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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]).

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).

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    $\begingroup$ I think by "For this convolution to be defined, both $f$ and $g$ must be integrable" you mean "If $f$ and $g$ are both integrable, then the convolution is defined", or maybe "The convolution is not always defined, but it is if $f$ and $g$ are both integrable" (which is what your link shows). For example, if one of $f$ or $g$ is $0$, then the convolution is defined regardless of the integrability of the other. $\endgroup$
    – LSpice
    Commented Dec 23, 2022 at 23:33
  • $\begingroup$ I was thinking that this could lead to infinity times zero. I will edit, though. Thanks $\endgroup$
    – Jacob Helwig
    Commented Dec 24, 2022 at 20:16
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    $\begingroup$ Re, measure-theoretic $0\cdot\infty$ is usually $0$, so that $\infty\chi_A$ and $0\chi_B$ are both correctly given integral $0$ when $A$ has measure $0$ and $B$ has measure $\infty$. $\endgroup$
    – LSpice
    Commented Dec 24, 2022 at 20:27

1 Answer 1

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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 of the book Jakowatz, Wahl, Eichel, Ghiglia, and Thompson - Spotlight-Mode Synthetic Aperture Radar: A Signal Processing Approach 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.)

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  • $\begingroup$ Please let me know if I have misunderstood how you arrived at your first step. First, the rotation property of the Fourier transform says that if I rotate the physical domain, then the frequency domain rotates equally: $$\mathcal F \mathcal Tf(x)[\xi] = \mathcal F \mathcal f(Rx)[\xi] = \mathcal F \mathcal f(x)[R\xi]= \mathcal T\mathcal F \mathcal f(x)[\xi]$$ Therefore: $$\mathcal F \mathcal Tf(x)[\xi] = \mathcal T\mathcal F \mathcal f(x)[\xi] = \mathcal F \mathcal f(x)[\xi]$$ The last equality is from assumption S. $\endgroup$
    – Jacob Helwig
    Commented Dec 22, 2022 at 23:29
  • $\begingroup$ Finally, from the uniqueness of the transform [source] $$\mathcal F \mathcal Tf(x)[\xi] = \mathcal F \mathcal f(x)[\xi]\iff \mathcal Tf=f, a.e.$$ $\endgroup$
    – Jacob Helwig
    Commented Dec 22, 2022 at 23:30
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    $\begingroup$ yes, that is how I understand it. $\endgroup$
    – Carlo Beenakker
    Commented Dec 23, 2022 at 9:17
  • $\begingroup$ Based on my understanding, for all $f$, the rotation property says that: $$ \mathcal T\mathcal F f =\mathcal F\mathcal T f $$ I also believe that the only way $\mathcal T\mathcal F f=\mathcal F f$ is if $f$ is complex valued, since if $f$ is real, $\mathcal F f$ will be Hermitian and therefore cannot have rotation symmetry. Both of these seem to not be true though. Suppose $f$ is real-valued and $\mathcal Tf=f$. Then: $$ \mathcal F \mathcal Tf = \mathcal F f $$ Also from the rotation property: $$ \mathcal F\mathcal T f = \mathcal T \mathcal Ff $$ $\endgroup$
    – Jacob Helwig
    Commented Dec 26, 2022 at 23:42
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    $\begingroup$ Hermitian is a concept that refers to an operator kernel $F(\mathbf{r}|\mathbf{r}')=\bar{F}(\mathbf{r}'|\mathbf{r})$; in the case you are considering the operator kernel is diagonal $F(\mathbf{r}|\mathbf{r}')=f(\mathbf{r})\delta(\mathbf{r}-\mathbf{r}')$; in that case Hermitian is the same as real, $f(\mathbf{r})=\bar{f}(\mathbf{r})$. $\endgroup$
    – Carlo Beenakker
    Commented Dec 28, 2022 at 7:07

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