Suppose that \(f, g\) are entire functions with \(|f(z)| \leq |g(z)|\) for all \(z \in \mathbb{C}\). Prove that there is a constant \(c \in \mathbb{C}\) such that \(f = cg\).
Solution
1. If \(g\) is not identically zero, the quotient \(f/g\) extends to a bounded entire function.
Let \(Z=\{z:g(z)=0\}\). The zeros of a nonzero entire function are isolated [@SS03]. On \(\mathbb C\setminus Z\), the quotient \(h=f/g\) is holomorphic and satisfies \(|h|\leq1\) by hypothesis. In a sufficiently small punctured disk around each point of \(Z\), this bound makes the singularity removable [@SS03]. The resulting local extensions agree with the same quotient off \(Z\), so they define one entire function \(H\). Continuity at the removed points preserves \(|H|\leq1\) on the whole plane.
2. The functions are proportional in every case.
Under the assumption of step 1, Liouville’s theorem makes \(H\) a constant \(c\) with \(|c|\leq1\) [@SS03]. Thus \(f=cg\) off \(Z\). At a point of \(Z\), the original inequality forces \(f=0=cg\), so the equality holds everywhere.
If instead \(g\) is identically zero, the same inequality forces \(f\) to be identically zero. Then \(f=cg\) with \(c=0\). These cases exhaust the possibilities.