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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 $f$.

Nandor
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  • I think it exists. You should probabaly modify the classical example of P. Du Bois Reymond of of a continuous periodic function which has Fourier series which diverges at some point. – an_ordinary_mathematician Feb 16 '24 at 13:23
  • What about the answer by @jjcale to this question: https://mathoverflow.net/questions/3764/does-there-exist-a-continuous-function-of-compact-support-with-fourier-transform? The function is $\frac{1/2-x}{\log(1/x)} 1_{[0,1/2]}$. – Dispersion Feb 16 '24 at 21:40
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    @Dispersion: $\lim_{n\to\infty}\int_{-n}^n\widehat{f}$ could still exist though, even when $\widehat{f}\notin L^1$. – Christian Remling Feb 16 '24 at 22:48
  • Ah yes you are right, I overlooked that we were taking the symmetric limit. – Dispersion Feb 16 '24 at 22:51

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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).

Iosif Pinelis
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Christian Remling
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  • Nice use of the uniform boundedness principle. Do you think an explicit example can be constructed here? – Iosif Pinelis Feb 16 '24 at 20:51
  • @IosifPinelis: Not sure, but it's certainly possible. Maybe one can just take one of the constructions of a continuous (and periodic) function with divergent Fourier series, and $f-c$ might still work in the current setting. – Christian Remling Feb 16 '24 at 21:02