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my question is:

Let $f\in S(\mathbb{R})$, with $f(0)=0$, then there exists $g\in S(\mathbb{R})$ such that: $$ f(x)=xg(x)\;\text{ for all }\;x\in \mathbb{R}.$$

I need to prove this.

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    $\begingroup$ Check that the function $g(x) := \begin{cases} \dfrac{f(x)}{x}, & x \ne 0, \\ 0, & x = 0 \end{cases}$ is from $S(\mathbb{R})$. $\endgroup$
    – njguliyev
    Oct 21 '13 at 13:47
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    $\begingroup$ The value in $0$ ought to be $f'(0)$ of course. @njguliyev typo'ed. $\endgroup$ Oct 21 '13 at 13:56
  • $\begingroup$ @DanielFischer, as usual... Thanks for correcting. $\endgroup$
    – njguliyev
    Oct 21 '13 at 15:06
  • $\begingroup$ I am trying to understand, thanks $\endgroup$
    – Wmmoreno
    Oct 23 '13 at 5:04
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There is a convenient way to define $g$ which is

$$ g(x)=\int_0^1f'(tx)dt\qquad \forall x\in \mathbb{R} $$

This handles $x\neq 0$ and $x=0$ simultaneously. Then it only remains to differentiate under the integral, etc...

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  • $\begingroup$ The only problem is to prove that we can differentiate under the integral. $\endgroup$ Oct 22 '13 at 13:35
  • $\begingroup$ @TZakrevskiy There is a theorem for that. $\endgroup$
    – Julien
    Oct 22 '13 at 13:47
  • $\begingroup$ Of course; but checking all its conditions - in my opinion - is quite a tedious work, especially for higher order derivatives. $\endgroup$ Oct 22 '13 at 14:00
  • $\begingroup$ @TZakrevskiy If the integral was, say, over $\mathbb{R}$, maybe. But it's over $[0,1]$ and $f$ is smooth. So this really is easy with the integral representation. $\endgroup$
    – Julien
    Oct 22 '13 at 14:25
  • $\begingroup$ I am trying to understand, thanks $\endgroup$
    – Wmmoreno
    Oct 23 '13 at 5:04
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Note that on $\Bbb R\setminus \{0\}$ the function $f(x)/x$ is $\mathcal C^\infty$ and it still satisfies the decreasing properties on infinity. The only thing to prove is that in zero our function $g(x)=f(x)/x$ is still $\mathcal C^\infty$.

First of all, obviously $g(0)=f'(0)$ by L'Hôpital's rule. Then again, you need to show that $$\frac{d^n}{dx^n}(f(x)/x)$$is continuous at zero. Leibnitz rule will give us $$\frac{d^n}{dx^n}(f(x)/x) = \sum_{k=0}^n (-1)^{n-k}(n-k)!\binom{n}{k}\frac{f^{(k)}(x)}{x^{n-k+1}}$$ $$=(-1)^{n}\frac{n!}{x^{n+1}}\sum_{k=0}^n \frac{f^{(k)}(x)(-x)^{k}}{k!}.$$ The sum - easy to recognise - is a Taylor developement of $f(0)$ in the point $x$, hence we can replace it by $f(0) - (-1)^{n+1}\frac{f^{(n+1)}(x)(-x)^{n+1}}{(n+1)!}+\mathcal O(x^{n+2}) $, which results in the following:

$$\frac{d^n}{dx^n}(f(x)/x)= \frac{n!}{x^{n+1}} \left( \frac{f^{(n+1)}(x)(-x)^{n+1}}{(n+1)!}+\mathcal O(x^{n+2})\right).$$ It's evident that the limit $x\to 0$ exist and is equal to $\frac{ f^{(n+1)}(0)}{n+1}$, which concludes the proof.

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  • $\begingroup$ I investigate the taylor expansion can not be applied if there are points of singularity in the domain of the function $\endgroup$
    – Wmmoreno
    Oct 21 '13 at 16:29
  • $\begingroup$ @Wmmoreno The term $\sum_{k=0}^n \frac{f^{(k)}(x)(-x)^{k}}{k!}$ is a Taylor expansion of $f$, which is $\mathcal C^\infty(\Bbb R)$, so no problems with singularities there. $\endgroup$ Oct 21 '13 at 16:37
  • $\begingroup$ I am trying to understand, thanks $\endgroup$
    – Wmmoreno
    Oct 21 '13 at 17:26

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