Let $n\in N$.
Let $\mu (n)$ be the classical Mobius function. In other words, it vanishes at square-full numbers, equals $+1$ if $n$ has an even number of distinct prime factors, and equals $-1$ is $n$ has an odd number of distinct prime factors.
Let $p_i$ be the $i$-th prime.
Let $p_{max} (n)$ be the largest prime factor of $n$.
Let $p_{nextmax} (n)$ be the first prime larger than $p_{max} (n)$.
Let $f(x)$ and $g(x)$ be integer functions of real variables. In other words, $f,g:R \to Z$.
Of interest is the following identity:
$$f(x)=\sum_{n p_{nextmax} (n) \leq x} \mu (n) g\left( \frac{x}{n} \right) $$
We notice that this sum runs over $n$, and roughly $n \leq \sqrt{x}$, not $n\leq x$, because of the condition $n p_{nextmax} (n) \leq x$. Thus, an easier variant of this question considers the identity:
$$f(x)=\sum_{n\leq \sqrt{x}} \mu (n) g\left( \frac{x}{n} \right) $$
Or, even more generally, the identity:
$$f(x)=\sum_{n\leq y\leq x} \mu (n) g\left( \frac{x}{n} \right) $$
Can the above identities be Mobius inverted? If so, what would the result be?
Thank you.