Vaughan's identity

For thresholds $U,V$, write ${\mu }_{\le U}(d)=\mu (d)1_{d\le U}$ and similarly for the other truncations. Vaughan's identity isIt decomposes a sum weighted by $\Lambda$ into short, Type I, and Type II bilinear sums.

For reduced $a/q$ with $|\alpha -a/q|\le q^{-2}$ and $1\le q\le N$, the displayed bound follows by splitting Vaughan identity with parameters $U=V=N^{2/5}$. Type I sums use short geometric-sum bounds; type II sums use the Cauchy-Schwarz inequality and rational spacing to control bilinear cancellation. A badly approximable number has a Dirichlet approximant with $q\asymp N^{1/2}$, yielding $O(N^{4/5}{\log }^{4}N)$ and hence $O(N^{9/10})$.

// Target: analytic-number-theory.bigb

The Von Mangoldt function is nonnegative and satisfies ${\sum }_{d\mid m}\Lambda (d)=\log m$. Therefore the middle short-divisor coefficient in the Vaughan identity has the displayed pointwise bound. Grouping its long-divisor term as ${\mu }_{>U}(m)b(n)$, with $b(n)={\sum }_{d\mid n,d>V}\Lambda (d)$, gives $0\le b(n)\le \log n$. This grouping permits bounded-coefficient Type II sums after division by $\log N$.

The displayed identity is exact for all positive integers and all positive cutoffs. Expand $(\mu -{\mu }_{\le U})\ast (\mu -{\mu }_{\le V})\ast 1$ and use $\mu \ast 1=\epsilon$. The two truncated terms are short sums; the negative term is a type I sum over divisors at most $UV$; the last term is a type II sum with both factor variables larger than a cutoff. This form of the Vaughan identity is specific to Möbius weights.

For $|f|\le 1$, if $|{\sum }_{n\le X}\mu (n)f(n)|\ge \delta X$ and $U+V\le \delta X/2$, the Vaughan identity for the Möbius function gives a type I sum or a type II sum of modulus at least $\delta X/4$. The first is ${\sum }_{d\le UV}{c}_d{\sum }_{k\le X/d}f(dk)$, with $c_d={\sum }_{bc=d,b\le U,c\le V}\mu (b)\mu (c)$. The second is ${\sum }_{d>V,w>U,dw\le X}{a}_d\mu (w)f(dw)$, with $a_d={\sum }_{c\mid d,c>V}\mu (c)$. Both coefficient sequences are bounded by the divisor function. Thus correlation forces either detection by small-modulus periodic indicators or a large multiplicatively organized bilinear sum, without claiming that $f$ itself is literally periodic or multiplicative.

For Dirichlet convolution, let $1(n)=1$ and $ϵ(n)=1_{n=1}$. Prime factorization gives $\mu \ast 1=ϵ$ and $\Lambda \ast 1=\log$. With subscripts denoting truncation at $U,V$,Rearranging proves the Vaughan identity.