# Log weight removal in general (weaker) prime number theorem

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 $$\limsup _{X\to \infty}\frac{\sum_{p\leq X} a_p}{X/\log X}\leq 1?$$

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Just use partial summation to remove the log. – Lucia Jun 13 '14 at 16:00

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