I'd like to know if the following statement is true:
For any $f \in \mathcal{C}^\infty(\mathbb{R})$ such that $f(0) = 0$, then $f(x)/x$ is smooth at $0$.
- if $f$ is analytic, then this is clear by simply Taylor-expanding it.
- hence, my question is for when $f$ is non-analytic. I am hoping that, maybe the fact that derivatives of $f$ look bizarre means that there won't be a singularity at $0$, even after dividing by $x$. (Which seems to be the case for the prototypical example of smooth but non-analytic function, the $\exp(-1/x)$ example).
I wonder if this can be done through some sort of polynomial approximation scheme, but am not at all versed in the topic.
It is a very bizarre question, and I am not a priori convinced it is true. However, this result seems to be used in a paper I am reading. Indeed:
I am working in the space of germs $\mathcal{C}_0^\infty(\mathbb{R})$ at $0$, which I treat as an $\mathbb{R}$-vector space. I consider the subspace:
$V =$ {$ax + K | a \in \mathcal{C}_0^\infty(\mathbb{R}), K \in \mathbb{R}$} $\subset \mathcal{C}_0^\infty(\mathbb{R})$
They claim, in Example 2 of §1 of this paper (whose DOI you might find useful if you want to download it; but they do not say much more), that $\dim_\mathbb{R}(\mathcal{C}_0^\infty(\mathbb{R})/V) = 0$, which seems to imply that $V = \mathcal{C}_0^\infty(\mathbb{R})$. Which in turn seems to indicate that any germ in $\mathcal{C}_0^\infty(\mathbb{R})$ can be written in the form of an element of $V$.
It is clearly true that any analytic germ can be put in this form. But since there definitely exist non-analytic germs, I don't see why that statement would hold; unless the question I asked holds.
Indeed, if for every non-analytic function $f$ such that $f(0) = 0$, we have that $f(x)/x$ is smooth at $0$, then this allows us to put any germ $f \in \mathcal{C}_0^\infty(\mathbb{R})$ in the form $f(x) = x \dfrac{f(x)-f(0)}{x} + f(0)$, so that $f \in V$ as required.
Hence: is this true?
