# SPDEs withα-Stable Lévy Noise: A Random Field Approach

Authors: Raluca M. Balan
Journal: International Journal of Stochastic Analysis
 (2014)

---

## Abstract

This paper is dedicated to the study of a nonlinear SPDE on a bounded domain inR
d, with zero initial conditions and Dirichlet boundary, driven by an α-stable Lévy noise Z with α
∈
(
0,2
), α
≠
1, and possibly nonsymmetric tails. To give a meaning to the concept of solution, we develop a theory of stochastic integration with respect to this noise. The idea is to first solve the equation with “truncated” noise (obtained by removing from Z the jumps which exceed a fixed value K), yielding a solution u
K, and then show that the solutions u
L
,
L
>
K coincide on the event t
≤
τ
K, for some stopping times τ
K converging to infinity. A similar idea was used in the setting of Hilbert-space valued processes. A major step is to show that the stochastic integral with respect to Z
K satisfies a pth moment inequality. This inequality plays the same role as the Burkholder-Davis-Gundy inequality in the theory of integration with respect to continuous martingales.


---

## Body

## 1. Introduction

Modeling phenomena which evolve in time or space-time and are subject to random perturbations are a fundamental problem in stochastic analysis. When these perturbations are known to exhibit an extreme behavior, as seen frequently in finance or environmental studies, a model relying on the Gaussian distribution is not appropriate. A suitable alternative could be a model based on a heavy-tailed distribution, like the stable distribution. In such a model, these perturbations are allowed to have extreme values with a probability which is significantly higher than in a Gaussian-based model.In the present paper, we introduce precisely such a model, given rigorously by a stochastic partial differential equation (SPDE) driven by a noise term which has a stable distribution over any space-time region and has independent values over disjoint space-time regions (i.e., it is a Lévy noise). More precisely, we consider the SPDE:(1)
L
u
(
t
,
x
)
=
σ
(
u
(
t
,
x
)
)
Z
˙
(
t
,
x
)
,
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions, where σ is a Lipschitz function, L is a second-order pseudo-differential operator on a bounded domain 𝒪
⊂
ℝ
d, and Z
˙
(
t
,
x
)
=
∂
d
+
1
Z
/
∂
t
∂
x
1
,
…
,
∂
x
d is the formal derivative of an α-stable Lévy noise with α
∈
(
0,2
),  α
≠
1. The goal is to find sufficient conditions on the fundamental solution G
(
t
,
x
,
y
) of the equation L
u
=
0 on ℝ
+
×
𝒪, which will ensure the existence of a mild solution of (1). We say that a predictable process u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} is a mild solution of (1) if for any t
>
0,  x
∈
𝒪,
(2)
u
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
a
.
s
.
We assume that G
(
t
,
x
,
y
) is a function in t, which excludes from our analysis the case of the wave equation with d
≥
3.To explain the connections with other works, we describe briefly the construction of the noise (the details are given in Section2). This construction is similar to that of a classical α-stable Lévy process and is based on a Poisson random measure (PRM) N on ℝ
+
×
ℝ
d
×
(
ℝ
∖
{
0
}
) of intensity d
t
d
x
ν
α
(
d
z
), where
(3)
ν
α
(
d
z
)
=
[
p
α
z
-
α
-
1
1
(
0
,
∞
)
(
z
)
+
q
α
(
-
z
)
-
α
-
1
1
(
-
∞
,
0
)
(
z
)
]
d
z
for some p,  q
≥
0 with p
+
q
=
1. More precisely, for any set B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
),
(4)
Z
(
B
)
=
∫
B
×
{
|
z
|
≤
1
}
‍
z
N
^
(
d
s
,
d
x
,
d
z
)
+
∫
B
×
{
|
z
|
>
1
}
‍
z
N
(
d
s
,
d
x
,
d
z
)
-
μ
|
B
|
,
where N
^
(
B
×
·
)
=
N
(
B
×
·
)
-
|
B
|
ν
α
(
·
) is the compensated process and μ is a constant (specified by Lemma 3). Here, ℬ
b
(
ℝ
+
×
ℝ
d
) is the class of bounded Borel sets in ℝ
+
×
ℝ
d and |
B
| is the Lebesgue measure of B.As the term on the right-hand side of (2) is a stochastic integral with respect to Z, such an integral should be constructed first. Our construction of the integral is an extension to random fields of the construction provided by Giné and Marcus in [1] in the case of an α-stable Lévy process {
Z
(
t
)
}
t
∈
[
0,1
]. Unlike these authors, we do not assume that the measure ν
α is symmetric.Since any Lévy noise is related to a PRM, in a broad sense, one could say that this problem originates in Itô’s papers [2, 3] regarding the stochastic integral with respect to a Poisson noise. SPDEs driven by a compensated PRM were considered for the first time in [4], using the approach based on Hilbert-space-valued solutions. This study was motivated by an application to neurophysiology leading to the cable equation. In the case of the heat equation, a similar problem was considered in [5–7] using the approach based on random-field solutions. One of the results of [6] shows that the heat equation:
(5)
∂
u
∂
t
(
t
,
x
)
=
1
2
Δ
u
(
t
,
x
)
+
∫
U
‍
f
(
t
,
x
,
u
(
t
,
x
)
;
z
)
N
^
(
t
,
x
,
d
z
)
+
g
(
t
,
x
,
u
(
t
,
x
)
)
has a unique solution in the space of predictable processes u satisfying sup
(
t
,
x
)
∈
[
0
,
T
]
×
ℝ
d
E
|
u
(
t
,
x
)
|
p
<
∞, for any p
∈
(
1
+
2
/
d
,
2
]. In this equation, N
^ is the compensated process corresponding to a PRM N on ℝ
+
×
ℝ
d
×
U of intensity d
t
d
x
ν
(
d
z
), for an arbitrary σ-finite measure space (
U
,
ℬ
(
U
)
,
ν
) with measure ν satisfying ∫
U
|
z
|
p
ν
(
d
z
)
<
∞. Because of this later condition, this result cannot be used in our case with U
=
ℝ
∖
{
0
} and ν
=
ν
α. For similar reasons, the results of [7] also do not cover the case of an α-stable noise. However, in the case α
>
1, we will be able to exploit successfully some ideas of [6] for treating the equation with “truncated” noise Z
K, obtained by removing from Z the jumps exceeding a value K (see Section 5.2).The heat equation with the same type of noise as in the present paper was examined in [8, 9] in the cases α
<
1 and α
>
1, respectively, assuming that the noise has only positive jumps (i.e., q
=
0). The methods used by these authors are different from those presented here, since they investigate the more difficult case of a non-Lipschitz function σ
(
u
)
=
u
δ with δ
>
0. In [8], Mueller removes the atoms of Z of mass smaller than 2
-
n and solves the equation driven by the noise obtained in this way; here we remove the atoms of Z of mass larger than K and solve the resulting equation. In [9], Mytnik uses a martingale problem approach and gives the existence of a pair (
u
,
Z
) which satisfies the equation (the so-called “weak solution”), whereas in the present paper we obtain the existence of a solution u for a given noise Z (the so-called “strong solution”). In particular, when α
>
1 and δ
=
1
/
α, the existence of a “weak solution” of the heat equation with α-stable Lévy noise is obtained in [9] under the condition
(6)
α
<
1
+
2
d
which we encounter here as well. It is interesting to note that (6) is the necessary and sufficient condition for the existence of the density of the super-Brownian motion with “α
-
1”-stable branching (see [10]). Reference [11] examines the heat equation with multiplicative noise (i.e., σ
(
u
)
=
u), driven by an α-stable Lévy noise Z which does not depend on time.To conclude the literature review, we should point out that there are many references related to stochastic differential equations withα-stable Lévy noise, using the approach based on Hilbert-space valued solutions. We refer the reader to Section 12.5 of the monograph [12] and to [13–16] for a sample of relevant references. See also the survey article [17] for an approach based on the white noise theory for Lévy processes.This paper is organized as follows.(i)
In Section2, we review the construction of the α-stable Lévy noise Z, and we show that this can be viewed as an independently scattered random measure with jointly α-stable distributions.
(ii)
In Section3, we consider the linear equation (1) (with σ
(
u
)
=
1) and we identify the necessary and sufficient condition for the existence of the solution. This condition is verified in the case of some examples.
(iii)
Section4 contains the construction of the stochastic integral with respect to the α-stable noise Z, for α
∈
(
0,2
). The main effort is dedicated to proving a maximal inequality for the tail of the integral process, when the integrand is a simple process. This extends the construction of [1] to the case random fields and nonsymmetric measure ν
α.
(iv)
In Section5, we introduce the process Z
K obtained by removing from Z the jumps exceeding a fixed value K, and we develop a theory of integration with respect to this process. For this, we need to treat separately the cases α
<
1 and α
>
1. In both cases, we obtain a pth moment inequality for the integral process for p
∈
(
α
,
1
) if α
<
1 and p
∈
(
α
,
2
) if α
>
1. This inequality plays the same role as the Burkholder-Davis-Gundy inequality in the theory of integration with respect to continuous martingales.
(v)
In Section6 we prove the main result about the existence of the mild solution of (1). For this, we first solve the equation with “truncated” noise Z
K using a Picard iteration scheme, yielding a solution u
K. We then introduce a sequence (
τ
K
)
K
≥
1 of stopping times with τ
K
↑
∞ a.s. and we show that the solutions u
L,  L
>
K coincide on the event t
≤
τ
K. For the definition of the stopping times τ
K, we need again to consider separately the cases α
<
1 and α
>
1.
(vi)
AppendixA contains some results about the tail of a nonsymmetric stable random variable and the tail of an infinite sum of random variables. Appendix B gives an estimate for the Green function associated with the fractional power of the Laplacian. Appendix C gives a local property of the stochastic integral with respect to Z (or Z
K).


## 2. Definition of the Noise

In this section we review the construction of theα-stable Lévy noise on ℝ
+
×
ℝ
d and investigate some of its properties.LetN
=
∑
i
≥
1
δ
(
T
i
,
X
i
,
Z
i
) be a Poisson random measure on ℝ
+
×
ℝ
d
×
(
ℝ
∖
{
0
}
), defined on a probability space (
Ω
,
ℱ
,
P
), with intensity measure d
t
d
x
ν
α
(
d
z
), where ν
α is given by (3). Let (
ε
j
)
j
≥
0 be a sequence of positive real numbers such that ε
j
→
0 as j
→
∞ and 1
=
ε
0
>
ε
1
>
ε
2
>
⋯. Let
(7)
Γ
j
=
{
z
∈
ℝ
;
ε
j
<
|
z
|
≤
ε
j
-
1
}
,
j
≥
1
,
Γ
0
=
{
z
∈
ℝ
;
|
z
|
>
1
}
.For any setB
∈
ℬ
b
(
ℝ
+
×
ℝ
d
), we define
(8)
L
j
(
B
)
=
∫
B
×
Γ
j
‍
z
N
(
d
t
,
d
x
,
d
z
)
=
∑
(
T
i
,
X
i
)
∈
B
Z
i
1
{
Z
i
∈
Γ
j
}
,
j
≥
0
.Remark 1.
The variableL
0
(
B
) is finite since the sum above contains finitely many terms. To see this, we note that E
[
N
(
B
×
Γ
0
)
]
=
|
B
|
ν
α
(
Γ
0
)
<
∞, and hence N
(
B
×
Γ
0
)
=
card
{
i
≥
1
;
(
T
i
,
X
i
,
Z
i
)
∈
B
×
Γ
0
}
<
∞.For anyj
≥
0, the variable L
j
(
B
) has a compound Poisson distribution with jump intensity measure |
B
|
·
ν
α
|
Γ
j; that is,
(9)
E
[
e
i
u
L
j
(
B
)
]
=
exp
{
|
B
|
∫
Γ
j
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
u
∈
ℝ
.
It follows that E
(
L
j
(
B
)
)
=
|
B
|
∫
Γ
j
‍
z
ν
α
(
d
z
) and Var
(
L
j
(
B
)
)
=
|
B
|
∫
Γ
j
‍
z
2
ν
α
(
d
z
) for any j
≥
0. Hence, Var
(
L
j
(
B
)
)
<
∞ for any j
≥
1 and Var
(
L
0
(
B
)
)
=
∞. If α
>
1, then E
(
L
0
(
B
)
) is finite. Define
(10)
Y
(
B
)
=
∑
j
≥
1
‍
[
L
j
(
B
)
-
E
(
L
j
(
B
)
)
]
+
L
0
(
B
)
.
This sum converges a.s. by Kolmogorov’s criterion since {
L
j
(
B
)
-
E
(
L
j
(
B
)
)
}
j
≥
1 are independent zero-mean random variables with ∑
j
≥
1
‍
Var
(
L
j
(
B
)
)
<
∞.From (9) and (10), it follows that Y
(
B
) is an infinitely divisible random variable with characteristic function:
(11)
E
(
e
i
u
Y
(
B
)
)
hh
=
exp
{
|
B
|
∫
ℝ
‍
(
e
i
u
z
-
1
-
i
u
z
1
{
|
z
|
≤
1
}
)
ν
α
(
d
z
)
}
,
hhhhhhhhhhhhhhlllhhhhhhhhhhhhhhhh
u
∈
ℝ
.
Hence, E
(
Y
(
B
)
)
=
|
B
|
∫
ℝ
‍
z
1
{
|
z
|
>
1
}
ν
α
(
d
z
) and Var
(
Y
(
B
)
)
=
|
B
|
∫
ℝ
‍
z
2
ν
α
(
d
z
).Lemma 2.
The family{
Y
(
B
)
;
B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
)
} defined by (10) is an independently scattered random measure; that is,(a)
for any disjoint sets
B
1
,
…
,
B
n
in
ℬ
b
(
ℝ
+
×
ℝ
d
)
,
Y
(
B
1
)
,
…
,
Y
(
B
n
)
are independent;
(b)
for any sequence
(
B
n
)
n
≥
1
of disjoint sets in
ℬ
b
(
ℝ
+
×
ℝ
d
)
such that
⋃
n
≥
1
‍
B
n
is bounded,
Y
(
⋃
n
≥
1
‍
B
n
)
=
∑
n
≥
1
‍
Y
(
B
n
)
a.s.Proof.
(a) Note that for any functionφ
∈
L
2
(
ℝ
+
×
ℝ
d
) with compact support K, we can define the random variable Y
(
φ
)
=
∑
j
≥
1
‍
[
L
j
(
φ
)
-
E
(
L
j
(
φ
)
)
]
+
L
0
(
φ
) where L
j
(
φ
)
=
∫
K
×
Γ
j
‍
φ
(
t
,
x
)
z N
(
d
t
,
d
x
,
d
z
). For any u
∈
ℝ, we have
(12)
E
(
e
i
u
Y
(
φ
)
)
hh
=
exp
{
∫
ℝ
+
×
ℝ
d
×
ℝ
‍
(
e
i
u
z
φ
(
t
,
x
)
-
1
hhhhhhhhhhhhhhhhh
-
i
u
z
φ
(
t
,
x
)
1
{
|
z
|
≤
1
}
e
i
u
z
φ
(
t
,
x
)
)
d
t
d
x
ν
α
(
d
z
)
∫
ℝ
+
×
ℝ
d
×
ℝ
‍
}
.
For any disjoint setsB
1
,
…
,
B
n and for any u
1
,
…
,
u
n
∈
ℝ, we have
(13)
E
[
exp
(
i
∑
k
=
1
n
‍
u
k
Y
(
B
k
)
)
]
=
E
[
exp
(
i
Y
(
∑
k
=
1
n
‍
u
k
1
B
k
)
)
]
=
exp
{
∫
ℝ
+
×
ℝ
d
×
ℝ
‍
(
∑
k
=
1
n
‍
u
k
1
B
k
(
t
,
x
)
e
i
z
∑
k
=
1
n
u
k
1
B
k
(
t
,
x
)
∑
k
=
1
n
‍
u
k
1
B
k
(
t
,
x
)
hhhhhhhhhhhhlhhhh
-
1
-
i
z
1
{
|
z
|
≤
1
}
hhhhhhhhhhhhlhhhh
×
∑
k
=
1
n
‍
u
k
1
B
k
(
t
,
x
)
)
d
t
d
x
ν
α
(
d
z
)
}
=
exp
{
∑
k
=
1
n
‍
|
B
k
|
∫
ℝ
‍
(
i
u
k
z
1
{
|
z
|
≤
1
}
e
i
u
k
z
-
1
hhhhhhhhhhhhhhhhhl
-
i
u
k
z
1
{
|
z
|
≤
1
}
e
i
u
k
z
)
ν
α
(
d
z
)
∑
k
=
1
n
‍
|
B
k
|
}
=
∏
k
=
1
n
‍
E
[
exp
(
i
u
k
Y
(
B
k
)
)
]
,
using (12) with φ
=
∑
k
=
1
n
‍
u
k
1
B
k for the second equality and (9) for the last equality. This proves that Y
(
B
1
)
,
…
,
Y
(
B
n
) are independent.
(b) LetS
n
=
∑
k
=
1
n
‍
Y
(
B
k
) and S
=
Y
(
B
), where B
=
⋃
n
≥
1
‍
B
n. By Lévy’s equivalence theorem, (
S
n
)
n
≥
1 converges a.s. if and only if it converges in distribution. By (13), with u
i
=
u for all i
=
1
,
…
,
k, we have
(14)
E
(
e
i
u
S
n
)
=
exp
{
|
⋃
k
=
1
n
‍
B
k
|
∫
ℝ
‍
(
e
i
u
z
-
1
-
i
u
z
1
{
|
z
|
≤
1
}
)
ν
α
(
d
z
)
}
.
This clearly converges to E
(
e
i
u
S
)
=
exp
{
|
B
|
∫
ℝ
‍
(
e
i
u
z
-
1
-
i
u
z
1
{
|
z
|
≤
1
}
)
ν
α
(
d
z
)
}, and hence (
S
n
)
n
≥
1 converges in distribution to S.Recall that a random variableX has an α
-stable distributionwith parameters α
∈
(
0,2
), σ
∈
[
0
,
∞
), β
∈
[
-
1,1
], and μ
∈
ℝ if, for any u
∈
ℝ,
(15)
E
(
e
i
u
X
)
=
exp
{
-
|
u
|
α
σ
α
(
1
-
i
sgn
(
u
)
β
tan
π
α
2
)
+
i
u
μ
}
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhl
if
α
≠
1
,
or
(16)
E
(
e
i
u
X
)
=
exp
{
-
|
u
|
σ
(
1
+
i
sgn
(
u
)
β
2
π
ln
|
u
|
)
+
i
u
μ
}
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhl
if
α
=
1
(see Definition 1.1
.
6 of [18]). We denote this distribution by S
α
(
σ
,
β
,
μ
).Lemma 3.
Y
(
B
) has a S
α
(
σ
|
B
|
1
/
α
,
β
,
μ
|
B
|
) distribution with β
=
p
-
q,
(17)
σ
α
=
∫
0
∞
‍
sin
x
x
α
d
x
=
{
Γ
(
2
-
α
)
1
-
α
cos
π
α
2
,
i
f
α
≠
1
,
π
2
,
i
f
α
=
1
,
μ
=
{
β
α
α
-
1
,
i
f
α
≠
1
,
β
c
0
,
i
f
α
=
1
,
and c
0
=
∫
0
∞
‍
(
sin
z
-
z
1
{
z
≤
1
}
)
z
-
2
d
z. If α
>
1, then E
(
Y
(
B
)
)
=
μ
|
B
|.Proof.
We first express the characteristic function (11) of Y
(
B
) in Feller’s canonical form (see Section XVII.2 of [19]):
(18)
E
(
e
i
u
Y
(
B
)
)
=
exp
{
i
u
b
|
B
|
+
|
B
|
∫
ℝ
‍
e
i
u
z
-
1
-
i
u
sin
z
z
2
M
α
(
d
z
)
}
with M
α
(
d
z
)
=
z
2
ν
α
(
d
z
) and b
=
∫
ℝ
‍
(
sin
z
-
z
1
{
|
z
|
≤
1
}
)
ν
α
(
d
z
). Then the result follows from the calculations done in Example XVII.3.(g) of [19].From Lemmas2 and 3, it follows that
(19)
Z
=
{
Z
(
B
)
=
Y
(
B
)
-
μ
|
B
|
;
B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
)
}
is an α-stable random measure, in the sense of Definition 3.3
.
1 of [18], with control measure m
(
B
)
=
σ
α
|
B
| and constant skewness intensity β. In particular, Z
(
B
) has a S
α
(
σ
|
B
|
1
/
α
,
β
,
0
) distribution.We say thatZ is an α-stable Lévy noise. Coming back to the original construction (10) of Y
(
B
) and noticing that
(20)
μ
|
B
|
=
-
|
B
|
∫
ℝ
‍
z
1
{
|
z
|
≤
1
}
ν
α
(
d
z
)
=
-
∑
j
≥
1
‍
E
(
L
j
(
B
)
)
,
hhhhhhhhhhhhhhhhhhhhhhhhhhl
if
α
<
1
,
μ
|
B
|
=
|
B
|
∫
ℝ
‍
z
1
{
|
z
|
>
1
}
ν
α
(
d
z
)
=
E
(
L
0
(
B
)
)
,
hhhhhhhhhhhhhhhhhhhhhll
if
α
>
1
,
it follows that Z
(
B
) can be represented as
(21)
Z
(
B
)
=
∑
j
≥
0
‍
L
j
(
B
)
=
:
∫
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
(
d
t
,
d
x
,
d
z
)
,
if
α
<
1
,
(22)
Z
(
B
)
=
∑
j
≥
0
‍
[
L
j
(
B
)
-
E
(
L
j
(
B
)
)
]
=
:
∫
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
^
(
d
t
,
d
x
,
d
z
)
,
if
α
>
1
.
Here N
^ is the compensated Poisson measure associated with N; that is, N
^
(
A
)
=
N
(
A
)
-
E
(
N
(
A
)
) for any relatively compact set A in ℝ
+
×
ℝ
d
×
(
ℝ
¯
∖
{
0
}
).In the caseα
=
1, we will assume that p
=
q so that ν
α is symmetric around 0, E
(
L
j
(
B
)
)
=
0 for all j
≥
1, and Z
(
B
) admits the same representation as in the case α
<
1.


## 3. The Linear Equation

As a preliminary investigation, we consider first equation (1) with σ
=
1:
(23)
L
u
(
t
,
x
)
=
Z
˙
(
t
,
x
)
,
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions. In this section 𝒪 is a bounded domain in ℝ
d or 𝒪
=
ℝ
d.By definition, the process{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} given by
(24)
u
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
Z
(
d
s
,
d
y
)
is a mild solution of (23), provided that the stochastic integral on the right-hand side of (24) is well defined.We define now the stochastic integral of a deterministic functionφ:
(25)
Z
(
φ
)
=
∫
0
∞
‍
∫
ℝ
d
‍
φ
(
t
,
x
)
Z
(
d
t
,
d
x
)
.Ifφ
∈
L
α
(
ℝ
+
×
ℝ
d
), this can be defined by approximation with simple functions, as explained in Section 3.4 of [18]. The process {
Z
(
φ
)
;
φ
∈
L
α
(
ℝ
+
×
ℝ
d
)
} has jointly α-stable finite dimensional distributions. In particular, each Z
(
φ
) has a S
α
(
σ
φ
,
β
,
0
)-distribution with scale parameter:
(26)
σ
φ
=
σ
(
∫
0
∞
‍
∫
ℝ
d
‍
|
φ
(
t
,
x
)
|
α
d
x
d
t
)
1
/
α
.More generally, a measurable functionφ
:
ℝ
+
×
ℝ
d
→
ℝ is integrable with respect to Z if there exists a sequence (
φ
n
)
n
≥
1 of simple functions such that φ
n
→
φ a.e., and, for any B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
), the sequence {
Z
(
φ
n
1
B
)
}
n converges in probability (see [20]).The next results show that conditionφ
∈
L
α
(
ℝ
+
×
ℝ
d
) is also necessary for the integrability of φ with respect to Z. Due to Lemma 2, this follows immediately from the general theory of stochastic integration with respect to independently scattered random measures developed in [20].Lemma 4.
A deterministic functionφ is integrable with respect to Z if and only if φ
∈
L
α
(
ℝ
+
×
ℝ
d
).Proof.
We write the characteristic function ofZ
(
B
) in the form used in [20]:
(27)
E
(
e
i
u
Z
(
B
)
)
hh
=
exp
{
∫
B
[
∫
ℝ
(
e
i
u
z
-
1
-
i
u
τ
(
z
)
)
ν
α
(
d
z
)
i
u
a
hhhhhhhhhhh
+
∫
ℝ
(
e
i
u
z
-
1
-
i
u
τ
(
z
)
)
ν
α
(
d
z
)
]
d
t
d
x
}
with a
=
β
-
μ, τ
(
z
)
=
z if |
z
|
≤
1 and τ
(
z
)
=
sgn
(
z
) if |
z
|
>
1. By Theorem 2.7 of   [20], φ is integrable with respect to Z if and only if
(28)
∫
ℝ
+
×
ℝ
d
‍
|
U
(
φ
(
t
,
x
)
)
|
d
t
d
x
<
∞
,
∫
ℝ
+
×
ℝ
d
‍
V
(
φ
(
t
,
x
)
)
d
t
d
x
<
∞
,
where U
(
y
)
=
a
y
+
∫
ℝ
‍
(
τ
(
y
z
)
-
y
τ
(
z
)
)
ν
α
(
d
z
) and V
(
y
)
=
∫
ℝ
‍
(
1
∧
|
y
z
|
2
)
ν
α
(
d
z
). Direct calculations show that, in our case, U
(
y
)
=
-
(
β
/
(
α
-
1
)
)
y
α if α
≠
1, U
(
y
)
=
0 if α
=
1, and V
(
y
)
=
(
2
/
(
2
-
α
)
)
y
α.The following result follows immediately from (24) and Lemma 4.Proposition 5.
Equation (23) has a mild solution if and only if for any t
>
0, x
∈
𝒪
(29)
I
α
(
t
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
s
,
x
,
y
)
α
d
y
d
s
<
∞
.
In this case, {
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} has jointly α-stable finite-dimensional distributions. In particular, u
(
t
,
x
) has a S
α
(
σ
I
α
(
t
)
1
/
α
,
β
,
0
) distribution.Condition (29) can be easily verified in the case of several examples.Example 6 (heat equation).
LetL
=
∂
/
∂
t
-
(
1
/
2
)
Δ. Assume first that 𝒪
=
ℝ
d. Then G
(
t
,
x
,
y
)
=
G
¯
(
t
,
x
-
y
), where
(30)
G
¯
(
t
,
x
)
=
1
(
2
π
t
)
d
/
2
exp
(
-
|
x
|
2
2
t
)
,
and condition (29) is equivalent to (6). In this case, I
α
(
t
)
=
c
α
,
d
t
d
(
1
-
α
)
/
2
+
1. If 𝒪 is a bounded domain in ℝ
d, then G
(
t
,
x
,
y
)
≤
G
¯
(
t
,
x
-
y
) (see page 74 of [11]) and condition (29) is implied by (6).Example 7 (parabolic equation).
LetL
=
∂
/
∂
t
-
ℒ where
(31)
ℒ
f
(
x
)
=
∑
i
,
j
=
1
d
‍
a
i
j
(
x
)
∂
2
f
∂
x
i
∂
x
j
(
x
)
+
∑
i
=
1
d
‍
b
i
(
x
)
∂
f
∂
x
i
(
x
)
is the generator of a Markov process with values in ℝ
d, without jumps (a diffusion). Assume that 𝒪 is a bounded domain in ℝ
d or 𝒪
=
ℝ
d. By Aronson estimate (see, e.g., Theorem 2.6 of [12]), under some assumptions on the coefficients a
i
j, b
i, there exist some constants c
1
,
c
2
>
0 such that
(32)
G
(
t
,
x
,
y
)
≤
c
1
t
-
d
/
2
exp
(
-
|
x
-
y
|
2
c
2
t
)
for all t
>
0 and x, y
∈
𝒪. In this case, condition (29) is implied by (6).Example 8 (heat equation with fractional power of the Laplacian).
LetL
=
∂
/
∂
t
+
(
-
Δ
)
γ for some γ
>
0. Assume that 𝒪 is a bounded domain in ℝ
d or 𝒪
=
ℝ
d. Then (see, e.g., Appendix B
.
5 of [12])
(33)
G
(
t
,
x
,
y
)
=
∫
0
∞
‍
𝒢
(
s
,
x
,
y
)
g
t
,
γ
(
s
)
d
s
=
∫
0
∞
‍
𝒢
(
t
1
/
γ
s
,
x
,
y
)
g
1
,
γ
(
s
)
d
s
,
where 𝒢
(
t
,
x
,
y
) is the fundamental solution of ∂
u
/
∂
t
-
Δ
u
=
0 on 𝒪 and g
t
,
γ is the density of the measure μ
t
,
γ, (
μ
t
,
γ
)
t
≥
0 being a convolution semigroup of measures on [
0
,
∞
) whose Laplace transform is given by
(34)
∫
0
∞
‍
e
-
u
s
g
t
,
γ
(
s
)
d
s
=
exp
(
-
t
u
γ
)
,
∀
u
>
0
.
Note that ifγ
<
1, g
t
,
γ is the density of S
t, where (
S
t
)
t
≥
0 is a γ-stable subordinator with Lévy measure ρ
γ
(
d
x
)
=
(
γ
/
Γ
(
1
-
γ
)
)
x
-
γ
-
1
1
(
0
,
∞
)
(
x
)
d
x.
Assume first that𝒪
=
ℝ
d. Then G
(
t
,
x
,
y
)
=
G
¯
(
t
,
x
-
y
), where
(35)
G
¯
(
t
,
x
)
=
∫
ℝ
d
‍
e
i
ξ
·
x
e
-
t
|
ξ
|
2
γ
d
ξ
.
Ifγ
<
1, then G
¯
(
t
,
·
) is the density of X
t, with  (
X
t
)
t
≥
0 being a symmetric (
2
γ
)-stable Lévy process with values in ℝ
d defined by X
t
=
W
S
t, with (
W
t
)
t
≥
0 a Brownian motion in ℝ
d with variance 2. By Lemma B.1 (Appendix B), if α
>
1, then (29) holds if and only if
(36)
α
<
1
+
2
γ
d
.
If𝒪 is a bounded domain in ℝ
d, then G
(
t
,
x
,
y
)
≤
G
¯
(
t
,
x
-
y
) (by Lemma 2.1 of [8]). In this case, if α
>
1, then (29) is implied by (36).Example 9 (cable equation inℝ).
LetL
u
=
∂
u
/
∂
t
-
∂
2
u
/
∂
x
2
+
u and 𝒪
=
ℝ. Then G
(
t
,
x
,
y
)
=
G
¯
(
t
,
x
-
y
), where
(37)
G
¯
(
t
,
x
)
=
1
4
π
t
exp
(
-
|
x
|
2
4
t
-
t
)
,
and condition (29) holds for any α
∈
(
0,2
).Example 10 (wave equation inℝ
d with d
=
1,2).
LetL
=
∂
2
/
∂
t
2
-
Δ and 𝒪
=
ℝ
d with d
=
1 or d
=
2. Then G
(
t
,
x
,
y
)
=
G
¯
(
t
,
x
-
y
), where
(38)
G
¯
(
t
,
x
)
=
1
2
1
{
|
x
|
<
t
}
,
if
d
=
1
,
G
¯
(
t
,
x
)
=
1
2
π
·
1
t
2
-
|
x
|
2
1
{
|
x
|
<
t
}
,
if
d
=
2
.
Condition (29) holds for any α
∈
(
0,2
). In this case, I
α
(
t
)
=
2
-
α
t
2 if d
=
1 and I
α
(
t
)
=
(
(
2
π
)
1
-
α
/
(
2
-
α
)
(
3
-
α
)
)
t
3
-
α if d
=
2.


## 4. Stochastic Integration

In this section we construct a stochastic integral with respect toZ by generalizing the ideas of [1] to the case of random fields. Unlike these authors, we do not assume that Z
(
B
) has a symmetric distribution, unless α
=
1.Letℱ
t
=
ℱ
t
N
∨
𝒩 where 𝒩 is the σ-field of negligible sets in (
Ω
,
ℱ
,
P
) and ℱ
t
N is the σ-field generated by N
(
[
0
,
s
]
×
A
×
Γ
) for all s
∈
[
0
,
t
],  A
∈
ℬ
b
(
ℝ
d
) and for all Borel sets Γ
⊂
ℝ
∖
{
0
} bounded away from 0. Note that ℱ
t
Z
⊂
ℱ
t
N where ℱ
t
Z is the σ-field generated by Z
(
[
0
,
s
]
×
A
),  s
∈
[
0
,
t
],  and  A
∈
ℬ
b
(
ℝ
d
).A processX
=
{
X
(
t
,
x
)
}
t
≥
0
,
x
∈
ℝ
d is called elementary if it is of the form
(39)
X
(
t
,
x
)
=
1
(
a
,
b
]
(
t
)
1
A
(
x
)
Y
,
where 0
≤
a
<
b, A
∈
ℬ
b
(
ℝ
d
), and Y is ℱ
a-measurable and bounded. A simple process is a linear combination of elementary processes. Note that any simple process X can be written as
(40)
X
(
t
,
x
)
=
1
{
0
}
(
t
)
Y
0
(
x
)
+
∑
i
=
0
N
-
1
‍
1
(
t
i
,
t
i
+
1
]
(
t
)
Y
i
(
x
)
with 0
=
t
0
<
t
1
<
⋯
<
t
N
<
∞ and Y
i
(
x
)
=
∑
j
=
1
m
i
‍
1
A
i
j
(
x
)
Y
i
j, where (
Y
i
j
)
j
=
1
,
…
,
m
i are ℱ
t
i-measurable and (
A
i
j
)
j
=
1
,
…
,
m
j are disjoint sets in ℬ
b
(
ℝ
d
). Without loss of generality, we assume that Y
0
=
0.We denote by𝒫 the predictable
σ
-field on Ω
×
ℝ
+
×
ℝ
d, that is, the σ-field generated by all simple processes. We say that a process X
=
{
X
(
t
,
x
)
}
t
≥
0
,
x
∈
ℝ
d is predictable if the map (
ω
,
t
,
x
)
↦
X
(
ω
,
t
,
x
) is 𝒫-measurable.Remark 11.
One can show that the predictableσ-field 𝒫 is the σ-field generated by the class 𝒞 of processes X such that t
↦
X
(
ω
,
t
,
x
) is left continuous for any ω
∈
Ω,  x
∈
ℝ
d and (
ω
,
x
)
↦
X
(
ω
,
t
,
x
) is ℱ
t
×
ℬ
(
ℝ
d
)-measurable for any t
>
0.Letℒ
α be the class of all predictable processes X such that
(41)
∥
X
∥
α
,
T
,
B
α
:
=
E
∫
0
T
‍
∫
B
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
<
∞
,
for all T
>
0 and B
∈
ℬ
b
(
ℝ
d
). Note that ℒ
α is a linear space.Let(
E
k
)
k
≥
1 be an increasing sequence of sets in ℬ
b
(
ℝ
d
) such that ⋃
k
‍
E
k
=
ℝ
d. We define
(42)
∥
X
∥
α
=
∑
k
≥
1
‍
1
∧
∥
X
∥
α
,
k
,
E
k
2
k
,
if
α
>
1
,
∥
X
∥
α
α
=
∑
k
≥
1
‍
1
∧
∥
X
∥
α
,
k
,
E
k
α
2
k
,
if
α
≤
1
.We identify two processesX and Y for which ∥
X
-
Y
∥
α
=
0; that is, X
=
Y
ν a.e., where ν
=
P
d
t
d
x. In particular, we identify two processes X and Y if X is a modification of Y; that is, X
(
t
,
x
)
=
Y
(
t
,
x
) a.s. for all (
t
,
x
)
∈
ℝ
+
×
ℝ
d.The spaceℒ
α becomes a metric space endowed with the metric d
α:
(43)
d
α
(
X
,
Y
)
=
∥
X
-
Y
∥
α
,
if
α
>
1
,
d
α
(
X
,
Y
)
=
∥
X
-
Y
∥
α
α
,
if
α
≤
1
.
This follows using Minkowski’s inequality if α
>
1 and the inequality |
a
+
b
|
α
≤
|
a
|
α
+
|
b
|
α if α
≤
1.The following result can be proved similarly to Proposition 2.3 of [21].Proposition 12.
For anyX
∈
ℒ
α there exists a sequence (
X
n
)
n
≥
1 of bounded simple processes such that ∥
X
n
-
X
∥
α
→
0 as n
→
∞.By Proposition 5.7 of [22], the α-stable Lévy process {
Z
(
t
,
B
)
=
Z
(
[
0
,
t
]
×
B
)
;
t
≥
0
} has a càdlàg modification, for any B
∈
ℬ
b
(
ℝ
d
). We work with these modifications. If X is a simple process given by (40), we define
(44)
I
(
X
)
(
t
,
B
)
=
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
Y
i
j
Z
(
(
t
i
∧
t
,
t
i
+
1
∧
t
]
×
(
A
i
j
∩
B
)
)
.
Note that, for any B
∈
ℬ
b
(
ℝ
d
), I
(
X
)
(
t
,
B
) is ℱ
t-measurable for any t
≥
0, and {
I
(
X
)
(
t
,
B
)
}
t
≥
0 is càdlàg. We write
(45)
I
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
(
d
s
,
d
x
)
.The following result will be used for the construction of the integral. This result generalizes Lemma 3.3 of [1] to the case of random fields and nonsymmetric measures ν
α.Theorem 13.
IfX is a bounded simple process then
(46)
sup
λ
>
0
λ
α
P
(
sup
t
∈
[
0
,
T
]
|
I
(
X
)
(
t
,
B
)
|
>
λ
)
hhhh
≤
c
α
E
∫
0
T
‍
∫
B
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
,
for any T
>
0 and B
∈
ℬ
b
(
ℝ
d
), where c
α is a constant depending only on α.Proof.
Suppose thatX is of the form (40). Since {
I
(
X
)
(
t
,
B
)
}
t
∈
[
0
,
T
] is càdlàg, it is separable. Without loss of generality, we assume that its separating set D can be written as D
=
∪
n
F
n where (
F
n
)
n is an increasing sequence of finite sets containing the points (
t
k
)
k
=
0
,
…
,
N. Hence,
(47)
P
(
sup
t
∈
[
0
,
T
]
|
I
(
X
)
(
t
,
B
)
|
>
λ
)
=
lim
n
→
∞
P
(
max
t
∈
F
n
|
I
(
X
)
(
t
,
B
)
|
>
λ
)
.
Fixn
≥
1. Denote by 0
=
s
0
<
s
1
<
⋯
<
s
m
=
T the points of the set F
n. Say t
k
=
s
i
k for some 0
=
i
0
<
i
1
<
⋯
<
i
N. Then each interval (
t
k
,
t
k
+
1
] can be written as the union of some intervals of the form (
s
i
,
s
i
+
1
]:
(48)
(
t
k
,
t
k
+
1
]
=
⋃
i
∈
I
k
‍
(
s
i
,
s
i
+
1
]
,
where I
k
=
{
i
;
i
k
≤
i
<
i
k
+
1
}. By (44), for any k
=
0
,
…
,
N
-
1 and i
∈
I
k,
(49)
I
(
X
)
(
s
i
+
1
,
B
)
-
I
(
X
)
(
s
i
,
B
)
=
∑
j
=
1
m
k
‍
Y
k
j
Z
(
(
s
i
,
s
i
+
1
]
×
(
A
k
j
∩
B
)
)
.
For anyi
∈
I
k, let N
i
=
m
k, and, for any j
=
1
,
…
,
N
i, define β
i
j
=
Y
k
j, H
i
j
=
A
k
j, and Z
i
j
=
Z
(
(
s
i
,
s
i
+
1
]
×
(
H
i
j
∩
B
)
). With this notation, we have
(50)
I
(
X
)
(
s
i
+
1
,
B
)
-
I
(
X
)
(
s
i
,
B
)
=
∑
j
=
1
N
i
‍
β
i
j
Z
i
j
,
∀
i
=
0
,
…
,
m
.
Consequently, for any l
=
1
,
…
,
m
(51)
I
(
X
)
(
s
l
,
B
)
=
∑
i
=
0
l
-
1
‍
(
I
(
X
)
(
s
i
+
1
,
B
)
-
I
(
X
)
(
s
i
,
B
)
)
=
∑
i
=
0
l
-
1
‍
∑
j
=
1
N
i
‍
β
i
j
Z
i
j
.
Using (47) and (51), it is enough to prove that for any λ
>
0,
(52)
P
(
max
l
=
0
,
…
,
m
-
1
|
∑
i
=
0
l
‍
∑
j
=
1
N
i
‍
β
i
j
Z
i
j
|
>
λ
)
≤
c
α
λ
-
α
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
α
d
x
d
s
.
First, note that(53)
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
α
d
x
d
s
=
∑
i
=
0
m
-
1
‍
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
.
This follows from the definition (40) of X and (48), since X
(
t
,
x
)
=
∑
i
=
0
N
-
1
‍
∑
i
∈
I
k
‍
1
(
s
i
,
s
i
+
1
]
(
t
)
∑
j
=
1
N
i
‍
β
i
j
1
H
i
j
(
x
).
We now prove (52). Let W
i
=
∑
j
=
1
N
i
‍
β
i
j
Z
i
j. For the event on the left-hand side, we consider its intersection with the event {
max
0
≤
i
≤
m
-
1
|
W
i
|
>
λ
} and its complement. Hence, the probability of this event can be bounded by
(54)
∑
i
=
0
m
-
1
‍
P
(
|
W
i
|
>
λ
)
+
P
(
max
0
≤
l
≤
m
-
1
|
∑
i
=
0
l
‍
W
i
1
{
|
W
i
|
≤
λ
}
|
>
λ
)
=
:
I
+
I
I
.
We treat separately the two terms.
For the first term, we note thatβ
¯
i
=
(
β
i
j
)
1
≤
j
≤
N
i is ℱ
s
i-measurable and Z
¯
i
=
(
Z
i
j
)
1
≤
j
≤
N
i is independent of ℱ
s
i. By Fubini’s theorem
(55)
I
=
∑
i
=
0
m
-
1
‍
∫
ℝ
N
i
‍
P
(
|
∑
j
=
1
N
i
‍
x
j
Z
i
j
|
>
λ
)
P
β
¯
i
(
d
x
¯
)
,
where x
¯
=
(
x
j
)
1
≤
j
≤
N
i and P
β
¯
i is the law of β
¯
i.
We examine the tail ofU
i
=
∑
j
=
1
N
i
‍
x
j
Z
i
j for a fixed x
¯
∈
ℝ
N
i. By Lemma 3, Z
i
j has a S
α
(
σ
(
s
i
+
1
-
s
i
)
1
/
α
|
H
i
j
∩
B
|
1
/
α
,
β
,
0
) distribution. Since the sets (
H
i
j
)
1
≤
j
≤
N
i are disjoint, the variables (
Z
i
j
)
1
≤
j
≤
N
i are independent. Using elementary properties of the stable distribution (Properties 1.2.1 and 1.2.3 of [18]), it follows that U
i has a S
α
(
σ
i
,
β
i
*
,
0
) distribution with parameters:
(56)
σ
i
α
=
σ
α
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
,
β
i
*
=
β
∑
j
=
1
N
i
|
x
j
|
α
|
H
i
j
∩
B
|
∑
j
=
1
N
i
‍
sgn
(
x
j
)
|
x
j
|
α
|
H
i
j
∩
B
|
.
By Lemma A.1 (Appendix A), there exists a constant c
α
*
>
0 such that
(57)
P
(
|
U
i
|
>
λ
)
≤
c
α
*
λ
-
α
σ
α
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
for any λ
>
0. Hence,
(58)
I
≤
c
α
*
λ
-
α
σ
α
∑
i
=
0
m
-
1
‍
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
=
c
α
*
λ
-
α
σ
α
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
α
d
x
d
s
.
We now treatI
I. We consider three cases. For the first two cases we deviate from the original argument of [1] since we do not require that β
=
0.
Case
1
(
α
<
1
). Note that
(59)
I
I
≤
P
(
max
0
≤
l
≤
m
-
1
M
l
>
λ
)
,
where {
M
l
=
∑
i
=
0
l
‍
|
W
i
|
1
{
|
W
i
|
≤
λ
}
,
ℱ
s
l
+
1
;
0
≤
l
≤
m
-
1
} is a submartingale. By the submartingale maximal inequality (Theorem 35.3 of [23]),
(60)
P
(
max
0
≤
l
≤
m
-
1
M
l
>
λ
)
≤
1
λ
E
(
M
m
-
1
)
=
1
λ
∑
i
=
0
m
-
1
‍
E
(
|
W
i
|
1
|
W
i
|
≤
λ
)
.
Using the independence betweenβ
¯
i and Z
¯
i it follows that
(61)
E
[
|
W
i
|
1
|
W
i
|
≤
λ
]
=
∫
ℝ
N
i
‍
E
[
|
∑
j
=
1
N
i
‍
x
j
Z
i
j
|
1
{
|
∑
j
=
1
N
i
x
j
Z
i
j
|
≤
λ
}
]
P
β
¯
i
(
d
x
¯
)
.
Let U
i
=
∑
j
=
1
N
i
‍
x
j
Z
i
j. Using (57) and Remark A.2 (Appendix A), we get
(62)
E
[
|
U
i
|
1
{
|
U
i
|
≤
λ
}
]
≤
c
α
*
σ
α
1
1
-
α
λ
1
-
α
(
s
i
+
1
-
s
i
)
×
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
.
Hence,
(63)
E
[
|
W
i
|
1
|
W
i
|
≤
λ
]
≤
c
α
*
σ
α
1
1
-
α
λ
1
-
α
(
s
i
+
1
-
s
i
)
×
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
.
From (59), (60), and (63), it follows that
(64)
I
I
≤
c
α
*
σ
α
1
1
-
α
λ
-
α
E
∫
0
T
‍
∫
B
|
X
(
s
,
x
)
|
α
‍
d
x
d
s
.
Case
2  (
α
>
1
). We have
(65)
I
I
≤
P
(
max
0
≤
l
≤
m
-
1
|
∑
i
=
0
l
‍
X
i
|
>
λ
2
)
+
P
(
max
0
≤
l
≤
m
-
1
Y
i
>
λ
2
)
=
:
I
I
′
+
I
I
′′
,
where X
i
=
W
i
1
{
|
W
i
|
≤
λ
}
-
E
[
W
i
1
{
|
W
i
|
≤
λ
}
∣
ℱ
s
i
] and Y
i
=
|
E
[
W
i
1
{
|
W
i
|
≤
λ
}
∣
ℱ
s
i
]
|.
We first treat the termI
I
′. Note that {
M
l
=
∑
i
=
0
l
‍
X
i
,
ℱ
s
l
+
1
;
0
≤
l
≤
m
-
1
} is a zero-mean square integrable martingale, and
(66)
I
I
′
=
P
(
max
0
≤
l
≤
m
-
1
|
M
l
|
>
λ
2
)
≤
4
λ
2
∑
i
=
0
m
-
1
‍
E
(
X
i
2
)
≤
4
λ
2
∑
i
=
0
m
-
1
‍
E
[
W
i
2
1
{
|
W
i
|
≤
λ
}
]
.
Let U
i
=
∑
j
=
1
N
i
‍
x
j
Z
i
j. Using (57) and Remark A.2 (Appendix A), we get
(67)
E
[
U
i
2
1
{
|
U
i
|
≤
λ
}
]
≤
2
c
α
*
σ
α
1
2
-
α
λ
2
-
α
(
s
i
+
1
-
s
i
)
×
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
.
As in Case 1, we obtain that
(68)
E
[
W
i
2
1
{
|
W
i
|
≤
λ
}
]
≤
c
α
*
σ
α
2
2
-
α
λ
2
-
α
(
s
i
+
1
-
s
i
)
×
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
,
and hence
(69)
I
I
′
≤
8
c
α
*
σ
α
1
2
-
α
λ
-
α
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
α
d
x
d
s
.
We now treatI
I
′′. Note that {
N
l
=
∑
i
=
0
l
‍
Y
i
,
ℱ
s
l
+
1
;
0
≤
l
≤
m
-
1
} is a semimartingale and hence, by the submartingale inequality,
(70)
I
I
′′
≤
2
λ
E
(
N
m
-
1
)
=
2
λ
∑
i
=
0
m
-
1
‍
E
(
Y
i
)
.
To evaluate E
(
Y
i
), we note that, for almost all ω
∈
Ω,
(71)
E
[
W
i
1
{
|
W
i
|
≤
λ
}
∣
ℱ
s
i
]
(
ω
)
=
E
[
∑
j
=
1
N
i
‍
β
i
j
(
ω
)
Z
i
j
1
{
|
∑
j
=
1
N
i
β
i
j
(
ω
)
Z
i
j
|
≤
λ
}
]
,
due to the independence between β
¯
i and Z
¯
i. We let U
i
=
∑
j
=
1
N
i
‍
x
j
Z
i
j with x
j
=
β
i
j
(
ω
). Since α
>
1, E
(
U
i
)
=
0. Using (57) and Remark A.2, we obtain
(72)
|
E
[
U
i
1
{
|
U
i
|
≤
λ
}
]
|
=
|
E
[
U
i
1
{
|
U
i
|
>
λ
}
]
|
≤
E
[
|
U
i
|
1
{
|
U
i
|
>
λ
}
]
≤
c
α
*
σ
α
α
α
-
1
λ
1
-
α
(
s
i
+
1
-
s
i
)
×
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
.
Hence, E
(
Y
i
)
≤
c
α
*
σ
α
(
α
/
(
α
-
1
)
)
λ
1
-
α
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
| and
(73)
I
I
′′
≤
c
α
*
σ
α
2
α
α
-
1
λ
-
α
E
∫
0
T
‍
∫
B
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
.
Case
3  (
α
=
1
). In this case we assume that β
=
0. Hence, U
i
=
∑
j
=
1
N
i
‍
x
j
Z
i
j has a symmetric distribution for any x
¯
∈
ℝ
N
i. Using (71), it follows that E
[
W
i
1
{
|
W
i
|
≤
λ
}
∣
ℱ
s
i
]
=
0 a.s. for all i
=
0
,
…
,
m
-
1. Hence, {
M
l
=
∑
i
=
0
l
‍
W
i
1
{
|
W
i
|
≤
λ
}
,
ℱ
s
l
+
1
;
0
≤
l
≤
m
-
1
} is a zero-mean square integrable martingale. By the martingale maximal inequality,
(74)
I
I
≤
1
λ
2
E
[
M
m
-
1
2
]
=
1
λ
2
∑
i
=
0
m
-
1
‍
E
[
W
i
2
1
{
|
W
i
|
≤
λ
}
]
.
The result follows using (68).We now proceed to the construction of the stochastic integral. IfY
=
{
Y
(
t
)
}
t
≥
0 is a jointly measurable random process, we define
(75)
∥
Y
∥
α
,
T
α
=
sup
λ
>
0
λ
α
P
(
sup
t
∈
[
0
,
T
]
|
Y
(
t
)
|
>
λ
)
.LetX
∈
ℒ
α be arbitrary. By Proposition 12, there exists a sequence (
X
n
)
n
≥
1 of simple functions such that ∥
X
n
-
X
∥
α
→
0 as n
→
∞. Let T
>
0 and B
∈
ℬ
b
(
ℝ
d
) be fixed. By linearity of the integral and Theorem 13,
(76)
∥
I
(
X
n
)
(
·
,
B
)
-
I
(
X
m
)
(
·
,
B
)
∥
α
,
T
α
≤
c
α
∥
X
n
-
X
m
∥
α
,
T
,
B
α
⟶
0
,
as n
,
m
→
∞. In particular, the sequence {
I
(
X
n
)
(
·
,
B
)
}
n is Cauchy in probability in the space D
[
0
,
T
] equipped with the sup-norm. Therefore, there exists a random element Y
(
·
,
B
) in D
[
0
,
T
] such that, for any λ
>
0,
(77)
P
(
sup
t
∈
[
0
,
T
]
|
I
(
X
n
)
(
t
,
B
)
-
Y
(
t
,
B
)
|
>
λ
)
⟶
0
.
Moreover, there exists a subsequence (
n
k
)
k such that
(78)
sup
t
∈
[
0
,
T
]
|
I
(
X
n
k
)
(
t
,
B
)
-
Y
(
t
,
B
)
|
⟶
0
a.s.
as k
→
∞. Hence, Y
(
t
,
B
) is ℱ
t-measurable for any t
∈
[
0
,
T
]. The process Y
(
·
,
B
) does not depend on the sequence (
X
n
)
n and can be extended to a càdlàg process on [
0
,
∞
), which is unique up to indistinguishability. We denote this extension by I
(
X
)
(
·
,
B
) and we write
(79)
I
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
(
d
s
,
d
x
)
.
If A and B are disjoint sets in ℬ
b
(
ℝ
d
), then
(80)
I
(
X
)
(
t
,
A
∪
B
)
=
I
(
X
)
(
t
,
A
)
+
I
(
X
)
(
t
,
B
)
a
.
s
.Lemma 14.
Inequality (46) holds for any X
∈
ℒ
α.Proof.
Let(
X
n
)
n be a sequence of simple functions such that ∥
X
n
-
X
∥
α
→
0. For fixed B, we denote I
(
X
)
=
I
(
X
)
(
·
,
B
). We let ∥
·
∥
∞ be the sup-norm on D
[
0
,
T
]. For any ɛ
>
0, we have
(81)
P
(
∥
I
(
X
)
∥
∞
>
λ
)
≤
P
(
∥
I
(
X
)
-
I
(
X
n
)
∥
∞
>
λ
ɛ
)
+
P
(
∥
I
(
X
n
)
∥
∞
>
λ
(
1
-
ɛ
)
)
.
Multiplying by λ
α and using Theorem 13, we obtain
(82)
sup
λ
>
0
λ
α
P
(
∥
I
(
X
)
∥
∞
>
λ
)
≤
ε
-
α
sup
λ
>
0
λ
α
P
(
∥
I
(
X
)
-
I
(
X
n
)
∥
∞
>
λ
)
+
(
1
-
ɛ
)
-
α
c
α
∥
X
n
∥
α
,
T
,
B
α
.
Let n
→
∞. Using (76) one can prove that sup
λ
>
0
λ
α
P
(
∥
I
(
X
n
)
-
I
(
X
)
∥
∞
>
λ
)
→
0. We obtain that sup
λ
>
0
λ
α
P
(
∥
I
(
X
)
∥
∞
>
λ
)
≤
(
1
-
ɛ
)
-
α
c
α
∥
X
∥
α
,
T
,
B
α. The conclusion follows letting ɛ
→
0.For an arbitrary Borel set𝒪
⊂
ℝ
d (possibly 𝒪
=
ℝ
d), we assume, in addition, that X
∈
ℒ
α satisfies the condition:
(83)
E
∫
0
T
‍
∫
𝒪
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
<
∞
,
∀
T
>
0
.
Then we can define I
(
X
)
(
·
,
𝒪
) as follows. Let 𝒪
k
=
𝒪
∩
E
k where (
E
k
)
k is an increasing sequence of sets in ℬ
b
(
ℝ
d
) such that ⋃
k
‍
E
k
=
ℝ
d. By (80), Lemma 14, and (83),
(84)
sup
λ
>
0
λ
α
P
(
sup
t
≤
T
|
I
(
X
)
(
t
,
𝒪
k
)
-
I
(
X
)
(
t
,
𝒪
l
)
|
>
λ
)
≤
c
α
E
∫
0
T
‍
∫
𝒪
k
∖
𝒪
l
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
⟶
0
,
as k
,
l
→
∞. This shows that {
I
(
X
)
(
·
,
𝒪
k
)
}
k is a Cauchy sequence in probability in the space D
[
0
,
T
] equipped with the sup-norm. We denote by I
(
X
)
(
·
,
𝒪
) its limit. As above, this process can be extended to [
0
,
∞
) and I
(
X
)
(
t
,
𝒪
) is ℱ
t-measurable for any t
>
0. We denote
(85)
I
(
X
)
(
t
,
𝒪
)
=
∫
0
t
‍
∫
𝒪
‍
X
(
s
,
x
)
Z
(
d
s
,
d
x
)
.
Similarly, to Lemma 14, one can prove that, for any X
∈
ℒ
α satisfying (83),
(86)
sup
λ
>
0
λ
α
P
(
sup
t
≤
T
|
I
(
X
)
(
t
,
𝒪
)
|
>
λ
)
≤
c
α
E
∫
0
T
‍
∫
𝒪
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
.


## 5. The Truncated Noise

For the study of nonlinear equations, we need to develop a theory of stochastic integration with respect to another processZ
K which is defined by removing from Z the jumps whose modulus exceeds a fixed value K
>
0. More precisely, for any B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
), we define
(87)
Z
K
(
B
)
=
∫
B
×
{
0
<
|
z
|
≤
K
}
‍
z
N
(
d
s
,
d
x
,
d
z
)
,
if
α
≤
1
,
(88)
Z
K
(
B
)
=
∫
B
×
{
0
<
|
z
|
≤
K
}
‍
z
N
^
(
d
s
,
d
x
,
d
z
)
,
if
α
>
1
.We treat separately the casesα
≤
1 and α
>
1.


### 5.1. The Caseα
≤
1

Note that{
Z
K
(
B
)
;
B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
)
} is an independently scattered random measure on ℝ
+
×
ℝ
d with characteristic function given by
(89)
E
(
e
i
u
Z
K
(
B
)
)
=
exp
{
|
B
|
∫
|
z
|
≤
K
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
∀
u
∈
ℝ
.We first examine the tail ofZ
K
(
B
).Lemma 15.
For any setB
∈
ℬ
b
(
ℝ
+
×
ℝ
d
),
(90)
sup
λ
>
0
λ
α
P
(
|
Z
K
(
B
)
|
>
λ
)
≤
r
α
|
B
|
,
where r
α
>
0 is a constant depending only on α (given by Lemma A.3).Proof.
This follows from Example 3.7 of  [1]. We denote by ν
α
,
K the restriction of ν
α to {
z
∈
ℝ
;
0
<
|
z
|
≤
K
}. Note that
(91)
ν
α
,
K
(
{
z
∈
ℝ
;
|
z
|
>
t
}
)
=
{
t
-
α
-
K
-
α
,
if
0
<
t
≤
K
,
0
,
if
t
>
K
,
and hence sup
t
>
0
t
α
ν
α
,
K
(
{
z
∈
ℝ
;
|
z
|
>
t
}
)
=
1. Next we observe that we do not need to assume that the measure ν
α
,
K is symmetric since we use a modified version of Lemma 2.1 of [24] given by Lemma A.3 (Appendix A).In fact, since the tail ofν
α
,
K vanishes if t
>
K, we can obtain another estimate for the tail of Z
K
(
B
) which, together with (90), will allow us to control its pth moment for p
∈
(
α
,
1
). This new estimate is given below.Lemma 16.
Ifα
<
1, then
(92)
P
(
|
Z
K
(
B
)
|
>
u
)
≤
α
1
-
α
K
1
-
α
|
B
|
u
-
1
,
∀
u
>
K
.
If α
=
1, then P
(
|
Z
K
(
B
)
|
>
u
)
≤
K
|
B
|
u
-
2 for all u
>
K.Proof.
We use the same idea as in Example 3.7 of [1]. For each k
≥
1, let Z
k
,
K
(
B
) be a random variable with characteristic function:
(93)
E
(
e
i
u
Z
k
,
K
(
B
)
)
=
exp
{
|
B
|
∫
{
k
-
1
<
|
z
|
≤
K
}
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
.
Since {
Z
k
,
K
(
B
)
}
k converges in distribution to Z
K
(
B
), it suffices to prove the lemma for Z
k
,
K
(
B
). Let μ
k be the restriction of ν
α to {
z
;
k
-
1
<
|
z
|
≤
K
}. Since μ
k is finite, Z
k
,
K
(
B
) has a compound Poisson distribution with
(94)
P
(
|
Z
k
,
K
(
B
)
|
>
u
)
=
e
-
|
B
|
μ
k
(
ℝ
)
∑
n
≥
0
‍
|
B
|
n
n
!
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
,
where μ
k
*
n denotes the n-fold convolution. Note that
(95)
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
=
[
μ
k
(
ℝ
)
]
n
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
,
where (
η
i
)
i
≥
1 are i.i.d. random variables with law μ
k
/
μ
k
(
ℝ
).
Assume first thatα
<
1. To compute P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
) we consider the intersection with the event {
max
1
≤
i
≤
n
|
η
i
|
>
u
} and its complement. Note that P
(
|
η
i
|
>
u
)
=
0 for any u
>
K. Using this fact and Markov’s inequality, we obtain that, for any u
>
K,
(96)
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
≤
P
(
|
∑
i
=
1
n
‍
η
i
1
{
|
η
i
|
≤
u
}
|
>
u
)
≤
1
u
∑
i
=
1
n
‍
E
(
|
η
i
|
1
{
|
η
i
|
≤
u
}
)
.
Note that P
(
|
η
i
|
>
s
)
≤
(
s
-
α
-
K
-
α
)
/
μ
k
(
ℝ
) if s
≤
K. Hence, for any u
>
K
(97)
E
(
|
η
i
|
1
{
|
η
i
|
≤
u
}
)
≤
∫
0
u
‍
P
(
|
η
i
|
>
s
)
d
s
=
∫
0
K
‍
P
(
|
η
i
|
>
s
)
d
s
≤
1
μ
k
(
ℝ
)
α
1
-
α
K
1
-
α
.
Combining all these facts, we get that for any u
>
K
(98)
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
≤
[
μ
k
(
ℝ
)
]
n
-
1
α
1
-
α
K
1
-
α
n
u
-
1
,
and the conclusion follows from (94).
Assume now thatα
=
1. In this case, E
(
η
i
1
{
|
η
i
|
≤
u
}
)
=
0 since η
i has a symmetric distribution. Using Chebyshev’s inequality this time, we obtain
(99)
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
≤
P
(
|
∑
i
=
1
n
‍
η
i
1
{
|
η
i
|
≤
u
}
|
>
u
)
≤
1
u
2
∑
i
=
1
n
‍
E
(
η
i
2
1
{
|
η
i
|
≤
u
}
)
.
The result follows as above using the fact that, for any u
>
K,
(100)
E
(
η
i
2
1
{
|
η
i
|
≤
u
}
)
≤
2
∫
0
u
‍
s
P
(
|
η
i
|
>
s
)
d
s
=
2
∫
0
K
‍
s
P
(
|
η
i
|
>
s
)
d
s
≤
1
μ
k
(
ℝ
)
K
.Lemma 17.
Ifα
<
1 then
(101)
E
|
Z
K
(
B
)
|
p
≤
C
α
,
p
K
p
-
α
|
B
|
f
o
r
a
n
y
p
∈
(
α
,
1
)
,
where C
α
,
p is a constant depending on α and p. If α
=
1, then
(102)
E
|
Z
K
(
B
)
|
p
≤
C
p
K
p
-
1
|
B
|
f
o
r
a
n
y
p
∈
(
1,2
)
,
where C
p is a constant depending on p.Proof.
Note that(103)
E
|
Z
K
(
B
)
|
p
=
∫
0
∞
‍
P
(
|
Z
K
(
B
)
|
p
>
t
)
d
t
=
p
∫
0
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
.
We consider separately the integrals for u
≤
K and u
>
K. For the first integral we use (90):
(104)
∫
0
K
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
r
α
|
B
|
∫
0
K
‍
u
-
α
+
p
-
1
d
u
=
r
α
|
B
|
1
p
-
α
K
p
-
α
.
For the second one we use Lemma 16: if α
<
1 then
(105)
∫
K
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
α
1
-
α
K
1
-
α
|
B
|
∫
K
∞
‍
u
p
-
2
d
u
=
α
(
1
-
α
)
(
1
-
p
)
|
B
|
K
p
-
α
,
and if α
=
1, then
(106)
∫
K
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
K
|
B
|
∫
K
∞
‍
u
p
-
3
d
u
=
|
B
|
1
2
-
p
K
p
-
1
.We now proceed to the construction of the stochastic integral with respect toZ
K. For this, we use the same method as for Z. Note that ℱ
t
Z
K
⊂
ℱ
t, where ℱ
t
Z
K is the σ-field generated by Z
K
(
[
0
,
s
]
×
A
) for all s
∈
[
0
,
t
] and A
∈
ℬ
b
(
ℝ
d
). For any B
∈
ℬ
b
(
ℝ
d
), we will work with a càdlàg modification of the Lévy process {
Z
K
(
t
,
B
)
=
Z
K
(
[
0
,
t
]
×
B
)
;
t
≥
0
}.IfX is a simple process given by (40), we define
(107)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
by the same formula (44) with Z replaced by Z
K. The following result shows that I
K
(
X
)
(
t
,
B
) has the same tail behavior as I
(
X
)
(
t
,
B
).Proposition 18.
IfX is a bounded simple process then
(108)
sup
λ
>
0
λ
α
P
(
sup
t
∈
[
0
,
T
]
|
I
K
(
X
)
(
t
,
B
)
|
>
λ
)
≤
d
α
E
∫
0
T
‍
∫
B
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
,
for any T
>
0 and B
∈
ℬ
b
(
ℝ
d
), where d
α is a constant depending only on α.Proof.
As in the proof of Theorem13, it is enough to prove that
(109)
P
(
max
l
=
0
,
…
,
m
-
1
|
∑
i
=
0
l
‍
∑
j
=
1
N
i
‍
β
i
j
Z
i
j
*
|
>
λ
)
≤
d
α
λ
-
α
∑
i
=
0
m
-
1
‍
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
,
where Z
i
j
*
=
Z
K
(
(
s
i
,
s
i
+
1
]
×
(
H
i
j
∩
B
)
). This reduces to showing that U
i
*
=
∑
j
=
1
N
i
‍
x
j
Z
i
j
* satisfies an inequality similar to (57) for any x
¯
∈
ℝ
N
i; that is,
(110)
P
(
|
U
i
*
|
>
λ
)
≤
d
α
*
λ
-
α
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
,
for any λ
>
0, for some d
α
*
>
0. We first examine the tail of Z
i
j
*. By (90),
(111)
P
(
|
Z
i
j
*
|
>
λ
)
≤
r
α
(
s
i
+
1
-
s
i
)
K
i
j
λ
-
α
,
where K
i
j
=
|
H
i
j
∩
B
|. Letting η
i
j
=
K
i
j
-
1
/
α
Z
i
j
*, we obtain that, for any u
>
0,
(112)
P
(
|
η
i
j
|
>
u
)
≤
r
α
(
s
i
+
1
-
s
i
)
u
-
α
,
∀
j
=
1
,
…
,
N
i
.
By Lemma A.3 (Appendix A), it follows that, for any λ
>
0,
(113)
P
(
|
∑
j
=
1
N
i
‍
b
j
η
i
j
|
>
λ
)
≤
r
α
2
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
b
j
|
α
λ
-
α
,
for any sequence (
b
j
)
j
=
1
,
…
,
N
i of real numbers. Inequality (110) (with d
α
*
=
r
α
2) follows by applying this to b
j
=
x
j
K
i
j
1
/
α.In view of the previous result and Proposition12, for any process X
∈
ℒ
α, we can construct the integral
(114)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
in the same manner as I
(
X
)
(
t
,
B
), and this integral satisfies (108). If in addition the process X
∈
ℒ
α satisfies (83), then we can define the integral I
K
(
X
)
(
t
,
𝒪
) for an arbitrary Borel set 𝒪
⊂
ℝ
d (possibly 𝒪
=
ℝ
d). This integral will satisfy an inequality similar to (108) with B replaced by 𝒪.The appealing feature ofI
K
(
X
)
(
t
,
B
) is that we can control its moments, as shown by the next result.Theorem 19.
Ifα
<
1, then for any p
∈
(
α
,
1
) and for any X
∈
ℒ
p,
(115)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
,
for any t
>
0 and B
∈
ℬ
b
(
ℝ
d
), where C
α
,
p is a constant depending on α, p. If 𝒪
⊂
ℝ
d is an arbitrary Borel set and we assume, in addition, that the process X
∈
ℒ
p satisfies
(116)
E
∫
0
T
‍
∫
𝒪
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
<
∞
,
∀
T
>
0
,
then inequality (115) holds with B replaced by 𝒪.Proof.
Consider the following steps.
Step
1. Suppose that X is an elementary process of the form (39). Then I
K
(
X
)
(
t
,
B
)
=
Y
Z
K
(
H
) where H
=
(
t
∧
a
,
t
∧
b
]
×
(
A
∩
B
). Note that Z
K
(
H
) is independent of ℱ
a. Hence, Z
K
(
H
) is independent of Y. Let P
Y denote the law of Y. By Fubini’s theorem,
(117)
E
|
Y
Z
K
(
H
)
|
p
=
p
∫
0
∞
‍
P
(
|
Y
Z
K
(
H
)
|
>
u
)
u
p
-
1
d
u
=
p
∫
ℝ
‍
(
∫
0
∞
‍
P
(
|
y
Z
K
(
H
)
|
>
u
)
u
p
-
1
d
u
)
P
Y
(
d
y
)
.
We evaluate the inner integral. We split this integral into two parts, for u
≤
K
|
y
| and u
>
K
|
y
|, respectively. For the first integral, we use (90). For the second one, we use Lemma 16. Therefore, the inner integral is bounded by
(118)
r
α
|
y
|
α
|
H
|
∫
0
K
|
y
|
‍
u
-
α
+
p
-
1
d
u
+
α
1
-
α
|
y
|
K
1
-
α
|
H
|
×
∫
K
|
y
|
∞
‍
u
p
-
2
d
u
=
C
α
,
p
′
K
p
-
α
|
y
|
p
|
H
|
,
E
|
Y
Z
K
(
H
)
|
p
≤
p
C
α
,
p
′
K
p
-
α
|
H
|
E
|
Y
|
p
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
2.   Suppose now that X is a simple process of the form (40). Then X
(
t
,
x
)
=
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
X
i
j
(
t
,
x
) where X
i
j
(
t
,
x
)
=
1
(
t
i
,
t
i
+
1
]
(
t
)
1
A
i
j
(
x
)
Y
i
j.
Using the linearity of the integral, the inequality|
a
+
b
|
p
≤
|
a
|
p
+
|
b
|
p, and the result obtained in Step 1 for the elementary processes X
i
j, we get
(119)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
E
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
|
I
K
(
X
i
j
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
∫
0
t
‍
∫
B
‍
|
X
i
j
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
3. Let X
∈
ℒ
p be arbitrary. By Proposition 12, there exists a sequence (
X
n
)
n of  bounded simple processes such that ∥
X
n
-
X
∥
p
→
0. Since α
<
p, it follows that ∥
X
n
-
X
∥
α
→
0. By the definition of I
K
(
X
)
(
t
,
B
) there exists a subsequence {
n
k
}
k such that {
I
K
(
X
n
k
)
(
t
,
B
)
}
k converges to I
K
(
X
)
(
t
,
B
) a.s. Using Fatou’s lemma and the result obtained in Step 2 (for the simple processes X
n
k), we get
(120)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
liminf
k
→
∞
E
|
I
K
(
X
n
k
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
liminf
k
→
∞
E
∫
0
t
‍
∫
B
‍
|
X
n
k
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
4. Suppose that X
∈
ℒ
p satisfies (116). Let 𝒪
k
=
𝒪
∩
E
k where (
E
k
)
k is an increasing sequence of sets in ℬ
b
(
ℝ
d
) such that ⋃
k
≥
1
‍
E
k
=
ℝ
d. By the definition of I
K
(
X
)
(
t
,
𝒪
), there exists a subsequence (
k
i
)
i such that {
I
K
(
X
)
(
t
,
𝒪
k
i
)
}
i converges to I
K
(
X
)
(
t
,
𝒪
) a.s. Using Fatou’s lemma, the result obtained in Step 3 (for B
=
𝒪
k
i) and the monotone convergence theorem, we get
(121)
E
|
I
K
(
X
)
(
t
,
𝒪
)
|
p
≤
liminf
i
→
∞
E
|
I
K
(
X
)
(
t
,
𝒪
k
i
)
|
p
≤
C
α
,
p
K
p
-
α
liminf
i
→
∞
E
∫
0
t
‍
∫
𝒪
k
i
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
𝒪
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.Remark 20.
Finding a similar moment inequality for the casesα
=
1 and p
∈
(
1,2
) remains an open problem. The argument used in Step 2 above relies on the fact that p
<
1. Unfortunately, we could not find another argument to cover the case p
>
1.


### 5.2.  The Caseα
>
1

In this case, the construction of the integral with respect toZ
K relies on an integral with respect to N
^ which exists in the literature. We recall briefly the definition of this integral. For more details, see Section 1.2.2 of [6], Section 24.2 of [25], or Section 8.7 of [12].Let𝔼
=
ℝ
d
×
(
ℝ
∖
{
0
}
) endowed with the measure μ
(
d
x
,
d
z
)
=
d
x
ν
α
(
d
z
) and let  ℬ
b
(
𝔼
) be the class of bounded Borel sets in 𝔼. For a simple process Y
=
{
Y
(
t
,
x
,
z
)
;
t
≥
0
,
(
x
,
z
)
∈
𝔼
}, the integral I
N
^
(
Y
)
(
t
,
B
) is defined in the usual way, for any t
>
0, B
∈
ℬ
b
(
𝔼
). The process I
N
^
(
Y
)
(
·
,
B
) is a (càdlàg) zero-mean square-integrable martingale with quadratic variation
(122)
[
I
N
^
(
Y
)
(
·
,
B
)
]
t
=
∫
0
t
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
N
(
d
s
,
d
x
,
d
z
)
and predictable quadratic variation
(123)
〈
I
N
^
(
Y
)
(
·
,
B
)
〉
t
=
∫
0
t
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
ν
α
(
d
z
)
d
x
d
s
.
By approximation, this integral can be extended to the class of all 𝒫
×
ℬ
(
ℝ
∖
{
0
}
)-measurable processes Y such that for any T
>
0 and B
∈
ℬ
b
(
𝔼
)
(124)
∥
Y
∥
2
,
T
,
B
2
≔
E
∫
0
T
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
ν
α
(
d
z
)
d
x
d
s
<
∞
.
The integral is a martingale with the same quadratic variations as above and has the isometry property: E
|
I
N
^
(
Y
)
(
t
,
B
)
|
2
=
∥
Y
∥
2
,
T
,
B
2. If, in addition, ∥
Y
∥
2
,
T
,
𝔼
<
∞, then the integral can be extended to 𝔼. By the Burkholder-Davis-Gundy inequality for discontinuous martingales, for any p
≥
1,
(125)
E
sup
t
≤
T
|
I
N
^
(
Y
)
(
t
,
𝔼
)
|
p
≤
C
p
E
[
I
N
^
(
Y
)
(
·
,
𝔼
)
]
T
p
/
2
.The previous inequality is not suitable for our purposes. A more convenient inequality can be obtained foranother stochastic integral, constructed for p
∈
[
1,2
] fixed, as suggested on page 293 of [6]. More precisely, one can show that, for any bounded simple process Y,
(126)
E
sup
t
≤
T
|
I
N
^
(
Y
)
(
t
,
𝔼
)
|
p
≤
C
p
E
∫
0
T
‍
∫
ℝ
d
‍
∫
ℝ
∖
{
0
}
‍
|
Y
(
t
,
x
,
z
)
|
p
ν
α
(
d
z
)
d
x
d
t
=
:
|
Y
|
p
,
T
,
𝔼
p
,
where C
p is the constant appearing in (125) (see Lemma 8.22 of [12]).By the usual procedure, the integral can be extended to the class of all𝒫
×
ℬ
(
ℝ
∖
{
0
}
)-measurable processes Y such that [
Y
]
p
,
T
,
𝔼
<
∞. The integral is defined as an element in the space L
p
(
Ω
;
D
[
0
,
T
]
) and will be denoted by
(127)
I
N
^
,
p
(
Y
)
(
t
,
𝔼
)
=
∫
0
t
‍
∫
ℝ
d
‍
∫
ℝ
∖
{
0
}
‍
Y
(
s
,
x
,
z
)
N
^
(
d
s
,
d
x
,
d
z
)
.
Its appealing feature is that it satisfies inequality (126).From now on, we fixp
∈
[
1,2
]. Based on (88), for any B
∈
ℬ
b
(
ℝ
d
), we let
(128)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
=
∫
0
t
‍
∫
B
‍
∫
{
|
z
|
≤
K
}
‍
X
(
s
,
x
)
z
N
^
(
d
s
,
d
x
,
d
z
)
,
for any predictable process X
=
{
X
(
t
,
x
)
;
t
≥
0
,
x
∈
ℝ
d
} for which the rightmost integral is well defined. Letting Y
(
t
,
x
,
z
)
=
X
(
t
,
x
)
z
1
{
0
<
|
z
|
≤
K
}, we see that this is equivalent to saying that p
>
α and X
∈
ℒ
p. By (126),
(129)
E
sup
t
≤
T
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
,
where C
α
,
p
=
C
p
α
/
(
p
-
α
). If, in addition, the process X
∈
ℒ
p satisfies (116) then (129) holds with B replaced by 𝒪, for an arbitrary Borel set 𝒪
⊂
ℝ
d.Note that (129) is the counterpart of (115) for the case α
>
1. Together, these two inequalities will play a crucial role in Section 6.Table1 summarizes all the conditions.Table 1
Conditions forI
K
(
X
)
(
t
,
B
) to be well defined.
α
<
1
α
>
1
B is bounded
X
∈
ℒ
α
X
∈
ℒ
p
for somep
∈
(
α
,
2
]
B
=
𝒪 is unbounded
X
∈
ℒ
α and X satisfies (83)
X
∈
ℒ
p and
X satisfies (116)
for somep
∈
(
α
,
2
]


## 5.1. The Caseα
≤
1

Note that{
Z
K
(
B
)
;
B
∈
ℬ
b
(
ℝ
+
×
ℝ
d
)
} is an independently scattered random measure on ℝ
+
×
ℝ
d with characteristic function given by
(89)
E
(
e
i
u
Z
K
(
B
)
)
=
exp
{
|
B
|
∫
|
z
|
≤
K
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
∀
u
∈
ℝ
.We first examine the tail ofZ
K
(
B
).Lemma 15.
For any setB
∈
ℬ
b
(
ℝ
+
×
ℝ
d
),
(90)
sup
λ
>
0
λ
α
P
(
|
Z
K
(
B
)
|
>
λ
)
≤
r
α
|
B
|
,
where r
α
>
0 is a constant depending only on α (given by Lemma A.3).Proof.
This follows from Example 3.7 of  [1]. We denote by ν
α
,
K the restriction of ν
α to {
z
∈
ℝ
;
0
<
|
z
|
≤
K
}. Note that
(91)
ν
α
,
K
(
{
z
∈
ℝ
;
|
z
|
>
t
}
)
=
{
t
-
α
-
K
-
α
,
if
0
<
t
≤
K
,
0
,
if
t
>
K
,
and hence sup
t
>
0
t
α
ν
α
,
K
(
{
z
∈
ℝ
;
|
z
|
>
t
}
)
=
1. Next we observe that we do not need to assume that the measure ν
α
,
K is symmetric since we use a modified version of Lemma 2.1 of [24] given by Lemma A.3 (Appendix A).In fact, since the tail ofν
α
,
K vanishes if t
>
K, we can obtain another estimate for the tail of Z
K
(
B
) which, together with (90), will allow us to control its pth moment for p
∈
(
α
,
1
). This new estimate is given below.Lemma 16.
Ifα
<
1, then
(92)
P
(
|
Z
K
(
B
)
|
>
u
)
≤
α
1
-
α
K
1
-
α
|
B
|
u
-
1
,
∀
u
>
K
.
If α
=
1, then P
(
|
Z
K
(
B
)
|
>
u
)
≤
K
|
B
|
u
-
2 for all u
>
K.Proof.
We use the same idea as in Example 3.7 of [1]. For each k
≥
1, let Z
k
,
K
(
B
) be a random variable with characteristic function:
(93)
E
(
e
i
u
Z
k
,
K
(
B
)
)
=
exp
{
|
B
|
∫
{
k
-
1
<
|
z
|
≤
K
}
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
.
Since {
Z
k
,
K
(
B
)
}
k converges in distribution to Z
K
(
B
), it suffices to prove the lemma for Z
k
,
K
(
B
). Let μ
k be the restriction of ν
α to {
z
;
k
-
1
<
|
z
|
≤
K
}. Since μ
k is finite, Z
k
,
K
(
B
) has a compound Poisson distribution with
(94)
P
(
|
Z
k
,
K
(
B
)
|
>
u
)
=
e
-
|
B
|
μ
k
(
ℝ
)
∑
n
≥
0
‍
|
B
|
n
n
!
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
,
where μ
k
*
n denotes the n-fold convolution. Note that
(95)
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
=
[
μ
k
(
ℝ
)
]
n
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
,
where (
η
i
)
i
≥
1 are i.i.d. random variables with law μ
k
/
μ
k
(
ℝ
).
Assume first thatα
<
1. To compute P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
) we consider the intersection with the event {
max
1
≤
i
≤
n
|
η
i
|
>
u
} and its complement. Note that P
(
|
η
i
|
>
u
)
=
0 for any u
>
K. Using this fact and Markov’s inequality, we obtain that, for any u
>
K,
(96)
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
≤
P
(
|
∑
i
=
1
n
‍
η
i
1
{
|
η
i
|
≤
u
}
|
>
u
)
≤
1
u
∑
i
=
1
n
‍
E
(
|
η
i
|
1
{
|
η
i
|
≤
u
}
)
.
Note that P
(
|
η
i
|
>
s
)
≤
(
s
-
α
-
K
-
α
)
/
μ
k
(
ℝ
) if s
≤
K. Hence, for any u
>
K
(97)
E
(
|
η
i
|
1
{
|
η
i
|
≤
u
}
)
≤
∫
0
u
‍
P
(
|
η
i
|
>
s
)
d
s
=
∫
0
K
‍
P
(
|
η
i
|
>
s
)
d
s
≤
1
μ
k
(
ℝ
)
α
1
-
α
K
1
-
α
.
Combining all these facts, we get that for any u
>
K
(98)
μ
k
*
n
(
{
z
;
|
z
|
>
u
}
)
≤
[
μ
k
(
ℝ
)
]
n
-
1
α
1
-
α
K
1
-
α
n
u
-
1
,
and the conclusion follows from (94).
Assume now thatα
=
1. In this case, E
(
η
i
1
{
|
η
i
|
≤
u
}
)
=
0 since η
i has a symmetric distribution. Using Chebyshev’s inequality this time, we obtain
(99)
P
(
|
∑
i
=
1
n
‍
η
i
|
>
u
)
≤
P
(
|
∑
i
=
1
n
‍
η
i
1
{
|
η
i
|
≤
u
}
|
>
u
)
≤
1
u
2
∑
i
=
1
n
‍
E
(
η
i
2
1
{
|
η
i
|
≤
u
}
)
.
The result follows as above using the fact that, for any u
>
K,
(100)
E
(
η
i
2
1
{
|
η
i
|
≤
u
}
)
≤
2
∫
0
u
‍
s
P
(
|
η
i
|
>
s
)
d
s
=
2
∫
0
K
‍
s
P
(
|
η
i
|
>
s
)
d
s
≤
1
μ
k
(
ℝ
)
K
.Lemma 17.
Ifα
<
1 then
(101)
E
|
Z
K
(
B
)
|
p
≤
C
α
,
p
K
p
-
α
|
B
|
f
o
r
a
n
y
p
∈
(
α
,
1
)
,
where C
α
,
p is a constant depending on α and p. If α
=
1, then
(102)
E
|
Z
K
(
B
)
|
p
≤
C
p
K
p
-
1
|
B
|
f
o
r
a
n
y
p
∈
(
1,2
)
,
where C
p is a constant depending on p.Proof.
Note that(103)
E
|
Z
K
(
B
)
|
p
=
∫
0
∞
‍
P
(
|
Z
K
(
B
)
|
p
>
t
)
d
t
=
p
∫
0
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
.
We consider separately the integrals for u
≤
K and u
>
K. For the first integral we use (90):
(104)
∫
0
K
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
r
α
|
B
|
∫
0
K
‍
u
-
α
+
p
-
1
d
u
=
r
α
|
B
|
1
p
-
α
K
p
-
α
.
For the second one we use Lemma 16: if α
<
1 then
(105)
∫
K
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
α
1
-
α
K
1
-
α
|
B
|
∫
K
∞
‍
u
p
-
2
d
u
=
α
(
1
-
α
)
(
1
-
p
)
|
B
|
K
p
-
α
,
and if α
=
1, then
(106)
∫
K
∞
‍
P
(
|
Z
K
(
B
)
|
>
u
)
u
p
-
1
d
u
≤
K
|
B
|
∫
K
∞
‍
u
p
-
3
d
u
=
|
B
|
1
2
-
p
K
p
-
1
.We now proceed to the construction of the stochastic integral with respect toZ
K. For this, we use the same method as for Z. Note that ℱ
t
Z
K
⊂
ℱ
t, where ℱ
t
Z
K is the σ-field generated by Z
K
(
[
0
,
s
]
×
A
) for all s
∈
[
0
,
t
] and A
∈
ℬ
b
(
ℝ
d
). For any B
∈
ℬ
b
(
ℝ
d
), we will work with a càdlàg modification of the Lévy process {
Z
K
(
t
,
B
)
=
Z
K
(
[
0
,
t
]
×
B
)
;
t
≥
0
}.IfX is a simple process given by (40), we define
(107)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
by the same formula (44) with Z replaced by Z
K. The following result shows that I
K
(
X
)
(
t
,
B
) has the same tail behavior as I
(
X
)
(
t
,
B
).Proposition 18.
IfX is a bounded simple process then
(108)
sup
λ
>
0
λ
α
P
(
sup
t
∈
[
0
,
T
]
|
I
K
(
X
)
(
t
,
B
)
|
>
λ
)
≤
d
α
E
∫
0
T
‍
∫
B
‍
|
X
(
t
,
x
)
|
α
d
x
d
t
,
for any T
>
0 and B
∈
ℬ
b
(
ℝ
d
), where d
α is a constant depending only on α.Proof.
As in the proof of Theorem13, it is enough to prove that
(109)
P
(
max
l
=
0
,
…
,
m
-
1
|
∑
i
=
0
l
‍
∑
j
=
1
N
i
‍
β
i
j
Z
i
j
*
|
>
λ
)
≤
d
α
λ
-
α
∑
i
=
0
m
-
1
‍
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
E
|
β
i
j
|
α
|
H
i
j
∩
B
|
,
where Z
i
j
*
=
Z
K
(
(
s
i
,
s
i
+
1
]
×
(
H
i
j
∩
B
)
). This reduces to showing that U
i
*
=
∑
j
=
1
N
i
‍
x
j
Z
i
j
* satisfies an inequality similar to (57) for any x
¯
∈
ℝ
N
i; that is,
(110)
P
(
|
U
i
*
|
>
λ
)
≤
d
α
*
λ
-
α
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
x
j
|
α
|
H
i
j
∩
B
|
,
for any λ
>
0, for some d
α
*
>
0. We first examine the tail of Z
i
j
*. By (90),
(111)
P
(
|
Z
i
j
*
|
>
λ
)
≤
r
α
(
s
i
+
1
-
s
i
)
K
i
j
λ
-
α
,
where K
i
j
=
|
H
i
j
∩
B
|. Letting η
i
j
=
K
i
j
-
1
/
α
Z
i
j
*, we obtain that, for any u
>
0,
(112)
P
(
|
η
i
j
|
>
u
)
≤
r
α
(
s
i
+
1
-
s
i
)
u
-
α
,
∀
j
=
1
,
…
,
N
i
.
By Lemma A.3 (Appendix A), it follows that, for any λ
>
0,
(113)
P
(
|
∑
j
=
1
N
i
‍
b
j
η
i
j
|
>
λ
)
≤
r
α
2
(
s
i
+
1
-
s
i
)
∑
j
=
1
N
i
‍
|
b
j
|
α
λ
-
α
,
for any sequence (
b
j
)
j
=
1
,
…
,
N
i of real numbers. Inequality (110) (with d
α
*
=
r
α
2) follows by applying this to b
j
=
x
j
K
i
j
1
/
α.In view of the previous result and Proposition12, for any process X
∈
ℒ
α, we can construct the integral
(114)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
in the same manner as I
(
X
)
(
t
,
B
), and this integral satisfies (108). If in addition the process X
∈
ℒ
α satisfies (83), then we can define the integral I
K
(
X
)
(
t
,
𝒪
) for an arbitrary Borel set 𝒪
⊂
ℝ
d (possibly 𝒪
=
ℝ
d). This integral will satisfy an inequality similar to (108) with B replaced by 𝒪.The appealing feature ofI
K
(
X
)
(
t
,
B
) is that we can control its moments, as shown by the next result.Theorem 19.
Ifα
<
1, then for any p
∈
(
α
,
1
) and for any X
∈
ℒ
p,
(115)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
,
for any t
>
0 and B
∈
ℬ
b
(
ℝ
d
), where C
α
,
p is a constant depending on α, p. If 𝒪
⊂
ℝ
d is an arbitrary Borel set and we assume, in addition, that the process X
∈
ℒ
p satisfies
(116)
E
∫
0
T
‍
∫
𝒪
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
<
∞
,
∀
T
>
0
,
then inequality (115) holds with B replaced by 𝒪.Proof.
Consider the following steps.
Step
1. Suppose that X is an elementary process of the form (39). Then I
K
(
X
)
(
t
,
B
)
=
Y
Z
K
(
H
) where H
=
(
t
∧
a
,
t
∧
b
]
×
(
A
∩
B
). Note that Z
K
(
H
) is independent of ℱ
a. Hence, Z
K
(
H
) is independent of Y. Let P
Y denote the law of Y. By Fubini’s theorem,
(117)
E
|
Y
Z
K
(
H
)
|
p
=
p
∫
0
∞
‍
P
(
|
Y
Z
K
(
H
)
|
>
u
)
u
p
-
1
d
u
=
p
∫
ℝ
‍
(
∫
0
∞
‍
P
(
|
y
Z
K
(
H
)
|
>
u
)
u
p
-
1
d
u
)
P
Y
(
d
y
)
.
We evaluate the inner integral. We split this integral into two parts, for u
≤
K
|
y
| and u
>
K
|
y
|, respectively. For the first integral, we use (90). For the second one, we use Lemma 16. Therefore, the inner integral is bounded by
(118)
r
α
|
y
|
α
|
H
|
∫
0
K
|
y
|
‍
u
-
α
+
p
-
1
d
u
+
α
1
-
α
|
y
|
K
1
-
α
|
H
|
×
∫
K
|
y
|
∞
‍
u
p
-
2
d
u
=
C
α
,
p
′
K
p
-
α
|
y
|
p
|
H
|
,
E
|
Y
Z
K
(
H
)
|
p
≤
p
C
α
,
p
′
K
p
-
α
|
H
|
E
|
Y
|
p
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
2.   Suppose now that X is a simple process of the form (40). Then X
(
t
,
x
)
=
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
X
i
j
(
t
,
x
) where X
i
j
(
t
,
x
)
=
1
(
t
i
,
t
i
+
1
]
(
t
)
1
A
i
j
(
x
)
Y
i
j.
Using the linearity of the integral, the inequality|
a
+
b
|
p
≤
|
a
|
p
+
|
b
|
p, and the result obtained in Step 1 for the elementary processes X
i
j, we get
(119)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
E
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
|
I
K
(
X
i
j
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∑
i
=
0
N
-
1
‍
∑
j
=
1
m
i
‍
∫
0
t
‍
∫
B
‍
|
X
i
j
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
3. Let X
∈
ℒ
p be arbitrary. By Proposition 12, there exists a sequence (
X
n
)
n of  bounded simple processes such that ∥
X
n
-
X
∥
p
→
0. Since α
<
p, it follows that ∥
X
n
-
X
∥
α
→
0. By the definition of I
K
(
X
)
(
t
,
B
) there exists a subsequence {
n
k
}
k such that {
I
K
(
X
n
k
)
(
t
,
B
)
}
k converges to I
K
(
X
)
(
t
,
B
) a.s. Using Fatou’s lemma and the result obtained in Step 2 (for the simple processes X
n
k), we get
(120)
E
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
liminf
k
→
∞
E
|
I
K
(
X
n
k
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
liminf
k
→
∞
E
∫
0
t
‍
∫
B
‍
|
X
n
k
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.
Step
4. Suppose that X
∈
ℒ
p satisfies (116). Let 𝒪
k
=
𝒪
∩
E
k where (
E
k
)
k is an increasing sequence of sets in ℬ
b
(
ℝ
d
) such that ⋃
k
≥
1
‍
E
k
=
ℝ
d. By the definition of I
K
(
X
)
(
t
,
𝒪
), there exists a subsequence (
k
i
)
i such that {
I
K
(
X
)
(
t
,
𝒪
k
i
)
}
i converges to I
K
(
X
)
(
t
,
𝒪
) a.s. Using Fatou’s lemma, the result obtained in Step 3 (for B
=
𝒪
k
i) and the monotone convergence theorem, we get
(121)
E
|
I
K
(
X
)
(
t
,
𝒪
)
|
p
≤
liminf
i
→
∞
E
|
I
K
(
X
)
(
t
,
𝒪
k
i
)
|
p
≤
C
α
,
p
K
p
-
α
liminf
i
→
∞
E
∫
0
t
‍
∫
𝒪
k
i
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
=
C
α
,
p
K
p
-
α
E
∫
0
t
‍
∫
𝒪
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
.Remark 20.
Finding a similar moment inequality for the casesα
=
1 and p
∈
(
1,2
) remains an open problem. The argument used in Step 2 above relies on the fact that p
<
1. Unfortunately, we could not find another argument to cover the case p
>
1.


## 5.2.  The Caseα
>
1

In this case, the construction of the integral with respect toZ
K relies on an integral with respect to N
^ which exists in the literature. We recall briefly the definition of this integral. For more details, see Section 1.2.2 of [6], Section 24.2 of [25], or Section 8.7 of [12].Let𝔼
=
ℝ
d
×
(
ℝ
∖
{
0
}
) endowed with the measure μ
(
d
x
,
d
z
)
=
d
x
ν
α
(
d
z
) and let  ℬ
b
(
𝔼
) be the class of bounded Borel sets in 𝔼. For a simple process Y
=
{
Y
(
t
,
x
,
z
)
;
t
≥
0
,
(
x
,
z
)
∈
𝔼
}, the integral I
N
^
(
Y
)
(
t
,
B
) is defined in the usual way, for any t
>
0, B
∈
ℬ
b
(
𝔼
). The process I
N
^
(
Y
)
(
·
,
B
) is a (càdlàg) zero-mean square-integrable martingale with quadratic variation
(122)
[
I
N
^
(
Y
)
(
·
,
B
)
]
t
=
∫
0
t
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
N
(
d
s
,
d
x
,
d
z
)
and predictable quadratic variation
(123)
〈
I
N
^
(
Y
)
(
·
,
B
)
〉
t
=
∫
0
t
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
ν
α
(
d
z
)
d
x
d
s
.
By approximation, this integral can be extended to the class of all 𝒫
×
ℬ
(
ℝ
∖
{
0
}
)-measurable processes Y such that for any T
>
0 and B
∈
ℬ
b
(
𝔼
)
(124)
∥
Y
∥
2
,
T
,
B
2
≔
E
∫
0
T
‍
∫
B
‍
|
Y
(
s
,
x
,
z
)
|
2
ν
α
(
d
z
)
d
x
d
s
<
∞
.
The integral is a martingale with the same quadratic variations as above and has the isometry property: E
|
I
N
^
(
Y
)
(
t
,
B
)
|
2
=
∥
Y
∥
2
,
T
,
B
2. If, in addition, ∥
Y
∥
2
,
T
,
𝔼
<
∞, then the integral can be extended to 𝔼. By the Burkholder-Davis-Gundy inequality for discontinuous martingales, for any p
≥
1,
(125)
E
sup
t
≤
T
|
I
N
^
(
Y
)
(
t
,
𝔼
)
|
p
≤
C
p
E
[
I
N
^
(
Y
)
(
·
,
𝔼
)
]
T
p
/
2
.The previous inequality is not suitable for our purposes. A more convenient inequality can be obtained foranother stochastic integral, constructed for p
∈
[
1,2
] fixed, as suggested on page 293 of [6]. More precisely, one can show that, for any bounded simple process Y,
(126)
E
sup
t
≤
T
|
I
N
^
(
Y
)
(
t
,
𝔼
)
|
p
≤
C
p
E
∫
0
T
‍
∫
ℝ
d
‍
∫
ℝ
∖
{
0
}
‍
|
Y
(
t
,
x
,
z
)
|
p
ν
α
(
d
z
)
d
x
d
t
=
:
|
Y
|
p
,
T
,
𝔼
p
,
where C
p is the constant appearing in (125) (see Lemma 8.22 of [12]).By the usual procedure, the integral can be extended to the class of all𝒫
×
ℬ
(
ℝ
∖
{
0
}
)-measurable processes Y such that [
Y
]
p
,
T
,
𝔼
<
∞. The integral is defined as an element in the space L
p
(
Ω
;
D
[
0
,
T
]
) and will be denoted by
(127)
I
N
^
,
p
(
Y
)
(
t
,
𝔼
)
=
∫
0
t
‍
∫
ℝ
d
‍
∫
ℝ
∖
{
0
}
‍
Y
(
s
,
x
,
z
)
N
^
(
d
s
,
d
x
,
d
z
)
.
Its appealing feature is that it satisfies inequality (126).From now on, we fixp
∈
[
1,2
]. Based on (88), for any B
∈
ℬ
b
(
ℝ
d
), we let
(128)
I
K
(
X
)
(
t
,
B
)
=
∫
0
t
‍
∫
B
‍
X
(
s
,
x
)
Z
K
(
d
s
,
d
x
)
=
∫
0
t
‍
∫
B
‍
∫
{
|
z
|
≤
K
}
‍
X
(
s
,
x
)
z
N
^
(
d
s
,
d
x
,
d
z
)
,
for any predictable process X
=
{
X
(
t
,
x
)
;
t
≥
0
,
x
∈
ℝ
d
} for which the rightmost integral is well defined. Letting Y
(
t
,
x
,
z
)
=
X
(
t
,
x
)
z
1
{
0
<
|
z
|
≤
K
}, we see that this is equivalent to saying that p
>
α and X
∈
ℒ
p. By (126),
(129)
E
sup
t
≤
T
|
I
K
(
X
)
(
t
,
B
)
|
p
≤
C
α
,
p
K
p
-
α
E
∫
0
T
‍
∫
B
‍
|
X
(
s
,
x
)
|
p
d
x
d
s
,
where C
α
,
p
=
C
p
α
/
(
p
-
α
). If, in addition, the process X
∈
ℒ
p satisfies (116) then (129) holds with B replaced by 𝒪, for an arbitrary Borel set 𝒪
⊂
ℝ
d.Note that (129) is the counterpart of (115) for the case α
>
1. Together, these two inequalities will play a crucial role in Section 6.Table1 summarizes all the conditions.Table 1
Conditions forI
K
(
X
)
(
t
,
B
) to be well defined.
α
<
1
α
>
1
B is bounded
X
∈
ℒ
α
X
∈
ℒ
p
for somep
∈
(
α
,
2
]
B
=
𝒪 is unbounded
X
∈
ℒ
α and X satisfies (83)
X
∈
ℒ
p and
X satisfies (116)
for somep
∈
(
α
,
2
]


## 6. The Main Result

In this section, we state and prove the main result regarding the existence of a mild solution of (1). For this result, 𝒪 is a bounded domain in ℝ
d. For any t
>
0, we denote
(130)
J
p
(
t
)
=
sup
x
∈
𝒪
∫
𝒪
‍
G
(
t
,
x
,
y
)
p
d
y
.Theorem 21.
Letα
∈
(
0,2
), α
≠
1. Assume that for any T
>
0
(131)
lim
h
→
0
∫
0
T
‍
∫
𝒪
‍
|
G
(
t
,
x
,
y
)
-
G
(
t
+
h
,
x
,
y
)
|
p
d
y
d
t
=
0
,
∀
x
∈
𝒪
,
(132)
lim
|
h
|
→
0
∫
0
T
‍
∫
𝒪
‍
|
G
(
t
,
x
,
y
)
-
G
(
t
,
x
+
h
,
y
)
|
p
d
y
d
t
=
0
,
∀
x
∈
𝒪
,
(133)
∫
0
T
‍
J
p
(
t
)
d
t
<
∞
,
for some p
∈
(
α
,
1
) if α
<
1, or for some p
∈
(
α
,
2
] if α
>
1. Then (1) has a mild solution. Moreover, there exists a sequence (
τ
K
)
K
≥
1 of stopping times with τ
K
↑
∞ a.s. such that, for any T
>
0 and K
≥
1,
(134)
sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
(
|
u
(
t
,
x
)
|
p
1
{
t
≤
τ
K
}
)
<
∞
.Example 22 (heat equation).
LetL
=
∂
/
∂
t
-
(
1
/
2
)
Δ. Then G
(
t
,
x
,
y
)
≤
G
¯
(
t
,
x
-
y
) where G
¯
(
t
,
x
) is the fundamental solution of L
u
=
0 on ℝ
d. Condition (133) holds if p
<
1
+
2
/
d. If α
<
1, this condition holds for any p
∈
(
α
,
1
). If α
>
1, this condition holds for any p
∈
(
α
,
1
+
2
/
d
], as long as α satisfies (6). Conditions (131) and (132) hold by the continuity of the function G in t and x, by applying the dominated convergence theorem. To justify the application of this theorem, we use the trivial bound (
2
π
t
)
-
d
p
/
2 for both G
(
t
+
h
,
x
,
y
)
p and G
(
t
,
x
+
h
,
y
)
p, which introduces the extra condition d
p
<
2. Unfortunately, we could not find another argument for proving these two conditions (In the case of the heat equation on ℝ
d, Lemmas A.2 and A.3 of [6] estimate the integrals appearing in (132) and (131), with p
=
1 in (131). These arguments rely on the structure of G
¯ and cannot be used when 𝒪 is a bounded domain.).Example 23 (parabolic equations).
LetL
=
∂
/
∂
t
-
ℒ where ℒ is given by (31). Assuming (32), we see that (133) holds if p
<
1
+
2
/
d. The same comments as for the heat equation apply here as well (Although in a different framework, a condition similar to (131) was probably used in the proof of Theorem 12.11 of [12] (page 217) for the claim lim
s
→
t
E
|
J
3
(
X
)
(
s
)
-
J
3
(
X
)
(
t
)
|
L
p
(
𝒪
)
p
=
0. We could not see how to justify this claim, unless d
p
<
2.).Example 24 (heat equation with fractional power of the Laplacian).
LetL
=
∂
/
∂
t
+
(
-
Δ
)
γ for some γ
>
0. By Lemma B.23 of [12], if α
>
1, then condition (133) holds for any p
∈
(
α
,
1
+
2
γ
/
d
), provided that α satisfies (36) (This condition is the same as in Theorem 12.19 of [12], which examines the same equation using the approach based on Hilbert-space valued solution.).
To verify conditions (131) and (132), we use the continuity of G in t and x and apply the dominated convergence theorem. To justify the application of this theorem, we use the trivial bound C
d
,
γ
t
-
d
p
/
(
2
γ
) for both G
(
t
+
h
,
x
,
y
)
p and G
(
t
,
x
+
h
,
y
)
p, which introduces the extra condition d
p
<
2
γ. This bound can be seen from (33), using the fact that 𝒢
(
t
,
x
,
y
)
≤
𝒢
¯
(
t
,
x
-
y
) where 𝒢 and 𝒢
¯ are the fundamental solutions of ∂
u
/
∂
t
-
Δ
u
=
0 on 𝒪 and ℝ
d, respectively. (In the case of the same equation on ℝ
d, elementary estimates for the time and space increments of G
¯ can be obtained directly from (35), as on page 196 of [26]. These arguments cannot be used when 𝒪 is a bounded domain.)The remaining part of this section is dedicated to the proof of Theorem21. The idea is to solve first the equation with the truncated noise Z
K (yielding a mild solution u
K) and then identify a sequence (
τ
K
)
K
≥
1 of stopping times with τ
K
↑
∞ a.s. such that, for any t
>
0, x
∈
𝒪, and L
>
K, u
K
(
t
,
x
)
=
u
L
(
t
,
x
) a.s. on the event {
t
≤
τ
K
}. The final step is to show that process u defined by u
(
t
,
x
)
=
u
K
(
t
,
x
) on {
t
≤
τ
K
} is a mild solution of (1). A similar method can be found in Section 9.7 of [12] using an approach based on stochastic integration of operator-valued processes, with respect to Hilbert-space-valued processes, which is different from our approach.Sinceσ is a Lipschitz function, there exists a constant C
σ
>
0 such that
(135)
|
σ
(
u
)
-
σ
(
v
)
|
≤
C
σ
|
u
-
v
|
,
∀
u
,
v
∈
ℝ
.
In particular, letting D
σ
=
C
σ
∨
|
σ
(
0
)
|, we have
(136)
|
σ
(
u
)
|
≤
D
σ
(
1
+
|
u
|
)
,
∀
u
∈
ℝ
.For the proof of Theorem21, we need a specific construction of the Poisson random measure N, taken from [13]. We review briefly this construction.Let(
𝒪
k
)
k
≥
1 be a partition of ℝ
d with sets in ℬ
b
(
ℝ
d
) and let (
U
j
)
j
≥
1 be a partition of ℝ
∖
{
0
} such that ν
α
(
U
j
)
<
∞ for all j
≥
1. We may take U
j
=
Γ
j
-
1 for all j
≥
1. Let (
E
i
j
,
k
,
X
i
j
,
k
,
Z
i
j
,
k
)
i
,
j
,
k
≥
1 be independent random variables defined on a probability space (
Ω
,
ℱ
,
P
), such that
(137)
P
(
E
i
j
,
k
>
t
)
=
e
-
λ
j
,
k
t
,
P
(
X
i
j
,
k
∈
B
)
=
|
B
∩
𝒪
k
|
|
𝒪
k
|
,
P
(
Z
i
j
,
k
∈
Γ
)
=
|
Γ
∩
U
j
|
|
U
j
|
,
where λ
j
,
k
=
|
𝒪
k
|
ν
α
(
U
j
). Let T
i
j
,
k
=
∑
l
=
1
i
‍
E
l
j
,
k for all i
≥
1. Then
(138)
N
=
∑
i
,
j
,
k
≥
1
‍
δ
(
T
i
j
,
k
,
X
i
j
,
k
,
Z
i
j
,
k
)
is a Poisson random measure on ℝ
+
×
ℝ
d
×
(
ℝ
∖
{
0
}
) with intensity d
t
d
x
ν
α
(
d
z
).This section is organized as follows. In Section6.1 we prove the existence of the solution of the equation with truncated noise Z
K. Sections 6.2 and 6.3 contain the proof of Theorem 21 when α
<
1 and α
>
1, respectively.


### 6.1. The Equation with Truncated Noise

In this section, we fixK
>
0 and we consider the equation:
(139)
L
u
(
t
,
x
)
=
σ
(
u
(
t
,
x
)
)
Z
˙
K
(
t
,
x
)
,
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions. A mild solution of (139) is a predictable process u which satisfies (2) with Z replaced by Z
K. For the next result, 𝒪 can be a bounded domain in ℝ
d or 𝒪
=
ℝ
d (with no boundary conditions).Theorem 25.
Under the assumptions of Theorem21, (139) has a unique mild solution u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
}. For any T
>
0,
(140)
sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
(
t
,
x
)
|
p
<
∞
,
and the map (
t
,
x
)
↦
u
(
t
,
x
) is continuous from [
0
,
T
]
×
𝒪 into L
p
(
Ω
).Proof.
We use the same argument as in the proof of Theorem 13 of [27], based on a Picard iteration scheme. We define u
0
(
t
,
x
)
=
0 and
(141)
u
n
+
1
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
for any n
≥
0. We prove by induction on n
≥
0 that (i)  u
n
(
t
,
x
) is well defined; (ii) K
n
(
t
)
:
=
sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
n
(
t
,
x
)
|
p
<
∞ for any T
>
0; (iii) u
n
(
t
,
x
) is ℱ
t-measurable for any t
>
0 and x
∈
𝒪; (iv) the map (
t
,
x
)
↦
u
n
(
t
,
x
) is continuous from [
0
,
T
]
×
𝒪 into L
p
(
Ω
) for any T
>
0.
The statement is trivial forn
=
0. For the induction step, assume that the statement is true for n. By an extension to random fields of Theorem 30, Chapter  IV of [28], u
n has a jointly measurable modification. Since this modification is (
ℱ
t
)
t-adapted (in the sense of (iii)), it has a predictable modification (using an extension of Proposition 3.21 of [12] to random fields). We work with this modification, that we call also u
n.
We prove that (i)–(iv) hold foru
n
+
1. To show (i), it suffices to prove that X
n
∈
ℒ
p, where X
n
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
n
(
s
,
y
)
). By (136) and (133),
(142)
E
∫
0
t
‍
∫
𝒪
‍
|
X
n
(
s
,
y
)
|
p
d
y
d
s
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
∫
0
t
‍
J
p
(
t
-
s
)
d
s
<
∞
.
In addition, if 𝒪
=
ℝ
d, we have to prove that X
n satisfies (83) if α
<
1, or (116) if α
>
1 (see Table 1). If α
<
1, this follows as above, since α
<
p and hence sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
(
t
,
x
)
|
α
<
∞; the argument for α
>
1 is similar.
Combined with the moment inequality (115) (or (129)), this proves (ii), since
(143)
E
|
u
n
+
1
(
t
,
x
)
|
p
≤
C
α
,
p
K
p
-
α
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
∫
0
t
‍
J
p
(
t
-
s
)
d
s
,
for any x
∈
𝒪. Property (iii) follows by the construction of the integral I
K.
To prove (iv), we first show the right continuity int. Let h
>
0. Writing the interval [
0
,
t
+
h
] as the union of [
0
,
t
] and (
t
,
t
+
h
], we obtain that E
|
u
n
+
1
(
t
+
h
,
x
)
-
u
n
+
1
(
t
,
x
)
|
p
≤
2
p
-
1
(
I
1
(
h
)
+
I
2
(
h
)
), where
(144)
I
1
(
h
)
=
E
|
∫
0
t
‍
∫
𝒪
‍
(
G
(
t
+
h
-
s
,
x
,
y
)
-
G
(
t
-
s
,
x
,
y
)
)
hhhhhhhlh
×
σ
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
∫
0
t
‍
∫
𝒪
‍
|
p
,
I
2
(
h
)
=
E
|
∫
t
t
+
h
‍
∫
𝒪
‍
G
(
t
+
h
-
s
,
x
,
y
)
σ
hhhhhhhhllh
×
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
∫
t
t
+
h
‍
∫
𝒪
‍
|
p
.
Using again (136) and the moment inequality (115) (or (129)), we obtain
(145)
I
1
(
h
)
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
t
‍
∫
𝒪
‍
|
G
(
s
+
h
,
x
,
y
)
-
G
(
s
,
x
,
y
)
|
p
d
y
d
s
,
I
2
(
h
)
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
h
‍
∫
𝒪
‍
G
(
s
,
x
,
y
)
p
d
y
d
s
.
It follows that both I
1
(
h
) and I
2
(
h
) converge to 0 as h
→
0, using (131) for I
1
(
h
) and the Dominated Convergence Theorem and (133) for I
2
(
h
), respectively. The left continuity in t is similar, by writing the interval [
0
,
t
-
h
] as the difference between [
0
,
t
] and (
t
-
h
,
t
] for h
>
0. For the continuity in x, similarly as above, we see that E
|
u
n
+
1
(
t
,
x
+
h
)
-
u
n
+
1
(
t
,
x
)
|
p is bounded by
(146)
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
t
‍
∫
𝒪
‍
|
G
(
s
,
x
+
h
,
y
)
-
G
(
s
,
x
,
y
)
|
p
d
y
d
s
,
which converges to 0 as |
h
|
→
0 due to (132). This finishes the proof of (iv).
We denoteM
n
(
t
)
=
sup
x
∈
𝒪
E
|
u
n
(
t
,
x
)
|
p. Similarly to (143), we have
(147)
M
n
(
t
)
≤
C
1
∫
0
t
‍
(
1
+
M
n
-
1
(
s
)
)
J
p
(
t
-
s
)
d
s
,
∀
n
≥
1
,
where C
1
=
C
α
,
p
K
p
-
α
D
σ
p
2
p
-
1. By applying Lemma 15 of Erratum to [27] with f
n
=
M
n, k
1
=
0, k
2
=
1, and g
(
s
)
=
C
J
p
(
s
), we obtain that
(148)
sup
n
≥
0
sup
t
∈
[
0
,
T
]
M
n
(
t
)
<
∞
,
∀
T
>
0
.
We now prove that{
u
n
(
t
,
x
)
}
n converges in L
p
(
Ω
), uniformly in (
t
,
x
)
∈
[
0
,
T
]
×
𝒪. To see this, let U
n
(
t
)
=
sup
x
∈
𝒪
E
|
u
n
+
1
(
t
,
x
)
-
u
n
(
t
,
x
)
|
p for n
≥
0. Using the moment inequality (115) (or (129)) and (135), we have
(149)
U
n
(
t
)
≤
C
2
∫
0
t
‍
U
n
-
1
(
s
)
J
p
(
t
-
s
)
d
s
,
where C
2
=
C
α
,
p
K
p
-
α
C
σ
p. By Lemma 15 of Erratum to [27], ∑
n
≥
0
‍
U
n
(
t
)
1
/
p converges uniformly on [
0
,
T
] (Note that this lemma is valid for all p
>
0.).
We denote byu
(
t
,
x
) the limit of u
n
(
t
,
x
) in L
p
(
Ω
). One can show that u satisfies properties (ii)–(iv) listed above. So u has a predictable modification. This modification is a solution of (139). To prove uniqueness, let v be another solution and denote H
(
t
)
=
sup
x
∈
𝒪
E
|
u
(
t
,
x
)
-
v
(
t
,
x
)
|
p. Then
(150)
H
(
t
)
≤
C
2
∫
0
t
‍
H
(
s
)
J
p
(
t
-
s
)
d
s
.
Using (133), it follows that H
(
t
)
=
0 for all t
>
0.


### 6.2. Proof of Theorem21: Case α
<
1

In this case, for anyt
>
0 and B
∈
ℬ
b
(
ℝ
d
), we have (see (21))
(151)
Z
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
(
d
s
,
d
x
,
d
z
)
.
The characteristic function of Z
(
t
,
B
) is given by
(152)
E
(
e
i
u
Z
(
t
,
B
)
)
=
exp
{
t
|
B
|
∫
ℝ
∖
{
0
}
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
hhhhhhhhhhhhhhhhhhhhhhl
∀
u
∈
ℝ
.
Note that {
Z
(
t
,
B
)
}
t
≥
0 is not a compound Poisson process since ν
α is infinite.We introduce the stopping times(
τ
K
)
K
≥
1, as on page 239 of [13]:
(153)
τ
K
(
B
)
=
inf
{
t
>
0
;
|
Z
(
t
,
B
)
-
Z
(
t
-
,
B
)
|
>
K
}
,
where Z
(
t
-
,
B
)
=
lim
s
↑
t
Z
(
s
,
B
). Clearly, τ
L
(
B
)
≥
τ
K
(
B
) for all L
>
K.We first investigate the relationship betweenZ and Z
K and the properties of τ
K
(
B
). Using construction (138) of N and definition (87) of Z
K, we have
(154)
Z
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
=
:
∑
j
,
k
≥
1
‍
Z
j
,
k
(
t
,
B
)
,
Z
K
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
|
Z
i
j
,
k
|
≤
K
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
.We observe that{
Z
j
,
k
(
t
,
B
)
}
t
≥
0 is a compound Poisson process with
(155)
E
(
e
i
u
Z
j
,
k
(
t
,
B
)
)
=
exp
{
t
|
𝒪
k
∩
B
|
∫
U
j
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
∀
u
∈
ℝ
.Note thatτ
K
(
B
)
>
T means that all the jumps of {
Z
(
t
,
B
)
}
t
≥
0 in [
0
,
T
] are smaller than K in modulus; that is, {
τ
K
(
B
)
>
T
}
=
{
ω
;
|
Z
i
j
,
k
(
ω
)
|
≤
K for  all  i
,
j
,
k
≥
1 for  which T
i
j
,
k
(
ω
)
≤
T and X
i
j
,
k
(
ω
)
∈
B
}. Hence, on {
τ
K
(
B
)
>
T
},
(156)
Z
(
[
0
,
t
]
×
A
)
=
Z
K
(
[
0
,
t
]
×
A
)
=
Z
L
(
[
0
,
t
]
×
A
)
,
for any L
>
K, t
∈
[
0
,
T
], and A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B. Using an approximation argument and the construction of the integrals I
(
X
) and I
K
(
X
), it follows that, for any X
∈
ℒ
α and for any L
>
K, a.s. on {
τ
K
(
B
)
>
T
}, we have
(157)
I
(
X
)
(
T
,
B
)
=
I
K
(
X
)
(
T
,
B
)
=
I
L
(
X
)
(
T
,
B
)
.The next result gives the probability of the event{
τ
K
(
B
)
>
T
}.Lemma 26.
For anyT
>
0 and B
∈
ℬ
b
(
ℝ
d
),
(158)
P
(
τ
K
(
B
)
>
T
)
=
exp
(
-
T
|
B
|
K
-
α
)
.
Consequently, lim
K
→
∞
P
(
τ
K
(
B
)
>
T
)
=
1 and lim
K
→
∞
τ
K
(
B
)
=
∞ a.s.Proof.
Note that{
τ
K
(
B
)
>
T
}
=
⋂
j
,
k
≥
1
‍
{
τ
K
j
,
k
(
B
)
>
T
}, where
(159)
τ
K
j
,
k
(
B
)
=
inf
{
t
>
0
;
|
Z
j
,
k
(
t
,
B
)
-
Z
j
,
k
(
t
-
,
B
)
|
>
K
}
.
Since ν
α
(
{
z
;
|
z
|
>
K
}
)
=
K
-
α and (
τ
K
j
,
k
(
B
)
)
j
,
k
≥
1 are independent, it is enough to prove that, for any j
,
k
≥
1,
(160)
P
(
τ
K
j
,
k
(
B
)
>
T
)
=
exp
{
-
T
|
B
∩
𝒪
k
|
ν
α
(
{
z
;
|
z
|
>
K
}
∩
U
j
)
}
.
Note that{
τ
K
j
,
k
(
B
)
>
T
}
=
{
ω
;
|
Z
i
j
,
k
(
ω
)
|
≤
K for  all i for  which T
i
j
,
k
≤
T and X
i
j
,
k
∈
B
} and (
T
n
j
,
k
)
n
≥
1 are the jump times of a Poisson process with intensity λ
j
,
k. Hence,
(161)
P
(
τ
K
j
,
k
(
B
)
>
T
)
hh
=
∑
n
≥
0
‍
∑
m
=
0
n
‍
∑
I
⊂
{
1
,
…
,
n
}
,
card
(
I
)
=
m
‍
P
(
T
n
j
,
k
≤
T
<
T
n
+
1
j
,
k
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
‍
{
X
i
j
,
k
∈
B
}
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
‍
{
|
Z
i
j
,
k
|
≤
K
}
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
c
‍
{
X
i
j
,
k
∉
B
}
)
hh
=
∑
n
≥
0
‍
e
-
λ
j
,
k
T
(
λ
j
,
k
T
)
n
n
!
hhhhhhh
×
[
1
-
P
(
X
1
j
,
k
∈
B
)
P
(
|
Z
1
j
,
k
|
>
K
)
]
n
hh
=
exp
{
-
λ
j
,
k
T
P
(
X
1
j
,
k
∈
B
)
P
(
|
Z
1
j
,
k
|
>
K
)
}
,
which yields (160).
To prove the last statement, letA
k
(
n
)
=
{
τ
K
(
B
)
>
n
}. Then P
(
lim
¯
K
A
K
(
n
)
)
≥
lim
¯
K
P
(
A
K
(
n
)
)
=
1 for any n
≥
1, and hence P
(
⋂
n
≥
1
‍
lim
¯
K
A
K
(
n
)
)
=
1. Hence, with probability 1, for any n, there exists some K
n such that τ
K
n
>
n. Since (
τ
K
)
K is nondecreasing, this proves that τ
K
→
∞ with probability 1.Remark 27.
The construction ofτ
K
(
B
) given above is due to [13] (in the case of a symmetric measure ν
α). This construction relies on the fact that B is a bounded set. Since Z
(
t
,
ℝ
d
) (and consequently τ
K
(
ℝ
d
)) is not well defined, we could not see why this construction can also be used when B
=
ℝ
d, as it is claimed in [13]. To avoid this difficulty, one could try to use an increasing sequence (
E
n
)
n of sets in ℬ
b
(
ℝ
d
) with ⋃
n
‍
E
n
=
ℝ
d. Using (157) with B
=
E
n and letting n
→
∞, we obtain that I
(
X
)
(
t
,
ℝ
d
)
=
I
K
(
t
,
ℝ
d
) a.s. on {
t
≤
τ
K
}, where τ
K
=
inf
n
≥
1
τ
K
(
E
n
). But P
(
τ
K
>
t
)
≤ P
(
lim
_
n
{
τ
K
(
E
n
)
>
t
}
)
≤ lim
_
n
P
(
τ
K
(
E
n
)
>
t
)
= lim
n
exp
(
-
t
|
E
n
|
K
-
α
)
=
0 for any t
>
0, which means that τ
K
=
0 a.s. Finding a suitable sequence (
τ
K
)
K of stopping times which could be used in the case 𝒪
=
ℝ
d remains an open problem.In what follows, we denoteτ
K
=
τ
K
(
𝒪
). Let u
K be the solution of (139), whose existence is guaranteed by Theorem 25.Lemma 28.
Under the assumptions of Theorem21, for any t
>
0, x
∈
𝒪, and L
>
K,
(162)
u
K
(
t
,
x
)
=
u
L
(
t
,
x
)
a
.
s
.
o
n
{
t
≤
τ
K
}
.Proof.
By the definition ofu
L and (157),
(163)
u
L
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
L
(
s
,
y
)
)
Z
L
(
d
s
,
d
y
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
L
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
a.s. on the event {
t
≤
τ
K
}. Using the definition of u
K and Proposition C.1 (Appendix C), we obtain that, with probability 1,
(164)
(
u
K
(
t
,
x
)
-
u
L
(
t
,
x
)
)
1
{
t
≤
τ
K
}
hh
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
hhhhhhhhhhhhl
×
(
σ
(
u
K
(
s
,
y
)
)
-
σ
(
u
L
(
s
,
y
)
)
)
hhhhhhhhhhhhl
×
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
.
LetM
(
t
)
=
sup
x
∈
𝒪
E
(
|
u
K
(
t
,
x
)
-
u
L
(
t
,
x
)
|
p
1
{
t
≤
τ
K
}
). Using the moment inequality (115) and the Lipschitz condition (135), we get
(165)
M
(
t
)
≤
C
∫
0
t
‍
J
p
(
t
-
s
)
M
(
s
)
d
s
,
where C
=
C
α
,
p
K
p
-
α
C
σ
p. Using (133), it follows that M
(
t
)
=
0 for all t
>
0.For anyt
>
0 and x
∈
𝒪, let Ω
t
,
x
=
⋂
L
>
K
‍
{
t
≤
τ
K
(
t
)
,
u
K
(
t
,
x
)
≠
u
L
(
t
,
x
)
}, where L and K are positive integers. Let Ω
t
,
x
*
=
Ω
t
,
x
∩
{
lim
K
→
∞
τ
K
=
∞
}.By Lemmas26 and 28, P
(
Ω
t
,
x
*
)
=
1.The next result concludes the proof of Theorem21.Proposition 29.
Under the assumptions of Theorem21, the process u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} defined by
(166)
u
(
ω
,
t
,
x
)
=
u
K
(
ω
,
t
,
x
)
,
i
f
ω
∈
Ω
t
,
x
*
,
t
≤
τ
K
(
ω
)
u
(
ω
,
t
,
x
)
=
0
,
i
f
ω
∉
Ω
t
,
x
*
is a mild solution of (1).Proof.
We first prove thatu is predictable. Note that
(167)
u
(
t
,
x
)
=
lim
K
→
∞
(
u
K
(
t
,
x
)
1
{
t
≤
τ
K
}
)
1
Ω
t
,
x
*
.
The process X
(
ω
,
t
,
x
)
=
1
{
t
≤
τ
K
}
(
ω
) is clearly predictable, being in the class 𝒞 defined in Remark 11. By the definition of Ω
t
,
x, since u
K
,
u
L are predictable, it follows that (
ω
,
t
,
x
)
↦
1
Ω
t
,
x
*
(
ω
) is 𝒫-measurable. Hence, u is predictable.
We now prove thatu satisfies (2). Let t
>
0 and x
∈
𝒪 be arbitrary. Using (157) and Proposition C.1 (Appendix C), with probability 1, we have
(168)
1
{
t
≤
τ
K
}
u
(
t
,
x
)
=
1
{
t
≤
τ
K
}
u
K
(
t
,
x
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
.
For the second last equality, we used the fact that processes X
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
K
(
s
,
y
)
)
1
{
s
≤
τ
K
} and Y
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
1
{
s
≤
τ
K
} are modifications of each other (i.e., X
(
s
,
y
)
=
Y
(
s
,
y
) a.s. for all s
>
0, y
∈
𝒪), and, hence, [
X
-
Y
]
α
,
t
,
𝒪
=
0 and I
(
X
)
(
t
,
𝒪
)
=
I
(
Y
)
(
t
,
𝒪
) a.s. The conclusion follows letting K
→
∞, since τ
K
→
∞ a.s.


### 6.3.  Proof of Theorem21: Case α
>
1

In this case, for anyt
>
0 and B
∈
ℬ
b
(
ℝ
d
), we have (see (22))
(169)
Z
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
^
(
d
s
,
d
x
,
d
z
)
.To introduce the stopping times(
τ
K
)
K
≥
1 we use the same idea as in Section 9.7 of [12].LetM
(
t
,
B
)
=
∑
j
≥
1
‍
(
L
j
(
t
,
B
)
-
E
L
j
(
t
,
B
)
) and P
(
t
,
B
)
=
L
0
(
t
,
B
), where L
j
(
t
,
B
)
=
L
j
(
[
0
,
t
]
×
B
) was defined in Section 2. Note that {
M
(
t
,
B
)
}
t
≥
0 is a zero-mean square-integrable martingale and {
P
(
t
,
B
)
}
t
≥
0 is a compound Poisson process with E
[
P
(
t
,
B
)
]
=
t
|
B
|
μ where μ
=
∫
|
z
|
>
1
‍
z
ν
α
(
d
z
)
=
β
(
α
/
(
α
-
1
)
). With this notation,
(170)
Z
(
t
,
B
)
=
M
(
t
,
B
)
+
P
(
t
,
B
)
-
t
|
B
|
μ
.We letM
K
(
t
,
B
)
=
P
K
(
t
,
B
)
-
E
[
P
K
(
t
,
B
)
]
=
P
K
(
t
,
B
)
-
t
|
B
|
μ
K, where
(171)
P
K
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
1
{
1
<
|
z
|
≤
K
}
N
(
d
s
,
d
x
,
d
z
)
and μ
K
=
∫
1
<
|
z
|
≤
K
‍
z
ν
α
(
d
z
). Recalling definition (88) of Z
K, it follows that
(172)
Z
K
(
t
,
B
)
=
M
(
t
,
B
)
+
P
K
(
t
,
B
)
-
t
|
B
|
μ
K
.For anyK
>
0, we let
(173)
τ
K
(
B
)
=
inf
{
t
>
0
;
|
P
(
t
,
B
)
-
P
(
t
-
,
B
)
|
>
K
}
,
where P
(
t
-
,
B
)
=
lim
s
↑
t
P
(
s
,
B
).Lemma26 holds again, but its proof is simpler than in the case α
<
1, since {
P
(
t
,
B
)
}
t
≥
0 is a compound Poisson process. By (138),
(174)
P
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
|
Z
i
j
,
k
|
>
1
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
,
P
K
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
1
<
|
Z
i
j
,
k
|
≤
K
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
.
Hence, on {
τ
K
(
B
)
>
T
}, for any L
>
K, t
∈
[
0
,
T
], and  A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B,
(175)
P
(
[
0
,
t
]
×
A
)
=
P
K
(
[
0
,
t
]
×
A
)
=
P
L
(
[
0
,
t
]
×
A
)
.Letb
K
=
μ
-
μ
K
=
∫
|
z
|
>
K
‍
z
ν
α
(
d
z
). Using (170) and (172), it follows that
(176)
Z
(
[
0
,
t
]
×
A
)
=
Z
K
(
[
0
,
t
]
×
A
)
-
t
|
A
|
b
K
=
Z
L
(
[
0
,
t
]
×
A
)
-
t
|
A
|
b
L
for any L
>
K, t
∈
[
0
,
T
], and  A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B. Let p
∈
(
α
,
2
] be fixed. Using an approximation argument and the construction of the integrals I
(
X
) and I
K
(
X
), it follows that, for any X
∈
ℒ
α and for any L
>
K, a.s. on {
τ
K
(
B
)
>
T
}, we have
(177)
I
(
X
)
(
T
,
B
)
=
I
K
(
X
)
(
T
,
B
)
-
b
K
∫
0
T
‍
∫
𝒪
‍
X
(
s
,
y
)
d
y
d
s
=
I
L
(
X
)
(
T
,
B
)
-
b
L
∫
0
T
‍
∫
𝒪
‍
X
(
s
,
y
)
d
y
d
s
.We denoteτ
K
=
τ
K
(
𝒪
). We consider the following equation:
(178)
L
u
(
t
,
x
)
=
σ
(
u
(
t
,
x
)
)
Z
˙
K
(
t
,
x
)
-
b
K
σ
(
u
(
t
,
x
)
)
,
hhhhhhhhhhhhhhhhllhl
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions. A mild solution of (178) is a predictable process u which satisfies
(179)
u
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
d
y
d
s
a
.
s
.
for any t
>
0, x
∈
𝒪. The existence and uniqueness of a mild solution of (178) can be proved similarly to Theorem 25. We omit these details. We denote this solution by v
K.Lemma 30.
Under the assumptions of Theorem21, for any t
>
0, x
∈
𝒪, and L
>
K,
(180)
v
K
(
t
,
x
)
=
v
L
(
t
,
x
)
a
.
s
.
o
n
{
t
≤
τ
K
}
.Proof.
By the definition ofv
L and (177), a.s. on the event {
t
≤
τ
K
}, v
L
(
t
,
x
) is equal to
(181)
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
Z
L
(
d
s
,
d
y
)
-
b
L
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
d
y
d
s
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
d
y
d
s
.
Using the definition ofv
K and Proposition C.1 (Appendix C), we obtain that, with probability 1,
(182)
(
v
K
(
t
,
x
)
-
v
L
(
t
,
x
)
)
1
{
t
≤
τ
K
}
=
1
{
t
≤
τ
K
}
(
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
-
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
d
y
d
s
∫
0
t
‍
∫
𝒪
‍
)
.
LettingM
(
t
)
=
sup
x
∈
𝒪
E
(
|
v
K
(
t
,
x
)
-
v
L
(
t
,
x
)
|
p
1
{
t
≤
τ
K
}
), we see that M
(
t
)
≤
2
p
-
1
(
E
|
A
(
t
,
x
)
|
p
+
E
|
B
(
t
,
x
)
|
p
) where
(183)
A
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
,
B
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
d
y
d
s
.
We estimate separately the two terms. For the first term, we use the moment inequality (129) and the Lipschitz condition (135). We get
(184)
sup
x
∈
𝒪
E
|
A
(
t
,
x
)
|
p
≤
C
∫
0
t
‍
J
p
(
t
-
s
)
M
(
s
)
d
s
,
where C
=
C
α
,
p
K
p
-
α
C
σ
p. For the second term, we use Hölder’s inequality |
∫
f
g
d
μ
|
≤
(
∫
‍
|
f
|
p
d
μ
)
1
/
p
(
∫
‍
|
g
|
q
d
μ
)
1
/
q with f
(
s
,
y
)
=
G
(
t
-
s
,
x
,
y
)
1
/
p  (
σ
(
v
K
(
s
,
y
)
)
)   -
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
} and g
(
s
,
y
)
=
G
(
t
-
s
,
x
,
y
)
1
/
q, where p
-
1
+
q
-
1
=
1. Hence,
(185)
|
B
(
t
,
x
)
|
p
≤
C
σ
p
K
t
p
/
q
hh
×
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
hhhhhhhh
×
|
v
K
(
s
,
y
)
-
v
L
(
s
,
y
)
|
p
1
{
s
≤
τ
K
}
d
y
d
s
,
where K
t
=
∫
0
t
‍
J
1
(
s
)
d
s
<
∞ (Since 𝒪 is a bounded set, J
1
(
s
)
≤
C
J
p
(
s
)
1
/
p where C is a constant depending on |
𝒪
| and p. Since p
>
1, ∫
0
t
‍
J
p
(
s
)
1
/
p
d
s
≤
c
t
(
∫
0
t
‍
J
p
(
s
)
d
s
)
1
/
p
<
∞ by (133). This shows that K
t
<
∞.). Therefore,
(186)
sup
x
∈
𝒪
E
|
B
(
t
,
x
)
|
p
≤
C
t
∫
0
t
‍
J
1
(
t
-
s
)
M
(
s
)
d
s
,
where C
t
=
C
σ
p
K
t
p
/
q. From (184) and (186), we obtain that
(187)
M
(
t
)
≤
C
t
′
∫
0
t
‍
(
J
p
(
t
-
s
)
+
J
1
(
t
-
s
)
)
M
(
s
)
d
s
,
where C
t
′
=
2
p
-
1
(
C
∨
C
t
). This implies that M
(
t
)
=
0 for all t
>
0.For anyt
>
0 and x
∈
𝒪, we let Ω
t
,
x
=
⋂
L
>
K
‍
{
t
≤
τ
K
,
v
K
(
t
,
x
)
≠
v
L
(
t
,
x
)
} where K and L are positive integers, and Ω
t
,
x
*
=
Ω
t
,
x
∩
{
lim
K
→
∞
τ
K
=
∞
}. By Lemma 30, P
(
Ω
t
,
x
*
)
=
1.Proposition 31.
Under the assumptions of Theorem21, the process u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} defined by
(188)
u
(
ω
,
t
,
x
)
=
v
K
(
ω
,
t
,
x
)
,
i
f
ω
∈
Ω
t
,
x
*
,
t
≤
τ
K
(
ω
)
,
u
(
ω
,
t
,
x
)
=
0
,
i
f
ω
∉
Ω
t
,
x
*
is a mild solution of (1).Proof.
We proceed as in the proof of Proposition29. In this case, with probability 1, we have
(189)
1
{
t
≤
τ
K
}
u
(
t
,
x
)
=
1
{
t
≤
τ
K
}
(
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
d
y
d
s
)
.
The conclusion follows letting K
→
∞, since τ
K
→
∞ a.s. and b
K
→
0.


## 6.1. The Equation with Truncated Noise

In this section, we fixK
>
0 and we consider the equation:
(139)
L
u
(
t
,
x
)
=
σ
(
u
(
t
,
x
)
)
Z
˙
K
(
t
,
x
)
,
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions. A mild solution of (139) is a predictable process u which satisfies (2) with Z replaced by Z
K. For the next result, 𝒪 can be a bounded domain in ℝ
d or 𝒪
=
ℝ
d (with no boundary conditions).Theorem 25.
Under the assumptions of Theorem21, (139) has a unique mild solution u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
}. For any T
>
0,
(140)
sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
(
t
,
x
)
|
p
<
∞
,
and the map (
t
,
x
)
↦
u
(
t
,
x
) is continuous from [
0
,
T
]
×
𝒪 into L
p
(
Ω
).Proof.
We use the same argument as in the proof of Theorem 13 of [27], based on a Picard iteration scheme. We define u
0
(
t
,
x
)
=
0 and
(141)
u
n
+
1
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
for any n
≥
0. We prove by induction on n
≥
0 that (i)  u
n
(
t
,
x
) is well defined; (ii) K
n
(
t
)
:
=
sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
n
(
t
,
x
)
|
p
<
∞ for any T
>
0; (iii) u
n
(
t
,
x
) is ℱ
t-measurable for any t
>
0 and x
∈
𝒪; (iv) the map (
t
,
x
)
↦
u
n
(
t
,
x
) is continuous from [
0
,
T
]
×
𝒪 into L
p
(
Ω
) for any T
>
0.
The statement is trivial forn
=
0. For the induction step, assume that the statement is true for n. By an extension to random fields of Theorem 30, Chapter  IV of [28], u
n has a jointly measurable modification. Since this modification is (
ℱ
t
)
t-adapted (in the sense of (iii)), it has a predictable modification (using an extension of Proposition 3.21 of [12] to random fields). We work with this modification, that we call also u
n.
We prove that (i)–(iv) hold foru
n
+
1. To show (i), it suffices to prove that X
n
∈
ℒ
p, where X
n
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
n
(
s
,
y
)
). By (136) and (133),
(142)
E
∫
0
t
‍
∫
𝒪
‍
|
X
n
(
s
,
y
)
|
p
d
y
d
s
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
∫
0
t
‍
J
p
(
t
-
s
)
d
s
<
∞
.
In addition, if 𝒪
=
ℝ
d, we have to prove that X
n satisfies (83) if α
<
1, or (116) if α
>
1 (see Table 1). If α
<
1, this follows as above, since α
<
p and hence sup
(
t
,
x
)
∈
[
0
,
T
]
×
𝒪
E
|
u
(
t
,
x
)
|
α
<
∞; the argument for α
>
1 is similar.
Combined with the moment inequality (115) (or (129)), this proves (ii), since
(143)
E
|
u
n
+
1
(
t
,
x
)
|
p
≤
C
α
,
p
K
p
-
α
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
∫
0
t
‍
J
p
(
t
-
s
)
d
s
,
for any x
∈
𝒪. Property (iii) follows by the construction of the integral I
K.
To prove (iv), we first show the right continuity int. Let h
>
0. Writing the interval [
0
,
t
+
h
] as the union of [
0
,
t
] and (
t
,
t
+
h
], we obtain that E
|
u
n
+
1
(
t
+
h
,
x
)
-
u
n
+
1
(
t
,
x
)
|
p
≤
2
p
-
1
(
I
1
(
h
)
+
I
2
(
h
)
), where
(144)
I
1
(
h
)
=
E
|
∫
0
t
‍
∫
𝒪
‍
(
G
(
t
+
h
-
s
,
x
,
y
)
-
G
(
t
-
s
,
x
,
y
)
)
hhhhhhhlh
×
σ
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
∫
0
t
‍
∫
𝒪
‍
|
p
,
I
2
(
h
)
=
E
|
∫
t
t
+
h
‍
∫
𝒪
‍
G
(
t
+
h
-
s
,
x
,
y
)
σ
hhhhhhhhllh
×
(
u
n
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
∫
t
t
+
h
‍
∫
𝒪
‍
|
p
.
Using again (136) and the moment inequality (115) (or (129)), we obtain
(145)
I
1
(
h
)
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
t
‍
∫
𝒪
‍
|
G
(
s
+
h
,
x
,
y
)
-
G
(
s
,
x
,
y
)
|
p
d
y
d
s
,
I
2
(
h
)
≤
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
h
‍
∫
𝒪
‍
G
(
s
,
x
,
y
)
p
d
y
d
s
.
It follows that both I
1
(
h
) and I
2
(
h
) converge to 0 as h
→
0, using (131) for I
1
(
h
) and the Dominated Convergence Theorem and (133) for I
2
(
h
), respectively. The left continuity in t is similar, by writing the interval [
0
,
t
-
h
] as the difference between [
0
,
t
] and (
t
-
h
,
t
] for h
>
0. For the continuity in x, similarly as above, we see that E
|
u
n
+
1
(
t
,
x
+
h
)
-
u
n
+
1
(
t
,
x
)
|
p is bounded by
(146)
D
σ
p
2
p
-
1
(
1
+
K
n
(
t
)
)
×
∫
0
t
‍
∫
𝒪
‍
|
G
(
s
,
x
+
h
,
y
)
-
G
(
s
,
x
,
y
)
|
p
d
y
d
s
,
which converges to 0 as |
h
|
→
0 due to (132). This finishes the proof of (iv).
We denoteM
n
(
t
)
=
sup
x
∈
𝒪
E
|
u
n
(
t
,
x
)
|
p. Similarly to (143), we have
(147)
M
n
(
t
)
≤
C
1
∫
0
t
‍
(
1
+
M
n
-
1
(
s
)
)
J
p
(
t
-
s
)
d
s
,
∀
n
≥
1
,
where C
1
=
C
α
,
p
K
p
-
α
D
σ
p
2
p
-
1. By applying Lemma 15 of Erratum to [27] with f
n
=
M
n, k
1
=
0, k
2
=
1, and g
(
s
)
=
C
J
p
(
s
), we obtain that
(148)
sup
n
≥
0
sup
t
∈
[
0
,
T
]
M
n
(
t
)
<
∞
,
∀
T
>
0
.
We now prove that{
u
n
(
t
,
x
)
}
n converges in L
p
(
Ω
), uniformly in (
t
,
x
)
∈
[
0
,
T
]
×
𝒪. To see this, let U
n
(
t
)
=
sup
x
∈
𝒪
E
|
u
n
+
1
(
t
,
x
)
-
u
n
(
t
,
x
)
|
p for n
≥
0. Using the moment inequality (115) (or (129)) and (135), we have
(149)
U
n
(
t
)
≤
C
2
∫
0
t
‍
U
n
-
1
(
s
)
J
p
(
t
-
s
)
d
s
,
where C
2
=
C
α
,
p
K
p
-
α
C
σ
p. By Lemma 15 of Erratum to [27], ∑
n
≥
0
‍
U
n
(
t
)
1
/
p converges uniformly on [
0
,
T
] (Note that this lemma is valid for all p
>
0.).
We denote byu
(
t
,
x
) the limit of u
n
(
t
,
x
) in L
p
(
Ω
). One can show that u satisfies properties (ii)–(iv) listed above. So u has a predictable modification. This modification is a solution of (139). To prove uniqueness, let v be another solution and denote H
(
t
)
=
sup
x
∈
𝒪
E
|
u
(
t
,
x
)
-
v
(
t
,
x
)
|
p. Then
(150)
H
(
t
)
≤
C
2
∫
0
t
‍
H
(
s
)
J
p
(
t
-
s
)
d
s
.
Using (133), it follows that H
(
t
)
=
0 for all t
>
0.


## 6.2. Proof of Theorem21: Case α
<
1

In this case, for anyt
>
0 and B
∈
ℬ
b
(
ℝ
d
), we have (see (21))
(151)
Z
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
(
d
s
,
d
x
,
d
z
)
.
The characteristic function of Z
(
t
,
B
) is given by
(152)
E
(
e
i
u
Z
(
t
,
B
)
)
=
exp
{
t
|
B
|
∫
ℝ
∖
{
0
}
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
hhhhhhhhhhhhhhhhhhhhhhl
∀
u
∈
ℝ
.
Note that {
Z
(
t
,
B
)
}
t
≥
0 is not a compound Poisson process since ν
α is infinite.We introduce the stopping times(
τ
K
)
K
≥
1, as on page 239 of [13]:
(153)
τ
K
(
B
)
=
inf
{
t
>
0
;
|
Z
(
t
,
B
)
-
Z
(
t
-
,
B
)
|
>
K
}
,
where Z
(
t
-
,
B
)
=
lim
s
↑
t
Z
(
s
,
B
). Clearly, τ
L
(
B
)
≥
τ
K
(
B
) for all L
>
K.We first investigate the relationship betweenZ and Z
K and the properties of τ
K
(
B
). Using construction (138) of N and definition (87) of Z
K, we have
(154)
Z
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
=
:
∑
j
,
k
≥
1
‍
Z
j
,
k
(
t
,
B
)
,
Z
K
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
|
Z
i
j
,
k
|
≤
K
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
.We observe that{
Z
j
,
k
(
t
,
B
)
}
t
≥
0 is a compound Poisson process with
(155)
E
(
e
i
u
Z
j
,
k
(
t
,
B
)
)
=
exp
{
t
|
𝒪
k
∩
B
|
∫
U
j
‍
(
e
i
u
z
-
1
)
ν
α
(
d
z
)
}
,
∀
u
∈
ℝ
.Note thatτ
K
(
B
)
>
T means that all the jumps of {
Z
(
t
,
B
)
}
t
≥
0 in [
0
,
T
] are smaller than K in modulus; that is, {
τ
K
(
B
)
>
T
}
=
{
ω
;
|
Z
i
j
,
k
(
ω
)
|
≤
K for  all  i
,
j
,
k
≥
1 for  which T
i
j
,
k
(
ω
)
≤
T and X
i
j
,
k
(
ω
)
∈
B
}. Hence, on {
τ
K
(
B
)
>
T
},
(156)
Z
(
[
0
,
t
]
×
A
)
=
Z
K
(
[
0
,
t
]
×
A
)
=
Z
L
(
[
0
,
t
]
×
A
)
,
for any L
>
K, t
∈
[
0
,
T
], and A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B. Using an approximation argument and the construction of the integrals I
(
X
) and I
K
(
X
), it follows that, for any X
∈
ℒ
α and for any L
>
K, a.s. on {
τ
K
(
B
)
>
T
}, we have
(157)
I
(
X
)
(
T
,
B
)
=
I
K
(
X
)
(
T
,
B
)
=
I
L
(
X
)
(
T
,
B
)
.The next result gives the probability of the event{
τ
K
(
B
)
>
T
}.Lemma 26.
For anyT
>
0 and B
∈
ℬ
b
(
ℝ
d
),
(158)
P
(
τ
K
(
B
)
>
T
)
=
exp
(
-
T
|
B
|
K
-
α
)
.
Consequently, lim
K
→
∞
P
(
τ
K
(
B
)
>
T
)
=
1 and lim
K
→
∞
τ
K
(
B
)
=
∞ a.s.Proof.
Note that{
τ
K
(
B
)
>
T
}
=
⋂
j
,
k
≥
1
‍
{
τ
K
j
,
k
(
B
)
>
T
}, where
(159)
τ
K
j
,
k
(
B
)
=
inf
{
t
>
0
;
|
Z
j
,
k
(
t
,
B
)
-
Z
j
,
k
(
t
-
,
B
)
|
>
K
}
.
Since ν
α
(
{
z
;
|
z
|
>
K
}
)
=
K
-
α and (
τ
K
j
,
k
(
B
)
)
j
,
k
≥
1 are independent, it is enough to prove that, for any j
,
k
≥
1,
(160)
P
(
τ
K
j
,
k
(
B
)
>
T
)
=
exp
{
-
T
|
B
∩
𝒪
k
|
ν
α
(
{
z
;
|
z
|
>
K
}
∩
U
j
)
}
.
Note that{
τ
K
j
,
k
(
B
)
>
T
}
=
{
ω
;
|
Z
i
j
,
k
(
ω
)
|
≤
K for  all i for  which T
i
j
,
k
≤
T and X
i
j
,
k
∈
B
} and (
T
n
j
,
k
)
n
≥
1 are the jump times of a Poisson process with intensity λ
j
,
k. Hence,
(161)
P
(
τ
K
j
,
k
(
B
)
>
T
)
hh
=
∑
n
≥
0
‍
∑
m
=
0
n
‍
∑
I
⊂
{
1
,
…
,
n
}
,
card
(
I
)
=
m
‍
P
(
T
n
j
,
k
≤
T
<
T
n
+
1
j
,
k
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
‍
{
X
i
j
,
k
∈
B
}
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
‍
{
|
Z
i
j
,
k
|
≤
K
}
)
hhhhhhhhhhhhhhhhhhhhhl
×
P
(
⋂
i
∈
I
c
‍
{
X
i
j
,
k
∉
B
}
)
hh
=
∑
n
≥
0
‍
e
-
λ
j
,
k
T
(
λ
j
,
k
T
)
n
n
!
hhhhhhh
×
[
1
-
P
(
X
1
j
,
k
∈
B
)
P
(
|
Z
1
j
,
k
|
>
K
)
]
n
hh
=
exp
{
-
λ
j
,
k
T
P
(
X
1
j
,
k
∈
B
)
P
(
|
Z
1
j
,
k
|
>
K
)
}
,
which yields (160).
To prove the last statement, letA
k
(
n
)
=
{
τ
K
(
B
)
>
n
}. Then P
(
lim
¯
K
A
K
(
n
)
)
≥
lim
¯
K
P
(
A
K
(
n
)
)
=
1 for any n
≥
1, and hence P
(
⋂
n
≥
1
‍
lim
¯
K
A
K
(
n
)
)
=
1. Hence, with probability 1, for any n, there exists some K
n such that τ
K
n
>
n. Since (
τ
K
)
K is nondecreasing, this proves that τ
K
→
∞ with probability 1.Remark 27.
The construction ofτ
K
(
B
) given above is due to [13] (in the case of a symmetric measure ν
α). This construction relies on the fact that B is a bounded set. Since Z
(
t
,
ℝ
d
) (and consequently τ
K
(
ℝ
d
)) is not well defined, we could not see why this construction can also be used when B
=
ℝ
d, as it is claimed in [13]. To avoid this difficulty, one could try to use an increasing sequence (
E
n
)
n of sets in ℬ
b
(
ℝ
d
) with ⋃
n
‍
E
n
=
ℝ
d. Using (157) with B
=
E
n and letting n
→
∞, we obtain that I
(
X
)
(
t
,
ℝ
d
)
=
I
K
(
t
,
ℝ
d
) a.s. on {
t
≤
τ
K
}, where τ
K
=
inf
n
≥
1
τ
K
(
E
n
). But P
(
τ
K
>
t
)
≤ P
(
lim
_
n
{
τ
K
(
E
n
)
>
t
}
)
≤ lim
_
n
P
(
τ
K
(
E
n
)
>
t
)
= lim
n
exp
(
-
t
|
E
n
|
K
-
α
)
=
0 for any t
>
0, which means that τ
K
=
0 a.s. Finding a suitable sequence (
τ
K
)
K of stopping times which could be used in the case 𝒪
=
ℝ
d remains an open problem.In what follows, we denoteτ
K
=
τ
K
(
𝒪
). Let u
K be the solution of (139), whose existence is guaranteed by Theorem 25.Lemma 28.
Under the assumptions of Theorem21, for any t
>
0, x
∈
𝒪, and L
>
K,
(162)
u
K
(
t
,
x
)
=
u
L
(
t
,
x
)
a
.
s
.
o
n
{
t
≤
τ
K
}
.Proof.
By the definition ofu
L and (157),
(163)
u
L
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
L
(
s
,
y
)
)
Z
L
(
d
s
,
d
y
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
L
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
a.s. on the event {
t
≤
τ
K
}. Using the definition of u
K and Proposition C.1 (Appendix C), we obtain that, with probability 1,
(164)
(
u
K
(
t
,
x
)
-
u
L
(
t
,
x
)
)
1
{
t
≤
τ
K
}
hh
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
hhhhhhhhhhhhl
×
(
σ
(
u
K
(
s
,
y
)
)
-
σ
(
u
L
(
s
,
y
)
)
)
hhhhhhhhhhhhl
×
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
.
LetM
(
t
)
=
sup
x
∈
𝒪
E
(
|
u
K
(
t
,
x
)
-
u
L
(
t
,
x
)
|
p
1
{
t
≤
τ
K
}
). Using the moment inequality (115) and the Lipschitz condition (135), we get
(165)
M
(
t
)
≤
C
∫
0
t
‍
J
p
(
t
-
s
)
M
(
s
)
d
s
,
where C
=
C
α
,
p
K
p
-
α
C
σ
p. Using (133), it follows that M
(
t
)
=
0 for all t
>
0.For anyt
>
0 and x
∈
𝒪, let Ω
t
,
x
=
⋂
L
>
K
‍
{
t
≤
τ
K
(
t
)
,
u
K
(
t
,
x
)
≠
u
L
(
t
,
x
)
}, where L and K are positive integers. Let Ω
t
,
x
*
=
Ω
t
,
x
∩
{
lim
K
→
∞
τ
K
=
∞
}.By Lemmas26 and 28, P
(
Ω
t
,
x
*
)
=
1.The next result concludes the proof of Theorem21.Proposition 29.
Under the assumptions of Theorem21, the process u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} defined by
(166)
u
(
ω
,
t
,
x
)
=
u
K
(
ω
,
t
,
x
)
,
i
f
ω
∈
Ω
t
,
x
*
,
t
≤
τ
K
(
ω
)
u
(
ω
,
t
,
x
)
=
0
,
i
f
ω
∉
Ω
t
,
x
*
is a mild solution of (1).Proof.
We first prove thatu is predictable. Note that
(167)
u
(
t
,
x
)
=
lim
K
→
∞
(
u
K
(
t
,
x
)
1
{
t
≤
τ
K
}
)
1
Ω
t
,
x
*
.
The process X
(
ω
,
t
,
x
)
=
1
{
t
≤
τ
K
}
(
ω
) is clearly predictable, being in the class 𝒞 defined in Remark 11. By the definition of Ω
t
,
x, since u
K
,
u
L are predictable, it follows that (
ω
,
t
,
x
)
↦
1
Ω
t
,
x
*
(
ω
) is 𝒫-measurable. Hence, u is predictable.
We now prove thatu satisfies (2). Let t
>
0 and x
∈
𝒪 be arbitrary. Using (157) and Proposition C.1 (Appendix C), with probability 1, we have
(168)
1
{
t
≤
τ
K
}
u
(
t
,
x
)
=
1
{
t
≤
τ
K
}
u
K
(
t
,
x
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
K
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
(
d
s
,
d
y
)
=
1
{
t
≤
τ
K
}
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
×
(
u
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
.
For the second last equality, we used the fact that processes X
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
K
(
s
,
y
)
)
1
{
s
≤
τ
K
} and Y
(
s
,
y
)
=
1
[
0
,
t
]
(
s
)
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
1
{
s
≤
τ
K
} are modifications of each other (i.e., X
(
s
,
y
)
=
Y
(
s
,
y
) a.s. for all s
>
0, y
∈
𝒪), and, hence, [
X
-
Y
]
α
,
t
,
𝒪
=
0 and I
(
X
)
(
t
,
𝒪
)
=
I
(
Y
)
(
t
,
𝒪
) a.s. The conclusion follows letting K
→
∞, since τ
K
→
∞ a.s.


## 6.3.  Proof of Theorem21: Case α
>
1

In this case, for anyt
>
0 and B
∈
ℬ
b
(
ℝ
d
), we have (see (22))
(169)
Z
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
N
^
(
d
s
,
d
x
,
d
z
)
.To introduce the stopping times(
τ
K
)
K
≥
1 we use the same idea as in Section 9.7 of [12].LetM
(
t
,
B
)
=
∑
j
≥
1
‍
(
L
j
(
t
,
B
)
-
E
L
j
(
t
,
B
)
) and P
(
t
,
B
)
=
L
0
(
t
,
B
), where L
j
(
t
,
B
)
=
L
j
(
[
0
,
t
]
×
B
) was defined in Section 2. Note that {
M
(
t
,
B
)
}
t
≥
0 is a zero-mean square-integrable martingale and {
P
(
t
,
B
)
}
t
≥
0 is a compound Poisson process with E
[
P
(
t
,
B
)
]
=
t
|
B
|
μ where μ
=
∫
|
z
|
>
1
‍
z
ν
α
(
d
z
)
=
β
(
α
/
(
α
-
1
)
). With this notation,
(170)
Z
(
t
,
B
)
=
M
(
t
,
B
)
+
P
(
t
,
B
)
-
t
|
B
|
μ
.We letM
K
(
t
,
B
)
=
P
K
(
t
,
B
)
-
E
[
P
K
(
t
,
B
)
]
=
P
K
(
t
,
B
)
-
t
|
B
|
μ
K, where
(171)
P
K
(
t
,
B
)
=
∫
[
0
,
t
]
×
B
×
(
ℝ
∖
{
0
}
)
‍
z
1
{
1
<
|
z
|
≤
K
}
N
(
d
s
,
d
x
,
d
z
)
and μ
K
=
∫
1
<
|
z
|
≤
K
‍
z
ν
α
(
d
z
). Recalling definition (88) of Z
K, it follows that
(172)
Z
K
(
t
,
B
)
=
M
(
t
,
B
)
+
P
K
(
t
,
B
)
-
t
|
B
|
μ
K
.For anyK
>
0, we let
(173)
τ
K
(
B
)
=
inf
{
t
>
0
;
|
P
(
t
,
B
)
-
P
(
t
-
,
B
)
|
>
K
}
,
where P
(
t
-
,
B
)
=
lim
s
↑
t
P
(
s
,
B
).Lemma26 holds again, but its proof is simpler than in the case α
<
1, since {
P
(
t
,
B
)
}
t
≥
0 is a compound Poisson process. By (138),
(174)
P
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
|
Z
i
j
,
k
|
>
1
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
,
P
K
(
t
,
B
)
=
∑
i
,
j
,
k
≥
1
‍
Z
i
j
,
k
1
{
1
<
|
Z
i
j
,
k
|
≤
K
}
1
{
T
i
j
,
k
≤
t
}
1
{
X
i
j
,
k
∈
B
}
.
Hence, on {
τ
K
(
B
)
>
T
}, for any L
>
K, t
∈
[
0
,
T
], and  A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B,
(175)
P
(
[
0
,
t
]
×
A
)
=
P
K
(
[
0
,
t
]
×
A
)
=
P
L
(
[
0
,
t
]
×
A
)
.Letb
K
=
μ
-
μ
K
=
∫
|
z
|
>
K
‍
z
ν
α
(
d
z
). Using (170) and (172), it follows that
(176)
Z
(
[
0
,
t
]
×
A
)
=
Z
K
(
[
0
,
t
]
×
A
)
-
t
|
A
|
b
K
=
Z
L
(
[
0
,
t
]
×
A
)
-
t
|
A
|
b
L
for any L
>
K, t
∈
[
0
,
T
], and  A
∈
ℬ
b
(
ℝ
d
) with A
⊂
B. Let p
∈
(
α
,
2
] be fixed. Using an approximation argument and the construction of the integrals I
(
X
) and I
K
(
X
), it follows that, for any X
∈
ℒ
α and for any L
>
K, a.s. on {
τ
K
(
B
)
>
T
}, we have
(177)
I
(
X
)
(
T
,
B
)
=
I
K
(
X
)
(
T
,
B
)
-
b
K
∫
0
T
‍
∫
𝒪
‍
X
(
s
,
y
)
d
y
d
s
=
I
L
(
X
)
(
T
,
B
)
-
b
L
∫
0
T
‍
∫
𝒪
‍
X
(
s
,
y
)
d
y
d
s
.We denoteτ
K
=
τ
K
(
𝒪
). We consider the following equation:
(178)
L
u
(
t
,
x
)
=
σ
(
u
(
t
,
x
)
)
Z
˙
K
(
t
,
x
)
-
b
K
σ
(
u
(
t
,
x
)
)
,
hhhhhhhhhhhhhhhhllhl
t
>
0
,
x
∈
𝒪
with zero initial conditions and Dirichlet boundary conditions. A mild solution of (178) is a predictable process u which satisfies
(179)
u
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
d
y
d
s
a
.
s
.
for any t
>
0, x
∈
𝒪. The existence and uniqueness of a mild solution of (178) can be proved similarly to Theorem 25. We omit these details. We denote this solution by v
K.Lemma 30.
Under the assumptions of Theorem21, for any t
>
0, x
∈
𝒪, and L
>
K,
(180)
v
K
(
t
,
x
)
=
v
L
(
t
,
x
)
a
.
s
.
o
n
{
t
≤
τ
K
}
.Proof.
By the definition ofv
L and (177), a.s. on the event {
t
≤
τ
K
}, v
L
(
t
,
x
) is equal to
(181)
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
Z
L
(
d
s
,
d
y
)
-
b
L
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
d
y
d
s
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
Z
K
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
v
L
(
s
,
y
)
)
d
y
d
s
.
Using the definition ofv
K and Proposition C.1 (Appendix C), we obtain that, with probability 1,
(182)
(
v
K
(
t
,
x
)
-
v
L
(
t
,
x
)
)
1
{
t
≤
τ
K
}
=
1
{
t
≤
τ
K
}
(
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
-
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
d
y
d
s
∫
0
t
‍
∫
𝒪
‍
)
.
LettingM
(
t
)
=
sup
x
∈
𝒪
E
(
|
v
K
(
t
,
x
)
-
v
L
(
t
,
x
)
|
p
1
{
t
≤
τ
K
}
), we see that M
(
t
)
≤
2
p
-
1
(
E
|
A
(
t
,
x
)
|
p
+
E
|
B
(
t
,
x
)
|
p
) where
(183)
A
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
Z
K
(
d
s
,
d
y
)
,
B
(
t
,
x
)
=
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
(
σ
(
v
K
(
s
,
y
)
)
)
-
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
}
d
y
d
s
.
We estimate separately the two terms. For the first term, we use the moment inequality (129) and the Lipschitz condition (135). We get
(184)
sup
x
∈
𝒪
E
|
A
(
t
,
x
)
|
p
≤
C
∫
0
t
‍
J
p
(
t
-
s
)
M
(
s
)
d
s
,
where C
=
C
α
,
p
K
p
-
α
C
σ
p. For the second term, we use Hölder’s inequality |
∫
f
g
d
μ
|
≤
(
∫
‍
|
f
|
p
d
μ
)
1
/
p
(
∫
‍
|
g
|
q
d
μ
)
1
/
q with f
(
s
,
y
)
=
G
(
t
-
s
,
x
,
y
)
1
/
p  (
σ
(
v
K
(
s
,
y
)
)
)   -
σ
(
v
L
(
s
,
y
)
)
1
{
s
≤
τ
K
} and g
(
s
,
y
)
=
G
(
t
-
s
,
x
,
y
)
1
/
q, where p
-
1
+
q
-
1
=
1. Hence,
(185)
|
B
(
t
,
x
)
|
p
≤
C
σ
p
K
t
p
/
q
hh
×
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
hhhhhhhh
×
|
v
K
(
s
,
y
)
-
v
L
(
s
,
y
)
|
p
1
{
s
≤
τ
K
}
d
y
d
s
,
where K
t
=
∫
0
t
‍
J
1
(
s
)
d
s
<
∞ (Since 𝒪 is a bounded set, J
1
(
s
)
≤
C
J
p
(
s
)
1
/
p where C is a constant depending on |
𝒪
| and p. Since p
>
1, ∫
0
t
‍
J
p
(
s
)
1
/
p
d
s
≤
c
t
(
∫
0
t
‍
J
p
(
s
)
d
s
)
1
/
p
<
∞ by (133). This shows that K
t
<
∞.). Therefore,
(186)
sup
x
∈
𝒪
E
|
B
(
t
,
x
)
|
p
≤
C
t
∫
0
t
‍
J
1
(
t
-
s
)
M
(
s
)
d
s
,
where C
t
=
C
σ
p
K
t
p
/
q. From (184) and (186), we obtain that
(187)
M
(
t
)
≤
C
t
′
∫
0
t
‍
(
J
p
(
t
-
s
)
+
J
1
(
t
-
s
)
)
M
(
s
)
d
s
,
where C
t
′
=
2
p
-
1
(
C
∨
C
t
). This implies that M
(
t
)
=
0 for all t
>
0.For anyt
>
0 and x
∈
𝒪, we let Ω
t
,
x
=
⋂
L
>
K
‍
{
t
≤
τ
K
,
v
K
(
t
,
x
)
≠
v
L
(
t
,
x
)
} where K and L are positive integers, and Ω
t
,
x
*
=
Ω
t
,
x
∩
{
lim
K
→
∞
τ
K
=
∞
}. By Lemma 30, P
(
Ω
t
,
x
*
)
=
1.Proposition 31.
Under the assumptions of Theorem21, the process u
=
{
u
(
t
,
x
)
;
t
≥
0
,
x
∈
𝒪
} defined by
(188)
u
(
ω
,
t
,
x
)
=
v
K
(
ω
,
t
,
x
)
,
i
f
ω
∈
Ω
t
,
x
*
,
t
≤
τ
K
(
ω
)
,
u
(
ω
,
t
,
x
)
=
0
,
i
f
ω
∉
Ω
t
,
x
*
is a mild solution of (1).Proof.
We proceed as in the proof of Proposition29. In this case, with probability 1, we have
(189)
1
{
t
≤
τ
K
}
u
(
t
,
x
)
=
1
{
t
≤
τ
K
}
(
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
Z
(
d
s
,
d
y
)
-
b
K
∫
0
t
‍
∫
𝒪
‍
G
(
t
-
s
,
x
,
y
)
σ
(
u
(
s
,
y
)
)
d
y
d
s
)
.
The conclusion follows letting K
→
∞, since τ
K
→
∞ a.s. and b
K
→
0.



---

*Source: 793275-2014-02-04.xml*