TITLE: basic exercise about Schwartz spaces
QUESTION [4 upvotes]: Let $f \in S(R)$ (the Schwartz space of rapidly decaying functions) such that $f(0)=0.$ Show that exists $g \in S(R) $such  that $f(x) = xg(x)$.
My try :
By the calculus fundamental theorem
$$ f(x) = \displaystyle\int_{0}^{1} \displaystyle\frac{d}{dt} f(tx) \ dt = \displaystyle\int_{0}^{1}  f^{'}(tx) x \ dt = x \displaystyle\int_{0}^{1}  f^{'}(tx)  \ dt.$$
Define $g(x) = \displaystyle\int_{0}^{1}  f^{'}(tx)  \ dt$. I know how prove that $g \in C^{\infty}(R).$
But to conclude that $g \in S(R)$ i need to show that $\displaystyle\sup_{x \in R} |x^{\alpha} g^{(\beta)}(x)| < \infty$ for all $(\alpha,\beta)\in N \times N.$
I dont know how to do that .
For example for the situation when $\alpha$ is arbitrary and $\beta = 0$ my best is this :
$$|x^{\alpha} g(x)| = |\displaystyle\int_{0}^{1}  f^{'}(tx)  \ dt|\leq \displaystyle\int_{0}^{1}|x^{\alpha } f^{'}(tx)| \ dt \leq |x|^{\alpha} \displaystyle\sup_{t \in [0,1]} |f^{'}(tx)|$$
someone can give me a hint ?
Thanks in advance.

REPLY [0 votes]: Hint: By the Mean Value Theorem for Integrals, there is $c\in(0,1)$ such that
$$\int_0^1|x^\alpha D^\beta f'(tx)|\;dt=|x^\alpha D^\beta f'(cx)|.$$
Solution:
By the hint,

$\displaystyle |x^\alpha g^{(\beta)}(x)|=\left|\int_0^1x^\alpha D^\beta f'(tx)\;dt\right|\leq \int_0^1|x^\alpha D^\beta f'(tx)|\;dt=\frac{1}{c^{|\alpha|}}|(cx)^\alpha D^\beta f'(cx)|$

and thus

 $\displaystyle\sup_{x\in\mathbb{R}^n}|x^\alpha g^{(\beta)}(x)|\leq \sup_{x\in\mathbb{R}^n}\frac{1}{c^{|\alpha|}}|(cx)^\alpha D^\beta f'(cx)|= \frac{1}{c^{|\alpha|}}\|f'\|_{\alpha,\beta}.$

Remark: The same argument can be used to prove that if $f\in\mathcal{S}(\mathbb R^n)$  and $f(0)=0$ then there are $g_1,...,g_n\in\mathcal{S}(\mathbb{R}^n)$ such that
$$f(x)=\sum_{j=1}^nx_jg_j(x),\qquad\forall\ x=(x_1,...,x_n)\in\mathbb R^n.$$
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