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!



Dirac delta composed with absolute value
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 $x^2$.


