# A Markov process for an infinite age-structured population

Dominika Jasińska  Instytut Matematyki, Uniwersytet Marii Curie-Skłodowskiej,
20-031 Lublin, Poland<EMAIL_ADDRESS>and  Yuri Kozitsky  Instytut
Matematyki, Uniwersytet Marii Curie-Skłodowskiej, 20-031 Lublin, Poland
<EMAIL_ADDRESS>
###### Abstract.

For an infinite system of particles arriving in and departing from a habitat
$X$ – a locally compact Polish space with a positive Radon measure $\chi$ – a
Markov process is constructed in an explicit way. Along with its location
$x\in X$, each particle is characterized by age $\alpha\geq 0$ – time since
arriving. As the state space one takes the set of marked configurations
$\widehat{\Gamma}$, equipped with a metric that makes it a complete and
separable metric space. The stochastic evolution of the system is described by
a Kolmogorov operator $L$, expressed through the measure $\chi$ and a
departure rate $m(x,\alpha)\geq 0$, and acting on bounded continuous functions
$F:\widehat{\Gamma}\to\mathds{R}$. For this operator, we pose the martingale
problem and show that it has a unique solution, explicitly constructed in the
paper. We also prove that the corresponding process has a unique stationary
state and is temporarily egrodic if the rate of departure is separated away
from zero.

###### Key words and phrases:

Age-structured population; Fokker-Planck equation; martingale problem; Markov
processes; resolvent

###### 1991 Mathematics Subject Classification:

60J25; 60J85; 92D25

##  1\. Introduction

In recent years, the stochastic dynamics of age-structured populations attract
considerable attention, mostly due to their various applications, see [6, 7,
8, 10, 13]. Finite populations of this kind are much more mathematically
tractable, in contrast to infinite ones where only few results are known. In
the present work, we construct a Markov process for an individual-based model
of an infinite population of point entities that arrive in and depart from a
continuous habitat $X\subseteq\mathds{R}^{d}$.

Let $X$ be a locally compact Polish space and $\chi$ a positive Radon measure
defined thereon. $X$ will serve as the habitat for possibly infinite
populations of entities (particles) that arrive in and depart from $X$ at
random. Along with the location $x\in X$, each entity is characterized by age
$\alpha\in[0+\infty)=:\mathds{R}_{+}$ – time of its presence in the
population. The probability that an entity appears at time $t$ in a given
compact $\Lambda\subset X$ is set to be $1-\exp(-t\chi(\Lambda))$. Similarly,
the probability that the entity located at $x$ departs at time $t$ is
$1-\exp(-tm(x,\alpha))$, where $\alpha$ is its age at the moment of departure.
Since we do not assume that $\chi(X)$ be finite, the total population can get
instantly infinite, even if it is finite in the initial state. That is, our
model is characterized by an arriving measure $\chi$ and a departing rate
$m(x,\alpha)$.

To describe microscopic states of the population we use so called _marked
configurations_ $\hat{\gamma}$, cf. [3, Definition 9.1.II, page 3]. Set

$\Gamma=\\{\gamma\subset X:|\gamma\cap\Lambda|<\infty,\ {\rm for}\ {\rm each}\
{\rm compact}\ \Lambda\subset X\\},$

where $|\gamma\cap\Lambda|$ is the number of the elements of $\gamma$
contained in $\Lambda$. Let $\mathcal{A}$ be the set of finite configurations
$a=\\{\alpha_{1},\dots,\alpha_{n}\\}\subset[0,+\infty)$, $n\in\mathds{N}_{0}$.
This set – equipped with a metric that makes it a complete and locally compact
metric space – serves as the space of marks for our model. That is, for $x\in
X$, by $a(x)\in\mathcal{A}$ we mean the ages of particles located at $x$. Then
a marked configuration $\hat{\gamma}$ is the pair $(\gamma,a)$ with
$\gamma\in\Gamma$ and $a:X\to\mathcal{A}$. In this case, we also write
$\gamma=p(\hat{\gamma})$ and call $\gamma$ ground configuration for
$\hat{\gamma}$. The set of all marked configurations $\widehat{\Gamma}$ is
equipped with a metric, in which it is complete and separable. Elements of
$\hat{\gamma}$ are pairs $(x,\alpha)=:\hat{x}$. They can be enumerated in
$\hat{\gamma}$, which allows one to consider sums
$\sum_{\hat{x}\in\hat{\gamma}}g(\hat{x})$ for continuous compactly supported
functions $g(\hat{x})=g(x,\alpha)$.

The evolution of the population is governed by the (backward) Kolmogorov
equation

$\frac{d}{dt}F_{t}=LF_{t},\qquad F_{t}|_{t=0}=F_{0},$ (1.1)

where $t$ is time, $F_{t}$ is a suitable (test) function of $\hat{\gamma}$ and
the Kolmogorov operator $L$ is given by the following expression

$\displaystyle(LF)(\hat{\gamma})$ $\displaystyle=$
$\displaystyle\sum_{(x,\alpha)\in\hat{\gamma}}\frac{\partial}{\partial\alpha}F(\hat{\gamma})+\sum_{(x,\alpha)\in\hat{\gamma}}m(x,\alpha)\left[F(\hat{\gamma}\setminus(x,\alpha))-F(\hat{\gamma})\right]$
$\displaystyle+$
$\displaystyle\int_{X}\left[F(\hat{\gamma}\cup(x,0))-F(\hat{\gamma})\right]\chi(dx).$

Here the first term describes aging, the second one corresponds to departing,
whereas the last term describes the appearance of new entities. In (1) and and
in the sequel, in expressions like $\hat{\gamma}\cup\hat{x}$ we consider
$\hat{x}=(x,\alpha)$ as a singleton configuration. The model parameters are
subject to the following

######  Assumption 1.1.

The departure rate $X\times\mathds{R}_{+}\ni(x,\alpha)\mapsto
m(x,\alpha)\in\mathds{R}_{+}$ is continuous and bounded, i.e., such that
$m(x,\alpha)\leq m_{*}$ for some $m_{*}>0$ and all $(x,\alpha)$. Moreover,
there exits $\varkappa:[0,1]\to\mathds{R}_{+}$ such that
$\varkappa(\epsilon)\to 0$ as $\epsilon\to 0$ and the following holds

$\forall x\in
X\qquad|m(x,\alpha)-m(x,\alpha^{\prime})|\leq\varkappa(|\alpha-\alpha^{\prime}|),\quad|\alpha-\alpha^{\prime}|\in(0,1).$
(1.3)

The arriving measure $\chi$ is just a positive Radon measure.

The result of the present work can be outlined as follows. We introduce a
Banach space $\mathcal{C}$ of bounded continuous functions
$F:\widehat{\Gamma}\to\mathds{R}$, in which we define $L$ as a closed and
densely defined linear operator that satisfies the conditions of the Hille-
Yosida theorem, and hence is the generator of a $C_{0}$-semigroup
$\\{S(t)\\}_{t\geq 0}$. Then the solution of (1.1) is obtained in the form
$F_{t}=S(t)F_{0}$. For a class of functions $\mathcal{F}_{\Theta}$, $F_{t}$
corresponding to $F_{0}\in\mathcal{F}_{\Theta}$ is obtained in an explicit
form. This allows us to explicitly construct the corresponding Markov
transition function $p^{\hat{\gamma}}_{t}$ and obtain finite-dimensional laws
of a Markov process $\mathcal{X}$ with values in $\widehat{\Gamma}$, which
describes the stochastic evolution of our model. Possible objects of this kind
are specified as stochastic processes that solve the martingale problem for
$L$. Then we show that this problem is well-posed, i.e., uniqueness holds. The
main ingredient of the proof here is showing that the corresponding Fokker-
Planck equation for $L$ has a unique solution, which we do by employing the
resolvent of $L$. Assuming that $m(x,\alpha)\geq m_{0}>0$, we also show that
the process $\mathcal{X}$ has a unique stationary state, explicitly
constructed in the paper, such that the laws of $\mathcal{X}(t)$ weakly
converge to this state as $t\to+\infty$.

##  2\. Preliminaries

In this work, we use the following standard notions and notations. For a
Polish (=separable and completely metrizable) space $E$, by $\mathcal{B}(E)$
we denote the corresponding Borel $\sigma$-field; $C_{\rm b}(E)$ (resp.
$B_{\rm b}(E)$) stands for the set of all bounded and continuous (resp.
bounded and measurable) functions $f:E\to\mathds{R}$. By $C_{\rm b}^{+}(E)$ we
denote the set of positive elements of $C_{\rm b}(E)$. A subset $C_{\rm
cs}(E)\subset C_{\rm b}(E)$ consists of continuous compactly supported
functions. A family of functions $\mathcal{F}$ is said to _separate_ the
points of $E$ if for each distinct $x,y\in E$, one finds $f\in\mathcal{F}$
such that $f(x)\neq f(y)$. By $\sigma\mathcal{F}$ we denote the smallest sub-
field of $\mathcal{B}(E)$ such that each $f\in\mathcal{F}$ is
$\sigma\mathcal{F}$-measurable; by $\mathcal{P}(E)$ we mean the set of all
probability Borel measures on $(E,\mathcal{B}(E))$. For a given measure $\mu$
and a suitable function $f$, we write $\mu(f)=\int fd\mu$. For a sequence
$\\{\mu_{n}\\}_{n\in\mathds{N}}\subset\mathcal{P}(E)$, by writing
$\mu_{n}\Rightarrow\mu\in\mathcal{P}(E)$ we mean its weak convergence, i.e.,
$\mu_{n}(f)\to\mu(f)$ for all $f\in C_{\rm b}(E)$. A family of functions
$\mathcal{F}$ is said to be _separating_ if $\mu_{1}(f)=\mu_{2}(f)$ holding
for all $f\in\mathcal{F}$ implies $\mu_{1}=\mu_{2}$ for any
$\mu_{1},\mu_{2}\in\mathcal{P}(E)$. If $\mathcal{F}$ separates the points of
$E$ and its linear span is an algebra with respect to pointwise operations,
then it is separating, see [5, Theorem 4.5, page 113]. A family of functions
$\mathcal{F}$ is said to be _convergence determining_ if $\mu_{n}(f)\to\mu(f)$
holding for all $f\in\mathcal{F}$ implies $\mu_{n}\Rightarrow\mu$. For a
suitable $A\subset E$, by $\mathds{1}_{A}$ we denote the indicator of $A$,
i.e., the functions such that $\mathds{1}_{A}(x)=1$ if $x\in A$ and
$\mathds{1}_{A}(x)=0$ otherwise.

###  2.1. The space of marks

As mentioned above, our populations dwell in a locally compact Polish space
$X$. By $\Lambda$ we always denote a compact subset of $X$. Each population
member is characterized by its compound trait
$\hat{x}=(x,\alpha)\in\widehat{X}=X\times\mathds{R}_{+}$. For a function,
$g:\widehat{X}\to\mathds{R}$, we use interchangeable writings $g(\hat{x})$ and
$g(x,\alpha)$. The space $\widehat{X}$ is equipped with the product topology
assuming that the topology of $\mathds{R}_{+}$ be defined by the metric which
we introduce now. For $\alpha\geq 0$, set
$\omega(\alpha)=\min\\{\alpha;1/\alpha\\}$, and then

${r}(\alpha,0)=\omega(\alpha),\quad{r}(\alpha,\alpha^{\prime})=\min\\{|\alpha-\alpha^{\prime}|;\omega(\alpha)+\omega(\alpha^{\prime})\\},$
(2.1)

where $|\beta|$ is the usual absolute value of $\beta\in\mathds{R}$.

######  Proposition 2.1.

The above introduced ${r}$ is a metric and $(\mathds{R}_{+},{r})$ is a compact
metric space.

###### Proof.

To prove the first part we just have to check the validity of the triangle
inequality

${r}(\alpha_{1},\alpha_{2})\leq{r}(\alpha_{1},\alpha_{3})+{r}(\alpha_{2},\alpha_{3}).$
(2.2)

This technical exercise is made in Appendix. To prove the compactness, we have
to show that: (a) ${r}$ is complete; (b) the space $(\mathds{R}_{+},{r})$ is
totally bounded. Assume that
$\\{\alpha_{n}\\}_{n\in\mathds{N}}\subset\mathds{R}_{+}$ is an ${r}$-Cauchy
sequence. Here one may have the following possibilities: (i) there exists
$\bar{\alpha}<\infty$ such that $\alpha_{n}\leq\bar{\alpha}$,
$n\in\mathds{N}$; (ii) the considered sequence contains a subsequence that
diverges in the usual sense. In case (i), $\\{\alpha_{n}\\}_{n\in\mathds{N}}$
contains a subsequence, say $\\{\alpha_{n_{k}}\\}_{k\in\mathds{N}}$, such that
$|\alpha_{n_{k}}-\alpha_{*}|\to 0$ as $k\to+\infty$ for some
$\alpha_{*}\leq\bar{\alpha}$. At the same time, for
$\varepsilon<2/\bar{\alpha}$, ${r}(\alpha_{n},\alpha_{m})<\varepsilon$ implies
$|\alpha_{n}-\alpha_{m}|<\varepsilon$, see (2.1), which means that
${r}(\alpha_{n},\alpha_{*})\to 0$ as $n\to+\infty$. In case (ii), the
divergent subsequence converges in ${r}$ to zero, which implies that the whole
sequence converges to zero in ${r}$. Hence, the latter metric is complete. To
prove (b), we set
$\tilde{B}_{\varepsilon}(\alpha)=\\{\alpha^{\prime}\in\mathds{R}_{+}:{r}(\alpha,\alpha^{\prime})<\varepsilon\\}$.
Fix $\varepsilon\in(0,1)$ and take the least $k\in\mathds{N}$ such that
$k+1>1/\varepsilon^{2}$. Then
$\mathds{R}_{+}=\cup_{j=0}^{k}\tilde{B}_{\varepsilon}(j\varepsilon)$, which
yields the property in question. ∎

Let us now compare $r$ with the absolute-value metrics of $\mathds{R}_{+}$. By
$C(\mathds{R}_{+},\mathcal{T}_{r})$ we will mean the sets of all bounded
$r$-continuous functions, whereas $C_{\rm
b}(\mathds{R}_{+},\mathcal{T}_{|\cdot|})$ is going to stand for the set of all
bounded $|\cdot|$-continuous functions.

######  Proposition 2.2.

$\mathcal{T}_{{r}}$ is coarser than $\mathcal{T}_{|\cdot|}$, and hence the
embedding
$(\mathds{R}_{+},\mathcal{T}_{|\cdot|})\hookrightarrow(\mathds{R}_{+},\mathcal{T}_{{r}})$
is continuous, whereas both latter topological spaces are Borel isomorphic.
Moreover,

$C(\mathds{R}_{+},\mathcal{T}_{{r}})=\\{u\in C_{\rm
b}(\mathds{R}_{+},\mathcal{T}_{|\cdot|}):\lim_{\alpha\to+\infty}u(\alpha)=u(0)\\}.$
(2.3)

###### Proof.

The validity of the first statement and (2.3) readily follows by the fact that
each $|\cdot|$-convergent sequence is also ${r}$-convergent, and each
${r}$-convergent sequence either converges in $|\cdot|$ to the same limit
$\alpha\neq 0$, or has two $|\cdot|$-accumulation points: $0$ and $\infty$.
Since the mentioned embedding is continuous and injective, it is also Borel-
measurable. By Kuratowski’s theorem [11], its inverse is also measurable and
thus is the isomorphism in question. This, in particular, means that the
corresponding Borel $\sigma$-fields coincide. ∎

######  Definition 2.3.

For a suitable $u\in C(\mathds{R}_{+},\mathcal{T}_{r})$, we introduce the map
$\mathds{R}_{+}\ni\alpha\mapsto u^{\prime}(\alpha)$ meaning the usual
derivative if $\alpha>0$, and the right-hand side one if $\alpha=0$. A given
$u$ is said to be continuously differentiable (on $\mathds{R}_{+}$) if
$u^{\prime}\in C(\mathds{R}_{+},\mathcal{T}_{r})$.

Let us consider the following functions

$u_{n}(\alpha)=\frac{\alpha^{2}}{1+n\alpha^{3}},\qquad\alpha\in\mathds{R}_{+},\quad
n\in\mathds{N}.$ (2.4)

It is clear that: (a) each $u_{n}$ is continuously differentiable, see
Definition 2.3; (b) $u_{n}$ is decreasing for $\alpha>(2/n)^{1/3}$. Moreover,
$u_{n}(\alpha)\leq 2^{2/3}/3n^{2/3}$ and

$u_{n}^{\prime}(\alpha)=\frac{2\alpha-n\alpha^{4}}{(1+n\alpha^{3})^{2}},\qquad\left|u_{n}^{\prime}(\alpha)\right|\leq
c/n^{1/3},$ (2.5)

the latter holding for some $c>0$ and all $\alpha\geq 0$.

Now let $\\{\sigma_{k}\\}_{k\in\mathds{N}}=:\varSigma\subset[0,+\infty)$ be
countable and such that: (i) $\sigma_{1}=0$; (ii) $\sigma_{k}<\sigma_{k+1}$
for all $k\in\mathds{N}$; $\sigma_{k}\to\bar{\sigma}<\infty$ as $n\to+\infty$.
Next, for $k,n\in\mathds{N}$, we set

$w_{k,n}(\alpha)=e^{-\sigma_{k}u_{n}(\alpha)}.$ (2.6)

Then $w_{k,n}$ is continuously differentiable and the following holds

$\left|w^{\prime}_{k,n}(\alpha)\right|\leq\frac{\bar{\sigma}c}{n^{1/3}}w_{k,n}(\alpha),$
(2.7)

where $c$ is the same as in (2.5).

Now let $a$ be a finite collection of points $\alpha_{l}\in\mathds{R}_{+}$.
That is, $a=\\{\alpha_{l}\\}_{1\leq l\leq m}$, $\alpha_{l}\leq\alpha_{l+1}$
for all $l$. For $\alpha\in a$, by $n_{a}(\alpha)\in\mathds{N}$ we will denote
the multiplicity of $\alpha$ in $a$, i.e., the number of elements of $a$
coinciding with this $\alpha$. We extend it to all $\alpha\geq 0$ by setting
$n_{a}(\alpha)=0$ whenever $\alpha$ is not in $a$. Two such $a$ and
$a^{\prime}$ are equal if they consist of exactly the same elements, with the
same multiplicities.

######  Proposition 2.4.

Let $a$ and $a^{\prime}$ be as just described. Then they are equal if

$\sum_{\alpha\in a}w_{k,n}(\alpha)=\sum_{\alpha\in
a^{\prime}}w_{k,n}(\alpha),$ (2.8)

holding for all $k,n\in\mathds{N}$.

###### Proof.

For $a$ as above and $\zeta\in\mathds{C}$, consider

$f_{n,a}(\zeta)=\sum_{\alpha\in a}\exp\left(-\zeta u_{n}(\alpha)\right),\qquad
n\in\mathds{N}.$

Each such $f$ is an exponential type entire function. By (2.6) and (2.8) we
have that $(f_{n,a}-f_{n,a^{\prime}})|_{\varSigma}=0$, holding for all
$n\in\mathds{N}$. Since $\varSigma$ has a limiting point, this implies
$f_{n,a}(\zeta)=f_{n,a^{\prime}}(\zeta)$ for all $\zeta\in\mathds{R}$ and
$n\in\mathds{N}$. Obviously, $\lim_{\zeta\to+\infty}f_{n,a}(\zeta)=n_{a}(0)$,
where $n_{a}(0)\geq 0$ is the multiplicity of $\alpha=0$ in $a$. Then the just
mentioned equality yields $n_{a}(0)=n_{a^{\prime}}(0)$, and also

$\sum_{\alpha\in a\setminus\\{0\\}}\exp\left(-\zeta
u_{n}(\alpha)\right)=\sum_{\alpha\in
a^{\prime}\setminus\\{0\\}}\exp\left(-\zeta u_{n}(\alpha)\right).$ (2.9)

Let $\alpha_{*}$ and $\alpha^{\prime}_{*}$ be the least positive elements of
$a$ and $a^{\prime}$, respectively. Take $n>2/\alpha^{3}_{\flat}$,
$\alpha_{\flat}:=\min\\{\alpha_{*};\alpha^{\prime}_{*}\\}$. Then, for such $n$
and all $\alpha>\alpha_{\flat}$, one has
$u_{n}(\alpha_{\flat})>u_{n}(\alpha)$. Now we multiply both sides of (2.9) by
$e^{\zeta u_{n}(\alpha_{\flat})}$ and pass to the limit $\zeta\to-\infty$.
This yields that $\alpha_{*}=\alpha^{\prime}_{*}$ and
$n_{a}(\alpha_{*})=n_{a^{\prime}}(\alpha^{\prime}_{*})$. Thereafter, we
subtract the coinciding terms from both sides of (2.9) and proceed to
comparing the remaining least elements of $a$ and $a^{\prime}$. This
eventually yields the equality to be proved. ∎

Let $\mathcal{A}$ be the set of all $a=\\{\alpha_{l}\\}_{1\leq l\leq m}$,
$m\in\mathds{N}_{0}$, $0\leq\alpha_{1}\leq\alpha_{2}\leq\cdots\leq\alpha_{m}$.
Define

$\displaystyle\rho(a,a^{\prime})=\sum_{k,n\in\mathds{N}}\frac{2^{-k-n}\rho_{k,n}(a,a^{\prime})}{1+\rho_{k,n}(a,a^{\prime})},$
(2.10) $\displaystyle\rho_{k,n}(a,a^{\prime}):=\left|\sum_{\alpha\in
a}w_{k,n}(\alpha)-\sum_{\alpha\in a^{\prime}}w_{k,n}(\alpha)\right|.$

By Proposition 2.4 it follows that $\rho$ is a metric on $\mathcal{A}$. Each
$a\in\mathcal{A}$ can be considered as a finite counting measure defined on
the compact space $(\mathds{R}_{+},r)$, for which $a(\Delta)=\sum_{\alpha\in
a}\mathds{1}_{\Delta}(a)=|a\cap\Delta|$, holding for all Borel subsets
$\Delta$. The weak topology of $\mathcal{A}$ is defined as the coarsest
topology that makes continuous all the maps $a\mapsto\sum_{\alpha\in a}w(a)$,
$w\in C(\mathds{R}_{+},\mathcal{T}_{r})$. In the weak topology, $\mathcal{A}$
is a closed subset of the space of all finite positive measures on
$(\mathds{R}_{+},\mathcal{T}_{r})$.

######  Proposition 2.5.

$(\mathcal{A},\rho)$ is a complete metric space. The corresponding metric
topology coincides with the weak topology that turns $\mathcal{A}$ into a
locally compact Polish space.

###### Proof.

As each $w_{k,n}$ is in $C(\mathds{R}_{+},\mathcal{T}_{r})$, the weak
convergence of a sequence $\\{a_{m}\\}_{m\in\mathds{N}}\subset\mathcal{A}$ to
a certain $a\in\mathcal{A}$ yields $\rho(a,a_{m})\to 0$, $m\to+\infty$. Assume
now that $\\{a_{m}\\}_{m\in\mathds{N}}$ is a $\rho$-Cauchy sequence. By taking
$\sigma=0$ we then get from the latter that, for some $m_{*}\in\mathds{N}$,
the cardinalities of all $a_{m}$, $m>m_{*}$, coincide. By Prohorov’s theorem
this yields that $\\{a_{m}\\}_{m\in\mathds{N}}$ contains a subsequence that
weakly converges to some $a$. Hence, the whole sequence converges in $\rho$ to
this $a$. Then the metric is complete and the corresponding metric topology is
exactly the weak topology of $\mathcal{A}$. The separability and local
compactness follow by the fact that $(\mathds{R}_{+},\mathcal{T}_{r})$ is
compact. ∎

###  2.2. Marked configuration spaces

Let ${\Gamma}$ be the set of all locally finite simple configurations on $X$.
That is, each ${\gamma}\in{\Gamma}$ is a subset of $X$ such that each compact
$\Lambda\subset X$ contains a finite number of the elements of ${\gamma}$. Let
now $\breve{\gamma}$ be the pair $({\gamma},n)$, ${\gamma}\in{\Gamma}$ and
$n:{\gamma}\to\mathds{N}$. The value of $n$ at a given $x\in{\gamma}$ can be
considered as the multiplicity of $x$ in $\breve{\gamma}$. That is,
$\breve{\gamma}$ is a configuration with multiple locations, for which
${\gamma}$ is the ground configuration. Sometimes, we will write
$n_{\breve{\gamma}}(x)$ to explicitly indicate that we mean the multiplicity
of $x$ in the mentioned $\breve{\gamma}$. By $\breve{\Gamma}$ we denote the
set of all such multiple configurations. For $\breve{\gamma}=(\gamma,n)$, we
write ${\gamma}=p(\breve{\gamma})$. The weak-hash (vague) topology of
$\breve{\Gamma}$ is defined as the coarsest topology that makes continuous all
the maps $\breve{\gamma}\mapsto\sum_{x\in p(\breve{\gamma})}n(x)g(x)$, $g\in
C_{\rm cs}(X)$. It is well-known, see e.g., [14, Lemma 1.2], that with this
topology $\breve{\Gamma}$ is a Polish space, whereas $\Gamma$ is a
$G_{\delta}$ subset of $\breve{\Gamma}$, by which it is also Polish. Following
[9] we will also consider $\breve{\gamma}$ as configurations of point
particles, in which distinct particles may have the same location. Such
particles can be enumerated, which allows one to write

$\sum_{x\in p(\breve{\gamma})}n(x)g(x)=\sum_{x\in\breve{\gamma}}g(x),$ (2.11)

where in the second sum we mean a certain enumeration of this sort. In the
same sense, we will write

$\sum_{{x}_{1}\in\breve{\gamma}}\sum_{{x}_{2}\in\breve{\gamma}\setminus
x_{1}}\cdots\sum_{x_{m}\in\breve{\gamma}\setminus\\{x_{1},\dots,x_{m-1}\\}}g(x_{1},\dots,x_{m}),\qquad
m\in\mathds{N},$

where in expressions like $\breve{\gamma}\setminus x$ we treat $x$ the
singleton $\\{x\\}$.

It is known, see, e.g., [14, page 397], that there exists a collection
$\\{v_{s}\\}_{s\in\mathds{N}}=:\mathcal{V}\subset C^{+}_{\rm cs}(X)$ of
suitable functions such that the metric

$d(\breve{\gamma},\breve{\gamma}^{\prime})=\sum_{s\in\mathds{N}}\frac{2^{-s}d_{s}(\breve{\gamma},\breve{\gamma}^{\prime})}{1+d_{s}(\breve{\gamma},\breve{\gamma}^{\prime})},\quad
d_{s}(\breve{\gamma},\breve{\gamma}^{\prime}):=\left|\sum_{x\in\breve{\gamma}}v_{s}(x)-\sum_{x\in\breve{\gamma}^{\prime}}v_{s}(x)\right|,$
(2.12)

is complete and consistent with the weak-hash topology of $\breve{\Gamma}$. In
the sequel, we will always mean this topology of $\breve{\Gamma}$. Obviously,
we can and will assume that $\mathcal{V}$ contains also the following
functions. Let $\delta$ be a complete metric of $X$ and $X^{\prime}$ a
countable dense subset of $X$. Each $x^{\prime}\in X^{\prime}$ has a countable
base of compact neighborhoods, which we denote by $D(x^{\prime})$. Each
$\Delta\in D(x^{\prime})$ contains balls $B_{q}(x^{\prime})=\\{x\in
X:\delta(x,x^{\prime})<q\\}$ with compact closures, where $q$ is a rational
number satisfying $q\leq q^{\prime}$ for a $\Delta$-specific
$q^{\prime}\in\mathds{Q}$. For $x^{\prime}\in X^{\prime}$, $\Delta\in
D(x^{\prime})$, $q\leq q^{\prime}$ and $\varsigma\in(0,1)\cap\mathds{Q}$, let
$v\in C^{+}_{\rm cs}(X)$ be such that: (a) $v(x)\equiv\varsigma$ for $x\in
B_{q}(x^{\prime})$; (b) $v(x)=0$ for $x\in X\setminus\Delta$. The countable
set of all such functions is supposed to be a part of $\mathcal{V}$, and hence
they are taken into account in (2.12). Since each $v_{s}$ has compact support,
for each compact $\Lambda\subset X$ and any two configurations,
$d_{s}(\breve{\gamma}\cap\Lambda,\breve{\gamma}^{\prime}\cap\Lambda)>0$ only
for finitely many $s$. Here
$\breve{\gamma}\cap\Lambda:=(p(\breve{\gamma})\cap\Lambda,n)$.

For $\gamma\in\Gamma$, let $a:\gamma\to\mathcal{A}$ be a map, for which we
denote

$|a(x)|=\sum_{\alpha\in a(x)}n_{a(x)}(\alpha).$ (2.13)

Then the pair $\hat{\gamma}=(\gamma,a)$ is a marked configuration whose ground
configuration is $\gamma$ and the mark map is $a$. By writing
$\hat{x}=(x,\alpha)\in\hat{\gamma}$ we will mean that $x\in\gamma$ and
$\alpha\in a(x)$. The configuration of marks
$a(x)=\\{\alpha_{1},\dots,\alpha_{|a(x)|}\\}$ yields the ages of the particles
located at $x\in\gamma$, whereas $|a(x)|$ is the total number of such
particles. In some cases, we write $a_{\hat{\gamma}}$ to indicate that $a$ is
defined on a given $\hat{\gamma}$. Let $\widehat{\Gamma}$ denote the set of
all marked configurations $\hat{\gamma}$. Let also
$\breve{p}:\widehat{\Gamma}\to\breve{\Gamma}$ be the map such that
$\breve{p}(\gamma,a)=(\gamma,|a|)$, where $|a|(x)=|a(x)|$ see (2.13). Then
$p\circ\breve{p}$ maps $\hat{\gamma}=(\gamma,a)$ into its ground configuration
$\gamma$. For brevity, by writing $p(\hat{\gamma})$ we will mean
$(p\circ\breve{p})(\hat{\gamma})$. Our aim now is to equip $\widehat{\Gamma}$
with a complete metric. Define

$\displaystyle\kappa(\hat{\gamma},\hat{\gamma}^{\prime})=\sum_{s,k,n\in\mathds{N}}\frac{2^{-(s+k+n)}\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})}{1+\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})},$
(2.14)
$\displaystyle\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime}):=\left|\sum_{x\in
p(\hat{\gamma})}v_{s}(x)\sum_{\alpha\in
a_{\hat{\gamma}}(x)}w_{k,n}(\alpha)-\sum_{x\in
p(\hat{\gamma}^{\prime})}v_{s}(x)\sum_{\alpha\in
a_{\hat{\gamma}^{\prime}}(x)}w_{k,n}(\alpha)\right|.$

Note that the latter can also be written as, cf. (2.11),

$\displaystyle\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})=\left|\sum_{\hat{x}\in\hat{\gamma}}g_{s,k,n}(\hat{x})-\sum_{\hat{x}\in\hat{\gamma}^{\prime}}g_{s,k,n}(\hat{x})\right|,$
(2.15) $\displaystyle g_{s,k,n}(x,\alpha):=v_{s}(x)w_{k,n}(\alpha).$

For a compact $\Lambda\subset X$, we write
$\hat{\gamma}\cap\Lambda=(p(\hat{\gamma})\cap\Lambda,a)$, where $a$ is the
restriction of $a$ from $\hat{\gamma}$ to $p(\hat{\gamma})\cap\Lambda$.

######  Proposition 2.6.

For each $\varepsilon>0$, one may find a compact $\Lambda_{\varepsilon}\subset
X$ such that, for any two configurations, the following holds

$\left|\kappa(\hat{\gamma},\hat{\gamma}^{\prime})-\kappa(\hat{\gamma}\cap\Lambda_{\varepsilon},\hat{\gamma}^{\prime}\cap\Lambda_{\varepsilon})\right|<\varepsilon.$
(2.16)

###### Proof.

Fix $\varepsilon>0$ and then pick $s_{*}\in\mathds{N}$ such that
$2^{s_{*}}>1/\varepsilon$. Now let $\Lambda_{\varepsilon}$ be covered by the
supports of $v_{s}$ with $s\leq s_{*}$. For such $s$ and all
$k,n\in\mathds{N}$, we have
$\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})=\kappa_{s,k,n}(\hat{\gamma}\cap\Lambda_{\varepsilon},\hat{\gamma}^{\prime}\cap\Lambda_{\varepsilon})$,
see (2.14). This clearly yields (2.16). ∎

Since $\sigma=0$ is in $\varSigma$, by (2.6) and (2.12) we have that

$d(\breve{p}(\hat{\gamma}),\breve{p}(\hat{\gamma}^{\prime}))\leq\kappa(\hat{\gamma},\hat{\gamma}^{\prime}).$
(2.17)

######  Proposition 2.7.

The metric space $(\widehat{\Gamma},\kappa)$ is complete.

###### Proof.

We begin by pointing out the following evident fact

$\tilde{\kappa}_{s}(\hat{\gamma},\hat{\gamma}^{\prime}):=\sum_{k,n\in\mathds{N}}\frac{2^{-(s+k+n)}\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})}{1+\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})}\leq{\kappa}(\hat{\gamma},\hat{\gamma}^{\prime}),$
(2.18)

holding for all $s\in\mathds{N}$ and $\hat{\gamma},\hat{\gamma}^{\prime}$. Let
now
$\\{\hat{\gamma}_{m}=(\gamma_{m},a_{m})\\}_{m\in\mathds{N}}\subset\widehat{\Gamma}$
be a $\kappa$-Cauchy sequence. By (2.17) the sequence
$\\{\breve{p}(\hat{\gamma}_{m})\\}_{m\in\mathds{N}}\subset\breve{\Gamma}$
converges to some $\breve{\gamma}$. Take now $x\in p(\breve{\gamma})$ and then
pick a compact $\Delta\subset X$ such that $\Delta\cap
p(\breve{\gamma})=\\{x\\}$. For this $\Delta$, we then set
$n_{m}(x)=\sum_{y\in
p(\breve{\gamma}_{m})\cap\Delta}n_{\breve{\gamma}_{m}}(y)$, $m\in\mathds{N}$.
From the convergence of $\\{\breve{p}(\hat{\gamma}_{m})\\}_{m\in\mathds{N}}$
to $\breve{\gamma}$, it follows that $n_{m}(x)\to n(x)$; hence, there exists
$m_{*}\in\mathds{N}$ such that $n_{m}(x)=n(x)$ for all $m>m_{*}$. Now we pick
$x^{\prime}\in X^{\prime}$ and $q\in Q$ such that $x\in B_{q/2}(x^{\prime})$
and the closure of $B_{q}(x^{\prime})$ lies in $\Delta$. Let now
$v_{s}\in\mathcal{V}$ be such that $v_{s}(y)=\varsigma\in(0,1)\cap\mathds{Q}$,
$y\in B_{q/2}(x^{\prime})$, and $v_{s}(y)=0$ for $y\in X\setminus
B_{q}(x^{\prime})$. For these $m_{*}$ and $s$, $\\{\hat{\gamma}_{m}\\}_{m\geq
m_{*}+1}$ is also a $\tilde{\kappa}_{s}$-Cauchy sequence, see (2.18), for
which we have

$\kappa_{s,k,n}(\hat{\gamma}_{m},\hat{\gamma}_{m+l})=\varsigma\left|\sum_{y\in
p(\hat{\gamma}_{m})\cap\Delta}\sum_{\alpha\in\hat{a}_{m}(y)}w_{\sigma_{k},n}(\alpha)-\sum_{y\in
p(\hat{\gamma}_{m+l})\cap\Delta}\sum_{\alpha\in\hat{a}_{m+l}(y)}w_{\sigma_{k},n}(\alpha)\right|,$

holding for all $k,n\in\mathds{N}$, $m>m_{*}$ and $l\in\mathds{N}$. Here
$\hat{a}_{m}:=a_{\hat{\gamma}_{m}}$, $\hat{a}_{m+l}:=a_{\hat{\gamma}_{m+l}}$.
Let us enumerate $\hat{x}=(x,\alpha)\in\hat{\gamma}_{m}\cap\Delta$ in such a
way that $\alpha_{p,m}\leq\alpha_{p+1,m}$ for all $p$. This yields
$\hat{\gamma}_{m}\cap\Delta=\\{(x_{1,m},\alpha_{1,m}),\dots(x_{n,m},\alpha_{n,m})\\}$
with $n=n_{m}(x)=n(x)$. Similarly, we have
$\hat{\gamma}_{m+l}\cap\Delta=\\{(x_{1,m+l},\alpha_{1,m+l}),\dots(x_{n,m+l},\alpha_{n,m+l})\\}$
with the same $n$. Then
$\\{\alpha_{1,m},\dots\alpha_{n,m}\\}=:a_{m}\in\mathcal{A}$, and also
$\\{\alpha_{1,m+l},\dots\alpha_{n,m+l}\\}=:a_{m+l}\in\mathcal{A}$, and the
latter equality can be rewritten as follows

$\kappa_{s,k,n}(\hat{\gamma}_{m},\hat{\gamma}_{m+l})=\varsigma\left|\sum_{p=1}^{n}w_{\sigma_{k},n}(\alpha_{p,m})-\sum_{p=1}^{n}w_{\sigma_{k},n}(\alpha_{p,m+l})\right|=\varsigma\rho_{k,n}(a_{m},a_{m+l}),$

see (2.10). By (2.18) and (2.10) we then get

$\rho(a_{m},a_{m+l})\leq(2^{s}/\varsigma)\tilde{\kappa}_{s}(\hat{\gamma}_{m},\hat{\gamma}_{m+l}).$

By Proposition 2.5 this yields the convergence of $\\{a_{m}\\}_{m>m_{*}}$ to
some $a(x)\in\mathcal{A}$, which holds for each $x\in\breve{\gamma}$. This
defines the map $a:p(\breve{\gamma})\to\mathcal{A}$, and hence the
configuration $\hat{\gamma}=(p(\breve{\gamma}),a)$. Our aim now is to prove
that $\kappa(\hat{\gamma}_{m},\hat{\gamma})\to 0$ as $m\to+\infty$.

Fix $\varepsilon>0$ and then pick a compact $\Lambda_{\varepsilon}\subset X$
such that (2.16) holds with $\varepsilon/3$ in the right-hand side. Let
$\Lambda_{\varepsilon}^{o}$ be its interior. Then pick compact mutually
disjoint $\Delta_{x}\subset\Lambda^{o}$, $x\in
p(\breve{\gamma})\cap\Lambda^{o}$ such that
$p(\breve{\gamma})\cap\Delta_{x}=\\{x\\}$. As $\Lambda_{\varepsilon}$ is
compact, $p(\breve{\gamma})\cap\Lambda_{\varepsilon}^{o}$ is finite. Let
$\\{x_{j}\\}_{j\leq J}$ be an enumeration of it. For brevity, we will write
$\Delta_{j}$ in place of $\Delta_{x_{j}}$, $j=1,\dots,J$. Similarly as above,
by the convergence of $\\{{\breve{p}(\hat{\gamma}_{m})}\\}_{m\in\mathds{N}}$
to $\breve{\gamma}$, one finds $m_{*}$ such that
$p(\hat{\gamma}_{m})\cap\Delta_{j}$ is a singleton and
$|\breve{p}(\hat{\gamma}_{m})\cap\Delta_{j}|=:n_{m}(x_{j})=n(x_{j})$, holding
for all $m>m_{*}$ and $j\leq J$. Now we repeat the construction just made in
each of $\Delta_{j}$. That is, we enumerate
$\hat{\gamma}_{m}\cap\Delta_{j}=\\{(x^{j}_{1,m},\alpha^{j}_{1,m}),\dots,(x^{j}_{n(x_{j}),m},\alpha^{j}_{n(x_{j}),m})\\}$,
and then set $a_{m}^{j}=\\{\alpha^{j}_{1,m},\dots,\alpha^{j}_{n(x_{j}),m}\\}$.
Then we set $\hat{\gamma}_{*,m}=(\gamma_{*,m},a_{*,m})$, where
$\gamma_{*,m}=p(\breve{\gamma})\cap\Lambda_{\varepsilon}^{o}=\\{x_{1},\dots,x_{J}\\}$
and $a_{*,m}(x_{j})=a^{j}_{m}$. In other words, the ground configuration of
$\hat{\gamma}_{*,m}$ is the part of the limiting configurations
$p(\hat{\gamma})$ contained in $\Lambda_{\varepsilon}^{o}$, whereas the marks
are taken from the corresponding part of $\hat{\gamma}_{m}$. By the triangle
inequality we then have

$\kappa(\hat{\gamma},\hat{\gamma}_{m})\leq\kappa(\hat{\gamma}\cap\Lambda^{o}_{\varepsilon},\hat{\gamma}_{*,m})+\kappa(\hat{\gamma}_{m}\cap\Lambda_{\varepsilon}^{o},\hat{\gamma}_{*,m})+\varepsilon/3.$
(2.19)

By (2.14), for each $s,k,n\in\mathds{N}$, we have

$\displaystyle\kappa_{s,k,n}(\hat{\gamma}\cap\Lambda^{o}_{\varepsilon},\hat{\gamma}_{*,m})$
$\displaystyle=$
$\displaystyle\left|\sum_{j=1}^{J}v_{s}(x_{j})\left[\sum_{\alpha\in
a^{j}_{m}}w_{\sigma_{k},n}(\alpha)-\sum_{\alpha\in
a(x_{j})}w_{\sigma_{k},n}(\alpha)\right]\right|$ $\displaystyle\leq$
$\displaystyle J\max_{j\leq J}\rho_{k,n}(a_{m}^{j},a(x_{j})).$

Likewise,

$\displaystyle\kappa_{s,k,n}(\hat{\gamma}_{m}\cap\Lambda^{o}_{\varepsilon},\hat{\gamma}_{*,m})$
$\displaystyle=$
$\displaystyle\bigg{|}\sum_{j=1}^{J}\left(\sum_{p=1}^{n(x_{j})}v_{s}(x^{j}_{p,m})\right)\sum_{\alpha\in
a^{j}_{m}}w_{\sigma_{k},n}(\alpha)$ $\displaystyle-$
$\displaystyle\sum_{j=1}^{J}v_{s}(x_{j})\sum_{\alpha\in
a^{j}_{m})}w_{\sigma_{k},n}(\alpha)\bigg{|}$ $\displaystyle\leq$
$\displaystyle
d_{s}(\breve{\gamma}_{m}\cap\Lambda^{o}_{\varepsilon},\breve{\gamma}\cap\Lambda^{o}_{\varepsilon}).$

Both latter estimates yield

$\kappa(\hat{\gamma}\cap\Lambda^{o}_{\varepsilon},\hat{\gamma}_{*,m})\leq
J\max_{j\leq J}\rho(a^{j}_{m},a(x_{j})),$
$\kappa(\hat{\gamma}_{m}\cap\Lambda_{\varepsilon}^{o},\hat{\gamma}_{*,m})\leq
d(\breve{\gamma}_{m},\breve{\gamma}).$

By the aforementioned convergence $\breve{\gamma}_{m}\to\breve{\gamma}$ and
$a^{j}_{m}\to a(x_{j})$, one can find $m_{\varepsilon}>m_{*}$ such that the
first two summands in (2.19) are smaller than $\varepsilon$ for
$m>m_{\varepsilon}$, which completes the proof. ∎

###  2.3. Measures and functions on configuration spaces

For $v_{s}$ and $w_{k,n}$ as in (2.14) we set

$\theta_{s,k,n}(x,\alpha)=\exp\left(-v_{s}(x)w_{k,n}(\alpha)\right)-1=\exp\left(-g_{s,k,n}(x,\alpha)\right)-1,$

see (2.15). Then $\theta_{s,k,n}(x,\alpha)\in C_{\rm cs}(\widehat{X})$ and
$\theta_{s,k,n}(x,\alpha)\in(-1,0]$. Let $\Theta$ be the subset of $C_{\rm
cs}(\widehat{X})$ consisting of

$\theta(\hat{x})=e^{-g(\hat{x})}-1,\qquad
g(x,\alpha)=\sum_{j}v_{s_{j}}(x)w_{k_{j},n_{j}}(\alpha),$ (2.20)

where the latter sum runs over a finite subset of $\mathds{N}^{3}$. That is,
each such $g$ is a finite sum of $g_{s,k,n}$ defined in (2.15). Note that
$\Theta$ is countable and closed under the map
$(\theta,\theta^{\prime})\mapsto(\theta\ast\theta^{\prime})$, where

$(\theta\ast\theta^{\prime})(\hat{x})=\theta(\hat{x})+\theta^{\prime}(\hat{x})+\theta(\hat{x})\theta^{\prime}(\hat{x})=-1+(1+\theta(x))(1+\theta^{\prime}(x)).$
(2.21)

Moreover, by (2.4), (2.6) and (2.20) it follows that

$g(x,\alpha)\leq g(x,0),$ (2.22)

holding for all $\alpha\geq 0$ and $x\in X$. Now for $\theta\in\Theta$, we set

$F^{\theta}(\hat{\gamma})=\prod_{\hat{x}\in\hat{\gamma}}(1+\theta(\hat{x}))=\exp\left(-\sum_{\hat{x}\in\hat{\gamma}}g(\hat{x})\right),\qquad\hat{\gamma}\in\widehat{\Gamma}.$
(2.23)

Then $F^{\theta}(\hat{\gamma})\in(0,1]$ for all
$\hat{\gamma}\in\widehat{\Gamma}$, and hence $F^{\theta}\in C_{\rm
b}(\widehat{\Gamma})$. The set of all such functions will be denoted by
$\mathcal{F}_{\Theta}$. For $\mu\in\mathcal{P}(\widehat{\Gamma})$, we then
have

$\mu(F^{\theta})=\int_{\widehat{\Gamma}}F^{\theta}(\hat{\gamma})\mu(d\hat{\gamma})\leq
1.$

The Poisson measure $\pi_{\varrho}$ with intensity measure $\varrho$ satisfies

$\pi_{\varrho}(F^{\theta})=\exp\left(\varrho(\theta)\right)=\exp\left(\int_{\widehat{X}}\theta(\hat{x})\varrho(d\hat{x})\right).$
(2.24)

For $\mu_{1},\mu_{2}\in\mathcal{P}(\widehat{\Gamma})$, their convolution is
defined by the expression

$(\mu_{1}\star\mu_{2})(F)=\int_{\widehat{\Gamma}^{2}}F(\hat{\gamma}_{1}\cup\hat{\gamma}_{2})\mu_{1}(d\hat{\gamma}_{1})\mu_{2}(d\hat{\gamma}_{2}),$
(2.25)

that ought to hold for all $F\in B_{\rm b}(\widehat{\Gamma})$. For
$F^{\theta}$ as in (2.23), it takes the form

$(\mu_{1}\star\mu_{2})(F^{\theta})=\mu_{1}(F^{\theta})\mu_{2}(F^{\theta}).$
(2.26)

Recall that a set $\mathcal{F}\subset C_{\rm b}(\widehat{\Gamma})$ is called
convergence determining if $\mu_{n}(F)\to\mu(F)$, $n\to+\infty$, implies
$\mu_{n}\Rightarrow\mu$, holding for each
$\\{\mu_{n}\\}_{n\in\mathds{N}}\subset\mathcal{P}(\widehat{\Gamma})$. It is
known, see [5, Theorem 4.5, page 113], that such $\mathcal{F}$ has this
property if it is closed under pointwise multiplication and is strongly
separating. The latter means that, for each $\hat{\gamma}\in\widehat{\Gamma}$
and $\epsilon>0$, there exists a finite family $\\{F_{j}\\}\subset\mathcal{F}$
such that

$\inf_{\hat{\gamma}^{\prime}\in\widehat{C}_{\epsilon}}\max_{j}\left|F_{j}(\hat{\gamma})-F_{j}(\hat{\gamma}^{\prime})\right|>0,\qquad\widehat{C}_{\epsilon}:=\widehat{\Gamma}\setminus\widehat{B}_{\epsilon}(\hat{\gamma}).$
(2.27)

Here
$\widehat{B}_{\epsilon}(\hat{\gamma})=\\{\hat{\gamma}^{\prime}:\kappa(\hat{\gamma},\hat{\gamma}^{\prime})<\epsilon\\}$,
see (2.14). Note that taking $\epsilon\geq 1$ does not make sense as
$\kappa(\hat{\gamma},\hat{\gamma}^{\prime})<1$ for all configurations.

######  Proposition 2.8.

The set $\mathcal{F}_{\Theta}$ is strongly separating and thus convergence
determining.

###### Proof.

By the very definition of $\Theta$, cf. (2.21), $\mathcal{F}_{\Theta}$ is
closed under pointwise multiplication. To prove (2.27), we note that, see
(2.23) and (2.15),

$\left|F^{\theta_{s,k,n}}(\hat{\gamma})-F^{\theta_{s,k,n}}(\hat{\gamma}^{\prime})\right|\geq\min\left\\{F^{\theta_{s,k,n}}(\hat{\gamma});F^{\theta_{s,k,n}}(\hat{\gamma}^{\prime})\right\\}\kappa_{s,k,n}(\hat{\gamma};\hat{\gamma}^{\prime})$
(2.28)

holding for all $\hat{\gamma},\hat{\gamma}^{\prime}\in\widehat{\Gamma}$. Now
we fix $\hat{\gamma}$ and $\epsilon\in(0,1)$ and then take $m$ such that
$2^{-m}<\epsilon/2$. For this $m$ and any
$\hat{\gamma^{\prime}}\in\widehat{C}_{\epsilon}$, by (2.14) we readily
conclude that

$\max_{(s,k,n):s+k+n\leq
m}\kappa_{s,k,n}(\hat{\gamma},\hat{\gamma}^{\prime})>0,$

which by (2.28) yields the proof. ∎

The important properties of the family $\mathcal{F}_{\Theta}$ are summarized
in the following statement.

######  Proposition 2.9.

The following is true:

  * (i)

$\mathcal{B}(\widehat{\Gamma})=\sigma\\{\mathcal{F}_{\Theta}\\}$;

  * (ii)

$B_{\rm b}(\widehat{\Gamma})$ is the $bp$-closure of the linear span of
$\mathcal{F}_{\Theta}$;

  * (iii)

$\mathcal{F}_{\Theta}$ is separating;

  * (iv)

$\mathcal{F}_{\Theta}$ is convergence determining.

The proof of (i) and (ii) is standard, see [4, Lemma 3.2.5 and Theorem 3.2.6,
page 43]. The proof of (iv) was done above, whereas (iii) is a direct
consequence of (iv), cf. [5, page 113].

######  Proposition 2.10.

Under Assumption 1.1, the Kolmogorov operator introduced in (1) has the
property $L:\mathcal{F}_{\Theta}\to C_{\rm b}(\widehat{\Gamma})$.

###### Proof.

We consider each of the summands in (1) – denoted by $L_{i}$, $i=1,2,3$ –
separately. Thus,

$\displaystyle(L_{3}F^{\theta})(\hat{\gamma})=F^{\theta}(\hat{\gamma})\int_{X}\theta(x,0)\chi(dx),$
(2.29)

i.e., it is just the multiplication operator by a $\theta$-dependent constant.
Next,

$(L_{2}F^{\theta})(\hat{\gamma})=-\sum_{x\in\hat{\gamma}}m(\hat{x})\theta(\hat{x})F^{\theta}(\hat{\gamma}\setminus\hat{x})=F^{\theta}(\hat{\gamma})\Psi_{2}(\hat{\gamma}),$
(2.30)

where

$\Psi_{2}(\hat{\gamma})=\sum_{\hat{x}\in\hat{\gamma}}\psi_{2}(\hat{x})=\sum_{\hat{x}\in\hat{\gamma}}m(\hat{x})\left(e^{g(\hat{x})}-1\right).$
(2.31)

Let us consider the following function

$\phi_{\tau}(\hat{x})=g(\hat{x})-\tau
m(\hat{x})\left(e^{g(\hat{x})}-1\right),\qquad\tau\geq 0.$

Since each $g_{s,k,n}(\hat{x})<1$, see (2.15), it follows that $g(\hat{x})\leq
J_{\theta}$ where $J_{\theta}$ is just the number of summands in the sum in
(2.20). At the same time, $m(\hat{x})\leq m_{*}<\infty$, see Assumption 1.1.
Taking this into account, we set

$\tau_{*}=1/m_{*}e^{J_{\theta}}.$ (2.32)

Then

$\phi_{\tau_{*}}(\hat{x})>0,\qquad\hat{x}\in\widehat{X}.$

Now by the simple inequality $\beta e^{-\tau\beta}\leq 1/e\tau$,
$\tau,\beta>0$, we have, see (2.30) and (2.31),

$\displaystyle(L_{2}F^{\theta})(\hat{\gamma})\leq\tau^{-1}_{*}\exp\left(-1-\sum_{\hat{x}\in\hat{\gamma}}\phi_{\tau_{*}}(\hat{x})\right)\leq
m_{*}e^{J_{\theta}-1},$ (2.33)

which yields the boundedness in question. The continuity of $L_{2}F^{\theta}$
follows by the continuity of $\Psi_{2}$, which in turn follows by the fact
that $\psi_{2}\in C_{\rm cs}(\widehat{X})$. Finally,

$\displaystyle(L_{1}F^{\theta})(\hat{\gamma})=\left(-\sum_{\hat{x}\in\hat{\gamma}}g^{\prime}(\hat{x})\right)F^{\theta}(\hat{\gamma})=:\Psi_{1}(\hat{\gamma})F^{\theta}(\hat{\gamma}).$

By (2.20) we have

$g^{\prime}(x,\alpha)=\sum_{j}v_{s_{j}}(x)w^{\prime}_{k_{j},n_{j}}(\alpha),$

which yields the continuity of $\Psi_{1}$. At the same time, by (2.7) we have

$|g^{\prime}(x,\alpha)|\leq\sum_{j}v_{s_{j}}(x)|w^{\prime}_{k_{j},n_{j}}(\alpha)|\leq\bar{\sigma}cg(x,\alpha),$
(2.34)

which yields

$|g^{\prime}(x,\alpha)|\leq\bar{\sigma}cJ_{\theta},$ (2.35)

and also

$\left|(L_{1}F^{\theta})(\hat{\gamma})\right|\leq\bar{\sigma}c/e.$

This completes the proof. ∎

We summarize the estimates obtained above in the following

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}|(LF^{\theta})(\hat{\gamma})|\leq\chi(|\theta(\cdot,0)|)+m_{*}e^{J_{\theta}-1}+\bar{\sigma}c/e.$
(2.36)

Note that

$\displaystyle\chi(|\theta(x,\alpha)|)$ $\displaystyle:=$
$\displaystyle\int_{X}|\theta(x,\alpha)|\chi(dx)$ $\displaystyle=$
$\displaystyle\int_{X}\left(1-e^{-g(x,\alpha)}\right)\chi(dx)\leq\int_{X}g(x,0)\chi(dx),$

see (2.22).

##  3\. The Kolmogorov Equation

###  3.1. Notations and useful estimates

For $\theta\in\Theta$, see (2.20), we set

$\displaystyle\theta_{t}(x,\alpha)$ $\displaystyle=$
$\displaystyle\theta(x,\alpha+t)\exp\bigg{(}M(x,\alpha)-M(x,\alpha+t)\bigg{)},$
(3.1) $\displaystyle M(x,\alpha)$ $\displaystyle=$
$\displaystyle\int_{0}^{\alpha}m(x,\beta)d\beta.$

Then $\widehat{X}\ni\hat{x}\mapsto\theta_{t}(\hat{x})$ is continuous and
compactly supported for all $t\geq 0$. Moreover, both maps
$t\mapsto\theta_{t}(\hat{x})$ and $\alpha\mapsto\theta_{t}(x,\alpha)$ are
continuously differentiable and the following holds

$\frac{\partial}{\partial
t}\theta_{t}(x,\alpha)=\frac{\partial}{\partial\alpha}\theta_{t}(x,\alpha)-m(x,\alpha)\theta_{t}(x,\alpha).$
(3.2)

Note that

$(\theta_{t})_{s}(\hat{x})=\theta_{t+s}(\hat{x}).$ (3.3)

Next, we define, cf. (2.20),

$g_{t}(\hat{x})=-\log(1+\theta_{t}(\hat{x})).$ (3.4)

By (3.1) it follows that

$|\theta_{t}(x,\alpha)|\leq|\theta(x,\alpha+t)|=1-e^{-g(x,\alpha+t)}\leq
1-e^{-J_{\theta}},$

where $J_{\theta}$ is the same as in (2.32). By (3.4) this yields

$g_{t}(\hat{x})\leq J_{\theta},\qquad t\geq 0,\ \ \hat{x}\in\widehat{X}.$
(3.5)

By (3.1) we also have

$|\theta^{\prime}_{t}(x,\alpha)|\leq 2m_{*}+|\theta^{\prime}(x,\alpha+t)|\leq
2m_{*}+J_{\theta},$

where $m_{*}$ is as in (2.32) and the estimate
$|\theta^{\prime}(x,\alpha)|\leq|g^{\prime}(x,\alpha)|\leq J_{\theta}$ was
used, see (2.20) and (2.35).

Now we define

$F^{\theta}_{t}(\hat{\gamma})=\exp\left[\int_{0}^{t}\left(\int_{X}\theta(x,\alpha)e^{-M(x,\alpha)}\chi(dx)\right)d\alpha\right]F^{\theta_{t}}(\hat{\gamma}),$
(3.6)

with $\theta_{t}$ as in (3.1). Clearly, $F^{\theta}_{t}\in C_{\rm
b}(\widehat{\Gamma})$ for all $t>0$ and $\theta\in\Theta$, and

$0<F^{\theta}_{t}(\hat{\gamma})\leq 1,\qquad\hat{\gamma}\in\widehat{\Gamma},$
(3.7)

and, for all $t,s\geq 0$, the following holds, see (3.3),

$F^{\theta}_{t+s}=\exp\left(\int_{0}^{s}\int_{X}\theta_{u}(x,0)\chi(dx)du\right)F^{\theta_{s}}_{t}.$
(3.8)

Let us prove that $LF^{\theta}_{t}\in C_{\rm b}(\widehat{\Gamma})$ for all
$\theta\in\Theta$ and $t\geq 0$. As in the proof of Proposition 2.10 we divide
$L$ into three parts. Similarly as in (2.29) we have

$\left|(L_{3}F_{t}^{\theta})(\hat{\gamma})\right|\leq\chi(|\theta(\cdot,t)|)\leq\chi(g(\cdot,0)),$

see (2.3) and (3.7). Since $g_{t}(\hat{x})$ satisfies (3.5) for all $t\geq 0$,
it follows that

$\left|(L_{2}F_{t}^{\theta})(\hat{\gamma})\right|\leq m_{*}e^{J_{\theta}-1},$

holding for all $t>0$, see (2.33). The estimate of
$\left|(L_{1}F_{t}^{\theta})(\hat{\gamma})\right|$ is obtained as follows.
Denote

$q_{t}(x,\alpha)=\exp\left(-\int_{\alpha}^{\alpha+t}m(x,\beta)d\beta\right).$

Then by (3.1) we have

$\Phi(g_{t}(x,\alpha))=q_{t}(x,\alpha)\Phi(g(x,\alpha+t)),\qquad\Phi(b):=1-e^{-b},\
\ b\geq 0,$ (3.9)

by which we get that $g_{t}(x,\alpha)\leq g(x,\alpha+t)$. Let us prove that

$g_{t}(x,\alpha)e^{-g_{t}(x,\alpha)}\geq
q_{t}(x,\alpha)g(x,\alpha+t)e^{-g(x,\alpha+t)}.$ (3.10)

By (3.9) this is equivalent to the fact that the function $b\mapsto
b/(e^{b}-1)$ is decreasing, which is obviously the case. Now we take the
$\alpha$-derivative from both sides of (3.9) and obtain

$\displaystyle
g^{\prime}_{t}(x,\alpha)=q_{t}(x,\alpha)g^{\prime}(x,\alpha+t)\exp\bigg{(}g_{t}(x,\alpha)-g(x,\alpha+t)\bigg{)}$
$\displaystyle+m(x,\alpha+t)(e^{g_{t}(x,\alpha)}-1)-m(x,\alpha)(e^{g_{t}(x,\alpha)}-1),$

that can be estimated as follows

$\displaystyle|g^{\prime}_{t}(x,\alpha)|=\bar{\sigma}cq_{t}(x,\alpha)g(x,\alpha+t)\exp\bigg{(}g_{t}(x,\alpha)-g(x,\alpha+t)\bigg{)}$
(3.11) $\displaystyle+2m_{*}(e^{g_{t}(x,\alpha)}-1)\leq
g_{t}(x,\alpha)e^{J_{\theta}}(\bar{\sigma}c+2m_{*}),$

where we used (2.34), (3.5) and (3.10). Now we proceed as in obtaining (2.36),
which eventually yields

$\left|(LF^{\theta}_{t})(\hat{\gamma})\right|\leq\chi(g(\cdot,0))+m_{*}e^{J_{\theta}-1}+(\bar{\sigma}c+2m_{*})e^{J_{\theta}}=:\ell_{\theta}.$
(3.12)

The key property of the latter estimate is that it is uniform in $t$. However,
it does depend on $\theta$.

Along with the estimates derived above, we will use also the following. For
$\theta\in\Theta$, the corresponding $g$ has the form as in (2.20). By (2.5)
and (2.6) we have that

$e^{-\bar{\sigma}2^{2/3}/3}w_{k,n}(0)\leq w_{k,n}(\alpha)\leq w_{k,n}(0),$

which means that, cf. (2.22),

$\bar{c}g(x,0)\leq g(x,\alpha)\leq
g(x,0),\qquad\bar{c}:=e^{-\bar{\sigma}2^{2/3}/3}.$ (3.13)

For $\Phi$ as in (3.9), we have $b\geq\Phi(b)\geq b-b^{2}/2$, $b\geq 0$, which
we use together with (3.13) to obtain the following

$\displaystyle g_{t}(\hat{x})\geq q_{t}(\hat{x})\Phi(\bar{c}g(x,0))\geq
e^{-m_{*}t}\Phi(\bar{c}g(x,0)/J_{\theta})$ (3.14) $\displaystyle\geq
e^{-m_{*}t}\frac{\bar{c}g(x,0)}{J_{\theta}}\left(1-\bar{c}g(x,0)/2J_{\theta}\right)$
$\displaystyle\geq
e^{-m_{*}t}\bar{c}_{\theta}g(x,0),\quad\bar{c}_{\theta}:=\bar{c}/2J_{\theta},$

where we have used the fact that $g(x,0)\leq J_{\theta}$, see (3.5).

###  3.2. The operator

We fix $\hat{\gamma}\in\widehat{\Gamma}$ and calculate the $t$-derivative of
(3.2). This yields

$\displaystyle\frac{\partial}{\partial t}F^{\theta}_{t}(\hat{\gamma})$
$\displaystyle=$
$\displaystyle\left(\int_{X}\theta(x,t)e^{-M(x,t)}\chi(dx)\right)F^{\theta}_{t}(\hat{\gamma})+\sum_{\hat{x}\in\hat{\gamma}}\frac{\partial\theta(\hat{x})}{\partial
t}F^{\theta}_{t}(\hat{\gamma}\setminus\hat{x})\qquad$ $\displaystyle=$
$\displaystyle\left(\int_{X}\theta_{t}(x,0)\chi(dx)\right)F^{\theta}_{t}(\hat{\gamma})+\sum_{\hat{x}\in\hat{\gamma}}\frac{\partial}{\partial\alpha_{x}}F^{\theta}_{t}(\hat{\gamma})$
$\displaystyle+$
$\displaystyle\sum_{\hat{x}\in\hat{\gamma}}m(\hat{x})\left[F^{\theta}_{t}(\hat{\gamma}\setminus\hat{x})-F^{\theta}_{t}(\hat{\gamma})\right]=(LF^{\theta}_{t})(\hat{\gamma}).$

This means that we have found a solution of the Kolmogorov equation for (1) in
the following sense. It is a map $t\mapsto F_{t}\in C_{\rm
b}(\widehat{\Gamma})$, which is pointwise in $\hat{\gamma}$ continuously
$t$-differentiable and such that the equality in (1.1) holds. Our aim now is
to solve (1.1) in a suitable Banach space. Recall that the paths
$t\mapsto\theta_{t}$ have the flow property (3.3), see also (3.8).

Below, by saying of a property of $\theta_{s}$, $s\geq 0$, holding for all
$\theta$, we attribute that this property to all $\theta_{s}$ given in (3.1)
with $\theta$ taken from $\Theta$.

######  Proposition 3.1.

For each $\theta\in\Theta$ and $s\geq 0$, it follows that
$F_{t}^{\theta_{s}}\to F^{\theta_{s}}$ as $t\to 0$ in the norm of $C_{\rm
b}(\widehat{\Gamma})$.

###### Proof.

For each $\hat{\gamma}$, $s\geq 0$ and $\theta\in\Theta$, by (3.2) it follows
that

$\displaystyle
F_{t}^{\theta_{s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})=\exp\left(-\int_{0}^{s}\int_{X}\theta_{u}(x,0)\chi(dx)du\right)\left[F^{\theta}_{t+s}(\hat{\gamma})-F^{\theta}_{s}(\hat{\gamma})\right]$
(3.16)
$\displaystyle=\exp\left(-\int_{0}^{s}\int_{X}\theta_{u}(x,0)\chi(dx)du\right)\int_{s}^{t+s}(LF_{u}^{\theta})(\hat{\gamma})du,$

which by (3.12) yields

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}\left|F_{t}^{\theta_{s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})\right|\leq
t\ell_{\theta}\exp\left(-\int_{0}^{s}\int_{X}\theta_{u}(x,0)\chi(dx)du\right).$
(3.17)

This completes the proof. ∎

The next statement is a refinement of the one just proved.

######  Proposition 3.2.

For each $\theta\in\Theta$ and $s\geq 0$, it follows that
$LF_{t}^{\theta_{s}}\to LF^{\theta_{s}}$ as $t\to 0$ in the norm of $C_{\rm
b}(\widehat{\Gamma})$.

###### Proof.

First of all we note that the equality in the first line of (3.16) allows one
to obtain the property in question by showing that $LF_{t+s}^{\theta}\to
LF^{\theta}_{s}$ in the same sense.

As in the proof of Proposition 2.10, we split $L$ into three parts and
consider each of them separately. Fix $\theta\in\Theta$ and then denote

$\dot{\eta}(u)=\int_{X}\theta_{u}(x,0)\chi(dx),\qquad\eta(t)=\int_{0}^{t}\dot{\eta}(u)du.$
(3.18)

Then

$\displaystyle\left|(L_{3}F_{t+s}^{\theta})(\hat{\gamma})-(L_{3}F^{\theta}_{s})(\hat{\gamma})\right|$
$\displaystyle\leq$
$\displaystyle\left|\dot{\eta}(t+s)e^{\eta(t+s)}-\dot{\eta}(s)e^{\eta(s)}\right|$
$\displaystyle+$
$\displaystyle|\dot{\eta}(s)|e^{\eta(s)}\left|F^{\theta_{t+s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})\right|$
$\displaystyle=:$ $\displaystyle I_{1}(t)+I_{2}(t),$

see (3.7). Then $I_{1}(t)\to 0$ as $t\to 0$ since $\dot{\eta}(t)e^{\eta(t)}$
is a continuous function of $t$. At the same time, $I_{2}(t)$ can be estimated
as in (3.17). This yields the proof for $L_{3}$.

Now we proceed to $L_{2}$, for which it follows that

$\displaystyle(L_{2}F_{t}^{\theta_{s}})(\hat{\gamma})-(L_{2}F^{\theta_{s}})(\hat{\gamma})=J_{t}(\hat{\gamma})+K_{t}(\hat{\gamma}),$
(3.20) $\displaystyle
J_{t}(\hat{\gamma})=\sum_{\hat{x}\in\hat{\gamma}}\frac{m(\hat{x})(-\theta_{t+s}(\hat{x}))}{1+\theta_{t+s}(\hat{x})}[F^{\theta_{s}}_{t}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})]$
$\displaystyle
K_{t}(\hat{\gamma})=F^{\theta_{s}}(\hat{\gamma})\sum_{\hat{x}\in\hat{\gamma}}\frac{m(\hat{x})(\theta_{s}(\hat{x})-\theta_{t+s}(\hat{x}))}{(1+\theta_{t+s}(\hat{x}))(1+\theta_{s}(\hat{x}))}.$

By (3.6) and (3.18) we have

$\displaystyle F^{\theta_{s}}_{t}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})$
$\displaystyle=$
$\displaystyle(e^{\eta(t+s)-\eta(s)}-1)F^{\theta_{t+s}}(\hat{\gamma})$
$\displaystyle+$ $\displaystyle
F^{\theta_{t+s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})=:\Upsilon_{1}(t,\hat{\gamma})+\Upsilon_{2}(t,\hat{\gamma}).$

By (2.3) and (3.1), (3.18) it follows that

$\left|e^{\eta(t+s)-\eta(s)}-1\right|\leq t\chi(g(\cdot,0)),$

which then yields

$\left|\Upsilon_{1}(t,\hat{\gamma})\right|\leq
t\chi(g(\cdot,0))\exp\left(-\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})\right).$
(3.22)

To estimate $\Upsilon_{2}$ we write

$h_{t}(\hat{x})=\min\\{g_{t+s}(\hat{x});g_{s}(\hat{x})\\}.$

Then by (3.11) and (3.14) we get

$\displaystyle\left|\Upsilon_{2}(t,\hat{\gamma})\right|\leq\exp\left(-\sum_{\hat{x}\in\hat{\gamma}}h_{t}(\hat{x})\right)\sum_{\hat{x}\in\hat{\gamma}}\left|g_{t+s}(\hat{x})-g_{s}(\hat{x})\right|$
(3.23) $\displaystyle\leq
e^{J_{\theta}}(\bar{\sigma}c+2m_{*})\exp\left(-\sum_{\hat{x}\in\hat{\gamma}}h_{t}(\hat{x})\right)\sum_{\hat{x}\in\hat{\gamma}}\int_{s}^{t+s}g_{u}(x,\alpha)du,$
$\displaystyle\leq
e^{J_{\theta}}(\bar{\sigma}c+2m_{*})\exp\left(-e^{-m_{*}t}\bar{c}_{\theta}\sum_{\hat{x}\in\hat{\gamma}}g(x,0)\right)\sum_{\hat{x}\in\hat{\gamma}}tg(x,0)$
$\displaystyle=:tC_{\theta}e^{-\bar{c}_{\theta}(t)\Psi_{0}(\hat{\gamma})}\Psi_{0}(\hat{\gamma}),\qquad\bar{c}_{\theta}(t):=e^{-m_{*}t}\bar{c}_{\theta},\quad\
\Psi_{0}(\hat{\gamma}):=\sum_{\hat{x}\in\hat{\gamma}}g(x,0).$

At the same time,

$\displaystyle
0\leq\sum_{\hat{x}\in\hat{\gamma}}\frac{m(\hat{x})(-\theta_{t+s}(\hat{x})}{1+\theta_{t+s}(\hat{x})}\leq
m_{*}e^{J_{\theta}}\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})=:m_{*}e^{J_{\theta}}\Psi_{1}(t,\hat{\gamma}).$

Thereafter, we have

$\displaystyle\left|J_{t}(\hat{\gamma})\right|\leq
tm_{*}e^{J_{\theta}}\chi(g(\cdot,0))\Psi_{1}(t,\hat{\gamma})e^{-\Psi_{1}(t,\hat{\gamma})}$
$\displaystyle+tC_{\theta}m_{*}e^{J_{\theta}}\Psi_{1}(t,\hat{\gamma})\Psi_{0}(\hat{\gamma})e^{-\bar{c}_{\theta}(t)\Psi_{0}(\hat{\gamma})}=:\Pi_{1}(t,\hat{\gamma})+\Pi_{2}(t,\hat{\gamma}),$

where $C_{\theta}$, $\bar{c}_{\theta}(t)$ and $\Psi_{0}$ are as in (3.23).
Then

$\Pi_{1}(t,\hat{\gamma})\leq tm_{*}e^{J_{\theta}-1}\chi(g(\cdot,0))\to
0,\qquad t\to+\infty.$

Let $t_{\theta}>0$ be the (unique) solution of
$\bar{c}_{\theta}e^{-m_{*}t}=2t^{1/3}$. Then for $t\leq t_{\theta}$, we have

$\displaystyle\Pi_{2}(t,\hat{\gamma})\leq
tC_{\theta}m_{*}e^{J_{\theta}}[\Psi_{0}(\hat{\gamma})]^{2}e^{-\bar{c}_{\theta}(t)\Psi_{0}(\hat{\gamma})}$
(3.24) $\displaystyle\leq
t^{1/3}C_{\theta}m_{*}e^{J_{\theta}-2}\exp\left(-(\bar{c}_{\theta}(t)-2t^{1/3})\Psi_{0}(\hat{\gamma})\right)\leq
t^{1/3}C_{\theta}m_{*}e^{J_{\theta}-2},$

which yields the convergence

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}|J_{t}(\hat{\gamma})|\to 0,\qquad t\to
0.$

Now we turn to $K_{t}(\hat{\gamma})$. First, by (3.2) we have

$\displaystyle\left|\theta_{s}(\hat{x})-\theta_{t+s}(\hat{x})\right|=\left|\int_{s}^{t+s}\left(\frac{\partial}{\partial\alpha}\theta_{u}(x,\alpha)-m(x,\alpha)\theta_{u}(x,\alpha)\right)du\right|.$
(3.25)

Next, by (3.4) and (3.11) it follows that

$\displaystyle\left|\frac{\partial}{\partial\alpha}\theta_{u}(x,\alpha)\right|=e^{-g_{u}(x,\alpha)}\left|g^{\prime}_{u}(x,\alpha)\right|\leq
e^{J_{\theta}}(\bar{\sigma}c+2m_{*})g_{u}(x,\alpha)$ $\displaystyle\leq
e^{J_{\theta}}(\bar{\sigma}c+2m_{*})g(x,0),$

and also

$\displaystyle\left|m(x,\alpha)\theta_{u}(x,\alpha)\right|\leq
m_{*}g_{u}(x,\alpha)\leq m_{*}g(x,0).$

The latter two estimates yield

$\displaystyle{\rm LHR}(\ref{JJ21b})\leq
e^{J_{\theta}}(\bar{c}+3m_{*})g(x,0).$

By (3.13) this yields

$\displaystyle K_{t}(\hat{\gamma})\leq
tm_{*}(\bar{\sigma}+3m_{*})e^{3J_{\theta}}\Psi_{0}(\hat{\gamma})e^{-\bar{c}\Psi_{0}(\hat{\gamma})}$
$\displaystyle\leq tm_{*}(\bar{\sigma}+3m_{*})e^{3J_{\theta}-1}/\bar{c}\to
0,\qquad t\to+\infty.$

By (3.20) this completes the proof for $L_{2}$.

Next, we write

$\displaystyle(L_{1}F^{\theta_{s}}_{t})(\hat{\gamma})-(L_{1}F^{\theta_{s}})(\hat{\gamma})=Q_{t}(\hat{\gamma})+R_{t}(\hat{\gamma}),$
(3.26) $\displaystyle
Q_{t}(\hat{\gamma})=\sum_{\hat{x}\in\hat{\gamma}}\theta^{\prime}_{t+s}(\hat{x})\left[F^{\theta_{s}}_{t}(\hat{\gamma}\setminus\hat{x})-F^{\theta_{s}}(\hat{\gamma}\setminus\hat{x})\right],$
$\displaystyle
R_{t}(\hat{\gamma})=\sum_{\hat{x}\in\hat{\gamma}}\left[\theta^{\prime}_{t+s}(\hat{x})-\theta^{\prime}_{s}(\hat{x})\right]F^{\theta_{s}}(\hat{\gamma}\setminus\hat{x}).$

Then

$\displaystyle
Q_{t}(\hat{\gamma})=-\sum_{\hat{x}\in\hat{\gamma}}g^{\prime}_{t+s}(\hat{x})\left[F_{t}^{\theta_{s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})\right]$
$\displaystyle+F^{\theta_{s}}(\hat{\gamma})\sum_{\hat{x}\in\hat{\gamma}}g^{\prime}_{t+s}(\hat{x})\left[e^{g_{s}(\hat{x})-g_{t+s}(\hat{x})}-1\right]=:Q^{(1)}_{t}(\hat{\gamma})+Q^{(2)}_{t}(\hat{\gamma}).$

By (3.11) and then by (3.2), (3.22), (3.23) we get

$\displaystyle\left|Q^{(1)}_{t}(\hat{\gamma})\right|\leq
e^{J_{\theta}}(\bar{\sigma}c+2m_{*})\left(\Upsilon_{1}(t,\hat{\gamma})+\Upsilon_{2}(t,\hat{\gamma})\right)\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})$
$\displaystyle\leq
t\chi(g(\cdot,0))e^{J_{\theta}}(\bar{\sigma}c+2m_{*})\left(\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})\right)\exp\left(-\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})\right)$
$\displaystyle+tC_{\theta}e^{J_{\theta}}(\bar{\sigma}c+2m_{*})e^{-\bar{c}_{\theta}(t)\Psi_{0}(\hat{\gamma})}[\Psi_{0}(\hat{\gamma})]^{2}=:\Xi^{(1)}_{t}(\hat{\gamma})+\Xi^{(2)}_{t}(\hat{\gamma}).$

Then

$\Xi^{(1)}_{t}(\hat{\gamma})\leq
t\chi(g(\cdot,0))e^{J_{\theta}-1}(\bar{\sigma}c+2m_{*})\to 0,\qquad t\to 0,$

and also, cf. (3.24),

$\displaystyle\Xi^{(2)}_{t}(\hat{\gamma})\leq
t^{1/3}C_{\theta}e^{J_{\theta}-2}(\bar{\sigma}c+2m_{*})\exp\left(-(\bar{c}_{\theta}(t)-2t^{1/3})\Psi_{0}(\hat{\gamma})\right)$
$\displaystyle\leq
t^{1/3}C_{\theta}m_{*}e^{J_{\theta}-2}(\bar{\sigma}c+2m_{*}),$

for $t\leq t_{\theta}$. The latter two estimates yield

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}|Q^{(1)}_{t}(\hat{\gamma})|\to 0,\qquad
t\to 0.$ (3.27)

Next, by (3.5), (3.11) and (3.13) we have

$\displaystyle\left|Q^{(2)}_{t}(\hat{\gamma})\right|\leq
e^{J_{\theta}}F^{\theta_{s}}(\hat{\gamma})\sum_{\hat{x}\in\hat{\gamma}}|g^{\prime}_{t+s}(\hat{x})|\left|g_{t+s}(\hat{x})-g_{s}(\hat{x})\right|$
(3.28) $\displaystyle\leq
e^{3J_{\theta}}(\bar{\sigma}c+2m_{*})^{2}F^{\theta_{s}}(\hat{\gamma})\sum_{\hat{x}\in\hat{\gamma}}g_{t+s}(\hat{x})\int_{s}^{t+s}g_{u}(\hat{x})du$
$\displaystyle\leq
tJ_{\theta}e^{3J_{\theta}}(\bar{\sigma}c+2m_{*})^{2}\Psi_{0}(\hat{\gamma})\exp\left(-\bar{c}\Psi_{0}(\hat{\gamma})\right)$
$\displaystyle\leq
tJ_{\theta}e^{3J_{\theta}}(\bar{\sigma}c+2m_{*})^{2}/\bar{c}\to 0,\qquad t\to
0,$

which together with (3.27) yields

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}|Q_{t}(\hat{\gamma}|\to 0,\qquad t\to
0.$ (3.29)

Now we turn to estimating $R_{t}$. By 3.1) we have

$\displaystyle\left|\theta_{t+s}^{\prime}(x,\alpha)-\theta_{s}^{\prime}(x,\alpha)\right|\leq|\theta^{\prime}(x,\alpha+t+s)|\left|q_{t+s}(x,\alpha)-q_{s}(x,\alpha)\right|$
(3.30)
$\displaystyle+\left|\theta^{\prime}(x,\alpha+t+s)-\theta^{\prime}(x,\alpha+s)\right|+\left|m(x,\alpha+t+s)-m(x,\alpha+s)\right||\theta_{t+s}(x,\alpha)|$
$\displaystyle+\left|m(x,\alpha+s)-m(x,\alpha)\right||\theta_{s}(x,\alpha)|=:\delta_{1}(t,\hat{x})+\delta_{2}(t,\hat{x})+\delta_{3}(t,\hat{x})+\delta_{4}(t,\hat{x}).$

By (3.11) we have

$\displaystyle\delta_{1}(t,\hat{x})\leq
e^{-g(x,\alpha+t+s)}|g^{\prime}(x,\alpha+t+s)|\int_{\alpha+s}^{\alpha+t+s}m(x,\beta)d\beta$
(3.31) $\displaystyle\leq tm_{*}e^{J_{\theta}}(\bar{\sigma}c+2m_{*})g(x,0).$

To estimate $\delta_{2}$, by (3.4) we first get

$\theta^{\prime}(\hat{x})=-g^{\prime}(\hat{x})e^{-g(\hat{x})},$

by which we then obtain

$\displaystyle\delta_{2}(t,\hat{x})$ $\displaystyle\leq$
$\displaystyle\left|g^{\prime}(x,\alpha+t+s)-g^{\prime}(x,\alpha+s)\right|e^{-g(x,\alpha+t+s)}$
$\displaystyle+$
$\displaystyle|g^{\prime}(x,\alpha+s)|\left|e^{-g(x,\alpha+t+s)}-e^{-g(x,\alpha+s)}\right|$
$\displaystyle\leq$
$\displaystyle\left|g^{\prime}(x,\alpha+t+s)-g^{\prime}(x,\alpha+s)\right|$
$\displaystyle+$
$\displaystyle|g^{\prime}(x,\alpha+s)|\left|g(x,\alpha+t+s)-g(x,\alpha+s)\right|$
$\displaystyle=:$
$\displaystyle\delta_{2,1}(t,\hat{x})+\delta_{2,2}(t,\hat{x}).$

Now we recall that $g(x,\alpha)$ is as in (2.20) with $w_{k,n}$ defined in
(2.6). Thus, we can write

$\displaystyle\delta_{2,1}(t,\hat{x})\leq\sum_{j}v_{s_{j}}(x)\int_{s}^{t+s}|w^{\prime\prime}_{k_{j},n_{j}}(\alpha+u)|du.$
(3.32)

For each $k$ and $n$, we have

$\displaystyle|w^{\prime\prime}_{k,n}(\alpha)|=\left|-\sigma_{k}u_{n}^{\prime\prime}(\alpha)e^{-\sigma_{k}u_{n}(\alpha)}+[\sigma_{k}u_{n}^{\prime}(\alpha)]^{2}e^{-\sigma_{k}u_{n}(\alpha)}\right|$
$\displaystyle\leq\bar{\sigma}|u_{n}^{\prime\prime}(\alpha)|+\left[\bar{\sigma}u_{n}^{\prime}(\alpha)\right]^{2}\leq\bar{C},$

holding for some $\bar{C}>0$ that is independent of $k,n$ and $\alpha$. The
latter conclusion follows by (2.7) and the fact that
$|u^{\prime\prime}_{n}(\alpha)|\leq 2\phi(n\alpha^{3})$ with
$\phi(\beta)=(1+\beta^{2})/(1+\beta)^{3}$, $\beta\geq 0$. Then by (3.32) we
get

$\delta_{2,1}(t,\hat{x})\leq t\bar{C}g(x,0).$ (3.33)

At the same time, by (2.34) and (2.35) it follows that

$\displaystyle\delta_{2,2}(t,\hat{x})\leq|g^{\prime}(x,\alpha+s)|\int_{s}^{t+s}|g^{\prime}(x,\alpha+u)|du\leq
t(\bar{\sigma}c)^{2}J_{\theta}g(x,0),$

which together with (3.33) yields

$\delta_{2}(t,\hat{x})\leq
t\left[\bar{C}+(\bar{\sigma}c)^{2}J_{\theta}\right]g(x,0).$ (3.34)

Finally, (1.3) and (3.4), (3.13) we have

$\displaystyle\delta_{3}(t,\hat{x})\leq\varkappa(t)g_{t}(\hat{x})\leq\varkappa(t)g(x,0).$

The same estimate holds true also for $\delta_{4}(t,\hat{x})$. Then by (3.30)
and (3.31), (3.34) we have that

$\left|\theta_{t}^{\prime}(x,\alpha)-\theta^{\prime}(x,\alpha)\right|\leq\omega(t)g(x,0),\qquad\omega(t)\to
0,\ \ t\to 0.$

holding for some continuous function $\omega$ and all $\hat{x}\in\widehat{X}$.
Now we use this in (3.26) and obtain, cf. (3.28)

$\displaystyle|R_{t}(\hat{\gamma})|\leq\omega(t)e^{J_{\theta}}F^{\theta}(\hat{\gamma})\Psi_{0}(\hat{\gamma})\leq\omega(t)e^{J_{\theta}}\Psi_{0}(\hat{\gamma})e^{-\bar{c}\Psi_{0}(\hat{\gamma})}\leq\omega(t)e^{J_{\theta}}/\bar{c},$
(3.35)

which together with (3.29) yields

$\sup_{\hat{\gamma}\in\widehat{\Gamma}}\left|(L_{1}F^{\theta}_{t})(\hat{\gamma})-(L_{1}F^{\theta})(\hat{\gamma})\right|\to
0,\qquad t\to 0.$

This completes the whole proof. ∎

###  3.3. The domain

We recall that $\Theta$ consists of the functions as in (2.20) and the
countable collection $\mathcal{F}_{\Theta}\subset C_{\rm b}(\widehat{\Gamma})$
consists of the functions introduced in (2.23). It has a number of useful
properties established in Propositions 2.8 and 2.9. Let $\mathcal{C}_{0}$ be
the linear span of the set $\\{F^{\theta_{s}}:s\in\mathds{Q}_{+},\
\theta\in\Theta\\}$, i.e., each $F\in\mathcal{C}_{0}$ is a finite linear
combination of $F^{\theta_{s}}$, with positive rational $s$ and $\theta_{s}$
given in (3.1) with all possible choices of $\theta\in\Theta$.

######  Remark 3.3.

The set $\mathcal{C}_{0}$ is countable. It enjoys all the properties mentioned
in Proposition 2.9.

Now we set

$\mathcal{C}=\overline{\mathcal{C}_{0}},$ (3.36)

i.e., $\mathcal{C}$ is the closure of $\mathcal{C}_{0}$ in the norm of $C_{\rm
b}(\widehat{\Gamma})$, which we denote $\|\cdot\|$. With this norm it is then
a separable Banach space.

For $\lambda>0$ and $\theta\in\Theta$ and $s\geq 0$, we define, cf. (3.6),

$\displaystyle F_{\lambda,\theta_{s}}(\hat{\gamma})$ $\displaystyle=$
$\displaystyle\int_{0}^{+\infty}e^{-\lambda
t}F_{t}^{\theta_{s}}(\hat{\gamma})dt$ $\displaystyle=$
$\displaystyle\int_{0}^{+\infty}\exp\left[-\lambda
t+\int_{0}^{t}\left(\int_{X}\theta_{s}(x,\alpha)e^{-M(x,\alpha)}\chi(dx)\right)d\alpha\right]F^{\theta_{t+s}}(\hat{\gamma})dt.$

Since $\theta_{s}(x,\alpha)\leq 0$ and $F_{t}^{\theta_{s}}$ satisfies (3.7),
the above integral converges for each $\hat{\gamma}$. By the dominated
convergence theorem and the boundedness $F^{\theta_{s}}_{t}(\hat{\gamma})\leq
1$ it follows that $F_{\lambda,\theta_{s}}\in C_{\rm b}(\widehat{\Gamma})$.
Moreover, it can also be understood as the Bochner integral in the latter
Banach space. Therefore, $F_{\lambda,\theta}$ can be approximated in
$\|\cdot\|$ by the Riemann integral sums centered at rational $t$, which means
that

$F_{\lambda,\theta_{s}}\in\mathcal{C},\qquad{\rm for}\ {\rm all}\ s\geq 0\
{\rm and}\ \lambda>0.$ (3.38)

At the same time, we also have

$0<F_{\lambda,\theta_{s}}(\hat{\gamma})<1/\lambda,\qquad\hat{\gamma}\in\widehat{\Gamma}.$
(3.39)

By the dominated convergence theorem one readily proves that the map
$(0,+\infty)\ni\alpha\mapsto
F_{\lambda,\theta_{s}}(\hat{\gamma}\setminus\hat{x}\cup(x,\alpha))\in\mathds{R}$
is continuously differentiable for each $\hat{\gamma}$ and
$\hat{x}\in\hat{\gamma}$.

######  Lemma 3.4.

For each $\lambda>0$, $s\geq 0$ and $\theta\in\Theta$, the following holds

$LF_{\lambda,\theta_{s}}=\lambda F_{\lambda,\theta_{s}}-F^{\theta_{s}}.$
(3.40)

###### Proof.

By (3.6) and (3.3) we have

$\displaystyle LF_{\lambda,\theta_{s}}=L\int_{0}^{+\infty}e^{-\lambda
t}F^{\theta_{s}}_{t}dt=\int_{0}^{+\infty}e^{-\lambda t}LF^{\theta}_{t}dt$
(3.41) $\displaystyle=\int_{0}^{+\infty}e^{-\lambda t}\frac{\partial}{\partial
t}F^{\theta_{s}}_{t}dt=-F^{\theta_{s}}+\lambda F_{\lambda,\theta_{s}},$

where we have taken into account the upper bound in (3.39). The commutation
$L\int=\int L$ can be justified by means of the Lebesgue dominated convergence
theorem. ∎

######  Lemma 3.5.

For each $\theta\in\Theta$ and $s\geq 0$, it follows that $\|\lambda
F_{\lambda,\theta_{s}}-F^{\theta_{s}}\|\to 0$ and $\|\lambda
LF_{\lambda,\theta_{s}}-LF^{\theta_{s}}\|\to 0$ as $\lambda\to+\infty$.

###### Proof.

In view of (3.39), $\\{\lambda F_{\lambda,\theta_{s}}:\lambda>0\\}$ is
bounded. By (3.3) we have

$\displaystyle\lambda
F_{\lambda,\theta_{s}}(\hat{\gamma})=\int_{0}^{+\infty}\exp\left(-t+\int_{0}^{\varepsilon
t}\left(\int_{X}\theta_{\alpha+s}(x,0)\chi(dx)\right)d\alpha\right)F^{\theta_{\varepsilon
t+s}}(\hat{\gamma})dt,\quad\varepsilon:=\lambda^{-1}.$

Then by (3.17) it follows that

$\displaystyle\left|\lambda
F_{\lambda,\theta_{s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})\right|\leq\int_{0}^{+\infty}e^{-t}\left|F_{\varepsilon
t}^{\theta_{s}}(\hat{\gamma})-F^{\theta_{s}}(\hat{\gamma})\right|dt\leq\ell_{\theta}/\lambda,$

which yields that $\|\lambda F_{\lambda,\theta_{s}}-F^{\theta_{s}}\|\to 0$ as
$\lambda\to+\infty$. In the same way, by Proposition 3.2 we have, cf. (3.35),
that

$\|\lambda
LF_{\lambda,\theta_{s}}-LF^{\theta_{s}}\|\leq\tilde{\omega}(1/\lambda)\to
0,\quad\lambda\to+\infty,$

holding for an appropriate continuous $\tilde{\omega}$ such that
$\tilde{\omega}(\epsilon)\to 0$ as $\epsilon\to 0$. ∎

Let $\mathcal{D}_{0}(L)$ denote the linear span of the set
$\\{F_{\lambda,\theta_{s}}:\lambda>0,s\geq 0,\theta\in\varTheta\\}$. By Lemma
3.5 and (3.36) it follows that

$\mathcal{C}_{0}\subset\overline{\mathcal{D}_{0}(L)},$ (3.42)

i.e., $\mathcal{C}_{0}$ is contained in the closure of $\mathcal{D}_{0}(L)$ in
the norm of $C_{\rm b}(\widehat{\Gamma})$. Hence, $\mathcal{D}_{0}(L)$ is a
dense subset of the Banach space $\mathcal{C}$, see (3.36). Define

$\|F\|_{L}=\|F\|+\|LF\|,\qquad F\in\mathcal{D}_{0}(L),$ (3.43)

where – as above – $\|\cdot\|$ is the norm of $C_{\rm b}(\widehat{\Gamma})$.

######  Definition 3.6.

By the domain of the Kolmogorov operator $L$, denoted by $\mathcal{D}(L)$, we
mean the closure of $\mathcal{D}_{0}(L)$ in the graph-norm introduced in
(3.43).

######  Lemma 3.7.

The operator $(L,\mathcal{D}(L))$ is closed and densely defined in
$\mathcal{C}$. Its resolvent set contains $(0,+\infty)$ and
$\mathcal{C}_{0}\subset\mathcal{D}(L)$.

###### Proof.

In view of (3.38) and (3.42), the $\|\cdot\|$-closure of $\mathcal{D}(L)$ is
$\mathcal{C}$. The closedness of $(L,\mathcal{D}(L))$ is immediate, and the
inclusion $\mathcal{C}_{0}\subset\mathcal{D}(L)$ follows by the second part of
Lemma 3.5. By (3.40) it follows that the resolvent of $L$, denoted
$R_{\lambda}(L)$, has the property

$R_{\lambda}(L)F^{\theta_{s}}=F_{\lambda,\theta_{s}},\qquad\lambda>0,\ \quad
s\geq 0,\quad\theta\in\Theta,$

by which and (3.39) we also have that the operator norm of $R_{\lambda}(L)$
satisfies $\|R_{\lambda}(L)\|\leq 1/\lambda$ as $F^{\theta_{s}}$ form a dense
subset of $\mathcal{C}$. This completes the whole proof. ∎

###  3.4. Solving the Kolmogorov equation

The result obtained in Lemma 3.7 allows one to solve the Kolmogorov equation
(1.1) in the following sense.

######  Theorem 3.8.

Let $(L,\mathcal{D}(L))$ and $\mathcal{C}$ be as in Lemma 3.7. Then, for each
$F\in\mathcal{D}(L)$, there exists a unique continuously differentiable map
$[0,+\infty)\ni t\mapsto F_{t}\in\mathcal{D}(L)\subset\mathcal{C}$, which
solves (1.1) with $F_{0}=F$. In particular, for $F=F^{\theta_{s}}$, with
$F^{\theta_{s}}$ as in (3.1), $s\geq 0$ and $\theta\in\Theta$, the solution
has the following explicit form, cf. (3.6)

$F_{t}(\hat{\gamma})=\exp\left[\int_{0}^{t}\left(\int_{X}\theta_{s}(x,\alpha)\chi(dx)\right)d\alpha\right]F^{\theta_{t+s}}(\hat{\gamma}).$
(3.44)

###### Proof.

By the celebrated Hille-Yosida theorem, see, e.g, [12, page 8],
$(L,\mathcal{D}(L))$ is the generator of a $C_{0}$-semigroup of bounded linear
operators $S(t):\mathcal{C}\to\mathcal{C}$ such that $\|S(t)\|=1$ and the
solution in question is $F_{t}=S(t)F$, the uniqueness of which is also a
standard fact, see [12, Theorem 1.3, page 102]. The validity of (3.44) follows
by the calculations as in (3.2). ∎

##  4\. The Result

###  4.1. Poisson measures and transition functions

We begin by recalling, see Proposition 2.9, that the class of functions
$\mathcal{F}_{\Theta}$, see (2.23), is separating, i.e., if
$\mu_{1}(F)=\mu_{2}(F)$ for all $F\in\mathcal{F}_{\Theta}$, then
$\mu_{1}=\mu_{2}$, that holds for each pair
$\mu_{1},\mu_{2}\in\mathcal{P}(\widehat{\Gamma})$. Next, for $t\geq 0$,
$\mu\in\mathcal{P}(\widehat{\Gamma})$ and $\theta\in\Theta$, we define
$\mu^{t}\in\mathcal{P}({\widehat{\Gamma}})$ by the relation

$\mu^{t}(F^{\theta})=\mu(F^{\theta_{t}}).$ (4.1)

Recall that, for a positive Radon measure $\varrho$ on $\widehat{X}$, the
Poisson measure $\pi_{\varrho}$ with intensity measure $\varrho$ is defined in
(2.24). For $t\geq 0$, we then introduce the Poisson measure
$\pi_{t}=\pi_{\varrho_{t}}$ by defining its intensity measure

$\varrho_{t}(d\hat{x})=\mathds{1}_{[0,t)}(\alpha)\exp\left(-M(\hat{x})\right)\chi(dx)d\alpha,$
(4.2)

where $\chi$ is the same as in (1), $d\alpha$ is the Lebesgue measure on
$\mathds{R}_{+}$ and $M$ is as in (3.1). Note that
$\pi_{0}(\\{\varnothing\\})=1$ since $\pi_{0}(F^{\theta})=1$ for all $\theta$.
Then we set, see (2.25),

$\mu_{t}=\pi_{t}\star\mu^{t},\qquad t\geq 0.$ (4.3)

Let $\delta_{\hat{\gamma}}$ be the Dirac measure centered at a given
$\hat{\gamma}\in\widehat{\Gamma}$. Then

$p^{\hat{\gamma}}_{t}:=\pi_{t}\star\delta^{t}_{\hat{\gamma}}$ (4.4)

is a transition function, cf. [5, page 156]. Indeed,
$p^{\hat{\gamma}}_{t}\in\mathcal{P}(\widehat{\Gamma})$,
$p^{\hat{\gamma}}_{0}=\delta_{\hat{\gamma}}$ and the measurability of the map
$(t,\hat{\gamma})\mapsto p^{\hat{\gamma}}_{t}(B)\in\mathds{R}$,
$B\in\mathcal{B}(\widehat{\Gamma})$, follows by the measurability of
$(t,\hat{\gamma})\mapsto\delta_{\hat{\gamma}}^{t}(B)\in\mathds{R}$ and the
continuity of $t\mapsto\pi_{t}(B)\in\mathds{R}$. In view of the separating
property of $\mathcal{F}_{\varTheta}$, see Proposition 2.9, item (iii), the
flow property of $\\{p^{\hat{\gamma}}_{t}\\}_{t\geq 0}$ can be obtained by
showing that

$p^{\hat{\gamma}}_{t+s}(F^{\theta})=\int_{\widehat{\Gamma}_{*}}p^{\hat{\gamma}^{\prime}}_{t}(F^{\theta})p^{\hat{\gamma}}_{s}(d\hat{\gamma}^{\prime}),\qquad
t,s\geq 0,\ \ \theta\in\varTheta.$ (4.5)

By (2.25) we have

$\displaystyle
p_{t}^{\hat{\gamma}^{\prime}}(F^{\theta})=\exp\left(\int_{X}\int_{0}^{t}\theta_{\alpha}(x,0)d\alpha\chi(dx)\right)\delta^{t}_{\hat{\gamma}^{\prime}}(F^{\theta})$
$\displaystyle=\exp\left(\int_{X}\int_{0}^{t}\theta_{\alpha}(x,0)d\alpha\chi(dx)\right)F^{\theta_{t}}(\hat{\gamma}^{\prime}),$

see also (4.1) and (3.1). By the latter formula and (4.2), (4.4) we then get

$\displaystyle{\rm RHS(\ref{Q12})}$ $\displaystyle=$
$\displaystyle\exp\left(\int_{X}\int_{0}^{t}\theta_{\alpha}(x,0)d\alpha\chi(dx)+\int_{X}\int_{0}^{s}\theta_{t+\alpha}(x,0)d\alpha\chi(dx)\right)\delta^{s}_{\hat{\gamma}}(F^{\theta_{t}})\qquad$
$\displaystyle=$
$\displaystyle\exp\left(\int_{X}\int_{0}^{t+s}\theta_{\alpha}(x,0)d\alpha\chi(dx)\right)F^{\theta_{t+s}}(\hat{\gamma})={\rm
LHS(\ref{Q12})}.$

As is known, cf. [5, Theorem 1.2.page 157], the transition function (4.4)
determines a Markov process, $\mathcal{X}$, with values in $\widehat{\Gamma}$,
the finite-dimensional distributions of which are given by the following
formula

$\displaystyle P(\mathcal{X}(s_{1})\in B_{1},\dots,\mathcal{X}(s_{n})\in
B_{n})$ (4.6)
$\displaystyle=\int_{\widehat{\Gamma}}\int_{B_{1}}\cdots\int_{B_{n-1}}p_{s_{n}-s_{n-1}}^{\hat{\gamma}_{n-1}}(B_{n})p_{s_{n-1}-s_{n-2}}^{\hat{\gamma}_{n-2}}(d\hat{\gamma}_{n-1})$
$\displaystyle\times\cdots\times
p_{s_{2}-s_{1}}^{\hat{\gamma}_{1}}(d\hat{\gamma}_{2})p_{s_{1}}^{\hat{\gamma}}(d\hat{\gamma}_{1})\mu(d\hat{\gamma}),$

holding for all $n\in\mathds{N}$, $0<s_{1}\leq s_{2}\leq\cdots\leq s_{n}$ and
$B_{i}\in\mathcal{B}(\widehat{\Gamma})$. Here
$\mu\in\mathcal{P}(\widehat{\Gamma})$ is the initial distribution of
$\mathcal{X}$. Our aim is to show that such a process is unique up to
modifications.

###  4.2. The statement

The process determined by (4.6) describes the stochastic evolution of the
population which we consider. To verify whether it is the only one, we have to
specify which processes of this kind can be associated to the model defined by
the Kolmogorov operator (1). As is standard, the corresponding specification
is made by their martingale property, see [5, Chgapter 4].

######  Definition 4.1.

Let $\mathcal{X}$ be a measurable process on some probability space
$(\Omega,\mathfrak{F},P)$ with values in $\widehat{\Gamma}$. Let also
$\\{\mathfrak{F}_{t}\\}_{t\geq 0}$ be a filtration such that $\mathcal{X}(t)$
and

$\int_{0}^{t}G(\mathcal{X}(u))du$

are $\mathfrak{F}_{t}$-measurable for all $t$ and $G\in B(\widehat{\Gamma})$.
We say that $\mathcal{X}$ is a solution of the martingale problem for
$(L,\mathcal{D}(L))$ if for each $F\in\mathcal{D}(L)$,

$\mathcal{M}(t):=F(\mathcal{X}(t))-\int_{0}^{t}(LF)(\mathcal{X}(u))du$

is a $\mathfrak{F}_{t}$-martingale. If there exists a solution of the
martingale problem for $(L,\mathcal{D}(L))$ and uniqueness holds, we say that
the problem is well-posed. In the same way, we define the martingale problem
for $(L,\mathcal{D}(L),\mu)$ if the initial distribution
$\mu\in\mathcal{P}(\widehat{\Gamma})$ is specified.

The process related to the transition function (4.4) solves the martingale
problem for $(L,\mathcal{D}(L))$. Its uniqueness will be shown by proving that
all other solutions have the same finite-dimensional marginals, i.e., they
coincide with those defined in (4.6). We are going also to show that the
solution is temporarily ergodic.

######  Definition 4.2.

Let the martingale problem for $(L,\mathcal{D}(L),\mu)$ be well-posed. Then
$\mu\in\mathcal{P}(\widehat{\Gamma})$ is said to be a stationary distribution
if for each $n$ and $0<s_{1}<s_{2}<\cdots<s_{n}$, the $n$-dimensional
marginals introduced in (4.6) corresponding to $t+s_{1},\dots,t+s_{n}$ are
independent of $t\geq 0$.

If $P$ is as in (4.6), then $\mu$ is stationary if and only if

$\mu=\int_{\widehat{\Gamma}}p_{s}^{\hat{\gamma}}\mu(d\hat{\gamma}),$

holding for all $t>0$. Now we can formulate our result.

######  Theorem 4.3.

The martingale problem for $(L,\mathcal{D}(L))$ is well-posed in the sense of
Definition 4.1. Its solution is defined by finite-dimensional marginals, see
(4.6), with the transition function defined in (4.4). If - in addition to
Assumption 1.1 \- the departure function satisfies $m(\hat{x})\geq m_{0}>0$,
holding for all $\hat{x}$ and some $m_{0}$, then there exists a unique
stationary distribution $\mu=\pi_{\varrho}$, which is the Poisson measure with
intensity measure

$\varrho(d\hat{x})=\exp\left(-M(\hat{x})\right)\chi(dx)d\alpha.$ (4.7)

Moreover, in this case the solution $\mathcal{X}$ of the martingale problem
for $(L,\mathcal{D}(L),\mu)$ is temporarily ergodic in the following sense.
Let $\mu_{t}\in\mathcal{P}(\widehat{\Gamma})$ be the law of $\mathcal{X}(t)$,
$t\geq 0$. Then $\mu_{t}\Rightarrow\pi_{\varrho}$ as $t\to+\infty$.

##  5\. The Proof

The proof of Theorem 4.3 is divided into the following steps:

  * (a)

Proving uniqueness.

  * (b)

Showing the stationarity and ergodicity if $m(\hat{x})\geq m_{*}>0$.

The realization of (a) is based on the Fokker-Planck equation for $L$, which
is a weak version of the forward Kolmogorov equation.

###  5.1. The Fokker-Planck equation

The Fokker-Planck equation is in a sense dual to the Kolmogorov equation
(1.1), see [1] for a general results on such equations. It describes the
evolution of states of our model and reads

$\mu_{t}(F)=\mu_{0}(F)+\int_{0}^{t}\mu_{s}(LF)ds,\qquad\mu_{0}\in\mathcal{P}(\widehat{\Gamma}).$
(5.1)

######  Definition 5.1.

By a solution of (5.1) we understand a map $\mathds{R}_{+}\ni
t\mapsto\mu_{t}\in\mathcal{P}(\widehat{\Gamma})$ possessing the following
properties: (a) for each $F\in C_{\rm b}(\widehat{\Gamma})$, the map
$\mathds{R}_{+}\ni t\mapsto\mu_{t}(F)\in\mathds{R}$ is measurable; (b) the
equality in (5.1) holds for all $F\in\mathcal{D}(L)$.

Note that, for each solution $\mu_{t}$ and $F\in\mathcal{D}(L)$, the map
$\mathds{R}_{+}\ni t\mapsto\mu_{t}(F)$ is Lipschitz-continuous. Hence, it is
almost everywhere differentiable, and its derivative is $\mu_{t}(LF)$.
Therefore

$\mu_{t_{2}}(F)=\mu_{t_{1}}(F)+\int_{t_{1}}^{t_{2}}\mu_{s}(LF)ds,\qquad 0\leq
t_{1}<t_{2}.$

It turns out that, for our model, one can construct solutions of (5.1)
explicitly, which we are going to realize now.

######  Lemma 5.2.

For each $\mu_{0}\in\mathcal{P}(\widehat{\Gamma})$, the map $t\mapsto\mu_{t}$
defined in (4.3) is a unique solution of (5.1).

###### Proof.

For each $\theta\in\Theta$, the map
$t\mapsto\mu^{t}(F^{\theta})=\mu_{0}(F^{\theta_{t}})$ is continuous (by the
dominated convergence theorem) and hence measurable. By (2.26) we then have

$\displaystyle\mu_{t}(F^{\theta})=\pi_{t}(F^{\theta})\mu^{t}(F^{\theta})=\pi_{t}(F^{\theta})\mu_{0}(F^{\theta_{t}})$
$\displaystyle\quad=\exp\left(\int_{0}^{t}\left[\int_{X}\theta(x,\alpha)e^{-M(x,\alpha)}\chi(dx)\right]d\alpha\right)\mu_{0}(F^{\theta_{t}}).$

Thus, the map $t\mapsto\mu_{t}(F^{\theta})$ is continuous and hence
measurable. Then the measurability of $t\mapsto\mu_{t}(F)$ for all $F\in
C_{\rm b}(\widehat{\Gamma})$ follows by claim (ii) of Proposition 2.9. Now we
turn to proving the equality in (5.1) for $F=F_{\lambda,\theta}$,
$\theta\in\varTheta$, see Definition 3.6. By (4.3) and (3.6) for $s,t\geq 0$
and $\theta\in\varTheta$, we have

$\displaystyle\mu_{t}(F^{\theta}_{s})=\exp\left(\int_{0}^{s}\left(\int_{X}\theta_{\alpha}(x,0)\chi(dx)\right)d\alpha\right)\pi_{t}(F^{\theta_{s}})\mu^{t}(F^{\theta_{s}})$
$\displaystyle=\exp\left(\int_{0}^{s}\left(\int_{X}\theta_{\alpha}(x,0)\chi(dx)\right)d\alpha+\int_{0}^{t}\left(\int_{X}\theta_{s+\alpha}(x,0)\chi(dx)\right)d\alpha\right)\mu_{0}(F^{\theta_{s+t}})$
$\displaystyle=\exp\left(\int_{0}^{s+t}\left(\int_{X}\theta_{\alpha}(x,0)\chi(dx)\right)d\alpha\right)\mu_{0}(F^{\theta_{s+t}})=\mu_{0}(F^{\theta}_{s+t}).$

Then by Fubini’s theorem and the latter fact we get

$\displaystyle\mu_{t}(F_{\lambda,\theta})-\mu_{0}(F_{\lambda,\theta})=\int_{0}^{+\infty}e^{-\lambda
s}\left[\mu_{t}(F^{\theta}_{s})-\mu_{0}(F^{\theta}_{s})\right]ds$
$\displaystyle\ \ =\int_{0}^{+\infty}e^{-\lambda
s}\mu_{0}(F^{\theta}_{t+s}-F^{\theta}_{s})ds=\int_{0}^{+\infty}e^{-\lambda
s}\int_{0}^{t}\frac{\partial}{\partial u}\mu_{0}(F^{\theta}_{s+u})dsdu$
$\displaystyle\ \ =\int_{0}^{+\infty}e^{-\lambda
s}\int_{0}^{t}\frac{\partial}{\partial
s}\mu_{u}(F^{\theta}_{s})dsdu=\int_{0}^{t}\mu_{u}\left(\int_{0}^{+\infty}e^{-\lambda
s}\frac{\partial}{\partial s}F^{\theta}_{s}ds\right)du$ $\displaystyle\ \
=\int_{0}^{t}\mu_{u}(LF_{\lambda,\theta})du,$

where we have used also (3.41). Now we prove uniqueness by applying arguments
similar to those used in [2, Lemma 2.11]. Assume that a map $t\mapsto\mu_{t}$
satisfies condition (a) Definition 5.1 and $F,G\in C_{\rm
b}(\widehat{\Gamma})$ are such that

$\mu_{t}(F)-\mu_{0}(F)=\int_{0}^{t}\mu_{s}(G)ds,$

holding for all $t\geq 0$. Then the map $t\mapsto\mu_{t}(F)$ is almost
everywhere differentiable and

$d\mu_{t}(F)=\mu_{t}(G)dt.$

Then integrating by parts we get

$\displaystyle-\lambda\int_{0}^{t}e^{-\lambda s}\mu_{s}(F)ds=e^{-\lambda
t}\mu_{t}(F)-\mu_{0}(F)-\int_{0}^{t}e^{-\lambda s}\mu_{s}(G)ds,$

which yields

$\mu_{0}(F)=e^{-\lambda t}\mu_{t}(F)+\int_{0}^{t}e^{-\lambda
s}\left[\lambda\mu_{s}(F)-\mu_{s}(G)\right]ds,$

holding for all $t,\lambda>0$. Passing here to the limit $t\to+\infty$, for
$F=F_{\lambda,\theta}$ and $G=LF_{\lambda,\theta}$, see (5.1), we arrive at

$\mu_{0}(F_{\lambda,\theta})=\int_{0}^{+\infty}e^{-\lambda s}\mu_{s}(\lambda
F_{\lambda,\theta}-LF_{\lambda,\theta})ds=\int_{0}^{+\infty}e^{-\lambda
s}\mu_{s}(F^{\theta})ds,$ (5.2)

see (3.40). Assume now that (5.1) has two solutions, $\mu_{t}$ and
$\tilde{\mu}_{t}$, satisfying the same initial condition
$\mu_{t}|_{t=0}=\tilde{\mu}_{t}|_{t=0}=\mu_{0}$. By (5.2) the Laplace
transforms of both maps $t\mapsto\mu_{t}(F^{\theta})$ and
$t\mapsto\tilde{\mu}_{t}(F^{\theta})$ coincide, which yields
$\mu_{t}(F^{\theta})=\tilde{\mu}_{t}(F^{\theta})$ holding for each $t$ and all
$F^{\theta}$, $\theta\in\Theta$. Then the uniqueness in question follows by
Proposition 2.9. This completes the whole proof. ∎

###  5.2. Completing the proof

The existence of a solution of the martingale problem for $(L,\mathcal{D}(L))$
was shown by the very construction of the finite-dimensional marginals of
$\mathcal{X}$ in (4.6). To prove uniqueness we use the following fact, see [5,
Proposition 4.2, page 184]. Given $\mu\in\mathcal{P}(\widehat{\Gamma})$, let
$\mathcal{X}$ and $\mathcal{X}^{\prime}$ be solutions of the martingale
problem for $(L,\mathcal{D}(L),\mu)$ whose one-dimensional marginals,
$\mu_{t}$ and $\mu^{\prime}_{t}$, coincide for all $t\geq 0$. Then all their
finite-dimensional marginals coincide and hence the problem is well-posed.
Clearly, both $\mu_{t}$ and $\mu^{\prime}_{t}$ solve the Fokker-Planck
equation with the initial condition $\mu$. Then they coincide by Lemma 5.2.
This yields well-posedness.

Now we show the stated ergodicity. If $m$ satisfies $m(\hat{x})\geq m_{0}>0$,
then $M(x,\alpha)\geq m_{0}\alpha$, see (3.1), which for $\theta$ as in (3.1)
yields

$\forall t>0\qquad|\theta_{t}(\hat{x})|\leq e^{-tm_{0}}.$

By the continuity of the map $t\mapsto\mu(F^{\theta_{t}})$ we then get

$\displaystyle\mu_{t}(F^{\theta})$ $\displaystyle=$
$\displaystyle\exp\left(\int_{0}^{t}\left[\theta_{\alpha}(x,0)\chi(dx)\right]d\alpha\right)\mu(F^{\theta_{t}})$
$\displaystyle\to$
$\displaystyle\exp\left(\int_{0}^{+\infty}\left[\int_{X}\theta_{\alpha}(x,0)\chi(dx)\right]d\alpha\right),\quad
t\to+\infty.$

By claim (iv) of Proposition 2.9 this yields
$\mu_{t}\Rightarrow\pi_{\varrho}$, see (4.7), holding for each initial
$\mu\in\mathcal{P}(\widehat{\Gamma})$. Clearly, $\mu_{t}=\pi_{\varrho}$ if
$\mu=\pi_{\varrho}$, which means that $\pi_{\varrho}$ is a stationary state.
If there exists another stationary state, say $\mu^{\prime}$, then (5.2) fails
to hold for $\mu=\mu^{\prime}$, which contradicts the convergence just
established. This completes the proof of Theorem 4.3.

## Acknowledgements

The research of both authors was financially supported by National Science
Centre, Poland, grant 2017/25/B/ST1/00051, that is cordially acknowledged by
them.

## Appendix

Here we prove that $\tilde{r}$ introduced in (2.1) satisfies (2.2). During the
whole proof, we deal with the function $\ell(\alpha)=\alpha+1/\alpha\geq
2=\ell(1)$, $\alpha>0$, which is decreasing for $\alpha\leq 1$ and increasing
for $\alpha\geq 1$.

First, consider the case of $\underline{\alpha_{2}=0}$, where
$\tilde{r}(\alpha_{1},\alpha_{2})=\omega(\alpha_{1})$. For $\alpha_{i}\leq 1$,
$i=1,3$, it follows that $\omega(\alpha_{i})=\alpha_{i}$ and
$\tilde{r}(\alpha_{1},\alpha_{3})=|\alpha_{3}-\alpha_{1}|$. Hence, (2.2) turns
into

$\alpha_{1}\leq|\alpha_{3}-\alpha_{1}|+\alpha_{3},$

which obviously holds true. For $\alpha_{3}>1$, we have the following two
possibilities: (a) $\alpha_{3}-\alpha_{1}\leq\alpha_{1}+1/\alpha_{3}$; (b)
$\alpha_{3}-\alpha_{1}>\alpha_{1}+1/\alpha_{3}$. In (a), we get

$\displaystyle\omega(\alpha_{1})=\alpha_{1}\leq\alpha_{3}-\alpha_{1}+1/\alpha_{3}=\tilde{r}(\alpha_{1},\alpha_{3})+\omega(\alpha_{3}),$
(5.4)

which yields the inequality in question for this case since
$2\alpha_{1}\leq\ell(\alpha_{3})$. Note that the equality in (5.4) is possible
only if $\alpha_{1}=\alpha_{3}=1$. In (b), we have
$\alpha_{1}\leq\alpha_{1}+2/\alpha_{3}$, which completes the proof of (2.2)
for this case. For $\alpha_{1}>1$ and $\alpha_{3}\leq 1$, we have the
following possibilities: (a)
$\alpha_{1}-\alpha_{3}\leq\alpha_{3}+1/\alpha_{1}$; (b)
$\alpha_{1}-\alpha_{3}>\alpha_{3}+1/\alpha_{1}$. In (a), we have

$\displaystyle\omega(\alpha_{1})=1/\alpha_{1}\leq\alpha_{1}=\alpha_{1}-\alpha_{3}+\alpha_{3},$

which yields (2.2) in this case. In case (b), we have

$\omega(\alpha_{1})=1/\alpha_{1}\leq\alpha_{3}+1/\alpha_{1}+\alpha_{3}.$

It remains to consider the cases: (i) $\alpha_{1}\geq\alpha_{3}>1$; (ii)
$1<\alpha_{1}<\alpha_{3}$. For $\alpha_{1}-\alpha_{3}\leq
1/\alpha_{1}+1/\alpha_{3}$, in (i) we have

$\omega(\alpha_{1})=1/\alpha_{1}\leq
1/\alpha_{3}\leq\alpha_{1}-\alpha_{3}+1/\alpha_{3},$

which proves (2.2) for this case. For
$\alpha_{1}-\alpha_{3}>1/\alpha_{1}+1/\alpha_{3}$, in (i) we have

$\omega(\alpha_{1})=1/\alpha_{1}\leq 1/\alpha_{1}+1/\alpha_{3}+1/\alpha_{3},$

which again yields (2.2). For $\alpha_{3}-\alpha_{1}\leq
1/\alpha_{1}+1/\alpha_{3}$, in (ii) we have to prove
$1/\alpha_{1}\leq\alpha_{3}-\alpha_{1}+1/\alpha_{3}$, which is equivalent to
$\ell(\alpha_{1})\leq\ell(\alpha_{3})$. The latter follows by
$\alpha_{3}\geq\alpha_{1}>1$ as $\ell$ is increasing. For
$\alpha_{3}-\alpha_{1}>1/\alpha_{1}+1/\alpha_{3}$, the proof of (2.2) f is
immediate.

Now we consider the case $\underline{0<\alpha_{1}<\alpha_{2}}$. If
$\underline{\alpha_{2}\leq 1}$ and $\alpha_{3}=0$, then
$\alpha_{2}-\alpha_{1}\leq\omega(\alpha_{1})+\omega(\alpha_{2})=\alpha_{1}+\alpha_{2}$.
For $\alpha_{3}\in(0,1]$, it follows that
$\tilde{r}(\alpha_{3},\alpha_{i})=|\alpha_{3}-\alpha_{i}|$. Then (2.2) turns
into the triangle inequality for $|\cdot|$. The same is true also for
$\alpha_{3}>1$ such that $\alpha_{3}-1/\alpha_{3}\leq 2\alpha_{1}$. For
$2\alpha_{1}<\alpha_{3}-1/\alpha_{3}\leq 2\alpha_{2}$, the right-hand side of
(2.2) is $\alpha_{3}-\alpha_{2}+\alpha_{1}+1/\alpha_{3}$. Then
$\alpha_{2}-\alpha_{1}\leq{\rm RHS}(\ref{QH})$ turns into
$2(\alpha_{2}-\alpha_{1})\leq\ell(\alpha_{3})$, which holds since
$2(\alpha_{2}-\alpha_{1})\leq 2<\ell(\alpha_{3})$ for $\alpha_{3}>1$. For
$2\alpha_{2}<\alpha_{3}-1/\alpha_{3}$, the right-hand side of (2.2) is
$\alpha_{1}+\alpha_{2}+2/\alpha_{3}$, which is bigger than
$\alpha_{2}-\alpha_{1}$.

Consider now $\underline{0<\alpha_{1}\leq 1<\alpha_{2}}$ and
$\underline{\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}}$. The latter means that
$\tilde{r}(\alpha_{2},\alpha_{1})=\alpha_{2}-\alpha_{1}$. For $\alpha_{3}=0$,
the right-hand side of (2.2) is $\alpha_{1}+1/\alpha_{2}$, and the latter
turns into $\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}$, which holds in this
case. The same is true for $2\alpha_{3}\leq\alpha_{2}-1/\alpha_{2}$. For
$\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{3}\leq 2\alpha_{1}$, the right-hand side
of (2.2) is $\alpha_{2}-\alpha_{3}+\alpha_{1}-\alpha_{3}$, which is bigger
than $\alpha_{2}-\alpha_{1}$ as $\alpha_{3}\leq\alpha_{1}$. For
$\alpha_{1}\leq\alpha_{3}\leq 1$, the right-hand side of (2.2) is
$\alpha_{2}-\alpha_{3}+\alpha_{3}-\alpha_{1}={\rm LHS}(\ref{QH})$, Next,
consider $1<\alpha_{3}\leq\alpha_{2}$, where
$\alpha_{3}-1/\alpha_{3}\leq\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}$. For
$\alpha_{2}-1/\alpha_{2}\leq\ell(\alpha_{3})$, the right-hand side of (2.2) is
$\alpha_{2}-\alpha_{3}+\alpha_{3}-\alpha_{1}={\rm RHS}(\ref{QH})$. The case of
$\alpha_{2}-1/\alpha_{2}>\ell(\alpha_{3})>2$ is impossible since
$\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}\leq 2$. For $\alpha_{3}>\alpha_{2}$
such that $\alpha_{3}-1/\alpha_{3}\leq 2\alpha_{1}$ and
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{2})$, we have
$\tilde{r}(\alpha_{3},\alpha_{1})=\alpha_{3}-\alpha_{1}$ and
$\tilde{r}(\alpha_{3},\alpha_{2})=\alpha_{3}-\alpha_{2}$. Then the right-hand
side of (2.2) is
$\alpha_{3}-\alpha_{1}+\alpha_{3}-\alpha_{2}=2\alpha_{3}-(\alpha_{2}+\alpha_{1})$,
which is bigger than $\alpha_{2}-\alpha_{1}$. The case of
$\alpha_{3}-1/\alpha_{3}>\ell(\alpha_{2})$ is impossible for
$\alpha_{3}-1/\alpha_{3}\leq 2\alpha_{1}$. For $\alpha_{3}>\alpha_{2}$ such
that $\alpha_{3}-1/\alpha_{3}>2\alpha_{1}$ and
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{2})$, we have
$\tilde{r}(\alpha_{3},\alpha_{1})=\alpha_{1}+1/\alpha_{3}$ and
$\tilde{r}(\alpha_{3},\alpha_{2})=\alpha_{3}-\alpha_{2}$. Then the right-hand
side of (2.2) is $\ell(\alpha_{3})-\alpha_{2}+\alpha_{1}$, which yields (2.2)
in the form $2(\alpha_{2}-\alpha_{1})\leq\ell(\alpha_{3})$. By
$\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}$, $\alpha_{2}>1$, we have that
$2(\alpha_{2}-\alpha_{1})\leq 2\sqrt{1+\alpha_{1}^{2}}$, whereas
$\alpha_{3}-1/\alpha_{3}>2\alpha_{1}$ yields
$\ell(\alpha_{3})>\ell(\alpha_{*})=2\sqrt{1+\alpha_{1}^{2}}$, which proves
(2.2) in this case. Here $\alpha_{*}$ is the positive solution of
$\alpha-1/\alpha=2\alpha_{1}$. For $\alpha_{3}>\alpha_{2}$ such that
$\alpha_{3}-1/\alpha_{3}>2\alpha_{1}$ and
$\alpha_{3}-1/\alpha_{3}>\ell(\alpha_{2})$, we have
$\tilde{r}(\alpha_{3},\alpha_{1})=\alpha_{1}+1/\alpha_{3}$ and
$\tilde{r}(\alpha_{3},\alpha_{2})=1/\alpha_{3}+1/\alpha_{2}$. Then (2.2) turns
to

$\alpha_{2}-\alpha_{1}\leq\alpha_{1}+2/\alpha_{3}+1/\alpha_{2},$

which holds as $\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}$.

Consider now $\underline{0<\alpha_{1}\leq 1<\alpha_{2}}$ and
$\underline{\alpha_{2}-1/\alpha_{2}>2\alpha_{1}}$. The latter means that
$\tilde{r}(\alpha_{1},\alpha_{2})=\alpha_{1}+1/\alpha_{2}$. For
$\alpha_{3}\in[0,\alpha_{1}]$, we have that
$\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{1}-\alpha_{3}$ and
$\tilde{r}(\alpha_{2},\alpha_{3})=\alpha_{3}+1/\alpha_{2}$. Hence, (2.2) turns
into equality. For $\alpha_{3}\in(\alpha_{1},1]$ such that
$\alpha_{2}-1/\alpha_{2}>2\alpha_{3}$, we have the right-hand side of (2.2) in
the following form $\alpha_{3}-\alpha_{1}+\alpha_{3}+1/\alpha_{2}$, which is
bigger that $\tilde{r}(\alpha_{1},\alpha_{2})$ since $\alpha_{1}<\alpha_{3}$.
For $\alpha_{3}\in(\alpha_{1},1]$ such that $\alpha_{2}-1/\alpha_{2}\leq
2\alpha_{3}$, we have that the right-hand side of (2.2) is
$\alpha_{3}-\alpha_{1}+\alpha_{2}-\alpha_{3}=\alpha_{2}-\alpha_{1}>\tilde{r}(\alpha_{1},\alpha_{2})$.
Consider now $\alpha_{3}>1$ such that $\alpha_{3}-1/\alpha_{3}\leq
2\alpha_{1}$, which means that $\alpha_{3}<\alpha_{2}$ and
$\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{3}-\alpha_{1}$. For
$\tilde{r}(\alpha_{2},\alpha_{3})=\alpha_{2}-\alpha_{3}$, the right-hand side
of (2.2) is
$\alpha_{2}-\alpha_{3}+\alpha_{3}-\alpha_{1}\geq\tilde{r}(\alpha_{2},\alpha_{3})$.
For $\tilde{r}(\alpha_{2},\alpha_{3})=1/\alpha_{3}+1/\alpha_{2}$, the right-
hand side of (2.2) is $\alpha_{3}-\alpha_{1}+1/\alpha_{2}+1/\alpha_{3}$;
hence, (2.2) turns into $2\alpha_{1}\leq\ell(\alpha_{3})$, which holds since
$\alpha_{1}\leq 1$ and $\ell(\alpha_{3})\geq 2$. For $\alpha_{3}>1$ such that
$\alpha_{3}-1/\alpha_{3}>2\alpha_{1}$, we have that
$\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{1}+1/\alpha_{3}$. Then (2.2) turns
into

$1/\alpha_{2}\leq 1/\alpha_{3}+\tilde{r}(\alpha_{2},\alpha_{3}),$ (5.5)

which clearly holds for $\alpha_{3}\leq\alpha_{2}$, and also for
$\alpha_{3}>\alpha_{2}$, where for
$\tilde{r}(\alpha_{2},\alpha_{3})=\alpha_{3}-\alpha_{2}$ it turns into
$\ell(\alpha_{2})\leq\ell(\alpha_{3})$ – which is true as $\ell(\alpha)$ is
increasing for $\alpha>1$. For
$\tilde{r}(\alpha_{2},\alpha_{3})=1/\alpha_{3}+1/\alpha_{2}$, the validity
(5.5) is immediate.

Let us consider now the case of $\underline{1<\alpha_{1}<\alpha_{2}}$ and
$\underline{\ell(\alpha_{1})\leq\alpha_{2}-1/\alpha_{2}}$, where
$\tilde{r}(\alpha_{2},\alpha_{1})=1/\alpha_{1}+1/\alpha_{2}$. For
$\alpha_{3}\leq 1$ such that $2\alpha_{3}\leq\alpha_{1}-1/\alpha_{1}$, we have
$\tilde{r}(\alpha_{3},\alpha_{i})=\alpha_{3}+1/\alpha_{i}$, $i=1,2$. Then
(2.2) obviously holds. For $\alpha_{3}\leq 1$ satisfying
$\alpha_{1}-1/\alpha_{1}<2\alpha_{3}\leq\alpha_{2}-1/\alpha_{2}$, we have
$\tilde{r}(\alpha_{3},\alpha_{1})=\alpha_{1}-\alpha_{3}$ and
$\tilde{r}(\alpha_{3},\alpha_{2})=\alpha_{3}+1/\alpha_{2}$. Hence, (2.2) turns
into equality in this case. The remaining case
$\alpha_{2}-1/\alpha_{2}<2\alpha_{3}\leq 2$ is impossible since
$\alpha_{2}-1/\alpha_{2}\geq\ell(\alpha_{1})>2$. For $\alpha_{3}>1$ such that
$2\alpha_{3}\leq\alpha_{1}-1/\alpha_{1}$, we have
$\tilde{r}(\alpha_{3},\alpha_{i})=1/\alpha_{3}+1/\alpha_{i}$, $i=1,2$. Then
(2.2) obviously holds. For $\alpha_{3}\in(1,\alpha_{1}]$ satisfying
$\alpha_{1}-1/\alpha_{1}<2\alpha_{3}\leq\alpha_{2}-1/\alpha_{2}$, (2.2) turns
into

$1/\alpha_{1}+1/\alpha_{2}\leq\alpha_{1}-\alpha_{3}+1/\alpha_{2}+1/\alpha_{3},$

which holds for $\alpha_{3}\leq\alpha_{1}$ as the function $\alpha-1/\alpha$
is increasing. For $\alpha_{3}\in(\alpha_{1},\alpha_{2}]$ such that
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{1})<\ell(\alpha_{3})<\alpha_{2}-1/\alpha_{2}$,
the right-hand side of (2.2) is
$\alpha_{3}-\alpha_{1}+1/\alpha_{3}+1/\alpha_{2}$, and hence the latter turns
into $\ell(\alpha_{1})<\ell(\alpha_{3})$. For
$\alpha_{3}\in(\alpha_{1},\alpha_{2}]$ satisfying
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{1})$ and
$\alpha_{2}-1/\alpha_{2}\leq\ell(\alpha_{3})$, we have (2.2) in the form
$1/\alpha_{2}+1/\alpha_{1}\leq\alpha_{2}-\alpha_{3}+\alpha_{3}-\alpha_{1}$,
which holds as $\ell(\alpha_{1})<\alpha_{2}-1/\alpha_{2}$. For
$\alpha_{3}\in(\alpha_{1},\alpha_{2}]$ satisfying
$\ell(\alpha_{1})<\alpha_{3}-1/\alpha_{3}$ and
$\alpha_{2}-1/\alpha_{2}\leq\ell(\alpha_{3})$, the right-hand side of (2.2) is
$1/\alpha_{1}+1/\alpha_{3}+\alpha_{2}-\alpha_{3}$ which is bigger than
$\tilde{r}(\alpha_{1},\alpha_{2})$ as $\alpha-1/\alpha$ is increasing. For
$\alpha_{3}>\alpha_{2}$ such that
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{2})$, the right-hand side of (2.2) is
$1/\alpha_{1}+1/\alpha_{3}+\alpha_{3}-\alpha_{2}$. Hence, (2.2) holds as
$\ell(\alpha_{2})<\ell(\alpha_{3})$. For
$\alpha_{3}-1/\alpha_{3}>\ell(\alpha_{2})$, (2.2) turns into
$1/\alpha_{1}+1/\alpha_{2}\leq
1/\alpha_{1}+1/\alpha_{3}+1/\alpha_{3}+1/\alpha_{2}$.

Finally, let us consider the case of $\underline{1<\alpha_{1}<\alpha_{2}}$ and
$\underline{\ell(\alpha_{1})>\alpha_{2}-1/\alpha_{2}}$, where
$\tilde{r}(\alpha_{2},\alpha_{1})=\alpha_{2}-\alpha_{1}$. For $\alpha_{3}\leq
1$ such that $2\alpha_{3}\leq\alpha_{1}-1/\alpha_{1}$, it follows that
$\tilde{r}(\alpha_{i},\alpha_{3})=\alpha_{3}+1/\alpha_{i}$, $i=1,2$. Then
(2.2) turns into $\alpha_{2}-1/\alpha_{2}\leq\ell(\alpha_{1})+2\alpha_{3}$,
which evidently holds in this case. For
$\alpha_{1}-1/\alpha_{1}<2\alpha_{3}\leq\alpha_{2}-1/\alpha_{2}$, we have that
$\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{1}-\alpha_{3}$ and
$\tilde{r}(\alpha_{2},\alpha_{3})=\alpha_{3}+1/\alpha_{2}$. Then (2.2) turns
into $\alpha_{2}-1/\alpha_{2}\leq 2\alpha_{1}$, which is the case for
$2\alpha_{1}>\ell(\alpha_{1})>\alpha_{2}-1/\alpha_{2}$. For $\alpha_{3}\leq 1$
satisfying $2\alpha_{3}>\alpha_{2}-1/\alpha_{2}$, it follows that
$\tilde{r}(\alpha_{i},\alpha_{3})=\alpha_{i}-\alpha_{3}$, $i=1,2$. Then (2.2)
turns into $2\alpha_{3}\leq 2\alpha_{1}$. which is obviously the case. For
$\alpha_{3}\in(1,\alpha_{1}]$ such that
$\ell(\alpha_{3})\leq\alpha_{1}-1/\alpha_{1}$, we have that
$\tilde{r}(\alpha_{i},\alpha_{3})=1/\alpha_{i}+1/\alpha_{3}$, $i=1,2$. Then
(2.2) amounts to $\alpha_{2}-1/\alpha_{2}\leq\ell(\alpha_{1})+2/\alpha_{3}$,
which obviously holds. For
$\alpha_{1}-1/\alpha_{1}<\ell(\alpha_{3})\leq\alpha_{2}-1/\alpha_{2}$, we have
that $\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{1}-\alpha_{3}$ and
$\tilde{r}(\alpha_{2},\alpha_{3})=1/\alpha_{2}+1/\alpha_{3}$. Then (2.2)
amounts to
$\alpha_{1}-\alpha_{3}+\alpha_{1}+1/\alpha_{3}\geq\alpha_{2}-1/\alpha_{2}$,
which is the case for
$\alpha_{1}-\alpha_{3}+\alpha_{1}+1/\alpha_{3}\geq\alpha_{1}+1/\alpha_{3}\geq\ell(\alpha_{1})>\alpha_{2}-1/\alpha_{2}$.
For $\alpha_{2}-1/\alpha_{2}<\ell(\alpha_{3})$, we have
$\tilde{r}(\alpha_{i},\alpha_{3})=\alpha_{i}-\alpha_{3}$, $i=1,2$. Then (2.2)
is $\alpha_{2}-\alpha_{1}\leq\alpha_{2}-\alpha_{3}+\alpha_{1}-\alpha_{3}$,
which obviously holds as $\alpha_{3}\leq\alpha_{1}$. Now we consider
$\alpha_{3}\in(\alpha_{1},\alpha_{2}]$. For
$\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{1})$, we have
$\tilde{r}(\alpha_{1},\alpha_{3})=\alpha_{3}-\alpha_{1}$ and
$\tilde{r}(\alpha_{2},\alpha_{3})=\alpha_{2}-\alpha_{3}$, which yields
equality in (2.2). Recall that $\alpha_{2}-1/\alpha_{2}<\ell(\alpha_{1})$;
hence, $\alpha_{3}-1/\alpha_{3}>\ell(\alpha_{1})$ is impossible for
$\alpha_{3}\leq\alpha_{2}$. It remains to consider $\alpha_{3}>\alpha_{2}$.
For $\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{1})$, we have that
$\tilde{r}(\alpha_{i},\alpha_{3})=\alpha_{3}-\alpha_{i}$, $i=1,2$. Then (2.2)
takes the form $\alpha_{2}-\alpha_{1}\leq 2\alpha_{3}-\alpha_{1}-\alpha_{2}$,
which obviously holds in this case. For
$\ell(\alpha_{1})<\alpha_{3}-1/\alpha_{3}\leq\ell(\alpha_{2})$, (2.2) amounts
to $2\alpha_{2}\leq\ell(\alpha_{3})+\ell(\alpha_{1})$, which holds as
$\ell(\alpha_{1})>\alpha_{2}-1/\alpha_{2}$ (assumed) and
$\ell(\alpha_{3})>\alpha_{2}+1/\alpha_{2}$ for $\alpha_{3}>\alpha_{2}$. For
$\alpha_{3}-1/\alpha_{3}>\ell(\alpha_{2})$, we have that
$\tilde{r}(\alpha_{3},\alpha_{i})=1/\alpha_{3}+1/\alpha_{i}$, $i=1,2$. Then
(2.2) turns into

$\alpha_{2}-\alpha_{1}\leq 2/\alpha_{3}+1/\alpha_{2}+1/\alpha_{1},$

which holds since $\ell(\alpha_{1})>\alpha_{2}-1/\alpha_{2}$. This completes
the whole proof.

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