For every $\varphi\in\mathcal S$, the <change of variables formula> gives the identity
$$
\widehat{(A^t)^*\varphi}(x)
=\frac1{|\det A|}\widehat\varphi(A^{-1}x).
$$
Using this, the distributional Fourier transform, and the pullback formula from part b,
$$
\begin{aligned}
\langle\widehat{A^*u},\varphi\rangle
&=\frac1{|\det A|}
\langle u,(A^{-1})^*\widehat\varphi\rangle\\
&=\langle u,\widehat{(A^t)^*\varphi}\rangle
=\langle\widehat u,(A^t)^*\varphi\rangle\\
&=\left\langle
\frac{((A^t)^{-1})^*\widehat u}{|\det A|},
\varphi\right\rangle.
\end{aligned}
$$
Therefore
$$
\boxed{\widehat{A^*u}
=\frac{((A^t)^{-1})^*\widehat u}{|\det A|}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 327 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-327/3/c/solution