Some general forms of sharp Sobolev inequalities

Meijun Zhu Department of Mathematics The University of British Columbia Vancouver, B.C. Canada V6T 1Z2 mzhu@math.ubc.ca

Abstract

In this paper, we establish some general forms of sharp Sobolev inequalities on the upper half space or any compact Riemannian manifold with smooth boundary. These forms extend some previous results due to Escobar [11], Li and Zhu [18].

0 Introduction

In the past several decades, the study of sharp Sobolev inequalities has attracted the attention of many mathematicians. Not only do those sharp type Sobolev inequalities play essential roles in the study of some problems arising from geometry and physics, but also those inequalities themselves indicate some rich and significant phenomena ( for example, under which circumstance the extremal functions for those inequalities exist).

The sharp Sobolev inequalities for functions vanishing on the boundary are well understood for manifolds with dimension greater than or equal to 3, see Aubin [3], Talenti [21], Brezis-Nirenberg [6], Lieb [19], Hebey and Vaugon [14]-[15], Hebey [13], and the references therein. However, it seems that those sharp Sobolev inequalities for functions which do not vanish on the boundary still need to be further studied, even though there are already some interesting results, see Lions [20], Brezis-Lieb [5], Escobar [11]-[12], Beckner [4], Adimurthi and Yadava [2], Carlen and Loss [8]-[9], Li and Zhu [17]-[18], and the references therein. One of the most interesting problems is to study the relations between the L^{2} norm of the gradient, the boundary L^{q} norm and the interior L^{p} norm. Here and throughout this paper we always set q = 2(n − 1)/(n − 2), p = 2n/(n − 2), where n ≥ 3 is the dimension of the manifold.

In this paper, we continue our previous work [17], [18] and give some sharp relations between the three terms we just mentioned. It turns out that some previous results are special cases, see Remark 0.2-0.3 and 0.5 below.

Denote ℝ_{+}^{n} = {(x′, x_{n}) : x′ = (x_{1}, x_{2}, ..., x_{n−1}), x_{n} > 0} as the upper half space, and D^{1,2}(ℝ_{+}^{n}) = {u : ∫_{ℝ_{+}^{n}} |∇u|^{2} < ∞, ∫_{ℝ_{+}^{n}} |u|^{p} < ∞}. Let 1/S = (n − 2)/2 · (2π^{n/2}/(Γ(n/2)))^{1/(n−1)} be the sharp constant corresponding to the trace inequality, Z_{0} = (n−2)C_{n}^{1/(n−1)}, where C_{n} = ∫_{ℝ^{n−1}}(\frac{1}{1+|x|^{2}} )^{n−1}dx = π^{(n−1)/2}Γ((n−1)/2)/Γ(n−1). One can check that

Z_{0} = \frac{1}{S}  (0.1)

For any ϵ > 0 and d ∈ ℝ, we define

u_{ϵ,d}(x) = (\frac{ϵ}{ϵ^{2} + |x′|^{2} + |x_{n} − ϵd|^{2}} )^{\frac{n−2}{2}} .  (0.2)

Considering three typical manifolds, we have corresponding theorems as follows. On the upper half space, we have

Theorem 0.1 For any Z ∈ (−1/S, Z_{0}), let

d_{z} = \frac{Z}{\sqrt{Z_{0}^{2} − Z^{2}}} ,  (0.3)

and S_{1}(Z) be given by

\frac{1}{S_{1}(Z)} = \frac{∫_{ℝ_{+}^{n}} |∇u_{ϵ,d_{z}}|^{2} + Z(∫_{∂ℝ_{+}^{n}} u_{ϵ,d_{z}}^{q})^{\frac{2}{q}}}{(∫_{ℝ_{+}^{n}} u_{ϵ,d_{z}}^{p})^{2/p}} .  (0.4)

Then

(∫_{ℝ_{+}^{n}} |u|^{p})^{\frac{2}{p}} ≤ S_{1}(Z)( ∫_{ℝ_{+}^{n}} |∇u|^{2} + Z(∫_{∂ℝ_{+}^{n}} |u|^{q})^{\frac{2}{q}} ), ∀u ∈ D^{1,2}(ℝ_{+}^{n}),  (0.5)

and equality holds if and only if u(x) = Cu_{ϵ,d_{z}} for some ϵ > 0, C ∈ ℝ.

For Z = Z_{0}, let 1/S_{1} = πn(n − 2)(Γ(n/2)/Γ(n))^{2/n} be the sharp constant of Sobolev inequality in ℝ^{n}. Then

(∫_{ℝ_{+}^{n}} |u|^{p})^{\frac{2}{p}} < S_{1}( ∫_{ℝ_{+}^{n}} |∇u|^{2} + Z_{0}(∫_{∂ℝ_{+}^{n}} |u|^{q})^{\frac{2}{q}} ), ∀u ∈ D^{1,2}(ℝ_{+}^{n}).  (0.6)

Remark 0.1 One can check that S_{1}(Z) → ∞ as Z → −1/S. An easy consequence of Theorem 0.1 is that S_{1}(Z) strictly decreases from ∞ to 1/S_{1} as Z goes from −1/S to Z_{0} = 1/S. The strictness of inequality (0.6) is due to the nonexistence of the extremal function.

Remark 0.2 For Z ≤ 0, this theorem was proved by Escobar in [11]. It seems to us that his method can not be applied to prove (0.5) for Z > 0.

Let Ω ⊂ ℝ^{n} (n ≥ 3) be a bounded domain, we have

Theorem 0.2 For any Z ∈ (−1/S, Z_{0}), let d_{z}, S_{1}(Z) be given as in Theorem 0.1. Then there exists a constant C(Z, Ω) such that

(∫_{Ω} |u|^{p})^{\frac{2}{p}} ≤ S_{1}(Z)( ∫_{Ω} |∇u|^{2} + Z(∫_{∂Ω} |u|^{q})^{\frac{2}{q}} ) + C(Z, Ω) ∫_{∂Ω} u^{2}, ∀u ∈ H^{1}(Ω). (0.7)

Remark 0.3 For Z = 0, (0.7) was proved by Y.Y.Li and M.Zhu in [18] and was partially proved by Adimurthi and S. L. Yadava in [2]. In this case, one can see that S_{1}(0) = 1/(2^{2/n}S_{1}).

Remark 0.4 It is interesting to give some upper bound estimates about the constant C(Z, Ω). For a general domain (except a ball), it is hard to say whether the extremal function for (0.7) exists or not.

Let (M, g) be any compact Riemannian manifold with smooth boundary and dimension n ≥ 3, we have

Theorem 0.3 For any Z ∈ (−1/S, Z_{0}), let d_{z}, S_{1}(Z) be given as in Theorem 0.1. Then there exists a constant D(Z, M) such that ∀u ∈ H^{1}(M)

\begin{matrix}(∫_{M} |u|^{p}dv_{g})^{\frac{2}{p}} \\ ≤ S_{1}(Z)( ∫_{M} |∇_{g}u|^{2}dv_{g} + Z(∫_{∂M} |u|^{q}ds_{g})^{\frac{2}{q}} ) + D(Z, M)( ∫_{∂M} u^{2}ds_{g} + ∫_{M} u^{2}dv_{g}). \end{matrix}

(0.8)

Remark 0.5 When Z = 0, (0.8) was proved by Y.Y.Li and M.Zhu in [18].

Remark 0.6 We do not know whether (0.7) and (0.8) still hold for Z = Z_{0} or not. We tend to believe that it is true, and give the following conjecture.

Conjecture Let Ω ⊂ ℝ^{n} be a bounded smooth domain. There exists some constant C(Ω) > 0 such that

(∫_{Ω} |u|^{p})^{\frac{2}{p}} ≤ S_{1}( ∫_{Ω} |∇u|^{2} + Z_{0}(∫_{∂Ω} |u|^{q})^{\frac{2}{q}} ) + C(Ω) ∫_{∂Ω} u^{2}, ∀u ∈ H^{1}(Ω). (0.9)

Let (M, g) be a compact Riemannian manifold with smooth boundary and dimension ≥ 3. There exists some constant D(M) such that ∀u ∈ H^{1}(M)

\begin{matrix}(∫_{M} |u|^{p}dv_{g})^{\frac{2}{p}} \\ ≤ S_{1}( ∫_{M} |∇_{g}u|^{2}dv_{g} + Z_{0}(∫_{∂M} |u|^{q}ds_{g})^{\frac{2}{q}} ) + D(M)( ∫_{∂M} u^{2}ds_{g} + ∫_{M} u^{2}dv_{g}). \end{matrix}

(0.10)

Since the upper half space is conformally equivalent to a ball, we know from Theorem 0.1 that (0.9) holds for some constant C(Ω) when Ω is a ball in ℝ^{n}.

Remark 0.7 It can be easily seen that if (0.10) held for some large constant D(M), Hebey and Vaugon’s inequality (see [14]) would be its corollary.

Remark 0.8 In [5], Brezis and Lieb proved that for any bounded domain Ω ⊂ ℝ^{n}, there exists a constant C, such that

||u||_{p,Ω} ≤ S_{1}^{1/2}||∇u||_{2,Ω} + C||u||_{q,∂Ω}, ∀u ∈ H^{1}(Ω).

They asked whether there exists a constant C_{1}, such that

||u||_{p,Ω}^{2} ≤ S_{1}||∇u||_{2,Ω}^{2} + C_{1}||u||_{q,∂Ω}^{2}, ∀u ∈ H^{1}(Ω).  (0.11)

It is easy to see that (0.11) follows directly from (0.9). Hence the answer to their question is affirmative when the domain is a ball.

The proof of Theorem 0.1 heavily depends on the conformal invariant property of the corresponding energy functionals between the upper half space and the unit ball. The key ingredient is to show that the infimum of the corresponding functional is attained under the assumption of small energy (see Proposition 1.1 below for precise statement). We use a new approach which combines some old ideas (blowup argument) with some new inequalities initiated by the work of Brezis and Lieb [5](see corollary 1.2 below). The proofs of Theorem 0.2 and 0.3 also involve this difficulty; we overcome it by using the same method. Some ingredients in the proofs of Theorem 0.2 and Theorem 0.3 have already appeared in our previous work [17],

[18] and [22].

This paper is organized as follows. In Section 1 we give the proof of Theorem 0.1. In Section 2 we prove Theorem 0.2 through an argument by contradiction . In Section 3 we sketch the proof of Theorem 0.3. In Section 4 we give some discussions concerning the conjecture and point out the obstacle of proving this conjecture by the current method. When I was working on this problem, I was informed by Y.Y.Li about the recent results of Carlen and Loss [9]. It turns out that Theorem 0.1 can also be derived from their results. This derivation is included in the appendix. It seems to me that their method can not be applied to general domains.

Throughout this paper, we use C_{0}, C, C_{1}, C_{2}, ..., to represent some various positive constants, ϵ, ϵ_{0}, ϵ_{1}, ..., δ_{0}, δ, δ_{1}, ..., to represent some various small positive constants. Without specific mention, we always pass to a limit up to some subsequence of ϵ or α.

Acknowledgment. I would like to thank Y.Y.Li for many valuable discussions and for informing me about the work of Carlen and Loss [9]. I would like to thank H.Brezis and Y.Y.Li for their interests in this work and their constant encouragements. I would like to thank E. Carlen and M. Loss for informing me of their work [8]. This work was completed when I was a Postdoctoral Fellow in the University of British Columbia. I would like to thank R. Froese, N. Ghoussoub and C. Gui for their supports.

1 The upper half space and unit ball

For any Z ∈ (−1/S, Z_{0}], we define

I_{Z}(u) = \frac{∫_{ℝ_{+}^{n}} |∇u|^{2} + Z(∫_{∂ℝ_{+}^{n}} |u|^{q})^{2/q}}{(∫_{ℝ_{+}^{n}} |u|^{p})^{2/p}} , ∀u ∈ D^{1,2}(ℝ_{+}^{n}) \ \{0\}

and

II_{Z}(u) = \frac{∫_{B_{1}} |∇u|^{2} + Z(∫_{∂B_{1}} |u|^{q})^{2/q} + \frac{n−2}{2} ∫_{∂B_{1}} u^{2}}{(∫_{B_{1}} |u|^{p})^{2/p}} , ∀u ∈ H^{1}(B_{1}) \ \{0\}.

Due to the conformal invariance, we know

inf_{u∈D^{1,2}(ℝ_{+}^{n})\{0}}I_{Z}(u) = inf_{u∈H^{1}(B_{1})\{0}}II_{Z}(u) := ξ_{Z}.  (1.1)

The key step in our proof of Theorem 0.1 is to establish the following proposition.

Proposition 1.1 If 0 < ξ_{Z} < 1/S_{1}, then inf I_{Z}(u) and inf II_{Z}(u) are achieved.

Proof. We prove this proposition by contradiction.

Suppose these infimums are not attained. For any 0 < ϵ < 1 and u ≠ 0, we define

II_{ϵ}(u) = \frac{∫_{B_{1}} |∇u|^{2} + Z(∫_{∂B_{1}} |u|^{q_{ϵ}})^{2/q_{ϵ}} + \frac{n−2}{2} ∫_{∂B_{1}} u^{2}}{(∫_{B_{1}} |u|^{p_{ϵ}})^{2/p_{ϵ}}}

where and throughout this paper, we set p_{ϵ} = p − ϵ, q_{ϵ} = q − ϵ/2. Denote

ξ_{ϵ} := inf_{u∈H^{1}(B_{1})\{0}}II_{ϵ}(u).

In order to derive a contradiction, we need several lemmas.

Lemma 1.1 As ϵ small enough,

0 ≤ ξ_{ϵ} ≤ ξ_{Z}.

Proof. ∀δ > 0, there exists a ū ∈ C^{∞}(B_{1}), such that

\frac{∫_{B_{1}} |∇ū|^{2} + Z(∫_{∂B_{1}} |ū|^{q})^{2/q} + \frac{n−2}{2} ∫_{∂B_{1}} ū^{2}}{(∫_{B_{1}} |ū|^{p})^{2/p}} ≤ ξ_{Z} + δ.

Also as ϵ → 0,

\frac{∫_{B1} |∇ū|^{2}+Z(∫_{∂B1} |ū|^{q})^{2/q}+ \frac{n−2}{2} ∫_{∂B1} ū^{2}}{(∫_{B1} |ū|^{p})^{2/p}}

≥ \frac{∫_{B_{1}} |∇ū|^{2}+Z(∫_{∂B_{1}} |ū|^{qϵ})^{2/qϵ}+ \frac{n−2}{2} ∫_{∂B_{1}} ū^{2}}{(∫_{B_{1}} |ū|^{pϵ})^{2/pϵ}} − δ

≥ ξ_{ϵ} − δ.

Combining above two inequalities together we know ξ_{ϵ} ≤ ξ_{Z}, as ϵ → 0. If Z ≥ 0, it is obvious that ξ_{ϵ} ≥ 0. If 0 > Z > −1/S, by sharp trace inequality, we know that for any fixed 0 < δ_{0} < Z/2 + 1/(2S),

∫_{ℝ_{+}^{n}} |∇u|^{2} + (Z − δ_{0})(∫_{∂ℝ_{+}^{n}} |u|^{q})^{2/q} ≥ 0, ∀ u ∈ D^{1,2}(ℝ_{+}^{n}).

It follows from the conformal invariance of the energy that

∫_{B_{1}} |∇u|^{2} + (Z − δ_{0})(∫_{∂B_{1}} |u|^{q})^{2/q} + \frac{n − 2}{2} ∫_{∂B_{1}} u^{2} ≥ 0, ∀u ∈ H^{1}(Ω).

Notice Z − δ_{0} < 0. We know from the above and Hölder inequality that as ϵ → 0, ξ_{ϵ} ≥ 0.

It follows from Lemma 1.1 and the standard variational method that as ϵ sufficiently small, there exists u_{ϵ} ≥ 0 with ||u_{ϵ}||_{p_{ϵ},B_{1}} = 1 such that

II_{ϵ}(u_{ϵ}) = inf_{u∈H^{1}(B_{1})\{0}}II_{ϵ}(u) = ξ_{ϵ}.

Next lemma is a slight extension of an inequality due to Brezis and Lieb [5].

Lemma 1.2 Let Ω be a bounded domain in ℝ^{n}, |Ω| = 1, 1 < r ≤ p, s = \frac{(n−1)r}{n}, then there exists a constant C(Ω, r), such that

(∫_{Ω} |f|^{r})^{1/r} ≤ S_{1}^{1/2}(∫_{Ω} |∇f|^{2})^{1/2} + C(Ω, r)(∫_{∂Ω} |f|^{s})^{1/s}, ∀ f ∈ H^{1}(Ω).  (1.2)

Further, as r close to 2n/(n − 2), we can choose C(Ω, r) independent of r.

Proof. We only need to show that (1.2) holds for any smooth function f. Let h be the solution of the following equation:

\{ \begin{matrix}−Δh = 0 & in Ω \\ h = f & on ∂Ω \end{matrix}

and u = f − h. Then u = 0 on ∂Ω. Therefore

||u||_{r,Ω} ≤ ||u||_{p,Ω} ≤ S_{1}^{1/2}||∇u||_{2,Ω}.  (1.3)

One can easily check that

||∇u||_{2,Ω}^{2} = ||∇f||_{2,Ω}^{2} − ||∇h||_{2,Ω}^{2}.  (1.4)

Also, by Minkowski inequality

||u||_{r,Ω} ≥ ||f||_{r,Ω} − ||h||_{r,Ω}.  (1.5)

Combining (1.3) and (1.4) with (1.5), we have

||f||_{r,Ω} − ||h||_{r,Ω} ≤ S_{1}^{1/2}(||∇f||_{2,Ω}^{2} − ||∇h||_{2,Ω}^{2})^{1/2} ≤ S_{1}^{1/2}||∇f||_{2,Ω}.  (1.6)

We claim:

||h||_{r,Ω} ≤ C(Ω, r)||f||_{s,∂Ω}

with C(Ω, r) independent of r as r close to 2n/(n − 2).

Lemma 1.2 directly follows from the claim and (1.6). Therefore, we are left to prove the claim.

Let ϕ be the solution of the following equation

\{ \begin{matrix}Δϕ = Y & in Ω \\ ϕ = 0 & on ∂Ω \end{matrix}

for some Y ∈ L^{t}(Ω). Then

∫_{Ω} hY = ∫_{∂Ω} h\frac{∂ϕ}{∂ν}.  (1.7)

By elliptic estimates, we have

||∂_{ij}ϕ||_{t,Ω} ≤ C_{t}||Y ||_{t,Ω}, ||∇ϕ||_{W^{1,t}(Ω)} ≤ C_{t}||Y ||_{t,Ω},

thus

||\frac{∂ϕ}{∂ν}||_{β,∂Ω} ≤ C_{t}||Y ||_{t,Ω}  (1.8)

where 1/β = n/((n−1)t)−1/(n−1). Also we can choose a uniform C_{t} if 2n/(n+2) ≤ t ≤ 2. It follows from (1.7) and (1.8) that

∫_{Ω} hY ≤ ||h||_{β′,∂Ω} · ||\frac{∂ϕ}{∂ν}||_{β,∂Ω} ≤ C_{t}||f||_{β′,∂Ω} · ||Y ||_{t,Ω}  (1.9)

where 1/β + 1/β′ = 1.

Let 1/r = 1 − 1/t, then 1/β′ = n/((n − 1)r), that is β′ = s. Therefore, by (1.9)

∫_{Ω} hY ≤ C_{t}||f||_{s,∂Ω} · ||Y ||_{t,Ω}.

Claim follows from the above directly.

A quick consequence of the above lemma is the following.

Corollary 1.1 Let Ω ∈ ℝ^{n} be a bounded domain with |Ω| = 1. There exists a constant C(Ω), such that as ϵ < 1/100,

||f||_{p_{ϵ},Ω} ≤ S_{1}^{1/2}||∇f||_{2.Ω} + C(Ω)||f||_{q_{ϵ},∂Ω}, ∀f ∈ H^{1}(Ω).  (1.10)

Proof. Let r = p_{ϵ} in Lemma 1.2, then

s = \frac{n − 1}{n} · p_{ϵ} = q − \frac{n − 1}{n} · ϵ < q_{ϵ}.

Corollary1.1 follows directly from Hölder inequality.

Later on, we will use this corollary in the following setting. We state it as another corollary.

Corollary 1.2 Let Ω ∈ ℝ^{n} be a bounded domain. ∀δ > 0, there exists a constant C(Ω, δ), such that as ϵ small enough,

||f||_{p_{ϵ},Ω}^{2} ≤ (S_{1} + δ)||∇f||_{2,Ω}^{2} + C(Ω, δ)||f||_{q_{ϵ},∂Ω}^{2}, ∀f ∈ H^{1}(Ω).  (1.11)

Proof. Without loss of generality, we assume {0} ∈ Ω and |Ω| = λ^{n}. Set Ω_{0} = Ω/λ and f_{0}(x) = λ^{2/(p_{ϵ}−2)}f(λx) for x ∈ Ω_{0}. Then due to Corollary 1.1, we have, for ϵ < 1/100,

||f_{0}||_{p_{ϵ},Ω_{0}} ≤ S_{1}^{1/2}||∇f_{0}||_{2.Ω_{0}} + C||f_{0}||_{q_{ϵ},∂Ω_{0}}.

By rescaling, we have

||f_{0}||_{p_{ϵ},Ω_{0}} = λ^{o(1)}||f||_{p_{ϵ},Ω}, ||∇f_{0}||_{2.Ω_{0}} = λ^{o(1)}||∇f||_{2.Ω}, ||f_{0}||_{q_{ϵ},∂Ω_{0}} = λ^{o(1)}||f||_{q_{ϵ},∂Ω}

where o(1) → 0 as ϵ → 0. Therefore, for any δ > 0, as ϵ < ϵ_{1} < 1/100 for some small ϵ_{1}, we have

||f||_{p_{ϵ},Ω} ≤ (S_{1}^{1/2} + \frac{δ}{2})||∇f||_{2.Ω} + C_{1}||f||_{q_{ϵ},∂Ω}.

Squaring both sides of the above and using Cauchy-Schwartz inequality we have our corollary.

We now continue the proof of Proposition 1.1. Since II_{ϵ}(u_{ϵ}) = ξ_{ϵ} and ||u_{ϵ}||_{p_{ϵ},B_{1}} = 1, we know that u_{ϵ} satisfies

\{ \begin{matrix}−Δu_{ϵ} & = ξ_{ϵ}u_{ϵ}^{p_{ϵ}−1} & in B_{1} \\ \frac{∂u_{ϵ}}{∂ν} & = −Z(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}u_{ϵ}^{q_{ϵ}−1} − \frac{n−2}{2}u_{ϵ} & on ∂B_{1}. \end{matrix}  (1.12)

Due to Cherrier [10], we know that u_{ϵ} is smooth up to the boundary. Hence we can assume u_{ϵ}(x_{ϵ}) = ||u||_{L^{∞}(B_{1})} for some x_{ϵ} ∈ \overline{B}_{1}. Since inf II_{Z} is not attained, we know that u_{ϵ}(x_{ϵ}) → ∞ as ϵ → 0.

Define

\{ \begin{matrix}μ_{ϵ} & = (u_{ϵ}(x_{ϵ}))^{− \frac{pϵ−2}{2}} , Ω_{ϵ} = \frac{B_{1}−x_{ϵ}}{μ_{ϵ}} , \\ v_{ϵ}(x) & = μϵ^{\frac{2}{pϵ−2}} u_{ϵ}(μ_{ϵ}x + x_{ϵ}) for x ∈ Ω_{ϵ}. \end{matrix}  (1.13)

Then v_{ϵ} satisfies

\{ \begin{matrix}−Δv_{ϵ} & = ξ_{ϵ}v_{ϵ}^{p_{ϵ}−1} & in Ω_{ϵ} \\ \frac{∂v_{ϵ}}{∂ν} & = −Z(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}v_{ϵ}^{q_{ϵ}−1} − \frac{n−2}{2}μ_{ϵ}v_{ϵ} & on ∂Ω_{ϵ}. \end{matrix}  (1.14)

Lemma 1.3

lim_{ϵ→0} ∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}} > 0.

Proof. We prove this lemma by contradiction. If not, there exists a subsequence of ϵ (we still denote it as ϵ), s.t.

∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}} → 0 as ϵ → 0.  (1.15)

From Lemma 1.1 we know: as ϵ small enough, 0 ≤ ξ_{ϵ} ≤ ξ_{Z} < 1/S_{1}, thus ||u_{ϵ}||_{H^{1}(B_{1})} ≤ C uniformly for some constant C. It follows that there exists a u_{0} ∈ H^{1}(B_{1}) such that ( after passing to a subsequence of ϵ)

u_{ϵ} → u_{0} weakly in H^{1}(B_{1}).

Using Corollary 1.2, we know for some small constant δ_{0} satisfying 1/(S_{1} + δ_{0}) > ξ_{Z} that

ξ_{ϵ} = ∫_{B_{1}} |∇u_{ϵ}|^{2} + Z(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}} + \frac{n−2}{2} ∫_{∂B_{1}} u_{ϵ}^{2}
≥ \frac{1}{S_{1}+δ_{0}} (∫_{B_{1}} |u_{ϵ}|^{p_{ϵ}})^{2/p_{ϵ}} + o_{ϵ}(1)
= \frac{1}{S_{1}+δ_{0}} + o_{ϵ}(1)

where o_{ϵ}(1) → 0 as ϵ → 0. We derive a contradiction when ϵ is small enough. Lemma 1.3 is established.

Set

T = lim_{ϵ→∞} \frac{dist(x_{ϵ}, ∂B_{1})}{μ_{ϵ}}.  (1.16)

(Recall the notation at the end of the introduction, we define the above limit by passing to a subsequence of ϵ.)

Lemma 1.4

T < ∞.

Proof. If T = ∞, we know that v_{ϵ} → v_{1} in C^{2}(B_{R}(0)) for any R > 1, where v_{1} satisfies

\{ \begin{matrix}−Δv_{1} & = ξ_{0}v_{1}^{p−1} in ℝ^{n} \\ v_{1}(0) & = 1, 0 ≤ v_{1} ≤ 1 \end{matrix}

and ξ_{0} = lim_{ϵ→0} ξ_{ϵ} (recall our notation: we take this limit up to some subsequence of ϵ). From [7], we know that v_{1}(x) = O(\frac{1}{|x|^{n−2}} ) as |x| → ∞ and v_{1}(x) ∈ D^{1,2}(ℝ^{n}). Multiplying the above equation by v_{1}, we have

∫_{ℝ^{n}} |∇v_{1}|^{2} = ξ_{0} ∫_{ℝ^{n}} v_{1}^{p}.

Also by the definition of S_{1}, we know

∫_{ℝ^{n}} |∇v_{1}|^{2} ≥ \frac{1}{S_{1}} (∫_{ℝ^{n}} v_{1}^{p})^{2/p}.

Notice ξ_{0} < 1/S_{1}. We conclude that

∫_{ℝ^{n}} v_{1}^{p} > 1.

On the other hand, note ||v_{ϵ}||_{p_{ϵ},Ω_{ϵ}} = μ_{ϵ}^{2/(p_{ϵ}−2)−n/p_{ϵ}}||u_{ϵ}||_{p_{ϵ},Ω} = μ_{ϵ}^{(n−2)ϵ/(p_{ϵ}(p_{ϵ}−2))} ≤ 1 and v_{ϵ} → v_{1} in C^{2}(B_{R}(0)), we have ∫_{ℝ^{n}} v_{1}^{p} ≤ 1. Contradiction. We complete the proof of Lemma 1.4.

Due to 0 ≤ ξ_{0} ≤ C, Lemma 1.3 and the fact that Ω_{ϵ} tends to ℝ_{T}^{n} = {x = (x′, x_{n}) : x_{n} > −T }, we know by the standard elliptic estimates that up to a subsequence, v_{ϵ} → v in C^{2}(B_{R} ∩ ℝ_{T}^{n}), and v satisfies

\{ \begin{matrix}−Δv & = ξ_{0}v^{p−1}, 0 ≤ v ≤ 1 & in ℝ_{T}^{n} \\ \frac{∂v}{∂ν} & = −Zav^{q−1} & on ∂ℝ_{T}^{n}, \end{matrix}  (1.17)

where a = lim_{ϵ→0}(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}. From [16] we know that v(x) = O(\frac{1}{|x|^{n−2}} ) as |x| → ∞ and v(x) ∈ D^{1,2}(ℝ_{T}^{n}). Therefore, multiplying (1.17) by v, we have

\frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Za ∫_{∂ℝ_{T}^{n}} v^{q}}{∫_{ℝ_{T}^{n}} v^{p}} = ξ_{0}.

Using rescaling, we know

(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1} = μ_{ϵ}^{( \frac{2qϵ}{pϵ−2} −(n−1))·(1− \frac{2}{qϵ})}(∫_{∂Ω_{ϵ}} v_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}.  (1.18)

Since (\frac{2q_{ϵ}}{p_{ϵ}−2} − (n − 1)) · (1 − \frac{2}{q_{ϵ}}) > 0 and μ_{ϵ} → 0, we know

\begin{aligned}(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q−1} & ≥ lim_{ϵ→0}(∫_{∂Ω_{ϵ}} v_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1} \\ & ≥ a. \end{aligned}

If Z ≤ 0, noticing ||v||_{p,ℝ_{T}^{n}} ≤ 1, we have

\begin{aligned}\frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Z(∫_{∂ℝ^{n}_{T}} v^{q})^{2/q}}{(∫_{ℝ^{n}_{T}} v^{p})^{2/p}} & ≤ \frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Z·a·∫_{∂ℝ^{n}_{T}} v^{q}}{∫_{ℝ^{n}_{T}} v^{p}} \\ & = ξ_{0} ≤ ξ_{Z}. \end{aligned}

Set v̄(x′, x_{n}) = v(x′, x_{n} − T), we know that inf I_{Z} is attained by v̄, contradiction. Thus we complete the proof of Proposition 1.1 in the case of Z ≤ 0.

Now we consider the case Z > 0. Using rescaling, we have

\begin{aligned}(∫_{ℝ_{T}^{n}} v^{p})^{1− \frac{2}{p}} & ≤ lim_{ϵ→0}(∫_{Ω_{ϵ}} v_{ϵ}^{p_{ϵ}})^{1− \frac{2}{pϵ}} \\ & ≤ lim_{ϵ→0} μϵ^{\frac{(n−2)ϵ}{pϵ}} . \end{aligned}  (1.19)

Let

a_{1} = lim_{ϵ→0} μ_{ϵ}^{( \frac{2qϵ}{pϵ−2} −(n−1))·(1− \frac{2}{qϵ})}, b = ||v||_{p,ℝ_{T}^{n}}^{p−2} .

Note as ϵ small enough

\frac{(n − 2)ϵ}{p_{ϵ}} > (\frac{2q_{ϵ}}{p_{ϵ} − 2} − (n − 1)) · (1 − \frac{2}{q_{ϵ}}) > 0,  (1.20)

we know b ≤ a_{1} ≤ 1.

In order to complete the proof of Proposition 1.1, we still need one more lemma.

Lemma 1.5

∫_{∂Ω_{ϵ}} v_{ϵ}^{q_{ϵ}} → ∫_{∂ℝ_{T}^{n}} v^{q} as ϵ → 0.  (1.21)

We relegate the proof of this lemma at the end of this Section and continue our proof of Proposition 1.1.

It follows from the above lemma and (1.18) that a = a_{1}||v||_{q,∂ℝ_{T}^{n}}^{2−q} . Therefore

\frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Z(∫_{∂ℝ^{n}_{T}} v^{q})^{2/q}}{(∫_{ℝ^{n}_{T}} v^{p})^{2/p}} ≤ \frac{∫_{ℝ^{n}_{T}} |∇v|^{2}}{∫_{ℝ^{n}_{T}} v^{p}} + \frac{Z·b·(∫_{∂ℝ^{n}_{T}} v^{q})^{2/q}}{∫_{ℝ^{n}_{T}} v^{p}}
≤ \frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Z·a·∫_{∂ℝ^{n}_{T}} v^{q}}{∫_{ℝ^{n}_{T}} v^{p}}
= ξ_{0} ≤ ξ_{Z}.

Set v̄(x′, x_{n}) = v(x′, x_{n} − T), we know that inf I_{Z} is attained by v̄, contradiction. This completes the proof of Proposition 1.1.

Now we are ready to give the proof of Theorem 0.1. Proof of Theorem 0.1. We first claim: if Z < Z_{0}, inf I_{Z} < 1/S_{1}.

For any d > 0, we define

v_{d}(x) = (\frac{1}{1 + |x′|^{2} + (x_{n} − d)^{2}} )^{\frac{n−2}{2}} .

Direct calculation shows

(∫_{∂ℝ_{+}^{n}} v_{d}^{q})^{2/q} = Cn^{\frac{n−2}{n−1}} (1 + d^{2})^{− \frac{n−2}{2}} , (∫_{ℝ_{+}^{n}} v_{d}^{p})^{2/p} = (D_{n}E_{n} − D_{n} ∫_{d}^{∞}(1 + t^{2})^{− \frac{n+1}{2}} dt)^{2/p},

and

\begin{aligned}∫_{ℝ_{+}^{n}} |∇v_{d}|^{2} & = (n − 2)^{2}C_{n}F_{n} − (n − 2)^{2}C_{n} ∫_{d}^{∞}(1 + t^{2})^{− \frac{n−1}{2}} dt \\ & −(n − 2)^{2}D_{n}E_{n} + (n − 2)^{2}D_{n} ∫_{d}^{∞}(1 + t^{2})^{− \frac{n+1}{2}} dt, \end{aligned}

where

C_{n} = ∫_{ℝ^{n−1}}(\frac{1}{1 + |x′|^{2}} )^{n−1}dx′, D_{n} = ∫_{ℝ^{n−1}}(\frac{1}{1 + |x′|^{2}} )^{n}dx′,

E_{n} = ∫_{−∞}^{+∞}(1 + t^{2})^{− \frac{n+1}{2}} dt, F_{n} = ∫_{−∞}^{+∞}(1 + t^{2})^{− \frac{n−1}{2}} dt

satisfying the relation

\frac{(n − 2)^{2}C_{n}F_{n} − (n − 2)^{2}D_{n}E_{n}}{(D_{n}E_{n})^{2/p}} = \frac{1}{S_{1}} .

It follows that as d large enough,

I_{Z}(v_{d}) = \frac{1}{S_{1}} + \frac{1}{(D_{n}E_{n})^{2/p}} · (ZCn^{\frac{n−2}{n−1}} − (n − 2)C_{n})d^{−(n−2)} + o(1)d^{−(n−2)}, (1.22)

where o(1) → 0 as d → ∞. Note Z < Z_{0}, ZCn^{\frac{n−2}{n−1}} − (n − 2)C_{n} < 0. Choosing d sufficiently large we establish the claim.

Due to Proposition 1.1, we know that for −1/S < Z < Z_{0} , inf I_{Z} is attained by some u_{z} ∈ D^{1,2}(ℝ_{+}^{n}). Without loss of generality we can assume that u_{z} satisfies ||u_{z}||_{p,ℝ_{+}^{n}} = 1 and u_{z} ≥ 0. Then one can easily see that u_{z} satisfies the following equation:

\{ \begin{matrix}−Δu_{z} = ξ_{Z}u_{z}^{p−1} & in ℝ_{+}^{n} \\ \frac{∂u_{z}}{∂ν} = −Z(∫_{∂ℝ_{+}^{n}} u_{z}^{q})^{2/q−1}u_{z}^{q−1} & on ∂ℝ_{+}^{n}. \end{matrix}  (1.23)

Let v_{z} = k_{z}u_{z} with k_{z} = (\frac{ξ_{Z}}{n(n−2)})^{(n−2)/4}. Then v_{z} satisfies:

\{ \begin{matrix}−Δv_{z} = n(n − 2)v_{1}^{p−1} & in ℝ_{+}^{n} \\ \frac{∂v_{z}}{∂ν} = −Z(∫_{∂ℝ_{+}^{n}} v_{z}^{q})^{2/q−1}v_{z}^{q−1} & on ∂ℝ_{+}^{n}. \end{matrix}  (1.24)

From [16] we know

v_{z} = (\frac{ϵ_{z}}{ϵ_{z}^{2} + |x′|^{2} + (t − t_{z})^{2}} )^{\frac{n−2}{2}} ,

where

t_{z} = (n − 2)^{−1}ϵ_{z}Z(∫_{∂ℝ_{+}^{n}} v_{z}^{q})^{2/q−1}

for some ϵ_{z} > 0. Direct computation yields

(∫_{∂ℝ_{+}^{n}} v_{z}^{q})^{2/q−1} = C_{n}^{− \frac{1}{n−1}} (1 + (\frac{t_{z}}{ϵ_{z}})^{2})^{1/2}.

Combining the above two identities together, we have

Z = \frac{(n − 2)Cn^{\frac{1}{n−1}} \frac{t_{z}}{ϵ_{z}}}{(1 + (\frac{t_{z}}{ϵ_{z}})^{2})^{1/2}} .

This gives the proof of Theorem 0.1 for −1/S < Z < Z_{0}.

If Z = Z_{0}, we will show that

inf I_{Z_{0}}(u) = \frac{1}{S_{1}}

and the infimum can not be attained through an argument by contradiction. It is well known that inf I_{Z_{0}}(u) ≤ 1/S_{1}. If inf I_{Z_{0}}(u) < 1/S_{1}, due to Proposition 1.1, we know that there exists a u_{0} ≥ 0 with ||u_{0}||_{p,ℝ_{+}^{n}} = 1 such that I_{Z_{0}}(u_{0}) = inf I_{Z_{0}}(u) := ξ_{Z_{0}}. It follows that u_{0} satisfies

\{ \begin{matrix}−Δu_{0} = ξ_{Z_{0}}u_{0}^{p−1} & in ℝ_{+}^{n} \\ \frac{∂u_{0}}{∂ν} = −Z_{0}(∫_{∂ℝ_{+}^{n}} u_{0}^{q})^{2/q−1}u_{0}^{q−1} & on ∂ℝ_{+}^{n}. \end{matrix}

It can be shown as the above that u_{0} = (ξ_{Z_{0}}/(n(n − 2)))^{−(n−2)/4}v_{Z_{0}}, where

v_{Z_{0}} = (\frac{ϵ_{0}}{ϵ_{0}^{2} + |x′|^{2} + (t − t_{0})^{2}} )^{\frac{n−2}{2}}

with

t_{0} = (n − 2)^{−1}ϵ_{0}Z_{0}(∫_{∂ℝ_{+}^{n}} v_{Z_{0}}^{q})^{2/q−1}

for some ϵ_{0} > 0. Therefore

Z_{0} = \frac{(n − 2)Cn^{\frac{1}{n−1}} \frac{t_{0}}{ϵ_{0}}}{(1 + (\frac{t_{0}}{ϵ_{0}} )^{2})^{1/2}} < (n − 2)Cn^{\frac{1}{n−1}} = Z_{0}.

Contradiction! The nonexistence of the extremal functions also follows from the same argument. This completes the proof of Theorem 0.1.

We are left to prove Lemma 1.5. In order to prove Lemma 1.5, we need the following inequality. One can find a proof of such inequality in Adam’s book [1](with a slight modification).

Lemma 1.6 Let Ω be a bounded domain in ℝ^{n}. For all γ > 0, ϵ > 0, there exists a constant Ĉ(γ, ϵ) depending on γ and ϵ such that

(∫_{∂Ω} |u|^{q_{ϵ}}ds)^{\frac{2}{qϵ}} ≤ γ ∫_{Ω} |∇u|^{2}dv + Ĉ(γ, ϵ)(∫_{Ω} |u|^{p_{ϵ}}dv)^{\frac{2}{pϵ}} , ∀ u ∈ H^{1}(Ω). (1.25)

Further, if ϵ is close to 0, one can choose Ĉ(γ, ϵ) independent of ϵ.

Proof of Lemma 1.5. Since ||u_{ϵ}||_{H^{1}} < C, we know that u_{ϵ} → u_{0} weakly in H^{1}(B_{1}) for some u_{0} ≥ 0. Noticing lim_{ϵ→0}(∫_{∂B_{1}} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1} = a > 0, we know that u_{0} satisfies

\{ \begin{matrix}−Δu_{0} & = ξ_{0}u_{0}^{p−1}, 0 ≤ u_{0} ≤ 1, & in B_{1} \\ \frac{∂u_{0}}{∂ν} & = −Zau_{0}^{q−1} − \frac{n−2}{2}u_{0} & on ∂B_{1}. \end{matrix}  (1.26)

Set u_{ϵ,1} = u_{ϵ} − u_{0}, then u_{ϵ,1} → 0 weakly in H^{1}(B_{1}). Since inf II_{Z} is not attained, ||u_{ϵ,1}||_{∞} → ∞. Let x_{ϵ,1} be the maximal point of u_{ϵ,1}, μ_{ϵ,1} = u_{ϵ,1}(x_{ϵ,1})^{(p_{ϵ}−2)/2} and

v_{ϵ,1}(x) = μϵ,1^{\frac{2}{pϵ−2}} u_{ϵ,1}(μ_{ϵ,1}x + x_{ϵ,1}) for x ∈ Ω_{ϵ,1} = \frac{B_{1} − x_{ϵ,1}}{μ_{ϵ,1}} .

As before, we know, that up to a subsequence, v_{ϵ,1} → v_{1} in C^{2}(B_{R} ∩ ℝ_{T_{1}}^{n}), and v_{1} satisfies

\{ \begin{matrix}−Δv_{1} & = ξ_{0}v_{1}^{p−1}, 0 ≤ v_{1} ≤ 1, v_{1}(0) = 1, & in ℝ_{T_{1}}^{n} \\ \frac{∂v_{1}}{∂ν} & = −Zav_{1}^{q−1} & on ∂ℝ_{T_{1}}^{n}, \end{matrix}  (1.27)

where

T_{1} = \overline{lim_{ϵ→∞}} \frac{dist(x_{ϵ,1}, ∂B_{1})}{μ_{ϵ,1}}.

For any R >> 1, define

u_{ϵ}^{(1)} = μ_{ϵ,1}^{− \frac{2}{pϵ−2}} v_{1}(μ_{ϵ,1}^{−1}(x − x_{ϵ,1})) · η(μ_{ϵ,1}^{−1}(x − x_{ϵ,1})), x ∈ B_{1},

where η(x) is a cutoff function with η(x) = 1 for x ∈ B_{R}(0) and η(x) = 0 in B_{2R}^{c}(0).

Due to (1.18) and (1.19), we only need to show that u_{0} = 0 (therefore u_{ϵ,1} = u_{ϵ}, x_{ϵ,1} = x_{ϵ} and T_{1} = T) and

||u_{ϵ,1} − u_{ϵ}^{(1)}||_{p_{ϵ},B_{1}}, ||u_{ϵ,1} − u_{ϵ}^{(1)}||_{q_{ϵ},∂B_{1}} = o_{ϵ}(1) + o_{R}(1)  (1.28)

where o_{ϵ}(1) → 0 as ϵ → 0 and o_{R}(1) → 0 as R → ∞.

It is easy to see that ||u_{ϵ,1}||_{p_{ϵ},B_{1}} ≥ C_{0} > 0 as ϵ small enough. Suppose that there exists some δ_{0} > 0, such that

||u_{ϵ,1} − u_{ϵ}^{(1)}||_{p_{ϵ},B_{1}} > δ_{0}.  (1.29)

Then we define u_{ϵ,2} = u_{ϵ} − u_{0} − u_{ϵ}^{(1)}. Easy to see that u_{ϵ,2} → 0 weakly in H^{1}(B_{1}). Since ||u_{ϵ,2}||_{p_{ϵ},B_{1}} ≥ δ_{0}, we know ||u_{ϵ,2}||_{∞} → ∞. Let x_{ϵ,2} be the maximal point of u_{ϵ,2}, μ_{ϵ,2} = u_{ϵ,2}(x_{ϵ,2})^{(p_{ϵ}−2)/2} and

v_{ϵ,2}(x) = μϵ,2^{\frac{2}{pϵ−2}} u_{ϵ,2}(μ_{ϵ,2}x + x_{ϵ,2}) for x ∈ Ω_{ϵ,2} = \frac{B_{1} − x_{ϵ,2}}{μ_{ϵ,2}} .

One observes that dist(x_{ϵ,1}, x_{ϵ,2}) > 100Rμ_{ϵ,1} for any fixed R as ϵ → 0. Also μ_{ϵ,1} ≤ μ_{ϵ,2}. Therefore, we know as before that, up to a subsequence, v_{ϵ,2} → v_{2} in C^{2}(B_{R} ∩ ℝ_{T_{2}}^{n}), and v_{2} satisfies

\{ \begin{matrix}−Δv_{2} & = ξ_{0}v_{2}^{p−1}, 0 ≤ v_{2} ≤ 1, v_{2}(0) = 1, & in ℝ_{T_{2}}^{n} \\ \frac{∂v_{2}}{∂ν} & = −Zav_{2}^{q−1} & on ∂ℝ_{T_{2}}^{n}, \end{matrix}  (1.30)

where

T_{2} = \overline{lim_{ϵ→∞}} \frac{dist(x_{ϵ,2}, ∂B_{1})}{μ_{ϵ,2}}.

Define

u_{ϵ}^{(2)} = μ_{ϵ,2}^{− \frac{2}{pϵ−2}} v_{2}(μ_{ϵ,2}^{−1}(x − x_{ϵ,2}))η(μ_{ϵ,2}^{−1}(x − x_{ϵ,2})), x ∈ B_{1},

where η(x) is the cutoff function as the above ( η(x) = 1 for x ∈ B_{R}(0) and η(x) = 0 in B_{2R}^{c}(0)).

Without loss of generality, we can assume that for any 0 < δ << 1, as ϵ → 0

||u_{ϵ,2} − u_{ϵ}^{(2)}||_{p_{ϵ},B_{1}} ≤ δ.  (1.31)

Otherwise we just keep this process going. Since ||u_{ϵ}||_{p_{ϵ}, B_{1}} = 1, we know this process must stop after several steps (depends on δ).

It is easy to check that ||u_{0} + u_{ϵ}^{(1)} + u_{ϵ}^{(2)}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}} = ||u_{0}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}} + ||u_{ϵ}^{(1)}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}} + ||u_{ϵ}^{(2)}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}} + o_{ϵ}(1) + o_{R}(1), and

||u_{0} + u_{ϵ}^{(1)} + u_{ϵ}^{(2)}||_{p_{ϵ},B_{1}}^{p_{ϵ}} = ||u_{0}||_{p_{ϵ},B_{1}}^{p_{ϵ}} + ||u_{ϵ}^{(1)}||_{p_{ϵ},B_{1}}^{p_{ϵ}} + ||u_{ϵ}^{(2)}||_{p_{ϵ},B_{1}}^{p_{ϵ}} + o_{ϵ}(1) + o_{R}(1). Combining with (1.31) and using Lemma 1.6, we have 1 − δ − o_{ϵ}(1) − o_{R}(1) ≤ ||u_{0}||_{p_{ϵ},B_{1}}^{p_{ϵ}} + ||u_{ϵ}^{(1)}||_{p_{ϵ},B_{1}}^{p_{ϵ}} + ||u_{ϵ}^{(2)}||_{p_{ϵ},B_{1}}^{p_{ϵ}} ≤ 1 + δ + o_{ϵ}(1) + o_{R}(1) and a^{\frac{q}{2−q}} −c(δ)−o_{ϵ}(1)−o_{R}(1) ≤ ||u_{0}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}}+||u_{ϵ}^{(1)}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}}+||u_{ϵ}^{(2)}||_{q_{ϵ},∂B_{1}}^{q_{ϵ}} ≤ a^{\frac{q}{2−q}} +c(δ)+o_{ϵ}(1)+o_{R}(1) where c(δ) → 0 as δ → 0.

Define

α_{0} = ||u_{0}||_{p_{ϵ}}^{p_{ϵ}}, β_{0} = ||u_{0}||_{q_{ϵ}}^{q_{ϵ}}/||u_{ϵ}||_{q_{ϵ}}^{q_{ϵ}},

and

α_{i} = ||u_{ϵ}^{(i)}||_{p_{ϵ}}^{p_{ϵ}}, β_{i} = ||u_{ϵ}^{(i)}||_{q_{ϵ}}^{q_{ϵ}}/||u_{ϵ}||_{q_{ϵ}}^{q_{ϵ}} for i = 1, 2.

Therefore

1 − δ − o_{ϵ}(1) − o_{R}(1) ≤ α_{0} + α_{1} + α_{2} ≤ 1 + δ + o_{ϵ}(1) + o_{R}(1)  (1.32)
1 − c′(δ) − o_{ϵ}(1) − o_{R}(1) ≤ β_{0} + β_{1} + β_{2} ≤ 1 + c′(δ) + o_{ϵ}(1) + o_{R}(1)  (1.33)

where c′(δ) → 0 as δ → 0.

It follows from (1.26), (1.27) and (1.30) that

||∇u_{0}||_{2,B_{1}}^{2} + Za||u_{0}||_{q,∂B_{1}}^{q} = ξ_{0}||u_{0}||_{p,B_{1}}^{p},

||∇v_{i}||_{2,ℝ_{T_{i}}^{+}}^{2} + Za||v_{i}||_{q,∂ℝ_{T_{i}}^{+}}^{q} = ξ_{0}||v_{i}||_{p,ℝ_{T_{i}}^{+}}^{p} for i = 1, 2.

Using (1.18), (1.19), (1.20) and the above, we have

||∇u_{0}||_{2,B_{1}}^{2} + Zβ_{0}^{1− \frac{2}{q}} ||u_{0}||_{q,∂B_{1}}^{2} = ξ_{0}α_{0}^{1− \frac{2}{p}} ||u_{0}||_{p,B_{1}}^{2} + o_{ϵ}(1),

||∇v_{i}||_{2,ℝ_{T_{i}}^{+}}^{2} + Zβ_{i}^{1− \frac{2}{q}} ||v_{i}||_{q,∂ℝ_{T_{i}}^{+}}^{2} ≤ ξ_{0}α_{i}^{1− \frac{2}{p}} ||v_{i}||_{p,ℝ_{T_{i}}^{+}}^{2} + o_{ϵ}(1) + o_{R}(1) for i = 1, 2.

Since inf I_{Z} ≥ ξ_{0} and α_{i} ≤ 1, we know that as ϵ → 0 and R → ∞,

β_{i}^{1− \frac{2}{q}} ≤ α_{i}^{1− \frac{2}{p}} for i = 0, 1, 2.  (1.34)

Noticing q < p and α_{1}, α_{2} are larger than some fixed number, we derive a contradiction due to (1.32), (1.33) and (1.34) when we choose δ suitable small and R suitable large. Thus (1.29) is false, that is: as ϵ → 0,

||u_{ϵ,1} − u_{ϵ}^{(1)}||_{p_{ϵ},B_{1}} → 0.

Using Lemma 1.6, we know that (1.28) holds. Similarly, we can show that u_{0} = 0. Lemma 1.5 is established.

2 Domain case

In this section, we assume that Ω is a smooth bounded domain in ℝ^{n} and give the proof of Theorem 0.2.

First we present a rough inequality with a slight larger constant than the sharp one S_{1}(Z) in Theorem 0.2. However, the case Z = Z_{0} is included.

Proposition 2.1 Let Z ∈ (−1/S, Z_{0}]. For any δ > 0, there exists C(δ) > 0 such that

(∫_{Ω} |u|^{p})^{\frac{2}{p}} ≤ (S_{1}(Z)+δ)( ∫_{Ω} |∇u|^{2}+Z(∫_{∂Ω} |u|^{q})^{\frac{2}{q}} )+C(δ) ∫_{∂Ω} u^{2}, ∀u ∈ H^{1}(Ω). (2.1)

When Z > 0, due to the positive L^{q} term in the right hand side of (2.1), we can not prove this inequality directly from Theorem 0.1 via the partition of unit, neither can we prove it by a similar argument used in [17] and [18]. Here, we again use blowup argument to prove this proposition.

Proof. We prove it by contradiction. Assume (2.1) is not true, that is, there exists δ_{1} > 0 such that ∀α > 1,

inf_{u∈H^{1}(Ω)} I_{Z,α}(u) := inf_{u∈H^{1}(Ω)} \frac{∫_{Ω} |∇u|^{2} + Z(∫_{∂Ω} |u|^{q})^{2/q} + α ∫_{∂Ω} u^{2}}{(∫_{Ω} |u|^{p})^{2/p}} := ξ_{Z,α} < \frac{1}{S_{1}(Z) + δ_{1}} .

(2.2)

Let ᾱ be some positive constant such that

\frac{1}{S} ||u||_{q,∂Ω}^{2} ≤ ||∇u||_{2,Ω}^{2} + ᾱ||u||_{2,∂Ω}^{2}, ∀u ∈ H^{1}(Ω).  (2.3)

The existence of such ᾱ was shown in [17].

Lemma 2.1 If Z ≥ 0, under condition (2.2), inf I_{Z,α} is achieved; If −1/S < Z < 0, for any fixed α > ᾱ, under condition (2.2), inf I_{Z,α} is achieved.

Proof. This can be proved in a similar way as that of Proposition 1.1. We sketch the proof here for readers’ convenience. For u ≠ 0, define

I_{ϵ}(u) = \frac{∫_{Ω} |∇u|^{2} + Z(∫_{∂Ω} |u|^{q_{ϵ}})^{2/q_{ϵ}} + α ∫_{∂Ω} u^{2}}{(∫_{Ω} |u|^{p_{ϵ}})^{2/p_{ϵ}}} .

If Z ≥ 0, or −1/S < Z < 0 and α > ᾱ, as before, we know I_{ϵ}(u) ≥ 0. The standard variational method shows that ∃ u_{ϵ} ≥ 0 with ||u_{ϵ}||_{p_{ϵ}} = 1 such that

I_{ϵ}(u_{ϵ}) = inf I_{ϵ}(u) := ξ_{ϵ}.

We want to show that ||u_{ϵ}||_{∞} ≤ C. Lemma 2.1 follows from this fact easily. Suppose ||u_{ϵ}||_{∞} → ∞ up to a subsequence. Easy to see that u_{ϵ} satisfies

\{ \begin{matrix}−Δu_{ϵ} & = ξ_{ϵ}u_{ϵ}^{p_{ϵ}−1} & in Ω \\ \frac{∂u_{ϵ}}{∂ν} & = −Z(∫_{∂Ω} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}u_{ϵ}^{q_{ϵ}−1} − αu_{ϵ} & on ∂Ω. \end{matrix}  (2.4)

By [10], we know that there exists a x_{ϵ} ∈ Ω such that u_{ϵ}(x_{ϵ}) = ||u_{ϵ}||_{∞} → ∞. Define

\{ \begin{matrix}μ_{ϵ} & = (v_{ϵ}(x_{ϵ}))^{− \frac{pϵ−2}{2}} , Ω_{ϵ} = \frac{Ω−x_{ϵ}}{μ_{ϵ}} , \\ v_{ϵ}(x) & = μϵ^{\frac{2}{pϵ−2}} u_{ϵ}(μ_{ϵ}x + x_{ϵ}) for x ∈ Ω_{ϵ}. \end{matrix}

Then v_{ϵ} satisfies

\{ \begin{matrix}−Δv_{ϵ} & = ξ_{ϵ}v_{ϵ}^{p_{ϵ}−1}, 0 ≤ v_{ϵ} ≤ 1, v(0) = 1, & in Ω_{ϵ} \\ \frac{∂v_{ϵ}}{∂ν} & = −Z(∫_{∂Ω} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}v_{ϵ}^{q_{ϵ}−1} − αμ_{ϵ}v_{ϵ} & on ∂Ω_{ϵ}. \end{matrix}  (2.5)

As in the proof of Proposition 1.1 (also we need to use (2.3) when Z < 0), we can show that when Z ≥ 0, or −1/S < Z < 0 and α > ᾱ,

0 ≤ lim_{ϵ→0}ξ_{ϵ} := ξ_{0} ≤ ξ_{Z,α} < \frac{1}{S_{1}(Z) + δ_{1}}  (2.6)

and

∫_{∂Ω} u_{ϵ}^{q_{ϵ}} ≥ C, T := \overline{lim_{ϵ→∞}} \frac{dist(x_{ϵ}, ∂Ω)}{μ_{ϵ}} < ∞.  (2.7)

It is easy to see α||u_{ϵ}||_{2,∂Ω} ≤ C. Combining this with (2.7) and the definition of μ_{ϵ}, we have

αμ_{α} ≤ \frac{αμ_{α}}{C} ∫_{∂Ω} u_{ϵ}^{q_{ϵ}} ≤ \frac{α}{C} ∫_{∂Ω} u_{ϵ}^{2} ≤ C.  (2.8)

Set C_{1} = lim_{ϵ→0} αμ_{ϵ}. By standard elliptic estimates, from (2.5), (2.7) and (2.8), we know that v_{ϵ} → v in C^{2}(B_{R} ∩ ℝ_{T}^{n}), and v satisfies

\{ \begin{matrix}−Δv & = ξ_{0}v^{p−1}, 0 ≤ v ≤ 1, v(0) = 1, & in ℝ_{T}^{n} \\ \frac{∂v}{∂ν} & = −Zav^{q−1} − C_{1}v & on ∂ℝ_{T}^{n}. \end{matrix}  (2.9)

where a = lim_{ϵ→0}(∫_{∂Ω} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}. If Z ≤ 0, notice that ||v||_{p,ℝ_{T}^{n}} ≤ 1 and (∫_{∂ℝ_{T}^{n}} v^{q})^{2/q−1} ≥ a (see (1.18) for details), we have

≤ \frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Za ∫_{∂ℝ^{n}_{T}} v^{q}}{∫_{ℝ^{n}_{T}} v^{p}} ≤ ξ_{0} < \frac{1}{S_{1}(Z)}.^{\frac{∫_{ℝ^{n}_{T}} |∇v|^{2}+Z(∫_{∂ℝ^{n}_{T}} v^{q})^{2/q}}{(∫_{ℝ^{n}_{T}} v^{p})^{2/p}}}

This contradicts to Theorem 0.1.

If Z > 0, slightly modifying the proof of Lemma 1.5 (we need to use Theorem 0.1 here), we can show that as ϵ → 0

∫_{∂Ω_{ϵ}} v_{ϵ}^{q_{ϵ}} → ∫_{∂ℝ_{T}^{n}} v^{q}.

Then following the proof of Proposition 1.1 closely, we can get

\frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q}}{(∫_{ℝ_{T}^{n}} v^{p})^{2/p}} ≤ ξ_{0} < \frac{1}{S_{1}(Z)}.

This again contradicts to Theorem 0.1. We thereby complete the proof of Lemma 2.1.

As α > ᾱ, without loss of generality, we can assume that inf I_{Z,α}(u) = I_{Z,α}(u_{α}) with u_{α} ≥ 0 and ||u_{α}||_{p,Ω} = 1. It is easy to see that u_{α} satisfies

\{ \begin{matrix}−Δu_{α} & = ξ_{Z,α}u_{α}^{p−1} & in Ω \\ \frac{∂u_{α}}{∂ν} & = −Z(∫_{∂Ω} u_{α}^{q})^{2/q−1}u_{α}^{q−1} − αu_{α} & on ∂Ω, \end{matrix}  (2.10)

and

α||u_{α}||_{2,∂Ω} ≤ C.  (2.11)

As in the proof of Lemma 1.3, due to ξ_{Z,α} < 1/(S_{1} + δ_{1}), we know that there exists a constant C > 0, such that

∫_{∂Ω} u_{α}^{q} ≥ C.  (2.12)

Due to Cherrier, we know that u_{α} is smooth up to the boundary. Let u_{α}(x_{α}) = ||u_{α}||_{∞} for some x_{α} ∈ Ω, and define

\{ \begin{matrix}μ_{α} & = (u_{α}(x_{α}))^{− \frac{p−2}{2}} , Ω_{α} = (Ω − x_{α})/μ_{α}, \\ v_{α}(x) & = μα^{\frac{2}{p−2}} u_{α}(μ_{α}x + x_{α}), x ∈ Ω_{α}. \end{matrix}  (2.13)

Then v_{α} satisfies

\{ \begin{matrix}−Δv_{α} & = ξ_{Z,α}v_{α}^{p−1}, 0 ≤ v_{α} ≤ 1, v(0) = 1, & in Ω_{α} \\ \frac{∂v_{α}}{∂ν} & = −Z(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1}v_{α}^{q−1} − αμ_{α}v_{α} & on ∂Ω_{α}. \end{matrix}  (2.14)

From (2.11), (2.12) and the definition of μ_{α}, we know that

αμ_{α} ≤ \frac{αμ_{α}}{C} ∫_{∂Ω} u_{α}^{q} = \frac{α}{C} ∫_{∂Ω} u_{α}^{2} ≤ C.

Set C_{2} = lim_{α→∞}αμ_{α} ≥ 0. Thus ||u_{α}||_{∞} → ∞. Also as in the proof of Proposition 1.1, due to ξ_{Z,α} < 1/(S_{1} + δ_{1}), we know

\overline{lim_{α→∞}} \frac{dist(x_{α}, ∂Ω)}{μ_{α}} = T < ∞.  (2.15)

By standard elliptic estimates, we know that v_{ϵ} → v in C^{2}(B_{R} ∩ ℝ_{T}^{n}), and v ≠ 0 satisfies

\{ \begin{matrix}−Δv & = ξ_{∞}v^{p−1}, 0 ≤ v ≤ 1, v(0) = 1, & in ℝ_{T}^{n} \\ \frac{∂v}{∂ν} & = −Za_{1}v^{q−1} − C_{2}v & on ∂ℝ_{T}^{n} \end{matrix}  (2.16)

where ξ_{∞} = lim_{α→∞} ξ_{Z,α} ≤ 1/(S_{1}(Z) + δ_{1}), a_{1} = lim_{α→∞}(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1}. If Z ≤ 0, one can easily see as before that

\frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q}}{(∫_{ℝ_{T}^{n}} v^{p})^{2/p}} ≤ \frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q} + C_{2} ∫_{∂ℝ_{T}^{n}} v^{2}}{∫_{ℝ_{T}^{n}} v^{p}} = ξ_{∞} < \frac{1}{S_{1}(Z)}.

If Z > 0, similarly, using Theorem 0.1, we can prove as in the proof of Lemma 1.5 that

lim_{α→∞}(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1} = (∫_{∂ℝ_{T}^{n}} v^{q})^{2/q−1},

thus we also have

\frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q}}{(∫_{ℝ_{T}^{n}} v^{p})^{2/p}} ≤ \frac{∫_{ℝ_{T}^{n}} |∇v|^{2} + Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q} + C_{2} ∫_{∂ℝ_{T}^{n}} v^{2}}{∫_{ℝ_{T}^{n}} v^{p}} = ξ_{∞} < \frac{1}{S_{1}(Z)}.

Both of the above two inequalities contradict to Theorem 0.1. Thus the proof of Proposition 2.1 is completed.

From now on, we begin to prove Theorem 0.2 through an argument by contradiction. Note that we assume Z < Z_{0}, thus 1/S_{1}(Z) < 1/S_{1}.

Suppose Theorem 0.2 is false, then for any α > ᾱ,

inf_{H^{1}(Ω)} I_{α}(u) = inf_{H^{1}(Ω)} \frac{∫_{Ω} |∇u|^{2} + Z(∫_{∂Ω} |u|^{q})^{2/q} + α ∫_{∂Ω} u^{2}}{(∫_{Ω} |u|^{p})^{2/p}} := ξ_{α} < \frac{1}{S_{1}(Z)}. (2.17)

From the proof of Proposition 2.1, we know that under assumption (2.17), inf I_{α}(u) is attained. Without loss of generality, we assume inf I_{α}(u) = I_{α}(u_{α}) with u_{α} ≥ 0 and ||u_{α}||_{p,Ω} = 1. It is easy to see that u_{α} satisfies

\{ \begin{matrix}−Δu_{α} & = ξ_{α}u_{α}^{p−1} & in Ω \\ \frac{∂u_{α}}{∂ν} & = −Z(∫_{∂Ω} u_{α}^{q})^{2/q−1}u_{α}^{q−1} − αu_{α} & on ∂Ω. \end{matrix}  (2.18)

Lemma 2.2 As α → ∞,

α||u_{α}||_{2,∂Ω}^{2} → 0, ξ_{α} → \frac{1}{S_{1}(Z)}.  (2.19)

Proof. From (2.17), Proposition 2.1 and ||u_{α}||_{p,Ω} = 1, we know that for any δ > 0, there exists a constant C(δ) such that

1 + \frac{δ}{S_{1}(Z)} ≥ (S_{1}(Z) + δ)ξ_{α}
= (S_{1}(Z) + δ)(||∇u_{α}||_{2,Ω}^{2} + Z||u_{α}||_{q,∂Ω}^{2} + α||u_{α}||_{2,∂Ω}^{2})
≥ 1 + (α(S_{1}(Z) + δ) − C(δ))||u_{α}||_{2,∂Ω}^{2}.

Thus

1 + \frac{δ}{S_{1}(Z)} ≥ (S_{1}(Z) + δ) lim sup_{α→∞} ξ_{α} ≥ 1 + (S_{1}(Z) + δ) lim sup_{α→∞} α∥u_{α}∥_{2,∂Ω}^{2},

1 + \frac{δ}{S_{1}(Z)} ≥ (S_{1}(Z) + δ) lim inf_{α→∞} ξ_{α} ≥ 1 + (S_{1}(Z) + δ) lim inf_{α→∞} α∥u_{α}∥_{2,∂Ω}^{2}.

Sending δ → 0, we have our lemma.

Since 1/S_{1}(Z) < 1/S_{1}, as in the proof of Lemma 1.3, we have the following.

Lemma 2.3 There exists a constant C > 0 such that

∫_{∂Ω} u_{α}^{q} ≥ C.  (2.20)

Remark 2.1 Condition Z < Z_{0} (thus 1/S_{1}(Z) < 1/S_{1}) is essential in the proof of (2.20). Actually, if Z = Z_{0}, we can show that ∫_{∂Ω} u_{α}^{q} → 0 as α → ∞. More details will be discussed in Section 4.

Since u_{α} satisfies (2.18), due to Cherrier, we know that u_{α} is smooth up to the boundary. Let u_{α}(x_{α}) = ||u||_{∞} for some x_{α} ∈ Ω, and define

\{ \begin{matrix}μ_{α} & = (u_{α}(x_{α}))^{− \frac{p−2}{2}} , Ω_{α} = (Ω − x_{α})/μ_{α}, \\ v_{α}(x) & = μα^{\frac{2}{p−2}} u_{α}(μ_{α}x + x_{α}) for x ∈ Ω_{α}. \end{matrix}  (2.21)

Then v_{α} satisfies

\{ \begin{matrix}−Δv_{α} & = ξ_{α}v_{α}^{p−1}, 0 ≤ v ≤ 1, v(0) = 1 & in Ω_{α} \\ \frac{∂v_{α}}{∂ν} & = −Z(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1}v_{α}^{q−1} − αμ_{α}v_{α} & on ∂Ω_{α}. \end{matrix}  (2.22)

Combining (2.19) with (2.20), as being shown in (2.8), we have

αμ_{α} → 0.  (2.23)

Also if we set \overline{lim}_{α→∞}dist(x_{α}, ∂Ω)/μ_{α} = T, due to 1/S_{1}(Z) < 1/S_{1}, as before, we know T < ∞.

By standard elliptic estimates we know that v_{α} → v in C^{3}(\overline{Ω_{α}} ∩ B_{R}(0)) for all R > 1. And again, using the argument in the proof of Lemma 1.5 (also we need to use Theorem 0.1 here), we know

lim_{α→∞} ∫_{Ω_{α}} |v_{α} − v|^{p} = lim_{α→∞} ∫_{Ω_{α}} |∇v_{α} − ∇v|^{2} = lim_{α→∞} ∫_{∂Ω_{α}} |v_{α} − v|^{q} = 0.  (2.24)

It follows that

lim_{α→∞}(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1} = (∫_{∂ℝ_{T}^{n}} v^{q})^{2/q−1},

and v satisfies

\{ \begin{matrix}−Δv = \frac{1}{S_{1}(Z)}v^{p−1}, 0 ≤ v(x) ≤ 1, v(0) = 1, & in ℝ_{T}^{n}, \\ \frac{∂v}{∂ν} = −Z(∫_{∂ℝ_{T}^{n}} v^{q})^{2/q−1}v^{q−1} & on ∂ℝ_{T}^{n}. \end{matrix}  (2.25)

If Z ≥ 0, it follows from [16] that

v = (\frac{1}{1 + c(n)(|x′|^{2} + |x_{n}|^{2})})^{\frac{n−2}{2}}  (2.26)

where c(n) = 1/(S_{1}(Z)(n − 2)n).

If Z < 0, due to 1/S_{1}(Z) < 1/(2^{2/n}S_{1}), one can check as in the proof of Lemma 1.4 that T = 0. Due to (2.24), we know ||v||_{p,ℝ_{+}^{n}} = 1. It follows from the proof of Theorem 0.1 that

v = (\frac{ϵ_{z}}{ϵ_{z}^{2} + |x′|^{2} + (x_{n} − ϵ_{z}d_{z})^{2}} )^{\frac{n−2}{2}}  (2.27)

where d_{z} satisfies (0.3), ϵ_{z} = (Z_{0}^{2} − Z^{2})/Z_{0}^{2}.

We are ready to give a L^{∞} estimate on v_{α} through the Moser iteration method as we did in [17] and [18]. First, Let’s recall that the conformal Laplacian operator L_{g} and the conformal boundary operator B_{g} corresponding to metric g are given by

\{ \begin{matrix}L_{g}ψ = Δ_{g}ψ − a(n)R_{g}ψ, \\ B_{g}ψ = \frac{∂_{g}ψ}{∂ν} + b(n)H_{g}ψ, \end{matrix}  (2.28)

where a(n) = \frac{n−2}{4(n−1)}, b(n) = \frac{n−2}{2}, R_{g} is the scalar curvature of Ω, and H_{g} is the mean curvature of ∂Ω with respect to the inner normal of ∂Ω (e.g., the unit ball in ℝ^{n} has positive mean curvature).

We write g_{0} as the standard Euclidean metric. Let v(x) be given by (2.26) or (2.27), and ĝ = v^{4/(n−2)}g_{0}, i.e. ĝ_{ij}dx^{i}dx^{j} = v^{4/(n−2)}dx^{i}dx^{i}. Then for all ψ ∈ C^{∞}(Ω_{α})

\{ \begin{matrix}Lĝ(ψ/v) = v^{−(n+2)/(n−2)}L_{g_{0}}(ψ) & in Ω_{α}, \\ Bĝ(ψ/v) = v^{−n/(n−2)}B_{g_{0}}(ψ) & on ∂Ω_{α}. \end{matrix}  (2.29)

Let ψ = v_{α} in (2.29) and write w_{α} = v_{α}/v, we have

\{ \begin{matrix}Δ_{g_{0}}v_{α} = v^{(n+2)/(n−2)}(Δĝw_{α} − a(n)Rĝw_{α}) & in Ω_{α}, \\ \frac{∂_{g_{0}}v_{α}}{∂ν} + b(n)H_{g_{0}}u_{α} = v^{n/(n−2)}(\frac{∂ĝw_{α}}{∂ν} + b(n)Hĝw_{α}) & on ∂Ω_{α}. \end{matrix}  (2.30)

Let ψ = v in (2.29), we get

\{ \begin{matrix}−a(n)Rĝv^{(n+2)/(n−2)} = Δ_{g_{0}}v \\ b(n)Hĝv^{n/n−2} = \frac{∂_{g_{0}}v}{∂ν} + b(n)H_{g_{0}}v. \end{matrix}  (2.31)

Combining (2.30), (2.31) with (2.22), we have

\{ \begin{matrix}−Δĝw_{α} = ξ_{α}w_{α}^{p−1} + Δ_{g_{0}}v/v^{(n+2)/(n−2)} · w_{α} & in Ω_{α}, \\ \frac{∂ĝw_{α}}{∂ν} = −Z(∫_{∂Ω} v_{α}^{q})^{2/q−1}w_{α}^{q−1} − (αμ_{α}/v^{2/(n−2)} + \frac{∂v}{∂ν} /v^{n/(n−2)})w_{α} & on ∂Ω_{α}. \end{matrix}

(2.32)

By a similar calculation to the proof of Lemma 2.3.1 in [22] ( see also [18]), we have, as α large enough, that

αμ_{α}/v^{2/(n−2)} + \frac{∂v}{∂ν} /v^{n/(n−2)} ≥ 0, x ∈ ∂Ω_{α}.

Thus w_{α} satisfies

\{ \begin{matrix}−Δĝw_{α} ≤ ξ_{α}w_{α}^{p−1} & in Ω_{α}, \\ \frac{∂ĝw_{α}}{∂ν} ≤ −Z(∫_{∂Ω} v_{α}^{q})^{2/q−1}w_{α}^{q−1} & on ∂Ω_{α}. \end{matrix}  (2.33)

Note Z may be a negative number here.

Define Θ_{α} = {y : y = x/|x|^{2}, x ∈ Ω_{α}}, y = x/|x|^{2}, W_{α}(y) = w_{α}(x) and g̃(y) = ĝ(x). Then we have

\{ \begin{matrix}−Δg̃W_{α} ≤ ξ_{α}W_{α}^{p−1} & in Θ_{α}, \\ \frac{∂g̃W_{α}}{∂ν} ≤ −Z(∫_{∂Ω} v_{α}^{q})^{2/q−1}W_{α}^{q−1} & on ∂Θ_{α}. \end{matrix}  (2.34)

If we write g̃(y) = g̃_{ij}(y)dy^{i}dy^{j}, due to g̃(y) = ĝ(x) = v^{4/(n−2)}g_{0}, we know g̃_{ij} = |x|^{4}v^{4/(n−2)}δ_{ij}. Thus, there exists a C > 0, such that 1/C ≤ g̃_{ij}(y) ≤ C for y ∈ Θ_{α} ∩ B_{1}(0).

Using these notations, we rewrite (2.24) in the following setting.

Lemma 2.4

lim_{α→∞} ∫_{Θ_{α}} |W_{α} − 1|^{p}dvg̃ = lim_{α→∞} ∫_{∂Θ_{α}} |W_{α} − 1|^{q}dsg̃ = 0.  (2.35)

Now we focus on proving the following proposition.

Proposition 2.2 There exists a constant C > 0 such that, ∀α > ᾱ,

v_{α} ≤ Cv for x ∈ Ω̄_{α},  (2.36)

where v is given by (2.26) or (2.27), depending on the value of Z.

Proof. We only need to show that (2.36) holds for α large, thus, without loss of generality, we can assume w_{α} = v_{α}/v satisfies (2.33). Due to the fact that v_{α} → v in C^{3}(\overline{Ω_{α}} ∩ B_{R}(0)) for all R > 1 , we only need to show that (2.36) holds for |x| large, that is to show W_{α} ≤ C for |y| small.

Let η be some smooth cutoff function with compact support in B_{1}(0). Multiplying (2.34) by W_{α}^{k}η^{2} for k > 1 and integrating by parts, we obtain (since ξ_{α}, Z(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1} ≤ C)

∫_{Θ_{α}} ∇g̃W_{α} · ∇g̃(W_{α}^{k}η^{2})dvg̃
≤ ξ_{α} ∫_{Θ_{α}} W_{α}^{p−1+k}η^{2}dvg̃ − Z(∫_{∂Ω_{α}} v_{α}^{q})^{2/q−1} · ∫_{∂Θ_{α}} W_{α}^{q−1+k}η^{2}dsĝ
≤ C ∫_{Θ_{α}} W_{α}^{p−1+k}η^{2}dvg̃ + C ∫_{∂Θ_{α}} W_{α}^{q−1+k}η^{2}dsĝ.

Direct computation yields:

∫_{Θ_{α}} ∇g̃W_{α} · ∇g̃(W_{α}^{k}η^{2})dvg̃
= \frac{4k}{(k+1)^{2}} ∫_{Θ_{α}} |∇g̃(W_{α}^{k+1/2}η)|^{2}dvg̃ + \frac{k−1}{(k+1)^{2}} ∫_{Θ_{α}} W_{α}^{k+1}Δg̃η^{2}dvg̃
−\frac{4k}{(k+1)^{2}} ∫_{Θ_{α}} W_{α}^{k+1}|∇g̃η|^{2}dvg̃ − \frac{k−1}{(k+1)^{2}} ∫_{∂Θ_{α}} W_{α}^{k+1}∇g̃η^{2} · νdsg̃.

We derive from the last two inequalities that

\begin{matrix}∫_{Θ_{α}} |∇g̃(W_{α}^{k+1/2}η)|^{2}dvg̃ \\ ≤ ∫_{Θ_{α}} W_{α}^{k+1}(|Δg̃η^{2}| + |∇g̃η|^{2})dvg̃ + \frac{k−1}{4k} ∫_{∂Θ_{α}} W_{α}^{k+1}∇g̃η^{2} · νdsg̃ \\ +\frac{C(k+1)^{2}}{4k} ∫_{Θ_{α}} W_{α}^{p−1+k}η^{2}dvg̃ + \frac{C(k+1)^{2}}{4k} ∫_{∂Θ_{α}} W_{α}^{q−1+k}η^{2}dsg̃. \end{matrix}  (2.37)

Set, for 0 < \overline{δ} < 1/2 (\overline{δ} will be chosen later),

R_{i} = (1 + \frac{1}{2^{i−1}} )\overline{δ}, i = 1, 2, 3, · · · .  (2.38)

we can choose some smooth cutoff function η_{i} satisfying

\{ \begin{matrix}η_{i}(y) = 1, |y| < R_{i+1}; η_{i}(y) = 0, |y| > R_{i}; \\ |∇g̃η_{i}| ≤ C2^{i}, |∇2 ; g̃η_{i}| ≤ C4^{i}. \end{matrix}

Taking η = η_{i} in (2.37) and using Sobolev embedding theorem (see Appendix A in [17]) we reach

\begin{aligned}& \{ ∫_{Θ_{α}∩B_{R_{i}}}(W_{α}^{(k+1)/2}η_{i})^{p}dvg̃\}^{2/p} + \{ ∫_{∂Θ_{α}∩B_{R_{i}}}(W_{α}^{(k+1)/2}η_{i})^{q}dsg̃\}^{2/q} \\ ≤ & 4^{i}C ∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{k+1}dvg̃ + 2^{i}C ∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{k+1}dsg̃ \\ & +\frac{C(k+1)^{2}}{k} ∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{p−1+k}dvg̃ + \frac{C(k+1)^{2}}{k} ∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{q−1+k}dsg̃. \end{aligned}  (2.39)

It follows from Lemma 2.4 that for any ϵ_{0} > 0, there exists a δ_{0} > 0 such that for any δ < δ_{0},

∫_{Θ_{α}∩B_{δ}(0)} W_{α}^{p}dvg̃ + ∫_{∂Θ_{α}∩B_{δ}(0)} W_{α}^{q}dsg̃ < ϵ_{0}.

By the standard Moser iteration, we know for any s̄ > p, there exists δ_{1} > 0, such that for any p ≤ s < s̄, δ < δ_{1},

∫_{Θ_{α}∩B_{δ}(0)} W_{α}^{s}dvg̃ + ∫_{∂Θ_{α}∩B_{δ}(0)} W_{α}^{s}dsg̃ < C(s̄).  (2.40)

Choose s_{0} ∈ (p, s̄) and s_{0} close to p. Let r_{0} = s_{0}/(p − 2), β = p(r_{0} − 1)/(2r_{0}) and \frac{t_{0}}{t_{0}−1} = \frac{q}{2β}. We can check β > 1. Also as s_{0} is close to p, β is close to 1. Therefore, we can make 2β < q and (q −2)t_{0} < s̄ after we choose a suitable s_{0}. Choose 2δ̄ < δ_{1}. By Hölder inequality, we know

∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{p−1+k}dvg̃ ≤ ( ∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{(k+1)r_{0}/(r_{0}−1)}dvg̃)^{(r_{0}−1)/r_{0}}(∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{s_{0}})^{\frac{1}{r_{0}}}

and

∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{q−1+k}dsg̃ ≤ ( ∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{(k+1)t_{0}/(t_{0}−1)}dsg̃)^{(t_{0}−1)/t_{0}}(∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{(q−2)t_{0}})^{\frac{1}{t_{0}}} .

Combining the above two inequalities with (2.40) we have

∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{p−1+k}dvg̃ ≤ C( ∫_{Θ_{α}∩B_{R_{i}}} W_{α}^{(k+1)r_{0}/(r_{0}−1)}dvg̃)^{(r_{0}−1)/r_{0}}  (2.41)

and

∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{q−1+k}dsg̃ ≤ C( ∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{(k+1)t_{0}/(t_{0}−1)}dsg̃)^{(t_{0}−1)/t_{0}}.  (2.42)

Also, from Hölder inequality,

∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{1+k}dsg̃ ≤ C( ∫_{∂Θ_{α}∩B_{R_{i}}} W_{α}^{(k+1)t_{0}/(t_{0}−1)}dsg̃)^{(t_{0}−1)/t_{0}}.  (2.43)

Set p_{0} = 2r_{0}/(r_{0} − 1) < p, p_{i} = βp_{i−1} = β^{i−1}p, q_{i} = p_{i}(r_{0} − 1)/r_{0} = 2β^{i}, for i ≥ 1. Taking k = q_{i} − 1 ( for i ≥ 1 ) in (2.39), and using (2.41), (2.42) and (2.43), we obtain

\begin{aligned}& ∥W_{α}∥_{p_{i+1},Θ_{α}∩B_{R_{i+1}}}^{q_{i}} + ∥W_{α}∥_{q̄_{i+1},∂Θ_{α}∩B_{R_{i+1}}}^{q_{i}} \\ ≤ & (4^{i}C + \frac{q_{i}^{2}C}{(q_{i}−1)})(∥W_{α}∥_{p_{i},Θ_{α}∩B_{R_{i}}}^{q_{i}} + ∥W_{α}∥_{q̄_{i},∂Θ_{α}∩B_{R_{i}}}^{q_{i}}), \end{aligned}

where q̄_{i} = q_{i} · \frac{t_{0}}{t_{0}−1} for i ≥ 1. Since β > 1, we have (a^{β} + b^{β})^{2/β} ≤ a + b. It follows that

\begin{aligned}& (∥W_{α}∥_{p_{i+1},Θ_{α}∩B_{R_{i+1}}}^{q_{i+1}} + ∥W_{α}∥_{q̄_{i+1},∂Θ_{α}∩B_{R_{i+1}}}^{q_{i+1}})^{1/q_{i+1}} \\ ≤ & (4^{i}C + \frac{q_{i}^{2}C}{q_{i}−1})^{1/q_{i}}(∥W_{α}∥_{p_{i},Θ_{α}∩B_{R_{i}}}^{q_{i}} + ∥W_{α}∥_{q̄_{i},∂Θ_{α}∩B_{R_{i}}}^{q_{i}})^{1/q_{i}}. \end{aligned}  (2.44)

As in [17], one can easily check that

∏_{i=1}^{∞}(4^{i}C + \frac{q_{i}^{2}C}{q_{i} − 1})^{1/q_{i}} ≤ C < ∞,

thus

∥W_{α}∥_{p_{i+1},Θ_{α}∩B_{R_{i+1}}} ≤ C(∥W_{α}∥_{p_{1},Θ_{α}∩B_{R_{1}}}^{2β} + ∥W_{α}∥_{q_{1},∂Θ_{α}∩B_{R_{1}}}^{2β})^{1/(2β)}
= C(∥W_{α}∥_{p,Θ_{α}∩B_{R_{1}}}^{2β} + ∥W_{α}∥_{q,∂Θ_{α}∩B_{R_{1}}}^{2β})^{1/(2β)}
≤ C_{1}.

Sending i to ∞, we have

∥W_{α}∥_{L^{∞}(Θ_{α}∩B\overline{δ})} ≤ C(\overline{δ}).  (2.45)

Therefore we complete the proof of Proposition 2.2.

Let Q_{α} ∈ ∂Ω be the closest point to x_{α}. By choosing an appropriate coordinate system centered at Q_{α}, we can assume without loss of generality that Q_{α} = 0, g_{ij}(0) = δ_{ij}, B_{1}^{+}(0) ⊂ Ω, {(x′, 0) : |x′| < 1} ⊂ ∂Ω .

Let R_{α} = 1/(αμ_{α}), h_{α} = g_{ij}(μ_{α}x)dx^{i}dx^{j} in B_{10R_{α}}^{+}(0), and

v̄_{α}(x) = μ_{α}^{(n−2)/2}u_{α}(μ_{α}x + x_{α}), for x ∈ B_{10R_{α}}^{+}(0).

It follows from (2.23) and (2.22) that R_{α} → ∞ as α → ∞, and v̄_{α} satisfies

\{ \begin{matrix}−Δ_{h_{α}} v̄_{α} = ξ_{α}v̄_{α}^{p−1} & in B_{10R_{α}}^{+}(0) \\ \frac{∂_{hα} v̄_{α}}{∂ν} = −Z(∫_{∂Ω} u_{α}^{q})^{2/q−1}v̄_{α}^{q−1} − αμ_{α}v̄_{α} & on \{(x′, 0) : |x′| < 10R_{α}\} \\ 0 < v̄_{α} ≤ μ_{α}^{(n−2)/2}u_{α}(0). \end{matrix} (2.46)

Clearly,

|h_{α}^{ij}(x) − δ^{ij}| ≤ C|μ_{α}x|, |Γ_{ij}^{k}(x)| ≤ Cμ_{α} in B_{10R_{α}}^{+}(0),  (2.47)

where Γ_{ij}^{k} is the Christoffel symbol of h_{α}.

As being explained before,

lim_{α→∞} ∥v̄_{α} − v_{T} ∥_{C^{2}(\overline{B_{R}^{+}(0)})} = 0, ∀R > 1,  (2.48)

where v_{T} = v(x′, x_{n} − T), v is given in (2.26) or (2.27). It is not difficult to see from Proposition 2.2 that as α > ᾱ,

v̄_{α}(x) ≤ \frac{C}{1 + |x|^{n−2}} for x ∈ \overline{B_{10R_{α}}^{+}}(0).  (2.49)

Note Z(∫_{∂Ω} u_{α}^{q})^{2/q−1} ≤ C, we can show, exactly in the same way as in [17], the following estimates on the first and second derivative of v̄_{α}.

Proposition 2.3 For all α ≥ ᾱ, x ∈ \overline{B_{R_{α}}^{+}(0)}, we have

|∇v̄_{α}(x)| ≤ \frac{C}{1 + |x|^{n−1}} , |∇^{2}v̄_{α}(x)| ≤ \frac{C}{1 + |x|^{n}} ,

where |∇^{2}v̄_{α}| = ∑_{i,j=1}^{n} |∂^{2}v̄_{α}/∂x^{i}∂x^{j}|, and C is some constant independent of α and x.

For n = 3, we need to obtain an appropriate lower bound of v̄_{α}.

Proposition 2.4 For n = 3, as α large enough,

v̄_{α}(x) ≥ \frac{1}{C(1 + |x|)}, ∀ x ∈ \overline{B_{R_{α}^{1/4}}^{+}(0)},

where C > 0 is some constant independent of α.

Proof. If Z ≤ 0, Proposition 2.4 can be provn exactly in the same way as that in [17], therefore we will focus on the case of Z > 0 here. The proof is slight different from that in [17]. Due to Z > 0, we know T > 0. Without loss of generality, we can assume T ≥ 1. In this case, we need to use more accurate boundary condition in (2.46). In the following, α is always assumed to be suitable large.

Let x̄ = (0, ..., 0, 1) and

G_{α}(x) = \frac{1}{|x − x̄|} − \frac{1}{R_{α}^{1/2}|x − x̄|^{1/2}} in B_{R_{α}^{1/3}}(x̄) \ B_{2}(x̄).

It is easy to see that

\frac{1}{2|x − x̄|} ≤ G_{α}(x) ≤ \frac{1}{|x − x̄|} in B_{R_{α}^{1/3}}(x̄) \ B_{2}(x̄).

As in [17], by using (2.47), one can check that Δ_{h_{α}}G_{α} ≥ 0 for x ∈ B_{R_{α}^{1/3}}(x̄)\B_{2T} (x̄).

Also, from (2.47), we know that for all x = (x′, 0), 1 < |x′| < R_{α}^{1/3}, there exists a constant C_{1} > 0 such that

\frac{∂_{h_{α}}}{∂ν} (G_{α}) ≤ −\frac{1}{C_{1}} · \frac{1}{|x − x̄|^{3}} ≤ −\frac{1}{C_{1}} · G_{α}^{3}.

We will use the maximum principle and Hopf lemma on A = {x ∈ ℝ_{+}^{n} : 2T < |x − x̄| < R_{α}^{1/3}}. Let Σ_{1} = ∂A ∩ {x_{n} = 0}, Σ_{2} = ∂A ∩ {|x − x̄| = 2T }, and Σ_{3} = ∂A ∩ {|x − x̄| = R_{α}^{1/3}}. Choose 0 < τ_{1} < 1 small enough such that τ_{1}G_{α} ≤ v̄_{α} on Σ_{2}. Note Z(∫_{∂Ω} u_{α}^{q})^{2/q−1} ≤ C (in the remains of the proof of Proposition 2.4, we always take C as the same positive constant), we choose τ_{2} < τ_{1} small enough such that 1/(C_{1}τ_{2}^{2}) ≥ C. Let H_{α} = τ_{2}G_{α} − max_{Σ_{3}}(τ_{2}G_{α}). One can check that v̄_{α} − H_{α} satisfies

Δ_{h_{α}}(v̄_{α} − H_{α}) ≤ 0  in A,
v̄_{α} − H_{α} ≥ 0  on Σ_{2} ∪ Σ_{3},
\frac{∂_{hα}(v̄_{α}−H_{α})}{∂ν} > C(H_{α}^{3} − v̄_{α}^{3}) on Σ_{1}.

It follows from the maximum principle and Hopf lemma that

v̄_{α} ≥ H_{α} in A.

Consequently, for all x ∈ B_{R_{α}^{1/4}}^{+}(0) \ B_{2T}^{+} (x̄),

v̄_{α}(x) ≥ H_{α}(x) ≥ \frac{τ_{2}}{2|x − x̄|} − \frac{τ_{2}}{R_{α}^{1/3}} ≥ \frac{τ_{2}}{4|x − x̄|}.

For |x − x̄| ≤ 2T, Proposition 2.4 follows from (2.48).

Let B_{R_{α}}^{+} = B_{R_{α}}(0) ∩ ℝ_{+}^{n}, Γ_{1} = ∂B_{R_{α}}^{+} ∩ ∂ℝ_{+}^{n}, Γ_{2} = ∂B_{R_{α}}^{+} ∩ ℝ_{+}^{n}. We always use dV for the volume element of the standard Euclidean metric, dS for the surface element of the standard Euclidean metric, ν for the unit outer normal vector of the corresponding surface with respect to the specified metrics, and “·” for the inner product under the standard Euclidean metric. As in [17], we have the following identity.

∫_{B_{Rα}^{+}} Δv̄_{α}(∇v̄_{α} · x)dV + \frac{n − 2}{2} ∫_{B_{Rα}^{+}} v̄_{α}Δv̄_{α}dV = J(R_{α}, v̄_{α}) + I(R_{α}, v̄_{α}), (2.50)

where

J(R_{α}, v̄_{α}) = \frac{1}{2} ∫_{Γ_{2}}\{|\frac{∂v̄_{α}}{∂ν} |^{2}|x| − |∂_{tan}v̄_{α}|^{2}|x| + (n − 2)\frac{∂v̄_{α}}{∂ν} v̄_{α}\}dS,  (2.51)

I(R_{α}, v̄_{α}) = \frac{1}{2} ∫_{Γ_{1}}\{2(∑_{i=1}^{n−1} x_{i} \frac{∂v̄_{α}}{∂x_{i}})\frac{∂v̄_{α}}{∂ν} + (n − 2)\frac{∂v̄_{α}}{∂ν} v̄_{α}\}dS.  (2.52)

Replacing Δv̄_{α} in (2.50) by

Δv̄_{α} = Δ_{h_{α}} v̄_{α} − (h_{α}^{ij} − δ^{ij})∂_{ij}v̄_{α} + h_{α}^{ij}Γ_{ij}^{k}∂_{k}v̄_{α},

we have

− ∫_{B_{Rα}^{+}}(x^{i}∂_{i}v̄_{α})Δ_{h_{α}} v̄_{α}dV − \frac{n−2}{2} ∫_{B_{Rα}^{+}} v̄_{α}Δ_{h_{α}} v̄_{α}dV
+ ∫_{B_{Rα}^{+}}(x^{k}∂_{k}v̄_{α})(h_{α}^{ij} − δ^{ij})∂_{ij}v̄_{α}dV − ∫_{B_{Rα}^{+}}(x^{l}∂_{l}v̄_{α})(h_{α}^{ij}Γ_{ij}^{k}∂_{k}v̄_{α})dV
+\frac{n−2}{2} ∫_{B_{Rα}^{+}} v̄_{α}(h_{α}^{ij} − δ^{ij})∂_{ij}v̄_{α}dV − \frac{n−2}{2} ∫_{B_{Rα}^{+}} v̄_{α}(h_{α}^{ij}Γ_{ij}^{k})∂_{k}v̄_{α}dV
= −J(R_{α}, v̄_{α}) − I(R_{α}, v̄_{α}).

Using equation (2.46), we get

A(h_{α}, v̄_{α}) = −J(R_{α}, v̄_{α}) − I(R_{α}, v̄_{α}),  (2.53)

where

A(h_{α}, v̄_{α}) = \frac{ξ_{α}}{p} ∫_{Γ_{2}} v̄_{α}^{p}|x|dS
+ ∫_{B_{Rα}^{+}}(x^{k}∂_{k}v̄_{α})(h_{α}^{ij} − δ^{ij})∂_{ij}v̄_{α}dV − ∫_{B_{Rα}^{+}}(x^{l}∂_{l}v̄_{α})(h_{α}^{ij}Γ_{ij}^{k}∂_{k}v̄_{α})dV
+\frac{n−2}{2} ∫_{B_{Rα}^{+}} v̄_{α}(h_{α}^{ij} − δ^{ij})∂_{ij}v̄_{α}dV − \frac{n−2}{2} ∫_{B_{Rα}^{+}} v̄_{α}(h_{α}^{ij}Γ_{ij}^{k})∂_{k}v̄_{α}dV.

By (2.47), we know

\begin{aligned}A(h_{α}, v̄_{α}) = & O(∫_{Γ_{2}} v̄_{α}^{p}|x|dS) \\ & +O( ∫_{B_{Rα}^{+}} μ_{α}|x|^{2}|∇v̄_{α}| |∇^{2}v̄_{α}|dV) + O( ∫_{B_{Rα}^{+}} μ_{α}|x||∇v̄_{α}|^{2}dV) \\ & +O( ∫_{B_{Rα}^{+}} μ_{α}|x|v̄_{α}|∇^{2}v̄_{α}|dV) + O( ∫_{B_{Rα}^{+}} μ_{α}v̄_{α}|∇v̄_{α}|dV) \end{aligned} (2.54)

We simplify I(R_{α}, v̄_{α}) by using equation (2.46). It is easy to see from (2.47) that

\frac{∂_{h_{α}} v̄_{α}}{∂ν} = \frac{∂v̄_{α}}{∂ν} + O(μ_{α}|x′| |∇v̄_{α}|), on Γ_{1}.

It follows that

\begin{aligned}2I(R_{α}, v̄_{α}) = & ∫_{Γ_{1}}\{2(∑_{i=1}^{n−1} x_{i} \frac{∂v̄_{α}}{∂x_{i}})\frac{∂_{hα} v̄_{α}}{∂ν} + (n − 2)\frac{∂_{hα} v̄_{α}}{∂ν} v̄_{α}\}dS \\ & +O(∫_{Γ_{1}}[μ_{α}|x′|^{2} |∇v̄_{α}|^{2} + μ_{α}|x′|v̄_{α}|∇v̄_{α}|]dS). \end{aligned}  (2.55)

Using the boundary condition in (2.46), we have

∫_{Γ_{1}}\{2(∑_{i=1}^{n−1} x_{i} \frac{∂v̄_{α}}{∂x_{i}})\frac{∂_{hα} v̄_{α}}{∂ν} + (n − 2)\frac{∂_{hα} v̄_{α}}{∂ν} v̄_{α}\}dS
= ∫_{Γ_{1}}\{−2(∑_{i=1}^{n−1} x_{i} \frac{∂v̄_{α}}{∂x_{i}})(αμ_{α}v̄_{α} + Z(∫_{∂Ω} u_{α}^{q})^{2/q−1}v̄_{α}^{q−1})
−(n − 2)αμ_{α}v̄_{α}^{2} − (n − 2)Z(∫_{∂Ω} u_{α}^{q})^{2/q−1}v̄_{α}^{q}}dS
= αμ_{α} ∫_{Γ_{1}} v̄_{α}^{2}dS − ∫_{∂Γ_{1}} αμ_{α}v̄_{α}^{2}|x|dS − \frac{2Z}{q} · (∫_{∂Ω} u_{α}^{q})^{2/q−1} ∫_{∂Γ_{1}} v̄_{α}^{q}|x|dS.

Thus

\begin{aligned}I(R_{α}, v̄_{α}) & = \frac{αμ_{α}}{2} ∫_{Γ_{1}} v̄_{α}^{2}dS + O( ∫_{∂Γ_{1}}(αμ_{α}v̄_{α}^{2}|x| + v̄_{α}^{q}|x|)dS) \\ & +O( ∫_{Γ_{1}}[μ_{α}|x′|^{2}|∇v̄_{α}|^{2} + μ_{α}|x′|v̄_{α} |∇v̄_{α}|]dS). \end{aligned}  (2.56)

Clearly,

J(R_{α}, v̄_{α}) = O( ∫_{Γ_{2}}(|x||∇v̄_{α}|^{2} + v̄_{α} |∇v̄_{α}|)dS).  (2.57)

We can rewrite (2.53) as the following Pohozaev type identity:

αμ_{α} ∫_{Γ_{1}} v̄_{α}^{2}dS = O(∫_{Γ_{2}} v̄_{α}^{p}|x|dS)

+O( ∫_{B_{Rα}^{+}} μ_{α}|x|^{2}|∇v̄_{α}| |∇^{2}v̄_{α}|dV) + O( ∫_{B_{Rα}^{+}} μ_{α}|x||∇v̄_{α}|^{2}dV)

+O( ∫_{B_{Rα}^{+}} μ_{α}|x|v̄_{α}|∇^{2}v̄_{α}|dV) + O( ∫_{B_{Rα}^{+}} μ_{α}v̄_{α}|∇v̄_{α}|dV)
+O( ∫_{Γ_{2}}(|x||∇v̄_{α}|^{2} + v̄_{α} |∇v̄_{α}|)dS)
+O( ∫_{∂Γ_{1}}(αμ_{α}v̄_{α}^{2}|x| + v̄_{α}^{q}|x|)dS)
+O( ∫_{Γ_{1}}[μ_{α}|x′|^{2}|∇v̄_{α}|^{2} + μ_{α}|x′|v̄_{α} |∇v̄_{α}|]dS).

(2.58)

We will derive a contradiction from (2.58) by showing that the left hand side is much larger than the right hand side as α tends to infinity.

Similarly as in [17], by using (2.48) and Proposition 2.4, we have

Lemma 2.5 For n ≥ 3, there exists some constant C > 0 independent of α, such that ∫_{Γ_{1}} v̄_{α}^{2}dS > 1/C for all α ≥ 1. Moreover for n = 3, ∫_{Γ_{1}} v̄_{α}^{2}dS ≥ (log R_{α})/C for all α ≥ 1.

Also, by using (2.49), Proposition 2.3 and some elementary calculations, we have

Lemma 2.6 The following estimates hold.

∫_{∂Γ_{1}}(αμ_{α}v̄_{α}^{2}|x| + v̄_{α}^{q}|x|)dS ≤ αμ_{α}R_{α}^{3−n},

∫_{Γ_{1}}(μ_{α}|x′|^{2}|∇v̄_{α}|^{2} + μ_{α}|x′|v̄_{α} |∇v̄_{α}|)dS ≤ \{ \begin{matrix}Cμ_{α} log R_{α}, & n = 3, \\ Cμ_{α}, & n ≥ 4, \end{matrix}

∫_{Γ_{2}}(|x||∇v̄_{α}|^{2} + v̄_{α} |∇v̄_{α}|)dS ≤ C(αμ_{α})^{n−2},
∫_{Γ_{2}} v̄_{α}^{p}|x|dS ≤ C(αμ_{α})^{n},

∫_{B_{Rα}^{+}}(μ_{α}|x|^{2}|∇v̄_{α}| |∇^{2}v̄_{α}| + μ_{α}|x||∇v̄_{α}|^{2})dV ≤ \{ \begin{matrix}Cμ_{α} log R_{α}, & n = 3, \\ Cμ_{α}, & n ≥ 4, \end{matrix}

∫_{B_{Rα}^{+}}(μ_{α}|x|v̄_{α}|∇^{2}v̄_{α}| + μ_{α}v̄_{α}|∇v̄_{α}|)dV ≤ \{ \begin{matrix}Cμ_{α} log R_{α}, & n = 3, \\ Cμ_{α}, & n ≥ 4. \end{matrix}

Proof of Theorem 0.2. From Lemma 2.5 and Lemma 2.6 we know that the left hand side is clearly much larger than the right hand side in (2.58) as α tends to infinity. Therefore we derive a contradiction basing on the assumption (2.17).

3 Compact manifold with boundary

Let (M, g) be a compact Riemannian manifold with smooth boundary ∂M and dimension n ≥ 3. In this section we sketch the proof of Theorem 0.3.

First we show a rough inequality as in Section 2.

Proposition 3.1 Let Z ∈ (−1/S, Z_{0}]. For any δ > 0, there exists D(δ) > 0 such that ∀u ∈ H^{1}(M),

\begin{matrix}(∫_{M} |u|^{p}dv_{g})^{\frac{2}{p}} \\ ≤ (S_{1}(Z) + δ)( ∫_{M} |∇u|^{2}dv_{g} + Z(∫_{∂M} |u|^{q}ds_{g})^{\frac{2}{q}} ) + D(δ)( ∫_{∂M} u^{2}ds_{g} + ∫_{M} u^{2}dv_{g}). \end{matrix}

(3.1)

Proof. We prove this proposition by contradiction. The proof is quite similar to that of Proposition 2.1, we sketch it below.

Assume that (3.1) is not true, that is, there exists some δ_{2} > 0 such that ∀α > 1,

\begin{aligned}inf_{u∈H^{1}(M)\\{0\}}I_{Z,α}(u) & := inf_{u∈H^{1}(M)\\{0\}} \frac{∫_{M} |∇_{g}u|^{2} + Z(∫_{∂M} |u|^{q})^{2/q} + α ∫_{∂M} u^{2} + α ∫_{M} u^{2}}{(∫_{M} |u|^{p})^{2/p}} \\ & := ξ_{Z,α} < \frac{1}{S_{1}(Z)+δ_{2}} . \end{aligned}

(3.2)

Lemma 3.1 ∀δ > 0, there exists a constant C(M, δ), such that ∀f ∈ H^{1}(M), as ϵ small enough,

||f||_{p_{ϵ},M}^{2} ≤ (S_{1} + δ)||∇f||_{2,M}^{2} + C(M, δ)(||f||_{2,M}^{2} + ||f||_{q_{ϵ},∂M}^{2} ).  (3.3)

Proof. This can be provn from Corollary 1.2 through the partition of unit. We omit the details here.

Lemma 3.2 As α large enough, if inf I_{Z,α} satisfies condition (3.2), inf I_{Z,α} is attained.

Proof. Due to [17], we know that there exists α_{1} < ∞ such that

||u||_{q,∂M}^{2} ≤ S||∇_{g}u||_{2,M}^{2} + α_{1}||u||_{2,∂M}^{2}, ∀u ∈ H^{1}(M).  (3.4)

For u ≠ 0, we define

I_{ϵ}(u) = \frac{∫_{M} |∇_{g}u|^{2} + Z(∫_{∂M} |u|^{q_{ϵ}})^{2/q_{ϵ}} + α ∫_{∂M} u^{2} + α ∫_{M} u^{2}}{(∫_{M} |u|^{p_{ϵ}})^{2/p_{ϵ}}} .

If α > α_{1}, we know as before that I_{ϵ}(u) ≥ 0. The standard variational method shows that ∃ u_{ϵ} ≥ 0 with ||u_{ϵ}||_{p_{ϵ}} = 1 such that

I_{ϵ}(u_{ϵ}) = inf I_{ϵ}(u) := ξ_{ϵ}.

Easy to see that u_{ϵ} satisfies

\{ \begin{matrix}−Δ_{g}u_{ϵ} & = ξ_{ϵ}u_{ϵ}^{p_{ϵ}−1} − αu_{ϵ} & in M \\ \frac{∂_{g}u_{ϵ}}{∂ν} & = −Z(∫_{∂M} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}u_{ϵ}^{q_{ϵ}−1} − αu_{ϵ} & on ∂M. \end{matrix}  (3.5)

Due to [14], we know that there exists α_{2} < ∞ such that

||u||_{p,M}^{2} ≤ S_{1}||∇_{g}u||_{2,M}^{2} + α_{2}||u||_{2,M}^{2}, ∀u ∈ H_{0}^{1}(M).  (3.6)

Set α_{0} = max{α_{1}, α_{2}}. We only need to show that as α > α_{0}, ||u_{ϵ}||_{∞} ≤ C. Suppose ||u_{ϵ}||_{∞} → ∞ up to a subsequence, by [10], we know that there exists a x_{ϵ} ∈ M such that u_{ϵ}(x_{ϵ}) = ||u||_{∞} → ∞. Set μ_{ϵ} = (u_{ϵ}(x_{ϵ}))^{− \frac{pϵ−2}{2}} .

Let (y^{1}, ⋯, y^{n−1}, y^{n}) denote some geodesic normal coordinates given by the exponential map exp_{x_{ϵ}}. In this coordinate system, the metric g is given by g_{ij}(y)dy^{i}dy^{j}.

For suitably small δ_{3} > 0 (independent of ϵ), we define v_{ϵ} in a neighborhood of z = 0 by

v_{ϵ}(z) = u_{ϵ}^{−1}(x_{ϵ})u_{ϵ}(exp_{x_{ϵ}}(μ_{ϵ}z)), z ∈ O_{ϵ} ⊂ ℝ^{n},

where

O_{ϵ} = {z ∈ ℝ^{n} : |z| < δ_{3}/μ_{ϵ}, exp_{x_{ϵ}}(μ_{ϵ}z) ∈ M}.  (3.7)

We write ∂O_{ϵ} = Γ_{ϵ}^{1} ∪ Γ_{ϵ}^{2}, where

Γ_{ϵ}^{1} = {z ∈ ∂O_{ϵ} : exp_{x_{ϵ}}(μ_{ϵ}z) ∈ ∂M}, Γ_{ϵ}^{2} = {z ∈ ∂O_{ϵ} : exp_{x_{ϵ}}(μ_{ϵ}z) ∈ M}.

Then v_{ϵ} satisfies

\{ \begin{matrix}−Δ_{g_{ϵ}}v_{ϵ} = ξ_{ϵ}v_{ϵ}^{p−1} − αμ_{ϵ}^{2}v_{ϵ}, & in O_{ϵ}, \\ \frac{∂_{gϵ}v_{ϵ}}{∂ν} = −Z(∫_{∂M} u_{ϵ}^{q_{ϵ}})^{2/q_{ϵ}−1}v_{ϵ}^{q_{ϵ}−1} − αμ_{ϵ}v_{ϵ}, & on Γ_{ϵ}^{1}, \\ v_{ϵ}(0) = 1,  0 ≤ v_{ϵ} ≤ 1, \end{matrix}  (3.8)

where g_{ϵ} denotes the metric on O_{ϵ} given by g_{ϵ} = g_{ij}(μ_{ϵ}z)dz^{i}dz^{j}. As in the proof of Lemma 1.1 (here we need to use (3.4)), we can show that as α > α_{0},

0 ≤ \overline{lim_{ϵ→0}}ξ_{ϵ} := ξ_{0} ≤ ξ_{Z,α} < \frac{1}{S_{1}(Z) + δ_{2}} .  (3.9)

We claim:

∫_{∂M} u_{ϵ}^{q_{ϵ}} ≥ C > 0.  (3.10)

If ∫_{∂M} u_{ϵ}^{q_{ϵ}} → 0 up to a subsequence, using Lemma 3.1, as in the proof of Lemma 1.3, we know ||u_{ϵ} − u_{0}||_{p_{ϵ},M} → 0 for some u_{0} ∈ H_{0}^{1}(M). It follows that

\frac{∫_{M} |∇_{g}u_{0}|^{2}dv_{g} + α ∫_{M} u_{0}^{2}dv_{g}}{(∫_{M} |u_{0}|^{p}dv_{g})^{2/p}} < \frac{1}{S_{1}} ,

this contradicts (3.6) as α > α_{2}.

Also as in the proof of Lemma 1.4, we can show that

T := \overline{lim_{ϵ→∞}} \frac{dist(x_{ϵ}, ∂M)}{μ_{ϵ}} < ∞.  (3.11)

Then we follow the proof of Lemma 2.1 closely and can derive a contradiction to Theorem 0.1. We thereby establish Lemma 3.2.

Due to Lemma 3.2, without loss of generality, we can assume that as α > α_{0}, inf I_{Z,α}(u) = I_{Z,α}(u_{α}) with u_{α} ≥ 0 and ||u_{α}||_{p,M} = 1. Then, we follow the proof of Proposition 2.1 closely and can complete the proof of Proposition 3.1. The only difference is to show ∫_{∂M} u_{α}^{q} ≥ C > 0. But this can be handled similarly to (3.10). We leave these details to interesting readers.

From now on, we begin to prove Theorem 0.3 through an argument by contradiction. Note that we assume Z < Z_{0}, thus 1/S_{1}(Z) < 1/S_{1}.

Suppose that Theorem 0.3 is false, then ∀α > 1,

\begin{aligned}inf_{H^{1}(M)\\{0\}} I_{α}(u) & = inf_{H^{1}(M)\\{0\}} \frac{∫_{M} |∇_{g}u|^{2}+Z(∫_{∂M} |u|^{q})^{2/q}+α ∫_{∂M} u^{2}+α ∫_{M} u^{2}}{(∫_{M} |u|^{p})^{2/p}} \\ & := ξ_{α} < \frac{1}{S_{1}(Z)}. \end{aligned} (3.12)

From the proof of Lemma 3.2, we know that as α > α_{0}, under (3.12), inf I_{α}(u) is attained. Without loss of generality, we can always assume α suitable large and inf I_{α}(u) = I_{α}(u_{α}) with u_{α} ≥ 0 and ||u_{α}||_{p,M} = 1. It is easy to see that u_{α} satisfies

\{ \begin{matrix}−Δ_{g}u_{α} & = ξ_{α}u_{α}^{p−1} − αu_{α} & in M \\ \frac{∂_{g}u_{α}}{∂ν} & = −Z(∫_{∂M} u_{α}^{q})^{2/q−1}u_{α}^{q−1} − αu_{α}, & on ∂M. \end{matrix}  (3.13)

Using Proposition 3.1, we have

Lemma 3.3 As α → ∞,

α||u_{α}||_{2,∂M}^{2} → 0, ξ_{α} → \frac{1}{S_{1}(Z)}.  (3.14)

Because of 1/S_{1}(Z) < 1/S_{1}, as in the proof of (3.10), we have the following.

Lemma 3.4 There exists a constant C > 0 such that

∫_{∂M} u_{α}^{q} ≥ C.  (3.15)

Since u_{α} satisfies (3.13), due to Cherrier, we know that u_{α} is smooth up to boundary. Let u_{α}(x_{α}) = ||u||_{∞} for some x_{α} ∈ \overline{M}. Set μ_{α} = (u_{α}(x_{α}))^{− \frac{p−2}{2}} . As before, from (3.14) and (3.15) we can show that

αμ_{α} → 0, as α → ∞.  (3.16)

Let (y^{1}, ⋯, y^{n−1}, y^{n}) denote some geodesic normal coordinates given by the exponential map exp_{x_{α}}. In this coordinate system, the metric g is given by g_{ij}(y)dy^{i}dy^{j}.

For a suitable small δ_{4} > 0 (independent of α), we define v_{α} in a neighborhood of z = 0 by

v_{α}(z) = u_{α}^{−1}(x_{α})u_{α}(exp_{x_{α}}(μ_{α}z)), z ∈ O_{α} ⊂ ℝ^{n},

where

O_{α} = {z ∈ ℝ^{n} : |z| < δ_{4}/μ_{α}, exp_{x_{α}}(μ_{α}z) ∈ M}.  (3.17)

We write ∂O_{α} = Γ_{α}^{1} ∪ Γ_{α}^{2}, where

Γ_{α}^{1} = {z ∈ ∂O_{α} : exp_{x_{α}}(μ_{α}z) ∈ ∂M}, Γ_{α}^{2} = {z ∈ ∂O_{α} : exp_{x_{α}}(μ_{α}z) ∈ M}.

Then v_{α} satisfies

\{ \begin{matrix}−Δ_{g_{α}}v_{α} = ξ_{α}v_{α}^{p−1} − αμ_{α}^{2}v_{α}, & in O_{α}, \\ \frac{∂_{gα}v_{α}}{∂ν} = −Z(∫_{∂M} u_{α}^{q_{α}})^{2/q_{α}−1}v_{α}^{q_{α}−1} − αμ_{α}v_{α}, & on Γ_{α}^{1}, \\ v_{α}(0) = 1,  0 ≤ v_{α} ≤ 1, \end{matrix}  (3.18)

where g_{α} denotes the metric on O_{α} given by g_{α} = g_{ij}(μ_{α}z)dz^{i}dz^{j}.

Also let \overline{lim}_{α→∞}dist(x_{α}, ∂M)/μ_{α} = T, due to 1/S_{1}(Z) < 1/S_{1}, as in Section 1, we know T < ∞.

By standard elliptic estimates, we know v_{α} → v in C^{3}(B_{R}(0) ∩ \overline{O}_{α}), where v(x) is given by (2.26) or (2.27) (depending on T > 0 or T = 0). Consequently, as before, we have the following lemma.

Lemma 3.5

lim_{α→∞} ∫_{O_{α}} |v_{α} − v|^{p} = lim_{α→∞} ∫_{O_{α}} |∇_{g_{α}}v_{α} − ∇_{g_{α}}v|^{2} = lim_{α→∞} ∫_{∂O_{α}} |v_{α} − v|^{q} = 0. (3.19)

As in [18], by using Lemma 3.5, we have

Proposition 3.2 There exists a constant C > 0 such that, ∀α > 1

v_{α} ≤ Cv, x ∈ M̄_{α}.  (3.20)

Then following the proof of Theorem 0.2 closely, by using Pohozaev identity, we derive a contradiction, thus complete the proof of Theorem 0.3. We refer [18] and [17] to interesting readers for more details.

4 Some further remarks

In this section, we give some details concerning Remark 2.1 and point out the obstacle by using the current method to prove the conjecture which we present in our introduction.

Assume Z = Z_{0}. Under condition (2.17), we know that inf I_{α}(u) = I_{α}(u_{α}) for some u_{α} ≥ 0, ||u_{α}||_{p,Ω} = 1 and u_{α} satisfies (2.18). In contrast to the case of Z < Z_{0}, here, we claim:

∫_{∂Ω} u_{α}^{q} → 0 as α → ∞.  (4.1)

We show (4.1) by contradiction. If not, as α → ∞,

∫_{∂Ω} u_{α}^{q} ≥ C > 0.  (4.2)

Define μ_{α}, Ω_{α}, and v_{α} as in (2.21), then v_{α} satisfies (2.22). As before, from (2.19) and (4.2), we know αμ_{α} → 0. Let \overline{lim}_{α→∞}dist(x_{α}, ∂Ω)/μ_{α} = T. If T < ∞, then v_{α} → v_{0} in C^{3}(\overline{Ω_{α}} ∩ B_{R}(0)) for all R > 1, where v_{0} satisfies

\{ \begin{matrix}−Δv_{0} = \frac{1}{S_{1}} v_{0}^{p−1} & in ℝ_{T}^{n}, \\ \frac{∂v_{0}}{∂ν} = −Z_{0}(∫_{∂ℝ_{T}^{n}} v_{0}^{q})^{2/q−1}v_{0}^{q−1} & on ∂ℝ_{T}^{n}, \\ v_{0}(0) = 1,  0 ≤ v_{0}(x) ≤ 1. \end{matrix}  (4.3)

However, a similar discussion as in the proof of Theorem 0.1 shows that (4.3) has no solution. Therefore T = ∞, and v_{α} → v_{1} in C^{3}(\overline{Ω_{α}} ∩ B_{R}(0)) for all R > 1, where v_{1} satisfies

\{ \begin{matrix}−Δv_{1} = \frac{1}{S_{1}} v_{1}^{p−1} in ℝ^{n}, \\ v_{1}(0) = 1,  0 ≤ v_{1}(x) ≤ 1. \end{matrix}  (4.4)

It follows that ||v_{1}||_{p,ℝ^{n}} = 1, therefore ||v_{α} − v_{1}||_{p,Ω_{α}} → 0.

Note ||∇v_{α} − ∇v_{1}||_{2,Ω_{α}} < C. Using Lemma 1.6 and property of v_{1}, we know that ∫_{∂Ω_{α}} v_{α}^{q} → 0. This contradicts to (4.2).

This discussion shows that we do need some new ideas to handle the extremal case Z = Z_{0} in the proof of Theorem 0.2 and 0.3.

5 Appendix

In this appendix, we present another proof of Theorem 0.1 based on a new result due to Carlen and Loss [9].

Proposition 5.1 (Carlen and Loss’s Theorem) For λ > 0

S(λ) = inf_{D^{1,2}(ℝ_{+}^{n})\\{0\}} \{\frac{||∇f||_{2,ℝ_{+}^{n}}}{||f||_{p,ℝ_{+}^{n}}} : \frac{||f||_{q,∂ℝ_{+}^{n}}}{||f||_{p,ℝ_{+}^{n}}} = λ\}  (5.1)

is attained.

Denote S(0) = 1/S_{1}^{1/2}. It is not difficult to see from [9] that S(λ) is a continuous function on [0, ∞).

Let II_{Z}(u) be given as in Section 1 and ξ_{Z} be given by (1.1). In order to prove Theorem 0.1, we only need to establish the following proposition, the other details can be carried out as in Section 1.

Proposition 5.2 For any Z ∈ (−1/S, Z_{0}), inf II_{Z} is attained.

Proof. For Z ≤ 0, this proposition was already proved in [9]. Consequently, a new proof of Escobar’s inequality was given by E. Carlen and M. Loss there. Here, we focus on the case of 0 < Z < Z_{0}.

It is well known that ξ_{Z} ≥ 1/(2^{2/n}S_{1}) for Z ≥ 0. The existence of minimizer of II_{Z} is equivalent to the existence of a extremal function for the following inequality

||∇u||_{2,ℝ_{+}^{n}}^{2} + Z||u||_{q,∂ℝ_{+}^{n}}^{2} ≥ ξ_{Z}||u||_{p,ℝ_{+}^{n}}^{2}, ∀u ∈ D^{1,2}(ℝ_{+}^{n}) \ {0},

i.e.

||u||_{p,ℝ_{+}^{n}}^{2} − \frac{Z}{ξ_{Z}} ||u||_{q,∂ℝ_{+}^{n}}^{2} ≤ \frac{1}{ξ_{Z}} ||∇u||_{2,ℝ_{+}^{n}}^{2}, ∀u ∈ D^{1,2}(ℝ_{+}^{n}) \ \{0\}.  (5.2)

Therefore, we only need to show

sup_{D^{1,2}(ℝ_{+}^{n})\\{0\}} \frac{||u||_{p,ℝ_{+}^{n}}^{2} − Z · ξ_{Z}^{−1}||u||_{q,∂ℝ_{+}^{n}}^{2}}{||∇u||_{2,ℝ_{+}^{n}}^{2}} = \frac{1}{ξ_{Z}}

and the supremum is attained.

From the definition of ξ_{Z}, it is not difficult to see that the supremum is less than or equals to 1/ξ_{Z}. Suppose

sup_{D^{1,2}(ℝ_{+}^{n})\\{0\}} \frac{||u||_{p,ℝ_{+}^{n}}^{2} − Z · ξ_{Z}^{−1}||u||_{q,∂ℝ_{+}^{n}}^{2}}{||∇u||_{2,ℝ_{+}^{n}}^{2}} = \frac{1}{τξ_{Z}}

for some τ > 1. Then

||u||_{p,ℝ_{+}^{n}}^{2} ≤ \frac{1}{τξ_{Z}} ||∇u||_{2,ℝ_{+}^{n}}^{2} + \frac{Z}{ξ_{Z}} ||u||_{q,∂ℝ_{+}^{n}}^{2}, ∀u ∈ D^{1,2}(ℝ_{+}^{n}) \ \{0\}.  (5.3)

From the definition of ξ_{Z}, we know that for all i ≥ 1, there exists u_{i}, such that

||u_{i}||_{p,ℝ_{+}^{n}}^{2} ≥ \frac{1}{ξ_{Z} + \frac{1}{i}} (||∇u_{i}||_{2,ℝ_{+}^{n}}^{2} + Z||u_{i}||_{q,∂ℝ_{+}^{n}}^{2})  (5.4)

and ||∇u_{i}||_{2,ℝ_{+}^{n}}^{2} = 1. Due to trace inequality, we know ||u_{i}||_{q,∂ℝ_{+}^{n}} ≤ C.

Combining (5.3) with (5.4), we have

(\frac{1}{ξ_{Z} + \frac{1}{i}} − \frac{1}{τξ_{Z}})||∇u_{i}||_{2,ℝ_{+}^{n}}^{2} ≤ (\frac{Z}{ξ_{Z}} − \frac{Z}{ξ_{Z} + \frac{1}{i}})||u_{i}||_{q,∂ℝ_{+}^{n}}^{2}.

Sending i to infinity, we have ||∇u_{i}||_{2,ℝ_{+}^{n}}^{2} → 0. Contradiction! Therefore τ = 1. To see the supremum is attained, one observes that

sup_{D^{1,2}(ℝ_{+}^{n})\\{0\}} \frac{||u||_{p,ℝ_{+}^{n}}^{2} − Z · ξ_{Z}^{−1}||u||_{q,∂ℝ_{+}^{n}}^{2}}{||∇u||_{2,ℝ_{+}^{n}}^{2}} = sup_{λ≥0}\{(1 − ξ_{Z}^{−1} · Zλ^{2})/S(λ)^{2}
= sup_{(ξ_{Z}^{−1}·Z)^{−1/2}≥λ≥0}{(1 − ξ_{Z}^{−1} · Zλ^{2})/S(λ)^{2}.

From our early calculation (see (1.22)), we can easily see that ξ_{Z} < 1/S_{1}, therefore the supremum can not be attained at λ = 0, that is

sup_{D^{1,2}(ℝ_{+}^{n})\\{0\}} \frac{||u||_{p,ℝ_{+}^{n}}^{2} − Z · ξ_{Z}^{−1}||u||_{q,∂ℝ_{+}^{n}}^{2}}{||∇u||_{2,ℝ_{+}^{n}}^{2}} = sup_{(ξ_{Z}^{−1}·Z)^{−1/2}≥λ>γ} \frac{(1 − ξ_{Z}^{−1} · Zλ^{2})}{S(λ)^{2}}

for some small γ > 0. The existence of a maximum follows from the continuity of S(λ) and Proposition 5.1.

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