Okay, here are the ingredients to this question.

Me: 60 years old. 39 years ago I took two semesters of Real Analysis using the Royden textbook. Rusty is an understatement. But I am still quite anal and OCD. I am also an electrical engineer, works in signal processing. DSP and Linear System Theory are important to me. I have also had two semesters (as a grad student) of Functional Analysis (using the Kreyszig text) and multiple courses in probability, random variables, and random processes (a.k.a. "stochastic processes").

Electrical Engineers (and I suspect many physicists) essentially treat $\delta(x)$ as a "function". But it isn't. One thing I remember from R.A. is that if

$$ f(x) = g(x) $$

almost everywhere in $E$, then

$$ \int_E f(x) \, dx \ = \ \int_E g(x) \, dx $$

problem is, of course, that electrical engineers (and their professors) like to think of

$$\begin{align} f(x) & = \delta(x) \\ g(x) & = 0 \end{align} $$

and that

$$ \int\limits_{-1}^{+1} f(x) \, dx = 1 \quad \ne \quad \int\limits_{-1}^{+1} g(x) \, dx = 0 $$

yet $f(x) = g(x)$ everywhere except at one single value of $x$ .

Now I have heard (or read) that the "Dirac delta function is not really a function but is a 'distribution' or a 'functional'." And I understand the meaning of the terms "distribution" in the context of random variables and "functional" in the context of metric spaces, normed spaces, etc. Is that the given usage of these two terms regarding $\delta(x)$?

Question 1: Is the usage of the notation

$$ \int\limits_{-\infty}^{+\infty} f(x) \, \delta(x) \, dx $$

a misnomer? There is no integration going on. It's just a linear functional that maps the function $f(x)$ to the number $f(0)$. Now EEs and maybe physicists will comfortably look at that as an integral that is the same as

$$ \int\limits_{-\infty}^{+\infty} f(0) \, \delta(x) \, dx = f(0)\int\limits_{-\infty}^{+\infty} \delta(x) \, dx = f(0) $$

But since $\delta(x)$ is not a function at all, what do mathematicians mean with that notation?

Question 2: How fatal is it for electrical engineers and physicists to consistently treat the Dirac delta function simply as a limit of "nascent deltas" such as

$$ \delta(x) \ \triangleq \ \lim_{\sigma \to 0^+} \frac{1}{ \sigma} \operatorname{rect} \left( \frac{x}{\sigma} \right) $$

where $ \operatorname{rect}(x) \triangleq \begin{cases} 1 \quad |x|<\frac{1}{2} \\ \frac{1}{2} \quad |x|=\frac{1}{2} \\ 0 \quad |x|>\frac{1}{2} \\ \end{cases} $


$$ \delta(x) \ \triangleq \ \lim_{\sigma \to 0^+} \frac{1}{\sigma} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x}{\sigma}\right)^2} $$

What is gonna kill us to simple-mindedly treat the Dirac delta as such a function? A function that is zero almost everywhere, yet it integrates to be equal to 1 (where the function that is zero everywhere integrates to be 0).

If we do that, within our own disciplines, what mathematical problem might crop up that kills us?

This is not exactly the same but smells a lot like this concern from Richard Hamming:

“Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane.”

I might ask the same question regarding the mathematician's and the engineer's understanding of the Dirac delta function. How might a mathematician answer that question?

  • 5
    $\begingroup$ The two uses of the word "distribution" are not the same. In the context of the delta function, "distribution" means a linear map on a space of functions with some properties. Here is an introduction. $\endgroup$
    – lulu
    Jul 2, 2016 at 23:28
  • $\begingroup$ thanks, @lulu . i sorta remember seeing that before. i think my questions remain, but i will read that article. $\endgroup$ Jul 2, 2016 at 23:30
  • $\begingroup$ In measure theory, $\delta_x$ is just the unit point mass at $x.$ $\endgroup$
    – zhw.
    Jul 3, 2016 at 0:23
  • $\begingroup$ .... Quite Anal? $\endgroup$
    – Enrico M.
    Mar 18, 2018 at 18:07
  • $\begingroup$ can't spell "Analysis" without "Anal", @VonNeumann. them shrinks knew that ever since Freud. $\endgroup$ Mar 19, 2018 at 5:45

1 Answer 1


First, a comment about the property you mention. It's true that if $f$ and $g$ are measurable functions and $f = g$ almost everywhere, then

$$ \int_Ef(x)dx = \int_Eg(x)dx $$ This won't hold if $f$ and $g$ are distributions, because as you say distributions aren't functions. So the integral notation is just that - it's notation. Now...

A1. The answer to your first question comes from wanting the theory of distributions to functionally "look like" the theory of nicer linear functionals, i.e. we were really good at manipulating integrals, so we wanted the new thing to operationally work the same. For physicists, who were used the the Riesz representation theorem, this meant wanting to think of linear functionals as being just integrals against a fixed function. So, hence the inner product/integral notation

$$ F(\varphi) = \langle F,\varphi\rangle = \int F(x)\varphi(x)dx $$ Note that in distribution theory, this is not a standard Lebesgue integral - it's just formal notation. The integral notation is ubiquitous though - it's not just linear functionals that are written this way, but also linear operators

$$ g(x) = \int k(x,y)f(y)dy $$ There is even a sort of generalization of the Riesz representation theorem called the Schwartz Kernel Theorem that says that any (nice enough) linear operator $g = Ky$ can be written using an "integral" like that, but where the kernel function $k(x,y)$ is possibly a distribution. The moral of the story is, you should extend your understanding of the integral notation to include other linear operations, not just integration of standard functions. Once you've proved that all the usual operations that you're used to, like integration by parts, make sense with distributions, you'll see that using the integral notation is very natural and 100% rigorous - as long as you remember that it's just notation for "apply the linear operation specified".

A2. It isn't fatal at all to think of the delta function in this way - in fact this is a preferred method to define the delta function and many other distributions. The rect function represents a sort of local averaging, and you can think of the delta function as being an "infinitely local" averaging (i.e. sampling). The one thing I would recommend is looking into "approximations to the identity" - the rect function construction is just one possible construction, and in order to show that the delta function is uniquely defined, one should show that any similar sequence of approximate deltas also gives the same result (e.g. triangles, Gaussians, etc). In other words, you could either define the delta function as "that linear functional such that $F(\varphi) = \varphi(0)$, in which case you need to show that this is a well-defined, bounded linear operation on some function space, or alternately you could define the delta function as "the limit of $\langle \delta_\epsilon,\varphi\rangle$ as $\epsilon\rightarrow 0$, in which case you still need to prove that this is a well-defined, linear operation on some function space. Either way, the result is the same.

  • 1
    $\begingroup$ Good answer! I’m in my MS at WWU right now :) We just got done doing distribution theory in Arpads class $\endgroup$
    – Prince M
    Mar 19, 2018 at 18:12
  • $\begingroup$ @PrinceM Awesome! He is a wonderful prof, enjoy your time there! $\endgroup$
    – icurays1
    Mar 19, 2018 at 18:25
  • $\begingroup$ icurays1, could you comment on the expression of the Dirac comb which is: $$\begin{align} \sum\limits_{k = -\infty}^{\infty} \delta(x-k) &= \sum\limits_{n = -\infty}^{\infty} e^{i 2 \pi n x} \\ &= 1+ 2 \sum\limits_{n = 1}^{\infty} \cos( 2 \pi n x) \\ \end{align}$$ do you math guys just say that these infinite series don't exist or that these "naked" dirac deltas (not clothed with an integral) are illegitimate expressions? what do mathematicians say about this? $\endgroup$ Sep 27, 2018 at 21:39
  • 1
    $\begingroup$ @robertbristow-johnson You should probably ask a separate question to get a complete answer, but the short answer is yes, the Dirac comb and its Fourier transform are perfectly well-defined mathematical objects. Those sums are "formal" expressions, meaning they only make sense if interpreted correctly, but this is all done properly in the theory of distributions (see the book by Friedlander, for instance). $\endgroup$
    – icurays1
    Sep 28, 2018 at 14:27
  • 1
    $\begingroup$ thanks, but i think i'll take the lazy route and just accept your short little answer. some of the mathematically more sophisticated electrical engineers are having a discussion about this now. my pragmatic feeling or response to anyone who admonishes me about misuse of anything Diracian is that i will simply define it, for all practical purposes, to be 1 Planck time wide and having an area of 1 (dimensionless). that's a good enough approximation and it's a regular function of time. $\endgroup$ Sep 28, 2018 at 18:54

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