TITLE: $f: [0,1] \to [0,1]$ is injective & $ f(2x-f(x))=x$ for $x \in [0,1]$ then $f(x)=x $
QUESTION [5 upvotes]: Let $f: [0,1] \to [0,1]$ be an injective function such that $ f(2x-f(x))=x$ for $x \in [0,1]$. I have to prove that $f(x)=x,  x\in [0,1]$.
My attempt:
let $$g(x)=2x-f(x)$$ then $$f(g(x))=x  \implies f$$ has an inverse, $$f^{-1}=2x-f(x) $$,
and I have no clue on how to proceed from here, the solution at the back of the back suggests defining a sequence $(x_{n})$ recursively as,   $(x_{n})=f(x_{n-1})$ but I couldn't understand it further.
Help  of any sort (hints or full solution, whatever seems possible) would be greatly appreciated

REPLY [2 votes]: $f(2x -f(x)) = x, y=f(x), f(2y-f(y))=y$ implies that $f(2f(x)-f^2(x))=f(x)$
f injective $2f(x)-f^2(x)= x, 2x=4f(x)- 2f^2(x), 2x-f(x) = 3f(x)-2f^2(x)$ this implies
$f(2x-f(x))= f(3f(x)-2f^2(x))=x$ f injective
$3f(x)-2f^2(x)=x$, since $2f(x)-f^2(x)= x, f(x)=x$
$3f(x)-2f^2(x)-2(2f(x)-f^2(x))=-f(x)=x-2x$, thus f(x)=x