# Thread: f(x)=x

1. ## f(x)=x

iIf $\displaystyle f:[0,1] \to [0,1]$ is an injection which satisfies $\displaystyle f(2x-f(x))=x$ then $\displaystyle f(x)=x$ for $\displaystyle x \in [0,1]$

2. Originally Posted by Chandru1
If $\displaystyle f:[0,1] \to [0,1]$ is an injection which satisfies $\displaystyle f(2x-f(x))=x$ then $\displaystyle f(x)=x$ for $\displaystyle x \in [0,1]$
Proof by contradiction: Suppose that $\displaystyle f(x)\ne x$ for some x, say $\displaystyle f(x) = x+\delta$ where $\displaystyle \delta\ne0.$ Then $\displaystyle x = f(2x-f(x)) = f(x-(f(x)-x)) = f(x-\delta)$, so that $\displaystyle f(x-\delta) = (x-\delta) + \delta$.

Now repeat the same argument with $\displaystyle x-\delta$ in place of x, to see inductively that $\displaystyle f(x-n\delta) = (x-n\delta) + \delta$ for n=1,2,3,... . But for n large enough, $\displaystyle x-n\delta$ will be negative (if $\displaystyle \delta>0$), or greater than 1 (if $\displaystyle \delta<0$). In either case that gives a contradiction.