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Question:

if $f:\mathbb{Z}^2\to \mathbb{R}$ is bounded ,and for any $x,y\in \mathbb{Z}$,we have $$4f(x,y)=f(x-1,y)+f(x,y-1)+f(x+1,y)+f(x,y+1)$$ show that $$f\equiv C$$ where $C$ is constant.

My try:let $x=y=0$,then we have $$4f(0,0)=f(-1,0)+f(0,-1)+f(1,0)+f(0,1)$$ and let $x=0,y=1$,then $$4f(0,1)=f(-1,1)+f(0,0)+f(1,1)+f(0,2)$$ let $x=1,y=0$, then $$4f(1,0)=f(0,0)+f(1,-1)+f(2,0)+f(1,1)$$ $$\cdots\cdots$$ Then I fell very ugly, so I can't works,maybe have other methods,and this problem is from a middle school student mathematics exercises

Thank you very much!

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  • $\begingroup$ In more technical terms, every bounded harmonic function on the grid is constant. $\endgroup$ – lhf Dec 12 '13 at 13:01
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    $\begingroup$ Related to math.stackexchange.com/questions/204365/…. $\endgroup$ – lhf Dec 12 '13 at 13:04
  • $\begingroup$ See also math.stackexchange.com/questions/154470/…. $\endgroup$ – lhf Dec 12 '13 at 13:12
  • $\begingroup$ The first link of @lhf is not only related, it completely solves the question. In short: since $f(x,y)$ is average of its neighbors, there can be no non-trivial extrema. Because $f$ is bounded, there is a sequence of $(x,y)$ approaching its suppremum and using harmonic property again one eventually shows that $f$ must be equal to that suppremum everywhere by first observing that this holds along that sequence. $\endgroup$ – Marek Dec 12 '13 at 13:49
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Apparently the solution below was the official solution to this problem, which appeared on the 2003 Bulgarian TST. I edited this from dgrosev's solution found here.

Assume on the contrary that $f(x,y)$ is a non constant bounded harmonic function (assume $0\leq f \leq 1$). Then there exist two consecutive points (along the $x$ or $y$ axes) in which $f$ has different values. Rotating the plane, if necessary, we may assume without loss of generality that $f(x_0+1,y_0) > f(x_0,y_0)$.

Let consider $g(x,y)=f(x+1,y)-f(x,y)$. Then $g$ is not identical to $0$ and $g$ is bounded. So if we take $M= \sup g(x,y) \implies 0< M < \infty$, it is easy to see that $g$ is also a harmonic function.

Let $\varepsilon >0$ and take $(x,y)$ so that: $g(x,y) > M-\varepsilon$. Because $g$ is harmonic, it follows: $g(x+1,y) > M-4\varepsilon$. Applying this argument several times implies that $g(x+n,y) > M-4^n \varepsilon$. The definition of $g$ shows that $$f(x+n,y)-f(x,y) = \sum_{k=0}^{n-1} g(x+k,y)\tag{1}$$

Note that LHS of $(1)$ is between $-1$ and $1$. However, if we take $n>\frac{2}{M}$, then take sufficiently small $\varepsilon$ and $(x,y)$ so that $f(x,y)>M-\varepsilon$, the RHS of $(1)$ is greater than $1$. This is a contradiction, and so $g\equiv 0$ and so $f$ is constant.

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