**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?



Contraction of probability measures
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.


