The chain $\textbf{absorbed off}$ of state-space $W$ is the Markov Chain whose transition probabilities are modified only so that $p(x,x)=1$ for $x\notin W$.
My question:
(1)what does 'transition probabilities are modified only so that $p(x,x)=1$ for $x\notin W$' mean? Why not $p(x,y)=1$ for $y\notin W$?
The definition is from http://mypage.iu.edu/~rdlyons/prbtree/book_online.pdf Chapter 2 P20.
(2)Maximum Principle: Let $W$ be a subset of states of a Markov chain on a finite or countable state space $V$. If $f: V\to \mathbb{R}$ is a harmonic on $W$ and the $\sup f$ on $V$ is got at some $x_0\in W$. Then $f$ is constant on all states accessible from $x_0$ in the chain absorbed off of $W$.
Here is my proof of the Maximum Principle, I am not sure if it is right?
Let $K:=\{y\in V: f(y)=\sup f\}$. For $x\in W\cap K$ and $y$ is accessbile from $x$ with $P(x,y)>0$. So $$f(x)=\sum_{x} P(x,y)f(y)$$ If $f(y)<f(x)$, then $f(x)=\sum_{x} P(x,y)f(y)<\sup f\sum_{x} P(x,y)=\sup f$ which is a contradiction. Hence, $f(y)=f(x)$ which implies $y\in K$.
I prove that "Then $f$ is constant on all states accessible from $x_0$" But how to show that "in the chain absorbed off of $W$"?
I am going to write a version of the maximum principle that is the useful one (in my opinion). Hope this will help you understand the maximum principle.
Theorem. Let $\mathrm{V}$ be the vertex set of a countable graph. Assume $p$ is an irreducible transition density on $\mathrm{V}$ (from every vertex, one can find a path to any other vertex with positive probability). If $f$ is a harmonic function relative to $p,$ then $f$ is either constant or else it does not attain neither maximum nor minimum.
Proof. Suffices to consider the case in which $f$ is not constant. Also, since both $f$ and $-f$ are harmonic relative to $p,$ suffices to show $f$ cannot attain a maximum. If $x_0$ were a maximiser of $f,$ then $f(x_0) = \sum\limits_{x:x \sim x_0} f(x) p(x_0, x),$ and then $f(x_0) = f(x)$ for every $x \sim x_0$ (every vertex that can be accessed from $x_0$ in one step). (To see this, notice that if $x' \sim x_0$ and $f(x_0) > f(x')$ then $$\begin{align*} \sum f(x) p(x_0, x) &= p(x_0, x') f(x') + \sum_{x \neq x'} p(x_0, x) f(x) \\ &< p(x_0, x') f(x_0) + \sum_{x \neq x'} p(x_0, x) f(x_0) = f(x_0), \end{align*}$$ contrary to the hypothesis of harmonicity.) Thus, if $f$ had a maximiser $x_0,$ this would entail that for every neighbour $x$ of $x_0,$ $f(x) = f(x_0),$ the hypothesis of irreducibility of $p$ (and an easy induction on the length of paths starting at $x_0$) shows at once $f$ had to be constant to begin with, which contradicts the hypothesis. Thus, $x_0$ cannot exists. Q.E.D.