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.

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    $\begingroup$ it's all a matter of approximation. If you take norm-1 continuous functions $f_n$ that converge punctually to (*), then $\int g(x) f_n(x) \to g(0)$ for every $g(x)$ continuous $\endgroup$
    – Exodd
    Jul 6 at 12:30
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    $\begingroup$ They are the same. It is common in physics and engineering to use sloppier definitions than in mathematics. $\endgroup$ Jul 6 at 12:51
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    $\begingroup$ A lot of distributions are given by a formula like $T_f(g) = \int f(x) g(x)\,\mathrm dx$. You may know $f \mapsto T_f$ as an embedding of $L^1$ into the space of distributions. Many definitions in distribution theory can be motivated by looking at what happens to $T_f$ (eg: Fourier transforms etc). Not every distribution is given by a formula of this form - the notion of "distribution" generalises the notion of "integrate against". Physicists still like to think of the delta distribution as "integrating against" a thing, as that makes it easier to think about, and you can use it to calculate. $\endgroup$ Jul 6 at 13:00
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    $\begingroup$ Just because someone has defined something as "the function having this property and that property" does not mean that such a function actually exists, in the strict sense of mathematical existence; that way lies the ontological argument. But the intuition and application of the concept is so useful and important that mathematicians worked hard to put the concept of the Dirac delta on a solid mathematical foundation. Their efforts involved the successful development of the theory of distributions. $\endgroup$
    – Lee Mosher
    Jul 6 at 13:38
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    $\begingroup$ And, by the way, if you read Dirac's book The Principles of Quantum Mechanics, where he proposes applications of the delta "function" to physics, I think you'll find that he was fully aware of the "distributional" nature of the delta. $\endgroup$
    – Lee Mosher
    Jul 6 at 13:41

1 Answer 1


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.

  • $\begingroup$ But if you look at how physicists use the delta function in practice, it is exactly the same as how mathematicians use it. Physicists often figure out how to use something correctly before mathematicians figure out a rigorous definition of it. Newton did this. $\endgroup$
    – Deane
    Jul 6 at 14:44
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    $\begingroup$ @Deane Perhaps my heavily edited answer is more satisfactory in this regard. No bashing of people was intended; I am aware that physicists don't actually get wrapped up in contradictions as a result of this. $\endgroup$
    – Ian
    Jul 6 at 15:04

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