Jacobian determinant of Lipschitz function

Question

Given a L-Lipschitz function $f:X \subseteq \mathbb{R}^n \to Y \subseteq \mathbb{R}^n$ is it true that $\det(J_xf) \leq L^n$?

Answer 1

Hadamard's inequality says the following: If the matrix $A\in{\mathbb R}^{n\times n}$ has column vectors ${\bf a}_k$ $\>(1\leq k\leq n)$ then $$|\det(A)|\leq |{\bf a}_1|\cdot|{\bf a}_2|\cdot\ldots\cdot|{\bf a}_n|\ .\tag{1}$$ The column vectors of $\bigl[df(x)\bigr]$ are the vectors $df(x).{\bf e}_k$. From the Lipschitz property of $f$ it easily follows that $$|df(x).{\bf e}_k|\leq L\>|{\bf e}_k|=L\ ,$$ so that $(1)$ implies $$\bigl|\det\bigl(df(x)\bigr)\bigr|\leq L^n\ .\tag{2}$$ On the other hand, for the map $f:=L\ {\rm id}$ we have equality in $(2)$. This shows that $(2)$ is optimal.