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Does there exist a compactly supported continuous function f on R, such that lim 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