HARNACK INEQUALITIES ON WEIGHTED GRAPHS AND SOME
APPLICATIONS FOR THE RANDOM CONDUCTANCE MODEL
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
A BSTRACT. We establish elliptic and parabolic Harnack inequalities on graphs with
unbounded weights. As an application we prove a local limit theorem for a continuous time random walk X in an environment of ergodic random conductances
taking values in [0, ∞) satisfying some moment conditions.
C ONTENTS
1. Introduction
1.1. The model
1.2. Harnack inequalities on graphs
1.3. The Method
1.4. Local Limit Theorem for the Random Conductance Model
2. Sobolev and Poincaré inequalities
2.1. Setup and Preliminaries
2.2. Local Poincaré and Sobolev inequality
2.3. The lemma of Bombieri and Giusti
3. Elliptic Harnack inequality
3.1. Mean value inequalities
3.2. From mean value inequalities to Harnack principle
3.3. Hölder continuity
4. Parabolic Harnack inequality
4.1. Mean value inequalities
4.2. From mean value inequalities to Harnack principle
4.3. On-diagonal estimates and Hölder-continuity
5. Local limit theorem
6. Improving the lower moment condition
Appendix A. Technical estimates
References
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2
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6
7
10
10
11
13
14
14
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20
23
26
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31
36
38
Date: December 19, 2013.
2000 Mathematics Subject Classification. 31B05, 39A12, 60J35, 60K37, 82C41.
Key words and phrases. Harnack inequality, Moser iteration, Random conductance model, local
limit theorem, ergodic.
1
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SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
1. I NTRODUCTION
Consider the operator La in divergence form,
d
X
La f (x) =
∂xi aij (·) ∂xj f (·) (x),
i,j=1
Rd ,
acting on functions on
where the symmetric positive definite matrix a = (aij (x))
is bounded, measurable and uniformly elliptic, i.e. for some 0 < λ1 , λ2 < ∞
λ1 |ξ|2 ≤
d
X
aij (x) ξi ξj ≤ λ2 |ξ|2 ,
∀ x, ξ ∈ Rd .
(1.1)
i,j=1
Then, a celebrated result of Moser [20] states that the elliptic Harnack inequality
(EHI) holds, that is there exits a constant CEH ≡ CEH (λ1 , λ2 ) < ∞ such that for any
positive function u > 0 which is La harmonic on the ball B(x0 , r), i.e. (La u)(x) = 0
for all x ∈ B(x0 , r), we have
max u ≤ CEH
B(x0 ,r/2)
min
B(x0 ,r/2)
u.
A few years later, Moser also proved that the parabolic Harnack inequality (PHI)
holds (see [21]), that is there exits a constant CPH ≡ CPH (λ1 , λ2 ) < ∞ such that
for any positive caloric function u > 0, i.e. any weak solution of the heat equation
(∂t − La )u(t, x) = 0 on [t0 , t0 + r2 ] × B(x0 , r), we have
max u(t, x) ≤ CPH
(t,x)∈Q−
min u(t, x),
(t,x)∈Q+
where Q− = t0 + 14 r2 , t0 + 12 r2 ×B x0 , 21 r and Q+ = t0 + 34 r2 , t0 +r2 ×B x0 , 21 r .
Both EHI and PHI have had a big influence on PDE theory and differential geometry, in particular they play a prominent role in elliptic regularity theory. One
application of Harnack inequalities – observed by Nash in [22] – is to show Hölder
regularity for solutions of elliptic or parabolic equations. For instance, the PHI implies the Hölder continuity of the fundamental solution pat (x, y) of the parabolic
equation
∂t − La pat (x, y) = 0,
pa0 (x, y) = δx (y).
Furthermore, the PHI is also one of the key tools in Aronson’s proof [4] of Gaussian
type bounds for pat (x, y).
The remarkable fact about these results is that they only rely on the uniform
ellipticity and not on the regularity of the matrix a. We refer to the monograph
[24] for more details on this topic.
1.1. The model. In this paper we will be dealing with a discretized version of the
operator La . We consider an infinite, connected, locally finite graph G = (V, E)
with vertex set V and edge set E. We will write x ∼ y if {x, y} ∈ E. A path of
length n between x and y in G is a sequence {xi : i = 0, . . . , n} with the property
that x0 = x, xn = y and xi ∼ xi+1 . Let d be the natural graph distance on G, i.e.
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
3
d(x, y) is the minimal length of a path between x and y. We denote by B(x, r) the
closed ball with center x and radius r, i.e. B(x, r) := {y ∈ V | d(x, y) ≤ r}.
The graph is endowed with the counting measure, i.e. the measure of A ⊂ V
is simply the number |A| of elements in A. For any non-empty, finite A ⊂ V and
p ∈ [1, ∞), we introduce space-averaged `p -norms on functions f : A → R by the
usual formula
!1
p
X
1
p
f f :=
:= max |f (x)|.
and
|f
(x)|
p,A
∞,A
x∈A
|A|
x∈A
For a given set B ⊂ V , we define the relative internal boundary of A ⊂ B by
∂B A := x ∈ A ∃ y ∈ B \ A s.th. {x, y} ∈ E
and we simply write ∂A instead of ∂V A. Throughout the paper we will make the
following assumption on G.
Assumption 1.1. For some d ≥ 2 the graph G satisfies the following conditions:
(i) volume regularity of order d, that is there exists Creg ∈ (0, ∞) such that
−1 d
Creg
r ≤ |B(x, r)| ≤ Creg rd
∀ x ∈ V, r ≥ 1.
(1.2)
(ii) relative isoperimetric inequality of order d, that is there exists Criso ∈ (0, ∞) such
that for all x ∈ V and r ≥ 1
|∂B(x,r) A|
Criso
≥
|A|
r
∀ A ⊂ B(x, r) s.th. |A| < 12 |B(x, r)|.
(1.3)
Remark 1.2. The Euclidean lattice, (Zd , Ed ), satisfies the Assumption 1.1.
Assume that the graph, G, is endowed with positive weights, i.e. we consider a
family ω = {ω(e) ∈ (0, ∞) : e ∈ E}. With an abuse of notation we also denote the
conductance matrix by ω, that is for x, y ∈ V we set ω(x, y) = ω(y, x) = ω({x, y}) if
{x, y} ∈ E and ω(x, y) = 0 otherwise. We also refer to ω(x, y) as the conductance of
the corresponding edge {x, y} and we call (V, E, ω) a weighted graph. Let us further
define measures µω and ν ω on V by
X
X
1
µω (x) :=
.
ω(x, y)
and
ν ω (x) :=
ω(x,
y)
y∼x
y∼x
For any fixed ω we consider a reversible continuous time Markov chain, Y = {Yt :
t ≥ 0}, on V with generator LωY acting on bounded functions f : V → R as
X
LωY f )(x) = µω (x)−1
ω(x, y) f (y) − f (x) .
(1.4)
y∼x
Pωx
We denote by
the law of the process starting at the vertex x ∈ V . The corresponding expectation will be denoted by Eωx . Setting pω (x, y) := ω(x, y)/µω (x),
this random walk waits at x an exponential time with mean 1 and chooses its next
position y with probability pω (x, y). Since the law of the waiting times does not
depend on the location, Y is also called the constant speed random walk (CSRW).
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SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Furthermore, Y is a reversible Markov chain with symmetrizing measure given by
µ. We recall that in [3] we mainly consider the variable speed random walk (VSRW)
X = {Xt : t ≥ 0}, which waits at x an exponential time with mean 1/µ(x), with
generator
X
ω(x, y) f (y) − f (x) = µω (x) · LωY f )(x).
Lω f )(x) =
y∈Zd
1.2. Harnack inequalities on graphs. In the case of uniformly bounded conductances, i.e. there exists 0 < λ1 , λ2 < ∞ such that
λ1 ≤ ω(x, y) ≤ λ2 ,
both the EHI and the PHI have been derived in the setting of an infinite, connected,
locally finite graphs by Delmotte in [15, 16] with constant CEH = CEH (λ1 , λ2 ) and
CPH = CPH (λ1 , λ2 ) depending on λ1 and λ2 only. It is obvious that for a given ball
B(x0 , n) the constants λ1 and λ2 can be replaced by
1
x,y∈B(x0 ,n) ω(x, y)
max
and
max
x,y∈B(x0 ,n)
ω(x, y).
Our main result shows that the uniform upper and lower bound can be replaced by
`p -bounds.
Theorem 1.3 (Elliptic Harnack inequality). For any x0 ∈ V and n ≥ 1 let B(n) =
B(x0 , n). Suppose that u > 0 is harmonic on B(n), i.e. LωY u = 0 on B(n). Then, for
any p, q ∈ (1, ∞) with
there exists CEH ≡ CEH
1 1
2
+ <
p q
d
ω
ω
µ , ν q,B(n) such that
p,B(n)
max u(x) ≤ CEH
x∈B(n/2)
min
x∈B(n/2)
u(x).
(1.5)
The constant CEH is more explicitly given by
κ CEH µω p,B(n) , ν ω q,B(n) = c1 exp c2 1 ∨ µω p,B(n) ν ω q,B(n)
for some positive ci = ci (d, p, q) and κ = κ(d, p, q).
Theorem 1.4 (Parabolic Harnack inequality). For any x0 ∈ V , t0 ≥ 0 and n ≥ 1
let Q(n) = [t0 , t0 + n2 ] × B(x0 , n). Suppose that u > 0 is caloric on Q(n), i.e.
∂t u − LωY u = 0 on Q(n). Then, for any p, q ∈ (1, ∞) with
there exists CPH = CPH
1 1
2
+ <
p q
d
ω
ω
µ , ν q,B(n) such that
p,B(n)
max u(t, x) ≤ CPH
(t,x)∈Q−
min u(t, x),
(t,x)∈Q+
(1.6)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
5
where Q− = t0 + 14 n2 , t0 + 12 n2 ×B x0 , 12 n and Q+ = t0 + 34 n2 , t0 +n2 ×B x0 , 12 n .
The constant CPH is more explicitly given by
κ κ
1 ∨ ν ω q,B(n)
CPH µω p,B(n) , ν ω q,B(n) = c1 exp c2 1 ∨ µω p,B(n)
for some positive ci = ci (d, p, q) and κ = κ(d, p, q).
Remark 1.5. Given a speed measure π ω : Zd → (0, ∞), one can also consider the
process, Z = {Zt : t ≥ 0} on V that is defined by a time change of the VSRW X, i.e.
Zt := Xat for t ≥ 0, where at := inf{s ≥ 0 : As > t} denotes the right continuous
inverse of the functional
Z t
π ω (Xs ) ds,
t ≥ 0.
At =
0
Its generator is given by
Lωπ f (x) = π ω (x)−1
X
ω(x, y) f (y) − f (x) .
(1.7)
y∈V
The maybe most natural choice for the speed measure is π ω = µω , for which we
get again the CSRW Y . Since we have the same harmonic functions w.r.t. Lωπ for
any choice of π, it is obvious that the EHI in Theorem 1.3 also holds for Lωπ . On
the other hand, for the PHI one can show the following statement along the lines
of the proof of Theorem 1.4: Let u > 0 be caloric w.r.t. Lωπ on Q(n), then for any
p, q, r ∈ (1, ∞) with
there exists CPH = CPH
1
1 r−1
1
2
+
+
<
(1.8)
r
p−1 r
q
d
ω
1
µ /(π ω )1− p , ν ω q,B(n) , π ω r,B(n) such that
p,B(n)
max u(t, x) ≤ CPH
(t,x)∈Q−
min u(t, x).
(t,x)∈Q+
In particular, for the VSRW X we have π ω ≡ 1 and choosing r = ∞ the condition
on p and q reads 1/(p − 1) + 1/q < 2/d in this case.
Notice that the Harnack constants CEH and CPH in Theorem 1.3 and Theorem 1.4
do depend on the ball under consideration, namely on its center point x0 as well as
on its radius n. As a consequence one cannot deduce directly Hölder continuity estimates from these results. This requires ergodicity w.r.t. translations as an additional
assumption (see Proposition 3.6 and Proposition 4.6 below).
Assumption 1.6. Assume that
µ̄ = sup lim sup µω x0 ∈V
n→∞
p,B(x0 ,n)
< ∞,
ν̄ = sup lim sup ν ω q,B(x0 ,n) < ∞.
x0 ∈V
n→∞
In particular, Assumption 1.6 implies for the Harnack constants appearing in Theo∗ := 2C
∗
rem 1.3 and Theorem 1.4 that CEH
EH (µ̄, ν̄) < ∞ and CPH := 2CPH (µ̄, ν̄) < ∞
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SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
do not depend on x0 and n. Furthermore, there exists sω (x0 ) ≥ 1 such that
∗
sup CEH (µω p,B(x0 ,n) , ν ω q,B(x0 ,n) ) ≤ CEH
< ∞,
n≥sω (x0 )
sup
n≥sω (x0 )
∗
CPH (µω p,B(x0 ,n) , ν ω q,B(x0 ,n) ) ≤ CPH
< ∞.
In other words, under Assumption 1.6 the results in Theorem 1.3 and Theorem 1.4 are Harnack inequalities becoming effective for balls with radius n large
enough.
Corollary 1.7. Suppose that Assumption 1.6 holds. Then, for any x0 ∈ V and
n ≥ sω (x0 ) the elliptic and parabolic Harnack inequalities in Theorem 1.3 and The∗ and
orem 1.4 hold with the Harnack constants in (1.5) and (1.6) replaced by CEH
∗ , respectively.
CPH
In the applications for the random conductance model discussed below the conductances will be stationary ergodic random variables, so under some moment conditions Assumption 1.6 will be satisfied by the ergodic theorem.
It is well known that a PHI is equivalent to Gaussian lower and upper bounds on
the heat kernel in many situations, for instance in the case of uniformly bounded
conductances on a locally finite graph, see [16]. Unfortunately, in our setting we
are not able to deduce off-diagonal Gaussian bounds from the PHI due to the dependence of the Harnack constant on the underlying ball. More precisely, in order to
get effective Gaussian off-diagonal bounds, one needs to apply the PHI on a number
of balls with radius n having a distance of order n2 . In general, the ergodic theorem
does not give the required uniform control on the convergence of space-averages
of stationary random variables over such balls (see [1]). However, we still obtain
some on-diagonal estimates (see Proposition 4.5 below).
1.3. The Method. Since the pioneering works [20, 21] Moser’s iteration technique
is by far the best-established tool in order to prove Harnack inequalities. Moser’s
iteration is based on two main ideas: the Sobolev-type inequality which allows to
control the `r norm with r = r(d) = d/(d − 2) > 1 in terms of the Dirichlet form,
and a control on the Dirichlet form of any harmonic (or caloric) function u. In the
uniformly elliptic case this is rather standard. In our case where the conductances
are unbounded from above and below, we need to work with a dimension dependent weighted Sobolev inequality, established in [3] by using Hölder’s inequality.
That is, the coefficient r(d) is replaced by
r(d, p, q) =
d − d/p
.
(d − 2) + d/q
For the Moser iteration we need r(d, p, q) > 1, of course, which is equivalent to
1/p + 1/q < 2/d appearing in statements of Theorem 1.3 and Theorem 1.4. As a
result we obtain a maximum inequality for u, that is an estimate for the `∞ -norm
of u on a ball in terms of the `α -norm of u on a slightly bigger ball for some α > 0.
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
7
Since the same holds for u−1 , to conclude the Harnack inequality we are left to link
the `α -norm and the `−α -norm of u. In the setting of uniformly elliptic conductances
(see [15]) this can be done by using the well-known John-Nirenberg inequality, that
is the exponential integrability of BMO functions. In this paper we establish this
link by an abstract lemma of Bombieri and Giusti [10] (see Lemma 2.4 below). In
order to apply this lemma, beside the maximum inequalities mentioned above, we
need to establish a weighted Poincaré inequality, which is classically obtained by
the Whitney covering technique (see e.g. [16, 5, 24]). However, in this paper we
can avoid the Whitney covering by using results from a recent work by Dyda and
Kassmann [17].
1.4. Local Limit Theorem for the Random Conductance Model. Our main motivation for deriving the above Harnack inequalities is the quenched local limit theorem for the random conductance model. Consider the d-dimensional Euclidean
lattice, (Vd , Ed ), for d ≥ 2. The vertex set, Vd , of this graph equals Zd and the
edge set, Ed , is given by the set of all non oriented nearest neighbour bonds, i.e.
Ed := {{x, y} : x, y ∈ Zd , |x − y| = 1}.
⊗ Ed be a measurable space. Let the graph (V , E ) be
d
Let (Ω, F) = RE
d
d
+ , B(R+ )
endowed with a configuration ω = {ω(e) : e ∈ Ed } ∈ Ω of conductances. We will
henceforth denote by P a probability measure on (Ω, F), and we write E to denote
the expectation with respect to P. A space shift by z ∈ Zd is a map τz : Ω → Ω
(τz ω)(x, y) := ω(x + z, y + z),
∀ {x, y} ∈ Ed .
(1.9)
The set τx : x ∈ Zd together with the operation τx ◦τy := τx+y defines the group of
space shifts. We will study the nearest-neighbor random conductance model. For any
fixed realization ω it is a reversible continuous time Markov chain, Y = {Yt : t ≥ 0},
on Zd with generator LωY defined as in (1.4). We denote by q ω (t, x, y) for x, y ∈ Zd
and t ≥ 0 the transition density (or heat kernel associated with LωY ) of the CSRW
with respect to the reversible measure µ, i.e.
Pωx Yt = y
ω
q (t, x, y) :=
.
µω (y)
As a consequence of (1.9) we have
q τz ω (t, x, y) = q ω (t, x + z, y + z).
Assumption 1.8. Assume that P satisfies the following conditions:
(i) P 0 < ω(e) < ∞ = 1 for all e ∈ Ed and E[µω (0)] < ∞.
(ii) P is ergodic with respect to translations of Zd , i.e. P ◦ τx−1 = P for all x ∈ Zd
and P[A] ∈ {0, 1} for any A ∈ F such that τx (A) = A for all x ∈ Zd .
Assumption 1.9. There exist p, q ∈ (1, ∞] satisfying 1/p + 1/q < 2/d such that
E ω(e)p < ∞ and E ω(e)−q < ∞
(1.10)
for any e ∈ Ed .
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SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Notice that under Assumptions 1.8 and 1.9 the spatial ergodic theorem gives that
for P-a.e. ω,
p
q
lim µω p,B(n) = E µω (0)p < ∞ and lim ν ω q,B(n) = E ν ω (0)q < ∞.
n→∞
n→∞
In particular, Assumption 1.6 is fulfilled in this case and therefore for P-a.e. ω the
EHI and the PHI do hold for LωY in the sense of Corollary 1.7, i.e. provided n ≥
sω (x0 ).
We are interested in the P almost sure or quenched long range behaviour, in
particular in obtaining a quenched functional limit theorem (QFCLT) or invariance
principle for the processes Y starting in 0 and a quenched local limit theorem for
Y . We first recall that the following QFCLT has been recently obtained as the main
result in [3].
Theorem 1.10 (QFCLT). Suppose that d ≥ 2 and Assumptions 1.8 and 1.9 hold.
(n)
Then, P-a.s. Yt := n1 Yn2 t converges (under P0ω ) in law to a Brownian motion on Rd
with a deterministic non-degenerate covariance matrix Σ2Y .
Proof. See Theorem 1.3 and Remark 1.5 in [3].
In this paper our main concern is to establish a local limit theorem for Y . It
roughly describes how the transition probabilities of the random walk Y can be
rescaled in order to get the Gaussian transition density of the Brownian motion,
which appears as the limit process in Theorem 1.10. Write
−1
1
exp − x · Σ2Y
x/2t
(1.11)
kt (x) ≡ ktΣY (x) := q
(2πt)d det Σ2Y
2
d
for the Gaussian
heat kernel with covariance matrix ΣY . For x ∈ R write bxc =
bx1 c, . . . bxd c .
Theorem 1.11 (Quenched Local Limit Theorem). Let T2 > T1 > 0 and K > 0 and
suppose that d ≥ 2 and Assumptions 1.8 and 1.9 hold. Then,
lim sup
sup nd q ω n2 t, 0, bnxc − a kt (x) = 0,
P -a.s.
n → ∞ |x|≤K t ∈[T ,T ]
1 2
with a := 1/ E[µω (0)].
Remark 1.12. Similarly to Remark 1.5 it is possible to state a local limit theorem
for the process Z, obtained as a time-change of the VSRW X in terms of a speed
measure π ω : Zd → (0, ∞) with generator Lωπ defined in (1.7). Indeed, if π ω (x) =
π τx ω (0) for every x ∈ Zd , if 0 < E[π ω (0)] < ∞ and if Assumptions 1.8 and 1.9
hold, we have a QFCLT also for the process Z with some non-degenerate covariance
matrix Σ2Z = E[π ω (0)]−1 Σ2X , where Σ2X denotes the covariance matrix in the QFCLT
for the VSRW X (see Remark 1.5 in [3]). If in addition there exists r > 1 such that
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
9
p, q and r satisfy (1.8) and E[π ω (0)r ] < ∞, writing qZω (t, x, y) for the heat kernel
associated with Lωπ and Z, we have
lim sup
sup nd qZω n2 t, 0, bnxc − a ktΣZ (x) = 0,
P -a.s.
n → ∞ |x|≤K t ∈[T ,T ]
1 2
with a := 1/ E[π ω (0)]. In particular, for the VSRW X with speed measure π = 1, we
have the quenched local limit theorem for the associated heat kernel pω (t, x, y) =
Pxω (Xt = y) if 1/(p − 1) + 1/q < 2/d.
Analogous results have been obtained in other settings, for instance for random
walks on supercritical percolation clusters, see [7]. In [14], the arguments of [7]
have been used in order to establish a general criterion for a local limit theorem
to hold, which is applicable in a number of different situations like graph trees
converging to a continuum random tree or local homogenization for nested fractals.
For the random conductance model a local limit theorem has been proven in
the case, where the conductances are i.i.d. random variables and uniformly elliptic
(see Theorem 5.7 in [7]). Later this has been improved for the VSRW under i.i.d.
conductances, which are only uniformly bounded away from zero (Theorem 5.14
in [6]). However, if the conductances are i.i.d. but have fat tails at zero, the QFCLT
still holds (see [2]), but due to a trapping phenomenon the heat kernel decay is subdiffusive, so the transition density does not have enough regularity for a local limit
theorem – see [8, 9]. Hence, it is clear that some moment conditions are needed.
In [18] Fontes and Mathieu consider a random walk under random ergodic conductances, for which they obtain sub-Gaussian decay of the annealed return probability. More precisely, in their example the conductances are given by ω(x, y) =
θω (x) ∧ θω (y), where {θω (x) : x ∈ Zd } is a family of i.i.d. random variables bounded
from above, i.e. p = ∞, and have a polynomial tail at zero, while the Gaussian
behaviour of the return probability fails once q < d/2. This makes us believe that at
least the moment condition on 1/ω(e) is optimal for general ergodic conductances.
However, in Section 6 below we will show that for some examples of conductances
our moment condition can be improved if it is replaced by a moment condition on
the heat kernel. While for the example in [18] no improvement can be achieved by
d
is sufficient if for inthis method (see Proposition 6.2), we obtain that q > 43 d+1
stance ω(x, y) = θω (x) ∨ θω (y) with {θω (x) : x ∈ Zd } as above (see Proposition 6.5).
Moreover, in the case of i.i.d. conductances uniformly bounded from above this
procedure gives the following result (see Proposition 6.3).
Theorem 1.13. Assume that the conductances {ω(e)
∈ Ed } are independent and
: e−q
uniformly bounded from above, i.e. p = ∞, and E ω(e)
< ∞ for some q > 1/4.
Then, the QFCLT in Theorem 1.10 and the quenched local limit theorem in Theorem 1.11 hold.
We shall prove Theorem 1.11 by using the approach in [7] and [14]. The
two main ingredients of the proof are the QFCLT in Theorem 1.10 and a Höldercontinuity estimate on the heat kernel, which at the end enables us to replace
10
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
the weak convergence given by the QFCLT by the pointwise convergence in Theorem 1.11.
When d ≥ 3 we can also establish a local limit theorem for the Green kernel.
Indeed, as a direct consequence from the PHI we get some on-diagonal bounds on
the heat-kernel (see Proposition 4.5 below) from which we obtain the existence of
the Green kernel g ω (x, y) defined by
Z ∞
ω
q ω (t, x, y) dt.
g (x, y) =
0
By integration we can deduce the following local central limit theorem for the Green
kernel from Theorem 1.11.
Theorem 1.14. Suppose that d ≥ 3 and Assumptions 1.8 and 1.9 hold. Then, for any
x 6= 0 we have
lim nd−2 g ω 0, bnxc − a gBM (x) = 0,
P -a.s.
n→∞
R∞
with a := 1/ E[µω (0)] and gBM (x) = 0 kt (x) dt denoting the Green kernel of a
Brownian motion with covariance matrix Σ2Y .
Finally, we remark that our result applies to a random conductance model given
by
ω(x, y) = exp φ(x) + φ(y) ,
{x, y} ∈ Ed ,
where for d ≥ 3, {φ(x) : x ∈ Zd } is the discrete massless Gaussian free field, cf. [11].
In this case the moment condition (1.10) holds for any p, q ∈ (0, ∞), of course.
The paper is organized as follows: In Section 2 we provide a collection of Sobolevtype inequalities needed to setup the Moser iteration and to apply the BombieriGiusti criterion. Then, Sections 3 and 4 contain the proof of the EHI and PHI in
Theorem 1.3 and 1.4, respectively. The local limit theorem for the random conductance model is proven in Section 5. Some examples for conductances allowing
an improvement on the moment conditions are discussed in Section 6. Finally, the
appendix contains a collection of some elementary estimates needed in the proofs.
Throughout the paper we write c to denote a positive constant which may change
on each appearance. Constants denoted Ci will be the same through each argument.
2. S OBOLEV AND P OINCAR É INEQUALITIES
2.1. Setup and Preliminaries.
For functions f : A → R, where either A ⊆ V or
p
A ⊆ E, the ` -norm f `p (A) will be taken with respect to the counting measure.
The corresponding scalar products in `2 (V ) and `2 (E) are denoted by h·, ·i`2 (V ) and
h·, ·i`2 (E) , respectively. Similarly, the `p -norm and scalar product w.r.t. to a mea sure ϕ : V → R will be denoted by f p
and h·, ·i 2
, i.e. hf, gi 2
=
` (V,ϕ)
` (V,ϕ)
` (V,ϕ)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
11
hf, g · ϕi`2 (V ) . For any non-empty, finite A ⊂ V and p ∈ [1, ∞), we introduce spaceaveraged norms on functions f : A → R by
!1
p
X
1
p
f :=
|f (x)| ϕ(x) .
p,A,ϕ
|A|
x∈A
are defined by ∇f : E → R and ∇∗ F : V → R
X
F (y, x) − F (x, y)
∇f (x, y) := f (y) − f (x),
and
∇∗ F (x) :=
The operators ∇ and
∇∗
x∼y
∇∗
for f : V → R and F : E → R. Mind that
is the adjoint of ∇, i.e. for all f ∈ `2 (V )
and F ∈ `2 (E) it holds h∇f, F i`2 (E) = hf, ∇∗ F i`2 (V ) . We define the products f · F
and F · f between a function, f , defined on the vertex set and a function, F , defined
on the edge set in the following way
f · F (x, y) := f (x) F (x, y),
and
F · f (x, y) := f (y) F (x, y).
Then, the discrete analog of the product rule can be written as
∇(f g) = g · ∇f + ∇g · f .
(2.1)
In contrast to the continuum setting, a discrete version of the chain rule cannot
be established. However, by means of the estimate (A.1), |∇f α | for f ≥ 0 can be
bounded from above by
1 α ∇f
≤ f α−1 · ∇f + ∇f · f α−1 ,
∀ α ∈ R.
(2.2)
1 ∨ |α|
On the other hand, the estimate (A.4) implies the following lower bound
∀ α ≥ 1.
2 ∇f α ≥ f α−1 · ∇f + ∇f · f α−1 ,
(2.3)
The Dirichlet form or energy associated to LωY is defined by
E ω (f, g) := hf, −LωY gi`2 (V,µω ) =
1
h∇f, ω∇gi`2 (E) ,
2
E ω (f ) ≡ E ω (f, f ).
(2.4)
For a given function η : B ⊂ V → R, we denote by Eηω2 (u) the Dirichlet form where
ω(x, y) is replaced by 21 (η 2 (x) + η 2 (y))ω(x, y) for {x, y} ∈ E. Mind that
Eηω2 (u) =
1
∇u, (η 2 · ω) ∇u `2 (E) .
2
2.2. Local Poincaré and Sobolev inequality. The main objective in this subsection
is to establish a version of a local Poincaré inequality.
Suppose that the graph (V, E) satisfies the condition (ii) in Assumption 1.1.
Then, by means of a discrete version of the co-area formula, the classical local
`1 -Poincaré inequality on V can be easily established, see e.g. [23, Lemma 3.3].
Moreover, the condition (i) in Assumption 1.1 ensures that balls in V have a regular volume growth. Mind that, due to [13, Théorème 4.1], the volume regularity of
12
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
balls B(x0 , n) and the local `1 -Poincaré inequality on V
u : V → R, that
X
n
inf u − a d ,B(x0 ,n) ≤ C1
a∈R
d−1
|B(x0 , n)|
implies for d ≥ 2 and any
u(x) − u(y).
(2.5)
x,y∈B(x0 ,n)
x∼y
This inequality is the starting point to prove a local `2 -Poincaré inequality on the
weighted graph (V, E, ω).
Proposition 2.1 (Local Poincaré inequality). Suppose d ≥ 2 and for any x0 ∈ V and
n ≥ 1, let B(n) ≡ B(x0 , n). Then, there exists CPI ≡ CPI (d) < ∞ such that for any
u: V → R
X
ω
2
n2
u − (u)B(n) 2
ν
≤
C
ω(x, y) u(x) − u(y) , (2.6)
d
PI
2,B(n)
,B(n)
2
|B(n)|
x,y∈B(n)
x∼y
where (u)B(n) := |B(n)|−1
P
x∈B(n) u(x).
Proof. Without lost of generality, let us assume that (u)B(n) = 0. First, we will
consider the case d = 2. Since inf a∈R ku − ak2,B(n) = ku − (u)B(n) k2,B(n) , the
assertion (2.6) follows immediately from (2.5) and an application of the CauchySchwarz inequality.
Now, assume that d > 2 and set α = 2(d − 1)/d and define ũα := |u|α sign(u).
Let us first establish a lower bound for the left-hand side of (2.5). Since α > 1, we
obtain in view of (A.6)
2 d−1
(2.7)
inf ũα − ãα d ,B(n) ≥ 2−α inf |u − a|α d ,B(n) = 2−α u2,Bd ,
a∈R
a∈R
d−1
d−1
αd
= 2 and (u)B(n) = 0. On the other hand, by
where we used in the last step that d−1
means of (A.1), Cauchy-Schwarz and Hölder inequality, we get that
X 1
ũα (x) − ũα (y)
|B(n)|
x,y∈B(n)
x∼y
1
≤ c ν ω 2d ,B(n)
2
1
|B(n)|
!1
2
X
2
ω(x, y) u(x) − u(y)
x,y∈B(n)
x∼y
α−1 |u|
2d
,B(n)
d−2
.
(2.8)
2d
Notice that (α − 1) d−2
= 2. Thus, by combining (2.7) and (2.8) with (2.5) and by
solving the resulting equation for kuk2,B(n) , we obtain (2.6).
In the later application we also need a weighted version of the local Poincaré
inequality.
Proposition 2.2 (Weighted local Poincaré inequality). Suppose d ≥ 2 and for x0 ∈ V
and n ≥ 1 let B(n) ≡ B(x0 , n). Further, let η be a non-negative, radially decreasing
weight with supp η ⊂ B(n), i.e. η(x) = Φ(d(x0 , x)) for some non-increasing and
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
13
non-negative càdlàg function Φ. Then, there exists M = M (d) < ∞ such that for any
u: V → R
Eηω2 ∧η2 (u)
2 ω
u − (u)B(n),η2 2
≤ CPI M n ν d ,B(n)
,
2,B(n),η 2
2
|B(n)|
P
P
2
where (u)B(n),η2 = x∈B(n) η 2 (x)u(x)
x∈B(n) η (x) and
2
1 X
Eηω2 ∧η2 (u) :=
min{η 2 (x), η 2 (y)} ω(x, y) u(x) − u(y) .
2
x,y∈V
Proof. Given the local Poincaré inequality in Proposition 2.1, this follows from the
same arguments as in Theorem 1 and Proposition 4 in [17, Corollary 5].
From the relative isoperimetric inequality in Assumption 1.1 one can deduce that
the following Sobolev inequality holds for any function u with finite support
u d
(2.9)
≤ CS1 ∇u`1 (E)
` d−1 (V )
(cf. Remark 3.3 in [3]). Note that (2.9) is a Sobolev inequality on an unweighted
graph, while for our purposes we need a version involving the conductances as
weights. Let
ρ = ρ(q, d) :=
qd
.
q(d − 2) + d
(2.10)
Notice that ρ(q, d) is monotone increasing in q and ρ(d/2, d) = 1. In [3] we derived
the following weighted Sobolev inequality by using (2.9) and the Hölder inequality.
Proposition 2.3 (Sobolev inequality). Suppose that the graph (V, E) has the isoperimetric dimension d ≥ 2 and let B ⊂ V be finite and connected. Consider a nonnegative function η with
supp η ⊂ B,
0 ≤ η ≤ 1
and
η ≡ 0 on
∂B.
Then, for any q ∈ [1, ∞], there exists C S ≡ C S (d, q) < ∞ such that for any u : V → R,
!
Eηω2 (u)
2
2
2 2
ω
(η u) + ∇η `∞(E) u 1,B,µω ,
≤ C S |B| d ν q,B
(2.11)
ρ,B
|B|
P
where Eηω2 (u) := 12 x,y η 2 (x) ω(x, y) (u(y) − u(x))2 .
Proof. See [3, Proposition 3.5].
2.3. The lemma of Bombieri and Giusti. In order to obtain the Harnack inequalities we will basically follow the approach in [15]. However, due to the lack of a
suitable control on the BMO norm and thus of a John-Nirenberg lemma we will use
the following abstract lemma by Bombieri and Giusti (cf. [10]):
Lemma 2.4. Let {Uσ : σ ∈ (0, 1]} be a collection of subsets of a fixed measure space
endowed with a measure m such that and Uσ0 ⊂ Uσ if σ 0 < σ. Fix 0 < δ < 1,
0 < α0 ≤ ∞ and let f be a positive function on U := U1 . Suppose that
14
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
(i) there exists γ > 0 such that for all δ ≤ σ 0 < σ ≤ 1 and 0 < α ≤ min{1, α0 /2},
Z
1
1
1 − 1 Z
α0
α
0 −γ
−1 α α0
α
α0
≤ CBG1 (σ − σ ) m[U ]
f dm ,
f dm
Uσ 0
Uσ
(ii) for all λ > 0,
m ln f > λ ≤ CBG2 m[U ] λ−1 ,
with some positive constants CBG1 and CBG2 . Then, there exists A = A(c1 , c2 , α0 , δ, γ)
such that
Z
f α0 dm ≤ A m[U ],
if α0 < ∞,
Uδ
and
sup f (x) ≤ A,
if α0 = ∞.
x∈Uδ
The constant A is more explicitly given by
3 A = exp c1 c2 + 2 CBG1 ∨ CBG2
for some positive c1 ≡ c1 (δ, γ) and c2 ≡ c2 (α0 ).
Proof. See Lemma 2.2.6 in [24].
3. E LLIPTIC H ARNACK INEQUALITY
In this section we prove Theorem 1.3.
3.1. Mean value inequalities.
Lemma 3.1. Consider a connected, finite subset B ⊂ V and a function η on V with
supp η ⊂ B,
0 ≤ η ≤ 1
and
η ≡ 0 on
∂B.
Further, let u > 0 be such that LωY u ≥ 0 on B. Then, there exists C2 < ∞ such that
for all α ≥ 1
Eηω2 (uα )
|B|
≤ C2
α4
(2α−1)2
2
∇η ∞ u2α .
` (E)
1,B,µω
(3.1)
Proof. Since LωY u ≥ 0, a summation by parts yields,
0 ≥
1 η 2 u2α−1 , −LωY u `2 (V,µω ) =
∇(η 2 u2α−1 ), ω ∇u `2 (E) .
2
(3.2)
As an immediate consequence of the product rule (2.1) and (A.2), we obtain
ω
α
2
2α−1
∇(η 2 u2α−1 ), ω ∇u `2 (E) ≥ 2 2α−1
E
(u
)
−
|∇η
|,
ω
|∇u|
·
u
. (3.3)
2
2
η
α
`2 (E)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
15
Since (A.5) is applicable after a suitable symmetrization, we find that
|∇η 2 |, ω |∇u| · u2α−1 `2 (E)
≤ 2 |∇η 2 |, ω uα · |∇uα | + |∇uα | · uα `2 (E)
= 4 uα · |∇η| + |∇η| · uα , ω η · |∇uα | `2 (E)
Thus, by applying the Young inequality |ab| ≤ 21 (εa2 + b2 /ε) and combining the
result with (3.3), we obtain
∇(η 2 u2α−1 ), ω ∇u `2 (E)
2
2α − 1
8
α
ω
∇η ∞ u2α .
(3.4)
≥ 2
(u
)
−
−
2ε
E
|B|
2
η
1,B,µω
` (E)
α2
ε
By choosing ε =
2α−1
4α2
and combining the result with (3.2), the assertion follows.
Proposition 3.2. For any x0 ∈ V and n ≥ 1, let B(n) ≡ B(x0 , n). Let u > 0 be such
that LωY u ≥ 0 on B(n). Then, for any p, q ∈ (1, ∞] with
1 1
2
+ <
p q
d
there exists κ = κ(d, p, q) ∈ (1, ∞) and C3 = C3 (d, q) such that for all β ∈ [2p∗ , ∞]
and for all 1/2 ≤ σ 0 < σ ≤ 1 we have
!
1 ∨ µω p,B(n) ν ω q,B(n) κ u
u
≤ C3
,
(3.5)
β,B(σ 0 n)
2p∗ ,B(σn)
(σ − σ 0 )2
where p∗ = p/(p − 1).
Proof. For fixed 1/2 ≤ σ 0 < σ ≤ 1, consider a sequence {B(σk n)}k of balls with
radius σk n centered at x0 , where
σk = σ 0 + 2−k (σ − σ 0 )
and
τk = 2−k−1 (σ − σ 0 ),
k = 0, 1, . . .
Mind that σk = σk+1 + τk and σ0 = σ. Further, we consider a cut-off function ηk
with supp ηk ⊂ B(σk n) having the property that ηk ≡ 1 on B(σk+1 n), ηk ≡ 0 on
∂B(σk n) and linear decaying on B(σk n) \ B(σk+1 n). This choice of ηk implies that
|ηk (x) − ηk (y)| ≤ 1/τk n for all (x, y) ∈ Ed .
Set αk = (ρ/p∗ )k . Since 1/p + 1/q < 2/d, it holds that ρ > p∗ . Hence, we have
that αk ≥ 1 for every k. Thus, by applying the Sobolev inequality (2.11) to uαk , we
obtain
(ηk uαk )2 ρ,B(σk n)
ω
!
µ Eηω2 (uαk )
2 p,B(σ
n)
k
k
u2αk +
.
≤ CS |B(σk n)| d ν ω q,B(σ n)
p∗ ,B(σk n)
k
|B(σk n)|
(τk n)2
16
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
On the other hand, by using the Hölder inequality with 1/p + 1/p∗ = 1 we have
Eηω2 (uαk )
2α αk 2 k
µω u k ≤ C2
.
p,B(σk n)
p∗ ,B(σk n)
|B(σk n)|
τk n
Since αk+1 p∗ = αk ρ, we obtain by combining these two estimates and using the
2 2d that
volume regularity which implies that |B(n)|/|B(σ 0 n)| ≤ Creg
u
2α
k+1 p∗ ,B(σk+1 n)
≤
ω
22k αk2 µω ν c
p,B(n)
q,B(n)
0
2
(σ − σ )
1
2αk
u
2α
k p∗ ,B(σk n)
(3.6)
for some c < ∞. First, for β ∈ [2p∗ , ∞) choose K < ∞ such that β ≤ 2αK p∗ . Then,
Jensen’s inequality implies kukβ,B(σK n) ≤ kuk2αK p∗ ,B(σK n) . By iterating (3.6) and
P
using the fact that ∞
k=0 k/αk < ∞ there exists C3 < ∞ independent of k such that
!1
K−1
Y µω p,B(n) ν ω q,B(n) 2αk u
u
≤ C3
.
β,B(σ 0 n)
2p∗ ,B(σn)
(σ − σ 0 )2
k=0
P
Thus, the claim is immediate with κ = 12 ∞
k=0 (1/αk ) < ∞. Now, let us consider
the case β = ∞. Mind that
1
max u(x) ≤ |B(σ 0 n)| 2αn p∗ u2αn p∗ ,B(σ0 n) ≤ c u2αn p∗ ,B(σn n) .
x∈B(σ 0 n)
By iterating the inequality (3.6) there exists C3 < ∞ independent of k such that
ω
ω
! 1
n
µ ν 2αk
Y
p,B(n)
q,B(n)
u
max u(x) ≤ C3
.
2p∗ ,B(σn)
0
2
(σ − σ )
x∈B(σ 0 n)
k=0
Finally, choosing κ as before the claim is immediate.
Corollary 3.3. Let u > 0 be such that LωY u = 0 on B(n). Then, for all α ∈ (0, ∞)
and 1/2 ≤ σ 0 < σ ≤ 1 there exists C4 < ∞ such that
!2p∗ κ
1 ∨ µω p,B(n) ν ω q,B(n) α u−1 max u(x)−1 ≤ C4
.
(3.7)
α,B(σn)
0
2
0
(σ − σ )
x∈B(σ n)
Proof. Since u > 0 is harmonic on B(n), the function f = u−α/2p∗ is sub-harmonic
on B(n), i.e. LωY f ≥ 0. Thus, (3.7) follows by applying (3.5) with β = ∞ to f . Corollary 3.4. Consider a non-negative u such that LωY u ≥ 0 on B(n). Then, for all
α ∈ (0, ∞) and 1/2 ≤ σ 0 < σ ≤ 1, there exists C5 < ∞ such that
! 0
1 ∨ µω p,B(n) ν ω q,B(n) κ u
max u(x) ≤ C5
(3.8)
α,B(σn)
(σ − σ 0 )2
x∈B(σ 0 n)
with κ0 = 1 ∨ 2pα∗ κ > 1.
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
17
Proof. The proof of this Corollary, based on arguments given originally in [24, Theorem 2.2.3], can be found in [3, Corollary 3.9]. Nevertheless, we will repeat it here
for the reader’s convenience.
In view of (3.5), for any α ≥ 2p∗ the statement (3.8) is an immediate consequence of Jensen’s inequality. It remains to consider the case α ∈ (0, 2p∗ ). For all
1/2 ≤ σ 0 < σ ≤ 1 we have from (3.5) with β = ∞,
!
1 ∨ µω p,B(n) ν ω q,B(n) κ u
.
(3.9)
max u(x) ≤ C3
2p∗ ,B(σn)
(σ − σ 0 )2
x∈B(σ 0 n)
In the sequel, let 1/2 ≤ σ 0 < σ ≤ 1 be arbitrary but fixed and set σk = σ−2−k (σ−σ 0 )
for any k ∈ N0 . Now, by Hölder’s inequality, we have for any α ∈ (0, 2p∗ )
u
2p∗ ,B(σ
k n)
θ
≤ uα,B(σ
k n)
1−θ
u
∞,B(σ
k n)
where θ = α/2p∗ . Hence, in view of (3.9) and the volume regularity which implies
2 2d , we obtain
that |B(σn)|/|B(σ 0 n)| ≤ Creg
u
∞,B(σ
k−1 n)
1−θ
θ
≤ 22κk J uα,B(σn) u∞,B(σ
k n)
,
(3.10)
κ
where we introduced J = c µω p,B(n) ν ω q,B(n) /(σ − σ 0 )2 to simplify notation.
Hence, by iteration, we get
Pi−1 (1−θ)k P
i
k
k=0
θ
2κ i−1
(k+1)(1−θ)
u(1−θ) . (3.11)
u
k=0
≤ 2
J u α,B(σn)
∞,B(σ n)
∞,B(σ 0 n)
i
2
As i tends to infinity, this yields u∞,B(σ0 n) ≤ 22κ/θ J 1/θ and (3.8) is immediate.
3.2. From mean value inequalities to Harnack principle.
Lemma 3.5. Consider a connected, finite subset B ⊂ V and a function η on V with
supp η ⊂ B,
0 ≤ η ≤ 1
and
η ≡ 0 on
∂B.
Further, let u > 0 be such that LωY u ≤ 0 on B. Then,
Eηω2 ∧η2 (ln u)
|B|
2
≤ 9 osr η 2 ∇η `∞(E) µω 1,B ,
(3.12)
where osr(η) := max η(y)/η(x) ∧ 1 | {x, y} ∈ E, η(x) 6= 0 .
Proof. Since u > 0 and LωY u ≤ 0 on B, a summation by parts yields
0 ≤
2 −1
1 η u , −LωY u `2 (V,µω ) =
∇(η 2 u−1 ), ω ∇u `2 (E) .
2
(3.13)
18
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
In view of the estimates (A.3) and (A.7), we obtain for any (x, y) ∈ E that
∇ η 2 u−1 (x, y) (∇u)(x, y)
1
min η 2 (x), η 2 (y) ∇u−1 (x, y) (∇u)(x, y) + 9 osr η 2 (∇η)2 (x, y)
2
2
1
≤ − min η 2 (x), η 2 (y) ∇ ln u (x, y) + 9 osr η 2 (∇η)2 (x, y). (3.14)
2
Thus, by combining the resulting estimate with (3.13) the claim (3.12) follows. ≤
Proof of Theorem 1.3. Consider a harmonic function u > 0 on B(n) ≡ B(x0 , n).
Pick ρ so that 3/4 ≤ ρ < 1 and set
d(x0 , x) − ρn +
∨ 0.
η(x) := 1 −
(1 − ρ) n
Obviously, the cut-off function η has the properties that supp η ⊂ B(n), η ≡ 1 on
B(ρn), η ≡ 0 on ∂B(n), k∇ηk`∞ (E) ≤ 1/(1 − ρ)n and osr(η 2 ) ≤ 4.
Now let f1 := u−1 eξ with ξ = ξ(u) := (ln u)B(n),η2 . Our aim is to apply
Lemma 2.4 to the function f1 with U = B(ρn), Uσ = B(σρn), δ = 1/2ρ, α0 = ∞
and m being the counting measure. Mind that Corollary 3.3 implies that the condition (i) of Lemma 2.4 is satisfied with γ = 4p∗ κ, where κ and p∗ are chosen as in
Proposition 3.2 and
2p∗ κ
CBG1 = c 1 ∨ µω p,B(n) ν ω q,B(n) .
To verify the second condition of Lemma 2.4, we apply the weighted local `2 Poincaré inequality in Proposition 2.2 to the function ln u. This yields
ln u − (ln u)B(n),η2 2
2,B(n),η 2
Eηω2 ∧η2 (ln u)
≤ CPI M n ν d ,B(n)
2
|B(n)|
ω
ω
≤ c ν d ,B(n) µ 1,B(n) ,
2 ω
2
where we used (3.12) in the last step. Hence,
1
ω
|B(n)| 2
ln u − ξ ν ω d
µ λ 1l{ln f1 > λ} 1,B(ρn) ≤
≤
c
.
2
,B(n)
1,B(n),η
1,B(n)
2
|B(ρn)|
This shows that the second hypothesis of Lemma 2.4 is satisfied by f1 . Thus, we can
apply Lemma 2.4 to f1 and, since Uδ = B(n/2), we conclude that
f1 ≤ Aω
⇐⇒
eξ ≤ Aω min u(x).
(3.15)
∞,B(n/2)
x∈B(n/2)
Analogously, using Corollary 3.4 one can verify that also the function f2 = u e−ξ
with ξ(u) := (ln u)B(n),η2 satisfies as well the conditions of Lemma 2.4. Hence,
f2 ≤ Aω
⇐⇒
max u(x) ≤ Aω eξ .
(3.16)
∞,B(n/2)
x∈B(n/2)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
19
By combining (3.15) and (3.16), we obtain the EHI. Finally, the explicit form of the
Harnack constant follows from the representation of the constant in Lemma 2.4. 3.3. Hölder continuity. As an immediate consequence from the elliptic Harnack
inequality we prove the Hölder-continutiy of harmonic functions.
Proposition 3.6. Suppose that Assumption 1.6 holds. Let x0 ∈ V and let sω (x0 ) and
∗ be as in Assumption 1.6. Further, let R ≥ s(x ) and suppose that u > 0 is a
CEH
0
harmonic function on B(x0 , R0 ) for some R0 ≥ 2R. Then, for any x1 , x2 ∈ B(x0 , R)
we have
θ
u(x1 ) − u(x2 ) ≤ c · R
max u,
R0 B(x0 ,R0 /2)
∗ /(2C ∗ − 1) / ln 2 and c only depending on C ∗ .
where θ = ln 2CEH
EH
EH
Proof. Set Rk = 2−k R0 and let Bk = B(x0 , Rk ) and note that Bk+1 ⊂ Bk . For
x ∈ Bk we define the function vk through
vk (x) =
u(x) − minBk u
∈ [0, 1],
maxBk u − minBk u
Mind that vk is harmonic on Bk and satisfies osc(vk , Bk ) = 1. (Here the oscillation
of a function u on A is defined by osc(u, A) = maxA u − minA u.) Replacing vk by
1 − vk if necessary we may assume that maxBk+1 vk ≥ 1/2.
Now for every k such that Rk ≥ sω (x0 ), an application of Theorem 1.3 yields
1
∗
≤ max vk ≤ CEH
min vk .
Bk+1
Bk+1
2
Moreover, for such k it follows that
maxBk+1 u − minBk+1 u
osc(u, Bk+1 ) =
osc(u, Bk )
osc(u, Bk )
maxBk+1 u − maxBk u
= 1+
− min vk osc(u, Bk ).
Bk+1
osc(u, Bk )
Hence,
osc(u, Bk+1 ) ≤ (1 − δ) osc(u, Bk )
−1
∗
with δ = 2CEH
. Now we choose k0 such that Rk0 ≥ R > Rk0 +1 . Then, by
iterating the above inequality on the chain of balls B1 ⊃ B2 ⊃ · · · ⊃ Bk0 we obtain
osc(u, Bk0 ) ≤ (1 − δ)k0 −1 osc(u, B1 ).
Note that B(x0 , R) ⊆ B(x0 , Rk0 ) and B1 = B(x0 , R0 /2). Since (1−δ)k0 ≤ c(R/R0 )θ ,
the claim follows.
4. PARABOLIC H ARNACK INEQUALITY
In this section we prove the parabolic Harnack inequality in Theorem 1.4.
20
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
4.1. Mean value inequalities.
Lemma 4.1. Suppose that Q = I × B, where I = [s1 , s2 ] is an interval and B is a
finite, connected subset of V . Consider a function η : V → R and a smooth function
ζ : R → R with
supp η ⊂ B,
0 ≤ η ≤ 1
and
η ≡ 0
supp ζ ⊂ I,
0 ≤ ζ ≤ 1
and
ζ(s1 ) = 0.
on
∂B,
Further, let u > 0 be such that ∂t u − LωY u ≤ 0 on Q. Then, for all α ≥ 1,
Z
Z
Eηω2 (uαt )
2α 2
ut dt
ζ(t)
dt ≤ 32 α2 ∇η `∞ (E) + ζ 0 L∞ (I)
1,B,µω
|B|
I
I
(4.1)
and
Z
2α 2
ut dt.
max ζ(t) (η uαt )2 1,B,µω ≤ 32 α2 ∇η `∞(E) + ζ 0 L∞ (I)
1,B,µω
t∈I
I
(4.2)
Proof. Since ∂t u − LωY u ≤ 0 on Q we have, for every t ∈ I,
2 2α−1
1
∂t η 2, u2α
=
η
u
,
∂
u
2
ω
t
t
t
t
` (V,µ )
`2 (V,µω )
2α
1 ∇(η 2 u2α−1
), ω ∇ut `2 (E) .
≤ η 2 u2α−1
, LωY ut `2 (V,µω ) = −
t
t
2
1
By choosing ε = 4α in (3.4) and exploiting the fact that α ≥ 1, we obtain
2α 1 ω α
ut .
E
(u
)
−
32
α
|B|
∇(η 2 u2α−1
),
ω
∇u
≥
2
2
t
t
t
1,B,µω
` (E)
α η
Hence,
Eηω2 (uαt )
2
α 2
∂t (η ut ) 1,B,µω +
≤ 32 α2 ∇η `∞(E) u2α
.
(4.3)
t
1,B,µω
|B|
By multiplying both sides of (4.3) with ζ(t), we get
Eηω2 (uαt )
∂t ζ(t) (η uαt )2 1,B,µω + ζ(t)
|B|
2
≤ 32 α2 ∇η `∞(E) u2α
+ ζ 0 L∞ (I) (η uαt )2 1,B,µω .
t
1,B,µω
Integrating this inequality over [s1 , s] for any s ∈ I, we obtain
Z s
Eηω2 (uαt )
α 2
ds
ζ(s) (η us ) 1,B,µω +
ζ(t)
|B|
s1
Z 0
2
2 u2α
≤ 32 α
∇η `∞(E) + ζ L∞ (I)
dt.
t
1,B,µω
(4.4)
I
Thus, by neglecting the first term on the left-hand side of (4.4), (4.1) is immediate,
whereas (4.2) follows once we neglect the second term on the left-hand side of
(4.4).
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
21
For any x0 ∈ V , t0 ≥ 0 and n ≥ 1, we write Q(n) ≡ [t0 , t0 + n2 ] × B(n) with
B(n) ≡ B(x0 , n). Further, we consider a family of intervals {Iσ : σ ∈ [0, 1]}, i. e.
h
i
Iσ := σ t0 + 1 − σ s0 , 1 − σ s00 + σ t0 + n2
interpolating between the intervals [t0 , t0 + n2 ] and [s0 , s00 ], where s0 ≤ s00 are chosen
such that given ε ∈ (0, 1/4) we have s0 − t0 ≥ εn2 and either t0 + n2 − s00 ≥ εn2 or
s00 = t0 + n2 . Moreover, set Q(σn) := Iσ × B(σn). In addition, let us introduce a
Lp -norm on functions u : R × V → R by
1
Z
p
p
1
u
:=
ut p,B,µω dt ,
p,I×B,µω
|I| I
where ut = u(t, .), t ∈ R.
Proposition 4.2. Let u > 0 be such that ∂t u − LωY u ≤ 0 on Q(n). Then, for any
p, q ∈ (1, ∞] with
1
1
2
+
<
(4.5)
p
q
d
there exists κ ≡ κ(d, p, q) and C6 ≡ C6 (d, q, ε) such that for all β ≥ 1 and for all
1/2 ≤ σ 0 < σ ≤ 1, we have
!
1 ∨ µω p,B(n) 1 ∨ ν ω q,B(n) κ u
u
≤ C6
. (4.6)
2β,Q(σ 0 n),µω
2,Q(σn),µω
(σ − σ 0 )2
Proof. Proceeding as in the proof of Proposition 3.2, consider a sequence {B(σk n)}k
of balls with radius σk n centered at x0 , where
σk = σ 0 + 2−k (σ − σ 0 )
τk = 2−k−1 (σ − σ 0 ),
and
k = 0, 1, . . .
and a sequence {ηk }k of cut-off functions in space
such
that supp ηk ⊂ B(σk n),
ηk ≡ 1 on B(σk+1 n), ηk ≡ 0 on ∂B(σk n) and ∇ηk `∞ (E) ≤ 1/τk n. Further, let
{ζk }k be a sequence of cut-off functions in time,
i.e. ζk ∈ C ∞ (R), supp ζk ⊂ Iσk ,
ζk ≡ 1 on Iσk+1 , ζk (σk t0 + (1 − σk )s0 ) = 0 and ζk0 L∞ ([t0 ,t0 +n2 ]) ≤ 1/ετk n2 .
Set αk = (1 + (ρ − p∗ )/ρ)k . Since 1/p + 1/q < 2/d, it holds that ρ > p∗ . Hence,
we have that αk ≥ 1 for every k ∈ N0 . By Hölder’s inequality, we have that
∗
2α 1+ ρ−p
ρ
u k ρ−p
ω
∗
1+
≤
ρ
,B(σk+1 n),µ
p∗
ω pρ∗
|B(σk n)| (ηk uαk )2 1− ρ
(ηk uαk )2 µ ω
t
t
1,B(σk n),µ
p,B(σk n) .
ρ,B(σk n)
|B(σk+1 n)|
2 2d , we obtain
Since by the volume regularity |B(σk n)|/|B(σk+1 n)| ≤ Creg
∗
2α 1+ ρ−p
p∗
p∗
αk 2 1− ρ
ρ
2
d ω ρ
u k ρ−p
max
≤
C
2
µ
(η
u
)
k t
ω
reg
∗
p,B(σk n)
1,B(σk n),µω
1+
ρ
,Q(σk+1 n),µ
t∈Iσk+1
×
1
|Iσk+1 |
Z
(ηk uαk )2 t
ρ,B(σ
Iσk+1
k
dt .
n)
22
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Now, in view of (2.11), the integrand can be estimated from above by
!
Eηω2 uαt k
1 2αk 2 ω
k
(ηk uαk )2 u
.
≤ CS n ν q,B(σ n)
+
t
1,B(σk n),µω
ρ,B(σk n)
k
|B(σk n)|
(τk n)2 t
On the other hand, we obtain from (4.1) that
Z
Z
Eηω2 uαt k
2α αk 2 τk k
u k dt.
dt ≤ 32
1+
t
1,B(σk n),µω
|B(σ
n)|
τ
n
ε
k
k
Iσ
Iσ
k+1
k
Combining the estimates above and using (4.2), we finally get
u
2α
,Q(σ
n),µω
k+1
≤
k+1
−1
cε
2α1k ω
ω pρ∗
22k αk2 u
1 ∨ µ p,B(n) ν q,B(n)
2αk ,Q(σk n),µω
(σ − σ 0 )2
for some c < ∞, where we used that τk ≤ 1 and |Iσk |/|Iσk+1 | = σk /σk+1 ≤ 2.
Now for any β ∈ [1, ∞) choose K <∞ such that 2β
≤
2αK . By iterating the
inequality above and using the fact that u 2β,Q(σ0 n) ≤ u2α ,Q(σ n) gives
K
K
ω
ω
! 1
K−1
1 ∨ µ p,B(n) 1 ∨ ν q,B(n) 2αk Y
u
u
≤
C
,
6
2β,Q(σ 0 n),µω
2,Q(σn),µω
(σ − σ 0 )2
k=0
(4.7)
P∞
where C6 < ∞ is a constant that is independent of k since k=0 k/αk < ∞ and
P
supk 2(k+1)/2αk < ∞. Finally, choosing κ = 21 ∞
k=0 1/αk < ∞ the claim follows. Corollary 4.3. Suppose that the assumptions of the Proposition 4.2 are satisfied. In
addition, assume that u > 0 solves ∂t u − Lω u = 0 on Q(n).
(i) Suppose that s0 − t0 ≥ εn2 and either t0 + n2 − s00 ≥ εn2 or s00 = t0 + n2 . Then,
for all α ∈ (0, ∞), there exists C7 = C7 (d, q, ε) and κ0 = 2(1 + κ)/α such that
! 0
1 ∨ µω p,B(n) 1 ∨ ν ω q,B(n) κ −1
u−1 max u(t, x)
≤ C7
.
α,Q(σn),µω
(σ − σ 0 )2
(t,x)∈Q(σ 0 n)
(4.8)
(ii) Suppose that s0 − t0 ≥ εn2 and t0 + n2 − s00 ≥ εn2 . Then, for all α ∈ (0, ∞),
there exists C8 = C8 (d, q, ε) and κ0 = (1 ∨ α2 )(1 + κ) such that
! 0
1 ∨ µω p,B(n) 1 ∨ ν ω q,B(n) κ u
max u(t, x) ≤ C8
.
α,Q(σn),µω
(σ − σ 0 )2
(t,x)∈Q(σ 0 n)
(4.9)
Proof. (i) We shall show that for β large enough there exists c < ∞ which is independent of n such that
!
1 ∨ ν ω 1,B(n) u−1 max u(t, x)−1 ≤ c
,
(4.10)
2β,Q(σ̄n),µω
(σ − σ 0 )2
(t,x)∈Q(σ 0 n)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
23
where σ̄ = (σ 0 + σ)/2. Once we have established (4.10), (4.8) follows by applying
(4.6) to the function f = u−α/2 with σ 0 replaced by σ̄. The proof of (4.10) uses
arguments similar to the ones given in [16]. Since u is caloric on Q(n), we have for
every t ∈ Iσ and x ∈ B(σn)
∂t u(t, x) = LωY u(t, ·)(x) ≥ −u(t, x),
which implies that u(t2 , x) ≥ e−(t2 −t1 ) u(t1 , x) for every t1 , t2 ∈ Iσ with t1 < t2 and
x ∈ B(σn). Now, choose (t∗ , x∗ ) ∈ Q(σ 0 n) in such a way that
u(t∗ , x∗ )−1 =
max
(t,x)∈Q(σ 0 n)
u(t, x)−1 .
Then, for any t ∈ [σ̄ t0 + (1 − σ̄) s0, σ 0 t0 + (1 − σ 0 ) s0 ] =: I 0 ⊂ Iσ̄ \ Iσ0 we have that
2
u(t, x∗ )−1 ≥ e−n u(t∗ , x∗ )−1 . This implies
Z
Z
−1 β
1
|I 0 | −n2
−1
u dt
≥
u(t,
x
)
dt
≥
e
u(t∗ , x∗ )−1 .
∗
t
β,B(σ̄n)
|B(σ̄n)|
|B(n)|
0
Iσ̄
I
Recall that s0 − t0 ≥ εn2 . Hence |I 0 | ≥ 12 ε(σ − σ 0 ) n2 . Choosing β = |B(n)|n2 ∨ α,
an application of the Cauchy-Schwarz inequality yields
1
−1 ω 2β
−1 1
1
0 β
u ν u max u(t, x)−1,
≥
σ
−
σ
≥
ω
1,B(σ̄n)
2β,Q(σ̄n),µ
β,Q(σ̄n)
c
(t,x)∈Q(σ 0 n)
where c = c(ε) ∈ (0, ∞). Here, we used that 1/µω 1,B(σ̄n) ≤ ν ω 1,B(σ̄n) . This
completes the proof of (4.10).
(ii) For every t ∈ [(1−σ 0 ) s00 +σ 0 (t0 +n2 ), (1− σ̄) s00 + σ̄ (t0 +n2 )] ⊂ Iσ̄ \Iσ0 we have
2
that u(t, x∗ ) ≥ e−n u(t∗ , x∗ ), where u(t∗ , x∗ ) = max(t,x)∈Q(σ0 n) u(t, x). Therefore,
by a computation similar to the one above and in view of (4.6), we obtain
!
1 ∨ µω p,B(n) 1 ∨ ν ω q,B(n) 1+κ u
max u(t, x) ≤ c
.
2,Q(σn),µω
(σ − σ 0 )2
(t,x)∈Q(σ 0 n)
In particular, this inequality implies (4.9) for every α ≥ 2. For α ∈ (0, 2) the proof
goes literally along the lines of the proof of Corollary 3.4.
4.2. From mean value inequalities to Harnack principle. From now on we denote by m the product measure m = dt × | · | on R+ × V , where | · | is still the
counting measure on V .
Lemma 4.4. For x0 ∈ V , t0 ≥ 0 and n ≥ 1, let Q(n) = [t0 , t0 + n2 ] × B(n) where
B(n) ≡ B(x0 , n). Further, let u > 0 be such that ∂t u − LωY u ≥ 0 on Q(n). Then, there
exists C9 = C9 (d) and ξ = ξ(u) such that for all λ > 0 and all ρ ∈ [1/2, 1),
h
i
m (t, z) ∈ I × B(ρn) : ln u(t, z) < −λ − ξ
≤ C9 ν ω d ,B(n) + µω 1,B(n) m I × B(ρn) λ−1 ,
2
24
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
and
h
i
m (t, z) ∈ I × B(ρn) : ln u(t, z) > λ − ξ
≤ C9 ν ω d ,B(n) + µω 1,B(n) m I × B(ρn) λ−1 .
2
Proof. Let η be defined as
η(x) =
d(x0 , x)
1−
n
.
+
In particular, η ≤ 1, maxx∼y |∇η(x, y)| ≤ 1/n and osr(η 2 ) ≤ 4. Moreover, note that
η ≥ 1 − ρ on B(ρn), and by the volume regularity of V we get
|B(n)| ≥ η 2 , 1 `2 (V ) ≥ 1 − ρ)2 |B(ρn)| ≥ c |B(n)|.
(4.11)
Since ∂t u − LωY u ≥ 0 on Q(n), we have
∂t η 2 , − ln ut `2 (V,µω ) = η 2 ut−1 , −∂t ut `2 (V,µω )
2 −1
1 ∇ η 2 u−1
, ω ∇ut `2 (E) .
η ut , −LωY ut `2 (V,µω ) =
t
2
In view of (3.14) and (4.11), we obtain
c1 1 ∇ η 2 u−1
, ω ∇ut `2 (E) ≤ −Eηω2 ∧η2 (ln ut ) + 2 η 2 , 1 `2 (V ) µω 1,B(n) .
t
2
n
Hence,
c1 ∂t η 2 , − ln ut `2 (V,µω ) ≤ −Eηω2 ∧η2 (ln ut ) + 2 η 2 , 1 `2 (V ) µω 1,B(n) . (4.12)
n
From now on the proof goes literally along the lines of [24, Lemma 5.4.1]. For
the reader’s convenience we provide the details here. Let us define
c1 w(t, x) := − ln u(t, x),
w(t, x) := w(t, x) − 2 µω 1,B(n) (t − t0 ),
n
2
η , wt `2 (V )
c1 µω :=
W (t) :=
,
W
(t)
W
(t)
−
(t − t0 ).
1,B(n)
hη 2 , 1i`2 (V )
n2
≤
With the notation introduced above, the inequality (4.12) reads
−1
c1 ∂t W (t) + η 2 , 1 `2 (V ) Eηω2 ∧η2 (wt ) ≤ 2 µω 1,B(n) .
n
Hence, applying the weighted local `2 -Poincaré inequality given in Proposition 2.2
and again (4.11), we find
−1 c1 wt − W (t)2
µω ∂t W (t) + CPI M n2 ν ω d ,B(n)
,
2 ≤
2,B(n),η
1,B(n)
2
2
n
where we used that |B(n)|/ η 2 , 1 `2 (V ) ≥ 1. This implies
−1 wt − W (t)2
≤ 0.
∂t W (t) + c2 n2 ν ω d ,B(n)
2,B(ρn)
2
(4.13)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
25
In particular, ∂t W (t) ≤ 0. Let ξ = ξ(u) = W (t0 ) and for any λ > 0 we define
Dt+ (λ) = z ∈ B(ρn) : w(t, z) > ξ + λ .
Then, we have for all t ∈ (t0 , t0 + n2 ] and z ∈ Dt+ (λ) that
w(t, z) − W (t) ≥ λ + ξ − W (t) > λ,
because ξ = W (t0 ) and ∂t W (t) ≤ 0. Using this in (4.13) we obtain
−1 ω
2
λ + ξ − W (t)2 D+ (λ) ≤ 0
∂t W (t) + c2 n |B(ρn)| ν d ,B(n)
t
2
which is equivalent to
+ λ + ξ − W (t)−1 .
D (λ) ≤ −c2 n2 |B(ρn)| ν ω d
∂
t
t
,B(n)
2
n2
Integrating from t0 to t0 +
gives
h
i
n2 |B(ρn)|
≤ c2 ν ω d ,B(n)
.
m (t, z) ∈ I × B(ρn) : w(t, z) > ξ + λ
2
λ
On the other hand, an application of the Markov inequality yields
h
i
n2 |B(ρn)|
c1 ≤ c1 µω 1,B(n)
.
m (t, z) ∈ I × B(ρn) : 2 µω 1,B(n) (t − s0 ) > λ2
n
λ
Since − ln u = w = w + nc12 µω 1,B(n) (t − t0 ), we obtain finally
h
i
m (t, z) ∈ I × B(ρn) : ln u(t, z) < −ξ − λ
h
i
≤ m (t, z) ∈ I × B(ρn) : w(t, z) > ξ + λ2
h
i
c1 + m (t, z) ∈ I × B(ρn) : 2 µω 1,B(n) (t − s0 ) > λ2
n
≤ c ν ω d ,B(n) + µω 1,B(n) n2 |B(ρn)| λ−1 ,
2
which proves the first inequality.
Working instead of Dt+ (λ) with the set Dt− (λ) =
z ∈ B(ρn) : w(t, z) < ξ − λ , the second inequality follows by a similar argument.
Proof of Theorem 1.4. Consider a caloric function u > 0 on space-time cylinder
Q(n) = [t0 , t0 + n2 ] × B(n) with B(n) = B(x0 , n) and fix some ρ ∈ [3/4, 1). Consider
the function f1 := u−1 e−ξ , where ξ = ξ(u) be as in Lemma 4.4. Our aim is to apply
Lemma 2.4 to the function f1 := u−1 e−ξ with U = [t0 , t0 + n2 ] × B(ρn),
Uσ = [σ t0 + (1 − σ) s0, t0 + n2 ] × B(σρn)
with
s0 = t0 +
3ρ
n2 ,
4ρ − 2
δ = 1/2ρ, α0 = ∞ and m = dt × | · |. In view of (4.8) the condition (i) of Lemma 2.4
is satisfied with γ = 2(1 + κ) and constant CBG1 given by
2(1+κ)
CBG1 = c 1 ∨ µω p,B(n)
1 ∨ ν ω q,B(n)
.
26
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Since ln f1 > λ is equivalent to ln u < −λ − ξ, the condition (ii) of Lemma 2.4
follows immediately from Lemma 4.4. Noting that Uδ = Q+ , an application of
Lemma 2.4 gives
max f1 (t, x) ≤ Aω
max u−1 (t, x) ≤ eξ Aω .
⇐⇒
(t,x)∈Q+
(t,x)∈Q+
(4.14)
ξ
Next, consider
ξ = ξ(u) be again as in Lemma 4.4.
the function f0 2 = u e , where
Now, set Uσ = σ t0 + (1 − σ) s , (1 − σ) s00 + σ(t0 + n2 ) × B(σρn) with
s0 = t0 +
ρ
n2
4ρ − 2
s00 = t0 +
and
ρ−1 2
n ,
2ρ − 1
where δ, α0 and m are as before. Analogously to the above procedure, the two
conditions of Lemma 2.4 are verified due to (4.9) and Lemma 4.4. Hence, since
Uδ = Q− , we conclude that
max f2 (t, x) ≤ Aω
(t,x)∈Q−
⇐⇒
max u(t, x) ≤ e−ξ Aω .
(t,x)∈Q−
(4.15)
By combining (4.14) and (4.15), the assertion is immediate.
4.3. On-diagonal estimates and Hölder-continuity. It is a well known fact that
the parabolic Harnack inequality is a powerful property, which has a number of
important consequences. We state here only a few of them, namely it provides
immediately on-diagonal estimates on the heat kernel and a quantitative Hölder
continuity estimate for caloric functions.
Proposition 4.5 (On-diagonal heat-kernel bounds). Let t > 0 and x1 ∈ V . Then,
(i) for any x2 ∈ V ,
q ω (t, x1 , x2 ) ≤
(ii) for any x2 ∈ B
1
2
√
t
µω
d
C Creg
CPH
√ ≤ ω PH
2d t− 2 ,
1
√
µ 1,B(x1 , t/2)
B(x1 , 2 t)
n
o
Creg
1
d − d2
√
≥
min
1,
2
t
,
2
2 B(x , 1 t)
CPH
CPH
1 2
kµkp,B(x1 ,√t) , kνkq,B(x1 ,√t) being the Harnack constant in Theorem 1.4.
q ω (t, x1 , x2 ) ≥
with CPH
Proof. We will proceed as in [16, Proposition 3.1]. Given x1 , x2 ∈√V and t > 0 we
apply the PHI to the caloric function q ω (·, ·, x2 ) on [t0 , t0 + t] × B( t) with t0 = 43 t.
√
This yields for any z ∈ B(x1 , 21 t)
q ω (t, x1 , x2 ) ≤ CPH q ω 23 t, z, x2 ,
with CPH = CPH kµkp,B(x1 ,√t) , kνkq,B(x1 ,√t) . Hence, by multiplying both sides
√
with µω (z), a summation over all z ∈ B(x1 , 21 t) gives
X
CPH
CPH
ω
√
√ ,
q ω (t, x1 , x2 ) ≤ ω P
Y
=
y
≤ ω
3t/2
x
2
1
√
µ B(x1 , 2 t)
µ B(x1 , 12 t)]
z∈B(x1 , t/2)
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
27
where we used in the first step both the symmetry and the definition of the transition density. The assertion (i) follows now from the volume
√ regularity assumption.
1
In order to prove (ii), suppose that x2 ∈ B(x1 , 2 t). Similar to the above
ω
considerations
√ an applicationωof1 Theorem 1.4 to ωthe caloric function q (·, ·, x21)√on
[0, t] × B(x1 , t) yields that q ( 2 t, z, x2 ) ≤ CPH q (t, x1 , x2 ) for any z ∈ B(x1 , 2 t)
and CPH as defined above. In particular,
X
√ q ω 12 t, z, x2 ≤ CPH B(x1 , 21 t) q ω (t, x1 , x2 ).
√
z∈B(x1 , t/2)
Next, consider the function
(
1,
u(τ, x) :=
P
if τ ∈ [0, t/2],
q ω (τ − t/2, x, z), if τ ∈ [t/2, t].
√
Notice that the function u is caloric on [0, t] × B(x1 , t) and by the parabolic Harnack inequality we have that
X
√ 2 1 = u 12 t, x2 ≤ CPH
q ω 12 t, z, x2 ≤ CPH
B(x1 , 21 t) q ω (t, x1 , x2 ),
√
z∈B(x1 , t/2)
√
z∈B(x1 , t/2)
where we used the definition of u and the symmetry of the transition density in the
second
step.
the volume regularity assumption because
√ Now, the claim
follows from
B(x1 , 1 t) ≤ Creg max 1, 2−d td/2 .
2
Proposition 4.6. Suppose that Assumption 1.6 holds. √
Let x0 ∈ V and let s(x0 ) and
∗ be as in Assumption 1.6. Further, let R ≥ s(x ) and T ≥ R. Set T := T + 1 and
CH
0
0
R02 := T0 and suppose that u > 0 is a caloric function on [0, T0 ] × B(x0 , R0 ). Then, for
any x1 , x2 ∈ B(x0 , R) we have
R θ
u(T, x1 ) − u(T, x2 ) ≤ c
u,
max
T 1/2 [3T0 /4,T0 ]×B(x0 ,R0 /2)
∗ /(2C ∗ − 1)))/ ln 2 and c only depending on C ∗ .
where θ = ln(2CPH
PH
PH
Proof. We will follow the arguments in [14, Lemma 6] (cf. also [7, Proposition 3.2]),
but some extra care is needed due to the special form of the Harnack constant. Set
Rk = 2−k R0 and let
Qk = [T0 − Rk2 , T0 ] × B(x0 , Rk ).
+
and let Q−
k and Qk be accordingly defined as
+
1 2
1
1 2
1
3 2
Q−
k = T0 − 4 Rk , T0 − 2 Rk × B(x0 , 2 Rk ), Qk = T0 − 4 Rk , T0 × B(x0 , 2 Rk ).
In particular, note that Qk+1 ⊂ Qk and Q+
k = Qk+1 . Setting
vk =
u − minQk u
,
maxQk u − minQk u
28
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
we have that vk is caloric, 0 ≤ vk ≤ 1 and osc(vk , Qk ) = 1. (Here osc(u, A) =
maxA u − minA u denotes the oscillation of u on A.) Replacing vk by 1 − vk if
necessary we may assume that maxQ− vk ≥ 21 .
k
Now for every k such that Rk ≥ s(x0 ) we apply the PHI in Theorem 1.4 and using
∗ we get
the definition of CPH
1
2
∗
≤ max vk ≤ CPH
min vk .
Q−
k
Q+
k
Since osc(u, Qk+1 ) = osc(u, Q+
k ), it follows that for such k,
osc(u, Qk+1 ) =
=
maxQ+ u − minQ+ u
k
k
osc(u, Qk )
osc(u, Qk )
maxQ+ u − maxQk u
k
− min vk osc(u, Qk ).
1+
osc(u, Qk )
Q+
k
Hence,
osc u, Qk+1
≤ (1 − δ) osc(u, Qk )
∗ )−1 . Now we choose k such that R
with δ = (2CPH
0
k0 ≥ R > Rk0 +1 . Then, we can
iterate the above inequality on the chain of boxes Q1 ⊃ Q2 ⊃ · · · ⊃ Qk0 to obtain
osc u, Qk0 ≤ (1 − δ)k0 −1 osc(u, Q1 ).
Note that B(x0 , R) ⊆ B(x0 , Rk0 ), T ∈ [T0 − Rk20 , T0 ] and Q1 = Q+
0 . Finally, since
(1 − δ)k0 ≤ c(R/T 1/2 )θ , the claim follows.
5. L OCAL LIMIT THEOREM
In this section, we prove the quenched local limit theorem for the random conductance model as formulated in Theorem 1.11. For this reason, we consider the ddimensional Euclidean lattice (Zd , Ed ) as the underlying graph and assume throughout this section that the random weights ω satisfy the Assumptions 1.8 and 1.9. Let
us recall that kt = ktΣY , defined in (1.11), denotes the Gaussian heat kernel with
diffusion matrix Σ2Y .
The proof of the local limit theorem is based on the approach in [7] and [14]. In
the sequel, we will briefly explain their strategy. For x ∈ Rd and δ > 0, let
Z
h
i
(n)
ω
J(t, n) ≡ J(t, x, δ, n) := P0 Yt ∈ C(x, δ) −
kt (y) dy.
(5.1)
C(x,δ)
Here, we denote by C(x, δ) = x + [−δ, δ]d the cube with center x and side length
2δ, i.e. C(x, δ) is a ball in Rd with respect to the supremums norm | · |∞ . Further,
for any n ≥ 1, we set C n (x, δ) := nC(x, δ) ∩ Zd .
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
29
Now, let us rewrite (5.1) as J(t, n) = J1 (t, n) + J2 (t, n) + J3 (t, n) + J4 (t, n), where
X J1 (t, n) :=
q ω n2 t, 0, z − q ω n2 t, 0, bnxc µω (z),
z∈C n (x,δ)
J2 (t, n) := µω C n (x, δ) q ω n2 t, 0, bnxc − n−d a kt (x) ,
J3 (t, n) := kt (x) µω C n (x, δ) n−d a − (2δ)d ,
Z
J4 (t, n) :=
kt (x) − kt (y) dy,
C(x,δ)
µω (0)
and a = 1/ E
. Thus, in order to conclude a quenched local limit theorem, it
remains to establish suitable upper bounds on the terms J1 , J3 and J4 uniformly for
t ∈ [T1 , T2 ] ⊂ (0, ∞) that are small compared to µω [C n (x, δ)]n−d . First, notice that
by means of the spatial ergodic theorem, cf. Lemma 5.1 below, µω [C n (x, δ)]n−d
converges P-a.s to E[µω (0)](2δ)d as n tends to infinity. Hence, we obtain that
limn→∞ |J3 (t, n)|/µω [C n (x, δ)]n−d = 0 and limn→∞ |J(t, n)|/µω [C n (x, δ)]n−d = 0
uniformly for all t ∈ [T1 , T2 ]. The latter follows from the quenched invariance principle, see Theorem 1.10. Further, mind that q ω (·, 0, ·) is a caloric function. Thus,
provided that the assumptions of Proposition 4.6 are satisfied, the Hölder continuity
estimate combined with the on-diagonal upper bounds of Proposition 4.5 implies
−d−θ
max nd q ω n2 t, 0, z − q ω n2 t, 0, bnxc ≤ c δ θ T1 2 2 .
z∈C n (x,δ)
This estimate allows to deduce that
lim lim nd
δ→0 n→∞
|J1 (t, n)|
ω
µ [C n (x, δ)]
=0
uniformly for t ∈ [T1 , T2 ]. The verification of the assumptions of Proposition 4.6
will be based on the ergodic theorem as well. Finally, a uniform upper bound of
|J4 (t, n)|/µω [C n (x, δ)]n−d can be deduced from the explicit form of kt (x).
For the proof Theorem 1.11 we will apply the general criterion in [14, Theorem 1], so we will just verify the required assumptions involving conditions on the
underlying sequence of rescaled graphs, the weights and a PHI on sufficiently large
balls.
Let us start with some immediate consequnces of the ergodic theorem.
Lemma 5.1. For any x ∈ Rd and δ > 0 we have that, for P-a.e. ω,
p
q
lim µω p,C n (x,δ) = E µω (0)p
and
lim ν ω q,C n (x,δ) = E ν ω (0)q .
n→∞
n→∞
rω (x, δ, ε)
In particular, for every ε > 0 we have
< ∞ for P-a.e. ω, where
n
p
rω (x, δ, ε) := inf n ≥ 1 : µω p,C n (x,δ) − E µω (0)p < ε
o
q
and ν ω q,C n (x,δ) − E ν ω (0)q < ε .
30
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Proof. The assertions of the lemma are an immediate consequence of the spatial
ergodic theorem [19, Theorem 2.8 in Chapter 6], i.e.
X
p
p
ω p
1
τx ω
lim µω p,C n (x,δ) = lim n
µ
(0)
=
E
µ (0) ,
n→∞ C (x, δ)
n→∞
x∈C n (x,δ)
n
provided we have shown
n that the sequence
C (x, δ) : n ∈ N is regular. For this
purpose, consider C 0, |x|∞ + δ : n ∈ N . Obviously, this sequence
of cubes
is increasing and have the property that C n (x, δ) ⊂ C n 0, |x|∞ + δ for all n ∈ N.
Moreover, there exists K ≡ K(x, δ) < ∞ such that
n
C 0, |x|∞ + δ ≤ K C n (x, δ),
∀ n ∈ N.
n
Thus, the sequence C (x, δ) : n ∈ N is regular.
Remark 5.2. Mind that B(bxnc, δn) : n ∈ N is also a regular sequence of boxes
for every given x ∈ Rd and δ > 0. Thus, the assertion of Lemma 5.1 also holds if we
replace C n (x, δ) by B(bnxc, δn).
Proof of Theorem 1.11. We just need to verify Assumption 1 and 4 in [14]. Then,
the assertion follows immediately from [14, Theorem 1].
Consider the sequence of rescaled Euclidean lattices ( n1 Zd , n1 Ed ) : n ∈ N on
the underlying metric space (Rd , | · − · |∞ ) and set α(n) = n, β(n) = nd / E[µω (0)]
and γ(n) = n2 . Recall that we denote by d(x, y) the graph distance for x, y ∈ n1 Zd .
Since,
α(n) |x − y|∞ ≤ d(x, y) ≤ d α(n) |x − y|∞ ,
∀ x, y ∈
1
n
Zd ,
and for all δ > 0
lim 1
n→∞ n
Zd ∩ C(0, δ) = C(0, δ)
with respect to the usual Hausdorff topology on non-empty compact subsets of
(Rd , | · − · |∞ ), the Assumption 1(a), (b) and Assumption 4(b) are satisfied. The
Assumption 1(c) requires that for every x ∈ Rd and δ > 0
Z
n
d
−1 ω
lim β(n) µ C (x, δ) = 2δ =
dy.
P -a.s.
n→∞
C(x,δ)
But, in view of Lemma 5.1 and the definition of β(n), this assumption is satisfied.
Further, the Assumption 1(d) requires that for any compact interval I ⊂ (0, ∞),
x ∈ Rd and δ > 0
Z
h
i
lim Pω0 n1 Yγ(n)t ∈ C(x, δ) =
kt (y) dy,
P -a.s.
n→∞
C(x,δ)
uniformly for all t ∈ I. As we have discussed above, this is a direct consequence of
Theorem 1.10.
It remains to verify the Assumption 4(c): For every x0 ∈ n1 Zd , n ≥ 1, there
∗
exists sn (x) < ∞ such that the parabolic Harnack inequality with constant CPH
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
31
holds on Q(r) ≡ [0, r2 ] × B(x0 , r) for all r ≥ sn (x0 ) and for every x ∈ Rd we have
α(n)−1 sn (bnxc) → 0 as n → ∞.
By the ergodic theorem we know that kµω kp,B(x0 ,n) and kν ω kq,B(x0 ,n) converge as
n tends to infinity for P-a.e. ω. In particular, by Lemma 5.1 and Remark 5.2, we
have that for all x ∈ Rd and ε > 0 that P-a.s. there exists rω (x, 1, ε) < ∞
ω
ω
1
1
µ ν ≤ ε + E µω (0)p p ,
≤ ε + E ν ω (0)q q
p,B(bnxc,n)
q,B(bnxc,n)
for all n ≥ rω (x, 1, ε). Since the Harnack constant appearing in Theorem 1.4 depends monotonically on kµω kp,B(bnxc,n) and kν ω kq,B(bnxc,n) , this implies that for
any r ≥ rω (x, 1, ε) the parabolic Harnack inequality
holds on Q(r) with constant
∗
ω
p
1/p
ω
q
1/q
CPH := CPH (ε + E[µ (0) ]) , (ε + E[ν (0) ])
. Thus, the Assumptiom 4(c) of
[14] is satisfied as well.
6. I MPROVING THE LOWER MOMENT CONDITION
It is naive to expect that for general ergodic distributions, one could formulate a
necessary condition for a local limit theorem based on the marginal distribution of
the conductances only. In fact, for any edge {x, y} ∈ Ed , the lower bound condition
E ω(x, y)−q < ∞
(6.1)
some q > 0 can be improved in several situations to a more general condition based
on the heat kernel of the VSRW
h
i
0
E pω (t, x, y)−q < ∞
(6.2)
or on the heat kernel of the CSRW
h
−q0 i
E µω (x) q ω (t, x, y) µω (y)
< ∞,
(6.3)
respectively, for some q 0 > 0.
Proposition 6.1. Suppose
thatd ≥ 2 and that Assumptions 1.8 and 1.9 hold with the
moment condition E ω(x, y)−q < ∞ replaced by
h
−q i
E pω (t, x, y)−q < ∞
or
E µω (x) q ω (t, x, y) µω (x)
< ∞
for any t > 0 and all x, y ∈ Zd such that {x, y} ∈ Ed . Then, the statements of
Theorem 1.10 and Theorem 1.11 hold.
Proof. We only consider the moment condition on the heat kernel q ω (t, x, y) of the
−1
P
CSRW. For x ∈ Zd and t > 0, set ν̃tω (x) := y∼x µω (x) q ω (t, x, y) µω (y) . Our
main objective is to establish a Sobolev inequality similar to the one in (2.11).
Indeed, one can show along the lines of the proof of this Sobolev inequality in [3,
Proposition 3.5] that
X
2
2 1
ω
(ηu)2 d ν̃ µω (x) q ω (t, x, y) µω (y) (ηu)(x) − (ηu)(y)
≤
c
|B|
t q,B
ρ,B
|B|
{x,y}∈E
32
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
(cf. equation (3.13) in [3]). Since the sum on the right hand side is smaller than
2
1 X ω
µ (x) q ω (t, x, y) µω (y) (ηu)(x) − (ηu)(y)
E qt (ηu) :=
2 x,y
and t−1 Eηq2t (u) % Eηω2 (u) as t & 0, cf. [12, p. 250], following the arguments in [3,
Proposition 3.5] we conclude that (2.11) holds with ν ω replaced by ν̃tω , but with a
constant depending now also on t. Finally, since ν̃tω (x) = ν̃tτx ω (0) for every x ∈ Zd ,
an application of the ergodic theorem gives that
q
lim ν̃tω q,B(n) = E ν̃tω (0)q < ∞.
n→∞
The statements of Theorem 1.10 and Theorem 1.11 now follow as before.
Of course, it is not easy to verify (6.3) (or (6.2)), because this would require
apriori lower bounds on the heat kernel. However, note that
∞
X
tn ω n
q ω (t, x, y) µω (y) = Pωx Yt = y = e−t
(p ) (x, y),
(6.4)
n!
n=0
P
where still pω (x, y) = ω(x, y)/µω (x) and (pω )n+1 (x, y) = z (pω )n (x, z) pω (z, y). In
particular, since (pω )n (0, x) = 0 for every x ∈ Zd with |x|∞ = 1 and even n, the
summation is only over odd n.
Our first result deals with an example, where (6.3) and (6.1) are equivalent, i.e.
no improvement can be achieved, see e.g. [18].
Proposition 6.2. Let either ω(x, y) = θω (x) ∧ θω (y) or ω(x, y) = θω (x) θω (y), where
{θω (x) : x ∈ Zd } is a family of stationary random variables, i.e. θω (x) = θτx ω (0) for
all x ∈ Zd . Then, for every {x, y} ∈ Ed and all q ≥ 1,
h
−q i
< ∞.
E ω(x, y)−q < ∞ ⇐⇒ E µω (x) q ω (t, x, y) µω (y)
Proof. It suffices to show that for every n odd and {x, y} ∈ Ed
µω (x)(pω )n (x, y) ≤ (2d)n ω(x, y).
(6.5)
Once the claim (6.5) is proven, the assertion of the Proposition is immediate. In
order to see this, mind that (6.4) implies the following trivial estimate
µω (x) q ω (t, x, y) µω (y) ≥ t e−t µω (x) pω (x, y) = t e−t ω(x, y).
On the other hand, using again (6.4) and (6.5) we obtain
X tk
µω (x) q ω (t, x, y) µω (y) ≤ e−t ω(x, y)
(2d)k ≤ e(2d−1)t ω(x, y).
k!
k odd
In what follows, we will prove the claim (6.5). For this purpose, we distinguish
two cases depending on the definition of the weights ω. To start with, suppose that
ω(x, y) = θω (x) ∧ θω (x) for {x, y} ∈ Ed . Since µω (x) ≤ 2d θω (x) we obtain that
µω (x) (pω )n (x, y) ≤ 2d θω (x).
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
33
On the other hand, by exploiting the detailed balance condition we get
µω (x) (pω )n (x, y) = µω (y) (pω )n (y, x) ≤ 2d θω (y).
Thus, (6.5) follows by combing these two estimates.
Let us now consider the case ω(x, y) = θω (x) θω (y). By introducing the abbreviaP
tion αω (x) = z∼x θ(z), we can rewrite
µω (x) = θω (x) αω (x)
and
pω (x, y) =
θω (y)
.
αω (x)
(6.6)
Thus, for n = 3, we have that
µω (x) (pω )3 (x, y) = θω (x) θω (y)
X X θω (z1 ) θω (z2 )
≤ (2d)2 ω(x, y),
ω (z ) αω (z )
α
2
1
z ∼x z ∼y
1
2
where we used that θω (z1 )/αω (z2 ) ≤ 1 and θω (z2 )/αω (z1 ) ≤ 1 since z1 ∼ z2 . This
implies (6.5) for n = 3. For general odd n the claim (6.5) follows inductively.
Note that in Proposition 6.2 no assumptions
have been made on the distribution
of the random variables θω (x) : x ∈ Zd . In general the computations can be
rather intricate, in particular due to the quotient in the definition of pω (x, y). For
simplicity we will now focus on the case with conductances bounded from above,
that is p = ∞.
Proposition 6.3. Assume that the conductances {ω(e) : e ∈ Ed } are independent and
uniformly bounded from above, i.e. P-a.s. ω(e) ≤ 1. Then, if (6.1) holds
for any q > 0,
0
−q
we have that (6.3) holds for all q < 2dq. In particular, if E ω(x, y)
< ∞ for some
q > 1/4 and all {x, y} ∈ Ed then there exists q 0 > d2 such that
h
−q0 i
E µω (x) q ω (t, x, y) µω (y)
< ∞.
For the proof we will need the following simple lemma.
Lemma 6.4. Let Z1 , Z2 , . . . , ZN be independent random variables such that
P 0 < Zi ≤ 1 = 1
and
ci (α) = E Zi−α < ∞,
for some α > 0. Then
h
−β i
< ∞
E Z1 + Z2 + . . . + ZN
for all β < N α.
Proof. Note that Z1 +Z2 +. . .+ZN ≥ maxi=1,...,N Zi , and Čebyšev’s inequality yields
h
P
max Zi
i=1,...,N
−β
> t
i
= P
h
max Zi < t
i=1,...,N
− β1
i
≤ t
− nα
β
n
Y
ci (α).
i=1
Since P[0 < Zi ≤ 1] = 1, the assertion follows by integrating the estimate above in
t over the interval [1, ∞).
34
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
γ4
γ3
x
y
γ1
γ2
F IGURE 1. Illustration of the 4 different disjoint nearest neighbour
paths {γi : i = 1, . . . 4} from x to y for d = 2.
Proof of Proposition 6.3. Suppose that (6.1) holds for any q > 0. Since the conductances are assumed to be bounded from above ω(x, y) ≤ 1 we have µω (x) ≤ 2d and
therefore (pω )n (x, y) ≥ (2d)−n ω n (x, y), where we define ω 1 (x, y) := ω(x, y) and
P
ω n+1 (x, y) = z ω n (x, z) ω(z, y). Thus, in view of (6.4) we get
µω (x) q ω (t, x, y) µω (y) ≥ c ω(x, y) + ω 3 (x, y) + ω 9 (x, y) ,
for some positive constant c ≡ c(d, t). Hence, it suffices to show that
h
−q0 i
E ω(x, y) + ω 3 (x, y) + ω 9 (x, y)
< ∞.
(6.7)
Note that for {x, y} ∈ Ed there are at least 2d different disjoint nearest neighbour
oriented paths {γi : i = 1, . . . , 2d} from x to say y, meaning that the sets of edges
along the paths (γi ) are disjoint: one direct path of length 1, 2(d − 1) paths of length
3 and one path of length 9 (see Figure 1). That is,
ω(x, y) + ω 3 (x, y) + ω 9 (x, y) ≥ Z1 + Z2 + . . . + Z2d ,
Q
where Zi = e∈γi ω(e) for i = 1, . . . , 2d. Using the independence we see that
E Zi−q < ∞ for i = 1, . . . , 2d. This implies (6.7) by Lemma 6.4.
Proposition 6.5. Let either ω(x, y) = θω (x) ∨ θω (y) or ω(x, y) = θω (x) + θω (y),
where {θω (x) : x ∈ Zd } is a family of uniformly bounded i.d.d. random variables,
that is p = ∞. Then, if (6.1) holds for any q > 0, we have that (6.3) holds for
d
and all
all q 0 < 32 (d + 1)q. In particular, if E ω(x, y)−q < ∞ for some q > 34 d+1
d
0
{x, y} ∈ Ed then there exists q > 2 such that
h
−q0 i
E µω (x) q ω (t, x, y) µω (y)
< ∞.
Lemma 6.6. Let X1 and X2 be i.i.d. random variables such that
P 0 < Xi ≤ 1 = 1,
and
c(α) = E Xi−α < ∞,
i = 1, 2,
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
35
for some α > 0. Then
h
−β i
E X12 X2 ∨ X1 X22
< ∞
for all β < 32 α.
Proof. First of all, note that for any s > 0 we have
i
h
1
1
P X12 X2 ∨ X1 X22 < s = P X1 < s 3 , X2 < s 3
h
i
1
1
+ P X1 ≥ s 3 , X2 < s 3 , X2 < sX1−2
h
i
1
1
−2
3
3
+ P X1 < s , X2 ≥ s , X1 < sX2
h
i h
i
h
i
1
1
1
= P X1 < s 3 P X2 < s 3 + 2 P X1 ≥ s 3 , X2 < sX1−2 ,
bα
where the second equality follows by a symmetry argument. Let us denote by E
−α
−1
b
the expectation w.r.t. dPα := c1 (α) X1 d P. Then, we obtain
i
i
h
h
1
−α
−2α
b
≤ c(α) s 3 α .
= c(α) Eα 1l{X1 ≥s1/3 } X1
E 1l{X1 ≥s1/3 } X1
Thus, Čebyšev’s inequality yields the following upper bound
i
h
h
i
2
1
P X1 ≥ s 3 , X2 < sX1−2 = E 1l{X1 ≥s1/3 } P X2−α > s−α X12α X1 ≤ c(α)2 s 3 α .
Hence, for any t > 0, again by Čebyšev’s inequality
i h
i
h
i
h
−β
− 1
− 1
P X12 X2 ∨ X1 X22
> t = P X1 < t 3β P X2 < t 3β
i
h
− 2α
− 1
−1
+ 2 P X1 ≥ t 3β , X2 < t β X1−2 ≤ 3 c(α)2 t 3β .
Proof of Proposition 6.5. Since
1 ω
θ (x) + θω (y) ≤ θω (x) ∨ θω (y) ≤ θω (x) + θω (y)
2
it is enough to consider the case ω(x, y) = θω (x) ∨ θω (y). Again it suffices to show
2
(6.7). First note that we have P θω (x) < s = P ω(x, y) < s for any edge {x, y} ∈
Ed and therefore
h
i
h
i1
ω −r
1
q 2
q
ω
− r1
−q
r
P θ (x) > t = P θ (x) < t
= P ω(x, y) > t
≤ t− 2r E ω(x, y)−q 2 .
In particular, E θω (x)−r < ∞ for all r < q/2. Let x = x0 , x1 , . . . , x2k , x2k+1 = y be
a disjoint nearest neighbour path from x to y. Then,
2k
Y
ω(xj , xj+1 ) =
j=0
2k
Y
θω (xj ) ∨ θω (xj+1 )
j=0
≥
k
Y
j=1
2k
Y
θω (xj ) · θω (xk ) ∨ θω (xk+1 ) ·
θω (xj ).
j=k+1
(6.8)
36
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
In particular, note that the right hand side does not depend on θω (x) and θω (y).
Moreover, by Lemma 6.6
Y
−β 2k
q
< ∞,
∀β < .
E
ω(xj , xj+1 )
j=0
3
In order to prove (6.7) we consider again 2d disjoint nearest neighbour paths from x
to y and define Zi , i = 1, . . . , 2d as above in the proof of Proposition 6.3. Then, using
(6.8) we have Zi ≥ Z̃i for i = 2, . . . , 2d, where {Z̃i : i = 2, . . . , d} is a collection of
independent random variables with E Z̃i−β < ∞ for all β < 3q . Furthermore, {Z̃i }
is independent of Z1 = ω(x, y). Thus,
ω(x, y) + ω 3 (x, y) + ω 9 (x, y) ≥ Z1 + Z̃2 + . . . + Z̃2d
and by Lemma 6.4 we conclude that (6.7) holds for all q 0 < q + (2d − 1)β, which
implies the claim.
A PPENDIX A. T ECHNICAL ESTIMATES
In this section we collect some technical estimates needed in the proofs. In a
sense some of them may be seen as a replacement for a discrete chain rule.
Lemma A.1. For a ∈ R we write ãα := |a|α · sign a for any α ∈ R \ {0}.
(i) For all a, b ∈ R and any α, β 6= 0
α
ã − b̃α ≤ 1 ∨ α ãβ − b̃β |a|α−β + |b|α−β .
β
(ii) For all a, b ≥ 0 and any α > 1/2
α2 α
α 2
a − b a2α−1 − b2α−1 .
≤ a −b
2α − 1
(A.1)
(A.2)
Moreover, if a, b > 0 then
2
ln a − ln b ≤ − a−1 − b−1 a − b .
(A.3)
(iii) For all a, b ≥ 0 and any α, β ≥ 0
aα + bα aβ − bβ ≤ 2 aα+β − bα+β .
(A.4)
(iv) For all a, b ≥ 0 and any α ≥ 1/2
a2α−1 + b2α−1 a − b ≤ 4 aα − bα aα + bα .
(A.5)
(v) For all a, b ∈ R and any α ≥ 1,
a − bα ≤ 2α−1 ãα − b̃α .
(A.6)
Proof. (i) First of all assume that a, b ≥ 0 or a, b ≤ 0. Then, for all α, β 6= 0,
Z |a|
α
α
α−β β
β
β−1 α−β
ã − b̃α = |a|α − |b|α = α
t
t
dt
≤
max
t
ã
−
b̃
.
β t∈[|b|,|a|]
|b|
Since t 7−→ tα−β for t ≥ 0 is monotone decreasing if α − β < 0 and monotone
increasing if α − β > 0, the maximum is attained at one of the boundary points. In
HARNACK INEQUALITIES ON WEIGHTED GRAPHS
37
particular, maxt∈[|b|,|a|] tα−β ≤ |a|α−β + |b|α−β . On the other hand, for a ≥ 0, b ≤ 0
or a ≤ 0, b ≥ 0, it holds that ãα − b̃α = |a|α + |b|α . Hence,
α
ã − b̃α ≤ |a|β + |b|β |a|α−β + |b|α−β = ãβ − b̃β |a|α−β + |b|α−β .
(ii) Notice that for all a, b ≥ and for any α > 0,
Z a
2
α
α 2
α−1
a −b
= α
t
dt .
b
Hence, for any α > 1/2, an application of the Cauchy-Schwarz inequality yields
Z a
2
Z a Z a
α2
α−1
2
α
t
dt ≤ α
dt
t2α−2 dt =
a − b a2α−1 − b2α−1 .
2α − 1
b
b
b
Similarly, for a, b > 0 we have
Z a
Z
2
−1
(ln a − ln b) =
t dt ≤
b
b
a
Z
dt
a
t−2 dt = −(a − b)(a−1 − b−1 ).
b
(iii) Since the assertion is trivial, if a = 0, we assumethat
a 6=
0 and set z =
α
β
b/a ≥ 0. Then, the left-hand side of (A.4) reads 1 + z
1 − z . Provided that
α ≥ 0, by distinguishing two cases, z ∈ [0, 1) or z ≥ 1, we obtain
1 + z α 1 − z β = 21 − z α+β − 1 − z α 1 + z β ≤ 21 − z α+β .
This completes the proof.
(iv) The assertion follows immediately from (A.1) and (A.4).
(v) Since the cases α = 1 is obvious, suppose that α > 1. To start with, let us
asuume that either a ≥ 0, b ≤ 0 or a ≤ 0, b ≥ 0. Then, by Jensens’s inequality
a − bα = |a| + |b|α ≤ 2α−1 |a|α + |b|α = 2α−1 ãα − b̃α .
It remains to consider the case when a, b ≥ 0 or a, b ≤ 0. Since the assertion
is trivial, if a = 0, we assume that a 6= 0. Further, we set z = b/a ≥ 0 and
f (z) = |1 − z|α /|1 − z α |. By distinguish the two cases, z ∈ [0, 1] and z ∈ (1, ∞),
an elementary computation shows that f (z) in monotone decreasing in [0, 1) and
monotone increasing in (1, ∞). Moreover, f (0) = 1 and limz→∞ f (z) = 1, i.e. f is
bounded by 1 on [0, ∞).
Lemma A.2. For all x, y > 0 and b ≥ a ≥ 0
 2

a2 1 1
b2
2

2
−
y−x + 9 2 b−a ,
b
a
2 y x
a
−
y−x ≤

y
x
b2 ,
a > 0,
(A.7)
a = 0.
Proof. First of all, the case a = 0 is trivial. Further, notice
that (A.7) holds true if
x = y or a = b where we use in the latter that y −1 −x−1 (y −x) ≤ 0. Therefore, we
assume in the sequel that x 6= y and 0 < a < b. Then, by using Young’s inequality
2ab ≤ ε a2 + b2 /ε with ε = y/x, we obtain the following apriori estimate
2
2
b
a2
y 2
x
−
y − x = b2 −
a + b2
+ a2 ≤ b − a .
(A.8)
y
x
x
y
38
SEBASTIAN ANDRES, JEAN-DOMINIQUE DEUSCHEL, AND MARTIN SLOWIK
Since
b2 a2
−
y
x
y−x
2
= a
1 1
−
y x
1 2
y−x +
b − a2 y − x ,
y
the assertion (A.7) is immediate without the need of the second
positive term pro
vided that b2 − a2 (y − x) y −1 ≤ −a2 y −1 − x−1 y − x /2. Thus, we are finally
left with the case
a2 −1
b2 − a2 y − x y −1 ≥ −
y − x−1 y − x .
(A.9)
2
Notice that this implies that x < y since a < b and the right-hand side of the
inequality above is positive. Now, an elementary computation, using the fact that
y −1 − x−1 = −x−1 (y − x)y −1 , reveals that (A.9) is equivalent to
b
0 < x−1 y − x ≤ 2 a−2 b2 − a2 ≤ 4 2 b − a .
a
Thus, by exploiting the fact that x < y, this estimate implies
2
2
a2 −2
b2
1
y−x ≤ 8 2 b−a
x
− a2 y −1 − x−1 y − x ≤
2
2
a
and the assertion (A.7) follows from (A.8) and (A.10).
(A.10)
Acknowledgement. We thank Martin Barlow, Takashi Kumagai and Pierre Mathieu
for useful discussions and valuable comments. M.S. gratefully acknowledge financial support from the DFG Forschergruppe 718 ”Analysis and Stochastics in Complex
Physical Systems”.
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R HEINISCHE F RIEDRICH -W ILHELMS U NIVERSIT ÄT B ONN
Current address: Endenicher Allee 60, 53115 Bonn
E-mail address: [email protected]
T ECHNISCHE U NIVERSIT ÄT B ERLIN
Current address: Strasse des 17. Juni 136, 10623 Berlin
E-mail address: [email protected]
T ECHNISCHE U NIVERSIT ÄT B ERLIN
Current address: Strasse des 17. Juni 136, 10623 Berlin
E-mail address: [email protected]