TITLE: Image of L^1 under the Fourier Transform
QUESTION [24 upvotes]: The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto.  It is known (if I remember correctly) that the range isn't closed, but is dense in $C_0$.  Everything I have read/heard says that the range is "difficult to describe".  
But, since $\mathcal{F}$ is injective and continuous, the image of $L^1$ must be Borel inside of $C_0$.  Is anything else known about its descriptive complexity?  If not, might this be an example of a natural set of high Borel rank?
This question may be relevant. Thanks for any insight/references.

REPLY [9 votes]: It seems that Mike's argument shows that in fact the range $R$ of the Fourier transform is $F_{\sigma\delta}$ in $C_0(\mathbb R)$. Define the $T_ng$ as above, 
$$T_ng(x)=\int_{\mathbb R} e^{-a_n\pi\vert t\vert}g(t) e^{2i\pi tx}dt\;.$$ Then a function $g\in C_0(\mathbb R)$ is in $R$ iff two things hold:
(i) all $T_ng$ are in $L^1$;
(ii) the sequence $(T_ng)$ is Cauchy in $L^1$.
Condition (i) can be written as follows:
$$ \forall n\;\exists N\in\mathbb N\; \left ( \forall d\in\mathbb R_+ \;:\;\int_{-d}^d \vert T_ng(x)\vert dx\leq N\right)$$
By dominated convergence, the condition under brackets is closed with respect to $g$; so (i) defines an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$.
Condition (ii) reads
$$\forall k\in\mathbb N\;\exists N \;\left(\forall p,q\geq N\; \forall d \;:\; \int_{-d}^d \vert T_pg(x)-T_qg(x)\vert dx\leq \frac 1k\right)$$
By dominated convergence again, the condition under brackets is closed wrt $g$, so (ii) defines an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$.
Altogether, $R$ is the intersection of two $F_{\sigma\delta}$ sets, hence an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$. I would be extremely surprised if it were better than that; i.e.   I  "conjecture" that it is not $G_{\delta\sigma}$.