I came across the following assertion: if $f\in PW_\infty([-a,a])$, i.e. the Bernstein space of functions in $L^\infty(\mathbb{R})$ which are the Fourier transform of a distribution supported on $[-a,a]$, then $$\int_0^{+\infty}\frac{\ln\frac{1}{|f(x)|}}{1+x^2}\mathrm{dx}<+\infty.$$ The authors say it's a classical result but I can't seem to find the proof or come with an idea to prove it. Any suggestions/references?
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5$\begingroup$ Koosis has written two volumes on exactly this integral, called The logarithmic integral, I think there's a good chance you'll find either the result or the tools needed there. $\endgroup$– Christian RemlingCommented Apr 1, 2022 at 19:47
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$\begingroup$ It should also follows from Krein's theorem, which gives this conclusion for entire functions $f,f^{\#}\in N$ (Nevanlinna class). $\endgroup$– Christian RemlingCommented Apr 1, 2022 at 19:51
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$\begingroup$ Another potentially helpful observation is that the compactly supported distribution has finite order, so $f=t^N\widehat{u}$ with $u\in C^2[-a,a]$ (say). Of course, the power is irrelevant inside the logarithm, and then you can apply the argument from my second comment to $\widehat{u}$. (Or refer to the fact that your property is known for Hardy space functions.) $\endgroup$– Christian RemlingCommented Apr 1, 2022 at 20:15
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$\begingroup$ Thank you for the references. $\endgroup$– pipenaussCommented Apr 2, 2022 at 21:15
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This is due to Wiener and Paley, (but can be also derived from the Jensen inequality and description of positive harmonic functions in a half-plane) and a more general formulation is this: If $f$ is entire, of exponential type, and $$\int\frac{\log^+|f(x)|}{1+x^2}dx<\infty,$$ then $$\int\frac{\log^-|f(x)|}{1+x^2}dx<\infty.$$ Here $a^+=\max\{ a,0\}$ and $a^-=(-a)^+$, so that $a=a^+-a^-$.
There is a whole book about these integrals:
P. Koosis, The logarithmic integral, 2 vols. Cambridge Univ. Press, 1988, 1992. (The fact that I stated is proved in section III G 2 of volume I.)
answered Apr 2, 2022 at 2:26
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$\begingroup$ Indeed, I found it in volume I, thanks again for the references. $\endgroup$ Commented Apr 2, 2022 at 21:16