Prove $F^* = \mu * F$

Let $$f: \mathbb{Q} \cap [0,1] \to K$$ and set $$F(n) = \sum_{k = 1}^n f(\frac k n)$$, $$F^*(n) = \sum_{k = 1, (k,n) = 1}^n f(\frac k n)$$.

Show that $$F^* = \mu * F$$ where $$*$$ is the Dirichlet product.

$$\mu * F (n) = \sum_{d|n} \mu(d) F(\frac n d) = \sum_{k = 1}^n\sum_{d|n} \mu(d) f(\frac k n)$$

Context

This is a previous step to obtain $$\mu(n) = \sum_{k = 1, (k,n) = 1} exp(2 \pi i\frac k n)$$ which can be concluded from the fact that $$F(n) = e(n) = \sum_{d | n} \mu(d)$$ using Möbius inversion formula. This result is also done in: The Möbius function is the sum of the primitive $n$th roots of unity..

References

Tom Apostol, Introduction to Analytic Number Theory, page 48.

• $$\sum_{k=1,gcd(k,n)=1}^n f(k/n) = \sum_{k=1}^n f(k/n)\sum_{d | gcd(k,n)} \mu(d)= \sum_{d | n} \mu(d) \sum_{k=1,d | k}^n f(k/n)$$ – reuns Apr 18 at 22:38

$$F^*(n) = \sum_{k = 1, (k,n) = 1}^n f\Big(\frac k n\Big) = \sum_{k = 1}^n f\Big(\frac k n\Big) \Big(\sum_{d | (k,n)} \mu(d)\Big) = \sum_{k = 1}^n \sum_{d|k,d|n} \mu(d)f\Big(\frac k n\Big) = \ldots$$

For each $$k$$ such that $$d|k$$ we write $$k = qd$$ and $$1 \le k \le n$$ is equivalent to $$1 \le q \le \frac n d$$ and we write:

$$\ldots = \sum_{d|n} \mu(d) \sum_{k = 1,d|k} f\Big(\frac k n\Big) = \sum_{d|n} \mu(d) \sum_{q = 1}^{n/d} f\Big(\frac {q}{n/d}\Big) = \mu * F(n)$$

Let's see how we prove the consequence:

Take $$f(x) = exp(\frac{2 \pi i x}{n})$$, then observe that $$F(n) = e(n)$$ where $$e(n) = 1$$ if $$n = 1$$ and $$0$$ otherwise. This is the neutral element for Dirichlet convolution. Therefore, $$\mu * F = \mu$$ and by the above $$\mu * F = F^*$$, so we conclude $$\mu = F^*$$. This proves the proposition.

Acknowledgements

Thanks to @reuns for his comment.