Consider the functional equation $$\Big(\frac{1}{x}-1\Big)f(x)+\Big(\frac{1}{x^{\phi-1}}-1\Big)f(x^\phi)=1$$ where $\phi$ is the golden ratio. I’m looking for a continuous function $f:[0,1)\to \mathbb R^+$ with $f(0)=0$ satisfying this equation. I’ve shown that this function is unique, so if I can find a single elementary function satisfying it, then I’ve found the only solution meeting these requirements.
QUESTION: Can anyone find the function $f$ in closed form? I’m not interested in integral or series representations.
How I know there’s a unique solution: To see why there is a unique continuous solution with $f(0)=0$, we can do a series of repeated substitutions into the original functional equation:
$$\Big(\frac{1}{x^\phi}-1\Big)f(x^{\phi})+\Big(\frac{1}{x^{(\phi-1)\phi}}-1\Big)f(x^{\phi^2})=1$$
$$\Big(\frac{1}{x^{\phi^2}}-1\Big)f(x^{\phi^2})+\Big(\frac{1}{x^{(\phi-1)\phi^2}}-1\Big)f(x^{\phi^3})=1$$ $$...$$
If we keep making the substitution $x\mapsto x^\phi$, we can treat this like a long system of equations in the variables $f(x),f(x^\phi),f(x^{\phi^2}),$ and so on. Through repeated substitution, we can solve for $f(x)$ in terms of $f(x^{\phi^n})$, which approaches $0$ as $n\to\infty$. The algebra is messy, but this leaves us with a different series representation for $f(x)$, showing that it is uniquely determined when we assume continuity and $f(0)=0$.
MOTIVATION: It turns out that the unique solution $f$ has the following series representation: $$f(x)=\sum_{n=1}^\infty x^{n+(\phi-1)\lfloor n (\phi-1)\rfloor}$$ and I’m trying to find a closed-form of this series (if not in terms of $x$, at least at some special values of $x$).
It’s a bit tricky to explain how I know $f$ satisfies this functional equation, but it can be proven from the following generalized identity: $$\frac{1-x}{x}\sum_{n\ge 1}x^n y^{\lfloor n\alpha\rfloor}+\frac{1-y}{y}\sum_{n\ge 1}y^n x^{\lfloor n/\alpha\rfloor}=1$$ which holds for all $x,y\in (0,1)$ and positive irrational $\alpha$. The functional equation for $f$ follows by setting $y=x^{\phi-1}$ and $\alpha=\phi-1$.
$\left(\dfrac{1}{x}-1\right)f(x)+\left(\dfrac{1}{x^{\phi-1}}-1\right)f(x^\phi)=1$
$\left(\dfrac{1}{(e^x)^{\phi-1}}-1\right)f((e^x)^\phi)=\left(1-\dfrac{1}{e^x}\right)f(e^x)+1$
$(e^{-\phi x}-e^{-x})f(e^{\phi x})=(e^{-x}-e^{-2x})f(e^x)+e^{-x}$
$(e^{-\phi\phi^x}-e^{-\phi^x})f(e^{\phi\phi^x})=(e^{-\phi^x}-e^{-2\phi^x})f(e^{\phi^x})+e^{-\phi^x}$
$(e^{-\phi^{x+1}}-e^{-\phi^x})f(e^{\phi^{x+1}})=(e^{-\phi^x}-e^{-2\phi^x})f(e^{\phi^x})+e^{-\phi^x}$
$f(e^{\phi^{x+1}})=\dfrac{e^{-\phi^x}-e^{-2\phi^x}}{e^{-\phi^{x+1}}-e^{-\phi^x}}f(e^{\phi^x})+\dfrac{e^{-\phi^x}}{e^{-\phi^{x+1}}-e^{-\phi^x}}$