Moreover, since z
2
= ¯ G0
[ ¯ G0
− 1], as noted in 1.15, we see that z
2
= 1 + exp[− log Z], i.e., r
2
= log Z, and so indeed equality holds in 3.16. The quenched LDP in Theorem 2.2, together with Varadhan’s lemma applied to 3.8, gives z
1
= 1 + exp[
−r
1
] with r
1
:= lim
N →∞
1 N
log F
1 N
X ≤ sup
Q ∈P
inv
Ý Z
d N
¨Z
Ý Z
d
π
1
Qd y log f y − I
que
Q «
X − a.s.,
3.20 where I
que
Q is given by 2.13–2.14. Without further assumptions, we are not able to reverse the inequality in 3.20. This point will be addressed in Section 4 and will require assumptions
1.10–1.12.
3.2 Proof of Theorem 1.3
To compare 3.20 with 3.16, we need the following lemma, the proof of which is deferred.
Lemma 3.2. Assume 1.1. Let Q
∗
= q
∗ ⊗N
with q
∗
as in 3.19. If m
Q
∗
∞, then I
que
Q
∗
I
ann
Q
∗
. With the help of Lemma 3.2 we complete the proof of the existence of the gap as follows. Since
log f is bounded from above, the function
Q 7→
Z
Ý Z
d
π
1
Qd y log f y − I
que
Q 3.21
is upper semicontinuous. Therefore, by compactness of the level sets of I
que
Q, the function in 3.21 achieves its maximum at some Q
∗∗
that satisfies r
1
= Z
Ý Z
d
π
1
Q
∗∗
d y log f y − I
que
Q
∗∗
≤ Z
Ý Z
d
π
1
Q
∗∗
d y log f y − I
ann
Q
∗∗
≤ r
2
. 3.22
If r
1
= r
2
, then Q
∗∗
= Q
∗
, because the function Q
7→ Z
Ý Z
d
π
1
Qd y log f y − I
ann
Q 3.23
has Q
∗
as its unique maximiser recall the discussion immediately after Lemma 3.1. But I
que
Q
∗
I
ann
Q
∗
by Lemma 3.2, and so we have a contradiction in 3.22, thus arriving at r
1
r
2
. In the remainder of this section we prove Lemma 3.2.
Proof. Note that q
∗
Z
d n
= X
x
1
,...,x
n
∈Z
d
p
n
x
1
+ · · · + x
n
¯ G0
− 1
n
Y
k=1
px
k
= p
2n
¯ G0
− 1 ,
n ∈ N,
3.24 and hence, by assumption 1.2,
lim
n →∞
log q
∗
Z
d n
log n = −α
3.25 565
and m
Q
∗
=
∞
X
n=1
nq
∗
Z
d n
=
∞
X
n=1
np
2n
¯ G0
− 1 .
3.26 The latter formula shows that m
Q
∗
∞ if and only if p·, · is strongly transient. We will show that m
Q
∗
∞ =⇒
Q
∗
= q
∗ ⊗N
6∈ R
ν
, 3.27
the set defined in 2.15. This implies Ψ
Q
∗
6= ν
⊗N
recall 2.16, and hence HΨ
Q
∗
|ν
⊗N
0, implying the claim because
α ∈ 1, ∞ recall 2.14. In order to verify 3.27, we compute the first two marginals of Ψ
Q
∗
. Using the symmetry of p ·, ·,
we have Ψ
Q
∗
a = 1
m
Q
∗
∞
X
n=1 n
X
j=1
X
x1,...,xn∈Z d
x j =a
p
n
x
1
+ · · · + x
n
¯ G0
− 1
n
Y
k=1
px
k
= pa P
∞ n=1
np
2n −1
a P
∞ n=1
np
2n
. 3.28
Hence, Ψ
Q
∗
a = pa for all a ∈ Z
d
with pa 0 if and only if
a 7→
∞
X
n=1
n p
2n −1
a is constant on the support of p·. 3.29
There are many p ·, ·’s for which 3.29 fails, and for these 3.27 holds. However, for simple
random walk 3.29 does not fail, because a 7→ p
2n −1
a is constant on the 2d neighbours of the origin, and so we have to look at the two-dimensional marginal.
Observe that q
∗
x
1
, . . . , x
n
= q
∗
x
σ1
, . . . x
σn
for any permutation σ of {1, . . . , n}. For a, b ∈ Z
d
, we have
m
Q
∗
Ψ
Q
∗
a, b = E
Q
∗
τ
1
X
k=1
1
κY
k
=a,κY
k+1
=b
=
∞
X
n=1 ∞
X
n
′
=1
X
x
1
,...,x
n+n′
q
∗
x
1
, . . . , x
n
q
∗
x
n+1
, . . . , x
n+n
′
n
X
k=1
1
a,b
x
k
, x
k+1
= q
∗
x
1
= a q
∗
x
1
= b +
∞
X
n=2
n − 1q
∗
{a, b} × Z
d n
−2
. 3.30
Since q
∗
x
1
= a = pa
2
¯ G0
− 1 +
∞
X
n=2
X
x
2
,...,x
n
∈Z
d
p
n
a + x
2
+ · · · + x
n
¯ G0
− 1 pa
n
Y
k=2
px
k
= pa
¯ G0
− 1
∞
X
n=1
p
2n −1
a 3.31
566
and q
∗
{a, b} × Z
d n
−2
= 1
n=2
papb ¯
G0 − 1
p
2
a + b + 1
n ≥3
papb ¯
G0 − 1
X
x
3
,...,x
n
∈Z
d
p
n
a + b + x
3
+ · · · + x
n n
Y
k=3
px
k
= papb
¯ G0
− 1 p
2n −2
a + b, 3.32
we find recall that ¯ G0
− 1 = P
∞ n=1
p
2n
Ψ
Q
∗
a, b = papb
h
∞
P
n=1
p
2n
ih
∞
P
n=1
np
2n
i
∞
X
n=1
p
2n −1
a
∞
X
n=1
p
2n −1
b +
∞
X
n=1
p
2n ∞
X
n=2
n − 1p
2n −2
a + b
. 3.33
Pick b = −a with pa 0. Then, shifting n to n − 1 in the last sum, we get
Ψ
Q
∗
a, −a pa
2
− 1 =
∞
P
n=1
p
2n −1
a
2
h
∞
P
n=1
p
2n
ih
∞
P
n=1
np
2n
i 0. 3.34
This shows that consecutive letters are not uncorrelated under Ψ
Q
∗
, and implies that 3.27 holds as claimed.
3.3 Proof of Theorem 1.6
