Fourier Transform – Function with Non Riemann-Integrable Transform

Question

Does there exist a compactly supported continuous function $f$ on $\mathbb R$, such that
$$
\lim_{n\to\infty}\int_{-n}^n\widehat{f}(x)\ dx
$$
does not exist?

Here $\widehat f$ is the Fourier transform of $f$.

Answer 1

Yes. We can adapt the functional analytic argument for the existence of continuous functions with divergent Fourier series.

Fix a $\varphi\in C[-2,2]$ with $0\le\varphi\le 1$, $\varphi(\pm 2)=0$, and $\varphi=1$ on $[-1,1]$, and consider the functionals $L_n: C[-2,2]\to\mathbb C$ $$ L_n(g) = \int_{-n}^n \widehat{\varphi g}\, dt = (D_n *(\varphi g))(0) =\int_{-2}^2 D_n(t)\varphi(t)g(t)\, dt . $$ Here $D_n(t)=\frac{\sin nt}{t}$ is the Dirichlet kernel.

For the (discontinuous) function $g(t)=\operatorname{sgn}(D_n(t))$ we have $\|g\|_{\infty}=1$, $$ |L_n(g)|\ge \int_{-1}^1 |D_n(t)|\, dt\gtrsim \log n . $$ Since we can get close to this situation also with continuous functions $g$, it follows that $\|L_n\|\to\infty$, and thus the uniform boundedness principle shows that there must be $f=\varphi g$ for which $\int_{-n}^n \widehat{f} = L_n(g)$ diverges (in fact, is unbounded).