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Harnack inequality for nondivergent linear elliptic operators on Riemannian manifolds : a self-contained… Mun, Byeongju 2013

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Harnack Inequality for NondivergentLinear Elliptic Operators onRiemannian ManifoldsA Self-contained ProofbyByeongju MunB.Sc., Sungkyunkwan University, 2011A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)September 2013c? Byeongju Mun 2013AbstractIn this paper, a self-contained proof is given to a well-known Harnack in-equality of second order nondivergent uniformly elliptic operators on Rie-mannian manifolds with the condition that M? [R(?)] > 0, following theideas of M. Safonov [5]. Basically, the proof consists of three parts: 1)Critical Density Lemma, 2) Power-Decay of the Distribution Functions ofSolutions, and 3) Harnack Inequality.iiPrefaceChapter 2. Preliminaries is a summary of the introductions on Riemannianmanifolds in Cabre? [1] and S. Kim [2].Chapter 3. Critical Density Lemma follows the results and the proofs alsoby Cabre? [1] and S. Kim [2] with more comments and explanations for clarity.In Chapter 4, The ideas of Safonov [5] is followed for a covering lemma onRiemannian manifolds and a power-decay property of the distribution func-tions of solutions.Chapter 5. Harnack Inequality follows the ideas of Cabre? [1], Caffarelli [4],and Caffarelli and Cabre? [3].iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Riemannian Geometry . . . . . . . . . . . . . . . . . . . . . 32.2 Differential Operators on Riemannian Manifolds . . . . . . . 62.3 Lemmas on Riemannian Manifolds . . . . . . . . . . . . . . . 73 Critical Density Lemma . . . . . . . . . . . . . . . . . . . . . . 114 Power Decay of the Distribution Functions of Solutions . 255 Harnack Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 32Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39AppendicesA Some Universal Constants . . . . . . . . . . . . . . . . . . . . 41ivAcknowledgementsDo homage to the Son ... (Psalm 2 : 12)Were it not for Jesus Christ , my writing would have been terrifying andcompleting this project would have been overwhelming. He has been theonly true fountain of comfort, happiness, strength, and wisdom for me; theAlmighty Redeemer, the Gracious Lord and the Good Sheperd of my lifeand study in Vancouver.I thank my joint-supervisors: Prof. Kim, Young-Heon for his enthusiasmand constant support for my study through which I could start and finishthis paper, and Prof. Tsai, Tai-Peng for his generous hospitality and kind-ness which provided me with a precious opportunity to study mathematicsin University of British Columbia. Furthermore, they much helped me tocarefully proofread this paper by suggesting improvements and detectingerrors.I also appreciate Prof. Chae, Dongho for his counsel and inspiring lectureswhich motivated me to seriously study Analysis, Prof. Lee, Jihoon for hiswarm favors and cheerfulness which helped me to broaden my perspectivestowards mathematics and problem-solving, Prof. Park, Jeong-Hyeong forparental hospitality and kind instruction which helped me to have interestsin Geometry, and Prof. Kang, Kyungkeun for his unusual guidance andthoughtful consideration which helped me not also to start and to have in-terests in studying this interesting area, Partial Differential Equations, butalso to reflect upon myself and my study.Moreover, I feel deep gratitude to Lee Yupitun who is our Graduate Pro-gram Coordinator and a special helper for graduate students in need, tomy warm friends?Alok and Subhashini, and Woo-Hang?who did not spareheartful advises for my study and showed our family deep hospitality, to myable colleagues?Dong-Seok, Dimitris Roxanos, Tatchai Titichetrakun, andYuwen Luo?who have been warm friends and encouraging helpers duringmy graduate study of mathematics, and to my families in Christ ?Irene,vAcknowledgementsStone and Erica, In-Young and Kevin, Jeong-Ah, Young-Tae, Junho andKyeong-Hye, Jae-Ho and Seon-Hee, Gui-Soon and Young-Sook, Soon-Gi,and Douglas and Florence?who have had edifying fellowship with our fam-ily in His Church, helped us in various aspects of life in Vancouver, andprayed for me to keep finishing my degree.Finally, I am very grateful to my wife, Sang-Eun for her patience and lov-ingkindness to wait for me and to take faithful care of John, our first sonwho was born during this project. It sincerely helped me to concentrate onmy work and study.viDedicationTo the only true triune God in Jesus Christ , our Savior and Lord ,to my families in His Church,in memory of my father, Yeon-Sik Mun,to my mother, Young-Sook Kim,and to my wife and our first son, Sang-Eun and John Giju.viiChapter 1IntroductionThere are many results about Harnack inequality on Riemannian manifolds.First, Yau, S-T [9] proved Harnack inequality for positive harmonic func-tions on Riemannian manifolds with nonnegative Ricci curvature. Also,Saloff-Coste [10] and Grigor?yan [11] obtained that the volume doublingproperty of measure on manifolds and a kind of weak Poincare? inequaltygive Harnack inequality for solutions of divergence parabolic equations onRiemannian manifolds. Especially, for the defintion of second-order, linear,nondivergent uniformly elliptic operators L, Cabre? [1] and Stroock [12] canreferred to, and the definition will be noted in Section 2.2.In this paper, a self-contained proof of Harnack inequality on Riemannianmanifolds with the condition thatM? [R(?)] > 0 (See Section 2.1), which isstronger?we cannot guarantee whether it is strictly or not?than nonnegativeRicci curvature condition, but weaker than nonnegative sectional curvature,is given. This result was earlier proven by S.Kim [2]; however, S.Kim onlygave the proofs of Critical Density Lemma part and refered to Cabre? [1] forthe remaining parts: Power-decay of distribution functions of solutions, andHarnack inequality.In fact, Cabre? [1] proved Harnack inequality for nondivergent elliptic opera-tors on Riemannian manifolds with nonnegative sectional curvatures, and S.Kim [2] proved a similar result with the condition thatM? [R(?)] > 0. Thedifference between the conditions on Riemannian manifolds needs differentproofs only for Critical Density Lemma part; in fact, the essense of S. Kim[2] improved from Cabre? [1] was the computationLdy(x) ?aLdy(x)with aL = (n? 1)?for any x ?M\ [Cut(y) ? {y}] under the geometric condition M? [R(?)] >0, and using it to get Critical Density Lemma. On the other hand, Cabre?usedD2dy(x)(?, ?) 61dy(x)|?|21Chapter 1. Introductionfor any x ? M\ [Cut(y) ? {y}] and any ? ? TxM under the condition ofnonnegative sectional curvature to get the lemma.The strategy used by Cabre? and S.Kim is basically based on the proof ofHarnack inequality in Rn by Caffarelli [4]. In Cabre??s paper, for the proofof Power-decay of distribution functions of solutions part, he followed Caf-farelli?s arguments and applied a kind of Caldero?n-Zygmund decompositionon Riemannian manifolds with nonnegative sectional curvature condition,using a result of M. Christ [6] which is highly nontrivial. S. Kim alsodirectly followed the arguments of Cabre? for that part since nonnegativeRicci curvature condition is sufficient to get the doubling-property of vol-ume (see Lemma 2.3) essential for the decomposition, and the conditionthat M? [R(?)] > 0 implies the nonnegativity of Ricci curvature (see Sec-tion 2.1).For the part, a more elementary approach using a covering lemma by Sa-fonov (see Lemma 4.2) is applied, and a simpler proof than those of Cabre?and S. Kim is given in this paper; this is a small improvement. In fact,there is a similar result using a covering lemma, by Aimar and Forzani,and Toledano [7] in a more general and abstract setting, e.g. homogeneousspaces. However, our proof is simpler and more direct, focusing on the caseof Riemannian manifolds.With respect to the parts of Critical Density Lemma and Power-decayproperty of distribution functions, Safonov [5] started with Alexandrov-Bakelman-Pucci estimate whose proof is well-known in Rn, and obtaineda Growth Lemma which is less restrictive for applications and a Double-section Lemma which make the proof for Harnack simple and direct. How-ever, it was difficult for us to get a lemma which is silimar to Alexandrov-Bakelman-Pucci estimate. Thus, except the part for Power-decay propertyof distribution functions, we basically followed the ideas and arugments ofCabre?, and S. Kim; that is, our paper can be regarded as a self-containedexposition of the results of Safonov, Cabre?, and S. Kim.2Chapter 2Preliminaries2.1 Riemannian GeometryNotationThe notation for some concepts on Riemannian manifolds is given as the fol-lowing. Let M be a smooth n-dimensional complete Riemannian manifoldwith a metric g. And the geodesic distance between points x and y on Mis denoted by d(x, y) or dy(x) or dx(y), the Riemannian measure of M bydV, the tangent space of M at x by TxM, and the Riemannian curvaturetensor by R(X,Y )Z. For convenience, the geodesic distance is sometimescalled just by distance.Since the geodesic parametrized by the arc-length and exponential map-ping on Riemannian manifolds will be often used later on, a summary ofthem including the concept of cut-points of a point x in M is given.Exponential FunctionsIf the exponential map expx : TxM ??M is considered, for any ? ? TxMwith |?| = 1, a function ?(t) = expx(t?) can be set. Then, ?(t) is thegeodesic parametrized by arc-length, that is, with unit-speed which satisfies?(0) = x and ??(0) = ?. Here, a constant t0 is defined byt0 = t0(x, ?) = sup { s > 0 | ?(t) is the minimal geodesic from x to ?(s)} .When t0 < ?, the point ?(t0) is called the cut point of x along ?(t), andthe set Cut(x) is defined byCut(x) = {cut point of x along ?(t) = expx(t?) | ? ? TxM and |?| = 1} .Then, it is well-known that Cut(x) has zero n-dimensional Riemannian mea-sure. An another set Gx is defined byGx = { t? | 0 6 t < d(x, expx(t0(x, ?)?)) for ? ? TxM with |?| = 1} .32.1. Riemannian GeometryThen, it is also well-known thatexpx : Gx ?? expx(Gx) is a diffeomorphism.Also, a concept of second-variation of vector-fields on Riemannian manifoldsis necessary to get a estimate of differential operators acted on the distancefunction under a geometric condition (see Lemma 2.1).Curvatures and Morse Index FormThe Riemannian curvature tensor is defined byR(X,Y )Z = ?X?Y Z ??Y?XZ ??[X,Y ]Z,where ? is the Levi-Civita connection. And, for a unit tangent vector v inTxM, the Ricci transformation R(v) : TxM ?? TxM is defined byR(v)X = R(X, v)v.For a given geodesic ? : [0, l] ?? M parametrized by the arc-length, theMorse index form I(V,W ) is defined byI(V,W ) =? l0{????V,???W ? ? g?R(??, V )W,???}dt,where V,W are piecewise smooth vector fields along ?.From now on, specific Riemannian manifolds satisfying the following condi-tion are considered in Chapter 3-5.Geometric AssumptionsIt is noted thatdp is smooth on M\ [Cut(p) ? {p}] .To express the condition of M under which Harnack inequalty is proven inthis paper, the Pucci?s extremal operator for a symmetric endomorphism Aon TxM is introduced:M? [A, ?,?] =M? [A] = ???j>0?j + ???j<0?j ,42.1. Riemannian Geometrywhere ?j are the eigenvalues of A. The sufficient condition for Harnackinequality found by S. Kim [2] was the following:M? [R(v)] > 0,for any x ?M and ? ? TxM with |?| = 1, where R(?) is the Ricci trans-formation on TxM in Section 2.1. It is noted that the condition is strongerthan the condition of nonnegative Ricci curvature, i.e. M? [R(v)] > 0 on Mimplies that M has nonnegative Ricci curvature. This can be easily checkedby the following:M? [R(v)] 6 tr (A(x) ?R(v))if A(x) is uniformly elliptic (see Section 2.2); especially, when A(x) = ?Id,nonnegative Ricci curvature condition for M is satisfied. Moreover, it is alsonoted that any Riemannian manifold with nonnegative sectional curvaturetrivially satisfies the conditionM? [R(v)] > 0,thus the condition is stronger than nonnegative Ricci curvature condition;however, weaker than nonnegative sectional curvature condition. The con-dition will be always assumed for any Riemannian manifold M in Chapter3-5.52.2. Differential Operators on Riemannian Manifolds2.2 Differential Operators on RiemannianManifoldsThe definitions of some elementary differential operators on Riemannianmanifolds are summarized.Gradient and Hessian of Functions on ManifoldsIt is noted that the Hessian of a function u at a point x in M is defined asan endomorphism of TxM byD2u ? ? = D??u ?? ? TxM,where D denotes the Levi-Civita connection in M and ?u(x) is the gradientof u at x.Second-order Nondivergent Linear Uniformly EllipticOperators on ManifoldsFirst, let A(x) be a positive definite symmetric endomorphism of TxM. Itis assumed that A(x) satisfies the uniformly ellipticity with some positiveconstants ? and ?:? |?|2 6 g ?A(x)?, ?? 6 ? |?|2 ?x ?M, ?? ? TxM,where |?|2 = g ??, ??.Next, a second-order, nondivergent, linear, uniformly elliptic operator L isdefined byLu = tr(A(x) ?D2u)= tr {? 7?? A(x)???u} ,where tr is the trace of endomorphism, ? is composition of endomorphisms,and D2u is the Hessian of a function u.62.3. Lemmas on Riemannian Manifolds2.3 Lemmas on Riemannian ManifoldsThis section is basically for some computations in the proof of Lemma 3.2.All the lemmas in this section directly refer to Cabre? [1] and S.Kim [2];however, some of the proofs are included for clarity.The following lemma provides a boundness of an elliptic operator L op-erated on the distance function under a geometric condition.Lemma 2.1. (S. Kim [2]) Let M satisfy M? [R(?)] > 0 on M (see Section2.1.) Let p be a point on M and x ?M\ [Cut(p) ? {p}]. Then, it is obtainedthatLdp(x) ?aLdp(x), where aL = (n? 1)?.Proof. Let ? : [0, l] ?? M be the minimal geodesic parametrized by arc-length from p = ?(0) to x = ?(l), and choose an orthnormal basis {j}nj=1 onTxM satisfying that 1 = ??(l) and {j}nj=1 are eigenvectors of D2dp. Here,by parallel transport along ?(t), {j}nj=1 can be extended to {j(t)}nj=1 with aparameter t ? [0, l]. If the Jacobi fields along ?(t), Vj(t) is defined, satisfying1) Vj(0) = 0 and Vj(l) = ej ,2)[Vj(t), ??(t)]= 0,then it is obtained that?D2dp(j), j?= ????Vj , Vj? (l) = I(Vj , Vj).Here, since a Jacobi field minimizes the Morse index form among all vectorfields along the same geodesic with the same boundary data, it is obtainedthatI(Vj , Vj) 6 I(tlj(t),tlj(t)).72.3. Lemmas on Riemannian ManifoldsThus, it can be computed thatLdp(x) =n?j=2ajj?D2dp(j), j?6n?j=2ajjI(Vj , Vj)6n?j=2ajjI(tlj(t),tlj(t))=n?j=2ajj? l0????1l????2?? l0(tl)2n?j=2ajj?R(??, j)j , ???6n?j=2ajj? l0????1l????2?? l0(tl)2M?[R(??)],and since M? [R(?)] > 06(n? 1)?l=(n? 1)?dp(x).This finishes the proof of the lemma.For the proof of Lemma 3.2, a computation of the Jacobian of the expo-nential mapping of the gradients of smooth functions on manifolds will benecessary.Lemma 2.2. (Cabre? [1]) Let v be a smooth function in an open set ? ofM. For the map ? : ? ??M defined by?(x) = expx?v(x),whenever ?v(x) ? Gx for some x ? ?, the following is satisfiedJac ?(x) = Jac expx(?v(x)) ??????detD2(v +d2y2)(x)?????,where y = ?(x) and Jac expx(?v(x)) denotes the Jacobian of the exponentialmapping evaluated at ?v(x) ? TxM.Let d be the exterior differentiation on M in the following proof.82.3. Lemmas on Riemannian ManifoldsProof. Set a geodesic ?(t) satisfying ?(0) = x and ? ?(0) = ? and a family ofgeodesics with a parameter s is considered by?s(t) = exp?(t) s?v(?(t)).Here, it is noted that ?0(t) = ?(t) and ?1(t) = exp?(t)?v(?(t)). Now, someJacobi fields are considered. First, a Jacobi field J(s) along expx s?v(x) isdefined byJ(s) =??t|t=0 ?s(t).Then, J(s) satisfiesJ(0) = ?, J(1) = d {expx?v(x)} ? ? and DsJ(0) = D2v(x) ? ?since Ds ???t |s=0= Dt???s |s=0= Dt?v(?(t)) = D2v(?(t)) ? ? ?(t), where d isthe exterior differentiation on M. Next, an another Jacobi field J?(s) alongexpx s?v(x) is set, satisfyingJ?(0) = ? and J?(1) = 0,and an another Jacobi field J?? also along expx s?v(x) byJ?? = J ? J?,then J?? naturally satisfies the following:J??(0) = 0 and DsJ??(0) = D2v(x) ? ? ?DsJ?(0),and alsod {expx?v(x)} ? ? = J(1) = J??(1) = d expx |?v(x) ?DsJ??(0).Here, consider an another family of geodesics ?s(t) = exp?(t) s exp?1?(t) y sat-isfying?0(t) = ?(t) and ?1(t) = y.Then, it is obtained that??t|t=0 ?s(t) = J?(s),and?DsJ?(0) = ?Ds??t|t=0 ? |s=0= Dt{? exp?1?(t) y}|t=0= Dt?(d2y2)(?(t)) |t=0= D2d2y2(x) ? ?.92.3. Lemmas on Riemannian ManifoldsThus, it follows thatDsJ??(0) = D2(v +d2y2)(x) ? ?Note. For Lemma 2.2, any curvature condition is not necessary, but ?v(x) ?Gx should be checked to use it.The doubling-property of measure under a geometric condition?nonnegativeRicci curvature?will be used to get a covering lemma(see Lemma 4.2).Lemma 2.3. (Gromov) Let M be an n-dimensional Riemannian manifoldwith nonnegative Ricci curvature. For any balls in M, M satisfies the volumedoubling property:|B2R(x)| ? 2n |BR(x)| .Proof. Since the proof of this lemma is purely geometric, Chavel [13] isreferred to.10Chapter 3Critical Density LemmaThe following theorem is called Critical Density Lemma since it gives animportant result that a sufficient measure?Critical Density?of a cuf-off setby a constant in a ball implies the lower-boundness of the function in alarger ball.Theorem 3.1. (modified Critical Density) Let u be a nonnegative smoothfunction in a ball BR(x0) and satisfy Lu 6 0 in the same ball. Then, u hasthe following property:???{u ? 1} ?BR8(x0)??????BR8(x0)???> ?1 =? infBR4(x0)u > ?1,where ?1 = 1? {n?12(C 12+ aL + ?)}n and ?1 =15732 + C 12.Note. For convenience, every universal constant which depends only on di-mension n and ellipticity constants ?, ? is collected in Appendix A.Remark. 1. C 12is to be given as C? with ? = 12 in the proof of Theorem 3.1,Step 1, pp20-22, and aL = (n? 1)? is given in Lemma 2.2.2. It suffices to check thatC 12> n?to guarantee that ?1 > 0, and it is trivial from the definition of C?.3. Theorem 3.1 can be restated as the following:For any 0 < ?1 615732 + C 12< 1, there exists a constant ?1 = ?1(n, ?,?)such that infBR4(x0)u 6 ?1 =????{u ? 1} ?BR8(x0)??????BR8(x0)???6 ?1.11Chapter 3. Critical Density LemmaThe proof of the theorem is given in the end of this section after somelemmas which are necessary to prove it. The following lemma is also calledCritical Density Lemma?only the radius of ball where the infimum is takenis different?, essential for Theorem 3.1, and the proof of it is due to Cabre? [1]and S. Kim [2] on Riemannian manifolds. Here, a detailed proof is includedfor clarity. In Rn, Safonov [5] proved a similar result named Growth Lemmausing ABP-estimate, and the lemma has a larger extent of constant ?1, thatis, ?1 ? (0, 1). This benefit comes from the application of the ABP-estimateon Rn.Lemma 3.2. (Critical Density) Let u be a nonnegative smooth function in aball B4R(x0) and infBR(x0) u 6 1. Then, the following inequality is satisfied:???BR4??? 61(n?)n?{u6 5732}?B3R(x0){(R2Lu+ aL + ?)+}n dV,where f+(x) = max{f(x), 0} and aL = (n? 1)?.Proof. From now on, BR will be used in every chapter to notify BR(x0) forsimplicity if there is no risk of confusion. First, a point y in BR4is arbitrarilychosen, and a continuous function wy is defined bywy(x) = R2u(x) +12d2y(x), .Step 1.A minimum of wy in B4R is achived at a point z0 in B3R.It is noted thatinfBRwy 6 R2 +(R+ R4)22= R2 +25R232=57R232< 2R2,and, since u > 0 in B4R, it is obtained thatwy(x) >(3R?R)22= 2R2 in B4R\B3R.Thus, it is concluded that the minimum of wy(x) in B3R is achieved at apoint of B3R, and that is also the minimum of wy(x) in B4R. That is,infB4Rwy = infB3Rwy = wy(z0) for some z0 in B3R.12Chapter 3. Critical Density LemmaSo, the claim of Step 1 is proved.Here, it is noted that for any y in BR4there exists such z0, and a set Aof such z0?s is defined byA ={z ? B3R | z is a minimum point of wy(x) in B3R for y ? BR4}.Step 2. The two points on M, y and z0, which were considered in Step 1have the following relation:y = expz0 ?(R2u)(z0) .From Step 1, it is easily noted that for ?z0 ? A,wy (z0) 6 wy (x) = R2u (x) +12d2y (x) ?x ? B4R.First, an arbitrary geodesic ? parametrized by the arc-length is considered,satisfying ?(0) = z0. Then, for any t,dy (?(t)) 6 t+ dy (z0) , thus,wy(z0) 6 R2u(?(t)) +12d2y(?(t)) 6 R2u(?(t)) +12{t+ dy(z0)}2 .Here, it is noted that both inequalities become equality when t = 0, that is,if a function f is set byf(t) = R2u(?(t)) +12{t+ dy(z0)}2 ? wy(z0),then f has its minimum when t = 0. Thus, if f is differentiated with respectto t at t = 0,0 6 f ?(t) |t=0 = g??(R2u)(?(t)), ??(t)?|t=0 + {t+ dy(z0)} |t=0= g??(R2u)(z0), ??(0)?+ dy(z0).? g??(R2u)(z0),???(0)?6 dy(z0).Here, since the geodesic ? was arbitrarily chosen only to satisfy ?(0) = z0with unit-speed, (3) is satisfied for any unit vector ? ?Mz0 instead of ???(0),that is,g??(R2u)(z0), ??6 dy(z0) ?? ?Mz0 with |?| = 1. (3.1)13Chapter 3. Critical Density LemmaNext, an another minimal geodesic ? parametrized by arc-length is con-sidered; however, joining z0 and y by the condition that ?(0) = z0 and?(dy(z0)) = y. Then, it is obtained thatdy(z0) = dy(?(t)) + t for 0 6 t 6 dy(z0), thus,wy(z0) 6 wy(?(t)) = R2u(?(t)) +12d2y(?(t)) = R2u(?(t)) +12{dy(z0)? t}2 .Here, it is also noted that this inequality become equality when t = 0, thatis, if a function h is defined byh(t) = R2u(?(t)) +12{dy(z0)? t}2 ? wy(z0),then h has its minimum when t = 0. Thus, by differentiation, it is computedthat0 6 h?(t) |t=0 = g??(R2u)(?(t)), ??(t)?|t=0 + {dy(z0)? t} |t=0= g??(R2u)(z0), ??(0)?? dy(z0).? g??(R2u)(z0), ??(0)?> dy(z0). (3.2)Here, since |??(0)| = 1, from (3.1) and (3.2) it is concluded thatg??(R2u)(z0), ??(0)?= dy(z0)? ?(R2u)(z0) = dy(z0)??(0).Then by the definition of exponential mapping on Riemannian manifold? expz0 ?(R2u)(z0) = ?(dy(z0)) = y.So, the claim of Step 2 is proved.If a smooth map ? : B3R ?? M is defined by ?(z) = expz?(R2u)(z),then we have proved that for any y ? BR4there is at least one z ? A suchthat ?(z) = y. Thus, by virtue of the area formula it is obtained that|BR4| 6?AJac ?(z) dV(z).14Chapter 3. Critical Density LemmaAnd, from Step 1, it is easily noted thatA ?{u 65732}?B3R.Hence, for the proof of this lemma, it suffices to showStep 3.Jac ?(z) 61(n?)n{(R2Lu(z) + aL + ?)+}n for any z ? A.Let z1 ? A and take y1 ? BR4such that wy1(z1) = infB3R wy1 , that is,y1 = ?(z1) = expz1 ?(R2u)(z1). Here, there are two different cases: 1) z1is not a cut point of y1, or 2) z1 is a cut point of y1.Case 1) This is the easier case. If z1 is not a cut point of y1?i.e. ?(R2u)(z1) ?Gz1?, then by Lemma 2.2 it is attained thatJac ?(z1) 6?????detD2(R2u+d2y12)(z1)?????= |detD2wy1(z1)|.Here, it is used thatJac expx(?) 6 1for any x ? M and ? ? Gx. Li [17] (see Bishop Comparison Theorem), S.Kim [2] can be referred to for details, and the sketch of it is the following:1)Let J(r, ?)d? be the area element of the geodesic sphere ?Br(x),2)J(r, ?)d? = rn?1A(r, ?)d? where A(r, ?) is the Jacobian of the map,expx at r? ? TxM,3)By the Laplace Comparison Theorem under nonnegative Ricci curvaturecondition (see Schoen[15] and Schoen and Yau[16]), it is obtained that?dp(x) 6n? 1dp(x),4)In Li[17], it is computed thatJ ?(r, ?)J(r, ?)= ?r, and by 3) ?r 6n? 1r,5)From 4) it is obtained that A(r, ?) is nondecreasing with respect to r,6)Since limr?0A(r, ?) = 1, it is concluded that A(r, ?) 6 1.15Chapter 3. Critical Density LemmaSince wy1 achieves its minimum at z1, D2wy1(z1) > 0. Therefore, by usingthe well-known inequalitydetA ?detB 6{tr (A ?B)n}nwhere A,B are symmetric and nonnegative,it is concluded thatJac ?(z1) 6 detD2wy1(z1)=1detA(z1)detA(z1) ? detD2wy1(z1)61?ndetA(z1) ? detD2wy1(z1)61(n?)n[tr{A(z1) ?D2wy1(z1)}]n=1(n?)n{Lwy1(z1)}n=1(n?)n{R2Lu(z1) + L(d2y12)(z1)}n61(n?)n{(R2Lu(z1) + aL + ?)+}n,where, in the last step, by Lemma 2.1 it is computed thatL(d2y12) = dy1Ldy1 + ?A?dy1 ,?dy1? 6 aL + ?|?dy1 |2,where?dy1(z) = ?exp?1z y1??exp?1z y1??if z 6= y1.Thus, the claim of Step 3 for the case of when z1 is not a cut point of y1 isproved.Case 2) When z1 is a cut point of y1, we can reduce this kind of criticalsituation to the previous non-critical case by using upper barrier techniquedue to Calabi [14] as the followings:(Upper Barrier Technique) Since y1 = expz1 ?(R2u)(z1), z1 is not acut point of ys = ?s(z1) := expz1 ?(sR2u)(z1) for 0 6 ?s < 1. By continu-ity, Jac ?(z1) = lims?1 Jac ?s(z1). As before,Jac ?s(z1) 6?????detD2(sR2u+d2ys2)(z1)?????.16Chapter 3. Critical Density LemmaSincelim infs?1?????detD2(sR2u+d2ys2)(z1)?????= lim infs?1?????detD2(R2u+d2ys2)(z1)?????= lim infs?1|detD2wys(z1)|,it only remains to prove thatlim infs?1|detD2wys(z1)| 61(n?)n{(R2Lu+ aL + ?)+}n.Here, since it cannot be guaranteed that D2wys(z1) is nonnegative, the aboveinequality between determinant and trace to D2wys(z1) is not directly ap-plied. By the way, sincedys(y1)? 0 as s? 1,by passing to the limit as s? 1, the inequality can be applied toD2wys(z1) +Ndys(y1)Id for some Ninstead of D2wys(z1), only if it is possible to check thatD2wys(z1) +Ndys(y1)Id is nonnegative.Here, let ?k2(k > 0) be a lower bound of sectional curvature along theminimal geodesic connecting z1 and y1. Then, the nonnegative definitenessof D2wys(z1) + Ndys(y1)Id is clear if it is noted that Hessian comparisontheorem (see Schoen [15], and Schoen and Yau [16]) states thatD2dys(z1) 6 k coth(k dys(z1)) Id 6 N Iduniformly in s ? (12 , 1) for some N , and an auxilary function is consideredbyR2u(z) +12{dys(z) + dys(y1)}2 = wys(z) + dys(y1)dys(z) +12dys(y1)2which is smooth near z1 and has a local minimum at z1, so that its Hessianat z1 is nonnegative. Thus, the following is obtained0 6 D2wys(z1) + dys(y1)D2dys(z1) 6 D2wys(z1) +Ndys(y1)Id.17Chapter 3. Critical Density LemmaNow the previously mentioned relation between determinant and trace isapplied to this nonnegative definite endomorphism for s ? (12 , 1), and it isobtained that0 6 lim infs?1|detD2wys(z1)|= lim infs?1??det{D2wys(z1) +Ndys(y1)Id}??61?nlim infs?1detA(z1) ? det{D2wys(z1) +Ndys(y1)Id}61?nlim infs?1[tr{A(z1) ?(D2wys(z1) +Ndys(y1)Id)}n]n=1(n?)nlim infs?1[tr {A(z1) ?D2wys(z1)}]nsince dys(y1) ?? 0 as s ?? 0=1(n?)nlim infs?1{Lwys(z1)}n=1(n?)nlim infs?1{R2Lu(z1) + L(d2ys2)(z1)}n,and by the same computation with the last line of Case 1)61(n?)n{(R2Lu(z1) + aL + ?)+}n.This finishes the proof of the theorem.Remark. 1. It is noted that any ball B3R+(x0) for any  > 0 instead ofB4R(x0) might be used since it makes no change in the proof; just the factthat a little bit bigger than B3R(x0) is sufficient.2. The condition that u > 0 in B4R\B3R is enough for the proof insteadof the condition that u > 0 in B4R. 3. Under an additional condition ofLu 6 0 a different statement of Lemma 3.2, which looks similar to the resultof Theorem 3.1, can be obtained:If u is nonnegative, smooth in B4R and infBR u 6 1, then(143)n|B3R| =???BR4??? 61(n?)n?{u6 5732}?B3R{(R2Lu+ aL + ?)+}n dV,and also if Lu 6 0 in B3R, which is a condition of Theorem 3.161(n?)n(aL + ?)n????{u 65732}?B3R???? .18Chapter 3. Critical Density LemmaThus, under the additional condition that Lu > 0 in B3R we get the followinginequality similar to Theorem 3.1:????{u 6 5732}?B3R??|B3R|>{n?12(aL + ?)}n????{u > 5732}?B3R??|B3R|6 1?{n?12(aL + ?)}n.Since L is linear and constants are independent of the radius of ball R, thisimplies thatCorollary 3.3. Let u be a nonnegative smooth function in a ball BR andsatisfy Lu 6 0 in the same ball. Then, u has the following property:???{u ? 1} ?BR2??????BR2???> ?0 =? infBR6u > ?0,where ?0 = 1? {n?12 (aL + ?)}n and ?0 =3257.Remark. 1. It is noted that any ball BR2 +for any  > 0 may be used insteadof BR.2. It is mentioned that the main difference between Theorem 3.1 and Corol-lary 3.3 is the following:1) Theorem 3.1: the radius of the ball where infima are taken is biggerthan the radius of the ball where measure is computed.2) Corollary 3.3: the radius of the ball where infima are taken is smallerthan the radius of the ball where measure is computed.Thus, to prove Theorem 3.1 it suffices to find a proper auxiliary functionv, so that v might reduce the radius of the ball on which the measure iscomputed, that is, where the integration is done. Then, the situation ofCorollary 3.3 might change into that of Theorem 3.1. Here, we can consideran application of Lemma 3.2 to a modified function u+v instead of u, wherev roughly satisfies the following:R2Lv + aL + ? 6 0 a.e. in B3R\B?R,19Chapter 3. Critical Density Lemmawhere B?R is a smaller ball than BR on which the infimum was taken. Then,it is obtained that1(n?)n?{u+v6 5732}?B3R(x0){(R2L(u+ v) + aL + ?)+}n dV,when Lu 6 0 in B4R=1(n?)n?{u+v6 5732}?B3R(x0){(R2Lv + aL + ?)+}n dV,=1(n?)n?{u+v6 5732}?B?R(x0){(R2Lv + aL + ?)+}n dV.It is also noted that in order to apply Lemma 3.2 to u+v, it would be betterfor v to basically satisfy the following:1) For u+ v > 0 in B4R\B3R, v > 0 in B4R\B3R when u > 0 in B4R\B3R,2) For infBR(u+ v) 6 1 in BR, v 6 0 in BR when infBRu 6 1 in BR.Now the above strategy is implemented in detail.Proof. of Theorem 3.1.Step 1. For 0 < ? < 1, there is a continuous function v? in B4R satisfyingthe following properties:1) v? is smooth in B4R\Cut(x0),2) R2Lv + aL + ? 6 0 a.e. in B3R\B?R,3) v > 0 in B4R\B3R,4) v 6 0 in BR,5) R2Lv 6 C? a.e. in B3R, and6) ? v? 6 C? in B4R,where C? depends only on ?.A function v? is defined byv?(x) = I?(dx0(x)R)20Chapter 3. Critical Density Lemmawhere I? is a smooth increasing function on R+ satisfying that1) I ??(0) = 0, (3.3)2) I?(r) = (32)? ? (3r)? for r > ?. (3.4)with ? to be chosen later.It is trivial that v? is continuous on B4R. Also, v? is smooth in B4R\Cut(x0)since dx0 is smooth in M\(Cut(x0)?{x0}) and I??(0) = 0. Thus, ?v? shouldbe bounded from above by some constant C1,?. From (3.4), 3) and 4) ofthe above claim are also trivial. For the remained 2) and 5), we need tocompute Lv? under the condition of the smoothness of v? in B4R\Cut(x0):Lv? =1RI ??(?)Ldx0 +1R2I ??? (?)g ?A?dx0 ,?dx0?=1R2I ??(?)?dx0Ldx0 +1R2I ??? (?)g ?A?dx0 ,?dx0? ,where for ? 6 ? < 3 it is noted thati) I ??(?) =?3(3?)?+1,ii) I ??? (?) = ??(? + 1)32(3?)?+2,iii) ? 6 g ?A?dx0 ,?dx0? 6 ?,iv) dx0Ldx0 6 aL.Thus, in B3R\ {B?R ? Cut(x0)}, it is also obtained thatLv? 61R2I ??(?)?dx0Ldx0 +1R2I ??? (?)g ?A?dx0 ,?dx0?61R2?32(3?)?+2aL ?1R2?(? + 1)32(3?)?+2?61R2?32(3?)?+2 (aL ? (? + 1)?)61R2?9(3?)?+2 (aL ? (? + 1)?) .Here, since 3? > 1 and aL?(?+1)? < 0 for a sufficient large ?, the last termcan be made smaller than ?(aL+?)R2 by choosing a large ?. Thus, 2) of theclaim is proven. Moreover, in B4R\Cut(x0), basically it is computed thatLv? 6aLR2sup0<?<4????I ??(?)?????+ sup0<?<4?R2??I ??? (?)?? = C2,? <?.21Chapter 3. Critical Density LemmaThus, 5) of the claim is obtained with C? = max {C1,?, C2,?, n?} where n?is taken to guarantee that C? > n? in Remark 2 of Theorem 3.1, and theproof of the claims of Step 1 is finished.Since the above v? is guaranteed to be smooth only in B4R\Cut(x0), butmight not be smooth in B4R itself, an approximating process by smoothfunctions to v? in B4R is necessary in order to apply Lemma 3.2.Step 2. There exist a smooth bump function ? such thati) 0 6 ? 6 1 in M,ii) ? ? 1 in B3R,iii) supp(?) ? B4R,and a sequence of smooth functions {wk} in M satisfying the followings:i) wk ?? ?v? uniformly in M,ii) D2wk ?? D2v? a.e. in B3R,iii) D2wk 6 C Id in M for some C independent of k.In regard to the proof of this step, Cabre? [1] is referred to for simplicitysince its arguments just consist of applications of an approximation of theidentity and partition of unity.Step 3.If Lemma 3.2 is applied to u+ wk, then Theorem 3.1 is proven.First, if u+wk is approximated byu+wk+k1+2kwith a sequence {k} convergingto 0, then it might assumed that u+wk satisfies the hypotheses of Theorem3.1:1) u+ wk be a nonnegative smooth function in a ball B4R(x0),2) infBR(x0)u+ wk 6 1.This is because a proper sequence {k} ?? 0 can be chosen by the following22Chapter 3. Critical Density Lemmasteps:1) For x ? BR, that is, wk(x) ?? v?(x) 6 0,a sufficiently large k1 is picked such that wk(x) > 0 for all k 6 k1,and a sequence is set by 1,k = supBRwk > wk(x) for k 6 k1and 1,k = 0 for k > k1=? infBRu+ wk + 1,k1 + 21,k61 + 21,k1 + 21,k= 1,2) For x ? B4R\B3R, that is, wk(x) ?? v?(x) > 0,a sufficiently large k2 is chosen such that wk(x) 6 0 for all k 6 k2,and a sequence is defined by 2,k = supB4R\B3R(?wk) > ?wk(x) for k 6 k1and 2,k = 0 for k > k2=?u+ wk + 2,k1 + 22,k> 0 in B4R\B3R,3) A sequence is set by k = max {1,k, 2,k} .It is also noted that1) For any  > 0, there is a sufficiently large k such that{u+ wk 65732}?B3R ?{u+ v? 65732+ }?B3R,2){(R2L(u+ wk) + aL + ?)+}nis uniformly bounded in Msince D2wk 6 C Id.Thus, by applying the Lebesgue dominated convergence theorem as k ???and Lemma 3.2, and with the conditions of v? in Step 1 and Lu 6 0, the23Chapter 3. Critical Density Lemmafollowing inequalities are computed:???BR4??? 61(n?)n?{u+v?65732}?B3R(x0){(R2L(u+ v?) + aL + ?)+}n dV=1(n?)n?{u+v?65732}?B3R(x0){(R2L(v?) + aL + ?)+}n dV=1(n?)n?{u+v?65732}?B?R(x0){(R2L(v?) + aL + ?)+}n dV61(n?)n?{u+v?65732}?B?R(x0){(C? + aL + ?)+}n dV6(C? + aL + ?n?)n ????{u+ v? 65732}?B?R(x0)????6(C? + aL + ?n?)n ????{u 65732+ C?}?B?R(x0)????Here, the inequality can be expressed like below:????{u 6 5732 + C?}?B?R??|B?R|>{n?4?(C? + aL + ?)}n????{u > 5732 + C?}?B?R??|B?R|6 1?{n?4?(C? + aL + ?)}n.Thus, since L is linear and all the constants are independent of the radiusof ball R, by letting ? = 12 the proof of (Theorem 3.1) is finished:If u is a nonnegative smooth function in a ball BR(x0) and satisfy Lu 6 0in the same ball. Then, u has the following property:???{u ? 1} ?BR8(x0)??????BR8(x0)???> ?1 =? infBR4(x0)u > ?1,where ?1 = 1? {n?12(C 12+ aL + ?)}n and ?1 =15732 + C 12.24Chapter 4Power Decay of theDistribution Functions ofSolutionsThis is a chapter to get a power-decay property of distribution functions ofsolutions of nondivergent uniformly elliptic partial differential operators.Theorem 4.1. (Power-decay) Let BR be a ball in M and L be as defined inSection 2.2. Let u be a nonnegative smooth function in a ball BR satisfyingLu 6 0 in the same ball. Then, for any nonnegative integer k, u has thefollowing property :???{u ?Mk} ?BR8??????BR8???> ?k =? infBR8u > 1,where M =1?1=5732+ C 12and ? =11 + 1??15n.The proof of this theorem is given in the end of this chapter, and AppendixA can be referred to for any universal constants.Now a covering lemma on Riemannian manifold is proven, and it will beused to prove Power decay of the distribution functions of nonnegative su-persolutions of L.Lemma 4.2. (Covering) Let M be a Riemannian manifold, A is an mea-surable set on M with respect to the Riemannian measure on M, and B0 =BR(x0) is a ball containing A on M. A positive number ? is chosen so that|A| < ?|B0| where | ? | denotes the Riemannian measure on M.A family of balls, S is set byS ={B | B ? B0 and|B ?A||B|> ?}where B are balls on M.25Chapter 4. Power Decay of the Distribution Functions of SolutionsAlso, an another set ? is defined by ? = ?B?SB. Then, A ? ? a.e. x and|?| ? (1 + c1)|A| where c1 = c1(?, n) = 1??5n .Proof. The proof of Safonov [5] in Rn is followed with languages of Rieman-nian geometry. A function ?A is defined by?A(x, r) =|Br(x) ?A||Br(x)|.Step 1.A ? ? a.e. x.First, it is mentioned that a similar covering lemma in metric spaces equippedwith a measure satisfying a doubling property, which implies the Lebesguedifferentiation theorem in those spaces, was proved in Coifman and Weiss[8]. Then, we can apply the Lebesgue differentiation theorem on Riemma-nian manifolds with nonnegative Ricci curvature, which is a metric spaceequipped with the doubling property of measure, to the charateristic func-tion of A, ?A. If x /? ?, then we can find a sequence of small balls Si whichcontain x, shrink to one point, and satisfy|Si ?A||Si|< ? < 1.Thus, by the Lebesque differentiation theorem on Riemannian manifoldsapplied to ?A,?A(x) = limdiam(Si)?01|Si|?Si?A,and also= limdiam(Si)?0|Si ?A||Si|< ?.for a.e. x /? ?. Thus, for a.e. x /? ?, x /? A, that is A ? ? a.e.x.For every Br(x) ? S , since |Br(x) ? A| ? ?|Br(x)| and |A| < ?|B0|, bythe continuity of ?A(x, r) as a function of x and r, a ball Br?(x?) can bechosen such that ?A(x?, r?) = ?. In detail, this can be done by finding aninterpolation ball between B0 = BR(x0) and Br(x). The interpolation ballBt with a parameter t is defined byBt = Btr+(1?t)R(?(t))26Chapter 4. Power Decay of the Distribution Functions of Solutionswhere ?(t) is the minimal geodesic conneting x0 and x, and satisfying?(0) = x0 and ?(1) = x. Here, it is noted that B0 = BR(x0), B1 = Br(x),andStep 2.B1 ? Bt ? B0 for any t ? [0, 1].This is becausefor p ? B1, d(p, ?(t)) 6 d(p, x) + d(x, ?(t))< r + d(x, ?(t)) since p ? B16 r + (1? t)d(x, x0) since ?(t) is the minimalgeodesic satisfying ?(0) = x0 and ?(1) = x< r + (1? t)(R? r) since Br(x) ? BR(x0)= tr + (1? t)R=? p ? Bt,andfor q ? Bt, d(q, x0) 6 d(q, ?(t)) + d(?(t), x0)< tr + (1? t)R+ d(?(t), x0) since q ? Bt6 tr + (1? t)R+ td(x, x0) since ?(t) is the minimalgeodesic satisfying ?(0) = x0 and ?(1) = x< tr + (1? t)R+ t(R? r) = R=? q ? B0.Here, an another function ? is set by?(t) =|Bt ?A||Bt|with the condition that ?(0) < ? and ?(1) > ?. Then, by the continuity of?, there exists a number 0 < t0 ? 1 such that ?(t0) = ?. Thus, the above x?and r? can be picked byx? = ?(t0) and r? = t0r + (1? t0)R,and a refinement of S , S? can be considered byS? ={Bt0 = Bt0r+(1?t0)R(?(t0)) | Br(x) ? S},27Chapter 4. Power Decay of the Distribution Functions of Solutionswhere ?(t) and t0 are taken by the above process for each ball Br(x) in S .Then, for every B ? S? it is trivial that|B ?A||B|= ?,so S? is called as a refinement of S . And a set ?? is defined by?? = ?B?S?B.Then, by the claim of Step 2 and a trivial fact that ?? ? ?, it is obtainedthat ? = ??. Thus, a modified way for the construction of ? is attained.Step 3.|??||A|> 1 +(1? ?)5n.With the above S? , a Vital Covering of A can be considered; that is, thereexist distjoint balls Bi = Bri(xi) ? S? such thatA ? ?iB5ri(xi).Here, by virtue of the doubling property of measure, it is computed that|A| 6 |?iB5ri(xi)| 6?i|B5ri(xi)| = 5n?i|Bri(xi)|.Then, since1) B?is are disjoint,2) B?is satisfy|Bi ?A||Bi|= ?,it is obtained that|???A| >?i|(???A) ?Bi| =?i|Bi ?A| = (1? ?)?i|Bi| >(1? ?)5n|A|.Thus, it follows that|??||A|= 1 +|??/A||A|> 1 +(1? ?)5n.This proves the theorem with a constant c1 = c1(n, ?) = 1??5n .28Chapter 4. Power Decay of the Distribution Functions of SolutionsNote. The condition of Riemannian manifold is considered only for theLebesgue Differentiation Theorem, and the doubling property of Rieman-nian measure. So, its proof is basically the same with that in Rn, andthis lemma essentially simplify our proof of Harnack inequality on Rieman-nian manifolds, comparing with the proof of Cabre? who used a Calderon-Zygmund decomposition theorem in Riemannian manifolds.The above covering lemma explains a uniform growth, which is essential forthe proof of Theorem 4.1, of the measure of cut-off sets by increasing lowerbounds. Thus, if it is iterated, uniform increments of the measures of cut-offsets is obtained successively. The following are details of that idea.Proof. of Theorem 4.1.For convenience, it is recalled thatM =1?1=5732+ C 12and ? =11 + 1??15n,where ?1 = 1? {n?12(C 12+ aL + ?)}n.When k = 0, this is trivial since???{u ? 1} ?BR8??????BR8???> 1 =????{u ? 1} ?BR8??? =???BR8????? infBR8u > 1.It suffices to prove that for any positive integer k,???{u ?Mk} ?BR8??????BR8???> ?k =????{u ?Mk?1} ?BR8??????BR8???> ?k?1because then it is concluded by induction that???{u ?Mk} ?BR8??????BR8???> ?k =? ? ? ? =????{u > 1} ?BR8??? =???BR8????? infBR8u > 1.29Chapter 4. Power Decay of the Distribution Functions of SolutionsFirst, it is assumed that???{u ?Mk} ?BR8??????BR8???> ?k.Here, let ? = BR8? {u ?Mk} for simplicity, then it is assumed that|?| > ?k???BR8??? .Next, a covering of ?, which consists of small balls contained in BR8, is con-sidered; moreover, simultaneously whose sufficient portions intersect with ?;that is, a colletion of such balls F is defined byF ={B ? BR8||? ?B||B|> ?1}, where ?1 is from Theorem 3.1.Then, by Theorem 3.1 and linearity of L, it is obtained that for any B ? FinfBu >Mk(15732 + C 12)= Mk?1 since M =5732+ C 12.Thus, if ?1 := ?B?FB is considered, by Lemma 4.2 it is noted that1) ? ? ?1 almost every x in BR2.2) ?1 ?{x ? BR8| u(x) >Mk?1}= {u ?Mk?1} ?BR8.3) |?1| > (1 +1? ?15n) |?| > (1 +1? ?15n)?k???BR8??? = ?k?1???BR8???since ? =11 + 1??15n.From 2) and 3), it is obtained that???{u ?Mk?1} ?BR8??????BR8???> ?k?1and finish the proof.30Chapter 4. Power Decay of the Distribution Functions of SolutionsA variant of Theorem 4.1 as a power decay expression is given for an appli-cation in the next section.Corollary 4.3. ( btd -expression) Let u be a nonnegative smooth functionin a ball BR satisfying Lu 6 0 in the same ball. Moreover, suppose thatinfBR8u 6 1. Then, for any number t > 1, u has the following property :???{u ? t} ?BR8??????BR8???6btd,where b =1?= (1 +1? ?15n) and d = logM1?.Note. Appendix A is referred to for any universal constants.Proof. The contrapositive statement of (Theorem 4.1) is considered:infBu 6 1 =???{u >Mk} ?B??|B|6 ?k,where B = BR8.Then, for any Mk 6 t < Mk+1,|{u > t} ?B||B|6??{u >Mk} ?B??|B|6 ?k =?k+1?= b ?k+1 with b =1?, and if let d = logM 1? ??1? = (Md), it is obtainedt < Mk+1 ??1?k+1= (Md)k+1 > td ?? ?k+1 <1td.Thus, it is computed|{u > t} ?B||B|6btd.and finish the proof.31Chapter 5Harnack InequalityThis is a chapter to prove Harnack inequality.Theorem 5.1. (Harnack Inequality) Let M be a smooth n-dimensional com-plete Riemannian manifold which satisfies M? [R(v)] > 0, and u be a non-negative smooth solution of Lu = 0 in a ball BR(x0) on M . Then u has thefollowing property:supBR2(x0)u ? C infBR2(x0)u,where C is a universal constant depending only on ?,?, n, and aL.The geometric assumption of M, M? [R(v)] > 0, is mentioned for clarity,and the proof of Harnack inequality is given in the end of this chapter aftersome auxiliary lemmas.Lemma 5.2. (a Lower Bound) Let u be a nonnegative smooth function ina ball BR(x0) and satisfy Lu = 0 in the same ball. Moreover, suppose thatinfBR8(x0) u 6 1. Then, for k > k? (see Remark of Lemma 5.2), u has thefollowing property:If there is x1 ? BR8(x0) such that 1) d(x1, ?BR8(x0)) > rk =P(LNk)dn,2) u(x1) > LNk?1,then supBrk (x1)u > LNk,where L = 2b1d = 2 + 2{2n?5(C 12+ aL + ?)}n, N =LL? 12> 1,P = 2R{2db1? bLd} 1n, k? is mentioned in the following remark.32Chapter 5. Harnack InequalityRemark. If, in fact, rk = P(LNk)dn> R8 , then the above lemma is meaningless;so, in the beginning it is necessary to check thatrk =P(LNk)bn6R8for sufficiently large k. (5.1)We check that (5.1) is true for some k?:rk =P(LNk)bn6R8? 2R(2db1? bLd) 1n 1(LNk)bn6R8? 162db1? bLd6 LdNkd? 162dbLdLd ? b6 LdNkd? 162dbLd 6 Nkd(Ld ? b).Now, since b, d, L, and N > 1 are constants and Ld? b > 0, it is trivial thatthere exists a large k? satisfies the claim.Proof. The idea of Caffarelli [4] is followed for the proof. Suppose that thereis a point x1 satisfying the hypotheses of this lemma, but supBrk (x1) u 6LNk, then a contradition finishes the proof. First, a function w is definedbyw(x) =LNk ? u(x)LNk ? LNk?1.Step 1.???{w > t} ?B rk8(x1)??????B rk8(x1)???6btdfor any positive number t.This trivially comes from Corollary 4.3 since w satisfies the following prop-erties:1) w(x) is nonnegative smooth function in Brk(x1),2) Lw 6 0 in Brk(x1),3) w(x1) =LNk ? u(x1)LNk ? LNk?16LNk ? LNk?1LNk ? LNk?1= 1 =? infBrk (x1)w 6 1.33Chapter 5. Harnack InequalityStep 2.??2R{2db1? bLd} 1n=??P 6 R{2db1? bLd} 1n, thus a contradiction happens.First, an estimate for the measure of the ball B rk8(x1) can be computed:???B rk8(x1)??? 6????{u 6LNk2}?B rk8(x1)????? ?? ?(1)+????{u >LNk2}?B rk8(x1)????? ?? ?(2)(5.2)Here, (1) and (2) can be computed by Step 1 and Corollary 4.3, respectively:(1) =???{w > L} ?B rk8(x1)??? 6bLd???B rk8???since B2 rk8(x1) ? BR(x0) and u(x) 6 LNk2 ?? w(x) =LNk?u(x)LNk?LNk?1 >LNk2LNk?LNk?1 =N2N?1 = L because N =LL? 12,(2) 6????{u >LNk2}?BR8(x0)???? 6b(LNk2 )d???BR8???since B rk8(x1) ? BR8(x0) and infBR8(x0) u 6 1.Next, it is obtained that(5.2) =? (1?bLd)???B rk8??? 6b(LNk2 )d???BR8??? ,=? (1?bLd)wn(rk8)n 62db(LNk)dwn(R8)n, where rk =P(LNk)dn,=? (1?bLd)Pn(LNk)d62db(LNk)dRn,=? P 6 R{2db1? bLd} 1n.Thus, the claim of Step 2 is proved, and a contradiction happens.34Chapter 5. Harnack InequalityNow it is concluded that if u satisfies all of the hypotheses of this lemmaand all constants are defined as above, then u should satisfy thatsupBrk (x1)u > LNk.This finishes the proof.Remark. 1. L may be an arbitrary constant satisfying1) L >12since N =LL? 12> 0,2) L > b1d since 1?bLd> 0.Thus, L can be defined by L = 2b1d = 2 + 2{ 2n?5(C 12+aL+?)}n > 2.2. P = 2R{2db1? bLd} 1ndepends on the radius of the ball; however, anotherconstants are independent of R. Thus, it is necessary to recognize the reasonwhy we can get a universal constant independent of R in the following lemmain spite of such P . This independency of constants is essential to obtainHarnack inequality.Lemma 5.3. (almost Harnack) Let u be a nonnegative smooth function ina ball BR and satisfy Lu = 0 in the same ball. Moreover, suppose thatinfBR8(x0) u 6 1. Then, u has the following property in B R16(x0):supB R16(x0)u 6 C0, where C0 is a universal constant.Proof. Since rk = P(LNk)dnin Lemma 5.2, a sufficiently large positive integerk0 > k? (see Remark of Lemma 5.2) can be chosen such that?k>k02rk 6R16. (5.3)A ClaimsupB R16(x0)u < LNk0?1 = C0.35Chapter 5. Harnack InequalitySuppose that the claim is false, that is, there exist a point xk0 ? B R16(x0)such that u(xk0) > LNk0?1.First, since Brk0 (xk0) ? B R16+R16(x0) = BR8(x0) by (5.3), in virtue of Lemma5.2, an another point xk0+1 ? Brk0 (xk0) ? BR8(x0) can be picked such that1) supBrk0(xk0 )u = u(xk0+1) > LNk0 ,2) Brk0+1(xk0+1) ? BR8(x0) by (5.3) again.In detail, 2) comes from the fact that, for any integer k > k0,d(x0, xk) 6 d(x0, xk0) +?k>k0d(xk, xk+1)6R16+?k>k02rk 6R16+R166R8.Next, the same process can be iterated to construct a sequence of points{xk}k>k0 in BR8(x0) such that, for any k > k0,1) supBrk (xk)u = u(xk+1) > LNk,2) Brk+1(xk+1) ? BR8(x0) also by (5.3).This implies thatu(xk) ?? ? in BR8(x0) as k ???;however, it is contradictory to the condition that u is smooth in BR(x0), sothat, u is bounded in BR8(x0). Thus, it is concluded that if u satisfies allthe condition in this lemma, thensupB R16(x0)u < C0 = LNk0?1, where k0 satisfies?k>k02rk =?k>k02P(LNk)dn6R16.Finally, to guarantee that C0 is a universal constant which is independentof u and the radius R, it is recalled thatL = 2b1d , N =LL? 12> 1, and P = 2R{2db1? bLd} 1n.36Chapter 5. Harnack InequalityRemark. The previous lemma is almost a Harnack inequality since it is easyto get a Harnack inequality from it by the following simple argument:Let u be a nonnegative smooth function in a ball BR and satisfy Lu = 0 inthe same ball. Here, an auxiliary function is defined byu?(x) =u(x)infB R16(x0) u+ for an arbitrarily given  > 0,then u?(x) naturally satisfies all the hypotheses of Lemma 5.3 sinceinfBR8 (x0)u? =infBR8 (x0)uinfB R16(x0) u+ 6 1,so that we havesupB R16(x0)u 6 C0{infB R16(x0)u+ }6 C0 infB R16(x0)u since  > 0 is arbitrary.Now the proof of a different statement of Harnack inequality, Theorem 5.1,is given below.Proof. of Theorem 5.1.It suffices to prove thatu(x) 6 C u(y) for any x and y in BR2(x0) for a universal constant C.Let x and y be any two points in BR2(x0), and consider the minimal geodesicfrom x to y , ?(t), which is parametrized by the arc-length satisfying?(0) = x, ?(l) = y, and l < R = the diameter of BR2(x0).Then, a colletion of small balls{Bj = B R2?16(xj) = B R32(xj)}can be con-structed by the following steps:1) A positive integer i is chosen such thatt0 = 0 < t1 =R32< t2 = 2R32< ? ? ? < ti = iR32< tm+1 = l 6 (i+ 1)R32, then iR32< l =? i <32lR< 32.2) Let x0 = x, xj = ?(tj), and xi+1 = ?(ti+1) = y on the geodesic curve.37Chapter 5. Harnack InequalityThen, the balls successively intersect each other since ?(t) is a unit-speedcurve, so that, by using the above remark of (Lemma 6.3) iteratively, it isconcluded thatu(x) = u(x0) 6 C0u(x1) 6 ? ? ? 6 C0iu(xi+1) = C0iu(y) < C032u(y).This finishes the proof with a universal constant C = C032.38Bibliography[1] Cabre?, Xavier, Nondivergent elliptic equations on manifolds with non-negative curvature. Communications on pure and applied mathematics,50 (1997), 623?665.[2] Kim, Seick, Harnack inequality for nondivergent elliptic operators onRiemannian manifolds. Pacific J. Math, 213 (2004), 281?293.[3] Caffarelli, Luis A. and Cabre?, Xavier, Fully nonlinear elliptic equations,volume 43 of American Mathematical Society Colloquium Publications.American Mathematical Society, Providence, RI, 1995.[4] Caffarelli, Luis A., Interior W 2,p estimates for solutions of the Monge-Ampere equation. The Annals of Mathematics, 131 (1990), 135?150.[5] Safonov, M. V., Harnack?s inequality for elliptic equations and the Ho?lderproperty of their solutions. Journal of Soviet Mathematics, 21 (1983),851?863.[6] Christ, Michael, A T (b) theorem with remarks on analytic capacity andthe Cauchy integral. Colloquium Mathematicum, 60/61 (1990), 601?628.[7] Aimar, H., Forzani, L., and Toledano, R., Ho?lder regularity of solu-tions of PDE?s: a geometric view. Communications in Partial DifferentialEquations, 26 (2001), 1145?1173.[8] Coifman, R. R. and Weiss, G., Analyse harmonique non-commutativesur certains espaces homogenes.(French) Etude de certaines inte?gralessingulieres. Lecture Notes in Mathematics, 242 (1971).[9] Yau, S.-T., Harmonic functions on complete Riemannian manifolds.Communications in Pure and Applied Mathematics, 28 (1975), 201?228.[10] Saloff-Coste, L., Uniformly elliptic operators on Riemannian manifolds.Journal of Differential Geometry, 2 (1992), 27?38.39[11] Grigor?yan, A. A., The heat equation on noncompact Riemannian man-ifolds (English). Math. USSR-Sb., 72(1) (1992), 47?77.[12] Stroock, D. W., Non-divergent form operators and variations on Yau?sexplosion criterion. Journal of Fourier Analysis and Applications, 4(4-5)(1998), 565?574.[13] Chavel, I., Riemannian Geometry: A Modern Introduction. CambridgeUniversity Press, 1993.[14] Calabi, E., An extension of E. Hopf?s maximum principle with an ap-plication to Riemannian geometry. Duke Math. J., 25 (1957), 45?56[15] Schoen, R., The effect of curvature on the behavior of harmonic func-tions and mappings, Nonlinear partial differential equations in differen-tial geometry IAS/Part City Math. Ser. 2. Amer. Math. Soc., Providence,RI, 1996[16] Schoen, R. and Yau, S.-T. Lectures on differential geometry. Interna-tional Press, Cambridge, MA, 1994[17] Li, Peter, Lecture notes on geometric analysis, Lecture Notes Series No.6, Seoul National University, Research Institute of Mathematics, GlobalAnalysis Research Center, Seoul, 199340Appendix ASome Universal ConstantsAll the constants depend only on dimension n and differential operator L:1) (Lemma 2.2: Upper Boundness of Ldp) aL = (n? 1)?,(in the proof of Theorem 3.1, Step 1, pp20-22) C 12= C? with ? =12,2) (Theorem 3.1: Critical Density)?1 = 1? {n?12(C 12+ aL + ?)}n, and ?1 =15732 + C 12,3) (Theorem 4.1: Power Decay)M =5732+ C 12, ? =11 + { 2n?5(C 12+aL+?)}n,4) (Corollary 4.3:btd-expression)b = 1 + {2n?5(C 12+ aL + ?)}n, d = log(5732+C 12)(1 + {2n?5(C 12+ aL + ?)}n),5) (Corollary 5.2: Lower Bound)L = 2 + 2{2n?5(C 12+ aL + ?)}n, N =LL? 12> 1, P = 2R(2db1? bLd) 1n,6) (Theorem 5.1: Harnack Inequality)C = L32N32(k0?1), where k0 satisfies?k>k01Nk6L64nd(2db1? bLd) 1d.41


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