Heat kernel and Harnack inequality on Riemannian manifolds

Transcription

1 Hea kernel and Harnack inequaliy on Riemannian manifolds Alexander Grigor yan UHK 11/02/2014 onens 1 Laplace operaor and hea kernel 1 2 Uniform Faber-Krahn inequaliy 3 3 Gaussian upper bounds 4 4 ean-value propery 7 5 Relaive Faber-Krahn inequaliy 9 6 On-diagonal lower esimae 11 7 Harnack inequaliy and Li-Yau esimae 11 8 Esimaes of derivaives of he hea kernel 13 1 Laplace operaor and hea kernel A weighed manifold is a couple, μ where is a Riemannian manifold and μ is a measure on wih smooh posiive densiy wih respec o he Riemannian measure μ 0 ; ha is, dμ = h 2 dμ 0 where h, h > 0. The weighed Laplace operaor Δ on is defined by Δ = 1 h 2 div h 2. Observe ha Δ is a symmeric operaor wih respec o measure μ. Indeed, for all f, g 0, Δf gdμ = div h 2 f gdμ 0 = h 2 f, g dμ 0 1

2 whence Δf gdμ = fδgdμ. Furhermore, he operaor Δ wih he domain 0 is admis he Friedrichs exension o a self-adjoin operaor in L 2, μ, which will also be denoed by Δ. The hea semigroup {exp Δ} 0 is defined by means of specral heory as a family of operaors exp Δ in L 2, μ, and he hea kernel p x, y is a unique smooh posiive funcion of > 0 and x, y such ha e Δ f x = p x, y f y dμ y, for all f L 2, μ. I is known ha p x, y exiss on any weighed manifold and coincides wih he minimal posiive fundamenal soluion of he hea equaion u = Δu on R +. Besides, he hea kernel saisfies he following properies. symmery: p x, y = p y, x. he semigroup ideniy: p +s x, y = p x, y dμ y 1. p x, z p s z, y dμ z. 1 Recall ha in R n wih he Lebesgue measure μ, Δ is he classical Laplace operaor Δ = n k=1, and is hea kernel is given by he Gauss-Weiersrass formula 2 x 2 k p x, y = 1 4π n/2 exp x y 2. 4 Explici formulas for he hea kernel exis also in hyperbolic spaces H n when μ is he Riemannian measure. For example in H 3 p x, y = 1 4π 3/2 r sinh r exp r For arbirary H n he formula looks complicaed, bu i implies he following esimae, for all > 0 and x, y H n : p x, y n r r exp λ r2 n/2 4 λr, 3 where λ = n 12 4 is he boom of he specrum of he Laplace operaor on H n. 2

3 2 Uniform Faber-Krahn inequaliy Le, μ be a weighed manifold. For any relaively compac open se Ω, denoe by λ 1 Ω he smalles eigenvalue of he weak Dirichle problem for Δ in Ω. Le Λ be a funcion on 0, +. We say ha a weighed manifold, μ saisfies he uniform Faber-Krahn inequaliy wih funcion Λ if, for any non-empy relaively compac open se Ω, λ 1 Ω Λ μ Ω. 4 For example, R n saisfies he Faber-Krahn inequaliy wih funcion Λ v = cv 2/n. Also, any aran-hadamard manifold of dimension n saisfies he Faber- Krahn inequaliy wih he same funcion Λ bu possibly wih a differen consan c. If K is a k-dimensional compac manifold hen he Riemannian produc = R m K saisfies he Faber-Krahn inequaliy wih funcion { v Λ v = c 2/n, v 1, v 2/m 5, v 1, where n = dim = k + m. Any n-dimensional manifold wih bounded geomery saisfies he Faber-Krahn inequaliy wih he funcion { v Λ v = c 2/n, v 1, v 2 6, v 1. Theorem 1 Assume ha, μ saisfies he Faber-Krahn inequaliy 4 wih a posiive coninuous decreasing funcion Λ such ha dv <. 7 vλ v 0 Then, for all > 0, sup p x, x 4 x γ/2, 8 where he funcion γ is defined by he ideniy = γ 0 dv vλv. 9 Approach o he proof. Assuming ha 4 holds, one deduces he following Nash ype inequaliy: for any non-zero funcion u D, 2 u 2 dμ 1 ε u 2 L 2Λ u 2 L 1, 10 ε u 2 L 2 for any ε 0, 1. Then one applies Nash s argumen: exending 10 o u = p x, and noicing ha u 2 dμ = uδudμ = 1 d 2 d u 2 L 11 2 and u L 1 1, one obains from 10 and 11 a differenial inequaliy for u 2 L 2 = p 2 x, x, whence he resul follows. 3

4 3 Gaussian upper bounds Fix some value D 0, + ] and consider he following weighed inegral of he hea kernel: d E, x := p 2 2 x, z x, z exp dμz. 12 D A priori, he value of E, x may be +. For example, in R n E, x = if D 2. Lemma 2 For all x, y, > 0, he following inequaliy holds p x, y E/2, xe/2, y exp d2 x, y. 13 Proof. For any poins x, y, z, le us denoe α = dy, z, β = dx, z and γ = dx, y. By he riangle inequaliy, α 2 + β γ2. y γ α x β z Figure 1: We have hen p x, y = p /2 x, zp /2 y, zdμz p /2 x, ze β2 D p/2 y, ze α2 D e γ2 dμz which was o be proved. p 2 /2x, ze 2β2 D dμz 1 2 = E/2, xe/2, y exp d2 x, y 1 p 2 /2y, ze 2α2 2 D dμy e γ2 Lemma 3 If D 2 hen for any x, he funcion E, x is decreasing in. In paricular, if E, x < for some = 0 hen E, x < for all > 0. Approach o he proof. The proof of he monooniciy of E, x amouns o verifying ha is ime derivaive is non-posiive. I is essenial for he proof ha d x, 1, which implies ha he funcion ξ, x = d2 x,y saisfies D, ξ + D 4 ξ

5 Theorem 4 Assume ha, for some x and for all > 0, p x, x γ, 15 where γ is an increasing posiive funcion on R + saisfying he doubling condiion: γ 2 Aγ for all > 0 16 for some consan A > 1. Then, for any D > 2 and all > 0, for some ε = ε D > 0 and = A,, D. E, x γε, 17 By puing ogeher Theorem 4 and Lemma 2, we obain he following resul. orollary 5 Assume ha, for some poins x, y and for all > 0, p x, x γ 1 and p y, y γ 2, 18 where γ 1 and γ 2 are increasing posiive funcion on R + boh saisfying 16. Then, for any D > 2 and all > 0, p x, y γ1 εγ 2 ε exp d2 x, y. 19 orollary 6 On any weighed manifold, μ and for any D > 2, E, x is finie for all > 0, x. oreover, he funcion E, x is coninuous and monoone decreasing. Skech of proof. By Theorem 4 and Lemma 3, i suffices o prove ha for any x here exis posiive consans and T such ha p x, x n/2, for all 0 < < T, 20 where n = dim. Fix a small relaively compac open se Ω conaining he poin x. By compacness argumen, he weighed manifold Ω, μ saisfies he following Faber-Krahn inequaliy: for all open ses U Ω, such ha μ U 1 2 μ Ω, λ 1 U cμ U 2/n, where c > 0 depends on Ω. Hence, by a sligh modificaion of Theorem 1, we obain ha he hea kernel p Ω of Ω, μ saisfies he esimae p Ω x, x n/2, for all 0, T, 5

6 where and T depend on Ω. onsider he funcion u, y = p x, y p Ω x, y and exend i o 0 by seing u, y 0. This funcion saisfies in R Ω he equaion u = Δu and hence i is -smooh in R Ω. In paricular, he funcion u, x is bounded on [0, T ], say u, x. Then we obain, for all 0 < < T, whence 20 follows. p x, x = p Ω x, x + u, x n/2 +, Theorem 7 On any weighed manifold, μ and for any D > 2, here exiss a posiive coninuous funcion Φ, x on R +, which is decreasing in and such ha he following inequaliy holds p x, y Φ, xφ, y exp λ min d2 x, y, 21 for all x, y and > 0, where λ min is he boom of he specrum of Δ on. Proof. Le us firs se Φ, x = E 1, x By orollary 6, his funcion is finie. By Lemma 3, he funcion Φ, x is decreasing in. By Lemma 2, we obain p x, y Φ, xφ, y exp d2 x, y. 23 This esimaes sill does no mach 21 because of he lack of he erm λ min. To handle i, le us find a posiive smooh funcion h saisfying on he equaion Δh + λh = 0 where λ = λ min. onsider he measure μ defined by d μ = h 2 dμ and he hea kernel p on he weighed manifold, μ. Applying 23 on, μ, we obain ha here exiss a funcion Φ, x decreasing in such ha p x, y Φ, x Φ, y exp Using he relaion beween he hea kernels d2 x, y p x, y = p x, yhxhye λ, and 24, we obain 21 wih Φ, x = Φ, xh x.. 24 Remark. As i follows from he consrucion of he funcion Φ, x and from he proof of orollary 6, for any compac se K here exis posiive consans and T such ha Φ, x n/4 for all x K and 0 < < T. 25 6

7 4 ean-value propery Here we presen an alernaive mehod of obaining Gaussian upper bounds, which avoids using E, x and, insead, is based on a cerain inegral esimae of he hea kernel and he mean-value propery. The following heorem shows ha he Gaussian exponenial erm appears naurally in he hea kernel upper esimaes on arbirary manifolds. Theorem 8 Le, μ be a weighed manifold and le A and B be wo μ-measurable ses on. Then p x, ydμxdμy μaμb exp d2 A, B, 26 4 A B where da, B is he geodesic disance beween ses A and B. To obain poinwise bounds from 26 one needs o combine i wih a mean-value propery. Definiion. We say ha he manifold admis he mean-value propery V if, for all > τ > 0, x and for any posiive soluion us, η of he hea equaion in he cylinder τ, ] Bx, τ, we have u, x τv x, τ τ Bx, τ us, zdμzds. 27 s,x u= u - -,x x _ Bx, Figure 2: Theorem 9 Assume ha he mean-value propery V holds on he manifold. Then, for all x and > 0, p x, x 7 V x,. 28

8 oreover, for all x, y, > 0, D > 2, p x, y V x, /2V y, /2 exp d2 x, y. 29 Theorem 9 admis a localized version. We say ha he manifold admis a resriced mean-value propery V xyτ 0, for some x, y and τ 0 R +, if he inequaliy 27 holds for all τ 0, τ 0 ] in each of he balls B x, τ and B y, τ. If admis V xyτ 0 hen similarly o he above heorem we obain p x, y V x, τv y, τ exp d2 x, y where τ = min/2, τ 0. Observe ha he propery V xyτ 0 holds on any manifold: for any given x, y, here exiss τ 0 such ha V xyτ 0 is rue. However, he consan in he mean-value inequaliy 27 depends on he cerain geomeric properies of he balls Bx, τ 0 and By, τ 0. Definiion. We say ha a weighed manifold, μ saisfies he volume doubling propery if, for all x and r > 0, V x, 2r V x, r. 30 I is known ha he volume doubling condiion implies he volume comparison condiion: for all 0 < r < R and x, V x, R V x, r N R, 31 r wih some N > 0. Assuming 31, one can improve he esimae 29 as follows. Theorem 10 Assume ha, μ saisfies he mean-value propery V and he volume comparison 31. Then, for all x, y and > 0, p x, y where d = dx, y. V x, V y, 1 + d N 1 exp d2 4 Noe for comparison ha on S n he hea kernel a he poles x, y admis he esimae p x, y c 1 + d n 1 exp d2, 0, n/2 4 which shows he sharpness of

9 5 Relaive Faber-Krahn inequaliy Le, μ be a geodesically complee weighed manifold. Definiion. We say ha, μ saisfies he relaive Faber-Krahn inequaliy if here exis posiive consans δ, c such ha, for any geodesic ball B x, r on and for any non-empy relaively compac open se Ω B x, r, λ 1 Ω c r 2 V x, r μ Ω δ. 33 For example, he relaive Faber-Krahn inequaliy holds in R n wih δ = 2/n since V x, r = cr n and hence 33 amouns o he uniform Faber-Krahn inequaliy 4 wih Λ v = cv 2/n. Theorem 11 If has non-negaive Ricci curvaure and μ is he Riemannian volume hen, μ saisfies he relaive Faber-Krahn inequaliy. Approach o he proof. The following key propery of manifolds of nonnegaive Ricci curvaure is used in he proof of his heorem. For any x, y le γ x,y : [0, L] be a shores geodesic beween x and y, where L = d x, y, γ x,y 0 = x and γ x,y L = y. For any x and 0 < s < 1, define a homohey Γ x s : by Γ x s y = γ x,y sl. Γ x s A A x y Γ x s y Figure 3: Homohey Γ x s Then here exiss c > 0 such ha for any 1 2 s 1 and any Borel se A, μ Γ x s A cμ A. 34 Furhermore, if a homohey wih he propery 34 can be defined on some manifold, μ hen his manifold saisfies he relaive Faber-Krahn inequaliy. The class of weighed manifolds wih he relaive Faber-Krahn inequaliy is much wider han hose wih non-negaive Ricci curvaure. In paricular, his class is sable under quasi-isomery. Anoher example of sabiliy: a conneced sum of k copies of he same manifold saisfying he relaive Faber-Krahn inequaliy also saisfies he relaive Faber-Krahn inequaliy. 9

10 Theorem 12 The relaive Faber-Krahn inequaliy implies he volume doubling and he mean value propery. ombining wih Theorem 9 or 10, we obain hea kernel bounds under he relaive Faber-Krahn inequaliy. Theorem 13 The following condiions are equivalen: a, μ saisfies he relaive Faber-Krahn inequaliy. b, μ saisfies he volume doubling propery and he hea kernel on, μ admis he esimae p x, x V x,, 35 for all x and > 0. c, μ saisfies he volume doubling propery and he hea kernel on, μ admis he esimae p x, y V x, exp c d2 x, y, 36 for all x, y and > 0. The consan c in he exponenial in 36 can be made arbirarily close o 1. In 4 fac, i can be aken exacly 1 a he expense of addiional facors as in he following 4 saemen. Theorem 14 Le, μ saisfy he relaive Faber-Krahn inequaliy. addiion ha for some N > 0 and for all 0 < r < R and x, Assume in Then, for all x, y and > 0, p x, y V x, V y, V x, R V x, r 1 + N R. 37 r N 1 d x, y exp d x, y Noe ha he volume comparison 37 follows from he relaive Faber-Krahn inequaliy 33 wih N = 2 δ. However, N in 37 does no have o be 2 δ. 10

11 6 On-diagonal lower esimae The following resul allows o obain an on-diagonal lower bound from an upper bound. Theorem 15 Assume ha for some poin x, V x, 2r V x, r for all r > 0 and Then, for all > 0, p x, x V x,, for all > 0. c p x, x V x,. 39 orollary 16 The following condiions are equivalen: a, μ saisfies he relaive Faber-Krahn inequaliy. b The hea kernel on, μ saisfies he esimaes p x, y V x, exp c d2 x, y. 40 for all x, y and > 0, and p x, x c V x,, 41 /2 for all x and > 0. orollary 17 If has non-negaive Ricci curvaure and μ is he Riemannian volume hen he hea kernel on, μ saisfies he upper bounds 36, 38 and he lower bound Harnack inequaliy and Li-Yau esimae In his secion we assume ha, μ is a geodesically complee weighed manifold. We say ha he hea kernel on, μ saisfies he Li-Yau esimae if, for all x, y and > 0, p x, y V x, exp c d2 x, y. 42 P. Li and S.-T. Yau proved his esimae on geodesically complee Riemannian manifolds wih non-negaive Ricci curvaure using he gradien esimaes see Theorem 22 below. 11

12 We say ha, μ saisfies he uniform parabolic Harnack inequaliy if, for any ball B z, r on and for any posiive soluion u, x of he hea equaion in he cylinder = 0, r 2 B z, r, he following holds: sup u, x inf u, x + where = 1 4 r2, 1 2 r2 B z, 1 2 r and + = 3 4 r2, r 2 B z, 1 2 r. r 2 3 /4r /2r 2 1 /4r 2-0 Bz, 1 /2r Bz,r Figure 4: ylinders + and I is well known ha he Harnack inequaliy holds for uniformly parabolic equaions in R n. The relaion o hea kernels is given by he following saemen. Theorem 18 A manifold, μ saisfies he Li-Yau esimae if and only if i saisfies he uniform Harnack inequaliy. To characerize manifolds wih he Harnack inequaliy, we need one more noion. We say ha a weighed manifold saisfies he weak Poincaré inequaliy if here exiss ε 0, 1 such ha for any ball B z, r and for any funcion u 1 B z, r, u s 2 dμ r 2 u 2 dμ. 43 inf s R Bz,εr The erm weak refers here o he facor ε < 1. Bz,r Theorem 19 A manifold, μ saisfies he Harnack inequaliy if and only if i saisfies he doubling volume propery and he Poincaré inequaliy. Hence, he Li-Yau esimae holds if and only if he doubling volume propery and he Poincaré inequaliy hold. Theorem 19 admis a localized version: Harnack inequaliy holds in all balls of radii R if and only if he doubling volume propery and Poincaré inequaliy hold in all balls of radii R. The proof of Theorem 19 uses he following resul, which is of is own ineres. 12

13 Theorem 20 If, μ saisfies he Poincaré inequaliy and he volume doubling propery hen i saisfies he relaive Faber-Krahn inequaliy. Noe ha he converse o Theorem 20 is no rue: i is possible o show ha a conneced sum of wo copies of R n, n 3, saisfies he relaive Faber-Krahn inequaliy bu no he Poincaré inequaliy. Using orollary 16, we obain ha he Poincaré inequaliy and he doubling volume propery imply he upper bound and he on-diagonal lower bound in 42. The off-diagonal lower bound requires addiional ools, which we do no ouch here and which are similar o oser s original proof of he Harnack inequaliy in R n. onnecion o he Ricci curvaure comes from he following saemen. Theorem 21 If has non-negaive Ricci curvaure and μ is he Riemannian volume hen, μ saisfies he Poincaré inequaliy and he volume doubling propery. In fac, boh Poincaré inequaliy and volume doubling propery come from he propery 34 of he homohey on such manifolds. learly, Theorems 21 and 20 imply Theorem 11. Successive applicaion of Theorems 21, 19, and 18 yields he following resul. Theorem 22 If has non-negaive Ricci curvaure and μ is he Riemannian volume hen he hea kernel on, μ saisfies he Li-Yau esimae 42. The above resuls are schemaically presened on he diagram: Homohey propery = Ricci 0 = Gradien esimaes Vol doubling & Poincaré Harnack inequaliy p x, y exp Relaive Faber-Krahn inequaliy Vol doubling & V inequaliy p x, y exp c p x, x V x, Vol doubling c d2 x,y V x, c d2 x,y V x, 8 Esimaes of derivaives of he hea kernel Fix some D 2, + ] and consider again he quaniy d E, x = p 2 2 x, y x, y exp dμy, D as well as E 1, x = d p 2 2 x, y x, y exp dμy D 13

14 and E 2, x = d Δp 2 2 x, y x, y exp dμy D where boh operaors and Δ ac on he variable y. ore generally, se d E n, x = n p 2 2 x, y x, y exp dμy D where { Δ n n/2, n even, = Δ n 1 2, n odd. The quaniy E n can be used o obain poinwise esimaes for he ime derivaive of he hea kernel. The following inequaliy is similar o Lemma 2. Lemma 23 For all x, y and > 0, we have n p x, y n E 2n /2, xe/2, y exp Approach o proof. Using he semigroup ideniy p x, y = p s x, zp s y, zdμz, we obain n n p x, y = = d2 x, y. n n p sx, zp s y, zdμz Δ n z p s x, zp s y, zdμz. Taking s = /2 and using auchy-schwarz inequaliy as in Lemma 2, we finish he proof. This mehod does no work for esimaing x p x, y as we would need z under he inegral. However, if one knows a priori ha x p x, z z p x, z hen one can obain in he same way a poinwise esimae for x p x, y. Our nex purpose is o obain he esimaes for E n, x. Theorem 24 For any x, he funcion E n x, is a finie, coninuous, and decreasing as long as D > 2. Theorem 25 Assume ha, for some x and all 0, T E x, 1 f 14

15 where f is some posiive L 1 loc funcion on 0, T. Define a funcion f n on 0, T by inducion as follows: f 0 = f, f n+1 = 0 f n τ dτ. Then, for all 0, T, where c = D 2 D/2+8. E n x, c n f n In fac, f n can be defined explicily by f n = For example, if f = α hen f n = n α+n. 0 τ n 1 f τ dτ. n 1! orollary 26 If, μ saisfies he relaive Faber-Krahn inequaliy 33 hen for all x, y and > 0 N+n+1 n p 1 + dx,y d x, y2 n x, y V n x, V y, exp, 4 where N is he exponen of he volume comparison condiion