Let $X$ be a Banach space and $g: X \to X$ a contraction, meaning that for all $x, y \in X$, we have that
$$||g(x) - g(y)|| ≤ L ||x - y||$$
for a constant $0 ≤ L < 1$. Now, consider the function $f: X \to X, f(x) = x + g(x)$.
First, I want to show that $f$ is bijective. Using that, I want to prove that $f$ is an homeomorphism. In case that $g$ is linear, I also want to concretely find the inverse function.
Now first, I see that $f$ is continuous because $g$ and the identity function are. In case $X = \mathbb{R}$, the bijectivity would already result from the fact that $f$ is strictly monotonically increasing. But I don't really have an approach for $X ≠ \mathbb{R}$. I know that $g$ has one but only one fixed point, according to the Banach fixed-point theorem; so there is exactly one $x_0 \in X$, so that $f(x_0) = 2 x_0$. Can we somehow use this?
For the second part, if proven that $f$ is bijective and considering we know that $f$ is continuous, we only would need to show that $f^{-1}$ is also continuous. But I'm not sure how we can do that. I must admit that I don't know either how we can determine $f^{-1}$ in case of $g$ being linear. $g$ doesn't need to be bijective after all, so we can't just solve $y = x + g(x)$ for x, can we?
Thanks in advance.
1) Injectivity:
Let $x_1,x_2\in X$ and $x_1+g(x_1)=x_2+g(x_2) \Leftrightarrow x_1-x_2=g(x_2)-g(x_1) \Rightarrow L\|x_2-x_1\|\ge\|g(x_2)-g(x_1)\|=\|x_2-x_1\|$. Therefore $x_1=x_2$, because $L<1$.
2) Surjectivity
Let $c\in X$ be an arbitrary element. We want to find $x\in X$ such that $f(x)=c$, i.e $x+g(x)=c \Leftrightarrow x=c-g(x):= r(x)$. For the function $r(x)=c-g(x)$ we have $\|r(x_1)-r(x_2)\|=\|c-g(x_1)-c+g(x_2)\|=\|g(x_1)-g(x_2)\|\leq L\|x_1-x_2\|$ which implies that $r: X\rightarrow X$ is also a contraction and by the Banach fixed-point theorem there exists an unique $x_0: x_0=r(x_0)$ or $x_0=c-g(x_0)$
To find the inverse of $f(x)$ you will need some more specific conditions on $g(x)$. Even if you had only $f(x)=g(x)$ it is still impossible to invert it without knowing something more about $g$.