The Dual Möbius Inversion Formula: Let $\mathcal D$ be a divisor closed set of natural numbers (i.e., if $d\in \mathcal D$ and $c \mid d$, then $c\in \mathcal D$.) Let $f$ and $g$ be two complex-valued functions on the natural numbers. Prove that
$$f(n) = \sum_{\substack{n\mid d \\ d\in \mathcal D}} g(d)$$
if and only if $$g(n) = \sum_{\substack{n\mid d \\ d\in \mathcal D}} \mu(d/n) f(d)$$
where $\mu$ is the Möbius function. Assume all series are absolutely convergent.
For the forward direction, I wrote
$$\begin{align*} \sum_{\substack{n\mid d \\ d\in \mathcal D}} \mu(d/n) f(d) &= \sum_{\substack{np = d \\ d\in \mathcal D}} \mu(d/n) \sum_{\substack{dq = k \\ k\in \mathcal D}} g(k)\\ &= \sum_{\substack{npq = k \\ k\in \mathcal D}} g(k)\mu(p) \end{align*}$$
How should I proceed?
I primarily want to use that $\sum_{d\mid n} \mu(d) = 1$ if and only if $n = 1$, and $\sum_{d\mid n} \mu(d) = 0$ otherwise.
Reference: An Introduction to Sieve Methods and Their Applications by
Alina Carmen Cojocaru and M. Ram Murty.
Best Answer
You've correctly recognized the one key step in all arithmetic Möbius inversion proofs: you want to recognize $\sum_{d \mid n} \mu(d)$ somewhere. Let's complete your half-proof. (The other direction is essentially identical).
$$\begin{align*} \sum_{\substack{n\mid d \\ d\in \mathcal D}} \mu(d/n) f(d) &= \sum_{\substack{np = d \\ d\in \mathcal D}} \mu(d/n) \sum_{\substack{dq = k \\ k\in \mathcal D}} g(k)\\ &= \sum_{\substack{npq = k \\ k\in \mathcal D}} g(k)\mu(p), \end{align*}$$ (where I inserted a missing $g(k)$ from the OP).
To be completely clear, this last summation is over all $p$ and $q$ such that $k = npq$ is in $\mathcal{D}$ — and $p, q$ are generic integers (not necessarily prime). We reorder the double summation. We instead sum over $k$ that are multiples of $n$ and such that $k = n\ell$ is in $\mathcal{D}$; and then over $p \mid \ell$:
$$ \sum_{\substack{p, q \\ npq = k \\ k \in \mathcal{D}}} g(k) \mu(p) = \sum_{\substack{k \\ n \mid k \\ k \in \mathcal{D}}} g(k) \sum_{\substack{p \mid \frac{k}{n}}} \mu(p).$$
This last sum is zero unless $n = k$, in which precisely one term survives on the right: $g(n)$. Thus the sums we have written are ultimately equal to $g(n)$. (Note also that if $n \not \in \mathcal{D}$, then every sum is trivially $0$).
Broadly, this is a typical summation reversal. In your notation of $np = d$, $dq = k$, and $k = npq$, we observe that the initial double sum has an outer sum over $p$ and an inner sum over $k$. The step I included changed the order so that the outer sum is over $k$ and the inner sum is over $p$ (where the variable names morally maintain their responsibilities).