I'm trying to solve the following problem
Suppose that $$\sum_{k=1}^{\infty}d_3(k)|f(kx)| \lt \infty,$$ where $d_3(k)$ denotes the number of factorizations of $k$ as a product of three numbers. Show that if $$ g(x)=\sum_{m=1}^{\infty}f(mx),$$ then $$f(x) = \sum_{n=1}^{\infty}\mu(n)g(nx)$$and conversly.
My probelm is that I seem to have proved it without using the condition about $d_3$. Here's how my proof goes: $\sum_{n=1}^{\infty}\mu(n)g(nx) = \sum_{n=1}^{\infty}\mu(n)\sum_{m=1}^{\infty}f(mnx)=\sum_{m,n}\mu(n)f(mnx)={\sum_{n=1}^{\infty}f(nx)\sum_{d|n}\mu(d)}=f(x)$
the converse goes very similarly, so my question is where exactly is the mistake in this proof?
With $\tau(m) = \sum_{d | m} 1$
$$ \sum_{n=1}^\infty \sum_{k=1}^\infty|\mu(n) f(knx)|\le\sum_{m=1}^\infty \tau(m) | f(mx)| \tag{1}$$ If $(1)$ converges, $g(x) = \sum_{k=1}^\infty f(kx)$ converges and $$\sum_{n=1}^\infty \mu(n) g(nx)=\sum_{n=1}^\infty \sum_{k=1}^\infty \mu(n) f(knx) = \sum_{m=1}^\infty f(mx) \sum_{d | m} \mu(d) = f(x)$$ since absolute convergence means the order of summation doesn't matter.