If $\mathcal{F}$ is the Fourier transform, what can be said about $\mathcal{F}(L^1(\mathbb{R})) \cap L^1(\mathbb{R})$?

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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 $L^1(\mathbb{R^n})$, the inverse Fourier transform can be applied to $\mathcal{F}(f)$, which yields a continuous function that is equal to $f$ almost everywhere.

Thus, it seems that the (inverse) Fourier transform also 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 that can be said about this these functions, about this map, or about this subset of $L^1(\mathbb{R})$?