Can we calculate for a multiple composition of Fourier transforms?

Question

I tried to find a formula for multiple composition of a Fourier transform (not a convolution). $FoFoFo...oF\{f(t)\} =$?

Can we also find a formula for multiple composition of Fourier for $f(t) \cdot g(t)$ ?

Answer 1

For your first question: you can actually view the Fourier transform as a rotation of $L^2(\mathbb{R})$ when put in a sufficiently nice orthonormal basis (the so-called Hermite Polynomials). There are four eigenspaces in this basis, each corresponding to the action of the Fourier transform rotating by an angle of $1, i,-1,-i$.

A more elementary way to answer your question: let's take the normalization $F\{f\}(k) = \int_{\mathbb{R}} f(t) e^{-2 \pi i k t} dt$. Recall that the inverse transform is the same as the forward transform, except the minus sign in the exponential is replaced with a plus sign. As an exercise, try to show that in this normalization, the following hold: $$ F\{f\} = \overline{F^{-1}\{\overline{f}\}} \\ \overline{F\{f\}} = F\{\overline{f_-}\} $$ where $f_-(t) = f(-t)$ is the 'rotation' of $f$. Now we have $$ F \circ F\{f\} = \overline{F^{-1}\{\overline{F\{f\}}\}} = \overline{F^{-1}\{F\{\overline{f_-}\}\}} = \overline{\overline{f_-}} = f_- $$ So we see that iterating $F$ twice gives the "rotation" $f \mapsto f_-$. Another two compositions and we have $F \circ F \circ F \circ F = \text{Id}$.

For your second question, you can use the convolution formula and the above to figure out what's going on.