# Bounded holomorphic function in unit disk

Does there exist a bounded holomorphic function in the unit disk $D$ such that $$f\left(1-\frac{1}{n}\right) = \frac{(-1)^n}{n}$$ for $n = 1,2,3,\dots$ ?

Sorry it's not homework so I've no idea how should I begin. I only noticed that $\{1-1/n\}$ has a limit point out of $D$. Also $f'(1)$ does not exist.

• The obvious candidate for such $f$ is $f(z)=(1-z)\cos\frac{\pi}{1-z}$, but that is not bounded – Hagen von Eitzen Aug 6 '18 at 6:08
• I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph. – Diger Aug 6 '18 at 6:41
• Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there. – zokomoko Aug 6 '18 at 6:58
• @HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here. – MonkeyKing Aug 6 '18 at 17:21

It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $\sum [1-|a_n|] <\infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $\{1-\frac 1 {2n}\}$ which do not satisfy the summability condition. Hence $f(z)\equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
• Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set. – Jyrki Lahtonen Aug 6 '18 at 7:52