用户: TravorLZH/通用的下界筛构造法

在本文中, 笔者将基于 Buchstab 迭代法给出一种基于 Selberg 筛法的下界筛构造, 并证明:

定理 0.1. 对于固定的 $0<\alpha <\beta$, 当 $z=D^{1/s},w=D^{1/s_0},\alpha \le s<s_0\le \beta$ 时, 倘若存在 $0<\Delta <1$ 使得: $$S(\mathcal{A},\mathcal{P},w)>XV(w)(1-\Delta )-\sum _{\begin{array}{c}d|P(w)\\ d<D\end{array}}{3}^{\omega (d)}|r(d)|$$则有: $$S(\mathcal{A},\mathcal{P},z)>XV(z)(1-{\Delta }^{\prime }+o(1))-\sum _{\begin{array}{c}d|P(z)\\ d<D\end{array}}{3}^{\omega (d)}|r(d)|$$其中: $${\Delta }^{\prime }={\left(\frac{s_0}{s}\right)}^{\kappa }\Delta +\kappa s^{-\kappa }{\int }_s^{s_0}{u}^{\kappa -1}\left[\frac{1}{{\sigma }_{\kappa }(u-1)}-1\right]\mathrm{d}u$$

1一个 $V(z)$ 的迭代公式

设 $p<p^{\prime }$ 为 $\mathcal{P}$ 中的两个相邻素数, 则有:

$$V(p^{\prime })-V(p)=\sum _{\begin{array}{c}d|P(p^{\prime })\\ p|d\end{array}}\mu (d)g(d)=-g(p)V(p)$$

不断使用这个恒等式, 我们就得到了 $V(z)$ 的迭代式:

定理 1.1. 当 $2\le w<z$ 时总有: $$V(z)=V(w)-\sum _{w\le p<z}g(p)V(p)$$特别地$$V(z)=1-\sum _{p<z}g(p)V(p)$$

2结论的证明

利用 Buchstab 迭代式, 可知:

$$\begin{array}{rl}S(\mathcal{A},\mathcal{P},z)& =S(\mathcal{A},\mathcal{P},w)-\sum _{w\le p<z}S({\mathcal{A}}_p,\mathcal{P},p)\\ & >X\left\{V(w)-\sum _{w\le p<z}g(p)V(p)\right\}-XV(w)\Delta \\ & -\sum _{w\le p<z}[S({\mathcal{A}}_p,\mathcal{P},p)-g(p)XV(p)]\\ & -\sum _{\begin{array}{c}d|P(w)\\ d<D\end{array}}{3}^{\omega (d)}|r(d)|\end{array}$$

现在根据 Selberg 上界筛, 有:

$$S({\mathcal{A}}_p,\mathcal{P},p)<g(p)XV(p)\left[\frac{1}{{\sigma }_{\kappa }(\log D_p/p)}+o(1)\right]+\sum _{\begin{array}{c}d|P(p)\\ d<D_p\end{array}}{3}^{\omega (d)}|r(pd)|$$

其中我们设置 $D_p=D/p$ 即可统一误差项:

$$\sum _{\begin{array}{c}d|P(w)\\ d<D\end{array}}{3}^{\omega (d)}|r(d)|+\sum _{w\le p<z}\sum _{\begin{array}{c}d|P(p)\\ d<D_p\end{array}}{3}^{\omega (d)}|r(d)|\le \sum _{\begin{array}{c}d|P(z)\\ d<D\end{array}}{3}^{\omega (d)}|r(d)|$$

因此剩下的任务就变成主项的估计了. 因为我们先前假设了:

$$\frac{V(w)}{V(z)}={\left(\frac{\log z}{\log w}\right)}^{\kappa }\left\{1+O\left(\frac{1}{\log w}\right)\right\}$$(1)

所以:

$$\begin{array}{rl}\frac{1}{XV(z)}& \sum _{w\le p<z}[S({\mathcal{A}}_p,\mathcal{P},p)-g(p)XV(p)]\\ & \lesssim \sum _{w\le p<z}g(p){\left(\frac{\log z}{\log p}\right)}^{\kappa }\left[\frac{1}{{\sigma }_{\kappa }(\log (D/p)/\log p)}-1\right]\\ & =s^{-\kappa }\sum _{w\le p<z}g(p){\left(\frac{\log D}{\log p}\right)}^{\kappa }\left[\frac{1}{{\sigma }_{\kappa }\left(\frac{\log D}{\log p}-1\right)}-1\right]\\ & \sim \kappa s^{-\kappa }{\int }_w^z{\left(\frac{\log D}{\log t}\right)}^{\kappa }\left[\frac{1}{{\sigma }_{\kappa }\left(\frac{\log D}{\log t}-1\right)}\right]\frac{\mathrm{d}t}{\log t}\\ & =\kappa s^{-\kappa }{\int }_s^{s_0}{u}^{\kappa -1}\left[\frac{1}{{\sigma }_{\kappa }(u-1)}-1\right]\mathrm{d}u\end{array}$$

又因为 (1) 意味着:

$$V(w)\Delta \sim V(z){\left(\frac{\log z}{\log w}\right)}^{\kappa }\Delta =V(z){\left(\frac{s_0}{s}\right)}^{\kappa }\Delta$$

所以合并起来我们就得到了结论.