How can we conclude exponential decay from this Lipschitz bound?

Question

Below I want to show that there is a $(\tilde c,\lambda)\in[0,\infty)\times(0,\infty)$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le\tilde ce^{-\lambda t}\operatorname W_d(\delta_x,\delta_y)\tag0$$ for all $x,y\in E$ and $t\in[0,t_0)$, which should be a rather simple conclusion.

Let $(E,d)$ be a complete separable metric space, $\mathcal M_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$, $\operatorname W_d$ denote the $1$-Wasserstein metric on $\mathcal M_1(E)$ induced by $d$, $(\kappa_t)_{\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ and $t_0>0$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,t_0)\tag1$$ for some $c\ge0$ and $$\operatorname W_d(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0})\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag2$$ for some $\alpha\in(0,1)$. We can extend both $1$ and $(2)$ to hold for $\delta_x,\delta_y$ replaced by arbitrary $\mu,\nu\in\mathcal M_1(E)$. From the extension of $(2)$, we easily obtain $$\operatorname W_d(\delta_x\kappa_{nt_0},\delta_y\kappa_{nt_0})\le\alpha^n\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }n\in\mathbb N_0\tag3.$$ Now, let $t\ge0$ and $x,y\in E$. Define $$n:=\left\lfloor\frac t{t_0}\right\rfloor$$ and $s:=t-nt_0\in[0,t_0)$. By the extension of $(1)$ and $(3)$, we obtain $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\alpha^n\operatorname W_d(\delta_x,\delta_y)\tag4.$$ But now I'm stuck. What I would like to achieve is bounding the left-hand side of $(4)$ by $$\tilde ce^{-\lambda t}\operatorname W_d(\delta_x,\delta_y)\tag5$$ for some suitably chosen $\lambda>0$. We can clearly define $\lambda:=\ln\alpha$ so that $\alpha^n=e^{\lambda n}$, but that's not sufficient, since it clearly does not hot $\alpha^n\le\alpha^{\frac t{t_0}}$.

So, what do we need to do? If it's important for the conclusion, feel free to assume that for all $t_0>0$ there are such $c$ and $\alpha$.

Answer 1

To generalize @0xbadf00d's proof to $t_0 \ne 1$, for $t>0$ we can write $t = n t_0 + r$ for some $n \in \mathbb{N}_0$ and $r \in [0,t_0)$. Then $(7)$ stays the same, i.e. $$W_d(\delta_x \kappa_t, \delta_y \kappa_t) \le \alpha^n W_d(\delta_x \kappa_r, \delta_y \kappa_r) \le c \alpha^n W_d(\delta_x,\delta_y)$$

for all $x,y \in E$. Setting $\bar c := \frac{c}{\alpha}$ and $\lambda := -\frac{\ln \alpha}{t_0}$ gives $W_d(\delta_x\kappa_t,\delta_y\kappa_t) \le \bar c e^{-\lambda t}W_d(\delta_x,\delta_y)$ for all $t$.

If we have that for all $t_0$ there exist such $c$ and $\alpha$, then this holds for all $t$ by simply applying the proof to an arbitrary $t_0$. In fact, since we showed that $W_d(\delta_x \kappa_{nt_0}, \delta_y \kappa_{nt_0}) \le \alpha^n W_d(\delta_x,\delta_y) < \alpha W_d(\delta_x,\delta_y)$ we have if such $c$ and $\alpha$ exist for some $t_0$ then they also exist for $n t_0$ and so we can conclude that they exist for all $T$ sufficiently large.

Answer 2

Partial answer:

I think the claim is easy to prove when $t_0=1$ (unless I'm missing something): From $(2)$, we easily deduce $$\operatorname W_d\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag6$$ for all $x,y\in\mathbb N$ and $n\in\mathbb N_0$. If $t>0$, we may write $t=n+r$ for some $n\in\mathbb N_0$ and $r\in[0,1)$ so that $$\operatorname W_d\left(\delta_x\kappa_t,\delta_y\kappa_t\right)\le\alpha^n\operatorname W_d\left(\delta_x\kappa_r,\delta_y\kappa_r\right)\le c\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag7$$ for all $x,y\in E$ by $(6)$ and $(1)$.

Now we only need to note that $$c\alpha^n=\frac c\alpha\alpha^{n+1}\le\frac c\alpha\alpha^t\tag8$$ (the last "$\le$" is actually a "$<$" as long as $c\ne0$) and hence we obtain $$\operatorname W_d\left(\mu\kappa_t,\nu\kappa_t\right)\le\tilde ce^{-\lambda t}\operatorname W_d(\mu,\nu)\tag9$$ for all $\mu,\nu\in\mathcal M_1(E)$, where $$\tilde c:=\frac c\alpha$$ and $$\lambda:=-\ln\alpha>0.$$