Question 1. Suppose that \(f , g\) are entire holomorphic functions with \(| f ( z ) | \leq | g ( z ) |\) for all \(z \in \mathbf { C }\) . Prove that there is a constant \(c \in \mathbf { C }\) so that \(f = c g\) ·
Solution
1. If \(g\equiv0\), the conclusion is immediate.
The inequality gives \(|f(z)|\le0\) for every \(z\), so \(f\equiv0\). Then \(f=cg\) for any constant \(c\), for example \(c=0\).
2. Otherwise, \(f/g\) extends holomorphically across every zero of \(g\).
Assume \(g\) is not identically zero. Let \(a\) be a zero of \(g\) of order \(m\). Write \[ g(z)=(z-a)^m u(z),\qquad u(a)\ne0. \] If \(f\) has a zero of order \(k\) at \(a\), write \(f(z)=(z-a)^k v(z)\) with \(v(a)\ne0\); if \(f\equiv0\) locally, then by the identity theorem \(f\equiv0\) globally and the conclusion is again immediate. For nearby \(z\ne a\), the hypothesis gives \[ |z-a|^k|v(z)|\le |z-a|^m|u(z)|. \] If \(k<m\), division by \(|z-a|^k\) and passage to \(z\to a\) would give \(|v(a)|\le0\), a contradiction. Hence \(k\ge m\).
Therefore the quotient \(h=f/g\), initially holomorphic away from the discrete zero set of \(g\), has a removable singularity at every zero of \(g\). Filling in those values gives an entire function \(h\).
3. The extended quotient is bounded and hence constant.
Where \(g(z)\ne0\), the original inequality gives \[ |h(z)|=\left|\frac{f(z)}{g(z)}\right|\le1. \] By continuity the same bound holds at the removable points. Thus \(h\) is a bounded entire function. Liouville’s theorem implies that \(h\equiv c\) for some \(c\in\mathbb C\). Consequently \[ f=cg \] on the whole plane, as required.