Levin & Peres 2017 Ex. 4.1: prove that $\overline{d}(t)=\sup_{\mu,\nu}\lVert\mu P^t-\nu P^t\rVert_{TV}$

Question

Definition Let $S$ be a finite set. For any probability distributions $\mu,\nu$ on $S$, $$\lVert\mu-\nu\rVert_{TV}:=\max_{A\subset S}|\mu(A)-\nu(A)|\text{.}$$ Let $P$ be a transition kernel with state space $S$. For any positive interger $t$, $$\overline{d}(t):=\max_{x,y\in S}\lVert P^t(x,\cdot)-P^t(y,\cdot)\rVert_{TV}\text{.}$$

Prove that $$\overline{d}(t)=\sup\lVert\mu P^t-\nu P^t\rVert_{TV}$$ where the supremum is taken over all distributions $\mu,\nu$ on $S$.

That $\mbox{LHS}\leq\mbox{RHS}$ is clear. For any distributions $\mu,\nu$ on $S$ and $A\subset S$, $$\begin{align*}|\mu P^t(A)-\nu P^t(A)|&=|\sum_x\mu(x)P^t(x,A)-\sum_x\mu(x)P^t(y,A)+\sum_x\nu(x)P^t(y,A)-\sum_x\nu(x)P^t(x,A)|\\ &=|\sum_x(\mu(x)-\nu(x))(P^t(x,A)-P^t(y,A))| \end{align*}$$ for any $y\in S$.

I don't know how to proceed any further.

Answer 1

Suppose by way of contradiction that $$\sup_{\mu,\nu \in \Delta(S)} \|\mu P^t - \nu P^t\|_{TV} > \sup_{x,y \in S} \|P^t(x,\cdot) - P^t(y,\cdot)\|_{TV},$$ where $\Delta(S)$ is the set of probability measures over $S$.

Let $\mu, \nu \in \Delta(S)$ and $A \subset S$ such that $$\mu P^t(A) - \nu P^t(A) > \sup_{x,y \in S} \|P^t(x,\cdot) - P^t(y,\cdot)\|_{TV}.$$ Note that $\mu \mapsto \mu P^t(A)$ is linear, hence it is maximized for $\mu = \delta_x$ for some $x \in S$, where $\delta_x$ is the Dirac measure. Similarly $\nu \mapsto \nu P^t(A)$ is minimized for $\nu = \delta_y$ for some $x \in S$. Hence $$\mu P^t(A) - \nu P^t(A) \leq P^t(x,A) - P^t(y,A) \leq \sup_{x,y\in S} \|P^t(x,\cdot) - P^t(y,\cdot)\|_{TV},$$ contradiction.