identity theorem to prove/disprove the existence of a holomorphic function

Question

Does there exist a holomorphic function $f$ the unit disk $\{z :|z|<1 \}$ 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.

Answer 1

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 $m\in\mathbb N_{>0}$ 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.

Answer 2

Conjecture: The function $f(z)=0$ suggested by @Desperado is the unique holomorphic function that satisfies $$|f(1/n)| \le \frac{1}{2^n}.$$