periodical function $f$ with period $1$ satisfying $f(x)+f(x+1/2)=f(2x)$ is zero. Let $f$ be continuously differentiable on $\Bbb R$, periodical with period $1$, and  $f(x)+f(x+1/2)=f(2x)$ for all $x\in\Bbb R$. Show that $f\equiv 0$.
A natural attempt is to use Fourire series. Let $f(x)=a_0/2+\sum_{n=1}^\infty(a_n\cos 2 n\pi x+b_n\sin 2n\pi x)$. Then after checking both side of $f(x)+f(x+1/2)=f(2x)$ , we find $a_0=0$, $2a_{2n}=a_n$, $2b_{2n}=b_n$. Then I have on idea to proceed on.

We are given that $f$ is $C^1$.
Then  $f'(x)+f'(x+\frac12)=2f'(2x)$.
As a continuous preriodic function, $f'$ reaches its maximum at some $a\in[0,1)$. Then so  it does also at $\frac a 2$ (and $\frac a2+\frac12$). By induction, this also happens at $2^{-n}a$, by continuity also at $0 $.
By the same argument, $f'$ reaches its minimum at $0$, hence is constant.
Then $f(x)=mx+b$, and only $m=0,b=0$ work.
