# Image of L^1 under the Fourier Transform

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.

• Isn't this space related to the Wiener Algebra? This is the space of $2\pi$-periodic functions with absolute summable Fourier coefficients. I vaguely remember that I heard somebody calling the image of $L^1$ under Fourier transform also "Wiener Algebra"...
– Dirk
Apr 9, 2013 at 8:46
• Dirk: yes. Also called the Fourier algebra Apr 9, 2013 at 18:28
• Ah, I see. Interesting that the Wikipedia pages of the Wiener algebra and the Fourier algebra do not mention each other...
– Dirk
Apr 10, 2013 at 6:37

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

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

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 $$$$T_n(g)(x) := \int_{\mathbb R} e^{-a_n\pi |t|} g(t)e^{2\pi itx} dt$$$$ is Cauchy in $$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

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

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

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

where we define

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

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 $$d\geq 1$$,

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

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

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

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 $G_{\delta\sigma}$.

• Yes, that's much cleaner. And I would also suspect it is not $G_{\delta\sigma}$. Apr 13, 2013 at 12:53