Everywhere it is specified (with good reason, since no such function could be defined) that the Dirac Delta function $\delta(x)$ is not a function, but a distribution.
I know that distributions are defined as continuous linear functionals $F[\varphi]$ acting on the space of test functions $F: T \to \Bbb{R} $, therefore distributions belong to its dual space $T^*$. Such functionals can be represented as integrals: $F = \int_{-\infty}^{+\infty}{f(t)\varphi(t)}dt $ where $f(t)$ is some function integrable on $\varphi$'s support.
In the Delta case, if we define the functional $\delta_{t_0}[\varphi]:= \varphi(t_0)$, no such $f$ can be found (except in a limit process of delta-approximating bump function), but we still define this hypotetic delta function as $\delta(t-t_0)$ implying that it only makes sense when it is inside an integral, in the same way (correct me if i'm wrong) as it is customary to write $P.V. (\frac{1}{x})$ inside some equation, without integration, knowing it will make sense (Cauchy's P.V.) once it is integrated.
So, why everywhere $\delta(t-t_0)$ (or Heaviside step, or anything else in that situation) is called a distribution? The functional $\delta_{t_0}(\varphi) = \int_{-\infty}^{+\infty}{\delta(t-t_0)\varphi(t)}dt $ should be the distribution, not its integral's kernel ($\delta_{t_0} \in T^*$ is in the space of distributions).
Could someone explain me the situation better?