Application of the $\delta$ distribution to a function which is not a Schwartz-Function?

Question

I come from the field of Physics, where lecturers roughly don't care that the $\delta$-distribution is not a function. In Physics, it is just used as if it were a function, in textbooks or lecture notes people will integrate over the $\delta$-function, multiplied with some other function, and are fine.

From my own studies, I however know that the $\delta$-Distribution is a non regular Distribution, that can be defined as a linear functional over the Schwartz space (the space of rapidly falling functions).

I understand that by defining distributions this way, they extend the functionals generated from Integration over a function in a nice way, while it is still possible to define operations like the Fourier-transform, multiplication with a function, or convolution in a nice and meaningful way.

What I don't understand however, is that physicists don't care what functions they apply the $\delta$-Distribution to. I have multiple times seen the $\delta$-Distribution be applied to polynomials or sinuses, which are clearly not elements of the Schwartz-space, as they are not falling. So my question is: How can one rigorously define the use of the $\delta$-Distribution, or of any other tempered Distribution in general, on a bigger class of functions, like polymials?

EDIT: I allready understand that physicists don't always care about rigorosity, yet I'd like to know wether there is a way to make the described procedure of applying tempered distributions to continuous functions rigorous.

Answer 1

Suppose that $u$ is a distibution with compact support and that $\varphi \in C^\infty$ (not necessarily having compact support). Let $\rho \in C_c^\infty$ be such that $\rho \equiv 1$ on a neighborhood of the support of $u$. Then we can set $$\langle u, \varphi \rangle := \langle u, \rho \varphi \rangle,$$ and this is independent of the choice of $\rho$.

Answer 2

As a measure, the Dirac $\delta$ measure can be applied to any function. Let some set $X$, then the powerset of $X$, symbolized by $\wp(X)$, is a $\sigma$-algebra and $$\delta_x(A):=\begin{cases}1,& x\in A\\0,&x\notin A\end{cases}\tag1$$ is a Dirac measure (for some chosen $x\in X$) on the measurable space $(X,\wp(X))$. That is, the triplet $(X,\wp(X),\delta_x)$ defines a measure space.

Now suppose that $\mathcal A$ is some $\sigma$-algebra in $E$ and $f\in E^X$. Then we says that $f$ is measurable when $f^{-1}(H)\in\wp(X)$ for every $H\in \mathcal A$, thus any function is measurable for this measure.

Then

$$\int_X f(y)\delta_x(y)=f(x)\tag2$$

for any chosen $f\in E^X$, so any function $f\in E^X$ is Bochner integrable respect to this measure, because $f=g$ almost everywhere where

$$g(t):=\begin{cases}f(t),& t=x\\0,&\text{otherwise}\end{cases}\tag3$$

and $g$ is simple.

But then $(2)$ can be used to define a distribution on the space of test functions $\mathcal D(X)$, that is, $\delta_x:\mathcal D(X)\to E,\, \varphi\mapsto\varphi(x)$ as a distribution, then in this context $\delta_x$ is defined just in this space. This is the difference between the two ways to see the Dirac $\delta$.