Fork me on GitHub
Math for the people, by the people.

User login

asymptotic estimates for real-valued nonnegative multiplicative functions

\documentclass{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
%%\usepackage{xypic}

\newtheorem*{thm*}{Theorem}
\newtheorem*{cor*}{Corollary}

\begin{document}
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.

\begin{thm*}
Let $f$ be a real-valued nonnegative multiplicative function such that the two following conditions are satisfied:

\begin{enumerate}
\item There exists $A \ge 0$ such that, for every $y \ge 0$, $\displaystyle \sum_{p \le y} f(p) \log p \le Ay$.
\item There exists $B \ge 0$ such that $\displaystyle \sum_p \sum_{k \ge 2} \frac{f(p^k)\log(p^k)}{p^k} \le B$.
\end{enumerate}

Then for all $x>1$, $\displaystyle \sum_{n \le x} f(n) \le (A+B+1) \frac{x}{\log x} \sum_{n \le x} \frac{f(n)}{n}$.
\end{thm*}

\begin{proof}
\begin{center}
$\begin{array}{ll}
\displaystyle \log x \sum_{n \le x} f(n) & \displaystyle =\sum_{n \le x} f(n)(\log x-\log n+\log n) \\
\\
& \displaystyle =\sum_{n \le x} f(n) \log\left(\frac{x}{n}\right)+\sum_{n \le x} f(n)\log n \\
\\
& \displaystyle \le \sum_{n \le x} f(n) \left(\frac{x}{n}\right)+\sum_{n \le x} f(n) \sum_{p^k \Vert n} \log (p^k) \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{p^k \le x} \log(p^k) \sum_{n \le x \text{ and }p^k \Vert n} f(n) \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{p^k \le x} \log(p^k) \sum_{n \le x \text{ and }p^k \Vert n} f(p^k)f\left(\frac{n}{p^k}\right) \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{p^k \le x} \log(p^k) \sum_{m \le \frac{x}{p^k}} f(p^k)f(m) \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{p \le x} f(p)\log p \sum_{m \le \frac{x}{p}} f(m)+\sum_{p \le x} \sum_{k \ge 2} f(p^k) \log(p^k) \frac{x}{p^k} \sum_{m \le \frac{x}{p^k}} \frac{f(m)}{m} \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{m \le x} f(m) \sum_{p \le \frac{x}{m}} f(p)\log p+x\sum_{m \le x} \frac{f(m)}{m} \sum_{p \le x} \sum_{k \ge 2} \frac{f(p^k)\log(p^k)}{p^k} \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+\sum_{m \le x} f(m) \left( \frac{Ax}{m} \right)+x\sum_{m \le x} \frac{f(m)}{m} B \\
\\
& \displaystyle \le x\sum_{n \le x} \frac{f(n)}{n}+Ax\sum_{n \le x} \frac{f(n)}{n}+Bx \sum_{n \le x} \frac{f(n)}{n} \\
\\
& \displaystyle \le (A+B+1)x\sum_{n \le x} \frac{f(n)}{n} \end{array}$
\end{center}

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

The theorem has an obvious corollary:

\begin{cor*}
If $f$ \PMlinkescapetext{satisfies} the conditions of the theorem, then for all $x>1$, $\displaystyle \sum_{n \le x} f(n)=O\left(\frac{x}{\log x} \sum_{n \le x} \frac{f(n)}{n}\right)$.
\end{cor*}
%%%%%
%%%%%
nd{document}