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Notations. Let $(X, \mathcal{B})$ be a separable Banach space, with its Borel sigma-algebra, $\|\cdot\|$ stands for the norm in $X$, $\mathcal{P}(X)$ - the set of all probability measures on $X$. Let $P(x,B)$ be a stochastic kernel with source and target spaces both equal to $(X, \mathcal{B})$. Let also

$$ L_b(X)=\left\{f\in C_b(X): \mathrm{Lip}(f):=\sup_{x_1,x_2\in X}\frac{|f(x_1)-f(x_2)|}{\|x_1-x_2\|}<\infty \right\},\\ \|f\|_L:=\|f\|_{C(X)}+\mathrm{Lip}(f). $$

By $\|\cdot\|_{TV}$, $\|\cdot\|^*_L$ we denote total variation and dual Lipschitz distances on $\mathcal{P}(X)$: $$ \|\mu_1-\mu_2\|_{TV}=\frac{1}{2}\sup_{f\in C_b(X),\\ \|f\|_{C(X)}\leq 1}\left|(f,\mu_1)-(f,\mu_2)\right|,\ \mu_1,\mu_2\in \mathcal{P}(X);\\ \|\mu_1-\mu_2\|^{*}_{L}=\sup_{f\in C_b(X),\|f\|_{L}\leq 1}|(f,\mu_1)-(f,\mu_2)|,\ \mu_1,\mu_2\in \mathcal{P}(X); $$ By $\|\cdot\|_K$ we denote Kantorovich distance on $$\mathcal{P}_1(X)=\{\mu\in\mathcal P(X):\int_X\|x-x_0\|\mu(dx)<\infty\mbox{ for some } x_0\in X\}:$$ $$ \|\mu_1-\mu_2\|_{K}=\sup_{f\in L_b(X),\\\mathrm{Lip}(f)\leq 1}|(f,\mu_1)-(f,\mu_2)|,\ \mu_1,\mu_2\in \mathcal{P_1}(X). $$

Question. If for all $x_1,x_2\in X$ we have $\|P(x_1,\cdot)-P(x_2,\cdot)\|_{TV}\leq \gamma$ for some $\gamma\in (0,1)$, then one can show that the map $$ G:\mathcal P(X)\to \mathcal P(X),\ G(\mu)(B)=\int_XP(x,B)\mu(dx) $$ is contraction $$ \|G(\mu_1)-G(\mu_2)\|_{TV}\leq \gamma \|\mu_1-\mu_2\|_{TV}. $$ Is it possible to obtain similar contraction result
1. in dual Lipschitz distance, if for all $x_1,x_2\in X$ $\|P(x_1,\cdot)-P(x_2,\cdot)\|^*_{L}\leq \gamma$ for some $\gamma\in (0,1)$?
2. in Kantorovich distance, if for all $x_1,x_2\in X$ $\|P(x_1,\cdot)-P(x_2,\cdot)\|^*_{K}\leq \gamma$ for some $\gamma\in (0,1)$?

If yes, please, could you give an idea of the proof or a reference?

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No. $P$ is not a contraction with respect to dual Lipschitz distance. As an example, take $X=\mathbb R$ and $P(x,\cdot)=\frac 12\delta_0+\frac 12\delta_{2x}$. This satisfies the contraction in $TV$ (because half of the mass is coupled in each step). But $\|G(\delta_0)-G(\delta_{\frac 14})\|_L^*= \|\delta_0-\delta_{\frac14}\|_L^*$ (as can be seen using the function $f(x)=\min(|x|,1)$).

The same argument shows that there is no contraction of the Kantorovich distance. If you replace $2x$ by $3x$, then the distances are actually increased.

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    $\begingroup$ Thank you very much for simple and illustrative example! I have realised that to have contraction in $\|\cdot\|_K$ we should have $\|P(x,\cdot)-P(y,\cdot)\|_K\leq \gamma \|x-y\|$ for all $x,y\in X$ with $\gamma\in (0,1)$. $\endgroup$ – Anton Jun 14 '16 at 15:58

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