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](https://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1) may be relevant. Thanks for any insight/references.



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 $F\_\sigma$.


