I'll get right to the point. For sequences $\{a_n\}$ and $\{b_n\}$ in $\mathbf C$, let $c_n = \sum_{d \mid n} a_db_{n/d}$. Set $A(x) = \sum_{n \leq x} a_n$, $B(x) = \sum_{n \leq x} b_n$, and $C(x) = \sum_{n \leq x} c_n$. If $A(x) = o(x)$ and $B(x) = o(x)$, must it be true that $C(x) = o(x)$, or is there a (simple) counterexample? And if there is a counterexample, is the conclusion valid if we also assume $A(x)$ or $B(x)$ is $O_\delta(x^{1-\delta})$ for some $\delta$ in $(0,1)$ but introduce no further assumptions?
This question came up when editing some course notes. When proving various relations are equivalent to the prime number theorem, the proof that $\sum_{n \leq x} \mu(n) = o(x)$ implies $\psi(x) \sim x$ appears to be basically the same everywhere I have looked and I'd prefer if there were a different argument. In the standard proof (whose basic structure goes back to Landau (see Section 1)), you reduce the problem to checking that $$ \sum_{dd' \leq x} \mu(d)(\log(d') - \tau(d') + 2\gamma) = o(x), $$ where $\tau(n) = \sum_{d \mid n} d$ and $\gamma$ is Euler's constant. The proof of this $o(x)$ estimate uses the following three properties:
$\sum_{n \leq x} \mu(n) = o(x)$,
$\sum_{n \leq x} (\log(n) - \tau(n) + 2\gamma) = O(\sqrt{x})$,
$|\mu(n)| \leq 1$ for all $a$.
Abstracting what makes the proof work, we can say for my question at the start that $A(x) = o(x)$ and $B(x) = o(x)$ implies $C(x) = o(x)$ if we add the conditions that $\{a_n\}$ is bounded (think $a_n = \mu(n)$) and $B(x) = O_\delta(x^{1-\delta})$ for some $\delta$ in $(0,1)$ (think $b_n = \log(n) - \tau(n) + 2\gamma$ and $\delta = 1/2$). I would prefer a proof of $C(x) = o(x)$ that is more symmetric in the assumptions on $\{a_n\}$ and $\{b_n\}$ while still being applicable to the intended example above ($a_n = \mu(n)$ and $b_n = \log n - \tau(n) + 2\gamma$).
I have not seen a result of the form "$A(x) = o(x)$ and $B(x) = o(x)$ implies $C(x) = o(x)$" anywhere, so I suspect it might not be true, but I have not seen (or created) a counterexample either.
For the simple counterexample:
Take $f$ increasing to $\infty$ such that $\tau(n)>f(n)$ for a positive proportion of integers.
($f(n)=\log\log(n+2)$ should work).
Let $$a(n)=b(n)= \frac1{\sqrt{f(n)}},\qquad a(d)b(n/d)\ge a(n)b(n)=\frac1{f(n)}$$
$a(n)=b(n)=o(1)$ so $A(n)=B(n)=o(n)$ but $$c(n)\ge \frac{\tau(n)}{f(n)}\ge 1$$ for a positive proportion of integers, therefore $C(n)\ne o(n)$.