Analysis Problem of the Day 5

The following problem appears as Problem 7 on the UCLA Fall 2018 Analysis Qual:

Problem 7: Let $f: \mathbb{C} \to \mathbb{C}$ be an entire function such that the function $U(z) = \log |f(z)|$ is Lebesgue area integrable, i.e.

$\int_{\mathbb{C}} |U(z)|dx dy < \infty.$

Prove that $f$ is constant.

Solution: One of the central insights of complex analysis is the connection between analytic functions and (sub)-harmonic functions. For instance, the real and complex parts of an analytic function are harmonic, and on a simply-connected domain, any real harmonic function is the real part of an analytic function that is unique up to an imaginary constant.

To that end, a classic fact is that for any entire function $f,$ $\log |f| = \text{Re}(\log f)$ is subharmonic (defining $\log |f(z)| = -\infty$ whenever $f(z)=0.$ ). Consequently, it suffices to prove that any integrable subharmonic function in the complex plane is constant. One can make a further simplification by recalling that the maximum of a family of subharmonic functions is subharmonic, and in particular, $U_+(z) = \log_+ |f(z)| := \max(\log |f(z)|,0)$ is a positive subharmonic function. This function is integrable by assumption, so by the definition of a subharmonic function, its value at any point is less than its average on a ball around that point, i.e.

$U_+(z_0) \leq \frac{1}{\pi r^2} \int_{|z-z_0|<r} U_+(z) d\mu(z), \quad z_0 \in \mathbb{C}.$

By our assumption of integrability, the integral on the right stays bounded for all $r,$ so as $r \to \infty,$ the right hand side tends towards $0.$ Since $z_0$ is arbitrary, it follows $U_+ \equiv 0.$ Now, we’re not quite done yet, since we want to show $U \equiv 0.$ But if $U_+$ vanishes, it follows that $\log |f(z)| \leq 0$ for all $z,$ i.e. $|f(z)| \leq 1.$ Thus, $f$ is a bounded entire function, so by Liouville’s theorem, $f$ must be constant, and in fact, $|f(z)| = 1,$ i.e. $f(z) = e^{i \theta}$ for some $\theta \in [0,2\pi).$

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