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.
Best Answer
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.