Proof. The proof is similar to that of Lemma 4.4 and is omitted here. We are now ready to prove the existence of
λ
p
and give a variational formula.
Corollary 6.6. For p ∈ 0, ∞, the annealed Lyapunov exponent λ
p
exists and is given by λ
p
= sup
α∈[0,γ]
inf
ββ
c r
−β + αL
sup p
β 6.21
for all γ 0 large enough.
Proof. Using Lemmas 6.1, 6.3, 6.4 and 6.5 one may proceed similarly to the quenched case.
Proof of Theorem 2.7. In order to derive the representation of Theorem 2.7, we distinguish two cases: First assume that L
sup p
does not have a zero in 0, β
c r
. From the fact that L
sup p
is continuous and increasing cf. Lemma 5.11, we infer that L
sup p
β 0 for all β ∈ −∞, β
c r
, whence taking α ↓ 0 in Corollary 6.6 yields λ
p
= −β
c r
. Otherwise, if such a zero exists, the properties of L
sup p
derived in Lemma 5.11 first imply the unique- ness of such a zero and then, in an analogous way to the proof of Theorem 2.1 and in combination
with 6.21, that λ
p
equals the zero of L
sup p
−·.
7 Further results
While in sections 4 and 6 we derived the existence of the corresponding Lyapunov exponents and gave formulae for them, we now concentrate on the proofs of the remaining results.
7.1 Quenched regime
We start with proving the result on the transition from zero to positive speed of the random walk in random potential, Proposition 2.5.
Proof of Proposition 2.5. a In order to deal with the explicit dependence of the respective quan- tities on h, we use the notation Y
h t
t ∈R
+
for a continuous-time random with generator κ∆
h
as well as
Z
n
for the set of discrete time simple random walk paths on Z starting in 1 and hitting 0 for the first time at time 2n + 1. Furthermore, J Y denotes the number of jumps of the process Y to the
right before it hits 0 and we may and do assume h 1. For β ≤ β
h c r
:= β
c r
= κ1 − p
1 − h
2
we then have
L
h
β = D
log E
1
exp n
Z
T
ξY
h s
+ β ds oE
= D
log X
n ∈N
X
Z ∈Z
n
1 + h 2
1 − h
2
4
n 2n
Y
k=0
κ κ − ξZ
k
− β E
= log1 + h + D
log E
1
exp n
Z
T
ξY
s
+ β ds + JY log1 − h
2
oE .
7.1
2323
Now in order to prove a, according to Corollary 2.3 it is enough to show lim sup
h ↓0
L
h
β
h c r
0. 7.2
By direct inspection or Lemma 5.5 we get that a.s. the expression log1 + h + log E
1
exp n
Z
T
ξY
s
+ β
h c r
ds + J Y log1 − h
2
o
= log X
n ∈N
X
z ∈Z
n
p 1 + h
2
2n+1
p 1
− h
2n
Y
k=0
κ p
1 − h
2
κ p
1 − h
2
− ξZ
k
7.3
is bounded from above uniformly in h ∈ [0, 12. Considering the right-hand side of 7.3 one
observes that for h ↓ 0 it converges to 7.3 evaluated at h = 0. Since furthermore L
β
c r
0, dominated convergence yields 7.2.
b Departing from 7.1 we write L
h
β
h c r
= D
log X
n ∈N
X
Z ∈Z
n
2
−2n+1
κ p
1 + h κ
p 1
− h − ξ1
2n
Y
k=1
κ p
1 − h
2
κ p
1 − h
2
− ξZ
k
E .
Now if L
1
β → ∞ for β ↑ β
1 c r
= κ, then L
h
β
h c r
→ ∞ for h ↑ 1 as well. Otherwise, if lim
β↑κ
L
1
β ∞, then dominated convergence yields L
h
β
h c r
→ L
1
κ = 〈log
κ −ξ1
〉 as h ↑ 1, and the claim follows.
Proof of Proposition 2.6. a We first show convexity. Writing u
κ
t, x to emphasise the depen- dence of the solution to 1.1 on
κ, we have for a random walk X
t t
∈R
+
with generator ∆
h
: λ
κ = lim
t →∞
1 t
log u
κ
t, x = lim
t →∞
1 t
log E exp
n Z
t
ξX
κs
ds o
= lim
t →∞
1 t
log E exp
n 1
κ Z
κt
ξX
s
ds o
= κ lim
t →∞
1 t
log E exp
n 1
κ Z
t
ξX
s
ds o
= κΨ1κ, 7.4
where Ψx := lim
t →∞
1 t
log E exp
n x
Z
t
ξX
s
ds o
for x ≥ 0. Note that the limit defining Ψx exists in [−1, 0] for all x ∈ R
+
due to Theorem 2.1 and 1.8. Hölder’s inequality now tells us that Ψ is convex and choosing
α :=
β x β x+1−β y
and γ :=
1−β y β x+1−β y
, we obtain the convexity of xΨ1 x in a similar manner:
β x + 1 − β yΨ 1
β x + 1 − β y = β x + 1 − β yΨ
α x
+ γ
y ≤ αβ x + 1 − β yΨ1x + γβ x + 1 − β yΨ1 y
= β xΨ1x + 1 − β yΨ1 y.
2324
In combination with 7.4 the convexity of κ 7→ λ
κ follows. To show that
λ κ is nonincreasing in κ ∈ 0, ∞ assume to the contrary that there is 0 κ
1
κ
2
such that λ
κ
1
λ κ
2
. The convexity of λ would then imply lim
κ→∞
λ κ = ∞ which is
impossible since we clearly have λ
κ ≤ 0 for all κ ∈ 0, ∞. b From Theorem 2.1 we deduce
λ κ ∈ [−β
c r
, 0], 7.5
and using 1.8 the claim follows. c 2.7 follows from 7.4 and the fact that lim
x ↓0
Ψx = 0, which is due to the boundedness of ξ from below. 2.8 follows using Ψ1 = λ
1 0, the fact that Ψx ∈ [−1, 0] due to Theorem 2.1 and 1.8, as well as 7.4 and the monotonicity of Ψ.
7.2 Annealed regime
