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!

  • $\begingroup$ 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

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 $C^1$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.