Image of L^1 under the Fourier Transform

Here is an attempt, somewhat rough around the edges but I think it works:

Claim: The range of the Fourier transform $\mathcal{F}:L^1(\mathbb R)\to C_0(\mathbb R)$ is a Borel set in $C_0(\mathbb R)$ of the form

\begin{equation} \bigcap_{k=1}^\infty \bigcup_{N=1}^\infty \bigcap_{m,n\geq N} E_{m,n,k} \end{equation}

where each $E_{m,n,k}$ is an $F_\sigma$.

Proof: Consider the cutoff functions $\{e^{-a\pi|t|}\}$ and fix a sequence $a_n\to 0$. It is a fact that a function $g\in C_0(\mathbb R)$ is the Fourier transform of some $f\in L^1$ if and only if the sequence \begin{equation} T_n(g)(x) := \int_{\mathbb R} e^{-a_n\pi |t|} g(t)e^{2\pi itx} dt \end{equation} is Cauchy in $L^1$, in which case if we put $f=\lim T_ng$ then $g =\widehat{f}$. Let $R$ denote the range of the Fourier transform in $C_0$. If we define

\begin{equation} E_{n,m,k} = \lbrace g\in C_0: T_ng, T_mg\in L^1 ,{||T_n(g)-T_m(g)||}_1 \leq\frac{1}{k} \rbrace \end{equation}

then $R$ has the claimed form.

To prove that $E_{m,n,k}$ is an $F_\sigma$, put $T_n(g)-T_m(g):=T_{mn}(g)$, explicitly

\begin{equation} T_{mn}g(x):= T_ng(x)-T_mg(x) =\int_{\mathbb R} h_{mn}(x,t)g(t) dt \end{equation}

where we define

\begin{equation} h_{mn}(x,t) = (e^{-a_m\pi|t|} - e^{-a_n\pi|t|} )e^{2\pi itx} \end{equation}

Note that by monotone convergence, $g\in E_{m,n,k}$ if and only if two conditions are satisfied: first, for fixed $n$, we need $T_ng\in L^1$. By monotone convergence this is equivalent to: There exists an integer $N$ such that for all integers $d\geq 1$,

\begin{equation} \left\| {\bf 1}_{[-d,d]}(x)T_ng(x)\right\|_1 \leq N. \end{equation}

By dominated convergence the set of all such $g$ obeying this for fixed $N,n,d$ is closed, so the set that obeys this for some $N$ and all $d$ (with $n$ held fixed) is an $F_\sigma$. Thus for fixed $m,n$ the set of $g$ for which $T_ng, T_mg\in L^1$ is an $F_\sigma$. Additionally, for all integers $d\geq 1$, we need the condition

\begin{equation} \left\| {\bf 1}_{[-d,d]}(x) T_{mn} g(x)\right\|_1 \leq \frac{1}{k}. \end{equation}

But similarly, the set of $g$ obeying this for fixed $m,n,k,d$ is closed in $C_0(\mathbb R)$, and we conclude that each $E_{m,n,k}$ is an $F_\sigma$.

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 $G_{\delta\sigma}$.