11footnotetext: Université de Lorraine, IECL, UMR 7502, Campus Scientifique,
B.P. 70239, Vandœuvre-lès-Nancy Cedex, F-54506, France22footnotetext: CNRS,
IECL, UMR 7502, Vandœuvre-lès-Nancy, F-54506, France33footnotetext: Inria,
TOSCA team, Villers-lès-Nancy, F-54600, France.
E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
# Uniform convergence of conditional distributions for absorbed one-
dimensional diffusions

Nicolas Champagnat1,2,3, Denis Villemonais1,2,3

###### Abstract

This article studies the quasi-stationary behaviour of absorbed one-
dimensional diffusions. We obtain necessary and sufficient conditions for the
exponential convergence to a unique quasi-stationary distribution in total
variation, uniformly with respect to the initial distribution. An important
tool is provided by one dimensional strict local martingale diffusions coming
down from infinity. We prove under mild assumptions that their expectation at
any positive time is uniformly bounded with respect to the initial position.
We provide several examples and extensions, including the sticky Brownian
motion and some one-dimensional processes with jumps.

Keywords: one-dimensional diffusions; absorbed process; quasi-stationary
distribution; uniform exponential mixing property; strict local martingales;
one dimensional processes with jumps.

2010 Mathematics Subject Classification. Primary: 60J60; 60J70; 37A25; 60B10;
60F99. Secondary: 60G44; 60J75.

##  1 Introduction

This article studies the quasi-stationary behaviour of general one-dimensional
diffusion processes in an interval $E$ of $\mathbb{R}$, absorbed at its finite
boundaries. When the process is absorbed, it is sent to some cemetary point
$\partial$. We recall that a _quasi-stationary distribution_ for a continuous-
time Markov process $(X_{t},t\geq 0)$ on the state space $E\cup\\{\partial\\}$
absorbed in $\partial$, is a probability measure $\alpha$ on $E$ such that

$\mathbb{P}_{\alpha}(X_{t}\in\cdot\mid
t<\tau_{\partial})=\alpha(\cdot),\quad\forall t\geq 0,$

where $\mathbb{P}_{\alpha}$ denotes the distribution of the process $X$ given
that $X_{0}$ has distribution $\alpha$, and

$\tau_{\partial}:=\inf\\{t\geq 0:X_{t}=\partial\\}.$

We refer to [15, 22, 19] for general introductions to the topic.

We consider a general diffusion $(X_{t})_{t\geq 0}$ on $[a,b)$ with
$a<b\leq+\infty$ absorbed at $\partial=a$ in a.s. finite time
$\tau_{\partial}$ and such that $b$ is an entrance or natural boundary. We are
interested in the existence of a quasi-stationary distribution $\alpha$ on
$(a,b)$ such that, for all probability measures $\mu$ on $(a,b)$,

$\displaystyle\left\|\mathbb{P}_{\mu}(X_{t}\in\cdot\mid
t<\tau_{\partial})-\alpha\right\|_{TV}\leq Ce^{-\gamma t},\ \forall t\geq 0$
(1.1)

where $\|\cdot\|_{TV}$ is the total variation norm and $C$ and $\gamma$ are
positive constants. We prove that this property is _equivalent_ to the
following condition:

(B)

The boundary $b$ is entrance for $X$ and there exists $t,A>0$ such that

$\mathbb{P}_{x}(\tau_{\partial}>t)\leq As(x),\quad\forall x>0,$

where $s$ is the scale function of the diffusion $X$.

In the sequel, without loss of generality, we will focus on the case of a
diffusion on natural scale on $E\cup\\{\partial\\}=[0,+\infty)$ and modify (B)
accordingly.

The exponential convergence (1.1) entails number of additional properties.
First, it implies the uniqueness of the quasi-stationary distribution $\alpha$
which attracts any initial distribution. Second, (1.1) implies [5, Theorem
2.1] that the following convergence is uniform in $x\in(a,b)$:

$\displaystyle\eta(x)=\lim_{t\rightarrow\infty}\frac{\mathbb{P}_{x}(t<\tau_{\partial})}{\mathbb{P}_{\alpha}(t<\tau_{\partial})}=\lim_{t\rightarrow+\infty}e^{\lambda_{0}t}\mathbb{P}_{x}(t<\tau_{\partial}).$
(1.2)

Note that we used the existence of a constant $\lambda_{0}>0$ such that

$\mathbb{P}_{\alpha}(t<\tau_{\partial})=e^{-\lambda_{0}t},\quad\forall t\geq
0,$

which is a well-known property of quasi-stationnary distributions. Moreover
[3, Prop. 2.3], $\eta$ belongs to the domain of the infinitesimal generator
$L$ of $X$ on the Banach space of bounded measurable functions on $[a,b)$
equipped with the $L^{\infty}$ norm and

$\displaystyle L\eta=-\lambda_{0}\eta.$

Third, it implies the existence and the exponential ergodicity of the
associated $Q$-process, defined as the process $X$ conditionned to never be
extinct (see [3, Thm. 3.1] for a precise definition). The convergence of the
conditional laws of $X$ to the law of the $Q$-process holds also uniformly in
total variation norm [5, Theorem 2.1], which also entails conditional ergodic
properties [5, Corollary 2.2].

The second main result of this paper is the fact that Condition (B) is closely
related to properties of strict local martingale diffusions $(Z_{t},t\geq 0)$
on $[0,\infty)$. We recall that $(Z_{t},t\geq 0)$ is called a strict local
martingale if $\mathbb{E}_{x}(Z_{t})<x$ for all $x,t>0$. We say that
$(Z_{t},t\geq 0)$ is a _strict martingale in the strong sense_ if, for all
$t>0$,

$\sup_{x>0}\mathbb{E}_{x}(Z_{t})<\infty.$ (1.3)

We prove, under a mild control on the oscillations of the speed measure near
$+\infty$, that the property (1.3) is true for all $t>0$. This new result has
its own interest in the theory of strict local martingales [20] and applies to
nearly all strict local martingale diffusions on $[0,\infty)$. For instance,
it is well-known that the solution on $(0,+\infty)$ to the SDE

$\displaystyle dZ_{t}=Z_{t}^{\alpha}dB_{t}$

is a strict local martingale if and only if $\alpha>1$. Our result implies
that, for all $\alpha>1$ and all $t>0$, (1.3) holds true, and hence that
$\mathbb{E}_{x}(Z_{t})<x$ for all $x>0$ is actually equivalent to
$\sup_{x>0}\mathbb{E}_{x}(Z_{t})<\infty$. This equivalence is also true for
example for the SDE

$\displaystyle dZ_{t}=Z_{t}\ln(Z_{t})^{\beta}dB_{t}$

which is a (strong) strict local martingale iff $\beta>1/2$.

We deduce from these results an explicit criterion ensuring Condition (B),
which shows that it is satisfied for nearly all diffusions coming down from
infinity and hitting 0 in a.s. finite time. However, in several cases,
Condition (B) is simple enough to be checked using elementary probability
tools, and to apply to many other settings (processes with jumps, piecewise
deterministic paths [3] or time-inhomogeneous processes [4]). Hence we also
provide a simple probabilistic criterion and show how it applies in an example
of one-dimensional process with jumps.

We also provide an explicit formula relating the speed measure of a diffusion
on natural scale and its quasi-stationary distribution. Conversely, we show
that, given a probability measure $\alpha$ on $(0,\infty)$ satisfying suitable
conditions and a positive constant $\lambda_{0}$, there exists a unique
diffusion on natural scale, with explicit speed measure, admitting $\alpha$ as
unique quasi-stationary distribution with associated absorbing rate
$\lambda_{0}$. We also use this explicit formula to compute the quasi-
stationary distribution of the sticky Brownian motion on $(-1,1)$, absorbed at
$\\{-1,1\\}$.

The usual tools to prove convergence as in (1.1) for conservative Markov
processes involve coupling arguments (Doeblin’s condition, Dobrushin coupling
constant, see e.g. [16]). However, a diffusion conditioned not to hit 0 before
a given time $t>0$ is a time-inhomogeneous diffusion process with a singular,
non-explicit drift for which these methods fail. For instance, the solution
$(X_{s})_{s\geq 0}$ to the SDE

$\displaystyle dX_{s}=\sigma(X_{s})dB_{s}+b(X_{s})ds$

with smooth coefficients and conditioned not to hit 0 up to a time $t>0$ has
the law of the solution $(X^{(t)}_{s})_{s\in[0,t]}$ to the SDE

$\displaystyle
dX^{(t)}_{s}=b(X^{(t)}_{s})ds+\sigma(X^{(t)}_{s})dB_{s}+\sigma(X^{(t)}_{s})\,\left[\nabla\ln\mathbb{P}_{\cdot}(t-s<\tau_{\partial})\right](X^{(t)}_{s})ds.$

Since $\mathbb{P}_{x}(t-s<\tau_{\partial})$ vanishes when $x$ converges to 0,
the drift term in the above SDE is singular and cannot in general be
quantified.

Hence, convergence of conditioned diffusion processes have been obtained up to
now using (sometimes involved) spectral theoretic arguments, which most often
require regularity of the coefficients (see for instance [2, 13, 10, 7, 17])
and are specific to time homogeneous one-dimensional diffusion processes. None
of the above references provide uniform convergence with respect to the
initial distribution. In particular, all the consequences of (1.1) listed
above are original results of this paper, even for solutions to SDEs with
regular coefficients. Moreover, the uniformity of the convergence with respect
to the initial distribution is important in applications, as discussed in [15,
Ex. 2]. Concerning the sole existence and uniqueness of a quasi-stationnary
distribution, in the few cases where our criterion is not satisfied, we refer
the reader to [13, 10, 7]. An interesting open problem is to prove that (1.1),
and hence Condition (B), actually holds true in these cases.

The paper is organized as follows. In Section 2, we precisely define the
absorbed diffusion processes and recall their construction as time-changed
Brownian motions. Section 3 contains several results on strict local
martingales, including the criterion ensuring (1.3). We give in Section 4 the
statements of all our main results on quasi-stationnary distributions
(Subsections 4.1 and 4.4) and on criteria ensuring Condition (B) (Subsections
4.2 and 4.3). We also study in Subsection 4.5 several examples including
sticky Brownian motion and diffusions with jumps. Finally, we give in Section
5 the proofs of the main results of Section 4.

##  2 Some reminders on diffusion processes

In this section, we recall the construction of general diffusion processes
$(X_{t},t\geq 0)$ on $[0,+\infty)$, absorbed at $\partial=0$. The typical
situation corresponds to stochastic population dynamics of continuous
densities with possible extinction.

The distribution of $X$ given $X_{0}=x\in[0,\infty)$ will be denoted
$\mathbb{P}_{x}$, and the semigroup of the process is given by
$P_{t}f(x)=\mathbb{E}_{x}[f(X_{t})]$ for all bounded measurable
$f:[0,\infty)\rightarrow\mathbb{R}$ and all $x\geq 0$.

A stochastic process $(X_{t},t\geq 0)$ on $[0,+\infty)$ is called a diffusion
if it has a.s. continuous paths in $[0,\infty)$, satisfies the strong Markov
property and is _regular_. By regular, we mean that for all $x\in(0,\infty)$
and $y\in[0,\infty)$, $\mathbb{P}_{x}(T_{y}<\infty)>0$, where $T_{y}$ is the
first hitting time of $y$ by the process $X$. Examples of such processes are
given by (weak) solutions of stochastic differential equations (see Section
4.5).

Given such a process, there exists a continuous and strictly increasing
function $s$ on $[0,\infty)$, called the _scale function_ , such that
$(s(X_{t\wedge T_{0}}),t\geq 0)$ is a local martingale [6]. The stochastic
process $(s(X_{t}),t\geq 0)$ is itself a diffusion process with identity scale
function. Since we shall assume that 0 is regular or exit and that
$T_{0}<\infty$ a.s., we necessarily have $s(0)>-\infty$ and
$s(\infty)=\infty$, and we can assume $s(0)=0$. Hence, replacing $(X_{t},t\geq
0)$ by $(s(X_{t}),t\geq 0)$, we can assume without loss of generality that
$s(x)=x$. Such a process is said to be on _natural scale_ and satisfies for
all $0<a<b<\infty$,

$\mathbb{P}_{x}(T_{a}<T_{b})=\frac{b-x}{b-a}.$

To such a process $X$, one can associate a unique locally finite positive
measure $m(dx)$ on $(0,+\infty)$, called the _speed measure_ of $X$, which
gives positive mass to any open subset of $(0,+\infty)$ and such that
$X_{t}=B_{\sigma_{t}}$ for all $t\geq 0$ for some standard Brownian motion
$B$, where

$\displaystyle\sigma_{t}=\inf\left\\{s>0:A_{s}>t\right\\},\text{ with
}A_{s}=\int_{0}^{\infty}L^{x}_{s}\,m(dx).$ (2.1)

and $L^{x}$ is the local time of $B$ at level $x$. Conversely, any such time
change of a Brownian motion defines a regular diffusion on $[0,\infty)$ [9,
Thm. 23.9].

We will use in the sequel the assumption that

$\displaystyle\int_{0}^{\infty}y\,m(dy)<\infty.$ (2.2)

Since the measure $m$ is locally finite on $(0,\infty)$, this assumption
reduces to local integrability of $y\,m(dy)$ near 0 and $+\infty$. These two
conditions are respectively equivalent to
$\mathbb{P}_{x}(\tau_{\partial}<\infty)=1$ for all $x\in(0,\infty)$ [9, Thm.
23.12], where $\tau_{\partial}:=T_{0}$ is the first hitting time of
$\partial=0$ by the process $X$, and to the existence of $t>0$ and $y>0$ such
that

$\inf_{x>y}\mathbb{P}_{x}(T_{y}<t)>0.$

This means that $\infty$ is an _entrance boundary_ for $X$ [9, Thm. 23.12], or
equivalently that the process $X$ _comes down from infinity_ [2].

We recall that $0$ is absorbing for the process $X$ iff
$\int_{0}^{1}\,m(dx)=\infty$ [9, Thm. 23.12], which is not necessarily the
case under the assumption that $\int_{0}^{1}y\,m(dy)<\infty$. Therefore, we
modify the definition of $X$ so that $\partial$ becomes an absorbing point as
follows:

$X_{t}=\begin{cases}B_{\sigma_{t}}&\text{if\ }0\leq t<\tau_{\partial}\\\
\partial&\text{if\ }t\geq\tau_{\partial}.\end{cases}$ (2.3)

Note that this could be done equivalently by replacing $m$ by
$m+\infty\delta_{0}$.

With this definition, $X$ is a local martingale. Note that $A_{s}$ is strictly
increasing since $m$ gives positive mass to any open interval, and hence
$\sigma_{t}$ is continuous and

$\sigma_{\tau_{\partial}}=T^{B}_{0},\quad\text{or,
equivalently,}\quad\tau_{\partial}=A_{T^{B}_{0}},$ (2.4)

where $T^{B}_{x}$ is the first hitting time of $x\in\mathbb{R}$ by the process
$(B_{t},t\geq 0)$. In particular, $X_{t}=B_{\sigma_{t}\wedge T_{0}^{B}}$,
$\forall t\geq 0$.

##  3 Strict local martingales

This section is devoted to the study of one-dimensional strict local
martingale diffusions on $[0,\infty)$ coming down from infinity and their link
with absorbed one-dimensional diffusion processes. We recall that a
nonnegative, real local martingale $(Z_{t},t\geq 0)$ with $Z_{0}=z\geq 0$ is
called strict if it is not a martingale, i.e. if $\mathbb{E}(Z_{t})<z$ for at
least one $z\in(0,\infty)$. If $(Z_{t},t\geq 0)$ is a diffusion on
$[0,\infty)$ on natural scale with speed measure $\widetilde{m}$, then

$(Z_{t},t\geq 0)\mbox{ is a strict local martingale }\Longleftrightarrow\
\int_{1}^{\infty}y\,\widetilde{m}(dy)<\infty,$ (3.1)

i.e. if $+\infty$ is an entrance boundary [11]. The notion of strict local
martingales is of great importance in the theory of financial bubbles. We
refer the interested reader to Protter [20] and references therein.

The goal of this section is to study a stronger notion of strict martingale.
We will say that a nonnegative, real local martingale $(Z_{t},t\geq 0)$ is a
_strict martingale in the strong sense_ if, for all $t>0$, there exists
$A_{t}<\infty$ such that

$\mathbb{E}_{z}(Z_{t})\leq A_{t},\ \forall z>0.$

###  3.1 Strict local martingales in a strong sense

The next result gives a sufficient criterion for a strict local martingale
diffusion process $Z$ on $[0,\infty)$ to be a strictly local martingale in the
strong sense.

######  Theorem 3.1.

Let $\widetilde{m}$ be a locally finite measure on $(0,\infty)$ giving
positive mass to any open subset of $(0,\infty)$, and let $Z$ be a diffusion
process on $[0,\infty)$ stopped at 0 on natural scale with speed measure
$\widetilde{m}$. Assume

$\int_{1}^{\infty}\frac{1}{x}\,\sup_{y\geq
x}\left(y\int_{y}^{\infty}\widetilde{m}(dz)\right)\,dx<\infty.$ (3.2)

Then, $(Z_{t})_{t\geq 0}$ is a strict local martingale in the strong sense.

Note that Fubini’s theorem implies that

$\displaystyle\int_{1}^{\infty}(z-1)\,\widetilde{m}(dz)$
$\displaystyle=\int_{1}^{\infty}\int_{y}^{\infty}\widetilde{m}(dz)\,dy\leq\int_{1}^{\infty}\frac{1}{x}\left(\sup_{y\geq
x}y\int_{y}^{\infty}\widetilde{m}(dz)\right)dx.$

Hence (3.2) and the fact that $\widetilde{m}$ is locally finite implies that

$\int_{1}^{\infty}y\,\widetilde{m}(dy)<\infty.$ (3.3)

######  Remark 1.

Although we do not know if all strict local martingale diffusions are strict
in the strong sense, the last result shows that it is true for most strict
local martingale diffusions. Indeed, when $\widetilde{m}$ has a density with
respect to Lebesgue’s measure $\Lambda$ on $(0,\infty)$, we recall that
$(Z_{t},t\geq 0)$ is (weak) solution to the SDE

$dZ_{t}=\sigma(Z_{t})dB_{t},$

where $(B_{t},t\geq 0)$ is a standard Brownian motion and
$\widetilde{m}(dx)=\sigma^{-2}(x)\Lambda(dx)$. In this case, the condition
(3.2) covers almost all the practical situations where (3.3) is satisfied. For
example, it is true if

$\displaystyle\frac{d\widetilde{m}}{d\Lambda}(x)\leq\frac{C}{x^{2}\log
x\cdots\log_{k-1}x\left(\log_{k}x\right)^{1+\epsilon}},$

or equivalently

$\displaystyle\sigma(x)\geq C^{\prime}x\sqrt{\log
x\cdots\log_{k-1}x\left(\log_{k}x\right)^{1+\epsilon}},$

for all $x$ large enough and for some $\epsilon>0$ and $k\geq 1$, where
$\log_{k}(x)=\log_{k-1}(\log x)$ (see [14, Ex. 2]). In other words, (3.2)
could fail only when $\widetilde{m}$ has strong oscillations close to
$+\infty$ and $y\,\widetilde{m}(dy)$ is nearly non-integrable at infinity.

###### Proof.

Since $(Z_{t},t\geq 0)$ is a positive local martingale, hence a
supermartingale, and a strong Markov process, we have for all $z\geq 1$ and
$t>0$,

$\displaystyle\mathbb{E}_{z}(Z_{t})$
$\displaystyle=\mathbb{E}_{z}(Z_{t}\mathbbm{1}_{t<T^{Z}_{1}})+\mathbb{E}_{z}(Z_{t}\mathbbm{1}_{T^{Z}_{1}<t})$
$\displaystyle\leq\mathbb{E}_{z}(Z_{t\wedge
T^{Z}_{1}}\mathbbm{1}_{t<T^{Z}_{1}})+\mathbb{E}_{z}[\mathbbm{1}_{T^{Z}_{1}<t}\mathbb{E}_{1}(Z_{t-T^{Z}_{1}})]$
$\displaystyle\leq\mathbb{E}_{z}(Z_{t\wedge T^{Z}_{1}})+1,$

where $T^{Z}_{1}$ is the first hitting time of 1 by $Z$. Hence, we only need
to prove that $\sup_{z\geq 1}\mathbb{E}_{z}(Z^{T_{1}}_{t})<\infty$, where
$Z^{T_{1}}$ is the diffusion process $Z$ stopped at time $T^{Z}_{1}$.

Since $Z^{T_{1}}$ is a diffusion process on $[1,\infty)$, stopped at 1, on
natural scale and with speed measure
$\nu(\cdot)=\widetilde{m}(\cdot\cap(1,\infty))$ satisfying

$\nu([1,2])<\infty,$

we can apply the result of [14, Thm. 4.1, Ex. 2]. This result ensures that,
under conditions (3.3) and (3.2), the probability density function of
$Z^{T_{1}}_{t}$ with respect to $\nu$, denoted by $p(x,y,t)$, is well defined
for all $t>0$ and there exists a constant $A^{\prime}_{t}>0$ such that

$\displaystyle\sup_{1\leq x,y<+\infty}p(x,y,t)\leq A^{\prime}_{t},\quad\forall
t>0.$

As a consequence, for all $t>0$,

$\displaystyle\mathbb{E}_{z}(Z^{T_{1}}_{t})$
$\displaystyle=\int_{1}^{\infty}y\,p(z,y,t)\,d\nu(y)\leq
A^{\prime}_{t}\int_{1}^{\infty}y\,d\widetilde{m}(y).\qed$

###  3.2 Links between strict local martingales and absorbed diffusions

The next result gives a change of measure linking the behavior near 0 and near
$+\infty$ of diffusion processes on $[0,\infty)$ on natural scale. Note that
it does not require that $X$ comes down from infinity nor that it hits 0 in
finite time.

######  Theorem 3.2.

Let $X$ be a diffusion process on $[0,+\infty)$ on natural scale stopped at 0
with speed measure $m$. We denote by $\mathbb{P}_{x}$ its law when $X_{0}=x$,
defined on the canonical space of continuous functions from $\mathbb{R}_{+}$
to itself, equipped with its canonical filtration $(\mathcal{F}_{t})_{t\geq
0}$. For all $x>0$, we define the following $h$-transform of $\mathbb{P}_{x}$:

$\mathchoice{{d\widetilde{\mathbb{Q}}_{x}\,\smash{\vrule
height=5.7778pt,depth=2.2119pt}}_{\,\mathcal{F}_{\tau}}}{{d\widetilde{\mathbb{Q}}_{x}\,\smash{\vrule
height=5.7778pt,depth=2.2119pt}}_{\,\mathcal{F}_{\tau}}}{{d\widetilde{\mathbb{Q}}_{x}\,\smash{\vrule
height=5.7778pt,depth=1.74722pt}}_{\,\mathcal{F}_{\tau}}}{{d\widetilde{\mathbb{Q}}_{x}\,\smash{\vrule
height=5.7778pt,depth=1.74722pt}}_{\,\mathcal{F}_{\tau}}}=\frac{X_{\tau}}{x}\mathchoice{{d\mathbb{P}_{x}\,\smash{\vrule
height=5.55557pt,depth=2.2119pt}}_{\,\mathcal{F}_{\tau}}}{{d\mathbb{P}_{x}\,\smash{\vrule
height=5.55557pt,depth=2.2119pt}}_{\,\mathcal{F}_{\tau}}}{{d\mathbb{P}_{x}\,\smash{\vrule
height=3.88889pt,depth=1.74722pt}}_{\,\mathcal{F}_{\tau}}}{{d\mathbb{P}_{x}\,\smash{\vrule
height=2.77777pt,depth=1.74722pt}}_{\,\mathcal{F}_{\tau}}},$

for all $(\mathcal{F}_{t})$-stopping time $\tau$ such that
$(X_{t\wedge\tau},t\geq 0)$ is a uniformly integrable martingale. Then, under
$(\widetilde{\mathbb{Q}}_{x})_{x>0}$, the process $Z_{t}:=1/X_{t}$ is a
diffusion process stopped at 0 on natural scale with speed measure

$\widetilde{m}(dz)=\frac{1}{z^{2}}(f*m)(dz),$ (3.4)

where $f*m$ is the pushforward measure of $m$ by the application $f(x)=1/x$.

In particular, if the measure $m(dx)$ is absolutely continuous w.r.t.
Lebesgue’s measure $\Lambda$ on $(0,+\infty)$, then $\widetilde{m}(dz)$ is
absolutely continuous w.r.t. $\Lambda$ and

$\frac{d\widetilde{m}}{d\Lambda}(z)=\frac{1}{z^{4}}\frac{dm}{d\Lambda}\left(\frac{1}{z}\right),\quad\forall
z>0.$ (3.5)

###### Proof.

The fact that $\widetilde{\mathbb{Q}}_{x}$, $x>0$, are probability measures
and that the process $Z=1/X$ is a diffusion on $[0,\infty)$ stopped at 0 on
natural scale under $(\widetilde{\mathbb{Q}}_{x})_{x>0}$ are proved in [18]
(see Lemma 2.5 for the last point). Hence, we only have to check that
$\widetilde{m}$ defined in (3.4) is its speed measure. By Green’s formula [9,
Lemma 23.10], this means that, for all $0<a<z<b<+\infty$ and for all bounded
measurable function $g:(0,\infty)\rightarrow\mathbb{R}$,

$\mathbb{E}^{\widetilde{\mathbb{Q}}_{1/z}}\left[\int_{0}^{T^{Z}_{a,b}}g(Z_{t})\,dt\right]=2\int_{a}^{b}\frac{(z\wedge
y-a)(b-z\vee y)}{b-a}g(y)\,\widetilde{m}(dy),$ (3.6)

where $T^{Z}_{a,b}:=\inf\\{t\geq 0:Z_{t}\in\\{a,b\\}\\}$.

On the one hand, the definition of $\widetilde{m}$ and the change of variable
formula entails that the right hand term of (3.6) is equal to

$2\int_{a}^{b}\frac{(z\wedge y-a)(b-z\vee
y)}{b-a}\frac{g(y)}{y^{2}}\,(f*m)(dy)\\\ \begin{aligned}
&=2\int_{1/b}^{1/a}\frac{(z\wedge\frac{1}{x}-a)(b-z\vee\frac{1}{x})}{b-a}x^{2}g(1/x)\,m(dx)\\\
&=2z\int_{1/b}^{1/a}\frac{(\frac{1}{a}-\frac{1}{z}\vee
x)(x\wedge\frac{1}{z}-\frac{1}{b})}{\frac{1}{a}-\frac{1}{b}}xg(1/x)\,m(dx)\\\
&=z\mathbb{E}_{1/z}\left[\int_{0}^{T_{1/b,1/a}}X_{t}g(1/X_{t})\,dt\right],\end{aligned}$

where the last equality comes from the fact that $X$ is a diffusion on natural
scale with speed measure $m$ under $(\mathbb{P}_{x})_{x\geq 0}$.

On the other hand, since $(X_{t\wedge T_{1/b,1/a}},t\geq 0)$ is a bounded
martingale under $\mathbb{P}_{1/z}$, we obtain by definition of
$(\widetilde{\mathbb{Q}}_{x})_{x>0}$

$\displaystyle\mathbb{E}^{\widetilde{\mathbb{Q}}_{1/z}}\left[\int_{0}^{T^{Z}_{a,b}}g(Z_{t})\,dt\right]$
$\displaystyle=\mathbb{E}^{\widetilde{\mathbb{Q}}_{1/z}}\left[\int_{0}^{T_{1/b,1/a}}g(1/X_{t})\,dt\right]$
$\displaystyle=z\mathbb{E}_{1/z}\left[X_{T_{1/b,1/a}}\int_{0}^{T_{1/b,1/a}}g(1/X_{t})\,dt\right].$

Now, Itô’s formula implies that, a.s. for all $t\geq 0$,

$X_{t}\int_{0}^{t}g(1/X_{s})\,ds=\int_{0}^{t}\left(\int_{0}^{s}g(1/X_{u})\,du\right)\,dX_{s}+\int_{0}^{t}X_{s}g(1/X_{s})\,ds.$

Since $(X_{t\wedge T_{1/b,1/a}},t\geq 0)$ is a bounded martingale under
$\mathbb{P}_{1/z}$, for all $n\geq 1$,

$\mathbb{E}_{1/z}\left[X_{n\wedge T_{1/b,1/a}}\int_{0}^{n\wedge
T_{1/b,1/a}}g(1/X_{s})\,ds\right]=\mathbb{E}_{1/z}\left[\int_{0}^{n\wedge
T_{1/b,1/a}}X_{s}g(1/X_{s})\,ds\right].$

Since $\mathbb{E}_{1/z}[T_{1/b,1/a}]<\infty$ (this is a general property of
diffusions), Lebesgue’s theorem implies (3.6), which ends the proof of Theorem
3.2. ∎

The nature (regular/exit or entrance) of the boundary points 0 and $\infty$
for $X$ and $Z$ are related.

######  Corollary 3.3.

With the previous notation, $X$ hits 0 in a.s. finite time iff $Z$ comes down
form infinity, and $X$ comes down from infinity iff $Z$ hits 0 in a.s. finite
time. In particular,

$\displaystyle(Z_{t})_{t\geq 0}\text{ is a strict local martingale}\
\Longleftrightarrow\ (X_{t})_{t\geq 0}\mbox{ hits 0 in a.s.\ finite time}.$

###### Proof.

We need to check that

$\int_{0}^{1}y\,m(dy)<\infty\quad\Longleftrightarrow\quad\int_{1}^{\infty}y\,\widetilde{m}(dy)<\infty$

and

$\int_{1}^{\infty}y\,m(dy)<\infty\quad\Longleftrightarrow\quad\int_{0}^{1}y\,\widetilde{m}(dy)<\infty.$

This follows from the definition (3.4) of $\widetilde{m}$ by a simple change
of variable. ∎

The next corollary gives an important relationship between the absorption
probability of $X$ under $\mathbb{P}_{x}$ and the expectation of $Z$ under
$\widetilde{\mathbb{Q}}_{x}$. We obtain in particular a condition on the
diffusion $X$ equivalent to the fact that $Z$ is a strict local martingale in
the strong sense.

######  Corollary 3.4.

With the previous notation, for all $t>0$ and $x>0$, we have

$\displaystyle\frac{\mathbb{P}^{X}_{x}(t<\tau_{\partial})}{x}=\mathbb{E}^{Z}_{1/x}(Z_{t}),$

where $\mathbb{P}^{X}_{x}$ and $\mathbb{P}^{Z}_{z}$ are the respective
distributions of the diffusion processes $X$ such that $X_{0}=x$ and $Z$ such
that $Z_{0}=z$. In particular, for any constant $A>0$ and $t>0$,

$\sup_{z>0}\mathbb{E}^{Z}_{z}(Z_{t})\leq A\ \Longleftrightarrow\
\mathbb{P}^{X}_{x}(t<\tau_{\partial})\leq Ax,\ \forall x\geq 0.$ (3.7)

###### Proof of Corollary 3.4.

We have, for all $x>0$ and all $t>0$,

$\mathbb{E}^{Z}_{1/x}(Z_{t})=\mathbb{E}^{\widetilde{\mathbb{Q}}_{x}}\left(\frac{1}{X_{t}}\right)=\frac{1}{x}\mathbb{P}^{X}_{x}(X_{t}>0).\qed$

##  4 Quasi-stationary distribution for one dimensional diffusions

We consider in this section a diffusion process $X$ on $[0,\infty)$ on natural
scale hitting 0 in a.s. finite time $\tau_{\partial}$. Recall that this is
equivalent to assuming

$\int_{0}^{1}y\,m(dy)<\infty$

where $m$ is the speed measure of $X$. We also set $X_{t}=0$ for all
$t>\tau_{\partial}$.

###  4.1 Exponential convergence to quasi-stationary distribution

The aim of this section is to establish that the following Condition (B) is
equivalent to the exponential convergence to a unique quasi-stationary
distribution of conditional distributions of $X$.

(B)

The process $X$ comes down from infinity and there exist two constants
$t_{1},A>0$ such that

$\mathbb{P}_{x}(\tau_{\partial}>t_{1})\leq Ax,\quad\forall x>0.$ (4.1)

Let us recall that $X$ comes down from infinity iff
$\int_{1}^{\infty}y\,m(dy)<\infty$.

######  Remark 2.

Note that it is well known that, for diffusion processes on natural scale on
$[0,\infty)$, $\mathbb{P}_{x}(\tau_{\partial}<\infty)>0$ for some $x>0$
implies that $\mathbb{P}_{x}(\tau_{\partial}<\infty)=1$ for all $x>0$ (it can
be also deduced from (3.1) and the results of Section 3.2). Hence (4.1)
implies that $\int_{0}^{1}y\,m(dy)<\infty$.

######  Remark 3.

The converse inequality in (4.1) is true for all diffusion process $X$ on
natural scale on $(0,+\infty)$: for all $t>0$, there exists a constant $a>0$
such that

$\mathbb{P}_{x}(t<\tau_{\partial})\geq ax,\quad\forall x\in(0,1).$

Indeed, by the strong Markov property at time $T_{1}$,

$\displaystyle\mathbb{P}_{x}(t<\tau_{\partial})\geq\mathbb{E}_{x}[\mathbbm{1}_{T_{1}<\tau_{\partial}}\mathbb{P}_{1}(t<\tau_{\partial})]=\mathbb{P}_{x}(T_{1}<\tau_{\partial})\mathbb{P}_{1}(t<\tau_{\partial})=x\mathbb{P}_{1}(t<\tau_{\partial}).$

The following result is proved in Section 5.

######  Theorem 4.1.

Assume that $X$ is a one-dimensional diffusion on natural scale on
$[0,\infty)$ with speed measure $m(dx)$ such that $\tau_{\partial}<\infty$
a.s. Then we have equivalence between

(i)

Assumption (B).

(ii)

There exist a probability measure $\alpha$ on $(0,\infty)$ and two constants
$C,\gamma>0$ such that, for all initial distribution $\mu$ on $(0,\infty)$,

$\displaystyle\left\|\mathbb{P}_{\mu}(X_{t}\in\cdot\mid
t<\tau_{\partial})-\alpha(\cdot)\right\|_{TV}\leq Ce^{-\gamma t},\ \forall
t\geq 0.$ (4.2)

In this case, $\alpha$ is the unique quasi-stationary distribution for the
process and there exists $\lambda_{0}>0$ such that
$\mathbb{P}_{\alpha}(t<\tau_{\partial})=e^{-\lambda_{0}t}$. Moreover, $\alpha$
is absolutely continuous with respect to $m$ and

$\displaystyle\frac{d\alpha}{dm}(x)=2\lambda_{0}\int_{0}^{\infty}(x\wedge
y)\,\alpha(dy).$ (4.3)

In addition,

$\int_{0}^{\infty}y\,\alpha(dy)<\infty.$

In the next result, we provide additional information on the function $\eta$
defined in (1.2).

######  Proposition 4.2.

Assume that $X$ is a one-dimensional diffusion on natural scale with speed
measure $m(dx)$ such that $\tau_{\partial}<\infty$ a.s. and such that (B) is
satisfied. Then the limit $\eta$ in (1.2) is given by the normalized right
hand side of (4.3):

$\displaystyle\eta(x)=\frac{\displaystyle{\int_{0}^{\infty}(x\wedge
y)\,\alpha(dy)}}{\displaystyle{\int_{0}^{\infty}\int_{0}^{\infty}(y\wedge
z)\,\alpha(dy)\,\alpha(dz)}}.$ (4.4)

Moreover, the function $\eta$ is non-decreasing, bounded, $\alpha(\eta)=1$ and
$\eta(x)\leq Cx$ for all $x\geq 0$ and some constant $C>0$.

######  Remark 4.

We deduce from the last results that the density of $\alpha$ with respect to
$m$ is proportional to $\eta$. Hence it is positive, bounded, (strictly)
increasing on $(0,+\infty)$, and differentiable at each point of
$(0,+\infty)$.

###  4.2 Conditions ensuring (B) using elementary probabilistic tools

The next result gives a condition implying (B). Its proof is particularly
simple. We give in the next section a slightly more general criterion which
makes use of the results on strict local martingales of Section 3.

######  Theorem 4.3.

With the previous notation, assume that $X$ comes down from infinity, hits 0
a.s. in finite time and for all $x\in(0,1)$,

$I(x):=\int_{0}^{x}y\,m(dy)\leq Cx^{\rho}$ (4.5)

for some constants $C>0$ and $\rho>0$. Then, for all $t>0$, there exists
$A_{t}<\infty$ such that

$\displaystyle\mathbb{P}_{x}(t<\tau_{\partial})\leq A_{t}\,x,\ \forall x>0.$

We immediately deduce the next corollary.

######  Corollary 4.4.

Assume that $X$ comes down from infinity and that $0$ is regular, i.e.
$\int_{0}^{1}m(dx)<\infty$. Then Condition (B) is satisfied.

######  Remark 5.

Equation (4.5) is satisfied if $m$ is absolutely continuous w.r.t. Lebesgue’s
measure $\Lambda$ on $(0,\infty)$ and if

$\frac{dm}{d\Lambda}(x)\leq\frac{C}{x^{\alpha}}$

in the neighborhood of $0$, where $\alpha<2$. This corresponds to a diffusion
process solution to the SDE

$dX_{t}=\sigma(X_{t})dB_{t}$

with $\sigma(x)\geq cx^{\alpha/2}$ in the neighborhood of $0$.

###### Proof.

The proof is based on the following lemma.

######  Lemma 4.5.

For all $k\geq 1$ and $x\in\mathbb{R}_{+}$, the function
$M_{k}(x)=\mathbb{E}_{x}\left(\tau_{\partial}^{k}\right)$ is bounded and
satisfies

$\displaystyle M_{k}(x)=2k\int_{0}^{\infty}(x\wedge
y)\,M_{k-1}(y)\,m(dy)\,<\infty.$ (4.6)

We admit this lemma for the moment and give its proof at the end of the
subsection.

First, notice that we can assume without loss of generality that
$\rho\not\in\mathbb{Q}$. We prove by induction that there exists a constant
$C_{k}>0$ such that for all $x\in(0,1)$,

$\displaystyle M_{k}(x)\leq C_{k}x^{(k\rho)\wedge 1}.$ (4.7)

This is enough to conclude since one can choose $k$ such that $k\rho>1$ and
use Markov’s inequality to obtain

$\mathbb{P}_{x}(t<\tau_{\partial})\leq\frac{M_{k}(x)}{t^{k}}\leq C_{t}x$

where the constant $C_{t}$ depends on $t$ but not on $x$.

Equation (4.7) is trivial for $k=0$. Let us assume it is true for $k\geq 0$
and let us denote by $\alpha_{k}$ the constant $(k\rho)\wedge 1$. For all
$x\in(0,1)$, by (4.6),

$\displaystyle M_{k+1}(x)$ $\displaystyle\leq
C\left(\int_{0}^{x}yM_{k}(y)m(dy)+x\int_{x}^{1}M_{k}(y)m(dy)+x\int_{1}^{\infty}M_{k}(y)m(dy)\right)$
$\displaystyle\leq
C\left(\int_{0}^{x}y^{1+\alpha_{k}}\,m(dy)+x\int_{x}^{1}y^{\alpha_{k}-1}\,y\,m(dy)+x\int_{1}^{\infty}m(dy)\right)$
$\displaystyle\leq
C\left[x^{\alpha_{k}}I(x)+x\left(I(1)+(1-\alpha_{k})\int_{x}^{1}y^{\alpha_{k}-2}I(y)\,dy\right)+x\int_{1}^{\infty}m(dy)\right]$
$\displaystyle\leq
Cx^{\alpha_{k}+\rho}+Cx+Cx\int_{x}^{1}y^{\alpha_{k}+\rho-2}\,dy,$

where the constant $C>0$ depends on $k$ and may change from line to line, and
the third inequality follows from integration by parts. Since
$\rho\not\in\mathbb{Q}$, we have

$\int_{x}^{1}y^{\alpha_{k}+\rho-2}\,dy\leq\begin{cases}1/(\alpha_{k}+\rho-1)&\text{if
}\alpha_{k}+\rho-1>0,\\\ x^{\alpha_{k}+\rho-1}/(1-\alpha_{k}-\rho)&\text{if
}\alpha_{k}+\rho-1<0.\end{cases}$

This concludes the induction. ∎

###### Proof of Lemma 4.5.

Let us define the functions $M_{k}(x)$ recursively from the formula (4.6) and
$M_{0}(x)=1$. Our aim is hence to prove that, for all $k\geq 0$, $M_{k}$ is
uniformly bounded and that, for all $x\in(0,+\infty)$, that
$M_{k}(x)=\mathbb{E}_{x}(\tau_{\partial}^{k})$.

An immediate induction procedure and (2.2) proves that, for all $k\geq 0$, the
function $M_{k}$ is bounded.

The Green function for diffusions on natural scale [9, Lemma 23.10] implies
that, for any $0<a<b<\infty$,

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{T_{a,b}}M_{k-1}(X_{t})\,dt\right)=2\int_{a}^{b}\frac{(x\wedge
y-a)(b-x\vee y)}{b-a}M_{k-1}(y)\,m(dy).$

Now, letting $a\rightarrow 0$ (by monotone convergence) and
$b\rightarrow\infty$ (by monotone convergence in the l.h.s and by Lebesgue’s
Theorem in the right hand side - remember that $M_{k-1}$ is bounded), we
obtain

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{\tau_{\partial}}M_{k-1}(X_{t})\,dt\right)=2\int_{0}^{\infty}(x\wedge
y)\,M_{k-1}(y)\,m(dy)$

and hence

$\displaystyle
M_{k}(x)=k\,\mathbb{E}_{x}\left(\int_{0}^{\tau_{\partial}}M_{k-1}(X_{t})\,dt\right).$

Let us now prove by induction that $M_{k}(x)=\mathbb{E}(\tau_{\partial}^{k})$
for all $k\geq 0$. The property is true for $k=0$. Assume that
$M_{k-1}(x)=\mathbb{E}_{x}(\tau_{\partial}^{k-1})$, then

$\displaystyle
M_{k}(x)=k\mathbb{E}_{x}\left(\int_{0}^{\tau_{\partial}}\mathbb{E}_{X_{t}}(\tau_{\partial}^{k-1})\,dt\right)=k\mathbb{E}_{x}\left(\int_{0}^{\tau_{\partial}}\left(\tau_{\partial}-t\right)^{k-1}\,dt\right)=\mathbb{E}_{x}(\tau_{\partial}^{k}),$

where we used the Markov property. ∎

###  4.3 Conditions ensuring (B) in relation with strict local martingales

The next result gives a sufficient condition for (B), which is a weaker
requirement than the one of Theorem 4.3. It is a direct consequence of Theorem
3.1, the definition (3.4) of $\widetilde{m}$ and the equivalence (3.7).

######  Theorem 4.6.

Assume that $X$ comes down from infinity and

$\int_{0}^{1}\frac{1}{x}\,\sup_{y\leq
x}\left(\frac{1}{y}\int_{(0,y)}z^{2}\,m(dz)\right)\,dx<\infty.$ (4.8)

Then, for all $t>0$, there exists $A_{t}<\infty$ such that

$\displaystyle\mathbb{P}_{x}(t<\tau_{\partial})\leq A_{t}\,x,\ \forall x>0.$

Note that Fubini’s theorem implies that

$\displaystyle\frac{1}{2}\int_{0}^{1/2}z\,m(dz)\leq\int_{0}^{1}z(1-z)\,m(dz)$
$\displaystyle=\int_{0}^{1}\frac{1}{x^{2}}\int_{0}^{x}z^{2}\,m(dz)$
$\displaystyle\leq\int_{0}^{1}\frac{1}{x}\left(\sup_{y\leq
x}\frac{1}{y}\int_{0}^{y}z^{2}\,m(dz)\right)dx.$

Hence (4.8) implies that $\int_{0}^{1}z\,m(dz)<\infty$, i.e. $X$ hits 0 in
a.s. finite time.

The last theorem is based (via Theorem 3.1) on abstract but powerful general
analytical results on the density of diffusion processes, which are hard to
extend to more general settings. But Condition (B) is simple enough to be
checked for other classes of models using elementary probability tools, as we
did in Theorem 4.3. We give in Subsection 4.5.3 a simple example of a one-
dimensional process with jumps where exponential convergence to the quasi-
stationary distribution can be proved by an easy extension of Theorem 4.3.

######  Remark 6.

As in Remark 1, in the case of a (weak) solution to the SDE
$dX_{t}=\sigma(X_{t})dB_{t}$, the condition (4.8) covers almost all the
practical situations where $\int_{0}^{\infty}y\sigma^{-2}(y)\,dy<\infty$. For
example, it is true if $X$ comes down from infinity and

$\displaystyle\sigma(x)\geq
Cx\sqrt{\log\frac{1}{x}\cdots\log_{k-1}\frac{1}{x}\left(\log_{k}\frac{1}{x}\right)^{1+\epsilon}}$

for all $x$ in a neighborhood of $0$ and for some $\epsilon,C>0$ and $k\geq
1$.

###  4.4 Consequences of the expression of $\alpha$ as a function of $m$

In Theorem 4.1, we proved that, under Assmuption (B), $\alpha$ is absolutely
continuous with respect to $m$ and that

$\displaystyle\frac{d\alpha}{dm}(x)=2\lambda_{0}\int_{0}^{\infty}(x\wedge
y)\alpha(dy),\ \forall x\in(0,+\infty).$ (4.9)

Our goal here is to provide a converse property, i.e. to give a necessary and
sufficient condition on a given positive measure $\alpha$, such that it is the
unique quasi-stationary distribution of a diffusion process on natural scale
on $[0,+\infty)$ with exponential convergence of the conditional laws.

We will say (with a slight abuse of notation) that a positive measure $m$ on
$(0,\infty)$ satisfies (B) if: it is locally finite, gives positive mass on
each open subset of $(0,\infty)$ and the diffusion process on $[0,\infty)$ on
natural scale with speed measure $m$ and stopped at 0 satisfies Condition (B).

######  Theorem 4.7.

Fix $\lambda_{0}>0$ and a probability measure $\alpha$ on $(0,+\infty)$ such
that the measure $(\frac{1}{x}\vee 1)\alpha(dx)$ satisfies (B). Then, there
exists a unique (in law) diffusion process $(X_{t})_{t\geq 0}$ on
$[0,+\infty)$ on natural scale a.s. absorbed at $0$ satisfying

$\displaystyle\left\|\mathbb{P}_{x}(X_{t}\in\cdot\mid
t<\tau_{\partial})-\alpha\right\|_{TV}\leq Ce^{-\gamma t},\quad\forall
x\in(0,+\infty),$

for some positive constants $C,\gamma$ and such that

$\displaystyle\mathbb{P}_{\alpha}(t<\tau_{\partial})=e^{-\lambda_{0}t},\quad\forall
t\geq 0.$

In addition, the speed measure of the process $X$ is given by

$\displaystyle m(dx)=\frac{\alpha(dx)}{2\lambda_{0}\int_{0}^{\infty}(x\wedge
y)\alpha(dy)},\quad\forall x\in(0,\infty).$

Conversely, if a speed measure $m$ satisfies (B), then the corresponding
quasi-stationary distribution $\alpha$ is such that $(\frac{1}{x}\vee
1)\alpha(dx)$ satisfies (B).

Because of the relation (4.9) and the equivalence between (B) and uniform
exponential convergence to a quasi-stationary distribution, this result
follows from the next lemma.

######  Lemma 4.8.

Let $\alpha$ be a probability measure on $(0,+\infty)$. Define the measure $m$
on $(0,+\infty)$ as

$\displaystyle m(dx)=\frac{\alpha(dx)}{\int_{0}^{\infty}(x\wedge
y)\alpha(dy)},\ \forall x\in(0,\infty).$ (4.10)

Then the measure $(\frac{1}{x}\vee 1)\alpha(dx)$ satisfies (B) if and only if
the measure $m$ satisfies (B).

###### Proof.

Note that, if $m_{1},m_{2}$ are two locally finite measures on $(0,+\infty)$
giving positive mass to each open subset of $(0,\infty)$, such that $m_{1}\leq
Cm_{2}$ for some positive constant $C>0$ and such that $m_{2}$ satisfies (B),
then $m_{1}$ also satisfies (B). This is a direct consequence of the
construction of diffusion processes given in Section 2 and of (2.1).

Assume first that $(\frac{1}{x}\vee 1)\alpha(dx)$ satisfies (B). Then, using
Remark 2,

$\displaystyle\int_{0}^{\infty}(1\vee y)\,\alpha(dy)<\infty.$ (4.11)

Thus

$\displaystyle\int_{0}^{\infty}(x\wedge
y)\alpha(dy)\leq\left(x\int_{0}^{\infty}\alpha(dy)\right)\wedge\left(\int_{0}^{\infty}y\,\alpha(dy)\right)<\infty.$
(4.12)

Hence $m\neq 0$, is locally finite and gives positive mass to each open subset
of $(0,\infty)$. Since

$\displaystyle\int_{0}^{\infty}(x\wedge
y)\alpha(dy)\geq\begin{cases}x\int_{1}^{\infty}\alpha(dy),&\text{ if }x\leq
1,\\\ \int_{1}^{\infty}\alpha(dy),&\text{ if }x>1,\end{cases}$

we have

$\frac{dm}{d\alpha}(x)=\left(\int_{0}^{\infty}(x\wedge
y)\alpha(dy)\right)^{-1}\leq\left(\int_{1}^{\infty}\alpha(dy)\right)^{-1}\left(\frac{1}{x}\vee
1\right).$

Therefore, if $(\frac{1}{x}\vee 1)\alpha(dx)$ satisfies (B), $m$ satisfies
(B).

Conversely, if $m$ satisfies (B), Theorem 4.1 implies that $\alpha$ is
proportional to the unique quasi-stationary distribution of $X$ with speed
measure $m$ and satisfies

$\int_{0}^{\infty}y\,\alpha(dy)<\infty.$

Since $\alpha$ is a finite measure, we deduce from (4.12) that

$\int_{0}^{\infty}(x\wedge y)\alpha(dy)\leq C(x\wedge 1)$

for some constant $C<\infty$. Hence

$\frac{dm}{d\alpha}(x)\geq C^{-1}\left(\frac{1}{x}\vee 1\right),$

i.e. $(\frac{1}{x}\vee 1)\alpha(dx)$ satisfies (B). ∎

###  4.5 Examples

####  4.5.1 On general diffusions

Let us first recall that our results also cover the case of general (i.e. not
necessarily on natural scale) diffusion processes $(Y_{t},t\geq 0)$ on $[a,b)$
such that $-\infty<a<b\leq+\infty$ with speed measure $m_{Y}$, which hits $a$
in a.s. finite time $\tau_{\partial}$, absorbed at $a$ and such that $b$ is a
natural or entrance boundary. Under these assumptions, there exists a
continuous and strictly increasing scale function
$s:[a,b)\rightarrow[0,\infty)$ of the process $Y$ such that $s(a)=0$ and
$s(b-)=+\infty$. Our results then apply to the process $X_{t}=s(Y_{t})$ which
is a diffusion on natural scale on $[0,\infty)$, whose speed measure is given
by

$\displaystyle m_{X}(dx)=s*m_{Y}(dx),$ (4.13)

where $s*m_{Y}$ is the pushforward measure of $m_{Y}$ through the function $s$
(this follows for instance from [21, Thm. VII.3.6]).

Hence $X$ satisfies (B) if and only if $Y$ satisfies the following Condition
(B’).

Condition (B’) Assume that $\int_{c}^{b}s(y)\,m_{Y}(dy)<\infty$ for some
$c\in(a,b)$ and there exist two constants $t_{1},A>0$ such that

$\mathbb{P}(t_{1}<\tau_{\partial}\mid Y_{0}=y)\leq As(y),\quad\forall y>a.$

Under this assumption, the process $Y$ satisfies (4.2) for a unique quasi-
stationary distribution $\alpha_{Y}$ on $(a,b)$ and $\alpha_{Y}$ satisfies

$\int_{a}^{b}s(y)\alpha_{Y}(dy)<\infty.$

Moreover

$\frac{d\alpha_{Y}}{dm_{Y}}(y)=2\lambda_{0}\int_{a}^{b}(s(y)\wedge
s(z))\alpha_{Y}(dz),$

where $\lambda_{0}$ is such that $\mathbb{P}(t<\tau_{\partial}\mid
Y_{0}\sim\alpha_{Y})=e^{-\lambda_{0}t}$. Theorem 4.6 entails the following
result.

######  Proposition 4.9.

With the previous notation, if $\int_{a}^{b}s(y)\,m_{Y}(dy)<\infty$ and

$\int_{0}^{1}\frac{1}{x}\left(\sup_{y\in(a,s^{-1}(x)]}\frac{1}{s(y)}\int_{a}^{y}s(z)^{2}\,m_{Y}(dz)\right)dx<\infty,$
(4.14)

then Condition (B’) is satisfied.

Let us also mention that our methods can be easily extended to diffusion
processes on a bounded interval $[a,b]$ where both boundary points are either
exit or regular (reflecting or absorbing). In the case where $b$ is exit or
regular absorbing (i.e. $\tau_{\partial}$ is the first hitting time of
$\\{a,b\\}$), we obtain two conditions of the form (4.14) in the neighborhood
of $a$ and $b$.

To illustrate the generality of the processes that are covered by our
criteria, we give two simple examples where the speed measure $m$ is singular
with respect to Lebesgue’s measure $\Lambda$. The case of speed measures
absolutely continuous w.r.t. Lebesgue’s measure (i.e. of SDEs) will be
discussed in the next subsection.

##### Example 1

We recall that a diffusion process on $\mathbb{R}$ with speed measure
$\Lambda+\delta_{0}$ is called a _sticky Brownian motion_ [8, 1]. We consider
a diffusion process on $[0,\infty)$ which comes down from infinity
($\int_{1}^{\infty}y\,m(dy)<\infty$) and which is a sticky Brownian motion on
$[0,1]$, “sticked” at the points $a_{1},a_{2},\ldots$, where $(a_{i})_{i\geq
1}$ is decreasing, converges to 0 and $a_{1}<1$, i.e.

$\mathchoice{{m\,\smash{\vrule
height=3.44444pt,depth=2.97502pt}}_{\,(0,1)}}{{m\,\smash{\vrule
height=3.44444pt,depth=2.97502pt}}_{\,(0,1)}}{{m\,\smash{\vrule
height=2.41112pt,depth=2.12502pt}}_{\,(0,1)}}{{m\,\smash{\vrule
height=1.72221pt,depth=2.12502pt}}_{\,(0,1)}}=\mathchoice{{\Lambda\,\smash{\vrule
height=5.46666pt,depth=2.97502pt}}_{\,(0,1)}}{{\Lambda\,\smash{\vrule
height=5.46666pt,depth=2.97502pt}}_{\,(0,1)}}{{\Lambda\,\smash{\vrule
height=3.82666pt,depth=2.12502pt}}_{\,(0,1)}}{{\Lambda\,\smash{\vrule
height=2.73332pt,depth=2.12502pt}}_{\,(0,1)}}+\sum_{i\geq 1}\delta_{a_{i}}.$

Assuming that there exist constants $C,\rho>0$ such that for all $j\geq 1$,

$\displaystyle\sum_{i\geq j}a_{i}\leq Ca^{\rho}_{j},$ (4.15)

then for all $x\in(0,1)$, defining $i_{0}:=\inf\\{j\geq 1:a_{j}<x\\}$,

$\int_{(0,x)}y\,m(dy)=\frac{x^{2}}{2}+\sum_{i\geq
i_{0}}a_{i}\leq\frac{x^{2}}{2}+Ca^{\rho}_{i_{0}}\leq\frac{x^{2}}{2}+Cx^{\rho},$

and we can apply Theorem 4.3.

For example, the choice $a_{i}=i^{-\frac{1}{1-\rho}}$, for all $i\geq 1$,
satisfies (4.15).

##### Example 2

Let $X$ be a sticky Brownian motion stopped at $-1$ and $1$. This means that
$X$ is a diffusion on natural scale with speed measure
$m(dx)=\Lambda(dx)+\delta_{0}(dx)$ on $(-1,1)$, absorbed at $-1$ and $1$. As
mentionned above, it is straightforward to adapt our results to processes on
$[-1,1]$. For the process of this example, the corresponding Assumption (B) is
clearly fulfilled since both boundaries $-1$ and $1$ are regular for $X$ (see
Corollary 4.4). Then the unique quasi-stationary distribution $\alpha$ of $X$
is

$\displaystyle\alpha(dx)=\frac{\gamma^{*}}{2}\sin\left(\gamma^{*}(1+x)\wedge(1-x)\right)m(dx),$

where $\gamma^{*}$ is the unique solution in $(0,\pi]$ of $\text{cotan
}\gamma=\gamma/2$.

Indeed, it satisfies the following adaptation of formula (4.3)

$\displaystyle\frac{d\alpha}{dm}(x)=\lambda_{0}\int_{-1}^{1}(x\wedge
y+1)(1-x\vee y)\,\alpha(dy),\ \forall x\in(-1,1).$ (4.16)

In particular, the measure $\alpha$ is absolutely continuous with respect to
Lebesgue’s measure $\Lambda$ on $(-1,1)\setminus\\{0\\}$ and satisfies

$\displaystyle\left(\frac{d\alpha}{d\Lambda}\right)^{\prime\prime}(x)=-2\lambda_{0}\frac{d\alpha}{d\Lambda}(x),\
\forall x\neq 0.$

Using the symmetry of the problem and the $0$ boundary conditions, we deduce
that there exist constants $a,b\in\mathbb{R}$ such that

$\displaystyle\alpha(dx)=a\delta_{0}(dx)+b\sin\left(\gamma(1+x)\wedge(1-x)\right)\Lambda(dx),$

where $\gamma=\sqrt{2\lambda_{0}}$. Note that the fact that $\alpha$ is a
positive measure implies that $\gamma\in(0,\pi]$. In addition, equality (4.16)
at $x=0$ entails

$\displaystyle
a=\lambda_{0}\left(2b\int_{0}^{1}(1-y)\sin(\gamma(1-y))dy+a\right)$

and hence

$\displaystyle(1-\gamma^{2}/2)a=b\left(\sin\gamma-\gamma\cos\gamma\right).$
(4.17)

In addition, $d\alpha/dm(x)$ is continuous at $0$ by Remark 4, hence

$\displaystyle a=b\sin\gamma.$ (4.18)

Finally, $\alpha$ is a probability measure, so that

$\displaystyle a+\frac{2b}{\gamma}(1-\cos\gamma)=1.$ (4.19)

Now, dividing (4.17) by (4.18), we obtain $\gamma=\gamma^{*}$. Equality (4.19)
becomes

$\displaystyle 1=a-b\sin\gamma+\frac{2b}{\gamma}.$

By (4.18), we deduce that $b=\gamma/2$ and finally that
$a=\frac{\gamma\sin\gamma}{2}$.

####  4.5.2 On processes solutions of stochastic differential equations

Let $Y$ be the solution of the general SDE on $[a,b)$ with $a<b\leq+\infty$

$dY_{t}=\sigma(Y_{t})dB_{t}+\mu(Y_{t})dt,$ (4.20)

absorbed at $a$, where $\mu/\sigma^{2}\in L^{1}_{\text{loc}}((a,b))$ (which
ensures the existence of a weak solution [9, Ch. 23], possibly exposive if
$b=+\infty$, or which can possibly hit $b$ if $b<\infty$). In this case, a
scale function of $Y$ is given by

$s(x)=\int_{c}^{x}\exp\left(-\int_{c}^{y}2\mu(z)\sigma^{-2}(z)dz\right)dy,\
\forall x\in(a,b)$ (4.21)

for any $c\in(a,b)$, and the speed measure by
$m_{Y}(dy)=2/(s^{\prime}(y)\sigma^{2}(y))\Lambda(dy)$. In the case where
$s(a+)>-\infty$, one can take $c=a$ in Formula (4.21), so that $s(a)=0$. If in
addition $s(b-)=\infty$ (which implies in particular that (4.20) is non-
explosive if $b=\infty$ or cannot hit $b$ in finite time if $b<\infty$), the
formula (4.21) can be used to check the condition (4.14) in Proposition 4.9.

In particular, our criteria apply to cases of speed measures with
discontinuous densities with respect to Lebesgue’s measure, i.e. to cases of
discontinuous $\sigma$. Such diffusions arise as rescaled solutions of SDEs
driven by their local times, including skew Brownian motions (see for instance
[12]). Hence our criteria also extend to one-dimensional SDEs driven by their
local time in the sense of [12].

Note that a major part of the existing literature [2, 13, 10] on the subject
only considers solutions to SDEs with absolutely continuous coefficient
$\sigma$. The reason behind this is that all the previously cited works are
based on Sturm-Liouville theory, which is particularly well suited for the
study of diffusion processes $(Y_{t},t\geq 0)$ on $[0,\infty)$ solution to a
SDE of the form

$dY_{t}=dB_{t}-q(Y_{t})dt.$

After a change of scale, this class of processes can only provide solutions to
SDEs of the form (4.20) with absolutely continuous $\sigma$.

####  4.5.3 A simple example with jumps

The following simple example shows how our method can be easily extended to
processes with jumps.

We consider a diffusion process $(X_{t},t\geq 0)$ on $[0,\infty)$ with speed
measure $m$ satisfying Assumption (B). Let us denote by $\mathcal{L}$ the
infinitesimal generator of $X$. We consider the Markov process
$(\widetilde{X}_{t},t\geq 0)$ with infinitesimal generator

$\widetilde{\mathcal{L}}f(x)=\mathcal{L}f(x)+(f(x+1)-f(x))\mathbbm{1}_{x\geq
1},$

for all $f$ in the domain of $\mathcal{L}$. In other words, we consider a
càdlàg process following a diffusion process with speed measure $m$ between
jump times, which occur at the jump times of an independent Poisson process
$(N_{t},t\geq 0)$ of rate $1$, with jump size $+1$ if the process is above 1,
and $0$ otherwise. We denote by $\widetilde{\tau}_{\partial}$ its first
hitting time of $0$.

The proof of the following proposition is postponed to Subsection 5.4.

######  Proposition 4.10.

Under the previous assumptions, there exist a unique probability measure
$\alpha$ on $(0,\infty)$ and two constants $C,\gamma>0$ such that, for all
initial distribution $\mu$ on $(0,\infty)$,

$\displaystyle\left\|\mathbb{P}_{\mu}(\widetilde{X}_{t}\in\cdot\mid
t<\widetilde{\tau}_{\partial})-\alpha(\cdot)\right\|_{TV}\leq Ce^{-\gamma t},\
\forall t\geq 0.$ (4.22)

In particular, the probability measure $\alpha$ is the unique quasi-stationary
distribution of the process.

##  5 Proof of the results of Section 4

The main part of Theorem 4.1, the uniform convergence in (1.2) and some of the
results of Proposition 4.2 are proved using the criterion developed in [3] for
general Markov processes. More precisely, the following condition (A) is
equivalent to Condition (ii) of Theorem 4.1 ([3, Thm. 2.1]), implies property
(1.2) and the fact that $\eta$ is bounded and $\alpha(\eta)=1$. ([3, Prop.
2.3]).

##### Assumption (A)

There exists a probability measure $\nu$ on $(0,\infty)$ such that

  * (A1)

there exist $t_{0},c_{1}>0$ such that for all $x>0$,

$\mathbb{P}_{x}(X_{t_{0}}\in\cdot\mid t_{0}<\tau_{\partial})\geq
c_{1}\nu(\cdot);$

  * (A2)

there exists $c_{2}>0$ such that for all $x>0$ and $t\geq 0$,

$\mathbb{P}_{\nu}(t<\tau_{\partial})\geq
c_{2}\mathbb{P}_{x}(t<\tau_{\partial}).$

Hence, we need first to prove that (B) is equivalent to (A) (in Subsection
5.1), second, to prove (4.4) and that $\eta(x)\leq Cx$ (in Subsection 5.2),
and third to prove (4.3) and that $\int y\,\alpha(dy)<\infty$ (in Subsection
5.3). We next give in Subsection 5.4 the proof of Proposition 4.10 on
processes with jumps.

###  5.1 Proof of the equivalence between (A) and (B)

Note that [3, Thm. 2.1] also assumes that

$\displaystyle\mathbb{P}_{x}(t<\tau_{\partial})>0,\quad\forall
x>0,\quad\forall t>0,$ (5.1)

so we need first to check that this is true for our diffusion processes. This
follows easily from the strong Markov and the regularity properties of $X$ (it
is for example enough to use that $\mathbb{P}_{x}(T_{x/2}<+\infty)>0$ and
$\mathbb{P}_{x/2}(T_{x}<+\infty)>0$).

We actually prove below that (A1) and (B) are equivalent and that these
properties imply (A2). In other words, for one-dimensional diffusion
processes, (4.2) is actually equivalent to (A1) alone.

We first prove that (A1) implies (B). If (A1) holds true, there exist $0<a<b$
such that

$\inf_{x>0}\mathbb{P}_{x}(X_{t_{1}}\in[a,b]\mid
t_{1}<\tau_{\partial})=\underline{c}>0.$ (5.2)

Now, for all $x>b$,

$\frac{\mathbb{P}_{x}(T_{b}<t_{1})}{\mathbb{P}_{b}(t_{1}<\tau_{\partial})}\geq\frac{\mathbb{P}_{x}(X_{t_{1}}\in[a,b])}{\mathbb{P}_{b}(t_{1}<\tau_{\partial})}\geq\mathbb{P}_{x}(X_{t_{1}}\in[a,b]\mid
t_{1}<\tau_{\partial}).$

Since we proved above that $\mathbb{P}_{b}(t_{1}<\tau_{\partial})>0$, we
deduce that $\inf_{x>b}\mathbb{P}_{x}(T_{b}<t_{1})>0$, i.e. that $\infty$ is
an entrance boundary for $X$. Equation (5.2) also implies that, for all $x<a$,

$\displaystyle\mathbb{P}_{x}(t_{1}<\tau_{\partial})$
$\displaystyle\leq\frac{\mathbb{P}_{x}(X_{t_{1}}\in[a,b])}{\underline{c}}$
$\displaystyle\leq\frac{\mathbb{E}_{x}(X_{t_{1}\wedge
T_{a}})}{a\underline{c}}=\frac{x}{a\underline{c}}.$

Hence (i) is proved.

The difficult part of the proof is the implication (B)$\Rightarrow$(A).

Step 1: the conditioned process escapes a neighborhood of 0 in finite time.
The goal of this step is to prove that there exists $\varepsilon,c>0$ such
that

$\mathbb{P}_{x}(X_{t_{1}}\geq\varepsilon\mid t_{1}<\tau_{\partial})\geq
c,\quad\forall x>0.$ (5.3)

To prove this, we first observe that, since $X$ is a local martingale, for all
$x\in(0,1)$,

$x=\mathbb{E}_{x}(X_{t_{1}\wedge
T_{1}})=\mathbb{P}_{x}(t_{1}<\tau_{\partial})\mathbb{E}_{x}(X_{t_{1}\wedge
T_{1}}\mid t_{1}<\tau_{\partial})+\mathbb{P}_{x}(T_{1}<\tau_{\partial}\leq
t_{1}).$

By the Markov property,

$\displaystyle\mathbb{P}_{x}(T_{1}<\tau_{\partial}\leq t_{1})$
$\displaystyle\leq\mathbb{E}_{x}\left[\mathbbm{1}_{T_{1}<\tau_{\partial}\wedge
t_{1}}\mathbb{P}_{1}(\tau_{\partial}\leq t_{1})\right]$
$\displaystyle\leq\mathbb{P}_{x}(T_{1}<\tau_{\partial})\mathbb{P}_{1}(\tau_{\partial}\leq
t_{1})$ $\displaystyle=x\mathbb{P}_{1}(\tau_{\partial}\leq t_{1}).$

Hence (4.1) entails

$\mathbb{E}_{x}(1-X_{t_{1}\wedge T_{1}}\mid t_{1}<\tau_{\partial})\leq
1-\frac{1}{A^{\prime}},$

where $A^{\prime}:=A/\mathbb{P}_{1}(t_{1}<\tau_{\partial})$. Note that,
necessarily, $A^{\prime}>1$. Markov’s inequality then implies that, for all
$x\in(0,1)$,

$\mathbb{P}_{x}(X_{t_{1}\wedge T_{1}}\leq\frac{1}{2A^{\prime}-1}\mid
t_{1}<\tau_{\partial})\leq\frac{1-1/A^{\prime}}{1-1/(2A^{\prime}-1)}=1-\frac{1}{2A^{\prime}}.$
(5.4)

Since $\mathbb{P}_{1/(2A^{\prime}-1)}(t_{1}<\tau_{\partial})>0$, there exists
$\varepsilon\in(0,1/(2A^{\prime}-1))$ such that

$\mathbb{P}_{1/(2A^{\prime}-1)}(t_{1}<T_{\varepsilon})>0.$ (5.5)

Hence, for all $x\in(0,1)$,

$\displaystyle\mathbb{P}_{x}(X_{t_{1}}\geq\varepsilon)$
$\displaystyle\geq\mathbb{P}_{x}(T_{1/(2A^{\prime}-1)}<t_{1})\mathbb{P}_{1/(2A^{\prime}-1)}(T_{\varepsilon}>t_{1})$
$\displaystyle\geq\mathbb{P}_{x}(X_{t_{1}\wedge T_{1}}\geq
1/(2A^{\prime}-1))\mathbb{P}_{1/(2A^{\prime}-1)}(T_{\varepsilon}>t_{1})$
$\displaystyle\geq\frac{\mathbb{P}_{x}(t_{1}<\tau_{\partial})}{2A^{\prime}}\mathbb{P}_{1/(2A^{\prime}-1)}(T_{\varepsilon}>t_{1})$

by (5.4). This ends the proof of (5.3) for $x<1$. For $x\geq
1>1/(2A^{\prime}-1)>\varepsilon$, the continuity and the strong Markov
property for $X$ entail

$\mathbb{P}_{x}(X_{t_{1}}>\varepsilon\mid
t_{1}<\tau_{\partial})\geq\mathbb{P}_{x}(X_{t_{1}}>\varepsilon)\geq\mathbb{P}_{x}(T_{\varepsilon}>t_{1})\geq\mathbb{P}_{1/(2A^{\prime}-1)}(T_{\varepsilon}>t_{1})>0.$

Hence (5.3) is proved.

Step 2: Construction of coupling measures for the unconditioned process.
Our goal is to prove that there exist two constants $t_{2},c_{1}>0$ such that,
for all $x\geq\varepsilon$,

$\mathbb{P}_{x}(X_{t_{2}}\in\cdot)\geq c_{1}\nu,$ (5.6)

where

$\nu=\mathbb{P}_{\varepsilon}(X_{t_{2}}\in\cdot\mid t_{2}<\tau_{\partial}).$

This kind of relations can be obtained with classical coupling arguments,
which we detail here for completeness. Fix $x\geq\varepsilon$ and construct
two independent diffusions $X^{\varepsilon}$ and $X^{x}$ with speed measure
$m(dx)$, and initial value $\varepsilon$ and $x$ respectively. Let
$\theta=\inf\\{t\geq 0:X^{\varepsilon}_{t}=X^{x}_{t}\\}$. By the strong Markov
property, the process

$Y^{x}_{t}=\begin{cases}X^{x}_{t}&\text{if }t\leq\theta,\\\
X^{\varepsilon}_{t}&\text{if }t>\theta\end{cases}$

has the same law as $X^{x}$. Since
$\theta\leq\tau^{x}_{\partial}:=\inf\\{t\geq 0:X^{x}_{t}=0\\}$, for all $t>0$,
$\mathbb{P}(\theta<t)\geq\mathbb{P}(\tau^{x}_{\partial}<t)$. Since $\infty$ is
an entrance boundary and $0$ is accessible for $X$, there exists $t_{2}>0$
such that

$\inf_{y>0}\mathbb{P}_{y}(\tau_{\partial}<t_{2})=c^{\prime}_{1}>0.$

Hence

$\mathbb{P}_{x}(X_{t_{2}}\in\cdot)=\mathbb{P}(Y^{x}_{t_{2}}\in\cdot)\geq\mathbb{P}(X^{\varepsilon}_{t_{2}}\in\cdot,\
\tau^{x}_{\partial}<t_{2})\geq
c^{\prime}_{1}\mathbb{P}_{\varepsilon}(X_{t_{2}}\in\cdot).$

Therefore, (5.6) is proved with
$c_{1}=c^{\prime}_{1}\mathbb{P}_{\varepsilon}(t_{2}<\tau_{\partial})$.

Step 3: Proof of (A1).
Using successively the Markov property, Step 2 and Step 1, we have for all
$x>0$

$\displaystyle\mathbb{P}_{x}(X_{t_{1}+t_{2}}\in\cdot\mid
t_{1}+t_{2}<\tau_{\partial})$
$\displaystyle\geq\mathbb{P}_{x}(X_{t_{1}+t_{2}}\in\cdot\mid
t_{1}<\tau_{\partial})$
$\displaystyle\geq\int_{\varepsilon}^{\infty}\mathbb{P}_{y}(X_{t_{2}}\in\cdot)\mathbb{P}_{x}(X_{t_{1}}\in
dy\mid t_{1}<\tau_{\partial})$ $\displaystyle\geq
c_{1}\int_{\varepsilon}^{\infty}\nu(\cdot)\mathbb{P}_{x}(X_{t_{1}}\in dy\mid
t_{1}<\tau_{\partial})$
$\displaystyle=c_{1}\nu(\cdot)\mathbb{P}_{x}(X_{t_{1}}\geq\varepsilon\mid
t_{1}<\tau_{\partial})\geq c_{1}c\nu(\cdot).$

This entails (A1) with $t_{0}=t_{1}+t_{2}$. Hence we have proved the
equivalence between (A1) and (B).

Step 4: Proof of (A2).
The general idea of the proof is close to the case of birth and death
processes in [3].

For all $0<a<b<\infty$, we have

$\mathbb{E}_{x}(T_{a}\wedge T_{b})=2\int_{a}^{b}\frac{(x\wedge y-a)(b-x\vee
y)}{b-a}\,m(dy),\quad\forall a<x<b.$

Hence,

$\mathbb{E}_{x}(T_{a}\wedge T_{b})\leq 2\int_{a}^{b}(x\wedge y-a)\,m(dy)\leq
2\int_{a}^{\infty}y\,m(dy).$

Since the process is non-explosive, the left hand side converges to
$\mathbb{E}_{x}(T_{a})$ when $b\rightarrow\infty$. But
$\int_{0}^{\infty}y\,m(dy)<\infty$ by assumption, so that, for all
$\varepsilon>0$, there exists $a_{\varepsilon}>0$ such that

$\displaystyle\sup_{x\geq
a_{\varepsilon}}\mathbb{E}_{x}(T_{a_{\varepsilon}})\leq\varepsilon.$

Therefore, $\sup_{x\geq a_{\varepsilon}}\mathbb{P}_{x}(T_{a_{\varepsilon}}\geq
1)\leq\varepsilon$ and, applying recursively the Markov property, $\sup_{x\geq
a_{\varepsilon}}\mathbb{P}_{x}(T_{a_{\varepsilon}}\geq k)\leq\varepsilon^{k}$.
Then, for all $\lambda>0$, there exists $y_{\lambda}\geq 1$ such that

$\sup_{x\geq y_{\lambda}}\mathbb{E}_{x}(e^{\lambda T_{y_{\lambda}}})<+\infty.$
(5.7)

The proof of the following lemma is postponed to the end of this section.

######  Lemma 5.1.

There exists $a>0$ such that $\nu([a,+\infty[)>0$ and, for all
$k\in\mathbb{N}$,

$\displaystyle\mathbb{P}_{a}(X_{kt_{0}}\geq a)\geq e^{-\rho kt_{0}},$

with $\rho>0$.

Take $a$ as in the previous lemma. From (5.7), one can choose $b>a$ large
enough so that

$\displaystyle\sup_{x\geq b}\mathbb{E}_{x}\left(e^{\rho T_{b}}\right)<\infty.$
(5.8)

We are also going to make use of two inequalities. Since

$\displaystyle\mathbb{P}_{a}(t<\tau_{\partial})\geq\mathbb{P}_{a}(T_{b}<s_{0})\mathbb{P}_{b}(t<\tau_{\partial})$

for all $s_{0}\geq 0$, we obtain the first inequality: $\forall t\geq 0$,

$\displaystyle\sup_{x\in[a,b]}\mathbb{P}_{x}(t<\tau_{\partial})=\mathbb{P}_{b}(t<\tau_{\partial})\leq
C\mathbb{P}_{a}(t<\tau_{\partial})=C\inf_{x\in[a,b]}\mathbb{P}_{x}(t<\tau_{\partial}),$
(5.9)

where $C$ is a positive constant. We recall that the regularity assumption
ensures that $\mathbb{P}_{a}(T_{b}<s_{0})>0$ for $s_{0}$ large enough. The
second inequality is an immediate consequence of the Markov property: $\forall
s<t$,

$\displaystyle\mathbb{P}_{a}(X_{\lceil s/t_{0}\rceil t_{0}}\geq
a)\mathbb{P}_{a}(t-s<\tau_{\partial})$ $\displaystyle=\mathbb{P}_{a}(X_{\lceil
s/t_{0}\rceil t_{0}}\geq
a)\inf_{x\in[a,\infty)}\mathbb{P}_{x}(t-s<\tau_{\partial})$
$\displaystyle\leq\mathbb{P}_{a}(t<\tau_{\partial}).$ (5.10)

In the following computation, we use successively (5.8), (5.9) and (5.10). For
all $x\geq b$, with a constant $C>0$ that may change from line to line,

$\displaystyle\mathbb{P}_{x}(t<\tau_{\partial})$
$\displaystyle\leq\mathbb{P}_{x}(t<T_{b})+\int_{0}^{t}\mathbb{P}_{b}(t-s<\tau_{\partial})\,\mathbb{P}_{x}(T_{b}\in
ds)$ $\displaystyle\leq Ce^{-\rho
t}+C\int_{0}^{t}\mathbb{P}_{a}(t-s<\tau_{\partial})\,\mathbb{P}_{x}(T_{b}\in
ds)$ $\displaystyle\leq Ce^{-\rho(\lceil
t/t_{0}\rceil-1)t_{0}}+C\mathbb{P}_{a}(t<\tau_{\partial})\int_{0}^{t}\frac{1}{\mathbb{P}_{a}(X_{\lceil
s/t_{0}\rceil t_{0}}\geq a)}\,\mathbb{P}_{x}(T_{b}\in ds)$ $\displaystyle\leq
C\mathbb{P}_{a}(t<\tau_{\partial})+C\mathbb{P}_{a}(t<\tau_{\partial})\int_{0}^{t}e^{\rho(s+t_{0})}\,\mathbb{P}_{x}(T_{b}\in
ds),$

where we used twice Lemma 5.1 in the last line. We deduce form (5.8) that, for
all $t\geq 0$,

$\displaystyle\sup_{x\geq b}\mathbb{P}_{x}(t<\tau_{\partial})\leq
C\mathbb{P}_{a}(t<\tau_{\partial}).$

Since $\nu([a,+\infty[)>0$, this ends the proof of (A2).

###### Proof of Lemma 5.1.

Fix $a>0$ such that $\nu([a,+\infty[)>0$. Step 3 of the previous proof and
(5.1) entail

$\displaystyle\mathbb{P}_{a}(X_{t_{0}}\geq a)\geq
c_{1}\nu([a,+\infty[)\mathbb{P}_{a}(t_{0}<\tau_{\partial})\stackrel{{\scriptstyle\text{def}}}{{=}}e^{-\rho
t_{0}}>0.$

Now, using Markov property,

$\displaystyle\mathbb{P}_{a}(X_{kt_{0}}\geq a)\geq\left(\inf_{x\geq
a}\mathbb{P}_{x}(X_{t_{0}}\geq a)\right)^{k}.$

Since $\inf_{x\geq a}\mathbb{P}_{x}(X_{t_{0}}\geq
a)=\mathbb{P}_{a}(X_{t_{0}}\geq a)$ by coupling arguments, the proof is
completed. ∎

###  5.2 Proof of Proposition 4.2

Since (A) is equivalent to (B), the convergence in (1.2) for the uniform
normis a direct consequence of Proposition [3, Prop 2.3]. This proposition
also entails that $\eta$ is bounded, positive on $(0,+\infty)$ and that
$\alpha(f)=1$. The fact that $\eta$ is non-decreasing comes from (1.2) and
from the fact that $\mathbb{P}_{x}(t<\tau_{\partial})$ is non-decreasing in
$x\geq 0$ by standard comparison arguments. The fact that $\eta(x)\leq Cx$
follows from assumption (B) since

$P_{t_{1}}\eta(x)\leq\|\eta\|_{\infty}\mathbb{P}_{x}(t_{1}\leq\tau_{\partial})\leq\|\eta\|_{\infty}Ax,\quad\forall
x\geq 0.$

It only remains to prove (4.4). For all measurable $f\geq 0$ and all $0\leq
a<x<b\leq\infty$,

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{T_{a}\wedge
T_{b}}f(X_{t})\,dt\right)=2\int_{a}^{b}\frac{(x\wedge y-a)(b-x\vee
y)}{b-a}\,f(y)\,m(dy).$ (5.11)

For $a=0$ and $b=\infty$, we obtain

$\displaystyle\mathbb{E}_{x}\left(\int_{0}^{\tau_{\partial}}f(X_{t})\,dt\right)=2\int_{0}^{\infty}(x\wedge
y)\,f(y)\,m(dy).$ (5.12)

For $f=\eta$, we deduce that

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{x}\left(\eta(X_{t})\mathbbm{1}_{t<\tau_{\partial}}\right)\,dt$
$\displaystyle=2\int_{0}^{\infty}(x\wedge y)\,\eta(y)\,m(dy).$

Since $\eta(0)=0$ and $L\eta=-\lambda_{0}\eta$, we have
$\mathbb{E}_{x}\left(\eta(X_{t})\mathbbm{1}_{t<\tau_{\partial}}\right)=P_{t}\eta(x)=e^{-\lambda_{0}t}\eta(x)$.
Then

$\displaystyle\frac{\eta(x)}{\lambda_{0}}=2\int_{0}^{\infty}(x\wedge
y)\,\eta(y)\,m(dy).$

This entails (4.4) provided we prove that $\alpha(dx)$ is proportional to
$\eta(x)m(dx)$ (observe that the normalizing constant is determined by the
condition $\alpha(\eta)=1$). This will be done in the next Subsection.

###  5.3 Proof of (4.3) and that $\int y\,\alpha(dy)<\infty$

Integrating (5.12) with respect to $\alpha(dx)$, we obtain

$\displaystyle\int_{0}^{\infty}\mathbb{E}_{\alpha}\left(f(X_{t})\mathbbm{1}_{t<\tau_{\partial}}\right)dt=2\int_{0}^{\infty}\int_{0}^{\infty}(x\wedge
y)\,f(y)\,m(dy)\,\alpha(dx).$

Since
$\mathbb{E}_{\alpha}\left(f(X_{t})\mathbbm{1}_{t<\tau_{\partial}}\right)=\alpha(f)\,e^{-\lambda_{0}t}$,
we deduce that

$\displaystyle\frac{\alpha(f)}{\lambda_{0}}=2\int_{0}^{\infty}\int_{0}^{\infty}(x\wedge
y)\,f(y)\,m(dy)\,\alpha(dx).$

This entails (4.3). We now prove that $\alpha(dx)$ is proportional to
$\eta(x)m(dx)$.

The following reversibility result for diffusions on natural scale is more or
less classical but we need a version with precise bounds on the test functions
in the case of a diffusion coming down from infinity and hitting 0 a.s. in
finite time. The proof is given at the end of the subsection for sake of
completeness.

######  Lemma 5.2.

Let $X$ be a diffusion on $[0,\infty)$ in natural scale with speed measure $m$
satisfying (2.2). Then it is reversible with respect to $m$ in the sense that,
for all bounded non-negative measurable functions $f$ on $(0,+\infty)$ and all
nonnegative measurable function $g$ on $(0,+\infty)$ such that $g(x)\leq Cx$
for some $C>0$,

$\int_{0}^{\infty}f(x)P_{t}g(x)m(dx)=\int_{0}^{\infty}g(x)P_{t}f(x)m(dx),\quad\forall
t\geq 0,$ (5.13)

where both sides are finite.

Applying this lemma for $g=\eta$, we obtain that, for all bounded measurable
non-negative $f$,

$\int_{0}^{\infty}f(x)\eta(x)m(dx)=e^{\lambda_{0}t}\int_{0}^{\infty}\eta(x)P_{t}f(x)m(dx)$

Now, it follows from (4.2) and (1.2) that

$e^{\lambda_{0}t}P_{t}f(x)=e^{\lambda_{0}t}\mathbb{P}_{x}(t<\tau_{\partial})\mathbb{E}_{x}(f(X_{t})\mid
t<\tau_{\partial})\rightarrow\eta(x)\alpha(f)$

when $t\rightarrow+\infty$, where the convergence is uniform in $x$. Since
$\int\eta\,dm<\infty$, Lebesgue’s theorem entails that

$\alpha(f)\int_{0}^{\infty}\eta^{2}(x)m(dx)=\int_{0}^{\infty}f(x)\eta(x)m(dx)$

for all bounded measurable $f$. Hence $\alpha\propto\eta\,dm$. Note also that
$f=\eta$ gives $\int_{0}^{\infty}\eta\,dm=\int_{0}^{\infty}\eta^{2}\,dm$.

It only remains to prove that $\int y\,\alpha(dy)<\infty$. Since $\eta$ is
bounded and

$\eta(x)\propto\int_{0}^{\infty}(x\wedge y)\,\alpha(dy),$

this follows from the monotone convergence theorem.

###### Proof of Lemma 5.2.

We start by proving that both sides of (5.13) are finite. This is obvious for
the r.h.s. because of (2.2). For the l.h.s., since the positive local
martingale $(X_{t},t\geq 0)$ is a supermartingale, we have
$P_{t}g(x)=\mathbb{E}_{x}[g(X_{t})\mathbbm{1}_{t<\tau_{\partial}}]\leq
C\mathbb{E}_{x}[X_{t}]\leq Cx$, which allows to conclude.

By Lebesgue’s theorem, it is enough to prove (5.13) for a $f,g$ in a dense
subset of the set of continuous functions on $(0,+\infty)$ with compact
support. Note that a function $g$ with compact support in $(0,+\infty)$
satisfies $g(x)\leq Cx$ for some $C>0$. For all $s\in[0,t]$, we define

$\psi(s)=\int_{0}^{\infty}P_{s}f(x)\,P_{t-s}g(x)\,m(dx).$

We use the characterization of the infinitesimal generator of diffusion
processes of [6, Thm. 2.81]: let $\mathcal{D}$ be the set of functions $f$
bounded continuous on $[0,\infty)$, such that the right derivative $f^{+}$ of
$f$ exists, is finite, continuous from the right and of bounded variation on
all compact intervals of $(0,\infty)$, and such that $df^{+}=h\,dm$, where
$df^{+}$ denotes the measure on $(0,\infty)$ such that

$f^{+}(y)-f^{+}(x)=df^{+}(x,y]$

and $h$ is some bounded continuous function on $[0,\infty)$ with $h(0)=0$.
Then, for all $f\in\mathcal{D}$,

$Lf(x):=\lim_{t\rightarrow
0}\frac{P_{t}f(x)-f(x)}{t}=\frac{df^{+}}{dm}(x),\quad\forall x\geq 0,$

where the convergence holds for the uniform norm on $[0,\infty)$.

So let $f,g\in\mathcal{D}$ have compact support. In particular, there exists
$C>0$ such that $P_{s}f(x)+P_{s}g(x)\leq Cx$ for all $x,s\geq 0$. Since
$f,g\in\mathcal{D}$, $P_{s}f$ and $P_{s}g$ also belong to $\mathcal{D}$ and
Lebesgue’s theorem entails that, for all $s\in[0,t]$

$\displaystyle\psi^{\prime}(s)$
$\displaystyle=\int_{0}^{\infty}LP_{s}f(x)\,P_{t-s}g(x)\,m(dx)-\int_{0}^{\infty}P_{s}f(x)\,LP_{t-s}g(x)\,m(dx)$
$\displaystyle=\int_{0}^{\infty}P_{t-s}g(x)\,d(LP_{s}f)^{+}(x)-\int_{0}^{\infty}P_{s}f(x)\,d(LP_{t-s}g)^{+}(x).$

The right-hand side is equal to zero by the integration by parts formula for
Stieljes integrals.

Since $\mathcal{D}$ is dense in the set of continuous functions on
$(0,\infty)$ with compact support, the proof is completed. ∎

###  5.4 Proof of Proposition 4.10

The proof is similar to the one of Theorem 4.1 (see Subsection 5.1) and we
only detail the steps that need to be modified. Our aim is to check that
conditions (A1) and (A2) hold.

Step 1. (A1) is satisfied.

Since $m$ satisfies the conditions of Theorem 4.1, we deduce that (A1) holds
for $X$. As a consequence, there exist two constants $c_{1}>0$ and $t_{0}>0$
and a probability measure $\nu$ on $(0,+\infty)$, such that, for all
$A\subset(0,\infty)$ measurable,

$\displaystyle\mathbb{P}_{x}(X_{t_{0}}\in A)\geq
c_{1}\nu(A)\mathbb{P}_{x}(t_{0}<\tau_{\partial}),\quad\forall
x\in(0,+\infty).$

By construction of $\widetilde{X}$, we have

$\displaystyle\mathbb{P}_{x}(\widetilde{X}_{t_{0}}\in A)\geq
e^{-t_{0}}\mathbb{P}_{x}(X_{t_{0}}\in A).$

Fix $x\in(0,1)$. Using the fact that $X_{t}=\widetilde{X}_{t}$ for all $t\leq
T_{1}$ under $\mathbb{P}_{x}$, we deduce that

$\displaystyle\mathbb{P}_{x}(t_{0}<\widetilde{\tau_{\partial}})$
$\displaystyle\leq\mathbb{P}_{x}(\\{t_{0}<\tau_{\partial}\\}\cup\\{\widetilde{X}_{t_{0}}\neq
X_{t_{0}}\\})$
$\displaystyle\leq\mathbb{P}_{x}(t_{0}<\tau_{\partial})+\mathbb{P}_{x}(T_{1}\leq\tau_{\partial})$
$\displaystyle\leq Ax+x$
$\displaystyle\leq\frac{A+1}{a}\mathbb{P}_{x}(t_{0}<\tau_{\partial}),$

where $a>0$ is the positive constant from Remark 3. As a consequence, for all
$A\subset(0,\infty)$ measurable,

$\displaystyle\mathbb{P}_{x}(\widetilde{X}_{t_{0}}\in
A)\geq\frac{c_{1}ae^{-t_{0}}}{A+1}\,\nu(A)\,\mathbb{P}_{x}(t_{0}<\widetilde{\tau_{\partial}}),$
(5.14)

which concludes the proof of (A1) for $\widetilde{X}$.

Step 2. (A2) is satisfied.

By construction of the process $\widetilde{X}$ from the process $X$ and an
independent Poisson process, it is clear that, for all $x\leq y$, all $t>0$
and $a\in(0,x)$, we have

$\displaystyle\mathbb{P}_{x}(\widetilde{T}_{a}\leq
t)\geq\mathbb{P}_{y}(\widetilde{T}_{a}\leq t),$

where $\widetilde{T}_{a}=\inf\\{t\geq 0,\ \widetilde{X}_{t}=a\\}$, and that
$\mathbb{P}_{x}(\widetilde{X}_{t}\geq a)\geq\mathbb{P}_{x}(X_{t}\geq a)$.
Hence we only need to prove that $\widetilde{X}$ satisfies (5.7), the rest of
the proof being the same as in Step 4 of Subsection 5.1.

Fix $\varepsilon>0$ and set $t_{\varepsilon}=-\log(1-\varepsilon/2)>0$. For
all $a>0$, we have, by independence of the Poisson process $(N_{t})_{t\geq
0}$,

$\displaystyle\inf_{x\in(a,+\infty)}\mathbb{P}_{x}(\widetilde{T}_{a}\leq
t_{\varepsilon})=\lim_{x\rightarrow\infty}\mathbb{P}_{x}(\widetilde{T}_{a}\leq
t_{\varepsilon})\geq\left(1-\frac{\varepsilon}{2}\right)\lim_{x\rightarrow\infty}\mathbb{P}_{x}(T_{a}\leq
t_{\varepsilon}).$

Since $X$ comes down from infinity, there exists $a_{\varepsilon}>0$ such that

$\lim_{x\rightarrow+\infty}\mathbb{P}_{x}(T_{a_{\varepsilon}}\leq
t_{\varepsilon})\geq 1-\frac{\varepsilon}{2}.$

Hence, for all $\varepsilon>0$ small enough, there exists $a_{\varepsilon}>0$
such that, for all $x\geq a_{\varepsilon}$,

$\displaystyle\mathbb{P}_{x}(1<\widetilde{T}_{a_{\varepsilon}})\leq\mathbb{P}_{x}(t_{\varepsilon}<\widetilde{T}_{a_{\varepsilon}})\leq
1-\left(1-\frac{\varepsilon}{2}\right)^{2}\leq\varepsilon.$

This entails (5.7) and (A2) as in Step 4 of the proof of Theorem 4.1.

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