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Suppose $f:\mathbb{R}\rightarrow\mathbb{C}$ is a uniformly bounded, uniformly continuous, smooth function with limits at $\pm\infty$.

In particular, $f$ has a distributional Fourier transform.

Question: Does there exist a sequence of functions $f_n$ such that:

  • $f_n\rightarrow f$ uniformly as $n\rightarrow\infty$,
  • for each $n$, the Fourier transform $\widehat{f}_n$ is compactly supported and continuous (or, failing this, just in $L^1(\mathbb{R})$)?

Thoughts: I think this can be done if one is happy for the $\widehat{f}_n$ to have distributional Fourier transforms. (For example one could define $f_n:=f*\phi_n$, where $\phi_n$ is defined by $\phi_n=n\phi(nx)$ for some fixed function $\phi$ with compactly supported $\widehat{\phi}$ and mass $1$. In that case, $\widehat{f}_n=\widehat{f}\widehat{\phi}_n$ would be a compactly supported distribution, but not necessarily continuous or $L^1$.)

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Unless $f$ vanishes at infinity (i.e., the limits at $\pm \infty$ are zero), this is impossible. To see this, note that if $\widehat{f_n}$ is in $L^1$, then $f_n \in C_0$ (continuous and vanishes at infinity), by Fourier inversion and the Riemann Lebesgue lemma.

But $C_0$ is closed with respect to uniform convergence, meaning that since $f_n \to f$ uniformly, you also have $f \in C_0$.

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