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.
 A: 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 $\{|z|<R\}$.
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.
