在完備黎曼流形上針對熱方程的哈納克不等式

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：陳瑞堂博士. Harnack Inequality for The Heat Equation on A Complete Riemannian Manifold. 研究生：張舜為. 中華民國一百零一年六月.

(2) 摘. 要. 設 ‫ ܯ‬是一個光滑且連通的完備非緊緻黎曼流形. 若 ‫ ܯ‬滿足體積倍增條件和弱 ‫ܮ‬ଶ 龐加萊不等式的話, 則針對底律雷特熱方程正解的哈納克不等式成立. 本論文主要探討這個定理. 基本上, 本論文可分成四個部分. 第一部分討論具有體積倍增條件和弱 ‫ܮ‬ଶ 龐加萊不等式的流形上的一些重要性質. 第二部分則利用這些性質證明納許不等式和索伯列夫不等式. 第三部分著重在底律雷特熱方程的 subsolutions 和 supersolutions 並分別從這兩種類型的解中萃取出均值不等式及逆赫爾德不等式. 最後一部分則運用了在前三個部份中所獲得的工具來完成本論文主要定理的證明. 關鍵字: 關鍵字哈納克不等式, 體積倍增條件, ‫ݏ‬-型覆蓋, 惠特尼型覆蓋, 弱 ‫ܮ‬ଶ 龐加萊不等式, 加權龐加萊不等式, 納許不等式, 索伯列夫不等式, 底律雷特熱方程, 均值不等式, 反向赫爾德不等式, 莫澤迭代法.. Abstract Let ‫ ܯ‬be a smooth connected complete non-compact Riemannian manifold. If ‫ܯ‬ satisfies the volume doubling condition (VDC) and the weak ‫ܮ‬ଶ Poincaré inequality (WPI), then the Harnack inequality for positive solutions to the Dirchlet heat equation holds on ‫ܯ‬. This is the main theorem in this paper. Basically, This paper can be seperated into four part. The first part discusses some important and useful properties on the manifold equipped with both VDC and WPI. The second part utilizes those properties to establish the Nash inequality and the Sobolev inequality. The third part focuses on subsolutions and supersolutions to the Dirichlet heat equation, and extracts the mean value inequality and the reverse Hölder inequality from them respectively. The last part applies all the tools obtained in previous parts to show the proof of the main theorem in this paper. Key Words: Harnack inequality, volume doubling condition, ‫ݏ‬-packing covering, Whitney-type covering, weak ‫ܮ‬ଶ Poincaré inequality, weighted Poincaré inequality, Nash inequality, Sobolev inequality, Dirichlet heat equation, mean value inequality, reverse Hölder inequality, Moser’s iteration..

(3) Harnack Inequality for The Heat Equation on A Complete Riemannian Manifold Shun-Wei Chang July 23, 2012.

(4) Abstract Let M be a smooth connected complete non-compact Riemannian manifold. If M satis…es the volume doubling condition (VDC) and the weak L2 Poincaré inequality (WPI), then the Harnack inequality for positive solutions to the Dirchlet heat equation holds on M . This is the main theorem in this paper. Basically, This paper can be seperated into four part. The …rst part discusses some important and useful properties on the manifold equipped with both VDC and WPI. The second part utilizes those properties to establish the Nash inequality and the Sobolev inequality. The third part focuses on subsolutions and supersolutions to the Dirichlet heat equation, and extracts the mean value inequality and the reverse Hölder inequality from them respectively. The last part applies all the tools obtained in previous parts to show the proof of the main theorem in this paper. Key Words. Harnack inequality, volume doubling condition, s-packing covering, Whitneytype covering, weak L2 Poincaré inequality, weighted Poincaré inequality, Nash inequality, Sobolev inequality, Dirichlet heat equation, mean value inequality, reverse Hölder inequality, Moser’s iteration..

(5) Contents 0 Introduction 1 Volume doubling condition and 1.1 Volume doubling condition . . 1.2 s-packing covering . . . . . . 1.3 Whitney-type covering . . . .. 1 two . . . . . . . . .. types of covering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2 Poincaré inequality and weighted Poincaré inequality. 1 1 2 5 13. 3 Nash inequality and Sobolev inequality 27 3.1 Nash inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.2 Heat kernel upper bound and Sobolev inequality . . . . . . . . . . . . . . . . 35 4 Subsolutions and supersolutions for the heat equation 50 4.1 Subsolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 4.2 Supersolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 5 Harnack inequality for the heat equation. 80. Bibliography. 87.

(6) 0. Introduction. This paper is a survey mainly about the Harnack inequality for positive solutions to the Dirichlet heat equation on a smooth connected complete non-compact Riemannian manifold which satis…es both the volume doubling condition and the weak L2 Poincaré inequality. In the beginning of this paper, we introduce the de…nition and give some propositions of the volume doubing condition, the weak L2 Poincaré inequality and two types of covering, the the s-packing covering and the Whitney type covering. The propositions of these two types of covering are very useful since we can get the Nash inequality and the weighted Poincaré inequality from them. Once we get the Nash inequality, then, applying the propositions of the Dirichlet heat kenel and the the spectrum theory for elliptic operators, we can further obtain the Sobolev inequality. The Sobolev inequality plays an extremely important role in Geometry Analysis. If the Sobolev inequality holds, we can get the mean value inequlity and the reverse Hölder inequality by applying Moser’s cerebrated iteration on the positive subsolution and supersolution to the Dirichlet heat equation respectively. Finally, by a powerful lemma, we can get the Harnarck ineaulaity for positive solutions to the Dirichlet heat equation from the mean value inequlity and the reverse Hölder inequality. Since that lemma is provided by the volume doubling condition and the weak L2 Poincaré inequality, and so are the last two inequalities, we can conclude that the Harnarck inequality for positive solutions to the Dirichlet heat equation holds on a complete non-compact Riemannian manifold which satis…es both the volume doubling condition and the weak L2 Poincaré inequality. This conclusion is the main goal of this paper.. 1. Volume doubling condition and two types of covering. Let M be a smooth connected complete non-compact Riemannian manifold. Let d be the canonical metric associated to the Riemannian structure of M . That is, d : M M ! R+ [ f0g is de…ned by d (x; y). the length of a geodesic curve joining x to y. for any x; y 2 M . In this case, by Hopf and Rinow’s theorem, we know that (M; d) forms a complete metric space and any bounded subset of M is precompact since M is complete. On the other hand, we let Bx (r) be the geodesic ball of center x 2 M and radius r > 0. That is, Bx (r) fy 2 M : d (y; x) < rg : Also, we denote by jBx (r)j the volume of Bx (r).. 1.1. Volume doubling condition. De…nition 1.1. M is said to have the volume doubling condition (VDC) if there exists a constant d0 1 such that jBx (2r)j d0 jBx (r)j (1.1) for any x 2 M and r > 0. In this case, d0 is called the controlling constant in VDC. 1.

(7) Proposition 1.1. Suppose that M has VDC. For any x; y 2 M , for any r d (x; y) < r, then r log2 d0 jBx (r)j d20 ; jBy (s)j s. s > 0, if (1.2). where d0 is the controlling constant in VDC. Proof. Since d (x; y). r, for any z 2 Bx (r), d (z; y). d (z; x) + d (x; y) < r+r = 2r:. That is, z 2 By (2r). Since z is arbitrary, we have Bx (r) jBx (r)j. By (2r) and thus. jBy (2r)j d0 jBy (r)j : (by (1.1)). (1). s > 0, there exists a positive integer k such that 2k 1 s. On the other hand, because r r 2k s. Therefore,. By 2k s. jBy (r)j. dk0 jBy (s)j (by (1.1)) 1+log2. d0 = d0. r s. r s. jBy (s)j (since 2k 1 s. log2 d0. r). jBy (s)j :. (2). Finally, pluging (2) into (1), we obtain jBx (r)j. d20. r s. log2 d0. jBy (s)j :. This …nishes the proof.. 1.2. s-packing covering. De…nition 1.2. Let. be an index set. For any s > 0, a collection of balls n o s E Bx : x 2 M; 2 2. is said to be an s-packing covering (s-PC) for M if it satis…es the following two conditions: 1. For any ; 2 , if 6= , then E \ E = ;; S 2. M 2E , where kE Bx ks for any k > 0. 2 2. Proposition 1.2. For any s > 0, there exists an s-PC for M .. 2.

(8) Proof. First we …x a point x0 2 M and let E0 = Bx0 2s . If there exists x1 2 M such that Bx1 2s \ Bx0 2s = ;, then we let E1 = Bx1 2s ; otherwise we let S = fE0 g. In the former case, if there exists x2 2 M such that Bx2 2s \ Bx0 2s [ Bx1 2s = ;, then we let E2 = Bx2 2s ; otherwise we let S = fE0 ; E1 g. Continue this process, we obtain a collection of balls S = fE0 ; E1 ; : : : ; En ; : : :g : For such S, it is clear that any two distinct balls in S are disjoint by construction. So S satis…es the …rst condition of an s-PC for M . On the other S Shand, if M is not contained in = E 2S 2E and thus x 2 = 2E = E 2S 2E , then there exists some x 2 M such that x 2 Bx (s) for all E 2 S. That is, d (x; x ) s for all E 2 S. In this case, we consider the ball Bx 2s . If there exists some E 2 S such that Bx 2s \ E 6= ;, then there exist some y 2 Bx 2s \ E for such E and thus d (x; x ). d (x; y) + d (y; x ) s s < + 2 2 = s:. So we get a contradiction since d (x; x ) s for all E 2 S. Hence, Bx 2s \ E = ; for all E 2 S. By construction of S, Bx 2s must be collected in S. Thus we get that S s x 2 Bx 2 SBx (s) E 2S 2E . This yields a contradiction again, hence M must be contained in E 2S 2E . Therefore, S also satis…es the second condition of an s-PC for M and thus S forms an s-PC for M . Proposition 1.3. For any s > 0, let S = E set, be an s-PC for M . For any z 2 M , de…ne. Bx. #f 2. N (z). s 2. : x 2 M;. 2. ,. is an index. : z 2 8E g :. If M has VDC, then d60 ;. sup N (z) z2M. where d0 is the controlling constant in VDC. Proof. For each z 2 M , let Then, for any. 2. z,. f 2. z. : z 2 8E g :. we have z 2 8E = Bx (4s) and thus d (z; x ) < 4s:. So, for any y 2 Bx. s 2. , d (y; z). d (y; x ) + d (x ; z) s < + 4s 2 < 8s:. That is, y 2 Bz (8s). Since y is arbitrary, we get Bx. s 2. Bz (8s) : 3. (1).

(9) This induces two results. First, since [ 2. is arbitrary, we have s 2. Bx. Bz (8s). z. and thus [. jBz (8s)j. =. 2. z. 2. z. [. =. X 2. Second, for any. 2. z,. s 2. Bx E. jE j : (by De…nition 1.2). z. s 2. by (1), we have x 2 Bx. (2). Bz (8s). So. d (x ; z) < 8s and thus jBz (8s)j = jE j. jBz (8s)j Bx 2s 8s. d20. log2 d0. (by (1.2)). s 2. = d60 : That is, d0 6 jBz (8s)j :. jE j Since. is arbitrary, we get X jE j 2. X 2. z. z. d0 6 jBz (8s)j. 6. = d0 jBz (8s)j. X 2. 1. z. = d0 6 jBz (8s)j # f 2 = d0 6 jBz (8s)j N (z) :. : z 2 8E g. Pluging (3) into (2), we obtain that jBz (8s)j. d0 6 jBz (8s)j N (z) :. That is, N (z). d60 :. Finally, since z is arbitrary, the proof of this proposition is …nished.. 4. (3).

(10) 1.3. Whitney-type covering. De…nition 1.3. Let a collection of balls E. be an index set. For any ball E of center x0 2 M and radius r > 0, Bx (r ) : x 2 E; r = 10 3 d (E ; @E) ;. 2. is said to be a Whitney-type covering (WC) for E if it satis…es the following two conditions: 1. For any ; 2 , if 6= , then E \ E = ;; S 2. E 2E , where kE Bx (kr ) for any k > 0. 2. Proposition 1.4. For any ball E of center x0 2 M and radius r > 0, there exists a WC for E. Proof. Let S=. Bz (rz ) : z 2 E; rz =. Note that rz =. d (z; @E) 1 + 103. :. d (z; @E) 1 + 103. is equivalent to rz = 10 3 d (Ez ; @E) : So, to construct a WC for E, we can pick balls from S. We start by picking a ball, say E0 , which has the largest possible radius in S. Actually, E0 must be the ball of center x0 and r radius 1+10 3 . That is, r E0 = Bx0 : 1 + 103 Indeed, for any ball in S of center y 2 En fx0 g, the radius d (y; @E) 1 + 103 r d (y; x0 ) = 1 + 103 r < : 1 + 103. ry =. 0 ;@E) and thus E0 2 S. Also, since r = d (x0 ; @E), the radius of E0 equals to d(x 1+103 Next, we show that there exists at least a ball which does not intersect E0 and has the maximal radius among all the radii of balls in SnE0 . To do so, we let be the least upper bound of the radii of balls that does not intersect E0 in SnE0 . Then, there exists an increasing sequence, say j j2N , of radii of balls that does not intersect E0 in SnE0 such that j ! as j ! 1. Note that, for each j 2 N, there exists at least a point, say zj , in E such that Bzj j 2 S. Since E is bounded, we can without lossing of generality (by extracting a subsequence if necessary) assume that zj ! z as j ! 1 for some z 2 E. Now we consider the ball Bz ( ). For such Bz ( ), …rst, since Bzj j \ E0 = ; for all j 2 N, by convergency, Bz ( ) \ E0 = ;. Second, Bz ( ) 2 SnE0 . To see this, …rst we prove that Bz ( ) 2 S. That is, we need to show that = d(z;@E) . For each j 2 N, let yj be a 1+103. 5.

(11) point such that d (zj ; yj ) = j and d Bzj j ; @E = d (yj ; @E). Then, again, since E is bounded, we can without lossing of generality (by extracting a subsequence if necessary) assume that yj ! y as j ! 1 for some y 2 E. For such y, note that d (z; y) =. lim d (zj ; yj ). j!1. =. lim. = so y 2 @Bz ( ). j. j!1. ;. Bz ( ) and thus we obtain that d (Bz ( ) ; @E) =. inf d (p; @E). p2Bz ( ). =. inf d (p; @E) p2Bz ( ). d (y; @E) (since y 2 Bz ( )) = lim d (yj ; @E) j!1. =. lim d Bzj. j!1. lim 103. =. j. ; @E. j. j!1 3. (1). = 10 :. On the other hand, for any " > 0, there exists some J 2 N such that Bz ( ) Bzj j + " for all j J. Indeed, since zj ! z and j ! as j ! 1, there exist J 2 N large enough such that d (z; zj ) < 2" " j < 2 for all j. J. So, for any y 2 Bz ( ), we have d (y; zj ). d (y; z) + d (z; zj ) " < + 2 " " < + j + 2 2 = j +". which indicates that y 2 Bzj j + " . Since y is arbitrary, we get Bz ( ) desired. Now, for the J 2 N, because d (Bz ( ) ; @E) =. j. + " as. inf d (p; @E). p2Bz ( ). inf. p2Bzj (. j +". ). = d Bzj. j. = d (zj ; @E) = d Bzj = 103 j for all j. Bzj. j. d (p; @E) (since Bz ( ). Bzj. j. +" ). + " ; @E j. +". ; @E. ". ". J. So, as j tends to 1, we get d (Bz ( ) ; @E) d (Bz ( ) ; @E) 6. 103 :. 103. ". Since " is arbitrary, (2).

(12) Finally, by (1) and (2), we obtain that d (Bz ( ) ; @E) = 103 which is equivalent to =. d (z; @E) : 1 + 103. Thus Bz ( ) 2 S. On the other hand, since Bz ( ) \ E0 = ;, we can conclude that Bz ( ) is a ball which does not intersect E0 in Sn fE0 g. We pick such Bz ( ) and thus, in this case, fE0 ; Bz ( )g S. For convenience, we write E1 = Bz ( ). Continue this process, we establish a collection of balls S = fE0 ; E1 ; : : : ; En ; : : :g : Such S must satis…es the …rst condition of a WC that S for E by its construction. So, to see d(z;@E) . S is a WC for E, we have to show that E 2E . For any z 2 E, let = E 2S 1+103 3 Then, it is clear that Bz ( ) 2 S and d (Bz ( ) ; @E) = 10 . Let N be the largest integer SN such that EN 2 S has radius rN . If Bz ( ) \ = ;, then, because the n=0 En radius of EN +1 , say rN +1 , is the least upper bound of radii of balls that does not intersect S N . This yields a contradiction n=0 En in Sn fE0 ; E1 ; : : : ; EN g, we obtain that rN +1 SN 6= ;. Let since N is the larest interger such that rN . Hence, Bz ( ) \ n=0 En SN n0 N . Note that y 2 Bz ( ) \ n=0 En , then y 2 Bz ( ) and y 2 En0 for some 0 frn gn2N forms a decreasing sequence of radii by the construction of S, we have rn0 rN and thus d (z; xn0 ). d (z; y) + d (y; xn0 ) < + rn0 2rn0 : S S It implies that z 2 Bxn0 (2rn0 ). because Bxn0 (2rn0 ) we get z 2 E 2S 2E E 2S 2E , S immediately. Fincally, since z is arbitrary, we obtain that E E 2S 2E as desired. Therefore, S forms a WC for E. Let E be a ball of center x0 2 M and radius r > 0. Also, let S = fE Bx (r ) : x 2 E; r = 10 3 d (E ; @E) ; 2 g be a WC for E, where is an index set. Then we have some important propositions as follows. Proposition 1.5. For any z 2 E, de…ne M (z). : z 2 102 E. 2. #. :. If M has VDC, then sup M (z) z2E. d13 0 ;. where d0 is the controlling constant in VDC. Proof. For each z 2 E, let Then, for any. 2. z,. z. =. 2. : z 2 102 E. :. since z 2 102 E = Bx (102 r ), we have d (z; x ) < 102 r : 7. (1).

(13) This gives us two results. First, because 103 r. = d (E ; @E) = inf d (p; @E) p2E. d (x ; @E) d (x ; z) + d (z; @E) < 102 r + d (z; @E) ; (by (1)) we have d (z; @E) > 103 r 102 r = 900r : That is, r <. 1 d (z; @E) : 900. Thus, for any x 2 E = Bx (r ), d (x; z) < = < <. d (x; x ) + d (x ; z) r + 102 r (by (1)) 101r 101 d (z; @E) 900 d (z; @E) :. That is, x 2 Bz (d (z; @E)). Since x is arbitrary, we get E. Bz (d (z; @E)) :. (2). Second, if we let y 2 @E such that d (x ; y) = d (x ; @E), then d (z; @E) =. < = = <. inf d (z; p). p2@E. d (z; y) d (z; x ) + d (x ; y) 102 r + d (x ; @E) (by (1)) 102 r + 103 + 1 r 1101r 211 r :. That is, r >2. 11. d (z; @E) :. (3). Note that 2 z is arbitrary in both (2) and (3), we can conclude that, for each 2 such that 102 E contains z, the corresponding ball of has radius at least 2 11 d (z; @E) and is contained in Bz (d (z; @E)). Also, since any two di¤erent balls in a WC are disjoint, the size of z is …nite. In fact, if we let 0 2 z be such that its corresponding ball E 0 = Bx 0 (r 0 ). 8.

(14) reaches the smallest volume among all balls in the collection fE : bound of the size of z can be estimated as follows. j. 2. z g,. then an upper. jBz (d (z; @E))j Bx 0 (r 0 ). zj. d (z; @E) r 0. d20. log2 d0. d20 211 = d13 0 :. log2 d0. (by (1.2)) (by (3)). Finally, since j. zj. = # 2 = M (z). : z 2 102 E. and z is arbitrary, the proof of this proposition is …nished. Proposition 1.6. For any E 2 S, let E 2 S, if 2E \ 6= ;, then Proof. First we note that, since of E = Bx0 (r),. be a geodesic curve joining x0 to x . For any r < 2r :. (1.3). is a geodesic curve joining x0 to x and x0 is the center d(. ; @E) = d (x ; @E) :. Now, to see the relation between r and r , we let y 2 2E \ contained in both 2E and , we get d (y; x ) < 2r. (1) . For such y, since y is (2). and d (y; @E) >. inf d (p; @E). p2. = d ( ; @E) = d (x ; @E) (by (1)) = 103 + 1 r > 103 r : Therefore, by (2) and (3), we have 103 r. 2r. < d (y; @E) d (y; x ) d (x ; @E) = 1 + 103 r :. It implies that r. 3 + 103 r 103 < 2r <. which …nishes the proof. 9. (3).

(15) Note that, by the construction of a WC for E, the …rst ball in S is the ball of center x0 and radius r0 = 1031+1 r. For such ball, we call it the central ball in S and denote it by E0 . For any E in S with E 6= E0 , let be a geodesic curve joining x0 to x . Since S is a WC, we get E0 \ E = ; and thus x 2 = E0 . In this case, we let p1 be the …rst point along (starting from x0 ) which does not belong to 2E0 . For such p1 , since p1 2 E and S is a WC for E, there exists a ball, say E 1 , in S such that p1 2 2E 1 . If 2E 1 also contains x , then we de…ne a string of balls L = fE. 0. = E0 ; E 1 ; E. = E g:. 2. If 2E 1 does not contain x , then we let p2 be the …rst point along (starting from x0 ) which does not belong to 2E 1 . For such p2 , similarly there exists a ball, say E 2 , in S such that p2 2 2E 2 . If 2E 2 contains x , then we de…ne a string of balls L = fE. 0. = E0 ; E 1 ; E 2 ; E. 3. = E g:. If 2E 2 does not contain x , then we continue this process until 2E n( ) 2 contains both pn( ) 2 and x for some n ( ) 2 N. The exsitence of n ( ) shall be proved in the next paragraph. In this case, we de…ne a string of balls n o L = E 0 = E0 ; E 1 ; : : : ; E n( ) 2 ; E n( ) 1 = E : To see n ( ) is …nite, we note that, for each E r. i. i. 2 L , since 2E i \. 6= ;,. 1 > r : (by (1.3)) 2. Thus n ( ) = jL j #. 2. 1 :r > r 2. :. Since 12 r > 0, by the construction of a WC for E, we know that the number of balls which are of radius larger than 21 r is …nite. That is, #. 1 :r > r 2. 2. < 1:. So n ( ) is …nite. Now, for any E 2 Sn fE0 g, if we consider a string of balls L for E , then we obtain two useful propositions as follows. Proposition 1.7. For any E 2 L , we have E Proof. Let center of E,. 104 E :. be the geodesic curve which induces L . Then, for any p 2 d (p; @E) = d (p; x ) + d (x ; @E) :. 10. (1.4) , since x0 is the (1).

(16) Because E 2 L , 2E must contains at least one point, say y, on contained in both 2E = Bx (2r ) and , we have. . For such y, since y is. d (y; x ) < 2r and thus d (y; x ) = d (y; @E) d (x ; @E) (by (1)) < d (y; @E) d (y; x ) + d (x ; @E) < 2r + 103 + 1 r = 103 + 3 r : This implies that d (x ; x ). d (x ; y) + d (y; x ) < 103 + 3 r + 2r = 103 + 5 r :. (3). Now, for any z 2 E , because d (z; x ). d (z; x ) + d (x ; x ) r + 103 + 5 r 2r + 103 + 5 r (by (1.3)) 103 + 7 r 104 r ;. < < = <. we have z 2 Bx (104 r ) = 104 E . Finally, since z is arbitrary, we get E desired. Proposition 1.8. For any E i ; E 1. 2 1 r 2. 8 1 E. r. i. i. i+1. E. 2 L , we have. 2r i ; 8E i ;. i+1. 3. max jE i j ; E. i+1. 104 E as. i+1. 8E i \ 8E. Proof. Let p be the …rst point along then we have. i+1. :. (starting from x0 ) which does not belong to 2E i , d (p; x i ) = 2r i :. Since, by the construction of L , p 2 2E d p; x. i+1. , we have. i+1. < 2r. i+1. :. Therefore, d x i; x. i+1. d (x i ; p) + d p; x < 2r i + 2r i+1 : 11. i+1. (1).

(17) It implies that 1 + 103 r. i+1. = d x i+1 ; @E d x i+1 ; x i + d (x i ; @E) < 2r i + 2r i+1 + 1 + 103 r = 2r i+1 + 3 + 103 r i+1. (by (1)). i. and thus r. 3 + 103 r 1 + 103 1003 r = 999 i < 2r i : <. i+1. i. Similarly, r. i. < 2r. i+1. :. Combining the last two inequalities, we get 1 r 2 as desired. Next, for any y 2 E. i+1. = Bx. i+1. d (y; x i ) < = < =. r. i. r. i+1. , because. i+1. 2r. (2). i. d y; x i+1 + d x i+1 ; x i r i+1 + 2r i + 2r i+1 (by (1)) 2r i + 3r i+1 2r i + 6r i (by (2)) 8r i ;. we have y 2 Bx i (8r i ) = 8E i . Since y is arbitrary, E. i+1. E. i. 8E i .. Similarly, 8E. i+1. :. Combining the last two results, we get 1 E 8. i. E. i+1. 8E. (3). i. as desired. Finally, applying (3), we obtain that E E. i i+1. 8E i \ 8E i+1 8E i \ 8E i+1. and thus max jE i j ; E. 8E i \ 8E. i+1. which …nished the proof. 12. i+1. ;.

(18) 2. Poincaré inequality and weighted Poincaré inequality. Let u 2 L1 (M ). For any x 2 M and r > 0, we denote uBx (r) the average of u on Bx (r). That is, Z 1 uBx (r) u: jBx (r)j Bx (r) De…nition 2.1. M is said to have the weak L2 Poincaré inequality (WPI) if there exists a constant P2 > 0 such that Z Z 2 2 jruj2 (2.1) P2 r u uBx (r) Bx (2r). Bx (r). for any x 2 M , r > 0 and u 2 C 1 (M ). In this case, P2 is called the controlling constant in WPI. Lemma 2.1. For any x0 2 M and r > 0, let E = Bx0 (r). Let S be a WC for E with the central ball E0 = Bx0 103r+1 . If M has WPI, then, for any E 2 S and any E i ; E i+1 2 L , we have ! 21 Z 1 p r i ; (2.2) 8 10P22 jruj2 u8E i u8E i+1 1 64E i jE i j 2. for any u 2 C 1 (M ), where P2 is the controlling constant in WPI and L is a string of balls joining E0 to E (which is obtained by the process in page 10). Proof. Let 2. I = u8E. u8E. i. 8E i \ 8E. i+1. i+1. ;. 2. then, since u8E. u8E. i. is constant on M ,. i+1. Z. 2. I = =. u8E Z. =. i \8E i+1. 2. u8E. i. i \8E i+1. i+1. 2. Z. u+u. i. i \8E i+1. u8E. 8E. 2. 1. 8E. u8E. 8E. 2. i+1. u8E. 8E. Z. u8E. i. Z. i \8E i+1. 8E. ". u 2. 2P2 (8r i ) = 128P2. Z. r2 i. 16E. 2. u + u. i. +. Z. u8E. i. u. 8E. u8E. i+1. jruj + 8r 2. i. i+1. 2. 2. 2. 16E. Z. i. i+1. 2. 2. u8E. i. u8E. jruj + r2 i+1 13. i+1. Z. i+1. 16E. Z. 16E. i+1. i+1. jruj. ! jruj. 2. !. :. 2. #. (by (2.1)) (1).

(19) Note that, following the proof in Proposition 1.8, we have d x and thus, for any y 2 16E. i+1. = Bx. i+1. ;x. 16r. i+1. d (y; x i ). < 2r i + 2ri+1. i. ,. i+1. d y; x i+1 + d x i+1 ; x i 16r i+1 + 2r i + 2ri+1 18r i+1 + 2r i 36r i + 2r i (by Proposition 1.8) 38r i 64r i :. < = = <. That is, y 2 Bx i (64r i ) = 64E i . Since y is arbitrary, 16E. 64E i :. i+1. Pluging this into (1), we obtain that I. r2 i. 128P2. Z. 64E. i. jruj2 + r2 i+1. = 128P2 r2 i + r2 i+1 Z. Z. 64E. i. Z. 64E. i. jruj2. !. jruj2. 128P2 r i + 4r i jruj2 (by Proposition 1.8) 64E i Z = 640P2 r2 i jruj2 : 2. 2. 64E. (2). i. On the other hand, applying Proposition 1.8 again, we have 2. I =. u8E. i. u8E. i. u8E. i. u8E. i+1. 2. u8E. i+1. 2. u8E. i+1. 8E i \ 8E. i+1. max jE i j ; E. i+1. jE i j :. (3). Combining (2) and (3), we …nally arrive at 2. u8E. i. u8E. i+1. r2 640P2 i jE i j. Z. 64E. i. jruj2. which is equivalent to u8E. i. u8E. i+1. 1 p r i 8 10P22 1 jE i j 2. So we get the proof. 14. Z. 64E. i. jruj2. ! 21. :.

(20) Lemma 2.2. For any. 2 (0; 1), let f : [0; 1) ! R be de…ned by 8 ; l 2 [0; ] < 1 1 l ; l 2 [ ; 1] : f (l) = : 1 0 ; l 2 [1; 1). Fix an x0 2 M and r > 0. Let E = Bx0 (r) and S = fE g for E. Let ' : M ! R be de…ned by. sup '2. 2 (1; 103 ], we have 222 inf '2 :. (2.3). E. E. 2 (1; 103 ], take y. Proof. For any E 2 S and. is an index set, be a WC. d (x; x0 ) r. ' (x) = f for all x 2 M . Then, for any E 2 S and. ,. 2. 1. and y. 2. on @ E so that. d (y 1 ; x0 ) = d (x ; x0 ) r d (y 2 ; x0 ) = d (x ; x0 ) + r. (1). :. For any x 2 E = Bx ( r ), since d (x; x ) < r , we have d (x; x0 ). d (x0 ; x ) + d (x ; x) < d (x0 ; x ) + r = d (y 2 ; x0 ) (by (1)). d (x; x0 ). d (x0 ; x ) d (x ; x) > d (x0 ; x ) r = d (y 1 ; x0 ) : (by (1)). and. That is, d (y 1 ; x0 ) < d (x; x0 ) < d (y 2 ; x0 ) : Since f is positive and decreasing on [0; 1), the last inequalities implies that 2. sup [' (x)]. =. x2 E. 2. f. d (x; x0 ) r. 2. sup f. d (x; x0 ) r. sup x2 E. =. x2 E. d (y 1 ; x0 ) r. f. 2. = [' (y 1 )]2. (2). and inf [' (x)]2 =. x2 E. =. d (x; x0 ) r. 2. f. 2. inf f. d (x; x0 ) r. inf. x2 E. x2 E. f. d (y 2 ; x0 ) r. = [' (y 2 )]2 : 15. 2. (3).

(21) Now, if d (y 2 ; x0 ) < r, then d (y 1 ; x0 ) < d (y 2 ; x0 ) < r. So, by the de…nition of f , we obtain that d (y 1 ; x0 ) r. ' (y 1 ) = f = 1. d (y 2 ; x0 ) r = ' (y 2 ) : = f. It implies that sup [' (x)]2. [' (y 1 )]2 (by (2)). x2 E. = [' (y 2 )]2 inf [' (x)]2 : (by (3)) x2 E. On the other hand, if d (y 2 ; x0 ). (4). r, we claim that sup ' (x). 222 inf ' (x) : x2 E. x2 E. To do so, we note that, for any l 2 [0; 1], f (l). 1 1. l. :. Thus ' (y 1 ) = f = f = f = f = f 1 1 1. d (y 1 ; x0 ) r d (x ; x0 ) r (by (1)) r r d (x ; @E) r r (103 + 1) r + r 1 r (1001 + ) r 1 r 1 (1001 + ) r (by (5)) r 1 2001r (since 103 ) r 1 211 r . r. 16. (5).

(22) Also, by a similar calculation, we have d (y 1 ; x0 ) r (1001 1. ' (y 2 ) = f = f =. 1. )r r. (1001. 1. )r r. 1 1. r : (since r. (since. d (y 1 ; x0 ) r. ). 103 ). Applying the last two estimates, we get [' (y 1 )]2. 222 r2 r2. 1. (1 )2 222 [' (y 2 )]2. and thus sup [' (x)]2. [' (y 1 )]2 (by (2)). x2 E. 222 [' (y 2 )]2 222 inf [' (x)]2 : (by (3)) x2 E. (6). Combining (4) and (6), we arrive at sup [' (x)]2. 222 inf [' (x)]2 : x2 E. x2 E. So we get the proof. Proposition 2.1. Fix x0 2 M and r > 0. For any 2 (0; 1), let f and ' be de…ned as the functions in Lemma 2.2. If M has both VDC and WPI, then there exists a constant Pw = Pw (d0 ; P2 ) > 0 such that Z Z 2 2 2 ju u' j ' Pw r jruj2 '2 ; (2.4) M. M. for any u 2 C 1 (M ), where d0 and P2 are the controlling constants in VDC and WPI, respectively. Also, in this inequality, Z 1 R u'2 : u' 2 ' M M. Proof. By the setting in Lemma 2.2, since S = fE g E. [ 2. 17. 2E. 2. is a WC for E = Bx0 (r), we have.

(23) and thus Z. E. ju. 2. u8E0 j '. XZ. 2. 2. 2E. 2E. 2 ju. XZ 2. = 2. u8E0 j2 '2. ju. XZ 2. ju. 2E. u8E j2 + ju8E 2. 2. u8E j ' +. = 2 (I + II) ;. u8E0 j2 '2. XZ 2. 2E. ju8E. 2. u8E0 j '. 2. ! (1). where E0 is the central ball of S and ( R P ju u8E j2 '2 I= 2 2E R P : ju8E u8E0 j2 '2 II = 2 2E. First, we estimate I. Note that, for any 2 , we have Z Z 2 2 2 ju u8E j ' sup ' ju u8E j2 8E 8E 8E Z sup '2 ju u8E j2 16E 8E Z 22 2 ju u8E j2 (by 2.3) 2 inf ' 16E 8E Z 2 22 2 2 inf ' P2 (8r ) jruj2 (by (2.1)) 16E 16E Z 228 P2 r2 jruj2 '2 Z 16E jruj2 '2 : 228 P2 r2 16E. Since. is arbitrary, we get I =. XZ 2. 2E. ju. u8E j2 '2. 8E. ju. u8E j2 '2. XZ 2. X. 28. 2 P2 r. 2. = 228 P2 r2. 2. Z. 16E. jruj2 '2. 16E. jruj2 '2 :. XZ 2. (2). By Proposition 1.5, we know that the overlapping number of 102 E with E 2 S is uniformly bounded. More precisely, accroding to that proposition, if we let M (z) = #. 2 18. : z 2 102 E.

(24) for all z 2 E, then, since M has VDC with the controlling constant d0 , we have d13 0 :. sup M (z) z2E. Thus,. XZ. 2. jruj '. 102 E. 2. Pluging this into (2), we obtain that 28. I. 2 P2 r. 2. d13 0. Z. E. XZ. 2. XZ. 228 P2 r2. 2. 2 228 d13 0 P2 r. jruj2 '2. 102 E. Z. E. (3). jruj2 '2. 16E. 2. jruj2 '2 :. jruj2 '2 :. (4). Next we estimate II. Note that, for any 2 , ju8E u8E0 j is constant on 8E . So we have XZ II = ju8E u8E0 j2 '2 2E. 2. XZ. 8E. 2. X. =. 2. X. =. 2. u8E0 j2 '2. ju8E. 2. Z. '2. ju8E. u8E0 j. ju8E. u8E0 j2 ' (8E ) ;. 8E. (5). where ' (8E ) is viewed as a notation which represents the integral of '2 on 8E . For any A M , let A : M ! R be de…ned by A. (x) =. 1 ;x 2 A : 0 ;x 2 =A. Then (5) =. X 2. =. Z X M. Now, for any E 2 S, let. ju8E. n L = E. 0. 2. 2. u8E0 j ' (8E ). ju8E. u8E0 j2. = E0 ; E 1 ; : : : ; E. Z. M. ' (8E ) jE j. n( ). ;E 2. 1 jE j E. E. (6). :. n( ) 1. =E. o. be a string of balls joining E0 to E (which is obtained by the method in page 10). Then we have ju8E. u8E0 j =. u8E. u8E. n( ) 1. 0. n( ) 1. X i=1. 19. u8E. i. u8E. i 1. :. (7).

(25) Also, for any E. i. 2 L , we have d (p; @E) = d (64E i ; @E). inf. p264E. i. = = = > > = = = = So, if we let y 2 @ (103 E ) and y. i. d (x i ; @E) 64r i 103 + 1 r i 64r 103 63 r i 2r i r (by (1.3)) 103 + 1 r 103 r d (x ; @E) 103 r d 103 E ; @E inf3 d (p; @E) :. i. p210 E. 2 @ (64E i ) such that. d (y ; @E) = inf p2103 E d (p; @E) ; d (y i ; @E) = inf p264E i d (p; @E) then d (y ; x0 ) = r = r. d (y ; @E) inf3 d (p; @E) p210 E. > r. inf. p264E. d (p; @E) i. = r d (y i ; @E) = d (y i ; x0 ) : Since f is decreasing on [0; 1), we get d (y ; x0 ) r d (y i ; x0 ) < f r = ' (y i ) :. ' (y ) = f. (8). Note that y is a point in 103 E such that y is closest to @E, that is, y is a piont in 103 E 0) such that y is farest from x0 . Thus ' (x) = f d(x;x must arrive at its in…mum at y on r 103 E since f is decreasing on [0; 1). That is, ' (y ) =. inf. ' (x). x2103 E. =. inf. x2103 E. ' (x) : (since f and d are continuous). Similarly, we have ' (y i ) =. inf. x264E. 20. ' (x) : i.

(26) So inf. x2103 E. ' (x) = ' (y ) < ' (y i ) (by (8)) = inf ' (x) x264E. i. and thus inf '2 < inf '2. 103 E. 64E. i. since f is positive on [0; 1). Applying this inequality, we get Z ' (8E ) 1 = '2 jE j jE j 8E Z 1 2 1 sup ' jE j 8E 8E = 8 sup '2 8E. 8 222 inf '2 (by (2.3)) 8E. 25. = 2 inf '2 8E. inf '2. 25. 2. 103 E. < 225 inf '2 : 64E. Note that E. i. i. is arbitrary, we can conclude the last inequality in the following form ' (8E ) < 225 inf 64E i jE j. '2. (9). 1. for all i = 1; 2; : : : ; n ( ). Now, let II = ju8E. ' (8E ) jE j. u8E0 j. then we get n( ) 1. II. 13. < 2. X. u8E. u8E. i. inf. i 1. 64E. i=1. n( ) 1 13. 2. X i=1 1 2. 1 p r 8 10P22 E. n( ) 1. 220 P2. X i=1. 1 2. 220 P2 r. r E. i=1. Z. 1 2. 1 1 2 i 1. i 1. 2. 64E. i 1. E. 64E. i 1. i 1. n( ) 1. X. 1 2. Z. i 1. 64E. (10). ;. 1 2. (by (7) and (9)). i 1. Z. i 1. '. 2. 1 2. jruj. jruj '2. i 1. 21. 2. ! 21. ! 21. jruj2 '2. ! 21. inf. 64E. i 1. '. 2. 1 2. (by (2.2)).

(27) and thus II 2. 2. 1. n( ) 1. < 4220 P22 r. E. X. 2. 1 1 2. E. i=1. 2. 240 P2 r2 4. X. X. E 2S. 1 2. 64E. i 1. Z. 1 jE j. 1 2. i 1. Z. 1 E. i=1. 64E. i 1. n( ) 1. = 240 P2 r2 4. Z. i 1. jruj2 '2. jruj2 '2. jruj2 '2. 64E. ! 21 32. ! 21. ! 12. 5. 104 E. Since the last inequality holds for any E 2 S, we have X. II 2. < 240 P2 r2. E. X. E 2S. E 2S. 2. = 240 P2 r2 4. 2 4. X. 1. 64E. Z. 1 jE j. E 2S. 1 2. jE j. E 2S. X. Z. 1 2. 104 E. 32 5. E. jruj2 '2. jruj2 '2. 64E. E. ! 21. i 1. 32 5. (by (1.4)). E. :. ! 21. 104 E. 104 E. 32 5. 32 5. X. E. E. E 2S. Note S is a WC for E, so any two di¤erent balls in S are disjoint. Therefore, X 1 E E 2S. and thus the last inequality becomes X. II 2. E. E 2S. 2. < 240 P2 r2 4. = 240 P2 r2 @ J P. E 2S. J. 1. E 2S. 0. where. Let g =. X. 104 E. . Since J. X. jE j. jE j J. 1 2. 1 2. 64E. 104 E. E 2S. 1. Z. Z. 64E. jruj2 '2. 12. A ;. jruj2 '2. g g2 M. =1. 1 2. 22. 2. 104 E. 32 5. (11). ! 12. :. 0 for all E 2 S, we have g R. ! 21. (12) 0. If g > 0, note that.

(28) and. g R. (. M. 1. g2 ) 2. > 0, we have Z. 1 2. g2. =. M. Z. R. R. g2. M. 1 2. g2 M. g g R 1 M g2 2 M Z sup g :. =. k k2 =1; >0. M. If g = 0, then the last inequality holds automaticlly. Therefore, we can conclude that Z. g. 2. sup k k2 =1; >0. M. That is,. 2 0 Z X 4 @ J M. Since. 12 3 21 A5. 104 E. E 2S. R:H:S: =. sup. X. sup. X. sup k k2 =1; >0. J. sup. 4. J 10 E. J. d16 0. k k2 =1; E 2S >0. =. d16 0. sup. 2 0 Z X 4 @ J M. E 2S. 104 E. M. 0 @. X. J. 104 E. E 2S. 1 j104 E j. 1 jE j 4 j10 E j. X. k k2 =1; E >0. we get. Z. 1. A :. 104 E. k k2 =1; E 2S >0. X. g :. M. Z. k k2 =1; E 2S >0. =. Z. 1 2. 1 J jE j j104 E j 2S. 12 3 12 A5. d16 0. sup. Z. 104 E. Z. (by (1.1)). 104 E. Z. 104 E. X. k k2 =1; E >0. 1 J jE j 4 j10 E j 2S. Now, for any z 2 E = Bx (r ), if y 2 104 E = Bx (104 r ), then d (y; z) = d (y; x ) + d (x ; z) < 104 r + r < 105 r : This implies y 2 Bz (105 r ) : Since y is arbitrary, we get 104 E. Bz 105 r 23. Z. 104 E. :. (13).

(29) and thus, for any positive Z 1 j104 E j 104 E. 2 C (M ), 1 4 j10 E j. Z. Bz (105 r. 5. ). 1 jBz (10 r )j = j104 E j jBz (105 r )j 5. =. jBz (10 r )j Bx (104 r ). < d60. =. 1 jBz (105 r )j. 1 d60 sup s>0 jBz (s)j. d60 M. (z) ;. Z. Z. Z. Bz (105 r. Bz (105 r. ). Bz (105 r. 1 jBz (105 r )j Z. 1 jBz (105 r )j Z. = d20 10log2 d0. Bz (105 r. 1 jBz (105 r )j. log2 d0. 105 r 104 r. d20. Z. ). Bz (105 r. (by (1.2)) ). ). ). Bz (s). (14). where M : M ! R is the maximal function de…ned by Z 1 M (z) = sup s>0 jBz (s)j Bz (s) for all z 2 M . Note that, in (14), z 2 E is arbitrary, we have ! Z Z Z 1 d60 M (z) dz dz < 4 j10 E j 104 E E E Z 6 = d0 M : E. Since ! Z ! Z 1 L:H:S: = 1dz j104 E j 104 E E Z jE j ; = j104 E j 104 E we get jE j j104 E j. Z. <. 104 E. d60. Z. E. 24. M :.

(30) Pluging this into (13), we arrive at 2 0 Z X 4 @ J M. 104 E. E 2S. 12 3 12 A5. d16 0. <. X. sup. J. Z. d60. k k2 =1; E 2S >0. d22 0. =. Z. sup k k2 =1; >0. M. E. X. J. M. E. M E 2S. 2 0 Z X @ J sup 4. d22 0. k k2 =1; >0. M. E 2S. 2 0 Z X @ J sup 4. d22 0. k k2 =1; >0. M. E. E 2S. 2 0 Z X 4 @ = Cd22 J 0 M. E. E. E 2S. 12 3 21 A5. 12 3 21 A5. Z. (M ). C. Z. 2. 1 2. M. 2. 1 2. M. 12 3 21. A5 ;. for some constant C = C (d0 ) > 0, where we apply the Hölder inequality and the maximal function property in the third and the fourth inequality, respectively. Note that the last inequality is equivalent to 0 12 0 12 Z Z X X @ @ C 2 d44 (15) J 104 E A J E A : 0 M. M. E 2S. E 2S. Thus,. Z X. II. M. =. 2 Z X M. <. Z. M. u8E0 j2. ju8E II 2. 2. 240 P2 r2 @. = 240 P2 r2. X. J. X. J. 104 E. E 2S. Z. M. 0 @. 240 P2 r2 C 2 d44 0. 104 E. E 2S. Z. M. 2 = 240 C 2 d44 0 P2 r. E. (by (6)). (by (10)). E. 0. ' (8E ) jE j. Z. M. 0 @. 0 @. 25. X. J. X. J. 12. A (by (11)) 12 A. E. E 2S. E 2S. E. 12. A (by (15)) 12. A :. (16).

(31) Again, since S is a WC for E, the intersection of any two distinct balls in E is empty. Thus 12 0 X X 2 @ J E J E A = E 2S. E 2S. X. =. J2. E. :. E 2S. Also, note that J =. 1 1. jE j 2. R. Z. 2. 64E. jruj '. X. J. 1 2. 2. 2. is constant on M , so XZ. =. E. M E 2S. E 2S. X. =. J2. J. 2. X. E 2S. Applying the last two results in (16), we get Z X 40 2 44 2 J2 (16) = 2 C d0 P2 r. Z. E. M. E 2S. =. E. M. J 2 jE j :. E. M E 2S. 2 = 240 C 2 d44 0 P2 r. X. E 2S. J 2 jE j. ! Z 1 2 = 240 C 2 d44 jruj2 '2 jE j (by (12)) 0 P2 r jE j 64E E 2S Z X 2 = 240 C 2 d44 P r jruj2 '2 2 0 X. E 2S. 2 240 C 2 d44 0 P2 r. E 2S 2 240 C 2 d57 0 P2 r. 64E. XZ. Z. E. jruj2 '2. 102 E. jruj2 '2 : (by (3)). That is, we have II. 40. 2. 2 C 2 d57 0 P2 r. Z. E. jruj2 '2 :. Now, combining (4) and (17), we arrive at Z ju u8E0 j2 '2 2 (I + II) E Z Z 2 2 28 13 2 40 2 57 2 2 d0 P2 r jruj ' + 2 C d0 P2 r jruj2 '2 ZE ZE 2 2 240 d57 jruj2 '2 + 240 C 2 d57 jruj2 '2 (since d0 0 P2 r 0 P2 r E E Z 240 d57 1 + C 2 P2 r 2 jruj2 '2 : 0 E. 26. (17). 1).

(32) Note that u' is constant on M , so u' 2 C 1 (M ). Thus, by (18), we have Z Z 2 2 40 57 2 2 ju' u8E0 j ' 2 d0 1 + C P2 r jru' j2 '2 E. E. = 0: (since u' is constant on M ). (19). Finally, applying the last inequalities, we get Z Z 2 2 ju u' j ' = ju u' j2 '2 (since ' 2 C0 (E) ) M ZE 2 ju u8E0 j2 + ju8E0 u' j2 '2 E Z Z 2 2 = 2 ju u8E0 j ' + ju8E0 u' j2 '2 E ZE jruj2 '2 + 0 (by (18) and (19)) 1 + C 2 P2 r 2 2 240 d57 0 Z E jruj2 '2 = 241 d57 1 + C 2 P2 r 2 0 ZE jruj2 '2 = 241 d57 1 + C 2 P2 r 2 0 M Z = Pw r 2 jruj2 '2 ; M. where Pw = 241 d57 1 + C 2 P2 0 is a positive constant dependent only on d0 and P2 . Thus the proof is …nished.. 3 3.1. Nash inequality and Sobolev inequality Nash inequality. For any s > 0, let us : M ! R be de…ned by us (y). uBy (s). for all y 2 M . Also, we denote by C01 ( ) the set of all C 1 functions de…ned on M and supported on , where is a domain of M . Lemma 3.1. Let E be a ball of center x0 2 M and raius r > 0 in M . If M has both VDC, then Z Z log d 2 d50 rs 2 0 2 juj us jEj M M. holds for any s 2 (0; r] and u 2 C01 (E), where d0 is the controlling constant in VDC. Proof. First we estimate supM jus j. For any y 2 M , if E \ By (s) = ;, then By (s). 27. EC ;.

(33) where E C is the complement of E in M . Therefore, Z 1 jus (y)j = u jBy (s)j By (s) Z 1 juj jBy (s)j By (s) Z 1 juj jBy (s)j E C = 0: (since u 2 C01 (E) ) Note that y is arbitrary with E \ By (s) = ;, so we get sup y2M E\By (s)=;. jus (y)j = 0:. (1). On the other hand, if E \ By (s) 6= ;, then there exists some z 2 E \ By (s). By such z, we have d (y; x0 ). d (y; z) + d (z; x0 ) < r+s. and thus jBx0 (r)j jEj = jBy (s)j jBy (s)j jBx0 (r + s)j jBy (s)j d20. r+s s. d20. 2r s. log2 d0. log2 d0. (since s. = d20 2log2 d0 = d30. r s. (by (1.2)). log2 d0. r s. r). log2 d0. :. It implies that Z 1 jus (y)j = u jBy (s)j By (s) Z 1 juj jBy (s)j By (s) Z 1 jEj = juj jEj jBy (s)j By (s) Z log2 d0 1 3 r d juj : (by (2)) jEj 0 s By (s) 28. (2).

(34) So, in this case, since y is arbitrary with E \ By (s) 6= ;, we get log d Z d30 rs 2 0 sup jus (y)j juj : jEj y2M By (s). (3). E\By (s)6=;. Combining (1) and (3), we get d30. sup jus (y)j. y2M. d30 Next we estimate Z. M. R. Z. r log2 d0 s. jEj. By (s). Z. r log2 d0 s. jEj M. jus j =. = =. = = =. M. juj. juj :. (4). u2s . Note that Z. Z 1 u (z) dz dy M jBy (s)j By (s) Z R ju (z)j dz By (s) dy jBy (s)j M Z Z 1 dy ju (z)j dz M Bz (s) jBy (s)j Z Z jBz (s)j 1 dy ju (z)j dz M Bz (s) jBy (s)j jBz (s)j Z Z s log2 d0 1 d20 dy ju (z)j dz (by (1.2)) s jBz (s)j M Bz (s) Z Z 1 2 d0 dy ju (z)j dz M Bz (s) jBz (s)j Z 2 d0 ju (z)j dz ZM juj : d20 M. That is,. Z. M. jus j. Z. d20. M. juj :. Finally, by previous arguments, we obtain that Z Z 2 jus j jus j us = M M Z sup jus j jus j M. d30. M Z log d 0 2 r s. jEj. =. d50. M. Z. r log2 d0 s. jEj. M. So the proof is …nished. 29. (5). !. d20. juj. M. 2. juj. Z. :. juj.

(35) Lemma 3.2. If M has both VDC and WPI, then Z Z 2 6 2 2 ju us j 8d0 1 + d0 P2 s jruj2 M. M. holds for any s > 0 and u 2 C01 (M ), where d0 and P2 are the controlling constants in VDC and WPI, respectively. Proof. Let S = fE g. be an s-PC for M . Since [ M 2E ;. 2. 2. we have Z. M. ju. Z. 2. us j. S. 2E. XZ 2. 2. us j2. 2E. ju. 2E. 2 ju. XZ 2. us j2. ju. XZ. = 2. u4E j2 +. ju. 2E. 2. = 2 (I + II) ; where. (. us j2. u4E j2 + ju4E XZ 2. 2E. ju4E. us j2. ! (1). R P ju u4E j2 I= 2 2E R P : us j2 ju4E II = 2 2E. First we estimate I. By a direct calculation, we have XZ I = ju u4E j2 2. 2E. XZ 2. =. 4E. XZ 2. X. u4E j2. ju. u. Bx (2s) 2. P2 (2s). 2. = 4P2 s2. XZ 2. = 4P2 s2. 2. Bx (4s). 4P2 s. jruj2 (by (2.1)). jruj2 :. 8E. 2. (2s). jruj2. Bx (4s). That is, I. Z. XZ. 2. uB x. XZ 2. 30. 8E. jruj2 :. (2).

(36) Next, we estimate II. For any y 2 2E = Bx (s), we have, for any z 2 By (s), d (z; x ). d (z; y) + d (y; x ) < s+s = 2s:. That is, z 2 Bx (2s) : Since z is arbitrary, we get By (s). Bx (2s) = 4E :. (3). On the other hand, for the given y, note that 2. jBy (s)j ju4E. us (y)j. 1 jBy (s)j. = jBy (s)j u4E. Z. 2. u. By (s). 2 Z 1 u = u4E jBy (s)j jBy (s)j By (s) 2 Z Z 1 u 1 u4E = jBy (s)j By (s) By (s) 2 Z 1 = (u4E u) : jBy (s)j By (s). Since, by Hölder’s inequality, Z. Z. u). (u4E. By (s). ! 21. 1. By (s). 1 2. = jBy (s)j. Z. Z. By (s). By (s). ju4E. ju4E uj2. uj2 ! 12. ! 21. ;. we have 1 jBy (s)j. Z. 2. (u4E. u). By (s). Z 1 jBy (s)j ju4E jBy (s)j By (s) Z = ju4E uj2 : By (s). Pluging this into (4), we get jBy (s)j ju4E. 2. us (y)j. Z. By (s). 31. ju4E. uj2. uj2. (4).

(37) and thus ju4E. 2. us (y)j. Z 1 ju4E uj2 jBy (s)j By (s) Z 1 ju u4E j2 (by (3)) jBy (s)j 4E Z 1 2 u uBx (2s) = jBy (s)j Bx (2s) Z 1 2 jruj2 (by (2.1)) P2 (2s) jBy (s)j Bx (4s) 2 Z 4P2 s = jruj2 : jBy (s)j 8E. Now, since y is arbitrary, we get XZ XZ 2 ju4E us j = 2E. 2. us (y)j2 dy. ju4E. 2E. 2. Z. XZ 2. = 4P2 s. 4P2 s2 2E jBy (s)j X Z. 2. 8E. 2. (5). jruj2 dy (by (5)). 8E. jruj. Z. 2. 2E. 1 dy: jBy (s)j. (6). Note that Z. 2E. 1 dy = jBy (s)j. Z. 1 dy Bx (s) jBy (s)j Z jBx (s)j 1 = dy jBy (s)j Bx (s) jBx (s)j Z 1 s log2 d0 d20 dy (by (1.2)) s Bx (s) jBx (s)j Z d20 = 1dy jBx (s)j Bx (s) = d20 :. (7). Thus XZ 2. 2E. ju4E. 2. us j. 4P2 s. 2. X Z 2. 4P2 s2. = 4d20 P2 s2 That is, II. 4d20 P2 s2. 8E. jruj2 d20 (by (7)). 8E. jruj2 :. XZ 2. 32. 1 dy (by (6)) jBy (s)j. 8E. XZ 2. Z. jruj. X Z 2. 2. 8E. jruj2 :. 2E. (8).

(38) So, by previous arguments, we obtain that Z ju us j2 2 (I + II) (by (1)) M XZ XZ 2 2 2 2 jruj + 8d0 P2 s 8P2 s 8E. 2. = 8 1 + d20 P2 s2. jruj2 :. 8E. 2. 8E. 2. XZ. jruj2 (by (2) and (8)) (9). Note that, by Proposition 1.3, we know that the overlapping number of 8E is uniformly bounded. More precisely, accroding to that proposition, if we let N (z) = # f 2. : z 2 8E g. for all z 2 M , then, since M has VDC with the controlling constant d0 , we have d60 :. sup N (z) z2M. Thus,. XZ 2. jruj. 8E. 2. d60. Z. jruj2 :. M. Pluging this into (9), we …nally get Z XZ 2 2 2 ju us j 8 1 + d0 P2 s M. 8E. 2. Z. 8d60 1 + d20 P2 s2. M. as desired.. jruj2 (by (9)). jruj2. Theorem 3.1. Let E be a ball of center x0 2 M and raius r > 0 in M . If M has both VDC and WPI, then there exists a constant CN = CN (d0 ; P2 ) > 0 such that 4 Z Z Z Z 1+ v2 v 1 CN r2 2 2 2 u juj jruj + 2 (3.1) u 2 r M M M M jEj v. holds for any u 2 C01 (E), where v = log2 d0 > 0, d0 and P2 are the controlling constant in VDC and WPI, respectively. The inequality in this theorem is called the Nash inequality (NI) and CN is called the controlling constant in NI. Proof. For any s 2 (0; r], by Lemma 3.1 and Lemma 3.2, we have Z Z 2 u 2 ju us j2 + jus j2 M M Z Z 2 = 2 ju us j + u2s M. 16d60 cs. 2. 1+. Z. M. = cs. 2. M. Z. M. d20. P2 s. 2. jruj + 2. jruj +. 2. Z. 2. jruj +. M 5 r v 2d0 s. Z. jEj. 2d50 rv jEj. s. 2. M. v. Z. M. 33. r log2 d0 s. 2d50. jEj. Z. M. 2. juj. juj 2. juj. ;. (1).

(39) where c = 16d60 1 + d20 P2 + 1. On the other hand, for any s 2 (r; 1), we have s2 r 2 > 1. Since c Z Z 2 2 2 u cs r u2 : M. 1, in this case we have (2). M. Combining (1) and (2), we obtain that, for any s > 0, Z Z Z Z 2 2d50 rv v 2 2 2 2 2 juj + cs r u2 s u cs jruj + jEj M M M M Z Z Z 2 5 v 1 2d0 r 2 2 v 2 jruj + 2 = cs juj : s u + r M jEj M M R Note that M u2 is independent R on2s, so we can minimize the R.H.S. in the last inequality to get a better upper bound for M u . To do so, we let ( R R A = c M jruj2 + r12 M u2 (3) 2 2d50 r v R B = jEj juj M and consider. v. h (s) = As2 + Bs on (0; 1). Then we have. h0 (s) = 2As. v 1. Bvs. and h00 (s) = 2A + Bv (v + 1) s. v 2. >0. for all s > 0. Since h (s) tends to 1 as s tends to 0+ and 1, the minima of h (s) appears at s0 with h0 (s0 ) = 0. To …nd such s0 , we solve h0 (s) = 0. Note that 0 = h0 (s) = 2As. Bvs. v 1. implies that 2As = Bvs so vB 2A. s= and thus we can choose h h. vB 2A. vB 2A. 1 v+2. 1 v+2. !. v 1. ;. 1 v+2. as a (better) upper bound for = A. vB 2A. = A. vB 2A. vB = 2. vB 2A. = B. vB 2A. = B. 2A vB 34. 2 v+2. +B. vB 2A. v+2 v v+2. +B. vB 2A. +B. vB 2A. v v+2. v v+2. v v+2. v +1 2 v +1 ; 2. R. M. v v+2. v v+2. v v+2. u2 . Note that.

(40) so Z. u2. vB 2A. h. M. = B. 1 v+2. v v+2. 2A vB. ! v +1 2. and thus Z. u. 2. v+2 v. v+2 2A v v +1 vB 2 v+2 v 1+ v2 A 2 v B +1 B v 2 v+2 2 2 v v +1 BvA v 2 #2 " Z Z Z 2 v v+2 2 v 2d50 rv 1 v 2 juj +1 (by (3)) c jruj + 2 u2 jEj r M v 2 M M 10 4 Z Z Z 2 v 1+ v2 2 v d v r 2 2c v 1 2 0 juj jruj + 2 +1 u2 2 v 2 r M M M jEj v 4 Z Z Z v 1 CN r2 2 jruj + 2 juj u2 ; 2 r M M M jEj v. B. M. = = =. = =. v+2 v. where. 1+ v2 2 10 2c v +1 2 v d0v : v 2 Recall that c and v are constants only involving d0 and P2 , so CN is also a constant involving nothing but d0 and P2 . Therefore, in conclusion, we get. CN =. Z. M. u. 2. 1+ v2. CN r2 jEj. 2 v. Z. M. 4 v. juj. Z. 1 jruj + 2 r M 2. Z. u2. M. with CN = CN (d0 ; P2 ) is constant as desired.. 3.2. Heat kernel upper bound and Sobolev inequality. Let E be a ball of center x0 2 M and raius r > 0 in M . Let H = H (x; t; y) be the Dirichlet heat kernel for E. That is, for each y 2 E, H is of C 1 (E R+ ) and satis…es the following conditions: 8 < H (x; t; y) @t H (x; t; y) = 0 on E R+ H (x; t; y) = 0 on @E R+ ; : limt!0+ H (x; t; y) = y (x) for any x 2 E. where is the Laplace-Beltrami operator which acts only on the x variable and y is the Dirac delta funtion concentrated at y. For such H, for any y 2 E and t 2 R+ , the domain of H ( ; t; y) can be extended by assigning H (x; t; y) = 0 for all x 2 E C . Thus, in this case, we have H ( ; t; y) 2 C01 (E) for 35.

(41) any y 2 E and t 2 R+ . Also, we can de…ne H ( ; ; y) = 0 if y 2 E C , so H can be viewed as a function de…ned on M R+ M . On the other hand, by elliptic theory, there exists eigenfunctions 1 ; 2 ; 3 ; : : : in C01 (E)\ such that C 1 E and corresponding eigenvalues 0 < 1 < 2 3 i. =. i i. for all i 2 N. Also, f i gi2N forms an orthonormal basis for L2 (E). Applying the eigenfunction expansion, we can express the Dirichlet heat kernel in the following form: 1 X H (x; t; y) = e i t i (x) i (y) : i=1. There are some signi…cant properties about the Dirichlet heat kernel. We shall just state them without proof. Proposition 3.1. Let E be a ball of center x0 2 M and raius r > 0 in M . Let H = H (x; t; y) be the Dirichlet heat kernel for E, then 1. H (x; t; y) 0 for any x; y 2 M and t > 0; R 2. M H (x; t; y) dy 1 for any x 2 M and t > 0;. 3. H (x; t; y) = H (y; t; x) for any x; y 2 M and t > 0; R 4. H (x; t + s; y) = M H (x; t; z) H (z; s; y) dz for any x; y 2 M and t; s > 0.. Theorem 3.2. Let E be a ball of center x0 2 M and raius r > 0 in M . Let H = H (x; t; y) be the Dirichlet heat kernel for E. If NI holds for all functions in C01 (E) on M , then H. rv jEj. vCN t. v 2. t. e r2. holds for any x; y 2 M and t > 0, where v is a positive constant and CN is the controlling contant in NI. The R.H.S. of the inequality in this theorem is called a Dirichlet heat kernel upper bound at t > 0. Proof. Fix y 2 M and de…ne u : M. R+ ! R by u (x; t) = H (x; t; y). for all x 2 M and t > 0. For any t > 0, since u 2 C01 (E), we have 4 Z Z Z Z 1+ v2 v CN r2 1 2 2 2 jruj + 2 u juj ; (by (3.1)) u 2 r M M M M jEj v. (1). where v is a positive constant and CN is the controlling contant in NI. Note that, by Proposition 3.1, Z Z juj = ju (x; t)j dx M M Z = jH (x; t; y)j dx M Z = H (x; t; y) dx M Z H (y; t; x) dx = M. 1:. 36.

(42) Pluging this into (1), we get Z u2. 1+ v2. Z. CN r2 jEj. M. 1 jruj + 2 r M. 2 v. 2. Z. u2. M. which is equivalent to Z. M. Z. 2. jruj. jEj v CN r2. 2. u. 1+ v2. 2. M. 1 + 2 r. Z. u2 :. (2). M. Since the last inequality holds and u 2 C01 (E) for all t > 0, Z Z 2 @t = @t u u2 M E Z @t u2 = E Z = 2 u@t u ZE = 2 u u EZ = 2 hru; rui (by Green’s identity) ZE = 2 jruj2 ZE jruj2 = 2 M. =. 2 Z Z 1+ v2 2 jEj v 2 2 u + 2 u2 (by (2)) CN r2 r M M Z Z 1+ v2 u2 A +B u2 ;. M. where. (. (3). M. 2. A= B=. 2jEj v CN r 2 2 r2. Now, de…ne f; g : (0; 1) ! R by f (t) =. Z. :. u2. M. and Bt. g (t) = e for all t 2 (0; 1). Then we have. Z. @t f (t) = @t. f (t). u2. M. A. Z. u. 2. 1+ v2. M. =. +B. Z. u2 (by (3)). M. 1+ v2. A [f (t)]. 37. + Bf (t). (4).

(43) and @t g (t) = @t e =. Bt. f (t). Be. Bt. f (t) + e. Bt. =. @t f (t) h i 2 Be Bt f (t) + e Bt A [f (t)]1+ v + Bf (t) (by (4)) Ae. Bt. [f (t)]1+ v. =. Ae. Bt. eBt g (t). =. Ae. 2Bt v. 2. 1+ v2. 2. [g (t)]1+ v :. That is, [g (t)]. 2 v. 1. @t g (t). Ae. 2Bt v. (5). :. So @t [g (t)]. 2 v. =. 2 2 [g (t)] v 1 @t g (t) v 2A 2Bt e v : (since v > 0 and by (5)) v. Note that the last inequality holds for any t > 0. So, for any t > 0, by the last inequality, we have Z t Z t 2 2A 2Bs @s [g (s)] v ds e v ds t t v 2 2 t. 2A v 2Bs e v v 2B. =. t 2. Bt A 2Bt e v ev B A Bt Bt 1 = ev ev B A Bt Bt ev B v At Bt ev: = v. =. On the other hand, since Z t @s [g (s)]. 2 v. 2 v. ds = [g (t)]. g. t 2. =. e. = e e. Bt. 2Bt v. f (t). [f (t)]. 2Bt v. [f (t)]. 2 v. e. 2 v. 2 v. e ;. together with the previous result, we get e. 2Bt v. [f (t)]. 2 v. 38. 2 v. t 2. At Bt ev v. Bt v. Bt 2. f. 2 v. t 2. f t 2. 2 v.

(44) and thus. v 2. v At. f (t). Bt. (6). e2:. Note that, by Proposition 3.1, Z. f (t) =. u2. ZM. =. ZM. =. ZM. =. [u (x; t)]2 dx H (x; t; y) H (x; t; y) dx H (y; t; x) H (x; t; y) dx. M. = H (y; 2t; y) : Pluging this into (6), we obtain that v At. H (y; 2t; y). v 2. e. Bt 2. and thus 2v At. H (y; t; y). 2v At. v 2. e v 2. Bt 4. Bt. e 2 : (since B > 0). (7). Note that y 2 M and t > 0 are both arbitrary in the last inequality. Finally, for any x; y 2 M and t > 0, by Proposition 3.1 again, Z t t H x; ; z H z; ; y dz H (x; t; y) = 2 2 M ! 21 Z Z 2 t t dz H z; ; y H x; ; z 2 2 M M 1. 2. dz. ! 21. 1. = [H (x; t; x)] 2 [H (y; t; y)] 2 2v At =. v 2. e. Bt 2. 2v CN r2 t 2 jEj v2. rv = jEj. vCN t. (by (7)) ! v2 v 2. e. 2r 2 t 2. t. e r2 ;. where the second inequality holds by Hölder’s inequality. So we get the proof. Let 1 ; 2 ; 3 ; : : : be eigenfunctions in C01 (E) \ C 1 E and 0 < correspoding eigenvalues such that i. = 39. i i. 1. <. 2. 3. be.

(45) for all i 2 N. Since f i gi2N forms an orthonormal basis for L2 (E), for any u 2 L2 (E), there exists fai = ai (u)gi2N such that 1 X u= ai i : i=1. In fact,. ai =. Z. u. i. M. for all i 2 N. In this case, we can de…ne u=. 1 X. ai (. i). i. i=1. so that form:. can be viewed as an operator on L2 (E). We write the last identity in the following (. ). 1 X. ai. i. 1 X. =. i=1. Then, by such point of view, for any c; span f 1 ; 2 ; 3 ; : : :g over R by setting (c. ). 1 X. ai. 2 R, we can de…ne an operator (c. bi. 1 X. i. i=1. bi (c +. (c. ). (c. ). bi. i). ) on. i. i=1. for any fbi gi2N R. Note that, when c = 0 and Beltrami operator as before. Also, (c ) and (c since 1 X. i i:. i=1. i. = 1, we get an "extended" Laplace) are the "inverse" to each other. = (c. ). i=1. 1 X. bi (c +. i). i. i=1. = =. 1 X i=1 1 X. bi (c + bi. i). (c +. i). i. i. i=1. for any fbi gi2N. R.. Theorem 3.3. Let E be a ball of center x0 2 M and raius r > 0 in M . Let H = H (x; t; y) be the Dirichlet heat kernel for E. Suppose that H. rv c jEj t. v 2. t. e r2. holds for any x; y 2 M and t > 0, where v and c are positive contants. If v > 2, then there exists a constant CS = CS (v; c) > 0 such that Z Z Z 1 v2 2 2v C 1 Sr 2 uv 2 jruj + 2 u2 (3.2) 2 r M M M jEj v holds for any u 2 C01 (E). The last inequlaity in this theorem is called the Sobolev inequality (SI) and CS is called the controlling constant in SI. 40.

(46) Proof. First, for convenience, we let (. v. rv c 2 jEj 1 r2. c1 = c2 =. (1). :. In this case, the assumption can be rewritten in the following form: H. v 2. c1 t. ec2 t :. (2). Now, let 1 ; 2 ; 3 ; : : : be eigenfunctions in C01 (E) \ C 1 E and 0 < be correspoding eigenvalues such that =. i. for all i 2 N. For any 2 R, we let (c2 since u 2 C01 (E) L2 (E), we have. 1. <. 2. i i. ) be de…ned as the operator in page ??. Then, 1 X. u=. ai i ;. i=1. where ai = ai (u) =. R. M. u. i. for all i 2 N, and thus (c2. ) u=. 1 X. ai (c2 +. i). i:. i=1. Note that Z Z 2 jruj + c2 u2 = M. M. =. Z. Z. u u + c2 u (c2. M. = =. Z. =. 1 X. ai. i. i=1 1 X. )u !". a2i (c2 +. M i=1 Z "X 1 M. Z. u2 (by Green’s identity). M. M. =. Z. M. Z. 1 X. aj (c2 +. (since f i gi2N is orthonormal). i). i). 1 2. i=1. i. #". 2. 1. j. j=1. ai (c2 +. (c2. j). #. 1 X. aj (c2 +. j=1. )2 u :. M. So, to show. Z. u. 2v v 2. 1. 2 v. jEj. is equivalent to show that. M. u. 2v v 2. 1. Z. CS r2. M. Z. 3. 2 v. 2 v. 2. M. CS r2 2. jEj v 41. Z. jruj + c2. M. (c2. Z. u2. M. 1. 2. )2 u :. j). 1 2. j. #.

(47) Let. 1. u e = (c2. then. 1. ) 2 u;. u = (c2. ). = (c2. ). 1 2. 1. (c2. 1 2. )2 u. u e. 1. since (c2 ) 2 and (c2 ) 2 are the inverse to each other by de…nition. Therefore, to show the last inequality, it is enough to show that Z. (c2. 1 2. ). M. u e. 2 v. 1. 2v v 2. CS r2 jEj. 2 v. Z. M. je uj2 :. (3). To see this, we apply the Marcinkiewicz interpolation theorem. For any x 2 M , we have (c2. ). 1 2. Z 1 X. u e (x) =. M. i=1. Z X 1. =. u e (y). (c2 +. i. (y) dy (c2 +. i). 1 2. i. (x). M i=1. Note that, by the property of the gamma distribution, Z 1 1 1 2e d = 1; 1 0. i. i). 1 2. i. (x). (y) u e (y) dy:. (4). 2. so (c2 +. i). 1 2. = (c2 +. i). Z. 1 2. 1. 1 1 2 1. 0. = (c2 + =. 1 2. i) 1. 1 2. 1 2. Z. 1. t. 1 2. 1 2. Z. 1. e d [(c2 +. i ) t]. 1 2. e. (c2 +. i )t. (c2 +. i ) dt. 0. e. (c2 +. i )t. dt. 0. holds for all i 2 N. Pluging this into (4), we obtain that " # Z X 1 1Z 1 1 1 (4) = t 2 e (c2 + i )t dt i (x) i (y) u e (y) dy 2 M i=1 0 1 1Z 1Z X 1 1 t 2 e (c2 + i )t i (x) i (y) u e (y) dydt = 2 0 M i=1 # Z "X 1 1Z 1 1 1 (c + )t = t 2 e (y) dydt: e 2 i i (x) i (y) u 2 0 M i=1. 42. (5).

(48) c2 t. Let G = G (x; t; y) = e. H (x; t; y), then, by the eigenfunction expansion of H, we have 1 X. c2 t. G (x; t; y) = e. it. e. i. (x). i. (y). i=1. 1 X. =. i=1 1 X. =. e(. c2. e. (c2 +. i )t. i )t. i. (x). i. (y). i. (x). i. (y) :. i=1. Pluging this into (5), we get 1. 1 2. (5) =. Z. 1. t. M. 0. That is, (c2. ). 1 2. 1. 1 2. u e (x) =. Next, for any T > 0, let ( L1 u e (x) = L2 u e (x) =. 1 2 1 2. Then. (c2. ). 1 2. RT t 0 1R1 t T 1. (c2. u e (z). n So, for such z, z 2 x 2 M : (c2 x 2 M : jL1 u e (x)j <. n where I = x 2 M : (c2 C. I Thus, jIj. ). 1 2. Z. 1. G (x; t; y) u e (y) dydt:. t. 1 2. ). 1 2. Z. M. 0. > 0, if z 2 x 2 M : jL1 u e (x)j <. For any. Z. 1 2. G (x; t; y) u e (y) dydt:. R G (x; t; y) u e (y) dydt M : 1 R 2 G (x; t; y) u e (y) dydt M. 1 2. (7). = L1 + L2 :. \ x 2 M : jL2 u e (x)j. 2. (6). , then. 2. = jL1 u e (z) + L2 u e (z)j (by (6)). jL1 u e (z)j + jL2 u e (z)j < : o 1 2 ) u e (x) < . Since z is arbitrary, we get 2. \ x 2 M : jL2 u e (x)j. IC;. 2. o u e (x) < , which is equivalent to. x 2 M : jL1 u e (x)j. x 2 M : jL1 u e (x)j. [ x 2 M : jL2 u e (x)j >. 2. 2. = A + B;. 43. +. 2. x 2 M : jL2 u e (x)j >. :. 2 (8).

(49) where. x 2 M : jL1 u e (x)j x 2 M : jL2 u e (x)j >. A= B=. 2. :. 2. Applying Hölder’s inequality (with respect to dy), we have jL2 u e (x)j =. = =. 1 2. 1. 1. Z. 1 2. t. T. 1 2. 1. 1 2. 1. 1 2. 1. 1 2. 1. 1 2. 1. 1 2. 1. where p; q 2 (1; v) and. Z. Z. M. 1. 1 2. t. T. Z. Z. M. 1. 1 2. t. T. Z. 1 2. t. 1. 1 2. t. sup [G (x; t; y)]. 1. Z. p 1 p. t. 1 p. sup G (x; t; y) y2M. 1. 1 2. t. +. 1 q. dt. 1 p. sup G (x; t; y). 1 q. q. q. je u (y)j dy. 1 q. dt. dt (by Proposition 3.1). je u (y)j dy. Z. 1 q. q. je u (y)j dy. dt. 1 q. q. je u (y)j dy. M. Z. M. M. Z. 1 q. y2M. T. je u (y)j dy. G (x; t; y) dy. M. 1 1 2. 1 q. q. M. Z. p 1 p. Z. M. sup [G (x; t; y)]. T. 1 p. [G (x; t; y)] dy. y2M. T. Z. 1 p. p. y2M. T. Z. G (x; t; y) je u (y)j dydt. M. 1. Z. Z. G (x; t; y) u e (y) dydt (by (7)). (8). dt;. = 1. Note that c2 t. sup G (x; t; y) = sup e y2M. y2M c2 t. = e. H (x; t; y). sup H (x; t; y) y2M v 2. e c2 t c1 t v = c1 t 2. ec2 t (by (2)) (9). Thus jL2 u e (x)j. =. 1 2. 1. 1 2. 1. 1 2. 1. Z. 1. 1 2. t. T. 1 q. c1. sup G (x; t; y) y2M. T. Z. Z. 1 q. 1. 1 2. t Z. M. c1 t. 1 q. v 2. Z. M. je u (y)j dy. 1 q. Z. Note that q < v. So 1 2. v <0 2q. 44. je u (y)j dy. je u (y)j dy. T. 1. t. 1 2. 1 q. dt (by (8)). 1 q. q. M. q. q. v 2q. dt:. dt (by (9)) (10).

(50) and thus Z. 1. 1. t. 1 2. v 2q. dt =. lim. 1. =. v 2q. 1 2. 1. =. v 2q T. 1 2. s!1. T. v 2q. 1 2. h. s. v 2q. t2. lim s. ! 1 2. v 2q. T. s!1 v 2q. 1. T2. 1 2. v 2q. i. :. Pluging this into (10), we get 1. 1 2. (10) =. c1 1. Z. 1 q. c1. M. 1. jL2 u e (x)j. 1. je u (y)j dy je uj. That is, 1 2. 1 q. q. M. 1 2. =. Z. 1 q. Z. 1 q. c1. 1 q. q. M. v 2q. 2q v. je uj. q. 1. v 2q. T2. q 1 q. 2q v. 1. 1 2. :. 1. q. v 2q. T2. T2. v 2q. :. Since the last inequality holds for any T > 0, we can choose an appropriate T in the beginning such that the right hand side is equal to 2 . Then, for such T , we get B = = 0 since x 2 M : jL2 u e (x)j > jIj. x 2 M : jL2 u e (x)j >. 2. is empty. Therefore,. 2. A + B (by (8)) = A =. x 2 M : jL1 u e (x)j q. 2. Z. M. 2. jL1 u e (x)jq dx: (by Chebyshev’s inequality). 45. (11).

(51) Note that, by Hölder’s inequality (with respect to dt), q. jL1 u e (x)j. 1. 1 2. = ". =. =. T. Z. 1 2. t. 1. 1 2. q. 1 2. q. 1 2. q. 1 2. q. Z. Z. T. 1 2. t. 0 T. G (x; t; y) u e (y) dydt. t. Z. Z. Z. Z. 1 2. T. q. G (x; t; y) je u (y)j dydt 1 p. 1 2. t. 0. q 2p. 1 2. t Z. dt. T. Z. 1 2. t. Since, by Hölder’s inequality agian, Z q = G (x; t; y) je u (y)j dy. Z. Z. 1 2. t. G (x; t; y) je u (y)j dy. dt. q. G (x; t; y) je u (y)j dy. (12). dt:. q. 1 q. G (x; t; y) dy. Z. M. c2 t. e. qc2 t p. Z. qc2 t p. H (x; t; y) dy. G (x; t; y) je u (y)jq dy q p. Z. M. Z. q p. H (x; t; y) dy. M. M. M. q. [G (x; t; y)] [G (x; t; y)] je u (y)j dy. M. e Z. G (x; t; y) je u (y)j dy dt. M. 0. M. = e. Z. q p. Z. =. q. 1 p. M. M. T. M. 0. Z. M. Z. q p. 1 q. 1 2. t. 0 T. #q. G (x; t; y) je u (y)j dydt. M. M. 0. T. q. M. 0. 1 2. =. Z. Z. G (x; t; y) je u (y)jq dy. M. G (x; t; y) je u (y)jq dy. G (x; t; y) je u (y)jq dy (by Proposition 3.1). G (x; t; y) je u (y)jq dy: (since. qc2 t > 0 for t > 0) p. Pluging this into (12), we get q. 1 2. (12). T. q 2p. Z. q. 1 2. q. jL1 u e (x)j. T. q 2p. M. jL1 u e (x)j dx. Z. 1 2. M. =. t. Z. M. Z. T. 1 2. t. 1 2. q. q. q 2p. T q. T 2p. G (x; t; y) je u (y)jq dydt:. Z. M. 0. Since x is arbitrary in the last inequality, q. 1 2. 0. That is,. Z. T. Z. Z. T. t. 0. T. 0. 46. t. 1 2. G (x; t; y) je u (y)jq dydt:. 1 2. Z. M. G (x; t; y) je u (y)jq dydtdx. M. G (x; t; y) je u (y)jq dydxdt:. Z Z M. (13).

(52) Note that Z Z Z Z q G (x; t; y) je u (y)j dydx = G (x; t; y) je u (y)jq dxdy M M ZM ZM = e c2 t H (x; t; y) dx je u (y)jq dy ZM ZM H (x; t; y) dx je u (y)jq dy (since c2 t > 0 for t > 0) ZM M je u (y)jq dy: (by Proposition 3.1) M. Pluging this into (13), we arrive at Z. M. 1 2. q. 1 2. q. q. =. 1 2. q. =. 1 2 1 2. q. q. jL1 u e (x)j dx. =. =. T. q 2p. Z. Z. T. 1 2. t. Z. M. 0. je u (y)jq dydt. q. Z. q 2p. M. je u (y)j dy T. M. je ujq T 2p T 2. M. je ujq T 2p + 2. M. je ujq T 2 : (since. Z. q. Z. t. 1 2. dt. 0. 1. q. Z. T. 1. 1 1 + = 1) p q. q. So q. jIj. 2. Z. M. q. jL1 u e (x)jq dx (by (11)) q. 1 2. 2. Z. q. M. je ujq T 2 :. (14). Recall that T was choosen such that 1. 1 2. Z. 1 q. c1. q. M. je uj. v. q. 1 q. 2q v. 1. q. thus T. 1 2. v 2q. 1 2. =. 2q. c1. 2. T. = = =. T T. q v 2q. 1 2. v 2q. 1 2. =. Z. je uj. M. It implies that q 2. v 2q. T2. 2. ;. 1 q. q. :. q2 q v. q2 q v. v. q2 q v. q 2q. 2 47. c1. Z. M. je uj. q. q v q. :.

(53) Pluging this into (14), we get q. (14). 2 = C 2. where. = 4. qv q v. qv v q. M. M. C. R. M. je uj. q. Z. q. 1 2 Z. 1 q. je uj. q. 3 vqvq 5. je uj. q. v. q2 q v. q 2q. c1. 2. Z. M. v v q. je uj. q. q v q. ;. That is,. 2. jIj ). n x 2 M : (c2. 1. 1 2. C=2. n Recall that I = x 2 M : (c2. 1 2. 1 2. 4. 1 2. ). v R. C. M. q v. 2q. je uj. 1. q. q. 1 q. (15). c1v :. 3 vqvq 5. :. o . So, actually, we arrive at. u e (x). 2. o. u e (x). 4. C. R. M. je uj. q. 1 q. 3 vqvq 5. :. (16). Note that the last inequality holds for 1 < q < v. Since v > 2, we can choose q1 and q2 such that 1 < q1 < 2 < q2 < v such that n x 2 M : (c2. ). 1 2. and n x 2 M : (c2. ). with. 1 2. u e (x). for i = 1; 2. Let. =. q1 2. q2 2 , q2 q 1. 1. 1 2. then 0 <. v. v. 2 4. C1. R. C2. R. qi 2qi. qi v. M. M. je uj. q1. je uj. q2. 1 q1. 1 q2. 3 vq1qv. 1. 5. 3 vq2qv. 2. 5. 1. c1v. < 1. Also, for such , we have. 1 + (1 q1 and. 4. o. u e (x). Ci = 2. 2. o. q1 + (1 q1 v. ). ). 48. v. 1 1 = q2 2. q1 v 2 = . q2 v 2v.

(54) Thus, by the Marcinkiewicz interpolation theorem or, more precisely, following the proof of the Marcinkiewicz interpolation theorem, there is some constant K = K (q1 ; q2 ) > 0 such that 1 Z Z h i v2v2 v2v2 2 1 2 1 2 u e : e (c2 ) u KC1 C2 M. M. To see. C1 C21. , we exactly choose. q1 = q2 =. 3 2 2+v 2. :. Then, for such q1 and q2 , we have q1 q2 2 2 q 2 q1 3 v 2 = 4 v 1 =. and thus C1 C21. 1. = 2. 1 2. 1. = 2. 1 2. 1. = 2. 1 2. 1 v. c1. 1 v. c1. ". v. 2q1. ". 2v. 3 2v. 3 6. # 43. 2v. v. #1 2+v 2v. v 2 4 + 2v. # 14. v+2 v 1. (v+2)2 8v(v 1). 4 + 2v v 2. 3. q2 v. q2. 2q2 ". v 2 v 1. 9(v 2) 8v(v 1). 6. 1 v. c1. # ". q1 v. q1. 1 v. = Rc1 ; where R is a constant determined only by v. Therefore, the last inequality becomes Z. (c2. 1 2. ). M. Thus we get Z. (c2. ). M. 1 2. u e. v 2 2v. 2v v 2. u e. Z. 1 v. KRc1. M. 1. 2v v 2. 2 v. 2. 2 v. 2. K R c1 K 2 R2 CS r. =. jEj. 2. 2 v. u e2. 1 2. :. Z. r2 c jEj Z M. u e2 M Z. 2 v. M. u e2 ;. u e2 (by (1)). where CS = cK 2 R2 :Since R is a constant dependent only on v, CS is a constant dependent on both c and v. So we obtain (3) and thus Z. u. 2v v 2. 1. 2 v. CS r2. M. holds by equivalency. Finally, since c2 =. jEj 1 , r2. 2 v. Z. M. 2. jruj + c2. Z. M. the proof is …nished. 49. u2.

(55) Applying Theorem 3.1, 3.2 and 3.3, we get the following result immediately. Corollary 3.1. Let E be a ball of center x0 2 M and raius r > 0 in M . If M has both VDC and WPI, then SI holds for all functions in C01 (E) on M . Moreover, the controlling constant in SI is dependent only on d0 and P2 in this case, where d0 and P2 are the controlling constants in VDC and WPI, respectively. Proof. First we note that the controlling constant in VDC can be enlarged without loss of generality such that d0 > 4. Then, since M has both VDC and WPI, by Theorem 3.1, there exists a constant CN = CN (d0 ; P2 ) > 0 such that Z. u. 1+ v2. 2. CN r2 2. jEj v. M. Z. 1 jruj + 2 r M 2. Z. u. Z. 2. M. M. 4 v. juj. holds for any u 2 C01 (E), where v = log2 d0 > 2, d0 and P2 are the controlling constant in VDC and WPI, respectively. That is, NI holds for all functions in C01 (E) on M . Thus, by Theorem 3.2, for the Dirichlet heat kernel H for E, we have rv jEj. H. vCN t. v 2. t. e r2. holds for any x; y 2 M and t > 0. Let c = vCN , then, by Theorem 3.3, since v > 2, there exists a constant CS = CS (v; c) > 0 such that Z. u. 2v v 2. 1. 2 v. CS r2 jEj. M. 2 v. Z. 1 jruj + 2 r M 2. Z. u2. M. holds for any u 2 C01 (E). That is, SI holds for all functions in C01 (E) on M . Note that v = log2 d0 and c = vCN = (log2 d0 ) CN (d0 ; P2 ), the controlling constant in SI, CS , is actually dependent only on d0 and P2 . So we get the proof.. 4. Subsolutions and supersolutions for the heat equation. De…nition 4.1. Let E be a ball of center x0 2 M and raius r > 0 in M and [a; b] R+ . Let u be a real-valued function de…ned on E [a; b]. u is said to be a subsolution (supersolution, solution) to the Dirichlet heat equation on E [a; b] if u ( ; t) 2 C01 (E) for any t 2 [a; b], u (x; ) 2 C 1 (a; b) for any x 2 E and u satis…es u on. @t u. ( ; =) 0. [a; b].. Remark 4.1. By De…nition 4.1, we know that if u is a solution to the Dirichlet heat equation, then u is both the subsolution and supersolution to the Dirichlet heat equation.. 50.

(56) 4.1. Subsolutions. Theorem 4.1. Let E be a ball of center x0 2 M and raius r > 0 in M . Let t0 > r2 and u be a positive subsolution to the Dirichlet heat equation on E [t0 r2 ; t0 ]. If SI holds for all functions in C01 (E) on M , then, for any 0 < 0 < 1 and p > 0, there exists a constant 0 CM = CM (CS ; ; ; p) > 0 such that sup u Q. CM. 0. 1 jEj r2. Z. u. 1 p. p. ;. (4.1). Q. where Q. E. r2 ; t0. t0. with E = Bx0 ( r) for all 2 (0; 1] and CS is the controlling constant in SI. The inequality in this theorem is called the mean value inequality (MVI) for subsolutions to the Dirichlet heat equation and CM is called the controlling contant in MVI. Proof. For any 0 < such that. and. 0. <. 0 < 1, let = . Let ; 8 r < (l) = 0 for l (l) = 1 for 0 l : 0 2 j j r. 8 <. (t) = 1 for t (t) = 0 for 0 : 0 2 j j 2 r2. respectively. Also, let ' : M. t0 l. : [0; 1) ! R be smooth functions 0. r. (1). 0 2. r. t0. r2 ;. (2). [0; 1) ! R be a test function de…ned by ' (x; t) =. (d (x; x0 )) (t). for all x 2 M and t 0. By a direct calculation, since u ( ; t) 2 C01 (E), for all t 2 2 [t0 r ; t0 ], and u is a subsolution to the Dirichlet heat equation, we have Z Z 2 r ' u ru = '2 u u (by Green’s identity) E ZE '2 u@t u E Z 1 = '2 @t u2 2 E Z 1 = @t '2 u2 u2 @t '2 2 E Z Z 1 2 2 = @t 'u + u2 '@t ': (3) 2 E E. 51.

(57) Note that, on the other hand, we have Z Z 2 r ' u ru = '2 ru + ur'2 ru E ZE = '2 jruj2 + 2'ur'ru ZE = j'ruj2 + 2 ('ru) (ur') + jur'j2 ZE Z 2 = j'ru + ur'j jur'j2 E ZE Z 2 = jr ('u)j jur'j2 : E. and thus. Z. 1 jr ('u)j + @t 2 E. Z. Z. 2 2. 'u. E. =. 2 2. 'u +. E. 2. ZE. ZE. jur'j +. Z. u2 '@t '. E. Z. u2 '@t '. E. u2 jr'j2 + '@t '. u2 E Z 6 2 r2. =. Z. 1 @t 2. E. 2. 4 2 + 2 r2 2 r2. (by (1) and (2)). u2 :. (5). E. The last inequality induces two important informations. First, since we have Z Z Z 1 1 2 2 2 jr ('u)j + @t @t 'u ' 2 u2 2 2 E E E Z 6 u2 : (by (5)) 2 r2 E That is, @t. Z. 12 2 r2. 2 2. 'u. E. Let. (4). E. So, combining (3) and (4), we get Z Z 2 jr ('u)j jur'j2 E. jur'j2. I. 0. Z. R. E. jr ('u)j2. 0,. u2 : E. 0 2. t0. r ; t0 ;. then, for any s 2 I 0 , by the last inequality, we have Z s Z Z s Z 12 2 2 @t 'u 2 r2 t0 r2 E t r2 Z 0t0 Z 12 2 r2 2 Zt0 r 12 = 2 2 u2 : r Q 52. u2 E. u2 E. (6).

(58) Note that Z. s. @t. t0. r2. Z. 2 2. 'u. Z. =. E. 'u. E. Z. =. ' 2 u2. E. Pluging this into (6), we obtain that Z ' 2 u2 E. Z. 2 2. t=s. t=s. ' 2 u2. t=t0. E. r2. : (by (2)). t=s. 12 2 r2. Z. u2 :. 12 2 r2. Z. u2 :. Q. Since s is arbitrary, we have sup I. 0. Z. 2 2. 'u. E. (7). Q. Second, if we integrate (5) directly on [t0 r2 ; t0 ], then we get Z t0 Z Z Z Z t0 Z 1 t0 6 2 2 2 jr ('u)j + 'u @t 2 r2 2 t0 r2 t0 r2 E E t r 2 Bx0 ( Z0 6 = 2 2 u2 : r Q. u2 r). (8). Note that Z. Z. t0. @t. t0. r2. 2 2. 'u. Z. =. 'u. ZE. E. =. Z. 2 2. ' 2 u2. E. t=t0. t=t0. ' 2 u2. E. r2. t=t0. (by (2)). 0;. so Z. t0. t0. r2. Z. E. Z. 2. jr ('u)j. t0. t0. 6 2 r2. Now, let v > 2 be …xed, then we have Z Z 4 2(1+ v2 ) ('u) = ('u)2 ('u) v E. E. Z. ('u). E. CS r2 2. jEj v. Z. 2v v 2. E. v 2 v. r2. Z. Z. 1 jr ('u)j + 2 E 2. Z. t0. @t r2. t0. Z. ' 2 u2. E. u2 : (by (8)). (9). Q. Z. 2. 2 v. (by Hölder’s inequality) ! Z 2 v 2 2 2 2 jr ('u)j + r ('u) ('u) (by (3.2)) ('u). E. E. 53.

(59) and thus, by (7), we get Z. t0. t0. 0 r2. Z. 1+ v2. 2(. ('u). Z. ). E. t0 0 r2. t0. CS r2 2. jEj v. CS r2. =. jEj where A=. Z. t0. t0. Since A =. Z. 0 r2. t0. Z. Z. t0. t0. t0. r2. Z. E. E. Z. 0 r2. Z. u. Z. 2 v. 2. Q. Z. 12 2 r2. 2 v. 2. 2. t0 0 r2. t0. 2 v. u2. 2. 2 v. E. Z. E. jr ('u)j2 + r. 2. ('u)2 (10). A;. Q. jr ('u)j2 + r. E. 0 r2. t0. E. 0. 12 2 r2. 2 v. 2. ! Z. jr ('u)j + r ('u) ('u) 2 E jEj v E ! v2 Z Z Z t0 2 2 jr ('u)j2 + r 2 ('u)2 sup ' u I. CS r2 jEj. Z. CS r2. 2. jr ('u)j + r 2. jr ('u)j + r. 2. Z. t0 0 r2. t0. 2. sup I. 0. ('u)2 :. 2. Z. Z. E. ('u)2 E !Z. t0. ' 2 u2. 1. t0. 0 r2. Z Z 6 2 2 12 0 2 u +r u2 r (by (7) and (9)) 2 r2 2 r2 Q Q Z Z 12 0 6 2 u + 2 2 u2 = 2 2 r Q r Q Z 6 (1 + 2 0 ) = u2 2 r2 Q Z 6 (1 + 2) u2 (since 0 < 0 < 1) 2 r2 Q Z 18 = 2 2 u2 ; r Q we have Z. t0. t0. 0 r2. Z. CS r2. 2(1+ v2 ). ('u). 2. jEj v. E. CS r2 jEj. =. 2 v. CS r2 jEj. 2 v. where c1 = 122 18v 54. 12 2 r2 12 2 r2 c1 2 r2 1 v+2. Z. u. 2. u. 2. 2 v. A (by (10)). Q. Z. Q. Z. Q. :. u2. 2 v. 18 2 r2. 1+ v2. ;. Z. u2. Q. (11).

(60) Note that Z. t0. t0. 0 r2. Z. Z. 2(1+ v2 ). ('u). t0. = =. Z 0t0 Zt0 Q. so (11) becomes. Z. Q. 0 r2. t. E. u. CS r2. 2(1+ v2 ). 2. 2. ('u)2(1+ v ) 0E. Z. 2. u2(1+ v ) (by (1)) 0E 2. u2(1+ v ) ; 0. c1 2 r2. jEj v. 0. 0 r2. Z. Z. u. 2. 1+ v2. (12). :. Q. Recall that (12) holds for all positive subsolution to the Dirichlet heat equation. For any q > 1, since uq. @t uq = div (ruq ) @t uq = div quq 1 ru quq 1 @t u = q ruq 1 ru + uq 1 div (ru) = q (q quq = quq 0;. quq 1 @t u. 1) uq 2 jruj2 + quq 1 u quq 1 @t u 1 u quq 1 @t u (since q > 1) 1 ( u @t u). uq is also a positive subsolution to the Dirichlet heat equation and thus uq also satis…es (12). That is, Z Z CS r2 c1 2q u u2q ; 2 2 r2 v Q 0 Q jEj where i =. 0 and set = 1 + v2 . Now, for the given 0 < 0 < < 1, we let = Pi i for all i 2 N. Note that, by this setting, we have j=1 2j and qi = i+1. =. =. i+1. 2i+1. 55. i. :. 0. = ,.

(61) Thus, applying Moser’s iteration on the last inequality with such setting, we obtain that Z. u2. Q. CS r2. i+1. jEj. i+1. =. 2 v. 2 2 i+1 r. 4i+1 c1 2 r2. CS r2 jEj. =. c1. 2 v. CS r2 jEj. c1 2 r2. 2 v. = K L. Z. Q. u2. i. u2. i. i. Z. Q. = K 1+. !. i. Z. 4(i+1). 4(i+1) 2. i. u2 i. 4(i+1) 4K L L. +. u2. Q. Z. Q. K L. !. 2. 4(i+1). +i. !. i. Z. 4i Z. 2. .. . K 1+. + +. L. +. 2. + +. i+1. !. i 1. u2. Q. Q. i. i. i 1. u2. i 1. +i. 2. i 1. 4(i+1). !. + +. !3. 2. 5. i+1. Z. Q. where. (. K= L=. u2 0. 0. !. i+1. ;. CS r 2 2. jEj v c1 2 r2. :. So, Z. Q. u i+1. 2. i+1. !. 1 i+1. K = K. 1 1 i+1 + i +. Pi+1. j=1. 1 j. +1. L. L. Pi. j=0. 56. 1 1 i+ i 1+. 1 j. Pi. 4. +1. j+1 j=0 j. 4 Z. (i+1) i i + i 1+. Q. +1. Z. Q. u2 :. u2 0. 0.

(62) letting i ! 1 in the last inequality, we get sup u Q. 2. P1. 1 j. P1. P1. 1 j. 0. L. j=0. where. j=0. Z. Now, for the given p > 0, if p Z. u2. v 4(1+ 2 ) :. 2, then Z. 2 p. up. Q. Q. Z. =. 2 p. 1. (by Hölder’s inequality). 1. Q. u. Z. 2 p. p. Q. Z. (13). 2. 1+ v2. v. c2 = CS2 c1 Z. j+1 j. 4 u2 Q Z 2 1 u2 = K 1 L 1 4( 1 ) Z Q 2 v v v 2 u2 (since = 1 + ) = K 2 L1+ 2 4(1+ 2 ) v Q ! v2 v Z 1+ 2 2 CS r2 c1 1+ v2 ) ( = 4 u2 2 2 r2 Q jEj v 2 v v v Z 1+ CS2 c1 2 4(1+ 2 ) = u2 2+v r 2 jEj Q Z c2 = 2+v 2 u2 ; r jEj Q K. j=1. 2 p. 1. 1. Q1. u. 2 p. p. r2 jEj. Q. 2 p. 1. (14). and thus sup u =. sup u2. Q. Q. 0. 0. ! 21. 1. c22 1+ v2. r jEj. 1 2. 1. c22 1+ v2 1. c22. =. 1+ v2. r jEj. 1 2. 1 r2 jEj. Z. u2. 1 2. (by (13)). Q. Z. u. p. 1 p. Q. Z. u. p. r2 jEj. 1 2. 1 p. (by (14)). 1 p. (15). :. Q. On the other hand, in the case of 0 < p < 2, we note that 1. sup u Q. 0. c22 1+ v2. 1. r jEj 2. Z. Q. 57. u. 2. 1 2. (by (13)).

(63) holds for any 0 <. 0. 1. So, for any 0 <. <. < Z. 1. c22. sup u Since Z. Q. u. 1. v. "1+ 2 r jEj 2. Q. Z. 1 2. 2. =. u. Q. 1 2. 2. :. +". 1 2. p 2 p. uu. Q. +". , we have. +". +". sup u Q. Z. 1 2. 2 p. Z. p 2. 1. sup u Q. u. Q. +". 1 2. p. +". u. 1 2. p. ;. Q. +". the previous inequality becomes 1. sup u. sup u. 1. v. "1+ 2 r jEj 2. Q. Set 0 = 0 , i+1 = i + "i+1 , "i = setting, we obtain that. Q. u. 1 2. p. :. Q. +". p . 2. and # = 1. 2i. Z. p 2. 1. c22. Iterating on the last inequality by such. sup u = sup u Q. Q. 0. 0 1. c22 1+ v2 21 1. c22. =. 1+ v2. sup u. 1. r jEj 2 Z. Q. u. 1. r jEj 2. = M N1. 1. !# Z. Q. 1 2. p. u. 1+ v2. 2. sup u Q. Q. sup u Q. 2. 1. !#. M N 1 4M N 2 = M 1+# N 1+2#. 1 2. p. sup u Q. sup u Q. 2. 2. 1. !#. !# 3# 5. !#2. .. . M 1+#+. +#. i. N 1+2#+. i. +(i+1)#. sup u Q. = M where. Pi. 8 <. j=0. #j. N. Pi. j=0 (j+1)#. j. sup u Q. 1. c22. M = 1+ v 1 2 rjEj 2 : N = 21+ v2 58. R. Q. u. p. i+1. 1 2. :. i+1. !#i+1. !#i+1 ;.

(64) In the last inequality, letting i ! 1, we obtain that sup u Q. M. 0. P1. j=0. #j. N. P1. j=0 (j+1)#. 1 (1 #)2. 1. = M1 # N 4 2 = M p N p2 " 1 c22 = 1 1+ v2 r jEj 2 1 p. v. c2 21+ 2. =. j. 1 2. up. Q. # p2. Z. 1 2 r jEj. 2+v p. Q. i+1. 1 Z. 4 p2. sup u. lim. i!1. !#i+1. u. p. 4 p2. v. 21+ 2 1 p. (16). :. Q. By (15) and (16), we know that, for any p > 0, sup u Q. where CM Recall that. 1 2 r jEj. CM. 0. Z. up. ;. Q. 8 1 1 < c 2 c p 21+ v2 2 2 = max v ; 2+v : 1+ 2 p v. 1 p. 4 p2. 9 = ;. :. 2. v. v 1+ c2 = CS2 c1 2 4(1+ 2 ) i1+ v2 v h 1 v 2 2 v v+2 2 = CS 12 18 4(1+ 2 ) 2. v. v v = 12 18 2 4(1+ 2 ) CS2 ;. and 0. =. :. So CM is just a constant dependent on CS ; 0 ; and p. Thus we get the proof.. 4.2. Supersolutions. Theorem 4.2. Let E be a ball of center x0 2 M and raius r > 0 in M . Let t0 > r2 and u be a positive supersolution to the Dirichlet heat equation on E [t0 r2 ; t0 ]. Fix v > 2 and let = 1 + v2 . If SI holds for all functions in C01 (E) on M , then, for any 0 < 0 < 1 p0 0 < p0 1, there exists a constant CRH = CRH (CS ; ; ; v; p; p0 ; r) > 0 such and 0 < p that ! p1 1 1 1 Z Z 0 p p0 p 1 p0 p u CRH u (4.2) jEj r2 e 0 e Q Q where. e Q. E. t0. r2 ; t0. (1. ) r2. with E = Bx0 ( r) for all 2 (0; 1] and CS is the controlling constant in SI. The inequality in this theorem is called the reverse Hölder inequality (RHI) and CRH is called the controlling constant in RHI. 59.

(65) Proof. For any 0 < such that. and. 0. <. 8 <. 0 < 1, let = . Let ; 8 r < (l) = 0 for l (l) = 1 for 0 l : 0 2 j j r. (t) = 0 for t (t) = 1 for 0 : 0 2 j j 2 r2. respectively. Also, let ' : M. t0 l. : [0; 1) ! R be smooth functions 0. r. (1). (1 ) r2 0 t0 (1 ) r2 ;. (2). [0; 1) ! R be the test function de…ned by ' (x; t) =. (d (x; x0 )) (t). for all x 2 M and t 0. Then, for any q 2 0; p0 , since u ( ; t) 2 C01 (E), for all t 2 [t0 r2 ; t0 ], and u is a supersolution to the Dirichlet heat equation, we have Z Z 2 q 1 r 'u ru = '2 uq 1 u (by Green’s identity) E ZE '2 uq 1 @t u E Z 1 = '2 quq 1 @t u q E Z 1 = '2 @t uq q E Z 1 @t '2 uq uq @t '2 = q E Z Z 1 2 2 q = @t 'u + uq '@t ': (1) q q E E Note that, on the other hand, Z Z 2 q 1 r 'u ru = E ZE = ZE = ZE =. we have '2 ruq. 1. + uq 1 r'2 ru. (q. 1) '2 uq. (q. 1) '2. 2. jruj2 + 2'uq 1 r'ru. q 4 q2 2 u q2 4. 1. 2. q. jruj2 + 2'u 2. 2 q q u2 q 2. q 2 q 4 (q 1) 2 4 q 2 2 r'ru 2 ' ru + 'u q2 q E Z 2 q q q 4 (q 1) q 'ru 2 + 'ru 2 u 2 r' = 2 q q 1 E. Since q. 'ru 2. 2. q. =. r 'u 2. =. r 'u 2. q. q. 2. 2. u 2 r' + uq jr'j2 60. q. q. 2u 2 r'r 'u 2. :. 1. r'ru. (2).

(66) and q q. 'ru. 1. q 2. q 2. u r'. h. q. =. q. 1 q. =. q. 1. r 'u. q 2. q 2. u r'. q. i. q. u 2 r'. q. q. u 2 r'r 'u 2. q. 1. uq jr'j2 ;. we get 4 (q 1) (2) = q2. Z. 2. q. r 'u 2. E. +. 2 q. q q q u 2 r'r 'u 2 1. 1 q. 1. uq jr'j2 :. (3). Since 2 q. 0. q 1. 2 q. =. q 1. q 2. u r' + r 'u. 2. q 2. 2. +2. we have 2 q. q q q u 2 r'r 'u 2 1. p0. Note that, since q. 1 2. 2 q. 2 q. 2. q. uq jr'j2 + r 'u 2. q 1. 2. q 1. q. q. u 2 r'r 'u 2 ; q 1 r 'u 2 2. uq jr'j2. 2. :. < 1, 4 (q 1) < 0; q2. so 4 (q 1) q2. (3) =. 4 (q 1) q2. Z " E. Z. E. r 'u. q 1 r 'u 2 2. 2 q. 1 2. 2. q 2. q 1. 2. q 1 r 'u 2 2. uq jr'j2. 1. 2. q. 1. uq jr'j2. q 2 2q + 2 q 2 : 2 u jr'j 2 (q 1). 2. That is, Z. E. 2 q 1. r 'u. ru. 4 (q 1) q2. Z. q 1 r 'u 2 2. E. q 2 2q + 2 q 2 : 2 u jr'j 2 (q 1). 2. Combining the last inequality and (1), we obtain that Z 2 q 1 4 (q 1) q 2 2q + 2 q 2 2 r 'u 2 u jr'j q2 2 2 (q 1) E It implies that Z q r 'u 2 E. 2. q 2 2q + 2 (q 1)2. So, if we set I=. Z. E. Z. E. q. u jr'j. r 'u. q 2. 1 @t q. q. 2. 2 (q q. 2. +. 61. 1). 2 (q. 1). Z. @t. @t. Z. 2 'u + q E. 'u +. E. '2 uq ;. Z. q. 2 q. E. Z. 2 q. (q. 1). uq '@t ':. E. Z. E. uq '@t ':. #.