Suppose I have a function $F : \mathbb{R}^n \rightarrow \mathbb{R}^n$ and that $F$ is $k$ times continuously differntiable.
Let $V$ be an open subset of $\mathbb{R}^n$ for which $F^{-1}$ is well defined.
I am interested in learning how to show
$F^{-1}: V \rightarrow \mathbb{R}^n$ is $k$ times continuously differntiable as well. I would greatly appreciate any comments. Thank you very much.
PS I forgot to add that I am also assuming the Jacobian of $F$ is non-singular on all of $F^{-1}(V)$.
On
As pointed in the comments, the result is false as stated, but true if you assume that ${\sf D}F(x)$ is non-singular at each point. In this case, the result follows from the Inverse Function Theorem. Here's a counter-example in the general case: $F(x) = x^3$ is a $C^\infty$ bijection, but $F^{-1}(x) = \sqrt[3]{x}$ is not differentiable at $x=0$ (note that the Inverse Function Theorem says that $F^{-1}$ is $C^\infty$ near enough each point $x \neq 0$).
You need to do part of the proof of the inverse function theorem, namely to prove that $F^{-1}$ is differentiable, with derivative $$DF^{-1}(x) = \big(DF(F^{-1}(x))\big)^{-1}.$$
Since $F$ is $C^1$, $DF^{-1}$ is the composition of continuous functions and hence is continuous.
You then can bootstrap: If $F$ is $C^2$, this formula tells us that $DF^{-1}$ is the composition of two $C^1$ functions, hence $C^1$. So $F^{-1}$ is $C^2$ as well. Continue by induction.
(For this point you can see the beginning of my lecture. For the actual proof of the bulk of the inverse function theorem, see the preceding lecture.)