The Han-Li conjecture in constant scalar curvature and constant boundary mean curvature problem on compact manifolds

Advances in Mathematics 358 (2019) 106854

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The Han-Li conjecture in constant scalar curvature and constant boundary mean curvature problem on compact manifolds Xuezhang Chen a,∗ , Yuping Ruan a,1 , Liming Sun b a

Department of Mathematics & IMS, Nanjing University, Nanjing 210093, PR China b Department of Mathematics, Johns Hopkins University, Baltimore, MD, 21218, USA

a r t i c l e

i n f o

Article history: Received 25 November 2018 Received in revised form 30 August 2019 Accepted 8 October 2019 Available online xxxx Communicated by YanYan Li MSC: primary 53C21, 58J32, 35J20 secondary 58J05, 35B33, 34B18

a b s t r a c t The Han-Li conjecture states that: Let (M, g0 ) be an ndimensional (n ≥ 3) smooth compact Riemannian manifold with boundary having positive (generalized) Yamabe constant and c be any real number, then there exists a conformal metric of g0 with scalar curvature 1 and boundary mean curvature c. Combining with Z.C. Han and Y.Y. Li’s results, we answer this conjecture affirmatively except for the case that n ≥ 8, the boundary is umbilic, the Weyl tensor of M vanishes on the boundary and has an interior non-zero point. © 2019 Elsevier Inc. All rights reserved.

Keywords: Mountain Pass Lemma Conformal Fermi coordinates Boundary Yamabe problem Positive mass theorem

* Corresponding author. E-mail addresses: [email protected] (X. Chen), [email protected] (Y. Ruan), [email protected] (L. Sun). 1 Current address: Department of Mathematics, University of Michigan, Ann Arbor, MI 48109, USA. https://doi.org/10.1016/j.aim.2019.106854 0001-8708/© 2019 Elsevier Inc. All rights reserved.

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

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Contents 1. 2. 3. 4. 5.

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The case of non-empty Z with PMT . . . . . . . . . . . . . . . . . . . . . . . . . Remaining cases in dimensions four and six . . . . . . . . . . . . . . . . . . . . n ≥ 7, umbilic boundary and non-zero Weyl tensor at a boundary point 5.1. Positive constant boundary mean curvature . . . . . . . . . . . . . . . 5.2. Non-positive constant boundary mean curvature . . . . . . . . . . . Appendix A. ............................................ References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1. Introduction On a smooth compact Riemannian manifold with boundary, the analogues of the Yamabe problem were initially studied P. Cherrier, then by J. Escobar more extensively. Over the past more than twenty years, such problems have been extensively investigated by numerous researchers. In some literatures, these problems are also referred to as Escobar problem. We give a brief summary for these problems. It is convenient to distinguish into three cases. The first case is concerned with the existence of conformal metrics with constant scalar curvature and zero boundary mean curvature. This problem was initially studied by J. Escobar [24] in the case of 3 ≤ n ≤ 5 or n ≥ 6 and the boundary has a non-umbilic point, later by S. Brendle-S. Chen [12] in the case of n ≥ 6 and the boundary is umbilic, assuming the validity of the Positive Mass Theorem (PMT). For recent associated curvature flows, readers are referred to [9,14,4] and the references therein. The second case is concerned with the existence of scalar-flat conformal metrics with constant boundary mean curvature under the condition that the corresponding generalized Yamabe constant is finite. It was also first studied by Escobar [20], subsequently in Escobar [22], and then by F. Marques [31] in the case of 3 ≤ n ≤ 5 or M is locally conformally flat with umbilic boundary or n ≥ 6 and the boundary has at least one non-umbilic point. Some remaining cases have been studied by Marques [30], S. Chen [13] with assuming PMT, S. Almaraz [1] etc. More recently, without the PMT, the remaining case was recently solved by M. Mayer and C. Ndiaye [32] through the barycenter technique, but in general the solution they obtained is not a minimizer of the associated energy functional. See [9,14,2] etc. for the related conformal curvature flows. The third case is concerned with the existence of conformal metrics with (non-zero) constant scalar curvature and (non-zero) constant boundary mean curvature. For variational methods, see [21,25,27,28,7,8,15,18] and the references therein. For flow approaches, see [15,16]. From now on, we focus on the last case. This paper can be regarded as a continuation of the first and third named authors’ paper [18]. Let (M, g0 ) be a smooth compact Riemannian manifold of dimension n ≥ 3 with boundary ∂M . The (generalized) Yamabe constant Y (M, ∂M, [g0 ]) is defined as

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

 Y (M, ∂M, [g0 ]) := inf

M

g∈[g0 ]

3

 Rg dμg + 2(n − 1) ∂M hg dσg ,  n−2 ( M dμg ) n

where Rg is the scalar curvature of M and hg is the mean curvature on ∂M of metric g. Define     4(n − 1) 2 2 |∇u|g0 + Rg0 u dμg0 + 2(n − 1) E[u] = hg0 u2 dσg0 . n−2 M

∂M

For c ∈ R, we consider a “free” functional  I[u] = E[u] − 4(n − 1)(n − 2)

2n n−2



u+ dμg0 − 4c M

2(n−1)

u+n−2 dσg0

(1.1)

∂M

for all u ∈ H 1 (M, g0 ), where u+ = max{u, 0}. Then we can verify that I ∈ C 2 (H 1 (M, g0 ); R). It is not difficult to check that any non-trivial critical point u of I solves ⎧ n+2 4(n − 1) ⎪ ⎨− Δg0 u + Rg0 u = 4n(n − 1)u+n−2 in M, n−2 (1.2) n ⎪ ∂u + n − 2 h u = cu n−2 ⎩ on ∂M, g0 + ∂νg0 2 where νg0 is the outward unit normal on ∂M with respect to g0 . When the Yamabe constant Y (M, ∂M, [g0 ]) is non-negative, a simple application of the maximum principle and the Hopf boundary point lemma yields that such u must be positive in M . The regularity theory in [19] shows that u is smooth in M . Thus if we let g = u4/(n−2) g0 , (1.2) implies that Rg = 4n(n −1) and hg = 2c/(n −2). For brevity, denote by Lg0 = − 4(n−1) n−2 Δg0 +Rg0 the conformal Laplacian and Bg0 = ∂ν∂g + n−2 h the boundary conformally covariant g0 2 0 operator, respectively. Both Lg0 and Bg0 have the following conformally covariant properties: Let g = u4/(n−2) g0 , then for any ϕ ∈ C ∞ (M ), there hold n+2

Lg0 (uϕ) = u n−2 Lg (ϕ)

n

and Bg0 (uϕ) = u n−2 Bg (ϕ).

(1.3)

In 1999, Zheng-Chao Han and Yan Yan Li [28] proposed the following conjecture. Conjecture (Han-Li). If Y (M, ∂M, [g0 ]) > 0, then given any c ∈ R, problem (1.2) is solvable. Compared with the previous two cases in Escobar Problem, besides the critical Sobolev exponents coming from both equations of (1.2) in the interior and on the boundary, the arbitrariness of the constant c leads to an additional difficulty in solving the Han-Li conjecture. Furthermore, they confirmed their conjecture if any of the following hypotheses is fulfilled:

4

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

(a) n ≥ 5 and ∂M admits at least one non-umbilic point; see [27]. (b) n ≥ 3 and (M, g0 ) is locally conformally flat with umbilic boundary ∂M ; see [28]. (c) n = 3; see [26]. Another natural motivation of the study of the Han-Li conjecture came from Escobar’s work [25] in 1990: Under what conditions can an Einstein manifold be conformally deformed to another Einstein manifold? A partial result of this problem was given in [25, Theorem 2.1], which demonstrates that an affirmative answer to the Han-Li conjecture is exactly a sufficient condition of this problem. As an application of the minimizing conformal metric obtained by Chen-Sun [18], Chen-Lai-Wang used this metric as a compatification for a Poincaré-Einstein manifold and then obtained a geometric inequality involving the generalized Yamabe constant and isoperimetric ratio; see [17, Theorem 7]. In [18], we studied the third case of Escobar problem and partially answered the Han-Li conjecture. We used subcritical approximations to find minimizers of E[u] with a suitable homogeneous constraint (equivalently, a quotient functional) and established that under some natural hypotheses on manifolds there exists a conformal metric with constant scalar curvature 1 and some positive constant boundary mean curvature. In particular, this constant allows to be very large. This provides us more evidences to the Han-Li conjecture. However, it is still difficult to solve this conjecture by seeking minimizers of the quotient functional, due to the reason that the Lagrange multiplier of Euler-Lagrange equation involves both the scalar curvature and the boundary mean curvature, and therefore it seems hard to obtain the arbitrariness of constant boundary mean curvature in the Han-Li conjecture. One way to get around the difficulty is to use the free functional (1.1), which was introduced by Z. C. Han and Y. Y. Li [27]. Their strategy was to find a non-trivial mountain pass critical point of I via the following Mountain Pass Lemma of Ambrosetti and Rabinowitz [6]. Mountain Pass Lemma (MPL). Let X be a Banach space and I ∈ C 1 (X; R). Suppose that I[0] = 0 and there exists 0 = u0 ∈ X such that I[u0 ] ≤ 0. Let Γ denote the set of continuous paths in X connecting 0 and u0 and define Imp := inf γ∈Γ supu∈γ I[u]. Suppose that Imp > 0 and that I satisfies the (P S) condition at the level Imp . Then Imp is a critical value of I. In general the (P S) condition for the associated functional I in (1.1) can not be satisfied due to the critical nonlinearities of I. However, Z. C. Han and Y. Y. Li proved that I satisfies a weak (P S) condition at any energy level below a certain threshold Sc , but it is enough to prove the existence result. Here Sc was given in [27], as well as will be defined in (2.6) below. The mountain pass structure of I can be verified through the following facts. Given any u ∈ H 1 (M, g0 ) with u+ ≡ 0, we define f (t) := I[tu] for t ∈ [0, ∞). It is not hard to show that f (0) = 0 and limt→∞ f (t) = −∞ and f (t) admits a unique maximum point t∗ (see also Section 2). Since Y (M, ∂M, [g0 ]) > 0, it is not hard to show that there exists

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0 > 0 such that I[u] ≥ 0 when u H 1 (M,g0 ) = r0 for small enough r0 > 0. For each u ∈ H 1 (M, g0 ) with u+ ≡ 0, there holds I[tu] < 0 for sufficiently large t > 0. We choose u0 = t0 u for some suitable t0 such that I[t0 u0 ] ≤ 0 and define Imp as in the statement of MPL. Indeed, with an improvement of the proof of the Han-Li’s [28, Lemma 1.2], we can verify that the (P S) condition for I below the mentioned energy level Sc is satisfied (see Lemma 2.1). Then, if there exists some u ∈ H 1 (M, g0 ) with u+ ≡ 0 such that max I[tu] < Sc ,

t∈[0,∞)

(1.4)

then 0 < 0 ≤ Imp < Sc . Hence, the existence of a nontrivial mountain pass critical point of I is guaranteed by the (P S) condition proved in Lemma 2.1 in Section 2. Consequently, in order to verify the Han-Li conjecture, the goal of this paper is to find some appropriate test functions satisfying (1.4). Since the functional I is conformally invariant, we sometimes work in conformal Fermi coordinates around a boundary point (see [30]) to simplify the analysis. Due to various geometric nature of the compact manifold, the construction of a test function has both a local and a global aspects. The developments listed in the beginning of this introduction deepen our understanding on how the test functions should be constructed. To state our main theorem, we let d = [(n − 2)/2] for n ≥ 3 and define as in [2,18]

Z = x0 ∈ ∂M ; lim sup dg0 (x, x0 )2−d |Wg0 (x)|g0 = 0 and x∈M,x→x0

 lim sup dg0 (x, x0 )1−d |˚ πg0 (x)|g0 = 0 ,

x∈∂M,x→x0

where Wg0 denotes the Weyl tensor of M , and πg0 , ˚ πg0 denote the second fundamental form on ∂M and its trace-free part, respectively. Combining with the aforementioned Han-Li’s existence results, we can solve Han-Li’s conjecture in the affirmative under the following natural geometric assumptions. Theorem 1.1. Let (M, g0 ) be a smooth compact Riemannian manifold of dimension n ≥ 3 with boundary ∂M and assume that Y (M, ∂M, [g0 ]) > 0, then the Han-Li conjecture is true, provided that one of the following hypotheses is satisfied: (i) (ii) (iii) (iv) (v)

3 ≤ n ≤ 7; n ≥ 8, Z is non-empty and M is spin; n ≥ 8, ∂M is umbilic and the Weyl tensor of M is non-zero at a boundary point; n ≥ 8, M is locally conformally flat with umbilic boundary; n ≥ 8, ∂M has at least one non-umbilic point.

Remark 1.2. For c = 0 in the Han-Li conjecture, Theorem 1.1 can directly follow from [24,12], as well as [4] from the viewpoint of a geometric flow. As mentioned earlier, the assertions (iv) and (v) are due to Z. C. Han-Y. Y. Li [28] and [27], respectively.

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Remark 1.3. Theorem 1.1 has established the Han-Li conjecture except for the case that n ≥ 8, ∂M is umbilic and the Weyl tensor of M vanishes on ∂M and has an interio non-zero point. We sketch the procedure of the proof of Theorem 1.1. When Z is non-empty, then by using the inverted coordinates near a point of Z, one can obtain an asymptotically flat manifold with non-compact boundary. It follows from the PMT in [3] that the mass of this manifold unless being isometric to the half space is strictly positive when 3 ≤ n ≤ 7. ¯(x ,) constructed in the first and third named authors’ Therefore the test function U 0 previous paper [18] (see also (3.5) and (4.6)) can be applied with some adaptation to all c ∈ R. In particular n = 3 will imply Z = ∂M , we give an alternative proof of [26] for n = 3. We mention that such ideas should be traced back to S. Brendle [10,11]. Thanks to [27], we can assume the boundary is umbilic for n ≥ 5. We can deduce from definition of Z that Z = ∂M when n = 5, and thus finish the proof of Han-Li conjecture for n = 5. Now we turn to the case Z = ∅ when n = 4, 6, 7. Since d = [(n − 2)/2] ≤ 2 in these cases, any boundary point x0 ∈ / Z will either be non-umbilic or have non-vanishing Weyl tensor at the same point. We construct some local test functions, which are modifications of ¯(x ,) , for the remaining cases in dimensions n = 4, 6 and prove (1.4) using them. Hence, U 0 we completely solve the Han-Li conjecture in dimensions 3 ≤ n ≤ 6, and dimension n = 7 except for the case that ∂M is umbilic and the Weyl tensor of M is non-zero at a boundary point. Furthermore, assuming that n ≥ 7, ∂M is umbilic and the Weyl tensor of M is non-zero at a boundary point, we succeed to prove the Han-Li conjecture. We need to point out that the case of dimensions n ≥ 7 is more subtle, partly due to the loss of the term log(ρ/) in deriving (1.4) as in dimensions n = 4, 6. Indeed, we combine the following two different types of test functions to achieve our goal. One is still to use the ¯(x ,) , we dig into the error correction term to obtain some new estimates. test function U 0 It boils down to prove an inequality (see (5.19)) for all dimensions n ≥ 7, which has independent interest. Possibly due to technical reasons, we are now only able to prove it when c is not less than a negative dimensional constant, which is given in Theorem 5.4. For the other range of c, we can construct another type of test function tailored to this case. In the work of F. Marques [31] and S. Almaraz [1] on the second case of Escobar problem, it is notable that their test functions have just one parameter and could not work here directly. Somewhat inspired by their ideas, with the use of three parameters κ0 , κ1 , κ2 , we explicitly construct a local test function defined in (5.29), still denoted by ¯(x ,) for simplicity. We define κ := (κ2 , κ1 , κ0 , 1) ∈ R4 . Through complex computations U 0 together with some delicate observations, we eventually arrive at ¯(x ,) ] = I[t∗ U ¯(x ,) ] = Sc + an κQκ |Rnanb |2 4 − bn |W |2 4 + h.o.t. max I[tU gx0 0 0

t∈[0,∞)

for small enough , where an , bn are two positive constants, Q is a real symmetric matrix, gx0 ∈ [g0 ], Rnanb is the Riemann curvature tensor, and W is the Weyl tensor of ∂M . See (5.35). Based on Lemma 5.7, our greatest challenging task is to find a vector κ

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above, such that κQκ < 0. After many attempts, finally we succeed in finding a “good” κ = −VS2 S1 (n −2)/2 with V = (2/3, −c/(n −2), 0, 1), where S1 , S2 are two non-singular matrices defined in (5.36) and (5.37). More details are referred to the end of Subsection 5.2. Consequently, these two test functions match so perfectly that we can completely solve the Han-Li conjecture in the aforementioned case. The main ingredients of our approach are as follows: Firstly, the test function constructed in our previous paper [18] and some necessary estimates therein play an important role in our proof. Another observation is that such estimates in [18] of the test function for positive c can be naturally extended to the ones for all real numbers c. Secondly, we used the precise expression (2.9) of the associated maximum t∗ with various test functions in the verification of condition (1.4). Thirdly, the conformal Fermi coordinate system [30] greatly simplifies the computations and plays the same role as the conformal normal coordinate system on the closed manifolds. Finally, similar to the resolution of the Yamabe problem, the construction of a global test function heavily relies on the PMT [3] with a non-compact boundary. In our forthcoming paper [16], we will employ a conformal curvature flow to study this constant scalar curvature and constant boundary curvature problem on a compact manifold. The paper is organized as follows. In Section 2, we collect some elementary properties of I and set up some notations. In Section 3, we establish the Han-Li conjecture when Z is non-empty and the PMT is valid, which together with Han-Li’s existence results covers a large class of the lower dimensions 3 ≤ n ≤ 7. See Theorem 3.4. In Section 4, the Han-Li conjecture is confirmed for the remaining cases of dimensions n = 4, 6. In Section 5, we employ two different types of test functions mainly according to the sign of c. In Subsection 5.1, we prove that problem (1.2) is solvable for any constant not less than a negative dimensional constant, under the condition that the boundary is umbilic and the Weyl tensor of M is non-zero at some boundary point. In Subsection 5.2, through constructing a local test function, we complete the proof of the Han-Li conjecture for the remaining case in dimension n = 7, as well as for n ≥ 8 with the same assumption. In Appendix A, we give detailed proofs of Lemma 2.1 and a technical result used in Subsection 5.1. Acknowledgments The authors would like to thank Professor Yan Yan Li for his fruitful discussions and encouragement. Part of this work was carried out during the first named author’s several visits to Rutgers University. He also would like to thank the Mathematics Department at Rutgers University for its hospitality and financial support. He was supported by NSFC (No.11771204), A Foundation for the Author of National Excellent Doctoral Dissertation of China (No.201417), the start-up grant of 2016 Deng Feng Program B (No.020314912201) at Nanjing University, the travel grants from AMS Fan fund and Hwa Ying foundation at Nanjing University. We thank the anonymous referee for careful reading and for helpful comments on the mathematics and the exposition.

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2. Preliminaries We collect some elementary properties of the standard bubble and I[u], as well as a criterion of seeking the (P S) condition for I, and introduce some notations at the end of this section. Let Rn+ = {y = (y 1 , · · · , y n ) ∈ Rn ; y n > 0} be the Euclidean half-space. Define Tc = −c/(n − 2) and W (y) = 

2−n 2

W (

−1

 y) =

 n−2 2

 2 + |y − Tc en |2

(2.1)

for any  > 0, where en is the unit direction vector in the n-th coordinate. Then W satisfies ⎧ n+2 ⎪ ⎨−ΔW = n(n − 2)Wn−2 , in Rn+ , n (2.2) ∂W ⎪ on Rn−1 . ⎩ n = (n − 2)Tc Wn−2 , ∂y Readers are referred to [29,25] for the classification theorem of all positive solutions to (2.2). Multiplying the equations in (2.2) by W , we integrate by parts to get 

 |∇W (y)| dy = n(n − 2) 2

Rn +

W (y)

2n n−2

 dy + c

Rn +

W (y)

2(n−1) n−2

dσ.

(2.3)

Rn−1

Throughout the paper, we set  A=



2n

W (y) n−2 dy

and

B=

Rn +

W (y)

2(n−1) n−2

dσ.

(2.4)

Rn−1

Then A and B are independent of . Define Sc :=

4(n − 1) n−2 

− 4c



 |∇W |2 dy − 4(n − 1)(n − 2)

Rn +

Rn + 2(n−1) n−2

W

dσ

Rn−1

=

4 n−2

2n

Wn−2 dy



 |∇W (y)|2 dy + 4(n − 2)

Rn +

2n

Wn−2 dy > 0

(2.5)

Rn +

by (2.3). Notice that Sc only depends on n, c by virtue of (2.3) and (2.5), more concretely, Sc = 8(n − 1)A +

4 cB. n−2

(2.6)

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Given any u ∈ H 1 (M, g0 ) with u+ ≡ 0, for t ∈ [0, ∞) we define f (t) := I[tu]



= t E[u] − 4(n − 1)(n − 2) 2

2n n−2

u+ dμg0 t

2n n−2



2(n−1)

− 4c

M

u+n−2 dσg0 t

2(n−1) n−2

.

∂M

We now claim that there exists a unique maximum point t∗ = t∗ (u) of f (t) in [0, ∞), namely max I[tu] =

t∈[0,∞)

1 E[u]t2∗ + 4(n − 2) n−1



2n

2n

u+n−2 dμg0 t∗n−2 ,

(2.7)

M

where t∗ satisfies n−2 E[u] = n(n − 2) 4(n − 1)



2n n−2

4 n−2

u+ dμg0 t∗ M

 +c

2(n−1)

2

u+n−2 dσg0 t∗n−2 .

(2.8)

∂M

To show this, we simplify f (t) as 2n

f (t) = at2 − θt n−2 + bt

2(n−1) n−2

,

t ∈ [0, ∞),

where a, θ > 0 and b ∈ R. Then f (0) = 0 and

lim f (t) = −∞.

t→∞

We compute n+2 n 2n n−2 2(n − 1) n−2 t t +b , n−2 n−2 4 2 2n(n + 2) n−2 2(n − 1)n n−2 f  (t) = 2a − θ t +b t . (n − 2)2 (n − 2)2

f  (t) = 2at − θ

Then there exists a unique t∗ ∈ (0, ∞) such that f  (t∗ ) = 0. Moreover, we obtain 4 2 8n 4(n − 1) n−2 t∗n−2 + b t∗ 2 2 (n − 2) (n − 2)

4 2n 2 n−2 θt∗ + 2a < 0. =− n−2 n−2

f  (t∗ ) = −θ

This implies that t∗ is the unique maximum point of f (t) in [0, ∞). ¯(x ,) to denote some qualified non-negative test function of For simplicity, we use U 0 (1.4), depending on small  > 0 and some suitable x0 ∈ ∂M , though it may vary in

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different sections. We also define n−2 ¯(x ,) ], A = n(n − 2) E[U E= 0 4(n − 1)





2n

¯ n−2 dμg , B = c U x0 (x0 ,) M

2(n−1)

¯ n−2 dσg . U x0 (x0 ,)

∂M

By (2.7), (2.8) and the assumption that Y (M, ∂M, [g0 ]) > 0, we obtain 2

t∗n−2 =

√

−B +

B 2 + 4EA , 2A

(2.9)

¯(x ,) . which is a positive constant depending on each U 0 The following lemma is a slightly improved version of Han-Li’s [28, Lemma 1.2] and provides the (P S) condition for I below the level Sc . Lemma 2.1 (Compactness). Suppose that Y (M, ∂M, [g0 ]) > 0. Let {ui ; i ∈ N} be a sequence of functions in H 1 (M, g0 ) satisfying I[ui ] → L < Sc and

v∈H

|I  [ui ](v)| →0 (M,g0 )\{0} v H 1 (M,g0 )

max 1

as i → ∞.

Then after passing to a subsequence, either (i) {ui } strongly converges in H 1 (M, g0 ) to some positive solution u of (1.2) or (ii) {ui } strongly converges to 0 in H 1 (M, g0 ). Traced back to Han-Li’s [28, Lemma 1.2], the original statement of Case (i) is that {ui } weakly converges in H 1 (M, g0 ) to some solution u of (1.2). Inspired by Han-Li’s idea, we present a proof in Appendix A to conclude Lemma 2.1. Convention. Let a, b, c, · · · range from 1 to n − 1 and i, j, k, · · · range from 1 to n. For small ρ > 0, let Bρ+ (0) = Bρ (0) ∩ Rn+ ;

∂ + Bρ+ (0) = ∂Bρ+ (0) ∩ Rn+ ;

Dρ (0) = ∂Bρ+ (0)\∂ + Bρ+ (0) and simplify Bρ+ (0), ∂ + Bρ+ (0), Dρ (0) as Bρ+ , ∂ + Bρ+ , Dρ without otherwise stated. Given any x0 ∈ ∂M and any integer N ≥ 1, it follows from F. Marques [30, Proposition 3.1] that there exists gx0 ∈ [g0 ] such that under gx0 -Fermi coordinates {(y 1 , · · · , y n ); y n > 0}, for small |y| there hold dμgx0 = (1 + O(|y|N ))dy and hgx0 = O(|y|N −1 ) near x0 .

(2.10)

+ In this paper, N = 2d + 2 is enough for our use. Let Ψx0 : B2ρ → M denote the map 1 induced by Fermi coordinates around x0 : set x = Ψx0 (y), y¯ = (y , · · · , y n−1 ) are geodesic normal coordinates on ∂M centered at x0 and expx0 (−y n νgx0 (x0 )) ∈ M for small y n > 0,

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then y = (¯ y , y n ) is so-called gx0 -Fermi coordinates around x0 . In particular, (gx0 )nn = 1 + and (gx0 )an = 0 in B2ρ . We write (gx0 )ij = exp(hij ) and set hij =

d  ∂ α hij α y + O(|y|d+1 ) := Hij + O(|y|d+1 ). α!

(2.11)

|α|=1

Denote by Ωρ = Ψx0 (Bρ+ ) the coordinate half-ball of radius ρ under the Fermi coordinates around x0 . Let ωn be the volume of the standard unit sphere S n . We adopt Einstein summation convention. 3. The case of non-empty Z with PMT In this section, we assume Z = ∅. For any x0 ∈ Z, Proposition 3.1 below shows that all derivatives of the metric around x0 vanish up to order d. For readers’ convenience, we present a proof of that, though it is probably well-known to the experts in this field, it can not be found in the literature. Proposition 3.1. For n ≥ 3 and x0 ∈ Z, let gx0 ∈ [g0 ] be the metric induced by the conformal Fermi coordinates around x0 , then there holds (gx0 )ij = δij + O(|y|d+1 ) near x0 . Proof. When n = 3, then d = 0 and the assertion is trivial. Now without loss of generality, we assume d ≥ 1. Under conformal gx0 -Fermi coordinates near x0 ∈ ∂M , it follows + from F. Marques [30, Proposition 3.1] that det gx0 = 1 + O(|y|2d+2 ) in B2ρ . If we write (gx0 )ij = exp(hij ) and hij =

d  ∂ α hij α y + O(|y|d+1 ) := Hij + O(|y|d+1 ), α!

|α|=1

+ then (gx0 )ij hij = O(|y|2d+2 ) in B2ρ . Moreover, the mean curvature hgx0 satisfies

hgx0 = −

1 1 (gx )ij ∂n (gx0 )ij = − ∂n (log det(gx0 )) = O(|x|2d+1 ). 2(n − 1) 0 2(n − 1)

Indeed we can prove a somewhat stronger result. πgx0 |gx0 = O(|y|k ) and hij = Hij + Claim. Suppose that |Wgx0 |gx0 = O(|y|k−1 ), |˚  k O(|y|k+1 ) near x0 , where Hij = |α|=1 (α!)−1 ∂ α hij (0)y α and 1 ≤ k ≤ d, then Hij = 0 near x0 . We prove it by induction on k. Suppose that k = 1. Since ∂n (gx0 )ab = − 12 (πgx0 )ab = O(|y|) near x0 ∈ ∂M , then ∂n (gx0 )ab (0) = 0, which implies ∂n hab (0) = 0. On the other hand, it follows from conformal Fermi coordinates that ∂c hab (0) = 0 for any 1 ≤ a, b, c ≤ n − 1. Thus we obtain

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

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Hab = 0 near x0 . One consequence of Fermi coordinates is Hin = 0 for 1 ≤ i ≤ n. Thus k = 1 is proved. Assume that the claim holds for k ≥ 1 and consider the case k + 1. Firstly from induction assumption, we obtain hij = Hij + O(|y|k+2 ),



where Hij =

|α|=k+1

∂ α hij (0) α y . α!

Notice that ∂n Hij =

¯  ∂ (α,n) hij (0) α¯ y , α ¯!

on {y n = 0},

(3.1)

|α|=k ¯

where α ¯ denotes a multi-index with components consisting of {1, 2, · · · , n − 1}. Also we α have ∂ ¯ ∂n (gx0 )ab = − 12 ∂ α¯ (πgx0 )ab = O(|y|) for all |¯ α| = k, by virtue of induction assumption for k + 1 that |˚ πgx0 |gx0 = O(|y|k+1 ) and the mean curvature hgx0 = O(|x|2d+1 ). Then ∂n Hab = 0 in a small neighborhood on ∂M of x0 . Since we are using Fermi coordinates, then all ∂n Hij = 0 on ∂M near x0 . + As in [11] (see also [2]), in B2ρ we define Aik = ∂i ∂m Hmk + ∂m ∂k Him − ΔHik −

1 ∂m ∂p Hmp δik n−1

and Zijkl = ∂i ∂k Hjl + ∂j ∂l Hik − ∂i ∂l Hjk − ∂j ∂k Hil +

1 (Ajl δik + Aik δjl − Ajk δil − Ail δjk ). n−2

(3.2)

Through direct computations (see also [5, formula (11) in the proof of Lemma 2.1]), we obtain the following expansion of the Weyl tensor Wijkl of gx0 Wijkl = −2Zijkl + O(|∂(h∂h)|2 ) + O(|y|k ).

(3.3)

By our assumption, it follows that |Wgx0 |gx0 = O(|y|k ) and |∂(h∂h)|2 = O(|y|2k ). This implies Zijkl = 0 near x0 , because (3.2) shows that Zijkl are only homogeneous polynomials of degree k − 1. Since we have already shown that ∂n Hij = 0 on {y n = 0}, then one can deduce that H = 0 from [12, Proposition 2.3]. Hence the induction step is complete. After finitely many steps, if k = d, we obtain the desired conclusion. 2 ¯(x ,) realizing (1.4). Denote by χ(y) = Next we shall construct some test function U 0 χ(|y|) a smooth cut-off function in Rn+ with χ = 1 in B1+ and χ = 0 in Rn+ \B2+ . For any ρ > 0, set χρ (y) = χ(|y|/ρ) for y ∈ Rn+ . Let G = Gx0 denote the Green’s function

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of the conformal Laplacian of metric gx0 with pole x0 ∈ ∂M , coupled with a boundary condition, namely ⎧ 4(n − 1) ⎪ ⎪ Δgx0 Gx0 + Rgx0 Gx0 = 0 , in M , ⎨− n−2 ⎪ 2 ∂Gx0 + h G = 0 , ⎪ on ∂M \{x0 } . ⎩ gx0 x0 n − 2 ∂νgx0

(3.4)

We assume that G is normalized such that limy→0 G(Ψx0 (y))|y|n−2 = 1, where Ψx0 is the conformal Fermi coordinates around x0 . In particular, G is positive in M \ {x0 }, see [4, Appendix B]. Define ¯(x ,) = [χρ W ] ◦ Ψ−1 +  n−2 2 (1 − χ ) ◦ Ψ−1 G U ρ x0 x0 0

(3.5)

¯(x ,) ]. for all c ∈ R. We shall calculate E[U 0 Before doing any calculation, let us point out a simple but vital estimate for W is that for all c ∈ R, there exists C = C(Tc , n) > 0 such that C −1 

n−2 2

( + |y|)2−n ≤ W (y) ≤ C

n−2 2

( + |y|)2−n , y ∈ Rn+ .

(3.6)

For a simple proof, it is not hard to show (3.6) for c ≥ 0. When c < 0, i.e. Tc > 0, it suffices to show 2 + |y − Tc en |2 ≥ C −1 ( + |y|)2 for some C = C(Tc , n) > 0. To see this, given any δ ∈ (0, 1), by Young’s inequality we have 2 + |y − Tc en |2 = 2 (1 + Tc2 ) + |y|2 − 2Tc y n  y |2 + (1 − δ)|y n |2 . ≥ 2 [1 + Tc2 (1 − δ −1 )] + |¯ Then by choosing δ = 2Tc2 /(1 + 2Tc2 ), we obtain 2 + |y − Tc en |2 ≥ min



1 1 2 , 1+2Tc2



(2 + |y|2 ) ≥ min



1 1 4 , 2(1+2Tc2 )



( + |y|)2

as desired. ¯(x ,) in (3.5), we shall separate it by To estimate the energy E of U 0  ¯(x ,) ; Ωρ ] = E[U 0 Ωρ

4(n − 1) ¯ 2 2 ¯ |∇U(x0 ,) |gx0 + Rgx0 U(x0 ,) dμgx0 n−2 

+ 2(n − 1) Ωρ ∩∂M

¯2 hgx0 U (x0 ,) dσgx0

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

14

and  ¯(x ,) ; M \Ωρ ] = E[U 0 M \Ωρ



4(n − 1) ¯ ¯2 |∇U(x0 ,) |2gx0 + Rgx0 U (x0 ,) dμgx0 n−2  ¯2 hgx0 U (x0 ,) dσgx0 .

+ 2(n − 1) ∂M \Ωρ

¯(x ,) varies, we keep the meanings of Throughout the paper, though the test function U 0 ¯ ¯ E[U(x0 ,) ; Ωρ ] and E[U(x0 ,) ; M \Ωρ ] unchanged for simplicity. We first mention that the estimate (3.6) enables us to extend some existing estimates in [18] for all non-positive constants c. Under conformal gx0 -Fermi coordinates around x0 ∈ ∂M , by (2.10) we have   ¯(x ,) ; Ωρ ] = E[U 0 Bρ+

  4(n − 1) |∇W |2gx0 + Rgx0 W2 dy + 2(n − 1) hgx0 W2 dσ n−2 Dρ

+ Cn−2 ρ2d+4−n   4(n − 1) = |∇W |2gx0 dy + (I1 + I2 )dy + Cn−2 ρ2d+4−n . n−2 Bρ+

Bρ+

Here, I1 = −

4(n − 1) ∂i W ∂k W hik + W2 ∂i ∂k hik n−2

= ∂i (W2 ∂k hik − 2W ∂k W hik ) + 2(W ∂i ∂k W − = ∂i (W2 ∂k hik − ∂k W2 hik ) +

n ∂i W ∂k W )hik n−2

2 n (W ΔW − |∇W |2 )trh, n n−2

where the third identity follows from (see [18, Equation (5.3)]) W ∂i ∂k W −

n 1 ∂i W ∂k W = n−2 n

 W ΔW −

 n |∇W |2 δik in Rn+ n−2

which equivalently states that W (y)4/(n−2) |dy|2 is Einstein, and I2 =

4(n − 1) ik (gx0 − δik + hik )∂i W ∂k W + (Rgx0 − ∂i ∂k hik )W2 n−2

Since trh = O(|y|2d+2 ) in Bρ+ , we estimate 

 I1 dy ≤ Bρ+

(W2 ∂k hik − ∂k W2 hik ) ∂ + Bρ+

yi dσ + Cn−2 ρ2d+4−n . |y|

(3.7)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

15

On the other hand, it follows from Proposition 3.1 that |gxik0 − δik + hik | ≤ C|y|2d+2 , and the expansion of scalar curvature in [18, Proposition 5.6] implies that |Rgx0 − ∂i ∂k hik | ≤ C|y|2d by taking Hij ≡ 0 there. Therefore, we are able to obtain  I2 dy ≤ Cn−2 ρ2d+4−n . Bρ+

Integrating by parts for the first term in (3.7) by using (2.2) and combining the above estimates for I1 and I2 lead to   2(n−1) 2n n−2 ¯(x ,) ; Ωρ ] ≤ 4n(n − 1) E[U W dy − 4(n − 1)T W n−2 dσ  c 0 Bρ+

+

4(n − 1) n−2

Dρ



i

∂i W W ∂ + Bρ+

y dσ + |y|



(W2 ∂j hij − ∂j W2 hij )

∂ + Bρ+

yi dσ |y|

n−2 2d+4−n

+ C

ρ

for 0 < 2 < ρ < ρ0 ≤ 1, where ρ0 , C are positive constants depending only on n, Tc , gx0 . On the other hand, using [18, (5.48)] together with the estimate that h = O(|y|d+1 ), we have ¯(x ,) ; M \Ωρ ] E[U 0  4(n − 1) ≤ n−2



−∂i W W + ∂j W W hij − 

n−2 2

 yi dσ |y|

(W ∂i G − G∂i W )

∂ + Bρ+

+ Cρ2d+4−n | log ρ|2 n−2 + Cρ2−n n−1 . The derivation of the above inequality is quite subtle. Basic idea behind this is to use integration by parts with the key fact that − 4(n−1) n−2 Δgx0 G + Rgx0 G = 0 in M \Ωρ and ¯ ¯ replace one U(x0 ,) in the integrand by U(x0 ,) − (n−2)/2 G, readers are referred to [18, pp. 43-46] for details. Consequently, we combine the above two estimates to obtain   2(n−1) 2n n−2 ¯(x ,) ] ≤ 4n(n − 1) E[U W dy − 4(n − 1)T W n−2 dσ  c 0 

Bρ+

(W2 ∂j hij +

+ ∂ + Bρ+

−

Dρ

4(n − 1) n−2  2 n−2

n yi ∂j W2 hij ) dσ n−2 |y|

 (W ∂i G − G∂i W ) ∂ + Bρ+

+ Cρ2d+4−n | log ρ|2 n−2 + Cn−1 ρ2−n .

yi dσ |y|

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16

By using the estimate of G in [18, (5.31)], we obtain (see also [18, (5.50)])  (W2 ∂j hij + ∂ + Bρ+

n yi ∂j W2 hij ) dσ n−2 |y|

4(n − 1) n−2 −  2 n−2

 (W ∂i G − G∂i W ) ∂ + Bρ+

yi dσ |y|

≤ − n−2 I(x0 , ρ) + Cn−1 ρ1−n , where I(x0 , ρ) is defined in [18, p. 34], more explicitly,  I (x0 , ρ) = − ∂ + Bρ+

  yi dσ |y|2−2n |y|2 ∂j hij − 2ny j hij |y|

4(n − 1) + n−2





|y|

2−n

∂i G − G∂i |y|

2−n

∂ + Bρ+

yi |y|

 dσ.

Hence, collecting the above calculations, we conclude that  ¯(x ,) ] ≤ 4n(n − 1) E[U 0

2n n−2

W

 dy − 4(n − 1)Tc

Bρ+

2(n−1) n−2

W

dσ

Dρ

− n−2 I(x0 , ρ) + Cρ2d+4−n | log ρ|2 n−2 + Cn−1 ρ2−n

(3.8)

for all c ∈ R. We shall use a notion of a mass associated to manifolds with boundary. Definition 3.2. Let (N, g) be a Riemannian manifold with a non-compact boundary ∂N . We say that N is asymptotically flat with order p > 0, if there exist a compact set N0 ⊂ N and a diffeomorphism F : N \N0 → Rn+ \B1+ (0) such that, in the coordinate chart defined by F (called asymptotic coordinates of N ), there holds 2 |gij (y) − δij | + |y||∂k gij (y)| + |y|2 |∂kl gij (y)| = O(|y|−p ), as |y| → ∞,

where i, j, k, l = 1, · · · , n, B1+ (0) = B1 (0) ∩ Rn+ . Provided that the following limit ⎡ ⎢ m(g) := lim ⎣



R→∞

{y∈Rn + ; |y|=R}

n  i,j=1

(gij,j − gjj,i )

yi dσ + |y|



{y∈Rn−1 ; |y|=R}

⎤ n−1  a=1

gna

ya ⎥ dσ ⎦ |y|

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

17

exists, we call it the mass of (N, g). The following positive mass type conjecture has been verified by Almaraz-Barbosa-de Lima [3], provided that 3 ≤ n ≤ 7 or n ≥ 8 and N is spin. Conjecture (Positive mass with a non-compact boundary). If (N, g) is asymptotically flat with order p > (n − 2)/2 and Rg ≥ 0, hg ≥ 0, then m(g) ≥ 0, equality holds if and only if (N, g) is isometric to (Rn+ , gRn ). 4/(n−2)

Proposition 3.3. Suppose that m(¯ gx0 ) > 0 for x0 ∈ Z, where g¯x0 = Gx0 gx0 is the ¯ metric on M \ {x0 } (see [18, Proposition 5.14]). Then U(x0 ,) satisfies (1.4) when  is sufficiently small. Proof. By (2.10) we have 



2n

¯ n−2 dμg = U x0 (x0 ,) M

n −n n−2 dμ (W ◦ Ψ−1 ) gx0 + O( ρ x0 ) 2n

Ωρ



2n

Wn−2 dy + O(n ρ2d+2−n ) + O(n ρ−n )

= Bρ+

= A + O(n ρ−n ).

(3.9)

Similarly we have 



2(n−1)

(W ◦ Ψ−1 x0 )

¯ n−2 dσg = U x0 (x0 ,)

2(n−1) n−2

dσgx0 + O(n−1 ρ1−n )

∂Ωρ ∩∂M

∂M



2(n−1) n−2

W

=

dσ + O(n−1 ρ2d+3−n ) + O(n−1 ρ1−n )

Dρ

= B + O(n−1 ρ1−n ).

(3.10)

Moreover, since m(¯ gx0 ) > 0, we obtain from [12, Proposition 4.3] that I(x0 , ρ) > C˜ > 0 for sufficiently small ρ > 0. From this and (3.8) with  small enough, we have  C˜ n−2 C˜ n−2 ¯ E[U(x0 ,) ] ≤ n(n − 2)A + cB −  = |∇W |2 dy − n−2 , (3.11) 4(n − 1) 2 2 Rn +

where the last identity follows from (2.3). ¯(x ,) . Thus, by Denote by t∗ > 0 the unique maximum point of (2.8) with u = U 0 (2.9) together with (3.9), (3.10), (3.11) and (2.3), we conclude that t∗ ∈ (0, 1) when  is sufficiently small. Therefore, for 0 <   ρ < ρ0 with small enough ρ0 , we obtain

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

18

¯(x ,) ] = I[t∗ U 0

1 ¯(x ,) ]t2∗ + 4(n − 2) E[U 0 n−1

4 < n−2





2n

M



|∇W (y)|2 dy + 4(n − 2) Rn +

2n

¯(x ,) ) n−2 dμg t∗n−2 (U x0 0 2n

W n−2 dy = Sc ,

Rn +

where the last inequality follows from (3.9), (3.11), Y (M, ∂M, [g0 ]) > 0 and t∗ ∈ (0, 1), ¯(x ,) satisfies (1.4). 2 namely, this U 0 It follows from [27] that if n ≥ 5 and ∂M admits one non-umbilic point, then the Han-Li conjecture is true. From this, a direct consequence of Proposition 3.3 is the following Theorem 3.4. The Han-Li conjecture is true, provided that one of the following assumptions is fulfilled: (i) n = 3, 5; (ii) n = 4 and ∂M has an umbilic point; (iii) n = 6, 7, the boundary is umbilic and the Weyl tensor of M vanishes at some boundary point; (iv) Z is non-empty and M is spin or M is locally conformally flat with umbilic boundary. Proof. Thanks to [27], we can assume ∂M is umbilic for n ≥ 5. Based on Proposition 3.3, it reduces to showing that there exists x0 ∈ Z such that m(¯ gx0 ) > 0. We would like to emphasize that the positive mass type theorem has been verified in [3, Theorem 1.3] and [24] for all cases (i)-(iv). We will verify these item by item. (i) If n = 3, then d = 0 and Z = ∂M . If n = 5, then d = 1 and we can deduce from definition of Z that Z = ∂M . (ii) If n = 4 and let x0 ∈ ∂M be an umbilic point, then d = 1 and lim sup dg0 (x, x0 )2−d |Wg0 (x)|g0 = lim sup dg0 (x, x0 )1−d |˚ πg0 (x)|g0 = 0. x→x0

x→x0

Thus, x0 ∈ Z. (iii) If n = 6, 7, then d = 2. Combining this and the assumption that the Weyl tensor of M is zero at some x0 ∈ ∂M , we have x0 ∈ Z. (iv) In this case, Z = ∅ and the positive mass type theorem also comes into play as in the previous three items. 2 Again from [27] together with Theorem 3.4, the remaining cases in lower dimension 3 ≤ n ≤ 7 are • n = 4 and ∂M admits at least one non-umbilic point;

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• n = 6, 7, ∂M is umbilic and the Weyl tensor Wg0 of M does not vanish everywhere on ∂M . 4. Remaining cases in dimensions four and six To solve these remaining cases, one needs an appropriate correction term to obtain a ¯(x ,) . As mentioned earlier, the test function constructed in [18] better test function U 0 will play an important role in solving these cases. For reader’s convenience, we collect some useful estimates therein and extend these estimates to all non-positive constant c with the help of the estimate (3.6). Recall that Hij is defined in (2.11). We can assume d ≥ 1 by virtue of Theorem 3.4. By [18, p. 31], see also [12, Appendix] and [13], one can prove the existence of a vector field V satisfying ⎧ 2n   n ⎪ ⎨  ∂i Wn−2 χρ Hij − ∂i Vj − ∂j Vi + 2 (divV )δij = 0, in Rn+ , n (4.1) i=1 ⎪ ⎩∂n Va = Vn = 0, on Rn−1 , where 1 ≤ i, j ≤ n, 1 ≤ a ≤ n − 1. Moreover, there holds |∂ β V (y)| ≤ C(n, Tc , |β|)

n−1 

d 

|∂ α hab |( + |y|)|α|+1−|β| , ∀ β.

(4.2)

a,b=1 |α|=1

We define symmetric trace-free 2-tensors S and T in Rn+ by Sij = ∂i Vj + ∂j Vi −

2 divV δij n

and

T =H −S.

(4.3)

Denote ψ = ∂i W Vi +

n−2 W divV. 2n

(4.4)

+ Thus, using (4.2) and the expression (2.1) of W , in B2ρ we have

|ψ(y)| ≤ C(n, Tc )

n−2 2

n−1 

d 

|∂ α hab |( + |y|)|α|+2−n .

(4.5)

a,b=1 |α|=1

We define a test function as ¯(x ,) (x) = [χρ (W + ψ)] ◦ Ψ−1 (x), U x0 0

(4.6)

where ψ is defined in (4.4). By definition (1.1) of the functional I, we need the expansions of the volumes of M and ∂M under conformal Fermi coordinates around x0 ∈ ∂M .

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

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Lemma 4.1. If 0 <   ρ < ρ0 for some sufficiently small ρ0 , there holds 

2n

¯ n−2 dμg = A + A2 + O(3 ) + O(n ρ−n ), U x0 (x0 ,)

M

where n(n + 2)2 A2 = n−2



4

Wn−2 ψ 2 dy = O(2 ). Bρ+

Proof. By (2.10) we obtain 



2n

¯ n−2 dμg = U x0 (x0 ,) M

n −n n−2 dμ ((W + ψ) ◦ Ψ−1 ) gx0 + O( ρ x0 ) 2n

Ωρ



(W + ψ) n−2 dy + O(n ρN −n ) + O(n ρ−n ). 2n

= Bρ+

+ It follows from (4.5) and (3.6) that |ψ(y)| ≤ C( + |y|)W (y) in B2ρ . Observe that



2n

(W + ψ) n−2 dy Bρ+



2n

Wn−2 dy +

= Bρ+

2n n−2



n+2

Wn−2 ψdy + Bρ+

n(n + 2) (n − 2)2



4

Wn−2 ψ 2 dy + O(3 ) Bρ+

:= A0 + A1 + A2 + O(3 ). It is easy to show A0 = A + O(n ρ−n ).

(4.7)

By definition (4.4) of ψ, we have n+2 2n 2n Wn−2 ψ = div(Wn−2 V ). n−2

Since Vn = 0 on Dρ , an integration by parts together with (4.2) gives 

2n



div(Wn−2 V )dy =

A1 = Bρ+

2n

Wn−2 Vi ∂ + Bρ+

yi dσ = O(n ρ1−n ). |y|

By collecting all the above estimates, the desired assertion follows. 2

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Lemma 4.2. If 0 <   ρ < ρ0 for some sufficiently small ρ0 , there holds  2(n−1) ¯ n−2 dσg = B + B1 + B2 + O(3 ) + O(n−1 ρ1−n ), U x0 (x0 ,) ∂M

where 2(n − 1) B1 = n−2



n n−2

W

n(n − 1) B2 = (n − 2)2

ψdσ,

Dρ

Proof. By (2.10) we have  2(n−1) ¯ n−2 dσg = U x0 (x0 ,)



((W + ψ) ◦ Ψ−1 x0 )



2

Wn−2 ψ 2 dσ = O(2 ). Dρ

2(n−1) n−2

dσg0 + O(n−1 ρ1−n )

∂Ωρ ∩∂M

∂M



=

(W + ψ)

2(n−1) n−2

dσ + O(n−1 ρN −n+1 ) + O(n−1 ρ1−n ).

Dρ

Notice that

 (W + ψ)

2(n−1) n−2



2(n−1) n−2

dσ =

Dρ

W Dρ

(n − 1)n + (n − 2)2

dσ + 

2(n − 1) n−2 2 n−2

W



n

Wn−2 ψdσ Dρ

ψ 2 dσ + O(3 )

Dρ

= B + B1 + B2 + O(3 ) + O(n−1 ρ1−n ). Hence we combine these two estimates to obtain the desired estimate. 2 We first assume that n = 4 and ∂M admits at least a non-umbilic point x0 , then it follows from F. Marques [30, Lemma 2.2] that there exists gx0 ∈ [g0 ] such that under gx0 -Fermi coordinates around x0 , it has the following expansion near x0 : (gx0 )ab = δab − 2πab y n + O(|y|2 ).

(4.8)

hab = −2πab y n + O(|y|2 ).

(4.9)

Then (4.8) implies

Since d = 1 when n = 4, (4.5) gives |ψ(y)| ≤ C

n−1 

|πab |( + |y|)W (y)

a,b=1

¯(x ,) is non-negative. By choosing ρ sufficiently small, U 0

+ in B2ρ .

(4.10)

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22

¯(x ,) in (4.6), we can estimate E[U ¯(x ,) ; M \Ωρ ], the energy By the expression of U 0 0 + outside Ωρ = Ψx0 (Bρ ). A direct calculation shows |∇W |2 = n−2 (n − 2)2

|¯ y |2 + |y n − Tc |2 . (2 + |y − Tc en |2 )n

(4.11)

Notice that   ¯(x ,) |2g ≤ C|∇U ¯(x ,) |2 ≤ C |∇χρ |2 (W + ψ)2 + χ2ρ |∇(W + ψ)|2 , |∇U 0 0 x0 it yields 

|∇χρ |2 (W + ψ)2 dy ≤ Cρ−2

+ B2ρ \Bρ+

 W2 dy ≤ Cn−2 ρ2−n ,

+ B2ρ \Bρ+





χ2ρ |∇(W + ψ)|2 dy ≤ C + B2ρ \Bρ+

|∇W |2 dy ≤ Cn−2 ρ2−n . + B2ρ \Bρ+

Similarly, we have 

 ¯2 Rgx0 U (x0 ,) dμgx0 ≤ C Ω2ρ \Ωρ

(W + ψ)2 dy ≤ Cn−2 ρ4−n , + B2ρ \Bρ+





¯2 hgx0 U (x0 ,) dσgx0 ≤ C Ψx0 (D2ρ \Dρ )

|¯ y |N −1 (W + ψ)2 dσ

D2ρ \Dρ

≤ Cn−2 ρ2+N −n . Hence, we obtain ¯(x ,) ; M \Ωρ ] ≤ Cn−2 ρ2−n . E[U 0

(4.12)

¯(x ,) ; Ωρ ], it follows from [18, Proposition 5.7] that with a suffiFor the energy E[U 0 ciently small ρ0 > 0, there holds  Bρ+

4(n − 1) |∇(W + ψ)|2gx0 + Rgx0 (W + ψ)2 dy n−2  + 2(n − 1)

hgx0 (W + ψ)2 dσ

Dρ



≤ 4n(n − 1) Bρ+

4 n−2

W

 W2

 n+2 2 ψ dy + n−2

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

4(n − 1) + n−2



yi ∂i W W dσ + |y|

∂ + Bρ+



− 4(n − 1)Tc

 (W2 ∂j hij − ∂j W2 hij )

∂ + Bρ+

2  Wn−2 W2 + 2W ψ +

23

yi dσ |y|

 n n−2 2 ψ2 − W2 Snn dσ 2 n−2 8(n − 1)

Dρ

 n−1 d   1 |∂ α hab |2 n−2 ( + |y|)2|α|+2−2n dy − λ∗ 2 a,b=1 |α|=1

+C

n−1 

d 

Bρ+

|∂ α hab |n−2 ρ|α|+2−n + Cn−2 ρ2d+4−n

(4.13)

a,b=1 |α|=1

for 0 < 2 < ρ < ρ0 ≤ 1, where λ∗ , ρ0 , C are positive constants depending only on n, Tc , gx0 . By (4.9) and (4.2), estimate (4.13) with n = 4 actually implies ⎛ n−2 ¯(x ,) ; Ωρ ] ≤ n(n − 2)A + c ⎜ E[U ⎝B + 2 0 4(n − 1) 1 − λ∗ |πgx0 |2gx0 2 2



⎞ n ⎟ Wn−2 ψdσ ⎠ + O(2 ρ2−n )

Dρ



( + |y|)4−2n dy,

(4.14)

Bρ+

n−1 2 where |πgx0 |2gx0 = a,b=1 πab > 0 at x0 by (4.9) and the fact that hgx0 = 0 at this non-umbilic point x0 . Notice that 

ρ ( + |y|)4−2n dy = O(log ). 

Bρ+

Therefore, combining (4.12) and (4.14), we can estimate ⎛ n−2 ¯(x ,) ] ≤ n(n − 2)A + c ⎜ E[U ⎝B + 2 0 4(n − 1) 1 − λ∗ |πgx0 |2gx0 2 2





⎞ n ⎟ Wn−2 ψdσ ⎠

Dρ

( + |y|)−4 dy + O(2 ρ−2 ).

(4.15)

Bρ+

Next by definition (1.1) of the functional I, we need the expansions of the volumes of M and ∂M under conformal Fermi coordinates around x0 ∈ ∂M . Therefore, putting (4.15), Lemmas 4.1 and 4.2 together, we obtain A =n(n − 2)A + O(2 ),

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

24

B = cB + cB1 + O(2 ), E ≤ n(n − 2)A + cB +

n−2 n−1 cB1

1 − λ∗ |πgx0 |2gx0 2 2



( + |y|)−4 dy + O(2 ρ−2 )

Bρ+

and B1 = O(), we obtain

4n(n − 2)2 A cB1 B + 4EA ≤ (cB + 2n(n − 2)A) + 2cB + n−1  − 2n(n − 2)A2 λ∗ |πgx0 |2gx0 ( + |y|)−4 dy + O(2 ρ−2 ). 2

2

Bρ+

Notice that 2n(n − 2)A + cB > 0 for all c ∈ R because of (2.3). Thus, (2.9) implies 2

c B1 (n − 1)(2n(n − 2)A + cB)  λ∗ |πgx0 |2gx0 2  − ( + |y|)−4 dy + O(2 ρ−2 ) 2(2n(n − 2)A + cB)

t∗n−2 ≤ 1 −

Bρ+

:= 1 +

2 ˜ B1 − C ∗ 2 log(ρ/) + O(2 ρ−2 ), n−2

(4.16)

where C ∗ is a positive constant depending on c, λ∗ , |πgx0 |2gx0 and ˜1 = B

−c n−2 B1 = O(). 2(n − 1) 2n(n − 2)A + cB

Plugging (4.16), (4.15) and (2.6) into (2.7) and using (2.6) and the assumption that Y (M, ∂M, [g0 ]) > 0, we conclude that ¯(x ,) ] max I[tU 0

t∈[0,∞)

=

1 ¯(x ,) ] + 4(n − 2) E[t∗ U 0 n−1



2n

¯(x ,) ) n−2 dμg (t∗ U x0 0 M

4c ˜1 − C ˜1 (nA + 1 cB) + 8nAB ˜ 2 log(ρ/) + O(2 ρ−2 ) B1 + 8B n−1 n−2 ˜ 2 log(ρ/) + O(2 ρ−2 ) = Sc − C ≤ Sc +

< Sc , ˜ Tc , |πg |2 ) > 0, when 0 <   ρ < ρ0 for some sufficiently small for some C˜ = C(n, x0 gx0 ˜1 . ρ0 , where the second identity follows from definitions of B1 and B

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

25

In the following, we turn to the dimensions n = 6, 7. Recall that when n = 6, 7, the remaining case is that the Weyl tensor of M is non-zero everywhere on the umbilic boundary ∂M . Indeed, we can achieve our goal when relaxing the assumption a little: Assume that n ≥ 6, ∂M is umbilic and the Weyl tensor Wg0 of M is non-zero at some x0 ∈ ∂M , then at x0 there hold ∂ α hab = 0 for all 0 ≤ |α| ≤ 1 and

n−1 



|∂ α hab |2 > 0.

(4.17)

a,b=1 |α|=2

Here the above second assertion in (4.17) follows from a contradiction argument by using (3.3) and the assumption that Wg0 (x0 ) = 0. Thus, it follows from (4.5) and (4.17) that + in B2ρ there holds |ψ(y)| ≤ C( + |y|)2 W (y).

(4.18)

In dimension n = 6, a local test function is also enough to our use and the argument can be similarly done as the one in dimension four. Suppose that n = 6, we still adopt ¯(x ,) in (4.6) except for replacing n = 4 by n = 6. Similarly, the same test function U 0 with the above refined estimate (4.18), we obtain 

2n

¯ n−2 dμg = A + O(4 ) + O(6 ρ−6 ), U x0 (x0 ,)

(4.19)

M



2(n−1)

¯ n−2 dσg = B + B1 + O(4 ) + O(6 ρ−6 ), U x0 (x0 ,)

(4.20)

∂M

where B1 =

2(n − 1) n−2



n

Wn−2 ψdσ Dρ

and then |B1 | ≤ C(n, Tc )2 . Similar to (4.15), by using (4.12) together with n = 6, (4.19), (4.20) and (4.13), we estimate n−2 ¯(x ,) ] E[U 0 4(n − 1)

⎛

⎜ ≤ n(n − 2)A + c ⎝B + 2



⎞ n n−2

W

⎟ ψdσ ⎠

Dρ

  n−1  1 |∂ α hab |2 4 ( + |y|)−6 dy + O(4 ρ−4 ). − λ∗ 2 |α|=2 a,b=1

Bρ+

(4.21)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

26

Notice that 

ρ ( + |y|)−6 dy = O(log ). 

Bρ+

¯(x ,) , one can proceed as the case n = 4 and find With these estimates and (2.8) for this U 0 2 ¯(x ,) ] are canceled out, eventually prove that U ¯(x ,) that all  -terms involved in I[t∗ U 0 0 satisfies (1.4). 5. n ≥ 7, umbilic boundary and non-zero Weyl tensor at a boundary point In this section, we assume that n ≥ 7, ∂M is umbilic and the Weyl tensor of M is ¯(x ,) defined non-zero at some x0 ∈ ∂M . In Subsection 5.1, we adopt the test function U 0 in [18] to prove the Han-Li conjecture in addition that the constant c is not less than a negative dimension constant. In Subsection 5.2, we explicitly construct a local test function to prove the Han-Li conjecture for all non-positive constant c. 5.1. Positive constant boundary mean curvature In this subsection, we will use the following test function constructed in [18] ¯(x ,) = [χρ (W + ψ)] ◦ Ψ−1 +  n−2 2 (1 − χ ) ◦ Ψ−1 G U ρ x0 x0 0

(5.1)

for all c ∈ R, where ψ is given in (4.4) and G is the Green’s function defined in (3.4). In contrast with those cases in dimensions four and six, due to the loss of the log ||-term from the first correction term in the following estimate of E, as well as of I, the situation becomes more complicated. We start with some elementary calculations and temporarily admit the following expansions: A = n(n − 2)A + n(n − 2)A˜ + O(6 ),

(5.2)

˜ + O(6 ), B = cB + cB

(5.3)

˜ + O(5 | log |), E = n(n − 2)A + cB + E

(5.4)

˜ ≤ C4 and |B| ˜ ≤ C2 and E ˜ ≤ C2 . where |A| A direct computation together with (2.3) yields ˜ + 4n(n − 2)AE ˜ B 2 + 4EA = (2n(n − 2)A + cB)2 + 2c2 B B ˜ 2 + 4n(n − 2)A(n(n ˜ + c2 B − 2)A + cB) + o(4 ) and

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

−B+

$

B 2 + 4EA

= 2n(n − 2)A + 2n(n − 2) −

27

˜ − cAB ˜+ AE

˜2 c2 B 4n(n−2)

+ [n(n − 2)A + cB]A˜

2n(n − 2)A + cB

˜ + 2n(n − 2)AE) ˜ 2 1 (c2 B B + o(4 ). 2 (2n(n − 2)A + cB)3

It follows from the above estimates and (2.9) that 2 n−2

t∗

˜ − cB ˜ E + =1+ 2n(n − 2)A + cB −

˜2 c2 B 4n(n−2)A

− n(n − 2)A˜

2n(n − 2)A + cB

˜ + 2n(n − 2)AE) ˜ 2 (c2 B B + o(4 ) 4n(n − 2)A(2n(n − 2)A + cB)3

:= 1 +

2 T∗ + o(4 ), n−2

whence T∗ ≤ C2 . Thus, we obtain t∗ = 1 + T ∗ +

n−4 2 T + o(4 ). 2(n − 2) ∗

Therefore, by (5.2), (5.4), (5.5) and (2.7), we conclude that ¯(x ,) ] max I[tU 0

0≤t<∞

=

1 ¯(x ,) ] + 4(n − 2) E[t∗ U 0 n−1



2n

¯(x ,) ) n−2 dμg (t∗ U x0 0 M

 4  2n − 6 2 ˜ + o(4 )) T∗ + o(4 ) (n(n − 2)A + cB + E = 1 + 2T∗ + n−2 n−2   2n 2n(n − 1) 2 4 T∗ + + 4(n − 2) 1 + T + o( ) (A + A˜ + O(5 )) n−2 (n − 2)2 ∗ 8 4 ˜ 8 ˜ = Sc + E+ (2n(n − 2)A + cB)T∗ + 4(n − 2)A˜ + ET∗ n−2 n−2 n−2

8(n − 3) 8n(n − 1) A T∗2 + o(4 ) + (n(n − 2)A + cB) + (n − 2)2 n−2 4(n − 1) ˜ ˜ − 4(n − 1)(n − 2)A˜ E − 4cB n−2 ˜ + 2n(n − 2)AE) ˜ 2 ˜ E ˜ − cB) ˜ ˜2 (c2 B B 4E( c2 B − + + n(n − 2)A n(n − 2)A(2n(n − 2)A + cB)2 2n(n − 2)A + cB

8n(n − 1) 8(n − 3) + A T∗2 + o(4 ) (n(n − 2)A + cB) + (n − 2)2 n−2

= Sc +

(5.5)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

28

4(n − 1) ˜ ˜ − 4(n − 1)(n − 2)A˜ E − 4cB n−2 4(n(n − 2)A + cB) ˜ ˜ 2 + o(4 ) (E − cB) + (2n(n − 2)A + cB)2

= Sc +

+ [2(n − 3)(n(n − 2)A + cB) + 2n(n − 1)(n − 2)A] = Sc +

˜ − cB) ˜ 2 (E (2n(n − 2)A + cB)2

˜ ˜ 2 4(n − 1) ˜ ˜ − 4(n − 1)(n − 2)A˜ + 2(n − 1)(E − cB) + o(4 ). (5.6) E − 4cB n−2 2n(n − 2)A + cB

We are now in a position to verify (5.2)-(5.4). Lemma 5.1. Assume that n ≥ 7 and ∂M is umbilic, then if 0 <   ρ < ρ0 for some sufficiently small ρ0 , the volume of M has the expansion 

2n

¯ n−2 dμg = A + A2 + O(ρ−6 6 ), U x0 (x0 ,)

(5.7)

M

where A2 =

n(n + 2) (n − 2)2



4

Wn−2 ψ 2 dy Bρ+

with |A2 | ≤ C4 . Proof. The proof follows from the same lines in Lemma 4.1 but with the estimate (4.18). 2 Lemma 5.2. Assume that n ≥ 7 and ∂M is umbilic, then if 0 <   ρ < ρ0 for some sufficiently small ρ0 , the volume of the boundary has 

2(n−1)

¯ n−2 dσg = B + B1 + B2 + O(ρ−6 6 ), U x0 (x0 ,)

∂M

where B1 =

B2 =

2(n − 1) n−2



n

Wn−2 ψdσ = Dρ

n(n − 1) (n − 2)2



1 2

 Dρ

2 n−2

W

ψ 2 dσ

Dρ

with |B1 | ≤ C(n, Tc )2 , |B2 | ≤ C(n, Tc )4 .

2(n−1) n−2

W

Snn dσ + O(n−1 ρ3−n ),

(5.8)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

Proof. By (2.10) we have  2(n−1) ¯ n−2 dσg = U (x0 ,)



((W + ψ) ◦ Ψ−1 x0 )

x0

2(n−1) n−2

29

dσgx0 + O(n−1 ρ1−n )

∂Ωρ ∩∂M

∂M



=

(W + ψ)

2(n−1) n−2

dσ + O(2d+2 ) + O(n−1 ρ1−n ).

Dρ

By choosing ρ small enough and (4.18) we have  (W + ψ)

2(n−1) n−2

 dσ =

Dρ

2(n−1) n−2

W

dσ +

Dρ

+

(n − 1)n (n − 2)2



2(n − 1) n−2 2 n−2

W



n

Wn−2 ψdσ Dρ

ψ 2 dσ + O(ρ−6 6 )

Dρ

:= B0 + B1 + B2 + O(ρ−6 6 ). Notice that B0 = B + O(n−1 ρ1−n ). Then it follows from (4.4) and (4.1) that n−1 2(n−1)  n 2(n − 1) n−2 1 2(n−1) W ψ = ∂a (W n−2 Va ) + W n−2 Snn n−2 2 a=1

on Dρ , an integration by parts gives 1 B1 = 2



2(n−1) n−2

W Dρ

=

1 2



−1



Snn dσ + ρ

2(n−1) n−2

W

Va y a dσ

∂Dρ 2(n−1) n−2

W

Snn dσ + O(ρ3−n n−1 )

Dρ

and |B1 | = O(2 ) by virtue of (4.2) and (4.17). 2 ¯(x ,) ] in this case. By adopting the However, it is a little bit tricky to estimate E[U 0 same notation in [18, Formula (5.33)], we decompose 4(n − 1) |∇(W + ψ)|2gx0 + Rgx0 (W + ψ)2 n−2 =

4  4 4(n − 1) 4(n − 1) |∇W |2 + n(n + 2)Wn−2 ψ 2 + Ji , n−2 n−2 i=1

(5.9)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

30

where Ji , 1 ≤ i ≤ 4 were defined in [18, p.33]. By [18, (5.34)] and Lemma 5.2, we obtain  J1 dy ≤ − Bρ+

8(n − 1) n−2

+C

n−1 



 ∂ + Bρ+

Dρ

d 

yi dσ |y|

|∂ α hab |n−2 ρ|α|+2−n + Cρ2d+4−n n−2

a,b=1 |α|=2

8(n − 1)c = n−2

(W2 ∂k hik − ∂k W2 hik )

∂n W ψdσ +



n n−2

W

 (W2 ∂k hik − ∂k W2 hik )

ψdσ + ∂ + Bρ+

Dρ

yi dσ |y|

+ O(n−2 ρ4−n )  yi (W2 ∂k hik − ∂k W2 hik ) dσ + O(n−2 ρ4−n ), = 4cB1 + |y|

(5.10)

∂ + Bρ+

n

n

where the first identity follows from (2.2): ∂n W = (n − 2)Tc Wn−2 = −cWn−2 on Dρ . Using an intermediate estimate in [18, (5.35)], we obtain 

%

1 J2 dy = − 4

Bρ+



:=−K

&'  (  2n n−2 Qik,l Qik,l dy − 2 W Tik Tik dy +

Bρ+

Bρ+

= −K + O(n−2 ρ6−n ) +

n+2 2(n − 2)

∂ + Bρ+



ξi

yi dσ − |y|

 ξn dσ

Dρ

2 W ∂n W Snn dσ + 4cB2 ,

(5.11)

Dρ

where the vector field ξ was defined in [18, Proposition 5.3] and we have used  ξi ∂ + Bρ+

yi dσ = O(n−2 ρ6−n ) |y|

and the exact expression of ξn (see [18, (5.28)]). Since ∂M is umbilic, using an independent estimate in [11, Corollary 12], we obtain an improved estimate of J3 and J4 (cf. [18, (5.37)]) J3 + J4 ≤C

n−1 

d 

(|∂ α hab |2 ( + |y|)2|α|+4−2n + |∂ α hab |( + |y|)|α|+d+3−2n )n−2

a,b=1 |α|=2

+ C( + |y|)2d+4−2n n−2 .

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

31

Thus, we have )

 (J3 + J4 )dy ≤ Bρ+

Cn−2 log ρ ,

if n = 7, 8,

C6 ,

if n ≥ 9.

(5.12)

Moreover, by (2.10) and (4.17) we have 

 hgx0 (W + ψ) dσ ≤ C

|y|2d+1 (W + ψ)2 dσ ≤ Cn−2 ρ2d+4−n .

2

Dρ

(5.13)

Dρ

Putting the above estimates (5.10)-(5.13) together, from (5.9) we obtain ¯(x ,) ; Ωρ ] E[U 0 4(n − 1) = n−2 −K +



ρ |∇W |2 dy + 4(n − 1)(n − 2)A2 + 4c(B1 + B2 ) + O(5 log ) 

Bρ+

n+2 2(n − 2)



 (W2 ∂k hik − ∂k W2 hik )

2 W ∂n W Snn dσ + ∂ + Bρ+

Dρ

yi dσ. (5.14) |y|

Using [18, (5.48)], when   ρ < ρ0 for some sufficiently small ρ0 we have ¯(x ,) ; M \Ωρ ] E[U 0  4(n − 1) ≤ n−2



−∂i W W + ∂j W W hij − 

n−2 2

 yi dσ |y|

(W ∂i G − G∂i W )

∂ + Bρ+

+ O(n−2 ρ4−n ).

(5.15)

Testing problem (2.2) with W and integrating over Bρ+ , we obtain 

 |∇W |2 dy − Bρ+

W  ∂i W 

∂ + Bρ+

 = n(n − 2)

yi dσ |y|

2n



Wn−2 dy − (n − 2)Tc

Bρ+

= n(n − 2)A + cB + O(n−1 ρ1−n ).

2(n−1) n−2

W

dσ

Dρ

(5.16)

Combining estimates (5.14) and (5.15), when  < ρ  ρ0 for some sufficiently small ρ0 , we conclude that

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

32

¯(x ,) ] E[U 0 =

⎛

4(n − 1) ⎜ ⎝ n−2

⎞





i

|∇W |2 dy −

Bρ+

∂i W W

∂ + Bρ+

y ⎟ dσ ⎠ |y|

+ 4(n − 1)(n − 2)A2 + 4c(B1 + B2 ) − K   n yi n+2 2 ∂j W2 hij ) dσ W ∂n W Snn dσ + (W2 ∂j hij + + 2(n − 2) n−2 |y| ∂ + Bρ+

Dρ

−

=

4(n − 1) n−2  2 n−2

 (W ∂i G − G∂i W ) ∂ + Bρ+

yi ρ dσ + O(5 log ) |y| 

4(n − 1) [n(n − 2)A + cB] + 4(n − 1)(n − 2)A2 + 4c(B1 + B2 ) n−2  n+2 ρ 2 −K + W ∂n W Snn dσ + O(5 log ). 2(n − 2) 

(5.17)

Dρ

Here the second identity follows from (5.16) and a rough estimate of [18, (5.50)], which indicates that the underlined terms are of order O(n−2 ). Comparing (5.7), (5.8), (5.17) and (5.2), (5.3), (5.4) respectively, we write A˜ = A2 ,

˜ = B1 + B2 , B

4(n − 1) ˜ ˜ −K + n+2 E = 4(n − 1)(n − 2)A2 + 4cB n−2 2(n − 2)

 2 W ∂n W Snn dσ. Dρ

˜ ≤ C4 , |B| ˜ ≤ C2 and E ˜ ≤ C2 as required. On the other hand, by Hence we have |A| (4.2) and (4.3), we estimate  2 K = O(4 ) and W ∂n W Snn dσ = O(4 ). Dρ

Consequently, keeping in mind that A2 = O(4 ) by Lemma 5.1 and B1 = O(2 ), B2 = O(4 ) by Lemma 5.2, we plug these estimates into (5.6) to obtain ¯(x ,) ] max I[tU 0

0≤t<∞

n+2 = Sc − K + 2(n − 2) ≤ Sc − K +

n+2 2(n − 2)

 2 W ∂n W Snn dσ + Dρ

˜ − cB) ˜ 2 2(n − 1)(E + o(4 ) 2n(n − 2)A + cB



2 W ∂n W Snn dσ + Λc2 B12 + o(4 ), Dρ

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

33

where Λ can be chosen as Λ=

2 . (n − 1)(2n(n − 2)A + cB)

(5.18)

In order to prove (1.4), it remains to show n+2 2(n − 2)

 2 W ∂n W Snn dσ + Λc2 B12 < K + o(4 ) Dρ

for sufficiently small  > 0. Recall definitions of B1 and K, it suffices to show n+2 2(n − 2) <

1 4

⎛

 2 W ∂n W Snn dσ + Dρ



Λ⎜ ⎝ 4

⎞2 ⎟ W ∂n W Snn dσ ⎠

Dρ



2n

Wn−2 Tik Tik dy + o(4 ).

Qik,l Qik,l dy + 2 Bρ+



Bρ+

Notice that Snn + Tnn = Hnn = 0 on Dρ , then the above inequality becomes

n+2 2(n − 2) <

1 4

⎛

 2 W ∂n W Tnn dσ + Dρ



Λ⎜ ⎝ 4

Bρ+

⎞2 ⎟ W ∂n W Tnn dσ ⎠

Dρ



Qik,l Qik,l dy + 2



2n

Wn−2 Tik Tik dy + o(4 )

(5.19)

Bρ+

for any c ∈ R, when 0 <   ρ < ρ0 for some sufficiently small ρ0 . Theorem 5.3. Assume that n ≥ 7, ∂M is umbilic and the Weyl tensor Wg0 of M is non-zero at some x0 ∈ ∂M . Then problem (1.2) is solvable for all non-negative constant c. Proof. Based on the above estimates, it reduces to showing (5.19) for all non-negative constant c. Without loss of generality we assume that Tnn ≡ 0 on Dρ , otherwise it is trivial. By Hölder’s inequality we have ⎛ ⎜ ⎝



Dρ

⎞2 ⎟ W ∂n W Tnn dσ ⎠ ≤



 W (−∂n W )dσ

Dρ

2 W (−∂n W )Tnn dσ,

Dρ n

since it follows from (2.2) that ∂n W = −cWn−2 on Dρ . Together with (5.18) and n ≥ 7, we obtain

34

X. Chen et al. / Advances in Mathematics 358 (2019) 106854



Λ 4

W (−∂n W )dσ ≤ Dρ

1 n+2 Λ cB ≤ < . 4 2(n − 1) 4(n − 2)

Then for any c ≥ 0, n+2 LHS of (5.19) ≤ 4(n − 2)

 2 W ∂n W Tnn dσ ≤ 0.

(5.20)

Dρ

On the other hand, thanks to estimate (3.6), (4.17) and the assumption that ∂M is umbilic, we apply Proposition A.2 to show 

∗ n−2



Qik,l Qik,l dy ≥ λ  Bρ+

( + |y|)6−2n dy ≥ Cλ∗ 4

(5.21)

Bρ+

for all ρ ≥ 2(1 + |Tc |), where λ∗ = λ∗ (n) > 0 and we have used (4.17) in the first inequality. Then (5.21) implies that 1 4



 Qik,l Qik,l dy + 2 Bρ+

2n

Wn−2 Tik Tik dy ≥ C1 4 ,

(5.22)

Bρ+

where C1 = C1 (n) > 0. Hence the estimate (5.19) follows from (5.20) and (5.22). 2 For any non-negative constant c, both the selection of Λ and the sign of c are very crucial in the above verification of estimate (5.19). However, at present we are not sure whether (5.19) is true for all negative constants c. Instead, we consider the inequality (5.19) on a spherical cap. This is realized by a pull-back of a stereographic projection from the spherical cap combined with an application of the sharp Sobolev trace inequality. Theorem 5.4. Let n ≥ 7 and c be a negative real number. Assume that ∂M is umbilic and the Weyl tensor Wg0 of M is non-zero at some x0 ∈ ∂M , then problem (1.2) is solvable for all c ∈ [−c0 , 0), where c0 is a positive dimensional constant. Proof. It suffices to prove the above inequality (5.19) when Bρ+ and Dρ are replaced by Rn+ and Rn−1 , respectively, because the integrals outside these domains are o(4 ), which can be verified by direct computations. Let π : S n (Tc en ) \ {Tc en + en+1 } → {y + Tc en ∈ Rn+1 ; y n+1 = 0}  Rn be the stereographic projection from the unit sphere S n (Tc en ) in Rn+1 centered at Tc en ; see [18, Fig. 1 on p. 9]. Let Σ := π −1 (Rn+ ) ⊂ S n denote a spherical cap equipped with the metric gΣ = 14 gS n , where gS n is the standard round metric of S n . Denote by ν = νgΣ 4/(n−2) the outward unit normal of gΣ on ∂Σ. We define T (x) = (W T ) ◦ π(x) as a symmetric 2-tensor on Σ. In particular, we have Tνν := T (νgΣ , νgΣ ) = Tnn on ∂Σ. Thus,

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

35

the quantities A and B in (2.4) are the volumes of (Σ, gΣ ) and ∂Σ with the induced metric of gΣ , respectively. We write (π −1 )∗ (gΣ ) = W (y) n−2 gRn 4

and define as in Brendle [11] Qij,k = W ∂k Tij +

2 (∂l W Til δjk + ∂l W Tjl δik − ∂i W Tjk − ∂j W Tik ) n−2

n−6

−1 )ij , = Wn−2 ∇Σ k (T ◦ π

where ∇Σ k denotes the covariant derivative with respect to the metric gΣ on Σ. Then by (2.2) we obtain 



|∇T |2gΣ dμgΣ ,

Qik,l Qik,l dy = 

Rn +

Σ



2n

|T |2gΣ dμgΣ ,

Wn−2 Tik Tik dy = Rn +

Σ



2 W ∂n W Tnn dσ



= −c

Rn−1

2 Tνν dσgΣ .

∂Σ

Thus, it suffices to show  Σ

1 (n + 2)c ( |∇T |2gΣ + 2|T |2gΣ )dμgΣ + 4 2(n − 2)

 |T |2gΣ dσgΣ

⎛ ⎞2  Λc2 ⎝ > Tνν dσgΣ ⎠ , 4

∂Σ

∂Σ

which implies the estimate (5.19). The sharp Sobolev trace inequality in [23, Theorem 1] (see also [29]) states that 4(n − 1) n−2



⎛ |∇ψ|2 dy ≥ Q(B n , S n−1 , [gRn ]) ⎝

Rn +

⎞ n−2 n−1

 |ψ|

2(n−1) n−2

dσ ⎠

(5.23)

Rn−1

for any ψ ∈ C ∞ (Rn+ ). Furthermore, the sharp constant is given by Q(B n , S n−1 , [gRn ]) =

1 n − 2 n−1 ωn−1 . 2

Given any ϕ ∈ H 1 (Σ, gΣ ), from the conformal invariance (1.3) of both LgΣ and BgΣ , we set ψ(y) = ((ϕ ◦ π −1 )W )(y) and use (5.23) to show

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

36

  Σ

  4(n − 1) |∇ϕ|2gΣ + RgΣ ϕ2 dμgΣ + 2(n − 1) hgΣ ϕ2 dσgΣ n−2 ∂Σ

⎛ ≥ Q(B n , S n−1 , [gRn ]) ⎝

⎞ n−2 n−1

 |ϕ|

2(n−1) n−2

dσgΣ ⎠

.

(5.24)

∂Σ

Moreover, it is easy to show that RgΣ = 4n(n − 1) and hgΣ = 2c/(n − 2) in virtue of (2.2). By choosing ϕ = |T |gΣ and using Kato’s inequality: |∇T |gΣ ≥ |∇|T |gΣ |gΣ , the above inequality (5.24) becomes   Σ

4(n − 1) |∇T |2gΣ + 4n(n − 1)|T |2gΣ n−2 ⎛

≥ Q(B , S n

n−1

, [gRn ]) ⎝



 dμgΣ +

4(n − 1)c n−2

 |T |2gΣ dσgΣ ∂Σ

⎞ n−2 n−1 2(n−1)

|T |gΣn−2 dσgΣ ⎠

.

∂Σ

Equivalently, 

  |∇T |2gΣ + n(n − 2)|T |2gΣ dμgΣ + c

Σ

 |T |2gΣ dσgΣ

∂Σ

⎛ n−2 Q(B n , S n−1 , [gRn ]) ⎝ ≥ 4(n − 1)

⎞ n−2 n−1



2(n−1) n−2 gΣ

|T |

dσgΣ ⎠

.

∂Σ

Consequently, we conclude that  Σ

1 (n + 2)c ( |∇T |2gΣ + 2|T |2gΣ )dμgΣ + 4 2(n − 2)  ≥

2 1 − 4 n(n − 2)



|T |2gΣ dσgΣ ∂Σ



|∇T

|2gΣ dμgΣ

+

Σ

⎛ +



1 Q(B n , S n−1 , [gRn ]) ⎝ 2n(n − 1)



2 n+2 − 2(n − 2) n(n − 2)

  c |T |2gΣ dσgΣ ∂Σ

⎞ n−2 n−1 2(n−1)

|T |gΣn−2 dσgΣ ⎠

∂Σ

⎞ n−2 ⎛ n−1

 2 2(n−1) 1 1 n−2 n + 2n − 4 n−1 n−2 ⎠ ⎝ ω cB n−1 ≥ + |T |gΣ dσgΣ . 4n(n − 1) n−1 2n(n − 2)

∂Σ

(5.25)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

37

On the other hand, by Hölder’s inequality we have ⎞2

⎛  Λ⎝ 4

W ∂n W Tnn dσ ⎠ ≤

Rn−1

Λ 4



 W ∂n W dσ

Rn−1

2 W ∂n W Tnn dσ

Rn−1

⎛ ⎞ n−2 n−1  2(n−1) n Λc2 n−1 n−2 ⎝ ⎠ B |T |gΣ dσgΣ . ≤ 4

(5.26)

∂Σ

In order to show (5.19), our goal is to find a constant c < 0 satisfying 1 1 n 2n 2 n−1 n2 + 2n − 4 n−1 1 n−2 n−1 ωn−1 cB + ≥ c B . 4n(n − 1) 2n(n − 2) n(n − 1)(n − 2) ωn

(5.27)

Assuming this claim temporarily, we obtain n n n 2n 2 n−1 1 1 Λc2 n−1 1 c2 B n−1 > B c B > , n(n − 1)(n − 2) ωn 2(n − 1) n(n − 2)A 4

1 where the second inequality follows from the fact that A > 2n+1 ωn for c < 0 and the third inequality follows from definition (5.18) of Λ and n(n − 2)A + cB > 0. Thus, (5.19) follows from (5.25) and (5.26). It remains to show the inequality (5.27). Let us collect some elementary facts on the c c geometric quantity B for c < 0. Recall that Tc = − n−2 and let cos r = $−T , then 1+T 2 c

r ∈ ( π2 , π) and B = B(r) = 21−n ωn−1 (sin r)n−1 . In terms of the new variable r we obtain 1

cB n−1 =

1 n n n − 2 n−1 (n − 2)2 n−1 ωn−1 cos r and c2 B n−1 = ωn−1 cos2 r sinn−2 r. n 2 2

(5.28)

Substituting (5.28) into (5.27), we assert that the inequality (5.19) holds for all r ∈ ( π2 , π) satisfying 1+

ωn−1 (n − 1)(n2 + 2n − 4) cos r ≥ 4 cos2 r sinn−2 r. n−2 ωn

Therefore, we conclude from the above inequality that (5.19) holds for all c ∈ [−c0 , 0), where c0 = c0 (n) > 0 is also determined by the above inequality. 2 Remark 5.5. Though we tend to believe that the test function (5.1) should be one of good choices of (1.4), all our attempts in this direction up to now have failed. As mentioned earlier, we will use a different type of test function in Subsection 5.2 to verify (1.4) under the hypothesis that n ≥ 7, ∂M is umbilic and the Weyl tensor of M is non-zero at a boundary point.

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

38

5.2. Non-positive constant boundary mean curvature We first present some preliminary results. Proposition 5.6. Suppose that n ≥ 7 and ∂M is umbilic, then under gx0 -Fermi coordi= 0, gxnn = 1 and nates around x0 ∈ ∂M , there hold gxan 0 0 1¯ c d n 2 gxab0 = δab + R acbd y y + Rnanb (y ) 3 1¯ 1 c d e n 2 c n 3 + R acbd,e y y y + Rnanb,c (y ) y + Rnanb,n (y ) 6 3   1 ¯ 1 ¯ ¯ Racbd,ef + Rachd Rbehf y c y d y e y f + 20 15   1 1 ¯ Rnanb,cd + Symab (Raced Rnenb ) (y n )2 y c y d + Rnanb,nc (y n )3 y c + 2 3 +

1 (Rnanb,nn + 8Rnanc Rncnb )(y n )4 + O(|y|5 ), 12

¯ abcd where the curvature quantities are with respect to metric gx0 and evaluated at x0 , R is the Riemannian curvature tensor of ∂M . Moreover at x0 there hold ¯ ab = 0, Rnn = R

−R,nn − 2(Rnanb )2 = 2Rnanb,ab ,

Rna,b = Rnn,a = Rnn,n = 0,

¯ g = −ΔR x0

1 |W gx0 |2gx0 . 6

Proof. For n ≥ 7, then d ≥ 2. Notice that hgx0 = O(|y|2d+1 ) near x0 and ∂M is umbilic, ¯ α πg (0) = 0 for all |α| ≤ 4. From this, the it yields |(πgx0 )ab | = O(|y|2d+1 ) and then ∇ x0 first assertion is a direct consequence of Marques [30, Lemma 2.3] (see also [1, Lemma 2.3]). Since det gx0 = 1 + O(|x|2d+2 ), then the coefficients in the expansion of det gx0 , which can be derived from [30, Lemma 2.2], vanish up to order 4. From this, the second assertion follows by the same lines in [30, Proposition 3.2]. 2 Let W and W denote the Weyl tensors of M and ∂M of metric gx0 respectively. We restate Almaraz [1, Lemma 2.5], which is crucial in the following test function construction. Lemma 5.7. Suppose that ∂M is umbilic and n ≥ 4. Then under gx0 -Fermi coordinates around x0 ∈ ∂M , Wijkl (x0 ) = 0 if and only if Rnanb (x0 ) = 0 = W abcd (x0 ). Theorem 5.8. Suppose that n ≥ 7, ∂M is umbilic and the Weyl tensor of M is non-zero at a boundary point, then problem (1.2) is solvable for all non-positive c.

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

39

Proof. We choose a test function as ¯(x ,) = [χρ (W + φ)] ◦ Ψ−1 , U x0 0

(5.29)

where the correction term is φ(y) = [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 )]

n−2 2

Rnanb y a y b (2 + |y − Tc en |2 )− 2

n

and κ0 , κ1 , κ2 are constants only depending on n, c to be determined later. Note that for + c ∈ R, (W + φ)(y) = (1 + O(|y|2 + 2 ))W (y) in B2ρ . Then for sufficiently small ρ > 0, there holds 1 ¯(x ,) (Ψx (y)) ≤ 2W (y) W (y) ≤ U 0 0 2

+ in B2ρ .

We remark that our test function includes the one used by Almaraz [1], both somewhat inspired by the idea of Marques [30]. In the following, we will use an elementary identity (see e.g., [31, p. 390]): Let q be a homogeneous polynomial of degree k and r > 0, then  Srn−2

r2 qdσ = k(n + k − 3)

 Δqdσ.

(5.30)

Srn−2

Using Δ2 (Rnanb Rncnd y a y b y c y d ) = 16(Rnanb )2 and (5.30), we have  Rnanb Rncnd y a y b y c y d dσ = Srn−2

2ωn−2 (Rnanb )2 n+2 r . (n − 1)(n + 1)

(5.31)

Recall that ∞

Γ(α)Γ(β) xα−1 dx = B(α, β) = α+β (1 + x) Γ(α + β)

0

for Re(α) > 0, Re(β) > 0. A similar argument in Section 4 gives ¯(x ,) ; M \Ωρ ] ≤ Cn−2 ρ2−n . E[U 0 ¯(x ,) ; Ωρ ]. Notice that Rnn = 0 at x0 by Proposition 5.6 and We turn to estimate E[U 0 then  Rnn ωn−2 rn = 0. Rnanb y a y b dσ = (5.32) n−1 Srn−2

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

40

Since n ≥ 7, we obtain 

2n

¯ n−2 dμg U x0 (x0 ,) Ωρ



(W + φ) n−2 dy + O(n ρN −n ) 2n

= Bρ+



2n n−2

=

W Bρ+





2n dy + n−2

n+2 n−2

W Bρ+

n(n + 2) φdy + (n − 2)2



4

Wn−2 φ2 dy + O(6 ) Bρ+

2n

Wn−2 dy + J + O(6 ),

= Bρ+

where J=

=

n(n + 2) (n − 2)2



4

Wn−2 φ2 dy Rn +

n(n + 2) 4  Rnanb Rncnd (n − 2)2



Rn +

=

y a y b y c y d [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 dy (1 + |y − Tc en |2 )n+2

2n(n + 2) (Rnanb )2 4 (n − 2)2 (n2 − 1)



Rn +

=

|¯ y |4 [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 dy (1 + |y − Tc en |2 )n+2

n(n + 2)ωn−2 4 2 n+1 B( n+3 2 , 2 )(Rnanb ) (n − 2)2 (n2 − 1) ∞ [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 n · dy n+1 (1 + (y n − Tc )2 ) 2 0

by using (5.31) and 2n n−2



n+2

Wn−2 φdy = 0 Bρ+

by virtue of (5.32). Similarly we obtain 

2(n−1)

¯ n−2 dσg U x0 (x0 ,) Ωρ ∩∂M



=

(W + φ) Dρ

2(n−1) n−2

dσ + O(n−1 ρN −n+1 )

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

 =

2(n−1) n−2

W

2(n − 1) n−2

dσ +

Dρ

n(n − 1) + (n − 2)2  =

2(n−1) n−2

W



2 n−2

W



n

Wn−2 φdσ Dρ

)

O(6 log ρ ) if n = 7

φ2 dσ +

Dρ

if n ≥ 8

O(6 )

)

O(6 log ρ ) if n = 7

ˆ+ dσ + B

O(6 )

Dρ

if n ≥ 8

,

where ˆ = n(n − 1) B (n − 2)2



2

Wn−2 φ2 dσ Rn−1



n(n − 1) 4  Rnanb Rncnd = (n − 2)2

Rn−1

2n 4 (Rnanb )2 = (n − 2)2 (n + 1)

y a y b y c y d (κ2 Tc2 − κ1 Tc + κ0 )2 dσ (1 + Tc2 + y¯2 )n+1 

|¯ y |4 (κ2 Tc2 − κ1 Tc + κ0 )2 dσ (1 + Tc2 + y¯2 )n+1

Rn−1

=

nωn−2 B (n − 2)2 (n + 1)

 n+3 2

 4 (κ2 Tc2 − κ1 Tc + κ0 )2 , n−1  (Rnanb )2 n−1 2 (1 + Tc2 ) 2

and 2(n − 1) n−2



n

Wn−2 φdσ = 0 Dρ

by virtue of (5.32). Observe that 

 ¯(x ,) |2 dμg = |∇U gx0 x0 0 Ωρ

|∇(W + φ)|2gx0 dy + O(n−2 ρN +2−n ) Bρ+





|∇(W + φ)| dy + 2

= Bρ+

(gxab0 − δab )∂a (W + φ)∂b (W + φ)dy

Bρ+

+ O(n−2 ρN +2−n ). By (2.2), an integration by parts gives 8(n − 1) n−2

 ∇W , ∇φdy Bρ+

8(n − 1) =− n−2

 Bρ+

8(n − 1) ΔW φdy − n−2

 φ Dρ

∂W dσ ∂y n

41

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

42

+

8(n − 1) n−2

 φ

∂W dσ ∂r

∂ + Bρ+

= O(n−2 ρ4−n ). From this we obtain

 |∇(W + φ)|2 dy

Bρ+



 |∇W | dy +

|∇φ|2 dy + O(n−2 ρ4−n ).

2

= Bρ+

Bρ+

Direct computations give |∇φ|2  = 4Rnena Rnenb y a y b D2 (2 + |y − Tc en |2 )−n + n(n − 4)Rnanb Rncnd y a y b y c y d D2 (2 + |y − Tc en |2 )−n−1 + Rnanb Rncnd y a y b y c y d (2κ2 (y n − Tc ) + κ1 )2 (2 + |y − Tc en |2 )−n − 2nRnanb Rncnd y a y b y c y d (2κ2 (y n − Tc ) + κ1 ) · D(y n − Tc )(2 + |y − Tc en |2 )−n−1

 − n2 2 Rnanb Rncnd y a y b y c y d D2 (2 + |y − Tc en |2 )−n−2 n−2 ,

where D = D(y n ) = κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 . From this, (5.30) and (5.31), we obtain  |∇φ|2 dy Bρ+

⎧ ⎪ ⎨ =

 4Rnena Rnenb

⎪ ⎩

(2

Bρ+

y a y b D(y)2 dy + |y − Tc en |2 )n 

+ n(n − 4)Rnanb Rncnd Bρ+

 + Rnanb Rncnd Bρ+

(2

y a y b y c y d D(y)2 dy + |y − Tc en |2 )n+1

y a y b y c y d (2κ2 (y n − Tc ) + κ1 )2 dy (2 + |y − Tc en |2 )n 

− 2nRnanb Rncnd Bρ+

y a y b y c y d (2κ2 (y n − Tc ) + κ1 )D(y)(y n − Tc ) dy (2 + |y − Tc en |2 )n+1

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

 −n2 2 Rnanb Rncnd Bρ+

⎧ ⎪ ⎪ ⎨ =

 4 ⎪ n−1 ⎪ ⎩ B+

a b c d

2

43

⎫ ⎪ ⎬

y y y y D(y) dy n−2 (2 + |y − Tc en |2 )n+2 ⎪ ⎭

|¯ y |2 [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 dy (1 + |y − Tc en |2 )n

ρ/

2n(n − 4) + n2 − 1 2 + 2 n −1

+ Bρ/

 + Bρ/

 −



|¯ y |4 [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 dy (1 + |y − Tc en |2 )n+1

|¯ y |4 [2κ2 (y n − Tc ) + κ1 ]2 dy (1 + |y − Tc en |2 )n

|¯ y |4 [2κ2 (y n −Tc )+κ1 ](y n −Tc )[κ2 (y n −Tc )2 +κ1 (y n −Tc )+κ0 ] dy (1+|y−Tc en |2 )n+1

4n n2 −1 + Bρ/

−

2n2 n2 − 1

 + Bρ/

⎫ ⎪ ⎪ ⎬

|¯ y |4 [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 dy 4 (Rnanb )2 . ⎪ (1 + |y − Tc en |2 )n+2 ⎪ ⎭

Notice that  + Bρ/

|¯ y |4 |y n |k dy (1 + |y − Tc en |2 )n

ωn−2 n+3 n−3 = B( 2 , 2 ) 2

∞ 0

 + Bρ/

∞ 0

n−3 2

dy n + O(ρ6−n n−6 ),

0 ≤ k ≤ 2,

|y n |k (1 + (y n − Tc )2 )

n−1 2

dy n + O(ρ6−n n−6 ),

0 ≤ k ≤ 4,

dy n + O(ρ6−n n−6 ),

0 ≤ k ≤ 4,

|¯ y | |y | dy (1 + |y − Tc en |2 )n+1 4

+ Bρ/

(1 + (y n − Tc )2 )

|¯ y |2 |y n |k dy (1 + |y − Tc en |2 )n

ωn−2 n+1 n−1 = B( 2 , 2 ) 2 

|y n |k

n k

ωn−2 n+3 n−1 = B( 2 , 2 ) 2

∞ 0

|y n |k (1 + (y n − Tc )2 )

n−1 2

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

44

 + Bρ/

=

|¯ y |4 |y n |k dy (1 + |y − Tc en |2 )n+2 ωn−2 n+3 n+1 B( 2 , 2 ) 2

∞ (1 +

0 n−1 B( n+1 2 , 2 )=

|y n |k (y n

− Tc

)2 )

n+1 2

dy n + O(ρ4−n n−4 ),

0 ≤ k ≤ 4.

2n n−1 B( n+3 2 , 2 ). n+1

Therefore, putting these facts together, we conclude that  |∇(W + φ)|2 dy Bρ+

 |∇W |2 dy + O(ρ4−n n−2 )

= Rn +

⎧ ∞ ⎨ [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ]2 n 2 n+3 n−1 + n B( 2 , 2 ) dy n−1 ⎩ (1 + (y n − Tc )2 ) 2 0

∞ n−3 + B( n+3 2 , 2 )

[2κ2 (y n − Tc ) + κ1 ]2 (1 + (y n − Tc )2 )

0

∞ −

n−1 2nB( n+3 2 , 2 )

n−3 2

dy n

[2κ2 (y n −Tc )+κ1 ][κ2 (y n −Tc )2 +κ1 (y n −Tc )+κ0 ](y n −Tc ) (1+(y n −Tc )2 )

n−1 2

0

∞ n+1 −n2 B( n+3 2 , 2 )

[κ2 (y −Tc ) +κ1 (y −Tc )+κ0 ] n

2

n

(1+(y n −Tc )2 )

n+1 2

0

2

dy n

⎫ ⎬

dy n

ωn−2 2 4 2 −1 (Rnanb )  .

⎭n

Next we need to estimate  (gxab0 − δab )∂a (W + φ)∂b (W + φ)dy Bρ+



 (gxab0

= Bρ+

− δab )∂a W ∂b W dy + 2 Bρ+



+

(gxab0 − δab )∂a W ∂b φdy

(gxab0 − δab )∂a φ∂b φdy.

Bρ+

By the symmetry of the half-ball, (5.30), (5.32) and Proposition 5.6, we estimate the first term (see also [1, Lemma A.1])

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

45

 (gxab0 − δab )∂a W ∂b W dy Bρ+

(n − 2)2 4  Rnanb,ab n2 − 1

=



+ Bρ/

+

(n − 2)2 4  (Rnanb )2 2(n − 1)  5

+ O( ) + Bρ/

|y n |2 |¯ y |4 dy (1 + |y − Tc en |2 )n 

+ Bρ/

|y n |4 |¯ y |2 dy (1 + |y − Tc en |2 )n

|y|7 dy (1 + |y − Tc en |2 )n

(n − 2)2 4 (n − 2)2 4 ρ  Rnanb,ab Θ1 +  (Rnanb )2 Θ2 + O(5 log ), 2 n −1 2(n − 1) 

:= where

 Θ1 = Rn +

|y n |2 |¯ y |4 dy (1 + |y − Tc en |2 )n

ωn−2 n+3 n−3 B( 2 , 2 ) = 2  Θ2 = Rn +

∞ 0

|y n |2 (1 + (y n − Tc )2 )

n−3 2

dy n ,

|y | |¯ y| dy (1 + |y − Tc en |2 )n n 4

2

ωn−2 n+1 n−1 B( 2 , 2 ) = 2

∞ 0

(y n )4 (1 + (y n − Tc )2 )

n−1 2

dy n .

For the second term, direct computations together with Proposition 5.6 give  (gxab0 − δab )∂a W ∂b φdy

2 Bρ+



n−2

(gxab0

=2 Bρ+

) · −

2

n−2 2

(2 − n)y a  2 − δab ) 2 n ( + |y n − Tc en |2 ) 2 Rnenb y e [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 ] n (2 + |y n − Tc en |2 ) 2

nRnenf y e y f y b 

n−2 2

[κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 ] n (2 + |y n − Tc en |2 ) 2 +1

dy

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

46



n−2

Rnanb (y n )2

=2 Rn +

) · −

2

n−2 2

(2 − n)y a  2 n (2 + |y n − Tc en |2 ) 2

Rnenb y e [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 ] n (2 + |y n − Tc en |2 ) 2

nRnenf y e y f y b 

n−2 2

[κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 2 ] n (2 + |y n − Tc en |2 ) 2 +1

dy

ρ + O(5 log )  2n(n − 2) 4 n−1 =− 2  ωn−2 (Rnanb )2 B( n+3 2 , 2 ) n −1 ∞ n 2 ρ (y ) [κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 ] n · dy + O(5 log ). n−1 n 2  (1 + (y − Tc ) ) 2 0

The third term can be estimated by  (gxab0 − δab )∂a φ∂b φdy = O(ρ8−n n−2 ). Bρ+

Thus, we obtain  (gxab0 − δab )∂a (W + φ)∂b (W + φ)dy Bρ+

=

(n − 2)2 ρ Rnanb,ab Θ1 4 + O(5 log ) n2 − 1  ⎧ ∞ ωn−2 4 ⎨ (n−2)2 (n+1) (y n )4 2 n n+1 n−1 + 2  (Rnanb ) B( 2 , 2 ) n−1 dy 4 n n −1 ⎩ (1+(y −Tc )2 ) 2 0

∞ 2 n−1 −2n(n − 2)B( n+3 2 , 2 )(Rnanb )

(y ) [κ2 (y −Tc ) +κ1 (y −Tc )+κ0 ] n 2

n

2

(1+(y n −Tc )2 ) 0

n

n−1 2

dy n

⎫ ⎬ ⎭

We also estimate  ¯2 Rgx0 U (x0 ,) dμgx0 Ωρ



 Rgx0 W2 dy + 2

= Bρ+

Bρ+

 Rgx0 W φdy + Bρ+

Rgx0 φ2 dy + O(n−2 ρN +6−n ).

.

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

47

¯ g = 1 |W g |2 at x0 and Rg = O(|y|2 ) by Proposition 5.6, it yields Notice that −ΔR x0 x0 gx0 x0 6  Rgx0 W2 dy



1 = 4 R,nn 2

Bρ+

+ Bρ/

−

|y n |2 dy (1 + |y − Tc en |2 )n−2

1 4 |W gx0 |2gx0 12(n − 1)  5

+ O( ) + Bρ/

 + Bρ/

|¯ y |2 dy (1 + |y − Tc en |2 )n−2

|y|3 dy (1 + |y − Tc en |2 )n−2

1 ρ 1 4  R,nn Θ3 − 4 |W gx0 |2gx0 Θ4 + O(5 log ), 2 12(n − 1)      Rgx0 W φdy = O |y|2 (2 + |y|2 ) W2 dy = O(5 ρ), =

Bρ+

Bρ+



 2

Rgx0 φ dy = Bρ+

  O |y|2 (2 + |y|2 )2 W2 dy = O(5 ρ3 ),

Bρ+

where  Θ3 = Rn +

|y n |2 dy (1 + |y − Tc en |2 )n−2

ωn−2 n−1 n−3 B( 2 , 2 ) = 2  Θ4 = Rn +

∞ 0

(y n )2 (1 + (y n − Tc )2 )

n−3 2

dy n ,

|¯ y |2 dy (1 + |y − Tc en |2 )n−2

ωn−2 n+1 n−5 B( 2 , 2 ) = 2

∞ 0

1 (1 + (y n − Tc )2 )

n−5 2

dy n .

Also we have 

 ¯2 hgx0 U (x0 ,) dσgx0 = Ωρ ∩∂M

Dρ/

O(N )|y|N −1 dσ = O(n−2 ρN +2−n ). (1 + |y − Tc en |2 )n−2

Notice that Θ1 =

n+1 Θ3 4(n − 2)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

48

and −R,nn − 2(Rnanb )2 = 2Rnanb,ab at x0 by Proposition 5.6, it yields 1 4(n − 2) Rnanb,ab Θ1 + R,nn Θ3 = −(Rnanb )2 Θ3 . n+1 2 Based on the above estimates, we conclude from (2.9) that ρ t∗ = 1 + K∗ 4 + O(5 log ),  where K∗ is a constant depending on n, Tc , M, gx0 . Then we claim that ¯(x ,) ] I∗ [U 0 2n



¯(x ,) ] − 4(n − 1)(n − 2)(t∗n−2 − 1) := (t2∗ − 1)E[U 0 2(n−1) n−2

+ 4(n − 2)Tc (t∗

M



2(n−1) n−2

¯ U (x0 ,) dσgx0

− 1) ∂M

.

 |∇W | dy − n(n − 2) 2

=

2n

¯ n−2 dμg U x0 (x0 ,)

Rn +

2n

Wn−2 dy

Rn +



/

2(n−1) n−2

+ (n − 2)Tc

W

dσ

Rn−1

8(n − 1) ρ K∗ 4 + O(5 log ) n−2 

ρ = O(5 log ),  where the last identity follows from (2.3). For convenience, noticing that Tc ≥ 0, we define π

2

sink θ cosl θdθ,

J(k, l) =

k, l ∈ N,

− arctan Tc

then the iteration formulae of J(k, l) are given by

J(k + 2, l) = J(k, l + 2) =

(−1)k+1 Tck+1 (l + k + 2)(1 +

Tc2 )

k+l+2 2

(−1)k Tck+1 (l + k + 2)(1 +

Tc2 )

k+l+2 2

+

k+1 J(k, l), l+k+2

k, l ∈ N,

(5.33)

+

l+1 J(k, l), l+k+2

k, l ∈ N.

(5.34)

Under the change of variables tan θ = z − Tc , we obtain

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

∞ I0 (k) = 0

(z − Tc )k (1 + (z − Tc )2 )

∞ I1 (k) = 0

(z − Tc )k (1 + (z − Tc )2 )

∞ I2 (k) = 0

n−3 2

n−1 2

(z − Tc )k (1 + (z − Tc )2 )

n+1 2

49

dz = J(k, n − 5 − k),

0 ≤ k ≤ 2,

dz = J(k, n − 3 − k),

0 ≤ k ≤ 4,

dz = J(k, n − 1 − k),

0 ≤ k ≤ 4.

Therefore, putting these facts together, together with (2.6) we conclude that ¯(x ,) ] I[t∗ U 0 ¯(x ,) ] ¯(x ,) ] + I∗ [U = I[U 0 0 ⎧ ⎪ 4(n − 1) ⎨ |∇W |2 dy + O(5 log( ρ )) = (n − 2) ⎪ ⎩n R+

⎡ + ⎣n2 B( n+3 , n−1 ) 2

∞

(κ2 (y n − Tc )2 + κ1 (y n − Tc ) + κ0 )2

2

(1 + (y n − Tc )2 )

0

∞ +

(2κ2 (y n − Tc ) + κ1 )2

n−3 B( n+3 2 , 2 )

(1 + (y n − Tc )2 )

0

∞ −

n−1 2nB( n+3 2 , 2 )

n−3 2

n−1 2

dy n

(2κ2 (y n −Tc )+κ1 )(κ2 (y n −Tc )2 +κ1 (y n −Tc )+κ0 )(y n −Tc ) (1+(y n −Tc )2 )

n−1 2

0

∞ n+1 −n2 B( n+3 2 , 2 )

(κ2 (y −Tc ) +κ1 (y −Tc )+κ0 ) 2

n

(1+(y n −Tc )2 )

2

n+1 2

dy n ⎦

ωn−2 4 ⎣ (n − 2)2 (n + 1) n−1  (Rnanb )2 B( n+1 + 2 2 , 2 ) n −1 4 ∞ 2 n−1 −2n(n − 2)B( n+3 2 , 2 )(Rnanb ) 0 2 ωn−2

−  (Rnanb ) 4

2

dy n

⎤ n

0

⎡

dy n

∞ n−3 B( n−1 2 , 2 ) 0

ωn−2 (Rnanb )2 4 n2 − 1

∞

(y n )4 (1+(y n −Tc )2 )

0

n−1 2

dy n

⎤⎫ ⎪ ⎬ (y n )2 (κ2 (y n −Tc )2 +κ1 (y n −Tc )+κ0 ) n⎦ dy n−1 ⎪ (1+(y n −Tc )2 ) 2 ⎭ (y n )2

(1 + (y n − Tc )2 )

n−3 2

dy n

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

50

⎡ 1 ⎢ 4 |W gx0 |2gx0 Θ4 + O(5 log( ρ )) − 4(n − 1)(n − 2) ⎣ − 12(n − 1)



2n

Wn−2 dy

Bρ+

∞ +

n(n+2)ωn−2 2 4 n+3 n+1 (n−2)2 (n2 −1) B( 2 , 2 )(Rnanb ) 

⎡ ⎢ + 4(n − 2)Tc ⎣

2

n

(1+(y n −Tc )2 )

n+1 2

0



2(n−1) n−2

W

dσ +

⎤ 2

⎥ dy n⎦ ⎤



 n+3 n−1 nωn−2 B (κ2 Tc2 −κ1 Tc +κ0 )2 4 (Rnanb )2 ⎥ 2 , 2

Dρ

= Sc +

(κ2 (y −Tc ) +κ1 (y −Tc )+κ0 ) n

 4 2 n−1 ωn−2 B( n+3 2 , 2 )κQκ  (Rnanb )

(n−2)2 (n+1)(1+Tc2 )

−

n−1 2

2 4 Θ4 12(n−1) |W gx0 |gx0 

⎦

+ O(5 log( ρ )), (5.35)

where κ = (κ2 , κ1 , κ0 , 1), . Q11 = = Q12 = = Q13 =

Q = (Qij )1≤i,j≤4 ,

/ 8 Tc5 I0 (2) + (n − 4)I1 (4) − (n − 1)I2 (4) + n−1 n−3 (1 + Tc2 ) 2

8n 4 J(4, n − 7) + J(2, n − 7) , (n − 2)(n + 1) n−3 . / 4 4 T c 4n n−1 (n−2)(n+1) (n − 3)I1 (3) − (n − 1)I2 (3) + n − 3 I0 (1) − (1 + Tc2 ) 2

8n 2 J(3, n − 6) + J(1, n − 6) , (n − 2)(n + 1) n−3 / . 4n Tc3 (n − 2)I1 (2) − (n − 1)I2 (2) + n−1 (n − 2)(n + 1) (1 + Tc2 ) 2 4n (n−2)(n+1)

8n J(2, n − 5), (n − 2)(n + 1)  4n  =− I1 (4) + 2Tc I1 (3) + Tc2 I1 (2) n+1  4n  =− J(4, n − 7) + 2Tc J(3, n − 6) + Tc2 J(2, n − 5) , n+1 . / 3 2 T c 4n = (n−2)(n+1) (n − 2)I1 (2) − (n − 1)I2 (2) + I0 (0) + n−1 n−3 (1 + Tc2 ) 2

8n 1 = J(2, n − 5) + J(0, n − 5) , (n − 2)(n + 1) n−3 . / 4n Tc2 = (n − 1)I1 (1) − (n − 1)I2 (1) − n−1 (n − 2)(n + 1) (1 + Tc2 ) 2 =

Q14

Q22

Q23

Qij = Qji ,

=

8n J(1, n − 4), (n − 2)(n + 1)

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

51

 4n  I1 (3) + 2Tc I1 (2) + Tc2 I1 (1) n+1  4n  J(3, n − 6) + 2Tc J(2, n − 5) + Tc2 J(1, n − 4) , =− n+1 . / 4n Tc nI1 (0) − (n − 1)I2 (0) + = n−1 (n − 2)(n + 1) (1 + Tc2 ) 2

Q24 = −

Q33

8n J(0, n − 3), (n − 2)(n + 1)  4n  I1 (2) + 2Tc I1 (1) + Tc2 I1 (0) =− n+1  4n  J(2, n − 5) + 2Tc J(1, n − 4) + Tc2 J(0, n − 3) , =− n+1  2 (I0 (2) + 2Tc I0 (1) + Tc2 I0 (0)) = − n−3  2n(n − 2) + (I1 (4) + 4Tc I1 (3) + 6Tc2 I1 (2) + 4Tc3 I1 (1) + Tc4 I1 (0)) n+1  2 (J(2, n − 7) + 2Tc J(1, n − 6) + Tc2 J(0, n − 5)) = − n−3 =

Q34

Q44

+ (J(4, n − 7) + 4Tc J(3, n − 6) + 6Tc2 J(2, n − 5)) + 4Tc3 J(1, n − 4)  2n(n − 2) + Tc4 J(0, n − 3)) n+1 by virtue of (5.34). By (5.35) and (1.4), it remains to prove the following Claim. There exists a vector κ = κ(n, c) ∈ R4 with its fourth element equal to 1 such that κQκ < 0. To this end, we choose two non-singular matrices 0

2 S1 = diag 1, 1, 1, − n−2

1 (5.36)

and ⎛

⎞ 1 0 0 −1 ⎜ 0 1 0 −2T ⎟ ⎜ c⎟ S2 = ⎜ 2 ⎟. ⎝ 0 0 1 −Tc ⎠ 0 0 0 1 Then we obtain a symmetric matrix Q =

(n−2)(n+1)   S2 S1 QS1 S2 , 8n

(5.37)

where

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

52

Q11 = J(4, n − 7) + Q12 = J(3, n − 6) +

4 n−3 J(2, n 2 n−3 J(1, n

− 7), − 6),

Q13 = J(2, n − 5), 4 Q14 = − n−3 J(2, n − 7) −

Q22 = J(2, n − 5) +

4Tc n−3 J(1, n

1 n−3 J(0, n

− 6),

− 5),

Q23 = J(1, n − 4), 2 Q24 = − n−3 J(1, n − 6) −

2Tc n−3 J(0, n

− 5),

Q33 = J(0, n − 3), Q34 = 0, Q44 =

2 n−3



 J(2, n − 7) + 2Tc J(1, n − 6) + Tc2 J(0, n − 5) .

For any a ∈ R satisfying 7a2 − 8a + 2 < 0, we choose V = (a, Tc , 0, 1) and find VQV 





4 2 J(2, n − 7) + 2aTc J(3, n − 6) + J(1, n − 6) = a J(4, n − 7) + n−3 n−3

4Tc 4 J(2, n − 7) − J(1, n − 6) + 2a − n−3 n−3

1 J(0, n − 5) + Tc2 J(2, n − 5) + n−3

2 2Tc J(1, n − 6) − J(0, n − 5) + 2Tc − n−3 n−3  2  J(2, n − 7) + 2Tc J(1, n − 6) + Tc2 J(0, n − 5) + n−3 (a − 1)2 Tc3 7a2 − 8a + 2 =− J(2, n − 7) < 0, n−3 + n − 3 (1 + Tc2 ) 2 n−3 2

where we have used (5.33) in the last identity.   Therefore, we can choose κ = − n−2 2 VS2 S1 whose fourth element equals 1, such that  κQκ < 0. This proves the Claim. 2 Appendix A Our first purpose is to prove that I satisfies (P S) condition for energy level below Sc . For clarity, we restate Lemma 2.1 here. Lemma A.1 (Compactness). Suppose that Y (M, ∂M, [g0 ]) > 0. Let {ui ; i ∈ N} be a sequence of functions in H 1 (M, g0 ) satisfying I[ui ] → L < Sc and |I  [ui ](v)| →0 0 )\{0} v H 1 (M,g0 )

max

v∈H 1 (M,g

as i → ∞.

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

53

Then after passing to a subsequence, either (i) {ui } strongly converges in H 1 (M, g0 ) to some positive solution u of (1.2) or (ii) {ui } strongly converges to 0 in H 1 (M, g0 ).  Proof. Since Y (M, ∂M, [g0 ]) > 0, we define u, v := 4(n−1) ∇u, ∇vg0 dμg0 + n−2 M   1 Rg0 uvdμg0 + 2(n − 1) ∂M hg0 uvdσg0 as an inner product of H (M, g0 ). The norm M defined by u = u, u1/2 is equivalent to u H 1 (M,g0 ) . Based on Han-Li [28, Lemma 1.2], we only need to show that if {ui } ⊂ H 1 (M, g0 ) weakly converges to some nontrivial solution u of (1.2) , then ui → u in H 1 (M, g0 ) as i → ∞. By the Sobolev embedding and trace inequalities, up to a subsequence we have ui  u

in H 1 (M, g0 );

ui → u

in L2 (M, g0 ) and L2 (∂M, g0 ).

For any v ∈ H 1 (M, g0 ) we have 1  I [ui − u]v 2



= ui − u, v − 4n(n − 1)

n+2

(ui − u)+n−2 vdμg0 −

 4(n−1) n−2 c

M

→ 0,

∂M

as i → ∞,

(A.1)

which implies that I  [ui − u] → 0 as i → ∞. By the Lebesgue dominant convergence theorem, we have 1 I[ui − u] = I[ui ] +

d I[ui − tu]dt dt

0

1 = I[ui ] −

I  [ui − tu]udt

0

1 →L−

I  [(1 − t)u]udt, as i → ∞.

0

Since u is a nontrivial smooth solution of (1.2), we obtain 2 d 22 I[tu] = 0, dt 2t=1 and then

n

(ui − u)+n−2 vdσg0

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

54

d I[tu] dt

⎡

= 2t ⎣E[u] − 4n(n − 1)t



4 n−2

u

2n n−2

dμg0

M

⎡

= 2t(1 − t n−2 ) ⎣E[u] + t n−2 4n(n − 1) 2

2

2 4(n − 1) n−2 ct − n−2



⎤

⎤

 u

2(n−1) n−2

dσg0 ⎦

∂M

u n−2 dμg0 ⎦ . 2n

M

Thus, d I[tu] = I  [tu]u ≥ 0, dt

∀ t ∈ [0, 1].

From these we conclude that lim I[ui − u] ≤ L < Sc .

(A.2)

i→∞

Therefore, thanks to (A.1) and (A.2), we can apply the same argument of Case (ii) in [28, Lemma 1.2] to ui − u and then obtain ui → u in H 1 (M, g0 ) as i → ∞. 2 Next we obtain the following refined estimate of [18, Proposition 5.5] (see also [12, Corollary 2.6]), which has been used in Subsection 5.1. Proposition A.2. Let (M, g0 ) be a compact Riemannian manifold of dimension n ≥ 3 with umbilic boundary ∂M and c ∈ R. Then there exists a positive dimensional constant λ∗ (independent of Tc ), such that ∗ n−2

λ 

n−1 



d 

|∂ hab | α

a,b=1 |α|=2



n 

≤ Bρ+ (0)

2

( + |y|)2|α|+2−2n dy

Bρ+ (0)

Qik,l Qik,l dy

i,k,l=1

for all ρ ≥ 2(1 + |Tc |). Proof. For r > 0, let Ur ⊂ Rn+ be an open ball of radius r/4 centered at (3r/2)en . Denote + by η(y/r) a smooth cut-off function such that η = 1 in Ur , η = 0 in B2r (0) \ Br+ . Since ∂M is umbilic, we can apply [12, Proposition 2.4] to show C(n)

n−1 

d 

a,b=1 |α|=2

for all r > 0.

 |∂ α hab |2 r2|α|−4+n ≤

n 

Ur i,j,k,l=1

|Zijkl (y)|2 dy

X. Chen et al. / Advances in Mathematics 358 (2019) 106854

55

Observe that n 

1 4

|Zijkl |2

i,j,k,l=1 n 

=

2 n−2

∂j (W−1 Qik,l )Zijkl +

i,j,k,l=1

n 

W−2 ∂k W Qil,j Zijkl .

i,j,k,l=1

Multiplying the above equation by η(y/r) and integrating over Rn+ , we obtain 1 4



n 

y |Zijkl | η( )dy = − r 2

i,j,k,l=1 Rn +



n 

i,j,k,l=1 Rn +

2 n−2

+

 y  W−1 Qik,l ∂j Zijkl η( ) dy r

  n i,j,k,l Rn +

y W−2 ∂k W Qil,j Zijkl η( )dy. r

Applying Hölder’s inequality, we have 

n 

|Zijkl |2 dy

Ur i,j,k,l=1

⎛ ≤ C

− n−2 2

(r + (1 + |Tc |))n−2 ⎝

n d  

⎞ 12 |∂ α hik |2 r2|α|−6+n ⎠

|α|=2 i,k=1

⎛ ⎜ ·⎝



n 

+ B2r (0)\Br+ (0)

⎛ ≤ Crn−3 ⎝

d 

⎞ 12 ⎟ |Qik,l (y)|2 dy ⎠

i,k,l=1

n 

|α|=2 i,k=1

⎞ 12 ⎛ ⎜ |∂ α hik |2 r2|α|−4+n ⎠ ⎝



+ B2r (0)\Br+ (0)

n 

⎞ 12 ⎟ |Qik,l (y)|2 dy ⎠

i,k,l=1

for all r ≥ (1 + |Tc |). For such , one can combine with [12, Proposition 2.4] to get the conclusion. 2 References [1] S. Almaraz, An existence theorem of conformal scalar-flat metrics on manifolds with boundary, Pacific J. Math. 248 (1) (2010) 1–22. [2] S. Almaraz, Convergence of scalar-flat metrics on manifolds with boundary under a Yamabe-type flow, J. Differential Equations 259 (2015) 2626–2694. [3] S. Almaraz, E. Barbosa, L. de Lima, A positive mass theorem for asymptotically flat manifolds with a non-compact boundary, Comm. Anal. Geom. 24 (4) (2016) 673–715.

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