Problem Statement
=================


Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption **I1**). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption **C**) is an integrable function (assumption **I2**) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption **S**). Finally, let assumption(s) **U** be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:


\begin{equation}
\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1}
\end{equation}


From the convolution theorem, the following is equivalent:


\begin{equation}
f\*\left(\mathcal Tg\right)=\mathcal T\left(f\*g\right).
\label{2}\tag{2}
\end{equation}


In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.


My question is to define **U** and show how these assumptions can be used to prove equation \eqref{1}.


Assumptions
===========


I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.


Assumption **I1** and **I2**
----------------------------


From the convolution theorem:


$$
\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f\*g.
$$


If $f$ and $g$ are integrable, then the convolution is defined ([[source]](https://math.stackexchange.com/questions/519034/show-that-convolution-of-two-measurable-functions-is-well-defined "Show that convolution of two measurable functions is well-defined")).


Assumption **C**
----------------


Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption **S**.


Assumption **S**
----------------


Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).



Proof of covariant convolution for a kernel function that is rotation symmetric in Fourier space
I would think that no extra assumption **U** is needed, assumption **S** suffices. The key thing to observe is that the Fourier transform of a rotated function is equal to a rotated version of the Fourier transform of that function, see for example [Appendix A: Rotation property of Fourier transforms](https://link.springer.com/content/pdf/bbm:978-1-4613-1333-5/1.pdf) of the book [Jakowatz, Wahl, Eichel, Ghiglia, and Thompson - Spotlight-Mode Synthetic Aperture Radar: A Signal Processing Approach](https://doi.org/10.1007/978-1-4613-1333-5) for a proof.


Denote by $R$ the operation that rotates the vector $x$, then assumption **S** that $\mathcal T\mathcal F f=\mathcal F f$ implies that $f(Rx)=f(x)$. Now apply this to the convolution,
$$(f\ast {\cal T}g)(x)=\int dz\, f(x-z)g(Rz)=\int dz\, f(Rx-Rz)g(Rz)$$
$$=\int dRz\, f(Rx-Rz)g(Rz)=(f\ast g)(Rx)={\cal T}(f\ast g)(x),$$
which is the desired identity. (When switching from $z$ to $Rz$ as integration variable I have used that the Jacobian for this transformation is unity.)


