On book Nonlinear Functional Analysis by K.Deimling, there is an exercise (Ex15.4 on P171) saying that:
If $F:\mathbb{R}^n\to\mathbb{R}^n$ is $C^1$ map and $\forall x \in \mathbb{R}^n$, $det JF(x)\neq 0$, then $F$ is a $C^1$ homeomorphism from $\mathbb{R}^n\to\mathbb{R}^n$ if and only if \begin{equation} \lim_{\|x\|\to\infty} \|F(x)\| = \infty \end{equation}
Now assume that $F$ satisfies the condition of the above proposition (i.e. $F$ is in $C^1(\mathbb{R}^n)$ with inversible Jacobian and $\lim_{\|x\|\to\infty} \|F(x)\| = \infty$), then its inverse is also in $C^1(\mathbb{R}^n)$. I am wondering that whether its inverse is Lipschitz.
My thought is $F^{-1}$ is in $C^1$, so it is Lipschitz if and only if its Jacobian has bounded norm, i.e.
\begin{equation} \exists M > 0, \text{ s.t. } \|JF^{-1}(x)\| = \|(JF(x))^{-1}\| \leq M, \forall x \end{equation}
And we know that
\begin{equation} \|JF^{-1}(x)\| \geq \frac{\|I\|}{\|JF(x)\|} \end{equation}
So, if we want to find counter-examples, we may need to find $F$ such that $JF(x)$ goes to $0$ when $x$ goes to someplace.
But I cannot find counter-examples in the 1-dimensional case, if $f'(x)$ goes to $0$, then it seems like the function $f$ cannot be bijective. It there a counter-example or we can prove that the inverse must be Lipschitz?
Update: $x\mapsto \ln x$ seems like a good candidate, its inverse is not Lipschitz, but the question is it is not $C^1$ on $\mathbb{R}$.