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I hope this question is well suited for this site; please excuse me if not.

I recently read that the value of $\delta(x^2)$ is an open question [1], with $\delta(x)$ the Dirac delta. Now I'm trying to get my head around what $\delta(|x|)$ might be, where $|x|$ is the absolute value of x. I know from [2] that $\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$, where $i$ indexes the roots $x_i$ of $g(x)$, which explains why $\delta(x^2)$ is difficult to define in $x=0$: $g'(x) = 2x = 0$.

Now, $|x|'$ is undefined in $x=0$ as well, BUT the case looks different from the one before. We do not exactly need $g'(x_i)$, only $|g'(x_i)|$. The limits from the left and from the right of $\lim_{x \to 0}|x|/x$ do exist, and both their absolute values is $1$. So, would it not make sense to define $|(|x|)'(0)|=|(|x|)'|(0)=1$ to obtain $\delta(|x|) = \delta(x)$? Is there such a thing as an "absolute value of the derivative" with appropriate properties such as "absolute differentiability"?

[1] http://www.gauge-institute.org/delta/DeltaFunction.pdf

[2] http://en.wikipedia.org/wiki/Dirac_delta_function#Composition_with_a_function

Thanks a lot!

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    $\begingroup$ since $\delta(x)=\delta(-x)$ it obviously only depends on the absolute value of $x$ $\endgroup$
    – Carlo Beenakker
    Commented Sep 2, 2013 at 9:59
  • $\begingroup$ Perhaps you first need to fix a definition of the delta function. For example: as a linear functional on WHICH space of test functions... $\endgroup$
    – Gerald Edgar
    Commented Sep 2, 2013 at 13:27
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    $\begingroup$ And while you're at it, please define $\delta(g(x))$ as well. In particular: Please clarify for what kind of $g$ this is defined in what ever framework you're using (it does not seem to be the usual notion of Schwartzian distributions). Judging from your reference [1] (and what it calls "open problems") I'm guessing that the question becomes vacuous and/or trivial if the relevant definitions are spelled out. $\endgroup$
    – Johannes Hahn
    Commented Sep 2, 2013 at 16:09
  • $\begingroup$ The site of "the gauge institute" seems to be a ... not so reliable site. $\endgroup$
    – paul garrett
    Commented Sep 2, 2013 at 22:31

1 Answer 1

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Let $\kappa:\mathbb R\rightarrow\mathbb R$ be a diffeomorphism with $\kappa(0)=0$. Mimicking the change of variable formula, we would like to have $$ \int \delta(\kappa(x))\vert \kappa'(x)\vert \phi(\kappa(x))dx=\int \delta(y)\phi(y) dy, $$ where the integrals should be replaced by brackets of duality. We shall in fact use that as a definition: let $\psi$ be a continuous function on $\mathbb R$. We define $$ \langle (\delta\circ \kappa)(x),\psi(x)\rangle=\langle \delta(y),\psi(\kappa^{-1}(y)) \vert \kappa'(\kappa^{-1}(y))\vert^{-1}\rangle=\psi(0)\vert \kappa'(0)\vert^{-1}. $$ When $\kappa$ is not a diffeomorphism, it is not so easy to define properly the pullback $\delta\circ \kappa$. However, in the case $\kappa(x)=\vert x\vert$ (not a diffeomorphism), you can somehow use an extension of the previous formula to define $\delta\circ \kappa=\delta$. Note that it won't be possible when $\kappa'(0)=0$ as it occurs for $x^2$.

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