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Let $f \colon \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map. Suppose that exists $c>0$ such that $||f(x)-f(y)||\geq c||x-y||$ for all $x,y \in \mathbb{R}^n$. Prove that $f$ is a diffeomorphism.

I know prove that $f \colon X \to X$, $f\in C^{0}$ such that $||f(x)-f(y)||\geq c||x-y||$ is a homeomorphism if $X$ is compact. It is a long and wearisome proof.

About the question, we have automatically that $f$ is injective, and it is open, cause $f$ is continous and injective, so we can consider a inverse, $h\colon f(\mathbb{R}^n) \to \mathbb{R}^n$, and it is continuous because $f$ is open. $A \subset \mathbb{R}^n$ open set $\implies$ $h^{-1}(A) = f(A)$ open.

My two difficulties are show that $f$ is surjective and $h$ is differentiable.

To show that $h$ is differentiable I was trying show that $D_{f}(x)= f^{'}(x)$ is invertible.(I don't know if it is true). I just prove that $||D_{f}(x)||\geq c$. It helps?

Thank you for any help.

Oddone
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  • For the differentiablility you could use the implicit function theorem. Of course, $f'(x)$ is not necessarily invertible, take $f(x)=x$... – Alex Jul 06 '16 at 23:40
  • I don't know how use it, I thought use the inverse function theorem. – Oddone Jul 06 '16 at 23:46
  • @Alex careful there. If $f:\Bbb R^n \to \Bbb R^n$, then at a fixed $x$, $f'(x)$ is an $n\times n$ matrix which is or isn't invertible. In the case of $f(x)=x$, $f'(x)$ is an invertible transformation for each fixed $x$. – Ben Grossmann Jul 06 '16 at 23:46
  • of course, you are right, thank you for the correction! – Alex Jul 06 '16 at 23:47
  • @Oddone in fact, part of the content of the inverse function theorem is that $h$ has derivative (Jacobian matrix) $[f'(x)]^{-1}$ at the point $f(x)$. – Ben Grossmann Jul 06 '16 at 23:49

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$f$ is surjective because $f(\mathbb R^n)$ is an open and closed subset of $\mathbb R^n$. You already mentioned that $f$ is an open map. To see that $f(\mathbb R^n)$ is closed, let $(y_n)$ be a sequence in $f(\mathbb R^n)$ converging to $y$. Then because $\|f^{-1}(y_n)-f^{-1}(y_m)\|\leq c^{-1}\|y_n-y_m\|$, $(f^{-1}(y_n))$ is a Cauchy sequence, hence converges to some $x\in\mathbb R^n$, and $f(x) =f(\lim\limits_{n\to\infty}f^{-1}(y_n))=\lim\limits_{n\to\infty}y_n=y$, the second to last equality by continuity. By connectedness, the only nonempty open and closed subset of $\mathbb R^n$ is $\mathbb R^n$.

You can show that $f'(x)$ is always invertible by contraposition. Let $c>0$. Suppose $f'(x)$ is not invertible for some $x$. Let $v\in\mathbb R^n$ be a unit vector such that $f'(x)v=0$. Then $\lim\limits_{h\to 0}\dfrac{\|f(x+hv)-f(x)\|}{|h|}=0$, so there exists $\delta>0$ such that $0<|h|<\delta$ implies $$\|f(x+hv)-f(x)\|<c|h|=c\|(x+hv)-x\|,$$ showing that the hypothesis doesn't hold.

Hence your $f$ does have $f'(x)$ invertible for all $x$, and the inverse function theorem therefore implies $f^{-1}$ is $C^1$.

Jonas Meyer
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