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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 $N^{0.9}/(2D) \le W \le N/D$.