# Is a continuously differentiable bi-Lipschitz map on $\mathbf R^d$ a $C^1$-diffeomorphism?

Question: Let $f:\mathbf R^d\to \mathbf R^d$ be a continuously differentiable map. Suppose that $f$ is bi-Lipschitz, that is, there exist a constant $C>1$, such that $$C^{-1}|x-y|\le |f(x)-f(y)| \le C|x-y|, \quad \forall x\in\mathbf R^d.$$ Is $f$ a $C^1$-diffeomorphism?

What I know is that $f:\mathbf R^d\to \mathbf R^d$ is bijective (see here), and then it is a Lipchitz homeomorphism whose inverse is also Lipschitz. But how to show the differentiability of the inverse $f^{-1}$?

Another way is using the inverse function theorem, in order to prove that $f$ is $C^1$-diffeomorphism, we need to show that the Jacobian determinant of $f$ is nowhere vanishing, i.e., $$\det(\nabla f(x))\ne0, \quad \forall x\in\mathbf R^d.$$ But I really don't know how to show this...

Now I even suspect that $f$ is not a $C^1$-diffeomorphism, but I can't find out any counterexample...

Could anyone give some hints or comments... TIA!

• I believe the result follows from terrytao.wordpress.com/2011/09/12/…. I missed the missing continuity of $f$ when I wrote my answer. – copper.hat Mar 27 '18 at 4:57
• I doubt that you can show that $f$ is $C^1$. I would guess differentiable is the best you can do. – copper.hat Mar 27 '18 at 5:08
• @copper.hat Oh, I'm sorry... I just missed the continuous differentiability of $f$. Now I add it... – Dreamer Mar 27 '18 at 5:13

Note that ${f(x+th)-f(x) \over t} \to {\partial f(x) \over \partial x} h$, and hence $\| {\partial f(x) \over \partial x} h \| \ge {1 \over C} \|h\|$ and hence the derivative is invertible.
We can use the inverse function theorem to conclude that the inverse is $C^1$.
• Oops, missed the fact that $f$ is just differentiable. – copper.hat Mar 27 '18 at 4:56