co.combinatorics - Discrete harmonic function on a planar graph

The answer is no.

I first describe the graph $G$. Let $N_i$ be a sequence of positive integers; we will choose $N_i$ later. Let $T$ be an infinite tree which has one root vertex, the root has $N_1$ children; the children of that root have $N_2$ children, those children have $N_3$ children and so forth. Let $V_0$ be the set containing the root, $V_1$ be the set of children of the root, $V_2$ the children of the elements of $V_1$, and so forth. To form our graph, take $T$ and add a sequence of cycles, one going through the vertices of $V_1$, one through $V_2$ and so forth. (In the way which is compatible with the obvious planar embedding of $T$.)

Every face of $G$ is either a triangle or a quadrilateral.

We will build a harmonic function $f$ on $G$ as follows: On the root, $f$ will be $0$. On $V_1$, we choose $f$ to be nonzero, but average to $0$. On $V_i$, for $i geq 2$, we compute $f$ inductively by the condition that, for every $u in V_{i-1}$, the function $f$ is constant on the children of $u$. Of course, we may or may not get a bounded function depending on how we choose the $N_i$. I will now show that we can choose the $N_i$ so that $f$ is bounded. Or, rather, I will claim it and leave the details as an exercise for you.

Let $a_i$ be a decreasing sequence of positive reals, approaching zero. Take $N_i = 6/(a_{i+1} - a_i)$. Exercise: If $f$ on $V_1$ is taken between $-1+a_1$ and $1-a_1$, then $f$ on $V_i$ will lie between $-1+a_i$ and $1-a_i$. In particular, $f$ will be bounded between $-1$ and $1$ everywhere.