If I am given that $d(L^* \mu_0, L^* \mu_1) \le \lambda d(\mu_0, \mu_1)$ for any two probability measure also given that $L^*\mu=\mu$, $\lambda \in (0,1)$.
(1) Then Could anyone tell me what will be $ d((L^*)^k \nu, \mu) \le ?$ for some probability measure $\nu$?
where $d(\mu, \nu)= \sup_{Lip(f)\le 1}| \int f d\mu- \int fd \nu| $
Thanks for helping! They actually asked to show, that $(L^*)^k \nu\to \mu $ exponentially fast in $d$ distance. I didn't understand what does this means.
Okay.
$1)$ Since $L^* \mu=\mu,$ we have
$$ d((L^*)^k \nu,\mu)=d((L^*)^k \nu,(L^*)^k\mu)\leq \lambda^k d(\nu,\mu), $$ which, indeed, is exponential decay (i.e. of theform $Ce^{-ck}$ for $C,c>0$).
$2)$ The authors defined $\mathcal{P}_q$ to mean the set of measures having finite $W_q$-distance to some $\delta$-measure. Note that $W_q=C_q^{\min\{1,1/q\}},$ so if one's finite, then the other is finite.
$3)$ This I didn't check up on, but it's pretty standard notation for push-forward. I.e. for some measure $\mu$ on some space $X$ and some map $f:X\to Y$, we define $f_* \mu(A)=\mu(f^{-1}(A))$ for all measurable subsets $A$ of $Y$.