# Global Implicit Function Theorem

Let $F:\mathbb R^2\rightarrow\mathbb R$ be a measurable function. Under what conditions on $F$ does there exist a function $\theta:\mathbb R^2\rightarrow\mathbb R$ such that

$F(x,\theta(z,x))=z$ for all $(x,y,z)\in\mathbb R^3$ with $F(x,y)=z$,

$\theta(\cdot,x):\mathbb R\rightarrow\mathbb R$ is continuous for all $x\in\mathbb R$,

$\theta(z,\cdot):\mathbb R\rightarrow\mathbb R$ is measurable for all $z\in\mathbb R.$

Local results can be found for example here, which allows for quite general functions $F$. References are welcome...

• Why not look at $G: \mathbb{R}^3 \to \mathbb{R}$ given by $G(x,y,z) = F(x,y)-z$ and seek $\theta(x,z)$ so that $G(x,\theta(x,z),z) = 0$ describes all roots of $G$ in $\mathbb{R}^3$? You now at least are only worrying about a singly global IFT. – Aaron Hoffman Apr 10 '13 at 12:15
• ok, you are right! – Andy Teich Apr 10 '13 at 12:42
• I changed the question! – Andy Teich Apr 10 '13 at 12:48

Theorem. Let $F:\mathbb R^n\rightarrow \mathbb R^n$ be a $C^1$ mapping such that $\forall x\in \mathbb R^n, \det F'(x)\not=0$. Then $F$ is a global $C^1$ diffeomorphism if $$\int_0^{+\infty}\inf_{\vert x\vert=r}\Vert(F'(x))^{-1}\Vert^{-1} dr=+\infty.$$