# Talk:Dirac delta function

Dirac delta function has been listed as one of the Mathematics good articles under the good article criteria. If you can improve it further, please do so. If it no longer meets these criteria, you can reassess it.
Article milestones
Date Process Result
September 29, 2010 Good article nominee Not listed
October 1, 2010 Good article nominee Listed
Current status: Good article

## The frequency domain

I would like to see some specifics on the Dirac Delta in the frequency domain. The mere substitution of "x" by "f" in the current definition formulae would be misleading. This clarification is important since some other articles depend on it. For instance, the statement The Fourier transform of a Dirac comb is also a Dirac comb here.MaskedAce (talk) 03:31, 30 September 2012 (UTC)

The Fourier transform is already covered in detail in the relevant section. It's not clear what you specifically believe is inadequate in the present treatment. Sławomir Biały (talk) 21:24, 30 September 2012 (UTC)

In the equation that follows the expression "The inverse Fourier transform of the tempered distribution f(ξ) = 1 is the delta function. Formally, this is expressed": When I naively try to do some calculus I obtain ${\displaystyle \int _{f=-\infty }^{\infty }e^{2\pi ift}df={\biggl [}{\frac {e^{2\pi ift}}{2\pi it}}{\biggr ]}_{f=-\infty }^{\infty }={\biggl [}{\frac {\cos {2\pi ft}+i\sin {2\pi ft}}{2\pi it}}{\biggr ]}_{f=-\infty }^{\infty }={\biggl [}{\frac {\sin {2\pi ft}}{2\pi t}}{\biggr ]}_{f=-\infty }^{\infty }=\lim _{f\to \infty }{\frac {2\sin {2\pi ft}}{2\pi t}}={\Bigg \{}{\begin{matrix}\lim _{f\to \infty }2f&{\textrm {as}}&t\to 0\\\in [-1,1]/(\pi t)&{\textrm {if}}&t\neq 0\end{matrix}}}$

Is the Dirac pulse amplitude defined to be ${\displaystyle 2\infty }$ so to speak? My question is if there is a reason that the definition doesn't contain a compensating factor of 1/2. I simply just don't understand and I believe that the article would benefit much if someone had any idea on how to explain the above in a pedagogical sense. Geo39geo (talk) 13:23, 12 February 2018 (UTC)

The value of the function ${\displaystyle \textstyle {\frac {2\sin {2\pi ft}}{2\pi t}}}$ at 0 is roughly the height of the hump; but did you think about the width of the hump? The integral of this function (over the whole real line) is equal to 1 (rather than 2), since ${\displaystyle \textstyle \int _{0}^{\infty }{\frac {\sin at}{t}}\,dt={\frac {\pi }{2}}}$ for all ${\displaystyle a>0.}$ Boris Tsirelson (talk) 18:53, 12 February 2018 (UTC)
Aha ok, I get it! This makes sense. Thank you very much for the succinct and fast reply. Geo39geo (talk) 22:10, 13 February 2018 (UTC)

## "The delta function only makes sense as a mathematical object when it appears inside an integral"

There has been a back and forth over whether the sentence in the lead "The delta function only makes sense as a mathematical object when it appears inside an integral" is a proper, accessible summary of the source, which says (Gel'fand & Shilov 1968, p. 1):

If the singular function occurs at all in the final result, it is only in an integrand where it is multiplied by some other sufficiently well-behaved function. There is therefore no actual necessity for answering the question of just what a singular function is per se; it is sufficient to know what is meant by the integral of a product of a singular function and a sufficiently 'good' function. For instance, rather than answer the question of what a delta function is, it is sufficient for our purposes to point out that for any sufficiently well-behaved function ${\displaystyle \phi (x)}$ we have ${\displaystyle \int _{-infty}^{\infty }\delta (x-x_{0})\phi (x)\,dx=\phi (x_{0}).}$

In addition, I have added a reference to Terry Gannon's article in the Princeton Companion, where he says in no uncertain terms

However, [the delta function] really only makes sense inside an integral.

Do we feel that the sources are being misrepresented in the context? If so, is there an equally accessible way to summarize the cited sources without that misrepresentation? Sławomir Biały (talk) 11:13, 27 July 2017 (UTC)

I note that an IP editor seems keen on having a discussion in article space. The purpose of these templates is not to engage in arguments in article space, but to indicate points that could be clarified or discussed, a request for clarification, not a mandate. Here is the long comment that I removed from the article:

on page 543 of The Princeton Companion to Mathematics Eq. 2 has a commutation relation equal a delta function , how is it that a delta function only makes sense inside an integral when this equation does not. Also, Green's functions are found by setting a differential equal to a Delta function without an integral. Third, Delta functions are use extensively in Quantum Mech. without being inside an integral. Isn't the state "makes sense" meaningless. What makes sense to me, might not make sense to you...

One has to consider the statement in the context of the article. We aren't talking about quantum mechanics or Green's functions. At the fundamental level to which such a statement is meant to apply, an equation like that of finding a Green's function, ${\displaystyle \Delta f=\delta }$, is merely shorthand for ${\displaystyle \int (\Delta f)\phi =\int \delta \phi }$ for all test functions ${\displaystyle \phi }$. I'm open to an equally accessible way to express this, but I think the current sentence is a fair summary for the target audience, which likely includes undergraduate engineers. For such readers, the sentence addresses the question of what the delta function "is". While pedants with a detailed knowledge of the subject may object to the niceties of the wording, they already know enough to not be misled by such pronouncements. Sławomir Biały (talk) 20:32, 27 July 2017 (UTC)

Many students are mislead by this statement and it hinders there ability to use delta function to solve physics problems. Wikipedia editors should at least make it clear that delta functions are (basically) never used "inside an integral" in practical problem solving (i.e. calculation). One student is convinced that the Kronecker delta "only make sense" when it appears inside a summation! — Preceding unsigned comment added by 131.252.127.172 (talk) 20:32, 29 July 2017 (UTC)

Ok. Would such students accept that the differential equation (for example) ${\displaystyle f''(x)=\delta (x)}$ is merely short-hand for the integral equation
${\displaystyle \int \varphi (x)f''(x)\,dx=\int \varphi (x)\delta (x)\,dx\qquad (=\varphi (0))}$
for all test functions ${\displaystyle \varphi }$? Sławomir Biały (talk) 21:09, 29 July 2017 (UTC)

Why do we insist on a test function? If "a vector" equal "b vector", then does not "a vector" dot "c vector" = "b vector" dot "c vector"? This test "c vector" (or function) at best just sits there through the calculation and at worst confuses even the smartest students.

For examples of how the Delta function is used in Quantum Mechanics:

"The Principles of Quantum Mechanics" P. A. M. Dirac Defines Dirac delta function and uses it to go from discrete basis to continuous basis. (Kronecker delta-> Dirac delta) p. 59

"Introduction to Quantum Mechanics" David J. Griffiths Uses Dirac delta to solve inhomogeneous Schroedinger eqn. by finding the Green's function. p.409

I see this disagreement in analogy with the ideas of imaginary numbers. At first people didn't think they made sense as mathematical objects. After all, x^2+1=0 has no root when you graph it. But later on when studying the cubic eqn. they saw the power of analysis using imaginary numbers. They could now find all three roots!

I have a feeling that most of the students reading this page are not Mathematicians rather Physicists and Engineers. These people want to know about the Dirac delta because they need to solve some problem that requires knowledge of it in their respected fields. They likely don't care about the mathematical niceties of distribution theory and whether it is Riemanan integrable or whether the limit exist etc. If they wanted to that know the niceties they would have taken Analysis courses. I gonna be bold and at the sametime very nonprecise in the next few statements. The Dirac delta maps a function onto itself. The Kronecker delta maps a vector onto itself. The Kronecker delta makes sense as mathematical object and so does the Dirac delta. Let's focus on what the Dirac delta is, not on what it is not. Let's focus on its power to solve some rather difficult problems without having to understand the details of Analysis and the theory of distributions.

131.252.127.172 — Preceding unsigned comment added by 131.252.127.172 (talk) 19:05, 30 July 2017 (UTC)

One doesn't need to understand the analytic details in order for the operational calculus to produce physically meaningful results. But to regard the delta function as a mathematical object requires test functions and integration. In the context of "a function that is equal to zero everywhere but the origin, where it is infinite, with total integral one", it is clearly important to point out that there is no such function, but the delta function can be made mathematically meaningful. This requires integrating against test functions. Sławomir Biały (talk) 21:03, 30 July 2017 (UTC)

First, you did not address the points I brought up in my last comment. Second, you did not explain why the Dirac delta cannot be made mathematically meaningful without a text function as you claim over and over. (And I repeatly ask you to explain.)

You are not even clear about what you take to be a mathematical meaningful object. (You just claim too much is being read into it) Does the identity element in a group have mathematical meaning to you? Does the identity matrix have mathematical meaning to you? In operational calculus the operator acting on the Green's function is equal to the delta function. Just like a group element and its inverse are equal to the identity element or the matrix and its inverse are equal to the identity matrix. Are we talking about some metaphysical ideas here or mathematics?

Why do you not back up your claims with clear/specific examples, like I do?

I can see you have dominated everyone in this talk page so you must feel very personal about what the Dirac delta means to you. Do you have any papers published where you use Dirac deltas? I am willing to learn... but let us not mislead students into believing that the Dirac delta does not have mathematical meaning.

131.252.127.172 — Preceding unsigned comment added by 131.252.127.172 (talk) 22:21, 30 July 2017 (UTC)

The article does not say that the delta function has no meaning. It says that "The delta function only makes sense as a mathematical object when it appears inside an integral." There is a lengthy quotation in a footnote, to Gelfand and Shilov: "There is therefore no actual necessity for answering the question of just what a singular function is per se." The parallel with the Kronecker delta is a false one. The Kronecker delta ${\displaystyle \delta _{ij}}$ is equal to one if ${\displaystyle i=j}$, and zero otherwise. But it is not the case that ${\displaystyle \delta (x)}$ is equal to zero for ${\displaystyle x\not =0}$ and equal to infinity for ${\displaystyle x=0}$: this is a meaningless statement. Sławomir Biały (talk) 23:09, 30 July 2017 (UTC)

You are so caught up on the fact the Dirac deltas are not functions that you will never see how useful (and powerful) they really are. And it seems like you are choosing to take young people down with you. :(

We already state on the page that the Dirac delta is not a function more times than I can count. Would taking the statement in question off the page really cause you so much discomfort that you are willing to argue to the bitter end to keep it up?

And I quote, "The delta function only makes sense as a mathematical object when it appears inside an integral."

            "In the context of "a function that is equal to zero everywhere but the origin, where it is infinite, with total integral one", it is clearly important to point out that there is no such function, but the delta function can be made mathematically meaningful."


I have given numerous examples of the mathematical meaning of Dirac delta outside the integral in my comments above. (You also have not answered the questions I ask you above.) There are many more in well accepted textbooks and published papers. You only have to look. Even your own cited books uses it outside an integral, and you still insist on being self righteous and use that quote to mislead people.

If you truly believe that Dirac deltas are not mathematical objects when they appear outside an integral, then you should submitted a paper to a peer reviewed journal where it can be judged by experts.

It is a shame a person like you has so much time and you choose to use it misleading young people on Wikipedia.

Don't bother replying if you are only gonna repeat that the "delta function as a mathematical object requires test functions and integration." It's getting tiresome. — Preceding unsigned comment added by 131.252.70.156 (talk) 23:59, 30 July 2017 (UTC)

### Comment and arbitrary break

Sławomir is correct. However, there is a formalism where the delta function can be handled without dragging along integrals that makes total sense (since it is once and for all rigorously established using integrals). There's a book in the GTM series,

written by a mathematician who throughout his career taught engineering and physics students (as well as students of pure mathematics), that highlights the soundness of delta function manipulation in the aforementioned formalism. Message: You can treat it engineering/physics style without worry. A mathematician should worry, it is his job. But for someone like Dirac (or bleaker copies or students of the non-mathematical variety), it would just be a total waste of time to worry. It isn't their job. YohanN7 (talk) 09:50, 31 July 2017 (UTC)

I'd add that just because the delta function needs to be regarded as a functional if we're talking about mathematics, doesn't mean that other attitudes about the delta function can't also be discussed. The overall perspective in such approaches is that one uses a particular realization of the delta function, say, as a limit of Gaussians (in quantum mechanics, esp) ${\displaystyle \delta (x)=\lim _{\sigma \to 0}\phi _{\sigma }(x)}$ where the limit is understood as "outside the integral". (I believe this is the attitude, for example, in the Griffiths textbook suggested above.) Very often such treatments will ask the student to "prove" identities regarding the delta function, where it is precisely this outside-the-integral-limit that allows for such a proof. So it really doesn't help students of physics and engineering either to pretend that the delta function can be treated mathematically independently of being thought of mathematically as an integrand or functional. This is not mere pedantry for its own sake: it is important for the basic mathematics to be correct. Sławomir Biały (talk) 12:47, 31 July 2017 (UTC)
No disagreement at all except maybe this: Had the basic mathematics been wrong, the formalism wouldn't have been around. Also, any responsible text or teacher will point out first thing that the delta function isn't really a function, at least not a function having all the advertised properties.
I guess the issue is this. Should the article
• Pretend that it is raining and don't mention formal use?
• Mention formal use and say it isn't rigorous?
• Mention formal use and say that the formal use can be backed up rigorously?
The first option ignores a big percentage of readers. The second option says (in effect) that most books are plain wrong. The third option is, to me, the verifiable (and IRL true) truth. I mean, even respected math texts use informal, but easily formalizable presentation. It the present case, we don't have "easily formalizable", but rather "with difficulty formalizable". The more reason so for allowing for it. YohanN7 (talk) 13:48, 31 July 2017 (UTC)
Please note, I haven't read the article in a while, just this talk thread. YohanN7 (talk) 13:17, 31 July 2017 (UTC)
I think an introductory section could be drafted illustrating one or two basic properties using the limit-outside-the-integral approach, and also the formal use in the applied setting. I think an example should be sufficient. Thoughts? Sławomir Biały (talk) 13:56, 31 July 2017 (UTC)
Sounds splendid. I'll see if I can find something "quotable" on the issue in that Vretblad book. YohanN7 (talk) 14:03, 31 July 2017 (UTC)
Something like
Regrettably, there is no ordinary real-(or complex)-valued function that satisfies these conditions. Condition 2 irrevocably implies that the integral in condition 3 must be zero. Nevertheless, using formal calculations involving the object δ, it is possible to arrive at results that are both physically meaningful and "correct"
is clearly citable or quotable verbatim. Since this issue is so "infected", it might be a good idea to have this quote (or some other like it) from an actual mathematician and not a physicist/engineer accompanying such an intro section. YohanN7 (talk) 14:51, 31 July 2017 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── I was thinking of something that starts with an example. The simplest (?) example is giving an object a kick with a delta function force ${\displaystyle p\delta (t)}$ (note that this is dimensionally consistent). The object then goes from rest to uniform motion with momentum p. Model this by a sustained force ${\displaystyle F=p/\Delta t}$ over a time interval ${\displaystyle \Delta t}$, i.e.,

${\displaystyle F(t)={\begin{cases}p/\Delta t&0

Taking the limit as ${\displaystyle \Delta t\to 0}$ gives the force ${\displaystyle F(t)=p\delta (t)}$, whose affect can be felt only if we understand this limit as taking place only outside the integral. Possibly it could also be mentioned very early on that this shows that the delta functions should be regarded as the derivative of the Heaviside function, provided the limit is "outside the integral". Sławomir Biały (talk) 16:16, 31 July 2017 (UTC)

Looks reasonable as a basic example. The equation
${\displaystyle {\frac {dH(t)}{dt}}=\delta (t)}$
is definitely a part of the basic formal (it being implicitly understood that it should be used in a context) tool-kit. This necessary qualification (where and how to use it like your "limit outside integral" (does interpretation as distributional derivative work b t w?)) is usually left out, or at least not repeatedly mentioned in the engineering/phys literature. It is omissions of this kind that accounts for most of the non-rigor in everyday practical usage. Brief mention of this is perhaps something that could go in? YohanN7 (talk) 12:46, 1 August 2017 (UTC)
In examples, we shouldn't emphasize the fully rigorous interpretation, no. But it is only with the "outside-the-integral" view that sources in this subject are able to prove "theorems" about the behavior of the delta function, and thus develop a context for the validity of the operational calculus. Most students are willing to accept these formal rules for manipulating integrals, and so they also accept the operational calculus. For example, Bracewell declares as his definition of the delta function the limit ${\displaystyle \lim _{\tau \to 0}\int _{-\infty }^{\infty }\tau ^{-1}\Pi \left({\frac {x}{\tau }}\right)dx}$. It is defined likewise in Schaum's Outline of Quantum Mechanics (for example), and probably lots of places if one really looks. Various properties of the delta function then follow from elementary manipulations of the integrals like the mean value theorem for integration. So this is not a mere matter of implicit understanding: typically sources do say explicitly that the delta function is the limit of something. I'd much rather present it this way in an example than treating the operational calculus as a magical black box. The only mathematically deep problem is specifying in what sense this limit-outside-the-integral is actually properly a limit, and in what sense all of the different limits are the same: that requires distributions, but can safely be omitted from initial examples. Sławomir Biały (talk) 13:54, 1 August 2017 (UTC)
I agree 100%, and perhaps you have misunderstood me. The delta is introduced the way you describe (or similarly, detail and rigor varies). It is, of course, only after this introduction it is "understood" how to properly use formulas like ${\displaystyle {\frac {dH(t)}{dt}}=\delta (t)}$ and similar ones. I understand that a mathematician wouldn't write this formula, even if facing the business end of a shoot gun, but a physicist is never happier than when writing them. In practice, these formulas (alas) serve more like a "table of integrals" of sorts, than as instructions to do proper math (taking limits). I think this is quite sound since it saves time, and always gives the right answer if the practitioner knows what he is doing. (See also new sections below by IP. The bra-ket notation is even more riddled with problems of this sort (and of ambiguities), but the end result is magically always the right one.) YohanN7 (talk) 07:07, 3 August 2017 (UTC)

### Dirac delta in 3D

It might be helpful to students if we added the three dimensional versions of the Dirac delta in different coordinates (spherical, polar, etc.), for easy reference.

### Application to Quantum Mechanics

This section isn't using the standard position and momentum basis that beginning QM students would easily recognize (i.e. use p instead of y). Minor note, one of the equations is not in bra-ket form. Can we add a statement in this section to better show the ease of changing basis with Bra-ket notation like "bra x" "ket psi" = psi(x) and using identity operator to change basis. The use of the Dirac delta is standard in QM to do this. This would be similar to the history section with the Fourier analysis but with the compact Dirac notation that is used in the physics community.

## Is Dirac delta function a function?

Hi all. Because Dirac delta is a distribution, it automatically means that it is a real-valued linear functional, right? But being a real-valued linear functional makes it a real-valued linear map. The latter makes Dirac delta a real-valued map. And a real-valued map is a function. So long story short, Dirac delta is a function. However, the article explicitly says it's not. It's like saying that a field is not a set. Could anyone fix it please? Thank you. Konstantin Pavlovskii (talk) 12:21, 9 August 2017 (UTC)

Hello D.Lazard,

I've decided to talk to you because apparently you reverted my recent contribution. I am extremely new to the Wikipedia and I haven't had enough time yet to get used to how things work here.

You wrote in the edit summary of your revert that "The article was not wrong" and that "the edit summary is mathematically wrong". I assume that the latter refers to the edit summary of my contribution. I think it would have been a good idea if you had left a note on the article's Talk page explaining the reason why you think my edit summary "is mathematically wrong" before actually reverting my contribution, exactly as Wikipedia recommends.

Because the contribution is already reverted by you, I'd still -- even more so -- appreciate a note from you on the article's Talk page explaining the reason why you think my edit summary is "mathematically wrong". This way you'd give me an opportunity to respond to your concern, so we could reach an editing consensus.

Thank you, I hope to talk to you soon. Konstantin Pavlovskii (talk) 10:16, 10 August 2017 (UTC)

The Dirac delta function is not a function, as clearly stated in the article about it. This fact is sufficiently known and accepted by all mathematicians, for not needing any discussion here. Thus, this is your assertion in the edit summary that is wrong, not the result of your edit. I apologise for not having done the distinction in my own edit summary. Nevertheless the revert was justified, because it is always useful to recall that the Dirac function is not a function, despite its name. As the point is not the distinction between a function and a distribution, I have edited the sentence for less emphasizing on this point. D.Lazard (talk) 12:42, 10 August 2017 (UTC)
To D.Lazard: Hello D.Lazard, what makes you think that a Dirac delta function that is a distribution, is not at the same time a function? Any distribution is a linear functional, so any distribution is a real-valued map. By definition any real-valued map is a function. I've already pointed it out in the Dirac delta function talk page the corresponding article that you refer to has been recently edited by the community to remove the incrorrect statement that the Dirac delta function is not a function. You could have a look at the new Dirac delta article yourself. I'd appreciate if you undid the harmful reverts of my contributions, please. And could please discuss my contributions on the corresponding talk pages first before reverting them. And thanks for that prompt reponse. Konstantin Pavlovskii (talk) 13:34, 10 August 2017 (UTC)
I think Konstantin's perspective is reasonable. I don't claim to fully agree with it, but I'm inclined to tone down the "not a function" aspect at least. Russian mathematicians even call it a "generalized function" ;-) I think we should clarify both the sense in which it is and is not a function, rather than relying on possibly hidden meanings of the word "function". Sławomir Biały (talk) 13:43, 10 August 2017 (UTC)
A distribution is a linear functional, that is, a map from a space of functions to a space of functions to the reals. Thus Dirac delta function is not a real-valued function, although it may be considered as a "function-valued" function. As, for everybody, a function, without further specification, means a real- or complex-valued function, you are wrong by saying that Dirac delta function is a function. D.Lazard (talk) 13:10, 10 August 2017 (UTC)
D.Lazarad, a linear functional is always a map from a vector space to its underlying field. In this particular case, it is a map from a vector space of real-valued functions of a real variable. The latter is a vector space whose underlying field is reals. Therefore the distribution (linear functional) used to define the Dirac delta function is a real-valued linear functional. And once again, a distribution has to map a vector space to its own underlying field not any other field. Therefore the Dirac delta function is a distribution, which is a real-valued functional, and -- you already know -- it automatically makes Dirac delta function a real-valued map (e.g., a real-valued function). Konstantin Pavlovskii (talk) 18:13, 10 August 2017 (UTC)
OK, Dirac delta function is a map from a unspecified vector space of infinite dimension to the reals. Moreover, the domain of this map depends on the context (either the functions, in the usual sense, that are defined for x = 0, or the functions that are defined everywhere, or the continuous functions, etc). Thus a correct formulation would be: "although they are generally considered as a generalization of real-valued functions of one variable, distributions and Dirac delta function are not such functions". In all the articles you have edited, Dirac delta function appear in contexts (for example Fourier transform), which are generally devoted to real-valued functions of a real or complex variable. In such a context, "function" means function of a real or complex variable. It is thus important to recall to non-experts that Dirac delta function is not a function in this usual sense. Thus I would agree to replace "is not a function" by something like "is not a function in the usual sense". But I disagree to remove the caveats. D.Lazard (talk) 19:00, 10 August 2017 (UTC)
First of all I did not understand why you brought up the question of dimensionality of a vector space the DDF maps to real and what follows it -- all these other things you have there up to ". Thus a". I certainly miss the reason you wrote them. What's helpful is to make sure that the users, should they have any doubt about what the Dirac delta function really is, could in one click reach a page where the fact that the DDF does not take any real value anywhere on the real line -- is expicitly written and I would strongly recommend against using word patterns that support the weird idea that the DDF is a misnomer, for example: "DDF is not a function, in <certain sense>". An engineer perceives this as you if you were saying that DDF is only a function to a certain extent (and not to the full extent). Sort of like a peanut being not exactly a nut. This pattern will continue to generate and suppport that "misnomer" that I even saw on someone's personal page here. And also could I ask you D.Lazard again to revert the reverts of my contributions that you reverted and bring any concerns you have about the reader confusing DDF with a function of a real variable to the corresponding talk pages before reverting the contributions. I will check that there are no vague statements in the articles that would make domain of the DDF appear real. Finally, as for your text suggestion, "generally considered" sounds like a pretty toxic term to me, but as I am a new user, let me focus on the fact that your text suggestion is obviously not true in the first place: "although they are <..> considered <...> generalization of real-valued functions of one variable <...> Dirac delta function are not such functions". Well, DDF is a real valued function of an exactly one variable -- of a function. It's a linear functional -- hence the name. You might want to consider consulting the article in question:namely, the definitions section. And as a reminder, please, revert your reverts of my contributions related to the DDF and bring your concerns to the corresp. Talk pages. Thank you. Konstantin Pavlovskii (talk) 20:43, 10 August 2017 (UTC)
This is just a note that I replied at User talk:D.Lazard's user talk page, but it's fine by me if anyone to refactor this discussion so it's all in one place if they want. Sławomir Biały (talk) 13:44, 10 August 2017 (UTC)
Done. Konstantin Pavlovskii (talk) 09:19, 13 August 2017 (UTC)

Just a note. Anyone first rigorously learning the notion of a function in a course on set theory will be very open-minded as to what constitutes a function. So it is good that the article is clear on what is its notion of "function". YohanN7 (talk) 14:03, 10 August 2017 (UTC)

Every article using Dirac delta function or talking about it supposes that the reader has some knowledge of calculus. The the reference to set theory is not relevant here. If one does not clearly state that Dirac delta function, is not really a function, a reader of these articles will surely think that it is a real-valued function of a real variable (the basis of calculus), which it is not. Thus not distinguishing Dirac delta function from usual functions is misleading and confusing. Thus even if, in some cases, it may make sense to consider Dirac delta function as a function, this should be avoided in WP. Moreover, any edit that would imply that Dirac delta is a function should include a reliable source supporting this. D.Lazard (talk) 14:53, 10 August 2017 (UTC)
It is rather strange to read that the reference to the set theory definition is "irrelevant". It is akin to saying that vectors that aren't arrows in 3 aren't vectors (because they aren't encountered in Calculus 101). Note that all I was saying is that it is good that the article now is trying to be precise on what it means by "function". YohanN7 (talk) 08:49, 11 August 2017 (UTC)
D.Lazard, when the article says that the Diract delta function is a function it does at the same time explicitly say that it is not a function of a real variable, and I strongly suggest to add another clarification: "therefore it is not defined for any real argument". There is no source of confusion. However, just saying that "Dirac delta function is not a function" is not true. As simple as that. Saying that it's not a function of a real variable is true, but it begs a question: "Eeeh, but a function of what kind of a variable then? Or is it not a function in principle?". There is no way around making a statement about the functionhood of the Dirac delta (in terms of the concept of a function -- weird to even mention that). And the principal answer to this question can't be "no it's not a function" because it's not true. So to me it looks like there's no other option but to explain straightforwardly to the reader whatever is already in the article plus that additional clarification that I suggested in one form or another. And (perhaps I'm getting too emotional here) there has never been an issue of a "Dirac delta function" being a "misnomer", this issue was born on the pages of the Wikipedia and traces back to at least 2002 when the community for some reason let that (mind slip?) happen that suddenly gave birth to the problem of a "misnomer". It's a pretty well-known fact that any distribution is a function and any measure is a function. Why whould anyone even expect of something defined to be a distribution (or a measure) to suddenly not be a function? This is what would sound wild to anyone (and is not true, above all).Konstantin Pavlovskii (talk) 18:50, 10 August 2017 (UTC)
The Dirac delta function was not defined to be a distribution, though - it was defined to be a real-valued function which is zero everywhere except the origin and integrates to 1 on the real line. Of course this function does not exist. This article has a challenge, therefore, that it is using the name for a non-existing object (the Delta function) to refer to an existing object (the Delta distribution). That is what the phrase "the Dirac delta function is not a function" is meant to suggest. Whether the Delta distribution is "really" a function, under the hood, is not the point. The ambivalence in this article between the Delta function and the Delta distribution has increased over time; older versions such as [1] had their own flaws but were more clear in the lede about the function nature of the Delta function. Also see my comment at [2]. — Carl (CBM · talk) 19:40, 10 August 2017 (UTC)
Hi Carl. I am not getting it, sorry, you say: <<using the name for a non-existing object>>. There is no such thing as the name of an non-existing object. It's math, objects don't appear and then pass away. Any of them either exists or doesn't in general, they are not parts of material world. When an object is non-existing, so are all its properties including the name, definition, etc. It sounds even more confusing to me as I read further: <<That is what the phrase "the Dirac delta function is not a function" is meant to suggest.>> Suggest what? You can't check whether or not an object is a function if it doesn't even exist. This statement doesn't make sense in the absense of the object that it's talking about. Could you please clarify the rest, I think I'm stuck. Thanks. Konstantin Pavlovskii (talk) 23:05, 10 August 2017 (UTC)
"The smallest odd multiple of 8" is a name for a nonexistent natural number. Here's a long list of nonexistent objects at MathOverflow [3]. So there are many names in mathematics for nonexistent objects. The prototypical "Dirac delta function" is defined as a real valued function on the real line that is zero except at the origin and integrates to 1 on the real line. We can prove that no such real valued function exists, of course. When most people write "the Dirac delta function is not a function", they are simply alluding to this fact, to point out that they are using the term "Dirac delta function" to mean something else. For example, this answer at MathOverflow [4] shows the usage: "δ isn't a function - yet sometimes it behaves like one." — Carl (CBM · talk) 01:21, 11 August 2017 (UTC)
Oh, I see it was a joke. Jesus I thought you were serious :) Of course you can make up whatever name you want and pretend that it's a name for something that's proven to not exist under ZFC, say. But the latter statement (that it's its name) is... could you tell me (a stupid physicist) what's its status in ZFC? As for the Dirac delta function, I think I've said enough, and the historical reasons for the cognitive slip that you think made it grow into a <<fake "misnomer">>... This slip now looks way more dissapointing to me than before. It so hard to avoid those misfortunate congnitive malfunctions big and small... they grow through the years and a couple decades later their falsehood just becomes a part of you. And one day you realize you're not quite the person you've been trying to be, not at all. Konstantin Pavlovskii (talk) 02:40, 11 August 2017 (UTC)

Konstantin Pavlovskii takes formal definitions too seriously. They often mix an idea and its implementation. Often an idea has many implementations. For instance, some sources define a natural number to be a finite ordinal in such a way that it is true that ${\displaystyle 5\in 6,}$ but it is evil to say so ("evil" according to nLab, not to me). See "Equivalent definitions of mathematical structures". Likewise, it is evil to say that delta function is a distribution and therefore a function. After all, in ZFC everything is a set; but it is evil to say "5 is a set". This happens outside math, too; it would be evil to say "the fifth bit of this song is 0" even though this could be true according to a given mp3 file. It is not serious, to bee too much serious; it is rather evil. "Some subjects are so serious that one can only joke about them." (Niels Bohr) Boris Tsirelson (talk) 14:03, 11 August 2017 (UTC)

Hi Boris. The article clearly identifies a particular class of implementations of the Dirac delta function in the very beginning by saying that <<the Dirac delta function, or δ function, is a generalized function, or distribution on the real number line...>>. Seeing this why don't you say that the article itself already "mixes an idea and its implementation". And then the same article goes on to say <<...[is a distribution] on the real number line that is zero everywhere except at zero>>. Jesus one of two things: either this functional operates on a trivial vector space on R (i.e. the R itself -- just {0} doesn't work because in this case it wouldn't talk about functional values "everywhere except zero") or the author doesn't know what they are talking about. Perhaps the latter since the Dirac delta function defined in the "Definitions" section doesn't take any values on the real number line whatsoever. So, a quick summary:
1. The article as of now does mix an idea and its implementation in the top section.
2. The article as of now does talk about the values that "the Dirac delta function", a "distribution on the real number line" takes for real arguments (it doesn't take them in any of the implementations of the Dirac delta described in the article).
3. Then, a few lines below the article states that the values it was just talking about (see #2) don't exist by saying that <<The Dirac delta function is not a function <...> of real variables>>
If the article picks the "distribution" implementation in the first sentence, then what's wrong with sticking to it until the end of the paragraph at least? And let's fix these statements in the top section that contradict each other. Otherwise it's just not a sound article: it discredits the Wikipedia (the best case scenario) or fools the reader into believing them all at once (the worst case scenario). When I suggest how to fix them, my contribution gets reverted, even though it explains things straightforwardly to the reader sticking to the implementation introduced in the beginning of the section. Even saying <<in the usual sense of real variables>> is ambigous. What does it mean? If someone asked what's a "function in the usual sense of real varibles?" I would answer "a real-valued function of a simply ordered set of real variables". Does anyone disagree? Is it what's actually meant by this sentence? Konstantin Pavlovskii (talk) 04:23, 12 August 2017 (UTC)
Hi Konstantin. Yes, the article does mix an idea and its implementation, since this is usual in mathematical literature. Yes, another article "Distribution (mathematics)" treats distributions as linear functionals. But it never calls these functionals "functions" (in spite of the uncontroversial fact that they are functions); this is also usual in mathematical literature. Unless otherwise stated, a reader usually interpret "function" (in this context) as a function on R or Rn.
Now think about a text that defines a natural number to be a finite ordinal in such a way that it is true that in fact 5 belongs to 6. Unless devoted to implementation details, such text never says that 5 belongs to 6 (in spite of the uncontroversial fact that it belongs).
A reader of Wikipedia, being often not a mathematician, usually is not interested in implementation details.
Think also about an equivalent definition of the space of distributions as the completion of the space of test functions w.r.t an appropriate topology. Now implementation of completion matters; usually you get that a distribution is an equivalence class of Cauchy sequences (or filters, etc). Now a distribution is not a function!
Well, frankly I am not sure it is not; it depends on very subtle implementation details; apriori it may happen that a single set from the ZFC universe is both a function (from something to something) and an equivalence class (of something). But you surely feel that this is a stupid matter. Who bothers whether a finite sequence of bits can be a legitimate sound file in one operation system and a legitimate executable in another system? Well, maybe some expert in computer crime could... but surely not a typical user of a computer.
We do not hide implementation details from the reader, but we give them DUE WEIGHT, and not more. Boris Tsirelson (talk) 04:50, 12 August 2017 (UTC)
On your first paragraph: OK. You understand that "unless otherwise stated" bit at least. Because the question of the functionhood of the Dirac delta function is controversial enough (not in Russia though) to ignite fierce discussions and is a well-known public misconception, I suggest simply clearing up this question in the top section once and for all, at the same time clearly stating the nature of the DDF functionhood right in the same sentence or right next to it. If you prefer a "light version" of the same fix that's fine with me personally (I'd call this an editing consensus, although see the note below), then just stating that the DDF is not a function of a real variable is enough I think, it is exactly the change that was introduced right after I brought up this topic. I didn't have major concerns about the new way it was worded <<The DDF is not a function, <an addition about a real variable> >>. However, this light version is not extremely appealing to me as to a person who frequently has to clear the air in terms of whether the DDF is a function (fullstop). A first-year engineering student is very likely (in my experience) to interpret "The DDF is not a function, <blah-blah-blah> " as "The DDF is not a function, I understand this part, I don't care about what's after the comma". This is because students don't know much about the functions that are not functions of a real variable, so they take that little piece after the comma as an additional clarification of whatever is before the comma rather then a (very serious!) reduction of scope of the statement. That's why saying "The DDF is not a function of a real variable" is much better in my view.
About the ordinals, there's no wide misconception among the public (with serious consequences) on whether or not in the implementation they use 5 is an element of 6. It's not a practical matter, so mentioning this explicitly is not needed, certainly not in the top section of an article that talks about a set of natural numbers. But if in the impelementation you describe 5 is indeed an element of 6, then saying explictly that it's not is not an option, it would be like selling your soul to the devil. Because it's not true and you would be confusing the more advanced readers. So just not mentioning this fact (even though it might be true) is totally acceptable and is not confusing to anyone.
I don't insist on specifying the details to the reader. I would merely want the top section of the article to not contradict itself and to clear up the question of the DDF functionhood to those students/engineers/physicists who are exposed to the publicly accepted view of the DDF as of a misnomer.
About the alternative implementations of a distribution. The Wikipedia is not trying to write about everything in the world and certainly not at once. So, in my view, picking a particular implementation (of the DDF and of a distribution and of a natural number and ...) is neccessary in the quick description of the DDF in the top section. Unneccessary generalizations and "meta-language" of ideas is confusing enough even for experienced readers. I remember reading Canadian Criminal Code that has a list of definitions of what "a public nuisance" or "a public intoxication" (or smth like that I don't recall) mean in this particular section. It's not an option for an article in the Wikipedia to resemble a text of a legislation, it would scare off the readers. It is for a reason that ideas and implementations get mixed in the mathematical literature and in any literature. For example, talking about the Soviet past, a Russian text would say "the Communist party", however the implementations of a communist party are numerous. Speaking about a relatively complicated mathematical concept in the "meta-language" of ideas won't let you use the terminology appropriate to the common implementation of the concept.
So selecting a common implementation and sticking to it in the top section is a must. Whoever is interested in the alternative implementations are advanced enough to not get confused with a top section written like that, it does no harm to them, at the same time, less experienced readers benefit from it because this way it is more understandable and uses the terms that they are used to.
So the program that I suggest is:
a) (a must) fix the mutually contradicting and vague (like <<in the usual sense of real variables>>) statements in the top section while sticking to the common implementations of the mathematical concepts throughout the top section, then
b) clarify the issue of the functionhood of the DDF in the top section with straightforward and unambigious sentences that preferably don't use the "the DDF is not a function <comma>" word pattern, replacing it with "the DDF is not a function of a real variable". The latter being done with the sole intent of benefiting your average engineer or physicist who is not used to the fact that commas in math routinely and greatly reduce the scope of the statements. Konstantin Pavlovskii (talk) 06:46, 12 August 2017 (UTC)

I have edited the lead in order to avoid excessive formalism introduced by Konstantin.pavlovskii, and to keep the important functional property of Delta function. IMO, this new lead is much more informative for readers that have never heard of distributions, and, nevertheless, remains mathematically correct. D.Lazard (talk) 10:42, 12 August 2017 (UTC)

Excellent! Thank you Daniel and Sławomir. Now it looks like a decent top section to me. Konstantin Pavlovskii (talk) 16:09, 12 August 2017 (UTC)

## Softplus-inspired (smoothed) infinite delta function

I've found it to be so:

y = ln(2) / ( abs(x) - ln(cosh(x)) ) - 1

I like it, because it grows really nice and it spire is sharp as hell! Does this has a name?

And its finite modification:

1 - ( abs(x) - ln(cosh(x)) ) / ln(2) — Preceding unsigned comment added by Xakepp35 (talkcontribs) 01:36, 15 February 2018 (UTC)

That's not a model of the delta function. Sławomir Biały (talk) 02:07, 15 February 2018 (UTC)