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

*

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