Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
Hints: Compute $f(2^{-k})$, use the identity theorem for holomorphic fuctions. –  t.b. Feb 15 '11 at 11:25
2  
"Complex Analysis" in unsuitable as a title. –  Aryabhata Feb 15 '11 at 17:20
    
Thanks for the edit. –  Chu Feb 16 '11 at 18:30

2 Answers 2

up vote 12 down vote accepted

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 $f(z)=z$.

share|improve this answer
3  
The question asks for a hint... –  Mariano Suárez-Alvarez Feb 15 '11 at 20:26
    
@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). –  GEdgar Nov 10 '11 at 12:58
2  
It was really a wonderful solution. –  srijan May 16 '12 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.