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Good day.

There is a question I have already encountered twice, in very different contexts, that is relatively simple looking, but both solutions I know involve some pretty advanced theorems from the respective fields. I would like to ask if someone could think of any different proofs.

So, here is the question: Let $f: \mathbb Z^2 \rightarrow \mathbb R$ satisfy $f(m,n) = \frac14(f(m+1,n) + f(m-1,n) + f(m,n+1) + f(m,n-1))$. i.e. $f$ is defined on a grid, at a value at a point is the average of its four neighbors. Assume that $f$ is bounded (in some versions, bounded from below only). Prove that $f$ is constant.

One proof of the question uses martingales, and relies, I think on the martingale convergence theorem. In the proof, you define a two-dimensional random walk $S_n$ and look at the process $f(S_n)$.

The other proof comes from functional analysis, and uses the Krein-Milman theorem on the set $K_1 = \{f \in L_{\infty}(\mathbb Z^2) | f(m,n) = \frac14(...), ||f||\leq1 \} $, after finding its extremal set (which is the constant functions $\{+1,-1\}$).

So, any other proofs of this seemingly simple question? I should suspect there is something related to complex analysis, since for analytic functions this would follow from the Liouville theorem. Also, perhaps a combinatorial solution?

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    $\begingroup$ Near duplicate of math.stackexchange.com/questions/51926/… . No elementary proofs there either, though. $\endgroup$ – David E Speyer Jun 27 '14 at 15:30
  • $\begingroup$ Can you think of a suitable title which includes some non-LaTeX text? Pure LaTeX titles cause some problems. $\endgroup$ – Peter Taylor Jun 27 '14 at 16:38
  • $\begingroup$ I left an elementary proof at the other question. $\endgroup$ – David E Speyer Jun 27 '14 at 16:57
  • $\begingroup$ Well, I'm not sure 'elementary' is the correct word to use here :). But it indeed does not seem to rely on any deep theorems, so in that sense it is elementary. $\endgroup$ – Aahz Jun 27 '14 at 17:02
  • $\begingroup$ Chebyshev's inequality, Newton's inequalities, generating functions and the inequality $\binom{2r}{r} \leq 2^{2r}/\sqrt{2r+1}$, which can be proved by induction on $r$. All of these are permissible in an IMO solution, so I figure they count as elementary. (I could make it sound more elementary by never using the word random walk, but that's pointless obscurity.) $\endgroup$ – David E Speyer Jun 27 '14 at 17:07
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This is the discrete equivalent of Liouville's principle for harmonic function: every bounded function with a null laplacian is constant.

Without loss of generality you can assume that $f(0,0)=0$ (if not so, replace $f$ with $f-f(0,0)$). Now consider the square $B_N=\{(m,n):|m|+|n|\leq N\}$. Over $B_N$, $f$ attains its maximum and minimum on $\partial B_N$, and $$\Gamma_N = \max_{(m,n)\in B_N}f(m,n)=\max_{(m,n)\in \partial B_N}f(m,n)$$ is non decreasing as a function of $N$. $\Gamma_N$ is also bounded by $\Gamma$, and satisfies the almost-midpoint-convexity condition: $$\Gamma_N\leq\frac{3\Gamma_{N+1}+\Gamma_{N-1}}{4}$$ since the values of $f$ over $\partial B_N$ are completely determined by the values of $f$ over $\partial B_{N+1}$ and $\partial B_{N-1}$. Since any non-decreasing, bounded, midpoint-convex function is constant, we would expect that the same applies here. We have: $$\Gamma_{N+1}-\Gamma_{N}\geq\frac{1}{3}(\Gamma_{N}-\Gamma_{N-1})\geq\frac{\Gamma_1}{3^N}$$ so for any $\epsilon>0$ we can find a point $(a_1,b_1)$ such that $f(a_1,b_1)\geq\frac{3}{2}\Gamma_1-\varepsilon>\frac{10}{7}\Gamma_1.$ This gives that the difference of the values of $f$ in two neighbour points cannot be bounded, hence the function itself cannot be bounded.

We have just used a Borel-Caratheodory-type argument to provide lower bounds for $\Gamma_N$.

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  • $\begingroup$ I do not quite follow. Can you explain why "over $B_N$, $f$ attains its maximum of $\partial B_N$", and why is $\Gamma_N$ non-decreasing in $N$? I do not see how those follow, not even intuitively. It seems to me that such a function, if not constant, may decrease in some directions and increase in other, as long as those average out. $\endgroup$ – Aahz Jun 27 '14 at 19:16
  • $\begingroup$ It is the same principle for which the maximum modulus of an olomorphic function is attained on the boundary of a ball: since any value of $f$ is the average of the values of $f$ in a neighbourhood, $\operatorname{argmax}_{z\in B}f(z)$ and $\operatorname{argmin}_{z\in B}$ cannot lie strictly inside $B$. Hence, for any compact ball $B$, $\operatorname{argmax}f$ lies on the boundary, and the function $\max_{z\in B_N}f(z)$ is non decreasing since $B_N\subset B_{N+1}$. $\endgroup$ – Jack D'Aurizio Jun 27 '14 at 19:57
  • $\begingroup$ I don't understand the end of your proof. I don't see why the difference cannot be bounded. In fact, a sequence starting with 0, 1 and with the same recurrence relation tends to 1.5. $\endgroup$ – Labo Aug 3 '17 at 18:11

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