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Which of the following functions belong to $S(R)=\{ f \epsilon C^\infty(R): |x^\alpha| \times |D^\beta f(x)| \leq C_{\alpha, \beta} \}$?

a) $f(x)=\frac{sin(x)}{x}$ b) $f(x)=1-e^{-x^{-2}}$ with $f(x)=1$ at $x=0$ c) $f(x)=x^5e^{-x^2}$ d) $f(x)=x^5e^{-|x|}$

I am not sure how to such the functions above are in $S(R^n)$. I do know that all the functions are $C^\infty(R)$ since sin(x), \frac{1}{x}, $e^x$ functions are in $C^\infty(R)$.

Username Unknown
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    Naively, $\phi \in C^{\infty}(\mathbb{R})$ will be in $\mathcal{S}(\mathbb{R})$ if, and only if, every derivative of $\phi$ goes to $0$ faster than every $\frac{1}{x^{\alpha}}$ when $x \to \infty$. Clearly, for a), $x^2 f(x)$ is unbounded, hence $f \notin \mathcal{S}(\mathbb{R})$. For b), $f'$ does not go to $0$ faster than $\frac{1}{x^4}$ hence $f \notin \mathcal{S}(\mathbb{R})$. However I believe that the $f$ of c) is in it, because every derivative has the form $p(x) e^{-x^2}$, where $p(x)$ is a polynomial, and the exponential will always kill any polynomial. – Amateur Jul 05 '14 at 04:35
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    As for d), are you sure $f \in C^{\infty}(\mathbb{R})$ ? $e^{-|x|}$ admits no derivative at $x=0$ so some sufficiently high derivative of $f$ might not exist at $x=0$. – Amateur Jul 05 '14 at 04:38
  • @Amateur Is there a formal proof for thes – Username Unknown Jul 07 '14 at 00:26

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