Liouville’s theorem states that every function $f: \mathbb{C} \to \mathbb{C}$ which is analytic everywhere and bounded is constant. I recently stumbled across the paper A Note on Liouville’s Theorem (A. Lenard, 1986), that proves Liouville’s theorem using only algebraic power series manipulation: no analysis techniques were used at all, no limits, derivatives, integrals involved. I was shocked this was even possible, but I couldn’t understand the paper. It took me a couple of (very discontinuous) days of work to grasp the cleverness of the proof; it’s really cool and I don’t want to put this understanding to waste, so here’s my attempt to present it in a more intuitive way with more details.

The proof

Our version of Liouville’s Theorem is as follows:

Theorem. If

$$f(z) = \sum_{n\geq 0} a_n z^n; \quad a_n \in \mathbb{C}$$

is a power series that converges everywhere in $\mathbb{C}$, and there exists $M > 0$ such that

$$\forall z \in \mathbb{C}, \enspace |f(z)| \leq M,$$

then $\forall n \in \mathbb{N} \setminus \{0\}, a_n = 0$. That is, $f$ is constant.

Proof. For this proof, we want to show that $a_n = 0$ for all $n \geq 1$ (so $f(z) = a_0$ is only the constant term). We will achieve this by using $f$ to construct a specific absolute sum, which we will then bound to show that $f$ must be constant.

Since $f$ converges everywhere on $\mathbb{C}$, it is also absolutely convergent on $\mathbb{C}$. Hence for any positive $r > 0$ we have

$$S(r) := \sum_{n \geq 0} |a_n| r^n < \infty.$$

Since the absolute sum $S(r)$ converges, if we take a tail of the sum starting at $m$, it must eventually tend towards 0 as $m \to \infty$:

$$\lim_{m\to\infty} \sum_{n \geq m} |a_n| r^n = 0.$$

In particular we can definitely pick some tail of the sequence that has value $\leq 1/S(r)$. Let $k = k(r)$ be the smallest positive integer which satisfies that criterion, so it is the smallest positive integer such that

$$S(r) \cdot \left(\sum_{n \geq k} |a_n| r^n\right) \leq 1.$$

Let $\omega = \exp(2 \pi i/k)$ be the $k$-th primitive root of unity. We know that¹

$$\frac 1{k} \sum_{j=0}^{k-1} \omega^{(n-m)j} = \begin{cases} 1 &\text{if } n \equiv m \pmod{k} \\ 0 &\text{otherwise.} \end{cases}$$

Remember this for later.

Let $F(r)$ be the arithmetic mean of $|f|^2 = f \overline{f}$ at each $r \cdot \omega^j$, where $j$ runs from $0$ to $k-1$. In other words,

$$F(r) := \frac 1{k} \sum_{j=0}^{k-1} \left|f(r \cdot \omega^j) \right|^2.$$

Since $F(r)$ is the average over some $z$-values of $|f(z)|^2$, and $|f(z)|^2 \leq M^2$ by assumption, we know $F(r) \leq M^2$. We can obtain a power series form for the mean $F(r)$:

$$\begin{align*} F(r) &= \frac 1{k} \sum_{j=0}^{k-1} \left|f(r \cdot \omega^j) \right|^2 \\ &= \frac 1k \sum_{j=0}^{k-1} f(r \omega^j)\cdot \overline{f(r \omega^j)} \\ &= \frac 1k \sum_{j=0}^{k-1} \left(\sum_{n \geq 0} a_n (r \omega^j)^n\right) \cdot \left(\sum_{m \geq 0} \overline{a_m (r \omega^j)^m}\right) \\ &= \frac 1k \sum_{j=0}^{k-1} \left(\sum_{n \geq 0} a_n r^n \omega^{jn} \right) \cdot \left(\sum_{m \geq 0} \overline{a_m} r^m (\overline{\omega})^{jm} \right). \\ \end{align*}$$

If we then apply $(\overline{\omega})^b = \omega^{-b}$, we get

$$\begin{align*} F(r) &= \frac 1k \sum_{j=0}^{k-1} \left(\sum_{n \geq 0} a_n r^n \omega^{jn} \right) \cdot \left(\sum_{m \geq 0} \overline{a_m} r^m \omega^{-j m} \right) \\ &= \frac 1k \sum_{j=0}^{k-1} \sum_{n \geq 0} \sum_{m \geq 0} a_n r^n \omega^{jn} \overline{a_m} r^m \omega^{-j m} \\ &= \frac 1k \sum_{j=0}^{k-1} \sum_{n \geq 0} \sum_{m \geq 0} a_n \overline{a_m} r^{n+m}\omega^{j(n-m)} \\ &= \sum_{n \geq 0} \sum_{m \geq 0} a_n \overline{a_m} r^{n+m} \frac 1k \sum_{j=0}^{k-1} \omega^{j(n-m)} \\ &= \sum_{n \geq 0} \sum_{m \geq 0} a_n \overline{a_m} r^{n+m} \cdot \begin{cases} 1 &\text{if } n \equiv m \pmod{k} \\ 0 &\text{otherwise} \end{cases} \\ &= \mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{n \equiv m \pmod{k}} a_n \overline{a_m} r^{n+m}. \end{align*}$$

If we consider all choices $n,m \in \mathbb{N} \cup \{0\}$ of the term $a_n \overline{a_m} r^{n + m}$, then the mean $F(r)$ sums up the terms corresponding to something like the black tiles in the following image:

We can break $F(r)$ up into two pieces: the diagonal where $n=m$, and the parts where $n \neq m$. That is, if we define the diagonal

$$A(r) := \sum_{n \geq 0} |a_n|^2 r^{2n},$$

we can write $F(r) = A(r) + G(r)$, where $G(r)$ represents the off-diagonal elements:

$$G(r) := \mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{\substack{n \equiv m \pmod{k}\\[0.25em] n \neq m}} a_n \overline{a_m} r^{n+m}.$$

This corresponds to the following image, where the diagonal $A(r)$ sums up the tiles in blue, and the off-diagonal $G(r)$ sums up the tiles in orange.

If we get a bound on the diagonal $A(r)$, then we can get a bound on $|a_n|$. Recall the definition of $k = k(r)$ again: it is the smallest positive integer such that

$$\begin{align*} 1 &\geq S(r) \cdot \left(\sum_{m \geq k} |a_m| r^m\right) = \left(\sum_{n\geq 0} |a_n| r^n\right)\left(\sum_{m \geq k} |a_m| r^m\right) = \sum_{n\geq 0} \sum_{m \geq k} |a_n| |a_m| r^{n+m}. \end{align*}$$

The sum on the RHS of the above equation corresponds to the tiles in purple.

Since $|a_n| |a_m| = |a_m| |a_n|$, we can notice that the above table can be mirrored around the line $n=m$. We can then notice that if we add together two copies of the RHS, like $2 \sum_{n\geq 0} \sum_{m \geq k} |a_n| |a_m| r^{n+m}$, then it can be depicted as covering the following tiles (some of them twice):

The algebraic way to see the above table is with two mirrored copies of the sum:

$$2 \sum_{n\geq 0} \sum_{m \geq k} |a_n| |a_m| r^{n+m} = \left(\sum_{n\geq 0} \sum_{m \geq k} |a_n| |a_m| r^{n+m}\right) + \left(\sum_{n\geq k} \sum_{m \geq 0} |a_n| |a_m| r^{n+m}\right).$$

Notice that $G(r)$, consisting of the off-diagonal elements (depicted in orange), is completely covered by the above!

So, if we bound each term in the off-diagonal sum $G(r)$ by its absolute value, it must be bounded above by the absolute sum corresponding to the purple tiles! Formally:

$$\begin{align*} |G(r)| &= \left|\mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{\substack{n \equiv m \pmod{k}\\[0.25em] n \neq m}} a_n \overline{a_m} r^{n+m} \right| \leq \mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{\substack{n \equiv m \pmod{k}\\[0.25em] n \neq m}} \left|a_n \overline{a_m} r^{n+m}\right| \\ &= {\color{orange} \mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{\substack{n \equiv m \pmod{k}\\[0.25em] n \neq m}} |a_n| |a_m| r^{n+m}} \\ &\leq {\color{purple} 2\left(\sum_{n\geq 0} \sum_{m \geq k} |a_n| |a_m| r^{n+m} \right)} \leq 2 \cdot 1 = 2. \end{align*}$$

This implies the diagonal term $A(r) = F(r) - G(r)$ must be bounded above:

$$A(r) = |A(r)| = |F(r) - G(r)| \leq |F(r)| + |G(r)| \leq M^2 + 2.$$

This also means that $A(r)$ also converges for all $r > 0$ (since every term in the series is positive and it’s bounded above.)

From here it’s trivial. Suppose for contradiction that some $a_q \neq 0$, $q \geq 1$. If we then pick

$$r = 1 + \sqrt[2q]{\frac{M^2 + 2}{|a_q|^2}},$$

then we violate the inequality we just proved,

$$A(r) = \sum_{n \geq 0} |a_n|^2 r^{2n} \geq |a_q|^2 r^{2q} > M^2 + 2,$$

contradiction. $\square$

Okay, wait, what?

When I first saw this proof I didn’t understand it at all. Even after understanding it and writing it up I still don’t really grasp it on a fundamental level. Here are some remarks that may or may not aid in this regard.

1. In the original paper, Lenard remarks that the definition of $F(r)$ as

   $$F(r) := \frac 1k \sum_{j=0}^{k-1} |f(r \cdot \omega^j)|^2$$

   is just a Riemann sum of the integral

   $$\frac{1}{2 \pi} \int_0^{2\pi} |f(r \cdot e^{i t})|^2 \,\mathrm dt.$$

   In fact, if you expand out $|f(r e^{i t})|^2 = f(r e^{i t}) \overline{f(r e^{i t})}$ in the above integral and compute the power series of the resulting function in $r$, it in fact equals our diagonal $A(r)$:

   $$\begin{align*} \frac{1}{2\pi} \int_0^{2\pi} |f(r \cdot e^{i t})|^2 \,\mathrm dt &= \frac{1}{2 \pi} \int_0^{2\pi} \left(\sum_{n\geq 0} \sum_{m \geq 0} a_n \overline{a_m} e^{i (n-m) t} r^{n+m} \right) \mathrm dt \\ &= \mathop{\sum_{n \geq 0} \sum_{m \geq 0}}\limits_{n=m} a_n \overline{a_m} r^{n+m} = \sum |a_n|^2 r^{2n} = A(r). \end{align*}$$

   So another way to see the sum of the off-diagonal elements, $G(r)$, is as the error between a Riemann sum and an actual integral. Our choice of $k$ is just large enough to make sure that the Riemann sum is close enough to make the proof work. (Another way to see that the sum $F(r)$ converges to the diagonal is to note that as $k \to \infty$, $n \equiv m \pmod{k}$ basically just means $n = m$.)

   You can see this if you try to follow the failure of the proof when you take out the assumption that $|f(z)| \leq M.$ The entire proof goes through up to and including proving $|G(r)| \leq 2$, but the bound fails for $A(r)$. This makes sense: $A(r)$ is the actually important quantity.

2. The expression $A(r) = \sum |a_n|^2 r^{2n}$ reminds me of Parseval’s identity, but I have no idea what the connection is (or if there even is one).

3. The fact that a proof even exists that purely uses power series methods makes one wonder whether or not Liouville’s theorem can be generalized beyond $\mathbb{C}$ in some manner. According to this math.stackexchange post, there is apparently a corresponding result to Liouville’s theorem over every algebraically closed normed field. (Does this mean there exists a purely power-series proof like the above that doesn’t use complex-number-specific constructions like the complex conjugate? That would be really cool, but is left as an exercise to the reader, because I have no idea.)

Postscript

I found this paper from the MathOverflow post “Liouville’s theorem with your bare hands”, which is itself worth a read.

Footnotes

1. You may have seen this before in the more general setting of series multisection. The intuition is that if $n \not\equiv m \pmod{k}$, then exponentiating each $\omega^j$ by $n-m$ just evenly spreads them out around the unit circle, so they add to 0. If on the other hand $n \equiv m \pmod{k}$, then we are exponentiating each $\omega^j$ by a multiple of $k$, so each of them lands 1 and the total sum of all the roots of unity is $k$, which we multiply out by $1/k$ to get the result. ↩