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Suppose we have a function $f$ defined on the 2D lattice points that takes values between $[0, 1]$. Furthermore, suppose that $f$ satisfies

$$f(a,b) = \frac{f(a-1,b) + f(a + 1,b) + f(a, b-1) + f(a, b+1)}4.$$

That is, the value of $f$ at $(a,b)$ is the average value of its neighbors. (This is also knows as the discrete harmonic property.) The problem is:

Prove that $f$ must be the constant function.

I have tried is to assume that $f$ is only defined on a finite grid of lattice points. Using an extremal argument, we can prove that $f$ must take its maximal value on the boundaries. However, this doesn't seem to lead anywhere in the infinite case. Any ideas?

Sandeep Silwal
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2 Answers2

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For several proofs see the end of Chapter 9 (Theorem 9.24) in the book "Probability on trees and networks" by Lyons and Peres. A very short proof uses the Krein Milman Theorem. The harmonic functions with values in [0,1] form a compact conves set, which is the closed convex hull of its extreme points. Every harmonic function is the average of four translates so an extreme harmonic function must be equal to these four translates, hence it is constant. This proof works in all discrete abelian groups. Another proof can be based on the recurrence of simple random walk $X_n$ in the 2D lattice. $f(X_n)$ is a bounded Martingale, hence it must converge almost surely; as the SRW visits every node infinitely often, this convergence forces $f$ to be constant.

Yuval Peres
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  • Does the proof using Krein-Milman also work for non-negative harmonic functions on $\mathbb{Z}^d$ (NN graph) with the constraint $f(0)=1$? (Using ad hoc Harnack-type inequality allows me to constrain $f(x)$ to be inside interval $[0,(2d)^{|x_1|+...+|x_d|}]$) – 5th decile Mar 10 '24 at 12:22
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One thing you can do is to first establish a mean value property (so that $f(x,y) = \frac{1}{4r}\sum\limits_{|a_1|+|a_2|=r} f(x+a_1,y+a_2)$). Use this to then show that $f(x,y)$ is equal to the average value over any "ball" (that is $\{(x+a,y+b)\colon |a|+|b|\leq r\}$).

Now consider the points $(x_1,y_1)$ and $(x_2,y_2)$. If we take larger and larger balls around these two points, they will intersect at a greater and greater proportion of their respective areas. Because we assume $f$ is bounded this means that the averages of $f$ over these two balls will converge to the same number, as the radius tends to infinity.

We conclude that

$$f(x_1,y_1)=f(x_2,y_2)$$

and so the function is constant.

Yankl
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    The average over the larger square does not follow so he cannot establish it. Averaging with transition probabilities for simple random walks works. – Yuval Peres Aug 10 '18 at 01:29
  • I think the different form of the averaging property holds. Namely, that the average value of the harmonic function over all points in a circle of radius $r$ centered at node $A$ deviates from $f(A)$ by no more than $cr$ for some global constant $c$ (we are assuming all values do not exceed 1 by absolute value). I think I saw that in one of the older articles on discrete harmonic functions but cannot recall right now which one. I wonder whether this is also true for a square centered at $A$... – JimT Nov 01 '22 at 03:15