Boundary behaviors for a class of continuous-state

nonlinear branching processes in critical cases

Shaojuan Ma111School of Mathematics and Information Science, North Minzu
University, Yinchuan, China. Email<EMAIL_ADDRESS>Xu Yang222School of
Mathematics and Information Science, North Minzu University, Yinchuan, China.
Email<EMAIL_ADDRESS>Corresponding author. and Xiaowen
Zhou333Department of Mathematics and Statistics, Concordia University,
Montreal, Canada. Email<EMAIL_ADDRESS>

Abstract. Using Foster-Lyapunov techniques we establish new conditions on non-
extinction, non-explosion, coming down from infinity and staying infinite,
respectively, for the general continuous-state nonlinear branching processes
introduced in Li et al. (2019). These results can be applied to identify
boundary behaviors for the critical cases of the above nonlinear branching
processes with power rate functions driven by Brownian motion and (or) stable
Poisson random measure, which was left open in Li et al. (2019). In
particular, we show that even in the critical cases, a phase transition
happens between coming down from infinity and staying infinite.

Mathematics Subject Classifications (2010): Primary 60F10, secondary 60C05.

Key words and phrases: Continuous-state branching process, nonlinear
branching, extinction, explosion, come down from infinity, stay infinite,
Foster-Lyapunov criteria, stochastic differential equation.

##  1 Introduction

Continuous-state branching processes (CSBPs in short) are nonnegative-valued
Markov processes satisfying the additive branching processes. They often arise
as time-population scaling limits of discrete-state branching processes, and
can also be obtained from spectrally positive Lévy processes via the Lamperti
transform. The introduction of CSBP allows the applications of stochastic
analysis, Lévy processes and stochastic differential equations (SDE in short)
techniques to its study. We refer to Li (2011), Li (2019) and Kyprianou (2012)
and references therein for comprehensive reviews on CSBPs.

Generalized versions of the CSBP have been proposed in recent years to
incorporate interactions between individuals and (or) between individuals and
the population. A class of CSBPs with nonlinear branching mechanism, obtained
by generalized Lamperti transform, is introduced in Li (2019). In Li et al.
(2019), a more general version of the nonlinear CSBP is proposed as the
solution to SDE

$X_{t}=x+\int_{0}^{t}a_{0}(X_{s})\mathrm{d}s+\int_{0}^{t}\int_{0}^{a_{1}(X_{s})}W(\mathrm{d}s,\mathrm{d}u)+\int_{0}^{t}\int_{(0,\infty)}\int_{0}^{a_{2}(X_{s-})}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u),$
(1.1)

where $x>0$, $a_{0}$ and $a_{1},a_{2}\geq 0$ are Borel functions on
$[0,\infty)$, $W(\mathrm{d}s,\mathrm{d}u)$ and
$\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$ denote a Gaussian white noise
and an independent compensated Poisson random measure, respectively. The model
of Li (2019) corresponds to solution $(X_{t})_{t\geq 0}$ to SDE (1.1) with
identical power rate functions $a_{i},i=0,1,2$.

These nonlinear CSBPs allow richer behaviors such as coming down from
infinity. Some extinction, extinguishing, explosion and coming down from
infinity properties are proved in Li (2019). By analyzing weighted occupation
times for spectrally positive Lévy process, asymptotic results on the speeds
of coming down from infinity and explosion are obtained in Foucart et al.
(2019) and Li and Zhou (2019), respectively, for nonlinear CSBP corresponding
to solution to SDE (1.1) with identical rate functions $a_{0}=a_{1}=a_{2}$.
Exponential ergodicity for the general continuous-state nonlinear branching
processes in Li et al. (2019) is studied by Li and Wang (2020) using coupling
techniques.

A version of SDE (1.1) with $a_{1}\equiv 0$ and power functions $a_{0}$ and
$a_{2}$ is considered earlier in Berestycki et al. (2015) where using the
Lamperti transform, a necessary and sufficient condition for extinction is
obtained and the pathwise uniqueness of solution is studied. Work on the
continuous-state logistic branching process can be found in Lambert (2005), Le
et al. (2013) and Le (2014).

Using a martingale approach, the extinction, explosion and coming down from
infinity behaviors are further discussed in Li et al. (2019) and some rather
sharp criteria in terms of $a_{0},a_{1},a_{2}$ and $\mu$ are obtained on
characterization of different kinds of boundary behaviors for the nonlinear
CSBPs as a Markov process. In Example 2.18 of Li et al. (2019) where
$a_{0},a_{1}$ and $a_{2}$ are taken to be power functions and $\tilde{M}$ is
taken to be an $\alpha$-stable Poison random measure with index
$\alpha\in(1,2)$. The above criteria are further expressed in terms of the
coefficients and the powers of functions $a_{i}$s and the stable index
$\alpha$. But for the critical cases, where the coefficients, the powers and
the index $\alpha$ satisfy certain linear equations, the martingale approach
fails and the corresponding boundary classification remains an open problem.

The main goal of this paper is to identify the exact boundary behaviors in the
above mentioned critical cases for the solution $(X)_{t\geq 0}$ to (1.1). For
this purpose, we adapt the Foster-Lyapunov approach and select logarithm type
test functions to obtain two new conditions under which the nonlinear CSBP
never becomes extinct and never explodes, respectively. Similarly, for the
boundary at infinity, we also find a Foster-Lyapunov condition with which we
can show that an interesting phase transition occurs between coming down from
infinity and staying infinite for different choices of coefficients and powers
of the power functions and differential values of the stable index $\alpha$.

The rest of the paper is arranged as follows. We introduce the generalized
CSBPs with nonlinear branching in more details and present the main theorem in
Section 2. The proofs of preliminary results on Foster-Lyapunov criteria and
the main theorem are deferred to Section 3.

##  2 Main results

Let $U$ be a Borel set on $(0,\infty)$. Given $\sigma$-finite measures $\mu$
and $\nu$ on $(0,\infty)$ such that

$(z\wedge
z^{2})\mu(\mathrm{d}z)\,\,\text{and}\,\,(1\vee\ln(1+z))\nu(\mathrm{d}z)\,\,\text{are
finite measures on $U$ and $(0,\infty)\setminus U$, respectively},$

we consider the following SDE that is a modification of (1.1):

$\displaystyle X_{t}$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!x+\int_{0}^{t}a_{0}(X_{s})\mathrm{d}s+\int_{0}^{t}\int_{0}^{a_{1}(X_{s})}W(\mathrm{d}s,\mathrm{d}u)$
(2.2)
$\displaystyle\\!\\!+\int_{0}^{t}\int_{U}\int_{0}^{a_{2}(X_{s-})}z\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)+\int_{0}^{t}\int_{(0,\infty)\setminus
U}\int_{0}^{a_{3}(X_{s-})}zN(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u),$

where $x>0$, $a_{0}$ and $a_{1},a_{2},a_{3}\geq 0$ are Borel functions on
$[0,\infty)$, $W(\mathrm{d}s,\mathrm{d}u)$ is a Gaussian white noise with
density $\mathrm{d}s\mathrm{d}u$,
$\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$ denotes a compensated Poisson
random measure on $(0,\infty)\times U\times(0,\infty)$ with density
$\mathrm{d}s\mu(\mathrm{d}z)\mathrm{d}u$, and
$N(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$ denotes a Poisson random measure on
$(0,\infty)\times((0,\infty)\setminus U)\times(0,\infty)$ with density
$\mathrm{d}s\nu(\mathrm{d}z)\mathrm{d}u$. We assume that
$W(\mathrm{d}s,\mathrm{d}u)$, $\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
and $N(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$ are independent of each other. A
similar SDE (2.2) is considered in Li et al. (2019) under the assumption that
$U=(0,\infty)$.

We only consider the solution of (2.2) before the minimum of their first times
of hitting zero and reaching infinity (that is $\tau_{0}^{-}$ and
$\tau_{\infty}^{+}$, which will be given in the following), respectively, i.e.
both zero and infinity are absorbing states for the solution. See Section 2 of
Li et al. (2019) for more details. By the same argument as Theorem 3.1 in Li
et al. (2019), we can show that SDE (2.2) has a pathwise unique solution if
functions $a_{0},a_{1},a_{2},a_{3}$ are locally Lipschitz on $(0,\infty)$. The
main purpose of this paper is to investigate the extinction, explosion and
coming down from infinity behaviors, and the uniqueness of solution to SDE
(2.2) is not required.

Throughout this paper we always assume that $a_{0},a_{1},a_{2},a_{3}$ are
bounded on any bounded interval of $[0,\infty)$ and that process
$(X_{t})_{t\geq 0}$ is defined on filtered probability space
$(\Omega,\mathscr{F},\mathscr{F}_{t},\mathbf{P})$ which satisfies the usual
hypotheses. We use $\mathbf{P}_{x}$ to denote the law of a process started at
$x$, and denote by $\mathbf{E}_{x}$ the associated expectation. For $a,b>0$ we
define first passage times

$\displaystyle\tau_{a}^{-}:=\inf\\{t>0:X_{t}<a\\},\quad\tau_{b}^{+}:=\inf\\{t>0:X_{t}>b\\}$

and

$\displaystyle\tau_{0}^{-}:=\inf\\{t>0:X_{t}=0\\},\quad\tau_{\infty}^{+}:=\lim_{n\to\infty}\tau_{n}^{+}$

with the convention $\inf\emptyset=\infty$. Let $C^{2}((0,\infty))$ denote the
space of twice continuously differentiable functions on $(0,\infty)$.

We next introduce several functions. For $u>0$ let

$\displaystyle\phi(u)$ $\displaystyle\\!\\!:=$
$\displaystyle\\!\\!-a_{0}(u)u^{-1}+\frac{1}{2}a_{1}(u)u^{-2}+a_{2}(u)\int_{U}z^{2}\mu(\mathrm{d}z)\int_{0}^{1}(u+vz)^{-2}(1-v)\mathrm{d}v$
$\displaystyle\\!\\!-a_{3}(u)\int_{(0,\infty)\setminus
U}z\nu(\mathrm{d}z)\int_{0}^{1}(u+vz)^{-1}\mathrm{d}v.$

For $\rho,z>0$ and $u>3$ let

$\displaystyle K_{\rho}(u,z):=\Big{(}\frac{\ln(u+z)}{\ln
u}\Big{)}^{-\rho}+\rho\frac{\ln(u+z)}{\ln u}-(\rho+1)>0$

and

$\displaystyle
H_{\rho}(u):=\frac{1}{2}a_{1}(u)u^{-2}+a_{2}(u)\int_{U}K_{\rho}(u,z)\mu(\mathrm{d}z)+a_{3}(u)\int_{(0,\infty)\setminus
U}K_{\rho}(u,z)\nu(\mathrm{d}z).$

Process $(X_{t})_{t\geq 0}$ becomes extinct if $\tau^{-}_{0}<\infty$; it
explodes if $\tau^{+}_{\infty}<\infty$; it stays infinite if
$\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}=0$ for all $t>0$ and all
large $a$; it comes down from infinity if
$\lim_{a\to\infty}\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}=1$ for
all $t>0$.

The following main result provides new criteria on non-extinction, non-
explosion, coming down from infinity and staying infinite for the solution to
SDE (1.1).

###### Theorem 2.1

For the solution $(X_{t})_{t\geq 0}$ to (2.2) we have

  * (i)

if $\phi(u)\leq 0$ for all small enough $u>0$, then
$\mathbf{P}_{x}\\{\tau_{0}^{-}=\infty\\}=1$ for all $x>0$, i.e. there is no
extinction;

  * (ii)

if $\phi(u)\geq 0$ for all large enough $u>0$, then
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\}=1$ for all $x>0$, i.e. there is
no explosion;

  * (iii)

if $\phi(u)\leq 0$ for all large enough $u>0$ and

$\limsup_{u\rightarrow\infty}H_{\rho}(u)<\infty$

for some constant $\rho>0$, then the process $(X_{t})_{t\geq 0}$ stays
infinite;

  * (iv)

if $\phi(u)\geq 0$ for all large enough $u>0$ and

$\displaystyle\lim_{u\rightarrow\infty}(\ln u)^{-\rho-2}H_{\rho}(u)=\infty$

for some constant $\rho>0$, then the process $(X_{t})_{t\geq 0}$ does not
explode and it comes down from infinity.

###### Remark 2.2

Since the process $(X_{t})_{t\geq 0}$ does not explode and comes down from
infinity under the assumptions of Theorem 2.1 (iv), then under additional
assumptions on functions $a_{i}$s, $(X_{t})_{t\geq 0}$ can be extended to a
Feller process defined on state space $[0,\infty]$ with $\infty$ as its
entrance boundary; see Foucart et al. (2020).

Until the end of this section we focus on the special case that
$U=(0,\infty)$, $a_{0},a_{1},a_{2}$ are power functions and $\mu(\mathrm{d}z)$
is an $\alpha$-stable measure, that is

$a_{i}(u)=b_{i}u^{r_{i}}~{}\text{for}~{}i=0,1,2~{}\text{with}~{}r_{0},r_{1},r_{2}\geq
0,b_{1},b_{2}\geq 0,b_{0}>0$ (2.3)

and

$\mu(\mathrm{d}z)=\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}1_{\\{z>0\\}}z^{-1-\alpha}\mathrm{d}z~{}\text{for
Gamma function}\,\Gamma\,\text{and}\,1<\alpha<2.$ (2.4)

By the properties of Gamma function we have

$\displaystyle\int_{0}^{\infty}z^{2}\mu(\mathrm{d}z)\int_{0}^{1}(u+vz)^{-2}(1-v)\mathrm{d}v=\Gamma(\alpha)u^{-\alpha}$

and then

$\displaystyle\phi(u)=-b_{0}u^{r_{0}-1}+\frac{1}{2}b_{1}u^{r_{1}-2}+\Gamma(\alpha)b_{2}u^{r_{2}-\alpha},\quad
u>0.$

We further estimate

$H_{\rho}(u)=\frac{1}{2}a_{1}(u)u^{-2}+a_{2}(u)\int_{0}^{\infty}K_{\rho}(u,z)\mu(\mathrm{d}z)$

for which we first estimate $\int_{0}^{\infty}K_{\rho}(u,z)\mu(\mathrm{d}z)$.
Note that for $y>0$ and $f(y):=y^{-\rho}+\rho y-(\rho+1)$, by Taylor’s formula
we have

$\displaystyle f(y)$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!f(1+y-1)=f(1+y-1)-f(1)-(y-1)f^{\prime}(1)$
$\displaystyle\\!\\!=$
$\displaystyle\\!\\!(y-1)^{2}\int_{0}^{1}f^{\prime\prime}(1+v(y-1))(1-v)\mathrm{d}v.$

Then by a change of variable,

$\displaystyle\\!\\!\int_{0}^{\infty}K_{\rho}(u,z)\mu(\mathrm{d}z)$
$\displaystyle\\!\\!=$
$\displaystyle\\!\\!\rho(\rho+1)\int_{0}^{\infty}\mu(\mathrm{d}z)\int_{0}^{1}\Big{(}\frac{\ln(u+vz)}{\ln
u}-1\Big{)}^{2}\Big{(}1+\frac{v\ln(u+vz)}{\ln
u}-v\Big{)}^{-\rho-2}(1-v)\mathrm{d}v$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!\rho(\rho+1)u^{-\alpha}\int_{0}^{\infty}\mu(\mathrm{d}z)\int_{0}^{1}\Big{(}\frac{\ln(1+vz)}{\ln
u}\Big{)}^{2}\Big{(}1+\frac{v\ln(1+vz)}{\ln
u}\Big{)}^{-\rho-2}(1-v)\mathrm{d}v.$

Observe that $\ln(1+z)\leq C(z\wedge\sqrt{z})$ for all $z>0$ and some constant
$C>0$, which implies that for $u>3$,

$\displaystyle\int_{0}^{\infty}K_{\rho}(u,z)\mu(\mathrm{d}z)$
$\displaystyle\\!\\!\leq$ $\displaystyle\\!\\!\rho(\rho+1)u^{-\alpha}(\ln
u)^{-2}\int_{0}^{\infty}(\ln(1+z))^{2}\mu(\mathrm{d}z)$ (2.5)
$\displaystyle\\!\\!\leq$ $\displaystyle\\!\\!C^{2}\rho(\rho+1)u^{-\alpha}(\ln
u)^{-2}\int_{0}^{\infty}(z\wedge z^{2})\mu(\mathrm{d}z).$ (2.6)

Moreover, it is elementary to see that

$\displaystyle\\!\\!\int_{0}^{\infty}K_{\rho}(u,z)\mu(\mathrm{d}z)$ (2.7)
$\displaystyle\\!\\!\geq$ $\displaystyle\\!\\!\rho(\rho+1)u^{-\alpha}(\ln
u)^{-2}\Big{(}1+\frac{\ln 3}{\ln
u}\Big{)}^{-\rho-2}\int_{1}^{2}(\ln(3/2))^{2}\mu(\mathrm{d}z)\int_{1/2}^{1}(1-v)\mathrm{d}v.$
(2.8)

###### Remark 2.3

In Section 2.5 of Li et al. (2019), the exact conditions are found for the
above mentioned model with polynomial rate functions to exhibit
extinction/non-extinction, explosion/non-explosion and coming-down-from-
infinity/staying-infinite behaviors, respectively, except for the critical
case that

$b_{0}=\frac{b_{1}}{2}+\Gamma(\alpha)b_{2}>0,\,r_{1}=r_{0}+1\,\,\text{
when}\,\,b_{1}>0\,\,\text{ and}\,\,r_{2}=r_{0}+\alpha-1\,\,\text{
when}\,\,b_{2}>0.$

Observe that in this critical case $\phi(u)=L(\ln)(u)=0$, where operator $L$,
to be defined in (3.3), denotes the generator of process $X$. This inspires us
to choose logarithm type test functions for the main proofs.

As the main goal of this paper, applying Theorem 2.1 together with
(2.5)–(2.7), we provide an answer to this open problem.

###### Corollary 2.4

Suppose that (2.3) and (2.4) hold with
$b_{0}=\frac{b_{1}}{2}+\Gamma(\alpha)b_{2}>0$, $r_{1}=r_{0}+1$ when $b_{1}>0$
and $r_{2}=r_{0}+\alpha-1$ when $b_{2}>0$. Then $\phi(u)=0$ for all $u>0$, and
we have for all $x>0$,

$\displaystyle\mathbf{P}_{x}\\{\tau_{0}^{-}=\infty\\}=1\quad\text{and}\quad\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\}=1.$

Moreover, process $(X_{t})_{t\geq 0}$ stays infinite if both $r_{1}\leq 2$
(when $b_{1}>0$) and $r_{2}\leq\alpha$ (when $b_{2}>0$), and it comes down
from infinity if either $r_{1}>2$ (when $b_{1}>0$) or $r_{2}>\alpha$ (when
$b_{2}>0$).

###### Remark 2.5

Note that in the critical cases, there is an interesting phase transition
between coming down from infinity and staying infinite. Intuitively, in these
cases the process comes down from infinity if the fluctuations are relatively
large and stays infinite otherwise.

Combining Corollary 2.4 and Example 2.18 of Subsection 2.5 in Li et al.
(2019), we recover the necessary and sufficient condition on the extinction of
solution to the SDE of Berestycki et al. (2015); see Theorem 1.1 there.

##  3 Proofs

Before presenting the proof of Theorem 2.1 we first prove some preliminary
Foster-Lyapunov criteria. Suppose that $g\in C^{2}((0,\infty))$ satisfies

$\displaystyle\sup_{z\geq 1,u\geq
v}\big{[}|g^{\prime}(u)|+|g^{\prime\prime}(u)|+|g(u+z)-g(u)|/\ln(1+z)\big{]}<\infty$
(3.1)

for all $v>0$. For $u>0$, put

$\displaystyle Lg(u)$ $\displaystyle\\!\\!:=$
$\displaystyle\\!\\!a_{0}(u)g^{\prime}(u)+\frac{1}{2}a_{1}(u)g^{\prime\prime}(u)+a_{2}(u)\int_{U}[g(u+z)-g(u)-zg^{\prime}(u)]\mu(\mathrm{d}z)$
(3.3) $\displaystyle\\!\\!+a_{3}(u)\int_{(0,\infty)\setminus
U}[g(u+z)-g(u)]\nu(\mathrm{d}z)$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!a_{0}(u)g^{\prime}(u)+\frac{1}{2}a_{1}(u)g^{\prime\prime}(u)+a_{2}(u)\int_{U}z^{2}\mu(\mathrm{d}z)\int_{0}^{1}g^{\prime\prime}(u+zv)(1-v)\mathrm{d}v$
(3.5) $\displaystyle\\!\\!+a_{3}(u)\int_{(0,\infty)\setminus
U}z\nu(\mathrm{d}z)\int_{0}^{1}g^{\prime}(u+zv)\mathrm{d}v$

by Taylor’s formula. By Itô’s formula,

$\displaystyle g(X_{t})$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!g(x)+\int_{0}^{t}Lg(X_{s})\mathrm{d}s+\int_{0}^{t}\int_{U}\int_{0}^{a_{2}(X_{s-})}[g(X_{s-}+z)-g(X_{s-})]\tilde{M}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$
$\displaystyle\\!\\!+\int_{0}^{t}\int_{(0,\infty)\setminus
U}\int_{0}^{a_{3}(X_{s-})}[g(X_{s-}+z)-g(X_{s-})]\tilde{N}(\mathrm{d}s,\mathrm{d}z,\mathrm{d}u)$

For $b>a>0$ let $\gamma_{a,b}:=\tau_{a}^{-}\wedge\tau_{b}^{+}$ and

$\displaystyle M_{t}^{g}:=g(X_{t})-g(x)-\int_{0}^{t}Lg(X_{s})\mathrm{d}s.$

Then under condition (3.1),

$\displaystyle t\mapsto M_{t\wedge\gamma_{a,b}}^{g}\quad\mbox{is a
martingale}$ (3.6)

for all $b>a>0$.

###### Lemma 3.1

Given $0<a<x<b<\infty$, for any function $g\in C^{2}((a,b))$ satisfying (3.1)
and constant $d_{a,b}>0$ satisfying

$\displaystyle Lg(u)\leq d_{a,b}g(u),\qquad u\in(a,b),$

we have

$\displaystyle\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}\leq
g(x){\mbox{\rm e}}^{d_{a,b}t},\qquad t\geq 0.$ (3.7)

Proof. It follows from (3.6) that

$\displaystyle\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}=g(x)+\mathbf{E}_{x}\Big{[}\int_{0}^{t\wedge\gamma_{a,b}}Lg(X_{s})\mathrm{d}s\Big{]}\leq
g(x)+d_{a,b}\int_{0}^{t}\mathbf{E}_{x}\big{[}g(X_{s\wedge\gamma_{a,b}})\big{]}\mathrm{d}s.$

By Gronwall’s lemma,

$\displaystyle\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}\leq
g(x){\mbox{\rm e}}^{d_{a,b}t},$

which ends the proof. $\Box$

###### Lemma 3.2

Let $(X_{t})_{t\geq 0}$ be the solution to SDE (2.2).

  * (i)

For any fixed $b>0$, if there exists a function $g\in C^{2}((0,\infty))$
strictly positive on $(0,b]$ satisfying (3.1) and $\lim_{u\to 0}g(u)=\infty$,
and there is a constant $d(b)>0$ such that $Lg(u)\leq d(b)g(u)$ for all
$0<u<b$, then $\mathbf{P}_{x}\\{\tau_{0}^{-}\geq\tau^{+}_{b}\\}=1$ for all
$0<x<b$.

  * (ii)

For any fixed $a>0$, if there exists a function $g\in C^{2}((0,\infty))$
strictly positive on $[a,\infty)$ satisfying (3.1) and
$\lim_{u\to\infty}g(u)=\infty$, and there is a constant $d(a)>0$ such that
$Lg(u)\leq d(a)g(u)$ for all $u>a$, then
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}\geq\tau_{a}^{-}\\}=1$ for all $x>a$.

  * (iii)

If there exists a function $g\in C^{2}((0,\infty))$ strictly positive on
$[u,\infty)$ for all large $u$ satisfying (3.1) and $\lim_{u\to\infty}g(u)=0$,
and for any large $a>0$, there is a constant $d(a)>0$ such that $Lg(u)\leq
d(a)g(u)$ for all $u>a$, then $(X_{t})_{t\geq 0}$ stays infinite.

Proof. We apply Lemma 3.1 for the proofs.

For part (i), (3.7) holds for all $0<a<b$ and with $d_{a,b}$ replaced by
$d(b)$. Then using Fatou’s lemma we have

$\displaystyle\mathbf{E}_{x}\big{[}\liminf_{a\to
0}g(X_{t\wedge\tau_{a}^{-}\wedge\tau_{b}^{+}})\big{]}\leq\liminf_{a\to
0}\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}\leq g(x){\mbox{\rm
e}}^{d(b)t}.$

Since $\lim_{u\to 0}g(u)=\infty$, then
$\mathbf{P}_{x}\\{\tau_{0}^{-}>t\wedge\tau_{b}^{+}\\}=1$ for all $t,b>0$.
Letting $t\to\infty$ we obtain
$\mathbf{P}_{x}\\{\tau_{0}^{-}\geq\tau_{b}^{+}\\}=1$, which gives the first
assertion.

For part (ii), (3.7) holds for all $b>a$ and with $d_{a,b}$ replaced by
$d(a)$. Then using Fatou’s lemma again we obtain

$\displaystyle\mathbf{E}_{x}\big{[}\liminf_{b\to\infty}g(X_{t\wedge\tau_{a}^{-}\wedge\tau_{b}^{+}})\big{]}\leq\liminf_{b\to\infty}\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}\leq
g(x){\mbox{\rm e}}^{d(a)t}.$

Since $\lim_{u\to\infty}g(u)=\infty$, then
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}>t\wedge\tau_{a}^{-}\\}=1$ for all $t>0$.
Letting $t\to\infty$ we get
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}\geq\tau_{a}^{-}\\}=1$, which implies the
second assertion.

For part (iii), given any large $a>0$, (3.7) holds for all $b>a$ and with
$d_{a,b}$ replaced by $d(a)$ again. We can also get

$\displaystyle\mathbf{E}_{x}\big{[}g(X_{\tau^{-}_{a}})1_{\\{\tau^{-}_{a}<t\wedge\tau_{\infty}^{+}\\}}\big{]}$
$\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\liminf_{b\to\infty}\mathbf{E}_{x}\big{[}g(X_{\tau^{-}_{a}})1_{\\{\tau^{-}_{a}<t\wedge\tau_{b}^{+}\\}}\big{]}$
$\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\liminf_{b\to\infty}\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}\leq
g(x){\mbox{\rm e}}^{d(a)t},$

which implies

$\displaystyle
g(a)\mathbf{P}_{x}\\{\tau^{-}_{a}<t\wedge\tau_{\infty}^{+}\\}\leq
g(x){\mbox{\rm e}}^{d(a)t}.$

Since $\lim_{u\to\infty}g(u)=0$, then for all $t,a>0$,

$\displaystyle\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau^{-}_{a}<t\wedge\tau_{\infty}^{+}\\}=0.$
(3.8)

Observe that
$\\{\tau^{-}_{a}\geq\tau_{\infty}^{+}\\}\subset\\{\tau^{-}_{a}=\infty\\}$.
Then combining (3.8) we have

$\displaystyle\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau^{-}_{a}<t\\}\leq\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau^{-}_{a}<t\wedge\tau_{\infty}^{+}\\}+\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau^{-}_{a}<t,\tau_{\infty}^{+}\leq\tau^{-}_{a}\\}=0$

for all $t>0$. Then the process stays infinite. $\Box$

The next lemma provides a condition that associates the probability of coming
down from infinity with the probability of non-explosion. Its proof is a
modification of Proposition 2.2 in Ren et al. (2019).

###### Lemma 3.3

Suppose that there exist a function $g(u)\in C^{2}((0,\infty))$ bounded and
strictly positive for all large $u$, satisfying (3.1) and
$\limsup_{u\to\infty}g(u)>0$, and a strictly positive function $d$ on
$(0,\infty)$ such that

$Lg(u)\geq d(a)g(u)\,\,\text{ for all large}\,\,u\,\,\text{
and}\,\,\lim_{a\to\infty}d(a)=\infty.$

Then for any $t>0$

$\displaystyle\lim_{a\to\infty}\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}\geq\liminf_{x\to\infty}\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\}.$

Consequently, process $(X_{t})_{t\geq 0}$ comes down from infinity if there is
no explosion.

Proof. The proof is a modification of that of Proposition 2.2 in Ren et al.
(2019). We present the details for completeness. By (3.6), for all large $a<b$

$\displaystyle\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}=g(x)+\mathbf{E}_{x}\Big{[}\int_{0}^{t\wedge\gamma_{a,b}}Lg(X_{s})\mathrm{d}s\Big{]}=g(x)+\int_{0}^{t}\mathbf{E}_{x}\Big{[}Lg(X_{s})1_{\\{s\leq\gamma_{a,b}\\}}\Big{]}\mathrm{d}s$

and then by integration by parts,

$\displaystyle\\!\\!\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}})\big{]}{\mbox{\rm
e}}^{-d(a)t}$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!g(x)+\int_{0}^{t}\mathbf{E}_{x}\big{[}g(X_{s\wedge\gamma_{a,b}})\big{]}\mathrm{d}({\mbox{\rm
e}}^{-d(a)s})+\int_{0}^{t}{\mbox{\rm
e}}^{-d(a)s}\mathrm{d}\big{(}\mathbf{E}_{x}\big{[}g(X_{s\wedge\gamma_{a,b}})\big{]}\big{)}$
$\displaystyle\\!\\!=$
$\displaystyle\\!\\!g(x)-d(a)\int_{0}^{t}\mathbf{E}_{x}\big{[}g(X_{s\wedge\gamma_{a,b}}){\mbox{\rm
e}}^{-d(a)s}\big{]}\mathrm{d}s+\int_{0}^{t}{\mbox{\rm
e}}^{-d(a)s}\mathbf{E}_{x}\big{[}Lg(X_{s})1_{\\{s\leq\gamma_{a,b}\\}}\big{]}\mathrm{d}s$
$\displaystyle\\!\\!\geq$
$\displaystyle\\!\\!g(x)-d(a)\int_{0}^{t}\mathbf{E}_{x}\big{[}g(X_{s\wedge\gamma_{a,b}})\big{]}{\mbox{\rm
e}}^{-d(a)s}\mathrm{d}s+d(a)\int_{0}^{t}{\mbox{\rm
e}}^{-d(a)s}\mathbf{E}_{x}\big{[}g(X_{s})1_{\\{s\leq\gamma_{a,b}\\}}\big{]}\mathrm{d}s,$

which implies that

$\displaystyle
g(x)\leq\mathbf{E}_{x}\big{[}g(X_{t\wedge\gamma_{a,b}}){\mbox{\rm
e}}^{-d(a)t}\big{]}+d(a)\mathbf{E}_{x}\Big{[}\int_{0}^{t}g(X_{\gamma_{a,b}}){\mbox{\rm
e}}^{-d(a)s}1_{\\{s>\gamma_{a,b}\\}}\mathrm{d}s\Big{]}.$

Letting $t\to\infty$ in the above inequality and using the dominated
convergence we obtain

$\displaystyle g(x)\leq
d(a)\mathbf{E}_{x}\Big{[}g(X_{\gamma_{a,b}})\int_{\gamma_{a,b}}^{\infty}{\mbox{\rm
e}}^{-d(a)s}\mathrm{d}s\Big{]}=\mathbf{E}_{x}\big{[}g(X_{\gamma_{a,b}}){\mbox{\rm
e}}^{-d(a)\gamma_{a,b}}\big{]}.$

It follows that

$\displaystyle g(x)$ $\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\mathbf{E}_{x}\Big{[}\lim_{b\to\infty}g(X_{\gamma_{a,b}}){\mbox{\rm
e}}^{-(\tau_{a}^{-}\wedge\tau^{+}_{\infty})d(a)}\big{(}1_{\\{\tau_{\infty}^{+}<\tau^{-}_{a}\\}}+1_{\\{\tau_{a}^{-}<t,\tau_{a}^{-}\leq\tau^{+}_{\infty}\\}}+1_{\\{t\leq\tau^{-}_{a}\leq\tau_{\infty}^{+}\\}}\big{)}\Big{]}$
$\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\limsup_{u\to\infty}g(u)\mathbf{P}_{x}\\{\tau_{\infty}^{+}<\infty\\}+g(a)\mathbf{P}_{x}\\{\tau_{a}^{-}<t,\tau_{a}^{-}\leq\tau^{+}_{\infty}\\}+g(a){\mbox{\rm
e}}^{-d(a)t}$ $\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\limsup_{u\to\infty}g(u)(1-\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\})+g(a)\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}+g(a){\mbox{\rm
e}}^{-d(a)t}.$

Letting $x\to\infty$ first,

$\displaystyle\limsup_{x\to\infty}g(x)$ $\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\limsup_{u\to\infty}g(u)\limsup_{x\to\infty}(1-\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\})$
$\displaystyle\\!\\!+g(a)\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}+g(a){\mbox{\rm
e}}^{-d(a)t}.$

Then letting $a\to\infty$, by the conditions in the lemma we have

$\displaystyle\limsup_{x\to\infty}g(x)$ $\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\limsup_{u\to\infty}g(u)\left(1-\liminf_{x\to\infty}\mathbf{P}_{x}\\{\tau_{\infty}^{+}=\infty\\}\right)$
$\displaystyle\\!\\!+\limsup_{a\to\infty}g(a)\limsup_{a\to\infty}\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}.$

Observing that $\lim_{x\to\infty}\mathbf{P}_{x}\\{\tau_{a}^{-}<t\\}$ is
increasing in $a$, the desired inequality then follows from the above
inequality. $\Box$

We are now ready to show the proofs of the main results.

Proof of Theorem 2.1. (i) Suppose that there is a constant $0<c_{1}<1$ so that
$\phi(u)\leq 0$ for all $0<u<c_{1}$. For $n\geq 1$ let $g_{n}(u)=1+\ln n+\ln
u^{-1}$. Then $g_{n}(u)>0$ for $0<u\leq n$ and $Lg_{n}(u)=\phi(u)$ by (3.3).
Thus $Lg_{n}(u)\leq 0$ for $0<u<c_{1}$. Since $a_{0},a_{1},a_{2},a_{3}$ are
bounded on $[c_{1},n]$, $Lg_{n}$ is bounded on $[c_{1},n]$. Now using Lemma
3.2(i) we obtain $\mathbf{P}_{x}\\{\tau_{0}^{-}\geq\tau^{+}_{n}\\}=1$ for all
$0<x<n$. Since the process is defined before the first time of hitting zero or
explosion, $\mathbf{P}_{x}\\{\tau_{0}^{-}=\infty$ or
$\tau_{\infty}^{+}=\infty\\}=1$. Letting $n\to\infty$ we prove the assertion.

(ii) Suppose that there is a constant $c_{2}>1$ so that $\phi(u)\geq 0$ for
all $u>c_{2}$. Let $g_{n}(u)=\ln u+\ln n+1$ for $n\geq 1$. Then $g_{n}(u)\geq
1$ for $u\geq n^{-1}$. It follows from (3.3) that $Lg_{n}(u)=-\phi(u)$ for all
$u\geq n^{-1}$. Then for all $n\geq 1$, $Lg_{n}(u)\leq 0$ for all $u\geq
c_{2}$ and $Lg_{n}$ is bounded on $[n^{-1},c_{2}]$, which gives
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}>\tau_{1/n}^{-}\\}=1$ for all $x>n^{-1}$ by
Lemma 3.2(ii). Letting $n\to\infty$ we have
$\mathbf{P}_{x}\\{\tau_{\infty}^{+}>\tau_{0}^{-}\\}=1$ for all $x>0$. The
assertion for (ii) then follows from the definition of the solution to SDE
(2.2).

(iii) Suppose that there exist constants $c_{3}>3$ and $c_{4}>0$ so that
$\phi(u)\leq 0$ and $H_{\rho}(u)\leq c_{4}$ for all $u>c_{3}$. Let $g\in
C^{2}((0,\infty))$ be a strictly positive function with $g(u)=(\ln u)^{-\rho}$
for $\rho>0$ and $u>3$.

Then for $u>3$,

$\displaystyle g(u+z)-g(u)$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!-\rho(\ln u)^{-\rho-1}[\ln(u+z)-\ln u]+(\ln
u)^{-\rho}K_{\rho}(u,z)$ $\displaystyle\\!\\!=$ $\displaystyle\\!\\!-\rho(\ln
u)^{-\rho-1}z\int_{0}^{1}(u+vz)^{-1}\mathrm{d}v+(\ln u)^{-\rho}K_{\rho}(u,z)$

and

$\displaystyle g^{\prime}(u)=-\rho(\ln
u)^{-\rho-1}u^{-1},~{}g^{\prime\prime}(u)=\rho(\ln
u)^{-\rho-1}u^{-2}+\rho(\rho+1)(\ln u)^{-\rho-2}u^{-2}.$

Consequently, for all $u>3$ and $z>0$,

$\displaystyle g(u+z)-g(u)-zg^{\prime}(u)$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!-\rho(\ln u)^{-\rho-1}\big{[}\ln(u+z)-\ln
u-zu^{-1}\big{]}+(\ln u)^{-\rho}K_{\rho}(u,z)$ $\displaystyle\\!\\!=$
$\displaystyle\\!\\!\rho(\ln
u)^{-\rho-1}z^{2}\int_{0}^{1}(u+zv)^{-2}(1-v)\mathrm{d}v+(\ln
u)^{-\rho}K_{\rho}(u,z).$

It follows that

$\displaystyle Lg(u)$ $\displaystyle\\!\\!=$ $\displaystyle\\!\\!\rho(\ln
u)^{-\rho-1}\phi(u)+\frac{1}{2}\rho(\rho+1)(\ln u)^{-2}a_{1}(u)u^{-2}g(u)$
$\displaystyle\\!\\!+g(u)a_{2}(u)\int_{U}K_{\rho}(u,z)\mu(\mathrm{d}z)+g(u)a_{3}(u)\int_{(0,\infty)\setminus
U}K_{\rho}(u,z)\nu(\mathrm{d}z)$ $\displaystyle\\!\\!\leq$
$\displaystyle\\!\\!\rho(\ln
u)^{-\rho-1}\phi(u)+[\rho(\rho+1)+1]g(u)H_{\rho}(u),\qquad u>3.$

Then $Lg(u)\leq c_{4}[\rho(\rho+1)+1]g(u)$ for all $u>c_{3}$. Thus,
$(X_{t})_{t\geq 0}$ stays infinite for all $x>0$ by Lemma 3.2(iii).

(iv) Let $g\in C^{2}((0,\infty))$ be a bounded and strictly positive function
with $g(u)=1+(\ln u)^{-\rho}$ for $\rho>0$ and $u>3$. It follows from the
argument in (iii) that for $u>3$,

$\displaystyle Lg(u)$ $\displaystyle\\!\\!=$ $\displaystyle\\!\\!\rho(\ln
u)^{-\rho-1}\phi(u)+\frac{1}{2}\rho(\rho+1)(\ln u)^{-\rho-2}a_{1}(u)u^{-2}$
$\displaystyle\\!\\!+(\ln
u)^{-\rho}a_{2}(u)\int_{U}K_{\rho}(u,z)\mu(\mathrm{d}z)+(\ln
u)^{-\rho}a_{3}(u)\int_{(0,\infty)\setminus U}K_{\rho}(u,z)\nu(\mathrm{d}z)$
$\displaystyle\\!\\!\geq$ $\displaystyle\\!\\!\rho(\ln
u)^{-\rho-1}\phi(u)+(\rho\wedge 1)(\ln u)^{-\rho-2}H_{\rho}(u).$

Then we can conclude the proof by the assumptions for this part together with
Theorem 2.1(ii) and Lemma 3.3. $\Box$

Acknowledgements. This work was supported by NSFC (Nos. 11772002 and
11771018), Major research project for North Minzu University (No. ZDZX201902)
and NSERC (RGPIN-2016-06704).

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