# Can we show that this transition semigroup preserves a certain Wasserstein space?

Let $$E$$ be a separable $$\mathbb R$$-Banach space, $$v:E\to[1,\infty)$$ be continuous, $$\rho(x,y):=\inf_{\substack{\gamma\:\in\:C^1([0,\:1],\:E)\\ \gamma(0)\:=\:x\\ \gamma(1)\:=\:y}}\int_0^1v\left(\gamma(t)\right)\left\|\gamma'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E,$$ $$(\Omega,\mathcal A,\operatorname P)$$ be a complete probability space, $$X:\Omega\times[0,\infty)\times E\to E$$ be a stochastic flow, $$X^x_t:=X(\;\cdot\;,t,x)\;\;\;\text{for }(t,x)\in[0,\infty)\times E$$ and $$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$

Assume $$\operatorname E[v(X^x_t)]\le cv^{\lambda(t)}(x)\;\;\;\text{for all }(t,x)\in[0,\infty)\times E\tag1$$ for some $$c>0$$ and decreasing $$\lambda:[0,\infty)\to[0,1]$$. By $$(1)$$, $$\operatorname E[\rho(X^x_t,X^y_t)]\le c\rho(x,y)\tag2$$ for all $$x,y\in E$$ and $$t\in[0,1]$$.

Let, $$\mathcal M_1$$ denote the set of probability measures on $$(E,\mathcal B(E))$$, $$\operatorname W_\rho$$ denote the Wasserstein metric associated with $$\rho$$ and $$\mathcal S^1:=\{\mu\in\mathcal M_1\mid\exists y\in E:(\mu\otimes\delta_y)\rho<\infty\}.$$ By $$(2)$$, $$\operatorname W_\rho(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_\rho(\delta_x,\delta_y)\tag3$$ for all $$x,y\in E$$ and $$t\in[0,1]$$.

Let $$t\ge0$$. Can we show that $$\kappa_t^\ast$$ is $$\mathcal S^1$$-preserving? Or even that $$\kappa_t^\ast\mathcal M_1\subseteq\mathcal S^1$$?

I'm quite sure that at least the $$\mathcal S^1$$-preserving claim is true. If $$\mu\in\mathcal M_1$$, then we need to show that there is a $$y\in E$$ with $$(\mu\kappa_t\otimes\delta_y)\rho_r<\infty$$. Maybe we can pick $$y=0$$.

EDIT 1: Assume $$\delta_x\kappa_t\in S^1$$ for all $$x\in E$$ and $$t\ge0$$.

EDIT 2: Assume there are nondecreasing $$v_i:[0,\infty)\to(1,\infty)$$ with $$v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)$$ for all $$x\in E$$ and $$rv_2(r)\le \alpha v_1^\beta(r)$$ for all $$r>0$$ for some $$\alpha\ge0$$ and $$\beta\ge1$$. Assume further that $$\operatorname E[V^\theta(X^x_t)]\le\eta v^{\beta\lambda(t)}(x)$$ for all $$x\in E$$ and $$t\ge0$$.

Then we easily see $$\rho(0,x)\le\alpha v^\beta(x)$$ for all $$x\in E$$. Now, since $$\lambda$$ is decreasing, it must hold $$\lambda(t)\to0$$ as $$t\to\infty$$ and hence $$\operatorname W_\rho(\mu\kappa_t,\delta_0)=\int\mu({\rm d}x)\operatorname E[\rho(0,X^x_t)]\le\alpha\eta\int\mu({\rm d}x)v^{\beta\lambda(t)}(x)\xrightarrow{t\to\infty}1\tag4$$ by monotone convergence for all $$\mu\in\mathcal M_1$$ and $$t\ge0$$. This should yield that $$\kappa_t^\ast$$ maps $$\mathcal M_1$$ to $$\mathcal S^1$$ for all $$t\ge0$$.

• In the equation between (2) and (3), I guess that $\rho_r$ is simply $\rho$ (i.e. $\mathcal{S}^1$ is the set of measures with finite first moment). Jun 29, 2020 at 12:13
• I do not see how you get (2) from (1). For example, if $\lambda\equiv 0$ and $v\equiv 1$, (1) is satisfied but (2) can be false. Jun 29, 2020 at 12:15
• @BenoîtKloeckner Regarding your first comment: Yes, it should be $\rho$ in the definition of $\mathcal S^1$. Regarding your second comment: $\lambda$ should be decreasing (not nonincreasing). So, $\lambda\equiv0$ is not a valid choice. Jun 29, 2020 at 13:08
• That $\lambda$ is decreasing actually changes nothing when $v\equiv 1$. I see no way (2) could follow from (1), and I cannot guess a variation of the hypotheses that would change this. (1) is about the local geometry at $x$ and $X_x^t$, while (2) needs control over a whole curve. Jun 29, 2020 at 13:19

There are some issues that I point out in comments, but assuming (3) you would get $$\mathcal{S}^1$$-preservation easily by convexity of Wasserstein distance, assuming that for at least one $$x\in E$$ you have $$\delta_x\kappa_t\in\mathcal{S}^1$$.

1. Convexity of $$\mathrm{W}_\rho$$ enables us to turn (3) into $$\mathrm{W}_\rho(\mu\kappa_t,\nu\kappa_t) \le c\mathrm{W}_\rho(\mu,\nu).$$ (Let indeed $$\mu,\nu\in\mathcal{S}^1$$, and for each $$t$$ and each $$(x,y)$$ choose (measurably) optimal transport plans $$\eta_{x,y}^t$$ between $$\delta_x\kappa_t$$ and $$\delta_y\kappa_t$$. Let $$\zeta$$ be an optimal transport plan from $$\mu$$ to $$\nu$$; Then $$\int \eta_{x,y}^t d\zeta(x,y)$$ is a transport plan from $$\mu\kappa_t$$ to $$\nu\kappa_t$$, so that \begin{align*} \mathrm{W}_\rho(\mu\kappa_t,\nu\kappa_t) &\le \iint \rho(x',y') d\eta_{x,y}^t(x',y') d\zeta(x,y) \\ &\le \int \mathrm{W}_\rho(\delta_x\kappa_t,\delta_y\kappa_t) d\zeta(x,y) \\ &\le \int c\mathrm{W}_\rho(\delta_x,\delta_y) d\zeta(x,y) = c\int \rho(x,y) d\zeta(x,y) = c\mathrm{W}_\rho(\mu,\nu) \end{align*} as claimed.)

2. Then if for some $$x\in E$$, $$\delta_x\kappa_t\in \mathcal{S}^1$$, for all $$\mu\in\mathcal{S}^1$$ we have $$\mathrm{W}_\rho(\mu\kappa_t,\delta_x\kappa_t)\le c\mathrm{W}_\rho(\mu,\delta_x) <\infty$$ and thus $$\mu\kappa_t\in\mathcal{S}^1$$, as you wished.

3. You do need the additional assumption on $$\delta_x\kappa_t$$, (1) or (2) are not enough. Take $$v\equiv 1$$ and let $$\kappa_t$$ be a Markov kernel sending $$\delta_x$$ to some distribution with infinite first moment, translated by $$x$$. Then you have obviously (1) and (2), but you do not have $$\mathcal{S}^1$$ preservation.

4. You cannot expect to have $$\mathcal{M}_1$$ sent into $$\mathcal{S}^1$$ without additional assumption: the trivial dynamics $$\delta_x\kappa_t=\delta_x$$ satisfies your assumption.

• Regarding 2.: I guess you conclude by the triangle inequality together with the assumption $\delta_x\kappa_t\in S^1$ which implies that $\operatorname W_\rho(\delta_x,\delta_x\kappa_t)<\infty$ and hence $\operatorname W_\rho(\mu\kappa_t,\delta_x)\le\operatorname W_\rho(\mu\kappa_t,\delta_x\kappa_t)+\operatorname W_\rho(\delta_x\kappa_t,\delta_x)<\infty$. Jun 29, 2020 at 13:42
• Please take note of my second edit. Under the additional assumptions, $\kappa_t^\ast$ should actually map $\mathcal M_1$ to $\mathcal S^1$. Jun 29, 2020 at 14:57
• I would like to know if you agree to my second edit. Am I missing something? Jun 30, 2020 at 14:09
• @0xbadf00d I do not know what $V^\theta$ is, and you seem to have produced your own answer about that. MO is not for others to check your details. Jul 2, 2020 at 18:10