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$C^k$ stands for $k$-times continuously differentiable ($k = \infty$ is included). The map $h : \mathbb R \to \mathbb R, h(x) = x^3$, is the standard example of a $C^\infty$-homeomorphism whose inverse is not even differentiable.

For $k < \infty$, does there exist a $C^k$-homeomorphism $h : \mathbb R \to \mathbb R$ whose inverse is $C^{k-1}$, but not $C^k$?

This seems to be a quite obvious question, but for $k > 1$ I did neither find any example in the literature nor via internet search. My own efforts were unsuccessful, candidates as $h(x) = x^n$ do not work.

Bonus question:

For $k < \infty$, does there exist a $C^\infty$-homeomorphism $h : \mathbb R \to \mathbb R$ whose inverse is $C^{k-1}$, but not $C^k$?

Paul Frost
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  • What about $\dfrac{x^3}{3}+x$? – Darsen Oct 16 '20 at 23:44
  • @Darsen For which $k$? – Paul Frost Oct 16 '20 at 23:50
  • $f(x)=\dfrac{x^3}{3}+x$ is clearly $C^\infty$. Now, $f'(x)=x^2+1$ is never $0$, so by the Inverse Function Theorem $f$ has a continuously differentiable inverse. However, $f''(x)=2x$ is $0$ at $x=0$, so if the inverse had second derivative, it would have to be infinity there. Therefore, the inverse is $C^1$, but not $C^2$. – Darsen Oct 16 '20 at 23:55
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    If $f$ is $C^k$ and $f^{-1}$ is differentiable, then $(f^{-1})^{\prime}(x)=1/f^{\prime}(f^{-1}(x))$, so $f^{-1}$ will be $C^k$ as well. What am I missing? – Thorgott Oct 16 '20 at 23:55
  • @Thorgott Ah, you are right. You should give an official answer. – Paul Frost Oct 16 '20 at 23:59
  • @Thorgott But you need your function and its derivatives to not be zero. – Darsen Oct 17 '20 at 00:04
  • The derivative is non-zero by the chain rule since $1=\operatorname{id}^{\prime}(x)=(f^{-1}\circ f)^{\prime}(x)=(f^{-1})^{\prime}(f(x))f^{\prime}(x)$. I don't see the issue with the function being $0$ (it has to be at one point, anyhow). – Thorgott Oct 17 '20 at 00:07
  • You're right. I meant to say nonzero first derivative. – Darsen Oct 17 '20 at 00:15

1 Answers1

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For $k>1$, there is no such map. The reason is that if $f\colon\mathbb{R}\rightarrow\mathbb{R}$ is a $C^k$-bijection with differentiable inverse, then the inverse is automatically $C^k$. This can be seen as a special instance of the inverse function theorem. If $f^{-1}$ is differentiable, this dictates we have $(f^{-1})^{\prime}(x)=1/f^{\prime}(f^{-1}(x))$ for all $x\in\mathbb{R}$. As composition of continuous functions, $(f^{-1})^{\prime}$ is continuous, so $f^{-1}$ is $C^1$. Then, as composition of $C^1$-functions, $(f^{-1})^{\prime}$ is $C^1$, hence $f^{-1}$ is $C^2$. Induction carries us all the way to $f^{-1}$ being $C^k$.

Thorgott
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