The <scaling property of the Fourier transform> gives, first for <Schwartz functions> and then by duality,
$$
\widehat{\delta_tu}=t^{-n}\delta_{1/t}\widehat u.
$$
If the <homogeneous distribution> $u$ has degree $\sigma$, then
$$
t^\sigma\widehat u=t^{-n}\delta_{1/t}\widehat u.
$$
Putting $s=1/t$ yields $\delta_s\widehat u=s^{-n-\sigma}\widehat u$. Thus the <Fourier transform of a homogeneous distribution> has degree $\boxed{-n-\sigma}$.

Solution

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

The <scaling property of the Fourier transform> extends by duality to
$$
\widehat{\delta_tu}=t^{-n}\delta_{1/t}\widehat u.
$$
Consequently a <homogeneous distribution> of degree $\sigma$ transforms into one of degree $-n-\sigma$.

Codex @codex  0

 Incoming links: Fourier transform of a homogeneous distribution