let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume $f(z) = f(2z)$, prove that f is constant

Question

1. $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume that $f(z) = f(2z)$ for all $z \in \mathbb{C}$. Prove that f is constant...

Then we are supposed to use this result to solve the second question which is ...

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be analytic throughout $\mathbb{C}$ and satisfy $f(2z) = 2f(z)$ for all z. Prove that there exists $c \in \mathbb{C}$ such that $f(z) = cz$ for all z.

Answer 1

Note that $f(z) = f({1 \over 2^n} z)$. Letting $n \to \infty$ gives $f(z) = f(0)$.

Answer 2

Working backward... $$ f(z) = f(z/2) = f(z/4) = f(z/8) \cdots$$

Let $\epsilon >0$ be given. Then there is a $delta >0$ such that $|z|<\delta$ implies $|f(z)-f(0)|<\epsilon$. So for any we can chose an $n$ such that $z/2^n$ is as small as we please, so $|f(z)-f(0)|$ is less than any $\epsilon>0$. Hence

$$ f(z) \equiv f(0) $$

For part (2), it is easy to see that $f(0)=2f(0)$ and hence $f(0)=0$. Let $g(z)=f(z)/z$. This has a removable singularity at $z=0$, so we can assume that $g$ is continuous.

$$g(2z)= f(2z)/(2z) = (2 f(z))/(2z) = f(z)/z= g(z)$$

Hence $g(z) = c$ a constant, and $f(z) = cz$

Answer 3

compute $$ f'(z)=\frac{1}{2}f'(2z)2=f'(2z) $$ apply the answer to the first question to conclude $f'(z)=C$ for all $z$, so $f(z)=Cz+D$ and $$ D=f(0)=2f(0)=2D \Rightarrow D=0 $$