Where the scaling property $\int_{-\infty}^\infty f(ax)dx=\frac1a \int_{-\infty}^\infty f(x)dx$ comes from in the theory of distributions? Where the scaling property $\int_{-\infty}^\infty f(ax)dx=\frac1a \int_{-\infty}^\infty f(x)dx$ comes from in the theory of distributions?
Basically, if the distributions are defined as linear maps $f:\psi \in {\mathcal {D}}(\mathbb {R} )\to\int_{-\infty}^\infty f(x)\psi(x)dx\in\mathbb{R}$, where the scaling property comes from? Or it is a separate postulate?
 A: So first, rigorously speaking, integrals are not defined for distributions. A distribution $\boldsymbol f$ (over $\Bbb R$) is indeed a linear form that to any $\varphi\in\mathcal D(\Bbb R)$ associates a value $\langle \boldsymbol f,\varphi\rangle\in\Bbb R$ (that is sometimes also denoted $\boldsymbol f[\varphi]$).

*

*Now, to make these linear forms behave as a generalization of functions, one associates to any function $f$ a distribution $\boldsymbol f$ defined by
$$
\langle \boldsymbol f,\varphi\rangle := \int_{\Bbb R} f(x)\,\varphi(x)\,\mathrm d x.
$$
Here I put a bold font to distinguish the function and the distribution, but in general one uses the same notation, and considers functions as a particular case of distributions (a bit as rational numbers are a particular case of real numbers, but they are formally usually initially build differently).


*But there are others distributions such as $\delta_0$ that are not functions, and for example its distributional derivative acts as $\langle\delta_0',\varphi\rangle = -\varphi'(0)$ which is not an integral. However, because of the fact that brackets denote the integral in the case of functions, we can think of them as being the analogue of the integral for other distributions.
Now that objects are more clear, you see that in principle, if $\boldsymbol f$ is a distribution, there is no meaning yet to $\boldsymbol f(ax)$ (and even no meaning either to $\boldsymbol f(x)$). Let me define $\tau_a$ the operation $\tau_a f(x) = f(ax)$. If we want to use the analogue of a change of variable formula for distributions, we should indeed define what it means first. To be consistent with the change of variable formula in integrals, it is defined by
$$\label{1}\tag{1}
\langle \tau_a\boldsymbol f,\varphi\rangle := \frac{1}{|a|}\langle \boldsymbol f,\tau_{1/a}\varphi\rangle,
$$
so that indeed, when $f$ is a function, this is consistent with the formula
$$
\int_{\Bbb R} f(ax)\,\varphi(x)\,\mathrm d x = \frac{1}{|a|}\int_{\Bbb R} f(x)\,\varphi(x/a)\,\mathrm d x.
$$
In the particular case when $\boldsymbol f$ is a compactly supported distribution, then one can take $\varphi = 1$ and define the analogue of the integral of $f$ by
$$
\int_{\Bbb R} \boldsymbol f := \langle \boldsymbol f,1\rangle,
$$
and your formula follows taking $\varphi = 1$ and $a>0$ in \eqref{1}, giving
$$
\int_{\Bbb R} \tau_a\boldsymbol f = \frac{1}{a} \int_{\Bbb R} \boldsymbol f
$$
since $\tau_{1/a}1 = 1$.
