Skip to main content
2 of 2
added top level tag
Ricardo Andrade
  • 6.2k
  • 5
  • 42
  • 69

Dirac delta composed with absolute value

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!

bers
  • 111
  • 1
  • 6