Why Schwarz space is given by the set of $f\in \mathcal C^\infty (\mathbb R^n)$ s.t. $$\sup_{x\in \mathbb R^n}|(1+|x|^N)\partial _x^\alpha f(x)|<\infty ,$$ where $\alpha \in \mathbb N^n$ and $N\in\mathbb N$.
To have $$\sup_{x\in \mathbb R^n}|x|^N|\partial _x^\alpha f(x)|<\infty,$$ it is not enough ? I have the impression that both definition are equivalent, no ?
On
Yes, the two conditions are equivalent as $|x| \to \infty$, but one may not forget that we are interested in dominant functions $g$ such that $$ \int_{\mathbb R^n} |g|\:dx <\infty $$ which is the case for $$ g(x)=\frac1{(1+|x|^N)}, $$ as $N$ is great enough and which is not the case for $$ g(x)=\frac1{|x|^N}. $$
In fact using $|x|^N$ is enough.
The seminorm $\sup|x|^N|f(x)|$ is not equivalent to the seminorm $\sup(1+|x|^N)|f(x)|$.
But that's irrelevant to the definition of the Schwarz space, which involves the family of seminorms for all $N$. The family of seminorms defined by $|x|^N$ is equivalent to the family of seminorms defined by $1+|x|^N$. So the two families of seminorms do define the same space.
This is clear, because on the one hand $|x|^N\le1+|x|^N$, while on the other hand $$\sup(1+|x|^N)|f(x)|\le\sup|f(x)|+\sup|x|^N|f(x)|.$$