If a Markov semigroup is eventually contractive, can we conclude that it admits a unique invariant measure?

Question

Let $E$ be a separable $\mathbb R$-Banach space, $\rho$ be a complete separable metric on $E$, $\operatorname W_\rho$ denote the Wasserstein metric of order $1$ associated to $\rho$, $\mathcal M_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$ and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W_\rho(\mu\kappa_t,\nu\kappa_t)\le ce^{-\lambda t}\operatorname W_\rho(\mu,\nu)\;\;\;\text{for all }\mu,\nu\in\mathcal M_1(\mu,\nu)\tag1$$ for some $c\ge0$ and $\lambda>0$.

Are we able to conclude that $(\kappa_t)_{t\ge0}$ has a unique invariant measure $\mu_\ast\in\mathcal M_1(E)$?

By $(1)$, the adjoint semigroup $(\kappa_t^\ast)_{t\ge0}$ is eventually contractive: Let $t_0\ge0$ with $$ce^{-\lambda t}<1\;\;\;\text{for all }t\ge t_0$$ and $t\ge t_0$. Since the Wasserstein space $$\mathcal S^1(E,\rho):=\left\{\mu\in\mathcal M_1(E):(\mu\otimes\delta_0)\rho<\infty\right\}$$ equipped with $\operatorname W_\rho$ is complete and hence we can apply Banach's fixed-point theorem yielding that there is a unique $\mu_\ast\in\mathcal S^1(E,\rho)$ with $$\mu_\ast\kappa_t=\mu_\ast\tag2.$$ Moreover, for any $\mu_0\in\mathcal S^1(E,\rho)$ and $$\mu_n:=\mu_{n-1}\kappa_t\;\;\;\text{for }n\in\mathbb N,$$ it holds $$\operatorname W_\rho(\mu_n,\mu_\ast)\xrightarrow{n\to\infty}0\tag3.$$

So, all what's left to prove is that $\mu_\ast$ does not depend on $t$, i.e. $\mu_\ast$ is invariant with respect to $\kappa_t$ for all $t\ge t_0$.

BTW: Is this all we can hope for or can we even conclude that $\mu_ast$ must be invariant with respect to $\kappa_t$ for all $t\ge\color{red}0$?

Answer 1

Note that your argument contains an implicit assumption that $\kappa_t \mu \in \mathcal{S}^1$ for every $\mu \in \mathcal{S}^1$ (otherwise the Banach fixed point theorem does not apply). I will also make that assumption. Also, I realized that I have written $\kappa_t \mu$ with $\mu$ on the right; sorry about that.

You have shown that for some fixed $t^* \ge t_0$, that $\kappa_{t^*}$ has an invariant measure $\mu_*$ which is unique in $\mathcal{S}^1$.

Let $t > 0$ be arbitrary. Then we have by the semigroup property that $$\kappa_{t^*} \kappa_t \mu_* = \kappa_{t+ t^*} \mu_* = \kappa_t \kappa_{t^*} \mu_* = \kappa_t \mu_*$$ which proves that $\kappa_t \mu_*$ is invariant for $\kappa_{t^*}$. By uniqueness, $\kappa_t \mu_* = \mu_*$. This proves that $\mu_*$ is invariant for $\kappa_t$.

If $t \ge t_0$, then your argument shows that $\mu_*$ is in fact the unique invariant measure in $\mathcal{S}^1$ for $\kappa_t$. Otherwise, for $t < t_0$, suppose $\mu' \in \mathcal{S}^1$ is another invariant measure for $\kappa_t$. Let $n$ a large enough integer so that $n t \ge t_0$; then $\mu' = \kappa_t^n \mu' = \kappa_{nt} \mu'$. Since $\kappa_{nt}$ has $\mu_*$ as its unique invariant measure, we have $\mu' = \mu_*$.

We have thus shown that for every $t$, $\mu_*$ is invariant for $\kappa_t$, and is the unique such measure in $\mathcal{S}^1$.