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I want to construct a holomorphic function in unit disk which is not identical zero and has infinitely many zeros in unit disk.

The hint said to find a holomorphic function $f:\bf{D}\rightarrow \bf{D}$ with simple zero at some point $a\in\bf{D}$, and non-zero otherwise. Then take products of such functions and make them converge.

It reminds me of the Möbius transformation $$f(z)=\frac{z-w}{1-\bar{w}z}$$ which maps $\bf{D}$ to $\bf{D}$ bijectively, with the zero $w$. Hence I suppose that $$f_n(z)=\frac{z-z_n}{1-z_nz} $$ where $$z_n=-1+\frac{1}{n^2}$$ for $n\in \mathbb{Z}$. Take the infinite product $$\prod_{n=1}^{\infty}f_n(z)=\prod_{n=1}^{\infty}\frac{z-z_n}{1-z_nz}=\prod_{n=1}^{\infty}\left(1+(1+z_n)\frac{z-1}{1-z_nz}\right)=\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}\frac{z-1}{1-\left(-1+\frac{1}{n^2}\right)z}\right)$$ Now I want to show this infinite product (absolutely) converges, and it suffices to show that $$\sum_{n=1}^{\infty}\frac{1}{n^2}\frac{z-1}{1-\left(-1+\frac{1}{n^2}\right)z}=\sum_{n=1}^{\infty}\frac{z-1}{(z+1)n^2-z}$$ (absolutely) converges. But I cannot prove this. Please tell me how to show it converges or not.

And even it may not converge... If my attempt is wrong, then what is the right way? Thank you!

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    $\begingroup$ How about $\sin\frac1{z-1}$? $\endgroup$ – Lord Shark the Unknown Sep 21 '17 at 1:31
  • $\begingroup$ @LordSharktheUnknown I know this but I want to follow this hint... $\endgroup$ – Aolong Li Sep 21 '17 at 1:32
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Clearly your product cannot converge since $\lim_{n \to \infty} \frac{z-z_n}{1-\overline{z_n} z} \ne 1$, If you want to repair it see Blaschke product

Now a simple solution is to construct an entire function and take $F(\frac{1}{1-z})$ :

  • For a sequence $(a_n), |a_n| < |a_{n+1}|, \sum_n \frac{1}{|a_n|} < \infty$ let $$F(z) = \prod_{n=1}^\infty (1-\frac{z}{a_n})$$ It converges to an entire function because for $|z| < \frac12 |a_N|$ $$\log \frac{F(z)}{\prod_{n=1}^{N-1} (1-\frac{z}{a_n})} =\sum_{n=N}^\infty \log (1-\frac{z}{a_n}) = \sum_{n=N}^\infty \mathcal{O}(\frac{z}{a_n})$$ is a locally uniformly convergent series of analytic functions, so it is analytic.

  • Thus $f(z) = F(\frac{1}{1-z})$ is analytic on $|z-1| > 0$ and it has some zeros at $\frac{1}{1-z} = a_n, z = 1-\frac{1}{a_n}$

  • If $\sum_n \frac{1}{|a_n|} = \infty$ then the same idea works with some regularization terms, see Weierstrass product

A good question would be to ask when $f$ is bounded on $|z| < 1$ so we can make it $\mathbb{D} \to \mathbb{D}$, ie. when $F$ is bounded on $\Re(z) < -1/2$. I'd say for this we need to look at $F(z) = \prod_{n=1}^\infty (1-\frac{z}{a_n})e^{z/a_n}$ with $\sum_n \frac{1}{|a_n|} = \infty$

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  • $\begingroup$ I think the function you construct is not from unit disk to unit disk. $\endgroup$ – Aolong Li Sep 21 '17 at 2:37
  • $\begingroup$ Thank you very much! This is really a great answer. $\endgroup$ – Aolong Li Sep 21 '17 at 4:48
  • $\begingroup$ But when we take $z_n$ as real numbers, the series converges to $1$. $\endgroup$ – Aolong Li Sep 21 '17 at 4:49

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