This is something I got wind of at MathOverflow; the question was shut down pretty quickly as off-topic, but I actually got some enjoyment out of thinking about this problem, which (obviously) I hadn’t seen before.
Suppose that for all we have
and . Then is constant.
What helped me is noticing that the equation (1) can be rewritten as
which is a way of say is “discrete harmonic”: for the discrete Laplacian. This gives some useful keywords with which one can do a Google search, if one is so inclined; for example, the topic crops up at various places in MO and Math.SE. We can think of the theorem above as a version of the Liouville theorem akin to the one for ordinary (continuous) harmonic functions, but using finite differences in place of derivatives. It turns out there is a bit of a literature on this type of thing; for the above theorem, see this for instance.
But we can prove this directly: what follows is my own write-up based on a nice idea of “orl” over at AoPS. Consider the space of such nonnegative discrete harmonic functions as a subspace of the product space , which is a locally convex TVS.
Put . Then is compact and convex.
Convexity is clear; we check compactness. By an easy consequence of equation (1), if is adjacent to on the -grid. This implies where is the norm, so
where the right side product is compact. Furthermore the space is defined as an equalizer in the category of Hausdorff spaces and so is closed in ; therefore , which is the intersection in of a closed subspace and the compact product subspace, is also compact.
If and , then , i.e., for all .
If then by inclusion (2).
A point in a convex cone in a locally convex TVS is conically extreme in if, whenever with , it follows that are nonnegative multiples of .
Suppose is an extreme point of . Then is conically extreme in .
Say with and . Assume and are both nonzero, so by corollary 1, and are both nonzero. Then putting , we have and and . But since is an extreme point of , this convex combination forces . Hence and .
We will show consists of a single point, the constant . Since is convex, and compact by applying Lemma 1, it will suffice by the Krein-Milman theorem to show has at one most extreme point . Such is conically extreme in by Proposition 1. If we define the operators and on by and , then is invariant under , and their inverses. Then, since we have
the conical extremality of forces and for some . Then from
we conclude
and hence . Since and , we conclude is constant, namely the constant , and we are done.