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. Suppose that f is an entire function. If for some integer $k \ge 0$, there are positive constants A and B such that $|f(z)| \le A + B|z|^k, \forall z \in \mathbb{C}$, then f is a polynomial of degree at most k.

Proof.

We proceed by mathematical induction on the integer $k$.

Base Case (k = 0)

If 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 $z \in \mathbb{C}$. 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.

Inductive Step. Assume the statement is true for some integer k -1 (where $k \ge 1$). That is, if an entire function g satisfies $|g(z)| \le C + D|z|^{k-1}$, then g is a polynomial of degree at most k -1. We must prove it is true for k. Let f be an entire function satisfying: $|f(z)| \le A + B|z|^k$

Define the function h(z) as $h(z) = \begin{cases} \frac{f(z)-f(0)}{z}, &z \ne 0 \\\\ f'(0), &z = 0 \end{cases}$

Claim 1: h(z) is entire.

Since f is entire, it has a Taylor series expansion at z = 0: $f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \cdots$ converges for every complex number z.

If a function is analytic on all of $\mathbb{C}$, then its Taylor series at any point (in particular at 0) has infinite radius of convergence.

Subtracting f(0) and dividing by z (for $z \ne 0$) gives: $h(z) = \frac{f(z)-f(0)}{z} = f'(0) + \frac{f''(0)}{2!}z + \frac{f'''(0)}{3!}z^2 + \cdots, \forall z \in \mathbb{C} \setminus \{ 0 \}$. 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).

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: $|h(z)| \le M + B|z|^{k-1}$ (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.

Outside the Unit Disk |z| > 1$: $$ \begin{aligned} |h(z)| &=|\frac{f(z)-f(0)}{z}| \\[2pt] &\text{By the Triangle inequality} \\[2pt] &\le \frac{|f(z)| + |f(0)|}{|z|} \\[2pt] &\text{By Induction Hypothesis} \\[2pt] &\le \frac{A + B|z|^k + |f(0)|}{|z|} \\[2pt] &=\frac{A + |f(0)|}{|z|} + B|z|^{k-1} \\[2pt] &\text{|z| > 1} \\[2pt] &\le A + |f(0)| + B|z|^{k-1} \end{aligned} $$

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

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 $z \ne 0, h(z) = \frac{f(z) - f(0)}{z} \implies f(z) = z \cdot h(z) + f(0)$. 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.