TITLE: If $f(2x)=2f(x)$ and $f'(0)=0$ then $f(x)=0$
QUESTION [8 upvotes]: Recently, when I was working on a functional equation, I encountered something like an ordinary differential equation with boundary conditions!

Theorem. If the following holds for all $x \in \mathbb R$ 
  $$\begin{align}
f(2x) &=2 f(x) \\
f'(0) &=0
\end{align}$$
  then $f(x)=0$ on $\mathbb R$.

Intuitively, it is evident for me that $f(x)=0$ but I cannot show this by a formal argument. In fact, I don't have any idea to work on it! :)
I will be thankful if you provide me a hint or help to show this with a nice formal proof.

REPLY [22 votes]: Setting $x=0$ in $f(2x)=2f(x)$, $f(0)=0$.  Now fix $x\neq 0$ and consider the values $f(x/2^n)$.  By induction, $f(x/2^n)=f(x)/2^n$ for all $n\in\mathbb{N}$.  But by the definition of the derivative, $$\frac{f(x/2^n)-f(0)}{x/2^n-0}=\frac{f(x)/2^n}{x/2^n}=\frac{f(x)}{x}$$ must converge to $f'(0)=0$ as $n\to\infty$.  It follows that $f(x)=0$.