Show $T(x) := x+sf(x)$ is a bijection with $f$ Lipschitz and $\vert s \vert \lt \frac{1}{L}$

Question

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a function and $L \gt 0$ such that

$\Vert f(x)-f(y)\Vert \leq L \Vert x-y\Vert\ $for all $x,y \in \mathbb{R}^n$

1. Show that $T(x) := x+sf(x)$ defines a bijection on $\mathbb{R}^n$ if $\vert s \vert \lt \frac{1}{L}$.
2. Prove that there is a constant $\tilde{L} \gt 0$, such that $\Vert T^{-1}(x)-T^{-1}(y)\Vert \leq \tilde{L} \Vert x-y \Vert\ $ for all $x,y \in \mathbb{R}^n$

I can prove that $T$ is injective but I am currently struggling with showing that it is also surjective. I don't really now how to approach 2. - I mean I know that the inverse of $T$ exists I thought about using the surjectivity in order to rewrite $x,y$ as $T(\tilde{x}),T(\tilde{y})$.

Answer 1

1. You want to prove that given $y\in\mathbb R^n$, there exists $x\in\mathbb R^n$ such that $T(x)=y$, i.e, $y-sf(x)=x$, i.e, $x$ is a fixed point of $y-sf(x)$. So consider the function $g_y(x)=y-sf(x)$. Prove that $g_y$ is $|s|L$-Lipschitz. Since $(\mathbb R^n,\|\|$) is complete and $|s|L<1$, by the Banach fixed-point theorem, $g_y$ has a unique fixed point. (this also shows that $T$ is also injective).

2. Notice that, as shown in 1., for $x\in\mathbb R^n$, $T^{-1}(x)$ is the unique fixed point of $g_x$. So given $x,y\in\mathbb R^n$, let $x_0=T^{-1}(x)$ and $y_0=T^{-1}(y)$. So you know that $x_0=g_x(x_0)$ and $y_0=g_y(y_0)$. So $\|T^{-1}(x)-T^{-1}(y)\|=\|g_x(x_0)-g_y(y_0)\|\cdots$