identity theorem to prove/disprove the existence of a holomorphic function Does there exist a holomorphic function $f$ the unit disk $\{z :|z|&lt;1 \}$ such that
$f(\frac{1}{n}) \leq \frac{1}{2^n} $ for all $ n \in \mathbb{N}$?
I know by identity theorem we can prove that $f$ not exists with $f(\frac{1}{n}) = \frac{1}{2^n} $ but not sure if this inequality can change the conclusion.
I guess the conjecture of @mjw is right, here a tentative proof. Let me now if there are mistakes.
The zero-function satisfies the conditions, and looking for a contradiction,
let $f$ be a non-zero holomorphic function satisfying the condition too. The function $f$must have a zero of order $m\in\mathbb N_{&gt;0}$ in $z=0$, hence
$g(z):=f(z)/z^m$ is also holomorphic and $g(0)\neq 0$. On the other hand, it holds
$$|g(0)|= \lim_{n\to \infty}|g(1/n)|=\lim_{n\to \infty} |n^m f(1/n)|\leq \lim_{n\to \infty}\frac{n^m}{2^n}=0$$
a contradiction.
Conjecture:  The function $f(z)=0$ suggested by @Desperado is the unique holomorphic function that satisfies $$|f(1/n)| \le \frac{1}{2^n}.$$