This question is about making consistent the definition and properties of the Dirac Delta function from two different sources.

In this online textbook, the following definition is given.

$$\delta_{\alpha}(x) = \begin{cases} \frac{1}{\alpha}, \quad |x| < \frac{\alpha}{2} \\ 0, \quad |x| > \frac{\alpha}{2}\end{cases}$$ Let $g: \mathbb{R} \to \mathbb{R}$ be a continuous function. We define $$ \int_{-\infty}^\infty g(x)\delta(x-x_0) dx=\lim_{\alpha \to 0} \int_{-\infty}^\infty g(x)\delta_{\alpha}(x-x_0) dx$$ Lemma: Let $g: \mathbb{R} \to \mathbb{R}$ be a continuous function. We have $$ \int_{-\infty}^\infty g(x) \delta(x - x_0) dx=g(x_0).$$

However, on the wikipedia article,

...This limit is meant in a weak sense: either that $$ \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \eta_\epsilon(x)f(x)dx = f(0)$$ for all continuous functions $f$ having compact support, or that this limit holds for all smooth functions $f$ with compact support.

where $\eta_\epsilon$ can be defined similarly to $\delta_{\alpha}$ in the previous case.

Why does the first reference not require compact support, while the second reference does? Is one of them incorrect?


1 Answer 1


Distributions are functionals which act linearly and continuously on a space of some class of test functions, such as compactly supported, or rapidly decreasing, or just all continuous functions as above. By analogy with the Riesz representation lemma, which says that certain classes of linear functionals are actually integrals, we may pretend that some other linear functionals, like the Dirac delta function, are also integrals.

The main reason to prefer smooth functions of compact support is that it makes integration by parts particularly easy to use, and therefore weak derivatives of distributions is easy to define.

For example, I would like to know what the derivative of the Dirac delta function (distribution) is.

$$\int^\infty_{-\infty} g(x)\delta'(x-x_0)\,dx = g(x)\delta(x-x_0)\big|^\infty_{-\infty}-\int^\infty_{-\infty} g'(x)\delta(x-x_0)\, dx $$

Since a priori $\delta$ is not a function, I do not know how to define its derivative. The left-hand side of the above equation is not defined, though I would like it to be true. So I simply define it to be true. I define the derivative of the delta function to be that functional which makes the above equation true. This is called a weak derivative.

Now if I chose $g(x)$ to be of compact support, or to at least converge sufficiently rapidly to zero at infinity, then I can conclude that the boundary term in my integration by parts vanishes. So I can define the derivative $\delta'(x-x_0)$ to be the functional which sends $g$ to $-g'(x_0).$

But if I assume that $g(x)$ is just any continuous function, then I have to carry that boundary term, and my weak derivative loses some good properties. This is a good reason to use test functions of compact support.

  • $\begingroup$ Hi! Thanks for your response. A few questions to test my understanding: 1. If we didn't require the derivative of $\delta$ to be defined, could we then remove the requirement that our test functions be compact? 2. Regarding your last paragraph, wouldn't the boundary term vanish regardless of how fast $g$ converges? Since $\delta(\infty)=0$? $\endgroup$
    – Student
    Commented Nov 8, 2017 at 5:49
  • 2
    $\begingroup$ @Student 1. Yes, I believe so. Well there may be other reasons to prefer a regular class of test functions: the larger your class of test functions, the smaller the possible distributions you include. Also they will have better Fourier transform properties. 2. Yes, that argument will apply for the delta function, but not for other functionals. $\endgroup$
    – ziggurism
    Commented Nov 8, 2017 at 6:00

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