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.

  • 7
    $\begingroup$ Hints: Compute $f(2^{-k})$, use the identity theorem for holomorphic fuctions. $\endgroup$
    – t.b.
    Feb 15, 2011 at 11:25
  • 2
    $\begingroup$ "Complex Analysis" in unsuitable as a title. $\endgroup$
    – Aryabhata
    Feb 15, 2011 at 17:20

2 Answers 2


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 $f(z)=z$.

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    $\begingroup$ The question asks for a hint... $\endgroup$ Feb 15, 2011 at 20:26
  • $\begingroup$ @Mariano: That rarely works here. Even when the OP asks for a hint, and some people provide hints, someone else will come along with a full solution (in this case just 3 hours later). $\endgroup$
    – GEdgar
    Nov 10, 2011 at 12:58
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    $\begingroup$ It was really a wonderful solution. $\endgroup$
    – Srijan
    May 16, 2012 at 0:22

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.


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