Dirac delta and its derivatives

For $a\in \mathrm{\Omega }$, the Dirac distribution at $a$ is the complex-linear functional ${\delta }_a(\phi )=\phi (a)$ on $\mathcal{D}(\mathrm{\Omega })$, with the bilinear convention of Distribution. It is continuous: the seminorm $q(\phi )=\mathrm{\mid }\phi (a)\mathrm{\mid }$ restricts on each ${\mathcal{D}}_K$ to a seminorm bounded by $p_0$ (and is zero if $a\mathrm{\notin }K$), so it is one of the admissible seminorms defining the test topology.

Its derivatives are those of Distributional derivative: $⟨{\mathrm{\partial }}^{\alpha }{\delta }_a,\phi ⟩=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\mathrm{\partial }}^{\alpha }\phi (a)\mathrm{.}$ They are distributions by that definition; on each fixed-support stage the absolute value is at most $p_{\mathrm{\mid }\alpha \mathrm{\mid }}(\phi )$. If $a\mathrm{\notin }K$ all derivatives of the test vanish at $a$, so the bound has zero left side. For $\alpha =0$ this is ${\delta }_a$; a first derivative evaluates the negative first derivative of the test. No conjugation and no choice are involved. On the empty open set there is no permitted point $a$, rather than a new Dirac distribution. The sharp order of these distributions requires a test witness and is proved in the assigned example later.