# Solving functional equations including $f\circ f(x)$ functions

Find all $$a\in\mathbb{R}$$ for which there exists a function $$f : \mathbb{R}\to\mathbb{R}$$ , such that

(i) $$f(f(x))=f(x)+x$$, for all $$x\in\mathbb{R}$$,

(ii) $$f(f(x)–x)=f(x)+ax$$, for all $$x\in\mathbb{R}$$

Normally in such functional equations, I'd put different values of $$x$$ (like $$x+1$$, $$x+2$$ etc) and try to get an equation for $$f(x)$$. But here I'm not getting anywhere by that method. I can't use the coefficient comparison method either, because $$f(x)$$ may not be a polynomial.

• Check out my edit to get a better feel for how $\rm\LaTeX$ works. Mar 2, 2021 at 3:52

NOTE: $$f^{-1}f(x) = f(f^{-1}(x)) =x$$ GIVEN:

(i) $$f(f(x))=f(x)+x$$, for all $$x$$ $$\in$$ $$ℝ$$,

(ii) $$f(f(x)–x)=f(x)+ax$$, for all $$x$$ $$\in$$ $$ℝ$$

Substitute f(x) as x in equation (i). you can verify the equation below is equivalent to equation (i) if you put $$x = f(x)$$ in the equation below. $$f(x)= x+ f^{-1}(x)$$

Then in RHS of equation 2, substitute what we have above: $$x = f(x)- f^{-1}(x)$$:

$$f(f(x)-(f(x)-f^{-1}(x)))=f(x)+ax \implies$$ $$x=f(x)+ax \implies f(x)=(1-a)x$$

Substitute and solve... I got a = golden ratio and its conjugate... Ask doubt if any below

• You are assuming here that the inverse function $f^{-1}$ is well defined (or even defined at all for all $x$). It is not clear why that is necessarily true.
– JimT
Feb 14, 2022 at 20:39
• is there a better way to solve this? @JimT
– Sid
Mar 28, 2022 at 13:34
• Proving that $f$ is injective is easy; if $f(x) = f(y)$ then it follows that $f(f(x)) = f(f(y))$ and then from the first recursive equation we obtain $x = y$. Thus, $f^{-1}$ is well defined but possibly not on the entire $\mathbb R$. Proving that $f$ is surjective does not seem easy,... of course, if that is true at all. But perhaps I am wrong and there exists a simple solution.
– JimT
Mar 29, 2022 at 16:59