Scaling property of the Fourier transform
= Scaling property of the Fourier transform
If $f_\lambda(x)=f(\lambda x)$ on $\mathbb R^n$ with $\lambda>0$, then
$$
\widehat {f_\lambda}(\xi)
=\lambda^{-n}\widehat f(\xi/\lambda),
$$
and consequently
$$
\|f_\lambda\|_p=\lambda^{-n/p}\|f\|_p,
\qquad
\|\widehat {f_\lambda}\|_q
=\lambda^{-n(1-1/q)}\|\widehat f\|_q.
$$