At page 3 of this article <https://arxiv.org/pdf/1706.03755.pdf> about the new proof of Halasz's Theorem, the authors claim that $$\sum\limits\_{(\log x)^4< p\leq x/2}\frac{\log p}{p\log(x/p)}\ll\log\log x.$$ It should be equivalent to $$\sum\limits\_{(\log x)^4< p\leq x/2}\frac1{p\log(x/p)}\ll\frac{\log\log x}{\log x}.$$


I know how to prove the bounds if $p$ runs up to a fixed power of $x$, but not all the way up to $x/2$.


Thank you.



Estimating a sum over prime numbers
Using the fact that


$$\sum\_{y\leq p\leq 2y} \frac{\log p}{p}\ll 1,$$


we get for any $A\geq 2$


$$\sum\_{x/2A\leq p\leq x/A} \frac{\log p}{p\log(x/p)}\ll \frac{1}{\log A}$$


Hence for any positive integer $l$ we have


$$\sum\_{x/2^{l+1}\leq p\leq x/2^l} \frac{\log p}{p\log(x/p)}\ll \frac{1}{l}.$$


Summing this over all $l$ with $2\leq 2^l\leq x$ we get


$$\sum\_{ p\leq x/2} \frac{\log p}{p\log(x/p)}\ll \sum\_{0\lt l\leq \log x} \frac{1}{l} \ll \log\log x,$$


as needed.


