# How prove this $f=C$ if $4f(x,y)=f(x-1,y)+f(x,y-1)+f(x+1,y)+f(x,y+1)$

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!

• In more technical terms, every bounded harmonic function on the grid is constant. – lhf Dec 12 '13 at 13:01
• Related to math.stackexchange.com/questions/204365/…. – lhf Dec 12 '13 at 13:04
• – lhf Dec 12 '13 at 13:12
• 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. – Marek Dec 12 '13 at 13:49

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.