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Let $f:\mathbb{R}\to \mathbb{R}$ be in $L^1$, with its Fourier transform $\widehat{f}$ also in $L^1$. What is a necessary and sufficient condition on $f$ so that $$\int_{-\infty}^x \widehat{f}(t) dt \geq 0$$ holds for every $x$?

Assume, if it helps, that $f$ is even, continuous and of compact support, with $f(0)=1$.

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  • $\begingroup$ If the question were to ensure that $\widehat{f}(t)\geq 0$, we would all know the answer -- it is necessary and sufficient for 𝑓 to be of the form 𝑔∗𝑔. Now, if I haven't miscalculated, $\int_{-\infty}^x \widehat{f}(t) dt$ is the Fourier transform of the distribution $-\frac{f(x)}{2\pi i x} + \frac{f(0)}{2} \delta(x)$, but I have no idea of what it would mean for that to be a self-convolution! $\endgroup$
    – H A Helfgott
    Commented Nov 26 at 11:41
  • $\begingroup$ That's related to your computation but did you try to introduce, for all $x$, the tempered distribution $g_x$ (sorry I am too lazy to compute it now) such that $\widehat{g_x} = \mathbf{1}_{(-\infty,x)}$ and then write your condition as " the integral of $\widehat{f}\widehat{g_x} = \widehat{f\star g_x}$ has to be non-negative for all $x$ " which in turn should be equivalent to $f\star g_x(0) \geq 0$, by Fourier inversion theorem (compact support of $f$ could be useful here to define the convolution product $f\star g_x$). I don't know if such a condition would be useful though ... $\endgroup$
    – Ayman Moussa
    Commented Nov 26 at 21:40
  • $\begingroup$ @AymanMoussa This gives me the condition $\frac{1}{2} + \widehat{\frac{i f}{2\pi x}}(t)\geq 0$ for every $t$. That's the same as what we had before, no? $\endgroup$
    – H A Helfgott
    Commented Nov 26 at 22:13
  • $\begingroup$ If your function is continuous and of compact support then you can use Whittaker–Shannon interpolation formula (or something like that) and you will see that the Fourier transform is written in terms of sinc function. $\endgroup$
    – Andronick Arutyunov
    Commented Nov 26 at 22:20
  • $\begingroup$ @AndronickArutyunov Aha? Details? Yes, assume it's continuous and of compact support. $\endgroup$
    – H A Helfgott
    Commented Nov 26 at 22:22

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