A differentiable injective function with Lipschitzian Inverse I'm having difficulty with the following question which was given to me following studying the inverse mapping theorem. 
Let $U\subseteq\mathbb{R}^{n}$
  be an open set and let $f:U\to\mathbb{R}^{n}$
  be injective and differentiable in $U$
 , assume also $f\left(U\right)$
 is an open set and let $g:f\left(U\right)\to U$
  be the inverse of $f$
 . Prove that if $g$
  is Lipschitzian then it is differentiable.
I'm assuming the main thing I'm missing is how to use the condition that $f(U)$ is open.
Help would be most appreciated!
 A: In fact, due to invariance of domain, if $U\subset\mathbb{R}^n$ is open and if $f:U\to\mathbb{R}^n$ is injective and continuous, then $f(U)$ is open.
Given $x\in U$, let us show that $g$ is differentiable at $y=f(x)$. Since $f$ is differentiable, there exist $L_x>0$ and  $r_x>0$, such that if $\|x'-x\|\le r_x$, then $x'\in U$ and $\|f(x')-y\|\le L_x\|x'-x\|$. Similarly, since $g$ is Lipschitz, there exist $L_y>0$ and $r_y>0$, such that if $\|y'-y\|\le r_y$, then $y'\in f(U)$ and $\|g(y')-x\|\le L\|y'-y\|$.
Let $r=\min(r_x,  L_x^{-1}r_y)$. If $\|x'-x\|\le r$, then for $y'=f(x')$, $\|y'-y\|\le r_y$, and hence
$$\|x'-x\|=\|g(y')-x\|\le L_y\|y'-y\|=L_y\|f(x')-y\|.$$
It implies that $f'(x)$, the Jacobi matrix of $f$ at $x$, is invertible, i.e. $\det f'(x)\ne 0$. Then it is easy to check that
$$\limsup_{y'\to y}\frac{\|g(y')-x- f'(x)^{-1}(y'-y)\|}{\|y'-y\|}\le\|f'(x)^{-1}\|\cdot\lim_{x'\to x}\frac{\|f(x)-y- f'(x)(x'-x)\|}{\|x'-x\|}=0,$$
i.e. $g$ is differentiable at $y$ and $g'(y)=f'(x)^{-1}$.
