Let e F = e
ψκ
F . Then, by 3.1 and 3.3, 3.6
≤ ≥
ZZ
Ω×Z
3
d µ
new ρ
η, x
V η, x F
2
η, x −
e
∓
K κ
12 κ
X
{a,b}
h F
η
a,b
, x − Fη, x
i
2
− e
∓
K κ
2 X
y : k y−xk=1
F η, y − Fη, x
2
= ZZ
Ω×Z
3
d µ
ρ
η, x
V η, x e
F
2
η, x −
e
∓
K κ
12 κ
X
{a,b}
h e
F η
a,b
, x − e
F η, x
i
2
− e
∓
K κ
2 X
y : k y−xk=1
h e
F η, y − e
F η, x
i
2
= e
∓
K κ
ZZ
Ω×Z
3
d µ
ρ
e
±
K κ
V e F
2
+ e
F A e
F .
3.7
Taking further into account that eF
2 L
2
µ
ρ
= kFk
2 L
2
µ
new ρ
, 3.8
and that e F
∈ DA if and only if F ∈ DA
new
, we get the claim.
3.2 Reduced key propositions
At this point we may combine the assertions in Lemmas 3.1–3.2 for the potentials V =
α e
−
1 κ
ψ
A e
1 κ
ψ
− A 1
κ ψ
3.9 and
V = α
κ S
T
φ 3.10
with ψ given by 2.10. Because of 2.25–2.26, the constant K in 3.3 may be chosen to be the
maximum of 2G and 2C p
T , resulting in K κ → 0 as κ → ∞. Moreover, from 2.27 and a Taylor
expansion of the r.h.s. of 3.9 we see that the potential in 3.9 is bounded for each κ and T ,
and the same is obviously true for the potential in 3.10 because of 2.4. In this way, using a moment inequality to replace the factor e
±Kκ
α by a slightly larger, respectively, smaller factor α
′
independent of T and κ, we see that the limits in Propositions 2.2–2.3 do not change when we
replace E
new ν
ρ
,0
by E
ν
ρ
,0
. Hence it will be enough to prove the following two propositions.
Proposition 3.3. For all α ∈ R,
lim sup
t, κ,T →∞
κ
2
t log E
ν
ρ
,0
exp
α
Z
t
ds e
−
1 κ
ψ
A e
1 κ
ψ
− A 1
κ ψ
Z
s
≤
α 6
ρ1 − ρG. 3.11
Proposition 3.4. For all α 0,
lim
t, κ,T →∞
κ
2
t log E
ν
ρ
,0
exp α
κ Z
t
ds S
T
φ Z
s
= 6
α
2
ρ1 − ρ
2
P
3
. 3.12
Proposition 3.3 has already been proven in [3], Proposition 4.4.2. Sections 4–5 are dedicated to the proof of the lower, respectively, upper bound in Proposition 3.4.
2102
4 Proof of Proposition 3.4: lower bound
In this section we derive the lower bound in Proposition 3.4. We fix α, κ, T 0 and use Lemma 3.1,
to obtain lim
t →∞
1 t
log E
ν
ρ
,0
exp α
κ Z
t
ds S
T
φ Z
s
= sup
F ∈DA
kFk L2
µρ =1
ZZ
Ω×Z
3
d µ
ρ
α κ
S
T
φ F
2
+ F A F .
4.1 In Section 4.1 we choose a test function. In Section 4.2 we compute and estimate the resulting
expression. In Section 4.3 we take the limit κ, T → ∞ and show that this gives the desired lower
bound.
4.1 Choice of test function
To get the desired lower bound, we use test functions F of the form F
η, x = F
1
ηF
2
x. 4.2
Before specifying F
1
and F
2
, we introduce some further notation. In addition to the counting mea- sure m on Z
3
, consider the discrete Lebesgue measure m
κ
on Z
3 κ
= κ
−1
Z
3
giving weight κ
−3
to each site in Z
3 κ
. Let l
2
Z
3
and l
2
Z
3 κ
denote the corresponding l
2
-spaces. Let ∆
κ
denote the lattice Laplacian on Z
3 κ
defined by ∆
κ
f x = κ
2
X
y ∈Z3κ
k y−xk=κ−1
f y − f x
. 4.3
Choose f ∈ C
∞ c
R
3
with k f k
L
2
R
3
= 1 arbitrarily, where C
∞ c
R
3
is the set of infinitely differen- tiable functions on R
3
with compact support. Define f
κ
x = κ
−32
f κ
−1
x ,
x ∈ Z
3
, 4.4
and note that k f
κ
k
l
2
Z
3
= k f k
l
2
Z
3 κ
→ 1 as κ → ∞. 4.5
For F
2
choose F
2
= k f
κ
k
−1 l
2
Z
3
f
κ
. 4.6
To choose F
1
, introduce the function e
φη = α
k f
κ
k
2 l
2
Z
3
X
x ∈Z
3
S
T
φ η, x f
2 κ
x. 4.7
Given K 0, abbreviate
S = 6T 1[ κ]
and U = 6K
κ
2
1[ κ]
4.8 recall 2.24. For
κ p
T K, define e
ψ: Ω → R by e
ψ = Z
U −S
ds T
s
e φ,
4.9 2103
where T
t t
≥0
is the semigroup generated by the operator L in 1.4. Note that the construction of e
ψ from e φ in 4.9 is similar to the construction of ψ from φ in 2.10. In particular,
− L e ψ = e
φ − T
U −S
e φ.
4.10 Combining the probabilistic representations of the semigroups
S
t t
≥0
generated by A in 2.8
and T
t t
≥0
generated by L in 1.4 with the graphical representation formulas 2.21–2.22, and using 4.4–4.5, we find that
e φη =
α k f k
2 l
2
Z
3 κ
Z
Z
3 κ
m
κ
d x f
2
x X
z ∈Z
3
p
S
κx, z[ηz − ρ] 4.11
and e
ψη = X
z ∈Z
3
hz[ ηz − ρ]
4.12 with
hz = α
k f k
2 l
2
Z
3 κ
Z
Z
3 κ
m
κ
d x f
2
x Z
U S
ds p
s
κx, z. 4.13
Using the second inequality in 2.20, we have ≤ hz ≤
C α
p T
, z
∈ Z
3
. 4.14
Now choose F
1
as F
1
= e
e ψ
−1 L
2
ν
ρ
e
e ψ
. 4.15
For the above choice of F
1
and F
2
, we have kF
1
k
L
2
ν
ρ
= kF
2
k
l
2
Z
3
= 1 and, consequently, kFk
L
2
µ
ρ
= 1. With F
1
, F
2
and e φ as above, and A as in 2.8, after scaling space by κ we ar-
rive at the following lemma.
Lemma 4.1. For F as in 4.2, 4.6 and 4.15, all α, T, K 0 and κ
p T
K, κ
2
ZZ
Ω×Z
3
d µ
ρ
α κ
S
T
φ F
2
+ F A F =
1 k f k
2 l
2
Z
3 κ
Z
Z
3 κ
d m
κ
f ∆
κ
f + κ
ke
e ψ
k
2 L
2
ν
ρ
Z
Ω
d ν
ρ
e φe
2 e ψ
+ e
e ψ
Le
e ψ
, 4.16
where e φ and e
ψ are as in 4.7 and 4.9.
4.2 Computation of the r.h.s. of 4.16
