# Approximating a bounded Borel function by a sequence of functions with compactly supported Fourier transform

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:

1. $$\displaystyle\lim_{i\rightarrow\infty}||f-f_i||_\infty = 0$$;
2. 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.
• Thanks for that. I've modified the question to the case I'm interested in, but I'm still a bit confused by your answer. I think it is true that every $f\in C_0(\mathbb{R})$ can be uniformly approximated by (smooth) functions $f_i$ with compactly supported Fourier transform. But if what you said is true then $f$ would have a holomorphic extension to $\mathbb{C}$, which shouldn't be true for arbitrary $f\in C_0(\mathbb{R})$, right? – geometricK Feb 26 '19 at 18:25
• That $(f_n)$ holomorphic converges uniformly on $\mathbb{R}$ doesn't imply the limit is holomorphic, only that the limit is continuous. For $f$ integrable continuous pick any $\hat{\phi} \in C^\infty_c, \int \phi = 1, \phi_n(x) = n\phi(nx)$ and let $f_n = f \ast \phi_n$ its Fourier transform is $\hat{f} \hat{\phi}(./n)$ and $f_n \to f$ uniformly. If $f$ is only bounded then $f_n \to f$ locally uniformly and it converges uniformly iff $f$ is uniformly continuous @ougoah – reuns Feb 26 '19 at 18:34
• And $f$ uniformly continuous is really the key, that $f$ is integrable and continuous isn't enough. The conclusion is that if $f_n \to f$ uniformly with $f_n$ uniformly continuous then $f$ is uniformly continuous. That $f$ is bounded isn't necessary, uniform continuity implies it has at most linear growth @ougoah – reuns Feb 26 '19 at 20:19