# The sum of contraction with identity map has a continuous inverse

In $(\mathbb R, d(x,y)=|x-y|)$, suppose $f:\mathbb R \to \mathbb R$ is a contraction, that is for all $x,y \in \mathbb R$, there exists a constant $A$ between 0 and 1 such that $|f(x)-f(y)| \leq A|x-y|$.

Then let $G(x)=x+f(x)$. I have already proved that $G:\mathbb R \to \mathbb R$ is a bijection and $G$ is continuous, how to show that the inverse of $G$ is also continuous?

I have tried to find connection between $| G^{-1}(x)-G^{-1}(y) |$ and $|x-y|$ but failed, any hint?

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There are some problems i guess, $G$ is a bijection where $f$ is only a contraction and what is $F$? – Mathematician Apr 23 '13 at 16:01
sorry,that should be G. Corrected. – Elvis Apr 23 '13 at 16:10
sinx is not bijective, but sinx+sin(sinx) might be. I used fixed point theorem to prove that G is surjective and it is easy to show that it is injective – Elvis Apr 23 '13 at 16:16
$\sin x$ is not a contraction, but $x+\sin x$ is a bijection. – Julián Aguirre Apr 23 '13 at 16:21
any hint on how to show inverse of G is continuous? – Elvis Apr 23 '13 at 16:24

$g$ is monotone, indeed $\forall x,y \in \mathbb R, x>y$ consider
$|f(x)-f(y)|<A(x-y)<(x-y)$ $\implies -(x-y)<f(x)-f(y)<(x-y)\implies g(x)-g(y)>0$ by using left part of the inequality. Now you have a continuous bijective monotone function from $\mathbb R \to \mathbb R$ what can you say about $g^{-1}$?