By <multiplication of a distribution by a smooth function>, for every <test function> $\psi$,
$$
\langle\rho\delta_0,\psi\rangle
=\langle\delta_0,\rho\psi\rangle
=\rho(0)\psi(0)=\psi(0)
=\langle\delta_0,\psi\rangle.
$$
Hence $\rho\delta_0=\delta_0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 327 / 2 / b / i / Solution

For a <distribution> $u$ and a <smooth function> $a$, the product $au$ is defined by $\langle au,\varphi\rangle=\langle u,a\varphi\rangle$. It obeys the <Leibniz rule>; in one dimension, $(au)'=a'u+au'$.

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