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.

You can inductively prove that
$$ f(x)
= \sum_{k=0}^{2^n - 1} f\left(\frac{x}{2^n} + \frac{k}{2^n}\right) $$
holds for all $x \in \Bbb{R}$ and $n \geq 1$. Since $f \in C^1$, this implies
$$ f'(x) = \sum_{k=0}^{2^n - 1} f'\left(\frac{x}{2^n} + \frac{k}{2^n}\right) \frac{1}{2^n} \xrightarrow[n\to\infty]{} \int_{0}^{1} f'(t) \, dt = f(1) - f(0) = 0. $$
So $f$ is constant, which then implies $f \equiv 0$ by the functional equation.

Edit (2017/08/20). Here are two remarks:


*

*We can give an alternative solution using Fourier series: Differentiating the functional equation, we have $f'(x) + f'(x+\frac{1}{2}) = 2f'(2x)$. So it follows that
$$ \hat{f'}(k)
:= \int_{0}^{1} f'(x)e^{-2\pi i k x} \, dx
= \int_{0}^{1} \frac{f'(\frac{x}{2}) + f'(\frac{x+1}{2})}{2} \, e^{-2\pi i k x} \, dx
= \hat{f'}(2k). $$
So by the Parseval's identity, for any $k \neq 0$ and $m \geq 1$ we have
$$ |\hat{f'}(k)|^2
= \frac{1}{m}\sum_{j=0}^{m-1} |\hat{f'}(2^j k)|^2
\leq \frac{1}{m} \sum_{n \in \mathbb{Z}} |\hat{f'}(n)|^2
= \frac{1}{m} \int_{0}^{1} |f'(x)|^2 \, dx $$
and letting $m\to\infty$ gives $\hat{f'}(k) = 0$. This tells that $f'$ is constant, from which we easily deduce $f' \equiv 0$ and consequently $f \equiv 0$.

*Without differentiability, we have non-trivial solutions such as
$$ f(x) = \sum_{n=1}^{\infty} \frac{\cos(2^n \pi x)}{2^n}. $$
Indeed, Weierstrass $M$-test tells that $f$ is continuous. Also, it follows that
$$ f(x) + f(x + \tfrac{1}{2})
= \sum_{n=1}^{\infty} \frac{\cos(2^n \pi x) + \cos(2^n \pi x + 2^{n-1}\pi)}{2^n} 
= \sum_{n=2}^{\infty} \frac{2\cos(2^n \pi x)}{2^n}
= f(2x).$$
So the differentiability condition is essential.
