Showing $\mathcal{F}(f\circ R_{\theta})=(\mathcal{F}f)\circ R_{\theta}$ If $R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ is the rotation in $\mathbb{R}^2$ through angle $\theta$ then for $f\in\mathcal{S}(\mathbb{R}^2)$ I get
$$(\mathcal{F}(f\circ R_\theta))(\xi)=\mathcal{F}(f(R_\theta(\xi))) = \int_{\mathbb{R}^2} f(x)e^{ix\cdot R_{\theta}(\xi)}\,dx$$
I don't immediately see why this would be equal to $(\mathcal{F}f)\circ R_\theta$ though?
 A: If $T : \Bbb{R}^n \to \Bbb{R}^n$ is an invertible linear transformation and $f \in \mathcal{S}(\Bbb{R}^n)$, then with the substitution $Tx = y$,
\begin{align*}
\mathcal{F}(f \circ T)(\xi)
&= \int_{\Bbb{R}^n} f(Tx)e^{i\langle x, \xi\rangle} \, dx \\
&= \frac{1}{|\det T|}\int_{\Bbb{R}^n} f(y)e^{i\langle T^{-1}y, \xi\rangle} \, dy \\
&= \frac{1}{|\det T|}\int_{\Bbb{R}^n} f(y)e^{i\langle y, T^{-\mathsf{T}}\xi\rangle} \, dy \\
&= |\det T|^{-1}(\mathcal{F}f)(T^{-\mathsf{T}} \xi),
\end{align*}
where $T^{-\mathsf{T}} = (T^{-1})^{\mathsf{T}} = (T^{\mathsf{T}})^{-1}$ is the transpose inverse of $T$. If $T$ is a rotation matrix, however, then $T = T^{-\mathsf{T}}$ and $|\det T| = 1$, hence we have $\mathcal{F} (f \circ T) = (\mathcal{F} f) \circ T$.
A: \begin{align}
& \int_{\mathbb{R}^2} f(x) \exp( ix\cdot R_{\theta}(\xi))\,dx \\[10pt]
= {} & \int_{\mathbb R^2} f(x) \exp( ix \cdot \zeta ) \, dx \qquad \text{where } \zeta = R_\theta(\xi) \\[10pt]
= {} & (\mathcal F f)(\zeta) = (\mathcal F f)(R_\theta(\xi)) = \Big( (\mathcal F f) \circ R_\theta \Big) (\xi)
\end{align}
