8
$\begingroup$

Is the following conjecture true?

Let $f:\mathbb Z^2\to \mathbb R$ be a discrete harmonic function. If $f$ has less than exponential growth then it is a polynomial.


Definitions:

  • Growth of $f$ is less then exponential if for every $a>0$ it is that $$ \lim_{\|x\|\to \infty} e^{-a\|x\|} f(x) = 0. $$
  • A function $f:\mathbb Z^2\to \mathbb R$ is discrete harmonic iff $$ f(x) = \dfrac{f(x_1 + 1, x_2)+f(x_1, x_2 + 1) + f(x_1 - 1, x_2) +f(x_1, x_2 - 1)}{4}. $$

Context: Given a discrete harmonic function $f:\mathbb Z^2\to \mathbb R$, it is known that

Does the conjecture follow?

$\endgroup$
4
  • $\begingroup$ What does "Let $n \in 2$ " mean in body of text that you labelled as "claim"? $\endgroup$ Commented Jan 4, 2023 at 21:43
  • $\begingroup$ @DionelJaime Hahaha, thanks for pointing out that typo. I originally wanted to ask the questions for any dimension $n\in \mathbb N$, but then I though that the case $n=2$ is perhaps most interesting. $\endgroup$ Commented Jan 4, 2023 at 22:51
  • 1
    $\begingroup$ I have no experience with discrete harmonic functions, but for harmonic functions $u:\Bbb C \to \Bbb R$ this is definitely wrong: Take the real part of any non-constant entire function of order less than one, e.g. $u(z) = \operatorname{Re}(\cos(\sqrt z)).$ $\endgroup$
    – Martin R
    Commented Jan 13, 2023 at 7:53
  • 1
    $\begingroup$ @orangeskid Yes, $f(x,y)=x$ is a harmonic polynomial...not contradicting the conjecture. $\endgroup$ Commented Jan 14, 2023 at 11:33

1 Answer 1

5
+50
$\begingroup$

Your conjecture seems wrong. It is possible to construct a counter-example in the form $f(x,y)=\sum a_n\cdot p_n(x,y)$ where each $p_n(x,y)$ is a discrete harmonic polynomial with degree $n$, and the coefficients $a_n$ tends to $0$ rapidly.

A concrete function is $$ f(x,y) = \sum_{n=0}^\infty \dfrac1{\color{red}{n^{2n}}}\bigg(\sum_{k=0}^n(-1)^k\binom{x{\color{red}{+n-k}}}{2(n-k)}\binom{y{\color{red}{+k}}}{2k}\bigg). $$


Updates:

I inserted the red $+n-k$ and $+k$ in the binomial coefficients.

A verification that the polynomial $p(x,y)=\sum\limits_{k=0}^n(-1)^k\tbinom{x+n-k}{2(n-k)}\tbinom{y+k}{2k}$ is harmonic:

For $a>0$ we have $$ \tbinom{x+a+1}{2a}-2\tbinom{x+a}{2a}+\tbinom{x+a-1}{2a} =\Big(\tbinom{x+a+1}{2a}-\tbinom{x+a}{2a}\Big) +\Big(\tbinom{x+a}{2a}-\tbinom{x+a-1}{2a}\Big) \\ =\tbinom{x+a}{2a-1}-\tbinom{x+a-1}{2a-1} =\tbinom{x+a-1}{2(a-1)}; $$ for $a=0$ we have $$ \tbinom{x+a+1}{2a}-2\tbinom{x+a}{2a}+\tbinom{x+a-1}{2a}=0. $$ So, $$ \big(p(x+1,y)-2p(x,y)+p(x-1,y)\big) +\big(p(x,y+1)-2p(x,y)+p(x,y-1)\big) \\ = \sum_{k=0}^n(-1)^k \Big( \tbinom{x+n-k+1}{2(n-k)}-2\tbinom{x+n-k}{2(n-k)}+\tbinom{x+n-k-1}{2(n-k)} \Big)\tbinom{y+k}{2k} \\ +\sum_{k=0}^n(-1)^k \tbinom{x+n-k}{2(n-k)} \Big( \tbinom{y+k+1}{2k}-2\tbinom{y+k}{2k}+\tbinom{y+k-1}{2k}\Big) \\ = \sum_{k=0}^{n-1}(-1)^k \tbinom{x+n-k-1}{2(n-k-1)} \tbinom{y+k}{2k} +\sum_{k=1}^n(-1)^k \tbinom{x+n-k}{2(n-k)} \tbinom{y+k-1}{2(k-1)}\Big) \\ = 0. $$

Concerning speed of growth.. $$ \Big|\sum\limits_{k=0}^n(-1)^k\tbinom{x+n-k}{2(n-k)}\tbinom{y+k}{2k}\Big| \le \sum\limits_{k=0}^n \Big|\tbinom{x+n-k}{2(n-k)}\Big|\cdot\Big|\tbinom{y+k}{2k}\Big|\\ <\sum\limits_{k=0}^n \frac{\big(|x|+n\big)^{2(n-k)}\big(|y|+n\big)^{2k} }{(2k)!(2n-k)!} <\frac{\big(|x|+|y|+2n\big)^{2n}}{(2n)!} $$ so for arbitrary positive integer $K>0$, $$ \big|f(x,y)\big| < \sum_{n=0}^\infty \frac{\big(|x|+|y|+2n\big)^{2n}}{n^{2n}(2n)!} <\sum_{n\le K}+\sum_{K<n\le|x|+|y|}+\sum_{n>|x|+|y|}\\ <\text{(some polynomial)} +\sum_{K<n\le|x|+|y|}\frac{\big(3(|x|+|y|)\big)^{2n}}{K^{2n}(2n)!} +\sum_{n>|x|+|y|}\frac{(3n)^{2n}}{n^{2n}{2n!}} \\ < \text{(some polynomial)} +e^{\frac3K(|x|+|y|)} + O(1). $$

$\endgroup$
6
  • $\begingroup$ Thank you for a very nice solution, I'm impressed by the simplicity of the idea! Could you recommend me a reference where it is shown that the polynomials in the form of products of two generalized combinatorial coefficients are discrete harmonic? $\endgroup$ Commented Jan 14, 2023 at 10:05
  • 1
    $\begingroup$ @PavelKocourek This seems wrong. If, for example, $f(x,y)=\binom{x}{2}\binom{y}{2}$, then $\frac{f(3,4)+f(5,4)+f(4,3)+f(4,5)}{4}=\frac{18+60+18+60}{4}=39$ is not equal to $f(4,4)=36$. $\endgroup$ Commented Jan 14, 2023 at 10:58
  • $\begingroup$ @EwanDelanoy Good point! So the sum over $k$ in the answer has to be discrete harmonic although its individual term don't have to be so. I'm wondering what would be an easy way to that the sum over $k$ is discrete harmonic for each $n$. $\endgroup$ Commented Jan 14, 2023 at 11:30
  • $\begingroup$ I like this answer, but I don't immediately see why $f(x, y)$ has less than exponential growth. Should this be obvious? Thanks a lot. $\endgroup$ Commented Jan 14, 2023 at 15:34
  • 1
    $\begingroup$ OK, I am going to add something about speed of growth... :-) $\endgroup$
    – G.Kós
    Commented Jan 14, 2023 at 16:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .