The Fourier transform gives a map of the Schwartz space to itself which turns out to be a linear homeomorphism of period 4.

However, when the domain is extended to $L^1(\mathbb{R})$, the situation is not as clean. The Fourier transform gives a bounded linear map $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$. If $\mathcal{F}(f)$ happens to be in $L^1(\mathbb{R})$, the inverse Fourier transform can be applied to $\mathcal{F}(f)$, which yields a continuous function that is equal to $f$ almost everywhere (everywhere if $f$ was continuous).

Thus, if we restrict $\mathcal{F}$ to $\mathcal{F}(L^1(\mathbb{R})) \cap L^1(\mathbb{R})$ (that is, $L^1$ functions which are also in the image of $\mathcal{F}$), it seems that the Fourier transform turns out to be a linear bijection of period 4 from $\mathcal{F}(L^1(\mathbb{R})) \cap L^1(\mathbb{R})$ to itself.

Is there anything else that can be said about this these functions, about this map, or about this subset $\mathcal{F}(L^1(\mathbb{R})) \cap L^1(\mathbb{R})$?

A [similar][1] question was asked a while back, about the image $\mathcal{F}(L^1(\mathbb{R}))$. For simplicity I used $\mathbb{R}$ as the domain, but of course I'm curious about the case for $\mathbb{R^n}$ as well.

[1]:https://mathoverflow.net/questions/126923/image-of-l1-under-the-fourier-transform