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

Is $F(x)$ continuous?

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 $F(x)$?

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

Is $F(x)$ uniformly bounded? And if so, is it continuous?

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

Is $F(x)$ continuous?