and, consequently, K
k,u 1
= e
2C αT
2 α
2
κ
2
Z
M
d r X
a ∈Z
3
Ξ
r
a Ξ
r
a − Ξ
M
a ≤ e
2C αT
2 α
2
κ
2
Z
M
d r X
a ∈Z
3
Ξ
r
a
2
. 5.69
Hence, taking into account 5.50, we arrive at 5.58.
5.2.4 Proof of Lemma 5.8
Proof. Using the same arguments as in 5.62–5.63, we can replace E
new ν
ρ
by E
ν
ρ
in formula 5.54, to obtain
E
6 k,u
t ≤ E
ν
ρ
exp 2
κ e
C αT
Z
t
d r Q
M
b φ
ξ
r κ
. 5.70
Because of 5.49, this yields exp
2 α
κ e
C αT
ρ t E
6 k,u
t ≤ E
ν
ρ
exp 2
α κ
e
C αT
Z
t
d r X
y ∈Z
3
Ξ
M
yξ
r κ
y .
5.71 Now, using the independent random walk approximation e
ξ of ξ see [3], Proposition 1.2.1, we find that
E
ν
ρ
exp 2
α κ
e
C αT
Z
t
d r X
y ∈Z
3
Ξ
M
yξ
r κ
y ≤
Z ν
ρ
dη Y
x ∈A
η
E
Y x
exp 2
α κ
e
C αT
Z
t
d r Ξ
M
Y
r κ
, 5.72
where A
η
is given by 5.12 and Y is simple random walk with step rate 1. Define vx, t = E
Y x
exp 2
α κ
e
C αT
Z
t
d r Ξ
M
Y
r κ
, x, t ∈ Z
3
× [0, ∞, 5.73
and write wx, t = vx, t
− 1. 5.74
Then we may bound 5.71 from above as follows: r.h.s. 5.71
≤ Z
ν
ρ
dη Y
x ∈Z
3
1 + ηxwx, t
= Y
x ∈Z
3
1 + ρwx, t
≤ exp ρ
X
x ∈Z
3
wx, t .
5.75
2118
By the Feynman-Kac formula, w is the solution of the Cauchy problem ∂
∂ t wx, t =
1 6
κ ∆wx, t +
2 α
κ e
C αT
Ξ
M
x 1 + wx, t
, w
·, 0 ≡ 0. 5.76
Therefore ∂
∂ r X
x ∈Z
3
wx, r = 2
α κ
e
C αT
X
x ∈Z
3
Ξ
M
x 1 + wx, r
. 5.77
Integrating 5.77 w.r.t. r over the time interval [0, t] and substituting the resulting expression into 5.75, we get
r.h.s. 5.71 ≤ exp
2
α κ
e
C αT
ρ Z
t
d r X
x ∈Z
3
Ξ
M
x 1 + wx, r
. 5.78
Since P
x ∈Z
3
Ξ
M
x = 1, this leads to E
6 k,u
t ≤ exp
2 α
κ e
C αT
ρ Z
t
d r X
x ∈Z
3
Ξ
M
xwx, r
. 5.79
An application of Lemma 2.6 to the expectation in the r.h.s. of 5.73 gives vx, t
≤ 1
− 2αe
C αT
GΞ
M ∞
−1
. 5.80
Next, using 5.47 and 5.50, we find that GΞ
M ∞
≤ G
6T +M κ
0 ≤ G
3K κ
2
0, 5.81
where the r.h.s. tends to zero as κ → ∞. Thus, if K 1 and κ κ
with κ
large enough not depending on the other parameters, then vx, t
≤ 2, and hence wx, t ≤ 1, for all x ∈ Z
3
and t
≥ 0, so that 5.76 implies that w ≤ b w, where
b w solves
∂ ∂ t
b wx, t =
1 6
κ ∆
b wx, t +
4 α
κ e
C αT
Ξ
M
x, b
w ·, 0 ≡ 0.
5.82 The solution of this Cauchy problem has the representation
b wx, t =
4 α
κ e
C αT
Z
t
d r X
y ∈Z
3
p
r κ
x, yΞ
M
y = 4
α κ
e
C αT
Z
t
d r Ξ
M +r
x. 5.83
Hence X
x ∈Z
3
Ξ
M
xwx, r ≤ 4
α κ
e
C αT
Z
r
d er
X
x ∈Z
3
Ξ
M
xΞ
M + er
x ≤
4 α
κ e
C αT
1 R
2 κ
Z
k+1R
κ
kR
κ
ds Z
k+1R
κ
kR
κ
d es
Z
r
d er p
12T 1[ κ]+
s+ es−2u+2M+er
κ
≤ 4
α κ
e
C αT
Z
∞
d er p
2M + er
κ
≤ 4C
α p
K κ
e
C αT
, 5.84
where we again use the second inequality of Lemma 2.4. Substituting 5.84 into 5.79, we arrive at the claim with D
α,T,K
= 8α
2
C e
2C αT
p K.
2119
5.2.5 Further reduction of Lemma 5.4
