Relation between Gevrey space $G^s$ and Schwartz space $S$

Question

I want to decide whether there is a relation between the Gevrey space $G^s(\mathbb{R}^m)\;(s>1)$ and the schwartz space of rapidly decreasing functions $S(\mathbb{R}^m)$. I know that polynomials are in Gevrey class but no non zero polynomial is in $S$. Thus $G^s$ can not be included in $S$. My difficulty is in the converse case. Thank you for your suggestions.

Answer 1

The Gevrey space is defined by a regularity condition, which can be summarized by a strong decay of the Fourier transform. A fonction $u$ belongs to $G^s$ if its Fourier transform satisfies for some $\rho>0$ $$ \vert\hat u(\xi)\vert\le C e^{-\rho \vert\xi\vert^{1/s}}. \tag 1$$ The Schwartz space $\mathscr S$ is defined by regularity and decay. As you pointed out, $G^s$ is not included in $\mathscr S$, but the converse is not true either and in fact you do even have $$ \mathscr S\not\subset \cup_{s>0} G^s. $$ To prove this, consider a fonction $f$ defined on $\mathbb R^n$ by $$ f(\xi)= e^{-\bigl(\ln(1+\vert\xi\vert^2)\bigr)^2}. $$ It can be shown that $f$ belongs to the Schwartz space and thus $u=\hat f$ is also in $\mathscr S$. However, obviously $\hat u$ cannot satisfy (1) for any $s>0$.

It is also interesting to note that for $s>1$, $G^s$ contains some smooth compactly supported functions, which is not the case for $s\le 1$.