Positivity of Ma-Trudinger-Wang curvature on Riemannian surfaces

Calc. Var.DOI 10.1007/s00526-013-0684-7 Calculus of Variations

Positivity of Ma-Trudinger-Wang curvature onRiemannian surfaces

Shi-Zhong Du · Qi-Rui Li

Received: 16 September 2013 / Accepted: 20 October 2013© Springer-Verlag Berlin Heidelberg 2013

Abstract In this paper we consider the optimal transportation on Riemannian surfaces whenthe cost function is squared distance. The main ingredient is the verification of MTW con-dition. It is known that MTW condition holds if Gauss curvature is sufficiently close to 1in C2 norm. In this paper we give an explicit condition on Gauss curvature such that MTWcondition is satisfied.

Mathematics Subject Classification (2000) 53C21

1 Introduction

For optimal transportation problem, the regularity of potential functions has been investi-gated in a number of papers in recent years. In the landmark work, the authors discoveredthe Ma–Trudinger–Wang curvature [25] (MTW curvature for short). It is now well-knownthe positivity of this curvature, namely the MTW condition, plays an essential role in theregularity theory. A geometric property of this condition was observed by Loeper [21], bywhich Loeper showed that a degenerate version of this condition, namely the weak MTWcondition, is necessary for the continuity of optimal maps. On Riemannian manifolds, regu-larity of potential functions is also closely related to the MTW condition [9,11,14,18,21–23].However the MTW condition involves a combination of derivatives of the cost function upto the fourth order, its verification is therefore very complicated.

Communicated by N. Trudinger.

S.-Z. DuShenzhen Graduate School, The Harbin Institute of Technology, Shenzhen 518055, Chinae-mail: [email protected]

Q.-R. Li (B)Centre for Mathematics and Its Applications,Australian National University, Canberra, ACT 0200, Australiae-mail: [email protected]

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S.-Z. Du, Q.-R. Li

For mass transfer problem on Riemannian manifold (M, g), the cost function

c (x, y) = 1

2d2

g (x, y) (1.1)

is of special interest. Here dg denotes the distance function with respect to Riemannian metricg. Suppose (M, g) is a connected, compact, smooth Riemannian manifold without boundary.Let μ , μ∗ be probability measures on M, both absolutely continuous with respect to thevolume measure dVg , and satisfying mass balance condition μ (M) = μ∗ (M). The optimaltransportation problem is to find a map T minimizing the total cost functional

C (s) =∫

Mc (x, s (x)) dμ (x) , (1.2)

over the set S of all the measure preserving maps

S =⎧⎨⎩s is Borel map onM|

∫

Mh (s (x)) dμ (x) =

∫

Mh (y) dμ∗ (y) ,∀h ∈ C0 (M)

⎫⎬⎭.

McCann [24] developed Brenier’s [1,2] polar decomposition theory and then proved thereexists a unique minimizer T of (1.2) among all the maps in S. McCann’s solution map T isin the form of

T (x) = expx (−∇u (x))

with potential function u. Assume μ = φdVg and μ∗ = φ∗dVg , then potential functionsatisfies a Monge–Ampère type equation

det(∇2

x c (x, exp (−∇u) (x)) − ∇2u (x))

=∣∣∣det

(∇2

x,yc (x, exp (−∇u) (x)))∣∣∣ φ (x)

φ∗ (exp (−∇u) (x)). (1.3)

When the manifold is Euclidean space and cost function is squared distance (1.1), the equation(1.3) can be reduced to the standard Monge–Ampère equation, and the regularity of solutionswas obtained by Caffarelli, Delanoë and Urbas [3–5,8,29].

For general cost functions, Ma et al. [25] introduced the following fourth order curvature

MT W kli j (x, y) := −

∑ (ci j,rs − ci j,pcp,qcq,rs

)cr,kcs,l , (1.4)

where ci, j := ∂2xi y j c (x, y), ci, j denotes the inverse of ci, j , and other quantities are defined

similarly. It is known that MT W kli j (x, y) is a (2, 2) form of x variable [17]. The non-negativity

of this curvature, namely (weak) MTW condition, is crucial for regularity of optimal map.We say a Riemannian manifold (M, g) satisfies the weak MTW condition if

MT W kli j ξ iξ jηkηl ≥ 0 (1.5)

for all vector ξ and covector η satisfying η · ξ = 0, and (M, g) satisfies the MTW conditionif

MT W kli j ξ iξ jηkηl ≥ C0 |ξ |2 |η|2 (1.6)

for all η · ξ = 0, where C0 is a positive constant. For general cost functions satisfyingconditions (1.6) or (1.5), the regularity of solutions has been extensively investigated in the

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Optimal mappings on surfaces

Euclidean space [15,19,21,20,25,27,28], and also on Riemannian manifolds [9,11,14,18,21–23].

The cost (1.1) satisfies the weak MTW but not MTW condition in Euclidean space. Aninteresting observation, also by Loeper [21], is that (1.1) satisfies the MTW condition onpositive constant sectional curvature spaces. The regularity of optimal transportation on suchspaces was then obtained in [9,22]. However positive sectional curvature condition itselfcannot imply the regularity. Kim [16] , and also Figalli et al. [12] found examples of positivelycurved manifolds which do not satisfy the MTW condition. It was proved that MTW conditionholds if g is a small C4-perturbation of the round sphere metric g0 by Figalli and Rifford [11]for n = 2 and by Figalli et al. [13] for n ≥ 2. Interestingly, the authors also proved that on aC4-perturbation of the round sphere S

n , all injectivity domains are uniformly convex [11,13]. Under a more intrinsic assumption that Riemannian curvature sufficiently close to 1 in C2

norm, Delanoë and Ge [9,10] showed MTW condition is satisfied. For two dimensional case,Figalli et al. [12] showed the MTW condition is stable under C4 -perturbation of metric if thenonfocal domains are uniformly convex. The regularity of optimal transportation was alsostudied on products of arbitrarily many round spheres with arbitrary sizes and dimensionsby Figalli et al. [14], and on Riemannian submersions and Riemannian products by Kimand McCann [18]. On the other hand, the stability of the MTW condition under Gromov–Hausdorff limits has been studied by Villani [30], and some interesting relation between theMTW condition and the geometry of the cut locus has also been investigated by Loeper,Figalli, Rifford and Villani [23,11–13].

In [9–13] the authors made MTW curvature link to various quantities which involvethe Jacobi fields along geodesics, then they study the behavior of MTW curvature by aclever choice of coordinate and by delicately estimating the corresponding quantities viaJacobi equations. Inspired by aforementioned works, we relate MTW condition to a quantityA, defined as (2.19) below, which involves the mean curvature of geodesic sphere and itsderivatives up to second order. Via the Ricatti equation satisfied by mean curvature, we findan explicit condition on Gauss curvature on Riemannian surface such that MTW conditionholds (apart away from cut-locus). It is natural that our condition involves the derivatives ofGauss curvature up to second order, as MTW curvature is a fourth order quantity of distancefunction. To the authors’ knowledge, this condition is the first explicit condition on Gausscurvature such that MTW condition holds.

Before stating our main theorem, we need some notions. Let Dx be a closed subset ofTxM given by

Dx = {v ∈ TxM|d (

expx v, x) = |v|}.

Then the injectivity domain Ix is defined as

Ix = exp int Dx ,

where int Dx stands for the interior of set Dx . If y ∈ Ix , let ey−x ∈ TxM be the unit vectorsuch that y = expx

(d (x, y) ey−x

). Our main result is the following

Theorem 1 Let M be a compact connected surface with positive Gauss curvature K . LetKM = maxM K , and L be any constant less than π√

KM. For all x, y ∈ M satisfying y ∈ Ix

and ρ = d (x, y) ≤ L, if condition

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S.-Z. Du, Q.-R. Li

2K (t) ≥ (ρ − t)2∣∣∇2 K

∣∣ (t) +⎡⎣ (ρ − t)

4 sin α0

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ 3

4

ρ∫

t

(ρ − τ) K (τ ) dτ + 8 + √3

8 sin α0

⎤⎦ (ρ − t) |∇K | (t) (1.7)

holds for all t ∈ [0, ρ], then

MT W kli j (x, y) ξ iξ jηkηl ≥ C0 |ξ |2 |η|2 , ∀η · ξ = 0

for some positive constant C0. Here K (t) , |∇K | (t) and∣∣∇2 K

∣∣ (t) take values at expy tex−y;and α0 is a constant given by

α0 = min

{arcsin

(sin

(√KM L

)√

KM L

),

π

16

}.

Remark 1 The condition (1.7) is invariant under scaling the metric. Namely, if (M, g) sat-isfies (1.7), then

(M, gλ)

also satisfies (1.7) for all λ > 0. Here gλ = λ2g is a rescaling ofg. Let us fix x, y and z = expy tex−y . We use superscript λ to denote the quantities corre-sponding to metric gλ. Clearly we have z = expλ

y tλeλx−y with tλ = λt , and also ρλ = λρ,

K λ = λ−2 K ,∣∣∇gλ K λ

∣∣gλ = λ−3 |∇K |,

∣∣∣∇2gλ K λ

∣∣∣gλ

= λ−4∣∣∇2 K

∣∣. It follows that the two

middle integral terms in the right hand side of (1.7) are invariant. For example, one maycheck

ρλ∫

tλ

(ρλ − τ

)K λ (τ ) dτ = λ−2

λρ∫

λt

(λρ − τ) K(λ−1τ

)dτ =

ρ∫

t

(ρ − τ) K (τ ) dτ.

Hence we get the invariance of condition (1.7) immediately.

We would like to point out the condition (1.7) implies MTW condition holds locally (apartaway from cut-locus), hence the optimal mapping is smooth if mass densities are smoothand supported in a sufficiently small geodesic ball. Our strategy of proving Theorem 1 isto introduce an auxiliary quantity A [see (2.19) below], and reduce the MTW condition tothe negativity of A. The quantity A involves the mean curvature H of geodesic spheres onsurface, so we make it link to the Gauss curvature K by employing the Ricatti equation(2.14) satisfied by H . Our main technique is to firstly differentiate A with respect to thetangential direction of the corresponding geodesic, then rewrite this derivative formula viaRicatti equation, and finally reformulate A by integrating this derivative formula back, see(4.21). The negativity of A under condition 1.7 is obtained by estimating (4.21) term by term.Some of our ideas and computation are inspired by [9–13]. For example the formulae (2.24)and (4.1)–(4.4) below are analogous to the formulae obtained in [10,12,13]. However weremark that the quantity A plays a crucial role in this paper, and our argument focuses on themean curvature of geodesic sphere and its Ricatti equation.

This paper is organized as follows: in Sect. 2 we show the MTW condition is equivalentto the negativity of a quantity A, which can be presented in terms of the second fundamentalform of geodesic spheres and their derivatives up to second order. Section 3 is devoted toobtain the initial data for A. In Sect. 4 we differentiate A along geodesic path and thenintegrate it to obtain a reformulation. We also get some estimates for the reformulation inthis section. The proof of Theorem 1 will be complete in Sect. 5.

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2 The MTW condition

Given x ∈ M, ξ, η, p ∈ TxM, let

(x, p) (ξ, η) := 1

2

d2

ds2

d2

dt2 d2 (expx (tξ) , expx (− (p + sη))

) |t=0,s=0 . (2.1)

It is known that (see [25,21])

MT W kli j (x, y) ξ iξ j ηk ηl = − (x, p) (ξ, η) ,

provided

y = expx (−p) and ηi = ci,k ηk .

Hence the MTW condition (1.6) is equivalent to

(x, p) (ξ, η) < −C0 |ξ |2 |η|2 ∀ξ⊥η. (2.2)

Recall that (see [21])

d2 (expx (tξ) , expx (−sη)

) = |ξ |2 t2 + |η|2 s2 − 2st |η| |ξ | cos θ

− K

3s2t2 |η|2 |ξ |2 sin2 θ + O

((s2 + t2)5/2

)(2.3)

where θ is the angle between vectors ξ and −η, and K = Sectx (ξ, η) is the sectionalcurvature. We conclude

(x, 0) (ξ, η) = −2

3K |ξ |2 |η|2 sin2 θ.

Therefore positive sectional curvature condition is necessary for MTW condition.If p �= 0, the situation becomes complicated. Inspired from [9–13] where authors repre-

sented the second derivatives of geodesic distance by quantities involving Jacobi fields, inthis paper, we relate (2.1) to the second fundamental form of the geodesic sphere. We needthe proposition below.

Proposition 1 Let d (x, y0) be the geodesic distance between point x and a fixed point y0.Assume x lies in the injectivity domain of y0, and γ : [0, 1] → M is the geodesic segmentstarting from y0 and ending at x. Then for any two vector fields X and Y , we have

Xd (x, y0) =⟨

X (x) ,γ

|γ | (1)

⟩

and

∇2X,Y d (x, y0) = ∂ Bρ(y0;x) (P X, PY )

where ∂ Bρ(y0;x) is the second fundamental form of the geodesic sphere ∂ Bρ (y0) , ρ =d (x, y0) and P = P∂ Bρ(y0;x) is the projection of TxM onto Tx∂ Bρ (y0).

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S.-Z. Du, Q.-R. Li

Proof The first formula is standard and can be found in [7]. Here we prove the second one.It is easy to see

X (Y d (x, y0)) = X

⟨Y,

γ

|γ |⟩

=⟨∇X Y,

γ

|γ |⟩+

⟨Y,∇X

(γ

|γ |)⟩

= (∇X Y ) d (x, y0) +⟨Y,∇X

(γ

|γ |)⟩

.

By the decomposition

X = P X + X⊥e0 and Y = PY + Y ⊥e0

where e0 = γ|γ | is the normal of ∂ Bρ (y0), we have

⟨Y,∇X

(γ

|γ |)⟩

=⟨PY + Y ⊥e0,∇P X+X⊥e0

e0

⟩

=⟨PY + Y ⊥e0,∇P X e0

⟩

= 〈PY,∇P X e0〉= ∂ Bρ(y0;x) (P X, PY ) .

The second formula follows by

∇2X,Y d (x, y0) = XY d (x, y0) − (∇X Y ) d (x, y0) .

��Let us set

ps = p + sη, ys = expx (−ps) , ρs = |ps | = d (x, ys) , (2.4)

and

Ps = P∂ Bρs (ys ;x).

By virtue of Proposition 1, we can make the following calculation

d2

dt2 |t=0d2 (expx (tξ) , ys

) = 2

(d

dt|t=0d

(expx (tξ) , ys

))2

+2ρsd2

dt2 |t=0d(expx (tξ) , ys

)

= 2

(⟨ξ,

ps

ρs

⟩)2

+ 2ρs ∂ Bρs (ys ;x) (Psξ, Psξ) .

Hence (2.1) can be rewritten as

(x, p) (ξ, η) = d2

ds2 |s=0

{〈ξ, ps〉2

ρ2s

+ ρs ∂ Bρs (ys ;x) (Psξ, Psξ)

}. (2.5)

To further process, we separate the discussion into two subsections below. It is clear from(2.5) that (x, p) (ξ, η) involves the derivatives of second fundamental form of geodesicsphere. To verify MTW condition, we first show the second fundamental form of geodesic

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sphere satisfies a Riccati type equation (2.13). This is standard, and can be found in e.g.[26]. To make the content more readable, we include it in the first subsection. In the secondsubsection, we introduce our key quantity A(ξ,η) (r) along geodesic segment joining y =expx (−p) and x . The MTW condition is then equivalent to the negativity of A(ξ,η) at r = |p|.2.1 Second fundamental form of geodesic spheres

Suppose dim M = n + 1. We choose an orthonormal frame {eα}nα=0 ∈ TxM such that

e0 = p

ρ, η = η0e0 + η1e1 and ξ = ξ0e0 + ξ1e1 + ξ2e2. (2.6)

We also introduce orthonormal basis {eα (ϕ)} with parameter ϕ, given by

(e0 (ϕ) , . . . , en (ϕ)) = (e0, . . . , en) Q (ϕ) ,

where

Q (ϕ) =(

B (ϕ) 00 In−1

)with B (ϕ) :=

(cos ϕ − sin ϕ

sin ϕ cos ϕ

). (2.7)

By parallel translating {eα (ϕ)} along geodesic expx (−te0 (ϕ)), we can define an orthonormalframe

{eα (t, ϕ)

}in the neighborhood of x .

Let ϕ (s) be the solution of the equation

tan ϕ (s) = sη1

ρ + sη0with s ∈ (−ε, ε) (2.8)

so that

e0 (ϕ (s)) = ps

ρs.

We introduce a family of geodesics γ (τ, s) : [0, 1] × (−ε, ε) → M, defined by

γ (τ, s) := expx ((1 − τ)(−ps)) .

For fixed s, γs (τ ) := γ (τ, s) is the geodesic segment starting from ys and ending at x . Recallwe have introduced the frame

{eα (t, ϕ)

}. By restricting this frame to the normal geodesic

γs (r) := γ(rρ−1

s , s), we define Eα (r, s) = eα (ρs − r, ϕs). Then we obtain n Jacobi fields

�i (r, s) along γs (r) , i = 1, ..., n, given by⎧⎨⎩

∇2E0

�i = Rm (E0,�i ) E0,

�i (0, s) = 0,

∇E0�i (0, s) = Ei (0, s) .

(2.9)

Suppose

�i (r, s) =n∑

α=0

�αi (r, s) Eα (r, s) .

It is clear, by using (2.9),

�0i ≡ 0

and {�

ji + S jk�

ki = 0,

�ji (0, s) = 0, �

ji (0, s) = δ

ji ,

(2.10)

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S.-Z. Du, Q.-R. Li

where Si j = Rm(E0, Ei , E0, E j

)takes value at γs (r), and we use a dot to designate the

derivatives w.r.t. E0 = 1ρs

ddτ

= ddr .

Since every Jacobi field can be viewed as a variation of a family of geodesics, we get

∇E0�i = ∇�i E0 for i = 1, . . . , n.

Hence

�ki kl�

lj = ⟨∇E0�i ,� j

⟩ =n∑

k=1

�ki �

kj , (2.11)

where i j (r, s) := ⟨∇Ei E0, E j⟩

is the second fundamental form of geodesic spheres

∂ Br (ys). Let us use � to denote the matrix(�i

j

)1≤i, j≤n

and denote the inverse of � by �−1

[before touching conjugate point, one may always assume � (r, s) is invertible for r > 0].Set

U = ��−1.

Then (2.11) can be rewritten as

i j = Ui j . (2.12)

It is easy to see from (2.10) that U satisfies Riccati equation:

U + U 2 + S = 0. (2.13)

We remark that U is symmetric and U is not defined at r = 0.

Remark 2 In the case of surfaces (i.e., n = 1), S and U are not matrices but functions. Letus use K and H to denote the Gauss curvature of the surface and the mean curvature of thecorresponding geodesic sphere respectively. Let θ be the function satisfying{

θ (r, s) + K (γs (r)) θ (r, s) = 0,

θ (0, s) = 0, θ (0, s) = 1.

Then S = K , U = H = θ/θ . Hence it follows from (2.13)

H + H2 + K = 0. (2.14)

2.2 The quantity A (r)

For fixed s, let ζs (r) := ζ (r, s) be a vector field defined on γs which satisfies ∇ .

γsζs = 0 and

ζs (ρs) = Psξ . Recall Ps denotes the projection of TxM onto Tx∂ Bρs (ys). With notationbefore, it is easy to see

Psξ = (e1 (ϕ (s)) , . . . , en (ϕ (s))) R(s)

where R(s) is a column vector given by

R (s) = (ξ1 cos ϕ(s) − ξ0 sin ϕ(s), ξ2, 0, . . . , 0)T . (2.15)

By (2.12), the term ∂ Bρs (ys ;x) (Psξ, Psξ) in (2.5) can be rewritten as

∂ Bρs (ys ;x) (Psξ, Psξ) = ∂ Bρs (ys ;x) (ζs, ζs)

= RT (s) U (ρs, s) R (s). (2.16)

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On the other hand, by ξ⊥η, it is clear

〈ξ, ps〉2 = ρ2ξ20 . (2.17)

So we obtain, by plugging (2.16) and (2.17) into (2.5),

(x, p) (ξ, η) = d2

ds2 |s=0

{ρ2

ρ2sξ2

0 + ρs RT (s) U (ρs, s) R (s)

}. (2.18)

We now introduce the quantity A(ξ,η), which plays a key role in our argument,

A(ξ,η)(r) := d2

ds2 |s=0

{ρ3

rρ2sξ2

0 + ρs RT (s) U

(r

ρρs, s

)R(s)

}where r ∈ (0, ρ]. (2.19)

It is easy to see

(x, p) (ξ, η) = A(ξ,η) (ρ) . (2.20)

Hence verification of the MTW condition is reduced to study the behavior of A(ξ,η) (r).Let us use a dot to designate the partial derivative ∂

∂r and a prime to designate the partialderivative ∂

∂s . For convenience, we omit subscript (ξ, η) and denote A(ξ,η) (r) by A (r) forsimplicity. In view of (2.19), straightforward calculation yields

A(r) =(

−2ξ20

r+ RT U R + r RT U R

)d2ρs

ds2 |s=0

+(

6ξ20

rρ+ 2r

ρRT U R + r2

ρRT U R

) (dρs

ds|s=0

)2

+2

((RT U R

)′ + r(

RT U R)′) dρs

ds|s=0 + ρ

(RT U R

)′′. (2.21)

Furthermore, by virtue of (2.4) and (2.8), one can check that

dϕ

ds|s=0 = η1

ρ,

d2ϕ

ds2 |s=0 = −2η0η1

ρ2 , (2.22)

and

dρs

ds|s=0 = η0,

d2ρs

ds2 |s=0 = η21

ρ. (2.23)

Plugging (2.22) and (2.23) into (2.21), we obtain

A (r) =(

−2ξ20

ρr+ 1

ρRT U R + r

ρRT U R

)η2

1

+(

6ξ20

rρ+ 2r

ρRT U R + r2

ρRT U R

)η2

0

+2

((RT U R

)′ + r(

RT U R)′)

η0 + ρ(

RT U R)′′

. (2.24)

Our expression (2.24), as well as the expression (4.1)–(4.4) for two dimensional case below,is analogous to the ones obtained in [10,12,13].

To verify the MTW condition, we need to study the behavior of A (r). Our strategy is tointegrate the differential identity satisfied by A (r) along geodesic path. In next section wewill first study its initial value.

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S.-Z. Du, Q.-R. Li

3 Initial value of A (r)

It seems from (2.24) that limr→0 A (r) might not exist. In this section, we show that singularterms appearing in (2.24) kill each other and the remaining terms make limr→0 A (r) = 0.To achieve this, one should first understand the asymptotic behavior of second fundamentalform of geodesic sphere.

By repeatedly using (2.10), we have for fixed s ∈ (−ε, ε)

�ki (0, s) = 0,

...�

ki (0, s) = −Ski ,

....�

ki (0, s) = −2∇0Ski ,(

�(5))k

i(0, s) = −3∇2

0 Ski + Skp Spi ,

and consequently

�(r, s) = I r − S

6r3 − ∇0S

12r4 + S2 − 3∇2

0 S

120r5 + O

(r6)

. (3.1)

Here S, ∇0S and ∇20 S are all calculated at ys . We deduce for all r > 0

r�−1 =(

�

r

)−1

= I + S

6r2 + ∇0S

12r3 + 7S2 + 9∇2

0 S

360r4 + O

(r5

). (3.2)

Combining (3.1) and (3.2), one gets for all r > 0

rU = I − S

3r2 − ∇0S

4r3 −

(S2

45+ ∇2

0 S

10

)r4 + O

(r5

). (3.3)

Summarizing the formulae above, we have the following proposition:

Proposition 2 Let H (r, s) denote the mean curvature of ∂ Br (ys) at the point γs (r). Then

r H (r, s) = n − Ric0

3r2 − ∇0 Ric0

4r3

−(

|S|245

+ ∇20 Ric0

10

)r4 + O

(r5

),

where Ric0 = Ric (E0, E0) is Ricci curvature and all the quantities on R.H.S. are computedat ys .

Moreover, given u ∈ Tγs (r)∂ Br (ys), we have

r (u, u) = g (u, u) − 1

3Su,ur2 − 1

4∇0Su,ur3

−(

1

45S2

u + 1

10∇2

0 Su,u

)r4 + O

(r5

),

where u ∈ Tys M is obtained by parallel translating u along γs , Su,u = Rm (E0, u, E0, u) ,

S2u = 〈Rm (E0, u, E0, ·) , Rm (E0, u, E0, ·)〉 and all the quantities on R.H.S. are computed

at ys .

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Combining (3.3) with (2.15), (2.22) and (2.23), we have(−2ξ2

0

ρr+ 1

ρRT U R + r

ρRT U R

)η2

1 = −2ξ20 η2

1

ρr+ O (r) ,

(6ξ2

0

rρ+ 2r

ρRT U R + r2

ρRT U R

)η2

0 = 6ξ20 η2

0

rρ+ O (r) ,

and

2

((RT U R

)′ + r(

RT U R)′)

η0 + ρ(

RT U R)′′

= ρ

r

d2

ds|s=0 (ξ1 cos ϕs − ξ0 sin ϕs)

2 + O(r)

= 2

ρr

{−ξ21 η2

1 + 2ξ0ξ1η0η1 + ξ20 η2

1

} + O(r).

By virtue of ξ⊥η, it follows from (2.24) that

limr→0

A (r) = 0. (3.4)

4 Verification of the MTW condition on surfaces

To find proper curvature condition, we check the sign of A (ρ). We have obtainedlimr→0 A (r) = 0 in previous section. In this section, we first derive the formula of d

dr A (r),i.e. (4.17), in the case of M is a surface, and then get a reformulation of A (r), namely(4.21), by integrating d

dr A (r) with its initial value (3.4). As mentioned in Remark 2, whendim M = 2, S, U are both functions, instead of matrices. Let us use K and H to denotethe Gauss curvature of the surface and mean curvature of corresponding geodesic sphererespectively, then S = K and U = H .

Throughout the remaining paper, we always assume ξ⊥η which gives

ξ0η0 = −ξ1η1.

When considering surfaces, in view of (2.15), (2.22) and (2.23), (2.24) can be rewritten as

A (r) = a00 (r) ξ20 + 2a01 (r) ξ0ξ1 + a11 (r) ξ2

1 (4.1)

where

a00 (r) = 6η20

ρr− 2η2

1

ρ

(1

r− H

), (4.2)

a01 (r) = −2η0η1

ρr H − 2η1 H ′, (4.3)

a11 (r) = −η21

ρH + η2

1

ρr H + 2η2

0

ρr H

+r2η20

ρH + 2rη0 H ′ + 2η0 H ′ + ρH ′′. (4.4)

123

S.-Z. Du, Q.-R. Li

Differentiating (4.1), we obtain

d

drA (r) = b00 (r) ξ2

0 + 2b01 (r) ξ0ξ1 + b11 (r) ξ21 ,

where

b00 = −6η20

ρr2 + 2η21

ρ

(1

r2 + H

), (4.5)

b01 = −2η0η1

ρH − 2η0η1

ρr H − 2η1 H ′ (4.6)

and

b11 = 2η20

ρH + 4η2

0 + η21

ρr H + 4η0 H ′

+r2η20

ρ

...H + 2rη0 H ′ + ρ H ′′

= −ρK ′′ + 2η20

ρH + 4η2

0 + η21

ρr H + 4η0 H ′

+r2η20

ρ

...H + 2rη0 H ′ − 2ρH ′2 − 2ρH H ′′. (4.7)

In the second equality of (4.7), we employ Ricatti equation (2.14).We next process the last term on the R.H.S. of (4.7). By virtue of (4.4), we have

− 2ρξ21 H H ′′ = 2ξ2

1 H

(−a11 − η2

1

ρH + η2

1

ρr H

+2η20

ρr H + r2η2

0

ρH + 2rη0 H ′ + 2η0 H ′

)

= 2H(−A + a00ξ

20 + 2a01ξ0ξ1

)

+2ξ21 H

(−η2

1

ρH + η2

1

ρr H + 2η2

0

ρr H

+r2η20

ρH + 2rη0 H ′ + 2η0 H ′

)

= −2HA + 12ξ20 η2

0

ρrH − 2ξ2

1 η21

ρH2

−4ξ20 η2

1

ρ

(1

r− H

)H − 8ξ0ξ1η0η1

ρr H H

+2ξ21 η2

1

ρr H H + 4ξ2

1 η20

ρr H H − 8ξ0ξ1η1 H H ′

+2ξ21 η2

0

ρr2 H H + 4ξ2

1 η0r H H ′ + 4ξ21 η0 H H ′. (4.8)

123


It follows that

b11ξ21 = −2HA − ρξ2

1 K ′′ + 12ξ20 η2

0

ρrH − 2ξ2

1 η21

ρH2

−4ξ20 η2

1

ρ

(1

r− H

)H + 2ξ2

1 η20

ρH

+(

2ξ21 η2

1

ρ+ 4ξ2

1 η20

ρ− 8ξ0ξ1η0η1

ρ

)r H H

−8ξ0ξ1η1 H H ′ + 4ξ21 η0 H H ′ + 4ξ2

1 η20 + ξ2

1 η21

ρr H

+2ξ21 η2

0

ρr2 H H + 4ξ2

1 η0 H ′ + 4ξ21 η0r H H ′

−2ρξ21 H ′2 + ξ2

1 η20

ρr2 ...

H + 2ξ21 η0r H ′. (4.9)

As a consequence of (4.5), (4.6) and (4.9), one obtains

dAdr

= −2HA − ρξ21 K ′′ − 6ξ2

0 η20

ρr2 + 2ξ20 η2

1

ρr2

+12ξ20 η2

0

ρrH − 2ξ2

1 η21

ρH2 − 4ξ2

0 η21

ρ

(1

r− H

)H

+(

2ξ21 η2

0

ρ+ 2ξ2

0 η21

ρ− 4ξ0ξ1η0η1

ρ

)H

+(

2ξ21 η2

1

ρ+ 4ξ2

1 η20

ρ− 8ξ0ξ1η0η1

ρ

)r H H

+ (4ξ2

1 η0 − 8ξ0ξ1η1)

H H ′ − 2ρξ21 H ′2

+(

4ξ21 η2

0

ρ+ ξ2

1 η21

ρ− 4ξ0ξ1η0η1

ρ

)r H

+2ξ21 η2

0

ρr2 H H + (

4ξ21 η0 − 4ξ0ξ1η1

)H ′

+4ξ21 η0r H H ′ + ξ2

1 η20

ρr2 ...

H + 2ξ21 η0r H ′. (4.10)

By virtue of (2.14), we have

H = −K − H2, (4.11)

H = −K − 2H H , (4.12)

and

...H = −K + 2H K + 4H2 H − 2H2. (4.13)

123

S.-Z. Du, Q.-R. Li

By plugging identities above into (4.10), one concludes

dAdr

= −2HA − ρξ21 K ′′ − ξ2

1 η20

ρr2 K

−(

4ξ21 η2

0

ρ+ 5ξ2

0 η20

ρ

)r K −

(2ξ2

1 η20

ρ+ 2ξ2

0 η21

ρ+ 4ξ2

0 η20

ρ

)K

+(

2ξ20 η2

1

ρ− 6ξ2

0 η20

ρ

) (1

r− H

)2

− 2ξ21 η2

0

ρr2

(H

r+ H

)2

+ (4ξ2

1 η0 + 8ξ20 η0

)H H ′ − 2ρξ2

1 H ′2 + 2ξ21 η0r H ′

+ (4ξ2

1 η0 + 4ξ20 η0 + 4ξ2

1 η0r H)

H ′. (4.14)

Let A (r) denote the terms in the last two rows on R.H.S. of (4.14). That is

A (r) = (4ξ2

1 η0 + 8ξ20 η0

)H H ′ − 2ρξ2

1 H ′2 + 2ξ21 η0r H ′

+ (4ξ2

1 η0 + 4ξ20 η0 + 4ξ2

1 η0r H)

H ′.

Using (4.11), (4.12) and (4.13) again, one gets

A (r) = −2ξ21 η0r K ′ + (

4ξ21 η0 + 8ξ2

0 η0)

H H ′ − 2ρξ21 H ′2

= −2ξ21 η0r K ′ − (

4ξ21 η0 + 4ξ2

0 η0)

K ′

−2ρξ21

[2η0r

ρ

(H

r+ H

)H ′ + H ′2

]

= −2ξ21 η0r K ′ − (

4ξ21 η0 + 4ξ2

0 η0)

K ′

+2ξ21 η2

0

ρr2

(H

r+ H

)2

−2ρξ21

[η0r

ρ

(H

r+ H

)+ H ′

]2

. (4.15)

Combining (4.15) and (4.14), we have

dAdr

= −2HA − ρξ21 K ′′ − ξ2

1 η20

ρr2 K − 2ξ2

1 η0r K ′

−(

4ξ21 η2

0

ρ+ 5ξ2

0 η20

ρ

)r K − (

4ξ21 η0 + 4ξ2

0 η0)

K ′

−(

2ξ21 η2

0

ρ+ 2ξ2

0 η21

ρ+ 4ξ2

0 η20

ρ

)K

+(

2ξ20 η2

1

ρ− 6ξ2

0 η20

ρ

) (1

r− H

)2

−2ρξ21

[η0r

ρ

(H

r+ H

)+ H ′

]2

. (4.16)

Without loss of generality, let us assume, by virtue of ξ⊥η,

ξ = cos αe0 + sin αe1, η = − sin αe0 + cos αe1.

123


Thus ξ and η correspond to an angle α. We set

Aα (r) := A(ξ,η) (r) .

Employing (4.16), one immediately obtains

dAα

dr= −2HAα + 1

ρIα (r) − 2

ρK

+2χα

ρ

(1

r− H

)2

− 2

ρsin2 α I Iα (r) . (4.17)

where

χα = cos2 α(1 − 4 sin2 α

),

Iα (r) = −ρ2 sin2 αK ′′ − sin4 αr2 K + 2 sin3 αrρ K ′

− sin2 α(4 + cos2 α

)r K + 4 sin αρK ′ (4.18)

and

I Iα (r) =[ρH ′ + sin α

(r K − r H

(1

r− H

))]2

. (4.19)

To analyze Aα (ρ), we will integrate (4.17). First of all, let us introduce B (r) and Cα (r),which are given by

{dBdr = −2HB − 2

ρK + 2

ρ

( 1r − H

)2,

limr→0 B (r) = 0,

and {dCα

dr = −2HCα + Iαρ

− 2ωα

ρ

( 1r − H

)2 − 2ρ

sin2 α I Iα,

limr→0 Cα (r) = 0,

where

ωα := 1 − χα = sin2 α(1 + 4 cos2 α).

Thus one gets

Aα(r) = B(r) + Cα(r).

Recall Remark 2. We have

H(t) = θ

θ(t)fort > 0,

with θ(t) satisfying{

θ + K θ = 0,

θ(0) = 0, θ (0) = 1.

The following lemmas can be easily established.

123

S.-Z. Du, Q.-R. Li

Lemma 1 We have

B(r) = − 2

ρ

(1

r− H

)

and

Cα(r) = 1

ρθ2(r)

r∫

0

θ2 Iαdt − 2 sin2 α

ρθ2(r)

r∫

0

θ2 I Iαdt

− 2ωα

ρθ2(r)

r∫

0

K θ2dt + 2ωα

ρ

(1

r− H

).

Proof Apparently we have

r∫

ε

H(t)dt =r∫

ε

θ

θdt = ln

θ(r)

θ(ε).

Hence

B(r) = limε→0

e−2∫ rε Hdt

⎧⎨⎩B(ε) +

r∫

ε

[− 2

ρK + 2

ρ

(1

t− H

)2]

e2∫ tε Hdτ dt

⎫⎬⎭

= − 2

ρθ2(r)

r∫

0

K θ2(t)dt + 2

ρθ2(r)

r∫

0

(θ

t− θ

)2

dt

= − 2

ρ

(1

r− H

)

and

Cα(r) = limε→0

e−2∫ rε Hdt

⎧⎨⎩Cα(ε) +

r∫

ε

[Iαρ

− 2 sin2 α

ρI Iα − 2ωα

ρ

(1

t− H

)2]

e2∫ tε Hdτ dt

⎫⎬⎭

= 1

ρθ2(r)

r∫

0

[θ2 Iα − 2 sin2 αθ2 I Iα − 2ωα

(θ

t− θ

)2]

dt

= 1

ρθ2(r)

r∫

0


ρθ2(r)

r∫

0

θ2 I Iαdt

− 2ωα

ρθ2(r)

r∫

0

K θ2dt + 2ωα

ρ

(1

r− H

).

In deriving the formulae above, we have used the integral identity

r∫

0

(θ

t− θ

)2

dt =r∫

0

K θ2dt − θ

(θ

r− θ

). (4.20)

��

123


Lemma 2 We have

1

r− H = 1

rθ

r∫

0

tθ K dt

and

θ (r) = 1 −r∫

0

θ K dt.

Proof To prove this lemma, it suffices to notice the identities

θ − r θ =r∫

0

d

dt

(θ − t θ

)dt =

r∫

0

tθ K dt

and

θ (r) = θ (0) +r∫

0

θ (t)dt = 1 −r∫

0

θ K dt.

��With the help of Lemmas 1 and 2, we obtain

ρθ2(r)Aα(r) =r∫

0


r∫

0

θ2 I Iαdt

−2ωα

r∫

0

K θ2dt − 2χα

θ(r)

r

r∫

0

tθ K dt. (4.21)

To estimate Iα , we need the following lemma.

Lemma 3 The function θt is non-increasing on (0, t0] if θ(t) ≥ 0 on the same interval (0, t0].

Consequently θ ≤ t holds on [0, t0].Proof Let f = θ

t and g = f − θ . We know θ(t) = t + O(t3). It is clear that{

f = − 1t g,

limt→0 f (t) = 1,

and {g = − 1

t g + K θ,

limt→0 g(t) = 0.

We thus conclude that, by noticing that θ(t) is nonnegative, g ≥ 0. This implies f is non-increasing. Furthermore, the initial value of f yields

θ

t≤ lim

t→0

θ

t= 1.

��

123

S.-Z. Du, Q.-R. Li

Lemma 4 One has

Iα(t) ≤ sin2 αpα(t) + 4 |sin α| q(t),

where

pα(t) = (ρ − t)2∣∣∇2 K

∣∣ (t) +⎡⎣cos2 α

β0(ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ (1 + |sin α|) |cos α|ρ∫

t

(ρ − τ) K (τ ) dτ + 2 |cos α|⎤⎦ (ρ − t) |∇K | (t) ,

q(t) = (ρ − t) |∇K | (t),

β0 = sin(√

KM L)

√KM L

,

and∣∣∇2 K

∣∣ (t), |∇K | (t) and K (t) take value at expy tex−y .

Proof Let us first define a geodesic spherical coordinate system {�, ϑ} at x , given byz (�, ϑ) = expx (�e0 (ϑ)), where

e0 (ϑ) = e0 cos ϑ + e1 sin ϑ,

e1 (ϑ) = −e0 sin ϑ + e1 cos ϑ,

and ei = −ei , e0, e1 ∈ TxM was introduced in Sect. 2.1. We also adopt ei (�, ϑ) to denotethe vector obtained by parallel translating ei (ϑ) along geodesic (�, ϑ).

It is not hard to see, in the coordinate {�, ϑ} , γ (r, s) can be represented as

γ (r, s) = (ρs − r, ϕ (s)) . (4.22)

In the sequel, let γ 0 = �−1◦ γ , γ 1 = ϑ−1◦ γ , and{�k

i j

}0≤i, j,k≤1

denote Christoffel symbols

under coordinate {�, ϑ}. It is readily to check that

�00i = �i

00 = 0,

�110 = H and �0

11 = −H θ2,

where H (t, ϕ) stands for the mean curvature of the geodesic sphere ∂ Bt (x), taking value atthe point (t, ϕ), and

θ (t, ϕ) := |∂ϑ (t, ϕ)| .Set J (r) := γ∗

( dds |s=0

). Then, in the local coordinate {�, ϑ}, we have, by employing

(4.22) (2.22) and (2.23),

J (r) = dρs

ds∂� + dϕ

ds∂ϑ

= sin αE0 + cos α

ρ∂ϑ, (4.23)

123


∇E0 J (r) = −∇∂�

(∂s γ

0∂� + ∂s γ1∂ϑ

)= −∂�∂s γ

0∂� − (∂�∂s γ

1 + ∂s γ1�1

01

)∂ϑ

= −cos α

ρH∂ϑ , (4.24)

and

∇J J (r) = ∇γ∗

(dds |s=0

)(

dρs

ds∂� + dϕ

ds∂ϑ

)

=[

d2ρs

ds2 +(

dϕ

ds

)2

�011

]∂�

+[

d2ϕ

ds2 + 2dϕ

ds

dρs

ds�1

01 +(

dϕ

ds

)2

�111

]∂ϑ

= −cos2 α

ρ2

(ρ − H |∂ϑ |2

)E0

+[

2 sin α cos α

ρ2

(1 − ρ H

)+ cos2 α

ρ2 �111

]∂ϑ . (4.25)

where E0, ∂ϑ , H take value at (ρ − r, 0).Since

K ′ = 〈∇K , J 〉 ,

K ′′ = ∇2 K (J, J ) + 〈∇K ,∇J J 〉and

K ′ = ∇2 K (J, E0) + ⟨∇K ,∇E0 J⟩,

by virtue of (4.23), (4.24), (4.25) and (4.18), we have

Iα (r) = − sin2 α∇2 K (ρ J − r sin αE0, ρ J − r sin αE0) + 4 sin α 〈∇K , ρ J − r sin αE0〉+2 sin3 αrρ

⟨∇K ,∇E0 J⟩ − sin2 α cos2 αr 〈∇K , E0〉 − ρ2 sin2 α 〈∇K ,∇J J 〉

= − sin2 α∇2 K (ρ J − r sin αE0, ρ J − r sin αE0) + 4 sin α 〈∇K , ρ J − r sin αE0〉+ sin2 α cos2 α

(ρ − r − H |∂ϑ |2

)〈∇K , E0〉

+(

2 sin3 α cos α (ρ − r) H − 2 sin3 α cos α − sin2 α cos2 α�111

)〈∇K , ∂ϑ 〉

= − sin2 α∇2 K (ρ J − r sin αE0, ρ J − r sin αE0) + 4 sin α 〈∇K , ρ J − r sin αE0〉+ sin2 α cos α (ρ − r)

⟨∇K , cos αE0 − sin α

∂ϑ

ρ − r

⟩

+ sin2 α cos α (ρ − r) H |∂ϑ |⟨∇K , sin α

∂ϑ

|∂ϑ | − |∂ϑ |ρ − r

cos αE0

⟩

− sin3 α cos α⟨∇K ,

(1 − (ρ − r) H

)∂ϑ

⟩− sin2 α cos2 α

⟨∇K , �111∂ϑ

⟩.

Letting

X = sin αE0 + cos α∂ϑ

ρ − r, Y = cos αE0 − sin α

∂ϑ

ρ − r,

123

S.-Z. Du, Q.-R. Li

Z = − |∂ϑ |ρ − r

cos αE0 + sin α∂ϑ

|∂ϑ | , W =(

1 − (ρ − r) H)

∂ϑ ,

we then obtain

Iα (r) = − sin2 α (ρ − r)2 ∇2 K (X, X) + 4 sin α (ρ − r) 〈∇K , X〉+ sin2 α cos α (ρ − r) 〈∇K , Y 〉+ sin2 α cos α (ρ − r) H |∂ϑ | 〈∇K , Z〉− sin3 α cos α 〈∇K , W 〉 − sin2 α cos2 α

⟨∇K , �111∂ϑ

⟩. (4.26)

The quantities on the R.H.S. of (4.26) are all calculated at (ρ − r, 0).For fixed ϕ, ∂ϑ(t, ϕ) is a Jacobi field along each geodesic ray (t, ϕ) emanating from x ,

with the initial condition limt→0 ∂ϑ(t, ϕ) = 0 and limt→0 ∇∂� ∂ϑ (t, ϕ) = e1(ϕ). We concludefrom Jacobi equation and Rauch comparison theorem that

∂ϑ (t, ϕ) = θ (t, ϕ) e1 (t, ϕ) and θ ≤ t. (4.27)

Consequently,

|X | ≤ 1, |Y | ≤ 1 and |Z | ≤ 1.

On the other hand, it is easy to see

1

2∂� |∂ϑ |2 = ⟨∇∂� ∂ϑ , ∂ϑ

⟩ = H |∂ϑ |2

which gives H = ∂�θ

θ. Since θ (0, ϕ) = 0 ∂�θ (0, ϕ) = 1 for all ϕ, by applying Lemma 2 to

θ , we have

∣∣∣H∣∣∣ |∂ϑ | ≤ 1 +

ρ−r∫

0

t K(expx (−te0)

)dt

= 1 +ρ∫

r

(ρ − τ) K(expx (− (ρ − τ) e0)

)dτ (4.28)

and

|W | ≤ρ−r∫

0

t2 K(expx (−te0)

)dt

≤ (ρ − r)

ρ∫

r

(ρ − τ) K(expx (− (ρ − τ) e0)

)dτ. (4.29)

Plugging (4.28) and (4.29) into (4.26), we reach the inequality

Iα (r) ≤ sin2 α (ρ − r)2∣∣∇2 K

∣∣ + 4 |sin α| (ρ − r) |∇K |+ sin2 α cos2 α |∇K | ∣∣�1

11

∣∣ θ + 2 sin2 α |cos α| (ρ − r) |∇K |

+ sin2 α (1 + |sin α|) |cos α|⎛⎝

ρ∫

r

(ρ − τ) K (τ ) dτ

⎞⎠ (ρ − r) |∇K | . (4.30)

123


We next estimate �111θ . By (4.27 ), it follows

�111 = |∂ϑ |−2 ⟨∇∂ϑ ∂ϑ , ∂ϑ

⟩

= e1θ + θ

2e1 〈e1, e1〉

= e1θ .

The last equality is due to 〈e1, e1〉 ≡ 1. On the other hand, we have

d

dt

(e1θ

)= e1e0θ + [

e0, e1]θ

= e1

(H θ

)+ ⟨∇e0 e1 − ∇e1 e0, e0

⟩e0θ

+ ⟨∇e0 e1 − ∇e1 e0, e1⟩e1θ

= e1

(H θ

)− ⟨∇e1 e0, e1

⟩e1θ

= e1

(H θ

)− H e1θ

= θG, (4.31)

where G := e1 H . In the third equality above, we employ⟨∇ei e j , e j

⟩ = 12 ei

∣∣e j∣∣2 = 0

and⟨∇e0 e1, e0

⟩ = e0 〈e1, e0〉 = 0. Noticing limt→0 e1θ = limt→0∂ϑ θ

θ= 0 [see (3.1)], we

integrate (4.31) and reach

�111 = e1θ =

t∫

0

Fdτ, (4.32)

where

F = θG.

Recall H satisfies the Ricatti equation, say

e0 H = −H2 − K .

Similar as in (4.31), one then concludes

d

dtG = e1e0 H + [

e0, e1]

H

= e1

(−H2 − K

)− H e1 H

= −3H G − e1 K .

Consequently

d

dtF = G

d

dtθ + θ

d

dtG

= G H θ + θ(−3H G − e1 K

)

= −2H F − θ e1 K . (4.33)

123

S.-Z. Du, Q.-R. Li

It is easy to see limt→0 F = limt→0 ∂ϑ H = 0. Solving (4.33), we get

F = limε→0

e−2∫ tε Hdτ

⎧⎨⎩F |t=ε −

t∫

ε

θ (e1 K ) e2∫ τε Hdξ dτ

⎫⎬⎭

= − 1

θ2

t∫

0

θ3 (e1 K ) dτ.

Plugging identity above into (4.32), one obtains

θ∣∣�1

11

∣∣ ≤ θ

t∫

0

1

θ2

τ∫

0

θ3 |∇K | dζdτ

≤ θ

⎛⎝

t∫

0

τ 2

θ2dτ

⎞⎠

⎛⎝

t∫

0

τ |∇K | dτ

⎞⎠

≤ t3

θ

t∫

0

τ |∇K | dτ

≤ t2

β0

t∫

0

τ |∇K | dτ.

The third inequality is derived by Lemma 3 (i.e., τ 2/θ2 is increasing.) and the fourth one isdue to Sturm’s comparison theorem (see [6]). Consequently, taking t = ρ − r , we get

θ∣∣�1

11

∣∣ ≤ (ρ − r)2

β0

ρ∫

r

(ρ − τ) |∇K | (τ ) dτ. (4.34)

The proof is complete by combining (4.34) and (4.30). ��

5 Proof of main result

In this section we prove our main result Theorem 1. It is sufficient to show Aα (ρ) < 0 forall α ∈ [0, π). Employing Lemma 4 and noticing

−2 sin2 α

ρθ2 (r)

r∫

0

θ2 I Iαdt ≤ 0,

(4.21) can be estimated in the form of

ρθ2 (ρ) Aα (ρ) ≤ sin2 α

ρ∫

0

θ2 pαdt − 2ωα

ρ∫

0

K θ2dt

+4 |sin α|ρ∫

0

θ2qdt − 2χα

θ (ρ)

ρ

ρ∫

0

tθ K dt. (5.1)

123


Recall that

χα = cos2 α(1 − 4 sin2 α

)and ωα = sin2 α

(1 + 4 cos2 α

).

We complete the proof of our main result by analyzing (5.1) case by case. Theorem 1 is adirect consequence of the following proposition.

Proposition 3 Suppose M is a compact connected surface with positive Gauss curvatureK . Let KM = maxM K , and L be any constant less than π√

KM. For all x, y ∈ M satisfying

y ∈ Ix and ρ = d (x, y) ≤ L, if the inequality

2K (t) ≥ (ρ − t)2∣∣∇2 K

∣∣ (t) +⎡⎣a (ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ b

ρ∫

t

(ρ − τ) K (τ ) dτ + c

⎤⎦ (ρ − t) |∇K | (t) (5.2)

holds for all t ∈ [0, ρ], then

MT W kli j (x, y) ξ iξ jηkηl ≥ 0, ∀η · ξ = 0. (5.3)

Furthermore, if the inequality (5.2) is strict, then (5.3) is strictly positive. Here K (t), |∇K | (t)and

∣∣∇2 K∣∣ (t) take values at expy tex−y; and a, b, c are respectively given by

a = maxi=1,2,3

ai , b = maxi=1,2,3

bi andc = maxi=1,2,3,4

ci ,

where

a1 = (sin α0)−1 max

[0,α0]

cos2 α

1 + 4 cos2 α,

a2 = (sin α0)−1 max[

α0, π6

]cos2 α

1 + 4 cos2 α,

a3 = (sin α0)−1 max[

π6 , π

2

] sin2 α cos2 α,

b1 = max[0,α0]

(1 + sin α) cos α

1 + 4 cos2 α,

b2 = max[α0, π

6

](1 + sin α) cos α

1 + 4 cos2 α,

b3 = max[π6 , π

2

][sin2 α(1 + sin α) cos α

],

c1 = 4

cos2 α0(1 − 4 sin2 α0

) ,

c2 = max[0,α0]

2 cos α

1 + 4 cos2 α,

c3 = max[α0, π

6

]2 sin α cos α + 4

sin α(1 + 4 cos2 α

) ,

c4 = max[π6 , π

2

](2 sin2 α cos α + 4 sin α

),

123

S.-Z. Du, Q.-R. Li

α0 = min

{arcsin

(sin(

√KM L)√

KM L

),π

l

}

and l > 6 is any constant.

Proof In view of (2.2), (2.20), (4.18), (4.19), (4.21), (4.26), and the compactness of M, itsuffices to show

Aα (ρ) ≤ 0 for any α ∈ [0, π). (5.4)

We split the proof into several cases.Case 1. α ∈ [0, α0) ∪ (π − α0, π)

In this case, sin α might be very small. To obtain (5.4), we use the fourth term on the R.H.S.of (5.1) to control the third. By choosing l > 6, χα has uniformly positive lower bound χπ/ l

for all α ∈ [0, α0) ∪ (π − α0, π). The condition (5.2) with a = b = 0 and c = c1 implies

2K (t) ≥ c1 (ρ − t) |∇K | (t)= 4

χα0

(ρ − t) |∇K | (t) ,

which yields, for α ∈ [0, α0) ∪ (π − α0, π),

4(ρ − t) |∇K | (t) ≤ 2χα K (t), ∀t ∈ [0, ρ] .

We therefore have

4 sin α

ρ∫

0

θ2qdt ≤ 4 sin α

ρ∫

0

(ρ − t)tθ |∇K | dt

≤ 2χα sin α

ρ∫

0

tθ K dt. (5.5)

The first inequality above is due to Lemma 3 which asserts θ ≤ t . Recall Sturm’s comparisontheorem (see [6]) gives

sin√

KM t√KM

≤ θ ≤ sin√

Kmt√Km

,

we conclude

sin α ≤ sin α0 ≤ sin√

KM L√KM L

≤ θ (L)

L.

Lemma 3 and inequality (5.5) hence imply

4 sin α

ρ∫

0

θ2qdt − 2χα

θ (ρ)

ρ

ρ∫

0

tθ K dt

≤(

ρ

θ (ρ)

θ (L)

L− 1

)2χαθ (ρ)

ρ

ρ∫

0

tθ K dt

≤ 0 whence α ∈ [0, α0) ∪ (π − α0, π) . (5.6)

123


On the other hand, by virtue of Lemma 4 and α0 < π6 , we find

sin2 αpα − 2ωα K

≤ ωα

{−2K + (ρ − t)2

∣∣∇2 K∣∣(

1 + 4 cos2 α)

+⎡⎣ cos2 α(

1 + 4 cos2 α)

sin α0(ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ (1 + sin α) |cos α|1 + 4 cos2 α

ρ∫

t

(ρ − τ) K (τ ) dτ

+ 2 |cos α|1 + 4 cos2 α

](ρ − t) |∇K |

}

≤ ωα

{−2K + (ρ − t)2∣∣∇2 K

∣∣

+⎡⎣a1 (ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ b1

ρ∫

t

(ρ − τ) K (τ ) dτ + c2

⎤⎦ (ρ − t) |∇K |

⎫⎬⎭

≤ 0.

The last inequality is due to (5.2) with a = a1, b = b1 and c = c2. Consequently

sin2 α

ρ∫

0

θ2 pαdt − 2ωα

ρ∫

0

K θ2dt ≤ 0. (5.7)

Plugging (5.6) and (5.7) into (5.1), one concludes

Aα (ρ) ≤ 0.

Case 2. α ∈ [α0,π6 ) ∪ ( 5π

6 , π − α0]In this case, we have χα ≥ 0. By virtue of Lemma 4, we have

sin2 αpα + 4 |sin α| q − 2ωa K

≤ ωa

{−2K + (ρ − t)2

∣∣∇2 K∣∣(

1 + 4 cos2 α)

+⎡⎣ cos2 α(

1 + 4 cos2 α)

sin α0(ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ (1 + sin α) |cos α|(1 + 4 cos2 α

)ρ∫

t

(ρ − τ) K (τ ) dτ

+ 2 sin α |cos α| + 4


)]

(ρ − t) |∇K |}

123

S.-Z. Du, Q.-R. Li

≤ ωa{−2K + (ρ − t)2

∣∣∇2 K∣∣

+⎡⎣a2 (ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ b2

ρ∫

t

(ρ − τ) K (τ ) dτ + c3

⎤⎦ (ρ − t) |∇K |

⎫⎬⎭

≤ 0.

We have used (5.2) for a = a2, b = b2 and c = c3 in the last inequality. Consequently, by(5.1),

Aα (ρ) ≤ −2χα

ρ2θ

ρ∫

0

tθ K dt ≤ 0.

Case 3. α ∈[

π6 , 5π

6

]By virtue of (4.20) and Lemma 2, one can see

r∫

0

K θ2dt ≥ θ2(

1

r− H

)= θ

r

r∫

0

tθ K dt.

Since χα ≤ 0 in this case, (5.1) becomes

ρθ2 (ρ) Aα (ρ) ≤ sin2 α

ρ∫

0

θ2 pαdt + 4 |sin α|ρ∫

0

θ2qdt − 2

ρ∫

0

K θ2dt.

In view of the condition (5.2), one concludes

sin2 αpα + 4 sin αq − 2K

≤ −2K + sin2 α (ρ − t)2∣∣∇2 K

∣∣ (t)

+⎡⎣ sin2 α cos2 α

sin α0(ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ sin2 α (1 + sin α) |cos α|ρ∫

t

(ρ − τ) K (τ ) dτ

+ 2 sin2 α |cos α| + 4 sin α](ρ − t) |∇K |

≤ −2K + (ρ − t)2∣∣∇2 K

∣∣ (t)

+⎡⎣a3 (ρ − t)

ρ∫

t

(ρ − τ) |∇K | (τ ) dτ

+ b3

ρ∫

t

(ρ − τ) K (τ ) dτ + c4

⎤⎦ (ρ − t) |∇K |

≤ 0.

123


The last inequality is by employing (5.2) for a = a3, b = b3 and c = c4. Consequently,

Aα (ρ) ≤ 0.

��We now complete the proof of main Theorem.

Proof of Theorem 1 With notation of Proposition 3, since

max {a1, a2} = (sin α0)−1 max[

0, π6

]cos2 α

1 + 4 cos2 α

= 1

5 sin α0

and

a3 = 1

4 sin α0,

we conclude a = 14 sin α0

.On the other hand, one has

max {b1, b2} = max[0, π

6

](1 + sin α) cos α

1 + 4 cos2 α

= max[0,1/2]

(1 + t)√

1 − t2

5 − 4t2

= exp

(max

[0,1/2]ln

(1 + t)√

1 − t2

5 − 4t2

)

=(1 + 1

2

) √1 − ( 1

2

)2

5 − 4( 1

2

)2

= 3√

3

16≈ 0.3248

and

b3 = max[π6 , π

2

] sin2 α (1 + sin α) cos α

= max[12 ,1

] t2 (1 + t)√

1 − t2

= exp

⎛⎝max[

12 ,1

] ln t2 (1 + t)√

1 − t2

⎞⎠

=(

1 + √33

8

)2 (1 + 1 + √

33

8

) √√√√1 −(

1 + √33

8

)2

≈ 0.7045.

There we deduce b < 34 .

123

S.-Z. Du, Q.-R. Li

Moreover, letting l = 16, we get

c1 = 4

cos2 α0(1 − 4 sin2 α0

)

≤ 4(cos2 π

16

) (1 − 4 sin2 π

16

)≤ 5.

We also have

c2 = max[0,α0]

2 cos α

1 + 4 cos2 α

= 2 cos α0

1 + 4 cos2 α0≤ 2,

c3 = max[α0, π

6

]2 sin α cos α + 4


)

≤ sin π3 + 4

sin α0(1 + 4 cos2 π

6

)

= 8 + √3

8 sin α0.

and

c4 = max[π6 , π

2

](2 sin2 α cos α + 4 sin α

)

≤ max[π6 , π

2

] sin 2α + 4

= 5.

We thus conclude that c ≤ max{

5, 8+√3

8 sin α0

}= 8+√

38 sin α0

.

Summarizing above, we already have a = 14 sin α0

, b < 34 and c ≤ 8+√

38 sin α0

, so Theorem 1follows from Proposition 3. ��Acknowledgments This paper was supported by the Australian Research Council. The first author waspartially supported by the National Natural Science Foundation of China (11101106). The second author wasalso supported by the Chinese Scholarship Council. The authors would like to show their appreciation toProfessor Xu-Jia Wang for his encouragement and instructive discussions.

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