= Solution
Set $L=\sqrt D$, and restrict attention to the set $\mathcal S$ of <squarefree integers> $d\leq L$ 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. Put
$$
h(t)=\prod_{p\mid t}(g(p)^{-1}-1),\qquad
k(t)=h(t)^{-1}=\prod_{p\mid t}\frac{g(p)}{1-g(p)},\qquad J=\sum_{t\in\mathcal S}k(t).
$$
In particular, the normalizing <sum> runs to \b[$\sqrt D$], as in the PDF; the converted TeX's $\sqrt d$ is a transcription error. Define $y_t=\sum_{\substack{m\in\mathcal S\\t\mid m}}g(m)\rho_m$. The finite multiples version of <Möbius inversion> is
$$
g(d)\rho_d=\sum_{\substack{t\in\mathcal S\\d\mid t}}\mu(t/d)y_t.
$$
It 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> give
$$
1\leq\left(\sum_{t\in\mathcal S}h(t)y_t^2\right)\left(\sum_{t\in\mathcal S}\frac{\mu(t)^2}{h(t)}\right)=\Sigma J.
$$
Equality holds at $y_t=\mu(t)/(Jh(t))$, which satisfies the constraint. Inverting, and writing $t=dv$, yields
$$
\boxed{\rho_d=\frac{\mu(d)}{J}\prod_{p\mid d}\frac1{1-g(p)}
\sum_{\substack{v\leq L/d\\v\text{ squarefree}\\(v,d)=1\\p\mid v\Rightarrow p\in\mathcal P}}\prod_{p\mid v}\frac{g(p)}{1-g(p)}.}
$$
For $d\notin\mathcal S$ set $\rho_d=0$. The formula has $\rho_1=1$ and gives \b[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 $v\leq L/d$, is a distinct term of $\mathcal S$. The total weight of these terms is
$$
\sum_{e\mid d}k(e)\sum_v k(v)=\prod_{p\mid d}(1-g(p))^{-1}\sum_v k(v).
$$
It is at most $J$, so the displayed formula gives $|\rho_d|\leq1$. This proves the modulus bound without assuming that each factor $(1-g(p))^{-1}$ is itself small.
Back to article page
