Existence of a bounded function on the disk with specified values

Question

I am trying to answer the question:

“Does there exist a bounded holomorphic function on the unit disk such that for all $n\in \mathbb{N}$, $f(1-\frac{1}{n}) = \frac{(-1)^n}{n}$?”

My solution goes roughly likes this: using Jensen’s formula we find that any holomorphic function with zeroes at $1-\frac{1}{n}$ cannot be bounded, so if there was such a function as the problem desired, it would be the only one, lest the difference between two such functions violate Poisson Jensen. But using some algebra tricks, given any one such function we mess it around to get another. This is a contradiction.

I just want to ask if there is a more straightforward way. Any other ideas are much appreciated.

Answer 1

Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-\frac{1}{n}$. By Blaschke condition, we must have $g \equiv 0$. But this implies $h(z)= \frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set $\{-1,1\}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.

Answer 2

Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-\frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $\frac{f}{1-z}$ has constant modulus and is therefore constant, so that $f\equiv z \mapsto a(1-z)$ for $a$ equal to $\pm 1$, independently of $z$ of course, which is a contradiction.