I need to integrate with $\delta$ against something that isn't a test function! In relation to ``How does integration over $\delta^{(n)}(x)$ work?,'' I need to evaluate $\int_{-a}^{a}f(x)\delta^{(n)}(x)\, dx$. However, while  my $f$ is smooth on its domain, it can't be a test function and doesn't shrink in any sense: $f(x)=\frac{\text{sinc}(a\cdot x)}{\text{Gaussian}(x)\cdot\text{sinc}(b\cdot x)}$. From what I have read about distributions, the only thing you can integrate over them is test functions (and the like). Does this create mathematical difficulty?
 A: The derivatives of the Dirac distribution only care about the behaviour of the function in arbitrarily small neighbourhoods of $0$. If you have two test functions $\varphi$ and $\psi$, and there is a neighbourhood $V$ of $0$ such that $\varphi\lvert_V = \psi\lvert_V$, then you have $\delta^{(n)}[\varphi] = \delta^{(n)}[\psi]$ for all $n \in \mathbb{N}$. Therefore it makes sense to extend the domain of $\delta^{(n)}$ to arbitrary smooth functions, without any growth condtitions. One then defines $\delta^{(n)}[f] := \delta^{(n)}[\varphi\cdot f]$, where $\varphi$ is any test function that is identically $1$ on some arbitrary neighbourhood of $0$. The result is of course $\delta^{(n)}[f] = (-1)^n \cdot f^{(n)}(0)$, it does not depend on the chosen cut-off function $\varphi$.
For $\delta^{(n)}$ individually, we could extend the domain further, it is only necessary that the $n^{\text{th}}$ derivative of $f$ exists at $0$. But when we're looking at extending not only $\delta^{(n)}$ but a large class of distributions, it is sensible to stop at smooth functions.
Systematically, let $\Omega$ be a (nonempty) open subset of $\mathbb{R}^n$ (or more generally a smooth manifold). We let $\mathscr{D}(\Omega)$ denote the space of smooth functions with compact support contained in $\Omega$, and $\mathscr{D}'(\Omega)$ the space of distributions on $\Omega$. Further, we let $\mathscr{E}(\Omega)$ denote the space of smooth functions on $\Omega$, and $\mathscr{E}'(\Omega)$ its topological dual. I omit a discussion of the (usual) topologies these spaces are endowed with and refer to the literature for that.
For a distribution $T \in \mathscr{D}'(\Omega)$, we say that $T$ vanishes on the open subset $W \subset \Omega$ if $T[\varphi] = 0$ for all $\varphi \in \mathscr{D}(\Omega)$ with $\operatorname{supp} \varphi \subset W$. We denote that by $T\lvert_{\mathscr{D}(W)} \equiv 0$. Since $\mathscr{D}(W)$ has a canonical inclusion into $\mathscr{D}(\Omega)$, that is only a slight abuse of notation. We define the support of $T$ as the complement of the largest open subset of $\Omega$ on which it vanishes,
$$\operatorname{supp} T = \Omega \setminus \bigcup \bigl\{ W \subset \Omega : W\text{ open and } T\lvert_{\mathscr{D}(W)} \equiv 0\bigr\}.$$
Then we can identify $\mathscr{E}'(\Omega)$ with the space of distributions on $\Omega$ with compact support, $\{ T \in \mathscr{D}'(\Omega) : \operatorname{supp} T \text{ is compact}\}$. For if $T$ is a distribution with compact support $K$, we can choose a $\varphi \in \mathscr{D}(\Omega)$ such that $\varphi^{-1}(\{1\})$ is a neighbourhood of $K$, and define
$$\tilde{T}[f] := T[\varphi\cdot f]$$
for $f \in \mathscr{E}(\Omega)$. The functional $\tilde{T}$ does not depend on the choice of $\varphi$, for if $\psi\in \mathscr{D}(\Omega)$ such that $\psi^{-1}(\{1\})$ is a neighbourhood of $K$, then $\varphi\cdot f - \psi\cdot f = (\varphi - \psi)\cdot f$ is, for every $f\in \mathscr{E}(\Omega)$, a test function that vanishes in a neighbourhood of $K$, so $\operatorname{supp} (\varphi - \psi)\cdot f \subset \Omega\setminus \operatorname{supp} T$, whence $T[(\varphi - \psi)\cdot f] = 0$. Since the map $\mu_{\varphi} \colon \mathscr{E}(\Omega) \to \mathscr{D}(\Omega);\; f \mapsto \varphi \cdot f$ is continuous for every $\varphi\in \mathscr{D}(\Omega)$, we have $\tilde{T} \in \mathscr{E}'(\Omega)$.
Conversely, since the inclusion $\mathscr{D}(\Omega) \hookrightarrow \mathscr{E}(\Omega)$ is continuous and has dense image, we have a canonical inclusion $\rho \colon \mathscr{E}'(\Omega) \hookrightarrow \mathscr{D}'(\Omega)$, so every element of $\mathscr{E}'(\Omega)$ "is" a distribution on $\Omega$, and since every neighbourood of $0$ in $\mathscr{E}(\Omega)$ constrains only the behaviour on some compact subset of $\Omega$, that distribution has compact support. One then verifies that $\rho(\tilde{T}) = T$ for $T \in \mathscr{D}'(\Omega)$ with compact support, and $\widetilde{\rho(S)} = S$ for $S \in \mathscr{E}(\Omega)$ to have the identification of $\mathscr{E}'(\Omega)$ with the space of distributions with compact support.
