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Let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $\omega\in\Omega$
  • $\delta_\omega$ denote the Dirac measureat $\omega$ on $(\Omega,\mathcal A)$
  • $E$ be a $\mathbb R$-Banach space
  • $\mathcal M$ denote the set of strongly $\mathcal A$-measurable $f:\Omega\to E$

If $f\in\mathcal M$, $$\langle\delta_\omega,f\rangle :=\int f\:{\rm d}\delta_\omega=f(\omega)$$ is well-defined. In that sense, $\delta_\omega$ can be thought of as an element of the algebraic dual space $\mathcal M^\ast$ of $\mathcal M$.


Now, assume

  • $(\Omega,\mathcal A)=(\mathbb R,\mathcal B(\mathbb R))$
  • $E=\mathbb R$

In the context of distribution theory, we can find the so-called scaling property of the Dirac delta function $$\langle\delta(a\;\cdot\;),f\rangle=\frac1{\left|a\right|}\left\langle\delta,f\left(\frac{\;\cdot\;}a\right)\right\rangle=\frac1{\left|a\right|}f(0),\tag1$$ where $a\in\mathbb R\setminus\left\{0\right\}$ and $f\in C^\infty(\mathbb R,\mathbb C)$.

My problem is that I don't understand how $\delta(a\;\cdot\;)$ is defined. Moreover, I'm aware of the usual proof of $(1)$ relying on the substitution rule. However, since the Dirac delta function is not a function, I don't understand why that rule is applicable.

It seems like one is assuming that there is a function $\delta$ such that the measure $\delta_x$ is Radon-Nikodým differentiable with respect to the Lebesuge measure $\lambda$ on $(\mathbb R,\mathcal B(\mathbb R))$ with $$\delta(\;\cdot\;-x)=\frac{{\rm d}\delta_x}{{\rm d}\lambda}\tag2$$ for all $x\in\mathbb R$. With that assumption, it's easy to see that $$\int f(y)\delta(a(y-x))\:\lambda({\rm d}y)=\frac1{|a|}\int f\left(\frac{\;\cdot\;}a\right)\delta(\;\cdot\;-ax)\:{\rm d}\lambda=\frac1{|a|}f(x)\tag3$$ for all $x\in\mathbb R$.

However, since $\delta_x$ is singular with respect to $\lambda$ for all $x\in\mathbb R$, $(2)$ is not well-defined. So, how can we state and prove $(1)$ in a rigorous way?

Please note that I know almost nothing about distribution theory. So, it would be great if there would be a purely measure theoretic answer.

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In fact (1) is the definition of $\delta(a\cdot)$. This is exactly how such things are "always" defined.

To save typing, let's say once and for all $\phi$ and $\psi$ are test functions on $\Bbb R$ and $u$ is a distribution. If $D$ is the derivative we note that $$\int(D\phi)\psi=-\int\phi D\psi,$$so we define $Du$ by $$(Du)(\psi)=-u(D\psi).$$If $\tau_a$ is the translation $\tau_a\phi(t)=\phi(t-a)$ then $$\int(\tau_a\phi)\psi=\int\phi\tau_{-a}\psi,$$hence the definition $$(\tau_au)(\psi)=u(\tau_{-a}\psi).$$Giving our dilation operators a name, say $a>0$ and define $\Delta_a \phi(t)=\phi(at)$. Then $$\int(\Delta_a\phi)\psi=\frac1{|a|}\int\phi\Delta_{1/a}\psi,$$hence the definition $$(\Delta_au)(\psi)=\frac1{|a|}u(\Delta_{1/a}\psi).$$

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    $\begingroup$ I think I've got it. In the concrete case of the question $\Delta_a\delta_x$ can even be extended to $\mathcal M$ (as the right-hand side of the definition is well-defined for any function of $\mathcal M$), right? (And note that you've missed to take the absolute value of $a$ in the denominator.) $\endgroup$ – 0xbadf00d Sep 30 '18 at 16:26

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