# Inverse function theorem: derive stronger form

Let $$E \subseteq \mathbb{R}^n$$ be an open set, and $$f: E \subseteq \mathbb{R}^n \to \mathbb{R}^n$$ be a $$C^r$$ function. If $$Df(x)$$ is invertible for some $$x \in E$$, there are open sets $$U,V$$ of $$\mathbb{R}^n$$ such that $$f:U \to V$$ is a bijection and $$f^{-1}$$ is a $$C^r$$ function.

Is there any way to prove this using either the theorem statement, or the proof of the statement below:

Let $$E \subseteq \mathbb{R}^n$$ be an open set, and $$f: E \subseteq \mathbb{R}^n \to \mathbb{R}^n$$ be a $$C^1$$ function. If $$Df(x)$$ is invertible for some $$x \in E$$, there are open sets $$U,V$$ of $$\mathbb{R}^n$$ such that $$f:U \to V$$ is a bijection and $$f^{-1}$$ is a $$C^1$$ function.

My definitions:

$$f: E \subseteq \mathbb{R}^n \to \mathbb{R}^n$$ is a $$C^1$$ function if all its component functions have continuous partial derivatives on $$E$$.

$$f: E \subseteq \mathbb{R}^n \to \mathbb{R}^n$$ is a $$C^r$$ function $$(r> 1)$$ if all partial derivatives of the component functions are of class $$C^{r-1}$$/

I also know: $$f \in C^1 \iff Df$$ continuous.

Yes, that's also correct. The proof is exactly the same as the proof of the inverse function theorem, almost no difference. Just in the end you will have to show that the inverse is $$C^r$$, this is done by induction.

Edit: Ok, I'll show the idea of the proof. Let's call the inverse $$g$$. We know that $$g$$ is differentiable and by the chain rule we also know how the derivative will look like:

$$D_g(y)=[D_f(g(y))]^{-1}$$

Note that we have a composition of three functions here: $$g$$, $$D_f$$ and the inverse operator. I assume you know that the inverse operator is $$C^\infty$$ and that composition of $$C^r$$ functions is $$C^r$$ as well. So all we have to do is talk about $$g$$ itself. So, by the inverse function theorem you know $$g$$ is continuous. But then we get that $$D_g$$ is continuous as a composition of $$3$$ continuous functions. That means $$g\in C^1$$. But now that means $$D_g$$ is $$C^1$$ as a composition of $$3$$ $$C^1$$ functions. That means $$g\in C^2$$. Continue by induction and you will get $$g$$ is actually $$C^r$$.

• Well, can you show that induction argument then please? This is exactly with what I struggle.
– user370967
Commented Sep 22, 2018 at 20:38
• Done. I didn't write all the details, just the general idea.
– Mark
Commented Sep 22, 2018 at 20:46
• Thank you. How do you define the $C^k$ classes for the inverse operator? This is a matrix map or a linear tranqformation map. I'm only familiar with $C^k$ classes of functions $\mathbb{R}^n \to \mathbb{R}^m$ (all the definitions I know are given in my post).
– user370967
Commented Sep 22, 2018 at 20:49
• Well, you can do a trick and think of $n\times n$ matrix as a vector in $R^{n^2}$: the first $n$ coordinates are the first row, the next $n$ are the second row and so on. And then you get functions that you can deal with.
– Mark
Commented Sep 22, 2018 at 20:58
• Thanks. Do you have a reference for this trick?
– user370967
Commented Sep 22, 2018 at 20:59