# basic exercise about Schwartz spaces

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 ?

• I think this is easier to prove if you use $g(x)=\{ \frac{f(x)}{x}$ if $x\neq 0, f'(0)$ if $x=0$ (You can probably show this is the same as your g but it's easier to just start over with this one) – mathematician Nov 20 '13 at 1:59
• @Leandro Tavares : yes, you know exactly what $g$ has to be (see above comment) so why not use i? – Stefan Smith Nov 20 '13 at 3:24

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 $x$.

• If i use your idea to show that $sup_{x \in R}|g^{'}(x) |< \infty$, i will have this : $|g^{'}(x)| \leq |f^{'}(x)x^{-1}| + |f(x)x^{-2}|$ (by the product rule and the triangular inequality) . How can i bound the right hand side of the inequality ? – math student Nov 20 '13 at 12:41
• @LeandroTavares $f$ is a Schwartz function, use the definition. – user103402 Nov 20 '13 at 12:44
• I might be missing something but I'm unsure about this answer. The boundedness of $\left|x^{k}f(x)\right|$ is only in issue if $k<0$ by the definition of $f\in {\cal S}(\mathbb{R})$. So, assume henceforth that $k<0$. In this case, the boundedness of $\left|x^kf(x)\right|$ away from $0$ is evident. Unfortunately, near $x=0$, there seems to be an issue. If $f(0)=0$ as assumed, then there is no issue if $k=-1$, i.e., $\left|\frac{f(x)}{x}\right|$ is bounded near $0$. However, how is $\left|x^kf(x)\right|$ even defined at $0$ if $k<-2$ and the order of vanishing of $f$ at $0$ is $1$? – Amitesh Datta Jul 6 '14 at 12:05

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 $$f(x)=\sum_{j=1}^nx_jg_j(x),\qquad\forall\ x=(x_1,...,x_n)\in\mathbb R^n.$$