$f_{n}\to f$ in $L^{1}\implies \hat{f_{n}}\to \hat{f}$ in $L^{1}$? Let $f\in L^{1}(\mathbb R)$ and it Fourier transform, $\hat{f} (y) : = \int _ {\mathbb R} f(x) e^{-2\pi i x\cdot y} dx ; y \in \mathbb R .$  Suppose there exists $f_{n},  \in L^{1}(\mathbb R)$ such that $\|f_{n}-f\|_{L^{1}(\mathbb R)}\to 0$ as $n\to \infty$, that is, $f_{n}$ converges to $f$ in $L^{1}(\mathbb R).$
Also, we  assume, $\hat{f_{n}}\in L^{1}(\mathbb R),$ for every $n\in \mathbb N.$

My Question: Can we expect $\hat {f_{n}}$ converges in $L^{1}(\mathbb R)$; if yes, is it converges to $\hat{f}$ ? Or, we can produce counter example ?

Thanks,
 A: No, you can't expect. You can approximate any $f\in L^1$ by functions $f_n$ in the Schwartz space $\mathscr S$ (even by smooth functions with compact support).
Then $\hat f_n \in \mathscr S \subseteq L^1$ but there is no reason to expect convergence of $\hat f_n$.
A: Here is a counterexample. Let $\phi\colon\mathbb{R}\to\mathbb{R}$ be $C^\infty$ supported on $[0,1]$, positive with $\int_0^1\phi(x)=\int_0^1\phi(x)^2\,dx=1$. Let
$$
f_n(x)=\sum_{k=1}^n2^{k}\phi(2^{2k}(x-k)).
$$
Then
$$
\int_{\mathbb{R}}f_n(x)\,dx=\sum_{k=1}^n2^{k}2^{-2k}<1.
$$
Moreover $f_n(x)\le f_{n+1}(x)$ for all $x\in\mathbb{R}$. It follows that $f_n$ converges pointwise and in $L^1$ to $f=\sum_{k=1}^\infty2^{k}\phi(2^{2k}(x-k))$. It is clear that $\hat f_n\in L^1$. But
$$
\int_{\mathbb{R}}|f(x)|^2\,dx=\sum_{k=1}^n2^{2k}2^{-2k}=\infty.
$$
That is, $f\not\in L^2$, hence $\hat f\not\in L^2$. Since $\hat f\in L^\infty$, this implies that $\hat f\not\in L^1$, because otherwise, by interpolation, we would have $\hat f\in L^2$.
