Non constant and bounded discrite harmonic function

Question

I'm preparing an exam and in a preparation sheet there is this exercise that I really don't know how to solve.

Find a function $ f : \mathbb{Z}^3 \to \mathbb{R} $ which is discrete harmonic on $ \mathbb{Z}^3 $ except at $ \{ (0,0,0) \} $, bounded but not constant.

I recall that discrete harmonic means $ \Delta f(x) = 0 $ where $$ \Delta f(x) = \frac{1}{\deg x } \sum_{y \sim x} (f(y) - f(x) ) $$ in our case we have

$$ \Delta f(x) = \frac{1}{6} \sum_{ y \sim x} (f(y)- f(x)) $$ or equivalently $$ f(x,y,z) = \frac{1}{6} \left( f(x+1,y,z) + f(x-1,y,z) + f(x,y+1,z) + f(x,y-1,z) + f(x,y,z+1) + f(x,y,z-1) \right) $$

and this last equality holds for all $ (x,y,z) \in \mathbb{Z}^3 $ except when $ (x,y,z)= (0,0,0) $ . The big problem is that an harmonic function has to verify the maximum principle, thus the maximum (and the minimum) has to be on the boundary. So I think actually that at $ (0,0,0) $ the value of $ f $ has to be maximal (or minimal). And then I think it has to oscillate in some way. Explicitly I'm not able to find it but I think that a such $f$ has to verify the following equation:

$$ \left\{\begin{matrix} \Delta f(x) = 0 & x \in \mathbb{Z}^3 \setminus \{ (0,0,0) \} \\ f(x) = F(x) & x \in \{ (0,0,0) \} \end{matrix}\right. $$

This discrete PDE is the discrete Laplace with Dirichlet boundary conditions $F$. I think actually that we can find an example

$$ \left\{\begin{matrix} \Delta f(x) = 0 & x \in \mathbb{Z}^3 \setminus \{ (0,0,0) \} \\ f(x) = \delta_{x=y} & x \in \{ (0,0,0) \} \end{matrix}\right. $$

where $y =(0,0,0)$, and thus we find as solution the discrete Harmonic Measure $H(x,y)$ i.e. $$ H(x,y) = \frac{1}{6} \sum_{z \sim x} H(z,y) $$ and the solution became $$ f(x) = H(x,y) $$ indeed for $x \in \mathbb{Z}^3 \setminus \{ (0,0,0) \} $ we have that $$ \Delta f(x) = \Delta H(x,y) = \frac{1}{6} \sum_{z \sim x} (H(z,y) - H(x,y)) $$ $$ = \left( \frac{1}{6} \sum_{z \sim x} H(z,y) \right) - H(x,y) = 0 $$ thus it is discrete harmonic on $\mathbb{Z}^3 \setminus \{ (0,0,0) \}$ moreover if $x= (0,0,0)$ we have that $$ \Delta f(x) = \Delta H(x,y) = \frac{1}{6} \sum_{z \sim x} (H(z,x) - H(x,y)) = \left( \frac{1}{6} \sum_{z \sim x} H(z,x) \right) - 1 $$ but actually I don't know if this Harmonic discrete function is Harmonic at $(0,0,0)$ unfortunately I think so. Maybe we have to modify the PDE.

Answer 1

Let $f(x):\mathbb Z^3 \to \mathbb [0,1]$ be defined as the probability that a symmetric 3D random walk returns to the origin $(0,0,0)$ from $x$. Thanks to the result due to Pólya (related post), $f(x)<1$ so long as $x$ is not at the origin itself. Since $f$ is a martingale everywhere outside of the absorption state $(0,0,0)$, considering that the random walk makes step with probability $1/6$ in each of the canonic directions, $$ f(x,y,z) = \frac{1}{6} \Big( f(x+1,y,z) + f(x-1,y,z) + f(x,y+1,z) + f(x,y-1,z) + f(x,y,z+1) + f(x,y,z-1) \Big), $$ and as you showed it means that $f$ is discrete harmonic except at the origin.

Interestingly, by the result of Pólya this construction of $f$ won't work in 1D or 2D, but it works 3D and in higher dimension.