Suppose \(f\) and \(g\) are entire functions such that \[ |f(z)|\le |g(z)| \] for every \(z\in\mathbb C\). Show that \[ f(z)=c\,g(z) \] for some constant \(c\in\mathbb C\).
Solution
1. If\[ g\equiv0, \]then\[ f\equiv0. \]
The hypothesis gives \[ \abs{f(z)} \leq \abs{g(z)} = 0 \] for every \(z\in\CC\). Hence \(f(z)=0\) for every \(z\).
2. If \(g\equiv0\), the required conclusion holds with\[ \boxed{c=0}. \]
By step 1, \[ f\equiv0=0\cdot g. \]
3. Assume henceforth that \(g\) is not identically zero, and let\[ Z=\{z\in\CC:g(z)=0\}. \]Then every point of \(Z\) is isolated.
The function \(g\) is entire and not identically zero. By the identity theorem, its zeros have no accumulation point in \(\CC\). Thus every zero is isolated.
4. On \(\CC\sm Z\), define\[ h(z)=\frac{f(z)}{g(z)}. \]Then \(h\) is holomorphic and satisfies\[ \abs{h(z)}\leq1. \]
On \(\CC\sm Z\), the denominator \(g(z)\) is nonzero, so the quotient of the entire functions \(f\) and \(g\) is holomorphic. The given inequality gives \[ \abs{h(z)} = \frac{\abs{f(z)}}{\abs{g(z)}} \leq 1. \]
5. Every point of \(Z\) is a removable singularity of \(h\).
Fix \(z_0\in Z\). By step 3, there is a punctured disk \[ 0<\abs{z-z_0}<r \] containing no zero of \(g\) other than \(z_0\). Thus \(h\) is holomorphic on this punctured disk. Step 4 gives \[ \abs{h(z)}\leq1 \] there, so \(h\) is bounded near \(z_0\). The removable singularity theorem therefore extends \(h\) holomorphically across \(z_0\).
6. The function \(h\) extends to an entire function\[ H:\CC\longrightarrow\CC \]such that\[ \abs{H(z)}\leq1 \]for every \(z\in\CC\).
Step 5 removes each isolated singularity at a zero of \(g\). The resulting function \(H\) is entire and agrees with \(h\) on \(\CC\sm Z\).
4. At a point \(z_0\in Z\), it follows by continuityof \(H\) from the same bound on the punctured neighborhood.
7. There is a constant \(c\in\CC\) such that\[ H\equiv c. \]
By step 6, the function \(H\) is bounded and entire. Liouville’s theorem therefore implies that \(H\) is constant.
8. One has\[ \boxed{ f(z)=c\,g(z) } \]for every \(z\in\CC\).
On \(\CC\sm Z\), steps 4 and 7 give \[ \frac{f(z)}{g(z)} = H(z) = c, \] so \[ f(z)=c\,g(z). \] The entire function \[ f-cg \] therefore vanishes on the nonempty open set \(\CC\sm Z\). By the identity theorem, it vanishes identically on \(\CC\).
9. Q.E.D.
Step 2 handles the case \(g\equiv0\), and step 8 handles the case in which \(g\) is not identically zero.