Scale-invariant boundary Harnack principle on
inner uniform domains in fractal-type spaces
Janna Lierl
October 8, 2013
Abstract
We prove a scale-invariant boundary Harnack principle for inner uniform domains in metric measure Dirichlet spaces. We assume the volume
doubling property and weak sub-Gaussian heat kernel bounds are satisfied, and the Dirichlet form is symmetric, strongly local, regular. We
make no assumptions on the pseudo-metric induced by the Dirichlet form,
hence the underlying space can be a fractal space.
2010 Mathematics Subject Classification: 31C25, 60J60, 60J45.
Keywords: Inner uniform domain, boundary Harnack principle, Dirichlet form,
metric measure space, fractal.
Acknowledgement: Research partially supported by NSF grant DMS 1004771.
Address: Malott Hall, Department of Mathematics, Cornell University, Ithaca,
NY 14853, United States.
Introduction
The boundary Harnack principle is a property of a domain that allows us to
compare the decay of different harmonic functions on the domain under Dirichlet
boundary condition. More precisely, a domain is satisfies a boundary Harnack
principle if the ratio of any two harmonic functions is bounded near some part
of the boundary of the domain where both functions vanish.
The boundary Harnack principle was introduced by Kemper [Kem72], followed by works of Dahlberg, Wu and Ancona [Dah77, Anc80, Wu78]. Whether
a given domain has this property depends on the geometry of its boundary as
well as on the precise formulation of the boundary Harnack principle.
The classical formulation is the global boundary Harnack principle. It applies to functions u, v on a domain D that are non-negative harmonic on an
open subset V ⊂ D and vanish continuously along V ∩ ∂D. Bass and Burdzy
([BB91]) proved that, if D is a twisted Hölder domain of order α ∈ (1/2, 1], then
the boundary Harnack inequality uv ≤ A1 = A1 (D, V, K) is satisfied on some
compact set K intersecting V ∩ ∂D.
In this paper, we are concerned with the scale-invariant (also called geometric) boundary Harnack principle. That is, V and K are replaced by two
1
concentric balls of radius A0 r and r, respectively, centered at a boundary point,
and the constant A1 = A1 (A0 ) is independent of the radius r.
The scale-invariant boundary Harnack principle is crucial in order to identify
the Martin boundary of a bounded domain with its topological boundary. Another interesting application are two-sided estimates of the Dirichlet heat kernel
as in [GSC11, LSCc]. In fact, the requirements on the domain in those works
are mostly due to the need for a geometric boundary Harnack principle.
A condition that is reasonably close to minimal requirements on the boundary, is the inner uniformity. This determines a very large class of domains that
are defined intrinsically using the inner metric of the domain. The boundary of
an inner uniform domain can be very rough (e.g. the Koch snowflake). However, the domain, or any part of it, must not have the shape of a cusp. This
condition affects the geometry of the domain both locally and globally and is
needed to ensure the scale-invariance of the result. A local boundary Harnack
principle (i.e. for small radius only) can be proved on domains that satisfy an
inner uniform condition only locally.
On uniform and inner uniform domains, respectively, the scale-invariant
boundary Harnack principle was first proved by Aikawa ([Aik01]) and Ancona
([Anc07]) for domains in Euclidean space. Gyrya and Saloff-Coste ([GSC11])
generalized Aikawa’s approach to uniform domains in (non-fractal) Dirichlet
spaces. They considered a metric measure space equipped with a symmetric
strongly local regular Dirichlet form that satisfies a parabolic Harnack inequality. Moreover, they deduced that the boundary Harnack principle holds on inner
uniform domains, by considering the inner uniform domain as a uniform domain
in a different metric space, namely the completion of the inner uniform domain
with respect to its inner metric.
In [LSCb], Saloff-Coste and the author proved the geometric boundary Harnack principle directly on inner uniform domains. There, the underlying metric
measure space is equipped with a local, regular (possibly non-symmetric) Dirichlet form and some technical assumptions (cf. [LSCa]) which provide control over
the non-symmetric Dirichlet form in terms of its symmetric strongly local part.
Furthermore, the volume doubling property and the Poincaré inequality are
assumed.
Both in [GSC11] and [LSCb], the case of fractal spaces was excluded by
putting certain hypotheses on the pseudo-metric induced by the Dirichlet form
and by assuming the classical space-time scaling Ψ(r) = r2 .
The aim of this paper is to give a proof of the geometric boundary Harnack
principle in a more general setting that includes fractal spaces. Applications of
this result to proving two-sided bounds for the Dirichlet heat kernel on inner
uniform domains will be presented in forthcoming papers [KLa, KLb].
In this paper, the underlying space (X, d, µ) is a metric measure space
equipped with a symmetric strongly local regular Dirichlet form (E, F ) that
satisfies volume doubling and weak two-sided sub-Gaussian bounds on the heat
kernel, w-HKE(Ψ). The space-time scaling is captured by a continuous strictly
increasing bijection Ψ : [0, ∞) → [0, ∞) which has the property that, for some
2
constants 1 < β1 ≤ β2 < ∞ and C > 0,
C −1
R
r
β1
≤
Ψ(R)
≤C
Ψ(r)
β2
R
r
∀0 < r ≤ R < ∞.
(1)
The main result of this paper is the following scale-invariant boundary Harnack principle.
Theorem 0.1. Let (X, d, µ, E, F ) be a metric measure Dirichlet space that satisfies volume doubling (VD), and weak sub-Gaussian heat kernel estimates wHKE(Ψ). Let Ω ⊂ X be an inner uniform domain.
e \ Ω, 0 <
Then there exist constants A0 , A1 ∈ (1, ∞) such that for any ξ ∈ Ω
r < R(Ω), and any two non-negative harmonic functions u and v on BΩ (ξ, A0 r)
e \ Ω, we have
with weak Dirichlet boundary condition along Ω
u(x)
v(x)
≤ A1
,
u(x′ )
v(x′ )
∀x, x′ ∈ BΩ (ξ, r).
Here BΩ (ξ, r) = {z ∈ Ω : dΩ (ξ, z) < r}, dΩ is the intrinsic distance of the
e the completion of Ω with respect to dΩ . R(Ω) depends on the
domain Ω, and Ω
inner uniformity constants and on the diameter of Ω, and can be chosen to be
infinity if Ω is unbounded.
For example, removing the bottom line in the Sierpinski gasket produces a
subset that is an inner uniform domain in the Sierpinski gasket. Hence, any two
harmonic functions on the Sierpinski gasket that vanish along the bottom line,
decay at comparable rates near the bottom line.
The structure of this paper follows the presentation of the non-fractal case in
[LSCb]. We begin by recalling well-known definitions and properties of Dirichlet
spaces and harmonic functions in Section 1. In Section 2 we state the geometric
hypotheses on the space, the volume doubling property and weak heat kernel
estimates, and relate them to further conditions, EHI, RΨ , CSA(Ψ), which will
be used throughout the paper. In Section 3 we discuss the (local) inner uniformity conditions and collect some properties. In addition, we present estimates
for the Dirichlet heat kernel on metric balls and for Green’s functions on balls
intersected with an inner uniform domain. Furthermore, we give detailed proofs
of a maximum principle and a representation formula for harmonic functions,
which were omitted in [LSCb]. The latter results are essential ingredients in the
proof of the geometric boundary Harnack principle (Theorem 4.2) in Section 4.
Finally, in Section 5, we give two applications: First, the identification of the
Martin boundary of a bounded inner uniform domain. Second, the construction of a harmonic profile on an unbounded inner uniform domain, which will
be important to obtain bounds on the Dirichlet heat kernel in the forthcoming
papers [KLa, KLb].
3
1
Dirichlet space and harmonic functions
1.1
Dirichlet space and harmonic functions
Let (X, d, µ, E, F ) be a metric measure Dirichlet space (MMD space). That
is, X is a connected, locally compact, separable, complete metric space, all
metric balls are relatively compact, and the metric d is geodesic. µ is a positive
Radon measure on X with full support. (E, F ) is a symmetric, strongly local,
regular Dirichlet form on L2 (X, µ), see [FŌT94]. We denote by (L, D(L)) the
infinitesimal generator of (E, F ).
There exists a measure-valued quadratic form Γ defined by
Z
1
f dΓ(u, u) = E(uf, u) − E(f, u2 ), ∀f, u ∈ F ∩ L∞ (X),
2
and extended to unbounded functions by setting Γ(u, u) = limn→∞ Γ(un , un ),
where un = max{min{u, n}, −n}. Using polarization, we obtain a bilinear form
Γ. In particular,
Z
E(u, v) = dΓ(u, v), ∀u, v ∈ F .
The quadratic form Γ(u, u) is called the energy measure of the form (E, F ) and
is denoted as µhui in [FŌT94].
For U ⊂ X open and connected, set
Floc (U ) = {f ∈ L2loc (U ) : ∀ compact K ⊂ U, ∃f ♯ ∈ F , f = f ♯ K a.e.},
2
where L2loc (U ) is the space of functions that
in
are locally
L (U ). For f, g ∈
♯ ♯ Floc (U ) we define Γ(f, g) locally by Γ(f, g) K = Γ(f , g ) K , where K ⊂ U is
compact and f ♯ , g ♯ are functions in F such that f = f ♯ , g = g ♯ a.e. on K.
Define
Z
Z
F (U ) = {u ∈ Floc (U ) :
|u|2 dµ +
dΓ(u, u) < ∞},
U
U
Fc (U ) = {u ∈ F (U ) : The essential support of u is compact in U },
1/2
Z
0
2
F (U ) = the closure of Fc (U ) in F for the norm E(u, u) + u dµ
.
Note that Fc (U ) is a linear subspace of F .
For a domain U in (X, d), define the inner metric dU on U by
dU (x, y) := inf length(γ)γ : [0, 1] → U continuous, γ(0) = x, γ(1) = y ,
where
length(γ) = sup
(
n
X
)
d(γ(ti ), γ(ti−1 )) : n ∈ N, 0 ≤ t0 < . . . < tn ≤ 1 .
i=1
e be the completion of U with respect to the inner metric dU .
Let U
4
Definition 1.1. Let V be an open subset of U . Set
0
e with
Floc
(U, V ) ={f ∈ L2loc (V, µ) : ∀ open A ⊂ V rel. compact in U
dU (A, U \ V ) > 0, ∃f ♯ ∈ F 0 (U ) : f ♯ = f µ-a.e. on A}.
Definition 1.2. Let V ⊂ U be open. A function u : V → R is harmonic on V ,
if
(i) u ∈ Floc (V ),
(ii) For any function φ ∈ Fc (V ), E(u, φ) = 0.
If in addition
0
u ∈ Floc
(U, V ),
then u is a harmonic function with Dirichlet boundary condition along Ve dU \ U ,
where Ve dU denotes the completion of V with respect to dU .
If u ∈ Floc (V ) satisfies E(u, φ) ≥ 0 for all φ ∈ Fc (V ) with φ ≥ 0, then u is
called superharmonic.
1.2
Dirichlet-type Dirichlet form
Definition 1.3. For an open set U ⊂ X, the Dirichlet-type Dirichlet form on
U is defined as
EUD (f, g) = E(f, g), f, g ∈ F 0 (U ).
D
D
Let (LD
U , D(LU )) be the infinitesimal generator and PU,t , t > 0, be the
semigroup associated with (EUD , F 0 (U )). If the semigroup admits a continuous
integral kernel, that is, a non-negative continuous map pD
U (t, x, y) : (0, ∞) × U ×
U → R such that, for any f ∈ L2 (U, µ),
Z
D
pD
∀t > 0, x ∈ U,
PU,t
f (x) =
U (t, x, y)f (y)µ(dy),
U
then pD
U (t, x, y) is called the Dirichlet heat kernel on U . Using the reasoning in
0
[Stu95, Proposition 2.3], one can show that the map y 7→ pD
U (t, x, y) is in F (U ).
The extended Dirichlet space F 0 (U )e is defined as the family of all measurable, almost everywhere finite functions u such that there exists an approximating sequence un ∈ F 0 (U ) that is EUD -Cauchy and u = lim un almost everywhere.
If (EUD , F 0 (U )) is transient then F 0 (U )e is complete by [FŌT94, Lemma 1.5.5].
1.3
Capacity
The potential theory for symmetric regular Dirichlet forms is developed in
[FŌT94, Chapter 2]. Here we recall some definitions and facts that we need
in the sequel.
Let U ⊂ X be open and relatively compact. Suppose (EUD , F 0 (U )) is transient.
5
For an arbitrary set A ⊂ U let
LA,U = {u ∈ Fe (U ) : u
e ≥ 1 a.e. on A}.
The 0-capacity of A in U can be defined as
(
inf u∈LA,U EUD (u, u), LA,U 6= ∅
CapU (A) =
+∞,
LA,U = ∅.
Note the set monotonicity of the capacity: If V ⊂ U is open and A ⊂ B ⊂ V ,
then CapU (A) ≤ CapV (B).
Let
LeA,U = {u ∈ Fe (U ) : u
e ≥ 1 q.e. on A},
where u
e is a quasi-continuous modification of u. If LA,U 6= ∅ then there exists
a unique element eA ∈ LA,U such that
CapU (A) = EUD (eA , eA ).
Moreover, eA is the unique element of Fe (U ) satisfying eeA = 1 q.e. on A and
EUD (eA , v) ≥ 0 for all v ∈ Fe (U ) with ve ≥ 0 q.e. on A. See [FŌT94, Theorem
2.1.5 and p.71].
Now assume that A is compact. By the 0-order counterpart of [FŌT94,
Lemma 2.2.6] (cf. also [FŌT94, Lemma 2.2.10]), eA is a (0-order) potential.
That is, eA = GU νA , where GU is the Green function on U and νA is a finite
measure supported on A which charges no set of zero capacity. νA is called the
equilibrium measure. In particular, we have
Z
D
D
CapU (A) = EU (eA , eA ) = EU (GU νA , eA ) = eeA dνA = νA (A).
2
Geometric hypotheses on the underlying space
Let (X, d, µ, E, F ) be a MMD space.
Definition 2.1. (X, µ) satisfies the volume doubling property VD if there exists
a constant CVD ∈ (0, ∞) such that for every x ∈ X, R > 0,
V (x, 2R) ≤ CVD V (x, R),
where V (x, R) = µ(B(x, R)) denotes the volume of the ball B(x, R).
Let Ψ : [0, ∞) → [0, ∞) be a continuous strictly increasing bijection that has
the following property. There are constants 1 < β1 ≤ β2 < ∞ and C > 0 such
that, for all 0 < r ≤ R < ∞,
C
−1
R
r
β1
Ψ(R)
≤
≤C
Ψ(r)
6
R
r
β2
.
(2)
Definition 2.2. (i) (E, F ) satisfies the weak two-sided heat kernel bounds wHKE(Ψ) if the heat kernel p(t, x, y) on X exists and if there are positive
constants c1 , c2 , c3 , c4 and ǫ > 0 so that
1/(β2 −1) !
c1
Ψ(d(x, y))
p(t, x, y) ≤
exp −c2
V (x, Ψ−1 (t))
t
for all t ∈ (0, ∞) and almost every x, y ∈ X, and
p(t, x, y) ≥
c3
,
V (x, Ψ−1 (t))
for all t ∈ (0, ∞) and almost every x, y ∈ X with
d(x, y) ≤ ǫΨ−1 (t).
(ii) (E, F ) satisfies the local lower estimate LLE(Ψ) if there exist c5 > 0 and
ǫ ∈ (0, 1) such that, for any ball B = B(x0 , R) ( X, the Dirichlet heat
pD
B (t, x, y) exists and satisfies
pD
B (t, x, y) ≥
c5
,
V (x0 , Ψ−1 (t))
(3)
for all 0 < t ≤ Ψ(ǫR), and almost every x, y ∈ B(x0 , ǫΨ−1 (t)).
The following theorem is proved in [BGK12, Theorem 3.1].
Theorem 2.3. Assume VD is satisfied. Then the following are equivalent:
(i) X satisfies w-HKE(Ψ).
(ii) X satisfies w-LLE(Ψ).
Further, under any of the conditions (i), (ii), (iii), the heat kernel p(t, x, y)
is a continuous function of (t, x, y) ∈ R+ × X × X and, for any open subset
U ⊂ X, the Dirichlet heat kernel pD
U (t, x, y) is a continuous function of (t, x, y) ∈
R+ × U × U .
Remark 2.4. Condition w-HKE(Ψ) is also equivalent to a weak parabolic Harnack inequality w-PHI(Ψ), see [BGK12].
Definition 2.5. Let V, U be open sets in X with V ⊂ V ⊂ U . A function
φ ∈ F is called a cutoff function for (V, U ) if 0 ≤ φ ≤ 1, φ = 0 on X \ U and
φ = 1 on V quasi-everywhere.
The following variants CSD and CSA of the cutoff Sobolev inequality were
introduced in [AB]. Note that in this paper we do not require cutoff functions
to be continuous.
7
Definition 2.6. Let D0 , D1 be open subsets of X with D0 ⊂ D0 ⊂ D1 , and
let U = D1 \ D0 . We say that property CSD(D0 ,D1 ,θ) holds if there exists a
cutoff function φ for D0 ⊂ D1 which satisfies, for all f ∈ F ,
Z
Z
Z
1
f 2 dΓ(φ, φ) ≤
φ2 dΓ(f, f ) + θ
f 2 dµ.
(4)
8
U
U
U
Definition 2.7. We say that condition CSA(Ψ) holds if there exists a constant
CS such that for every x ∈ X, R > 0, r > 0 the condition CSD(B(x, R),
B(x, R + r), CS Ψ(r)−1 ) holds.
Definition 2.8. (E, F ) satisfies the elliptic Harnack inequality EHI, if there
exist constants C > 0 and δ ∈ (0, 1) such that, for any ball B(x, R) ( X and
any non-negative bounded function u ∈ F that is harmonic on B(x, R), we have
ess sup u(z) ≤ C ess inf u(z).
z∈B(x,δR)
z∈B(x,δR)
Definition 2.9. (E, F ) satisfies the weak Poincaré inequality PI(Ψ) on X if
there exist constants P0 and κ ≥ 1 such that for any ball B = B(x, R) ⊂ X,
and any f ∈ Fc (B),
Z
Z
2
(f − fB ) dµ ≤ P0 Ψ(R)
dΓ(f, f ).
B
B(x,κR)
R
Here fB = µ(B)−1 B f dµ.
Definition 2.10. (E, F ) satisfies the resistance condition RΨ if there exist
constants K, C > 1 such that for any B(x, (K + 1)R) ( X,
C −1
−1
Ψ(R)
Ψ(R)
≤ CapB(x,KR) (B(x, R))
.
≤C
V (x, R)
V (x, R)
Theorem 2.11. Assume that (E, F ) satisfies VD and w-HKE(Ψ). Then (E, F )
also satisfies EHI, and CSA(Ψ), PI(Ψ), and RΨ for all K > 1.
The classical proof of the Poincaré inequality (cf. [SC92]) uses the semigroup
associated with the Neumann-type form
EUN (f, g) = E(f, g),
f, g ∈ F (U ),
defined on an open set U ⊂ X. A proof in [BBK06] claimed that the Neumann
semigroup admits a kernel; it is not clear to the author that this is the case.
Here we give a proof that does not rely on the existence of the Neumann kernel.
Proposition 2.12. Assume that CSA(Ψ) holds. Then the Neumann-type form
(EUN , F (U )) is a strongly local Dirichlet form on L2 (U, µ|U ).
8
Proof. It is clear from the definition that (EUN , F (U )) is a strongly local Markovian symmetric form. It suffices to show that (EUN , F (U )) is closed. We follow
[GSC11, Proposition 2.50]. Let (ui ) be a Cauchy sequence in F (U ) for the norm
1/2
R
R
. Let u ∈ L2 (U, µ) be its limit in L2 (U, µ).
N (u) = U u2 dµ + U dΓ(u, u)
Let V ⊂ U be open and relatively compact in U . By CSA(Ψ) there exists a
cutoff function ψ for V ⊂ U . Hence, by the cutoff Sobolev inequality (4),
Z
Z
u2i dΓ(ψ, ψ)
ψ 2 dΓ(ui , ui ) + 2
E(ψui , ψui ) ≤ 2
U
U
Z
Z
2
≤ (2 + )
u2i dµ.
dΓ(ui , ui ) + C
8 U
U
Hence (ψui ) is Cauchy in F for the E1 -norm and converges to ψu in F . It
follows that u ∈ Floc (U ).
Since Γ(ui , ui ) − Γ(uj , uj ) ≤ Γ(ui − uj , ui − uj ), the sequence Γ(ui , ui ) of
Radon measures on U is Cauchy in the total variation norm. Let ν be the
Radon measure on U with finite total variation that is the limit of the sequence
Γ(ui , ui ). Since ν coincides
R with Γ(u, u) on any compact subset of U , we must
have ν = Γ(u, u). Hence U dΓ(u, u) < ∞ and u ∈ F (U ). Now we see that ui
tends to u in the norm N , hence the form (EUN , F (U )) is closed.
Let PUN (t), t > 0, be the strongly continuous contraction semigroup associated with the Neumann-type form (EUN , F (U )).
Lemma 2.13. Let B = B(x, R) ( X and ǫB = B(x, ǫR), ǫ ∈ (0, 1). Then
the Dirichlet semigroup on ǫB is dominated by the Neumann semigroup on B.
That is,
PtB,N f ≥ PtǫB,D f
∀0 ≤ f ∈ L2 (ǫB), t > 0.
(5)
Proof. Let f ∈ L2 (ǫB), f ≥ 0. Then
(PtǫB,D f − PtB,N f )+ ≤ PtǫB,D f.
Hence, by [GH08, Lemma 4.4], (PtǫB,D f − PtB,N f )+ ∈ F (B). Since PtǫB,D f and
hence (PtǫB,D f − PtB,N f )+ can be approximated by functions in Fc (ǫB), we find
that (PtǫB,D f − PtB,N f )+ ∈ F 0 (ǫB).
Now we follow [Ouh96, Proof of Theorem 3.3]. Let u = PtǫB,D f and v =
B,N
Pt f . Then
D
N
N
(u − v, (u − v)+ ) + EǫB
(v, (u − v)+ )
EǫB
(u, (u − v)+ ) = EǫB
N
N
≥ EǫB
(v, (u − v)+ ) = EB
(v, (u − v)+ ).
Rewriting,
Z
+
LD
ǫB u(u − v) dµ ≤
9
Z
+
LN
B v(u − v) dµ,
and
Z Z d
d
+
u (u − v) dµ ≤
v (u − v)+ dµ.
dt
dt
R
d
Hence, the function g(t) = ((u − v)+ )2 dµ satisfies dt
g ≤ 0 and g(0) = 0.
Therefore (u − v)+ = 0, which proves the claim.
Proof of Theorem 2.11. The elliptic Harnack inequality EHI follows from the
weak parabolic Harnack inequality w-PHI(Ψ) which is equivalent to w-HKE(Ψ),
see [BGK12, Theorem 3.1].
Let us show that (E, F ) is conservative. Then the implication VD, wHKE(Ψ) ⇒ CSA(Ψ) follows easily by adapting the arguments of [AB, Theorem
5.5, Lemma 5.3] to the present setting. The proof of [AB, Lemma 5.3] refers to
[GH], where conservativeness of the form is assumed.
In the case that X is bounded, it is clear that (E, F ) is conservative because
the constant function 1 is in F .
In the case when X is unbounded, it is proved in [GT12, Theorem 7.4] that
w-HKE(Ψ) is equivalent to EHI together with an exit time estimate E(Ψ). By
[GT12, Lemma 7.3], condition E(Ψ) implies the conservativeness. Note that
the heat kernel bounds which we denote by w-HKE(Ψ), are the same as UEweak
together with NLEweak in the notation of [GT12]. To see that these notions are
β 1−1
2
equivalent, it suffices to verify that Φ(d(x, y), t) ≍ Ψ(d(x,y))
, where Φ is
t
as in [GT12, (3.34)]. But this follows from [GT12, Lemma 3.19].
Next, we show PI(Ψ). We follow the line of reasoning in the proof of [SC02,
Theorem 5.5.1]. Let PtB,N , t > 0, be the Neumann semigroup on the ball
B = B(x, 1ǫ R), and PtǫB,D , t > 0, the Dirichlet semigroup on the ball ǫB =
B(x, R), where ǫ ∈ (0, 1) is the parameter in w-LLE(Ψ). Recall that w-HKE(Ψ)
is equivalent to w-LLE(Ψ) by Theorem 2.3. By Lemma 2.13,
PtB,N f ≥ PtǫB,D f
∀0 ≤ f ∈ L2 (ǫB), t > 0.
(6)
Let f ∈ F (B), y ∈ B(x, ǫR). Let φ be a cutoff function that is 1 on B(x, ǫR) and
has compact support in B(x, R). Applying (5) and w-LLE(Ψ) with t = Ψ(R),
we obtain
B,N
B,N
[f − PΨ(R)
f (y)]2 (y)
PΨ(R)
B,N
ǫB,D
φ[f − PΨ(R)
f (y)]2 (y)
≥ PΨ(R)
Z
B,N
2
pD
=
ǫB (Ψ(R), y, z)φ|f (z) − PΨ(R) f (y)| dµ(z)
B
Z
B,N
2
≥
pD
ǫB (Ψ(R), y, z)|f (z) − PΨ(R) f (y)| dµ(z)
B(x,ǫR)
Z
c5
B,N
≥
|f (z) − PΨ(R)
f (y)|2 dµ(z)
V (x, R) B(x,ǫR)
Z
c5
|f (z) − fB(x,ǫR) |2 dµ(z),
≥
V (x, R) B(x,ǫR)
10
where fB(x,ǫR) is the mean of f over the ball B(x, ǫR).
Integrating over B(x, ǫR) and using volume doubling yields
Z
B,N
B,N
[f − PΨ(R)
f (y)]2 (y)dµ(y)
PΨ(R)
B(x,ǫR)
Z
V (x, ǫR)
|f (z) − fB(x,ǫR) |2 dµ(z)
V (x, R) B(x,ǫR)
Z
|f (z) − fB(x,ǫR) |2 dµ(z).
≥ c(ǫ)
≥ c5
(7)
B(x,ǫR)
The next computation uses the fact that the Neumann semigroup PtB,N , t > 0,
is conservative, i.e. PtB,N 1 = 1 for all t > 0, and an L2 -contraction.
Z
B,N
B,N
[f − PΨ(R)
f (y)]2 (y)dµ(y)
PΨ(R)
B(x,ǫR)
Z
h
i2
h
i2
B,N
B,N
B,N
B,N
PΨ(R)
(f 2 ) − 2 PΨ(R)
f (y) + PΨ(R)
=
1(y) PΨ(R)
f (y) dµ(y)
B(x,ǫR)
B,N
f k2B(x,ǫR),2 = −
≤ kf k2B(x,ǫR),2 − kPΨ(R)
Z
Z
Ψ(R)
B,N
hLN
f, PsB,N f ids = 2
B Ps
0
0
Z
N
dΓ(f, f ).
≤ 2Ψ(R)EB (f, f ) = 2Ψ(R)
=2
Z
Ψ(R)
0
Ψ(R)
∂
kP B,N f k2B(x,ǫR),2ds
∂s s
N
EB
(PsB,N f, PsB,N f )ds
(8)
B
N
To see the last inequality, observe that s → EB
(PsB,N f, PsB,N f ) is a nonincreasing function. This can be proved by noting that
N
B,N
1/2 B,N
EB
(PsB,N f, PsB,N f ) = hLN
f, PsB,N f i = k(LN
Ps f k22 .
B Ps
B)
From (7) and (8) we obtain
Z
Z
2
|f (z) − fB(x,ǫR) |2 dµ(z) ≤
dΓ(f, f ).
Ψ(R)
c(ǫ)
B(x,ǫR)
B
We have proved that VD and w-HKE(Ψ) imply PI(Ψ).
Now RΨ for all K > 1 follows from VD and CSA(Ψ) together with PI(Ψ)
by the same arguments as in [BB04, Lemma 5.1].
3
3.1
Inner uniform domains and Green’s function
estimates
(Inner) uniformity
For the rest of this paper, we let (X, d, µ, E, F ) be a MMD space that satisfies
VD and w-HKE(Ψ).
11
In this Section we discuss inner uniformity conditions and some properties
of inner uniform domains. Section 3.1 is nearly identical to the corresponding
section in [LSCb]. We include it for the convenience of the reader and due to
its importance for the understanding of the main result.
Let Ω ⊂ X be open and connected. Recall that the inner metric on Ω is
defined as
dΩ (x, y) = inf length(γ)γ : [0, 1] → Ω continuous, γ(0) = x, γ(1) = y .
e be the completion of Ω with respect to dΩ , and ∂ e Ω = Ω\Ω
e
Let Ω
the boundary.
Ω
e
e
For x ∈ Ω, we will consider the inner balls BΩ
e (x, r) := {y ∈ Ω : dΩ (x, y) < r}
and BΩ (x, r) := BΩ
e (x, r) ∩ Ω. We always assume that the radius r of an the
inner ball is minimal in the sense that BΩ
e (x, r) 6= BΩ
e (x, s) for all s < r.
e dΩ be its completion with respect to the metric
For an open set B ⊂ Ω, let B
e dΩ \ B be its boundary. Let ∂Ω B = ∂ e B ∩ Ω be the part
dΩ , and let ∂Ω
eB = B
Ω
of the boundary that lies in Ω.
For a set V ⊂ Ω let diamΩ (V ) be the diameter of V with respect to the inner
metric dΩ .
Definition 3.1. (i) Let γ : [α, β] → Ω be a rectifiable curve in Ω and let
c ∈ (0, 1), C ∈ (1, ∞). We call γ a (c, C)-uniform curve in Ω if
δΩ γ(t) ≥ c · min d γ(α), γ(t) , d γ(t), γ(β) ,
for all t ∈ [α, β], (9)
and if
length(γ) ≤ C · d γ(α), γ(β) .
The domain Ω is called (c, C)-uniform if any two points in Ω can be joined
by a (c, C)-uniform curve in Ω.
(ii) Inner uniformity is defined analogously by replacing the metric d on X
with the inner metric dΩ on Ω.
(iii) The notion of (inner) (c, C)-length-uniformity
is defined analogously by
replacing d(γ(a), γ(b)) by length(γ [a,b] ).
The following proposition is taken from [GSC11, Proposition 3.3]. See also
[Mar, Lemma 2.7].
Proposition 3.2. Assume that (X, d) is a complete, locally compact length
metric space with the property that there exists a constant D such that for any
r > 0, the maximal number of disjoint balls of radius r/4 contained in any ball
of radius r is bounded above by D. Then any connected open subset U ⊂ X is
uniform if and only if it is length-uniform.
Lemma 3.3. Let Ω be a (cu , Cu )-inner uniform domain in (X, d). For every
e
ball B = BΩ
e (x, r) in (Ω, dΩ ) with minimal radius, there exists a point xr ∈ B
e \ Ω) ≥ cu r/8.
with dΩ (x, xr ) = r/4 and d(xr , Ω
12
Proof. See [GSC11, Lemma 3.20].
The following lemma is crucial for the proof of the boundary Harnack principle on inner uniform domains rather than uniform domains. A version of this
lemma was already used in [Anc07] to prove a boundary Harnack principle on
inner uniform domains in Euclidean space.
e → Ω be the natural projection onto the completion Ω of Ω with
Let p : Ω
e and any ball D = B(p(x), r), let
respect to the metric d in X. For any x ∈ Ω
′
−1
D be the connected component of p (D ∩ Ω) that contains x. It follows that
e contains x.
D′ ∩ Ω is the connected component of D ∩ Ω whose closure in Ω
Lemma 3.4. Suppose µ has the volume doubling property on X. Let Ω be a
(cu , Cu )-inner uniform domain in (X, d). Then there exists a positive constant
e
CΩ such that for any ball D = B(p(x), r) with x ∈ Ω,
′
BΩ
e (x, r) ⊂ D ⊂ BΩ
e (x, CΩ r).
The constant CΩ depends only on the volume doubling constant CVD and on the
inner uniformity constants cu , Cu of Ω.
Proof. See [LSCb, Lemma 3.8].
Remark 3.5. In the above lemma, the hypothesis that the whole domain
Ω is inner uniform can be relaxed to the hypothesis that any two points in
BΩ (x, CΩ r) can be connected by a path that is inner uniform in Ω. See [LSCb,
Remark 3.9(ii)].
Fix cu ∈ (0, 1) and Cu ∈ (1, ∞). Let A3 = 2(12 + 12Cu ), A0 = A3 + 7.
Definition 3.6. For ξ ∈ ∂Ω
e Ω, let Rξ be the largest radius so that
(i) 26Rξ /cu ≤ diamΩ (Ω)/2 if Ω is a bounded domain.
(ii) Any two points in BΩ
e (ξ, (A0 ∨12/cu )Rξ ) can be connected by a curve that
is (cu , Cu )-inner uniform in Ω.
The condition Rξ > 0 for some boundary point ξ can be understood as a
local inner uniformity condition.
Remark 3.7. An inner uniform domain Ω clearly satisfies Rξ > 0 for all bounde \ Ω. However, the converse is not true unless the domain is
ary points ξ ∈ Ω
bounded.
3.2
Green’s function estimates
Recall that (X, d, µ, E, F ) is a MMD space that satisfies VD and w-HKE(Ψ).
Theorem 3.8. Let B = B(a, R) be a ball with B(a, KR) ( X for some K > 1.
13
(i) For any fixed ǫ ∈ (0, 1) there are constants c2 , C2 ∈ (0, ∞) such that for
any x, y ∈ B(a, R), t ≥ ǫΨ(R), the Dirichlet heat kernel pD
B is bounded
above by
pD
B (t, x, y) ≤
C2
exp(−c2 t/Ψ(R)).
V (x, R)
(ii) There exist constants c3 , C3 ∈ (0, ∞) such that for any x, y ∈ B(a, R),
t > 0, the Dirichlet heat kernel pD
B is bounded above by
1/(β2 −1) !
Ψ(d(x, y))
C3
D
.
exp −c3
pB (t, x, y) ≤
V (x, Ψ−1 (t) ∧ R)
t
All the constants ci , Ci above depend only on K, CVD , the constants in wHKE(Ψ, and β1 , β2 in (2). They are independent of a and R.
Proof of Theorem 3.8. (ii) follows from the upper bound in w-HKE(Ψ) and the
set monotonicity of the kernel pD
B . (i) follows from changing notation in [HSC01,
Lemma 5.13 part 3].
Lemma 3.9. For any relatively compact open set V ⊂ X, the Green function
0
y 7→ GV (x, y) is in Floc
(V, V \ {x}) for any fixed x ∈ V .
Proof. We follow the line of reasoning in [GSC11, Proof of Lemma 4.7]. Since
we do not have a carré du champ, we use CSA(Ψ) to control the energy of cutoff
functions.
Let Ω ⊂ V be open and relatively compactSin V with d(Ω,Sx) > 0. Pick
finitely many balls Bi = B(zi , si ) so that Ω ⊂ i Bi and K = i B(zi , 4si ) ⊂
X \ {x}. For each i, there exists by CSA(Ψ) a continuous cut-off function
ψi
P
with ψ = 1 on Bi and ψi = 0 on X \ B(zi , 2si ). Let ψ(y) = min{1, i ψi (y)}
for all y ∈ K, and f ♯ = ψGV (x, ·). Then f ♯ = GV (x, ·) on Ω. It remains to
show that f ♯ is in F 0 (V ).
0
Recall that the map y 7→ pD
V (t, x, y) is in F (V ). The heat kernel upper
bounds of Theorem 3.8 imply that ψGV (x, ·) ∈ L2 (X, µ). Indeed, pD
V (t, x, y) ≤
pD
(t,
x,
y)
with
B
=
B(x,
R)
and
R
is
chosen
large
enough
so
that
V
⊂ B. By
B
Theorem 3.8, there are constants c, C(B) ∈ (0, ∞) such that for all t ≥ Ψ(R)
and x, y ∈ V ,
−ct/Ψ(R)
pD
,
V (t, x, y) ≤ C(B)e
(10)
and there are constants c′ , C ′ ∈ (0, ∞) depending on V , K, β1 , β2 , and the
constants in VD and w-HKE(Ψ) such that for all 0 < t < Ψ(R) and x, y ∈ V ∩K,
′
′ −c /t
pD
.
V (t, x, y) ≤ C e
(11)
R∞
This shows that the integral ψGV (x, ·) = 0 ψpD
V (t, x, ·)dt converges at 0 and
∞ in L2 (X, µ). Hence ψGV (x, ·) is in L2 (X, µ).
14
Rb
2
For fixed 0 < a < b < ∞, set g = a pD
V (t, x, ·)dt and observe that ψg, ψ g ∈
0
F (V ). Thus, applying CSA(Ψ), we obtain
Z
Z
Z
2
g 2 dΓ(ψ, ψ)
ψ dΓ(g, g) + 2
dΓ(ψg, ψg) ≤ 2
K∩V
K∩V
V
Z
XZ
dΓ(g, g) + C
g 2 dΓ(ψi , ψi )
≤2
V
≤ C′
Z
i
B(zi ,2si )∩V
g(−Lg) dµ + C sup Ψ(2si )−1
Z
g 2 dµ
Z
Z
′
D
D
≤C
g 2 dµ
g pV (a, x, ·) − pV (b, x, ·) dµ + C
V
V
Z
Z
≤ C′
g 2 dµ.
(a,
x,
·)dµ
+
C
gpD
V
i
V
V
V
V
The constants C and C ′ change from line to line and depend on Ω, Ψ and the
constant CS from CSA(Ψ). Now, observe that (10)-(11) imply that
!2
Z
Z
Z
b
g 2 dµ =
V
V
a
pD
V (t, x, ·)dt
dµ
tends to 0 when a, b tend to infinity or when a, b tend to 0 (this is indeed
the argument we used above to Rshow that GV (x, ·) is in L2 (X, dµ)). The same
estimates (10)-(11) imply that V gpD
V (a, x, ·)dµ tends to 0 when a, b tend to
infinity
or
when
a,
b
tend
to
0.
This
implies that the integral ψGV (x, y) =
R∞
0
ψ 0 pD
V (t, x, ·)dt converges in F (V ) as desired.
Lemma 3.10. Let B(z, 2R) ( X.
(i) There is a constant C1 depending only on β1 , β2 , and the constants in VD
and w-HKE(Ψ), such that for any x, y ∈ B(z, R) with d(x, y) ≥ (ǫR) for
some ǫ > 0,
Ψ(R)
.
(12)
GB(z,R) (x, y) ≤ C1
V (x, R)
(ii) Fix θ ∈ (0, 1). There is a constant C2 depending only on θ, β1 , β2 , and
the constants in VD and w-HKE(Ψ), such that
∀x, y ∈ B(z, θR),
GB(z,R) (x, y) ≥ C2
Ψ(R)
.
V (x, R)
(13)
Proof. The lower bound (13) follows from integrating (3) from t = Ψ θǫ R to
t = Ψ 2 θǫ R and applying VD and (2).
For the upper bound (12) we again use the heat kernel estimates of Theorem
3.8,

1/(β2 −1) Ψ(d(x,y))
C3

,
t < 12 Ψ(R),
exp
−c
3
V (x,Ψ−1 (t))
t
pD
(t,
x,
y)
≤
B
 C2
t ≥ 12 Ψ(R),
V (x,R) exp(−c2 t/Ψ(R)),
15
where B = B(z, R). Therefore,
Z ∞
Z ∞
D
pB (t, x, y)dt ≤
1
2 Ψ(R)
1
2 Ψ(R)
Ψ(R)
C2
exp(−c2 t/Ψ(R))dt ≤ C
,
V (x, R)
V (x, R)
and
Z
0
1
2 Ψ(R)
pD
B (t, x, y)dt
1/(β2 −1) !
Ψ(d(x, y))
C3
dt
exp −c3
≤
V (x, Ψ−1 (t))
t
0
1/(β2 −1) !
Z 12 Ψ(R)
C3
Ψ(ǫR)
V (x, R)
≤
dt.
exp −c3
V (x, R) 0
V (x, Ψ−1 (t))
t
Z
1
2 Ψ(R)
Due to VD and (2), the integrand is bounded by a constant independent of R.
Hence, the right hand side is less or equal to C VΨ(R)
(x,R) . Finally,
GB(z,R) (x, y) ≤
Z
0
1
2 Ψ(R)
pD
B (t, x, y)dt +
Z
∞
1
2 Ψ(R)
pD
B (t, x, y)dt
≤C
Ψ(R)
.
V (x, R)
Lemma 3.11. Fix θ ∈ (0, 1). Let U ⊂ X be open.
(i) Let B(z, 2R) ( X. Then there is a constant C1 depending only on θ, β1 ,
β2 , and the constants in VD and w-HKE(Ψ), such that
GBU (z,R) (x, y) ≤ GU∩B(z,R) (x, y) ≤ C1
Ψ(R)
,
V (x, R)
(14)
for all x, y ∈ U ∩ B(z, R) with d(x, y) ≥ (θR).
(ii) Let z ∈ U , R > 0, δ ∈ (0, 1/3). Suppose that U ∩ B(z, 2R) ( U and
any two points in BU (z, δR) can be connected by a (cu , Cu )-inner uniform
curve in U . Then there is a constant C2 depending only on cu , Cu , θ, β1 ,
β2 , and the constants in VD and w-HKE(Ψ), such that
GBU (z,R) (x, y) ≥ C2
Ψ(R)
,
V (x, R)
(15)
for all x, y ∈ BU (z, δR) with d(x, X \ U ), d(y, X \ U ) ∈ (θR, ∞) and
dU (x, y) ≤ δR/Cu .
Proof. We follow the line of reasoning of [GSC11, Lemma 4.9]. (i) Set B =
B(z, R), W = U ∩ B(z, R). The upper bound (14) follows easily from Lemma
3.10 and the monotonicity inequality GW ≤ GB . (ii) By assumption, there is an
ǫ1 > 0 such that, for any x, y ∈ BU (z, δR) satisfying d(x, X \ U ), d(y, X \ U ) >
16
θR, there is a path in U from x to y of length less than Cu dU (x, y) ≤ δR that
stays at distance at least ǫ1 R from X \ U . Since x, y ∈ BU (z, δR) and δ < 1/3,
this path is contained in BU (z, R) ∩ {ζ ∈ U : d(ζ, X \ U ) > ǫ1 R}. Using this
path, the Harnack inequality easily reduces the lower bound (15) to the case
when y satisfies d(x, y) = ηR for some arbitrary fixed η ∈ (0, ǫ1 ) small enough.
Pick η > 0 so that, under the conditions of the Lemma, the ball Bx = B(x, 2ηR)
is contained in BU (z, R). Let W = BU (z, R). Then the monotonicity property
of Green functions implies that GW (x, y) ≥ GBx (x, y). Lemma 3.10 and the
volume doubling property then yield GW (x, y) ≥ C2 VΨ(R)
(x,R) . This is the desired
lower bound.
3.3
Maximum principle
Let Ω ⊂ X be open. In this subsection, we do not assume inner uniformity.
Proposition 3.12 (Maximum principle). Suppose (EΩD , F 0 (Ω)) is transient. Let B = BΩ (x, r) be an inner ball such that BΩ (x, 2r) ( Ω.
(i) Let f be a non-negative harmonic function on BΩ (x, 2r) with Dirichlet
e
boundary condition along ∂Ω
e Ω ∩ BΩ
e (x, 2r). Then f has a modification f
which is continuous on BΩ (x, 3r/2) and satisfies
sup fe = sup fe.
B
∂Ω B
(ii) Let f be a non-negative harmonic function on Ω \ BΩ (x, r) with Dirichlet
e
boundary condition along ∂Ω
e Ω \ BΩ
e (x, r). Then f has a modification f
which is continuous on Ω \ BΩ (x, 3r/2) and satisfies
sup
Ω\BΩ (x,2r)
fe =
sup
∂Ω BΩ (x,2r)
fe.
Proof. Since the proof of (ii) is very similar to that of (i), we only show (i). Let
f be as required in (i). Then there is a function f ♯ ∈ F 0 (Ω) such that f = f ♯
a.e. on BΩ (x, 3r/4). Replacing f by f ♯ if necessary, it suffices to consider the
case when f ∈ F 0 (Ω). Recall from Theorem 2.11 that EHI holds. By a standard
argument, (e.g., [SC02, Theorem 5.4.7]) the elliptic Harnack inequality implies
the Hölder continuity of harmonic functions. Hence, f admits a modification fe
which is continuous on BΩ (x, 3r/2), in particular on the part of the boundary
∂Ω B. Let (Xt ), t > 0, be the diffusion process associated with (EΩD , F 0 (Ω)) and
let F 0 (Ω)e be the extended Dirichlet space with respect to (EΩD , F 0 (Ω)). Set
U = Ω \ B and let
σU = inf{t > 0|Xt ∈ U }
be the first hitting time of the set U for the process X. Let
HU fe(z) = Ez (fe(XσU ); σU < ∞).
17
It is immediate from the definition that HU fe satisfies the maximum principle.
We show that HU fe is a quasi-continuous modification of f . Since (EΩD , F 0 (Ω))
is transient, its extended Dirichlet space admits the orthogonal decomposition
F 0 (Ω)e = F 0 (Ω)e,Ω\U ⊕ HU ,
F 0 (Ω)e,Ω\U = u ∈ F 0 (Ω)e : u
e = 0 q.e. on U .
Let PHU : F 0 (Ω)e → HU be the orthogonal projection. By [FŌT94, Theorem
4.3.2], HU fe is a quasi-continuous modification of PHU f .
Since f , hence also −f , is harmonic on U , and by the 0-order analogue of
[FŌT94, Lemma 2.2.6 (iii) ⇒ (ii)], we have
E(f, v) = 0,
∀v ∈ F 0 (Ω)e,Ω\U .
That is, f ∈ HU . Hence, f = PHU f . We have proved that f admits the
quasi-continuous modification HU fe which satisfies the maximum principle.
3.4
Representation formula for harmonic functions
Let Ω ⊂ X be an open subset. Suppose (EΩD , F 0 (Ω)) is transient.
An essential step in the proof of the boundary Harnack principle is the
reduction to Green functions estimates, see Section 4.1, through a representation
formula for harmonic functions. On Euclidean space, a representation formula
was used for this purpose by Aikawa ([Aik01], [ALM03, p.331]). In [GSC11] it
is claimed that the representation extends to (non-fractal) symmetric Dirichlet
spaces. Note that classic potential theoretic references require the domain of
integration to be compact (see, e.g., [GH13, Lemma 6.2]), which is not the case
in (16) below. Thus, we include a full proof.
e \ Ω with Rξ > 0. Suppose that any
Proposition 3.13. Let r > 0 and ξ ∈ Ω
two points in BΩ (ξ, (1 + 2CΩ )r) can be connected by a path that is (cu , Cu )inner uniform in Ω. Let f be non-negative harmonic on BΩ (ξ, (1 + 2CΩ )r) with
Dirichlet boundary condition along (∂Ω
e Ω) ∩ BΩ
e (ξ, (1 + 2CΩ )r). Then f admits
e
a modification f which satisfies
Z
fe(x) =
GBΩ (ξ,2CΩ r) (x, y)dνf (y), ∀x ∈ BΩ (ξ, r).
(16)
∂Ω BΩ (ξ,r)
e → Ω is the natural projection. By CSA(Ψ) there exists
Proof. Recall that p : Ω
a cutoff function ψ for B(p(ξ), r) in B(p(ξ), 2r) which satisfies the cutoff Sobolev
inequality. Let B ′ be the connected component of p−1 (Ω ∩ B(p(ξ), 2r)) which
′
′
contains ξ. By Lemma 3.4, we have B ′ ⊂ BΩ
e (ξ, 2CΩ r). Let BΩ = B ∩ Ω. Set
g := f ψ1BΩ′ .
We claim that g ∈ F 0 (BΩ (ξ, 2CΩ r)). Since f satisfies Dirichlet boundary condi♯
0
tion on (∂Ω
e Ω ∩ BΩ (ξ, (1 + 2CΩ )r), there exists a function f ∈ F (Ω) such that
18
f = f ♯ on BΩ (ξ, (1/2 + 2CΩ )r). In particular, g = f ♯ ψ1BΩ′ . Hence, there is a
R
sequence (fn ) in F ∩ C0 (Ω) such that E(f ♯ − fn , f ♯ − fn ) + (f ♯ − fn )2 dµ → 0.
Let
[
e \ Ω) < 1 d(Vn , Ω
e \ Ω)}.
Vn =
supp(fk ), Ωn = {y ∈ Ω : dΩ (y, Ω
n
k≤n
′
Let φn ∈ F be a cutoff function that is 1 on Vn ∩BΩ
and 0 on Ωn+1 ∪(B(p(ξ), 2r)\
′
BΩ ). Let
gn := fn ψφn .
Note that gn ∈ Fc (Ω). Write
gn − g = (fn − f ♯ )ψ1BΩ′ + fn ψ(φn − 1BΩ′ ).
′
′
Observe that the support of fn ψ is contained in (Vn ∩ BΩ
) ∪ (B(p(ξ), 2r) \ BΩ
),
♯
where (φn − 1BΩ′ ) vanishes. Hence, gn − g = (fn − f )ψ1BΩ′ . In particular, (gn )
converges to g in the L2 -norm. Next, we show that (gn ) is a Cauchy sequence
for the E1 -norm. Observe that for m > n,
fn ψ(φn − φm ) = 0,
and hence
gn − gm = (fn − fm )ψφm .
Therefore, applying CSA(Ψ) and using the fact that ψ, φm ≤ 1,
E(gn − gm , gn − gm ) = E((fn − fm )ψφm , (fn − fm )ψφm )
Z
Z
≤ 2 dΓ(fn − fm , fn − fm ) + 4 (fn − fm )2 dΓ(ψ, ψ)
Z
+ 4 (fn − fm )2 ψ 2 dΓ(φm , φm )
Z
≤ CE1 (fn − fm , fn − fm ) + 4 (fn − fm )2 ψ 2 dΓ(φm , φm ).
By definition of (fn ), E1 (fn − fm , fn − fm ) is a Cauchy sequence. The second
integral on the right hand side is equal to
Z
Z
(fn − fm )2 ψ 2 φm dΓ(φm , φm ) + (fn − fm )2 ψ 2 (1 − φm )dΓ(φm , φm ) = 0
due to strict locality. Indeed, φm = 0 on the support of (fn − fm )2 ψ 2 (1 − φm )
and φm = 1 on the support of (fn − fm )2 ψ 2 φm .
Since (E, F ) is closed, the sequence (gn ) converges in the E1 -norm to its
L2 -limit g and g ∈ F 0 (BΩ (ξ, 2CΩ r)).
Let Y = BΩ (ξ, 2CΩ r), A = {x ∈ Ω : dΩ (ξ, x) ≤ r} and F = ∂Ω BΩ (ξ, r). Let
χ be a solution to the Dirichlet problem
(
χ is harmonic on Y \ A
χ − g ∈ F 0 (Y \ A).
19
Existence of χ is guaranteed by [GT12, Lemma 7.1]. Let
(
g on A
ĝ :=
χ on Y \ A.
Following the line of arguments given in [GH13, Proof of Lemma 6.4], it can
be shown that ĝ is superharmonic on Y \ F .
Hence, EYD (g, v) ≥ 0 for all v ∈ F 0 (Y )e such that ve ≥ 0 on F . Now it
follows from the 0-order version of [FŌT94, Lemma 2.2.6] that ĝ is a (0 order)
potential for some positive Radon measure ν supported on F . By [FŌT94,
Theorem 2.2.5],
Z
ve(x)ν(dx), ∀v ∈ F 0 (Y )e .
EYD (g, v) =
∂Ω B(ξ,r)
Applying this to v = GY φ for any test function φ ∈ C0 (Y ), we obtain
Z
Z Z
D
ĝ(x)φ(x)dµ(x) = EY (g, v) =
GY (y, x)φ(x)dµ(x)dν(y)
Y
Y
ZF Z
=
GY (y, x)dν(y) φ(x)dµ(x).
Y
Hence,
ĝ(x) =
Z
GY (y, x)dν(y)
F
for a.e. x ∈ Y.
F
Since f = ĝ on BΩ (ξ, r), the assertion follows for almost every x ∈ BΩ (ξ, r).
Since f is harmonic, it satisfies EHI, hence admits a continuous modification fe.
Also, the Green function is continuous (see, e.g., [GH13, Lemma 5.2]). Hence,
the assertion follows.
4
4.1
Boundary Harnack Principle
Reduction to Green functions estimates
Let (X, d, µ, E, F ) be a MMD space that satisfies VD and w-HKE(Ψ).
Let Ω ⊂ X be open and connected. Fix cu ∈ (0, 1) and Cu ∈ (1, ∞). Let
A3 = 2(12 + 12Cu), A0 = A3 + 7. We will assume later that Ω is locally (cu , Cu )inner uniform near a boundary point ξ in the sense of Definition 3.6, that is,
Rξ > 0. We do not require the inner uniformity of the whole domain Ω.
Theorem 4.1. There exists a constant A′1 ∈ (1, ∞) such that for any ξ ∈ ∂Ω
eΩ
with Rξ > 0 and any
0 < r < inf{Rξ′ : ξ ′ ∈ ∂Ω
e Ω ∩ BΩ
e (ξ, 7Rξ )},
we have
GY ′ (x, y)
GY ′ (x, y ′ )
≤ A′1
,
′
GY ′ (x , y)
GY ′ (x′ , y ′ )
20
for all x, x′ ∈ BΩ (ξ, r) and y, y ′ ∈ ∂Ω BΩ (ξ, 6r). Here Y ′ = BΩ (ξ, A0 r). The
constant A′1 depends only on cu , Cu , β1 , β2 , and the constants in VD and
w-HKE(Ψ).
The proof of this theorem is the content of Section 4.2 below. It is based
on the estimates for the Green’s functions in Section 3.2 and the maximum
principles of Section 3.3.
Theorem 4.2. There exists a constant A1 ∈ (1, ∞) such that for any ξ ∈ ∂Ω
eΩ
with Rξ > 0, any
0<r<
A0
inf{Rξ′ : ξ ′ ∈ ∂Ω
e Ω ∩ BΩ
e (ξ, 7Rξ )},
2CΩ
and any two non-negative harmonic functions u, v on BΩ (ξ, (1 + 2CΩ )r) with
Dirichlet boundary condition along (∂Ω
e Ω) ∩ BΩ
e (ξ, (1 + 2CΩ )r), we have
u(x)
v(x)
≤ A1
,
′
u(x )
v(x′ )
Ω
for all x, x′ ∈ BΩ (ξ, 2C
A0 r). The constant A1 depends only on cu , Cu , β1 , β2 ,
and the constants in VD and w-HKE(Ψ). The constant CΩ depends only on
cu , Cu and the constant in VD.
2CΩ
′
′
Proof. Fix ξ ∈ ∂Ω
e Ω and r > 0 as in the theorem. Let r = A0 r and Y =
′
BΩ (ξ, A0 r ). By the representation formula of Proposition 3.13, there exists a
Borel measure νu such that
Z
u(x) =
GY ′ (x, y)dνu (y)
(17)
∂Ω BΩ (ξ,r)
for all x ∈ BΩ (ξ, r), where GY ′ is the Green’s function corresponding to the
Dirichlet form (EYD′ , D(EYD′ )).
By Theorem 4.1, there exists a constant A′1 ∈ (1, ∞) such that for all x, x′ ∈
BΩ (ξ, r′ ) and all y, y ′ ∈ ∂Ω BΩ (ξ, 6r′ ), we have
′
GY ′ (x, y)
′ GY ′ (x, y )
≤
A
.
1
GY ′ (x′ , y)
GY ′ (x′ , y ′ )
For any (fixed) y ′ ∈ ∂Ω BΩ (ξ, 6r′ ), we find that
Z
1
GY ′ (x, y ′ )
u(x)
≤
GY ′ (x′ , y)dνu (y)
A′1
GY ′ (x′ , y ′ ) ∂Ω BΩ (ξ,6r)
=
GY ′ (x, y ′ )
u(x′ ) ≤ A′1 u(x).
GY ′ (x′ , y ′ )
We get a similar inequality for v. Thus, for all x, x′ ∈ BΩ (ξ, r′ ),
v(x)
GY ′ (x, y ′ )
1 u(x)
≤ A′1
≤
.
′
′
′
′
A1 u(x )
GY ′ (x , y )
v(x′ )
21
(18)
Corollary 4.3 (Carleson estimate). Let ξ ∈ ∂Ω
e Ω with Rξ > 0 and r ∈ (0, Rξ /3).
Let u be a non-negative harmonic function on BΩ (ξ, A0 r) with weak Dirichlet
boundary condition along (∂Ω
e Ω) ∩ BΩ
e (ξ, 6r). Then, for any x ∈ BΩ (ξ, r/2),
u(y) ≤ Cu(xr )
∀y ∈ BΩ (x, r/2),
where xr is as in Lemma 3.3. The constant C depends only on cu , Cu , β1 , β2 ,
and the constants in VD and w-HKE(Ψ).
Proof. This can be proved by applying the boundary Harnack principle of Theorem 4.2 to u and a Green function on a suitably chosen ball, and then using
the Green function estimates of Lemma 3.11. The proof is omitted since it is
similar to the proof of [GSC11, Theorem 4.17].
4.2
Proof of Theorem 4.1
We follow closely [Aik01], [GSC11] and [LSCb]. Let Ω be as above and fix
ξ ∈ ∂Ω
e Ω with Rξ > 0.
Definition 4.4. For η ∈ (0, 1) and any open U ⊂ X, define the capacity width
wη (U ) by
(
)
CapB(x,2r) B(x, r) \ U
wη (U ) = inf r > 0 : ∀x ∈ U,
≥η .
CapB(x,2r) B(x, r)
Note that wη (U ) is a decreasing function of η ∈ (0, 1) and an increasing
function of the set U .
Lemma 4.5. There are constants A7 , η ∈ (0, ∞) depending only on cu , Cu , β1 ,
β2 , and the constants in VD and w-HKE(Ψ), such that for all 0 < r < 2Rξ ,
wη ({y ∈ Y ′ : dΩ (y, ∂Ω
e Ω) < r}) ≤ A7 r.
Proof. We follow [GSC11, Lemma 4.12]. Let Yr = {y ∈ Y ′ : dΩ (y, ∂Ω
e Ω) < r}
and y ∈ Yr . Since r < cu diamΩ (Ω)/12, there exists a point x ∈ Ω such that
dΩ (x, y) = 4r/cu . There exists an inner uniform curve connecting y to x in
Ω. Let z ∈ ∂Ω BΩ (y, 2r/cu ) be a point on this curve and note that dΩ (y, z) =
2r/cu ≤ dΩ (x, y) − dΩ (y, z) ≤ dΩ (x, z). Hence,
dΩ (z, ∂Ω
e Ω) ≥ cu min{dΩ (y, z), dΩ (z, x)} = 2r.
So for any y ∈ Yr there exists a point z ∈ ∂Ω BΩ (y, 2r/cu ) with dΩ (z, ∂Ω
e Ω) = 2r.
Thus, B(z, r) ⊂ B(y, A7 r) \ Yr if A7 = 2/cu + 1. The capacity of B(y, A7 r) \ Yr
in B(y, 2A7 r) is larger than the capacity of B(z, r) in B(y, 2A7 r), which is larger
than the capacity of B(z, r) in B(z, 3A7 r). The latter capacity is comparable to
V (z, r)/Ψ(r), since RΨ holds for all K > 1 by Theorem 2.11. Hence, wη (Yr ) ≤
A7 r for some η > 0.
22
Fix η ∈ (0, 1) small enough so that the conclusion of Lemma 4.5 applies and
write w(U ) := wη (U ) for the capacity width of an open set U ⊂ Ω.
The following lemma relates the capacity width to the harmonic measure ω.
We write f ≍ g to indicate that C1 g ≤ f ≤ C2 g, for some constants C1 , C2 that
only depend on cu , Cu , β1 , β2 , and the constants in VD and w-HKE(Ψ).
For any open set B ⊂ X let ∂X B := B \ B denote the boundary of B with
respect to the metric d on X.
Lemma 4.6. There is a constant a1 depending only on β1 , β2 , and the constants
in VD and w-HKE(Ψ), such that for any non-empty open set U ⊂ X and any
x ∈ U , r > 0, we have
ωU∩B(x,r) (x, U ∩ ∂X B(x, r)) ≤ exp(2 − a1 r/w(U )).
Proof. We follow [Aik01, Lemma 1] and [GSC11, Lemma 4.13]. We may assume
that r/w(U ) > 2. For any κ ∈ (0, 1), we can pick w(U ) ≤ s < w(U ) + κ so that
CapB(y,2s) (B(y, s) \ U )
CapB(y,2s) (B(y, s))
≥η
∀y ∈ U.
Fix y ∈ U and let E = B(y, s) \ U . Let νE be the equilibrium measure of E in
B = B(y, 2s). We claim that there exists A2 > 0 such that
GB νE ≥ A2 η
on B(y, s).
(19)
Let F = B(y, s) and νF be the equilibrium measure of F in B. Then, by EHI,
we have for any z with d(y, z) = 3s/2,
GB (z, ζ) ≍ GB (z, y) ∀ζ ∈ B(y, s).
Hence,
GB νF (z) =
Z
GB (z, ζ)νF (dζ) ≍ GB (z, y)νF (F )
F
and
GB νE (z) =
Z
GB (z, ζ)νE (dζ) ≍ GB (z, y)νE (E).
E
Moreover, since νF (F ) = CapB (F ), RΨ together with Lemma 3.10 yield that
GB νF (z) ≃ 1. Hence, by the choice of s, for any z ∈ ∂X B(y, 3s/2),
GB νE (z) ≍
GB νE (z)
νE (E)
CapB (E)
≍
≍
≥ η.
GB νF (z)
νF (F )
CapB (F )
This proves (19).
Let x ∈ U . For simplicity, write ω(·) = ωU∩B(x,r) (·, U ∩ ∂X B(x, r)). Let k
be the integer such that 2kw(U ) < r < 2(k + 1)w(U ), and pick s > w(U ) so
close to w(U ) that 2ks < r. We claim that
sup
{ω} ≤ (1 − A2 η)j
U∩B(x,r−2js)
23
(20)
for j = 0, 1, . . . , k with A2 , η as in (19). Note that for j = k, (20) yields the
inequality stated in this Lemma:
r
ω(x) ≤ (1 − A2 η)k ≤ exp log (1 − A2 η) 2w(U )
≤ e2 exp(−a1 r/w(U )),
with a1 = −(log(1 − A2 η))/2.
Inequality (20) is proved by induction, starting with the trivial case j = 0.
Assume that (20) holds for j − 1. By the maximum principle of [GH13, Lemma
4.1(2)], it suffices to prove
sup
{ω} ≤ (1 − A2 η)j .
(21)
U∩∂X B(x,r−2js)
Let y ∈ U ∩ ∂X B(x, r − 2js). Then B(y, 2s) ⊂ B(x, r − 2(j − 1)s) so that
the induction hypothesis implies that
ω ≤ (1 − A2 η)j−1
on U ∩ B(y, 2s).
Since ω vanishes (quasi-everywhere) on (∂X U ) ∩ B(x, r) ⊃ (∂X U ) ∩ B(y, 2s),
the mean value property implies that
ω(b) ≤ (1 − A2 η)j−1 ωU∩B(y,2s) (b, U ∩ ∂X B(y, 2s))
for any b ∈ U ∩ B(y, 2s). To estimate
u = ωU∩B(y,2s) (·, U ∩ ∂X B(y, 2s)),
on U ∩ B(y, 2s), we compare it to
v = 1 − GB(y,2s) νE ,
where, as above, νE denotes the equilibrium measure of E = B(y, 2s) \ U in
B(y, 2s). Both functions are harmonic in U ∩ B(y, 2s) and u ≤ v on ∂X (U ∩
B(y, 2s)) quasi-everywhere (in the limit sense). By (19), this implies
u = ωU∩B(y,2s) (·, U ∩ ∂X B(y, 2s)) ≤ v ≤ 1 − A2 η
on U ∩ B(y, 2s). Hence,
ω ≤ (1 − A2 η)j
on U ∩ B(y, 2s).
Since this holds for any y ∈ U ∩ ∂X B(x, r − 2js), (21) is proved.
Lemma 4.7. There exist constants A2 , A3 ∈ (0, ∞) depending only on cu , Cu ,
β1 , β2 , and the constants in VD and w-HKE(Ψ), such that for any 0 < r < Rξ
and any x ∈ BΩ (ξ, r), we have
V (ξ, r)
G
(x, ξ16r ).
ωBΩ (ξ,2r) x, ∂Ω BΩ (ξ, 2r) ≤ A2
Ψ(r) BΩ (ξ,CΩ A3 r)
Here ξ16r is any point in Ω with dΩ (ξ, ξ16r ) = 4r and
d(ξ16r , X \ Ω) = d(ξ16r , X \ Y ′ ) ≥ 2cu r.
24
Proof. We follow [LSCb, Lemma 4.7]. Let A3 = 2(12+12Cu) so that all (cu , Cu )inner uniform paths that connect two points in BΩ (ξ, 12r) stay in BΩ (ξ, A3 r/2).
Recall that Y ′ = BΩ (ξ, A0 r), where A0 = A3 + 7. For any z ∈ BΩ (ξ, A3 r), set
G′ (z) = GBΩ (ξ,A3 r) (z, ξ16r ).
Let s = min{cu r, 5r/Cu }. As BΩ (ξ16r , s) ⊂ BΩ (ξ, A3 r) \ BΩ (ξ, 2r), the maximum principle yields
∀y ∈ BΩ (ξ, 2r),
G′ (y) ≤
G′ (z).
sup
z∈∂Ω BΩ (ξ16r ,s)
Lemma 3.11 and the volume doubling condition yield
G′ (z) ≤ C
sup
z∈∂Ω BΩ (ξ16r ,s)
Ψ(r)
,
V (ξ, r)
for some constant C > 0. Hence, there exists ǫ1 > 0 such that
∀y ∈ BΩ (ξ, 2r),
Write BΩ (ξ, 2r) =
Uj =
Let Vj =
S
k≥j
S
j≥0
′
ǫ1
V (ξ, r) ′
G (y) ≤ e−1 .
Ψ(r)
Uj ∩ BΩ (ξ, 2r), where
x ∈ Y : exp(−2
j+1
V (ξ, r) ′
j
G (x) < exp(−2 ) .
) ≤ ǫ1
Ψ(r)
Uk ∩ BΩ (ξ, 2r). We claim that
wη (Vj ) ≤ A4 r exp −2j /σ
(22)
for some constants A4 , σ ∈ (0, ∞).
Suppose x ∈ Vj . Observe that for z ∈ ∂Ω BΩ (ξ16r , s), by the inner uniformity
of the domain, the length of the Harnack chain of balls in BΩ (ξ, A3 r) \ {ξ16r }
e \ Y ′ )) for some constants
connecting x to z is at most A5 log(1 + A6 r/dΩ (x, Ω
A5 , A6 ∈ (0, ∞). Therefore there are constants ǫ2 , ǫ3 , σ > 0 such that
!σ
e \ Y ′)
V (ξ, r) ′
V (ξ, r) ′
dΩ (x, Ω
j
exp(−2 ) > ǫ1
G (x) ≥ ǫ2
G (z)
Ψ(r)
Ψ(r)
r
!σ
e \ Y ′)
dΩ (x, Ω
≥ ǫ3
.
r
The last inequality is obtained by applying Lemma 3.11 with R = A3 r and
δ = 5/A3 . Now we have that for any x ∈ Vj ,
−1/σ
′
e
dΩ (x, ∂Ω
exp(−2j /σ)r ∧ 2r.
e Ω) = dΩ (x, Ω \ Y ) ≤ ǫ3
25
This together with Lemma 4.5 yields (22).
Let R0 = 2r and for j ≥ 1,
Rj =
j
6 X 1
2− 2
π
k2
k=1
!
r.
Then Rj ↓ r and
X
∞
6a1 −2
a1 (Rj−1 − Rj )
j+1
j+1
j
exp 2
−
exp 2
−
=
j exp(2 /σ)
A4 r exp(−2j /σ)
A4 π 2
j=1
j=1
∞
X
3a1
−2
j
j
exp(2
/σ)
exp 2j+1 −
≤
CΩ A4 π 2
j=1
∞
X
< C < ∞.
Let ω0 = ω(·, ∂Ω BΩ (ξ, 2r), BΩ (ξ, 2r)) and
n
o
(
0 (x)
sup VΨ(r)ω
(ξ,r)G′ (x) : x ∈ Uj ∩ BΩ (ξ, Rj ) ,
dj =
0,
(23)
if Uj ∩ BΩ (ξ, Rj ) 6= ∅,
if Uj ∩ BΩ (ξ, Rj ) = ∅.
Since the sets Uj ∩ BΩ (ξ, 2r) cover BΩ (ξ, 2r) and BΩ (ξ, r) ⊂ BΩ (ξ, Rk ) for each
k, to prove Lemma 4.7, it suffices to show that
sup dj ≤ A2 < ∞
j≥0
where A2 is as in Lemma 4.7.
We proceed by iteration. Since ω0 ≤ 1, we have by definition of U0 ,
d0 =
Ψ(r)ω0 (x)
≤ ǫ1 e2 .
′ (x)
V
(ξ,
r)G
U0 ∩BΩ (ξ,2r)
sup
Let j > 0. For x ∈ Uj−1 ∩ BΩ (ξ, Rj−1 ), we have by definition of dj−1 that
ω0 (x) ≤ dj−1
V (ξ, r) ′
G (x).
Ψ(r)
Also, ω0 ≤ 1. Therefore the maximum principle yields that, for x ∈ Vj ∩
BΩ (ξ, Rj−1 ),
ω0 (x) ≤ ω(x, Vj ∩ ∂Ω BΩ (ξ, Rj−1 ), Vj ∩ BΩ (ξ, Rj−1 )) + dj−1
V (ξ, r) ′
G (x). (24)
Ψ(r)
For x ∈ Vj ∩ BΩ (ξ, Rj ), let D = B(p(x), CΩ−1 (Rj−1 − Rj )) and let D′ be the
connected component of p−1 (D ∩ Ω) that contains x. Then by Lemma 3.4,
D′ ∩ Ω ⊂ BΩ (x, Rj−1 − Rj ) ⊂ BΩ (ξ, Rj−1 ),
26
hence D′ ∩ Vj ∩ ∂Ω BΩ (ξ, Rj−1 ) = ∅. Thus, the first term on the right hand side
of (24) is not greater than
Rj−1 − Rj
Rj−1 − Rj
′
′
ω x, Vj ∩ D ∩ ∂X B p(x),
, Vj ∩ D ∩ B p(x),
2CΩ
2CΩ
a1 Rj−1 − Rj
a1 Rj−1 − Rj
≤ exp 2 −
≤ exp 2 −
2CΩ wη (Vj ∩ D′ )
2CΩ wη (Vj )
Rj−1 − Rj
a1
j
≤ exp 2 − ǫ6 j −2 exp(2j /σ) ,
exp(2 /σ)
≤ exp 2 −
2CΩ A4
r
by Lemma 4.6, monotonicity of U 7→ wη (U ) and (22). Here ǫ6 =
Moreover, by definition of Uj ,
ǫ1
3a1
π 2 A4 C Ω .
V (ξ, r) ′
G (x) ≥ exp(−2j+1 )
Ψ(r)
for x ∈ Uj . Hence, for x ∈ Uj ∩ BΩ (ξ, Rj ), (24) becomes
V (ξ, r) ′
G (x)
ω0 (x) ≤ exp 2 − ǫ6 j −2 exp(2j /σ) + dj−1
Ψ(r)
V (ξ, r) ′
G (x).
≤ ǫ1 exp 2 + 2j+1 − ǫ6 j −2 exp(2j /σ) + dj−1
Ψ(r)
(ξ,r) ′
Dividing both sides by VΨ(r)
G (x) and taking the supremum over x ∈ Uj ∩
BΩ (ξ, Rj ),
dj ≤ ǫ1 exp 2 + 2j+1 − ǫ6 j −2 exp(2j /σ) + dj−1 ,
and hence for every integer i > 0,
∞
X
exp 2j+1 −
di ≤ ǫ1 e 1 +
2
j=1
3a1
−2
j
j exp(2 /σ)
= ǫ1 e2 (1 + C) < ∞
π 2 A4 CΩ
by (23).
Proof of Theorem 4.1. We follow [GSC11, Theorem 4.5] and [Aik01, Lemma 3].
Recall that A0 = A3 + 7 = 2(12 + 12Cu ) + 7. Fix ξ ∈ ∂Ω
e Ω with Rξ > 0. Let
′
0 < r < {Rξ′ : ξ ′ ∈ ∂Ω
e Ω ∩ BΩ
e (ξ, 7Rξ )} and set Y = BΩ (ξ, A0 r). Note that
any two points in BΩ (ξ, 12r) can be connected by a (cu , CU )-inner uniform path
that stays in BΩ (ξ, A3 r/2).
Fix x∗ ∈ BΩ (ξ, r), y ∗ ∈ ∂Ω BΩ (ξ, 6r) such that c1 r ≤ dΩ (x∗ , ∂Ω
e Ω) ≤ r and
6c0 r ≤ dΩ (y ∗ , ∂Ω
e Ω) ≤ 6r, for some constants c0 , c1 ∈ (0, 1) depending on cu and
Cu . Existence of x∗ and y ∗ follows from the inner uniformity of Ω. It suffices
to show that for all x ∈ BΩ (ξ, r) and y ∈ ∂Ω BΩ (ξ, 6r) we have
GY ′ (x, y) ≍
GY ′ (x∗ , y)
GY ′ (x, y ∗ ).
GY ′ (x∗ , y ∗ )
27
(25)
Fix y ∈ ∂Ω BΩ (ξ, 6r), and call u (v, respectively) the left(right)-hand side of
(25), viewed as a function of x. Then u is positive and harmonic in Y ′ \ {y},
whereas v is positive and harmonic in Y ′ \ {y ∗ }. Both functions vanish quasieverywhere on the boundary of Y ′ .
Since y ∗ ∈ ∂Ω BΩ (ξ, 6r) and 6c0 r ≤ dΩ (y ∗ , ∂Ω
e Ω) ≤ 6r, it follows that the
ball BΩ (y ∗ , 3c0 r) is contained in BΩ (ξ, 9r) \ BΩ (ξ, 3r). Let z ∈ ∂Ω BΩ (y ∗ , c0 r).
By a repeated use of Harnack inequality (a finite number of times, depending
only on cu and Cu ), one can compare the value of v at z and at x∗ , so that by
Lemma 3.11 (notice that dΩ (x∗ , y) ≥ c1 r) and the volume doubling property,
v(z) ≤ C1 v(x∗ ) = C1 GY ′ (x∗ , y) ≤ C1′
Ψ(r)
.
V (ξ, r)
Now, if y ∈ BΩ (y ∗ , 2c0 r), then by Lemma 3.11 (notice that dΩ (z, y) ≤ 3r ≤
and z, y ∈ BΩ (ξ, A0 r/6)) and the volume doubling property,
u(z) = GY ′ (z, y) ≥ C2
A0 r
6Cu
Ψ(r)
,
V (ξ, r)
so that we have u(z) ≥ C3 v(z) in this case for some C3 > 0. If instead y ∈
Ω \ BΩ (y ∗ , 2c0 r), then we can connect z and x∗ by a path of length comparable
to r that stays away (at scale r) from both ∂Ω
e Ω and the point y. Hence, the
Harnack inequality implies that u(z) ≍ u(x∗ ) = v(x∗ ) ≍ v(z) in this case. This
shows that we always have
u(z) ≥ ǫ3 v(z) ∀z ∈ ∂Ω BΩ (y ∗ , c0 r).
By the maximum principle, we obtain
u ≥ ǫ3 v
on Y ′ \ BΩ (y ∗ , c0 r).
Since BΩ (ξ, r) ⊂ Y ′ \ BΩ (y ∗ , c0 r), we have proved that u ≥ ǫ3 v on BΩ (ξ, r),
that is,
GY ′ (x∗ , y)
GY ′ (x, y) ≥ ǫ3
(26)
GY ′ (x, y ∗ )
GY ′ (x∗ , y ∗ )
for all x ∈ BΩ (ξ, r) and y ∈ ∂Ω BΩ (ξ, 6r). This is one half of (25).
We now focus on the other half of (25), that is,
ǫ4 GY ′ (x, y) ≤
GY ′ (x∗ , y)
GY ′ (x, y ∗ ),
GY ′ (x∗ , y ∗ )
(27)
for all x ∈ BΩ (ξ, r) and y ∈ ∂Ω BΩ (ξ, 6r).
For x ∈ BΩ (ξ, 2r) and z ∈ BΩ (ξ, 9r)\BΩ (ξ, 3r), Lemma 3.11 and the volume
doubling condition yield
GY ′ (x, z) ≤ C1
28
Ψ(r)
.
V (ξ, r)
Regarding GY ′ (x, z) as harmonic function of x, the maximum principle gives
GY ′ (·, z) ≤ C1
Ψ(r)
ω(·, ∂Ω BΩ (ξ, 2r), BΩ (ξ, 2r))
V (ξ, r)
on BΩ (ξ, 2r).
Using Lemma 4.7 (note that A0 > A3 ) and the Harnack inequality (to move
from ξ16r to y ∗ ), we get for x ∈ BΩ (ξ, r) and z ∈ BΩ (ξ, 9r) \ BΩ (ξ, 3r), that
GY ′ (x, z) ≤ C1 A2
Ψ(r) V (ξ, r)
GY ′ (x, ξ16r ) ≤ C4 GY ′ (x, y ∗ ),
V (ξ, r) Ψ(r)
(28)
for some constant C4 > 0. Fix x ∈ BΩ (ξ, r) and y ∈ ∂Ω BΩ (ξ, 6r). If we have
∗ ∗
∗
∗
dΩ (y, ∂Ω
e Ω) ≥ c0 r/2, then GY ′ (x, y) ≍ GY ′ (x, y ) and GY ′ (x , y) ≍ GY ′ (x , y )
by the Harnack inequality, so that (27) follows. Hence we now assume that
′
y ∈ ∂Ω BΩ (ξ, 6r) satisfies dΩ (y, ∂Ω
e Ω) < c0 r/2. Let ξ ∈ ∂Ω
e Ω be a point such
′
′
that dΩ (y, ξ ) < c0 r/2. It follows that y ∈ BΩ (ξ , r). Also,
BΩ (ξ ′ , 2r) ⊂ BΩ (y, 3r) ⊂ BΩ (ξ, 9r) \ BΩ (ξ, 3r).
We apply inequality (28) to get GY ′ (x, z) ≤ C4 GY ′ (x, y ∗ ) for any z ∈ BΩ (ξ ′ , 2r).
Regarding GY ′ (x, y) as harmonic function of y, we obtain
GY ′ (x, y) ≤ C4 GY ′ (x, y ∗ ) ω(y, ∂Ω BΩ (ξ ′ , 2r), BΩ (ξ ′ , 2r)).
(29)
Let us apply Lemma 4.7 with ξ replaced by ξ ′ . This yields
V (ξ ′ , r)
′
GBΩ (ξ′ ,CΩ A3 r) (y, ξ16r
)
Ψ(r)
V (ξ, r)
′
≤ A′2
, y),
GY ′ (ξ16r
Ψ(r)
ω(y, ∂Ω BΩ (ξ ′ , 2r), BΩ (ξ ′ , 2r)) ≤ A2
(30)
′
′
′
where ξ16r
∈ Ω is any point such that dΩ (ξ16r
, ξ ′ ) = 4r and dΩ (ξ16r
, ∂Ω
e Ω) ≥
2cu r. Observe that we have used the volume doubling property as well as the set
monotonicity of the Green function, and that BΩ (ξ ′ , A3 r) ⊂ BΩ (ξ, A0 r) because
A0 = A3 + 7 and dΩ (ξ, ξ ′ ) ≤ 7r. Now, (29) and (30) give
GY ′ (x, y) ≤ C5
V (ξ, r)
′
, y)GY ′ (x, y ∗ ).
GY ′ (ξ16r
Ψ(r)
(31)
′
′
By construction, dΩ (ξ16r
, y) ≥ d(ξ16r
, ξ ′ ) − dΩ (ξ ′ , y) ≥ 2r and dΩ (x∗ , y) ≥
∗
dΩ (ξ, y) − dΩ (ξ, x ) ≥ 5r. Using the inner uniformity of Ω, we find a chain
′
of balls, each of radius ≍ r and contained in Y ′ \ {y}, going from x∗ to ξ16r
,
so that the length of the chain is uniformly bounded in terms of cu , Cu . Ap′
, y) ≍ GY ′ (x∗ , y).
plying the Harnack inequality repeatedly thus yields GY ′ (ξ16r
∗
∗
As Lemma 3.11 gives GY ′ (x , y ) ≍ Ψ(r)/V (ξ, r), inequality (31) implies (27).
This completes the proof.
29
5
Applications
5.1
Martin boundary
Theorem 5.1. Let (X, d, µ, E, F ) be a MMD space that satisfies VD and wHKE(Ψ). Let Ω ⊂ X be a bounded inner uniform domain in X. Then the
e and each
Martin compactification relative to (E, F ) of Ω is homeomorphic to Ω
e
boundary point ξ ∈ Ω \ Ω is minimal.
Proof. The assertion can be proved along the line of [ALM03, Theorem 1.1]
using the boundary Harnack principle of Theorem 4.2.
5.2
Existence of a harmonic profile
On domains that satisfy the (scale-invariant) boundary Harnack principle, the
latter is crucial to obtain precise estimates for the Dirichlet heat kernel. See
[GSC11, LSCc], where the case of regular Dirichlet spaces that induce a nondegenerate metric and satisfy a parabolic Harnack inequality, is treated.
In the case of unbounded domains, an important technique to obtain Dirichlet heat kernel estimates is the h-transform of the Dirichlet form by a harmonic
profile h for (E, F ) on the domain U . In this section, we prove the existence of
a harmonic profile on inner uniform domains in a MMD space.
In forthcoming papers [KLa, KLb], the harmonic profile, as well as the geometric boundary Harnack principle of Theorem 4.2, will be used to prove twosided bounds for the Dirichlet heat kernel in inner uniform domains on fractal
spaces.
Definition 5.2. A function h on an unbounded domain U is called a harmonic
profile if
(i) h is harmonic on U ,
0
(ii) h satisfies Dirichlet boundary condition along ∂Ue U , that is, h ∈ Floc
(U ),
(iii) h > 0 in U .
Remark 5.3. Due to the maximum principle, a harmonic profile can exist
only if the domain is unbounded domain. On bounded domains, Dirichlet heat
kernel estimates can be obtained by using a ground state transform instead of
an h-transform.
Proposition 5.4. Let (X, d, µ, E, F ) be a MMD space that satisfies VD and
w-HKE(Ψ). Let U be an unbounded inner uniform domain in X. Then there
exists a harmonic profile for (E, F ) on U . It is unique up to multiplication by a
scalar.
Proof. Uniqueness can be deduced from the boundary Harnack principle and a
classical argument, see, e.g., [Anc80, Section 6]. To show existence, we follow
[GSC11, Section 4.3]. Consider a fixed point x0 ∈ U and a sequence of inner
30
balls Bi = BUe (x0 , ri ) ∩ U so that ri ↑ ∞. For each i, let xi be a point such that
dU (xi , x0 ) = ri /2 (such a sequence exists because U is unbounded). We will
define the harmonic profile as the limit of a sequence of non-negative functions
hi . Set
GBi (xi , x)
,
hi (x) =
GBi (xi , x0 )
for x ∈ Bi and h(x) = 0 for x ∈ U \ Bi . Recall from Lemma 3.9 that GBi (·, x) ∈
Floc (Bi \ {x}). Hence, for each i, hi belongs to Floc (Bi \ {x}) and is harmonic
on Bi \ {xi }.
By a standard argument ([?, p.587]), bounded non-negative harmonic functions are Hölder continuous (with parameters depending only on the constant
in the elliptic Harnack inequality). Hence, and because hi (xi ) = 1 for all i,
the family {hi : i ≥ i0 } is equicontinuous on U ∩ BUe (x0 , ri0 /3 ). There exists
a subsequence of the sequence (hi ) that converges uniformly in any compact
subset of U . Replacing the original sequence by such a subsequence, we set
h(x) = lim hi (x)
i→∞
∀x ∈ U.
(32)
Let us show that hi converges to h locally in F (U ). It is clear that hi converges
to h locally in L2 (U ). By the Leibniz rule for the energy measure Γ, for any
function φ ∈ Cc (U ) ∩ F and i, j large enough, we have
Z
(hi − hj )2 dΓ(φ, φ) + E(hi − hj , φ2 (hi − hj )).
E(φ(hi − hj ), φ(hi − hj )) =
U
The last term on the right hand side is 0 because φ2 (hi − hj ) is in Cc (U ) ∩ F
and hi − hj is harmonic on an open subset of U containing the compact support
of φ. Since hi converges to h locally uniformly in U , this proves convergence
locally in F (U ).
It easily follows that h is harmonic, since for any φ ∈ Cc (U ) ∩ F ,
E(h, φ) = E(lim hi , φ) = lim E(hi , φ) = 0.
By the Harnack inequality and because h(x0 ) = 1, it follows that h > 0 in U .
0
(U ).
It remains to show that h is in Floc
The first step is to show that the sequence (hi ) is Cauchy not only in
e , dµ). Let K be a compact set in X and V = K ∩U .
L2loc (U, dµ) but also in L2loc (U
We need to show that (hi ) is Cauchy in L2 (V, µV ).
Choose a ball B(x0 , r) that contains K. Let ξ0 a point in [B(x0 , (A0 + 1)r) \
B(x0 , A0 r)]∩U . Let g(x) = GU∩B(x0 ,(A0 +1)r) (ξ0 , x) be the corresponding Green
function. Note that g is continuous and positive in B(x0 , A0 r)∩U and harmonic
on B(x0 , A0 r) ∩ U with Dirichlet boundary condition along ∂Ue U . Applying the
boundary Harnack principle to any of the pairs hi , g with i large enough yields
sup
U∩K
hi
≤ A1
g
31
for all i.
For any η ∈ (0, 1) small enough, let Vη = {x ∈ V : d(x, X \ U ) ≥ η} and
note that Vη is a compact subset of U . For i, j large enough, we have
Z
Z
Z
2
2
|hi − hj | dµ + 2A1
|hi − hj | dµ ≤
g 2 dµ.
V
Vη
V \Vη
R
As V \Vη g 2 dµ tends to 0 with η and (hi ) converges to h in L2loc (U, µ), this indeed
shows that (hi ) is Cauchy in L2 (V, µ).
The second and last step is to show that for any open set V of U which is
relatively compact in X, (hi ) is Cauchy in F (V ). Since hi ∈ F 0 (U, V ) for large
0
i, this implies that h ∈ F 0 (U, V ) and thus h ∈ Floc
(U ) as desired.
To this end, let B = B(x0 , r) be a ball that contains V . By CSA, there
0
exists a cutoff function φ for B. Since hi ∈ Floc
(Bi , Bi \ {xi }), we have φhi ∈
0
0
F (Bi ) ⊂ F (U ) for large i. We show that φhi is Cauchy in F 0 (U ). We have
Z
(hi − hj )2 dΓ(φ, φ) + E(hi − hj , φ2 (hi − hj )).
E(φ(hi − hj ), φ(hi − hj )) =
U
The last term on the right hand side is zero, because hi − hj is harmonic in
Bi \ {xi } and φ2 (hi − hj ) ∈ F 0 (U ) can be approximated by functions with
compact support contained in Bi \ {xi }. Hence, applying CSA(Ψ),
Z
Z
(hi − hj )2 dΓ(φ, φ)
dΓ(hi − hj , hi − hj ) ≤ E(φ(hi − hj ), φ(hi − hj )) =
U
B(x0 ,2r)
Z
Z
1
≤
dΓ(hi − hj , hi − hj ) + CS Ψ(r)−1
(hi − hj )2 dµ.
8 B(x0 ,2r)
B(x0 ,2r)
Therefore,
Z
8
dΓ(hi − hj , hi − hj ) ≤ CS Ψ(r)−1
9
B(x0 ,2r)
Z
(hi − hj )2 dµ.
B(x0 ,2r)
e ), this shows that it is also Cauchy in F (V ).
Since (hi ) is Cauchy in L2loc (U
Acknowledgement
I thank Naotaka Kajino and Laurent Saloff-Coste for many helpful discussions.
Further, I thank Naotaka Kajino for his very careful reading and his comments
on an earlier version of this paper.
References
[AB]
Sebastian Andres and Martin T. Barlow.
Energy inequalities for cutoff functions and some applications.
preprint,
http://arxiv.org/pdf/1202.0722v2.pdf.
32
[Aik01]
Hiroaki Aikawa. Boundary Harnack principle and Martin boundary
for a uniform domain. J. Math. Soc. Japan, 53(1):119–145, 2001.
[ALM03] Hiroaki Aikawa, Torbjörn Lundh, and Tomohiko Mizutani. Martin
boundary of a fractal domain. Potential Anal., 18(4):311–357, 2003.
[Anc80]
Alano Ancona. Erratum: “Principe de Harnack à la frontière et
théorème de Fatou pour un opérateur elliptique dans un domaine lipschitzien” [Ann. Inst. Fourier (Grenoble) 28 (1978), no. 4, 169–213;
MR 80d:31006]. In Potential theory, Copenhagen 1979 (Proc. Colloq.,
Copenhagen, 1979), volume 787 of Lecture Notes in Math., page p.
28. Springer, Berlin, 1980.
[Anc07]
Alano Ancona. Sur la théorie du potentiel dans les domaines de John.
Publ. Mat., 51(2):345–396, 2007.
[BB91]
Richard F. Bass and Krzysztof Burdzy. A boundary Harnack principle
in twisted Hölder domains. Ann. of Math. (2), 134(2):253–276, 1991.
[BB04]
Martin T. Barlow and Richard F. Bass. Stability of parabolic Harnack
inequalities. Trans. Amer. Math. Soc., 356(4):1501–1533 (electronic),
2004.
[BBK06] Martin T. Barlow, Richard F. Bass, and Takashi Kumagai. Stability
of parabolic Harnack inequalities on metric measure spaces. J. Math.
Soc. Japan, 58(2):485–519, 2006.
[BGK12] Martin T. Barlow, Alexander Grigor’yan, and Takashi Kumagai. On
the equivalence of parabolic Harnack inequalities and heat kernel estimates. J. Math. Soc. Japan, 64(4):1091–1146, 2012.
[Dah77] Björn E. J. Dahlberg. Estimates of harmonic measure. Arch. Rational
Mech. Anal., 65(3):275–288, 1977.
[FŌT94] Masatoshi Fukushima, Yōichi Ōshima, and Masayoshi Takeda.
Dirichlet forms and symmetric Markov processes, volume 19 of de
Gruyter Studies in Mathematics. Walter de Gruyter & Co., Berlin,
1994.
[GH]
Alexander Grigor′ yan and Jiaxin Hu. Upper bounds of heat kernels
on doubling spaces. preprint.
[GH08]
Alexander Grigor′yan and Jiaxin Hu. Off-diagonal upper estimates
for the heat kernel of the Dirichlet forms on metric spaces. Invent.
Math., 174(1):81–126, 2008.
[GH13]
A. Grigor’yan and J. Hu. Heat kernels and Green functions on metric
measure spaces. to appear in Canad. J. Math., 2013.
[GSC11] Pavel Gyrya and Laurent Saloff-Coste. Neumann and Dirichlet heat
kernels in inner uniform domains. Astérisque, (336):viii+144, 2011.
33
[GT12]
Alexander Grigor’yan and Andras Telcs. Two-sided estimates of heat
kernels on metric measure spaces. Ann. Probab., 40(3):1212–1284,
2012.
[HSC01] W. Hebisch and L. Saloff-Coste. On the relation between elliptic
and parabolic Harnack inequalities. Ann. Inst. Fourier (Grenoble),
51(5):1437–1481, 2001.
[Kem72] John T. Kemper. A boundary Harnack principle for Lipschitz domains
and the principle of positive singularities. Comm. Pure Appl. Math.,
25:247–255, 1972.
[KLa]
Naotaka Kajino and Janna Lierl. Dirichlet heat kernel estimates in
unbounded inner uniform domains of fractal-type spaces. In preparation.
[KLb]
Naotaka Kajino and Janna Lierl. Neumann and Dirichlet heat kernel
estimates in inner uniform domains of local resistance Dirichlet spaces.
In preparation.
[LSCa]
Janna Lierl and Laurent Saloff-Coste. Parabolic Harnack inequality for time-dependent non-symmetric Dirichlet forms. Submitted,
arxiv.org/pdf/1204.1926v1.pdf.
[LSCb]
Janna Lierl and Laurent Saloff-Coste.
Scale-invariant boundary Harnack principle in inner uniform domains.
Submitted,
arxiv.org/pdf/1110.2763v2.pdf.
[LSCc]
Janna Lierl and Laurent Saloff-Coste. The Dirichlet heat kernel in
inner uniform domains: local results, compact domains and nonsymmetric forms. Submitted, http://arxiv.org/pdf/1210.4586.pdf.
[Mar]
O. Martio. Injectivity theorems in plane and space. Ann. Acad. Sci.
Fenn. Ser. A I Math., 4(2):383–401.
[Mos61] Jürgen Moser. On Harnack’s theorem for elliptic differential equations.
Comm. Pure Appl. Math., 14:577–591, 1961.
[Ouh96] El-Maati Ouhabaz. Invariance of closed convex sets and domination
criteria for semigroups. Potential Anal., 5(6):611–625, 1996.
[SC92]
L. Saloff-Coste. A note on Poincaré, Sobolev, and Harnack inequalities. Internat. Math. Res. Notices, (2):27–38, 1992.
[SC02]
Laurent Saloff-Coste. Aspects of Sobolev-type inequalities, volume 289
of London Mathematical Society Lecture Note Series. Cambridge University Press, Cambridge, 2002.
[Stu95]
Karl-Theodor Sturm. Analysis on local Dirichlet spaces. II. Upper
Gaussian estimates for the fundamental solutions of parabolic equations. Osaka J. Math., 32(2):275–312, 1995.
34
[Wu78]
Jang Mei G. Wu. Comparisons of kernel functions, boundary Harnack
principle and relative Fatou theorem on Lipschitz domains. Ann. Inst.
Fourier (Grenoble), 28(4):147–167, vi, 1978.
35