Let Let \(f:{\mathbb C}\rightarrow {\mathbb C}\) be an entire function. Assume the existence of a non-negative integer \(m\), and of positive constants \(L\) and \(R\), such that for all \(z\) with \(|z|>R\) the inequality \[|f(z)| \leq L |z|^m\] holds. Prove that \(f\) is a polynomial of degree \(\leq m\).
Let \(f:{\mathbb C}\rightarrow {\mathbb C}\) be an entire function. Suppose that there exists a real number \(M\) such that for all \(z\in {\mathbb C}, \Re(f) \leq M\). Prove that \(f\) must be a constant.
Solution
\[ \abs{f^{(n)}(z)} &= \abs{ {1\over 2\pi i} \oint_\gamma {f(\xi) \over (\xi - z)^{n+1}} \dxi } \\ &\leq {1\over 2\pi i} \oint_\gamma { \abs{ f(\xi) } \over \abs{\xi - z}^{n+1}} \dxi \\ &\leq {1\over 2\pi i } \oint_\gamma {LR^m \over R^{n+1} } \dxi \\ &= {L\over 2\pi i} R^{m-(n+1)} \cdot 2\pi R \\ &= LR^{m-n} \\ &\convergesto{R\to\infty} 0 \qquad \iff m-n<0 \iff n>m ,\] so \(f\) is a polynomial of degree at most \(m\).
Now if \(f\) is entire, \(g(z) \da e^{f(z)}\) is entire and \[ \abs{g(z)} = \abs{e^{f(z)}} = e^{\Re(f)} \leq e^M ,\] so \(g\) is an entire bounded function and thus constant by Liouville, making \(f\) constant. Why this is true: \[ e^{f} = C \implies e^f \cdot f' = 0 \implies f'\equiv 0 ,\] since \(e^f\) is nonvanishing, and \(f'\equiv 0\) implies \(f\) is constant.