A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 ·...

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Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 237418, 5 pages http://dx.doi.org/10.1155/2013/237418 Research Article A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian on Riemannian Manifolds Peihe Wang 1 and Ying Li 2 1 School of Mathematical Sciences, Qufu Normal University, Shandong, Qufu 273165, China 2 College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China Correspondence should be addressed to Peihe Wang; [email protected] Received 8 December 2012; Revised 19 February 2013; Accepted 9 March 2013 Academic Editor: Wenming Zou Copyright © 2013 P. Wang and Y. Li. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. e paper starts with a discussion involving the Sobolev constant on geodesic balls and then follows with a derivation of a lower bound for the first eigenvalue of the Laplacian on manifolds with small negative curvature. e derivation involves Moser iteration. 1. Introduction e Laplacian is one of the most important operator on Riemannian manifolds, and the study of its first eigenvalue is also an interesting subject in the field of geometric analysis. In general, people would like to estimate the first eigenvalue of Laplacian in terms of geometric quantities of the manifolds such as curvature, volume, diameter, and injectivity radius. In this sense, the first interesting result is that of Lichnerowicz and Obata, which proved the following result in [1]: let be an -dimensional compact Riemannian manifold without boundary with Ric () ≥ ( − 1), then the first eigenvalue of Laplacian on will satisfy that 1 () ≥ , and the inequality becomes equality if and only if ≅ . e above result implies that the first eigenvalue of the Laplacian will have a lower bound less than if the Ricci curvature of manifolds involved has a lower bound −1 except on a small part where the Ricci curvature satisfied that Ric () ≥ 0. Now a natural question arises: what is the lower bound of the first eigenvalue of Laplacian on such a manifold? In [2], Petersen and Sprouse gave a lower bound under the assumption that the bad part of the manifolds is small in the sense of -norm, where is a constant larger than half of the dimension of the manifold. In this paper, we are interested in the lower bound of the first eigenvalue under the global pinching of the Ricci curvature and we obtain a universal estimate of this lower bound on a certain class of manifolds. 2. A Sobolev Constant on the Geodesic Ball e Sobolev inequality is one of the most important tools in geometric analysis, and the Sobolev constant plays an important part in the study of this field. In this section, we will obtain a general Sobolev constant only depending on the dimension of the manifold on the geodesic ball with small radius. Definition 1. Let () ⊆ be a geodesic ball with radius ; we define the Sobolev constant () on it to be the infimum among all the constant such that the inequality |||| 2 2/(−2) ≤ ||∇|| 2 2 holds for all ∈ 1.2 0 ( ()). Definition 2. Let () ⊆ be a geodesic ball with radius ; we define the isoperimetric constant 0 () on it to be the supremum among all the constant such that the inequality Area (Ω) ≥ Vol (Ω) 1−(1/) holds for all Ω⊆ () with smooth boundary. For any fixed point and radius , Croke proves that the equality () = 4(( − 1)/( − 2)) 2 0 () −2 holds [3], but one expects the constant () to be independent on the location of the point , under some assumptions. In what follows, we will give an upper bound to () independent of the point . Let be an -dimensional Riemannian manifold, is the unit tangent bundle of , and : → is the canonical projective map. Ω ⊆ , ∈ Ω, () is the

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Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 237418 5 pageshttpdxdoiorg1011552013237418

Research ArticleA Global Curvature Pinching Result of the First Eigenvalue ofthe Laplacian on Riemannian Manifolds

Peihe Wang1 and Ying Li2

1 School of Mathematical Sciences Qufu Normal University Shandong Qufu 273165 China2 College of Science University of Shanghai for Science and Technology Shanghai 200093 China

Correspondence should be addressed to Peihe Wang peihewanghotmailcom

Received 8 December 2012 Revised 19 February 2013 Accepted 9 March 2013

Academic Editor Wenming Zou

Copyright copy 2013 P Wang and Y Li This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The paper starts with a discussion involving the Sobolev constant on geodesic balls and then follows with a derivation of a lowerbound for the first eigenvalue of the Laplacian onmanifolds with small negative curvatureThe derivation involves Moser iteration

1 Introduction

The Laplacian is one of the most important operator onRiemannian manifolds and the study of its first eigenvalue isalso an interesting subject in the field of geometric analysisIn general people would like to estimate the first eigenvalueof Laplacian in terms of geometric quantities of themanifoldssuch as curvature volume diameter and injectivity radius Inthis sense the first interesting result is that of Lichnerowiczand Obata which proved the following result in [1] let 119872119899be an 119899-dimensional compact Riemannian manifold withoutboundary with Ric (119872) ge (119899 minus 1) then the first eigenvalueof Laplacian on 119872

119899 will satisfy that 1205821(119872) ge 119899 and the

inequality becomes equality if and only if119872119899 cong 119878119899

The above result implies that the first eigenvalue of theLaplacian will have a lower bound less than 119899 if the Riccicurvature of manifolds involved has a lower bound 119899 minus 1

except on a small part where the Ricci curvature satisfiedthat Ric (119872) ge 0 Now a natural question arises what is thelower bound of the first eigenvalue of Laplacian on such amanifold In [2] Petersen and Sprouse gave a lower boundunder the assumption that the bad part of the manifolds issmall in the sense of 119871119901-norm where 119901 is a constant largerthan half of the dimension of the manifold In this paper weare interested in the lower bound of the first eigenvalue underthe global pinching of the Ricci curvature and we obtain auniversal estimate of this lower bound on a certain class ofmanifolds

2 A Sobolev Constant on the Geodesic Ball

The Sobolev inequality is one of the most important toolsin geometric analysis and the Sobolev constant plays animportant part in the study of this field In this section wewill obtain a general Sobolev constant only depending on thedimension of the manifold on the geodesic ball with smallradius

Definition 1 Let 119861119901(119877) sube 119872 be a geodesic ball with radius

119877 we define the Sobolev constant 119862119904(119877) on it to be the

infimum among all the constant 119862 such that the inequality||119891||2

2119899(119899minus2)le 119862||nabla119891||

2

2holds for all 119891 isin 119882

12

0(119861119901(119877))

Definition 2 Let 119861119901(119877) sube 119872 be a geodesic ball with radius

119877 we define the isoperimetric constant 1198620(119877) on it to be the

supremum among all the constant 120572 such that the inequalityArea (120597Ω) ge 120572Vol (Ω)1minus(1119899) holds for all Ω sube 119861

119901(119877) with

smooth boundaryFor any fixed point 119901 and radius 119877 Croke proves that the

equality119862119904(119877) = 4((119899minus1)(119899minus2))

2

1198620(119877)minus2 holds [3] but one

expects the constant 119862119904(119877) to be independent on the location

of the point 119901 under some assumptions In what follows wewill give an upper bound to 119862

119904(119877) independent of the point

119901Let 119872 be an 119899-dimensional Riemannian manifold 119878119872

is the unit tangent bundle of 119872 and 120587 119878119872 rarr 119872 isthe canonical projective map Ω sube 119872 120585 isin 119878Ω 120574

120585(119905) is the

2 Abstract and Applied Analysis

normalized geodesic from 120587(120585) with the initial velocity 120585 Wedefine some notations as follows

120591 (120585) = sup 120591 gt 0 | 120574120585(119905) isin Ω forall119905 isin (0 120591) (1)

119862(120585) is the arc length from 120587(120585) to the cut locus pointalong 120574

120585(119905) Consider

119880Ω = 120585 isin 119878Ω | 119862 (120585) ge 120591 (120585)

119880119909= (120587 | 119880Ω)

minus1

(119909) 120596119909=120583119909(119880119909)

119888119899minus1

120596 = inf119909isinΩ

120596119909

(2)

where 120583119909is the standard surface measure of the unit sphere

119888119899minus1

is denoted to be the area of the unit sphere 119878119899minus1

Definition 3 Using the Notation above 120596 = inf119909isinΩ

120596119909is

called the visibility angle ofΩIf the manifold has Inj (119872) ge 119894 which ensures that any

minimal geodesic starting from any point in 119861119901(1198942) will

reach the boundary 120597119861119901(1198942) before it reaches its cut locus

then the visibility angle of 119861119901(1198942) for any point 119901 which we

denote by 120596(1198942) satisfies 120596(1198942) = 1

Lemma 4 Let 119872119899 be a closed Riemannian manifold withInj (119872) ge 119894 then for any 119901 isin 119872 the following Sobolevinequality holds on 119861

119901(1198942) ||119891||2

2119899(119899minus2)le 119862||nabla119891||

2

2 where

119891 isin 11988212

0(119861119901(1198942)) and 119862 = 119862(119899)

Proof Croke proved the following inequality [4]

Area (120597Ω)Vol (Ω)1minus(1119899)

ge119888119899minus1

(1198881198992)1minus(1119899)

1205961+(1119899)

(3)

whereΩ sube 119861119901(1198942) 120597Ω isin 119862

infin and120596 is just the visibility angleof the domainΩ

As discussed above we will have120596(Ω) = 1 ifΩ sube 119861119901(1198942)

then according to Crokersquos inequality we obtain 1198620(1198942) ge

119888119899minus1

(1198881198992)1minus(1119899) The relation between 119862

0(1198942) and 119862

119904(1198942)

tells us that 119862119904(1198942) le 119862(119899) where 119862(119899) is a constant only

depending on the dimension 119899

Proposition 5 Let119872119899 be a closed 119899-dimensional Riemannianmanifold with Inj (119872) ge 119894 then for all 119901 isin 119872 Vol (119861

119901(1198942))

ge 119862119899119894119899 where119862

119899is a constant only depending on the dimension

119899

Proof Also take the inequality of Croke

Area (120597Ω)Vol (Ω)1minus(1119899)

ge119888119899minus1

(1198881198992)1minus(1119899)

1205961+(1119899)

(4)

then the result can easily be derived from the fact that120596(1198942) = 1 and Area (120597119861

119901(119903)) = 119889Vol (119861

119901(119903))119889119903 after we

integrate both sides of the inequality

3 The First Eigenfunction and Eigenvalue

Let 119872119899 be a closed 119899-dimensional Riemannian manifoldsuppose that120582

1(119872) is the first eigenvalue of the Laplacian and

119906 is the first eigenfunction In other words they will satisfythat Δ119906 + 120582

1(119872)119906 = 0 By linearity we can assume that

minus1 le 119906 le 1 and inf119909isin119872

119906 = minus1 for the linearity For theconvenience we call it the normalized eigenfunctionNextwewill study some properties of the normalized eigenfunctionand the eigenvalue

Lemma 6 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 Then a constant1198621(119899 119894) gt 0 can be found such that 120582

1(119872) le 119862

1(119899 119894)

Proof One of the theorems of Yau and Schoen [1] shows that1205821(119872) le 119864

1198991198892

le 1198641198991198942

≜ 1198621(119899 119894) if Ric (119872) ge 0 where 119889 is

the diameter of the manifold and 119864119899is a constant depending

only on 119899We will now introduce some notation Let Ric

minus(119909) denote

the lowest eigenvalue of the Ricci curvature tensor at 119909 Fora function 119891(119909) on 119872

119899 we denote 119891+(119909) = max119891(119909) 0

Notice that a Riemannian manifold satisfies Ric ge 119899 minus 1 ifand only if ((119899 minus 1) minus Ric

minus)+equiv 0

The well-known Myers theorem shows that a closedmanifold with Ric ge 119899 minus 1 would have a bounded diameter119889 le 120587 In other words one can deduce that 119889 le 120587 if one has(1Vol(119872)) int

119872

((119899 minus 1) minus Ricminus)+119889vol = 0 We will show next

a result analogous to the one in [5] which we will use in ourestimation of the eigenvalueThe proof follows identically soit will be omitted (the reader can refer to the aforementionedarticle)

Lemma 7 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 then for any 120575 gt 0 there exists1205980= 1205980(119899 120575) gt 0 such that if

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

0(119899 120575) (5)

then the diameter will satisfy 119889 lt 120587 + 120575 In particular thereexists 120598

1= 1205981(119899) such that if

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (6)

then the diameter will satisfy 119889 lt 2120587 This fact together withthe volume comparison theorem implies that Vol (119872) le

1198622(119899) where 119862

2(119899) is also a constant only dependent of 119899

Now we can get a rough lower bound for the first eigen-value

Lemma 8 For 119899 isin N let 1205981= 1205981(119899) gt 0 as above and suppose

that119872119899 is a closed manifold with

Ric (119872) ge 0

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899)

(7)

Abstract and Applied Analysis 3

then there exists a constant 1198623(119899) gt 0 such that 120582

1(119872) ge

1198623(119899)

Proof The proof mainly belongs to Li and Yau [6] Let 119906 bethe normalized eigenfunction of119872 set V = log (119886 + 119906) where119886 gt 1 Then we can easily get that

ΔV =minus1205821(119872) 119906

119886 + 119906minus |nablaV|

2

(8)

Denote that 119876(119909) = |nablaV|2(119909) and we then have by theRicci identity on manifolds with Ric (119872) ge 0

Δ119876 = 2V2

119894119895+ 2V119895V119895119894119894ge 2V2

119894119895+ 2 ⟨nablaV nablaΔV⟩ (9)

For the term V2119894119895 we have

sum

119894119895

V2

119894119895ge(ΔV)2

119899ge1

119899(1198762

+21205821(119872) 119906

119886 + 119906) (10)

and for the term ⟨nablaV nablaΔV⟩ we have

⟨nablaV nablaΔV⟩ = minus1198861205821(119872)

119886 + 119906119876 minus ⟨nablaV nabla119876⟩ (11)

Therefore assume 1199090isin 119872 to be the maximum of 119876 then at

1199090 we have

0 ge2

119899119876 (1199090) + (

41205821(119872)

119899minus2 (119899 + 2) 119886120582

1(119872)

119899 (119886 minus 1)) (12)

Therefore

119876 (119909) le 119876 (1199090) le

(119899 + 2) 1198861205821(119872)

119886 minus 1 (13)

Denote 120574 to be the minimizing unit speed geodesic join-ing themaximum andminimumpoints of 119906 then integrating11987612 along 120574 one will get

log( 119886

119886 minus 1) le log(119886 +max 119906

119886 minus 1) le 119889(

(119899 + 2) 1198861205821(119872)

119886 minus 1)

12

(14)

Let 119905 = (119886 minus 1)119886 then for any 119905 isin (0 1) we have (119899 +2)1205821(119872) ge 119905(119889

minus2

(log (1119905))2)Considering the maximum of the right hand and the

upper bound of the diameter derived in Lemma 7 we candeduce that a positive constant 119862

3(119899) can be found such that

1205821(119872) ge

4119890minus2

(119899 + 2) 1198892ge 1198623(119899) (15)

where 119889 is the diameter of the manifold

Corollary 9 If the manifold one discussed satisfies all theconditions in Lemma 8 and its injectivity radius satisfiesInj (119872) ge 119894 and if one let 119906 to be the normalized eigenfunctionthen there exists a constant 119862

4(119899 119894) gt 0 such that |nabla119906|2 le

1198624(119899 119894)

Proof Set 119886 = 2 in the (13) from above Then applyingLemma 6 one obtains

1

9|nabla119906|2

le |nablaV|2

(119909) le 21198621(119899 119894) (119899 + 2) (16)

therefore

|nabla119906|2

le 1198624(119899 119894) (17)

Proposition 10 Let 119872119899 be a closed 119899-dimensional Rieman-nian manifold 119906 the first eigenfunction of the Laplacian and1205821(119872) the corresponding eigenvalue thenΔ|119906|+120582

1(119872)|119906| ge 0

holds in the sense of distribution Moreover if 119872119899 is compactwith boundary then the same conclusion holds for its Neumannboundary value problem

Proof From the definition we know that Δ119906 + 1205821(119872)119906 = 0

holds on119872 Denote

119872+

= 119909 isin 119872 | 119906 (119909) gt 0

119872minus

= 119909 isin 119872 | 119906 (119909) lt 0

1198720

= 119909 isin 119872 | 119906 (119909) = 0

(18)

According to the maximum principle of elliptic equation andthe discussion about nodal set and nodal regions in [1] wecan conclude that 120597119872+ = 120597119872

minus

= 1198720 is a smooth manifold

with dimension 119899 minus 1For all 120601 isin 119862

infin

(119872) 120601 ge 0 integrating by parts we thenhave

int119872

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int119872+

|119906| Δ120601 + 1205821(119872) 120601 |119906| + int

119872minus

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int120597119872+

(119906120597120601

120597119899+minus 120601

120597119906

120597119899+) + int119872+

120601 (Δ119906 + 1205821(119872) 119906)

minus int120597119872minus

(119906120597120601

120597119899minusminus 120601

120597119906

120597119899minus) minus int119872minus

120601 (Δ119906 + 1205821(119872) 119906)

= int120597119872minus

120601120597119906

120597119899minusminus int120597119872+

120601120597119906

120597119899+ge 0

(19)

where 119899+ and 119899minus denote the outward normal direction withrespect to the boundaries of 119872+ and 119872minus respectively Notethat 120597119906120597119899minus ge 0 on 120597119872

minus and 120597119906120597119899+

le 0 on 120597119872+ for the

definition of119872minus and119872+ This completes the proof

When119872 has boundary we can apply the same reasoningexcept that the test function will require 120601 isin 119862

infin

0(119872) This

gives the proofAs long as the given manifold is compact one knows

that the first normalized eigenfunction is then determinedThis indicates that the first normalized eigenfunction ofthe Laplacian has a close relation with the geometry of

4 Abstract and Applied Analysis

the manifold In particular one would hope to bound the1198712-norm of first normalized eigenvalue of Laplacian from

below by the geometric quantities In this sense we have thefollowing result

Theorem 11 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 If 119906 is thenormalized eigenfunction of the Laplacian then there exists aconstant 119862

5(119899 119894) gt 0 such that int

119872

1199062

ge 1198625(119899 119894)

Proof We use Moser iteration to get the result From Propo-sition 10 we know that Δ|119906| + 120582

1(119872)|119906| ge 0 holds on 119872 in

the sense of distribution Set V = |119906| and take the point 119901 isin 119872

such that 119906(119901) = minus1For 119886 ge 1 denote 119877 = 1198942 120601 is a cut-off function on

119861119901(119877) then we have by integrating by parts

1205821(119872)int

119861119901(119877)

1206012

V2119886

= minusint119861119901(119877)

1206012

V2119886minus1

ΔV

= 2int119861119901(119877)

120601V2119886minus1

+ (2119886 minus 1) int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

ge 2int119861119901(119877)

120601V2119886minus1

+ 119886int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(20)

However using the identity

int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

= int119861119901(119877)

1003816100381610038161003816nabla1206011003816100381610038161003816

2

V2119886

+ 2119886int119861119901(119877)

120601V2119886minus1

⟨nabla120601 nablaV⟩

+ 1198862

int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(21)

we have

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

minus int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

(22)

therefore using the Sobolev inequality in Lemma 4

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

+ int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

ge1

119862(int119861119901(119877)

(120601V119886

)2119899(119899minus2)

)

(119899minus2)119899

=1

119862

1003817100381710038171003817120601V1198861003817100381710038171003817

2

2119899(119899minus2)

(23)

Let

120601 =

1 in 119861119901(120588)

120588 + 120590 minus 119903

120590 in

119861119901(120588 + 120590)

119861119901(120588)

0 in119861119901(119877)

119861119901(120588 + 120590)

(24)

Putting 120601 into the inequality above and we then have bysplitting the integral into three parts and using the values of120601 on each of them

V2119886119899(119899minus2)119861

119901(120588)

le [119862(1205821(119872) 119886 +

1

1205902)]

12119886

V2119886119861119901(120588+120590)

(25)

where we denote ||119891||119901Ω

= (intΩ

119891119901

)1119901 only for emphasizing

the integral domainSet

119886119895 1198860= 1 1198861=

119899

119899 minus 2 119886

119895= (

119899

119899 minus 2)

119895

120590119895 1205900=119877

4 1205901=119877

8 120590

119895=

119877

2minus(2+119895)

120588119895 120588minus1= 119877 120588

119895= 119877 minus

119895

sum

119904=0

120590119897

(26)

And putting 119886119895 120590119895 120588119895 into (25) we can derive after

iteration that

V2119886119895+1 119861119901(120588119895)le [119862(120582

1(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119886119895119861119901(120588119895minus1)

le

119895

prod

119904=0

[119862(1205821(119872) 119886

119904+

1

1205902119904

)]

12119886119904

V2119861119901(119877)

(27)

Let 119895 rarr +infin then

Vinfin119861119901(1198772)

le

infin

prod

119895=0

[119862(1205821(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119861119901(119877) (28)

The product can be estimated as follows

infin

prod

119895=0

[119862(1205821(119872) 119886

119895+

1

1205902

119895

)]

12119886119895

le

infin

prod

119895=0

[119862(1205821(119872) +

16

1198772)]

12119886119895

41198952119886119895

(29)

The right hand will converge to a fixed number by usingthe fact thatprodinfin

119895=0119861120583119895

= 119861120583(120583minus1) and the factsuminfin

119895=0119895120583minus119895 is finite

for some 119861 isin 119877 120583 gt 1 From 1205821(119872) le 119862

1(119899 119894) we can find a

positive constant 1198625(119899 119894) gt 0 such that

Vinfin119861119901(1198772)

le 119862minus12

5(119899 119894) V

2119861119901(119877)

le 119862minus12

5(119899 119894) V

2119872

(30)

Therefore

int119872

1199062

ge 1198625(119899 119894) (31)

Abstract and Applied Analysis 5

4 The Lower Bound of the First Eigenvalue

Using the same notation as above we can state the followingresult

Theorem 12 For 119899 isin N 119894 120575 isin R+ there is an 120598 = 120598(119899 119894 120575) gt

0 such that any closed manifold 119872119899 with Ric (119872) ge 0

Inj (119872) ge 119894 and

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598 (32)

will satisfy that 1205821(119872) ge 119899 minus 120575

Proof Assume that 119906 is the normalized eigenfunction ofLaplacian on 119872

119899 let 119876(119909) = |nabla119906|2

+ (1205821(119872)119899)119906

2 directcomputation shows that

1

2Δ119876 = 119906

2

119894119895+ 119906119895119906119894119894119895+ 119877119894119895119906119894119906119895+1205821(119872)

119899|nabla119906|2

+1205821(119872)

119899119906Δ119906

ge1 minus 119899

1198991205821(119872) |nabla119906|

2

+ 119877119894119895119906119894119906119895

(33)

Integrating both sides on119872119899 we have

0 ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

119877119894119895119906119894119906119895119889vol

ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

Ricminus|nabla119906|2

119889vol

=1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + (119899 minus 1) int119872

|nabla119906|2

119889vol

minus int119872

[(119899 minus 1) minus Ricminus] |nabla119906|

2

119889vol

(34)

therefore

1205821(119872) ge 119899 minus

Vol (119872)

1205821(119872) int

119872

|119906|2

119889vol1

Vol (119872)

times int119872

((119899 minus 1) minus Ricminus)+|nabla119906|2

119889vol

(35)

if we suppose that

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (36)

If 1205981is the one obtained in Lemma 7 then one has

1205821(119872) ge 119899 minus

11986221198624

11986231198625

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol

(37)

Finally if one chooses

0 lt 120598 = 120598 (119899 119894 120575) le min1205981 12057511986231198625

11986221198624

(38)

then 1205821(119872) ge 119899 minus 120575 as long as (1Vol (119872)) int

119872

((119899 minus 1) minus

Ricminus)+119889vol le 120598 and this proves the theorem

Acknowledgments

The authors owe great thanks to the referees for theircareful efforts to make the paper clearer Research of thefirst author was supported by STPF of University (noJ11LA05) NSFC (no ZR2012AM010) the Postdoctoral Fund(no 201203030) of Shandong Province and PostdoctoralFund (no 2012M521302) of China Part of this work was donewhile the first author was staying at his postdoctoral mobileresearch station of QFNU

References

[1] S T Yau and R Schoen Lectures in Differential GeometryScientific Press 1988

[2] P Petersen andC Sprouse ldquoIntegral curvature bounds distanceestimates and applicationsrdquo Journal of Differential Geometryvol 50 no 2 pp 269ndash298 1998

[3] D Yang ldquoConvergence of Riemannian manifolds with integralbounds on curvature IrdquoAnnales Scientifiques de lrsquoEcole NormaleSuperieure Quatrieme Serie vol 25 no 1 pp 77ndash105 1992

[4] I Chavel Riemannian Geometry A Mordern IntroductionCambridge University Press Cambridge UK 2000

[5] C Sprouse ldquoIntegral curvature bounds and bounded diameterrdquoCommunications in Analysis andGeometry vol 8 no 3 pp 531ndash543 2000

[6] P Li and S T Yau ldquoEstimates of eigenvalues of a compactRiemannianmanifoldrdquo inAMS Proceedings of Symposia in PureMathematics pp 205ndash239 American Mathematical SocietyProvidence RI USA 1980

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 2: A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 · Introduction The Laplacian is one of the most important operator on ... curvature

2 Abstract and Applied Analysis

normalized geodesic from 120587(120585) with the initial velocity 120585 Wedefine some notations as follows

120591 (120585) = sup 120591 gt 0 | 120574120585(119905) isin Ω forall119905 isin (0 120591) (1)

119862(120585) is the arc length from 120587(120585) to the cut locus pointalong 120574

120585(119905) Consider

119880Ω = 120585 isin 119878Ω | 119862 (120585) ge 120591 (120585)

119880119909= (120587 | 119880Ω)

minus1

(119909) 120596119909=120583119909(119880119909)

119888119899minus1

120596 = inf119909isinΩ

120596119909

(2)

where 120583119909is the standard surface measure of the unit sphere

119888119899minus1

is denoted to be the area of the unit sphere 119878119899minus1

Definition 3 Using the Notation above 120596 = inf119909isinΩ

120596119909is

called the visibility angle ofΩIf the manifold has Inj (119872) ge 119894 which ensures that any

minimal geodesic starting from any point in 119861119901(1198942) will

reach the boundary 120597119861119901(1198942) before it reaches its cut locus

then the visibility angle of 119861119901(1198942) for any point 119901 which we

denote by 120596(1198942) satisfies 120596(1198942) = 1

Lemma 4 Let 119872119899 be a closed Riemannian manifold withInj (119872) ge 119894 then for any 119901 isin 119872 the following Sobolevinequality holds on 119861

119901(1198942) ||119891||2

2119899(119899minus2)le 119862||nabla119891||

2

2 where

119891 isin 11988212

0(119861119901(1198942)) and 119862 = 119862(119899)

Proof Croke proved the following inequality [4]

Area (120597Ω)Vol (Ω)1minus(1119899)

ge119888119899minus1

(1198881198992)1minus(1119899)

1205961+(1119899)

(3)

whereΩ sube 119861119901(1198942) 120597Ω isin 119862

infin and120596 is just the visibility angleof the domainΩ

As discussed above we will have120596(Ω) = 1 ifΩ sube 119861119901(1198942)

then according to Crokersquos inequality we obtain 1198620(1198942) ge

119888119899minus1

(1198881198992)1minus(1119899) The relation between 119862

0(1198942) and 119862

119904(1198942)

tells us that 119862119904(1198942) le 119862(119899) where 119862(119899) is a constant only

depending on the dimension 119899

Proposition 5 Let119872119899 be a closed 119899-dimensional Riemannianmanifold with Inj (119872) ge 119894 then for all 119901 isin 119872 Vol (119861

119901(1198942))

ge 119862119899119894119899 where119862

119899is a constant only depending on the dimension

119899

Proof Also take the inequality of Croke

Area (120597Ω)Vol (Ω)1minus(1119899)

ge119888119899minus1

(1198881198992)1minus(1119899)

1205961+(1119899)

(4)

then the result can easily be derived from the fact that120596(1198942) = 1 and Area (120597119861

119901(119903)) = 119889Vol (119861

119901(119903))119889119903 after we

integrate both sides of the inequality

3 The First Eigenfunction and Eigenvalue

Let 119872119899 be a closed 119899-dimensional Riemannian manifoldsuppose that120582

1(119872) is the first eigenvalue of the Laplacian and

119906 is the first eigenfunction In other words they will satisfythat Δ119906 + 120582

1(119872)119906 = 0 By linearity we can assume that

minus1 le 119906 le 1 and inf119909isin119872

119906 = minus1 for the linearity For theconvenience we call it the normalized eigenfunctionNextwewill study some properties of the normalized eigenfunctionand the eigenvalue

Lemma 6 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 Then a constant1198621(119899 119894) gt 0 can be found such that 120582

1(119872) le 119862

1(119899 119894)

Proof One of the theorems of Yau and Schoen [1] shows that1205821(119872) le 119864

1198991198892

le 1198641198991198942

≜ 1198621(119899 119894) if Ric (119872) ge 0 where 119889 is

the diameter of the manifold and 119864119899is a constant depending

only on 119899We will now introduce some notation Let Ric

minus(119909) denote

the lowest eigenvalue of the Ricci curvature tensor at 119909 Fora function 119891(119909) on 119872

119899 we denote 119891+(119909) = max119891(119909) 0

Notice that a Riemannian manifold satisfies Ric ge 119899 minus 1 ifand only if ((119899 minus 1) minus Ric

minus)+equiv 0

The well-known Myers theorem shows that a closedmanifold with Ric ge 119899 minus 1 would have a bounded diameter119889 le 120587 In other words one can deduce that 119889 le 120587 if one has(1Vol(119872)) int

119872

((119899 minus 1) minus Ricminus)+119889vol = 0 We will show next

a result analogous to the one in [5] which we will use in ourestimation of the eigenvalueThe proof follows identically soit will be omitted (the reader can refer to the aforementionedarticle)

Lemma 7 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 then for any 120575 gt 0 there exists1205980= 1205980(119899 120575) gt 0 such that if

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

0(119899 120575) (5)

then the diameter will satisfy 119889 lt 120587 + 120575 In particular thereexists 120598

1= 1205981(119899) such that if

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (6)

then the diameter will satisfy 119889 lt 2120587 This fact together withthe volume comparison theorem implies that Vol (119872) le

1198622(119899) where 119862

2(119899) is also a constant only dependent of 119899

Now we can get a rough lower bound for the first eigen-value

Lemma 8 For 119899 isin N let 1205981= 1205981(119899) gt 0 as above and suppose

that119872119899 is a closed manifold with

Ric (119872) ge 0

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899)

(7)

Abstract and Applied Analysis 3

then there exists a constant 1198623(119899) gt 0 such that 120582

1(119872) ge

1198623(119899)

Proof The proof mainly belongs to Li and Yau [6] Let 119906 bethe normalized eigenfunction of119872 set V = log (119886 + 119906) where119886 gt 1 Then we can easily get that

ΔV =minus1205821(119872) 119906

119886 + 119906minus |nablaV|

2

(8)

Denote that 119876(119909) = |nablaV|2(119909) and we then have by theRicci identity on manifolds with Ric (119872) ge 0

Δ119876 = 2V2

119894119895+ 2V119895V119895119894119894ge 2V2

119894119895+ 2 ⟨nablaV nablaΔV⟩ (9)

For the term V2119894119895 we have

sum

119894119895

V2

119894119895ge(ΔV)2

119899ge1

119899(1198762

+21205821(119872) 119906

119886 + 119906) (10)

and for the term ⟨nablaV nablaΔV⟩ we have

⟨nablaV nablaΔV⟩ = minus1198861205821(119872)

119886 + 119906119876 minus ⟨nablaV nabla119876⟩ (11)

Therefore assume 1199090isin 119872 to be the maximum of 119876 then at

1199090 we have

0 ge2

119899119876 (1199090) + (

41205821(119872)

119899minus2 (119899 + 2) 119886120582

1(119872)

119899 (119886 minus 1)) (12)

Therefore

119876 (119909) le 119876 (1199090) le

(119899 + 2) 1198861205821(119872)

119886 minus 1 (13)

Denote 120574 to be the minimizing unit speed geodesic join-ing themaximum andminimumpoints of 119906 then integrating11987612 along 120574 one will get

log( 119886

119886 minus 1) le log(119886 +max 119906

119886 minus 1) le 119889(

(119899 + 2) 1198861205821(119872)

119886 minus 1)

12

(14)

Let 119905 = (119886 minus 1)119886 then for any 119905 isin (0 1) we have (119899 +2)1205821(119872) ge 119905(119889

minus2

(log (1119905))2)Considering the maximum of the right hand and the

upper bound of the diameter derived in Lemma 7 we candeduce that a positive constant 119862

3(119899) can be found such that

1205821(119872) ge

4119890minus2

(119899 + 2) 1198892ge 1198623(119899) (15)

where 119889 is the diameter of the manifold

Corollary 9 If the manifold one discussed satisfies all theconditions in Lemma 8 and its injectivity radius satisfiesInj (119872) ge 119894 and if one let 119906 to be the normalized eigenfunctionthen there exists a constant 119862

4(119899 119894) gt 0 such that |nabla119906|2 le

1198624(119899 119894)

Proof Set 119886 = 2 in the (13) from above Then applyingLemma 6 one obtains

1

9|nabla119906|2

le |nablaV|2

(119909) le 21198621(119899 119894) (119899 + 2) (16)

therefore

|nabla119906|2

le 1198624(119899 119894) (17)

Proposition 10 Let 119872119899 be a closed 119899-dimensional Rieman-nian manifold 119906 the first eigenfunction of the Laplacian and1205821(119872) the corresponding eigenvalue thenΔ|119906|+120582

1(119872)|119906| ge 0

holds in the sense of distribution Moreover if 119872119899 is compactwith boundary then the same conclusion holds for its Neumannboundary value problem

Proof From the definition we know that Δ119906 + 1205821(119872)119906 = 0

holds on119872 Denote

119872+

= 119909 isin 119872 | 119906 (119909) gt 0

119872minus

= 119909 isin 119872 | 119906 (119909) lt 0

1198720

= 119909 isin 119872 | 119906 (119909) = 0

(18)

According to the maximum principle of elliptic equation andthe discussion about nodal set and nodal regions in [1] wecan conclude that 120597119872+ = 120597119872

minus

= 1198720 is a smooth manifold

with dimension 119899 minus 1For all 120601 isin 119862

infin

(119872) 120601 ge 0 integrating by parts we thenhave

int119872

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int119872+

|119906| Δ120601 + 1205821(119872) 120601 |119906| + int

119872minus

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int120597119872+

(119906120597120601

120597119899+minus 120601

120597119906

120597119899+) + int119872+

120601 (Δ119906 + 1205821(119872) 119906)

minus int120597119872minus

(119906120597120601

120597119899minusminus 120601

120597119906

120597119899minus) minus int119872minus

120601 (Δ119906 + 1205821(119872) 119906)

= int120597119872minus

120601120597119906

120597119899minusminus int120597119872+

120601120597119906

120597119899+ge 0

(19)

where 119899+ and 119899minus denote the outward normal direction withrespect to the boundaries of 119872+ and 119872minus respectively Notethat 120597119906120597119899minus ge 0 on 120597119872

minus and 120597119906120597119899+

le 0 on 120597119872+ for the

definition of119872minus and119872+ This completes the proof

When119872 has boundary we can apply the same reasoningexcept that the test function will require 120601 isin 119862

infin

0(119872) This

gives the proofAs long as the given manifold is compact one knows

that the first normalized eigenfunction is then determinedThis indicates that the first normalized eigenfunction ofthe Laplacian has a close relation with the geometry of

4 Abstract and Applied Analysis

the manifold In particular one would hope to bound the1198712-norm of first normalized eigenvalue of Laplacian from

below by the geometric quantities In this sense we have thefollowing result

Theorem 11 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 If 119906 is thenormalized eigenfunction of the Laplacian then there exists aconstant 119862

5(119899 119894) gt 0 such that int

119872

1199062

ge 1198625(119899 119894)

Proof We use Moser iteration to get the result From Propo-sition 10 we know that Δ|119906| + 120582

1(119872)|119906| ge 0 holds on 119872 in

the sense of distribution Set V = |119906| and take the point 119901 isin 119872

such that 119906(119901) = minus1For 119886 ge 1 denote 119877 = 1198942 120601 is a cut-off function on

119861119901(119877) then we have by integrating by parts

1205821(119872)int

119861119901(119877)

1206012

V2119886

= minusint119861119901(119877)

1206012

V2119886minus1

ΔV

= 2int119861119901(119877)

120601V2119886minus1

+ (2119886 minus 1) int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

ge 2int119861119901(119877)

120601V2119886minus1

+ 119886int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(20)

However using the identity

int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

= int119861119901(119877)

1003816100381610038161003816nabla1206011003816100381610038161003816

2

V2119886

+ 2119886int119861119901(119877)

120601V2119886minus1

⟨nabla120601 nablaV⟩

+ 1198862

int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(21)

we have

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

minus int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

(22)

therefore using the Sobolev inequality in Lemma 4

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

+ int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

ge1

119862(int119861119901(119877)

(120601V119886

)2119899(119899minus2)

)

(119899minus2)119899

=1

119862

1003817100381710038171003817120601V1198861003817100381710038171003817

2

2119899(119899minus2)

(23)

Let

120601 =

1 in 119861119901(120588)

120588 + 120590 minus 119903

120590 in

119861119901(120588 + 120590)

119861119901(120588)

0 in119861119901(119877)

119861119901(120588 + 120590)

(24)

Putting 120601 into the inequality above and we then have bysplitting the integral into three parts and using the values of120601 on each of them

V2119886119899(119899minus2)119861

119901(120588)

le [119862(1205821(119872) 119886 +

1

1205902)]

12119886

V2119886119861119901(120588+120590)

(25)

where we denote ||119891||119901Ω

= (intΩ

119891119901

)1119901 only for emphasizing

the integral domainSet

119886119895 1198860= 1 1198861=

119899

119899 minus 2 119886

119895= (

119899

119899 minus 2)

119895

120590119895 1205900=119877

4 1205901=119877

8 120590

119895=

119877

2minus(2+119895)

120588119895 120588minus1= 119877 120588

119895= 119877 minus

119895

sum

119904=0

120590119897

(26)

And putting 119886119895 120590119895 120588119895 into (25) we can derive after

iteration that

V2119886119895+1 119861119901(120588119895)le [119862(120582

1(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119886119895119861119901(120588119895minus1)

le

119895

prod

119904=0

[119862(1205821(119872) 119886

119904+

1

1205902119904

)]

12119886119904

V2119861119901(119877)

(27)

Let 119895 rarr +infin then

Vinfin119861119901(1198772)

le

infin

prod

119895=0

[119862(1205821(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119861119901(119877) (28)

The product can be estimated as follows

infin

prod

119895=0

[119862(1205821(119872) 119886

119895+

1

1205902

119895

)]

12119886119895

le

infin

prod

119895=0

[119862(1205821(119872) +

16

1198772)]

12119886119895

41198952119886119895

(29)

The right hand will converge to a fixed number by usingthe fact thatprodinfin

119895=0119861120583119895

= 119861120583(120583minus1) and the factsuminfin

119895=0119895120583minus119895 is finite

for some 119861 isin 119877 120583 gt 1 From 1205821(119872) le 119862

1(119899 119894) we can find a

positive constant 1198625(119899 119894) gt 0 such that

Vinfin119861119901(1198772)

le 119862minus12

5(119899 119894) V

2119861119901(119877)

le 119862minus12

5(119899 119894) V

2119872

(30)

Therefore

int119872

1199062

ge 1198625(119899 119894) (31)

Abstract and Applied Analysis 5

4 The Lower Bound of the First Eigenvalue

Using the same notation as above we can state the followingresult

Theorem 12 For 119899 isin N 119894 120575 isin R+ there is an 120598 = 120598(119899 119894 120575) gt

0 such that any closed manifold 119872119899 with Ric (119872) ge 0

Inj (119872) ge 119894 and

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598 (32)

will satisfy that 1205821(119872) ge 119899 minus 120575

Proof Assume that 119906 is the normalized eigenfunction ofLaplacian on 119872

119899 let 119876(119909) = |nabla119906|2

+ (1205821(119872)119899)119906

2 directcomputation shows that

1

2Δ119876 = 119906

2

119894119895+ 119906119895119906119894119894119895+ 119877119894119895119906119894119906119895+1205821(119872)

119899|nabla119906|2

+1205821(119872)

119899119906Δ119906

ge1 minus 119899

1198991205821(119872) |nabla119906|

2

+ 119877119894119895119906119894119906119895

(33)

Integrating both sides on119872119899 we have

0 ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

119877119894119895119906119894119906119895119889vol

ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

Ricminus|nabla119906|2

119889vol

=1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + (119899 minus 1) int119872

|nabla119906|2

119889vol

minus int119872

[(119899 minus 1) minus Ricminus] |nabla119906|

2

119889vol

(34)

therefore

1205821(119872) ge 119899 minus

Vol (119872)

1205821(119872) int

119872

|119906|2

119889vol1

Vol (119872)

times int119872

((119899 minus 1) minus Ricminus)+|nabla119906|2

119889vol

(35)

if we suppose that

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (36)

If 1205981is the one obtained in Lemma 7 then one has

1205821(119872) ge 119899 minus

11986221198624

11986231198625

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol

(37)

Finally if one chooses

0 lt 120598 = 120598 (119899 119894 120575) le min1205981 12057511986231198625

11986221198624

(38)

then 1205821(119872) ge 119899 minus 120575 as long as (1Vol (119872)) int

119872

((119899 minus 1) minus

Ricminus)+119889vol le 120598 and this proves the theorem

Acknowledgments

The authors owe great thanks to the referees for theircareful efforts to make the paper clearer Research of thefirst author was supported by STPF of University (noJ11LA05) NSFC (no ZR2012AM010) the Postdoctoral Fund(no 201203030) of Shandong Province and PostdoctoralFund (no 2012M521302) of China Part of this work was donewhile the first author was staying at his postdoctoral mobileresearch station of QFNU

References

[1] S T Yau and R Schoen Lectures in Differential GeometryScientific Press 1988

[2] P Petersen andC Sprouse ldquoIntegral curvature bounds distanceestimates and applicationsrdquo Journal of Differential Geometryvol 50 no 2 pp 269ndash298 1998

[3] D Yang ldquoConvergence of Riemannian manifolds with integralbounds on curvature IrdquoAnnales Scientifiques de lrsquoEcole NormaleSuperieure Quatrieme Serie vol 25 no 1 pp 77ndash105 1992

[4] I Chavel Riemannian Geometry A Mordern IntroductionCambridge University Press Cambridge UK 2000

[5] C Sprouse ldquoIntegral curvature bounds and bounded diameterrdquoCommunications in Analysis andGeometry vol 8 no 3 pp 531ndash543 2000

[6] P Li and S T Yau ldquoEstimates of eigenvalues of a compactRiemannianmanifoldrdquo inAMS Proceedings of Symposia in PureMathematics pp 205ndash239 American Mathematical SocietyProvidence RI USA 1980

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 · Introduction The Laplacian is one of the most important operator on ... curvature

Abstract and Applied Analysis 3

then there exists a constant 1198623(119899) gt 0 such that 120582

1(119872) ge

1198623(119899)

Proof The proof mainly belongs to Li and Yau [6] Let 119906 bethe normalized eigenfunction of119872 set V = log (119886 + 119906) where119886 gt 1 Then we can easily get that

ΔV =minus1205821(119872) 119906

119886 + 119906minus |nablaV|

2

(8)

Denote that 119876(119909) = |nablaV|2(119909) and we then have by theRicci identity on manifolds with Ric (119872) ge 0

Δ119876 = 2V2

119894119895+ 2V119895V119895119894119894ge 2V2

119894119895+ 2 ⟨nablaV nablaΔV⟩ (9)

For the term V2119894119895 we have

sum

119894119895

V2

119894119895ge(ΔV)2

119899ge1

119899(1198762

+21205821(119872) 119906

119886 + 119906) (10)

and for the term ⟨nablaV nablaΔV⟩ we have

⟨nablaV nablaΔV⟩ = minus1198861205821(119872)

119886 + 119906119876 minus ⟨nablaV nabla119876⟩ (11)

Therefore assume 1199090isin 119872 to be the maximum of 119876 then at

1199090 we have

0 ge2

119899119876 (1199090) + (

41205821(119872)

119899minus2 (119899 + 2) 119886120582

1(119872)

119899 (119886 minus 1)) (12)

Therefore

119876 (119909) le 119876 (1199090) le

(119899 + 2) 1198861205821(119872)

119886 minus 1 (13)

Denote 120574 to be the minimizing unit speed geodesic join-ing themaximum andminimumpoints of 119906 then integrating11987612 along 120574 one will get

log( 119886

119886 minus 1) le log(119886 +max 119906

119886 minus 1) le 119889(

(119899 + 2) 1198861205821(119872)

119886 minus 1)

12

(14)

Let 119905 = (119886 minus 1)119886 then for any 119905 isin (0 1) we have (119899 +2)1205821(119872) ge 119905(119889

minus2

(log (1119905))2)Considering the maximum of the right hand and the

upper bound of the diameter derived in Lemma 7 we candeduce that a positive constant 119862

3(119899) can be found such that

1205821(119872) ge

4119890minus2

(119899 + 2) 1198892ge 1198623(119899) (15)

where 119889 is the diameter of the manifold

Corollary 9 If the manifold one discussed satisfies all theconditions in Lemma 8 and its injectivity radius satisfiesInj (119872) ge 119894 and if one let 119906 to be the normalized eigenfunctionthen there exists a constant 119862

4(119899 119894) gt 0 such that |nabla119906|2 le

1198624(119899 119894)

Proof Set 119886 = 2 in the (13) from above Then applyingLemma 6 one obtains

1

9|nabla119906|2

le |nablaV|2

(119909) le 21198621(119899 119894) (119899 + 2) (16)

therefore

|nabla119906|2

le 1198624(119899 119894) (17)

Proposition 10 Let 119872119899 be a closed 119899-dimensional Rieman-nian manifold 119906 the first eigenfunction of the Laplacian and1205821(119872) the corresponding eigenvalue thenΔ|119906|+120582

1(119872)|119906| ge 0

holds in the sense of distribution Moreover if 119872119899 is compactwith boundary then the same conclusion holds for its Neumannboundary value problem

Proof From the definition we know that Δ119906 + 1205821(119872)119906 = 0

holds on119872 Denote

119872+

= 119909 isin 119872 | 119906 (119909) gt 0

119872minus

= 119909 isin 119872 | 119906 (119909) lt 0

1198720

= 119909 isin 119872 | 119906 (119909) = 0

(18)

According to the maximum principle of elliptic equation andthe discussion about nodal set and nodal regions in [1] wecan conclude that 120597119872+ = 120597119872

minus

= 1198720 is a smooth manifold

with dimension 119899 minus 1For all 120601 isin 119862

infin

(119872) 120601 ge 0 integrating by parts we thenhave

int119872

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int119872+

|119906| Δ120601 + 1205821(119872) 120601 |119906| + int

119872minus

|119906| Δ120601 + 1205821(119872) 120601 |119906|

= int120597119872+

(119906120597120601

120597119899+minus 120601

120597119906

120597119899+) + int119872+

120601 (Δ119906 + 1205821(119872) 119906)

minus int120597119872minus

(119906120597120601

120597119899minusminus 120601

120597119906

120597119899minus) minus int119872minus

120601 (Δ119906 + 1205821(119872) 119906)

= int120597119872minus

120601120597119906

120597119899minusminus int120597119872+

120601120597119906

120597119899+ge 0

(19)

where 119899+ and 119899minus denote the outward normal direction withrespect to the boundaries of 119872+ and 119872minus respectively Notethat 120597119906120597119899minus ge 0 on 120597119872

minus and 120597119906120597119899+

le 0 on 120597119872+ for the

definition of119872minus and119872+ This completes the proof

When119872 has boundary we can apply the same reasoningexcept that the test function will require 120601 isin 119862

infin

0(119872) This

gives the proofAs long as the given manifold is compact one knows

that the first normalized eigenfunction is then determinedThis indicates that the first normalized eigenfunction ofthe Laplacian has a close relation with the geometry of

4 Abstract and Applied Analysis

the manifold In particular one would hope to bound the1198712-norm of first normalized eigenvalue of Laplacian from

below by the geometric quantities In this sense we have thefollowing result

Theorem 11 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 If 119906 is thenormalized eigenfunction of the Laplacian then there exists aconstant 119862

5(119899 119894) gt 0 such that int

119872

1199062

ge 1198625(119899 119894)

Proof We use Moser iteration to get the result From Propo-sition 10 we know that Δ|119906| + 120582

1(119872)|119906| ge 0 holds on 119872 in

the sense of distribution Set V = |119906| and take the point 119901 isin 119872

such that 119906(119901) = minus1For 119886 ge 1 denote 119877 = 1198942 120601 is a cut-off function on

119861119901(119877) then we have by integrating by parts

1205821(119872)int

119861119901(119877)

1206012

V2119886

= minusint119861119901(119877)

1206012

V2119886minus1

ΔV

= 2int119861119901(119877)

120601V2119886minus1

+ (2119886 minus 1) int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

ge 2int119861119901(119877)

120601V2119886minus1

+ 119886int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(20)

However using the identity

int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

= int119861119901(119877)

1003816100381610038161003816nabla1206011003816100381610038161003816

2

V2119886

+ 2119886int119861119901(119877)

120601V2119886minus1

⟨nabla120601 nablaV⟩

+ 1198862

int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(21)

we have

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

minus int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

(22)

therefore using the Sobolev inequality in Lemma 4

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

+ int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

ge1

119862(int119861119901(119877)

(120601V119886

)2119899(119899minus2)

)

(119899minus2)119899

=1

119862

1003817100381710038171003817120601V1198861003817100381710038171003817

2

2119899(119899minus2)

(23)

Let

120601 =

1 in 119861119901(120588)

120588 + 120590 minus 119903

120590 in

119861119901(120588 + 120590)

119861119901(120588)

0 in119861119901(119877)

119861119901(120588 + 120590)

(24)

Putting 120601 into the inequality above and we then have bysplitting the integral into three parts and using the values of120601 on each of them

V2119886119899(119899minus2)119861

119901(120588)

le [119862(1205821(119872) 119886 +

1

1205902)]

12119886

V2119886119861119901(120588+120590)

(25)

where we denote ||119891||119901Ω

= (intΩ

119891119901

)1119901 only for emphasizing

the integral domainSet

119886119895 1198860= 1 1198861=

119899

119899 minus 2 119886

119895= (

119899

119899 minus 2)

119895

120590119895 1205900=119877

4 1205901=119877

8 120590

119895=

119877

2minus(2+119895)

120588119895 120588minus1= 119877 120588

119895= 119877 minus

119895

sum

119904=0

120590119897

(26)

And putting 119886119895 120590119895 120588119895 into (25) we can derive after

iteration that

V2119886119895+1 119861119901(120588119895)le [119862(120582

1(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119886119895119861119901(120588119895minus1)

le

119895

prod

119904=0

[119862(1205821(119872) 119886

119904+

1

1205902119904

)]

12119886119904

V2119861119901(119877)

(27)

Let 119895 rarr +infin then

Vinfin119861119901(1198772)

le

infin

prod

119895=0

[119862(1205821(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119861119901(119877) (28)

The product can be estimated as follows

infin

prod

119895=0

[119862(1205821(119872) 119886

119895+

1

1205902

119895

)]

12119886119895

le

infin

prod

119895=0

[119862(1205821(119872) +

16

1198772)]

12119886119895

41198952119886119895

(29)

The right hand will converge to a fixed number by usingthe fact thatprodinfin

119895=0119861120583119895

= 119861120583(120583minus1) and the factsuminfin

119895=0119895120583minus119895 is finite

for some 119861 isin 119877 120583 gt 1 From 1205821(119872) le 119862

1(119899 119894) we can find a

positive constant 1198625(119899 119894) gt 0 such that

Vinfin119861119901(1198772)

le 119862minus12

5(119899 119894) V

2119861119901(119877)

le 119862minus12

5(119899 119894) V

2119872

(30)

Therefore

int119872

1199062

ge 1198625(119899 119894) (31)

Abstract and Applied Analysis 5

4 The Lower Bound of the First Eigenvalue

Using the same notation as above we can state the followingresult

Theorem 12 For 119899 isin N 119894 120575 isin R+ there is an 120598 = 120598(119899 119894 120575) gt

0 such that any closed manifold 119872119899 with Ric (119872) ge 0

Inj (119872) ge 119894 and

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598 (32)

will satisfy that 1205821(119872) ge 119899 minus 120575

Proof Assume that 119906 is the normalized eigenfunction ofLaplacian on 119872

119899 let 119876(119909) = |nabla119906|2

+ (1205821(119872)119899)119906

2 directcomputation shows that

1

2Δ119876 = 119906

2

119894119895+ 119906119895119906119894119894119895+ 119877119894119895119906119894119906119895+1205821(119872)

119899|nabla119906|2

+1205821(119872)

119899119906Δ119906

ge1 minus 119899

1198991205821(119872) |nabla119906|

2

+ 119877119894119895119906119894119906119895

(33)

Integrating both sides on119872119899 we have

0 ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

119877119894119895119906119894119906119895119889vol

ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

Ricminus|nabla119906|2

119889vol

=1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + (119899 minus 1) int119872

|nabla119906|2

119889vol

minus int119872

[(119899 minus 1) minus Ricminus] |nabla119906|

2

119889vol

(34)

therefore

1205821(119872) ge 119899 minus

Vol (119872)

1205821(119872) int

119872

|119906|2

119889vol1

Vol (119872)

times int119872

((119899 minus 1) minus Ricminus)+|nabla119906|2

119889vol

(35)

if we suppose that

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (36)

If 1205981is the one obtained in Lemma 7 then one has

1205821(119872) ge 119899 minus

11986221198624

11986231198625

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol

(37)

Finally if one chooses

0 lt 120598 = 120598 (119899 119894 120575) le min1205981 12057511986231198625

11986221198624

(38)

then 1205821(119872) ge 119899 minus 120575 as long as (1Vol (119872)) int

119872

((119899 minus 1) minus

Ricminus)+119889vol le 120598 and this proves the theorem

Acknowledgments

The authors owe great thanks to the referees for theircareful efforts to make the paper clearer Research of thefirst author was supported by STPF of University (noJ11LA05) NSFC (no ZR2012AM010) the Postdoctoral Fund(no 201203030) of Shandong Province and PostdoctoralFund (no 2012M521302) of China Part of this work was donewhile the first author was staying at his postdoctoral mobileresearch station of QFNU

References

[1] S T Yau and R Schoen Lectures in Differential GeometryScientific Press 1988

[2] P Petersen andC Sprouse ldquoIntegral curvature bounds distanceestimates and applicationsrdquo Journal of Differential Geometryvol 50 no 2 pp 269ndash298 1998

[3] D Yang ldquoConvergence of Riemannian manifolds with integralbounds on curvature IrdquoAnnales Scientifiques de lrsquoEcole NormaleSuperieure Quatrieme Serie vol 25 no 1 pp 77ndash105 1992

[4] I Chavel Riemannian Geometry A Mordern IntroductionCambridge University Press Cambridge UK 2000

[5] C Sprouse ldquoIntegral curvature bounds and bounded diameterrdquoCommunications in Analysis andGeometry vol 8 no 3 pp 531ndash543 2000

[6] P Li and S T Yau ldquoEstimates of eigenvalues of a compactRiemannianmanifoldrdquo inAMS Proceedings of Symposia in PureMathematics pp 205ndash239 American Mathematical SocietyProvidence RI USA 1980

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 · Introduction The Laplacian is one of the most important operator on ... curvature

4 Abstract and Applied Analysis

the manifold In particular one would hope to bound the1198712-norm of first normalized eigenvalue of Laplacian from

below by the geometric quantities In this sense we have thefollowing result

Theorem 11 Let 119872119899 be a closed 119899-dimensional Riemannianmanifold with Ric (119872) ge 0 and Inj (119872) ge 119894 If 119906 is thenormalized eigenfunction of the Laplacian then there exists aconstant 119862

5(119899 119894) gt 0 such that int

119872

1199062

ge 1198625(119899 119894)

Proof We use Moser iteration to get the result From Propo-sition 10 we know that Δ|119906| + 120582

1(119872)|119906| ge 0 holds on 119872 in

the sense of distribution Set V = |119906| and take the point 119901 isin 119872

such that 119906(119901) = minus1For 119886 ge 1 denote 119877 = 1198942 120601 is a cut-off function on

119861119901(119877) then we have by integrating by parts

1205821(119872)int

119861119901(119877)

1206012

V2119886

= minusint119861119901(119877)

1206012

V2119886minus1

ΔV

= 2int119861119901(119877)

120601V2119886minus1

+ (2119886 minus 1) int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

ge 2int119861119901(119877)

120601V2119886minus1

+ 119886int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(20)

However using the identity

int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

= int119861119901(119877)

1003816100381610038161003816nabla1206011003816100381610038161003816

2

V2119886

+ 2119886int119861119901(119877)

120601V2119886minus1

⟨nabla120601 nablaV⟩

+ 1198862

int119861119901(119877)

1206012

V2119886minus2

|nablaV|2

(21)

we have

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

minus int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

(22)

therefore using the Sobolev inequality in Lemma 4

1205821(119872) 119886int

119861119901(119877)

V2119886

1206012

+ int119861119901(119877)

V21198861003816100381610038161003816nabla120601

1003816100381610038161003816

2

ge int119861119901(119877)

1003816100381610038161003816nabla (120601V119886

)1003816100381610038161003816

2

ge1

119862(int119861119901(119877)

(120601V119886

)2119899(119899minus2)

)

(119899minus2)119899

=1

119862

1003817100381710038171003817120601V1198861003817100381710038171003817

2

2119899(119899minus2)

(23)

Let

120601 =

1 in 119861119901(120588)

120588 + 120590 minus 119903

120590 in

119861119901(120588 + 120590)

119861119901(120588)

0 in119861119901(119877)

119861119901(120588 + 120590)

(24)

Putting 120601 into the inequality above and we then have bysplitting the integral into three parts and using the values of120601 on each of them

V2119886119899(119899minus2)119861

119901(120588)

le [119862(1205821(119872) 119886 +

1

1205902)]

12119886

V2119886119861119901(120588+120590)

(25)

where we denote ||119891||119901Ω

= (intΩ

119891119901

)1119901 only for emphasizing

the integral domainSet

119886119895 1198860= 1 1198861=

119899

119899 minus 2 119886

119895= (

119899

119899 minus 2)

119895

120590119895 1205900=119877

4 1205901=119877

8 120590

119895=

119877

2minus(2+119895)

120588119895 120588minus1= 119877 120588

119895= 119877 minus

119895

sum

119904=0

120590119897

(26)

And putting 119886119895 120590119895 120588119895 into (25) we can derive after

iteration that

V2119886119895+1 119861119901(120588119895)le [119862(120582

1(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119886119895119861119901(120588119895minus1)

le

119895

prod

119904=0

[119862(1205821(119872) 119886

119904+

1

1205902119904

)]

12119886119904

V2119861119901(119877)

(27)

Let 119895 rarr +infin then

Vinfin119861119901(1198772)

le

infin

prod

119895=0

[119862(1205821(119872)119886119895+

1

1205902

119895

]

12119886119895

V2119861119901(119877) (28)

The product can be estimated as follows

infin

prod

119895=0

[119862(1205821(119872) 119886

119895+

1

1205902

119895

)]

12119886119895

le

infin

prod

119895=0

[119862(1205821(119872) +

16

1198772)]

12119886119895

41198952119886119895

(29)

The right hand will converge to a fixed number by usingthe fact thatprodinfin

119895=0119861120583119895

= 119861120583(120583minus1) and the factsuminfin

119895=0119895120583minus119895 is finite

for some 119861 isin 119877 120583 gt 1 From 1205821(119872) le 119862

1(119899 119894) we can find a

positive constant 1198625(119899 119894) gt 0 such that

Vinfin119861119901(1198772)

le 119862minus12

5(119899 119894) V

2119861119901(119877)

le 119862minus12

5(119899 119894) V

2119872

(30)

Therefore

int119872

1199062

ge 1198625(119899 119894) (31)

Abstract and Applied Analysis 5

4 The Lower Bound of the First Eigenvalue

Using the same notation as above we can state the followingresult

Theorem 12 For 119899 isin N 119894 120575 isin R+ there is an 120598 = 120598(119899 119894 120575) gt

0 such that any closed manifold 119872119899 with Ric (119872) ge 0

Inj (119872) ge 119894 and

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598 (32)

will satisfy that 1205821(119872) ge 119899 minus 120575

Proof Assume that 119906 is the normalized eigenfunction ofLaplacian on 119872

119899 let 119876(119909) = |nabla119906|2

+ (1205821(119872)119899)119906

2 directcomputation shows that

1

2Δ119876 = 119906

2

119894119895+ 119906119895119906119894119894119895+ 119877119894119895119906119894119906119895+1205821(119872)

119899|nabla119906|2

+1205821(119872)

119899119906Δ119906

ge1 minus 119899

1198991205821(119872) |nabla119906|

2

+ 119877119894119895119906119894119906119895

(33)

Integrating both sides on119872119899 we have

0 ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

119877119894119895119906119894119906119895119889vol

ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

Ricminus|nabla119906|2

119889vol

=1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + (119899 minus 1) int119872

|nabla119906|2

119889vol

minus int119872

[(119899 minus 1) minus Ricminus] |nabla119906|

2

119889vol

(34)

therefore

1205821(119872) ge 119899 minus

Vol (119872)

1205821(119872) int

119872

|119906|2

119889vol1

Vol (119872)

times int119872

((119899 minus 1) minus Ricminus)+|nabla119906|2

119889vol

(35)

if we suppose that

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (36)

If 1205981is the one obtained in Lemma 7 then one has

1205821(119872) ge 119899 minus

11986221198624

11986231198625

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol

(37)

Finally if one chooses

0 lt 120598 = 120598 (119899 119894 120575) le min1205981 12057511986231198625

11986221198624

(38)

then 1205821(119872) ge 119899 minus 120575 as long as (1Vol (119872)) int

119872

((119899 minus 1) minus

Ricminus)+119889vol le 120598 and this proves the theorem

Acknowledgments

The authors owe great thanks to the referees for theircareful efforts to make the paper clearer Research of thefirst author was supported by STPF of University (noJ11LA05) NSFC (no ZR2012AM010) the Postdoctoral Fund(no 201203030) of Shandong Province and PostdoctoralFund (no 2012M521302) of China Part of this work was donewhile the first author was staying at his postdoctoral mobileresearch station of QFNU

References

[1] S T Yau and R Schoen Lectures in Differential GeometryScientific Press 1988

[2] P Petersen andC Sprouse ldquoIntegral curvature bounds distanceestimates and applicationsrdquo Journal of Differential Geometryvol 50 no 2 pp 269ndash298 1998

[3] D Yang ldquoConvergence of Riemannian manifolds with integralbounds on curvature IrdquoAnnales Scientifiques de lrsquoEcole NormaleSuperieure Quatrieme Serie vol 25 no 1 pp 77ndash105 1992

[4] I Chavel Riemannian Geometry A Mordern IntroductionCambridge University Press Cambridge UK 2000

[5] C Sprouse ldquoIntegral curvature bounds and bounded diameterrdquoCommunications in Analysis andGeometry vol 8 no 3 pp 531ndash543 2000

[6] P Li and S T Yau ldquoEstimates of eigenvalues of a compactRiemannianmanifoldrdquo inAMS Proceedings of Symposia in PureMathematics pp 205ndash239 American Mathematical SocietyProvidence RI USA 1980

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 · Introduction The Laplacian is one of the most important operator on ... curvature

Abstract and Applied Analysis 5

4 The Lower Bound of the First Eigenvalue

Using the same notation as above we can state the followingresult

Theorem 12 For 119899 isin N 119894 120575 isin R+ there is an 120598 = 120598(119899 119894 120575) gt

0 such that any closed manifold 119872119899 with Ric (119872) ge 0

Inj (119872) ge 119894 and

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598 (32)

will satisfy that 1205821(119872) ge 119899 minus 120575

Proof Assume that 119906 is the normalized eigenfunction ofLaplacian on 119872

119899 let 119876(119909) = |nabla119906|2

+ (1205821(119872)119899)119906

2 directcomputation shows that

1

2Δ119876 = 119906

2

119894119895+ 119906119895119906119894119894119895+ 119877119894119895119906119894119906119895+1205821(119872)

119899|nabla119906|2

+1205821(119872)

119899119906Δ119906

ge1 minus 119899

1198991205821(119872) |nabla119906|

2

+ 119877119894119895119906119894119906119895

(33)

Integrating both sides on119872119899 we have

0 ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

119877119894119895119906119894119906119895119889vol

ge1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + int119872

Ricminus|nabla119906|2

119889vol

=1 minus 119899

1198991205821(119872)int

119872

|nabla119906|2

119889vol + (119899 minus 1) int119872

|nabla119906|2

119889vol

minus int119872

[(119899 minus 1) minus Ricminus] |nabla119906|

2

119889vol

(34)

therefore

1205821(119872) ge 119899 minus

Vol (119872)

1205821(119872) int

119872

|119906|2

119889vol1

Vol (119872)

times int119872

((119899 minus 1) minus Ricminus)+|nabla119906|2

119889vol

(35)

if we suppose that

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol le 120598

1(119899) (36)

If 1205981is the one obtained in Lemma 7 then one has

1205821(119872) ge 119899 minus

11986221198624

11986231198625

1

Vol (119872)int119872

((119899 minus 1) minus Ricminus)+119889vol

(37)

Finally if one chooses

0 lt 120598 = 120598 (119899 119894 120575) le min1205981 12057511986231198625

11986221198624

(38)

then 1205821(119872) ge 119899 minus 120575 as long as (1Vol (119872)) int

119872

((119899 minus 1) minus

Ricminus)+119889vol le 120598 and this proves the theorem

Acknowledgments

The authors owe great thanks to the referees for theircareful efforts to make the paper clearer Research of thefirst author was supported by STPF of University (noJ11LA05) NSFC (no ZR2012AM010) the Postdoctoral Fund(no 201203030) of Shandong Province and PostdoctoralFund (no 2012M521302) of China Part of this work was donewhile the first author was staying at his postdoctoral mobileresearch station of QFNU

References

[1] S T Yau and R Schoen Lectures in Differential GeometryScientific Press 1988

[2] P Petersen andC Sprouse ldquoIntegral curvature bounds distanceestimates and applicationsrdquo Journal of Differential Geometryvol 50 no 2 pp 269ndash298 1998

[3] D Yang ldquoConvergence of Riemannian manifolds with integralbounds on curvature IrdquoAnnales Scientifiques de lrsquoEcole NormaleSuperieure Quatrieme Serie vol 25 no 1 pp 77ndash105 1992

[4] I Chavel Riemannian Geometry A Mordern IntroductionCambridge University Press Cambridge UK 2000

[5] C Sprouse ldquoIntegral curvature bounds and bounded diameterrdquoCommunications in Analysis andGeometry vol 8 no 3 pp 531ndash543 2000

[6] P Li and S T Yau ldquoEstimates of eigenvalues of a compactRiemannianmanifoldrdquo inAMS Proceedings of Symposia in PureMathematics pp 205ndash239 American Mathematical SocietyProvidence RI USA 1980

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: A Global Curvature Pinching Result of the First Eigenvalue of the Laplacian … · 2014-04-25 · Introduction The Laplacian is one of the most important operator on ... curvature

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Fast Local Laplacian Filters: Theory and Applications · Fast Local Laplacian Filters: Theory and Applications • 3 Local Laplacian filtering. Paris et al. [2011] introduced local

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Prokaryotic cells divide by pinching in two

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Analysis of Hand Strength for Pinching Activityeprints.utem.edu.my/5161/1/Analysis_Of_Hand_Strenght_For... · 2015-05-28 · Analysis of Hand Strength for Pinching Activity Thesis

Analysis of Hand Strength for Pinching Activityeprints.utem.edu.my/5161/1/Analysis_Of_Hand_Strenght_For... · 2015-05-28 · Analysis of Hand Strength for Pinching Activity Thesis

Guia Thesis Complex Laplacian

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Research Article A Global Curvature Pinching Result of the ...downloads.hindawi.com/journals/aaa/2013/237418.pdfA Global Curvature Pinching Result of the First Eigenvalue of the Laplacian

Research Article A Global Curvature Pinching Result of the ...downloads.hindawi.com/journals/aaa/2013/237418.pdfA Global Curvature Pinching Result of the First Eigenvalue of the Laplacian