Can continuous functions be made smooth by changing the smooth structure on the domain? Suppose $M$ is a smooth manifold and $f : M \to \mathbb R$ is a continuous function.
The function $f$ may not be smooth, but does there exist another smooth structure $M'$ (on the same topological manifold $M$) such that $f : M' \to \mathbb R$ is a smooth function?
For example, if $M = \mathbb R$ and $f (x) = x^{\frac 13}$ then $f$ is not smooth w.r.t. the usual smooth structure on $\mathbb R$. However if we take the smooth structure provided by the chart $\varphi (x) = x^{\frac 13}$ then in these coordinates $f \circ \varphi^{-1} (t) = t$, so $f : (\mathbb R, \varphi) \to \mathbb R$ is a smooth function.
As another example, $g (x) = \lvert x \rvert$ is not smooth, but if we take the chart $\psi (x) = \operatorname{sign} (x) \sqrt{\lvert x \rvert}$ then $\psi^{-1} (t) = t \lvert t \rvert $, so $g \circ \psi^{-1} (t) = t^2$, which means $g$ is a smooth function on $(\mathbb R, \psi)$.
 A: Usually not. Let's take $M = \mathbb{R}$ and let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be some continuous nowhere differentiable function. If $M$ has an atlas under which $f$ becomes a smooth map then for any $p \in \mathbb{R}$ you can find a homeomorphism $\phi \colon I \rightarrow U_p$ where $I \subseteq \mathbb{R}$ is an open interval and $U_p$ is an open interval around $p$ such that $f|_{U_p} \circ \phi = g \colon I \rightarrow \mathbb{R}$ is smooth. But then $f|_{U_p} = g \circ \phi^{-1}$ is the composition of a smooth map and a homeomorphism between open intervals. But any such homeomorphism must be monotone and any monotone function is differentiable almost everywhere which then implies that $f|_{U_p}$ must be differentiable almost everywhere, a contradiction.
A: HINT:
For $M=\mathbb{R}$ : consider a continuous function $f$ from $\mathbb{R}$ to $\mathbb{R}$ that is not monotonic on any open interval, for this see nowhere monotonic continuous function.  Such a function cannot be of class $C^1$, even locally. 
$\bf{added:}$ The point is that "not monotonic on any interval" is preserved under restrictions to open intervals of the domain and under compositions local homeomorphism ( charts) that are monotonic. So under the composition with any chart such a map cannot be $C^1$. 
For $M=\mathbb{R}^n$, take $F(x_1, \ldots, x_n) =f(x_1)$, where $f$ is as above. This requires some arguing. 
$\bf{added:}$ The point is that for a non-constant $C^1$ map $F$ from  a smooth manifold $X$ to $\mathbb{R}$,  there exists $\phi \colon (-\epsilon,\epsilon) \to X$ an homeomorphism on the image such that $F\circ \phi$ is strictly increasing ( think of a path around a point with direction the gradient at that point). But there no such path for our function $F$. 
$\bf{Note:}$ it is the case that a function from $\mathbb{R}$ to $\mathbb{R}$ that has no derivative at any point is not monotonic on any interval (since any monotonic function has a derivative almost everywhere - this is a non-trivial result--Lebesgue theorem on monotone functions). However, there also exist functions that are not monotonic on any interval and having a derivative at every point--see link 
