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This is theorem 9.24 in Rudin, known as the inverse function theorem:

Suppose $f$ is a continuously differentiable map of an open set $E \subseteq \mathbb{R}^n$ into $\mathbb{R}^n$, $f'(a)$ is invertible for some $a \in E$ and $b= f(a)$. Then

(a) There exists open sets $U,V$ in $\mathbb{R}^n$ with $a \in U, b \in V$ such that $f: U \to V$ is a bijection.

(b)$f^{-1}$ is continuously differentiable.

My question: Can we add the following to the conclusion of the inverse function theorem?

(c) $f'(x)$ is invertible for all $x \in U$?

Looking at Rudin's proof, I think we can, but maybe I'm not too sure.

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  • $\begingroup$ what does "$f'(a)$ is invertible" mean? I don't know what it means for a point to be invertible. $\endgroup$ – mathworker21 Sep 20 '18 at 8:10
  • $\begingroup$ If you consider $f'(a)$ as a linear transformation, that it is an isomorphism. If you consider it as a matrix, that it is non-singular. $f'(a)$ means the same thing as $Df(a)$, if you are familiar with that notation. $\endgroup$ – user370967 Sep 20 '18 at 8:11
  • $\begingroup$ yes but that is just chain rule applied to $f^{-1}\circ f=\operatorname{id}$ using (b) to see $f'(x)$ has inverse $(f^{-1})'(f(x))$ $\endgroup$ – user10354138 Sep 20 '18 at 8:14
  • $\begingroup$ If you use the chain rule, you use that $f^{-1}$ is differentiable at a point which is what has to be proven. $\endgroup$ – user370967 Sep 20 '18 at 8:18
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Yes, (although perhaps for a smaller neighborhood than $U$). Note that $f$ is continuously differentiable. Since $f'(a)$ is invertible, $\det f'(a)\neq0$. But then, by continuity, $\det f'(x)\neq0$ when $x$ is close enough to $a$, and this means that, for those $x$'s, $f'(x)$ is invertible.

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  • $\begingroup$ And can we keep the requirement that $V=f(U)$ is open then? This is my main problem here. $\endgroup$ – user370967 Sep 20 '18 at 8:13
  • $\begingroup$ @Math_QED Yes, that will still be true. $\endgroup$ – José Carlos Santos Sep 20 '18 at 8:14
  • $\begingroup$ @JoséCarlosSantos how do you know that? $\endgroup$ – mathworker21 Sep 20 '18 at 8:14
  • $\begingroup$ I think the $U$ Rudin defines in his proof works to show this. I'd be glad if someone can verify if this works? $\endgroup$ – user370967 Sep 20 '18 at 8:14
  • $\begingroup$ @Math_QED Because the fact that $f'(x)$ is invertible for each $x$ implies that the restriction of $f$ to $U$ is an open mapping. $\endgroup$ – José Carlos Santos Sep 20 '18 at 8:15

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