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Vaughan's decomposition 2

In the previous post we estimated the first term on the right hand side of: \[ \mathbb E_{N^{0.9} < n \le N   }\mu (n)\bar f(n) = - \mathbb E_{N^{0.9}<n\le N  } \sum _{b\le U,c\le V , bc|n } \mu (b) \mu (c) \bar f(n)  + \mathbb E_{N^{0.9}<n\le N  } \sum _{b> U,c> V , bc|n } \mu (b) \mu (c) \bar f(n) . \] In this post we estime the second term. Let us consider the parameter $d=n/b $ and eliminate $n$ from the expression. Noting that from $bc \le n=db $ we get $d\ge c >V$: \[ (secondterm) = \frac 1{N-N^{0.9} } \sum _{V<d \le N/U }\sum _{N^{0.9}/d <b\le N/d }1_{b>U} \sum _{c>v,c|d }\mu (b) \mu (c)  \bar f(db). \]

Again we take the crude bound $| \sum _{c>V,c|d } \mu (c) |\le t(d) $. The previous quantity is bound in magnitude by: \[ \le \frac 1{N-N^{0.9} } \sum _{V<d \le N/U } t(d) |\sum _{N^{0.9}/d <b\le N/d } 1_{b>U}\mu (b)  \bar f (db) | . \] We interpret this as the inner product of $(t(d) /\sqrt d )_d $ and $\left( \sqrt d |\sum _{N^{0.9}/d <b\le N/d } 1_{b>U}\mu (b)  \bar f (db) | \right)_d $. The Cauchy-Schwarz inequality then gives \[ (previous)^2\le \frac 1{(N-N^{0.9} )^2} \left( \sum _{V<d \le N/U} t(d)^2 /d \right) \left( \sum _{V<d \le N/U} d \left| \sum _{N^{0.9}/d <b\le N/d } 1_{b>U}\mu (b)  \bar f (db)\right| ^2\right) \] We use the bound $\sum _{V<d \le N/U} t(d)^2 /d \le \sum _{1\le d\le N} t(d)^2/d \ll (\log N)^4$ from the previous post to get the bound: \[ \ll \frac{(\log N)^4}{N^2} \left( \sum _{V<d \le N/U} d \left| \sum _{N^{0.9}/d <b\le N/d } 1_{b>U}\mu (b)  f (db)\right| ^2\right) . \] We take the dyadic decomposition $[1,2]\cup (2,4]\cup \cdots $ of the interval $V<d \le N/U$ to find a $V<D\le N/U$ such that \[ (the-right-hand-side ) \ll \frac{(\log N )^5}{N^2} D \sum _{D<d \le 2D }  \left| \sum _{N^{0.9}/d <b\le N/d } 1_{b>U}\mu (b)  f (db)\right| ^2  \]

Next we partition the interval $(N^{0.9}/d , \le N/d] $ to find a $W\in (N^{0.9}/d , \le N/d] $ such that: \[ (previous) \ll \frac{(\log N )^6}{N^2} D \sum _{D<d \le 2D }  \left| \sum _{W <b\le 2W } 1_{b>U}\mu (b)  f (db)\right| ^2 . \] Note that $N^{0.9}/(2D) \le W \le N/D $.