TITLE: For holomorphic $f$, $f(\frac{z}{2})= \frac{1}{2}f(z) \Longrightarrow f(z) = z$
QUESTION [6 upvotes]: Let $f$ be a holomorphic function on the open unitary disk $\mathbb{D}$ and continuous on $\mathbb{\overline{D}}$. If $f(\frac{z}{2})= \frac{1}{2}f(z)$ for all $z\in \mathbb{\overline{D}}$ and $f(1)=1$, then $f(z)=z$ for all $z\in \mathbb{\overline{D}}$.
Got this as homework. Any hints would be highly appreciated.

REPLY [4 votes]: Another way of looking at this is that you are given that $f(z) - z = 0$ for $z = 1$. Then use the condition that $f(z/2) = {1 \over 2}f(z)$ to inductively show that $f(z) - z$ also has zeroes at $z = 2^{-n}$ for all positive integers $n$. Thus the zeroes of $f(z) - z$ have an accumulation point at $z = 0$, which is only possible if $f(z) - z = 0$ for all $z$ since nonzero analytic functions can only have isolated zeroes.