Let $f\in L^\infty(\mathbb{R})$ be a bounded Borel function.
Q: Does there exist a sequence of functions $(f_i)_{i\in\mathbb{N}}$ in $L^\infty(\mathbb{R})$ such that:
- $\displaystyle\lim_{i\rightarrow\infty}||f-f_i||_\infty = 0$;
- for each $i$, $f_i$ has compactly supported (distributional) Fourier transform?
Remark: I think condition $2$ forces the approximating functions $f_i$ to be smooth.
I understand that such a sequence $(f_i)$ does exist if $L^\infty$ were replaced by $L^1\cap L^2$, but as I understand, the Fourier transform of a bounded Borel function is a distribution. Since I am not familiar with the theory of distributions, the purpose of this question is to find out whether such an approximation can still be done in this case.
Modified Q: As MaoWao pointed out, the answer to the question as stated is no. So I'd like to make a modification, namely to replace $L^\infty$ in the question by $C_b(\mathbb{R})$, the space of bounded continuous functions.
This is not possible. By the Paley-Wiener-Schwartz theorem, the functions $f_j$ have holomorphic extensions to the complex plane. Hence the uniform limit (on $\mathbb{R}$) is necessarily continuous.
Even if you only know that each $f_j$ is smooth, the uniform limit will still be continuous, so this is enough to find counterexamples.