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Let $\kappa>0$ be given. What is the function $f:\mathbb{R}\to [0,\infty)$ with $\int_\mathbb{R} f(x) dx = 1$ such that $$\int_\mathbb{R} |x| f(x) dx + \kappa \int_{|t|\geq T}\left| \frac{\widehat{f}(t)}{t} \right| dt$$ is minimal?


Commentary. The motivation of this question resides in its applications to bounds in analytic number theory -- for example, explicit versions of the prime number theorem. The question is essentially equivalent to the one I have asked in Best smoothing for the Prime Number Theorem? and Optimizing a smoothing function with the Prime Number Theorem in mind . The main difference is that now I have a plausible candidate: we could work with $$f(x) = \frac{1/\sigma}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2}.$$ Since then $$\int_{\mathbb{R}} |x| f(x) dx = 2 \int_0^\infty \frac{x/\sigma}{\sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2} dx = - \frac{2\sigma}{\sqrt{2\pi}} \int_0^\infty \left(e^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2}\right)' dx = \sqrt{\frac{2}{\pi}} \cdot\sigma$$ and, by $\widehat{f}(t) = e^{-2 \pi^2 \sigma^2 t^2}$, $$\frac{\partial}{\partial \sigma}\int_{|t|\geq T} \left|\frac{\widehat{f}(t)}{t}\right| dt = 2\frac{\partial}{\partial \sigma} \int_{t\geq T} \frac{e^{-2\pi^2 \sigma^2 t^2}}{t} dt = - 2 \int_{t\geq T} 4 \pi^2 \sigma t e^{-2\pi^2\sigma^2 t^2} dt = - 2\frac{e^{-2\pi^2 \sigma^2 T^2}}{\sigma} ,$$ we see that the optimal value of $\sigma$ is the one for which $\sigma = \sqrt{2\pi} \kappa \cdot e^{-2 \pi^2 \sigma^2 T^2}$. We write $\sigma = \frac{\sqrt{2 \log T + \eta}}{2 \pi T}$, and see we must solve for $\frac{\sqrt{2 \log T + \eta}}{2\pi} = \sqrt{2\pi} \kappa e^{-\eta/2}$, so $\eta = -2 \log \frac{\sqrt{\log T}}{2\pi^{3/2}\kappa} + O\left(\frac{1}{\log T}\right) = - \log \log T + 2 \log(2 \pi^{3/2} \kappa) + O\left(\frac{1}{\log T}\right)$.

Then $$\begin{aligned}\int_\mathbb{R} |x| f(x) dx + \kappa \int_{|t|\geq T}\left| \frac{\widehat{f}(t)}{t} \right| dt &= \sqrt{\frac{2}{\pi}} \sigma + 2 \kappa \int_{t\geq T} \frac{e^{-2\pi^2 \sigma^2 t^2}}{t} dt\\ &= \sqrt{\frac{2}{\pi}} \sigma - \kappa \textrm{Ei}(-2\pi^2 \sigma^2 T^2) = \sqrt{\frac{2}{\pi}} \sigma- \kappa \textrm{Ei}(- \log T - \eta/2) \\ &\sim \frac{\sqrt{\log(2\pi^{3/2} \kappa T) - \frac{1}{2} \log \log T}}{\pi^2 T} + \kappa \sqrt{\frac{2}{\pi}} \frac{e^{-\eta/2}}{T (\log T + \frac{\eta}{2})}\\ &= \frac{1}{\pi^2 T} \left(\sqrt{\log(2\pi^{3/2} \kappa T)} - \frac{\log \log T}{4 \sqrt{\log 2\pi^{3/2} \kappa T}} + \frac{1}{\sqrt{2\log T}} + \dotsc\right). \end{aligned}$$ Can one do better?

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  • $\begingroup$ Right, like the one I propose. $\endgroup$ Oct 31, 2022 at 17:04
  • $\begingroup$ What is the relationship between $f(x)$ and $\widehat{f}(t)$? $\endgroup$ Oct 31, 2022 at 21:54
  • $\begingroup$ You mean, how the Fourier transform is defined? $\hat{f}(t) = \int_{\mathbb{R}} f(x) e^{-2\pi i x t} dx$. $\endgroup$ Nov 1, 2022 at 1:28
  • $\begingroup$ Thanks, I was a bit confused because your earlier questions seemed to be based on the Mellin transform. $\endgroup$ Nov 1, 2022 at 2:22
  • $\begingroup$ ... which is a Fourier transform with a change of variables, basically. $\endgroup$ Nov 1, 2022 at 2:23

1 Answer 1

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If you care about the correct order of magnitude only (i.e., don't mind a bounded constant factor separated from both $0$ and infinity), the problem is fairly straightforward.

Scaling By playing with $uf(ux)$ instead of $f$ and choosing $u=1/\kappa$, one can immediately see that the answer for the minimum is $\kappa F(\kappa T)$ where $F$ is the answer with $\kappa=1$, so I'll assume $\kappa=1$ from now on.

The case $T\gtrsim 1$. Take any smooth non-negative even $\psi$ supported on $[-2,2]\setminus[-1,1]$ and such that $\int_{\mathbb R}\psi=1$ and put $\psi_T(t)=T^{-1}\psi(t/T)$. Then the inverse Fourier transform $\Psi_T$ of $\psi_T$ satisfies $\Psi_T(x)\ge 1-2T|x|$. Thus, integrating $$ (2T)^{-1}=(2T)^{-1}\int_{\mathbb R}f \le \int_{\mathbb R}[|x|+(2T)^{-1}\Psi_T]f = \\ \int_{\mathbb R}|x|f+(2T)^{-1}\int_{\mathbb R}\psi_T \widehat f \lesssim \Phi(f) $$ where $\Phi(f)$ is your functional to be minimized if $T\gtrsim 1$, so in this regime $\Phi(T)\gtrsim T^{-1}$. To get a matching upper bound, just take any positive $f_0$ with integral $1$, Fourier transform supported on $[-1,1]$ and some minimally decent decay like $|f_0(x)|\lesssim \frac 1{1+|x|^3}$ and put $f(x)=Tf_0(Tx)$, so that the Fourier transform of $f$ vanishes outside $[-T,T]$ and we are left with $\int_{\mathbb R}|x|f=T^{-1}\int_{\mathbb R}|x|f_0\approx T^{-1}$.

Now we want to investigate what happens as $T\to 0$.

Note that if $\int_{|x|>T^{-1/2}}|f|\ge T^{1/4}$, say, then $\Phi(f)\ge T^{-1/4}$ already. Otherwise $$ |\widehat f(t)|\ge 1-T^{-1/2}t-T^{1/4}>1/2 $$ when $t\in (T,T^{3/4})$, say, so the Fourier part of $\Phi(f)$ is $\gtrsim \log(1/T)$.

Hence $\Phi(f)\gtrsim \log(1/T)$ in this case. To have a matching upper bound just take any fixed smooth fast decaying function.

The final answer is that, up to a constant factor, the minimum is $\frac 1T$ if $\kappa T\ge 1$ and $\kappa\log\frac e{\kappa T}$ if $\kappa T<1$.

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  • $\begingroup$ Nice. I do care about the leading-order term, though, and not just up to a constant factor. Assume $T$ large. $\endgroup$ Nov 1, 2022 at 1:20
  • $\begingroup$ Also, don't you mean $\frac{1}{\kappa} F\left(\frac{x}{\kappa}\right)$ rather than $\kappa F(\kappa x)$? $\endgroup$ Nov 1, 2022 at 2:05
  • $\begingroup$ So, it would make sense to choose $f$ so that (a) $\int_{\mathbb{R}} f(x) dx = 1$, (b) $\widehat{f}$ has support contained in $[-1,1]$, (b) \int_{\mathbb{R}} f(x) |x| dx = 1$. (Not that that necessarily gives an optimal solution to the original problem!) Surely this must be known? $\endgroup$ Nov 1, 2022 at 2:32
  • $\begingroup$ @HAHelfgott "I do care about the leading-order term, though". Well, suppose that I tell you that it is $C/T$ with some crazy transcendental constant $C$. What shall it change? If you want to beat some particular value, just let me know which one. Otherwise what's the point to care about a numerical constant? As to the minimization of $\int_f(x)|x|dx$ over non-negative $f$ with Fourier transform supported on $[-1,1]$, that is an interesting problem, indeed. I don't know the answer off hand and suspect that it is not explicit. $\endgroup$
    – fedja
    Nov 1, 2022 at 18:14
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    $\begingroup$ @HAHelfgott The answer to your question depends on what exactly you mean by "decays as fast as...". If you do not want to scale the decay rate with $T$, then there is no problem with keeping $f$ compactly supported on $[-1,1]$. However, if $e^{-|x|}$ actually means $e^{-T|x|}$, then there is trouble, indeed. In all honesty, it will be better if you just explain the full setup of what you are doing and trying to achieve (as far as I could understand from your comments, you want to compute something fast and $f$ should be reasonable from both mathematical and computational views, so what is it?) $\endgroup$
    – fedja
    Nov 4, 2022 at 0:22

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