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Apr 24, 2019 at 4:26 comment added Terry Tao The reason is spectral theory - traditionally expressed via complex analysis (e.g. Perron's formula), but also expressible via Fourier analysis (e.g. Parseval formula) or from the Gelfand theory of Banach algebras. One can for instance start from the reproducing formula $\mu(n) \log n = - \sum_{d|n} \mu(d) \Lambda(n/d)$ (traditionally Selberg's symmetry formula is used instead). On the Fourier side, this type of formula can be used to show $\Lambda$ either has a "large Fourier coefficient" ($\mu$ pretends to be $n^{it}$) or $\mu$ has mean zero; this is a special case of Halasz's theorem.
Apr 24, 2019 at 3:38 comment added kodlu there is a nice discussion of pretentiousness in Andrew Granville's paper at citeseerx.ist.psu.edu/viewdoc/…
Apr 24, 2019 at 3:15 comment added mathworker21 @TerryTao Vague question: Why does it suffice, to prove the prime number theorem, to show that $\mu(n)$ does not pretend to be like $n^{it}$? In other words, why is that pretention the only obstruction?
Apr 22, 2019 at 13:04 vote accept Q_p
Apr 21, 2019 at 23:12 history edited kodlu CC BY-SA 4.0
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Apr 21, 2019 at 18:20 history edited David E Speyer CC BY-SA 4.0
Changed o(1) to 0.
Apr 21, 2019 at 17:49 comment added Terry Tao In particular, if $t$ a non-zero real, then $\lim_{s \to 1^+} \sum_{n=1}^\infty \frac{n^{it}}{n^s}$ can be shown to converge to a finite value, whereas $\lim_{x \to \infty} \sum_{n \leq x} \frac{n^{it}}{n}$ is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
Apr 21, 2019 at 3:16 comment added Yemon Choi @NateEldredge Or magic guarantees like (weak) compactness...
Apr 21, 2019 at 3:03 comment added Nate Eldredge Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
Apr 21, 2019 at 2:48 history answered kodlu CC BY-SA 4.0