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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 $L^1(\mathbb{R})$, 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 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.

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juan arroyo
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    In recent years this thing has been called the Lebesgue-Fourier algebra of ${\mathbb R}$ (because it's the intersection of $L^1({\mathbb R})$ with the Fourier algebra $A({\mathbb R})\cong L^1(\widehat{\mathbb R})$). I'm not sure what you're hoping to say about such functions. Of course, since they are $C_0$ and integrable, they belong to every $L^p$ – Yemon Choi Feb 17 '16 at 20:35
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    Also, why the "descriptive-set-theory" tag? – Yemon Choi Feb 17 '16 at 20:36
  • I'm not sure, I was wondering if there is anything else that characterizes them.The post I linked to included that tag, I suppose it can be removed. – juan arroyo Feb 17 '16 at 20:39

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