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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!

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

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  • $\begingroup$ Oops, missed the fact that $f$ is just differentiable. $\endgroup$ – copper.hat Mar 27 '18 at 4:56

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