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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.

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  • $\begingroup$ do you want upper bounds or asymptotics? $\endgroup$ – Dr. Pi Oct 4 '20 at 0:12
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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.

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