Understanding this characterization of Schwartz's space

Question

While reading up on tempered distributions I came across this definition of Schwartz's space:
$S(\mathbb{R}^n) = \{ f \in \mathbb{C}^\infty : \underset{x \in \mathbb{R}^n}{sup} \: \underset{\lvert \alpha \rvert < k}{sup} (1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert < \infty \:\: \forall k \in \mathbb{N}_0\}$
Where $\alpha$ is a multi-index of dimension $n$.
I have trouble understanding the nested $sup$s. Am I supposed to read them as one big $sup$, where i try to maximize with respect to $x$ and $\alpha$ simultaneously, or is the order important in a way?
I tend to the further, because I cannot see how it would be possible to attempt to maximize consecutively.
Thanks.

Answer 1

There are two nested suprema, and as written, they are evaluated consecutively. So if $f\in C^\infty$ and $x\in\mathbb R^n$, you have $$ \sup_{x\in\mathbb R^n}\sup_{|\alpha|<k}(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert = \sup_{x\in\mathbb R^n}a(x), $$ where $a(x)=\sup_{|\alpha|<k}(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert$.

But, fortunately enough, the order of two sups does not matter and you can also regard it as one big sup. That is, $$ \sup_{x\in\mathbb R^n}a(x) = \sup_{|\alpha|<k}b(\alpha) = \sup C, $$ where $$ b(\alpha)=\sup_{x\in\mathbb R^n}(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert $$ and $$ C=\{(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert;x\in\mathbb R^n,|\alpha|<k\}. $$ Proving this is a fairly simple exercise using only the definition of a supremum.