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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.

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    $\begingroup$ $$\sum_{k=1,gcd(k,n)=1}^n f(k/n) = \sum_{k=1}^n f(k/n)\sum_{d | gcd(k,n)} \mu(d)= \sum_{d | n} \mu(d) \sum_{k=1,d | k}^n f(k/n)$$ $\endgroup$ – reuns Apr 18 at 22:38
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$$F^*(n) = \sum_{k = 1, (k,n) = 1}^n f\Big(\frac k n\Big) = \sum_{k = 1}^n f\Big(\frac k n\Big) \Big(\sum_{d | (k,n)} \mu(d)\Big) = \sum_{k = 1}^n \sum_{d|k,d|n} \mu(d)f\Big(\frac k n\Big) = \ldots$$

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.

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