Let $d$ be a number $\geq 0$, and $\Omega$ an open subset of $\mathbb{R}^n$. One calls d$^{\text{th}}$ Gevrey class in $\Omega$, and denotes by $G_d(\Omega)$, the space of $C^{\infty}$ functions $f$ such that, to every compact subset $K$ of $\Omega$, there is a constant $A(f,K)>0$ such that, for all $p \in \mathbb{N}^n$, $$\sup_{x \in K}\bigg|\left(\frac{\partial}{\partial x}\right)^p f(x)\bigg| \leq A(f,K)^{|p|+1}(p!)^d.$$
Question. What are the elements of $G_0(\Omega)$ and of $G_1(\Omega)$?
I thought of the following: Let any $f \in G_0(\Omega)$ , we have for all $K \subset \Omega$ compact there exists a constant $A(f,K)>0$ such that $$\sup_{x \in K}\bigg|\left(\frac{\partial}{\partial x}\right)^p f(x)\bigg| \leq A(f,K)^{|p|+1})$$ and hence $$|f|_{m,K} \leq \sup_{|p|\leq m} A(f,K)^{|p|+1}=:B_0(p)$$ with $m \in \mathbb{N}$, since in particular $f \in C^{\infty}(\Omega)$.
Similarly, given any $f \in G_1(\Omega)$, we have for all $K \subset \Omega$ compact, there exists a constant $A(f,K)>0$ so that $$|f|_{m,K} \leq A(f,K)^{|p|+1}p!=:B_1(p),$$ with $m \in \mathbb{N}$.
In this way, given $m \in \mathbb{N}$ , $p \in \mathbb{N}^n$ e $K \subset \Omega$ compact we concluded that if $f \in G_0(\Omega)$, then $|f|_{m,K}$ "bounded" by $B_0(p)$. And if $f \in G_1(\Omega)$, so $|f|_{m,K}$ is "bounded" for $B_1(p)$. Here, $$|f|_{m,K}=\sup_{|p|\leq m} \Bigg( \sup_{x \in K}\bigg|\left(\frac{\partial}{\partial x}\Bigg)^p f(x)\bigg| \right), \; \forall \; f \in C^{\infty}(\Omega).$$
Can I say then that the elements of $G_0(\Omega)$ and $G_1(\Omega)$ are characterized by these "boundedness"?