I want to prove that there is no holomorphic function $f(z)$ on the open unit disk that satisfies: $$ f(\frac{1}{n}) = 2^{-n} \quad \quad \forall n\in\{2,3,4,\cdots\} $$
I recall that if a holomorphic function accumulates zeros, then that function is equal to zero, but I am having trouble using that fact to show that no function satisfies the above statement.
Thank you!
A variation on some of the other solutions:
Lemma. The analytic function $g(z)= f(z/2) -f^2(z)$ satisfies $g (\frac{1}{n})=0$ for every positive integer $n$.
Proof. At $z=\frac{1}{n}$, $ f(z/2)= f(\frac{1}{2n}) = 2^{-2n} = (2^{-n})^2 = f(z)^2$.
From the Lemma and the Identity Theorem, it follows that $g(z)$ is the constant $0$ at each $z$ in the unit disk. That is, $ f(z/2)= f^2(z)$ there.
This functional relation is too good to be true! Playing the right side against the left we find that the radius $R$ of convergence of the Taylor expansion of $f$ must satisfy $2R= R$ so $R= \infty$. (Thus $f$ is entire.) Since $f(0)=0$ there is a maximal open disk $\Delta$ centered at $z=0$ on which $|f|<1$. Once again the functional equation implies $\Delta =2\Delta$ so $\Delta=\infty$. Thus $f$ is a bounded entire function. It must be 0.