# Dirac Delta's Ill-defineness Property

This problem arises from the following property of Dirac $\delta-$function: $$\delta(f(x))=\sum_{a_i\in Z(f)}\frac{\delta(x-a_i)}{|\frac{df}{dx}(a_i)|}$$ where $Z(f):=\{x\in dom(f)|\,f(x)=0\}$, the zero-set of $f$.

Here, many people use the Taylor expansion of $f(x)$ to prove the property. Specifically, they expand $f(x)$ around those zero point of $f$, for example, $f(x)=f(a_i)+f'(a_i)(x-a_i)+\mathcal{O}((x-a_i)^2)$, where $f(a_i)=0$, and say that $\mathcal{O}((x-a_i)^2)$ is higher order infinitesimal to $(x-a_i)$ therefore they are neglegible, which sounds reasonable.

However, when $f(x)$ is the power of $x$, for example, $f(x)=x^2$. Then the above form tell us $\delta(x^2)=\frac{\delta(x-0)}{2\cdot0}$ which is ill-defined. Similarly to $x^n$.

One may define the extended version such that $\frac{1}{0}:=\infty$, then the whole $\delta(x^2)$ still make sense. However, if one tries to evaluate the following integral:

$$\int_{-\infty}^{\infty}x\delta(x^2)dx$$ by the above formula, it will give us $$\int_{-\infty}^{\infty}x\delta(x^2)dx=\int_{-\infty}^{\infty}x\frac{\delta(x)}{2\cdot 0}dx=\frac{0}{2\cdot 0}$$ which is an ill-defined result.

On the other hand, notice $xdx=\frac{1}{2}d(x^2)$

$$\int_{-\infty}^{\infty}x\delta(x^2)dx=\frac{1}{2}\int_{-\infty}^{\infty}\delta(x^2)d(x^2)=\frac{1}{2}$$ which I think this is just a coincidence. (the reason is one can have the integrand to be $x^3\delta(x^2)$, again use $xdx=\frac{1}{2}d(x^2)$, this time the first way generate the same wield formula, while the second way gives you $0$).

Can anyone explain it to me? (Notice, whenever the denominator is not zero, everything looks fine)

• Related: math.stackexchange.com/q/2481114/11127 and links therein. – Qmechanic Mar 10 '18 at 14:22
• @Qmechanic See you again, sir! Thanks for your links. I think it is very helpful! However, I just wondered where you get the general formula for $\delta$. Are there any paper or book relate to that? – Hamio Jiang Mar 10 '18 at 17:46

## 1 Answer

The formula $$\begin{equation} \delta(f(x))=\sum_{a_i\in Z(f)}\frac{\delta(x-a_i)}{\left|\frac{df}{dx}(a_i)\right|} \tag{1} \label{1} \end{equation}$$ is only a formula for expressing the composition between a Dirac distribution and a function belonging to a well defined class of ordinary function. Its "ill-definiteness" simply due to the fact that the formula is valid not valid for the function $$f(x)=x^2$$. Precisely, equation \eqref{1} can be proved only under the following two hypotheses (, §1.9 pp. 22-23), both implicitly used in all the proposed answers to question "Dirac Delta Function of a Function":

1. $$f\in C^1$$: if $$f\notin C^1$$, the real number $$|\,f^\prime(a_i)|$$ does not exists finite, thus the right hand term of \eqref{1} does not make sense.
2. The zeros of $$f$$ on its domain must be simple and isolated, and thus $$Z(f):=\{x\in \mathrm{dom}(f)|\,f(x)=0\}$$ must be countable: even this fact is implicitly used in all the proposed answers to [question "Dirac Delta Function of a Function"]. The first statement on the zeros of $$f$$ is tied to the requirement of point 1: if $$a_i\in Z(f)$$ is not simple for some $$I$$, then $$|\,f^\prime(a_i)|=0$$ and again \eqref{1} does not make sense. The second statement is related to both the structure of \eqref{1}, for a infinite sum in mathematical analysis has a meaning only for a countable set of terms, and to a more general expression of $$\delta(f(x))$$ (which is discussed below).

The general definition of $$\delta(f(x))$$ in $$\mathscr{D}^\prime$$

The problem of defining of $$\delta(f(x))$$ for a general $$f$$ is a particular instance of the problem of defining the composition of an ordinary function with a distribution: this in turn is one of the motivating problems which gave rise to several "nonlinear theories" of generalized functions. The approach followed by Vladimirov (, §1.9 p. 22) for defining this composition in In $$\mathscr{D}^\prime$$ is the following one: $$\delta(f(x))=\lim_{\varepsilon\to0+}\omega_\varepsilon(f(x))\,\text{ in }\mathscr{D}^\prime\iff \langle\delta(f),\varphi\rangle=\lim_{\varepsilon\to0+}\!\int\limits_{Z(f)\cap[a,b]}\!\!\!\!\omega_\varepsilon(f(x))\varphi(x)\mathrm{d}x\tag{2}\label{2} \quad\forall\varphi\in\mathscr{D}([a,b])$$ where $$\mathscr{D}([a,b])$$ is the space of $$C^\infty$$ functions whose support is contained in any interval $$[a,b]$$ of interest, and $$\begin{equation} \omega_\varepsilon(x)= \begin{cases} C_\varepsilon e^{-\frac{\varepsilon^2}{\varepsilon^2-x^2}} & |x|\leq\varepsilon\\ &\\ 0 & |x|\geq\varepsilon \end{cases} \qquad C_\varepsilon=\frac{1}{\varepsilon \int\limits_{|\xi|<1}e^{-\frac{1}{1-\xi^2}}\mathrm{d}\xi}\qquad\forall\varepsilon>0 \end{equation}$$

is the standard $$\delta$$-sequence converging to the Dirac distribution (I made explicit the domain of integration in this formula respect to the one found in reference  in order to show clearly what is the rôle of $$Z(f)$$).

• If $$f(x)$$ satisfies requirements 1 and 2, the limit formula \eqref{2} can be used to deduce formula \eqref{1} from the (smooth) change of variables for distributions (, §1.9 p. 22) and from the lemma of "piecewise sewing" (, §1.5 pp. 13-15).
• If $$f(x)$$ does not satisfies the requirement 1 and 2, the the limit \eqref{2} is not equal to formula \eqref{1} and may or may not exists, i.e $$\delta(f(x))$$ may or may not be a distribution.

• Finally, let's consider the case where $$Z(f)$$ is not made of only isolated points: if, for example $$f(x)=0$$ for all $$x$$ in a given interval $$[a,b]$$, the limit \eqref{2} is $$\infty$$ since $$\omega_\varepsilon(f(x))$$ goes to $$\infty$$ for $$\varepsilon\to0$$ on a set of finite measure $$b-a$$, thus again does not defines a distribution.

 Vladimirov, V. S. (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR 2012831, Zbl 1078.46029.

• Thanks for your answer, but I have one question. As you mentioned, the zeros of $f$ on its domain must be simple and isolated. I understand the definition of simple root if $f$ is a polynomial, but I do not understand what it means for a general function, for example, $f(x)=\cos(x)-1$, the derivative of $f(x)$ is $-\sin(x)$ which is $0$ when $x=0$. – Hamio Jiang Mar 10 '18 at 17:44
• Hamio, you're welcome. Regarding you question, a simple zero is precisely a point $a_1$ for which $f(a_1)=0$ and $f^{(1)}(a)\neq0$ while a multiple zero of order $N$ is point $a_N$ for which $f(a_N)=0$, $f^{(i)}(a)=0$ for all $i=1,\dots,N-1$ and $f^{(1N}(a)\neq0$. In the example you gave, $f(x)=\cos(x)-1$, $x=0$ is a second order zero for $f$ as in it is in the case $f(x)=x^2$: obviously this concept make sense only for differentiable functions and, in some sense, means that a given function behaves like a (McLaurin-Taylor) polynomial in the neighborhood of a given point. – Daniele Tampieri Mar 10 '18 at 19:11