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.
To show the decay of derivatives, it's easier to use the formula $g(x)=x^{-1}f(x)$ (which is valid away from $0$). Indeed, the $n$th derivative of $g$ is $$\sum_{k=0}^n \binom{n}{k} (x^{-1})^{(k)} f^{(n-k)}(x)$$ Multiplying this by $x^{\beta}$, you get a finite sum where each term is some derivative of $f$ multiplied by some power of $x$.