Let $a(\cdot), b(\cdot)$ be non-negative multiplicative functions supported on square-free integers (that is, $a(p^k) = b(p^k) = 0$ for all primes $p$ and $k \geq 2$). Consider the summatory functions


$$\displaystyle A(x) = \sum\_{n \leq x} a(n), B(x) = \sum\_{n \leq x} b(n).$$


Suppose that $A(x), B(x)$ satisfy asymptotic formulas of the form


$$\displaystyle A(x) \sim c\_1 x (\log x)^{k\_1}, B(x) \sim c\_2 x (\log x)^{k\_2}.$$


What can be said in general about the order of magnitude of the summatory function


$$\displaystyle AB(x) = \sum\_{n \leq x} a\_n b\_n?$$


By Cauchy-Schwarz, one has that


$$\displaystyle AB(x) \leq \left(\sum\_{n \leq x} a\_n^2 \right)^{1/2} \left(\sum\_{n \leq x} b\_n^2 \right)^{1/2},$$


but this is not a good upper bound in some cases. For example if $b\_n = 1$ when $n$ is square-free and zero otherwise and $a\_n = 2^{\omega(n)}$ when $n$ is square-free and zero otherwise, one sees that


$$\displaystyle AB(x) = A(x) = \sum\_{\substack{n \leq x \\ n \text{ square-free}}} 2^{\omega(n)} \sim c\_1 x \log x,$$


while


$$\displaystyle \sum\_{\substack{n \leq x \\ n \text{ square-free}}} a\_n^2 = \sum\_{\substack{n \leq x \\ n \text{ square-free}}} 4^{\omega(n)} \sim c\_1^\prime x (\log x)^3.$$


Thus Cauchy-Schwarz gives an upper bound of $O(x (\log x)^{3/2})$, which is the wrong order of magnitude.


The case I am most interested in is when $k\_2 = 0$ in the notation above: that is, when $b\_n$ is constant on average. Can one guarantee an upper bound of the correct order of magnitude, at least when $a\_n$ is sufficiently uniform? One can take $a\_n = \lambda^{\omega(n)}$ with $\lambda > 1$ for $n$ square-free, as this is the case I am most interested in.



Estimating a sum of the shape $\sum_{n \leq x} a(n) b(n)$
If for every $q>1 $ the function $b(n)^q$ has average $O\_q(1)$ then by H"{o}lder one can get $$ \sum\_{n<x} \lambda^{\omega(n)} b(n) \ll\_\epsilon x (\log x)^{\lambda-1+\epsilon}$$ for every fixed $\epsilon >0$. Apart from the $(\log x)^\epsilon$ term, this is close to the best one can hope in general.


If you know the stronger (compared to the bounded average of $b^q$) assumption that $b(p)=1+o(1)$ on the primes then for $\lambda>1 $ you can use (1.82) of Iwaniec-Kowalski to get directly $$ \sum\_{n<x} \lambda^{\omega(n)} b(n)\ll x (\log x)^{\lambda-1}, $$ which is best possible.


