Let's say I have a sequence of functions $f_n\in L^1(\Bbb{R})$ such that
- $f_n\rightarrow f$ in $L^1(\Bbb{R})$, and
- $\hat{f}_n\in L^1(\Bbb{R})$ for all $n$ (where hat is the Fourier transform), and
- $\hat{f}_n\rightarrow g$ for some $g\in L^1(\Bbb{R})$.
Then it should be the case that $g = \hat{f}$, correct? Certainly it suffices to show that $\hat{f}_n\rightarrow \hat{f}$ in $L^1$. By the boundedness of $\mathcal{F}:L^1\rightarrow L^\infty$, we have trivially that
$$ \|\hat{f}_n-\hat{f}\|_\infty\leq\|f_n-f\|_1 $$ and so $\hat{f}_n\rightarrow \hat{f}$ pointwise almost everywhere (actually, everywhere since $f_n$ are continuous by virtue of $\hat{f}_n\in L^1$). From here I can think of a few complicated directions, but can you think of an elegant proof?
Let $g_n = \hat f_n$. We have that $g_n \to g$ in $L^1$, so by the proof that $L^1$ is a Banach space, there is a subsequence $(g_{{n_k}})$ converging to $g$ a.e. Now, $(g_{n_k})$ is a subsequence of $(g_n)$, so it must also converge to $\hat f$ a.e. as so does $(g_n)$. Thus, $g = \hat f$ a.e. By continuity, we get $g = \hat f$ on $\Bbb R$.