Functional equation $\big(\frac{1}{x}-1\big)f(x)+\big(\frac{1}{x^{\phi-1}}-1\big)f(x^\phi)=1$

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 $$\alpha=\phi-1$$.

• Is there any reason to expect it to have an elementary representation? Dec 27 '19 at 14:04
• Can you maybe explain to me the idea you used to prove uniqueness and to obtain the series you got? Dec 27 '19 at 14:09
• @Fimpellizieri No, no particular reason. I’m just hopeful. :) Dec 27 '19 at 14:10

$$\left(\dfrac{1}{x}-1\right)f(x)+\left(\dfrac{1}{x^{\phi-1}}-1\right)f(x^\phi)=1$$
$$\left(\dfrac{1}{(e^x)^{\phi-1}}-1\right)f((e^x)^\phi)=\left(1-\dfrac{1}{e^x}\right)f(e^x)+1$$
$$(e^{-\phi x}-e^{-x})f(e^{\phi x})=(e^{-x}-e^{-2x})f(e^x)+e^{-x}$$
$$(e^{-\phi\phi^x}-e^{-\phi^x})f(e^{\phi\phi^x})=(e^{-\phi^x}-e^{-2\phi^x})f(e^{\phi^x})+e^{-\phi^x}$$
$$(e^{-\phi^{x+1}}-e^{-\phi^x})f(e^{\phi^{x+1}})=(e^{-\phi^x}-e^{-2\phi^x})f(e^{\phi^x})+e^{-\phi^x}$$
$$f(e^{\phi^{x+1}})=\dfrac{e^{-\phi^x}-e^{-2\phi^x}}{e^{-\phi^{x+1}}-e^{-\phi^x}}f(e^{\phi^x})+\dfrac{e^{-\phi^x}}{e^{-\phi^{x+1}}-e^{-\phi^x}}$$