What's the relation between Lipschitz constant and the determinant of Jacobian matrix? [closed]

Question

For a vector fuction $y=f(x)$ where $x, y$ are vectors. What's the relation between Lipschitz constant and the determinant of Jacobian matrix?

Answer 1

No relation in general. Fix $a >0$ and consider $f:\mathbb{R}^2 \to \mathbb{R}^2$ defined by $f(x,y) = (x,ay)$. If $a \leq 1$ then the Lipschitz number of $f$ is $1$ but the determinant of its Jacobian is everywhere $a$.

So Lipschitz number can be arbitrarily larger than the determinant of the Jacobian. The converse is also true; consider $g:\mathbb{R}^n \to \mathbb{R}^n$ defined by $g(\vec{x}) = 2\vec{x}$. The Lipschitz number is $2$ but the determinant of the Jacobian is $2^n$.

I guess the Lipschitz number is always at least the $n$th root of the absolute value of the determinant of the Jacobian, for maps from $\mathbb{R}^n$ to $\mathbb{R}^n$.

Answer 2

The absolute value of the determinant of the Jacobian matrix of the transformation $f$ shows how much $f$ changes the $n$-volume, locally. The Lipschitz constant of $f$ shows how much $f$ changes the lengths, at the most.

Nik Weaver mentioned that the Lipschitz number is always at least the $n$th root of the absolute value of the determinant of the Jacobian quite plausible.

The purpose here is to provide a proof of this statement. Indeed, approximating locally $f$ by an affine transformation, we may assume that $f$ is a linear transformation given by an $n\times n$ real matrix $A$. Then the mentioned statement can be rewritten as $$|\det A|^{1/n}\overset{\text{(?)}}\le\|A\|, \tag{1} $$ where $\|A\|$ is the spectral norm of $A$, so that $\|A\|^2=\|A^T A\|=c_1$, where $c_1\ge\cdots\ge c_n$ are the (nonnegative) eigenvalues of $A^T A$. We have $$|\det A|^2=\det(A^T A)=c_1\cdots c_n\le c_1^n=\|A\|^{2n}, $$ so that (1) follows.

This reasoning also shows that the equality in (1) holds (iff $c_1=\cdots=c_n$, that is) iff $A$ is a real multiple of an orthogonal matrix.

On the other hand, we can have $\det A=0$ but $\|A\|=1$ (say).

Answer 3

The absolute of Jacobian determinant is bounded by $L^n$: $|\det{J_f(x)}|\le L^n$. It comes from the Lipschitz property and the Hadamard's inequality:

Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a $L$-Lipschitz function.

1. by Lipschitz property: $$|J_f(x)\cdot \mathbf{e}_k|\le L|\mathbf{e}_k|=L\ \tag{1},$$ where $\mathbf{e}_k \>(k=1,2,\cdots,n)$ is the $k$th canonical basis vector in $\mathbb{R}^n$.
2. by Hadamard's inequality, for a matrix $A\in\mathbb{R}^{n\times n}$ with column vectors $\mathbf{a}_k$, it follows: $$|\det{A}|\le \prod_{k=1}^n|\mathbf{a}_k|\ \tag{2}$$ Let $A=J_f(x)$, given by $(1)$ and $(2)$: $$|\det J_f(x)|\le\prod_{k=1}^n|J_f(x)\cdot\mathbf{e}_k|\le\prod_{k=1}^n L|\mathbf{e}_k|=L^n$$

Answer 4

Only an inequality holds, with Lip constant bounding the Jacobian.