Proof. By flatness,

$$ \begin{align*}e^{A^t(\bar y)} \le e^{\lvert A^t(\bar y)-A^t(\bar x)\rvert} e^{A^t(\bar x)}\le e^{B\omega\circ d(x,y)} e^{A^t(\bar x)} \le \underbrace{e^{B\omega(\operatorname{diam} \Omega)}}_{C} e^{A^t(\bar x)}\end{align*} $$

and similarly $e^{A^t(\bar y)}\ge e^{-B\omega \circ d(x,y)} e^{A^t(\bar x)}$ .

Proof.

Here we need the normalization assumption to ensure these two measures are both of the same mass. Fix $t\in \mathbb {N}$ , $x,y\in \Omega $ and observe that $\mathscr {L}^{*t}\delta _x = (e_t)_*( e^{A^t} \,{d} m_x^t)$ , where $e_t:\Omega ^t\to \Omega $ is the projection to the last coordinates. We seek an efficient transport plan between $\mathscr {L}^{*t}\delta _x$ and $\mathscr {L}^{*t}\delta _y$ , and we will construct it as $(e_t,e_t)_{*}\Pi $ , where $\Pi $ is a transport plan between $e^{A^t} \,{d} m_x^t$ and $e^{A^t} \,{d} m_y^t$ . What we are given by the coupling $\mathsf {P}$ is a transport plan $\Pi _{x,y}^t$ between $m_x^t$ and $m_y^t$ , and we will modify it to take into account the $e^{A^t}$ factors. Define a function

$$ \begin{align*} a : \Omega^t\times \Omega^t &\to \mathbb{R} \\ (\bar x,\bar y) &\mapsto \min ( e^{A^t(\bar x)} , e^{A^t(\bar y)} ) \end{align*} $$

so that $a \,{d} \Pi _{x,y}^t$ is a positive measure whose marginals are less than $e^{A^t} \,{d} m_x^t$ and $e^{A^t} \,{d} m_y^t$ , respectively. There must thus exist some positive measure $\Lambda $ on $\Omega ^t\times \Omega ^t$ such that

$$ \begin{align*}\Pi := a \,{d} \Pi_{x,y}^t + \Lambda\end{align*} $$

is a probability measure with marginals exactly $e^{A^t} \,{d} m_x^t$ and $e^{A^t} \,{d} m_y^t$ . We want to bound above the $\omega $ -cost of $\Pi $ ; the basic idea is that the first term will be small by the decay hypothesis, the second one will be small because $\Lambda $ has small mass.

By Lemma 3, for $\Pi _{x,y}^t$ almost all $(\bar x,\bar y)$ and for some constant B,

$$ \begin{align*}a(\bar x,\bar y) \ge e^{A^t(\bar x)} e^{-B \omega\circ d(x,y)};\end{align*} $$

using that A is normalized, it comes that the total mass of $a\,{d}\Pi _{x,y}^t$ is at least $e^{-B\omega \circ d(x,y)}$ . Since $\Omega $ has finite diameter, up to enlarging B, this total mass can be bounded from below both by a constant $e^{-B}\in (0,1)$ and by $1-B\omega \circ d(x,y)$ . The total mass of $\Lambda $ is therefore bounded above as follows:

$$ \begin{align*} \int \boldsymbol{1} \,{d} \Lambda \le \min( B\omega\circ d(x,y) , 1-e^{-B}).\end{align*} $$

We introduce a new modulus of continuity

$$ \begin{align*}\omega' = \min ( K\omega, \omega(\operatorname{diam}\Omega) ),\end{align*} $$

where K is a positive constant to be specified later on (independently of $x,y$ ). We have $\omega '\circ d(x,y) \ge \omega \circ d(x,y)$ for all $x,y\in \Omega $ and $\omega '\le K\omega $ , so that $\omega \circ d$ and $\omega '\circ d$ are Lipschitz-equivalent metrics on $\Omega $ , and as a consequence, $\operatorname {W}_\omega $ and $\operatorname {W}_{\omega '}$ are Lipschitz-equivalent (with the same constants).

We decompose the cost as

$$ \begin{align*}\int \omega'\circ d(x_t,y_t) \,{d}\Pi(\bar x,\bar y) &= \int \omega'\circ d(x_t,y_t) a(\bar x,\bar y) \,{d}\Pi_{x,y}^t(\bar x,\bar y)\\ &\quad+ \int \omega'\circ d(x_t,y_t) \,{d}\Lambda(\bar x,\bar y).\end{align*} $$

For the first term, we get from $\omega $ -decay of $\mathsf {P}_A$ (with D only depending on A):

$$ \begin{align*} \int \omega'\circ d(x_t,y_t) a(\bar x,\bar y) \,{d}\Pi_{x,y}^t(\bar x,\bar y) &\le K \int \omega\circ d(x_t,y_t) e^{A^t(\bar x)} \,{d}\Pi_{x,y}^t(\bar x,\bar y) \\ &\le DK\cdot F(t,\omega \circ d(x,y)) \\ &\le DK\cdot F(t,\omega' \circ d(x,y)). \end{align*} $$

For the second term, we distinguish two cases. If $\omega \circ d(x,y)\ge \omega (\operatorname {diam} \Omega )/K$ , then $\omega '\circ d(x,y)= \omega (\operatorname {diam} \Omega ) = \omega '(\operatorname {diam} \Omega )$ and we bound the mass of $\Lambda $ by $1-e^{-B}$ , so that

$$ \begin{align*} \int \omega'\circ d(x_t,y_t) \,{d}\Lambda(\bar x,\bar y) &\le (1-e^{-B}) \omega'(\operatorname{diam} \Omega) \\ &\le (1-e^{-B})\omega'\circ d(x,y). \end{align*} $$

If $\omega \circ d(x,y)\le \omega (\operatorname {diam} \Omega )/K$ , then $\omega '\circ d(x,y)=K\omega \circ d(x,y)$ and we bound the mass of $\Lambda $ by $B\omega \circ d(x,y)$ :

$$ \begin{align*} \int \omega'\circ d(x_t,y_t) \,{d}\Lambda(\bar x,\bar y) &\le B\omega\circ d(x,y) \omega'(\operatorname{diam} \Omega) \\ &\le \frac{B\omega(\operatorname{diam} \Omega)}{K} \omega'\circ d(x,y). \end{align*} $$

Choosing K large enough to ensure ${B\omega (\operatorname {diam} \Omega )}/{K}\le 1-e^{-B}$ , we get in both cases

(1) $$ \begin{align} \int \omega'\circ d(x_t,y_t) \,{d}\Pi(\bar x,\bar y) \le DK\cdot F(t,\omega' \circ d(x,y)) + (1-e^{-B}) \omega'(d(x,y)). \end{align} $$

Since $F(t,r)\lesssim r$ , the previous step implies in particular that $\Pi $ , as a restricted coupling at time t, is $\omega '$ -Hölder; but $\omega \le \omega '\le K\omega $ on $[0,\operatorname {diam}\Omega ]$ so that $\Pi $ is also $\omega $ -Hölder. Then the claim follows from Lemma 2.

Step 5. There exist $\theta _1\in (0,1)$ and $k_1\in \mathbb {N}$ such that for all r, all $x,y\in \Omega $ such that $\omega '\circ d(x,y)\ge r$ and all $t\ge k_1\tau (r/2^{k_1})$ ,

$$ \begin{align*}\operatorname{W}_{\omega'}(\mathscr{L}^{*t}\delta_x,\mathscr{L}^{*t}\delta_y) \le \theta_1 \omega'\circ d(x,y).\end{align*} $$

We choose any $\theta _1 \in (1-e^{-B},1)$ and $k_1$ large enough to ensure $DK/2^{k_1} + (1-e^{-B}) \le \theta _1$ , and then apply equation (1) (note that $k_1\tau (r/2^{k_1})\ge \tau (r)+\tau (r/2)+\cdots +\tau (r/2^{k_1})$ ).

Step 6. There exist $\theta \in (0,1)$ and $k_2\in \mathbb {N}$ such that for all r, all $\mu ,\nu \in \operatorname {\mathcal {P}}(\Omega )$ such that $\operatorname {W}_{\omega '}(\mu ,\nu ) = r$ and all $t\ge k_2\tau (r/k_2)$ ,

$$ \begin{align*}\operatorname{W}_{\omega'}(\mathscr{L}^{*t}\mu,\mathscr{L}^{*t}\nu) \le \theta \operatorname{W}_{\omega'}(\mu,\nu).\end{align*} $$

Choose any $\theta \in (\theta _1,1)$ and let $\eta>0$ be small enough to ensure $\theta _1+C\eta \le \theta $ , where C is the constant of Step 4. Let $\mu ,\nu $ be any two probability measures and let $\Pi \in \Gamma (\mu ,\nu )$ be optimal for $\operatorname {W}_{\omega '}(\mu ,\nu )=:r$ . Define $s:=\eta r$ and $E:=\{(x,y) \mid \omega '\circ d(x,y) \ge s \}$ . For all $t\ge k_1 \tau (s/2^{k_1})$ , using Lemma 2.9, we obtain

$$ \begin{align*} \operatorname{W}_{\omega'}(\mathscr{L}^{*t}\mu,\mathscr{L}^{*t}\nu ) &\le \int \operatorname{W}_{\omega'}(\mathscr{L}^{*t}\delta_x,\mathscr{L}^{*t}\delta_y ) \,{d}\Pi(x,y) \\ &\le \int_E \operatorname{W}_{\omega'}(\mathscr{L}^{*t}\delta_x,\mathscr{L}^{*t}\delta_y ) \,{d}\Pi(x,y)\\ &\quad+ \int_{\Omega\times\Omega\setminus E} \mkern-36mu \operatorname{W}_{\omega'}(\mathscr{L}^{*t}\delta_x,\mathscr{L}^{*t}\delta_y ) \,{d}\Pi(x,y) \\ &\le \theta_1 \int_E \omega'\circ d(x,y) \,{d}\Pi(x,y) + C \int_{\Omega\times\Omega\setminus E} \mkern-36mu \omega'\circ d(x,y) \,{d}\Pi(x,y) \\ &\le \theta_1 \operatorname{W}_{\omega'}(\mu,\nu) + C \eta r \\ &\le \theta \operatorname{W}_{\omega'}(\mu,\nu). \end{align*} $$

It suffices to choose $k_2\ge 2^{k_1}/\eta $ .

We deduce that the $\theta $ decay time $\tau ^{\omega '}_\theta (r)$ of $\mathscr {L}^*$ with respect to $\operatorname {W}_{\omega '}$ is at most $k_2\tau (r/k_2)$ .

Then for all $n\in \mathbb {N}$ ,

$$ \begin{align*}\tau^{\omega'}_{\theta^n}(r) \le k_2\tau(r/k_2)+k_2\tau(\theta r/k_2) + \cdots + k\tau(\theta^{n-1} r/k_2) \end{align*} $$

and taking n large enough to ensure $\theta ^n\le {1}/{2K}$ , we get

$$ \begin{align*}\tau^{\omega'}_{{1}/{2K}}(r) \le k_2 n \tau(\theta^{n-1} r/k_2) \le k \tau(r/k) \quad \text{for some }k.\end{align*} $$

Now since $\operatorname {W}_\omega \le \operatorname {W}_{\omega '} \le K\operatorname {W}_\omega $ , the decay time for $\mathscr {L}^*$ with respect to $\operatorname {W}_\omega $ satisfies $\tau ^\omega _{1/2} \le \tau ^{\omega '}_{{1}/{2K}}$ , and we are done.

Proof. Both points are essentially obvious: $e^{A(x_1)}\,{d}\pi _{x,y}(x_1,y_1)$ is absolutely continuous with respect to $\pi _{x,y}$ and $\int _{E_{x,y}} e^{A(x_1)}\,{d}\pi _{x,y}\ge p \min (e^A)>0$ . For the claim to make sense though, one has to observe that $(e^{A}\pi _{x,y})_{x,y}$ is a Markov transition kernel, which follows from the normalization hypothesis:

$$ \begin{align*}\int \boldsymbol{1} e^{A(x_1)} \,{d} \pi_{x,y}(x_1,y_1) = \int \boldsymbol{1} e^{A(x_1)} \,{d} m_x(x_1) =\boldsymbol{1}.\\[-40pt]\end{align*} $$

Proof. It suffices to consider the case $t=1$ , then conclude by induction. However, this follows from the definition: denoting by $\unicode{x3bb} _1$ the contraction factor of $\pi _{x,y}$ on $E_{x,y}$ ,

then we take $\unicode{x3bb} = p\unicode{x3bb} _1 + (1-p)$ .

Proof. Let A be a bounded, normalized potential. Then for all modulus of continuity $\omega $ , some $\unicode{x3bb} \in (0,1)$ and all $x,y,t$ :

$$ \begin{align*} \int \omega\circ d(x_t,y_t) \,{d}(e^{A^t}\Pi_{x,y}^t)(\bar x,\bar y) &\le \omega \bigg( \int d(x_t,y_t) \,{d}(e^{A^t}\Pi_{x,y}^t)(\bar x,\bar y) \bigg) \\ &\le \omega(\unicode{x3bb}^t d(x,y)). \end{align*} $$

It only remains to observe that for all $\alpha '<\alpha $ , $\omega _{\alpha +\beta \log }(\unicode{x3bb} ^t r) \lesssim \unicode{x3bb} ^{\alpha 't} \omega _{\alpha +\beta \log }(r)$ and that

$$ \begin{align*}\omega_{\beta\log}(\unicode{x3bb}^t r) = \frac{\omega_{\beta\log}(r)}{(1+ t\cdot \omega_{\beta\log}(r)^{1/\beta} \log(1/\unicode{x3bb}))^\beta},\end{align*} $$

which provides exactly the polynomial decay of degree $\beta $ .

Proof. The natural coupling is Markovian, with transition kernel

$$ \begin{align*}\pi_{x,y}=\sum_j \frac1k\delta_{(x^{\eta(j)},y^{\sigma(j)})},\end{align*} $$

which is non-dilating and contracting with positive probability thanks to the hypothesis that $\mathsf {M}$ is weakly contracting. Corollary 8 provides the conclusion.