Since $\delta_{1/t}\varphi(x)=\varphi(x/t)$ maps the <Schwartz space> continuously to itself, the formula
$$
\langle\delta_tu,\varphi\rangle=t^{-n}\langle u,\delta_{1/t}\varphi\rangle
$$
defines a continuous <linear functional> on $\mathcal S$, hence a <tempered distribution>. For a <locally integrable function> $u$, the <change of variables formula> $y=tx$ gives
$$
\int_{\mathbb R^n}u(tx)\varphi(x)\,dx
=t^{-n}\int_{\mathbb R^n}u(y)\varphi(y/t)\,dy,
$$
so this <dilation of a distribution> agrees with ordinary function dilation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 327 / 1 / b / i / Solution

{title2=$\delta_tu$}

For $t>0$, the dilation $\delta_tu$ of a distribution on $\mathbb R^n$ is defined by
$$
\langle\delta_tu,\varphi\rangle=t^{-n}\langle u,\varphi(\mathord\cdot/t)\rangle.
$$
For a regular distribution this agrees with the function $x\mapsto u(tx)$.


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 Incoming links: Dilation of a distribution