Existence of continuous functions defined on $x\geq 0$ such that $f(4x)-f(2x)-f(x)=0$

Question

How to show that there exists a non-zero continuous function defined on $x\geq 0$ such that $f(4x)-f(2x)-f(x)=0$?

By setting $a_n=f(2^nx)$, then $f(x)=(-\frac{1+\sqrt{5}}{2})^n f(x/2^n)$. Then what happens next?

Answer 1

There is $f(x) = \phi^{\log_2(x)}$, with $f(0) = 0$, where $\phi = \frac{(1+\sqrt{5})}{2}$, the golden ratio.

So we have: $$f(4x)-f(2x)-f(x) = \phi^{\log_2(x)} (\phi^2 - \phi - 1) = 0$$

Answer 2

Define $\,f(x):=\phi^y\,g(z)\,$ where $$ \phi:=\frac{1+\sqrt{5}}2,\quad y:=\log_2(x) \quad\text{ and } \quad z:=x/2^{\lfloor y\rfloor}. \tag{1} $$ Now $$ f(2x)=\phi^{y+1}g(z),\quad f(4x)=\phi^{y+2}g(z),\quad\text{ and } \tag{2}$$ $$ f(4x)-f(2x)-f(x) = \phi^y(y^2-y-1)g(z) = 0. \tag{3} $$ If the function $\,g(z)\,$ is continuous on $\,[1,2]\,$ and $\,g(2)=\phi\, g(1)\,$ then $\,f(x)\,$ is continuous for all $\,x>0.$