Let \(f\) and \(g\) be nonconstant entire functions, and suppose \[ f\circ g=f. \] Prove that \[ g\in\operatorname{Aut}(\mathbb C). \]
Solution
The identity \(f\circ g=f\) forces every iterate of \(g\) to preserve \(f\). This makes the family of iterates normal on all of \(\mathbb C\). A nonlinear entire function has nonempty Julia set, so \(g\) must be affine, hence an automorphism of the plane.
1. For every integer \(n\ge1\),\[ f\circ g^n=f. \]
The case \(n=1\) is the hypothesis. If \(f\circ g^n=f\), then \[ f\circ g^{n+1} =(f\circ g^n)\circ g =f\circ g =f. \] The result follows by induction.
2. The family of iterates\[ \{g^n:n\ge1\} \]is normal in a neighborhood of every point of \(\mathbb C\).
Fix \(z_0\in\mathbb C\). Choose a disk \(U\) about \(z_0\) whose closure is compact. Then \[ K=f(\overline U) \] is compact.
Since \(f\) is nonconstant entire, it is unbounded. We may therefore choose three points \(a_1,a_2,a_3\in\mathbb C\) such that the values \[ f(a_1),\quad f(a_2),\quad f(a_3) \] are distinct and all lie outside \(K\).
For \(z\in U\), no iterate \(g^n(z)\) can equal any \(a_j\). Indeed, if \(g^n(z)=a_j\), then 1 gives \[ f(z)=f(g^n(z))=f(a_j), \] but \(f(z)\in K\) while \(f(a_j)\notin K\), a contradiction.
Thus every function \(g^n|_U\) omits the same three values \(a_1,a_2,a_3\). By Montel’s theorem, the family \[ \{g^n|_U:n\ge1\} \] is normal. Since \(z_0\) was arbitrary, the iterates of \(g\) form a normal family on all of \(\mathbb C\).
3. The function \(g\) cannot be a nonlinear entire function.
Recall the standard Fatou–Julia theorem for entire iteration: if an entire map is nonlinear (that is, either a polynomial of degree at least two or a transcendental entire function), then its Julia set is nonempty. Equivalently, its family of iterates fails to be normal in a neighborhood of at least one point of \(\mathbb C\).
But 2 shows that the iterates of \(g\) are normal in a neighborhood of every point. Hence \(g\) is not nonlinear. Since \(g\) is nonconstant, it must have the form \[ g(z)=az+b \] with \(a\neq0\).
4. Therefore \(g\in\operatorname{Aut}(\mathbb C)\).
For \[ g(z)=az+b, \qquad a\neq0, \] the inverse map is \[ g^{-1}(w)=\frac{w-b}{a}, \] which is entire. Thus \(g\) is a biholomorphic self-map of \(\mathbb C\), i.e. \[ g\in\operatorname{Aut}(\mathbb C). \]