# Bounded Holomorphic function on Right half plane.

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!

• What is the "right upper half plane"? – WimC Jan 1 '16 at 8:38
• probably $x>0, y>0$. – p Groups Jan 1 '16 at 8:41
• Yes, pGroups got a point. Thanks. – nicksohn Jan 1 '16 at 8:44
• $\sqrt{n}$ is not in there? – WimC Jan 1 '16 at 8:46
• OMG, I lose my mind, I fixed this, Right upper half -> Right half. Sorry confusing you. – nicksohn Jan 1 '16 at 8:58

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.

• I edited my deleted answer, but now I see that it matches your answer. (+1) – robjohn Jan 1 '16 at 10:03
• 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....? – nicksohn Jan 1 '16 at 10:03
• @nicksohn: I undeleted my answer. Look at the Szegö Theorem. – robjohn Jan 1 '16 at 10:04
• Ah, Uniqueness Theorem, Right? – nicksohn Jan 1 '16 at 10:12
• @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). – WimC Jan 1 '16 at 10:52

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.

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