For $F\in C^1(\mathbb{R}^n)$, if its Jacobian is singular at least one point, could its inverse be Lipschitz?

Question

Consider map $F \in C^1(\mathbb{R}^n)$, if there exists $x_0$ such that Jacobian $JF(x_0)$ is singular, we know that the inverse $F^{-1}$ may exist (e.g. $x\mapsto x^3$) and if it exists, it cannot be also in $C^1(\mathbb{R}^n)$ (because it is not differentiable at $x_0$). But the question is, in this case, could the inverse be Lipschitz?

I can neither prove that the inverse is definitely not Lipschitz nor provide an example. Any hints? Thanks!

Answer 1

The inverse is not Lipschitz. Assume:

• $F\colon\Bbb R^n\to \Bbb R^n$ has an inverse $G\colon\Bbb R^n\to \Bbb R^n$;
• $F(0)=0$;
• $DF(x_0)$ is not invertible. Then $G$ isn't Lipschitz near $0$.

Proof. Let $v\in\Bbb R^n$ be such that $DF(x_0)\cdot v=0$ (the existence of such $v$ follows from the assumption on $DF(x_0)$. This means that the directional derivative at $0$ in the direction $v$ is zero, that is $$0= \lim_{h\to 0}\frac{F(0+hv)-F(0)}{h}=\lim_{h\to 0}\frac{F(hv)}{h}.\tag{*}$$ Let $y_h:=F(hv)$. Then $y_h\to 0$ as $h\to 0$ and $hv=G(y_h)$. From (*) we get $$ \lim_{h\to 0}\frac{\|G(y_h)-G(0)\|}{\|y_h-0\|} = \lim_{h\to 0}\frac{|h|}{\|F(hv)\|}\cdot \|v\|\to\infty.$$