Sufficient conditions on $f_n\longrightarrow f$ so that $\hat{f_n}\longrightarrow\hat f$ in $L^1$ sense

Question

Let $f:\mathbb{R}\longrightarrow\mathbb{C}$ be a Schwartz function, and suppose $(f_n)_n\subset\mathcal{C}_c^{\infty}$ is such that $f_n\xrightarrow{n\to\infty}f$ in the $L^1(\mathbb{R})$ norm.

Is it true that then, $\hat{f_n}\xrightarrow{n\to\infty}\hat{f}$ in the $L^1(\mathbb{R})$ sense as well? (Here $\hat f$ denotes the Fourier transform of $f$.)

[EDIT: it appears to be false in general, as necessary conditions for $\hat{f_n}\xrightarrow{n\to\infty}\hat{f}$ in $L^1(\mathbb{R})$ include that $(f_n)_n$ converge uniformly to $f$. What would be sufficient conditions to make that happens, if there are any?]

Any help is greatly appreciated, thanks in advance!

Answer 1

If $\hat{f_n}\to \hat f$ in $L^1$, then $$f_n(0)=\int_{\mathbb R}\hat{f_n}\to \int_{\mathbb R}\hat f=f(0).$$ Thus it suffices to find $f_n$ and $f$ such that $f_n\to f$ in $L^1$ but not pointwise to construct a counterexample. I leave that to you.

Answer 2

Your question is equivalent to asking whether Fourier transform $\mathscr{F}$ is bounded from $L^1$ to $L^1$. We know it is bounded from $L^1$ to $L^\infty$. But it is not bounded from $L^1$ to $L^1$. If it is bounded, by interpolation then it is bounded from $L^1$ to $L^p$ for all $p$, which is not possible.

It is useful to use scaling invariance to check the boundedness of an operator. If $\|\hat{f}\|_1\leq C\|f\|_1$ for all $f$ and some constant $C$. Then substituting $f$ by $f_{\lambda}(x):=f(\lambda x)$. Then you will get:

$$ \|\hat{f}\|_1\leq\lambda^{-d}C\|f\|_1 $$ ($d$ is the dimension $\mathbb{R}^d$) Letting $\lambda\rightarrow\infty$, then we have $f=0$.