It is given that $f :[0,1] \rightarrow [0,1] $ and it is bijective.
If $f(2x-f(x))=x$ , find all such f.
Is my solution correct?
My attempt
$f(x)$ is bijective. thus there exists g(x) which is the inverse of f(x).
$f(2x- f(x)) = x $
$=>f(x)+g(x) = 2x$
Assume $f(y)\neq y$ for some $y$.
Then , $f(y) = y+ d$ and $g(y) = y-d$ for some $d$.
Now, $f(y+d) \neq y+d$ as $f(y)=y+d$ and $f$ is bijective.
Then, $f(y+d) = y+d+h$ for some $h$.
Therefore $g(y+d) = y+d-h$ but $g(y+d) = y$.
Thus $d=h$, which implies $f(y+d)=y + 2d$. => $f(f(y)) = y + 2d$
By iteration we get, $f^n(y)=y+nd$. There exists n such that
If $d>0, y+nd >1$.
If $d<0 , y+nd<0$.
Thus we arrive at a contradiction.
Best Answer
We have the following functional equation $$f(2x-f(x))=x,\ \ \ \ \ (*)$$ for given function $f:[0,1]\rightarrow [0,1]$.
$\bf{Proof\ 1}$: Assume that that the function $f$ is one-to-one (injective), then we have: by replacing $x:=f(x)$ in $(*)$ $$f(2f(x)-f(f(x)))=f(x)$$ and so $$f^{2}(x)=2f(x)-x$$ thus by induction, we obtain that $$f^{n}(x)=nf(x)-(n-1)x-x=n(f(x)-x)+x$$ for all natural $n$, now since $f^{n}(x)\in [0,1]$ for any fixed $x$ and all natural $n$, thus $$(n(f(x)-x)+x)\in [0,1]$$ for all $x\in [0,1]$ and any natural $n$, therefore it implies that $f(x)=x$ for all $x\in [0,1]$.
$\bf{Proof\ 2}$: Note that from relation $(*)$ implies that the function $f$ is surjective and the function $g:[0,1]\rightarrow [0,1]$ with $g(x):=2x-f(x)$ is well-define and $f(g(x))=x$ for all $x\in [0,1]$, an so the function $g$ is a right inverse for the function $f$. Now, we have $$g^{2}(x)=2g(x)-f(g(x))=4x-2f(x)-x=3x-2f(x);$$ $$g^{3}(x)=g^{2}(g(x))=3g(x)-2f(g(x))=4x-3f(x);$$ therefore, by induction, we get that $$g^{n}(x)=nx-(n-1)f(x)=n(x-f(x))+f(x)$$ for any fixed $x\in [0,1]$ and all natural $n$. Now since $g^{n}(x)\in [0,1]$ for any fixed $x$ and all natural $n$, thus $$n(x-f(x))+f(x)\in [0,1]$$ for all $x\in [0,1]$ and any natural $n$, therefore it implies that $f(x)=x$ for all $x\in [0,1]$.
$\bf{Therefore}$: Let $f:[0,1]\rightarrow [0,1]$ be a function satisfying the functional equation $$f(2x-f(x))=x,$$ then $f(x)=x$ for all $x\in [0,1]$.