- Find all entire analytic functions satisfying \(| f ( z ) | \leq | e ^ { z } |\) for all \(z \in \mathbb { C }\)
Solution
The functions are exactly \(\boxed{f(z)=ce^z,\ |c|\leq1}\).
1. The quotient by the exponential is a bounded entire function.
Since \(e^{-z}\) is entire, \(h(z)=f(z)e^{-z}\) is entire. The given inequality implies \[ |h(z)|=|f(z)|/|e^z|\leq1 \] for every \(z\), using \(e^z\ne0\). Liouville’s theorem therefore gives \(h=c\) for a constant \(c\) with \(|c|\leq1\) [@SS03]. Thus \(f(z)=ce^z\).
2. Every such constant gives an admissible function.
For \(|c|\leq1\), the function \(ce^z\) is entire and \(|ce^z|=|c||e^z|\leq|e^z|\) everywhere. Hence every listed function satisfies the hypothesis, completing both directions of the classification.