# For holomorphic $f$, $f(\frac{z}{2})= \frac{1}{2}f(z) \Longrightarrow f(z) = z$

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.

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Hints: Compute $f(2^{-k})$, use the identity theorem for holomorphic fuctions. – t.b. Feb 15 '11 at 11:25
"Complex Analysis" in unsuitable as a title. – Aryabhata Feb 15 '11 at 17:20
Thanks for the edit. – Sak Feb 16 '11 at 18:30

Since $f$ is holomorphic it has a unique power series representation about origin, $$f(z)=\sum_{k=0}^{\infty} c_k z^k.$$ Since $f\left(\dfrac{z}{2}\right)=\dfrac{1}{2} f(z)$ we obtain $$\frac{c_k}{2^k}=\frac{1}{2}c_k,\quad k=0,1,\ldots.$$ From this we obtain $c_k=0$ if $k\neq 1$. Therefore $f(z)=c_1z$. We know that $f(1)=1$. Thus $c_1=1$ and $f(z)=z$.
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.