# Is the Dirac Delta "Function" really a function?

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I am given to understand that the Dirac delta function is strictly not a function in the conventional sense and it is a "functional or a distribution".

The part which I can not understand why the Delta "function" makes sense only when it acts on another function and that too only inside an integral and how is a "functional" or "distribution" different from a function.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user AchiralSarkar
asked Jan 24, 2013
retagged Jun 16, 2014
See this answer.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Did

This is a ridiculous question to import here given its purely mathematical content, please close it.

No not until at least 3 people with > 500 rep vote to close, PhysicsOverflow intentionally refrains from adopting the anti-mathematics policies enforced on some other physics Q&A sites, that is why we have a mathematics category in the Q&A section to store questions about math useful to and applied by physicists. (In fact I am slowly imorting questions, that have been wrongly moved out of sight of the physicists that are interested in them...) The Dirac Delta function (or more exactly distribution ;-) ...) and its properties are heavily applied in physics.

@Dilaton

I agree with @physicsnewbie, this is a pure mathematics question, which

• does not need a physicist's point-of-view answer.
• is not motivated by physics
• does not explicitly mention any reference to physics or an application to physics

In fact, whether it's a function or a distribution hardly matters from a physicist's point of view...

Physical Physics, Mathematical Physics, and Physical Mathematics are all on-topic, but Mathematical Mathematics is not, and should not be, because they'll be better off elsewhere (assuming it's high-level enough).

Lots of things are "of interest to physicsists" but that doesn't mean that we can import every question related to nearly all of Mathematics, typesetting in TeX, and whatever. I thought we decided that this is an appropriate way to determine the appropriateness of a pure maths question?

@Dimension10

No I do not 100% agree the answer on the other physics Q&A site, the question is definitifely of interest to physicists. We will see what the community says, but my personal first order approach is  what is of interest for out community is on-topic. The answer there already contains too much of the specific philosphy applied there to be directly applicable to PhysicsOverflow IMHO.

BTW I am not randomly importing math questions ;-), but the once that have wrongly been migrated away and that were appreciated (upvoted and answered) by another physics community and that (or their answers) seem interesting enough for PO.

@Dilaton I'm not sure what you're saying, are you saying that even questions about LaTeX typesetting should be on-topic because they are of interest of physicists? I always thought that the Mathematics section was for what I call "Physical Mathematics", Mathematics from a physicist's point of view (in the same way that Mathematical Physics is Physics from a Mathematics point of view).

I guess I'll retract my close vote now, though, after Ron's answer, which seems to be more of a physicist's point-of-view answer.

Mathematicians dont give a damn about the Dirac Delta, and actually I have met many of them who hadnt ever heard about it. It is physicists stuff.

It is not the last time we will have such discussions, since by definition physics is the science that tries to emulate nature by means of mathematical models.

I trend to think like Dilaton in that sense. At least up to some extent, many things that are of interest to physicists are not off-topic. OK, a question about importing a picture to Latex should be closed, but "How do I write h bar?" should be right.

@Eduardo Do you think physicists care a damn about the Dirac delta function "really" being a function? There's nothing wrong importing questions that are of relevance to physicists, unlike this one.

@physicsnewbie of course everybody can always express his opinion and point of view, but if the majority of other people who cares about the issue disagrees with you, it is probably a good idea to accept this. You already have made your point of view completely clear, so ...

In addition, I am not 100% sure if you can really judge what kind of maths question  (in particular theoretical/fundamental/mathematical) physicists would appreciate and consider as useful (?) ...

## 6 Answers

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First you should confront the question why should I think of the $\delta$-function as a function at all? If you are trying to imagine it as a real-valued function of real inputs, which just happens to be $0$ just about everywhere, then you are off to a bad (but very common) start. You can define $\delta$ as a symbol with certain properties relating to combining it with an actual function and some other symbols (e.g. $\int$), and this really suffices for most purposes, so why insist on trying to cram such an interesting object into a limited definition of "function?"

So instead, let's take a different approach. Let $f : \mathbb{R} \to \mathbb{C}$ be a generic function from the reals to the complexes. Consider the set of all1 such functions, and call it $L$ for lack of a better letter. $L$ is a set just like $\mathbb{R}$, and so we can define maps (read: functions) from it to $\mathbb{C}$ as well. The $\delta$-function is one such beast, defined by \begin{align} \delta : L & \to \mathbb{C} \\ f & \mapsto f(0). \end{align} Thus it is a function, but not of real numbers. It is a function of functions of reals, which is sometimes called a functional.

So what about the integrals? Well you can also approach this in a limiting fashion. One way is to note that $$\lim_{\sigma\to0} \int\limits_\mathbb{R} f(x) \frac{1}{\sqrt{2\pi\sigma^2}} \mathrm{e}^{-x^2/2\sigma^2} \mathrm{d}x = f(0).$$ Exchange the limit and the integral2, and you see that there is a "function" - or rather a limit of a sequence of functions from $L$ that is itself not a member of $L$ - whose values seem to be given by $$\delta(x) = \lim_{\sigma\to0} \frac{1}{\sqrt{2\pi\sigma^2}} \mathrm{e}^{-x^2/2\sigma^2}.$$ This is what a distribution is, with terminology suggestive of the probability distributions one so often integrates against (though I could be mistaken on the etymology). Note though that we really weren't allowed to switch that limit and integral while we still called that Gaussian-looking thing a member of $L$. After all, taking the pointwise limit first produces something that vanishes everywhere but a point, and such an object will cause the Lebesgue integral we were using to vanish as well.

In any event, the integral was there from the very beginning. You can think of this as overbearing notation for what we really wanted to say: "Give the value that results when $\delta$ acts on $f$." The integral notation has another advantage, though, and that is in connection with inner product spaces. Secretly, we constructed $L$ to be a vector space over $\mathbb{R}$. Then the set of linear maps from $L$ to $\mathbb{C}$ form its dual space $L^*$. For every $g \in L$ there is a corresponding $g^* \in L^*$, which can conveniently be represented in this integral notation as the complex conjugate of $g$.3 The inner product of $f$ and $g$ is $$\langle f | \underbrace{g}_{g\in L} \rangle = \int\limits_\mathbb{R} f(x) \underbrace{g^*}_{g,g^*\in L}(x) \mathrm{d}x,$$ and so you can identify \begin{align} \underbrace{g^*}_{g^*\in L^*} : L & \to \mathbb{C} \\ f & \mapsto \int\limits_\mathbb{R} f\underbrace{g^*}_{g,g^*\in L}. \end{align}

Now for every $g \in L$ there is a corresponding dual member that you can write as the complex conjugate of $g$ for the purposes of such integration, but the converse is not true.4 $\delta$ is an example of a member of $L^*$ that has no actual function in $L$ we can complex conjugate and integrate against to replicate its behavior.

1 In practice this is often too much. It's better to restrict attention to, e.g., all square-integrable functions from $\mathbb{R}$ to $\mathbb{C}$.

2 Beware! A very dangerous thing to do!

3 Yes, we are about to thoroughly abuse the two meanings of $*$ - be on the lookout.

4 It won't be in general unless $L$ is finite-dimensional, but in that case you have Kronecker deltas and finite sums rather than Dirac deltas and integrals.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Chris White
answered Jan 24, 2013 by (80 points)
I don't believe the term distribution as used for $\delta$ has anything etymologically to do with the term distribution used in probability.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user KCd

The terminology is deliberately the same. A probablilty distribution is a distribution (in the sense of a positve linear functional) that is nonnegatvie on nonnegative function and integrates to 1.

+ 6 like - 0 dislike

The reason that we try to make sense of the $\delta$-function inside of an integral is because its defining characteristic is given in terms of an integral. That is, the $\delta$-function is zero everywhere other than the origin, and

$$\int_{\mathbb{R}} \delta(x) dx = 1$$

Heuristically, the $\delta$-function concentrates all its mass at the origin. Of course, no actual function $f(x)$ enjoys this property, since if $f(x)$ was zero everywhere other than the origin, it would have integral $0$, even if $f(0) = + \infty$.

That being said, there are situations, such as in the study of electromagnetism, that we would like to talk of positive mass existing at a point. The language of integral calculus is indispensable, but it does not classically allow for such constructions. Thus, the $\delta$-function emerges as a way of allowing our theory of integration to make sense of these point masses.

One way to remedy the fact that the $\delta$-function is not a function is to reinterpret it as a distribution, as Chris explained above. Another option is to think of it as a measure. If you haven't studied measure theory, I'll avoid the technical details, only to mention that a measure is a way of assigning a size to a set. The integral above ends with $dx$, which corresponds to the measure that assigns to every set its "obvious" size. The size of $[0,4]$ is $4$, the size of a point set is $0$, and this can be extended to most "messy" sets in a sensible way. When we used the measure $dx$, it was impossible for our integral to detect point masses, since a point was assigned zero size, and hence was inconsequential with respect to integration.

However, we can define a measure $\delta_0$ that assigns a set size $1$ if it contains $0$, and assigns it a size of $0$ otherwise. If we integrate using this measure, all mass is concentrated at the origin, and indeed we have

$$\int_{\mathbb{R}} f(x) d\delta_0= f(0) =  \int_{\mathbb{R}} f(x)\delta(x) dx"$$

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Isaac Solomon
answered Jan 24, 2013 by (60 points)
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There are several ways to look at this. Personally, I think it is more clear to say that the Dirac $\delta$ function is actually a function -- that's what the name says. Unfortunately, the Dirac $\delta$ function does not exist. There are closely related objects, however, which do exist, and which let us do most of the things that we wish we could do with the $\delta$ function if it actually existed. Understanding both of the previous two sentences is key to understanding what is going on.

Because the $\delta$ function does not exist, many people redefine the term "$\delta$ function" to mean something that does exist, so that they can still talk "as if" the $\delta$ function existed, even though it does not. I find this somewhat revisionist, but it is the most common approach I have seen among people who use the $\delta$ function in their day to day work, so you have to expect it when you look at the literature.

The idea is that any statement involving the $\delta$ function is actually an abbreviation of a different statement, or family of statements, each of which only involves objects that actually exist. For example, the equation $$\int f \delta\,dx = f(0)$$ can be viewed as an abbreviation for $$\lim_{n \to \infty} \int f\phi_n\,dx = f(0)$$ where $(\phi_n)$ is a particular sequence of actually-existing functions. Thus the $\delta$ function is replaced by the $\delta$ distribution. The answer by Chris White has more details.

On the other hand, $$\int f \delta\,dx = f(0)$$ can also be viewed as an abbreviation for $$\int f\,d\delta = f(0)$$ where the "$\delta$" on the right-hand side is not the $\delta$ function, it's the Dirac measure. This is explained in more detail by Isaac Solomon. Again, "$\int f\delta\,dx$" is a purely formal abbreviation because (in the jargon of measure theory) the $\delta$ measure is not absolutely continuous to Lebesgue measure and so it has no Radon-Nikodym derivative; if this derivative existed, it would be the $\delta$ function.

One reason to continue writing $$\int f \delta\,dx = f(0)$$ is that it does not commit us to either of these two interpretations; we can switch back and forth between them whenever it is convenient. It is also convenient for setting up computational problems, in the same way that some people set up integrals by drawing diagrams labeled with infinitesimals while at the same time accepting that infinitesimals don't exist.

Another example of a non-existent but still useful mathematical object is the field with one element. There is no field with one element - so this object does not, strictly speaking, exist. But it has nevertheless been useful as a way of thinking about results that involve objects that do exist.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Carl Mummert
answered Jan 29, 2013 by (50 points)
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Not really. $δ$ is not a pointwise defined object. It's a distribution and defined in terms of how it acts on test functions.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user B. S.
answered Jan 24, 2013 by (40 points)
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The easiest way to understand this is by making a lattice and taking a limit. On a lattice, a function is a discrete thing, and a delta-function is a simple construction. Then you can ask about limits as you make the lattice small.

There are many functions you get depending on the condition for convergence. If you insist that the value at every point has to converge to a fixed value as the lattice gets small, you get an ordinary function. These are the functions you study in grade school, they can be defined directly as functions on R, without taking an explicit limit (the limit is implicit, however, in the definition of R).

But you can also allow the values be diverge, so long as the averaged or integrated vaue over any region is convergent. This defines distributions. In this way of taking a limit, a delta function is nonzero only at a point, and it's integrated value is 1.

To make everything differentiable, so that you can define the derivative of a delta function, you need to define the average using weights on the lattice which converge to infinitely differentiable functions. The result is that a distribution is defined by it's linear action on infinitely differentiable functions, and the derivatives of distributions are defined by integration by parts, by the action of the distribution on the derivative of the function. There are lots of crazy distributions, for example, the derivative of a particular instance of a random walk, which are everywhere divergent, but whose integral over an interval is convergent.

You always should have a lattice or some sort of other regulator in mind when talking about the real numbers in the context of geometry.

Distributions don't exhaust all meaningful generalized functions. There are functions which are diverging in the continuum limit too fast to be distributions, the integral over a region is also divergent! An example of this is the continuum limit of a 1-d field whose correlation function goes like 1/|x-y|^2, where x and y are two positions. This has a log-divergence when you integrate it over x and y, so it doesn't become well-defined even when smeared out. I have heard that this shows up in the scaling limit for boundary correlators in the Ising model, but I don't know the details. This is not such a big problem, it just means you need to be a little more careful when defining the limiting process in this case.

answered Jun 16, 2014 by (7,720 points)
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$\delta$ is a function to exactly the same extent that $\infty$ is a number.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user TonyK
answered Mar 27, 2014 by (0 points)
Huh? In what exact sense could this be?

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Dirk
@Dirk: (i) A beginner treats $\delta$ as a function and $\infty$ as a number. (ii) A more sophisticated mathematician learns that neither of these assumptions is justified. (iii) A professional mathematician knows how to define $\delta$ as a function, and $\infty$ as a number, so that both definitions make sense.

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user TonyK
Ah I see. You meant that the individual perception of $\delta$ and $\infty$ may be "exactly the same"… It was not meant as a statement about the mathematical role of either of the two, right?

This post imported from StackExchange Mathematics at 2014-06-16 11:24 (UCT), posted by SE-user Dirk

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