A distribution of order k+1 is sum of two.

Question

Let $\lambda$ be a compactly supported distribution of order $k+1$ then how to prove $\lambda=\delta'+\mu$ for some compactly supported distributions of order $k$?

Answer 1

Let $\lambda \in \def\E{\mathcal E}\E'(\def\R{\mathbb R}\R)$ of order $k+1$. Given any $\varphi \in \E(\R)$, define $\bar\varphi \in \E(\R)$ by $\bar\varphi(t) = \int_0^t \varphi(s)\, ds$. Let $$ \delta(\varphi) = -\lambda(\bar\varphi) $$ Then $\delta$ is of order $k$, as $$ \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|} \let\phi\varphi\abs{\delta(\phi)} = \abs{\lambda(\bar\varphi)} \le C\norm{\bar\varphi}_{k+1} \le C\norm{\varphi}_{k}. $$ Moreover we have $$ \delta'(\phi) = -\delta(\phi') = \lambda(\overline{\phi'}) = \lambda\bigl(\phi -\phi(0)\bigr) = \lambda(\phi) - \phi(0) \cdot \lambda(1). $$ That is, if we let $\mu(\phi) = \phi(0) \cdot \lambda(1)$, then $\lambda = \delta' + \mu$.