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The question is:

Let $f$ be holomorphic function in unit disc, which is bounded and have zeros in points $1-1/n$ for natural $n$. Then $f$ is equal to zero.

Identity theorem does not help because point 1 is not inside unit disc. I suppose it follows from the fact that any holomorphic function in unit disc is either unbounded or may be extended on larger disc, but I can not prove it. Is it correct?

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    $\begingroup$ There are bounded holomorphic functions on the unit disk that cannot be extended to a larger disk. Any bounded holomorphic function with infinitely many zeros in the disk that is $\not\equiv 0$ for example. $\endgroup$ Feb 13, 2014 at 15:39
  • $\begingroup$ This is an application of the identity theorem. $\endgroup$
    – user21467
    Feb 13, 2014 at 16:09
  • $\begingroup$ But limit point is 1, which is not inside of unit disc $\endgroup$
    – evgeny
    Feb 13, 2014 at 16:15
  • $\begingroup$ Oh, I see. Yes, I read too quickly. $\endgroup$
    – user21467
    Feb 13, 2014 at 17:21

1 Answer 1

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This follows from Blaschke's condition: the sequence $(a_n)$ is the zero sequence of a bounded holomorphic function on the unit disc if and only if

$$ \sum_n (1-|a_n|) < \infty. $$

The simplest proof is via Jensen's formula.


As noted, this doesn't follow from the identity theorem: there are unbounded holomorphic functions on the unit disc whose zero sequence is $(1-1/n)$.

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