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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-327/1/b/ii/solution