Prove that $f$ is a $C^1$ diffeomorphism with this condition this is an exercise that I cannot solve:
Suppose that $f \in C^
1
(\mathbb{R}^n
, \mathbb{R}^n
)$ is regular
everywhere and that there exists $k>0$ such that $\|f(x_1)−f(x_2)\|
\geq k\|x_1−x_2\|$, for
all $x_1, x_2\in \mathbb{R}^n$.
Prove that $f$ is a $C^1$
diffeomorphism.
I just know that $f$ is injective and open. How can I continue?
Thanks.
 A: Since $f:\mathbb{R}^n\rightarrow Ran(f)$ is injective, the inverse $f^{-1}:Ran(f)\rightarrow \mathbb{R}^n$ exists. Now, for each $x\in \mathbb{R}^n$ there exist open neighborhoods $U_x$ of $x$ and $V_{f(x)}$ of $f(x)$ such that $f:U_x\rightarrow V_{f(x)}$ is a $C^1$-diffeomorphism. That is, the local inverse of $f$, which we can denote by $g_x:V_{f(x)}\rightarrow U_x$, is continuously differentiable. But the restriction of $f^{-1}$ to $V_{f(x)}$ is $g_x$, so it must be differentiable in $f(x)$. Since $x$ was arbitrary, $f^{-1}$ is continuously differentiable on the whole range of $f$.
So now we only have to show that $Ran(f)=\mathbb{R}^n$. We will show that $Ran(f)$ is both open and closed in $\mathbb{R}^n$. Our claim then follows from the fact that $\mathbb{R}^n$ is connected. 
First, note that $Ran(f)=\bigcup_{x\in \mathbb{R}^n}V_{f(x)}$, where $V_{f(x)}$ is as above. Since the inverse function theorem guarantees that $V_{f(x)}$ is open in $\mathbb{R}^n$, $Ran(f)$ is also open as a union of open sets. Secondly, $f$ is a closed map. To see this, let $A$ be a closed set, $(y_n)$ a convergent sequence in $f(A)$ and $y$ its limit. Then this also a Cauchy sequence, hence $(f^{-1}(y_n))$ is a Cauchy sequence as well and $x:=\lim_{n\to \infty}f^{-1}(y_n)$ exists. Since $A$ is closed, we must have $x\in A$. A simple continuity argument shows $y=f(x)\in f(A)$. This proves that $f$ is a closed map and since $\mathbb{R}^n$ is closed, $Ran(f)$ is closed as well.
