I'm trying to understand the proof of the Maximum Principle for discrete Markov chains:
--- start of proof ---
Let $W$ be a set of states of a Markov chain on finite or countable state space $V$. If $f: V \rightarrow \mathbb{R}$ is a function that is harmonic on W and the supremum of $f$ on $V$ is achieved at some element $x_0$ in the chain absorbed off of $W$.
$f(x) = \sum_{x \sim y} p(x,y) f(y)$
Prove that $f$ is a constant on all states accessible from $x_0$ in the chain absorbed off at $W$.
The proof of this goes like:
Let $K = \{ y \in V; f(y) = sup\ f \}$
If $x \in W \cap K $ and $p(x,y) > 0$ then $y \in K$ because $f$ is harmonic.
Hence the conclusion follows.
--- end of proof ---
What does the author mean by "because $f$ is harmonic"? Would a more detailed proof involve the mean-value property of harmonic funcions and proving by contradiction using the closed sets argument on graphs?
It would. The more detailed argument is: we have $$ \sum_{y} p(x,y) f(y) = f(x) = f(x)\left(\sum_{y} p(x,y)\right) = \sum_{y} p(x,y) f(x) $$ because the probabilities $p(x,y)$ add up to $1$, so $$ \sum_{y} p(x,y) (f(x) - f(y)) = 0. $$ In this expression, because $f(x) = \sup f$, each difference $f(x) - f(y)$ is nonnegative, and therefore each term is nonnegative. If we had a single positive term, then we couldn't make the sum equal $0$.
Therefore we have $p(x,y) (f(x)-f(y)) = 0$ for all $y$, which means that for all $y$ with $p(x,y) > 0$, $f(x) - f(y) = 0$. That is, for all $y$ with $p(x,y)>0$, $f(y) = f(x) = \sup f$.