# Dimension of the space of discrete harmonic functions $f:\mathbb Z^n\to \mathbb R$

Fix $$n\in \mathbb N$$. Let $$F(n)$$ be the space of discrete harmonic functions $$f:\mathbb Z^n\to \mathbb R$$. What is the minimal $$d \in \mathbb N$$ (if there is one) such that there exists $$B\subset F(n)$$ with $$|B|=d$$ that every $$f\in F(n)$$ can be written as a linear combination of functions from $$B$$ "subject to a translation", i.e. there is $$m\in \mathbb N$$ and $$(a^1,f^1,x^1),\ldots,(a^m,f^m,x^m) \in \mathbb R \times B \times \mathbb Z^n$$ such that $$f(x) = \sum_{j=1}^m a^j f^m(x-x^m) \quad \forall x\in \mathbb Z^n. \tag{*}$$

A function $$f:\mathbb Z^n\to \mathbb R$$ is harmonic iff $$2nf(x) = \sum_{i=1}^n [f(x-e_i)+f(x+e_i)],$$ for all $$x\in \mathbb Z^n$$, where $$(e_1,\ldots,e_n)$$ is the canonical basis of $$\mathbb Z^n$$. Note that $$f$$ being discrete harmonic can be interpreted as that $$f$$ is a martingale w.r.t. a symmetric random walk on $$\mathbb Z^n$$.

Case $$n=1$$: It is easy to see that $$f(x)$$ has to have a constant growth, thus $$F(1)$$ is the space of all the linear functions $$f:\mathbb Z\to \mathbb R$$.

Case $$n=2$$: Analogously, $$F(2)$$ also contains all the linear functions $$f:\mathbb Z^2\to \mathbb R$$. However, the function $$f^2_{12}(x) = (x_1)^2 - (x_2)^2$$ is also discrete harmonic. Does $$f^2_{12}$$ together with the basis of the space of linear functions $$f:\mathbb Z^2\to \mathbb R$$ span the whole $$F(2)$$? Clearly not, as @NeitherNor pointed out, there are other linearly independent functions in $$F(2)$$ and as @RyszardSzwarc brought up, their translations are also in $$F(2)$$, thus the complicated definition of "basis" $$B$$.

Case $$n\geq 3$$: Besides the linear functions also the functions $$f^3_{ij}=(x_i)^2 - (x_j^2), i\neq j, \ i,j \in \{1,2,3\}$$ are harmonic...

• Case $n=2$: also $x_1x_2$ and $x_1x_2(x_1^2-x_2^2)$ are discrete harmonic. Indeed, there are many more I forgot, but: for all $k>0$, there are 2 linearly independent polynomial harmonic functions of order $k$. Anyways, this doesn't mean that there are no non-polynomial ones additionally... Commented Jan 3, 2023 at 19:46
• If $\sum_{i=1}^n z_i+z_i^{-1} = 2n$ then $f(x)= \prod_{i=1}^n z_i^{x_i}$ is solution Commented Jan 3, 2023 at 23:30
• @reuns Won’t this imply that $z_i=1$ for all $i$? Commented Jan 3, 2023 at 23:38
• When $f(x)$ is a solution so is $f(x+k)$ for any $k\in\mathbb{Z}^n.$ Commented Jan 4, 2023 at 0:23
• @NeitherNor In the example of reuns you cannot take $z_2=0,$ as then $f(z)$ is not well defined. Instead you can take $z_2=-1.$ Commented Jan 4, 2023 at 5:50

The answer is : the dimension $$d$$ is infinite for $$n\ge 2.$$
At the beginning I will put the problem into a more general framework of finitely generated groups, although it will not help solving it. Let $$G=\mathbb{Z}^n.$$ The group is generated by $$e_1,e_2,\ldots, e_n.$$ The generators lead to a natural distance on $$G$$ $$\|x\|=|x_1|+|x_2|+\ldots +|x_n|$$ Consider the operator acting on functions defined on $$G$$ by $$(Af)(x)={1\over 2n}\sum_{j=1}^n[f(x-e_j)+f(x+e_j)]={1\over 2n}\sum_{\|y\|=1} f(x+y)$$ Observe that $$A$$ is a convolution operator on the group $$G$$ with the function $$\varphi={1\over 2n}\sum_{j=1}^n[\delta_{e_j}+\delta_{-e_j}]={1\over 2n}\sum_{\|y\|=1}\delta_y$$ i.e. the normalized indicator function of elements with norm $$1.$$
Concerning the problem, we want to determine the dimension of the eigenspace of $$A$$ corresponding to the eigenvalue $$1,$$ i.e. $$Af=f.$$ We do not impose any norm conditions on $$f.$$ Some solutions are multiplicative (called usually the characters of the group, if $$|f|=1$$). Namely if $$f(a+b)=f(a)f(b)$$ then $$f(x)=(Af)(x)={1\over 2n}\sum_{j=1}^n[f(x)f(e_j)+f(x)f(e_j)^{-1}]\\ = {1\over 2n}\sum_{j=1}^n [f(e_j)+f(e_j)^{-1}]\,f(x)$$ This provides a solution (found by @reuns) $${1\over 2n}\sum_{j=1}^n[z_j+z_j^{-1}]=1,\quad z_j:=f(e_j)$$
We are going to show that the dimension of harmonic functions is infinite. Consider $$n=2.$$ For $$z>1$$ let $$f_z$$ denote the function corresponding to $$z_2=-z$$ and $$z_1>1,$$ i.e. $$z_1=2+{1\over 2}(z+z^{-1})+\sqrt{[2+\textstyle{1\over 2}(z+z^{-1})]^2-1}\ge 1+(z+z^{-1})>z=|z_2|$$ Observe that for $$w>z>1$$ we have $$w_1>z_1>1.$$ Let $$z(k)>z(k-1)>\ldots >z(1)>1.$$ We claim that the functions $$f_{z(j)}$$ are linearly independent. The corresponding parameters are $$z_1(j)$$ and $$z_2(j).$$ Assume $$a_1f_{z(1)}+a_2f_{z(2)}+\ldots +a_kf_{z(k)}=0$$ With no loss of generality we may assume $$a_k\neq 0.$$ Substituting $$x=je_1$$ gives $$a_1z_1(1)^j+a_2z_1(2)^j+\ldots +a_kz_1(k)^j=0$$ The left hand side is equal asymptotically to $$a_kz_1(k)^j$$ when $$j\to \infty.$$ The translation is irrelevant as the translation of $$f_z$$ is a multiple of $$f_z.$$ Therefore $$a_k=0,$$ a contradiction. The proof is made for $$n=2,$$ but it can be adapted to any $$n\ge 2.$$
• That makes me wonder – is there infinitely many solutions that are neither multiplicative nor a product of a multilicative solution with some more elementary solution, i.e $x_1x_2(x_1^2-x_1^2)$ won't count, but $x_1^2-x_1^2$ would. Commented Jan 4, 2023 at 15:49
• I had in mind a solution in terms of elements of the "bases" $B$, sorry for being vague. Commented Jan 4, 2023 at 17:13