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The question is : Prove that there exists a holomorphic function $f$ on the open unit disc $\{z \in \mathbb{C} : |z| <1\}$ with the properties that $f(0) = 0$ and $f(1-1/n)=1$ for every integer $n$ greater than $1$.

My first idea was to use the general version of the Weierstrass factorization theorem to say that there exists a holomorphic function, $g(z)$ in $\mathbb{D}$ with zeros at $z_n = 1-1/n$ and no other zeros. Thus $g(z)+1$ is a candidate for our function. However we need to ensure that $g(0) = -1$. I don't know how.

Note: If such a function exists then it will be unbounded as for bounded holomorphic functions in $\mathbb{D}$ with zeros at $z_n$, $\sum (1-|z_n|)$ must converge.

Sourav D
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What about $$ f(z) = 1 + \frac{1}{\pi z} \sin\left(\frac{\pi}{1 - z}\right) $$

Seub
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  • That is indeed an elegant example. Did you just come up with the function just using the periodic property of sine function or is there a more systematic process? – Sourav D Dec 17 '13 at 02:11
  • Well, $\sin$ is a handy example of a holomorphic function with a nice infinite set of zeros. So first I thought of $\sin(\pi/(1-z))$ (so that $1/(1-z) = n$ when $z = 1 - 1/n$). Then I realized that we can correct this so that it satisfies $f(0) = 1$. Oh wait I got it wrong, I'll edit. – Seub Dec 17 '13 at 02:21
  • So no, no systematic process. Although from a practical point of view, it often works to cook something up with $\sin$ in exercises ;) – Seub Dec 17 '13 at 02:25
  • Yes it indeed does. Thank you for the insight. By the way, your first solution was an example of the function zero at the points $1-1/n$ and $-1$ at the origin. Its an equivalent solution. :) – Sourav D Dec 17 '13 at 02:32
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Actually, it is much simpler. Take $g = f-1.$ then that function has to have zeros at your $1-1/n$ and $-1$ at $0.$ Take the Weierstrass function $f$. It will have zeros where it is supposed to and some random value $v$ at $0.$ Now, just divide by $-v.$

Igor Rivin
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  • I am sorry, but can you elaborate a little bit. From the Mittag-Leffler's theorem I can definitely guarantee the existence of such a holomorphic function but additionally I can only specify the poles (none in this case) and the principal part of the Laurent series of the function. How does that help in specifying the value at $z=0$. – Sourav D Dec 17 '13 at 02:00
  • Mittag-leffler actually is not the right thing, see the edit. – Igor Rivin Dec 17 '13 at 02:06
  • What if $v$ is zero... (then you can work things out, agreed) – Seub Dec 17 '13 at 02:07
  • Thank you. I am sorry to have got stuck at such a simple observation. I was thinking of composing with some transformation to force the origin to go to zero and all that complicated stuff. – Sourav D Dec 17 '13 at 02:09
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    @Seub notice that the weierstrass function ONLY has zeros at the prescribed points -- I did think of that :) – Igor Rivin Dec 17 '13 at 02:10
  • Oh, I see. I'm wondering: can you improve Wierstrass' theorem to prescribe arbitrary values on a closed discrete set? – Seub Dec 17 '13 at 02:11
  • Yes, this is the general Mittag-Leffler theorem, but notice that the OP's set is NOT discrete (though it is discrete in the interior...) – Igor Rivin Dec 17 '13 at 02:12
  • Yes it is discrete! – Seub Dec 17 '13 at 02:13
  • See: http://math.stackexchange.com/questions/581113/specifying-a-holomorphic-function-by-a-sequence-of-values – Igor Rivin Dec 17 '13 at 02:15
  • Thanks for the link! I assume it works for any closed discrete set in an open set. – Seub Dec 17 '13 at 02:16
  • Yes, I believe so, though the general machine spawned by this is the whole subject of sheaves (see Grauert and Remmert, etc). – Igor Rivin Dec 17 '13 at 02:18