# Understanding why the domain of a distribution is defined to be smooth functions with compact support

I'm trying to understand the motivation for the definition of a distribution and am probably missing something basic. Why is the domain of distribution / generalized function defined to be the set of smooth functions with compact set support instead of the set of functions with some other property and compact support?

In particular why aren't "competitors" like the following used as the domain

• the set of differentiable functions from $$\mathbb{R} \to \mathbb{R}$$ with compact support
• the set of piecewise linear functions with finitely many pieces with only removable discontinuities with compact support.
• the set of piecewise constant functions with finitely many pieces with compact support

I'm wondering how you would motivate the actual definition ... and in particular how you would motivate it by considering a couple of simpler ideas and then striking them down as unworkable.

A distribution / generalized function is defined as a map from $$T \to \mathbb{R}$$ where $$T$$, the set of test functions, is defined as follows

$$T \stackrel{def}{=} \left\{ x \in \mathbb{R} \to \mathbb{R} \text { where x is smooth and has compact support }\right\} \tag{1}$$

I'm picturing the dirac delta function as a motivating example for a distribution that isn't a function $$\mathbb{R} \to \mathbb{R}$$ .

Here's a strawman definition of $$\delta$$ as a function $$\mathbb{R} \to \mathbb{R} \cup \left\{ \infty \right\}$$ . ✱✱ is used here to mark an unacceptable definition.

$$✱✱ \;\;\;\delta(x) \stackrel{def}{=} \text{if } x = 0 \text{ then } \infty \text{ else } 0 \tag{2}$$

The definition fails to capture that $$\int_{-\infty}^{\infty} \delta(x) = 1$$ .

But the distributional definition of the Dirac delta is:

$$\left\langle \delta, \varphi \right\rangle \stackrel{def}{=} \varphi \left(0\right) \tag{3a}$$ $$\delta\left(\varphi\right) \stackrel{def}{=} \varphi \left(0\right) \tag{3b}$$

where (3a) is conventional notation and (3b) emphasizes that $$\delta$$ is a higher order function.

And there is where I get stuck. I don't know what arguments to make to suggest that $$T$$ as a domain for a generalized function is the natural/correct choice.

The Wikipedia article on distributions offers the following paragraph which briefly describes what you can say about distributions whose domain is a subset of $$T$$, which suggests that you are doing something nontrivial when you pick $$T$$ .

Distribution theory reinterprets functions as linear functionals acting on a space of test functions. Standard functions act by integration against a test function, but many other linear functionals do not arise in this way, and these are the "generalized functions". There are different possible choices for the space of test functions, leading to different spaces of distributions. The basic space of test function consists of smooth functions with compact support, leading to standard distributions. Use of the space of smooth, rapidly (faster than any polynomial increases) decreasing test functions (these functions are called Schwartz functions) gives instead the tempered distributions, which are important because they have a well-defined distributional Fourier transform. Every tempered distribution is a distribution in the normal sense, but the converse is not true: in general the larger the space of test functions, the more restrictive the notion of distribution. On the other hand, the use of spaces of analytic test functions leads to Sato's theory of hyperfunctions; this theory has a different character from the previous ones because there are no analytic functions with non-empty compact support.

Emphasis mine.

• The derivative of a distribution is defined via $\langle f',\phi\rangle=-\langle f,\phi'\rangle$ and to make sense of it we need $\phi'$ to be in the domain. Dec 29, 2018 at 23:03

As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($$c^{\infty}$$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.