$f\in \mathcal{C}^{\infty}_{c}(\mathbb{R})$ such that $f(0)=0$ there exists $g\in\mathcal{C}^{\infty}_{c}(\mathbb{R})$ such that $f(x)=xg(x)$ 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?
 A: 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 $g\in C^\infty_c(\mathbb{R}).$
A: If $f$ vanishes outside $[a,b]$, so does $g$. As a quotient of smooth functions, $g$ is smooth for $x\neq0$. At $x=0$ you have
$$
f(x)=f(0)+f’(0)x+\frac 1 {2}f’’(0)x^2+\dots=f’(0)x+\frac 1{2}f”(0)x^2+o(x^2)
$$
by the local Taylor formula. Actually you can expand up to any order $n$ with remainder $o(x^n)$. Dividing by $x$ you get
$$
g(x)=f’(0)+\frac 1 {2}f’’(0)x+o(x)
$$
(Or the corresponding expansion to order $n$) which is infinitely differentiable at zero with $g(0)=f’(0)$ etc.
The reason is: suppose the above relation holds ($n=1$). Setting $x=0$ you obtain $g(0)=f’(0)$. Then, dividing throughout by $x$ and taking the limit as $x\to 0$ you obtain, by definition of “little o” and of derivative
$$
g’(0)=f’’(0)/2.
$$
Using the corresponding expansion with $n=2$, and the previous values and dividing throughout by $x^2$ you get in the limit
$$
g’’(0)=f’’’(0)/3
$$
Etc.
