# Diffeomorphism, from class $C^1$ to $C^k$

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.

• You're right, it follows directly from the inverse function theorem. – Pedro Mar 25 '17 at 2:19
• @Pedro Thank you. I have edited the question to include my attempt at a proof. Could you please offer more insight? – khalatnikov Mar 26 '17 at 7:44

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 $Y$.
• Thank you very much. I think the IVT states that if $f$ has a derivative, then $f$ is locally invertible if and only if its derivative is locally invertible. Hence, by induction we obtain that if $f$ is $C^k$ at some point and is locally invertible, then it's a $C^k$ local diffeomorphism. The induction step is what I was missing. – khalatnikov Mar 26 '17 at 3:37
• I think that in the case where $f$ is a $C^1$ diffeomorphism between some open subsets of $\mathbb{R}^n$, your first statement becomes quite nontrivial, if we have to start we the following version of the inverse function theorem 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''. Can you offer some insight on how to prove the strong version of IVT that you quoted in this case? We can understand $C^k$ as meaning that partial derivatives of orders up to $k$ exist and are continuous. – khalatnikov Mar 26 '17 at 9:33