I assume this question has been considered before, but I can't find an literature on it. Let $\mu(n)$ denote the usual Mobius function and define:

$F(x) : = \sum_{n=1}^{\infty} \frac{\mu(n)}{n}e(nx)$

where $e(x):= e^{2\pi i x}$.

The Prime Number Theorem is equivalent to the statement that $F(0)=0$. More generally, one can show that $F(x)$ is uniformly bounded. This follows from partial summation and an old theorem of Davenport.

Two further questions naturally follow:

  1. Is $F(x)$ continuous?
  2. Davenport's theorem is ineffective due to the possible existence of Siegel zeros. Can one obtain an unconditional effective uniform bound on $F(x)$?
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    $\begingroup$ Maybe a very stupid question: is $F$ Lipschitz? $\endgroup$ – Sina Baghal Jul 12 at 2:45
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    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$ – GH from MO Jul 12 at 3:01
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    $\begingroup$ @SinaBaghal Maybe I am wrong and this a stupid answer, but I think if you look at the linked question, Davenport inequality would imply a Lipschitz like condition. $\endgroup$ – polfosol Jul 12 at 15:23

I can answer the first question. Davenport proved that $$S(N,x):=\sum_{n\leq N}\mu(n)e(nx)\ll\frac{N}{(\log N)^2},$$ with an upper bound independent of $x\in\mathbb{R}$. Therefore, by partial summation we see that $F(x)$ converges uniformly, and it equals $$F(x)=\sum_{n=1}^\infty\frac{S(n,x)}{n(n+1)}.$$ Uniform convergence implies continuity, since the terms in $F(x)$ are continuous (i.e. the partial sums of $F(x)$ are continuous).


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