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 $t\ge0$.

  • $\begingroup$ 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). $\endgroup$ Jun 29, 2020 at 12:13
  • $\begingroup$ 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. $\endgroup$ Jun 29, 2020 at 12:15
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Jun 29, 2020 at 13:08
  • $\begingroup$ 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. $\endgroup$ Jun 29, 2020 at 13:19

1 Answer 1


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.

  • 1
    $\begingroup$ 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 $\operatorname W_\rho(\mu\kappa_t,\delta_x)\le\operatorname W_\rho(\mu\kappa_t,\delta_x\kappa_t)+\operatorname W_\rho(\delta_x\kappa_t,\delta_x)<\infty$. $\endgroup$
    – 0xbadf00d
    Jun 29, 2020 at 13:42
  • $\begingroup$ Please take note of my second edit. Under the additional assumptions, $\kappa_t^\ast$ should actually map $\mathcal M_1$ to $\mathcal S^1$. $\endgroup$
    – 0xbadf00d
    Jun 29, 2020 at 14:57
  • $\begingroup$ I would like to know if you agree to my second edit. Am I missing something? $\endgroup$
    – 0xbadf00d
    Jun 30, 2020 at 14:09
  • $\begingroup$ @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. $\endgroup$ Jul 2, 2020 at 18:10

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