where τ
r
is the exit time for the process X from the box B
r
. Let
{λ
ω i
r, i ∈ [1, B
r
]} be the set of eigenvalues of −G
ω r
labelled in increasing order, and {ψ
ω,r i
, i ∈ [1, B
r
]} the corresponding eigenfunctions with due normalization in L
2
B
r
, π
ω
.
3.2 Proof of the upper bound.
In this last part, we will complete the proof of Theorem 1.1 by giving the proof of the upper bound 1.13.
Proof of Theorem 1.1. Assume that the origin belongs to
C
ξ
. By lemma 3.1, we have P
ω
X t = 0 ≤ 2
t Z
t t
2
P
ω
X s = 0ds = 2
t E
ω
Z
t t
2
1
{X s=0}
ds
.
The additive functional A
ξ
being a continuous increasing function of the time, so by operating a variable change by setting s = A
ξ −1
u i.e. u = A
ξ
s, we get E
ω
Z
t t
2
1
{X s=0}
ds
=
E
ω
Z
t t
2
1
{X s=0}
ϕX sds
=
E
ω
Z
A
ξ
t A
ξ
t2
1
{X
ξ
u=0}
du
, which is bounded by
E
ω
Z
t A
ξ
t2
1
{X
ξ
u=0}
du
,
since A
ξ
t ≤ t. Therefore, for
ε ∈ 0, 1 P
ω
X t = 0 ≤ 2
t E
ω
Z
t A
ξ
t2
1
{A
ξ
t2≥t
ε
}
1
{X
ξ
u=0}
du
+
2 t
E
ω
Z
t A
ξ
t2
1
{A
ξ
t2≤t
ε
}
1
{X
ξ
u=0}
du
≤
2 t
Z
t t
ε
P
ω
X
ξ
u = 0du + 2
t Z
t
P
ω
A
ξ
t2 ≤ t
ε
du 2079
and using lemma 2.3, P
ω
X t = 0 ≤ c
t Z
t t
ε
u
−d2
du + 2
t P
ω
A
ξ
t2 ≤ t
ε
t ≤
c t
ε
d 2
−ε+1
+ 2P
ω
A
ξ
t2 ≤ t
ε
3.7 It remains to estimate the second term in the right-hand side of the last inequality, i.e.
P
ω
A
ξ
t2 ≤ t
ε
or more simply P
ω
A
ξ
t ≤ 2
ǫ
t
ε
, but we can neglect the constant 2
ε
in the calculus as one will see in 3.15. For each
λ ≥ 0, Chebychev inequality gives
P
ω
A
ξ
t ≤ t
ε
= P
ω
A
ξ
t ≤ t
ε
; t τ
r
+ P
ω
A
ξ
t ≤ t
ε
; τ
r
≤ t ≤ P
ω
e
−λA
ξ
t
≥ e
−λt
ε
; t τ
r
+ P
ω
τ
r
≤ t ≤ e
λt
ε
E
ω
h e
−λA
ξ
t
; t τ
r
i + P
ω
τ
r
≤ t. 3.8
From the Carne-Varopoulos inequality, it follows that P
ω
τ
r
≤ t ≤ C t r
d −1
e
−
r2 4t
+ e
−ct
, 3.9
where C and c are numerical constants, see Appendix C in [15]. With our choice of r such that t
∼ r
2
log r
−b
b 1, we get that P
ω
τ
r
≤ t decays faster than any polynomial as t tends to +
∞. Thus Theorem 1.1 will be proved if we can check, for a particular choice of
λ 0 that may depend on t, that
lim sup
t →+∞
log e
λt
ε
E
ω
h e
−λA
ξ
t
; t τ
r
i log t
≤ − d
2 .
3.10 That will be true if e
λt
ε
E
ω
h e
−λA
ξ
t
; t τ
r
i decays faster than any polynomial in t as t tends
to + ∞.
The Dirichlet form of −L
ω r
on L
2
B
r
, π
ω
endowed with the usual scalar product see Lemma 3.1, can be written as
E
ω,r
f , f = ¬
−L
ω r
f , f ¶
ω
= 1
2 X
b ∈B
r+1
d f b
2
ω
b
,
2080
where d f b = f y − f x and the sum ranges over b = x, y ∈ B
ω r+1
. By the min-max Theorem see [12] and 3.4, we have
λ
ω 1
r = inf
f 6≡0
E
ω,r
f , f + λ P
x ∈C
ξ r
f
2
xπ
ω
x π
ω
f
2
. 3.11
where C
ξ r
is the largest connected component of C
ξ
∩ B
r
, and the infimum is taken over functions with Dirichlet boundary conditions. Recall that
λ
ω 1
r is the first eigenvalue of −G
ω r
. To estimate the decay of the first term in the right-hand side of 3.8, we will also need to
estimate the first eigenvalue λ
ω 1
r. Recall that µ denotes an arbitrary positive constant.
Lemma 3.4 Under assumption 1.7, for any d
≥ 2 and γ 0, we have Q − a.s. for r large enough,
λ
ω 1
r ≥ 8d
−1
r
−
d γ
+µ
log n
−5
, 3.12
for λ proportional to r
−
d γ
+µ
. Proof. For some arbitrary
µ 0, let r be large enough so that 3.2 holds. Let h be a hole that intersects the box B
r
, and for notational ease we will use the same notation for h ∩ B
r
. Define
∂ h to be the outer boundary of h, i.e. the set of sites in C
ξ r
which are adjacent to some vertex in h. Let us associate to each hole h a fixed site h
∗
∈ C
ξ r
situated at the outer boundary of h and for x
∈ h call κx, h
∗
a self-avoiding path included in h with end points x and h
∗
, and let |κx, h
∗
| denote the length of such a path. Now let f
∈ L
2
B
r
, π
ω
and let B
ω
h denote the set of the bonds of h. For each x ∈ h, write f x =
X
b ∈κx,h
∗
d f b + f h
∗
and, using Cauchy-Schwarz f
2
x ≤ 2|κx, h
∗
| X
b ∈κx,h
∗
|d f b|
2
+ 2 f
2
h
∗
. In every path
κx, h
∗
, we see each bond only one time. Multiply the last inequality by π
ω
x
2081
and sum over x ∈ h to obtain
X
x ∈h
f
2
xπ
ω
x ≤ 2
X
x ∈h
|κx, h
∗
| X
b ∈κx,h
∗
|d f b|
2
π
ω
x + 2 X
x ∈h
f
2
h
∗
π
ω
x ≤ 4d max
x ∈h
|κx, h
∗
| max
b ∈B
ω
h
1 ω
b
h X
b ∈B
ω
h
|d f b|
2
ω
b
+ 2 X
x ∈h
f
2
h
∗
π
ω
x, 3.13
which, by virtue of lemma 2.1, 1.7, 3.2 and since π
ω
h
∗
≥ ξ, is bounded by 4d r
d γ
+µ
log r
5
X
b ∈B
ω
h
|d f b|
2
ω
b
+ 4d
ξ h f
2
h
∗
π
ω
h
∗
, Thus,
X
x ∈h
f
2
xπ
ω
x ≤ 4d r
d γ
+µ
log r
5
X
b ∈B
ω
h
|d f b|
2
ω
b
+ 4d
ξ h f
2
h
∗
π
ω
h
∗
. Let
C
c r
ξ denote the complement of C
ξ r
in the box B
r
and sum over h to obtain X
x ∈C
c r
ξ
f
2
xπ
ω
x ≤ 8d r
d γ
+µ
log r
5
E
ω,r
f , f + 8d
2
ξ h
X
x ∈C
ξ r
f
2
xπ
ω
x, where in the last term, we multiply by 2d since we may associate the same h
∗
to 2d different holes. Then
X
x ∈B
r
f
2
xπ
ω
x ≤ 1 + 8d
2
log r
5 2
ξ
−1
X
x ∈C
ξ r
f
2
xπ
ω
x + 8d r
d γ
+µ
log r
5
E
ω,r
f , f . ≤ 8d
2
log r
5
ξ
−1
X
x ∈C
ξ r
f
2
xπ
ω
x + 8d r
d γ
+µ
log r
5
E
ω,r
f , f .
So, according to 3.11 and for λ = dξ
−1
r
−
d γ
+µ
, we get λ
ω 1
r ≥ 8d
−1
r
−
d γ
+µ
log r
−5
. 3.14
2082
Let us get back to the proof of the upper bound. Let λ = dξ
−1
r
−
d γ
+µ
; mr := 8d
−1
r
−
d γ
+µ
log r
−5
. For f
≡ 1, observe that
R
ω,r t
f 0 = E
ω
h e
−λA
ξ
t
; t τ
r
i =
X
i
e
−λ
ω i
rt
¬ 1,
ψ
ω,r i
¶ ψ
ω,r i
0, and
R
ω,r t
f
2
0π
ω
0 ≤ X
x ∈B
r
R
ω,r t
f
2
xπ
ω
x =
X
i
e
−2λ
ω i
rt
¬ 1,
ψ
ω,r i
¶
2
≤ e
−2λ
ω 1
rt
X
x
1
2
xπ
ω
x ≤ 2
d+2
d r
d
e
−2λ
ω 1
rt
. Then, for large enough t and by 3.14, we have
e
λt
ε
E
ω
h e
−λA
ξ
t
; t τ
r
i ≤ 2
d+2
d
1 2
e
λt
ε
e
−tλ
ω 1
r
r
d 2
≤ 2
d+2
d
1 2
r
d 2
exp {λt
ε
− tmr} ≤ 2
d+2
d
1 2
r
d 2
e
−
t 2
mr
, 3.15
since ε 1.
By our choice of t ∼ r
2
log r
−b
b 1, we deduce
e
λt
ε
E
ω
h e
−λA
ξ
t
; t τ
r
i ≤ 2
d+2
d
1 2
r
d 2
exp §
−[16dlog r
b+5
]
−1
r
2
r
−
d γ
+µ
ª ≪ t
−
d 2
if γ
d 2
− µ ,
3.16 which yields 3.10.
In conclusion, as µ is arbitrary and according to 3.9–3.10, we obtain
lim sup
t →+∞
log P
ω
A
ξ
t ≤ t
ε
log t ≤ −
d 2
for γ
d 2
, 2083
and finally, by 3.7 lim
ε→1
lim sup
t →+∞
log P
ω
X t = 0 log t
≤ − d
2 for
γ d
2 ,
which gives 1.13. We conclude that for any sufficiently small
ξ, then Q
ξ
−a.s. 1.11 is true and since Q
∪
ξ0
{0 ∈ C
ξ
ω}
= 1, it remains true Q-a.s.
3.3 Proof of the discrete-time case.
