Does the Mobius Inversion Theorem hold when the sum function is over multiples instead of divisors.
Formally, are the following two expressions equivalent:
$$ f(n) = \sum_{k:n|k}h(k) * \mu\left(\frac{k}{n}\right) $$
and
$$ h(n) = \sum_{k:n|k}f(k) $$
I encountered this while solving a problem which had $h$ as:
$$ h(k) = \left\lfloor\frac{\alpha}{k}\right\rfloor \left\lfloor\frac{\beta}{k}\right\rfloor $$
where $\alpha$ and $\beta$ are constants
Specifically in that problem I had the following expression: $$ F(g) = \sum_{k:g|k}\mu\left(\frac{k}{g}\right)h(k)\tag{1} $$
Then I assumed that Mobius Inversion Theorem holds when sum function is over multiples, to get:
$$ h(g) = \sum_{k:g|k}F(k) $$
which gives me:
$$ F(g) = \left\lfloor\frac{n}{g}\right\rfloor\left\lfloor\frac{m}{g}\right\rfloor - \sum_{k:g|k, k \neq g}F(k)\tag{2} $$
I wrote this code(C++) to check whether $(1)$ and $(2)$ are equivalent, and it seems that they are.
Let's assume that all functions considered are decaying fast enough (we will be a bit more explicit with what "fast enough" is below) that we need not worry about convergence and the validity of reordering the sums. For such functions, let's define
$$S(f) \colon n \mapsto \sum_{k : n\mid k} f(k) = \sum_{m = 1}^{\infty} f(m\cdot n)$$
and
$$M(f) \colon n \mapsto \sum_{k : n\mid k} \mu\biggl(\frac{k}{n}\biggr) f(k) = \sum_{m = 1}^{\infty} \mu(m) f(m\cdot n).$$
Then we have indeed $M(S(f)) = f = S(M(f))$. The key is, as for the usual Möbius inversion, that
$$\sum_{d \mid n} \mu(d) = \begin{cases} 1 &, n = 1 \\ 0 &, n > 1. \end{cases}$$
With that, we calculate
\begin{align} M(S(f))(n) &= \sum_{m = 1}^{\infty} S(f)(mn) \\ &= \sum_{m = 1}^{\infty} \sum_{k = 1}^{\infty} \mu(k)f(kmn) \\ &= \sum_{r = 1}^{\infty} \Biggl(\sum_{k \mid r} \mu(k)\Biggr) f(rn) \\ &= f(n) \end{align}
and similarly
\begin{align} S(M(f))(n) &= \sum_{k = 1}^{\infty} \mu(k)M(f)(kn) \\ &= \sum_{k = 1}^{\infty} \sum_{m = 1}^{\infty} \mu(k)f(kmn) \\ &= \sum_{r = 1}^{\infty} \Biggl(\sum_{k \mid r} \mu(k)\Biggr) f(rn) \\ &= f(n). \end{align}
In your example, we have $h(k) = 0$ for $k > c := \min\: \{\lvert\alpha\rvert,\, \lvert\beta\rvert\}$, and hence also $F(k) = 0$ for $k > c$, so we are effectively dealing with finite sums and no convergence problems arise. In general, the reordering is certainly legitimate if
$$\sum_{k = 1}^{\infty} \sum_{m = 1}^{\infty} \lvert \mu(k) f(kmn)\rvert = \sum_{r = 1}^{\infty} \Biggl(\sum_{k \mid r} \lvert \mu(k)\rvert\Biggr) \lvert f(rn)\rvert < +\infty$$
for every $n$, and that is the case if
$$\sum_{n = 1}^{\infty} \tau(n)\lvert f(n)\rvert < +\infty,$$
where $\tau(n)$ is the number of divisors of $n$.
We note that essentially the same computation shows that - under suitable conditions - the generalised Möbius inversion formula
$$h(x) = \sum_{m = 1}^{\infty} f\bigl(x\cdot c(m)\bigr) \iff f(x) = \sum_{m = 1}^{\infty} \mu(m) h\bigl(x \cdot c(m)\bigr)$$
holds for every non-constant completely multiplicative function $c$. The case $c(m) = \dfrac{1}{m}$ is famous.