Let $\Omega \subseteq \mathbb{R}^n$ be neighborhood of $0$ and let $f : \Omega \to R$ be a smooth function, when does there exist a diffeomorphism $c : U \to \Omega$ with $c(0) = 0$ such that $f \circ c$ is real analytic in a neighborhood of $0$? Or more generally the same question, but now for $f : \Omega \to \mathbb{R}^m$?
I had thought of the following examples, which show it is possible for some $f$ and impossible for some other $f$:
If the derivative of $f$ is non-vanishing at $0$, we can use the submersion theorem to find a coordinate transformation such that $f(x_1, ..., x_n) = x_1$. Thus, in this case, there exists a coordinate transformation making $f$ real-analytic.
For $$f(x) = \begin{cases} e^{-1/x} & \text{x > 0} \\ 0 & \text{otherwise} \end{cases}$$, it is clear that for every coordinate transformation $c$ around $0$, all the derivatives of $f \circ c$ will vanish. However, either for $x > 0$ or $x < 0$, $f\circ c(x)$ will be non-zero. So, $f \circ c$ is also not real analytic.
In case the codomain of $f$ is $\mathbb{R}^m$ and $Df$ has constant rank, we can find coordinates such that $f(x_1, ..., x_n) = (x_1, ..., x_k, 0, ..., 0)$. So, $f$ can be made real analytic.
Although, I have the above examples, I struggle to see how one might answer the question in general.
