Are all discrete harmonic function $f:\mathbb Z^2\to \mathbb R$ with less than exponential growth polynomials?

Question

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.

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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 $$ f(x) = \dfrac{f(x_1 + 1, x_2)+f(x_1, x_2 + 1) + f(x_1 - 1, x_2) +f(x_1, x_2 - 1)}{4}. $$

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

• if $f$ is bounded, then it is constant: A stronger version of discrete "Liouville's theorem"
• if the gradient of $f$ is bounded, then $f$ is linear: Harmonic functions on $\mathbf{Z}^2$
• analogously: if all the $k$-th gradients of $f$ are bounded, then $f$ is a polynomial of at most order $k$.

Does the conjecture follow?

Answer 1

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

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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 $K>0$, $$ \big|f(x,y)\big| < \sum_{n=0}^\infty \frac{\big(|x|+|y|+2n\big)^{2n}}{n^{2n}(2n)!} <\sum_{n\le K}+\sum_{K<n\le|x|+|y|}+\sum_{n>|x|+|y|}\\ <\text{(some polynomial)} +\sum_{K<n\le|x|+|y|}\frac{\big(3(|x|+|y|)\big)^{2n}}{K^{2n}(2n)!} +\sum_{n>|x|+|y|}\frac{(3n)^{2n}}{n^{2n}{2n!}} \\ < \text{(some polynomial)} +e^{\frac3K(|x|+|y|)} + O(1). $$