2
$\begingroup$

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 $t\ge0$.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Benoît Kloeckner Jun 29 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. $\endgroup$ – Benoît Kloeckner Jun 29 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 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. $\endgroup$ – Benoît Kloeckner Jun 29 at 13:19
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 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 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 at 14:57
  • $\begingroup$ I would like to know if you agree to my second edit. Am I missing something? $\endgroup$ – 0xbadf00d Jun 30 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. $\endgroup$ – Benoît Kloeckner Jul 2 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.