Proof. We first show that E φ
i
6 Cω
i
, where ω
i
= i if α 2 and ω
i
= i log i if α = 2, where E denotes expectation w.r.t. the field
{Γ
u
, u ∈ Z
2
}. Thanks to Lemma A.1 given in the Appendix, from 3.18 we have
E φ
i
6 c
1
X
a i
X
b i
ab − a
−α
. 3.20
Next we estimate P
b i
b − a
−α
6 c
2
i − a
1 −α
, so that, using the Riemann integral, we obtain E
φ
i
6 c
2
X
a i
ai − a
1 −α
= c
2
i
2 −α
X
a i
a i
1
− a
i
1 −α
6 c
3
i
3 −α
Z
1 −
1 i
y1 − y
1 −α
d y 6 c
3
i
3 −α
Z
1 1
i
y
1 −α
d y 6 c
4
ω
i
. Hence, for C large we can estimate
P φ
i
2C ω
i
6 Pφ
i
− Eφ
i
Cω
i
6 C ω
i −4
E h
φ
i
− Eφ
i 4
i ,
3.21 where we use P to denote the law of the variables
{Γ
u
}. The proof then follows from the Borel– Cantelli Lemma and the following estimate to be established below: There exists C
∞ such that, for all i
∈ N, E
h φ
i
− Eφ
i 4
i 6
C i
2
. 3.22
To prove 3.22 we write E
h φ
i
− Eφ
i 4
i =
X
a
X
b
X
u ∼a
X
v ∼b
Φu, v Gu, v , 3.23
where the sums are over a = a
1
, . . . , a
4
, b = b
1
, . . . , b
4
such that a
k
i 6 b
k
, u
∼ a stands for
the set of u = u
1
, . . . , u
4
such that u
k
∈ F
a
k
, and we have defined, for u
∼ a, v ∼ b: Φu, v =
4
Y
k=1
b
k
− a
k
ρ
u
k
,v
k
, G
u, v =
4
Y
k=1
Γ
u
k
Γ
v
k
− E[Γ
u
k
Γ
v
k
]
. From the independence assumption on the field
{Γ
u
} we know that Gu, v = 0 unless for every
k = 1, . . . , 4 there exists a k
′
= 1, . . . , 4 with k 6= k
′
and {u
k
, v
k
} ∩ {u
k
′
, v
k
′
} 6= ;. Moreover, when this condition is satisfied using 1.18 we can easily bound G
u, v from above by constant C. By
symmetry we may then estimate X
a
X
b
X
u ∼a
X
v ∼b
Φu, v Gu, v
6 C
X
a
X
b
X
u ∼a
X
v ∼b
Φu, v χ ∀k ∃k
′
6= k : {u
k
, v
k
} ∩ {u
k
′
, v
k
′
} 6= ; 6
3 C X
a
X
b
X
u ∼a
X
v ∼b
Φu, v
h χ u
1
= u
2
; u
3
= u
4
+ + χ u
1
= u
2
; v
3
= v
4
+ χ v
1
= v
2
; v
3
= v
4
i .
3.24 2601
We claim that each of the three terms in the summation above is of order i
2
as i grows. This will prove the desired estimate 3.22. The first term in 3.24 satisfies
X
a
X
b
X
u ∼a
X
v ∼b
Φu, v χ u
1
= u
2
; u
3
= u
4
6 Ai
2
, 3.25
where Ai :=
X
a
1
i
X
b
1
i
X
b
2
i
b
1
− a
1
b
2
− a
1
X
u
1
∈F
a1
X
v
1
∈F
b1
X
v
2
∈F
b2
ρ
u
1
,v
1
ρ
u
1
,v
2
. Similarly the third term in 3.24 is bounded above by Bi
2
, with Bi :=
X
a
1
i
X
a
2
i
X
b
1
i
b
1
− a
1
b
1
− a
2
X
u
1
∈F
a1
X
u
2
∈F
a2
X
v
1
∈F
b1
ρ
u
1
,v
1
ρ
u
2
,v
1
. Finally, the middle term in 3.24 is less than or equal to the product AiBi. Therefore, to prove
3.22 it suffices to show that Ai 6 C i and Bi 6 C i. Using Lemma A.1 we see that X
u
1
∈F
a1
X
v
1
∈F
b1
X
v
2
∈F
b2
ρ
u
1
,v
1
ρ
u
1
,v
2
6 C a
1
b
1
− a
1 −1−α
b
2
− a
1 −1−α
. This bound yields
Ai 6 C X
a
1
i
a
1
i − a
1 2
−2α
, for some new constant C, where we have used the fact that
X
b
2
i
b
2
− a
1 −α
6 Ci
− a
1 1
−α
. Using the Riemann integral we obtain
X
a
1
i
a
1
i − a
1 2
−2α
6 C i
4 −2α
Z
1 −1i
x1 − x
2 −2α
d x 6
C i
4 −2α
Z
1 1
i
x
2 −2α
d x 6 2 α − 3C i .
This proves that Ai = Oi. Similarly, from Lemma A.1 we see that X
u
1
∈F
a1
X
u
2
∈F
a2
X
v
1
∈F
b1
ρ
u
1
,v
1
ρ
u
2
,v
1
6 C b
−1 1
a
1
a
2
b
1
− a
1 −1−α
b
1
− a
2 −1−α
. Therefore
Bi 6 C X
b
1
i
b
−1 1
X
a
1
i
a
1
b
1
− a
1 −α
2
6 C
′
i
2
X
b
1
i
b
−1 1
b
1
− i + 1
2 −2α
6 C
′′
i ,
2602
where we have used the estimate X
a
1
i
a
1
b
1
− a
1 −α
6 i
X
a
1
i
b
1
− a
1 −α
6 C i b
1
− i + 1
1 −α
, and the fact that, for
α 2, we have X
b
1
i
b
−1 1
b
1
− i + 1
2 −2α
6 i
−1 ∞
X
k=1
k
−2
= Ci .
We remark that a proof of Theorem 1.6 could be obtained by application of the variational principle 1.6 as in the proof of Theorem 1.5. To see this one can start from the network Γ,
ρ introduced at the beginning of this section and choose a trial function that is constant in each F
a
. Then, for any non–decreasing sequence f
, f
1
, . . . such that f = 0 and f
k
0 eventually, one has R
n
x A
n
f where
A
n
f = f
−2 n
n
X
a=0 ∞
X
b=a+1
f
b
− f
a 2
X
u ∈F
a
Γ
u
X
v ∈F
b
Γ
v
|v − u|
−2−α
. 3.26
One then choose f
k
= log1 + k for α 2 and f
k
= logloge + k for α = 2 and the desired conclusions will follow from suitable control of the fluctuations of the random sum appearing in
3.26. Here the analysis is slightly more involved than that in the proof of Theorem 1.5 and it requires estimates as in 3.24 above. Moreover, one needs a fifth moment assumption with this
approach instead of the fourth moment condition 1.18. Under this assumption, and using Lemma A.1, it is possible to show that there a.s. exists a constant c such that
X
u ∈F
a
Γ
u
X
v ∈F
b
Γ
v
|v − u|
−2−α
6 c a b
− a
−1−α
, a
b . 3.27
Once this estimate is available the proof follows from simple calculations.
4 Proof of Proposition 1.7 and Theorem 1.8
