Entire Functions of Polynomial Growth
An entire function with a bounded k-th derivative, or growth at most |z|^k, is a polynomial of degree at most k.
1 Statement
Theorem 1.1 (Polynomial growth). Let \(f\) be entire and \(k \geq 0\) an integer. The following are equivalent:
- (a)
- \(f\) is a polynomial of degree at most \(k\);
- (b)
- \(f^{(k)}\) is bounded on \(\C \);
- (c)
- there are constants \(A, B \geq 0\) with \(\abs {f(z)} \leq A + B \abs {z}^k\) for all \(z \in \C \).
The case \(k = 0\) of (c) \(\Rightarrow \) (a) is Liouville’s theorem.
2 Setting
We use two standard facts about an entire function \(f\) [1, ch. 4–5]: it equals its Taylor series \(f(z) = \sum _{n \geq 0} a_n z^n\) on all of \(\C \), with \(a_n = f^{(n)}(0)/n!\), and its derivatives are again entire.
Lemma 2.1 (Cauchy’s estimate). If \(f\) is entire and \(M(R) = \max _{\abs {z} = R} \abs {f(z)}\), then \(\abs {a_n} \leq M(R) / R^n\) for every \(R > 0\) and \(n \geq 0\).
Proof. By Cauchy’s integral formula for derivatives on the circle \(\abs {z} = R\),
3 Idea of the proof
Growth of \(f\) controls its Taylor coefficients through Lemma 2.1. If \(f\) grows no faster than \(\abs {z}^k\), every coefficient beyond \(a_k\) is crushed to zero as \(R \to \infty \).
4 Proof
(c) \(\Rightarrow \) (a). For \(R \geq 1\) we have \(M(R) \leq A + B R^k\), so for \(n > k\), Lemma 2.1 gives
Hence \(a_n = 0\) for all \(n > k\), and \(f(z) = \sum _{n=0}^{k} a_n z^n\).
(b) \(\Rightarrow \) (a). The function \(f^{(k)}\) is entire and bounded, so it satisfies (c) with exponent \(0\). By the case already proved, it is a constant \(c\). Then \(f^{(n)} \equiv 0\) for \(n > k\), so \(a_n = f^{(n)}(0)/n! = 0\) for \(n > k\).
(a) \(\Rightarrow \) (b) and (c). If \(f(z) = \sum _{n=0}^{k} a_n z^n\), then \(f^{(k)} = k!\, a_k\) is constant. For (c), note that \(\abs {z}^n \leq 1 + \abs {z}^k\) for \(0 \leq n \leq k\), since one of \(1\) and \(\abs {z}^k\) is at least \(\abs {z}^n\). So \(\abs {f(z)} \leq \sum _{n} \abs {a_n} (1 + \abs {z}^k)\).
5 Remarks
Corollary 5.1 (Sublinear growth). If \(f\) is entire and \(f(z)/z \to 0\) as \(z \to \infty \), then \(f\) is constant.
Proof. By Lemma 2.1 with \(n = 1\), \(\abs {a_1} \leq M(R)/R \to 0\), and for \(n \geq 2\), \(\abs {a_n} \leq M(R)/R^n \leq M(R)/R \to 0\) once \(R \geq 1\).
Remark 5.2. Nothing like this holds for real functions. The function \(\sin x\) is smooth on \(\R \) and all of its derivatives are bounded by \(1\), but it is not a polynomial. Its complex extension is unbounded in every direction except the real axis: \(\abs {\sin (iy)} = \abs {\sinh y}\). Bounding a function on the real line says nothing about the complex plane, and Liouville’s theorem needs the whole plane.
Remark 5.3. Read contrapositively, (c) \(\Rightarrow \) (a) says that an entire function that is not a polynomial, such as \(e^z\), outgrows every power: for each \(k\) there is a sequence \(z_j \to \infty \) with \(\abs {f(z_j)} / \abs {z_j}^k \to \infty \).