Suppose a continuous function $f:[0,1] \to \mathbb{R}$ satisfies the following equation for all $z \in \left(0,\frac{1}{2}\right)$,
$$\int\_z^{2z} [f(x)-f(z)] dx = 0.$$
It is clear that a constant function $f(x)=c$ satisfies it. I would like to prove that there are no other such continous functions.


Note: this is a missing part of a larger proof I'm working on. I've already verified that if $f$ is a polynomial then it must be constant. Any hints on how to prove it for arbitrary continuous functions would be appreciated.



Functional equation $\int_z^{2z} [f(x)-f(z)] dx = 0$
Let $p$ be a zero of $2^{p+1}-p-2$ other than $-1$ and $0$ (e.g. one is approximately $2.54536493037426+10.7539751752688 i$). Then
the real and imaginary parts of $f(x) = x^p$ satisfy the equation.
Note that (with $f(0)=0$) $f$ is continuous on $[0,\infty)$ if $\text{Re}(p) > 0$.


