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let $x\in\mathbb{R}-\mathbb{Z}$ and $e(x)=e^{2\pi ix}$. If we have this sum $$\left|\overset{q}{\underset{h=1}{\sum}^{*}}e\left(h\, x\right)\underset{\underset{p\equiv h\,\textrm{mod}\, q}{p\leq N}}{\sum}\log p\right|$$where ${\sum}^{*}$ is the sum with the condition $\left(h,q\right)=1$ can we affirm that exist a $h^{*}\in\left[1,q\right],\,\left(h^{*},q\right)=1$ such that $$\left|\overset{q}{\underset{h=1}{\sum}^{*}}e\left(h\, x\right)\underset{\underset{p\equiv h\,\textrm{mod}\, q}{p\leq N}}{\sum}\log p\right|\leq\left|\overset{q}{\underset{h=1}{\sum}^{*}}e\left(h\, x\right)\right|\underset{\underset{p\equiv h^{*}\,\textrm{mod}\, q}{p\leq N}}{\sum}\log p$$ so there exists a $h^{*}$ that maximizes this sum? Thank you.

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  • $\begingroup$ A bit more background would be nice... $\endgroup$ Oct 24, 2014 at 7:48

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No. If $q$ is not square-free and $x=1/q$, then the right hand side of your inequality is zero, while the left hand side is surely not.

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  • $\begingroup$ Thank you! Do you think can be true with some hypothesis? For example $q$ prime and $x=a/q$, with $(a,q)=1$. $\endgroup$
    – peppo
    Oct 24, 2014 at 8:50
  • $\begingroup$ @peppo: I don't know, but my feeling is that even for $q$ prime the $h$-sum on the right exhibits too much cancellation to make the inequality true. In some sense $x=a/q$ is the worst case in this regard, you might have better success with $x$ badly approximable (I am not optimistic though). $\endgroup$
    – GH from MO
    Oct 25, 2014 at 14:43

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