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Dirichlet forms and associated heatkernels on the Cantor set induced byrandom walks on trees

Jun KigamiGraduate School of Informatics

Kyoto UniversityKyoto 606-8501, Japan

e-mail:kigami@i.kyoto-u.ac.jpAbstract

Transient random walk on a tree induces a Dirichlet form on its Martinboundary, which is the Cantor set. The procedure of the inducementis analogous to that of the Douglas integral on S 1 associated with theBrownian motion on the unit disk. In this paper, those Dirichlet formson the Cantor set induced by random walks on trees are investigated.Explicit expressions of the hitting distribution (harmonic measure) andthe induced Dirichlet form on the Cantor set are given in terms of theeffective resistances. An intrinsic metric on the Cantor set associated withthe random walk is constructed. Under the volume doubling property of with respect to the intrinsic metric, asymptotic behaviors of the heatkernel, the jump kernel and moments of displacements of the processassociated with the induced Dirichlet form are obtained. Furthermore,relation to the noncommutative Riemannian geometry is discussed.

1 Introduction

Transient random walks eventually go to innity, which is not just a singlepoint but a collection of possible behaviors of random walks as the time tends tothe innity. A rigorous way to describe this innity is the Martin boundary,where all the boundary values of harmonic functions lie. In certain cases, atransient random walk naturally induces a (Hunt) process (or equivalently aDirichlet form) on its Martin boundary. In this paper, we are going to studysuch an induced (Hunt) process in the case of a random walk on a tree, whoseMartin boundary is known to coincide with the Cantor set.

A well-known example of such an induced Dirichlet form is the Douglasintegral

D (, ) = 4 20 20 (() ( ))( () ( ))sin2 ( 2 ) (d) (d )

on the circle S 1 = {z|z = ei , R }, where (d) = d/ 2 is the uniformdistribution on S 1 . In this case the process which induces the Douglas integralon its Martin boundary S 1 is (reected) Brownian motion on the unit disk

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D = {z R , |z| < 1}, which is not even a random walk. We use, however, thisexample to illustrate the procedure of inducement of Dirichlet forms on Martinboundaries. The Dirichlet from associated with the Brownian motion on D isthe (half of) Dirichlet integral

E (u, v) = 12 D ux vx + uy vy dxdy

dened for u, v H 1(D ) = {u|u, ux , uy L2(D )}, where D is identied with{(x, y)|x, y R , x2 + y2 < 1}and the derivatives are in the sense of distributions.Note that the Brownian motion starting from 0 D will hit the boundary S 1uniformly due to the symmetry. In this case, the uniform distribution on S 1is called the hitting distribution (or harmonic measure) of S 1 starting from 0.The key stone of the bridge between the Dirichlet integral on D and the Douglas

integral on S 1 is the Poisson integral which gives harmonic functions on D fromboundary values on S 1 . Let H be the operation of applying the Poisson integral.Namely, we dene H () by

(H ())( re i ) = 20 1 r 21 2r cos( ) + r 2 ( ) (d ).Then

Dom( D ) = {|L2(S 1 , ), HH 1(D )} and D(, ) = E (H ,H )for any , Dom( D).The above example shows us the essence of obtaining Dirichlet forms onMartin boundaries from random walks. Suppose that we have a Markov processor a random walk on a space X , that its Martin boundary M is well-denedand that we have the following ingredients (a), (b) and (c):(a) the Dirichlet form E associated with the original process or the randomwalk(b) the hitting distribution of M starting from certain point in X (c) the map H which transforms functions on M into harmonic functions onX .Then the induced form E M on the Martin boundary M is given by

E M (, ) = E (H ,H )for ,

F M , where

F M =

{

|

L2(M, ), H

Dom(

E )

}.

The theory of the Martin boundary for a transient random walk originatedwith the classical works of Doob [4] and Hunt [6]. Since then, it has beendeveloped by many authors and, as a result, all the necessary ingredients forconstructing an induced form ( E M , F M ) are already well-established. For ex-ample, one can nd them in [16].

In this paper, we are going to focus on the case of a transient random walkon a tree and study the induced form ( E M , F M ) on its Martin boundary. The

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original motivation of this work comes from the study of traces of the Brownian

motion on the Sierpinski gasket. By removing the line segment on the bottomof the Sierpinski gasket, one can associate a random walk on a binary tree withthe Brownian motion. We will present the details on this idea in a separatepaper[10].

Due to Cartier[2], the Martin boundary of a transient random walk on a treeis known to be (homeomorphic to) the Cantor set = {1, 2}N = {i1 i2 . . . |i j {1, 2} for j N }. Hereafter we use to denote the Martin boundary in placeof M . Let us x what we mean by a random walk and a tree at this point.Random walk : Let (V, E ) be a non-directed graph, i.e. V is the set of vertices and E = {(x, y)|x= y T, x and y are connected by an edge } is theset of edges. We assume that {y|(x, y ) E } is a nite set for any x V . Foreach (x, y ) E , we assign a conductance C (x, y) > 0 which satises C (x, y ) =C (y, x ). Dene p(x, y ) = C (x, z )/

(x,z )E C (x, y ). p(x, y ) gives the probability

of transition from x to y in the unit time.Tree : A non-directed graph ( T, E ) is a tree if and only if it is connected andthere exists no non-trivial cyclic path. We always x a reference point T and assign conductances on ( T, E ) as described above.

In this framework, assuming the transience of the random walk, we aregoing to obtain explicit expressions of the hitting distribution starting fromthe reference point T and the induced form ( E , F ) in terms of effectiveresistances, prove that ( E , F ) is a regular Dirichlet form, and construct anintrinsic metric on the Martin boundary with respect to the random walk.Then, assuming the volume doubling property of with respect to the intrinsicmetric, we are going to determine asymptotic behaviors of the heat kernel, the jump kernel and moments of displacements of the process generated by the

regular Dirichlet form ( E , F ).All the results in this paper are obtained in the generality of the aboveframework. In this introduction, however, we are going to present statementsin the case of the binary tree for the sake of simplicity. Let T = m 0{1, 2}m ,where {1, 2}0 = {} and {1, 2}m = {w1 . . . wm |w1 , . . . , w m {1, 2}} and letE = {(w,wi ), (wi,w)|w T, i {1, 2}}. (T, E ) is called the (complete innite)binary tree. See Figure 1. For w T , we set rwi = C (w,wi ) 1 for i = 1 , 2,T w = {wv|v T } and w = {wi1 i2 . . . |i j {1, 2} for any j N }. T w is asubtree of T = T and w is a subset of the Martin boundary = . Thequadratic form E w associated with the subtree T w is given by

E w (f, f ) =vT w i=1 ,2

1r vi

(f (v) f (vi)) 2 .

The effective resistance between w and w with respect to the subtree T w isgiven by

Rw = (min{E w (f, f )|f (w) = 1 , the support of f is a nite set} 1 .Under these notations, assuming the transiency of the random walk, we haveobtained the following results.

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1 2

11 12 21 22

The binary tree T

w

w = wi

w2w1

r w = 1C (w ,w )

r w 1 rw 2

Local structure of T

Figure 1: Random walk on the binary tree T

Explicit expression of the hitting distribution (Section 3): The hittingdistribution is characterized by () = 1 and

( wi ) = Rw

r wi + Rwi ( w )

for any wT and any i {1, 2}.Explicit expression of (E , F ) (Section 4): For w T , let (u)w be theaverage of u on w with respect to , i.e. (u)w = w u() (d)/ ( w ). ThenE

(, ) =wT (

()w 1 ()w 2

(()w 1 ()w 2

r w 1 + Rw 1 + r w 2 + Rw 2(1.1)

for any , F = {|E (,) < +}. In particular, ( E , F ) is shown tobe a regular Dirichlet form on L2( , ).Wavelet base consisting of eigenfunctions (Section 4): Dene

w = ( w 2) w 1 ( w 1) w 2 ( w 1)2 ( w 2) + ( w 2)2 ( w 1)

for any w T , where A is the characteristic function of A . Then{1,w |w T } is a complete orthonormal system of L2( , ) consisting of theeigenfunctions of the non-negative self-adjoint operator L associated with theDirichlet form (

E ,

F ). Furthermore, dene Dw = Rw ( w ) for any w

T .

Then L w = ( D w ) 1w for any wT . Such a wavelet base on the Cantorset consisting of eigenfunctions has been observed in Kozyrev[11] and Pearson-Bellissard[13]. In fact, the example in [13] is shown to be a special case of ourframework in Section 12. Also it is noteworthy that Dw is expressed as theGromov product of certain metric related to effective resistances. See (4.1).Construction of intrinsic metric (Section 5): For , with = ,the conuence [ , ] T is 1 . . . m , where = 12 . . . , = 1 2 . . . ,

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m = min {k|k = k} 1. Dene d(, ) = D[, ]. Then d(, ) is a metric on and it is thought of as the intrinsic metric with respect to the random walk.Moreover, we show that has the volume doubling property with respect to d,i.e. (B (, 2r )) c (B (, r )) for any and any r > 0, where c is indepen-dent of and r and B(, r ) = { | , d(, ) < r }, if and only if there existc > 0 and (0, 1) such that ( wi ) ( w ) and Dwi 1 ...i m cm Dw forany wT and any i, i1 , . . . , i m {1, 2}.Asymptotic behavior of the heat kernel (Section 6): Assume that hasthe volume doubling property with respect to d. Dene

p(t ,, ) =n 0

e [ ]n 1 t e [ ]n t ( []n )

[ ]n ( )

for any t > 0 and any ,

, where []n = 1 . . . n for = 12 . . .

. Then p(t ,, ) is continuous on (0 , + ] and is the heat kernel(fundamental solution) of the heat equation ut = L u. Namely,(e L t u0)() = p(t ,, )u0( ) (d )

for any u0 L2( , ). Moreover, p(t ,, )min

td(, ) ( [, ])

, 1

(B (, t)) (1.2)

for any t (0, 1] and any , , where the notation is dened at theend of introduction. This heat kernel estimate is a variant of that studied inChen-Kumagai[3] and satises a typical asymptotic behavior of jump processes.From (1.2), we also have asymptotic behaviors of moments of displacement:

E (d(, X t ) )

t if > 1,t(| log t|+ 1) if = 1,t if 0 < < 1

(1.3)

for any and any t (0, 1], where E () is the expectation with respectto the Hunt process associated with the Dirichlet form ( E , F ) on L2( , ). InSection 14, (1.3) is shown to hold in general for jump processes which satisfythe heat kernel estimate (1.2).Generalization and inverse problem (Sections 9 and 10): We investigate

the following class of quadratic forms Qs on L2( , ) given by

Q(, ) =wT

aw ((),w 1 (),w 2)(( ),w 1 (),w 2), (1.4)

where aw > 0 for any w T , is a Borel regular probability measure on and ( u),v = ( v ) 1 v u()(d). We show an equivalent condition forQ being a regular Dirichlet form. Moreover, we consider when a quadratic5

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form Q given by (1.4) is induced by a transient random walk on T . Let w =aw ( w )/ (( w 1)( w 2)) for w T . Dene two conditions (A) and (B) asfollows:(A) w < wi for any (w, i ) T {1, 2} and limm []m = + for any (B) w < wi for any (w, i ) T {1, 2} and lim m []m = + for -a.e. .Then we prove

(A) Q is induced by a random walk on T (B).Furthermore, we show that neither (A) nor (B) is equivalent to Qbeing inducedby a random walk on T in Section 11.Relation to noncommutative Riemannian geometry (Section 12): In [13],

Pearson and Bellissard have constructed noncommutative Riemannian geometryon the Cantor set including Laplacians and Dirichlet forms from an ultra-metric.In fact, their Dirichlet forms belong to the class described by (1.4). As anapplication of the inverse problem, we prove that Dirichlet forms derived fromself-similar ultra-metrics on the binary tree are induced by random walks andobtain heat kernel estimate (1.2) and moments of displacement (1.3). Thisextends the result of [13] where the authors have studied self-similar ultra-metrics whose similarity ratios are equal.

Behind all those results, the theory of resistance forms, which is brieyreviewed in Section 13, plays an important role. For example, if a random walkon a graph ( V, E ) is transient, we may add the innity I to the vertices V and consider a natural resistance form on V {I } associated with the randomwalk on V . See Section 2 for details. By using this fact, the Martin kernelis expressed by the resistance metric (effective resistance) associated with theresistance form on V {I }. This idea is essentially the key of obtaining theresults of this paper.

Finally we introduce several conventions in the notation in this paper.(1) Let f and g be real valued functions on a set A. We write f (x)g(x) onA if and only if there exist c1 , c2 > 0 such that c1g(x) f (x) c2g(x) for anyxA.(2) Let X be a set. We dene (X ) = {u|u : X R }.

2 Weighted graphs as random walks

Although our main subject is a random walk on a tree, we introduce and studymore general framework of weighted graphs in this section. The most importantresult is (2.3), where the Martin kernel is expressed by the resistance metric.This relation (2.3) is the foundation of all the theories in the following sections.Except that, most of the denitions and results are classical and can be foundin [16] for example.

First we dene weighted graphs and associated notions.

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Denition 2.1. (1) A pair ( V, C ) is called a weighted graph if and only if V

is a countable set and C : V V [0, ) satises that C (x, y ) = C (y, x ) forany x, y V and that C (x, x ) = 0 for any x V . The points in V are calledvertices of the graph ( V, C ). Two vertices x, y V are said to be connected if and only if C (x, y ) > 0. The neighborhood N x (V, C ) of x V is dened byN x (V, C ) = {y|C (x, y ) > 0}. For x0 , x1 , . . . , x n V , p = ( x0 , x1 , . . . , x n ) iscalled a path if and only if C (x i , x i+1 ) > 0 for any i = 0 , . . . , n 1. We denethe length | p| of a path p = ( x0 , x1 , . . . , x n ) by n . If x0 = x and xn = y, then apath ( x0 , . . . , x n ) is called a path between x and y. A path p = ( x0 , . . . , x n ) iscalled simple if and only if x i= xj for any i = j .(2) A weighted graph ( V, C ) is called irreducible if and only if a path betweenx and y exists for any x, y V .(3) A weighted graph ( V, C ) is called locally nite if and only if N x (V, C ) is anite set for any x

V .

In this paper, we always assume that V is an innite set and a weightedgraph ( V, C ) is irreducible and locally nite. A weighted graph gives a reversibleMarkov chain on V .

Denition 2.2. Let C (x) = yV C (x, y) and let p(x, y ) = C (x, y )/C (x). Forn = 0, 1, 2, . . . , we dene p(n ) (x, y ) for x, y V inductively by p(0) (x, y) = xy ,where xy is the Diracs delta, and p(n +1) (x, y ) =

zV

p(n ) (x, z ) p(z, y).

Dene G(x, y ) =

n 0 p

(n ) (x, y ), which may be innite. G(x, y ) is called the

Green function of ( V, C ). A weighted graph ( V, C ) is said to be transient if andif G(x, y) < + for any x, y V .

We regard p(x, y ) as the transition probability from x to y in the unit time.Let ({Z n }n 0 , {Qx}xV ) be the associated random walk (or the Markov chain)on V . Then p(n ) (x, y ) = Qx (Z n = y), i.e. p(n ) (x, y ) is the probability of thetransition from x to y at time n . Since p(x, y )C (x) = p(y, x )C (y), this Markovchain is reversible. One can easily see that ( V, C ) is transient if G(x0 , y0) < + for a single pair of points ( x0 , y0).

Next we dene the notions of Laplacian, harmonic functions and resistanceform associated with a (locally nite irreducible) weighted graph ( V, C ).

Denition 2.3. (1) The Laplacian L : (V ) (V ) associated with ( V, C ) isdened by(Lu )(x) =

yV

C (x, y )(u(y) u(x))

for any u (V ) and any x X . We say u (V ) is harmonic on V withrespect to ( V, C ) if (Lu )(x) = 0 for any xV . We use H(V, C ), H+ (V, C ) andH (V, C ) to denote the collection of harmonic functions, non-negative harmonic

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functions, bounded harmonic functions on V with respect to ( V, C ) respectively.

(2) Dene

F = {u|u : V R ,x,y V

C (x, y )(u(x) u(y)) 2 < + } and

E (u, v ) = 12

x,y V

C (x, y )(u(x) u(y))( v(x) v(y))

for any u, v F . For a reference point V , we deneE (u, v) = E (u, v) + u()v()

for any u, v F . Dene C 0(V ) = {u|u (V ), #(supp( u)) < +}, wheresupp( u) is the support of u and #( A) is the number of elements in A.Proposition 2.4. (1) (E , F ) is a resistance form on V .(2) If uC 0(V ) and v F , then

E (u, v ) = xV

u(x)(Lv)(x). (2.1)

See Section 13 for the denition of resistance forms.

Denition 2.5. (E , F ) dened in the last proposition is called the resistanceform associated with the weighted graph ( V, C ).The following theorem is one of the most essential results on the type prob-

lem of random walks. It has originated with Yamasaki in [17, 18]. See [15,

Theorem 4.8] for details.Theorem 2.6. The following conditions are equivalent:(Tr1) ( V, C ) is transient.(Tr2) 1 /(C 0(V ))E , where (C 0(V ))E is the closure of C 0(V ) with respect to

E .(Tr3) ( C 0(V )) E = F .(Tr4) sup {u(x)2 / E (u, u )|uC 0(V )}< + for any xV .By the above theorem, if ( V, C ) is transient, we may introduce a point I ,

which is thought of as the point of innity, and construct a new resistance formon V {I }. Later in (2.3), the Martin kernel of ( V, C ) is shown to be describedby the resistance metric associated with this new resistance form on V {I }.Proposition 2.7. Assume that (V, C ) is transient. Dene F = ( C 0(V ))E +R = {f + a|f (C 0(V )) E , aR }. Let I /V and dene u(I ) = a if u = f + a for f (C 0(V )) E and aR . Denote V = V {I }. Then (1) (E , F ) is a resistance form on V .(2) There exists g : V V [0, ) such that the following conditions (a) ,(b) and (c) are satised:(a) g(x, y ) = g(y, x ) for any x, y V .

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(b) For any x V , let gx(y) = g(x, y ). Then gx is a unique element in (C 0(V )) E which satises E (g

x, u) = u(x) for any u(C

0(V )) E .(c) For any x, y V ,g(x, y ) =

R(x, I ) + R(y, I ) R(x, y )2

, (2.2)

where R(, ) is the resistance metric on V associated with (E , F ) given by

R(x, y ) = max(u(x) u(y))2

E (u, u )uF , E (u, u ) = 0 .

Note that the right-hand side of (2.2) is the Gromov product associated withthe metric space ( V {I }, R).Remark. There is no weighted graph ( V , C ) whose associated resistance formis (E , F ). Otherwise we have F , where is the characteristic functionof the innity I . Then there exists u (C 0(V ))E such that = u + 1.Since u 1 on V and (T, C ) is transient, this contradicts to the fact that1 /(C 0(V )) E .Remark. Let R(, ) be the resistance metric associated with the resistance form(E , F ) on V . Then R(, ) and R(, ) are called the limit resistance and theminimal resistance respectively in [14]. Also, they are called the free resistanceand the wired resistance respectively in [7]. For the use of the terminology freeand wired, see [12].

Proof. (1) We will verify the conditions (RFi) for i = 1 , . . . , 5 in Denition 13.4.

(RF1) and (RF2) are immediate. Let x be the characteristic function of a singlepoint xV . Since x C 0(V ) for any xV , we have (RF3). For x = y V ,since F F , the supremum in (RF4) with F = F is nite. If x V andy = I , then the supremum is also nite by the condition (Tr4) of Theorem 2.6.Finally, let uF . Then u = f + a for some f (C 0(V )) E and aR . Thenf + a a(C 0(V )) E . Hence u(I ) = ( f + a)( I ) = a = u(I ). This shows (RF5)for (E , F ). Thus ( E , F ) is a resistance form on V .Applying Theorem 13.2 to the resistance form ( E , F ) on V with B = {I },we obtain the statements (2) and (3). Note that in this case, B is a single point.The remark after Theorem 13.2 shows that R B = R.

The element I is considered as the point of innity.

Denition 2.8. Assume that ( V, C ) is transient. When it is necessary to specify(V, C ), we use I (V,C ) to denote the innity I associated with ( E , F ) and writeRx (V, C ) instead of R(x, I ), which is the resistance metric between x and I (V,C )with respect to ( E , F ). If (V, C ) is not transient, then we dene Rx (V, C ) = .

According to the use of terminology in Section 13, g(, ) is the I -Greenfunction associated with ( E , F ). In light of the following theorem, we callg(, ) the symmetrized Green function of the weighted graph ( V, C ).

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I

x y

r xI

r xy

r yI

Trace to {x,y,I }

I

x y

z

R I = g (x, y )

R x Ry

After the Y transform

Figure 2: Calculation of g(x, y) by means of Y transform

Proof of Theorem 2.9. By Lemma 2.10-(4), it follows that Gy(C 0(V )) E and

E (u, G y /C (y)) = u(y) for any u (C 0(V )) E . Proposition 2.7-(b) shows thatG(x, y )/C (y) = Gy (x)/C (y) = g(x, y ).The next lemma is technically useful in the following sections. One can refer

to Theorem 13.3 for the denition of a trace of a resistance form.

Lemma 2.11. Assume that (V, C ) is transient. Set U = {x,y,I } for x, y X with x= y. Let (E|U , F |U ) be the trace of the resistance form (E , F ) on U .Suppose that

E|U (u, u ) = cxI (u(x)

u(I )) 2 + cyI (u(y)

u(I ))2 + cxy (u(x)

u(y))2

for any u : U R . Dene rxI = 1/c xI , r yI = 1 /c yI and rxy = 1 /c xy . Then

g(x, y ) =

r xI r yIr xI + r yI + r xy if cxI cyI cxy > 0,

0 if cxy = 0 ,r yI if cxI = 0 ,r xI if cyI = 0 .

This lemma is proven by using the Y transform to E|U as illustrated inFigure 2. See [9, Lemma 2.1.15] for the Y transform.Proof. Assume cxy cyI cxI > 0. Let Rx = rxI r xy /R and Ry = ryI r xy /R andR I = r xI r yI /R , where R = r xI + ryI + r xy . Dene a resistance form

E on four

points {x,y, I ,z } by

E (u, u ) = p= x,y,I

1R p

(u( p) u(z)) 2 .

Then by the Y transform, ( E|U , F |U ) is equal to the trace of resistance formE to U . Hence R(x, I ) = Rx + R I , R(y, I ) = Ry + R I and R(x, y) = Rx + Ry .

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Using (2.2), we immediately obtain g(x, y ) = R I . If any of cxy , cxI and cyI is

0, then direct calculation easily shows the proposition.In the rest of this section, we introduce the notion of the Martin boundary

of a transient weighted graph and related results originally studied in 1960s.See [16, Section 24] for the references and details.

Denition 2.12. Assume ( V, C ) is transient. Dene

K x 0 (x, y) = G(x, y )G(x0 , y)

for any x0 ,x ,y T . K x 0 (x, y ) is called the Martin kernel of ( V, C ).Using (2.2), we have

K x 0 (x, y) =

g(x, y )

g(x0 , y) =

R(x, I ) + R(y, I )

R(x, y )

R(x0 , I ) + R(y, I ) R(x0 , y) , (2.3)where R is the resistance metric with respect to ( E , F ).Proposition 2.13. Assume (V, C ) is transient. Then there exists a unique minimal compactication V of V (up to homeomorphism) such that K x 0 is ex-tended to a continuous function from V V to R . V is independent of the choice of x0 . Moreover, there exists a V \V -valued random variable Z such that Qx limn Z n = Z = 1 for any xX .

Denition 2.14. Assume that ( V, C ) is transient.

V is called the Martin

compactication of V . Dene M (V, C ) = V \V , which is called the Martinboundary of ( V, C ). Dene a probability measure x on M (V, C ) by x (B ) = Qx (Z B )for any Borel set B M (V, C ). x is called the hitting distribution on theMartin boundary M (V, C ) starting from x.

Note that x s for x V are mutually absolutely continuous and satisfy x = yV p(x, y ) y . The following theorem is the fundamental result on theMartin boundary and representation of harmonic functions on a weighted graph.Theorem 2.15. Assume that (V, C ) is transient.(1) K x 0 (, y)H+ (V, C ) H (V, C ) for any x0 V and any y

V .

(2) For any h

H+ (V, C ), there exists a Borel regular measure h on M (V, C )

such that h(x) = M (V,C ) K x 0 (x, y) h (dy).

(3) For any hH (V, C ), there exists f L (M (V, C ), x 0 ) such that h(x) = M (V,C ) K x 0 (x, y )f (y) x 0 (dy).

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3 Transient trees and their Martin boundaries

In this section, we introduce the notion of a tree, which is a special case of weighted graphs. The main goals are Theorems 3.8 and 3.13, where the hittingdistribution on the Martin boundary and the Martin kernel are expressed interms of resistance metrics of sub-trees.

Denition 3.1. (1) A pair ( T, C ) is called a tree if and only if it satises thefollowing conditions (Tree1) and (Tree2):(Tree1) ( T, C ) is an irreducible and locally nite weighted graph.(Tree2) For any x, y T , there exists a unique simple path between x and y.(2) Let ( T, C 1) and ( T, C 2) be trees. Then ( T, C 1) and ( T, C 2) is said to havea common graph structure if and only if C 1(x, y) = 0 implies C 2(x, y ) = 0, andvise versa for any x, y T . We write ( T, C 1) G (T, C 2) if (T, C 1) and ( T, C 2)have a common graph structure.

Fix a countably innite set T . Then the relation G is an equivalence relationon {(T, C )|(T, C ) is a tree}. We use G(T, C ) to denote the equivalence class of a tree ( T, C ) with respect to G . The equivalence class G(T, C ) determines thestructure of T as a non-directed graph.

Hereafter in this section, ( T, C ) is always a tree.

Denition 3.2. (1) For x, y T , the unique simple path between x and yis called the geodesic between x and y and denoted by xy. The path distance|x, y | between x and y is dened by the length of the geodesic between x and y.(2) For xT , dene x : T T by

x (y) =xn 1 if x = y and xy = ( x0 , . . . , x n 1 , xn ),x if y = x.

If x= y, x (y) is called the predecessor of y with respect to x. Also we deneS x (y) = N y (T, C )\{x (y)}. The points in S x (y) are called successors of y withrespect to x.(3) An innite path ( x0 , x1 , . . . ) T N is called a geodesic ray originatedfrom x V if and only if x0 = x and (x0 , x1 , . . . , x n ) = x0xn for any n 1. Two geodesic rays ( x0 , x1 , . . . ) and ( y0 , y1 , . . . ) is equivalent if and only if (xm , xm +1 , . . . ) = ( yk , yk+1 , . . . ) for some m and k. Dene ( T, C ) as the col-lection of the equivalence classes of geodesic rays. We write

T = T ( T, C ).

Easily, ( T, C ) is identied with the collections of geodesic rays originatedfrom a xed point xT .

Lemma 3.3. Let xT . Dene x (T, C ) by the collection of the geodesic rays

originated from x. Then ( T, C ) is naturally identied with x (T, C ).

We need the notion of sub-tree T xy and associated collection of geodesic rays xy (T, C ) to describe ner structure of x (T, C ).

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Denition 3.4. We write z xy if and only if xy = ( x0 , . . . , x n ) and z = xifor some i = 0 , . . . , n . Dene T

xy = {z|y xz} and xy (T, C ) = {(x0 , x1 , . . . )|(x0 , x1 , . . . ) x (T, C ), xm = y for some m 0}.

Dene T xy = T xy xy (T, C ).

By Lemma 3.3, we always identify ( T, C ) as x (T, C ). If no confusionmay occur, we write , x and xy in place of (T, C ), x (T, C ) and xy (T, C )respectively hereafter.

Proposition 3.5. Dene

O= {U |U

T , for any x U ,

there exists y T such that xy and T

xy U .}

Then O gives a topology of T and ( T ,O) is compact. Moreover, the closure of T is T .The restriction of O to T is the discrete topology on T . ( T ,O) is called theend compactication of T .Remark. Note that the notions introduced in this section so far only depend onthe equivalence class G(T, C ). In particular, if ( T, C 1) and ( T, C 2) are trees andG(T, C 1) = G(T, C 2), then ( T, C 1) is naturally identied with ( T, C 2) andthe end compactications of ( T, C 1) and ( T, C 2) are the same.

The following fundamental theorem on the Martin boundary of a tree is due

to Cartier [2]. See [16] for details.Theorem 3.6. Assume (T, C ) is transient. Then the Martin compactication T of T coincides with the end compactication T . For any hH

(T, C ), there exists a unique L

(M (T, C ), x 0 ) such that

h(x) = M (T,C ) K x 0 (x, y )(y) x 0 (dy) for any x V . Moreover, h(xn ) () as n x 0 -almost every ( T, C ), where (x0 , x1 , . . . ) is the geodesic ray originated from x0 representing ( T, C ).

By the above theorem, we identify the Martin boundary M (T, C ) with hereafter. Let O be the relative topology of Oon . Then ( , O ) is compact.Recall that x is the hitting distribution on starting from xV dened by x (B ) = Qx (Z B ) for a Borel set B . We are going to give an expressionof x ( xy ) by means of resistance metrics of sub-trees ( T xy , C xy ) dened below.Denition 3.7. Let x, y T with x= y. Dene rxy = 1 /C (x (y), y). Let C xybe the restriction of C onto T xy . We write R xy = Ry (T xy , C xy ) and xy = r xy + Rxy .

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Remark. If (T xy , C xy ) is not transient, then Rxy = xy = . Moreover, ( T xy , C xy )is not transient if and only if ( T

xz , C

xz ) is not transient for all z S x (y).Theorem 3.8. Assume that (T, C ) is transient. Then

x ( xy ) =R x x ( y )

xy x ( x x (y ) ) if (T

x x (y ) , C

x x (y ) ) is transient,

0 otherwise (3.1)

for any y T \{x}.By the inductive use of this theorem, if xy = ( x0 , . . . , x n ), then

( xy ) =R xx 0 R

xx 1 R

xx n 1xx 1

xx 2

xx n

if (T xy , C xy ) is transitive,

0 if (T xy , C xy ) is not transitive.

Lemma 3.9. Assume (T, C ) is transient. Then

x ( xy ) = Rx (T, C )/xy

for any y N x (T, C ).Note that Rx (T, C ) = (yN x (T,C ) 1/ xy ) 1 .Proof. If (T xy , C xy ) is not transient, then x ( xy ) = Qx (Z xy ) = 0. Assumethat ( T xy , C xy ) is transient. Let F (x, y) = Qx (Z n = y for some n 0). By [16,(1.13)-(b)] and Theorem 2.9,

F (x, y ) = G(x, y )G(y, y)

= g(x, y )g

(y, y) (3.2)

Using [16, (26.5)], we have

x (T xy ) = F (x, y )(1 F (x, y ))

1 F (x, y )F (y, x ) =

(g(x, x ) g(x, y ))g(x, y )g(x, x )g(y, y) g(x, y )2

. (3.3)

We consider the trace of ( E , F ) to the three points {x,y,I }, where I = I (T,C )as in Lemma 2.11. By using the same notations,g(x, x ) = Rx (T, C ) =

rxI (r xy + r yI )R

,

g(y, y) = Ry (T, C ) = ryI (r xy + r xI )

R ,

g(x, y ) = rxI r yI

R ,

where R = rxy + r xI + r yI . (3.3) and these imply

x (T xy ) = rxI

R =

Rx (T, C )r xy + r y

= Rx (T, C )

xy.

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Proof of Theorem 3.8. Let J xy = {Z n T xy for sufficiently large n}. Then x (

xy ) = Qx (J

xy ). Set I

xz = {Z n T

xz for any n 0}, where z = x (y).By the Markov property,

x ( xy ) = Qx (J xy ) = Qx (J xz )Qz (J xy I xz |I xz ) = x ( xz )Qz (J xy I xy |I xz ).Let ({Z n }n 0 , {Qw}wT xz ) be the Markov chain associated with ( T xz , C xz ) andlet z is the associated hitting distribution. ThenQz (J xy I xz |I xz ) = Qz (Z n T

xy for sufficiently large n ) = z (

xy ).

Applying Lemma 3.9 to ( T xz , C xz ), we obtain z (xy ) = Rxz / xy . Thus we have

(3.1).

As an application of Theorem 3.8, we may identify the support of x , whichis the Poisson boundary of ( T, C ).

Denition 3.10. Dene

T x = {y|y T, (T xy , C xy ) is transient }and let C x be the restriction of C to T x .

( T x , C x) is naturally identied with {(x0 , x1 , . . . )|(x0 , x1 , . . . ) x , xm T x for any m 0}. In this sense, ( T x , C x) is regarded as a subset of .Corollary 3.11. The support of x is ( T x , C x).

Next we give an expression of the Martin kernel in terms of resistance metrics

of sub-trees.Denition 3.12. Assume that ( T, C ) is transient. For x = y T , dene

xy =R xy xy

if (T xy , C xy ) is transient,

1 otherwise.

Theorem 3.13. Let x0 , x T with x0 = x and let x0x = ( x0 , x1 , . . . , x n ),where xn = x. For any y T , dene k(x0 ,x ,y ) as the unique k which satises y T x 0x k T xx k . Then,K x 0 (x, y ) =

(xx 0

xx 1 xx k 1

1x 0x k +1 x 0x n , (3.4)where k = k(x0 , x ,y ).

The rest of this section is devoted to proving Theorem 3.13.

Lemma 3.14. Assume that (T, C ) is transient. Let z T . If x N z (T, C )and y T zx , then K z (x, y ) = 1 / xz . (3.5)

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I y

z x

w

g (z, y )-1

r yI

r xy

r zw rxw

r wI

I y

z x

O

r yI

r zx

Rxz rxI rxy

I y

z

w

g (z, y )-2

r yI

r zw

r wI rwy

I y

z

w

g (z, y )-3

r zw

g(z, y)

I y

x

g (x, y )-1

r yI

r xyRxI

I y

x

g (x, y )-2

g(x, y)

Figure 3: Calculation of g(x, y ) and g(z, y)

Proof. We have three cases, namely,Case 1: both ( T xz , C xz ) and ( T zx , C zx ) are transient.Case 2: (T xz , C xz ) is transient while ( T zx , C zx ) is not.Case 3: (T zx , C zx ) is transient while ( T xz , C xz ) is not.First we consider Case 1. Let

cxI (u(x) u(I )) 2 + cyI (u(y) u(I ))2 + cxy (u(x) u(y)2be the trace of the resistance form associated with ( T zx , C zz ). Assume thatcxI cyI cxy > 0. Set r xI = 1 /c xI , r yI = 1 /c yI and r xy = 1 /r xy . Now our networkcorresponds to O at the upper left corner of Figure 3. We will follow thearrows in Figure 3 to calculate g(x, y ) and g(z, y). To obtain g(x, y), wecalculate the combined resistance RxI of rxz + Rxz and rxI and get g(x, y )-1 from O. Applying Y transform to g(x, y )-1, we have g(x, y )-2. ThenLemma 2.11 gives g

(x, y ). After performing these processes,

g(x, y) = rxI r yI (Rxz + r xz )

(r yI + r xy )(Rxz + r xz + r xI ) + ( Rxz + r xz )r xI .

To get g(z, y), we rst apply the Y transform to three points network

{z,x,I } in O and obtain g(z, y)-1. To set rwy = rxw + rxy gives g(z, y)-2.Finally applying the Y transform to three point network {I ,w,y }, we get

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g(x, y)-3. Then Lemma 2.11 gives g(z, y). After performing these processes,

g(z, y) = rxI r yI Rxz

(r yI + r xy )(Rxz + r xz + r xI ) + ( Rxz + r xz )r xI .

Since K z (x, y) = g(x, y )/g (z, y), we have (3.5). If one of cxI , cyI and cxy = 0,then the corresponding edge in O of Figure 3 is disconnected. The calculationis considerably easier and one can conrm (3.5) as well.

In Case 2, rxI and ryI are innite and the corresponding edges in O of Figure 3 are disconnected. Hence it is easy to obtain (3.5).

In Case 3, Rxz is innite and the corresponding edge in O of Figure 3 isdisconnected. Therefore g(x, z ) = g(x, y ) and this implies (3.5).

Proof of Theorem 3.13. Let z = x0 . We use induction in |z, x |. Assume |z, x | =1. Then y

T zx

or y

T z

x. If y

T z

x, then Lemma 3.14 shows (3.4). If y

T x

z ,

then K z (x, y ) = ( K x (z, y)) 1 = zz and so (3.4) follows. Next assume that (3.4)holds if |z, x | m. Let |z, x | = m +1 and let zx = ( x0 , x1 , . . . , x m , xm +1 ). Thenby Lemma 3.14,

K z (x, y ) = K z (xm , y)K x m (x, y ) = K z (xm , y) (xx m 1 if y T x mx ,x mx if y T xx m .Note that x mx = zx and T x mx = T zx . Hence using the hypothesis of the induction,we obtain (3.4). Thus we have completed the proof.

4 Induced form on the Martin boundary

In this section, we present a structure theorem (Theorem 4.4) of the Dirichletform on the Cantor set induced by a transient random walk on a tree. Through-out this section, ( T, C ) is a transient tree and T is a reference point.The trace on the Martin boundary ( E , F ) is dened as follows.Denition 4.1. Fix a reference point T . Write T # = T \{}. Dene alinear map H : L1( , ) (T ) by

H (f )(x) = x K (x, y )f (y) (dy)for any xX . Moreover, dene

F = {f |f L2( , ), H (f )F}and

E (f, g ) = E (H (f ), H (g))for any f , g F .

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Note that H (f )H(T, C ) since K (, y)H(T, C ) for any y .We are going to study the quadratic form ( E , F ). Since the support of is (T , C ), we will consider (T , C ) in place of (T, C ). Equivalently, we

assume that ( T xy , C xy ) is transient for any x, y T hereafter. As a result, theMartin boundary is equal to the Poisson boundary and every edge in T must hasat least one successor, i.e. S x (y) = for any x, y T . (Note that if S x (y) = ,then ( T xy , C xy ) is recurrent.) For ease of notation, we omit writing in notations.For example, we use , ,T x , C x , R x , x and x instead of , , T x , C x , R x , xand x .

The values Dx and x dened below play an essential role in this paper. Forexample, {x |x T }{0} will be identied with the collection of eigenvaluesof the self-adjoint operator associated with ( E , F ).Denition 4.2. Dene Dx = ( x )Rx and x = 1 /D x for any xT . The map

x x from T to (0, ) is called the eigenvalue map associated with ( T, C ).By (2.2) and Theorem 5.2-(1), we will see thatD x =

R(x, I ) + R(, I ) R(x, )2

, (4.1)

whose right-hand side coincides with the Gromov product of the metric R.

Lemma 4.3. For any xT and any y S (x),D yD x

= Ryy

< 1 (4.2)

and D x

Dy = ry ( y ) (4.3)

In particular, Dx > D y and y > x for any xT and any y S (x).Proof. By Theorem 3.8,

DyD x

= Rx Ryy Rx

= Ryy

= Ryr y + Ry

< 1

Again by Theorem 3.8, ( y )y = ( x )Rx . This immediately implies (4.3).Since every (T y , C y ) is assumed to be transient, it follows that ( y ) > 0.

The next theorem shows that E| has a simple expression by means of {x}xT and .Theorem 4.4. For any f L1( , ),

E (H (f ), H (f )) =xT

x2 ( x ) y,z S (x )

( y ) ( z )((f )y (f )z 2=

xT

Rx2

y,z S (x )

1y z((f )y (f )z 2

(4.4)

where (f )y = ( y ) 1 y f (x) (dx).19

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Remark. By Theorem 3.8, we see that

x2 ( x )

( y ) ( z ) = Rx

2y z

for any xT and any y, z S (x).The statement of the above theorem includes E (H (f ), H (f )) < + if andonly if the right-hand side of (4.4) is nite.

Denition 4.5. (1) Dene |x| = |, x | for any xT . Let x = ( x0 , . . . , x n ).Then dene [ x]m = xm for any m = 0 , 1, . . . , n = |x|.(2) Let = ( x0 , x1 , x2 , . . . ). Dene []n = xn for n 0.(3) For , with = , dene N (, ) = max {n|[]n = [ ]n } and[, ] = []N (, ) . We use [, ]m to denote [[, ]]m .Using Theorem 4.4, we may realize aspects of the nature of the quadratic

form (E , F ) as follows.Theorem 4.6. (1) (E , F ) is a regular Dirichlet form on L2( , ).(2) Dene J : ( ) \ [0, ) by

J (, ) = 12

+N (, ) 1

m =0

[, ]m +1 [, ]m ( [, ]m +1 ) ) (4.5)

for any , with = , where = {(, )| }. Then

F = u u

L2( , ),

J (, )(u()

u( )) 2 (d) (d ) < +

and, for any u, v F ,

E (u, v ) = J (, )(u() u( ))( v() v( )) (d) (d ).J (, ) is called the jump kernel of (E , F ).(3) Let L be the non-negative self-adjoint operator on L2( , ) associated with (E , F ) on L2( , ). Dene

E x = =yS (x )

ay y ,yS (x )

ay / y = 0

for xT . Then E x is contained in the domain of L and L = x for any E x . Moreover, let {x,i }i =1 ,..., #( S (x )) 1 be a orthonormal base of E x with respect to L2( , )-inner product. Then { ,x,i |xT, i = 1 , . . . , #( S (x)) 1}is a complete orthonormal system of L2( , ).Remark. By Lemma 4.3, J (, ) > 0 for any (, )( ) \.

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We will prove Theorem 4.6 in Section 9 as a special case of generalized

framework. Let us forget ( T, C ) for the moment and think T as a non-directgraph. Namely, we x T and choose an equivalence class with respect toG . Then we have : T T , , S (x), x and so on as we remarked afterProposition 3.5. Let be a Borel regular probability measure on whichsatises ( x ) > 0 for any x T and let : T [0, ). Dene a quadraticform

Q(u, v) =xT

(x)2( x ) y,z S (x )

( y )( z )(( u)y, (u)z, )(( v)y, (v)z, ),

where ( f )x, = ( x ) 1 x f (y)(dy). Set D = {f |f L2( , ), Q(f, f )

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I (T xz ,C xz ) z x y I (T xy ,C xy )

I (T 0 ,C 0 )

Rxz rzx rxy Rxy

x

z = R xz + r zx y = Rxy + r xy

Figure 4: Calculation of M y /M x

Theorem 4.8. For any (x, y )T # T ,K (x, y ) = x K ((x), y) + (1 x )

1 ( x )

bT x (y).

Proof. Let x = ( x0 , . . . , x n 1 , xn ) where x0 = and xn = x. Write z =xn 1(= (x)). If x / T x , then Theorem 3.13 yields K (x, y) = x K (z, y).Suppose x T x . Again by Theorem 3.13,K (x, y ) = ( xz ) 1K (z, y) = x K (z, y) +

((xz )

1 x K (z, y)= x K (z, y) +

Rxz + r zxRxz

RzxR zx + r zx K (z, y)

= x K (z, y) + (1 x )Rxz + R zx + r zx

Rxz(x

xx 1 xx n 2 ) 1

Now Lemma 4.7 implies K (x, y) = x K (z, y) + (1 x )/ ( x ).Corollary 4.9. For any f L1( , ),

H (f )(x) =(f ) = fd if x = ,x H (f )((x)) + (1 x )(f )x if x = . (4.9)

Moreover, let f, u L1( , ). If H (f )(x) = H (u)(x) for any x T , then f () = u() for -almost every .Proof. (4.9) is direct from Theorem 4.8. Using (4.9) inductively, we see that(f )x = ( u)x for any x T . Therefore, f () = u() for -almost every .

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Denition 4.10. Dene W m = {x|x T, |x| = m} for m 0. For f L

1

( , ), dene [f ]m = xW m (f )x x , E 0(f ) = 0 and, for m 1,

E m (f ) =xmk =1 W k

1r x(H (f )((x)) H (f )(x) 2 .

Corollary 4.9 implies the following lemma.

Lemma 4.11. For any m 1 and any f L1( , ),

E m (f ) = E m 1(f ) +xW m

r x(x )2(H (f )((x)) (f )x 2

= E m 1(f ) +xW m 1 yS (x )

r y(y )2

(H (f )(x) (f )y

2

Denition 4.12. Let (E x , F x ) be the resistance form associated with ( T x , C x )for xT . DeneRk,x = (inf {E x (u, u )|u : T x R , u(x) = 1 , u(y) = 0 if |y| |x|+ k + 1}) 1

for k 1 and R 0,x = 0.Lemma 4.13. (1) limk Rk,x = Rx .(2) For any k 1 and any xT ,

1Rk,x

=yS (x )

1r y + Rk 1,y

Proof. (1) Note that ( T x , C x ) is transient. Dene F ,x = ( C 0(T x )) E x +1, whereE x (u, v) = E (u, v ) + u(x)v(x). Write I x = I (T x ,C x ) . Recall that ( E x , F ,x ) is aresistance form on T x{I x} and that

Rx = (min{E x (u, u )|uF ,x , u(x) = 1 , u(I x ) = 0 } 1 .Hence Rx Rk,x . There exists F ,x such that (x) = 1 and (I x ) = 0and E x (, ) = 1 /R x . Since (I x ) = 0 , (C 0(T x ))E x . Hence there ex-ists {n }n 1 C 0(T x ) such that n (x) = 1 and E x (n , n ) 1/R x as n . Choose kn so that supp( n ) {y|y T x , |y| |x| + kn + 1}. ThenE x (n , n ) 1 Rk n ,x . Since kn as n and Rk,x Rk+1 ,x for any k,it follows that Rx limk Rk,x Rx .(2) This follows by applying the formula of combined resistance.Lemma 4.14. Let 1 m n .

ming:gL 1 ( , ) ,[g ]m =[ f ]m E n (g) = E m (f ) + xW m

Rn m,x(Rx )2 (H (f )(x) (f )x 2 (4.10)

for any f L1( , ).

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Proof. We use an inductive argument. If [ f ]n = [g]n , then E n (f ) = E n (g) by(4.9). Hence (4.10) holds for n = m. Assume that we have (4.10) for m. ChooseuL

1( , ) so that [ u]m 1 = [f ]m 1 . By (4.9) and the induction hypothesis,

ming:gL 1 ( , ) ,[g ]m =[ u ]m E n (g) = E m (u)+ xW m 1 yS (x )

Rn m,y(Ry )2 (H (u)(y)(u)y 2

= E m 1(u) +xW m 1 yS (x )

r y + Rn m,y(y )2 (H (u)(x) (u)y 2

= E m 1(f ) +xW m 1 yS (x )

r y + Rn m,y(y )2 (H (f )(x) (u)y 2

Now, we consider the minimum of the last expression subject to the constraint

[u]m 1 = [f ]m 1 , i.e.(f )x =

yS (x )

Rxy

(u)y

for any xW m 1 . The method of Lagrange multipliers along with Lemma 4.13-(2) shows

minu :[u ]m 1 =[ f ]m 1

ming :[g]m =[ u ]m E (g)

= E m 1(f ) +xW m 1

1(Rx )2 yS (x ) 1r y + Rn m,y 1(H (f )(x) (f )x 2

=

E m 1(f ) +

xW m 1

Rn m +1 ,y

(Rx )2 (H (f )(x)

(f )x

2 .

Lemma 4.15. For any f L1( , ),(f )xRx

(H (f )(x) (f )x ) yS (x )

(f )yRy

(H (f )(y) (f )y )

= Rx

2y,z S (x )

1y z((f )y (f )z 2 . (4.11)

Proof. Note that ( f )x =

yS (x )

(Rx / y )( f )y . Also by (4.9), if y

S (x), then

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H (f )(y) (f )y = y (H (f )(x) (f )y ). By those relations,(f )xRx

(H (f )(x) (f )x ) yS (x )

(f )yRy

(H (f )(y) (f )y )

= yS (x ) 1y (f )y (H (f )(x) (f )x ) yS (x ) (f )yy (H (f )(x) (f )y )= Rx yS (x ) 1y (f )y

2

+yS (x )

1y

(( f )y )2

=yS (x )

1y

Rx(y )2

(( f )y )2 Rxy,z S (x ) ,y = z

1y z

(f )y (f )z

= RxyS (x )

1y zS (x ) ,z = y

1z (( f )y )

2

Rx y,z S (x ) ,y = z1

y z (f )y (f )z

=Rx2

y,z S (x )

1y z((f )y (f )z 2 .

Lemma 4.16. Let f L1( , ). For any m 0,

xW m

(f )xRx

(H (f )(x) (f )x ) =xm 1j =0 W j

Rx2

y,z S (x )

1y z((f )y (f )z 2 .

(4.12)

For m = 0 , the right-hand side of (4.12) is considered as 0.Proof. Since H (f )() = ( f ) , we have (4.12) for m = 0. Suppose that (4.12)holds for m = 0 , 1, . . . , n 1. Then Lemma 4.15 suffices to show (4.12) form = n.Lemma 4.17. Let f L1( , ). Then

E m (f )+xW m

1Rx(H (f )(x)(f )x 2 = xm 1j =0 W j Rx2 y,z S (x ) 1y z((f )y (f )z 2 .

Proof. Dene V m = (mj =1 W j )(xW m {I x}), where I x = I (T x ,C x ) . For u (V m ), dene

E m (u, u ) =xmj =1 W m

1r x(u(x) u((x)) 2 + xW m 1Rx (u(x) u(I x ))2 .

Then ( E m , (V m )) is a resistance form on V m . Let Lm be the associated Lapla-cian, i.e. non-positive denite symmetric operator from (V m ) to (V m ) whichsatises E (u, v) = xV m u(x)(Lm v)(x). Dene F : V m R by F (x) =

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H (f )(x) for x mj =0 W j and F (I x ) = ( f )x for x W m . By (4.9), for anyxW

m ,

(Lm F )(x) = 1Rx

(F (I x ) F (x)) + 1r x

(F ((x)) F (x))=

1Rx

(( f )x H (f )(x)) + 1r x

(H (f )((x)) H (f )(x)) = 0 .In the same manner, it follows that ( Lm F )(x) = 0 for any x

mj =0 W m . There-

fore,

E m (F, F ) = xW m

F (I x )(Lm F )( I x ) = xW m

(f )xRx

(H (f )(x) (f )x ),

which equals to the left-hand side of (4.12). On the other hand,

E m (F, F ) = E m (f ) +xW m

1Rx(H (f )(x) (f )x 2

by denition. Combining these facts with Lemma 4.16, we immediately deducethe desired equation.

Proof of Theorem 4.4. By Lemma 4.14,

E n (f ) E m (f ) +xW m

Rn m,x(Rx )2 (H (f )(x) (f )x 2

for any n m 1. Applying Lemma 4.13-(1) and Lemma 4.17, we haveE (H (f ), H (f ))

xm 1j =0 W j

Rx2

y,z S (x )

1y z((f )y (f )z 2 E m (f )

for any m 1. Since E m (f ) E (H (f ), H (f )) as m , we immediatelyverify the theorem.

5 Intrinsic metric and volume doubling prop-erty

In the last section, we have introduced {Dx}xT and shown that D x n is mono-tonically decreasing if = ( x0 , x1 , . . . ) . In this section, we construct anultra-metric d on where the diameter of x equals to Dx for any x T andprovide a simple condition which is equivalent to the volume doubling propertyof with respect to this ultra-metric d. The ultra-metric d will turn out tobe suitable for describing asymptotic behaviors of the Hunt process associatedwith the regular Dirichlet form ( E , F ) on L2( , ) in the next section.

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In this section, ( T, C ) is a transient tree. We assume that ( T xy , C xy ) is tran-

sient for any x, y T . We x a reference point X and use the same notationas in the previous sections. To avoid nonessential complications, we further as-sume that #( S (x)) 2 for any x T . Even without this assumption, thestatements in this and following sections, except Corollary 6.9, hold with minormodication. This assumption implies the following proposition.

Proposition 5.1. ( , O ) has no isolated point and is a Cantor set, i.e. com-pact, perfect and totally disconnected.To describe the next theorem, we use the notations from Denition 2.8.

Recall that F = ( C 0(T ))E + R and that g is the symmetrized Green functionof (T, C ).Theorem 5.2. (1) g(, x) = Dx for any x

T .(2) xW n (

x )D x 0 as n .(3) D[]n 0 as n for -almost every .Notation. We write D n, = D []n .

We have an example of ( T, C ) where lim n D []n > 0 for some inSection 11.Proof. (1) Dene (x) = g(, x). Then is a {, I }-harmonic function withboundary value () = R and (I ) = 0. Let ( E m , F m ) be the trace of ( E , F )on T m {I }, where T m = mn =0 W m and let Lm be the associated discreteLaplacian on W m {I }. Note that |T m {I } is also a {, I }-harmonic functionwith respect to ( E m , F m ). Hence, for xW m ,

Lm (x) = ((x)) (x)

r x+

() (x)Rx

= 0

Since () = 0, we obtain

(x)/ ((x)) = Dx /D (x ) (5.1)

by (4.2). As () = R = D , (5.1) implies (x) = D x on W m for any m 0inductively.(2) Let ( E m , (T m )) be the resistance form on a nite set T m associated with(T m , C |T m ) and let H m be the associated (discrete) Laplacian. Then by (4.3)

E m

(, ) =

xT m(x)(H

m)(x)

= RyS ( )

D y D r y xT m

D xD (x ) Dx

r x

= R xT m

D x ( x ).

(5.2)

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Note that E m (, ) E (, ) = R as m . Hence by (5.2), we havedesired result.(3) Dene n : R by n = xW n Dx x . By (4.2), n is positive andmonotonically decreasing as n . By (2),

n () (d) = xT m D x ( x ) 0as n . This immediately implies that n () 0 as n for -almostevery . Since n () = Dn, , we have completed our proof.

Now we dene an ultra-metric on the Cantor set by means of D x .

Denition 5.3. Dene

d(, ) = D [, ] if = ,0 if = .

Proposition 5.4. (1) d(, ) is a metric on . Moreover it is an ultra-metric,i.e.max{d(, ), d(, )} d(, ). (5.3)

for any , , .(2) max d(, ) = R for any .(3) Dene B (, r ) = { |d(, ) < r } for any and r > 0. B(, r ) = []nif and only if D[]n < r D []n 1 .(4) The identity from ( , d) to ( , O ) is continuous. Moreover, ( , d) is homeomorphic to ( ,O

) if and only if D[]n

0 as n

for any

.

Proof. (1) It is enough to show (5.3), which implies the triangle inequality.Other properties of metric are immediate. If T [, ] T [, ], then multiple ap-plications of (4.2) show that d(, ) < d (, ). Hence we have (5.3). Otherwise,d(, ) = d(, ) and (5.3) holds.(2) For any , we may choose so that [ , ] = . Then d(, ) =D = R .(3) Fix . Then d(, ) {D m, |m 0}. Moreover, d(, ) < D n 1, d(, ) Dn, []n . Therefore, B(, r ) = []n if and only if Dn, < r Dn 1, .(4) Note that { x}xT is a fundamental system of neighborhoods of O . Forany x T # , choose x . If r = D x , then B(, r ) = x . Hence x is openwith respect to ( , d) for any x

T . This implies the continuity of the identityfrom ( , d) to ( , O ). Assume that Dn, 0 as n for any .Let U be an open set with respect to ( , d). For any U , B(, r ) U forsome r > 0. By the assumption, there exists n such that Dn, < r and hence []n U . Therefore, U is an open set with respect to O . This shows that( , d) and ( , O ) are homeomorphic.Finally assume Dn, D > 0 for some . If 0 < r < D , thenB (, r ) = {}. Therefore {} is an open set with respect to ( , d). On the

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other hand, {} / O by Proposition 5.1. Hence ( , d) is not homeomorphicto ( , O ).The following theorem gives necessary and sufficient condition for to havethe volume doubling property with respect to the metric d.

Theorem 5.5. has the volume doubling property with respect to d, i.e., there exists c > 0 such that (B (x, 2r )) c (B (x, r )) for any x and any r > 0,if and only if the following two conditions (EL) and (D) hold:(EL) : There exists c1 (0, 1) such that c1 ( x )/ ( (x ) ) for any xT # .(D): For any n 0 and any ,

D []n + m D []n ,where m 1 and (0, 1) are independent of n and .

By the condition (D), D []n 0 for any if has the volume doublingproperty with respect to d. Also the number of neighboring vertices is shownto be uniformly bounded under the volume doubling property as follows.

Proposition 5.6. If (EL) hold, then supxT,y S (x ) ( y )/ ( x ) < 1 and sup xT #( S (x)) < + .Proof. If #( S (x)) > 1, then c1 zS (x ) ,z = y ( z )/ ( x ) = 1 ( y )/ ( x ).Hence ( y )/ ( x ) 1 c1 . Moreover, c1#( S (x)) yS (x ) ( y )/ ( x ) =1. This shows #( S (x)) 1/c 1 .

The following alternative denition of the volume doubling property is some-times useful.

Proposition 5.7. has the volume doubling property with respect to d if and only if there exist c > 0 and (0, 1) such that (B (, r )) c (B (,r ) for any and any r(0, R ].Proof of Theorem 5.5. Assume the volume doubling property. Then,

(B (, Dn, + )) c (B (, Dn, / 2 + /2))Choose so that Dn, + < Dn 1, and Dn, / 2 + /2 D n, . Proposition 5.4-(3) shows ( []n ) c (B (, Dn, )) = c ( []n +1 ). If xT and any y S (x),then x = []n and y = []n +1 for some . Hence 1 /c ( y )/ ( x ).Thus we have (EL) . By Proposition 5.6, there exists (0, 1) such that (

y)/ (

x)

for any x

T and any y

S (x). Choose m

1 so that

m < 1/c . For any , we have (B (, Dn + m, )) = ( []n + m +1 ) m ( []n +1 ) = m (B (, Dn, ).

Using the volume doubling property, we obtain

1c

(B (, 2D n + m, )) (B (, Dn + m, )) m (B (, Dn, ).

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Hence D n + m, D n, / 2. Thus we have (D).Conversely, we assume (EL) and (D). If Dn, < r D0, = R , then wemay choose k n such that D k, < r D k 1, . Since D k + m, D k, < r ,we have ( []k + m ) (B (,r )). On the other hand, by (EL) , ( []k + m ) (c1)m ( []k ) = ( c1)m (B (, r )). Thus

(B (, r ) c (B (,r )) (5.4)for any r > D n, , where c = ( c1) (m +1) . By Proposition 5.7, we have thevolume doubling property of with respect to d.

6 Asymptotic behaviors of the process

In this section, we study the asymptotic behavior of the heat kernel (transitiondensity), the jump kernel J (, ) and moments of displacement associated withthe Dirichlet form ( E , F ) on L2( , ) under the assumption that is volumedoubling with respect to d. As in the last section, ( T, C ) is a transient tree,(T xy , C xy ) is assumed to be transient for any x, y T . We x a reference point. Moreover, we continue to assume that #( S (x)) 2 for any x T . All thestatements except Corollary 6.9, however, will hold without this assumption.

Making use of Theorem 4.6, we have a formal expression of a heat kernelassociated with the Dirichlet form ( E , F ) on L2( , ) as follows:

p(t ,, ) = 1 +xT

e x t#( S (x )) 1

j =1

x,j ()x,j ( ). (6.1)

By Lemma 6.1 below,

#( S (x )) 1

j =1

x,j ()x,j ( ) =yS (x )

1 ( y )

y () y ( ) 1

( x ) x () x ( ).

Combining this with (6.1), we obtain

p(t ,, ) =

1 +

n =0

1 ( []n +1 )

1 ( []n )

e [ ]n t if = ,

N (, )

n =0

1 ( [, ]n )

(e [, ]n 1 t e [, ]n t if = ,

(6.2)

where we dene [] 1 = 0 and write [ , ]n = [[, ]]n . If we allowas a value, p(t ,, ) is well-dened on (0 , ) 2 through (6.2). The value may occuron the diagonal. Also, by (6.2), p(t ,, ) is continuous on (0 , )(( ) \ ,where is the diagonal.30

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Proof. If = , then

p(t ,, ) =n 0

e [ ]n 1 t e [ ]n t ( []n )

[ ]n ( ) (6.5)

Note that this is an innite sum of non-negative functions. Using (6.5), weobtain (6.4) by a routine but careful calculation. The fact that { pt}t> 0 is aMarkovian transition function is immediate from (6.4).

By this proposition, if lim n D []n = 0 for any , then ( pt u)() =(T t u)() for -a.e. , where {T t }t> 0 is the strongly continuous semigroupon L2( , ) associated with the Dirichlet form ( E , F ) on L2( , ). Moreover,we have the following theorem.Theorem 6.3. If limn D []n = 0 for any , then there exists a Hunt process ({X t }t> 0 , {P } ) on whose transition density is p(t ,, ), i.e.

E (f (X t )) = p(t ,, )f ( ) (d ) for any and any Borel measurable bounded function f : R , where E () is the expectation with respect to P .

Note that the Hunt process ( {X t}t> 0 , {P } ) is naturally associated withthe Dirichlet form ( E , F ) on L2( , ).The essence of the proof of Theorem 6.3 is to show that the transition func-tion { pt}t> 0 is a Feller transition function. Namely we are going to prove that pt (C ())

C () and

|| pt u

u

||

0 as t

0 for any u

C (), where C ()is the collection of continuous functions on .

Lemma 6.4. Let Cm = {xW m ax x |ax R for any xW m } and let C =m 0Cm . Then || pt u u|| 0 as t 0 for any uC.Proof. For any uCm , by using an inductive argument, it follows that

u = c +m

k=1 xW k

#( S (x )) 1

i =1

bx,i x,i .

(See Lemma 9.2 for details.) This implies

pt u = c +m

k=1 xW k

#( S (x )) 1

i =1e x t bx,i x,i .

Hence there exists c > 0 such that || pt u u|| c(1 e t ).

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Proof of Theorem 6.3. Let , and let N = N (, ). Then by (6.5),

| p(t ,, ) p(t ,, )|=

n>N

e [ ]n 1 t e [ ]n t ( []n )

[ ]n ( ) + e [ ]n 1 t e [xi ]n t

( []n ) [ ]n ( ))

Since limn D []n = 0 for any , we have

| pt, pt, |d = 2e [, ] tHence if u is a bounded Borel measurable function on , then

|( pt u)()

( pt u)( )

| 2e [, ] t

||u

|| .

Again by the fact that lim n D []n = 0 for any , we see pt uC (). Inparticular pt (C ()) C ().Let uC () and x > 0. Then there exist m 0 and um Cm such that||u um || < / 3. By Lemma 6.4,

|| pt u u|| || pt u pt um || + || pt um um || + ||u um || < for sufficiently small t > 0. Hence || pt u u|| 0 as t 0. Thus { pt }t> 0 isa Feller transition function. Then by [5, Theorem A.2.2], (see also [1, Theo-rem I.9.4]), we obtain the desired statement.

Notation. We write (m, ) = ( []m ) for any and any m 0.Without any further assumptions, p(t ,, ) satises the following estimates.

Proposition 6.5. (1) For any and any t > 0, p(t ,,)

1e

1 (B (, t))

(6.6)

(2) If 0 < t d(, ), then p(t ,, )

td(, ) ( [, ])

. (6.7)

Proof. (1) If Dn, < t Dn 1, for some n 1, then t/D m, 1 for m =0, 1, . . . , n

1. Hence

p(t ,,) 1 +n 1

m =0

1 (m + 1 , )

1 (m, )

e 1 1

e (n, ).

By Proposition 5.4-(3), it follows that (n, ) = (B (, t)). Hence we have(6.6) for t(0, R ]. For t R , B (, t) = and hence p(t ,,) 1 e 1 =e 1 / (B (, t)).

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(2) Write N = N (, ), n = []n , D n = Dn, and n = ( []n ). Then

d(, ) = DN . By letting f (t) = p(t ,, ), (6.2) implies

f (t) =N

n =0

e n 1 t e n t n

,

f (t) =N

n =0

n e n t n 1e n 1 t n

,

f (t) =n

n =0

(n 1)2e n 1 t (n )2e n t n

,

where 1 = 0. Since n 1 t n t DN /D n 1, we see that f (t) 0 andf (t) 0 for any t[0, D N ]. Hencef (DN )t f (t) f (0) t (6.8)

for any t[0, D N ]. (6.8) along with

f (0) =N 1

n =0n

1 n

1 n +1

+ N N

1DN N

= 1

d(, ) ( [, ])

shows (6.7) for any t(0, d(, )].

The volume doubling property of leads us to two-sided estimates of p(t ,, )and J (, ). Note that under the volume doubling property of , we havelimn D []n = 0 for any by Theorem 5.5.Theorem 6.6. Suppose has the volume doubling property with respect to d.

(1) p(t ,, ) is continuous on (0, ) . Dene q (t ,, ) =

td(, ) ( [, ])

if 0 < t d(, ),1

(B (, t)) if t > d (, ).

Then p(t ,, )q (t ,, ) (6.9)

on (0, ) .(2) For any (, )( ) \ ,J (, )

1d(, ) ( [, ])

. (6.10)

The proof of this theorem is given at the end of this section.The heat kernel estimate (6.9) can be thought of as a generalized version

of the counterpart in [3], where is supposed to satisfy the uniform volumedoubling property: i.e. (B (x, r )) f (r ) for any r > 0 and f (r ) has thedoubling property. In this paper, however, we do not require the uniform volumedoubling property.

The following proposition gives an alternative expression for q (t ,, ).

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Proposition 6.7. For any t > 0 and any , ,q (t ,, ) = min t

d(, ) ( [, ]), 1

(B (, t)).

The above proposition is immediate from the following lemma.

Lemma 6.8. t (B (, t)) d(, ) ( [, ]) if and only if t d(, ).Proof. Assume 0 < t D = R . Then Dn, < t D n 1, for some n 1. ByProposition 5.4-(3), B(, t) = []n . If t d(, ), then N (, ) n 1 andhence t ( []n ) d(, ) ( [, ]). Otherwise, t > d (, ) implies N (, ) n.Therefore, t ( []n ) > d (, ) ( [, ]).

Next assume t > D . Then B (, t) = and t > d (, ). Hence t ( []n ) >d(, ) ( [, ]).

Next we have estimate of moments of the displacement. In the next corollary,the assumption that #( S (x)) 2 for any xT is essential.Corollary 6.9. Suppose that has the volume doubling property with respect to d. Then

E (d(, X t ) )

t if > 1,t(| log t|+ 1) if = 1t if 0 < < 1.

(6.11)

for any and any t(0, 1].Note that we have an extra log-term when = 1. This is due to the slow

polynomial decay of the off-diagonal part of p(t ,, ) with respect to the spacevariable . See the discussion after Theorem 14.1 for details. Corollary 6.9 is aspecial case of Theorem 14.1, which shows that the above behavior of momentsof displacement occurs whenever p(t ,, ) enjoys certain type of heat kernelestimate which is typical in certain class of jump processes.

Proof. Let us verify all the assumptions of Theorem 14.1 with X = , = and(r ) = r. Theorem 6.3 suffices to show the assumption (1) in Section 14. ByProposition 5.6, there exist c1 , c2 , c3 (0, 1) and R > 0 such that c1 (B (, r )) (B (, c3r ) c2 (B (, r )) for any and any r (0, R]. Hence we have(14.1). Theorem 6.6 implies (14.2). Therefore, we may apply Theorem 14.1 andobtain the above corollary.

The rest of this section is devoted to showing Theorem 6.6.

Proof of Theorem 6.6-(1). Combining (6.1) and Lemma 6.1, we have

p(t ,, ) = 1+n 0

e [ ]n t 1

( []n +1 ) [ ]n +1 ( )

1 ( []n )

[ ]n ( ) . (6.12)

Note that every term in the sum is continuous on (0 , ) .

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On the other hand, by Theorem 5.5, the volume doubling property implies

that there exist (0, 1) and c > 0 such that ( []n ) n and Dn, c nfor any and any n 0. Hence the absolute value of each term in (6.12)is no greater than c n e c

1 n T on [T,) . Since the innite sum of these terms is convergent, the innite sum in (6.12) is uniformly convergent on[T,) by the Weierstrass M-test. Therefore, p(t ,, ) is continuous on(0, ) .Proof of Theorem 6.6-(2). We use the notations in the proof of Proposition 6.5.By Theorem 5.5, there exist (0, 1) and M N such that

m n

n + m (EL)

Dn + M D n (D)for any n and m. Choosing suitable (0, 1) and c > 0, we also have

Dn + m cm Dn (D)for any n and m. Note that , c and M are independent of .The upper estimate for 0 < t d(, ) is immediate by Proposition 6.5-(2).Lower estimate for 0 < t d (, ): Assume that M N + 1. By (EL) ,

f (DN ) N

n = N M +1

n e n D N n 1e n 1 D N n

N

n = N M +1 N n n e

n D N n 1e n 1 D N N

M 1

N DN 1e

DN DN M

e D N

D N M .

(6.13)

By (D), we see D N /D N M < 1. Hencef (DN )

M 1(e 1 e ) N D N

C N DN

. (6.14)

Next we consider the case where N + 1 M . Since Dm +1 , < D m, for any and any m 0, there exists (0, 1) such that Dm +1 , D m, forany and any m M . Note that N 1 M and hence DN D N 1 .Therefore

f (D N ) N e 1 N 1e N 1 D N

N

1

N D N 1e

DN D N 1

e D N D N 1

e 1 e N DN

.

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Using this and (6.14), we have the desired lower estimate for 0 < t d(, )from (6.8).Upper estimate for d (, ) < t : By (6.2), p(t ,, ) p(t ,,). Hence it isenough to show that p(t ,,) c/ (B (, t)). Suppose Dm < t D m +1 . Then

1+m 1

n =0

1 n +1

1 n

e n t 1+m 1

n =0

1 n +1

1 n

= 1 m

= 1

(B (, t)) (6.15)

Let n m. Then by (D), t/D n Dm /D n c 1 m n . Also by (EL) ,1/ n +1 1/ n 1/ n +1 (n +1 m ) / m . These imply

n = m

1 n +1

1 n

e n t m

k=0

k e c1 k

C (B (, t))

. (6.16)

Since C is independent of and m, (6.15) and (6.16) yield the upper estimatefor d(, ) < t D0 . If t > D 0 , then (B (, t)) = 1. Since

p(t ,,) = 1 +

n =0

1 n +1

1 n

e n t 1 +

n =0 (n +1) e c

1 n ,

we have the upper estimate in this case.Lower estimate for d (, ) < t : If D 0 < t , then B (, t) = and

p(t ,, ) 1 0

(1 e 0 t ) 1 e 1 = 1e 1 (B (, t))

.

Assume d(, ) < t D 0 . Then Dm < t D m 1 for some m {1, 2, . . . , N }.Suppose M N . Then we may choose k {0, . . . , m 1}so that m k + M N . By (EL) , it follows that m M n for any n = k, k +1 , . . . , k + M . Hence

p(t ,, ) k + M

n = k+1

e n 1 t e n t n

M e k t e k + M t m

M e b e a m

,

where a = t/D k+ M and b = t/D k . Using (D), we see that b 1 a andb a . Since min{e b e a |b 1 a, b a }> 0, it follows that p(t ,, ) C / (B (, t)), where C is independent of and t.Finally assume N + 1 M . As in the proof of the lower estimate for0 < t d(, ), we have Dm D m 1 , where (0, 1) is independent of

and t . Hence p(t ,, )

e m 1 t e m t m

.

Since t/D m 1 1 t/D m and t/D m 1 t/D m , the similar argument asabove shows p(t ,, ) C / (B (, t)).

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Proof of Theorem 6.6-(3). We continue to use the notations in the proof of

Proposition 6.5. By (4.5),

J (, ) = 12

0 +N 1

m =0

m +1 m m +1

,

where N = |x|. Note that (EL) , (D) and (D) in the proof of Theorem 6.6-(2)hold under the volume doubling property of . First we show the lower estimate.Assume that N M + 1, where M is the constant appearing in (D). By (EL) , N i M N for 0 i M . Hence, by (D),

J (, ) M

2

N 1

m = N M

m +1 m N

M

2N N M

N

M (1 )N 2 N = M (1 )2d(, ) ( [, ]) .

In case N M + 1, we may apply the same discussion as in the lower estimatefor d(, ) < t in the proof of Theorem 6.6-(2) and obtain the lower estimate inthis case.

Next by (D), it follows that N i c i N . This implies

J (, ) 12

N

m =0

m N

c(1 + + . . . + N )N 2 N

c2(1 )d(, )( [, ])

.

Thus we have shown the upper estimate.

7 Completion with respect to resistance metric

As in the last section, ( T, C ) is a transient tree, ( E , F ) and R are the associatedresistance form and resistance metric on T respectively.Let T be the completion of T with respect to the resistance metric. Then by

[9, Theorem 2.3.10], ( E , F ) can be thought of as a resistance form on T . WriteT R = T \T . Then by Theorem 13.3, we have a resistance form ( E|T R , F|T R )on T R which is the trace of the resistance from ( E , F ) on T R . The naturalquestion is if T R = and ( E , F ) = ( E|T R , F|T R ) or not. In particular, if thisis true, then ( E , F ) is a resistance form on . In the followings, we establisha sufficient condition in Theorem 7.1 and a necessary condition in Theorem 7.2for being T

R = and (

E|T R,

F|T R) = (

E ,

F ). Using these theorems, we will

show that both T R = and T R = can occur by examples in the next section.

Theorem 7.1. Let rx = C (x, (x)) 1 . If m 1(max xW m rx ) < +, where W m = {x|x T, |x| = m}, then there exists a homeomorphism between T and T which is the identity on T . Moreover, (E , F ) coincides with (E| , F| )which is the trace of (E , F ) on , where is identied with T \T through the homeomorphism. In particular, (E , F ) is a resistance form on .38

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Proof of Theorem 7.1. Dene r m = max xW m rx . Let . For m > n ,

R([]n , []m ) =m

i = n +1

r []i m

i= n +1

r i

Therefore, {[]n }n 0 is an R-Cauchy sequence. Denote the R-limit of {[]n }n 0by ( ) T . Since R([]n , [ ]n ) r[]N ( , )+1 for n > N (, ), : T is injective. Similar argument shows that () T \T . Let {xn }n 0 be anR-Cauchy sequence and let x be its R-limit in T . Assume x /T . Then thereexists a subsequence {yn }n 0 of {xn }n 0 such that |yn | as n . Forany m 0, there exists zm W m such that yn T zm for sufficiently large n.Note that = ( z0 , z1 , . . . ) is a geodesic ray and that R(, z) i = m +1 r i forany z T zm . Those facts imply x = . Hence is bijective. Now, extend : T

T by (x) = x for x

T . It is routine to see that is continuous.Thus is a homeomorphism between T and T and |T is identity. Hereafter,we identify with T \T .Now, let f F . Then H (f ) F . Note that H (f ) is extended to acontinuous function on T = T . Hence Theorem 3.6 shows that H (f )(x) =H (u)(x) for any x T , where u = H (f )| . Making use of Corollary 4.9, wehave ( u)x = ( f )x for any x T . This implies u = f for -almost everywhere.Hence f F| = {h| |hF}.Let f F| . Let h(f ) be the -harmonic function with boundary value f ,i.e. h(f )F attains the following minimum:

min{E (v, v)|v F , v| = f }.

Since h(f ) is harmonic on T and is bounded, Theorem 3.6 shows that h(f )(x) =H (u)(x) for some u L ( , ) and h(f )([]n ) u(x) for -almost every . Note that h(f ) is continuous on T and h(f )| = f . It follows that u = f and hence h(f ) = H (f ). Therefore, f F and hence F| = F . FinallyE| (f, f ) = E (h(f ), h(f )) = E (H (f ), H (f )) = E (f, f ) for any f F = F| .Thus ( E| , F| ) is identied with ( E , F ).

Next we state a necessary condition.

Theorem 7.2. If (E , F ) is a resistance form on and the associated resis-tance metric gives the same topology as O , then R[]n 0 as n for any .Proof of Theorem 7.2. Using [8, Theorem 9.4], we see that p(t ,, ) is jointlycontinuous and hence it is bounded. (Since p(t ,, ) is continuous except thediagonal , the heat kernel given in [8, Theorem 9.4] is equal to p(t ,, ).) By [8,Corollary 9.6], tp(t ,,) 0 as t 0 for any . Using Proposition 6.5-(1),we have t/ (B (, t)) 0 as t 0. If D n, < t Dn 1, , then B (, t) = []n .Hence t/ (B (, t)) Dn, / ( []n ) = R[]n . Since n as t 0, we haveR []n 0 as n .

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8 Example: binary tree

In this section, we are going to illustrate the results obtained in the previoussections by examples which are (innite complete) binary trees. Our examplesare divided into two classes. The rst class is the collection of self-similartrees, where the volume doubling property is automatic under the assumptionof transience. The other class is homogeneous trees, through which we willexplore various phenomena when the volume doubling property fails.

Denition 8.1. Let W m = {1, 2}m for m 0, where W 0 = {}. Dene W =m 0W m . We denote ( w1 , . . . , w m ) as w1 . . . wm . For w1 . . . wm W \W 0 , de-ne (w1 . . . wm ) = w1 . . . wm 1 and S (w1 . . . wm ) = {w1 . . . wm 1, w1 . . . wm 2}.Assume that C : W W [0, ) satises C (w, v) = C (v, w) and C (w, v) > 0if and only if (w) = v or (v) = w. (W , C ) is called the (innite complete)binary tree.

For binary trees, we always choose as the reference point. Then, thenotions (w), S (w) and W m are consistent with those dened in Sections 3 and4. For any ( W , C ), the collection of innite geodesic rays originated from , , is identied with the Cantor set {1, 2}N . As a standard metric on , weintroduce d(, ).Denition 8.2. Dene d(, ) = 2 N (, ) for any = and d(, ) = 0if = .

It is easy to see that d(, ) is a distance on .First we consider a kind of self-similar binary tree ( W , C S ).Denition 8.3. Let r1 , r 2 > 0. For w W , dene C S (w,wi ) = ( r w r i )

1,where r w = rw 1 r w m for any w = w1 . . . wm W .

Theorem 8.4. (W , C S ) is transient if and only if r1r 2 / (r 1 + r 2) < 1. In particular, if r1 = r2 = r , then (W , C S ) is transient if and only if 0 < r < 2.

Proof. Let ({X n }n 0 , {P w}wW ) be the random walk associated with ( W , C S ).Then P (|X n +1 | = |X n | + 1 ||X n | 1) = ( r 1 + r2)/ (r 1 + r2 + r1r 2) = p1 andP (|X n +1 | = |X n | 1||X n | 1) = r1r 2 / (r 1 + r2 + r1r 2) = p2 . Therefore, wemay associate a random walk ( {Z n }n 0 , {Qk}kN) on N = {0, 1, . . .}such thatQk (Z n +1 = Z n + 1 |Z n 1) = p1 and Qk (Z n +1 = Z n 1|Z n 1) = p2 if Z n 1and Qk (Z n +1 = 1 |Z n = 0) = 1. This random walk is transient if and only if p1 > p 2 r 1r 2 / (r 1 + r 2) < 1.By applying the results in the previous sections, we obtain the following

statements.

Lemma 8.5. Assume that (W , C S ) is transient.

(1) R =r 1 + r 2

r 1r 2 1 1

.

(2) Rw = rw R for any wW .

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(3) Let 1 = r2 / (r 1 + r 2) and 2 = r1 / (r 1 + r 2). Then ( w ) = w1 w m for any w = w1 . . . wm W .(4) Dw = ( w )Rw = ( r 1r 2 / (r 1 + r 2)) |w | R for any wW .(5) has the volume doubling property with respect to d(, ), where d(, ) has been given in Denition 5.3.Proof. (1) Considering the self-similarity, R satises

1r 1(R + 1)

+ 1

r 2(R + 1) =

1R

.

(2) This is obvious from the self-similarity.(3) Use Theorem 3.8.(4) Combine (2) and (3).(5) One can easily verify the conditions (EL) and (D) in Theorem 5.5.

The above lemma shows that d(, ) = D[, ] = ( r 1r 2 / (r 1 + r2)) N (, ) R .Hence we have d(, ) = d(, )R for any , , where

= logr 1 + r 2

r 1r 2/ log 2.

Combining those with Theorem 6.6, we have the following heat kernel estimate.

Theorem 8.6. Assume that (W , C S ) is transient. Let p(t ,, ) be the associ-ated heat kernel. Dene

q (t ,, ) =

td(, ) ( [, ])

if 0 < t d(, ) ,1

(B(, t1/ )) if t > d (, ) .

Then p(t ,, )q (t ,, ) (8.1)

on (0, ) , where B(, r ) = { |d(, ) < r }. In particular, if r 1 = r2 ,then = 1 log r/ log2 and we may replace q in (8.1) by

q (t ,, ) =t

d(, )+1 if 0 < t d(, ) ,

t 1/ if t > d (, ) .

Theorem 8.7. (E , F ) is a resistance form on and the associated resistance metric gives the same topology as O if and only if 0 < r 1 < 1 and 0 < r 2 < 1.Proof. Assume that 0 < r 1 < 1 and 0 < r 2 < 1. Let r = max {r 1 , r 2}. Then0 < r < 1 and max xW m rx rm . By Theorem 7.1, ( E , F ) is a resistanceform on .

Conversely say r1 1. Let = 1111 . . .. Then R[]n = ( r 1)n R . NowTheorem 7.2 yields the desired conclusion.

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Next we study a class of homogeneous binary trees.

Denition 8.8. Let r(i) > 0 for any i 0. Dene C H (w,wi ) = r(|w|) 1 forany wW and any i {1, 2}.Theorem 8.9. (W , C H ) is transient if and only if n 0 2 (n +1) r (n) < + .Proof. Let us reduce W m to an single point m N {0}. Then the resultingweighted graph is ( N {0}, C ), where C (m, m + 1) = 2 m +1 /r (m). Hence R =R0 = n 0 r (n)/ 2(n +1) . Since the transience is equivalent to the condition thatR0 < + , we have the claim of the theorem.

By the homogeneity of the tree, we can easily see that ( x ) = ( y ) forany x, y W m . Also the same method as in the proof of the above theoremgives Rw . As a consequence, we have the followings.Lemma 8.10. Assume that (W , C H ) is transient. Then for any wW

,

( w ) = 2 | w | , Rw =n 0

r (|w|+ n)2n +1

and Dw =n | w |

r (n)2n +1

. (8.2)

Hereafter, we always assume that ( W , C H ) is transient. By (8.2), Dw onlydepends on |w|. For n 0, we dene Dn = Dw and n = 1 /D n for w W n .Note that {D n }n 0 is strictly decreasing and lim n D n = 0. Conversely, givena strictly decreasing sequence {Dn }n 0 with limn Dn = 0, we may constructan associated {r (n)}n 0 by letting r (n) = 2 n +1 (Dn Dn +1 ). By (6.2), we have

p(t ,,) = 1 +

n =0

2n e n t (8.3)

and for = ,

p(t ,, ) =N (, )

n =02n (e n 1 t e n t )

= 1 +N (, ) 1

n =02n e n t 2N (, ) e N ( , ) t

(8.4)

From these, if p(T,,) < + for some T > 0, then p(t ,, ) is continuouson [T,) . Choosing an appropriate decreasing sequence {Dn }n 0 , wemay obtain a variety of heat kernels with interesting behaviors. For instance,we have an example where pt, () = p(t,, ) is getting more and more regularas t log 2 as follows.Example 8.11. Let D n = ( n + 1) 1 for n 0. Then

p(t ,, ) = (1 e t )2N +1 e (N +1) t 1

2e t 1 if t = log 2,

(N + 1)(1 e t ) if t = log 2,(8.5)

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where N = N (, ). Dene pt, by pt, ( ) = p(t ,, ). Then by (8.2) and (8.5),

for q 1, pt,

Lq

( , ) if and only if (1 1/q )log2 < t . (We regard 1 /q as0 if q = .) In other words, if 0 < t log 2, then pt, Lq( , ) for 1 q 0,

12 0 e s 2f (s/t ) ds p(t ,,) 0 e s 2f (s/t ) ds.

By using the above proposition, an asymptotic behavior of p(t ,,) as t

0

may be determined even if the volume doubling property fails as in the nextexample.

Example 8.13. Let Dn = ( n + 1) 2 for n 0. Then n = ( n + 1) 2 . Inthis case, does not have the volume doubling property with respect to d.Proposition 8.12 implies that

p(t ,,) 0 e s 2 s/t ds.on (0, ) . Now, for any c > 0,

0 ec s/t s ds = 1 + c t exp c24t c2 t e y 2 dy.Hence

p(t ,,)1 + 1 t exp

(log 2)2

4t

on (0, ). Since (B (, t)) 12 1/t , the degree of divergence of p(t ,,)as t 0 is actually much higher than what is expected by the formula whichholds with the volume doubling property.

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9 Generalization and jump kernel

In this section, as a generalization of ( E , F ), we study a class of Dirichletforms and/or closed forms on the space of innite geodesic rays of a tree asa non-directed graph.

Let T be a countably innite set. To give an a priori structure of a treeas a non-directed graph, we x an equivalence class with respect to G . This isequivalent to choose T and : T T such that () = and, for anyxT \{}, (n ) (x) = for some n 1, where (n ) = . . . n -times . All the notionsassociated with the structure of a tree as a non-directed graph can be derivedfrom and , for example, T # = T \{},

S (x) = {y|(y) = x},T x = {y|y T,

(n )

(y) = x for some n 0}, = {(x i )i 0|x0 = , (x i +1 ) = xi for any i 0}, x = {(x i )i 0|(x i ) i 0 , xm = x for some m 0}.

To avoid unnecessary technical complexity, we assume that 2 #( S (x)) < + for all x T . The space is equipped with the canonical topology O whichgenerated by the basis of open sets { x |xT }. Note that ( , O ) is compact.Denition 9.1. (1) Let : T [0, ) and let MP (), where

MP () = {| is a Borel regular probability measure on ( x ) > 0 for any xT }.

Then for = ( , ), we dene

D = u uL2( , ),

xT

(x)2( x ) y,z S (x )

( y )( z )((u),y (u),z )2 < + ,where ( u),x = ( x ) 1 x ud, andQ (u, v ) =

xT

(x)2( x ) y,z S (x )

( y )( z )((u),y (u),z )(( v),y (v),z )for any u, v

D .

(2) Dene

E x, = f f =yS (x )

ay y , f (y)(dy) = 0 .Comparing with (4.4), it is apparent that Q is a generalization of E . Weare going to show that ( Q , D ) has properties which are analogous to ( E , F ).

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To obtain an alternative expression of Q by using an integral kernel, wedene a transformation and .Denition 9.4. For any x T , dene [x]n = xn for n = 1 , 2, . . . , M , whereM = |x| and (x0 , x1 , . . . , x M 1 , xM ) is the geodesic between and x. (Hencex0 = and xM = x.) Let be a Borel regular probability measure on . Denea linear map : (T ) (T ) and : (T ) (T ) by

( ())( x) = 12

([x]0) +|x | 1

m =0

([x]m +1 ) ([x]m )( [x ]m +1 ) ) (9.3)

for any (T ) and any xT , and

( (J ))( x) = 2 J (x)( x ) + 2

|x | 1

m =0J ([x]m )((

[x ]m ) ( [x ]m +1 ) (9.4)for any J (T ) and any xT .

Simple calculation shows that is the inverse of .

Lemma 9.5. = = Identity on (T ).Denition 9.6. For J : T [0, ), dene LJ : ( ) \ [0, ) byLJ (, ) = J ([, ]) for any , with = ,

DJ, = u uL2( , ), 2 LJ (, )(u() u( )) 2(d)(d ) < + and, for any u, v DJ, ,

QJ, (u, v) = 2 LJ (, )(u() u( ))( v() v( ))(d)(d ).Theorem 9.7. Let : T [0, ) and let MP () . Write = ( , ). Then (Q , D ) is a regular Dirichlet from on L2( , ) if and only if ( ())( x) 0 for any x T . Moreover, assume that ( ())( x) 0 for any x T . Then D = D ( ) , and Q (u, v) = Q ( ) , (u, v ) for any u, v D .Remark. Dene + (T ) = {u|u : T [0, )}. Then ()(x) 0 for any xT if and only if (+ (T )).

Before proving Theorem 9.7, we state an immediate corollary which followsby (9.4) and Lemma 9.5.

Corollary 9.8. Let J : T [0, ) and let M p() . Then (QJ, , DJ, ) is a regular Dirichlet from on L2( , ). Moreover, (QJ, , DJ, ) = ( Q , D ), where = ( (J ), ).Next two lemmas are needed to prove Theorem 9.7.

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Lemma 9.9. Let J : T [0, ), let MP () and let x T . Then for any E x, and any u L

2

( , ), LJ (, )(() ( ))( u() u( )) is -integrable on and

LJ (, )(() ( ))( u() u( ))(d)(d ) = x (, u) , (9.5)where x = ( (J ))( x) and (, u) = ()u()(d).Proof. Dene Y x = y,z S (x ) ,y = z y z . Then LJ = xT J (x) Y x . LetK u,v (, ) = LJ (, )(u() u( ))( v() v( ) and let = yS (x ) ay y .Since ()(d) = 0, it follows that yS (x ) ay ( y ) = 0. We devide{(, )|() = ( )}into three regions Y x , x ( \ x ) and ( \ x ) x . Forthe rst part, since L J (, ) = J (x) on Y x , K ,u (, ) is integrable on Y x and

Y x K ,u (, )=y,z S (x )

(ay az ) ( z ) y u()(d) ( y ) z u( )(d )= 2 ( x )J (x)(, u) .

For the second region, let U x,m = [x ]m \ [x ]m +1 . Then \ x = |x | 1m =0 U x,m .

Note that LJ (, ) = J ([x]m ) on x U x,m . Hence K ,u (, ) is integrable oneach x U x,m . Now

x U x,m

K ,u (, )(d)(d ) =

x U x,m

J ([x]m )()u()(d)( )

= J ([x]m )(( [x ]m ) ( [x ]m +1 )(, u) .Hence

x \ x K ,u (, )(d)(d ) = |x | 1m =0 J ([x]m )(( [x ]m ) ( [x ]m +1 ))(, u) .The third part is the same as the second part. As a whole, we obtain (9.5).

Lemma 9.10. Let J : T [0, ) and let MP () . Then (QJ, , DJ, ) is a regular Dirichlet form on L2( , ). Moreover,

DJ, = a00 + xT M (x )

n =1ax,n x,n (a0)2 +

xT (1 + x )

M (x )

n =1(ax,n )2 < + ,

(9.6)where x = ( (J ))( x), and

QJ, (u, v ) =xT

xM (x )

n =1ax,n bx,n (9.7)

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for any u = a00 +

xT

M (x )n =1 ax,n x,n , v = b00 +

xT

M (x )n =1 bx,n x,n

DJ, .Proof. By [5, Example 1.2.4], it follows that ( QJ, , DJ, ) is a Dirichlet form.Note that {0 ,x,n |xT, n = 1 , . . . ,