On the distance to the stationary distribution A Markov Chain $M$ has only one stationary distribution $q$.
For distribution $p$, with $D_{TV}(p,Mp)=x$, can we bound $D_{TV}(p,q)$?
Clearly, $x=0$ implies $D_{TV}(p,q)=0$. Does general bound hold?
We may write $q=\lim_{n\mapsto  \infty} \frac{p+Mp+M^2p+\cdots}{n}$.
 A: It is impossible to bound $D_{TV}(p,q)$ in terms of $x=D_{TV}(p,Mp)$ without further assumptions on the chain, like expansion. This is due to the phenomenon known as metastability.
Rich examples are discussed in [1], [2] and [3], for instance.
The simplest example is a chain on two states $a,b$ with transition probabilities
$M(a,b)=M(b,a)=\epsilon$ and $M(a,a)= M(b,b)=1-\epsilon$.
Then $q$ such that $q(a)=q(b)=1/2$ is the unique stationary measure, but $p$ with $p(a)=1$ satisfies
$D_{TV}(p,Mp)=\epsilon$ yet  $D_{TV}(p,q)=1/2.$
[1] Olivieri, Enzo, and Maria Eulália Vares. Large deviations and metastability. No. 100. Cambridge University Press, 2005.
[2] Capocaccia, D., Cassandro, M. and Olivieri, E., 1974. A study of metastability in the Ising model. Communications in Mathematical Physics, 39(3), pp.185-205.
[3] Levin, David A., Malwina J. Luczak, and Yuval Peres. "Glauber dynamics for the mean-field Ising model: cut-off, critical power law, and metastability." Probability Theory and Related Fields 146, no. 1 (2010): 223-265.
