# Schwarz distribution as a derivative of a continuous function

Every distribution $t\in\mathcal{D}'(\mathbb{R}^n)$ of compact support is of the form

$$t(x) = \sum_{|\alpha|<m}\partial^\alpha t_\alpha(x),$$ where $\alpha$ is multi-index, $m$ - some constant and $t_\alpha \in C(\mathbb{R}^n)$.

It is not true that every distribution $t\in\mathcal{D}'(\mathbb{R}^n)$ is of the above form (the sum may be infinite).

However, I suppose that every Schwarz distribution $t\in\mathcal{S}(\mathbb{R}^n)$ is of the above form, i.e. $$t(x) = \sum_{|\alpha|<m}\partial^\alpha t_\alpha(x),$$ where $\alpha$ is multi-index, $m$ - some constant and $t_\alpha \in C(\mathbb{R}^n)$. Am I right? Can you provide a reference.

• See math.stackexchange.com/questions/1963159/… it is the same as saying every distribution with compact support has finite order – reuns Dec 15 '16 at 12:01
• All tempered distributions are of finite order, hence you hypothesis is true. – TZakrevskiy Dec 16 '16 at 17:13