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Resistance forms, quasisymmetric maps and heat

kernel estimates

Jun Kigami

Author address:

Graduate School of Informatics, Kyoto University, Kyoto 606-8501, Japan

E-mail address: kigami@i.kyoto-u.ac.jp

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Contents

1. Introduction 1

Part 1. Resistance forms and heat kernels 72. Topology associated with a subspace of functions 83. Basics on resistance forms 10

4. the Green function 145. Topologies associated with resistance forms 176. Regularity of resistance forms 217. Annulus comparable condition and local property 228. Trace of resistance form 259. Resistance forms as Dirichlet forms 2810. Transition density 30

Part 2. Quasisymmetric metrics and volume doubling measures 3911. Semi-quasisymmetric metrics 4012. Quasisymmetric metrics 4313. Relations of measures and metrics 4514. Construction of quasisymmetric metrics 50

Part 3. Volume doubling measures and heat kernel estimates 5515. Main results on heat kernel estimates 5616. Example: the -stable process on R 6117. Basic tools in heat kernel estimates 6418. Proof of Theorem 15.6 6819. Proof of Theorems 15.10, 15.11 and 15.13 71

Part 4. Random Sierpinski gaskets 7520. Generalized Sierpinski gasket 7621. Random Sierpinski gasket 8122. Resistance forms on Random Sierpinski gaskets 83

23. Volume doubling property 8724. Homogeneous case 9225. Introducing randomness 97

Bibliography 99Assumptions, Conditions and Properties in Parentheses 101List of Notations 102Index 104

v

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Abstract

Assume that there is some analytic structure, a differential equation or astochastic process for example, on a metric space. To describe asymptotic behaviorsof analytic objects, the original metric of the space may not be the best one. Everynow and then one can construct a better metric which is somehow intrinsic withrespect to the analytic structure and under which asymptotic behaviors of theanalytic objects have nice expressions. The problem is when and how one can findsuch a metric.

In this paper, we consider the above problem in the case of stochastic processesassociated with Dirichlet forms derived from resistance forms. Our main concernsare following two problems:(I) When and how can we find a metric which is suitable for describing asymptoticbehaviors of the heat kernels associated with such processes?(II) What kind of requirement for jumps of a process is necessary to ensure goodasymptotic behaviors of the heat kernels associated with such processes?

Note that in general stochastic processes associated with Dirichlet forms havejumps, i. e. paths of such processes are not continuous.

The answer to (I) is for measures to have volume doubling property with respectto the resistance metric associated with a resistance form. Under volume doubling

property, a new metric which is quasisymmetric with respect to the resistancemetric is constructed and the Li-Yau type diagonal sub-Gaussian estimate of theheat kernel associated with the process using the new metric is shown.

About the question (II), we will propose a condition called annulus comparablecondition, (ACC) for short. This condition is shown to be equivalent to the existenceof a good diagonal heat kernel estimate.

As an application, asymptotic behaviors of the traces of 1-dimensional -stableprocesses are obtained.

In the course of discussion, considerable numbers of pages are spent on thetheory of resistance forms and quasisymmetric maps.

1991 Mathematics Subject Classification. Primary 30L10, 31E05, 60J35; Secondary 28A80,43A99, 60G52.

Key words and phrases. resistance form, Green function, quasisymmetric map, volume dou-bling property, jump process, heat kernel.

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1. INTRODUCTION 1

1. Introduction

Originally, the main purpose of this paper is to give answers to the following twoquestions on heat kernels associated with Dirichlet forms derived from resistanceforms. Such Dirichlet forms roughly correspond to Hunt processes for which everypoint has positive capacity.

(I) When and how can we find metrics which are suitable for describing asymptoticbehaviors of heat kernels?

(II) What kind of requirement for jumps of processes and/or Dirichlet forms isnecessary to ensure good asymptotic behaviors of associated heat kernels?

Eventually we are going to make these questions more precise. For the moment,let us explain what a heat kernel is. Assume that we have a regular Dirichlet form(E, D) on L2(X, ), where X is a metric space, is a Borel regular measure on X, Eis a nonnegative closed symmetric form on L2(X, ) and

Dis the domain of

E. Let

L be the Laplacian associated with this Dirichlet form, i.e. Lv is characterizedby the unique element in L2(X, ) which satisfies

E(u, v) = X

u(Lv)d

for any u F. A nonnegative measurable function p(t,x,y) on (0, +) X2 iscalled a heat kernel associated with the Dirichlet form (E, D) on L2(X, ) if

u(t, x) =

X

p(t,x,y)u(y)(dy)

for any (t,x,y) (0, +) X2 and any initial value u L2(X, ), where u(t, x) isthe solution of the heat equation associated with the Laplacian L:

u

t = Lu.

The heat kernel may not exist in general. However, it is know to exist in manycases like the Brownian motions on Euclidean spaces, Riemannian manifolds andcertain classes of fractals.

If the Dirichlet form (E, D) has the local property, in other words, the corre-sponding stochastic process is a diffusion, then one of the preferable goals on anasymptotic estimate of a heat kernel is to show the so-called Li-Yau type (sub-)Gaussian estimate, which is

(1.1) p(t,x,y) c1Vd(x, t1/)

exp

c2

d(x, y)

t

1/(1),

where d is a metric on X, Vd(x, r) is the volume of a ball Bd(x, r) ={

y|d(x, y) < r

}and 2 is a constant. It is well-known that the heat kernel of the Brownianmotion on Rn is Gaussian which is a special case of (1.1) with d(x, y) = |xy|, = 2and Vd(x, r) = r

n. Li and Yau have shown in [42] that, for a complete Riemannianmanifold with non-negative Ricci curvature, (1.1) holds with = 2, where d is thegeodesic metric and Vd(x, r) is the Riemannian volume. In this case, (1.1) is calledthe Li-Yau type Gaussian estimate. Note that Vd(x, t1

/) may have inhomogeneitywith respect to x in this case. For fractals, Barlow and Perkins have shown in [9]that the Brownian motion on the Sierpinski gasket satisfies sub-Gaussian estimate,that is, (1.1) with d(x, y) = |x y|, = log5/ log2 and Vd(x, r) = r, where

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2

= log3/ log 2 is the Hausdorff dimension of the Sierpinski gasket. Note thatVd(x, r) is homogeneous in this particular case. Full generality of (1.1) is realized,

for example, by a certain time change of the Brownian motion on [0, 1], whose heatkernel satisfies (1.1) with > 2 and inhomogeneous Vd(x, r). See [38] for details.

There have been extensive studies on the conditions which are equivalent to(1.1). For Riemannian manifolds, Grigoryan [23] and Saloff-Coste [47] have in-dependently shown that the Li-Yau type Gaussian estimate is equivalent to thePoincare inequality and the volume doubling property. For random walks onweighted graphs, Grigoryan and Telcs have obtained several equivalent conditionsfor general Li-Yau type sub-Gaussian estimate, for example, the combination ofthe volume doubling property, the elliptic Harnack inequality and the Poincare in-equality in [25, 26]. Similar results have been obtained for diffusions. See [31] and[10] for example.

The importance of the Li-Yau type (sub-)Gaussian estimate (1.1) is that itdescribes asymptotic behaviors of an analytical object, namely, p(t,x,y) in termsof geometrical objects like the metric d and the volume of a ball Vd(x, r). Suchan interplay of analysis and geometry makes the study of heat kernels interesting.In this paper, we have resistance forms on the side of analysis and quasisymmetricmaps on the side of geometry. To establish a foundation in studying heat kernelestimates, we first need to do considerable works on both sides, i.e. resistance formsand quasisymmetric maps. Those two subjects come to the other main parts of thispaper as a consequence.

The theory of resistance forms has been developed to study analysis on low-dimensional fractals including the Sierpinski gasket, the 2-dimensional Sierpinskicarpet, random Sierpinski gaskets and so on. Roughly, a symmetric non-negativedefinite quadratic form Eon a subspace F of real-valued functions on a set X iscalled a resistance form on X if it has the Markov property and

min{E(u, u)|u F, u(x) = 1 and u(y) = 0}exists and is positive for any x = y X. The reciprocal of the above minimum,denoted by R(x, y), is known to be a metric (distance) and is called the resistancemetric associated with (E, F). See [36] for details. Note that unlike the Dirichletforms, a resistance form is defined without referring to any measure on the spaceand hence it is not necessarily a Dirichlet form as it is. In Part 1, we are going toestablish fundamental notions on resistance forms, for instance, the existence andproperties of the Green function with an infinite set as a boundary, regularity ofa resistance form, traces, construction of a Dirichlet form from a resistance form,the existence and continuity of heat kernels. Assume that (X, R) is compact forsimplicity and let be a Borel regular measure on (X, R). In Section 9, a regularresistance form (

E,F

) is shown to be a regular Dirichlet form on L2(X, ). Wealso prove that the associated heat kernel p(t,x,y) exists and is continuous on(0, +) X X in Section 10. Even if (X, R) is not compact, we are going toobtain a modified versions of those statements under mild assumptions. See Part Ifor details.

The notion of quasisymmetric maps has been introduced by Tukia and Vaisalain [51] as a generalization of quasiconformal mappings in the complex plane. Soonits importance has been recognized in wide areas of analysis and geometry. Therehave been many works on quasisymmetric maps since then. See Heinonen [32] andSemmes [48] for references. In this paper, we are going to modify the resistance

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1. INTRODUCTION 3

metric R quasisymmetrically to obtain a new metric which is more suitable fordescribing asymptotic behaviors of a heat kernel. The key of modification is to

realize the following relation:(1.2) Resistance Volume (Distance),where Volume is the volume of a ball and Distance is the distance with respectto the new metric. With (1.5), we are going to show that the mean exit time froma ball is comparable with r and this fact will lead us to Li-Yau type on-diagonalestimate of the heat kernel described in (1.6). Quasisymmetric modification ofa metric has many advantages. For example, it preserves the volume doublingproperty of a measure. In Part 2, we will study quasisymmetric homeomorphisms ona metric space. In particular, we are going to establish relations between propertiessuch as (1.2) concerning the original metric D, the quasisymmetrically modifiedmetric d and the volume of a ball Vd(x, r) = (Bd(x, r)) and show how to constructa metric d which is quasisymmetric to the original metric D and satisfy a desiredproperty like (1.2).

Let us return to question (I). We will confine ourselves to the case of diffusionprocesses for simplicity. The lower part of the Li-Yau type (sub-)Gaussian estimate(1.1) is known to hold only when the distance is geodesic, i.e. any two points areconnected by a geodesic curve. This is not the case for most of metric spaces. So,we use an adequate substitute called near diagonal lower estimate, (NDL),d for

short. We say that (NDL),d holds if and only if

(1.3)c3

Vd(x, t1/) p(t,x,y)

for any x, y which satisfy d(x, y) c4t. For upper estimates, the Li-Yau type(sub-)Gaussian upper estimate of order , (LYU),d for short, is said to hold if and

only if

(1.4) p(t,x,y) c5Vd(x, t1/)

exp

c6

d(x, y)

t

1/(1).

Another important property is the doubling property of a heat kernel, (KD) forshort, that is,

(1.5) p(t,x,x) c7p(2t,x,x).Note that p(t,x,x) is monotonically decreasing with respect to t. It is easy to seethat the Li-Yau type (sub-)Gaussian heat kernel estimate together with the volumedoubling property implies (KD). Let p(t,x,y) be the heat kernel associated with adiffusion process. Now, the question (I) can be rephrased as follows:

Question When and how can we find a metric d under which p(t,x,y) satisfies(LYU),d , (NDL),d and (KD)?

In Corollary 15.12, we are going to answer this if the Dirichlet form associatedwith the diffusion process is derived from a resistance form. Roughly speaking, weobtain the following statement.

Answer The underlying measure has the volume doubling property with respectto the resistance metric R if and only if there exist > 1 and a metric d which isquasisymmetric with respect to R such that (LYU),d , (NDL),d and (KD) hold.

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4

Of course, one can ask the same question for general diffusion process with a heatkernel. Such a problem is very interesting. In this paper, however, we only consider

the case where the process is associated with a Dirichlet form induced by a resistanceform.

Next, we are going to explain the second problem, the question (II). Recently,there have been many results on asymptotic behaviors of a heat kernel associatedwith a jump process. See [12, 15, 16, 5] for example. They have dealt with aspecific class of jump processes and studied a set of conditions which is equivalentto certain kind of (off-diagonal) heat kernel estimate. For example, in [15], theyhave shown the existence of jointly continuous heat kernel for a generalization of-stable process on an Ahlfors regular set and given a condition for best possibleoff-diagonal heat kernel estimate. In this paper, we will only consider the followingLi-Yau type on-diagonal estimate, (LYD),d for short,

(1.6) p(t,x,x)

1

Vd(x, t1/)which is the diagonal part of (1.1). Our question is

Question When and how can we find a metric d with (LYD),d for a given (jump)process which possesses a heat kernel?

In this case, the when part of the question includes the study of the requirementon jumps. In this paper, again we confine ourselves to the case where Dirichletforms are derived from resistance forms. Our proposal for a condition on jumps isthe annulus comparable condition, (ACC) for short, which says that the resistancebetween a point and the complement of a ball is comparable with the resistancebetween a point and an annulus. More exactly, (ACC) is formulated as

(1.7) R(x, BR(x, r)c)

R(x, AR(x,r, (1 + )r))

on (x, r) X (0, +) for some > 0, where R is a resistance metric, BR(x, r)is a resistance ball and AR(x,r, (1 + )r) = BR(x, (1 + )r)\BR(x, r) is an annulus.(BR(x, s) is the closure of a ball BR(x, r) with respect to the resistance metric.) Ifthe process in question has no jump, i.e. is a diffusion process, then the quantitiesin the both sides of (1.7) coincide and hence (ACC) holds. As our answer to theabove question, we obtain the following statement in Theorem 15.11:

Theorem 1.1. The following three conditions are equivalent:(C1) The underlying measure has the volume doubling property with respect toR and (ACC) holds.(C2) The underlying measure has the volume doubling property with respect toR and the so-called Einstein relation:

Resistance Volume Mean exit timeholds for the resistance metric.(C3) (ACC) and (KD) is satisfied and there exist > 1 and a metric d which isquasisymmetric with respect to R such that (LYD),d holds.

See [26, 50] on the Einstein relation, which is known to be implied by theLi-Yau type (sub-)Gaussian heat kernel estimate.

Our work on heat kernel estimates is largely inspired by the previous two papers[6] and [41]. In [6], the strongly recurrent random walk on an infinite graph has

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1. INTRODUCTION 5

been studied by using two different metrics, one is the shortest path metric d onthe graph and the other is the resistance metric R. In [6], the condition R(), that

is,R() R(x, y)Vd(x, d(x, y)) d(x, y)has been shown to be essentially equivalent to the random walk version of (1.1).Note the resemblance between (1.2) and R(). The metric d is however fixed intheir case. In [41], Kumagai has studied the (strongly recurrent) diffusion processassociated with a resistance form using the resistance metric R. He has shownthat the uniform volume doubling property with respect to R is equivalent to thecombination of natural extensions of (LYU),d and (NDL),d with respect to R. Seethe remark after Theorem 15.10 for details. Examining those results carefully fromgeometrical view point, we have realized that quasisymmetric change of metrics(implicitly) plays an important role. In this respect, this paper can be though ofan extension of those works.

There is another closely related work. In [39], a problem which is very similarto our question (I) has been investigated for a heat kernel associated with a self-similar Dirichlet form on a self-similar set. The result in [39] is also quite similarto ours. It has been shown that the volume doubling property of the underlyingmeasure is equivalent to the existence of a metric with (LYD),d . Note that the

results in [39] include higher dimensional Sierpinski carpets where the self-similarDirichlet forms are not induced by resistance forms. The processes studied in [39],however, have been all diffusions

Finally, we present an application of our results to an -stable process on R for (1, 2]. Define

E()(u, u) =R2

(u(x) u(y))2|x y|1+ dxdy

and F() = {u|E()(u, u) < +} for (1, 2) and (E(2), F(2)) is the ordinaryDirichlet form associated with the Brownian motion on R. Then (E(), F()) is aresistance form for (1, 2] and the associated resistance metric is c|x y|1.If = 2, then the corresponding process is not a diffusion but has jumps. Letp()(t,x,y) be the associated heat kernel. (We will show the existence of the heatkernel p()(t,x,y) in Section 16.) It is well known that p()(t,x,x) = ct1/. Let(E()|K, F()|K) be the trace of (E(), F()) onto the ternary Cantor set K. Letp(

)K (t,x,y) be the heat kernel associated with the Dirichlet form on L

2(K, ) in-

duced by (E()|K, F()|K), where is the normalized Hausdorff measure of K. ByTheorem 15.13, we confirm that (ACC) holds and obtain

p()K (t,x,x) t,

where =log 2

(1)log3+log2 . See Section 16 for details.This paper consists of four parts. In Part 1, we will develop basic theory of

resistance forms regarding the Green function, trace of a form, regularity and heatkernels. This part is the foundation of the discussion in Part 3. Part 2 is devotedto studying quasisymmetric homeomorphisms. This is another foundation of thediscussion in Part 3. After preparing those basics, we will consider heat kernelestimates in Part 3. Finally in Part 4, we consider estimates of heat kernels onrandom Sierpinski gaskets as an application of the theorems in Part 3.

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6

The followings are conventions in notations in this paper.(1) Let f and g be functions with variables x1, . . . , xn. We use f g for any(x1, . . . , xn) A if and only if there exist positive constants c1 and c2 such that

c1f(x1, . . . , xn) g(x1, . . . , xn) c2f(x1, . . . , xn)for any (x1, . . . , xn) A.(2) The lower case c (with or without a subscript) represents a constant which isindependent of the variables in question and may have different values from placeto place (even in the same line).

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Part 1

Resistance forms and heat kernels

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In this part, we will establish basics of resistance forms such as the Greenfunction, harmonic functions, traces and heat kernels. In the previous papers [ 36,

34, 37], we have established the notions of the Green function, harmonic functionsand traces if a boundary is a finite set. One of the main subjects is to extend thoseresults to the case where a boundary is an infinite set. In fact, we should determinewhat kind of an infinite set can be regarded as a proper boundary in the first place.To do so, we introduce a new topology determined by the domain of a resistanceform and show that the closed set with respect to this new topology can be regardedas a boundary. Moreover, we will establish the existence of jointly continuous heatkernel associated with the Dirichlet form derived from a resistance form underseveral mild assumptions, which do not include the ultracontractivity. Recall thatif the transition semigroup associated with a Dirichlet form is ultracontractive, thenthere exists a (pointwise) integral kernel of the transition semigroup, which will givea heat kernel after necessary modifications. See [18, Lemma 2.1.2], [22, Lemma 3.2]and [5, Theorem 2.1] for details.

The followings are basic notations used in this paper.

Notation. (1) For a set V, we define (V) = {f|f : V R}. If V is afinite set, (V) is considered to be equipped with the standard inner product ( , )Vdefined by (u, v)V =

pV u(p)v(p) for any u, v (V). Also |u|V =

(u, u)V for

any u (V).(2) Let V be a finite set. The characteristic function VU of a subset U V isdefined by

VU(q) =

1 if q U,0 otherwise.

If no confusion can occur, we write U instead ofVU. IfU = {p} for a point p V,

we write p instead of {p}. If H : (V)

(V) is a linear map, then we setHpq = (Hq)(p) for p, q V. Then (Hf)(p) = qV Hpqf(q) for any f (V).(3) Let (X, d) be a metric space. Then

Bd(x, r) = {y|y X, d(x, y) < r}for x X and r > 0.

2. Topology associated with a subspace of functions

In this section, we will introduce an operation B BF from subsets of a spaceX to itself associated with a linear subspace Fof real valued functions (X). Thisoperation will turn out to be essential in describing whether a set can be treatedas a boundary or not. More precisely, we will show in Theorems 4.1 and 4.3 thata set B can be a proper boundary if and only if BF = B. Also the importance of

the condition that BF = B is revealed in Theorem 6.3 where we have equivalentconditions for regularity of resistance forms.

Definition 2.1. Let Fbe a linear subspace of (X) for a set X. For a subsetB X, define

F(B) = {u|u F, u(x) = 0 for any x B}.and

BF =

uF(B)

u1(0)

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2. TOPOLOGY ASSOCIATED WITH A SUBSPACE OF FUNCTIONS 9

The following lemma is immediate from the definition.

Lemma 2.2. LetF

be a linear subspace of (X) for a set X.(1) For any B X, B BF, F(B) = F(BF) and (BF)F = BF.(2) XF = X.(3) F = if and only if {u(x)|u F} = R for any x X.

The above lemma suggests that the operation B BF satisfies the axiom ofclosure and hence it defines a topology on X. Indeed, this is the case ifFis stableunder the unit contraction.

Definition 2.3. (1) For u : X R, define u : X [0, 1] by

u(p) =

1 if u(p) 1,u(p) if 0 < u(p) < 1,

0 if u(p)

0.

u is called the unit contraction of u.(2) Let Fbe a linear subspace of (X) for a set X. F is said to be stable underthe unit contraction if and only if u Ffor any u F.

In the case of Dirichlet forms, the condition that the domain is stable underthe unit contraction is one of the equivalent conditions of the Markov property. See[21, Section 1.1] for details. (In [21], their terminology is that the unit contractionoperates on F in place of that Fis stable under the unit contraction.)

Theorem 2.4. LetFbe a linear subspace of (X) for a set X. Assume that{u(x)|u F} = R for any x X and that F is stable under the unit contraction.Define

CF=

{B

|B

X, BF = B

}.

ThenCF satisfies the axiom of closed sets and it defines a topology of X. Moreover,the T1-axiom of separation holds for this topology, i.e. {x} is a closed set for anyx X, if and only if, for any x, y X with x = y, there exists u F such thatu(x) = u(y).

As a topology given by a family of real-valued function, the notion of finetopology has been introduced in classical axiomatic potential theory by Brelot [13].In the case of resistance forms, we will see that the topology given by CF coincideswith the fine topology associated with the cone of nonnegative functions in thedomain of resistance form in Theorem 5.7. Our proof of Theorem 5.7, however,depends essentially on Theorem 4.3, where the condition BF = B has alreadyplayed a crucial role. So the coincidence of

CR and the fine topology does give us

small help in studying resistance forms. See the comments after Theorem 5.7 fordetails.

The rest of this section is devoted to proving the above theorem.

Lemma 2.5. Under the assumptions of Theorem 2.4, if B CF and x X\B,then there exists u F such that u F(B), u(x) = 1 and 0 u(y) 1 for anyy X.

Proof. Since BF = B, there exists v F(B) such that v(x) = 0. Letu = v/v(x). Then u satisfies the required properties.

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10

Proof of Theorem 2.4. First we show that CF satisfies the axiom of closedsets. Since F(X) = {0}, XF = X. Also we have F = by Lemma 2.2-(3). LetBi CF for i = 1, 2 and let x (B1 B2)

c

, where Ac

is the complement of A inX, i.e. Ac = X\A. By Lemma 2.5, there exists ui F(Bi) such that ui = ui andui(x) = 1. Let v = u1 + u2 1. Then v(x) = 1 and v(y) 0 for any y B1 B2.If u = v, then u F(B1 B2) and u(x) = 1. Hence B1 B2 CF. Let B CFfor any . Set B = B. If x / B, then there exists such thatx / B . We have u F(B) F(B) satisfying u(x) = 0. Hence x / BF. Thisshows B CF. Thus CF satisfies the axiom of closed sets.

Next define Ux,y = f(x)

f(y)

f F. We will show that Ux,y = R2 if thereexists u Fsuch that u(x) = u(y). Suppose that u(x) = 0. Considering u/u(x),we see that

1a

Ux,y , where a = 1. Since there exists v F with v(y) = 0, itfollows that

b1

Ux,y for some b R. Now we have five cases.Case 1: Assume that a

0. Considering the operation u

u, we have

10 Ux,y .Also b1 Ux,y. Since Ux,y is a linear subspace ofR2, Ux,y = R2.

Case 2: Assume that b 0. By similar arguments as Case 1, we have Ux,y = R2.Case 3: Assume that b 1. The u-operation shows that 11 Ux,y. Since

11

,

1a

is independent, Ux,y = R

2.

Case 4: Assume that a (0, 1) and b (0, 1). Then

1a

,b1

is independent.

Hence Ux,y = R2.

Case 5: Assume that a > 1 and b (0, 1). The u-operation shows 11 Ux,y .Then

11

,b1

is independent and hence Ux,y = R2.

Thus Ux,y = R2 in all the cases. Exchanging x and y, we also deduce the sameconclusion even if u(x) = 0. In particular, the fact that Ux,y = R2 implies that

y / {x}F

. Hence if there exists u Fsuch that u(x) = u(y) for any x, y X withx = y, then {x} CF for any x X. The converse direction is immediate.

3. Basics on resistance forms

In this section, we first introduce definition and basics on resistance forms.

Definition 3.1 (Resistance form). Let X be a set. A pair (E, F) is called aresistance form on X if it satisfies the following conditions (RF1) through (RF5).(RF1) Fis a linear subspace of (X) containing constants and Eis a non-negativesymmetric quadratic form on F. E(u, u) = 0 if and only if u is constant on X.(RF2) Let be an equivalent relation on Fdefined by u v if and only if u vis constant on X. Then (F/, E) is a Hilbert space.(RF3) Ifx

= y, then there exists u

Fsuch that u(x)

= u(y).

(RF4) For any p, q X,sup

|u(p) u(q)|2E(u, u)

u F, E(u, u) > 0is finite. The above supremum is denoted by R(E,F)(p, q).(RF5) u F and E(u, u) E(u, u) for any u F, where u is defined in Defini-tion 2.3.

Note that the definition of resistance forms does not require any measure onthe space X at all. Being combined with a measure , a resistance form may induce

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3. BASICS ON RESISTANCE FORMS 11

a Dirichlet form on L2(X, ) and the associated process, semigroup and Laplacian.See Section 9.

By (RF3) and (RF5) along with Theorem 2.4, the axiom of closed sets holdsfor CF and the associated topology satisfies the T1-separation axiom.

Proposition 3.2. Assume that u F for any u F. Then (RF3) in theabove definition is equivalent to the following conditions:(RF3-1) FF = F for any finite subset F X.(RF3-2) For any finite subset F X and any v (F), there exists u Fsuchthat u|F = v.

Proof. (RF3) (RF3-1) By Theorem 2.4, (RF3) implies that {x}F = {x}for any x X. Let F be a finite subset of X. Again by Theorem 2.4, FF =(xF{x})F = xF{x}F = F.(RF3-1) (RF3-2) Let F be a finite subset of X. Set Fx = F\{x} for x F.Since (Fx)

F = Fx, there exists ux F

such that ux|Fx

0 and ux(x) = 1. For any

v (F), define u = xF v(x)ux. Then u|F = v and u F.(RF3-2) (RF3) This is obvious.

Remark. In the previous literatures [36, 34, 37], (RF3-2) was employed as apart of the definition of resistance forms in place of the current (RF3).

By the results in [36, Chapter 2], we have the following fact.

Proposition 3.3. Let (E, F) be a resistance form on X. The supremum in(RF4) is the maximum and R(E,F) is a metric on X.

Definition 3.4. Let (E, F) be a resistance form on X. R(E,F) is called theresistance metric on X associated with the resistance form (E, F).

One of the most simple examples of resistance forms is weighted graphs.

Example 3.5 (Weighted graph). Let V be an (infinite) countable set and letH = {Hxy}x,yX satisfy the following three conditions (WG1), (WG2) and (WG3).

(WG1) Hxy = Hyx 0 and Hxx = 0 for any x, y X.(WG2) N(x) = {y|Hxy > 0} is a finite set for any x X.(WG3) For any x, y X, there exist x1, . . . , xn X such that x1 = x, xn = yand Hxixi+1 > 0 for any i = 1, . . . , n 1.Then (V, H) is called a (locally finite irreducible) weighted graph. (The condi-tion (WG3) is called the locally finiteness and the condition (WG4) is called theirreducibility.) Define

F(V,H) = uu (V), x,yV Hxy(u(x) u(y))

2 < +

and, for u, v F(V,H),

E(V,H)(u, v) = 12

x,yV

Hxy(u(x) u(y))(v(x) v(y)).

Then (E(V,H), F(V,H)) is a resistance form on V. There exists a random walk on Vassociated with the weight graph (V, H). Namely define x =

yN(x) Hxy and

P(x, y) = Hxy/x. We give the transition probability from x to y in the unit timeby P(x, y). This random walk is called the random walk associated with (V, H).

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Relations between (E(V,H), F(V,H)) and the random walk associated with (V, H)have been one of the main subjects in the theory of random walks and discrete

potential theory. See [52, 53], [43] and [49] for example. In particular, as wementioned in the introduction, asymptotic behaviors of the heat kernel associatedwith the random walk has been studied in [6].

Note that in the above example, the set V is countable. This is not the case forgeneral resistance forms. For example, we have the resistance forms on R associatedwith -stable process defined in the introduction and the resistance forms on the(random) Sierpinski gaskets in Part 4.

If no confusion can occur, we use R to denote R(E,F). By (RF4), we immediatelyobtain the following fact.

Proposition 3.6. Let (E, F) be a resistance form on X and let R be the as-sociated resistance metric. For any x, y X and any u F,

(3.1) |u(x) u(y)|2

R(x, y)E(u, u).In particular, u Fis continuous with respect to the resistance metric.

Next we introduce the notion of Laplacians on a finite set and harmonic func-tions with a finite set as a boundary. See [36, Section 2.1] for details, in particular,the proofs of Proposition 3.8 and 3.10.

Definition 3.7. Let V be a non-empty finite set. Recall that (V) is equippedwith the standard inner-product (, )V . A symmetric linear operator L : (V) (V) is called a Laplacian on V if it satisfies the following three conditions:

(L1) L is non-positive definite,(L2) Lu = 0 if and only if u is a constant on V,(L3) Lpq 0 for all p = q V.

We use LA(V) to denote the collection of Laplacians on V.By [36, Proposition 2.1.3], we have the next proposition, which says that a

resistance form on a finite set corresponds to a Laplacian.

Proposition 3.8. Let V be a non-empty finite set and let L be a symmetriclinear operator form (V) to itself. Define a symmetric bilinear form EL on (V)by EL(u, v) = (u,Lv)V for any u, v (V). Then, EL is a resistance form on Vif and only if L LA(V).

Using the standard inner-product (, )V , we implicitly choose the uniform dis-tribution, i.e. the sum of all the Dirac masses on the space, as our measure on V.This is why we may relate Laplacians with resistance forms on a finite set withoutmentioning any measure explicitly. (Recall the comment after Definition 3.1.)

Definition 3.9. Let V be a finite set and let L LA(V). The resistance form(EL, (V)) on V is called the resistance form associated with L.

The harmonic function with a finite set as a boundary is defined as the energyminimizing function.

Proposition 3.10. [36, Lemma 2.3.5] Let (E, F) be a resistance form on Xand let V be a finite subset of X. Let (V). Then there exists a unique u Fsuch that u|V = and u attains the following minimum:

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

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3. BASICS ON RESISTANCE FORMS 13

Moreover, the map from to u is a linear map from (V) to F. Denote this mapby hV. Then there exists a Laplacian L LA(V) such that(3.2) EL(, ) = E(hV(), hV()).

In Lemma 8.2, we are going to extend this proposition when V is an infiniteset.

Definition 3.11. hV() defined in Proposition 3.10 is called the V-harmonicfunction with the boundary value . Also we denote the above L LA(V) in (3.2)by L(E,F),V.

To construct concrete examples of resistance forms, we often start from a se-quence of resistance forms on finite sets which has certain compatibility as follows.

Definition 3.12. Let {Vm}m0 be a sequence of finite sets and let Lm LA(Vm) for m 0. {(Vm, Lm)}m0 is called a compatible sequence if and only ifL(ELm+1 ,(Vm+1)),Vm = Lm for any m 0, i.e.

ELm(u, u) = min{ELm+1(v, v)|v (Vm+1), u = v|Vm}for any u (Vm).

Combining the results in [36, Sections 2.2 and 2.3], in particular [36, Theo-rems 2.2.6 and 2.3.10], we obtain the following theorem on construction of a resis-tance form from a compatible sequence.

Theorem 3.13. Let{Vm}m0 be a sequence of finite sets and letLm LA(Vm)for m 0. Assume thatS = {(Vm, Lm)}m0 is a compatible sequence. Let V =m0Vm. Define

FS = {u|u (V), limm+

ELm(u|Vm , u|Vm) < +}and

ES(u, v) = limm+

ELm(u|Vm , v|Vm)for any u, v FS. Then(ES, FS) is a resistance form on V and L(ES,FS),Vm =Lm for any m 0. Moreover, let RS be the resistance metric associated with(ES, FS) and let (X, R) be the completion of (V, RS). Then there exists a uniqueresistance form (E, F) on X such that, for any u F, u is a continuous functionon X, u|V FS and E(u, u) = ES(u|V , u|V). In particular, R coincides with theresistance metric associated with (E, F).

We are going to use this theorem to construct resistance forms in Example 5.5and Section 22.

In contrast with the above theorem, if (X, R) is separable, the resistance form

(E, F) is always expressed as a limit of a compatible sequence. More precisely, thefollowing fact has been shown in [36, Section 2.3].

Theorem 3.14. Let (E, F) be a resistance form on a set X and let R be theassociated resistance metric. Let{Vm}m1 be an increasing sequence of finite sub-sets of X. Assume that V = m1Vm is dense in X. Set Lm = L(E,F),Vm whereL(E,F),Vm is defined in Definition 3.11. Then {(Vm, Lm)}m1 is a compatible se-quence. Moreover,

F= {u|u C(X, R), limm+

ELm(u|Vm , u|Vm) < +},

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where C(X, R) is the collection of real-valued continuous functions with respect tothe resistance metric, and

E(u, v) = limm+

ELm(u|Vm , v|Vm)for any u, v F.

By using this theorem, we have the following fact, which is used in Section 5.

Proposition 3.15. Let (E, F) be a resistance form on a set X and let R bethe resistance metric. Assume that (X, R) is separable. For u : X R, define u+ :R [0, +) by u+(x) = max{u(x), 0}. Then u+ FandE(u+, u+) E(u, u) forany u F.

Proof. Since (X, R) is separable, we may choose a sequence {Vm}m0 of sub-sets of X so that Vm Vm+1 for any m 0 and V = m0Vm is dense in (X, R).Let Lm as in Theorem 3.14. Then

ELm(u, u) = x,yVm(Lm)xy(u(x) u(y))

2/2

for any u (Vm). Since (u+(x) u+(y))2 (u(x) u(y))2, it follows thatELm(u+|Vm , u+|Vm) ELm(u|Vm.u|Vm). Hence by Theorem 3.14, we see that u+ Fand E(u+, u+) E(u, u).

4. the Green function

In this section, we study the Green function associated with an infinite set asa boundary. In the course of discussion, we will show that a set B is a suitableboundary if and only if BF = B. Conditions ensuring BF = B are given in thenext section. For example, if B is compact with respect to the resistance metric,then BF will be shown to coincide with B.

Throughout this section, (E, F) is a resistance form on a set X and R is theassociated resistance metric. The next theorem establishes the existence and basic

properties of the Green function with an infinite set as a boundary.

Theorem 4.1. Let B X be non-empty. Then (E, F(B)) is a Hilbert spaceand there exists a unique gB : X X R that satisfies the following condition(GF1):(GF1) Define gxB(y) = gB(x, y). For any x X, gxB F(B) and E(gxB, u) = u(x)for any u F(B).

Moreover, gB satisfies the following properties (GF2), (GF3) and (GF4):(GF2) gB(x, x) gB(x, y) = gB(y, x) 0 for any x, y X. gB(x, x) > 0 if andonly if x / BF.(GF3) Define R(x, B) = gB(x, x) for any x X. If x / BF, then

R(x, B) =

min{E(u, u)|u F(B), u(x) = 1}

1

= sup |u(x)|2E(u, u) u F(B), u(x) = 0.(GF4) For any x,y,z X, |gB(x, y) gB(x, z)| R(y, z).

By (GF2), if B = BF, then gxB 0 for any x BF\B. Such a set B is not agood boundary.

We will prove this and the next theorem at the same time.

Definition 4.2. The function gB(, ) given in the above theorem is called theGreen function associated with the boundary B or the B-Green function.

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4. THE GREEN FUNCTION 15

As we have remarked after Definition 3.1, no measure is required to define aresistance form. Thus the definition of the Green function is also independent of a

choice of measures. After introducing a measure and constructing the Dirichletform on L2(X, ) induced by a resistance form in Section 8, we may observe that theGreen function gB(x, y) defined above coincides with the Green function associatedwith the Dirichlet form. See Theorem 10.10 and Corollary 10.11 for details. Forexample, gB(x, y) is shown to be the integral kernel of the nonnegative self-adjointoperator associated with the Dirichlet form.

The next theorem assures another advantage of being B = BF. Namely, ifB = BF, we may reduce B to a one point, consider the shorted resistanceform (E, FB) and obtain a expression of the Green function (4.1) by the shortedresistance metric RB(, ). In the case of weighted graphs, such an identification ofa subset of domain is called Rayleighs shorting method. See [19] for details.

Theorem 4.3. Let (

E,

F) be a resistance form and let B

X be non-empty.

Suppose that BF = B. SetFB = {u|u F, u is a constant on B}

andXB = {B}(X\B). Then(E, FB) is a resistance form on XB. Furthermore,if RB(, ) is the resistance metric associated with (E, FB), then

(4.1) gB(x, y) =RB(x, B) + RB(y, B) RB(x, y)

2

for any x, y X. In particular, R(x, B) = RB(x, B) for any x X\B.Remark. In [45, Section 3], V. Metz has shown (4.1) in the case where B is a

one point.

The proofs of the those two theorems are divided into several parts.

Note that by Proposition 5.1 B is closed with respect to R if BF

= B.

Proof of the first half of Theorem 4.1. Let x B and let F(x) =F({x}). By (RF2), (E, F(x)) is a Hilbert space. Note that F(B) F(x). If {um}m1 is a Cauchy sequence in F(B), there exists the limit u F(x). Fory B,

|um(y) u(y)|2 R(x, y)E(um u, um u).Letting m +, we see that u(y) = 0. Hence u F(B). This shows that(E, F(B)) is a Hilbert space. For any z X and any u F(B), |u(z)|2 R(x, y)E(u, u). The map u u(z) is continuous linear functional and hence thereexists a unique z F(B) such that E(z, u) = u(z) for any u F(B). De-fine gB(z, w) = z(w). Since E(z, w) = z(w) = w(z), we have (GF1) andgB(z, w) = gB(w, z). If z

BF, then u(z) = 0 for any u

F(B). Hence

gB(z, z) = gzB(z) = 0. Conversely, assume gB(z, z) = 0. Since gB(z, z) = E(gzB, gzB),(RF1) implies that gzB is constant on X. On the other hand, g

zB(y) = 0 for any

y B. Hence gzB 0. For any u F(B), u(z) = E(gzB, u) = 0. Therefore,z BF.

Definition 4.4. Let B X be non-empty. If x / BF, we define xB =gxB/gB(x, x).

Note that gB(x, x) > 0 if and only if x / BF by the above proof. Hence xB iswell-defined.

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Lemma 4.5. LetB X be non-empty and let x / BF. ThenxB is the uniqueelement which attains the following minimum:

min{E(u, u)|u F(B), u(x) = 1}.In particular, (GF3) holds.

Proof. Let u F(B) with u(x) = 1. Since

E(u xB, xB) =E(u xB, gxB)

gB(x, x)=

(u(x) 1)gB(x, x)

= 0,

we haveE(u, u) = E(u xB, u xB) + E(xB, xB)

Hence E(u, u) E(xB, xB) and if the equality holds, then u = xB. Now,

E(xB, xB) =E(gxB, gxB)gB(x, x)2

=1

gB(x, x).

This suffices for (GB3).

Lemma 4.6. Let B X be non-empty. Then gB(x, x) gB(x, y) 0 for anyx, y X.

Proof. If x BF, then gxB 0. Otherwise, define v = xB. Then by (RF5),E(xB, xB) E(v, v). The above lemma shows that Bx = v. Hence 0 xB 1.

Combining Lemma 4.6 and the results from Proof of the first half of Theo-rem 4.1, we have (GF2).

So far, we have obtained (GF1), (GF2) and (GF3). Before showing (GF4), weprove Theorem 4.3.

Proof of Theorem 4.3. (RF1), (RF2) and (RF5) are immediate by the def-

inition ofFB

. To show (RF3), let x and y X with x = y. We may assume y = Bwithout loss of generality. Set Bx = B {x}. Since (Bx)F = Bx, there existsu F(Bx) such that u(y) = 0. Hence we obtain (RF3). To see (RF4), note that

sup

|u(x) u(y)|2E(u, u)

u FB, E(u, u) > 0 R(E,F)(x, y)because FB F. Hence we have (RF4). To prove (4.1), it is enough to show thecase where B is a one point. Namely we will show that

(4.2) g{z}(x, y) =R(x, z) + R(y, z) R(x, y)

2

for any x,y,z X. We write g(x, y) = g{z}(x, y). The definition of R(, ) alongwith Lemma 4.5 shows that g(x, x) = R(x, z). Also by Lemma 4.5, if u(y) =g(x, y)/g(x, x), then u is the

{x, z

}-harmonic function whose boundary values are

u(z) = 0 and u(x) = 1. Let V = {x,y,z}. Then by Proposition 3.10, there existsa Laplacian L LA(V) with (3.2). Note that

EL(u|V , u|V) = min{EL(v, v)|v (V), v(x) = 1, v(z) = 0}.Therefore, (Lu)(y) = 0. Set L = (Lpq)p,qV . Hereafter we assume that Lpq > 0for any p, q V with p = q. (If this condition fails, the proof is easier.) LetRpq = (Lpq)

1. Solving Lu(y) = 0, we have

(4.3) u(y) =Lxy

Lxy + Lyz=

RyzRxy + Ryz

.

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5. TOPOLOGIES ASSOCIATED WITH RESISTANCE FORMS 17

Define Rx = RxyRxz/R, Ry = RyxRyz/R and Rz = RzxRzy/R, where R =Rxy + Ryz + Rzx. By the -Y transform, R(p, q) = Rp + Rq for any p and q with

p = q. (See [36, Lemma 2.1.15] for -Y transform.) Hence

(4.4)R(x, z) + R(y, z) R(x, y)

2= Rz

Since g(x, x) = R(x, z), (4.3) implies

g(x, y) = g(x, x)u(y) = R(x, z)u(y) =Rxz(Rxy + Ryz)

Ru(y) = Rz.

By (4.4), we have (4.2).

Proof of (GF4) of Theorem 4.1. Let K = BF. Note that gB(x, y) =gK(x, y). By (4.1),

|gB(x, y) gB(x, z)| |R(y, K) R(z, K)| + |RK(x, y) RK(x, z)|2

RK(y, z) R(y, z).

5. Topologies associated with resistance forms

We continue to assume that (E, F) is a resistance form on a set X and R isthe associated resistance metric. In the previous section, we have seen that thecondition for a subset B X being a good boundary is that BF = B. Note thatby Theorem 2.4, CF gives a topology on X which satisfies T1-axiom of separation.Lemma 2.2 implies that BF = B if and only if B is a closed set with respect tothe CF-topology. Therefore, to see when BF = B occurs is to consider the relationbetween the topologies induced by the resistance metric R and CF. Furthermore,there exists a classical notion of fine topology associated with a cone of non-negative functions introduced by Brelot in [13]. In this section, we will studyrelations of those topologies on X. The topologies induced by the resistance metricand CF are called the R-topology and the CF-topology respectively.

First we show that the CF-topology is coarser than the R-topology in general.Proposition 5.1. BF is a closed set with respect to the resistance metric R.

In other word, the CF-topology is coarser (i.e. weaker) than that given by the R-topology.

Proof. Let

{xn

}n1

BF. Assume limn+ R(x, xn) = 0. If u

F(B),

then u(x) = limn+ u(xn) = 0 for any u F(B). Hence x BF. Later in Example 5.5, we have a resistance form on {0, 1, . . .} where there exists

an R-closed set which is not CF-closed.Next, we study a sufficient condition ensuring that BF = B.

Definition 5.2. (1) Let (X, d) be a metric space. For a non-empty subset ofB X, define

Nd(B, r) = min{#(A)|A B yABd(y, r)}

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for any r > 0, where #(A) is the number of the elements of A.(2) Let (E, F) be a resistance form on X and let R be the associated resistancemetric. For any subsets U, V X, define

R(U, V) = inf{R(x, y)|x U, y V}.

Hereafter in this section, we use N(B, r) to denote NR(B, r).The following theorem plays an important role in proving heat kernel estimates

in Part III.

Theorem 5.3. Let (E, F) be a resistance form on X. Let B be a non-emptysubset of X and let x X\B. If N(B, R(x, B)/2) < +, then x / BF and

R(x, B)

4N(B, R(x, B)/2) R(x, B) R(x, B).

The key idea of the following proof has been extracted from [6, Lemma 2.4]and [41, Lemma 4.1].

Proof. Write uy = x{y} for any x, y X. Then, by (GF4),

uy(z) = uy(z) uy(y) R(y, z)R(y, x)

.

If y B, x X\B and z BR(x, R(x, B)/2), then uy(z) 1/2. Supposethat n = N(B, R(x, B)/2) is finite. We may choose y1, . . . , yn B so thatB ni=1BR(yi, R(x, B)/2). Define v(z) = mini=1,...,n uyi(z) for any z X. Thenv F, v(x) = 1 and v(z) 1/2 for any z B. Letting h = 2(v 1/2), we see that0 h(z) 1 for any z X, h(x) = 1 and h F(B). Hence x / BF. Moreover,

E(h, h) 4E(v, v) 4n

i=1E(uyi , uyi) 4

ni=1

1

R(x, yi) 4n

R(x, B) .

Therefore,

R(x, B) =

min{E(u, u)|u F(B), u(x) = 1}1 R(x, B)4n

.

Corollary 5.4. Let (E, F) be a resistance form on X. If B is compact withrespect to the resistance metric associated with (E, F), then BF = B.

In general, even if (X, R) is locally compact, BF= B may happen for anR-closed set B as you can see in the next example.

Example 5.5. Let X =N

{0} and let Vm = {1, . . . , m}{0}. Define a linearoperator Lm : (Vm) (Vm) by

(Lm)ij =

2 if |i j| = 1 or |i j| = m,1 if {i, j} = {0, k} for some k {1, . . . , m}\{1, m},4 if i = j and i {1, m},5 if i = j and i {1, . . . , m}\{1, m},(m + 2) if i = j = 0,0 otherwise.

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5. TOPOLOGIES ASSOCIATED WITH RESISTANCE FORMS 19

1 2 3 1 2 3 4

0 0

2 2 2 2 2

2 1 2 2 1 1 2

L3

L4

The italic number between vertices i and j is the value of (Lm)i,j .

Figure 1. L3 and L4

See Figure 1 for L3 and L4. Lm is a Laplacian on Vm and {(Vm, Lm)}m1 isa compatible sequence. Let S = {(Vm, Lm)}m1. Then by Theorem 3.13, we havea resistance form (ES, FS) on V = m1Vm. Note that V = X. We use E andF instead of ES and FS respectively for ease of notation. Let R be the resistancemetric associated with (E, F). Hereafter in this example, we only consider thetopology induced by the resistance metric R. Using the fact that R(i, j) = Rm(i, j)for i, j

Vm, where Rm is the resistance metric with respect to

ELm , we may

calculate R(i, j) for any i, j X. As a result,R(0, j) = 1

3for any j 1,

R(i, j) = 23

(1 2|ij|) if i, j 1.Since 1/3 R(i, j) 2/3 for any i, j X with i = j, any one point set {x} isclosed and open. In particular, (X, R) is locally compact. Let B = N. Since B isthe complement of a open set {0}, B is closed. Define (X) by (0) = 1 and(x) = 0 for any x B. Since Em(|Vm , |Vm) = m+2 + as m +, we seethat / F. Therefore if u F(B), then u(0) = 0. This shows that BF = B {0}.

Now following Brelot [13], we introduce the notion of fine topology associatedwith a cone of extended nonnegative valued functions.

Definition 5.6. (1) Define R++ = [0, +) {+} and f : R++ [0, 1]by f(x) = 1 (x + 1)1. (Note that f(+) = 1.) We give a metric d on R++ byd(a, b) = |f(x) f(y)|.(2) Let X be a set and let {f|f : X R++}. Assume that is a cone, i.e.a1f1 + a2f2 if a1, a2 R and f1, f2 and that + , where + means aconstant function whose value is +. (We use a convention that 0 + = 0.) LetO be a topology on X, i.e. O is a family of sets satisfying the axiom of open sets.The coarsest topology which is finer than O and for which all the functions in arecontinuous is called the fine topology associated with and O. We use OF(O, )

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to denote the family of open sets with respect to the fine topology associated withO and . In particular, if O = {, X}, then we write OF() = OF(O, ).

Considering the resistance form (E, F), the adequate candidate of is F++defined by

F++ = {u|u F, u(x) 0 for any x X} {+}Define OF = {U|X\U CF}. Also define OR as the collection of open sets with

respect to the R-topology. By Proposition 5.1, we have OF OR.Theorem 5.7. Assume that (X, R) is separable.

(1) OF(OR, F++) = OR.(2) OF(F++) = OF.

By Theorem 5.7, we realize an odd situation where the fine topology is coarserthan the R-topology. Because of this and the fact that we need Theorem 4.3, wherethe condition BF = B has already been used, to prove Theorem 5.7, we can take

small advantage of the axiomatic potential theory developed in [13].To prove the above theorem, we need the following lemma.

Lemma 5.8. Assume that (X, R) is separable. Then for anyx X, there existsu Fsuch that u(x) = 0 and u(y) > 0 for any y X\{x}.

Proof. First note that by (RF2), (Fx, E) is a Hilbert space, where Fx ={u|u(x) = 0, u F}.

For any y X, define uy(z) = gy{x}(z). Then by (GF2) and (GF4), ifR(y, z) 0 for any z BR(y, R(y, x)).

Now fix > 0 and let {xn}n1 be a dense subset of X\BR(x, ). Choosen > 0 so that n1 nR(xn, x) < +. Define vn = nm=1 nuxm . Then

n1

E(vn vn+1, vn vn+1) =

n1

n+1R(xn+1, x) < +. This im-

plies that {vn}n1 is a Cauchy sequence in the Hilbert space (Fx, E). Hencev =

n1 nuxn Fx. By the above argument, for any z X\BR(x, ), there

exists xn such that uxn(z) > 0. Therefore, v(z) > 0 for any z X\BR(x, ). Weuse v to denote v. Let un = v1/n. Choose n so that

n1

nE(un, un) < +.

Then similar argument as

n1 nuxn, we see that

n1 nun belongs to Fx. Letu =

n1 nun. Then u(z) > 0 for any z X\{x}.

Proof of Theorem 5.7. First we show that

(5.1) OF(F++) OF OR.Note that

OF(

F++) is generated by u

1((

, c)) and u1((c, +

)) for any c

R

and any u F++. Let u F and let c R. Set B = {x|u(x) c}. Definev(x) = max{u(x) c, 0}. Then Proposition 3.15 shows that v F. Furthermore,v F(B) and v(x) > 0 for any x / B. Hence BF = B and so B CF. Thereforeu1((, c)) OF. Using similar argument, we also obtain that u1((c, +)) OF. Thus we have shown (5.1)

Since OF(OR, F++) is finer than OR, (5.1) implies OF(OR, F++) = OR.Hence we have shown (1).

To prove (2), let B CF. By Theorem 4.3, (E, FB) is a resistance form on XB.Let RB be the resistance metric associated with (E, FB) on XB. Then RB(x, y)

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6. REGULARITY OF RESISTANCE FORMS 21

R(x, y) for any x, y X\B and RB(x, B) infzB R(x, z). Therefore, since (X, R)is separable, (XB, RB) is separable. Hence applying Lemma 5.8 to (E, FB) withx = B, we obtain u F

B

which satisfies u(y) > 0 for any y X\B. Since u F++,

u1(0) = B is closed under OF(F++). This yields that OF OF(F++). 6. Regularity of resistance forms

Does a domain Fof a resistance form Econtain enough many functions? Thenotion of regularity of a resistance form will provide an answer to such a question.As you will see in Definition 6.2, a resistance form is regular if and only if the domainof the resistance form is large enough to approximate any continuous function witha compact support. It is notable that the operation B BF plays an importantrole again in this section.

Let (E, F) be a resistance form on a set X and let R be the associated resistancemetric on X. We assume that (X, R) is separable in this section.

Definition 6.1. (1) Let u : X R. The support of u, supp(u) is definedby supp(u) = {x|u(x) = 0}, where U is the closure of U X with respect to theresistance metric. We use C0(X) to denote the collection of continuous functionson X whose support are compact.(2) Let K be a subset of X and let u : X R. We define the supremum norm ofu on K, ||u||,K by

||u||,K = supxK

|u(x)|.We write | | | | = | | | |,X if no confusion can occur.

Definition 6.2. The resistance form (E, F) on X is called regular if and onlyifF C0(X) is dense in C0(X) in the sense of the supremum norm | | | |.

The regularity of a resistance form is naturally associated with that of a Dirich-let form. See Section 9 for details. The following theorem gives a simple criteriafor regularity.

Theorem 6.3. Assume that(X, R) is locally compact. The following conditionsare equivalent:(R1) (E, F) is regular.(R2) BF = B for anyR-closed subset B. In other words, the R-topology coincideswith the CF-topology.(R3) If B is R-closed and Bc is R-compact, then BF = B.(R4) If K is a compact subset of X, U is an R-open subset of X, K U and Uis R-compact, then there exists F such that supp() U, 0 (y) 1 forany y X and |K 1.

Combining the above theorem with Corollary 5.4, we obtain the following result.

Corollary 6.4. If (X, R) is R-compact, then (E, F) is regular.In general, even if (X, R) is locally compact, (E, F) is not always regular. Recall

Example 5.5.Hereafter, the R-topology will be always used when we will consider a resistance

form unless we say otherwise. For example, an open set means an R-open set.To prove Theorem 6.3, we need the following lemma, which can be proven by

direct calculation.

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Lemma 6.5. If u, v F C0(X), then uv F C0(X) and

E(uv,uv)

2||

u||

2

E(v, v) + 2

||v||

2

E(u, u).

Proof of Theorem 6.3. (R1) (R2) Let x / B. Choose r > 0 so thatB(x, r) is compact and B B(x, r) = . Then there exists f C0(X) such that0 f(y) 1 for any y X, f(x) = 1 and supp(f) B(x, r). Since (E, F) isregular, we may find v F C0(X) such that ||v f|| 1/3. Define u = 3v 1.Then u(x) = 1 and u|B 0. Hence x / BF.(R2) (R3) This is obvious.(R3) (R4) By (R3), (Uc)F = Uc. Hence, for any x K, we may choose rx sothat B(x, rx) U and xUc(y) 1/2 for any y B(x, rx). Since K is compact,K ni=1B(xi, rxi) for some x1, . . . , xn K. Let v =

ni=1

xiUc . Then v(y) 1/2

for any y K and supp(v) U. If = 2v, then u satisfies the desired properties.(R4) (R1) Let u C0(X). Set K = supp(u). Define UK = {u|K : u F C0(X)}. Then by (R4) and Lemma 6.5, we can verify the assumptions of theStone Weierstrass theorem for UK with respect to | | | |,K. (See, for example,[54] on the Stone Weierstrass theorem.) Hence, UK is dense in C(K) with respectto the supremum norm on K. For any > 0, there exists u F C0(X) suchthat ||u u||,K < . Let V = K {x : |u(x)| < }. Suppose that x K andthat there exists {xn}n=1,2,... Vc such that R(xn, x) 0 as x +. Then|u(xn)| for any n and hence |u(x)| . On the other hand, since xn Kc,u(xn) = 0 for any n and hence u(x) = 0. Since x K, this contradict to the factthat ||u u||,K < . Therefore, V is open. We may choose a open set U so thatK U, U is compact and U V. Let be the function obtained in (R4). Definev = u. Then by Lemma 6.5, v F C0(X). Also it follows that ||uv|| .This shows that F C0(X) is dense in C0(X) with respect to the norm | | | |.

7. Annulus comparable condition and local property

A resistance form may have a long-distance connection and/or a jump in gen-eral. For instance, let us modify a given resistance form by adding a new resistorbetween two separate points. The modified resistance form has a jump inducedby the added resistor. Such a jump naturally appears in the associated prob-abilistic process. In this section, we introduce the annulus comparable condition,(ACC) for short, which ensures certain control of such jumps, or direct connectionsbetween two separate points. For example, Theorems in Section 15 will show that(ACC) is necessary to get the Li-Yau type on-diagonal heat kernel estimate.

We need the following topological notion to state (ACC).

Definition 7.1. Let (X, d) be a metric space. (X, d) is said to be uniformlyperfect if and only if there exists > 0 such that Bd(x, (1 + )r)\Bd(x, r) = forany x X and r > 0 with X\Bd(x, r) = .

In [51], the notion of uniformly perfect is called homogeneously dense.Note that if (X, d) is connected, then it is uniformly perfect. Uniformly per-

fectness ensures that widths of gaps in the space are asymptotically of geometricprogression. For example, the ternary Cantor set is uniformly perfect. In general,any self-similar set where the contractions are similitudes is uniformly perfect asfollows.

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7. ANNULUS COMPARABLE CONDITION AND LOCAL PROPERTY 23

Example 7.2. Let (X, d) be a complete metric space and let N 2. For anyi = 1, . . . , N , Fi : X X is assumed to be a contraction, i.e.

supx=yRn

d(Fi(x), Fi(y))d(x, y)

< 1.

Then there exists a unique non-empty compact set K X which satisfies K =Ni=1Fi(K). See [36] for details. K is called the self-similar set associated with{Fi}i=1,...,N. If (X, d) is Rn with the Euclidean metric and every Fi is a similitude,i.e. Fi(x) = riAix + ai, where ri (0, 1), Ai is an orthogonal matrix and ai Rn,then K is uniformly perfect.

Next, we have an example which is complete and perfect but not uniformlyperfect.

Example 7.3. Let X = [0, 1]. If F1(x) = x2/3 and F2(x) = x/3 + 2/3, the

self-similar set associated with

{F1, F2

}is a (topological) Cantor set, i.e. complete,

perfect and compact. We denote the n-th iteration ofF1 by (F1)n, i.e. (F1)0(x) = xfor any x R and (F1)n+1 = (F1)n F1. Define an = (F1)n(1/3) and bn =(F1)

n(2/3). Then by inductive argument, we see that (an, bn) K = for anyn 0. Now, an = 3e2n+1 log 3 and bn = 3e2n+1(log 3(log2)/2). Hence bn/an =(3/

2)2

n+1

. This shows that K is not uniformly perfect.

In this section, (E, F) is a regular resistance form on X and R is the associatedresistance metric. We assume that (X, R) is separable and complete.

Definition 7.4. A resistance form (E, F) on X is said to satisfy the annuluscomparable condition, (ACC) for short, if and only if (X, R) is uniformly perfectand there exists > 0 such that

(ACC) R(x, BR

(x, r)c)

Rx, BR(x, (1 + )r) BR(x, r)cfor any x X and any r > 0 with BR(x, r) = X.

Remark. It is obvious that

R(x, BR(x, r)c) Rx, BR(x, (1 + )r) BR(x, r)c.

So the essential requirement of (ACC) is the opposite inequality up to a constantmultiplication.

The annulus comparable condition holds if (X, R) is uniformly perfect and(E, F) has the local property defined below.

Definition 7.5. (E, F) is said to have the local property if and only if E(u, v) =0 for any u, v Fwith R(supp(u), supp(v)) > 0.

Proposition 7.6. Assume that (E, F) has the local property and that BR(x, r)is compact for any x X and any r > 0. If BR(x, (1 + )r) BR(x, r)c= , then

R(x, BR(x, r)c) = R

x, BR(x, (1 + )r) BR(x, r)c

.

In particular, we have (ACC) if (X, R) is uniformly perfect.

Proof. Let K = BR(x, (1 + )r) BR(x, r)c. Recall that xK(y) = gK(x,y)gK(x,x)and that E(xK, xK) = R(x, K)1. By Theorem 6.3, there exists Fsuch thatsupp() BR(x, (1 + /2)r), 0 (y) 1 for any y X and (y) = 1 for any

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y BR(x, r). By Lemma 6.5, if 1 = xK and 2 = xK(1 ), then 1 and 2belong to F. Since supp(2) Br(x, (1 + )r)c, the local property implies

E(xK, xK) = E(1, 1) + E(2, 2) E(1, 1).Note that 1(y) = 0 for any y BR(x, r)c and that 1(x) = 1. Hence, E(1, 1) E(xB, xB), where B = B(x, r)c. On the other hand, since K B, E(xB, xB) E(xK, xK). Therefore, we have

R(x, B)1 = E(xB, xB) = E(xK, xK) = R(x, K)1.

There are non-local resistance forms which satisfy (ACC), for example, the -stable process on R and their traces on the Cantor set. See Sections 16. In the nextsection, we will show that if the original resistance form has (ACC), then so do itstraces, which are non-local in general.

To study non-local cases, we need the doubling property of the space.

Definition 7.7. Let (X, d) be a metric space.(1) (X, d) is said to have the doubling property or be a doubling space if and onlyif

(7.1) supxX,r>0

Nd(Bd(x, r), r) < +

for any (0, 1), where Nd(B, r) is defined in Definition 5.2. (2) Let be a Borelregular measure on (X, d) which satisfies 0 < (Bd(x, r)) < + for any x X andany r > 0. is said to have the volume doubling property with respect to d or bevolume doubling with respect to d, (VD)d for short, if and only if there exists c > 0such that

(VD)d (Bd(x, 2r))

c(Bd(x, r))

for any x X and any r > 0.Remark. (1) It is easy to see that (7.1) holds for all (0, 1) if it holds for

some (0, 1). Hence (X, d) is a doubling space if (7.1) holds for some (0, 1).(2) If is (VD)d, then, for any > 1, (Bd(x,r)) (Bd(x, r)) for any x Xand any r > 0.

One of the sufficient condition for the doubling property of a space is theexistence of a measure which has the volume doubling property. The followingtheorem is well-known. See [32] for example.

Proposition 7.8. Let (X, d) be a metric space and let be a Borel regularmeasure on (X, R) with 0 < (BR(x, r)) < + for any x X and any r > 0. If is (VD)d, then (X, d) has the doubling property.

The next proposition is straightforward from the definitions.

Proposition 7.9. If a metric space (X, d) has the doubling property, then anybounded subset of (X, d) is totally bounded.

By the above proposition, if the space is doubling and complete, then everybounded closed set is compact.

Now we return to (ACC). The following key lemma is a direct consequence ofTheorem 5.3.

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8. TRACE OF RESISTANCE FORM 25

Lemma 7.10. Assume that (X, R) has the doubling property and is uniformlyperfect. Then, for some > 0,

(7.2) Rx, BR(x, r)c BR(x, (1 + )r) rfor any x X and any r > 0 with BR(x, r) = X.

Proof. Set B = BR(x, (1 + )r) BR(x, r)c. Choose so that B = for anyx X and any r > 0 with BR(x, r) = X. Then, r R(x, B) (1 + )r. This andthe doubling property of (X, R) imply

N(B, R(x, B)/2) N(B,r/2) N(BR(x, (1 + 2)r), r/2) c,where c is independent of x and r. Using Theorem 5.3, we see

r

8c R(x, B) (1 + )r.

By the above lemma, (ACC) turns out to be equivalent to (RES) defined belowif (X, R) is a doubling space.

Definition 7.11. A resistance form (E, F) on X is said to satisfy the resistanceestimate, (RES) for short, if and only if

(RES) R(x, BR(x, r)c) r

for any x X and any r > 0 with BR(x, r) = X.Theorem 7.12. Assume that (X, R) has the doubling property. Then (X, R)

is uniformly perfect and (RES) holds if and only if (ACC) holds.

Proof of Theorem 7.12. If (ACC) holds, then (7.2) and (ACC) immedi-ately imply (RES). Conversely, (RES) along with (7.2) shows (ACC).

Corollary 7.13. If (E, F) has the local property, (X, R) has the doublingproperty and is uniformly perfect, then (RES) holds.

8. Trace of resistance form

In this section, we introduce the notion of the trace of a resistance form on asubset of the original domain. This notion is a counterpart of the notion of tracesin the theory of Dirichlet forms, which has been extensively studied in [21, Section6.2], for example.

Throughout this section, (E, F) is a resistance form on Xand R is the associatedresistance distance. We assume that (X, R) is separable and complete.

Definition 8.1. For a non-empty subset Y

X, define

F|Y =

{u

|Y : u

F}.

Lemma 8.2. Let Y be a non-empty subset of (X, R). For any u F|Y, thereexists a unique u Fsuch that u|Y = u and

E(u, u) = min{E(v, v)|v F, v|Y = u}.This lemma is an extension of Proposition 3.10. In fact, if Y is a finite set,

then we have F|Y = (Y) by (RF3-2). Hence Lemma 8.2 holds in this case byProposition 3.10. The unique u is thought of as the harmonic function with theboundary value u on Y.

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Proof of Lemma 8.2. Let p Y. Replacing u by u u(p), we may assumethat u(p) = 0 without loss of generality. Choose a sequence {vn}n1 F sothat vn|Y = u and limn+ E(vn, vn) = inf{E(v, v)|v F, v|Y = u}. Let C =supn E(vn, vn). By Proposition 3.2, if v = vn, then(8.1) |v(x) v(y)|2 CR(x, y)and

(8.2) |v(x)|2 CR(x, p)Let {Vm}m1 be an increasing sequence of finite subsets of X. Assume that V =m1Vm is dense in X. (Since (X, R) is separable, such {Vm}m1 does exists.) By(8.1) and (8.2), the standard diagonal construction gives a subsequence {vni}i1which satisfies {vni(x)}i1 is convergent as i + for any x V = m1Vm.Define u(x) = limi+ vni(x) for any x V. Since u satisfies (8.1) and (8.2)on V

with v = u

, u

is extended to a continuous function on X. Note that

the extended function also satisfies (8.1) and (8.2) on X with v = u. Set Em =EL(E,F),Vm . Then, by Theorem 3.14,(8.3) Em(vn, vn) E(vn, vn) Cfor any m 1 and any n 1. Define M = inf{E(v, v)|v F, v|Y = u}. For any > 0, if n is large enough, then (8.3) shows Em(vn, vn) M + for any m 1.Since vn|Vm u|Vm as n +, it follows that Em(u, u) M + for anym 1. Theorem 3.14 implies that u Fand E(u, u) M.

Next assume that ui F, ui|Y = u and E(ui, ui) = M for i = 1, 2. SinceE((u1 + u2)/2, (u1 + u2)/2) E(u1, u1), we have E(u1, u2 u1) 0. Similarly,E(u2, u1u2) 0. Combining those two inequalities, we obtain E(u1u2, u1u2) =0. Since u1 = u2 on Y, we have u1 = u2 on X.

The next definition is an extension of the notion ofhV defined in Definition 3.11.

Definition 8.3. Define hY : F|Y Fby hY(u) = u, where u and u are thesame as in Lemma 8.2. hV(u) is called the Y-harmonic function with the boundaryvalue u. For any u, v F|Y, define E|Y(u, v) = E(hY(u), hY(v)). (E|Y , F|Y) iscalled the trace of the resistance form (E, F) on Y.

Making use of the harmonic functions, we construct a resistance form on asubspace Y of X, which is called the trace.

Theorem 8.4. Let Y be a non-empty subset of X. ThenhY : F|Y F islinear and (E|Y , F|Y) is a resistance form on Y. The associated resistance metricequals to the restriction of R on Y

. If Y is closed and (

E,

F) is regular, then

(E|Y, F|Y) is regular.We denote the restriction of R on Y Y by R|Y.The following lemma is essential to prove the above theorem.

Lemma 8.5. LetY be a non-empty subset of X. Define

HY = {u|u F, E(u, v) = 0 for any v F(Y)},where F(Y) is defined in Definition 2.1. Then, for any f F|Y, u = hY(f) if

and only if u HY and u|Y = f.

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8. TRACE OF RESISTANCE FORM 27

By this lemma, HY = Im(hY) is the space of Y-harmonic functions and F=HY + F(Y), i.e. F is the direct sum of HY and F(Y). Moreover, E(u, v) = 0 forany u HY and any v F(Y). The counterpart of this fact has been know forDirichlet forms. See [21, Section 6.2] for details. In the case of weighted graphs(Example 3.5) such a decomposition is known as the Roydens decomposition. See[49, Theorem (6.3)] for details.

Proof. Let f = hY(f). Ifv Fand v|Y = f, thenE(t(v f) + f, t(v f) + f) E(f, f)

for any t R. Hence E(v f, f) = 0. This implies that f HY . Converselyassume that u HY and u|Y = f. Then, for any v Fwith v|Y = f,

E(v, v) = E((v u) + u, (v u) + u) = E(v u, v u) + E(u, u) E(u, u).Hence by Lemma 8.2, u = hY(f).

Proof of Theorem 8.4. By Lemma 8.5, ifrY : HY F|Y is the restrictionon Y, then rY is the inverse of hY. Hence hY is linear. The conditions (RF1)through (RF4) for (E|Y, F|Y) follow immediately from the counterparts for (E, F).About (RF5),

E|Y(u, u) = E(hY(u), hY(u)) E(hY(u), hY(u)) E(hY(u), hY(u)) = E|Y(u, u).

The rest of the statement is straightforward.

In the rest of this section, the conditions (ACC) and (RES) are shown to bepreserved by the traces under reasonable assumptions.

Theorem 8.6. Let (E, F) be a regular resistance form on X and let R bethe associated resistance metric. Assume that (E, F) satisfies (RES). If Y is aclosed subset of X and (Y, R|Y) is uniformly perfect, then (RES) holds for the trace(E|Y, F|Y).

By Theorem 7.12, we immediately have the following corollary.

Corollary 8.7. Let(E, F) be a regular resistance form on X and let R be theassociated resistance metric. Assume that (X, R) has the doubling property. Let Ybe a closed subset of X and assume that (Y, R|Y) is uniformly perfect. If (ACC)holds for (E, F), then so does for the trace (E|Y, F|Y).

Notation. Let (E, F) be a resistance form on X and let R be the associ-ated resistance metric. For a non-empty subset Y of X, we use R

Y

to denotethe resistance metric associated with the trace (E|Y, F|Y) on Y. Also we writeBYR (x, r) = BR(x, r) Y for any x Y and r > 0.

Note that RY(x, y) = R(x, y) for any x, y Y by Theorem 8.4. Hence RY =R|Y.

Proof of Theorem 8.6. Note that RY(x, Y\BYR (x, r)) = R(x, BR(x, r)c Y). Hence if (RES) holds for (E, F) then,(8.4) RY(x, Y\BYR (x, r)) R(x, BR(x, r)c) c1r.

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On the other hand, since (Y, RY) is uniformly perfect, there exists > 0 such thatBYR (x, (1 + )r)\BYR (x, r) = for any x Y and r > 0 with BYR (x, r) = Y. Lety B

YR (x, (1 + )r)\B

YR (x, r). Then

(8.5) (1 + )r RY(x, y) RY(x, Y\BYR (x, r)).Combining (8.4) and (8.5), we obtain (RES) for (E|Y , F|Y).

9. Resistance forms as Dirichlet forms

In this section, we will present how to obtain a regular Dirichlet form from aregular resistance form and show that every single point has a positive capacity. Asin the previous sections, (E, F) is a resistance form on X and R is the associatedresistance metric on X. We continue to assume that (X, R) is separable, completeand locally compact.

Let be a Borel regular measure on (X, R) which satisfies 0 < (BR(x, r)) 0. Note that C0(X) is a dense subset of L

2(X, ) bythose assumptions on .

Definition 9.1. For any u, v F L2(X, ), define E1(u, v) byE1(u, v) = E(u, v) +

X

uvd.

By [36, Theorem 2.4.1], we have the following fact.

Lemma 9.2. (F L2(X, ), E1) is a Hilbert space.Since F C0(X) F L2(X, ), the closure of F C0(X) is a subset of

F L2(X, ).Definition 9.3. We use

Dto denote the closure of

F C

0(X) with respect to

the inner product E1.Note that if (X, R) is compact, then D = F.Theorem 9.4. If (E, F) is regular, then (E|DD, D) is a regular Dirichlet form

on L2(X, ).

See [21] for the definition of a regular Dirichlet form.For ease of notation, we write Einstead of EDD.Definition 9.5. The Dirichlet form (E, D) on L2(X, ) is called the Dirichlet

form induced by a resistance form (E, F).Proof. (E, D) is closed form on L2(X, ). Also, since C0(X) is dense in

L2(X, ), the assumption thatF

C0(X) is dense in C0(X) shows thatD

is densein L2(K, ). Hence (E, D) is a regular Dirichlet form on L2(X, ) with a coreF C0(X).

Hereafter in this section, (E, F) is always assumed to be regular. Next we studycapacity of points associated with the Dirichlet form constructed above.

Lemma 9.6. Letx X. Then there exists cx > 0 such that|u(x)| cx

E1(u, u)

for any u D. In other words, the map u u(x) from D to R is bounded.

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9. RESISTANCE FORMS AS DIRICHLET FORMS 29

Proof. Assume that there exists a sequence {un}n1 Fsuch that un(x) = 1and E1(un, un) 1/n for any n 1. By (3.1),

|un(x) un(y)| R(x, y)n

R(x, y)Hence un(y) 1/2 for any y BR(x, 1/4). This implies that

||un||22 B(x,1/4)

u(y)2d (BR(x, 1/4))/4 > 0

This contradicts the fact that E1(un, un) 0 as n +. Lemma 9.7. If K is a compact subset of X, then the restriction map K :

D C(K) defined by K(u) = u|K is a compact operator, where D and C(K) areequipped with the norms

E1(, ) and | | | |,K respectively.

Proof. Set D = supx,yKR(x, y). Let

Ube a bounded subset of

D, i.e. there

exists M > 0 such that E1(u, u) M for any u U. Then by (3.1),|u(x) u(y)|2 R(x, y)M

for any x, y X and any u U. Hence U is equicontinuous. Choose x K. ByLemma 9.6 along with (3.1),

u(x)2 2|u(x) u(x)|2 + 2|u(x)|2 2DM + 2c2xMfor any u U and any x K. This shows that U is uniformly bounded on K.By the Ascoli-Arzela theorem, {u|K}uU is relatively compact with respect to thesupremum norm. Hence K is a compact operator.

Definition 9.8. For an open set U X, define the E1-capacity of U, Cap(U),by

Cap(U) = inf{E1(u, u)|u D, u(x) 1 for any x U}.If {u|u D, u(x) 1 for any x U} = , we define Cap(U) = +. For anyA X, Cap(A) is defined by

Cap(A) = inf{Cap(U)|U is an open subset of X and A U}.Theorem 9.9. For any x X, 0 < Cap({x}) < +. Moreover, if K is a

compact subset of X, then 0 < infxKCap({x}).The proof of Theorem 9.9 requires the following lemma.

Lemma 9.10. For any x X, there exists a unique g D such thatE1(g, u) = u(x)

for any u D. Moreover, let = g/g(x). Then, is the unique element in{u|u D, u(x) 1} which attains the following minimummin{E1(u, u)|u D, u(x) 1}.

Proof. The existence ofg follows by Lemma 9.6. Assume that E1(f, u) = u(x)for any u D. Since E1(f g, u) = 0 for any u D, we have f = g. Now, ifu Dand u(x) = a > 1, then E1(u a,) = u(x)/g(x) 1/g(x) = 0. Hence,

E1(u, u) = E1(u a,u a) + E1(a, a) E1(, ).This immediately shows the rest of the statement.

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Definition 9.11. We denote the function g and in Lemma 9.10 by gx1 andx1 respectively.

Proof of Theorem 9.9. Fix x X. By the above lemma, for any open setU with x U,

Cap(U) = min{E1(u, u)|u D, u(y) 1 for any y U} min{E1(u, u)|u D, u(x) 1} E1(x1 , x1 ) =

1

gx1 (x).

Hence 0 < 1/gx1 (x) < Cap({x}) < Cap(U) < +.Let K be a compact subset ofX. By Lemma 9.7, there exists cK > 0 such that

||u||,K cK

E1(u, u) for any u D. Now, for x K,

gx1 (x) = E1(gx1 , gx1 ) = supuD,u=0

E1(gx1 , u)2E1(u, u) = supuD,u=0

u(x)2

E1(u, u) (cK)2

Hence Cap({x}) 1/gx1 (x) (cK)2. Following [21, Section 2.1], we introduce the notion of quasi continuous func-

tions.

Definition 9.12. A function u : X R is called quasi continuous if and onlyif, for any > 0, there exists V X such that Cap(V) < and u|X\V is continuous.

Theorem 9.9 implies that every quasi continuous function is continuous.

Proposition 9.13. Any quasi continuous function is continuous on X.

Proof. Let u be a quasi continuous function. Let x X. Since (X, R) islocally compact, BR(x, r) is compact for some r > 0. By Theorem 9.9, we maychoose > 0 so that infyB

R(x,r)

Cap(

{y

}) > . There exists V

X such that

Cap(V) < and u|X\V is continuous. Since V BR(x, r) = , u is continuous atx. Hence u is continuous on X.

10. Transition density

In this section, without ultracontractivity, we establish the existence of jointlycontinuous transition density (i.e. heat kernel) associated with the regular Dirichletform derived from a resistance form.

As in the last section, (E, F) is a resistance form on X and R is the associatedresistance metric. We assume that (X, R) is separable, complete and locally com-pact. is a Borel regular measure on X which satisfies 0 < (BR(x, r)) < +for any x X and any r > 0. We continue to assume that (E, F) is regular. ByTheorem 9.4, (

E,D

) is a regular Dirichlet form on L2(X, ), whereD

is the closureofF C0(X) with respect to the E1-inner product.

Let H be the nonnegative self-adjoint operator associated with the Dirichletform (E, D) on L2(X, ) and let Tt be the corresponding strongly continuous semi-group. Since Ttu D for any u L2(X, ), we always take the continuous versionof Ttu. In other words, we may naturally assume that Ttu is continuous for anyt > 0.

Let M = (, {Xt}t>0, {Px}xX) be the Hunt process associated with the reg-ular Dirichlet form (E, D) on L2(X, ). By [21, Chapter 7], such a Hunt processis determined outside a set of capacity zero in general. Note that Cap(A) = 0 if

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10. TRANSITION DENSITY 31

and only if A = by Theorem 9.9. Hence, the Hunt process M is determined forevery x X in our case. Moreover, by [21, Theorem 4.2.1], every exceptional set isempty. (See [21, Section 4.1] for the definition of exceptional sets and its relationto other notions like polar sets, semi-polar sets and negligible sets.) Let ptbe the transition semigroup associated with the Hunt process M. For non-negative-measurable function u,

(ptu)(x) = Ex(u(Xt))

for any x X. Let Bbe the Borel -algebra of (X, R). We say that u is Borel mea-surable, if and only ifu1((a, b]) Bfor any a, b R. Combining Proposition 9.13and [21, Theorem 4.2.3], we have the following statement.

Proposition 10.1. For any nonnegative u L2(X, ), (ptu)(x) = (Ttu)(x)for any t > 0 and any x X.

Definition 10.2. Let U be an open subset of X. Define

DU =

{u

|u

D, u|Uc 0}. Also we define EU = E|DUDU.Note that if U is compact, then DU = F(Uc).Combining the results in [21, Section 4.4], we have the following facts.

Theorem 10.3. Let U be the restriction of on U, i.e. U(A) = (A U)for any Borel set U. Then (EU, DU) is a regular Dirichlet form on L2(U, U).Moreover, if MU = (U, X

Ut , P

Ux ) is the associated Hunt process, then

PUx (XUt A) = P(Xt A, t < U)

for any Borel set A and any x U, where U is the exit time of U defined byU() = inf{t > 0|Xt() X\U}.

Moreover, if pUt is the transition semigroup associated with MU, hen

(10.1) (pUt u)(x) = EUx (u(X

Ut )) = Ex({t

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Theorem 10.4. Assume thatBR(x, r) is compact for any x X andr > 0. LetU be a non-empty open subset of X. Then there exits pU(t,x,y) : (0, + X X [0, +) which satisfies the following conditions:(TD1) pU(t,x,y) is continuous on (0, +) X X. Define pt,xU (y) = pU(t,x,y).Then pt,xU DU for any (t, x) (0, +) X.(TD2) pU(t,x,y) = pU(t,y,x) for any (t,x,y) (0, +) X X.(TD3) For any non-negative (Borel)-measurable function u and any x X,(10.2) (pUt u)(x) =

X

pU(t,x,y)u(y)(dy)

(TD4) For any t,s > 0 and any x, y X,(10.3) pU(t + s,x,y) =

X

pU(t,x,z)pU(s,y,z)(dz).

Furthermore, let A be a Borel subset of X which satisfies 0 < (A) < +. DefineR(x, A) = supyA R(x, y) for any x X. Then(10.4) pU(t,x,x) 2R(x, A)

t+

2

(A)

for any x X and any t > 0.The proof of the upper heat kernel estimate (10.4) is fairly simple. Originally,

the same result has been obtained by more complicated discussion in [6] and [41].A simplified argument, which is essentially the same as ours, for random walks canbe found in [8].

Remark. In fact, we have the following inequality which is slightly better than(10.4). For any > 0,

(10.5) pU(t,x,x) 1 + 1R(x, A)t +

1 + (A) .

This inequality implies that

limt+

pU(t,x,x) 1(X)

.

for any x X.Definition 10.5. pU(t,x,y) is called the transition density and/or the heat

kernel associated with the Dirichlet form (EU, DU) on L2(X, ).Corollary 10.6. Assume that BR(x, r) is compact for any x X and any

r > 0. Let U be a non-empty open subset of X. Then

limt0

tpU(t,x,x) = 0

for any x X.Proof. Choose A = BR(x, r). By (10.4), it follows that tpU(t,x,x) 3r for

sufficiently small t.

Now we begin our proof of Theorem 10.4 which consists of several lemmas.First we deal with the case where U is compact.

Lemma 10.7. If U is compact, then we have pU(t,x,y) : (0, +) X Xwhich satisfies (TD1), (TD2), (TD3) and (TD4).

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10. TRANSITION DENSITY 33

Proof. Let HU be the non-negative self-adjoint operator on L2(U, U) associ-

ated with the Dirichlet form (EU, DU). Then by Lemma 9.7, HU has compact resol-vent. Hence, there exists a complete orthonormal system {n}n1 ofL

2

(U, U) and{n}n1 [0, +) such that n Dom(HU) DU, HUn = nn, n n+1and limn+ n = +.Claim 1:

(10.6)n1

1

n + 1< +

Proof of Claim 1: By Lemma 9.6, for any x U, there exists gx1,U DU such thatE1(gx1,U, u) = u(x) for any u DU. Since

n(x) = E1(gx1,U, n) = (n + 1)U

gx1,UndU,

we have gx1,U = n1 n(x)n+1 n in L2(U, U). Hence(10.7) gx1,U(x) = E1(gx1,U, gx1,U) =

n1

n(x)2

n + 1

On the other hand, by the same argument as in the proof of Theorem 9.9, thereexists cU > 0 such that

|E1(u, gx1,U)| |u(x)| | |u||,K cU

E1(u, u)for any u DU, where K = U. This implies that E1(gx1,U, gx1,U) cU. Combiningthis with (10.7), we see that gx1,U(x) is uniformly bounded on U. Hence integrating

(10.7) with respect to x, we obtain (10.6) by the monotone convergence theorem.Claim 2: ||n||

Dn for any n 2, where D = supx,yU R(x, y).

Proof of Claim 2: By (3.1),

(10.8) |n(x) n(y)|2 E(n, n)R(x, y) = nR(x, y).We have two cases. First if U = X, then n(y) = 0 for any y Uc. Hence (10.8)implies the claim. Secondly, if U = X, then (X, R) is compact. It follows that 1 =0 and 1 is constant on X. Hence

X

n(x)(dx) = 0 for any n 2. For any x X,we may find y X so that n(x)n(y) 0. Since |n(x)|2 |n(x) n(y)|2,(10.8) yields the claim.Claim 3:

n1 e

ntn(x)n(y) converges uniformly on [T, +)XX for anyT > 0.Proof of Claim 3: Note that ea 2/a2 for any a > 0. This fact with Claim 2shows that |entn(x)n(y)| 2/(nt2). Using Claim 1, we immediately obtainClaim 3.

Now, let pU(t,x,y) = n1 entn(x)n(y). By Claim 3, p is continuouson (0, +) X X. Also, pU is the integral kernel of the strongly continuoussemigroup {TUt }t>0 associated with the Dirichlet form (EU, DU) on L2(U, U). LetA be a Borel set. Then

(TUt A)(x) =

X

pU(t,x,y)A(y)(dy).

In particular TUt A is continuous on X. By [21, Theorem 4.2.3], pUt A is a quasi

continuous version of TUt A. Since any quasi continuous function is continuousby Proposition 9.13, we have (TUt A)(x) = (p

Ut A)(x) for any x X. Letting

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pU(t,x,y) = pU(t,x,y), we have (TD3). The rest of the desired properties arestraightforward.

The following facts are well-known in general setting. See [21] for example. Inthis particular situation, they can be checked by the eigenfunction expansion of theheat kernel above.

Lemma 10.8. Assume that U is compact.(1) For any t > 0 and any x, y X,

pUt

(t,x,y) = E(pt/2,xU , pt/2,yU ).

(2) For any t,s > 0 and any x X,

E(pt,xU , ps,xU ) 2

t + spU

t + s

2, x , x.

Lemma 10.9. If U is compact, then (10.4) holds for any Borel subset A of Xwhich satisfies 0 < (A) < +.

Proof. SinceA

pU(t,x,y)(dy) X

pU(t,x,y)(dy) 1, there exists y A such that pU(t,x,y) 1/(A). By this fact along with Lemma 10.8-(2),

1

2pU(t,x,x)

2 pU(t,x,y)2 + |pU(t,x,x) pU(t,x,y)|2

1(A)2

+ R(x, A)E(pt,xU , pt,xU ) 1

(A)2+

R(x, A)

tpU(t,x,x)

Solving this with respect to pU(t,x,x), we have

pU(t,x,x) R(x, A)t

+ 2

(A)2+

R(x, A)2

t2

12 2R(x, A)

t+

2(A)

.

Remark. To get (10.5), we only need to use

pU(t,x,x)2 (1 + )pU(t,x,y)2 +

1 +

1

|pU(t,x,x) pU(t,x,y)|2

in place of the inequality with = 1 in the above proof.

Thus we have shown Theorem 10.4 if U is compact.

Proof of Theorem 10.4. IfU is compact, then we have completed the proof.Assume that U is not compact. Fix x X and set Un = BR(x, n) U for anyn = 1, 2, . . .. Note that Un is compact. Write pn(t,x,y) = pUn(t,x,y).Claim 1 pn(t,x,y) pn+1(t,x,y) for any x, y X and any n 1.Proof of Claim 1. Let n = Un . Then n n+1 for any n. Hence

(pUnt u)(x) = Ex(t

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10. TRANSITION DENSITY 35

Let A be a Borel subset of X which satisfies 0 < (A) < +. By (TD4) and(10.4), we have

(10.9) pn(t,x,y) pn(t,x,x)pn(t,y,y)

2R(x, A)

t+

2

(A)

12

2R(y, A)

t+

2

(A)

12

for any x X, any t > 0 and any n. Hence pn(t,x,y) is uniformly bounded andmonotonically nondecreasing as n +. This shows that pn(t,x,y) convergesas n +. If p(t,x,y) = limn+pn(t,x,y), then p(t,x,y) satisfies the sameinequality as (10.9). In particular (10.4) holds for p(t,x,x). Also, we immediatelyverity (TD2) and (TD4) for p(t,x,y) from corresponding properties of pn(t,x,y).About (TD3), let u be a non-negative Borel-measurable function. Then by (TD3)for pn(t,x,y),

(pUnt u)(x) = Ex({t 0 and any x X. Using (TD3) and (TD4), we see

pt,xU = pUt/2(p

t/2,xU ) = Tt(p

t/2,xU )

for any t > 0 and any x, y X, where {Tt}t>0 is the strongly continuous semigroupassociated with the Dirichlet form (EU, DU) on L2(X, ). Hence pt,xU D for anyt > 0 and any x X.

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At the end of this section, we give a fundamental relations between the Greenfunction, Laplacian and the transition density. By Theorem 10.3, (EU, DU) is aregular Dirichlet form on L

2

(U, |U). (DU and EU are defined in Definition 10.2.)Theorem 10.