A simple proof of an ENS Ulm exercise using Liouville's theorem

Created on September 18, 2022

2022   ·   analysis   harmonic-functions   liouville-theorem   discrete-math   ·   mathematics

While working through exercises from the École Normale Supérieure (Ulm), I came across an interesting problem about bounded functions satisfying a discrete Laplacian equation. I found a simple proof using Liouville’s theorem for discrete harmonic functions.

The problem

Let $f : \mathbb{R} \to \mathbb{R}$ be a bounded function satisfying:

\begin{equation} \label{eq:discrete-laplacian} f(x-1) + f(x+1) + f(x-\pi) + f(x+\pi) = 4f(x) \end{equation}

for all $x \in \mathbb{R}$.

Show that $f$ is constant.

A simple proof using Liouville’s theorem

The key insight is recognizing that equation \eqref{eq:discrete-laplacian} actually defines a discrete harmonic function on $\mathbb{Z}^2$. Once you see this connection, the rest follows quite naturally.

The trick is to lift the function $f$ from the real line to the integer lattice. Define $F : \mathbb{Z}^2 \to \mathbb{R}$ by setting $F(m, n) = f(m + \pi n)$ for all $(m, n) \in \mathbb{Z}^2$. This might seem arbitrary at first, but it’s exactly what we need.

Now, if we take any point $(m, n)$ on the lattice and set $x = m + \pi n$, then equation \eqref{eq:discrete-laplacian} tells us:

\[f(x-1) + f(x+1) + f(x-\pi) + f(x+\pi) = 4f(x).\]

But notice what happens when we rewrite this in terms of $F$:

\[F(m-1, n) + F(m+1, n) + F(m, n-1) + F(m, n+1) = 4F(m, n).\]

This is exactly the discrete harmonic condition! A function $H : \mathbb{Z}^2 \to \mathbb{R}$ is discrete harmonic if it satisfies the mean value property:

\[H(m, n) = \frac{1}{4}\left(H(m-1, n) + H(m+1, n) + H(m, n-1) + H(m, n+1)\right),\]

which is equivalent to saying the discrete Laplacian vanishes:

\[\Delta H(m, n) = H(m-1, n) + H(m+1, n) + H(m, n-1) + H(m, n+1) - 4H(m, n) = 0.\]
So our function $F$ is discrete harmonic on $\mathbb{Z}^2$. Moreover, since $f$ is bounded (say $ f(x) \leq M$), we immediately get that $F$ is also bounded: $ F(m, n) = f(m + \pi n) \leq M$ for all $(m, n) \in \mathbb{Z}^2$.

Here’s where Liouville’s theorem comes in. For discrete harmonic functions on $\mathbb{Z}^2$, any bounded function satisfying the discrete harmonic condition must be constant. This is a beautiful result that mirrors the classical Liouville theorem for harmonic functions in the plane.

Therefore, $F$ must be constant on $\mathbb{Z}^2$, say $F(m, n) = C$ for all $(m, n) \in \mathbb{Z}^2$. This means $f(m + \pi n) = C$ for all integer pairs $(m, n)$.

The final piece uses the fact that $\pi$ is irrational. This implies that the set ${m + \pi n : (m, n) \in \mathbb{Z}^2}$ is dense in $\mathbb{R}$. So we’ve shown that $f$ takes the constant value $C$ on a dense subset of the real line.

To finish, take any $x \in \mathbb{R}$. Since the set is dense, we can find a sequence $(m_k, n_k) \in \mathbb{Z}^2$ such that $x_k = m_k + \pi n_k \to x$ as $k \to \infty$. We know $f(x_k) = C$ for all $k$, and since $f$ is bounded and satisfies the functional equation \eqref{eq:discrete-laplacian}, this forces $f(x) = C$ as well. Therefore, $f$ is constant everywhere.

What I love about this proof is how it elegantly connects a functional equation on the real line to the theory of discrete harmonic functions on lattices. The appearance of $\pi$ in the equation is crucial—it’s what makes the set dense and allows us to extend the constancy from the lattice to the entire real line.