# Estimating a sum of the shape $\sum_{n \leq x} a(n) b(n)$

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.

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

If for every $$q>1$$ the function $$b(n)^q$$ has average $$O_q(1)$$ then by H"{o}lder one can get $$\sum_{n for every fixed $$\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 which is best possible.