12.2 Operations on distributions

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If $T$ and $S$ are distributions of order $m$ on ${\mathbb{R}}^n$ , then clearly $T+S$ and $aT$ are distributions of this order for all $a\in \mathbb{R}$ . Thus ${\mathcal{D}}^{\prime m}({\mathbb{R}}^n)$ is a vector space.

Exercise

Prove that $T+S$ is linear and continuous. Similarly for $aT$

The product $ST$ of two distributions is not a distribution. For example, if we were to define $(ST)(\phi )=S(\phi )T(\phi )$ , this is not linear in $\phi .$ . However, if $\alpha$ is a $C^m$ function on ${\mathbb{R}}^n$ and $T$ is a distribution of order $m$ then $\alpha$ $T$ can be defined as a distribution of order $m,$ , by setting

$$(\alpha T)(\phi )=T(\alpha \phi ),$$

since $\alpha \phi \in {\mathcal{D}}^m({\mathbb{R}}^n)$ for all $\phi \in {\mathcal{D}}^m({\mathbb{R}}^n)$ ). Note that $\alpha$ need not be a test function for this construction – it works even if the function $\alpha$ does not have compact support.

If $T$ is a regular distribution on ${\mathbb{R}}^n,T=T_f$ , then $\alpha$ $T_f=T_{\alpha f}$ . For

$$\begin{array}{l}\alpha T_f(\phi )=T_f(\alpha \phi )\\ ={\int }_{{\mathbb{R}}^n}\cdots \int \phi \alpha f{\mathrm{d}}^{n}x\\ =T_{\alpha f}(\phi ).\end{array}$$

The operation of multiplying the regular distribution $T_f$ by $\alpha$ is equivalent to simply multiplying the corresponding density function $f$ by $\alpha$. In this case $\alpha$ need only be a locally integrable function.

Example 12.3

The distribution $\delta$ defined in Example 12.2 is a distribution of order zero, since it is well-defined on the space of continuous test functions, ${\mathcal{D}}^0(\mathbb{R})$ . For any continuous function $\alpha (x)$ we have

$$\alpha \delta (\phi )=\delta (\alpha \phi )=\alpha (0)\phi (0)=\alpha (0)\delta (\phi ).$$

Thus

$$\alpha \delta =\alpha (0)\delta .\text{\hspace{1em}(12.3)}$$

In terms of the ‘delta function’ this identity is commonly written as

$$\alpha (x)\delta (x)=\alpha (0)\delta (x),$$

since

$${\int }_{-\infty }^{\infty }\alpha (x)\delta (x)\phi (x)dx=\alpha (0)\phi (0)={\int }_{-\infty }^{\infty }\alpha (0)\delta (x)\phi (x)dx.$$

For the delta function at an arbitrary point $a$ , these identities are replaced by

$$\alpha {\delta }_a=\alpha (a){\delta }_a,\alpha (x)\delta (x-a)=\alpha (a)\delta (x-a).\text{\hspace{1em}(12.4)}$$

Setting $\alpha (x)=x$ results in the useful identities

$$x\delta =0,x\delta (x)=0.\text{\hspace{1em}(12.5)}$$

Exercise

Extend these identities to the $n$-dimensional delta function,

$$\alpha {\delta }_{\mathbf{a}}=\alpha (\mathbf{a}){\delta }_{\mathbf{a}},\alpha (\mathbf{x}){\delta }^n(\mathbf{x}-\mathbf{a})=\alpha (\mathbf{a})\delta (\mathbf{x}-\mathbf{a}).$$

Differentiation of distributions

Let $T_f$ be a regular distribution where $f$ is a differentiable function. Standard results in real analysis ensure that the derivative $f^{\prime }=\mathrm{d}f/\mathrm{d}x$ is a locally integrable function. Let $\phi$ be any test function from ${\mathcal{D}}^1(\mathbb{R})$ . Using integration by parts

$$\begin{array}{l}{T}_{f^{\prime }}(\phi )={\int }_{-\infty }^{\infty }\phi (x)\frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{d}x\\ ={\left[\phi f\right]}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }\frac{\mathrm{d}\phi }{\mathrm{d}x}f(x)\mathrm{d}x\\ =T_f(-{\phi }^{\prime }),\end{array}$$

since $\phi (\pm \infty )=0$ . We can extend this identity to general distributions, by defining the derivative of a distribution $T$ of order $m\ge 0$ on $\mathbb{R}$ to be the distribution $T^{\prime }$ of order $m+1$ given by

$$T^{\prime }(\phi )=T(-{\phi }^{\prime })=-T({\phi }^{\prime }).\text{\hspace{1em}(12.6)}$$

The derivative of a regular distribution then corresponds to taking the derivative of the density function. Note that the order of the distribution increases on differentiation, for ${\phi }^{\prime }\in {\mathcal{D}}^m(\mathbb{R})$ implies that $\phi \in {\mathcal{D}}^{m+1}(\mathbb{R})$ ). In particular, if $T$ is a distribution of order 0 then $T^{\prime }$ is a distribution of order 1.

To prove that $T^{\prime }$ is continuous (linearity is obvious), we use the fact that in the definition of convergence to order $m+1$ of a sequence of functions ${\phi }_n\to \phi$ it is required that all derivatives up to and including order $m+1$ converge uniformly on a compact subset $K$ of

$\mathbb{R}$. In particular, ${\phi }_n^{\prime }(x)\to {\phi }^{\prime }(x)$ for all $x\in K$ , and

$$T^{\prime }({\phi }_n)=T(-{\phi }_n^{\prime })\to T(-{\phi }^{\prime })=T^{\prime }(\phi ).$$

It follows that every distribution of any order is infinitely differentiable.

If $T$ is a distribution of order greater or equal to 0 on ${\mathbb{R}}^n$ , we may define its partiall derivatives in a similar way,

$$\frac{\partial T}{\partial x_k}(\phi )=-T\left(\frac{\partial \phi }{\partial x_k}\right).$$

As for distributions on $\mathbb{R},$ , any such distribution is infinitely differentiable. For higher derivatives it follows that

$$D_{\underline{m}}T(\phi )=(-1{)}^{m}T(D_{\underline{m}}\phi )\text{ where }m=|\underline{m}|=\sum _i{m}_i.$$

Exercise

Show that

$$\frac{{\partial }^{2}T}{\partial x_i\partial x_j}=\frac{{\partial }^{2}T}{\partial x_j\partial x_j}.$$

Example 12.4

Set $\theta (x)$ to be the Heaviside step function

$$\theta (x)=\{\begin{array}{ll}1& \text{ if }x\ge 0,\\ 0& \text{ if }x<0.\end{array}$$

This is evidently a locally integrable function, and generates a regular distribution $T_{\theta }$ . For any test function $\phi \in {\mathcal{D}}^1(\mathbb{R})$

$$\begin{array}{l}{T}_{{\theta }^{\prime }}(\phi )=T_{\theta }(-{\phi }^{\prime })=-{\int }_{-\infty }^{\infty }{\phi }^{\prime }(x)\theta (x)\mathrm{d}x\\ =-{\int }_0^{\infty }\frac{\mathrm{d}\phi }{\mathrm{d}x}\mathrm{d}x\\ =\phi (0)\text{ since }\phi (\infty )=0\\ =\delta (\phi ).\end{array}$$

Thus we have the distributional equation, valid only over ${\mathcal{D}}^1(\mathbb{R})$ ,

$$T_{\theta }^{\prime }=\delta .$$

This is commonly written in terms of ‘functions’ as

$$\delta (x)={\theta }^{\prime }(x)=\frac{\mathrm{d}\theta (x)}{\mathrm{d}x}.$$

Intuitively, the step at $x=0$ is ‘infinitely steep’.

Example 12.5

The derivative of the delta distribution is defined as the distribution ${\delta }^{\prime }$ of order 1, which may be applied to any test function $\phi \in {\mathcal{D}}^1(\mathbb{R})$

$${\delta }^{\prime }(\phi )=\delta (-{\phi }^{\prime })=-{\phi }^{\prime }(0).$$

Expressed in terms of the delta function, this reads

$${\int }_{-\infty }^{\infty }{\delta }^{\prime }(x)\phi (x)\mathrm{d}x=-{\phi }^{\prime }(0),$$

for an arbitrary function differentiable on a neighbourhood of the origin $x=0$ . Continuing to higher derivatives, we have

$${\delta }^{\prime \prime }(\phi )={\phi }^{\prime \prime }(0)$$

or in Dirac’s notation

$${\int }_{-\infty }^{\infty }{\delta }^{\prime \prime }(x)\phi (x)dx={\phi }^{\prime \prime }(0).$$

For the $m$th derivative,

$${\delta }^{(m)}(\phi )=(-1{)}^m{\phi }^{(m)}(0),{\int }_{-\infty }^{\infty }{\delta }^{(m)}(x)\phi (x)\mathrm{d}x=(-1{)}^m{\phi }^{(m)}(0).$$

For the product of a differentiable function $\alpha$ and a distribution $T$ we obtain the usua Leibnitz rule,

$$(\alpha T{)}^{\prime }=\alpha T^{\prime }+{\alpha }^{\prime }T,$$

for

$$\begin{array}{rl}(\alpha T{)}^{\prime }(\phi )& =\alpha T(-{\phi }^{\prime })\\ & =T(-\alpha {\phi }^{\prime })\\ & =T((-\alpha \phi {)}^{\prime }+{\alpha }^{\prime }\phi )\\ & =T^{\prime }(\alpha \phi )+{\alpha }^{\prime }T(\phi )\\ & =\alpha T^{\prime }(\phi )+{\alpha }^{\prime }T(\phi ).\end{array}$$

Example 12.6

From Examples 12.3 and 12.5 we have that

$$(x\delta {)}^{\prime }=0^{\prime }=0$$

and

$$(x\delta {)}^{\prime }=x{\delta }^{\prime }+x^{\prime }\delta =x{\delta }^{\prime }+\delta .$$

Hence

$$x{\delta }^{\prime }=-\delta .$$

We can also derive this equation by manipulating the delta function in natural ways,

$$x{\delta }^{\prime }(x)=(x\delta (x){)}^{\prime }-x^{\prime }\delta (x)=0^{\prime }-1.\delta (x)=-\delta (x).$$

Exercise

Verify the identity $x{\delta }^{\prime }=-\delta$ by applying both sides as distributions to an arbitrary test function $\varphi$($x$).

Change of variable in $\delta$-functions

In applications of the mathematics of delta functions it is common to consider ‘functions’ such as $\delta (f(x))$ . While this is not an operation that generalizes to all distributions, there is a sense in which we can define this concept for the delta distribution for many functions $f$ . Firstly, if $f:\mathbb{R}\to \mathbb{R}$ is a continuous monotone increasing function such that $f(\pm \infty )=\pm \infty$ and we adopt Dirac’s notation then, assuming integrals can be manipulated by the standard rules for change of variable,

$$\begin{array}{l}{\int }_{-\infty }^{\infty }\phi (x)\delta (f(x))\mathrm{d}x={\int }_{-\infty }^{\infty }\phi (x)\delta (y)\frac{\mathrm{d}y}{f^{\prime }(x)}\text{where}y=f(x)\\ ={\int }_{-\infty }^{\infty }\frac{\phi (f^{-1}(y))}{f^{\prime }(f^{-1}(y))}\delta (y)\mathrm{d}y\\ =\frac{\phi (a)}{f^{\prime }(a)}\text{where}f(a)=0.\end{array}$$

If $f(x)$ is monotone decreasing then the range of integration is inverted to ${\int }_{-\infty }^{\infty }$ resulting in a sign change. The general formula for a monotone function $f$ of either direction, having a unique zero at $x=a$ , is

$${\int }_{-\infty }^{\infty }\phi (x)\delta (f(x))dx=\frac{\phi (a)}{|f^{\prime }(a)|}.\text{\hspace{1em}(12.7)}$$

Symbolically, we may write

$$\delta (f(x))=\frac{1}{|f^{\prime }(a)|}\delta (x-a),$$

or in terms of distributions,

$$\delta \circ f=\frac{1}{|f^{\prime }(a)|}{\delta }_a.\text{\hspace{1em}(12.8)}$$

Essentially this equation can be taken as the definition of the distribution $\delta \circ f$ . Setting $f(x)=-x$ , it follows that $\delta (x)$ is an even function $\delta (-x)=\delta (x)$

If two test functions $\phi$ and $\psi$ agree on an arbitrary neighbourhood $[-ϵ,ϵ]$ of the origin $x=0$ then

$$\delta (\phi )=\delta (\psi )=\phi (0)=\phi (\psi ).$$

Hence the distribution $\delta$ can be regarded as being a distribution on the space of functions $\mathcal{D}([-ϵ,ϵ])$ , since essentially it only samples values of any test function $\phi$ in a neighbourhood of the origin. Thus it is completely consistent to write

$$\delta (\phi )={\int }_{-ϵ}^{ϵ}\phi (x)\delta (x)dx.$$

This just reiterates the idea that $\delta (x)=0$ for all $x\ne 0.$

If $f(x)$ has zeros at $x=a_1,a_2,\dots$ . and $f$ is a monotone function in the neighbourhood of each $a_i.$ , then a change of variable to $y=f(x)$ gives, on restricting integration to a smal

neighbourhood of each zero,

$${\int }_{-\infty }^{\infty }\phi (x)\delta (f(x))dx=\sum _i\frac{\phi \left(a_i\right)}{\left|f^{\prime }\left(a_i\right)\right|}.$$

Hence

$$\delta (f(x))=\sum _i\frac{1}{|f^{\prime }(a_i)|}\delta (x-a_i),\text{\hspace{1em}(12.9)}$$

or equivalently

$$\delta \circ f=\sum _i\frac{1}{|f^{\prime }(a_i)|}{\delta }_{a_i}.$$

Example 12.7

The function $f=x^2-a^2=(x-a)(x+a)$ is locally monotone at both its zeros $x=\pm a$ , provided $a\ne 0$ . In a small neighbourhood of $x=a$ the function $f$ may be approximated by the monotone increasing function $2a(x-a)$ , while in a neighbourhood of $x=-a$ it is monotone decreasing and approximated by $-2a(x+a)$ . Thus

$$\delta (x^2-a^2)=\delta (2a(x-a))+\delta (-2a(x+a))=\frac{1}{2a}(\delta (x-a)+\delta (x+a)),$$

in agreement with Eq. (12.9).

Problems

Problem 12.5

In the sense of convergence defined in Problem 12.4 show that if $T_n\to T$ then $T_n^{\prime }\to T^{\prime }$

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

$$f_n(x)=-\frac{2n^{3}x}{\sqrt{\pi }}{\mathrm{e}}^{-n^2{x}^2}\to {\delta }^{\prime }(x).$$

Problem 12.6

Evaluate

(a) ${\int }_{-\infty }^{\infty }{\mathrm{e}}^{at}\sin (bt){\delta }^{(n)}(t)\mathrm{d}t$ for $n=0,1,2$.

(b) ${\int }_{-\infty }^{\infty }(\cos t+\sin t){\delta }^{(n)}(t^3+t^2+t)\mathrm{d}t\mathrm{for}n=0,1.$

Problem 12.7

Show the following identities:

$$\delta ((x-a)(x-b))=\frac{1}{b-a}(\delta (x-a)+\delta (x-b)).\text{\hspace{1em}(a)}$$ $$\text{ (b) }\frac{\mathrm{d}}{\mathrm{d}x}\theta (x^2-1)=\delta (x-1)-\delta (x+1)=2x\delta (x^2-1).$$

(c) $\frac{\mathrm{d}}{\mathrm{d}x}\delta (x^2-1)=\frac{1}{2}({\delta }^{\prime }(x-1)+{\delta }^{\prime }(x+1)).$

$${\delta }^{\prime }(x^2-1)=\frac{1}{4}({\delta }^{\prime }(x-1)-{\delta }^{\prime }(x+1)+\delta (x-1)+\delta (x+1)).\text{\hspace{1em}(d)}$$

Problem 12.8

Show that for a monotone function $f$($x$) such tha $f(\pm \infty )=\pm \infty \mathrm{with}f(a)=0$

$${\int }_{-\infty }^{\infty }\phi (x){\delta }^{\prime }(f(x))dx=-\frac{1}{f^{\prime }(x)}\frac{d}{dx}\left(\frac{\phi (x)}{|f^{\prime }(x)|}\right){|}_{x=a}.$$

For a general function $f(x)$ that is monotone on a neighbourhood of all its zeros, find a genera formula for the distribution ${\delta }^{\prime }\circ f$

Problem 12.9

Show the identities

$$\frac{\mathrm{d}}{\mathrm{d}x}(\delta (f(x)))=f^{\prime }(x){\delta }^{\prime }(f(x))$$

and

$$\delta (f(x))+f(x){\delta }^{\prime }(f(x))=0.$$

Hence show that $\varphi (x,y)=\delta (x^2-y^2)$ is a solution of the partial differential equation

$$x\frac{\partial \varphi }{\partial x}+y\frac{\partial \varphi }{\partial y}+2\varphi (x,y)=0.$$