Is a Schwartz space closed under fractional Fourier transforms?

Question

The Fourier transform is a linear automorphism of a Schwartz space. A (1/n)-fractional Fourier transform is an operator whose nth iterate is the Fourier transform. Note that this might not be unique.

For all n, is there a 1/n fractional Fourier transform on a Schwartz space, that is a linear automorphism? My gut says yes but the math is working out unpleasantly.

Answer 1

$\mathcal{F}^4 = \text{Id}$ implies the eigenspace decomposition of the Fourier transform $$T_m[\phi] = \frac14 \sum_{k=0}^3 i^{-mk} \mathcal{F}^k[\phi], \qquad T_m[T_m[\phi]]=T_m[\phi],\qquad T_n[T_m[\phi]]=0, \qquad \phi = \sum_{m=0}^3 T_m[\phi]\\ \mathcal{F}[T_m[\phi]] = i^m T_m[\phi]$$ From there the (fractional) iterates of $\mathcal{F}$ have an obvious expression

$$\mathcal{F}^a[\phi] = \sum_{m=0}^3 e^{a m i \pi / 2} T_m[\phi]$$

Which leads to $$\mathcal{F}^{a+b}[\phi]= \mathcal{F}^a[\mathcal{F}^b[\phi]], \qquad \mathcal{F}[\phi] = \underbrace{\mathcal{F}^{1/n}\circ \ldots \circ \mathcal{F}^{1/n}}_n[\phi]$$

Everything is clearly continuous on the Schwartz space.