# $f\in \mathcal{C}^{\infty}_{c}(\mathbb{R})$ such that $f(0)=0$ there exists $g\in\mathcal{C}^{\infty}_{c}(\mathbb{R})$ such that $f(x)=xg(x)$

distribution-theorygeneral-topology

Let $$f\in \mathcal{C}^{\infty}_{c}(\mathbb{R})$$ such that $$f(0)=0$$ I want to show that there exists $$g\in\mathcal{C}^{\infty}_{c}(\mathbb{R})$$ such that $$f(x)=xg(x)$$. For this I tried to consider the following function
$$g(x)=\frac{f(x)}{x} \qquad g(0)=f'(0)$$
But I cannot show that $$g\in \mathcal{C}^{\infty}_{c}(\mathbb{R})$$. Does this function check the hypotheses?

Define $$g:\mathbb{R}\to\mathbb{R}$$ by $$g(x) = \begin{cases} \dfrac{f(x)}{x} & (x\neq 0), \\ f'(0) & (x=0). \\ \end{cases}$$ This function is obviously continuous for $$x\neq 0.$$ At $$x=0$$ we have $$\lim_{x\to 0} g(x) = \lim_{x\to 0} \frac{f(0+x)-f(0)}{x} = f'(0) = g(0)$$ so $$g$$ is also continuous there. Also, $$\operatorname{supp}g \subseteq \operatorname{supp}f$$ so $$g\in C_c(\mathbb{R}).$$
Now, as reuns suggests, we can notice that $$g(x) = \int_0^1 f'(xt) \, dt.$$ (Remember to check both the case $$x\neq 0$$ and $$x=0$$.)
Derivatives of $$g$$ are given by $$g^{(k)}(x) = \int_0^1 t^k f^{(k+1)}(xt) \, dt$$ which by construction are defined for all $$x$$ and all $$k\in\mathbb{N}$$ since $$f\in C^\infty(\mathbb{R}).$$ Thus $$g\in C^\infty(\mathbb{R}).$$
So, since $$g\in C_c(\mathbb{R})$$ and $$g\in C^\infty(\mathbb{R})$$ we have $$g\in C^\infty_c(\mathbb{R}).$$