# Exercise 5, Chapter 8. Complex analysis, Stein

Suppose that $$F:\mathbb{H}\rightarrow\mathbb{C}$$ is holomorphic and bounded where $$\mathbb{H}$$ is the upper-half plane. Also suppose that $$F(z)$$ vanishes when $$z=ir_n$$, $$n=1,2,\ldots,$$ where $$\{r_n\}$$ is a bounded sequence of positive numbers. Prove that if $$\sum_{n=1}^\infty r_n=\infty$$ then $$F=0$$.

Progress: Since $$\{r_n\}$$ is bounded has a subsequence convergent $$\{r_{n_k}\}$$ to $$r_0\in [0,\infty)$$. If $$r_0\in (0,\infty)\subset \mathbb{H}$$ then $$F=0$$. So I think the problem is when $$r_0=0$$. In that case, for simplicity, I considered that $$\{r_n\}$$ converges to $$0$$. I tried to use the Schwarz lemma, but I get an inequality that doesn't seem to help: $$F(z)\leq M\left| \frac{r_{n_0}-r_n}{r_{n_0}+r_n}\right|,\forall |z|\leq r_n,z\in \mathbb{H}$$ where $$M$$ is a bound for $$F$$ and $$r_{n_0}> r_n$$ are sufficiently small.

• You can use Jensen's formula. To apply the formula as is, first map $\mathbb{H}$ to the unit disc. You will need to use the equivalence of the convergence of $\sum_n\log(1-x_n)$ with that of $\sum_n x_n$.
– plop
Jun 9, 2021 at 3:00
• I looks like in Chapter 5, Section 2, of that book they did pretty much this work. You can essentially apply the contrapositive of the theorem therein.
– plop
Jun 9, 2021 at 3:14
• Sorry but I don't follow you. That theorem is about functions of order $\rho$ (for example). The Jensen's Formula gives me an identity of the zeros inside a circle or radius $r<1$. I already computed it and don't see how it helps me. @plop
– Luz
Jun 10, 2021 at 1:14
• For other readers: This other question is about the same exercise.
– plop
Jun 10, 2021 at 12:54

Assume that $$|F|\leq M$$ is a positive bound for $$F$$.
To use Jensen's formula let's map $$\mathbb{H}$$ to the unit disc $$\mathbb{D}$$ using $$g(z)=\frac{i-z}{i+z}$$ We will be applying the formula to $$f=F\circ g^{-1}:\mathbb{D}\to\mathbb{C}$$. The roots of $$f$$ are now $$\frac{1-r_n}{1+r_n}$$. I am assuming already the reduction that you made of assuming that $$r_n\to0$$. In addition, let's re-order the roots to have them decreasing.
If $$f$$ is not identically zero, we can replace $$f(z)$$ by $$f(z)/z^m$$, for some $$m$$, and assume that $$f(0)\neq0$$.
From Jensen's formula, we have that, for $$R<1$$ and chosen such that no zeros of $$f$$ are on $$|z|=R$$,$$\log|f(0)|+\sum_{k=1}^{N}\log\left|R^{-1}\frac{1+r_n}{1-r_n}\right|=\frac{1}{2\pi}\int_{0}^{2\pi}\log|f(Re^{it})|dt\leq \log(M)$$ Here $$N$$ the number of roots of $$f$$ in $$\{|z|.
Let's take limit as $$R\to1^-$$. We get that $$\log|f(0)|+\sum_{n=1}^{\infty}\log\left(\frac{1+r_n}{1-r_n}\right)<\log(M)$$
On the other hand, for out positive $$r_n$$, the series $$\sum_{k=1}^{n}\log\left(\frac{1+r_n}{1-r_n}\right)=\sum_{k=1}^{n}\log\left(1+\frac{2r_n}{1-r_n}\right)$$ converges if and only if $$\sum_{k=1}^{n}\frac{2r_n}{1-r_n}$$ converges, which does so if and only if $$\sum_{k=1}^{n} r_n$$ converges. Since this is not the case, then the assumption that $$z=0$$ was a zero of finite order was not true.