I'm starting studying differential topology and I somehow arrived to the following (which I feel is wrong).
If $M,N$ are $C^r$ manifolds and $f:M\to N$ is a $C^r$ diffeomorphism then $f$ is actually a $C^\infty$ diffeomorphism (*) ($r>0$)
I couldn't find an error in my proof but the statement looks wrong because this would imply that a differentiable function $f:V\to W$ between open subsets of $\mathbb{R}$ with continuous derivative actually has all derivatives and this doesn't look right.
So the question is:
Is (*) right or is there a counterexample?
The definition of $C^\infty$ we are using is: A $C^\infty$ map is a function $f:M\to N$ such that for each $p\in M$ there is an atlas $(U,x)$ of $M$, an atlas $(V,y)$ of $N$ such that $p\in U$, $f(p)\in V$ and $y\circ f\circ x^{-1}$ is a $C^\infty$ map (between open subsets of some $\mathbb{R}^n$). A $C^\infty$ diffeomorphism is a bijective function $f:M\to N$ such that both $f$ and $f^{-1}$ are $C^\infty$ maps.
If your manifolds are only endowed with $C^r$ atlases, it doesn't make sense to talk about derivatives of higher order than $r$ - since you can't differentiate the transition maps enough, different charts will not necessarily agree on these derivatives.
If you require $M,N$ to be $C^\infty$ so that the question makes more sense, the answer is in the negative. A simple counterexample is the $C^1$ diffeomorphism $f:\mathbb R \to \mathbb R$ given by $f(x) = x(|x|+1)$, which has continuous derivative $f(x)=2|x|+1$ but is not even $C^2$.