# Uniformly approximating a function by functions with nice Fourier transform

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 $$L^1$$.)

## 1 Answer

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 $$f \in C_0$$.