I'm trying to find a function $f$ that fulfills the following property: The area under the curve starting at some point $x_0$ with a width of $x_0$ should always be the same for all $x_0$. In other words: When I evaluate the integral starting from $x_0$ with a width of $x_0$ it should be the same as the integral starting from $\alpha x_0$ with a width of $\alpha x_0$, for all $x_0$. Setting $\alpha=2$ for simplicity gives:
$$ \int_{x_0}^{2x_0} f(x) dx = \int_{2x_0}^{4x_0} f(x) dx $$
Expressing this equation by the antiderivate $F$, with $\frac{d}{dx} F(x) = f(x)$ gives the following recurrence equation:
$$ F(4x_0) - 2 F(2x_0) + F(x_0) = 0 $$
I'm currently totally confused how to approach this problem, and I'm not even sure whether it has a solution (intuitively I was expecting some simple solution based on an exponential function), let alone, how to find all possible solutions to such a problem. Any hint in the right direction is highly appreciated.
Hint: The condition is for all $x$, we have $$\int_x^{2x}f(t)dt=C$$ Or, $$\frac{d}{dx}\int_x^{2x}f(t)dt=\frac{d}{dx}\left(\int_0^{2x}-\int_0^x\right)f(t)dt=2f(2x)-f(x)=0$$ Then solve the function equation $f(x)=2f(2x)$ to get required $f$.