s dx = 1 for allk ∈ N, then by continuity, and up to considering a subsequence, there existsrk > 0 such thatQ(rk) =∫Brk|wk(x, 0)|2∗α(s)|x|s dx =12for all k ∈ N.Define the rescaled sequence vk(x, y) := rn−α2k wk(rkx, rky) for k ∈ N and(x, y) ∈ Rn+1+ , in such a way that (vk)k∈N is also a minimizing sequencefor S(n, α, γ, s). Indeed, it is easy to check that vk ∈ Xα(Rn+1+ ) and that‖wk‖2 = ‖vk‖2,limk→∞(kα∫Rn+1+y1−α|∇vk|2dxdy − γ∫Rn|vk(x, 0)|2|x|α dx)= S(n, α, γ, s)(3.15)and ∫Rn|vk(x, 0)|2∗α(s)|x|s dx =∫Rn|wk(x, 0)|2∗α(s)|x|s dx = 1.Moreover, we have that∫B1|vk(x, 0)|2∗α(s)|x|s dx =12for all k ∈ N. (3.16)In addition, ‖vk‖2 = S(n, α, γ, s) + o(1) as k →∞, so (3.10) yields that(vk)k∈N is bounded in Xα(Rn+1+ ). Therefore, without loss of generality,there exists a subsequence -still denoted vk- such thatvk ⇀ v in Xα(Rn+1+ ) and vk(., 0)→ v(., 0) in Lqloc(Rn) for all 1 ≤ q < 2∗α.(3.17)We shall show that the weak limit of the minimizing sequence is notidentically zero, that is v 6≡ 0. Indeed, suppose v ≡ 0. It follows from (3.17)thatvk ⇀ 0 in Xα(Rn+1+ ) and vk(., 0)→ 0 in Lqloc(Rn) for every 1 ≤ q < 2∗α.(3.18)243.2. Proof of Theorem 3.1For δ > 0, define B+δ := {(x, y) ∈ Rn+1+ : |(x, y)| < δ}, Bδ := {x ∈ Rn : |x| <δ} and let η ∈ C∞0 (B+1 ) be a cut-off function such that η ≡ 1 in B+12and0 ≤ η ≤ 1 in Rn+1+ .We use η2vk as test function in (3.14) to get thatkα∫Rn+1+y1−α∇vk.∇(η2vk)dxdy − γ∫Rnvk(x, 0)(η2vk(x, 0))|x|α dx= S(n, α, γ, s)∫Rn|vk(x, 0)|2∗α(s)−1(η2vk(x, 0))|x|s dx+ o(1).(3.19)Simple computations yield |∇(ηvk)|2 = |vk∇η|2 +∇vk.∇(η2vk), so that wehavekα∫Rn+1+y1−α|∇(ηvk)|2dxdy − kα∫Rn+1+y1−α∇vk.∇(η2vk)dxdy= kα∫Rn+1+y1−α|vk∇η|2dxdy= kα∫Ey1−α|∇η|2|vk|2dxdy,where E := supp(|∇η|). Since α ∈ (0, 2), y1−α is an A2-weight, and sinceE is bounded, we have that the embedding H1(E, y1−α) ↪→ L2(E, y1−α) iscompact (See [5] and [41]). It follows from (3.18)1 thatkα∫Ey1−α|∇η|2|vk|2dxdy → 0 as k →∞.Therefore,kα∫Rn+1+y1−α|∇(ηvk)|2dxdy = kα∫Rn+1+y1−α∇vk.∇(η2wk)dxdy + o(1).By plugging the above estimate into (3.19), we get that‖ηvk‖2 = kα∫Rn+1+y1−α|∇(ηvk)|2dxdy − γ∫Rn|ηvk(x, 0)|2|x|α dx= S(n, α, γ, s)∫Rn|vk(x, 0)|2∗α(s)−2(|ηvk(x, 0)|2)|x|s dx+ o(1)= S(n, α, γ, s)∫B1|vk(x, 0)|2∗α(s)−2(|ηvk(x, 0)|2)|x|s dx+ o(1).(3.20)253.2. Proof of Theorem 3.1Note that in the last equality we used the fact that supp(η(x, 0)) ⊂ B1. Wenow apply Ho¨lder’s inequality with exponents 2∗α(s)2∗α(s)−2 and2∗α(s)2 to get that∫B1|vk(x, 0)|2∗α(s)−2(|ηvk(x, 0)|2)|x|s dx =∫B1|vk(x, 0)|2∗α(s)−2|x|s2∗α(s)−22∗α(s)|ηvk(x, 0)|2|x|s 22∗α(s)dx≤(∫B1|vk(x, 0)|2∗α(s)|x|s dx) 2∗α(s)−22∗α(s)(∫B1|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s).It then follows from (3.16) that∫B1|vk(x, 0)|2∗α(s)−2(|ηvk(x, 0)|2)|x|s dx ≤(12)1− 22∗α(s)(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s).Plugging the above inequality into (3.20), we get that‖ηvk‖2 = kα∫Rn+1+y1−α|∇(ηvk)|2dxdy − γ∫Rn|ηvk(x, 0)|2|x|α dx≤ S(n, α, γ, s)21− 22∗α(s)(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)+ o(1).On the other hand, it follows from the definition of S(n, α, γ, s) thatS(n, α, γ, s)(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)≤ ‖ηvk‖2≤ S(n, α, γ, s)21− 22∗α(s)(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)+ o(1).Note that S(n,α,γ,s)21− 22∗α(s)< S(n, α, γ, s) for s ∈ (0, α), which yields that∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx = o(1). (3.21)By straightforward computations and Ho¨lder’s inequality, we get that(∫B1|vk(x, 0)|2∗α(s)|x|s dx) 12∗α(s)=(∫B1|ηvk(x, 0) + (1− η)vk(x, 0)|2∗α(s)|x|s dx) 12∗α(s)≤(∫B1|ηvk(x, 0)|2∗α(s)|x|s dx) 12∗α(s)+(∫B1|(1− η)vk(x, 0)|2∗α(s)|x|s dx) 12∗α(s)≤(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 12∗α(s)+ C(∫B1|vk(x, 0)|2∗α(s)dx) 12∗α(s).263.2. Proof of Theorem 3.1From (3.18)2, and the fact that 2∗α(s) < 2∗α, we obtain∫B1|vk(x, 0)|2∗α(s)dx→ 0 as k →∞.Therefore,(∫B1|vk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)≤(∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)+ o(1).(3.22)It then follows from (3.21) and (3.22) thato(1) =∫Rn|ηvk(x, 0)|2∗α(s)|x|s dx =∫B1|vk(x, 0)|2∗α(s)|x|s dx+ o(1).This contradicts (3.16) and therefore v 6≡ 0.We now conclude by proving that vk converges weakly in Rn+1+ to v, andthat ∫Rn|v(x, 0)|2∗α(s)|x|s dx = 1.Indeed, for k ∈ N, let θk = vk − v, and use the Brezis-Lieb Lemma (see [10]and [70]) to deduce that1 =∫Rn|vk(x, 0)|2∗α(s)|x|s dx =∫Rn|v(x, 0)|2∗α(s)|x|s dx+∫Rn|θk(x, 0)|2∗α(s)|x|s dx+o(1),which yields that both∫Rn|v(x,0)|2∗α(s)|x|s dx and∫Rn|θk(x,0)|2∗α(s)|x|s dx are in the interval [0, 1]. (3.23)The weak convergence θk ⇀ 0 in Xα(Rn+1+ ) implies that‖vk‖2 = ‖v + θk‖2 = ‖v‖2 + ‖θk‖2 + o(1).273.2. Proof of Theorem 3.1By using (3.14) and the definition of S(n, α, γ, s), we get thato(1) = ‖vk‖2 − S(n, α, γ, s)∫Rn|vk(x, 0)|2∗α(s)|x|s dx=(‖v‖2 − S(n, α, γ, s)∫Rn|v(x, 0)|2∗α(s)|x|s dx)+(‖θk‖2 − S(n, α, γ, s)∫Rn|θk(x, 0)|2∗α(s)|x|s dx)+ o(1)≥ S(n, α, γ, s)(∫Rn|v(x, 0)|2∗α(s)|x|s dx) 22∗α(s)−∫Rn|v(x, 0)|2∗α(s)|x|s dx+ S(n, α, γ, s)(∫Rn|θk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)−∫Rn|θk(x, 0)|2∗α(s)|x|s dx+ o(1).Set nowA :=(∫Rn|v(x, 0)|2∗α(s)|x|s dx) 22∗α(s)−∫Rn|v(x, 0)|2∗α(s)|x|s dx,andB :=(∫Rn|θk(x, 0)|2∗α(s)|x|s dx) 22∗α(s)−∫Rn|θk(x, 0)|2∗α(s)|x|s dx.Note that since 2∗α(s) > 2, we have a22∗α(s) ≥ a for every a ∈ [0, 1], andequality holds if and only if a = 0 or a = 1. It then follows from (3.23)that both A and B are non-negative. On the other hand, the last inequalityimplies that A+B = o(1), which means that A = 0 and B = o(1), that is∫Rn|v(x, 0)|2∗α(s)|x|s dx =(∫Rn|v(x, 0)|2∗α(s)|x|s dx) 22∗α(s).Hence,either∫Rn|v(x, 0)|2∗α(s)|x|s dx = 0 or∫Rn|v(x, 0)|2∗α(s)|x|s dx = 1.283.2. Proof of Theorem 3.1The fact that v 6≡ 0 yields ∫Rn |v(x,0)|2∗α(s)|x|s dx 6= 0, and ∫Rn |v(x,0)|2∗α(s)|x|s dx = 1,which yields thatkα∫Rn+1+y1−α|∇v|2dxdy − γ∫Rn|v(x, 0)|2|x|α dx = S(n, α, γ, s).Without loss of generality we may assume v ≥ 0 (otherwise we take |v| in-stead of v), and we then obtain a positive extremal for S(n, α, γ, s) in thecase s ∈ (0, α).Suppose now that s = 0 and γ ≥ 0. By a result in [19], extremals existfor S(n, α, γ, s) whenever s = 0 and γ = 0. Hence, we only need to showthat there exists an extremal for S(n, α, γ, 0) in the case γ > 0. First notethat in this case, we have thatS(n, α, γ, 0) < S(n, α, 0, 0). (3.24)Indeed, if w ∈ Xα(Rn+1+ ) \ {0} is an extremal for S(n, α, 0, 0), then byestimating the functional at w, and using the fact that γ > 0, we obtainS(n, α, γ, 0) = infu∈Xα(Rn+1+ )\{0}‖u‖2Xα(Rn+1+ )− γ ∫Rn |u(x,0)|2|x|α dx(∫Rn |u(x, 0)|2∗αdx)22∗α≤‖w‖2Xα(Rn+1+ )− γ ∫Rn |w(x,0)|2|x|α dx(∫Rn |w(x, 0)|2∗αdx)22∗α<‖w‖2Xα(Rn+1+ )(∫Rn |w(x, 0)|2∗αdx)22∗α= S(n, α, 0, 0).Now we show that S(n, α, γ, 0) is attained whenever S(n, α, γ, 0) <S(n, α, 0, 0). Indeed, let (wk)k∈N ⊂ Xα(Rn+1+ )\{0} be a minimizing sequencefor S(n, α, γ, 0). Up to multiplying by a positive constant, we assume thatlimk→∞(kα∫Rn+1+y1−α|∇wk|2dxdy − γ∫Rn|wk(x, 0)|2|x|α dx)= S(n, α, γ, 0)(3.25)and ∫Rn|wk(x, 0)|2∗αdx = 1. (3.26)293.2. Proof of Theorem 3.1The sequence(‖wk‖Xα(Rn+1+ ))k∈Nis therefore bounded, and there exists asubsequence - still denoted wk- such that wk ⇀ w weakly in Xα(Rn+1+ ). Theweak convergence implies that‖wk‖2Xα(Rn+1+ ) = ‖wk − w‖2Xα(Rn+1+ )+ ‖w‖2Xα(Rn+1+ )+ o(1),and∫Rn|w(x, 0)|2|x|α dx =∫Rn|(w − wk)(x, 0)|2|x|α dx+∫Rn|wk(x, 0)|2|x|α dx+ o(1).The Brezis-Lieb Lemma ([10, Theorem 1]) and (3.26) yield that∫Rn|(wk − w)(x, 0)|2∗αdx ≤ 1,for large k, henceS(n, α, γ, 0) = ‖wk‖2Xα(Rn+1+ ) − γ∫Rn|wk(x, 0)|2|x|α dx+ o(1)≥ ‖wk − w‖2Xα(Rn+1+ ) + ‖w‖2Xα(Rn+1+ )− γ∫Rn|w(x, 0)|2|x|α dx+ o(1)≥ S(n, α, 0, 0)(∫Rn|(wk − w)(x, 0)|2∗αdx)22∗α+ S(n, α, γ, 0)(∫Rn|w(x, 0)|2∗αdx)22∗α + o(1)≥ S(n, α, 0, 0)∫Rn|(wk − w)(x, 0)|2∗αdx+ S(n, α, γ, 0)∫Rn|w(x, 0)|2∗αdx+ o(1).Use the Brezis-Lieb Lemma again to get thatS(n, α, γ, 0) ≥ (S(n, α, 0, 0)− S(n, α, γ, 0))∫Rn|(wk − w)(x, 0)|2∗αdx+ S(n, α, γ, 0)∫Rn|wk(x, 0)|2∗αdx+ o(1)= (S(n, α, 0, 0)− S(n, α, γ, 0))∫Rn|(wk − w)(x, 0)|2∗αdx+ S(n, α, γ, 0) + o(1).303.2. Proof of Theorem 3.1Since S(n, α, γ, 0) < S(n, α, 0, 0), we get that wk(., 0)→ w(., 0) in L2∗α(Rn),that is ∫Rn|w(x, 0)|2∗αdx = 1.The lower semi-continuity of I then implies that w is a minimizer for S(n, α,γ, 0). Note that |w| is also an extremal in Xα(Rn+1+ ) for S(n, α, γ, 0), there-fore there exists a non-negative extremal for S(n, α, γ, s) in the case γ > 0and s = 0, and this completes the proof of the case when s = 0 and γ ≥ 0.Now we consider the case when γ < 0.Claim 3.5. If γ ≤ 0, then S(n, α, γ, 0) = S(n, α, 0, 0), hence, there are noextremals for S(n, α, γ, 0) whenever γ < 0.Indeed, we first note that for γ ≤ 0, we have S(n, α, γ, 0) ≥ S(n, α, 0, 0).On the other hand, if we consider w ∈ Xα(Rn+1+ )\{0} to be an extremal forS(n, α, 0, 0) and define for δ ∈ R, and x¯ ∈ Rn, the function wδ := w(x−δx¯, y)for x ∈ Rn and y ∈ R+, then by a change of variable, we getS(n, α, γ, 0) ≤ Iδ : =‖wδ‖2Xα(Rn+1+ ) − γ∫Rn|wδ(x,0)|2|x|α dx(∫Rn |wδ(x, 0)|2∗αdx)22∗α=‖w‖2Xα(Rn+1+ )− γ ∫Rn |w(x,0)|2|x+δx¯|α dx(∫Rn |w(x, 0)|2∗αdx)22∗α,so thatS(n, α, γ, 0) ≤ limδ→∞Iδ =‖w‖2Xα(Rn+1+ )(∫Rn |w(x, 0)|2∗αdx)22∗α= S(n, α, 0, 0).Therefore, S(n, α, γ, 0) = S(n, α, 0, 0). Since there are extremals for S(n, α,0, 0) (see [19]), there is none for S(n, α, γ, 0) whenever γ < 0. This establishes(2) and completes the proof of Theorem 3.3.Back to Theorem 3.1, since the non-negative α-harmonic function w is aminimizer for S(n, α, γ, s) in Xα(Rn+1+ )\{0}, which exists from Theorem 3.3,then u := Tr(w) = w(., 0) ∈ Hα20 (Rn) \ {0} and by (2.12), u is a minimizerfor µγ,s,α(Rn) in Hα20 (Rn) \ {0}. Therefore, (1) and (2) of Theorem 3.1hold. For (3), let u∗ be the Schwarz symmetrization of u. By the fractionalPolya-Szego¨ inequality [57], we have‖(−∆)α2 u∗‖2L2(Rn) ≤ ‖(−∆)α2 u‖2L2(Rn).313.3. Proof of Theorem 3.2Furthermore, it is clear (Theorem 3.4. of [51]) that∫Rn|u|2|x|αdx ≤∫Rn|u∗|2|x|α dx and∫Rn|u|2∗α(s)|x|s dx ≤∫Rn|u∗|2∗α(s)|x|s dx.Combining the above inequalities and the fact that γ ≥ 0, we get thatµγ,s,α(Rn) ≤‖(−∆)α2 u∗‖2L2(Rn) − γ∫Rn|u∗|2|x|α dx(∫Rn|u∗|2∗α(s)|x|s dx)22∗α(s)≤‖(−∆)α2 u‖2L2(Rn) − γ∫Rn|u|2|x|αdx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s)= µγ,s,α(Rn).This implies that u∗ is also a minimizer and achieves the infimum of µγ,s,α(Rn).Therefore the equality sign holds in all the inequalities above, that isγ∫Rn|u|2|x|αdx = γ∫Rn|u∗|2|x|α dx and∫Rn|u|2∗α(s)|x|s dx =∫Rn|u∗|2∗α(s)|x|s dx.From Theorem 3.4. of [51], in the case of equality, it follows that u = |u| = u∗if either γ 6= 0 or if s 6= 0. In particular, u is positive, radially symmetricand decreasing about origin. Hence u must approach a limit as |x| → ∞,which must be zero.3.3 Proof of Theorem 3.2We shall now use the existence of extremals for the fractional Hardy-Sobolevtype inequalities, established in Section 3.2, to prove that there exists a non-trivial weak solution for (3.6). The energy functional Ψ associated to (3.6)is defined as follows:Ψ(w) =12‖w‖2− 12∗α∫Rn|u|2∗αdx− 12∗α(s)∫Rn|u|2∗α(s)|x|s dx for w ∈ Xα(Rn+1+ ),(3.27)where again u := Tr(w) = w(., 0). Fractional trace Hardy, Sobolev andHardy-Sobolev inequalities yield that Ψ ∈ C1(Xα(Rn+1+ )). Note that a weaksolution to (3.6) is a non-trivial critical point of Ψ.Throughout this section, we use the following notation for any sequence(wk)k∈N ∈ Xα(Rn+1+ ):uk := Tr(wk) = wk(., 0) for all k ∈ N.We split the proof in three parts:323.3. Proof of Theorem 3.23.3.1 Existence of a suitable Palais-Smale sequenceWe first verify that the energy functional Ψ satisfies the conditions of theMountain Pass Lemma leading to a minimax energy level that is below asuitable threshold. The following is standard.Lemma 3.6 (Ambrosetti and Rabinowitz [3]). Let (V, ‖ ‖) be a Banachspace and Ψ : V → R a C1−functional satisfying the following conditions:(a) Ψ(0) = 0,(b) There exist ρ,R > 0 such that Ψ(u) ≥ ρ for all u ∈ V , with ‖u‖ = R,(c) There exists v0 ∈ V such that lim supt→∞Ψ(tv0) < 0.Let t0 > 0 be such that ‖t0v0‖ > R and Ψ(t0v0) < 0, and definecv0(Ψ) := infσ∈Γsupt∈[0,1]Ψ(σ(t)),whereΓ := {σ ∈ C([0, 1], V ) : σ(0) = 0 and σ(1) = t0v0}.Then, cv0(Ψ) ≥ ρ > 0, and there exists a Palais-Smale sequence at levelcv0(Ψ), that is there exists a sequence (wk)k∈N ∈ V such thatlimk→∞Ψ(wk) = cv0(Ψ) and limk→∞Ψ′(wk) = 0 strongly inV ′.Moreover, we have that cv0(Ψ) ≤ supt≥0Ψ(tv0).We now prove the following.Proposition 3.7. Suppose 0 ≤ γ < γH(α) and 0 ≤ s < α, and considerΨ defined in (3.27) on the Banach space Xα(Rn+1+ ). Then, there existsw ∈ Xα(Rn+1+ ) \ {0} such that w ≥ 0 and 0 < cw(Ψ) < c?, wherec? = min{α2nS(n, α, γ, 0)nα ,α− s2(n− s)S(n, α, γ, s)n−sα−s}, (3.28)and a Palais-Smale sequence (wk)k∈N in Xα(Rn+1+ ) at energy level cw(Ψ),that is,limk→∞Ψ(wk) = cw(Ψ) and limk→∞Ψ′(wk) = 0 strongly in (Xα(Rn+1+ ))′.(3.29)333.3. Proof of Theorem 3.2Proof of Proposition 3.7. In the sequel, we will use freely the following ele-mentary identities involving 2∗α(s):12 − 12∗α =α2n ,2∗α2∗α−2 =nα ,12 − 12∗α(s) =α−s2(n−s) and2∗α(s)2∗α(s)−2 =n−sα−s .First, we note that the functional Ψ satisfies the hypotheses of Lemma3.6, and that condition (c) is satisfied for any w ∈ Xα(Rn+1+ ) \ {0}. Indeed,it is standard to show that Ψ ∈ C1(Xα(Rn+1+ )) and clearly Ψ(0) = 0, sothat (a) of Lemma 3.6 is satisfied. For (b), note that by the definition ofS(n, α, γ, s), we have thatS(n, α, γ, 0)(∫Rn|u|2∗αdx)22∗α ≤ ‖w‖2andS(n, α, γ, s)(∫Rn|u|2∗α(s)|x|s dx)22∗α(s) ≤ ‖w‖2.Hence,Ψ(w) ≥ 12‖w‖2 − 12∗αS(n, α, γ, 0)−2∗α2 ‖w‖2∗α − 12∗α(s)S(n, α, γ, s)−2∗α(s)2 ‖w‖2∗α(s)=(12− 12∗αS(n, α, γ, 0)−2∗α2 ‖w‖2∗α−2 − 12∗α(s)S(n, α, γ, s)−2∗α(s)2 ‖w‖2∗α(s)−2)‖w‖2.(3.30)Since s ∈ [0, α), we have that 2∗α − 2 > 0 and 2∗α(s) − 2 > 0. Thus, by(3.10), we can find R > 0 such that Ψ(w) ≥ ρ for all w ∈ Xα(Rn+1+ ) with‖w‖Xα(Rn+1+ ) = R. Regarding (c), note thatΨ(tw) =t22‖w‖2 − t2∗α2∗α∫Rn|u|2∗αdx− t2∗α(s)2∗α(s)∫Rn|u|2∗α(s)|x|s dx,hence limt→∞Ψ(tw) = −∞ for any w ∈ Xα(Rn+1+ ) \ {0}, which means thatthere exists tw > 0 such that ‖tww‖Xα(Rn+1+ ) > R and Ψ(tw) < 0, for t ≥ tw.Now we show that there exists w ∈ Xα(Rn+1+ )\{0} such that w ≥ 0 andcw(Ψ) <α2nS(n, α, γ, 0)nα . (3.31)From Theorem 3.3, we know that there exists a non-negative extremal w inXα(Rn+1+ ) for S(n, α, γ, 0) whenever γ ≥ 0. By the definition of tw, and thefact that cw(Ψ) > 0, we obtain343.3. Proof of Theorem 3.2cw(Ψ) ≤ supt≥0Ψ(tw) ≤ supt≥0f(t),wheref(t) =t22‖w‖2 − t2∗α2∗α∫Rn|u|2∗αdx ∀t > 0.Simple computations yield that f(t) attains its maximum at the point t˜ =(‖w‖2∫Rn |u|2∗αdx) 12∗α−2. It then follows thatsupt≥0f(t) = (12− 12∗α) ‖w‖2(∫Rn |u|2∗αdx)22∗α2∗α2∗α−2=α2n ‖w‖2(∫Rn |u|2∗αdx)22∗αnα .Since w is an extremal for S(n, α, γ, 0), we get thatcw(Ψ) ≤ supt≥0f(t) =α2nS(n, α, γ, 0)nα .We now need to show that equality does not hold in (3.31). Indeed, otherwisewe would have that 0 < cw(Ψ) = supt≥0Ψ(tw) = supt≥0f(t). Consider t1 > 0(resp., t2 > 0) where supt≥0Ψ(tw) (resp., supt≥0f(t)) is attained. We get thatf(t1)− t2∗α(s)12∗α(s)∫Rn|w(x, 0)|2∗α(s)|x|s dx = f(t2),which means that f(t1) > f(t2) since t1 > 0. This contradicts the fact thatt2 is a maximum point of f(t), hence the strict inequality in (3.31) holds.To finish the proof of Proposition 3.7, we can assume without loss thatα− s2(n− s)S(n, α, γ, s)n−sα−s <α2nS(n, α, γ, 0)nα .Let now w in Xα(Rn+1+ )\{0} be a positive minimizer for S(n, α, γ, s), whoseexistence was established in Section 3.2, and setf¯(t) =t22‖w‖2 − t2∗α(s)2∗α(s)∫Rn|u|2∗α(s)|x|s dx.353.3. Proof of Theorem 3.2As above, we havecw(Ψ) ≤ supt≥0f(t) = (12− 12∗α(s)) ‖w‖2(∫Rn|u|2∗α(s)|x|s dx)22∗α(s)2∗α(s)2∗α(s)−2=α− s2(n− s)S(n, α, γ, s)n−sα−s .Again, if equality holds, then 0 < cw(Ψ) ≤ supt≥0Ψ(tw) = supt≥0f¯(t), and ift1, t2 > 0 are points where the respective suprema are attained, then acontradiction is reached sincef¯(t1)− t2∗α12∗α∫Rn|u|2∗αdx = f¯(t2).Therefore,0 < cw(Ψ) < c? = min{α2nS(n, α, γ, 0)nα ,α− s2(n− s)S(n, α, γ, s)n−sα−s}.Finally, the existence of a Palais-Smale sequence at that level follows imme-diately from Lemma 3.6.3.3.2 Analysis of the Palais-Smale sequencesWe now study the concentration properties of weakly null Palais-Smale se-quences. For δ > 0, we shall write B+δ := {(x, y) ∈ Rn+1+ : |(x, y)| < δ} andBδ := {x ∈ Rn : |x| < δ} .Proposition 3.8. Let 0 ≤ γ < γH(α) and 0 < s < α. Assume that(wk)k∈N is a Palais-Smale sequence of Ψ at energy level c ∈ (0, c?). Ifwk ⇀ 0 in Xα(Rn+1+ ) as k →∞, then there exists a positive constant0 = 0(n, α, γ, c, s) > 0 such that for every δ > 0, one of the followingholds:1. lim supk→∞∫Bδ|uk|2∗αdx = lim supk→∞∫Bδ|uk|2∗α(s)|x|s dx = 0;2. lim supk→∞∫Bδ|uk|2∗αdx and lim supk→∞∫Bδ|uk|2∗α(s)|x|s dx ≥ 0.The proof of Proposition 3.8 requires the following two lemmas.363.3. Proof of Theorem 3.2Lemma 3.9. Let (wk)k∈N be a Palais-Smale sequence as in Proposition3.8. If wk ⇀ 0 in Xα(Rn+1+ ), then for any ω ⊂⊂ Rn \ {0} and any D ⊂⊂Rn+1+ \ {0}, there exists a subsequence of (wk)k∈N, still denoted by (wk)k∈N,such thatlimk→∞∫ω|uk|2|x|α dx = limk→∞∫ω|uk|2∗α(s)|x|s dx = 0 (3.32)andlimk→∞∫D∗|uk|2∗αdx = limk→∞∫Dy1−α|∇wk|2dxdy = 0, (3.33)where uk := wk(., 0) for all k ∈ N, and D∗ := {(x, y) ∈ D : y = 0} ⊂⊂Rn \ {0}.Proof of Lemma 3.9. Fix ω ⊂⊂ Rn \ {0}, and note that the following frac-tional Sobolev embedding is compact:Hα20 (Rn) ↪→ Lq(ω) for every 1 ≤ q < 2∗α.Using the trace inequality (2.13), and the assumption wk ⇀ 0 in Xα(Rn+1+ ),we get thatuk → 0 strongly for every 1 ≤ q < 2∗α.On the other hand, the fact that |x|−1 is bounded on ω ⊂⊂ Rn \{0} impliesthat there exist constants C1, C2 > 0 such that0 ≤ limk→∞∫ω|uk|2|x|α dx ≤ C1 limk→∞∫ω|uk|2dxand0 ≤ limk→∞∫ω|uk|2∗α(s)|x|s dx ≤ C2 limk→∞∫ω|uk|2∗α(s)dx.Since s ∈ (0, α), we have that 1 ≤ 2, 2∗α(s) < 2∗α. Thus, (3.32) holds.To show (3.33), we let η ∈ C∞0 (Rn+1+ ) be a cut-off function such thatη∗ := η(., 0) ∈ C∞0 (Rn \ {0}), η ≡ 1 in D and 0 ≤ η ≤ 1 in Rn+1+ . We firstnote thatkα∫Rn+1+y1−α|∇(ηwk)|2dxdy = kα∫Rn+1+y1−α|η∇wk|2dxdy + o(1). (3.34)Indeed, apply the following elementary inequality for vectors X,Y in Rn+1,∣∣|X + Y |2 − |X|2∣∣ ≤ C(|X||Y |+ |Y |2),373.3. Proof of Theorem 3.2with X = y1−α2 η∇wk and Y = y 1−α2 wk∇η, to get for all k ∈ N, that∣∣y1−α|∇(ηwk)|2 − y1−α|η∇wk|2∣∣ ≤ C (y1−α|η∇wk||wk∇η|+ y1−α|wk∇η|2) .By Ho¨lder’s inequality, we get∣∣∣∣∣∫Rn+1+y1−α|∇(ηwk)|2dxdy −∫Rn+1+y1−α|η∇wk|2dxdy∣∣∣∣∣≤ C3[‖wk‖Xα(Rn+1+ ) (∫supp(∇η)y1−α|wk|2dxdy) 12 +∫supp(∇η)y1−α|wk|2dxdy]≤ C4[(∫supp(∇η)y1−α|wk|2dxdy) 12 +∫supp(∇η)y1−α|wk|2dxdy].(3.35)Since the embedding H1(supp(∇η), y1−α) ↪→ L2(supp(∇η), y1−α) is com-pact, and wk ⇀ 0 in Xα(Rn+1+ ), we get that∫supp(∇η)y1−α|wk|2dxdy = o(1),which gives∫Rn+1+y1−α|∇(ηwk)|2dxdy =∫Rn+1+y1−α|η∇wk|2dxdy + o(1).Thus, (3.34) holds.Now recall that the sequence (wk)k∈N has the following property:limk→∞Ψ′(wk) = 0 strongly in (Xα(Rn+1+ ))′. (3.36)Since η2wk ∈ Xα(Rn+1+ ) for all k ∈ N, we can use it as a test function in(3.36) to get thato(1) = 〈Ψ′(wk), η2wk〉= kα∫Rn+1+y1−α〈∇wk,∇(η2wk)〉dxdy − γ∫Rnη2∗|uk|2|x|α dx−∫Rnη2∗|uk|2∗αdx−∫Rnη2∗|uk|2∗α(s)|x|s dx.383.3. Proof of Theorem 3.2Regarding the first term, we havekα∫Rn+1+y1−α〈∇wk,∇(η2wk)〉dxdy = kα∫Rn+1+y1−α|η∇wk|2dxdy+ kα∫Rn+1+y1−αwk〈∇(η2),∇wk〉dxdy.From Ho¨lder’s inequality, and the fact that wk → 0 in L2(supp(|∇η|), y1−α),it follows that as k →∞,∣∣∣∣∣kα∫Rn+1+y1−α〈∇wk,∇(η2wk)〉dxdy − kα∫Rn+1+y1−α|η∇wk|2dxdy∣∣∣∣∣=∣∣∣∣∣kα∫Rn+1+y1−αwk〈∇(η2),∇wk〉dxdy∣∣∣∣∣≤ kα∫Rn+1+y1−α|wk||∇(η2)||∇wk|dxdy≤ C∫supp(|∇η|)y1−α|wk||∇wk|dxdy≤ C‖wk‖Xα(Rn+1+ )(∫supp(|∇η|)y1−α|wk|2dxdy) 12= o(1).Thus, we have proved thatkα∫Rn+1+y1−α〈∇wk,∇(η2wk)〉dxdy = kα∫Rn+1+y1−α|η∇wk|2dxdy + o(1).Using the above estimate coupled with (3.34), we obtaino(1) = 〈Ψ′(wk), η2wk〉= kα∫Rn+1+y1−α|η∇wk|2dxdy − γ∫Kη2∗|uk|2|x|α dx−∫Rnη2∗|uk|2∗αdx−∫Kη2∗|uk|2∗α(s)|x|s dx+ o(1)= kα∫Rn+1+y1−α|∇(ηwk)|2dxdy −∫Rnη2∗|uk|2∗αdx+ o(1)≥ ‖ηwk‖2 −∫Rnη2∗|uk|2∗αdx+ o(1) as k →∞,(3.37)393.3. Proof of Theorem 3.2where K = supp(η∗). Therefore,‖ηwk‖2 ≤∫Rn|η∗uk|2|uk|2∗α−2dx+ o(1) as k →∞. (3.38)By Ho¨lder’s inequality, and using the definition of S(n, α, γ, 0), we then getthat‖ηwk‖2 ≤(∫Rn|η∗uk|2∗αdx) 22∗α(∫Rn|uk|2∗αdx) 2∗α−22∗α+ o(1)≤ S(n, α, γ, 0)−1‖ηwk‖2(∫Rn|uk|2∗αdx) 2∗α−22∗α+ o(1).(3.39)Thus, 1− S(n, α, γ, 0)−1(∫Rn|uk|2∗αdx) 2∗α−22∗α ‖ηwk‖2 ≤ o(1). (3.40)In addition, it follows from (3.29) thatΨ(wk)− 12〈Ψ′(wk), wk〉 = c+ o(1),that is,(12− 12∗α)∫Rn|uk|2∗αdx+ (12− 12∗α(s))∫Rn|uk|2∗α(s)|x|s dx = c+ o(1), (3.41)from which follows that∫Rn|uk|2∗αdx ≤ 2nαc+ o(1) as k →∞. (3.42)Plugging (3.42) into (3.40), we obtain that[1− S(n, α, γ, 0)−1(2nαc)αn]‖ηwk‖2 ≤ o(1) as k →∞.On the other hand, by the upper bound (3.28) on c, we have thatc <α2nS(n, α, γ, 0)nα .This yields that 1− S(n, α, γ, 0)−1(2nα c)αn > 0, and therefore,limk→∞‖ηwk‖2 = 0.403.3. Proof of Theorem 3.2Using (2.13) and (3.10), we obtain thatlimk→∞kα∫Rn+1+y1−α|∇(ηwk)|2dxdy = 0.It also follows from the definition of S(n, α, γ, 0) thatlimk→∞∫Rn|η∗uk|2∗αdx = 0.Since η∣∣D≡ 1 and η∗∣∣D∗≡ 1, the last two equality yield (3.33).Lemma 3.10. Let (wk)k∈N be Palais-Smale sequence as in Proposition 3.8.For any δ > 0, setθ := lim supk→∞∫Bδ|uk|2∗αdx; ζ := lim supk→∞∫Bδ|uk|2∗α(s)|x|s dx andµ := limk→∞∫B+δy1−α|∇wk|2dxdy − γ∫Bδ|uk|2|x|α dx,(3.43)where uk := Tr(wk) = wk(., 0) for all k ∈ N. If wk ⇀ 0 in Xα(Rn+1+ ) ask →∞, then the following hold:1. θ22∗α ≤ S(n, α, γ, 0)−1µ and ζ22∗α(s) ≤ S(n, α, γ, s)−1µ.2. µ ≤ θ + ζ.Proof of Lemma 3.10. First note that it follows from Lemma 3.9 that θ, ζand µ are well-defined and are independent of the choice of δ > 0. Let nowη ∈ C∞0 (Rn+1+ ) be a cut-off function such that η∗ := η(., 0) ∈ C∞0 (Rn \ {0}),η ≡ 1 in B+δ , and 0 ≤ η ≤ 1 in Rn+1+ .1. Since ηwk ∈ Xα(Rn+1+ ), we get from the definition of S(n, α, γ, s)S(n, α, γ, 0)(∫Rn|η∗uk|2∗αdx)22∗α≤ kα∫Rn+1+y1−α|∇(ηwk)|2dxdy − γ∫Rn|η∗uk|2|x|α dx.(3.44)413.3. Proof of Theorem 3.2On the other hand, from the definition of η, it follows thatkα∫Rn+1+y1−α|∇(ηwk)|2dxdy − γ∫Rn|η∗uk|2|x|α dx=∫B+δy1−α|∇wk|2dxdy − γ∫Bδ|uk|2|x|α dx+∫supp(η)\B+δy1−α|∇wk|2dxdy − γ∫supp(η∗)\Bδ|η∗uk|2|x|α dx,and(∫Bδ|uk|2∗αdx)22∗α ≤ (∫Rn|η∗uk|2∗αdx)22∗α .Note that supp(η) \ B+δ ⊂⊂ Rn+1+ \ {0} and supp(η∗) \ Bδ ⊂⊂ Rn \ {0}.Therefore, taking the upper limits at both sides of (3.44), and using Lemma3.9, we get thatS(n, α, γ, 0)(∫Bδ|uk|2∗αdx)22∗α ≤∫B+δy1−α|∇wk|2dxdy− γ∫Bδ|uk|2|x|α dx+ o(1),as k →∞, which givesθ22∗α ≤ S(n, α, γ, 0)−1µ.Similarly, we can prove thatζ22∗α(s) ≤ S(n, α, γ, s)−1µ.2. Since η2wk ∈ Xα(Rn+1+ ) and 〈Ψ′(wk), η2wk〉 = o(1) as k →∞, wehave423.3. Proof of Theorem 3.2o(1) = 〈Ψ′(wk), η2wk〉= kα∫Rn+1+y1−α〈∇wk,∇(η2wk)〉dxdy − γ∫Rn|η∗uk|2|x|α dx−∫Rnη2∗|uk|2∗αdx−∫Rnη2∗|uk|2∗α(s)|x|s dx=(kα∫Rn+1+y1−α|η∇wk|2dxdy − γ∫Rn|η∗uk|2|x|α dx)−∫Rnη2∗|uk|2∗αdx−∫Rnη2∗|uk|2∗α(s)|x|s dx+ kα∫Rn+1+y1−αwk〈∇(η2),∇wk〉dxdy.(3.45)By Ho¨lder’s inequality, and the fact that wk → 0 in L2(supp(|∇η|), y1−α),we obtain that∣∣∣∣∣kα∫Rn+1+y1−αwk〈∇(η2),∇wk〉dxdy∣∣∣∣∣ ≤ kα∫Rn+1+y1−α|wk||∇(η2)||∇wk|dxdy≤ C∫supp(|∇η|)y1−α|wk||∇wk|dxdy≤ C‖wk‖Xα(Rn+1+ )‖wk‖L2(supp(|∇η|),y1−α)≤ o(1) as k →∞.Plugging the above estimate into (3.45), and using (2.13), we get thato(1) = 〈Ψ′(wk), η2wk〉=(kα∫Rn+1+y1−α|∇(ηwk)|2dxdy − γ∫Rn|η∗uk|2|x|α dx)−∫Rnη2∗|uk|2∗αdx−∫Rnη2∗|uk|2∗α(s)|x|s dx+ o(1)≥(∫B+δy1−α|∇wk|2dxdy − γ∫Bδ|uk|2|x|α dx)−(∫Bδ|uk|2∗αdx)−(∫Bδ|uk|2∗α(s)|x|s dx)433.3. Proof of Theorem 3.2−∫supp(η∗)\Bδ(γ|η∗uk|2|x|α dx+ η2∗|uk|2∗αdx+η2∗|uk|2∗α(s)|x|s)dx+ o(1).Noting that supp(η∗)\Bδ ⊂⊂ Rn \{0}, and taking the upper limits on bothsides, we get that µ ≤ θ + ζ.Proof of Proposition 3.8. It follows from Lemma 3.10 thatθ22∗α ≤ S(n, α, γ, 0)−1µ ≤ S(n, α, γ, 0)−1θ + S(n, α, γ, 0)−1ζ,which givesθ22∗α (1− S(n, α, γ, 0)−1θ2∗α−22∗α ) ≤ S(n, α, γ, 0)−1ζ. (3.46)On the other hand, by (3.41), we haveθ ≤ 2nαc.Substituting the last inequality into (3.46), we get that(1− S(n, α, γ, 0)−1(2nαc)αn )θ22∗α ≤ S(n, α, γ, 0)−1ζ.Recall that the upper bounded (3.28) on c implies that1− S(n, α, γ, 0)−1(2nαc)αn > 0.Therefore, there exists δ1 = δ1(n, α, γ, c) > 0 such that θ22∗α ≤ δ1ζ. Similarly,there exists δ2 = δ2(n, α, γ, c, s) > 0 such that ζ22∗α(s) ≤ δ2θ. These twoinequalities yield that there exists 0 = 0(n, α, γ, c, s) > 0 such thateither θ = ζ = 0 or {θ ≥ 0 and ζ ≥ 0}. (3.47)It follows from the definition of θ and ζ thateither lim supk→∞∫Bδ|uk|2∗αdx = lim supk→∞∫Bδ|uk|2∗α(s)|x|s dx = 0;or lim supk→∞∫Bδ|uk|2∗αdx ≥ 0 and lim supk→∞∫Bδ|uk|2∗α(s)|x|s dx ≥ 0.443.3. Proof of Theorem 3.23.3.3 End of proof of Theorem 3.4We shall first eliminate the possibility of a zero weak limit for the Palais-Smale sequence of Ψ, then we prove that the nontrivial weak limit is indeeda weak solution of Problem (3.6). In the sequel (wk)k∈N will denote thePalais-Smale sequence for Ψ obtained in Proposition 3.8.First we show thatlim supk→∞∫Rn|uk|2∗αdx > 0. (3.48)Indeed, otherwise limk→∞∫Rn |uk|2∗αdx = 0, which once combined with the factthat 〈Ψ′(wk), wk〉 → 0 yields‖wk‖2 =∫Rn|uk|2∗α(s)|x|s dx+ o(1) as k →∞.By combining this estimate with the definition of S(n, α, γ, s), we obtain(∫Rn|uk|2∗α(s)|x|s dx) 22∗α(s)≤ S(n, α, γ, s)−1‖wk‖2≤ S(n, α, γ, s)−1∫Rn|uk|2∗α(s)|x|s dx+ o(1),which implies that(∫Rn|uk|2∗α(s)|x|s dx) 22∗α(s)[1− S(n, α, γ, s)−1(∫Rn|uk|2∗α(s)|x|s dx)2∗α(s)−22∗α(s)]≤ o(1).It follows from (3.28) and (3.41) that as k →∞,∫Rn|uk|2∗α(s)|x|s dx = 2cn− sα− s + o(1)and(1− S(n, α, γ, s)−1(2cn− sα− s)α−sn−s ) > 0.Hence,limk→∞∫Rn|uk|2∗α(s)|x|s dx = 0. (3.49)453.3. Proof of Theorem 3.2Using that limk→∞∫Rn |uk|2∗αdx = 0, in conjunction with (3.49) and (3.41), weget that c+ o(1) = 0, which contradicts the fact that c > 0. This completesthe proof of (3.48).Now, we show that for small enough  > 0, there exists another Palais-Smale sequence (vk)k∈N for Ψ satisfying the properties of Proposition 3.8,which is also bounded in Xα(Rn+1+ ) and satisfies∫B1|vk(x, 0)|2∗αdx =  for all k ∈ N. (3.50)For that, consider 0 as given in Proposition 3.8. Let β = lim supk→∞∫Rn |uk|2∗αdx, which is positive by (3.48). Set 1 := min{β, 02 } and fix  ∈ (0, 1). Upto a subsequence, there exists by continuity a sequence of radii (rk)k∈N suchthat∫Brk|uk|2∗αdx =  for each k ∈ N. Let nowvk(x, y) := rn−α2k wk(rkx, rky) for x ∈ Rn and y ∈ R+.It is clear that∫B1|vk(x, 0)|2∗αdx =∫Brk|uk|2∗αdx =  for all k ∈ N. (3.51)It is easy to check that (vk)k∈N is also a Palais-Smale sequence for Ψ thatsatisfies the properties of Proposition 3.8.We now show that (vk)k∈N is bounded in Xα(Rn+1+ ). Indeed, since(vk)k∈N is a Palais-Smale sequence, there exist positive constants C1, C2 > 0such thatC1 + C2‖vk‖ ≥ Ψ(vk)− 12∗α(s)〈Ψ′(vk), vk〉≥(12− 12∗α(s))‖vk‖2 +(12∗α− 12∗α(s))∫Rn|vk(x, 0)|2∗αdx≥(12− 12∗α(s))‖vk‖2.(3.52)The last inequality holds since 2 < 2∗α(s) < 2∗α. Combining (3.52) with(3.10), we obtain that (vk)k∈N is bounded in Xα(Rn+1+ ).It follows that there exists a subsequence – still denoted by vk – suchthat vk ⇀ v in Xα(Rn+1+ ) as k →∞. We claim that v is a nontrivial weaksolution of (3.6). Indeed, if v ≡ 0, then Proposition 3.8 yields thateither lim supk→∞∫B1|vk(x, 0)|2∗αdx = 0 or lim supk→∞∫B1|vk(x, 0)|2∗αdx ≥ 0.463.3. Proof of Theorem 3.2Since  ∈ (0, 02 ), this is in contradiction with (3.51), thus, v 6≡ 0.To show that v ∈ Xα(Rn+1+ ) is a weak solution of (3.6), consider any ϕ ∈C∞0 (Rn+1+ ), and writeo(1) = 〈Ψ′(vk), ϕ〉= kα∫Rn+1+y1−α〈∇vk,∇ϕ〉dxdy − γ∫Rnvk(x, 0)ϕ|x|α dx−∫Rn|vk(x, 0)|2∗α−2vk(x, 0)ϕdx−∫Rn|vk(x, 0)|2∗α(s)−2vk(x, 0)ϕ|x|s dx.(3.53)Since vk ⇀ v in Xα(Rn+1+ ) as k →∞, we have that∫Rn+1+y1−α〈∇vk,∇ϕ〉dxdy →∫Rn+1+y1−α〈∇v,∇ϕ〉dxdy, ∀ϕ ∈ C∞0 (Rn+1+ ).In addition, the boundedness of vk in Xα(Rn+1+ ) yields thatvk(., 0), |vk(., 0)|2∗α−2vk(., 0) and |vk(., 0)|2∗α(s)−2vk(., 0)are bounded in L2(Rn, |x|−α), L2∗α2∗α−1 (Rn) and L2∗α(s)2∗α(s)−1 (Rn, |x|−s), respec-tively. Therefore, we have the following weak convergence:vk(., 0) ⇀ v(., 0) in L2(Rn, |x|−α)|vk(., 0)|2∗α−2vk(., 0) ⇀ |v(., 0)|2∗α−2v(., 0) in L2∗α2∗α−1 (Rn)|vk(., 0)|2∗α(s)−2vk(., 0) ⇀ |v(., 0)|2∗α(s)−2v(., 0) in L2∗α(s)2∗α(s)−1 (Rn, |x|−s).Thus, taking limits as k →∞ in (3.53), we obtain that0 = 〈Ψ′(v), ϕ〉= kα∫Rn+1+y1−α〈∇v,∇ϕ〉dxdy − γ∫Rnv(x, 0)ϕ|x|α dx−∫Rn|v(x, 0)|2∗α−2v(x, 0)ϕdx−∫Rn|v(x, 0)|2∗α(s)−2v(x, 0)ϕ|x|s dx.Hence v is a weak solution of (3.6).47Chapter 4Mass and AsymptoticsAssociated to FractionalHardy-Schro¨dingerOperators in CriticalRegimes4.1 IntroductionThroughout this chapter, we shall assume that0 < α < n and 0 ≤ γ < γH(α) = 2α Γ2(n+α4)Γ2(n−α4), (4.1)which is the best fractional Hardy constant on Rn (see below). We may alsosometimes use the following notations for β+(γ) and β−(γ) introduced inRemark 1.4:β+ := β+(γ) and β− := β−(γ).Our main focus will be on the case when α < 2, that is when (−∆)α2is not a local operator. We shall study problems on bounded domains, butwill start by recalling the properties of (−∆)α2 on the whole of Rn, where itcan be defined on the Schwartz class S(Rn) (the space of rapidly decayingC∞ functions on Rn) via the Fourier transform,(−∆)α2 u = F−1(|2piξ|αF(u)).Here F(u) is the Fourier transform of u. For α ∈ (0, 2), the fractional Sobolevspace Hα20 (Rn) is defined as the completion of C∞c (Rn) under the norm‖u‖2Hα20 (Rn)=∫Rn|2piξ|α|Fu(ξ)|2dξ =∫Rn|(−∆)α4 u|2dx.484.1. IntroductionAs we mentioned before in (2.17), the fractional Hardy inequality in Rn thenstates thatγH(α) := inf∫Rn |(−∆)α4 u|2dx∫Rn|u|2|x|αdx; u ∈ Hα20 (Rn) \ {0} = 2αΓ2(n+α4 )Γ2(n−α4 ) ,which means that the fractional Hardy-Schro¨dinger operator Lγ,α is positivewhenever (4.1) is satisfied. In this case, a Hardy-Sobolev type inequalityholds for Lγ,α. It states that if 0 ≤ s < α < n, and 2∗α(s) = 2(n−s)n−α , thenµγ,s,α(Rn) is finite and strictly positive, where the latter is the best constantµγ,s,α(Rn) := infu∈Hα20 (Rn)\{0}∫Rn |(−∆)α4 u|2dx− γ ∫Rn |u|2|x|αdx(∫Rn|u|2∗α(s)|x|s dx)22∗α(s). (4.2)Note that any minimizer for (4.2) leads –up to a constant– to a variationalsolution of the following borderline problem on Rn,{(−∆)α2 u− γ u|x|α = u2∗α(s)−1|x|s in Rnu ≥ 0 ; u 6≡ 0 in Rn. (4.3)Indeed, a function u ∈ Hα20 (Rn) is said to be a weak solution to (4.3) ifu ≥ 0, u 6≡ 0 and for any ϕ ∈ Hα20 (Rn), we haveCn,α2∫(Rn)2(u(x)− u(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy =∫Rn(γu|x|α +u2∗α(s)−1|x|s )ϕ dx.Unlike the case of the Laplacian (α = 2), no explicit formula is known forthe best constant µγ,s,α(Rn) nor for the extremals where it is achieved. Wetherefore try to describe their asymptotic profile whenever they exist. Thiswas considered in Ghoussoub-Shakerian [39]; where Theorem 3.1 is proved.Note that the cases when γ = 0 are by now well known. Indeed, itwas stated in [19] that the infimum in µ0,0,α(Rn) is attained. Actually, afunction u˜ ∈ Hα20 (Rn) \ {0} is an extremal for µ0,0,α(Rn) if and only if thereexist x0 ∈ Rn, k ∈ R \ {0} and r > 0 and such thatu˜(x) = k(r2 + |x− x0|2)− (n−α)2 for all x ∈ Rn.Asymptotic properties of the positive extremals of µ0,s,α(Rn) (i.e., whenγ = 0 and 0 < s < α) were given by Y. Lei [48], Lu-Zhu [54], and Yang-Yu494.1. Introduction[71]. The latter proved that an extremal u¯(x) for µ0,s,α(Rn) must have thefollowing behaviour: There is C > 0 such thatC−1(1 + |x|2)− (n−α)2 ≤ u¯(x) ≤ C (1 + |x|2)− (n−α)2 for all x ∈ Rn. (4.4)Recently, Dipierro-Montoro-Peral-Sciunzi [21] found a similar control of theextremal for µγ,0,α(Rn) (i.e., when 0 < γ < γH(α) and s = 0). Our firstresult is an improvement of their estimate since it gives the exact asymptoticbehaviour of the extremal of µγ,s,α(Rn) in the general case. For that, weconsider the functionΨn,α(β) := 2αΓ(n−β2 )Γ(α+β2 )Γ(n−β−α2 )Γ(β2 ). (4.5)Theorem 4.1. Assume 0 ≤ s < α < 2, n > α and 0 ≤ γ < γH(α). Then,any positive extremal u ∈ Hα20 (Rn) for µγ,s,α(Rn) satisfies u ∈ C1(Rn \ {0})andlimx→0|x|β−(γ)u(x) = λ0 and lim|x|→∞ |x|β+(γ)u(x) = λ∞, (4.6)where λ0, λ∞ > 0 and β−(γ) (resp., β+(γ)) is the unique solution in(0, n−α2)(resp., in(n−α2 , n− α)) of the equation Ψn,α(t) = γ. In particular, thereexist C1, C2 > 0 such thatC1|x|β−(γ) + |x|β+(γ) ≤ u(x) ≤C2|x|β−(γ) + |x|β+(γ) for all x ∈ Rn \ {0}.Remark 4.2. Note that a direct consequence of Theorem 4.1 is (4.4) andthe corresponding control by Dipierro-Montoro-Peral-Sciunzi [21].Also note that if α = 2, that is when the fractional Laplacian is theclassical Laplacian, the best constant in the Hardy inequality is then γH(2) =(n−2)24 . The best constant associated with the Hardy-Sobolev inequality isµγ,s,2(Rn) := infu∈D1,2(Rn)\{0}∫Rn |∇u|2dx− γ∫Rn|u|2|x|2 dx(∫Rn|u|2?(s)|x|s dx)22∗(s),where s ∈ [0, 2), 2∗(s) := 2(n−s)n−2 , 0 ≤ γ < γH(2) = (n−2)24 and D1,2(Rn)is the completion of C∞c (Ω) with respect to the norm ‖u‖2 =∫Rn |∇u|2dx.504.1. IntroductionThe extremals for µγ,s,2(Rn) are then explicit and are given by multiples ofthe functions u(x) = −n−22 U(x ) for  > 0, whereU(x) =1(|x|(2−s)σ−(γ)n−2 + |x|(2−s)σ+(γ)n−2)n−22−sfor Rn \ {0},andσ±(γ) =n− 22±√(n− 2)24− γ.Note that the radial function u(x) = |x|−β is a solution of Lγ,2(u) = 0 onRn \ {0} if and only ifβ ∈ {σ−(γ), σ+(γ)}. (4.7)Back to the case 0 < α < 2, we now turn to when Ω is a smooth boundeddomain in Rn with 0 in its interior. The best constant in the correspondingfractional Hardy-Sobolev inequality is then,µγ,s,α(Ω) := infu∈Hα20 (Ω)\{0}Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ω|u|2|x|αdx(∫Ω|u|2∗α(s)|x|s dx)22∗α(s),where Hα20 (Ω) is the closure of C∞c (Ω) with respect to the norm‖u‖2Hα20 (Ω)=Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy =∫Rn|(−∆)α4 u|2dx.In Proposition 4.15, we note that –just like the case when α = 2– we haveµγ,s,α(Ω) = µγ,s,α(Rn), and therefore (4.3) restricted to Ω, with Dirichletboundary condition has no extremal, unless Ω is essentially Rn. We thereforeresort to a setting popularized by Brezis-Nirenberg [11] by considering thefollowing boundary value problem:(−∆)α2 u− γ u|x|α =u2∗α(s)−1|x|s + λu in Ωu ≥ 0 in Ωu = 0 in Rn \ Ω,(4.8)where 0 < λ < λ1(Lγ,α) and λ1(Lγ,α) is the first eigenvalue of the operatorLγ,α = (−∆)α2 − γ|x|α with Dirichlet boundary condition, that isλ1 := λ1(Lγ,α) = infu∈Hα20 (Ω)\{0}Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ωu2|x|α∫Ω u2dx.514.1. IntroductionOne then considers the quantityµγ,s,α,λ(Ω) = infu∈Hα20 (Ω)\{0}Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ωu2|x|α dx− λ∫Ωu2dx(∫Ωu2∗α(s)|x|s dx) 22∗α(s),and uses the fact that compactness is restored as long as µγ,s,α,λ(Ω) <µγ,s,α(Rn); see Proposition 4.10 and also [11] for more details. This type ofcondition is now classical in borderline variational problems; see Aubin [4]and Brezis-Nirenberg [11].When α = 2, i.e., in the case of the standard Laplacian, the minimizationproblem µγ,s,α,λ(Ω) has been extensively studied, see for example Lieb [50],Chern-Lin [17], Ghoussoub-Moradifam [35] and Ghoussoub-Robert [36]. Thenon-local case has also been the subject of several studies, but in the absenceof the Hardy term, i.e., when γ = 0. In [65], Servadei proved the existence ofextremals for µ0,0,α,λ(Rn), and completed the study of problem (4.8) whichhas been initiated by Servadei-Valdinoci [63, 64]. Recently, it has beenshown by Yang-Yu [71] that there exists a positive extremal for µ0,s,α,λ(Rn)when s ∈ [0, 2). In this chapter, we consider the remaining cases.In the spirit of Jannelli [45], who dealt with the Laplacian case, weobserve that problem (4.8) is deeply influenced by the value of the parameterγ. Roughly speaking, if γ is sufficiently small then µγ,s,α,λ(Ω) is attained forany 0 < λ < λ1. This is essentially what was obtained by Servadei-Valdinoci[64] when s = γ = 0 and n ≥ 2α via local arguments. This is, however notthe case, when γ is closer to γH(α), which amounts to dealing with lowdimensions: see for instance Servadei-Valdinoci [63]. In this context of lowdimension, the local arguments generally fail, and it is necessary to use globalarguments via the introduction of a notion of mass in the spirit of Schoen[61]. In the present case, and as in the work of Ghoussoub-Robert [37], wedefine a notion of mass for the operator Lγ,α−λI, which again turns out tobe critical for this non-local case. The mass is defined via the following keyresult.Theorem 4.3. Let Ω be a bounded smooth domain in Rn (n > α) andconsider, for 0 < α < 2, the boundary value problem(−∆)α2H −(γ|x|α + a(x))H = 0 in Ω \ {0}H > 0 in Ω \ {0}H = 0 in Rn \ Ω,(4.9)where a(x) ∈ C0,τ (Ω) for some τ ∈ (0, 1). Assuming the operator (−∆)α2 −( γ|x|α + a(x)) coercive, there exists then a threshold −∞ < γcrit(α) < γH(α)524.1. Introductionsuch that for any γ with γcrit(α) < γ < γH(α), there exists a unique solutionto (4.9) (in the sense of Definition 4.6) H : Ω→ R, H 6≡ 0, and a constantc ∈ R such thatH(x) =1|x|β+(γ) +c|x|β−(γ) + o(1|x|β−(γ))as x→ 0.We define the fractional Hardy-singular internal mass of Ω associated to theoperator Lγ,α to bemαγ,a(Ω) := c ∈ R.We then prove the following existence result, which complements those in[65] and [71] to the case when γ > 0.Theorem 4.4. Let Ω be a smooth bounded domain in Rn(n > α) such that0 ∈ Ω, and let 0 ≤ s < α, 0 ≤ γ < γH(α).Then, there exist extremals forµγ,s,α,λ(Ω) under one of the following two conditions:1. 0 ≤ γ ≤ γcrit(α) and 0 < λ < λ1(Lγ,α),2. γcrit(α) < γ < γH(α), 0 < λ < λ1(Lγ,α) and mαγ,λ(Ω) > 0.The idea of studying how critical behavior occurs while varying a param-eter γ on which an operator Lγ,α continuously depends goes back to [45], whoconsidered the classical Hardy-Schro¨dinger operator Lγ,2 := −∆− γ|x|2 , andshowed the existence of extremals for any λ > 0 provided 0 ≤ γ ≤ (n−2)24 −1.In this case, γcrit(2) =(n−2)24 − 1. The definition of the mass and the coun-terpart of Theorem 4.4 for the operator Lγ,2 was established by Ghoussoub-Robert [37]. The complete picture can be described as follows.Hardy term Dimension Singularity Analytic. cond. Ext.0 ≤ γ ≤ γcrit(α) n ≥ 2α s ≥ 0 λ > 0 Yesγcrit(α) < γ < γH(α) n ≥ 2α s ≥ 0 mαγ,λ(Ω) > 0 Yes0 ≤ γ < γH(α) α < n < 2α s ≥ 0 mαγ,λ(Ω) > 0 YesEven though the constructions and the methods are heavily inspired bythe work of Ghoussoub-Robert [37] on the Laplacian case, the fact that theoperator is nonlocal here induces several fundamental difficulties that hadto be overcome. First, the construction of the mass in the local case usesa precise classification of singularities for solutions of corresponding ellipticequations, that follows from the comparison principle stating that behavior534.2. The fractional Hardy-Schro¨dinger operator Lγ,α on Rnin a domain is governed by the behavior on its boundary. In the nonlocalcase, this fails since one needs to consider the whole complement of thedomain, and not only its boundary. We were able to bypass this difficultyby using sharp regularity results available for the fractional Laplacian. An-other difficulty we had to face came from the test-functions estimates in thepresence of the mass. In the classical local case, one estimates the associ-ated functional on a singular test-function, counting on the mass to appearafter suitable integrations by parts. In the nonlocal context, this strategyfails. We overcome this difficulty by looking at the integral on the boundaryof a domain as a limit of integrals on the domain after multiplying by acut-off functions whose support converge to the boundary. This process iswell-defined in the nonlocal context and proves to be efficient in tackling theestimates involving the mass.4.2 The fractional Hardy-Schro¨dinger operatorLγ,α on RnIn this section, we study the local behavior of solutions of the fractionalHardy-Schro¨dinger operator Lγ,α := (−∆)α2 − γ|x|α on Rn. The most basicsolutions for Lγ,αu = 0 on Rn are of the form u(x) = |x|−β, and a straight-forward computation yields (see [34])(−∆)α2 |x|−β = Ψn,α(β)|x|−β−α in the sense of S ′(Rn) when 0 < β < n− α,whereΨn,α(β) := 2αΓ(n−β2 )Γ(α+β2 )Γ(n−β−α2 )Γ(β2 ). (4.10)Recall that the best constant in the fractional Hardy inequalityγH(α) := µ0,α,α(Rn) = inf∫Rn |(−∆)α4 u|2dx∫Rn|u|2|x|αdx; u ∈ Hα20 (Rn) \ {0}is never achieved (see Fall [29]), is equal to Ψn,α(n−α2 ) = 2α Γ2(n+α4)Γ2(n−α4)(seeHerbst and Yafaev [43, 69]), and it converges to the best classical Hardyconstant γH(2) =(n−2)24 whenever α→ 2.We summarize some properties of the function β 7→ Ψn,α(β) which will beused freely in this section. They are essentially consequences from knownproperties of Gamma function Γ.544.2. The fractional Hardy-Schro¨dinger operator Lγ,α on RnProposition 4.5 (Frank-Lieb-Seiringer [34]). The following properties hold:1. Ψn,α(β) > 0 for all β ∈ (0, n− α).2. The graph of Ψn,α in (0, n−α) is symmetric with respect to n−α2 , thatis,Ψn,α(β) = Ψn,α(n− α− β) for all β ∈ (0, n− α).3. Ψn,α is strictly increasing in (0,n−α2 ), and strictly decreasing in (n−α2 ,n− α).4. Ψn,α(n− α2)= γH(α).5. limβ↘0Ψn,α(β) = limβ↗n−αΨn,α(β) = 0.6. For any γ ∈ (0, γH(α)), there exists a unique β−(γ) ∈ (0, n−α2 ) suchthat Ψn,α(β−(γ)) = γ.7. For any 0 < β ≤ n− α, we have that(−∆)α2 |x|−β = Ψn,α(β)|x|−α−β + cn,α1{β=n−α}δ0 in S ′(Rn), (4.11)where we define Ψn,α(n− α) = 0 and cn,α > 0 is a constant.In particular, for 0 < β < n− α,((−∆)α2 − γ|x|α)|x|−β = 0 in S ′(Rn) if and only if β ∈ {β+(γ), β−(γ)},where 0 < β−(γ) < n−α2 is as in Proposition 4.5 and β+(γ) := n − α −β−(γ) ∈(n−α2 , n− α). In particular, it follows from Proposition 4.5 thatβ−(γ), β+(γ) are the only solutions to Ψn,α(β) = γ in (0, n − α). Since0 < β−(γ) < n−α2 < β+(γ) < n − α, we get that x 7→ |x|−β−(γ) is locallyin Hα20 (Rn). It is the“small” or variational solution, while x 7→ |x|−β+(γ)is the“large” or singular solution. We extend β−(γ), β+(γ) to the wholeinterval [0, γH(α)] by definingβ−(0) := 0, β+(0) := n− α, and β−(γH(α)) = β+(γH(α)) = n− α2,(4.12)which is consistant with Proposition 4.5.554.2. The fractional Hardy-Schro¨dinger operator Lγ,α on RnWe now proceed to define a critical threshold γcrit(α) as follows. Assumingfirst that n > 2α, then n−α2 <n2 < n− α and therefore, by Proposition 4.5,there exists γ¯(α) ∈ (0, γH(α)) such thatn2< β+(γ) < n− α if γ ∈ (0, γ¯(α))β+(γ) =n2if γ = γ¯(α)n− α2< β+(γ) <n2if γ ∈ (γ¯(α), γH(α)).We then setγcrit(α) :=γ¯(α) if n > 2α0 if n = 2α−1 if n < 2α.(4.13)One can easily check that for γ ∈ [0, γH(α)), we have thatγ ∈ (γcrit(α), γH(α)) ⇔ β+(γ) < n2⇔ x 7→ |x|−β+(γ) ∈ L2loc(Rn).We now introduce the following terminology in defining a notion of solutionon a punctured domain.Definition 4.6. Let Ω be a smooth domain (not necessarily bounded) of Rn,n > 1. Let f be a function in L1loc(Ω \ {0}). We say that u : Ω → R is asolution to {(−∆)α2 u = f in Ω \ {0}u = 0 in ∂Ω,provided1. For any η ∈ C∞c (Rn \ {0}), we have that ηu ∈ Hα20 (Ω);2.∫Ω|u(x)|1+|x|n+α dx <∞;3. For any ϕ ∈ C∞c (Ω \ {0}), we have thatCn,α2∫Rn∫Rn(u(x)− u(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy =∫Rnf(x)ϕ(x) dx.Note that the third condition is consistent thanks to the two preceding it.If Ω is bounded, the second hypothesis rewrites as u ∈ L1(Ω).564.3. Profile of solutions4.3 Profile of solutionsThroughout this chapter, we shall frequently use the following fact:Proposition 4.7. A measurable function u : Rn → R belongs to Hα20 (Rn) ifand only if∫Rn |u|2?α(0) dx < +∞ and ∫(Rn)2 |u(x)−u(y)|2|x−y|n+α dxdy < +∞.The proof consists of approximating u by a compactly supported func-tion satisfying the same properties. Then, by convoluting with a smoothmollifier, this approximation is achieved by a smooth compactly supportedfunction. The rest is classical and the details are left to the reader.To prove Theorem 4.1, we shall use a similar argument as in Dipierro-Montoro-Peral-Sciunzi [21]. The main idea is to transform problem (4.3)into a different nonlocal problem in a weighted fractional space by using arepresentation introduced in Frank-Lieb-Seiringer [34].Lemma 4.8 (Ground State Representation [34]; Formula (4.3)). Assume0 < α < 2, n > α, 0 < β < n−α2 . For u ∈ C∞c (Rn\{0}), we let vβ(x) =|x|βu(x) in Rn\{0}. Then,Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy = Ψn,α(β)∫Rnu2(x)|x|α dx+Cn,α2∫Rn∫Rn|vβ(x)− vβ(y)|2|x− y|n+αdx|x|βdy|y|β .Let now u ∈ Hα20 (Rn) be a positive weak solution to (4.3). Then by (4.4)and Remark 4.4 in [21], we haveCn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy = γ∫Rnu2(x)|x|α dx+∫Rnu2∗α(s)(x)|x|s dx.Set v(x) = |x|β−(γ)u(x) on Rn\{0}. It follows from Lemma 4.8 and the574.3. Profile of solutionsdefinition of β−(γ) thatCn,α2∫Rn∫Rn|v(x)− v(y)|2|x− y|n+αdx|x|β−(γ)dy|y|β−(γ)=Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy −Ψn,α(β−(γ))∫Rnu2(x)|x|α dx= γ∫Rnu2(x)|x|α dx+∫Rnu2∗α(s)(x)|x|s dx−Ψn,α(β−(γ))∫Rnu2(x)|x|α dx=∫Rnu2∗α(s)(x)|x|s dx=∫Rnv2∗α(s)(x)|x|s+β−(γ)2∗α(s)dx.For 0 < β < n−α2 , define the space Hα2,β0 (Rn) as the completion of C∞c (Rn \{0}) with respect to the norm‖φ‖Hα2 ,β0 (Rn):=(∫Rn∫Rn|φ(x)− φ(y)|2|x− y|n+αdx|x|βdy|y|β) 12.Many of the properties of the space Hα2,β0 (Rn) were established in [22]. ByLemma 4.8, Remark 4.4 in [21] and [1], we have that v ∈ Hα2,β0 (Rn). Now,we introduce the operator (−∆β)α2 , whose action on a function w is givenvia the following duality: For φ ∈ Hα2,β0 (Rn),〈(−∆β)α2w, φ〉 = Cn,α2∫Rn∫Rn(w(x)− w(y))(φ(x)− φ(y))|x− y|n+αdx|x|βdy|y|β .This means that v is a weak solution to(−∆β−(γ))α2 v =v2∗α(s)−1|x|s+β−(γ)2∗α(s) in Rn, (4.14)in the sense that for any φ ∈ H α2 ,β−(Rn), we have thatCn,α2∫Rn∫Rn(v(x)− v(y))(φ(x)− φ(y))|x− y|n+αdx|x|β−dy|y|β− =∫Rnv2∗α(s)−1|x|s+β−2∗α(s)φ dx.The following proposition gives a regularity result and a Harnack inequalityfor weak solutions of (4.14).584.3. Profile of solutionsProposition 4.9. Assume 0 < s < α < 2, n > α and 0 < β < n−α2 , and letv ∈ Hα2,β0 (Rn) be a non-negative, non-zero weak solution to the problem(−∆β)α2 v =v2∗α(s)−1|x|s+β2∗α(s) in Rn.Then, v ∈ L∞(Rn) and there exist constants R > 0 and C > 0 such thatC ≤ v(x) in BR(0).Proof. The statement that v(x) ≥ C in BR(0) is essentially the Harnackinequality for super-harmonic functions associated to the nonlocal operator(−∆β)α2 , which is just Theorem 3.4 in Abdellaoui-Medina-Peral-Primo [2].See also the proof of Lemma 3.10 in [2] and also [21]. We now show thatv ∈ L∞(Rn) by using a similar argument as in [21]. For any p ≥ 1 andT > 0, define the functionφp,T (t) ={tp if 0 ≤ t ≤ TpT p−1(t− T ) + T p if t > T .It is easy to check that the function φp,T (t) has the following properties:• φp,T (t) is convex and Lipschitz in [0,∞).• φp,T (t) ≤ tp for all t ≥ 0.• tφ′p,T (t) ≤ 2pφp,T (t) for all t ≥ 0, sincetφ′p,T (t) ={pφp,T (t) if 0 < t < TpT p−1t if t > T .• If T2 > T1 > 0, then φp,T1(t) ≤ φp,T2(t) for all t ≥ 0.Since φp,T (t) is convex and Lipschitz, then as noted in [49],(−∆β)α2 φp,T (v) ≤ φ′p,T (v)(−∆β)α2 v in Rn. (4.15)Since φp,T (t) is Lipschitz and φp,T (0) = 0, then φp,T (v) ∈ Hα2,β0 (Rn). By theweighted fractional Hardy-Sobolev inequality, the ground state representa-tion formula, Lemma 4.8, and (4.2), we get that there exists some constantC0 > 0 which only depends on n, α, s and β such that[∫Rn|φp,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)≤ C02∫Rn∫Rn|φp,T (v(x))− φp,T (v(y))|2|x− y|n+αdx|x|βdy|y|β .(4.16)594.3. Profile of solutionsSince φp,T (t) ≥ 0 for all t ≥ 0, we get from (4.15) that∫(Rn)2|φp,T (v(x))− φp,T (v(y))|2|x− y|n+αdx|x|βdy|y|β =∫Rnφp,T (v)(−∆β)α2 φp,T (v)dx≤∫Rnφp,T (v)φ′p,T (v)(−∆β)α2 vdx=∫Rnφp,T (v)φ′p,T (v)v2∗α(s)−1|x|s+β2∗α(s)dx≤ 2p∫Rn|φp,T (v)|2 v2∗α(s)−2|x|s+β2∗α(s)dx.Note that the last inequality holds, since tφ′p,T (t) ≤ 2pφ(t) for all t ≥ 0. By(4.16), we have[∫Rn|φp,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)≤ pC0∫Rn|φp,T (v)|2 v2∗α(s)−2|x|s+β2∗α(s) dx. (4.17)Letting p1 =2∗α(s)2, then[∫Rn|φp1,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)≤ p1C0∫Rn|φp1,T (v)|2v2∗α(s)−2|x|s+β2∗α(s) dx. (4.18)For m > 0, a simple computation and Ho¨lder’s inequality yield thatp1C0∫Rn|φp1,T (v)|2v2∗α(s)−2|x|s+β2∗α(s)dx= p1C0∫v(x)≤m|φp1,T (v)|2v2∗α(s)−2|x|s+β2∗α(s) dx+ p1C0∫v(x)>m|φp1,T (v)|2v2∗α(s)−2|x|s+β2∗α(s)dx≤ p1C0m2∗α(s)−2∫v(x)≤m|φp1,T (v)|2|x|s+β2∗α(s) dx+ p1C0∫v(x)>m|φp1,T (v)|2|x|2(s+β2∗α(s))2∗α(s)v2∗α(s)−2|x|s+β2∗α(s)−2(s+β2∗α(s))2∗α(s)dx604.3. Profile of solutions≤ p1C0m2∗α(s)−2∫Rn|φp1,T (v)|2|x|s+β2∗α(s) dx+ p1C0[∫v(x)>m|φp1,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)[∫v(x)>mv2∗α(s)|x|s+β2∗α(s) dx]α−sn−s≤ p1C0m2∗α(s)−2∫Rn|φp1,T (v)|2|x|s+β2∗α(s) dx+ p1C0[∫Rn|φp1,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)[∫v(x)>mv2∗α(s)|x|s+β2∗α(s) dx]α−sn−s.Recall that v ∈ Hα2,β0 (Rn), hence∫Rnv2∗α(s)|x|s+β2∗α(s) dx < ∞. Thus, we cantake a large M0  1 and fix it in such a way thatp1C0[∫v(x)>M0v2∗α(s)|x|s+β2∗α(s) dx]α−sn−s≤ 12.Since φp1,T (t) ≤ tp1 for all t ≥ 0, then by (4.18) and the fact that p1 =2∗α(s)2,we get[∫Rn|φp1,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)≤ 2p1C0M2∗α(s)−20∫Rn|φp1,T (v)|2|x|s+β2∗α(s) dx≤ 2p1C0M2∗α(s)−20∫Rn|v|2p1|x|s+β2∗α(s) dx= 2p1C0M2∗α(s)−20∫Rnv2∗α(s)|x|s+β2∗α(s) dx. (4.19)Let C1 = 2C0M2∗α(s)−20 . By taking T → ∞ in (4.19) and applying Fatou’slemma, we get that[∫Rnvp12∗α(s)|x|s+β2α∗(s) dx] 22α∗(s)≤ p1C1∫Rnv2∗α(s)|x|s+β2∗α(s) dx <∞.Define now recursively the sequence {pk}∞k=2 as follows:2pk+1 + 2∗α(s)− 2 = pk2∗α(s) for all k ≥ 1. (4.20)614.3. Profile of solutionsUsing (4.17) and (4.20), we have[∫Rn|φpk+1,T (v)|2∗α(s)|x|s+β2∗α(s) dx] 22∗α(s)≤ pk+1C0∫Rn|φpk+1,T (v)|2v2∗α(s)−2|x|s+β2∗α(s)dx≤ pk+1C0∫Rnv2pk+1v2∗α(s)−2|x|s+β2∗α(s)dx= C0pk+1∫Rnvpk2∗α(s)|x|s+β2∗α(s)dx. (4.21)We also have used the fact that φpk+1,T (t) ≤ tpk+1 for all t ≥ 0. By takingT →∞ in (4.21) and applying Fatou’s lemma, we get that[∫Rnvpk+12∗α(s)|x|s+β2∗α(s)dx] 22∗α(s)≤ C0pk+1∫Rnvpk2∗α(s)|x|s+β2∗α(s)dx for all k ≥ 1.Hence, by (4.20), we obtain that[∫Rnvpk+12∗α(s)|x|s+β2∗α(s) dx] 12∗α(s)(pk+1−1)≤ (C0pk+1)12(pk+1−1)[∫Rnvpk2∗α(s)|x|s+β2∗α(s) dx] 12(pk+1−1)= (C0pk+1)12(pk+1−1)[∫Rnvpk2∗α(s)|x|s+β2∗α(s) dx] 12∗α(s)(pk−1).For k ≥ 1, setIk :=[∫Rnvpk2∗α(s)|x|s+β2∗α(s) dx] 12∗α(s)(pk−1)and Dk = (C0pk+1)12(pk+1−1) .We have Ik+1 ≤ DkIk for all k ≥ 1, andln Ik+1 ≤ lnDk + ln Ik ≤k∑j=1lnDj + ln I1 ≤k∑j=1lnC0 + ln pj+12(pj+1 − 1) + ln I1.It follows from (4.20) that pk+1 = pk1(p1− 1) + 1 for all k ≥ 0. This coupled624.3. Profile of solutionswith the fact that p1 > 1 yieldln Ik+1 ≤k∑j=1lnC02pj1(p1 − 1)+k∑j=1ln[pj1(p1 − 1) + 1]2pj1(p1 − 1)+ ln I1≤k∑j=1lnC02pj1(p1 − 1)+k∑j=1ln pj+112pj1(p1 − 1)+ ln I1 < C2 <∞.For any fix R ≥ 1, we then have[∫|x|≤Rvpk2∗α(s)|x|s+β2∗α(s) dx] 12∗α(s)(pk−1)≤ Ik ≤ eC2 =: C3 for all k ≥ 1.Since s+ β2∗α(s) > 0, we then get[∫|x|≤Rvpk2∗α(s) dx] 12∗α(s)pk≤ C3Rs+β2∗α(s)2∗α(s)pk for all k ≥ 1.Since limk→∞pk =∞, we have‖v‖L∞(BR(0)) = limk→∞[∫|x|≤Rvpk2∗α(s) dx] 12∗α(s)pk≤ C3,and finally, that ‖v‖L∞(Rn) ≤ C3.Proof of Theorem 4.1. Let v(x) = |x|β−(γ)u(x) in Rn\{0}, by the discussionbefore at the beginning of section 3, we know that v ∈ Hα2,β0 (Rn) is a positiveweak solution to (4.14). We deduce from Proposition 4.9 that for all R > 0,there exist some constant C > 1 such that C−1 ≤ v(x) ≤ C in BR(0). Sincev(x) = |x|β−(γ)u(x) in Rn\{0}, thenC−1|x|β−(γ) ≤ u(x) ≤C|x|β−(γ) in BR(0)\{0}. (4.22)In order to prove the asymptotic behavior at zero, it is enough to show thatlimx→0|x|β−(γ)u(x) exists. To that end, we proceed as follows:Claim 1: u ∈ C1(Rn \ {0}).634.3. Profile of solutionsThis is consequence of regularity theory and we only sketch the proof. Firstwe define f0(x) := γ|x|−αu+ u2∗α(s)−1|x|−s, so that for any ω ⊂⊂ Rn \ {0},we have that (−∆)α/2u = f0 in ω in the sense that u ∈ Hα20 (Rn) andCn,α2∫(Rn)2(u(x)− u(y))(φ(x)− φ(y))|x− y|n+α dxdy =∫ωf0ϕdx for all ϕ ∈ C∞c (ω).It follows from (4.22) that f0 ∈ L∞(ω). Since u ≥ 0 and f0 ∈ L∞(ω), itfollows from Remark 2.5 (see also Theorem 2.1) in Jin-Li-Xiong [46] thatthere exists τ > 0 such that u ∈ C0,τloc (Rn \ {0}). Then, using recursivelyTheorem 2.1 in Jin-Li-Xiong [46], we get that u ∈ C1(Rn \{0}). This provesthe claim.Claim 2: There exists C > 0 such that |x|β−(γ)+1|∇u(x)| ≤ C for all x ∈B1(0) \ {0}.If not, then there exists a sequence (xi)i∈N ∈ B1(0) \ {0} such thatlimi→+∞|xi|β−(γ)+1|∇u(xi)| = +∞.For simplicity, we write β− := β−(γ). It follows from from Claim 1 thatlimi→+∞xi = 0.We define ri := |xi| and we setui(x) := rβ−i u(rix) for all x ∈ Rn \ {0}.It is easy to see that ui ∈ Hα20 (Rn), ui ≥ 0 for all i ∈ N and (−∆)α/2ui = fi inω ⊂⊂ Rn \{0} where fi(x) := γ|x|−αui+ r(2∗α(s)−2)(n−α2 −β−)i u2∗α(s)−1i |x|−s forall x ∈ Rn \ {0}. Using the apriori bound of Remark 2.5 (see also Theorem2.1) in Jin-Li-Xiong [46], we get that there exists τ > 0 such that for anyR > 1, there exists C(R) > 0 such that ‖ui‖C0,τ (BR(0)−BR−1 (0)) ≤ C(R) forall i ∈ N. Using recursively Theorem 2.1 of [46] as in Step 1, we get thatfor any ω ⊂⊂ Rn \ {0}, there exists C(ω) > 0 such that ‖ui‖C1(ω) ≤ C(ω).Taking ω large enough and estimating |∇ui( xi|xi|)|, we get a contradiction,which proves Claim 2.Set now h(x) := u2∗α(s)−2|x|s , so that (−∆)α/2u − γ|x|αu = h(x)u in Rn. Itfollows from Claims 1 and 2, that h ∈ C1(Rn \ {0}), and for some C > 0,|h(x)|+ |x| · |∇h(x)| ≤ C|x|θ−α for all x ∈ B1(0) \ {0},644.4. Analytic conditions for the existence of extremalswhere θ := (2∗α(s) − 2)(n−α2 − β−) > 0. It then follows from Lemma 4.14below that there exists λ0 > 0 such thatlimx→0|x|β−u(x) = λ0 > 0.In order to deal with the behavior at infinity, let w be the fractional Kelvintransform of u, that is,w(x) = |x|α−nu(x∗) := |x|α−nu(x|x|2)in Rn\{0}.By Lemma 2.2 and Corollary 2.3 in [32], we have that w ∈ Hα20 (Rn). Asimple calculation gives us that w is also a positive weak solution to (4.3).Indeed, we have(−∆)α2w(x) = 1|x|n+α((−∆)α2 u)( x|x|2)= γw(x)|x|α +w2∗α(s)−1(x)|x|s .Arguing as in the first part of the proof, we get that there exists λ∞ > 0such thatlimx→0|x|β−(γ)w(x) = λ∞ > 0.Coming back to u, this implies thatlim|x|→∞|x|β+(γ)u(x) = λ∞ > 0.This ends the proof of Theorem 4.1.4.4 Analytic conditions for the existence ofextremalsLet a ∈ C0,τ (Ω) for some τ ∈ (0, 1), and define the functional JΩa : Hα20 (Ω) −→R byJΩa (u) :=Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ωu2|x|αdx−∫Ω au2dx(∫Ω|u|2∗α(s)|x|s dx) 22∗α(s),in such a way thatµγ,s,α,a(Ω) := inf{JΩa (u) : u ∈ Hα20 (Ω) \ {0}}.We now prove the following proposition, which gives analytic conditionsfor the existence of extremals for µγ,s,α,a(Ω).654.4. Analytic conditions for the existence of extremalsProposition 4.10. Let Ω be a bounded domain in Rn (n > α) such that0 ∈ Ω, and assume that 0 ≤ γ < γH(α) and 0 ≤ s ≤ α.1. If µγ,s,α,a(Ω) < µγ,s,α(Rn), then there are extremals for µγ,s,α,a(Ω) inHα20 (Ω).2. If a(x) is a constant λ, with 0 < λ < λ1(Lγ,α) and if s < α, thenµγ,s,α,a(Ω) > 0.Proof. Let (uk)k∈N ⊂ Hα20 (Ω)\{0} be a minimizing sequence for µγ,s,α,a(Ω),that is,JΩa (uk) = µγ,s,α,a(Ω) + o(1) as k →∞.Up to multiplying by a constant, we may assume that∫Ω|uk|2∗α(s)|x|s dx = 1 (4.23)Cn,α2∫Rn∫Rn|uk(x)− uk(y)|2|x− y|n+α dxdy−∫Ω(γ|x|α + a)u2kdx = µγ,s,α,λ(Ω)+o(1)(4.24)as k → +∞. By (4.23), we have∫Ωu2kdx ≤ C < ∞ for all k. Since 0 ≤γ < γH(α), the fractional Hardy inequality combined with (4.24) yields that‖uk‖Hα20 (Ω)≤ C for all k. It then follows that there exists u ∈ Hα20 (Ω) suchthat, up to a subsequence, such that (uk) goes to u weakly in Hα20 (Ω) andstrongly in L2(Ω) as k →∞.We first show that∫Ω|u|2∗α(s)|x|s dx = 1. Define θk = uk − u for all k ∈ N. Itfollows from the boundedness in Hα20 (Ω) that, up to a subsequence, we havethat θk ⇀ 0 weakly in Hα20 (Ω), strongly in L2(Ω) as k →∞, and θk(x)→ 0for a.e. x ∈ Ω as k → +∞. Hence, by the Brezis-Lieb lemma (see [10] and[70]), we get that∫Rn∫Rn|uk(x)− uk(y)|2|x− y|n+α dxdy =∫Rn∫Rn|θk(x)− θk(y)|2|x− y|n+α dxdy+∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy + o(1),1 =∫Ω|uk|2∗α(s)|x|s dx =∫Ω|θk|2∗α(s)|x|s dx+∫Ω|u|2∗α(s)|x|s dx+ o(1),664.4. Analytic conditions for the existence of extremals∫Ωu2k|x|αdx =∫Ωθ2k|x|αdx+∫Ωu2|x|αdx+ o(1),and ∫Ωu2kdx =∫Ωu2dx+ o(1), as k →∞.Thus, we haveµγ,s,α,a(Ω) =[Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy −∫Ω(γ|x|α + a)u2dx]+[Cn,α2∫Rn∫Rn|θk(x)− θk(y)|2|x− y|n+α dxdy − γ∫Ωθ2k|x|αdx]+ o(1)(4.25)as k → +∞. The definition of µγ,s,α,a(Ω) and Hα20 (Ω) ⊂ Hα20 (Rn) yieldµγ,s,α,a(Ω)(∫Ω|u|2∗α(s)|x|s dx) 22∗α(s)≤ Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy−∫Ω(γ|x|α + a)u2dx,andµγ,s,α(Rn)(∫Ω|θk|2∗α(s)|x|s dx) 22∗α(s)≤ Cn,α2∫Rn∫Rn|θk(x)− θk(y)|2|x− y|n+α dxdy− γ∫Ωθ2k|x|αdx.(4.26)Summing these two inequalities and using (4.23) and (4.25), and passing tothe limit k →∞, we obtainµγ,s,α(Rn)(1−∫Ω|u|2∗α(s)|x|s dx) 22∗α(s)≤ µγ,s,α,a(Ω)1− (∫Ω|u|2∗α(s)|x|s dx) 22∗α(s) .Finally, the fact that µγ,s,α,a(Ω) < µγ,s,α(Rn) implies that∫Ω|u|2∗α(s)|x|s dx = 1.It remains to show that u is an extremal for µγ,s,α,a(Ω). For that, note thatsince∫Ω|u|2∗α(s)|x|s dx = 1, the definition of µγ,s,α,a(Ω) yields thatµγ,s,α,a(Ω) ≤ Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy −∫Ω(γ|x|α + a)u2dx.674.5. The fractional Hardy singular interior mass of a domain in the critical caseThe second term in the right-hand-side of (4.25) is nonnegative due to (4.26).Therefore, we get thatµγ,s,α,a(Ω) =Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy −∫Ω(γ|x|α + a)u2dx.This proves the first claim of the Proposition.Now assume that λ ∈ (0, λ1(Lγ,α)) and 0 ≤ γ < γH(α), then for all u ∈Hα20 (Ω) \ {0},JΩλ (u) =Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy −∫Ω(γ|x|α + λ)u2dx(∫Ω|u|2∗α(s)|x|s dx) 22∗α(s)≥(1− λλ1(Lγ,α)) Cn,α2∫Rn∫Rn|u(x)−u(y)|2|x−y|n+α dxdy − γ∫Ωu2|x|αdx(∫Ω|u|2∗α(s)|x|s dx) 22∗α(s)≥(1− λλ1(Lγ,α))(1− γγH(α))µ0,s,α,0(Ω)=(1− λλ1(Lγ,α))(1− γγH(α))µ0,s,α,0(Rn) > 0.Therefore, µγ,s,α,λ(Ω) > 0.4.5 The fractional Hardy singular interior massof a domain in the critical caseIn this section, we define the fractional Hardy singular interior mass of adomain by proving Theorem 4.3. We shall need the following five lemmae.Lemma 4.11. Assume 0 < β ≤ n − α, and let η ∈ C∞c (Ω) be a cut-offfunction such that 0 ≤ η(x) ≤ 1 in Ω, and η(x) ≡ 1 in Bδ(0), for someδ > 0 small. Then x 7→ η(x)|x|−β ∈ Hα20 (Ω) and there exists fβ ∈ L∞loc(Rn)with fβ(x) ≥ 0 on Bδ(0) and fβ ∈ C1(Bδ(0)) such that(−∆)α2 (η|x|−β) = Φn,α(β)|x|−αη|x|−β + fβ in D′(Ω \ {0}), (4.27)684.5. The fractional Hardy singular interior mass of a domain in the critical casein the sense that, if vβ(x) := η(x)|x|−β, then for all ϕ ∈ C∞c (Ω \ {0}),Cn,α2∫Rn∫Rn(vβ(x)− vβ(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy = Φn,α(β)∫Ωvβϕ|x|αdx+∫Ωfβϕ(x) dx.Moreover, if β < n−α2 , then vβ ∈ Hα20 (Ω) and equality (4.27) holds in theclassical sense of Hα20 (Ω).Proof. When β < n−α2 , it follows from Proposition 4.7 that x 7→ η(x)|x|−β ∈Hα20 (Ω). In the general case, for ϕ ∈ C∞c (Ω \ {0}), straightforward compu-tations yieldCn,α2∫Rn∫Rn(vβ(x)− vβ(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy = 〈(−∆)α2 |x|−β , ηϕ〉+∫Ωfβϕdx,wherefβ(x) := Cn,α lim→0∫|x−y|>η(x)− η(y)|x− y|n+α ·1|y|β dy for all x ∈ Rn.Note that fβ ∈ L∞loc(Rn), and for x ∈ Bδ(0), we have thatfβ(x) := Cn,α∫Rn1− η(y)|x− y|n+α ·1|y|β dy ≥ 0,yielding that fβ ∈ C1(Bδ(0)). Since ϕ ≡ 0 around 0, the lemma is a conse-quence of (4.11).Lemma 4.12 (A comparison principle via coercivity). Suppose Ω be abounded smooth domain in Rn, 0 < α < 2, γ < γH(α) and a(x) ∈ C0,τ (Ω)for some τ ∈ (0, 1). Assume that the operator (−∆)α2 − ( γ|x|α + a(x)) iscoercive. Let u be a function in Hα20 (Rn) that satisfies (−∆)α2 u−(γ|x|α + a(x))u ≥ 0 in Ωu ≥ 0 on ∂Ω,in the sense that u ≥ 0 in Rn \ Ω andCn,α2∫Rn∫Rn(u(x)− u(y))(v(x)− v(y))|x− y|n+α dxdy−∫Ω(γ|x|α + a(x))uvdx ≥ 0,for all v ∈ Hα20 (Ω) with v ≥ 0 a.e. in Ω. Then, u ≥ 0 in Ω.694.5. The fractional Hardy singular interior mass of a domain in the critical caseProof. Let u−(x) = −min(u(x), 0) be the negative part of u. It follows fromProposition 4.7 that u− ∈ Hα20 (Ω). We can therefore use it as a test functionto get〈Lu, u−〉 : = Cn,α2∫Rn∫Rn(u(x)− u(y))(u−(x)− u−(y))|x− y|n+α dxdy− γ∫Ωuu−|x|α dx−∫Ωa(x)uu−dx ≥ 0.LetΩ+ := {x : u(x) ≥ 0} and Ω− := {x : u(x) < 0}.Straightforward computations yield0 ≤ −〈Lu−, u−〉 − Cn,α2∫Ω−∫Ω+(u(x)− u(y))u−(y)|x− y|n+α dxdy+Cn,α2∫Ω+∫Ω−(u(x)− u(y))u−(x)|x− y|n+α dxdy,which yields via coercivityc‖u−‖2Hα20 (Rn)≤ 〈Lu−, u−〉 ≤ 0.Thus, u− ≡ 0, and therefore, u ≥ 0 on Ω.Lemma 4.13. Assume that u ∈ Hα20 (Ω) is a weak solution of(−∆)α2 u−(γ +O(|x|τ )|x|α)u = 0 in Hα20 (Ω),for some τ > 0. If u 6≡ 0 and u ≥ 0, then there exists a constant C > 0 suchthatC−1 ≤ |x|β−(γ)u(x) ≤ C for x→ 0, x ∈ Ω.Proof. We use the weak Harnack inequality to prove the lower bound. In-deed, using Theorem 3.4 and Lemma 3.10 in [2], we get that there existsC1 > 0 such that for δ1 > 0 small enough,u(x) ≥ C1|x|−β−(γ) in Bδ1 .The other inequality goes as in the iterative scheme used to prove Proposi-tion 4.9.704.5. The fractional Hardy singular interior mass of a domain in the critical caseLemma 4.14 (See Fall-Felli [30]). Consider an open subset ω ⊂ Ω with0 ∈ ω, and a function h ∈ C1(ω) such that for some τ > 0,|h(x)|+ |x| · |∇h(x)| ≤ C|x|τ−α for all x ∈ ω \ {0}.Let u ∈ Hα20 (Ω) be a weak solution of(−∆)α2 u− γ|x|αu = h(x)u in ω ⊂ Ω,in the sense that for all ϕ ∈ C∞c (ω),Cn,α2∫Rn∫Rn(u(x)− u(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy−γ∫Ωuϕ|x|αdx =∫Ωh(x)uϕdx.Assume further that there exists C > 0 such thatC−1 ≤ |x|β−(γ)u(x) ≤ C for x→ 0, x ∈ Ω.Then, there exists l > 0 such thatlimx→0|x|β−(γ)u(x) = l.Proof. This result is an extension of Theorem 1.1 proved by Fall-Felli [30],who showed that under these conditions, one haslimτ→0|τx|n−α2 −√(n−α2 )2+µ u(τx) = ψ(0,x|x|)in C1loc(B1(0)) \ {0} (4.28)where µ ∈ R and ψ : Sn+1+ := {θ ∈ Sn+1 θ1 > 0} → R are respectively aneigenvalue and an eigenfunctions for the problem{ −div(θ1−α1 ∇ψ) = µθ1−α1 ψ in Sn+1+− limθ1→0 θ11−α∂νψ(θ1, θ′) = γkαψ(0, θ′) for θ′ ∈ ∂Sn+1+ ,(4.29)where kα is a positive constant defined in Section 2.3. We refer to [30] forthe explicit definition of this eigenvalue problem, in particular the relevantspaces used via the Caffarelli-Silvestre classical representation [12]. It thenfollows from the pointwise control (4.22) thatβ−(γ) :=n− α2−√(n− α2)2+ µ,714.5. The fractional Hardy singular interior mass of a domain in the critical caseand by Proposition 2.3 in Fall-Felli [30], that µ is the first eigenvalue of theeigenvalue problem (4.29). Then, using classical arguments, we get that thecorresponding eigenspace is one-dimensional and is spanned by any positiveeigenfunction of (4.29) (no matter the value of µ, it must necessarily be thefirst eigenvalue).We are left with proving that ψ(0, x|x|) is independant of x. In view of theremarks above, this amonts to prove the existence of a positive eigenfunctionthat is constant on the boundary.We now exhibit such an eigenfunction by following the argument in Propo-sition 2.3 in [30]. First, use ([29], Lemma 3.1) to obtain Γ ∈ C0([0,+∞) ×Rn) ∩ C2((0,+∞)× Rn) such that−div(t1−α∇Γ) = 0 in (0,+∞)× Rn− limt→0 t1−α∂νΓ(t, x) = kα γ|x|α for x ∈ Rn = ∂((0,+∞)× Rn)Γ(0, x) = |x|−β−(γ) for x ∈ Rn = ∂((0,+∞)× Rn).(4.30)Moreover, Γ is in the relevant function space, Γ > 0 and satisfiesΓ(z) = |z|−β−(γ)Γ(z|z|)for all z ∈ (0,+∞)× Rnwhere |z| = √t2 + |x|2 if z = (t, x). In particular, we have that Γ(z) =|z|−β−(γ)ψ0(θ) for θ := z|z| and some ψ0 ∈ C0(Sn+1+ ) ∩ C2(Sn+1+ ). Following[30], we get that ψ0 is an eigenfunction for the problem (4.29). Moreover,ψ0 > 0. Therefore, ψ0 corresponds to the first eigenvalue and spans thecorresponding eigenspace. Finally, we remark that for θ ∈ ∂Sn+1+ , we havethatψ0(0, θ) = Γ(0, θ) = |θ|−β−(γ) = 1.Since the eigenspace is one-dimensional, there exists l ∈ R such that ψ =l · ψ0. Therefore ψ(0, x|x|) = l for all x ∈ B1(0) \ {0} ⊂ Rn. It then followsfrom (4.28) thatlimx→0|x|β−(γ)u(x) = l > 0,which complete the proof of Lemma 4.14.Proof of Theorem 4.3. We first prove the existence of a solution. For δ > 0small enough, let η ∈ C∞c (Ω) be a cut-off function as in Lemma 4.11 such724.5. The fractional Hardy singular interior mass of a domain in the critical casethat η(x) ≡ 1 in Bδ(0). Set β := β+(γ) ≤ n− α in (4.27) and definef(x) : = −((−∆)α2 −(γ|x|α + a(x)))(η|x|−β+(γ))= −fβ+(γ) +aη|x|β+(γ) in Ω \ {0},in the distribution sense. In particular, f ∈ C1(Bδ(0)\{0}) and there existsa positive constant C > 0 such that|f(x)|+ |x| · |∇f(x)| ≤ C|x|−β+(γ) for x 6= 0 close to 0. (4.31)In the sequel, we write β+ := β+(γ) and β− := β−(γ). Note that theassumption γ > γcrit(α) implies that β+ <n2 <n+α2 . Thus, using (4.31)and the fact that β+ <n+α2 , we get that f ∈ L2nn+α (Ω). Since L2nn+α (Ω) =(L2nn−α (Ω))′ ⊂ (H α20 (Ω))′ , there exists g ∈ H α20 (Ω) such that((−∆)α2 −(γ|x|α + a(x)))g = f weakly in Hα20 (Ω).SetH(x) :=η(x)|x|β+ + g(x) for all x ∈ Ω \ {0}. (4.32)Thanks to (4.27), H : Ω→ R is a solution to{(−∆)α2H −(γ|x|α + a(x))H = 0 in Ω \ {0}H = 0 in Rn \ Ω,(4.33)in the sense of Definition 4.6. The idea is to now write f as the difference oftwo positive C1 functions. The decomposition f = |f | − 2f− does not workhere since the resulting functions are not necessarily C1. To smooth out thefunctions x 7→ |x| and x 7→ x−, we considerϕ1(x) :=√1 + x2 and ϕ2(x) := ϕ1(x)− x for all x ∈ R.It is clear that ϕ1, ϕ2 ∈ C1(R) and there exists C > 0 such that0 ≤ ϕi(x) ≤ C(1 + |x|) , |ϕ′i(x)| ≤ C and x = ϕ1(x)− ϕ2(x), (4.34)for all x ∈ R and i = 1, 2. Define fi := ϕi ◦ f for i = 1, 2. In particular,f = f1 − f2. Let g1, g2 ∈ Hα20 (Ω) be solutions to(−∆)α2 gi −(γ|x|α + a(x))gi = fi weakly in Hα20 (Ω) (4.35)734.5. The fractional Hardy singular interior mass of a domain in the critical casefor i = 1, 2. Since f1, f2 ≥ 0, Lemma 4.12 yields g1, g2 ≥ 0. Also((−∆)α2 −(γ|x|α + a(x)))(g − (g1 − g2)) = f − (f1 − f2) = 0.It follows from coercivity that g = g1− g2. Assuming g1 6≡ 0, it follows fromLemma 3.10 in [2] that there exists K ′ > 0 such that g1(x) ≥ K ′|x|−β− inBδ(0) \ {0}.Since g1 ∈ Hα20 (Ω), it follows from (4.35) and Theorem 2.1 of Jin-Li-Xiong[46] that g1 ∈ C0,τloc (Ω \ {0}) for some τ > 0. Arguing as in the proof ofTheorem 4.1, we get that g1 ∈ C1(Ω \ {0}). Settingh(x) :=f1(x)g1(x)for x close to 0,we have that h ∈ C1(Bδ(0)). Now use (4.31) and (4.34) to get thatf1(x) ≤ C(1 + |f(x)|) ≤ C|x|−β+ = C|x|−β− |x|α−(β+−β−)|x|−α≤ K1|x|−α+(α−(β+−β−))g1(x).Using the fact that γ > γcrit(α) if and only if α − (β+ − β−) > 0, we getthat |h(x)| ≤ C|x|τ−α for x→ 0 where τ := α− (β+ − β−) > 0. Therefore,we have that(−∆)α2 g1 − γ +O(|x|(α−(β+−β−)))|x|α g1 = 0 weakly in Hα20 (Ω),with g1 ≥ 0 and g1 6≡ 0. It then follows from Lemma 4.13 that there existsc > 0 such that c−1 ≤ |x|β−g1(x) ≤ c for x ∈ Ω, x 6= 0 close to 0. Arguingas in Claim 2 in the proof of Theorem 4.1, we get that there exists C > 0such thatc−1 ≤ |x|β−g1(x) ≤ c and |x|β−+1|∇g1(x)| ≤ C for all x ∈ Bδ(0). (4.36)We now deal with the differential of h. With the controls (4.31), (4.34) and(4.36), we get that|x| · |∇h(x)| ≤ C|x|τ−α for x ∈ Bδ/2(0) \ {0}.Now, writing (−∆)α2 g1− γ|x|α g1 = h(x)g1 in Ω and using Lemma 4.14, we getthat |x|−β−g1(x) has a finite limit as x → 0. Note that this is also clearly744.6. Existence of extremalsthe case if g1 ≡ 0. The same holds for g2. Therefore, there exists a constantc ∈ R such that |x|−β−g(x)→ c as x→ 0. In other words,H(x) =1|x|β+ +c|x|β− + o(1|x|β−)as x→ 0,and there exists C > 0 such that |g(x)| ≤ C|x|−β− for all x ∈ Ω.We now prove that H > 0 in Ω \ {0}. Indeed, from the above asymptoticexpansion we have that H(x) > 0 for x → 0, x 6= 0. Since χH ∈ Hα20 (Rn)for all χ ∈ C∞c (Rn \{0}), it follows from Proposition 4.7 that H− ∈ Hα20 (Ω \B(0)) for some  > 0 small. We then test (4.33) against H−, and arguingas in the proof of Lemma 4.12 we get that H− ≡ 0, and then H ≥ 0. SinceH 6≡ 0, H ∈ C1(Ω\{0}), it follows from the Harnack inequality (see Lemma3.10 in [2]) that H > 0 in Ω \ {0}. This proves the existence of a solution uto Problem (4.9) with the relevant asymptotic behavior.We now deal with uniqueness. Assume that there exists another solutionu′ satisfying the hypothesis of Theorem 4.3. We define u¯ := u − u′. Thenu¯ : Ω→ R is a solution to{(−∆)α2 u¯−(γ|x|α + a(x))u¯ = 0 in Ω \ {0}u¯ = 0 in Rn \ Ω,in the sense of Definition 4.6. Since |u¯(x)| ≤ C|x|−β− for all x ∈ Ω whereC > 0 is some uniform constant, then by using Proposition 4.7 one concludesthat u¯ ∈ Hα20 (Ω) is a weak solution to(−∆)α2 u¯−(γ|x|α + a(x))u¯ = 0 in Ω,that is, for all ϕ ∈ Hα20 (Ω),Cn,α2∫(Rn)2(u¯(x)− u¯(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy−∫Rn(γ|x|α + a(x))u¯ϕ dx = 0.Taking ϕ := u¯ and using the coercivity, we get that u¯ ≡ 0, and then u ≡ u′,which yields the uniqueness.4.6 Existence of extremalsThis section is devoted to prove the main result, which is Theorem 4.4. Bychoosing a suitable test function, we estimate the functional JΩa (u), and754.6. Existence of extremalswe show that the condition µγ,s,α,a(Ω) < µγ,s,α(Rn) holds under suitableconditions on the dimension or on the mass of the domain. Recall thatProposition 4.10 implies that it is this strict inequality that guarantees theexistence of extremals for µγ,s,α,a(Ω).We fix a ∈ C0,τ (Ω), τ ∈ (0, 1) and η ∈ C∞c (Ω) such thatη ≡ 1 in Bδ(0) and η ≡ 0 in Rn \B2δ(0) with B4δ(0) ⊂ Ω. (4.37)Let U ∈ Hα20 (Rn) be an extremal for µγ,s,α,0(Rn). It follows from Theorem4.1 that, up to multipliying by a nonzero constant, U satisfies for someκ > 0,(−∆)α2 U − γ|x|αU = κU2∗α(s)−1|x|s weakly in Hα20 (Rn). (4.38)Moreover, U ∈ C1(Rn \ {0}), U > 0 andlim|x|→∞|x|β+U(x) = 1. (4.39)SetJΩa (u) :=Cn,α2∫(Rn)2(u(x)−u(y))2|x−y|n+α dxdy −∫Ω(γ|x|α + a)u2 dx(∫Ω|u|2∗α(s)|x|s dx) 22∗α(s)=A(u)B(u)22∗α(s),(4.40)whereA(u) := 〈u, u〉 −∫Ωau2 dx and B(u) :=∫Ω|u|2∗α(s)|x|s dx (4.41)with〈u, v〉 := Cn,α2∫(Rn)2(u(x)− u(y))(v(x)− v(y))|x− y|n+α dxdy (4.42)−∫Rnγ|x|αuv dx for u, v ∈ Hα20 (Rn).Consideru(x) := −n−α2 U(−1x) for x ∈ Rn \ {0}.It follows from Proposition 4.7 that ηu ∈ Hα20 (Ω).764.6. Existence of extremals4.6.1 General estimates for ηuWe define the following bilinear form Bη on Hα20 (Rn) as follows: For anyϕ,ψ ∈ Hα20 (Rn),Bη(ϕ,ψ) := 〈ηϕ, ψ〉 − 〈ϕ, ηψ〉=Cn,α2∫(Rn)2η(x)− η(y)|x− y|n+α (ϕ(y)ψ(x)− ϕ(x)ψ(y)) dxdy.(4.43)This expression makes sense since η ≡ 1 around 0 and η ≡ 0 around ∞.Note that〈ηu, ηu〉 = 〈u, η2u〉+ β+−β−Bη(uβ+−β−2,ηuβ+−β−2). (4.44)It follows from (4.38) and the definition of u that〈u, ϕ〉 = κ∫Rnu2∗α(s)−1|x|s ϕdx for all ϕ ∈ Hα20 (Rn).By a change of variable, we get as → 0,〈u, η2u〉 = κ∫Rnη2u2∗α(s)|x|s dx= κ∫Rnu2∗α(s)|x|s dx+O(∫Rn\Bδ(0)u2∗α(s)|x|s dx)= κ∫RnU2∗α(s)|x|s dx+O(∫Rn\B−1δ(0)U2∗α(s)|x|s dx).With (4.39), we get that〈u, η2u〉 = κ∫RnU2∗α(s)|x|s dx+O(2∗α(s)2(β+−β−))= κ∫RnU2∗α(s)|x|s dx+ o(β+−β−).(4.45)We now deal with the second term of (4.44). First note thatBη(uβ+−β−2,ηuβ+−β−2)=Cn,α2∫(Rn)2(η(x)− η(y))2|x− y|n+αu(x)β+−β−2· u(y)β+−β−2dxdy.774.6. Existence of extremalsIt follows from (4.39) and the pointwise control of Theorem 4.1 that thereexists C > 0 such that for any x ∈ Rn \ {0}, we have thatlim→0u(x)β+−β−2= S(x) :=1|x|β+ and∣∣∣∣∣ u(x)β+−β−2∣∣∣∣∣ ≤ C|x|β+ . (4.46)Since η(x) = 1 for all x ∈ Bδ(0) and β+(γ) < n, Lebesgue’s convergencetheorem yieldslim→0Bη(uβ+−β−2,ηuβ+−β−2)=Cn,α2∫(Rn)2(η(x)− η(y))2|x− y|n+α S(x)S(y) dxdy= Bη(S, ηS).(4.47)By plugging together (4.44), (4.45) and (4.47), we get as → 0,〈ηu, ηu〉 = κ∫RnU2∗α(s)|x|s dx+Bη(S, ηS)β+−β− + o(β+−β−). (4.48)Arguing as in the proof of (4.45), we obtain as → 0,∫Rn(ηu)2∗α(s)|x|s dx =∫RnU2∗α(s)|x|s dx+ o(β+−β−). (4.49)As an immediate consequence, we getProposition 4.15. Suppose that 0 ≤ s < α < n, 0 < α < 2 and 0 ≤ γ <γH(α). Then,µγ,s,α,0(Ω) = µγ,s,α(Rn).Proof. It follows from the definition of µγ,s,α(Ω) that µγ,s,α,0(Ω) ≥ µγ,s,α(Rn).We now show the reverse inequity. Using the estimates (4.48) and (4.49)above, we have as → 0,JΩ0 (U) =κ∫RnU2∗α(s)|x|s dx(∫RnU2∗α(s)|x|s dx) 22∗α(s)+O(β+−β−)= JRn0 (U) +O(β+−β−) = µγ,s,α(Rn) +O(β+−β−).Letting → 0 yields µγ,s,α,0(Ω) ≤ µγ,s,α(Rn) from which follows thatµγ,s,α,0(Ω) = µγ,s,α(Rn).784.6. Existence of extremals4.6.2 The test functions for the non-critical caseWe now estimate J(ηu) when 0 ≤ γ ≤ γcrit(α), that is in the case whenβ− ≥ n2 . Note that since β− + β+ = n− α, we have thatβ+ − β− > α when γ < γcrit(α) and β+ − β− = α if γ = γcrit(α). (4.50)We start with the following:Proposition 4.16. Let 0 ≤ s < α < 2, 0 ≤ γ ≤ γcrit(α) and n ≥ 2α. Then,as → 0,∫Ωa(ηu)2dx ={α(∫Rn U2dx)a(0) + o(α) if 0 ≤ γ < γcrit(α)ωn−1a(0)α ln(−1) + o(α ln ) if γ = γcrit(α).Proof of Proposition 4.16. We write∫Ωa(ηu)2dx =∫Bδau2dx+∫B2δ\Bδa(ηu)2dx= α∫B−1δa(x)U2dx+O(β+−β−).Assume that γ < γcrit(α). Since β+ >n2 and U ∈ C1(Rn\{0}) satisfies (4.6),we get that U ∈ L2(Rn) and therefore, Lebesgue’s convergence theorem andthe assumption β+(γ)− β−(γ) > α yield∫Ωa(ηu)2 dx = α(∫RnU2dx)a(0) + o(α) as → 0.If now γ = γcrit(α), then lim|x|→∞ |x|n2U(x) = 1 and β+−β− = α. Therefore∫Ω(ηu)2dx = ωn−1a(0)α ln(−1) + o(α ln ) as → 0.This proves Proposition 4.16.Plugging together (4.48), (4.49) and Proposition 4.16 then yields, as → 0,JΩa (ηu) =κ∫RnU2∗α(s)|x|s dx(∫RnU2∗α(s)|x|s dx) 22∗α(s)− a(0)∫Rn U2 dx(∫RnU2∗α(s)|x|s dx) 22∗α(s)α + o(α)= JRn0 (U)− a(0)∫Rn U2 dx(∫RnU2∗α(s)|x|s dx) 22∗α(s)α + o(α), (4.51)794.6. Existence of extremalswhen γ < γcrit(α), andJΩa (ηu) = JRn0 (U)− a(0)ωn−1(∫RnU2∗α(s)|x|s dx) 22∗α(s)α ln1+ o(α ln1), (4.52)when γ = γcrit(α).4.6.3 The test function for the critical caseHere, we assume that γ > γcrit(α). It follows from Theorem 4.3 that thereexists H : Ω \ {0} → R such thatH ∈ C1(Ω \ {0}) , ξH ∈ Hα20 (Ω) for all ξ ∈ C∞c (Rn \ {0}),(−∆)α2H −(γ|x|α + a))H = 0 weakly in Ω \ {0}H > 0 in Ω \ {0}H = 0 in ∂Ωand limx→0 |x|β+H(x) = 1.Here the solution is in the sense of Definition 4.6. In other words, the secondidentity means that for any ϕ ∈ C∞c (Ω \ {0}), we have thatCn,α2∫(Rn)2(H(x)−H(y))(ϕ(x)− ϕ(y))|x− y|n+α dxdy−∫Rn(γ|x|α + a)Hϕdx = 0.(4.53)Note that this latest identity makes sense since H ∈ L1(Ω) (since β+ < n).Let now η be as in (4.37). Following the construction of the singular functionH in (4.32), there exists g ∈ Hα20 (Ω) such thatH(x) :=η(x)|x|β+ + g(x) for x ∈ Ω \ {0},where(−∆)α2 g −(γ|x|α + a)g = f, (4.54)with f ∈ L∞(Ω) and f ∈ C1(Bδ(0) \ {0}). It follows from (4.31) that thereexists c > 0 such that|f(x)| ≤ C|x|β+ for x ∈ Ω \ {0} and |∇f(x)| ≤C|x|β++1 for all x ∈ Bδ/2(0) \ {0}.(4.55)804.6. Existence of extremalsWe also have thatg(x) =mαγ,a(Ω)|x|β− +o(1|x|β−)as x→ 0, and |g(x)| ≤ C|x|−β− for all x ∈ Ω.(4.56)Define the test function asT(x) = ηu(x) + β+−β−2 g(x) for all x ∈ Ω \ {0},whereu(x) := −n−α2 U(−1x) for x ∈ Rn \ {0},and U ∈ Hα20 (Rn) is such that U > 0, U ∈ C1(Rn \ {0}) and satisfies (4.38)above for some κ > 0 and also (4.39). It is easy to see that T ∈ Hα20 (Ω) forall  > 0.This subsection is devoted to computing the expansion of JΩa (T) whereJΩa is defined in (4.40), (4.41) and (4.42). For simplicity, we set S(x) :=1|x|β+for x ∈ Rn \ {0}. In particular, it follows from (4.11) that we have that(−∆)α2 S − γ|x|αS = 0 weakly in Rn \ {0}, (4.57)in the sense that 〈S, ϕ〉 = 0 for all ϕ ∈ C∞c (Rn \ {0}). First note thatlim→0Tβ+−β−2= H in L∞loc(Ω \ {0}).Therefore, since |−β+−β−2 T(x)| ≤ C|x|−β+ for x ∈ Ω \ {0} with 2β+ < n,Lebesgue’s theorem yields as → 0,∫ΩaT 2 dx = β+−β−∫ΩaH2 dx+ o(β+−β−),Since T = ηu + β+−β−2 g, we have thatA(T) = 〈T, T〉 − β+−β−∫ΩaH2 dx+ o(β+−β−)= 〈ηu, ηu〉+ 2β+−β−2 〈ηu, g〉+ β+−β−〈g, g〉− β+−β−∫ΩaH2 dx+ o(β+−β−)814.6. Existence of extremalsWe are now going to estimate these terms separately. First, Formula (4.43)and (4.48) yield, as → 0A(T) = κ∫RnU2∗α(s)|x|s dx+ 2β+−β−2 〈u, ηg〉+ β+−β−M + o(β+−β−),(4.58)whereM := Bη (S, ηS) + 2Bη(uβ+−β−2, g)+ 〈g, g〉 −∫ΩaH2 dx.As to the second term of (4.58), we have〈u, ηg〉 = κ∫Rnu2∗α(s)−1 ηg|x|s dx.We set θ :=∫Rnu2∗α(s)−1 ηg|x|s dx. It is easy to check that, since ηg ∈ Hα20 (Ω),(u) is bounded in Hα20 (Rn) and goes to 0 weakly as → 0, we have thatlim→0θ = 0. (4.59)Therefore we can rewrite (4.58) asA(T) = κ∫RnU2∗α(s)|x|s dx+ 2κβ+−β−2 θ + β+−β−M + o(β+−β−)(4.60)as → 0.We now estimate M. First, we writeBη(uβ+−β−2, g)=Cn,α2∫(Rn)2F(x, y) dxdy,whereF(x, y) :=η(x)− η(y)|x− y|n+α(uβ+−β−2(y)g(x)− uβ+−β−2(x)g(y)).Remembering that η ≡ 1 in Bδ(0) and η ≡ 0 in B2δ(0)c and using (4.56),we get that∣∣F(x, y)1|x|<δ/2∣∣ ≤ C1|x|<δ/21|y|>δ|x|−β+ |y|−(n+α+β−) ∈ L1((Rn)2).824.6. Existence of extremalsSimilarly, we have a bound on F on {|x| > 3δ}. By symmetry, this yieldsalso a bound on {|y| < δ/2} ∪ {|y| > 3δ}. We are then left with getting abound on A := [B3δ(0) \Bδ/2(0)]2.For (x, y) ∈ A, we have that|F(x, y)| ≤ C |x− y||x− y|n+α ·(uβ+−β−2(y)− uβ+−β−2(x))g(x)+ C|x− y||x− y|n+α ·(uβ+−β−2(x)(g(x)− g(y)))≤ C|x− y|1−α−n(∣∣∣∣∣ uβ+−β−2 (y)− uβ+−β−2 (x)∣∣∣∣∣+ |g(x)− g(y)|).As noticed in the proof of Theorem 4.3, it follows from elliptic theory thatg ∈ C1(Ω \ {0}). Therefore, there exists C > 0 such that |g(x) − g(y)| ≤C|x− y| for all (x, y) ∈ A.Setting u˜ := −β+−β−2 u, it follows from (4.38) that(−∆)α2 u˜ − γ|x|α u˜ = κ2∗α(s)−22(β+−β−) u˜2∗α(s)−1|x|s weakly in Hα20 (Rn).It then follows from (4.46) and arguments similar to the Proof of Theorem4.1 (see Remark 2.5 and Theorem 2.1 of Jian-Li-Xiong [46]) that (u˜) isbounded in C1loc(Rn \ {0}). Therefore, there exists C > 0 such that |u˜(x)−u˜(y)| ≤ C|x− y| for all (x, y) ∈ A. Then, we get|F(x, y)| ≤ C|x− y|2−α−n ∈ L1(A).Therefore, (F) is uniformly dominated on (Rn)2. Noting that uβ+−β−2(x)→S(x) as → 0 for all x ∈ Rn \ {0}, Lebesgue’s theorem yieldslim→0Bη(uβ+−β−2, g)= Bη (S, g) . (4.61)Here again, note that Bη(S, g) makes sense. Therefore, we get that M =M + o(1) as → 0 whereM := Bη(S, ηS) + 2Bη (S, g) + 〈g, g〉 −∫ΩaH2 dx. (4.62)834.6. Existence of extremalsWe now estimate B(T). Note first that since p > 2, there exists C(p) > 0such that∣∣|x+ y|p − |x|p − p|x|p−2xy∣∣ ≤ C(p) (|x|p−2y2 + |y|p) for all x, y ∈ R.We therefore get thatB(T) =∫Rn∣∣∣ηu + β+−β−2 g∣∣∣2∗α(s)|x|s dx=∫Rn(ηu)2∗α(s)|x|s dx+ 2∗α(s)β+−β−2∫Rnu2∗α(s)−1 η2∗α(s)−1g|x|s dx+O(β+−β−∫Rnu2∗α(s)−2 η2∗α(s)−2g2|x|s dx+ 2∗α(s)2(β+−β−)∫Rn|g|2∗α(s)|x|s dx).Since η ≡ 1 around 0, we get that∫Rnu2∗α(s)−1 η2∗α(s)−1g|x|s dx =∫Rnu2∗α(s)−1 ηg|x|s dx+O(∫Ω\Bδ(0)u2∗α(s)−1 g|x|s dx)= θ + o(β+−β−2),as → 0. Therefore, in view of (4.49), we deduce thatB(T) =∫RnU2∗α(s)|x|s dx+ 2∗α(s)β+−β−2 θ + o(β+−β−), (4.63)as → 0. Plugging (4.60), (4.59) and (4.63) into (4.40), we get thatJΩa (T) =κ∫RnU2∗α(s)|x|s dx(∫RnU2∗α(s)|x|s dx) 22∗α(s)1 + Mκ∫RnU2∗α(s)|x|s dxβ+−β− + o(β+−β−)= JRn0 (U)1 + Mκ∫RnU2∗α(s)|x|s dxβ+−β− + o(β+−β−) (4.64)as → 0, where M is defined in (4.62) and JRn0 is as in (4.40).We now express M in term of the mass. Note that in the classical(pointwise) setting, an integration by parts yield that Bη(ϕ,ψ) defined in(4.43) is an integral on the boundary of a domain. Hence, the mass appearsby simply integrating by part independently the singular function H. The844.6. Existence of extremalscentral remark we make here is that the integral on the boundary on adomain (defined in the local setting) can be seen as the limit of an integral onthe domain via multiplication by a cut-off function with support convergingto the boundary –which happened to be defined in the nonlocal setting.Therefore, despite the nonlocal aspect of our problem, we shall be able toapply the same strategy as in the local setting.We shall be performing the following computations in the same order asthe ones above made to get A(T). The constant M will therefore appearnaturally in the two settings.Let χ ∈ C∞(Rn) such that χ ≡ 0 in B1(0) and χ ≡ 1 in Rn \ B2(0). Fork ∈ N \ {0}, define χk(x) := χ(kx) for x ∈ Rn, so thatχk(x) = 0 for |x| < 1kand χk(x) = 1 for |x| > 2k.In particular, (χk)k is bounded in L∞(Rn) and χk(x) → 1 as k → +∞ fora.e. x ∈ Rn. Since χkH ∈ Hα20 (Ω), then by the very definition of H (see(4.53)), we have that0 = 〈H,χkH〉 −∫RnaHχkH dx= 〈ηS + g, χkηS + χkg〉 −∫RnχkaH2 dx= 〈ηS, χkηS〉+ 〈ηS, χkg〉+ 〈χkηS, g〉+ 〈g, χkg〉 −∫RnχkaH2 dx= 〈S, χkη2S〉+Bη(S, χkηS) + 〈S, ηχkg〉+Bη(S, χkg) + 〈S, χkηg〉+Bχkη(S, g) + 〈g, χkg〉 −∫RnχkaH2 dx.Since aH2 ∈ L1(Ω) (this is a consequence of 2β+ < n) and S is a solutionto (4.57), we get that0 = Bη(S, χkηS) +Bη(S, χkg) +Bχkη(S, g) + 〈g, χkg〉 −∫RnaH2 dx+ o(1),as k → +∞. We now estimate these terms separately.Our first claim is thatlimk→+∞〈χkg, g〉 = 〈g, g〉. (4.65)854.6. Existence of extremalsIndeed,‖χkg − g‖2Hα20 (Rn)=Cn,α2∫(Rn)2|(1− χk)(x)g(x)− (1− χk)(y)g(y)|2|x− y|n+α dxdy≤ Cn,α∫(Rn)2|1− χk(x)|2 |g(x)− g(y)|2|x− y|n+α dxdy+ Cn,α∫(Rn)2g(y)2|χk(x)− χk(y)|2|x− y|n+α dxdy.The first integral goes to 0 as k → +∞ with Lebesgue’s convergence theormesince g ∈ Hα20 (Rn). For the second term, we use the change of variableX = kx, Y = ky and the control of g(x) by |x|−β− . This proves that(χkg)→ g in Hα20 (Rn) as k → +∞. The claim follows and (4.65) is proved.We now writeBη(S, χkηS) =Cn,α2(∫(Rn)2χk(x)F˜ (x, y) dxdy +∫(Rn)2Gk(x, y) dxdy),whereF˜ (x, y) :=η(x)− η(y)|x− y|n+α (S(y)(ηS)(x)− S(x)(ηS)(y)) ,andGk(x, y) :=(η(x)− η(y))(χk(x)− χk(y))(ηS)(y)S(x)|x− y|n+α .As in the proof of (4.61) and (4.47), F˜ ∈ L1((Rn)2) and Lebesgue’s conver-gence theorem yieldslimk→+∞Cn,α2∫(Rn)2χk(x)F˜ (x, y) dxdy = Bη(S, ηS).Arguing as in the proof of (4.61), we get the existence of G ∈ L1((Rn)2)such that |Gk(x, y)| ≤ G(x, y) for all (x, y) ∈ (Rn)2 such that |x| < δ/2or |x| > 3δ. By symmetry, a similar control also holds for (x, y) ∈ (Rn)2such that |y| < δ/2 or |y| > 3δ. Moreover, for δ > 0 small enough, wehave that Gk(x, y) = 0 for (x, y) ∈ (Rn)2 such that |x| > δ/2 and |y| > δ/2(this is due to the definition of χk). Therefore, since limk→+∞(χk(x) −χk(y)) = 0 for a.e. (x, y) ∈ (Rn)2, Lebesgue’s convergence theorem yields∫(Rn)2 Gk(x, y) dxdy → 0 as k → +∞. We can then conclude thatlimk→+∞Bη(S, χkηS) = Bη(S, ηS).864.6. Existence of extremalsSimilar arguments yieldlimk→+∞Bη(S, χkg) = Bη(S, g).Therefore, we get that0 = Bη(S, ηS) +Bη(S, g) +Bχkη(S, g) + 〈g, g〉 −∫RnaH2 dx+ o(1),as k → +∞. We also have thatBχkη(S, g) =Cn,α2∫(Rn)2χk(x)η(x)− η(y)|x− y|n+α (S(y)g(x)− S(x)g(y)) dxdy+Cn,α2∫(Rn)2η(y)χk(x)− χk(y)|x− y|n+α (S(y)g(x)− S(x)g(y)) dxdy.As above, the first integral of the right-hand-side goes to Bη(S, g) as k →+∞. We now deal with the second integral. Using that β+ + β− = n − α,the change of variables X = kx and Y = ky yield∫(Rn)2η(y)χk(x)− χk(y)|x− y|n+α (S(y)g(x)− S(x)g(y)) dxdy =∫(Rn)2Fk(X,Y )dXdY,whereFk(X,Y ) := η(Yk)χ(X)− χ(Y )|X − Y |n+α(1|Y |β+ g(Xk)k−β− − 1|X|β+ g(Yk)k−β−).Note that there exists C > 0 such that |g(x)| ≤ C|x|−β− for all x ∈ Ω\{0}.Since χ(X) = 0 for |X| < 1 and χ(X) = 1 for |X| > 2, arguing as in theproof of (4.61), we get that |Fk(X,Y )| is uniformly bounded from above bya function in L1((Rn)2) for (X,Y ) ∈ (Rn)2 such that X 6∈ B3(0) \ B1/2(0)or Y 6∈ B3(0) \B1/2(0).There exists C > 0 such that |η(X) − η(Y )| ≤ C|X − Y | for all (X,Y ) ∈[B3(0) \B1/2(0)]2. Therefore, for such (X,Y ), we have that|Fk(X,Y )| ≤ C|X − Y |1−α−n∣∣∣∣( 1|Y |β+ − 1|X|β+)g(Xk)k−β−∣∣∣∣+C|X − Y |1−α−n 1|X|β+∣∣∣∣g(Xk)k−β− − g(Yk)k−β−∣∣∣∣≤ C|X − Y |2−α−n+C|X − Y |1−α−n∣∣∣∣g(Xk)k−β− − g(Yk)k−β−∣∣∣∣ .874.6. Existence of extremalsDefine gk(X) := g(Xk)k−β− for X ∈ kΩ. It follows from (4.54) and (4.55)that(−∆)α2 gk −(γ|X|α + k−αa(k−1X))gk = fk weakly in Hα/20 (kΩ),wherefk(X) := k−β−−αf(k−1X) so that |fk(X)| ≤ Ck−(α−(β+−β−))|X|−β+ ,for all X ∈ kΩ. Here again, elliptic regularity yields that (gk) is bounded inC1loc(Rn \ {0}). Therefore, there exists C > 0 such that|gk(X)− gk(Y )| ≤ C|X − Y |,for all (X,Y ) ∈ [B3(0) \B1/2(0)]2. Therefore, we get that|Fk(X,Y )| ≤ C|X − Y |2−α−n for all (X,Y ) ∈ [B3(0) \B1/2(0)]2.Therefore, since α < 2, (Fk) is also dominated on this domain, and then on(Rn)2. Finally, it follows from the definition (4.56) of the mass thatlimk→+∞Fk(X,Y ) = mαγ,a(Ω)χ(X)− χ(Y )|X − Y |n+α(1|Y |β+ |X|β− −1|X|β+ |Y |β−),for a.e. (X,Y ) ∈ (Rn)2. Therefore, Lebesgue’s convergence theorem yields0 = Bη(S, ηS) + 2Bη(S, g) +K ·mαγ,a(Ω) + 〈g, g〉 −∫RnaH2 dx,whereK :=Cn,α2∫(Rn)2χ(X)− χ(Y )|X − Y |n+α(1|Y |β+ |X|β− −1|X|β+ |Y |β−)dXdY.Without loss of generality, we can assume that χ is radially symetrical andnondecreasing. Therefore, we get that K > 0. With (4.62), we then get thatM = −K ·mαγ,a(Ω) with K > 0.Plugging this identity in (4.64) yieldsJΩa (T) = JRn0 (U)1− Kκ∫RnU2∗α(s)|x|s dx·mαγ,a(Ω)β+−β− + o(β+−β−) .(4.66)4.6.4 Proof of Theorem 4.4Theorem 4.4 is now a direct consequence of (4.51), (4.52), (4.66) and Propo-sition 4.10.88Chapter 5Existence Results forNon-linearly PerturbedFractionalHardy-Schro¨dinger Problems5.1 IntroductionThroughout this chapter, we shall use the following notations:β+ := β+(γ), β− := β−(γ) and mαγ,λ := mαγ,λ(Ω).Recall that the mass mαγ,λ(Ω) is defined in Theorem 1.5, and parametersβ+(γ) and β−(γ) are introduced in Section 4.2; see also Remark 1.4.We consider the following perturbed problem associated with the oper-ator Lγ,α on bounded domains Ω ⊂ Rn with 0 ∈ Ω:(−∆)α2 u− γ u|x|α − λu =u2∗α(s)−1|x|s + huq−1 in Ωu ≥ 0 in Ω,u = 0 in Rn \ Ω,(5.1)where 0 ≤ s < α < 2, n > α, 2∗α(s) := 2(n−s)n−α , and λ, γ ∈ R. We also assumethat h ∈ C0(Ω), h ≥ 0, and q ∈ (2, 2∗α) with 2∗α := 2∗α(0).When (h ≡ 0), Problem (5.1) has been studied in both the local and non-local setting. See [17, 36, 37] and [38] and the references therein. In [44],Jaber considered the local version of the problem in the Riemannian contextbut in the absence of the Hardy term (i.e., γ = 0). A similar problem, withthe second order operator replaced by the fourth order Paneitz operator,was studied by Esposito-Robert [26] (see also Djadli- Hebey-Ledoux [23]).895.1. IntroductionBy using ideas from [38] and [44], we now investigate the role of thelinear perturbation (i.e., λu), the non-linear perturbation (i.e., huq−1), aswell as the geometry of the domain (the mass mαγ,λ) on the existence of apositive solution of (5.1).As in Jaber [44], our main tool here to investigate the existence of so-lutions is the Mountain Pass Lemma of Ambrosetti- Rabinowitz [3] (seeLemma 3.6). We shall use the extremal of µγ,s,α(Rn) (defined in (1.2)) andits profile at zero and infinity to build appropriate test-functions for thefunctional under study.Our analysis shows that the existence of a solution for problem (5.1)depends only on the non-linear perturbation when the operator Lγ,α is non-critical (i.e., 0 ≤ γ ≤ γcrit(α)). The critical case ( i.e., γcrit(α) < γ < γH(α))is more complicated and depends on other conditions involving both theperturbation and the global geometry of the domain. More precisely, when0 ≤ γ ≤ γcrit(α), the competition is between the linear and non-linearperturbations, and since q > 2, the non-linear term dominates. In the criticalcase, this competition is more challenging as it is between the mass and thenon-linear perturbation. In this situation, there exists a threshold qcrit ∈(2, 2∗α), where the dominant factor switches from the non-linear perturbationto the mass. The transition at 2∗α is most interesting. We shall establish thefollowing result.Theorem 5.1 (Shakerian [66]). Let Ω be a smooth bounded domain inRn(n > α) such that 0 ∈ Ω, and let 2∗α(s) := 2(n−s)n−α , 0 ≤ s < α, −∞ <λ < λ1(Lγ,α), and 0 ≤ γ < γH(α). We consider 2 < q < 2∗α, h ∈ C0(Ω) andh ≥ 0. Then, there exists a positive solution u ∈ Hα20 (Ω) to (5.1) under oneof the following conditions:(1) 0 ≤ γ ≤ γcrit(α) and h(0) > 0.(2) γcrit(α) < γ < γH(α) andh(0) > 0 if q > qcritc1h(0) + c2mαγ,λ > 0 if q = qcritmαγ,λ > 0 if q < qcrit,where c1, c2 are two positive constant and can be computed explicitly (seeSection 5.4), and qcrit = 2∗α − 2β+−β−n−α ∈ (2, 2∗α).Remark 5.2. When the operator Lγ,α is non-critical, our condition de-pends only on the non-linear perturbation h (i.e., h(0) > 0), and not on thepositivity of λ, which was the case in the non-perturbed case.905.2. The Palais-Smale condition below a critical thresholdFor γ < γH(α), the same argument as in the beginning of Section 3.2coupled with the fractional Hardy inequality yield that‖|u‖| =(Cn,α2∫Rn∫Rn|u(x)− u(y)|2|x− y|n+α dxdy − γ∫Ωu2|x|α dx) 12is a well defined norm on Hα20 (Ω), and is equivalent to the norm ‖u‖H α20 (Ω).We shall consider the following functional Φ : Hα20 (Ω) → R whose criticalpoints are solutions for (5.1): For u ∈ Hα20 (Ω), letΦ(u) =12‖|u‖|2 − λ2∫Ωu2dx− 12∗α(s)∫Ωu2∗α(s)+|x|s dx−1q∫Ωhuq+dx,where u+ = max(0, u) is the non-negative part of u. Note that any criticalpoint of the functional Φ(u) is essentially a variational solution of (5.1).Indeed, we have for any v ∈ Hα20 (Ω),〈Φ′(u), v〉 = Cn,α2∫Rn∫Rn(u(x)− u(y))(v(x)− v(y))|x− y|n+α dxdy−∫Rn(γu|x|α + λu+u2∗α(s)−1+|x|s + huq−1+ )v dx.5.2 The Palais-Smale condition below a criticalthresholdIn this section, we prove the followingProposition 5.3. If c < α−s2(n−s)µγ,s,α(Rn)n−sα−s , then every Palais-Smale se-quence (uk)k∈N for Φ at level c has a convergent subsequence in Hα20 (Ω).Proof of Proposition 5.3. Assume c < α−s2(n−s)µγ,s,α(Rn)n−sα−s and let(uk)k∈N ∈Hα20 (Ω) be a Palais-Smale sequence for Φ at level c, that is Φ(uk)→ c andΦ′(uk)→ 0 in (Hα20 (Ω))′, (5.2)where (Hα20 (Ω))′ denotes the dual of Hα20 (Ω).We first prove that (uk)k∈N is bounded in Hα20 (Ω). One can use uk ∈ Hα20 (Ω)915.2. The Palais-Smale condition below a critical thresholdas a test function in (5.2) to get that‖|uk‖|2−λ∫Ωu2kdx =∫Ω(uk)2∗α(s)+|x|s dx+∫Ωh(uk)q+dx+o(‖|uk‖|) as k →∞.(5.3)On the other hand, from the definition of Φ, we deduce that‖|uk‖|2 − λ∫Ωu2kdx = 2Φ(uk) +22∗α(s)∫Ω(uk)2∗α(s)+|x|s +2q∫Ωh(uk)q+dx. (5.4)It follows from the last two identities that as k →∞,2Φq(uk) =(1− 22∗α(s))∫Ω(uk)2∗α(s)+|x|s dx+(1− 2q)∫Ωh(uk)q+dx+ o(‖|uk‖|).(5.5)This coupled with the Palais-Smale condition Φ(uk)→ c, and the fact thath ≥ 0 yield2c =(1− 22∗α(s))∫Ω(uk)2∗α(s)+|x|s dx+(1− 2q)∫Ωh(uk)q+dx+ o(1)≥(1− 22∗α(s))∫Ω(uk)2∗α(s)+|x|s dx+ o(1) as k →∞.Thus, (1− 2q)∫Ωh(uk)q+dx = O(1) as k →∞.We finally obtain‖|uk‖|2 − λ∫Ωu2kdx ≤ O(1) + o(‖|uk‖|) as k →∞.Using that λ < λ1(Lγ,α) and γ < γH(α), we get that0 <(1− γγH(α))(1− λλ1(Lγ,α))‖uk‖2Hα20 (Ω)≤(1− λλ1(Lγ,α))‖|uk‖|2≤ ‖|uk‖|2 − λ∫Ωu2kdx ≤ O(1) + o(‖|uk‖|) as k →∞.925.2. The Palais-Smale condition below a critical thresholdWe then deduce that (uk)k∈N is bounded in Hα20 (Ω), which implies that thereexists u ∈ Hα20 (Ω) such that, up to a subsequence,(1) uk ⇀ u weakly in Hα20 (Ω).(2) uk → u strongly in Lp1(Ω) for all p1 ∈ [2, 2∗α).(3) uk → u strongly in Lp2(Ω, |x|−sdx) for all p2 ∈ [2, 2∗α(s)).(5.6)We now claim that, up to a subsequence, we have‖|uk − u‖|2 =∫Ω(uk − u)2∗α(s)+|x|s dx+ o(1) as k →∞, (5.7)andα− s2(n− s)‖|uk − u‖|2 ≤ c+ o(1) as k →∞. (5.8)Indeed, straightforward computations yieldo(1) = 〈Φ′(uk)− Φ′(u), uk − u〉= ‖|uk − u‖|2 − λ∫Ω(uk − u)2−∫Ω(uk − u)((uk)2∗α(s)−1+ − u2∗α(s)−1+)|x|s dx+∫Ωh(uk − u)[(uk)q−1+ − uq−1+]dx as k →∞.(5.9)We first write∫Ω(uk − u)((uk)2∗α(s)−1+ − u2∗α(s)−1+)|x|s dx =∫Ω(uk)2∗α(s)+|x|s dx−∫Ωuku2∗α(s)−1+|x|s dx−∫Ωu(uk)2∗α(s)−1+|x|s dx+∫Ωu2∗α(s)+|x|s dx.It now follows from integral theory thatlimk→∞∫Ωuku2∗α(s)−1+|x|s dx =∫Ωu2∗α(s)+|x|s dx = limk→∞∫Ωu(uk)2∗α(s)−1+|x|s dx.So, we get that∫Ω(uk − u)((uk)2∗α(s)−1+ − u2∗α(s)−1+)|x|s dx =∫Ω(uk)2∗α(s)+|x|s dx−∫Ωu2∗α(s)+|x|s dx.935.2. The Palais-Smale condition below a critical thresholdIn order to deal with the right hand side of the last identity, we use thefollowing basic inequality:∣∣∣(uk)2∗α(s)+ − u2∗α(s)+ − (uk − u)2∗α(s)+ ∣∣∣ ≤ c(u2∗α(s)−1+ |uk−u|+(uk−u)2∗α(s)−1+ |u|),for some constant c > 0. We multiply both sides of the above inequality by|x|−s and take integral over Ω, and then use (5.6) to get thatlimk→∞∫Ω(uk)2∗α(s)+ − (uk − u)2∗α(s)+|x|s dx =∫Ωu2∗α(s)+|x|s dx.We therefore have∫Ω(uk−u)((uk)2∗α(s)−1+ − u2∗α(s)−1+)|x|s dx =∫Ω(uk − u)2∗α(s)+|x|s dx+o(1) as k →∞.In addition, the embeddings (5.6) yield that∫Ω(uk − u)2 = o(1) as k →∞,and∫Ωh(uk−u)[(uk)q−1+ − uq−1+]dx =∫Ω(uk−u)q+dx+o(1) = o(1) as k →∞.Plugging back the last three estimates into (5.9) gives (5.7). On the otherhand, since u is a weak solution of (5.1) then Φ(u) ≥ 0, and since Φ(uk)→ cas k →∞, it follows that α−s2(n−s)‖uk−u‖2 ≤ c+ o(1). This proves the claim.We now show thatlimk→∞uk = u in Hα20 (Ω). (5.10)Indeed, test the inequality (1.5) on uk − u, and use (5.6) and Proposition4.15 to obtain that∫Ω(uk − u)2∗α(s)+|x|s dx ≤ µγ,s,α(Rn)−2∗α(s)2 ‖|uk − u‖|2∗α(s) + o(1). (5.11)Combining this with (5.7), we get‖|uk − u‖|2(1− µγ,s,α(Rn)−2∗α(s)2 ‖|uk − u‖|2∗α(s)−2 + o(1))≤ o(1).945.3. Mountain pass geometry and existence of a Palais-Smale sequenceIt then follows from the last inequality and (5.8) that(1− µγ,s,α(Rn)−2∗α(s)2(2(n− s)α− s c) 2∗α(s)−22+ o(1))‖|uk − u‖|2 ≤ o(1).Note that the assumption c < α−s2(n−s)µγ,s,α(Rn)n−sα−s implies that(2(n− s)α− s c) 2∗α(s)−22< µγ,s,α(Rn)2∗α(s)2 ,and therefore (1− µγ,s,α(Rn)−2∗α(s)2(2(n− s)α− s c) 2∗α(s)−22)> 0.Thus, ‖|uk − u‖| → 0 as k →∞, and this proves (5.10).Finally, we have that Φ(u) = c, since the functional is continuous on Hα20 (Ω).5.3 Mountain pass geometry and existence of aPalais-Smale sequenceProposition 5.4. For every w ∈ Hα20 (Ω) \ {0} with w ≥ 0, there exists anenergy level c, with0 < c ≤ supt≥0Φ(tw), (5.12)and a Palais-Smale sequence (uk)k for Φ at level c, that isΦ(uk)→ c and Φ′(uk)→ 0 in (Hα20 (Ω))′.Proof. We show that the functional Φ satisfies the hypotheses of the moun-tain pass lemma 3.6. It is standard to show that Φ ∈ C1(Hα20 (Ω)) and clearlyΦ(0) = 0, so that (a) of Lemma 3.6 is satisfied.For (b), we show that 0 is a strict local minimum. Indeed, by the defi-nition of λ1(Lγ,α) and µγ,s,α(Ω), we have thatλ1(Lγ,α)∫Ω|w|2dx ≤ ‖|w‖|2 and µγ,s,α(Rn)(∫Ω|w|2∗α(s)|x|s dx)22∗α(s) ≤ ‖|w‖|2.955.4. Proof of Theorem 5.1In addition, it follows from (5.6)2 that there exists a positive constant S > 0such thatS(∫Ωh|w|qdx) 2q ≤ ‖|w‖|2.Hence,Φ(w) ≥ 12‖|w‖|2 − 12λλ1(Lγ,α)‖|w‖|2 − 1qS−q2 ‖|w‖|q− 12∗α(s)µγ,s,α(Rn)−2∗α(s)2 ‖|w‖|2∗α(s)= ‖|w‖|2(12(1− λλ1(Lγ,α))− 1qS−q2 ‖|w‖|q−2− 12∗α(s)µγ,s,α(Rn)−2∗α(s)2 ‖|w‖|2∗α(s)−2).Since λ < λ1(Lγ,α), q ∈ (2, 2∗α) and s ∈ [0, α), we have that 1− λλ1(Lγ,α) > 0,q − 2 > 0 and 2∗α(s) − 2 > 0, respectively. Thus, we can find R > 0 suchthat Φ(w) ≥ ρ for all w ∈ Hα20 (Ω) with ‖w‖H α20 (Ω) = R.Regarding (c), we haveΦ(tw) =t22‖|w‖|2 − t2λ2∫Ωw2dx− t2∗α(s)2∗α(s)∫Ωw2∗α(s)+|x|s dx−tqq∫Ωhwq+dx,hence limt→∞Φ(tw) = −∞ for any w ∈ Hα20 (Ω) \ {0} with w+ 6≡ 0, whichmeans that there exists tw > 0 such that ‖tww‖Hα20 (Ω)> R and Φ(tw) < 0,for t ≥ tw. In other words,0 < ρ ≤ inf{Φ(w); ‖w‖Hα20 (Ω)= R} ≤ c = infγ∈Fsupt∈[0,1]Φ(γ(t)) ≤ supt≥0Φ(tw),where F is the class of all path γ ∈ C([0, 1];Hα20 (Ω)) with γ(0) = 0 andγ(1) = tww.The rest follows from the Ambrosetti-Rabinowitz lemma.5.4 Proof of Theorem 5.1This section is devoted to prove Theorem 5.1. Since Φ satisfies the Palais-Smale condition only up to level α−s2(n−s)µγ,s,α(Rn)n−sα−s , we need to check which965.4. Proof of Theorem 5.1conditions on γ, q, h and the mass of the domain Ω guarantee that thereexists a w ∈ Hα20 (Ω) such thatsupt≥0Φ(tw) <α− s2(n− s)µγ,s,α(Rn)n−sα−s .We shall use the test functions ηu(resp., T) constructed in Section 4.6 forthe case when the operator Lγ,α is non-critical (resp., critical) to obtain thegeneral condition of existence.One can summarize the definition and properties of the test-functions asfollows (see also Section 4.6):v ={U := ηu if 0 ≤ γ ≤ γcrit(α)T := U + β+−β−2 g(x) if γcrit(α) < γ < γH(α),(5.13)where the cut-off function η ∈ C∞0 (Ω) verifies (4.37) andu(x) := −n−α2 U(−1x) for x ∈ Rn \ {0},and U ∈ Hα20 (Rn) is such that U > 0, U ∈ C1(Rn \ {0}) and satisfies (4.38)for some κ > 0 and also (4.39). Also, the function g satisfiesg(x) =mαγ,λ|x|β− + o(1|x|β−)as x→ 0, and |g(x)| ≤ C|x|−β− for all x ∈ Ω.We refer the readers to Subsection 4.6.3 for the definition and properties ofg(x) in detail.We now prove the following proposition which plays a crucial role in theproof of Theorem 5.1.Proposition 5.5. There exists τl > 0 for l = 1, .., 5 such that1) If 0 ≤ γ ≤ γcrit(α), thensupt≥0Φ(tv) = Υ− τ1h(0)n−qn−α2 + o(n−qn−α2 ); (5.14)2) If γcrit(α) < γ < γH(α), thensupt≥0Φ(tv) = Υ +−τ2h(0)n−qn−α2 + o(n−qn−α2 ) if q > qcrit−(τ3h(0) + τ4mαγ,λ)β+−β− + o(β+−β−) if q = qcrit−τ5mαγ,λβ+−β− + o(β+−β−) if q < qcrit,(5.15)where Υ := α−s2(n−s)µγ,s,α(Rn)n−sα−s and qcrit := 2∗α − 2β+−β−n−α ∈ (2, 2∗α).975.4. Proof of Theorem 5.1Proof of Proposition 5.5. We expand Φ(tv) in the following way:Φ(tv) =t22I − t2∗α(s)2∗α(s)J − tqqK as → 0,whereI := ‖|v‖|2 − λ‖v‖2L2(Ω), J :=∫Ω|v|2∗α(s)|x|s dx and K :=∫Ωh|v|qdx.Here is a summary of the estimates obtained in Section 4.6 which will beused freely in this section:Let EU :=∫RnU2∗α(s)|x|s dx. There exist positive constant c1, c2, c3 such thatI =κEU − c1λα + o(α) if 0 ≤ γ < γcrit(α)κEU − c2λα ln −1 + o(α ln −1) if γ = γcrit(α)κEU − c3mαγ,λβ+−β− + 2κβ+−β−2 θ + o(β+−β−) if γcrit(α) < γ < γH(α),(5.16)as → 0, andJ ={EU + o(β+−β−) if 0 ≤ γ ≤ γcrit(α)EU + 2∗α(s)β+−β−2 θ + o(β+−β−) if γcrit(α) < γ < γH(α),(5.17)as  → 0. Here, θ :=∫Rnu2∗α(s)−1 ηg|x|s dx and we have lim→0θ = 0; see (4.59).We are then left with estimating K.Estimate for K : We will consider two following cases.Case 1: 0 ≤ γ ≤ γcrit(α). We split K into two integrals as followsK =∫Ωh|U|qdx =∫Bδh|U|qdx+∫Ω\Bδh|U|qdx.985.4. Proof of Theorem 5.1We start by estimating the first term:∫Bδh|U|qdx = qα−n2∫Bδh(x)|U(x)|qdx= n−qn−α2∫B δh(X)|U(X)|qdX= n−qn−α2 h(0)∫Rn|U(X)|qdX+ n−qn−α2∫Rn\B δh(X)|U(X)|qdX.Note that we used the change of variable x = X . From the asymptotic(4.39) and the fact that q > 2, it then follows that∫Bδh|U|qdx = n−qn−α2 h(0)∫Rn|U(X)|qdX +O(qβ+−β−2 )= n−qn−α2 h(0)∫Rn|U(X)|qdX + o(β+−β−).Following the same argument that we treat the second integral in the lastterm yields that∫Ω\Bδh|U|qdx = O(qβ+−β−2 ) = o(β+−β−).Therefore,∫Ωh|U|qdx = n−qn−α2 h(0)[∫Rn|U(X)|qdX]+ o(β+−β−). (5.18)It now follows from (4.50) thatβ+ − β− ≥ α if 0 ≤ γ ≤ γcrit(α).On the other hand, the condition 2 < q < 2∗α implies that0 < n− qn− α2< α.Combining the last two inequalities, we then get thatn− qn− α2< β+ − β−,995.4. Proof of Theorem 5.1and thereforeK = n−q n−α2 h(0)[∫Rn|U(X)|qdX]+ o(n−qn−α2 ), (5.19)when 0 ≤ γ ≤ γcrit(α).Case 2: γcrit(α) < γ < γH(α). In order to estimate K in the criticalcase, we need the following inequality: For q > 2, there exists C = C(q) > 0such that||X + Y |q − |X|q| − qXY |X|q−2| ≤ C(|X|q−2Y 2 + |Y |q) for all X,Y ∈ R.We writeK =∫Ωh|T|qdx =∫Bδh|U + β+−β−2 g(x)|qdx+O(qβ+−β−2 ),where the last term came from the fact that∫Ω\Bδh|T|qdx = O(qβ+−β−2 ) = o(β+−β−).Let now X = U and Y = β+−β−2 g(x) in the above inequality. Takingintegral from both sides then leads us to∫Bδh|U + β+−β−2 g(x)|qdx =∫Bδh|U|qdx+ qβ+−β−2∫Bδhg|U|q−1dx+R,whereR = O(β+−β−∫Bδh|U|q−2g2dx+ qβ+−β−2∫Bδhgqdx)= O(qβ+−β−2 ) = o(β+−β−).(5.20)Regarding the second term, we haveqβ+−β−2∫Bδhg|U|q−1dx = O((β+−β−)+(n−qn−α2)) +O(qβ+−β−2 )= o(β+−β−) for all q ∈ (2, 2∗α).(5.21)Combining (5.18), (5.20) and (5.21), we get that there exist a constant C >0 such thatK = Ch(0)n−q n−α2 + o(n−qn−α2 ) + o(β+−β−) if γcrit(α) < γ < γH(α).1005.4. Proof of Theorem 5.1We point out that the situation in the critical case is more delicate, andunlike the non-critical case, we have that bothβ+ − β− and n− qn− α2are in the interval (0, α).Therefore, there is a competition between the terms β+−β− and n−qn−α2 .In order to find the threshold, namely qcrit, we equate the exponents of the terms, and solve the equation for q to get thatqcrit = 2∗α − 2β+ − β−n− α .One should note that qcrit ∈ (2, 2∗α), since α > β+ − β− > 0 in the criticalcase. This implies thato(n−qn−α2 ) + o(β+−β−) = o(n−qn−α2 ) if q > qcrito(n−qn−α2 ) + o(β+−β−) = o(n−qn−α2 ) = o(β+−β−) if q = qcrito(n−qn−α2 ) + o(β+−β−) = o(β+−β−) if q < qcrit.We finally obtainK =c4h(0)n−q n−α2 + o(n−qn−α2 ) if q > qcritc5h(0)β+−β− + o(β+−β−) if q = qcrito(β+−β−) if q < qcrit.(5.22)for some c4, c5 > 0, and as long as γcrit(α) < γ < γH(α).We now defineI0 := lim→0I = κ∫RnU2∗α(s)|x|s dx and J0 := lim→0 J =∫RnU2∗α(s)|x|s dx,and it is easy to check thatlim→0K = 0 for all cases.In the next step, we claim that, up to a subsequence of (u)>0, there existsT0 := T0(n, s, α) > 0 such thatsupt≥0Φ(tv) = d(Ψ(v))2∗α(s)2∗α(s)−2 − Tq0qK + o(K), (5.23)1015.4. Proof of Theorem 5.1whereΨ(v) =‖|v‖|2 − λ‖v‖2L2(Ω)(∫Ω|v|2∗α(s)|x|s dx) 22∗α(s)=IJ22∗α(s).The proof of this claim goes exactly as Step II in [44, Proposition 3]. Weomit it here.Let us now compute Ψ(v). It follows from (4.51), (4.52) and (4.66) thatthere exist positive constants c6, c7, c8 such thatΨ(v) = µγ,s,α(Rn)(1 + Θ,γ)(5.24)whereΘ,γ =−c6λα + o(α) if 0 ≤ γ < γcrit(α)−c7λα ln −1 + o(α ln −1) if γ = γcrit(α)−c8mαγ,λβ+−β− + o(β+−β−) if γcrit(α) < γ < γH(α).We are now going to estimate supt≥0Φ(tv). This will be done again by con-sidering two cases:Case 1: 0 ≤ γ ≤ γcrit(α). In this case, plugging (5.19) and (5.24) into(5.23) implies that there exist constants c9, c10, c11, c12 > 0 such thatsupt≥0Φ(tv) =α− s2(n− s)µγ,s,α(Rn)n−sα−s− c9λα − c10h(0)n−qn−α2 + o(α) + o(n−qn−α2 ),when 0 ≤ γ < γcrit(α), andsupt≥0Φ(tv) =α− s2(n− s)µγ,s,α(Rn)n−sα−s− c11λα ln(−1)− c12h(0)n−qn−α2 + o(α ln(−1)) + o(n−qn−α2 ),when γ = γcrit(α).Recall that α > n− q n−α2 , since q > 2. This implies thato(α) + o(n−qn−α2 ) = o(n−qn−α2 ).Thus, there exist a positive constant τ1 such that, for every 0 ≤ γ ≤ γcrit(α),we havesupt≥0Φ(tv) = Υ− τ1h(0)n−qn−α2 + o(n−qn−α2 ),1025.4. Proof of Theorem 5.1where Υ := α−s2(n−s)µγ,s,α(Rn)n−sα−s .Case 2: γcrit(α) < γ < γH(α). The critical case needs a careful anal-ysis as a new phenomena happens in this situation. We shall show thatthere is a competition between the geometry of the domain, the mass (i.e.,β+−β−mαγ,λ), and the non-linear perturbation (i.e., n−q n−α2 h(0)). Indeed, itfollows from plugging (5.22) and (5.24) into (5.23) that there exist constantsc12, c13, c14 > 0 such thatsupt≥0Φ(tv) = Υ− c12mαγ,λβ+−β− + o(β+−β−)+−c13h(0)n−qn−α2 + o(n−qn−α2 ) if q > qcrit−c14h(0)β+−β− + o(β+−β−) if q = qcrito(β+−β−) if q < qcrit.Following our analysis in the critical case of estimating K, one can thensummarize the competition results as follows.Competitive Terms q > qcrit q = qcrit q < qcrit.n−qn−α2 h(0) Dominate Equally Dominate ×β+−β−mαγ,λ × Equally Dominate DominateTherefore, we finally deduce that there exists τl > 0 for l = 2, .., 5 such thatsupt≥0Φ(tv) = Υ +−τ2h(0)n−qn−α2 + o(n−qn−α2 ) if q > qcrit−(τ3h(0) + τ4mαγ,λ)β+−β− + o(β+−β−) if q = qcrit−τ5mαγ,λβ+−β− + o(β+−β−) if q < qcrit,where Υ := α−s2(n−s)µγ,s,α(Rn)n−sα−s .Remark 5.6. We point out that the value qcrit corresponds to the valueq = 4 obtained in [44, Proposition 3]. 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