A remark on trigonometric sums

Summary Let <InlineEquation ID=IE"1"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"2"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"3"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"4"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"5"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"6"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"7"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"8"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"9"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"10"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"11"><EquationSource Format="TEX"><!CDATA<InlineEquation ID=IE"12"><EquationSource Format="TEX"><!CDATA$]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>$]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation>]]></EquationSource></InlineEquation> S(x,\alpha\mid X_p):=\sum_{\substack{p_1p_2<x\\ p_1<p_2}} X_{p_1}X_{p_2}e^{2\pi i\alpha p_1p_2},\qquad \pi_2(x)=\sum_{\substack{p_1p_2<x\ p_1<p_2}} 1, $$ where $p$, $p_1$, $p_2$ run over the prime numbers. It is proved that $$ \max_{\substack{|X_p|\le 1\\ p}} \frac{{S(x,\alpha,X_p)}}{{\pi_2(x)}} =\Delta(x,\alpha)\to 0\qquad(x\to\infty) $$ for almost all irrational $\alpha$.