Function on $\mathbb Z^2$ whose value equals the average of values at adjacent points $\Rightarrow$ function is constant

Question

This is a reference request. I am not asking for a proof.

If I remember correctly, there is a theorem that states that if a bounded [criterion added after editing] function $f:\mathbb Z^2\to\mathbb R$ satisfies, for all $(x,y)\in\mathbb Z^2$, $$f(x,y)=\frac{f(x+1,y)+f(x-1,y)+f(x,y+1)+f(x,y-1)}{4},$$ then $f$ is constant. As I recall, the proof I saw is non-elementary and uses ~~Alaoglu’s compactness theorem~~ the Krein–Milman theorem.

Can anyone help me locate an author/text/name? Thanks in advance.

False, any linear function $f(x,y) = Ax+By+C$ also satisfy the given requirement. — achille hui, Aug 18 '15 at 06:22
I remember a Putnam problem stating that if $f:\Bbb R^2\to\Bbb R$ sums to zero on any square in $\Bbb R^2$ then $f=0$ identically. (Same is true of equilateral triangles I believe. Fun problem.) — anon, Aug 18 '15 at 06:30

score 2 · Answer 1 · answered Aug 18 '15 at 06:22

2

You do not remember correctly. Try $f(x,y) = x$. Perhaps there was an assumption that $f$ is bounded?

answered Aug 18 '15 at 06:22

Robert Israel

470,583

Yes! Now you say it, I do remember boundedness. – triple_sec Aug 18 '15 at 06:59

Batominovski · Answer 2 · 2015-08-18T07:06:39.200

1

Although I didn't try to prove it, I think that all polynomial solutions to this functional equation (without the boundedness condition) are of the form $f(x,y)=a+bx+cy+dxy+e\left(x^2-y^2\right)$ for all $x,y\in\mathbb{Z}$, where $a,b,c,d,e$ are fixed real numbers. There are other solutions like $f(x,y)=\text{Re}\left(c\alpha^x\beta^y\right)$ for all $x,y\in\mathbb{Z}$, where $\alpha,\beta,c\in\mathbb{C}$ are such that $\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}=4$.

edited Aug 18 '15 at 07:06

answered Aug 18 '15 at 06:53

Batominovski

50,341

Thank you for pointing out these counterexamples. Boundedness is indispensable for the result, indeed. – triple_sec Aug 18 '15 at 17:12
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There are other polynomial solutions. e.g. $x^3-3xy^2$, $3x^2y-y^3$ and $x^3y-xy^3$... – achille hui Aug 18 '15 at 19:22
@achille hui, nice examples. As I said, I didn't try to proof my claim. Do you think it is possible to characterize all polynomial solutions? – Batominovski Aug 18 '15 at 19:24
I don't know how to characterize all polynomial solutions. In fact, I'm not sure whether there are other polynomial solution linear independent from what we have. – achille hui Aug 18 '15 at 19:32
Interestingly, if $(x+\text{i}y)^n=a(x,y)+\text{i},b(x,y)$, where $a(x,y),b(x,y)\in\mathbb{Z}[x,y]$, then $f=a$ and $f=b$ are solutions for $n=0,1,2,3$. For $n=4$, $f=b$ is a solution, but $f=a$ is not. However, for $n\geq5$, $f=a$ and $f=b$ are both not solutions. – Batominovski Aug 18 '15 at 20:10

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

Found it. It is the “discrete version” of Liouville’s theorem. I was wrong about Alaoglu: the functional-analytic approach uses the Krein–Milman theorem (although proofs based on martingale theory are available, too). See more here.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 18 '15 at 13:14

triple_sec

23,935

Function on $\mathbb Z^2$ whose value equals the average of values at adjacent points $\Rightarrow$ function is constant

3 Answers3