Let $U$ be an open subset of $\sR^n$ such that $0\in U$ . Then the delta distribution is the mapping \begin{eqnarray*} \delta : \cD(U) &\to & \sC \\ u &\mapsto & u(0). \end{eqnarray*} Claim The delta distribution is a distribution of zeroth order, i.e., $\delta\in \cD'^0(U)$ .

Proof. With obvious notation, we have \begin{eqnarray*} \delta(u+v)&=&(u+v)(0)=u(0)+v(0) = \delta(u) + \delta(v),\\ \delta(\alpha u) &=& (\alpha u)(0)=\alpha u(0)=\alpha \delta(u), \end{eqnarray*}so $\delta$ is linear. To see that $\delta$ is continuous, we use condition (3) on this this page. Indeed, if $K$ is a compact set in $U$ , and $u\in \cD_K$ , then $$ |\delta(u)| = |u(0)| \le ||u||_\infty,$$ where $||\cdot||_\infty$ is the supremum norm.