A few minor definitions are needed to understand what distributions represent. We define a set $X$ and function $\phi: X\to\mathbb{R}$ for the rest of this post.

The first two definitions are very simple.

**Definition:**Suppose $\phi$ is in $L_p(\mathbb{R}^n)$ and $X$ is open. We say $\phi$ is

*locally integrable*if, for all compact subsets $A$ of $X$,

$$\int_A |\phi(x)|\;dx< \infty.$$The space of all such functions is called $L_p^{loc}$.

The formal definition of compactness can be found here. For those who haven't studied real analysis, a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.

**Definition:**The

*support*of $\phi$, written supp($\phi$), is the closure of the set of points in X where f is non-zero. That is,

$$\operatorname{supp}(\phi) = \{x\in X \,|\, \phi(x)\ne 0\}.$$(Topologists use a slightly different definition.)

From here, a slightly more specific property can be considered:

**Definition:**A function $\phi$ is said to have compact support if $supp(\phi)$ is compact.

It's hard to come up with a compactly supported function without specifying that the complement of the support is zero. As a result, most easily representable test functions, even the continuous and infinitely differentiable ones, are defined piecewise. We consider a few examples.

Note that compact support can also be interpreted as the function vanishing outside a compact set; continuous functions are always nonzero on an open set, so taking the closure in the definition of support is necessary.

One of the simplest examples of a compactly supported function is $\chi_A(x)$, where $A$ is a compact set and

$$\chi_A(x)=\left\{\begin{array}{ll}1&x\in A\\0& x \notin A.\end{array}\right.$$This is the identity on $A$ and zeros out everything else. In fact, the composition of $\chi_A(x)$ with any function on $x$ will have compact support as well. Here are a couple examples:

Note that compact support can also be interpreted as the function vanishing outside a compact set; continuous functions are always nonzero on an open set, so taking the closure in the definition of support is necessary.

One of the simplest examples of a compactly supported function is $\chi_A(x)$, where $A$ is a compact set and

$$\chi_A(x)=\left\{\begin{array}{ll}1&x\in A\\0& x \notin A.\end{array}\right.$$This is the identity on $A$ and zeros out everything else. In fact, the composition of $\chi_A(x)$ with any function on $x$ will have compact support as well. Here are a couple examples:

(

This example leads well into the last definition.

*Test yourself! Is H(x) from the previous post compactly supported? Are B-splines?*)This example leads well into the last definition.

**Definition:**A function $\phi$ is a

*test function*if it has compact support and is infinitely differentiable (

*i.e*., in $C^\infty$). We refer to the space of all test functions on a set $X$ as $\mathcal{D}(X)$.

This is a crucial definition! It's weird for a function to have compact support but to also be infinitely differentiable, so let's generate a couple examples. Consider

$$\psi(x)=\left\{\begin{array}{ll}e^{-\frac{1}{1-x^2}}&|x|<1\\0& |x|\geq 1.\end{array}\right.$$This is a lot smoother than the previous function, and looks like a bump:

A slightly more complicated example would be

$$u_A(x)=\int_{\mathbb{R}^n}\chi_A(x-y)u(y)\;dy$$where $u(x)$ is a locally integrable function in $\mathbb{R}^n$ (the technique used to generate this example is called Sobolev's mollification method). If you're familiar with convolution already, it should not be difficult to prove this function is compactly supported. It looks like someone built a sandcastle shaped like a regular $\chi$ function and a wave rolled over it:

Many operations, such as translation and scaling, preserve infinite differentiability and compact support. Linear combinations of test functions and products of test functions are also test functions themselves.

(

*Test yourself! Can test functions be analytic?*)

So, our point---these definitions are necessary in order to understand what distributions are. We'll go into this in detail next week.

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