- Deduce from Hadamard’s theorem that if \(F\) is entire and of growth order \(\rho\) that is non-integral, then F has infinitely many zeros.
Solution
Suppose, toward a contradiction, that the entire function \(F\) has only finitely many zeros. Hadamard’s factorization theorem then reduces to \[ F(z)=e^{P(z)}Q(z), \] where \(P\) and \(Q\) are polynomials and \(Q\) records the finitely many zeros of \(F\).
If \(P\) is constant, then \(F\) is a polynomial and therefore has growth order \(0\).
If \(P\) has degree \(d\ge1\), then \(F\) has growth order exactly \(d\). Indeed, the upper bound \[ |F(z)|\le A e^{B|z|^d} \] is immediate from polynomial growth of \(P\) and \(Q\). Conversely, write the leading term of \(P\) as \(cz^d\), \(c\ne0\), and choose an angle \(\theta\) so that \(ce^{id\theta}=|c|\). Along \(z=re^{i\theta}\), \[ \Re P(z)=|c|r^d+O(r^{d-1}), \] while \(\log|Q(z)|=O(\log r)\) away from the finitely many zeros of \(Q\). Hence \[ \log|F(re^{i\theta})|=|c|r^d+O(r^{d-1}), \] which rules out every growth-order bound with exponent \(<d\).
Thus an entire function with only finitely many zeros has growth order either \(0\) or a positive integer. Since the given order \(\rho\) is non-integral, this is impossible. Therefore \(F\) has infinitely many zeros.