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In Mean values of multiplicative functions over function fields the mention a proof of Halasz inequality in one of their future pre-prints. In fact here is an a proof from 1999.

I would like to see a proof in any way shape or form, as it might lead to a proof of the prime number theorem.


There are various statements floating around here are two:

[1,3] Let $f : \mathbb{N} \to \mathbb{C}$ be multiplicative function such that $|f(n)| \leq 1$. Then if we establish for all real numbers $|t| < T$ that: $$ \sum_{p < x} \frac{1 - \mathrm{Re}[\, f(p) p^{-it} \,]}{p} \geq M$$ then the average value of $f(n)$ is quite small: $$ \frac{1}{x} \sum_{n < x} f(n) \ll (1+M)e^{-M} + \frac{1}{\sqrt{T}} $$ even the "cheap" version stated in the blog might already imply the prime number theorem.

[2] If the mean value of $f$ is "large" in absolute value, then $f(n)$ pretends to be $n^{it}$ for some "small" real $t$. Here, pretending is defined in terms of Kullback-Liebler distance:

$$ \mathbb{D}(f, g; x)^2 = \sum_{p \leq x} \frac{1 - \mathrm{Re}\big[\,f(p)\overline{g(p)}\,\big]}{p} $$

All of the proofs seem long and technical, and the new proof may not be much simpler. I still might like to know if the weaker version implies PNT

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    $\begingroup$ I do not see any connection to the Kullback-Liebler . Maybe we should call it "Mochizuki norm" instead? $\endgroup$ – aosjckajsd Jul 6 '16 at 19:22
  • $\begingroup$ @aosjckajsd are you joking? $\endgroup$ – john mangual Jul 6 '16 at 20:27
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    $\begingroup$ Yes! But you should be up-voting me instead of asking questions. $\endgroup$ – aosjckajsd Jul 6 '16 at 20:30
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Heuristically (and in fact rigorously with a bit of work) we have $$ \sum_{p \leq x} \frac{1 + \Re(p^{it})}{p} = \log\log x + \Re \log \zeta(1 + \frac{1}{\log x} + it) + O(1) $$ Therefore if $\zeta$ has a zero at $1 + iu$ then the hyp othesis of Halasz's theorem fails and it doesn't imply the Prime number theorem. So you get the Prime Number Theorem from Halasz + non-vanishing of $\zeta(s)$ on the $1$-line.

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    $\begingroup$ Can the ignorant masses explain why they are down-voting a correct answer? $\endgroup$ – aosjckajsd Jul 6 '16 at 23:00
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It has been over a year since I asked. The rumor it was being worked on (in fact clearly stated by the authors). I note that two papers came out yesterday:

  1. A more intuitive proof of a sharp version of Halász's theorem
  2. A new proof of Halász's Theorem, and its consequences

For both papers, authors are: Andrew Granville, Adam J Harper, Kannan Soundararajan.

I was looking for any proof of Halasz theorem, completing a develoment of the "pretentious" approach to number theory. Here's the lemma they build it from: an averaging formula.

$$ \int_{-T}^T \left|\frac{a_n\,\Lambda(n)}{n^{1+it}} \right|^2 \, dt \ll \sum_{T^2 \leq n \leq x} \frac{|a_n|^2\Lambda(n)}{n}$$

I am not a specialist in analytic number theory and these averages look rather technical and in need of interpretation. They say the proof splits into a combinaorial part and an analytic part.

Combinatorics and Real Analysis are very broad areas with a wide range of techniques available. At least, in 6 short pages, we have the result.

The combinatorial lemma looks like many we have seen before: $$ \sum_{n \leq x} f(n) = \frac{1}{\log x} \sum_{n\leq x} f(n) \big( \log n + \log x/ n \big) = \frac{1}{\log x} \sum_{mp \leq x} f(m)f(p) \log x + O \left( \frac{x}{\log x} \right) $$
This is a double convolution which gets turned into a triple convolution. However, it is a "triple convolution" after discarding many primes and only approximately. $$ (\log x)^4 < p < \frac{x}{2} $$ They do this to make the proof work. For me, an outsider, it takes a bit of thought too appreciate why these substitutions are OK.

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