# Dirac delta

So I've been told that the Dirac delta functional is a distribution, but I don't see why that's the case. I had an introduction to distributions in my calculus IV course, but as I remember it, a distribution involves and integral containing a the product of a function from the Schwartz space and a function that is "continuous and slowly increasing" (I don't know the actual English expression for them). How does the Dirac delta fit into that?

Unless my understanding of the terminology is incorrect, you're confusing the terms distribution and distribution function. From Wikipedia, "a probability distribution identifies either the probability of each value of a random variable (when the variable is discrete), or the probability of the value falling within a particular interval (when the variable is continuous)."

A distribution function F(x), however, is defined as F(x) = P(X$$\leq$$x) - the probability that the random variable X takes a value less than or equal to x. So the Dirac delta is a distribution, not a distribution function. In fact, the distribution function of the Dirac delta is the Heaviside step function.

I don't think I'm confusing these concepts. The distributions I'm talking about are functionals, not functions. And the Heaviside function is the ant-derivative of the dirac delta in the sense of distributions. I think the correct term for what I'm talking about is "temperate distribution".

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arildno
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It is rather informal on the technical distribution concept, but more rigorous in establishing the connection between the Dirac function(al) and the integral representation of it.

Wow that's great. I read the first post and what got me confused is something rather different than what you describe arildno. My (math) teacher taught me that writing "the delta" inside of an integral is a useful but ultimately wrong "notation abuse" of the convolution of distributions. What got me confused is that, if you can't write the delta inside the integral, how do you define it as a distribution?

arildno
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Wow that's great. I read the first post and what got me confused is something rather different than what you describe arildno. My (math) teacher taught me that writing "the delta" inside of an integral is a useful but ultimately wrong "notation abuse" of the convolution of distributions. What got me confused is that, if you can't write the delta inside the integral, how do you define it as a distribution?
A distribution is a linear functional, Amok.
There might be some other technicalities here, but that is essentially what it is.

Thus, a distribution D on some function space satisfies the condition of linearity, most importantly D(a*f+b*g)=a*D(f)+b*D(g), where f and g are any two functions in the function space, and a and b scalars.

jambaugh
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Here's a way to understand the delta "function".

Consider functions on some interval as vectors in an abstract vector space. We can then consider linear functionals which are linear mappings from functions to numbers. These form the dual space. If a vector space is finite dimensional and has an inner product we can express every functional using the inner product with another vector:

$$f \mapsto \langle g,f\rangle$$
The integral of the product of two functions over an interval forms a good inner product and we can express many of the linear functionals via:
$$f \mapsto \int_{x_1}^{x_2} g(x)f(x)dx$$
But since the function space is infinite dimensional there are "more" dual vectors i.e. linear functionals than there are vectors i.e. functions. For example we cannot express the evaluation mapping:
$$f \mapsto f(0)$$
as an integral of f with another function. So we invent a dummy function name to hold the place of that missing function in the integral notation. And that is the delta "function".

It allows us to continue using the integral inner product to express those linear functionals which are not actually expressible as integrals.
$$f \mapsto f(0)=\langle \delta , f\rangle = \int_{x_1}^{x_2} \delta(x)f(x)dx$$