Is the Jacobian bounded a.e?

Question

In this paper first Corollary, I find:

let $\Phi$ a map from $\mathbb R^d$, $d ≥ 1$, onto itself which is bi- Lipschitz continuous, has an almost everywhere constant Jacobian, and maps the unit ball onto the unit cube.

Is the jacobian of $\Phi^{-1}$ bounded almost everywhere?

Thanks for your help.

Answer 1

Yes, you can find on Evans, measure theory and fine properties of functions, that:

If $f, g: \mathbb{R}^n \to \mathbb{R}^n$ are locally lipschitz continous, and if $Y = \{ x \in \mathbb{R}^n\ :\ g(f(x))\ =\ x \}$ then $Dg(f(x))Df(x)=I$ for $L^n$ almost every $x \in Y$.

In your case you have $g = f^{-1}$ thus $Y = \mathbb{R}^n$ and so $D(f^{-1}(f(x))D(f(x))= I$ for almost every $x \in \mathbb{R}^n$. And the claim follows.