# What really is a Dirac delta function?

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Yesterday a friend asked me what a Dirac delta function really is. I tried to explain it but eventually confused myself. It seems that a Dirac delta is defined as a function that satisfies these constraints:

$$\delta (x-x') = 0 \quad\text{if}\quad x \neq x'$$

$$\delta (x-x') = \infty \quad\text{if}\quad x = x'$$

$$\int_{-\infty} ^{+\infty} \delta(x-x') dx = 1$$

I have seen approximation of the dirac delta function as an infinitely peaked Gaussian. I have also seen interpretation of Dirac delta function as a Fourier Transform which is widely adopted in study of quantum theory.

So what really is a Dirac delta function? Is it a function? Or is it some kind of limit for a large class of functions? I am really confused.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Zhengyan Shi
I personally like the standard definition in terms of dual spaces as $\delta_{x_0}(f) = f(x_0)$, I find it the easiest to understand.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Gennaro Tedesco
@GennaroTedesco Totally agreed. But I think the equally rigorous notion of distributions as sequences alongside it conveys vital intuition for physical / signal theoretic problems. Then the latter is also, as you say, much more complicated, so it's fun to understand that one can use the two equivalent notions as needed and reflect that all the complicated sequence stuff really is as simple as an abstract, simple linear functional.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
@HDE: While true, the physics context suggests the theory of Schwartz distributions is the right answer, rather than other notions of generalized function (e.g. measure theory) that could be used instead.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Hurkyl
This has been asked before on Physics, and we migrated it to Math then too.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Chris White

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It is a distribution. The easiest, cleanest way to think of it is as a linear functional $\mathscr{H}\to\mathbb{R}$ on the Hilbert space $\mathscr{H}$ of functions $\mathbb{R}^N\to\mathbb{R}$ that are $\mathbf{L}^2(\mathbb{R}^N)$. Input a function $f\in\mathbf{L}^2(\mathbb{R}^N)$, and DiracDelta spits out $f(0)$. It's a manifestly linear operator.

Historically it was intuitively motivated by the need for a function $\delta:\mathbb{R}^N\to\mathbb{R}$, in the everyday sense, with precisely the properties you state. It's also motivated by the Riesz Representation Theorem: a Hilbert space is complete, so every continuous linear functional can be represented as an inner product in the space: for every continuous linear functional $f:\mathbf{L}^2(\mathbb{R}^N)$ there exists a unique function $f_0:\mathbb{R}^N\to\mathbb{R}$ in the everyday sense such that:

$$\langle f_0,\,g\rangle = \int_{\mathbb{R}^N} f_0(U)\,g(U)\,\mathrm{d} U\tag{1}$$

In finite dimensional vector spaces over $\mathbb{R}$ kitted with inner product, all linear functionals are also continuous. Thus, for example, the one forms make up all the linear functionals that there are: there are no linear functionals that cannot be written as the action $X\to\omega(X)$ of a one form $\omega$ on the vectors in a finite dimensional Hilbert space. In physicist's speak: all linear functionals can be written as vectors with their indices lowered by contraction with the metric tensor $g$.

In infinite dimensional spaces, the notion of continuous linear functional is strictly more precise than simply linear functional: there are always linear functionals which are not continuous and indeed DiracDelta is one. See, for example, my answer here (as well as the MathOverflow threads linked in it) on rigged Hilbert spaces for more information. So we can discuss distributions through this notion of rigged Hilbert space: we kit the Hilbert space with a stronger topology than simply the Hilbert norm topology so that the new space's topology is precisely strong enough, but no stronger, to deem all linear functionals to be continuous with respect to this newer, stronger topology. We speak of the "topological dual space" of all continuous linear functionals to a Hilbert space as distinct from the larger, "algebraic dual" of linear, but not needfully continuous functionals. It is the former that a Hilbert space, by definition, is equivalent to, not the latter. The Riesz Representation Theorem in this context is then the assertion that this definition of topological self duality is the same as the Hilbert space definition as a complete, inner product space.

So there is no everyday function with the properties you state precisely because the notion of linear functional is strictly more general than the notion of continuous linear functional in infinite dimension Hilbert space.

We can also think of distributions as sequences of everyday functions. This is the approach you allude to in the thinking of DiracDelta as a a sequence of ever peakier Gaussians. This is the approach that M. J. Lighthill, "An Introduction to Fourier Analysis and Generalised Functions" uses. It's a cute little book: rigorous, highly readable but quite off-beat (it uses nomenclature "good functions" for the Schwartz functions that I have not seen elsewhere, for example) and a little dated. But it is still a valid conception of the notion of distribution.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
answered Oct 3, 2015 by (485 points)
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The delta function is an instruction to switch the order of limits.

Here's how to think about it. Let's say we have an infinite sequence of functions $\delta_1$, $\delta_2$, ..., which satisfy the following properties:

1) $\int\delta_i(x)dx=1$

2) The integral of $\delta_i$ over any patch not containing $0$ approaches $0$ as $i\rightarrow\infty$

Then you can say the following: if $f$ is any continuous function, $\lim_{i\rightarrow\infty}\int \delta_i(x)f(x)dx = f(0)$. You CAN'T say that $\int\lim_{i\rightarrow\infty} \delta_i(x)f(x)dx = f(0)$, because the limit of $\delta_i$ might not exist, or it might not be integrable, or it might integrate to zero. This is, in fact, exactly what happens; the $\delta$ function is NOT a real function. It's an instruction. Whenever you see $\int \delta(x)f(x)$, replace $\delta$ with a sequence of functions converging as described above, and flip the limit and the integral. In practice, this amounts to the rules you gave above.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Jahan Claes
answered Oct 3, 2015 by (40 points)
It's a point that's annoyingly often left implicit, but $\lim \int \delta_i(x) f(x) \, dx = \int \lim \delta_i(x) f(x) \, dx$ is true if the right hand limit is meant as a limit of distributions.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Hurkyl
+1 Great summary of the Lighthill approach

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
All correct, but I always wondered if these kind of definitions only define the scalar products (i. e. the integrals) against the $\delta$ rather than the object $\delta$ itself. Can one reconstruct the $\delta$ as element on itself back from the limits?

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Gennaro Tedesco
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This will not be a purely mathematical answer. The way that I think about a Dirac Delta Function is that it is an infinitely tall spike that has infinitesimal width, with it's area defined to be 1. When you integrate over a delta function you get the area under the "curve", which is by definition 1. It also has an awesome property when you do $$\int \delta(x-a)f(x) = f(a)$$ This is because the delta function is 0 everywhere except where the argument in the delta function is 0.

This post imported from StackExchange Physics at 2015-10-05 20:47 (UTC), posted by SE-user Carl
answered Oct 4, 2015 by (60 points)

Correct.

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