Suppose \(f\) and \(g\) are entire and \(\abs{f(z)}\leq \abs{g(z)}\) for all \(z\in\CC\). Show that \(f=cg\) for some constant \(c\).
Solution
If \(g\equiv0\), then the inequality forces \(f\equiv0\), so take \(c=0\).
Assume \(g\not\equiv0\) and set \(h=f/g\) away from the zeros of \(g\). There \(\abs{h}\leq1\). If \(a\) is a zero of \(g\), then \(h\) is bounded on a punctured neighborhood of \(a\), so \(a\) is a removable singularity of \(h\). Thus \(h\) extends to an entire function on \(\CC\) with \(\abs h\leq1\) everywhere. By Liouville’s theorem, \(h\equiv c\) for some \(c\in\CC\) with \(\abs c\leq1\). Hence \[ f=cg. \]