0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Aaron Hoffman Apr 10 '13 at 12:15
  • $\begingroup$ ok, you are right! $\endgroup$ – Andy Teich Apr 10 '13 at 12:42
  • $\begingroup$ I changed the question! $\endgroup$ – Andy Teich Apr 10 '13 at 12:48
2
$\begingroup$

Let me quote the simplest and most classical result for a global inverse function theorem, due to Hadamard and Plastock (see L. Nirenberg, Topics in Nonlinear functional analysis, Courant LN,6, 2001).

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 $$ \int_0^{+\infty}\inf_{\vert x\vert=r}\Vert(F'(x))^{-1}\Vert^{-1} dr=+\infty. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.