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It seems to me that it would be more natural to consider Dirac Delta as a piecewise-defined function, as described here, with the scaling rule $\delta (ax)=\delta(x)$. This way we keep all the integral transforms properties of delta function, and its derivatives as long as its argument is the integration variable. But at the same time, we can define arbitrary functions on delta function, including rising it to any positive power, evaluate it at zero, and consider it outside of integrals.

It seems to me absolutely natural that a derivative of scaling-invariant function, like $\operatorname{sign}x$ also should be scaling-invariant. This, in particular, allows to generalize sign function to dual numbers while keeping its properties.

So, the question is: why another convention, that is $\delta(a x)=\frac1{|a|}\delta(x)$ was adopted, which does not even allow to ascribe to it a value at zero?

Anixx
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    Maybe ask this question on Math Stack Exchange? – paul garrett Feb 01 '23 at 15:55
  • $\delta (ax)=\delta(x)$ goes against the defining property of $\delta$-function. Consider the standard approximation: $\frac1{2h}1_{[-h,h]}(ax)=\frac1{2h}1_{[-h/|a|,h/|a|]}(x)=\frac1{|a|}\frac1{2h/|a|}1_{[-h/|a|,h/|a|]}(x)$, from which the standard scaling rule falls out. It is not meant to be scaling invariant. That would also be inconsistent with the image of scaling in operational calculus, or with $\delta$'s interpretation as a measure/distribution. Getting raised to powers and evaluated at zero is not what $\delta$ was designed to do either, and surely nobody cared about dual numbers. – Conifold Feb 03 '23 at 04:08
  • @Conifold well, this answer https://math.stackexchange.com/a/4628238/2513 seemingly says that if delta is defined as a measure, the scaling rule that comes is $\delta(ax)=\delta(x)$. – Anixx Feb 03 '23 at 11:27
  • No, it says that it comes out as $\delta_0(aX) = \delta_0(X)$, where $X$ is a set, and $\delta_0$ is the unit atom measure at $0$. Then $\delta_0(x)$ makes no sense with $x$ as a point, but if you define $\delta(x)$ to be its symbolic (pseudo) "density" relative to the Lebesgue measure you'll end up with the same old standard scaling rule $\delta(a x)=\frac1{|a|}\delta(x)$. – Conifold Feb 04 '23 at 05:00
  • @Conifold "Then $δ_0(x)$ makes no sense with x as a point" - why? I think, the question (and my own answer to it) describes exactly this function. – Anixx Feb 04 '23 at 10:41
  • Because measures are functions of sets, not of points. Whatever you describe is not a measure, since you stick $x$ into it, nor is it $δ_0$ discussed by your link, which is a measure. You just took the scaling rule for Dirac's measure $δ_0$ on sets and confused it with the scaling rule for Dirac's "function" $δ$ on points (your link clearly distinguishes them). Scaling rules do not translate from measures to their densities by simply replacing the set variable by the point variable. That is what your link explains, and that is where the $\frac1{|a|}$ factor comes from. – Conifold Feb 04 '23 at 13:12
  • @Conifold I did not "take" the measure, I do not know measure theory. The measure was given in answer by LL – Anixx Feb 04 '23 at 15:59
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    One does not need to know measure theory to mistake a formula for measures for a formula for functions. Indeed, not knowing measure theory is what makes such a mistake more likely. Looking into measure theory, specifically how scaling of measures relates to scaling of densities, should help you dispose of your question. Also, all those algebraic things you wanted to do with Dirac's function can be unproblematically done with Dirac's measure instead. It is a function on subsets of reals, defined on every subset, that takes value 1 if the subset contains the origin and 0 otherwise. – Conifold Feb 05 '23 at 08:04
  • @Conifold Well, how would one define integration on Dirac measure that would satisfy the rules for $\bar{\delta}(x)$? Such as $$\int_{-\infty }^{\infty } \left( \begin{cases} f(\pi \overline{\delta} (0)) & x=0 \ 0 & x\neq 0 \ \end{cases} \right) , dx=f'(0)+\int_0^{\infty } f''(x) , dx=f'(\pi \overline{\delta} (0))?$$ This is a generalization of integration property of Dirac distribution, but more extended, for instance, it allows to integrate $\bar{\delta}(x)^2$. – Anixx Feb 05 '23 at 10:26

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We want the usual change of variables formulas to work.

$$ \int_{-1}^1 f(x)\delta(x)\;dx = f(0) . $$ What if we change variables $x=ay$ where $a>0$? Then $y = x/a$ and $dx = a\, dy$ and so for change of variables we want $$ f(0) = \int_{-1}^1 f(x)\delta(x)\;dx = \int_{-1/a}^{1/a} f(ay)\delta(ay) a\;dy . $$ But if we use the incorrect $\delta(ay) = \delta(y)$ we would get $$ f(0) = \int_{-1/a}^{1/a} f(ay)\delta(ay) a\;dy= \int_{-1/a}^{1/a} f(ay)\delta(y) a\;dy= f(0) a . $$

Gerald Edgar
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