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.

  • $\begingroup$ 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? $\endgroup$ – geometricK Feb 26 '19 at 18:25
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    $\begingroup$ 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 $\endgroup$ – reuns Feb 26 '19 at 18:34
  • $\begingroup$ OK I've accepted the combination of answer + comment from reuns. $\endgroup$ – geometricK Feb 26 '19 at 19:36
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    $\begingroup$ 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 $\endgroup$ – reuns Feb 26 '19 at 20:19
  • $\begingroup$ @reuns You are of course right, I corrected the answer. $\endgroup$ – MaoWao Feb 26 '19 at 22:43

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