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)

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 ?

**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,