In my book there is a theorem which says that-
> Let a function $f:[a,b] \to [a,b]$ be a differentiable function. Then $f$ is contraction iff $\exists \alpha \in (0,1)$ such that $|f'(x)|≤\alpha$ $\forall x \in [a,b]$.
My question: I can prove the same question for any interval $I$, irrespective of closed interval, though one part is obvious, any if $f$ is contraction then $|f'(x)|≤\alpha \forall x \in I$
For converse part I use Lagrange's mean value theorem Let us take any $x,y \in I$ then applying mean value theorem on $[x,y]$ $\exists c\in (x,y)$ such that $|f(x)-f(y)|≤f'(c)|x-y|$ From hypothesis $f'(c)≤\alpha$, where $\alpha\in (0,1)$ Therefore we have $|f(x)-f(y)|≤\alpha|x-y|$, $\alpha\in (0,1)$ $\Rightarrow$ $f$ is contraction on $I$.
Can anybody tell me that I did something wrong in proof or my my proof is not wrong and proposition claimed by me is true. Thanks in advance