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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.

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  • $\begingroup$ It seems to me that the condition $f(0,0)=1$ is redundant, if $f$ is "uniquely determined from its values on the boundary". $\endgroup$ – Han de Bruijn May 3 '16 at 18:52
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    $\begingroup$ @HandeBruijn It is redundant indeed. If you don't like the normalization $f(0,0)=1$, replace this condition with "$f$ nonzero". Then it can be shown that $f(0,0)$ is nonzero. And maximize $\frac{f(1,0)}{f(0,0)}$ instead of $f(1,0)$ $\endgroup$ – Ewan Delanoy May 3 '16 at 19:00
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    $\begingroup$ It seems clear a priori that the maximum is attained at some $f$ whose boundary support is a single point: $f(1,0)$ and $f(0,0)$ are both positively-weighted averages of the values of $f$ on $\partial C_n$, and we're asking in effect for the largest ratio between weights evaluated at the same boundary point. Showing that the maximal ratio occurs at $(n,0)$ looks harder. $\endgroup$ – Noam D. Elkies May 6 '16 at 5:25

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