Characterization of $\mathscr{S}(\mathbb R^n)$? Consider the vector space $$\displaystyle\mathscr{S}(\mathbb R^n)=\{f\in C^\infty(\mathbb R^n): \lim_{|x|\to \infty} |x^\alpha \partial^\beta \phi(x)|=0, \forall \alpha, \beta\in\mathbb N_0^n\}.$$
I'd like to show $f\in C^\infty(\mathbb R^n)$ is an element of $\mathscr{S}(\mathbb R^n)$ if and only if $$\displaystyle \sup_{x\in\mathbb R^n} |x^\alpha \partial^\beta f(x)|<\infty.$$
I proved the implication $(\Rightarrow)$ as follows:
Let $\varepsilon>0$. There exists $M>0$ such that $|x|>M\Rightarrow |x^\alpha \partial^\beta f(x)|<\varepsilon$. Now suppose $x\in\mathbb R^n$ is such that $|x|\leq M$. Then, $$|x^\alpha \partial^\beta f(x)|\leq |x|^{|\alpha|}|\partial^\beta f(x)|\leq M^{|\alpha|}\sup_{x\in \overline{B(0, M)}}|\partial^\beta f(x)|.$$ Setting $C_{\alpha, \beta}=\max\{\varepsilon, M^{|\alpha|} \displaystyle\sup_{x\in \overline{B(0, M)}}|\partial^\beta f(x)|\}$ we find $|x^\alpha \partial^\beta f(x)|\leq C_{\alpha, \beta}$ for all $x\in\mathbb R^n$ what implies $\displaystyle\sup_{x\in\mathbb R^n} |x^\alpha \partial^\beta f(x)|<\infty$.
However, I wasn't manage to prove the implication $(\Leftarrow)$. Any hint?
 A: Hint. The point for $(\Leftarrow)$ is, that you have boundedness for all $\alpha$. Let $\alpha, \beta$ be given, we have 
$$ |x^\alpha \partial^\beta f(x)| \le |x|^{|\alpha|}|\partial^\beta f(x)| = \frac{|x|^{|\alpha| + 1}|\partial^\beta f(x)|}{|x|} $$
If we can bound the numerator, we are done.
A: Since it's bounded for any multiindex add $1$ to $\alpha_i$ to show that $|x^{\alpha} \partial^\beta f(x)|\leq \frac{M_i}{|x_i|}$ for every $i$. Selecting the largest $M_i$ to be $M$ we get $|x^{\alpha} \partial^\beta f(x)|\leq \min_i\frac{M_i}{|x_i|}\leq\frac{M}{\max_i|x_i|}$ for all $x\neq0$. But if $|x|\to\infty$ then $\max_i|x_i|\to\infty$.
A: Well, using @martini 's idea I wrote a proof: 
Let $\varepsilon>0$ be given and fix $\alpha, \beta\in\mathbb N_0^n$. If $p\in\mathbb N$ then using the multinomial theorem: 
$$ \begin{align}
\displaystyle |x|^{2p} |\partial^\beta f(x)|^2&=(x_1^2+\ldots+x_n^2)^p |\partial^\beta f(x)|^2\\
&=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}x^{2\upsilon}|\partial^\beta f(x)|^2\\
&=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}|x^\upsilon|^2|\partial^\beta f(x)|^2\\
&=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}|x^\upsilon \partial^\beta f(x)|^2\\
&\leq \sum_{|\upsilon|=p} \frac{p!}{\upsilon!}C_{\upsilon, \beta}^2\\
&\leq (\max_{|\upsilon|=p} C_{\upsilon, \beta})^2\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}\\
&=n^p (\max_{|\upsilon|=p} C_{\upsilon, \beta})^2\\
&=C_{p, \beta}^2,
\end{align}$$
Consequently: $$|x|^p |\partial^\beta f(x)|\leq C_{p, \beta}.$$
Applying this with $p=|\alpha|+1\in\mathbb N$ we find: 
$$\begin{align*}
\displaystyle |x^\alpha \partial^\beta f(x)|&\leq \frac{|x|^{|\alpha|+1}}{|x|}|\partial^\beta f(x)|\\
&\leq \frac{C_{\alpha, \beta}}{|x|}
\end{align*}$$ 
Now choosing $\displaystyle M=\frac{C_{\alpha, \beta}}{\varepsilon}$ we see that $|x|>M\Rightarrow |x^\alpha \partial^\beta f(x)|<\varepsilon$. Therefore, $$\lim_{|x|\to +\infty} |x^\alpha \partial^\beta f(x)|=0.$$
Obs: Above I also used the general fact $$\sum_{|\upsilon|=p}\frac{p!}{\upsilon!}=n^p,$$ whenever $\upsilon\in\mathbb N_0^n$. This follows expanding $(1+\ldots+1)^p$ with the multinomial theorem.
