TITLE: Fourier Series with Mobius coefficients
QUESTION [16 upvotes]: 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:

Is $F(x)$ continuous?
Davenport's theorem is ineffective due to the possible existence of Siegel zeros. Can one obtain an unconditional effective uniform bound on $F(x)$?

REPLY [14 votes]: 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).