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Consider a Dirichlet series $\sum_n a_n n^{-s}$ with desirable analytic properties (e.g., analytic extension to $\Re s>0$); one example would be $a_n=\mu(n)$. Say we want to estimate $\sum_{n\leq x} a_n$, and that we are given that the Riemann hypothesis has been verified for all non-trivial zeros $\rho$ with $|\Im \rho|\leq T$.

Then we want to choose a smoothing function $\eta:[0,\infty)\to \mathbb{R}$ close to $1_{[0,1]}$ so that we can estimate $\sum_{n} a_n \eta(n/x)$ by analytic means and then add the error term $\sum_{n\leq x} a_n |\eta(n/x)-1| + \sum_{n>x} a_n |\eta(n/x)|$, which, if $a_n$ is bounded (in $L^\infty$ or even $L^1$ norm), should be at most about $x \int_0^\infty |\eta(t)-1_{[0,1]}(t)| dt$. In the estimation of $\sum_{n\leq x} a_n \eta(n/x)$, the main term will be very roughly of the form $$\kappa x \int_T^\infty |M\eta(1+it)| dt,$$ where $\kappa$ is a constant depending on the series and $M\eta$ is the Mellin transform of $\eta$.

The question then is how to minimize $$I = \int_0^\infty |\eta(t)-1_{[0,1]}(t)| dt + \kappa \int_T^\infty |M\eta(1+it)| dt,$$ and what that minimum shall be.

Can one get the best $\eta$ and the best $I$? Or at least get the minimum of $I$ within a constant factor?

The problem of choosing $\eta$ has of course been considered in the literature (for $\sum_n a_n n^{-s} = \zeta(s)$: I am aware of Büthe and Faber-Kadiri) but not in ways that I find completely satisfactory. Let us consider one plausible candidate for $\eta$.

Let $\eta$ be given by $$\eta(t) = \frac{1}{2} \left(1 - \mathrm{erf}\left(\frac{\log t}{\delta/T}\right)\right),$$ where we can choose $\delta$. This choice has the advantage that the Mellin transform $M\eta$ of $\eta$ is simply $$M\eta(s) = e^{\left(\frac{\delta s}{2 T}\right)^2}.$$ If we set $\delta$ of size close $\sqrt{8 \log T}$, we obtain a bound on $I$ with leading-order term $$\frac{\sqrt{(8/\pi) \log T}}{T}.$$

Can one do better than this? Can one do better than $O(\sqrt{\log T}/T)$? It is not hard to show that one cannot do better than $O(1/T)$ (see the answer to Fourier optimization problem related to the Prime Number Theorem).

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  • $\begingroup$ In the definition of I, shouldn't the first integral by on $[0,1]$ since you sum only for $n\leq x$? $\endgroup$
    – Lior Silberman
    Commented Nov 18 at 0:50
  • $\begingroup$ Fixed that - one has to sum for all $n$, and the integral had to be written a bit differently. $\endgroup$
    – H A Helfgott
    Commented Nov 18 at 3:34
  • $\begingroup$ It is clear that the $\sqrt{\log T}$ is coming from the decay $e^{(cs/T)^2}=O(e^{-(ct/T)^2})$ in the Mellin transform of $M\eta(s)$ (really the decay $O(e^{-t^2/4})$ of the Mellin transform of $e^{-(\log x)^2}$, to simplifty). One can just choose a function whose Mellin transform $M\eta$ decays even faster - as long as $(\eta-1_{[0,1]})(x)$ decays faster than any power, at least as for $x\to 0$, so that the Mellin transform still converges for $\Re s\leq 1+\epsilon$. $\endgroup$
    – H A Helfgott
    Commented Nov 18 at 5:01
  • $\begingroup$ Under that condition, $M\eta(1+it)$ cannot decrease as fast as $\exp(-e^{\epsilon |t|})$, no matter how small $\epsilon$ is (see mathoverflow.net/questions/433831/…), so we are not going to get $I\ll (\log \log T)/T$ in this way. I wonder what is best? $\endgroup$
    – H A Helfgott
    Commented Nov 18 at 5:04
  • $\begingroup$ That is, basically, how fast can the Fourier transform of a function $f$ that decays faster than any exponential (for $x\to \infty)$) decay? $\endgroup$
    – H A Helfgott
    Commented Nov 18 at 5:09

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