The Fourier Transform is in general terms a continuous mapping, as well as it's inverse when it exists. Is there any abstract result that translates convergence of the transforms of a sequence of functions into convergence of the original sequence when the inverse does not exist but convergence takes place to a function in the image of the initial space under the Fourier Transform?

Assume for instance that $f_{n}$ is a sequence in $L^{1}$ and that $\hat{f}_{n} \to \hat{f}$ uniformly, for some $f \in L^{1}$. Can something be said about the convergence of $f_{n}$ to $f$ in $L^{1}$?

That is, if for given spaces $X,Y$, the function $\mathcal{F}:X \to Y$ is continuous (I'm thinking for instance in $X = L^{p}$ and $Y=L^{p'}$, $p \in (1,2]$ ) and the image under the Fourier Transform of a sequence $f_{n}$ converges in $Y$ to a function $\hat{f} \in \mathcal{F}(X)$, does $f_n$ converge to $f$ in $X$?