I recently came across dirac delta function and trying to learn about it has led me to learn that it is a distribution/generalized function and is not a ordinary function. But most of the explanations as mathematically abstract. So I am trying to understand it using examples. The common example given while explaining the dirac delta function is that it helps express in a mathematically-correct form such idealized concepts as the density of a material point, a point charge or a point dipole, the (space) density of a simple or double layer, the intensity of an instantaneous source, etc. I am wondering how the distribution helps mathematically represent concept such as density. The sentence "On the other hand, the concept of a generalized function reflects the fact that in reality a physical quantity cannot be measured at a point; only its mean values over sufficiently small neighbourhoods of a given point can be measured. Thus, the technique of generalized functions serves as a convenient and adequate apparatus for describing the distributions of various physical quantities. Hence generalized functions are also called distributions." (Ref:https://www.encyclopediaofmath.org/index.php/Generalized_function) means that the mass is "continuous" (in continuum approximation sense), but when talking about "density at a point", in reality we consider the only the mass closely "distributed" around that point. This is why density is a "distribution".

a)So is density not continuous?

b)How does the integral: $$ \int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a) $$ help explain the concept of density? is the test function $f(x)$ giving the continuous density through out the one dimensional space and the delta function help us get the density at a point a?

C)Why not directly get the value of the trial function at point 'a' using f(a) instead of using the round about way of delta fucntion?

Note: Here is a good notes that I found useful -> https://math.mit.edu/~stevenj/18.303/delta-notes.pdf

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    $\begingroup$ For intuition, you can think of the delta function as being just a smooth nonnegative function which is zero everywhere except for an extremely narrow spike near the origin. The spike is tall enough that the area under the curve is $1$. $\endgroup$
    – littleO
    Aug 2, 2019 at 15:00

1 Answer 1


a) If you tried to define what "density" is and what "continuous" for it means, you would find that it makes no sense. As you use it, it defines a measure, that is, it assigns every interval (and thus every Borel set) some positive value. One can show that this is equivalent to a linear continuous functional on the space of continuous test functions, either compactly supported or fast falling. In that later sense, continuity is built-in.

b) It does not explain anything, it is just a definition of the left side by the right, as the left side does not have a meaning in itself. You can use it as another symbolic way to denote the application of the delta distribution to a test function. The non-symbolic content is the right side, it defines what the left side means.

c) You are right, one could say that this is just the evaluation operator in $a$ and write $E_a(f)=f(a)$. But Dirac introduced this notation and since then it is traditional. It is also compatible with the idea of an approximate identity (of the convolution of $L^1$ functions) which approximate delta in a distributional sense. To recap, any function $\phi$ which is non-negative, and integrable with integral $1$ defines an approximate identity (at $x=0$) with the sequence of functions $\phi_n(x)=n\phi(nx)$. So it is common to write $\phi_n\to\delta_0$ for $n\to\infty$ (in the topology of distributions).

See Carl Offner: "A little harmonic analysis" for a deeper trieatment of these ideas.

  • $\begingroup$ a) So instead of giving a value for density at a point, like a function, we get a value in a small area around that point, i.e. in a "interval" by a "compactly supported functions". These compactly supported functions make up the test function space and give thus give value fot the density, maintaining "continuity", at every point in space. Is this right ? $\endgroup$
    Aug 5, 2019 at 6:19
  • $\begingroup$ b) For this example, the test function space has "functions" which give the value of the density "in interval" around different points in space ? $\endgroup$
    Aug 5, 2019 at 6:25
  • $\begingroup$ a) By making the test functions under consideration arbitrarily small you can localize the behavior of the distribution, you find that if the support does not include $a$, the value you get is zero. This leads to declaring that $\{a\}$ is the support of $δ_a$. But again, at $a$ you get no value of $δ_a$ as function. $\endgroup$ Aug 5, 2019 at 6:27
  • $\begingroup$ b) There is no reason to expect that such a density exists. If it exists, one speaks of "regular distributions", most of the interesting ones are not regular. $\endgroup$ Aug 5, 2019 at 6:32
  • $\begingroup$ @GRANZER : As an exercise contemplate if you can fit the "dipole" $\delta'$ into the density picture. And what exactly $x\delta'(x)$ amounts to. $\endgroup$ Aug 5, 2019 at 16:22

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