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.
Asked
Active
Viewed 119 times
2
-
1Check that the function $g(x) := \begin{cases} \dfrac{f(x)}{x}, & x \ne 0, \ 0, & x = 0 \end{cases}$ is from $S(\mathbb{R})$. – njguliyev Oct 21 '13 at 13:47
-
2The value in $0$ ought to be $f'(0)$ of course. @njguliyev typo'ed. – Daniel Fischer Oct 21 '13 at 13:56
-
@DanielFischer, as usual... Thanks for correcting. – njguliyev Oct 21 '13 at 15:06
-
I am trying to understand, thanks – Wmmoreno Oct 23 '13 at 05:04
2 Answers
1
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...
Julien
- 44,791
-
The only problem is to prove that we can differentiate under the integral. – TZakrevskiy Oct 22 '13 at 13:35
-
-
Of course; but checking all its conditions - in my opinion - is quite a tedious work, especially for higher order derivatives. – TZakrevskiy Oct 22 '13 at 14:00
-
@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. – Julien Oct 22 '13 at 14:25
-
0
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.
TZakrevskiy
- 22,980
-
I investigate the taylor expansion can not be applied if there are points of singularity in the domain of the function – Wmmoreno Oct 21 '13 at 16:29
-
@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. – TZakrevskiy Oct 21 '13 at 16:37
-