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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?

Thanks in advance.

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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
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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
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@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

1 Answer 1

The implicit function theorem, assuming it's assumptions are satisfied, says there is a locally unique solution to the equation you wrote down. So you can expect exactly what the theorem claims, not more, not less, cause you get exactly what the theorem claims. In Siminores example the assumptions of the theorem are violated.

Oh, and I don't know of a Lipschitz version of the theorem and doubt there is one.

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There are many extensions of the implicit & inverse function theorems, but they involve more technical machinery. For example, see Clarke's book "Optimization and Nonsmooth Analysis", look at the corollary in Section 7.1.3 on the implicit function theorem. Ultimately, something needs to be invertible in some sense. –  copper.hat Aug 21 '12 at 16:26
    
@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

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