Lemma 2. i Let 1
u γ and f ∈ H
u
such that f ≤ Ψ f . Then for all t ≥ 0,
f t ≤ C
u α1 − uα
u α − λ − u
2
α
2
e
u αt
− C λ
u α − λ − u
2
α
2
e
αt
. ii If
λ = 14 and f ∈ H
1
is such that f ≤ Φ f , then for all t ≥ 0,
f t ≤ e
t 2
. Before proving Lemma 2, we conclude the proof of Theorem 3ii. For 0
λ 14, from 7, we may apply Lemma 2i applied to 1
u βα, C = 1. We get that f
n 1
t ≤ C
u
e
αut
for some C
u
0. The monotone convergence theorem implies that f
1
t = lim
n →∞
f
n 1
t exists and is bounded by C
u
e
αut
. Therefore f
1
solves the integral equation 4 and is equal to x
a
for some a
≥ 0. From what precedes, we get x
a
t ≤ C
u
e
αut
, however, since αu β, the only possibility is
a = 0 and f
1
t = e
αt
. Similarly, if
λ = 14, from Lemma 2ii, f
1
t ≤ e
t 2
. This proves that f
1
is finite, and we thus have f
1
= x
a
for some a ≥ 0. Again, the only possibility is a = 0 since x
a
t ≤ e
t 2
implies a = 0. Proof of Lemma 2. i. The fixed points of the mapping Ψ are the functions h
a,b
such that h
a,b
t = ae
αt
+ be
β t
+ C u
α1 − uα u
α − λ − u
2
α
2
e
u αt
, with a+b+C
u α1−uα
u α−λ−u
2
α
2
= C. The only fixed point in H
u
is h
∗
:= h
a
∗
,0
with a
∗
= −Cλuα−λ−u
2
α
2
. Let
C
u
denote the set of continuous functions in H
u
, note that Ψ is also a mapping from C
u
to C
u
. Now let g
∈ C
u
and for k ≥ 1, g
k
= Ψg
k −1
. We first prove that for all t ≥ 0 , lim
k
g
k
t = h
∗
t. If 1
u γ then uα1 − uα λ and
u α1−uα
u α−λ−u
2
α
2
is positive. We deduce easily that if g t ≤ Le
u αt
then g
1
t = Ψgt ≤ C e
u αt
+
L λ
u α1−uα
e
u αt
− 1 ≤ L
1
e
u αt
, with L
1
= C +
L λ
u α1−uα
. By recursion, we obtain that lim sup
k
g
k
t ≤ L
∞
e
u αt
, with L
∞
= Cuα1 − uαuα − λ − u
2
α
2
∞. From Arzela-Ascoli’s theorem, g
k k
∈N
is relatively compact in C
u
and any accumulation point converges to h
∗
since h
∗
is the only fixed point of Ψ in C
u
. Now since f
∈ H
u
, there exists a constant L 0 such that for all t ≥ 0, f t ≤ g
t := Le
u αt
. The monotonicity of the mapping Ψ implies that Ψ f
≤ Ψg = g
1
. By assumption, f ≤ Ψ f thus by
recursion f ≤ lim
n
g
n
= h
∗
. ii. The function x
t = e
t 2
is the only fixed point of Φ in H
1
. Moreover, if gt ≤ C e
t 2
then we also have Φgt
≤ C e
t 2
. Then, if g is continuous, arguing as above, from Arzela-Ascoli’s theorem, Φ
k
g
k ∈N
converges to x . We conclude as in i.
2.2 Proof of Theorem 3i
We define f
p
t = E
λ
[Y t
p
]. As above, we often drop the parameter λ in E
λ
and other objects depending on
λ.
Lemma 3. Let p ≥ 2, there exists a polynomial Q
p
with degree p such that for all t 0,
2020
i If λ ∈ 0, pp + 1
−2
, then f
p
t = Q
p
e
αt
. ii If
λ ≥ pp + 1
−2
, then f
p
t = ∞, Note that if such polynomial Q
p
exists then Q
p
x ≥ 1 for all x ≥ 1. Note also that λ ∈ 0, pp+1
−2
implies that p γ = βα where γ was defined by 8, and thus pα β 1. Hence Lemma 3
implies Theorem 3i since E[N
p
] = R
f
p
te
−t
d t. Let
κ
p
X denote the p
th
cumulant of a random variable X whose moment generating function is defined in a neighborhood of 0: ln Ee
θ X
= P
p ≥0
κ
p
X θ
p
p. In particular κ X = 0, κ
1
X = EX and
κ
2
X = VarX . Using the exponential formula E exp
X
ξ
i
∈Φ
h ξ
i
, Z
i
= expλ Z
∞
Ee
hx,Z
− 1d x, 9
valid for all non-negative function h and iid variables Z
i
, i ∈ N, independent of Φ = {ξ
i
}
i ∈N
a Poisson point process of intensity
λ, we obtain that for all p ≥ 1, κ
p
X
i: ξ
i
≤t
h ξ
i
, Z
i
= λ
Z
t
Eh
p
x, Zd x. 10
Due to this last formula, it will be easier to deal with the cumulant g
p
t = κ
p
Y t. By recursion, we will prove the next lemma which implies Lemma 3.
Lemma 4. Let p ≥ 2, there exists a polynomial R
p
with degree p, positive on [1, ∞ such that, for all
t 0,
i If λ ∈ 0, pp + 1
−2
, then f
p
t ∞ and g
p
t = R
p
e
αt
. ii If
λ ≥ pp + 1
−2
, then f
p
t = ∞, Proof of Lemma 4. In §2.1, we have computed f
p
for p = 1 and found R
1
x = x. Let p ≥ 2 and assume now that the statement of the Lemma 4 holds for q = 1,
· · · , p − 1. We assume first that f
p
t ∞, we shall prove that necessarily λ ∈ 0, pp + 1
−2
and g
p
t = R
p
e
αt
. Without loss of generality we assume that 0
λ 14. From Fubini’s theorem, using the linearity of cumulants in 3 and 10, we get
g
p
t = λ Z
t
Z
∞
E[Y x + s
p
]e
−s
dsd x =
λ Z
t
e
x
Z
∞ x
f
p
se
−s
dsd x, 11
note that Fubini’s Theorem implies the existence of f
p
s for all s 0. From Jensen inequality f
p
t ≥ g
1
t
p
= e
p αt
and the integral R
∞ x
e
p αs
e
−s
dsd x is finite if and only if p α 1. We may thus
assume that p α 1. We now recall the identity: EX
p
= P
π
Q
I ∈π
κ
|I|
X , where the sum is over all set partitions of
{1, · · · , p}, I ∈ π means I is one of the subsets into which the set is partitioned, and |I| is the cardinal of I. This formula implies that EX
p
= κ
p
X + Σ
p −1
κ
1
X , · · · , κ
p −1
X , where 2021
Σ
p −1
x
1
, · · · , x
p −1
is a polynomial in p − 1 variables with non-negative coefficients and each of its monomial
Q
k ℓ=1
x
n
ℓ
i
ℓ
satisfies P
ℓ
n
ℓ
i
ℓ
= p. Using the recurence hypothesis, we deduce from 11 that there exists a polynomial ˜
R
p
x = P
p k=1
r
k
x
k
of degree p with r
p
0 such that g
p
t = λ Z
t
e
x
Z
∞ x
g
p
se
−s
+ ˜ R
p
e
αs
e
−s
dsd x
=
p
X
k=1
λr
k
k α1 − kα
e
k αt
+ λ Z
t
e
x
Z
∞ x
g
p
se
−s
dsd x, 12
recall that p α 1. Now we take the derivative of this last expression, multiply by e
−t
and take the derivative again. We get that g
p
is a solution of the differential equation: x
′′
− x
′
+ λx = −
p
X
k=1
λr
k
e
k αt
, 13
with initial condition x0 = 0. Thus necessarily g
p
t = ae
αt
+ be
β t
+ ϕt, where ϕt is a particular solution of the differential equation 13. Assume first that
λ 6= pp + 1
−2
, then it is easy to check that p + 1
λ − pα and pp + 1
−2
− λ are different from 0 and have the same sign. Looking for a function
ϕ of the form ϕt = P
p k=1
c
k
e
k αt
gives c
k
= −λr
k
k
2
α
2
− kα + λ
−1
= λr
k
k − 1
−1
k + 1λ − kα
−1
. If λ pp + 1
−2
then p α β and the leading term in g
p
is c
p
e
p αt
. However, if
λ pp + 1
−2
, c
p
0 and thus g
p
t 0 for t large enough. This is a contradiction with Equation 11 which asserts that g
p
t is positive. We now check that if 0
λ pp + 1
−2
then f
p
t is finite. We define f
n p
t = E[minY t, n
p
]. We use the following identity,
N
X
i=1
y
i p
=
N
X
i=1 p
−1
X
k=0
p − 1
k y
k+1 i
N
X
j 6=i
y
i
p −k−1
. Then from 3 we get,
Y t − 1
p d
= 14
X
ξ
i
≤t
Y
i
ξ
i
+ D
i p
+ X
ξ
i
≤t p
−2
X
k=0
p − 1
k Y
i
ξ
i
+ D
i k+1
X
ξ
j
6=ξ
i
≤t
Y
j
ξ
j
+ D
j
p −k−1
. The recursion hypothesis implies that there exists a constant C such that f
k
t = Q
k
e
αt
≤ C e
k αt
for all 1 ≤ k ≤ p − 1. Thus, the identity Y t
p
= Y t − 1
p
− P
p −1
k=0 p
k
−1
p −k
Y t
k
gives f
n p
t ≤ E[minY t − 1, n
p
] +
p −1
X
k=0
p k
C e
k αt
≤ E[minY t − 1, n
p
] + C
1
e
p αt
. From the recursion hypothesis, if 1
≤ k ≤ p − 1, Z
t
E[Y x + D
k
]d x = Z
t
e
x
Z
∞ x
f
k
se
−s
dsd x = ˜ Q
k
e
αt
≤ C e
k αt
2022
for some constant C 0. We take the expectation in 14 and use Slyvniak’s theorem to obtain
f
n p
t ≤ C
1
e
p αt
+ λ Z
t
e
x
Z
∞ x
f
n p
se
−s
dsd x + λ
Z
t p
−2
X
k=0
p − 1
k E[Y
i
x + D
i k+1
]E
X
ξ
j
≤t
Y
j
ξ
j
+ D
j p
−k−1
d x
≤ C
1
e
p αt
+ λ Z
t
e
x
Z
∞ x
f
n p
se
−s
dsd x + λ
p −2
X
k=0
p − 1
k ˜
Q
k+1
e
αt
E[Y t − 1
p −k−1
] ≤ C
2
e
p αt
+ λ Z
t
e
x
Z
∞ x
f
n p
se
−s
dsd x So finally for a suitable choice of C,
| f
n p
t ≤ C e
p αt
+ λ Z
t
e
x
Z
∞ x
f
n p
se
−s
dsd x. 15
From Lemma 2, f
n p
t ≤ C
′
e
p αt
, and, by the monotone convergence theorem, g
p
t ≤ f
p
t ≤ C
′
e
p αt
. From what precedes: g
p
t = ae
αt
+ be
β t
+ ϕt, with ϕt = P
p k=1
c
k
e
k αt
, with c
p
0. If b
0, since λ pp + 1
−2
then p α β and the leading term in g
p
is be
β t
which is in contradiction with g
p
t ≤ C
′
e
p αt
. If b 0, this is a contraction with Equation 11 which asserts that g
p
t is positive. Therefore b = 0 and g
p
t = ae
αt
+ ϕt = R
p
e
αt
. It remains to check that if
λ = pp + 1
−2
then for all t 0, f
p
t = ∞. We have proved that, for all λ pp+1
−2
, g
p
t = u
p
λp−1
−1
p+1λ−pα
−1
e
p αt
+S
p −1
e
αt
, where S
p −1
is a polynomial of degree at most p
−1 and u
p
λ 0. Note that lim
λ↑pp+1
−2
p + 1λ − pα = 0. A closer look at the recursion shows also that u
p
λ is a sum of products of terms in λ and λℓ − 1
−1
ℓ + 1λ − ℓα
−1
, with 2
≤ ℓ ≤ p − 1. In particular, we deduce that lim
λ↑pp+1
−2
u
p
λ 0. Similarly, the coefficients of S
p −1
are equal to sums of products of integers and terms in λ and λℓ − 1
−1
ℓ + 1λ − ℓα
−1
, with 2
≤ ℓ ≤ p − 1. Thus they stay bounded as λ goes to pp + 1
−2
and we obtain, for all t 0,
lim inf
λ↑pp+1
−2
f
p
t ≥ lim
λ↑pp+1
−2
g
p
t = ∞. 16
Now, for all t 0, the random variable Y t is stochastically non-decreasing with λ. Therefore
E
λ
[Y t
p
] is non-decreasing and 16 implies that E
1 4
[Y t
p
] = ∞. The proof of the recursion is complete.
2.3 Proof of Theorem 3iii
