I think that the proof of the following statement should be an exercise in inverse function theorem. Can someone please offer some insight?
Prove that a $C^1$ diffeomorphism $f$ with $f$ of class $C^k$ is a
$C^k$ diffeomorphism.
I think some clarifications are in order.
A mapping $f$ between open sets of $\mathbb{R}^n$, $f:u\to v$, is called a
homeomorphism if it is bijective and if $f$ and its inverse
mapping $f^{-1}$ are continuous. A bijective mapping is a $C^k$
diffeomorphism if $f$ and $f^{-1}$ are of class $C^k$.
I have attempted the following argument, but I don't know how to finish it.
The function $f:u\to v$ is a $C^1$ diffeomorphism between open sets
$u$ and $v$ in $\mathbb{R}^n$. By hypothesis, both $f$ and $f^{-1}$ are
$C^1$, i.e., both have continuous partial derivatives of order
1. Let $(x^1,\ldots,x^n)$ be the standard coordinates on
$\mathbb{R}^n$. We say that $f$ is a $C^k$ function if all partial
derivatives of $f$ of orders up to $k$ exist and are
continuous. Hence, it suffices to prove that $f^{-1}$ has
continuous partial derivatives of all orders up to $k$.
The inverse function theorem states that a differentiable function
$f$ is invertible in some neighborhood of a point $p\in u$ if and
only if its Jacobian matrix at $p$ is invertible.
Best Answer
The inverse function theorem states that if $f$ is a $C^k$ function, and $(Df)(a)$ (the derivative of $f$ at $a$) is invertible, then $f$ is invertible on a neighbourhood of $a$. Moreover, this inverse is of class $C^k$.
If $f:X\rightarrow Y$ is a $C^1$ diffeomorphism, $f^{-1}$ is of class $C^1$ by definition. It follows that $(Df)(a)$ is invertible for all $a\in X$ because you can show that $((Df)(a))^{-1} = (Df^{-1})(f(a))$.
Now we know that $f$ is of class $C^k$ and its derivative is invertible on the entire domain, so $f$ is a $C^k$ diffeomorphism from $X$ onto $Y$.