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?

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

distribution-theorygeneral-topology

## Best Answer

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

reunssuggests, 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 $g\in C^\infty_c(\mathbb{R}).$