# What smoothness properties does an implicit function inherit?

Consider the relation $f(x,y)=0$, with $f:\mathbb{R}^n\times\mathbb{R}^m\rightarrow \mathbb{R}^n$. The (standard) implicit value theorem gives you conditions for the existence of a function $g:B\rightarrow \mathbb{R}^m$ such that $f(x,g(x))=0$ for all $x$ in some open ball $B$ around a given point $a\in\mathbb{R}^n$. In addition, if $g$ exists then it inherits certain smoothness properties from $f$, i.e. if $f\in\mathcal{C}^k$ then $g\in\mathcal{C}^k$.

Suppose you already know there that there exists a $g$ such that $f(x,g(x))=0$ for all $x$ in a given open set A. What properties smoothness properties can $g$ be expected to have? More specifically if $f\in\mathcal{C}^k$ is $g\in\mathcal{C}^k$ and if $f$ is globally Lipschitz continuous is $g$ globally Lipschitz continuous?

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Let $f(x,y)=1-y^2$ and $g(x)=-1$ when $x <0$, $g(x)=+1$ when $x \geq 0$. Then $f(x,g(x))=0$ for every $x$. – Siminore Aug 21 '12 at 16:16
Generally some sort of invertibility is required. This is what constrains $g$ sufficiently to give it desirable properties. – copper.hat Aug 21 '12 at 16:28
Take $f(x,y) = x-\arctan y$. Then $f$ is smooth and globally Lipschitz, but an inverse function is not globally Lipschitz. – copper.hat Aug 21 '12 at 16:47
@copper.hat thanks, you're right in the end I'm getting requirements on the invertibility of different Jacobians. – jkn Aug 21 '12 at 18:13
For example, assume $f\in\mathcal{C}^1$. Then $f(x,g(x))=0$, so $f'(x,g(x)) = 0$, $D_x f(x,g(x)) = \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}=0\Rightarrow \frac{\partial g}{\partial x} = \left(\frac{\partial f}{\partial g}\right)^{-1}\frac{\partial f}{\partial x}$. Hence $g\in\mathcal{C}^1$ if and only if $\frac{\partial f}{\partial g}$ is invertible. – jkn Aug 21 '12 at 18:58

@copper.hat Right you are. But as far as I can tell, you wont have that 'ivertibilty in some sense' in general if $f$ is merely Lipschitz. – user20266 Aug 21 '12 at 16:46