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

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 $$t\ge\color{red}0$$?

• That's fine, but it only proves uniqueness in the space of measures with a finite first moment. – R W Jun 27 '20 at 17:59
• @RW Yes, this space is $\mathcal S^1(E,\rho)$. I would like to show that there is a $\mu_\ast\in\mathcal S^1(E,\rho)$ with $\mu_\ast\kappa_t=\mu_\ast$ for all $t\ge t_0$. – 0xbadf00d Jun 27 '20 at 18:02

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 $$\mathcal{S}^1$$.

• Thank you for your answer. (a) For your uniqueness argument in your second argument to work, we need that $\kappa_t^\ast\mu_\ast\in\mathcal S^1$. It's not clear to me why this is the case. Is it a general fact that a Markov kernel is $\mathcal S^1$-preserving? (b) Just to be sure: The result we've obtained is that $(\kappa_t)_{t\ge0}$ has a unique invariant measure in $\mathcal S^1$, but there might be other invariant measures in $\mathcal M_1\setminus\mathcal S^1$, right? – 0xbadf00d Jun 28 '20 at 5:09
• @0xbadf00d: Good point. In fact, consider the following counterexample: let $\nu$ be a probability measure not in $\mathcal{S}^1$ and take $\kappa_t f(x) = \int f\,d\nu$ for all $t$. Then $\kappa_t \mu = \nu$ for every $\mu$ and every $t>0$, so (1) is trivially satisfied (the LHS is always zero). Indeed, this is also a gap in your argument, because $\kappa_t$ has a unique invariant measure and it is not in $\mathcal{S}^1$. I think we need to assume that $\kappa_t$ takes $\mathcal{S}^1$ into $\mathcal{S}^1$ for any of this to work. – Nate Eldredge Jun 28 '20 at 12:04
• @0xbadf00d: I agree that even with this assumption, nothing appears to rule out the possibility of another invariant measure which is not in $\mathcal{S}^1$. I don't know of an example where this happens, however. – Nate Eldredge Jun 28 '20 at 13:07