# analytic in open unit disk,corresponding to a bounded sequence and a bounded functional sequence

Let $f$ be analytic in open unit disk, we need to show there exist $\{z_n\}$ with $|z_n|<1$ and $|z_n|\rightarrow 1$ then $f(z_n)$ is bounded.

could any one give me Hints for this one?

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What are your tools? Is it from a book or a homework? – AD. Sep 7 '12 at 5:46
"then" doesn't make sense there. Do you mean "and"? – Jonas Meyer Sep 7 '12 at 5:48

## 1 Answer

Hint: If no such sequence exists, $f$ has only finitely many zeros in the open unit disk. Use the Maximum Modulus principle on $p(z)/f(z)$ for a suitable polynomial $p$.

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+1 (Assuming $z_n\to1$ was a typo). – AD. Sep 7 '12 at 6:19
I do not understand the connection between non-existance of such sequence and finitely many $0$'s. – miosaki Sep 7 '12 at 6:38
@miosaki: Try thinking about the contrapositive of the first sentence of this answer: If there are infinitely many zeros, then can you find such a sequence? – Jonas Meyer Sep 7 '12 at 6:47
@JonasMeyer If there are infinitely many zeros of $f$ in the open unit disk then by the theorem "every bounded infinite set of $\mathbb{R}^n$ has a limit point in $\mathbb{R}^n$" and hence by Identity Theorem of analytic function we get $f\equiv 0$ thats what I can say. But here limit point may not be in the open unit disk right? Then I can not say $f\equiv 0$ – miosaki Sep 7 '12 at 7:14
If there are infinitely many zeros, they have a limit point. If that limit point is in the open disk, $f = 0$ everywhere. Otherwise it's on the circle $|z|=1$, and you take $z_n$ to be a sequence of those zeros. – Robert Israel Sep 7 '12 at 15:37