I have a question on the following proof, that $\delta_0$ is not a regular distribution. We define $\delta_0$ as the linear function on test function with
$$
\delta_0(\varphi) = \varphi(0)
$$
for every test function $\varphi$ (note it is not defined as some function with $\delta(0) = 1, \delta(x) = 0$ for $x \ne 1$, so this property could not be used here).
Proof: Suppose there is a representation
$$
\delta_0(\varphi) = \int_{\Omega} f(x) \varphi(x) ~\mathrm dx
$$
with a function $f \in L_{\mathrm{loc}}^1(\Omega)$. The restriction of $\delta_0$ on $\Omega\setminus \{ 0 \}$ is the Zero-Distribution, i.e. those distribution that assigns each $\varphi$ the value $0$. Therefore, the function which represents $\delta_0$ is $f \equiv 0$ almost everywhere on $\Omega$, and therefore $\delta_0$ would be the Zero-Distribtion as well. Contradiction. $\square$
The proof could be found here, p. 95 (in german).
My Question: What is the restriction of a distribution? Distribution are linear functionals, so if I restrict the domain, I restrict the set of test functions, but here some subsets, maybe $\Omega \subseteq \mathbb R$, is restricted. And also if I restrict the domain of integration, then the set of test functions is altered too, so who said that if I get a result for some set of test functions, I get the same if I consider another set of test functions?
Best Answer
For $U \subset V$, we have a natural (continuous) injection
$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$
Its transpose,
$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$
is called the restriction of the distributions on $V$ to distributions on $U$. For regular distributions, that corresponds to the restriction of the locally integrable function representing the distribution to $U$.
The fundamental lemma, that for any $f\in L^1_\text{loc} \Omega$ and $W\subset \Omega$ we have
$$\bigl(\forall \varphi \in \mathscr{D}(W)\bigr)\left(\int_W f(x)\varphi(x)\,dx = 0\right) \implies f\lvert_W = 0 \text{ a.e.},$$
makes the two restrictions compatible.
So here, looking at only test functions with support in $\Omega\setminus\{0\}$ yields the result that $f$ would be zero almost everywhere in $\Omega\setminus\{0\}$, hence almost everywhere in $\Omega$.
For an alternative proof that the Dirac distribution is not regular, consider a test function $\varphi\in\mathscr{D}(\Omega)$ with $\varphi(0) = 1$, and note that then
$$1 = \int_\Omega f(x)\varphi(nx)\,dx$$
for all $n\in\mathbb{N}\setminus\{0\}$ violates the dominated convergence theorem.