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

Loreno Heer
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    What do you mean for $L^n$? if $L$ is the Lipschitz constant, then it depends if $L$ is or not greater than $1$. Let's suppose it is, then yes, your statement is true. As you can see from http://en.wikipedia.org/wiki/Lipschitz_continuity – Oscar Mar 18 '15 at 16:40
  • @Oscar made a thinking-mistake, should be just $L$ the lipschitz constant. Do you know a counterexample in case $L > 1$? – Loreno Heer Mar 18 '15 at 18:25
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    @Oscar I thought for a second I made a mistake and $\leq L$ would suffice, but got an answer now and it is indeed $\leq L^n$... – Loreno Heer Mar 18 '15 at 21:44

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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.

Christian Blatter
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