Reproducing the delta function in physics by using the definition adopted in the theory of distributions I used to study physics, where the delta function on the real line is defined as a function $\delta$ that satisfies
$$\delta(x)=\begin{cases}\infty&,x=0\\
0&,x\neq0
\end{cases}\tag{$*$}$$
and
$$\int_{-\infty}^\infty\delta(x)dx=1.\tag{$**$}$$
Now I'm studying mathematics for peace of mind, and surprised to know that mathematicians define this exotic function in a different way. Precisely, the theory of distributions defines the delta function as the continuous linear map $\delta:C_0^\infty(\mathbb{R})\to\mathbb{C}$ given by $\delta(f)=f(0)$. Unlike the one defined in the field of physics, this definition talks about a linear functional on a vector space of functions. I was wondering how these two definitions are related to each other. Is it possible to start with the definition in mathematics to obtain results similar to $(*)$ and $(**)$, and vice versa? After all, these two definitions bear the same name; they must have somthing to do with each other.
I have been looking over several threads on the delta function, but haven't found any of them answering my question. It would be great if someone could tell me more about it. By the way, I'm studying [Georgiev2015_Book_TheoryOfDistributions, Springer] and [Duistermaat-Kolk2010_Book_Distributions, Birkhäuser], and I just acquainted myself with the definition of a distribution on an open subset of $\mathbb{R}^n$. If you happpen to know any chapters in these two books that shed light on this topic, please be sure to inform me. Thank you.
 A: The physicists' "definition" is strictly speaking contradictory, because the function defined by (*) has (Lebesgue) integral $0$, i.e. it does not satisfy (**). Even if you don't know how to integrate the function defined by (*), it still tells you that $c\delta$ for all $c>0$ are all the same function. Thus if (*) is literally true then we do not have the freedom to normalize $\delta$ to ensure that (**) holds.
The rigorous meaning behind this definition arises by passing to an "approximate identity". Namely, if you have a sequence of nonnegative integrable functions $f_n$ with $\int_{-\infty}^\infty f_n(x) dx = 1$ and $\lim_{n \to \infty} \int_{-\delta}^\delta f_n(x) dx = 1$ for all $\delta>0$, then $\lim_{n \to \infty} \int_{-\infty}^\infty f_n(x) g(x) dx = g(0)$ for any test function* $g$. Thus we say that the distributions $d_n(g)=\int_{-\infty}^\infty f_n(x) g(x) dx$ converge in distribution to $\delta$.
Now the physicists' "definition", taken on its face, is essentially taking this same limit but interpreting it pointwise. This interpretation is mathematically problematic; in particular, this is a situation where $f_n$ converges pointwise to $f$ but $\int f_n$ doesn't converge to $\int f$, which is where the contradiction I mentioned in the first paragraph is coming from.
With all this said, when you actually look at calculations, physicists are not really using this definition. Instead they are in effect using the mathematicians' definition even if they don't write it down.
*The test function requirement can be relaxed considerably.
