# Extension of Inverse Function Theorem from $\mathbb{R}$ to $\mathbb{R^n}$

Consider the Inverse Function Theorem in $$\mathbb{R}$$:

Let $$O ∈\mathbb{R}$$ be open for $$f:O → \mathbb{R}$$. If $$f$$ is continuously differentiable, and for a $$x_0 ∈O$$, $$f'\left(x_0\right) \ne 0$$. Then there exist an open interval $$\boldsymbol{I}$$ about $$x_0$$ and open interval $$\boldsymbol{J}$$ about the image of $$f\left(x_0\right)$$ such that $$f: \boldsymbol{I} → \boldsymbol{J}$$ is one to one and onto. Futhermore, $$f^{-1}: \boldsymbol{J} → \boldsymbol{I}$$ is continuously differentiable, and if $$y ∈\boldsymbol{J}$$, $$x ∈\boldsymbol{I}$$ such that $$f\left(x\right) = y$$, then $$\begin{split} (f^{-1})'(y) = \frac{1}{f'(x)} \end{split}$$

Compared with the Inverse Function Theorem in $$\mathbb{R^n}$$:

Let $$O ∈\mathbb{R^n}$$ be open for $$\boldsymbol{F}:O → \mathbb{R^n}$$. If $$\boldsymbol{F}$$ is continuously differentiable, and for a $$\boldsymbol{x_*} ∈O$$, $$\boldsymbol{DF}\left(\boldsymbol{x_*}\right)$$ is invertible. Then there exist a neighborhood $$\boldsymbol{I}$$ about $$\boldsymbol{x_*}$$ and an neighborhood $$\boldsymbol{J}$$ about the image of $$\boldsymbol{F}\left(\boldsymbol{x_*}\right)$$ such that $$\boldsymbol{F}: \boldsymbol{I} → \boldsymbol{J}$$ is one to one and onto. Futhermore, $$\boldsymbol{F}^{-1}: \boldsymbol{J} → \boldsymbol{I}$$ is continuously differentiable, and if $$\boldsymbol{y} ∈\boldsymbol{J}$$, $$\boldsymbol{x} ∈\boldsymbol{I}$$ such that $$\boldsymbol{F}\left(\boldsymbol{x}\right) = \boldsymbol{y}$$, then $$\begin{split} \boldsymbol{DF}^{-1}(\boldsymbol{y}) = [\boldsymbol{DF}(\boldsymbol{x})]^{-1} \end{split}$$

I noticed that for the Inverse Function Theorem in $$\mathbb{R^n}$$, it is possible for all $$\boldsymbol{x} ∈O$$ $$(O ∈\mathbb{R^n})$$ to satisfy the requirements of the theorem, yet still have $$\boldsymbol{F}: O → \mathbb{R^n}$$ be not one to one. For example: $$$$\boldsymbol{F}(x,y) = \left(e^x\cos \left(y\right),e^x\sin \left(y\right)\right)$$$$ $$\boldsymbol{F}$$ is continuously differentiable, and for any $$\boldsymbol{x} ∈\mathbb{R^n}$$, the determinant of $$\boldsymbol{DF}\left(\boldsymbol{x}\right)$$ does not equal to $$0$$. Yet, $$\boldsymbol{F}$$ is not one to one (e.g. the points $$\left(0,0\right)$$ and $$\left(0,2π\right)$$ produce the same result). This, I presume, is due to the fact that the Inverse Function Theorem is only true for the functon $$\boldsymbol{F}$$ on its restrictions (from $$\boldsymbol{I}$$ to $$\boldsymbol{J}$$).

Question:

1) Could this also happen for the Inverse Function Theorem in $$\mathbb{R}$$? That is, could $$f: O → \mathbb{R}$$ be continuously differentiable, and $$f'\left(x\right) \ne 0$$ for all $$x ∈\mathbb{R}$$, yet still have $$f: O → \mathbb{R}$$ be not one to one? If no, why are they different?

2) Could this also happen to the onto property of the function?

To be explicit, I am taking continuous differentiability here to mean that the the function is continuous, and has a continuous first derivative.

If $f^\prime (x) \neq 0$ for all $x\in O$ by continuity it must be the case that $f^\prime (x)>0$ for all $x\in O$ or $f^\prime (x) <0$ for all $x\in O$.

In either case, $f$ is strictly monotonic and hence must be one-to-one.

This function may not be onto. Consider any affine function defined on an open interval that is not the entire real line.

I don't have a good explanation for why the difference, but my suspicion is that, in multiple dimensions, you can have saddle/inflection points in some dimensions but still have the Jacobian be invertible. However, this is decidedly not the case in one variable.

• That doesn't satisfy the requirement that $f'(x) \ne 0$ for all $x$ – bli00 Dec 12 '16 at 3:25
• Yes, of course. Was being sloppy. Apologies. – Theoretical Economist Dec 12 '16 at 3:26
• @thestateofmay Answer edited. – Theoretical Economist Dec 12 '16 at 3:38