Is a differentiable function the product of differentiable functions subject to certain conditions? Honestly this is just a case of me not being great at constructing 'pathological' differentiable functions and I am not sure how to search for what I am looking for. This is related to an algebraic method of constructing cotangents that I have seen before, but it has been a long time so I don't remember the details.
Let $D$ be the set of functions $f$ on $\Bbb{R}$ which are differentiable at $0$ and have $f(0)=0$.
Suppose $f_0\in D$ has $f_0'(0)=0$.
Let $D\cdot D$ be the set of finite sums of products of elements of $D$. (i.e. the product ideal)
Is it the case that $f_0\in D\cdot D$? If not is there a counterexample?
It is easy to show that $f_0\in C\cdot D$ where $C$ is the set of continuous functions $f$ with $f(0)=0$ but I can't see if it is true in the restricted case.
Any help or references would be appreciated
 A: Every function   in $D$ satisfies $\limsup_{x \to 0} f(x)/x <\infty$,
so every function
$$h(x)=\sum_{i=1}^n f_i g_i$$ in $D \cdot D$ must satisfy
$$\limsup_{x \to 0} h(x)/x^2 <\infty \,,$$
so in particular the function $x^{3/2}$ proposed by Ancar is not in $ D \cdot D$.
A: Sorry about the confusion earlier. This should work assuming you are only interested in the question can $f_0$ be $gh$ with $g,h\in D$.
Take $f_0(x)=|x|\cdot x$. Clearly $f_0\in D$ and $\lim_{x\to 0} \frac{f_0(x)}{x}=\lim_{x\to 0}|x|=0$. Now assume $f_0=gh$ for some $g,h\in D$.
Fix $\varepsilon>0$ such that $2\varepsilon(|g'(0)|+|h'(0)|+2\varepsilon)<1$. Then on some small interval $(-\delta,\delta)$ we must have that $\frac{g(x)}{x}=g'(0)+\alpha(x)$ with $|\alpha(x)|<\varepsilon$ and similarly $\frac{h(x)}{x}=h'(0)+\beta(x)$ with $|\beta(x)|<\varepsilon$.
Multiplying, we get that $\frac{|x|}{x}=g'(0)h'(0)+\alpha(x)h'(0)+\beta(x)g'(0)+\alpha(x)\beta(x)$ for all $x\in(-\delta,\delta)$. But note that the LHS can take the values $\pm 1$ while the range of values on the RHS is less than $1$ by choice of $\varepsilon$, contradiction.
