Let $\mathscr{D}(\mathcal{O})$ be the space of basic functions, $\phi$, which is equipped with convergence, and where $\phi$ vanish outside of the set $\mathcal{O}$.
Let $\mathscr{D}'(\mathcal{O})$ be the space of generalized functions, which is also equipped with convergence, and where $(f_k,\phi_k)\longrightarrow 0 $, as $k\to\infty$.
Theorem (Vladimirov page 13): For a linear functional $f$ on $\mathscr{D}(\mathcal{O})$, to belong to the space of generalized functions $\mathscr{D}'(\mathcal{O})$, it is necessary and sufficient that for any open set $\mathcal{O}'\subset \mathcal{O}$ (where $\mathcal{O}'$ is also dense in $\mathcal{O}$), that there exists a number $K=K(\mathcal{O'})$ and $m=m(\mathcal{O}')$ such that
$$|(f,\phi)|\leq K ||\phi||_{C^m(\bar{\mathcal{O}'})'}$$
where $\phi$ are the basic functions, $\phi \in \mathscr{D}(\mathcal{O'})$, $|(f,\phi)|$ is the value of the integral.
At first, I would have thought that $C^m$ is the space of continuous functions with $m$-norm. However, this is not true. In the book by Vladimirov, $m$ is said to be the order of the function $f$ on the set $\mathcal{O}$.
Vldimirov further elaborates saying:
"The order of the delta function is for example 0, while the order of the generalized function $(f,\phi)=\sum_{k\geq 1} \mathcal{O}^{(k)}(k)$ is infinite. "
As an additional note, the order of the singular generalized function $\mathscr{P}\frac1x$ which forms the principal value of the integral of the function $\frac1x$ is on the other hand 1.
What is the best definition of order of a function in the given context of the generalized functions in distribution theory? I cannot find any other definition of the order here than in this theorem, and it seems more to be a integer that defines some property of the set $\mathcal{O}$ on which the basic functions are defined in.
Thanks
For $f\in L^\infty_{\text{loc}}$ the corresponding distribution is defined as $$ \varphi \mapsto \int f(x) \, \varphi(x) \, dx. $$
If $\operatorname{supp}\varphi \subseteq K$ we have $$ \left| \int f(x) \, \varphi(x) \, dx \right| = \left| \int_K f(x) \, \varphi(x) \, dx \right| \leq \int_K |f(x)| \, |\varphi(x)| \, dx \leq \int_K |f(x)| \, \sup_{x\in K}|\varphi(x)| \, dx \\ = \int_K |f(x)| \, dx \cdot \sup_{x\in K}|\varphi(x)| = C_K \|\varphi\|_{0}, $$ where $C_K = \int_K |f(x)| \, dx.$
Since we don't need to include any derivatives of $\varphi$ in the bound $C_k \|\varphi\|_{0}$, the distribution has order zero.