Fourier rotation theorem in higher dimensions

Question

Let $F(\mu, \nu)$ denote the Fourier transform of $f(x,y)$, then the (2D) Fourier rotation theorem says that the Fourier transform of a rotated function $f(x \cos \theta + y \sin \theta, -x \sin \theta + y \cos \theta)$ is $$F(\mu \cos \theta + \nu \sin \theta, -\mu \sin \theta + \nu \cos \theta)$$ which is the rotated version of $F(\mu, \nu)$ by the same angle $\theta$.

I wonder if this theorem holds in higher dimensions. Thanks.

Answer 1

It does, as explained here. Since your question is not really a duplicate, I'll put an edited version of that answer below.

Suppose $A:\mathbb R^n\to\mathbb R^n$ is a linear transformation. Then $\mathcal F(f\circ A)$ is computed as $$\mathcal F(f\circ A)(\xi)=\int_{\mathbb R^n} f(Ax)e^{-2\pi i \xi\cdot x}\,dx = \frac{1}{|\det A|}\int_{\mathbb R^n} f(y)e^{-2\pi i \xi\cdot A^{-1}y}\,dy $$ by the change of variables $x=A^{-1}y$. Also, $\xi\cdot A^{-1}y = ((A^{-1})^* \xi )\cdot y$ by the definition of adjoint. Therefore, $$\mathcal F(f\circ A)=\frac{1}{\det A} \mathcal{F}(f)\circ (A^{-1})^* $$

If the matrix $A$ is orthogonal (this includes rotations), then $(A^{-1})^* =A$ and $|\det A|=1$. In this special case, the Fourier transform commutes with composition: $$\mathcal F(f\circ A)= \mathcal{F}(f)\circ A$$