TITLE: Mean value of arithmetic function
QUESTION [7 upvotes]: Suppose we define a mean value of arithmetic function $G(f)$ as $$ G(f)=\lim_{x \rightarrow \infty} \frac{1}{x \log{x}} \sum_{n \leq x} f(n) \log{n},$$ and suppose now for an arithmetic function $f$, $G(f)$ exist and is equal to $A$, how to use this result to show that the ordinary mean value of arithmetic function $M(f)=\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} f(n)$ also exists?

REPLY [5 votes]: Via Abel's summation formula:
$$\sum_{n\le x} (f(n)\log n)\frac{1}{\log n}=\left(\sum_{n\le x}f(n)\log n\right)\frac{1}{\log x}+\int_2^x \left(\sum_{m\le u} f(m)\log m\right)\frac{du}{u\log^2 u}.$$
Divide by $x$ and subtract, obtain:
$$M_x(f)-G_x(f)=\frac{1}{x}\int_2^xG_u(f)\frac{du}{\log u}=\frac{\mathrm{Li}(x)}{x}\left(A+O(1)\right)\to0.$$