Random walks on ultra-metric spaces

Transcription

1 Random walks on ultra-metric spaces Alexander Grigor yan Nankai University and Bielefeld University Mini-course, September 2018, AMSS CAS, Beijing Based on a series of joint works with A.Bendikov, Eryan Hu, Ch.Pittet, W.Woess

2 Contents 1 Ultra-metric spaces 3 2 Markov operators 10 3 Heat kernel and isotropic Dirichlet form 19 4 Laplacian and Green function 23 5 Isotropic semigroup in Q p 29 6 Ultra-metric product spaces 34 7 Heat kernels on more general spaces 39 8 α-regular ultra-metric spaces 51 1

3 9 Example: jump measure on products Operators of Vladimirov and Taibleson Semi-bounded jump kernels Example: degenerated jump kernel Approach to the proof Lemma of growth Weak Harnack inequality 93 2

4 1 Ultra-metric spaces Let (X, d) be a metric space. The metric d is called ultra-metric if it satisfies the ultra-metric inequality d(x, y) max{d(x, z), d(z, y)}, (1) that is obviously stronger than the usual triangle inequality. In this case (X, d) is called an ultra-metric space. A well-known example of an ultra-metric distance is given by the p-adic norm. Given a prime p, the p-adic norm on Q is defined as follows: if x = p n a, where a, b, n Z and a, b are not divisible by p, then b x p := p n. If x = 0 then x p := 0. The p-adic norm on Q satisfies the ultra-metric ( ) inequality. Indeed, if y = p m c and m n then x + y = d pm p n m a + c. b d Since the denominator bd is not divisible by p, it follows that } x + y p p m = max { x p, y p. 3

5 Hence, Q with the metric x y p is an ultra-metric space, and so is its completion Q p the field of p-adic numbers. The next example of an ultra-metric space is the product Q n p = n times {}}{ Q p... Q p where the ultra-metric is given by the vector p-norm (x 1,..., x n ) p := max i=1,...,n x i p. Various constructions of Markov processes on Q p and on more general locally compact Abelian groups carrying an ultra-metric, were developed by Steven Evans 1989, Albeverio and Karwowski 1991, Kochubei 2001, Del Muto and Figà-Talamanca 2004, Rodríges-Vega and Zúňiga-Galindo 2008 and many others, by means of Fourier transform on such groups. 4

6 Analysis on Q p was developed by Taibleson 1975 and by Vladimirov and Volovich 1989, also using Fourier transform. They have introduced a class of pseudo-differential operators on Q p and on Q n p, in particular, a p-adic Laplacian. Vladimirov et al. studied the corresponding p-adic Schrödinger equation. Among other results, they explicitly computed (as series expansions) certain heat kernels as well as the Green function of the p-adic Laplacian. In this work we define a natural class of random walks on an ultrametric space (X, d) that satisfies in addition the following conditions: it is separable, proper (that is, all balls are compact), and non-compact. Our construction is very easy, takes full advantage of ultra-metric property and uses no Fourier Analysis. In the case of Q p this class of processes coincides with the one constructed by Albeverio and Karwowski, and their generators coincide with the operators of Taibleson and Vladimirov. 5

7 Let us first discuss some properties of ultra-metric balls B r (x) = {y X : d(x, y) r}, where x X and r > 0. The ultra-metric property (1) implies that any two metric balls of the same radius are either disjoint or identical. Indeed, let two balls B r (x) and B r (y) have a non-empty intersection: z B r (x) B r (y). Then d(x, z) r and d(y, z) r whence it follows d(x, y) r. Consider now an arbitrary point z B r (x). We have d(x, z) r and d(x, y) r whence d(y, z) r and z B r (y). Therefore, B r (x) B r (y) and, similarly, B r (y) B r (x) whence B r (x) = B r (y). Consequently, a collection of all distinct balls of the same radius r forms a partition of X, which is a key property for our construction. 6

8 Other properties of ultra-metric spaces: Every point inside a ball is its center. Indeed, if y B r (x) then the balls B r (y) and B r (x) have a nonempty intersection whence B r (x) = B r (y). Consequently, the distance from any point y B r (x) to the complement B r (x) c is larger than r. Every ball is open and closed as a set. Indeed, any ball B r (x) is closed by definition, but it is also open because any y B r (x) has a neighborhood B r (y) B r (x). Consequently, the topological boundary B r (x) is empty. 7

9 Any ultrametric space X is totally disconnected, that is, any nonempty connected subset S of X is an one-point set. Indeed, if S contains two distinct points, say x and y, set r = 1 d(x, y) 2 and observe that S is covered by two disjoint open sets B r (x), B r (x) c both having non-empty intersection with S. Hence, S is disconnected. Consequently, X cannot carry any non-trivial diffusion process. Any two balls B r1 (x) and B r2 (y) of arbitrary radii r 1, r 2 > 0 are either disjoint or one of them contains the other. Indeed, let r 1 r 2. If the balls B r1 (x) and B r2 (y) are not disjoint then also the balls B r1 (x) and B r1 (y) are not disjoint, whence B r1 (x) = B r1 (y) B r2 (y). 8

10 Any triangle {x, y, z} X is isosceles; moreover, the largest two sides of the triangle are equal. Indeed, if d(y, z) is smallest among all three distances then we obtain d(x, y) max (d(x, z), d(y, z)) = d(x, z) and similarly d(x, z) d(x, y) whence d(x, y) = d(x, z). For any x X, a set M = {d(x, y) : y X} has no accumulation point in (0, + ); in particular, M is countable. Let r (0, ) be an accumulation point of M, i.e. {r n } M\{r} such that r n r. Choose y n X such that d(x, y n ) = r n. By compactness of balls, we can assume that {y n } converges, say y n y. Then d(x, y) = r. Since d(y, y n ) 0, it follows that r = r n, which contradicts to the choice of {r n }. For example, in Q p we have M = {p m } m Z. 9

11 2 Markov operators Let μ be a Radon measure with full support on an ultra-metric space X. Then μ (B r (x)) is finite and positive for all x X and r > 0. Let also μ (X) =. Define a family {Q r } r>0 of averaging operators acting on f L (X): Q r f(x) = 1 f dμ. (2) μ (B r (x)) B r (x) Clearly, Q r is a Markov operator. Let σ(r) be a cumulative probability distribution function on (0, ) that is strictly monotone increasing, left-continuous, and σ (0+) = 0, σ ( ) = 1. 10

12 The following convex combination of Q r is also a Markov operator: P f = 0 Q r f dσ(r) It determines a discrete time Markov chain {X n } n N on X with the following transition rule: X n+1 is μ-uniformly distributed in B r (X n ) where the radius r > 0 is chosen at random according to the distribution σ. We refer to P as an isotropic Markov operator associated with (d, μ, σ). 11

13 Example. Consider X = Q p with the p-adic distance d(x, y) = x y p. Every x Q p has a presentation in p-adic numeral system: x =... a k... a 2 a 1 a 0.a 1 a 2... a N = k= N a k p k, where N N and each a k is a p-adic digit: a k {0, 1,..., p 1}. Then x p = p l provided a l 0 and a k = 0 for all k < l. Consider a ball B r (x) of radius r = p m, where m Z. For any y =... b k... b 2 b 1 b 0.b 1 b 2... b N B r (x) we have x y p p m, that is, the first non-zero a k b k occurs for k m; that is, b k = a k for k < m and b k are arbitrary for k m, so that y =... b m+2 b m+1 b m a m 1 a m 2 a m 3... Since b m can take p values, any ball B r (x) of radius r = p m consists of p disjoint balls of radii p (m+1) that are determined by the value of b m. 12

14 Let μ be the Haar measure on Q p with the normalization condition μ (B 1 (x)) = 1. Then we obtain that μ (B p m(x)) = p m. If p m r < p (m 1) then B r (x) = B p m(x) which implies μ (B r (x)) = p m r. The Markov chain {X n } with the transition operator P has the following transition rule from X n to X n+1. One chooses at random r > 0 and, hence, m as above, then changes all the digits a k of X n with k m to b k, where all b k are uniformly and independently distributed in {0, 1,..., p 1}: X n =... a m+2 a m+1 a m a m 1 a m 2 a m 3... X n+1 =... b m+2 b m+1 b m a m 1 a m 2 a m

15 The averaging operator Q r on an ultra-metric space X has some unique features arising from ultra-metric properties. We have 1 Q r f(x) = 1 Br (x)fdμ = q r (x, y)f(y)dμ(y), μ (B r (x)) where the kernel q r (x, y) = X 1 μ (B r (x)) 1 B r (x)(y) = X 1 μ (B r (y)) 1 B r (y)(x) is symmetric in x, y because B r (y) = B r (x) for any y B r (x). As a Markov operator with symmetric kernel, Q r extends to a bounded self-adjoint operator in L 2 (X, μ). Claim. Q r is an orthoprojector in L 2 (X, μ) and spec Q r [0, 1]. Proof. For any ball B of radius r > 0, any point x B is a center of B. The value Q r f(x) is the average of f in B and, hence, is the same for all x B; that is, Q r f = const on B. A second application of Q r to Q r f does not change this constant, whence we obtain Q 2 r = Q r. Therefore, Q r is an orthoprojector. It follows that spec Q r [0, 1]. 14

16 Note that general symmetric Markov operators have spectrum in [ 1, 1] and the negative part of the spectrum may be non-empty. For example, the stochastic symmetric matrix ( 1 ) has eigenvalues 1 and The averaging operator Q r in R n is also Markov and symmetric, but it has a non-empty negative part of the L 2 -spectrum (and, hence, is not an orthoprojector). For example, the averaging operator in R has the Fourier transform Q 1 f(x) = 1 2 Q 1 f (ξ) = 1 3 x+1 x 1 sin 2πξ 2πξ f (t) dt f (ξ) sin 2πξ so that its L 2 -spectrum consists of all values (ξ R) and, hence, 2πξ it has a negative part. In fact, min spec Q

17 Summary of Lecture 1 Let (X, d) be an ultra-metric space, that is, d is a metric that satisfies d (x, y) max (d (x, z), d (y, z)). Then all metric balls B r (x) = {y X : d (x, y) r} (where x X and r > 0) are open and closed. Also, any point in a ball is its center. Assume further that all balls are compact, while X is not. Let μ be a Radon measure on X with full support. Consider for any r > 0 the averaging operator 1 Q r f (x) = f dμ μ (B r (x)) B r (x) that is a Markov operator in L. Then Q r extends to L 2 (X, μ) as an orthoprojector, and Im Q r consists of all L 2 functions that are constant on any ball of radius r. Consequently, spec Q r [0, 1]. It follows also that the family of subspaces {Im Q r } r>0 is monotone decreasing in r in the sense of inclusion. 16

18 It is easy to prove that the family of operators {Q r } r>0 is right continuous in the strong operator topology and that s- lim r 0 Q r = id and s- lim r Q r = 0 (for the last claim we need μ (X) = ). Hence, {Q r } r>0 is a resolution of identity. By a standard convention in spectral theory, a resolution of identity is a family {E λ } λ>0 of orthoprojectors that is monotone increasing in λ, left-continuous and E 0+ = 0, E = id. We obtain such a family by setting E λ = Q 1/λ, but it will be more convenient to work directly with {Q r }. Let σ be a cumulative probability distribution on (0, ). It determines a Markov operator P by P := Q r dσ(r) = σ(r)dq r. 0 0 We obtain in the right hand side explicitly the spectral resolution of P, where the spectral projectors Q r are themselves Markov operators! 17

19 Consequently, P is a self-adjoint operator in L 2 and spec P [0, 1] because 0 σ(r) 1. By the functional calculus, for any ϕ C [0, 1], ϕ (P ) = 0 ϕ (σ (r)) dq r. In particular, the power P t is well-defined for any t > 0 and satisfies P t = 0 σ t (r)dq r = 0 Q r dσ t (r). (3) Setting P 0 = id, we obtain a family {P t } t 0 that is a symmetric strongly continuous Markov semigroup in L 2 (X) (note that P t P s = P t+s!). Definition. The semigroup {P t } t 0 is referred to as an isotropic heat semigroup on X (that is determined by the triple (d, μ, σ)). Let {X t } t 0 be continuous time Markov process with the transition semigroup {P t }. Since the n-step transition operator of the discrete time Markov chain {X n } is P n, we see that the discrete time Markov chain {X n } embeds into the continuous time Markov process {X t }. 18

20 3 Heat kernel and isotropic Dirichlet form Proposition 1 The isotropic heat semigroup P t has a density with respect to μ, that is, P t f(x) = p t (x, y)f(y)dμ(y), where the heat kernel p t (x, y) is a continuous function given by X p t (x, y) = d(x,y) dσ t (r) μ(b r (x)). (4) Proof. Since by (3) and Q r has the kernel P t = q r (x, y) = 0 Q r dσ t (r) 1 μ (B r (x)) 1 B r (x)(y), 19

21 it follows that P t has the kernel p t (x, y) = = 0 d(x,y) q r (x, y) dσ t (r) = dσ t (r) μ (B r (x)). 0 1 μ (B r (x)) 1 B r (x)(y)dσ t (r) As it is well known, any symmetric strongly continuous Markov semigroup in L 2 (X) is associated with a Dirichlet form. In particular, the Dirichlet form (E, F) associated with {P t } is given by 1 E (f, f) = lim t 0 t (f P t f, f) L 2 1 = lim t 0 2t X X (f(x) f(y)) 2 p t (x, y)dμ(x)dμ(y), (5) where the limit always exists in [0, + ], and the domain F consists of functions f L 2 (X) where the limit is finite. 20

22 Proposition 2 The Dirichlet form (E, F) associated with {P t } is a jump type Dirichlet form E(f, f) = 1 (f(x) f(y)) 2 J(x, y)dμ(x)dμ(y) (6) 2 with the jump kernel X X J(x, y) = d(x,y) 1 d ln σ(r). (7) μ (B r (x)) Besides, (E, F) is regular (that is, F C 0 (X) is dense both in C 0 (X) and in F). We refer to this Dirichlet form (E, F) as an isotropic Dirichlet form. Proof. Indeed, comparing (5) and (6), as well as using (4), we obtain 1 J(x, y) = lim t 0 t p 1 tσ t 1 (r)dσ(r) t(x, y) = lim t 0 t d(x,y) μ (B r (x)) = d(x,y) σ 1 dσ(r) μ (B r (x)). 21

23 The regularity of (E, F) follows from the fact that, for any ball B, the indicator function 1 B is continuous in X (because B = ) and 1 B F. Indeed, let B = B ρ (z). For f = 1 B we have by (6) and (7) E(f, f) = J(x, y)dμ(x)dμ(y) B ρ (z) Bρ c(z) 1 = d ln σ(r)dμ(x)dμ(y) {x B ρ (z),y Bρ c(z),r d(x,y) } μ (B r (x)) ( ) dμ(x) = dμ(y)d ln σ(r) ρ y B r (x)\b ρ (z) x B ρ (z) μ (B r (z)) μ (B ρ (z)) = dμ(y)d ln σ(r) μ (B r (z)) = ρ ρ y B r (z)\b ρ (z) μ (B r (z) \ B ρ (z)) μ (B ρ (z)) d ln σ(r) μ (B r (z)) μ (B ρ (z)) ln 1 σ(ρ) <. 22

24 4 Laplacian and Green function Let L be the generator of (E, F) that is a positive definite self-adjoint operator in L 2 (X). We refer to L as an isotropic Laplacian. Since the heat semigroup of (E, F) is given by { e tl}, it follows that t 0 e tl = P t and, hence, L = ln P = 0 ln σ(r)dq r. (8) Denote by C the space of functions f L 2 (X) satisfying the condition that r > 0 such that f const on any ball of radius r. Theorem 3 The space C is dense in L 2 (X), is contained in the domain dom(l) of the Laplacian L, and, for any f C, we have Lf (x) = (f(x) f(y)) J(x, y)dμ(y). (9) X 23

25 The spectrum of L is given by { spec L = ln 1 } σ (r) : r Λ {0}, (10) where Λ = {d(x, y) : x, y X, x y}. Furthermore, L has a complete system of eigenfunctions of the form f = 1 μ(b ) 1 B 1 μ(b) 1 B where B is any ball in X and B is any maximal ball such that B B. The eigenvalue of f is λ = ln 1 where r is the largest radius of B. σ(r) The identity (9) follows from (8) by integration by parts, where one should watch the singularity of ln σ (r) near r = 0. By (8), the spectrum of L is determined by the values of ln σ (r) at those r where dq r does not vanish, which occurs exactly at r Λ. Observe that, for any x B, there exists the maximal ball B containing x and such that B B: in fact, B = B r (x) where r is the largest value in (0, r) of d (x, ). 24

26 The Green function g(x, y) on X X is defined by g(x, y) = 0 p t (x, y)dt. It is known that if g finite (which means g (x, y) < for all x y) then g is in some sense inverse to L: the minimal non-negative solution to Lu = f (where f 0) is given by u(x) = g(x, y)f(y)dμ(y). M Also, it is known that {X t } t 0 is transient of and only if g is finite. Proposition 4 We have g(x, y) = d(x,y) 1 μ(b r (x)) d 1 ln σ(r). (11) 25

27 Proof. Using (4), we obtain g(x, y) = = 0 d(x,y) tσ t 1 (r) μ(b r (x)) dσ(r)dt ( ) d ln σ(r) tσ t (r) dt μ(b r (x)). d(x,y) 0 Since for any a > 0 it follows that and g(x, y) = d(x,y) 0 0 te at dt = 1 a 2, tσ t (r) dt = 1 (ln σ (r)) 2 d ln σ(r) μ(b r (x)) (ln σ (r)) 2 = d(x,y) 1 μ(b r (x)) d 1 ln σ(r). 26

28 Example. Assume that the space (X, d, μ) is α-regular, that is, for all x X and r > 0, μ (B r (x)) r α, where α > 0 (in fact, α has to be the Hausdorff dimension of (X, d)). Choose σ as ( follows: σ (r) = exp ( ) ) c β r where c, β > 0 (Fréchet distribution) By (4) we obtain p t (x, y) = d(x,y) t α/β tσ t (r) d ln σ (r) μ (B r (x)) exp ( cβ d(x,y)/t s 1/β β t d(x,y) ) exp ( tcβ r α β 1 dr r β ) s α β 1 ds 1 t α/β ( 1 + ) (α+β) d(x, y), t 1/β 27

29 so that t p t (x, y). (12) (t 1/β α+β + d(x, y)) The jump kernel admits then the estimate p t (x, y) J (x, y) = lim t 0 t For the Green function, we have g(x, y) = d(x,y) d 1 ln σ(r) μ(b r (x)) d(x,y) d(x, y) (α+β). { dr β, α β r = α d(x, y) (α β), α > β. Recall for comparison that the symmetric stable process in R n of the index β (0, 2) (generated by ( Δ) β/2 ) has the heat kernel while p t (x, y) t (t 1/β + x y ) n+β, J(x, y) = c n,β x y (n+β) and g(x, y) = c n,β x y (n β) (if n > β). 28

30 5 Isotropic semigroup in Q p Set X = Q p with the p-adic distance d(x, y) = x y p and with the Haar measure μ normalized so that μ (B 1 (x)) = 1. We already know that μ (B r (x)) = p n if p n r < p n+1, (13) where n Z. Fix some β > 0 and set ( ( p ) ) β σ(r) = exp. r Knowing μ (B r (x)) exactly enables us to make a precise computation of J(x, y) as follows. By (7) we have J(x, y) = d(x,y) 1 βr β 1 dr d ln σ(r) = pβ μ (B r (x)) x y p μ (B r (x)). 29

31 Let x y p = p k for some k Z. Using (13), we obtain p k β r β 1 dr μ (B r (x)) = n k = n k p n+1 p n p n+1 p n = ( 1 p β) = β r β 1 dr μ (B r (x)) dr β n k p n = n k 1 p β 1. 1 p (1+β) x y 1+β p ( 1 1 p n p 1 ) nβ p (n+1)β 1 p = ( 1 p β) p k(1+β) n(1+β) 1 p (1+β) Hence, we obtain the identity J(x, y) = pβ 1 1. (14) 1 p (1+β) x y 1+β p The jump kernel (14) arises from a completely different point of view. 30

32 As a locally compact abelian group, Q p has the dual group, that is again Q p, which allows to define Fourier transform. The Fourier transform f f of a function f on Q p is defined by f(ξ) = e 2πi{xξ} f(x)dμ(x), Q p where ξ Q p and {xξ} is the fractional part of the p-adic number xθ, that is, {xξ} Q. It is known that f f is a linear isomorphism of the space C 0 of locally constant functions on Q p with compact support. Using the Fourier transform, Vladimirov and Volovich introduced in 1989 the following class of fractional derivatives D β on functions on Q p. Definition. For any β > 0, the operator D β is defined on functions f C 0 by D β f(ξ) = ξ β f(ξ), p ξ Q p. (15) 31

33 They showed that D β can be written as singular integral operator D β f(x) = pβ 1 f(x) f(y) dμ(y). (16) 1 p (1+β) Q p x y 1+β p Comparison with (14) shows that D β coincides ( with the isotropic Laplacian L that corresponds to σ(r) = exp ( ) ) p β. More precisely, we have r D β = L in C 0 so that D β is essentially self-adjoint in L 2 (Q p ). Theorem 5 The operator D β generates a heat semigroup in L 2 (Q p ) that admits a continuous heat kernel p t (x, y) satisfying the estimate p t (x, y) t. (17) (t 1/β + x y p ) 1+β The Green function of D β is finite if and only if β < 1, and in this case it is given by g(x, y) = 1 p β x y (1 β) 1 p (1 β) p. (18) 32

34 Proof. Since D β = L, we can apply all the previous results. The heat kernel estimate (17) follows from (12) because Q p is α-regular with α = 1. The identity (18) for the Green function follows by exact integration in (11) similarly to the computation of J(x, y). Despite the fact that the above statement is a simple consequence of the previous results, we call it Theorem for the following reason: without knowing the theory of isotropic heat semigroup, the question of estimating the heat kernel of D β was very difficult and it remained open for many years. In fact, the full estimate (17) was obtained for the first time in our work in 2014 by using the isotropic Laplacian. In contrast to that, the identity (18) for the Green function was proved by Vladimirov and Volovich directly from (15). 33

35 6 Ultra-metric product spaces Let {(X i, d i )} n i=1 be a finite sequence of ultra-metric spaces. Define their ultra-metric product (X, d) by X = X 1... X n and d(x, y) = max 1 i n d i(x i, y i ). where x = (x 1,..., x n ) X and y = (y 1,... y n ) Y. Then (X, d) is again an ultra-metric space, and balls in X are products of balls in X i : B r (x) = n i=1 B (i) r (x i ). If there is a Radon measure μ i on each (X i, d i ), then we consider on (X, d) the product measure μ = μ i. Given a probability distribution σ on (0, ) as above, we obtain an isotropic semigroup P t on the product space X. 34

36 For example, consider Q n p that is the ultra-metric product of n copies of Q p, with the p-adic metric d (x, y) = x y p = max 1 i n x i y i p. The product of the normalized Haar measures μ on Q p is the normalized Haar measure μ n on Q n p. Hence, if p m r < p (m 1) where m Z then, for any x Q n p, μ n (B r (x)) = n i=1 μ ( B (i) r (x i ) ) = p nm r n. Fix any β > 0 and consider the distribution function ( ( p ) ) β σ(r) = exp. r As in the one-dimensional case, computing J(x, y) from (7), that is, J(x, y) = d(x,y) 1 d ln σ(r), μ (B r (x)) 35

37 and using the exact values of μ (B r (x)), one obtains J(x, y) = pβ 1 x y (n+β) 1 p (n+β) p. (19) Similarly, (11) yields, in the case n > β, that and (4) implies p t (x, y) g(x, y) = 1 p β x y (n β) 1 p (n β) p, t (t 1/β + x y p ) = 1 n+β t n/β ( 1 + x y ) (n+β) p. t 1/β Hence, the jump kernel, Green function and the heat kernel for the isotropic Markov process in Q n p match the same quantities for the symmetric stable process of index β in R n (apart from the values of constants and the range of β because β (0, 2) in R n and β (0, ) in Q n p). 36

38 Summary of Lecture 2 Let (X, d) be a proper, separable ultra-metric space and μ be a Radon measure on X with full support. For any distribution function σ on (0, ), we have constructed an isotropic heat semigroup {P t } t 0 in L 2 (X), and obtained the following identity for its heat kernel: p t (x, y) = d(x,y) dσ t (r) μ (B r (x)). The associated Dirichlet form (E, F) is regular and has the jump kernel J (x, y) = d(x,y) 1 d ln σ (t). μ (B r (x)) Fix α, β > 0, assume that (X, d, μ) α-regular, that is, and choose σ as follows: μ (B r (x)) r α, σ (r) = exp ( ( p ) ) β. r 37

39 We have proved that and p t (x, y) 1 t α/β ( 1 + ) (α+β) d (x, y). (20) t 1/β J (x, y) d (x, y) (α+β). (21) Moreover, in Q n p the isotropic jump kernel satisfies the identity and (20) holds with α = n. J (x, y) = c n,β d (x, y) (n+β) Suppose now that (E, F) is a general (not isotropic) regular Dirichlet form of jump type on the ultra-metric space. How to characterize those Dirichlet forms whose heat kernels satisfy (21)? It is true that if the jump kernel (not isotropic) satisfies (21) then the heat kernel satisfies (20)? Even in Q n p these questions are highly non-trivial. 38

40 7 Heat kernels on more general spaces Let (X, d) be a a separable, proper metric space and μ be a Radon measure on X with full support. Let (E, F) be a regular Dirichlet form in L 2 (X, μ) and {P t } t 0 is the associated heat semigroup. One of the most discussed problems is obtaining estimates of the corresponding heat kernel p t (x, y) (as well as its existence). There are very few situations when the heat kernel can be computed exactly and explicitly. In R n with the Lebesgue measure, the classical Dirichlet form E (f, f) = f 2 dx, R n has the generator L = Δ and the Gauss-Weierstrass heat kernel ( ) 1 x y 2 p t (x, y) = exp, n/2 (4πt) 4t that is the normal distribution at any time t. 39

41 For the symmetric stable process of index 1, generated by Δ, the heat kernel is the Cauchy distribution with the parameter t, that is, with some c n > 0. p t (x, y) = c n t ( t2 + x y 2) n+1, 2 In R n with measure dμ = e x 2 dx, the Dirichlet form E (f, f) = f 2 dμ R n has the generator L = Δ 2x and the Mehler heat kernel ( ) 1 2x y e 2t x 2 y 2 p t (x, y) = exp nt. n/2 (2π sinh 2t) 1 e 4t In the hyperbolic space H 3, the Laplace-Beltrami operator has the heat kernel ) 1 r p t (x, y) = ( (4πt) n/2 sinh r exp r2 4t t where r = d (x, y). 40

42 For many application quantitate properties of the heat kernels are important, so it becomes essential to have at least good estimates. Let us recall some results about heat kernel bounds assuming that the space (X, d, μ) is α-regular, that is, where necessarily α = dim H X. μ (B r (x)) r α, Let first X be a Riemannian manifold with the geodesic distance d and Riemannian measure μ. For the heat kernel of the local Dirichlet form E (f, f) = f 2 dμ the following is known: it satisfies the two-sides Gaussian estimates p t (x, y) c ( ) 1 d 2 (x, y) exp c tα/2 2 t (where c 1, c 2 > 0) if and only if the following Poincaré inequality holds: for any ball B = B r (x 0 ) and any f C 1 (B), ( ) 2 f f dμ Cr 2 f 2 dμ, (22) εb X 41 B

43 where f = fdμ and the constants C and ε (0, 1] are the same for εb all balls and functions. For example, (22) holds in R n and, moreover, on all manifolds of nonnegative Ricci curvature. However, it fails on the following manifold that is a connected sum of two copies of R n, because it has a bottleneck. 42

44 Development of Analysis on fractal spaces has brought into life sub- Gaussian estimates of heat kernels of local Dirichlet forms. This is the estimate of the form p t (x, y) c 1 t α/β exp ( c 2 ( d β (x, y) t ) ) 1 β 1, (23) where β is a new parameter that is called the walk dimension of the corresponding diffusion process. For example, the walk dimension of a diffusion process on a manifold, satisfying the Gaussian estimate, is clearly β = 2. One can show that (23) implies β 2. It was proved in 1990s by M.Barlow, R.Bass et al. that, on a large class of fractals (like unbounded Sierpinski gasket and carpet), there is a diffusion process whose heat kernel satisfies the sub-gaussian estimate (23) with β > 2. 43

45 For example, on the Sierpinski gasket α = ln 3 and ln 2 β = ln , on ln 2 the Sierpinski carpet α = ln 8 and ln 3 β Let us discuss a possibility of the heat kernel estimates (23) on a general metric measure space X. If (23) is true for some diffusion on X then X has to be α-regular and μ has to be comparable to the Hausdorff measure of dimension α. In particular, α = dim H X so that α is an invariant of the metric space (X, d). 44

46 To describe the nature of β, consider for any β > 0 the following quadratic form in L 2 (X, μ): E α,β (f, f) = X X (f (x) f (y)) 2 d (x, y) α+β dμ (x) dμ (y). It was proved by AG, Jiaxin Hu and K.S. Lau in 2003 that the walk dimension β admits the following characterization: β = sup {β > 0 : E α,β extends to a regular Dirichlet form}. (24) Consequently, β is also an invariant of the metric structure (X, d) alone! The identity (24) holds under the hypothesis that a diffusion on X satisfies the sub-gaussian estimate. However, the right hand side makes sense on an arbitrary α-regular metric space, so we can take now (24) as a new definition of the walk dimension β. It is valid for any α-regular metric space independently of heat kernels. It follows from (24) that always β 2 because, for any β < 2, Lipschitz functions with compact support are in the domain of E α,β. 45

47 If X is a Riemannian manifold then one can deduce from (24) that β = 2. On fractals, as we know, typically β > 2. Let us ask what is the walk dimension β of an ultra-metric space? As we know, on an α-regular ultra-metric space, the isotropic Dirichlet form E with the distribution function σ (r) = exp( (c/r) β ) with arbitrary β > 0 has the jump kernel J (x, y) d (x, y) (α+β). Since this jump kernel is comparable with the jump kernel of E α,β, we have E α,β (f, f) E (f, f). Since E is a regular Dirichlet form (Proposition 2), it follows that E α,β is also a regular Dirichlet form for any β > 0, which implies β =! Hence, in the family of all α-regular metric spaces, manifolds and ultrametric spaces are extremal cases: for the manifolds we have β = 2, while for the ultra-metric spaces β =. 46

48 However, these two extremal classes of metric spaces have something in common: both manifolds and ultra-metric spaces possess a priori rich classes of test functions with controlled energy: on manifolds these are usual bump or tent functions, while on ultra-metric spaces these are indicators of balls, as we have seen. The presence of such test functions is very essential for the proofs of heat kernel estimates as all known techniques for obtaining off-diagonal upper bounds make use of such test functions. In the setting of general metric spaces, one has to make an additional assumption about existence of good test functions. 47

49 To conclude the discussion about general metric spaces, let us mention the following result of AG and T.Kumagai 2008: if the heat kernel of a conservative Dirichlet form (E, F) satisfies the estimate of the form p t (x, y) c 1 t α/β Φ ( c 2 d (x, y) t 1/β for some positive α and β then either E is strongly local or Φ (s) (1 + s) (α+β). ) Since on ultra-metric spaces strongly local Dirichlet forms do not exist, we obtain that the only possible estimate of the above type is a stable-like estimate p t (x, y) 1 t α/β ( 1 + ) (α+β) d(x, y). (25) t 1/β Our next purpose is to characterize those ultra-metric space and Dirichlet forms (not necessarily isotropic) when this estimate holds. 48

50 The following necessary conditions for (25) are known: the α-regularity: for any metric ball B r (x), we have μ (B r (x)) r α (V ) (consequently, α = dim H X and μ H α ). the jump kernel estimate: for all x, y X, J(x, y) d(x, y) (α+β). (J) Z.-Q.Chen and T.Kumagai proved in 2003 that, on general metric spaces (with a certain mild restriction on the metric), if 0 < β < 2 then (V )+(J) (25). However, if the walk dimension β of the space in question is larger than 2, then the value of β in (J) can be > 2. In this case, on top of (V ) and (J) we need one more condition that ensures the existence of good test functions. 49

51 Such a condition was established in 2016 independently by Z.-Q. Chen, T. Kumagai, Jian Wang: condition CSJ (cutoff Sobolev inequality for jumps); AG, Jiaxin Hu, Eryan Hu: condition Gcap (generalized capacity condition). A common result of these works: (V ) + (J) + (Gcap) (25). We will show that, in the setting of ultra-metric spaces, the third condition is not needed. 50

52 8 α-regular ultra-metric spaces Let (X, d) be a separable, proper ultra-metric space and let μ be an α-regular Radon measure on X. The results below were proved by A.Bendikov, AG, Eryan Hu Theorem 6 Let J be a symmetric non-negative function on X X such that J(x, y) d(x, y) (α+β) (J) for some β > 0. Then the quadratic form E (f, f) = (f(x) f(y)) 2 J(x, y)dμ(x)dμ(y) X X determines a regular Dirichlet form in L 2 (X, μ). Its heat kernel p t (x, y) exists, is continuous in (t, x, y), Hölder continuous in (x, y) and satisfies the stable-like estimate p t (x, y) 1 t α/β ( 1 + ) (α+β) d(x, y) (26) t 1/β for all x, y X and t > 0. Consequently, (V )+(J) (26). 51

53 Next, let us relax pointwise upper and lower estimates of J(x, y) in (J). We slightly change a setup and assume that we are given a symmetric Radon measure j on X X of the form dj = J(x, dy)dμ(x). Both j and J are referred to as jump measures. Definition. We say that J satisfies the β-poincaré inequality if, for any ball B = B r (x 0 ) and any function f L 2 (B), f f 2 dμ Cr β (f(x) f(y)) 2 J(x, dy)dμ(x) (P I) εb where f = εb B B fdμ and C and ε (0, 1] are constants. Definition. We say that J satisfies the β-tail condition if, for any ball B r (x), B r (x) c J(x, dy) Cr β. (T J) If dj = J(x, y)dμ (x) dμ (y) and X is α-regular then the following implications hold: J(x, y) cd(x, y) (α+β) (P I) J(x, y) cd(x, y) (α+β) (T J) 52

54 Theorem 7 (Main Theorem) Let (X, d, μ) be α-regular ultra-metric space and let J(x, dy) be a jump measure on X X that satisfies (T J). Then the quadratic form E (f, f) = (f(x) f(y)) 2 J(x, dy)dμ(x) X X extends to a regular Dirichlet form (E, F) in L 2 (X, μ). If in addition J satisfies (P I) then the heat kernel p t (x, y) of (E, F) exists, is continuous in (t, x, y), Hölder continuous in (x, y) and satisfies for all x, y X and t > 0 the following weak upper estimate p t (x, y) C t α/β ( 1 + d(x, y) t 1/β ) β, (W UE) and the near-diagonal lower estimate p t (x, y) c provided d(x, y) δt 1/β. (NLE) t α/β Moreover, under the standing assumption (T J), we have (P I) (W UE) + (NLE). (27) 53

55 Equivalence (27) is analogous to the aforementioned result that, on α- regular manifolds, the Poincaré inequality for the Dirichlet integral is equivalent to the two-sided Gaussian estimates of the heat kernel. An analogue of the condition (T J) is in this case the locality of the Dirichlet form. Note that the exponent β in (W UE) does not match the exponent (α + β) in the optimal heat kernel bound (26). There are examples showing that, under (T J) and (P I), one cannot guarantee any estimate of the form with γ > β. p t (x, y) C t α/β ( 1 + ) γ d(x, y) t 1/β In the same way, the lower bound (NLE) cannot be improved to any estimate of the form p t (x, y) c ( ) γ d(x, y) 1 + t α/β t 1/β with any, even very large, γ. 54

56 9 Example: jump measure on products Here we give an example showing that the estimates (W UE) and (NLE) of Theorem 7 cannot be improved assuming only (T J) and (P I). Let {(X i, d i, μ i )} n i=1 be a sequence of ultra-metric measure spaces such that X i is α i -regular, where α 1,..., α n is a prescribed sequence of positive reals. For example, we can take X i = Q p and d i (x, y) = x y 1/α i p. Since Q p with x y p is 1-regular, it follows that (X i, d i ) is α i -regular. Fix β > 0 and consider on each X i the isotropic Dirichlet form (E i, F i ) associated with σ (r) = exp( (c/r) β ), so that its jump kernel J i satisfies and its heat kernel p (i) t J i (x, y) d i (x, y) (α i+β) satisfies p (i) t (x, y) 1 t α i/β ( 1 + d ) (αi +β) i(x, y). (28) t 1/β 55

57 Consider now the product space X = X 1... X n with the ultra-metric d(x, y) = max 1 i n d i(x i, y i ) and the product measure μ = μ 1... μ n. Then X is α-regular with α = α α n. Let L i be the generator of E i. We apply L i to functions f = f (x 1,..., x n ) on X by considering f as a function of x i only (like partial derivatives in R n ). Consider the operator acting on functions on X. L = L L n Proposition 8 The operator L is essentially self-adjoint, it generates a heat semigroup { e tl} in t 0 L2 (X) and its heat kernel satisfies the estimate p t (x, y) 1 t α/β n i=1 ( 1 + d ) (αi +β) i (x i, y i ). (29) t 1/β 56

58 Proof. Main idea: the operators L i commute, whence e tl = e tl 1 e tl 2... e tl n. This implies that e tl has the heat kernel p t (x, y) = n p (i) t (x i, y i ). Substituting the estimates (28) for p (i) t, we obtain (29). Let us verify that p t (x, y) satisfies both (W UE) and (NLE). p t (x, y) C t α/β i=1 Indeed, for any pair x, y, choosing i so that d (x, y) = d (x i, y i ), we obtain ( ) (αi +β) 1 + C t α/β ( 1 + d (x, y) t 1/β d (x, y) t 1/β ) β. If d (x, y) t 1/β then also d i (x i, y i ) t 1/β for all i whence p t (x, y) c t. α/β 57

59 The Dirichlet form (E, F) generated by L has the form n E (f, f) = (Lf, f) L 2 (X) = (L i f, f) L 2 (X) = n... i... X 1 i=1 i=1 X n E i (f, f) dμ 1... i...dμ n where E i (f, f) = X i [f(x 1,..., y i,..., x n ) f(x 1,..., x i,..., x n )] 2 J i (x i, y i )dμ i (x i )dμ i (y i ). It follows that (E, F) is a jump type Dirichlet form with the following jump measure (not jump kernel!) J(x, dy) = n δ x1 (dy 1 )... δ xi 1 (dy i 1 ) J i (x i, y i )dμ i (y i )δ xi+1 (dy i+1 )... δ xn (dy n ), i=1 where δ xk (dy k ) is a unit measure on X k sitting at x k. 58

60 It is easy to check that J satisfies (T J): B r (x) c J (x, dy) = n i=1 B (i) r (x i ) c J i (x i, y i ) dμ i (y i ) Cr β. Since the heat kernel on X satisfies (W UE) and (NLE), we conclude by Theorem 7, that the Poincaré inequality (P I) is also satisfied on X. Consider the range of x, y, t such that Then (29) yields d 1 (x 1, y 1 ) > t 1/β and d i (x i, y i ) t 1/β for i = 2,..., n. p t (x, y) 1 t α/β = 1 t α/β ( 1 + d 1 (x 1, y 1 ) ( 1 + ) (α1 +β) t 1/β ) (α1 +β) d(x, y) t 1/β Since α 1 can be chosen arbitrarily small, we see that (W UE) is optimal.. 59

61 Similarly, consider the range of x, y such that Then d(x, y) d i (x i, y i ) and d i (x i, y i ) d j (x j, y j ) for all i, j. p t (x, y) 1 t α/β 1 t α/β n i=1 ( 1 + ( 1 + d i (x i, y i ) t 1/β ) (αi +β) ) (α+nβ) d(x, y). t 1/β Since n can be chosen arbitrarily large, while α and β are fixed, we see that one cannot ensure any lower bound of the form p t (x, y) c t α/β In this sense, (NLE) is optimal. ( 1 + ) N d(x, y). t 1/β 60

62 10 Operators of Vladimirov and Taibleson Let us slightly modify the above example. For any i = 1,..., n, let X i be Q p with the p-adic metric d i (x, y) = x y p. That is, in the terminology of the previous example, all α i = 1. Fix β > 0 and consider the fractional derivative D β i acting in X i. On the product space Q n p = X 1... X n we have the operator V β = n D β i, i=1 that is called Vladimirov operator. The operator V β was introduced by Vladimirov and Volovich and was considered as a free Hamiltonian in their theory of p-adic Quantum Mechanics. Since D β i coincides with the isotropic Laplacian L i on X i = Q p, we obtain from Proposition 8 the following. 61

63 Corollary 9 The operator V β is essentially self-adjoint, and the heat semigroup exp ( tv β) has the heat kernel p t (x, y) that satisfies for all t > 0 and x, y Q n p the estimate p t (x, y) 1 t n/β n i=1 ( 1 + x ) i y i (1+β) p. t 1/β Corollary 10 If (n 1)/2 < β < n then the Green function of V β exists and satisfies the estimate g(x, y) x y (n β) p. (30) The estimate (30) was known before only for a very special case n = 3, β = 2 and when all the components x i y i are the same. Another natural way of constructing a Markov operator on Q n p is to use the Fourier transform in Q n p that is defined for functions f on Q n p by f (ξ) = e 2πi x,ξ f (x) dμ (x), where ξ Q n p and x, ξ = n k=1 {x kξ k }. Q n p 62

64 Definition. For any β > 0 the Taibleson operator T β is defined on functions f C 0 (Q n p) by T β f(ξ) = ξ β p f(ξ), ξ Q n p. Clearly, in the case n = 1 the operator T β coincides with D β and V β. In the case n > 1, the operators T β and V β are different! In particular, this can be seen from the following result. Theorem 11 The operator T β is essentially self-adjoint, it generates a heat semigroup in L 2 (Q n p) that admits a continuous heat kernel p t (x, y) satisfying the estimate p t (x, y) 1 t n/β ( 1 + x y ) (n+β) p. t 1/β The Green function of T β is finite if and only if β < n, and in this case it satisfies the identity g(x, y) = c n,p x y (n β) p. Proof. Everything follows from the observation that T β coincides on C 0 with the isotropic Laplacian L associated with the distribution function σ(r) = exp( (p/r) β ). For the proof of T β = L, we compare eigenfunctions and eigenvalues of T β and L and show that they coincide. 63

65 11 Semi-bounded jump kernels Let (X, d, μ) be α-regular ultra-metric space and (E, F) be a regular Dirichlet form with a jump kernel J(x, y). Consider two conditions: and J(x, y) Cd(x, y) (α+β) (J ) J(x, y) cd(x, y) (α+β). (J ) Theorem 12 If (J ) and (P I) are satisfied then the heat kernel satisfies for all x, y X and t > 0 the optimal upper bound p t (x, y) C ( ) (α+β) d(x, y) 1 + (UE) t α/β t 1/β and the near-diagonal lower bound p t (x, y) c t α/β provided d(x, y) δt 1/β. (NLE) In fact, we have (J ) + (P I) (UE) + (NLE). 64

66 Theorem 13 If (J ) and (T J) are satisfied then the heat kernel satisfies for all x, y X and t > 0 the optimal lower bound p t (x, y) and the weak upper bound p t (x, y) c t α/β C t α/β ( 1 + ( 1 + ) (α+β) d(x, y) (LE) t 1/β ) β d(x, y). (W UE) t 1/β Moreover, under the standing assumption (T J), we have (J ) (W UE) + (LE). Clearly, Theorems 12 and 13 imply that which is equivalent to Theorem 6. (J ) + (J ) (UE) + (LE), 65

67 12 Example: degenerated jump kernel Here we construct an example of a jump kernel J(x, y) on X = Q p that satisfies (J ) and (P I) but not (J ). In fact, J vanishes on large subsets. Let J be a symmetric kernel on X X and let Φ be an increasing positive function on (0, ). We say that J satisfies Φ-Poincaré inequality if, for any ball B X of radius r and for any f L 2 (B), (f(x) f(y)) 2 dμ(x)dμ(y) Φ (r) (f(x) f(y)) 2 J(x, y)dμ(x)dμ(y). B B B B Lemma 14 The above inequality is equivalent to ( ) 2 Φ(r) f f dμ (f(x) f(y)) 2 J(x, y)dμ(x)dμ(y), (31) B 2μ(B) B B where f = fdμ. B Note that if μ (B) r α and Φ (r) = r α+β then (31) coincides with the β-poincaré inequality. 66

68 Proof. We have (f(x) f(y)) 2 dμ(x)dμ(y) = B B B B = 2μ (B) B ( = 2μ (B) ( f(x) 2 2f(x)f(y) + f(y) 2) dμ(x)dμ(y) ( f 2 dμ 2 B B f 2 dμ f 2 μ (B) ) 2 fdμ ) and ( ) 2 f f dμ = f 2 dμ 2f fdμ + f 2 μ (B) = B B B B f 2 dμ f 2 μ (B). Hence, we obtain (f(x) f(y)) 2 ( ) 2 dμ(x)dμ(y) = 2μ (B) f f dμ, B B B whence the claim follows. 67

69 Set Φ (r) = r α+β with α = 1. We need to construct on Q p a jump kernel that satisfies const Φ-Poincaré inequality, vanishes on large subsets and such that 1 J(x, y) Φ (d(x, y)). For simplicity, we construct J not on Q p but on a discrete subset of Q p. Let M Q p be the set of p-adic fractions.x 1 x 2..., that is, M is the set of sequences x = {x i } i=1, where x i F p and x i = 0 for large enough i. The set M has the additive group structure as follows: x + y = {x i + y i } i=1, where the sum x i + y i is understood in F p. Recall that x p = p n if x n 0 and x i = 0 for all i > n. The distance function on M is d(x, y) = x y p, and balls are defined by B r (x) = {y M : d(x, y) r}. 68

70 Define a function S on M by S (x) = x i F p, i=1 and consider the following subset N of M M: N = {(x, y) M M : S (x) = 0 and S (y) = 1 or S (x) = 1 and S (y) = 0}. Proposition 15 Let p 3. For the jump kernel J(x, y) = 1 N c(x, y) Φ (d(x, y)), the following inequality holds for any ball B of radius r and any function f on B: (f(x) f(y)) 2 5Φ(r) (f(x) f(y)) 2 J(x, y) (32) (x,y) B B (x,y) B B 69

71 Proof. We have (x,y) (B B) N c (f(x) f(y)) 2 (f(x) f(y)) 2 Φ (r) Φ (d(x, y)) (x,y) (B B) N c = Φ (r) (f(x) f(y)) 2 J(x, y) (x,y) B B We will prove that (f (x) f (y)) 2 4 (x,y) (B B) N which will then imply (32). (x,y) (B B) N c (f (x) f (y)) 2, (33) For simplicity, let p = 3. Observe first the following: any two points x, y M form with the point z = (x + y) an equilateral triangle. Indeed, we have z x = 2x y = x y (since 2 = 1 mod 3), whence z x 3 = x y 3 and in the same way z y 3 = x y 3. 70

72 Consequently, if x, y B then also z B since x is a center of B. The second observation is that if (x, y) N then both (x, z) and (y, z) belong to N c. Indeed, by the definition of z we have S (z) = (S (x) + S (y)). Since (x, y) N, we have S (x) + S (y) = 1 whence S (z) = 1 = 2. Consequently, any pair (, z) belongs to N c. Combining the above observations, we conclude that if (x, y) (B B) N then (x, z) (B B) N c, and the same is true for (y, z). Next, we have (f (x) f (y)) 2 2 (f (x) f (z)) (f (y) f (z)) 2 71

73 and (f (x) f (y)) 2 2 (f (x) f (z)) 2 (x,y) (B B) N +2 (x,y) (B B) N (x,y) (B B) N (f (y) f (z)) 2. Observe that the mapping (x, y) (x, z) = (x, (x + y)), is injective because the pair (x, z) allows to recover the pair (x, y) uniquely by y = (x + z). Therefore, (f (x) f (z)) 2 (f (x) f (z)) 2, (x,y) (B B) N (x,z) (B B) N c The same applies to the sum of (f (y) f (z)) 2, and we obtain (f (x) f (y)) 2 4 (f (x) f (z)) 2, (x,y) (B B) N (x,z) (B B) N c thus proving (33). 72

74 13 Approach to the proof We outline most essential parts of the proofs of Theorems 6, 7, 12, 13. Let (X, d, μ) be an α-regular ultra-metric space and (E, F) be a jump type Dirichlet form with the jump kernel J(x, y). We write dj = J(x, y)dμ(x)dμ (y) = J(x, dy)dμ(x). Assuming that J satisfies the β-tail condition and the β-poincaré inequality f f 2 dμ Cr β B r we need to prove the weak upper estimate ( p t (x, y) 1 + B r (x) c J(x, dy) Cr β (T J) B r B r (f(x) f(y)) 2 J(x, dy)dμ(x), (P I) C t α/β 73 d(x, y) t 1/β ) β (W UE)

75 and the near-diagonal lower estimate p t (x, y) for some δ > 0. If in addition J satisfies c t α/β provided d (x, y) δt 1/β, (NLE) J (x, y) Cd (x, y) (α+β) (J ) then heat kernel should satisfy the optimal upper estimate p t (x, y) C ( ) (α+β) d(x, y) 1 +, (UE) t α/β t 1/β and if in addition J (x, y) cd (x, y) (α+β) (J ) then heat kernel should satisfy the optimal lower estimate p t (x, y) c ( ) (α+β) d(x, y) 1 +. (LE) t α/β t 1/β There are also issues with the existence of the heat kernel and its Hölder contnuity, as well as the opposite implications. 74

76 The proof is very long and consists of many steps. We outline the structure of the proof and some most essential moments. Overall, the proof uses the same techniques as in general metric spaces but the presense of an ultra-metric brings some simplifications. For any open set Ω X, consider the function space F (Ω) that is the closure of F C 0 (Ω) in F. Then (E, F (Ω)) is a regular Dirichlet form in L 2 (Ω) that corresponds to a Markov process killed outside Ω. It is important, that in ultra-metric space satisfying (T J), for any ball B = B r (x), 1 B F (B) because 1 B C 0 (B) and E (1 B, 1 B ) Cμ (B) r β. Denote by P Ω t the heat semigroup of (E, F(Ω)) and by G Ω = the Green operator. It is known that P Ω t and G Ω are increasing in Ω. 0 We say that a function u F is superharmonic in Ω if E (u, ϕ) 0 for any non-negative ϕ F (Ω). A function u is subharmonic if u is superharmonic. Finally, u is harmonic if u is super- and subharonic. 75 P Ω t dt

77 Step 1. (P I) implies the Nash inequality: for any f F L 1 (X), f 2(1+ν) L 2 CE(f, f) f 2ν L1, (34) where ν = β/α. The latter implies the existence of the heat kernel and the diagonal upper estimate, for all t > 0 and almost all x, y X, p t (x, y) Ct α/β. (DUE) One of the consequences of (DUE) is the following estimate of the meat exit time from balls: for any ball B of radius r, G B 1 Cr β. (35) In the case α > β it is simple (while the case α β requires more care): G B 1 G1 B = 0 P t 1 B dt r β P t 1 dt + p t (x, y) dμ (y) dt 0 r β B ( ) r β + C t α/β dt dμ r β + Cr ( α r β) 1 α/β = Cr β. B r β 76

78 One more consequence of the Nash inequality (34) is the Faber-Krahn inequality: for any measurable set E X of finite measure and any f F such that f = 0 a.e. outside E, we have Indeed, by Cauchy-Schwarz inequality, E (f, f) cμ (E) ν f 2 L 2. (36) so by (34) E (f, f) c f 2(1+ν) L 2 f 2 L 1 μ (E) f 2 L 2 f 2ν L 1 c f 2 L 2 μ (E) ν. Step 2. This is the largest and most technical part of the proof. One obtains a weak Harnack inequality for harmonic functions of (E, F), where the main ingrdient of the proof is Lemma of growth. We give some details below. The weak Harnack inequality implies an oscillation inequality for harmonic functions and, consequently, the Hölder continuity of harmonic functions. 77

79 The mean exit time estimate (35) implies G B f L Cr β f L, which allows to extends oscillation inequality to solutions u of Lu = f with bounded functions f. Considering a function u (t, ) = P t ϕ as solution to Lu = t u and estimating t u L by means of (DUE), we obtain the oscillation inequality and the Höder continuity for P t f and, hence, also for the heat kernel. Step 3. Here one obtains the lower bound for mean exit time: G B 1 cr β in B (37) that is, in fact, a consequence of the Lemma of growth. The function u = G B 1 is superharmonic in B; hence, by a corollary of a Lemma of growth, it satisfies inf B ( u c B ) 1 1 u dμ. On the other hand, using φ = 1 B F (B), we obtain 1 φ2 dμ = (φ, u u ) = E(GB φ, φ2 φ2 ) = E(u, u u ). B 78

80 Next one uses the following general inequality (Lemma 19 below): E(u, φ2 ) 3E (φ, φ). u Since by (T J) E (φ, φ) Cr α β, we obtain 1 u dμ Cr β, whence (37) follows. The estimates (35) and (37) yield B G B 1 r β in B. This implies the following survival estimate: P B t 1 ε in B, provided t 1/β δr, (S) with some ε, δ > 0. Indeed, (S) follows from a general inequality Pt B 1 GB 1 t. G B 1 L 79

81 Step 4. Here we prove (NLE). For any ball B = B r (x), assuming t 1/β δr, we have, using the semigroup identity and (S), p 2t (x, x) = p t (x, y) 2 dμ (y) p t (x, y) 2 dμ (x) X 1 μ (B) ( Choosing r = δ 1 r 1/β, we obtain B B p t (x, y) dμ (x)) 2 p t (x, x) ct α/β. By the oscillation inequality from the second step, ( P B t 1 ) 2 μ (B) p t (x, x) p t (x, y) Ct α/β ( d(x, y) t 1/β Hencee, if d (x, y) δt 1/β with small enough δ, then whence (N LE) follows. p t (x, x) p t (x, y) c 2 t a/β, 80 ) θ. ε2 μ (B) r α.

82 Step 5. Here we prove (W UE). The main difficulty is in obtaining the following estimate: for any ball B of radius r and any t > 0, P t 1 B c C t r β. (T P ) If this is already known then we have, by setting r = d (x, y) /2, p 2t (x, y) = p t (x, z) p t (z, y) dμ (z) X ( ) p t (x, z) p t (z, y) dμ (z) B r (x) c + B r (y) c (sup p t ) P t 1 Br (x) c + (sup p t ) P t 1 Br (y) c Ct α/β t r. β Since by (DUE) also p t (x, y) Ct α/β, it follows ( p 2t (x, y) Ct α/β min 1, t r β 81 ) t α/β ( 1 + r t 1/β ) β.

83 However, the main difficulty here lies in proving (T P ) which itself a multi-step procedure that is based on reiterating of the suvival estimate (S). Indeed, (S) implies, for t 1/β δr, that P t 1 B c 1 P t 1 B 1 P B t 1 1 ε, which gives (T P ) provided t 1/β = δr. A certain bootstrapping argument allows to extend this to all t. Step 6. In the case when J satisfies (J ), one can extend the argument of Step 5 to prove the optimal upper estimate (UE), which requires additional techniques. One uses the truncated jump kernel the heat kernel q (ρ) t estimate J (ρ) = min (J, ρ), (x, y) associated with J (ρ), and the following general p t (x, y) q (ρ) t (x, y) + 2t sup J (x, y ). {x,y X:d(x,y ) ρ} 82

84 For the truncated heat kernel one obtains the estimate ( (x, y) Ct α/β exp 4ρ β t c min( q (ρ) t which together with (J ) allows to obtain (UE). d (x, y), ρ ) ρ t ), 1/β Step 7. In the case when J satisfies (J ), one uses the following general result: assuming that conditions (S) and (N LE) are satisfied, the following estimate holds for all t > 0, x, y X: p t (x, y) c t α/β min(1, {tμ (B t 1/β (y)) essinf x B t 1/β (x) y B t 1.β (y) J(x, y )}). (38) Hence, if r := d (x, y) δt 1/β then d (x, y ) Cr and, hence, J (x, y ) cr (α+β) which implies p t (x, y) min (1, t1+α/β α/β r α+β c t ) 1 t α/β ( 1 + r t 1/β ) (α+β). 83