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Let $E$ be a separable $\mathbb R$-Banach space, $\rho_r$ be a metric on $E$ for $r\in(0,1]$ with $\rho_r\le\rho_s$ for all $0<r\le s\le1$, $\rho:=\rho_1$, $$d_{r,\:\delta,\:\beta}:=1\wedge\frac{\rho_r}\delta+\beta\rho\;\;\;\text{for }(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$$ and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$.

Assume we arbe able to show that for all $n\in\mathbb N$ there is a $\alpha\in[0,1)$ and $(r,\delta,\beta)\in[0,1]\times(0,\infty)\times(0,1)$ with$^1$ $$\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x,\delta_y\right)\tag1$$ for all $x,y\in E$, where $\delta_x$ denotes the Dirac measure on $(E,\mathcal B(E))$ at $x\in E$. Why are we able to conclude that there is a $(c,\lambda\in[0,\infty)^2$ with $$\operatorname W_\rho\left(\nu_1\kappa_t,\nu_2\kappa_t\right)\le ce^{-\lambda t}\operatorname W_\rho\left(\nu_1,\nu_2\right)\tag2$$ for all $\nu_1,\nu_2\in\mathcal M_1(E)$ and $t\ge0$?

It's clear to me that if $\kappa$ is any Markov kernel on $(E,\mathcal B(E))$ and $d$ is any metric on $E$ such that there is a $\alpha\ge0$ with $\operatorname W_d\left(\delta_x\kappa,\delta_y\kappa\right)\le\alpha\operatorname W_d\left(\delta_x,\delta_y\right)$ for all $x,y\in E$, then this extends to $\operatorname W_d(\mu\kappa,\nu\kappa)\le\alpha\operatorname W_d(\mu,\nu)$ for all $\mu,\nu\in\mathcal M_1(E)$. Moreover, it's clear that $\operatorname W_d\left(\delta_x,\delta_y\right)=d(x,y)$.

Note that for any choice of $(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$, it holds $$\beta\rho\le d_{r,\:\delta,\:\beta}\le\left(\frac1\delta+\beta\right)\rho.\tag3$$

Remark: The desired claim seems to be used in the proof of Theorem 3.4 in https://arxiv.org/pdf/math/0602479.pdf.


$^1$ If $(E,d)$ is a complete separable metric space and $\mathcal M_1(E)$ is the space of probability measures on $\mathcal B(E)$, then the Wasserstein metric $\operatorname W_d$ on $\mathcal M_1(E)$ satisfies the identity $$\operatorname W_d(\mu,\nu)=\sup_{\substack{f\::\:E\:\to\:\mathbb R\\|f|_{\operatorname{Lip}(d)}\:\le\:1}}(\mu-\nu)f\;\;\;\text{or all }\mu,\nu\in\mathcal M_1(E),$$ where $$|f|_{\operatorname{Lip}(d)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{d(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $\mu f:=\int f\:{\rm d}\mu$ for $\mu$-integrable $f:E\to\mathbb R$.

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I can answer assuming some regularity on the Markov semigroup, which I would expect to be satisfied in most cases. Specifically, assume local (in time) Lipschitz continuity on your Markov semigroup, i.e. $$\forall s_0>0, \exists C>0, \forall s\in[0,s_0], \forall \mu_1,\mu_2 : \mathrm{W}(\mu_1\kappa_s,\mu_2\kappa_s)\le C\mathrm{W}(\mu_1,\mu_2)$$ (I do not precise for which metric, since the two metrics under consideration are Lipschitz-equivalent and so only the constant $C$ would change when passing from one to the other.)

Using convexity of Wasserstein distance, every Lipschitz/contraction bound we have on Dirac masses is also true for arbitrary measures (I guess that is what you mean at the end of your question, although an $\alpha$ appears to be missing).

For any $t_0$, using (1) with $n=1$ iteratively and the double inequality (3): \begin{align*} \mathrm{W}_\rho(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0}) &\le \frac1\beta \mathrm{W}_{d_{r,\delta,\beta}}(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0}) \\ &\le \frac{\alpha^{t_0}}{\beta} \mathrm{W}_{d_{r,\delta,\beta}}(\delta_x,\delta_y) \\ &\le \alpha^{t_0}\Big(\frac{1}{\beta\delta}+1\Big) \mathrm{W}_\rho(\delta_x,\delta_y) \end{align*} Since $\alpha\in(0,1)$, this is what you needed.

(Side note: this kind of computation shows that any decay of the form $$ d(T^n(x),T^n(y)) \le f(n) d(x,y)$$ where $d$ is any metric, $T$ is any Lipschitz dynamical system, and $f(n) \to 0$ as $n\to \infty$ (or even $f(n)<1$ for some $n$), actually imply exponential decay. This is pretty basic, but seems to be sometimes overlooked.)

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  • $\begingroup$ Thank you very much for your answer. (a) How do you define the notion of local Lipschitz continuity of a transition semigroup? (b) I guess you've assumed $t_0=n$ to obtain the second inequality in your third displayed equation, right? I'm not sure what your goal is at this point. Is it to establish an inequality of the form as in your first displayed equation (with $\eta=\alpha^{t_0}((\beta\delta)^{-1}+1)$)? (c) Why is it sufficient to take $t_0$ large enough (and why can we do that)? How precisely do we need to choose $\lambda$ in $(2)$? $\endgroup$ – 0xbadf00d Jun 11 at 4:56
  • $\begingroup$ @0xbadf00d (a) for all $s_0$, there exists $C$ such that for all $s\in[0,s_0]$ $\mathrm{W}(\mu_1\kappa_s,\mu_2\kappa_s) \le C \mathrm{W}(\mu_1,\mu_2)$; (b) no, I use (1) with $1$ in the role of $n$ (hence the power $t_0$ for the $\alpha$); yes my goal is to prove the first displayed equation, this is why I wrote "this $t_0$ is given by"; (c) $\alpha<1$ so $\alpha^{t_0}\to 0$; we can choose $t_0$ freely here because all three inequalities are true for all $t_0$. For $\lambda$, recall $\eta<1$ and $k\sim t$. $\endgroup$ – Benoît Kloeckner Jun 11 at 8:47
  • $\begingroup$ Could it be the case that you've mistakenly mixed things up in the third equation? If you put $d:=d_{r,\:δ,\:\beta}$, then assumption $(1)$ essentially is that there is a $t_0\ge0$ such that $$\text W_d(δ_xκ_{t_0},δ_yκ_{t_0})\le\alpha\text W_d(δ_x,δ_y)\;\;\;\text{for all }x,y\in E\tag4.$$ Now, by $(3)$, $$\text W_\rho\le\frac1\beta\text W_d\le\left(\frac1{\betaδ}+1\right)\text W_\rho\tag5$$ and hence, by $(4)$, $$\text W_\rho(δ_xκ_{t_0},δ_yκ_{t_0})\le\alpha\left(\frac1{\betaδ}+1\right)\text W_{\rho}(δ_x,δ_y)\;\;\;\text{for all }x,y\in E\tag6.$$ $\endgroup$ – 0xbadf00d Jun 11 at 8:49
  • $\begingroup$ (a): You didn't specify any metric associated to the Wasserstein distance. Or will the Lipschitz continuity hold for all metrics as soon as it holds for one metric? Is this notion somehow related to the usual notion of Lipschitz continuity when we consider the semigroup as a mapping from $[0,\infty)$ into a suitable space? (b): I'm sorry, I don't get what you mean by "with $1$ in the role of $n$". What is your $t_0$ then? Isn't $(6)$ the instance of your first equation we are seeking for? Otherwise it seems like a Lipschitz constant is missing in your second inequality of the third equation. $\endgroup$ – 0xbadf00d Jun 11 at 10:19
  • $\begingroup$ Applying your second equation to $(6)$ would yield $$\text W_\rho(δ_xκ_t,δ_yκ_t)\le C\left[\alpha\left(\frac1{\betaδ}+1\right)\right]^{\left\lfloor\frac t{t_0}\right\rfloor}\text W_{\rho}(δ_x,δ_y)\;\;\;\text{for all }x,y\in E\tag7.$$ (c) I don't get why these inequalities should be true for all $t_0$, since your assumption was that there is a particular $t_0$ such that your first equation holds. Or did you use my assumption from the question here (with $t_0=n$)? $\endgroup$ – 0xbadf00d Jun 11 at 10:20
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Building up on Benoît Kloeckner's answer, consider the following simplified scenerio: Let $(E,d)$ be a complete separable metric space, $(\kappa_t)_{\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,1)\tag{10}$$ for some $c\ge0$ and $$\operatorname W_d(\delta_x\kappa_1,\delta_y\kappa_1)\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11}$$ for some $\alpha\in(0,1)$.

From $(11)$, we easily deduce $$\operatorname W_d\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{12}$$ for all $x,y\in\mathbb N$ and $n\in\mathbb N_0$. If $t>0$, we may write $t=n+r$ for some $n\in\mathbb N_0$ and $r\in[0,1)$ so that $$\operatorname W_d\left(\delta_x\kappa_t,\delta_y\kappa_t\right)\le\alpha^n\operatorname W_d\left(\delta_x\kappa_r,\delta_y\kappa_r\right)\le c\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{13}$$ for all $x,y\in E$ by $(12)$ and $(10)$.

Now we only need to note that $$c\alpha^n=\frac c\alpha\alpha^{n+1}\le\frac c\alpha\alpha^t\tag{14}$$ (the last "$\le$" is actually a "$<$" as long as $c\ne0$) and hence we obtain $$\operatorname W_d\left(\mu\kappa_t,\nu\kappa_t\right)\le\tilde ce^{-\lambda t}\operatorname W_d(\mu,\nu)\tag{15}$$ for all $\mu,\nu\in\mathcal M_1(E)$, where $$\tilde c:=\frac c\alpha$$ and $$\lambda:=-\ln\alpha.$$

Remark

I'd still be interested in the question whether this result still holds when $(10)$ and $(11)$ are replaced by the following assumption: There is a $t_0>0$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,t_0)\tag{10'}$$ and $$\operatorname W_d(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0})\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11'}$$ for some $\alpha\ge0$.

(The original statement in this answer is the particular case $t_0=1$.)

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