12.1 Test functions and distributions

中文

Spaces of test functions

The support of a function $f:{\mathbb{R}}^n\to \mathbb{R}$ is the closure of the region of ${\mathbb{R}}^n$ where $f(\mathbf{x})\ne 0$ We will say a real-valued function $f$ on ${\mathbb{R}}^n$ has compact support if the support is a closed bounded set; that is, there exists $R>0$ such that $f(x_1,\dots ,x_n)=0\mathrm{for}|\mathbf{x}|\ge R$ . A function $f$ is said to be $C^m$ if all partial derivatives of order $m$,

$$D_{\underline{m}}f=\frac{{\partial }^{m}f}{\partial x_1^{m_1},\dots ,\partial x_n^{m_n}},$$

exist and are continuous, where $\underline{m}=(m_1,\dots ,m_n)$ and $\begin{array}{r}m=|\underline{m}|\equiv {\sum }_{i=1}^n{m}_i\end{array}$ . We adopt the convention that $D_{(0,0,\dots ,0)}f=f$. A function $f$ is said to be $C^{\infty }$ , or infinitely differentiable, if it is $C^m$ to all orders $m=1,2,\dots$ . We set ${\mathcal{D}}^m({\mathbb{R}}^n)$ to be the vector space of all $C^m$ functions on ${\mathbb{R}}^n$ with compact support, called the space of test functions of order $m$

Exercise

Show that ${\mathcal{D}}^m({\mathbb{R}}^n)$ is a real vector space.

The space of infinitely differentiable test functions, ${\mathcal{D}}^{\infty }({\mathbb{R}}^n)$ , is often denoted simply as $\mathcal{D}({\mathbb{R}}^n)$ and is called the space of test functions,

$$\mathcal{D}({\mathbb{R}}^n)=\bigcap _{n=1}^{\infty }{\mathcal{D}}^m({\mathbb{R}}^n).$$

To satisfy ourselves that this space is not empty, consider the function $f:\mathbb{R}\to \mathbb{R}$ defined by

$$f(x)=\{\begin{array}{ll}{\mathrm{e}}^{-1/x}& \text{ if }x>0,\\ 0& \text{ if }x\le 0.\end{array}$$

This function is infinitely differentiable everywhere, including the point $x=0$ where all derivatives vanish both from the left and the right. Hence the function $\phi :\mathbb{R}\to \mathbb{R}$

$$\phi (x)=f(-(x-a))f(x+a)=\{\begin{array}{ll}\exp \left(\frac{2a}{x^2-a^2}\right)& \text{ if }|x|<a,\\ 0& \text{ if }|x|\ge a,\end{array}$$

is everywhere differentiable and has compact support $[-a,a]$ . There is a counterpart in ${\mathbb{R}}^n$ ,

$$\phi (\mathbf{x})=\{\begin{array}{ll}\exp \left(\frac{2a}{|\mathbf{x}{|}^2-a^2}\right)& \text{ if }|\mathbf{x}|<a,\\ 0& \text{ if }|\mathbf{x}|\ge a,\end{array}$$

where

$$|\mathbf{x}|=\sqrt{(x_1{)}^2+(x_2{)}^2+\cdots +(x_n{)}^2}.$$

A sequence of functions ${\phi }_n\in \mathcal{D}({\mathbb{R}}^n)$ is said to converge to order $m$ to a function $\phi \in \mathcal{D}({\mathbb{R}}^n)$ if the functions ${\phi }_n$ and $\phi$ all have supports within a common bounded set and

$$D_{\underline{k}}{\phi }_n(\mathbf{x})\to D_{\underline{k}}\phi (\mathbf{x})$$

uniformly for all $x$, for all $k$ of orders $k=0,1,\dots ,m$ . If we have convergence to order $m$ for all $m=0,1,2,...$ . then we simply say ${\phi }_n$ converges to $\phi ,$ , written ${\phi }_n\to \phi$

Example 12.1

Let $\phi :\mathbb{R}\to \mathbb{R}$ be any differentiable function having compact support on $K$ in $\mathbb{R}$. The sequence of functions

$${\phi }_n(x)=\frac{1}{n}\phi (x)\sin nx$$

are all differentiable and have common compact support $K$. Since $|\phi (x)\sin nx|<1$ it is evident that these functions approach the zero function uniformly as $n\to \infty$ , but their derivatives

$${\phi }_n^{\prime }(x)=\frac{1}{n}{\phi }^{\prime }(x)\sin nx+\phi (x)\cos nx\nrightarrow 0.$$

This is an example of a sequence of functions that converge to order 0 to the zero function, but not to order 1.

To define this convergence topologically, we can proceed in the following manner. For every compact set $K\subset {\mathbb{R}}^n$ let ${\mathcal{D}}^m(K)$ be the space of $C^m$ functions of compact support within $K$. This space is made into a topological space as in Example 10.25, by defining a norm

$$\Vert f{\Vert }_{K,m}=\underset{\mathbf{x}\in K}{\sup }\sum _{|\underline{k}|\le m}\left|D_{\underline{k}}f(\mathbf{x})\right|.$$

On ${\mathcal{D}}^m({\mathbb{R}}^n)$ we define a set $U$ to be open if for every $f\in U$ there exists a compact set $K$ and a real $a>0$ such that $f\in K$ and

$$\{g\in K\mid \Vert g-f{\Vert }_{K,m}<a\}\subseteq U.$$

It then follows that a sequence $f_k\in {\mathcal{D}}^m({\mathbb{R}}^n)$ converges to order $m$ to a function $f\in {\mathcal{D}}^m({\mathbb{R}}^n)$

if and only if $f_n\to f$ with respect to this topology. A similar treatment gives a topology on $\mathcal{D}({\mathbb{R}}^n)$ leading to convergence in all orders (see Problem 12.2).

Distributions

In this chapter, when we refer to ‘continuity’ of a functional $S$ on a space such as $\mathcal{D}({\mathbb{R}}^n)$ ), we will mean that whenever $f_n\to f$ in some specified sense on $\mathcal{D}({\mathbb{R}}^n)$ we have $S(f_n)\to S(f)$. A distribution of order $m$ on ${\mathbb{R}}^n$ is a linear functional $T$ on $\mathcal{D}({\mathbb{R}}^n)$ ),

$$T(a\phi +b\psi )=aT(\phi )+bT(\psi ),$$

which is continuous to order $m$; that is, if ${\phi }_k\to \phi$ is any sequence of functions in $\mathcal{D}({\mathbb{R}}^n)$ convergent to order $m$ then $T({\phi }_k)\to T(\phi )$ . A linear functional $T$ on $\mathcal{D}({\mathbb{R}}^n)$ that is continuous with respect to sequences ${\phi }_i$ in $\mathcal{D}({\mathbb{R}}^n)$ that are convergent to all orders will simply be referred to as a distribution on ${\mathbb{R}}^n$ . In this sense of continuity, the space of distributions of order $m$ on ${\mathbb{R}}^n$ is the dual space of ${\mathcal{D}}^m({\mathbb{R}}^n)$ (see Section 10.9), and the space of distributions is the dual space of $\mathcal{D}({\mathbb{R}}^n)$ . Accordingly, these are denoted ${\mathcal{D}}^{\prime m}({\mathbb{R}}^n)$ ) and ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ respectively.

Note that a distribution $T$ of order $m$ is also a distribution of order $m^{\prime }$ for all $m^{\prime }>m$ For, if ${\phi }_i$ is a convergent sequence of functions in ${\mathcal{D}}^{m^{\prime }}({\mathbb{R}}^n)$ , then ${\phi }_i$ and all its derivatives up to order $m^{\prime }$ converge uniformly to a function $\phi \in {\mathcal{D}}^{m^{\prime }}({\mathbb{R}}^n)$ ). In particular, it is also a sequence in ${\mathcal{D}}^m({\mathbb{R}}^n)$ converging to order $m<m^{\prime }$ to $\phi$ . Therefore a linear functional $T$ of order $m$ , having the property $T({\phi }_i)\to T(\phi )$ for all convergent sequences ${\phi }_i$ in ${\mathcal{D}}^m({\mathbb{R}}^n)$ , automatically has this property for all convergent sequences in ${\mathcal{D}}^{m^{\prime }}({\mathbb{R}}^n)$ . This is a curious feature, characteristic of dual spaces: given a function that is $C^m$ we can only conclude that it is $C^{m^{\prime }}$ for $m^{\prime }\le m$ , yet given a distribution of order $m$ we are guaranteed that it is a distribution of order $m^{\prime }$ for all $m^{\prime }\ge m$

Regular distributions

A function $f:{\mathbb{R}}^n\to \mathbb{R}$ is said to be locally integrable if it is integrable on every compact subset $K\subset {\mathbb{R}}^n$ . Set $T_f:\mathcal{D}({\mathbb{R}}^n)\to \mathbb{R}$ to be the continuous linear functional defined by

$$T_f(\phi )={\int }_{{\mathbb{R}}^n}\phi f\mathrm{d}{\mu }^n={\int }_{{\mathbb{R}}^n}\cdots {\int }_{{\mathbb{R}}^n}\phi (\mathbf{x})f(\mathbf{x})\mathrm{d}{x}_1\dots \mathrm{d}{x}_n.$$

The integral always exists, since every test function $\phi$ vanishes outside some compact set. Linearity is straightforward by elementary properties of the integral operator,

$$T_f(a\phi +b\psi )=a{T}_f(\phi )+b{T}_f(\psi ).$$

Continuity of $T_f$ follows from the inequality (11.11) and Lebesgue’s dominated convergence theorem 11.11,

$$\begin{array}{l}|T_f({\phi }_i)-T_f(\phi )|=|{\int }_{{\mathbb{R}}^n}\cdots \int f(\mathbf{x})({\phi }_i(\mathbf{x})-\phi (\mathbf{x})){\mathrm{d}}^{n}x|\\ \le {\int }_{{\mathbb{R}}^n}\cdots \int |f(\mathbf{x})({\phi }_i(\mathbf{x})-\phi (\mathbf{x}))|{\mathrm{d}}^{n}x\\ \le {\int }_{{\mathbb{R}}^n}\cdots \int |f(\mathbf{x})||({\phi }_i(\mathbf{x})-\phi (\mathbf{x}))|{\mathrm{d}}^{n}x\\ \to 0,\end{array}$$

since the sequence of integrable functions $f{\phi }_i$ is dominated by the integrable function $\left({\sup }_i|{\phi }_i|\right)|f|$ . Hence $T_f$ is a distribution and the function $f$ is called its density. In fact $T_f$ is a distribution of order 0, since only convergence to order 0 is needed in its definition.

Two locally integrable functions $f$ and $g$ that are equal almost everywhere give rise to the same distribution, $T_f=T_g$ . Conversely, if $T_f(\phi )=T_g(\phi )$ for all test functions $\phi$ then the density functions $f$ and $g$ are equal a.e. An outline proof is as follows: let $I^n$ be any product of closed intervals $I^n=I_1\times I_2\times \cdot \cdot \cdot \times I_n$ , and choose a test function $\phi$ arbitrarily close to the unit step function ${\chi }_{I^n}$ . Then ${\int }_{I^n}(f-g)\mathrm{d}{\mu }^n=0$ , which is impossible for all $I^n\mathrm{if}f-g$ has non-vanishing positive part, $(f-g{)}^{+}>0$ , on a set ofpositive measure. This argument may readily be refined to show that $f-g=0$ a.e. Hence the density $f$ is uniquely determined by $T_f$ except on a set of measure zero. By identifying $f$ with $T_f:$ , locally integrable functions can be thought ofas distributions. Not all distributions, however, arise in this way; distributions having a density $T=T_f$ are sometimes referred to as regular distributions, while those not corresponding to any locally integrable function are called singular.

Example 12.2

Define the distribution ${\delta }_a$ on $\mathcal{D}(\mathbb{R})$ by

$${\delta }_a(\phi )=\phi (a).$$

In particular, we write $\delta$ for ${\delta }_0$ , so that $\delta (\phi )=\phi (0)$ . The map ${\delta }_a:\mathcal{D}(\mathbb{R})\to \mathbb{R}$ is obviously linear, ${\delta }_a(b\phi +c\psi )=b\phi (a)+c\psi (a)=b{\delta }_a(\phi )+c{\delta }_a(\psi )$ , and is continuous since ${\phi }_n\to \phi ⟹{\phi }_n(a)\to \phi (a)$ . Hence ${\delta }_a$ is a distribution, but by the reasoning at the beginning of this chapter it cannot correspond to any locally integrable function. It is therefore a singular distribution. Nevertheless, physicists and engineers often maintain the density notation and write

$$\delta (\phi )\equiv {\int }_{-\infty }^{\infty }\phi (x)\delta (x)dx=\phi (0).\text{\hspace{1em}(12.2)}$$

In writing such an equation, the distribution $\delta$ is imagined to have the form $T_{\delta }$ for a density function $\delta (x)$ concentrated at the point $x=0$ and having an infinite value there as in Eq. (12.1), such that

$${\int }_{-\infty }^{\infty }\delta (x)\mathrm{d}x=1.$$

Using a similar convention, the distribution ${\delta }_a$ may be thought of as representing the density function ${\delta }_a(x)$ such that

$${\int }_{-\infty }^{\infty }\phi (x){\delta }_a(x)\mathrm{d}x=\phi (a)$$

for all test functions $\phi .$ . It is common to write ${\delta }_a(x)=\delta (x-a)$ , for on performing the ‘change of variable $x=y+a$’

$${\int }_{-\infty }^{\infty }\phi (x)\delta (x-a)dx={\int }_{-\infty }^{\infty }\phi (y+a)\delta (y)dy=\phi (a).$$

The $n$-dimensional delta function may be similarly defined by

$${\delta }_{\mathbf{a}}^n(\phi )=\phi (\mathbf{a})=\phi (a_1,\dots ,a_n)$$

and can be written

$${\delta }_{\mathbf{a}}^n(\phi )\equiv T_{{\delta }_{\mathbf{a}}^n}(\phi )={\int }_{{\mathbb{R}}^n}\cdots \int \phi (\mathbf{x}){\delta }^n(\mathbf{x}-\mathbf{a}){\mathrm{d}}^{n}x=\phi (\mathbf{a})$$

where

$${\delta }^n(\mathbf{x}-\mathbf{a})=\delta (x_1-a_1)\delta (x_2-a_2)\dots \delta (x_n-a_n).$$

Although it is not in general possible to define the product of distributions, no problems arise in this instance because the delta functions on the right-hand side depend on separate and independent variables.

Problems

Problem 12.1

Construct a test function such that $\varphi (x)=1\mathrm{for}|x|\le 1$ and $\varphi (x)=0$ for $|x|\ge 2$

Problem 12.2

For every compact set $K\subset {\mathbb{R}}^n$ let $\mathcal{D}(K)$ be the space of $C^{\infty }$ functions of compact support within $K$ . Show that if all integer vector $\underline{k}$ are set out in a sequence where $N(\underline{k})$ denotes the position of $\underline{k}$ in the sequence, then

$$\Vert f{\Vert }_K=\underset{\mathbf{x}\in K}{\sup }\sum _{|\underline{k}|}\frac{1}{2^{N(\underline{k})}}\frac{|D_{\underline{k}}f(\mathbf{x})|}{1+|D_{\underline{k}}f(\mathbf{x})|}$$

is a norm on $\mathcal{D}(K)$ . Let a set $U$ be defined as open in $\mathcal{D}({\mathbb{R}}^n)$ if it is a union of open ball $\{g\in K\mid \Vert g-f{\Vert }_K<a\}$ . Show that this is a topology and sequence convergence with respect to this topology is identical with convergence of sequences of functions of compact support to all orders.

Problem 12.3

Which of the following is a distribution?

$$T(\varphi )=\sum _{n=1}^m{\lambda }_n{\varphi }^{(n)}(0)({\lambda }_n\in \mathbb{R}).$$ $$\text{(b)}T(\varphi )=\sum _{n=1}^m{\lambda }_n\varphi (x_n)({\lambda }_n,x_n\in \mathbb{R}).$$ $$\text{ (c) }T(\varphi )={\left(\varphi (0)\right)}^2.$$

(d) $T(\varphi )=\sup \varphi .$

(e) $T(\varphi )={\int }_{-\infty }^{\infty }|\varphi (x)|\mathrm{d}x.$

Problem 12.4

We say a sequence ofdistributions $T_n$ converges to a distribution $T$ , written $T_n\to T$ if $T_n(\varphi )\to T(\varphi )$ for all test functions $\varphi \in \mathcal{D}$ (this is sometimes called weak convergence). If a sequence of continuous functions $f_n$ converges uniformly to a function $f(x)$ on every compact subset of $\mathbb{R}$, show that the associated regular distributions $T_{f_n}\to T_f$

In the distributional sense, show that we have the following convergences:

$$f_n(x)=\frac{n}{\pi (1+n^2{x}^2)}\to \delta (x),$$ $$g_n(x)=\frac{n}{\sqrt{\pi }}{\mathrm{e}}^{-n^2{x}^2}\to \delta (x).$$