# Fourier Series with Mobius coefficients

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 $$F(x)$$?
• Maybe a very stupid question: is $F$ Lipschitz? – Sina Baghal Jul 12 at 2:45
• Please use a high-level tag like "nt.number-theory". I added this tag now. – GH from MO Jul 12 at 3:01
• @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. – 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).