# Existence of function satisfying $f(f'(x))=x$ almost everywhere

My project is to Study the existence of a continuous function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ differentiable almost everywhere satisfying $$f\circ f'(x)=x$$ almost everywhere $$x \in \mathbb{R}$$

I began the study by supposing $$f\in C ^ 1(\mathbb{R})$$, I have shown that f does not exist.

After, I found some difficulties when we assume only f differentiable on $$\mathbb{R}$$, I had an answer using Darboux's theorem https://math.stackexchange.com/questions/3312572/questions-about-the-existence-of-a-function?noredirect=1#comment6815760_3312572.

Now, I want to attack the initial problem. Previous arguments do not work!

Do you have any suggestions for me?

I have already asked the question https://math.stackexchange.com/questions/3313126/existence-of-function-satisfying-ffx-x-almost-everywhere, but the subject will be closed for a reason that I do not understand

I think, we need other non-classical arguments

Looking for a solution of the form $$f(x)=ax^b$$, $$x>0$$, one finds $$a = \phi^{-\phi/(\phi+1)}, ~~~ b=\phi$$ where $$\phi=\frac{\sqrt{5}+1}{2}$$ is the Golden ratio.
• But we must have $f∘f′(x)=x$ a.e $x\in \mathbb{R}$ ( for x<0, f(x)=?) If the question is to Study the existence of a continuous function $f : \mathbb{R_+^*} \rightarrow \mathbb{R_+^*}$ differentiable almost everywhere satisfying $f\circ f'(x)=x$ almost everywhere $x \in \mathbb{R_+^*}$ The answer is yes, there is even a $C^1$ function $f : \mathbb{R_+^*} \rightarrow \mathbb{R_+^*}$ satisfying $\forall x\in\mathbb{R_+^*},\mbox{ } f\circ f'(x)=x.$ we can take $f(x)=ax^b,\forall x>0$ with $b=(1+\sqrt 5)/2$ and $a=\exp(\frac{-b\ln b}{b+1})$ – Tina Aug 4 '19 at 14:53
• But if we take $f(x)=a|x|^b$, then it will be differentiable everywhere except $0$, and the condition will be satisfied for all $\mathbb R$. So $f(x)=a|x|^b$ with $a$, $b$ as above is a solution to the OP's problem. – FusRoDah Aug 4 '19 at 18:07
• Almost, @FusRoDah 1 . The function $f(x)=a|x|^b$ is even, so $f(f^\prime(x))=|x|$. Instead, choose $f(x)=ax^b$ for $x>0$ and $f(x)=-a(-x)^b$ if $x<0$. – Marc Chamberland Aug 4 '19 at 19:14
• Almost, @Marc Chamberland. Your function is odd, its derivative is even, so the composition is even, and we get $|x|$ again. – Pietro Majer Aug 5 '19 at 6:41
• There is a solution of the form $\alpha(-x)^\beta$ for $x<0$, but with another $\alpha$ and $\beta$, and it has $\beta<0$, so this solution is unbounded near $0$. – Pietro Majer Aug 5 '19 at 7:41