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.

  • $\begingroup$ 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 $\sum_n x_n$. $\endgroup$
    – plop
    Jun 9, 2021 at 3:00
  • $\begingroup$ 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. $\endgroup$
    – plop
    Jun 9, 2021 at 3:14
  • $\begingroup$ 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 $\endgroup$
    – Luz
    Jun 10, 2021 at 1:14
  • 1
    $\begingroup$ For other readers: This other question is about the same exercise. $\endgroup$
    – plop
    Jun 10, 2021 at 12:54

1 Answer 1


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.


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