The Schwartz space is defined as $\mathcal{S} = \left\{ f \in C^\infty : \|f\|_{(N,\alpha)} < \infty \text{ for all } N,\alpha \right\}$, where the family of semi-norms $\|\cdot\|_{(N,\alpha)}$ is given by $$ \|f\|_{(N,\alpha)} = \sup_{x \in \mathbb{R}^n} (1+|x|)^N |\partial^\alpha f(x)|. $$
In Folland's Real Analysis, he states that "clearly, $C_c^\infty \subset \mathcal{S}$. It is not immediately clear to me why this is true. Of course, any $C_c^\infty$ function vanishes at infinity, but I don't see why this would have to be faster than any power of $|x|$.
I first thought to consider $\lim_{x \to \pm\infty} (1+|x|)^N| \partial^\alpha f(x)|$, which is of indeterminate type $\infty\cdot 0$. Recasting as a fraction $(1+|x|)^N/|\partial^\alpha f(x)|^{-1}$ and applying L'Hopital's rule, the quantity $|\partial^\alpha f(x)|^{-1}$ will always tend to $\pm\infty$ no matter how many derivatives we take, but eventually $(1+|x|)^{N}$ would become finite even as $x \to \pm\infty$ and hence, the limit is finite. Is this reasoning correct?
You wrote:
The set $C_c^\infty$ is the space of compactly supported infinitely differentiable functions. "Compactly supported" implies it's $0$ outside of some compact set. All compact sets are bounded. Thus if $f\in C_c^\infty,$ then for some number $M$ you have for every $x>M$ or $x<-M,$ $f(x) = 0.$ If $f(x)$ reaches $0$ and then stays there as $x$ approaches $+\infty$ or $-\infty,$ then certainly it goes to $0$ faster than anything that doesn't do that.