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Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.

  1. Suppose that $F:\mathbb{H}\to\mathbb{C}$ is holomorphic and bounded. Also, suppose $F(z)$ vanishes when $z=ir_n$, $n=1,2,3,\ldots,$ where $\{r_n\}$ is a bounded sequence of positive numbers. Prove that if $\sum r_n=\infty$ then $F=0$.
  2. If $\sum r_n<\infty$, it is possible to construct a bounded function on the upper half-plane with zeros precisely at the points $ir_n$.

There is something weird about the first part. If the sequence $\{ir_n\}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $\Bbb H$ and $F$ is zero. Am I missing something here?

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If $\{ir_n\}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if $\{ir_n\}$ accumulates at $0$ and the convergence $r_n\to 0$ is "slow" enough to make $\sum_n r_n =\infty$, then $F$ must be $0$. Put differently, if $F\neq 0$ is bounded on the upper half plane and $\{z_n\}$ are zeros of $F$, then it must be that $\Im(z_n)\to 0 $ fast enough to make $$ \sum_n \Im (z_n)<\infty.$$

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  • $\begingroup$ what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane? $\endgroup$ – UserA Dec 16 '18 at 20:36
  • $\begingroup$ That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = \frac{aw+b}{cw+d}$ or something like this. $\endgroup$ – Myeonghyeon Song Dec 16 '18 at 20:38
  • $\begingroup$ Use the map $T:\Bbb D\to \Bbb H$ defined by $z\mapsto i\frac{1-z}{1+z}$? $\endgroup$ – UserA Dec 16 '18 at 20:40
  • $\begingroup$ I guess that will work :) $\endgroup$ – Myeonghyeon Song Dec 16 '18 at 20:43
  • $\begingroup$ what about the second part? $\endgroup$ – UserA Dec 16 '18 at 20:46
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The sequence $(ir_n)$ accumulates at a point of $\mathbb{C}$, but not necessarily at a point of $\mathbb{H}$. Indeed, if $r_n\to 0$ then they accumulate only at $0$, which is not in $\mathbb{H}$.

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