---
abstract: |
    Using a coupling for the weighted sum of independent random variables and the explicit expression of the transition semigroup of Ornstein-Uhlenbeck processes driven by compound Poisson processes, we establish the existence of a successful coupling and the Liouville theorem for general Ornstein-Uhlenbeck processes. Then we present the explicit coupling property of Ornstein-Uhlenbeck processes directly from the behaviour of the corresponding symbol or characteristic exponent. This approach allows us to derive gradient estimates for Ornstein-Uhlenbeck processes via the symbol.

    **Keywords:** Ornstein-Uhlenbeck processes; coupling property; Liouville theorem; gradient estimates.

    **MSC 2010:** 60J25; 60J75.
author:
- 'René L. Schilling Jian Wang'
title: '**On the Coupling Property and the Liouville Theorem for Ornstein-Uhlenbeck Processes**'
---

[^1] [^2]

Main Results {#section1}
============

Let $(X^x_t)_{t{\geqslant}0}$ be an $n$-dimensional Ornstein-Uhlenbeck process, which is defined as the unique strong solution of the following stochastic differential equation $$\label{ou1}
    dX_t = AX_t\,dt + B\,dZ_t,\qquad X_0=x\in{\mathds R}^n.$$ Here $A$ is a real $n\times n$ matrix, $B$ is a real $n\times d$ matrix and $Z_t$ is a Lévy process in ${\mathds R}^d$; note that we allow $Z_t$ to take values in a proper subspace of ${\mathds R}^d$. It is well known that $$X_t^x
    =e^{tA}x + \int_0^t e^{(t-s)A}B\,dZ_s.$$ The characteristic exponent or symbol $\Phi$ of $Z_t$, defined by $${\mathds E}\bigl(e^{i{\langle\xi,Z_t\rangle}}\bigr)
    =e^{-t\Phi(\xi)},\quad \xi\in{\mathds R}^d,$$ enjoys the following Lévy-Khintchine representation: $$\label{ou2}
     \Phi(\xi)
     =\frac{1}{2}{\langleQ\xi,\xi\rangle} +i{\langleb,\xi\rangle} +\int_{z\neq 0} \Bigl(1-e^{i{\langle\xi,z\rangle}}+i{\langle\xi,z\rangle}{\mathds 1}_{B(0,1)}(z)\Bigr)\nu(dz),$$ where $Q=(q_{j,k})_{j,k=1}^d$ is a positive semi-definite matrix, $b\in{{\mathds R^d}}$ is the drift vector and $\nu$ is the Lévy measure, i.e.a $\sigma$-finite measure on ${\mathds R}^d\setminus\{0\}$ such that $\int_{z\neq 0}(1\wedge |z|^2)\,\nu(dz)<\infty$. For every $\varepsilon>0$, define ${\nu}_\varepsilon$ on ${\mathds R}^d$ as follows: $${\nu}_\varepsilon(C)
    =
    \begin{cases}
        \nu(C),                           & \text{if\ \ } \nu({\mathds R}^d)<\infty;\\
        \nu(C\setminus \{z: |z|<\varepsilon\}), & \text{if\ \ } \nu({\mathds R}^d)=\infty.
    \end{cases}$$

Let $(Y_t)_{t{\geqslant}0}$ be a Markov process on ${\mathds R}^n$ with transition function $P_t(x,\cdot)$. Then, according to [@Li; @T; @SW], we say that $(Y_t)_{t{\geqslant}0}$ admits a *successful coupling* (also: enjoys the *coupling property*) if for any $x,y\in{\mathds R}^n$, $$\label{prex1}
    \lim_{t\rightarrow\infty}\|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}=0,$$ where $\|\cdot\|_{{\mathrm{Var}}}$ stands for the total variation norm. If a Markov process admits a successful coupling, then it also has the Liouville property, i.e. every bounded harmonic function is constant; in this context a function $f$ is harmonic, if $Lf=0$ where $L$ is the generator of the Markov process. See [@CG; @CW] and the references therein for this result and more details on the coupling property.

Let $A$ be an $n\times n$ matrix. We say that an eigenvalue $\lambda$ of $A$ is *semisimple* if the dimension of the corresponding eigenspace is equal to the algebraic multiplicity of $\lambda$ as a root of characteristic polynomial of $A$. Note that for symmetric matrices $A$ all eigenvalues are real and semisimple. Recall that for any two bounded measures $\mu$ and $\nu$ on $({\mathds R}^d,{\mathscr{B}}({\mathds R}^d))$, $\mu\wedge\nu:=\mu-(\mu-\nu)^+$, where $(\mu-\nu)^{\pm}$ refers to the Jordan-Hahn decomposition of the signed measure $\mu-\nu$. In particular, $\mu\wedge\nu=\nu\wedge\mu$, and $\mu\wedge
\nu\,({\mathds R}^d)=\frac{1}{2}\big[\mu({\mathds R}^d)+\nu({\mathds R}^d)-\|\mu-\nu\|_{{\mathrm{Var}}}\big].$

One of our main results is the following

\[th1\] Let $P_t(x,\cdot)$ be the transition probability of the Ornstein-Uhlenbeck process $\{X_t^x\}_{t{\geqslant}0}$ given by . Assume that ${\operatorname{Rank}}(B)=n$ (which implies $n{\leqslant}d$), and that there exist $\varepsilon,\delta>0$ such that $$\label{th2233}
    \inf_{z\in{\mathds R}^d,|z|{\leqslant}\delta}\nu _\varepsilon\wedge (\delta_z*\nu_\varepsilon)({\mathds R}^d)>0.$$

If the real parts of all eigenvalues of $A$ are non-positive and if all purely imaginary eigenvalues are semisimple, then there exists a constant $C=C(\varepsilon,\delta,\nu,A,B)>0 $ such that for all $x,y\in{\mathds R}^n$ and $t>0$, $$\label{th21}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}
    {\leqslant}\frac{C(1+|x-y|)}{\sqrt{t}}\wedge2.$$

As a consequence of Theorem \[th1\], we immediately obtain the following result which partly answers the following question about Liouville theorems for non-local operators from [@PZ page 458]: *A challenging task would be to apply other probabilistic techniques, based on ... coupling to non-local operators*.

Under the conditions of Theorem \[th1\], the Ornstein-Uhlenbeck process $\{X_t^x\}_{t{\geqslant}0}$ admits a successful coupling and has the Liouville property.

\[remarkth1\] (1) If $A=0$, $d=n$ and $B={\operatorname{id}}_{{\mathds R}^n}$, then $X_t$ is just a Lévy process on ${\mathds R}^n$. The condition is one possibility to guarantee sufficient jump activity such that the Lévy process $X_t$ admits a successful coupling. To see that is sharp, we can use the example in [@SW Remark 1.2].

\(2) Let $Z_t$ be a (rotationally symmetric) $\alpha$-stable Lévy process $Z_t$, $0<\alpha<2$, and denote by $X_t$ the $n$-dimensional Ornstein-Uhlenbeck process driven by $Z_t$, i.e.$$dX_t = AX_t\,dt + dZ_t.$$ If at least one eigenvalue of $A$ has positive real part, then $X_t$ does not have the coupling property. Indeed, according to [@PZ Example 3.4 and Theorem 3.5], we know that $X_t$ does not have the Liouville property, i.e. there exists a bounded harmonic function which is not constant. According to [@Li Theorem 21.12] or [@CG Theorem 1 and its second remark], $X_t$ does not have the coupling property. This example indicates that the non-positivity of the real parts of the eigenvalues of $A$ is also necessary.

In [@SW Theorem 4.1 and Corollary 4.2] we show that Lévy processes which have the strong Feller property admit the coupling property. A similar conclusion, however, does not hold for general Ornstein-Uhlenbeck processes. Consider, for instance, the one-dimensional Ornstein-Uhlenbeck process given by $$dX_t=X_t\,dt+dZ_t,\qquad X_0=x\in{\mathds R},$$ where $Z_t$ is an $\alpha$-stable Lévy process $Z_t$ on ${\mathds R}$. According to [@PZ2 Theorem 1.1] (or [@NS Theorem A]) and [@PZ2 Proposition 2.1], we know that $X_t$ has the strong Feller property. However, the argument used in Remark \[remarkth1\] shows that this process fails to have the coupling property.

Recently, F.-Y. Wang [@wang1] has studied the coupling property of an Ornstein-Uhlenbeck process $X_t$ defined by . Assume that ${\operatorname{Rank}}(B)=n$ and ${\langleAx,x\rangle}{\leqslant}0$ holds for $x\in{\mathds R}^n$. In [@wang1 Theorem 3.1] it is proved that is satisfied for some constant $C>0$, whenever the Lévy measure of $Z_t$ satisfies $\nu(dz){\geqslant}\rho_0(z)dz$ such that $$\label{wang22}
    \int_{\{|z-z_0|{\leqslant}\varepsilon\}}
    \frac{dz}{\rho_0(z)}<\infty$$ holds for some $z_0\in{\mathds R}^d$ and some $\varepsilon>0$.

Let us compare F.-Y. Wang’s result with our Theorem \[th1\].

\[improvement\] Assume that holds for some $\rho_0\in L^1_{\mathrm{loc}}({\mathds R}^d\setminus\{0\})$, some $z_0\in{\mathds R}^d$ and some $\varepsilon>0$. Then, there exist a closed subset $F\subset \overline{B(z_0,\varepsilon)}=\{z\in{\mathds R}^d: |z-z_0|{\leqslant}\varepsilon\}$ and a constant $\delta>0$ such that $$\inf_{x\in{\mathds R}^d, |x|{\leqslant}\delta}\int_F\big(\rho_0(z)\wedge\rho_0(z-x)\big)\,dz>0.$$

We postpone the technical proof of Proposition \[improvement\] to Section \[subsec-appendix2\] in the appendix. Proposition \[improvement\] shows that Theorem \[th1\] improves [@wang1 Theorem 3.1], even if the Lévy measure $\nu$ of $Z_t$ has an absolutely continuous component as we will see in the following example.

Let $C_{3/4}$ be a Smith-Volterra-Cantor set in $[0,1]$ with Lebesgue measure ${\operatorname{Leb}}(C_{3/4})=3/4$, i.e. $C_{3/4}$ is a perfect set with empty interior, see e.g. [@AB Chapter 3, Section 18]. Consider the following one-dimensional Ornstein-Uhlenbeck process $$dX_t=-X_t\,dt + dZ_t,\qquad X_0=x\in{\mathds R},$$ where $Z_t$ is a real-valued Lévy process with Lévy measure $\nu(dz)={\mathds 1}_{C_{3/4}}(z)\,dz$. We will see that we can use Theorem \[th1\] to show the coupling property of the process $X_t$ while the criterion from [@wang1 Theorem 3.1] fails.

Let $\delta \in (0, 1/8)$ and $z\in [-\delta, \delta]$. Then $$\begin{aligned}
        \nu _\varepsilon\wedge (\delta_z*\nu_\varepsilon)({\mathds R})
        &=\int\Bigl({\mathds 1}_{C_{3/4}}(x)\wedge{\mathds 1}_{C_{3/4}}(x+z)\Bigr)dx\\
        &={\operatorname{Leb}}\bigl( C_{3/4}\cap (C_{3/4}-z)\bigr)\\
        &={\operatorname{Leb}}(C_{3/4}) + {\operatorname{Leb}}(C_{3/4}-z) - {\operatorname{Leb}}\bigl( C_{3/4}\cup (C_{3/4}-z)\bigr)\\
        &{\geqslant}\frac 64 - {\operatorname{Leb}}[-|z|,1+|z|]{\geqslant}\frac 14.
    \end{aligned}$$ This shows that the conditions of Theorem \[th1\] are satisfied.

On the other hand, since $C_{3/4}$ contains no intervals, we see that for all $z_0\in{\mathds R}$ and $\varepsilon>0$, $$\int_{\{|z-z_0|{\leqslant}\varepsilon\}}\frac{dz}{{\mathds 1}_{C_{3/4}}(z)}=\infty$$ (here we use the convention $\frac 10 = +\infty$). This means that does not hold.

Now we are going to estimate $\|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}$ for large values of $t$ with the help of the characteristic exponent $\Phi(\xi)$ of the Lévy process $Z_t$. We restrict ourselves to the case where $Q=0$ in , i.e. to Lévy process $(Z_t)_{t{\geqslant}0}$ without a Gaussian part. For $t, \rho>0$, define $$\varphi_t(\rho)
    :=\sup_{|\xi|{\leqslant}\rho}\int_0^t{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds,$$ where $M^\top$ denotes the transpose of the matrix $M$.

\[coup\] Let $P_t(x,\cdot)$ be the transition function of the Ornstein-Uhlenbeck process $\{X_t^x\}_{t{\geqslant}0}$ on ${\mathds R}^n$ given by . Assume that there exists some $t_0 > 0$ such that $$\label{coup1}
       \liminf\limits_{|\xi|\rightarrow\infty}\frac{\int_0^{t_0}{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds}{\log (1+|\xi|)} >2n+2.$$ If $$\label{coup2}
    \int \exp\left(-\int_0^t{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)ds\right) |\xi|^{n+2} \,d\xi
    = \mathsf{O}\left(\varphi_t^{-1}(1)^{2n+2}\right)
    \quad\text{as\ \ } t\to\infty,$$ then there exist $t_1,C>0$ such that for any $x,y\in{\mathds R}^n$ and $t{\geqslant}t_1$, $$\label{coup3}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}
    {\leqslant}C|e^{tA}(x-y)|\,\varphi^{-1}_t(1).$$ In particular, when $$\label{coup4}\xi\mapsto\int_0^\infty{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top
e^{sA^\top}\xi\big)\,ds \quad\textrm{ is locally bounded},$$ we only need the condition to get .

Note that is, e.g. satisfied, if the real parts of all eigenvalues of $A$ are negative and $$\limsup_{|\xi|\rightarrow0}\frac{{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top
\xi\big)}{|\xi|^\kappa}<\infty$$ for some constant $\kappa>0$.

The remaining part of this paper is organized as follows. In Section \[section2\] we first present the proof of Theorem \[th1\], where a coupling for the weighted sum of independent random variables and the explicit expression of the transition semigroup of Ornstein-Uhlenbeck processes driven by a compound Poisson process are used. Then, we follow the approach of our recent paper [@SSW] to prove Theorem \[coup\]. As a byproduct, we also derive explicit gradient estimates for Ornstein-Uhlenbeck processes, cf. the Appendix \[subsec-appendix1\].

Proofs of Theorems {#section2}
==================

We begin with the proof of Theorem \[th1\].

The proof is split into six steps. *Step 1*. For any $\varepsilon>0$, let $(Z_t^\varepsilon)_{t{\geqslant}0}$ be a compound Poisson process on ${\mathds R}^d$ whose Lévy measure is $\nu_\varepsilon$. Then, $(Z_t^\varepsilon)_{t{\geqslant}0}$ and $(Z_t-Z_t^\varepsilon)_{t{\geqslant}0}$ are independent Lévy processes. It follows, in particular, that the random variables $$X_t^{\varepsilon,x}:=e^{tA}x+\int_0^t e^{(t-s)A}B\,dZ_s^\varepsilon$$ and $$X_t^x-X_t^{\varepsilon,x}:=\int_0^t e^{(t-s)A}B\,d(Z_s-Z_s^\varepsilon)$$ are independent for any $\varepsilon>0$ and $t{\geqslant}0$.

*Step 2*. Denote by $\mu_{\varepsilon,t}$ the law of random variable $$X_t^{\varepsilon,0} := X_t^{\varepsilon,x}-e^{tA}x = \int_0^te^{(t-s)A}B\,dZ_s^\varepsilon.$$ We will compute $\mu_{\varepsilon,t}$, which coincides with the law of $\int_0^te^{sA}B\,dZ_s^\varepsilon$, cf. Lemma \[lemmaproofth11\] below. Our argument follows the proof of [@PZ2 Theorem 1.1], which is motivated by [@SA Theorem 27.7].

The law of the compound poisson process $Z_t^\varepsilon$ is given by $$e^{-C_\varepsilon t}\bigg[\delta_0+\sum_{k=1}^\infty \frac{(C_\varepsilon t)^k}{k!}\,\bar{\nu}_\varepsilon^{*k}\bigg],$$ where $C_\varepsilon=\nu_\varepsilon({\mathds R}^d)$, $\bar{\nu}_\varepsilon=\nu_\varepsilon/C_\varepsilon$ and $\bar{\nu}_\varepsilon^{*k}$ is the $k$-fold convolution of $\bar{\nu}_\varepsilon$.

Construct a sequence $(\xi_i)_{i{\geqslant}1}$ of iid random variables which are exponentially distributed with intensity $C_\varepsilon$, and introduce a further sequence $(U_i)_{i{\geqslant}1}$ of iid random variables on ${\mathds R}^d$ with law $\bar{\nu}_\varepsilon$. We will assume that the random variables $(U_i)_{i{\geqslant}1}$ are independent of the sequence $(\xi_i)_{i{\geqslant}1}$. It is not difficult to check that the random variable $$\label{proofs0}
    0\cdot{\mathds 1}_{\{\xi_1>t\}} +\sum_{k=1}^\infty{\mathds 1}_{\{\xi_1+\cdots+\xi_k{\leqslant}t<\xi_1+\cdots+\xi_{k+1}\}} \Big(e^{\xi_1A}BU_1+\cdots+e^{(\xi_1+\cdots+\xi_k)A}BU_k\Big)$$ also has the probability distribution $\mu_{\varepsilon,t}$.

Using we find for any $f\in B_b({\mathds R}^n)$, $$\label{proofs1}
    {\mathds E}f\bigl(X_t^{\varepsilon,x}\bigr)
    =\int f\bigl(e^{tA }x+z\bigr)\,\mu_{\varepsilon,t}(dz)
    =f\bigl(e^{tA}x\bigr)\,e^{-C_\varepsilon t}+Hf(x),
$$ where $$\begin{aligned}
&Hf(x)\\
&:={\mathds E}f\left(\sum_{k=1}^\infty{\mathds 1}_{\{\xi_1+\cdots+\xi_k{\leqslant}t<\xi_1+\cdots+\xi_{k+1}\}} \Bigl(e^{tA}x+e^{\xi_1A}BU_1+\cdots+e^{(\xi_1+\cdots+\xi_k)A}BU_k\Bigr)\right)\\
&=\sum_{k=1}^\infty{\mathds E}f\left({\mathds 1}_{\{\xi_1+\cdots+\xi_k{\leqslant}t<\xi_1+\cdots+\xi_{k+1}\}} \Bigl(e^{tA}x+e^{\xi_1A}BU_1+\cdots+e^{(\xi_1+\cdots+\xi_k)A}BU_k\Bigr)\right)\\
&=\sum_{k=1}^\infty \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}{\;\;\times}\\
&\qquad\quad{\times}\;\int\limits_{{\mathds R}^d}\cdots\int\limits_{{\mathds R}^d} f\bigl(e^{tA}x+e^{t_1A}By_1+\cdots+e^{(t_1+\cdots+t_k)A}By_k\bigr) \,\bar{\nu}_\varepsilon(dy_1)\cdots \bar{\nu}_\varepsilon(dy_k)\\
&=\sum_{k=1}^\infty \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}\;\;\times\\
&\qquad\quad\times\;\int\limits_{{\mathds R}^n} f\bigl(e^{tA}x+z\bigr)\,\mu_{t_1,\cdots,t_k}(dz).\end{aligned}$$ Here $\mu_{t_1,\cdots,t_k}$ is the probability measure on ${\mathds R}^n$ which is the image of the $k$-fold product measure $\bar{\nu}_\varepsilon\times \cdots\times \bar{\nu}_\varepsilon$ under the linear transformation $J_{t_1,\ldots,t_k}$ (independent of $\varepsilon$) acting from $({\mathds R}^{d})^k$ into ${\mathds R}^n$: $$J_{t_1,\ldots,t_k}(y_1,\ldots,y_k)
    =e^{t_1A}B y_1 + \cdots +e^{(t_1+\cdots+t_k)A}B y_k,$$ for $y_i\in{\mathds R}^d$ and $i=1,\cdots,k$.

*Step 3*. Let $P_t(x,\cdot)$ and $P_t$ be the transition function and the transition semigroup of the Ornstein-Uhlenbeck process $(X^x_t)_{t{\geqslant}0}$. Similarly, we denote by $P^\varepsilon_t(x,\cdot)$ and $P^\varepsilon_t$ the transition function and the transition semigroup of $(X_t^{\varepsilon,x})_{t{\geqslant}0}$, and by $Q^\varepsilon_t(x,\cdot)$ and $Q^\varepsilon_t$ the transition function and the transition semigroup of $(X_t^x-X_t^{\varepsilon,x})_{t{\geqslant}0}$. By the independence of the processes $(X_t^{\varepsilon,x})_{t{\geqslant}0}$ and $(X_t^x-X_t^{\varepsilon,x})_{t{\geqslant}0}$, we get $$\label{proofs2}\begin{aligned}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}
    &= \sup_{\|f\|_\infty{\leqslant}1}\big|P_tf(x)-P_tf(y)\big|\\
    &=\sup_{\|f\|_\infty{\leqslant}1} \big|P_t^\varepsilon Q_t^\varepsilon f(x)-P_t^\varepsilon Q_t^\varepsilon f(y)\big|\\
    &{\leqslant}\sup_{\|h\|_\infty{\leqslant}1} \big|P^\varepsilon_th(x)-P^\varepsilon_th(y)\big|.
\end{aligned}$$ Furthermore, it follows from that $$\label{proofs3}\begin{aligned}
   &\sup_{\|h\|_\infty{\leqslant}1} \big|P^\varepsilon_th(x)-P^\varepsilon_th(y)\big|\\
    &{\leqslant}2e^{-C_\varepsilon t}+\sum_{k=1}^\infty \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}\;\;\times\\
    &\qquad\times\sup_{\|h\|_\infty{\leqslant}1} \bigg|\int\limits_{{\mathds R}^n} h\big(e^{tA}x+z\big)\,\mu_{t_1,\cdots,t_k}(dz) - \int\limits_{{\mathds R}^n} h\big(e^{tA}y+z\big)\,\mu_{t_1,\cdots,t_k}(dz)\biggr|\\
    &= 2e^{-C_\varepsilon t}+\sum_{k=1}^\infty \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}\;\;\times\\
    &\qquad\times\sup_{\|h\|_\infty{\leqslant}1} \bigg|\int\limits_{{\mathds R}^n} h\big(e^{tA}(x-y)+z\big)\,\mu_{t_1,\cdots,t_k}(dz) - \int\limits_{{\mathds R}^n} h(z)\,\mu_{t_1,\cdots,t_k}(dz)\bigg|\\
    &{\leqslant}2e^{-C_\varepsilon t}+\sum_{k=1}^\infty\mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}\;\;\times\\
&\qquad\times\|\delta_{e^{tA}(x-y)}*\mu_{t_1,\cdots,t_k}-\mu_{t_1,\cdots,t_k}\|_{{\mathrm{Var}}}.
\end{aligned}$$

*Step 4*. For any $a\in{\mathds R}^n$, $a\neq 0$, let $R_a$ be the non-degenerate rotation such that $R_a a=|a|e_1$. Then, by [@SW Lemma 3.2], $$\begin{aligned}
    \big\|\delta_{e^{tA}(x-y)}*\mu_{t_1,\cdots,t_k} &-\mu_{t_1,\cdots,t_k}\big\|_{{\mathrm{Var}}}\\
    &=\big\|\delta_{|e^{tA}(x-y)|e_1}*\big(\mu_{t_1,\cdots,t_k}\circ R_{e^{tA}(x-y)}^{-1}\big)-\mu_{t_1,\cdots,t_k}\circ R_{e^{tA}(x-y)}^{-1}\big\|_{{\mathrm{Var}}}.\end{aligned}$$ Since $\mu_{t_1,\cdots,t_k}$ is the law of the random variable $$\sum_{i=1}^ke^{(t_1+\cdots+t_i)A}BU_i,$$ $\mu_{t_1,\cdots,t_k}\circ R_{e^{tA}(x-y)}^{-1}$ is the law of the random variable $$\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i\big).$$

To estimate $\big\|\delta_{|e^{tA}(x-y)|e_1}*\big(\mu_{t_1,\cdots,t_k}\circ R_{e^{tA}(x-y)}^{-1}\big)-\mu_{t_1,\cdots,t_k}\circ R_{e^{tA}(x-y)}^{-1}\big\|_{{\mathrm{Var}}}$, we will use the Mineka and Lindvall-Rogers couplings for random walks. The remainder of this part is based on the proof of [@SW Proposition 3.3]. In order to ease notations, we set ${\mathsf{n}}:=\bar{\nu}_\varepsilon$ and ${\mathsf{n}}^{a}:=\delta_a*\bar\nu_\varepsilon$ for any $a\in{\mathds R}^d$.

Since ${\operatorname{Rank}}(B)=n$, there exists a real $d\times n$ matrix $\bar{B}$ such that $B\bar{B}={\operatorname{id}}_{{\mathds R}^n}$, see e.g. [@Bern Theorem 2.6.1, Page 35]. For any $i{\geqslant}1$, let $(U_i,\Delta U_i)\in {\mathds R}^d \times {\mathds R}^d$ be a pair of random variables with the following distribution $${\mathds P}\big((U_i,\Delta U_i)\in C\times D\big)
    =
    \begin{cases}
    \qquad \frac 12 ({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i})(C), & \text{if\ \ } D=\{a_i\};\\
    \qquad \frac 12 ({\mathsf{n}}\wedge{\mathsf{n}}^{a_i})(C),  & \text{if\ \ } D= \{-a_i\};\\
    \big({\mathsf{n}}- \frac 12 ({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i}+{\mathsf{n}}\wedge{\mathsf{n}}^{a_i})\big)(C), & \text{if\ \ } D=\{0\};
    \end{cases}$$ where $C\in{\mathscr{B}}({\mathds R}^d)$, $a_i=\bar B\, e^{(t-(t_1+\cdots+t_i))A}\,(x-y)$ and $D$ is any of the following three sets: $\{-a_i\}$, $\{0\}$ or $\{a_i\}$. Again by [@SW Lemma 3.2], $$\begin{aligned}
    {\mathds P}\big(\Delta U_i=-a_i\big)
    &=\frac{1}{2}\big({\mathsf{n}}\wedge\big(\delta_{a_i}*{\mathsf{n}})\big)({\mathds R}^d)\\
    &=\frac{1}{2}\big({\mathsf{n}}\wedge\big(\delta_{-a_i}*{\mathsf{n}})\big)({\mathds R}^d)\\
    &={\mathds P}(\Delta U_i=a_i).\end{aligned}$$ It is clear that the distribution of $U_i$ is ${\mathsf{n}}$. Let $U_i'=U_i+\Delta U_i$. We claim that the distribution of $U_i'$ is also ${\mathsf{n}}$. Indeed, for any $C\in\mathscr{B}({\mathds R}^d)$, $$\begin{aligned}
    &{\mathds P}(U_i'\in C)\\
    &={\mathds P}(U_i-a_i\in C, \Delta U_i=-a_i)
        + {\mathds P}(U_i+a_i\in C, \Delta U_i=a_i)
        +{\mathds P}(U_i \in A, \Delta U_i=0)\\
    &= \frac 12\left(\delta_{-a_i}*({\mathsf{n}}\wedge{\mathsf{n}}^{a_i})\right)(C)
        +\! \frac12\left(\delta_{a_i}*({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i})\right)(C)
       \! +\! \left(\!\!{\mathsf{n}}-\!\! \frac 12\,\big({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i}+{\mathsf{n}}\wedge{\mathsf{n}}^{a_i}\big)\!\!\right)(C)\\
    &={\mathsf{n}}(C),\end{aligned}$$ where we have used that $$\delta_{a_i}*({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i})={\mathsf{n}}\wedge{\mathsf{n}}^{a_i}\quad\textrm{ and }\quad\delta_{-a_i}*({\mathsf{n}}\wedge{\mathsf{n}}^{a_i})={\mathsf{n}}\wedge{\mathsf{n}}^{- a_i}.$$

Without loss of generality, we can assume that the pairs $(U_i,U_i')$ are independent for all $i{\geqslant}1$. Now we construct the coupling $$(S_k,S_k')_{k{\geqslant}1}
    =\left(\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i\big),
        \sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU'_i\big)\right)_{k{\geqslant}1}$$ of $$S_k:=\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i\big).$$ Since $U'_i-U_i=\Delta U_i$ is either $\pm a_i$ or $0$, we know that $$\begin{aligned}
(S_k-& S_k')_{k{\geqslant}1}\\
    &=\left(\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU'_i\big) -\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i\big)\right)_{k{\geqslant}1}\\
    &=\left(\sum_{i=1}^kR_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}B(U'_i-U_i)\big)\right)_{k{\geqslant}1}\end{aligned}$$ is a random walk on ${\mathds R}^n$ whose steps are symmetrically (but not necessarily identically) distributed and take only the values $\pm |e^{tA}(x-y)| e_1$ and $0$.

Set $S^j_{k}=\sum_{i=1}^k\eta^j_{i}$ and $S^{j\,\prime}_{k}=\sum_{i=1}^k\eta^{j\,\prime}_{i}$ for $1{\leqslant}j{\leqslant}n$, where $$(\eta^{1}_{i},\ldots,\eta^{n}_{i})=R_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i\big)$$ and $$(\eta_{i}^{1\,\prime},\ldots,\eta^{n\,\prime}_{i})=R_{e^{tA}(x-y)}\big(e^{(t_1+\cdots+t_i)A}BU_i^{\prime}\big).$$ Then $(S^1_k-S^{1\,\prime}_k)_{k{\geqslant}1}$ is a random walk on ${\mathds R}$ whose steps are independent and attain the values $-|e^{tA}(x-y)|$, $0$ and $|e^{tA}(x-y)|$ with probabilities $\frac 12(1-p_i)$, $p_i$ and $\frac 12(1-p_i)$, respectively; the values of the $p_i$ are given by $$\begin{aligned}
    p_i
    :&={\mathds P}(\eta^{1\,\prime}_i-\eta^1_{i}=0)\\
    &= \left({\mathsf{n}}- \tfrac 12 ({\mathsf{n}}\wedge{\mathsf{n}}^{-a_i}+{\mathsf{n}}\wedge{\mathsf{n}}^{a_i})\right)({\mathds R}^d)\\
    &= 1-{\mathsf{n}}\wedge{\mathsf{n}}^{-a_i}({\mathds R}^d).\end{aligned}$$ Since $S^j_{k}=S^{j\,\prime}_{k}$ for $2{\leqslant}j{\leqslant}n$, we get $$\label{proofs6}
  \|\delta_{e^{tA}(x-y)}*\mu_{t_1,\cdots,t_k}-\mu_{t_1,\cdots,t_k}\|_{{\mathrm{Var}}}
    {\leqslant}2\,{\mathds P}(T^S>k),$$ where $$T^S=\inf\{i{\geqslant}1\::\: S^1_{i}=S^{1\,\prime}_{i}+|e^{tA}(x-y)|\}.$$

*Step 5*. Since the real parts of all eigenvalues of $A$ are non-positive and since all purely imaginary eigenvalues are semisimple, we know from [@Bern Proposition 11.7.2, Page 438] that $C_A:=\sup_{t{\geqslant}0}\|e^{tA}\|<\infty$. In particular, when $t{\geqslant}t_1+\cdots+t_i$, $$\big|e^{(t-(t_1+\cdots+t_i))A}(x-y)\big|{\leqslant}C_A|x-y|.$$ From we get that for all $i{\geqslant}1$ and $x, y\in{\mathds R}^n$ with $|x-y|{\leqslant}\delta(C_A\|\bar{B}\|)^{-1}$, $$\label{proofcon}\begin{aligned}
    \frac 12(1-p_i)
    &= \frac{1}{2}\big({\mathsf{n}}\wedge\big(\delta_{-a_i}*{\mathsf{n}})\big)({\mathds R}^d)\\
    &{\geqslant}\frac{1}{2}\inf_{|a|{\leqslant}C_A\|\bar{B}\||x-y|}{\mathsf{n}}\wedge (\delta_a*{\mathsf{n}})({\mathds R}^d)\\
    &{\geqslant}\frac{1}{2}\inf_{|a|{\leqslant}\delta}{\mathsf{n}}\wedge (\delta_a*{\mathsf{n}})({\mathds R}^d)\\
    &=:\frac{1}{2}\,\gamma(\delta)>0.
\end{aligned}$$

We will now estimate ${\mathds P}(T^S>k)$. Let $V_i$, $i{\geqslant}1$, be independent symmetric random variables on ${\mathds R}$, whose distributions are given by $${\mathds P}(V_i=z)
    = \begin{cases}
        \frac 12(1-p_i),     &\text{if\ \ } z=-|e^{tA}(x-y)|;\\
        \frac 12(1-p_i),     &\text{if\ \ } z=|e^{tA}(x-y)|;\\
        \qquad p_i,          &\text{if\ \ } z=0.
    \end{cases}$$ Set $Z_k:=\sum_{i=1}^k V_i$. We have seen earlier that $$T^S=\inf\{k{\geqslant}1\::\: Z_k=|e^{tA}(x-y)|\}.$$

For any $k{\geqslant}1$, let $$\eta=\eta(k):=\#\big\{i\::\: i{\leqslant}k\textrm{ and }V_i\neq 0\big\}$$ and set $\tilde{Z}_k :=\sum_{i=1}^k\tilde{V}_i$, where $\tilde{V}_i$ denotes the $i$th $V_j$ such that $V_j\neq 0$. Then, $\tilde{Z}_k$ is a symmetric random walk with iid steps which are either $-|e^{tA}(x-y)|$ or $|e^{tA}(x-y)|$ with probability $1/2$. Define $$T^{\tilde{Z}}:=\inf\{k{\geqslant}1\::\: \tilde{Z}_k=|e^{tA}(x-y)|\}.$$ By , $$\label{lll1}\begin{aligned}
    {\mathds P}(T^S>k)
    &={\mathds P}\left(T^S>k,\; \eta{\geqslant}\frac12\,\gamma(\delta)k\right)
        +{\mathds P}\left(T^S>k,\, \eta{\leqslant}\frac12\, \gamma(\delta)k\right)\\
    &{\leqslant}{\mathds P}\left(T^{\tilde{Z}}> \frac12\,\gamma(\delta)k\right)
        +{\mathds P}\bigg(\eta{\leqslant}\frac{1}{2}\sum_{i=1}^k(1-p_i)\bigg)\\
    &{\leqslant}{\mathds P}\left(T^{\tilde{Z}}>\frac12\,\gamma(\delta)k\right)
        +{\mathds P}\bigg(\Big|\eta-\sum_{i=1}^k(1-p_i)\Big|{\geqslant}\frac{1}{2}\sum_{i=1}^k(1-p_i)\bigg).
\end{aligned}$$

Note that $$\eta=\eta(k)=\sum_{i=1}^k\zeta_i,$$ where $\zeta_i = {\mathds 1}_{\{V_i\neq 0\}}$, ${1{\leqslant}i{\leqslant}k}$, are independent random variables with ${\mathds P}(\zeta_i = 0) = p_i$ and ${\mathds P}(\zeta_i=1)=1-p_i$. Chebyshev’s inequality shows that $$\label{lll2}\begin{aligned}
    {\mathds P}\bigg(\Big|\eta-\sum_{i=1}^k(1-p_i)\Big|{\geqslant}\frac{1}{2}\sum_{i=1}^k(1-p_i)\bigg)
    &{\leqslant}\frac{4{\mathrm{Var}}(\eta)}{\Big(\sum_{i=1}^k(1-p_i)\Big)^2}\\
    &=\frac{4\sum_{i=1}^kp_i(1-p_i)}{\Big(\sum_{i=1}^k(1-p_i)\Big)^2}\\
    &{\leqslant}\frac{4(1-\gamma(\delta))\sum_{i=1}^k(1-p_i)}{\Big(\sum_{i=1}^k(1-p_i)\Big)^2}\\
    &{\leqslant}\frac{4(1-\gamma(\delta))}{\gamma(\delta)k}.
\end{aligned}$$ For the second and the last inequality we have used .

On the other hand, by Lemma \[lemmaproofth13\] below, $$\begin{aligned}
    {\mathds P}\bigg(T^{\tilde{Z}}>\frac{\gamma(\delta)k}{2}\bigg)
    &={\mathds P}\bigg(\max_{i{\leqslant}\big[\frac{\gamma(\delta)k}{2}\big]}\tilde{Z}_i<  |e^{tA}(x-y)|\bigg)\\
    &{\leqslant}2\,{\mathds P}\left(0{\leqslant}\tilde{Z}_{\big[\frac{\gamma(\delta)k}{2}\big]} {\leqslant}|e^{tA}(x-y)|\right).\end{aligned}$$ From the construction above, we know that $(\tilde{Z}_k)_{k{\geqslant}1}$ is a symmetric random walk with iid steps with values $\pm |e^{tA}(x-y)|$. Using the central limit theorem we find for sufficiently large values of $k{\geqslant}k_0$ and some constant $C=C(k_0)$ $$\label{lll3}\begin{aligned}
    {\mathds P}\left(T^S>\frac12\,\gamma(\delta)k\right)
    &= 2\,{\mathds P}\left(0{\leqslant}\frac{Z_k}{|e^{tA}(x-y)|\sqrt{\big[\frac{\gamma(\delta)k}{2}\big]}} {\leqslant}{\left[\frac{\gamma(\delta)k}{2}\right]}^{-1/2}\right)\\
    &{\leqslant}\frac{C}{\sqrt{2\pi}} \int_{0}^{  {\left[\frac{\gamma(\delta)k}{2}\right]}^{-1/2}} e^{-u^2/2}\,du\\
    &{\leqslant}\frac{C_{\gamma(\delta)}}{\sqrt k}.
\end{aligned}$$

Combining , and gives for all $x,y\in{\mathds R}^n$ with $|x-y| {\leqslant}\delta(C_A\|\bar{B}\|)^{-1}$, $t{\geqslant}t_1+\cdots+t_k$ and $k{\geqslant}k_0$ that $${\mathds P}\big(T^S>k\big)
    {\leqslant}\frac{C_{\gamma(\delta)}}{\sqrt k}+\frac{4(1-\gamma(\delta))}{\gamma(\delta)k}.$$ Finally, yields for all $x,y\in{\mathds R}^n$ with $|x-y|{\leqslant}\delta(C_A\|\bar{B}\|)^{-1}$, $t{\geqslant}t_1+\cdots+t_k$ and $k{\geqslant}1$, that $$\label{proofs7}
  \|\delta_{e^{tA}(x-y)}*\mu_{t_1,\cdots,t_k}-\mu_{t_1,\cdots,t_k}\|_{{\mathrm{Var}}}
    {\leqslant}\frac{C_{1,\delta,{\mathsf{n}}}}{\sqrt k}.$$

*Step 6*. If we combine , and , we obtain that for all $x, y\in{\mathds R}^n$ with $|x-y|{\leqslant}\delta(C_A\|\bar{B}\|)^{-1}$, $$\begin{aligned}\label{proofs8}
 \|&P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}\\
 &{\leqslant}2e^{-C_\varepsilon t}
  +C_{1,\delta,{\mathsf{n}}}\sum_{k=1}^\infty\frac{1}{\sqrt{k}} \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t<t_1+\cdots+t_{k+1}} \!\!\!\!\!\!\!\!C_\varepsilon^{k+1}e^{-C_\varepsilon(t_1+\cdots+t_{k+1})}\,dt_1\cdots dt_{k+1}\\
    &{\leqslant}2e^{-C_\varepsilon t}+C_{1,\delta,{\mathsf{n}}}e^{-C_\varepsilon t}\sum_{k=1}^\infty \frac{C_\varepsilon^{k+1}}{\sqrt{k}} \mathop{\int\cdots\int}\limits_{t_1+\cdots+t_k{\leqslant}t}\!\!dt_1\cdots dt_{k}\\
    &{\leqslant}2e^{-C_\varepsilon t}+C_{1,\delta,{\mathsf{n}}}C_\varepsilon \sum_{k=1}^\infty\frac{C_\varepsilon^k t^k}{\sqrt{k}\,k!}e^{-C_\varepsilon t}\\
    &{\leqslant}2e^{-C_\varepsilon t}+\frac{\sqrt{2}C_{1,\delta,{\mathsf{n}}} C_\varepsilon (1-e^{-C_\varepsilon t})}{\sqrt{C_\varepsilon t}}\\
    &{\leqslant}\frac{C_{2,\epsilon,\delta,{\mathsf{n}}}}{\sqrt{t}},
\end{aligned}$$ where the penultimate inequality follows as in [@SW Proposition 2.2].

For any $x,$ $y\in{\mathds R}^n$, set $k=\left[\frac{C_A\|\bar{B}\||x-y|}{\delta}\right]+1$. Pick $x_0, x_1, \ldots, x_k\in{\mathds R}^n$ such that $x_0=x$, $x_k=y$ and $|x_i-x_{i-1}|{\leqslant}\delta(C_A\|\bar{B}\|)^{-1}$ for $1{\leqslant}i{\leqslant}k$. By , $$\begin{aligned}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}
    &{\leqslant}\sum_{i=1}^k\|P_t(x_i,\cdot)-P_t(x_{i-1},\cdot)\|_{{\mathrm{Var}}}\\
    &{\leqslant}\frac{C_{\epsilon,\delta,{\mathsf{n}},A,B}(1+|x-y|)}{\sqrt{t}},\end{aligned}$$ which finishes the proof of .

The following two lemmas have been used in the proof of Theorem \[th1\] above. For the sake of completeness we include their proofs.

\[lemmaproofth12\] Let $B\in{\mathds R}^{n\times d}$ and $(Z_t)_{t{\geqslant}0}$ be a $d$-dimensional Lévy process with characteristic exponent $\Phi$ as in . Then, $(Z^B_t)_{t{\geqslant}0}:=(BZ_t)_{t{\geqslant}0}$ is a Lévy process on (a subspace of) ${\mathds R}^n$, and the corresponding characteristic exponent is $${\mathds R}^n\ni\xi \mapsto \Phi_B(\xi):=\Phi(B^\top\xi).$$ The Lévy triplet $(Q_B,b_B,\nu_B)$ of $(Z^B_t)_{t{\geqslant}0}$ is given by $ Q_B=BQB^\top$, $\nu_B(C) = \nu\{y : By\in C\}$ and $$b_B=Bb+\int_{x\neq 0} Bx\,\big({\mathds 1}_{\{z\in{\mathds R}^d:|z|{\leqslant}1\}}(Bx) - {\mathds 1}_{\{z\in{\mathds R}^d:|z|{\leqslant}1\}}(x)\big)\, \nu(dx).$$

For all $\xi\in{\mathds R}^n$ and $t{\geqslant}0$, we have $${\mathds E}(e^{i{\langle\xi,Z^B_t\rangle}})
        ={\mathds E}(e^{i{\langle\xi,BZ_t\rangle}})
        ={\mathds E}(e^{i{\langleB^\top\xi,Z_t\rangle}})
        =e^{-t\Phi(B^\top\xi)}.$$ The assertion follows from and some straightforward calculations.

\[lemmaproofth11\] Let $A\in{\mathds R}^{n\times n}$, $B\in{\mathds R}^{n\times d}$ and $(Z_t)_{t{\geqslant}0}$ be a $d$-dimensional Lévy process with the characteristic exponent $\Phi$ as in . For all $t>0$ the random variables $\int_0^t e^{(t-s)A}B\,dZ_s$ and $\int_0^t e^{sA}B\,dZ_s$ have the same probability distribution. Furthermore, both random variables are infinitely divisible, and the characteristic exponent (log-characteristic function) is given by $${\mathds R}^n\ni\xi \mapsto \Phi_t(\xi):=\int_0^t\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds.$$

We first assume that $n=d$ and $B={\operatorname{id}}_{{\mathds R}^d}$. For any $t>0$, we can use Lemma \[lemmaproofth12\] and follow the proof of [@SA (17.3)] to deduce $${\mathds E}\left[\exp\left(i{\left\langle\xi,\int_0^t e^{(t-s)A}\,dZ_s\right\rangle}\right)\right]
        =\exp\left[-\int_0^t\Phi(e^{(t-s)A^\top}\xi)\,ds\right]$$ for all $\xi\in{\mathds R}^d$. Similarly, for every $\xi\in{\mathds R}^d$, $${\mathds E}\left[\exp\left(i{\left\langle\xi,\int_0^t e^{sA}\,dZ_s\right\rangle}\right)\right]
        =\exp\left[-\int_0^t\Phi(e^{sA^\top}\xi)\,ds\right].$$ Since $$\exp\left[-\int_0^t\Phi(e^{(t-s)A^\top}\xi)\,ds\right]
        =\exp\left[-\int_0^t\Phi(e^{sA^\top}\xi)\,ds\right],$$ it follows that $\int_0^t e^{(t-s)A}B\,dZ_s$ and $\int_0^t e^{sA}B\,dZ_s$ have the same law.

Now replace in the preceding calculations $A$ with $\frac 1k\,A$, $k{\geqslant}1$, and set $Y_k := \int_0^t e^{s\,\frac 1k\, A}\,dZ_s$. Denote by $Y_k^{(j)}$, $1{\leqslant}j{\leqslant}k$, independent copies of $Y_k$. It is straightforward to see that $\sum_{j=1}^k Y_k^{(j)}$ and $\int_0^t e^{sA}\,dZ_s$ have the same law. This proves the infinite divisibility.

If $n\neq d$, we consider, as in Lemma \[lemmaproofth12\], the Lévy process $(Z^B_t)_{t{\geqslant}0}:=(BZ_t)_{t{\geqslant}0}$ on (a subspace of) ${\mathds R}^n$. Then, for any $\xi\in{\mathds R}^n$, $${\mathds E}\left[\exp\left(i{\left\langle\xi,\int_0^t e^{(t-s)A}B\,dZ_s\right\rangle}\right)\right]
        ={\mathds E}\left[\exp\left(i{\left\langle\xi,\int_0^t
        e^{(t-s)A}\,dZ^B_s\right\rangle}\right)\right],$$ and the claim follows from the first part of our proof.

The following result presents the upper estimate for the distribution of the maximum of a symmetric random walk, by using the reflection principle. Since we could not find a precise reference in the literature, we include the complete proof for the readers’ convenience.

\[lemmaproofth13\] Consider a random walk $(S_i)_{i{\geqslant}1}$ on ${\mathds Z}$ with iid steps, which attain the values $-1$, $1$ and $0$ with probabilities $(1-r)/2$, $(1-r)/2$ and $r$ $(0{\leqslant}r<1)$, respectively. Then for any positive integers $a$ and $k$, we have $$\label{lemmaproofth1311}
    2{\mathds P}(S_k> a)
    {\leqslant}{\mathds P}\Big(\max_{i{\leqslant}k}S_i{\geqslant}a\Big)
    {\leqslant}2{\mathds P}(S_k{\geqslant}a)$$ and $$2{\mathds P}\big(0<S_k<a\big)
    {\leqslant}{\mathds P}\Big(\max_{1{\leqslant}i{\leqslant}k}S_i < a\Big){\leqslant}2{\mathds P}\big(0{\leqslant}S_k{\leqslant}a\big).$$

Fix any positive integer $a$ and define $\tau :=\tau_a := \inf\{i{\geqslant}1\::\: S_i=a\}$. Since the random walk has iid steps, it is obvious that $(S_{i+\tau}-S_\tau)_{i{\geqslant}0}$ and $(S_i)_{i{\geqslant}0}$ are independent random walks having the same law. Observing that $S_\tau = a$ and $\left\{\max_{i{\leqslant}k}S_i{\geqslant}a\right\} = \{\tau {\leqslant}k\}$ we find, therefore, $$\begin{aligned}
    {\mathds P}\Bigl(\max_{i{\leqslant}k}S_i{\geqslant}a\Bigr)
    &={\mathds P}\Bigl(\max_{i{\leqslant}k}S_i{\geqslant}a,\; S_k{\geqslant}a\Bigr) + {\mathds P}\Bigl(\max_{i{\leqslant}k}S_i{\geqslant}a,\; S_k < a\Bigr)\\
    &={\mathds P}(S_k{\geqslant}a)+{\mathds P}(\tau{\leqslant}k,S_k<S_\tau)\\
    &={\mathds P}(S_k{\geqslant}a)+{\mathds P}(\tau{\leqslant}k,S_k>S_\tau)\\
    &={\mathds P}(S_k{\geqslant}a)+{\mathds P}(S_k>a).
    \end{aligned}$$ From this we conclude that $$2{\mathds P}(S_k{\geqslant}a)
        {\geqslant}{\mathds P}\Bigl(\max_{i{\leqslant}k}S_i{\geqslant}a\Bigr)
        {\geqslant}2{\mathds P}(S_k>a).$$ Since ${\mathds P}(S_k{\geqslant}0)={\mathds P}(S_k{\leqslant}0){\geqslant}1/2$, we see $$\begin{aligned}
        {\mathds P}\Bigl(\max_{i{\leqslant}k}S_i< a\Bigr)
        &=1-{\mathds P}\Bigl(\max_{i{\leqslant}k}S_i{\geqslant}a\Bigr)\\
        &{\leqslant}1-2{\mathds P}(S_k>a)\\
        &{\leqslant}2\big({\mathds P}(S_k{\geqslant}0)-{\mathds P}(S_k>a)\big)\\
        &=2{\mathds P}(0{\leqslant}S_k{\leqslant}a);
    \end{aligned}$$ the other inequality follows similarly if we use ${\mathds P}(S_k>0)={\mathds P}(S_k<0){\leqslant}1/2$.

Next, we turn to the proof of Theorem \[coup\].

*Step 1*. As in the proof of Lemma \[lemmaproofth11\] we may, without loss of generality, assume that $n=d$ and $B={\operatorname{id}}_{{\mathds R}^d}$. For $t>0$, denote by $\mu_t$ the law of $X_t^0:=\int_0^t e^{(t-s)A}\,dZ_s$. According to Lemma \[lemmaproofth11\], the law $\mu_t$ is an infinitely divisible probability distribution, and the characteristic exponent of $\mu_t$ is given by $$\Phi_t(\xi):=\int_0^t\Phi\big(e^{sA^\top}\xi\big)\,ds.$$ Since the driving Lévy process $(Z_t)_{t{\geqslant}0}$ has no Gaussian part, the Lévy [triplet]{} $(0,b_t,\nu_t)$ of $\Phi_t$ is given by, cf. [@SAT Theorem 3.1], $$\begin{gathered}
        \nu_t(C)= \int_0^t\nu(e^{-sA}C)\,ds,\qquad C\in\mathscr{B}({\mathds R}^d\setminus\{0\}),\\
        b_t= \int_0^t e^{sA}b\,ds + \int_{z\neq 0} \int_0^t e^{sA}z\Big({\mathds 1}_{\{|z|{\leqslant}1\}} \big(e^{sA}z\big) - {\mathds 1}_{\{|z|{\leqslant}1\}}(z) \Big)\,ds\,\nu(dz).
    \end{gathered}$$ For every $r>0$, let $\{\mu_t^r, t{\geqslant}0\}$ be the family of infinitely divisible probability measures on ${\mathds R}^d$ whose Fourier transform is of the form $\widehat{\mu}^r_t(\xi)=\exp(-\Phi_{t,r}(\xi))$, where $$\Phi_{t,r}(\xi) = \int_{|z|{\leqslant}r} \left(1-e^{i{\langle\xi,z\rangle}}+i{\langle\xi,z\rangle}\right)\,\nu_t(dz)$$ with $\nu_t$ as above.

Set $h(t):=1\big/\varphi_t^{-1}(1)$. Following the proof of [@SSW Propostion 2.2], the conditions and ensure that there exists $t_1>0$ such that for all $t{\geqslant}t_1$, the measure $\mu^{h(t)}_t$ has a density $p^{h(t)}_t\in C^{n+2}_b({\mathds R}^d)$; moreover, $$\label{proofcoup1}
       \big|\nabla p^{h(t)}_t(y)\big|
       {\leqslant}c(n,\Phi)\,h(t)^{-(n+1)}\big(1+h(t)^{-1}|y|\big)^{-(n+1)}$$ holds for all $y\in{\mathds R}^d$.

*Step 2*. For $r>0$ and $\xi\in{\mathds R}^d$, define $$\Psi_{t,r}(\xi)
    := \Phi_t(\xi)-\Phi_{t,r}(\xi)
    = \int_{|z|>r}\left(1-e^{i{\langle\xi,z\rangle}}\right)\,\nu_t(dz) - i{\left\langle\xi,\int_{1<|z|{\leqslant}r}z\,\nu_t(dz)-b_t\right\rangle}.$$ Since $\Psi_{t,r}$ is given by a Lévy-Khintchine formula, it is the characteristic exponent of some $d$-dimensional infinitely divisible random variable. Let $\{{\mathsf{\pi}}_t^{r}, t{\geqslant}0\}$ be the family of infinitely divisible measures whose Fourier transforms are of the form $\widehat{{\mathsf{\pi}}}^{r}_t(\xi)=\exp(-\Psi_{t,r}(\xi))$. Clearly, $\mu_t=\mu_t^r*{\mathsf{\pi}}_t^{r}$ for all $t,r>0$.

Let $P_t(x,\cdot)$ and $P_t$ be the transition function and the transition semigroup of the Ornstein-Uhlenbeck process $\{X_t^x\}_{t{\geqslant}0}$ given by . For all $f\in B_b({\mathds R}^d)$ we have $$\begin{aligned}
        P_tf(x)
        &=\int f\bigl(e^{tA}x+z\bigr)\,\mu_t(dz)\\
        &=\int f\bigl(e^{tA}x+z\bigr)\,\mu^r_t*{\mathsf{\pi}}_t^{r}(dz)\\
        &=\iint f\bigl(e^{tA}x+z_1+z_2\bigr)\,{\mathsf{\pi}}_t^{r}(dz_1)\,\mu^r_t(dz_2).
    \end{aligned}$$

Taking $r=h(t)$ we get, using the conclusions of step 1, that for all $t{\geqslant}t_1$ and $x\in{\mathds R}^d$, $$\begin{aligned}
        P_tf(x)
        &= \int p^{h(t)}_t(z_2)\,dz_2 \int f\bigl(e^{tA}x+z_1+z_2\bigr)\,{\mathsf{\pi}}_t^{h(t)}(dz_1)\\
        &= \int p^{h(t)}_t\bigl(z_2-e^{tA}x\bigr)\,dz_2 \int f(z_1+z_2)\,{\mathsf{\pi}}_t^{h(t)}(dz_1).
    \end{aligned}$$ If $\|f\|_\infty{\leqslant}1$, then $$\bigg\|\int f(z_1+\cdot)\,{\mathsf{\pi}}_t^{h(t)}(dz_1)\bigg\|_\infty
        {\leqslant}\|f\|_\infty\, {\mathsf{\pi}}_t^{h(t)}({\mathds R}^d) {\leqslant}1.$$

*Step 3*. For all $x,y\in{\mathds R}^d$, $$\label{proofcoup2}\begin{aligned}
    \|P_t(x,&\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}\\
    &= \sup_{\|f\|_\infty{\leqslant}1}\big|P_tf(x)-P_tf(y)\big|\\
    &= \sup_{\|f\|_\infty{\leqslant}1} \bigg|\int p^{h(t)}_t\bigl(z_2-e^{tA}x\bigr)\,dz_2 \int f(z_1+z_2)\,{\mathsf{\pi}}_t^{h(t)}(dz_1)\\
    &\qquad\qquad\quad\mbox{}-\int p^{h(t)}_t\bigl(z_2-e^{tA}y\bigr)\,dz_2 \int f(z_1+z_2)\,{\mathsf{\pi}}_t^{h(t)}(dz_1)\bigg|\\
    &{\leqslant}\sup_{\|g\|_\infty{\leqslant}1} \bigg|\int g(z)p^{h(t)}_t\bigl(z-e^{tA}x\bigr)\,dz
    - \int g(z)p^{h(t)}_t\bigl(z-e^{tA}y\bigr)\,dz\bigg|\\
    &= \sup_{\|g\|_\infty{\leqslant}1} \bigg|\int g(z)\Big(p^{h(t)}_t\bigl(z-e^{tA}x\bigr)-p^{h(t)}_t\bigl(z-e^{tA}y\bigr)\Big)\, dz\bigg|\\
    &= \int \Big|p^{h(t)}_t\bigl(z-e^{tA}x\bigr)-p^{h(t)}_t\bigl(z-e^{tA}y\bigr)\Big|\,dz.
    \end{aligned}$$

With the argument used in the proof of [@SSW Theorem 3.1], follows from and .

*Step 4*. By assumption , $$\varphi_\infty(\rho):=\sup_{|\xi|{\leqslant}\rho}\int_0^\infty{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds$$ is finite on $(0,\infty)$; in particular, $\varphi_\infty^{-1}(1)\in(0,\infty]$. On the other hand, for any $t{\geqslant}t_0$, according to , $$\begin{aligned}
\int \exp\left(-\int_0^t{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds\right)& |\xi|^{n+2} \,d\xi\\
       &{\leqslant}\int \exp\left(-\int_0^{t_0}{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds\right) |\xi|^{n+2} \,d\xi\\
       &=:C(t_0)<\infty. \end{aligned}$$ Since the function $t\mapsto \varphi^{-1}_t(1)$ is decreasing on $(0,\infty]$, holds. This finishes the proof.

Appendix {#sec-appendix}
========

Gradient Estimates for Ornstein-Uhlenbeck Processes {#subsec-appendix1}
---------------------------------------------------

Motivated by [@SSW Theorem 1.3], we have the following results for gradient estimates of an Ornstein-Uhlenbeck process. This is the counterpart of Theorem \[coup\]. For $t, \rho>0$, define $$\varphi(\rho)
    :=\sup_{|\xi|{\leqslant}\rho}{\ensuremath{\operatorname{Re}}}\Phi\bigl(B^\top \xi\bigr)\quad\textrm{ and }\quad \varphi_t(\rho)
    :=\sup_{|\xi|{\leqslant}\rho}\int_0^t{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds,$$ where $\Phi$ is the characteristic exponent of the driving Lévy process $(Z_t)_{t{\geqslant}0}$ from .

\[strong\] Let $P_t(x,\cdot)$ be the transition function of the $n$-dimensional Ornstein-Uhlenbeck process $\{X_t^x\}_{t{\geqslant}0}$ given by . Assume that $$\label{strong1}
        \liminf_{|\xi|\rightarrow\infty} \frac{{\ensuremath{\operatorname{Re}}}\Phi\bigl(B^\top \xi\bigr)}{\log (1+|\xi|)} =\infty.$$ If for any $C>0$, $$\label{strong2}
        \int \exp\left[-C t{\ensuremath{\operatorname{Re}}}\Phi\bigl(B^\top\xi\bigr)\right] |\xi|^{n+2} \,d\xi
        = \mathsf{O}\left(\varphi^{-1}\Big(\frac{1}{t}\Big)^{2n+2}\right) \qquad\text{as\ \ } t\to 0,$$ then there exists $c>0$ such that for all $t>0$ and $f\in{B}_b({\mathds R}^n)$, $$\label{strong3}
        \|\nabla P_t f\|_\infty{\leqslant}c\|f\|_\infty \,\varphi^{-1}\Big(\frac{1}{t\wedge1}\Big).$$ If, in addition, $$\xi\mapsto\int_0^\infty{\ensuremath{\operatorname{Re}}}\Phi\big(B^\top e^{sA^\top}\xi\big)\,ds\quad \textrm{ is locally bounded},$$ then there exist $t_1,c>0$ such that for $t{\geqslant}t_1$ and $f\in{B}_b({\mathds R}^n)$, $$\label{strong4}
        \|\nabla P_t f\|_\infty
        {\leqslant}c\,\|f\|_\infty \bigg[\|e^{tA}\|\,\varphi_t^{-1}(1)\bigg],$$ where $\|M\|=\sup_{|x|{\leqslant}1}|Mx|$ denotes the norm of the matrix of $M$.

To illustrate the power of Theorem \[strong\], we consider

\[examplegradient\] Let $\mu$ be a finite nonnegative measure on the unit sphere ${\mathds{S}}\subset {\mathds R}^n$ and assume that $\mu$ is nondegenerate in the sense that its support is not contained in any proper linear subspace of ${\mathds R}^n$. Let $\alpha\in(0,2)$, $\beta\in (0,\infty]$ and assume that the Lévy measure $\nu$ satisfies $$\nu(C)
    {\geqslant}\int_0^{r_0}\int_{{\mathds{S}}}{\mathds 1}_C(s\theta)s^{-1-\alpha} \,ds\, \mu(d\theta)
    + \int_{r_0}^\infty\int_{{\mathds{S}}}{\mathds 1}_C(s\theta)s^{-1-\beta}\,ds\, \mu(d\theta)$$ for some constant $r_0>0$ and all $C\in {\mathscr{B}}({\mathds R}^n\setminus\{0\})$. Consider the following Ornstein-Uhlenbeck process $X_t$ on ${\mathds R}^n$ given by $$dX_t=AX_t\,dt+dZ_t,$$ where $(Z_t)_{t{\geqslant}0}$ is a Lévy process on ${\mathds R}^n$ with the Lévy measure $\nu$. By Theorem \[strong\] there exists a constant $c>0$ such that for all $t>0$ and $f\in B_b({\mathds R}^n)$, $$\|\nabla P_tf\|_\infty
        {\leqslant}c\,\|f\|_\infty \, (t\wedge 1)^{-1/\alpha}.$$ Furthermore, if the real parts of all eigenvalues of $A$ are negative, then there exists a constant $c>0$ such that for all $t>0$ and $f\in B_b({\mathds R}^n)$, $$\|\nabla P_tf\|_\infty
        {\leqslant}c\,\|f\|_\infty \,\frac{\|e^{tA}\|\quad}{(t\wedge 1)^{1/\alpha}}.$$

Recently, F.-Y. Wang [@W2 Theorem 1.1] has presented explicit gradient estimates for Ornstein-Uhlenbeck processes, by assuming that the corresponding Lévy measure has absolutely continuous (*with respect to Lebesgue measure*) lower bounds. Since lower bounds of Lévy measure in Example \[examplegradient\] could be much irregular, Theorem \[strong\] is more applicable than [@W2 Theorem 1.1].

Assuming the conditions and , we can mimic the proof of [@SSW Theorem 3.2] to show that there exist $t_1, C>0$ such that for all $x,y\in{\mathds R}^n$ and $t{\leqslant}t_1$, $$\label{proofstrong}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}
    {\leqslant}C\,|e^{tA}(x-y)|\,\varphi^{-1}\Big(\frac{1}{t}\Big).$$ Thus we can apply to find for all $f\in B_b({\mathds R}^n)$ with $\|f\|_\infty= 1$, $$\label{proofstrong}\begin{aligned}
    |\nabla P_tf(x)|
    &{\leqslant}\limsup_{y\to x}\frac{|P_tf(x)-P_tf(y)|}{|y-x|}\\
    &{\leqslant}\limsup_{y\to x}\frac{\sup_{\|w\|_\infty{\leqslant}1}|P_tw(x)-P_tw(y)|}{|y-x|}\\
    &{\leqslant}\limsup_{y\to x}\frac{\|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}}{|y-x|}\\
    &{\leqslant}C\,\|e^{tA}\|\,\varphi^{-1}\Big(\frac{1}{t}\Big)\\
    &{\leqslant}\Big[C\,\sup_{s{\leqslant}t_1}\|e^{sA}\|\Big]\,\varphi^{-1}\Big(\frac{1}{t}\Big).
\end{aligned}$$ Because of the Markov property of the semigroup $P_t$, the function $$t\mapsto \sup_{f\in B_b({\mathds R}^n),\, \|f\|_\infty =1}{ \|\nabla P_t f\|_\infty}$$ is deceasing. Combining this and yields .

The assertion follows if we combine the above argument with : there exist $t_2,C>0$ such that for all $x,y\in{\mathds R}^n$ and $t{\geqslant}t_2$, $$\begin{gathered}
    \|P_t(x,\cdot)-P_t(y,\cdot)\|_{{\mathrm{Var}}}{\leqslant}C\,|e^{tA}(x-y)|\,\varphi_t^{-1}(1).
\qedhere\end{gathered}$$

Proof of Proposition \[improvement\] {#subsec-appendix2}
------------------------------------

Because of , we can choose a closed subset $F\subset \overline{B(z_0, \varepsilon)}$ such that $0\notin F$ and $$\int_F \frac{dz}{\rho_0(z)} < \infty.$$ By the Cauchy-Schwarz inequality, we have $$\left(\int_F \rho_0(z)\,dz\right)^{-1}
        {\leqslant}\frac{1}{{\operatorname{Leb}}(F)^2} \int_F \frac{dz}{\rho_0(z)} < \infty.$$ Hence, $$K:=\int_F\rho_0(z)\,dz>0.$$

Since $F$ is a compact set and $0\notin F$, there exists some $\delta_0>0$ such that $
        0\notin F+\overline{B(0,\delta_0)},
    $ where $F+\overline{B(0,\delta_0)}:=\{a+b:a\in F, |b|{\leqslant}\delta_0\}.$ Since $\rho_0$ is locally integrable, we know that $$K{\leqslant}\int_{F+\overline{B(0,\delta_0)}}\rho_0(z)\,dz<\infty.$$ The remainder of the proof is now similar to the argument which shows that the shift $x\mapsto \|f(\cdot - x)-f\|_{L^1}$, $f\in L^1({\mathds R}^d,{\operatorname{Leb}})$, is continuous, see e.g. [@STR Lemma 6.3.5] or [@RSCC Theorem 14.8]: choose $\chi\in C_c^\infty ({\mathds R}^d)$ such that ${\operatorname{supp}}\chi\subset F+\overline{B(0,\delta_0)}$ and $$\int_{F+\overline{B(0,\delta_0)}}|\rho_0(z)-\chi(z)|\, dz{\leqslant}\frac{K}{4}.$$ Therefore, for any $x\in{\mathds R}^d$ with $|x|{\leqslant}\delta_0$, we obtain $$\begin{aligned}
    \int_F & |\rho_0(z)-\rho_0(z-x)|\,dz\\
    &{\leqslant}\int_F|\rho_0(z)-\chi(z)|\,dz
        +\int_F|\chi(z)-\chi(z-x)|\,dz
        +\int_F|\rho_0(z-x)-\chi(z-x)|\,dz\\
    &= \int_F|\rho_0(z)-\chi(z)|\,dz
        +\int_F|\chi(z)-\chi(z-x)|\,dz
        +\int_{F+x}|\rho_0(z)-\chi(z)|\,dz\\
    &{\leqslant}2\int_{F+\overline{B(0,\delta_0)}}|\rho_0(z)-\chi(z)|\,dz
        +\int_F|\chi(z)-\chi(z-x)|\,dz\\
    &{\leqslant}\frac{K}{2}+\int_F|\chi(z)-\chi(z-x)|\,dz.
    \end{aligned}$$ By the dominated convergence theorem we see that $$x\mapsto\int_F|\chi(z)-\chi(z-x)|\,dz$$ is continuous on ${\mathds R}^d$. Therefore, there exists $0<\delta{\leqslant}\delta_0$ such that $$\sup_{x\in{\mathds R}^d, |x|{\leqslant}\delta}\int_F|\chi(z)-\chi(z-x)|\,dz{\leqslant}\frac{K}{4}$$ and, in particular, $$\sup_{x\in{\mathds R}^d, |x|{\leqslant}\delta}\int_F|\rho_0(z)-\rho_0(z-x)|\,dz{\leqslant}\frac{3K}{4}.$$ Using $2(a\wedge b)=a+b-|a-b|$ for all $a,b{\geqslant}0$, we get $$\begin{aligned}
        \inf_{x\in{\mathds R}^d, |x|{\leqslant}\delta} & \int_F\big(\rho_0(z)\wedge\rho_0(z-x)\big)\,dz\\
        &=\frac{1}{2}\inf_{x\in{\mathds R}^d, |x|{\leqslant}\delta} \bigg[\int_F\big(\rho_0(z)+\rho_0(z-x)\big)\,dz
            - \int_F\big|\rho_0(z)-\rho_0(z-x)\big|\,dz\bigg]\\
        &{\geqslant}\frac{1}{2}\int_F\rho_0(z)\,dz
            - \frac{1}{2}\sup_{x\in{\mathds R}^d, |x|{\leqslant}\delta}\int_F\big|\rho_0(z)-\rho_0(z-x)\big|\,dz\\
        &{\geqslant}\frac{K}{8}>0.
    \end{aligned}$$ This finishes the proof.

Financial support through DFG (grant Schi 419/5-1) and DAAD (PPP Kroatien) (for René L. Schilling) and the Alexander-von-Humboldt Foundation and the Natural Science Foundation of Fujian $($No. 2010J05002$)$ (for Jian Wang) is gratefully acknowledged.

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[^1]: *R. Schilling:* TU Dresden, Institut für Mathematische Stochastik, 01062 Dresden, Germany. `rene.schilling@tu-dresden.de`

[^2]: *J. Wang:* School of Mathematics and Computer Science, Fujian Normal University, 350007, Fuzhou, P.R. China *and* TU Dresden, Institut für Mathematische Stochastik, 01062 Dresden, Germany. `jianwang@fjnu.edu.cn`
