Extended Liouville's Theorem: Polynomial Growth of Entire Functions

Sometimes people don’t want to hear the truth because they don’t want their illusions destroyed, Friedrich Nietzsche.

Extended Liouville’s Theorem. Let $f(z)$ be an entire function. Suppose there exist an integer $k \ge 0$ and positive constants $A$ and $B$ such that: $|f(z)| \le A + B|z|^k, \quad \forall z \in \mathbb{C}$. Then, $f(z)$ is a polynomial of degree at most $k$.

Proof.

We proceed by mathematical induction on the non-negative integer $k$.

Step 1. Base Case (k = 0)

Suppose k = 0, $|f(z)| \le A + B|z|^0 = A + B \le A, \forall z \in \mathbb{C}$. This means |f(z)| is bounded by a constant for all $z \in \mathbb{C}$. In other words, f is a bounded entire function.

By the standard Liouville’s Theorem (which states that every bounded entire function is constant), $f(z)$ must be constant. A constant is a polynomial of degree 0. Thus, the statement holds for k = 0.

Step 2. Inductive Step. Assume the statement is true for some integer k -1 (where $k \ge 1$). That is, any entire function g satisfying $|g(z)| \le C + D|z|^{k-1}$ is a polynomial of degree at most k -1.

We must prove that it is true for k. Let f be an entire function satisfying: $|f(z)| \le A + B|z|^k$

We define a new reduced function h(z) as $h(z) = \begin{cases} \frac{f(z)-f(0)}{z}, &z \ne 0 \\\\ f'(0), &z = 0 \end{cases}$

💡 Intuitively, we are “removing” the constant term and dividing by z. This lowers the degree of a polynomial by 1 — so we expect this to reduce the growth order from k to k − 1.

Claim 1: $h(z)$ is an entire function.

Since f is entire, it has a Taylor series expansion centered at $0$ that converges for all $z \in \mathbb{C}$: $f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \cdots$.

If a function is analytic on all of $\mathbb{C}$ (that is an entire function, e.g., polynomials, $e^x, sin(z), cos(z)$), then its Taylor series at any point (in particular at 0) has infinite radius of convergence.

For a Taylor series centered at a point $z_0$, the radius of convergence is the distance from $z_0$ to the nearest singularity. Since entire functions have no singularities in ℂ, there are no finite distances to singularities. By the Radius of Convergence Theorem in complex analysis, if a function is analytic everywhere, its Taylor series at any point must have an infinite radius of convergence.

Subtracting f(0) and dividing by z (for $z \ne 0$) yields: $h(z) = \frac{f(z)-f(0)}{z} = f'(0) + \frac{f''(0)}{2!}z + \frac{f'''(0)}{3!}z^2 + \cdots, \forall z \in \mathbb{C} \setminus \{ 0 \}$. This power series converges for all z, making $h(z)$ analytic everywhere (entire).

Note: $h(z)=\sum _{k=0}^{\infty }\frac{f^{(k+1)}(0)}{(k+1)!}z^k$. This is a new power series with coefficients $a_k=\frac{f^{(k+1)}(0)}{(k+1)!}.$ The radius of convergence of a power series does not change when you drop the first term, divide by z, or reindex because these operations do not affect the growth rate of the coefficients and since the original series had radius $R = \infty$, the new series also has radius $R = \infty$

Claim 2: Establish the bound for $h(z)$. We aim to show that $h(z)$ satisfies the condition for the induction hypothesis (degree k-1).

Case 1: Inside the Unit Disk ($|z| \le 1$).

Since h(z) is continuous and the closed unit disk $\overline{D}(0,1)$ is compact (closed and bounded), hence |h(z)| is bounded by some maximum constant M on this disk. We can trivially say: $|h(z)| \le M \le M + B|z|^{k-1}$ (since $B|z|^{k-1}$ is non-negative, this inequality holds safely).

Recall: In $\mathbb{R}^n$ or $\mathbb{C}^n$, the Heine–Borel theorem states that a set is compact if and only if it is closed and bounded. Futhermore, a continuous function on a compact set is bounded and attains its maximum and minimum.

Case 2: Outside the Unit Disk ($|z| > 1$).

$$ \begin{aligned} |h(z)| &=|\frac{f(z)-f(0)}{z}| \\[2pt] &\text{By the Triangle inequality} \\[2pt] &\le \frac{|f(z)| + |f(0)|}{|z|} \\[2pt] &\text{By Induction Hypothesis} \\[2pt] &\le \frac{A + B|z|^k + |f(0)|}{|z|} \\[2pt] &=\frac{A + |f(0)|}{|z|} + B|z|^{k-1} \\[2pt] &\text{Since |z| > 1, we know that } \frac{1}{|z|} < 1 \\[2pt] &\le A + |f(0)| + B|z|^{k-1} \end{aligned} $$

Combining both cases, there exist new constants A’ and B’ such that $\forall z \in \mathbb{C}, |h(z)| \le A' + B'|z|^{k-1}$.

Step 3. Conclusion.

By our Inductive Hypothesis, since $h(z)$ grows at most like $|z|^{k-1}$, $h(z)$ must be a polynomial of degree at most $k-1$. Let $h(z) = c_0 + c_1 z + \dots + c_{k-1} z^{k-1}$.

Recall the definition of $h(z)$ for $z \ne 0, h(z) = \frac{f(z) - f(0)}{z} \implies f(z) = z \cdot h(z) + f(0)$.

Substituting the polynomial expression for $h(z)$: $f(z) = f(0) + c_0 z + c_1 z^2 + \dots + c_{k-1} z^k$

Multiplying a polynomial of degree at most k -1 by z increases the degree by exactly 1. In other words, f(z) is a polynomial with a degree of at most k. The induction is complete. $\square$

Analytic Rigidty

The intuition behind this theorem.

Ordinary Liouville’s Theorem states that an entire function is bounded, then it is constant. While a real-valued $C^\infty$ function can “wiggle” as much as it wants while staying beneath a polynomial growth curve, an analytic function cannot. Its global behavior is strictly tied to its local power series; if it doesn’t grow fast enough at infinity, it is forced to truncate its own power series, becoming a polynomial.

In other words, entire functions cannot “pretend” to grow like polynomials — unless they actually are polynomials.

Conceptual Interpretation

Entire functions come in two broad types:

1. Polynomials — grow like powers of |z|.
2. Transcendental entire functions (like $e^z, \sin(z)$) — grow faster than any polynomial.

Extended Liouville’s Theorem tells us:

If an entire function does not grow faster than a fixed power of |z|, then it must belong to the first class.

There is no middle ground.

A Powerful Corollary

If an entire function satisfies $|f(z)| \le C (1+|z|)^k$, then it is a polynomial of degree at most k.

This is just a rephrasing of the same hypothesis.