# 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

The formula $$$$\delta(f(x))=\sum_{a_i\in Z(f)}\frac{\delta(x-a_i)}{\left|\frac{df}{dx}(a_i)\right|} \tag{1} \label{1}$$$$ is only a formula for expressing the composition between a Dirac distribution and a function belonging to a well defined class of ordinary function. Its "ill-definiteness" simply due to the fact that the formula is valid not valid for the function $$f(x)=x^2$$. Precisely, equation \eqref{1} can be proved only under the following two hypotheses ([1], §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], §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 $$$$\omega_\varepsilon(x)= \begin{cases} C_\varepsilon e^{-\frac{\varepsilon^2}{\varepsilon^2-x^2}} & |x|\leq\varepsilon\\ &\\ 0 & |x|\geq\varepsilon \end{cases} \qquad C_\varepsilon=\frac{1}{\varepsilon \int\limits_{|\xi|<1}e^{-\frac{1}{1-\xi^2}}\mathrm{d}\xi}\qquad\forall\varepsilon>0$$$$

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 [1] 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], §1.9 p. 22) and from the lemma of "piecewise sewing" ([1], §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.

[1] 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