# Weak Uniqueness for the Stochastic Heat Equation Driven by a Multiplicative
Stable Noise

Sayantan Maitra

###### Abstract.

We consider the stochastic heat equation

($\star$) $\displaystyle\frac{\partial Y_{t}(x)}{\partial
t}=\frac{1}{2}\Delta_{x}Y_{t}(x)+Y_{t-}(x)^{\beta}\dot{L}^{\alpha},$

with $t\geq 0$, $x\in\mathbb{R}$ and $L^{\alpha}$ being an $\alpha$-stable
white noise without negative jumps. When $\alpha\in(1,2)$ and
$\beta\in(\frac{1}{\alpha},1)$ we prove that weak uniqueness holds for
($\star$ ‣ Weak Uniqueness for the Stochastic Heat Equation Driven by a
Multiplicative Stable Noise) using the approximating duality approach
developed by Mytnik[Myt98].

###### Key words and phrases:

Weak uniqueness; Stochastic heat equation; Stable noise; Martingale problem;
Approximating duality

###### 2020 Mathematics Subject Classification:

60H15; 35R60; 60J68

Indian Statistical Institute, Bangalore Centre, 8th Mile Mysore Road,
Bengaluru - 560059, Karnakata.
E-mail<EMAIL_ADDRESS>
Weak Uniqueness for the Stochastic Heat Equation Driven by a Multiplicative
Stable Noise

##  1\. Introduction

In this paper we study the stochastic equation

(1.1) $\displaystyle\frac{\partial Y_{t}(x)}{\partial t}$
$\displaystyle=\frac{1}{2}\Delta_{x}Y_{t}(x)+Y_{t-}(x)^{\beta}\dot{L}^{\alpha},\qquad
t\geq 0,x\in\mathbb{R}$

where $L^{\alpha}$ is a stable noise of index $\alpha$ without negative jumps.
We show that its solutions are unique in law when $1<\alpha<2$, $0<\beta<1$
and $1<\alpha\beta$ (see Theorem 1.2). Proving uniqueness in law, also called
weak uniqueness, is an important step for establishing that a model of an
underlying system of interacting particles converges to a limit.

When $\alpha=2$ the above equation is similar to the following SPDE

(1.2) $\displaystyle\frac{\partial Y_{t}(x)}{\partial
t}=\frac{1}{2}\Delta_{x}Y_{t}(x)+Y_{t}(x)^{\beta}\dot{W}_{t,x}$

where $\dot{W}$ is the Gaussian space-time white noise. When
$\beta=\frac{1}{2}$ this describes the density process of the super-Brownian
motion (SBM) in $\mathbb{R}$ which can be obtained as the scaling limit of
interacting branching Brownian motions. The weak uniqueness of (1.2) with
$\beta=\frac{1}{2}$ follows from the martingale problem formulation of SBM
using duality (see [Per02, Theorem II.5.1]). In the $\beta\in(\frac{1}{2},1)$
case, the weak existence of (1.2) was proved by Mueller, Perkins [MP92] and
the weak uniqueness was established by Mytnik [Myt98]. The question of
pathwise uniqueness of (1.2) for $\beta>\frac{3}{4}$ was settled by Mytnik and
Perkins in 2011 [MP11]. Some negative results are also known: [MMP14] proved
pathwise non-uniqueness of solutions to (1.2) for $\beta\in(0,\frac{3}{4})$
and [BMP10] showed pathwise non-uniqueness for $\beta\in(0,\frac{1}{2})$ with
an added non-trivial drift.

Let us now consider (1.1) where $x\in\mathbb{R}^{d}$. For $\alpha\neq 2$,
Mueller [Mue98] proved a certain short time (strong) existence of solution to
(1.1) under the relations
$d<\frac{2(1-\alpha)}{\alpha\beta-(1-\alpha)},\alpha\in(0,1)$. The weak
existence was shown by [Myt02] under the relations
$0<\alpha\beta<\frac{2}{d}+1$, $1<\alpha<\min(2,\frac{2}{d}+1)$. When $d=1$
and $\alpha\beta=1$ it is known that (1.1) describes the density of the super-
Brownian motion with $\alpha$-stable branching mechanism; see [MP03] for more
details. The weak uniqueness for this case was proved in [Myt02] while the
same for the general case was left open (see [Myt02, Remark 5.9]). As stated
earlier in our main result we resolve the question for the case $d=1$,
$1<\alpha<2$, $0<\beta<1$ and $1<\alpha\beta$. We note that the weak
uniqueness for (1.1) when $d=1$, $1<\alpha<2$, $\alpha\beta<1$ still remains
open.

It is known that pathwise uniqueness implies weak uniqueness for (1.1). But as
the coefficients of the noise term in (1.1) is not Lipschitz, standard
techniques such as Grönwall’s inequality cannot be used to prove pathwise
uniqueness of solutions. However, more recently Yang and Zhou [YZ17] have
established pathwise uniqueness for (1.1) in the regime
$\frac{2(\alpha-1)}{(2-\alpha)^{2}}<\beta<\frac{1}{\alpha}+\frac{\alpha-1}{2}$
and $d=1$. This region partially overlaps with that stated in Theorem 1.2 when
$\beta>\frac{1}{\alpha}$.

In the next subsection we precisely define our model and state the main
theorem.

###  1.1. Model and Main Result

To define our model and state the main result we need to introduce the
following notations. Let $M_{F}\equiv M_{F}(\mathbb{R})$ be the set of all
non-negative finite measures on the real line, $\mathbb{R}$, with the topology
of weak convergence. We denote the space of all cadlag paths in $M_{F}$ as
$D\equiv D([0,\infty),M_{F})$. This space is equipped with the topology of
weak convergence and $\mathcal{B}(D)$ denotes the Borel $\sigma$-algebra on
$D$. Similarly, $\mathcal{B}(R)$ denotes the $\sigma$-algebra of all Borel
measurable subsets of $\mathbb{R}$ and for $A\in\mathcal{B}(\mathbb{R})$ we
use $|A|$ for the Lebesgue measure of $A$. Also let
$\mathcal{S}\equiv\mathcal{S}(\mathbb{R})$ be the space of all smooth rapidly-
decreasing functions defined on $\mathbb{R}$ whose derivatives of all orders
are also rapidly-decreasing and
$\mathcal{S}_{+}\equiv\mathcal{S}(\mathbb{R})_{+}$ be its subset containing
all non-negative functions. For $\mu\in M_{F}$ and $\varphi\in\mathcal{S}$ we
denote $\langle\mu,\varphi\rangle=\int_{\mathbb{R}}\varphi\,d\mu$.

######  Definition 1.1.

We say a stochastic process $Y\in D([0,\infty),M_{F})$ on some probability
space $(\Omega,\mathcal{F},\mathbb{P})$ solves (1.1) if for all
$\psi\in\mathcal{S}(\mathbb{R})$ and $t\geq 0$,

(1.3) $\displaystyle\langle Y_{t},\psi\rangle=\langle
Y_{0},\psi\rangle+\int_{0}^{t}\langle
Y_{s},\frac{1}{2}\Delta\psi\rangle\,ds+\int_{0}^{t}\int_{\mathbb{R}}(Y_{s-}(x))^{\beta}\psi(x)L^{\alpha}(\,ds\,dx),$

where $L^{\alpha}$ is a stable martingale measure (in Walsh’s sense) on
$[0,\infty)\times\mathbb{R}$ of index $\alpha>0$ without negative jumps. More
precisely, for each $A\in\mathcal{B}(\mathbb{R})$ with $|A|<\infty$,
$\\{L^{\alpha}_{t}(A):=L^{\alpha}([0,t]\times A)\\}_{t\geq 0}$ is a martingale
and

(1.4) $\displaystyle\mathbb{E}\exp{(-\lambda
L_{t}(A))}=\exp{(\lambda^{\alpha}t|A|)},$

for all $t\geq 0$, $\lambda\geq 0$.

We need to recall some notions of existence and uniqueness for solutions to
(1.1) that will be used in this article.

  * •

(1.1) is said to admit a weak solution with initial condition $Y_{0}$ if there
exists a filtered probability space
$(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq 0},\mathbb{P})$ and an
$\\{\mathcal{F}_{t}\\}$-adapted pair $(Y,L^{\alpha})$ such that $L^{\alpha}$
satisfies (1.4) and (1.3) holds.

  * •

Weak uniqueness holds for (1.1) if whenever the pairs $(Y,L^{\alpha})$ and
$(\tilde{Y},\tilde{L}^{\alpha})$ satisfy (1.3) with the same initial
condition, they have the same finite dimensional distributions.

Our main result is the following.

######  Theorem 1.2.

Assume that $1<\alpha<2$ and $\frac{1}{\alpha}<\beta<1$ and
$Y_{0}=\varphi\in\mathcal{S}(\mathbb{R})_{+}$. Then weak uniqueness holds for
solutions to (1.3), i.e. if $(Y,L^{\alpha})$ and $(\tilde{Y},L^{\alpha})$ are
both weak solutions of (1.3) and $Y_{0}=\tilde{Y}_{0}$, then $Y$ and
$\tilde{Y}$ have the same finite dimensional distributions.

We will now describe our approach for proving this result.

###  1.2. Proof Strategy

An approach for showing weak uniqueness of stochastic equations is the
following. First one shows that solutions to (1.3) are equivalently also
solutions of an appropriate martingale problem. Then it is enough to show that
any two solutions to the martingale problem have the same one-dimensional
distributions (cf. [EK86, Theorem 4.4.2]). We may define the (local)
martingale problem as follows. For $\psi\in\mathcal{S}_{+}$ and $t\geq 0$ let

(1.5) $\displaystyle M^{Y}_{t}(\psi)=e^{-\langle
Y_{t},\psi\rangle}-e^{-\langle Y_{0},\psi\rangle}-\int_{0}^{t}e^{-\langle
Y_{s-},\psi\rangle}\left(-\langle Y_{s-},\frac{1}{2}\Delta\psi\rangle+\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle\right)\,ds$

where $Y_{t}$s are the coordinate maps on $D$, i.e $Y_{t}(\omega)=\omega(t)$
for $\omega\in D$. A probability measure $\mathbb{P}$ on $(D,\mathcal{B}(D))$
is said to be a solution of the (local) martingale problem for (1.5) if for
all $\psi\in\mathcal{S}_{+}$ $\\{M^{Y}_{t}(\psi)\\}_{t}$ is a (local)
martingale under $\mathbb{P}$. We know from [Myt02, Proposition 4.1] that a
solution to the local martingale problem (1.5) exists with stopping times

(1.6) $\displaystyle\gamma^{Y}(k):=\inf\left\\{s\geq
0\middle|\int_{0}^{s}\frac{1}{2}\lVert
Y_{r}\rVert^{\alpha\beta}_{\alpha\beta}\,dr>k\right\\},\quad k\in\mathbb{N}.$

In light of the above discussion we can rephrase Theorem 1.2 into the
equivalent result concerning the martingale problem (1.5).

######  Theorem 1.3.

Under the assumptions of Theorem 1.2, any two solutions of (1.5) have same
one-dimensional distributions.

Motivated by [Myt98] we use an approximating duality argument and would like
to show the following.

######  Theorem 1.4.

For $\varphi,\psi\in\mathcal{S}_{+}$, there exists a sequence of processes
$\\{Z^{(n)}\\}_{n}$ with $Z^{(n)}_{0}=\psi$ for all $n\geq 1$ such that for
each $t\geq 0$

(1.7) $\displaystyle\mathbb{E}\exp(-\langle
Y_{t},\psi\rangle)=\lim_{n\to\infty}\mathbb{E}\exp(-\langle\varphi,Z^{(n)}_{t}\rangle)$

for each solution $Y$ of (1.5) with $Y_{0}=\varphi$ which is independent of
$Z^{(n)}$’s.

By the virtue of [Myt96, Theorem 1.3] this will imply Theorem 1.3.

Now we shall discuss a formal strategy of how one would prove Theorem 1.4.
Suppose $Y$ solves (1.5) and $Z$ solves SPDE given below

(1.8) $\displaystyle\frac{\partial Z_{t}(x)}{\partial
t}=\frac{1}{2}\Delta_{x}Z_{t}(x)+Z_{t-}(x)^{\frac{1}{\beta}}\dot{L}^{\alpha\beta},\qquad
Z_{0}=\psi$

or equivalently the local martingale problem

(1.9) $\displaystyle
M^{Z}_{t}(\varphi)=e^{-\langle\varphi,Z_{t}\rangle}-e^{-\langle\varphi,Z_{0}\rangle}-\int_{0}^{t}e^{-\langle\varphi,Z_{s-}\rangle}\left(-\langle\frac{1}{2}\Delta\varphi,Z_{s-}\rangle+\langle\varphi^{\alpha\beta},Z_{s-}^{\alpha},\rangle\right)\,ds$

is an $\mathcal{F}^{Z}_{t}$-local martingale. Then one could try to establish
the below exponential duality relation

(1.10) $\mathbb{E}\exp(-\langle
Y_{t},\psi\rangle)=\mathbb{E}\exp(-\langle\varphi,Z_{t}\rangle).$

However this duality relationship will only hold formally as the required
integrability conditions will fail to hold (see [EK86, Theorem 4.4.11]). Thus
one uses the approximate duality technique. In this approach we construct an
approximating sequence $Z^{(n)}$ to $Z$ using the framework of [Myt98] to
prove the Theorem 1.4.

However there are two key difficulties to overcome. First, we require a moment
estimate of the solutions of (1.5) and the second difficulty is the fact that
$M^{Y}(\psi)$, as defined above, are only local martingales. A moment estimate
was shown in [YZ17, Lemma 2.4] for very general initial condition but we prove
an improved estimate in Proposition 3.1 when
$Y_{0}\in\mathcal{S}(\mathbb{R})_{+}$ and for the range of $\alpha$ and
$\beta$ in Theorem 1.2. From this we can show that $M^{Y}(\psi)$ is indeed a
martingale.

######  Remark 1.5.

Note that the condition $\alpha\beta>1$ is crucial for our argument and the
technique of approximate duality. Consequently the case when $\alpha\beta<1$
is not covered by this method.

Layout of the paper. We briefly sketch the construction of $Z^{(n)}$ in the
next section. The moment estimates and the proof that $M^{Y}(\psi)$ is a
martingale can be found in Section 3. We have split the proof of Theorem 1.4
into Propositions 4.1, 4.2 and 4.3 and have stated them in Section 4. In this
section we also finish the proof of Theorem 1.4 assuming these three results.
Their proofs can be found in Section 5. Finally, Appendix A contains some
auxiliary results that are used in various places of this paper.

We use the notations $c$, $C$, $C_{1}$ etc. to denote constants whose value
may change from one line to the next. Wherever necessary we will denote their
dependence on the relevant parameters.

Acknowledgments. This paper is part of my Ph.D. thesis. I would like to thank
my advisor Siva Athreya for proposing this problem to me and for numerous
helpful and motivating conversations. I also want to specially thank Leonid
Mytnik for several useful discussions and pointers on techniques used in the
proof including the moment estimate derived in Proposition 3.1.

##  2\. Preliminaries

This section contains notations that are used throughout the paper, some
useful results regarding the mild forms of (1.3) and the construction of the
approximating sequence $Z^{(n)}$.

Let $p_{t}(x)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{x^{2}}{2t}\right)$ for
all $t\geq 0,x\in\mathbb{R}$. For any function $f:\mathbb{R}\to\mathbb{R}$, we
will denote $P_{t}f(x)=\int_{\mathbb{R}}p_{t}(x-y)f(y)\,dy$, $\lVert
f\rVert_{\infty}=\sup_{x\in\mathbb{R}}|f(x)|$ and $\lVert
f\rVert_{p}=\left(\int_{\mathbb{R}}|f(x)|^{p}\,dx\right)^{1/p}$ for $p\geq 1$.
As in [YZ17] define a measure on $\mathbb{R}$,

(2.1) $\displaystyle
m_{0}(dz)=\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}z^{-1-\alpha}\mathbf{1}\\{z>0\\}\,dz.$

We first show that the solution $Y_{t}$ in (1.3) of Definition 1.1 can be
written in the following equivalent mild forms.

######  Proposition 2.1.

Let $Y$ be a solution as in Definition 1.1 and $Y_{0}=\varphi$. Then

  * (a)

For $t\geq 0,x\in\mathbb{R}$,

(2.2) $\displaystyle
Y_{t}(x)=P_{t}\varphi(x)+\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-y)Y_{s}(y)^{\beta}\,L^{\alpha}(ds\,dy).$

  * (b)

There exists a Poisson random measure (PRM) $N$ on
$(0,\infty)^{2}\times\mathbb{R}$ with intensity $ds\,m_{0}(dz)dx$ such that

(2.3) $\displaystyle
Y_{t}(x)=P_{t}\varphi(x)+\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}z\,p_{t-s}(x-y)Y_{s}(y)^{\beta}\,\tilde{N}(ds\,dz\,dy),$

where $\tilde{N}(ds\,dz\,dy)=N(ds\,dz\,dy)-ds\,m_{0}(dz)dx$.

  * (c)

On an enlarged probability space there exists a PRM $N_{0}$ on
$(0,\infty)^{2}\times\mathbb{R}\times(0,\infty)$ with intensity
$ds\,m_{0}(dz)\,dy\,dv$ such that, for all $t\geq 0$ and a.e.
$x\in\mathbb{R}$,

(2.4) $\displaystyle
Y_{t}(x)=P_{t}\varphi(x)+\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{s}(y)^{\alpha\beta}}z\,p_{t-s}(x-y)\tilde{N}_{0}(\,ds\,dz\,dy\,dv),$

where $\tilde{N}_{0}$ is defined similarly as $\tilde{N}.$

###### Proof.

(a) This can be shown by an argument similar to the one in the proof of
Theorem 1.1(a) in [MP03]. The only difference here is to show that for each
$t>0$,

(2.5)
$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}(t-s)^{-\alpha/2}Y_{s}(y)^{\alpha\beta}\,dy\,ds\leq\left(\sup_{s\leq
t}\lVert
Y_{s}\rVert_{\alpha\beta}^{\alpha\beta}\right)\int_{0}^{t}(t-s)^{-\alpha/2}\,ds<\infty\text{
a.s.}$

This follows from the facts that $\int_{0}^{t}\lVert
Y_{s}\rVert^{\alpha\beta}_{\alpha\beta}\,ds<\infty$ and $s\mapsto\lVert
Y_{s}\rVert_{\alpha\beta}=\langle
Y_{s}^{\alpha\beta},1\rangle^{\frac{1}{\alpha\beta}}$ is a cadlag map.

The claim in part (b) follows from the above and [MP03, Theorem 1.1(a)], while
(b) and [YZ17, Proposition 2.1] imply part (c). ∎

######  Remark 2.2.

Yang and Zhou used the condition stated in [YZ17, Assumption 1.4] to show that
the stochastic integral in (2.4) is well-defined. We here observe that our
proof of (2.4) above does not require this assumption.

As mentioned before we need to construct an approximating sequence
$\\{Z^{(n)}\\}_{n}$ to $Z$ described in (1.8). We shall use the construction
given in [Myt02, §3]. For completeness we only present the sketch below.

Define $Z^{(n)}_{0}(\,dx)=\psi(x)\,dx$ and let
$b_{n}=\frac{\alpha\beta}{\Gamma(2-\alpha\beta)}n^{\alpha\beta-1}$. For
$\mu\in M_{F}$ let $V^{n}_{\cdot}(\mu)$ be the unique non-negative solution to
the partial differential equation (PDE)

(2.6) $\displaystyle
v_{t}=P_{t}\mu-\int_{0}^{t}P_{t-s}(b_{n}v_{s}^{\alpha})\,ds$

where $(P_{t}\mu)(x)=\int_{\mathbb{R}}p_{t}(x-y)\mu(\,dy)$. We have collected
some properties of the above PDE in Appendix A.2.

The idea behind this $Z^{(n)}$ is as follows. $Z^{(n)}$ evolves according to
the PDE (2.6), jumps after a random time by a random height at a random
location (denoted in the following by $\gamma^{n}(T^{n}_{k})$, $S^{n}_{i}$ and
$U^{n}_{i}$ respectively) and so on. More precisely, let
$\tilde{T}^{Z,n}_{i}:=\tilde{T}^{n}_{i}\sim\operatorname{Exp}(n^{\alpha\beta}\frac{(\alpha\beta-1)}{\Gamma(2-\alpha\beta)})$,
$i\in\mathbb{N}$, be iid random varibales and
$T^{n}_{i}:=\sum_{k=1}^{i}\tilde{T}^{n}_{k}$. The jump heights are given by
iid $[\frac{1}{n},0)$-valued random variables $\\{S^{n}_{i}\mid
i\in\mathbb{N}\\}$ defined by

$\displaystyle\mathbb{P}(S^{n}_{i}\geq
b)=\frac{\int_{b\vee(1/n)}^{\infty}\lambda^{-\alpha\beta-1}\,d\lambda}{\int_{1/n}^{\infty}\lambda^{-\alpha\beta-1}\,d\lambda},\qquad
b\geq 0.$

So $A^{n}_{t}:=\sum_{k=1}^{\infty}S^{n}_{k}\mathbf{1}(T^{n}_{k}\leq t)$ is the
process that jumps by height $S^{n}_{i}$ at time $T^{n}_{i}$ for all
$i\in\mathbb{N}$. For $0\leq t\leq T^{n}_{1}$ define the time change

$\displaystyle\gamma^{n}(t)=\inf\left\\{s\geq
0\middle|\frac{1}{2}\int_{0+}^{s}\lVert
V^{n}_{r}(\mu)\rVert^{\alpha}_{\alpha}\,dr>t\right\\}.$

Also let $\\{G(f,\cdot)\mid f\in L^{\alpha}(\mathbb{R})_{+}\\}$ be a
collection of probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$
such that for $E\in\mathcal{B}(\mathbb{R})$

$\displaystyle G(f,E)=\frac{\int_{E}f(x)^{\alpha}\,dx}{\lVert
f\rVert_{\alpha}^{\alpha}}$

Lastly, let $U^{n}_{1}$ be a $\mathbb{R}$-valued random variable defined by
the relation

$\mathbb{P}(U^{n}_{1}\in
A\mid\mathcal{F}^{A^{n}}_{T^{n}_{1}})=G(Z^{(n)}_{\gamma^{n}(T^{n}_{1})-},A)$

for $A\in\mathcal{B}(\mathbb{R})$.

Then we can define the approximating sequence $Z^{(n)}$ in the (random)
interval $[0,\gamma^{n}(T^{n}_{1})]$.

$\displaystyle Z^{(n)}_{t}:=\begin{cases}V^{n}_{t}(Z^{(n)}_{0}),\qquad 0\leq
t<\gamma^{n}(T^{n}_{1}),\\\
Z^{(n)}_{\gamma^{n}(T^{n}_{1})-}+S^{n}_{1}\delta_{U^{n}_{1}},\qquad
t=\gamma^{n}(T^{n}_{1}).\end{cases}$

For $t>\gamma^{n}_{1}(T^{n}_{1})$, $Z^{(n)}$ is defined inductively:

$\displaystyle
Z^{(n)}_{t}:=\begin{cases}V^{n}_{t-\gamma^{n}(T^{n}_{k})}(Z^{(n)}_{\gamma^{n}(T^{n}_{k})}),\qquad
t\in[\gamma^{n}(T^{n}_{k}),\gamma^{n}(T^{n}_{k+1})),\\\
Z^{(n)}_{\gamma^{n}(T^{n}_{k})-}+S^{n}_{k+1}\delta_{U^{n}_{k+1}},\qquad
t=\gamma^{n}(T^{n}_{k+1}),\end{cases}$

where

$\displaystyle\gamma^{n}(t)=\inf\left\\{s\geq
0\middle|T^{n}_{k}+\frac{1}{2}\int_{0+}^{s-\gamma^{n}(T^{n}_{k})}\lVert
V_{r}^{n}(Z^{(n)}_{T^{n}_{k}})\rVert_{\alpha}^{\alpha}\,dr>t\right\\},\qquad
T^{n}_{k}\leq t<T^{n}_{k+1},$

and

$\displaystyle\mathbb{P}(U^{n}_{k+1}\in
A\mid\mathcal{F}^{A^{n}}_{T^{n}_{k+1}}):=G(Z^{(n)}_{\gamma^{n}(T^{n}_{k+1})-},A),\qquad
A\in\mathcal{B}(\mathbb{R}).$

This completes the construction of $Z^{(n)}$. It is known that $Z^{(n)}$
solves a local martingale problem as described by the following lemma.

######  Lemma 2.3.

Let $\eta:=\frac{\alpha\beta(\alpha\beta-1)}{\Gamma(2-\alpha\beta)}$,
$g(r,y):=\int_{0+}^{r}(e^{-\lambda y}-1+\lambda
y)\lambda^{-\alpha\beta-1}\,d\lambda$. For all $\varphi\in\mathcal{S}_{+}$ and
$n\geq 1$

(2.7) $\displaystyle
M^{Z,n}_{t}(\varphi)=e^{-\langle\varphi,Z^{(n)}_{t}\rangle}-e^{-\langle\varphi,Z^{(n)}_{0}\rangle}$
$\displaystyle-\int_{0}^{t}e^{-\langle\varphi,Z^{(n)}_{s-}\rangle}\left(-\langle\frac{1}{2}\Delta\varphi,Z^{(n)}_{s}\rangle+\langle\varphi(\cdot)^{\alpha\beta}-\eta
g\left(1/n,\varphi(\cdot)\right),(Z^{(n)}_{s-}(\cdot))^{\alpha}\rangle\right)\,ds$

is an $(\mathcal{F}^{Z^{(n)}}_{t})$-local martingale with stopping times

(2.8) $\displaystyle\gamma^{Z,n}(k):=\gamma^{n}(k)=\inf\left\\{s\geq
0\middle|\int_{0}^{s}\frac{1}{2}\lVert
Z^{(n)}_{r}\rVert^{\alpha}_{\alpha}\,dr>k\right\\},\quad k\in\mathbb{N}.$

###### Proof.

See [Myt02, Lemma 3.7]. ∎

We conclude the section with a result describing the behaviour of the
compensator of $N^{Z,n}=N^{n}$ which is the counting measure tracking the
jumps of $Z^{(n)}$.

######  Lemma 2.4.

The compensator of $N^{n}$ is, for
$B_{1}\in\mathcal{B}([0,\infty)),B_{2}\in\mathcal{B}(\mathbb{R})$

(2.9) $\displaystyle\hat{N}^{n}(t,B_{1}\times B_{2})=\eta\int_{0}^{t\wedge
T^{*}_{n}}\,dr\int_{B_{1}}\,d\lambda\int_{B_{2}}\,dx\frac{\left(Z^{(n)}_{r-}(x)\right)^{\alpha}}{\lVert
Z^{(n)}_{r-}\rVert_{\alpha}^{\alpha}}\mathbf{1}(\lambda>1/n)\lambda^{-\alpha\beta-1}$

where

(2.10) $\displaystyle T^{*}_{n}=\inf\\{t\geq
0\mid\gamma^{n}(t)=\infty\\}=\inf\\{t\geq 0\mid A^{n}_{t}-b_{n}t=0\\}.$

###### Proof.

See [Myt02, Lemma 3.5]. ∎

##  3\. Moment Estimate and Martingale Problem

In this section we will establish the key moment estimate for solutions $Y$ of
(1.3) and also show that $M^{Y}(\psi)$ defined in (1.5) is a martingale for
all $\psi\in\mathcal{S}_{+}$. The following improves the estimate presented in
[YZ17, Lemma 2.4].

######  Proposition 3.1.

Let $1<q<\alpha$. Then for all $t\in[0,T]$ we have

(3.1) $\displaystyle\sup_{x\in\mathbb{R}}\mathbb{E}(Y_{t}(x)^{q})\leq
C_{1}e^{C_{2}t}$

for some constants $C_{1}=C_{1}(T),C_{2}=C_{2}(T)>0$.

###### Proof.

From (2.4) we have, for $t\in[0,T]$,

(3.2) $\displaystyle
Y_{t}(x)=P_{t}\varphi(x)+\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}zp_{t-r}(x-y)\tilde{N}_{0}(\,dr\,dz\,dy\,dv).$

Let us define

$\displaystyle\tau_{N}=\inf\left\\{t\mid\int_{0}^{t}\lVert
Y_{r}\rVert_{\alpha\beta}^{\alpha\beta}\,dr\geq N\right\\},$

when $N\in\mathbb{N}$ and from [PZ07, Lemma 8.21] recall that the quadratic
variation of

$\int_{0}^{s}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}zp_{t-r}(x-y)\tilde{N}_{0}(\,dr\,dz\,dy\,dv)$

equals

$\int_{0}^{s}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}N_{0}(\,dr\,dz\,dy\,dv).$

By Burkholder-Davis-Gundy inequality and the fact that $q<2$ we have

$\displaystyle I:=$
$\displaystyle\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}zp_{t-r}(x-y)\tilde{N}_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q}\mathbf{1}(t\leq\tau_{N})\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}zp_{t-r}(x-y)\mathbf{1}(r\leq\tau_{N})\tilde{N}_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q}\right]$
$\displaystyle\leq$ $\displaystyle
c\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q/2}\right]$
$\displaystyle\leq$ $\displaystyle
c\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q/2}\right]$
(3.3)
$\displaystyle+c\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q/2}\right].$

Fix $p\in(\alpha,2)$. Applying Jensen’s inequality to the above (noting that
$p/q>1$) we have

$\displaystyle I\leq$ $\displaystyle
c\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{p/2}\right]^{q/p}$
$\displaystyle+c\mathbb{E}\left[\left\lvert\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{2}p_{t-r}(x-y)^{2}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q/2}\right]$
$\displaystyle\leq$ $\displaystyle
c\mathbb{E}\left[\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{p}p_{t-r}(x-y)^{p}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right]^{q/p}$
(3.4)
$\displaystyle+c\mathbb{E}\left[\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}z^{q}p_{t-r}(x-y)^{q}\mathbf{1}(r\leq\tau_{N})N_{0}(\,dr\,dz\,dy\,dv)\right],$

where the second inequality above is due a fact about random sums (see the
proof of [PZ07, Lemma 8.22]). Now we use the definition of the PRM $N_{0}$,
integrate out $z$ and use the inequality $u^{q/p}\leq u+1$ for $u\geq 0$.

$\displaystyle I=$ $\displaystyle
c\mathbb{E}\left[\int_{0}^{t}\int_{\mathbb{R}}Y_{r}(y)^{\alpha\beta}p_{t-r}(x-y)^{p}\mathbf{1}(r\leq\tau_{N})\,dr\,dy\right]^{q/p}$
$\displaystyle+c\mathbb{E}\left[\int_{0}^{t}\int_{\mathbb{R}}Y_{r}(y)^{\alpha\beta}p_{t-r}(x-y)^{q}\mathbf{1}(r\leq\tau_{N})\,dr\,dy\right]$
(3.5) $\displaystyle\leq$ $\displaystyle
c+c\mathbb{E}\int_{0}^{t}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})Y_{r}(y)^{\alpha\beta}\mathbf{1}(r\leq\tau_{N})\,dr\,dy.$

Note that $\left(P_{t}\varphi(x)\right)^{q}\leq P_{t}(\varphi^{q})(x)$. From
(3.2) and (3) we have

$\displaystyle\mathbb{E}\left[Y_{t}(x)^{q}\mathbf{1}(t<\tau_{N})\right]$
$\displaystyle\leq$ $\displaystyle
cP_{t}(\varphi^{q})(x)+c+c\mathbb{E}\int_{0}^{t}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})Y_{r}(y)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\,dr\,dy$
(3.6) $\displaystyle=$ $\displaystyle
cP_{t}(\varphi^{q})(x)+c+c\int_{0}^{t}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})\mathbb{E}\left[Y_{r}(y)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dr\,dy.$

by applying Fubini in the last line. Use the definition of $p_{t}(x)$ to get

$\displaystyle\mathbb{E}\left[Y_{t}(x)^{q}\mathbf{1}(t<\tau_{N})\right]$
$\displaystyle\leq$ $\displaystyle
cP_{t}(\varphi^{q})(x)+c+c\int_{0}^{t}\,dr((t-r)^{-\frac{p-1}{2}}+(t-r)^{-\frac{q-1}{2}})\int_{\mathbb{R}}p_{t-r}(x-y)\mathbb{E}\left[Y_{r}(y)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dy$
(3.7) $\displaystyle\leq$ $\displaystyle
cP_{t}(\varphi^{q})(x)+c+c\int_{0}^{t}\,dr(t-r)^{-\frac{p-1}{2}}\int_{\mathbb{R}}p_{t-r}(x-y)\mathbb{E}\left[Y_{r}(y)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dy$

where in the last line we have used the fact that $(t-r)^{-\frac{q-1}{2}}\leq
C_{T}(t-r)^{-\frac{p-1}{2}}$. When $q=\alpha\beta$ this becomes

$\displaystyle\mathbb{E}\left[Y_{t}(x)^{\alpha\beta}\mathbf{1}(t<\tau_{N})\right]$
(3.8) $\displaystyle\leq$ $\displaystyle
cP_{t}(\varphi^{\alpha\beta})(x)+c+c\int_{0}^{t}\,dr(t-r)^{-\frac{p-1}{2}}\int_{\mathbb{R}}p_{t-r}(x-y)\mathbb{E}\left[Y_{r}(y)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dy.$

Let $s\in[0,T]$ be such that $s\geq t$. Apply $P_{s-t}$ to both sides and use
Fubini,

$\displaystyle\mathbb{E}\left[P_{s-t}\left(Y_{t}^{\alpha\beta}\right)(x)\mathbf{1}(t<\tau_{N})\right]$
$\displaystyle\leq$ $\displaystyle
cP_{s}(\varphi^{\alpha\beta})(x)+c+c\int_{0}^{t}\,dr\,(t-r)^{-\frac{p-1}{2}}\int_{\mathbb{R}}p_{s-t}(x-y)\int_{\mathbb{R}}p_{t-r}(y-z)\mathbb{E}\left[Y_{r}(z)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dz\,dy$
$\displaystyle=$ $\displaystyle
cP_{s}(\varphi^{\alpha\beta})(x)+c+c\int_{0}^{t}\,dr(t-r)^{-\frac{p-1}{2}}\int_{\mathbb{R}}p_{s-r}(x-z)\mathbb{E}\left[Y_{r}(z)^{\alpha\beta}\mathbf{1}(r<\tau_{N})\right]\,dz$
(3.9) $\displaystyle=$ $\displaystyle
cP_{s}(\varphi^{\alpha\beta})(x)+c+c\int_{0}^{t}\,dr\,(t-r)^{-\frac{p-1}{2}}\mathbb{E}\left[P_{s-r}\left(Y_{r}^{\alpha\beta}\right)(x)\mathbf{1}(r<\tau_{N})\right]$

Since the above holds for every $t\in[0,s]$, by Lemma A.1 there exist
constants $c_{1}(s),c_{2}(s)>0$ such that for all $t\leq s$,

(3.10)
$\displaystyle\mathbb{E}\left[P_{s-t}\left(Y_{t}^{\alpha\beta}\right)(x)\mathbf{1}(t<\tau_{N})\right]\leq
c_{1}(s)e^{c_{2}(s)t}.$

Observe that, in the proof of Lemma A.1, the constants $c_{1}(s)$ and
$c_{2}(s)$ depend only on $s$ and $p$ (i.e. they are independent of the
function $f$) and moreover, they are non-decreasing in $s$. So we have
$c_{1}(s)\leq c_{1}(T)$ and $c_{2}(s)\leq c_{2}(T)$ and (3.10) gives

(3.11)
$\displaystyle\mathbb{E}\left[P_{s-t}\left(Y_{t}^{\alpha\beta}\right)(x)\mathbf{1}(t<\tau_{N})\right]\leq
c_{1}(T)e^{c_{2}(T)t}$

for all $t\leq s\leq T$. Now replace $s$ by $t$ in the above. We get,

(3.12)
$\displaystyle\mathbb{E}\left[\left(Y_{t}(x)\right)^{\alpha\beta}\mathbf{1}(t<\tau_{N})\right]\leq
c_{1}(T)e^{c_{2}(T)t}$

for each $t\in[0,T]$. Take $N\to\infty$ and we obtain the required result. ∎

We here observe that the previous moment estimate can be utilized to show that
the stochastic integrals appearing in (2.2), (2.3) and (2.4) are martingales.
For this we recall the notion of a _class DL_ process (see [RY99, Definition
IV.1.6]).

A real valued and adapted stochastic process $X$ is said to be of _class DL_
if for every $t>0$, the set

$\\{X_{\tau}:\tau\leq t\text{ is a stopping time}\\}$

is uniformly integrable. And we know from [RY99, Proposition IV.1.7] that a
local martingale $X$ is a martingale if and only if it is of class DL. For
practical purposes it is enough to show that there is an $\epsilon>0$ such
that

$\sup_{\tau\leq t}\mathbb{E}(|X_{\tau}|^{1+\epsilon})<\infty$

where the supremum is taken over all stopping times $\tau\leq t$.

######  Lemma 3.2.

The stochastic integrals

(3.13)
$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-y)Y_{s}(y)^{\beta}\,L^{\alpha}(ds\,dy)=\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{s}(y)^{\alpha\beta}}z\,p_{t-s}(x-y)\tilde{N}_{0}(\,ds\,dz\,dy\,dv)$

are martingales with respect to $\mathcal{F}^{Y}$.

###### Proof.

Since the integrals in (3.13) are stochastic integrals with respect to
martingale measure $L^{\alpha}$ and the compensated PRM $\tilde{N}_{0}$, they
are clearly $\mathcal{F}^{Y}$-local martingales. Therefore we have to show
that

(3.14)
$\displaystyle\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{s}(y)^{\alpha\beta}}z\,p_{t-s}(x-y)\tilde{N}_{0}(\,ds\,dz\,dy\,dv)$

is of class DL.

Let $\tau\leq t$ be a stopping time. We omit some of the calculations since
they are similar to the ones found in the previous proof. Proceeding as in (3)
and (3) we obtain,

$\displaystyle\mathbb{E}\left[\left\lvert\int_{0}^{\tau}\int_{0}^{\infty}\int_{\mathbb{R}}\int_{0}^{Y_{r}(y)^{\alpha\beta}}zp_{t-r}(x-y)\tilde{N}_{0}(\,dr\,dz\,dy\,dv)\right\rvert^{q}\right]$
$\displaystyle\leq$ $\displaystyle
c+c\mathbb{E}\int_{0}^{\tau}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})Y_{r}(y)^{\alpha\beta}\,dr\,dy$
$\displaystyle\leq$ $\displaystyle
c+c\int_{0}^{t}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})\mathbb{E}(Y_{r}(y)^{\alpha\beta})\,dr\,dy$
$\displaystyle\leq$ $\displaystyle
c+c_{1}\int_{0}^{t}\int_{\mathbb{R}}(p_{t-r}(x-y)^{p}+p_{t-r}(x-y)^{q})e^{C_{2}r}\,dr\,dy$

applying (3.1) at the end. The RHS of the above is finite as

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}p_{t-r}(x-y)^{p}\,dy\,dr\leq
c_{2}\int_{0}^{t}(t-r)^{\frac{1-p}{2}}\int_{\mathbb{R}}p_{t-r}(x-y)\,dy\,dr<\infty$

and it is independent of the stopping times $\tau$. This shows that (3.14) is
in class DL and is therefore a martingale with respect to $\mathcal{F}^{Y}$. ∎

Next we observe that $M^{Y}_{t}(\psi)$ defined in (1.5) is a martingale. This
will be crucial for simplifying our approximate duality argument in the proof
of Proposition 4.1.

######  Proposition 3.3.

For each $\psi\in\mathcal{S}_{+}$, the local martingale $M^{Y}(\psi)$ is in
fact a martingale with respect to $\mathcal{F}^{Y}$.

###### Proof.

Recall that

$\displaystyle M^{Y}_{t}(\psi)=e^{-\langle Y_{t},\psi\rangle}-e^{-\langle
Y_{0},\psi\rangle}-\int_{0}^{t}I(Y_{s-},\psi)\,ds$

is an $\mathcal{F}^{Y}_{t}$-local martingale, where

$I(Y_{s-},\psi)=e^{-\langle Y_{s-},\psi\rangle}\left(-\langle
Y_{s-},\frac{1}{2}\Delta\psi\rangle+\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle\right)$

To show that $M^{Y}_{t}(\psi)$ is a martingale we show that it is in class DL,
i.e. for each $t>0$,

(3.15) $\displaystyle\sup_{\tau\leq
t}\mathbb{E}\left(|M^{Y}_{\tau}(\psi)|^{1+\epsilon}\right)<\infty$

for some $\epsilon>0$. The supremum ranges over all $\mathcal{F}^{Y}$-stopping
times $\tau$ that are bounded by $t$.

From the expression above it is enough to prove

(3.16) $\displaystyle\sup_{\tau\leq
t}\mathbb{E}\left(\left|\int_{0}^{\tau}I(Y_{s-},\psi)\,ds\right|^{1+\epsilon}\right)<\infty.$

Fix a stopping time $\tau\leq t$. By Jensen’s inequality

$\displaystyle\left|\int_{0}^{\tau}I(Y_{s-},\psi)\,ds\right|^{1+\epsilon}=$
$\displaystyle\left|\int_{0}^{\tau}e^{-\langle
Y_{s-},\psi\rangle}\left(-\langle Y_{s-},\frac{1}{2}\Delta\psi\rangle+\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle\right)\,ds\right|^{1+\epsilon}$
$\displaystyle=$
$\displaystyle\tau^{1+\epsilon}\left|\frac{1}{\tau}\int_{0}^{\tau}e^{-\langle
Y_{s-},\psi\rangle}\left(-\langle Y_{s-},\frac{1}{2}\Delta\psi\rangle+\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle\right)\,ds\right|^{1+\epsilon}$
$\displaystyle\leq$ $\displaystyle\tau^{\epsilon}\int_{0}^{\tau}\left|-\langle
Y_{s-},\frac{1}{2}\Delta\psi\rangle+\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle\right|^{1+\epsilon}\,ds$ (3.17)
$\displaystyle\leq$ $\displaystyle
C_{\epsilon}t^{\epsilon}\int_{0}^{t}\left(|\langle
Y_{s-},\frac{1}{2}\Delta\psi\rangle|^{1+\epsilon}+|\langle
Y_{s-}^{\alpha\beta},\psi^{\alpha}\rangle|^{1+\epsilon}\right)\,ds.$

Let $0<\epsilon<\frac{1}{\beta}-1$. Again apply Jensen’s inequality, Funbini’s
Theorem and Proposition 3.1.

$\displaystyle\mathbb{E}\int_{0}^{t}|\langle
Y_{s-},\frac{1}{2}\Delta\psi\rangle|^{1+\epsilon}\,ds=$
$\displaystyle\frac{1}{2^{1+\epsilon}}\mathbb{E}\int_{0}^{\tau}\left|\int_{\mathbb{R}}Y_{s-}(x)\Delta\psi(x)\,dx\right|^{1+\epsilon}\,ds$
$\displaystyle\leq$
$\displaystyle\frac{\lVert\Delta\psi\rVert_{L^{1}}^{1+\epsilon}}{2^{1+\epsilon}}\mathbb{E}\int_{0}^{t}\left|\frac{1}{\lVert\Delta\psi\rVert_{L^{1}}}\int_{\mathbb{R}}Y_{s-}(x)|\Delta\psi(x)|\,dx\right|^{1+\epsilon}\,ds$
$\displaystyle\leq$
$\displaystyle\frac{\lVert\Delta\psi\rVert_{L^{1}}^{\epsilon}}{2^{1+\epsilon}}\int_{0}^{t}\int_{\mathbb{R}}\mathbb{E}Y_{s-}(x)^{1+\epsilon}|\Delta\psi(x)|\,dx\,ds$
$\displaystyle\leq$ $\displaystyle
C_{T}\frac{\lVert\Delta\psi\rVert_{L^{1}}^{\epsilon}}{2^{1+\epsilon}}\int_{0}^{t}\int_{\mathbb{R}}e^{c_{2}s}|\Delta\psi(x)|\,dx\,ds$
(3.18) $\displaystyle=$ $\displaystyle
C_{T}\frac{\lVert\Delta\psi\rVert_{L^{1}}^{1+\epsilon}}{2^{1+\epsilon}}e^{c_{2}t}.$

Similarly,

$\displaystyle\mathbb{E}\int_{0}^{t}|\langle
Y_{s-}^{\alpha\beta},\frac{1}{2}\Delta\psi\rangle|^{1+\epsilon}\,ds\leq$
$\displaystyle\lVert\psi^{\alpha}\rVert_{L^{1}}^{\epsilon}\int_{0}^{t}\int_{\mathbb{R}}\mathbb{E}Y_{s-}(x)^{\alpha\beta(1+\epsilon)}\psi(x)^{\alpha}\,dx\,ds$
$\displaystyle=$ $\displaystyle
C_{T}\frac{\lVert\psi^{\alpha}\rVert_{L^{1}}^{1+\epsilon}}{2^{1+\epsilon}}\int_{0}^{t}e^{c_{2}s}\,ds$
(3.19) $\displaystyle=$ $\displaystyle
C_{T,\psi,\alpha,\beta,\epsilon}e^{c_{2}t}$

Plugging (3) and (3) in (3) we get,

(3.20)
$\displaystyle\mathbb{E}\left(\left|\int_{0}^{\tau}I(Y_{s-},\psi)\,ds\right|^{1+\epsilon}\right)\leq
C_{T,\psi,\alpha,\beta,\epsilon}e^{c_{2}t}.$

Taking supremum over all $\tau\leq t$ gives (3.16). ∎

We now show the above result holds for
$\psi:\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$ satisfying certain
assumptions.

######  Proposition 3.4.

Let $T\in(0,\infty)$ If $Y$ is a solution to the martingale problem (1.5) and
$\psi:[0,T]\times\mathbb{R}\to\mathbb{R}$ is such that

  * (i)

The map $[0,T]\ni s\mapsto\psi_{s}\in L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$
is continuous.

  * (ii)

$\sup_{s\leq T}\left\lVert\frac{\partial}{\partial
s}\psi_{s}\right\rVert_{\frac{\alpha\beta}{\alpha\beta-1}}<\infty$.

  * (iii)

The map $[0,T]\to L^{\infty}(\mathbb{R})$,
$s\mapsto\frac{\partial^{2}}{\partial x^{2}}\psi_{s}$ is continuous.

Then,

(3.21) $\displaystyle\tilde{M}^{Y}_{t}(\psi)=e^{-\langle
Y_{t},\psi_{t}\rangle}-e^{-\langle
Y_{0},\psi_{0}\rangle}-\int_{0}^{t}\tilde{I}(Y_{s-},\psi_{s})\,ds$

is an $\mathcal{F}^{Y}_{t}$ martingale, where

(3.22) $\displaystyle\tilde{I}(Y_{s-},\psi_{s})=e^{-\langle
Y_{s-},\psi_{s}\rangle}\left[-\langle
Y_{s-},\frac{1}{2}\partial^{2}_{xx}\psi_{s}+\partial_{s}\psi_{s}\rangle+\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle\right].$

This result is probably already known, but we could not find a self-contained
proof in the literature. Therefore we present its proof in the Appendix A.3.

##  4\. Overview of the Proof of Theorem 1.4

In this section we describe our plan for proving Theorem 1.4. Our proof
follows the argument in [Myt98] and will be split into various propositions
which we state in the following. At the end of this section we establish the
theorem assuming these results.

The first proposition describes the behaviour of $Y$ when coupled with the
solutions of the evolution equations used to construct $Z^{(n)}$. In what
follows we denote by $\mathbb{E}_{Y}$ the expectation with respect to to $Y$.
In particular, under $\mathbb{E}_{Y}$ we treat all the random variables used
to construct $Z^{(n)}$ in Section 2 as non-random owing to our assumption of
independence.

######  Proposition 4.1.

Let $Y$ be a solution to the martingale problem (1.5). Then for each $n\geq 1$
and $\mu\in M_{F}$,

(4.1) $\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-t},V^{n}_{t}(\mu)\rangle}\right]=\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},V^{n}_{0}(\mu)\rangle}\right]+\mathbb{E}_{Y}\left[\int^{t}_{0}\tilde{\mathcal{I}}(Y_{(T-r)-},V^{n}_{r}(\mu))\,dr\right]$

where

$\tilde{\mathcal{I}}(Y_{(T-r)-},V^{n}_{r}(\mu))=e^{-\langle
Y_{(T-r)-},V^{n}_{r}(\mu)\rangle}\frac{1}{2}\left\\{-\langle
Y^{\alpha\beta}_{(T-r)-},\left(V^{n}_{r}(\mu)\right)^{\alpha}\rangle+\langle
Y_{(T-r)-},b_{n}\left(V^{n}_{r}(\mu)\right)^{\alpha}\rangle\right\\}.$

In the next proposition we describe the relationship between $Y$ and the jumps
of $Z^{(n)}$. Define

(4.2) $\displaystyle\tau^{n}(t):=\frac{1}{2}\int_{0}^{t}\lVert
Z^{(n)}_{r-}\rVert^{\alpha}_{\alpha}\,dr$

and observe from (2.8) that $\tau^{n}$ is the inverse of $\gamma^{n}$:
$\tau^{n}(\gamma^{n}(t))=t$ and vice-versa.

######  Proposition 4.2.

If $Y$ is a solution to the martingale problem (1.5), independent of
$Z^{(n)}$s, then for all $t\in[0,T]$,

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-t},Z^{(n)}_{t}\rangle}\right]$ (4.3) $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z^{(n)}_{r-})\,dr+\int_{0}^{\tau^{n}(t)}\int_{\mathbb{R}}\int_{\mathbb{R}_{+}}\theta_{n}(s,x,\lambda)N^{n}(\,ds,\,dx,\,d\lambda)\right]$

where

$\theta_{n}(s,x,\lambda)=e^{-\langle
Y_{T-\gamma^{n}(s)},Z^{(n)}_{\gamma^{n}(s)-}\rangle}\left(e^{-\lambda
Y_{T-\gamma^{n}(s)}(x)}-1\right)$

In the last proposition before we prove our main result we show that the
previous result holds at the stopping time
$\Upsilon^{n}_{k}(t):=\gamma^{n}(k)\wedge t$.

######  Proposition 4.3.

If $Y$ is a solution to the martingale problem (1.5), independent of
$Z^{(n)}$s, then for each $m\in\mathbb{N}$ and $t\in[0,T]$,

$\displaystyle\mathbb{E}[\exp(-\langle
Y_{T-\Upsilon^{n}_{m}(t)},Z_{\Upsilon^{n}_{m}(t)}\rangle)]=$
$\displaystyle\mathbb{E}[\exp\left(-\langle Y_{T},Z_{0}\rangle\right)]$ (4.4)
$\displaystyle-\frac{\eta}{2}\mathbb{E}\left[\int^{\Upsilon^{n}_{m}(t)}_{0}e^{-\langle
Y_{(T-r)-},Z^{(n)}_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),(Z^{(n)}_{r-})^{\alpha}\rangle\,dr\right]$

We note that Propositions 4.1 and 4.2 are used to prove Proposition 4.3. Now
we present the proof of Theorem 1.4 assuming that the above propositions hold.
We will prove them in the next section.

###### Proof of Theorem 1.4.

Let $Y$ and $Z^{(n)}$ be as in the statement of Theorem 1.4. Let $k_{n}=\ln
n$. We first show that for each $t\in[0,T]$,

(4.5) $\displaystyle\lim_{n\to\infty}\lvert\mathbb{E}[\exp(-\langle
Y_{T-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle)]-\mathbb{E}[\exp(-\langle
Y_{T},Z^{(n)}_{0}\rangle)]\rvert=0.$

Note that for all $1<p<2$ and $\lambda\geq 0$,

$e^{-\lambda}-1+\lambda\leq\frac{\lambda^{p}}{p}.$

Also by our assumptions on $\alpha$ and $\beta$ we have
$1<\frac{\alpha(\beta+1)}{2}<2$. So,

$\displaystyle g\left(\frac{1}{n},Y_{T-s}(x)\right)$
$\displaystyle=\int_{0+}^{1/n}\left(e^{-\lambda Y_{T-s}(x)}-1+\lambda
Y_{T-s}(x)\right)\lambda^{-\alpha\beta-1}\,d\lambda$
$\displaystyle\leq\frac{2}{\alpha(\beta+1)}\int_{0+}^{1/n}\left(\lambda
Y_{T-s}(x)\right)^{\frac{\alpha(\beta+1)}{2}}\lambda^{-\alpha\beta-1}\,d\lambda$
$\displaystyle=\frac{2}{\alpha(\beta+1)}Y_{T-s}(x)^{\frac{\alpha(\beta+1)}{2}}\int_{0+}^{1/n}\lambda^{\frac{\alpha(\beta+1)}{2}-\alpha\beta-1}\,d\lambda$
(4.6)
$\displaystyle=\frac{2}{\alpha(\beta+1)}Y_{T-s}(x)^{\frac{\alpha(\beta+1)}{2}}\frac{2}{\alpha-\alpha\beta}n^{-\frac{\alpha-\alpha\beta}{2}}.$

Eq. (4.3) and the above calculation gives us

$\displaystyle\left|\mathbb{E}\exp\left(-\langle
Y_{T-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle\right)-\mathbb{E}e^{-\langle
Y_{T},Z_{0}\rangle}\right|$ $\displaystyle=$
$\displaystyle\left\lvert\frac{\eta}{2}\mathbb{E}\left[\int^{\Upsilon^{n}_{k_{n}}(t)}_{0+}e^{-\langle
Y_{(T-s)-},Z^{(n)}_{s}\rangle}\langle
g(1/n,Y_{(T-s)-}(\cdot)),\left(Z^{(n)}_{s-}\right)^{\alpha}\rangle\,ds\right]\right\rvert$
$\displaystyle=$
$\displaystyle\frac{\eta}{2}\mathbb{E}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}g(1/n,Y_{T-s}(x))\,dx\,ds\right]$
(4.7) $\displaystyle\leq$
$\displaystyle\frac{\eta}{2}\frac{2}{\alpha(\beta+1)}\frac{2}{\alpha-\alpha\beta}n^{-\frac{\alpha-\alpha\beta}{2}}\mathbb{E}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}Y_{T-s}(x)^{\frac{\alpha(\beta+1)}{2}}\,dx\,ds\right]$

using the fact that $Z^{(n)}_{s-}(\cdot)\geq 0$ and $Y_{(T-s)-}(\cdot)\geq 0$
for the second equality. Now use the estimate from Proposition 3.1 with
$\bar{p}=\frac{\alpha(\beta+1)}{2}$. We have by Fubini

$\displaystyle\mathbb{E}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}Y_{T-s}(x)^{\frac{\alpha(\beta+1)}{2}}\,dx\,ds\right]=$
$\displaystyle\mathbb{E}_{Z}\mathbb{E}_{Y}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}Y_{T-s}(x)^{\bar{p}}\,dx\,ds\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}_{Z}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}\mathbb{E}_{Y}\left(Y_{T-s}(x)^{\bar{p}}\right)\,dx\,ds\right]$
$\displaystyle\leq$ $\displaystyle
C\mathbb{E}_{Z}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\int_{\mathbb{R}}Z^{(n)}_{s-}(x)^{\alpha}e^{C(T-s)}\,dx\,ds\right]$
$\displaystyle=$ $\displaystyle
C\mathbb{E}_{Z}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\lVert
Z^{(n)}_{s-}\rVert^{\alpha}_{\alpha}e^{C(T-s)}\,ds\right]$ (4.8)
$\displaystyle\leq$ $\displaystyle
Ce^{CT}\mathbb{E}_{Z}\left[\int_{0}^{\Upsilon^{n}_{k_{n}}(t)}\lVert
Z^{(n)}_{s-}\rVert^{\alpha}_{\alpha}\,ds\right]\leq C_{T}k_{n}$

The last inequality follows from the definition of $\gamma^{n}$ (see (2.8)).
Plugging this in (4) gives

$\displaystyle\left|\mathbb{E}\exp\left(-\langle
Y_{T-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle\right)-\mathbb{E}e^{-\langle
Y_{T},Z_{0}\rangle}\right|\leq
C_{\alpha.\beta,T}n^{-\frac{\alpha-\alpha\beta}{2}}k_{n}$

Since $k_{n}=\ln n$ the RHS converges to $0$ as $n\to\infty$. This proves
(4.5).

Next we show that

(4.9) $\displaystyle\lim_{n\to\infty}\lvert\mathbb{E}\exp(-\langle
Y_{0},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle)-\mathbb{E}\exp(-\langle
Y_{t},\psi\rangle)\rvert=0$

for all $t\in[0,T]$ where $Z^{(n)}_{0}=\psi$. Observe that by (4.5) it is
enough to show

(4.10) $\displaystyle\lim_{n\to\infty}\lvert\mathbb{E}\exp(-\langle
Y_{t-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle)-\mathbb{E}\exp(-\langle
Y_{t},\psi\rangle)\rvert=0.$

By definition of $\Upsilon^{n}$ and $T^{*}_{n}$

$\displaystyle\lvert\mathbb{E}\exp(-\langle
Y_{t-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle)-\mathbb{E}\exp(-\langle
Y_{t},\psi\rangle)\rvert$ $\displaystyle=$
$\displaystyle\lvert\mathbb{E}\left[\exp(-\langle
Y_{t-\Upsilon^{n}_{k_{n}}(t)},Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}\rangle)-\exp(-\langle
Y_{t},\psi\rangle);\Upsilon^{n}_{k_{n}}(t)<t\right]\rvert$ $\displaystyle\leq$
$\displaystyle\mathbb{P}(\Upsilon^{n}_{k_{n}}(t)<t)$ $\displaystyle=$
$\displaystyle\mathbb{P}(\gamma^{n}(k_{n})<t)=\mathbb{P}(T^{*}_{n}>k_{n}).$

As $\mathbb{P}(T^{*}_{n}<\infty)=1$ (see [Myt02, eq. (3.14)]), we have

$\mathbb{P}(T^{*}_{n}>k_{n})\to\infty\text{ as }n\to\infty.$

This proves (4.10).

Now define $\tilde{Z}^{(n)}_{t}:=Z^{(n)}_{\Upsilon^{n}_{k_{n}}(t)}$. The above
gives,

$\displaystyle\lim_{n\to\infty}\lvert\mathbb{E}[e^{-\langle
Y_{0},\tilde{Z}^{(n)}_{t}\rangle}]-\mathbb{E}[e^{-\langle
Y_{t},\psi\rangle}]\rvert=0,\qquad t\in[0,T]$

and we are done with the approximate dual processes being $\tilde{Z}^{(n)}$. ∎

##  5\. Proof of key Propositions

We will prove the three propositions required for the proof of Theorem 1.4 in
this section. For Proposition 4.1 we start by verifying (4.1) for measures
having densities and then prove it for the case of general measures.

###### Proof of Proposition 4.1.

Let $\varphi_{l}\in\mathcal{S}(\mathbb{R})_{+}$, $l\in\mathbb{N}$, be such
that $\mu_{l}(dx):=\varphi_{l}(x)\,dx\implies\mu(dx)$ as $l\to\infty$. Since
$n$ is fixed in this proof, let $v_{l}(\cdot)=V^{n}_{\cdot}(\mu_{l})$ solve

$\displaystyle\partial_{t}v_{l}(t)$
$\displaystyle=\frac{1}{2}\partial^{2}_{xx}v_{l}(t)-\frac{1}{2}b_{n}v_{l}(t)^{\alpha}$
(5.1) $\displaystyle v_{l}(0)$ $\displaystyle=\varphi_{l}.$

Fix $l,k\in\mathbb{N}$. Let $\psi(s,x):=v_{l}(T-s,x)=V^{n}_{T-s}(\mu_{l})(x)$.
Lemma A.2 says that $\psi$ satisfies the conditions of Proposition 3.4.

From (3.22) recall that

$\displaystyle\tilde{I}(Y_{s-},\psi_{s})=e^{-\langle
Y_{s-},\psi_{s}\rangle}\left[-\langle
Y_{s-},\frac{1}{2}\Delta\psi_{s}\rangle+\frac{1}{2}\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle-\langle
Y_{s-},\frac{\partial}{\partial s}\psi_{s}\rangle\right].$

Then by Proposition 3.4 for each $k\in\mathbb{N}$ and $t\in[0,T]$,

$\displaystyle\mathbb{E}_{Y}\left[\tilde{M}^{Y}_{T-t}(\psi)\right]=\mathbb{E}_{Y}\left[\tilde{M}^{Y}_{T}(\psi)\right]$

which implies,

(5.2) $\displaystyle\mathbb{E}_{Y}\exp\left(-\langle
Y_{T-t},\psi_{T-t}\rangle\right)=\mathbb{E}_{Y}\left[e^{-\langle
Y_{T}\psi_{T}\rangle}-\int_{T-t}^{T}\tilde{I}(Y_{s-},\psi_{s})\right].$

From the above definition of $\tilde{I}(Y,\psi)$ and (5) we get,

$\displaystyle\int_{T-t}^{T}\tilde{I}(Y_{s-},\psi_{s})\,ds$
$\displaystyle=\int_{T-t}^{T}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left\\{-\langle
Y_{s-},\frac{1}{2}\Delta\psi_{s}\rangle+\frac{1}{2}\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle-\langle
Y_{s-},\frac{\partial}{\partial s}\psi_{s}\rangle\right\\}\,ds$
$\displaystyle=\int^{T}_{T-t}e^{-\langle
Y_{s-},v_{l}(T-s)\rangle}\left\\{-\langle Y_{s-},\frac{1}{2}\Delta
v_{l}(T-s)\rangle+\frac{1}{2}\langle
Y_{s-}^{\alpha\beta},v_{l}(T-s)^{\alpha}\rangle-\langle
Y_{s-},\frac{\partial}{\partial s}v_{l}(T-s)\rangle\right\\}\,ds$
$\displaystyle=-\int^{0}_{t}e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}\left\\{-\langle Y_{(T-r)-},\frac{1}{2}\Delta
v_{l}(r)\rangle+\frac{1}{2}\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle+\langle
Y_{(T-r)-},\frac{\partial}{\partial r}v_{l}(r)\rangle\right\\}\,dr,$
$\displaystyle=-\int_{t}^{0}e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}\left\\{\frac{1}{2}\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle+\langle
Y_{(T-r)-},\frac{\partial}{\partial r}v_{l}(r)-\frac{1}{2}\Delta
v_{l}(r)\rangle\right\\}\,dr$ (5.3) $\displaystyle=-\int_{t}^{0}e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}\left\\{\frac{1}{2}\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-},\frac{1}{2}b_{n}v_{l}(r)^{\alpha}\rangle\right\\}\,dr,$

using the substitution $r=T-s$ for the third equality.

By (5.2) and (5),

(5.4) $\displaystyle\mathbb{E}_{Y}\exp\left(-\langle
Y_{T-t},v_{l}(t)\rangle\right)=\mathbb{E}_{Y}\exp\left(-\langle
Y_{T},v_{l}(0)\rangle\right)+\mathbb{E}_{Y}\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},v_{l}(r))\,dr$

where

$\tilde{\mathcal{I}}(Y_{(T-r)-},v_{l}(r))=e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}\frac{1}{2}\left(\langle
Y_{(T-r)-},b_{n}\left(v_{l}(r)\right)^{\alpha}\rangle-\langle
Y^{\alpha\beta}_{(T-r)-},\left(v_{l}(r)\right)^{\alpha}\rangle\right)$

We now have to check whether this holds when
$\mu:=w-\lim_{l\to\infty}\mu_{l}$.

Let $v(r)=V^{n}_{r}(\mu)$,

$\displaystyle
R_{l}:=\mathbb{E}_{Y}\int^{t}_{0}\tilde{\mathcal{I}}(Y_{(T-r)-},V^{n}_{r}(\mu_{l}))\,dr=\mathbb{E}_{Y}\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},v_{l}(r)))\,dr,$

and

$\displaystyle
R:=\mathbb{E}_{Y}\int^{t}_{0}\tilde{\mathcal{I}}(Y_{(T-r)-},V^{n}_{r}(\mu))\,dr=\mathbb{E}_{Y}\int^{t}_{0}\tilde{\mathcal{I}}(Y_{(T-r)-},v(r))\,dr.$

We only have to prove $R_{l}\to R$ as $l\to\infty$. In $R_{l}-R$ adding and
subtracting the term

$e^{-\langle Y_{(T-r)-},v(r)\rangle}\left(b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle\right)$

we have

$\displaystyle|R_{l}-R|\leq\frac{1}{2}(I_{1}^{l}+I_{2}^{l}),$

where

$\displaystyle I_{1}^{l}=$
$\displaystyle\mathbb{E}_{Y}\left|\int_{0}^{t}\left(e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}-e^{-\langle
Y_{(T-r)-},v(r)\rangle}\right)\left(b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle\right)\,dr\right|$
$\displaystyle I_{2}^{l}=$
$\displaystyle\mathbb{E}_{Y}\left|\int_{0}^{t}e^{-\langle
Y_{(T-r)-},v(r)\rangle}\left(b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}-v(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}-v(r)^{\alpha}\rangle\right)\,dr\right|.$

To prove $|R_{l}-R|\to 0$ as $l\to\infty$ we need to show

(i): $I_{1}^{l}\to 0$; and

(ii): $I_{2}^{l}\to 0$ as $l\to\infty$.

Proof of (i):  Let $1<q<\frac{1}{\beta}$ and $p>1$ be such that
$\frac{1}{p}+\frac{1}{q}=1$. Note that

$\displaystyle I_{1}^{l}$
$\displaystyle\leq\int_{0}^{t}\mathbb{E}_{Y}\left|\left(e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}-e^{-\langle
Y_{(T-r)-},v(r)\rangle}\right)\left(b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle\right)\right|\,dr$
$\displaystyle\leq\int_{0}^{t}\mathbb{E}_{Y}\left(\left|e^{-\langle
Y_{(T-r)-},v_{l}(r)\rangle}-e^{-\langle
Y_{(T-r)-},v(r)\rangle}\right|^{p}\right)^{1/p}\mathbb{E}_{Y}\left(\left|b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle\right|^{q}\right)^{1/q}\,dr$
(5.5) $\displaystyle=:\int_{0}^{t}I^{l}_{11}(r)I^{l}_{12}(r)\,dr$

using Hölder’s inequality in the second line. Here $I^{l}_{11}(r)$ and
$I^{l}_{12}(r)$ denote the first and second terms of the integrand in the
above.

Since$I^{l}_{11}(r)\to 0$ for each $r\in[0,t]$ and $|I^{l}_{11}(r)|\leq 2$ for
all $l$ and $r$, by the dominated convergence theorem, to prove (i) as above
we only have to show that $I^{l}_{12}(r)\leq C<\infty$ for some constant $C$
independent of $l$ and $r$.

$\displaystyle I^{l}_{12}(r)$
$\displaystyle=\mathbb{E}_{Y}\left(\left|b_{n}\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle\right|^{q}\right)^{1/q}$
$\displaystyle\leq b_{n}\mathbb{E}_{Y}\left(\left|\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle\right|^{q}\right)^{1/q}+\mathbb{E}_{Y}\left(\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle^{q}\right)^{1/q}$ (5.6)
$\displaystyle:=b_{n}I_{121}^{l}(r)+I_{122}^{l}(r)$

using Minkowski’s inequality.

For all $r<t$,

$\displaystyle I_{121}^{l}(r)=$
$\displaystyle\mathbb{E}_{Y}\left(\left|\langle
Y_{(T-r)-},v_{l}(r)^{\alpha}\rangle\right|^{q}\right)^{1/q}$ $\displaystyle=$
$\displaystyle\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}\mathbb{E}_{Y}\left[\left(\int_{\mathbb{R}}\frac{1}{\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}}v_{l}(r,x)^{\alpha}Y_{T-r}(x)\,dx\right)^{q}\right]^{1/q}$
$\displaystyle\leq$ $\displaystyle\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}\left[\mathbb{E}_{Y}\left(\frac{1}{\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}}\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}Y_{T-r}(x)^{q}\,dx\right)\right]^{1/q}$
$\displaystyle=$ $\displaystyle\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha-\alpha/q}\left[\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}\mathbb{E}_{Y}(Y_{T-r}(x)^{q})\,dx\right]^{1/q}$
$\displaystyle\leq$ $\displaystyle C_{T}\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha-\alpha/q}\left[\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}e^{c_{2}(T-r)}\,dx\right]^{1/q}$
$\displaystyle\leq$ $\displaystyle C_{T}\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha-\alpha/q}e^{c_{2}T}\left[\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}\,dx\right]^{1/q}$
(5.7) $\displaystyle=$ $\displaystyle C_{T}\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}e^{c_{2}T}.$

Here we have used Jensen’s inequality and Proposition 3.1 (applicable by our
assumption that $q<\alpha$) in the first and second inequalities respectively.
Now let us use a notation from Fleischmann [Fle88]:

$\lVert v\rVert_{\mathbf{L}^{\alpha,T}}:=\sup_{0\leq t\leq T}\lVert
v(t)\rVert_{L^{\alpha}(\mathbb{R})}$

By [Fle88, Proposition A2], we have $v_{l}\to v$ in $\mathbf{L}^{\alpha,T}$
and this implies that for large enough $l\in\mathbb{N}$, $\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}\leq\lVert
v\rVert_{\mathbf{L}^{\alpha,T}}+1$ for all $r\in[0,T]$. Therefore (5) gives us

(5.8) $\displaystyle I_{121}^{l}(r)\leq C_{T}(\lVert
v\rVert_{\mathbf{L}^{\alpha,T}}+1)^{\alpha}e^{c_{2}T},$

when $l$ is large.

For the term $I^{l}_{122}$ we again proceed as in the calculation (5). Note
that, as $\alpha\beta q<\alpha$ by our assumption, we can again apply
Proposition 3.1 in the following. Let $r<t$.

$\displaystyle I^{l}_{122}(r)=$ $\displaystyle\mathbb{E}_{Y}\left(\langle
Y_{(T-r)-}^{\alpha\beta},v_{l}(r)^{\alpha}\rangle^{q}\right)^{1/q}$
$\displaystyle\leq$ $\displaystyle\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}\left[\mathbb{E}_{Y}^{A}\left(\frac{1}{\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}}\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}Y_{T-r}(x)^{\alpha\beta
q}\,dx\right)\right]^{1/q}$ $\displaystyle=$ $\displaystyle\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha-\alpha/q}\left[\int_{\mathbb{R}}v_{l}(r,x)^{\alpha}\mathbb{E}_{Y}^{A}(Y_{T-r}(x)^{\alpha\beta
q})\,dx\right]^{1/q}$ (5.9) $\displaystyle\leq$ $\displaystyle C_{T}\lVert
v_{l}(r)\rVert_{L^{\alpha}(\mathbb{R})}^{\alpha}e^{c_{2}T}\leq C_{T}(\lVert
v\rVert_{\mathbf{L}^{\alpha,T}}+1)^{\alpha}e^{c_{2}T}$

for large $l$. We can observe that (5.8) and (5) together show that
$I^{l}_{12}\leq C$ where $C$ is independent of $l$ nd $r$. Thus (i) is proved.

Proof of (ii):  First note that $v_{l}\to v$ in $\mathbf{L}^{\alpha,T}$
implies the following almost everywhere convergence along a sub-sequence:
there exists a sequence $(l_{i})_{i}$ of natural numbers such that

$v_{l_{i}}(r,x)\to v(r,x)\text{ as }i\to\infty$

for a.e. $(r,x)\in[0,T]\times\mathbb{R}$

Also observe that since the stochastic integration part in (2.2) is a
martingale (see Lemma 3.2) we have $\mathbb{E}Y_{t}(x)=P_{t}\varphi(x)$. By
Proposition 3.1,

$\displaystyle I^{l_{i}}_{2}=$
$\displaystyle\mathbb{E}_{Y}\left|\int_{0}^{t}e^{-\langle
Y_{(T-r)-},v(r)\rangle}\left(b_{n}\langle
Y_{(T-r)-},v_{l_{i}}(r)^{\alpha}-v(r)^{\alpha}\rangle-\langle
Y_{(T-r)-}^{\alpha\beta},v_{l_{i}}(r)^{\alpha}-v(r)^{\alpha}\rangle\right)\,dr\right|$
$\displaystyle\leq$
$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}\left|v_{l_{i}}(r,x)^{\alpha}-v(r,x)^{\alpha}\right|\cdot\mathbb{E}_{Y}\left(b_{n}Y_{(T-r)-}(x)+Y^{\alpha\beta}_{(T-r)-}(x)\right)\,dr\,dx$
$\displaystyle\leq$ $\displaystyle
C_{T}\int_{0}^{t}\int_{\mathbb{R}}\left[b_{n}P_{T-r}\varphi(x)+C_{1}e^{C_{2}(T-r)}\right]\left|v_{l_{i}}(r,x)^{\alpha}-v(r,x)^{\alpha}\right|\,dr\,dx$
(5.10) $\displaystyle\leq$ $\displaystyle
C_{T}(b_{n}C_{3}+C_{1}e^{C_{2}T})\int_{0}^{t}\int_{\mathbb{R}}\left|v_{l_{i}}(r,x)^{\alpha}-v(r,x)^{\alpha}\right|\,dr\,dx$

using the fact that $P_{T-r}\varphi(x)\leq C$ (independent of $r$, $T$ and
$x$) as $\varphi$ is bounded (recall that $\varphi\in\mathcal{S}_{+}$).

Now we can choose $\mu_{l}$’s in such a way that for each $l\geq 1$,
$\mu_{l}-\mu,\mu_{l}-\mu_{l+1}\in M_{F}(\mathbb{R})$ (e.g. if
$\mu=\delta_{0}$, choose $\varphi_{l}\geq 0$ to be test functions converging
to $\delta_{0}$ as distributions, and $\mu_{l}(\,dx)=\varphi_{l}(x)\,dx$). By
[Myt02, Lemma 2.6(a)] we thus know that the convergence $v_{l}(r,x)\to v(r,x)$
is monotone decreasing. This proves that

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}\left|v_{l_{i}}(r,x)^{\alpha}-v(r,x)^{\alpha}\right|\,dr\,dx\to
0$

as $i\to\infty$ and hence proves (ii). ∎

Next we prove Proposition 4.2. For the proof we will need to understand how
$Y$ behaves when $Z^{(n)}$ jumps. Since $n$ is fixed in this proof, we drop it
to simplify the following notations. $Z=Z^{(n)}$, $V=V^{n}$, $S=S^{n}$,
$U=U^{n}$, $T_{l}=T^{n}_{l}$, $\tau=\tau^{n}$, $N=N^{n}$,
$\hat{N}=\hat{N}^{n}$, $\gamma(s):=\gamma^{n}(s)$ and
$\gamma_{l}:=\gamma^{n}(T^{n}_{l})$ for $l\in\mathbb{N}$. Also recall the
notation

$\displaystyle\theta(s,x,\lambda):=\theta_{n}(s,x,\lambda)=e^{-\langle
Y_{T-\gamma^{n}(s)},Z^{(n)}_{\gamma^{n}(s)-}\rangle}\left(e^{-\lambda
Y_{T-\gamma^{n}(s)}(x)}-1\right).$

###### Proof of Proposition 4.2.

Fix $t\in[0,T]$ and let

$\displaystyle\theta_{j}:=\theta(T_{j},U_{j},S_{j})=e^{-\langle
Y_{T-\gamma_{j}},Z_{\gamma_{j}}\rangle}-e^{-\langle
Y_{T-\gamma_{j}},Z_{\gamma_{j}-}\rangle}=e^{-\langle
Y_{T-\gamma_{j}},Z_{\gamma_{j}-}\rangle}\left(e^{-S_{j}Y_{T-\gamma_{j}}(U_{j})}-1\right).$

Suppose we show that on the event $\\{\gamma_{l}\leq t<\gamma_{l+1}\\}$ we
have,

(5.11) $\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-t},Z_{t}\rangle}\right]=\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l}\theta_{i}\right],$

then we can write

$\displaystyle\sum_{i=1}^{l}\theta_{i}=\int_{0}^{\tau(t)}\int_{\mathbb{R}}\int_{\mathbb{R}_{+}}\theta(s,x,\lambda)N(\,ds,\,dx,\,d\lambda),$

since for $\gamma_{l}\leq t<\gamma_{l+1}$ by definition (see (2.8) and (4.2))
$\tau(t)\in[T_{l},T_{l+1})$. Replace the above in (5.11) and we obtain (4.2).
So, to complete the proof of (4.2) we need to establish (5.11).

We will prove this by induction on $l=0,1,2,\ldots$ and use (4.1) repeatedly
in the following. Note that (5.11) for $t=0$ is trivial. When
$0=\gamma_{0}<t<\gamma_{1}$, by our convention $l=0$. In this case (5.11) is

(5.12) $\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-t},Z_{t}\rangle}\right]=\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr\right]$

and this follows directly from (4.1).

Now assume that (5.11) holds on the event $\\{\gamma_{l}\leq
s<\gamma_{l+1}\\}$. We first show that

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-\gamma_{l+1}},Z_{\gamma_{l+1}}\rangle}\right]$ (5.13) $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{\gamma_{l+1}}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l+1}\theta_{i}\right].$

By definition of $\theta_{l+1}$ and induction hypothesis,

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-\gamma_{l+1}},Z_{\gamma_{l+1}}\rangle}\right]=\mathbb{E}_{Y}[\theta_{l+1}]+\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-\gamma_{l+1}},Z_{\gamma_{l+1}-}\rangle}\right]$ $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}[\theta_{l+1}]+\lim_{\begin{subarray}{c}{s\uparrow\gamma_{l+1}}\\\
{\gamma_{l}\leq s<\gamma_{l+1}}\end{subarray}}\mathbb{E}_{Y}e^{-\langle
Y_{T-s},Z_{s}\rangle}$ $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}[\theta_{l+1}]+\lim_{\begin{subarray}{c}{s\uparrow\gamma_{l+1}}\\\
{\gamma_{l}\leq s<\gamma_{l+1}}\end{subarray}}\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{s}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l}\theta_{i}\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}_{Y}[\theta_{l+1}]+\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{\gamma_{l+1}}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l}\theta_{i}\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{\gamma_{l+1}}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l+1}\theta_{i}\right].$

This proves (5).

The last step of the induction is to prove (5.11) when $l$ is replaced with
$l+1$ and $\gamma_{l+1}<t<\gamma_{l+2}$. We use (4.1) with $T-\gamma_{l+1}$,
$t-\gamma_{l+1}$ instead of $T$, $t$ and then apply (5) to get,

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-t},Z_{t}\rangle}\right]=\mathbb{E}_{Y}\left[\exp\left(-\langle
Y_{T-t},V_{t-\gamma_{l+1}}(Z_{\gamma_{l+1}})\rangle\right)\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}_{Y}\left[\exp\left(-\langle
Y_{T-\gamma_{l+1}},V_{0}(Z_{\gamma_{l+1}})\rangle\right)\right]+\mathbb{E}_{Y}\left[\int_{0}^{t-\gamma_{l+1}}\tilde{\mathcal{I}}(Y_{(T-\gamma_{l+1}-r)-},V_{r}(Z_{\gamma_{l+1}}))\,dr\right]$
$\displaystyle=$ $\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}+\int_{0}^{\gamma_{l+1}}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\sum_{i=1}^{l+1}\theta_{i}\right]$
(5.14)
$\displaystyle+\mathbb{E}_{Y}\left[\int_{\gamma_{l+1}}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr\right],$

which is the required expression. This completes the induction argument and
proves (5.11).

∎

For the proof of our final proposition, we continue to suppress $n$ and use
the notations introduced before the previous proof. Define

(5.15) $\displaystyle
M_{s}=\int^{s}_{0}\int_{\mathbb{R}}\int_{0}^{\infty}\theta(r,x,\lambda)[N(\,dr,\,dx,\,d\lambda)-\hat{N}(\,dr,\,dx,\,d\lambda)]$

and note that $Y$-a.s. $M$ is an $\mathcal{F}^{Z^{(n)}}$-martingale.

###### Proof of Proposition 4.3.

We recall that

$\displaystyle\eta=\frac{\alpha\beta(\alpha\beta-1)}{\Gamma(2-\alpha\beta)}\text{
and }g(r,y)=\int_{0+}^{r}(e^{-\lambda y}-1+\lambda
y)\lambda^{-\alpha\beta-1}\,d\lambda,\quad r,y\geq 0.$

Since for all $y\geq 0$,

$\displaystyle y^{\alpha\beta}=\eta\int_{0+}^{\infty}(e^{-\lambda y}-1+\lambda
y)\lambda^{-\alpha\beta-1}\,d\lambda=\eta
g(1/n,y)+\eta\int_{1/n}^{\infty}(e^{-\lambda
y}-1)\lambda^{-\alpha\beta-1}\,d\lambda+b_{n}y,$

we can write

$\displaystyle\mathbb{E}_{Y}\left[\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr\right]=\mathbb{E}_{Y}\left[\int^{t}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\frac{1}{2}\langle
b_{n}Y_{(T-r)-}-Y^{\alpha\beta}_{(T-r)-},Z_{r-}^{\alpha}\rangle\,dr\right]$
$\displaystyle=$
$\displaystyle-\frac{\eta}{2}\mathbb{E}_{Y}\left[\int^{t}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),Z_{r-}^{\alpha}\rangle\,dr\right]$ (5.16)
$\displaystyle-\frac{\eta}{2}\mathbb{E}_{Y}\left[\int^{t}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle\int_{1/n}^{\infty}(e^{-\lambda
Y_{(T-r)-}(\cdot)}-1)\lambda^{-\alpha\beta-1}\,d\lambda,Z_{r-}^{\alpha}\rangle\,dr\right].$

Let

$\displaystyle h(r)=$ $\displaystyle e^{-\langle
Y_{T-r},Z_{r-}\rangle}\int_{\mathbb{R}}\frac{\left(Z_{r-}(x)\right)^{\alpha}}{\lVert
Z_{r-}\rVert_{\alpha}^{\alpha}}\int_{1/n}^{\infty}\left(e^{-\lambda
Y_{T-r}(x)}-1\right)\lambda^{-\alpha\beta-1}\,d\lambda\,dx$
$\displaystyle\beta(r)=$ $\displaystyle\frac{1}{2}\lVert
Z_{r-}\rVert^{\alpha}_{\alpha}.$

Then by definition $\gamma(t)=\inf\\{s\geq 0\mid\int_{0}^{s}\beta(s)\,ds>t\\}$
and also recall from (4.2) that $\gamma(\tau(s))=s$. Applying [EK86, Exercise
6.12], for any $s\geq 0$, $Y$-a.s. we have

$\displaystyle\frac{\eta}{2}\int^{s}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle\int_{1/n}^{\infty}(e^{-\lambda
Y_{(T-r)-}(\cdot)}-1)\lambda^{-\alpha\beta-1}\,d\lambda,Z_{r-}^{\alpha}\rangle\,dr$
$\displaystyle=$
$\displaystyle\eta\int^{s}_{0}h(r)\beta(r)\,dr=\eta\int^{\gamma(\tau(s))}_{0}h(r)\beta(r)\,dr=\eta\int^{\tau(s)}_{0}h(\gamma(r))\,dr$
$\displaystyle=$ $\displaystyle\eta\int^{\tau(s)}_{0}e^{-\langle
Y_{T-\gamma(r)},Z_{\gamma(r)-}\rangle}\int_{\mathbb{R}}\frac{\left(Z_{\gamma(r)-}(x)\right)^{\alpha}}{\lVert
Z_{\gamma(r)-}\rVert_{\alpha}^{\alpha}}\int_{1/n}^{\infty}\left(e^{-\lambda
Y_{T-\gamma(r)}(x)}-1\right)\lambda^{-\alpha\beta-1}\,d\lambda\,dx\,dr$
$\displaystyle=$
$\displaystyle\eta\int^{\tau(s)}_{0}\int_{\mathbb{R}}\int_{0}^{\infty}\theta(r,x,\lambda)\frac{\left(Z_{\gamma(r)-}(x)\right)^{\alpha}}{\lVert
Z_{\gamma(r)-}\rVert_{\alpha}^{\alpha}}\mathbf{1}(\lambda>1/n)\lambda^{-\alpha\beta-1}\,dr\,dx\,d\lambda$
(5.17) $\displaystyle=$
$\displaystyle\int^{\tau(s)}_{0}\int_{\mathbb{R}}\int_{0}^{\infty}\theta(r,x,\lambda)\hat{N}(\,dr,\,dx,\,d\lambda)$

using Lemma 2.4 in the last line.

Combining (5.15) and the calculations in (5), (5) we get,

$\displaystyle\mathbb{E}_{Y}\left[\int_{0}^{t}\tilde{\mathcal{I}}(Y_{(T-r)-},Z_{r-})\,dr+\int_{0}^{\tau(t)}\int_{\mathbb{R}}\int_{\mathbb{R}_{+}}\theta(s,x,\lambda)N(\,ds,\,dx,\,d\lambda)\right]$
(5.18) $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}\left[M_{\tau(t)}-\frac{\eta}{2}\int_{0}^{t}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),Z_{r-}^{\alpha}\rangle\,dr\right]$

We can now use (5) to rewrite (4.2).

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle Y_{T-t},Z_{t}\rangle}\right]$
(5.19)
$\displaystyle=\mathbb{E}_{Y}\left[M_{\tau(t)}\right]-\mathbb{E}_{Y}\left[\frac{\eta}{2}\int_{0}^{t}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),Z_{r-}^{\alpha}\rangle\,dr\right]$

Recall the notation $\Upsilon_{m}(t)=\gamma(m)\wedge t$ and observe that for
any $m\in\mathbb{N}$, $\tau(\Upsilon_{m}(t))=\tau(t)\wedge m$. We localize the
above as follows.

$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T-\Upsilon_{m}(t)},Z_{\Upsilon_{m}(t)}\rangle}\right]$ $\displaystyle=$
$\displaystyle\mathbb{E}_{Y}\left[e^{-\langle
Y_{T},Z_{0}\rangle}\right]+\mathbb{E}_{Y}\left[M_{\tau(t)\wedge m}\right]$
(5.20)
$\displaystyle-\frac{\eta}{2}\mathbb{E}_{Y}\left[\int^{\Upsilon_{m}(t)}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),Z_{r-}^{\alpha}\rangle\,dr\right]$

Apply $\mathbb{E}_{Z}$ to the above. As

$\mathbb{E}_{Z}\mathbb{E}_{Y}(M_{\tau^{Z}(t)\wedge
m})=\mathbb{E}_{Y}\mathbb{E}_{Z}(M_{\tau^{Z}(t)\wedge m})=0$

we have,

$\displaystyle\mathbb{E}\left[e^{-\langle
Y_{T-\Upsilon_{m}(t)},Z_{\Upsilon_{m}(t)}\rangle}\right]=$
$\displaystyle\mathbb{E}\left[e^{-\langle Y_{T},Z_{0}\rangle}\right]$ (5.21)
$\displaystyle-\frac{\eta}{2}\mathbb{E}\left[\int^{\Upsilon_{m}(t)}_{0}e^{-\langle
Y_{(T-r)-},Z_{r-}\rangle}\langle
g(1/n,Y_{(T-r)-}(\cdot)),Z_{r-}^{\alpha}\rangle\,dr\right].$

This is the required expression. ∎

##  Appendix A

###  A.1. A Grönwall-type Lemma

######  Lemma A.1.

Let $\gamma\in(0,1)$ and $f:[0,T]\to[0,\infty)$ be a function such that
$\int_{0}^{T}f(r)\,dr<\infty$ and for all $t\in[0,T]$

(A.1) $\displaystyle f(t)\leq c+c\int_{0}^{t}(t-r)^{-\gamma}f(r)\,dr$

for some constant $c>0$. Then there exists constants $c_{1}c_{2}>0$ (depending
only on $T$ and $\gamma$) such that, for $t\in[0,T]$

(A.2) $\displaystyle f(t)\leq c_{1}e^{c_{2}t}.$

###### Proof.

Let $k>0$ be the smallest integer such that $\gamma<\frac{k}{k+1}$ and
$t\in[0,T]$. We apply (A.1) and use the substitution $v=\frac{r-u}{t-u}$,

$\displaystyle f(t)\leq$ $\displaystyle
c+c\int_{0}^{t}(t-r)^{-\gamma}\,dr+c\int_{0}^{t}\int_{0}^{r}(t-r)^{-\gamma}(r-u)^{-\gamma}f(u)\,du\,dr$
$\displaystyle=$ $\displaystyle
c_{1}(t)+c\int_{0}^{t}\,duf(u)\int_{u}^{t}\,dr(t-r)^{-\gamma}(r-u)^{-\gamma}$
$\displaystyle=$ $\displaystyle
c_{1}(t)+c\int_{0}^{t}f(u)(t-u)^{1-2\gamma}\,du\int_{0}^{1}\,dv(1-v)^{-\gamma}v^{-\gamma}$
(A.3) $\displaystyle=$ $\displaystyle
c_{1}(t)+c^{\prime}_{1}\int_{0}^{t}f(u)(t-u)^{1-2\gamma}\,du$

where $c_{1}(t)=c+c\frac{t^{1-\gamma}}{1-\gamma}$ and
$c^{\prime}_{1}=B(1-\gamma,1-\gamma)$ with $B$ here denoting the Beta
function. Again applying (A.1) to (A.1) we have

$\displaystyle f(t)\leq$ $\displaystyle
c_{2}(t)+c^{\prime}_{2}\int_{0}^{t}f(u)(t-u)^{2-3\gamma}\,du,$

where $c_{2}(t)=c_{1}(t)+c^{\prime}_{1}\frac{t^{2-2\gamma}}{2-2\gamma}$ and
$c^{\prime}_{2}=c^{\prime}_{1}B(1-\gamma,2-2\gamma)$. Continuing this process
for $k$ steps we get,

$\displaystyle f(t)\leq$ $\displaystyle
c_{k}(t)+c^{\prime}_{k}\int_{0}^{t}f(u)(t-u)^{k-(k+1)\gamma}\,du$ (A.4)
$\displaystyle\leq$ $\displaystyle
c_{k}(t)+c^{\prime}_{k}T^{k-(k+1)\gamma}\int_{0}^{t}f(u)\,du.$

where the last step is obtained by our assumption on $k$. Note that $c_{k}(t)$
is an increasing function of $t$. Also note that $f$ is non-negative and
integrable on $[0,T]$ by hypothesis. Therefore we can apply the standard
Grönwall’s inequality (see [Eva10, Appendix B2]) and have,

$\displaystyle f(t)\leq c_{1}(t)\exp(c_{2}(T)t)$

for all $t\in[0,T]$. ∎

###  A.2. Norm Estimates for Solutions of the Evolution Equation

This section contains some useful properties of the solutions to the PDE

$\displaystyle\frac{\partial}{\partial t}v(t,x)$
$\displaystyle=\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}v(t,x)-\frac{b_{n}}{2}v(t,x)^{\alpha},\quad x\in\mathbb{R},t\in[0,T],$
(A.5) $\displaystyle v(0,\cdot)$ $\displaystyle=\varphi.$

where $T$ is arbitrary but finite. When
$\varphi\in\mathcal{S}(\mathbb{R})_{+}$, [Isc86, Theorem A] guarantees that
this equation admits a unique solution.

######  Lemma A.2.

If $v=v(t,x)$ solves the PDE (A.2) and
$\varphi\in\mathcal{S}(\mathbb{R})_{+}$, then $v$ satisfies all the hypotheses
of Proposition 3.4.

###### Proof.

(a) It follows from [Fle88, Proposition A2] and proof of [Myt02, Lemma 2.1(c)]
that $s\mapsto v(s)\in L^{1}\cap L^{2}$ is continuous.

(b) We first prove that

(A.6) $\displaystyle\sup_{s\leq T}\left\lVert\frac{\partial^{2}}{\partial
x^{2}}v(s)\right\rVert_{\infty}=\left\lVert\frac{\partial^{2}}{\partial
x^{2}}v\right\rVert_{L^{\infty}([0,T]\times\mathbb{R})}<\infty.$

Note that as $\varphi\in\mathcal{S}_{+}$, by [Isc86, Theorem A], the solution
$v:[0,T]\to C_{0}(\mathbb{R}_{+})_{+}$ (continuous, non-negative functions
vanishing at infinity) is a continuous map. Therefore by (A.2), to show (A.6)
it is enough to prove that

(A.7) $\displaystyle\sup_{s\leq t}\lVert w(s)\rVert_{\infty}<\infty$

where we have used the notation $w(s)=\dot{v}(s)=\frac{\partial}{\partial
s}v(s).$

From the proof of [Isc86, Theorem A] it follows that $w$ must satisfy the PDE

(A.8) $\displaystyle w(t)=P_{t}(\tilde{\varphi})-\frac{\alpha
b_{n}}{2}\int_{0}^{t}P_{t-s}(v(s)^{\alpha-1}w(s))\,ds$

where $\tilde{\varphi}=\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\varphi-\frac{b_{n}}{2}\varphi^{\alpha}$. This gives us,

(A.9) $\displaystyle\lVert
w(t)\rVert_{\infty}\leq\lVert\tilde{\varphi}\rVert_{\infty}+\frac{\alpha
b_{n}}{2}\int_{0}^{t}\lVert(v(s)^{\alpha-1}\rVert_{\infty}\lVert
w(s)\rVert_{\infty}\,ds,$

from which using Grönwall’s inequality (see [Eva10, Appendix B2]) we obtain

(A.10) $\displaystyle\lVert
w(t)\rVert_{\infty}\leq\lVert\tilde{\varphi}\rVert_{\infty}\exp\left(\frac{\alpha
b_{n}}{2}\int_{0}^{t}\lVert v(s)^{\alpha-1}\rVert_{\infty}\,ds\right)<\infty.$

As $v$ is continuously differentiable (see the proof of [Isc86, Theorem A]),
$s\mapsto w(s,\cdot)$ is continuous. This fact along with the above gives us
(A.7).

Next we show that the map $[0,T]\to L^{\infty}(\mathbb{R})$,
$t\mapsto\frac{\partial^{2}}{\partial x^{2}}v(t)$ is continuous, i.e.

(A.11) $\displaystyle\left\lVert\frac{\partial^{2}}{\partial
x^{2}}v(s)-\frac{\partial^{2}}{\partial x^{2}}v(t)\right\rVert_{\infty}\to
0\text{ as }s\to t\text{ in }[0,T].$

Similarly as above, since $v\in C_{0}(\mathbb{R})_{+}$, by (A.2) it is enough
to show that

(A.12) $\displaystyle\lVert w(t)-w(s)\rVert_{\infty}\to 0\text{ as }s\to t$

and we use (A.8) for this purpose.

Let $0\leq s<t\leq T$. Let $f_{r}=v(r)^{\alpha-1}w(r)$. Then from (A.8)

$\displaystyle w(t)-w(s)=$ $\displaystyle
P_{t}(\tilde{\varphi})-P_{s}(\tilde{\varphi})-\frac{\alpha
b_{n}}{2}\left[\int_{0}^{t}P_{t-r}(f_{r})\,dr-\int_{0}^{s}P_{s-r}(f_{r})\,dr\right]$
$\displaystyle=$ $\displaystyle
P_{t}(\tilde{\varphi})-P_{s}(\tilde{\varphi})-\frac{\alpha
b_{n}}{2}\left[\int_{s}^{t}P_{t-r}(f_{r})\,dr+\int_{0}^{s}\left(P_{t-r}(f_{r})-P_{s-r}(f_{r})\right)\,dr\right]$
(A.13) $\displaystyle=$ $\displaystyle
P_{t}(\tilde{\varphi})-P_{s}(\tilde{\varphi})-\frac{\alpha
b_{n}}{2}\left[\int_{s}^{t}P_{t-r}(f_{r})\,dr+\int_{0}^{s}\left(P_{t-s}(P_{s-r}(f_{r}))-P_{s-r}(f_{r})\right)\,dr\right]$

using that fact $P_{t-r}f_{r}=P_{t-s}(P_{s-r}f_{r})$.

As $\varphi\in\mathcal{S}(\mathbb{R})_{+}$, by our definition
$\tilde{\varphi}\in\mathcal{S}(\mathbb{R})$. Therefore when $s\to t$, $\lVert
P_{t}\tilde{\varphi}-P_{s}\tilde{\varphi}\rVert_{\infty}\to 0$. For the second
term in (A.2), note that if we can prove that

$\sup_{r\leq t}\lVert P_{t-r}(f_{r})\rVert_{\infty}<\infty,$

it will follow that $\int_{s}^{t}P_{t-r}(f_{r})\,dr\to 0$ in
$L^{\infty}(\mathbb{R})$ as $s\to t$. We have, for $x\in\mathbb{R}$

$\displaystyle\left|P_{t-r}(f_{r})(x)\right|\leq\left|\int_{\mathbb{R}}p_{t-r}(x-y)f_{r}(y)\,dr\right|\leq\lVert
f_{r}\rVert_{\infty}=\lVert v(r)^{\alpha-1}w(r)\rVert_{\infty}<\infty.$

since know $v(r)\in C_{0}(\mathbb{R})$ and we have already shown that
$\sup_{r\leq t}\lVert w(r)\rVert_{\infty}<\infty$. Similarly, the third term
in (A.2) can be shown to be converging to $0$ in $L^{\infty}(\mathbb{R})$ as
$s\to t$. This proves (A.12) and hence (A.11).

(c) Let $\rho=\frac{\alpha\beta}{\alpha\beta-1}$. To show $\sup_{s\leq
t}\lVert\dot{v}(s)\rVert_{\rho}<\infty$ we again use (A.8). Note that

(A.14) $\displaystyle\int_{\mathbb{R}}|w(t,x)|^{\rho}\,dx\leq
C\left[\int_{\mathbb{R}}|P_{t}(\tilde{\varphi})|^{\rho}\,dx+\frac{\alpha
b_{n}}{2}\int_{\mathbb{R}}\left|\int_{0}^{t}P_{t-s}(v(s)^{\alpha-1}w(s))\,ds\right|^{\rho}\,dx\right].$

Using Jensen inequality,

$\displaystyle\int_{\mathbb{R}}|P_{t}(\tilde{\varphi})|^{\rho}\,dx=$
$\displaystyle\int_{\mathbb{R}}\left|\int_{\mathbb{R}}p_{t}(x-y)\tilde{\varphi}(y)\,dy\right|^{\rho}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}}\int_{\mathbb{R}}p_{t}(x-y)|\tilde{\varphi}(y)|^{\rho}\,dy\,dx$
$\displaystyle=$
$\displaystyle\lVert\tilde{\varphi}\rVert_{\infty}^{\rho}<\infty.$

By definition of $P_{t-s}$ and using Jensen’s inequality once more we have,

$\displaystyle\int_{\mathbb{R}}\left|\int_{0}^{t}P_{t-s}(v(s)^{\alpha-1}w(s))\,ds\right|^{\rho}\,dx=$
$\displaystyle\int_{\mathbb{R}}\left|\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-y)v(s,y)^{\alpha-1}w(s,y)\,dy\,ds\right|^{\rho}\,dx$
$\displaystyle\leq$ $\displaystyle
t^{\rho-1}\int_{\mathbb{R}}\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-y)|v(s,y)^{\alpha-1}w(s,y)|^{\rho}\,dy\,ds\,dx.$

Using Fubini’s Theorem, as all terms are non-negative, integrating out $x$ in
the above we have

$\displaystyle\int_{\mathbb{R}}\left|\int_{0}^{t}P_{t-s}(v(s)^{\alpha-1}w(s))\,ds\right|^{\rho}\,dx$
$\displaystyle\leq
t^{\rho-1}\int_{0}^{t}\int_{\mathbb{R}}|v(s,y)^{\alpha-1}w(s,y)|^{\rho}\,dy\,ds$
(A.15) $\displaystyle\leq C_{1}\int_{0}^{t}\lVert
v(s)\rVert_{\infty}^{\rho(\alpha-1)}\lVert w(s)\rVert_{\rho}^{\rho}\,ds.$

Using (A.2) in (A.14) we have

(A.16) $\displaystyle\lVert w(t)\rVert_{\rho}^{\rho}\leq
C\lVert\tilde{\varphi}\rVert_{\infty}^{\rho}+C\int_{0}^{t}\lVert
v(s)\rVert_{\infty}^{\rho(\alpha-1)}\lVert w(s)\rVert_{\rho}^{\rho}\,ds.$

Again by Grönwall’s inequality

(A.17) $\displaystyle\lVert w(t)\rVert_{\rho}^{\rho}\leq
C\lVert\tilde{\varphi}\rVert_{\infty}^{\rho}\exp\left(\int_{0}^{t}\lVert
v(s)\rVert_{\infty}^{\rho(\alpha-1)}\,ds\right)<\infty$

and the required result follows as in the previous part. ∎

###  A.3. Proof of Proposition 3.4

Since the proof is a little long we carry it out in two steps. The first shows
that a solution of the weak form (1.3) also satisfies a time-dependent version
as described in (A.18). The proof follows the argument of [Shi94, Theorem
2.1].

######  Lemma A.3.

Let $T>0$ be fixed and assume that $Y$ satisfies (1.3) and the following
conditions hold for $\psi:[0,T]\times\mathbb{R}\to[0,\infty)$.

  * (i)

The map $[0,T]\ni s\mapsto\psi_{s}\in L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$
is continuous.

  * (ii)

$\sup_{s\leq T}\lVert\frac{\partial}{\partial
s}\psi_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}}<\infty$, and

  * (iii)

$s\mapsto\frac{\partial^{2}}{\partial x^{2}}\psi_{s}$ is continuous in
$L^{\infty}(\mathbb{R})$, i.e. $\lVert\frac{\partial^{2}}{\partial
x^{2}}\psi_{s}-\frac{\partial^{2}}{\partial x^{2}}\psi_{t}\rVert_{\infty}\to
0$ as $|s-t|\to 0$.

Then for each $t\in[0,T]$, we have

(A.18) $\displaystyle\langle Y_{t},\psi_{t}\rangle=\langle
Y_{0},\psi_{0}\rangle+\int_{0}^{t}\langle
Y_{s},\left(\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}+\frac{\partial}{\partial
s}\right)\psi_{s}\rangle\,ds+\int_{0}^{t}\int_{\mathbb{R}}(Y_{s-}(x))^{\beta}\psi_{s}(x)L^{\alpha}(\,ds\,dx).$

###### Proof.

Fix $0\leq t\leq T$ and let $\Delta=\\{0=t_{0}<t_{1}<\cdots<t_{N}=t\\}$ be a
partition of $[0,t]$. For all $s\in[t_{i-1},t_{i}]$, denote
$\pi_{\Delta}(s)=t_{i-1}$ and $\bar{\pi}_{\Delta}(s)=t_{i}$. Then we have

$\displaystyle\langle Y_{t},\psi_{t}\rangle-\langle Y_{0},\psi_{0}\rangle$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{N}(\langle
Y_{t_{i}},\psi_{t_{i}}-\psi_{t_{i-1}}\rangle-\langle
Y_{t_{i}}-Y_{t_{i-1}},\psi_{t_{i-1}}\rangle)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{N}\left[\int_{t_{i-1}}^{t_{i}}\langle
Y_{\bar{\pi}_{\Delta}(s)},\dot{\psi}_{s}\rangle\,ds+\int_{t_{i-1}}^{t_{i}}\langle
Y_{s},\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\psi_{\pi_{\Delta}(s)}\rangle\,ds+\int_{t_{i-1}}^{t_{i}}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{\pi_{\Delta}(s)}(x)L^{\alpha}(\,ds,\,dx)\right]$
(A.19) $\displaystyle=$ $\displaystyle\int_{0}^{t}\left(\langle
Y_{\bar{\pi}_{\Delta}(s)},\dot{\psi}_{s}\rangle+\langle
Y_{s},\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\psi_{\pi_{\Delta}(s)}\rangle\right)\,ds+\int_{0}^{t}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{\pi_{\Delta}(s)}(x)L^{\alpha}(\,ds,\,dx)$

To prove the lemma we have to show that, as $|\Delta|\to 0$

  * (a)

$\int_{0}^{t}\langle
Y_{\bar{\pi}_{\Delta}(s)},\dot{\psi}_{s}\rangle\,ds\to\int_{0}^{t}\langle
Y_{s},\dot{\psi}_{s}\rangle\,ds$ a.s. ,

  * (b)

$\int_{0}^{t}\langle Y_{s},\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\psi_{\pi_{\Delta}(s)}\rangle\,ds\to\int_{0}^{t}\langle
Y_{s},\frac{1}{2}\frac{\partial^{2}}{\partial x^{2}}\psi_{s}\rangle\,ds$ a.s.
, and

  * (c)

$\int_{0}^{t}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{\pi_{\Delta}(s)}(x)L^{\alpha}(\,ds,\,dx)\to\int_{0}^{t}\int_{\mathbb{R}^{d}}(Y_{s-}(x))^{\beta}\psi_{s}(x)L^{\alpha}(\,ds\,dx)$
in probability.

For (a) and (b), we need to show that the integrand converges pointwise (i.e.
for each $s$) and that the dominated convergence theorem (DCT) can be applied.

(a) Recall that $s\mapsto Y_{s}$ is right continuous measure-valued a.s. and
by the definition of weak convergence we have, for each $s\in[0,t]$

$\displaystyle|\langle
Y_{\bar{\pi}_{\Delta}(s)}-Y_{s},\dot{\psi}_{s}\rangle|\to 0,\text{ a.s.}$

as $\dot{\psi}_{s}$ is bounded and continuous (in the space variable).

By Hölder’s inequality, as $\alpha\beta>1$, we have a.s.

$\displaystyle|\langle
Y_{\bar{\pi}_{\Delta}(s)}-Y_{s},\dot{\psi}_{s}\rangle|\leq$
$\displaystyle\int_{\mathbb{R}}\lvert
Y_{s}(x)\rvert\lvert\dot{\psi}_{s}(x)\rvert\,dx+\int_{\mathbb{R}}\lvert
Y_{\bar{\pi}_{\Delta}(s)}(x)\rvert\lvert\dot{\psi}_{s}(x)\rvert\,dx$
$\displaystyle\leq\lVert
Y_{s}\rVert_{\alpha\beta}\lVert\dot{\psi}_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}}+\lVert
Y_{\bar{\pi}_{\Delta}(s)}\rVert_{\alpha\beta}\lVert\dot{\psi}_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}}.$

Therefore, a.s.

$\displaystyle\int_{0}^{t}|\langle
Y_{\bar{\pi}_{\Delta}(s)}-Y_{s},\dot{\psi}_{s}\rangle|\,ds$
$\displaystyle\leq(\sup_{s\leq
t}\lVert\dot{\psi}_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}})\left[\int_{0}^{t}\lVert
Y_{s}\rVert_{\alpha\beta}\,ds+\int_{0}^{t}\lVert
Y_{\bar{\pi}_{\Delta}(s)}\rVert_{\alpha\beta}\,ds\right]$
$\displaystyle=(\sup_{s\leq
t}\lVert\dot{\psi}_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}})\left[\left(\int_{0}^{t}\lVert
Y_{s}\rVert_{\alpha\beta}\,ds\right)^{\frac{\alpha\beta}{\alpha\beta}}+\left(\int_{0}^{t}\lVert
Y_{\bar{\pi}_{\Delta}(s)}\rVert_{\alpha\beta}\,ds\right)^{\frac{\alpha\beta}{\alpha\beta}}\right]$
$\displaystyle\leq(\sup_{s\leq
t}\lVert\dot{\psi}_{s}\rVert_{\frac{\alpha\beta}{\alpha\beta-1}})\left[\left(t^{\alpha\beta}\frac{1}{t}\int_{0}^{t}\lVert
Y_{s}\rVert_{\alpha\beta}^{\alpha\beta}\,ds\right)^{\frac{1}{\alpha\beta}}+\left(t^{\alpha\beta}\frac{1}{t}\int_{0}^{t}\lVert
Y_{\bar{\pi}_{\Delta}(s)}\rVert_{\alpha\beta}^{\alpha\beta}\,ds\right)^{\frac{1}{\alpha\beta}}\right],$

using Jensen in the last line. The quantity above is finite by assumption (ii)
and the fact that $Y\in
L^{\alpha\beta}_{loc}(\mathbb{R}_{+}\times\mathbb{R})$. This implies that
$s\mapsto|\langle Y_{\bar{\pi}_{\Delta}(s)}-Y_{s},\dot{\psi}_{s}\rangle|$ is
a.s. integrable on $[0,t]$.

(b) Fix $s\in[0,t]$. Then

$\displaystyle|\langle Y_{s},\frac{\partial^{2}}{\partial
x^{2}}(\psi_{s}-\psi_{\pi_{\Delta}(s)})\rangle|$
$\displaystyle\leq\left(\sup_{x\in\mathbb{R}}|\frac{\partial^{2}}{\partial
x^{2}}(\psi_{s}-\psi_{\pi_{\Delta}(s)})(x)|\right)\langle
Y_{s},1\rangle,\text{ a.s.}$

We know that $Y_{s}$ is a finite measure, i.e. $\langle
Y_{s},1\rangle<\infty$. Thus the RHS above converges to $0$ by our assumption
(iii).

Let us introduce a new stopping time: for $k\in\mathbb{N}$,

$\sigma_{k}=\inf\\{s\geq 0\mid\langle Y_{s},1\rangle>k\\}.$

For $s\leq\sigma_{k}\wedge t$ we have, a.s.

$\displaystyle|\langle Y_{s},\frac{\partial^{2}}{\partial
x^{2}}(\psi_{s}-\psi_{\pi_{\Delta}(s)})\rangle|$
$\displaystyle\leq\left(\sup_{x\in\mathbb{R}}|\frac{\partial^{2}}{\partial
x^{2}}(\psi_{s}-\psi_{\pi_{\Delta}(s)})(x)|\right)\langle Y_{s},1\rangle$
$\displaystyle\leq 2k\left(\sup_{\stackrel{{\scriptstyle s\leq
t}}{{x\in\mathbb{R}}}}|\frac{\partial^{2}}{\partial
x^{2}}\psi_{s}(x)|\right)<\infty$

by hypothesis. As $\sigma_{k}\to\infty$ as $k\to\infty$ the above is true for
all $s\leq t$. Thus we can apply DCT to obtain (b).

(c) Recall the notations introduced in the beginning of Section 2. We have

(A.20) $\displaystyle
L^{\alpha}(\,ds,\,dx)=\int_{0}^{\infty}z\tilde{N}(\,ds,\,dz,\,dx)$

where $N(\,ds,\,dz,\,dx)$ is a PRM on $(0,\infty)^{2}\times\mathbb{R}$ with
intensity $\,ds\,m_{0}(dz)\,dx$.

Note that

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))L^{\alpha}(\,ds,\,dx)$
$\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\tilde{N}(\,ds,\,dz,\,dx)$
(A.21)
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\tilde{N}(\,ds,\,dz,\,dx).$

Here we note that $\int_{0}^{1}z^{2}m_{0}(dz)<\infty$ and
$\int_{1}^{\infty}zm_{0}(dz)<\infty$. Using Kunita’s inequality [App09,
Theorem 4.4.23] for the first term, Fubini and Proposition 3.1 we have

$\displaystyle\mathbb{E}\left(\left|\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\tilde{N}(\,ds,\,dz,\,dx)\right|^{2}\right)$
$\displaystyle\leq$ $\displaystyle
C\mathbb{E}\left(\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\left|Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\right|^{2}\,ds\,m_{0}(dz)\,dx\right)$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{t}\int_{\mathbb{R}}\left|\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x)\right|^{2}\mathbb{E}(Y_{s-}(x)^{2\beta})\,ds\,dx$
$\displaystyle\leq$ $\displaystyle
C\left(\sup_{s}\|\psi_{\pi_{\Delta}(s)}-\psi_{s}\|_{L^{2}(\mathbb{R})}\right)\int_{0}^{t}e^{c_{2}s}\,ds$
(A.22) $\displaystyle=$ $\displaystyle
C\left(\sup_{s}\|\psi_{\pi_{\Delta}(s)}-\psi_{s}\|_{L^{2}(\mathbb{R})}\right)e^{c_{2}t}\to
0$

as $|\Delta|\to 0$. The last line follows from our assumption (i) of
continuity of the map $s\mapsto\psi_{s}\in L^{2}(\mathbb{R})$, which implies
uniform continuity of the same on $[0,t]$. For the second term in (A.3) we
proceed as in the previous calculation.

$\displaystyle\mathbb{E}\left(\left|\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\tilde{N}(\,ds,\,dz,\,dx)\right|\right)$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\left(\left|\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))zN(\,ds,\,dz,\,dx)\right|\right)$
$\displaystyle\hskip
56.9055pt+\mathbb{E}\left(\left|\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\,ds\,m_{0}(dz)\,dx\right|\right)$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\left|Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\right|N(\,ds,\,dz,\,dx)$
$\displaystyle\hskip
56.9055pt+\mathbb{E}\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\left|Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\right|\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle
2\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\mathbb{E}\left|Y_{s-}(x)^{\beta}(\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x))z\right|\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle
c_{\alpha}\int_{0}^{t}\int_{\mathbb{R}}\mathbb{E}(Y_{s-}(x)^{\beta})|\psi_{\pi_{\Delta}(s)}(x)-\psi_{s}(x)|\,ds\,dx$
$\displaystyle\leq$ $\displaystyle
c_{\alpha}\left(\sup_{s}\lVert\psi_{s}-\psi_{\pi_{\Delta}(s)}\rVert_{L^{1}(\mathbb{R})}\right)\int_{0}^{t}e^{c_{2}s}\,ds$
(A.23) $\displaystyle=$ $\displaystyle
c_{\alpha,\beta}\left(\sup_{s}\lVert\psi_{s}-\psi_{\pi_{\Delta}(s)}\rVert_{L^{1}(\mathbb{R})}\right)e^{c_{2}t},$

where we have used Proposition 3.1 for the third inequality. By assumption (i)
the RHS above converges to $0$ as $|\Delta|\to 0$. The calculations (A.3),
(A.3)and (A.3) together prove (c). ∎

In the last lemma we show how to turn the time dependent weak form of our SPDE
(A.18), which was proved in the previous result, into a martingale. This will
complete the proof of Proposition 3.4.

######  Lemma A.4.

If (A.18) holds for some smooth $\psi:[0,T]\times R\to\mathbb{R}$ and $0\leq
t\leq T$, then

(A.24) $\displaystyle M^{Y}_{t}(\psi)=e^{-\langle
Y_{t},\psi_{t}\rangle}-e^{-\langle
Y_{0},\psi_{0}\rangle}-\int_{0}^{t}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(-\langle
Y_{s-},\frac{1}{2}\Delta\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle+\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle\right)\,ds$

is an $\mathcal{F}^{Y}$-martingale.

###### Proof.

The proof is an application of Ito’s formula. Using the representation (A.20)
and some algebraic manipulations we have,

$\displaystyle\langle Y_{t},\psi\rangle=$ $\displaystyle\langle
Y_{0},\psi_{0}\rangle+\int_{0}^{t}\langle
Y_{s},\frac{1}{2}\Delta\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle\,ds+\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)z\tilde{N}(\,ds,\,dz,\,dx)$
$\displaystyle=$ $\displaystyle\langle
Y_{0},\psi_{0}\rangle+\int_{0}^{t}\langle
Y_{s},\frac{1}{2}\Delta\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle\,ds+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)z\tilde{N}(\,ds,\,dz,\,dx)$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)zN(\,ds,\,dz,\,dx)-\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)z\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle\langle
Y_{0},\psi_{0}\rangle+\int_{0}^{t}\left[\langle
Y_{s},\frac{1}{2}\Delta\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle-\int_{1}^{\infty}z\langle
Y_{s-}(\cdot)^{\beta},\psi_{s}\rangle\,m_{0}(dz)\right]\,ds$ (A.25)
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)z\tilde{N}(\,ds,\,dz,\,dx)+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}Y_{s-}(x)^{\beta}\psi_{s}(x)zN(\,ds,\,dz,\,dx)$

Since $\int_{1}^{\infty}z\,m_{0}(dz)=\frac{\alpha}{\Gamma(2-\alpha)}$ the
above can be written formally as

$\displaystyle d\langle Y_{t},\psi\rangle=$ $\displaystyle\left(\langle
Y_{t},\frac{\Delta}{2}\psi_{t}+\frac{\partial}{\partial
t}\psi_{t}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{t-}^{\beta},\psi_{t}\rangle\right)\,dt$ (A.26)
$\displaystyle+\int_{0}^{1}\int_{\mathbb{R}}Y_{t-}(x)^{\beta}\psi_{t}(x)z\tilde{N}(\,dt,\,dz,\,dx)+\int_{1}^{\infty}\int_{\mathbb{R}}Y_{t-}(x)^{\beta}\psi_{t}(x)zN(\,dt,\,dz,\,dx).$

Ito’s formula as given in [App09, Theorem 4.4.7] can now be applied with
$f(x)=e^{-x}$ and $G(t)=\left(\langle
Y_{t},\frac{\Delta}{2}\psi_{t}+\frac{\partial}{\partial
t}\psi_{t}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{t-}^{\beta},\psi_{t}\rangle\right)$. We have from (A.3),

$\displaystyle e^{-\langle Y_{t},\psi_{t}\rangle}-e^{-\langle
Y_{0},\psi_{0}\rangle}=f(\langle Y_{t},\psi_{t}\rangle)-f(\langle
Y_{0},\psi_{0}\rangle)$ $\displaystyle=$
$\displaystyle\int_{0}^{t}f^{\prime}(\langle
Y_{s-},\psi_{s}\rangle)G(s)\,ds+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}\left[f(\langle
Y_{s-},\psi_{s}\rangle+Y_{s-}(x)^{\beta}\psi_{s}(x)z)-f(\langle
Y_{s-},\psi_{s}\rangle)\right]N(\,dt,\,dz,\,dx)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\left[f(\langle
Y_{s-},\psi_{s}\rangle+Y_{s-}(x)^{\beta}\psi_{s}(x)z)-f(\langle
Y_{s-},\psi_{s}\rangle)\right]\tilde{N}(\,dt,\,dz,\,dx)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}\left[f(\langle
Y_{s-},\psi_{s}\rangle+Y_{s-}(x)^{\beta}\psi_{s}(x)z)-f(\langle
Y_{s-},\psi_{s}\rangle)-Y_{s-}(x)^{\beta}\psi_{s}(x)zf^{\prime}(\langle
Y_{s-},\psi_{s}\rangle)\right]\,ds\,m_{0}(dz)\,dx$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}e^{-\langle Y_{s-},\psi_{s}\rangle}\left(\langle
Y_{s-},\frac{\Delta}{2}\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{s-}^{\beta},\psi_{s}\rangle\right)\,ds$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)N(\,ds,\,dz,\,dx)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\tilde{N}(\,ds,\,dz,\,dx)$
$\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle-\int_{0}^{t}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(\langle
Y_{s-},\frac{\Delta}{2}\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{s-}^{\beta},\psi_{s}\rangle\right)\,ds$
$\displaystyle+\left[\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\tilde{N}(\,ds,\,dz,\,dx)\right.$
$\displaystyle\left.+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\tilde{N}(\,ds,\,dz,\,dx)\right]$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\,ds\,m_{0}(dz)\,dx$
(A.27) $\displaystyle+\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx$

adding and subtracting the term
$\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\,ds\,m_{0}(dz)\,dx$
in the last line. Note that the term in the square bracket above is

$\displaystyle
M_{t}(\psi)=\int_{0}^{t}\int_{0}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\tilde{N}(\,ds,\,dz,\,dx)$

and it is an $\mathcal{F}^{Y}$-martingale. We consider the last term in the
RHS of (A.3). Recall the definition of $m_{0}$ from (2.1) and the fact that
for $y\geq 0$,

$\displaystyle\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}\int_{0+}^{\infty}\left(e^{-\lambda
y}-1+\lambda y\right)\lambda^{-\alpha-1}\,d\lambda=y^{\alpha}.$

From these we can get,

$\displaystyle\int_{0}^{t}\int_{0}^{1}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle\int_{0}^{t}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left[\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}\int_{0}^{\infty}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)z^{-1-\alpha}\,dz\right]\,ds\,dx$
$\displaystyle-\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx$
$\displaystyle=$ $\displaystyle\int_{0}^{t}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}Y_{s-}(x)^{\alpha\beta}\psi_{s}(x)^{\alpha}\,ds\,dx$
(A.28)
$\displaystyle-\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx.$

To finish the proof use the result of (A.3) in (A.3). By algebraic
manipulations we have,

$\displaystyle e^{-\langle Y_{t},\psi_{t}\rangle}-e^{-\langle
Y_{0},\psi_{0}\rangle}$ $\displaystyle=$
$\displaystyle-\int_{0}^{t}e^{-\langle Y_{s-},\psi_{s}\rangle}\left(\langle
Y_{s-},\frac{\Delta}{2}\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{s-}^{\beta},\psi_{s}\rangle\right)\,ds+M_{t}(\psi)$
$\displaystyle+\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1\right)\,ds\,m_{0}(dz)\,dx+\int_{0}^{t}e^{-\langle
Y_{s-},\psi_{s}\rangle}\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle\,ds$
$\displaystyle-\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(e^{-Y_{s-}(x)^{\beta}\psi_{s}(x)z}-1+Y_{s-}(x)^{\beta}\psi_{s}(x)z\right)\,ds\,m_{0}(dz)\,dx,$
$\displaystyle=$ $\displaystyle-\int_{0}^{t}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(\langle
Y_{s-},\frac{\Delta}{2}\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle-\frac{\alpha}{\Gamma(2-\alpha)}\langle
Y_{s-}^{\beta},\psi_{s}\rangle+\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle\right)\,ds+M_{t}(\psi)$
$\displaystyle-\int_{0}^{t}\int_{1}^{\infty}\int_{\mathbb{R}}e^{-\langle
Y_{s-},\psi_{s}\rangle}Y_{s-}(x)^{\beta}\psi(x)z\,ds\,m_{0}(dz)\,dx,$ (A.29)
$\displaystyle=$ $\displaystyle-\int_{0}^{t}e^{-\langle
Y_{s-},\psi_{s}\rangle}\left(\langle
Y_{s-},\frac{\Delta}{2}\psi_{s}+\frac{\partial}{\partial
s}\psi_{s}\rangle+\langle
Y_{s-}^{\alpha\beta},\psi_{s}^{\alpha}\rangle\right)\,ds+M_{t}(\psi)$

again using the fact that
$\int_{1}^{\infty}z\,m_{0}(dz)=\frac{\alpha}{\Gamma(2-\alpha)}$. ∎

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