arXiv:2110.15436v5 [math.DG] 15 Nov 2021

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SOLVING THE YAMABE PROBLEM ON CLOSED MANIFOLDS BY

ITERATION SCHEMES AND PERTURBATION METHODS

JIE XU

Abstract. We introduces a double iterative scheme and a perturbation method to solve the Yam-

abe equation gu := −4(n−1)n−2

∆gu + Sgu = λun+2

n−2 on closed manifolds (M, g) with dimM > 3.Thus g admits a conformal change to a constant scalar curvature metric. In contrast to the tra-ditional method, the Yamabe problem is completely solved in five cases classified by the sign ofthe scalar curvature Sg and the sign of the first eigenvalue of conformal Laplacian. In contrastto the traditional proof, we focus on local analysis; both Weyl tensor and positive mass theoremare not required, and the proof is essentially dimension independent. To treat the most difficultcase λ > 0, we construct the lower and upper solutions to solve a perturbed Yamabe equation

guβ + βuβ = (λβ − κ)un+2

n−2

β with κ > 0, β < 0 and some specific choices of λ′

βs, then take a limitby letting β → 0.

1. Introduction

In this paper, we will solve the Yamabe problem by a double iterative method inspired by [23], inwhich the Yamabe equation can be solved by an iterative method in a small enough bounded, opensubset of Rn equipped with some metric g and with Dirichlet boundary conditions. This iterativemethod is developed for hyperbolic operators [11], [12] and elliptic operators [23], [27] with a longhistory in PDE theory dating back to [21, 22]. Instead of a global analysis in terms of calculus ofvariation, this article applies the local analysis developed especially in [23] and [26] to the Yamabeequation. The local analysis allows us to prove our results in all dimensions by one method, incontrast to previous proofs that separately treated low dimensions and locally conformally flatmanifolds. When λ < 0, a pair of global sub- and super-solutions are constructed; when λ > 0, aperturbation method is applied and a local test function is good enough for the existence of the

solution of a perturbed Yamabe equation −4(n−1)n−2 ∆gu+(Sg + β) u = λu

n+2n−2 on any closed manifolds

M with dimM > 3 and β < 0, we then show the existence of Yamabe equation by letting β → 0.In particular, the local analysis gives us the flexibility to choose λ in perturbed Yamabe equationsuch that λ < λ(Sn). Here λ(Sn) is the Yamabe invariant with respect to the standard sphere.

In 1960, Yamabe proposed the following generalization of the classical uniformization theoremfor surfaces:

The Yamabe Conjecture. Given a compact Riemannian manifold (M,g) of dimension n > 3,there exists a metric conformal to g with constant scalar curvature.

Let Sg be the scalar curvature of g, and let Sg be the scalar curvature of the conformal metric

g = e2fg. Set e2f = up−2, where p = 2nn−2 and u > 0. Then

Sg = u1−p

(

−4 · n− 1

n− 2∆gu+ Sgu

)

. (1)

Setting a = 4 · n−1n−2 > 0, we have that g = up−2g has constant scalar curvature λ if and only if u > 0

satisfies the Yamabe equation− a∆gu+ Sgu = λup−1, (2)

1

http://arxiv.org/abs/2110.15436v10

2 J. XU

where ∆g = −d∗d is negative definite. was solved in several steps by important work of Yamabe,Trudinger, Aubin and Schoen, with the final cases solved in 1984. A thorough presentation of thehistorical development of the solution is in [18]. There are also results for manifolds with boundary[4, 6, 7, 20] and open manifolds [3, 10, 14] with certain restrictions.

For a more precise statement of the cases, let η1 be the first eigenvalue of the conformal Laplaciang := −a∆g + Sg on (M,g). The main theorem is the following:

Theorem. Let (M,g) be a closed Riemannian manifold with dimM > 3. Then

(i) If η1 = 0, (2) has a real, positive solution u ∈ C∞(M) with λ = 0;(ii) If η1 < 0 and Sg 6 0 everywhere or Sg > 0 somewhere, (2) has a real, positive solution

u ∈ C∞(M) with some λ < 0;(iii) If η1 > 0 and Sg 6 0 somewhere or Sg > 0 everywhere, (2) has a real, positive solution

u ∈ C∞(M) with some λ > 0.

These cover all cases of the Yamabe problem. Case (i) is a trivial eigenvalue problem; Case (ii)is solved in Theorem 4.1 and 4.2 below; Case (iii) is solved in Theorem 4.4 and 4.6. Theorem 4.3,in which a perturbed Yamabe equation is solved, plays a central role in Case (iii).

Historically, the PDE (2) is analyzed as the Euler-Lagrange equation of the functional

λ(M) = infu 6=0

ˆ

u

´

M a|∇gu|2 + Sgu2dVolg

(´

M updVolg)

2p

= infu 6=0

Q(u) (3)

and the Yamabe Conjecture reduced to proving a minimizer exists. After results showing theexistence the solution provided λ(M) < λ(Sn), the problem is equivalent to the nontrivial task offinding a test function φ such Q(φ) < λ(Sn) in (3).

There are some critical technical differences between our approach and the classical one. Ina small enough domain Ω with radius r, a local test function φ(ǫ, x) is chosen to show that theYamabe quotient Q(φ(ǫ, x)) satisfies

Q(φ(ǫ, x)) 6 (1 +O(r2))(aT +O(ǫ)).

with best Sobolev constant T for Euclidean Sobolev embedding Lp → H1. Since we want λ(M) <aT , the key is to control the smallness of O(r2). In the classical approach, a local test functionφ ∈ C∞

c (Ω) and a particular Riemannian metric within the conformal class were carefully chosen toshow that Q(φ) < λ(Sn) = aT when n > 6 and the manifold is not locally conformally flat; Schoensuccessfully applied the positive mass theorem to the remaining cases. In this article, especially forthe cases when λ > 0, we apply a direct analysis, which can be found in Appendix A, to show thata perturbed functional

Qβ(u) =

´

M a|∇gu|2 + (Sg + β) u2dVolg(´

M updVolg) 2

p

, β < 0.

satisfies Qβ(φ) < aT locally in Ω for all metric g with n > 3, provided that Ω is small enough.This results in the existence of a local solution (36) of the form Lu = b(x)up−1 + f(x, u) on Ω withtrivial Dirichlet boundary condition. We then apply monotone iteration scheme, e.g. [3, 16], toshow the existence of a positive solution of the perturbed Yamabe equation

− a∆guβ + (Sg + β) uβ = (λβ − κ) up−1β onM. (4)

Here λβ = infu 6=0Qβu and κ > 0 is some fixed constant. The local analysis and monotone iterationscheme give us the flexibility to choose the coefficient, especially the introduction of κ, ahead ofthe nonlinear term; the introduction of κ cannot be achieved by classical method. A limitingargument shows u = limβ→0 uβ exists for a subsequence of uβ and solves the Yamabe equation(2). The inequality Qβ(φ) < aT holds for all n > 3, and is independent of the Weyl tensor precisely

SOLVING THE YAMABE PROBLEM ON CLOSED MANIFOLDS BY ITERATION SCHEMES AND PERTURBATION METHODS3

because of the introduction of β. This method bypasses the subcritical exponent analysis of uniformboundedness of the sequence ϕs when s→ p, where ϕs solves gϕs = λsϕ

s−1s for s ∈ (2, p).

In particular, we constructed sub-solutions u− and super-solutions u+ of the Yamabe equation (2)or the perturbed Yamabe equation (72), then applied monotone iterative method–see e.g. [3, 16]–to get a solution, this is one layer of iteration. The essential difficulty is to show that u− 6≡ 0.Historically the cases when η1 < 0 is easy to handle. We applied local analysis in [23] to give adifferent way to construct the sub-solutions/super-solutions, this is the second layer of iteration.For the cases when η1 > 0, barely no one constructed the sub-solution/super-solution before, basedon the best understanding of us. The full power of this iteration scheme can be seen in solving theboundary Yamabe problem on compact manifolds with boundary. The size of the local domain wechoose is also important: in particular, the scalar curvature at each point tells us the deviation ofthe volume of small enough geodesic ball centered at the same point from the volume of Euclideanball with same radius. Weyl tensor cannot provide this type of information.

The paper is organized as follows. In §2, the preliminaries and required results for doubleiteration scheme are listed. In §3, we constructed local solutions on a small enough single chartof (M,g) with Dirichlet boundary conditions for different cases classified by sgn(Sg) and sgn(η1):when η1 < 0, the local analysis in [23] is applied, see Proposition 3.1 and 3.2; when η1 > 0, a localcalculus of variation is applied, see Proposition 3.3. When η1 > 0, the region in which Sg < 0 playsa central role in the local analysis.

In §4, sub-solutions u− and super-solutions u+ are constructed in various cases: (i) Theorem 4.1and 4.2 are related to the case η1 < 0. As discussed before, they are historically easy case, herea new construction of sub- and super- solutions is applied. (ii) Theorem 4.4 and 4.6 are relatedto the case η1 > 0. For the case η1 > 0, Theorem 4.3 plays a central role for the existence ofsolution of a perturbed Yamabe equation. The constructions of sub- and super- solutions rely onthe local analysis given in Proposition 3.3. To handle the cases when Sg 6 0 everywhere andSg > 0 everywhere, we shows in Theorem 4.5 and Corollary 4.1 that they are equivalent to thecases Sg > 0 somewhere, η1 < 0 and Sg < 0 somewhere, η1 > 0, respectively. Note that the lasttwo results mentioned above relates to the question: whether a function f can be the prescribedscalar curvature of some metric under conformal change?

2. The Preliminaries

In this section, we list definitions and results required for our analysis.

Let Ω be a connected, bounded, open subset of Rn with smooth boundary ∂Ω equipped with someRiemannian metric g that can be extended smoothly to Ω. We call (Ω, g) a Riemannian domain.Furthermore, let (Ω, g) be a compact manifold with smooth boundary extended from (Ω, g). Let(M,g) be a closed manifold with dimM > 3. Let (M ′, g) be a general compact manifold withinterior M ′ and smooth boundary ∂M ′. Throughout this article, we denote the space of smoothfunctions with compact support by C∞

c , smooth functions by C∞, and continuous functions by C0.We define two equivalent versions of the Lp norms and two equivalent versions of the Sobolev

norms on (Ω, g). We also define global Lp norms and Sobolev norms on (M,g).

Definition 2.1. Let (Ω, g) be a Riemannian domain. Let (M,g) be a closed Riemannian n-manifoldwith volume density dω with local expression dVolg. Let u be a real valued function. Let 〈v,w〉g and

|v|g = 〈v, v〉1/2g denote the inner product and norm with respect to g.

4 J. XU

(i) For 1 6 p <∞,

Lp(Ω) is the completion of

u ∈ C∞c (Ω) : ‖u‖pp :=

ˆ

Ω|u|pdx <∞

,

Lp(Ω, g) is the completion of

u ∈ C∞c (Ω) : ‖u‖pp,g :=

ˆ

Ω|u|p dVolg <∞

,

Lp(M,g) is the completion of

u ∈ C∞(M) : ‖u‖pp,g :=ˆ

M|u|p dVolg <∞

.

(ii) For ∇u the Levi-Civita connection of g, and for u ∈ C∞(Ω) or u ∈ C∞(M),

|∇ku|2g := (∇α1 . . .∇αku)(∇α1 . . .∇αku). (5)

In particular, |∇0u|2g = |u|2 and |∇1u|2g = |∇u|2g.

(iii) For s ∈ N, 1 6 p <∞,

W s,p(Ω) =

u ∈ Lp(Ω) : ‖u‖pW s,p(Ω) :=

ˆ

Ω

s∑

j=0

∣

∣Dju∣

∣

pdx <∞

, (6)

W s,p(Ω, g) =

u ∈ Lp(Ω, g) : ‖u‖pW s,p(Ω,g) =

s∑

j=0

ˆ

Ω

∣

∣∇ju∣

∣

p

gdVolg <∞

,

W s,p(M,g) =

u ∈ Lp(M,g) : ‖u‖pW s,p(M,g) =

s∑

j=0

ˆ

M

∣

∣∇ju∣

∣

p

gdω <∞

, .

Here |Dju|p :=∑

|α|=j|∂αu|p in the weak sense. Similarly, W s,p0 (Ω) is the completion of C∞

c (Ω)

with respect to the W s,p-norm. In particular, Hs(Ω) := W s,2(Ω) and Hs(Ω, g) := W s,2(Ω, g),Hs(M,g) :=W s,2(M,g) are the usual Sobolev spaces. We similarly define Hs

0(Ω),Hs0(Ω, g).

(iv) We define theW s,p-type Sobolev space on (M ′, g) the same as in (iii) when s ∈ N, 1 6 p <∞.

The elliptic regularity for Riemannian domain (Ω, g) and closed manifold (M,g), and for W s,p-type and Hs-type, are listed below, respectively.

Theorem 2.1. Let (Ω, g) be a Riemmannian domain. Let (M,g) be a closed Riemannian manifoldand (M ′, g) be a compact Riemannian manifold with smooth boundary.

(i)[24, Ch. 5, Thm. 1.3] Let L be a second order elliptic operator of the form Lu = −∆gu+Xuwhere X is a first order differential operator with smooth coefficients on M ′. For f ∈ L2(M ′, g), asolution u ∈ H1

0 (Ω, g) to Lu = f in M ′ with u ≡ 0 on ∂M ′ belongs to H2(M ′, g), and

‖u‖H2(M ′,g) 6 C∗(

‖f‖L2(M ′,g) + ‖u‖H1(M ′,g)

)

. (7)

C∗ = C∗(L,M ′, g) depends on L and (M ′, g).

(ii)[13, Vol. 3, Ch. 17]Let −∆g be the Laplacian with respect to Riemannian metric g on Ω, andlet u ∈ H1

0 (Ω, g) be a weak solution of −∆gu = f .(Interior Regularity) If f ∈W s,p(Ω, g), then u ∈W s+2,p(Ω, g) . Also, if u ∈ Lp(Ω, g), then

‖u‖W s+2,p(Ω,g) 6 D1(‖−∆gu‖W s,p(Ω,g) + ‖u‖Lp(Ω,g)), (8)

for some D1 > 0 where D1 only depends on −∆g, Ω and ∂Ω.(Schauder Estimates) If f ∈ Cs,α(Ω), then u ∈ Cs+2,α(Ω). Also, if u ∈ C0,α(Ω), then

‖u‖Cs+2,α(Ω) 6 D2(‖−∆gu‖Cs,α(Ω) + ‖u‖C0,α(Ω)), (9)

for some D2 > 0 where D2 only depends on −∆g, Ω and ∂Ω.


(iii) [18, Thm. 2.4]Let −∆g be the Laplace-Beltrami operator, and let u ∈ H1(M,g) be a weaksolution of −∆gu = f . If f ∈ W s,p(M,g), then u ∈ W s+2,p(M,g). Also, if u ∈ Lp(M,g), thenabove estimates in (8) and (9) also hold when replacing (Ω, g) by (M,g).

Sobolev embeddings, both on (Ω, g) and (M,g), play an important role in this article.

Theorem 2.2. [1, Ch. 4] (Sobolev Embeddings) Let Ω ∈ Rn be a bounded, open set with smooth

boundary ∂Ω and equipped with any Riemannian metric g.(i) For s ∈ N and 1 6 p 6 p′ <∞ such that

1

p− s

n6

1

p′, (10)

W s,p(Ω) continuously embeds into Lp′(Ω) with the following estimates:

‖u‖Lp′ (Ω,g) 6 K‖u‖W s,p(Ω,g). (11)

(ii) For s ∈ N, 1 6 p <∞ and 0 < α < 1 such that

1

p− s

n6 −α

n, (12)

Then W s,p(Ω) continuously embeds in the Holder space C0,α(Ω) with the following estimates:

‖u‖C0,α(Ω) 6 K ′‖u‖W s,p(Ω,g). (13)

(iii) The above results and estimates also hold when applying to the closed manifold (M,g).

A more precise version of Sobolev embedding is provided by Gagliardo-Nirenberg interpolationinequality.

Proposition 2.1. [2, Thm. 3.70] Let M be either Rn, or a compact Riemannian manifold with

or without boundary. Let q, r, l be real numbers with 1 6 q, r, l 6 ∞, and let j,m be integers with0 6 j < m. Define α by solving

1

l=j

n+ α

(

1

r− m

n

)

+1− α

q, (14)

as long as l > 0. If α ∈[

jm , 1

]

, then there exists a constant C0, depending only on n,m, j, q, r, α

and on the manifold, such that for all u ∈ C∞c (M) (with

´

M u = 0 in the compact case withoutboundary),

‖∇ju‖Lℓ(Ω,g) 6 C0‖∇mu‖αLr(Ω,g)‖u‖1−αLq(Ω,g). (15)

(If r = nm−j 6= 1, then (14) is not valid for α = 1.)

To see how different choices of λ affect the behaviors of solutions of Yamabe equation −a∆gu+Sgu = λup−1 in (Ω, g) and u ≡ c > 0 on ∂Ω, we need both strong and weak maximum principles.

Theorem 2.3. (i) [8, Cor. 3.2] (Weak Maximum Principle) Let L be a second order elliptic operatorof the form

Lu = −∑

|α|=2

−aα(x)∂αu+∑

|β|=1

−bβ(x)∂βu+ c(x)u

where aα, bβ, c ∈ C∞(Ω) are smooth and bounded real-valued functions on the bounded domainΩ ⊂ R

n. Let u ∈ C2(Ω). Denote u− := min(u, 0), and u+ := max(u, 0). Then we have

Lu > 0, c(x) > 0 ⇒ infΩu = inf

∂Ωu−;

Lu 6 0, c(x) > 0 ⇒ supΩu = sup

∂Ωu+.

(16)

6 J. XU

(ii) [8, Thm. 3.1] Let u, L and its coefficients be the same as in (i) above. Then we have

Lu > 0, c(x) = 0 ⇒ infΩu = inf

∂Ωu;

Lu 6 0, c(x) = 0 ⇒ supΩu = sup

∂Ωu.

(17)

(iii) [8, Thm. 3.5] (Strong Maximum Principle) Let u ∈ C2(Ω). Assume that ∂Ω ∈ C∞. Let Lbe a second order uniformly elliptic operator as above. If Lu > 0, c(x) > 0, and if u attains anonpositive minimum over Ω in an interior point, then u is constant within Ω; If Lu 6 0, c(x) > 0,and if u attains a nonnegative maximum over Ω in an interior point, then u is constant within Ω.

(iv) [8, Ch. 8] All weak and strong maximum principles above hold when u ∈ H1(Ω, g) or u ∈H1(M,g), respectively, provided that L is uniformly elliptic with some other restrictions.

Let’s denote the conformal Laplacian by

gu := −a∆gu+ Sgu, u ∈ C∞c (Ω) or u ∈ C∞(M). (18)

The next two results from Kazdan and Warner [16] for closed manifold, and from Escobar [7] forcompact manifold with boundary below, respectively, show that the sign of λ is determined by thesign of η1.

Theorem 2.4. (i) [16, Lemma 2.5] Let (M,g) be a closed manifold. Let η1 be the first nonzeroeigenvalue of the conformal Laplacian g. Then the sign of λ, the constant scalar curvature withrespect to g = up−2g, must be the same as the sign of η1.

(ii) [7, Lemma 1.1] Let (M ′, g) be a compact manifold with boundary. Then the same conclusionabove holds on (M ′, g) with the boundary condition ∂u

∂ν+n−22 hgu = 0, where hg is the mean curvature

with respect to ∂M ′.

Remark 2.1. Note that η1 satisfies the PDE

−a∆gϕ+ Sgϕ = η1ϕ.

It is well-known that the first eigenfunction ϕ > 0 on closed manifolds M , or on M ′ on compactmanifold with smooth boundary and the same boundary condition in Theorem 2.4(ii).

The key step that converts our local solutions on Riemannian domain to manifolds is the applica-tion of monotone iteration method, in terms of obtaining solutions of Yamabe equation/perturbedYamabe equation provided the existences of subsolution and supersolution. Here we cite a resultdue to Kazdan and Warner, see [15] and [16].

Theorem 2.5. [16, Lemma 2.6] Let (M,g) be a closed manifold with dimM > 3. Let h,H ∈C∞(M) for some p > n = dimM . Let m > 1 be a constant. If there exists functions u−, u+ ∈C0(M) ∩H1(M,g) such that

−a∆gu− + hu− 6 Hum− in (M,g);

−a∆gu+ + hu+ > Hum+ in (M,g),(19)

hold weakly, with 0 6 u− 6 u+ and u− 6≡ 0, then there is a u ∈W 2,p(M,g) satisfying

− a∆gu+ hu = Hum in (M,g). (20)

In particular, u ∈ C∞(M).

Remark 2.2. (i)In the proof of Kazdan and Warner [15], [16], they constructed the monotoneiteration steps as

−a∆guj+1 + kuj+1 = −f(x, uj) + kuj , j ∈ Z>0.

where

f(x, u) := hu−Hum, k = supx∈M,u−(x)6u(x)6u+(x)

∂f(x, u)

∂u, k > 0, u0 := u+.


The requirement of u−, u+ ∈ W 2,p(M,g) for p large enough implies that u−, u+ ∈ C0(M), due toSobolev embedding in Theorem 2.2(iii). Starting with u0 = u+, the function f(x, u0) ∈ Lp(M,g),hence u1 ∈W 2,p(M,g) by elliptic regularity. Inductively with f(x, uj) ∈ Lp(M,g), elliptic regular-ity in Theorem 2.1(iii) shows that uj+1 ∈ W 2,p(M,g), hence uj+1 ∈ C0(M) and thus f(x, uj+1) ∈Lp(M,g) by the same Sobolev embedding argument. It follows that Theorem 2.5 still holds if westart with u−, u+ ∈ C0(M). In addition, u−, u+ ∈ H1(M,g) are required, since then the sub-solution and super-solution at least hold in the weak sense and a maximum principle in weak senseapplies. We see this by claiming that (−a∆g + k)(u − v) > 0 with u, v ∈ H1(M) ∩ C0(M) im-plies u > v. We choose w = max0, v − u, then w ∈ H1(M) and w > 0. Observe also that(−a∆g + k)w 6 0. Thus

0 >

ˆ

Mw(−a∆gw + kw)dω =

ˆ

M

(

a|∇gw|2 + kw2)

dω > 0 ⇒ w = 0.

Thus we must have u > v.

(ii)The relation u− 6 . . . 6 uj+1 6 uj 6 . . . 6 u+ is proved by taking subtraction of two adjacentiterations above, applying mean value theorem and maximum principle inH1-spaces. It then followsfrom [16] that u ∈ W 2,p(M,g) solves (20). Therefore u ∈ C1,α(M) by Sobolev embedding. Whenh,H ∈ C0(M), the requirement of u− > 0 can be replaced by

u− > 0 and u− 6≡ 0.

If u− satisfies the restrictions above, it follows that u has the same property. Choosing M =maxx∈Mh−Hum−1, 0, (20) implies that

−a∆gu+Mu > −a∆gu+(

h−Hum−1)

u = 0.

By strong maximum principle in Theorem 2.3, we conclude that if u(x0) = 0 for some x0 ∈M thenu ≡ 0, contradicting to u > u− and u− 6≡ 0. Hence u > 0 on M .

(iii) In later sections, Theorem 2.5 are used by taking h = Sg or h = Sg + β with some β < 0,

H = λ or H = λ− κ with some κ > 0 and m = p− 1 = n+2n−2 with u−, u+ ∈ C0(M) ∩H1(M,g).

Theorem 2.6. [19, Thm. 7] Let (M , g) be a compact manifold with smooth boundary, let rinjbe the injectivity radius of M , and let hg be the minimum of the mean curvature of ∂M . ChooseK ≥ 0 such that Ricg > −(n− 1)K. If λ1 is the first nonzero eigenvalue

−∆gu = λ1u inM ′, u ≡ 0 on ∂M ′.

Then

λ1 >1

γ

(

1

4(n − 1)r2inj(log γ)2 − (n− 1)K

)

, (21)

where

γ = max

exp[1 +(

1− 4(n− 1)2r2injK)

12 ], exp[−2(n − 1)hgrinj]

. (22)

Remark 2.3. We apply Theorem 2.6 above for Riemannian domain (Ω, g), for which (Ω, g) isconsidered to be a compact manifold with boundary. In Riemannian normal coordinates centeredat P ∈ Ω, g agrees with the Euclidean metric up to terms of order O(r2). Thus if Ω is a g-geodesicball of radius r, the mean curvature of ∂Ω is close (n − 1)/r, the mean curvature of a Euclideanr-ball in R

n. In (22), as r → 0, K can be taken to be unchanged (since g is independent of r),rinj → 0, and h·rinj → n−1. Thus γ → e2, the right hand side of (21) goes to infinity as r → 0, andλ1 → ∞. If Ω is a general, small enough Riemannian domain with a small enough injectivity radius,then Ω sits inside a g-geodesic ball Ω′′ of small radius. By the Rayleigh quotient characterizationof λ1, we have λ1,Ω′′ 6 λ1,Ω. Thus for all Riemannian domains (Ω, g), λ−1

1 → 0 as the radius of Ωgoes to zero.

8 J. XU

3. The Local Analysis of Yamabe Equation

In this section, we consider the solvability of the Yamabe equation

−a∆gu+ (Sg + β) u = λup−1 in (Ω, g);

u ≡ c on ∂Ω(23)

with some constant c > 0 and β = 0 in Proposition 3.1 and 3.2, due to an iterative scheme from[23]. We then consider the case c = 0 and β < 0 in Proposition 3.3, due to a local analysis incalculus of variations. The small enough Riemannian domain (Ω, g) is chosen to be a subset of theclosed manifold (M,g). Note that all results are dimensional specific, i.e. different dimensions ofmanifolds may lead to different choices of λ.

In this section, we introduce local analysis of Yamabe equation/perturbed Yamabe equation for(i) Sg < 0 everywhere and λ < 0 in Proposition 3.1; (ii) Sg > 0 somewhere and λ < 0 in Proposition3.2; (iii) Sg < 0 somewhere and λ > 0 in Proposition 3.3. In the proofs of Proposition 3.1 and 3.2,we refer to Theorem 2.3, Theorem 2.8 and Theorem 3.1 in [23].

The first result is for the case Sg < 0 everywhere and λ < 0. Note that Sg < 0 implies´

M Sg < 0, hence η1 in (18) is negative. Therefore by Theorem 2.4 we search for the solution ofYamabe equation with λ < 0.

Proposition 3.1. [23] Let (Ω, g) be Riemannian domain in Rn, n > 3, with C∞ boundary, and

with Volg(Ω) and the Euclidean diameter of Ω sufficiently small. Assume Sg < 0 on Ω. Then forsome λ < 0, (23) has a real, positive, smooth solution u ∈ C∞(Ω) such that infx∈Ω u = c > 0.

Proof. As in [23], this result is proved by applying the following iterative scheme

au0 − a∆gu0 = f0 in (Ω, g), u0 ≡ c on ∂Ω;

auk − a∆guk = auk−1 − Sguk−1 + λup−1k−1 in (Ω, g), uk ≡ c on ∂Ω, k ∈ N.

(24)

Setting uk = uk − c, k ∈ Z>0, the above iterations are equivalent to

au0 − a∆gu0 = f0 − ac in (Ω, g), u0 ≡ 0 on ∂Ω;

auk − a∆guk = auk−1 − Sguk−1 + λup−1k−1 − ac in (Ω, g), uk ≡ 0 on ∂Ω, k ∈ N.

(25)

The solvability of each linear elliptic PDE in (24) and (25) is achieved by Lax-Milgram [17, Ch. 6].Thus uk, uk ∈ H1

0 (Ω, g)∩H2(Ω, g),∀k, due to elliptic regularity in Theorem 2.1(i). With appropriatechoice of f0 > 0, f0 ∈ C∞(M), we conclude that u0, u0 ∈ C∞(Ω). By maximum principle in Theorem2.3

‖u0‖H2(Ω,g) 6 1, u0 > 0 in Ω, |u0| 6 CMn ,∀x ∈ Ω. (26)

The upper bound for |u0| is obtained by appropriate choice of f0 and repeated use of W s,p-typeelliptic regularity in Theorem 2.1(ii) and Sobolev embedding in Theorem 2.2(i). We refer [23,Thm. 3.1] for details. Choosing λ here such that

|λ|Cp−2Mn

6 inf(−Sg) ⇒ |λ| supup−20 6 inf(−Sg) ⇒ −Sgu0 + λup−1

0 > 0. (27)

With (27) and the same argument in Theorem 2.3 and Theorem 3.1 of [23], we conclude that

u1, u1 ∈ C∞(Ω), ‖u1‖H2(Ω,g) 6 1, u1 > 0 in Ω, |u1| 6 CMn ,∀x ∈ Ω.

Inductively, we assume

uk−1, uk−1 ∈ C∞(Ω), ‖uk−1‖H2(Ω,g) 6 1, uk−1 > 0 in Ω, |uk−1| 6 CMn ,∀x ∈ Ω, k ∈ N. (28)

By choosing λ the same as in (27),

|λ|Cp−2Mn

6 inf(−Sg) ⇒ |λ| supup−2k−1 6 inf(−Sg) ⇒ −Sguk−1 + λup−1

k−1 > 0. (29)


It follows that

uk, uk ∈ C∞(Ω), ‖uk‖H2(Ω,g) 6 1, uk > 0 in Ω, |uk| 6 CMn ,∀x ∈ Ω. (30)

Note that CMn is uniform in k. In addition, (24) implies that

auk − a∆guk > auk−1

based on the choice of λ in (27). If the infimum of uk is attained on ∂Ω, then infx∈Ω uk = c. If theinfimum of uk is attained at some interior point x0, then we have −a∆guk(x0) 6 0. It follows that

a infx∈Ω

uk = auk(x0) > auk(x0)− a∆guk(x0) > auk−1(x0) > a infx∈Ω

uk−1.

If infx∈Ω u0 = c, which can be done by setting infx∈Ω f0 = c, then infx∈Ω uk = c,∀k. HenceCMn > c.

Therefore the sequence uk is uniformly bounded in H2-norm and is convergent to some u ∈H2(Ω, g) which solves (23). By Sobolev embedding and elliptic regularity again, it follows thatu ∈ C∞(Ω) and u > 0. Due to strong maximum principle in Theorem 2.3(iii) we conclude thatu > 0 since u ≡ c > 0 on ∂Ω. Observe that

|λ|up−2k 6 −Sg,∀x ∈ Ω ⇒ −Sgu+ λup−1

> 0 ⇒ −a∆gu > 0.

By Theorem 2.3(ii) we conclude that u achieves its minimum at ∂Ω. Note that in this proof, weneed to choose λ small enough twice so that (30) holds, uniformly for all k.

Next result corresponds to the case Sg > 0 somewhere in M and λ < 0, the proof is essentiallythe same as in [23]. In the next proposition, we assume, without loss of generality, that Sg 6 a

2 .This can be done by scaling the metric g. We start with this metric.

Proposition 3.2. [23] Let (Ω, g) be Riemannian domain in Rn, n > 3, with C∞ boundary, and

with Volg(Ω) and the Euclidean diameter of Ω sufficiently small. Assume Sg(x) 6a2 ,∀x ∈M , and

Sg > 0 within the small enough closed domain Ω. Then for some λ < 0, (23) has a real, positive,smooth solution u ∈ C∞(Ω) such that supx∈Ω u = c > 0.

Proof. We apply the same iterative scheme as in (24) and (25) as above. In this case we choosef0 = 0 in (24), thus we have

au0 − a∆gu0 = 0 in (Ω, g), u ≡ c > 0 on ∂Ω.

It follows immediately from the PDE above that −∆gu 6 0, therefore u0 > 0 by weak maximumprinciple in Theorem 2.3(i). Then by strong maximum principle in Theorem 2.3(ii), we concludethat supx∈Ω u0 = c. For the first iteration k = 1 in (24), we choose λ so that

|λ|cp−2 6a

26 a+ inf

Ω(−Sg) ⇒ |λ| supup−2

0 6 a− Sg ⇒ au0 − Sgu0 + λup−10 > 0. (31)

By the same argument as in Theorem 2.3, Theorem 2.8, Theorem 3.1 of [23] and Proposition 3.1above, we conclude that (26) holds with 0 6 u0 6 c. Inductively we assume that (28) holds with0 6 uk−1 6 c. observe that since Sg > 0, λ < 0 and uk−1 > 0, it follows from the kth iteration of(24) that

auk − a∆guk 6 auk−1 − Sguk−1 + λup−1k−1 6 auk−1.

If the supremum is attained at the boundary, then we have supx∈Ω uk = c. If the supremum isattained in some interior point x0, at this point we have −a∆guk(x0) > 0 hence

a supx∈Ω

uk = auk(x0) 6 auk(x0)− a∆guk(x0) 6 auk−1(x0) 6 a supx∈Ω

uk−1 = ac.

Choosing the same λ as in (31), it follows that

auk−1 − Sguk−1 + λup−1k−1 > 0.

10 J. XU

Therefore (30) holds with 0 6 uk 6 c. Due to the same argument as above, we conclude that thesmooth sequence uk ⊂ C∞(Ω) converges to some u ∈ H2(Ω, g). Since u ≡ c on ∂Ω, it is not atrivial solution. By bootstrapping strategy with respect to Sobolev inequality (11), (13) and ellipticregularity (8), we conclude that u ∈ C0,α(Ω) and hence Schauder estimates (9) in Theorem 2.1(iii)says that u ∈ C∞(Ω). In particular, since each uk ∈ [0, c], it follows that u ∈ [0, c].

With the choice of λ we have

Sg > 0,−λ > 0, u > 0 ⇒ Sgu− λup−1 > 0 ⇒ −a∆gu 6 0,

on (Ω, g) and it follows that u achieves its supremum at ∂Ω by Theorem 2.3(ii).

Remark 3.1. Consider the PDE

−a∆gu+ Sgu− λup−1 = 0 in (Ω, g).

For λ < 0 satisfies Theorem 2.3 of [23] and (31) above, we have

|λ|cp−26a

2⇒ −λcp−1

6a

2c⇒ λcp−1

> −a2c. (32)

(32) we be used later in Section 4.

For the case when Sg < 0 somewhere and λ > 0, we need the following theorem, due to Wang[26], which gives the existence of the positive solution of the following Dirichlet problem

Lu := −∑

i,j

∂i (aij(x)∂ju) = b(x)up−1 + f(x, u) in Ω;

u > 0 in Ω, u = 0 on ∂Ω.

(33)

Here p− 1 = n+2n−2 is the critical exponent with respect to the H1

0 -solutions of (33). The solution of

(33) is the minimizer of the functional

J(u) =

ˆ

Ω

1

2

∑

i,j

aij(x)∂iu∂ju− b(x)

pup+ − F (x, u)

dx, (34)

where u+ = maxu, 0 and F (x, u) =´ u0 f(x, t)dt. Set

A(D) = essinfx∈Ddet(aij(x))

|b(x)|n−2,∀D ⊂ Ω;

T = infu∈H1

0 (Ω)

´

Ω|Du|2dx(´

Ω|u|pdx)

2p

;

K = infu 6=0

supt>0

J(tu),K0 =1

nT

n2 (A(Ω))

12 ;

TL = infu 6=0

´

Ω aij∂iu∂judx(´

Ω b(x)|u|p+1dx) 2

p

.

(35)

Theorem 3.1. [26, Thm. 1.1, Thm. 1.4] Let Ω be a bounded smooth domain in Rn, n > 3. Let

Lu = −∑i,j ∂i (aij(x)∂ju) be a second order elliptic operator in divergence form. Let Volg(Ω) and

the diameter of Ω sufficiently small. Let b(x) 6= 0 be a nonnegative bounded measurable function.Let f(x, u) be measurable in x and continuous in u. Assume

(1) There exist c1, c2 > 0 such that c1|ξ|2 6∑

i,j aij(x)ξiξj 6 c2|ξ|2,∀x ∈ Ω, ξ ∈ Rn;

(2) limu→+∞f(x,u)up−1 = 0 uniformly for x ∈ Ω;

(3) limu→0f(x,u)

u < λ1 uniformly for x ∈ Ω, where λ1 is the first eigenvalue of L;


(4) There exists θ ∈ (0, 12),M > 0, σ > 0, such that F (x, u) =´ u0 f(x, t)dt 6 θuf(x, u) for any

u >M , x ∈ Ω(σ) = x ∈ Ω, 0 6 b(x) 6 σ.Furthermore, we assume that f(x, u) > 0, f(x, u) = 0 for u 6 0. We also assume that aij(x) ∈C0(Ω). If

K < K0 or K <1

nT

n2L , (36)

then the Dirichlet problem (33) possesses a solution u which satisfies J(u) 6 K.

We now apply Theorem 3.1 to show the following result. Unfortunately we cannot show

−a∆gu+ Sgu = λup−1 in Ω, u = 0 on ∂Ω

directly. Instead we perturb Sg by Sg + β for some β < 0, and show the existence of the positivesolution of

−a∆gu+ (Sg + β)u = λup−1 in Ω, u = 0 on ∂Ω.

Proposition 3.3. Let (Ω, g) be Riemannian domain in Rn, n > 3, with C∞ boundary, and with

Volg(Ω) and the Euclidean diameter of Ω sufficiently small. Let β < 0 be any constant. AssumeSg < 0 within the small enough closed domain Ω. Then for any λ > 0, the following Dirichletproblem

− a∆gu+ (Sg + β) u = λup−1 in Ω, u = 0 on ∂Ω. (37)

has a real, positive, smooth solution u ∈ C∞(Ω)∩H10 (Ω, g) in a very small domain Ω that vanishes

at ∂Ω.

Proof. We show this by showing that (37) can be converted to a special case in Theorem 3.1, andthus a positive solution is admitted. Without loss of generality, we may assume 0 ∈ Ω. UnderW s,p(Ω, g)-norms, (37) is the Euler-Lagrange equation of the functional

J∗(u) =

ˆ

Ω

(

1

2a√

det(g)gij∂iu∂ju−√

det(g)

pλup+ −

ˆ u

0

√

det(g)(x)(−Sg(x)− β)tdt

)

dx. (38)

It is straightforward to check that

d

dt

∣

∣

∣

∣

t=0

J∗(u+ tv) =

ˆ

Ω

((

−a∂i(√

det(g)gij∂ju))

v −√

det(g)λup−1+ v −

√

det(g)(−Sg − β)uv)

dx

=

ˆ

Ω

(

−a 1√

det(g)

(

∂i(√

det(g)gij∂ju))

− λup−1+ + (Sg + β) u

)

v√

det(g)dx

= 〈−a∆gu− λup−1+ + (Sg + β) u, v〉g.

Hence u > 0 in Ω minimizes J∗(u) if and only if (37) admits a positive solution. Note that J∗u isjust a special expression of (34) with

aij(x) = a√

det(g)gij(x), b(x) = λ√

det(g)(x), f(x, u) =√

det(g)(x) (−Sg(x)− β) u. (39)

We check that hypotheses in Theorem 3.1 are satisfied.

First of all, b(x) > 0 on Ω clearly. As mentioned in Remark 2.3, we can choose Ω small enoughso that it is a geodesic ball centered at some point p ∈ M . At the point p, gij = δij and it followsthat gij is a small purterbation of δij within small enough Ω. Thus the first hypothesis in Theorem3.1 holds. With the expressions if b and f in (39), we have

limu→+∞

f(x, u)

up−1= lim

u→+∞

√

det(g)(x)(−Sg(x) + β)u2−p = 0.

Hence the second hypothesis is satisfied.

12 J. XU

Hypothesis (3) is for the positivity of J(u). In the general case (33), we choose ϕ > 0 be thefirst eigenfunction of the Dirichlet problem Lϕ = λ1ϕ in Ω, ϕ = 0 on ∂Ω, we observe that if u > 0solves (33),

λ1

ˆ

Ωuϕdx =

ˆ

Ωu · Lϕdx =

ˆ

ΩLu · ϕdx =

ˆ

Ω

(

b(x)up−1 + f(x, u))

ϕdVolg

=

ˆ

Ω

(

b(x)up−1 +f(x, u)

u· u)

ϕdx >

ˆ

Ω

f(x, u)

u· uϕdx.

It follows that if limu→0f(x,u)

u > λ1 then it is impossible to get a positive solution u. In our casefor J∗, we choose ϕ > 0 and λ1 > 0 be the first eigenfunction and first nonzero eigenvalue of −∆g,respectively. It follows from (37) that

aλ1

ˆ

ΩuϕdVolg =

ˆ

Ωu · (−a∆gϕ)dVolg =

ˆ

Ω−a∆gu · ϕdx =

ˆ

Ω

(

λup−1 + (−Sg − β)u)

ϕdVolg

>

ˆ

Ω(−Sg − β) · uϕdVolg.

It follows that Hypothesis (3) holds if supx∈Ω(−Sg)− β 6 aλ1. For later use, we further require

a

n−(

n− 2

2n+

1

2

)

(

supx∈Ω

(−Sg)− β

)

· λ−11 > 0. (40)

by making λ1 large enough. Due to the argument of first eigenvalue in Remark 2.3, this can beachieved when Ω is small enough. Fix this Ω from now on. It follows that for any β′ < 0 withβ < β′, the same Ω applies.

For the last hypothesis, Wang [26] mentioned that if b(x) > 0 a.e. in Ω then (4) in Theorem 3.1

is not needed. Here our b(x) =√

det(g)λ > 0 as always since in a small geodesic ball,√

det(g) ≈1 +O(diam(Ω)2) and λ is chosen to be positive.

Lastly, we check (36) by checking K < K0. This can be done by showing that there exists someu > 0 in Ω, u ≡ 0 on ∂Ω such that supt>0J

∗(tu) < K0. Assume such an u does exist, then we have

J∗(tu) =

ˆ

Ω

(

t21

2a√

det(g)gij∂iu∂ju− tp√

det(g)λ

pup − t2

1

2

√

det(g) (−Sg(x)− β) u2

)

dx.

It follows that J∗(tu) < 0 when t large enough. Hence supt>0 J∗(tu) > 0 is achieved for some finite

t0 > 0. If t0 = 0, then supt>0 J∗(tu) = 0 and thus K 6 0, again K < K0 holds trivially. Assume

now t0 > 0. In this case, we have

d

dt

∣

∣

∣

∣

t=t0

J∗(tu) = 0

⇒ t0

ˆ

Ωa√

det(g)gij∂iu∂judx− tp−10

ˆ

Ω

√

det(g)λupdx− t0

ˆ

Ω

√

det(g)(−Sg(x)− β)u2dx = 0.

(41)

Denote

V1 =

ˆ

Ωa√

det(g)gij∂iu∂judx, V2 =

ˆ

Ω

√

det(g)(−Sg(x)− β)u2dx,W =

(ˆ

Ω

√

det(g)λupdx

)1p

.

Since t0 > 0, (41) implies

tp−20 =

V1 − V2W p

⇒ t0 =(V1 − V2)

1p−2

Wp

p−2

. (42)


By (42), J∗(t0u) is of the form

J∗(t0u) =(V1 − V2)

2p−2

2W2pp−2

V1 −(V1 − V2)

pp−2

Wp2

p−2

· 1pW p − (V1 − V2)

2p−2

2W2pp−2

V2.

Note that p = 2nn−2 hence we have

2

p− 2=n− 2

2,

p

p− 2=n

2, p− p2

p− 2= p

(

1− p

p− 2

)

= −n⇒ 2p

p− 2= n =

p2

p− 2− p.

In terms of n, J∗(t0u) becomes

J∗(t0u) =

(

1

2− 1

p

)

(V1 − V2)n2

W n=

1

n

(V1 − V2)n2

W n

=1

n

´

Ω a√

det(g)gij∂iu∂judx−´

Ω

√

det(g) (−Sg − β) u2dx(

´

Ω

√

det(g)λupdx)

2p

n2

.

(43)

The inequality K < K0 reduces to find some u ∈ H10 (Ω), u > 0 such that (43) is less than K0. We

check the expression of K0 first. by aij , b in (39), we have

A(Ω) = essinfΩdet(aij(x))

|b(x)|n−2= essinfΩ

det(

a√

det(g)gij)

(

√

det(g)λ)n−2 = essinfΩ

(

√

det(g))nan det(gij)

(

√

det(g)λ)n−2

= essinfΩanλ2−n

√

det(g)2det(gij) = λ2−nan.

It follows that

K0 =1

nT

n2A(Ω)

12 =

1

nλ

2−n2 a

n2 T

n2 . (44)

Compare expressions in (43) and (44), showing K < K0 is equivalent to find some u > 0 in Ω suchthat

J0 :=

´

Ω a√

det(g)gij∂iu∂judx−´

Ω

√

det(g) (−Sg − β)u2dx(

´

Ω

√

det(g)λupdx)

2p

< λ2−nn aT

⇔´

Ω

√

det(g)gij∂iu∂judx− 1a

´

Ω

√

det(g) (−Sg − β) u2dx(

´

Ω

√

det(g)updx) 2

p

< T.

(45)

Claim: When both Ω and ǫ are small enough, the function uǫ,Ω(x) =ϕΩ(x)

(ǫ+|x|2)n−22

satisfies

‖∇guǫ,Ω‖2L2(Ω,g) +1a

´

Ω (Sg + β)u2ǫ,Ω√

det(g)dx

‖uǫ,Ω‖2Lp(Ω,g)

< T. (46)

This claim is proved in Appendix A.

Finally we conclude thatK < K0.

Note that this result is peculiar for our special PDE in (38) with small enough Ω. The regularityof u follows exactly the same as in [Sec, 1][5]. We would like to denote here that when u ∈ L∞(Ω)then up−1 ∈ Lp(Ω, g) for all p, and by bootstrapping method, it follows that u ∈ C∞(Ω), providedthat ∂Ω is smooth. In addition, u ∈ C1,α(Ω), the constant α is determined by the nonlinear termup−1.

14 J. XU

Remark 3.2. The proof of Theorem A.1 in Appendix A plays a central role for the validity ofProposition 3.3. This proof is essentially due to Brezis and Nirenberg [5], but they are not the same.The proof in Theorem A.1 is a sharp estimate in the sense that, for example, if any negative term

with order ǫ4−n2 on numerator and any positive term of order ǫ

4−n2 on denominator is dropped when

n > 5, then the estimate fails. A detailed analysis with respect to Beta and Gamma functions arerequired to connect the best Sobolev constant, the area of unit sphere, and K2 defined in AppendixA.

Remark 3.3. The quantity Qǫ,Ω we estimate in Appendix A is a perturbation of Yamabe quotient,and hence is not conformally invariant. We show that Qǫ,Ω < T for all n > 3, no matter the ambientmanifold is locally conformally flat or not.

4. Yamabe Equation on Closed Manifolds

In this section, we apply the sub-solution and super-solution technique in Theorem 2.5 andRemark 2.2 to solve Yamabe equation for five cases:

(A). Sg 6 0 everywhere;(B). Sg > 0 somewhere and the conformal Laplacian g admits a negative first eigenvalue η1;(C). Sg < 0 somewhere and the conformal Laplacian g admits a positive first eigenvalue η1;(D). Sg > 0 everywhere;(E). η1 = 0.

When η1 = 0 in case (E), we can solve Yamabe problem trivially with λ = 0, this is just aneigenvalue problem. Note that generically zero is not an eigenvalue of conformal Laplacian g, see[9].

For the rest four cases above, the sub-solutions and super-solutions will be constructed due toPropositions 3.1 through 3.3, respectively. By Theorem 2.4(i), we know that (A) and (B) areequivalent to find a solution of (2) with λ < 0; (C) and (D) are equivalent to find a solution of(2) with λ > 0. In contrast to the classical calculus of variation approach–in which they findthe minimizer of (3) first, and conclude that the desired λ is automatically determined by somefunctional with the minimizer of (3)–this article determines an appropriate choice of λ first, andthen conclude that the Yamabe equation admits a real, smooth, positive solution associated withthe choice of λ.

Recall the eigenvalue problem of conformal Laplacian on (M,g).

− a∆gϕ+ Sgϕ = η1ϕ on (M,g). (47)

The first nonzero eigenvalue is characterized by

η1 = infϕ 6=0

´

M

(

a|∇gϕ|2 + Sgϕ2)

dω´

M ϕ2dω:= inf

ϕ 6=0L(ϕ). (48)

Observe that L(ϕ) = L(|ϕ|), we may characterize η1 by nonnegative functions, for details, see e.g.[16]. By bootstrapping and maximum principle, we conclude that the first eigenfunction ϕ > 0 issmooth.

The first result concerns the case globally when Sg < 0 everywhere. Note that taking u = 1 ineither (3) or (48), we have λ(M) < 0, or η1 < 0, respectively.

Theorem 4.1. Let (M,g) be a closed manifold with n = dimM > 3. Assume the scalar curvatureSg to be negative everywhere on M . Then there exists some λ < 0 such that the Yamabe equation(2) has a real, positive, smooth solution.


Proof. Due to Theorem 2.5 and Remark 2.2, we construct a subsolution u− ∈ C0(M) ∩H1(M,g)and a super-solution u+ ∈ C0(M) ∩H1(M,g).

First of all, we choose a finite cover Uα, φα of (M,g). Pick up a small enough Riemanniandomain (Ω, g) ⊂ (φα(Uα), g). We can definitely take Ω ⊂ φα(Uα) to be a small enough closed ball.Take any constant c > 0.

We construct the sub-solution first. By Proposition 3.1, there exists some λ ∈ (η1, 0) such thatu1 ∈ C∞(Ω) solves the Yamabe equation (23) locally with β = 0 and u1 ≡ c > 0 on ∂Ω. Fix the λ.Extend u1 to u− by setting

u−(x) =

u1(x), x ∈ Ω;

c, x ∈M\Ω,Clearly u− ∈ C0. Observe that u−− c is the extension of u1− c by zero outside Ω. Due to extensiontheorem, it follows that u1 − c ∈ H1(φα(Uα), g) since u1 − c ∈ H1

0 (Ω, g). To see this, we canapproximate by u1 − c by a sequence of smooth functions vk with compact support in Ω, thenextend vk by zero to the φα(Uα), then take the limit of vk with H1(φα(Uα), g)-norm. It followsthat the limit u1 − c ∈ H1(φα(Uα), g). Since M is compact, we can extend u1 − c by zero to thewhole manifold. Thus we conclude that

u− ∈ C0(M) ∩H1(M,g).

In addition, u− > 0 and u− 6≡ 0. We check that u− is a sub-solution of Yamabe equation. WithinΩ, u− gives an equality of (2), outside Ω, we observe that

− a∆gu− + Sgu− − λup−1− = −a∆gc+ Sgc− λcp−1 = c(Sg − λcp−2). (49)

As we mentioned in Proposition 3.1, CMn > c and thus

|λ|cp−2 6 |λ|Cp−2Mn

6 inf(−Sg).It follows that Sg − λcp−2 6 0,∀x ∈ M , hence the left side of (25) is nonpositive for x ∈ M\Ω. Inconclusion, u− is a sub-solution of Yamabe equation in the weak sense.

The super-solution is constructed as follows. For the fixed λ, pick a constant C such that

infx∈M

C > max(

infx∈M Sgλ

)1

p−1

, supx∈M

u−.

Denoteu+ := C.

Immediately, u+ ∈ C∞(M). Since λ < 0, we have

C >

(

infx∈M Sgλ

) 1p−2

⇒ Sg > λCp−2 ⇒ SgC > λCp−1

⇒− a∆gC + SgC > λCp−1 ⇒ −a∆gu+ + Sgu+ > λup−1+ .

Thus u+ is a super-solution of the Yamabe equation. Due to the choice of C, we conclude that0 6 u− 6 u+ on M . It then follows from Theorem 2.5 that there exists some real, positive functionu ∈ C∞(M) that solves (2).

Remark 4.1. Note that u1 above, which is constructed in Proposition 3.1 achieves its minimumc on ∂Ω, and hence u− = maxu1, c. It is easy to see that the maximum of two sub-solutions isagain a sub-solution in the weak sense.

Next result deals with the case Sg > 0 somewhere and conformal Laplacian g has a negativefirst eigenvalue. In the next theorem, we require Sg(x) ∈ [−a

4 ,a2 ],∀x ∈ M . This can be done by

scaling the original metric g.

16 J. XU

Theorem 4.2. Let (M,g) be a closed manifold with n = dimM > 3. Assume the scalar curvatureSg(x) ∈ [−a

4 ,a2 ], x ∈ M , which is positive somewhere on M . Let the first eigenvalue of conformal

Laplacian g satisfying η1 < 0. Then there exists some λ < 0 such that the Yamabe equation (2)has a real, positive, smooth solution.

Proof. Since η1 < 0, we solve the Yamabe problem for some λ < 0. We construct both the sub-solution and super-solution again here.

We construct the super-solution first. Pick up some c > 0 which will be determined later.Since Sg > 0 somewhere, we take a small enough Riemannian domain (Ω, g) in which Sg > 0 andmaxx∈M Sg is achieved in the interior of Ω. As above, we can take Ω ⊂ φα(Uα) to be a small enoughclosed ball for some cover Uα, φα of (M,g). By Proposition 3.2, there exists some λ < 0 when|λ| is small enough such that u2 ∈ C∞(Ω) solves (24) locally in (Ω, g) with u ≡ c > 0 on ∂Ω. SinceSg > −a

4 , we further choose λ such that

a

46 |λ|cp−2

6a

2⇒ −a

4> λcp−2 ⇒ Sg > λcp−2 ⇒ Sgc > λcp−1,∀x ∈M. (50)

Note that the choice of λ in (50) is compliant with the choice of λ in (31) in Proposition 3.2, dueto Remark 3.3. Note also that we need |λ| to be small enough so that the requirement in Theorem2.3 of [23] is also satisfied. In addition, let’s require |λ| 6 |η1|, where η1 is the first eigenvalue ofg on M in this situation. This can be done by choose an appropriate c. Setting

u+(x) =

u2(x), x ∈ Ω;

c, x ∈M\Ω,Clearly u+ > 0 on M . By the same argument in Theorem 4.1, we conclude that u+ ∈ C0(M) ∩H1(M,g) is a super-solution of Yamabe equation in the weak sense. To see this, we check thatwithin Ω, u+ = u2 and the equality of Yamabe equation holds. When outside Ω, we have

−a∆gu+ + Sgu+ − λup−1+ = Sgc− λcp−1 > 0,

due to (50).

For the sub-solution. We apply the result of eigenvalue problem, which says that there existssome function ϕ > 0 that solves

−a∆gϕ+ Sgϕ = η1ϕ on (M,g).

Observe that we have chosen λ such that |λ| 6 |η1| ⇔ η1 6 λ 6 0. Scaling ϕ by δϕ with δ < 1small enough such that

supx∈M

δϕ 6 min infx∈M

u+, 1.

Denoteu− := δϕ.

Observe that 0 6 u− 6 1 implies that 0 6 up−1− 6 u− since p − 1 = n+2

n−2 > 1. Note that λ < 0, wethen have

− a∆gϕ+ Sgϕ = η1ϕ 6 λϕ⇒ −a∆gδϕ+ Sgδϕ 6 λδϕ

⇒− a∆gu− + Sgu− 6 λu− 6 λup−1− .

Hence u− > 0 is a sub-solution of the Yamabe equation and u− 6≡ 0. Obviously u− ∈ C∞(M).In addition, 0 6 u− 6 u+ on M . It then follows from Theorem 2.5 that there exists some real,positive function u ∈ C∞(M) that solves (2).

Remark 4.2. Note that u2 above, which is constructed in Proposition 3.2 achieves its maximumc on ∂Ω, and hence u+ = minu2, c. It is easy to see that the minimum of two super-solutions isagain a super-solution in the weak sense.


Remark 4.3. The case when Sg 6 0 everywhere–slightly different from the hypothesis in Theorem4.1–can be converted to the case when Sg > 0 somewhere and η1 < 0. This can be done by applyingCorollary 4.1 below. Note the sign of first eigenvalue of g is a conformal invariant.

Remark 4.4. Historically the cases when η1 < 0 is easy to handle. On one hand, there are manyother ways to construct sub-solutions and super-solutions, for instance, see [16] on closed manifoldsand [3] on non-compact manifolds. On the other hand, when Sg 6 0 or

´

M Sg < 0, the functional(3) satisfies λ(M) < 0 and hence λ(M) < λ(Sn) holds trivially. Thus by Yamabe, Trudinger andAubin’s result, the Yamabe equation has a positive solution.

Next we show the case when Sg < 0 somewhere and the conformal Laplacian g has a positivefirst eigenvalue. This means we are looking for solutions of Yamabe equation with λ > 0. Let β < 0be any constant. Denote

λβ = infu 6=0,u∈H1(M)

´

M a|∇gu|2dω +´

M (Sg + β)u2dω(´

M updω) 2

p

. (51)

Recall that

λ(M) = infu 6=0,u∈H1(M)

´

M a|∇gu|2dω +´

M Sgu2dω

(´

M updω)

2p

> 0. (52)

It is crucial to assume Sg < 0 somewhere in order to construct a sub-solution within this region.We cannot work on the Yamabe equation directly. Instead, we solve

−a∆gu+ (Sg + β)u = λup−1

for some specific choice of λ with any β < 0. Before showing this, we need the following lemma,which indicates that a small perturbation of conformal Laplacian will not change the sign of λβ in(51) provided that λ(M) > 0.

Lemma 4.1. Let η1 > 0 be the first eigenvalue of g on close manifolds (M,g) with dimM > 3.Let β < 0 be a constant. Then λβ defined in (51) is also positive provided that |β| is small enough.

Proof. It is immediate that if η1 > 0 then λ(M) > 0. It is due to the characterization of η1

η1 = infϕ 6=0

´

M

(

a|∇gϕ|2 + Sgϕ2)

dω´

M ϕ2dω

as well as the characterization of λ(M) in (52).Due to the characterization of λβ in (51), we conclude that for each ǫ > 0, there exists a function

u0 such that´

M

(

a|∇gu0|2 + (Sg + β) u20)

dω(´

M up0dω) 2

p

6 λβ + ǫ.

It follows that

0 < λ(M) 6

´

M

(

a|∇gu0|2 + Sgu20

)

dω(´

M up0dω) 2

p

6

´

M

(

a|∇gu0|2 + (Sg + β) u20)

dω(´

M up0dω) 2

p

− β

´

M u20dω(´

M up0dω) 2

p

6 λβ − β

(´

M up0dω)

2p ·Volg(M)

p−2p

(´

M up0dω) 2

p

+ ǫ = λβ − βVolg(M)p−2p + ǫ

Since ǫ can be arbitrarily small, it follows that

λβ > λ(M) + βVolg(M)p−2p .

Thus λβ > 0 if |β| is small enough.

18 J. XU

We now show the existence of the solution of the perturbed Yamabe equation with a special choiceof constant in front of the nonlinear term up−1. One of the advantages from the local analysis isthe flexibility of the choice of this constant, without knowing λ(M) < λ(Sn). We would like tomention here that if we already know λ(M) < λ(Sn), then we can apply the analysis developed byAubin and Trudinger to show the existence result directly, see [18, §4]. With λ(M) < λ(Sn), nolocal analysis developed here is required anymore.

Theorem 4.3. Let (M,g) be a closed manifold with n = dimM > 3. Let Sg be the scalar curvaturewith respect to g which is negative somewhere on M . Assume the first eigenvalue of conformalLaplacian g satisfying η1 > 0. Let β < 0 be a small negative constant such that λβ > 0 andη1 + β > 0, and let κ > 0 small enough so that λβ − κ > 0. Then the following PDE

− a∆gu+ (Sg + λ) u = (λβ − κ) up−1 inM (53)

has a real, positive, smooth solution with λs given in (51).

Proof. We construct both the sub-solution and super-solution again here.

Let β < 0 be a small enough negative constant such that the first eigenvalue η1 of the conformalLaplacian admits a positive solution φ ∈ C∞(M) satisfying

− a∆gϕ+ (Sg + β)ϕ = η1ϕ+ βϕ inM. (54)

Note that any scaling δϕ also solves (54). For the given λβ and κ, we want

(η1 + β) infM

(δϕ) > 2p−2 (λβ − κ) supM

(

δp−1ϕp−1)

⇔ (η1 + β)

2p−2 (λβ − κ)> δp−2 supM ϕp−1

infM ϕ.

For fixed η1, λβ, ϕ, κ, β, this can be done by letting δ small enough. We denote φ = δϕ. It followsthat

−a∆gφ+ (Sg + β)φ = (η1 + β)φ inM ;

(η1 + β) infMφ > 2p−2 (λβ − κ) sup

Mφp−1

> 2p−2 (λβ − κ)φp−1 > (λβ − κ)φp−1 inM. (55)

Set

β′ = (η1 + β) supM

φ− 2p−2 (λβ − κ) infMφp−1 > (η1 + β)φ− 2p−2 (λβ − κ)φp−1 pointwisely. (56)

Thus we have

− a∆gφ+ (Sg + β)φ = (η1 + β)φ > 2p−2 (λβ − κ)φp−1 > (λβ − κ)φp−1 inM pointwisely. (57)

For sub-solution, we apply Proposition 3.3 within the small enough region (Ω, g) in which Sg < 0.Since Sg is smooth onM hence it’s bounded below. Thus based on hypothesis (3) in Theorem 3.1, wechoose Ω small enough in which the first eigenvalue λ1 of −∆gu is larger than a (supx∈M (−Sg) + β).By the result of (3.3), we have u3 ∈ C0(Ω) ∩H1

0 (Ω, g), u3 > 0 in Ω solves (37) for λβ − κ > 0, i.e.

−a∆gu3 + (Sg + β)u3 = (λβ − κ)up−13 in Ω, u3 = 0 on ∂Ω;

⇔ −a∆g

(u32

)

+ (Sg + β)u32

= 2p−2 (λβ − κ)(u32

)p−1in Ω, u3 = 0 on ∂Ω

(58)

Define

u−(x) =

u3(x), x ∈ Ω

0, x ∈M\Ω .


By the same argument in Theorem 4.1, u− ∈ C0(M)∩H1(M,g). We show that u− is a sub-solution

of (53). In local coordinates, the outward normal derivative ν = − ∇u3|∇u3|

along ∂Ω, thus locally

taking some Ω′ ⊃ Ω and any test function v > 0 we haveˆ

Ω′

(

a∇gu− · ∇gv + (Sg + β) u−v − λup−1− v

)

dVolg

=

ˆ

Ω

(

−a∆gu3 + (Sg + β)u3 − λup−13

)

vdVolg +

ˆ

∂Ω

∂u3∂ν

vdS

=

ˆ

∂Ω(∇u3 · ν) vdS =

ˆ

∂Ω∇u3 ·

(

− ∇u3|∇u3|

)

vdS =

ˆ

∂Ω−|∇u3|vdS 6 0.

This argument holds on M also since u− is trivial outside Ω. Thus we conclude

−a∆gu− + (Sg + β)u− − (λβ − κ) up−1− 6 0 weakly inM.

We now construct the super-solution. Pick up γ ≪ 1 such that

0 < 20 (λβ − κ) γ + 2γ ·(

supM

|Sg|+ |β|)

γ <β′

2, 31 (λβ − κ) (φ+ γ)p−2γ <

β′

2. (59)

This is dimensional specific. Set

V = x ∈ Ω : u3(x) > φ(x), V ′ = x ∈ Ω : u3(x) < φ(x),D0 = x ∈ Ω : u3(x) = φ(x),

D′0 = x ∈ Ω : dist(x,D0) < δ,D′′

0 =

x ∈ Ω : dist(x,D0) >δ

2

,

∂D′0 = x ∈ Ω : dist(x,D0) = δ, ∂D′′

0 =

x ∈ Ω : dist(x,D0) =δ

2

.

The constant δ > 0 is chosen so that

supx∈D′

0

|u3(x)− φ(x)| < γ.

If φ > u3 pointwisely, then φ is a super-solution. If not, a good candidate of super-solution willbe maxu3, φ in Ω and φ outside Ω, this is an H1 ∩ C0-function. Let ν be the outward normal

derivative of ∂V along D0. If ∂u3∂ν = ∂φ

∂ν on D0 then the super-solution has been constructed.

However, this is in general not the case. If not, then ∂u3−∂φ∂ν 6= 0, which follows that 0 is a regular

point of the function u3 − φ and hence D0 is a smooth submanifold of Ω. Define

Ω1 = V ∩D′′0 ,Ω2 = V ′ ∩D′′

0 ,Ω3 = D′0. (60)

Clearly Ωi, i = 1, 2, 3 are open sets,⋃

iΩi = Ω and Ω1 ∩ Ω2 = ∅. Note that D0,D′0 will never

intersect ∂Ω. Without loss of generality, we may assume that all Ωi, i = 1, 2, 3 are connected, sinceif not, then we do local analysis in all components where u3 − φ changes sign and combine resultstogether. For any 0 < ǫ≪ δ

4 , set

D1 = x ∈ V : dist(x, ∂D′0) < ǫ,D′

1 = x ∈ V ′ : dist(x, ∂D′0) < ǫ,

D2 = x ∈ V : dist(x, ∂D′′0 ) < ǫ,D′

2 = x ∈ V ′ : dist(x, ∂D′′0 ) < ǫ,

E1 = ∂D1 ∩ Ω1 ∩ Ω3, E2 = ∂D2 ∩ Ω1, E′1 = ∂D′

1 ∩Ω2 ∩ Ω3, E′2 = ∂D′

2 ∩ Ω2;

F1 = (Ω1 ∩ Ω3) \ (D1 ∪D2) , F2 = (Ω2 ∩ Ω3) \(

D′1 ∪D′

2

)

.

In conclusion, D1,D2, E1, E2, F1 are subsets of V , in which u3 > φ; D′1,D

′2, E

′1, E

′2, F2 are subsets

of V ′, in which u3 < φ. Note that ∂F1 = E1 ∪ E2, ∂F2 = E′1 ∪ E′

2. We now construct appropriate

20 J. XU

partition of unity subordinate to Ωi, i = 1, 2, 3. Choose a local coordinate system xi on Ω, weconsider the PDE

−∑

i,j

gij(u3 − φ− γ)∂2v1∂xi∂xj

− 2∑

i,j

gij(

∂u3∂xj

− ∂φ

∂xj

)

∂v1∂xi

−∑

i,j,k

(u3 − φ− γ)gijΓkij

∂v1∂xk

= 0 in F1;

v1 = 0 on E2; v1 = 1 on E1.

(61)

This seemingly complicated PDE (61) is a linear, second order elliptic PDE. For each fixed ǫ > 0,this differential operator is uniformly elliptic. Since two parts of boundaries do not intersect, (61)has a unique smooth solution v1 due to [13, Vol. 3]. Since the zeroth order term of this differentialoperator is zero, it follows from maximum principle that both maximum and minimum of v1 areachieved at ∂F1, it follows that

0 6 v1 6 1 in F1.

Let ψ be the standard mollifier defined on B0(1). Note that ψ can be defined as a radial functionin terms of |x|. Scaling this we get

ψǫ(x) =1

ǫnϕ(x

ǫ

)

, x ∈ B0(ǫ).

Define

v′′1 := min1,maxv1, 0, v′1(x) := v′′1 ∗ ψǫ(x),∀x ∈ Ω1 ∩ Ω3;

χ1(x) :=

1, x ∈ Ω1\Ω3

v′1(x), x ∈ Ω1 ∩Ω3

0, x ∈ Ω\Ω1

.(62)

Since v1 is only nontrivial within F1, we conclude that χ1(x) = 0 for all x ∈ ∂D′′0 ∩ Ω1 since E2 has

an ǫ-gap away from ∂D′′0 due to construction. By the same argument, χ1(x) = 1 for all x ∈ Ω1\Ω3.

In particular, χ1(x) = 1 on ∂D′0∩Ω1. Thus χ1 is continuous. Actually χ1 is smooth. For instantce,

we pick up a point x ∈ ∂D′′0 ∩ Ω1, then we have

∂αv′1(x) =

ˆ

Bx(ǫ)v′′1(y)∂

αψǫ(x− y)dy ≡ 0

since v′′1 (y) ≡ 0 on Bx(ǫ) due to the construction. The smoothness at ∂D′0 ∩Ω1 is almost the same.

Lastly, χ1 ∈ [0, 1]. For example, we pick up a point x ∈ Ω1 ∩Ω3, and have

1− χ1(x) = 1− v′1(x) =

ˆ

B0(ǫ)1 · ψǫ(y)dy −

ˆ

B0(ǫ)v′′1 (x− y)ψǫ(y)dy

=

ˆ

B0(ǫ)(1− v′′1 (x− y)) · ψǫ(y)dy > 0.

The last inequality is immediate from the construction of v′′1 . By a similar argument, we can seethat χ1(x) > 0. We then consider the PDE

−∑

i,j

gij∂2v2∂xi∂xj

−∑

i,j,k

gijΓkij

∂v2∂xk

= 0 in F2;

v2 = 0 on E′2; v2 = 1 on E′

1.

(63)


By the same argument as above, we conclude that (63) has a unique smooth solution v2 ∈ [0, 1] inF2. Define

v′′2 := min1,maxv2, 0, v′2(x) := v′′2 ∗ ψǫ(x),∀x ∈ Ω2 ∩ Ω3;

χ2(x) :=

1, x ∈ Ω2\Ω3

v′1(x), x ∈ Ω2 ∩Ω3

0, x ∈ Ω\Ω2

.(64)

In the same manner, we conclude that χ2 ∈ [0, 1] is smooth on Ω. Lastly we define

χ3(x) = 1− χ1(x)− χ2(x),∀x ∈ Ω. (65)

Observe that the smooth function χ3 ≡ 1 in Ω3\(

Ω1 ∪ Ω2

)

, and vanishes outside Ω3. We useχ1, χ2, χ3 defined in (62), (64) and (65) as a smooth partition of unity subordinate to Ωi, i = 1, 2, 3.Define

u = χ1u3 + χ2φ+ χ3 (φ+ γ) . (66)

We show that u ∈ C∞(Ω) is a classical super-solution of Yamabe equation in Ω. This verificationis classified in five different cases. Note that

Ω =

3⋃

i=1

Ωi =(

Ω1\Ω3

)

∪(

Ω2\Ω3

)

∪(

Ω3\(

Ω1 ∪ Ω2

))

∪ (Ω1 ∩ Ω3) ∪ (Ω2 ∪ Ω3) . (67)

In first set of (67), we have

u = χ1u3 = u3 in Ω1\Ω3 ⇒ −a∆gu+ (Sg + β) u− (λβ − κ) (u)p−1 = 0 in Ω1\Ω3.

due to (58). Similarly, in the second set of (67), we have

u = χ2φ = φ in Ω2\Ω3 ⇒ −a∆gu+ (Sg + β) u− (λβ − κ) (u)p−1 > 0 in Ω2\Ω3.

due to (57). In the third set of (67), we have

u = χ3 (φ+ γ) = φ+ γ in Ω3\ (Ω1 ∩ Ω2)

⇒− a∆gu+ (Sg + β) u− (λβ − κ) (u)p−1 = −a∆gφ+ (Sg + β) (φ+ γ)− (λβ − κ) (φ+ γ)p−1

> − (Sg + β)φ+ (η1 + β)φ+ (Sg + β) (φ+ γ)− 2p−2 (λβ − κ)φp−1 − 2p−2 (λβ − κ) γp−1

> β′ + (Sg + β) γ − 16 (λβ − κ) γ > 0;

⇒− a∆gu+ (Sg + β) u− (λβ − κ) (u)p−1 > 0 in Ω3\(

Ω1 ∪ Ω2

)

.

The last line is due to the choices of β′, γ in (59) and the observation that 2p−2 = 24

n−2 6 16,∀n > 3.In the fourth set of (67), we have

u = χ1u3 + χ3 (φ+ γ) in Ω1 ∩Ω3.

Note that in Ω1 ∩ Ω3 ⊂ Ω, Sg < 0, λ > 0, note that φ > u3 − γ and u3 > φ in Ω1 ∩ Ω3, hence wehave

u = χ1u3 + χ3 (φ+ γ) < χ1u3 + χ3 (u3 + γ) = u3 + χ3γ;

⇒ (Sg + β) u > (Sg + β)u3 + (Sg + β)χ3γ;

− (λβ − κ) (u)p−1> − (λβ − κ) (u3 + χ3γ)

p−1> − (λβ − κ) up−1

3 −(

2p−1 − 1)

(λβ − κ) up−23 χ3γ

> − (λβ − κ) up−13 −

(

2p−1 − 1)

(λβ − κ) (φ+ γ)p−2χ3γ

> − (λβ − κ) up−13 − 31 (λβ − κ) (φ+ γ)p−2χ3γ.

Here we use the inequality (a+ b)p−1 6 ap−1 +(

2p−1 − 1)

ap−2b when 0 6 b 6 a. Here a = u3, b =

χ3γ in Ω1 ∩ Ω3. The constant 31 is the largest possible value of(

2p−1 − 1)

for all n > 3.

22 J. XU

The hard part is −a∆gu. Define the commutator between −a∆g and any smooth function f by

[−a∆g, f ]u = −a∆g(fu)− f (−a∆gu) = −a∆g(fu) + f (a∆gu) . (68)

Immediately, [−a∆g, 1] = 0. In local coordinates, we see that

−a∆gu = −a∆g (χ1u3)− a∆g (χ3 (φ+ γ))

= χ1 (−a∆gu3) + [−a∆g, χ1]u3 + χ3 (−a∆g(φ+ γ)) + [−a∆g, χ3] (φ+ γ)

:= V1 + V2 +W1 +W2.

For V1 +W1, we use u3 < φ+ γ in Ω1 ∩ Ω3 and have

V1 +W1 = χ1 (−a∆gu3) + χ3 (−a∆gφ)

= χ1

(

− (Sg + β)u3 + (λβ − κ)up−13

)

+ χ3 (− (Sg + β)φ+ (η1 + β)φ)

> −χ1 (Sg + β) u3 + χ1 (λβ − κ)up−13 − χ3 (Sg + β) u3 + χ3 (λβ − κ) up−1

3

+ χ3

(

(Sg + β) γ + (η1 + β)φ− (λβ − κ) up−13

)

> − (Sg + β)u3 + (λβ − κ) up−13

+ χ3

(

(Sg + β) γ + (η1 + β)φ− (λβ − κ) (φ+ γ)p−1)

> − (Sg + β)u3 + (λβ − κ) up−13

+ χ3

(

(Sg + β) γ + (η1 + β)φ− 2p−2 (λβ − κ)φp−1 − 2p−2 (λβ − κ) γp−1)

> − (Sg + β)u3 + (λβ − κ) up−13

+ χ3

(

(Sg + β) γ + (η1 + β)φ− 2p−2 (λβ − κ)φp−1 − 16 (λβ − κ) γ)

.

To check V2 +W2, we recall that χ1 + χ3 = 1 in Ω1 ∩ Ω3, hence

0 = [−a∆g, 1] = [−a∆g, χ1] + [−a∆g, χ3] ⇒ [−a∆g, χ1] = −[−a∆g, χ3].

It follows that

V2 +W2 = [−a∆g, χ1]u3 + [−a∆g, χ3] (φ+ γ) = [−a∆g, χ1] (u3 − φ− γ)

= χ1 (a∆g) (u3 − φ− γ)− a∆g (χ1 (u3 − φ− γ))

= −∑

i,j

gij(u3 − φ− γ)∂2χ1

∂xi∂xj−∑

i,j

2gij(

∂u3∂xj

− ∂φ

∂xj

)

∂χ1

∂xi

−∑

i,j,k

gijΓkij(u3 − φ− γ)

∂χ1

∂xk.

Next we show thatV2 +W2 > −Lǫ′ (69)

in Ω1∩Ω3 for some constant L > 0 which is independent of ǫ′. Here ǫ′ > 0 can be arbitrarily small.We verify (71) in three types of points. Set

G1 := x ∈ Ω : dist(x,E1) 6 ǫ, G2 := x ∈ Ω : dist(x,E2) 6 ǫ.For points F1\ (G1 ∪G2), i.e. points in Ω1 ∩Ω3 that is at least distance ǫ from either E1 or E2, wehave

χ1(x) =

ˆ

B0(ǫ)v1(x− y)ψǫ(y)dy.

Since v1 ∈ C∞, it follows that

|∂αχ1(x)− ∂αv1(x)| 6 ǫ′,∀|α| 6 2.


Here ǫ′ǫ→0−−→ 0. Hence in this case,

V2 +W2 > −∑

i,j

gij(u3 − φ− γ)∂2v1∂xi∂xj

−∑

i,j

2gij(

∂u3∂xj

− ∂φ

∂xj

)

∂v1∂xi

−∑

i,j,k


∂v1∂xk

− Lǫ′

= −Lǫ′.Here L depends on u3, φ, g

ij ,Γkij and is independent of ǫ′. When x in Ω1∩Ω3∩G1 or Ω1 ∩Ω3∩G2,

we conclude that∂αχ1 = o(1),∀α,∀x ∈ Ω1 ∩ Ω3 ∩ (G1 ∪G2) .

This is due to the smoothness of χ1 and ∂αχ1 = 0 on ∂D′0 ∩Ω1 and ∂D′′

0 ∩ Ω1. To be precise, whenx ∈ Ω1 ∩ Ω3 ∩G1, u3 − φ− γ → 0 when ǫ→ 0; furthermore, the first order derivative satisfies

∂χ1

∂xj=

ˆ

Bx(ǫ)v′1(y)

∂ψǫ(x− y)

∂xjdy =

ˆ

Bx∩Ω1∩Ω3∩G1

∂v1∂yj

ψǫ(x− y)dy.

The boundary terms on ∂D1 are killed since on ∂D1 we have v1 = 1. The second order derivativeis of the form

∂2χ1

∂xixj=

ˆ

Bx(ǫ)v′1(y)

∂2ψǫ(x− y)

∂xixjdy =

ˆ

Bx(ǫ)v′1(y)(−1)2

∂2ψǫ(x− y)

∂yiyjdy

=

ˆ

Bx(ǫ)∩Ω1∩Ω3

∂2v1(y)

∂yiyjψǫ(x− y)dy −

ˆ

∂(Bx(ǫ)∩∂D1)

∂v1∂xi

νjψǫ(x− y)dS

The first term above can be controlled by the PDE (61); for the the second term above, we observe

that u3 − φ − γ = O(ǫ) and ∂v1∂xi

= O(γ) due to the construction of PDE, thus by the fact thatǫ≪ γ,

−gij(u3 − φ− γ)

ˆ

∂(Bx(ǫ)∩∂D1)

∂v1∂xi

νjψǫ(x− y)dS → 0

as ǫ→ 0. When x ∈ Ω1∩Ω3∩G2, the first order derivatives are controlled exactly as just discussed.Note that u3 − φ− γ < 0 here, the second order derivative will have

∂2χ1

∂xixj=

ˆ

Bx(ǫ)v′1(y)

∂2ψǫ(x− y)

∂xixjdy =

ˆ

Bx(ǫ)v′1(y)(−1)2

∂2ψǫ(x− y)

∂yiyjdy

=

ˆ

Bx(ǫ)∩Ω1∩Ω3

∂2v1(y)

∂yiyjψǫ(x− y)dy −

ˆ

∂(Bx(ǫ)∩∂D2)

∂v1∂xi

νjψǫ(x− y)dS

>

ˆ

Bx(ǫ)∩Ω1∩Ω3

∂2v1(y)

∂yiyjψǫ(x− y)dy.

The last inequality is due to the fact that v1 is descending to zero, therefore the derivative isnegative when very close to ∂D2. It follows that the boundary part above is positive. Within the2ǫ-strip of ∂D′

0 and ∂D′′0 , we have

V2 +W2 = −∑

i,j

gij(u3 − φ− γ)∂2χ1

∂xi∂xj−∑

i,j

2gij(

∂u3∂xj

− ∂φ

∂xj

)

∂χ1

∂xi

−∑

i,j,k


∂χ1

∂xk= o(1) > −Lǫ′.

24 J. XU

Therefore (71) holds. It follows that

−a∆gu = V1 + V2 +W1 +W2 > − (Sg + β) u3 + (λβ − κ) up−13

+ χ3

(


− Lǫ′.(70)

Combining all estimates together, we have

− a∆gu+ (Sg + β) u− (λβ − κ) up−1> − (Sg + β) u3 + (λβ − κ) up−1

3

+ χ3

(


− Lǫ′

+ (Sg + β) u3 + (Sg + β)χ3γ − (λβ − κ) up−13 − 31 (λβ − κ) (φ+ γ)p−2χ3γ

= χ3

(

(η1 + β)φ− 2p−2 (λβ − κ)φp−1 − 16 (λβ − κ) γ + 2 (Sg + β) γ − 31 (λβ − κ) (φ+ γ)p−2γ)

− Lǫ′

> χ3

(

β′ + 2 (Sg + β) γ − 31 (λβ − κ) (φ+ γ)p−2γ)

− Lǫ′.

Due to the choice of γ in (59) and the fact that ǫ′ can be arbitrarily small by making ǫ arbitrarilysmall, we conclude that

−a∆gu+ (Sg + β) u− (λβ − κ) up−1> 0 in Ω1 ∩ Ω3.

In the fifth set of (67), we haveu = χ2φ+ χ3 (φ+ γ) .

Repeat the same argument above in replacing u3 by φ, we conclude that u is a super-solution inΩ2 ∩ Ω3. The argument is essentially the same thus we just sketch here. In Ω2 ∩Ω3,

u = φ+ χ3γ 6 φ+ γ;

(Sg + β) u > (Sg + β)φ+ (Sg + β) γ;

− (λβ − κ) (u)p−1> −2p−2 (λβ − κ)φp−1 − 16 (λβ − κ) γ.

The Laplacian part can be estimated as

−a∆gu = −a∆g (χ2φ)− a∆g (χ3 (φ+ γ))

= χ2 (−a∆gφ) + [−a∆g, χ2]φ+ χ3 (−a∆g (φ+ γ)) + [−a∆g, χ3] (φ+ γ)

> −a∆gφ+ [−a∆g, χ2] (−γ) = −a∆gφ− a∆g (−γχ2) .

We observe that −a∆g (−γχ2) > −Lǫ′ with arbitrarily small ǫ′ by the same reasoning above, dueto the construction in (63) and (64). The rest are the same as above, hence we conclude that

−a∆gu+ (Sg + β) u− (λβ − κ) (u)p−1> 0 in Ω2 ∩Ω3.

Hence u is a super-solution of Yamabe equation (2) in Ω pointwisely. By the setup of Ωi, i = 1, 2, 3,it is immediate to check that

u > u3 in Ω.

Since u = φ near ∂Ω, we define

u+ :=

u, in Ω;

φ, inM\Ω. (71)

It follows that u+ ∈ C∞(M) is a super-solution of (54) and we conclude that 0 6 u− 6 u+. Lastly,by Theorem 2.5 and Remark 2.2, we conclude that there exists a solution u ∈ C∞(M) that solvesthe perturbed Yamabe equation (53) with the fixed λs > 0 given in (51). By maximum principleand the fact that u > u− 6≡ 0, we conclude that u > 0 on M .

Remark 4.5. Recall that the local PDE (37) holds for all λ > 0. This is crucial since it gives usthe flexibility to choose the constant associate with up−1 term, hence we can introduce the extraconstant κ > 0 here.


We now apply Theorem 4.3 to show the existence of the solution of Yamabe equation with η1 > 0.Note that from perturbed Yamabe equation (54) to the Yamabe equation, we need to pass the limitby letting β → 0−. Yamabe tried to let us−1 → up−1 when s→ p, and the uniform boundedness ofthe sequence is not that easy. Due to the local construction of our perturbed Yamabe equations in(53), we obtain the uniformly positive upper and lower bounds of ‖uβ‖Lp(M,g) when β ∈ [β0, 0], withsome fixed β0 < 0. With the introduction of the extra positive constant κ > 0, a similar argumentdue to Trudinger and Aubin then applies to show the uniform boundedness of ‖uβ‖Lr(M,g) forall β ∈ [β0, 0] for some r > p. In contrast to classical method, we obtain this without usingλ(M) < λ(Sn).

Theorem 4.4. Let (M,g) be a closed manifold with n = dimM > 3. Let κ > 0 be some smallenough positive constant. Assume the scalar curvature Sg < 0 somewhere on M and the firsteigenvalue η1 > 0. Then there exists some λ > 0 such that the Yamabe equation (2) has a real,positive, smooth solution.

Proof. By Theorem 4.3, we have a sequence of real, positive, smooth solutions uβ when β < 0and |β| is small enough, i.e.

− a∆guβ + (Sg + β) uβ = (λβ − κ) up−1β inM. (72)

We show first that λβ is bounded above, and is increasing and continuous when β → 0−. It thenfollows that we can choose κ uniformly so that

λβ − κ > 0,∀β ∈ [β0, 0].

We may assume´

M dω = 1 for this continuity verification, since otherwise only an extra term withrespect to Volg will appear. Recall that

λβ = infu 6=0,u∈H1(M)

´

M a|∇gu|2dω +´

M (Sg + β)u2dω(´

M updω) 2

p

.

It is immediate that if β1 < β2 < 0 then λβ1 6 λβ2 . For continuity we assume 0 < β2 −β1 < γ. Foreach ǫ > 0, there exists a function u0 such that

´

M a|∇gu0|2dω +´

M (Sg + β1)u20dω

(´

M up0dω) 2

p

< λβ1 + ǫ.

It follows that

λβ2 6

´

M a|∇gu0|2dω +´

M (Sg + β2) u20dω

(´

M up0dω)

2p

6

´

M a|∇gu0|2dω +´

M (Sg + β1) u20dω

(´

M up0dω) 2

p

+(β2 − β1)

´

M u20dω(´

M up0dω) 2

p

6 λβ1 + ǫ+ β2 − β1 < λβ1 + ǫ+ β2 − β1.

Since ǫ is arbitrarily small, we conclude that

0 < β2 − β1 < γ ⇒ |λβ2 − λβ1 | 6 2γ.

By the argument in Appendix A, we conclude that λβ < aT , where T is the best Sobolev constantfor the embedding Lp(Rn) → H1(Rn). Fix some β0 < 0 with η1,β0 > 0, we have

λβ0 6 λβ 6 aT,∀β ∈ [β0, 0], limβ→0−

λβ − κ := λ. (73)

26 J. XU

Next we show ‖uβ‖H1(M) 6 C uniformly when β ∈ [β0, 0]. Pairing the PDE (72) with uβ, it followsthat

a‖∇guβ‖2L2(M,g) = (λβ − κ) ‖uβ‖pLp(M,g) −ˆ

M(Sg + β) u2βdω

6 (λβ − κ) ‖uβ‖pLp(M,g) +

(

supM

|Sg|+ |β0|)

‖uβ‖2L2(M,g).

(74)

Thus it suffices to show‖uβ‖Lp(M,g) 6 C ′,∀β ∈ [β0, 0].

Denote the corresponding local solutions of (37) by uβ, i.e.−a∆guβ + (Sg + β) uβ = (λβ − κ) up−1

β in Ω, uβ = 0 on ∂Ω

with fixed domain Ω. Again we have

a‖∇guβ‖2L2(Ω,g) = (λβ − κ) ‖uβ‖pLp(Ω,g) −ˆ

M(Sg + β)u2βdVolg

⇒ (λβ − κ) ‖uβ‖pLp(Ω,g) 6 a‖∇guβ‖2L2(Ω,g) +

(

supM

|Sg|+ |β|)

‖uβ‖2L2(Ω,g).

(75)

Recall the fuctional J(u) in (34) and the quantity K0 in (35), due to Theorem 1.1 of [26], eachsolution uβ satisfies

J(uβ) 6 K0 ⇒a

2‖∇guβ‖2L2(Ω,g) −

(λβ − κ)

p‖uβ‖pLp(Ω,g) −

1

2

ˆ

M(Sg + β) u2βdVolg 6 K0. (76)

Note that since λβ is bounded above and below when β ∈ [β0, 0], the quantity K0 is uniformlybounded above, and we denote it by K0 again. Apply the estimate (75) into (76), we have

a

2‖∇guβ‖2L2(Ω,g) 6 K0 +

1

p

(

a‖∇guβ‖2L2(Ω,g) +

(

supM

|Sg|+ |β|)

‖uβ‖2L2(Ω,g)

)

+

(

supM

|Sg|+ |β|)

‖uβ‖2L2(Ω,g)

6 K0 +a(n− 2)

2n‖∇guβ‖2L2(Ω,g)

+

(

n− 2

2n+ 1

)(

supM

|Sg|+ |β|)

· λ−11 ‖∇guβ‖2L2(Ω,g);

⇒(

a

n−(

n− 2

2n+ 1

)(

supM

|Sg|+ |β|)

· λ−11

)

‖∇guβ‖2L2(Ω,g) 6 K0.

We have already chosen Ω small enough so that (40) holds for β0, thus for all β ∈ [β0, 0], (40) holds.It follows from above that there exists a constant C ′

0 such that

‖∇guβ‖2L2(Ω,g) 6 C ′0,∀β ∈ [β0, 0].

Apply (75) with the other way around, we conclude that

(λβ − κ) ‖uβ‖pLp(Ω,g) 6 a‖∇guβ‖2L2(Ω,g) +

(

supM

|Sg|+ |β|)

‖uβ‖2L2(Ω,g)

6

(

a+

(

supM

|Sg|+ |β|)

λ−11

)

‖∇guβ‖2L2(Ω,g).

We conclude that‖uβ‖pLp(Ω,g) 6 C ′

1,∀β ∈ [β0, 0]. (77)


Recall the construction of super-solution of each uβ in Theorem 4.3, we have

0 < uβ 6 u+,β =

uβ, in Ω

φ, inM\Ω .

where uβ is of the formuβ = χ1uβ + χ2φ+ χ3(φ+ γ).

It follows from the uniform upper bound of (77) that

‖uβ‖pLp(M,g) 6 ‖u+,β‖pLp(M,g) 6 E0

(

‖uβ‖pLp(Ω,g) + ‖φ‖pLp(M,g)

)

.

Recall in determining φ we require

(η1 + β)

2p−2 (λβ − κ)> δp−2 supM ϕp−1

infM ϕ.

Since when β ∈ [β0, 0], we have λβ ∈ [λβ0 , aT ], we can choose a fixed scaling δ for all β, thus‖φ‖pLp(M,g) is uniformly bounded. It follows that there exists a constant C ′ such that

‖uβ‖pLp(M,g) 6 C ′,∀β ∈ [β0, 0].

Due to Holder’s inequality, we conclude that

‖uβ‖L2(M,g) 6 ‖uβ‖2p

Lp(M,g)·Volg(M)

p−2p .

Thus by (74) we conclude that

‖uβ‖H1(M,g) 6 C,∀β ∈ [β0, 0]. (78)

According to the definition of λβ, we apply the solution of (72) with the form (74), it follows thatfor all β ∈ [β0, β].

λβ 6

´

M a|∇guβ |2dω +´

M (Sg + β) u2βdω(

´

M upβdω)

2p

= (λβ − κ) ·‖uβ‖pLp(M,g)(

´

M upβdω)

2p

⇒ λβλβ − κ

6 ‖uβ‖p− 2

p

Lp(M,g) ⇒ ‖uβ‖Lp(M,g) > 1.

The last cancellation is due to the fact that λβ > λβ −κ > λβ0 −κ > 0,∀β ∈ [β0, 0]. It follows that

‖uβ‖H1(M,g) 6 C, 1 6 ‖uβ‖Lp(M,g) 6 C ′,∀β ∈ [β0, 0]. (79)

Due to uniform boundedness of Lp-norms of uβ, we can scale the metric g 7→ γg, γ < 1 uniformlyfor all β ∈ [β0, 0] such that

0 < C ′′ 6 ‖uβ‖Lp(M,g) 6 1,∀β ∈ [β0, 0]. (80)

We still denote the new metric by g. This scaling does not affect the local estimates in AppendixA since β < 0, and hence does not affect the local solvability of the PDE in Proposition 3.3. Thisscaling does not affect the fact that λβ0 6 λβ 6 aT since again β < 0 and hence does not affectthe choice of κ. We now apply the analysis, essentially due to Trudinger and Aubin, to show thatthere exists some r > p, such that

‖uβ‖Lr(M,g) 6 K,∀β ∈ [β0, 0]. (81)

28 J. XU

Let δ > 0 be some constant. Pairing (72) with u1+2δβ and denote wβ = u1+δ

β , it follows thatˆ

Ma∇guβ · ∇g

(

u1+2δβ

)

dω +

ˆ

M(Sg + β) u2+2δ

β dω = (λβ − κ)

ˆ

Mup+2δβ dω;

⇒1 + 2δ

1 + δ2

ˆ

Ma|∇gwβ|2dω = (λβ − κ)

ˆ

Mw2βu

p−2β dω −

ˆ

M(Sg + β)w2

βdω.

According to the sharp Sobolev inequality on closed manifolds [18, Thm. 2.3, Thm. 3.3], it followsthat for any ǫ > 0,

‖wβ‖2Lp(M,g) 6 (1 + ǫ)1

T‖∇gwβ‖2L2(M,g) + C ′

ǫ‖wβ‖2L2(M,g)

= (1 + ǫ)1

aT· 1 + δ2

1 + 2δ

(

1 + 2δ

1 + δ2

ˆ

Ma|∇gwβ|2dω

)

+ C ′ǫ‖wβ‖2L2(M,g)

= (1 + ǫ)λβ − κ

aT· 1 + δ2

1 + 2δ

ˆ

Mw2βu

p−2β dω + Cǫ‖wβ‖2L2(M,g)

6 (1 + ǫ)λβ − κ

aT· 1 + δ2

1 + 2δ‖wβ‖2Lp(M,g)‖uβ‖

pLp(M,g) + Cǫ‖wβ‖2L2(M,g)

6 (1 + ǫ)λβ − κ

aT· 1 + δ2

1 + 2δ‖wβ‖2Lp(M,g) + Cǫ‖wβ‖2L2(M,g).

Note that Cǫ is uniform over all β. The last two lines above are due to Holder’s inequality and(80). Since λβ 6 aT , λβ − κ < aT ; thus we can choose ǫ, δ small enough such that

(1 + ǫ)λβ − κ

aT· 1 + δ2

1 + 2δ< 1.

It follows that

‖wβ‖2Lp(M,g) 6 K1‖wβ‖2L2(M,g) 6 K2‖uβ‖1+δLp(M,g) 6 K3,∀β ∈ [β0, 0].

Note that ‖wβ‖Lp(M,g) = ‖uβ‖1+δLp(1+δ)(M,g)

, it follows that (81) holds. Since r > p, by bootstrapping

method we conclude that‖uβ‖C2,α(M) 6 K′,∀β ∈ [β0, 0]. (82)

Due to Arzela-Ascoli, we pass the limit by letting β → 0− on both sides of (72), it follows that

−a∆gu+ Sgu = λup−1 onM.

Another bootstrapping method shows that u ∈ C∞(M) and u > 0 since u is the limit of a positivesequence. Lastly we show that u > 0. Due to uniform boundedness in (82) and lower bound ofLp-norms in (80), we conclude that

‖u‖Lp(M,g) > C ′′ > 0.

It follows from maximum principle that u > 0 on M .

Remark 4.6. With the introduction of κ > 0 in (72), we can apply this method for all cases, evenwhen the manifold is conformal to the standard sphere Sn, since we still have λβ−κ < λ(Sn) = aT .

Lastly we show the case when Sg > 0 everywhere. Due to [16], this amounts to say η1 > 0.Therefore we consider the solution of Yamabe equation when λ > 0. We show this by showing thatthere exists some g = e2fg = up−2g for some positive u ∈ C∞(M) such that Sg < 0 somewhere.Obviously, Sg cannot be too negative too often, actually due to Theorem 2.4(i) and taking ϕ = 1 in(48), a necessary condition is

´

M Sg > 0. Consequently we can apply Theorem 4.3 to obtain somemetric with positive constant scalar curvature.


Theorem 4.5. Let (M,g) be a closed manifold with n = dimM > 3. Assume the scalar curvatureSg > 0 everywhere on M . Then there exists some smooth function H ∈ C∞(M) which is negativesomewhere such that H is the scalar curvature with respect to some conformal change of g.

Proof. We show this by constructing the smooth function H with desired property for which thefollowing PDE

− a∆gu+ Sgu = Hup−1 in (M,g) (83)

admits a smooth, real, positive solution u. Without loss of generality, we can assume supx∈M(Sg) 61 for some g after scaling.

To construct H, we first construct a smooth function F ∈ C∞(M) such that: (i) F is verynegative at some point in M ; (ii)

´

M Fdω = 0; and (iii) ‖F‖Hs−2(M,g) is very small simultaneously.

Here s = n2 +1 if n is even and s = n+1

2 if n is odd. For simplicity, we discuss the even dimensionalcase, the odd dimensional case is exactly the same except the order s.

Pick up a normal ball Br in M with radius r < 1 centered at some point q ∈ M . Choose afunction

f(q) = −C < −1, f 6 0, |f | 6 C, f ∈ C∞(M), f ≡ 0 onM\Br. (84)

Such an f can be chosen as any smooth bump function or cut-off function in partition of unitymultiplying by −1. For f , we have

|∂αf | 6 C ′

r|α|,∀|α| 6 n

2− 1 = s− 2. (85)

By (85), we have∣

∣

∣

∣

ˆ

Mfdω

∣

∣

∣

∣

=

∣

∣

∣

∣

ˆ

Br

fdVolg

∣

∣

∣

∣

6 CVolg(Br) 6 C · Γrn. (86)

Here C is the upper bound of |f |. The hypothesis of the statement says that Sg > 0 everywehere,hence Γ depends only on the dimension of M for every geodesic ball centered at any point of M ,due to Bishop-Gromov comparison inequality. Specifically, the volume of a geodesic ball of radius

r ≪ 1 centered at p is given by V olg(Br) =(

1− Sg(0)6(n+2)r

2 +O(r4))

V ole(Br). Thus Γ is fixed when

(M,g) is fixed. Now we defineF := f + ǫ (87)

where 0 < ǫ < C is determined later. First we show that for any C, we can choose ǫ arbitrarilysmall so that

´

M Fdω = 0. Set

ǫ :=1

Volg(M)

ˆ

Br

−fdVolg ⇒ˆ

MFdω =

ˆ

M(f + ǫ)dω =

ˆ

Br

fdVolg + ǫ

ˆ

Mdω = 0.

As we have shown in (86),

ǫ 61

Volg(M)

∣

∣

∣

∣

ˆ

Br

fdVolg

∣

∣

∣

∣

61

Volg(M)CΓrn = O(rn).

Thus if we choose r small enough then ǫ is small enough.Secondly we show that for any C,C ′, we can make ‖F‖Hs−2(M,g) arbitrarily small. By Bishop-

Gromov inequality and (85), we observe that

‖F‖2Hs−2(M,g) =

ˆ

M|f + ǫ|2dω +

∑

16|α|6s−2

ˆ

Br

|∂αf |2dVolg

6 C0rn +

∑

16|α|6s−2

ˆ

Br

(C ′)2

rn−2dVolg 6 C0r

n +∑

16|α|6s−2

(C ′)2

rn−2Γrn 6 (C ′)2C ′

0r2

30 J. XU

for some C ′0 depends only on the the order s − 2 = n

2 − 1, the lower bound of Ricci curvature ofM , and the diameter of (M,g). Therefore, for any C,C ′, we can choose r small enough so that‖F‖2Hs−2(M,g) is arbitrarily small.

It follows that this smooth function F we constructed satisfies all requirements mentioned above.Now we construct u ∈ C∞(M). Let u′ be the solution of

− a∆gu′ = F in (M,g). (88)

Due to Lax-Milgram [17, Ch. 6] and hypothesis (i) of F , there exists such an u′ that solves (88).Since F ∈ C∞(M), standard elliptic regularity implies that u′ ∈ C∞(M). Recall the Sobolevembedding in Theorem 2.2(iii) and Hs-type elliptic regularity in Theorem 2.1(i), we have fixedconstants K ′ and C∗∗ such that

|u′| 6 ‖u′‖C0,β(M) 6 K ′‖u′‖Hs(M,g),1

2− s

n=

1

2− 1

2− 1

n< 0;

‖u′‖Hs(M,g) 6 C∗(

‖F‖Hs−2(M,g) + ‖u′‖L2(M,g)

)

6 C∗∗‖F‖Hs−2(M,g).

Note that the last inequality comes from the PDE (88) by pairing u′ on both sides, and applyPoincare inequality. However, we should have

´

M u′dω = 0 to apply Poincare inequality, this canbe done by taking u′ 7→ u′ + ǫ if necessary. Thus K ′ depends only on n, α, s,M, g and C∗∗ dependsonly on s,−a∆g, λ1,M, g, all of which are fixed once M,g, s are fixed.

Note that f(q) = −C for some C > 1 at the point q ∈ M . First choose r small enough so thatǫ < 1

2 , it follows that

F (q) = f(q) + ǫ 6 −C2. (89)

Now we further shrink r such that

C∗∗K ′‖F‖Hs−2(M,g) 6 C∗∗K ′ · C ′(

C ′0

) 12 r 6

C

8.

Thus by Sobolev embedding and elliptic regularity, it follows that

|u′| 6 C∗∗K ′‖F‖Hs−2(M,g) 6C

8⇒ u′ ∈ [−C

8,C

8],∀x ∈M. (90)

We now define u to be

u := u′ +C

4. (91)

By (90), it is immediate to see that u > 0 on M and u ∈ C∞(M) due to definition and smoothnessof u′. In particular u ∈ [C8 ,

3C8 ]. Furthermore, u solves (88). Finally, we define our H to be

H := u1−p (−a∆gu+ Sgu) ,∀x ∈M. (92)

Clearly H ∈ C∞(M). By −a∆gu = F , (89), (90) and (91), we observe that

H(q) = u(q)1−p (−a∆gu(q) + Sg(q)u(q)) = u(q)1−p (F (q) + Sg(q)u(q)) 6 u(q)1−p

(

−C2+

3C

8

)

< 0.

Based on our constructions of u and H, (92) is equivalent to say that the following PDE

−a∆gu+ Sgu = Hup−1 in (M,g)

has a real, positive, smooth solution. It follows that H is the scalar curvature of the metric up−2gon M .

Remark 4.7. We see from above that

H(q) 6 u(q)1−p

(

−C2+

3C

8

)

< 0.


Furthermore, we observe from above that

|u′| 6 C∗∗K ′‖F‖Hs−2(M,g) 6 C∗∗K · C ′(

C ′0

) 12 r.

It follows that we can make |u′| < ǫ′ for any ǫ′ small enough. Thus taking

u = u′ + 2ǫ′ ⇒ |u| 6 3ǫ′.

With fixed C, we can make H(q) arbitrarily large since u1−p = u−n+2n−2 would be arbitrarily large

with arbitrarily small ǫ′.

An immediate analogy of Theorem 4.5 is that when Sg 6 0 everywhere and hence η1 < 0, thereexists some H ′ ∈ C∞(M), positive somewhere, that is the prescribed scalar curvature of some gunder conformal change.

Corollary 4.1. Let (M,g) be a closed manifold with n = dimM > 3. Let Sg 6 0 everywhere onM be the scalar curvature with respect to g. Then there exists some smooth function H ′ ∈ C∞(M)which is positive somewhere such that H ′ is the scalar curvature with respect to some conformalchange of g.

Proof. Taking f > 0 with properties as in Theorem 4.5 and do the same argument.

With the help of Theorem 4.5, we can solve the Yamabe equation for the case Sg > 0 everywhere.Note that this amounts to say η1 > 0 due to (48).

Theorem 4.6. Let (M,g) be a closed manifold with n = dimM > 3. Assume the scalar curvatureSg > 0 everywhere on M . Then there exists some λ > 0 such that the Yamabe equation (2) has areal, positive, smooth solution.

Proof. After scaling we can, without loss of generality, assume supM Sg 6 1. We start with thismetric. By Theorem 4.4, we conclude that there exists some smooth function H < 0 somewheresuch that it is the scalar curvature with respect to the metric g1 = up−2g for some smooth u > 0on M . By Theorem 2.4(i) the first eigenvalue for the conformal Laplacian −a∆g1 +H is positive.

Consequently by Theorem 4.3, there exists some λ > 0 such that (2) has some real, positivesolution v ∈ C∞(M), i.e. λ is the scalar curvature of the metric g = vp−2g1. Combining bothconformal changes together, we see that

g = vp−2g1 = vp−2up−2g = (uv)p−2g.

Clearly uv > 0 and smooth on M . We now conclude that g is associated with the positive constantscalar curvature λ under the conformal change g = (uv)p−2g.

Acknowledgement

The author would like to thank his advisor Prof. Steven Rosenberg for his support and mentor-ship.

Appendix A. Proof of The Claim

In this section, we show that the claim (46) in Proposition 3.3 holds. The key is a very careful

analysis in terms of local expression of√

det(g) and gij . This estimate is sharp in the sense thatif we drop any negative term in numerator or any positive term in denominator with some specificorder then we cannot detect the desired upper bound, see Remark 3.2. Note that the quantity Qǫ,Ω

is not a Yamabe quotient because of the perturbed term β. Without loss of generality, we chooseΩ = B0(r) be a ball of radius r and centered at 0. We fix the radius r due to the discussion inProposition 3.3. We interchangeably use Ω and B0(r) when there will be no confusion.

32 J. XU

When n = 3, We set Ω = B0(1) and thus we choose the cut-off function as

ϕΩ(x) = cos

(

π|x|2

)

.

When n > 4, we choose ϕΩ(x) = ϕΩ (|x|) to be any radial bump function supported in Ω suchthat ϕΩ(x) ≡ 1 on B0

(

r2

)

. Therefore, the test function we choose in Ω is

uǫ,Ω(x) =ϕΩ(x)

(ǫ+ |x|2)n−22

, n > 3. (93)

We will simply denote uǫ,Ω = u when there would be no confusion.

When n > 4, denote

K1 = (n − 2)2ˆ

Rn

|y|2(1 + |y|2)n dy;

K2 =

(ˆ

Rn

1

(1 + |y|2)n dy)

2p

;

K3 =

ˆ

Rn

1

(1 + |y|2)n−2dy.

We will keep using K1,K2 when n = 3, but K3 is not integrable when n = 3. The best Sobolevconstant T satisfies

T =K1

K2, n > 3. (94)

We use the notation T for all cases n > 3.

Theorem A.1. Let Ω = B0(r) for small enough r with dimΩ > 3. Let β < 0 be any negativeconstant and T be the best Sobolev constant in (94). The quantity Qǫ,Ω satisfies

Qǫ,Ω :=‖∇guǫ,Ω‖2L2(Ω,g) +

1a

´

Ω (Sg + β) u2ǫ,Ω√

det(g)dx


< T. (95)

with the test functions given in (93).

The theorem will be proved in three cases: (i) n > 5; (ii) n = 4; (iii) n = 3.

Proof. Let Ω be a small enough geodesic ball centered at 0 with radius r, as discussed above. It iswell known that [18, §5], [7, §3] on normal coordinates xi within a small geodesic ball of radiusr, we have:

gij = δij − 1

3

∑

r,s

Rijrsxrxs +O(

|x|3)

;

√

det(g) = 1− 1

6

∑

i,j

Rijxixj +O(

|x|3)

.(96)

Here Rij are the coefficients of Ricci curvature tensors. In local expression of√

det(g), the term∑

i,j Rijxixj = Ric(x, x). Since the Ricci curvature tensor is a symmetric bilinear form, there exists

an orthonormal principal basis v1, . . . , vn on tangent space of each point in the geodesic ball suchthat its corresponding eigenvalues λi satisfies

∑

i

λi = Sg(0).


Let y1, . . . , yn be the corresponding normal coordinates with respect to the principal basis, theRicci curvature tensor is diagonalized at the center 0, thus in a very small ball we have

Rij(y) = δijRij(0) + o(1) = δijRij(0) +∑

|α|=1

∂αRij(0)yα +O(|y|2)

⇒Sg(y) = Rijgij = Sg(0) +

∑

i,|α|=1

∂αRii(0)yα +O(|y|2);

Rij(y) = δijRij(0) +∑

|α|=1

∂αRij(0)yα +O(|y|2)

⇒√

det(g) = 1− 1

6

∑

i

Rii(0) +∑

|α|=1

∂αRii(0)yα

y2i −1

6

∑

i 6=j,|α|=1

∂αRij(0)yαyiyj + o(|y|3).

(97)

The smallness o(1) above only depends on the smallness of r and is no larger than O(r) due to theTaylor expansion of the smooth functions Rij(y) near y = 0:

Rij(y) = Rij(0) +∑

|α|=1

∂αRij(0)yα +O(|y|2).

Since ϕΩ(y) = ϕΩ(|y|) is radial; note also that in normal coordinates yi, we also have

s2 = |y|2 ⇒ gss ≡ 1 ⇒ |∇guǫ,Ω|2 = |∂su|2.We apply (97) and have

‖∇gu‖2L2(Ω,g) =

ˆ

Ω|∇gu|2

√

det(g)dy

6

ˆ

Ω

1− 1

6

∑

i

Rii(0) +∑

|α|=1

∂αRii(0)yα

y2i

|∂su|2dy

−ˆ

Ω

1

6

∑

i 6=j,|α|=1

∂αRij(0)yαyiyj + o(|y|3)

|∂su|2dy

6

ˆ

Ω|∂su|2dy −

1

6n

ˆ

ΩSg(0)|y|2|∂su|2dy + C1

ˆ

Ω|y|3|∂su|2dy

:= ‖∂su‖2L2(Ω) +A1 +A2;

1

a

ˆ

Ω(Sg + β)u2

√

det(g)dx 61

a

ˆ

Ω(Sg(0) + β) |u|2dy + 1

a

ˆ

Ω

∑

i,|α|=1

∂αRii(0)yα|u|2dy

+ C2

ˆ

Ω|y|2|u|2dy + C3

ˆ

Ω|y|3|u|2dy

:=1

a

ˆ

Ω(Sg(0) + β) |u|2dy +B1 +B2 +B3;

‖u‖pLp(Ω,g) =

ˆ

Ω|u|p√

det(g)dx =

ˆ

Ω|u|pdx− 1

6n

ˆ

ΩSg(0)|y|2|u|pdy

+ C4

ˆ

Ω|y|3|u|pdy := ‖u‖pLp(Ω) + C1 + C2.

(98)

34 J. XU

Right sides of inequalities in (98) are all Euclidean norms and derivatives. We show that when ǫsmall enough, the test function in (93) satisfies

‖∇guǫ,Ω‖2L2(Ω,g) +1a

´

Ω (Sg + β) u2ǫ,Ω√

det(g)dx


6‖∂suǫ,Ω‖2L2(Ω) +

´

Ω Sg(0)u2ǫ,Ωdx+A1 +A2 +B1 +B2 +B3

(

‖uǫ,Ω‖pLp(Ω) + C1 +C2

) 2p

< T.

We show the inequality (95) in three cases, classified by dimensions. Note that by (93), u = uǫ,Ω isradial, thus we have

∂suǫ,Ω = ∂s

(

ϕΩ(s)

(ǫ+ |s|2)n−22

)

=∂sϕΩ(|y|)

(ǫ+ |y|2)n−22

− (n− 2)ϕΩ(|y|)y(ǫ+ |y|2)n

2

.

Thus the terms ‖∂su‖2L2(Ω), A1, A2 and A3 can be estimated as

‖∂su‖2L2(Ω) =

ˆ

Ω

|∂sϕΩ|2(ǫ+ |y|2)n−2

dy − 2(n− 2)

ˆ

Ω

ϕΩ (∂sϕΩ · y)(ǫ+ |y|2)n−1

dy + (n− 2)2ˆ

Ω

ϕ2Ω|y|2

(ǫ+ |y|2)n dy

= (n− 2)2ˆ

Rn

|y|2(ǫ+ |y|2)n dy +

ˆ

Ω

|∂sϕΩ|2(ǫ+ |y|2)n−2

dy − 2(n − 2)

ˆ

Ω

ϕΩ (∂sϕΩ · y)(ǫ+ |y|2)n−1

dy

+ (n− 2)2ˆ

Ω

(

ϕ2Ω − 1

)

|y|2(ǫ+ |y|2)n dy − (n− 2)2

ˆ

Rn\Ω

|y|2(ǫ+ |y|2)n dy

:= (n− 2)2ˆ

Rn

|y|2(ǫ+ |y|2)n dy +A0,1 +A0,2 +A0,3 +A0,4;

A1 = −Sg(0)6n

ˆ

Ω

|y|2 |∂sϕΩ|2(ǫ+ |y|2)n−2

dy +Sg(0)(n − 2)

3n

ˆ

Ω

ϕΩ|y|2 (∂sϕΩ · y)(ǫ+ |y|2)n−1

dy

− Sg(0)(n − 2)2

6n

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)n dy

:= A1,1 +A1,2 +A1,3;

A2 = C1

ˆ

Ω|y|3|∂suǫ,Ω|2dy = C1

ˆ

Ω

|y|3 |∂sϕΩ|2(ǫ+ |y|2)n−2

dy − 2C1(n− 2)

ˆ

Ω

ϕΩ|y|3 (∂sϕΩ · y)(ǫ+ |y|2)n−1

dy

+ C1(n − 2)2ˆ

Ω

ϕ2Ω|y|5

(ǫ+ |y|2)n dy

:= A2,1 +A2,2 +A2,3.


The other two terms above can be estimated as

1

a

ˆ

Ω(Sg(0) + β)u2ǫ,Ωdx =

1

a

ˆ

Ω(Sg(0) + β)

ϕ2Ω

(ǫ+ |y|2)n−2dy =

1

a

ˆ

Rn

(Sg(0) + β)1

(ǫ+ |y|2)n−2dy

− 1

a

ˆ

Rn\Ω(Sg(0) + β)

1

(ǫ+ |y|2)n−2dy

+1

a(Sg(0) + β)

ˆ

Ω

ϕ2Ω − 1

(ǫ+ |y|2)n−2dy

:=1

a

ˆ

Rn

(Sg(0) + β)1

(ǫ+ |y|2)n−2dy +B0,1 +B0,2;

‖uǫ,Ω‖pLp(Ω) =

ˆ

Ω

ϕpΩ

(ǫ+ |y|2)n dy =

ˆ

Rn

1

(ǫ+ |y|2)ndy −ˆ

Rn\Ω

1

(ǫ+ |y|2)n dy

+

ˆ

Ω

ϕpΩ − 1

(ǫ+ |y|2)n dy

:=

ˆ

Rn

1

(ǫ+ |y|2)ndy +C0,1 + C0,2.

Case I: n > 5. Fix r to be small enough, independent of the choice of ǫ. Claim that

‖∂suǫ,Ω‖2L2(Ω) +1

a

ˆ

ΩSg(0)u

2ǫ,Ωdx+A1 +A2 +B1 +B2 +B3

< (n− 2)2ˆ

Rn

|y|2(ǫ+ |y|2)n dy +

(n− 2)β

4(n − 1)ǫ−

4−n2

ˆ

Rn

1

(1 + |y|2)n−2dy +O

(

ǫ5−n2

)

− ǫ4−n2Sg(0)(n − 2)

6

n− 2

n

ˆ

BP

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

= (n− 2)2ˆ

Rn

|y|2(ǫ+ |y|2)n dy +

(n− 2)β

4(n − 1)ǫ−

4−n2

ˆ

Rn

1

(1 + |y|2)n−2dy +O

(

ǫ5−n2

)

+Σ2;

‖uǫ,Ω‖pLp(Ω)+ C1 + C2 =

ˆ

Rn

1

(ǫ+ |y|2)n dy +O(

ǫ3−n2

)

+Σ1.

(99)

Recall that β < 0 is some fixed constant, independent of ǫ.

36 J. XU

Let’s estimate the rest Bi, Ci, Ai,j , Bi,j , Ci,j except the crucial term A1,3. Let Ω′ = B0

(

rǫ−12

)

.

Since ϕΩ ≡ 1 on B0

(

r2

)

, |∂sϕΩ| 6 Lr for some fixed constant L, we observe that

A0,1 = O(1), A0,2 = O(1), A0,3 6 0, A0,4 6 0;

A1,1 = O(1), |A1,2| = O(1), A2,1 = O(1), |A2,2| = O(1);

A2,3 = C1(n− 2)2ˆ

Ω

ϕ2Ω|y|5

(ǫ+ |y|2)n dy = ǫ5−n2 C1(n− 2)2

ˆ

Ω′

ϕ2Ω|y|5

(1 + |y|2)n dy 6 L1ǫ5−n2

⇒A2,3 = O(

ǫ5−n2

)

;

|B1| 6

∣

∣

∣

∣

∣

∣

1

a

ˆ

Ω

∑

i,|α|=1

∂αRii(0)yα|u|2dy

∣

∣

∣

∣

∣

∣

6 L2

ˆ

Ω

|y|(ǫ+ |y|2)n−2

dy

6 L2ǫ5−n2

ˆ

Ω′

|y|(1 + |y|2)n−2

dy ⇒ B1 = O(

ǫ5−n2

)

;

B2 = C2

ˆ

Ω

ϕ2Ω|y|2

(ǫ+ |y|2)n−2dy = ǫ

6−n2 C2

ˆ

Ω′

ϕ2Ω|y|2

(1 + |y|2)n−2dy 6 L3ǫ

6−n2

⇒B2 = O(

ǫ6−n2

)

;B3 6 O(

ǫ7−n2

)

;

B0,1 = O(1), B0,2 = O(1);C0,1 = O(1), C0,2 = O(1);

C1 + C2 = − 1

6n

ˆ

ΩSg(0)|y|2|u|pdy + C4

ˆ

Ω|y|3|u|pdy

= −Sg(0)6n

ˆ

Ω

|y|2ϕpΩ

(ǫ+ |y|2)n dy +O(

ǫ3−n2

)

= −ǫ 2−n2Sg(0)

6n

ˆ

B0

(

( r2)ǫ

−12

)

|y|2(1 + |y|2)n dy +O (1) +O

(

ǫ3−n2

)

:= Σ1 +O(

ǫ3−n2

)

.

Here Li, i = 1, . . . , 3 are constants independent of r, ǫ.For the crucial terms A1,3, we show that

A1,3 +1

a

ˆ

Rn

(Sg(0) + β)1

(ǫ+ |y|2)n−2dy

< βǫ4−n2

ˆ

Rn

1

(1 + |y|2)n−2dy − ǫ

4−n2Sg(0)(n − 2)

6· n− 2

n

ˆ

B0

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

+O(1)

:=(n − 2)β

4(n − 1)ǫ4−n2

ˆ

Rn

1

(1 + |y|2)n−2dy +Σ2 +O(1).

(100)


The term A1,3 can be bounded as

A1,3 = −Sg(0)(n − 2)2

6n

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)n dy = −ǫ 4−n2Sg(0)(n − 2)2

6n

ˆ

Ω′

ϕ2Ω|y|4

(1 + |y|2)n dy

= −ǫ 4−n2Sg(0)(n − 2)2

6n

ˆ

Ω′

ϕ2Ω|y|2

(

|y|2 − nn−2

)

(1 + |y|2)n dy − ǫ4−n2Sg(0)(n − 2)2

6n

ˆ

Ω′

ϕ2Ω|y|2 · n

n−2

(1 + |y|2)n dy

6 −ǫ 4−n2Sg(0)

12n

ˆ

Rn

|y|2 ·2(n − 2)2

(

|y|2 − nn−2

)

(1 + |y|2)n dy

− ǫ4−n2Sg(0)(n − 2)2

6n

ˆ

Ω′

ϕ2Ω|y|2 · n

n−2

(1 + |y|2)n dy := Γ1 + Γ2.

The above inequality holds since the radius of Ω′ is much larger than 1, which follows that |y|2 −n

n−2 > 0 for y ∈ Ω′c. It is direct to check that for Euclidean Laplacian −∆, the function 1(1+|y|2)n−2

satisfies

∆

(

1

(1 + |y|2)n−2

)

=2(n − 2)2

(

|y|2 − nn−2

)

(1 + |y|2)n .

Using this, we estimate Γ1 as

Γ1 = −ǫ 4−n2Sg(0)

12n

ˆ

Rn

|y|2 ·2(n− 2)2

(

|y|2 − nn−2

)

(1 + |y|2)n dy

= −ǫ 4−n2Sg(0)

12n

ˆ

Rn

|y|2 ·∆(

1

(1 + |y|2)n−2

)

dy = −ǫ 4−n2Sg(0)

12n

ˆ

Rn

(

∆|y|2)

· 1

(1 + |y|2)n−2dy

= −ǫ 4−n2Sg(0)

6

ˆ

Rn

1

(1 + |y|2)n−2dy.

For Γ2, we see that

Γ2 = −ǫ 4−n2Sg(0)(n − 2)2

6n

ˆ

Ω′

ϕ2Ω|y|2 · n

n−2

(1 + |y|2)n dy

= −ǫ 4−n2Sg(0)(n − 2)

6

ˆ

B0

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy − ǫ

4−n2Sg(0)(n − 2)

6

ˆ

Ω′\B0

(

r2ǫ−

12

)

ϕ2Ω|y|2

(1 + |y|2)n dy

= −ǫ 4−n2Sg(0)(n − 2)

6· n− 2

n

ˆ

B0

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

− ǫ4−n2Sg(0)(n − 2)

6· 2n

ˆ

B0

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy +O (1)

< −ǫ 4−n2Sg(0)(n − 2)

6· n− 2

n

ˆ

B0

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

− ǫ4−n2Sg(0)

6· 2(n − 2)

n

ˆ

Rn

|y|2(1 + |y|2)n dy +O (1)

:= Σ2 + Γ2,1 +O (1) .

38 J. XU

Consider the integration´

Rn|y|2

(1+|y|2)ndy in Γ2,1. Recall that ωn is the area of the unit n− 1-sphere.

We check with change of variables s = tan(θ) in some middle step thatˆ

Rn

|y|2(1 + |y|2)n dy = ωn

ˆ ∞

0

sn+1

(1 + s2)nds = ωn

ˆ π2

0

tann+1(θ) sec2(θ)

(1 + tan2(θ))ndθ

= ωn

ˆ π2

0sin2(

n2+1)−1 cos2(

n2−1)−1 dθ = ωnB

(n

2+ 1,

n

2− 1)

= ωnΓ(

n2 + 1

)

Γ(

n2 − 1

)

Γ(n).

Here B(a, b) is the Beta function with parameters a, b and Γ(a) is the Gamma function withparameter a. The term below satisfies

1

a

ˆ

Rn

Sg(0)1

(ǫ + |y|2)n−2dy = ǫ

4−n2

1

a· Sg(0)

ˆ

Rn

1

(1 + |y|2)n−2dy.

Check the integration´

Rn

|y|2

(1+|y|2)ndy above, we have

ˆ

Rn

|y|2(1 + |y|2)n dy = ωnB

(n

2,n

2− 2)

= ωnΓ(

n2

)

Γ(

n2 − 2

)

Γ(n− 2).

By the standard relation Γ(z + 1) = zΓ(z) we conclude thatˆ

Rn

|y|2(1 + |y|2)n dy =

n(n− 4)

4(n − 1)(n − 2)

ˆ

Rn

1

(1 + |y|2)n−2dy. (101)

Note that 1a = n−2

4(n−1) . Applying (101) as well as the estimates of Γ1 and Γ2 above, we conclude

that

A1,3 +1

a(Sg(0) + β)

ˆ

Rn

1

(ǫ+ |y|2)n−2dy 6 Γ1 + Γ2 + ǫ

4−n2

1

a· (Sg(0) + β)

ˆ

Rn

1

(1 + |y|2)n−2dy

= −ǫ 4−n2Sg(0)

6

ˆ

Rn

1

(1 + |y|2)n−2dy +Σ2

+n(n− 4)

4(n − 1)(n − 2)·(

−ǫ 4−n2Sg(0)

6· 2(n− 2)

n

)ˆ

Rn

1

(1 + |y|2)n−2dy

+ ǫ4−n2

n− 2

4(n− 1)· (Sg(0) + β)

ˆ

Rn

1

(1 + |y|2)n−2dy +O(1)

=(n− 2)β

4(n − 1)ǫ4−n2

ˆ

Rn

1

(1 + |y|2)n−2dy +Σ2 +O(1).


It follows that (100) holds. Therefore we conclude that the estimates in (99) hold by combining allestimates above together. Recall K1,K2 defined above, we apply estimates in (99), we get

Qǫ,Ω 6

(n− 2)2´

Rn|y|2

(ǫ+|y|2)ndy + (n−2)β

4(n−1) ǫ4−n2

´

Rn1

(1+|y|2)n−2 dy +O(

ǫ5−n2

)

+Σ2

(

´

Rn1

(ǫ+|y|2)ndy +O

(

ǫ3−n2

)

+Σ1

) 2p

=(n− 2)2ǫ

2−n2

´

Rn

|y|2

(1+|y|2)ndy + (n−2)β

4(n−1) ǫ4−n2

´

Rn1

(1+|y|2)n−2 dy +O(

ǫ5−n2

)

+Σ2

(

ǫ−n2

´

Rn1

(1+|y|2)ndy +O

(

ǫ3−n2

)

+Σ1

) 2p

=K1 + ǫ (n−2)β

4(n−1)K3 +O(

ǫ32

)

+ ǫn−22 Σ2

K2 +O(

ǫ32

)

+ n−2n ǫ

n2 Σ1K

− 2n−2

2

.

The last term above is due to the Taylor expansion of the function f(x) = x2p . To control the last

terms in numerator and denominator above, respectively, we first recall that the ratio K1K2

is thesquare of the reciprocal of the best Sobolev constant of

‖u‖Lp(Rn) 6 E‖Du‖L2(Rn) ⇒ infEE−2 = T =

K1

K2.

Aubin and Talenti [2] showed that

T− 12 = π−

12n−

12

(

1

n− 2

) 12

(

Γ(n)

Γ(

n2

)

) 1n

.

Recall that β < 0, independent of ǫ, thus Qǫ,Ω can be estimated as

Qǫ,Ω 6

K1 + ǫ (n−2)β4(n−1)K3 +O

(

ǫ32

)

+ ǫn−22 Σ2

K2 +O(

ǫ32

)

+ n−2n ǫ

n2 Σ1K

− 2n−2

2

6K1 + ǫ

n−22 Σ2

K2 +n−2n ǫ

n2 Σ1K

− 2n−2

2

+O(

ǫ32

)

− ǫ · (n− 2)|β|4(n− 1)

K3

K2.

The last inequality is due to the fact that both ǫn2Σ1K

− 2n−2

2 and ǫn−22 Σ2 are bounded above and

below by constant multiples of ǫ. Lastly we must show that

K1 + ǫn−22 Σ2

K2 +n−2n ǫ

n2Σ1K

− 2n−2

2

6K1

K2. (102)

By the best Sobolev constant, we observe that

K1

K2= T ⇒ K1 = K2T.

40 J. XU

Thus (102) is equivalent to

K1 + ǫn−22 Σ2

K2 +n−2n ǫ

n2Σ1K

− 2n

2

6K1

K2⇔ K1 + ǫ

n−22 Σ2

K2 +n−2n ǫ

n2Σ1K

− 2n−2

2

− K1

K26 0

K22T + ǫ

n−22 Σ2K2 −K2

2T − n−2n K2Tǫ

n2Σ1K

− 2n−2

2(

K2 +n−2n ǫ

n2 Σ1K

− 2n−2

2

)

K2

6 0

⇔ǫn−22 Σ2K2 −

n− 2

nK2Tǫ

n2 Σ1K

− 2n−2

2 6 0

⇔ǫn−22

(

−ǫ 4−n2Sg(0)(n − 2)

6

n− 2

n

ˆ

BP

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

)

K2

− n− 2

nK2Tǫ

n2

(

−ǫ 2−n2Sg(0)

6n

ˆ

BP

(

r2ǫ−

12

)

|y|2(1 + |y|2)n dy

)

K− 2

n−2

2 6 0

⇔(n− 2)− TK− 2

n−2

2

n6 0.

Recall that the best Sobolev constant T is of the form

T = πn(n− 2)

(

Γ(

n2

)

Γ(n)

)2n

For the term K− 2

n−2

2 , we have

K− 2

n−2

2 =

(ˆ

Rn

1

(1 + |y|2)n dy)− 2

n

=

(

ωn

ˆ ∞

0

rn−1

(1 + r2)ndy

)− 2n

:= (ωnD1)− 2

n

Here ωn is the surface area of the (n− 1)-sphere. It is well known that

ωn = n · πn2

Γ(

n2 + 1

) = n · 2πn2

nΓ(

n2

) =2π

n2

Γ(

n2

) ⇒ ω− 2

nn =

(

Γ(

n2

)

2πn2

)2n

=(

Γ(n

2

)) 2n2−

2nπ−1.

For D1, we use change of variables r = tan θ, θ ∈ [0, π2 ], we have

D1 =

ˆ ∞

0

rn−1

(1 + r2)ndy =

ˆ π2

0

tann−1(θ)

(1 + tan2(θ))nsec2(θ)dθ =

ˆ π2

0

sinn−1(θ) secn+1(θ)

sec2n(θ)dθ

=

ˆ π2

0sinn−1(θ) cosn−1 θdθ = 21−n

ˆ π2

0sinn−1(2θ)dθ = 2−n

ˆ π

0sinn−1(x)dx.

We use 2θ = x in the last equality. It is well-known that the last integration above is symmetricwith respect to x = π

2 and is connected to the Gamma function. To be precise,

D1 = 2−n

ˆ π

0sinn−1(x)dx = 21−n

ˆ π2

0sinn−1(x)dx = 21−n

√πΓ(

n−1+12

)

2Γ(

n−12 + 1

)

⇒ D− 2

n1 =

(

2nπ−12Γ(

n2 + 1

2

)

Γ(

n2

)

) 2n

= 22π−1nΓ

(

n

2+

1

2

)2n

Γ(n

2

)− 2n.


Packing terms together, we have

TK− 2

n−2

2

n= n−1πn(n− 2)

(

Γ(

n2

)

Γ(n)

) 2n

·(

Γ(n

2

))2n2−

2nπ−1

· 22π− 1n

(

Γ

(

n

2+

1

2

)) 2n (

Γ(n

2

))− 2n

= (n− 2)22−2nπ−

1n

(

Γ

(

n

2+

1

2

)

Γ(n

2

)

)2n

(Γ(n))−2n

The Legendre duplication formula for Gamma functions says

Γ

(

z +1

2

)

Γ (z) = 21−2z√πΓ(2z).

Take z = n2 above, we have

TK− 2

n−2

2

n= (n− 2)22−

2nπ−

1n

(

Γ

(

n

2+

1

2

)

Γ(n

2

)

) 2n

(Γ(n))−2n

= (n− 2)22−2nπ−

1n

(

21−2n2√πΓ(n)

)2n(Γ(n))−

2n

= (n− 2).

Hence we conclude that

(n− 2)− TK− 2

n−2

2

n= (n − 2) − (n − 2) = 0 ⇒ K1 + ǫ

n−22 Σ2

K2 +n−2n ǫ

n2Σ1K

− 2n−2

2

6K1

K2.

Therefore (102) holds. It follows that when ǫ is small enough,

Qǫ,Ω 6K1

K2O(

ǫ32

)

− ǫ · (n− 2)|β|4(n − 1)

K3

K2<K1

K2= T.

Thus the inequality in (95) holds when n > 5.

Case II: n = 4. In this case, we keep notations to be the same as n > 5 cases. Again we fix r.Most estimates of Ai,j, Bi, Bi,j, Ci, Ci,j with n = 4 are the same as above:

A0,1 = O(1), A0,2 = O(1), A0,3 6 0, A0,4 6 0;

A1,1 = O(1), |A1,2| = O(1), A2,1 = O(1), |A2,2| = O(1);A2,3 = O(

ǫ12

)

;

|B1| = O(

ǫ12

)

;B2 = O (ǫ) ;B3 = O(

ǫ32

)

;

C0,1 = O(1), C0,2 = O(1);C1 > 0, C2 = O(

ǫ−12

)

.

Here C1 > 0 due to the fact that Sg(0) < 0. We do not need B0,1 and B0,2 since the key term

1

a· (Sg(0) + β)

ˆ

Ω

ϕ2Ω

(ǫ+ |y|)2)2 dy

is not integrable. Note that the term A1,3 with n = 4 can be estimated as

A1,3 = −Sg(0)(n − 2)2

6n

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)n dy = −Sg(0)6

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)4dy

6 −Sg(0)6

ˆ

Ω

ϕ2Ω

(ǫ+ |y|2)2 dy

42 J. XU

since the ratio |y|2

ǫ+|y|2< 1 as always. Thus with a = n−2

4(n−1) =16 when n = 4, we can see

A1,3 +1

a· (Sg(0) + β)

ˆ

Ω

ϕ2Ω

(ǫ+ |y|)2)2 dy 6β

6

ˆ

Ω

ϕ2Ω

(ǫ+ |y|)2)2 dy.

Applying exactly the same calculation as in [5, Lemma. 1.1], we conclude thatˆ

Ω

ϕ2Ω

(ǫ+ |y|)2)2 dy > L5|log ǫ|

with some positive constant L5 independent of ǫ. It follows that

A1,3 +1

a· (Sg(0) + β)

ˆ

Ω

ϕ2Ω

(ǫ+ |y|)2)2 dy 6 −L5|β||log ǫ|. (103)

Applying (103) and all other estimates above, we have

Qǫ,Ω 6

4´

Rn

|y|2

(ǫ+|y|2)4dy − L5|β||log ǫ|+O(1) +O

(

ǫ12

)

(

´

Rn1

(ǫ+|y|2)4dy +O(1) +O

(

ǫ−12

)) 12

64ǫ−1

´

Rn

|y|2

(1+|y|2)2dy − L5|β||log ǫ|+O (1)

(

ǫ−2´

Rn1

(1+|y|2)ndy +O

(

ǫ−12

)) 12

6

K1 − L5|β|ǫ|log ǫ|+O(

ǫ32

)

K2 +O(

ǫ32

)

=K1

K2+O (ǫ)− L5|β|

K2ǫ|log ǫ|.

It follows that when ǫ small enough, we have

Qǫ,Ω < T.

Thus (95) holds for n = 4.

Case III : n = 3. This case is inspired by the argument in Brezis and Nirenberg [5, Lemma 1.3],but not the same. We have to do some conformal change so that

Sg(0)

8+π2

4= 0 on B0(1). (104)

This can be done due to Theorem 4.5 and Remark 4.7. Thus

Sg(0) + β

8+π2

4< 0 on B0(1)

since β < 0. Note that the first eigenvalue λ1 of −∆g is almost the same as the first eigenvalue of

Euclidean Laplacian when the geodesic ball is small enough. Since Sgλ−11 is scaling invariant, we

can scale the metric, i.e. scale the radius of the ball without changing the relation in (104) as wellas the relation in (40) with a small enough β. Thus it suffices to set Ω = B0(1). We would liketo point out that the choice of Sg(0) in (104) is compatible with the choice of Sg(0) in (40). Weestimate A1 first. Recall that

A1 = −Sg(0)18

ˆ

Ω

|y|2 |∂sϕΩ|2ǫ+ |y|2 dy +

Sg(0)

9

ˆ

Ω

ϕΩ|y|2 (∂sϕΩ · y)(ǫ+ |y|2)2 dy − Sg(0)

18

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)3 dy

:= A1,1 +A1,2 +A1,3.


For A1,1, we have

A1,1 = −Sg(0)18

ˆ

Ω

|y|2 |∂sϕΩ|2ǫ+ |y|2 dy =

π2

9ω3 ·

π2

4

ˆ 1

0

sin2(

πs2

)

s4

ǫ+ s2ds 6

π2

9ω3 ·

π2

4

ˆ 1

0sin2

(πs

2

)

s2ds

= ω3 ·π4

36

(

1

6+

1

π2

)

.

For A1,2, we have

A1,2 =Sg(0)

9

ˆ

Ω

ϕΩ|y|2 (∂sϕΩ · y)(ǫ+ |y|2)2 dy =

2π2

9ω3π

4

ˆ 1

0

sin(πs)s5

(ǫ+ s2)2ds 6

2π2

9ω3π

4

ˆ 1

0sin(πs)sds

= ω3π3

18· 1π= ω3

π2

18.

For A1,3, we have

A1,3 = −Sg(0)18

ˆ

Ω

ϕ2Ω|y|4

(ǫ+ |y|2)3dy =π2

9ω3

ˆ 1

0

cos2(

πs2

)

s6

(ǫ+ s2)3ds 6

π2

9ω3

ˆ 1

0cos2

(πs

2

)

ds

= ω3π2

18.

Here L6 > 0 is some constant independent of ǫ, r. For A2, C1, C2, C0,1, C0,2, a simple estimate showsthat

A2 = O(

ǫ12

)

, C1 > 0, C2 = O(1), C0,1 = O(1), C0,2 = O(1).

Based on Brezis and Nirenberg [5], the term ‖∂su‖2L2(Ω) can be written as

‖∂su‖2L2(Ω) = ω3π2

4

ˆ 1

0

sin2(

πs2

)

s2

ǫ+ s2ds+ 3ω3ǫ

ˆ 1

0

cos2(

πs2

)

s2

(ǫ+ s2)3ds := E1 + E2.

Due to Brezis and Nirenberg again, we have

E1 = ω3π2

4

ˆ 1

0sin2

(πs

2

)

ds+O(ǫ).

For the term E2, we estimate as

E2 = 3ω3ǫ

ˆ 1

0

cos2(

πs2

)

s2

(ǫ+ s2)3ds

= 3ω3ǫ

ˆ ∞

0

s2

(ǫ+ s2)3ds− 3ω3ǫ

ˆ ∞

1

s2

(ǫ+ s2)3ds− 3ω3ǫ

ˆ 1

0

sin2(

πs2

)

s2

(ǫ+ s2)3ds

:= 3ω3ǫ

ˆ ∞

0

s2

(ǫ+ s2)3ds+ E2,1 + E2,2.

Using polar coordinates s = ǫ12 tan(θ), the term E2,1 can be computed as

E2,1 = −3ω3ǫ

ˆ ∞

1

s2

(ǫ+ s2)3ds = −3ω3ǫ

− 12

ˆ π2

arctan(

ǫ−12

)

tan2(θ) sec2(θ)

sec6(θ)dθ

= −3ω3ǫ− 1

2

ˆ π2

arctan(

ǫ−12

)

1

4dθ + o(1)

= −3

4ω3ǫ

− 12

(π

2− arctan

(

ǫ−12

))

+ o(1) = −3

4ω3 + o(1).

44 J. XU

The last inequality is due to the Taylor expansion of arctan(x) at x = +∞, which is

arctan(x)

∣

∣

∣

∣

x=+∞

=π

2− 1

x+

1

3x3+ o

(

1

x3

)

.

Again by Brezis and Nirenberg, we know that

3π

16=

ˆ ∞

0

s4

(1 + s2)3ds = 3

ˆ ∞

0

s2

(1 + s2)3ds. (105)

Thus with (105), the Taylor expansion of sin(x) at origin, we have

E2,2 = −3ω3ǫ

ˆ 1

0

sin2(

πs2

)

s2

(ǫ+ s2)3ds = −3ω3ǫ

π2

4

ˆ 1

0

s4

(ǫ+ s2)3ds+O(ǫ)

> −9π3

64ω3 +O(ǫ).

By a direct calculation, it follows that

A1 + E2,1 + E2,2 6 O(ǫ).

By (105), we can see that

3ω3ǫ

ˆ ∞

0

s2

(ǫ+ s2)3ds = 3ω3ǫ

− 12

ˆ ∞

0

s2

(1 + s2)3ds = ω3ǫ

− 12

ˆ ∞

0

s4

(1 + s2)3ds = ǫ−

12K1.

Exactly following Brezis and Nirenberg [5, Lemma 1.3], we have

‖u‖2L6(Ω) = ǫ−12K2 +O

(

ǫ12

)

;

Sg(0) + β

8

ˆ

Ω

ϕ2Ω(y)

ǫ+ |y|2dy =Sg(0) + β

8ω3

ˆ 1

0cos2

(πs

2

)

ds+O(

ǫ12

)

.

Lastly, note thatˆ 1

0cos2

(πs

2

)

ds =

ˆ 1

0sin2

(πs

2

)

ds.

Thus the quantity Qǫ,Ω can be estimated as

Qǫ,Ω 6

ǫ−12K1 + ω3

´ 10 cos2

(

πs2

)

ds(

π2

4 +Sg(0)+β

8

)

+O(

ǫ12

)

ǫ−12K2 +O

(

ǫ12

)

6K1

K2+ǫ12

K2ω3

ˆ 1

0cos2

(πs

2

)

ds

(

π2

4+Sg(0) + β

8

)

+O(ǫ).

By (104), we know that π2

4 +Sg(0)+β

8 < 0 hence

Qǫ,Ω < T

When ǫ is small enough. All cases are now proven.

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Department of Mathematics and Statistics, Boston University, Boston, MA, USA

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arXiv:2110.15436v5 [math.DG] 15 Nov 2021

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