Analysis Problem of the Day 40

Today’s problem appeared as Problem 11 on the UCLA Fall 2023 Analysis Qual:

Problem 11: Let $f$ be an entire function that is not a polynomial. Show that

$\frac{1}{\log R} \int_{0}^{2\pi} \log_+ |f(Re^{i\theta})| d\theta \to \infty$

as $R \to \infty,$ where $\log_+ x = \max(\log x, 0).$

Solution: Call the above integral $\frac{1}{\log R}I^+_f(R).$ Recall that by Jensen’s formula, if $f(0) \not = 0,$ then

$\log |f(0)| = \frac{1}{2\pi} \int_0^{2\pi }\log|f(Re^{i\theta})| d\theta + \sum_k \log \frac{|a_k|}{R},$

where $\{a_k\}$ are the zeros of $f$ in $\{|z| \leq R\}.$ In particular, if $f$ is a polynomial and $f(0)\not = 0,$ the left hand side is bounded and the sum on the right tends to 0 as $R \to \infty.$ Moreover, $|f(z)| \to \infty$ as $|z| \to \infty$ since $f$ is a polynomial, so $\log_+ (|f(Re^{i\theta})|)=\log(|f(Re^{i\theta})|)$ for large $R.$ Thus, $I^+_f(R) \to \log |f(0)|,$ i.e. $\frac{I^+(R)}{\log R} \to 0$ as $R \to \infty.$ If $f(z) = z^k p(z)$ where $p(0) \not = 0,$ the integrand has an extra $\log |z^k| = k \log |z|$ term, which contributes a $k$ to the integral. Thus, $\frac{I^+(R)}{\log R} \to k$ as $R \to \infty.$

Now, suppose $f$ is not a polynomial. By Great Picard, there exists an $|\alpha| \leq 1$ such that $f_\alpha(z):=f(z)-\alpha$ has infinitely many zeros. Define $g(z):=f(z) B(z),$ where $B$ is a finite Blaschke product that cancels out all the zeros $\{a_k\}$ of $f$ in $B(0,R),$ so that $\log|g(Re^{i\theta}|=\log|f(Re^{i\theta})|.$ In particular $g(0) \not =0,$ so since $\log |g|$ is harmonic, one gets

$I_g(R):=\frac{1}{2\pi} \int_0^{2\pi} \log|g(Re^{i\theta})| d \theta = \log |g(0)| = \log |f_\alpha(0)|+ \sum_k \log \frac{R}{|a_k|},$

with $a_k$ as above. In particular, the sum above satisfies

$\log |f_\alpha(0)|+ \sum_k \log \frac{R}{|a_k|} \geq \log |f_\alpha(0)| + \sum_{|a_k| \leq R^{\frac12}} \log \frac{R}{|a_k|} \geq \log |f_\alpha(0)| + \frac{N_R}{2} \log R.$

It follows that $\frac{I_g(R)}{\log R} \geq \frac{N_R}{2} \to \infty$ as $R \to \infty.$ Finally,

$\log_+ |f(z)-\alpha| \leq \log_+ |f(z)|+\log 2,$

$I^+_f(R) \geq I^+_{f_\alpha}(R) - \log 2 = I^+_{g}(R) - \log 2 \geq I_g(R)-\log 2,$

and we are done.

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