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Compactness of conformal metrics with constantQ-curvature. I

YanYan Li∗ and Jingang Xiong†

Abstract

We establish compactness for nonnegative solutions of the fourth order constantQ-curvature equations on smooth compact Riemannian manifolds of dimension≥ 5. If theQ-curvature equals−1, we prove that all solutions are universally bounded. If theQ-curvature is1, assuming that Paneitz operator’s kernel is trivial and itsGreen functionis positive, we establish universal energy bounds on manifolds which are either locallyconformally flat (LCF) or of dimension≤ 9. By assuming a positive mass type theoremfor the Paneitz operator, we prove compactness inC4. Positive mass type theorems havebeen verified recently on LCF manifolds or manifolds of dimension≤ 7, when the Yamabeinvariant is positive. We also prove that, for dimension≥ 8, the Weyl tensor has to vanish atpossible blow up points of a sequence of solutions. This implies the compactness result indimension≥ 8 when the Weyl tensor does not vanish anywhere. To overcome difficultiesstemming from fourth order elliptic equations, we develop ablow up analysis procedurevia integral equations.

Contents

1 Introduction 2

2 Preliminaries 82.1 Paneitz operator in conformal normal coordinates . . . . .. . . . . . . . . . . 82.2 Two Pohozaev type identities . . . . . . . . . . . . . . . . . . . . . . .. . . . 13

3 Blow up analysis for integral equations 15

∗Supported in part by NSF grants DMS-1065971 and DMS-1203961.†Supported in part by Beijing Municipal Commission of Education for the Supervisor of Excellent Doctoral

Dissertation (20131002701).

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http://arxiv.org/abs/1506.00739v2

Y. Y. Li & J. Xiong

4 Expansions of blow up solutions of integral equations 33

5 Blow up local solutions of fourth order equations 41

6 Global analysis, and proof of Theorem 1.2 50

7 Proof of Theorems 1.1, Theorem 1.3, Theorem 1.4 56

8 Proof of Theorem 1.5 57

A Local estimates for solutions of linear integral equations 58

B Riesz potentials of some functions 60

1 Introduction

On a compact smooth Riemannian manifold(M, g) of dimension≥ 3, the Yamabe problem,which concerns the existence of constant scalar curvature metrics in the conformal class ofg,was solved through the works of Yamabe [57], Trudinger [54],Aubin [2] and Schoen [49].Different proofs of the Yamabe problem in the casen ≤ 5 and in the case(M, g) is locallyconformally flat are given by Bahri and Brezis [4] and Bahri [3]. The problem is equivalent tosolving the Yamabe equation

− Lgu = Sign(λ1)un+2n−2 , u > 0 onM, (1)

whereLg := ∆g −(n−2)4(n−1)

Rg, ∆g is the Laplace-Beltrami operator associated withg, Rg isthe scalar curvature, andSign(λ1) denotes the sign of the first eigenvalueλ1 of the conformalLaplacian−Lg. The sign ofλ1 is conformally invariant, i.e., it is the same for every metric inthe conformal class ofg.

If λ1 < 0, there exists a unique solution of (1). Ifλ1 = 0, the equation is linear andsolutions are unique up to multiplication by a positive constant. If λ1 > 0, non-uniquenesshas been established; see Schoen [50] and Pollack [46]. If(M, g) is the standard unit sphere,all solutions are classified by Obata [44] and there is no uniformL∞ bound for them. Schoen[52] established a uniformC2 bound for all solutions ifM is locally conformally flat but notconformal to the sphere. The uniformC2 bound was established in dimensionsn ≤ 7 by Li-Zhang [38] and Marques [43] independently. Forn = 3, 4, 5, see works of Li-Zhu [40], Druet[17, 18] and Li-Zhang [37]. For8 ≤ n ≤ 24, the answer is positive provided that the positivemass theorem holds in these dimensions; see Li-Zhang [38, 39] for 8 ≤ n ≤ 11, and Khuri-Marques-Schoen [32] for12 ≤ n ≤ 24. On the other hand, the answer is negative in dimensionn ≥ 25; see Brendle [6] forn ≥ 52, and Brendle-Marques [7] for25 ≤ n ≤ 51.

2

Compactness of conformal metrics with constantQ-curvature

In this paper, we are interested in a fourth order analogue ofthe Yamabe problem. Namely,the constantQ-curvature problem. Let us recall the conformally invariant Paneitz operator andthe correspondingQ-curvature, which are defined as1

Pg = ∆2g − divg(anRgg + bnRicg)d+

n− 4

2Qg (2)

Qg = −1

2(n− 1)∆gRg +

n3 − 4n2 + 16n− 16

8(n− 1)2(n− 2)2R2g −

2

(n− 2)2|Ricg|

2, (3)

whereRg andRicg denote the scalar curvature and Ricci tensor ofg respectively, andan =(n−2)2+4

2(n−1)(n−2), bn = − 4

n−2. The self-adjoint operatorPg was discovered by Paneitz [45] in 1983,

andQg was introduced later by Branson [5]. Paneitz operator is conformally invariant in thesense that

• If n = 4, for any conformal metricg = e2wg, w ∈ C∞(M), there holds

Pg = e−4wPg and Pgw +Qg = Qge4w. (4)

• If n = 3 or n ≥ 5, for any conformal metricg = u4

n−4 g, 0 < u ∈ C∞(M), there holds

Pg(φ) = u−n+4n−4Pg(uφ) ∀ φ ∈ C∞(M). (5)

Hence, finding constantQ-curvature in the conformal class ofg is equivalent to solving

Pgw +Qg = λe4w onM (6)

if n = 4, andPgu = λu

n+4n−4 , u > 0 onM, (7)

if n = 3 or n ≥ 5, whereλ is a constant.Whenn = 4, there is a Chern-Gauss-Bonnet type formula involving theQ-curvature; see

Chang-Yang [11]. The constantQ-curvature problem has been studied by Chang-Yang [10],Djadli-Malchiodi [15], Li-Li-Liu [34] and references therein. Bubbling analysis and compact-ness for solutions have been studied by Druet-Robert [19], Malchiodi [42], Weinstein-Zhang[56] among others.

Whenn ≥ 5, the constantQ-curvature problem is a natural extension of the Yamabe prob-lem. However, the lack of maximum principle for fourth orderelliptic equations makes theproblem much harder. The first eigenvalues of fourth order self-adjoint elliptic operators arenot necessarily simple and the associated eigenfunctions may change signs. We might not be

1If n = 4, 1

2Qg is defined as theQ-curvature in some papers.

3

Y. Y. Li & J. Xiong

able to divide the study of (7) into three mutually exclusivecases by linking the constantλ tothe sign of the first eigenvalue of the Paneitz operator. Up tonow, the existence of solutionshas been obtained withλ = 1, roughly speaking, under the following three types of assump-tions. The first one is on the equation. Assuming, among others, the coefficients of the Paneitzoperator are constants, Djadli-Hebey-Ledoux [14] proved some existence results, where theydecompose the operator as a product of two second order elliptic operators and use the max-imum principle of second order elliptic equations. This assumption is fulfilled, for instance,when the background metric is Einstein. The second one is on the geometry and topology ofthe manifolds. Assuming that the Poincare exponent is lessthan(n−4)/2, Qing-Raske [47, 48]proved the existence result on locally conformally flat manifolds of positive scalar curvature.The last one is purely geometric. Assuming that there existsa conformal metric of nonnegativescalar curvature and semi-positiveQ-curvature, Gursky-Malchiodi [22] recently proved the ex-istence result forn ≥ 5. By their condition, the scalar curvature was proved to be positive. Ina very recent preprint, Hang-Yang [24] replaced the positive scalar curvature condition by thepositive Yamabe invariant (which is equivalent toλ1 > 0). More precisely, (7) admits a solutionwith λ = 1 if

λ1(−Lg) > 0, Qg ≥ 0 andQg > 0 somewhere onM, (8)

whereλ1(−Lg) is the first eigenvalue of−Lg defined above. See also Hang-Yang [23] forn = 3.

Each of the above three types of assumptions implies that

KerPg = 0 and the Green’s functionGg of Pg is positive. (9)

In fact,Pg is coercive in Djadli-Hebey-Ledoux [14], Qing-Raske [47, 48] and Gursky-Malchiodi[22]. We refer to the latest paper Gursky-Hang-Lin [21] for further discussions on these con-ditions. If (9) holds andλ1 > 0, there exists a positive mass type theorem forGg, providedM is locally conformally flat orn = 5, 6, 7, but not conformal to the standard sphere; seeHumbert-Raulot [28], Gursky-Malchiodi [22] and Hang-Yang[25].

Starting from this paper, we study the compactness of solutions of the constantQ-curvatureequation forn ≥ 5. For positive constantQ-curvature problem, there are non-compact exam-ples. If (M, g) is a sphere, the constantQ-curvature metrics are not compact inC4 due to thenon-compactness of the conformal diffeomorphism group of the sphere. Recently, Wei-Zhao[55] produced non-compact examples on manifolds of dimensionn ≥ 25 not conformal to thestandard sphere.

Theorem 1.1. Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 5, butnot conformal to the standard sphere. Assume (9). For1 < p ≤ n+4

n−4, let 0 < u ∈ C4(M) be a

solution ofPgu = c(n)up onM, (10)

4


wherec(n) = n(n + 2)(n − 2)(n − 4). Suppose that one of the following conditions is alsosatisfied:

i) λ1(−Lg) > 0 and(M, g) is locally conformally flat,

ii) λ1(−Lg) > 0 andn = 5, 6, 7,

iii) (M, g) is locally conformally flat or5 ≤ n ≤ 9, and the positive mass type theorem holdsfor the Paneitz operator,

iv) The Weyl tensor ofg does not vanish anywhere, i.e.,|Wg|2 > 0 onM .

Then there exists a constantC > 0, depending only onM, g, and a lower bound ofp− 1, suchthat

‖u‖C4(M) + ‖1/u‖C4(M) ≤ C. (11)

The assumption (9) in the theorem can be replaced by (8), as explained above. The positivemass type theorem for Paneitz operator in dimension8, 9 is understood as in Remark 2.1. Thecase5 ≤ n ≤ 9 for positive constantQ-curvature equation shows some similarity to3 ≤ n ≤ 7for the Yamabe equation with positive scalar curvature.

The following situations, included in Theorem 1.1, were proved before. IfM is locally con-formally flat andp = n+4

n−4, (11) was established by Qing-Raske [47, 48] with the assumptions

thatλ1 > 0 and the Poincare exponent is less than(n − 4)/2, and by Hebey-Robert [26, 27]with C depending on theH2 norm ofu, where they assumed thatPg is coercive.

Neitherλ1(−Lg) > 0 nor the positive mass type theorem for Paneitz operator is assumed,we have an energy bound of solutions.

Theorem 1.2. Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 5.Assume (9), and assume that eithern ≤ 9 or (M, g) is locally conformally flat.

Let0 < u ∈ C4(M) be a solution of(10). Then

‖u‖H2(M) ≤ C,

whereC > 0 depends only onM, g, and a lower bound ofp− 1.

Next, we establish Weyl tensor vanishing results.

Theorem 1.3. Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 8.Assume (9). Letui be a sequence of positive solutions of

Pgui = c(n)upii ,

wherepi ≤n+4n−4

, pi →n+4n−4

asi → ∞. Suppose that there is a sequence ofXi → X ∈ M suchthatui(Xi) → ∞. Then the Weyl tensor has to vanish atX, i.e.,Wg(X) = 0.

5

Y. Y. Li & J. Xiong

Furthermore, ifn = 8, 9, there existsX ′i → X such that, for alli,

|Wg(X′i)|

2 ≤ C

(log ui(X′i))

−1, if n = 8,

ui(X′i)

− 2n−4 , if n = 9,

whereC > 0 depends only onM andg.

Theorem 1.4.In addition to the assumptions in Theorem 1.3 withn ≥ 10, we assume that thereexist a neighborhoodΩ of X and a constantb > 0 such that

ui(X) ≤ b · distg(X,Xi)− 4

pi−1 ∀ X ∈ Ω, (12)

Xi is a local maximum point ofui, supΩui ≤ bui(Xi). (13)

Then, for sufficiently largei,

|Wg(Xi)|2 ≤ C

ui(Xi)− 4

n−4 log ui(Xi), if n = 10,

ui(Xi)− 4

n−4 , if n ≥ 11,

whereC > 0 depends only onM, g, distg(X, ∂Ω) and b.

The rates of decay of|Wg(Xi)| in Theorem 1.3 and Theorem 1.4 correspond to the Yamabeproblem casen = 6, 7 andn ≥ 8 respectively; see theorem 1.3 and theorem 1.2 in [38]. Condi-tion (12) and (13) can often be reduced to, by some elementaryconsideration, in applications.

In a subsequent paper, we will establish compactness results analogous to those establishedin 8 ≤ n ≤ 24 for the Yamabe equation by Li-Zhang [38, 39] and Khuri-Marques-Schoen [32].The present paper provides analysis foundations.

For the negative constantQ-curvature equation, we have

Theorem 1.5.Let(M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 5. Thenfor any1 < p <∞, there exists a positive constantC, depending only onM, g andp, such thatevery nonnegativeC4 solution of

Pg(u) = −up onM (14)

satisfies‖u‖C4(M) ≤ C.

The proofs of Theorems 1.1, Theorem 1.3 and Theorem 1.4 make use of important ideas forthe proof of compactness of positive solutions of the Yamabeequation, which were outlined firstby Schoen [50, 51, 52], as well as methods developed through the work Li [35], Li-Zhu [40],Li-Zhang [37, 38, 39], and Marques [43]. Our main difficulty now stems from the fourth order

6


equation, which we explain in details. To understand the profile of possible blow up solutions,it is natural to scale the solutions in local coordinates centered at local maximum points. By theLiouville theorem in Lin [41], one can conclude that these solutions are close to some standardbubbles in small geodesic balls, whose sizes become smallerand smaller as solutions blowingup; see e.g., Proposition 6.1. Then we need to answer two questions:

(i) Do these blow up points accumulate?

(ii) If not, how do these solutions behave in geodesic balls with some fixed size?

For the first one, we may scale possible blow up points apart and look at them individually.It turns out that we end up with the situation of question (ii). After scaling we need to carryout local analysis. In the Yamabe case, properties of secondorder elliptic equations, whichinclude the maximum principle, comparison principle, Harnack inequality and Bocher theoremfor isolated singularity, were used crucially. Now we don’thave these properties for fourthorder elliptic equations. This leads to an obstruction to using fourth order equations to developlocal analysis.

We observe that along scalings the bounds of Green’s function are preserved. In view ofGreen’s representation, we develop a blow up analysis procedure for integral equations andanswer the above two questions completely in dimensions less than10. This is inspired byour recent joint work with Jin [31] for a unified treatment of the Nirenberg problem and itsgeneralizations, which in turn was stimulated by our previous work on a fractional Nirenbergproblem [29, 30]. The approach of the latter two papers were based on the Caffarelli-Silvestreextension developed in [8]. Our analysis is very flexible andcan easily be adapted to deal withhigher order and fractional order conformally invariant elliptic equations. The organization ofthe paper is shown in the table of Contents.

Notations. Lettersx, y, z denote points inRn, and capital lettersX, Y, Z denote points onRiemannian manifolds. Denote byBr(x) ⊂ Rn the ball centered atx with radiusr > 0. Wemay writeBr in replace ofBr(0) for brevity. ForX ∈ M , Bδ(X) denotes the geodesic ballcentered atX with radiusδ. Throughout the paper, constantsC > 0 in inequalities may varyfrom line to line and are universal, which means they depend on given quantities but not onsolutions. f = O(k)(rm) denotes any quantity satisfying|∇jf(r)| ≤ Crm−j for all integers1 ≤ j ≤ k, wherek is a positive integer andm is a real number.|Sn−1| denotes the area of thestandardn− 1-sphere. Here are specified constants used throughout the paper:

• c(n) = n(n+ 2)(n− 2)(n− 4) appears in constantQ-curvature equation,

• αn = 12(n−2)(n−4)|Sn−1|

appears in the expansion of Green’s functions,

• cn = c(n) · αn = n(n+2)2|Sn−1|

.

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Y. Y. Li & J. Xiong

Added note on June 1, 2015:Theorem 1.1 was announced by the first named author in histalk at the International Conference on Local and Nonlocal Partial Differential Equations, NYUShanghai, China, April 24-26, 2015; while the part of the theorem for general manifolds ofdimensionn = 5, 6, 7 and for locally conformally flat manifolds of dimensionn ≥ 5 was an-nounced in his talk at the Conference on Partial Differential Equations, University of Sussex,UK, September 15-17, 2014. We noticed that two days ago an article was posted on the arXiv,[Gang Li, A compactness theorem on Branson′s Q-curvature equation, arXiv:1505.07692v1[math.DG] 28 May 2015], where a compactness result in dimension n = 5, under the assump-tion thatRg > 0 andQg ≥ 0 but not identically equal to zero, was proved independently.

Acknowledgments:J. Xiong is grateful to Professor Jiguang Bao and Professor Gang Tian fortheir supports.

2 Preliminaries

2.1 Paneitz operator in conformal normal coordinates

Let (M, g) be a smooth Riemannian manifold (with or without boundary) of dimensionn ≥ 5,andPg be the Paneitz operator onM . For any pointX ∈M , it was proved in [33], together withsome improvement in [9] and [20], that there exists a positive smooth functionκ (with control)onM such that the conformal metricg = κ

−4n−4 g satisfies, ing-normal coordinatesx1, . . . , xn

centered atX,det g = 1 in Bδ

for someδ > 0. We refer such coordinates as conformal normal coordinates. Notice thatdet g = 1+O(|x|N) will be enough for our use ifN is sufficiently large. Since one can viewxas a tangent vector ofM atX, thusdet g(x) = 1+O(|x|2). It follows thatκ(x) = 1+O(|x|2).In particular,

κ(0) = 1, ∇κ(0) = 0. (15)

In the g-normal coordinates,

Rij(0) = 0, SymijkRij,k(0) = 0,

R,i(0) = 0, ∆gR(0) = −1

6|Wg(0)|

2,

where the Ricci tensorRij , scalar curvatureR, and Weyl tensorW are with respect tog. Wealso have

∆g = ∆+ ∂lgkl∂k + (gkl − δkl)∂kl =: ∆ + d

(1)k ∂k + d

(2)kl ∂kl,

8


and∆2g = ∆2 + f

(1)k ∂k + f

(2)kl ∂kl + f

(3)kls∂kls + f

(4)klst∂klst,

where

f(1)k : = ∆d

(1)k + d(1)s ∂sd

(1)k + d

(2)st ∂std

(1)k = O(1),

f(2)kl : = ∂kd

(1)l +∆d

(2)kl + d

(1)k d

(1)l + d(1)s ∂sd

(2)kl + d

(2)sl ∂sd

(1)k + d

(2)st ∂std

(2)kl = O(1),

f(3)kls : = 2d(1)s δkl + ∂sd

(2)kl + 2d(1)s d

(2)kl + d

(2)st ∂td

(2)kl = O(|x|),

f(4)klst : = 2d

(2)kl δ

st + d(2)kl d

(2)st = O(|x|2).

Now the second term of the Paneitz operatorPg can be expressed as

−divg(anRgg + bnRicg)d = −∂l((anRgst + bnRst)gskgtl∂k) =: f

(5)k ∂k + f

(6)kl ∂kl,

where

f(5)k : = −∂l

(

(anRgst + bnRst)gskgtl

)

= O(1),

f(6)kl : = −(anRgst + bnRst)g

skgtl = O(|x|).

By abusing notations, we relabelf (1)k asf (1)

k + f(5)k , andf (2)

kl asf (1)kl + f

(6)kl . Hence,

E(u) : = Pgu−∆2u

=n− 4

2Qgu+ f

(1)k ∂ku+ f

(2)kl ∂klu+ f

(3)kls∂klsu+ f

(4)klst∂klstu, (16)

where

f(1)k (x) = O(1), f

(2)kl (x) = O(1), f

(3)kls (x) = O(|x|), f

(4)klst(x) = O(|x|2). (17)

We point out that each term off (1)k takes up to three times derivatives ofg totally, each term of

f(2)kl (x) takes twice, each term off (3)

kls (x) takes once, and no derivative ofg is taken in any termof f (4)

klst. Hence, we see that

‖f (1)k ‖L∞(Bδ) + ‖f (2)

kl ‖L∞(Bδ) + ‖∇f (3)kls‖L∞(Bδ) + ‖∇2f

(4)klst‖L∞(Bδ)

≤ C∑

k≥1,2≤k+1≤4

‖∇kg‖lL∞(Bδ)(18)

9

Y. Y. Li & J. Xiong

Lemma 2.1. In the g-normal coordinates, we have, for any smooth radial function u,

Pgu =∆2u+1

2(n− 1)R,kl(0)x

kxl(c∗1u′

r+ c∗2u

′′)−4

9(n− 2)r2

∑

kl

(Wikjl(0)xixj)2(u′′ −

u′

r)

+n− 4

24(n− 1)|Wg(0)|

2u+ (ψ5(x)

r2+ ψ3(x))u

′′ − (ψ5(x)

r3+ψ′3(x)

r)u′ + ψ1(x)u

+O(r4)u′′ +O(r3)u′ +O(r2)u,

wherer = |x|, ψk(x), ψ′k(x) are homogeneous polynomials of degreek, and

c∗1 =2(n− 1)

(n− 2)−

(n− 1)(n− 2)

2+ 6− n, c∗2 = −

n− 2

2−

2

n− 2. (19)

Proof. Sincedet g = 1 andu is radial, we have∆2gu = ∆2u. The rest of the proof is same as

that of Lemma 2.8 of [22]. It suffices to expand the coefficients of lower order terms ofPg inTaylor series to a higher order so that(ψ5(x)

r2+ ψ3(x))u

′′ − (ψ5(x)r3

+ψ′3(x)

r)u′ + ψ1(x)u appears.

If KerPg = 0, thenPg has unique Green functionGg, i.e.,PgGg(X, ·) = δX(·) for everyX ∈ M , whereδX(·) is the Dirac measure atX on manifolds(M, g). It is easy to check thatKerPg = 0 is conformally invariant.

Proposition 2.1([22], [24]). Let (M, g) be a smooth compact Riemannian manifold of dimen-sionn ≥ 5, on whichKerPg = 0. Then there exists a small constantδ > 0, depending onlyon (M, g), such that ifdet g = 1 in the normal coordinatex1, . . . , xn centered atX, theGreen’s functionG(X, expX x) ofPg has the expansion, forx ∈ Bδ(0),

• If n = 5, 6, 7, orM is flat in a neighborhood ofX,

G(X, expX x) =αn

|x|n−4+ A+O(4)(|x|),

• If n = 8,

G(X, expX x) =αn

|x|n−4−

αn1440

|W (X)|2 log |x|+O(4)(1),

• If n ≥ 9,

G(X, expX x) =αn

|x|n−4

(

1 + ψ4(x))

+O(4)(|x|9−n),

whereαn = 12(n−2)(n−4)|Sn−1 |

, A is a constant,W (X) is the Weyl tensor atX, andψ4(x) ahomogeneous polynomial of degree4.

10


Corollary 2.1. Suppose the assumptions in Proposition 2.1. Then in the normal coordinatecentered atX we have

G(expX x, expX y) =αn(1 +O(4)(|x|2) +O(4)(|y|2))

|x− y|n−4+ a+O(4)(|x− y|6−n),

wherex, y ∈ Bδ, x− y = (x1 − y1, . . . , xn − yn), |x− y| =√

∑ni=1(xi − yi)2, a is a constant

anda = 0 if n ≥ 6.

Proof. We only prove the case thatM is non-locally formally flat. DenoteX = expX x andY = expX y for x, y ∈ Bδ, whereδ > 0 depends only on(M, g). ForX 6= X, we can findgX = v

4n−4 g such that in thegX-normal coordinate centered atX there holddet gX = 1 and

v(Y ) = 1+O(4)(distgX(X, Y )2). LetGgX be the Green’s function ofPgX . By Proposition 2.1,

GgX (X, Y ) = αndistgX (X, Y )4−n + A+O(4)(distgX(X, Y )6−n),

whereA is a constant andA = 0 if n ≥ 6. By the conformal invariance of the Paneitz operator,we have the transformation law

G(X, Y ) = GgX(X, Y )v(X)v(Y ) = GgX (X, Y )v(Y ).

SincegX = v4

n−4 g andv(Y ) = 1 +O(4)(distgX (X, Y )2), we obtain

distgX (X, Y ) = distgX (expX x, expX y)

= (1 +O(4)(|x− y|2))distg(expX x, expX y)

= (1 +O(4)(|x− y|2))(1 +O(4)(|x|2) +O(4)(|y|2))|x− y|,

whereg is viewed as a Riemannian metric onBδ because of the exponential mapexpX .Therefore, we get

G(expX x, expX y) = αn1 +O(4)(|x|2) +O(4)(|y|2)

|x− y|n−4+O(4)(|x− y|6−n).

If X = X, it follows Proposition 2.1. We complete the proof.

The following positive mass type theorem for Paneitz operator was proved through [28],[22] and [24].

Theorem 2.1.Let (M, g) be a compact manifold of dimensionn ≥ 5, andX ∈ M be a point.Let g be a conformal metric ofg such thatdet g = 1 in the g-normal coordinatex1, . . . , xncentered atX. Suppose also thatλ1(−Lg) > 0 and(9) holds. Ifn = 5, 6, 7, or (M, g) is locallyconformally flat, then the constantA in Proposition 2.1 is nonnegative, andA = 0 if and onlyif (M, g) is conformal to the standardn-sphere.

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Y. Y. Li & J. Xiong

Remark 2.1. Suppose the assumptions in Theorem 2.1. IfW (X) = 0, it follows from Proposi-tion 2.1 of [24] that, in theg-normal coordinates centered atX, the Green’s functionG of Pghas the expansion

G(X, expX x) =

α8|x|−4 + ψ(θ) + log |x|O(4)(|x|), n = 8,

α9|x|−5(1 +R,ij(X)xixj |x|2

384) + A+O(4)(|x|), n = 9,

wherex = |x|θ, ψ is a smooth function ofθ, andA is constant. In dimensionn = 8, 9, wesay the positive mass type theorem holds for Paneitz operator if

∫

Sn−1 ψ(θ) dθ > 0 andA > 0respectively.

Let

Uλ(x) :=

(

λ

1 + λ2|x|2

)n−42

, λ > 0,

which is the unique positive solution of∆2u = c(n)un+4n−4 in R

n, n ≥ 5, up to translations byLin [41]. By Lemma 2.1, in theg-normal coordinates we have

PgUλ = c(n)Un+4n−4

λ + fλUλ, (20)

wherefλ(x) is a smooth function satisfying thatλ−k|∇kxfλ(x)|, k = 0, 1, . . . , 5, is uniformly

bounded inBδ independent ofλ ≥ 1. Indeed, by direct computations

∂rUλ = (4− n)λn2 (1 + λ2r2)

2−n2 r

∂2rrUλ = (4− n)(2− n)λn+42 (1 + λ2r2)

−n2 r2 + (4− n)λ

n2 (1 + λ2r2)

2−n2 .

Inserting them to the expression in Lemma 2.1, (20) follows.

Corollary 2.2. Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 5, onwhichKerPg = 0. Then there exists a small constantδ > 0, depending only on(M, g), suchthat if det g = 1 in the normal coordinatex1, . . . , xn centered atX, then

Uλ(x) = c(n)

∫

Bδ

G(expX x, expX y)Uλ(y)n+4n−4 + c′λ(x)Uλ(y) dy + c′′λ(x),

whereδ > 0 depends only onM, g, andc′λ, c′′λ are smooth functions satisfying

λ−k|∇kc′λ(x)| ≤ C, |∇kc′′λ(x)| ≤ Cλ4−n2 ,

for k = 0, 1, . . . , 5 and someC > 0 independent ofλ ≥ 1.

12


Proof. Let η(x) = η(|x|) be a smooth cutoff function satisfying

η(t) = 1 for t < δ/2, η(t) = 0 for t > δ.

By the Green’s representation formula, we have

(Uλη)(x) =

∫

Bδ

G(expX x, expX y)Pg(Uλη)(y) dy.

Making use of (20) and Lemma 2.1, we see thatc′λ =fλc(n)

and proof is finished.

2.2 Two Pohozaev type identities

For r > 0, define in Euclidean space

P(r, u) :=

∫

∂Br

n− 4

2

(

∆u∂u

∂ν− u

∂

∂ν(∆u)

)

−r

2|∆u|2

− xk∂ku∂

∂ν(∆u) + ∆u

∂

∂ν(xk∂ku) dS,

(21)

whereν = xr

is the outward normal to∂Br.

Proposition 2.2. Let 0 < u ∈ C4(Br) satisfy

∆2u+ E(u) = Kup in Br,

whereE : C4(Br) → C0(Br) is an operator,p > 0, r > 0 andK ∈ C1(Br). Then

P(r, u) =

∫

Br

(xk∂ku+n− 4

2u)E(u) dx+N (r, u),

where

N (r, u) :=(n

p+ 1−n− 4

2)

∫

Br

Kup+1 dx+1

p+ 1

∫

Br

xk∂kKup+1 dx

−r

p + 1

∫

∂Br

Kup+1 dS.

(22)

Proof. A similar Pohozaev identity withoutE(u) was derived in [16]. We present the proof forcompleteness. For anyu ∈ C4(Br), by Green’s second identity we have

∫

Br

u∆2u dx =

∫

Br

(∆u)2 dx+

∫

∂Br

u∂

∂ν(∆u)−

∂u

∂ν∆u dS

13

Y. Y. Li & J. Xiong

and∫

Br

xk∂ku∆2u dx =

∫

Br

∆(xk∂ku)∆u dx+

∫

∂Br

xk∂ku∂

∂ν(∆u)−

∂

∂ν(xk∂ku)∆u dS.

Using Green’s first identity, we have∫

Br

∆(xk∂ku)∆u dx = 2

∫

Br

(∆u)2 dx+1

2

∫

Br

xk∂k(∆u)2 dx

=4− n

2

∫

Br

(∆u)2 dx+

∫

∂Br

r

2(∆u)2 dS.

Therefore, we obtain

n− 4

2

∫

Br

u∆2u dx+

∫

Br

xk∂ku∆2u dx

=n− 4

2

∫

∂Br

u∂

∂ν(∆u)−

∂u

∂ν∆u dS

+

∫

∂Br

r

2|∆u|2 + xk∂ku

∂

∂ν(∆u)−

∂

∂ν(xk∂ku)∆u dS.

By the equation ofu we get

P(r, u) =

∫

Br

(xk∂ku+n− 4

2u)E(u) dx−

∫

Br

(xk∂ku+n− 4

2u)Kup dx.

Since∫

Br

xk∂kuKup dx =

1

p+ 1

∫

Br

Kxk∂kup+1 dx

= −n

p + 1

∫

Br

Kup+1 dx−1

p+ 1

∫

Br

xk∂kKup+1

+r

p+ 1

∫

Br

Kup+1 dS,

we complete the proof.

Lemma 2.2. For G(x) = |x|4−n + A +O(4)(|x|), whereA is constant. Then

limr→0

P(r, G) = −(n− 4)2(n− 2)A|Sn−1|.

The following proposition is a special case of Proposition 2.15 of [31].

14


Proposition 2.3. For R > 0, let 0 ≤ u ∈ C1(BR) be a solution of

u(x) =

∫

BR

K(y)u(y)p

|x− y|n−4dy + hR(x),

wherep > 0, andhR(x) ∈ C1(BR), ∇hR ∈ L1(BR). Then(

n− 4

2−

n

p+ 1

)∫

BR

K(x)u(x)p+1 dx−1

p+ 1

∫

BR

x∇K(x)u(x)p+1 dx

=n− 4

2

∫

BR

K(x)u(x)phR(x) dx+

∫

BR

x∇hR(x)K(x)u(x)p dx

−R

p + 1

∫

∂BR

K(x)u(x)p+1 dS.

3 Blow up analysis for integral equations

In the section, the idea of dealing with integral equation isinspired by [31], but we have toconsider general integral kernels and remainder terms. We will useA1, A2, A3 to denote positiveconstants, andτi∞i=1 to denote a sequence of nonnegative constants satisfyinglimi→∞ τi = 0.Set

pi =n + 4

n− 4− τi. (23)

Let Gi(x, y)∞i=1 be a sequence of functions onB3 ×B3 satisfying

Gi(x, y) = Gi(y, x), Gi(x, y) ≥ A−11 |x− y|4−n,

|∇lxGi(x, y)| ≤ A1|x− y|4−n−l, l = 0, 1, . . . , 5

Gi(x, y) = cn1 +O(4)(|x|2) +O(4)(|y|2)

|x− y|n−4+ ai +O(4)(

1

|x− y|n−6)

(24)

for all x, y ∈ B3, wherecn = n(n+2)2|Sn−1|

is the constant given towards the end of the introduction,

f = O(4)(rm) denotes any quantity satisfying|∇jf(r)| ≤ A1rm−j for all integers1 ≤ j ≤ 4,

andai is a constant andai = 0 if n ≥ 6. Let Ki∞i=1 ∈ C∞(B3) satisfy

limi→∞

Ki(0) = 1, Ki ≥ A−12 , ‖Ki‖C5(B3) ≤ A2. (25)

Let hi∞i=1 be a sequence of nonnegative functions inC∞(B3) satisfying

maxBr(x)

hi ≤ A2 minBr(x)

hi

5∑

j=1

rj|∇jhi(x)| ≤ A2‖hi‖L∞(Br(x))

(26)

15

Y. Y. Li & J. Xiong

for all x ∈ B2 and0 < r < 1/2.

Givenpi, Gi, Ki, andhi satisfying (23)-(26), let0 ≤ ui ∈ L2nn−4 (B3) be a solution of

ui(x) =

∫

B3

Gi(x, y)Ki(y)upii (y) dy + hi(x) in B3. (27)

It follows from [36] and Proposition A.2 thatui ∈ C4(B3). In the following we will alwaysassumeui ∈ C4(B3).

We say thatui blows up if‖ui‖L∞(B3) → ∞ asi→ ∞.

Definition 3.1. We say a pointx ∈ B3 is an isolated blow up point ofui if there exist0 < r < dist(x, ∂B3), C > 0, and a sequencexi tending tox, such that,xi is a local maximumof ui, ui(xi) → ∞ and

ui(x) ≤ C|x− xi|−4/(pi−1) for all x ∈ Br(xi).

Let xi → x be an isolated blow up ofui. Define

ui(r) =1

|∂Br|

∫

∂Br(xi)

ui dS, r > 0, (28)

andwi(r) = r4/(pi−1)ui(r), r > 0.

Definition 3.2. We sayxi → x ∈ B3 is an isolated simple blow up point, ifxi → x is an isolatedblow up point, such that, for someρ > 0 (independent ofi) wi has precisely one critical pointin (0, ρ) for large i.

Lemma 3.1. Givenpi, Gi, Ki andhi satisfying(23)-(26), let 0 ≤ ui ∈ C4(B3) be a solutionof (27). Suppose that0 is an isolated blow up point ofui with r = 2, i.e., for some positiveconstantA3 independent ofi,

ui(x) ≤ A3|x|−4/(pi−1) for all x ∈ B2. (29)

Then for any0 < r < 1/3 we have

supB2r\Br/2

ui ≤ C infB2r\Br/2

ui,

whereC is a positive constant depending only onn,A1, A2, A3.

16


Proof. For every0 < r < 1/3, set

wi(x) = r4/(pi−1)ui(rx).

By the equation ofui, we have

wi(x) =

∫

B3/r

Gi,r(x, y)Ki(ry)wi(y)pi dy + hi(x) x ∈ B3/r,

whereGi,r(x, y) = rn−4Gi(rx, ry) for r > 0

andhi(x) := r4/(pi−1)hi(rx). Since0 is an isolated blow up point ofui,

wi(x) ≤ A3|x|−4/(pi−1) for all x ∈ B3. (30)

SetΩ1 = B5/2 \ B1/4, Ω2 = B2 \ B1/2 andVi(y) = Ki(ry)wi(y)pi−1. Thuswi satisfies the

linear equation

wi(x) =

∫

Ω1

Gi,r(x, y)Vi(y)wi(y) dy + hi(x) for x ∈ B5/2 \B1/4,

where

hi(x) = hi(x) +

∫

B3/r\Ω1

Gi,r(x, y)Ki(ry)wi(y)pi dy.

By (30) and (25),‖Vi‖L∞(Ω1) ≤ C(n,A1, A2, A3) < ∞. SinceKi andwi are nonnegative, by(24) onGi and (26) onhi we havemaxΩ2

hi ≤ C(n,A1, A2)minΩ2h. Applying Proposition

A.1 towi givesmaxΩ2

wi ≤ CminΩ2

wi,

whereC > 0 depends only onn,A1, A2 andA3. Rescaling back toui, the lemma follows.

Proposition 3.1. Suppose that0 ≤ ui ∈ C4(B3) is a solution of(27) and all assumptions inLemma 3.1 hold. LetRi → ∞ with Rτi

i = 1 + o(1) and εi → 0+, whereo(1) denotes somequantity tending to0 asi→ ∞. Then we have, after passing to a subsequence (still denotedasui, τi and etc . . .),

‖m−1i ui(m

−(pi−1)/4i ·)− (1 + | · |2)(4−n)/2‖C3(B2Ri

(0)) ≤ εi,

ri := Rim−(pi−1)/4i → 0 as i→ ∞,

wheremi = ui(0).

17

Y. Y. Li & J. Xiong

Proof. Let

ϕi(x) = m−1i ui(m

−(pi−1)/4i x) for |x| < 3m

(pi−1)/4i .

By the equation ofui, we have,

ϕi(x) =

∫

B3m

(pi−1)/4i

Gi(x, y)Ki(y)ϕi(y)pi dy + hi(x), (31)

whereGi(x, y) = Gi,m

−pi−1

4i

(x, y), Ki(y) = Ki(m−

pi−1

4i y) andhi(x) = m−1

i hi(m−

pi−1

4i x).

First of all,max∂B1 hi ≤ max∂B1 ui ≤ A3, by (26) we have

hi → 0 in C5loc(R

n) asi→ ∞. (32)

Secondly, since0 is an isolated blow up point ofui,

ϕi(0) = 1, ∇ϕi(0) = 0, 0 < ϕi(x) ≤ A3|x|−4/(pi−1). (33)

For anyR > 0, we claim that‖ϕi‖C4(BR) ≤ C(R) (34)

for sufficiently largei.Indeed, by Proposition A.2 and (33), it suffices to prove thatϕi ≤ C in B1. If ϕi(xi) =

supB1ϕi → ∞, set

ϕi(z) = ϕi(xi)−1ϕi(ϕi(xi)

−(pi−1)/4z + xi) ≤ 1 for |z| ≤1

2ϕi(xi)

(pi−1)/4.

By (33),ϕi(zi) = ϕi(xi)

−1ϕi(0) → 0

for zi = −ϕi(xi)(pi−1)/4xi. Sinceϕi(xi) ≤ A3|xi|−4/(pi−1), we have|zi| ≤ A4/(p1−1)3 . Hence,

we can findt > 0 independent ofi such that such thatzi ∈ Bt. Applying Proposition A.1 toϕiin B2t (sinceϕi satisfies a similar equation to (31)), we have

1 = ϕi(0) ≤ Cϕi(zi) → 0,

which is impossible. Hence,ϕi ≤ C in B1.It follows from (34) that there exists a functionϕ ∈ C4(Rn) such that, after passing subse-

quence,ϕi(x) → ϕ in C3

loc(Rn) asi→ ∞. (35)

18


Thirdly, for everyR > 0, let

gi(R, x) :=

∫

B3m

(pi−1)/4i

\BR

Gi(x, y)Ki(y)ϕi(y)pi dy.

SinceKi andϕi are nonnegative, a simple computation using (24) gives that, for anyx ∈ BR−1,

|∇kgi(R, x)| ≤ Cgi(R, x), k = 1, . . . , 5.

Note thatgi(R, x) ≤ ϕi(x) ≤ C(R). It follows that, after passing to a subsequence,

gi(R, x) → g(R, x) ≥ 0 in C4(BR−1) asi→ ∞. (36)

By (24) and (25), we have

Gi(x, y) → cn1

|x− y|n−4∀ x 6= y

andKi(y) → Ki(0) = 1. Combining (32), (35) and (36) together, by (31) we have that for anyfixedR > 0 andx ∈ BR−1

g(R, x) = ϕ(x)− cn

∫

BR

ϕ(y)n+4n−4

|x− y|n−4dy. (37)

By (37), g(R, x) is non-increasing inR. For any fixedx and|y| ≥ R >> |x|, by (24) wehave

Gi,m

−(pi−1)/4i

(x, y)

Gi,m

−(pi−1)/4i

(0, y)=Gi,|y|m

−(pi−1)/4i

( x|y|, y|y|)

Gi,|y|m

−(pi−1)/4i

(0, y|y|)= 1 +O(

|x|

|y|).

Hence,gi(R, x) = (1 +O( |x|R))gi(R, 0), which implies

limR→∞

g(R, x) = limR→∞

g(R, 0) := c0 ≥ 0. (38)

SendingR to ∞ in (37), it follows from Lebesgue’s monotone convergence theorem that

ϕ(x) = cn

∫

Rn

ϕ(y)n+4n−4

|x− y|n−4dy + c0 x ∈ R

n.

We claim thatc0 = 0. If not,

ϕ(x)− c0 = cn

∫

Rn

ϕ(y)n+4n−4

|y|n−4dy > 0,

19

Y. Y. Li & J. Xiong

which implies that

1 = ϕ(0) ≥ cn

∫

Rn

cn+4n−4

0

|x− y|n−4= ∞.

This is impossible.The use of monotonicity in the above argument is taken from [31].It follows from the classification theorem in [12] or [36] that

ϕ(x) =(

1 + |x|2)−n−4

2 ,

where we have used thatϕ(0) = 1 and∇ϕ(0) = 0.The proposition follows immediately.

Since passing to subsequences does not affect our proofs, inthe rest of the paper we willalways chooseRi → ∞ with Rτi

i = 1 + o(1) first, and thenεi → 0+ as small as we wish(depending onRi) and then choose our subsequenceuji to work with. Sincei ≤ ji andlimi→∞ τi = 0, one can ensure thatR

τjii = 1 + o(1) as i → ∞. In the sequel, we will still

denote the subsequences asui, τi and etc.

Remark 3.1. By checking the proof of Proposition 3.1, together with the fact∇2(1+ |x|2)−n−42

is negatively definite near zero and theC2 convergence in a fixed neighborhood of zero, thefollowing statement holds. Let0 ≤ ui ∈ C4(B3) be a solution of(27)and satisfy(29). Supposethat ui(0) → ∞ as i → ∞, ∇ui(0) = 0 andmaxB3 ui ≤ bui(0) for some constantb ≥ 1independent ofi. Then, after passing to a subsequence,0 must be a local maximum point ofuifor i large. Namely,0 is an isolated blow up point ofui after passing to a subsequence.

Proposition 3.2. Under the hypotheses of Proposition 3.1, there exists a constant C > 0,depending only onn,A1, A2 andA3, such that,

ui(x) ≥ C−1mi(1 +m(pi−1)/2i |x|2)(4−n)/2, |x| ≤ 1.

In particular, for anye ∈ Rn, |e| = 1, we have

ui(e) ≥ C−1m−1+((n−4)/4)τii .

20


Proof. By change of variables and using Proposition 3.1, we have forri ≤ |x| ≤ 1,

ui(x) ≥ C−1

∫

|y|≤ri

ui(y)pi

|x− y|n−4dy

≥ C−1mi

∫

|z|≤Ri

(

m−1i ui(m

−(pi−1)/4i z)

)pi

|m(pi−1)/4i x− z|n−4

dz

≥ C−1mi

∫

|z|≤Ri

U1(z)pi

|m(pi−1)/4i x− z|n−4

dz

≥1

2C−1mi

∫

Rn

U1(z)n+4n−4

|m(pi−1)/4i x− z|n−4

dz

=1

2C−1miU1(m

(pi−1)/4i x).

(39)

Recall that

Uλ(z) =

(

λ

1 + λ2|z|2

)(n−4)/2

, λ > 0.

The proposition follows immediately.

Lemma 3.2.Suppose the hypotheses of Proposition 3.1 and in addition that0 is also an isolatedsimple blow up point with the constantρ > 0. Then there existδi > 0, δi = O(R−4

i ), such that

ui(x) ≤ Cui(0)−λi |x|4−n+δi, for all ri ≤ |x| ≤ 1,

whereλi = (n− 4− δi)(pi − 1)/4− 1 andC > 0 depends only onn,A1, A3 andρ.

Proof. We divide the proof into several steps.Step 1. From Proposition 3.1, we see that

ui(x) ≤ Cmi

(

1

1 + |m(pi−1)/4i x|2

)n−42

≤ CmiR4−ni for all |x| = ri = Rim

−(pi−1)/4i . (40)

Let ui(r) be the average ofui over the sphere of radiusr centered at0. It follows from theassumption of isolated simple blow up points and Proposition 3.1 that

r4/(pi−1)ui(r) is strictly decreasing forri < r < ρ. (41)

21

Y. Y. Li & J. Xiong

By Lemma 3.1, (41) and (40), we have, for allri < |x| < ρ,

|x|4/(pi−1)ui(x) ≤ C|x|4/(pi−1)ui(|x|)

≤ Cr4/(pi−1)i ui(ri)

≤ CR4−n2

i ,

where we usedRτii = 1 + o(1). Thus,

ui(x)pi−1 ≤ CR−4

i |x|−4 for all ri ≤ |x| ≤ ρ. (42)

Step 2. Let

Liφ(y) :=

∫

B3

Gi(x, z)Ki(z)ui(z)pi−1φ(z) dz.

Thusui = Liui + hi.

Note that for4 < µ < n and0 < |x| < 2,∫

B3

Gi(x, y)|y|−µ dy ≤ A1

∫

Rn

1

|x− y|n−4|y|µdy

= A1|x|4−n

∫

Rn

1

||x|−1x− |x|−1y|n−4|y|µdy

= A1|x|−µ+4

∫

Rn

1

||x|−1x− z|n−4|z|µdz

≤ C( 1

n− µ+

1

µ− 4

)

|x|−µ+4,

where we did the change of variablesy = |x|z. By (42), one can properly choose0 < δi =O(R−4

i ) such that∫

ri<|y|<ρ

Gi(x, y)Ki(y)ui(y)pi−1|y|−δi dy ≤

1

4|x|−δi , (43)

and∫

ri<|y|<ρ

Gi(x, y)Ki(y)ui(y)pi−1|y|4−n+δi dy ≤

1

4|x|4−n+δi , (44)

for all ri < |x| < ρ.SetMi := 4nA2

1max∂Bρ ui + 2maxBρhi,

fi(x) :=Miρδi |x|−δi + Am−λi

i |x|4−n+δi ,

22


and

φi(x) =

fi(x), ri < |x| < ρ,

ui(x), otherwise,

whereA > 1 will be chosen later.By (43) and (44), we have forri < |x| < ρ.

Liφi(x) =

∫

B3

Gi(x, y)Ki(y)ui(y)pi−1φi(y) dy

=

(∫

|y|≤ri

+

∫

ri<|y|<ρ

+

∫

ρ≤|y|<3

)


≤ A1

∫

|y|≤ri

ui(y)pi

|x− y|n−4dy +

fi4+

Mi

2n−4,

where we used, in view of (24),∫

ρ≤|y|<3


=

∫

ρ≤|y|<3

Gi(x, y)Ki(y)ui(y)pi dy

≤ A212n+4

∫

ρ≤|y|<3

Gi(ρx

|x|, y)Ki(y)ui(y)

pi dy

≤ A212n+4max

∂Bρ

ui ≤ 24−nMi.

By change of variables and using Proposition 3.1, we have, similar to (39),

∫

|y|≤ri

ui(y)pi

|x− y|n−4dy = mi

∫

|z|≤Ri

(

m−1i ui(m

−(pi−1)/4i z)

)pi

|m(pi−1)/4i x− z|n−4

dz

≤ 2mi

∫

|z|≤Ri

U1(z)pi

|m(pi−1)/4i x− z|n−4

dz

≤ Cmi

∫

Rn

U1(z)n+4n−4

|m(pi−1)/4i x− z|n−4

dz

= CmiU1(m(pi−1)/4i x),

where we usedR(n−4)τi = 1 + o(1). Since|x| > ri, we see

miU1(m(pi−1)/4i x) ≤ Cm

1−(pi−1)(n−4)/4i |x|4−n

≤ Cm−λii |x|4−n+δi.

23

Y. Y. Li & J. Xiong

Therefore, we conclude that

Liφi(x) + hi(x) ≤ φi(x) for all ri ≤ |x| ≤ ρ, (45)

providedA is large independent ofi.

Step 3. Note that

lim inf|x|→r+i

fi(x) > Am−λii R4−n+δi

i m(pi−1)(n−4−δi)/4i = AR4−n+δi

i mi.

In view of (40), we may chooseA large such that

lim inf|x|→r+i

(fi(x)− ui(x)) > 0. (46)

We claim thatui(x) ≤ φi(x). (47)

Indeed, if not, let

1 < ti := inft > 1, tφi(x) ≥ ui(x) for all ri ≤ |x| ≤ ρ <∞.

By (46),ti > 1, together withfi > ui on∂Bρ, we can find a sufficient small open neighborhoodof ∂Bri ∪ ∂Bρ in which tiφi > ui. By the continuity there existsyi ∈ Bρ \ Bri such that

0 = tiφi(yi)− ui(yi) ≥ Li(tiφi − ui)(yi) + (ti − 1)hi(yi) > 0.

We derived a contradiction and thus (47) is valid.

Step 4. By (26), we havemaxBρhi ≤ A2max∂Bρ

hi ≤ A2max∂Bρui. Hence,

Mi ≤ Cmax∂Bρ

ui.

For ri < θ < ρ,

ρ4/(pi−1)Mi ≤ Cρ4/(pi−1)ui(ρ)

≤ Cθ4/(pi−1)ui(θ)

≤ Cθ4/(pi−1)Miρδiθ−δi + Am−λi

i θ4−n+δi.

Chooseθ = θ(n, ρ, A1, A2, A3) sufficiently small so that

Cθ4/(pi−1)ρδiθ−δi ≤1

2ρ4/(pi−1).

24


Hence, we haveMi ≤ Cm−λi

i .

It follows from (47) that

ui(x) ≤ φi(x) ≤ Cm−λii |x|−δi + Am−λi

i |x|4−n+δi ≤ Cm−λii |x|4−n+δi .

We complete the proof of the lemma.

Lemma 3.3. Under the assumptions in Lemma 3.2, fork < n we have

Ik[ui](x) ≤ C

mn−2k+4

n−4+o(1)

i , if |x| < ri,

m−1+o(1)i |x|k−n, if ri ≤ |x| < 1,

where

Ik[ui](x) =

∫

B1

|x− y|k−nui(y)pi dy.

Proof. Making use of Proposition 3.1 and Lemma 3.2, we have

Ik[ui](x) =

∫

Bri

ui(y)pi

|x− y|n−kdy +

∫

B1\Bri

ui(y)pi

|x− y|n−kdy

≤ Cmn−2k+4

n−4+o(1)

i

∫

BRi

U1(z)pi

|m(pi−1)/4i x− z|n−k

dz

+ Cm−n+4

n−4+o(1)

i

∫

B1\Bri

1

|x− y|n−k|y|n+4dy.

If |x| < ri, we see that

∫

BRi

U1(z)pi


dz ≤ C

by Lemma B.1, and

∫

B1\Bri

1

|x− y|n−k|y|n+4dy ≤

∫

B1\Bri

1

|y|2n−k+4dy ≤ C(n)R

−(n−k+4)i m

2(n−k+4)n−4

+o(1)

i .

Hence,Ik[ui](x) ≤ Cmn−2k+4

n−4+o(1)

i .

25

Y. Y. Li & J. Xiong

If ri < |x| < 1, then|m(pi−1)/4i x| ≥ 1. It follows from Lemma B.1 that

∫

BRi

U1(z)pi


dz ≤

∫

BRi

1

|m(pi−1)/4i x− z|n−k(1 + |z|)n+4+o(1)

dz

≤ C|m(pi−1)/4i x|k−n.

By change of variablesz = m(pi−1)/4i y,

∫

B1\Bri

1

|x− y|n−k|y|n+4dy = m

2(n−k+4)n−4

+o(1)

i

∫

Bm

(pi−1)/4i

\BRi

1

|m(pi−1)/4i x− z|n−k|z|n+4

dz

≤ Cm2(n−k+4)

n−4+o(1)

i |m(pi−1)/4i x|k−n.

Thus

Ik[ui](x) ≤ Cmn−2k+4

n−4+o(1)

i |m(pi−1)/4i x|k−n = m

−1+o(1)i |x|k−n.

Therefore, the proof of the lemma is completed.

Lemma 3.4. Under the assumptions in Lemma 3.2, we have

τi = O(ui(0)−2/(n−4)+o(1)).

Consequently,mτii = 1 + o(1).

Proof. Forx ∈ B1, we write equation (27) as

ui(x) = cn

∫

B1

Ki(y)ui(y)pi

|x− y|n−4dy + bi(x), (48)

wherebi(x) := Q′i(x) +Q′′

i (x) + hi(x),

Q′i(x) :=

∫

B1

(Gi(x, y)− cn|x− y|4−n)Ki(y)ui(y)pi dy (49)

and

Q′′i (x) :=

∫

B3\B1

Gi(x, y)Ki(y)ui(y)pi dy.

Notice that

|Gi(x, y)− cn|x− y|4−n| ≤C|x|2

|x− y|n−4+ |ai|+ C|x− y|6−n.

|∇x(Gi(x, y)− cn|x− y|4−n)| ≤C|x|2

|x− y|n−3+

C|x|

|x− y|n−4+ C|x− y|5−n.

26


Hence,

|Q′i(x)| ≤ C(|x|2ui(x) + |ai|‖u

pii ‖L1(B1) + I6[u

pii ](x)),

|∇Q′i(x)| ≤ C(|x|2I3 + |x|I4 + I5)[u

pii ](x),

whereIk[upii ](x) =

∫

B1|x− y|k−nui(y)pi dy.

By Lemma 3.2, we haveui(x) ≤ Cm−λii for all x ∈ B3/2 \B1/2. Hence,Q′′

i (x) + hi(x) ≤

ui(x) ≤ Cm−1+o(1)i for anyx ∈ ∂B1 . It follows from (26) that

maxB2

hi(x) ≤ Cmin∂B1

hi(x) ≤ Cm−1+o(1)i .

and|∇hi(x)| ≤ Cmax

B2

hi(x) ≤ Cm−1+o(1)i for all x ∈ B1.

Sinceui is nonnegative, by (24) it is easy to check that

|Q′′i (x)|+ |∇Q′′

i (x)| ≤ Cm−1+o(1)i for all x ∈ B1.

Applying Proposition 2.3 to (48), we have

τi

∫

B1

ui(x)pi+1 −A2

∫

B1

|x|ui(x)pi+1 dx

≤ C(

∫

B1

(|Q′i(x)|+ |x||∇Q′

i(x))ui(x)pi +m

−1+o(1)i

∫

B1

upii +

∫

∂B1

upi+1i ds

)

. (50)

By Proposition 3.1 and change of variables,

∫

B1

ui(x)pi+1 dx ≥ C−1

∫

Bri

mpi+1i

(1 + |m(pi−1)/4i y|2)(n−4)(pi+1)/2

dy

≥ C−1mτi(1−n/4)i

∫

Ri

1

(1 + |z|2)(n−4)(pi+1)/2dz

≥ C−1mτi(1−n/4)i ,

By Proposition 3.1, Lemma 3.2, we have∫

B1

upii ≤ Cm−1+o(1)i ,

∫

B1

|x|supi+1i ≤ Cm

−2s/(n−4)+o(1)i , for − n < s < n,

27

Y. Y. Li & J. Xiong

and∫

∂B1

upi+1i ds ≤ Cm

−2n/(n−4)+o(1)i .

It follows from Lemma 3.3 that∫

B1

(|Q′i(x)|+ |x||∇Q′

i(x)|)ui(x)pi+1 dx ≤ Cm

−2/(n−4)+o(1)i .

Therefore, we complete the proof.

Lemma 3.5. For −4 < s < 4, we have, asi→ ∞,

m1+ 2s

n−4

i

∫

Bri

|y|sui(y)pi dy →

∫

Rn

|z|s(1 + |z|2)−n+42 dz

and

m1+ 2s

n−4

i

∫

B1\Bri

|y|sui(y)pi dy → 0.

Proof. By a change of variablesy = m−(pi−1)/4i z, we have

∫

Bri

|y|sui(y)pi dy = m

−(pi−1)(s+n)

4+pi

i

∫

BRi

|z|s(m−1i ui(m

−(pi−1)/4i z))pi dz

By Lemma 3.4,m−

(pi−1)(s+n)

4+pi

i = (1 + o(1))m−1− 2s

n−4

i . In view of Proposition 3.1 and−4 <s < 4, it follows from Lebesgue’s dominated convergence theoremthat

∫

BRi

|z|s(m−1i ui(m

−(pi−1)/4i z))pi dz →

∫

Rn

|z|s(1 + |z|2)−n+42 dz.

Hence, the first convergence result in the lemma follows.By Lemma 3.2,

∫

ri≤|y|<1

|y|sui(y)pi dy ≤ Cm−λipi

i

∫

ri≤|y|<1

|y|s|y|(4−n+δi)pi dy

≤ Cm−λipii m

(pi−1)((n−4−δi)pi−s−n)

4i R

n+s−(n−4−δi)pii

= Cm−

(pi−1)(s+n)

4+pi

i Rn+s−(n−4−δi)pii ,

where0 < δi = O(R−4i ) andλi = (n − 4 − δi)(pi − 1)/4 − 1. Sincem

−(pi−1)(s+n)

4+pi

i =

(1 + o(1))m−1− 2s

n−4

i andn + s − (n − 4 − δi)pi → s − 4 < 0 asi → ∞, we have the secondconvergence result in the lemma.

In conclusion, the lemma is proved.

28


Proposition 3.3. Under the assumptions in Lemma 3.2, we have

ui(x) ≤ Cu−1i (0)|x|4−n, for all |x| ≤ 1.

Proof. For |x| ≤ ri, the proposition follows immediately from Proposition 3.1and Lemma 3.4.We shall show first that

sup|e|=1

ui(ρe)ui(0) ≤ C. (51)

If not, then along a subsequence we have, for some unit vectorsei,

limi→∞

ui(ρei)ui(0) = +∞.

Sinceui(x) ≤ A3|x|−4/(pi−1) in B2, it follows from Proposition A.1 that for any0 < ε < 1there exists a positive constantC(ε), depending only onn,A1, A2, A3 andε, such that

supB3/2\Bε

ui ≤ C(ε) infB3/2\Bε

ui. (52)

Letϕi(x) = ui(ρei)−1ui(x). Then for|x| ≤ 1,

ϕi(x) =

∫

B3

Gi(x, y)Ki(y)ui(ρei)pi−1ϕi(y)

pi dy + hi(x),

wherehi(x) = ui(ρei)−1hi(x). Sinceϕi(ρei) = 1, by (52)

‖ϕi‖L∞(B3/2\Bε) ≤ C(ε) for 0 < ε < 1. (53)

By (26), we have that for anyx ∈ B1

hi(x) ≤ A2hi(ρei) ≤ A2.

Besides, by Lemma 3.2,ui(ρei)

pi−1 → 0 (54)

as i → ∞. Because of (24)-(26), by applying Proposition A.2 toϕi we conclude that thereexistsϕ ∈ C3(B1 \ 0) such thatϕi → ϕ in C3

loc(B1 \ 0) after passing to a subsequence.Let us write the equation ofϕi as

ϕi(x) =

∫

B1


pi dy + bi(x), (55)

wherebi(x) :=∫

B3\B1Gi(x, y)Ki(y)ui(ρei)

pi−1ϕi(y)pi dy + hi(x). By (53), there existsb ∈

C3(B1) such thatbi(x) → b(x) ≥ 0 in C3

loc(B1) (56)

29

Y. Y. Li & J. Xiong

after passing to a subsequence. Therefore,∫

B1


pi dy = ϕi(x)− bi(x) → ϕ(x)− b(x)

in C3loc(B1 \ 0). DenoteΓ(x) := ϕ(x)− b(x). For any|x| > 0 and0 < ε < 1

2|x|, in view of

(53) and (54) we have

Γ(x) = limi→∞

∫

Bε


pi dy

= limi→∞

(

Gi(x, 0)

∫

Bε

Ki(y)ui(ρei)pi−1ϕi(y)

pi dy +O(mi)

∫

Bε

|y|ui(y)pi dy

)

= limi→∞

(Gi(x, 0)

∫

Bε


pi dy +O(m− 2

n−4

i ))

=: G∞(x, 0)a(ε), (57)

where we used Lemma 3.5 in the third identity,a(ε) is a bounded nonnegative function ofε,

G∞(x, 0) = cn|x|4−n + a +O′(|x|6−n)

by (24), a ≥ 0 anda = 0 if n ≥ 6. Clearly,a(ε) is nondecreasing inε, so limε→0 a(ε) existswhich we denote asa. Sendingε→ 0, we obtain

Γ(x) = aG∞(x, 0).

Since0 is an isolated simple blow point ofui∞i=1, we havern−42 ϕ(r) ≥ ρ

n−42 ϕ(ρ) for 0 < r <

ρ. It follows thatϕ is singular at0, and thus,a > 0. Hence,

limi→∞

∫

B1/8


pi dy ≥ a(ε) ≥ a > 0.

However,∫

B1/8


pi dy

≤ Cui(ρei)−1

∫

B1/8

ui(y)pi dy

≤C

ui(ρei)ui(0)→ 0 asi→ ∞,

where we used Lemma 3.5 in the last inequality. This is a contradiction.

30


Without loss of generality, we may assume thatρ ≤ 1/2. It follows from Proposition A.1and (51) that Proposition 3.3 holds forρ ≤ |x| ≤ 1. To establish the inequality in the Proposi-tion for ri ≤ |x| ≤ ρ, we only need to rescale and reduce it to the case of|x| = 1. Suppose thecontrary that there exists a subsequencexi satisfying|xi| ≤ ρ andlimi→∞ ui(xi)ui(0)|xi|n−4 =+∞.

Setri := |xi|, ui(x) = r4/(pi−1)i ui(rix). Thenui satisfies

ui(x) =

∫

B3

Gi,ri(x, y)Ki(riy)ui(y)pi dy + hi(x) for x ∈ B2,

wherehi(x) =∫

B3/ri\B3

Gi,ri(x, y)Ki(riy)ui(y)pi dy + r

4/(pi−1)i hi(rix). One can easily check

that ui and the above equation satisfy all hypotheses of Proposition 3.3 forui and its equation.It follows from (51) that

ui(0)ui(xiri) ≤ C.

It follows (using Lemma 3.4) that

limi→∞

ui(xi)ui(0)|xi|n−4 <∞.

This is again a contradiction.Therefore, the proposition is proved.

Proposition 3.4. Under the assumptions in Lemma 3.2, we have

|∇kui(x)| ≤ Cu−1i (0)|x|4−n−k, for all ri ≤ |x| ≤ 1,

wherek = 1, . . . , 4.

Proof. Since0 is an isolated blow up point inB2, by Proposition A.1 we see that Proposition3.3 holds for all|x| ≤ 3

2. For anyri ≤ |x| < 1, let

ϕi(z) =( |x|

4

)4

pi−1ui(x+

|x|

4z).

By the equation ofui, we have

ϕi(z) =

∫

y:|x+|x|4y|≤3

Gi(z, y)Ki(y)ϕi(y)pi−1ϕi(y) dy + hi(z),

whereGi(z, y) = ( |x|4)n−4Gi(x+

|x|4z, x+ |x|

4y), Ki(y) = Ki(x+

|x|4y), andhi(z) = ( |x|

4)

4pi−1hi(x+

|x|4z). Since0 is an isolated blow up point ofui, we haveϕi(z)pi−1 ≤ Api−1

2 for all |z| ≤ 1.Sinceϕi, Gi, Ki andhi are nonnegative, by Proposition A.2 we have

|∇kϕi(0)| ≤ C(‖ϕi‖L∞(B1) + ‖hi‖C4(B1)).

31

Y. Y. Li & J. Xiong

This gives

(|x|

4)k|∇kui(x)| ≤ C‖ui‖L∞(B |x|

4

(x)) + Cm−1i

≤ Cui(0)−1|x|4−n.

Therefore, the proposition follows.

Corollary 3.1. Under the hypotheses of Lemma 3.2, we have∫

B1

|x|sui(x)pi+1 ≤ Cui(0)

−2s/(n−4), for − n < s < n,

Proof. Making use of Proposition 3.1, Lemma 3.4 and Proposition 3.3, the corollary followsimmediately.

By Proposition 3.3 and its proof, we have the following corollary.

Corollary 3.2. Under the assumptions in Lemma 3.2, if we letTi(x) = T ′i (x) + T ′′

i (x), where

T ′i (x) := ui(0)

∫

B1

Gi(x, y)Ki(y)ui(y)pi dy

and

T ′′i (x) := ui(0)

∫

B3\B1

Gi(x, y)Ki(y)ui(y)pi dy + ui(0)hi(x).

then, after passing a subsequence,

T ′i (x) → aG∞(x, 0) in C3

loc(B1 \ 0)

andT ′′i (x) → h(x) in C3

loc(B1)

for someh(x) ∈ C3(B2), whereG∞ is the limit of a subsequence ofGi in C3loc(B1 \ 0),

a =

∫

Rn

(

1

1 + |y|2

)n+42

dy. (58)

Consequently, we have

ui(0)ui(x) → aG∞(x, 0) + h(x) in C3loc(B1 \ 0).

32


Proof. Similar to that in the proof of Proposition 3.3, we setϕi(x) = ui(0)ui(x), which satisfies

ϕi(x) =

∫

B3

Gi(x, y)Ki(y)ui(0)1−piϕi(y)

pi dy + ui(0)hi(x)

=:

∫

B1

Gi(x, y)Ki(y)ui(0)1−piϕi(y)

pi dy + T ′′i (x) = T ′

i (x) + T ′′i (x).

We have all the ingredients as in the proof of Proposition 3.3. Hence, we only need to evaluatethe positive constanta. By (57) and Lemma 3.5, we have

a = limε→0

limi→∞

ui(0)

∫

Bε

Ki(y)ui(y)pi dy =

∫

Rn

(

1

1 + |y|2

)n+42

dy.

4 Expansions of blow up solutions of integral equations

In this section, we are interested in stronger estimates than that in Proposition 3.3. To makestatements closer to the main goal of the paper, we restrict our attention to a specialKi. Namely,givenpi, Gi, andhi satisfying (23), (24) and (26) respectively,κi satisfying (25) withKi re-placed byκi, let 0 ≤ ui ∈ C4(B3) be a solution of

ui(x) =

∫

B3

Gi(x, y)κi(y)τiupii (y) dy + hi(x) in B3. (59)

We also assume∇κi(0) = 0. (60)

Suppose that0 is an isolated simple blow up point ofui with ρ = 1, i.e.,

ui(x) ≤ A3|x|−4/(pi−1) for all x ∈ B2. (61)

andr4/(pi−1)ui(r) has precisely one critical point in(0, 1).Let us first introduce a non-degeneracy result.

Lemma 4.1. Letv ∈ L∞(Rn) be a solution of

v(x) = cnn + 4

n− 4

∫

Rn

U1(y)8

n−4 v(y)

|x− y|n−4dy.

Then

v(z) = a0

(

n− 4

2U1(z) + z · ∇U1(z)

)

+

n∑

j=1

aj∂jU1(z),

wherea0, . . . , an are constants.

33

Y. Y. Li & J. Xiong

Proof. Sincev ∈ L∞(Rn), by using Lemma B.1 iteratively a finite number of times we obtain

|v(x)| ≤ C(1 + |x|)4−n. (62)

LetF : Rn → Sn \ N,

F (x) =

(

2x

1 + |x|2,1− |x|2

1 + |x|2

)

denote the inverse of the inverse of the stereographic projection andh(F (x)) := v(x)JF (x)−n−4

2n ,whereJF = ( 2

1+|x|2)n is the Jacobian determinant ofF andN is the north pole. It follows from

(62) thath ∈ L∞(Sn). Let ξ = F (x) andη = F (y). Then

|ξ − η| =2|x− y|

√

(1 + |x|2)(1 + |y|2)and dη =

(

2

1 + |y|2

)n

dy

are respectively the distance betweenξ andη in Rn+1 and the surface measure ofSn. It followsthat

h(ξ) = 2−4n(n− 2)(n+ 2)(n+ 4)αn

∫

Sn

|ξ − η|4−nh(η) dη. (63)

By the regularity theory for Riesz potentials,h ∈ C∞(Sn). Note that the Paneitz operator

PgSn = ∆2gSn

−n2 − 2n− 4

2∆gSn +

n(n− 2)(n+ 2)(n− 4)

16

with respect to the standard metricgSn onSn satisfiesPgSnφ = |JF |−n+4

2n ∆2(|JF |n−42n φ F ) for

anyφ ∈ C∞(Sn). By the integral equation ofv we have

PgSnh = 2−4n(n− 2)(n+ 2)(n+ 4)h.

Let Y (k) be a spherical harmonics of degreek ≥ 0. We have

PgSnY(k) = 2−4(2k + n + 2)(2k + n)(2k + n− 2)(2k + n− 4)Y (k).

Hence,h must be a spherical harmonics of degree one. Transformingh back, we complete theproof.

In view of Corollary 2.2, we assume in this and next section that

Uλ(x) =

∫

B3

Gi(x, y)Uλ(y)n+4n−4 + c′λ,i(y)Uλ(y) dy + c′′λ,i(x) ∀ λ ≥ 1, x ∈ B3 (64)

34


wherec′λ,i, c′′λ,i ∈ C5(B3) satisfy

Θi :=5∑

k=0

‖λ−k∇kc′λ,i‖L∞(B2) ≤ A2, (65)

and‖c′′λ,i‖C5(B2) ≤ A2λ4−n2 , respectively.

Lemma 4.2. Let 0 ≤ ui ∈ C4(B3) be a solution of(59) and0 is an isolated simple blow uppoint ofui with some constantρ, sayρ = 1. Suppose(64)holds and letΘi be defined in(65).Then we have

|ϕi(z)− U1(z)| ≤ C

maxτi, m−2i , if 5 ≤ n ≤ 7,

maxτi,Θim−2i logmi, m

−2i , if n = 8,

maxτi,Θim− 8

n−4

i , m−2i , if n ≥ 9,

∀ |z| ≤ mpi−1

4i ,

whereϕi(z) = 1miui(m

−pi−1

4i z),mi = ui(0), andC > 0 depends only onn,A1, A2 andA3.

Proof. For brevity, setℓi = mpi−1

4i . By the equation ofui, we have

ϕi(z) =

∫

Bℓi

Gi,ℓ−1i(z, y)κi(y)

τiϕi(y)pi dy + hi(z), (66)

whereGi,ℓ−1i(z, y) = ℓ4−ni Gi(ℓ

−1i x, ℓ−1

i y), κi(z) = κi(ℓ−1i z), andhi(z) = m−1

i hi(ℓ−1i z) with

hi(x) =

∫

B3\B1

Gi(x, y)κi(y)τiui(y)

pi dy + hi(x).

Since0 is an isolated simple blow up point ofui, by Proposition 3.3 we have

ui(x) ≤ Cm−1i |x|4−n for |x| < 1. (67)

It follows thathi(x) ≤ Cm−1i for x ∈ B1 andhi(z) ≤ Cm−2

i for z ∈ Bℓi .Notice thatUℓi(x) ≤ Cm−1

i for 1 ≤ |x| ≤ 3. Let z = ℓix. By (64) withλ = ℓi we have for|z| ≤ ℓi

U1(z) =

∫

Bℓi

Gi,ℓ−1i(z, y)(U1(y)

n+4n−4 +m

− 8n−4

i c′ℓi,i(ℓ−1i y)U1(y)) dy +O(m−2

i )

=

∫

Bℓi

Gi,ℓ−1i(z, y)(κi(y)

τiU1(y)pi + Ti(y)) dy +O(m−2

i ), (68)

35

Y. Y. Li & J. Xiong

where we usedmτii = 1 + o(1), and

Ti(y) := U1(y)n+4n−4 − κi(y)

τiU1(y)pi +m

− 8n−4

i c′ℓi,i(ℓ−1i y)U1(y).

Here and throughout this section,O(m−2i ) denotes some functionfi satisfying‖∇kfi‖B(1−ε)ℓi

≤

C(ε)m−2− 2k

n−4

i for smallε > 0 andk = 0, . . . , 5.In the following, we adapt some arguments from Marques [43] for the Yamabe equation;

see also the proof of Proposition 2.2 of Li-Zhang [38]. Let

Λi = max|z|≤ℓi

|ϕi − U1|.

By (67), for any0 < ε < 1 andεℓi ≤ |z| ≤ ℓi, we have|ϕi(z) − U1(z)| ≤ C(ε)m−2i , where

we usedmτii = 1 + o(1). Hence, we may assume thatΛi is achieved at some point|zi| ≤ 1

2ℓi,

otherwise the proof is finished. Set

vi(z) =1

Λi(ϕi(z)− U1(z)).

It follows from (66) and (68) thatvi satisfies

vi(z) =

∫

Bℓi

Gi,ℓ−1i(z, y)(bi(y)vi(y) +

1

ΛiTi(y)) dy +

1

ΛiO(m−2

i ), (69)

where

bi = κτiiϕpii − Upi

1

ϕi − U1.

SinceGi,ℓ−1

i(z, y) ≤ A1|z − y|4−n

and

|Ti(y)| ≤ Cτi(| logUi|+ | log κi|)(1 + |y|)−pi(n−4) +Θim− 8

n−4

i (1 + |y|)4−n, (70)

we obtain∫

Bℓi

Gi,ℓ−1i(z, y)|Ti(y)| dy ≤ C(τi +Θiαi) for |z| ≤

ℓi2,

where

αi =

m−2i , if 5 ≤ n ≤ 7,

m−2i logmi, if n = 8,

m− 8

n−4

i , if n ≥ 9.

(71)

36


Sinceκi(x) is bounded andϕi ≤ CU1, we see

|bi(y)| ≤ CU1(y)pi−1 ≤ C(1 + |y|)−7.5, y ∈ Bℓi. (72)

Noticing that‖vi‖L∞(Bℓi) ≤ 1, by Lemma B.1 we have

∫

Bℓi

Gi,ℓ−1i(z, y)|bi(y)vi(y)| dy ≤ C(1 + |z|)−minn−4,3.5.

Hence, we get

vi(z) ≤ C((1 + |z|)−minn−4,3.5 +1

Λi(τi +Θiαi +m−2

i )) for |z| ≤ℓi2. (73)

Suppose the contrary that1Λi

maxτi,Θiαi, m−2i → 0 asi → ∞. Sincev(zi) = 1, by (73)

we see that|zi| ≤ C.

Differentiating the integral equation (69) up to three times, together with (70) and (72), we seethat theC3 norm of vi on any compact set is uniformly bounded. By Arzela-Ascoli theoremlet v := limi→∞ vi after passing to a subsequence. Using Lebesgue’s dominatedconvergencetheorem, we obtain

v(z) = cn

∫

Rn

U1(y)8

n−4 v(y)

|z − y|n−4dy.

It follows from Lemma 4.1 that

v(z) = a0(n− 4

2U1(z) + z · ∇U1(z)) +

n∑

j=1

aj∂jU1(z),

wherea0, . . . , an are constants. Sincev(0) = 0 and∇v(0) = 0, v has to be zero. However,v(zi) = 1. We obtain a contradiction.

Therefore,Λi ≤ C(τi + αi) and the proof is completed.

Lemma 4.3. Under the same assumptions in Lemma 4.2, we have

τi ≤ C

m−2i , if 5 ≤ n ≤ 7,

maxΘim−2i logmi, m

−2i , if n = 8,

maxΘim− 8

n−4

i , m−2i , if n ≥ 9.

37

Y. Y. Li & J. Xiong

Proof. The proof is also by contradiction. Recall the definition ofαi in (71). Suppose thecontrary that1

τimaxΘiαi, m

−2i → 0 asi→ ∞. Set

vi(z) =ϕi(z)− U1(z)

τi.

It follows from Lemma 4.2 that|vi(z)| ≤ C in Bℓi, whereℓi = mpi−1

4i . As (69), we have

vi(z) =

∫

Bℓi


1

τiTi(y)) dy +

1

τiO(m−2

i ), (74)

where


1

ϕi − U1,

and

Ti(y) := U1(y)n+4n−4 − κi(y)

τiU1(y)pi +m

− 8n−4


By the estimates (72) and (70) forbi andTi respectively, we conclude from the integralequation that‖vi‖C3 is uniformly bounded over any compact set. It follows thatvi → v inC2loc(R

n) for somev ∈ C3(Rn).Multiplying both sides of (74) bybi(z)φ(z), whereφ(z) = n−4

2U1(z) + z · ∇U1(z), and

integrating overBℓi, we have, using the symmetry ofGi,ℓ−1i

in y andz,

∫

Bℓi

bi(z)vi(z)

(

φ(z)−

∫

Bℓi

Gi,ℓ−1i(z, y)bi(y)φ(y) dy

)

dz

=1

τi

∫

Bℓi

Ti(z)

∫

Bℓi

Gi,ℓ−1i(z, y)bi(y)φ(y) dy dz +

1

τiO(m−2

i )

∫

Bℓi

bi(z)φ(z) dz.

As i→ ∞, we have

∫

Bℓi

Gi,ℓ−1i(z, y)bi(y)φ(y) dy → cn

∫

Rn

U1(y)8

n−4φ(y)

|z − y|n−4dz = φ(z),

1

τiO(m−2

i )

∫

Bℓi

bi(z)φ(z) dz → 0 by the contradiction hypothesis,

andTi(z)

τi→ (logU1(z))U1(z)

n+4n−4 .

38


Hence, by Lebesgue’s dominated convergence theorem we obtain

limi→∞

1

τi

∫

Bℓi

Ti(z)

∫

Bℓi

Gi,ℓ−1i(z, y)bi(y)φ(y) dy dz

=

∫

Rn

φ(z)(logU1(z))U1(z)n+4n−4dz = 0.

This is impossible, because

∫

Rn

φ(z)(logU1(z))U1(z)n+4n−4dz

=(n− 4)2|Sn−1|

4

∫ ∞

0

(r2 − 1)rn−1

(1 + r2)n+1log(1 + r2) dr

=(n− 4)2|Sn−1|

2

∫ ∞

1

(r2 − 1)rn−1

(1 + r2)n+1log r dr > 0,

where we used

∫ 1

0

(r2 − 1)rn−1

(1 + r2)n+1log(1 + r2) dr = −

∫ ∞

1

(s2 − 1)sn−1

(1 + s2)n+1(log(1 + s2)− log s2) ds

by the change of variabler = 1s.

We obtain a contradiction and thusτi ≤ αi. Therefore, the lemma is proved.

Proposition 4.1. Under the hypotheses in Lemma 4.2, we have

|ϕi(z)− U1(z)| ≤ C

m−2i , if 5 ≤ n ≤ 7,

maxΘim−2i logmi, m

−2i , if n = 8,

maxΘim− 8

n−4

i , m−2i , if n ≥ 9,

∀ |z| ≤ mpi−1

4i .

Proof. It follows immediately from Lemma 4.2 and Lemma 4.3.

Proposition 4.2. Under the hypotheses in Lemma 4.2, we have, for every|z| ≤ mpi−1

4i ,

|ϕi(z)− U1(z)| ≤ C

maxΘim−2i m

2n−4

i (1 + |z|)−1, m−2i , if n = 8,

maxΘim−2i m

2(n−8)n−4

i (1 + |z|)8−n, m−2i , if n ≥ 9.

39

Y. Y. Li & J. Xiong

Proof. Letαi be defined in (71). We may assume thatm−2i

Θiαi→ 0 asi→ ∞ for n ≥ 8; otherwise

the proposition follows immediately from Proposition 4.1.Set

α′i =

m−2i m

2n−4

i , if n = 8,

m−2i m

2(n−8)n−4

i , if n ≥ 9,

and

vi(z) =ϕi(z)− U1(z)

Θiα′i

, |z| ≤ mpi−1

4i .

Since m−2i

Θiαi→ 0, it follows from Proposition 4.1 that|vi| ≤ C. Since0 < ϕi ≤ CU1, we

only need to prove the proposition when|z| ≤ 12ℓi, whereℓi = m

pi−1

4i . Similar to (69),vi now

satisfies

vi(z) =

∫

Bℓi


1

Θiα′i

Ti(y)) dy +1

Θiα′i

O(m−2i ),

where


1

ϕi − U1

andTi(y) := U1(y)

n+4n−4 − κi(y)

τiU1(y)pi +m

− 8n−4


Noticing that

|Ti(y)| ≤ Cτi(| logU1|+ | log κi|)(1 + |y|)−4−n +m− 8

n−4

i Θi(1 + |y|)4−n,

we have

1

Θiα′i

∫

Bℓi

Gi,ℓ−1i(z, y)|Ti(y)| dy ≤ C

∫

Bℓi

1

|z − y|n−4(1 + |y|)4m2

n−4

i

dy

≤ C

∫

Bℓi

1

|z − y|n−4(1 + |y|)5dy

≤ C(1 + |z|)−1,

if n = 8, and1

Θiα′i

∫

Bℓi

Gi,ℓ−1i(z, y)|Ti(y)| dy ≤ C(1 + |z|)8−n

if n ≥ 9, where we used Lemma B.1. Thus

|vi(z)| ≤ C((1 + |z|)−3.5 + (1 + |z|)−1)

40


for n = 8, and|vi(z)| ≤ C((1 + |z|)−3.5 + (1 + |z|)8−n)

for n ≥ 9. If n = 8, 9, 10, 11, the conclusion follows immediately from multiplying bothsidesof the above inequalities byα′

i. If n ≥ 12, the above estimate gives|vi(z)| ≤ C(1 + |z|)−3.5.Plugging this estimate to the term

∫

Gi,ℓ−1i(z, y)bi(y)vi(y) dy yields|vi(z)| ≤ C(1 + |z|)8−n as

long asn ≤ 14. Repeating this process, we complete the proof.

Corollary 4.1. Under the hypotheses in Lemma 4.2, we have, for very|z| ≤ mpi−1

4i ,

|∇k(ϕi − U1)(z)|

≤ C(1 + |z|)−k

m−2i , if 5 ≤ n ≤ 7,

maxΘim−2i m

2n−4

i (1 + |z|)−1, m−2i , if n = 8,

maxΘim−2i m

2(n−8)n−4

i (1 + |z|)8−n, m−2i , if n ≥ 9.

wherek = 1, 2, 3, 4.

Proof. Considering the integral equation ofvi = ϕi − U1, the conclusion follows from LemmaB.1. Indeed, ifk < 4, we can differentiate the integral equation ofvi directly and then useLemma B.1. Ifk = 4, we can use a standard technique (see the proof of Proposition A.2) forproving the higher order regularity of Riesz potential sincevi and the coefficients are ofC1.

5 Blow up local solutions of fourth order equations

In the previous two sections, we have analyzed the blow up profiles of the blow up local so-lutions of integral equations. In this section, we will assume that those blow up solutions alsosatisfy differential equations, which is only used to checkthe Pohozaev identity in Proposi-tion 2.2. It should be possible to completely avoid using differential equations after improvingCorollary 2.1. This is the case on the sphere; see our joint work with Jin [31]. On the other hand,as mentioned in the Introduction, without extra information fourth order differential equationsthemselves are not enough to do blow up analysis for positivelocal solutions.

Proposition 5.1. In addition to the hypotheses in Lemma 4.2, assume thatui also satisfies

Pgiui = c(n)κτii upii in B3, (75)

wheredet gi = 1,B3 is a normal coordinates chart ofgi at 0 and‖gi‖C10(B3) ≤ A1.

41

Y. Y. Li & J. Xiong

(i) If either n ≤ 9 or gi is flat, then

lim infr→0

P(r,Γ) ≥ 0, (76)

whereΓ is a limit ofui(0)ui(x) along a subsequence.

(ii) If n ≥ 8, then|Wgi(0)|2 ≤ C∗Giβi withC∗ > 0 depending only onn,A1, A2, A3, where

Gi :=∑

k≥1, 2≤k+l≤4

Θi‖∇kgi‖

lL∞(B3)

+∑

k≥1, 6≤k+l≤8

‖∇kgi‖lL∞(B3)

, (77)

and

βi :=

(logmi)−1, if n = 8,

m− 2

n−4

i , if n = 9,

m− 4

n−4

i logmi, if n = 10,

m− 4

n−4

i , if n ≥ 11.

(78)

(iii) (76)holds ifn ≥ 10 and|Wgi(0)|

2 > C∗Giβi. (79)

Proof. It follows from Corollary 3.2 that after passing a subsequence

lim ui(0)ui(x) =: Γ(x),

whereΓ(x) is inC3(B1 \ 0). We will still denote the subsequence asui.Notice that for every0 < r < 1

m2iP(r, ui) → P(r,Γ) asi→ ∞. (80)

By Proposition 2.2,

P(r, ui) =

∫

Br

(xk∂kui +n− 4

2ui)E(ui) +N (r, ui),

whereE(ui) is as in (16) withg andu replaced bygi andui respectively, i.e.,

E(ui) : = Pgiui −∆2ui

=n− 4

2Qgiui + f

(1)i,k ∂kui + f

(2)i,kl∂klui + f

(3)i,kls∂klsui + f

(4)i,klst∂klstui,

f(1)i,k (x) = O(1), f

(2)i,kl(x) = O(1), f

(3)i,kls(x) = O(|x|), f

(4)i,klst(x) = O(|x|2), (81)

42


and

N (r, ui) =c(n)τipi + 1

∫

Br

(n− 4

2κτii + xk∂kκiκ

τi−1i )upi+1

i −r

pi + 1

∫

∂Br

c(n)κτii upi+1i .

By Proposition 3.3, for0 < r < 1 we have, for someC > 0 independent ofi andr,

m2iN (r, ui) ≥ −

m2i r

pi + 1

∫

∂Br

c(n)κτii upi+1i ≥ −Cr−nm1−pi

i ,

where we used the facts thatκi(x) = 1+O(|x|2) and|∇κi(x)| = O(|x|) withO(·) independentof i. Hence, we have

lim infi→∞

m2iN (r, ui) ≥ 0. (82)

Throughout this section, without otherwise stated, we useC to denote some constants inde-pendent ofi andr.

If gi is flat, then we complete the proof becauseE(ui) = 0.

Now we assumegi is not flat. By a change of variablesz = ℓix with ℓi = mpi−1

4i , we have

Ei(r) : = m2i

∫

Br

(xk∂kui +n− 4

2ui)E(ui) dx

= m2im

2+(4−n)pi−1

4i

∫

Bℓir

(

zk∂kϕi +n− 4

2ϕi

)

·

(n− 4

2ℓ−4i Qgi(ℓ

−1i z)ϕi +

4∑

j=1

ℓ−4+ji f

(j)i (ℓ−1

i z)∇jϕi

)

dz,

whereϕi(z) = m−1i ui(m

−pi−1

4i z), f (1)

i (ℓ−1i z)∇1ϕi = f

(1)i,k (ℓ

−1i z)∂kϕi and f (j)

i (ℓ−1i z)∇jϕi is

defined in the same fashion forj 6= 1. Define

Ei(r) : = m2im

2+(4−n)pi−1

4i

∫

Bℓir

(

zk∂kU1 +n− 4

2U1

)

·

(n− 4

2ℓ−4i Qgi(ℓ

−1i z)U1 +

4∑

j=1

ℓ−4+ji f

(j)i (ℓ−1

i z)∇jU1

)

dz.

Notice thatm2+(4−n)

pi−1

4i = 1 + o(1), andQg = O(1). By Proposition 4.1, Proposition 4.2,

43

Y. Y. Li & J. Xiong

Corollary 4.1, (81) and (18), we have

|Ei(r)− Ei(r)| (83)

≤ Cm2im

− 4n−4

i

∫

Bℓir

4∑

j=0

|∇j(ϕi − U1)|(z)(1 + |z|)2−n+j dz

≤ C∑

k≥1, 2≤k+l≤4


r2, if n = 5, 6, 7,

maxΘir, r2, if n = 8, 9,

maxΘi log(rmi), r2, if n = 10,

maxΘim2(n−10)

n−4

i , r2, if n ≥ 11.

(84)

Now we estimateEi(r).If n = 5, 6, 7, we have

Ei(r) = m2+(n−4)τii

∫

Br

(xk∂kUℓi +n− 4

2Uℓi)E(Uℓi) dx

= m2+(n−4)τii

∫

Br

(xk∂kUℓi +n− 4

2Uℓi)(Pgi −∆2)Uℓi dx

= O(1)m2i

∫

Br

|xk∂kUℓi +n− 4

2Uℓi |Uℓi,

(85)

where we have used(Pgi −∆2)Uℓi = O(1)Uℓi because of (20), and

||x|k∇kxUℓi(x)| ≤ C(n, k)Uℓi(x) for k ∈ N.

Hence,|Ei(r)| ≤ Cr8−n ≤ Cr. (86)

Therefore, (76) follows from (82), (84) and (86) whenn = 5, 6, 7.If n ≥ 8, by Lemma 2.1 we have

Ei(r) =−2

nγi

∫

Bℓir

(s∂sU1 +n− 4

2U1)(c

∗1s∂sU1 + c∗2s

2∂ssU1) dz

−32(n− 1)γi3(n− 2)n2

∫

Bℓir

(s∂sU1 +n− 4

2U1)s

2(∂ssU1 −∂sU1

s) dz

+ (n− 4)γi

∫

Bℓir

(s∂sU1 +n− 4

2U1)U1 dz +O(α′′

i )∑

k≥1, 6≤k+l≤8


,

44


where we used the symmetry so that those terms involving homogeneous polynomials of odd

degrees are gone,s = |z|, γi =m

2(n−8)n−4 +(n−4)τi

i |Wgi(0)|2

24(n−1)≥ 0,

α′′i =

∫

Br

|x|2Uℓi(x)2 dx = O(1)

r10−n, if n = 8, 9,

log rmi, if n = 10,

m2(n−10)

n−4

i , if n ≥ 11.

andc∗1, c∗2 are given in (19). By direct computations,

r∂rU1 +n− 4

2U1 =

n− 4

2

1− r2

(1 + r2)n−22

,

c∗1r∂rU1 + c∗2r2∂rrU1 = (4− n)

(c∗1 + c∗2)r2

(1 + r2)n−22

+ (4− n)(2− n)c∗2r

4

(1 + r2)n2

= (4− n)(c∗1 + c∗2)r

2 + (c∗1 + (3− n)c∗2)r4

(1 + r2)n2

,

∂rrU1 −∂rU1

r= (n− 4)(n− 2)(1 + r2)

−n2 r2.

Thus

Ei(r) =(n− 4)2

nγi|S

n−1|Ji +O(α′′i )

∑

k≥1, 6≤k+l≤8


,

where

Ji :=

∫ ℓir

0

(1− s2)[n2+ (c∗1 + c∗2 + n)s2 + (c∗1 + (3− n)c∗2 −

16(n−1)3n

+ n2)s4]sn−1

(1 + s2)n−1ds.

If n = 8, we have−(c∗1+(3−n)c∗2+n2) = (2n−12)+ 14

3−4 = 14

3. Since

∫ ℓir

0s13

(1+s2)7ds→ ∞

as ℓi → ∞, Hence,Ji → ∞ as i → ∞. For n ≥ 9, we notice that for positive integers2 < m+ 1 < 2k,

∫ ∞

0

tm

(1 + t2)kdt =

m− 1

2k −m− 1

∫ ∞

0

tm−2

(1 + t2)kdt.

If ℓir = ∞, we have

Ji =

−2n

n− 4− (c∗1 + c∗2 + n)

8n

(n− 6)(n− 4)

− (c∗1 + (3− n)c∗2 −16(n− 1)

3n+n

2)

12n(n+ 2)

(n− 8)(n− 6)(n− 4)

∫ ∞

0

sn−1

(1 + s2)n−1ds.

45

Y. Y. Li & J. Xiong

We compute the coefficients of the integral,

−2n

n− 4− (c∗1 + c∗2 + n)

8n

(n− 6)(n− 4)

− (c∗1 + (3− n)c∗2 −16(n− 1)

3n+n

2)

12n(n+ 2)

(n− 8)(n− 6)(n− 4)

=2n

n− 4

− 1 + (n(n− 2)

2− 8)

4

(n− 6)+ (

3n

2+

16(n− 1)

3n− 12)

6(n+ 2)

(n− 8)(n− 6)

=2n

n− 4

2n2 − 5n− 26

n− 6+

9(n+ 2)

(n− 6)+

32(n− 1)(n+ 2)

n(n− 8)(n− 6)

≥4n(n2 + 2n− 4)

(n− 4)(n− 6)> 0.

Therefore, for any0 < r < 1 and sufficiently largei (the largeness ofi may depend onr), wehave

Ji ≥ 1/C(n) > 0. (87)

In conclusion, we obtain

Ei(r) ≥

1C|Wgi(0)|

2 logmi +O(α′′i )∑

k≥1, 6≤k+l≤8 ‖∇kgi‖lL∞(B3)

, if n = 8,

1C|Wgi(0)|

2m2(n−8)n−4

i +O(α′′i )∑

k≥1, 6≤k+l≤8 ‖∇kgi‖lL∞(B3)

, if n ≥ 9.(88)

Combing (82), (84) and (88) together, we see that, forn ≥ 8,

m2iP(r, ui) = m2

iN (r, ui) + (Ei(r)− Ei(r)) + Ei(r)

≥ m2iN (r, ui) +

1

2Ei(r)− Cr (89)

whereCr can be set to zero whenn ≥ 9. If n = 8, 9, by sendingi → ∞ in (89) we haveP(r,Γ) ≥ −Cr. Thus (76) follows and the conclusion (i) is proved.

If n ≥ 10 and|Wgi(0)|2 satisfies (79) for largeC∗ > 0, by (88) we see that(Ei(r)−Ei(r))+

Ei(r) ≥ 0. Hence, the conclusion (iii) follows.Since|P(r,Γ)| ≤ C, it follows from (89) that for largei, Ei(r) ≤ C. In view of (88) and

the definition ofα′′, the conclusion (ii) follows.

Proposition 5.2. Givenpi, Gi, andhi satisfying(23), (24) and (26) respectively,κi satisfying(25) with Ki replaced byκi, let 0 ≤ ui ∈ C4(B3) solve both(59) and (75), and assume(64)holds. Suppose that0 is an isolated blow up point ofui with (61)holds. Then0 is an isolatedsimple blow up point, if one of the three cases happens:

46


• gi is flat;

• n ≤ 9;

• n ≥ 10 and (79)holds.

Proof. By Proposition 3.1,r4/(pi−1)ui(r) has precisely one critical point in the interval0 < r <

ri, whereRi → ∞ ri = Riui(0)−

pi−1

4 as in Proposition 3.1. Suppose the contrary that0 is notan isolated simple blow up point and letµi be the second critical point ofr4/(pi−1)ui(r). Thenwe must have

µi ≥ ri, limi→∞

µi = 0. (90)

Setvi(x) = µ

4/(pi−1)i ui(µix), x ∈ B3/µi .

By the assumptions of Proposition 3.1,vi satisfies

vi(x) =

∫

B3/µi

Gi(x, y)κi(y)τivi(y)

pi dy + hi(x)

|x|4/(pi−1)vi(x) ≤ A3, |x| < 2/µi → ∞,

limi→∞

vi(0) = ∞,

r4/(pi−1)vi(r) has precisely one critical point in0 < r < 1,

andd

dr

r4/(pi−1)vi(r)

∣

∣

∣

r=1= 0,

whereGi = Gi,µi, κi(y) = κi(µiy), hi(x) = µ4/(pi−1)i hi(µix) andvi(r) = |∂Br|−1

∫

∂Brvi.

Therefore,0 is an isolated simple blow up point ofvi.Claim. We have

vi(0)vi(x) →acn

|x|n−4+ acn in C3

loc(Rn \ 0). (91)

wherea > 0 is given in (58).First of all, by Proposition 3.3 we havehi(e) ≤ vi(e) ≤ Cvi(0)

−1 for anye ∈ Sn−1, whereC > 0 is independent ofi. It follows from the assumption (26) onhi that

vi(0)hi(x) ≤ C for all |x| ≤ 2/µi

and‖∇(vi(0)hi)‖L∞(B 1

9µi

) ≤ µi‖vi(0)hi‖L∞(B 14µi

) ≤ Cµi. (92)

47

Y. Y. Li & J. Xiong

Hence, for some constantc0 ≥ 0, we have, along a subsequence,

limi→∞

‖vi(0)hi(x)− c0‖L∞(Bt) = 0, ∀ t > 0. (93)

Secondly, by Corollary 3.2 and Proposition 3.3 we have, up toa subsequence,

vi(0)

∫

Bt


pi dy →acn

|x|n−4in C3

loc(Bt \ 0) for anyt > 0, (94)

where we used thatGi(x, 0) → cn|x|4−n. Notice that for anyx ∈ Bt/2

Q′′i (x) :=

∫

B3/µi\Bt


pi dy ≤ C(n,A1)max∂Bt

vi.

Sincemax∂Bt vi ≤ Ct4−nvi(0)−1, we have as in the proof of (35), after passing to a subse-

quence,vi(0)Q

′′i (x) → q(x) in C3

loc(Bt) asi→ ∞

for someq ∈ C3(Bt). For any fixed largeR > t+ 1, it follows from (94) that

vi(0)

∫

t≤|y|≤R


pi dy → 0

as i → ∞, since the constanta is independent oft. By the assumption (24) onGi, for anyx ∈ Bt and|y| > R, we have

|∇xGi(x, y)| ≤ A1|x− y|3−n ≤A1

R− t|x− y|4−n ≤

A21

R − tGi(x, y).

Therefore, we have|∇q(x)| ≤ A21

R−tq(x). By sendingR → ∞, we have|∇q(x)| ≡ 0 for any

x ∈ Bt. Thus,q(x) ≡ q(0) for all x ∈ Bt.

Sinced

dr

r4/(pi−1)vi(0)vi(r)

∣

∣

∣

r=1= vi(0)

d

dr

r4/(pi−1)vi(r)

∣

∣

∣

r=1= 0,

we have, by choosing, for example,t = 2 and sendingi to ∞, that

q(0) + c0 = acn > 0. (95)

Therefore, (91) is proved.It follows from (91) and Lemma 2.2 that

lim infi→∞

vi(0)2P(r, vi) = −(n− 4)2(n− 2)a2c2n|S

n−1| < 0 for all 0 < r < 1.

48


On the other hand, by (75)vi satisfies

Pgivi = c(n)κτii vpii in B3/µi ,

wheregi(z) = gi(µiz). It is easy to see that (64) is still correct withGi replaced byGi. If n ≤ 9or gi is flat, it follows from Proposition 5.1 that

lim infr→0

lim infi→∞

vi(0)2P(r, vi) ≥ 0. (96)

If n ≥ 10, by (64), we have

Uλ(x) =

∫

B3/µi

Gi,µi(x, y)Uλ(y)n+4n−4 + µ4

i c′λ/µi,i

(µiy)Uλ(y) dy + µn−42

i c′′λ/µi,i(µix)

=

∫

B3

Gi(x, y)Uλ(y)n+4n−4 + c′λ,i(y)Uλ(y) dy + c′′λ,i(x) ∀ λ ≥ 1, x ∈ B3,

wherec′λ,i(y) := µ4i c

′λ/µi,i

(µiy) and

c′′λ,i(x) =

∫

B3/µi\B3

Gi,µi(x, y)Uλ(y)n+4n−4 + µ4

i c′λ/µi,i

(µiy)Uλ(y) dy + µn−42

i c′′λ/µi,i(µix).

By the assumptions forc′λ,i andc′′λ,i, we have

Θi :=

5∑

i=0

‖λ−k∇kc′λ,i‖L∞(B3) ≤ µ4iΘi,

and‖c′′λ,i‖C5(B2) ≤ CA2λ4−n2 , whereC > 0 depends only onn,A1, A2. Clearly, we have

|Wgi(0)|2 = µ4

i |Wgi(0)|2. Hence (79) is satisfied. By Proposition 5.1, we also have (96). We

obtain a contradiction.Therefore,0 must be an isolated simple blow up point ofui and the proof is completed.

Lemma 5.1. Let 0 ≤ ui ∈ C4(B3) solve both(59) and (75) with n ≥ 10, and assume(64)holds. Forµi → 0, let

vi(x) = µ4

pi−1

i ui(µix).

Suppose that0 is an isolated blow up point ofvi and (79) holds. Then0 is also an isolatedsimple blow up point.

Proof. From the end of proof of Proposition 5.2, we see that the condition (79) is preserved

under the scalingvi(x) = µ4

pi−1

i ui(µix). Hence, the lemma follows from Proposition 5.2.

49

Y. Y. Li & J. Xiong

6 Global analysis, and proof of Theorem 1.2

Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 5. Suppose thatKerPg = 0 and the Green’s functionGg of Pg is positive. Consider the equation

Pgu = c(n)up, u ≥ 0 onM, (97)

where1 < p ≤ n+4n−4

.

Proposition 6.1. Assume the above. For any givenR > 0 and0 ≤ ε < 1n−4

, there exist positiveconstantsC0 = C0(M, g,R, ε),C1 = C1(M, g,R, ε) such that, for any smooth positive solutionof (97) with

maxM

u(X) ≥ C0,

thenn+4n−4

− p < ε and there exists a set of finite distinct points

S (u) := Z1, . . . , ZN ⊂M

such that the following statements are true.(i) EachZi is a local maximum point ofu and

Bri(Zi) ∩ Brj (Zj) = ∅ for i 6= j,

where ri = Ru(Zi)(1−p)/4, andBri(Zi) denotes the geodesic ball inB2 centered atZi with

radiusri(ii) For eachZi,

∥

∥

∥

∥

∥

1

u(Zi)u

(

expZi

(

y

u(Zi)(p−1)/4

))

−

(

1

1 + |y|2

)n−42

∥

∥

∥

∥

∥

C4(B2R)

< ε.

(iii) u(X) ≤ C1distg(x, Z1, . . . , ZN)−4/(p−1) for all X ∈M .

Proof. The proof is standard by now.

Proposition 6.2. If eithern ≤ 9 or (M, g) is locally conformally flat, then, forε > 0, R > 1and any solution(97)withmaxM u > C0, we have

|Z1 − Z2| ≥ δ∗ > 0 for anyZ1, Z2 ∈ S (u), Z1 6= Z2,

whereδ∗ depends only onM, g.

50


Proof. Suppose the contrary, then there exist a sequence0 ≤ n+4n−4

− pi < ε andui satisfying(97) withp = pi,

maxM

ui(X) > C0,

anddistg(Z1i, Z2i) = min

1≤k,l≤Ni, k 6=ldistg(Zki, Zli) → 0 (98)

asi → ∞, whereS (ui) = (Z1i, . . . , ZNii) be the local maximum points ofui as selected byProposition 6.1. Without loss of generality, we may assume

ui(Z1i) ≥ ui(Z2i).

SinceBRui(Z1i)−(pi−1)/4(Z1i) andBRui(Z2i)−(pi−1)/4(Z2i) have to be disjoint, we have, because of(98), thatui(Z1i) → ∞ andui(Z2i) → ∞.

Let x1, . . . , xn be the conformal normal coordinates centered atZ1i. We write (97) as

Pgiui = c(n)κτii upii onM, (99)

wheregi = κ−4n−4

i g, ui = κiui, κi > 0, κi(Z1i) = 1, ∇gκi(Z1i) = 0, andτi = n+4n−4

− pi. Sincedistg(Z1i, Z2i) → 0, for largei we letz2i ∈ R

n such thatexpZ1iz2i = Z2i, and let

ϑi := |z2i| → 0.

We will sit in the conformal normal coordinates chartBt atZ1i, wheret > 0 is independent ofi, and writef(expZ1,i

x) simply asf(x). Set

ϕi(x) = ϑ4/(pi−1)i ui(ϑix) for |x| ≤ t/ϑi.

By the equation (99), we have

Pgiϕi(x) = c(n)κi(x)τiϕi(x)

pi in Bt/ϑi , (100)

whereκi(x) = κi(ϑix), gi(x) = gi(ϑix). Using the Green representation for (99),

ui(x) = c(n)

∫

Bt

Gi(x, y)κi(y)τiui(y)

pi dy + hi(x),

whereGi(x, y) = Ggi(expZ1,ix, expZ1,i

y) and

hi(x) =

∫

M\expZ1,iBt

Ggi(expZ1,ix, Y )κi(Y )

τiui(Y ) dvolgi(Y ).

51

Y. Y. Li & J. Xiong

Hence,ϕi also satisfies

ϕi(x) =

∫

Bt/ϑi

Gi,ϑi(x, y)Ki(ϑiy)ϕi(y)pi dy + hi(x) for all x ∈ Bt/ϑi , (101)

whereGi,ϑi(x, y) = ϑn−4i G(ϑix, ϑiy) andhi = ϑ

4/(pi−1)i hi(ϑiy).

By proposition 6.1, we have

ui(x) ≤ C∗1 |x|

−4/(pi−1) for all |x| ≤ 3ϑi/4,

ui(x) ≤ C∗1 |x− z2i|

−4/(pi−1) for all |x− z2i| ≤ 3ϑi/4,

whereC∗1 depending only onC1,M andg. Hence,

ϕi(x) ≤ C∗1 |x|

−4/(pi−1) for all |x| ≤ 3/4,

ϕi(x) ≤ C∗1 |x− ϑ−1

i z2i|−4/(pi−1) for all |x− ϑ−1

i z2i| ≤ 3/4.(102)

Setξi = ϑ−1i z2i. We claim that, after passing to a subsequence,

ϕi(0), ϕi(ξi) → ∞ asi→ ∞. (103)

It is clear thatϕi(0) andϕi(ξi) are bounded from below by some positive constant independentof i. If there exists a subsequence (still denoted asϕi) such thatlim

i→∞ϕi(0) = ∞ butϕi(ξi) stays

bounded, we have that0 is an isolated blow up point forϕi in B3/4 wheni is large; see Remark3.1. Using equation (101) and (102), by the same proof of (34)we havesupB1/2(ξi)

ϕi < ∞.It follows from Proposition 3.3 and Proposition A.1 thatlim

i→∞ϕi(ξi) = 0, but this is impossible

sinceϑi > Rui(Z2i)−(pi−1)/4 and thusϕi(ξi) ≥ 1

CR for someC > 0 depending only onM ,

g, R and ε. On the other hand, if there exists a subsequence (still denoted asϕi) such thatϕi(0) andϕi(ξi) remain bounded, we know from a similar argument as above thatϕi is locallybounded. The same proof of Proposition 3.1 yields that afterpassing to a subsequenceϕi → ϕin C3

loc(Rn) for someϕ satisfying

ϕ(x) = cn

∫

Rn

ϕ(y)n+4n−4

|x− y|n−4dy for all x ∈ R

n,

∇ϕ(0) = 0, ∇ϕ(z) = limi→∞∇ϕi(ξi) = limi→∞ϑiϕi(ξi)∇κi(z2i)

κi(z2i)= 0, where|z| = 1 is the limit

of ξi up to passing a subsequence. This contradicts to the Liouville theorem in [12] or Li [36].Hence, (103) is proved.

Since∇ϕi(0) = 0, it follows from the first inequality of (102) and (103) that0 is an isolatedblow up point ofϕi. Sincen ≤ 9 or (M, g) is locally conformally flat, by Proposition 5.2 we

52


conclude that0 is an isolated simple blow up point ofϕi. It follows from Corollary 3.2 thatfor all x ∈ B1/2

ϕi(0)

∫

B1/2

Gi,ϑi(x, y)ϕi(y)piKi(ϑiy) dy → acn|x|

4−n (104)

andϕi(0)(Q

′′i (x) + hi(x)) → h(x) ≥ 0 in C3

loc(B1/2),

wherea > 0 is given in (58),h(x) ∈ C5(B1/2) and

Q′′i (x) =

∫

Bt/ϑ\B1/2

Gi,ϑi(x, y)ϕi(y)piKi(ϑiy) dy x ∈ B1/2.

Note that

ϕi(0)Q′′i (x) ≥

1

Cϕi(0)

∫

B1/2(ξi)

ϕi(y)pi dy.

It follows from (102), (103) and the proof of (34) that there exists a constantC > 0, dependingonly onM, g,R andε such thatϕi(x) ≤ Cϕi(ξi) for all |x− ξi| ≤

12. It follows from the proof

of Proposition 3.1 that there exist a constantλ and an pointx0 ∈ Rn with 1 ≤ λ ≤ C and

|x0| ≤ C such that for any fixedR > 0 we have

limi→∞

∥

∥

∥

∥

1

ϕi(ξi)ϕi(ξi + ϕi(ξi)

−(pi−1)/4x)− Uλ(x− x0)

∥

∥

∥

∥

C4(BR)

= 0

By changing of variablesy = ξi + ϕi(ξi)−(pi−1)/4x, we have

ϕi(0)

∫

B1/2(ξi)

ϕi(y)pi dy

= ϕi(0)ϕi(ξi)pi−

(pi−1)n

4

∫

Bϕi(ξi)/2(0)

(

1

ϕi(ξi)ϕi(ξi + ϕi(ξi)

−(pi−1)/4x)

)pi

dx

≥1

Cϕi(0)ϕi(ξi)

pi−(pi−1)n

4

∫

Rn

Uλ(x− x0)pi dx

≥1

Cϕi(0)ϕi(ξi)

pi−(pi−1)n

4

∫

Rn

(1 + |x|2)n+42 dx,

where we used1 ≤ λ ≤ C and|x0| ≤ C. Sinceui(Z1i) ≥ ui(Z2i), we haveϕi(0) ≥ 1Cϕi(ξi)

for someC depending only onM, g. By Lemma 3.4, we haveϕi(ξi)1+pi−(pi−1)n

4 = 1 + o(1).Therefore, we obtain

limi→∞

ϕi(0)Q′′i (x) ≥

1

C

∫

Rn

(1 + |ξ|2)n+42 dξ =: a0 > 0. (105)

53

Y. Y. Li & J. Xiong

In conclusion,ϕi(0)ϕi(x) → acn|x|

4−n + h(x) in C3loc(B1/2 \ 0)

for some nonnegative bounded function inC3(B1/2) with h(0) ≥ a0. It follows from Lemma2.2 that

lim infr→0

lim infi→∞

ϕ2iP(r, ϕi) < −(n− 4)2(n− 2)aa0cn|S

n−1|.

On the other hand, notice thatϕi satisfies (99). We also have Corollary 2.2. It follows fromProposition 5.1 that

lim infr→0

lim infi→∞

ϕi(0)2P(r, ϕi) ≥ 0.

We arrive at a contradiction. Therefore, (98) is not valid and the proposition follows.

Theorem 1.2 is a part of the following theorem.

Theorem 6.1. Let ui ∈ C4(M) be a sequences of positive solutions ofPgui = c(n)upii onM ,where0 ≤ (n + 4)/(n − 4) − pi → 0 as i → ∞. Assume(9). If eithern ≤ 9 or (M, g) islocally conformally flat, then

‖ui‖H2(M) ≤ C,

whereC > 0 depending only onM, g. Furthermore, after passing to a subsequence,ui isuniformly bounded or has only isolated simple blow up pointsand the distance between anytwo blow up points is bounded below by some positive constantdepending only onM, g.

Remark 6.1. On (M, g), we say a pointX ∈ M is an isolated blow up point forui if thereexists a sequenceXi ∈ M , where eachXi is a local maximum point forui andXi → X,

such thatui(Xi) → ∞ as i → ∞ and ui(X) ≤ Cdistg(X,Xi)− 4

pi−1 in Bδ(Xi) for someconstantsC, δ > 0 independent ofi. Under the assumptions thatui is a positive solution ofPgui = c(n)upii with 0 ≤ n+4

n−4− pi → 0, KerPg = 0 and that the Green’s functionGg of

Pg is positive, it is easy to see that ifXi → X ∈ M is an isolated blow up point ofui,then in the conformal normal coordinates centered atXi, 0 is an isolated blow up point of

ui(expXix), where the exponential map is with respect to conformal metric gi = κ

−4n−4

i g,κi > 0 is under control onM , and ui = κiui; see Remark 3.1. Since in Theorem 6.1 andthe sequel those assumptions will always be assumed, the notation of isolated simple blow uppoints on manifolds is understood in conformal normal coordinates.

Proof of Theorem 6.1.The last statement follows immediately from Proposition 6.1, Proposi-tion 6.2 and Proposition 5.2. Consequently, it follows fromCorollary 3.1 and Proposition 3.3

and Proposition A.1 that∫

Mu

2nn−4

i dvolg ≤ C. By the Green’s representation and standard esti-mates for Riesz potential, we have theH2 estimates.

Now we consider thatn ≥ 10.

54


Proposition 6.3. Let (M, g) be a smooth compact Riemannian manifold of dimensionn ≥ 10.Assume (9). Letui be a sequence of positive solutions ofPgu = c(n)upi, wherepi ≤ n+4

n−4,

pi →n+4n−4

asi → ∞. Suppose that there is a sequenceXi → X ∈ M such thatui(Xi) → ∞.For any smallε > 0 andR > 1, let S (ui) denote the set selected as in Proposition 6.1 forui.If |Wg(X)|2 ≥ ε0 > 0 onM for some constantε0, then there existsδ∗ > 0 depending only onM, g andε0 such thatBδ∗(X) ∩ S (ui) contains precisely one point.

Proof. Let δ > 0 such that|Wg(X)|2 ≥ ε0/2 for X ∈ Bδ(X). Assume the contrary of theproposition, then for a subsequence ofui (still denoted asui) there exist distinct pointsX1i, X1i ∈ S (ui) such thatX1i, X1i → X. Definefi : S (ui) → (0,∞) by

fi(X) := minX′∈S (ui)\X

distg(X′, X).

LetRi → ∞ satisfyingRifi(X1i) → 0.Claim. There exists a subsequence ofi→ ∞ such that one can findX ′

i ∈ S (ui)∩Bδ/9(X)satisfying

fi(X′i) ≤ (2Ri + 1)fi(X1i)

and

minX∈S (ui)∩BRifi(X

′i)(X

′i)fi(X) ≥

1

2fi(X

′i).

Indeed, suppose the contrary, then there existsI ∈ N such that for anyi ≥ I,X ′i in the claim

can not been selected. Sincefi(X1i) ≤ (2Ri+1)fi(X1i), by the contradiction hypothesis, theremust existX2i ∈ S (ui) ∩ BRifi(X1i)(X1i) such thatfi(X2i) <

12fi(X1i). We can defineXli ∈

S (ui), l = 3 . . . , satisfyingfi(Xli) <12fi(X(l−1)i) and0 < distg(Xli, X(l−1)i) < Rifi(X(l−1)i)

inductively as follows. OnceXli, l ≥ 2, is defined, we have, for2 ≤ m ≤ l, that

distg(Xmi, X(m−1)i) < Rifi(X(m−1)i) < Ri2−1fi(X(m−2)i) < · · · < Ri2

2−mfi(X1i),

which implies

distg(Xli, X1i) ≤l∑

m=2

distg(Xmi, X(m−1)i) < Rifi(X1i)

l∑

m=2

22−m < 2Rifi(X1i),

andfi(Xli) ≤ distg(Xli, X1i) + fi(X1i) ≤ (2Ri + 1)fi(X1i).

so X ′i := Xli satisfiesX ′

i ∈ S (ui) ∩ Bδ/9(X) and the first inequality of the claim. Bythe contradiction hypothesis, there must existX(l+1)i ∈ S (ui) ∩ BRifi(Xli)(Xli) such thatfi(X(l+1)i) <

12fi(Xli). But S (ui) is a finite set and we can not work for alll ≥ 2. Therefore,

the claim follows.By the claim, we can follow the proof of Proposition 6.2 withZ1i replaced byX ′

i. We thenderive a contradiction to Proposition 5.1. Therefore, we complete the proof.

55

Y. Y. Li & J. Xiong

7 Proof of Theorems 1.1, Theorem 1.3, Theorem 1.4

Proof of Theorem 1.3.For n = 8, 9, sinceui(Xi) → ∞, it follows from Proposition 6.2 andTheorem 6.1 thatX is an isolated simple blow up points ofui. Then Theorem 1.3 followsfrom item (ii) of Proposition 5.1. Whenn ≥ 10, we may argue by contradiction. Suppose that|Wg(X)|2 > 0. By Proposition 6.3, Proposition 5.2, and in view of the proof of (34) under (33),that there exists a sequence ofX ′

i → X which is an isolated simple blow up point ofui; seeRemark 6.1. By item (ii) of Proposition 5.1, we have|Wg(X

′i)|

2 → 0. It gives|Wg(X)|2 = 0.We obtain a contradiction. Hence, we complete the proof.

Proof of Theorem 1.4.Suppose the contrary, then, after passing to a subsequence,

|Wg(Xi)|2 >

1

|o(1)|

ui(Xi)− 4

n−4 log ui(Xi), if n = 10,

ui(Xi)− 4

n−4 , if n ≥ 11.(106)

For anyε > 0 andR > 1, let S (ui) = Z1i, . . . , ZNii be the set selected as in Proposition 6.1for ui, whereNi ∈ N+. Let, without loss of generality,

distg(Xi, Z2i) = infZji∈S (ui),Zji 6=Xi

distg(Xi, Zji).

If there exists a constantδ∗ > 0 independent ofi such thatdistg(Xi, Z2i) ≥ δ∗, thenXi ∈S (ui) for largei. It follows from item (iii) of Proposition 6.1 and Proposition 5.2 thatXi → Xhas to be an isolated simple blow up point ofui, using the fact that (106) guarantees (79). Byitem (ii) of Proposition 5.1, we obtain an opposite side inequality of (106). Contradiction.

If distg(Xi, Z2i) → 0 as i → ∞. Let x1, . . . , xn be the conformal normal coordinatescentered atXi. Defineϕi as that in the proof of Proposition 6.2 withZ1i replaced byXi. SincesupΩ ui ≤ bui(Xi), we must haveϕi(0) → ∞ by the Liouville theorem in [12] or [36]; see theproof of (103). Because of (12) and (13),0 has to be an isolated blow up point ofϕi; seeRemark 3.1. It follows from the contradiction hypothesis (106), which guarantees (79), as inthe proof of Lemma 5.1 that0 has to be an isolated simple blow up point ofϕi for i large.Then we we arrive at a contradiction by item (ii) of Proposition 5.1 again.


Proof of Theorem 1.1.If n ≥ 8 and |Wg|2 > 0 on M , Theorem 1.1 is a direct corollary of

Theorem 1.3. Hence, we only need to considern ≤ 9 or (M, g) is locally conformally flat.By Proposition 6.1, it suffices to consider thatp is close ton+4

n−4. Suppose the contrary that

there exists a sequences of positive solutionsui ∈ C4(M) of Pgui = c(n)upii on M , wherepi → (n+4)/(n− 4) asi→ ∞, such thatmaxM ui → ∞. By Theorem 6.1, letXi → X ∈M

56


be an isolated simple blow up point ofui; see Remark 6.1. It follows from Proposition 5.1that, in thegX-normal coordinates centered atX,

lim infr→0

P(r, c(n)G) ≥ 0,

wheregX a conformal metric ofg with det gX = 1 in an open ballBδ of thegX-normal coordi-nates,G(x) = GgX

(X, expX x) andGgXis the Green’s function ofPgX . On the other hand, if

n = 5, 6, 7 or (M, g) is locally conformally flat, by Theorem 2.1 and Lemma 2.2 we have

P(r, c(n)G) < −A for smallr,

whereA > 0 depends only onM, g. We obtain a contradiction.If n = 8, 9, by Theorem 1.3 we haveWg(X) = 0. In view of Remark 2.1, we have

limr→0

P(r, G) =

−2−∫

Sn−1 ψ(θ), n = 8,

−52A, n = 9,

whereψ(θ) andA are as in Remark 2.1. If the positive mass type theorem holds for Paneitzoperator in dimensionn = 8, 9, we obtainlimr→0P(r, G) < 0. Again, we derived a contradic-tion.

Therefore,ui must be uniformly bounded and the proof is completed.

8 Proof of Theorem 1.5

This section will not use previous analysis and thus is independent. The proof of Theorem 1.5is divided into two steps.

Step 1. Lp estimate. Letu ≥ 0 be a solution of (14). Integrating both sides of (14) and usingHolder inequality, we have

∫

M

updvolg =

∣

∣

∣

∣

−

∫

M

uPg(1)dvolg

∣

∣

∣

∣

≤n− 4

2‖u‖Lp(M)‖Qg‖Lp′(M),

where 1p′+ 1

p= 1. It follows that‖u‖p−1

Lp(M) ≤n−42‖Qg‖Lp′(M).

Step 2. If KerPg = 0, there exist a unique Green function ofPg. If the kernel ofPgis non-trivial, since the spectrum of Paneitz operator is discrete, there exists a small constantε > 0 such that the kernel ofPg − ε is trivial. Let Gg be the Green function of the operatorPg − ε, whereε ≥ 0. Then there exists a constantδ > 0, depending onlyM, g andε, suchthat, for everyX ∈ M , we haveG(X, Y ) > 0 for Y ∈ Bδ(X) and|Gg(X, Y )| ≤ C(δ, ε) forY ∈M \ Bδ(X). Rewrite the equation ofu as

Pgu− εu = −(up + εu).

57

Y. Y. Li & J. Xiong

It follows from the Green representation theorem that

u(X) = −

∫

M

Gg(X, Y )(up + εu)(Y )dvolg(Y )

= −

∫

Bδ(X)


−

∫

M\Bδ(X)


≤ −

∫

M\Bδ(X)


≤ Cmax‖u‖pLp(M), ‖u‖Lp(M) ≤ C.

By the arbitrary choice ofX, we have‖u‖L∞ ≤ C. The higher order estimate follows from thestandard linear elliptic partial differential equation theory; see Agmon-Douglis-Nirenberg [1].


A Local estimates for solutions of linear integral equations

LetΩ2 ⊂⊂ Ω1 be a bounded open set inRn, n ≥ 5. Forx, y ∈ Ω1, letG(x, y) satisfy

G(x, y) = G(y, x), G(x, y) ≥ A−11 |x− y|4−n,

|∇lxGi(x, y)| ≤ A1|x− y|4−n−l, l = 0, 1, . . . , 5

G(x, y) = cn1 +O(4)(|x|2) +O(4)(|y|2)

|x− y|n−4+O(4)(

1

|x− y|n−6)

(107)

and let0 ≤ h ∈ C4(Ω1) satisfy

supΩ2

h ≤ A2 infΩ2

h and4∑

j=1

rj |∇jh(x)| ≤ A2‖h‖L∞(Br(x)) (108)

for all x ∈ Ω2 and0 < r < dist(Ω2, ∂Ω1). We recall some local estimates for solutions of theintegral equation

u(x) =

∫

Ω1

G(x, y)V (y)u(y) dy + h(x) for x ∈ Ω1. (109)

58


Proposition A.1. LetG, h satisfy(107)and (108), respectively. Let0 ≤ V ∈ L∞(Ω1), and let0 ≤ u ∈ C0(Ω1) be a solution of(109). Then we have

supΩ2

u ≤ C infΩ2

u,

whereC > 0 depends only onn,A1, A2, Ω1, Ω2 and‖V ‖L∞(Ω1).

Proof. It follows from some simple modification of the proof of Proposition 2.3 of [31]. In fact,the third line of (107) is not needed.

Proposition A.2. Suppose the hypotheses in Proposition A.1. Thenu ∈ C3(B2) and

‖u‖C3(Ω2) ≤ C‖u‖L∞(Ω1),

whereC > 0 depends only onn,A1, A2, the volume ofΩ2, dist(Ω2, ∂Ω1) and‖V ‖L∞(Ω1).If V ∈ C1(Ω1), thenu ∈ C4(Ω2) and for anyΩ3 ⊂⊂ Ω2 we have

‖∇4u‖L∞(Ω3) ≤ C‖u‖L∞(Ω1),

whereC > 0 depends only onn,A1, A2, the volume ofΩ2, dist(Ω2, ∂Ω1), dist(Ω3, ∂Ω2) and‖V ‖C1(B3).

Proof. Let f := V u. If k < 4, making use of (107), (108) and (109) we have

∇ku(x) =

∫

Ω1

∇kxG(x, y)f(y) dy +∇kh(x) for x ∈ Ω2,

and thus

|∇ku(x)| ≤ A1‖f‖L∞(Ω1)

∫

Ω1

|x− y|n−4+k dy + |∇kh(x)|

≤ C(‖u‖L∞(Ω1) + ‖h‖L∞(Ω1)).

Sinceu andV are nonnegative, we have0 ≤ h(x) ≤ u(x). We proved the first conclusion.Let Ω3 ⊂⊂ Ω2. Without loss of generality, we may assume∂Ω2 ∈ C1. If V ∈ C1(Ω1), we

havef ∈ C1(Ω2). By the third line of (107), we see

∇xG(x, y) = −∇yG(x, y) + (O(3)(|x|)− O(3)(|y|))|x− y|4−n +O(3)(|x− y|5−n).

We have forx ∈ Ω3 and1 ≤ j ≤ n,

∇xj∇3u(x) =

∫

Ω2

∇xj∇3xG(x, y)(f(y)− f(x)) dy − f(x)

∫

∂Ω2

∇3yG(x, y)νj dS(x)

+ f(x)O(

∫

Ω2

|x− y|1−n dy) +

∫

Ω1\Ω2

∇xj∇3xG(x, y)f(y) dy +∇xj∇

3h(x),

whereν = (ν1, . . . , νn) denotes the outward normal to∂Ω2. By (107) and (108), the prooffollows immediately. Hence, we complete the proof.

59

Y. Y. Li & J. Xiong

B Riesz potentials of some functions

Lemma B.1. For 0 < α < n, we have

∫

Rn

1

|x− y|n−α(1 + |y|)µdy

≤ C(n, α, µ)

(1 + |x|)α−µ, if α < µ < n,

(1 + |x|)α−n log(2 + |x|), if µ = n,

(1 + |x|)α−n, if µ > n.

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Y.Y. LiDepartment of Mathematics, Rutgers University,110 Frelinghuysen Road, Piscataway, NJ 08854, USA

Email: [email protected]

J. XiongSchool of Mathematical Sciences, Beijing Normal UniversityBeijing 100875, China

Email: [email protected]

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