the results will follow once we prove that, for Γ, ρ, the effective resistance from 0 to Q
c 0,2n
= {u ∈ Z
d
: kuk
∞
n} satisfies the desired bounds. From now on we consider the cases d = 1 and d = 2 separately.
3.1 Proof of Theorem 1.5
Set d = 1. We further reduce the network Γ, ρ introduced above by collapsing into a single node
ev each pair {v, −v}. This gives a network on {0, 1, 2, . . . } where across each pair 0 6 i j there are now e
Γ
i
e Γ
j
wires, where e Γ
i
:= Γ
i
+ Γ
−i
i 6= 0 and e
Γ := Γ
recall that by Γ we now mean the
original Γ plus 1. Each of these wires has a resistance at least
ρ
i, j
and thus we further reduce the network by assigning each wire the same resistance
ρ
i, j
. We shall call e Γ, ρ this new network and
e R
n
0 its effective resistance from 0 to Q
c 0,2n
. An application of the variational formula 1.6 to the network e
Γ, ρ yields the upper bound e
R
n −1
= e C
n
0 6 1
f
2 n
n
X
i=0 ∞
X
j=i+1
e Γ
i
e Γ
j
j − i
−1−α
f
j
− f
i 2
, 3.1
for any sequence { f
i
}
i 0
such that f
i
= f i, f being a non–decreasing function on [0, ∞ taking the value 0 at the origin only. Next, we choose f as
f x := Z
x
g
α
td t, g
α
t :=
1 + Z
t
1
∧ s
2
s
1+ α
ds
−1
. 3.2
Note that f satisfies the differential equation f
′
t
2
1 +
Z
t
1
∧ s
2
s
1+ α
ds
= f
′
t . 3.3
Moreover, f is increasing on [0, ∞, f 0 = 0 and f
k
= f k behaves as
f
k
∼
log k if
α = 1 , k
α−1
if 1 α 2 ,
k log k
if α = 2 ,
k if
α 2 . 3.4
Here f
k
∼ a
k
means that there is a constant C 1 such that C
−1
a
k
6 f
k
6 C a
k
, for all k C. Since g
α
is non–increasing we have the concavity bounds f
j
− f
i
6 g
α
i j − i , j i 0 .
3.5 Let us first prove the theorem for the easier case
α 2. We point out that here we do not need condition 1.17 and a finite first moment condition suffices. Indeed, set
ξ
i
:= P
j i
j − i
1 −α
e Γ
j
. These random variables are identically distributed and have finite first moment since
α 2. Note that e
Γ
i
and ξ
i
are independent so that E[Γ
i
ξ
i
] ∞. From the ergodic theorem it follows that there exists a constant C such that P–a.s.
n
X
i=0
e Γ
i
ξ
i
6 C n ,
2597
for all n sufficiently large. Due to 3.1, we conclude that e C
n
0 6 n
−2
P
n i=0
e Γ
i
ξ
i
6 C n
−1
and the desired bound e
R
n
0 c n follows. The case 1 6 α 6 2 requires more work. Thanks to our choice of f , we shall prove the following deterministic estimate.
Lemma 3.1. There exists a constant C ∞ such that for any α 1
X
i
:=
∞
X
j=i+1
j − i
−1−α
f
j
− f
i 2
6 C g
α
i , i
∈ N . 3.6
Let us assume the validity of Lemma 3.1 for the moment and define the random variables ξ
i
:=
∞
X
j=i+1
e Γ
j
j − i
−1−α
f
j
− f
i 2
. Let us show that P-a.s.
ξ
i
satisfies the same bound as X
i
in Lemma 3.1. Set Λ λ := log E[e
λe Γ
i
]. From assumption 1.17 we know Λ
λ ∞ for all λ 6 ǫ for some ǫ 0. Moreover, Λλ is convex and Λ
λ 6 c λ for some constant c, for all λ 6 ǫ. Therefore, using Lemma 3.1 we have, for some new constant C:
E [e
a
i
ξ
i
] = Y
j i
exp
Λa
i
j − i
−1−α
f
j
− f
i 2
6 Y
j i
exp
c a
i
j − i
−1−α
f
j
− f
i 2
= exp [c a
i
X
i
] 6 exp [C a
i
g
α
i] , 3.7
provided the numbers a
i
0 satisfy a
i
j − i
−1−α
f
j
− f
i 2
6 ǫ for all j i 0. Note that the last
requirement is satisfied by the choice a
i
:= ǫg
α
i
2
since, using 3.5: a
i
j − i
−1−α
f
j
− f
i 2
6 a
i
j − i
1 −α
g
α
i
2
6 a
i
g
α
i
2
, for j
i, α 1. This will be our choice of a
i
for 1 6 α 6 2. From 3.7 we have
P ξ
i
2c
1
ǫ
−1
g
α
i 6 exp −2a
i
c
1
ǫ
−1
g
α
iE[e
a
i
ξ
i
] 3.8
6 exp
−2c
1
ǫ
−1
a
i
g
α
i exp C a
i
g
α
i 6 e
−c
1
ǫ g
α
i
−1
, if c
1
is large enough. Clearly, g
α
i 6 Clog i
−1
for 1 6 α 6 2 and i large enough. Therefore, if c
1
is sufficiently large, the left hand side in 3.8 is summable in i ∈ N and the Borel Cantelli lemma
implies that P–a.s. we have ξ
i
6 c
2
g
α
i, c
2
:= 2c
1
ǫ
−1
, for all i i , where i
is an a.s. finite random number. Next, we write
n
X
i=0 ∞
X
j=i+1
e Γ
i
e Γ
j
j − i
−1−α
f
j
− f
i 2
6
i
X
i=0
e Γ
i
ξ
i
+ c
2 n
X
i=1
e Γ
i
g
α
i . 3.9
The first term is an a.s. finite random number. The second term is estimated as follows. First, note that
n
X
i=1
g
α
i 6 f
n
, 3.10
2598
since by concavity f
n
=
n −1
X
j=0
f
j+1
− f
j n
−1
X
j=0
g
α
j + 1 =
n
X
j=1
g
α
i . Then we estimate
P
n
X
i=1
e Γ
i
g
α
i 2c
3
f
n
6 e
−2c
3
f
n
n
Y
i=1
E [e
e Γ
i
g
α
i
] = e
−2c
3
f
n
e
P
n i=1
Λg
α
i
. If i is large enough so that g
α
i 6 ǫ we can estimate Λg
α
i 6 c g
α
i. Using 3.10 we then have, for c
3
large enough P
n
X
i=1
e Γ
i
g
α
i 2c
3
f
n
6 e
−c
3
f
n
. Since f
n
log n for all α 1 we see that, if c
3
is sufficiently large, the Borel Cantelli lemma implies that the second term in 3.9 is P–a.s. bounded by 2c
3
f
n
for all n n for some a.s. finite
random number n . Recalling 3.1 we see that there exists an a.s. positive constant c
0 such that e
R
n
0 c f
n
. The proof of Theorem 1.5 is thus complete once we prove the deterministic estimate in Lemma 3.1.
Proof of Lemma 3.1. We only need to consider the cases α ∈ [1, 2]. We write
X
i
=
2i
X
j=i+1
f
j
− f
i 2
j − i
1+ α
+ X
j 2i
f
j
− f
i 2
j − i
1+ α
. 3.11
We can estimate the first term of the right–hand side by using the concavity of f and equation 3.3:
2i
X
j=i+1
f
j
− f
i 2
j − i
1+ α
6 g
2 α
i
i
X
k=1
k
2
k
1+ α
6 C g
α
i 3.12
for some positive constant C. As far as the second term is concerned, first observe that
f
j
− f
i
j − i is non-increasing in j by the concavity of f and so is the general term of the series. As a consequence
X
j 2i
f
j
− f
i 2
j − i
1+ α
6 Z
+∞ 2i
f x − f i
2
x − i
1+ α
d x . 3.13
In the case α = 1 we get, for any i 1,
X
j 2i
f
j
− f
i 2
j − i
1+ α
6 Z
+∞ 2i
1 x
− i ln
1 + x 1 + i
2
d x 6 1
i Z
+∞ 2i
1
x i
− 1 ln
x i
2
d x i
6 2g
α
i Z
+∞ 2
ln t t
− 1
2
d t . 3.14 In the case
α 1 we have, X
j 2i
f
j
− f
i 2
j − i
1+ α
6 2
1+ α
Z
+∞ 2i
f
2
x x
1+ α
d x , 3.15
2599
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
