# asymptotic estimates for real-valued nonnegative multiplicative functions

Note that, within this entry, $p$ always refers to a prime, $k$, $m$, and $n$ always refer to positive integers, and $\log$ always refers to the natural logarithm   .

###### Theorem.
1. 1.

There exists $A\geq 0$ such that, for every $y\geq 0$, $\displaystyle\sum_{p\leq y}f(p)\log p\leq Ay$.

2. 2.

There exists $B\geq 0$ such that $\displaystyle\sum_{p}\sum_{k\geq 2}\frac{f(p^{k})\log(p^{k})}{p^{k}}\leq B$.

Then for all $x>1$, $\displaystyle\sum_{n\leq x}f(n)\leq(A+B+1)\frac{x}{\log x}\sum_{n\leq x}\frac{% f(n)}{n}$.

###### Proof.

$\begin{array}[]{ll}\displaystyle\log x\sum_{n\leq x}f(n)&\displaystyle=\sum_{n% \leq x}f(n)(\log x-\log n+\log n)\\ \\ &\displaystyle=\sum_{n\leq x}f(n)\log\left(\frac{x}{n}\right)+\sum_{n\leq x}f(% n)\log n\\ \\ &\displaystyle\leq\sum_{n\leq x}f(n)\left(\frac{x}{n}\right)+\sum_{n\leq x}f(n% )\sum_{p^{k}\|n}\log(p^{k})\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{p^{k}\leq x}\log(p^{k})% \sum_{n\leq x\text{ and }p^{k}\|n}f(n)\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{p^{k}\leq x}\log(p^{k})% \sum_{n\leq x\text{ and }p^{k}\|n}f(p^{k})f\left(\frac{n}{p^{k}}\right)\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{p^{k}\leq x}\log(p^{k})% \sum_{m\leq\frac{x}{p^{k}}}f(p^{k})f(m)\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{p\leq x}f(p)\log p\sum_% {m\leq\frac{x}{p}}f(m)+\sum_{p\leq x}\sum_{k\geq 2}f(p^{k})\log(p^{k})\frac{x}% {p^{k}}\sum_{m\leq\frac{x}{p^{k}}}\frac{f(m)}{m}\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{m\leq x}f(m)\sum_{p\leq% \frac{x}{m}}f(p)\log p+x\sum_{m\leq x}\frac{f(m)}{m}\sum_{p\leq x}\sum_{k\geq 2% }\frac{f(p^{k})\log(p^{k})}{p^{k}}\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+\sum_{m\leq x}f(m)\left(\frac% {Ax}{m}\right)+x\sum_{m\leq x}\frac{f(m)}{m}B\\ \\ &\displaystyle\leq x\sum_{n\leq x}\frac{f(n)}{n}+Ax\sum_{n\leq x}\frac{f(n)}{n% }+Bx\sum_{n\leq x}\frac{f(n)}{n}\\ \\ &\displaystyle\leq(A+B+1)x\sum_{n\leq x}\frac{f(n)}{n}\end{array}$

Dividing the inequality  $\displaystyle\log x\sum_{n\leq x}f(n)\leq(A+B+1)x\sum_{n\leq x}\frac{f(n)}{n}$ by $\log x$ yields the desired result. ∎

The theorem has an obvious corollary:

###### Corollary.

If $f$ the conditions of the theorem, then for all $x>1$, $\displaystyle\sum_{n\leq x}f(n)=O\left(\frac{x}{\log x}\sum_{n\leq x}\frac{f(n% )}{n}\right)$.

Title asymptotic estimates for real-valued nonnegative multiplicative functions AsymptoticEstimatesForRealvaluedNonnegativeMultiplicativeFunctions 2013-03-22 16:08:42 2013-03-22 16:08:42 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Theorem msc 11N37 AsymptoticEstimate DisplaystyleSum_nLeXTaunaO_axlogX2a1ForAGe0