so that, for 1 α 2, there are two positive constants C and C
′
such that X
j 2i
f
j
− f
i 2
j − i
1+ α
6 2
1+ α
Z
+∞ 2i
C
2
x
2 α−2
x
1+ α
d x 6 2
1+ α
C
2
2 − α
2i
α−2
6 C
′
g
α
i 3.16
and, for α = 2, there are two positive constants C and C
′
such that X
j 2i
f
j
− f
i 2
j − i
1+ α
6 8
Z
+∞ 2i
C
2
x log
2
x d x 6
8C
2
log 2i 6
C
′
g
α
i . 3.17
3.2 Proof of Theorem 1.6
To prove Theorem 1.6 we shall make a series of network reductions that allow us to arrive at a nearest–neighbor, one–dimensional problem. We start from the network Γ,
ρ defined at the beginning of this section. We write F
a
= {u ∈ Z
2
: kuk
∞
= a}, a ∈ N. The next reduction is obtained by collapsing all nodes u
∈ F
a
into a single node for each a ∈ N. Once all nodes in each F
a
are identified we are left with a one-dimensional network with nodes a
∈ {0, 1, . . . }. Between nodes a and b we have a total of
P
u ∈F
a
Γ
u
P
v ∈F
b
Γ
v
wires, with a wire of resistance ρ
u,v
for each u ∈ F
a
and v
∈ F
b
. Finally, we perform a last reduction that brings us to a nearest–neighbor one–dimensional network. To this end we consider a single wire with resistance
ρ
u,v
between node a and node b, with a
b − 1. This wire is equivalent to a series of b − a wires, each with resistance ρ
u,v
b − a. That is we can add b
− a − 1 fictitious points to our network in such a way that the effective resistance does not change. Moreover the effective resistance decreases if each added point in the
series is attached to its corresponding node a + i, i = 1, . . . , b − a − 1, in the network. If we repeat
this procedure for each wire across every pair of nodes a b − 1 then we obtain a nearest–neighbor
network where there are infinitely many wires in parallel across any two consecutive nodes. In this new network, across the pair i
− 1, i we have a resistance R
i −1,i
such that φ
i
:= R
−1 i
−1,i
= X
a i
X
b i
X
u ∈F
a
X
v ∈F
b
b − a Γ
u
Γ
v
ρ
−1 u,v
. 3.18
Moreover, the reductions described above show that R
n
x
n+1
X
i=1
R
i −1,i
. Therefore Theorem 1.6 now follows from the estimates on R
i −1,i
given in the next lemma.
Lemma 3.2. There exists a positive constant c such that P–almost surely, for i sufficiently large
R
i,i+1
c i
−1
if α 2 ,
i log i
−1
if α = 2 .
3.19
2600
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
