Fourier transform of $f\circ A$ where $A$ linear and $f\in L^1(\mathbb{R}^n)$

Question

Let $A: \mathbb{R}^n \to \mathbb{R}^n$ be a linear map and $f\in L^1(\mathbb{R}^n)$. I would like to express the fourier transform of $f_A:= f\circ A$ in terms of the Fourier transform $\hat{f}$ of $f$.

Also for which $A$ do we have $\hat{f_A} = \hat{f} \circ A$, other than for $A=id$?

Answer 1

First off, you should require that $A$ is invertible. Because if $f\in L^1$ (or $L^2$) and $A$ is not invertible, there is no guarantee that $f\circ A\in L^1$ (or $L^2$): trivially, if $A=0$ and $f(0)\neq 0$, then $$\int_{\mathbb{R}^n}|f\circ A|^p(x)dx=\int_{\mathbb{R}^n}|f(0)|^pdx=\infty,\quad \forall p\in [1,\infty) $$ If $A$ is invertible, then by a change of variables \begin{align*}\widehat{f\circ A}(\xi)&=\int_{\mathbb{R}^n}f(Ax)e^{-ix\xi}dx=|\det A^{-1}|\int_{\mathbb{R}^n}f(x)e^{-iA^{-1}x\cdot \xi}dx=\\ &=|\det A^{-1}|\int_{\mathbb{R}^n}f(x)e^{-ix\cdot (A^{-1})^*\xi}dx=|\det A^{-1}|\hat{f}\left((A^{-1})^*\xi\right) \end{align*} Where $A^*$ is the transpose of $A$. Thus $$\widehat{f\circ A}=\hat{f}\circ A \iff A=(A^{-1})^* $$ and the latter condition is satisfied iff the matrix $A$ is orthogonal.

Answer 2

If you're not given that $A$ is invertible the function $f_A$ need not have a Fourier transform in the classical sense. For example if $A=0$ then $f_A=f(0)$, and non-zero constants don't have Fourier transforms.