space of schwartz, problem my question is:
Let $f\in S(\mathbb{R})$, with $f(0)=0$, then there exists $g\in S(\mathbb{R})$ such that: 
$$ f(x)=xg(x)\;\text{ for all }\;x\in \mathbb{R}.$$
I need to prove this.
 A: There is a convenient way to define $g$ which is

$$
g(x)=\int_0^1f'(tx)dt\qquad \forall x\in \mathbb{R}
$$

This handles $x\neq 0$ and $x=0$ simultaneously. Then it only remains to differentiate under the integral, etc...
A: Note that on $\Bbb R\setminus \{0\}$ the function $f(x)/x$ is $\mathcal C^\infty$ and it still satisfies the decreasing properties on infinity. The only thing to prove is that in zero our function $g(x)=f(x)/x$ is still $\mathcal C^\infty$.
First of all, obviously $g(0)=f'(0)$ by L'Hôpital's rule. Then again, you need to show that $$\frac{d^n}{dx^n}(f(x)/x)$$is continuous at zero. Leibnitz rule will give us
$$\frac{d^n}{dx^n}(f(x)/x) = \sum_{k=0}^n (-1)^{n-k}(n-k)!\binom{n}{k}\frac{f^{(k)}(x)}{x^{n-k+1}}$$
$$=(-1)^{n}\frac{n!}{x^{n+1}}\sum_{k=0}^n \frac{f^{(k)}(x)(-x)^{k}}{k!}.$$
The sum - easy to recognise - is a Taylor developement of $f(0)$ in the point $x$, hence we can replace it by $f(0) - (-1)^{n+1}\frac{f^{(n+1)}(x)(-x)^{n+1}}{(n+1)!}+\mathcal O(x^{n+2}) $, which results in 
the following:
$$\frac{d^n}{dx^n}(f(x)/x)= \frac{n!}{x^{n+1}} \left(  \frac{f^{(n+1)}(x)(-x)^{n+1}}{(n+1)!}+\mathcal O(x^{n+2})\right).$$ It's evident that the limit $x\to 0$ exist and is equal to $\frac{ f^{(n+1)}(0)}{n+1}$, which concludes the proof.
