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How to prove $\delta [g(x)]= \sum_i \frac{\delta (x-x_i)}{|g'(x_i)|} $

This is given as a property I came upon on MathWorld - Delta Function. This is very helpful in proving other properties of the delta function such as $\delta(ax) = \frac{1}{a}\delta(x)$

Ian Hsiao
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  • Also is there another way to prove $\delta(x^2-a^2) = \frac{1}{2a} (\delta(x+a)+\delta(x-a))$ without this so-called property? – Ian Hsiao Dec 19 '21 at 08:32
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    When $g$ has $C^1$ inverse, then integration by substitution easily proves the relation. When $g$ has isolated zeros and has local $C^1$ inverses near each zero, then you can split the integral (against a test function, of course) over the intervals near zeros of $g$ and then apply the previous observation to each of the integrals. – Sangchul Lee Dec 19 '21 at 08:36
  • https://math.stackexchange.com/a/2389360/168433 – md2perpe Dec 19 '21 at 08:55
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    What is your definition of $\delta(g(x))$ – reuns Dec 19 '21 at 08:55
  • @md2perpe do you have a more intuitive example? – Ian Hsiao Dec 19 '21 at 09:20
  • @reuns so $g(x)$ is a random nice function, and $\delta(x)$ is the typical Dirac delta function, hopefully that answers your concern? – Ian Hsiao Dec 19 '21 at 09:22
  • No. I'm asking for the definition of $\delta(g(x))$. – reuns Dec 19 '21 at 09:45
  • Note that it is quite tricky to rigorously define $\delta(\cdot)$, because the idea of the Dirac delta and integral of function does not get along. Mathematicians introduce the notion of distribution to overcome this difficulty, but then distribution in general do not behave well under "composition" (if such a thing is well defined at all). So, your input on the definition of $\delta(g(x))$ is very important for proving the property. – Sangchul Lee Dec 19 '21 at 10:25
  • @reuns sorry for the undefined term (I took it for granted because that's what appeared in my textbook), I think in some books it is written as $(fg)$, so $f(g(x))\equiv (fg)$ – Ian Hsiao Dec 19 '21 at 14:12
  • @IanHsiao. Composition is usually written with a small circle: $(f\circ g)(x) := f(g(x)).$ The problem here is that $\delta$ is not a function but a distribution, which is a functional on a space of very nice "test functions," so the composition $\delta\circ g$ cannot be defined the usual way. – md2perpe Dec 19 '21 at 14:38
  • @md2perpe okay, thanks for the clearification! – Ian Hsiao Dec 20 '21 at 05:27

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