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Question

Let $n\in N$, and let $S$ be a finite set in $Z^n$. What is the dimension $d(n,S)$ of the space $F(n,S)$ of bounded functions $f:\mathbb Z^n\to \mathbb R$ that are discrete harmonic on $Z^n\setminus S$?

Definition

A function $f:\mathbb Z^n\to \mathbb R$ is discrete harmonic iff $$ 2nf(x) = \sum_{i=1}^n [f(x-e_i)+f(x+e_i)], \tag{*}\label{H} $$ for all $x\in \mathbb Z^n$, where $(e_1,\ldots,e_n)$ is the canonical basis of $\mathbb Z^n$.

Note that the condition \eqref{H} can be interpreted as that $f$ is a martingale w.r.t. a symmetric random walk on $\mathbb Z^n$.

The case $n=3, S=\{0\}$

This question was motivated by the question about existence of a bounded non-constant function $f:\mathbb Z^3 \to \mathbb R$ that is discrete harmonic except at the origin $(0,0,0)$. An example of $f\in F(3,\{0\})$ is $f_1(x)$ equal to the probability that a symmetric 3D random walk starting from $x$ hits the origin. It can then be asked if all functions in $F(3,\{0\})$ can be obtained as linear combinations of $f_1(x)$ and the constant function $f_0(x)=1$.

The case $S=\{0\}$

For $n=1$ and $n=2$ the question is closely related to the related question with $S=\emptyset$. A random walk in 1D and 2D is guaranteed to return to the origin. This suggest that $d(1,\{0\})=d(2,\{0\})=1$.

For $n\geq 3$ it is no more guaranteed that the random walk will be absorbed in at the origin and so there is at least one more dimension associated the "value of never returning". Can we conclude that $d(n,\{0\}) = 2$ for $n\geq 3$?

The case $n=1$

Consider $S=\{a,b\}$. If $b=a+1$, then $f(a)=f(a-1)=\ldots$ and $f(b)=f(b+1)=\ldots$, but $f(a),f(b)$ can be arbitrary. What is more, $f(x)$ has to be linear between the points $a$ and $b$. For the general case of $S\subset \mathbb Z$ with $|S|\geq 2$, $f$ is constant below $\min(S)$ and above $\max(S)$, and is linear in between the points of $S$.

Pavel Kocourek
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    It's not true that you can have two different constants in one dimension: If two different constants meet, there are two points at which the function isn't discrete harmonic. If you have two different constants on $\mathbb N$ and $-\mathbb N$ and the value at the origin is the average, then the function is discrete harmonic at the origin but not at its two neighbours. In either case, there are two points at which the function isn't discrete harmonic. – joriki Jan 03 '23 at 19:29
  • @joriki Thanks a lot for your valuable comment, you are absolutely right, I edited the discussion accordingly. – Pavel Kocourek Jan 18 '23 at 12:58

1 Answers1

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I try to answer to this interesting question.

Consider the discrete PDE $$ \left\{\begin{matrix} \Delta f(x) = \rho & \text{ on } G \\ f(x) = F(x)& \text{ on } \partial G \end{matrix}\right. $$ where $G$ is a graph. Then the problem is linear, i.e. if we have solutions for$f_1$ and $f_2$ for data $(\rho_1,F_1)$ and $(\rho_2,F_2)$ then $\alpha f_1 + \beta f_2 $ is the unique solution for $(\alpha \rho_1 + \beta \rho_2, \alpha F_1+ \beta F_2)$. We are interested in solutions of the data $(0,F)$. Which says that $f$ is harmonic on $G$. Let $y \in \partial G$ $$ \left\{\begin{matrix} \Delta f(x) = 0 & \text{ on } G \\ f(x) = \delta_{x=y}& \text{ on } \partial G \end{matrix}\right. $$ the discrete harmonic measure $H(x,y)$ is the unique solutions to this problem. Consider $S \subset \partial G$ and the PDE $$ \left\{\begin{matrix} \Delta f(x) = 0 & \text{ on } G \\ f(x) = \delta_{x \in S}& \text{ on } \partial G \end{matrix}\right. $$ we have that $H(x,S)= \sum_{ y \in S} H(x,y)$ is the unique solutions by linearity. In particular every solution $f$ to the general equation with data $(0,F)$ is of the form $$ f(x) = \sum_{y \in \partial G} H(x,y) F(y).$$

If we want to give probabilistic interpretation to this, if we consider $X:=(X_n)_{n \geq 0}$ to be a simple random walk on $\mathbb{Z}^d$ which start at $x $ and let $\tau $ to be the first time that $X$ visits the boundary then $H(x,S)= \mathbb{P}_x(X_{\tau} \in S)$, i.e. is the probability that the simple random walk starting at $x$ exits in $S$. And for the general solution is $$f(x) = \mathbb{E}[ F(X_{\tau}) ].$$

Consider the case $G= \mathbb{Z}^n \setminus S$ and $\partial G = S$, where $S$ is finite. For $n=1,2$ we have that simple random walks are recurrent, and so $f$ is fully determined by its values on $S$. Hence $d(1,S)=d(2,S) = |S|$. In contrast, for $n\geq 3$ there is a positive probability that a given random walk will never hit any of the points in $S$, and thus we can also choose a constant associate with the random walk never hitting $S$, and so $d(n,S)=|S|+1$.

Conclusion: $$d(n,S) = \begin{cases} |S| & \text{if } \ n\leq 2 \\ |S|+1 & \text{otherwise}. \end{cases}$$

Pavel Kocourek
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3m0o
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  • Thank you for a very nice discussion of the problem. I'm convinced by your argument except for a few details:
    1. I think that after the sentence "We are interested in solutions of the data $(0,F)$" you wanted to replace $\rho$ by $0$, is that right?
    2. Did you mean $H(x,S)\in \mathop{span}({H(x,y),y\in S})$ instead of $H(x,S=\sum_{y\in S} H(x,y))$ ?
    3. Your conclusion disagrees with my observations for case $n=1$.
     – Pavel Kocourek Jan 18 '23 at 12:47
  • For 1. yes is a typo, I will edit. For 2. I don't understand where you say that I mean $H(x,S)$. – 3m0o Jan 18 '23 at 12:56
  • I am sorry, my wrong, I understand now why you had $H(x,S)=\sum_{y\in S} H(x,y)$ and not a linear combination, I confused $H(x,S)$ with $f(x)$. – Pavel Kocourek Jan 18 '23 at 13:02
  • For 3. yes, my conclusion disagrees, seems more likely to be $d(1,S)=d(2,S) = \left| S \right| $ since $H(x,y) $ coincide with the constant function. But I need to check my hypothesis. – 3m0o Jan 18 '23 at 13:16
  • Consider $n=1$ or $n=2$ and $X_t$ the random process starting from $x$. It will hit the boundary $S$ with certainty, that is why I think that $f(x)= \sum_{y\in S}P(X_\tau = y) f(y)$...and $f$ on $S$ can be chosen arbitrarily. – Pavel Kocourek Jan 18 '23 at 13:35
  • What I still don't understand is how to argue that for $n\geq 3$ there should be unique parameter associated with the process never stopping. What if there were different regions in which the random walk might be wondering? However, I think that your discussion answers this as well, all we need is to consider the discrete harmonic functions on $\Bbb Z^n$ (with no boundary), and it is known that all such functions, if bounded, are constant. – Pavel Kocourek Jan 18 '23 at 13:42
  • I would like to accept your answer, but I would first like us to agree on the conclusion. Please let me know if you can find time to edit the conclusion or if you give me permission to do so. Thanks in advance! – Pavel Kocourek Jan 19 '23 at 17:00
  • Sure I agree with you on conclusion! On these days I'm studying for some exams so if you want you can edit by yourself or just wait some days. – 3m0o Jan 19 '23 at 23:38
  • Interestingly enough, the fact that discrete harmonic, bounded functions are constant is related to the fact that $\mathbb{Z}^n$ is an amenable group. In fact, for non-amenable groups $G$, and for every probability $\mu$ on $G$, there always exists bounded $\mu$-harmonic functions that are non constant. But if we restrict the premises we have the following version: Suppose that for every $g \in G$, we have that exists $x_1,\ldots,x_n \in \operatorname{supp}(\mu) $ such that $g=x_1\cdot \ldots \cdot x_n$, and suppose that $f$ is $\mu$-harmonic and attains the maximum, then $f$ is constant. – 3m0o Jan 19 '23 at 23:55