Let \(f(z)\) be a nowhere zero analytic function on the entire complex plane \(\mathbb{C}\) and write \(u(z) = \log|f(z)|\). Assume \(|u|\) is Lebesgue integrable: \(\int_{\mathbb{C}} |u(z)|\,dx\,dy < +\infty\), where \(z = x + iy\). Prove \(f\) is constant. What is the value of \(f\)?
Solution
Because \(f\) is entire and nowhere zero and \(\mathbb C\) is simply connected, \(f\) has an entire logarithm. In particular \[ u(z)=\log|f(z)| \] is harmonic on all of \(\mathbb C\).
Let \[ L=\int_{\mathbb C}|u(z)|\,dA(z)<\infty. \] For any \(a\in\mathbb C\) and \(R>0\), the mean-value property gives \[ |u(a)| \le \frac1{\pi R^2} \int_{D(a,R)}|u(z)|\,dA(z) \le \frac{L}{\pi R^2}. \] Letting \(R\to\infty\) yields \(u(a)=0\). Thus \[ |f(z)|=1 \qquad(z\in\mathbb C). \] Liouville’s theorem, or the open mapping theorem, now shows that \(f\) is constant. Its value can be any constant of modulus one: \[ \boxed{f(z)\equiv c,\qquad |c|=1.} \]