# The zero-noise limit of SDEs with $L^{\infty}$ drift

Ulrik Skre Fjordholm  ,  Markus Musch  Department of Mathematics, University
of Oslo, PO Box 1053 Blindern, N-316 Oslo, Norway  and  Andrey Pilipenko
Institute of Mathematics, National Academy of Sciences of Ukraine,
Tereshchenkivska str. 3, 01601, Kiev, Ukraine

###### Abstract.

We study the zero-noise limit for autonomous, one-dimensional ordinary
differential equations with discontinuous right-hand sides. Although the
deterministic equation might have infinitely many solutions, we show, under
rather general conditions, that the sequence of stochastically perturbed
solutions converges to a unique distribution on classical solutions of the
deterministic equation. We provide several tools for computing this limit
distribution.

##  1\. Introduction

Consider a scalar, autonomous ordinary differential equation (ODE) of the form

$\begin{split}\frac{dX}{dt}(t)&=a(X(t))\qquad\text{for }t>0,\\\
X(0)&=0\end{split}$ (1.1)

where $a\colon{\mathbb{R}}\rightarrow{\mathbb{R}}$ is Borel measurable. (The
initial data $X(0)=0$ can be translated to an arbitrary point
$x_{0}\in{\mathbb{R}}$, if needed.) If the drift $a$ is non-smooth then
uniqueness of solutions might fail — this is the _Peano phenomenon_. To
distinguish physically reasonable solutions from non-physical ones, we add
stochastic noise to the equation, with the aim of letting the noise go to
zero. Thus, we consider a stochastic differential equation

$\begin{split}dX_{\varepsilon}(t)&=a(X_{\varepsilon}(t))dt+\varepsilon
dW(t),\\\ X_{\varepsilon}(0)&=0.\end{split}$ (1.2)

where $W(t)$ is a one-dimensional Brownian motion on a given probability space
$(\Omega,{\mathcal{F}},{\mathbb{P}})$, and $\varepsilon>0$. By the
Zvonkin–Veretennikov theorem [28, 29], equation (1.2) has a unique strong
solution.

In this paper we consider the following problem:

> _Identify the limit $\lim_{\varepsilon\to 0}X_{\varepsilon}$, and show that
> it satisfies (1.1)._

Somewhat informally, the challenges are:

  * •

determining whether the sequence $\\{X_{\varepsilon}\\}_{\varepsilon}$ (or a
subsequence) converges, and in what sense;

  * •

identifying the limit(s), either by a closed form expression or some defining
property;

  * •

proving that the limit solves (1.1) by passing to the limit in the (possibly
discontinuous) term $a(X_{\varepsilon})$.

The problem originated in the 1981 paper by Veretennikov [27], and was treated
extensively in the 1982 paper by Bafico and Baldi [2]. Only little work has
been done on this problem since then, despite its great interest. The original
work of Bafico and Baldi dealt with the Peano phenomenon for an autonomous
ordinary differential equation. They considered continuous drifts which are
zero at some point and are non-Lipschitz continuous on at least one side of
the origin. In their paper they show that the $\varepsilon\to 0$ limit of the
probability measure that represents the solution of the stochastic equation is
concentrated on at most two trajectories. Further, they compute explicitly
some limit probability measures for specific drifts. Unfortunately, since the
result of Bafico and Baldi relies on the direct computation of the solution of
an elliptic PDE, it only works in one dimension. In one dimension this
elliptic PDE reduces to a second-order boundary value problem for which an
explicit solution can be computed. Therefore, there is little hope that this
approach will also work in higher dimensions.

The only other work that is known to us dating back to the previous century is
the paper by Mathieu from 1994 [18]. In 2001 Grandinaru, Herrmann and Roynette
published a paper [12] which showed some of the results of Bafico and Baldi
using a large deviations approach. Herrmann did some more work on small-noise
limits later on together with Tugaut [13, 14, 15].

Yet another approach to Bafico and Baldi’s original problem was presented by
Delarue and Flandoli in [8]. They apply a careful argument based on exit
times. Noteworthy it also works in arbitrary dimension but with a very
specific right-hand side, in contrast to the original assumption of a general
continuous function; see also Trevisian [26]. We also point out the recent
paper by Delarue and Maurelli [9], where multidimensional gradient dynamics
with Hölder type coefficients was perturbed by a small Wiener noise.

The 2008 paper by Buckdahn, Ouknine and Quincampoix [7] shows that the the
zero noise limit is concentrated on the set of all Filippov solutions of
(1.1). Since this set is potentially very large, this result is of limited use
to us.

Even less work was done for zero noise limits with respect to partial
differential equations. To our best knowledge the only paper published so far
is Attanasio and Flandoli’s note on the linear transport equation [1].

A new approach was proposed by Pilipenko and Proske when the drift in (1.1)
has Hölder-type asymptotics in a neighborhood of $x=0$ and the perturbation is
a self-similar noise [21]. They used space-time scaling and reduce a solution
of the small-noise problem to a study of long time behaviour of a stochastic
differential equation with a fixed noise. This approach can be generalized to
multidimensional case and multiplicative Levy-noise perturbations [22, 17, 19,
23].

###  1.1. Uniqueness of classical solutions

If the drift $a=a(x)$ is continuous then the question of existence and
uniqueness of solutions of (1.1) is well established. If $a$ is continuous
then it’s known since Peano that there always exists at least one solution (at
least for small times). Binding [4] found that the solution is unique if and
only if $a$ satisfies a so-called Osgood condition at all zeros $x_{0}$ of
$a$:

$\int_{x_{0}-\delta}^{x_{0}}\frac{1}{a(z)\wedge
0}\,dz=-\infty,\qquad\int_{x_{0}}^{x_{0}+\delta}\frac{1}{a(z)\vee
0}\,dz=+\infty$ (1.3)

for all $\delta\in(0,\delta_{0})$ for some $\delta_{0}>0$. (Here and in the
remainder we denote $\alpha\wedge\beta\coloneqq\min(\alpha,\beta)$ and
$\alpha\vee\beta\coloneqq\max(\alpha,\beta)$.) The unique solution starting at
$x$ is then given by

$X(t;x)=\begin{cases}x&\text{if }a(x)=0\\\ A^{-1}(t)&\text{if }a(x)\neq
0\end{cases}$ (1.4)

(at least for small $t$), where $A(y)\coloneqq\int_{x}^{y}1/a(z)\,dz$ and
$A^{-1}$ is its inverse function.

If $a$ is discontinuous — say, $a\in L^{\infty}({\mathbb{R}})$ — then the
question of existence and uniqueness is much more delicate. The paper [10]
gives necessary and sufficient conditions for the uniqueness of _Filippov
solutions_ of (1.1). We remark here that the extension to Filippov solutions
might lead to non-uniqueness, even when the classical solution is unique. To
see this, let $E\subset{\mathbb{R}}$ be measure-dense, i.e. a set for which
both $U\cap E$ and $U\setminus E$ have positive Lebesgue measure for any
nonempty, open set $U\subset{\mathbb{R}}$ (see [25] for the construction of
such a set), and let $a=1+\mathbbm{1}_{E}$. Then (1.4) is the unique classical
solution for any starting point $x\in{\mathbb{R}}$, whereas any function
satisfying $\frac{d}{dt}X(t)\in[1,2]$ for a.e. $t>0$ will be a Filippov
solution. We will show that even in cases such as this one, the stochastically
perturbed solutions converge to the classical solution, and not just any
Filippov solution, as was shown in [7].

###  1.2. Main result

We aim to prove that the distribution of solutions $X_{\varepsilon}$ of (1.2)
converges to a distribution concentrated on either a single solution of the
deterministic equation (1.1), or two “extremal” solutions. Based on the
discussion in the previous section, we can divide the argument into cases
depending on whether $a$ is positive, negative or changes sign in a
neighbourhood, and in each case, whether an Osgood-type condition such as
(1.3) holds. The case of negative drift is clearly analogous to a positive
drift, so we will merely state the results for negative drift, without proof.

Under the sole assumption $a\in L^{\infty}({\mathbb{R}})$, the sequence
$\\{X_{\varepsilon}\\}_{\varepsilon}$ is weakly relatively compact in
$C([0,T])$, for any $T>0$. (Indeed, by (1.2), $X_{\varepsilon}-\varepsilon W$
is uniformly Lipschitz, and $\varepsilon W\overset{P}{\to}0$ as
$\varepsilon\to 0$. See e.g. [3] for the full argument.) Hence, the problems
are to characterize the distributional limit of any convergent subsequence, to
determine whether the entire sequence converges (i.e., to determine whether
the limit is unique), and to determine whether the sense of convergence can be
strengthened.

Without loss of generality we will assume that the process starts at $x=0$. If
$a(0)=0$ but $a$ does not satisfy the Osgood condition (1.3) at $x=0$, then
both $\psi_{-}$ and $\psi_{+}$ are classical solutions of (1.1) (along with
infinitely many other solutions), where

$\psi_{\pm}(t)\coloneqq A_{\pm}^{-1}(t),\qquad\text{where
}A_{\pm}(x)\coloneqq\int_{0}^{x}\frac{1}{a(z)}\,dz\text{ for
}x\in{\mathbb{R}}_{\pm}.$ (1.5)

Generally, the functions $\psi_{\pm}$ are defined in a neighborhood of 0. We
have assumed that $a$ is bounded, so $\psi_{\pm}$ cannot blow up in finite
time, but they can reach singular points $R_{\pm}$ where $A_{\pm}$ blow up. If
$t_{\pm}\in(0,\infty]$ are the times when $\psi_{\pm}(t_{\pm})=R_{\pm}$ then
we set $\psi_{\pm}(t)\equiv R_{\pm}$ for all $t\geqslant t_{\pm}$. We aim to
prove that the distribution of $X_{\varepsilon}$ converges to a distribution
concentrated on the two solutions $\psi_{-},\ \psi_{+}$, and to determine the
weighting of these two solutions.

######  Theorem 1.1.

Let $a\in L^{\infty}({\mathbb{R}})$ satisfy $a\geqslant 0$ a.e. in
$(-\delta_{0},\delta_{0})$ for some $\delta_{0}>0$, and

$\int_{0}^{\delta_{0}}\frac{1}{a(z)}dz<\infty.$ (1.6)

Then, for any $T>0$, $X_{\varepsilon}$ converges uniformly in probability to
$\psi_{+}$:

$\big{\|}X_{\varepsilon}-\psi_{+}\big{\|}_{C([0,T])}\overset{P}{\to}0\qquad\text{as
}\varepsilon\to 0.$ (1.7)

An analogous result holds for _negative_ drifts, with obvious modifications.

The proof of Theorem 1.1 for strictly positive drifts $a$ is given in Section
3, while the general case is considered in Section 5. The final theorem
applies also to signed drifts:

######  Theorem 1.2.

Let $a\in L^{\infty}({\mathbb{R}})$ satisfy

$-\int_{\alpha}^{0}\frac{1}{a(z)\wedge
0}\,dz<\infty,\qquad\int_{0}^{\beta}\frac{1}{a(z)\vee 0}\,dz<\infty$ (1.8)

for some $\alpha<0<\beta$ (compare with (1.3)). Let
$\\{\varepsilon_{k}\\}_{k}$ be some sequence satisfying $\varepsilon_{k}>0$
and $\lim_{k\to\infty}\varepsilon_{k}=0$, and define

$p_{k}\coloneqq\frac{s_{\varepsilon_{k}}(0)-s_{\varepsilon_{k}}(\alpha)}{s_{\varepsilon_{k}}(\beta)-s_{\varepsilon_{k}}(\alpha)}\in[0,1],\qquad
s_{\varepsilon}(r)\coloneqq\int_{0}^{r}\exp\Bigl{(}-\frac{2}{\varepsilon^{2}}\int_{0}^{z}a(u)\,du\Bigr{)}\,dz.$
(1.9)

Then $\\{P_{\varepsilon_{k}}\\}_{k}$ is weakly convergent if $\\{p_{k}\\}_{k}$
converges. Defining $p\coloneqq\lim_{k}p_{k}$ and
$P\coloneqq\operatorname{wlim}_{k}P_{\nu_{k}}$, we have

$P=(1-p)\delta_{\psi_{-}}+p\delta_{\psi_{+}}.$ (1.10)

The proof is given in Section 4, where we also provide tools for computing
$p$.

###  1.3. Outline of the paper

We now give an outline of the rest of this manuscript. In Section 2 we give
several technical results on convergence of SDEs with respect to perturbations
of the drift; the relation between the solution and its exit time; and the
distribution of the solution of an SDE. The goal of Section 3 is to prove
Theorem 1.1 in the case where $a>0$, and in Section 5 we extend to the case
$a\geqslant 0$. In Section 4 we prove Theorem 1.2 and provide several results
on sufficient conditions for convergence. Finally, we give some examples in
Section 6.

##  2\. Technical results

In this section we list a few technical results. The first two results are
comparison principles. In order to prove them we use approximations by SDEs
with smooth coefficients and the classical results on comparison. Since we do
not suppose that the drift is smooth or even continuous, the results are not
standard.

######  Theorem 2.1.

Let $\\{a_{n}\colon{\mathbb{R}}\rightarrow{\mathbb{R}}\\}_{n\geqslant 0}$ be
uniformly bounded measurable functions such that $a_{n}\to a_{0}$ pointwise
a.e. as $n\to\infty$. Let $X_{n}$ be a solution to the SDE

$X_{n}(t)=x_{n}+\int_{0}^{t}a_{n}(X_{n}(s))ds+W(t),\qquad t\in[0,T].$

Then $\\{X_{n}\\}_{n}$ converges uniformly in probability:

$\bigl{\|}X_{n}(t)-X_{0}(t)\bigr{\|}_{C([0,T])}\overset{P}{\to}0\qquad\text{as
}n\to\infty.$

For a proof, see e.g. [20, Theorem 2.1].

######  Theorem 2.2.

Let $a_{1},a_{2}\colon{\mathbb{R}}\rightarrow{\mathbb{R}}$ be locally bounded
measurable functions satisfying $a_{1}\leqslant a_{2}$ and let $x_{1}\leqslant
x_{2}$. Let $X_{1},X_{2}$ be solutions to the equations

$\displaystyle X_{i}(t)=x_{i}+\int_{0}^{t}a_{i}(X_{i}(s))ds+W(t),\qquad
i=1,2.$

Then

$X_{1}(t)\leqslant X_{2}(t)\qquad\forall\ t\geqslant 0$

with probability 1.

The proof is given in Appendix A.

######  Lemma 2.3.

Let $\\{f_{n}\\}_{n\geqslant 1}\subset C([0,T])$ be a uniformly convergent
sequence of non-random continuous functions and let $f_{0}\in C([0,T])$ be a
strictly increasing function. Set
$\tau^{x}_{n}\coloneqq\inf\bigl{\\{}t\geqslant 0:f_{n}(t)=x\bigr{\\}}$ for
every $n\geqslant 0$, and assume that

$\tau^{x}_{n}\to\tau^{x}_{0}\qquad\text{for every
}x\in\big{(}f_{0}(0),f_{0}(T)\bigr{)}\cap{\mathbb{Q}}.$

Then

$f_{n}\to f_{0}\qquad\text{in }C([0,T])\text{ as }n\to\infty.$

###### Proof.

Let $\mathcal{T}\coloneqq f_{0}^{-1}({\mathbb{Q}})$, and note that this is a
dense subset of $[0,T]$, since $f_{0}^{-1}$ is continuous. Let
$t\in\mathcal{T}$ be arbitrary and let $x\coloneqq f_{0}(t)\in{\mathbb{Q}}$.
By assumptions of the lemma we have
$t=\tau_{0}^{x}=\lim_{n\to\infty}\tau_{n}^{x}.$ Moreover, since
$f_{n}(\tau^{x}_{n})=x$ for sufficiently large $n$, we have

$f_{0}(t)=x=\lim_{n\to\infty}f_{n}(\tau^{x}_{n})=\lim_{n\to\infty}f_{n}(t),$
(2.1)

the last step following from the fact that $f_{n}$ converges uniformly and
$\tau^{x}_{n}\to\tau^{x}_{0}=t$ as $n\to\infty$. Thus, $\\{f_{n}\\}_{n}$
converges pointwise to $f_{0}$ on a dense subset of $[0,T]$. But
$\\{f_{n}\\}_{n}$ is uniformly convergent by assumption, so necessarily
$f_{n}\to f_{0}$ uniformly. ∎

######  Corollary 2.4.

Let $\\{\xi_{n}\\}_{n\geqslant 1}$ be a sequence of continuous stochastic
processes $\xi_{n}\colon[0,\infty)\to{\mathbb{R}}$ that is locally uniformly
convergent with probability $1$. Let $\xi_{0}$ be a strictly increasing
continuous process satisfying $\xi_{0}(0)=0$ and
$\lim_{t\to\infty}\xi_{0}(t)=\infty$. Set
$\tau_{n}^{x}\coloneqq\inf\\{t\geqslant 0:\xi_{n}(t)\geqslant x\\}$ and assume
that

$\tau_{n}^{x}\overset{P}{\to}\tau_{0}^{x}\qquad\text{for every
}x\in[0,\infty)\cap{\mathbb{Q}}.$

Then

$\xi_{n}\to\xi_{0}\qquad\text{locally uniformly with probability }1.$

###### Proof.

Enumerate the positive rational numbers as
${\mathbb{Q}}\cap(0,\infty)=\\{x_{n}\\}_{n}$. Select a sequence
$\\{n^{1}_{k}\\}_{k}$ such that

$\lim_{k\to\infty}\tau^{x_{1}}_{n^{1}_{k}}=\tau^{x_{1}}_{0}\qquad\text{${\mathbb{P}}$-a.s.}$

Then select a sub-subsequence $\\{n^{2}_{k}\\}_{k}$ of $\\{n^{1}_{k}\\}_{k}$
such that

$\lim_{k\to\infty}\tau^{x_{2}}_{n^{2}_{k}}=\tau^{x_{2}}_{0}\qquad\text{${\mathbb{P}}$-a.s.,}$

and so on. Then

${\mathbb{P}}\Bigl{(}\forall\
j\in{\mathbb{N}}\quad\lim_{k\to\infty}\tau^{x_{j}}_{n^{k}_{k}}=\tau^{x_{j}}_{0}\Bigr{)}=1.$

From Lemma 2.3 it follows that

${\mathbb{P}}\Bigl{(}\lim_{k\to\infty}\xi_{n^{k}_{k}}=\xi_{0}\quad\text{uniformly
in }[0,T]\Bigr{)}=1$

for any $T>0$. This yields the result. ∎

Assume that $a,\sigma\colon{\mathbb{R}}\to{\mathbb{R}}$ are bounded measurable
functions, $\sigma$ is separated from zero. It is well known that the
stochastic differential equation

$d\xi(t)=a(\xi(t))dt+\sigma(\xi(t))dW(t),\qquad t\geqslant 0,$

has a unique (weak) solution, which is a continuous strong Markov process,
i.e., $\xi$ is a diffusion process.

Denote $L\coloneqq
a(x)\frac{d}{dx}+\frac{1}{2}\sigma^{2}(x)\frac{d^{2}}{dx^{2}}$ and let $s$ and
$m$ be a scale function and a speed measure of $\xi,$ see details in [24,
Chapter VII]. Define the hitting time of $\xi$ as
$\tau^{y}\coloneqq\inf\\{t\geqslant 0:\xi(t)=y\\}$. Recall that $s$ and $m$
are well-defined up to constants, and $s$ is a non-degenerate $L$-harmonic
function, i.e.,

$Ls=0,$ (2.2)

in particular

$s(x)\coloneqq\int_{y_{1}}^{x}\exp\left(-\int_{y_{2}}^{y}\frac{2a(z)}{\sigma(z)^{2}}dz\right)dy,$
(2.3)

and

$m(dy)=\frac{2}{s^{\prime}(y)\sigma(y)^{2}}dy$ (2.4)

for any choices of $y_{1},y_{2},$ see [24, Chapter VII, Exercise 3.20].

######  Theorem 2.5.

Let $x_{1}<x_{2}$ be arbitrary.

  1. (i)

[24, Chapter VII, Proposition 3.2 and Exercise 3.20]

$\displaystyle{\mathbb{P}}^{x}\big{(}\tau^{x_{1}}\wedge\tau^{x_{2}}<\infty\big{)}=1\qquad$
$\displaystyle\forall\ x\in[x_{1},x_{2}]$ and
$\displaystyle{\mathbb{P}}^{x}\bigl{(}\tau^{x_{1}}<\tau^{x_{2}}\bigr{)}=\frac{s(x_{2})-s(x)}{s(x_{2})-s(x_{1})}\qquad$
$\displaystyle\forall\ x\in[x_{1},x_{2}],$

  2. (ii)

[24, Chapter VII, Corollary 3.8] For any $I=(x_{1},x_{2})$, $x\in I$ and for
any non-negative measurable function $f$ we have

${\mathbb{E}}^{x}\biggl{(}\int_{0}^{\tau^{x_{1}}\wedge\tau^{x_{2}}}\\!\\!f(\xi(t))dt\biggr{)}=\int_{x_{1}}^{x_{2}}\\!G(x,y)f(y)m(dy),$
(2.5)

where $G=G_{I}$ is a symmetric function such that

$G_{I}(x,y)=\frac{(s(x)-s(x_{1}))(s(x_{2})-s(y))}{s(x_{2})-s(x_{1})},\qquad
x_{1}\leqslant x\leqslant y\leqslant x_{2}.$

######  Remark 2.6.

  1. (i)

The function
$\tilde{u}(x)\coloneqq{\mathbb{E}}^{x}\Bigl{(}\int_{0}^{\tau^{x_{1}}\wedge\tau^{x_{2}}}f(\xi(t))dt\Bigr{)}$
from the left-hand side of (2.5) is a solution to

$\begin{cases}L\tilde{u}(x)=-f(x),&x\in(x_{1},x_{2})\\\
\tilde{u}(x_{1})=\tilde{u}(x_{2})=0.\end{cases}$

The function $G$ from (2.5) is the corresponding Green function, in the sense
that $\tilde{u}(x)$ can be written as the right-hand side of (2.5).

  2. (ii)

If we take $f(x)=1$ in (2.5), then we get a formula for the expectation of the
exit time $u(x)\coloneqq{\mathbb{E}}^{x}(\tau^{x_{1}}\wedge\tau^{x_{2}})$,
$x\in[x_{1},x_{2}]$. In particular,

$u(x)=-\int_{x_{1}}^{x}2\Phi(y)\int_{x_{1}}^{y}\frac{dz}{\sigma(z)^{2}\Phi(z)}dy+\int_{x_{1}}^{x_{2}}2\Phi(y)\int_{x_{1}}^{y}\frac{dz}{\sigma(z)^{2}\Phi(z)}dy\frac{\int_{x_{1}}^{x}\Phi(y)dy}{\int_{x_{1}}^{x_{2}}\Phi(y)dy},$

where
$\Phi(x)=\exp\left(-\int_{x_{1}}^{x}\frac{2a(z)}{\sigma(z)^{2}}dz\right).$

Finally, the following result will be quite useful when taking limits
$\sigma=\sigma_{\varepsilon}(x)\coloneqq\varepsilon\to 0$ in terms such as $s$
and $u$ above.

######  Lemma 2.7.

Let $\alpha<\beta$ and $\varepsilon\neq 0$, let $f,g\in L^{1}((\alpha,\beta))$
with $f>0$ almost everywhere, and let

$g_{\varepsilon}(y)\coloneqq\int_{y}^{\beta}\exp\left(-\int_{y}^{z}\frac{f(u)}{\varepsilon^{2}}\,du\right)\frac{f(z)}{\varepsilon^{2}}g(z)\,dz,\qquad
y\in[\alpha,\beta].$

Then $g_{\varepsilon}\to g$ as $\varepsilon\to 0$ in $L^{1}((\alpha,\beta))$
and pointwise a.e. in $y\in(\alpha,\beta)$. The same is true if

$g_{\varepsilon}(y)\coloneqq\int_{\alpha}^{y}\exp\left(-\int_{z}^{y}\frac{f(u)}{\varepsilon^{2}}\,du\right)\frac{f(z)}{\varepsilon^{2}}g(z)\,dz,\qquad
y\in[\alpha,\beta].$

The proof is given in Appendix A. Note that this lemma provides a positive
answer to the question raised by Bafico and Baldi in [2, Remark b in Section
6] on whether [2, Proposition 3.3] still holds under the sole assumption of
$\int_{0}^{r}1/a(z)dz<+\infty$.

##  3\. Positive drifts

This section is dedicated to the proof of Theorem 1.1. In order to prove the
theorem, we first prove the following:

######  Theorem 3.1.

Let $a\in L^{\infty}({\mathbb{R}})$ and assume that there exist positive
constants $\delta_{0},c_{-}>0$ such that

$a(x)\geqslant c_{-}\quad\text{for a.e. }x\in(-\delta_{0},\infty).$ (3.1)

Then we have the uniform convergence in probability

$\|X_{\varepsilon}-\psi_{+}\|_{C([0,T])}\overset{P}{\to}0\quad\text{as
}\varepsilon\to 0\text{ for all }T>0.$ (3.2)

###### Proof of Theorem 3.1.

The proof consists of these steps:

  1. 1.

Show weak relative compactness of $\\{X_{\varepsilon}\\}_{\varepsilon}$.

  2. 2.

Show that $\bar{X}_{0}$ is strictly increasing, where $\bar{X}_{0}$ is a limit
point of $\\{X_{\varepsilon}\\}_{\varepsilon}$.

  3. 3.

Reduce to proving convergence of the hitting times
$\tau^{\varepsilon}\to\tau$, see Lemma 2.3.

Step 1: For any $T>0$ the family $\\{X_{\varepsilon}\\}_{\varepsilon\in(0,1]}$
is weakly relatively compact in $C([0,T])$ (see e.g. [3]). Since $\psi_{+}$ is
non-random, the convergence statement (3.2) is equivalent to the weak
convergence

$X_{\varepsilon}\Rightarrow\psi_{+}\qquad\text{ in }C([0,T])\text{ as
}\varepsilon\to 0.$

for any $T>0$. To prove the latter, it suffices to verify that if
$\\{X_{\varepsilon_{k}}\\}_{k}$ is any convergent subsequence, then $\psi_{+}$
is its limit.

Step 2: Assume that $X_{\varepsilon_{k}}\Rightarrow\bar{X}_{0}$ as
$k\to\infty$. Since

$X_{\varepsilon_{k}}(t)=\int_{0}^{t}a(X_{\varepsilon_{k}}(s))\,ds+\varepsilon_{k}W(t)\qquad\forall\
t\in[0,T],$

and $\varepsilon_{k}W\overset{P}{\to}0$, Slutsky’s theorem implies that also

$\int_{0}^{\cdot}a(X_{\varepsilon_{k}}(s))\,ds\Rightarrow\bar{X}_{0}\qquad\text{in
}C([0,T]).$ (3.3)

By Skorokhod’s representation theorem [3, Theorem 1.6.7], we may assume that
the convergence in (3.3) happens almost surely. Since $c_{-}\leqslant
a\leqslant c_{+}$ (for some $c_{+}>0$), we conclude that

$c_{-}\leqslant\frac{\bar{X}_{0}(t_{2})-\bar{X}_{0}(t_{1})}{t_{2}-t_{1}}\leqslant
c_{+}\qquad\forall\ t_{1},t_{2}\in[0,T],\text{ almost surely.}$

In particular, $\bar{X}_{0}$ is strictly increasing.

Step 3: Notice that assumption (3.1) implies that
$\lim_{t\to\infty}\psi_{+}(t)=+\infty.$ Define

$\tau_{\varepsilon}^{x}\coloneqq\inf\\{t\geqslant
0\,:\,X_{\varepsilon}(t)=x\\},\qquad\tau_{0}^{x}\coloneqq\inf\\{t\geqslant
0\,:\,\psi_{+}(t)=x\\}=A(x)$

where $A(x)\coloneqq\int_{0}^{x}a(z)^{-1}\,dz$ (cf. (1.4)). By Corollary 2.4
it is enough to show convergence in probability of $\tau_{\varepsilon}$:

$\tau_{\varepsilon}^{x}\overset{P}{\to}A(x)\qquad\text{as }\varepsilon\to
0\text{ for every }x\in{\mathbb{Q}}\cap[0,\infty).$ (3.4)

To check (3.4) it is sufficient to verify that

$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{E}}(\tau_{\varepsilon}^{x})=A(x)$ $\displaystyle\text{for any
}x\in{\mathbb{Q}}\cap[0,\infty),$ (3.5a) $\displaystyle\lim_{\varepsilon\to
0}\mathrm{Var}(\tau_{\varepsilon}^{x})=0$ $\displaystyle\text{for any
}x\in{\mathbb{Q}}\cap[0,\infty).$ (3.5b)

We prove these properties under less restrictive conditions on $a$, given in
the lemma below.

######  Lemma 3.2.

Let $R,\delta>0$ and let $a\in L^{\infty}({\mathbb{R}})$ satisfy $a>0$ a.e. in
$(-\delta,R)$. Assume that the Osgood-type condition

$\int_{0}^{R}\frac{1}{a(z)}\,dz<\infty$ (3.6)

is satisfied. Denote $A(r)\coloneqq\int_{0}^{r}a(z)^{-1}\,dz$ for $r\in[0,R]$.
Then

$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}^{x}\big{(}\tau^{-\delta}_{\varepsilon}>\tau^{R}_{\varepsilon}\big{)}=1$
$\displaystyle\forall\ 0\leqslant x\leqslant R,$ (3.7a)
$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{E}}^{x}\big{(}\tau^{-\delta}_{\varepsilon}\wedge\tau^{r}_{\varepsilon}\big{)}=A(r){-A(x)}$
$\displaystyle\forall\ 0\leqslant x<r\leqslant R.$ (3.7b) Moreover, if
$a(x)\geqslant c_{-}$ for $x\in(-\infty,-\delta)$ for some constant $c_{-}>0$,
then also $\displaystyle{\lim_{\varepsilon\to
0}{\mathbb{E}}^{0}(\tau^{r}_{\varepsilon})=A(r)}$ $\displaystyle\forall\
0<r\leqslant R,$ (3.7c) and if $a(x)\geqslant c_{-}>0$ for all
$x\in{\mathbb{R}}$, then $\displaystyle{\lim_{\varepsilon\to
0}\mathrm{Var}^{0}(\tau^{r}_{\varepsilon})=0}$ $\displaystyle\forall\
0<r\leqslant R.$ (3.7d)

We finalize the proof of Theorem 3.1 and then prove the claims of Lemma 3.2
separately. Define the function

$\tilde{a}(x):=\begin{cases}a(x)&\text{if }x>-\delta,\\\ c_{-}&\text{if
}x\leqslant-\delta,\end{cases}$

and denote the solution to the corresponding stochastic differential equation
by $\tilde{X}_{\varepsilon}$. It follows from Lemma 3.2 that

$\|\tilde{X}_{\varepsilon}-\psi_{+}\|_{C([0,T])}\overset{P}{\to}0\qquad\text{as
}\varepsilon\to 0\text{ for all }T>0.$

Uniqueness of the solution yields
${\mathbb{P}}\bigl{(}\tilde{X}_{\varepsilon}(t)=X_{\varepsilon}(t)\text{ for
}t\leqslant\tau_{\varepsilon}^{-\delta}\bigr{)}=1.$ It is easy to see that
${\mathbb{P}}(\tau_{\varepsilon}^{-\delta}=\infty)\to 1$ as $\varepsilon\to
0.$ This completes the proof of Theorem 3.1. ∎

###### Proof of (3.7a) in Lemma 3.2.

By Theorem 2.5(i), we can write

${\mathbb{P}}^{x}(\tau^{r}_{\varepsilon}<\tau^{-\delta}_{\varepsilon})=\frac{s_{\varepsilon}(x)}{s_{\varepsilon}(r)}\geqslant\frac{s_{\varepsilon}(0)}{s_{\varepsilon}(r)}$

for every $x\in[0,r]$, where (cf. (2.3))

$s_{\varepsilon}(x)\coloneqq\int_{-\delta}^{x}e^{-B(y)/\varepsilon^{2}}\,dy,\qquad
B(y)\coloneqq 2\int_{-\delta}^{y}a(z)dz.$ (3.8)

We have

$s_{\varepsilon}(0)=\int_{-\delta}^{0}e^{-B(y)/\varepsilon^{2}}\,dy\geqslant\delta
e^{-B(0)/\varepsilon^{2}}$ (3.9)

since $B$ is nondecreasing. For sufficiently small $\varepsilon>0$ we can find
$y_{\varepsilon}>0$ such that $B(y_{\varepsilon})=B(0)+\varepsilon$. Note that
$y_{\varepsilon}\to 0$ as $\varepsilon\to 0$. Again using the fact that $B$ is
nondecreasing, we can estimate

$\displaystyle s_{\varepsilon}(r)$
$\displaystyle=s_{\varepsilon}(0)+\int_{0}^{r}e^{-B(y)/\varepsilon^{2}}\,dy\leqslant
s_{\varepsilon}(0)+y_{\varepsilon}e^{-B(0)/\varepsilon^{2}}+(r-y_{\varepsilon})e^{-B(y_{\varepsilon})/\varepsilon^{2}}$
$\displaystyle\leqslant
e^{-B(0)/\varepsilon^{2}}\Bigl{(}s_{\varepsilon}(0)+y_{\varepsilon}+re^{-1/\varepsilon}\Bigr{)}.$

Using (3.9), we get

${\mathbb{P}}^{x}(\tau^{r}_{\varepsilon}<\tau^{-\delta}_{\varepsilon})\geqslant\frac{s_{\varepsilon}(0)e^{B(0)/\varepsilon^{2}}}{s_{\varepsilon}(0)e^{B(0)/\varepsilon^{2}}+y_{\varepsilon}+re^{-1/\varepsilon}}\geqslant\frac{\delta}{\delta+y_{\varepsilon}+re^{-1/\varepsilon}}.$

Since $y_{\varepsilon}+re^{-1/\varepsilon}\to 0$ as $\varepsilon\to 0$, we
conclude that
${\mathbb{P}}^{x}(\tau^{r}_{\varepsilon}<\tau^{-\delta}_{\varepsilon})\to 1$
as $\varepsilon\to 0$. ∎

###### Proof of (3.7b) in Lemma 3.2.

We will show that for any $r\in(0,R]$ and $x\in[0,r]$, we have
$\lim_{\varepsilon\to
0}{\mathbb{E}}^{x}\big{(}\tau^{-\delta}_{\varepsilon}\wedge\tau^{r}_{\varepsilon}\big{)}=\int_{x}^{r}a(z)^{-1}dz.$
It follows from Theorem 2.5(ii) with $x_{1}=-\delta$, $x_{2}=r$, $f\equiv 1$,
$s=s_{\varepsilon}$ (cf. (3.8)) and $m=m_{\varepsilon}$ (cf. (2.4)) that for
any $\delta>0$ and $x\in[0,r]$,

$\displaystyle{\mathbb{E}}^{x}\big{(}\tau^{-\delta}_{\varepsilon}\wedge\tau^{r}_{\varepsilon}\big{)}=\int_{-\delta}^{r}G_{\varepsilon}(x,y)\,m_{\varepsilon}(dy)$
(3.10)
$\displaystyle=\int_{-\delta}^{x}G_{\varepsilon}(y,x)\,m_{\varepsilon}(dy)+\int_{x}^{r}G_{\varepsilon}(x,y)\,m_{\varepsilon}(dy)$
$\displaystyle=\int_{-\delta}^{x}\frac{s_{\varepsilon}(y)(s_{\varepsilon}(r)-s_{\varepsilon}(x))}{s_{\varepsilon}(r)}\,m_{\varepsilon}(dy)+\int_{x}^{r}\frac{s_{\varepsilon}(x)(s_{\varepsilon}(r)-s_{\varepsilon}(y))}{s_{\varepsilon}(r)}\,m_{\varepsilon}(dy)$
$\displaystyle=\int_{-\delta}^{x}\underbrace{\frac{s_{\varepsilon}(y)}{s_{\varepsilon}(r)}}_{\eqqcolon\,p_{\varepsilon}(y)}(s_{\varepsilon}(r)-s_{\varepsilon}(x))\,m_{\varepsilon}(dy)+\underbrace{\frac{s_{\varepsilon}(x)}{s_{\varepsilon}(r)}}_{=\,p_{\varepsilon}(x)}\int_{x}^{r}(s_{\varepsilon}(r)-s_{\varepsilon}(y))\,m_{\varepsilon}(dy)$
$\displaystyle=\int_{-\delta}^{x}p_{\varepsilon}(y)\left[\int_{x}^{r}\exp\left(-\int_{-\delta}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)dz\right]\frac{2}{\varepsilon^{2}}\exp\left(\int_{-\delta}^{y}\frac{2a(z)}{\varepsilon^{2}}dz\right)dy$
$\displaystyle\quad+p_{\varepsilon}(x)\int_{x}^{r}\left[\int_{y}^{r}\exp\left(-\int_{-\delta}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)dz\right]\frac{2}{\varepsilon^{2}}\exp\left(\int_{-\delta}^{y}\frac{2a(z)}{\varepsilon^{2}}dz\right)dy$
$\displaystyle=\int_{-\delta}^{x}p_{\varepsilon}(y)\int_{x}^{r}\exp\left(-\int_{y}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)\frac{2}{\varepsilon^{2}}\,dzdy$
$\displaystyle\quad+p_{\varepsilon}(x)\int_{x}^{r}\int_{y}^{r}\exp\left(-\int_{y}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)\frac{2}{\varepsilon^{2}}\,dzdy$
$\displaystyle={\int_{-\delta}^{x}p_{\varepsilon}(y)\int_{y}^{r}\exp\left(-\int_{y}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)\frac{2a(z)}{\varepsilon^{2}}\frac{\mathbbm{1}_{(x,r)}(z)}{a(z)}\,dzdy}$
$\displaystyle\quad+p_{\varepsilon}(x)\int_{x}^{r}\int_{y}^{r}\exp\left(-\int_{y}^{z}\frac{2a(u)}{\varepsilon^{2}}du\right)\frac{2a(z)}{\varepsilon^{2}}\frac{1}{a(z)}\,dzdy$
$\displaystyle=I_{\varepsilon}+\mathit{II}_{\varepsilon}.$

By Theorem 2.5(i) we have
$p_{\varepsilon}(x)={\mathbb{P}}^{x}(\tau_{\varepsilon}^{-\delta}>\tau_{\varepsilon}^{r})$,
and (3.7a) in Lemma 3.2 implies that $\lim_{\varepsilon\to
0}p_{\varepsilon}(x)=1$ for every $x\in[0,r]$. Letting $f(z)=2a(z)$ and
$g(z)=\frac{1}{a(z)}\mathbbm{1}_{(x,r)}(z)$ for $z\in[0,r]$, we see that the
$z$-integral in $\mathit{II}_{\varepsilon}$ can be written as

${\int_{y}^{r}\exp\left(-\int_{y}^{z}\frac{f(u)}{\varepsilon^{2}}du\right)\frac{f(z)}{\varepsilon^{2}}g(z)\,dz.}$

Note that $f,g\in L^{1}([0,r])$, by (3.6). Thus, we can apply Lemma 2.7 with
$\alpha=0$, $\beta=r$ to get

$g_{\varepsilon}(y)\coloneqq\int_{y}^{r}\exp\left(-\int_{y}^{u}\frac{2a(z)}{\varepsilon^{2}}dz\right)\frac{2}{\varepsilon^{2}}\,du\to
g(y)$

in $L^{1}([0,r])$ and pointwise a.e. as $\varepsilon\to 0$, so that

$\mathit{II}_{\varepsilon}\to\int_{x}^{r}g(y)\,dy=\int_{x}^{r}\frac{1}{a(y)}\,dy.$

A similar manipulation will hold for $I_{\varepsilon}$, with the same
functions $f$ and $g$, yielding

$I_{\varepsilon}\to\int_{-\delta}^{x}\frac{1}{a(y)}\mathbbm{1}_{(x,r)}(y)\,dy=0.$

Putting these together gives

$\lim_{\varepsilon\to
0}{\mathbb{E}}^{x}\big{(}\tau^{-\delta}_{\varepsilon}\wedge\tau^{r}_{\varepsilon}\big{)}=\lim_{\varepsilon\to
0}I_{\varepsilon}+\mathit{II}_{\varepsilon}=\int_{x}^{r}\frac{1}{a(y)}\,dy.$

This concludes the proof. ∎

###### Proof of (3.7c) in Lemma 3.2.

For any $x\in[0,r)$, note that
$\lim_{\delta\to+\infty}{\mathbb{E}}^{x}(\tau^{-\delta}_{\varepsilon}\wedge\tau^{r}_{\varepsilon})={\mathbb{E}}^{x}(\tau^{r}_{\varepsilon})$.
Using (3.10) and the assumption $a\geqslant c_{-}>0$ it is easy to obtain the
uniform estimates for expectations and to see that $\lim_{\varepsilon\to
0}{\mathbb{E}}^{0}(\tau^{r}_{\varepsilon})=A(r).$ ∎

###### Proof of (3.7d) in Lemma 3.2.

Let $X_{\varepsilon}$ solve (1.2) and define
$Y_{\varepsilon}(t)=\varepsilon^{-2}X_{\varepsilon}(\varepsilon^{2}t)$.
Substitution into (1.2) then gives

$Y_{\varepsilon}(t)=\int_{0}^{t}a\big{(}\varepsilon^{2}Y_{\varepsilon}(s)\big{)}\,ds+B(t)$
(3.11)

where $B(t)=\varepsilon^{-1}W(\varepsilon^{2}t)$ is another Brownian motion.
Applying the same scaling to $\tau$, we see that if $\pi^{n}_{\varepsilon}$ is
the exit time of $Y_{\varepsilon}$ from $(-\infty,n]$ then
$\pi^{n}_{\varepsilon}=\varepsilon^{-2}\tau^{\varepsilon^{2}n}_{\varepsilon}$.
To this end, fix $x>0$, let $n=\varepsilon^{-2}x$ (assumed for simplicity to
be an integer) and define the increments
$\zeta^{1}_{\varepsilon}=\pi^{1}_{\varepsilon}$,
$\zeta^{2}_{\varepsilon}=\pi^{2}_{\varepsilon}-\pi^{1}_{\varepsilon}$,
$\dots$,
$\zeta^{n}_{\varepsilon}=\pi^{n}_{\varepsilon}-\pi^{n-1}_{\varepsilon}$. The
strong Markov property ensures that
$\zeta^{1}_{\varepsilon},\dots,\zeta^{n}_{\varepsilon}$ are independent random
variables. Hence,

$\displaystyle\mathrm{Var}(\tau^{x}_{\varepsilon})$
$\displaystyle=\varepsilon^{4}\mathrm{Var}(\pi^{n}_{\varepsilon})=\varepsilon^{4}\mathrm{Var}\Biggl{(}\sum_{k=1}^{n}\zeta^{k}_{\varepsilon}\Biggr{)}$
$\displaystyle=\varepsilon^{4}\sum_{k=1}^{n}\mathrm{Var}(\zeta^{k}_{\varepsilon}).$

Hence, if we can bound $\mathrm{Var}(\zeta^{k}_{\varepsilon})$ by a constant
independent of $\varepsilon$, then
$\mathrm{Var}(\tau^{x}_{\varepsilon})\leqslant\varepsilon^{4}Cn=Cx\varepsilon^{2}\to
0$, and we are done. To this end, note first the naive estimate
$\mathrm{Var}(\zeta^{k}_{\varepsilon})\leqslant{\mathbb{E}}((\zeta^{k}_{\varepsilon})^{2})$.
Next, we invoke the comparison principle Theorem 2.2 between $Y_{\varepsilon}$
and

$Z_{\varepsilon}(t)\coloneqq\int_{0}^{t}c_{-}\,dt+B(t)=c_{-}t+B(t),$

yielding $Z_{\varepsilon}(t)\leqslant Y_{\varepsilon}(t)$ for all $t\geqslant
0$, almost surely. Hence,
$\pi^{n}_{\varepsilon}\leqslant\tilde{\pi}^{n}_{\varepsilon}$, where
$\tilde{\pi}^{n}_{\varepsilon}$ is the exit time of $Z_{\varepsilon}$, and
correspondingly,
$\zeta^{k}_{\varepsilon}\leqslant\tilde{\zeta}^{k}_{\varepsilon}$ for
$k=1,\dots,n$. Since $(\tilde{\zeta}^{k}_{\varepsilon})_{k=1}^{n}$ are
identically distributed, we get

${\mathbb{E}}\big{(}(\zeta^{k}_{\varepsilon})^{2}\big{)}\leqslant{\mathbb{E}}\big{(}(\tilde{\zeta}^{k}_{\varepsilon})^{2}\big{)}={\mathbb{E}}\big{(}(\tilde{\zeta}^{1}_{\varepsilon})^{2}\big{)}={\mathbb{E}}\big{(}(\tilde{\pi}^{1}_{\varepsilon})^{2}\big{)}.$

To estimate the latter, we have (letting
$p_{t}=\mathrm{Law}(B_{t})=\frac{1}{\sqrt{2\pi t}}e^{-|\cdot|^{2}/(2t)}$)

$\displaystyle{\mathbb{P}}\big{(}\tilde{\pi}^{1}_{\varepsilon}>t\big{)}$
$\displaystyle={\mathbb{P}}\big{(}\tilde{\pi}^{1}_{\varepsilon}>t,\
c_{-}t+B_{t}<1\big{)}+\underbrace{{\mathbb{P}}\big{(}\tilde{\pi}^{1}_{\varepsilon}>t,\
c_{-}t+B_{t}\geqslant 1\big{)}}_{=\;0}$
$\displaystyle\leqslant{\mathbb{P}}\big{(}c_{-}t+B_{t}<1\big{)}={\mathbb{P}}\big{(}B_{t}<1-c_{-}t\big{)}$
$\displaystyle=\int_{-\infty}^{1-c_{-}t}\frac{1}{\sqrt{2\pi
t}}\exp\biggl{(}-\frac{|x|^{2}}{2t}\biggr{)}\,dx$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{(1-c_{-}t)/\sqrt{t}}\exp\biggl{(}-\frac{|y|^{2}}{2}\biggr{)}\,dy.$

It follows that

${\mathbb{E}}((\tilde{\pi}^{1}_{\varepsilon})^{2})=\int_{0}^{\infty}2t{\mathbb{P}}(\tilde{\pi}^{1}_{\varepsilon}>t)\,dt\leqslant\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}2t\int_{-\infty}^{(1-c_{-}t)/\sqrt{t}}\exp\left(-\frac{|y|^{2}}{2}\right)\,dy\,dt<\infty,$

and we are done. ∎

Using the above theorem and standard comparison principles, we extend the
result to drifts satisfying an Osgood-type condition:

######  Lemma 3.3.

Let $a\in L^{\infty}({\mathbb{R}})$ satisfy $a>0$ a.e. in
$(-\delta_{0},\infty)$ for some $\delta_{0}>0$. Assume that for all $R>0$,

$\int_{0}^{R}\frac{1}{a(z)}dz<\infty.$

Then, for any $T>0$, $X_{\varepsilon}$ converges to $\psi_{+}$:

$\big{\|}X_{\varepsilon}-\psi_{+}\big{\|}_{C([0,T])}\overset{P}{\to}0\qquad\text{as
}\varepsilon\to 0\text{ for all }T>0$ (3.12)

(where $\psi_{+}$ is the maximal solution (1.5)).

###### Proof.

As in the proof of Theorem 3.1 we know that
$\\{X_{\varepsilon}\\}_{\varepsilon}$ is weakly relatively compact, so it has
some weakly convergent subsequence $\\{X_{\varepsilon_{k}}\\}_{k}$. Due to
Skorokhod’s representation theorem [3, Theorem 1.6.7] there exists a sequence
of copies $\tilde{X}_{\varepsilon_{k}}$ of $X_{\varepsilon_{k}}$ that satisfy
the corresponding SDEs with Wiener processes $B_{\varepsilon_{k}}$ and such
that $\\{\tilde{X}_{\varepsilon_{k}}\\}_{k}$ converges almost surely to some
continuous non-decreasing process $\tilde{X}$:

${\mathbb{P}}\Bigl{(}\lim_{k\to\infty}\|\tilde{X}_{\varepsilon_{k}}-\tilde{X}\|_{C([0,T])}=0\quad\forall\
T>0\Bigr{)}=1.$ (3.13)

The limit process is non-decreasing, so without loss of generality we may
assume that function $a$ is such that $a(x)=c_{-}$ for all
$x\in(-\infty,-\delta_{0}),$ where $c_{-}>0$ is a constant. Define
$a_{n}\coloneqq a+\nicefrac{{1}}{{n}}$, let $\tilde{X}_{n,\varepsilon}$ be the
corresponding stochastic process and let $X_{n}$ denote the solution of the
corresponding deterministic problem. It holds for all $n\in{\mathbb{N}}$ that
$a_{n}\geqslant\nicefrac{{1}}{{n}}$, thus the result above holds for $a_{n}$.

Let $\pi^{x}$, $\pi^{x}_{\varepsilon_{k}}$, $\pi^{x}_{n,\varepsilon_{k}}$,
$\tau^{x}_{n}$ and $\tau^{x}$ be the hitting times of $\tilde{X}$,
$\tilde{X}_{\varepsilon_{k}}$, $\tilde{X}_{n,\varepsilon_{k}}$, $X_{n}$ and
$\psi_{+}$, respectively. By the comparison principle Theorem 2.2, we know
that

$\tilde{X}_{n,\varepsilon_{k}}\geqslant\tilde{X}_{\varepsilon_{k}},\qquad\text{or
equivalently,}\qquad\pi^{x}_{n,\varepsilon_{k}}\leqslant\pi^{x}_{\varepsilon_{k}}\;\forall\
x$ (3.14)

(cf. Lemma 2.3). It follows from Theorem 3.1 that
$\tilde{X}_{n,\varepsilon_{k}}\to X_{n}$ a.s. as $k\to\infty$, which together
with (3.13) and (3.14) implies

$X_{n}\geqslant\tilde{X},\qquad\text{or
equivalently,}\qquad\tau^{x}_{n}\leqslant\pi^{x}\;\forall\ x.$ (3.15)

The lower semi-continuity of a hitting time with respect to its process also
implies that $\pi^{x}\leqslant\liminf_{k\to\infty}\pi^{x}_{\varepsilon_{k}}$
a.s. for any $x\geqslant 0$. Hence, for any $x\geqslant 0$,

$\displaystyle A(x)$
$\displaystyle=\lim_{n\to\infty}A_{n}(x)=\lim_{n\to\infty}\tau_{n}^{x}\leqslant{\mathbb{E}}(\pi^{x})$
$\displaystyle\leqslant{\mathbb{E}}\Bigl{(}\liminf_{k\to\infty}\pi_{\varepsilon_{k}}^{x}\Bigr{)}\leqslant\liminf_{k\to\infty}{\mathbb{E}}\bigl{(}\pi_{\varepsilon_{k}}^{x}\bigr{)}=A(x),$

the last equality following from (3.7b) in Lemma 3.2. Hence,
${\mathbb{E}}(\pi^{x})=A(x)$ for all $x\geqslant 0$, and since
$\pi^{x}\geqslant\tau_{n}^{x}\to A(x)$ as $n\to\infty$, we conclude that
$\pi^{x}=A(x)$ almost surely for every $x\geqslant 0$, so Corollary 2.4
implies that $\tilde{X}=A^{-1}=\psi_{+}$ almost surely. Since $\psi_{+}$ is
non-random, we have the uniform convergence in probability

${\mathbb{P}}\biggl{(}\lim_{k\to\infty}\|X_{\varepsilon_{k}}-\psi_{+}\|_{C([0,T])}=0\quad\forall\
T>0\biggr{)}=1.$

And finally, since the limit $\psi_{+}$ is unique, we can conclude that the
entire sequence $\\{X_{\varepsilon}\\}_{\varepsilon}$ converges. ∎

We are now ready to prove Theorem 1.1 under the additional condition that
$a>0$ a.e. in $(-\delta_{0},0)$:

###### Proof of Theorem 1.1 for positive $a$.

The case when $\int_{0}^{R}\frac{dx}{a(x)\vee 0}<\infty$ for any $R>0$ (and
hence, in particular, $a>0$ a.e. in $(-\delta_{0},\infty)$) has been
considered in Lemma 3.3. Thus, we can assume that there is some $R>0$ such
that $a>0$ a.e. on $(-\delta_{0},R)$, and for any (small) $\delta>0$,

$\int_{0}^{R-\delta}\frac{dx}{a(x)}<\infty\quad\text{but}\quad\int_{0}^{R+\delta}\frac{dx}{a(x)\vee
0}=\infty.$ (3.16)

Recall that

$\psi_{+}(x)=\begin{cases}A^{-1}(x),&x\in[0,A(R)),\\\ R,&x\geqslant
A(R).\end{cases}$

(Note that $A(R)$ may be equal to $\infty.$) The proof of the theorem consists
of the following steps:

  1. 1.

Prove the theorem for the stopped process
$X_{\varepsilon}(\cdot\wedge\tau^{R}_{\varepsilon})$

  2. 2.

Prove the theorem for nonnegative drifts

  3. 3.

Extend to possibly negative drifts.

Step 1. Set $\widehat{a}_{m}(x)\coloneqq a(x)\mathbbm{1}_{x\leqslant
R-\nicefrac{{1}}{{m}}}+\mathbbm{1}_{x>R-\nicefrac{{1}}{{m}}}$ for
$m\in{\mathbb{N}}$, and note that $\widehat{a}_{m}$ satisfies the conditions
of Lemma 3.3. Let $\widehat{X}_{m,\varepsilon}$ denote the solution to the
corresponding SDE, $\widehat{X}_{m}$ its limit, and
$\widehat{\tau}_{m,\varepsilon}^{x},\ \widehat{\tau}_{m}^{x}$ the
corresponding hitting times. It follows from the uniqueness of a solution that

${\mathbb{P}}\Bigl{(}\widehat{\tau}_{m,\varepsilon}^{R-\nicefrac{{1}}{{m}}}=\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\Bigr{)}=1\quad\text{and}\quad{\mathbb{P}}\Bigl{(}\widehat{X}_{m,\varepsilon}(t)=X_{\varepsilon}(t)\quad\forall\
t\leqslant\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\Bigr{)}=1.$

Thus, by Lemma 3.3,

$\begin{split}\sup_{t\in[0,T]}\big{|}X_{\varepsilon}\bigl{(}t\wedge\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\bigr{)}-A^{-1}\big{(}t\wedge\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\big{)}\big{|}&\overset{P}{\to}0\qquad\text{as
}\varepsilon\to 0\text{ for all }T>0,\\\
\sup_{t\in[0,T]}\big{|}\widehat{X}_{m,\varepsilon}\bigl{(}t\wedge\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\bigr{)}-A^{-1}\bigl{(}t\wedge\widehat{\tau}_{\varepsilon}^{R-\nicefrac{{1}}{{m}}}\bigr{)}\big{|}&\overset{P}{\to}0\qquad\text{as
}\varepsilon\to 0\text{ for all }T>0,\end{split}$ (3.17)

for every $m\in{\mathbb{N}}$.

Let $\overline{X}_{0}$ be a limit point of
$\\{X_{\varepsilon}\\}_{\varepsilon}$ and
$X_{\varepsilon_{k}}\Rightarrow\overline{X}_{0}$ as $k\to\infty.$ It follows
from (3.17) that
$\overline{X}_{0}(\cdot\wedge\tau^{R-\nicefrac{{1}}{{m}}}_{m})=A^{-1}(\cdot\wedge\tau^{R-\nicefrac{{1}}{{m}}}_{m})$,
and since $m$ is arbitrary, we have
$\overline{X}_{0}(\cdot\wedge\tau^{R})=A^{-1}(\cdot\wedge\tau^{R})$, that is,
$\overline{X}_{0}(\cdot\wedge\tau^{R})=\psi_{+}(\cdot\wedge\tau^{R})$. In
particular, the entire sequence of stopped processes converges, by uniqueness
of the limit.

Step 2. Assume next, in addition to (3.16), that $a\geqslant 0$ a.e. in
${\mathbb{R}}$. Any limit point of $\\{X_{\varepsilon}\\}_{\varepsilon}$ is a
non-decreasing process, so to prove the theorem it suffices to verify that for
any $\delta>0$ and $M>0$

$\limsup_{k\to\infty}{\mathbb{P}}\bigl{(}\tau^{R+\delta}_{\varepsilon_{k}}<M\bigr{)}=0$

Set $a_{n}\coloneqq a+\nicefrac{{1}}{{n}}$ and let $X_{n,\varepsilon}$ denote
the solution to the corresponding SDE. It follows from comparison Theorem 2.2
that for any $M>0$

$\limsup_{k\to\infty}{\mathbb{P}}\bigl{(}\tau^{R+\delta}_{\varepsilon_{k}}<M\bigr{)}\leqslant\liminf_{n\to\infty}\limsup_{k\to\infty}{\mathbb{P}}\bigl{(}\tau^{R+\delta}_{n,\varepsilon_{k}}<M\bigr{)}.$

Theorem 3.1 implies that $\lim_{\varepsilon\to
0}X_{n,\varepsilon}=X_{n}=A^{-1}_{n},$ so the right hand side of the above
inequality equals zero for any $M$. This concludes the proof if $a$ is non-
negative everywhere.

Step 3. In the case that $a$ takes negative values, we consider the processes
$X_{\varepsilon}^{+}$ satisfying the corresponding SDEs with drift
$a^{+}(x)\coloneqq a(x)\vee 0$. We have already proved in Step 2 that

$\displaystyle\bigl{\|}X_{\varepsilon}^{+}-\psi_{+}\bigr{\|}_{C([0,T])}\overset{P}{\to}0$
$\displaystyle\text{as }\varepsilon\to 0\;\forall\ T>0$ (since $a^{+}$ has the
same deterministic solution $\psi_{+}$ as $a$ does), and in Step 1 that
$\displaystyle\bigl{\|}X_{\varepsilon}\big{(}\cdot\wedge\tau^{R}_{0}\big{)}-\psi_{+}\bigr{\|}_{C([0,T])}\overset{P}{\to}0$
$\displaystyle\text{as }\varepsilon\to 0\;\forall\ T>0.$

Theorem 2.2 yields $X_{\varepsilon}^{+}(t)\geqslant X_{\varepsilon}(t)$.
Therefore, any (subsequential) limit of
$\\{X_{\varepsilon}^{+}\\}_{\varepsilon}$ is greater than or equal to a limit
of $\\{X_{\varepsilon}\\}_{\varepsilon}$, and if $\bar{X}_{0}$ is a limit
point of $\\{X_{\varepsilon}\\}_{\varepsilon}$ then

${\mathbb{P}}\Bigl{(}\bar{X}_{0}(t)=\psi_{+}(t)\ \forall\
t\leqslant\tau^{R}_{0}\text{ and }\bar{X}_{0}(t)\leqslant R\ \forall\
t>\tau^{R}_{0}\Bigr{)}=1.$

On the other hand, it can be seen that any limit point $\bar{X}_{0}$ of
$\\{X_{\varepsilon}\\}_{\varepsilon}$ satisfies

${\mathbb{P}}\Bigl{(}\exists\
t\geqslant\tau^{0}_{R}:\bar{X}_{0}(t)<R\Bigr{)}=0.$

Thus we have equality, $\bar{X}_{0}(t)=\psi_{+}(t)$ for all $t\geqslant 0$
almost surely. This concludes the proof for the case $a(x)>0$ for
$x\in(-\delta_{0},0)$. The case $a(x)\geqslant 0$ for $x\in(-\delta_{0},0)$
will be considered in §5. ∎

##  4\. Velocity with a change in sign

In this section we consider the repulsive case and prove Theorem 1.2. We also
provide several tools for computing the zero noise probability distribution.

###  4.1. Convergence in the repulsive case

######  Lemma 4.1.

Let $\alpha<0<\beta$, assume that $a\in L^{\infty}({\mathbb{R}})$ satisfies
the “repulsive Osgood condition” (1.8), and define $p_{\varepsilon}$ by

$p_{\varepsilon}\coloneqq\frac{-s_{\varepsilon}(\alpha)}{s_{\varepsilon}(\beta)-s_{\varepsilon}(\alpha)},\qquad
s_{\varepsilon}(r)\coloneqq\int_{0}^{r}e^{-B(z)/\varepsilon^{2}}\,dz,\qquad
B(z)\coloneqq 2\int_{0}^{z}a(u)\,du.$ (4.1)

Then

$\limsup_{\varepsilon\to
0}{\mathbb{E}}^{0}\big{(}\tau_{\varepsilon}^{\alpha}\wedge\tau_{\varepsilon}^{\beta}\big{)}\leqslant\int_{\alpha}^{\beta}\frac{1}{|a(x)|}\,dx<\infty.$

If $p_{\varepsilon_{k}}\to p$ as $k\to\infty$, then

${\mathbb{E}}^{0}\big{(}\tau_{\varepsilon_{k}}^{\alpha}\wedge\tau_{\varepsilon_{k}}^{\beta}\big{)}\to{(1-p)}\int_{\alpha}^{0}\frac{-1}{a(z)}\,dz+{p}\int_{0}^{\beta}\frac{1}{a(z)}\,dz\qquad\text{as
}k\to\infty.$

###### Proof.

By (2.2), (2.4), and (2.5) with $f=1$ we can write

$\displaystyle{\mathbb{E}}^{0}\big{(}\tau_{\varepsilon}^{\alpha}\wedge\tau_{\varepsilon}^{\beta}\big{)}=\int_{\alpha}^{0}\frac{(s_{\varepsilon}(y)-s_{\varepsilon}(\alpha))(s_{\varepsilon}(\beta)-s_{\varepsilon}(0))}{s_{\varepsilon}(\beta)-s_{\varepsilon}(\alpha)}\frac{2e^{B(y)/\varepsilon^{2}}}{\varepsilon^{2}}\,dy$
$\displaystyle\qquad+\int_{0}^{\beta}\frac{(s_{\varepsilon}(0)-s_{\varepsilon}(\alpha))(s_{\varepsilon}(\beta)-s_{\varepsilon}(y))}{s_{\varepsilon}(\beta)-s_{\varepsilon}(\alpha)}\frac{2e^{B(y)/\varepsilon^{2}}}{\varepsilon^{2}}\,dy$
$\displaystyle\quad={(1-p_{\varepsilon})}\int_{\alpha}^{0}(s_{\varepsilon}(y)-s_{\varepsilon}(\alpha))\frac{2e^{B(y)/\varepsilon^{2}}}{\varepsilon^{2}}\,dy+{p_{\varepsilon}}\int_{0}^{\beta}(s_{\varepsilon}(\beta)-s_{\varepsilon}(y))\frac{2e^{B(y)/\varepsilon^{2}}}{\varepsilon^{2}}\,dy$
$\displaystyle\quad={(1-p_{\varepsilon})\int_{\alpha}^{0}\int_{\alpha}^{y}\frac{2e^{(B(y)-B(z))/\varepsilon^{2}}}{\varepsilon^{2}}\,dz\,dy+p_{\varepsilon}\int_{0}^{\beta}\int_{y}^{\beta}\frac{2e^{(B(y)-B(z))/\varepsilon^{2}}}{\varepsilon^{2}}\,dz\,dy}$
$\displaystyle\quad=(1-p_{\varepsilon})\int_{\alpha}^{0}\int_{\alpha}^{y}\frac{2\exp\Bigl{(}{\textstyle-\int_{z}^{y}\frac{2a(u)}{\varepsilon^{2}}du}\Bigr{)}}{\varepsilon^{2}}\,dz\,dy$
$\displaystyle\qquad+p_{\varepsilon}\int_{0}^{\beta}\int_{y}^{\beta}\frac{2\exp\Bigl{(}{\textstyle-\int_{z}^{y}\frac{2a(u)}{\varepsilon^{2}}du}\Bigr{)}}{\varepsilon^{2}}\,dz\,dy$
$\displaystyle\quad=(1-p_{\varepsilon})\int_{\alpha}^{0}\int_{\alpha}^{y}\exp\Bigl{(}{\textstyle-\int_{z}^{y}\frac{2a(u)}{\varepsilon^{2}}du}\Bigr{)}\frac{2a(z)}{\varepsilon^{2}}\frac{1}{a(z)}\,dz\,dy$
$\displaystyle\qquad+p_{\varepsilon}\int_{0}^{\beta}\int_{y}^{\beta}\exp\Bigl{(}{\textstyle-\int_{z}^{y}\frac{2a(u)}{\varepsilon^{2}}du}\Bigr{)}\frac{2a(z)}{\varepsilon^{2}}\frac{1}{a(z)}\,dz\,dy.$

Setting $f(z)=2\mathop{\rm sign}(z)a(z)$ and $g(z)=\frac{1}{a(z)}$ in Lemma
2.7, we find that the above two integrals with $\varepsilon=\varepsilon_{k}$
converge to

$\int_{\alpha}^{0}\frac{-1}{a(z)}\,dz\qquad\text{and}\qquad\int_{0}^{\beta}\frac{1}{a(z)}\,dz$

respectively, as $k\to\infty$. This concludes the proof. ∎

We can now prove the main theorem in the repulsive case.

###### Proof of Theorem 1.2.

Let $X_{\varepsilon_{k}^{\prime}}$ be any weakly convergent subsequence of
$\\{X_{\varepsilon_{k}}\\}_{k}$, and let $\tau_{\varepsilon_{k}^{\prime}}$ and
$\tau$ be the hitting times of $X_{\varepsilon_{k}^{\prime}}$ and its limit,
respectively. By Lemma 4.1 we have for any $\alpha<0<\beta$

${\mathbb{E}}^{0}(\tau^{\alpha}\wedge\tau^{\beta})\leqslant\liminf_{k\to\infty}{\mathbb{E}}^{0}\bigl{(}\tau^{\alpha}_{\varepsilon_{k}}\wedge\tau^{\beta}_{\varepsilon_{k}}\bigr{)}={(1-p)A(\alpha)+pA(\beta)}.$

Consequently,
${\mathbb{P}}^{0}\bigl{(}\tau^{\alpha}\wedge\tau^{\beta}=\infty\bigr{)}=0$, so
${\mathbb{P}}^{0}(\tau^{\alpha}<\tau^{\beta})=\lim_{k\to\infty}{\mathbb{P}}^{0}(\tau^{\alpha}_{\varepsilon_{k}^{\prime}}<\tau^{\beta}_{\varepsilon_{k}^{\prime}})={1-p}$
and ${\mathbb{P}}^{0}(\tau^{\alpha}>\tau^{\beta})={p}$. Using Theorem 1.1 and
the strong Markov property, the probability of convergence once the process
escapes $(\alpha,\beta)$ at $x=\beta$ is one:

$\lim_{k\to\infty}{\mathbb{P}}^{0}\Bigl{(}\bigl{\|}X_{\varepsilon_{k}^{\prime}}(\cdot-\tau_{\beta})-\psi_{+}(\cdot-A(\beta))\bigr{\|}_{C([0,T])}>\delta\bigm{|}\tau^{\alpha}>\tau^{\beta}\Bigr{)}=1,$

for any sufficiently small $\delta>0$, and likewise for those paths escaping
at $x=\alpha$. Passing $\alpha,\beta\to 0$ yields

$\displaystyle\lim_{\delta\to
0}\lim_{k\to\infty}{\mathbb{P}}^{0}\Bigl{(}\|X_{\varepsilon_{k}^{\prime}}-\psi_{-}\|_{C([0,T])}>\delta\Bigr{)}={1-p},$
$\displaystyle\lim_{\delta\to
0}\lim_{k\to\infty}{\mathbb{P}}^{0}\Bigl{(}\|X_{\varepsilon_{k}^{\prime}}-\psi_{+}\|_{C([0,T])}>\delta\Bigr{)}={p}.$

Since this is true for any weakly convergent subsequence
$\varepsilon_{k}^{\prime}$, and the limit is unique, the entire sequence
$\varepsilon_{k}$ must converge. ∎

###  4.2. Probabilities in the repulsive case

Theorem 1.2 gives a concrete condition for convergence of the sequence of
perturbed solutions, as well as a characterization of the limit distribution.
In this section we give an explicit expression for the probabilities in the
limit distribution, and an equivalent condition for convergence.

Consider the integral

$B(x)\coloneqq\int_{0}^{x}a(y)\,dy$

and denote $B_{\pm}=B\bigr{|}_{{\mathbb{R}}_{\pm}}$. Select any
$\alpha>0,\beta>0$ such that the function $\mu\colon[0,\beta)\to(\alpha,0]$
defined by $\mu=B_{-}^{-1}\circ B_{+}$ is well-defined — that is,

$B_{+}(x)=B_{-}(\mu(x)),\quad\forall\ x\in[0,\beta).$

Clearly, $B_{\pm}$ are Lipschitz continuous. Since $a$ is strictly positive
(negative) for $x>0$ ($x<0$), the inverses of $B_{\pm}$ are absolutely
continuous (see e.g. [6, Exercise 5.8.52]), so $\mu$ is also absolutely
continuous. We now rewrite the probability of choosing the left/right extremal
solutions $X^{\pm}$ in terms of $\mu$.

######  Theorem 4.2.

Let $a\in L^{\infty}({\mathbb{R}})$ satisfy (1.8) and let
$\mu\colon[0,\beta)\to(\alpha,0]$ be as above. Then
$\\{p_{\varepsilon}\\}_{\varepsilon}$ converges if either the derivative
$\mu^{\prime}(0)$ exists, or if $\mu^{\prime}(0)=-\infty$. In either case, we
have

$\lim_{\varepsilon\to
0}p_{\varepsilon}={\frac{-\mu^{\prime}(0)}{1-\mu^{\prime}(0)}}.$ (4.2a)
Moreover, the derivative $\mu^{\prime}(0)$ exists if and only if the limit
$\lim_{u\downarrow 0}\frac{B_{-}^{-1}(u)}{B_{+}^{-1}(u)}$ exists, and we have
the equality: $\mu^{\prime}(0)=\lim_{u\downarrow
0}\frac{B_{-}^{-1}(u)}{B_{+}^{-1}(u)}.$ (4.2b)

To prove the theorem we will need the following lemmas:

######  Lemma 4.3.

Let $\alpha<0<\beta$. Define $p_{\varepsilon}$ as in (4.1) and
$p_{\varepsilon}^{\prime}$ similarly, where $\alpha,\beta$ are exchanged with
any $\alpha^{\prime}<0<\beta^{\prime}.$ Then $\lim_{\varepsilon\to
0}p_{\varepsilon}^{\prime}/p_{\varepsilon}=1$. In particular,
$p_{\varepsilon_{k}}$ converges to some $p$ as $k\to\infty$ if and only if
$p_{\varepsilon_{k}}^{\prime}$ converges to $p$.

The proof follows from the following observation: Since $B$ is strictly
increasing, then for any positive $r_{1}<r_{2}$ or negative $r_{1}>r_{2}$,

$\lim_{\varepsilon\to
0}\frac{\int_{r_{1}}^{r_{2}}e^{-B(z)/\varepsilon^{2}}\,dz}{\int_{0}^{r_{1}}e^{-B(z)/\varepsilon^{2}}\,dz}=0.$

Next, we prove a technical lemma:

######  Lemma 4.4.

Let $0<a\in L^{\infty}([0,\beta])$ and $f\in L^{1}({\mathbb{R}})$, and for
$\varepsilon>0$ and $x\in[0,\beta)$ define

$\displaystyle
B(x)=2\int_{0}^{x}a(y)\,dy,\qquad\nu_{\varepsilon}(x)=e^{-B(x)/\varepsilon^{2}}\mathbbm{1}_{[0,\beta]}(x),$
$\displaystyle\bar{\nu}_{\varepsilon}=\int_{0}^{\beta}\nu_{\varepsilon}(y)\,dy,\qquad
f_{\varepsilon}(x)=\frac{1}{\bar{\nu}_{\varepsilon}}\int_{0}^{\beta}f(x+y)\nu_{\varepsilon}(y)\,dy.$

Then $f_{\varepsilon}(x)\to f(x)$ as $\varepsilon\to 0$ if and only if $x$ is
a Lebesgue point of $f$.

###### Proof.

Let $x\in[0,\beta)$. For $s\in(0,\beta-x)$, let

$F(s)=\int_{0}^{s}|f(x+y)-f(x)|\,dy,\qquad
C_{s}=\sup_{y\in(0,s)}\tfrac{F(y)}{y}.$

Then $C_{s}\to 0$ as $s\to 0$ if and only if $x$ is a Lebesgue point. We
estimate

$\displaystyle|f_{\varepsilon}(x)-f(x)|$
$\displaystyle=\frac{1}{\bar{\nu}_{\varepsilon}}\biggl{|}\int_{0}^{\beta}(f(x+y)-f(x))\nu_{\varepsilon}(y)\,dy\biggr{|}$
$\displaystyle\leqslant\underbrace{\frac{1}{\bar{\nu}_{\varepsilon}}\int_{0}^{s}|f(x+y)-f(x)|\nu_{\varepsilon}(y)\,dy}_{=\,I_{1}}+\underbrace{\frac{1}{\bar{\nu}_{\varepsilon}}\int_{s}^{\beta}|f(x+y)-f(x)|\nu_{\varepsilon}(y)\,dy}_{=\,I_{2}}.$

For the first term we integrate by parts several times to get

$\displaystyle I_{1}$
$\displaystyle=F(s)\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}-\frac{1}{\bar{\nu}_{\varepsilon}}\int_{0}^{s}F(y)\nu_{\varepsilon}^{\prime}(y)\,dy\leqslant
F(s)\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}-\frac{C_{s}}{\bar{\nu}_{\varepsilon}}\int_{0}^{s}y\nu_{\varepsilon}^{\prime}(y)\,dy$
$\displaystyle=F(s)\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}-\frac{C_{s}}{\bar{\nu}_{\varepsilon}}s\nu_{\varepsilon}(s)+\frac{C_{s}}{\bar{\nu}_{\varepsilon}}\int_{0}^{s}\nu_{\varepsilon}(y)\,dy$
$\displaystyle\leqslant
F(s)\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}+\frac{C_{s}}{\bar{\nu}_{\varepsilon}}\int_{0}^{\beta}\nu_{\varepsilon}(y)\,dy$
$\displaystyle=F(s)\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}+C_{s}.$

For the second term we estimate

$\displaystyle I_{2}$ $\displaystyle\leqslant
2\|f\|_{L^{1}}\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}.$

If we can find $s=s_{\varepsilon}$ such that both $s_{\varepsilon}\to 0$ and
$\frac{\nu_{\varepsilon}(s_{\varepsilon})}{\bar{\nu}_{\varepsilon}}\to 0$ as
$\varepsilon\to 0$, then both $I_{1}$ and $I_{2}$ vanish in the
$\varepsilon\to 0$ limit, and we can conclude the result. Below we explain the
existence of such a choice.

Since $B$ is increasing and Lipschitz continuous, with $B(0)=0$ and
$\|B\|_{\mathrm{Lip}}\leqslant 2\|a\|_{L^{\infty}}<\infty$, there is some
$\kappa<s$ satisfying $B(\kappa)=\tfrac{1}{2}B(s)$, and
$\kappa\geqslant\frac{1}{2\|B\|_{\mathrm{Lip}}}B(s)$. Moreover, since
$\nu_{\varepsilon}$ is decreasing we have

$\bar{\nu}_{\varepsilon}=\int_{0}^{\beta}\nu_{\varepsilon}(y)\,dy\geqslant\kappa\nu_{\varepsilon}(\kappa)=\kappa
e^{-B(\kappa)/\varepsilon^{2}}=\kappa e^{-B(s)/(2\varepsilon^{2})},$

so

$\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}\leqslant\frac{1}{\kappa}e^{-B(s)/(2\varepsilon^{2})}\leqslant
2\|B\|_{\mathrm{Lip}}\frac{e^{-B(s)/(2\varepsilon^{2})}}{B(s)}.$

Now choose $s=s_{\varepsilon}$ such that $B(s_{\varepsilon})=\varepsilon$.
(Such a number exists for sufficiently small $\varepsilon>0$.) Then
$s_{\varepsilon}\to 0$ as $\varepsilon\to 0$, and

$\frac{\nu_{\varepsilon}(s)}{\bar{\nu}_{\varepsilon}}\leqslant
2\|B\|_{\mathrm{Lip}}\frac{e^{-1/(2\varepsilon)}}{\varepsilon}\to 0$

as $\varepsilon\to 0$. This finishes the proof. ∎

###### Proof of Theorem 4.2.

We have

$p_{\varepsilon}=\frac{{-s_{\varepsilon}(\alpha)}}{s_{\varepsilon}(\beta)-s_{\varepsilon}(\alpha)}={\frac{-\frac{s_{\varepsilon}(\alpha)}{s_{\varepsilon}(\beta)}}{1-\frac{s_{\varepsilon}(\alpha)}{s_{\varepsilon}(\beta)}}}.$

By Lemma 4.3 we may assume $\mu(\beta)=\alpha$, so

$\displaystyle s_{\varepsilon}(\alpha)$
$\displaystyle=\int_{0}^{\alpha}e^{-B(\mu^{-1}(x))/\varepsilon^{2}}\,dx=\int_{0}^{\beta}e^{-B(x)/\varepsilon^{2}}\mu^{\prime}(x)\,dx.$

Thus,

$\frac{s_{\varepsilon}(\alpha)}{s_{\varepsilon}(\beta)}=\frac{1}{\bar{\nu}_{\varepsilon}}\int_{0}^{\beta}\nu_{\varepsilon}(x)\mu^{\prime}(x)\,dx$

where

$\nu_{\varepsilon}(x)=e^{-B(x)/\varepsilon^{2}},\qquad\bar{\nu}_{\varepsilon}=\int_{0}^{\beta}e^{-B(z)/\varepsilon^{2}}\,dz.$

From Lemma 4.4 with $f(x)\coloneqq\mu^{\prime}(x)$ it now follows that
$p_{\varepsilon}$ converges if either $0$ is a Lebesgue point for
$\mu^{\prime}$, or $\lim_{x\to 0}\mu^{\prime}(x)={-\infty}$. In the former
case, we notice that $0$ is a Lebesgue point for $\mu^{\prime}$ if the
following limit exists:

${\lim_{h\downarrow
0}}\frac{\int_{0}^{h}\mu^{\prime}(z)\,dz}{h}={\lim_{h\downarrow
0}}\frac{\mu(h)-\mu(0)}{h}.$

The right hand side of the last equation is the usual definition of the
derivative.

To prove (4.2b) notice that

$\lim_{h\downarrow 0}\frac{\mu(h)-\mu(0)}{h}=\lim_{h\downarrow
0}\frac{\mu(h)}{h}=\lim_{h\downarrow 0}\frac{B_{-}^{-1}\circ
B_{+}(h)}{h}=\lim_{u\downarrow 0}\frac{B_{-}^{-1}(u)}{B_{+}^{-1}(u)}.$

∎

###  4.3. Repulsive, regularly varying drifts

Although Theorem 1.2 provides an explicit expression (4.2) of the limit
probabilities, the limit (4.2b) might be difficult to evaluate in practice. It
is clearly easier to study existence of the limits

$\lim_{x\downarrow 0}\frac{a(-x)}{a(x)}$ (4.3)

or

$\lim_{x\downarrow 0}\frac{B(-x)}{B(x)}$ (4.4)

than that for the inverse functions in (4.2). We will show that the limit in
(4.2) can easily be calculated using (4.3) or (4.4) if $a$ or $B$ are
regularly varying at $0$.

Recall that a positive, measurable function $f\colon[0,\infty)\to(0,\infty)$
is _regularly varying_ of index $\gamma$ at $+\infty$ if
$\lim_{x\to\infty}\frac{f(\lambda x)}{f(x)}=\lambda^{\rho}$ for all
$\lambda>0$. It is regularly varying of index $\rho$ at $0$ if the function
$x\mapsto f(1/x)$ is a regularly varying function of index $-\rho$ at
$+\infty$. The set of regularly varying functions of index $\rho$ (at
$+\infty$) is denoted by $R_{\rho}.$ It is well known that if $f\in R_{\rho}$,
then $f(x)=x^{\rho}\ell(x)$ for some slowly varying function $\ell$, i.e. some
$\ell\colon[0,\infty)\to(0,\infty)$ for which
$\lim_{x\to\infty}\frac{\ell(\lambda x)}{\ell(x)}=1$ for all $\lambda>0$.

We first consider the case when $B$ is regularly varying, and then the case
when $a$ is. Note that the latter implies the former, but not vice versa.

######  Proposition 4.5.

Assume that the functions $x\mapsto B_{\pm}(\pm x)$ are regularly varying of
index $\rho>0$ at 0, and that the limit $c\coloneqq\lim_{x\downarrow
0}B_{-}(-x)/B_{+}(x)$ exists (or equals $\infty$). Then
$\\{p_{\varepsilon}\\}_{\varepsilon>0}$ converges, and

$p\coloneqq\lim_{\varepsilon\to
0}p_{\varepsilon}=\frac{{c^{-1/\rho}}}{1+c^{-1/\rho}}.$ (4.5)

If the functions $x\mapsto B_{\pm}(\pm x)$ are regularly varying of different
indices $\rho_{\pm},$ then

$p\coloneqq\lim_{\varepsilon\to
0}p_{\varepsilon}=\begin{cases}1&\rho_{+}<\rho_{-}\\\
0&\rho_{+}>\rho_{-}.\end{cases}$

###### Proof.

It follows from [5, Exercise 14, p. 190] that if
$f_{1},f_{2}\colon(0,\infty)\to(0,\infty)$ are non-decreasing, regularly
varying functions at $0$ of index $\rho>0$, then

$\lim_{x\to 0}\frac{f_{1}(x)}{f_{2}(x)}=1\qquad\text{if and only
if}\qquad\lim_{x\to 0}\frac{f_{1}^{-1}(x)}{f_{2}^{-1}(x)}=1,$

where $f_{1}^{-1},f_{2}^{-1}$ are inverse functions. Write now
$f_{1}(x)=B_{-}(-x)$, $f_{2}(x)=cB_{+}(x)$. Then

$\lim_{x\to 0}\frac{f_{1}(x)}{f_{2}(x)}=1.$ (4.6)

The inverse function for $x\mapsto cB_{+}(x)$ is $x\mapsto B_{+}^{-1}(x/c),$
and $B_{+}^{-1}$ is regularly varying of index $1/\rho$ (see [5, Theorem
1.5.12]), so

$B_{+}^{-1}(x/c)=(x/c)^{1/\rho}\ell_{1}(x/c)\sim(x/c)^{1/\rho}\ell_{1}(x)=c^{-1/\rho}B_{+}^{-1}(x)\qquad\text{as
}x\to 0$

(where equivalence is meant in the sense of slowly varying functions). Hence,
(4.6) yields

$\lim_{x\to 0}\frac{B_{-}^{-1}(x)}{B_{+}^{-1}(x)}=c^{-1/\rho}.$ (4.7)

The same computation can be easily performed in reverse, so (4.6) and (4.7)
are equivalent, and the result now follows from Theorem 4.2 if $B_{\pm}$ are
of the same index.

If $x\mapsto B_{\pm}(\pm x)$ are regularly varying of different indices
${\rho_{\pm}},$ then the inverse functions are regularly varying functions of
indices $\frac{1}{\rho_{\pm}},$ and the result is obvious. ∎

######  Proposition 4.6.

Assume that both $x\mapsto a(\pm x)$ (for $x\geqslant 0$) are regularly
varying at $0$ with index $\rho>0$, and that the limit
$c\coloneqq\lim_{x\downarrow 0}\frac{-a(-x)}{a(x)}$ exists. Then
$\\{p_{\varepsilon}\\}_{\varepsilon>0}$ converges, and

$p\coloneqq\lim_{\varepsilon\to
0}p_{\varepsilon}=\frac{{c^{-1/(1+\rho)}}}{1+c^{-1/(1+\rho)}}.$

If the functions $x\mapsto a(\pm x)$ are regularly varying of different
indices $\rho_{\pm},$ then

$p\coloneqq\lim_{\varepsilon\to 0}p_{\varepsilon}=\begin{cases}1&\text{if
}\rho_{+}<\rho_{-}\\\ 0&\text{if }\rho_{+}>\rho_{-}.\end{cases}$

###### Proof.

It follows from the Karamata theorem, see [5, Theorem 1.6.1], that for $x>0$,

$\displaystyle B(x)$
$\displaystyle=2\int_{0}^{x}a(y)\,dy=2\int_{1/x}^{\infty}a(1/z)z^{-2}\,dz=2\int_{1/x}^{\infty}\ell(z)z^{-2-\rho}\,dz$
$\displaystyle\sim\frac{2\ell(1/x)x^{1+\rho}}{1+\rho}\sim\frac{xa(x)}{1+\rho}$

as $x\to 0$, and likewise for $x<0$. Thus, $x\mapsto B(\pm x)$ are regularly
varying of index $1+\rho$. Letting

$c\coloneqq\lim_{x\downarrow 0}\frac{-a(-x)}{a(x)},$

we can now apply Proposition 4.5 with $1+\rho$ in place of $\rho$ and get the
desired result.

The case when $a(\pm x)$ are regularly varying with different indices can be
considered similarly, cf. Proposition 4.5. ∎

Finally, we provide a result which simplifies the computation of the limit
distribution for severely oscillating drifts.

######  Proposition 4.7.

Let $a\colon{\mathbb{R}}\to{\mathbb{R}}$ satisfy $xa(x)\geqslant 0$ for all
$x\in{\mathbb{R}}$, and assume that it is of the form

$a(x)=b(x)+|x|^{\gamma}g(\tfrac{1}{x}),$

where $\gamma>0$, $b$ is regularly varying at $0$ of order $\rho<\gamma+1$,
and $g\in L^{\infty}({\mathbb{R}})$ is such that its antiderivative
$G(x)=\int_{0}^{x}g(y)\,dy$ also lies in $L^{\infty}({\mathbb{R}})$. Assume
also that the limit $c\coloneqq\lim_{x\downarrow 0}\frac{-b(-x)}{b(x)}$
exists. Then $\\{p_{\varepsilon}\\}_{\varepsilon>0}$ converges, and

$p\coloneqq\lim_{\varepsilon\to
0}p_{\varepsilon}=\frac{{c^{1/(1+\rho)}}}{1+c^{1/(1+\rho)}}.$

###### Proof.

We claim first that

$\int_{0}^{x}y^{\gamma}g(\tfrac{1}{y})\,dy=O(x^{1+\gamma})=o(x^{\rho})\qquad\text{as
}x\downarrow 0.$

Indeed,

$\displaystyle\biggl{|}\int_{0}^{x}y^{\gamma}g(\tfrac{1}{y})\,dy\biggr{|}$
$\displaystyle=\biggl{|}\int_{1/x}^{\infty}z^{-2-\gamma}g(z)\,dz\biggr{|}$
$\displaystyle=\biggl{|}-x^{2+\gamma}G(\tfrac{1}{x})-(\rho+2)\int_{1/x}^{\infty}z^{-3-\gamma}G(z)\,dz\biggr{|}$
$\displaystyle\leqslant
x^{2+\gamma}\|G\|_{L^{\infty}}+(\rho+2)\|G\|_{L^{\infty}}\int_{1/x}^{\infty}z^{-3-\gamma}\,dz$
$\displaystyle=\Bigl{(}1+\frac{\rho+2}{\gamma+2}\Bigr{)}x^{2+\gamma}\|G\|_{L^{\infty}}.$

It follows that the antiderivative $B(x)=\int_{0}^{x}a(y)\,dy$ equals a
regularly varying function of order $1+\rho$, plus a term of order
$o(x^{1+\rho})$. Following the same procedure as in the proof of Proposition
4.6 yields the desired result. ∎

##  5\. Proof of Theorem 1.1

We have already proven the Theorem in Section 3 if $a>0$ a.e. in a small
neighborhood of 0. We will only prove the result for $a$ such that $a\geqslant
0$ for negative $x$ and $\int_{0}^{R}\frac{dy}{a(y)\vee 0}<\infty$ for all
$R>0$. The general case, i.e., $a(x)\geqslant 0$ for a.e.
$x\in(-\delta_{0},0)$ and $\int_{0}^{\delta_{0}}\frac{dy}{a(y)}<0$, is
considered similarly to the reasoning in Section 3.

It follows from the comparison theorem that for any $x>0$ we have the
inequality $X_{\varepsilon}(t)\leqslant X_{\varepsilon}^{x}(t)$ for
$t\geqslant 0$ with probability 1, where $X_{\varepsilon}^{x}$ is a solution
of (1.2) that started from $x$, $X_{\varepsilon}^{x}(0)=x.$ Since $a$ is a.e.
positive on $(0,x)$, we have already seen that
$\\{X_{\varepsilon}^{x}(t)\\}_{\varepsilon}$ converges to
$\psi_{+}(\psi_{+}^{-1}(x)+t)$ as $t\to\infty.$ Thus, any limit point of
$\\{X_{\varepsilon}(t)\\}_{\varepsilon}$ must be less than or equal to
$\psi_{+}(\psi_{+}^{-1}(x)+t)$ for any $x>0$, almost surely. Therefore, any
limit point of $\\{X_{\varepsilon}(t)\\}_{\varepsilon}$ does not exceed
$\psi_{+}(t).$

Define the function

$a_{n}(x):=\begin{cases}a(x)&\text{if }x\geqslant 0\\\
-\tfrac{1}{n}a(-\tfrac{x}{n})&\text{if }x<0,\end{cases}$

and denote the corresponding solutions to stochastic differential equations by
$X_{n,\varepsilon}(t).$ Let us apply Theorem 4.2 to the sequence
$\\{X_{n,\varepsilon}\\}_{\varepsilon}.$ Calculate the limit (4.2b):

$B_{n,+}(x)=\int_{0}^{x}a(y)dy,\qquad
B_{n,-}(x)=\int_{0}^{x}a(y/n)dy/n=B_{+}(x/n).$

Thus,

$(B_{n,-})^{-1}(u)=n(B_{n,+})^{-1}(u),\qquad\lim_{u\downarrow
0}\frac{(B_{n,-})^{-1}(u)}{(B_{n,+})^{-1}(u)}=n,$

and we get convergence

$P_{X_{n,\varepsilon}}\Rightarrow\frac{1}{n+1}\delta_{-n\psi_{+}(n^{-2}t)}+\frac{n}{n+1}\delta_{\psi_{+}(t)}\qquad\text{as
}\varepsilon\to 0.$

By the comparison theorem we have the inequality
$X_{n,\varepsilon}(t)\leqslant X_{\varepsilon}(t),t\geqslant 0$ with
probability 1. Therefore, any limit point of
$\\{X_{\varepsilon}\\}_{\varepsilon}$ equals $\psi_{+}$ with probability at
least $\frac{n}{n+1}.$ We conclude that the limit of
$\\{X_{\varepsilon}\\}_{\varepsilon}$ exists and equals $\psi_{+}$ almost
surely. The limit is non-random, so we have convergence in probability, as in
(1.7). This finishes the proof of Theorem 1.1.

##  6\. Examples

######  Example 6.1.

For some fixed $\rho\in(0,1)$ we consider the function

$a(x)\coloneqq\mathop{\rm
sign}(x)|x|^{\rho}\bigl{(}1+\tfrac{1}{2}\phi\bigl{(}\tfrac{1}{x}\bigr{)}\bigr{)}\qquad\text{where
}\phi(y)\coloneqq\sum_{n\in{\mathbb{Z}}}\mathbbm{1}_{[2n-1,2n)}-\mathbbm{1}_{[2n,2n+1)},$

defined for all $x\neq 0$. Using Proposition 4.7 with $b(x)=\mathop{\rm
sign}(x)|x|^{\rho}$, $\gamma=\rho$ and $g(y)=\tfrac{1}{2}\mathop{\rm
sign}(y)\phi(y)$, we get $c=1$, and that $p_{\varepsilon}\to\frac{1}{2}$. We
also see that $a$ satisfies the repulsive condition (1.8) of Theorem 1.2, so
we conclude that

$P_{\varepsilon}\Rightarrow\tfrac{1}{2}\delta_{\psi_{-}}+\tfrac{1}{2}\delta_{\psi_{+}}\qquad\text{as
}\varepsilon\to 0$

where $\psi_{\pm}$ are the maximal classical solutions.

Figure 1 shows an ensemble of approximate solutions for the above drift. We
used noise sizes $\varepsilon=\frac{3^{-i}}{e},\ i=-2,\dots,-9$, and computed
150 samples of the solution with the Euler–Maruyama scheme with a step size
$\Delta t=2.5\times 10^{-3}$ up to time $t=0.5$. The left-hand figure shows
all sample paths (vertical axis) as a function of time (horizontal axis),
where bigger $\varepsilon$ were given a lighter shades of grey. The sample
paths with the smallest $\varepsilon$ are depicted in red. The right-hand
figure shows the cumulative distribution function of the samples at the final
time $t=0.5$ using the smallest value for $\varepsilon$. We can clearly see
that the solution is concentrated on the extreme sample paths
$\psi_{-},\psi_{+}$, each with probability $\tfrac{1}{2}$.

Figure 1. Sample paths (left) and cumulative distribution function (right) for
Example 6.1.

######  Example 6.2.

Let $a(x)=x^{\beta}$, $x>0,$ where $\beta\in(0,1).$ We claim that we can
continuously extend $a$ to the set $(-\infty,0]$ such that

  1. (a)

$-a(-x)\leqslant a(x)<0$ for all $x<0$;

  2. (b)

$\int_{-1}^{0}\frac{1}{a(x)}dx=-\infty,$ i.e., the Osgood condition is not
satisfied to the left of zero;

  3. (c)

$P_{X_{\varepsilon}}\Rightarrow\tfrac{1}{2}\delta_{\psi_{+}}+\tfrac{1}{2}\delta_{0}$
as $\varepsilon\to 0,$ i.e., the limit process with probability $\tfrac{1}{2}$
moves like the maximal positive solution
$\psi_{+}(t)=((1-\beta)t)^{\frac{1}{1-\beta}},t\geqslant 0$ and stays at 0
forever with probability $\tfrac{1}{2}$ too.

This example is not covered by the theory in the previous sections, and should
therefore be read as a demonstration of the complex behaviours that can occur
in the zero noise limit. Note also that the zero-noise limit is not only
concentrated on the maximal solution $\psi_{+}$, but also on the trivial
solution $\psi_{-}\equiv 0$.

Before we construct the extension, let us provide some simple preliminary
analysis. If a function $a\colon{\mathbb{R}}\to{\mathbb{R}}$ satisfies the
linear growth condition, then the family $\\{X_{\varepsilon}\\}_{\varepsilon}$
is weakly relatively compact. If additionally the function $a$ is continuous,
then any limit point of $\\{X_{\varepsilon}\\}_{\varepsilon}$ satisfies (1.1).
Both conditions (a) and (b) yield that any solution to (1.1), and hence any
limit point of $\\{X_{\varepsilon}\\}_{\varepsilon}$ has a form

$X_{0}(t)=\begin{cases}0,&t\leqslant\tau\\\
((1-\beta)(t-\tau))^{\frac{1}{1-\beta}},&t>\tau,\end{cases}$ (6.1)

where $\tau\in[0,\infty].$ Our aim is to find an extension of $a$ such that

${\mathbb{P}}(\tau=0)={\mathbb{P}}(\tau=\infty)=\tfrac{1}{2}$ (6.2)

for any limit point $X_{0}$ having representation (6.1).

Let $A=\cup_{k\geqslant
1}[-\frac{1}{2^{k}},-\frac{1}{2^{k}}+\frac{1}{4^{k}}].$ Set

$\displaystyle\tilde{a}(x)$ $\displaystyle\coloneqq\mathop{\rm
sign}(x)a(|x|)\mathbbm{1}_{x\notin A}=\begin{cases}x^{\beta},&x>0\\\
-|x|^{\beta},&x\leqslant 0,\ x\notin A\\\ 0,&x\leqslant 0,\ x\in
A,\end{cases}$ $\displaystyle\bar{a}(x)$ $\displaystyle\coloneqq\mathop{\rm
sign}(x)a(|x|)=\mathop{\rm sign}(x)|x|^{\beta}=\begin{cases}x^{\beta},&x>0\\\
-|x|^{\beta},&x\leqslant 0.\end{cases}$

Define $a$ on $(-\infty,0)$ to be any negative, continuous function such that
$\int_{-\delta}^{0}\frac{1}{a(x)}dx=-\infty$ for any $\delta>0$, and

$\bar{a}(x)\leqslant a(x)\leqslant\tilde{a}(x)\text{ for all
}x\in(-\infty,0).$

It is clear that there exists a function $a$ satisfying these properties.
Introduce the transformed process

$Y_{\varepsilon}(t)\coloneqq\varepsilon^{\frac{-2}{1+\beta}}X_{\varepsilon}\bigl{(}\varepsilon^{\frac{2(1-\beta)}{1+\beta}}t\bigr{)}.$

It can be seen (see [22] for a more general case) that

$dY_{\varepsilon}(t)=a_{\varepsilon}(Y_{\varepsilon}(t))dt+dw_{\varepsilon}(t),$
(6.3)

where
$w_{\varepsilon}(t)=\varepsilon^{\frac{-(1-\beta)}{1+\beta}}w\bigl{(}\varepsilon^{\frac{2(1-\beta)}{1+\beta}}t\bigr{)}$
is a Wiener process, and

$a_{\varepsilon}(y)=\varepsilon^{\frac{-2\beta}{1+\beta}}a\bigl{(}\varepsilon^{\frac{2}{1+\beta}}y\bigr{)}.$
(6.4)

Notice that $a_{\varepsilon}(y)=a(y)$ for all $y\in(0,\infty)$ and for all
$y<0$ such that $\varepsilon^{\frac{2}{1+\beta}}y\notin A$. For all other
$y<0$ we have the inequality $-|y|^{\beta}\leqslant a(y)<0$, by the choice of
the function $a$. We have convergence
$a_{\varepsilon}(y)\to\bar{a}(y)=\mathop{\rm sign}(y)|y|^{\beta}$ in Lebesgue
measure on any interval $y\in[-R,R]$. Observe also that

$\int_{0}^{x}a_{\varepsilon}(y)dy\geqslant\int_{0}^{x}\hat{a}(y)dy\qquad\forall\varepsilon>0,\
\forall x<0$

where

$\hat{a}(x)=\begin{cases}0,\ &x\in[-\tfrac{3}{2}\cdot 2^{n},-2^{n}]\text{ for
some }n\in{\mathbb{Z}}\\\ -|x|^{\beta}&\text{otherwise.}\end{cases}$

In particular, the last estimate yields

$\sup_{\varepsilon\in(0,1]}\lim_{R\to+\infty}\int_{-\infty}^{-R}\exp\biggl{(}-2\int_{0}^{x}a_{\varepsilon}(y)\,dy\biggr{)}dx=0.$

Set

$\sigma^{X}_{\varepsilon}(p)\coloneqq\inf\\{t\geqslant
0:X_{\varepsilon}(t)=p\\},$
$\sigma^{Y}_{\varepsilon}(p)\coloneqq\inf\\{t\geqslant
0:Y_{\varepsilon}(t)=p\\}.$

The observations above and formulas of Theorem 2.5 yield that for any $R>0$,
and for any sequences $\\{R^{\pm}_{\varepsilon}\\}$ such that
$\lim_{\varepsilon\to 0}R^{\pm}_{\varepsilon}=\pm\infty$ we have

$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}\Bigl{(}\sigma^{Y}_{\varepsilon}(R)<\sigma^{Y}_{\varepsilon}(-R)\mid
Y_{\varepsilon}(0)=0\Bigr{)}=\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{Y}_{\varepsilon}(-R)<\sigma^{Y}_{\varepsilon}(R)\mid
Y_{\varepsilon}(0)=0\Big{)}$ $\displaystyle={}$
$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{Y}_{\varepsilon}(R^{+}_{\varepsilon})<\sigma^{Y}_{\varepsilon}(R^{-}_{\varepsilon})\mid
Y_{\varepsilon}(0)=0\Big{)}=\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{Y}_{\varepsilon}(R^{-}_{\varepsilon})<\sigma^{Y}_{\varepsilon}(R^{+}_{\varepsilon})\mid
Y_{\varepsilon}(0)=0\Big{)}$ $\displaystyle={}$ $\displaystyle\tfrac{1}{2}.$

Hence, for any $\delta^{\pm}>0$ we have

$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{X}_{\varepsilon}\big{(}R\varepsilon^{\frac{2}{1+\beta}}\big{)}<\sigma^{X}_{\varepsilon}\bigl{(}-R\varepsilon^{\frac{2}{1+\beta}}\big{)}\mid
X_{\varepsilon}(0)=0\Big{)}$ (6.5) $\displaystyle={}$
$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{X}_{\varepsilon}\bigl{(}-R\varepsilon^{\frac{2}{1+\beta}}\big{)}<\sigma^{X}_{\varepsilon}\big{(}R\varepsilon^{\frac{2}{1+\beta}}\big{)}\mid
X_{\varepsilon}(0)=0\Big{)}$ $\displaystyle={}$
$\displaystyle\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{X}_{\varepsilon}(\delta^{+})<\sigma^{X}_{\varepsilon}(\delta^{-})\mid
X_{\varepsilon}(0)=0\Big{)}=\lim_{\varepsilon\to
0}{\mathbb{P}}\Big{(}\sigma^{X}_{\varepsilon}(\delta^{-})<\sigma^{X}_{\varepsilon}(\delta^{+})\mid
X_{\varepsilon}(0)=0\Big{)}$ $\displaystyle={}$ $\displaystyle\tfrac{1}{2}.$

Hence, if $X_{0}$ is a limit point of $\\{X_{\varepsilon}\\}$ having
representation (6.1), then ${\mathbb{P}}(\tau=\infty)\geqslant\tfrac{1}{2}.$
It also follows from Theorem 2.5 that for any $R>0$

$\sup_{\varepsilon>0}{\mathbb{E}}\Big{(}\sigma^{Y}_{\varepsilon}(R)\wedge\sigma^{Y}_{\varepsilon}(-R)\mid
Y_{\varepsilon}(0)=0\Big{)}<\infty.$

Thus,

$\sigma^{X}_{\varepsilon}\big{(}R\varepsilon^{\frac{2}{1+\beta}}\big{)}\wedge\sigma^{X}_{\varepsilon}\bigl{(}-R\varepsilon^{\frac{2}{1+\beta}}\big{)}\overset{{\mathbb{P}}}{\to}0\qquad\text{as
}\varepsilon\to 0$ (6.6)

if $X_{\varepsilon}(0)=0.$

Let $\bar{X}_{\varepsilon}$ be a solution to

$d\bar{X}_{\varepsilon}(t)=\bar{a}\big{(}\bar{X}_{\varepsilon}(t)\big{)}dt+\varepsilon
dw(t)$

and define

$\bar{Y}_{\varepsilon}(t)\coloneqq\varepsilon^{\frac{-2}{1+\beta}}\bar{X}_{\varepsilon}\bigl{(}\varepsilon^{\frac{2(1-\beta)}{1+\beta}}t\bigr{)}.$

Then (cf. (6.3), (6.4))

$d\bar{Y}_{\varepsilon}(t)=\bar{a}(\bar{Y}_{\varepsilon}(t))dt+dw_{\varepsilon}(t).$

In particular, if $\bar{X}_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}}$
for all $\varepsilon>0,$ where $R$ is a constant, then all processes
$\bar{Y}_{\varepsilon}$ have the same distribution independent of
$\varepsilon.$

Notice that for any $R>0$,

${\mathbb{P}}\Bigl{(}X_{\varepsilon}(t)=\bar{X}_{\varepsilon}(t),\
t\in[0,\sigma^{X}_{\varepsilon}(0)]\mid
X_{\varepsilon}(0)=\bar{X}_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}}\Bigr{)}=1.$
(6.7)

and

$\displaystyle p_{R}$
$\displaystyle\coloneqq{\mathbb{P}}\Bigl{(}\sigma^{X}_{\varepsilon}(0)=\infty\mid
X_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}}\Bigr{)}$ (6.8)
$\displaystyle={\mathbb{P}}\Bigl{(}X_{\varepsilon}(t)>0,t\geqslant 0\mid
X_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}}\Bigr{)}$
$\displaystyle={\mathbb{P}}\Bigl{(}\bar{X}_{\varepsilon}(t)>0,t\geqslant
0\mid\bar{X}_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}}\Bigr{)}$
$\displaystyle={\mathbb{P}}\Bigl{(}\bar{Y}_{\varepsilon}(t)>0\mid\bar{Y}_{\varepsilon}(0)=R\Bigr{)}\to
1\qquad\text{as }R\to\infty.$

It follows from [22] that if
$\bar{X}_{\varepsilon}(0)=R\varepsilon^{\frac{2}{1+\beta}},\varepsilon>0$,
then

$\bar{X}_{\varepsilon}\Rightarrow
p_{R}\delta_{\psi_{+}}+(1-p_{R})\delta_{\psi_{-}}\qquad\text{as
}\varepsilon\to 0.$ (6.9)

Hence, (LABEL:eq:1566), (6.6), (6.7), (6.8), and (6.9) yield that for any
limit point $X_{0}$ of $\\{X_{\varepsilon}\\}$ we have
${\mathbb{P}}(\tau=0)\geqslant\tfrac{1}{2}.$ This concludes the proof of the
convergence
$P_{X_{\varepsilon}}\Rightarrow\tfrac{1}{2}\delta_{\psi_{+}}+\tfrac{1}{2}\delta_{0}$
as $\varepsilon\to 0.$

Figure 2 shows the same type of simulation as in Example 6.1. From the figure
it is clear that for small $\varepsilon$, the samples split in two groups of
equal size, one moving along $\psi_{+}$ and the other remaining around the
origin. As the noise decreases, the left-going samples concentrate around the
trivial solution $X\equiv 0$.

Figure 2. Sample paths (left) and cumulative distribution function (right) for
Example 6.2.

##  Appendix A Appendix

###### Proof of Theorem 2.2.

For a sequence of numbers $0<\varepsilon_{n}\to 0$, let
$a_{i,n}=a_{i}*\omega_{\varepsilon_{n}}$, where
$\omega_{\varepsilon}(z)=\varepsilon^{-1}\omega\big{(}z\varepsilon^{-1}\big{)}$
and $\omega\in C_{c}^{\infty}({\mathbb{R}})$ is a nonnegative mollifier. Let
$X_{i,n}$ be the unique solution of

$dX_{i,n}=a_{i,n}(X_{i,n})dt+dW,\qquad i=1,2,\ n\in{\mathbb{N}}.$ (A.1)

For the smoothened drift functions it still holds $a_{1,n}\leqslant a_{2,n}$.
Therefore, it follows from the classic comparison theorem that
$X_{1,n}\leqslant X_{2,n}$ (see e.g. the comparison theorem in [16]).

The application of Theorem 2.1 completes the proof in the case when
$a_{1},a_{2}\in L^{\infty}({\mathbb{R}})$. If $a_{1},a_{2}$ are only locally
bounded, then we approximate $X_{1},X_{2}$ by solutions to SDEs with drifts
$a_{i,M}\coloneqq a_{i}\mathbbm{1}_{[-M,M]}$. It follows from [29, Remark 3b,
p. 145] that

${\mathbb{P}}\bigl{(}X_{i}(t)=X_{i,M}(t)\;\forall\
t\leqslant\tau_{i,M}\bigr{)}=1,\qquad i=1,2,$

where $\tau_{i,M}=\inf\\{t\geqslant 0:|X_{i}(t)|\geqslant M\\}.$ We have
already proved that $X_{1,M}(t)\leqslant X_{2,M}(t)$ almost surely. This
completes the proof of the theorem. ∎

###### Proof of Lemma 2.7.

We assume that $f$ is positive; the negative case follows similarly. Denote
$B(z)\coloneqq\int_{\alpha}^{z}f(u)\,du$. Then $B$ is absolutely continuous
and invertible, and since $B^{\prime}(z)=f(z)>0$ for a.e. $z$, the inverse
$B^{-1}$ is also absolutely continuous (see e.g. [6, Exercise 5.8.52]). Hence,
we can write

$\displaystyle g_{\varepsilon}(y)$
$\displaystyle=\int_{y}^{\beta}e^{-(B(z)-B(y))/\varepsilon^{2}}\frac{B^{\prime}(z)}{\varepsilon^{2}}g(z)\,dz$
$\displaystyle=\int_{B(y)}^{B(\beta)}\frac{e^{-(v-B(y))/\varepsilon^{2}}}{\varepsilon^{2}}g\big{(}B^{-1}(v)\big{)}\,dv$

(where we made the change of variables $v=B(z)$). The function $[0,\infty)\ni
v\mapsto\frac{e^{-v/\varepsilon^{2}}}{\varepsilon^{2}}$ is an approximate
identity and therefore

$g_{\varepsilon}(y)\to g(B^{-1}(B(y)))=g(y)\qquad\text{as }\varepsilon\to 0$

in $L^{1}((\alpha,\beta))$, and pointwise whenever $v=B(y)$ is a Lebesgue
point for $v\mapsto g\big{(}B^{-1}(v)\big{)}$; see e.g. [11, Theorems 8.14,
8.15]. But $B$ and $B^{-1}$ are absolutely continuous, so these points
coincide with the Lebesgue points for $g$. ∎

## Acknowledgements

U. S. Fjordholm was partially supported by the Research Council of Norway
project INICE, project no. 301538. A. Pilipenko acknowledges the support by
the National Research Foundation of Ukraine (project 2020.02/0014 “Asymptotic
regimes of perturbed random walks: on the edge of modern and classical
probability”) and the Senter for internasjonalisering av utdanning (SIU),
within the project Norway–Ukrainian Cooperation in Mathematical Education,
project number CPEA-LT-2016/10139.

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