- #1
- 7,007
- 10,463
Hi, everyone:
I was wondering if anyone knew of any extensions to the inverse function theorem
to this effect:
If f is a differentiable map, and df(x) is non-zero, then the IFT guarantees
there is a nhood (neighborhood) U_x containing x , such that [itex] f|(U_x) [/itex]
a diffeom. [itex] U_x-->f(U_x) [/itex].
Now, under what conditions on f , can we be guaranteed to have that f
has a global differentiable inverse?, i.e, f has a global inverse [itex] f^-1 [/itex] and
[itex] f^-1 [/itex] is differentiable . I imagine df(x) not 0 for all x is necessary, but not
sure that it is sufficient.
For simple cases like f(x)=[itex] x^2 [/itex] this is true in any interval [a,b]
not containing zero, and it may be relatively easy to do a proof of the claim
above for maps f:IR->IR . But I have no idea how well this would generalize
to maps f:[itex] R^n\to\mathbb{R^n} [/itex]
Any Ideas?
Thanks.
I was wondering if anyone knew of any extensions to the inverse function theorem
to this effect:
If f is a differentiable map, and df(x) is non-zero, then the IFT guarantees
there is a nhood (neighborhood) U_x containing x , such that [itex] f|(U_x) [/itex]
a diffeom. [itex] U_x-->f(U_x) [/itex].
Now, under what conditions on f , can we be guaranteed to have that f
has a global differentiable inverse?, i.e, f has a global inverse [itex] f^-1 [/itex] and
[itex] f^-1 [/itex] is differentiable . I imagine df(x) not 0 for all x is necessary, but not
sure that it is sufficient.
For simple cases like f(x)=[itex] x^2 [/itex] this is true in any interval [a,b]
not containing zero, and it may be relatively easy to do a proof of the claim
above for maps f:IR->IR . But I have no idea how well this would generalize
to maps f:[itex] R^n\to\mathbb{R^n} [/itex]
Any Ideas?
Thanks.