Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 27 / 3

Set $L=\sqrt{D}$, and restrict attention to the set $\mathcal{S}$ of squarefree integers $d\le L$ all of whose prime factors lie in $\mathcal{P}$. Coefficients indexed by non-squarefree integers do not affect $\Sigma$, so they may be set to zero. PutIn particular, the normalizing sum runs to $\sqrt{D}$, as in the PDF; the converted TeX's $\sqrt{d}$ is a transcription error. Define $y_t={\sum }_{\begin{array}{c}m\in \mathcal{S}\\ t\mid m\end{array}}g(m){\rho }_m$. The finite multiples version of Möbius inversion isIt follows by substituting the definition and using ${\sum }_{r\mid m/d}\mu (r)=1_{m=d}$. Taking $d=1$ gives $1={\sum }_t\mu (t)y_t$. Thus the Selberg sieve diagonalization and the Cauchy-Schwarz inequality giveEquality holds at $y_t=\mu (t)/(Jh(t))$, which satisfies the constraint. Inverting, and writing $t=dv$, yieldsFor $d\notin \mathcal{S}$ set ${\rho }_d=0$. The formula has ${\rho }_1=1$ and gives minimum $\Sigma =1/J$.

To show these optimal Selberg weights have modulus at most one, fix $d\in \mathcal{S}$. Each integer $ev$, with $e\mid d$, $(v,d)=1$ and $v\le L/d$, is a distinct term of $\mathcal{S}$. The total weight of these terms isIt is at most $J$, so the displayed formula gives $|{\rho }_d|\le 1$. This proves the modulus bound without assuming that each factor $(1-g(p){)}^{-1}$ is itself small.

Write $f(n)=3^{\omega (n)}$, where $\omega$ is the prime omega function. If $p^a\Vert n$, its contribution to the Dirichlet convolution $(f\ast \Lambda )(n)$ isSince $3(a-1+1/3)\ge a$, summing over the prime factors proves the log-weighted convolution bound for three to the prime omega, $f(n)\log n\le 3(f\ast \Lambda )(n)$. On $\sqrt{D}\le n\le D$, $\log n\ge \frac{1}{2}\log D$, henceThis is the required inequality. The standard Chebyshev estimate $\psi (t)\ll t$ bounds its right side by $D(\log D{)}^{-1}{\sum }_{m\le D}f(m)/m$. To bound the latter sum, use multiplicativity and extend to all numbers with prime factors at most $D$:The Mertens second theorem states ${\sum }_{p\le D}1/p=\log \log D+O(1)$. Taking logarithms of the product, with a summable $O(p^{-2})$ remainder, gives $3\log \log D+O(1)$. Therefore the product is $O({\log }^{3}D)$, proving the summatory bound for three to the prime omega in the required range: