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On a fractional Nirenberg problem, part I: blow up analysis and compactness of solutions Tianling Jin∗, YanYan Li†, Jingang Xiong‡ June 22, 2018

Contents 1

Introduction

2

Preliminaries 7 2.1 A weighted Sobolev space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Weak solutions of degenerate elliptic equations . . . . . . . . . . . . . . . . . . . . 9 2.3 Local Schauder estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3

Proof of Theorem 1.5

20

4

Local analysis near isolated blow up points

24

5

Estimates on the sphere and proofs of main theorems

41

A Appendix A.1 A Kazdan-Warner identity . . . . . A.2 A proof of Lemma 4.4 . . . . . . . A.3 Two lemmas on maximum principles A.4 Complementarities . . . . . . . . .

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49 49 50 51 54

1 Introduction The Nirenberg problem concerns the following: For which positive function K on the standard sphere (Sn , gSn ), n ≥ 2, there exists a function w on Sn such that the scalar curvature (Gauss ∗ Supported in part by a University and Louis Bevier Dissertation Fellowship at Rutgers University and by Rutgers University School of Art and Science Excellence Fellowship. † Supported in part by NSF grant DMS-0701545 and DMS-1065971 and by Program for Changjiang Scholars and Innovative Research Team in University in China. ‡ Supported in part by CSC project for visiting Rutgers University and NSFC No. 11071020.

1

curvature in dimension n = 2) Rg of the conformal metric g = ew gSn is equal to K on Sn ? The problem is equivalent to solving −∆gSn w + 1 = Ke2w , and

n+2

−∆gSn v + c(n)R0 v = c(n)Kv n−2 ,

on S2 , on Sn for n ≥ 3,

n−2

where c(n) = (n−2)/(4(n−1)), R0 = n(n−1) is the scalar curvature of (Sn , gSn ) and v = e 4 w . The first work on the problem is by D. Koutroufiotis [65], where the solvability on S2 is established when K is assumed to be an antipodally symmetric function which is close to 1. Moser [78] established the solvability on S2 for all antipodally symmetric functions K which is positive somewhere. Without assuming any symmetry assumption on K, sufficient conditions were given in dimension n = 2 by Chang and Yang [30] and [31], and in dimension n = 3 by Bahri and Coron [6]. Compactness of all solutions in dimensions n = 2, 3 can be found in work of Chang, Gursky and Yang [29], Han [55] and Schoen and Zhang [88]. In these dimensions, a sequence of solutions can not blow up at more than one point. Compactness and existence of solutions in higher dimensions were studied by Li in [68] and [69]. The situation is very different, as far as the compactness issues are concerned: In dimension n ≥ 4, a sequence of solutions can blow up at more than one point, as shown in [69]. There have been many papers on the problem and related ones, see, e.g., [1, 2, 3, 7, 9, 10, 16, 17, 25, 26, 29, 30, 31, 32, 27, 34, 35, 33, 37, 44, 47, 55, 56, 58, 61, 62, 67, 75, 77, 83, 84, 93, 95, 96]. In [54], Graham, Jenne, Mason and Sparling constructed a sequence of conformally covariant elliptic operators, {Pkg }, on Riemannian manifolds for all positive integers k if n is odd, and for k ∈ {1, · · · , n/2} if n is even. Moreover, P1g is the conformal Laplacian −Lg := −∆g + c(n)Rg and P2g is the Paneitz operator. The construction in [54] is based on the ambient metric construction of [49]. Up to positive constants P1g (1) is the scalar curvature of g and P2g (1) is the Q-curvature. Prescribing Q-curvature problem on Sn was studied extensively, see, e.g., [8, 41, 42, 43, 50, 91, 92]. Making use of a generalized Dirichlet to Neumann map, Graham and Zworski [53] introduced a meromorphic family of conformally invariant operators on the conformal infinity of asymptotically hyperbolic manifolds. Recently, Chang and Gonz´alez [28] reconciled the way of Graham and Zworski to define conformally invariant operators Pσg of non-integer order σ ∈ (0, n2 ) and the localization method of Caffarelli and Silvestre [22] for factional Laplacian (−∆)σ on the Euclidean space Rn . These lead naturally to a fractional order curvature Rσg := Pσg (1), which will be called σ-curvature in this paper. A typical example is that standard conformal spheres (Sn , [gSn ]) are the conformal infinity of Poincar´e disks (Bn+1 , gBn+1 ). In this case, σ-curvature can be expressed in the following explicit way. Let g be a representative in the conformal class [gSn ] and 4 write g = v n−2σ gSn , where v is positive and smooth on Sn . Then the σ-curvature for (Sn , g) can be computed as n+2σ Rgσ = v − n−2σ Pσ (v), (1.1) where Pσ is an intertwining operator and Γ(B + Pσ = Γ(B +

1 2 1 2

+ σ) , − σ)

B=

s

−∆gSn +

n−1 2

2

,

(1.2)

Γ is the Gamma function and ∆gSn is the Laplace-Beltrami operator on (Sn , gSn ). The operator Pσ can be seen more concretely on Rn using stereographic projection. The stereographic projection 2

from Sn \{N } to Rn is the inverse of n

n

F : R → S \ {N },

x 7→

2x |x|2 − 1 , 1 + |x|2 |x|2 + 1

,

where N is the north pole of Sn . Then (Pσ (φ)) ◦ F = |JF |−

n+2σ 2n

(−∆)σ (|JF |

where |JF | =

n−2σ 2n

2 1 + |x|2

(φ ◦ F )),

n

for φ ∈ C ∞ (Sn )

(1.3)

,

and (−∆)σ is the fractional Laplacian operator (see, e.g., page 117 of [86]). When σ ∈ (0, 1), Pavlov and Samko [81] showed that Z v(ξ) − v(ζ) dvolgSn (ζ) (1.4) Pσ (v)(ξ) = Pσ (1)v(ξ) + cn,−σ |ξ − ζ|n+2σ n S for v ∈ C 2 (Sn ), where cn,−σ =

22σ σΓ( n+2σ ) 2 n π2

Γ(1−σ)

and

R

Sn

is understood as lim

R

. ε→0 |x−y|>ε

For the σ-curvatures on general manifolds we refer to [53], [28], [52] and references therein. Corresponding to the Yamabe problem, fractional Yamabe problems for σ-curvatures are studied in [51], [52] and [82], and fractional Yamabe flows on Sn are studied in [64]. From (1.1), we consider n+2σ

Pσ (v) = c(n, σ)Kv n−2σ ,

on Sn ,

(1.5)

where c(n, σ) = Pσ (1), and K > 0 is a continuous function on Sn . When K = 1, (1.5) is the Euler-Lagrange equation for a functional associated to the fractional Sobolev inequality on Sn (see [8]), and all positive solutions must be of the form vξ0 ,λ (ξ) =

2λ 2 + (λ2 − 1)(1 − cos distgSn (ξ, ξ0 ))

n−2σ 2

,

ξ ∈ Sn

(1.6)

for some ξ0 ∈ Sn and positive constant λ. This classification can be found in [74], [36] and [70]. In general, (1.5) may have no positive solution, since if v is a positive solution of (1.5) with K ∈ C 1 (Sn ) then it has to satisfy the Kazdan-Warner type condition Z 2n h∇gSn K, ∇gSn ξiv n−2σ dξ = 0. (1.7) Sn

Consequently if K(ξ) = ξn+1 +2, (1.5) has no solutions. The proof of (1.7) is provided in Appendix A.1. In this and a subsequent paper [63], we study (1.5) with σ ∈ (0, 1), a fractional Nirenberg problem. Throughout the paper, we assume that σ ∈ (0, 1) without otherwise stated. Definition 1.1. For d > 0, we say that K ∈ C(Sn ) has flatness order greater than d at ξ if, in some local coordinate system {y1 , · · · , yn } centered at ξ, there exists a neighborhood O of 0 such that K(y) = K(0) + o(|y|d ) in O. 3

Theorem 1.1. Let n ≥ 2, and K ∈ C 1,1 (Sn ) be an antipodally symmetric function, i.e., K(ξ) = K(−ξ) ∀ ξ ∈ Sn , and be positive somewhere on Sn . If there exists a maximum point ξ0 of K at which K has flatness order greater than n − 2σ, then (1.5) has at least one positive C 2 solution. For 2 ≤ n < 2 + 2σ, K ∈ C 1,1 (Sn ) has flatness order greater than n − 2σ at every maximum point. When σ = 1, the above theorem was proved by Escobar and Schoen [46] for n ≥ 3. Theorem 1.2. Let n ≥ 2. Suppose that K ∈ C 1,1 (Sn ) is a positive function satisfying that for any critical point ξ0 of K, in some geodesic normal coordinates {y1 , · · · , yn } centered at ξ0 , there exist some small neighborhood O of 0 and positive constants β = β(ξ0 ) ∈ (n − 2σ, n), γ ∈ (n − 2σ, β] such that K ∈ C [γ],γ−[γ](O) (where [γ] is the integer part of γ) and K(y) = K(0) +

n X j=1

aj |yj |β + R(y),

in O,

Pn where aj = aj (ξ0 ) 6= 0, j=1 aj 6= 0, R(y) ∈ C [β]−1,1 (O) satisfies P[β] s −β+s → 0 as y → 0. If s=0 |∇ R(y)||y| X (−1)i(ξ) 6= (−1)n , P n ξ∈Sn such that ∇gSn K(ξ)=0, j=1 aj (ξ)<0 where i(ξ) = #{aj (ξ) : ∇gSn K(ξ) = 0, aj (ξ) < 0, 1 ≤ j ≤ n},

then (1.5) has at least one C 2 positive solution. Moreover, there exists a positive constant C depending only on n, σ and K such that for all positive C 2 solutions v of (1.5), 1/C ≤ v ≤ C

and kvkC 2 (Sn ) ≤ C.

For n = 3, σ = 1, the existence part of the above theorem was established by Bahri and Coron [6], and the compactness part were given in Chang, Gursky and Yang [29] and Schoen and Zhang [88]. For n ≥ 4, σ = 1, the above theorem was proved by Li [68]. We now consider a class of functions K more general than that in Theorem 1.2, which is modified from [68]. Definition 1.2. For any real number β > 1, we say that a sequence of functions {Ki } satisfies condition (∗)′β for some sequence of constants L(β, i) in some region Ωi , if {Ki } ∈ C [β],β−[β](Ωi ) satisfies [∇[β] Ki ]C β−[β] (Ωi ) ≤ L(β, i), and, if β ≥ 2, that

|∇s Ki (y)| ≤ L(β, i)|∇Ki (y)|(β−s)/(β−1) ,

for all 2 ≤ s ≤ [β], y ∈ Ωi , ∇Ki (y) 6= 0. Note that the function K in Theorem 1.2 satisfies (∗)′β condition.

4

Remark 1.1. For 1 ≤ β1 ≤ β2 , if {Ki } satisfies (∗)′β2 for some sequences of constants {L(β2 , i)} in some regions Ωi , then {Ki } satisfies (∗)′β1 for {L(β1 , i)}, where β2 −s β1 −s β2 −1 − β1 −1 β −β 2 1 L(β2 , i) max max k∇Ki kL∞ (Ωi ) , diam(Ωi ) , if [β2 ] = [β1 ] 2≤s≤[β1 ] β1 −s β2 −[β1 ]−1 β2 −s L(β1 , i) = β2 −1 − β1 −1 β2 −1 1+[β1 ]−β1 , , k∇Ki kL∞ (Ωi ) diam(Ωi ) L(β2 , i) max max k∇Ki kL∞ (Ωi ) 2≤s≤[β1 ] if [β2 ] > [β1 ] in the corresponding regions.

2n

The following theorem gives a priori bounds of solutions in L n−2σ norm. Theorem 1.3. Let n ≥ 2, and K ∈ C 1,1 (Sn ) be a positive function. If there exists some constant d > 0 such that K satisfies (∗)′(n−2σ) for some constant L > 0 in Ωd := {ξ ∈ Sn : |∇g0 K(ξ)| < d}, then for any positive solution v ∈ C 2 (Sn ) of (1.5), kvk

2n

L n−2σ (Sn )

≤ C,

(1.8)

where C depends only on n, σ, inf Sn K > 0, kKkC 1,1(Sn ) , L, and d. The above theorem was proved by Schoen and Zhang [88] for n = 3 and σ = 1, and by Li [68] for n ≥ 4 and σ = 1. Denote H σ (Sn ) by the closure of C ∞ (Sn ) under the norm Z vPσ (v) dvolg0 . Sn

The estimate (1.8) for the solution v is equivalent to kvkH σ (Sn ) ≤ C. However, the estimate (1.8) is not sufficient to imply L∞ bound for v on Sn . For instance, Z Z 2n vξn−2σ dvolg0 , = (ξ) dvol g 0 0 ,λ Sn

Sn

n−2σ 2

→ ∞ as λ → ∞. Furthermore, a sequence of solutions vi may blow up at but vξ0 ,λ (ξ0 ) = λ more than one point, and it is the case when σ = 1 (see [69]). The following theorem shows that the latter situation does not happen when K satisfies a little stronger condition. Theorem 1.4. Let n ≥ 2. Suppose that {Ki } ∈ C 1,1 (Sn ) is a sequence of positive functions with uniform C 1,1 norm and 1/A1 ≤ Ki ≤ A1 on Sn for some A1 > 0 independent of i. Suppose also that {Ki } satisfying (∗)′β condition for some constants β > n − 2σ, L, d > 0 in Ωd . Let {vi } ∈ C 2 (Sn ) be a sequence of corresponding positive solutions of (1.5) and vi (ξi ) = maxSn vi for some ξi . Then, after passing to a subsequence, {vi } is either bounded in L∞ (Sn ) or blows up at exactly one point in the strong sense: There exists a sequence of M¨obius diffeomorphisms {ϕi } from n−2σ Sn to Sn satisfying ϕi (ξi ) = ξi and | det dϕi (ξi )| 2n = vi−1 (ξi ) such that kTϕi vi − 1kC 0 (Sn ) → 0, where Tϕi vi := (v ◦ ϕi )| det dϕi |

n−2σ 2n

. 5

as i → ∞,

For n = 3, σ = 1, the above theorem was established by Chang, Gursky and Yang in [29] and by Schoen and Zhang in [88]. For n ≥ 4, σ = 1, the above theorem was proved by Li in [68]. M¨obius diffeomorphisms ϕ from Sn to Sn are those given by ϕ = φ ◦ F where φ is a M¨obius transformation from Rn ∪ {∞} to Rn ∪ {∞} generated by translations, multiplications by nonzero constant and the inversion x → x/|x|2 . Our local analysis of solutions of (1.5) relies on a localization method introduced by Caffarelli and Silvestre in [22] for the factional Laplacian (−∆)σ on the Euclidean space Rn , through which (1.5) is connected to a degenerate elliptic differential equation in one dimension higher (see section 2). The proofs of Theorem 1.3 and Theorem 1.4 make use of blow up analysis of solutions of (1.5), which is an adaptation of the analysis for σ = 1 developed in [88] and [68]. Our blow up analysis requires a Liouville type theorem. For the definitions of weak solutions and the space Hloc (t1−2σ , Rn+1 + ) in the following Liouville type theorem we refer to Definition 2.1 and the beginning of section 3. n+1 Theorem 1.5. Let U ∈ Hloc (t1−2σ , Rn+1 and U 6≡ 0, be a weak solution of + ), U (X) ≥ 0 in R+

(

div(t1−2σ ∇U (x, t))

= 0,

− lim t1−2σ ∂t U (x, t) = U t→0

in Rn+1 + ,

n+2σ n−2σ

(x, 0),

x ∈ Rn .

(1.9)

Then U (x, 0) takes the form Nσ cn,σ 2

2σ

n−2σ 4σ

λ 1 + λ2 |x − x0 |2

n−2σ 2

where λ > 0, x0 ∈ Rn , cn,σ is the constant in (1.5) and Nσ is the constant in (2.4). Moreover, Z U (x, t) = Pσ (x − y, t)U (y, 0) dy Rn

for (x, t) ∈ Rn+1 + , where Pσ (x) is the kernel given in (2.2). n+2σ

n+2σ Remark 1.2. If we replace U n−2σ (x, 0) by U p (x, 0) for 0 ≤ p < n−2σ in (1.9), then the only nonnegative solution of (1.9) is U ≡ 0. Moreover, for p < 0, (1.9) has no positive solution. These can be seen from the proof of Theorem 1.5 with a standard modification (see, e.g., the proof of n+2σ Theorem 1.2 in [24]). For σ ∈ (1/2, 1) and 1 < p < n−2σ , this result has been proved in [40].

Remark 1.3. We do not make any assumption on the behavior of U near ∞. If we assume that n+2σ U ∈ H(t1−2σ , Rn+1 + ), the theorem in the case of p = n−2σ follows from [36] and [70]. When 1 σ = 2 , the above theorem can be found in [59], [60], [73], [80] and [72]. Given the pages needed to present the proofs of all the results, we leave the proofs of Theorem 1.1 and the existence part of Theorem 1.2 to the subsequent paper [63]. The needed ingredients for a proof of the existence part of Theorem 1.2 are all developed in this paper. With these ingredients, the existence part of Theorem 1.2 follows from a perturbation result and a degree argument which are given in [63].

6

The present paper is organized as the following. In section 2 we derive some properties for solutions of fractional Laplacian equations. In particular we prove that local Schauder estimates hold for positive solutions. In section 3, using the method of moving spheres, we establish Theorem 1.5. This Liouville type theorem and the local Schauder estimates are used in the blow up analysis of solutions of (1.5). In section 4 we establish accurate blow up profiles of solutions of (1.5) near isolated blow up points. In fact most of the estimates hold also for subcritical approximations to such equations as well including in bounded domains of Rn . In section 5, we provide H σ (Sn ) norm a priori estimates, at most one isolated simple blow up point, and L∞ (Sn ) norm a priori estimates for solutions of (1.5) under appropriate hypotheses on K. The proofs of Theorem 1.2, 1.3 and 1.4 are given in this section. In the Appendix we provide a Kazdan-Warner identity, Lemma 4.4 that is in the same spirit of the classical Bˆocher theorem, two lemmas on maximum principles and some complementarities.

2 Preliminaries 2.1 A weighted Sobolev space Let σ ∈ (0, 1), X = (x, t) ∈ Rn+1 where x ∈ Rn and t ∈ R. Then |t|1−2σ belongs to the Muckenhoupt A2 class in Rn+1 , namely, there exists a positive constant C, such that for any ball B ⊂ Rn+1 Z Z 1 1 |t|1−2σ dX |t|2σ−1 dX ≤ C. |B| B |B| B

Let D be an open set in Rn+1 . Denote L2 (|t|1−2σ , D) as the Banach space of all measurable functions U , defined on D, for which kU kL2 (|t|1−2σ ,D) :=

Z

D

|t|

1−2σ

2

U dX

21

< ∞.

We say that U ∈ H(|t|1−2σ , D) if U ∈ L2 (|t|1−2σ , D), and its weak derivatives ∇U exist and belong to L2 (|t|1−2σ , D). The norm of U in H(|t|1−2σ , D) is given by kU kH(|t|1−2σ ,D) :=

Z

D

|t|

1−2σ

2

U (X) dX +

Z

D

|t|

1−2σ

2

|∇U (X)| dX

21

.

It is clear that H(|t|1−2σ , D) is a Hilbert space with the inner product Z |t|1−2σ (U V + ∇U ∇V ) dX. hU, V i := D

Note that the set of smooth functions C ∞ (D) is dense in H(|t|1−2σ , D). Moreover, if D is a domain, i.e. a bounded connected open set, with Lipschitz boundary ∂D, then there exists a linear, bounded extension operator from H(|t|1−2σ , D) to H(|t|1−2σ , Rn+1 ) (see, e.g., [39]). Let Ω be an open set in Rn . Recall that H σ (Ω) is the fractional Sobolev space defined as |u(x) − u(y)| σ 2 2 H (Ω) := u ∈ L (Ω) : ∈ L (Ω × Ω) n |x − y| 2 +σ 7

with the norm kukH σ (Ω) :=

Z

u2 dx + Ω

Z Z Ω

Ω

|u(x) − u(y)|2 dx dy |x − y|n+2σ

1/2

.

The set of smooth functions C ∞ (Ω) is dense in H σ (Ω). If Ω is a domain with Lipschitz boundary, then there exists a linear, bounded extension operator from H σ (Ω) to H σ (Rn ). Note that H σ (Rn ) with the norm k · kH σ (Rn ) is equivalent to the following space u ∈ L2 (Rn ) : |ξ|σ F (u)(ξ) ∈ L2 (Rn ) with the norm

k · kL2 (Rn ) + k|ξ|σ F (·)(ξ)kL2 (Rn ) where F denotes the Fourier transform operator. It is known that (see, e.g., [76]) there exists C > 0 n+1 depending only on n and σ such that for U ∈ H(t1−2σ , Rn+1 + ) ∩ C(R+ ), kU (·, 0)kH σ (Rn ) ≤ CkU kH(t1−2σ ,Rn+1 ) . Hence by a standard density argument, every U ∈ H(t1−2σ , Rn+1 + ) has a + σ n well-defined trace u := U (·, 0) ∈ H (R ). We define H˙ σ (Rn ) as the closure of the set Cc∞ (Rn ) of compact supported smooth functions under the norm kukH˙ σ (Rn ) = k|ξ|σ F (u)(ξ)kL2 (Rn ) . Then there exists a constant C depending only on n and σ such that kuk

2n

L n−2σ (Rn )

≤ CkukH˙ σ (Rn )

for all u ∈ Cc∞ (Rn ).

(2.1)

For any u ∈ H˙ σ (Rn ), set U (x, t) = Pσ [u] :=

Z

Rn

Pσ (x − ξ, t)u(ξ) dξ,

where Pσ (x, t) = β(n, σ) with constant β(n, σ) such that

R

Rn

(x, t) ∈ Rn+1 := Rn × (0, +∞), +

(2.2)

t2σ (|x|2 + t2 )

n+2σ 2

2 1−2σ Pσ (x, 1) dx = 1. Then U ∈ C ∞ (Rn+1 , K) + ), U ∈ L (t

2 1−2σ , Rn+1 for any compact set K in Rn+1 + , and ∇U ∈ L (t + ). Moreover, U satisfies (see [22])

div(t1−2σ ∇U ) = 0

in Rn+1 + ,

(2.3)

k∇U kL2 (t1−2σ ,Rn+1 ) = Nσ kukH˙ σ (Rn ) ,

(2.4)

+

and − lim t1−2σ ∂t U (x, t) = Nσ (−∆)σ u(x), t→0

in Rn

(2.5)

in distribution sense, where Nσ = 21−2σ Γ(1 − σ)/Γ(σ). We refer U = Pσ [u] in (2.2) to be the extension of u for any u ∈ H˙ σ (Rn ). For a domain D ⊂ Rn+1 with boundary ∂D, we denote ∂ ′ D as the interior of D ∩ ∂Rn+1 in + n+1 n ′′ ′ R = ∂R+ and ∂ D = ∂D \ ∂ D. 8

Proposition 2.1. Let D = Ω × (0, R) ⊂ Rn × R+ , R > 0 and ∂Ω be Lipschitz. (i) If U ∈ H(t1−2σ , D) ∩ C(D ∪ ∂ ′ D), then u := U (·, 0) ∈ H σ (Ω), and kukH σ (Ω) ≤ CkU kH(t1−2σ ,D) where C is a positive constant depending only on n, σ, R and Ω. Hence every U ∈ H(t1−2σ , D) has a well-defined trace U (·, 0) ∈ H σ (Ω) on ∂ ′ D. Furthermore, there exists Cn,σ > 0 depending only on n and σ such that kU (·, 0)k

2n

L n−2σ (Ω)

≤ Cn,σ k∇U kL2 (t1−2σ ,D)

for all U ∈ Cc∞ (D ∪ ∂ ′ D).

(2.6)

(ii) If u ∈ H σ (Ω), then there exists U ∈ H(t1−2σ , D) such that the trace of U on Ω equals to u and kU kH(t1−2σ ,D) ≤ CkukH σ (Ω) where C is a positive constant depending only on n, σ, R and Ω. Proof. The above results are well-known and here we just sketch the proofs. For (i), by the pre˜ ∈ H(t1−2σ , Rn+1 ) such that viously mentioned result on the extension operator, there exists U ˜ = U in D and U ˜ kH(t1−2σ ,Rn+1 ) ≤ CkU kH(t1−2σ ,D) . kU σ n Hence by the previously mentioned result on the trace from H(t1−2σ , Rn+1 + ) to H (R ), we have

˜ (·, 0)kH σ (Rn ) ≤ CkU ˜ k 1−2σ n+1 ≤ CkU kH(t1−2σ ,D) . kukH σ (Ω) ≤ kU H(t ,R ) +

For (2.6), we extend U to be zero in the outside of D and let V be the extension of U (·, 0) as in (2.2). The inequality (2.6) follows from (2.1), (2.4) and k∇V kL2 (t1−2σ ,Rn+1 ) ≤ k∇U kL2(t1−2σ ,Rn+1 ) +

+

where Lemma A.3 is used in the above inequality. For (ii), since ∂Ω is Lipschitz, there exists u ˜ ∈ H σ (Rn ) such that u ˜ = u in Ω and k˜ ukH σ (Rn ) ≤ CkukH σ (Ω) . Then U = Pσ [u], the extension of u˜, satisfies (ii).

2.2 Weak solutions of degenerate elliptic equations 2n

n+2σ (∂ ′ D) and b ∈ L1loc (∂ ′ D). Consider Let D be a domain in Rn+1 with ∂ ′ D 6= ∅. Let a ∈ Lloc +

(

div(t1−2σ ∇U (X)) = 0 − lim t1−2σ ∂t U (x, t) = a(x)U (x, 0) + b(x) t→0+

in D on ∂ ′ D.

(2.7)

Definition 2.1. We say that U ∈ H(t1−2σ , D) is a weak solution (resp. supersolution, subsolution) of (2.7) in D, if for every nonnegative Φ ∈ Cc∞ (D ∪ ∂ ′ D) Z Z t1−2σ ∇U ∇Φ = (resp. ≥, ≤) aU Φ + bΦ. (2.8) D

∂′ D

9

We denote QR = BR × (0, R) where BR ⊂ Rn is the ball with radius R and centered at 0. 2n

n

Proposition 2.2. Suppose that a(x) ∈ L 2σ (B1 ) and b(x) ∈ L n+2σ (B1 ). Let U ∈ H(t1−2σ , Q1 ) be a weak solution of (2.7) in Q1 . There exists δ > 0 depending only on n and σ such that if n ka+ kL 2σ < δ, then there exists a constant C depending only on n, σ and δ such that (B ) 1

kU kH(t1−2σ ,Q1/2 ) ≤ C(kU kL2 (t1−2σ ,Q1 ) + kbk Consequently, if a ∈ Lp (B1 ) for p >

n 2σ ,

2n

L n+2σ (B1 )

).

then C depends only on n, σ, kakLp (B1 ) .

Proof. Let η ∈ Cc∞ (Q1 ∪ ∂ ′ Q1 ) be a cut-off function which equals to 1 in Q1/2 and supported in Q3/4 . By a density argument, we can choose η 2 U as a test function in (2.8). Then we have, by Cauchy-Schwarz inequality, Z Z Z 1−2σ 2 2 1−2σ 2 2 a+ (ηU )2 + bη 2 U dx. t |∇η| U dX + 2 t η |∇U | dX ≤ 4 ∂ ′ Q1

Q1

Q1

By H¨older inequality and Proposition 2.1, Z a+ (ηU )2 dx ≤ δkηU k2

2n

L n−2σ (∂ ′ Q1 )

∂ ′ Q1

By Young’s inequality ∀ ε > 0, Z bη 2 U (·, 0) dx ≤ εkηU k2

2n

L n−2σ (∂ ′ Q1 )

∂ ′ Q1

≤ δC(n, σ)k∇(ηU )k2L2 (t1−2σ ,Q1 )

+ C(ε)kbk2

2n

L n+2σ (∂ ′ Q1 )

≤ εC(n, σ)k∇(ηU )k2L2 (t1−2σ ,Q1 ) + C(ε)kbk2

2n

L n+2σ (∂ ′ Q1 )

.

The first conclusion follows immediately if δ is sufficient small. n If a ∈ Lp (B1 ), we can choose r small such that kakL 2σ < δ for any ball Br (x0 ) ⊂ B1 . (Br (x0 )) n−2σ ˆ ˆ(x) = r2σ a(rx + x0 ) and ˆb(x, t) = Then U (x, t) = r 2 U (rx + x0 , rt) satisfies (2.7) with a n+2σ ˆ , we have < δ, applying the above result to U r 2 b(rx + x0 ) in Q1 . Since kˆ ak n L 2σ (B1 )

kU kH(t1−2σ ,B1/2 ×(0,r/2)) ≤ C(kU kL2 (t1−2σ ,Q1 ) + kbk

2n

L n+2σ (B1 )

)

where C depends only on n, σ, kakL∞ (B1 ) . This, together with the fact that (2.7) is uniformly elliptic in B1 × (r/4, 1), finishes the proof. n

Proposition 2.3. Suppose that a(x) ∈ L 2σ (B1 ). There exists δ > 0 which depends only on n and 2n n σ such that if ka+ kL 2σ < δ, then for any b(x) ∈ L n+2σ (B1 ), there exists a unique solution in (B1 ) H(t1−2σ , Q1 ) to (2.7) with U |∂ ′′ Q1 = 0. Proof. We consider the bilinear form Z Z t1−2σ ∇U ∇V dX − B[U, V ] :=

∂ ′ Q1

Q1

aU V dx,

U, V ∈ A

where A := {U ∈ H(t1−2σ , Q1 ) : U |∂ ′′ Q1 = 0 in trace sense}. By Proposition 2.1, it is easy to verify that B[·, ·] is bounded and coercive provided δ is sufficiently small. Therefore the proposition follows from the Riesz representation theorem. 10

Lemma 2.1. Suppose U ∈ H(t1−2σ , D) is a weak supersolution of (2.7) in D with a ≡ b ≡ 0. If U (X) ≥ 0 on ∂ ′′ D in trace sense, then U ≥ 0 in D. Proof. Use U − as a test function to conclude that U − ≡ 0. The following result is a refined version of that in [90]. Such De Giorgi-Nash-Moser type theorems for degenerated equations with Dirichlet boundary conditions have been established in [48]. n Proposition 2.4. Suppose a, b ∈ Lp (B1 ) for some p > 2σ . 1−2σ (i) Let U ∈ H(t , Q1 ) be a weak subsolution of (2.7) in Q1 . Then ∀ ν > 0

sup U + ≤ C(kU + kLν (t1−2σ ,Q1 ) + kb+ kLp (B1 ) )

Q1/2

where U + = max(0, U ), and C > 0 depends only on n, σ, p, ν and ka+ kLp (B1 ) . (ii) Let U ∈ H(t1−2σ , Q1 ) be a nonnegative weak supersolution of (2.7) in Q1 . Then for any 0 < µ < τ < 1, 0 < ν ≤ n+1 n we have inf U + kb− kLp (B1 ) ≥ CkU kLν (t1−2σ ,Qτ ) Qµ

where C > 0 depends only on n, σ, p, ν, µ, τ and ka− kLp (B1 ) . (iii) Let U ∈ H(t1−2σ , Q1 ) be a nonnegative weak solution of (2.7) in Q1 . Then we have the following Harnack inequality sup U ≤ C( inf U + kbkLp(B1 ) ), Q1/2

Q1/2

(2.9)

where C > 0 depends only on n, σ, p, kakLp (B1 ) . Consequently, there exists α ∈ (0, 1) depending only on n, σ, p, kakLp(B1 ) such that any weak solution U (X) of (2.7) is of C α (Q1/2 ). Moreover, kU kC α(Q1/2 ) ≤ C(kU kL∞ (Q1 ) + kbkLp(B1 ) ) where C > 0 depends only on n, σ, p, kakLp (B1 ) . Proof. The proofs are modifications of those in [90], where the method of Moser iteration is used. Here we only point out the changes. Let k = kb+ kLp (B1 ) if b+ 6≡ 0, otherwise let k > 0 be any number which is eventually sent to 0. Define U = U + + k and, for m > 0, let ( U if U < m, Um = k+m if U ≥ m. Consider the test function β

φ = η 2 (U m U − k β+1 ) ∈ H(t1−2σ , Q1 ), for some β ≥ 0 and some nonnegative function η ∈ Cc1 (Q1 ∪ ∂ ′ Q1 ). Direction calculations yield β

2 that, with setting W = U m U, Z Z Z 1 b+ 2 2 )η W . (a+ + t1−2σ |∇η|2 W 2 + 4 t1−2σ |∇(ηW )|2 ≤ 16 1 + β Q1 k ∂ ′ Q1 Q1

11

(2.10)

By H¨older’s inequality and the choice of k, we have Z b+ 2 2 (a+ + )η W ≤ (ka+ kLp (B1 ) + 1)kη 2 W 2 kLp′ (B1 ) k ∂ ′ Q1 p n where p′ = p−1 < n−2σ . Choose 0 < θ < 1 such that inequality gives that, for any ε > 0,

kη 2 W 2 kLp′ (B1 ) ≤ εkηW k2

1 p′

=θ+

+ ε−

2n

L n−2σ (B1 )

1−θ θ

(1−θ)(n−2σ) . n

The interpolation

kη 2 W 2 kL1 (B1 ) .

By the trace embedding inequality in Proposition 2.1, there exists C > 0 depending only on n, σ such that Z 2 t1−2σ |∇(ηW )|2 . ≤C kηW k 2n L n−2σ (B1 )

Q1

By Lemma 2.3 in [90], there exist δ > 0 and C > 0 both of which depend only on n, σ such that Z Z 1 δ kη 2 W 2 kL1 (B1 ) ≤ ε θ t1−2σ |∇(ηW )|2 + ε− θ t1−2σ η 2 W 2 . Q1

Q1

By choosing ε small, the above inequalities give that Z Z t1−2σ |∇(ηW )|2 ≤ C(1 + β)δ/θ Q1

Q1

t1−2σ (η 2 + |∇η|2 )W 2

where C depends only on n, σ and ka+ kLp (B1 ) . Then the proof of Proposition 3.1 in [90] goes through without any change. This finishes the proof of (i) for ν = 2. Then (i) also holds for any ν > 0 which follows from standard arguments. For part (ii) we choose k = kb− kLp (B1 ) if b− 6≡ 0, otherwise let k > 0 be any number which is eventually sent to 0. Then we can show that there exists some ν0 > 0 for which (ii) holds, by exactly the same proof of Proposition 3.2 in [90]. Finally −β use the test function φ = U η 2 with β ∈ (0, 1) to repeat the proof in (i) to conclude (ii) for 0 < ν ≤ n+1 n . Part (iii) follows from (i), (ii) and standard elliptic equation theory. Remark 2.1. Harnack inequality (2.9), without lower order term b, has been obtained earlier in [23] using a different method. The above proofs can be improved to yield the following result. n

n and U ∈ H(t1−2σ , Q1 ) is a Lemma 2.2. Suppose a ∈ L 2σ (B1 ), b ∈ Lp (B1 ) with p > 2σ weak subsolution of (2.7) in Q1 . There exists δ > 0 which depends only on n and σ such that if n ka+ kL 2σ < δ, then (B ) 1

kU + (·, 0)kLq (∂ ′ Q1/2 ) ≤ C(kU + kH(t1−2σ ,Q1 ) + kb+ kLp (B1 ) ). where C > 0 depends only on n, p, σ, δ, and q = min

n(p−1) 2(n+1) n−2σ , (n−2σ)p

·

2n n−2σ

.

Remark 2.2. Analogues estimates were established for −∆u = a(x)u in [15] (see Theorem 2.3 there) and for −div(|∇u|p−2 ∇u) = a(x)|u|p−2 u in [4] (see Lemma 3.1 there). 12

older Proof of Lemma 2.2. We start from (2.10), where we choose β = min n2 , 2(2σp−n) (n−2σ)p . By H¨ inequality and Proposition 2.1, Z b+ 2 2 n )η W ≤ δkη 2 W 2 kL n−2σ + kη 2 W 2 kLp′ (B1 ) (a+ + (B1 ) k ∂ ′ Q1 Z t1−2σ |∇(ηW )|2 + Cn,σ,p kU kH(t1−2σ ,Q1 ) . ≤ C(n, σ)δ Q1

By Poincare’s inequality in [48], we have Z t1−2σ |∇η|2 W 2 ≤ Cn,σ,p kU kH(t1−2σ ,Q1 ) . Q1

If δ is sufficiently small, the the above together with (2.10) imply that Z t1−2σ |∇(ηW )|2 ≤ Cn,σ,p kU kH(t1−2σ ,Q1 ) . Q1

Hence it follows from H¨older inequality and Proposition 2.1 that, by sending m → ∞, Z kU (·, 0)kLq (∂ ′ Q1/2 ) ≤ Cn,σ,p t1−2σ |∇(ηW )|2 ≤ Cn,σ,p kU kH(t1−2σ ,Q1 ) . Q1

This finishes the proof. Corollary 2.1. Suppose that K ∈ L∞ (B1 ), U ∈ H(t1−2σ , Q1 ) and U ≥ 0 in Q1 satisfies, for some 1 ≤ p ≤ (n + 2σ)(n − 2σ), (

div(t1−2σ ∇U (X)) = 0 − lim+ t1−2σ ∂t U (x, t) = K(x)U (x, 0)p t→0

in Q1 on ∂ ′ Q1 .

′ ∞ Then (i) U ∈ L∞ loc (Q1 ∪ ∂ Q1 ), and hence U (·, 0) ∈ Lloc (B1 ). (ii) There exist C > 0 and α ∈ (0, 1) depending only on n, σ, p, kukL∞ (B3/4 ) , kKkL∞ (B3/4 ) such that U ∈ C α (Q1/2 ) and

kU kH(t1−2σ ,Q1/2 ) + kU kC α(Q1/2 ) ≤ C. n+2

Note that the regularity of solution of −∆u = u n−2 was proved by Trudinger in [89]. 2n

Proof of Corollary 2.1. By Proposition 2.1, U (·, 0) ∈ H σ (B1 ) ⊂ L n−2σ (B1 ). Thus U (·, 0)p−1 ∈ n L 2σ (B1 ). Then part (i) follows from Lemma 2.2 and Proposition 2.4. Part (ii) follows from Proposition 2.2 and Proposition 2.4.

13

2.3 Local Schauder estimates 2n

n+2σ Let Ω be a domain in Rn , a ∈ Lloc (Ω) and b ∈ L1loc (Ω). We say u ∈ H˙ σ (Rn ) is a weak solution of (−∆)σ u = a(x)u + b(x) in Ω

if for any φ ∈ C ∞ (Rn ) supported in Ω, Z Z σ σ 2 2 (−∆) u(−∆) φ = a(x)uφ + b(x)φ. Rn

Ω

Then by (2.5), u ∈ H˙ σ (Rn ) is a weak solution of (−∆)σ u =

1 a(x)u + b(x) Nσ

in B1

if and only if U = Pσ [u], the extension of u defined in (2.2), is a weak solution of (2.7) in Q1 . For α ∈ (0, 1), C α (Ω) denotes the standard H¨older space over domain Ω. For simplicity, we use α C (Ω) to denote C [α],α−[α] (Ω) when 1 < α ∈ / N (the set of positive integers). In this part, we shall prove the following local Schauder estimates for nonnegative solutions of fractional Laplace equation. Theorem 2.1. Suppose a(x), b(x) ∈ C α (B1 ) with 0 < α 6∈ N. Let u ∈ H˙ σ (Rn ) and u ≥ 0 in Rn be a weak solution of (−∆)σ u = a(x)u + b(x), in B1 . Suppose that 2σ + α is not an integer. Then u ∈ C 2σ+α (B1/2 ). Moreover, kukC 2σ+α (B1/2 ) ≤ C( inf u + kbkC α(B3/4 ) ) B3/4

(2.11)

where C > 0 depends only on n, σ, α, kakC α (B3/4 ) . Remark 2.3. Replacing the assumption u ≥ 0 in Rn by u ≥ 0 in B1 , estimate (2.11) may fail (see [66]). Without the sign assumption of u, (2.11) with inf B3/4 u substituted by kukL∞ (Rn ) holds, which is proved in [21], [20] and [19] in a much more general setting of fully nonlinear nonlocal equations. The following proposition will be used in the proof of Theorem 2.1. Proposition 2.5. Let a(x), b(x) ∈ C k (B1 ), U (X) ∈ H(t1−2σ , Q1 ) be a weak solution of (2.7) in Q1 , where k is a positive integer. Then we have k X i=0

k∇ix U kL∞ (Q1/2 ) ≤ C(kU kL2 (t1−2σ ,Q1 ) + kbkC k (B1 ) ),

where C > 0 depends only on n, σ, k, kakC k (B1 ) .

14

Proof. We know from Proposition 2.4 that U is H¨older continuous in Q8/9 . Let h ∈ Rn with |h| sufficiently small. Denote U h (x, t) = U(x+h,t)−U(x,t) . Then U h is a weak solution of |h| (

div(t1−2σ ∇U h (X)) = 0 − lim t1−2σ ∂t U h (x, t) = a(x + h)U h + ah U + bh t→0+

in Q8/9 on ∂ ′ Q8/9 .

(2.12)

By Proposition 2.2 and Proposition 2.4, kU h kH(t1−2σ ,Q2/3 ) + kU h kC α (Q2/3 ) ≤ C(kU h kL2 (t1−2σ ,Q3/4 ) + kbkC 1 (B1 ) ) ≤ C(k∇U kL2 (t1−2σ ,Q4/5 ) + kbkC 1(B1 ) )

≤ C(kU kL2 (t1−2σ ,Q1 ) + kbkC 1(B1 ) )

for some α ∈ (0, 1) and positive constant C > 0 depending only on n, σ, kakC 1 (B1 ) . Hence ∇x U ∈ H(t1−2σ , Q2/3 ) ∩ C α (Q2/3 ), and it is a weak solution of (

div(t1−2σ ∇(∇x U ) = 0 − lim+ t1−2σ ∂t (∇x U ) = a∇x U + U ∇x a + ∇x b t→0

in Q2/3 on ∂ ′ Q2/3 .

Then this Proposition follows immediately from Proposition 2.2 and Proposition 2.4 for k = 1. We can continue this procedure for k = 2, 3, · · · (by induction). To prove Theorem 2.1 we first obtain Schauder estimates for solutions of the equation ( div(t1−2σ ∇U (X)) = 0 in QR 1−2σ − lim t ∂t U (x, t) = g(x) on ∂ ′ QR .

(2.13)

t→0+

Theorem 2.2. Let U (X) ∈ H(t1−2σ , Q2 ) be a weak solution of (2.13) with R = 2 and g(x) ∈ C α (B2 ) for some 0 < α 6∈ N. If 2σ + α is not an integer, then U (·, 0) is of C 2σ+α (B1/2 ). Moreover, we have kU (·, 0)kC 2σ+α (B1/2 ) ≤ C(kU kL∞ (Q2 ) + kgkC α(B2 ) ), where C > 0 depends only on n, σ, α. This theorem together with Proposition 2.4 implies the following Theorem 2.3. Let U (X) ∈ H(t1−2σ , Q1 ) be a weak solution of (2.7) with D = Q1 and a(x), b(x) ∈ C α (B1 ) for some 0 < α 6∈ N. If 2σ + α is not an integer, then U (·, 0) is of C 2σ+α (B1/2 ). Moreover, we have kU (·, 0)kC 2σ+α(B1/2 ) ≤ C(kU kL∞ (Q1 ) + kbkC α (B1 ) ), where C > 0 depends only on n, σ, α, kakC α (B1 ) . Proof. From Proposition 2.4, U is H¨older continuous in Q3/4 . Theorem 2.3 follows from bootstrap arguments by applying Theorem 2.2 with g(x) := a(x)U (x, 0) + b(x).

15

Proof of Theorem 2.2. Our arguments are in the spirit of those in [18] and [71]. Denote C as various constants that depend only on n and σ. Let ρ = 12 , Qk = Qρk (0), ∂ ′ Qk = Bk , k = 0, 1, 2, · · · . (Note that we have abused notations a little bit. Only in this proof we refer Qk , Bk as Qρk , Bρk .) We also denote M = kgkC α(B2 ) . From Proposition 2.4 we have already known that U is H¨older continuous in Q0 . First we assume that α ∈ (0, 1) Step 1: We consider the case of 2σ + α < 1. Let Wk be the unique weak solution of (which is guaranteed by Proposition 2.3) 1−2σ ∇Wk (X)) = 0 in Qk div(t 1−2σ − lim+ t ∂t Wk (x, t) = g(0) − g(x) on ∂ ′ Qk (2.14) t→0 W (X) = 0 on ∂ ′′ Q k

k

Let Uk = Wk + U in Qk and hk+1 = Uk+1 − Uk in Qk+1 , then

kWk kL∞ (Qk ) ≤ CM ρ(2σ+α)k .

(2.15)

Indeed (2.15) follows by applying Lemma 2.1 to the equation of ρ−2σk Wk (ρk x) ± (t2σ − 3)M ραk in Q0 . Hence by weak maximum principle again we have khk+1 kL∞ (Qk ) ≤ CM ρ(2σ+α)k . By Proposition 2.5, we have, for i = 0, 1, 2, 3 k∇ix hk+1 kL∞ (Qk+2 ) ≤ CM ρ(2σ+α−i)k .

(2.16)

Similarly apply Proposition 2.5 to U0 , we have k∇ix U0 kL∞ (Q2 ) ≤ C(kU0 kL∞ (Q1 ) + M ) ≤ C(kU kL∞ (Q0 ) + M ) For any given point z near 0, we have |U (z, 0) − U (0, 0)| ≤ |Uk (0, 0) − U (0, 0)| + |U (z, 0) − Uk (z, 0)| + |Uk (z, 0) − Uk (0, 0)| = I1 + I2 + I3

Let k be such that ρk+4 ≤ |z| ≤ ρk+3 . By (2.15),

I1 + I2 ≤ CM ρ(2σ+α)k ≤ CM |z|2σ+α .

For I3 , by (2.16) and (2.17), I3 ≤ |U0 (z, 0) − U0 (0, 0)| +

k X j=1

|hj (z, 0) − hj (0, 0)|

k X k∇x hj kL∞ (Qk+3 ) ≤ C|z| k∇x U0 kL∞ (Qk+3 ) + j=1

k X ≤ C|z| kU kL∞ (Q0 ) + M + M ρ(2σ+α−1)j j=1

≤ C|z| kU kL∞ (Q0 ) + M (1 + |z|2σ+α−1 ) . 16

(2.17)

Thus, for 2σ + α < 1, we have |U (z, 0) − U (0, 0)| ≤ C M + kU kL∞ (Q0 ) |z|2σ+α .

which finishes the proof of Step 1.

Step 2: For 1 < 2σ + α < 2, the arguments in Step 1 imply that k∇x U (·, 0)kL∞ (B1 ) ≤ C kU kL∞ (Q0 ) + M .

(2.18)

Apply (2.18) to the equation of Wk we have, together with (2.15),

k∇x Wk (·, 0)kL∞ (Bk+1 ) ≤ CM ρ(2σ+α−1)k By (2.16) and (2.17), |∇x Uk (z, 0) − ∇x Uk (0, 0)| ≤ |∇x U0 (z, 0) − ∇x U0 (0, 0)| +

k X j=1

|∇x hj (z, 0) − ∇x hj (0, 0)|

k X k∇2x hj kL∞ (Qk+3 ) ≤ C|z| k∇2x U0 kL∞ (Qk+3 ) + j=1

≤ C|z| kU kL∞ (Q0 ) + M + M

Hence

k X

ρ(2σ−2+α)j

j=1

≤ C|z| kU kL∞ (Q0 ) + M (1 + |z|2σ+α−2 ) .

|∇x U (z, 0) − ∇x U (0, 0)| ≤ |∇x Wk (0, 0)| + |∇x Wk (z, 0)| + |∇x Uk (z, 0) − ∇x Uk (0, 0)| ≤ CM ρ(2σ+α−1)k + C|z| kU kL∞ (Q0 ) + M (1 + |z|2σ+α−2 ) ≤ C M + kU kL∞(Q0 ) |z|2σ+α−1 .

which finishes the proof of Step 2.

Step 3: For 2σ + α > 2, the arguments in Step 2 imply that k∇2x U (·, 0)kL∞ (B1 ) ≤ C kU kL∞ (Q0 ) + M ,

Apply (2.19) to the equation of Wk we have, together with (2.15),

k∇2x Wk (·, 0)kL∞ (Bk+1 ) ≤ CM ρ(2σ+α−2)k

17

(2.19)

By (2.16) and (2.17), |∇2x Uk (z, 0) − ∇2x Uk (0, 0)| ≤ |∇2x U0 (z, 0) − ∇2x U0 (0, 0)| +

≤ C|z| k∇3x U0 kL∞ (Qk+3 ) +

k X j=1

k X j=1

|∇2x hj (z, 0) − ∇2x hj (0, 0)|

k∇3x hj kL∞ (Qk+3 )

k X ρ(2σ+α−3)k ≤ C|z| kU kL∞ (Q0 ) + M + M j=1

Hence

≤ C|z| kU kL∞ (Q0 ) + M (1 + |z|2σ+α−3 ) .

|∇2x U (z, 0) − ∇2x U (0, 0)|

≤ |∇2x Wk (0, 0)| + |∇2x Wk (z, 0)| + |∇2x Uk (z, 0) − ∇x Uk (0, 0)| ≤ CM ρ(2σ+α−2)k + C|z| kU kL∞ (Q0 ) + M (1 + |z|2σ+α−3 ) ≤ C M + kU kL∞(Q0 ) |z|2σ+α−2 .

which finishes the proof of Step 3. This finishes the proof of Theorem 2.2 for α ∈ (0, 1). For the case that α > 1, we may apply ∇x to (2.13) [α] times, as in the proof of Proposition 2.5, and repeat the above three steps. Theorem 2.2 is proved. Proof of Theorem 2.1. Since u ∈ H˙ σ (Rn ) is nonnegative, its extension U ≥ 0 in Rn+1 and U ∈ + H(t1−2σ , Q1 ) is a weak solution of (2.7) in Q1 . The theorem follows immediately from Theorem 2.3 and Proposition 2.4. Remark 2.4. Another way to show Theorem 2.1 is the following. Let u ∈ H˙ σ (Rn ) and u ≥ 0 in Rn be a solution of (−∆)σ u = g(x), in B1 where g ∈ C α (B1 ). Let η be a nonnegative smooth cut-off function supported in B1 and equal to 1 in B7/8 . Let v ∈ H˙ σ (Rn ) be the solution of (−∆)σ v = η(x)g(x),

in Rn

where ηg is considered as a function defined in Rn and supported in B1 , i.e., v is a Riesz potential of ηg Z Γ( n−2σ η(y)g(y) 2 ) dy. v(x) = 2σ n/2 2 π Γ(σ) Rn |x − y|n−2σ Then if 2σ + α and α are not integers, we have (see, e.g., [86])

kvkC 2σ+α (B1/2 ) ≤ C(kvkL∞ (Rn ) + kηgkC α (Rn ) ) ≤ CkgkC α (B1 ) . 18

Let w = u − v which belongs to H˙ σ (Rn ) and satisfies (−∆)σ w = 0,

in B7/8 .

˜ = W + kvkL∞ (Rn ) ≥ 0 in Rn+1 . Notice that W ˜ Let W = Pσ [w] be the extension of w, and W + is a nonnegative weak solution of (2.7) with a ≡ b ≡ 0 and D = Q1 . By Proposition 2.5 and Proposition 2.4, we have kw + kvkL∞ (Rn ) kC 2σ+α (B1/2 )

˜ kL2 (t1−2σ ,Q ) ≤ C inf W ˜ ≤ C( inf u + kvkL∞ (Rn ) ). ≤ CkW 7/8 Q3/4

Q3/4

Hence kukC 2σ+α(B1/2 ) ≤ kvkC 2σ+α (B1/2 ) + kwkC 2σ+α (B1/2 ) ≤ C( inf u + kgkC α (B1 ) ). B3/4

Using bootstrap arguments as that in the proof of Theorem 2.3, we conclude Theorem 2.1. Remark 2.5. Indeed, our proofs also lead to the following. If we only assume that a(x), b(x), g(x) ∈ L∞ (B1 ), and let U , u be those in Theorem 2.2 and in Theorem 2.1 respectively, then the estimates kU (·, 0)kC 2σ (B1/2 ) ≤ C1 (kU kL∞ (Q1 ) + kgkL∞(B1 ) ) kukC 2σ (B1/2 ) ≤ C2 ( inf u + kbkL∞ (B3/4 ) ) B3/4

hold provided σ 6= 1/2 , where C1 > 0 depends only on n, σ, α and C2 > 0 depends only on n, σ, α, kakL∞ (B3/4 ) . For σ = 21 , we have the following log-Lipschitz property: for any y1 , y2 ∈ B1/4 , y1 6= y2 , |U (y1 , 0) − U (y2 , 0)| ≤ C1 (kU kL∞ (Q1 ) − kgkL∞ (B1 ) log |y1 − y2 |), |y1 − y2 | |u(y1 ) − u(y2 )| ≤ −C2 log |y1 − y2 |( inf u + kbkL∞(B3/4 ) ) B3/4 |y1 − y2 | where C1 > 0 depends only on n, σ and C2 > 0 depends only on n, σ, kakL∞ (B3/4 ) . Next we have Lemma 2.3. (Lemma 4.5 in [23]) Let g ∈ C α (B1 ) for some α ∈ (0, 1) and U ∈ L∞ (Q1 ) ∩ H(t1−2σ , Q1 ) be a weak solution of (2.13). Then there exists β ∈ (0, 1) depending only on n, σ, α such that t1−2σ ∂t U ∈ C β (Q1/2 ). Moreover, there exists a positive constant C > 0 depending only on n, σ and β such that kt1−2σ ∂t U kC β (Q1/2 ) ≤ C(kU kL∞ (Q1 ) + kgkC α (B1 ) ).

19

Proposition 2.6. Suppose that K ∈ C 1 (B1 ), U ∈ H(t1−2σ , Q1 ) and U ≥ 0 in Q1 is a weak solution of ( div(t1−2σ ∇U ) = 0, in Q1 (2.20) 1−2σ p − lim t ∂t U (x, t) = K(x)U (x, 0), on ∂ ′ Q1 , t→0

n+2σ where 1 ≤ p ≤ n−2σ . Then there n, σ, p, kU kL∞ (Q1 ) , kKkC 1(Q1 ) such

∇x U

exist C > 0 and α ∈ (0, 1) both of which depend only on that

and t1−2σ ∂t U

are of C α (Q1/2 )

and k∇x U kC α (Q1/2 ) + kt1−2σ ∂t U kC α (Q1/2 ) ≤ C. Proof. We use C and α to denote various positive constants with dependence specified as in the ′ proposition, which may vary from line to line. By Corollary 2.1, U ∈ L∞ loc (Q1 ∪ ∂ Q1 ) and kU kC α (Q8/9 ) ≤ C. With the above, we may apply Theorem 2.3 to obtain U (·, 0) ∈ C 1,σ (B7/8 ) and kU (·, 0)kC 1,σ (B7/8 ) ≤ C. Hence we may differentiate (2.20) with respect to x (which can be justified from the proof of Proposition 2.5) and apply Proposition 2.4 to ∇x U to obtain k∇x U kC α (Q1/2 ) ≤ C. Finally we can apply Lemma 2.3 to obtain kt1−2σ ∂t U kC α (Q1/2 ) ≤ C.

3 Proof of Theorem 1.5 n+1 ∞ We first introduce some notations. We say that U ∈ L∞ loc (R+ ) if U ∈ L (QR ) for any R > 0.

1−2σ Similarly we say U ∈ Hloc (t1−2σ , Rn+1 , QR ) for any R > 0. + ) if U ∈ H(t + In the following BR (X) is denoted as the ball in Rn+1 with radius R and center X, and BR (X) n+1 + + as BR (X) ∩ R+ . We also write BR (0), BR (0) as BR , BR for short respectively. We start with a Lemma, which is a version of the strong maximum principle.

Proposition 3.1. Suppose U (X) ∈ H(t1−2σ , Dε ) ∩ C(B1+ ∪ B1 \ {0}) and U > 0 in B1+ ∪ B1 \ {0} is a weak supersolution of (2.7) with a ≡ b ≡ 0 and D = Dε := B1+ \ Bε+ for any 0 < ε < 1, then lim inf U (x, t) > 0.

(x,t)→0

20

Proof. For any δ > 0, let Vδ = U + Then V is also a weak supersolution in D

δ − min U. + |(x, t)|n−2σ ∂ ′′ B0.8 2

δ n−2σ

. Applying Lemma 2.1 to Vδ in D

2 . For any (x, t) ∈ ciently small δ, we have Vδ ≥ 0 in D n−2σ δ 0, i.e., U (x, t) ≥ min∂ ′′ B+ U .

+ B0.8 \{0},

2

δ n−2σ

for suffi-

we have limδ→0 Vδ (x, t) ≥

0.8

The proof of Theorem 1.5 uses the method of moving spheres and is inspired by [73], [72] and [24]. For each x ∈ Rn and λ > 0, we define, X = (x, 0), and n−2σ λ2 (ξ − X) λ UX,λ (ξ) := , ξ ∈ Rn+1 U X+ (3.1) + \{X}, |ξ − X| |ξ − X|2 the Kelvin transformation of U with respect to the ball Bλ (X). We point out that if U is a solution ¯ ∈ ∂Rn+1 \ Bε+ , for every x of (1.9), then Ux¯,λ is a solution of (1.9) in Rn+1 + , λ > 0, and ε > 0. + n+1 By Corollary 2.1 any nonnegative weak solution U of (1.9) belongs to L∞ loc (R+ ), and hence by

Proposition 2.4, U is H¨older continuous and positive in Rn+1 + . By Theorem 2.2, U (·, 0) is smooth in Rn . From classical elliptic equations theory, U is smooth in Rn+1 + . Lemma 3.1. For any x ∈ Rn , there exists a positive constant λ0 (x) such that for any 0 < λ < λ0 (x), + UX,λ (ξ) ≤ U (ξ), in Rn+1 (3.2) + \Bλ (X). Proof. Without loss of generality we may assume that x = 0 and write Uλ = U0,λ . Step 1. We show that there exist 0 < λ1 < λ2 which may depend on x, such that Uλ (ξ) ≤ U (ξ), ∀ 0 < λ < λ1 , λ < |ξ| < λ2 . 2

λ ξ + For every 0 < λ < λ1 < λ2 , ξ ∈ ∂ ′′ Bλ2 , we have |ξ| 2 ∈ Bλ . Thus we can choose λ1 = λ1 (λ2 ) 2 small such that n−2σ 2 λ λ ξ Uλ (ξ) = U |ξ| |ξ|2 n−2σ λ1 ≤ sup U ≤ inf+ U ≤ U (ξ) λ2 ∂ ′′ Bλ B+ 2

λ2

Hence Uλ ≤ U

on ∂ ′′ (Bλ+2 \Bλ+ )

for all λ2 > 0 and 0 < λ < λ1 (λ2 ). We will show that Uλ ≤ U on (Bλ+2 \Bλ+ ) if λ2 is small and 0 < λ < λ1 (λ2 ). Since Uλ satisfies (1.9) in Bλ+2 \ Bλ+1 , we have div(t1−2σ ∇(Uλ − U )) = 0, lim t1−2σ ∂t (Uλ − U ) t→0

=U

in n+2σ n−2σ

Bλ+2 \Bλ+ ;

n+2σ

(x, 0) − Uλn−2σ (x, 0), 21

on ∂ ′ (Bλ+2 \Bλ+).

(3.3)

Let (Uλ − U )+ := max(0, Uλ − U ) which equals to 0 on ∂ ′′ (Bλ+2 \Bλ+ ). Hence, by a density argument, we can use (Uλ − U )+ as a test function in the definition of weak solution of (3.3). We will make use of the narrow domain technique from [11]. With the help of the mean value theorem, we have Z t1−2σ |∇(Uλ − U )+ |2 + + Bλ \Bλ

=

Z2

n+2σ

Bλ2 \Bλ

Z

≤C

n+2σ

(Uλn−2σ (x, 0) − U n−2σ (x, 0))(Uλ − U )+ 4σ

Bλ2 \Bλ

≤C

Z

≤C

Z

((Uλ − U )+ )2 Uλn−2σ

Bλ2 \Bλ

+ + Bλ \Bλ

((Uλ − U )+ ) t

1−2σ

2

2n n−2σ

! n−2σ n + 2

|∇(Uλ − U ) |

Z

2n n−2σ

Bλ2 \Bλ

! Z

U

Uλ

2n n−2σ

Bλ2

! 2σ n

! 2σ n

where Proposition 2.1 is used in the last inequality and C is a positive constant depending only on n and σ. We fix λ2 small such that ! 2σ Z n 2n

C

U n−2σ

< 1/2.

Bλ2

Then ∇(Uλ − U )+ = 0 in Bλ+2 \Bλ+. Since (Uλ − U )+ = 0 on ∂ ′′ (Bλ+2 \Bλ+), (Uλ − U )+ = 0 in Bλ+2 \Bλ+. We conclude that Uλ ≤ U on (Bλ+2 \Bλ+) for 0 < λ < λ1 := λ1 (λ2 ). Step 2. We show that there exists λ0 ∈ (0, λ1 ) such that ∀ 0 < λ < λ0 Uλ (ξ) ≤ U (ξ), |ξ| > λ2 , ξ ∈ Rn+1 + .

Let φ(ξ) =

λ2 |ξ|

n−2σ

inf U , which satisfies

∂ ′′ Bλ2

(

div(t1−2σ ∇φ) = 0, − limt→0 t1−2σ ∂t φ(x, t) = 0,

in Rn+1 \ Bλ+2 + x ∈ Rn \ Bλ2 ,

and φ(ξ) ≤ U (ξ) on ∂ ′′ Bλ2 . By the weak maximum principle Lemma 2.1, n−2σ λ2 inf U, ∀ |ξ| > λ2 , ξ ∈ Rn+1 U (ξ) ≥ + . ∂ ′′ Bλ2 |ξ| 1

Let λ0 = min(λ1 , λ2 ( ′′inf U/ sup U ) n−2σ ). Then for any 0 < λ < λ0 , |ξ| ≥ λ2 , we have ∂ B λ2

Uλ (ξ) ≤ (

B λ2

λ0 λ2 λ n−2σ λ2 ξ ) U ( 2 ) ≤ ( )n−2σ sup U ≤ ( )n−2σ ′′inf U ≤ U (ξ). ∂ B λ2 |ξ| |ξ| |ξ| |ξ| B λ2

Lemma 3.1 is proved. 22

With Lemma 3.1, we can define for all x ∈ Rn , + ¯ λ(x) = sup{µ > 0 : UX,λ ≤ U in Rn+1 + \Bλ , ∀ 0 < λ < µ}.

¯ By Lemma 3.1, λ(x) ≥ λ0 (x).

¯ Lemma 3.2. If λ(x) < ∞ for some x ∈ Rn , then UX,λ(x) ≡ U. ¯ ¯ = λ(0). ¯ Proof. Without loss of generality we assume that x = 0 and write Uλ = U0,λ and λ By the ¯ definition of λ, Uλ¯ ≥ U in Bλ+ ¯ \{0},

¯ and therefore, for all 0 < ε < λ, ( div(t1−2σ ∇(Uλ − U )) = 0, in Bλ+ \Bε+ ; − lim t1−2σ ∂t (Uλ − U ) ≥ 0 on ∂ ′ (Bλ+ \Bε+).

(3.4)

t→0

We argue by contradiction. If Uλ¯ is not identically equal to U , applying the Harnack inequality Proposition 2.4 to (3.4), we have Uλ¯ > U in Bλ¯ \{{0} ∪ ∂ ′′ Bλ¯ }, and in view of Proposition 3.1, lim inf (Uλ¯ (ξ) − U (ξ)) > 0. ξ→0

So there exist ε1 > 0 and c > 0 such that Uλ¯ (ξ) > U (0) + c, ∀ 0 < |ξ| < ε1 . Choose ε2 small such that ¯ n−2σ λ c (U (0) + c) > U (0) + . ¯ + ε2 2 λ ¯<λ<λ ¯ + ε2 , Thus for all 0 < |ξ| < ε1 and λ Uλ (ξ) =

¯ 2 ¯ n−2σ ¯ n−2σ λ λ ξ λ ≥ ¯ Uλ¯ (U (0) + c) ≥ U (0) + c/2. 2 λ λ λ + ε2

Choose ε3 small such that for all 0 < |ξ| < ε3 , U (0) > U (ξ) − c/4. Hence for all 0 < |ξ| < ε3 and ¯<λ<λ ¯ + ε2 , λ Uλ (ξ) > U (ξ) + c/4. ¯ − δ}. Then there For δ small, which will be fixed later, denote Kδ = {ξ ∈ Rn+1 : ε3 ≤ |ξ| ≤ λ + exists c2 = c2 (δ) such that Uλ¯ (X) − U (X) > c2 in Kδ . ¯ < λ< By the uniform continuous of U on compact sets, there exists ε4 ≤ ε2 such that for all λ ¯ λ + ε4 Uλ − Uλ¯ > −c2 /2 in Kδ . 23

Hence Uλ − U > c2 /2 in Kδ .

¯ − δ ≤ |ξ| ≤ λ}. Using the narrow domain technique Now let us focus on the region {ξ ∈ Rn+1 :λ + as that in Lemma 3.1, we can choose δ small (notice that we can choose ε4 as small as we want) such that ¯ − δ ≤ |ξ| ≤ λ}. Uλ ≥ U in {ξ ∈ Rn+1 :λ +

¯<λ<λ ¯ + ε4 In conclusion there exists ε4 such that for all λ

Uλ ≥ U in {ξ ∈ Rn+1 : 0 < |ξ| ≤ λ} + ¯ which contradicts with the definition of λ. Proof of Theorem 1.5. It follows from the same arguments in [72], with the help of Lemma 3.2, that: ¯ ¯ (i) Either λ(x) = ∞ for all x ∈ Rn or λ(x) < ∞ for all x ∈ Rn ; (Lemma 2.3 in [72]) n ¯ (ii) If for all x ∈ R , λ(x) = ∞ then U (x, t) = U (0, t), ∀ (x, t) ∈ Rn+1 + ; (Lemma 11.3 in [72]) ¯ (iii) If λ(x) < ∞ for all x ∈ Rn , then by Lemma 11.1 in [72] u(x) := U (x, 0) = a

λ 2 1 + λ |x − x0 |2

n−2σ 2

(3.5)

where λ > 0, a > 0 and x0 ∈ Rn . We claim that (ii) never happens, since this would imply, using (1.9), that n+2σ

U (x, t) = U (0) − U (0) n−2σ

t2σ 2σ

which contradicts to the positivity of U . Then (iii) holds. We are only left to show that V := U − Pσ [u] ≡ 0 where u(x) is given in (3.5) and belongs to H˙ σ (Rn ). Hence, V satisfies ( div(t1−2σ ∇V ) = 0, in Rn+1 + V = 0 on ∂Rn+1 + . By Lemma 3.2, we know that Vλ¯ can be extended toR a smooth function near 0. Multiplying the above equation by V and integrating by parts, it leads to Rn+1 t1−2σ |∇V |2 = 0. Hence we have V ≡ 0. + n−2σ Finally a = Nσ cn,σ 22σ 4σ follows from (1.3) with φ = 1 and (2.5).

4 Local analysis near isolated blow up points The analysis in this and next section adapts the blow up analysis developed in [88] and [68] to give accurate blow up profiles for solutions of degenerate elliptic equations. For σ = 21 , similar results have been proved in [57] and [45], where equations are elliptic.

24

Let Ω ⊂ Rn (n ≥ 2) be a domain, τi ≥ 0 satisfy limi→∞ τi = 0, pi = (n + 2σ)/(n − 2σ) − τi , and Ki ∈ C 1,1 (Ω) satisfy, for some constants A1 , A2 > 0, that 1/A1 ≤ Ki (x) ≤ A1 for all x ∈ Ω, kKi kC 1,1 (Ω) ≤ A2 .

(4.1)

Let ui ≥ 0 in Rn and ui ∈ L∞ (Ω) ∩ H˙ σ (Rn ) satisfying

(−∆)σ ui = c(n, σ)Ki (x)upi i ,

in Ω.

(4.2)

We say that {ui } blows up if kui kL∞ (Ω) → ∞ as i → ∞.

Definition 4.1. Suppose that {Ki } satisfies (4.1) and {ui } satisfies (4.2). We say a point y ∈ Ω is an isolated blow up point of {ui } if there exist 0 < r < dist(y, Ω), C > 0, and a sequence yi tending to y, such that, yi is a local maximum of ui , ui (yi ) → ∞ and ui (y) ≤ C|y − yi |−2σ/(pi −1)

for all y ∈ Br (yi ).

Let yi → y be an isolated blow up of ui , define Z 1 ui , ui (r) = |∂Br | ∂Br (yi )

r > 0,

(4.3)

and w j (r) = r2σ/(pi −1) ui (r),

r > 0.

Definition 4.2. We say yi → y ∈ Ω is an isolated simple blow up point, if yi → y is an isolated blow up point, such that, for some ρ > 0 (independent of i) w i has precisely one critical point in (0, ρ) for large i. In this section, we are mainly concerned with the profile of blow up of {ui }. And under certain conditions, we can show that isolated blow up points have to be isolated simple blow up points. Let ui ∈ C 2 (Ω) ∩ H˙ σ (Rn ) and ui ≥ 0 in Rn satisfy (4.2) with Ki satisfying (4.1). Without loss of generality, we assume throughout this section that B2 ⊂ Ω and yi → 0 as i → ∞ is an isolated blow up point of {ui } in Ω. Let Ui = Pσ [ui ] be the extension of ui (see (2.2)). Then we have div(t1−2σ ∇Ui ) = 0, in Rn+1 + , (4.4) ∂U (x, t) i − lim t1−2σ for any x ∈ Ω, = c0 Ki (x)Ui (x, 0)pi , t→0 ∂t where c0 = Nσ c(n, σ) with Nσ = 21−2σ Γ(1 − σ)/Γ(σ). Lemma 4.1. Suppose that ui ∈ C 2 (Ω) ∩ H˙ σ (Rn ) and ui ≥ 0 in Rn satisfies (4.2) with {Ki } satisfying (4.1), and yi → 0 is an isolated blow up point of {ui }, i.e., for some positive constants A3 and r¯ independent of i, |y − yi |2σ/(pi −1) ui (y) ≤ A3 ,

for all y ∈ Br¯ ⊂ Ω.

Denote Ui = Pσ [ui ], and Yi = (yi , 0). Then for any 0 < r < inequality Ui ≤ C inf sup

1 3 r,

+ + B2r (Yi )\Br/2 (Yi )

+ + (Yi ) B2r (Yi )\Br/2

we have the following Harnack Ui ,

where C is a positive constant depending only on n, σ, A3 , r¯ and sup kKi kL∞ (Br (yi )) . i

25

(4.5)

Proof. For 0 < r < 3r¯ , set Vi (Y ) = r2σ/(pi −1) Ui (Yi + rY ),

in Y ∈ B3+ .

It is easy to see that div(s1−2σ ∇Vi ) = 0,

in B3+ ,

and − lim s1−2σ ∂s Vi (y, s) = c0 K(yi + ry)Vi (y, 0)pi , s→0

on ∂ ′ B3+ .

Since yi → 0 is an isolated blow up point of ui , Vi (y, 0) ≤ A3 |y|−2σ/(pi −1) ,

for all y ∈ B3 .

Lemma 4.1 follows after applying Proposition 2.4 and the standard Harnack inequality for uniform elliptic equation together to Vi in the domain Q2 \ Q1/2 . Proposition 4.1. Suppose that ui ∈ C 2 (Ω) ∩ H˙ σ (Rn ) and ui ≥ 0 in Rn satisfies (4.2) with Ki ∈ C 1,1 (Ω) satisfying (4.1). Suppose also that yi → 0 be an isolated blow up point of {ui } with (4.5). Then for any Ri → ∞, εi → 0+ , we have, after passing to a subsequence (still denoted as {ui }, {yi }, etc. ...), that −(pi −1)/2σ

km−1 i ui (mi

· +yi ) − (1 + ki | · |2 )(2σ−n)/2 kC 2 (B2Ri (0)) ≤ εi , −(pi −1)/2σ

Ri mi

where mi = ui (yi ) and ki = Ki (yi )1/σ /4.

→0

as i → ∞,

Proof. Let −(pi −1)/2σ

φi (x) = m−1 i ui (mi It follows that

x + yi ),

for x ∈ Rn .

−(pi −1)/2σ

x + yi )φpi i ,

(−∆)σ φi (x) = c(n, σ)Ki (mi 0 < φi ≤ A3 |x|−2σ/(pi −1) ,

(pi −1)/2σ

|x| < rmi

,

(4.6)

and φi (0) = 1,

∇φi (0) = 0.

Let Φi = Pσ [φi ] be the extension of φi (see (2.2)). Then Φ satisfies pi −1 div(t1−2σ ∇Φi (x, t)) = 0, |x, t| < r¯mi 2σ , pi −1 − lim t1−2σ ∂t Φi (x, t) = Nσ c(n, σ)Ki (m− 2σ x + yi )Φi (x, 0)pi , i

t→0

pi −1

|x| < r¯mi 2σ .

By the weak maximum principle we have, for any 0 < r < 1, 1 = φi (0) = Φi (0, 0) ≥ min Φi . It ′′ follows from Lemma 4.1 that

Φi ≤ C. Φi ≤ C min max φi ≤ max ′′ ′′ ∂Br

∂ Br

∂ Br

26

∂ Br

Namely, max φi ≤ C B1

for some C > 0 depending on n, σ, A1 , A2 , A3 . This and (4.6) implies that for any R > 1 max φi ≤ C(R) BR

for some C(R) > 0 depending on n, σ, A1 , A2 , A3 and R. Then by Corollary 2.1 there exists some α ∈ (0, 1) such that for every R > 1, kΦi kH(t1−2σ ,QR ) + kΦi kC α (QR ) ≤ C1 (R), where α and C1 (R) are independent of i. Bootstrap using Theorem 2.1, we have, for every 0 < β < 2 with 2σ + β 6∈ N, kφi kC 2σ+β (BR ) ≤ C2 (R, β) where C2 (R, β) is independent of i. Thus, after passing to a subsequence, we have, for some nonα negative functions Φ(X) ∈ Hloc (t1−2σ , Rn+1 ) ∩ Cloc (Rn+1 ) and φ ∈ C 2 (Rn ), 1−2σ , Rn+1 + ), Φi ⇀ Φ weakly in Hloc (t α/2 n+1 Φi → Φ in Cloc (R+ ), 2 φi → φ in Cloc (Rn ).

It follows that

Φ(·, 0) ≡ φ,

and Φ satisfies (

φ(0) = 1,

∇φ(0) = 0,

div(t1−2σ ∇Φ) = 0 − lim t1−2σ ∂t Φ(x, t) = c0 KΦ(x, 0)(n+2σ)/(n−2σ)

in Rn+1 , on ∂ ′ Rn+1 ,

t→0

with K = lim Ki (yi ). By Theorem 1.5, we have i→∞

φ(x) = (1 + lim ki |x|2 )(2σ−n)/2 , i→∞

where ki = Ki (yi )1/σ /4. Proposition 4.1 follows immediately. Note that since passing to subsequences does not affect our proofs, we will always choose Ri → ∞ first, and then εi → 0+ as small as we wish (depending on Ri ) and then choose our subsequence {ui } to work with. Proposition 4.2. Under the hypotheses of Proposition 4.1, there exists some positive constant C = C(n, σ, A1 , A2 , A3 ) such that, (pi −1)/σ

ui (y) ≥ C −1 mi (1 + ki mi

|y − yi |2 )(2σ−n)/2 ,

In particular, for any e ∈ Rn , |e| = 1, we have −1+((n−2σ)/2σ)τi

ui (yi + e) ≥ C −1 mi where τi = (n + 2σ)/(n − 2σ) − pi . 27

.

|y − yi | ≤ 1.

−(pi −1)/2σ

Proof. Denote ri = Ri mi

. It follows from Proposition 4.1 that ri → 0 and

ui (y) ≥ C −1 mi Ri2σ−n ,

for all |y − yi | = ri .

By the Harnack inequality Lemma 4.1, we have Ui (Y ) ≥ C −1 mi Ri2σ−n ,

for all |Y − Yi | = ri ,

where Ui = Pσ [ui ] is the extension of ui , Y = (y, s) with s ≥ 0, and Yi = (yi , 0). Set 3 Ψi (Y ) = C −1 Ri2σ−n rin−2σ mi (|Y − Yi |2σ−n − ( )2σ−n ), 2

ri ≤ |Y − Yi | ≤

3 . 2

Clearly, Ψi satisfies div(s1−2σ ∇Ψi ) = 0 = div(s1−2σ ∇Ui ),

ri ≤ |Y − Yi | ≤

on ∂ ′′ Bri ∪ ∂ ′′ B3/2 ,

Ψi (Y ) ≤ Ui (Y ),

− lim+ s1−2σ ∂s Ψi (y, s) = 0 ≤ − lim+ s1−2σ ∂s Ui (y, s), s→0

3 , 2

s→0

ri ≤ |y − yi | ≤

3 . 2

By the weak maximum principle Lemma 2.1 applied to Ui − Ψi , we have Ui (Y ) ≥ Ψi (Y ) for all ri ≤ |Y − Yi | ≤

3 . 2

Therefore, Proposition 4.2 follows immediately from Proposition 4.1. Lemma 4.2. Under the hypotheses of Proposition 4.1, and in addition that yi → 0 is also an isolated −2σ+o(1) ), such that simple blow up point with the constant ρ, there exist δi > 0, δi = O(Ri ui (y) ≤ C1 ui (yi )−λi |y − yi |2σ−n+δi ,

for all ri ≤ |y − yi | ≤ 1,

where λi = (n − 2σ − δi )(pi − 1)/2σ − 1 and C1 is some positive constant depending only on n, σ, A1 , A3 and ρ. Proof. From Proposition 4.1, we see that ui (y) ≤ Cui (yi )Ri2σ−n

for all |y − yi | = ri .

(4.7)

Let ui (r) be the average of ui over the sphere of radius r centered at yi . It follows from the assumption of isolated simple blow up and Proposition 4.1 that r2σ/(pi −1) ui (r)

is strictly decreasing for ri < r < ρ.

By Lemma 4.1, (4.8) and (4.7), we have, for all ri < |y − yi | < ρ, |y − yi |2σ/(pi −1) ui (y) ≤ C|y − yi |2σ/(pi −1) ui (|y − yi |) 2σ/(pi −1)

≤ ri

≤ CRi 28

ui (ri )

2σ−n +o(1) 2

,

(4.8)

where o(1) denotes some quantity tending to 0 as i → ∞. Applying Lemma 4.1 again, we obtain −2σ+o(1)

Ui (Y )pi −1 ≤ O(Ri

)|Y − Yi |−2σ

for all ri ≤ |Y − Yi | ≤ ρ.

Consider operators ( L(Φ) = div(s1−2σ ∇Φ(Y )), Li (Φ) = − lim+ s1−2σ ∂s Φ(y, s) − c0 Ki upi i −1 (y)Φ(y, 0), s→0

(4.9)

in B2+ , on ∂ ′ B2+ .

Clearly, Ui > 0 satisfies L(Ui ) = 0 in B2+ and Li (Ui ) = 0 on ∂ ′ B2+ . For 0 ≤ µ ≤ n − 2σ, a direct computation yields L(|Y − Yi |−µ − εs2σ |Y − Yi |−(µ+2σ) ) n ε(µ + 2σ)(n − µ)s2σ o = s1−2σ |Y − Yi |−(µ+2) − µ(n − 2σ − µ) + |Y − Yi |2σ and Li (|Y − Yi |−µ − εs2σ |Y − Yi |−(µ+2σ) ) = |Y − Yi |−(u+2σ) (2εσ − c0 Ki upi i −1 |Y − Yi |2σ ). −2σ+o(1)

It follows from (4.9) that we can choose εi = O(Ri −2σ+o(1) ) > 0 such that for ri < |y − yi | < ρ, O(Ri

) > 0, and then choose δi =

Li (|Y − Yi |−δi − εi s2σ |Y − Yi |−(δi +2σ) ) ≥ 0,

Li (|Y − Yi |2σ−n+δi − εi s2σ |Y − Yi |−n+δi ) ≥ 0 and for ri < |Y − Yi | < ρ, L(|Y − Yi |−δi − εi s2σ |Y − Yi |−(δi +2σ) ) ≤ 0,

L(|Y − Yi |2σ−n+δi − εi s2σ |Y − Yi |−n+δi ) ≤ 0. Set Mi = 2 max∂ ′′ Bρ+ Ui , λi = (n − 2σ − δi )(pi − 1)/2σ − 1 and Φi =Mi ρδi (|Y − Yi |−δi − εi s2σ |Y − Yi |−(δi +2σ) )

+ 2Aui (yi )−λi (|Y − Yi |2σ−n+δi − εi s2σ |Y − Yi |−n+δi ),

where A > 1 will be chosen later. By the choice of Mi and λi , we immediately have Φi (Y ) ≥ Mi ≥ Ui (Y ) for all |Y − Yi | = ρ. Φi ≥ AUi (Yi )Ri2σ−n+δi ≥ AUi (Yi )Ri2σ−n

for all |Y − Yi | = ri .

Due to (4.9), we can choose A to be sufficiently large such that Φi ≥ U i

for all |Y − Yi | = ri .

Therefore, applying maximum principles in section A.3 to Φi − Ui in Bρ \Bri , it yields U i ≤ Φi

for all ri ≤ |Y − Yi | ≤ ρ. 29

For ri < θ < ρ, the same arguments as that in (4.9) yield ρ2σ/(pi −1) Mi ≤ Cρ2σ/(pi −1) ui (ρ) ≤ Cθ2σ/(pi −1) ui (θ)

≤ Cθ2σ/(pi −1) {Mi ρδi θ−δi + Aui (yi )−λi θ2σ−n+δi }.

Choose θ = θ(n, σ, ρ, A1 , A2 , A3 ) sufficiently small so that Cθ2σ/(pi −1) ρδi θ−δi ≤

1 2σ/(pi −1) ρ . 2

It follows that Mi ≤ Cui (yi )−λi . Then Lemma 4.2 follows from the above and the Harnack inequality. Below we are going to improve the estimate in Lemma 4.2. First, we prove a Pohozaev type identity. + + Proposition 4.3. Suppose that K ∈ C 1 (B2R ). Let U ∈ H(t1−2σ , B2R ) and U ≥ 0 in B2R be a weak solution of ( + div(t1−2σ ∇U ) = 0, in B2R (4.10) + 1−2σ p − lim t ∂t U (x, t) = K(x)U (x, 0), on ∂ ′ B2R , t→0

where p > 0. Then Z

+ ∂ ′ BR

B ′ (X, U, ∇U, R, σ) +

where B ′ (X, U, ∇U, R, σ) = and B ′′ (X, U, ∇U, R, σ) =

Z

+ ∂ ′′ BR

t1−2σ B ′′ (X, U, ∇U, R, σ) = 0,

(4.11)

n − 2σ KU p+1 + hX, ∇U iKU p 2

R ∂U 2 n − 2σ ∂U U − |∇U |2 + R| | . 2 ∂ν 2 ∂ν

+ Proof. Let Ωε = BR ∩ {t > ε} for small ε > 0. Multiplying (4.10) by hX, ∇U i and integrating by parts in Ωε , we have, with notations ∂ ′ Ωε = interior of Ωε ∩ {t = ε}, ∂ ′′ Ωε = ∂Ωε \ ∂ ′ Ωε and ν = unit outer normal of ∂Ωε , Z Z ∂U 2 1−2σ | t1−2σ R| t ∂t U hX, ∇U i + − ∂ν ′′ ′ ∂ Ωε Z Z ∂ Ωε 1 t1−2σ |∇U |2 + t1−2σ X · ∇(|∇U |2 ) = 2 Ωε Ωε Z Z (4.12) n − 2σ 1 1−2σ 2 1−2σ 2 = − t |∇U | + t R|∇U | 2 2 ∂ ′′ Ωε Ωε Z 1 − t2−2σ |∇U |2 . 2 ∂ ′ Ωε

30

Multiplying (4.10) by U and integrating by parts in Ωε , we have Z Z Z t1−2σ U ∂t U + t1−2σ |∇U |2 = −

∂ ′′ Ωε

∂ ′ Ωε

Ωε

t1−2σ

∂U U. ∂ν

(4.13)

By Corollary 2.1 and Proposition 2.6, there exists some α ∈ (0, 1) such that U , ∇x U , and t1−2σ ∂t U belong to C α (Br+ ) for all r < 2R. With this we can send ε → 0 as follows. By (4.10), −t1−2σ ∂t U (x, t) → K(x)U p (x, 0) uniformly in B3R/2 as t → 0.

Hence (4.11) follows by sending ε → 0 in (4.12) and (4.13). Lemma 4.3. Under the assumptions in Lemma 4.2, we have τi = O(ui (yi )−2/(n−2σ)+o(1) ), and thus ui (yi )τi = 1 + o(1). Proof. Since Ui satisfies (4.4) and div(y − yi ) = n, we have, using integration by part, Z 1 B ′ (Y, Ui , ∇Ui , 1, σ) c0 ∂ ′ B1+ (Yi ) Z n − 2σ div(y − yi )Ki U pi +1 = 2n ∂ ′ B1+ (Yi ) Z 1 hy − yi , ∇y Uipi +1 iKi + pi + 1 ∂ ′ B1+ (Yi ) Z i h n − 2σ hy − yi , ∇y Ki iUipi +1 + hy − yi , ∇y Uipi +1 iKi =− 2n ∂ ′ B+ (Yi ) Z 1 Z n − 2σ 1 pi +1 + K i Ui + hy − yi , ∇y Uipi +1 iKi + 2n p + 1 ′ i ∂B1 (yi ) ∂ B1 (Yi ) Z τi (n − 2σ)2 hy − yi , ∇y Uipi +1 iKi = 2n(2n − τi (n − 2σ)) ∂ ′ B1+ (Yi ) Z Z n − 2σ n − 2σ hy − yi , ∇y Ki iUipi +1 + Ki Uipi +1 − 2n 2n ∂ ′ B1+ (Yi ) ∂B1 (yi ) and Z

∂ ′ B1+ (Yi )

= −n

Z

hy − yi , ∇y Uipi +1 iKi

∂ ′ B1+ (Yi )

Ki Uipi +1 −

Z

∂ ′ B1+ (Yi )

hy − yi , ∇y Ki iUipi +1 +

Z

∂B1 (yi )

Combining the above two, together with Proposition 4.3, we conclude that Z nZ τi Uipi +1 ≤C(n, σ, A1 , A2 ) |y − yi |Uipi +1 ∂ ′ B1+ (Yi )

Ki Uipi +1 .

∂ ′ B1+ (Yi )

+

Z

∂B1 (yi )

Uipi +1 +

Z

∂ ′′ B1+ (Yi )

31

o t1−2σ |B ′′ (Y, Ui , ∇Ui , 1, σ)| .

(4.14)

Since Ui = ui on ∂ ′ B1 (Yi ) = B1 (yi ) × {0}, it follows from Proposition 4.2 that Z Z upi i +1 Uipi +1 = B1 (yi )

∂ ′ B1 (Yi )

≥C

−1

≥C

−1

mpi i +1

Z

B1 (yi )

(pi −1)/2σ

(1 + |mi Z

τ (n/2σ−1) mi i

B

m

≥

(y − yi )|2 )(n−2σ)(pi +1)/2 1 2 (n−2σ)(p i +1)/2 (1 + |z| ) (pi −1)/2σ

(4.15)

i

τ (n/2σ−1) , C −1 mi i (p −1)/2σ

(y − yi ) in the second inequality. where we used change of variables z = mi i By Proposition 2.6 and Lemma 4.2, it is easy to see that the last two integral terms of right−2+o(1) ). By Proposition 4.1, we have handed side of (4.14) are in O(mi Z Z |y − yi |upi i +1 |Y − Yi |Uipi +1 = Bri (yi )

∂ ′ Bri (Yi )

≤C

Z

Bri (yi )

|y − yi |mpi i +1

(p −1)/2σ

(1 + |mi i Z

−2/(n−2σ)+o(1)

≤ Cmi ≤

BRi

−2/(n−2σ)+o(1) . Cmi

By Lemma 4.2 and that Ri → ∞, we have Z Z |Y − Yi |Uipi +1 = ∂ ′ B1 (Yi )\∂ ′ Bri (Yi )

(y − yi )|2 )(n−2σ)(pi +1)/2 |z| (1 + |z|2 )n+o(1)

B1 (yi )\Bri (yi )

|y − yi |upi i +1

−λi (pi +1) n+1+(2σ−n+δi )(pi +1) ri −2/(n−2σ)+o(1) ). o(mi

≤ mi =

(4.16)

(4.17)

Combining (4.14), (4.15), (4.16), (4.17) and that τi = o(1), we complete the proof. Proposition 4.4. Under the assumptions in Lemma 4.2, we have 2σ−n ui (y) ≤ Cu−1 , i (yi )|y − yi |

for all |y − yi | ≤ 1.

Our proof of this Proposition makes use of the following Lemma 4.4. Let n ≥ 2. Suppose that for all ε ∈ (0, 1), U ∈ H(t1−2σ , B1+ \ Bε+ ) and U > 0 in B1+ \ Bε+ be a weak solution of (

div(t1−2σ ∇U ) = 0 − lim t1−2σ ∂t U (x, t) = 0, t→0

32

in B1+ \ Bε+ , in B1 \ Bε+ .

(4.18)

Then U (X) = A|X|2σ−n + H(X), where A is a nonnegative constant and H(X) ∈ H(t1−2σ , B1+ ) satisfies ( div(t1−2σ ∇H) = 0 in B1+ , 1−2σ − lim t ∂t H(x, t) = 0, in B1 .

(4.19)

t→0

The proof of Lemma 4.4 is provided in Appendix A.2. Proof of Proposition 4.4. For |y − yi | < ri , it follows from Proposition 4.1 that ui (y) ≤ Cmi

1 (pi −1)/2σ

1 + |mi

−1− n−2σ 2σ τi

≤ Cmi

(y − yi )|2

!(n−2σ)/2

(4.20)

|y − yi |2σ−n

2σ−n ≤ Cm−1 , i |y − yi |

where Lemma 4.3 is used the last inequality. n+1 Suppose |y − yi | ≥ ri . Let e ∈ R+ with |e| = 1, and set Vi (Y ) = Ui (Yi + e)−1 Ui (Y ). Then Vi satisfies ( div(s1−2σ ∇Vi ) = 0, in B2+ , for y ∈ B2+ . − lim s1−2σ ∂s Vi (y, s) = c(n, σ)KUi (Yi + e)pi −1 Vipi , s→0

Note that Ui (Yi + e) → 0 by Lemma 4.2, and for any r > 0 Vi (Y ) ≤ C(n, σ, A1 , r),

for all r < |y − yi | ≤ 1

(4.21)

which follows from Lemma 4.1. It follows that {Vi } converges to some positive function V in + + ∞ α (B 3/2 \ {0}) for some α ∈ (0, 1), and V satisfies Cloc (B3/2 ) ∩ Cloc (

in B1+ for y ∈ B1+ \ {0}.

div(s1−2σ ∇V ) = 0, − lim s1−2σ ∂s V (y, s) = 0 s→0

Hence lim r2σ/(pi +1) v¯i (r) = rn−2σ v¯(r), where v(y) = V (y, 0). Since ri → 0 and yi → 0 is i→∞

an isolated simple blow up point of {ui }, it follows from Lemma 4.1 that r(n−2σ)/2 V (r) is almost decreasing for all 0 < r < ρ, i.e., there exists a positive constant C (which comes from Harnack inequality in Lemma 4.1) such that for any 0 < r1 ≤ r2 < ρ, (n−2σ)/2

r1

(n−2σ)/2

V (r1 ) ≥ Cr2

V (r2 ).

Therefore, V has to have a singularity at Y = 0. Lemma 4.4 implies V (Y ) = A|Y |2σ−n + H(Y ), 33

(4.22)

where A > 0 is a constant and H is as in Lemma 4.4. We first establish the inequality in Proposition 4.4 for |Y − Yi | = 1. Namely, we prove that Ui (Yi + e) ≤ CUi−1 (Yi )

(4.23)

Suppose that (4.23) does not hold, then along a subsequence we have lim Ui (Yi + e)Ui (Yi ) = ∞.

(4.24)

i→∞

By integration by parts (using Ωε and sending ε → 0, as in the proof of Proposition 4.3), we obtain Z div(s1−2σ ∇Vi ) 0=− B1+

=

Z

∂ ′′ B1+

s

1−2σ ∂Vi

∂ν

−1

+ c(n, σ)Ui (Yi + e)

(4.25)

Z

∂ ′ B1+

KUipi .

By Lemma 4.3 and similar computation in (4.16) and (4.17), we see that Z KUipi ≤ CUi (Yi )−1 . ∂ ′ B1+

Due to (4.24), lim Ui (Yi + e)−1

i→∞

Z

∂ ′ B1+

KUipi = 0.

A direct computation yields with (4.21) (again using Ωε and sending ε → 0) Z Z ∂Vi ∂ lim s1−2σ = lim s1−2σ (A|Y |2σ−n + H(Y )) + + i→∞ ∂ ′′ B i→∞ ∂ ′′ B ∂ν ∂ν 1 1 Z s1−2σ < 0, = A(2σ − n) ∂ ′′ B1+

which contradicts to (4.25). Thus we proved (4.23). By Lemma 4.1, we have established the inequality in Proposition 4.4 for ρ ≤ |Y − Yi | ≤ 1. By a standard scaling argument, we can reduce the case of ri ≤ |Y − Yi | < ρ to |Y − Yi | = 1. We refer to [68] (page 340) for details. Proposition 4.2 and 4.4 give a clear picture of ui near the isolated simple blow up point. By the estimates there, it is easy to see the following result. Lemma 4.5. We have Z

|y−yi |≤ri

|y − yi |s ui (y)pi +1 −2s/(n−2σ) ), O(ui (yi ) −2n/(n−2σ) = O(ui (yi ) log ui (yi )), −2n/(n−2σ) o(ui (yi ) ), 34

−n < s < n, s = n, s > n,

and Z

ri <|y−yi |≤1

|y − yi |s ui (y)pi +1 −2s/(n−2σ) ), o(ui (yi ) = O(ui (yi )−2n/(n−2σ) log ui (yi )), O(ui (yi )−2n/(n−2σ) ),

−n < s < n, s = n, s > n.

Proof. The first estimate in the above Lemma follows from Proposition 4.1 and Lemma 4.3, and the second one follows from Proposition 4.4 and Lemma 4.3. For later application, we replace Ki by Ki (x)Hi (x)τi in (4.2) and consider (−∆)σ ui (x) = c(n, σ)Ki (x)Hi (x)τi upi i (x),

in B2 ,

(4.26)

where {Hi } ∈ C 1,1 (B2 ) satisfies A−1 4 ≤ Hi (y) ≤ A4 ,

for all y ∈ B2 ,

and kHi kC 1,1 (B2 ) ≤ A5

(4.27)

for some positive constants A4 and A5 . Lemma 4.6. Suppose that {Ki } satisfies (4.1) and (∗)β condition with β < n for some positive constants A1 , A2 , {L(β, i)}, and that {Hi } satisfies (4.27) with A4 , A5 . Let ui ∈ H˙ σ (Rn )∩C 2 (B2 ) and ui ≥ 0 in Rn be a solution of (4.26). If yi → 0 is an isolated simple blow up point of {ui } with (4.5) for some positive constant A3 , then we have τi ≤Cui (yi )−2 + C|∇Ki (yi )|ui (yi )−2/(n−2σ)

+ C(L(β, i) + L(β, i)β−1 )ui (yi )−2β/(n−2σ) ,

where C > 0 depends only on n, σ, A1 , A2 , A3 , A4 , A5 , β and ρ. Proof. Using Lemma 4.3 and arguing the same as in the proof of Lemma 4.3, we have Z pi +1 τi −2 τi ≤ Cui (yi ) + C hy − yi , ∇y (Ki Hi )iui B1 (yi ) Z ≤ Cui (yi )−2 + Cτi |y − yi |upi i +1 B1 (yi ) Z +C hy − yi , ∇Ki iHiτi upi i +1 . B1 (yi )

35

Making use of Lemma 4.5, we have Z hy − yi , ∇Ki iHiτi upi i +1 B1 (yi ) Z ≤ C|∇Ki (yi )| |y − yi |upi i +1 B1 (yi ) Z |y − yi ||∇Ki (y) − ∇Ki (yi )|upi i +1 +C B1 (yi )

≤ C|∇Ki (yi )|ui (yi )−2/(n−2σ) Z |y − yi ||∇Ki (y) − ∇Ki (yi )|upi i +1 . +C B1 (yi )

Recalling the definition of (∗)β , a directly computation yields |∇Ki (y) − ∇Ki (yi )| ≤

[β] nX s=2

|∇s Ki (yi )||y − yi |s−1 + [∇[β] Ki ]C β−[β] (B1 (yi )) |y − yi |β−1 [β]

o

(4.28)

nX o ≤ CL(β, i) |∇Ki (yi )|(β−s)/(β−1) |y − yi |s−1 + |y − yi |β−1 . s=2

By Cauchy-Schwartz inequality, we have L(β, i)|∇Ki (yi )|(β−s)/(β−1) |y − yi |s

(4.29)

≤ C(|∇Ki (yi )||y − yi | + (L(β, i) + L(β, i)β−1 )|y − yi |β ).

Hence, by Lemma 4.5 we obtain Z |y − yi ||∇Ki (y) − ∇Ki (yi )|upi i +1

(4.30)

B1 (yi )

−2/(n−2σ)

≤ C|∇Ki (yi )|ui (yi )

β−1

+ C(L(β, i) + L(β, i)

−2β/(n−2σ)

)ui (yi )

.

Lemma 4.6 follows immediately. Lemma 4.7. Under the hypotheses of Lemma 4.6, |∇Ki (yi )| ≤ Cui (yi )−2 + C(L(β, i) + L(β, i)β−1 )ui (yi )−2(β−1)/(n−2σ) , where C > 0 depends only on n, σ, A1 , A2 , A3 , A4 , A5 , β and ρ. Proof. Choose a cutoff function η(Y ) ∈ Cc∞ (B1/2 ) satisfying η(Y ) = 1,

|Y | ≤

1 and η(Y ) = 0, 4

36

|Y | ≥

1 . 2

Let Ui (Y ) be the extension of ui (y), namely, ( div(s1−2σ ∇Ui ) = 0, − lim s1−2σ ∂s U (y, s) = c0 Ki (y)Hiτi Uipi , s→0

in Rn+1 + y ∈ B2 .

(4.31)

Multiplying (4.31) by η(Y − Yi )∂yj Ui (y, s), j = 1, · · · , n, and integrating by parts over B1 , we obtain Z div(s1−2σ ∇Ui )η∂yj Ui 0= B1+

=− =

1 2

Z

B1+

s

1−2σ

Z

+ + B1/2 \B1/4

c0 − pi + 1

∇Ui ∇(η∂yj Ui ) + c0

Z

∂ ′ B1+ (Yi )

ηKi Hiτi ∂yj Ui Uipi

s1−2σ (|∇Ui |2 ∂yj η − 2∇Ui ∇η∂yj Ui )

Z

∂ ′ B1+

∂yj (Ki Hiτi η)Uipi +1 .

By Proposition 4.4, we have Ui (Y ) ≤ CUi (Yi )−1 , and

Z

+ + B1/2 \B1/4

for all 1/2 ≥ |Y | ≥ 1/4

s1−2σ |∇Ui |2 ≤ CUi (Yi )−2 .

Therefore by Lemma 4.5 we conclude that Z τi pi +1 −2 ∂yj Ki Hi ui ≤ Cui (yi ) + Cτi . B1

Hence

Z τi pi +1 ∂j Ki (yi ) Hi ui − Cui (yi )−2 − Cτi B1 Z |∂j Ki (yi ) − ∂j Ki (y)|Hiτi upi i +1 ≤ B1

Summing over j, then making use of (4.28), (4.29) and Lemma 4.5, we have 1 |∇Ki (yi )| ≤ Cui (yi )−2 + Cτi + |∇Ki (yi )| 2 + C(L(β, i) + L(β, i)β−1 )ui (yi )−2(β−1)/(n−2σ) . Then Lemma 4.7 follows from Lemma 4.6. Lemma 4.8. Under the assumptions of Lemma 4.6 we have τi ≤ Cui (yi )−2 + C(L(β, i) + L(β, i)β−1 )ui (yi )−2β/(n−2σ) . 37

(4.32)

Proof. It follows immediately from Lemma 4.6 and Lemma 4.7. Corollary 4.1. In addition to the assumptions of Lemma 4.6, we further assume that one of the following two conditions holds: (i) β = n − 2σ and L(β, i) = o(1), and (ii) Then for any 0 < δ < 1 we have

β > n − 2σ and L(β, i) = O(1).

lim ui (yi )2

i→∞

Z

Bδ (yi )

(y − yi ) · ∇(Ki Hiτi )upi i +1 = 0.

Proof. Z pi +1 τi (y − yi ) · ∇(Ki Hi )ui Bδ (yi ) Z Z p +1 p +1 ≤ (y − yi ) · ∇Ki Hiτi ui i + τi (y − yi ) · ∇Hi Hiτi −1 Ki ui i Bδ (yi ) Bδ (yi ) Z |y − yi |upi i +1 ≤ C|∇Ki (yi )| Bδ (yi ) Z Z +C |y − yi ||∇Ki (y) − ∇Ki (yi )|upi i +1 + τi |y − yi |upi i +1 . Bδ (yi )

Bδ (yi )

The corollary follows immediately from Lemma 4.7, (4.30) and Lemma 4.8. Proposition 4.5. Suppose that {Ki } satisfies (4.1) and (∗)n−2σ condition for some positive constants A1 , A2 , L independent of i, and that {Hi } satisfies (4.27) with A4 , A5 . Let ui ∈ H˙ σ (Rn ) ∩ C 2 (B2 ) be a solution of (4.26). If yi → 0 is an isolated blow up point of {ui } with (4.5) for some positive constant A3 , then yi → 0 is an isolated simple blow up point. Proof. Due to Proposition 4.1, r2σ/(pi −1) ui (r) has precisely one critical point in the interval 0 < pi −1 r < ri , where ri = Ri ui (yi )− 2σ as before. Suppose yi → 0 is not an isolated simple blow up point and let µi be the second critical point of r2σ/(pi −1) ui (r). Then we see that µi ≥ ri ,

lim µi = 0.

i→∞

Without loss of generality, we assume that yi = 0. Set 2σ/(pi −1)

φi (y) = µi

ui (µi y),

y ∈ Rn .

Clearly, φi satisfies ˜ i (y)H ˜ τi (y)φpi (y), (−∆)σ φi (y) = K i i |y|2σ/(pi −1) φi (y) ≤ A3 , lim φi (0) = ∞, i→∞

38

|y| < 1/µi ,

(4.33)

r2σ/(pi −1) φi (r) has precisely one critical point in 0 < r < 1, and

o d n 2σ/(pi −1) r = 0, φi (r) dr r=1

R ˜ i (y) = Ki (µi y), H ˜ i (y) = Hi (µi y) and φ (r) = |∂Br |−1 φ. where K i ∂Br i Therefore, 0 is an isolated simple blow up point of φi . Let Φi (Y ) be the extension of φi (y) in the upper half space. Then Lemma 4.1, Proposition 4.4, Lemma 4.4 and elliptic equation theory together imply that α 2 Φi (0)Φi (Y ) → G(Y ) = A|Y |2σ−n + H(Y ) in Cloc (Rn+1 \ {0}) ∩ Cloc (Rn+1 + + ).

and 2 φi (0)φi (y) → G(y, 0) = A|y|2σ−n + H(y, 0) in Cloc (Rn \{0})

as i → ∞, where A > 0, H(Y ) satisfies ( div(s1−2σ ∇H) = 0 − lim s1−2σ ∂s H(y, s) = 0 s→0

(4.34)

in Rn+1 + for y ∈ Rn .

Note that G(Y ) is nonnegative, we have lim inf |Y |→∞ H(Y ) ≥ 0. It follows from the weak maximum principle and the Harnack inequality that H(y) ≡ H ≥ 0 is a constant. Since o o d n 2σ/(pi −1) d n 2σ/(pi −1) φi (r) = φi (0) = 0, r φi (0)φi (r) r dr dr r=1 r=1

we have, by sending i to ∞ and making use of (4.34), that A = H > 0.

We are going to derive a contradiction to the Pohozaev identity Proposition 4.3, by showing that for small positive δ Z lim sup Φi (0)2 i→∞

and lim sup Φi (0)2 i→∞

Z

B ′ (Y, Φi , ∇Φi , δ, σ) ≤ 0,

(4.35)

s1−2σ B ′′ (Y, Φi , ∇Φi , δ, σ) < 0.

(4.36)

∂ ′ Bδ+

∂ ′′ Bδ+

And thus Proposition 4.5 will be established. By Proposition 2.6, it is easy to verify (4.36) by that Z lim sup Φi (0)2 s1−2σ B ′′ (Y, Φi , ∇Φi , δ, σ) ∂ ′′ Bδ+

i→∞

=

Z

∂ ′′ Bδ+

s

1−2σ

(n − 2σ)2 2 B (Y, G, ∇G, δ, σ) = − A 2 ′′

39

Z

∂ ′′ B1+

t1−2σ < 0,

which shows (4.36). On the other hand, via integration by parts, we have Z B ′ (Y, Φi , ∇Φi , δ, σ) ∂ ′ Bδ+

Z Z n − 2σ τi pi +1 ˜ τi φpi ˜ iH ˜ ˜ hy, ∇φi iK + = Ki Hi φi i i 2 Bδ Bδ Z Z n − 2σ ˜ τi φpi +1 − n ˜ τi φpi +1 ˜ iH ˜ iH = K K i i i i 2 p + 1 i B Bδ Zδ Z 1 ˜ τi )iφpi +1 + δ ˜ iH ˜ τi φpi +1 ˜ iH hy, ∇(K K − i i i i pi + 1 Bδ pi + 1 ∂Bδ Z 1 ˜ τi )iφpi +1 + Cφi (0)−(pi +1) . ˜ iH ≤− hy, ∇(K i i pi + 1 Bδ

˜ i } satisfies (∗)n−2σ with where Proposition 4.4 is used in the last inequality. It is easy to see that {K L(β, i) = o(1). Therefore, (4.35) follows from Corollary 4.1. Proposition 4.6. Suppose the assumptions in Proposition 4.5 except the (∗)n−2σ condition for Ki . Then |∇Ki (yi )| → 0, as i → ∞. Proof. Suppose that contrary that |∇Ki (yi )| → d > 0.

(4.37)

Without loss of generality, we assume yi = 0. There are two cases. Case 1. 0 is an isolated simple blow up point. In this case, we argue as in the proof of Lemma 4.7 and obtain Z τi pi +1 ∇Ki Hi ui ≤ Cu−2 i (0) + Cτi . B1

It follows from the mean value theorem, Lemma 4.3 and Lemma 4.5 that Z |∇Ki (y) − ∇Ki (0)|Hiτi upi i +1 + o(1) = o(1). |∇Ki (0)| ≤ C B1

Case 2. 0 is not an isolated simple blow up point. In this case we argue as the proof of Proposition 4.5. The only difference is that we cannot derive (4.35) from Corollary 4.1, since (∗)n−2σ condition for Ki is not assumed. Instead, we will use the condition (4.37) to show (4.35). Let µi , φi , Φi , K˜i and H˜i be as in the proof of Proposition 4.5. The computation at the end of the proof of Proposition 4.5 gives Z B ′ (Y, Φi , ∇Φi , δ, σ) ∂ ′ Bδ+

≤−

1 pi + 1

Z

Bδ

˜ τi )iφpi +1 + Cφi (0)−(pi +1) . ˜ iH hy, ∇(K i i 40

R ˜ τi )iφpi +1 . Using Lemma 4.3 and arguing the ˜ iH Now we estimate the integral term Bδ hy, ∇(K i i same as in the proof of Lemma 4.3, we have Z −2 ˜ i (y)|H τi φpi +1 |y||∇K τi ≤ Cφi (0) + C i i Bδ

−2

≤ Cφi (0)

+ Cµi φi (0)−2/(n−2σ) .

By (4.32), Z

˜ τi φpi +1 ≤ Cφi (yi )−2 + Cτi . ˜ iH ∇K i i

Bδ

It follows that

˜ i (0)| ≤ C |∇K

Z

Bδ

˜ i (y) − ∇K ˜ i (0)|φpi +1 + Cφi (0)−2 + Cτi |∇K i

≤ Cµi φi (0)−2/(n−2σ) + Cφi (0)−2 + Cτi . ˜ i (0)| = µi |∇Ki (0)| ≥ (d/2)µi , we have Since |∇K µi ≤ Cφi (0)−2 + Cτi . It follows that τi ≤ Cφi (0)−2 Therefore,

Z

Bδ

and µi ≤ Cφi (0)−2 .

pi +1 τi −2−2/(n−2σ) ˜ ˜ hy, ∇(Ki Hi )iφi ≤ Cφi (0)

and (4.35) follows immediately.

5 Estimates on the sphere and proofs of main theorems Consider Pσ (v) = c(n, σ)Kv p ,

on Sn ,

(5.1)

n+2σ ] and K satisfies where p ∈ (1, n−2σ

A−1 1 ≤ K ≤ A1 ,

on Sn ,

(5.2)

and kKkC 1,1 (Sn ) ≤ A2 .

(5.3)

Proposition 5.1. Let v ∈ C 2 (Sn ) be a positive solution to (5.1). For any 0 < ε < 1 and R > 1, there exist large positive constants C1 , C2 depending on n, σ, A1 , A2 , ε and R such that, if max v ≥ C1 , n S

41

n+2σ then n−2σ − p < ε, and there exists a finite set ℘(v) ⊂ Sn such that (i). If P ∈ ℘(v), then it is a local maximum of v and in the stereographic projection coordinate system {y1 , · · · , yn } with P as the south pole,

kv −1 (P )v(v −

(p−1) 2σ

(P )y) − (1 + k|y|2 )(2σ−n)/2 kC 2 (B2R ) ≤ ε,

(5.4)

where k = K(P )1/σ /4. (ii). If P1 , P2 belonging to ℘(v) are two different points, then BRv(P1 )−(p−1)/2σ (P1 ) ∩ BRv(P2 )−(p−1)/2σ (P2 ) = ∅. (iii). v(P ) ≤ C2 {dist(P, ℘(v))}−2σ/(p−1) for all P ∈ Sn . Proof. Given Theorem 1.5, Remark 1.2 and the proof of Proposition 4.1, the proof of Proposition 5.1 is similar to that of Proposition 4.1 in [68] and Lemma 3.1 in [88], and is omitted here. We refer to [68] and [88] for details. Proposition 5.2. Assume the hypotheses in Proposition 5.1. Suppose that there exists some constant d > 0 such that K satisfies (∗)n−2σ for some L in Ωd = {P ∈ Sn : |∇K(P )| < d}. Then, for ε > 0, R > 1 and any solution v of (5.1) with maxSn v > C1 , we have |P1 − P2 | ≥ δ ∗ > 0,

for any P1 , P2 ∈ ℘(v) and P1 6= P2 ,

where δ ∗ depends only on n, σ, δ, ε, R, A1 , A2 , L2 , d. Proof. Suppose the contrary, then there exists sequences of {pi } and {Ki } satisfying the above assumptions, and a sequence of corresponding solutions {vi } such that lim |P1i − P2i | = 0,

i→∞

where P1i , P2i ∈ ℘(vi ), and |P1i − P2i | =

min

P1 ,P2 ∈℘(vi ) P1 6=P2

(5.5)

|P1 − P2 |.

Since BRvi (P1i )−(pi −1)/2σ (P1i ) and BRvi (P2i )−(pi −1)/2σ (P2i ) have to be disjoint, we have, because of (5.5), that vi (P1i ) → ∞ and vi (P2i ) → ∞. Therefore, we can pass to a subsequence (still denoted as vi ) with Ri → ∞, εi → 0 as in Proposition 4.1 (εi depends on Ri and can be chosen as small as we need in the following arguments) such that, for y being the stereographic projection coordinate with south pole at Pji , j = 1, 2, we have −(pi −1)/2σ

km−1 i vi (mi

y) − (1 + kji |y|2 )(2σ−n)/2 kC 2 (B2Ri (0)) ≤ εi ,

(5.6)

where mi = vi (0), kji = Ki (qji )1/σ , j = 1, 2; i = 1, 2, · · · In the stereographic coordinates with P1i being the south pole, the equation (5.1) is transformed into (−∆)σ ui (y) = c(n, σ)Ki (y)Hiτi (y)upi i (y), y ∈ Rn , (5.7) where

ui (y) =

Hi (y) =

2 1 + |y|2 2 1 + |y|2

(n−2σ)/2 (n−2σ)/2

42

vi (F (y)), (5.8) ,

and F is the inverse of the stereographic projection. Let us still use P2i ∈ Rn to denote the stereographic coordinates of P2i ∈ Sn and set ϑi = |P2i | → 0. For simplicity, we assume P2i is a local maximum point of ui . Since we can always reselect a sequence of points as in the proof of Proposition 5.1 to substitute for P2i . From (ii) in Proposition 5.1, there exists some constant C depending only on n, σ, such that ϑi >

1 max{Ri ui (0)−(pi −1)/2σ , Ri ui (P2i )−(pi −1)/2σ }. C

Set

2σ/(pi −1)

wi (y) = ϑi

ui (ϑi y),

(5.9)

in Rn .

It is easy to see that wi which is positive in Rn , satisfies ˜ i (y)H ˜ τi (y)wi (y)pi , (−∆)σ wi (y) = c(n, σ)K i

in Rn

(5.10)

and wi (y) ∈ C 2 (Rn ),

lim inf wi (y) < ∞, |y|→∞

˜ i (y) = Ki (ϑi y), H ˜ i (y) = Hi (ϑi y). where K By Proposition 5.1, ui satisfies ui (y) ≤ C2 |y|−2σ/(pi −1) ui (y) ≤ C2 |y − P2i |

for all |y| ≤ ϑi /2

−2σ/(pi −1)

for all |y − P2i | ≤ ϑi /2.

In view of (5.9), we therefore have lim wi (0) = ∞,

|y − |P2i |

−1

|y|

P2i |

i→∞ 2σ/(pi −1) 2σ/(pi −1)

lim wi (|P2i |−1 P2i ) = ∞

i→∞

wi (y) ≤ C2 , |y| ≤ 1/2,

wi (y) ≤ C2 , |y − |P2i |−1 P2i | ≤ 1/2.

After passing a subsequence, if necessary, there exists a point P ∈ Rn with |P | = 1 such that |P2i |−1 P2i → P as i → ∞. Hence 0 and P are both isolated blow up points of wi . If |∇Ki (0)| ≤ d/2, then 0 is an isolated simple blow up point of wi because of the (∗)n−2σ condition and Proposition 4.5. If |∇Ki (0)| ≥ d/2, arguing as in the proof of Proposition 4.6 we can conclude that 0 is an isolated simple blow up point of wi . Similarly, P is also an isolated simple blow up point of wi . By Proposition 4.4, wi (0)wi (y) ≤ Cε , for all ε ≤ |y| ≤ 1/2,

where Cε is independent of i. Let Wi be the extension of wi . Due to Proposition 5.1, Harnack inequality Lemma 4.1, and the choice of P1i , P2i , there exists an at most countable set ℘ ⊂ Rn such that inf{|x − y| : x, y ∈ ℘, x 6= y} ≥ 1, and 0 \ ℘) lim Wi (0)Wi (Y ) = G(Y ), in Cloc (Rn+1 +

i→∞

G(Y ) > 0, Y ∈ Rn+1 \ ℘. + 43

Let ℘1 ⊂ ℘ contain those points near which G is singular. Clearly, 0, P ∈ ℘1 . Since pi > 1, it follows from (5.10) that ( div(s1−2σ ∇G) = 0, in Rn+1 + , 1−2σ − lim s ∂s G(y, s) = 0, for all y ∈ Rn \ ℘1 . s→0

By Lemma 4.4 and maximum principle, there exist positive constants N1 , N2 and some nonnegative function H satisfying ( div(s1−2σ ∇H) = 0, in Rn+1 + , 1−2σ − lim s ∂s H(y, s) = 0, for all y ∈ Rn \ {℘1 \ {0, P }} s→0

such that G(Y ) = N1 |Y |2σ−n + N2 |Y − P |2σ−n + H(Y ),

Y ∈ Rn+1 \ {℘1 }. +

Applying Proposition 2.6 to H, it is not difficult to verify (4.36) with Φi replaced by Wi . On the other hand, we can establish (4.35) with Φi replaced by Wi if |∇Ki (0)| ≤ d/2, because (∗)n−2σ ˜ i and thus Corollary 4.1 holds. If |∇Ki (0)| ≥ d/2, we can condition with L = o(1) holds for K apply the argument in the proof of Proposition 4.6 to conclude that ϑi , τi ≤ wi (0)−2 , and hence (4.35) also holds for Wi . Proposition 5.2 is established. Consider Pσ (v) = c(n, σ)Ki vipi

on Sn ,

vi > 0, on Sn , n + 2σ pi = − τi , τi ≥ 0, τi → 0. n − 2σ

(5.11)

Theorem 5.1. Suppose Ki satisfies the assumption of K in Proposition 5.2. Let vi be solutions of (5.11), we have kvi kH σ (Sn ) ≤ C, (5.12) where C > 0 depends only on n, σ, A1 , A2 , L, d. Furthermore, after passing to a subsequence, either {vi } stays bounded in L∞ (Sn ) or {vi } has only isolated simple blow up points and the distance between any two blow up points is bounded blow by some positive constant depending only on n, σ, A1 , A2 , L, d. Proof. The theorem follows immediately from Proposition 5.2, Proposition 4.6, Proposition 4.5, Proposition 4.1 and Lemma 4.5. Proof of Theorem 1.3. It follows immediately from Theorem 5.1. In the next theorem, we impose a stronger condition on Ki such that {ui } has at most one blow up point.

44

Theorem 5.2. Suppose the assumptions in Theorem 5.1. Suppose further that {Ki } satisfies (∗)n−2σ condition for some sequences L(n − 2σ, i) = o(1) in Ωd,i = {q ∈ Sn : |∇g0 Ki | < d} or {Ki } satisfies (∗)β condition with β > n − 2σ in Ωd,i . Then, after passing to a subsequence, either {vi } stays bounded in L∞ (Sn ) or {vi } has precisely one isolated simple blow up point. Proof. The strategy is the same as the proof of Proposition 5.2. We assume there are two isolated blow up points. After some transformation, we can assume that they are in the same half sphere. The condition of {Ki } guarantees that Corollary 4.1 holds for ui , where ui is as in (5.8). Hence (4.35) holds for Ui , which is the extension of ui . Meanwhile (4.36) for Ui is also valid, since the distance between these blow up points is uniformly lower bounded which is due to Proposition 5.2. Proof of Theorem 1.4. By Theorem 5.2, we only need to show the latter case of theorem. After passing a subsequence, ξi → ξ is the only isolated simple blow up point of vi . For simplicity, assume that ξi is identical to the south pole and K(ξi ) = 1. Let F : Rn → Sn be the inverse of stereographic projection defined at the beginning of the paper. Define, for any λ > 0, ψλ : x 7→ λx,

∀x ∈ Rn .

2

Set ϕi = F ◦ ψλi ◦ F −1 with λi = vi (ξi )− n−2σ . Then Tϕi vi satisfies n+2σ

Pσ (Tϕi vi ) = c(n, σ)K ◦ ϕi Tϕi vin−2σ , Let ui (x) = and u˜i (x) = Note that | det dϕi (F (x))| Hence, u ˜i (x) = λ

n−2σ 2

2 1 + |x|2

2 1 + |x|2

n−2σ 2n

=

n−2σ 2

n−2σ 2

x ∈ Rn

vi ◦ F (x),

Tϕi vi ◦ F (x),

x ∈ Rn .

n −n 2 2 n λ i 1 + |λi x|2 1 + |x|2

ui (λi x) for any x ∈ Rn and 0 < ui ≤ 2 u ˜i (x) →

on Sn .

2 1 + |x|2

n−2σ 2

,

n−2σ 2

n−2σ 2n

.

. Arguing as before, we see that

2 in Cloc (Rn ).

2 Therefore, vi → 1 in Cloc (Sn \ {N }), where N is the north pole of Sn . Since Tϕi vi is uniformly bounded near the north pole, it follows from H¨older estimates that there exists a constant α ∈ (0, 1) such that Tϕi vi → f in C α (Bδ (N )) for small constant δ > 0 and some function f ∈ C α (Bδ (N )). It is clear that f = 1. Therefore, we complete the proof.

Theorem 5.3. Let vi be positive solutions of (5.11). Suppose that {Ki } ⊂ C ∞ (Sn ) satisfies (5.3), and for some point P0 ∈ Sn , ε0 > 0, A1 > 0 independent of i and 1 < β < n, that {Ki } is bounded in C [β],β−[β](Bε0 (q0 )),

45

Ki (P0 ) ≥ A1

and

(β)

|y| ≤ ε0 ,

Ki (y) = Ki (0) + Qi (y) + Ri (y),

(β)

where y is the stereographic projection coordinate with P0 as the south pole, Qi (y) satisfies (β) (β) Qi (λy) = λβ Qi (y), ∀λ > 0, y ∈ Rn , and Ri (y) satisfies [β] X s=0

|∇s Ri (y)||y|−β+s → 0

uniformly for i as y → 0. (β) Suppose also that Qi → Q(β) in C 1 (Sn−1 ) and for some positive constant A6 that A6 |y|β−1 ≤ |∇Q(β) (y)|, and

R

Rn

|y| ≤ ε0 ,

∇Q(β) (y + y0 )(1 + |y|2 )−n dy

(β) (y + y0 )(1 + |y|2 )−n dy Rn Q

R

!

6= 0,

(5.13)

∀ y 0 ∈ Rn .

(5.14)

If P0 is an isolated simple blow up point of vi , then vi has to have at least another blow up point. Proof. Suppose the contrary, P0 is the only blow up point of vi . We make a stereographic projection with P0 being the south pole to the equatorial plane of Sn , with its inverse π. Then the Eq. (5.11) is transformed to n+2σ

(−∆)σ ui = c(n, σ)Ki (y)uin−2σ , with ui (y) =

2 1 + |y|2

(n−2σ)/2

in Rn ,

(5.15)

vi (π(y)).

Let yi → 0 be the local maximum point of ui . It follows from Lemma 4.7 that |∇Ki (yi )| = O(ui (yi )−2 + ui (yi )−2(β−1)/(n−2σ) ). First we establish |yi | = O(ui (yi )−2/(n−2σ) ).

(5.16)

Since we have assumed that vi has no other blow up point other than P0 , it follows from Proposition 4.4 and Harnack inequality that for |y| ≥ ε > 0, ui (y) ≤ C(ε)|y|2σ−n ui (yi )−1 . By Proposition A.1 we have Z 2n

Rn

∇Ki uin−2σ = 0.

It follows that for ε > 0 small we have Z 2n ≤ C(ε)ui (yi )−2n/(n−2σ) . n−2σ ∇Ki (y + yi )ui (y + yi ) Bε

46

(5.17)

Using our hypotheses on ∇Q(β) and Ri we have Z 2n (β) (1 + oε (1))∇Qi (y + yi )ui (y + yi ) n−2σ ≤ C(ε)ui (yi )−2n/(n−2σ) . Bε

(2/(n−2σ))(β−1)

, where mi = ui (yi ), we have Multiplying the above by mi Z 2n 2/(n−2σ) (β) n−2σ y + y˜i )ui (y + yi ) (1 + oε (1))∇Qi (mi Bε

≤ C(ε)ui (yi )(2/(n−2σ))(β−1−n)

2/(n−2σ)

yi . Suppose (5.16) is false, namely, y˜i → +∞ along a subsequence. Then it where y˜i = mi follows from Proposition 4.1 (we may choose Ri ≤ |˜ yi |/4) that Z 2n 2/(n−2σ) (β) y + y˜i )ui (y + yi ) n−2σ (1 + oε (1))∇Qi (mi |y|≤Ri m−2/(n−2σ) i Z 2n n−2σ −2/(n−2σ) (β) −1 yi |β−1 . = z + yi ) (1 + oε (1))∇Qi (z + y˜i ) mi ui (mi ∼ |˜ |z|≤Ri

On the hand, it follows from Lemma 4.5 that Z 2n 2/(n−2σ) (β) y + y˜i )ui (y + yi ) n−2σ (1 + oε (1))∇Qi (mi Ri m−2/(n−2σ) ≤|y|≤ε i Z 2n 2/(n−2σ) β−1 ≤C y| + |˜ yi |β−1 ui (y + yi ) n−2σ |mi Ri m−2/(n−2σ) ≤|y|≤ε i ≤ o(1)|˜ yi |β−1 .

It follows that

(2/(n−2σ))(β−1−n)

|˜ yi |β−1 ≤ C(ε)mi

,

which implies that −(2/(n−2σ))(n/(β−1))

|yi | ≤ C(ε)mi

−2/(n−2σ)

= o(mi

This contradicts to that y˜i → ∞. Thus (5.16) holds. We are going to find some y0 such that (5.14) fails. It follows from Kazdan-Warner condition Proposition A.1 that Z hy, ∇Ki (y + yi )iui (y + yi )2n/(n−2σ) = 0.

).

(5.18)

Rn

Since P0 is an isolated simple blow up point and the only blow up point of vi , we have for any ε > 0, Z 2n/(n−σ) −2n/(n−2σ) hy, ∇K (y + y )iu (y + y ) . i i i i ≤ C(ε)ui (yi ) Bε

47

It follows from Lemma (4.5) and expression of Ki that Z (β) 2n/(n−2σ) hy, ∇Qi (y + yi )iui (y + yi ) Bε

≤ C(ε)ui (yi )−2n/(n−2σ) Z |y||y + yi |β−1 ui (y + yi )−2n/(n−2σ) + oε (1) ≤

Bε C(ε)ui (yi )−2n/(n−2σ)

+ oε (1) ≤

Z

(|y|β + |y||yi |β−1 )ui (y + yi )−2n/(n−2σ)

Bε C(ε)ui (yi )−2n/(n−2σ)

+ oε (1)ui (yi )−2β/(n−2σ) ,

where we used (5.16) in the last inequality. Multiplying the above by ui (yi )2β/(n−2σ) , due to β < n we obtain Z (β) 2β/(n−2σ) 2n/(n−2σ) lim ui (yi ) hy, ∇Qi (y + yi )iui (y + yi ) = oε (1). i→∞

(5.19)

Bε

2

Let Ri → ∞ as i → ∞. We assume that ri := Ri ui (yi )− n−2σ → 0 as we did in Proposition 4.1. By Lemma 4.5, we have Z (β) ui (yi )2β/(n−2σ) hy, ∇Qi (y + yi )iui (y + yi )2n/(n−2σ) ri ≤|y|≤ε Z (5.20) 2β/(n−2σ) β β−1 2n/(n−2σ) ≤ lim ui (yi ) (|y| + |y||yi | )ui (y + yi ) →0 i→∞ ri ≤|y|≤ε as i → ∞. Combining (5.19) and (5.20), we conclude that Z (β) lim ui (yi )2β/(n−2σ) hy, ∇Qi (y + yi )iui (y + yi )2n/(n−2σ) = oε (1). i→∞ Br i

2

It follows from changing variable z = ui (yi ) n−2σ y, applying Proposition 4.1 and then letting ε → 0 that Z hz, ∇Q(β) (z + z0 )i(1 + k|z|2 )−n = 0, (5.21) Rn

where z0 = limi→∞ ui (yi )2/(n−2σ) yi and k = limi→∞ Ki (yi )1/σ . On the other hand, from (5.17) Z ∇Ki (y + yi )ui (y + yi )2n/(n−2σ) = 0.

(5.22)

Rn

Arguing as above, we will have Z

Rn

∇Q(β) (z + z0 )(1 + k|z|2 )−n = 0. 48

(5.23)

It follows from (5.21) and (5.23) that Z Q(β) (z + z0 )(1 + k|z|2 )−n dz Rn Z = β −1 hz + z0 , ∇Q(β) (z + z0 )i(1 + k|z|2 )−n dz

(5.24)

Rn

= 0.

Therefore, (5.14) does not hold for y0 =

√ kz0 .

Theorem 5.4. Let σ ∈ (0, 1) and n ≥ 3. Suppose that K ∈ C 1,1 (Sn ), for some constant A1 > 0, 1/A1 ≤ Ki (ξ) ≤ A1

for all ξ ∈ Sn .

Suppose also that for any critical point ξ0 of K, under the stereographic projection coordinate system {y1 , · · · , yn } with ξ0 as south pole, there exist some small neighborhood O of 0, a positive constant L, and β = β(ξ0 ) ∈ (n − 2σ, n) such that k∇[β] KkC β−[β](O) ≤ L and

(β)

K(y) = K(0) + Q(ξ0 ) (y) + R(ξ0 ) (y) (β)

(β)

(β)

in O,

where Qξ0 (y) ∈ C [β]−1,1 (Sn−1 ) satisfies Qξ0 (λy) = λβ Qξ0 (y), ∀λ > 0, y ∈ Rn , and for some positive constant A6 A6 |y|β−1 ≤ |∇Q(β) (y)|, y ∈ O, and

R

Rn

R

∇Q(β) (y + y0 )(1 + |y|2 )−n dy (β)

2 −n

!

6= 0,

∀ y 0 ∈ Rn ,

(y + y0 )(1 + |y| ) dy P s −β+s = 0. and Rξ0 (y) ∈ C [β]−1,1 (O) satisfies limy→0 [β] s=0 |∇ R|ξ0 (y)|y| Then there exists a positive constant C ≥ 1 depending on n, σ, K such that for any solution v of (1.5) 1/C ≤ v ≤ C, on Sn . Rn

Q

Proof. It follows directly from Theorem 5.2 and Theorem 5.3. Proof of the compactness part of Theorem 1.2. It is easy to check that, if K satisfies the condition in Theorem 1.2, then it must satisfy the condition in the above theorem. Therefore, we have the lower and upper bounds of v. The C 2 norm bound of v follows immediately.

A

Appendix

A.1 A Kazdan-Warner identity In this section we are going to show (1.7), which is a consequence of the following

49

Proposition A.1. Let K > 0 be a C 1 function on S n , and let v be a positive function in C 2 (S n ) satisfying n+2σ Pσ (v) = Kv n−2σ , on S n . (A.1) Then, for any conformal Killing vector field X on S n , we have Z 2n (∇X K)v n−2σ dVgSn = 0.

(A.2)

Sn

Let ϕt : S n → S n be a one parameter family of conformal diffeomorphism (in this case they are M¨obius transformations), depending on t smoothly, |t| < 1, and ϕ0 = identity. Then d X := (ϕt )−1 is a conformal Killing vector field on S n . (A.3) dt t=0

Proof. The proof is standard (see, e.g., [12] for a Kazdan-Warner identity for prescribed scalar curvature problems) and we include it here for completeness. Since Pσ is a self-adjoint operator, (A.1) has a variational formulation: Z Z 2n n − 2σ 1 vPσ (v) dVgSn − Kv n−2σ dVgSn . I[v] := 2 Sn 2n Sn

Let X be a conformal Killing vector field, then there exists {ϕt } satisfying (A.3). Let vt := (v ◦ ϕt )wt where wt is given by

4

gt := ϕ∗t gSn = wtn−2σ gSn . Then

1 I[vt ] = 2

Z

vPσ (v) dVgSn

Sn

n − 2σ − 2n

Z

Sn

2n

n−2σ dV K(ϕ−1 gSn . t (x))v

It follows from (A.1) that Z 2n d n − 2σ d ′ 0 = I [v] =− (∇X K)v n−2σ dVgSn . vt = I[vt ] dt t=0 dt 2n Sn t=0

A.2 A proof of Lemma 4.4 The classical Bˆocher theorem in harmonic function theory states that a positive harmonic function u in a punctured ball B1 \ {0} must be of the form ( −a log |x| + h(x), n = 2, u(x) = a|x|2−n + h(x), n ≥ 3, where a is a nonnegative constant and h is a harmonic function in B1 . + We are going to establish a similar result, Lemma 4.4, in our setting. Denote BR = {X : |X| < + ′ + ′′ ′ + R, t > 0}, ∂ BR = {(x, t) : |x| < R} and ∂ B = ∂BR \ ∂ BR . 50

Proof of Lemma 4.4. We adapt the proof of the Bˆocher theorem given in [5]. Define R 1−2σ U (x, t)dSr ∂ ′′ Br+ t R A[U ](r) = t1−2σ dSr ∂ ′′ Br+

where r = |(x, t)| > 0 and dSr is the volume element of ∂ ′′ Br . By direct computation we have d A[U ](r) = dr Let f (r) =

R

∂ ′′ Br+

Z

∂ ′′ Br+

t1−2σ ∇U (x, t) · (x,t) r dSr R . 1−2σ t dSr ∂ ′′ Br+

t1−2σ ∇U (x, t) ·

(x, t) dSr . r

Since U satisfies (4.18), by integration by parts we have f (r1 ) = f (r2 ), ∀ 0 < r1 , r2 < 1. Notice that

Z

∂ ′′ Br+

t

1−2σ

dSr = r

n+1−2σ

Z

∂ ′′ B1+

t1−2σ dS1 .

Thus there exists a constant b such that d A[U ](r) = br−n−1+2σ . dr So there exist constants a and b such that A[U ](r) = a + br2σ−n . Since we have the Harnack inequalities for U as in the proof of Lemma 4.1, the rest of the arguments are rather similar to those in [5] and are omitted here. We refer to [5] for details.

A.3 Two lemmas on maximum principles Lemma A.1. Let Q1 = B1 × (0, 1) ⊂ Rn+1 + , then there exists ε = ε(n, σ) such that for all |a(x)| ≤ ε|x|−2σ , if U ∈ H(t1−2σ , Q1 ), U ≥ 0 on ∂ ′′ Q1 , and Z Z aU (·, 0)ϕ for all 0 ≤ ϕ ∈ Cc∞ (Q1 ). t1−2σ ∇U ∇ϕ ≥ B1

Q1

Then U ≥0

in Q1 .

Proof. By a density argument, we can use U − as a test function. Hence we have Z Z 1−2σ − 2 |a|(U − (·, 0))2 . t |∇U | ≤ B1

Q1

51

(A.4)

We extend U − to be zero outside of Q1 and still denote it as U − . Then the trace U − (·, 0) ∈ H˙ σ (Rn ). Since kU − (·, 0)k2H˙ σ (Rn ) =

Z

Rn+1 +

t1−2σ |∇Pσ ∗ U − (·, 0)|2 ≤

we have kU − (·, 0)k2H˙ σ (Rn ) ≤

Z

B1

Z

Rn+1 +

t1−2σ |∇U − |2 ,

|a|(U − (·, 0))2 .

By Hardy’s inequality (see, e.g., [94]) Z C(n, σ) |x|−2σ (U − (·, 0))2 ≤ kU − (·, 0)k2H˙ σ (Rn ) Rn

Γ((n+2σ)/4) where C(n, σ) = 22σ Γ((n−2σ)/4) is the best constant. Hence if ε < C(n, σ), U − (·, 0) ≡ 0 and − hence by (A.4), U ≡ 0 in Q1 .

Lemma A.2. Let a(x) ∈ L∞ (B1 ). Let W ∈ C(Q1 ) ∩ C 2 (Q1 ) satisfying ∇x W ∈ C(Q1 ), t1−2σ ∂t W ∈ C(Q1 ), and 1−2σ ∇W ) ≥ 0 in Q1 −div(t 1−2σ − lim t ∂t W (x, t) ≥ a(x)W (x, 0) on ∂ ′ Q1 (A.5) t→0 W > 0 in Q1 . If U ∈ C(Q1 ) ∩ C 2 (Q1 ) satisfying ∇x U ∈ C(Q1 ), t1−2σ ∂t U ∈ C(Q1 ), and 1−2σ ∇U ) ≥ 0 in Q1 −div(t − lim t1−2σ ∂t U (x, t) ≥ a(x)U (x, 0) on ∂ ′ Q1 t→0 U ≥ 0 in ∂ ′′ Q1 .

(A.6)

Then U ≥ 0 in Q1 .

Proof. Let V = U/W . Then 1−2σ 1−2σ ∇V ) − 2t1−2σ ∇VW∇W − div(t W ∇W )V −div(t V − lim t1−2σ ∂t V + W − lim t1−2σ ∂t W (x, t) − a(x)W (x, 0) t→0 t→0 V

≥0 ≥0 ≥0

in Q1 on ∂ ′ Q1

(A.7)

′′

in ∂ Q1 .

We are going to show that V ≥ 0 in Q1 . If not, then we choose k such that inf Q1 v < k ≤ 0. Let Vk = V − k

and Vk− = max(−Vk , 0).

Multiplying Vk− to (A.7), we have Z Z t1−2σ |∇Vk− |2 ≤ 2 Q1

Q1

52

t1−2σ W −1 Vk− ∇Vk− ∇W.

(A.8)

Case 1: Suppose 1 − 2σ ≤ 0. Denote Γk = Supp(∇Vk− ). Then by the H¨older inequality and the bounds of ∇x W , t1−2σ ∂t W , 21 Z Z 21 Z t1−2σ W −1 Vk− ∇Vk− ∇W ≤ C t1−2σ |∇Vk− |2 2 . t1−2σ |Vk− |2 Q1

Q1

Hence it follows from (A.8) that Z

Γk

Z

t1−2σ |∇Vk− |2 ≤ C

Q1

Γk

t1−2σ |Vk− |2 .

(A.9)

Since Vk− = 0 on ∂ ′′ Q1 , by Lemma 2.1 in [90], Z

t

1−2σ

Q1

|Vk− |2(n+1)/n

n n+1

≤C

Z

Q1

t1−2σ |∇Vk− |2 .

(A.10)

By (A.9), (A.10) and H¨older inequality, Z

Γk

t1−2σ ≥ C.

This yields a contradiction when k → inf Q1 v, since ∇V = 0 on the set of V ≡ inf Q1 V . Case 2: Suppose 1 − 2σ > 0. Denote Γk = Supp(Vk− ). Then by H¨older inequality and the bounds of ∇x W , t1−2σ ∂t W , Z Z − 2 1−2σ t1−2σ W −1 Vk− ∇Vk− ∇W t |∇Vk | ≤ 2 Q1 Q1 Z Vk− ∇Vk− ≤C Q1 Z Z − 2 1/2 1−2σ t2σ−1 |Vk− |2 )1/2 . t |∇Vk | ) ( ≤ C( Q1

Q1

Hence Z

Q1

Since

Vk−

t1−2σ |∇Vk− |2

Z

Q1

Z

t1−2σ |∇Vk− |2 ≤ C

Q1

Q1

t2σ−1 |Vk− |2 .

′′

= 0 on ∂ Q1 , by the proof of Lemma 2.3 in [90], for any β > −1, Z Z t1−2σ |∇Vk− |2 . tβ |Vk− |2 ≤ C(β) Q1

Q1

In the following we choose β = σ − 1. Hence, Z Z Z tσ−1 |Vk− |2 ≤ C t1−2σ |∇Vk− |2 Z

Γk

Q1

Q1

Q1

i.e.

Z

t1−2σ |∇Vk− |2

t1−2σ |∇Vk− |2

Z

Γk

tσ−1 |Vk− |2 ≤ C 53

Z

Γk

t1−2σ |∇Vk− |2

Z

t1−2σ |∇Vk− |2

Z

Q1

t2σ−1 |Vk− |2 ,

Γk

t2σ−1 |Vk− |2 .

Fixed ε > 0 sufficiently small which will be chosen later. By the strong maximum principle inf Q1 V has to be attained only on ∂ ′ Q1 , then we can choose k sufficiently closed to inf Q1 V such that Γk ⊂ B1 × [0, ε]. Then Z Z ε−σ tσ−1 |Vk− |2 . t2σ−1 |Vk− |2 ≤ C Γk

Γk

Choose ε small enough such that ε−σ > C + 1. It follows that Z Z − 2 1−2σ t |∇Vk | t2σ−1 |Vk− |2 = 0. Γk

Γk

Hence one of them has to be zero, which reaches a contradiction immediately.

A.4 Complementarities Lemma A.3. Let u(x) ∈ Cc∞ (Rn ) and V (·, t) = Pσ (·, t) ∗ u(·). For any U ∈ Cc∞ (Rn+1 ∪ ∂Rn+1 + + ) with U (x, 0) = u(x), Z Z Rn+1 +

t1−2σ |∇V |2 ≤

Rn+1 +

t1−2σ |∇U |2 .

+ + Proof. Let 0 ≤ η(x, t) ≤ 1, Supp(η) ⊂ B2R , η = 1 in BR and |∇η| ≤ 2/R. In the end we will let + . Since div(t1−2σ ∇V ) = 0, then R → ∞ and hence we may assume that U is supported in BR/2

0= =

Z

Rn+1 +

Z

Rn+1 +

t1−2σ ∇V ∇(η(U − V )) t

1−2σ

η∇U ∇V −

Z

Rn+1 +

t

1−2σ

2

η|∇V | −

Z

+ + B2R \BR

t1−2σ V ∇η∇V

+ where we used η(U − V ) = 0 on the boundary of B2R in the first equality. + + Note that for (x, t) ∈ B2R \BR Z t2σ |V (x, t)| = β(n, σ) n+2σ u(ξ) dξ Rn (|x − ξ|2 + t2 ) 2 Z (|x|2 + t2 )σ ≤ β(n, σ) n+2σ |u(ξ)| dξ Rn (|x|2 /4 + t2 ) 2 n

≤ C(n, σ)(|x|2 + t2 )− 2 kukL1

+ . where in the first inequality we have used that U is supported in BR/2

54

Direct computations yield Z t1−2σ V ∇η∇V + + B2R \BR ≤

Z

≤

Z

+ + B2R \BR

+ + B2R \BR

t

1−2σ

t

1−2σ

|∇V |

2

|∇V |

2

!1/2

Z

+ + B2R \BR

t

1−2σ

2

V |∇η|

2

!1/2

!1/2

· C(n, σ)|u|L1 (Rn ) (Rn+2−2σ−2−2n )1/2 → 0 as R → ∞ R where we used (2.4) that Rn+1 t1−2σ |∇V |2 < ∞. Therefore, we have +

Z

Rn+1 +

Finally, by H¨older inequality, Z

t

1−2σ

Rn+1 +

Z 1−2σ |∇V | ≤ t ∇U ∇V . Rn+1 + 2

t1−2σ |∇V |2 ≤

Z

Rn+1 +

t1−2σ |∇U |2 .

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Tianling Jin Department of Mathematics, Rutgers University 110 Frelinghuysen Road, Piscataway, NJ 08854, USA Email: [email protected] YanYan Li Department of Mathematics, Rutgers University 110 Frelinghuysen Road, Piscataway, NJ 08854, USA Email: [email protected] Jingang Xiong School of Mathematical Sciences, Beijing Normal University Beijing 100875, China and Department of Mathematics, Rutgers University 110 Frelinghuysen Road, Piscataway, NJ 08854, USA Email: [email protected]/[email protected]

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