discrete harmonic functions

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 $(m, n) \in L = \mathbb{Z}^2$ we have

(1)$f(m, n) = \frac1{4}(f(m+1, n) + f(m-1, n) + f(m, n+1) + f(m, n-1))$

and $f \geq 0$. Then $f$ is constant.

What helped me is noticing that the equation (1) can be rewritten as

$([f(m+1, n) - 2f(m, n) + f(m-1, n)] + [f(m, n+1) - 2f(m, n) + f(m, n-1)]) = 0$

which is a way of say $f$ is “discrete harmonic”: $(\Delta_x^2 + \Delta_y^2)(f) = 0$ 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 $H$ of such nonnegative discrete harmonic functions as a subspace of the product space $\mathbb{R}^L$, which is a locally convex TVS.

Put $K_\lambda \coloneqq \{g \in H: g(0) \leq \lambda\}$. Then $K_\lambda$ is compact and convex.

Convexity is clear; we check compactness. By an easy consequence of equation (1), $g(y) \leq 4g(x)$ if $y$ is adjacent to $x$ on the $\mathbb{Z}^2$-grid. This implies $g(x) \leq 4^{{|x|}}g(0)$ where ${|x|}$ is the $l_1$ norm, so

(2)$K_\lambda \subseteq \prod_{x \in L} [0, 4^{{|x|}}\lambda]$

where the right side product is compact. Furthermore the space $H$ is defined as an equalizer in the category of Hausdorff spaces and so is closed in $\mathbb{R}^L$; therefore $K_\lambda$, which is the intersection in $\mathbb{R}^L$ of a closed subspace $H$ and the compact product subspace, is also compact.

If $g \in H$ and $g(0) = 0$, then $g = 0$, i.e., $g(x) = 0$ for all $x \in L$.

If $g(x) = 0$ then $g \in K_0 = \{0\}$ by inclusion (2).

A point $x$ in a convex cone $C$ in a locally convex TVS is *conically extreme in* $C$ if, whenever $y, z \in C$ with $x = y + z$, it follows that $y, z$ are nonnegative multiples of $x$.

Suppose $f$ is an extreme point of $K= \{g \in H: g(0) = 1\}$. Then $f$ is conically extreme in $H$.

Say $f = g + h$ with $g, h \in H$ and $g(0) = \lambda, h(0) = \mu$. Assume $g$ and $h$ are both nonzero, so by corollary 1, $\lambda = g(0)$ and $\mu = h(0)$ are both nonzero. Then putting $g' = \frac1{\lambda}g, h' = \frac1{\mu}h$, we have $g'(0) = 1$ and $h'(0) = 1$ and $f = \lambda g' + \mu h'$. But since $f$ is an extreme point of $\{g \in H: g(0) = 1\}$, this convex combination forces $g' = f = h'$. Hence $g = \lambda f$ and $h = \mu f$.

We will show $K = \{g \in H: g(0) = 1\}$ consists of a single point, the constant $g \equiv 1$. Since $K$ is convex, and compact by applying Lemma 1, it will suffice by the Krein-Milman theorem to show $K$ has at one most extreme point $f$. Such $f$ is conically extreme in $H$ by Proposition 1. If we define the operators $T$ and $S$ on $\mathbb{R}^L$ by $T(g)(x) = g(x + (1, 0))$ and $S(g)(x) = g(x + (0, 1))$, then $H$ is invariant under $T, S$, and their inverses. Then, since we have

$f = \frac1{4}(T(f) + T^{-1}(f) + S(f) + S^{-1}(f))$

the conical extremality of $f$ forces $T(f) = \lambda f$ and $S(f) = \mu f$ for some $\lambda, \mu \gt 0$. Then from

$f = \frac1{4}(\lambda + \lambda^{-1} + \mu + \mu^{-1})f$

we conclude

$4 = \lambda + \lambda^{-1} + \mu + \mu^{-1}$

and hence $\lambda = 1, \mu = 1$. Since $T(f) = f$ and $S(f) = f$, we conclude $f: L \to \mathbb{R}$ is constant, namely the constant $f \equiv 1$, and we are done.

Revised on January 8, 2015 at 02:29:50
by
Todd Trimble