Theorem 12.4.2: Properties of the Fourier Transform

(1)
$\ha{f*g}=\ha{f}\cd\ha{g}$ ^[1]
(2)
If $f_{a}(x)\define f(x+a)$ then $\ha{f_a}(\xi)=e^{2\pi ia}\ha{f}(\xi)$
(3)
If $T\in GL(n,\R)$ then $\ha{f\circ T}(\xi)=\abs{\det T}\inv\ha{f}((T\inv)^{*}\xi)$^[2]. In particular if $T\in SO(n,\R)$ then $\ha{f\circ T}(\xi)=\ha{f}(T\xi)$.^[3]

Proof.

(1)
\begin{align*}\ha{f*g}(\xi)=\integral{\R^n}{}\integral{\R^n}{}e^{-2\pi i x\cd\xi}f(x-y)g(y)dydx=\integral{}{}\integral{}{}e^{-2\pi i(x-y)\cd\xi}f(x-y)e^{-2\pi iy\cd \xi}g(y)dxdy=\ha{f}(\xi)\integral{}{}e^{-2\pi i y\cd\xi}g(y)dy=\ha{f}(\xi)\ha{g}(\xi)\end{align*}

by Fubini and translation invariance of Lebesgue measure.
(2)
Change of variable
(3)
Let $y=Tx$ then $x=T\inv y$ and $g(x)\define f(T(x))$ so
\begin{align*}\ha{f\circ T}(\xi)=\integral{\R^n }{}e^{-2\pi i\xi\cd x}g(x)dx=\integral{\R^n }{}e^{-2\pi i\xi\cd T\inv(y)}f(y)\abs{\det T}\inv dy=\abs{\det T}\inv \integral{}{}e^{-2\pi i((T\inv)^*\xi)\cd y}f(y)dy\end{align*}

$\square$

The Eigenspace