Proof of Lemma 4.5. By an analogue essentially inverse procedure as in the proof of Lemma 4.4, we first get
4.82 = X
x
n k
t
i
, 2
≤n≤N
i+1
−1, 1
≤k≤n N
i+1
Y
n=2 n
−1
Y
k=1
ϕ
n
x
n −1
k
t
i
, x
n k+1
t
i
× det h
F
N
i+1
+1−l,1
x
N
i+1
k
t
i+1
− x
l 1
t
i
, a, b i
1 ≤k,l≤N
i+1
. 4.83
Now we insert by linearity the factor Q
N
i+1
n=l+1
ϕ
n
x
n −1
l
t
i
, x
n l+1
t
i
to terms F
N
i+1
+1−l,1
x
N
i+1
k
t
i+1
− x
l 1
t
i
, a, b as well as the sum over the corresponding variables. The sums are carried out by using 4.71, from which we get the r.h.s. of 4.82.
Lemma 4.6. Let us define
φ
n
x, y = ϕ
n
x, y, φ
n
x
n −1
n
, y = v
y n
. 4.84
Then v
x
n 1
n
det
ϕ
n
x
n −1
k
, x
n l
1 ≤k,l≤n
= det
φ
n
x
n −1
k
, x
n l
1 ≤k,l≤n
4.85 Proof of Lemma 4.6. It is a consequence of the fact that both determinants are zero if the variables
x
j i
do not interlace and when they do, the matrices are upper-triangular with diagonal equal to zero and with equal entries in the first n
− 1 rows. The only difference is for the last row, where the matrix in l.h.s. of 4.85 has entries 1 and r.h.s. of 4.85 has entries v
x
n l
n
.
5 Asymptotic analysis
5.1 Flat initial conditions
To prove Theorem 2.2 we need the uniform convergence of the kernel in bounded sets as well as bounds uniform in T . These results are provided in the following Propositions 5.1, 5.2, 5.3.
Let us define the rescaled and conjugate kernel by K
resc T
u
1
, s
1
; u
2
, s
2
= Kn
1
, t
1
, x
1
; n
2
, t
2
, x
2
T
1 3
e
t
2
2L+R2
2
x
2
e
t
1
2L+R2
2
x
1
5.1 where n
i
= nu
i
, t
i
= tu
i
, and x
i
= [−2n
i
+ v t
i
− s
i
T
1 3
]. 5.2
Proposition 5.1 Uniform convergence in a bounded set. Fix u
1
, u
2
, then for any fixed ℓ 0, the
rescaled kernel K
resc T
converges uniformly for s
1
, s
2
∈ [−ℓ, ℓ]
2
as lim
T →∞
K
resc T
u
1
, s
1
; u
2
, s
2
= S
−1 v
K
A
1
S
−1 h
u
1
, S
−1 v
s
1
; S
−1 h
u
2
, S
−1 v
s
2
, 5.3
with K
A
1
the kernel of the Airy
1
process, see 2.25, and S
v
, S
h
are defined in 2.13. 1408
Proof of Proposition 5.1. First we consider the term coming from the second integral in 3.23, namely
−T
1 3
2 πi
I
Γ
1
dz e
Rt
1
1−z+L t
1
1−z
e
Rt
2
z+L t
2
z
z
n
1
+n
2
+x
2
1 − z
n
1
+n
2
+x
1
+1
e
t
2
2L+R2
2
x
2
e
t
1
2L+R2
2
x
1
. 5.4
Define the functions Hz
= Rz + L
z − R2 − 2L lnz, g
z = πθ + θ Hz, g
1
z, u = −uπ
′
θ + 1Hz + u1 − π
′
θ lnz1 − z, g
2
z, u, s = u
2
π
′′
θ [Hz + lnz1 − z] + s lnz, 5.5
from which we then set f
z = g
1 − z − g z,
f
1
z = g
1
1 − z, u
1
− g
1
z, u
2
− g
1
12, u
1
+ g
1
12, u
2
, f
2
z = g
2
1 − z, u
1
, s
1
− g
2
z, u
2
, s
2
− g
2
12, u
1
, s
1
+ g
2
12, u
2
, s
2
, f
3
z = − ln1 − z. 5.6
With these notations we get 5.4 =
−T
1 3
2 πi
I
Γ
1
dze
T f z+T
2 3
f
1
z+T
1 3
f
2
z+ f
3
z
. 5.7
The function f z has a double critical point at z = 12 and the contribution for large T will be
dominated by the one close z = 1 2. Thus we need to do series expansions around the critical point.
Computations leads to f
z =
1 3
κ z − 12
3
+ O z − 12
4
, f
1
z = −u
1
− u
2
κ
1
z − 12
2
+ O z − 12
3
, f
2
z = −2s
1
+ s
2
z − 12 + O z − 12
2
, f
3
z = ln2 + O z − 12 5.8
with κ
= 88L + Rπθ + θ , κ
1
= R + 4Lπ
′
θ + 1 + 41 − π
′
θ . 5.9
First we choose Γ
1
to be a steep descent path
3
for f z. Important for the later analysis is that
the chosen steep descent path is, close to the critical point, the steepest descent one. We consider Γ
1
= γ ∨ γ
c
∨ ¯γ, where γ = {12 + e
−iπ3
ξ, 0 ≤ ξ ≤ 12}, ¯γ its image with respect to complex conjugation, and
γ
c
= {1 − 12e
i φ
, π3 ≤ φ ≤ 2π − π3}. We also have f
z = S
R
zRπθ + θ + S
L
zLπθ + θ , with S
R
z = 1 − 2z + 1
2 lnz
1 − z, S
L
z = 1
1 − z
− 1
z − 2 lnz1 − z.
5.10
3
For an integral I = R
γ
dze
t f z
, we say that γ is a steep descent path if 1 Re f z is maximum at some z
∈ γ: Re f z
Re f z for z ∈ γ \ {z
}, and 2 Re f z is monotone along γ except at its maximum point z and, if
γ is closed, at a point z
1
where the minimum of Re f is reached.
1409
On γ, simple computations leads to
dReS
R
z d
ξ = −
8 ξ
2
1 + 2ξ
2
1 + ξ
2
+ 2ξ
2
1 − ξ
2
+ 2ξ
2
, dReS
L
z d
ξ = −
64 ξ
2
1 + 2ξ
2 2
− 12ξ
4
1 + ξ
2
+ 2ξ
2 2
1 − ξ
2
+ 2ξ
2 2
5.11 which are both strictly less than 0 for
ξ ∈ 0, 12. Now consider the part of γ
c
with φ ∈ [π3, π].
Then dReS
R
z d
φ =
− 4 sin
φ1 − cosφ 5
− 4 cosφ ,
dReS
L
z d
φ =
− 32 sin
φ1 − cosφ2 − cosφ 5 − 4 cosφ
2
5.12 which are both strictly less than 0 for
φ ∈ [π3, π. For the piece of γ
c
with φ ∈ π, 2π − π3],
5.12 is strictly positive, which is right since when φ increases we go closer to the critical point.
Therefore the chosen Γ
1
is a steep descent path for f z.
Take any δ 0 and set Γ
δ 1
= {z ∈ Γ
1
||z − 12| ≤ δ}. Then, if in 5.7 we integrate only along Γ
δ 1
instead of integrating along Γ
1
, the error made is just of order O e
−cT
for some c 0 more exactly, c
∼ δ
3
for δ small. Thus we now consider the integral on Γ
δ 1
only. There, we can use the above series expansions to obtain
−2T
1 3
2 πi
Z
Γ
δ 1
dze
1 3
κ T z
−12
3
+u
2
−u
1
κ
1
T
2 3
z−12
2
−2s
1
+s
2
z−12
×e
O Tz−12
4
,T
2 3
z−12
3
,T
1 3
z−12
2
,z −12
. 5.13
The difference between 5.13 and the same integral without the error term can be bounded by applying
|e
x
− 1| ≤ |x|e
|x|
to O · · · . Thus, this error term can be bounded by
2T
1 3
2 π
Z
Γ
δ 1
dz e
1 3
c κ
T z −12
3
+u
2
−u
1
c
1
κ
1
T
2 3
z−12
2
−2c
2
s
1
+s
2
z−12
×O T z − 12
4
, T
2 3
z − 12
3
, T
1 3
z − 12
2
, z − 12
5.14 for some c
, c
1
, c
2
which can be taken as close to 1 as needed by setting δ small enough. Then, by
the change of variable T
1 3
z − 12 = w one gets that this error term is of order O T
−13
what is needed is just c
0. It remains to consider the leading term, namely 5.13 without the error terms. By extending
the integral to infinity by continuing the two small straight segments forming Γ
δ 1
, the error made is of order
O e
−cT
. Thus we obtained that 5.4 is, up to an error O e
−cT
, T
−13
uniform for s
1
, s
2
∈ [−ℓ, ℓ]
2
, equal to −2T
1 3
2 πi
Z
γ
∞
dze
1 3
κ T z
−12
3
+u
2
−u
1
κ
1
T
2 3
z−12
2
−2s
1
+s
2
z−12
, 5.15
1410
where γ
∞
is a path going from e
i π3
∞ to e
−iπ3
∞. By the change of variable w = κ T
1 3
z −12, we get
5.15 = −1
2 πi
Z
γ
∞
dw 2
κ
1 3
e
1 3
w
3
+u
2
−u
1
w
2
κ
1
κ
2 3
−2s
1
+s
2
wκ
1 3
5.16 = S
−1 v
Ai S
−2 h
u
2
− u
1 2
+ S
−1 v
s
1
+ s
2
×e
2 3
S
−3 h
u
2
−u
1 3
+S
−1 v
S
−1 h
u
2
−u
1
s
1
+s
2
with S
v
and S
h
defined in 2.13. Here we used the Airy function representation −1
2 πi
Z
γ
∞
dve
v
3
3+av
2
+bv
= Aia
2
− b exp2a
3
3 − ab. 5.17
To finish the proof, we need to consider the term coming from the first integral in 3.23, namely −
T
1 3
2 πi
I
Γ
dw 1
w
x
1
−x
2
+1
w 1
− w
n
2
−n
1
e
Rw+Lwt
1
−t
2
e
t
2
2L+R2
2
x
2
e
t
1
2L+R2
2
x
1
. 5.18
This can be rewritten as 5.18 =
−T
1 3
2 πi
I
Γ
dw w
e
T
2 3
p w−p
12+T
1 3
p
1
w−p
1
12
5.19 with
p w = u
2
− u
1
π
′
θ + 1Hw − u
2
− u
1
1 − π
′
θ lnw1 − w, p
1
w = −u
2 2
− u
2 1
π
′′
θ 2
[Hw + lnw1 − w] − s
2
− s
1
lnw, 5.20
where Hw is the function defined in 5.5. Remark that we need to do the analysis only for u
2
u
1
. The function p has critical point at w = 1
2. The series expansions of p and p
1
around w = 1
2 are p
w = p
12 + κ
1
u
2
− u
1
w − 12
2
+ O w − 12
3
, p
1
w = p
1
12 + 2s
1
− s
2
w − 12 + O w − 12
2
. 5.21
We choose as path Γ = {
1 2
e
i φ
, φ ∈ −π, π]}. This is a steep descent path for p
. In fact, for w ∈ Γ
, ReHw = R
2 + 2L cosφ + R2 − 2L ln2, 5.22
Re − lnw1 − w = ln2 − ln |1 − w| = 2 ln2 −
1 2
ln5 − 4 cosφ,
which are decreasing when cos φ decreases. Thus, we can integrate only on Γ
δ
= {w ∈ Γ ||w −
1 2| ≤ δ} and, for a small δ, the error term is just of order O e
−cT
2 3
with c 0 c ∼ δ
2
as δ ≪ 1.
The integral over Γ
δ
is then given by −2T
1 3
2 πi
Z
Γ
δ
dwe
κ
1
u
2
−u
1
w−12
2
T
2 3
+2s
1
−s
2
w−12T
1 3
×e
O w−12
3
T
2 3
,w −12
2
T
1 3
,w −12
. 5.23
1411
As above, we use |e
x
−1| ≤ |x|e
|x|
, to control the difference between 5.23 and the same expression without the error terms. By taking
δ ≪ 1 and the change of variable ˜ w = w
− 12T
1 3
, we get that this difference is of order
O T
−13
uniformly for s
1
, s
2
in a bounded set. Once we have taken away the error terms in 5.23, we extend the integral to 1
2 ± i∞. By this we make only an error of order
O e
−cT
2 3
. The integration path can be deformed to 12 + i
R
without passing through any poles, therefore by setting w = 1
2 + i y T
−13
we get −
1 π
Z
R
d y e
−κ
1
u
2
−u
1
y
2
+2s
1
−s
2
= − 1
p πκ
1
u
2
− u
1
exp
− s
2
− s
1 2
κ
1
u
2
− u
1
= −
S
−1 v
Æ 4
πu
2
− u
1
S
−1 h
exp −
s
2
− s
1 2
S
−2 v
4u
2
− u
1
S
−1 h
. 5.24
Since all the error terms in the series expansions are uniform for s
1
, s
2
∈ [−ℓ, ℓ]
2
, the result of the Proposition is proven.
Proposition 5.2 Bound for the diffusion term of the kernel. For any s
1
, s
2
∈
R
and u
2
− u
1
0 fixed, the bound e
t
2
2L+R2
2
x
2
e
t
1
2L+R2
2
x
1
T
1 3
2 πi
I
Γ
dw 1
w
x
1
−x
2
+1
w 1
− w
n
2
−n
1
e
Rw+Lwt
1
−t
2
≤ const e
−|s
1
−s
2
|
5.25 holds for T large enough and const independent of T . The const is uniform in s
1
, s
2
but not in u
2
− u
1
. Proof of Proposition 5.2. From the analysis in Proposition 5.1, we just need a bound for
|s
2
− s
1
| ≥ ℓ, ℓ 0 fixed. We start with 5.19 but to obtain a decaying bound for large |s
2
− s
1
| we consider another path Γ
. Consider an
ǫ with 0 ǫ ≪ 1 and set Γ = {w = ρe
i φ
, φ ∈ [−π, π}, with
ρ =
1 2
+
s
2
−s
1
T
−13
u
2
−u
1
κ
1
, if |s
2
− s
1
| ≤ ǫT
1 3
,
1 2
+
ǫ u
2
−u
1
κ
1
, if s
2
− s
1
≥ ǫT
1 3
,
1 2
−
ǫ u
2
−u
1
κ
1
, if s
2
− s
1
≤ −ǫT
1 3
. 5.26
We have
d d
φ
Rew −
1 2
lnw = −ρ sinφ,
d d
φ
Re1 w + 2 lnw = −
4 ρ
sin φ, and
d d
φ
Re − lnw1 −
w = −
ρ sinφ 1
−2ρ cosφ+ρ
2
. Thus Γ is a steep descent path for p
w. Moreover, since on Γ we have
Relnw = ln ρ is a constant, Γ
is also a steep descent path for p w plus the term of p
1
w proportional to s
2
− s
1
. Let, for a small δ 0 fixed, Γ
δ
= {w = ρe
i φ
, φ ∈ −δ, δ}. Then
5.19 = e
T
2 3
p ρ−p
12+T
1 3
p
1
ρ−p
1
12
5.27 ×
O e
−cT
2 3
+ −T
1 3
2 πi
Z
Γ
δ
dw w
e
T
2 3
p w−p
ρ+T
1 3
p
1
w−p
1
ρ
for some c 0 for small δ, c ∼ δ
2
. On Γ
δ
the s
i
-dependent term in Rep
1
w − p
1
ρ is equal to zero and the rest is of order
O φ
2
. Therefore the last integral can be bounded by T
1 3
2 π
Z
δ −δ
d φ
ρ e
− 1
2 T
2 3
u
2
−u
1
π
′
θ +1Rρ+Lρ+1−π
′
θ ρ1−ρ
2
φ
2
+O T
2 3
φ
4
,T
1 3
φ
2
. 5.28
1412
For δ small enough, and T large enough, the terms O T
2 3
φ
4
and O T
1 3
φ
2
are both controlled by the first term in the exponential. Then, by the change of variable T
1 3
φ = ψ one sees that r.h.s. of 5.28 is bounded by a constant, uniformly in T .
What remains is therefore to bound the first term in the r.h.s. of 5.27. By the choice in 5.26 of ρ, |ρ − 12| ≤ ǫu
2
− u
1
κ
1
≪ 1 for ǫ small enough which can be still chosen. Series expansion for
ρ close to 12 leads to p
ρ − p 12 = −2s
2
− s
1
ρ − 12T
1 3
1 + O ρ − 12 + κ
1
u
2
− u
1
ρ − 12
2
T
2 3
1 + O ρ − 12. 5.29
By 5.26 we obtain the bounds p
ρ − p 12 = −
s
2
− s
1 2
u
2
− u
1
κ
1
1 + O ǫ, if |s
2
− s
1
| ≤ ǫT
1 3
, 5.30
p ρ − p
12 = − |s
2
− s
1
|ǫT
1 3
u
2
− u
1
κ
1
1 + O ǫ, if |s
2
− s
1
| ≥ ǫT
1 3
. Combining the above result we have
|5.25| ≤ h
O e
−cT
2 3
+ O 1 ih
e
−
s2−s1 2
u2−u1κ1
1+O ǫ
+ e
−
|s2−s1|ǫT 1
3 u2−u1κ1
1+O ǫ
i .
5.31 Thus by taking an
ǫ small enough and then T large enough the bound 5.31 implies the statement to be proven, since for any
α 0, there exists a C
α
∞ such that e
−αs
2
−s
1 2
≤ C
α
e
−|s
2
−s
1
|
.
Proposition 5.3 Bound on the main term of the kernel. Let u
1
, u
2
be fixed. Then, for any s
1
, s
2
∈ [−ℓ, ∞
2
, the bound −T
1 3
2 πi
I
Γ
1
dz e
Rt
1
1−z+L t
1
1−z
e
Rt
2
z+L t
2
z
z
n
1
+n
2
+x
2
1 − z
n
1
+n
2
+x
1
+1
e
t
2
2L+R2
2
x
2
e
t
1
2L+R2
2
x
1
≤ const e
−s
1
+s
2
5.32 holds for T large enough, where const is a constant independent of T .
Proof of Proposition 5.3. Let ˜ ℓ be a constant independent from T , which can still be chosen large if
needed. For s
1
, s
2
∈ [−ℓ, ˜ℓ]
2
, the result is a consequence of the estimates in the proof of Propo- sition 5.1. Therefore we can consider just s
1
, s
2
∈ [−ℓ, ∞
2
\ [−ℓ, ˜ℓ]
2
. Introduce the notation ˜
s
i
= s
i
+ ℓ + ˜ℓT
−23
, which then belongs to [˜ ℓT
−23
, ∞.
The integral to be bounded is −T
1 3
2 πi
I
Γ
1
dze
T f z+T
2 3
f
1
z+T
1 3
f
2
z+ f
3
z
5.33 where f
1
z and f
3
z are given in 5.6, and f z and f
2
z are just slight modifications of the functions in 5.6, namely
f z = πθ + θ H1 − z − Hz + ˜s
1
ln21 − z − ˜s
2
ln2z, f
2
z = g
2
1 − z, u
1
, −ℓ − ˜ℓ − g
2
z, u
2
, −ℓ − ˜ℓ
5.34 −g
2
12, u
1
, −ℓ − ˜ℓ + g
2
12, u
2
, −ℓ − ˜ℓ.
1413
We put ˜ s
1
and ˜ s
2
in f z, because they are not restricted to be of order T
−23
as it was the case in Proposition 5.1.
First we need to find a steep descent path for f z. We choose it as Γ
1
= {1 − ρe
i φ
, φ ∈ [−π, π}
with 0 ρ ≤ 12, chosen as follows,
ρ = ¨
1 2
− ˜s
1
+ ˜ s
2
κ
1 2
, ˜ s
1
+ ˜ s
2
≤ ǫ,
1 2
− ǫκ
1 2
, ˜
s
1
+ ˜ s
2
≥ ǫ, 5.35
for some small ǫ 0 to be fixed later. Recall that the ˜s
i
0. To see that Γ
1
is a steep descent path, we consider f z term by term. First consider φ ∈ [0, π], the
case φ ∈ [−π, 0] is obtained by symmetry. The term proportional to Rπθ + θ satisfies
d d
φ Re1
− 2z +
1 2
lnz 1 − z = −
ρ3 − 8ρ cosφ + 4ρ
2
sinφ 1
− 2ρ cosφ + ρ
2
≤ 0 5.36
for all 0 ρ ≤ 12, with equality only at φ = 0, π. The term proportional to Lπθ + θ satisfies
d d
φ Re1
1 − z − 1z − 2 lnz1 − z = − 1 − 2ρ cosφ + 2ρ
2 2
− ρ
2
sinφ 1 − 2ρ cosφ + ρ
2 2
ρ ≤ 0 5.37
for all 0 ρ ≤ 12, with equality only at φ = 0, π. Finally, Reln1 − z is constant on Γ
1
and −Reln2z = − ln2|z| is strictly decreasing while moving on Γ
1
with |φ| increasing.
For a small δ 0, Γ
δ 1
= {1 − ρe
i φ
, φ ∈ −δ, δ}. We also define
Q ρ = exp
Re T f
1 − ρ + T
2 3
f
1
1 − ρ + T
1 3
f
2
1 − ρ
. 5.38
Since Γ
1
is a steep descent path of f z, the integral over Γ
1
\ Γ
δ 1
is bounded by Q
ρO e
−cT
5.39 for some c
0 independent of T . The contribution of the integral over Γ
δ 1
is bounded by Q
ρ −T
1 3
2 πi
Z
Γ
δ 1
dze
T f z− f
1−ρ+T
2 3
f
1
z− f
1
1−ρ+T
1 3
f
2
z− f
2
1−ρ+ f
3
z
5.40 The series expansion around
φ = 0 is Re f
1 − ρe
i φ
− f 1 − ρ = −γ
1
φ
2
1 + O φ 5.41
with γ
1
= ˜
s
2
ρ 21
− ρ
2
+ πθ + θ 1 − 2ρ
1 − ρ
2
R
ρ3 − 2ρ 4
+ L1
− ρ + 2ρ
2
3 ρ1 − ρ
,
5.42 and
Re f
1
1 − ρe
i φ
− f
1
1 − ρ = γ
2
φ
2
1 + O φ, 5.43
with γ
2
= u
2
− u
1
κ
1
+ O ρ − 12. 5.44
1414
Finally, Re f
2
1 − ρe
i φ
− f
2
1 − ρ = O φ
2
. Thus, by the change of variable z = 1 − ρe
i φ
, the above estimates, and by setting
γ = γ
1
+ γ
2
T
−13
, we get 5.40 = Qρ
T
1 3
ρ 2
π1 − ρ Z
δ −δ
d φe
−γφ
2
T 1+ O φ1+O T
−13
. 5.45
By choosing δ small enough independent of T and then T large enough, the factors with the error
terms can be replaced by 1 2, thus
5.40 ≤ Qρ T
1 3
ρ 2
π1 − ρ Z
δ −δ
d φe
−γφ
2
T 2
≤ Qρ 1
p 2
πγT
1 3
. 5.46
Remark that, the worse case is when γ becomes small, and this happens when ρ → 12, i.e., it is
the case of small values of ˜ s
1
+ ˜s
2
. But even in this case, γ
1
T
1 3
∼ s
1
+ s
2
+ 2ℓ + 2˜ℓ
1 2
≥ 2˜ℓ
1 2
. Since
γ
2
is of order one, γT
1 3
= γ
1
T
1 3
+ γ
2
0 for ˜ℓ large enough. So, for by setting ˜ℓ large enough, 5.40
≤ constQρ. This estimate, combined with 5.39, implies that the Proposition will be proven by showing that Q
ρ ≤ const e
−s
1
+s
2
. Since 1 −ρ is close to 12, we can apply the series
expansion of f
i
around z = 1 2. The expansion of f
1
is in 5.8, while the one of f
2
is the same as in 5.8 with s
1
+ s
2
= −2ℓ − 2˜ℓ. Finally, f
z =
1 3
κ z − 12
3
1 + O z − 12
2
− ˜s
1
+ ˜s
2
z − 121 + O z − 12. 5.47
First consider ˜ s
1
+ ˜ s
2
≤ ǫ. Then, with ρ chosen as in 5.35, we get Q
ρ = e
− 2
3 T ˜
s
1
+˜ s
2 3
2
κ
−12
T 1+ O
p ǫ
e
u
2
−u
1
κ
1
˜ s
1
+˜ s
2
T
2 3
κ
−1
1+O p
ǫ
×e
−2ℓ+˜ℓ˜s
1
+˜ s
2
κ
−12
T
1 3
1+O p
ǫ
= e
− 2
3 s
1
+s
2
+2ℓ+2˜ℓ
3 2
κ
−12
1+O p
ǫ
e
u
2
−u
1
κ
1
s
1
+s
2
+2ℓ+2˜ℓκ
−1
1+O p
ǫ
×e
−2ℓ+˜ℓs
1
+s
2
+2ℓ+2˜ℓκ
−12
T
−13
1+O p
ǫ
. 5.48
Recall that s
1
+ s
2
+ 2ℓ + 2˜ℓ ≥ 2˜ℓ ≫ 1 for ˜ℓ ≫ 1. Therefore by choosing ˜ℓ large enough depending only on the coefficients
κ ,
κ
1
, u
1
, u
2
which are however fixed, all the terms are controlled by the first one, i.e.,
Q ρ ≤ e
− 1
3 s
1
+s
2
+2ℓ+2˜ℓ
3 2
κ
−12
≤ e
− 1
3 s
1
+s
2 3
2
κ
−12
. 5.49
Since this decays more rapidly that exp −s
1
+ s
2
, the Proposition holds for ˜ s
1
+ ˜ s
2
≤ ǫ. The last case is ˜
s
1
+ ˜ s
2
≥ ǫ. In this case, with ρ chosen as in 5.35, we obtain Q
ρ = e
T κ
−12
1+O p
ǫ p
ǫǫ3−˜s
1
+˜ s
2
e
u
2
−u
1
κ
1
κ
−1
ǫT
2 3
1+O p
ǫ
×e
−4ℓ+˜ℓκ
−12
ǫT
1 3
1+O p
ǫ
. 5.50
But now, ǫ3 − ˜s
1
+ ˜ s
2
≤ −
2 3
˜ s
1
+ ˜s
2
, thus the first term in the exponential is, up to a positive constant,
− p
ǫT
1 3
s
1
+ s
2
+ 2ℓ + 2˜ℓ, which dominates the second term ∼ ǫT
2 3
≤ s
1
+ s
2
+ 2ℓ + 2˜ℓ, and it also dominates the third term. Therefore, for any choice of
ǫ and ˜ℓ made before, we can take T large enough such that
Q ρ ≤ e
−
1 3
p ǫT
1 3
s
1
+s
2
, 5.51
1415
which ends the proof of the Proposition. Proof of Theorem 2.2. The proof of Theorem 2.2 is the complete analogue of Theorem 2.5 in [3].
The results in Propositions 5.1,5.3,5.4, and 5.5 in [3] are replaced by the ones in Proposi- tion 5.1, 5.2, 5.3. The strategy is to write the Fredholm series of the expression for finite T and,
by using the bounds in Propositions 5.2 and 5.3, see that it is bounded by a T -independent and integrable function. Once this is proven, one can exchange the sumsintegrals and the T
→ ∞ limit by the theorem of dominated convergence. For details, see Theorem 2.5 in [3].
5.2 Sketch of the result 2.23
