Does there exist a function $f$ which is holomorphic on $B_0(2)$ (open disc of radius 2 in the complex plane) such that $f(1/n)=1/(n+1) \forall n \in \mathbb{N}$? At the moment I'm thinking not but a proof is seeming elusive. Any hints would be appreciated.


Let $a_n:=\frac 1n$. Such a function should satisfy $f(a_n)=\frac{a_n}{1+a_n}$. Let $g(z):=f(z)-\frac z{z+1}$ on $B(0,1)$, the open ball. What can you say about the zeros of $g$?

If the question was

Does there exist a function $f$ which is holomorphic on $B_0( 1)$ (open disc of radius 1 in the complex plane) such that $f(1/n)=1/(n+1) \forall n \in \mathbb{N}$?

there would not be any contradiction, since $f(z):=z/(z+1)$ would do the job.

However, since we asked for the ball $B_0(2)$, the only function which can do the job is $z\mapsto z/(z+1$ for $|z|<1$, but this function cannot be continuous at $-1$, hence it cannot be holomorphic on the ball $B_0(2)$.

  • $\begingroup$ Perfect, the first part of the question was to show that a holomorphic function has isolated zeros, so this must be the right approach. Obvious now I look at it. $\endgroup$ – user61496 Feb 22 '13 at 11:17
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    $\begingroup$ @DavideGiraudo I am dealing with an almost identical case , and I am at the point where we can see that as $\alpha_n\to0, g(\alpha_n)=f(\alpha_n) -\frac{\alpha_n}{\alpha_n+1}=0$ , hence $f(z)=\frac{z}{z+1}$ by the identity theorem . Shouldn't there be a contradiction that would prove such a fuction cannot exist? $\endgroup$ – helplessKirk Aug 24 '14 at 22:52
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    $\begingroup$ @helplessKirk Here the key point is that we consider the ball centered at the origin and of radius 2. If we had for example considered the radius $1/2$, there wouldn't be a contradiction. $\endgroup$ – Davide Giraudo Aug 25 '14 at 15:17
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    $\begingroup$ @DavideGiraudo So it has to do with where the zeros lie (and maybe if a maximum is achieved inside)? In my case we consider $D(0,1) $ and we need to prove that this function cannot exist except (maybe) for some finite cases. Thanks for answering! $\endgroup$ – helplessKirk Aug 25 '14 at 15:42
  • $\begingroup$ Can someone explain what role the radius plays? I can only think of focusing 0. I would appreciate a clarification. $\endgroup$ – Meitar Dec 16 '15 at 19:44

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