Are $C^{k}$ manifolds the same as $C^{\infty}$ manifolds? This is a theorem of Hassler Whitney:

For $0<k<\infty$ and any $n$-dimensional $C^k$ manifold the maximal
  atlas contains a $C^\infty$ atlas on the same underlying set.

It seems to me that this theorem says every $C^{k}$ manifold can be thought of as if it were a $C^{\infty}$ manifold proceeding like this:


*

*Start with a given a $C^{k}$ atlas $\mathcal{A}$ on a topological
manifold $M$

*Consider the maximal $C^{k}$ atlas $\overline{\mathcal{A}}$
containing $\mathcal{A}$ (i.e. the atlas containing every
$C^{k}$ chart on $M$ compatible with $\mathcal{A}$)

*Extract a $C^{\infty}$ subatlas $\mathcal{A}_{\infty}$ of
$\overline{\mathcal{A}}$ (it can be done because of Whitney's
theorem)

*Now make $M$ a $C^{\infty}$ manifold  considering the atlas
$\mathcal{A}_{\infty}$
After this maneuver we ended up with two differentiable structures over the same underlying topological manifold $M$: one of $C^{k}$ type given by $\mathcal{A}$ and the other of $C^{\infty}$ type given by $\mathcal{A}_{\infty}$.
My questions are:


*

*Up to what extent are this two differentiable manifolds,
$(M,\mathcal{A})$ and $(M,\mathcal{A}_{\infty})$, the same?

*Is it true that the maximal atlas $\overline{\mathcal{A}_{\infty}}$
generated by $\mathcal{A}_{\infty}$ is the same as
$\overline{\mathcal{A}}$ with all the non $C^{\infty}$ charts
removed?

*Are there examples of $C^{k}$ manifolds for which exists a
$C^{k}$ atlas such that none of its charts is $C^{r}$ for some
$r>k$?

*When studying functions defined on a $C^{k}$ manifold $M$ we
can consider differentials up to order $k$, with this limit $k$
imposed by the $C^{k}$ differentiable structure. Is the theorem of Whitney
telling us that this restriction is artificial since we can get a
$C^{\infty}$ atlas for $M$?
Thanks.
 A: Let $f:\mathbb R^n\to \mathbb R^n$ be a $C^k$ diffeomorphism which is not $C^{k+1}$ and consider the $C^k$ manifold $M$ whose underlying topological space is $\mathbb R^n$ and whose atlas consists of $f$ and the identity map $Id:\mathbb R^n\to \mathbb R^n:x\mapsto x$ .
If you discard one of the maps you obtain two different $C^\infty$ manifolds $M_{Id}$ and $M_{f}$ whose atlas consists of respectively the single homeomorphism $Id$ and the single homeomorphism $f$.
These different manifolds have the same underlying $C^k$ manifold, namely $M$.
This proves that passing from a $C^k$ manifold to a $C^\infty$ manifold is indeed always possible, but not in a canonical way.  The same is true for passing from $C^k$ to $C^{k+1}$.
Remark
The existence of a $C^k$ diffeomorphism $f$ which is not $C^{k+1}$ is probably easy to prove but the only reference I could find (by browsing the web for two minutes...) is for the case  $n=2$ in this probably much too sophisticated article .
A: (1) The identity map from the $C^k$-manifold to its $C^\infty$-refinement is a $C^k$-diffeomorphism. 
(2) I don't understand your question. 
(3) This question does not make sense to me. What does it mean for a map to be more differentiable than the maximum-definable level of differentiability? 
(4) You're asking a vague question here.  Could you be more specific. 
