Harmonic function on a binary tree

Question

Let $T$ be a binary tree (each of the vertices have degree 3). Let $P$ be the transition matrix of a symmetric random walk on $T$. Let $0$ be any vertex of $T$ and think of $0$ as the root of three disjoint sub-trees of $T$.

I want to show that any solution to $2a_{n+1} - 3a_n + a_{n-1}=0$ defines a harmonic function $\alpha$ on a given subtree, by $\alpha(x) = a_n$, where $n$ is the distance between $0$ and $x$.

According to the definition, the function $f: V \to \mathbb{R}$ is harmonic on $U$ (w.r.t. the transition matrix $P$) if \begin{align} f(u) = \sum_{u \in V} p_{u,v} f(v), \qquad u \in U. \end{align} However, I do not understand this definition well enough to apply it on the above recurrence relation $a_n = \frac{2}{3}a_{n+1} + \frac{1}{3}a_{n-1}$. Any help is appreciated!

Answer 1

The transition matrix is symmetric, with three entries equal to $1/3$ in each row, and the rest of them being zero. It is the adjacency matrix of the graph multiplied by $1/3$.

So, the definition of harmonicity says that each value must be the average of the three neighboring values. Let $x$ be any vertex different from $0$, and let $n$ be its distance to $0$. Then exactly one of the neighbors of $x$ lies on the geodesic from $x$ to $0$, and thus has distance $n-1$. The other two have distance $n+1$, since they are joined to $0$ by the simple path that consists of the edge going to $x$, and then the geodesic from $x$ to $0$.

Thus, $a_n = \frac{2}{3}a_{n+1} + \frac{1}{3}a_{n-1}$ is precisely saying that the value at $x$ is the average of the neighboring values.

The above does not apply to $x=0$, which was the reason for excluding it and considering the subtrees of $T\setminus \{0\}$.