# How can we prove the scaling property of the Dirac delta function rigorously?

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.

## 1 Answer

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 $$(\Delta_au)(\psi)=\frac1{|a|}u(\Delta_{1/a}\psi).$$

• 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.) Sep 30, 2018 at 16:26