Doeblin Condition of Markov Chains Let $P: X \times X \to R_+$ be the transition kernel of a Markov chain with state space $X$. A version of Doeblin condition is: there exists a constant $0 < \alpha < 1$ such that
$$
\|P(x, \cdot) - P(y, \cdot)\|_{TV} < 1 - \alpha
$$
for every $(x, y) \in X \times X$.
It is intuitively easy to see that this implies for any two probability measures $\mu, \nu$ on $X$ we have
$$
\|\mu P - \nu P\|_{TV} \le (1 - \alpha) \|\mu - \nu\|_{TV}
,
$$
but how to prove this rigorously?
Here $\|\|_{TV}$ is the total variation distance, and $\mu P$ is the probability measure defined by $\mu P(A) := \int_X \mu(dx) P(x, A)$.
If the Doeblin condition is replaced by $\alpha$-minorisation condition, I can see the proof.
I had this question while reading the paper Optimal approximating Markov chains
for Bayesian inference (https://arxiv.org/pdf/1508.03387.pdf).
 A: Here is one way to think about it, although I should put here the disclaimer that it may not be as rigorous as you'd like. It's also possible that there are much simpler arguments available, this is just what came to mind.
You can use the characterization of total variation distance in terms optimal couplings. For probability distributions $Q_{1}$ and $Q_{2}$ on $X$, a coupling of $Q_{1}$ and $Q_{2}$ is a joint distribution $\pi$ on $X \times X$ with marginals $Q_{1}$ and $Q_{2}$. That is, $\pi(B, X) = Q_{1}(B)$ and $\pi(X,B) = Q_{2}(B)$ for any measurable $B \subseteq X.$
We then have the characterization
$$\Vert Q_{1} - Q_{2} \rVert_{TV} = \inf_{\pi}\,\mathbb{E}_{\pi}[1_{\{Z_{1} \neq Z_{2}\}}]\qquad (*)$$
where the infimum is taken over all couplings $\pi$ of $Q_{1}$ and $Q_{2}$ and the random pair $(Z_{1}, Z_{2}) \in X \times X$ is distributed according to $\pi.$
The Doeblin condition gives that there is some $\alpha > 0$ such that for any pair $x,y \in X$, there is a coupling $\rho_{xy}$ of the distributions $P(x,\cdot)$ and $P(y,\cdot)$ satisfying $\mathbb{E}_{\rho_{xy}}[1_{\{Z_{1} \neq Z_{2}\}}] < 1 - \alpha.$ Fix such a coupling for each $x,y$, so that we have the family $\{\rho_{xy}\}_{x,y \in X}$. We only insist that $\rho_{xx}$ is the optimal coupling of $P(x,\cdot)$ with itself for each $x \in X$, so that $\mathbb{E}_{\rho_{xx}}[1_{\{Z_{1} \neq Z_{2}\}}] = 0$.
Now, let $\mu$ and $\nu$ be probability distributions on $X$. Observe that any coupling $\pi$ of $\mu$ and $\nu$ and the family of couplings $\{\rho_{xy}\}$ give rise to a single coupling of $\mu P$ and $\nu P$. The coupling is as follows:

*

*Draw $(W_{1}, W_{2})$ according to $\pi$.

*Conditioned on the outcome of (1), independently draw $(Y_{1}, Y_{2})$ according to $\rho_{W_{1}W_{2}}.$

*Set $(Z_{1}, Z_{2}) = (Y_{1}, Y_{2}).$
Denote the distribution of $(Z_{1},Z_{2})$ by $\rho\pi$. It is not too hard to see that $\rho\pi$ is indeed a coupling of $\mu P$ and $\nu P$.
Now, we see that
$$\lVert \mu P - \nu P \rVert_{TV} \leq \inf_{\pi} \mathbb{E}_{\rho\pi}[1_{\{Z_{1} \neq Z_{2}\}}] \qquad (**)$$
where the infimum is over all couplings $\pi$ of $\mu$ and $\nu.$ We get an inequality instead of the equality from (*) because we are taking an infimum over a smaller set. It remains to show that the right-hand side is at most $(1-\alpha)\lVert \mu - \nu\rVert_{TV}.$
To see this, let $\pi$ be any coupling of $\mu$ and $\nu$. In the probability space of the coupling $\rho\pi$, we claim:
$$1_{\{Z_{1} \neq Z_{2}\}} = 1_{\{Y_{1} \neq Y_{2}\}}1_{\{W_{1} \neq W_{2}\}}$$
as random variables. This holds because of our requirements on $\rho_{xx}$: if $W_{1} = W_{2}$, then necessarily $Y_{1} = Y_{2}$ and hence $Z_{1} = Z_{2}$.
Now, by "conditioning on the outcome of $(W_{1},W_{2})$" (this is the part which is not completely rigorous, but I believe it can be made rigorous),
$$\mathbb{E}_{\rho\pi}[1_{\{Z_{1} \neq Z_{2}\}}] = \mathbb{E}_{\rho\pi}[1_{\{Y_{1} \neq Y_{2}\}}1_{\{W_{1} \neq W_{2}\}}] = \mathbb{E}_{\pi}[1_{\{W_{1} \neq W_{2}\}}\mathbb{E}_{\rho_{W_{1}W_{2}}}[1_{\{Y_{1} \neq Y_{2}\}}\,|\,(W_{1},W_{2})]]$$
where we use that $1_{\{Y_{1} \neq Y_{2}\}}$ is conditionally independent from the outcome of $(W_{1},W_{2})$.
The inner conditional expectation $\mathbb{E}_{\rho_{W_{1}W_{2}}}[1_{\{Y_{1} \neq Y_{2}\}}\,|\,(W_{1},W_{2})]$ is less than $1-\alpha$ by our assumptions on the family $\{\rho_{xy}\}$, so in all we get
$$\mathbb{E}_{\rho\pi}[1_{\{Z_{1} \neq Z_{2}\}}] < (1-\alpha)\mathbb{E}_{\pi}[1_{\{W_{1}\neq W_{2}\}}]$$
and so, applying an infimum as in (**),
$$\lVert \mu P - \nu P \rVert_{TV} \leq \inf_{\pi}\mathbb{E}_{\rho\pi}[1_{\{Z_{1} \neq Z_{2}\}}] \leq (1-\alpha)\inf_{\pi}\mathbb{E}_{\pi}[1_{\{W_{1}\neq W_{2}\}}] = (1-\alpha)\lVert \mu - \nu \rVert_{TV}$$
as desired.
