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
4.1 Proof of Proposition 1.7
The proof of Proposition 1.7 is based on the following technical lemma related to renewal theory:
Lemma 4.1. Given δ 1, define the probability kernel
q
k
= cδk
−δ
k ∈ N ,
4.1 c
δ being the normalizing constant 1 P
k 1
k
−δ
and define recursively the sequence f n as f 0 = 1 ,
f n = P
n −1
k=0
f kq
n −k
, n
∈ N . If 1
δ 2, then lim
n ↑∞
n
2 −δ
f n = Γ2 − δ
Γδ − 1 .
4.2 2603
Proof. Let {X
i
}
i 1
be a family of IID random variables with PX
i
= k = q
k
, k ∈ N. Observe now
that PX
i
k = P
∞ s=k
q
s
∼ c k
1 −δ
since δ 1. In particular, if 1 δ 2 we can use Theorem B of
[10] and get 4.2 with un instead f n, where un is defined as follows: Consider the random walk S
n
on the set N ∪ {0}, starting at 0, S
= 0, and defined as S
n
= X
1
+ X
2
+ · · · + X
n
for n 1. Given n
∈ N define un as un := E
{m 0 : S
m
= n}
=
∞
X
m=0
PS
m
= n . Trivially u0 = 1, while the Markov property of the random walk S
n
gives for n 1 that un =
∞
X
m=1 n
−1
X
k=0
PS
m −1
= k, S
m
= n =
∞
X
m=1 n
−1
X
k=0
PS
m −1
= kq
n −k
=
n −1
X
k=0
ukq
n −k
. Hence, f n and un satisfy the same system of recursive identities and coincide for n = 0, thus
implying that f n = un for each n ∈ N.
We have now all the tools in order to prove Proposition 1.7: Proof of Proposition 1.7. We shall exhibit a finite energy unit flux f
·, · from x to infinity in the
network S, ϕ. To this end we define f ·, · as follows
f x
i
, x
k
=
f iq
k −i
if 0 6 i k ,
− f x
k
, x
i
if 0 6 k i ,
otherwise , 4.3
where f m, q
m
are defined as in the previous lemma for some δ ∈ 1, 2 that will be fixed below.
Since ϕ Cϕ
p, α
, the energy E f dissipated by the flux f ·, · is
E f =
∞
X
n=0 ∞
X
k=n+1
f x
n
, x
k 2
ϕ|x
n
− x
k
| 6
c
∞
X
n=0 ∞
X
k=n+1
r
p, α
x
k
, x
n
f
n
q
k −n
2
, 4.4
where r
p, α
x, y := 1ϕ
p, α
|x − y|. Hence, due to the previous lemma we obtain that E f 6 c
∞
X
n=0 ∞
X
k=n+1
r
p, α
x
k
, x
n
1 + n
2 δ−4
k − n
−2δ
. In order to prove that the energy
E f is finite P–a.s., it is enough to show that EE f is finite for some
δ ∈ 1, 2. To this end we observe that, due to assumption 1.22 and since r
p, α
x
k
, x
n
= 1
∨ x
n
− x
k 1+
α
, for suitable constants c
1
, c
2
, E
E f 6 c
1 ∞
X
n=0
1 + n
2 δ−4
∞
X
u=1
1 + E
|x
u
− x |
1+ α
u
−2δ
6 c
2 ∞
X
n=1
1 + n
2 δ−4
∞
X
u=1
u
1+ α−2δ
. 4.5
2604
Hence, the mean energy is finite if 2 δ−4 −1 and 1+α−2δ −1. In particular, for each α ∈ 0, 1
one can fix δ ∈ 1, 2 satisfying the conditions above. This concludes the proof of the transience of
S, ϕ for P–a.a. S. It remains to verify assumption 1.22 whenever P is a renewal point process such that Ex
1
− x
1+ α
∞. To this end we observe that, by convexity, x
u
− x
1+ α
= u
1+ α
1 u
u −1
X
k=0
x
k+1
− x
k 1+
α
6 u
1+ α
1 u
u −1
X
k=0
x
k+1
− x
k 1+
α
. Since by the renewal property x
k+1
− x
k k 0
is a sequence of i.i.d. random variables, the mean of the last expression equals u
1+ α
E x
1
− x
1+ α
= cu
1+ α
. Therefore, 1.22 is satisfied.
4.2 Proof of Theorem 1.8
