Analysis Problem of the Day 34

Today’s problem combines Problems 1 and 2 from Chapter 5 of Stein-Shakarchi’s “Complex Analysis”:

Problem: a) Let $f \in H(\mathbb{D})$ be bounded non-constant. If $\{a_n\}$ are the zeros of $f$ counted up to multiplicity, show that $\sum_n (1-|a_n|)<\infty.$

b) Show that for $|z| \leq r<1$ and $\alpha \in \mathbb{D},$

$\left|\frac{\alpha+|\alpha|z}{(1-\overline{\alpha}z)\alpha} \right| \leq \frac{1+r}{1-r}.$

c) Show that for any sequence $\{a_n\}$ of zeros as in a), the function $f: \mathbb{D} \to \mathbb{D}$ given by the infinite product

$f(z) = \prod_{n=1}^\infty \frac{z-\alpha}{1-\overline{\alpha}z} \frac{|\alpha|}{\alpha}$

converges normally and its zeros are precisely $\{a_n\}$ on $\mathbb{D}.$

Solution: a) We use Jensen’s formula. Without loss of generality, suppose $f(0)=1,$ since one can scale and divide out by the zeros at the origin. Then, since $f$ is bounded, $\log|f|$ is bounded above, so using Jensen’s, i.e.

$1=\log |f(0)|=\frac{1}{2\pi}\int_0^{2\pi} \log|f(re^{i\theta})|d\theta + \sum_{n} \log \frac{|a_n|}{r},$

so we conclude that $\sum_n \log |a_n|$ is bounded by sending $r \to 1$ (which can be done while avoiding the zeros of $f$ since the zeros are discrete). Thus, the sum above converges absolutely, and since $\sum_n \log |a_n| = \sum_n \log(1-(1-|a_n|)) = \sum_n (1-|a_n|) + O(\sum (1-|a_n|)^2),$ it follows that $\sum_n (1-|a_n|)< \infty.$

b) This computation follows directly from triangle inequality, as

$\left|\frac{\alpha+|\alpha|z}{(1-\overline{\alpha}z)\alpha} \right| \leq \frac{|\alpha|(1+|z|)}{|\alpha|(1-|\overline{\alpha} z|)} \leq \frac{1+r}{1-|\alpha|r} \leq \frac{1+r}{1-r}$

for $\alpha \in \mathbb{D}.$

c) Recall that a sufficient condition for the normal convergence of the product $\prod_n (1+f_n)$ of holomorphic functions $f_n$ is for the sum $\sum_n |f_n|$ to be finite. We thus compute

$\left|\frac{z-\alpha}{1-\overline{\alpha}z} \frac{|\alpha|}{\alpha} - 1\right| = \left|\frac{\alpha+|\alpha|z}{(1-\overline{\alpha}z)\alpha} (1-|\alpha|)\right| \leq \frac{1+r}{1-r} (1-\alpha) \leq C(1-\alpha)$

for $|z| \leq r<1.$ Thus, since the condition in a) holds, the sum satisfies $\sum_{n=1}^\infty |f_n| \leq C \sum_n (1-|a_n|) < \infty,$ so the product converges normally on $|z|<1.$ Finally, by a standard application of the argument principle, if the partial products are nonvanishing in a neighborhood of some $z_0$ and converge uniformly to a nonconstant function, the limit cannot vanish anywhere on the neighborhood. This implies that the zeros of the infinite product are precisely the zeros of its terms, which are given precisely by the sequence $\{a_n\}.$ Finally, as a product of Blaschke factors (which are all bounded in norm by 1), the infinite product is bounded by 1.

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