I need a hint for the following exercise: Prove that there exists no $f:\mathbb{D}\to\mathbb{C}$ that is holomorphic and such that $f(\frac{1}{n})=\frac{1}{2^n}$ for all $n=2,3,\dots$
Normally I'd try to apply the Identity theorem, but the exponential is messing things up for me; I can't find a nice function. Also, I tried composing f with some known holomorphic functions, but no result. Any ideas?
On
By analyticity,
$$(*)\quad f^{(m)}(0) = m! \lim_{z\to 0} \frac{ f(z) - \sum_{k=0}^{m-1} f^{(k)}(0) \frac{z^k}{k!}}{z^m}.$$
Now use induction to show that $f^{(m)}(0)=0$ for all $m$.
Base: $f(0)=0$.
Induction: $f^{(k)}(0)=0$ for $k=0,\dots, m-1$. Then by $(*)$, $$f^{(m)}(0)= m! \lim_{z\to 0} \frac{f(z)}{z^m}= m! \lim_{n\to\infty} f(1/n)n^m=m!\lim_{n\to\infty} 2^{-n} n^m=0.$$
Hint: Suppose there is such an $f.$ Then $f(0)=0$ but $f\not \equiv 0.$ Therefore $f(z) = z^mg(z)$ for some $m\in\mathbb N$ and holomorphic $g$ in $\mathbb D $ with $g(0)\ne 0.$ This implies $f(1/n)$ is on the order of $(1/n)^m$ as $n\to \infty.$ Is that possible?