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I am reading the proof in Buck's Advanced Calculus of the inverse function theorem, on p. 359. The way he proves it is to show that $(DF)_{p_0}^{-1}$ satisfies

$$F^{-1}(p_0 + h) - F^{-1}(p_0) = (DF)_{p_0}^{-1}(h) + o(h).$$

Therefore $(DF)_{p_0}^{-1}$ must be the differential of $F^{-1}$ at the point $p_0$, and since $(DF)^{-1}_p$ is a rational function of the entries of $(DF)_p$ (with denominator = the Jacobian, which is nonzero), $F^{-1}$ must be $C^1$, since $F$ is assumed to be $C^1$.

My Question: This has only shown that

$$F \in C^1 \Rightarrow F^{-1}\in C^1.$$

It has not shown the result that if $F$ is $C^n$, then $F^{-1}$ is $C^n$. If we want to maintain the structure of this proof, is it easy enough to get that extra result? Or is this stronger result more easily shown with another style of proof altogether? (I am aware there are other methods, based on the contraction principle, or a fixed point theorem.)

Eric Auld
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  • This can be shown using the definition of the inverse as the adjugate times the inverse of the determinant. – Alex Youcis Aug 22 '13 at 04:50
  • @AlexYoucis Can you clarify? I am aware of what you mean, but I'm not sure how that can be leveraged to show $F^{-1}\in C^n$. – Eric Auld Aug 22 '13 at 04:53
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    Because the inverse of the Jaobian, is the Jacobian of the inverse. From the fact you know it's C^1, you get the Jacobian exists. But, since you can express the entries of the Jacobian of $F^{-1}$ as rational functions of the entries of the Jacobian of $F$, you get that they are in the same differentiability class. – Alex Youcis Aug 22 '13 at 05:30
  • Thank you, that is helpful. – Eric Auld Aug 22 '13 at 05:37
  • Same deal as with the other post. – Alex Youcis Aug 22 '13 at 06:03

1 Answers1

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You know that the Jacobian of $F^{-1}$ (which exists since $F$ is $C^1$) is the inverse of the Jacobian of $F$. Since the inverse of the Jacobian of $F^{-1}$ has entries which can be expressed as rational functions in the entries of $\text{Jac}_F$, you can see that $F$ and $F^{-1}$ are in the same differentiability class.

Alex Youcis
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