TITLE: Log weight removal in general (weaker) prime number theorem
QUESTION [5 upvotes]: Let $a_n$ be a sequence of non-negative numbers.
Assume that $$\limsup _{X\to \infty}\frac{\sum_{p\leq X} a_p\log p}{X}\leq 1.$$
Can we prove that $$\limsup _{X\to \infty}\frac{\sum_{p\leq X} a_p}{X/\log X}\leq 1?$$

REPLY [5 votes]: Here is a direct argument avoiding partial summation. Let $Y:=X/\log^2 X$, then
$$ \sum_{p\leq X}a_p = \sum_{p\leq Y}a_p + \sum_{Y<p\leq X}a_p\leq \sum_{p\leq Y}a_p\frac{\log p}{\log 2} + \sum_{Y<p\leq X}a_p\frac{\log p}{\log Y} $$
$$ \leq (1+o(1))\left(\frac{Y}{\log 2}+\frac{X}{\log Y}\right) = (1+o(1))\frac{X}{\log X}.$$