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mathoverflow.net/q/149621
math.stackexchange.com/q/849284
math.stackexchange.com/q/51926
mathnet.ru/PresentFiles/17995/presentation.pdf
有界函数 $f:\mathbb{Z}^2\to\mathbb R$ 满足任意一点的 $f$ 值等于其相邻四点的 $f$ 值的平均值：
$$f(m,n) = \frac{f(m+1,n) + f(m-1,n) + f(m,n+1) + f(m,n-1)}{4}$$则 $f$ 为常数。
Normalization and SetupAssume $f$ is bounded, meaning there exists $M > 0$ such that $|f(m, n)| \le M$ for all $(m, n)$. Without loss of generality, replace $f$ by $f - f(0,0)$ so that $f(0,0) = 0$.
Define the $l_1$ balls
\[
B_N = \{(m, n) \in \mathbb{Z}^2 : |m| + |n| \le N\},
\]
and set
\[
\Gamma_N = \max_{(m, n) \in B_N} f(m, n).
\]
Discrete Maximum PrincipleThe discrete maximum principle states that a discrete harmonic function cannot attain a strict maximum at an interior point of a domain. Thus, the maximum on $B_N$ is attained on the boundary $\partial B_N = \{(m, n) : |m| + |n| = N\}$. Since $B_N \subset B_{N+1}$, we have $\Gamma_N \le \Gamma_{N+1}$, so $\{\Gamma_N\}$ is non-decreasing. (The same holds for the minimum, but with a non-increasing sequence.)
Midpoint-Convexity and Growth EstimateConsider a point $(m, n) \in \partial B_N$. The neighbors fall into $B_{N-1}$ or $B_{N+1}$.
- For "interior" boundary points (not vertices), two neighbors are in $B_{N-1}$ and two in $B_{N+1}$, so
\[
f(m, n) \le \frac{2\Gamma_{N+1} + 2\Gamma_{N-1}}{4} = \frac{\Gamma_{N+1} + \Gamma_{N-1}}{2}.
\] - For vertex points (e.g., $(N, 0)$), three neighbors are in $B_{N+1}$ and one in $B_{N-1}$, so
\[
f(m, n) \le \frac{3\Gamma_{N+1} + \Gamma_{N-1}}{4}.
\]
Since $\Gamma_{N+1} \ge \Gamma_{N-1}$ and the maximum may be at a vertex, the controlling bound is the weaker one:
\[
\Gamma_N \le \frac{3\Gamma_{N+1} + \Gamma_{N-1}}{4}.
\]
Rearranging gives
\[
\Gamma_{N+1} - \Gamma_N \ge \frac{1}{3} (\Gamma_N - \Gamma_{N-1}).
\]
Let $d_N = \Gamma_N - \Gamma_{N-1} \ge 0$ (with $\Gamma_0 = 0$, so $d_1 = \Gamma_1$). Then $d_{N+1} \ge \frac{1}{3} d_N$, and by induction,
\[
d_{N+1} \ge \frac{\Gamma_1}{3^N}.
\]
Thus,
\[
\Gamma_N = \sum_{k=1}^N d_k \ge \sum_{k=1}^N \frac{\Gamma_1}{3^{k-1}} = \Gamma_1 \frac{1 - (1/3)^N}{1 - 1/3} = \frac{3}{2} \Gamma_1 \left(1 - \frac{1}{3^N}\right).
\]
If $\Gamma_1 > 0$, then $\Gamma_N \ge \frac{3}{2} \Gamma_1 (1 - 1/3^N)$, approaching $\frac{3}{2} \Gamma_1$ as $N \to \infty$.
ConclusionAssume $\Gamma_1 > 0$ (otherwise, by monotonicity, $\Gamma_N = 0$ for all $N$, so $f \le 0$; applying to $-f$ shows $f = 0$). The lower bound suggests $\Gamma_N$ approaches a finite limit, but this assumes equality can be achieved at each step. However, achieving equality requires the maximum on $\partial B_N$ to be at a vertex with neighbors attaining the exact maxima, leading to inconsistencies in the harmonic condition at non-vertex points where the maximum is also forced.
For example, assuming equality propagates the maximum to certain points, but at non-vertex boundary points (where the neighbor distribution is 2-2), the harmonic average cannot match the required value, as it is bounded above by $(\Gamma_{N+1} + \Gamma_{N-1})/2 < \Gamma_N$ under the equality assumptions. This contradiction implies the actual growth must exceed the lower bound, forcing $\Gamma_N \to \infty$, contradicting boundedness.
Thus, $\Gamma_1 = 0$, so $f \equiv 0$ (shifting back, $f$ is constant). |
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Czhang271828
posted 2024-10-6 12:26
