# Stable limit theorems for additive functionals of one-dimensional diffusion
processes

Loïc Béthencourt Sorbonne Université, CNRS, Laboratoire de Probabilités,
Statistique et Modélisation, F-75005 Paris, France. Email:
<EMAIL_ADDRESS>I would like to
thank Nicolas Fournier for all the fruitful discussions and advice.

###### Abstract

We consider a positive recurrent one-dimensional diffusion process with
continuous coefficients and we establish stable central limit theorems for a
certain type of additive functionals of this diffusion. In other words we find
some explicit conditions on the additive functional so that its fluctuations
behave like some $\alpha$-stable process in large time for $\alpha\in(0,2]$.

2010 Mathematics Subject Classification: 60J60, 60F05.

Keywords and phrases: One-dimensional diffusion processes, Stable central
limit theorem, Stable processes, Local times.

##  1 Introduction and main result

Consider a one-dimensional diffusion process with continuous coefficients $b$
and $\sigma$, i.e. a continuous adapted process $(X_{t})_{t\geq 0}$ satisfying
the SDE

$X_{t}=\int_{0}^{t}b(X_{s})\mathrm{d}s+\int_{0}^{t}\sigma(X_{s})\mathrm{d}B_{s},$
(1)

where $(B_{t})_{t\geq 0}$ is a Brownian motion. Without loss of generality, we
assume $X_{0}=0$, even if it means changing $b$ and $\sigma$. If $\sigma$ does
not vanish, then weak existence and uniqueness in law hold for (1), see
Kallenberg [16, Chapter 23 Theorem 23.1].

######  Assumption 1.

The functions $b,\sigma:\mathbb{R}\to\mathbb{R}$ are continuous and for all
$x\in\mathbb{R}$, $\sigma(x)>0$. Moreover $b$ and $\sigma$ are such that the
process $(X_{t})_{t\geq 0}$ is positive recurrent in the sense of Harris.

We introduce the scale function $\mathfrak{s}$ of $(X_{t})_{t\geq 0}$ defined
by

$\mathfrak{s}(x)=\int_{0}^{x}\exp\left(-2\int_{0}^{v}b(u)\sigma^{-2}(u)\mathrm{d}u\right)\mathrm{d}v,$
(2)

which is a $C^{2}$, strictly increasing function solving
$\frac{1}{2}s^{\prime\prime}\sigma^{2}+s^{\prime}b=0$. We also introduce the
speed measure density

$m(x)=\sigma^{-2}(x)\exp\left(2\int_{0}^{x}b(v)\sigma^{-2}(v)\mathrm{d}v\right)=[\sigma^{2}(x)\mathfrak{s}^{\prime}(x)]^{-1}$
(3)

######  Remark 2.

Assume $b$ and $\sigma$ are continuous and $\sigma$ does not vanish. Then we
have the equivalence between the two following propositions, see for instance
Revuz-Yor [23, Chapter VII].

  1. 1.

$(X_{t})_{t\geq 0}$ is positive recurrent in the sense of Harris.

  2. 2.

$\lim_{x\to\pm\infty}\mathfrak{s}(x)=\pm\infty$ and
$\kappa^{-1}:=\int_{\mathbb{R}}m(x)\mathrm{d}x<\infty$.

Moreover, in this case, the measure $\mu(\mathrm{d}x)=\kappa m(x)\mathrm{d}x$
is the unique invariant probability measure for the process $(X_{t})_{t\geq
0}$.

The ergodic theorem for Harris recurrent processes, see Azema-Duflo-Revuz [1]
or Revuz-Yor [23, Chapter X], tells us that for
$f\in\mathrm{\bm{L}}^{1}(\mu)$, a.s.

$\frac{1}{t}\int_{0}^{t}f(X_{s})\mathrm{d}s\underset{t\to\infty}{\longrightarrow}\mu(f),$
(4)

where $\mu(f)=\int_{\mathbb{R}}f\mathrm{d}\mu$. The convergence in (4) can be
seen as a strong law of large numbers for the additive functional
$\int_{0}^{t}f(X_{s})\mathrm{d}s$ of the process $(X_{t})_{t\geq 0}$. Then it
is very natural to study its fluctuations, i.e. to describe the asymptotic
behavior of $\frac{1}{t}\int_{0}^{t}(f(X_{s})-\mu(f))\mathrm{d}s$. In this
paper, we give simple conditions on $f$ for these asymptotic fluctuations to
be $\alpha$-stable for some $\alpha\in(0,2]$. The conditions on $f$ are
entirely prescribed by the coefficients $b$ and $\sigma$.

We recall that $\ell:[0,\infty)\mapsto(0,\infty)$ is said to be slowly varying
if for every $\lambda>0$, $\ell(\lambda x)/\ell(x)\to 1$ as $x$ goes to
infinity. The use of such functions is justified by the fact that domains of
attraction of stable laws involve slowly varying functions.

######  Assumption 3.

The function $f:\mathbb{R}\mapsto\mathbb{R}$ is locally bounded and Borel.
Moreover, there exists $\alpha>0$, $(f_{+},f_{-})\in\mathbb{R}^{2}$ and a
continuous slowly varying function $\ell:[0,\infty)\mapsto(0,\infty)$ such
that

$[\sigma(x)\mathfrak{s}^{\prime}(x)]^{-2}|\mathfrak{s}(x)|^{2-1/\alpha}\ell(|\mathfrak{s}(x)|)f(x)\underset{x\to\pm\infty}{\longrightarrow}f_{\pm},$

and $|f_{+}|+|f_{-}|>0$. If $f\in\mathrm{\bm{L}}^{1}(\mu)$, we impose
$\mu(f)=0$. Finally, if $\alpha\geq 2$, we impose $f$ to be continuous.

In the critical stable regime $\alpha=1$, we will add a mild assumption which
ensures that the set $\mathrm{\bm{L}}^{1}(\mu)$ is big enough.

######  Assumption 4.

There exists $\lambda>0$ such that
$|\mathfrak{s}|^{\lambda}\in\mathrm{\bm{L}}^{1}(\mu)$.

Under Assumption 3, we set
$\rho=\int_{1}^{\infty}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x$
which can be infinite, and we define the diffusive constant as follows:

  * •

If $\alpha>2$ or $\alpha=2$ and $\rho<\infty$,
$\sigma_{\alpha}^{2}=4\kappa\int_{\mathbb{R}}\mathfrak{s}^{\prime}(x)\left(\int_{x}^{\infty}f(v)[\sigma^{2}(v)\mathfrak{s}^{\prime}(v)]^{-1}\mathrm{d}v\right)^{2}\mathrm{d}x$.

  * •

If $\alpha=2$ and $\rho=\infty$,
$\sigma_{2}^{2}=4\kappa(f_{+}^{2}+f_{-}^{2})$.

  * •

If $\alpha\in(0,2)$, $\sigma_{\alpha}^{\alpha}=\frac{\kappa
2^{\alpha-2}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}(|f_{+}|^{\alpha}+|f_{-}|^{\alpha})$.

We also introduce, for $\xi\in\mathbb{R}$, the complex number

  * •

$z_{\alpha}(\xi)=1-i\frac{\mathrm{sgn}(f_{+})|f_{+}|^{\alpha}+\mathrm{sgn}(f_{-})|f_{-}|^{\alpha}}{|f_{+}|^{\alpha}+|f_{+}|^{\alpha}}\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)$
if $\alpha\in(0,2)\setminus\\{1\\}$;

  * •

$z_{1}(\xi)=1+i\frac{f_{+}+f_{-}}{|f_{+}|+|f_{-}|}\frac{2}{\pi}\mathrm{sgn}(\xi)\left[\log\left(\frac{\pi|\xi|}{2(|f_{+}|+|f_{-}|)}\right)+2\gamma+\log(2)+\frac{\log|f_{+}|f_{+}+\log|f_{-}|f_{-}}{f_{+}+f_{-}}\right]$,
where $\gamma$ is the Euler constant.

Finally, for a family of processes $((Y_{t}^{\epsilon})_{t\geq
0})_{\epsilon>0}$ valued in $\mathbb{R}$, we say that
$(Y_{t}^{\epsilon})_{t\geq 0}\overset{f.d.}{\longrightarrow}(Y_{t}^{0})_{t\geq
0}$ if for all $n\geq 1$ and all $0<t_{1}<\cdots<t_{n}$, the vector
$(Y_{t_{i}}^{\epsilon})_{1\leq i\leq n}$ converges in law to
$(Y_{t_{i}}^{0})_{1\leq i\leq n}$ in $\mathbb{R}^{n}$. We are now ready to
state our main theorem.

######  Theorem 5.

Suppose Assumptions 1 and 3. Let $(X_{t})_{t\geq 0}$ be a solution of (1),
$(W_{t})_{t\geq 0}$ a Brownian motion and $(S_{t}^{(\alpha)})_{t\geq 0}$ a
$\alpha$-stable process such that $\mathbb{E}[\exp(i\xi
S_{t}^{(\alpha)})]=\exp(-t|\xi|^{\alpha}z_{\alpha}(\xi))$.

  1. (i)

If $\alpha>2$, or $\alpha=2$ and $\rho<\infty$,

$\left(\epsilon^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}W_{t}\right)_{t\geq
0}\qquad\hbox{as $\epsilon\to 0$}.$

  2. (ii)

If $\alpha=2$ and $\rho=\infty$, we set
$\rho_{\epsilon}=\int_{1}^{1/\epsilon}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x$
and we have

$\left(|\epsilon/\rho_{\epsilon}|^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{2}W_{t}\right)_{t\geq
0}\qquad\hbox{as $\epsilon\to 0$}.$

  3. (iii)

If $\alpha\in(0,2)\setminus\\{1\\}$,

$\left(\epsilon^{1/\alpha}\ell(1/\epsilon)\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}\qquad\hbox{as $\epsilon\to 0$}.$

  4. (iv)

If $\alpha=1$ and Assumption 4 holds, there exists a deterministic family
$(\xi_{\epsilon})_{\epsilon>0}$ of real numbers such that

$\left(\epsilon\ell(1/\epsilon)\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s-\xi_{\epsilon}t\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{1}S_{t}^{(1)}\right)_{t\geq
0}\qquad\hbox{as $\epsilon\to 0$}.$

Moreover $\xi_{\epsilon}\underset{\epsilon\to
0}{\sim}\kappa(f_{+}+f_{-})\ell(1/\epsilon)\zeta_{\epsilon}$, where
$\zeta_{\epsilon}=-\int_{1/\epsilon}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}$ if
$f\in\mathrm{\bm{L}}^{1}(\mu)$ and
$\zeta_{\epsilon}=\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)}$
otherwise.

For the reader more familiar with infinitesimal generators than with
characteristic functions, let us mention the following remark.

######  Remark 6.

When $\alpha\in(0,2)$, the process $(\sigma_{\alpha}S_{t}^{(\alpha)})_{t\geq
0}$ of Theorem 5 is a Lévy process with Lévy measure

$\nu(\mathrm{d}x)=\left(c_{+}x^{-1-\alpha}\bm{1}_{\\{x>0\\}}+c_{-}|x|^{-1-\alpha}\bm{1}_{\\{x<0\\}}\right)\mathrm{d}x,$

where

$c_{+}=\lambda_{\alpha}\left[\bm{1}_{\\{f_{+}>0\\}}|f_{+}|^{\alpha}+\bm{1}_{\\{f_{-}>0\\}}|f_{-}|^{\alpha}\right],\quad
c_{-}=\lambda_{\alpha}\left[\bm{1}_{\\{f_{+}<0\\}}|f_{+}|^{\alpha}+\bm{1}_{\\{f_{-}<0\\}}|f_{-}|^{\alpha}\right]$

and $\lambda_{\alpha}=\frac{\kappa
2^{\alpha-2}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}$.
Its infinitesimal generator $\mathcal{L}^{(\alpha)}$ is such that for all
$\phi\in C^{\infty}_{c}(\mathbb{R})$,

  * •

$\mathcal{L}^{(\alpha)}\phi(x)=\int_{\mathbb{R}\setminus\\{0\\}}[\phi(x+z)-\phi(x)]\nu(dz)$
if $\alpha\in(0,1)$,

  * •

$\mathcal{L}^{(\alpha)}\phi(x)=\int_{\mathbb{R}\setminus\\{0\\}}[\phi(x+z)-\phi(x)-z\phi^{\prime}(x)]\nu(dz)$
if $\alpha\in(1,2)$,

  * •

if $\alpha=1$,
$\mathcal{L}^{(\alpha)}\phi(x)=a\phi^{\prime}(x)+\int_{\mathbb{R}\setminus\\{0\\}}[\phi(x+z)-\phi(x)-z\bm{1}_{\\{|z|\leq
1\\}}\phi^{\prime}(x)]\nu(dz)$, where

$a=-(f_{+}+f_{-})\left[2\gamma+\log(2)+\kappa\log(\kappa)+\frac{\log|f_{+}|f_{+}+\log|f_{-}|f_{-}}{f_{+}+f_{-}}+\kappa\frac{\pi}{2}A\right],$

$\gamma$ is the Euler constant and
$A=\int_{0}^{1}\frac{\sin(x)-x}{x^{2}}dx+\int_{1}^{\infty}\frac{\sin(x)}{x^{2}}dx$.

One can refer to Bertoin [3, page 24] to understand how to go from the Lévy-
Khintchine triplet to the generator and to Sato [25, Chapter 3.14] to go from
the Lévy-Khintchine triplet to the characteristic function.

Observe that in the Lévy regime $\alpha\in(0,2)$, we have a convergence in
finite dimensional distribution of a continuous process towards a
discontinuous process (stable processes possess many jumps). Hence we cannot
hope to obtain convergence in law as a process, for example for the Skorokhod
distance. Observe also that if $f_{-}=0$, the stable process has only positive
jumps. Thus the limiting process is completely asymmetric although the process
$(f(X_{t}))_{t\geq 0}$ may visit the whole space infinitely often.

When $f\in\mathrm{\bm{L}}^{1}(\mu)$ and $\mu(f)\neq 0$, we can use Theorem 5
with the function $f-\mu(f)$ provided it satisfies Assumption 3.

When $\alpha\geq 2$, the stable process obtained in the limit is a Brownian
motion, which is the only $2$-stable process. A standard strategy to show
central limit theorems for the additive functional
$\int_{0}^{t}f(X_{s})\mathrm{d}s$ is to solve the Poisson equation
$\mathcal{L}g=f$, where $\mathcal{L}$ is the infinitesimal generator of
$(X_{t})_{t\geq 0}$ with domain $D_{\mathcal{L}}$. One can refer to Jacod-
Shiryaev [13, Chapter VIII], Pardoux-Veretennikov [22] or Cattiaux, Chafaï and
Guillin [7]. Assume $f$ is continuous and define the function

$g(x)=2\int_{0}^{x}\mathfrak{s}^{\prime}(v)\int_{v}^{\infty}f(u)[\sigma^{2}(u)\mathfrak{s}^{\prime}(u)]^{-1}\mathrm{d}u\mathrm{d}v.$
(5)

Then $g$ is a $C^{2}$ function solving the Poisson equation
$2bg^{\prime}+\sigma^{2}g^{\prime\prime}=-2f$. Hence, applying the Itô formula
with $g$, we express the additive functional as a martingale plus some
remainder, and hope to obtain the result using a central limit theorem for
martingales. This will be the strategy used when $\alpha\geq 2$, and we will
actually prove the following more general result.

######  Theorem 7.

Suppose Assumption 1 and let $(X_{t})_{t\geq 0}$ be the solution of (1). Let
also $f$ be a continuous function in $\mathrm{\bm{L}}^{1}(\mu)$ such that
$\mu(f)=0$ and $g$ be defined by (5). If
$g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$, then

$\left(\epsilon^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\gamma W_{t}\right)_{t\geq
0}\qquad\hbox{as $\epsilon\to 0$},$

where $(W_{t})_{t\geq 0}$ is a Brownian motion and
$\gamma^{2}=\int_{\mathbb{R}}[g^{\prime}\sigma]^{2}\mathrm{d}\mu$. Moreover,
if $g$ is bounded, the finite dimensional convergence can be replaced by a
convergence in law of continuous processes, with the topology of uniform
convergence on compact time intervals.

Note that in the Lévy regime, we do not require $f$ to be regular whereas we
need $f$ to be continuous in the diffusive regime, so we can apply the Itô
formula with $g$.

### References

The central limit theorem for Markov processes has a long history which goes
back to Langevin [17]. Roughly, he studied a one-dimensional particle, the
velocity of which is subject to random shocks and a restoring force $F(v)=-v$,
and showed that the position of the particle behaves like a Brownian motion,
when $t$ tends to infinity.

Probabilistic techniques to obtain such results can be found in Jacod-Shiryaev
[13, Chapter VIII]: if there exists a solution to the Poisson equation having
few properties, then we have a central limit theorem. This strategy is applied
in Pardoux-Veretennikov [22] for very general multidimensional diffusion
processes. However, [22] does not completely include the diffusive regime of
Theorem 5: we have a little less assumptions but our result only holds in
dimension one. It seems that the critical diffusive regime $\alpha=2$ was not
much studied. Finally, one can refer to Cattiaux, Chafaï and Guillin [7] for a
detailed state of the art of the techniques used to show such theorems.

Regarding the Lévy regime, we could not find many results for Markov processes
and this does not seem to be well developed. Through the prism of ergodic
theory, Gouëzel [11] proved stable limit theorems for a certain class of maps.
Applied to a positive recurrent Markov chain $(X_{n})_{n\geq 0}$ with
countable state space, its results tells us that if the function $f$ is such
that, up to a slowly varying function,

$\mathbb{P}_{i}\left(\sum_{k=0}^{\tau_{i}-1}f(X_{k})>x\right)\underset{x\to\infty}{\sim}c_{+}|x|^{-\alpha}\quad\text{and}\quad\mathbb{P}_{i}\left(\sum_{k=0}^{\tau_{i}-1}f(X_{k})<x\right)\underset{x\to-\infty}{\sim}c_{-}|x|^{-\alpha},$
(6)

where $\tau_{i}=\inf\\{k>0,X_{k}=i\\}$, then we have

$\frac{1}{n^{\alpha}}\sum_{k=0}^{n-1}f(X_{k})\overset{d}{\longrightarrow}S^{\alpha},$

where $S^{\alpha}$ is a stable random variable. This is more or less obvious
in this case since we have a sum of i.i.d blocks in the domain of attraction
of stable laws and we can thus apply classical stable central limit theorems.

Similar results with more tangible assumptions were proved by Jara, Komorowski
and Olla [14] for Markov chains with general state space. They assume the
function $f$ is such that, roughly, $\mu(\\{|f|>x\\})\simeq|x|^{-\alpha}$,
where $\mu$ is the invariant measure of the chain and a spectral gap
condition, the latter one being more or less equivalent to return times having
exponential moments. Since the return times are small, it is reasonable to
think that we have
$\mathbb{P}_{i}(\sum_{k=0}^{\tau_{i}-1}f(X_{k})>x)\simeq\mu(\\{f>x\\})$, at
least if we consider a countable state space. They actually give two proofs,
one using a martingale approximation and a second one with a renewal method
involving a coupling argument.

Mellet, Mischler and Mouhot [20] showed that for a linearized Boltzmann
equation with heavy-tailed invariant distribution, the rescaled distribution
of the position converges to the solution of the fractional heat equation,
i.e. the position process behaves like a stable process. Their work is closely
related to [14] although their proof is entirely analytic.

The method proposed in this article is rather powerful and gives a simple
condition on the additive functional for stable limit theorems to occur. The
assumption made on the function $f$ does not seem to be a time-continuous
equivalent of (6) and we do not assume anything on the return times, nor on
$\mu(\\{|f|>x\\})$. We will see in Section 4 that, indeed, the index $\alpha$
is not always prescribed by $\mu(\\{|f|>x\\})$.

The strategy is the following: we classically express the process
$(X_{t})_{t\geq 0}$ as a Brownian motion $(W_{t})_{t\geq 0}$ changed in time
and in space. Then we write
$\int_{0}^{t}f(X_{s})\mathrm{d}s=\int_{0}^{\rho_{t}}\phi(W_{s})\mathrm{d}s$
for a certain function $\phi$, which roughly looks like
$\int_{0}^{\tau_{t}}|W_{s}|^{1/\alpha-2}\mathrm{d}s$ (up to asymmetry and
principal values issues) in large time, where $(\tau_{t})_{t\geq 0}$ is the
generalized inverse of the local time at $0$ of $(W_{t})_{t\geq 0}$. This
process is known to be a stable process, see Itô-McKean [12, page 226], Jeulin
and Yor [15] and Biane and Yor [4]. The computations are thus tractable and
the method is very robust.

This method was proposed by Fournier and Tardif [10], studying a kinetic model
and the purpose of this paper is to extend this method to general one-
dimensional diffusions and to more general additive functionals. The model
they were studying was the following:

$V_{t}=B_{t}-\frac{\beta}{2}\int_{0}^{t}\frac{V_{s}}{1+V_{s}^{2}}\mathrm{d}s,\qquad
X_{t}=\int_{0}^{t}V_{s}\mathrm{d}s,$ (7)

where $(B_{t})_{t\geq 0}$ is a Brownian motion and $\beta>1$. The processes
$(V_{t})_{t\geq 0}$ and $(X_{t})_{t\geq 0}$ are respectively the velocity and
the position of a one-dimensional particle subject to the restoring force
$F(v)=-\frac{\beta}{2}\frac{v}{1+v^{2}}$ and random shocks. Lebeau and Puel
[18] showed that, when $\beta\in(1,5)\setminus\\{2,3,4\\}$, the rescaled
distribution of the position $X_{t}$ converges to the solution of the
fractional heat equation. Their work is analytical and relies on a deep
spectral analysis leading to an impressive result. The diffusive regime
($\beta>5$) and critical diffusive regime ($\beta=5$) are treated by
Nasreddine-Puel [21] and Cattiaux-Nasreddine-Puel [8]. We stress that all
these papers are P.D.E papers.

To summarize, if the restoring force field is strong enough ($\beta\geq 5$),
the rescaled position process resembles a Brownian motion and we say that
there is a normal diffusion limit. On the other hand, when the force is weak
($\beta\in(1,5)$), the position resembles a stable process and there is an
anomalous diffusion limit (or fractional diffusion limit).

Actually, physicists discovered that atoms diffuse anomalously when they are
cooled by a laser. See for instance Castin, Dalibard and Cohen-Tannoudji [6],
Sagi, Brook, Almog and Davidson [24] and Marksteiner, Ellinger and Zoller
[19]. A theoretical study (see Barkai, Aghion and Kessler [2]) modeling the
motion of atoms precisely by (7) proved with quite a high level of rigor the
observed phenomenons.

In the Lévy regime, the index $\alpha$ is not prescribed by the function
$f=\text{id}$ and the invariant measure $\mu$. Indeed observe that we have
$\alpha=(\beta+1)/3$ whereas one can check, see Section 4, that

$\mu(\\{|f|>x\\})\simeq|x|^{1-\beta}.$

In this situation, the return times are large and the dynamics are more
complex.

Then, using probabilistic techniques, Fournier and Tardif [10] treated all
cases of (7) for a larger class of symmetric forces. Naturally, we recover
their result and enlarge it to asymmetrical forces, see Section 4. In a
companion paper [9], they also prove the result in any dimension and the proof
is much more involved.

### Plan of the paper

In Section 2, we recall some facts about slowly varying functions, local times
and generalize some results on stable processes found in [12], [15] and [4].
Section 3 is dedicated to the proof of Theorem 5 and Theorem 7. Finally, we
apply our result to several models in Section 4, illustrating some remarks
made in the previous section.

### Notations

Throughout the paper, we will use the function
$\mathrm{sgn}_{a,b}(x)=a\bm{1}_{\\{x>0\\}}+b\bm{1}_{\\{x<0\\}}$ where
$(a,b)\in\mathbb{R}^{2}$. The function $\mathrm{sgn}$ will denote the usual
sign function, i.e. $\mathrm{sgn}(x)=\mathrm{sgn}_{1,-1}(x)$.

##  2 Preliminaries

###  2.1 Slowly varying functions

In this section, we define slowly varying functions and give the properties of
these functions that will be useful to us. One can refer to Bingham, Goldie
and Teugels [5] for more details.

######  Definition 8.

A measurable function $\ell:[0,\infty)\mapsto(0,\infty)$ is said to be slowly
varying if for every $\lambda>0$, $\ell(\lambda x)/\ell(x)\longrightarrow 1$
as $x\to\infty$.

Famous examples of slowly varying functions are powers of logarithm, iterated
logarithms or functions having a strictly positive limit at infinity. It holds
that for any $\theta>0$, $x^{\theta}\ell(x)\to\infty$ and
$x^{-\theta}\ell(x)\to 0$, as $x\to\infty$. We will need the following lemma.

######  Lemma 9.

Let $\ell$ be a continuous slowly varying function.

  1. (i)

For each $0<a<b$, we have $\sup_{\lambda\in[a,b]}\left|\ell(\lambda
x)/\ell(x)-1\right|\underset{x\to\infty}{\longrightarrow}1.$

  2. (ii)

(Potter’s bound). For any $\delta>0$ and $C>1$, there exists $x_{0}>0$, such
that for all $x,y\geq x_{0}$,

$\frac{\ell(y)}{\ell(x)}\leq
C\left(\Big{|}\frac{y}{x}\Big{|}^{\delta}\vee\Big{|}\frac{x}{y}\Big{|}^{\delta}\right).$

  3. (iii)

The functions $L:x\mapsto\int_{1}^{x}\frac{\mathrm{d}v}{v\ell(v)}$ and
$M:x\mapsto\int_{1}^{x}(\int_{v}^{\infty}\frac{\mathrm{d}u}{u^{3/2}\ell(u)})^{2}\mathrm{d}v$
are slowly varying, as well as
$N:x\mapsto\int_{x}^{\infty}\frac{\mathrm{d}v}{v\ell(v)}$ if
$\int_{1}^{\infty}\frac{\mathrm{d}v}{v\ell(v)}<\infty$.

###### Proof.

Points (i) and (ii). See [5, pages 6 and 25].

Point (iii). Let us start with the function $L$. For $\lambda>0$, we have

$L(\lambda x)=\int_{1}^{\lambda
x}\frac{\mathrm{d}v}{v\ell(v)}=L(x)+\int_{x}^{\lambda
x}\frac{\mathrm{d}v}{v\ell(v)}=L(x)+\int_{1}^{\lambda}\frac{\mathrm{d}u}{u\ell(xu)}$

where we used the substitution $v=xu$ in the last equality. By (i),
$\sup_{u\in[1,\lambda]}|l(x)/l(ux)-1|\longrightarrow 0$ as $x\to\infty$, so
that
$\int_{1}^{\lambda}\frac{\mathrm{d}u}{u\ell(xu)}\sim\log(\lambda)/\ell(x)$ as
$x$ tends to infinity and it remains to show that
$\ell(x)L(x)\longrightarrow\infty$ at infinity. Let $A>1$ and consider $x>A$:
we have

$\ell(x)L(x)=\ell(x)\int_{1}^{x}\frac{\mathrm{d}v}{v\ell(v)}=\ell(x)\int_{1/x}^{1}\frac{\mathrm{d}u}{u\ell(xu)}\geq\int_{1/A}^{1}\frac{\ell(x)\mathrm{d}u}{u\ell(xu)}\underset{x\to\infty}{\longrightarrow}\log(A),$

by Point (i) again. Now let $A$ to $\infty$ and the result follows. Regarding
the function $N$, we assume
$\int_{1}^{\infty}\frac{\mathrm{d}v}{v\ell(v)}<\infty$ and we write, for
$\lambda>0$,

$N(\lambda x)=N(x)-\int_{x}^{\lambda
x}\frac{\mathrm{d}v}{v\ell(v)}=N(x)-\int_{1}^{\lambda}\frac{\mathrm{d}u}{u\ell(xu)}$

and, as previously, we only need to show that
$\ell(x)N(x)\longrightarrow\infty$. We write

$\ell(x)N(x)=\ell(x)\int_{x}^{\infty}\frac{\mathrm{d}v}{v\ell(v)}=\int_{1}^{\infty}\frac{\ell(x)\mathrm{d}u}{u\ell(xu)}\geq\int_{1}^{A}\frac{\ell(x)\mathrm{d}u}{u\ell(xu)}$

for any $A>1$. But once again,
$\int_{1}^{A}\frac{\ell(x)\mathrm{d}v}{v\ell(xv)}\longrightarrow\log(A)$ and
the result holds letting $A$ to infinity. Finally we show the result for $M$.
We have, for $\lambda>0$,

$M(\lambda x)=M(x)+\int_{x}^{\lambda
x}\left(\int_{v}^{\infty}\frac{\mathrm{d}u}{u^{3/2}\ell(u)}\right)^{2}\mathrm{d}v=M(x)+\int_{1}^{\lambda}\left(\int_{y}^{\infty}\frac{\mathrm{d}z}{z^{3/2}\ell(xz)}\right)^{2}\mathrm{d}y$

A little study shows that
$\int_{1}^{\lambda}\left(\int_{y}^{\infty}\frac{\mathrm{d}z}{z^{3/2}\ell(xz)}\right)^{2}\mathrm{d}y\sim\int_{1}^{\lambda}\left(\int_{y}^{\infty}\frac{\mathrm{d}z}{z^{3/2}\ell(x)}\right)^{2}\mathrm{d}y=\frac{4\log(\lambda)}{\ell^{2}(x)}$.
It remains to show that $\ell^{2}(x)M(x)\longrightarrow\infty$. Let $A>1$ and
$x>A$, we have

$\ell^{2}(x)M(x)=\int_{1/x}^{1}\left(\int_{y}^{\infty}\frac{\ell(x)\mathrm{d}z}{z^{3/2}\ell(xz)}\right)^{2}\mathrm{d}y\geq\int_{1/A}^{1}\left(\int_{y}^{\infty}\frac{\ell(x)\mathrm{d}z}{z^{3/2}\ell(xz)}\right)^{2}\mathrm{d}y\underset{x\to\infty}{\longrightarrow}4\log(A)$

and the result follows by letting $A$ to $\infty$. ∎

###  2.2 Brownian local times

Local times have been widely studied in Revuz-Yor [23] to which we refer for
much more details. Let $(W_{t})_{t\geq 0}$ be a Brownian motion. We introduce
the local time of $(W_{t})_{t\geq 0}$ at $x\in\mathbb{R}$, which is the
process $(L_{t}^{x})_{t\geq 0}$ defined by

$L_{t}^{x}=|W_{t}-x|-|x|-\int_{0}^{t}\mathrm{sgn}(W_{s}-x)\mathrm{d}W_{s}.$

The process $(L_{t}^{x})_{t\geq 0}$ is continuous and non-decreasing and the
random non-negative measure $\mathrm{d}L_{t}^{x}$ on $[0,\infty)$ is a.s.
carried by the set $\\{t\geq 0,W_{t}=x\\}$, which is a.s. Lebesgue-null. We
will heavily use the occupation times formula, see [23, Chapter 6 Corollary
1.6 page 224], which tells that a.s., for every $t\geq 0$ and for all Borel
function $\varphi:\mathbb{R}\to\mathbb{R}_{+}$,

$\int_{0}^{t}\varphi(W_{s})\mathrm{d}s=\int_{\mathbb{R}}\varphi(x)L_{t}^{x}\mathrm{d}x.$

We will also use the fact that the map $x\mapsto L_{t}^{x}$ is a.s. Hölder of
order $\theta$, for $\theta\in(0,1/2)$, uniformly on compact time intervals,
see [23, Corollary 1.8 page 226], i.e. that

$\sup_{[0,T]\times\mathbb{R}}|x|^{-\theta}\left|L_{t}^{x}-L_{t}^{0}\right|<\infty\quad\text{a.s.}$
(8)

We now introduce $\tau_{t}=\inf\\{u\geq 0,L_{u}^{0}>t\\}$, the right-
continuous generalized inverse of $(L_{t}^{0})_{t\geq 0}$. Some properties of
$(\tau_{t})_{t\geq 0}$ will be needed. For all $t>0$,
$\mathbb{P}(\tau_{t_{-}}<\tau_{t})=0$, which means that $(\tau_{t})_{t\geq 0}$
has no deterministic jump time and a.s., for all $t\geq 0$,
$L_{\tau_{t}}^{0}=t$ since $(L_{t}^{0})_{t\geq 0}$ is continuous. We will also
use that a.s., for all $t\geq 0,W_{\tau_{t}}=0$ and that a.s., $\tau_{0}=0$.

Finally, the processes $(\tau_{t},W_{t})_{t\geq 0}$ and
$\left(L_{\tau_{t}}^{x}\right)_{x\in\mathbb{R},t\geq 0}$ enjoy the following
scaling property: for all $c>0$

$(\tau_{t},W_{t})_{t\geq
0}\overset{d}{=}(c^{-2}\tau_{ct},c^{-1}W_{c^{2}t})_{t\geq
0}\quad\text{and}\quad\left(L_{\tau_{t}}^{x}\right)_{x\in\mathbb{R},t\geq
0}\overset{d}{=}\left(c^{-1}L_{\tau_{ct}}^{xc}\right)_{x\in\mathbb{R},t\geq
0}$ (9)

###  2.3 Stable processes

There is a huge litterature on stable processes and we will mainly refer to
Sato [25]. In this section we generalize some results on stable processes
found in Itô-McKean [12, page 226], Jeulin-Yor [15] and Biane-Yor [4], where
they worked in a symmetric or completely asymmetric framework. Let us first
recall a classic result on stable processes, see Sato [25, Theorem 14.15 page
86 and Definition 14.16 page 87].

######  Theorem 10.

Let $\alpha\in(0,2)\setminus\\{1\\}$ and let $(X_{t})_{t\geq 0}$ be a
$\alpha$-strictly stable process, i.e. a Lévy process enjoying the scaling
property $(X_{t})_{t\geq 0}\overset{d}{=}(c^{-1/\alpha}X_{ct})_{t\geq 0}$ for
all $c>0$. Then there exist $\nu\geq 0$ and $\beta\in[-1,1]$ such that for all
$\xi\in\mathbb{R}$

$\mathbb{E}\left[\exp(i\xi X_{t})\right]=\exp\left(-\nu
t|\xi|^{\alpha}\left(1-i\beta\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right).$

If $(X_{t})_{t\geq 0}$ is a $1$-stable process, i.e. a Lévy process such that
for all $c>0$, there exists $k_{c}\in\mathbb{R}$ with $(X_{t})_{t\geq
0}\overset{d}{=}(c^{-1}X_{ct}+k_{c}t)_{t\geq 0}$, then there exist $\nu\geq
0$, $\beta\in[-1,1]$ and $\tau\in\mathbb{R}$ such that for all
$\xi\in\mathbb{R}$

$\mathbb{E}\left[\exp(i\xi X_{t})\right]=\exp\left(-\nu
t|\xi|\left(1+i\beta\frac{2}{\pi}\mathrm{sgn}(\xi)\log|\xi|\right)+it\tau\xi\right).$

In any case, if $(X_{t})_{t\geq 0}$ has only positive jumps, i.e. if its Lévy
measure is carried by $(0,\infty)$, then $\beta=1$.

We now consider a Brownian motion $(W_{t})_{t\geq 0}$, its local time
$(L_{t}^{0})_{t\geq 0}$ at 0, and the generalized inverse $(\tau_{t})_{t\geq
0}$ of its local time. We have the following result.

######  Lemma 11.

Let $\alpha\in(0,1)$, $(a,b)\in\mathbb{R}^{2}$ and define the process
$K_{t}=\int_{0}^{t}\mathrm{sgn}_{a,b}(W_{s})|W_{s}|^{1/\alpha-2}\mathrm{d}s$.
Then $(S_{t})_{t\geq 0}=(K_{\tau_{t}})_{t\geq 0}$ is strictly $\alpha$-stable
and for all $\xi\in\mathbb{R}$ and all $t\geq 0$, we have

$\mathbb{E}\left[\exp\left(i\xi
S_{t}\right)\right]=\exp\left(-c_{\alpha,a,b}t|\xi|^{\alpha}\left(1-i\beta_{\alpha,a,b}\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right),$

where
$c_{\alpha,a,b}=\frac{2^{\alpha-2}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}(|a|^{\alpha}+|b|^{\alpha})$
and
$\beta_{\alpha,a,b}=\frac{\mathrm{sgn}(a)|a|^{\alpha}+\mathrm{sgn}(b)|b|^{\alpha}}{|a|^{\alpha}+|b|^{\alpha}}$.

When $\alpha\in(0,1)$, stable processes have finite variations and no drift
part meaning they are pure jump processes and should be seen this way:
$S_{t}=\sum_{s\in[0,t]}\int_{\tau_{s-}}^{\tau_{s}}\mathrm{sgn}_{a,b}(W_{u})|W_{u}|^{1/\alpha-2}\mathrm{d}u$.
The jumping times are the ones of $(\tau_{t})_{t\geq 0}$ and the size of a
jump is equal to the integral of the function
$\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}$ over the corresponding excursion of
the Brownian motion.

While $(K_{t})_{t\geq 0}$ is well-defined when $\alpha\in(0,1)$ since
$\int_{0}^{T}|W_{s}|^{1/\alpha-2}\mathrm{d}s<\infty$ a.s., this is not the
case when $\alpha\in[1,2)$ and we have to work a little more.

######  Lemma 12.

Let $\alpha\in(1,2)$, $(a,b)\in\mathbb{R}^{2}$ and define
$K_{t}=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}(L_{t}^{x}-L_{t}^{0})\mathrm{d}x$.
Then $(S_{t})_{t\geq 0}=(K_{\tau_{t}})_{t\geq 0}$ is strictly $\alpha$-stable
and for all $\xi\in\mathbb{R}$ and all $t\geq 0$, we have

$\mathbb{E}\left[\exp\left(i\xi
S_{t}\right)\right]=\exp\left(-c_{\alpha,a,b}t|\xi|^{\alpha}\left(1-i\beta_{\alpha,a,b}\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right),$

where
$c_{\alpha,a,b}=\frac{2^{\alpha-2}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}(|a|^{\alpha}+|b|^{\alpha})$
and
$\beta_{\alpha,a,b}=\frac{\mathrm{sgn}(a)|a|^{\alpha}+\mathrm{sgn}(b)|b|^{\alpha}}{|a|^{\alpha}+|b|^{\alpha}}$.

The case $\alpha=1$ is a little more tedious.

######  Lemma 13.

Let $(a,b)\in\mathbb{R}$ and define the process
$K_{t}=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{-1}(L_{t}^{x}-L_{t}^{0}\bm{1}_{\\{|x|\leq
1\\}})\mathrm{d}x$. Then $(S_{t})_{t\geq 0}=(K_{\tau_{t}})_{t\geq 0}$ is
$1$-stable. For all $\xi\in\mathbb{R}$ and all $t\geq 0$, we have

$\mathbb{E}\left[\exp\left(i\xi
S_{t}\right)\right]=\exp\left(-c_{a,b}t|\xi|\left(1+i\beta_{a,b}\frac{2}{\pi}\log(|\xi|)\mathrm{sgn}(\xi)\right)+it\tau_{a,b}\xi\right),$

where $c_{a,b}=\frac{\pi}{2}(|a|+|b|)$, $\beta_{a,b}=\frac{a+b}{|a|+|b|}$,
$\tau_{a,b}=-(a+b)\left[2\gamma+\log(2)+\frac{\log|a|a+\log|b|b}{a+b}\right]$
and $\gamma$ is the Euler constant

To handle the proof, we need first to make the following general remark.

######  Remark 14.

Let $\phi:\mathbb{R}\to\mathbb{R}$ be a measurable function such that for all
$T>0$, $\int_{0}^{T}|\phi(W_{s})|\mathrm{d}s<\infty$ a.s. The process
$Z_{t}=\int_{0}^{\tau_{t}}\phi(W_{s})\mathrm{d}s$ is a Lévy process with
respect to the filtration $(\mathcal{F}_{\tau_{t}})_{t\geq 0}$. Indeed, for
all $0<s<t$,
$Z_{t}-Z_{s}=\int_{\tau_{s}}^{\tau_{t}}\phi(W_{u})\mathrm{d}u=\int_{0}^{\tau_{t}-\tau_{s}}\phi(W_{u+\tau_{s}})\mathrm{d}u=\int_{0}^{\bar{\tau}_{t-s}}\phi(\bar{W}_{u})\mathrm{d}u$,
where $(\bar{W}_{u})_{u\geq 0}=(W_{u+\tau_{s}})_{u\geq 0}$ is a Brownian
motion independent of $\mathcal{F}_{\tau_{s}}$ thanks to the strong Markov
property and since $\bar{W}_{0}=W_{\tau_{s}}=0$ a.s., and where
$(\bar{\tau}_{u})_{u\geq 0}$ is the generalized inverse of its local time.
Hence the increments of $(Z_{t})_{t\geq 0}$ are independent and stationary.

###### Proof of Lemma 11.

Let us first show that $\left(S_{t}\right)_{t\geq 0}$ is strictly
$\alpha$-stable. It is a Lévy process by Remark 14. For all $c>0$, we have
from (9)

$\begin{split}S_{t}&\overset{d}{=}\int_{0}^{c^{-2}\tau_{ct}}\mathrm{sgn}_{a,b}(c^{-1}W_{c^{2}s})|c^{-1}W_{c^{2}s}|^{1/\alpha-2}\mathrm{d}s\\\
&=c^{2-1/\alpha}\int_{0}^{c^{-2}\tau_{ct}}\mathrm{sgn}_{a,b}(W_{c^{2}s})|W_{c^{2}s}|^{1/\alpha-2}\mathrm{d}s\\\
&=c^{-1/\alpha}\int_{0}^{\tau_{ct}}\mathrm{sgn}_{a,b}(W_{u})|W_{u}|^{1/\alpha-2}\mathrm{d}u,\end{split}$

which equals $c^{-1/\alpha}S_{ct}$. Hence $\left(S_{t}\right)_{t\geq 0}$ is
strictly $\alpha$-stable. We now focus on showing the explicit form of its
characteristic function and introduce first the processes

$C_{t}=\int_{0}^{\tau_{t}}\bm{1}_{\\{W_{s}>0\\}}|W_{s}|^{1/\alpha-2}\mathrm{d}s,\qquad
D_{t}=\int_{0}^{\tau_{t}}\bm{1}_{\\{W_{s}<0\\}}|W_{s}|^{1/\alpha-2}\mathrm{d}s$

and $H_{t}=C_{t}-D_{t}$. We have $S_{t}=aC_{t}+bD_{t}$. It is clear that
$(C_{t})_{t\geq 0}$, $(D_{t})_{t\geq 0}$ and $(H_{t})_{t\geq 0}$ are also
strictly stable processes because e.g. $(C_{t})_{t\geq 0}$ is nothing but
$(S_{t})_{t\geq 0}$ when $a=1$ and $b=0$. It is also known from Biane-Yor [4]
that for all $\xi\in\mathbb{R}$, for all $t\geq 0$ that

$\mathbb{E}\left[\exp(i\xi
H_{t})\right]=\exp\left(-tc_{\alpha}|\xi|^{\alpha}\right)\>\>\>\text{with}\>\>\>c_{\alpha}=\frac{2^{\alpha-1}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}.$
(10)

Using the occupation times formula, we can write

$C_{t}=\int_{0}^{\infty}|x|^{1/\alpha-2}L_{\tau_{t}}^{x}\mathrm{d}x\quad\text{and}\quad
D_{t}=\int_{-\infty}^{0}|x|^{1/\alpha-2}L_{\tau_{t}}^{x}\mathrm{d}x.$

But it is well known, see [15, page 215], that for a fixed $t$, the processes
$(L_{\tau_{t}}^{x})_{x\geq 0}$ and $(L_{\tau_{t}}^{x})_{x\leq 0}$ are
independent and thus $C_{t}$ and $D_{t}$ are independent. It is also clear
that they have the same law. Since $(C_{t})_{t\geq 0}$ has only positive
jumps, Theorem 10 tells us that its characteristic function can be written

$\mathbb{E}\left[\exp(i\xi
C_{t})\right]=\exp\left(-c_{\alpha}^{\prime}t|\xi|^{\alpha}\left(1-i\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right),$

for some $c_{\alpha}^{\prime}>0$. Now since $\mathbb{E}\left[\exp(i\xi
H_{t})\right]=\mathbb{E}\left[\exp(i\xi
C_{t})\right]\mathbb{E}\left[\exp(-i\xi
D_{t})\right]=\exp\left(-2c_{\alpha}^{\prime}t|\xi|^{\alpha}\right)$ it is
clear from (10) that $c_{\alpha}^{\prime}=c_{\alpha}/2$. Now having in mind
that $S_{t}=aC_{t}+bD_{t}$, and $C_{t}$ and $D_{t}$ are i.i.d variables, we
easily complete the proof. ∎

###### Proof of Lemma 12.

To show that $(S_{t})_{t\geq 0}$ is a Lévy process, we first introduce, for
$\eta>0$,

$\begin{split}K_{t}^{\eta}&=\int_{0}^{t}\mathrm{sgn}_{a,b}(W_{s})|W_{s}|^{1/\alpha-2}\bm{1}_{\\{|W_{s}|\geq\eta\\}}\mathrm{d}s-\left(\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\bm{1}_{\\{|x|\geq\eta\\}}\mathrm{d}x\right)L_{t}^{0}\\\
&=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\bm{1}_{\\{|x|\geq\eta\\}}(L_{t}^{x}-L_{t}^{0})\mathrm{d}x,\end{split}$

where we used the occupation times formula. We show that
$(K_{t}^{\eta})_{t\geq 0}$ converges a.s., uniformly on compact time intervals
to $(K_{t})_{t\geq 0}$. For $\eta>0$, we have

$\begin{split}\sup_{[0,T]}\left|K_{t}^{\eta}-K_{t}\right|&\leq\int_{\mathbb{R}}\left|\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\bm{1}_{\\{|x|\geq\eta\\}}-\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\right|\times\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|\mathrm{d}x\\\
&\leq\left(\sup_{[0,T]\times\mathbb{R}}|x|^{-\theta}\left|L_{t}^{x}-L_{t}^{0}\right|\right)\int_{\mathbb{R}}\bm{1}_{\\{|x|<\eta\\}}|x|^{\theta+1/\alpha-2}\mathrm{d}x,\end{split}$

where we have fixed $\theta\in(0,1/2)$ such that $1/\alpha-2+\theta>-1$, which
is possible since $\alpha<2$. We conclude using that
$\sup_{[0,T]\times\mathbb{R}}|x|^{-\theta}\left|L_{t}^{x}-L_{t}^{0}\right|<\infty$
a.s., see (8).

From Remark 14, and since a.s. for all $t\geq 0$, $L_{\tau_{t}}^{0}=t$, we
know that $(S_{t}^{\eta})_{t\geq 0}=(K^{\eta}_{\tau_{t}})_{t\geq 0}$ is a Lévy
process for all $\eta>0$. Hence $(S_{t})_{t\geq 0}$ is a Lévy process, as a
limit of Lévy processes. We now show that it has the appropriate scaling
property. By (9), we have, for any $c>0$,

$\begin{split}S_{t}&=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}(L_{\tau_{t}}^{x}-t)\mathrm{d}x\\\
&\overset{d}{=}\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}(c^{-1}L_{\tau_{ct}}^{cx}-t)\mathrm{d}x\\\
&=c^{-1/\alpha}\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}(L_{\tau_{ct}}^{x}-ct)\mathrm{d}x,\end{split}$

which equals $c^{-1/\alpha}S_{ct}$.

As in the previous proof, we consider the processes

$C_{t}=\int_{0}^{\infty}|x|^{1/\alpha-2}(L_{\tau_{t}}^{x}-t)\mathrm{d}x,\qquad
D_{t}=\int_{-\infty}^{0}|x|^{1/\alpha-2}(L_{\tau_{t}}^{x}-t)\mathrm{d}x$

and $H_{t}=C_{t}-D_{t}$. For a fixed $t\geq 0$, $C_{t}$ and $D_{t}$ are
independent and identically distributed random variables and it is clear that
they only have positive jumps and thus their characteristic function is equal
to
$\exp\left(-c_{\alpha}^{\prime}t|\xi|^{\alpha}\left(1-i\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right)$
for some $c_{\alpha}^{\prime}>0$ by Theorem 10. But once again we know from
Biane-Yor [4] that the characteristic function of $H_{t}$ is equal to
$\exp\left(-tc_{\alpha}|\xi|^{\alpha}\right)$ so once again it comes that
$c_{\alpha}^{\prime}=c_{\alpha}/2$ and we can conlude as in the previous
proof. ∎

###### Proof of Lemma 13.

As previously, we can show that $(S_{t})_{t\geq 0}$ is a Lévy process by
approximation. Then it is enough to show that the characteristic function of
$S_{t}$ has the stated form. To this end, we approach $(S_{t})_{t\geq 0}$ by
the $\alpha$-stable processes from Lemma 12, when $\alpha\in(1,2)$. We denote
by $(K_{t}^{\alpha})_{t\geq 0}$ and $(S_{t}^{\alpha})_{t\geq 0}$ the processes
from Lemma 12. One can show, using the Hölder property of the local times,
recalled in (8), that for all $T>0$, a.s.,

$\sup_{[0,T]}\left|K_{t}^{\alpha}+\left(\int_{|x|\geq
1}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\mathrm{d}x\right)L_{t}^{0}-K_{t}\right|\underset{\alpha\downarrow
1}{\longrightarrow}0.$

Since $\int_{|x|\geq
1}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\mathrm{d}x=\frac{(a+b)\alpha}{\alpha-1}$,
it comes that for all $t\geq 0$ and all $\xi\in\mathbb{R}$,

$\lim_{\alpha\downarrow 1}\mathbb{E}\left[\exp\left(i\xi
S_{t}^{\alpha}\right)\right]e^{i\xi
t\frac{(a+b)\alpha}{\alpha-1}}=\mathbb{E}\left[\exp\left(i\xi
S_{t}\right)\right].$

But we know from Lemma 12 that

$\mathbb{E}\left[\exp\left(i\xi S_{t}^{\alpha}\right)\right]e^{i\xi
t\frac{(a+b)\alpha}{\alpha-1}}=\exp\left(-c_{\alpha,a,b}t|\xi|^{\alpha}+it\xi\left(c_{\alpha,a,b}\beta_{\alpha,a,b}|\xi|^{\alpha-1}\tan\left(\frac{\alpha\pi}{2}\right)+\frac{(a+b)\alpha}{\alpha-1}\right)\right),$

where
$c_{\alpha,a,b}\beta_{\alpha,a,b}=\frac{2^{\alpha-2}\pi}{\alpha\sin(\alpha\pi/2)}\left(\frac{\alpha^{\alpha}}{\Gamma(\alpha)}\right)^{2}(\mathrm{sgn}(a)|a|^{\alpha}+\mathrm{sgn}(b)|b|^{\alpha})\underset{\alpha\downarrow
1}{\to}\frac{\pi}{2}(a+b)$. The function
$h_{a,b}(\alpha)=c_{\alpha,a,b}\beta_{\alpha,a,b}$ defined for $\alpha>0$ is
$C^{1}$ in a neighborhood of $1$ and a careful computation shows that

$h_{a,b}^{\prime}(1)=\frac{\pi}{2}(a+b)\left[1+\log(2)+2\gamma+\frac{\log|a|a+\log|b|b}{a+b}\right],$

where $\gamma$ is the Euler constant, which appears from the fact that
$\Gamma^{\prime}(1)=-\gamma$. Hence we have
$c_{\alpha,a,b}\beta_{\alpha,a,b}\underset{\alpha\downarrow
1}{=}\frac{\pi}{2}(a+b)+h_{a,b}^{\prime}(1)(\alpha-1)+o(\alpha-1)$. Now using
that $|\xi|^{\alpha-1}\underset{\alpha\downarrow
1}{=}1+(\alpha-1)\log|\xi|+o(\alpha-1)$ and
$\tan\left(\frac{\alpha\pi}{2}\right)\underset{\alpha\downarrow
1}{=}\frac{-2}{\pi(\alpha-1)}+o(1)$, it comes that

$c_{\alpha,a,b}\beta_{\alpha,a,b}|\xi|^{\alpha-1}\tan\left(\frac{\alpha\pi}{2}\right)+\frac{(a+b)\alpha}{\alpha-1}=(a+b)-\frac{2}{\pi}h_{a,b}^{\prime}(1)-(a+b)\log|\xi|+o(1),$

and thus we have

$\mathbb{E}\left[\exp\left(i\xi S_{t}^{\alpha}\right)\right]e^{i\xi
t\frac{\alpha}{\alpha-1}}\underset{\alpha\downarrow
1}{\to}\exp\left(-c_{a,b}t|\xi|\left(1+i\beta_{a,b}\frac{2}{\pi}\log(|\xi|)\mathrm{sgn}(\xi)\right)+it\tau_{a,b}\xi\right),$

where $c_{a,b}=\frac{\pi}{2}(|a|+|b|)$, $\beta_{a,b}=\frac{a+b}{|a|+|b|}$ and
$\tau_{a,b}=-(a+b)\left[2\gamma+\log(2)+\frac{\log|a|a+\log|b|b}{a+b}\right]$.
∎

###  2.4 Inverting time-changes

We remind some classical convergence results enabling to treat the convergence
of the generalized inverses of time changes.

######  Lemma 15.

For all $n\geq 1$, let $(a_{t}^{n})_{t\geq 0}$ from $[0,+\infty)$ to itself be
a continuous, increasing and bijective function for all $n\geq 1$ and consider
its inverse $(r_{t}^{n})_{t\geq 0}$. Assume $(a_{t}^{n})_{t\geq 0}$ simply
converges, as $n\to\infty$, to a non-decreasing function $(a_{t})_{t\geq 0}$
such that $\lim_{t\to+\infty}a_{t}=\infty$. Consider $r_{t}=\inf\\{u\geq
0,a_{u}>t\\}$ its right-continuous generalized inverse and $J=\\{t\geq
0,r_{t_{-}}<r_{t}\\}$. For all $t\in[0,+\infty)\setminus J$, we have
$\lim_{n\to+\infty}r_{t}^{n}=r_{t}$.

##  3 Main proofs

From now on, we will always suppose at least that Assumptions 1 and 3 holds.
In Subsection 3.1, we characterize the integrability of $f$ and
$g^{\prime}\sigma$. In Subsection 3.2, we represent the process
$(X_{t/\epsilon})_{t\geq 0}$ with a time-changed Brownian motion
$(W_{\tau_{t}^{\epsilon}})_{t\geq 0}$ which enables us to state that
$\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s$ is equal in law to
$H_{\tau_{t}^{\epsilon}}=\int_{0}^{\tau_{t}^{\epsilon}}\phi_{\epsilon}(W_{s})\mathrm{d}s$.

In Subsection 3.3, we first check that $A_{t}^{\epsilon}$, inverse of
$\tau_{t}^{\epsilon}$, converges to the local time at $0$ of $(W_{t})_{t\geq
0}$ so that $\tau_{t}^{\epsilon}$ converges towards $\tau_{t}$. We also
slightly prepare the proof of the diffusive and critical diffusive regime.
Then, as
$\epsilon^{1/\alpha}\ell(1/\epsilon)\phi_{\epsilon}(x)\to\mathrm{sgn}_{f_{+},f_{-}}(x)|x|^{1/\alpha-2}$,
we show that $H_{\tau_{t}^{\epsilon}}$ converges to $K_{\tau_{t}}$ a.s. for
fixed a fixed $t$, which is enough to conclude. We will work a little bit more
in the critical Lévy regime.

In Subsection 3.4, we adopt the martingale strategy writing
$\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s$ as a local martingale plus some
remainder and we use classical central limit theorems for local martingales to
conclude.

###  3.1 Integrability of $f$ and $g^{\prime}\sigma$

In this subsection, we characterize with the index $\alpha$ the integrability
of $f$ and $g^{\prime}\sigma$ with respect to the measure $\mu$. We first
introduce the functions $\psi:\mathbb{R}\to(0,\infty)$ and
$\phi:\mathbb{R}\to\mathbb{R}$ defined by

$\psi=\left(\mathfrak{s}^{\prime}\circ\mathfrak{s}^{-1}\right)\times\left(\sigma\circ\mathfrak{s}^{-1}\right)\quad\text{and}\quad\phi=(f\circ\mathfrak{s}^{-1})/\psi^{2}.$
(11)

Assumption 3 precisely tells us that $\phi$ is a regular varying with index
$1/\alpha-2$. Indeed, since $\mathfrak{s}$ is an increasing bijection by
Assumption 1 and Remark 2, we have

$|x|^{2-1/\alpha}\ell(|x|)\phi(x)\underset{x\to\pm\infty}{\longrightarrow}f_{\pm}.$
(12)

We make the following proposition

######  Proposition 16.

$f\in\mathrm{\bm{L}}^{1}(\mu)$ is equivalent to $\alpha>1$ or $\alpha=1$ and
$\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}<\infty$.

###### Proof.

If $\alpha>1$, then $2-1/\alpha>1$. Since $\ell$ is a slowly varying function,
we can find $\delta>0$ with $2-1/\alpha-\delta>1$ such that
$|x|^{\delta}\ell(|x|)\to\infty$ as $|x|\to\infty$. Thus we have by (12) that

$|\phi(x)|\underset{x\to\pm\infty}{=}o\left(|x|^{\delta+1/\alpha-2}\right)$

and the function $\phi$ is integrable with respect to the Lebesgue measure on
$\mathbb{R}$. But using the substitution $x=\mathfrak{s}(v)$, we have

$\int_{\mathbb{R}}|\phi(x)|\mathrm{d}x=\int_{\mathbb{R}}\frac{|f(v)|}{\sigma^{2}(v)\mathfrak{s}^{\prime}(v)}\mathrm{d}v=\frac{1}{\kappa}\int_{\mathbb{R}}|f|\mathrm{d}\mu<\infty.$
(13)

If $\alpha=1$ and $\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}<\infty$, then
(12) tells us that $|x|\ell(|x|)\phi(x)\to f_{\pm}$, and thus $\phi$ is
integrable with respect to the Lebesgue measure on $\mathbb{R}$ which tells us
by (13) that $f$ is integrable with respect to the Lebesgue measure.

We can make a similar reasoning to show that $\alpha<1$ or $\alpha=1$ and
$\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}=\infty$ implies that
$f\notin\mathrm{\bm{L}}^{1}(\mu)$. ∎

We now focus on the function $g$ defined by (5) when $\alpha\geq 2$ and recall
that

$g(x)=2\int_{0}^{x}\mathfrak{s}^{\prime}(v)\int_{v}^{\infty}f(u)[\sigma^{2}(u)\mathfrak{s}^{\prime}(u)]^{-1}\mathrm{d}u\mathrm{d}v=2\kappa^{-1}\int_{0}^{x}\mathfrak{s}^{\prime}(v)\mu(f\bm{1}_{(v,\infty)})\mathrm{d}v.$

By Assumption 3, Proposition 16 and since $\alpha\geq 2$, we have $\mu(f)=0$
and thus $\mu(f\bm{1}_{(v,\infty)})=-\mu(f\bm{1}_{(-\infty,v)})$ so that

$g(x)=-2\int_{0}^{x}\mathfrak{s}^{\prime}(v)\int_{-\infty}^{v}f(u)[\sigma^{2}(u)\mathfrak{s}^{\prime}(u)]^{-1}\mathrm{d}u\mathrm{d}v.$
(14)

We now express the function $g\circ\mathfrak{s}^{-1}$, using the substitutions
$v=\mathfrak{s}^{-1}(y)$ and $u=\mathfrak{s}^{-1}(z)$:

$g(\mathfrak{s}^{-1}(x))=2\int_{0}^{x}\int_{y}^{\infty}f(\mathfrak{s}^{-1}(z))[\sigma(\mathfrak{s}^{-1}(z))\mathfrak{s}^{\prime}(\mathfrak{s}^{-1}(z))]^{-2}\mathrm{d}z\mathrm{d}y=2\int_{0}^{x}\int_{y}^{\infty}\phi(z)\mathrm{d}z\mathrm{d}y.$
(15)

Using (14), we also get

$g(\mathfrak{s}^{-1}(x))=-2\int_{0}^{x}\int_{-\infty}^{y}\phi(z)\mathrm{d}z\mathrm{d}y.$
(16)

We are now ready to state the proposition characterizing the integrability of
$g^{\prime}\sigma$ with respect to $\mu$.

######  Proposition 17.

If $\alpha>2$, or $\alpha=2$ and
$\rho=\int_{1}^{\infty}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x<\infty$,
then $g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$.

###### Proof.

From (15) and (16), we get that

$\displaystyle(g\circ\mathfrak{s}^{-1})^{\prime}(x)$
$\displaystyle=2\int_{x}^{\infty}\phi(v)\mathrm{d}v=-2\int_{-\infty}^{x}\phi(v)\mathrm{d}v.$
(17)

Assume $\alpha>2$, then $2-1/\alpha>3/2$. From Assumption 3 and since $\ell$
is a slowly varying function, we can find $\delta>0$ such that
$2-1/\alpha-\delta>3/2$ and

$|\phi(x)|\underset{x\to\pm\infty}{=}o\left(|x|^{\delta+1/\alpha-2}\right),$

and thus

$|(g\circ\mathfrak{s}^{-1})^{\prime}(x)|^{2}\underset{x\to\pm\infty}{=}o\left(|x|^{2(\delta+1/\alpha-1)}\right).$

Since $2(\delta+1/\alpha-1)<-1$,
$(g\circ\mathfrak{s}^{-1})^{\prime}\in\mathrm{\bm{L}}^{2}(\mathrm{d}x)$. Now
using the fact that
$(g\circ\mathfrak{s}^{-1})^{\prime}=[\mathfrak{s}^{\prime}\circ\mathfrak{s}^{-1}]^{-1}\times
g^{\prime}\circ\mathfrak{s}^{-1}$ and the substitution $x=\mathfrak{s}(v)$, we
have

$\int_{\mathbb{R}}\frac{|g^{\prime}(\mathfrak{s}^{-1}(x))|^{2}}{[\mathfrak{s}^{\prime}(\mathfrak{s}^{-1}(x))]^{2}}\mathrm{d}x=\int_{\mathbb{R}}\frac{|g^{\prime}(v)|^{2}}{\mathfrak{s}^{\prime}(v)}\mathrm{d}v=\frac{1}{\kappa}\int_{\mathbb{R}}|g^{\prime}\sigma|^{2}\mathrm{d}\mu,$

whence the result. Now assume $\alpha=2$ and
$\rho=\int_{1}^{\infty}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x<\infty$.
Assumption 3 tells us that
$\phi(x)\underset{x\to\infty}{\sim}f_{\pm}\frac{1}{|x|^{3/2}\ell(|x|)}$.
Integrating and using (17), we get

$(g\circ\mathfrak{s}^{-1})^{\prime}(x)\underset{x\to\infty}{\sim}2f_{+}\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}.$

Hence the fact that $\rho<\infty$ tells us precisely that
$(g\circ\mathfrak{s}^{-1})^{\prime}\in\mathrm{\bm{L}}^{2}(\mathbb{R}_{+},\mathrm{d}x)$.
Using a similar computation involving (17), we find that
$(g\circ\mathfrak{s}^{-1})^{\prime}\in\mathrm{\bm{L}}^{2}(\mathbb{R}_{-},\mathrm{d}x)$.
We conclude that $g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$ as when
$\alpha>2$. ∎

###  3.2 Scale function and speed measure

In this subsection, we classically represent the solution $(X_{t})_{t\geq 0}$
to (1) as a function of a time-changed Brownian motion. Roughly, the process
$(\mathfrak{s}(X_{t}))_{t\geq 0}$ is a continuous local martingale and the
Dubins-Schwarz theorem tells us $(\mathfrak{s}(X_{t}))_{t\geq 0}$ is a time-
changed Brownian motion. The next lemma enables us to represent
$X_{t/\epsilon}$ and $\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s$ through the
functions $\psi$, $\phi$, defined by (11), and $\mathfrak{s}$.

######  Lemma 18.

Let $\epsilon>0$ and $a_{\epsilon}>0$. We consider a Brownian motion
$(W_{t})_{t\geq 0}$ and we define

$A_{t}^{\epsilon}=\epsilon
a_{\epsilon}^{-2}\int_{0}^{t}\psi^{-2}\left(W_{s}/a_{\epsilon}\right)\mathrm{d}s$

and its inverse $(\tau_{t}^{\epsilon})_{t\geq 0}$, which is a.s. a continuous
and strictly increasing bijection from $[0,+\infty)$ in itself. Let us set

$X_{t}^{\epsilon}=\mathfrak{s}^{-1}\left(W_{\tau_{t}^{\epsilon}}/a_{\epsilon}\right)\quad\text{and}\quad
F_{t}^{\epsilon}=H_{\tau_{t}^{\epsilon}}^{\epsilon}\quad\text{where}\quad
H_{t}^{\epsilon}=a_{\epsilon}^{-2}\int_{0}^{t}\phi\left(W_{s}/a_{\epsilon}\right)\mathrm{d}s.$

For $(X_{t})_{t\geq 0}$ the unique solution to (1), we have

$\left(\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s,X_{t/\epsilon}\right)_{t\geq
0}\overset{d}{=}\left(F_{t}^{\epsilon},X_{t}^{\epsilon}\right)_{t\geq 0}.$

We introduce the unusual degree of freedom $a_{\epsilon}$ that will allow us
to make converge the time-change $A_{t}^{\epsilon}$ without normalization.

###### Proof.

We recall that the function $\psi$ and $\phi$ are defined by
$\psi=\left(\mathfrak{s}^{\prime}\circ\mathfrak{s}^{-1}\right)\times\left(\sigma\circ\mathfrak{s}^{-1}\right)$
and $\phi=(f\circ\mathfrak{s}^{-1})/\psi^{2}$. Let us consider the function
$\psi_{\epsilon}(w)=\epsilon^{-1/2}a_{\epsilon}\psi(w/a_{\epsilon})$, so that
$A_{t}^{\epsilon}=\int_{0}^{t}\psi_{\epsilon}^{-2}(W_{s})\mathrm{d}s$. We set
$Y_{t}^{\epsilon}=W_{\tau_{t}^{\epsilon}}$ which classicaly solves, see [16,
page 452],

$Y_{t}^{\epsilon}=\int_{0}^{t}\psi_{\epsilon}\left(Y_{s}^{\epsilon}\right)\mathrm{d}B_{s}^{\epsilon}$

where $(B_{t}^{\epsilon})_{t\geq 0}$ is a Brownian motion. Next, we consider
the function $\varphi_{\epsilon}(y)=\mathfrak{s}^{-1}(y/a_{\epsilon})$. We can
apply Itô’s formula to $X_{t}^{\epsilon}=\varphi_{\epsilon}(Y_{t}^{\epsilon})$
since $\mathfrak{s}^{-1}$ is $C^{2}$ and we get

$X_{t}^{\epsilon}=\int_{0}^{t}\varphi_{\epsilon}^{\prime}(Y_{s}^{\epsilon})\psi_{\epsilon}\left(Y_{s}^{\epsilon}\right)\mathrm{d}B_{s}^{\epsilon}+\frac{1}{2}\int_{0}^{t}\varphi_{\epsilon}^{\prime\prime}(Y_{s}^{\epsilon})\psi_{\epsilon}^{2}\left(Y_{s}^{\epsilon}\right)\mathrm{d}s.$

The functions are such that:

$\varphi_{\epsilon}^{\prime}(y)\psi_{\epsilon}(y)=\frac{1}{a_{\epsilon}\mathfrak{s}^{\prime}(\mathfrak{s}^{-1}(y/a_{\epsilon}))}\times\frac{a_{\epsilon}\psi(y/a_{\epsilon})}{\epsilon^{1/2}}=\frac{\sigma(\mathfrak{s}^{-1}(y/a_{\epsilon}))}{\epsilon^{1/2}}=\frac{\sigma(\varphi_{\epsilon}(y))}{\epsilon^{1/2}}$

and,

$\varphi_{\epsilon}^{\prime\prime}(y)\psi_{\epsilon}^{2}(y)=\frac{-1}{a_{\epsilon}^{2}}\frac{\mathfrak{s}^{\prime\prime}(\mathfrak{s}^{-1}(y/a_{\epsilon}))}{[\mathfrak{s}^{\prime}(\mathfrak{s}^{-1}(y/a_{\epsilon}))]^{3}}\times\frac{a_{\epsilon}^{2}\psi^{2}(y/a_{\epsilon})}{\epsilon}=\frac{2b(\mathfrak{s}^{-1}(y/a_{\epsilon}))}{\epsilon}=\frac{2b(\varphi_{\epsilon}(y))}{\epsilon}.$

Indeed remember that $\mathfrak{s}$ solves
$\frac{1}{2}\mathfrak{s}^{\prime\prime}\sigma^{2}+\mathfrak{s}^{\prime}b=0$.
Finally we have

$X_{t}^{\epsilon}=\epsilon^{-1}\int_{0}^{t}b(X_{s}^{\epsilon})\mathrm{d}s+\epsilon^{-1/2}\int_{0}^{t}\sigma(X_{s}^{\epsilon})\mathrm{d}B_{s}^{\epsilon}$

Now taking equation (1), we write

$X_{t/\epsilon}=\int_{0}^{t/\epsilon}b(X_{s})\mathrm{d}s+\int_{0}^{t/\epsilon}\sigma(X_{s})\mathrm{d}B_{s}=\epsilon^{-1}\int_{0}^{t}b(X_{s/\epsilon})\mathrm{d}s+\epsilon^{-1/2}\int_{0}^{t}\sigma(X_{s/\epsilon})\mathrm{d}\hat{B}_{s}^{\epsilon},$

where $(\hat{B}_{t}^{\epsilon})_{t\geq
0}=(\epsilon^{1/2}B_{t/\epsilon})_{t\geq 0}$ is a Brownian motion. Thus,
$\forall\epsilon>0$, the processes $(X_{t/\epsilon})_{t\geq 0}$ and
$(X_{t}^{\epsilon})_{t\geq 0}$ are solutions of the same SDE, for which we
have uniqueness in law, driven by different Brownian motion,
$(\hat{B}_{t}^{\epsilon})_{t\geq 0}$ and $(B_{t}^{\epsilon})_{t\geq 0}$. Thus
they are equal in law : $(X_{t/\epsilon})_{t\geq
0}\overset{d}{=}(X_{t}^{\epsilon})_{t\geq 0}$. As
$\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s=\epsilon^{-1}\int_{0}^{t}f(X_{s/\epsilon})\mathrm{d}s$,
we have

$\left(\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s,X_{t/\epsilon}\right)_{t\geq
0}\overset{d}{=}\left(\epsilon^{-1}\int_{0}^{t}f(X_{s}^{\epsilon})\mathrm{d}s,X_{t}^{\epsilon}\right)_{t\geq
0}.$

Using the substitution $u=\tau_{s}^{\epsilon}\Leftrightarrow
s=A_{u}^{\epsilon}$, we have
$\mathrm{d}s=\psi_{\epsilon}^{-2}(W_{u})\mathrm{d}u$, and

$\epsilon^{-1}\int_{0}^{t}f(X_{s}^{\epsilon})\mathrm{d}s=\epsilon^{-1}\int_{0}^{t}f\circ\varphi_{\epsilon}\left(W_{\tau_{s}^{\epsilon}}\right)\mathrm{d}s=\epsilon^{-1}\int_{0}^{\tau_{t}^{\epsilon}}\frac{f\circ\varphi_{\epsilon}(W_{u})}{\psi_{\epsilon}^{2}(W_{u})}\mathrm{d}u=a_{\epsilon}^{-2}\int_{0}^{\tau_{t}^{\epsilon}}\phi(W_{u}/a_{\epsilon})\mathrm{d}u=F_{t}^{\epsilon}$

which ends the proof. ∎

We end this section with the following proposition that we will use several
times later on.

######  Proposition 19.

Let $\epsilon>0$ and $a_{\epsilon}>0$ such that $a_{\epsilon}\to 0$ when
$\epsilon\to 0$. Let also $(W_{t})_{t\geq 0}$ be a Brownian motion and
$(L_{t}^{0})_{t\geq 0}$ its local time at $0$. For every
$\varphi\in\mathrm{\bm{L}}^{1}(\mu)$, we have a.s., for all $T>0$,

$\sup_{[0,T]}\left|\kappa
a_{\epsilon}^{-1}\int_{0}^{t}\frac{\varphi\circ\mathfrak{s}^{-1}(W_{s}/a_{\epsilon})}{\psi^{2}(W_{s}/a_{\epsilon})}\mathrm{d}s-\mu(\varphi)L_{t}^{0}\right|\underset{\epsilon\to
0}{\longrightarrow}0.$

###### Proof.

Denote $Y_{t}^{\epsilon}=\kappa
a_{\epsilon}^{-1}\int_{0}^{t}\frac{\varphi\circ\mathfrak{s}^{-1}(W_{s}/a_{\epsilon})}{\psi^{2}(W_{s}/a_{\epsilon})}\mathrm{d}s$.
Using the occupation times formula, we get

$Y_{t}^{\epsilon}=\kappa
a_{\epsilon}^{-1}\int_{\mathbb{R}}\frac{\varphi\circ\mathfrak{s}^{-1}(x/a_{\epsilon})}{\psi^{2}(x/a_{\epsilon})}L_{t}^{x}\mathrm{d}x=\kappa\int_{\mathbb{R}}\frac{\varphi(v)\mathfrak{s}^{\prime}(v)}{\psi^{2}(\mathfrak{s}(v))}L_{t}^{a_{\epsilon}\mathfrak{s}(v)}\mathrm{d}v=\int_{\mathbb{R}}\varphi(v)L_{t}^{a_{\epsilon}\mathfrak{s}(v)}\mu(\mathrm{d}v),$

where we used the substitution $x=a_{\epsilon}\mathfrak{s}(v)$. We recall that
$\mu(\mathrm{d}x)=\kappa[\sigma^{2}(x)\mathfrak{s}^{\prime}(x)]^{-1}\mathrm{d}x$
and that
$\psi=(\mathfrak{s}^{\prime}\circ\mathfrak{s}^{-1})\times(\sigma\circ\mathfrak{s}^{-1})$.
Hence we have

$\sup_{[0,T]}\left|Y_{t}^{\epsilon}-\mu(\varphi)L_{t}^{0}\right|\leq\int_{\mathbb{R}}|\varphi(v)|\sup_{[0,T]}\left|L_{t}^{a_{\epsilon}\mathfrak{s}(v)}-L_{t}^{0}\right|\mu(\mathrm{d}v).$

But for all $v\in\mathbb{R}$,
$\sup_{[0,T]}|L_{t}^{a_{\epsilon}\mathfrak{s}(v)}-L_{t}^{0}|\underset{\epsilon\to
0}{\longrightarrow}0$ a.s. (see [23] Corollary 1.8, page 226), and
$\sup_{[0,T]}|L_{t}^{a_{\epsilon}\mathfrak{s}(v)}-L_{t}^{0}|\leq
2\sup_{[0,T]\times\mathbb{R}}L_{t}^{x}<\infty$ a.s.. We can conclude by
dominated convergence since $\varphi\in\mathrm{\bm{L}}^{1}(\mu)$ by
assumption. ∎

###  3.3 The Lévy regime

In this section, we first show two lemmas, that will enable us to conclude.
The first lemma says that if in Lemma 18 we choose carefully $a_{\epsilon}$,
the time change $A_{t}^{\epsilon}$ converges to the local time of the Brownian
motion at 0. It also recovers a weak version of the ergodic theorem and
prepares some useful results for the diffusive case.

######  Lemma 20.

Let $(W_{t})_{t\geq 0}$ be a Brownian motion and $(L_{t}^{0})_{t\geq 0}$ its
local time at 0. For all $\epsilon>0$, we consider the processes
$(A_{t}^{\epsilon})_{t\geq 0}$, $(\tau_{t}^{\epsilon})_{t\geq 0}$ and
$(X_{t}^{\epsilon})_{t\geq 0}$ from Lemma 18 with the choice
$a_{\epsilon}=\epsilon/\kappa$.

  1. (i)

We have a.s., for all $T>0$,

$\sup_{[0,T]}\left|A_{t}^{\epsilon}-L_{t}^{0}\right|\underset{\epsilon\to
0}{\longrightarrow}0.$

  2. (ii)

For all $t>0$, a.s., $\tau_{t}^{\epsilon}\underset{\epsilon\to
0}{\longrightarrow}\tau_{t}$, the generalized inverse of $(L_{t}^{0})_{t\geq
0}$.

  3. (iii)

Suppose Assumption 4. For any slowly varying function $\gamma$, we have a.s.,
for all $T>0$,

$\sup_{[0,T]}\gamma(1/\epsilon)\left|A_{t}^{\epsilon}-L_{t}^{0}\right|\underset{\epsilon\to
0}{\longrightarrow}0.$

  4. (iv)

If $\alpha=2$ and
$\rho=\int_{1}^{\infty}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x=\infty$,
we have a.s., for all $T>0$,

$\sup_{[0,T]}\left|T_{t}^{\epsilon}-\sigma_{2}^{2}L_{t}^{0}\right|\underset{\epsilon\to
0}{\longrightarrow}0\quad\text{where}\quad
T_{t}^{\epsilon}=\kappa^{2}|\epsilon\rho_{\epsilon}|^{-1}\int_{0}^{t}\chi(W_{s}/a_{\epsilon})\mathrm{d}s,$

where $\chi=[(g\circ\mathfrak{s}^{-1})^{\prime}]^{2}$ with $g$ defined by (5)
and
$\rho_{\epsilon}=\int_{1}^{1/\epsilon}\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}\mathrm{d}x$.

  5. (v)

For all $\varphi\in\mathrm{\bm{L}}^{1}(\mu)$, for all $t>0$, we have a.s.

$\int_{0}^{t}\varphi(X_{s}^{\epsilon})\mathrm{d}s\underset{\epsilon\to
0}{\longrightarrow}\mu(\varphi)t.$

###### Proof.

Point (i). The first point is just the application of Proposition 19 with
$\varphi=1$.

Point (ii). The second point follows immediately from the first and Lemma 15,
since for all $t\geq 0$, as recalled in Subsection 2.2,
$\mathbb{P}(\tau_{t-}<\tau_{t})=0$.

Point (iii). Using the same scheme as in the proof of Proposition 19, we have

$\sup_{[0,T]}\gamma(1/\epsilon)\left|A_{t}^{\epsilon}-L_{t}^{0}\right|=\sup_{[0,T]}\left|\kappa\gamma(1/\epsilon)\int_{\mathbb{R}}\frac{L_{t}^{\epsilon
y/\kappa}-L_{t}^{0}}{\psi^{2}(y)}\mathrm{d}y\right|\leq\int_{\mathbb{R}}\gamma(1/\epsilon)\sup_{[0,T]}\left|L_{t}^{\epsilon\mathfrak{s}(v)/\kappa}-L_{t}^{0}\right|\mu(\mathrm{d}v).$

Now we use (8) with some $\theta\in(0,\frac{1}{2}\wedge\lambda)$, where
$\lambda$ refers to the constant of Assumption 4, so that
$|\mathfrak{s}|^{\lambda}\in\mathrm{\bm{L}}^{1}(\mu)$. We thus have
$\gamma(1/\epsilon)\sup_{[0,T]}|L_{t}^{\epsilon\mathfrak{s}(v)/\kappa}-L_{t}^{0}|\leq
C|\mathfrak{s}(v)|^{\theta}\gamma(1/\epsilon)\epsilon^{\theta}$, for some
random $C$. Since $\gamma$ is slowly varying, we conclude that
$\gamma(1/\epsilon)\sup_{[0,T]}|L_{t}^{\epsilon\mathfrak{s}(v)/\kappa}-L_{t}^{0}|\longrightarrow
0$ as $\epsilon\to 0$, for each fixed $v$. Moreover

$\sup_{\epsilon\in(0,1)}\gamma(1/\epsilon)\sup_{[0,T]}\left|L_{t}^{\epsilon\mathfrak{s}(v)/\kappa}-L_{t}^{0}\right|\leq
C\left(\sup_{\epsilon\in(0,1)}\gamma(1/\epsilon)\epsilon^{\theta}\right)|\mathfrak{s}(v)|^{\theta}$

and we can conclude by dominated convergence since
$|\mathfrak{s}|^{\theta}\in\mathrm{\bm{L}}^{1}(\mu)$.

Point (iv). Step 0 : We first estimate asymptotically the function $\chi$. Let
us recall from (17) that

$(g\circ\mathfrak{s}^{-1})^{\prime}(x)=2\int_{x}^{\infty}\phi(v)\mathrm{d}v=-2\int_{-\infty}^{x}\phi(v)\mathrm{d}v.$

Let us define the function $h$ on $(0,\infty)$:

$h(x)=\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{2}.$

By (12), we have
$|x|^{3/2}\ell(|x|)\phi(x)\underset{x\to\pm\infty}{\longrightarrow}f_{\pm}$
and thus

$\left(\int_{x}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{-1}\int_{x}^{\infty}\phi(v)dv\underset{x\to\infty}{\longrightarrow}f_{+}$

and

$\left(\int_{|x|}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v)}\right)^{-1}\int_{-\infty}^{x}\phi(v)dv\underset{x\to-\infty}{\longrightarrow}f_{-}.$

All in all, recalling that $\chi=[(g\circ\mathfrak{s}^{-1})^{\prime}]^{2}$, we
have
$[h(|x|)]^{-1}\chi(x)\underset{x\to\pm\infty}{\longrightarrow}4f_{\pm}^{2}$
from which

$\int_{-x}^{x}\chi(z)\mathrm{d}z\underset{x\to\infty}{\sim}4(f_{+}^{2}+f_{-}^{2})\int_{1}^{x}h(v)\mathrm{d}v.$
(18)

We used that $\rho=\int_{1}^{\infty}h(v)\mathrm{d}v=\infty$ by assumption.
Moreover there exists $A>0$ such that for all $x$, we have $\chi(x)\leq
Ah(|x|)$. Now we write for $\delta>0$ fixed,
$T_{t}^{\epsilon}=D_{t}^{\epsilon,\delta}+E_{t}^{\epsilon,\delta}$ where

$D_{t}^{\epsilon,\delta}=\int_{0}^{t}\kappa^{2}|\epsilon\rho_{\epsilon}|^{-1}\chi(\kappa
W_{s}/\epsilon)\bm{1}_{\\{|W_{s}|>\delta\\}}\mathrm{d}s\>\>\>\text{ and
}\>\>\>E_{t}^{\epsilon,\delta}=\int_{0}^{t}\kappa^{2}|\epsilon\rho_{\epsilon}|^{-1}\chi(\kappa
W_{s}/\epsilon)\bm{1}_{\\{|W_{s}|\leq\delta\\}}\mathrm{d}s.$

Step 1 : We show that $\sup_{[0,T]}|D_{t}^{\epsilon,\delta}|\to 0$ as
$\epsilon\to 0$. We have $\chi(\kappa W_{s}/\epsilon)\leq
Ah(\kappa|W_{s}|/\epsilon)$. Now notice that in this case, we have a.s.,

$\displaystyle\bm{1}_{\\{|W_{s}|>\delta\\}}h(\kappa|W_{s}|/\epsilon)$
$\displaystyle=\bm{1}_{\\{|W_{s}|>\delta\\}}\left(\int_{\kappa|W_{s}|/\epsilon}^{\infty}\frac{\mathrm{d}u}{u^{3/2}\ell(u)}\right)^{2}$
$\displaystyle=\bm{1}_{\\{|W_{s}|>\delta\\}}\frac{\epsilon}{\kappa|W_{s}|}\left(\int_{1}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(\kappa|W_{s}|v/\epsilon)}\right)^{2}$
$\displaystyle\leq\bm{1}_{\\{|W_{s}|>\delta\\}}\frac{\epsilon}{\kappa\delta}\left(\int_{1}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(\kappa|W_{s}|v/\epsilon)}\right)^{2}.$

We know from Lemma 9-(i) (with $a=\delta$ and $b=\sup_{[0,T]}|W_{s}|$) that
the following uniform convergence holds a.s.

$\sup_{[0,T]}\bm{1}_{\\{|W_{s}|>\delta\\}}\left|\frac{\ell(v/\epsilon)}{\ell(\kappa|W_{s}|v/\epsilon)}-1\right|\underset{\epsilon\to
0}{\longrightarrow}0.$

Hence there exists a random $M>0$ such that for any $s\in[0,T]$ and
$\epsilon\in(0,1)$, $|W_{s}|>\delta$ implies that
$1/\ell(\kappa|W_{s}|v/\epsilon)\leq M/\ell(v/\epsilon)$. Consequently,

$\sup_{[0,T]}\bm{1}_{\\{|W_{s}|>\delta\\}}|\chi(\kappa
W_{s}/\epsilon)|\leq\sup_{[0,T]}\bm{1}_{\\{|W_{s}|>\delta\\}}Ah(\kappa|W_{s}|/\epsilon)\leq\frac{\epsilon
AM^{2}}{\kappa\delta}\left(\int_{1}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v/\epsilon)}\right)^{2}.$

We know from Potter’s bound (Lemma 9-(ii)) that there exists $x_{0}>1$ such
that for all $x,y\geq x_{0}$,

$\frac{\ell(y)}{\ell(x)}\leq
2\left(\Big{|}\frac{y}{x}\Big{|}^{1/4}\vee\Big{|}\frac{x}{y}\Big{|}^{1/4}\right).$

Hence, if $\epsilon\in(0,1/x_{0})$, we have

$\left(\int_{1}^{\infty}\frac{\mathrm{d}v}{v^{3/2}\ell(v/\epsilon)}\right)^{2}\leq\frac{4}{\ell^{2}(1/\epsilon)}\left(\int_{1}^{\infty}\frac{|v|^{1/4}\vee|v|^{-1/4}\mathrm{d}v}{v^{3/2}}\right)^{2}.$

All in all, there exists a random constant that we name $M$ again, such that

$\sup_{[0,T]}|D_{t}^{\epsilon,\delta}|\leq\frac{MT}{\delta\ell^{2}(1/\epsilon)\rho_{\epsilon}}.$

It holds that $\ell^{2}(1/\epsilon)\rho_{\epsilon}\longrightarrow\infty$, see
the end of the proof of Lemma 9-(iii) and use that
$\rho_{\epsilon}=M(1/\epsilon)$, and thus for each $\delta>0$,
$\sup_{[0,T]}|D_{t}^{\epsilon,\delta}|\longrightarrow 0$ as $\epsilon\to 0$.

Step 2 : We concentrate on $E_{t}^{\epsilon,\delta}$. Using the occupation
times formula, we have

$E_{t}^{\epsilon,\delta}=\left(\int_{-\delta}^{\delta}\frac{\kappa^{2}\chi(\kappa
x/\epsilon)\mathrm{d}x}{|\epsilon\rho_{\epsilon}|}\right)L_{t}^{0}+\int_{-\delta}^{\delta}\frac{\kappa^{2}\chi(\kappa
x/\epsilon)(L_{t}^{x}-L_{t}^{0})\mathrm{d}x}{|\epsilon\rho_{\epsilon}|}=r_{\epsilon,\delta}L_{t}^{0}+R_{t}^{\epsilon,\delta},$

the last equality standing for a definition. We have, using a substitution and
(18),

$r_{\epsilon,\delta}=\frac{\kappa}{\rho_{\epsilon}}\int_{-\kappa\delta/\epsilon}^{\kappa\delta/\epsilon}\chi(z)\mathrm{d}z\underset{\epsilon\to
0}{\sim}\frac{4\kappa(f_{+}^{2}+f_{-}^{2})}{\rho_{\epsilon}}\int_{1}^{\kappa\delta/\epsilon}h(x)\mathrm{d}x\underset{\epsilon\to
0}{\sim}4\kappa(f_{+}^{2}+f_{-}^{2})=\sigma_{2}^{2},$

where we used that $\rho_{\epsilon}=\int_{1}^{1/\epsilon}h(x)\mathrm{d}x$ and
that the function $M(x)=\int_{1}^{x}h(v)\mathrm{d}v$ is slowly varying by
Lemma 9-(iii).

Step 3 : We conclude. Recalling that
$T_{t}^{\epsilon}=r_{\epsilon,\delta}L_{t}^{0}+R_{t}^{\epsilon,\delta}+D_{t}^{\epsilon,\delta}$,
we proved that a.s., for all $T>0$ and all $\delta>0$,

$\limsup_{\epsilon\to
0}\sup_{[0,T]}\left|T_{t}^{\epsilon}-\sigma_{2}^{2}L_{t}^{0}\right|\leq\limsup_{\epsilon\to
0}\sup_{[0,T]}|R_{t}^{\epsilon,\delta}|.$

But

$\sup_{[0,T]}|R_{t}^{\epsilon,\delta}|\leq
r_{\epsilon,\delta}\sup_{[0,T]\times[-\delta,\delta]}|L_{t}^{x}-L_{t}^{0}|,$

which implies

$\limsup_{\epsilon\to
0}\sup_{[0,T]}\left|T_{t}^{\epsilon}-\sigma_{2}^{2}L_{t}^{0}\right|\leq\sigma_{2}^{2}\sup_{[0,T]\times[-\delta,\delta]}|L_{t}^{x}-L_{t}^{0}|.$

Now, we let $\delta$ to $0$ which completes the proof.

Point (v). Using the definition of $X_{t}^{\epsilon}$, and making the
substitution $u=\tau_{s}^{\epsilon}\Leftrightarrow s=A_{u}^{\epsilon}$, we
have

$\int_{0}^{t}\varphi(X_{s}^{\epsilon})\mathrm{d}s=\epsilon
a_{\epsilon}^{-2}\int_{0}^{\tau_{t}^{\epsilon}}\frac{\varphi\circ\mathfrak{s}^{-1}(W_{s}/a_{\epsilon})}{\psi^{2}(W_{s}/a_{\epsilon})}\mathrm{d}s=\kappa
a_{\epsilon}^{-1}\int_{0}^{\tau_{t}^{\epsilon}}\frac{\varphi\circ\mathfrak{s}^{-1}(W_{s}/a_{\epsilon})}{\psi^{2}(W_{s}/a_{\epsilon})}\mathrm{d}s.$

Let $T=\sup_{\epsilon\in(0,1)}\tau_{t}^{\epsilon}$ which is a.s. finite from
the second point. We have

$\begin{split}\left|\int_{0}^{t}\varphi(X_{s}^{\epsilon})\mathrm{d}s-\mu(\varphi)t\right|\leq&\sup_{[0,T]}\left|\kappa
a_{\epsilon}^{-1}\int_{0}^{u}\frac{\varphi\circ\mathfrak{s}^{-1}(W_{s}/a_{\epsilon})}{\psi^{2}(W_{s}/a_{\epsilon})}\mathrm{d}s-\mu(\varphi)L_{u}^{0}\right|\\\
&+\mu(\varphi)\left|L_{\tau_{t}^{\epsilon}}^{0}-t\right|.\end{split}$

The first term on the right hand side goes to $0$ by Proposition 19 and the
second goes to $0$ since $\tau^{\epsilon}_{t}\to\tau_{t}$ by Point 2,
$L_{\tau_{t}}^{0}=t$ a.s., and $(L_{t}^{0})_{t\geq 0}$ is continuous. ∎

Let us state two immediate consequences of the previous lemma.

######  Remark 21.

Point (5) of the preceding Lemma recovers a weak version of the ergodic
theorem for the process $(X_{t})_{t\geq 0}$. Although it is straightforward
since the process is Harris recurrent, this provides a self-contained proof of
this weak version, which we only need. For all
$\varphi\in\mathrm{\bm{L}}^{1}(\mu)$ and all $t\geq 0$,
$\epsilon\int_{0}^{t/\epsilon}\varphi(X_{s})\mathrm{d}s\overset{d}{=}\int_{0}^{t}\varphi(X_{s}^{\epsilon})\mathrm{d}s$,
hence for all $t\geq 0$

$\epsilon\int_{0}^{t/\epsilon}\varphi(X_{s})\mathrm{d}s\overset{\mathbb{P}}{\longrightarrow}\mu(\varphi)t.$

######  Remark 22.

Point (4) prepares the critical diffusive regime. Using arguments similar to
those in the proof of point (5), we easily see that if $\alpha=2$ and
$\rho=\infty$, then for every $t\geq 0$, a.s.,
$T_{\tau_{t}^{\epsilon}}^{\epsilon}\underset{\epsilon\to
0}{\longrightarrow}\sigma_{2}^{2}t$. Now one can also see that as
$|\epsilon/\rho_{\epsilon}|\int_{0}^{t/\epsilon}[g^{\prime}(X_{s})\sigma(X_{s})]^{2}\mathrm{d}s\overset{d}{=}T_{\tau_{t}^{\epsilon}}^{\epsilon}$,
we have

$|\epsilon/\rho_{\epsilon}|\int_{0}^{t/\epsilon}[g^{\prime}(X_{s})\sigma(X_{s})]^{2}\mathrm{d}s\overset{\mathbb{P}}{\longrightarrow}\sigma_{2}^{2}t.$

The next result is the last preliminary before proving the main result in the
Lévy case.

######  Lemma 23.

Let $(W_{t})_{t\geq 0}$ be a Brownian motion, and $(L_{t}^{0})_{t\geq 0}$ its
local time in 0. Let also $(K_{t})_{t\geq 0}$ be the process defined in
section 2.3, with $(a,b)=(f_{+},f_{-})$. For all $\epsilon>0$, we consider the
process $(H_{t}^{\epsilon})_{t\geq 0}$ from Lemma 18 with the choice
$a_{\epsilon}=\epsilon/\kappa$.

  1. (i)

If $\alpha\neq 1$, then a.s., for all $T>0$,

$\sup_{[0,T]}\left|\epsilon^{1/\alpha}\ell(1/\epsilon)H_{t}^{\epsilon}-\kappa^{1/\alpha}K_{t}\right|\underset{\epsilon\to
0}{\longrightarrow}0.$

  2. (ii)

If $\alpha=1$, there exists $(\xi_{\epsilon})_{\epsilon>0}$ such that, a.s.,
for all $T>0$,

$\sup_{[0,T]}\left|\epsilon\ell(1/\epsilon)H_{t}^{\epsilon}-\xi_{\epsilon}L_{t}^{0}-\kappa
K_{t}\right|\underset{\epsilon\to 0}{\longrightarrow}0,$

where $\xi_{\epsilon}\underset{\epsilon\to
0}{\sim}-\kappa(f_{+}+f_{-})\ell(1/\epsilon)\int_{1/\epsilon}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}$
if $f\in\mathrm{\bm{L}}^{1}(\mu)$ and $\xi_{\epsilon}\underset{\epsilon\to
0}{\sim}\kappa(f_{+}+f_{-})\ell(1/\epsilon)\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)}$
otherwise.

###### Proof.

Technical estimates. We first show some technical estimates that we will use
in this proof.

(i) For all $x\in\mathbb{R}^{*}$, it holds (and this is the case for all
$\alpha\in(0,2)$) that

$\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)\underset{\epsilon\to
0}{\longrightarrow}\kappa^{1/\alpha}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}.$
(19)

Indeed we have by (12), since $\ell$ is slowly varying and since
$(a,b)=(f_{+},f_{-})$,

$\forall
x>0,\quad\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)\underset{\epsilon\to
0}{\longrightarrow}\kappa^{1/\alpha}ax^{1/\alpha-2},$ $\forall
x<0,\quad\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)\underset{\epsilon\to
0}{\longrightarrow}\kappa^{1/\alpha}b|x|^{1/\alpha-2}.$

(ii) For all $\delta>0$, there exist $A>0$ and $\epsilon_{0}>0$ (and this is
the case for all $\alpha\in(0,2)$) such that for all $x\in\mathbb{R}$ and all
$\epsilon\in(0,\epsilon_{0})$,

$\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa x/\epsilon\right)\leq
A(|x|^{1/\alpha-2}+1)(|x|^{\delta}+|x|^{-\delta})$ (20)

Indeed, we start using (12) again and that $\phi$ and $\ell$ are locally
bounded: there exists $A>0$ such that for all $x\in\mathbb{R}$,
$\ell(|x|)|\phi(x)|\leq A(|x|^{1/\alpha-2}+1)$. Hence we have

$\epsilon^{1/\alpha-2}\ell(1/\epsilon)|\phi\left(\kappa x/\epsilon\right)|\leq
A(|x|^{1/\alpha-2}+1)\frac{\ell(1/\epsilon)}{\ell(\kappa|x|/\epsilon)},$

the value of $A$ being allowed to change. We know from Potter’s bound (Lemma
9-(ii)) that there exists $x_{0}$ such that for all $x,y\geq x_{0}$,

$\frac{\ell(y)}{\ell(x)}\leq
2\left(\Big{|}\frac{y}{x}\Big{|}^{\delta}\vee\Big{|}\frac{x}{y}\Big{|}^{\delta}\right).$

Thus, for $\epsilon\in(0,1/x_{0})$ we have

$\epsilon^{1/\alpha-2}\ell(1/\epsilon)|\phi\left(\kappa
x/\epsilon\right)|\bm{1}_{\\{\kappa|x|\geq\epsilon x_{0}\\}}\leq
A(|x|^{1/\alpha-2}+1)(|x|^{\delta}\vee|x|^{-\delta}).$

The function $\ell$ is strictly positive and bounded from below on every
compact set and thus there exists $\ell_{0}>0$ such that for all
$x\in[0,x_{0}]$, $\ell(x)\geq\ell_{0}$. Now if $\kappa|x|<\epsilon x_{0}$, we
have $\ell(\kappa|x|/\epsilon)\geq\ell_{0}$ and $1\leq(\kappa|x|/\epsilon
x_{0})^{-\delta}$. Hence

$\epsilon^{1/\alpha-2}\ell(1/\epsilon)|\phi\left(\kappa
x/\epsilon\right)|\bm{1}_{\\{\kappa|x|<\epsilon x_{0}\\}}\leq
A(|x|^{1/\alpha-2}+1)|x|^{-\delta}\sup_{\epsilon\in(0,1)}\epsilon^{\delta}\ell(1/\epsilon).$

We proved (20).

Point (i) when $\bm{\alpha\in(0,1)}$. Let us remind that
$K_{t}=\int_{0}^{t}\mathrm{sgn}_{a,b}(W_{s})|W_{s}|^{1/\alpha-2}ds$. We have
$\epsilon^{1/\alpha}\ell(1/\epsilon)H_{t}^{\epsilon}=\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\int_{0}^{t}\phi\left(\kappa
W_{s}/\epsilon\right)\mathrm{d}s$ and

$\sup_{[0,T]}\left|\epsilon^{1/\alpha}\ell(1/\epsilon)H_{t}^{\epsilon}-\kappa^{1/\alpha}K_{t}\right|\leq\int_{0}^{T}\left|\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
W_{s}/\epsilon\right)-\kappa^{1/\alpha}\mathrm{sgn}_{a,b}(W_{s})|W_{s}|^{1/\alpha-2}\right|\mathrm{d}s$

which tends to 0 a.s. by dominated convergence. Indeed, since
$\\{t\in[0,T],W_{t}=0\\}$ is a.s. Lebesgue-null, we have by (19) that a.s.,
for a.e. $s\in[0,T]$,

$|\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
W_{s}/\epsilon\right)-\kappa^{1/\alpha}\mathrm{sgn}_{a,b}(W_{s})|W_{s}|^{1/\alpha-2}|\to
0$

Next we can dominate $\epsilon^{1/\alpha-2}\ell(1/\epsilon)|\phi(\kappa
W_{s}/\epsilon)|$ using (20) by

$A(|W_{s}|^{1/\alpha-2}+1)(|W_{s}|^{\delta}+|W_{s}|^{-\delta})$

which is a.s. integrable on $[0,T]$ if $\delta>0$ is small enough so that
$1/\alpha-2-\delta>-1$.

Point (i) when $\bm{\alpha\in(1,2)}$. We notice that
$\mu(f)=\kappa\int_{\mathbb{R}}\phi(z)\mathrm{d}z$, which equals $0$ by
Assumption 3. Indeed,

$\int_{\mathbb{R}}\phi(z)\mathrm{d}z=\int_{\mathbb{R}}\frac{f(\mathfrak{s}^{-1}(z))}{\sigma^{2}(\mathfrak{s}^{-1}(z))\mathfrak{s}^{\prime}(\mathfrak{s}^{-1}(z))^{2}}\mathrm{d}z=\int_{\mathbb{R}}\frac{f(v)}{\sigma^{2}(v)\mathfrak{s}^{\prime}(v)}\mathrm{d}v=\frac{\mu(f)}{\kappa}.$
(21)

We can now use the occupation times formula to write

$\begin{split}\epsilon^{1/\alpha}\ell(1/\epsilon)H_{t}^{\epsilon}&=\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\int_{0}^{t}\phi\left(\kappa
W_{s}/\epsilon\right)\mathrm{d}s\\\
&=\int_{\mathbb{R}}\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)L_{t}^{x}\mathrm{d}x\\\
&=\int_{\mathbb{R}}\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)(L_{t}^{x}-L_{t}^{0})\mathrm{d}x,\end{split}$

since $\int_{\mathbb{R}}\phi(z)\mathrm{d}z=0$ as noted above. Thus, recalling
that
$K_{t}=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}(L_{t}^{x}-L_{t}^{0})\mathrm{d}x$
in this case, we have

$\displaystyle\sup_{[0,T]}\left|\epsilon^{1/\alpha}\ell(1/\epsilon)H_{t}^{\epsilon}-\kappa
K_{t}\right|$ $\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}}\left|\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)-\kappa^{1/\alpha}\mathrm{sgn}_{a,b}(x)|x|^{1/\alpha-2}\right|\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|\mathrm{d}x.$

We conclude by dominated convergence. First, by (19), the integrand converges
to $0$ for each $x\in\mathbb{R}^{*}$.

We next use (20) to dominate
$R_{\epsilon}(x)=|\kappa^{2}\epsilon^{1/\alpha-2}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)|\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|$ by

$A(|x|^{1/\alpha-2}+1)(|x|^{\delta}+|x|^{-\delta})|\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|\leq
M(|x|^{1/\alpha-2}+1)(|x|^{\delta}+|x|^{-\delta})|\min(1,|x|^{\theta}),$

for some random constant $M>0$ and $\theta\in(0,1/2)$ fixed. We used (8) and
the fact that
$\sup_{[0,T]\times\mathbb{R}}\left|L_{t}^{x}-L_{t}^{0}\right|<\infty$ a.s.
This bound is integrable on $\mathbb{R}$ if we choose $\delta>0$ and
$\theta\in(0,1/2)$ such that $1/\alpha-2+\delta<-1$ and
$1/\alpha-2+\theta-\delta>-1$. This is possible because $-3/2<1/\alpha-2<-1$.

Point (ii): Convergence. Setting
$\xi_{\epsilon}=\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\int_{-1}^{1}\phi\left(\kappa
x/\epsilon\right)\mathrm{d}x$ and using the occupation times formula, we have

$\epsilon\ell(1/\epsilon)H_{t}^{\epsilon}-\xi_{\epsilon}L_{t}^{0}=\int_{|x|\geq
1}\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)L_{t}^{x}\mathrm{d}x+\int_{|x|\leq
1}\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\phi\left(\kappa
x/\epsilon\right)(L_{t}^{x}-L_{t}^{0})\mathrm{d}x$

Then, remembering that
$K_{t}=\int_{\mathbb{R}}\mathrm{sgn}_{a,b}(x)|x|^{-1}(L_{t}^{x}-L_{t}^{0}\bm{1}_{\\{|x|\leq
1\\}})\mathrm{d}x$, we write

$\displaystyle\sup_{[0,T]}\left|\epsilon
H_{t}^{\epsilon}-\xi_{\epsilon}L_{t}^{0}-\kappa K_{t}\right|$
$\displaystyle\leq$ $\displaystyle\int_{|x|\leq
1}\left|\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)-\kappa\mathrm{sgn}_{a,b}(x)|x|^{-1}\right|\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|\mathrm{d}x$
$\displaystyle+\int_{|x|>1}\left|\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)-\kappa\mathrm{sgn}_{a,b}(x)|x|^{-1}\right|L_{T}^{x}\mathrm{d}x,$

Again, we conclude by dominated convergence. By (19), the integrand in both
terms converges to $0$ for each $x\in\mathbb{R}^{*}$.

We first use (20) to dominate
$R_{\epsilon}(x)=|\kappa^{2}\epsilon^{-1}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)|\sup_{[0,T]}\left|L_{t}^{x}-L_{t}^{0}\right|$, for all $|x|\leq
1$, by

$M(|x|^{-1}+1)(|x|^{\delta}+|x|^{-\delta})|\times|x|^{\theta},$

for some random constant $M>0$ and $\theta\in(0,1/2)$ fixed. We used (8) and
the fact that
$\sup_{[0,T]\times\mathbb{R}}\left|L_{t}^{x}-L_{t}^{0}\right|<\infty$ a.s.
This bound is integrable on $[-1,1]$ if we choose $\delta>0$ and
$\theta\in(0,1/2)$ such that $-1+\theta-\delta>-1$, which is possible:
$\theta=1/4$, $\delta=1/8$.

Now we dominate the quantity
$P_{\epsilon}(x)=|\epsilon^{-1}\ell(1/\epsilon)\phi(\kappa
x/\epsilon)|L_{T}^{x}$ for all $|x|>1$. The second integral is actually
supported by the compact set $\\{|x|\in[1,S_{T}]\\}$ where
$S_{T}=\sup_{[0,T]}|W_{s}|$ since $L_{T}^{x}=0$ as soon as $|x|>S_{T}$. Using
(20) with $\delta=1/2$,

$P_{\epsilon}(x)\leq A(|x|^{-1}+1)(|x|^{1/2}+|x|^{-1/2})L_{T}^{x},$

which is a.s. bounded and thus integrable on $\\{|x|\in[1,S_{T}]\\}$.

Point (ii): Asymptotics of $\xi_{\epsilon}$. First remember from Proposition
16 that $f\in\mathrm{\bm{L}}^{1}(\mu)$ if and only if
$\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}<\infty$.

Case 1: Let us first suppose
$\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}=\infty$ and show that
$\xi_{\epsilon}\underset{\epsilon\to
0}{\sim}\kappa(f_{+}+f_{-})\ell(1/\epsilon)\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)}$.
For $\epsilon\in(0,\kappa)$

$\xi_{\epsilon}=\kappa\ell(1/\epsilon)\int_{-\kappa/\epsilon}^{\kappa/\epsilon}\phi(x)\mathrm{d}x=\kappa\ell(1/\epsilon)\int_{-\kappa/\epsilon}^{-1}\phi(x)\mathrm{d}x+\kappa\ell(1/\epsilon)\int_{-1}^{1}\phi(x)\mathrm{d}x+\kappa\int_{1}^{\kappa/\epsilon}\phi(z)\mathrm{d}z.$
(22)

The middle term is
$o(\ell(1/\epsilon)\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)})$ and
will not contribute. Assume e.g. $f_{+}\neq 0$, then
$\phi(x)\underset{x\to\infty}{\sim}f_{+}/[x\ell(x)]$ and it comes that

$\kappa\int_{1}^{\kappa/\epsilon}\phi(x)\mathrm{d}x\underset{\epsilon\to
0}{\sim}\kappa
f_{+}\int_{1}^{\kappa/\epsilon}\frac{\mathrm{d}x}{x\ell(x)}\underset{\epsilon\to
0}{\sim}\kappa f_{+}\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)},$

where we used the fact that $L(x)=\int_{1}^{x}\frac{\mathrm{d}v}{v\ell(v)}$ is
a slowly varying function by Lemma 9-(iii). If $f_{-}\neq 0$, then
$\kappa\int_{-\kappa/\epsilon}^{-1}\phi(x)\mathrm{d}x\underset{\epsilon\to
0}{\sim}\kappa f_{-}\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)}$ as
previously. If next $f_{-}=0$, we have
$\phi(x)\underset{x\to-\infty}{=}o(|x\ell(|x|)|^{-1})$, then
$\kappa\int_{-\kappa/\epsilon}^{-1}\phi(x)\mathrm{d}x\underset{\epsilon\to
0}{=}o(\int_{1}^{1/\epsilon}\frac{\mathrm{d}x}{x\ell(x)})$ because
$\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}=\infty$.

Now, if $f_{+}=0$, then necessarily $f_{-}\neq 0$ and we can proceed as
previously.

Case 2: $\int_{1}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}<\infty$. We must show
$\xi_{\epsilon}\underset{\epsilon\to
0}{\sim}-\kappa(f_{+}+f_{-})\ell(1/\epsilon)\int_{1/\epsilon}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}$.
In this case, $f\in\mathrm{\bm{L}}^{1}(\mu)$ and we impose $\mu(f)=0$ so that
$\int_{\mathbb{R}}\phi(x)\mathrm{d}x=0$ by (21) and thus

$\xi_{\epsilon}=-\kappa\ell(1/\epsilon)\int_{-\infty}^{-\kappa/\epsilon}\phi(x)\mathrm{d}x-\kappa\ell(1/\epsilon)\int_{\kappa/\epsilon}^{\infty}\phi(x)\mathrm{d}x.$

Let us treat the case $f_{+}\neq 0$ and $f_{-}=0$, the other cases being
treated similarly. Since $f_{+}\neq 0$, we can integrate the equivalence
$\phi(x)\underset{x\to\infty}{\sim}f_{+}/[x\ell(x)]$ and get that
$\int_{\kappa/\epsilon}^{\infty}\phi(x)\mathrm{d}x\underset{x\to\infty}{\sim}f_{+}\int_{1/\epsilon}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}$,
where we used that $N(x)=\int_{x}^{\infty}\frac{\mathrm{d}x}{x\ell(x)}$ is
slowly varying by Lemma 9. Next
$\int_{-\infty}^{-\kappa/\epsilon}\phi(x)\mathrm{d}x\underset{x\to-\infty}{=}o(\int_{1/\epsilon}^{\infty}\frac{\mathrm{d}x}{x\ell(x)})$
since $\phi(x)\underset{x\to-\infty}{=}o(|x\ell(|x|)|^{-1})$. ∎

We are now ready to give the proof of Theorem 5-(iii)-(iv).

###### Proof of Theorem 5-(iii)-(iv)..

Consider a Brownian motion $(W_{t})_{t\geq 0}$, its local time
$(L_{t}^{0})_{t\geq 0}$ at $0$ and $\tau_{t}=\inf\\{u\geq 0,L_{u}^{0}>t\\}$.
Consider also the process $(K_{t})_{t\geq 0}$ defined in Subsection 2.3 for
$(a,b)=(f_{+},f_{-})$.

For each $\epsilon>0$, consider the processes $(A_{t}^{\epsilon})_{t\geq}$,
$(\tau_{t}^{\epsilon})_{t\geq}$ and $(H_{t}^{\epsilon})_{t\geq}$ from Lemma 18
with the choice $a_{\epsilon}=\epsilon/\kappa$. We know from this Lemma that

$\left(\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{d}{=}\left(H_{\tau_{t}^{\epsilon}}^{\epsilon}\right)_{t\geq 0}$
(23)

And we know from Lemma 20-(ii) that
$\tau_{t}^{\epsilon}\longrightarrow\tau_{t}$ a.s. for each $t$ fixed. Thus
$T=\sup_{\epsilon\in(0,1)}\tau_{t}^{\epsilon}$ is a.s. finite.

Point (iii). We aim at showing that
$(\epsilon^{1/\alpha}\ell(1/\epsilon)\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s)_{t\geq
0}\overset{f.d.}{\longrightarrow}(\sigma_{\alpha}S_{t}^{(\alpha)})_{t\geq 0}$.
By (23) it is enough to show that a.s.,

$\Delta_{t}(\epsilon)=\left|\epsilon^{1/\alpha}\ell(1/\epsilon)H_{\tau_{t}^{\epsilon}}^{\epsilon}-\kappa^{1/\alpha}K_{\tau_{t}}\right|\underset{\epsilon\to
0}{\longrightarrow}0,$

for each fixed $t\geq 0$. Indeed, setting
$S_{t}^{(\alpha)}=\sigma_{\alpha}^{-1}\kappa^{1/\alpha}K_{\tau_{t}}$, Lemmas
11 and 12 tell us that $(S_{t}^{(\alpha)})_{t\geq 0}$ is a stable process such
that

$\mathbb{E}[\exp(i\xi
S_{t}^{(\alpha)})]=\exp\left(-c_{\alpha,a,b}t|\sigma_{\alpha}^{-1}\kappa^{1/\alpha}\xi|^{\alpha}\left(1-i\beta_{\alpha,a,b}\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)\right)\right)=\exp(-t|\xi|^{\alpha}z_{\alpha}(\xi)),$

where
$z_{\alpha}(\xi))=1-i\beta_{\alpha,a,b}\tan\left(\frac{\alpha\pi}{2}\right)\mathrm{sgn}(\xi)$
as in the statement.

We have

$\Delta_{t}(\epsilon)\leq\sup_{[0,T]}\left|\epsilon^{1/\alpha}\ell(1/\epsilon)H_{s}^{\epsilon}-\kappa^{1/\alpha}K_{s}\right|+\kappa^{1/\alpha}\left|K_{\tau_{t}^{\epsilon}}-K_{\tau_{t}}\right|.$

The first term goes to $0$ from Lemma 23-(i) and the second by continuity of
$(K_{t})_{t\geq 0}$.

Point (iv). We want to show that with $(\xi_{\epsilon})_{\epsilon>0}$ defined
in Lemma 23-2,

$\left(\epsilon\ell(1/\epsilon)\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s-\xi_{\epsilon}t\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{1}S_{t}^{(1)}\right)_{t\geq
0}.$

Once again, by (23), it is enough to show that

$\Delta_{t}(\epsilon)=\left|\epsilon\ell(1/\epsilon)H_{\tau_{t}^{\epsilon}}^{\epsilon}-\xi_{\epsilon}t-\kappa
K_{\tau_{t}}\right|\underset{\epsilon\to 0}{\longrightarrow}0,$

for each fixed $t\geq 0$. Indeed, setting $S_{t}^{(1)}=\sigma_{1}^{-1}\kappa
K_{\tau_{t}}$, Lemma 13 tells us that $(S_{t}^{(1)})_{t\geq 0}$ is a stable
process such that

$\displaystyle\mathbb{E}[\exp(i\xi S_{t}^{(1)})]$
$\displaystyle=\exp\left(-c_{a,b}t|\sigma_{1}^{-1}\kappa\xi|\left(1+i\beta_{a,b}\frac{2}{\pi}\log(\sigma_{1}^{-1}\kappa|\xi|)\mathrm{sgn}(\xi)\right)+it\tau_{a,b}\sigma_{1}^{-1}\kappa\xi\right)$
$\displaystyle=\exp\left(-t|\xi|\left(1+i\beta_{a,b}\frac{2}{\pi}\log(\sigma_{1}^{-1}\kappa|\xi|)\mathrm{sgn}(\xi)-i\tau_{a,b}\sigma_{1}^{-1}\kappa\,\mathrm{sgn}(\xi)\right)\right)$
$\displaystyle=\exp(-t|\xi|z_{1}(\xi)),$

where
$z_{1}(\xi)=1+i\beta_{a,b}\frac{2}{\pi}\log(\sigma_{1}^{-1}\kappa|\xi|)\mathrm{sgn}(\xi)-i\tau_{a,b}\sigma_{1}^{-1}\kappa\,\mathrm{sgn}(\xi)$
as in the statement.

We write

$\begin{split}\Delta_{t}(\epsilon)&\leq\left|\epsilon\ell(1/\epsilon)H_{\tau_{t}^{\epsilon}}^{\epsilon}-\xi_{\epsilon}L_{\tau_{t}^{\epsilon}}^{0}-\kappa
K_{\tau_{t}^{\epsilon}}\right|+\xi_{\epsilon}\left|A_{\tau_{t}^{\epsilon}}^{\epsilon}-L_{\tau_{t}^{\epsilon}}^{0}\right|+\kappa\left|K_{\tau_{t}^{\epsilon}}-K_{\tau_{t}}\right|\\\
&\leq\sup_{[0,T]}\left|\epsilon\ell(1/\epsilon)H_{t}^{\epsilon}-\xi_{\epsilon}L_{t}^{0}-\kappa
K_{t}\right|+\sup_{[0,T]}\xi_{\epsilon}\left|A_{t}^{\epsilon}-L_{t}^{0}\right|+\kappa\left|K_{\tau_{t}^{\epsilon}}-K_{\tau_{t}}\right|.\end{split}$

The first term goes to $0$ from Lemma 23-(ii) and the last one by continuity
of $(K_{t})_{t\geq 0}$. Now remember that
$\xi_{\epsilon}=\kappa\ell(1/\epsilon)\int_{-\kappa/\epsilon}^{\kappa/\epsilon}\phi(x)\mathrm{d}x=\gamma(1/\epsilon)$
where $\gamma(x)=\kappa\ell(x)\int_{-\kappa x}^{\kappa x}\phi(v)\mathrm{d}v$.
Using the notations of Lemma 9 and the asymptotic estimate of
$\xi_{\epsilon}$, see Lemma 23-(ii), we have

$\gamma(x)\underset{x\to\infty}{\sim}-\kappa(f_{+}+f_{-})l(x)N(\kappa
x)\quad\text{if $f\in\mathrm{\bm{L}}^{1}(\mu)$},$

and

$\gamma(x)\underset{x\to\infty}{\sim}\kappa(f_{+}+f_{-})l(x)L(\kappa
x)\quad\text{if $f\notin\mathrm{\bm{L}}^{1}(\mu)$}.$

In any case, $\gamma$ is equivalent to the product of of two slowly varying
functions, and thus, is a slowly varying function. Hence, Lemma 20-(iii) tells
us that the second term goes to $0$. ∎

###  3.4 The diffusive regime and critical diffusive regime

This case is standard and we use the strategy explained in the introduction.
It consists in solving the Poisson equation $\mathcal{L}g=f$, see Jacod-
Shiryaev [13, Chapter VIII section 3f]. We recall the function $g$ defined by

$g(x)=2\int_{0}^{x}\mathfrak{s}^{\prime}(v)\int_{v}^{\infty}f(u)[\sigma^{2}(u)\mathfrak{s}^{\prime}(u)]^{-1}\mathrm{d}u\mathrm{d}v,$

which is a $C^{2}$ function solving
$2bg^{\prime}+\sigma^{2}g^{\prime\prime}=-2f$. Also since $(X_{t})_{t\geq 0}$
is a regular positive recurrent diffusion by Assumption 1, we classically
deduce that $X_{t}$ tends in law to $\mu$ as $t$ goes to infinity, see
Kallenberg [16, Theorem 23.15]. We first show Theorem 7.

###### Proof of Theorem 7..

Using Itô formula with the function $g$, we have

$\int_{0}^{t}f(X_{s})\mathrm{d}s=\int_{0}^{t}g^{\prime}(X_{s})\sigma(X_{s})\mathrm{d}B_{s}-g(X_{t}).$

Thus
$\epsilon^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s=M_{t}^{\epsilon}-Y_{t}^{\epsilon}$
where
$M_{t}^{\epsilon}=\epsilon^{1/2}\int_{0}^{t/\epsilon}g^{\prime}(X_{s})\sigma(X_{s})\mathrm{d}B_{s}$
is a local martingale and $Y_{t}^{\epsilon}=\epsilon^{1/2}g(X_{t/\epsilon})$.
Since $g$ is continuous, $g(X_{t/\epsilon})$ converges in law as $\epsilon\to
0$ and thus $Y_{t}^{\epsilon}$ converges to $0$ in probability, for each fixed
$t$. Moreover if $g$ is bounded, we have a.s.
$\sup_{[0,T]}Y_{t}^{\epsilon}\underset{\epsilon\to 0}{\longrightarrow}0$.

We now show that $(M_{t}^{\epsilon})_{t\geq 0}$ converges in in law as a
continuous process to $(\sigma_{\alpha}W_{t})_{t\geq 0}$. It is enough (see
Jacod-Shiryaev [13, Theorem VII-3.11 page 473]) to show that for each $t\geq
0$, $\langle
M^{\epsilon}\rangle_{t}\overset{\mathbb{P}}{\to}\sigma_{\alpha}^{2}t$ as
$\epsilon\to 0$. But $\langle
M^{\epsilon}\rangle_{t}=\epsilon\int_{0}^{t/\epsilon}[g^{\prime}(X_{s})\sigma(X_{s})]^{2}\mathrm{d}s$
and the result follows from Remark 21 since
$g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$ by assumption.

All in all we have proved that
$(\epsilon^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s)_{t\geq 0}$ converges
in finite dimensional distributions to $(\sigma_{\alpha}W_{t})_{t\geq 0}$ and
as a continuous process if $g$ is bounded. ∎

###### Proof of Theorem 5-(i)-(ii)..

The first point is covered by Theorem 7 since
$g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$ by Proposition 17. In the case
$\alpha=2$ and $\rho=\infty$, we can use exactly the same proof: we write
$|\epsilon/\rho_{\epsilon}|^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s=M_{t}^{\epsilon}-Y_{t}^{\epsilon}$,
where
$M_{t}^{\epsilon}=|\epsilon/\rho_{\epsilon}|^{1/2}\int_{0}^{t/\epsilon}g^{\prime}(X_{s})\sigma(X_{s})\mathrm{d}B_{s}$
is a local martingale and
$Y_{t}^{\epsilon}=|\epsilon/\rho_{\epsilon}|^{1/2}g(X_{t/\epsilon})$. First
$Y_{t}^{\epsilon}$ tends to $0$ in probability for each $t$ fixed, as
previously, and since $|\epsilon/\rho_{\epsilon}|\to 0$. Finally it only
remains to show that for each $t\geq 0$,

$\langle
M^{\epsilon}\rangle_{t}=|\epsilon/\rho_{\epsilon}|\int_{0}^{t/\epsilon}[g^{\prime}(X_{s})\sigma(X_{s})]^{2}\mathrm{d}s\overset{\mathbb{P}}{\longrightarrow}\sigma_{2}^{2},$

which is the case by Remark 22. ∎

######  Remark 24.

We stress that it is plausible we can show the limit theorem in the Lévy
regime, at least when $f_{+}=-f_{-}$, using the martingale strategy, i.e.
writing $\int_{0}^{t}f(X_{s})\mathrm{d}s=M_{t}-g(X_{t})$ where $(M_{t})_{t\geq
0}$ is a local martingale with bracket
$\int_{0}^{t}[g^{\prime}(X_{s})\sigma(X_{s})]^{2}\mathrm{d}s$. Using similar
arguments as in Lemma 23, we can show that, correctly rescaled, this bracket
behaves like a stable subordinator. Hence $M_{t}$ might behave like a Brownian
motion subordinate by a stable subordinator, which is known to be a stable
process.

##  4 Applications

Diffusions with heavy tailed invariant measure : Consider the following
stochastic differential equation

$X_{t}=-\frac{\theta+1}{2}\int_{0}^{t}\mathrm{sgn}(X_{s})|X_{s}|^{\theta}\mathrm{d}s+B_{t},$
(24)

where $\theta>0$ and $(B_{t})_{t\geq 0}$ is a Brownian motion. This model
includes the Ornstein-Uhlenbeck process ($\theta=1$). Following equations (2)
and (3), the invariant measure and the scale function are given by

$\mu(\mathrm{d}x)=\kappa
e^{-|x|^{\theta+1}}\mathrm{d}x,\qquad\mathfrak{s}(x)=\int_{0}^{x}e^{|v|^{\theta+1}}\mathrm{d}v.$

One can check that
$\mathfrak{s}(x)\underset{x\to\pm\infty}{\sim}\mathrm{sgn}(x)e^{|x|^{\theta+1}}/[(\theta+1)|x|^{\theta}]$.
Hence, as soon as $f$ is a continuous function such that there exists
$\alpha>0$ and $(f_{+},f_{-})\in\mathbb{R}^{2}$ satisfying

$|x|^{(1/\alpha-2)\theta}e^{-|x|^{\theta+1}/\alpha}f(x)\underset{x\to\pm\infty}{\longrightarrow}f_{\pm},$

and $|f_{+}|+|f_{-}|>0$, we can apply Theorem 5 with $\ell\equiv 1$, provided
$\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s$ is replaced by
$\int_{0}^{t/\epsilon}[f(X_{s})-\mu(f)]\mathrm{d}s$ when $\alpha>1$. We can
also generalize using slowly varying functions.

Roughly, the return times of $(X_{t})_{t\geq 0}$ have exponential moments and
in this situation the asymptotic behavior of the function $f$ is prescribed by
the invariant measure $\mu$: $f$ is such that there exists a slowly varying
function $L$ and $c_{+},c_{-}>0$ such that

$L(x)|x|^{\alpha}\mu(\\{f>x\\})\underset{x\to\infty}{\longrightarrow}c_{+}\quad\text{and}\quad
L(x)|x|^{\alpha}\mu(\\{f<x\\})\underset{x\to-\infty}{\longrightarrow}c_{+}.$
(25)

A toy kinetic model : Here we slightly generalize the results of [21], [8],
[18] and [10]. Consider a one-dimensional particle with position
$X_{t}\in\mathbb{R}$ and velocity $V_{t}\in\mathbb{R}$ subject to random
shocks and a force field
$F(v)=\frac{\beta}{2}\frac{\Theta^{\prime}(v)}{\Theta(v)}$ for some $\beta>1$
and a function $\Theta:\mathbb{R}\mapsto(0,+\infty)$ of class $C^{1}$. The
Newton equations describing the motion of the particle are

$V_{t}=B_{t}+\int_{0}^{t}F(V_{s})\mathrm{d}s,\qquad
X_{t}=\int_{0}^{t}V_{s}\mathrm{d}s,$ (26)

where $(B_{t})_{t\geq 0}$ is a Brownian motion modeling the random shocks. The
invariant measure and the scale function are given by

$\mu(\mathrm{d}x)=\kappa[\Theta(x)]^{\beta}\mathrm{d}x,\qquad\mathfrak{s}(x)=\int_{0}^{x}[\Theta(v)]^{-\beta}\mathrm{d}v.$

We make the following assumption on the function $\Theta$ :

######  Assumption 25.

There exists $(c_{-},c_{+})\in\mathbb{R}_{+}^{2}$ such that
$|x|\Theta(x)\longrightarrow c_{\pm}$ as $x\to\pm\infty$ and $c_{+}+c_{-}>0$.
If $c_{+}=0$ (respectively $c_{-}=0$), we moreover impose there exists $A>0$
such that for all $x\geq A$, $\Theta^{\prime}(x)<0$ (respectively for all
$x\leq-A$, $\Theta^{\prime}(x)>0$).

Obviously, $\lim_{x\to\pm\infty}\mathfrak{s}(x)=\pm\infty$ and since
$\beta>1$, $\mu(\mathbb{R})<\infty$. Now take $f=\text{id}$, we want to find
$\alpha>0$ such that Assumption 3 holds. Let us show that, with
$\alpha=(\beta+1)/3$, we have

$\displaystyle|\mathfrak{s}^{\prime}(x)|^{-2}|\mathfrak{s}(x)|^{2-1/\alpha}x\underset{x\to\infty}{\longrightarrow}(\beta+1)^{1/\alpha-2}c_{+}^{\beta/\alpha},$
(27)
$\displaystyle|\mathfrak{s}^{\prime}(x)|^{-2}|\mathfrak{s}(x)|^{2-1/\alpha}x\underset{x\to-\infty}{\longrightarrow}-(\beta+1)^{1/\alpha-2}c_{-}^{\beta/\alpha}.$
(28)

Observe that we have the following estimations

$|x|^{\beta}|\mathfrak{s}^{\prime}(x)|^{-1}\underset{x\to\pm\infty}{\longrightarrow}c_{\pm}^{\beta}\quad\text{and}\quad|x|^{\beta+1}|\mathfrak{s}(x)|^{-1}\underset{x\to\pm\infty}{\longrightarrow}(\beta+1)c_{\pm}^{\beta}.$
(29)

Hence if $c_{+}>0$ and $c_{-}>0$, the result is immediate. We consider e.g.
the case $c_{+}=0$. First we notice that
$\mathfrak{s}^{\prime\prime}(x)=-\beta\Theta^{\prime}(x)[\Theta(x)]^{-\beta-1}$
and thus Assumption 25 tells us that there exists $A>0$, such that
$\mathfrak{s}^{\prime}$ is increasing on $[A,\infty)$. Hence for all $x\geq
A$, we have

$0\leq\mathfrak{s}(x)\leq\mathfrak{s}(A)+\mathfrak{s}^{\prime}(x)(x-A)\leq\mathfrak{s}(A)+\mathfrak{s}^{\prime}(x)x,$

and thus for all $x\geq A$, since $2-1/\alpha>0$ (because $\beta>1$),

$|\mathfrak{s}^{\prime}(x)|^{-2}|\mathfrak{s}(x)|^{2-1/\alpha}x\leq|\mathfrak{s}^{\prime}(x)|^{-2}|\mathfrak{s}(A)+\mathfrak{s}^{\prime}(x)x|^{2-1/\alpha}x\underset{x\to\infty}{\sim}|\mathfrak{s}^{\prime}(x)|^{-1/\alpha}|x|^{3-1/\alpha}$

But since $\alpha=(\beta+1)/3$, we have $3-1/\alpha=\beta/\alpha$ and thus
$|\mathfrak{s}^{\prime}(x)|^{-1/\alpha}|x|^{3-1/\alpha}=|x^{\beta}[\mathfrak{s}^{\prime}(x)]^{-1}|^{1/\alpha}$
which goes to $0$ by (29).

We can apply Theorem 5: with $\alpha=(\beta+1)/3$, $m=\mu(\text{id})$,
$\ell\equiv 1$, $f_{+}=(\beta+1)^{1/\alpha-2}c_{+}^{\beta/\alpha}$ and
$f_{-}=-(\beta+1)^{1/\alpha-2}c_{-}^{\beta/\alpha}$,

  1. (i)

If $\beta>5$, $\left(\epsilon^{1/2}(X_{t/\epsilon}-mt/\epsilon)\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}W_{t}\right)_{t\geq 0}.$

  2. (ii)

If $\beta=5$,
$\left(|\epsilon/\log\epsilon|^{1/2}(X_{t/\epsilon}-mt/\epsilon)\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{2}W_{t}\right)_{t\geq 0}.$

  3. (iii)

If $\beta\in(2,5)$,
$\left(\epsilon^{1/\alpha}(X_{t/\epsilon}-mt/\epsilon)\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

  4. (iv)

If $\beta=2$,
$\left(\epsilon(X_{t/\epsilon}-\xi_{\epsilon}t/\epsilon)\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

  5. (v)

If $\beta\in(1,2)$, $\left(\epsilon^{1/\alpha}X_{t/\epsilon}\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

Remark that $\alpha$ ranges from $2/3$ to infinity. The additional assumption
that $\Theta$ is eventually monotonic when $c_{+}=0$ (or $c_{-}=0$) is not
optimal, the true assumption is (27) and (28).

We stress that this model is the starting point of this paper as it was
explained in the references section. Considering the symmetric function
$\Theta(v)=(1+v^{2})^{-1/2}$ and reasoning on the law of $(X_{t},V_{t})$,
which is the solution of the associated kinetic Fokker-Planck equation, a
series of P.D.E papers (Nasreddine-Puel [21] : $\beta>5$, Cattiaux-Nasreddine-
Puel [8] : $\beta=5$ and Lebeau-Puel [18] :
$\beta\in(1,5)\setminus\\{2,3,4\\}$) answered these questions, finding some
symmetric stable processes at the limit. Then, using probabilistic techniques,
Fournier and Tardif [10] treated all the cases (even when $\beta\in(0,1]$),
still in a symmetric context. We thus recover their results and extend them to
asymmetrical forces.

When $\beta\in(0,1)$, the process $(V_{t})_{t\geq 0}$ is null recurrent and
Fournier and Tardif [10] shows the rescaled process
$(\epsilon^{1/2}V_{t/\epsilon})_{t\geq 0}$ converges in finite dimensional
distribution to a symmetrized Bessel process and
$(\epsilon^{3/2}X_{t/\epsilon})_{t\geq 0}$ to an integrated symmetrised Bessel
process (their proof actually holds for $\beta\in(-1,1)$). Although we can
extend their result to asymmetrical forces, this phenomenon seems specific to
this equation and we were not able to extend it to general diffusions.

Remark that for this model, the return times do not have exponential moments
and the exponent $\alpha$ is not prescribed by the invariant measure. Indeed
we have

$|x|^{\beta-1}\mu(\\{f>x\\})\underset{x\to\infty}{\longrightarrow}\kappa
c_{+}^{\beta}/(\beta-1)\quad\text{and}\quad|x|^{\beta-1}\mu(\\{f<x\\})\underset{x\to\infty}{\longrightarrow}\kappa
c_{-}^{\beta}/(\beta-1)$

and $\alpha=(\beta+1)/3$, while one might expect $\alpha=\beta-1$. Observe
that if $\beta\in(1,2)$, we have $(\beta+1)/3>\beta-1$ and if $\beta>2$, we
have $(\beta+1)/3<\beta-1$. The dynamics are a little bit more complicated and
one must take into account the fact that the return times can be long.

SDE without drift : Consider the following SDE

$X_{t}=\int_{0}^{t}\left(1+|X_{s}|\right)^{\beta/2}\mathrm{d}B_{s},$ (30)

with $\beta>1$. We have $\mathfrak{s}(x)=x$ (indeed $(X_{t})_{t\geq 0}$ is a
martingale) and $\mu(\mathrm{d}x)=\kappa(1+|x|)^{-\beta}\mathrm{d}x$. Set for
instance $f(x)=x/(1+|x|)^{1-\gamma}$ with $\gamma>\beta-2$, then $\mu(f)=0$
and one can see that, with $\alpha=1/(\gamma+2-\beta)>0$, $f_{+}=1$ and
$f_{-}=-1$, we have

$|x|^{2-1/\alpha-\beta}f(x)\underset{x\to\pm\infty}{\longrightarrow}f_{\pm}.$

Hence we can apply Theorem 5 with $\ell\equiv 1$.

  1. (i)

If $\beta-2<\gamma<\beta-3/2$, then
$\left(\epsilon^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}W_{t}\right)_{t\geq 0}.$

  2. (ii)

If $\gamma=\beta-3/2$, then
$\left(|\epsilon/\log\epsilon|^{1/2}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{2}W_{t}\right)_{t\geq 0}.$

  3. (iii)

If $\beta-3/2<\gamma<\beta-1$, then
$\left(\epsilon^{1/\alpha}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

  4. (iv)

If $\gamma=\beta-1$, then
$\left(\epsilon\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

  5. (v)

If $\gamma>\beta-1$, then
$\left(\epsilon^{1/\alpha}\int_{0}^{t/\epsilon}f(X_{s})\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{\alpha}S_{t}^{(\alpha)}\right)_{t\geq
0}.$

Now if $\gamma\leq\beta-2$, one can see that the function $g$, solution of the
Poisson equation, is such that

$g^{\prime}(x)=\int_{x}^{\infty}f(v)\sigma^{-2}(v)\mathrm{d}v\underset{x\to\infty}{\sim}x^{\gamma-\beta+1}\quad\text{and}\quad
g^{\prime}(x)=\int_{-\infty}^{x}f(v)\sigma^{-2}(v)\mathrm{d}v\underset{x\to-\infty}{\sim}-|x|^{\gamma-\beta+1}.$

Hence $g^{\prime}\in\mathrm{\bm{L}}^{2}(\mathrm{d}x)$ since
$2(\gamma-\beta+1)<-1$ and thus $g^{\prime}\sigma\in\mathrm{\bm{L}}^{2}(\mu)$
(since $\mu(\mathrm{d}x)=\sigma^{-2}(x)\mathrm{d}x$). We can apply Theorem 7
which tells us that the diffusive regime holds for $\gamma\leq\beta-2$.

Once again, the return times are slow and the index $\alpha$ is not the
expected one. Indeed we have for $\gamma>0$

$\mu(\\{f>x\\})\underset{x\to\infty}{\sim}C|x|^{(1-\beta)/\gamma}\quad\text{and}\quad\mu(\\{f<x\\})\underset{x\to-\infty}{\sim}C|x|^{(1-\beta)/\gamma}$

for some constant $C$.

Assume $\beta=5/4$ and let $f$ be a function going to $0$ at $\pm\infty$ such
that $\mu(f)\neq 0$. If $\alpha=4/3$, we have

$|x|^{2-1/\alpha-\beta}(f(x)-\mu(f))\underset{x\to\pm\infty}{\longrightarrow}-\mu(f),$

and thus we can apply Theorem 5-(iii) which tells us that

$\left(\epsilon^{3/4}\int_{0}^{t/\epsilon}[f(X_{s})-\mu(f)]\mathrm{d}s\right)_{t\geq
0}\overset{f.d.}{\longrightarrow}\left(\sigma_{4/3}S_{t}^{(4/3)}\right)_{t\geq
0}.$

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