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Does there exist a Bounded Holomorphic function defined on Right half plane which have all $\sqrt{n}$ as root for all natural number $n$?

I guess It is a just $0$ function.

But How Could I approach this one?

(I've been trying to use Blascke product.)

Thanks!

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  • $\begingroup$ What is the "right upper half plane"? $\endgroup$ – WimC Jan 1 '16 at 8:38
  • $\begingroup$ probably $x>0, y>0$. $\endgroup$ – p Groups Jan 1 '16 at 8:41
  • $\begingroup$ Yes, pGroups got a point. Thanks. $\endgroup$ – nicksohn Jan 1 '16 at 8:44
  • $\begingroup$ $\sqrt{n}$ is not in there? $\endgroup$ – WimC Jan 1 '16 at 8:46
  • $\begingroup$ OMG, I lose my mind, I fixed this, Right upper half -> Right half. Sorry confusing you. $\endgroup$ – nicksohn Jan 1 '16 at 8:58
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If $f$ is a bounded non-constant holomorphic function on the disc then $\{1-|z| \mid f(z)=0\}$ is summable (see here). The function $$z\mapsto\frac{1+z}{1-z}$$ maps the unit disc onto the right half plane. The pre-image of $\sqrt{n}$ is $$\frac{\sqrt{n}-1}{\sqrt{n}+1} = 1 -\frac2{\sqrt{n}+1}$$ so a bounded function on the unit disc with those roots must be identically zero. Therefore also a bounded function on the right half plane with roots $\sqrt{n}$ must be identically zero.

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  • $\begingroup$ I edited my deleted answer, but now I see that it matches your answer. (+1) $\endgroup$ – robjohn Jan 1 '16 at 10:03
  • $\begingroup$ Thanks! But Why "a bounded function on the unit disc with those roots must be identically zero."? I know preimage of $\sqrt{n}$ goes to $1$. So....? $\endgroup$ – nicksohn Jan 1 '16 at 10:03
  • $\begingroup$ @nicksohn: I undeleted my answer. Look at the Szegö Theorem. $\endgroup$ – robjohn Jan 1 '16 at 10:04
  • $\begingroup$ Ah, Uniqueness Theorem, Right? $\endgroup$ – nicksohn Jan 1 '16 at 10:12
  • $\begingroup$ @nicksohn The numbers $(\sqrt{n}-1)/(\sqrt{n}+1)$ converge too slowly to $1$ to be the roots of a bounded non-zero function on the disc ($2/(\sqrt{n}+1)$ is not summable). $\endgroup$ – WimC Jan 1 '16 at 10:52
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The function $$ z\mapsto\frac{z-1}{z+1} $$ maps the right half plane to the unit circle. The points $\sqrt{n}$ get mapped to $$ \frac{\sqrt{n}-1}{\sqrt{n}+1} $$ Since $$ \sum_{n=1}^\infty\left(1-\left|\frac{\sqrt{n}-1}{\sqrt{n}+1}\right|\right)=\sum_{n=1}^\infty\frac2{\sqrt{n}+1}=\infty $$ the Szegö Theorem says that the only bounded function on the unit disk with those roots vanishes identically.

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  • $\begingroup$ Right upper half -> Right half. Sorry confusing you. $\endgroup$ – nicksohn Jan 1 '16 at 9:15
  • $\begingroup$ Thanks! Szegö Theorem look fine, But Somewhat.. Too difficult to me for now. I'll check it very soon. $\endgroup$ – nicksohn Jan 1 '16 at 10:08
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    $\begingroup$ @nicksohn : Are you sure you didn't learn about the condition when studying Blaschke products? $\endgroup$ – DisintegratingByParts Jan 2 '16 at 22:24

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