Formal proof that Schwartz space is space of rapidly decreasing functions Everybody says that the Schwartz space is a space of rapidly decreasing functions, or of functions that rapidly vanish, but I am baffled with proving it formally. I can come up with nice reasons why it is true, but not with a solid formal proof.

What I want to prove is
  $$f\in S(\mathbb{R}^n)\implies\lim_{|x|\to\infty}f(x)=0$$
  or more precisely 
  $\forall\varepsilon>0,\exists K\subset\mathbb{R}^n$ compact, that $x\in(\mathbb{R}^n\setminus K)\implies|f(x)|<\varepsilon$.

The Schwartz space is 
$$S(\mathbb{R}^n)=\{f:\mathbb{R}^n\to\mathbb{R}^n\mid f\in C^\infty(\mathbb{R}^n),\; \|f\|_{\alpha,\beta}<\infty\}$$
where $\alpha,\beta\in\mathbb{N}_0^n$ are multi-indexes and 
$$\|f\|_{\alpha,\beta}=\sup_{x\in\mathbb{R}^n}|x^\alpha \partial_\beta f(x)|.$$
 A: 
Proposition. Let $f\in C^\infty(\mathbb R^n)$. The following statements are equivalent.
(a) $\displaystyle\|f\|_{\alpha,\beta}:=\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|<\infty$ for all $\alpha,\beta\in\mathbb{N}^n$.
(b) $\displaystyle\sup_{x\in\mathbb{R}^n}\|x\|^k|D^\beta f(x)|<\infty$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(c) $\displaystyle\lim_{|x|\to\infty}\|x\|^kD^\beta f(x)=0$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.
(d) $\displaystyle\lim_{|x|\to\infty} x^\alpha D^\beta f(x)=0$ for all $\alpha,\beta\in\mathbb{N}^n$.

As a consequence we have the following

Corollary. Let $f\in C^\infty(\mathbb R^n)$. Then, $f$ is a Schwartz function if and only if $f$ is a rapidly decreasing function.

Proof of the Corollary: To say that  $f$ is in the Schwartz space means that $f$ satisfies the condition $(a)$ of the proposition. To say that $f$ is rapidly decreasing means that each $D^\beta f$ tends to $0$ faster than $\|x\|^{-k}$ for all $k\geq 0$ as $|x|\to\infty$ which is  precisely the condition $(c)$ of the proposition. $\blacksquare$
Proof of the Proposition:
$\text{(a)}\Rightarrow\text{(b)}$ Let $e_1,...,e_n$ be the canonical basis of $\mathbb{R}^n$. Then
$$\|x\|^k=\left(\sum_{j=1}^n x_j^2 \right)^{k/2}\leq C_{k,n}\sum_{j=1}^n(x_j^2)^{k/2}=C_{k,n}\sum_{j=1}^n|x_j|^k=C_{k,n}\sum_{j=1}^n|x^{ke_j}|$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$, where $C_{k,n}$ is a positive constant that depends on $k$ and $n$. Thus,
$$\begin{align}
\sup_{x\in\mathbb R^n}\|x\|^k|D^\beta f(x)|&\leq\sup_{x\in\mathbb R^n}\left(C_{k,n}\sum_{j=1}^n|x^{ke_j}|\right)|D^\beta f(x)|=C_{k,n}\sup_{x\in\mathbb R^n}\sum_{j=1}^n|x^{ke_j}D^\beta f(x)|\\
&\leq C_{k,n}\sum_{j=1}^n\|f\|_{ke_j,\beta}<\infty.
\end{align}$$
$\text{(b)}\Rightarrow\text{(c)}$ There exists a constant $M$ such that $\|x\|^{k+1}|D^{\beta}f(x)|\leq M$ for all $x\in\mathbb R^n$ and thus
$$\left|\|x\|^{k}D^{\beta}f(x)\right|\leq\frac{M}{|x|},\qquad\forall\ x \neq 0.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(c)}\Rightarrow\text{(d)}$ Let $\alpha=(\alpha_1,...,\alpha_n)$ be a multi-index and $k=\alpha_1+\cdots+\alpha_n$. Then
$$|x^\alpha|=\prod_{j=1}^n|x_j^{\alpha_j}|=\prod_{j=1}^n(x_j^{2})^{\alpha_j/2}\leq \prod_{j=1}^n\left(\sum_{r=1}^nx_r^2\right)^{\alpha_j/2}=\|x\|^{k}$$
for all $x=(x_1,...,x_n)\in\mathbb{R}^n$. Thus,
$$|x^\alpha D^\beta f(x)|\leq\left|\|x\|^kD^\beta f(x)\right|,\qquad\forall \ x  \in\mathbb R^n.$$
Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.
$\text{(d)}\Rightarrow\text{(a)}$ There exists a constant $M$ such that
$$|x^\alpha D^\beta f(x)|\leq 1,\quad\forall\ x\notin \overline{B(0,M)}$$
and thus
$$\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|\leq\left\{1,\max_{x\in \overline{B(0,M)}}|x^\alpha D^\beta f(x)|\right\}<\infty.\blacksquare$$
A: My "informal" proof: It must hold for all $\alpha,\beta\in\mathbb{N}_0^n$. So if $\sup |x^\alpha f(x)|$ has to be bounded, then the limit of $f(x)$ in infinity has to be zero, because $|x^\alpha|$ in infinity is infinity. So if $f(x)$ is schwartz and $x^\alpha f(x)$ shall be bounded, then the only way to achieve this is to "put $f(x)$ down" to zero in infinity(because if $\lim_{x\to\infty} f(x)=c$, then $\lim_{x\to\infty}f(x)x^\alpha=\lim_{x\to\infty}f(x)\lim_{x\to\infty}x^\alpha=c.\infty=\infty$). If $f(x)$ is zero in infinity and $f(x)$ is smooth enough, then we can use L'hospital's rule $|\alpha|$ times and the resulting limit will be zero.
