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 Question: How to show $\mathcal{F}L^{1}$ is a complete with respect to the above norm ?
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,
The answer depends on how you understand the line $$\mathcal{F}L^1 = \{f\in L^{\infty}(\mathbb R):\hat{f}\in L^{1}(\mathbb R)\} \tag{1}$$ What does it mean to take the Fourier transform of an $L^\infty$ function? Obviously, the integral defining the transform has no reason to converge.
Version 1. What you actually mean is $$\mathcal{F}L^1 = \{\check g : g\in L^{1}(\mathbb R)\} \tag{2}$$ (Since $\check g$ is automatically in $L^\infty$, I do not include this as a condition.)
Then your proof works as is. There is not much to prove, actually. The correspondence $g\mapsto \check g$ is an isometry from $L^1$ onto $\mathcal{F}L^1$, and isometries preserve completeness.
Version 2. Every element $f\in L^\infty$ is a tempered distribution; we know how to take Fourier transform of those. The transform $\hat f$ is in general a tempered distribution, but if it happens to be represented by an $L^1$ function, then we say $f\in \mathcal{F}L^1$.
In this version you need to know that the inversion formula works for tempered distributions. But this is a lot easier than it sounds: all we do is pass hats from distributions to test functions, and since test functions are all in $L^1\cap \mathcal FL^1$, inversion works.