Say that an arithmetic function f has a strong logarithmic mean value A, and write L(f) = A, if f satises an estimate of the form
\begin{align} \sum\limits_{n=1}^{x}\frac{f(n)}{n} = Alog(x) + B + o(1) \end{align}
It is clear to see that f has a logarithmic mean value. I am having trouble proving that f also has an actual mean value. Was hoping someone may be able to lead me in the correct direction for a proof.
Thanks.
I have found a solution to my own problem but I would like it to be critiqued by anyone willing to read it. We do the following.
\begin{equation} \sum_{n=1}^{x}f(n)=\sum_{n=1}^{x}\frac{f(n)}{n}n =x\sum_{n=1}^{x}\frac{f(n)}{n} -\int_{1}^{x}\sum_{n=1}^{t}\frac{f(n)}{n}dt \end{equation}
By summation by parts. From here we turn (1) into the following.
\begin{equation} \sum_{n=1}^{x}f(n)=x(Alog(x)+B+o(1)) - \int_{1}^{x}Alog(t)+B+o(1)dt \end{equation}
The integral above can be solved for easily and we get what is required to finish our proof.
\begin{equation} Axlog(x)+B+o(1)-(A(x(log(x)-1))+B(x-1)+o(x))=Ax+B+o(x) \end{equation} Therefore,
\begin{equation} \lim_{x->\infty}\frac{1}{x}\sum_{n=1}^{x}f(n)=\lim_{x->\infty}A+\frac{B}{x}+\frac{o(x)}{x}=A+0+0=A \end{equation}
In conclusion, if f has a strong logarithmic mean value, it must also have a regular mean value. Moreover, the values are both A.