Let $|{}\cdot{}|$ be the Euclidean norm on $\mathbb{R}^{n}$ and $f(x) = x/(1+|x|)$, $x \in \mathbb{R}^n$. Prove $f$ is a $C^{1}$ diffeomorphism on $\mathbb{R}^n - \{0\}$.
Here is my sketch of proof:
- $f$ is bijective:
Let $x_1$, $x_2 \in \mathbb{R}^n$:
\begin{align} f(x_1) &= f(x_2) \\ \frac{x_1}{1+|x_1|} &= \frac{x_2}{1+|x_2|} \\ x_1\cdot\frac{1+|x_2|}{1+|x_1|} &= x_2 \end{align}
So $x_1$ and $x_2$ are colinear, and $x_1 = 0 \implies x_2 = 0$. Let $x_1$, $x_2 \neq 0$. Consider $g: \mathbb{R}^{+}_{0} \to \mathbb{R}$ such that $g(u) = 1/(1+u)$. Clearly $g$ is injective, and $|x_1|$, $|x_2| \in \mathbb{R}^{+}_{0}$, then:
\begin{align} \frac{1}{1+|x_1|} &= \frac{1}{1+|x_2|} \iff |x_1| = |x_2| \\ \frac{x_2}{1+|x_1|} &= \frac{x_2}{1+|x_2|} \\ \frac{x_2}{1+|x_1|} &= \frac{x_1(1+|x_2|)}{(1+|x_1|)(1+|x_2|)}\\ x_1 &= x_2 \end{align} Then, $f$ is injective.
Let $y \in \mathbb{R}^{n}-{0}$:
\begin{align} y &= \frac{x}{1+|x|}\\ (1+|x|)y &= x \tag 1 \end{align}
Applying the norm on (1), we have: \begin{align} |(1+|x|)y| &= |x|\\ |y|+|x||y| &= |x|\\ \frac{|y|}{1-|y|} &= |x| \tag 2 \end{align}
Replacing (2) in (1): $x = \frac{y}{1-|y|}$ Then $f$ is surjective. Therefore, $f$ is bijective.
- $f$ is $C^{1}$:
\begin{align} f(x) &= \frac{x}{1+|x|} \\ &= \frac{(x_1, x_2, \ldots, x_n)}{1+|x|} \\ &= \left(\frac{x_1}{1+|x|}, \ldots, \frac{x_n}{1+|x|}\right)\\ \end{align} Then, $f_i(x) = \dfrac{x_i}{1+|x|}$, $i=1,\ldots,n$ and $\displaystyle\frac{\partial{f_i(x)}}{\partial{x_j}} = \frac{1}{1+|x|}$ if $i=j$
$\displaystyle\frac{\partial{f_i(x)}}{\partial{x_j}} = 0$ if $i \neq j$
Then, all partial derivatives are continuous, so we conclude $f$ is $C^{1}$.
- Diffeomorphism
Clearly $Jf(x) = \frac{1}{1+|x|} I_n$. Then, $\det(Jf(x)) = \left(\frac{1}{1+|x|}\right)^n \neq 0$. By the inverse function theorem, we conclude $f^{-1}$ is $C^{1}$. Therefore $f$ is a diffeomorphism.
Is my proof correct? I suspect the injective part is wrong or not well-supported. Finally, I don’t see why it is necessary to work on $\mathbb{R}-{0}$, it makes sense if it would be $\mathbb{R}^{n}-\{x\}$ such that $|x|=1$.