Finite State Space (Markov Chain): Show $\alpha_{m+n}\leq\alpha_{n}\alpha_{m}$ It is a question in Durrett's book probability: theory and examples, Chapter 6, Exercise 6.6.3.
Assume the state space is finite. For any transition matrix $p$, define 
$\alpha_n = \sup_{i,j} \frac{1}{2}\sum_{k}|p^n(i,k)-p^n(j,k)|.$ The $1/2$ is there because for any $i$ and $j$ we can defince random variables $X$ and $Y$ so that $P(X=k)=p^n(i,k), P(Y=k)=p^n(j,k)$, and $P(X \neq Y)=\frac{1}{2}\sum_{k}|p^n(i,k)-p^n(j,k)|. $
Show that $\alpha_{m+n}\leq\alpha_{n}\alpha_{m}$. 
Anyone has any idea? It seems that I can easily prove that $\alpha_{m+n}\leq 2\alpha_{n}\alpha_{m}$, but I just cannot improve more. 

The idea is to use the coupling strategy used in the proof for the convergence of irreducible aperiodic Markov chains to their stationary distribution. Since it is difficult (and not at all straightforward) to manipulate these expressions algebraicly, we can instead interpret them as probabilities and exploit the Markov property to obtain the bound. To form the coupling, define the transition matrix $\mathbb{Q}$ on the state space $S^2$ by
\begin{align*}
q((x_1,y_1),(x_2,y_2)) = \begin{cases}
p(x_1,x_2)p(y_1,y_2) \hspace{1em} &\text{if } x_1 \neq y_1 \\
p(x_1,x_2) &\text{if } x_1 = y_1, x_2 = y_2 \\
0 &\text{otherwise.}
\end{cases}
\end{align*}
In words, all we have done is ``coupled'' two independent copies of the Markov chain defined by $p$, and set the two copies equal to each other as soon as they match. For example, a realization of the coupling could look like
\begin{align*}
(1,8), (2,6), (3,9), (4,4), (8,8), (5,5), (8,8), \dots
\end{align*}
Now suppose $(X_n,Y_n)$ evolves according to $q$. At this point it suffices to show that for any starting point $(i,j) \in S^2$, we have
\begin{align*}
P_{(i,j)}(X_{n+m}\neq Y_{n+m}) \leq \sup_{i,j} P_{(i,j)}(X_{n} \neq Y_n) \sup_{i,j} P_{(i,j)} (X_m \neq Y_m).
\end{align*}
Now notice that if the coupling doesn't occur by time $m+n$, then it must not have occured by times $m$ or $n$ either. Letting $\mathcal{F}_n$ denote the usual filtration, the left hand side becomes
\begin{align*}
P_{(i,j)}(X_{n+m} \neq Y_{n+m}) &= E_{(i,j)}( E_{(i,j)} (\textbf{1}_{\{X_{n+m}\neq Y_{n+m}\}} \mid \mathcal{F}_n)) \\
&= E_{(i,j)}(E_{(i,j)}(\textbf{1}_{\{X_m \neq Y_m\}} \circ \theta_n \mid \mathcal{F}_n) \textbf{1}_{\{X_n \neq Y_n\}}) \\
&= E_{(i,j)}(\textbf{1}_{\{X_n \neq Y_n\}} E_{(X_n,Y_n)}(\textbf{1}_{\{X_m \neq Y_m\}}) ),
\end{align*}
where the first equality is from the tower property, the second follows from the observation made above, and the third by the markov property. Now note that on the set $\{X_n \neq Y_n\}$, the inner expectation is bounded by
\begin{align*}
E_{(X_n,Y_n)}(\textbf{1}_{\{X_m \neq Y_m\}}) \leq \sup_{i,j} P_{(i,j)}(X_m \neq Y_m).
\end{align*}
Now, taking another sup over the pairs $(i,j)$, we obtain the inequality
\begin{align*}
\sup_{i,j} P_{(i,j)}(X_{n+m} \neq Y_{n+m}) \leq \sup_{(i,j)} P_{(i,j)}(X_n \neq Y_n) \sup_{(i,j)} P_{(i,j)} (X_m \neq Y_m),
\end{align*}
i.e. $\alpha_{n+m}\leq \alpha_n \alpha_m$, as desired.
