 Solution

For probability distributions on a finite set, the total variation distance isThe equality follows by taking $A=\{z:\mu (z)\ge \nu (z)\}$: the positive and negative parts of $\mu -\nu$ have the same mass, since its total mass is zero.

Use the stationary distribution identity $\pi P^t=\pi$. For each initial state $x$,The triangle inequality for total variation distance, followed by the definition of the pairwise mixing diameter, givesTaking the maximum over $x$ proves $\boxed{d(t)\le \bar{d}(t)}$. This particular inequality only needs a stationary distribution; it does not require an irreducible Markov chain or an aperiodic Markov chain.