e show that has the same distribution as $x_t$ in , and has the same distribution as $y_t$ in . Thus, for any $t$, the process $(x_t,y_t)$ defined by is a valid coupling for and .

[One step contraction]{} \[ss:step\_gaussian\]

\[l:gaussian\_contraction\] Let $f$ be as defined in Lemma \[l:fproperties\] with parameters $\epsilon$ satisfying $\epsilon \leq \frac{\Rq}{\aq\Rq^2 + 1}$. Let $x_t$ and $y_t$ be as defined in . If we assume that $\E{\lrn{y_0}_2^2} \leq 8\lrp{R^2 + \beta^2/m}$ and $T\leq \min\lrbb{\frac{\epsilon^2}{\beta^2}, \frac{\epsilon}{6 L\sqrt{R^2 + \beta^2/m}}}$, then $$\begin{aligned}
        \E{f(x_T - y_T)}
        \leq e^{-\lambda T} \E{f(x_0 - y_0)} + 3T (L+\LN^2) \epsilon
        \end{aligned}$$

For ease of reference: $m,L,\LR, R$ are from Assumption \[ass:U\_properties\], $\cm, \beta$ are from Assumption \[ass:xi\_properties\], $\aq, \Rq, \LN, \lambda$ are defined in .

For notational convenience, for the rest of this proof, let us define $z_t := x_t - y_t$ and $\nabla_t := \nabla U(x_t) - \nabla U(y_t)$, $\Delta_t := \nabla U(y_0) - \nabla U(y_t)$ $N_t := N(x_t) - N(y_t)$.

It follows from that $$\begin{aligned}
        d z_t = - \nabla_t dt + \Delta_t dt + 2 \cm \gamma_t \gamma_t^T dV_t + \lrp{N_t + N(y_t) - N(y_0)} dW_t
        \numberthis \label{e:coupled_difference_sde}
        \end{aligned}$$

Using Ito’s Lemma, the dynamics of $f(z_t)$ is given by $$\begin{aligned}
        &d f(z_t)\\
        =& {\lin{\nabla f(z_t), dz_t}}
        + {2\cm^2\tr\lrp{\nabla^2 f(z_t) \lrp{\gamma_t \gamma_t^T}}} dt
        + {\frac{1}{2}\tr\lrp{\nabla^2 f(z_t) \lrp{N_t+ N(y_t) - N(y_0)}^2}} dt\\
        =& \underbrace{-\lin{\nabla f(z_t), \nabla_t}}_{\circled{1}} dt + \underbrace{\lin{\nabla f(z_t), \Delta_t}}_{\circled{2}} dt  +  \underbrace{\lin{\nabla f(z_t), 2 \cm \gamma_t \gamma_t^T dV_t + \lrp{N_t + N(y_t) - N(y_0)} dW_t }}_{\circled{3}}\\
        &\quad + \underbrace{2\cm^2\tr\lrp{\nabla^2 f(z_t) \lrp{\gamma_t \gamma_t^T}}}_{\circled{4}} dt
        + \underbrace{\frac{1}{2}\tr\lrp{\nabla^2 f(z_t