Let $n>0$, and let $S_n$ denote the discrete square $S_n=[|-n,n|]^2$ (so $S_n$ has $(2n+1)^2$ elements). Let $K_n$ denote the set of four corner points $\lbrace (\pm n,\pm n)\rbrace$, and $C_n=S_n\setminus K_n$. let $f$ be a nonnegative function on $C_n$, such that $f$ is harmonic on $S_{n-1}$ (i.e. $f(x,y)=\frac{f(x,y-1)+f(x,y+1)+f(x-1,y)+f(x+1,y)}{4}$ for any $(x,y)\in S_{n-1}$.) Also, normalize $f$ so that $f(0,0)=1$. It is well known that $f$ is uniquely determined from its values on the boundary $\partial C_n$ of $C_n$ (a form of the so-called "maximum principle").
Is it true that under those conditions, $f(1,0)$ is maximal iff $f$ is zero on $\partial C_n$ except on $(n,0)$ ? I have checked this for small values of $n$.
This can be explained quite intuitively by saying that if we want to maximize $f(1,0)$, we must concentrate all the boundary mass at the nearest point from $(1,0)$. I do not see how to turn this idea into a rigorous proof however.