Consider the space of all Fourier transforms of $L^{1}(\mathbb R),$ that is,
$$\mathcal{F}L^{1}=\mathcal{F}L^{1}(\mathbb R):= \{f\in L^{\infty}(\mathbb R):\hat{f}\in L^{1}(\mathbb R)\},$$
with the norm, $\|f\|\_{\mathcal{F}(L^{1})}=\|\hat{f}\|\_{L^{1}(\mathbb R)}.$
(By uniqueness theorem and using the fact that Fourier transform is a linear, one can deduce that, this is actually a norm)



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> **My Questions**:(1) Can we expect $\mathcal{F}L^{1}$ is a complete with respect to the norm $\|\cdot\|\_{\mathcal{F}L^{1}}$ ? (2) Put, $A=\{f\in \mathcal{F}L^{1}: |f|\in \mathcal{F}L^{1}\}.$ Can we expect $A$ is closed in $\mathcal{F}L^{1}$ ? (3) (Bit Vague question) Can you suggests some proper closed subsets of $\mathcal{F}L^{1}$; can we expect its characterization ?
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**My attempt**: Suppose $\{f\_{n}\}\_{n\in \mathbb N}$ is a Cauchy sequence in $\mathcal{F}L^{1},$
that is, there is $N\in \mathbb N$, such that, $\|\hat{f\_{n}}-\hat{f\_{m}}\|\_{L^{1}(\mathbb R)}\to 0$, for every $n, m \geq N$; and since $(L^{1}(\mathbb R), \|\cdot\|\_{L^{1}(\mathbb R)})$ is complete, there is a $g\in L^{1}(\mathbb R)$, such that, $\|\hat{f\_{n}}-g\|\_{L^{1}(\mathbb R)}\to 0$ as $n\to \infty.$ We must find, $h\in \mathcal{F}L^{1}$ such that $\|f\_{n}-h\|\_{\mathcal{F}L^{1}}\to 0$ as $n\to \infty.$ Now my guess work is that, we should take, $h:=\check{g}$; but then the problem is: can we expect, $\hat{h}=g$ (I know this one can expect, if both $f$ and $\hat{f}$ both are in $L^{1}(\mathbb R)$, by inversion formula); or am I missing some thing ?


Thanks,



Closed sets in the space of Fourier transforms $\mathcal{F}L^{1}$
(1) The space is obviously complete since you defined the norm in such a way that $\mathcal{F}$ is an isometry $L^1 \cong \mathcal{F}L^1$ (maybe an isometry up to constants depending on your definition of the fourier transform). Consequently the answer to (3) is that the closed sets of $\mathcal{F}L^1$ are exactly the fourier transforms of closed sets of $L^1$.


