Let \(\mathcal{A}=\{a(n)\}_{n=1}^{\infty}\) be a
sequence of non-negative integers and \[|\mathcal{A}|=\sum_{n=1}^{\infty} a(n)<\infty
.\] Let \(\mathcal{P}\) be a set
of primes and let \(z \in \mathbb{R}, z \geq
2\). Let \[P(z)=\prod_{\substack{p \in
\mathcal{P} \\ p< z}} p .\] We define the sieving function as
\[S(\mathcal{A}, \mathcal{P}, z)=\sum_{(n,
P(z))=1} a(n)\] Since \[\sum_{d \mid
m} \mu(d)= \begin{cases}1 & \text { if } m=1 \\ 0 & \text {
otherwise }\end{cases}\] we obtain \[S(\mathcal{A}, \mathcal{P}, z)=\sum_{d \mid P(z)}
\mu(d)\left|\mathcal{A}_{d}\right|\] where \[\left|\mathcal{A}_{d}\right|=\sum_{d\mid n} a(n)
\quad \text { converges. }\] Assume that for every \(n \geq 1\), we have a multiplicative
function \(g(d)\) such that \[0 \leqslant g(p)<1\] for all \(p \in \mathcal{P}\).
Then for \(d \mid P(z)\) square-free
integers define \[r(d)=\left|\mathcal{A}_{d}\right|-\sum_{n} a(n)
g_{n}(d)\] and it follows that \[\begin{aligned}
S(\mathcal{A}, \mathcal{P}, z) & =\sum_{d \mid P(z)}
\mu(d)\left(\sum_{n} a(n) g_{n}(d)+r(d)\right) \\
& =\sum_{n} a(n) \prod_{p \mid P(z)}\left(1-g_{n}(p)\right)+\sum_{d
\mid P(z)} \mu(d) r(d)\\
&=\sum_{n} a(n) V_{n}(z)+R(z)
\end{aligned}\] where \[V_{n}(z)=\prod_{p \mid
P(z)}\left(1-g_{n}(p)\right) \ \text { and } \ R(z)=\sum_{d \mid P(z)}
\mu(d) r(d)\] However, \(R(z)\)
sometimes causes "trouble" as it is too large to estimate; in this
sense, it is much larger than \(V_{n}(z)\). To reduce error, we replace the
möbius function with \(\lambda^{ \pm}\)
truncated function such that \[\lambda^{+}(1)=\lambda^{-}(1)=1\] and for
every \(m \geq 2\), \[\sum_{d \mid m} \lambda^{+}(d) \geqslant 0 .
\quad \sum_{d \mid m} \lambda^{-}(d) \leqslant 0 .\] Moreover, if
\(\exists D>0\) such that \(\lambda^{+}(d)=0\) for all \(d \geq D\), then we call D a support
level.
Suppose that \(\lambda^{+}\) and \(\lambda^{-}\) have \(D\) as a support level. Then, \[\sum_{(n, P(z))=1} a(n) \leqslant \sum_{n} a(n)
\sum_{d\mid P(z)} \lambda^{+}(d)\] and by previous \[\sum_{n=1}^{\infty} a(n) G_{n}\left(z,
\lambda^{-}\right)+R^{-} \leq S(\mathcal{A}, \mathcal{P}, z) \leq
\sum_{n=1}^{\infty} a(n) G_{n}\left(z, \lambda^{+}\right)+R^{+}\]
where
\[G_{n}\left(z, \lambda^{
\pm}\right)=\sum_{d \mid P(z)} \lambda^{ \pm}(d) g_{n}(d) \text { and }
R^{ \pm}(z)=\sum_{\substack{d \mid P(z) \\ d<D}} \lambda^{ \pm}(d)
r(d)\] We then define \(\lambda_{1}^{
\pm}(d)\) over the range \(P_{1}\), which \(P=P_1 \cup \mathcal{Q}\) and for \(d \mid P(z)\) it can be uniquely factored
as \(d_1d_2\) which \(d_{1}\) is relatively prime to the product
of all primes in \(\mathcal{Q}\) ,
denoted as \(Q\). Now we define \[\lambda^{ \pm}(d)=\lambda_{1}^{ \pm}(d)
\mu\left(d_{2}\right)\] for \(d\mid
P(z)\), otherwise it is 0; and it is easy to verify that it
satisfies the previous properties of \(\mathrm{d}^{ \pm}\), and it has support
level \(DQ\) . It follows that \[G_{n}\left(z, \lambda^{ \pm}\right)=G_{n}\left(z,
\lambda_{1}^{ \pm}\right) \prod_{q\mid
Q(z)}\left(1-g_{n}(q)\right)\] where \[G_{n}\left(z, \lambda_{1}^{
\pm}\right)=\sum_{d_{1} \mid P_1(z)} \lambda_{1}^{
\pm}\left(d_{1}\right) g_{n}\left(d_{1}\right).\] and we replace
\(R^{ \pm}\) by \(\pm R(D Q, \mathcal{P}, z)\) for \(\left|\lambda_{1}^{ \pm}(d)\right| \leqslant
1\), \[R(D Q, \mathcal{P}, z)=\sum_{d
\mid P(z)}|r(d)| .\] We then construct the truncated function
based on the subset of integers from 1 to \(D\).
Let \(\beta>0, D>0\) be real
numbers. Let \(D^{+}\) be the set
consisting of 1 of all square-free numbers \(d\) with \[d=p_{1} p_{2} \cdots p_{k}\] such that
\[p_{k}<\cdots<p_{2}<p_{1}<D\]
and \[p_{m}<\left(\frac{D}{p_{1} p_{2}
\cdots p_{m}}\right)^{\frac{1}{\beta}}\] for all odd integers
\(m\). Let \(D^{-}\) be the set consisting of 1 and all
square-free numbers \(d\) with \[d=p_{1} \cdots p_{k}\] such that \[p_{k}<\cdots<p_{2}<p_{1}<D\]
and \[p_{m}<\left(\frac{D}{p_{1} \cdots
p_{m}}\right)^{1 / \beta}\] for all even integers. Let \[P(D)=\prod_{\substack{p \in \mathcal{P} \\ p <
D}} p\] and define
\[\lambda^{+}(d)= \begin{cases}\mu(d)
& \text { if } d \in D^{+} \text {and } d \mid P(D) \\ 0 & \text
{ otherwise }\end{cases}\] and \[\lambda^{-}(d)= \begin{cases}\mu(d) & \text {
if } d \in D^{-} \text {and } d \mid P(D) \\ 0 & \text { otherwise
}\end{cases}\] It follows from the definition that these
functions are truncation functions. Recall that for \(g(d)\) multiplicative function, we have
\[V(z)=\prod_{\substack{p \in \mathcal{P} \\
p<z}}(1-g(p))=\sum_{p \mid P(z)} \mu(d) g(d) .\] By induction
over the number of primes in an interval, we get \[\sum_{\substack{p \in \mathcal{P} \\ w \leqslant
p<z}} g(p) V(p)=V(w)\] Combining with the construction of
\(\lambda \pm(d)\), we obtain the
following Lemma.
Lemma. Let \(\beta\) be a set of
primes. For \(\beta>1\) and \(2 \leqslant z \leqslant D\). Let \[y_{m}=y_{m}\left(\beta, D, p_{1} \cdots
p_{m}\right)=\left(\frac{D}{p_{1} \cdots p_{m}}\right)^{1 / \beta}
.\] Let \(\lambda^{ \pm}(d)\) be
the truncated functions we defined previously, let \[G\left(z, \lambda^{ \pm}\right)=\sum_{d\mid P(z)}
\lambda^{ \pm}(d) g(d) .\] Let \[\operatorname{T_n}(D, z)=
\sum_{\substack{p_1,...,p_n\in\mathcal{P}\\ y_n\leq
p_n<...<p_1<z \\ p_m<y_m \ \forall m<n,m\equiv
n(\operatorname{mod} 2)}} g(p_1...p_n)V(p_n)\] Then \[G\left(z,
\lambda^{+}\right)=V(z)+\sum_{\substack{n=1 \\ n\equiv
1(\operatorname{mod} 2)}}^{\infty} \operatorname{T_n}(D, z)\] and
\[G\left(z,
\lambda^{-}\right)=V(z)-\sum_{\substack{n=1 \\ n \equiv 0(\bmod
2)}}^{\infty} \operatorname{T_n}(D, z)\] Moreover, \[\operatorname{T_n}(D, z) \geqslant 0\] for
all \(n \geq 1\), and \[G\left(z, \lambda^{-}\right) \leqslant V(z)
\leqslant G\left(z, \lambda^{+}\right) .\] If \(\beta \leqslant \frac{\log D}{\log z}=s\),
then \[\operatorname{T_n}(D, z)=0 \text { for
} n \leqslant s-\beta \text {. }\] Now we consider for \(\beta=2\) from now on. From the
combinatorial identity of \(V(z)\), we
obtain \[T_{1}(D, z)=V\left(D^{1 /
3}\right)-V(z) .\] Let \(z \geq
2\) and choose \(D \geqslant z\)
if \(n\) is odd, \(D \geq z^{2}\) if \(n\) is even, so \[s=\frac{\log D}{\log z} \geqslant \begin{cases}1
& \text { if } n \text { is odd. } \\ 2 & \text { if } n \text {
is even. }\end{cases}\] Let \(n \geq
2\). If \(n\) is even, or if
\(n\) is odd and \(s \geqslant 3\), then \[T_{n}(D, z)=\sum_{\substack{p \in \mathcal{P} \\
p<z}} g(p) T_{n-1}\left(\frac{D}{p}, p\right) .\] If \(n\) is odd and \(1 \leqslant S \leqslant 3\), then \[\operatorname{T_n}(D, z)=\sum_{\substack{p \in
\mathcal{P} \\ p < D^{\frac{1}{3}}}} g(p) T_{n-1}\left(\frac{D}{p},
p\right)\] If there exists some \(K>1\) such that \[\prod_{\substack{p \in \mathcal{P} \\ u \leqslant
p<z}}(1-g(p))^{-1} \leqslant K \frac{\log z}{\log u}\] for all
\(u\) such that \(1<U<Z\). Let \(\Phi\) be a continuous, increasing function
on the interval \([w, z]\) which \(w\) is any real number in \((1, z)\). Then the step function \[\begin{aligned}
S(u) & =\sum_{u \leqslant p(z} g(p) V(p) \\
& =V(u)-V(z) \\
& \leqslant\left(K \frac{\log z}{\log u}-1\right) V(z) .
\end{aligned}\] Let \[w \leqslant
p_{k}<p_{k-1}<\cdots<p_{1}<z\] which \(p_{i} \in \mathcal{P}\) for all \(i=1, \ldots, k\).
Then \[\begin{aligned}
\sum_{\substack{p \in \mathcal{P} \\
w \leq p<z}} g(p) V(p) \Phi(p)= & \sum_{i=1}^{k}
g\left(p_{i}\right) V\left(p_{i}\right) \Phi\left(p_{i}\right) \\
= & S\left(p_{k}\right)
\Phi(w)+S\left(p_{k}\right)\left(\Phi\left(p_{w}\right)-\Phi(w)\right)
\\
& +\sum_{i=1}^{k-1}
S\left(p_{i}\right)\left(\Phi\left(p_{i}\right)-\Phi\left(p_{i+1}\right)\right)
\\
= & S(w) \Phi(w)+\int_{w}^{z} S(u) d \Phi(u) \\
= & S(z) \Phi(z)-\int_{w}^{z} \Phi(u) d S(u) \\
\leq & (K-1) V(z) \Phi(z)-K V(z) \int_{w}^{z} \Phi(u)
d\left(\frac{\log z}{\log u}\right).
\end{aligned}\] For \(n=1\) and
\(1 \leqslant s \leqslant 3\), we have
\[\begin{aligned}
\frac{T_{1}(D, z)}{V(z)} &
=\frac{V\left(D^{\frac{1}{3}}\right)}{V(z)}-1 \\
& =\prod_{D^{\frac{1}{3}} \leq p<z}(1-g(p))^{-1}-1 \\
& \leqslant 3 K \frac{\log D}{\log z}-1 \\
& <\left(\frac{3}{s}-1\right)+3(K-1)
\end{aligned}\] Let \(n \geq
2\). If \(n\) is even or if
\(n\) is odd and \(s \geq 3\), then \[T_{n}(D, z)=\sum_{\substack{p \in \mathcal{P} \\
p<z}} g(p) T_{n-1}\left(\frac{D}{p}, p\right)\] To bound \(T_{n}(D, z)\), it is natural to assume that
there is a sequence of smooth functions times \(V(z)\). Let \(\Psi_{n}\) be a sequence of smooth
functions, and define \[\Phi_{n}(u)=\Psi_{n}\left(\frac{\log D}{\log
n}-1\right)\] and claim that \[T_{n}(D, z)<V(z) \Psi_{n}(s) .\] then
by the induction hypothesis, it follows that \[\begin{aligned}
T_{n}(D, z) & <\sum_{\substack{p \in \mathcal{P} \\
p<z}} g(p) V(p) \Psi_{n-1}\left(\frac{\log D}{\log p}-1\right) \\
& =\sum_{\substack{p \in \mathcal{P} \\
p<z}} g(p) V(p) \Phi_{n}(p) \\
& =(K-1) V(z) \Psi_{n-1}(s-1)-\frac{K V(z)}{s} \int_{1}^{z}
\Phi_{n-1}(u) d \frac{\log D}{\log u} \\
& =(K-1) V(z) \Psi_{n-1}(s-1)+\frac{K V(z)}{s} \int_{s}^{\infty}
\Psi_{n-1}(t-1) d t
\end{aligned}\] Similarly, for \(n \geq
3\) odd and \(1 \leqslant s \leqslant
3\), then \[T_{n}(D, z)<(K-1) V(z)
\Psi_{n-1}(2)+K V(z) \int_{3}^{\infty} \Psi_{n-1}(t-1) d t\] then
by choosing an appropiate \(K\), and
constructing \(\Psi_{n}\) carefully,
the induction can be proved. Let \[\Psi_{n}(s)=f_{n}(s)+h_{n}(s)\] which
\(f_n \ \text{and} \ h_n\) are
constructed as follows. For \(s \geqslant
1\). let \(R_{n}(s)\) be the
open convex region of Euclidean space consisting of all points \(\left(t_{1}, \ldots, t_{n}\right) \in
\mathbb{R}^{n}\) s.t. \[\begin{aligned}
& 0<t_{n}<\cdots<t_{1}<\frac{1}{s}, \\
& t_{1}+\cdots+t_{n}+2 t_{n}>1 .
\end{aligned}\] and \[t_{1}+\ldots+t_{m}+2 t_{m}<1 \ \ \ \ \text{if}
\ m<n \ \text{and} \ m \equiv n(\operatorname{mod} 2).\] For
\(n \geqslant 1\) and \(s \geqslant 1\), we define the function
\(f_{n}(s)\) by the integral \[s f_{n}(s)=\int \cdots \int_{R_{n}(s)} \frac{d
t_{1} \cdots d t_{n}}{\left(t_{1} \cdots t_{n}\right) t_{n}}\] We
construct \(h(s)\) for \(s \geqslant 1\) as follows: \[h(s)=\left\{\begin{array}{lll}
e^{-2} & \text { for } & 1 \leq s \leq 2 \\
e^{-s} & \text { for } & 2 \leq s \leq 3 \\
3 s^{-1} e^{-s} & \text { for } & s \geq 3
\end{array}\right.\] For \(s \geqslant
2\), let \[H(s)=\int_{s}^{\infty}
h(t-1) d t .\] Let \[\alpha=\frac{H(2)}{2 h(2)}=\frac{e^{2}
H(2)}{2}=1-\frac{1}{2 e}+\frac{3 e^{2}}{2} \int_{0}^{\infty} t^{-1}
e^{-t} d t\] We define the number \[\tau=\alpha+5(K-1)+11 e^{-8}\] and the
function \[h_{n}(s)=(K-1) \tau^{n} e^{10}
h(s)\] and in this case we let \(K\) satisfy \[1<K<1+\frac{1}{200} .\] And after
some calculation, we obtain the bound for \(T_{n}(D, z)\), and for \(s \geqslant 1\) \[F(s)=1+\sum_{\substack{n=1 \\ n\equiv 1
(\operatorname{mod}2)}}^{\infty}
f_{n}(s)=1+O\left(e^{-s}\right),\] and for \(s \geqslant 2\) \[f(s)=1-\sum_{\substack{n=1 \\ n \equiv
0(\operatorname{mod} 2)}}^{\infty} f_{n}(s)=1+O\left(e^{-s}\right)
.\] Moreover, let \(0<\varepsilon<\frac{1}{200},
K=1+\varepsilon\), and \[G\left(z,
\lambda^{ \pm}\right)=\sum_{d \mid P(z)} \lambda^{ \pm}(d) g(d)
.\] Then \[G\left(z,
\lambda^{+}\right)<V(z)\left(F(s)+\varepsilon
e^{14-\varepsilon}\right)\] and \[G\left(z,
\lambda^{-}\right)>V(z)\left(f(s)-\varepsilon
e^{14-\varepsilon}\right)\] Combining with the fact that this
\(\lambda^{ \pm}\) is actually \(\lambda_{1}^{ \pm}\) defined on \(\mathcal{P}_{1}\), we obtain the following
theorem.
Theorem(Jurkat-Richert). Let \(\mathcal{A}=\{a(n)\}_{n=1}^{\infty}\) such
that \[a(n) \geqslant 0 \quad \forall n \in
\mathbb{N} \text { and }|\mathcal{A}|=\sum_{n=1}^{\infty}
a(n)<\infty\] Let \(P\) be a
set of primes, for \(z \geq 2\), let
\[P(z)=\prod_{\substack{p \in \mathcal{P} \\
p<z}} p .\] Let \[S(\mathcal{A},
\mathcal{P}, z)=\sum_{\substack{n=1 \\(n, P(z))=1}}^{\infty}
a(n)\] For every \(n \geqslant
1\), let \(g_{n}(d)\) be a
multiplicative function s.t. \[0<g_{n}(p)<1 \quad \forall p \in P
.\] Define \(r(d)\) by \[\left|\mathcal{A}_{d}\right|=\sum_{d=1}^{\infty}
a(n)=\sum_{n=1}^{\infty} a(n) g(d)+r(d)\] Let \(\mathcal{Q}\) be a finite subset of \(\mathcal{P}\), and let \(Q\) be the product of primes in \(\mathcal{Q}\). Suppose that, for some \(\varepsilon\) satisfying \(0<\varepsilon<\frac{1}{200}\), the
inequality \[\prod_{\substack{p \in
\mathcal{P} \backslash \mathcal{Q} \\ u \leq
p<z}}^{\pi}\left(1-g_{n}(p)\right)^{-1}<(1+\varepsilon) \frac{\log
z}{\log u}\] holds for all \(1<u<z\). Then for any \(D \geq z\) there is the upper bound \[S(\mathcal{A}, \mathcal{P},
z)<\left(F(s)+\varepsilon e^{14-\varepsilon} \right)X+R\] and
for \(D \geqslant z^{2}\) there is the
lower bound \[S(\mathcal{A}, \mathcal{P},
z)>\left(f(s)-\varepsilon e^{14-\varepsilon}\right) X+R\]
where \[\begin{aligned}
s & =\frac{\log D}{\log z}, \\
X & =\sum_{n=1}^{\infty} a(n) \prod_{p\mid P(z)}(1-g(p)).
\end{aligned}\] and the remainder term is \[R=\sum_{\substack{d \mid P(z) \\ d <
DQ}}|r(d)| .\] If there is a multiplicative function \(g(d)\) s.t. \(g_{n}(d)=g(d)\) for all \(n\), then \[X=V(z)|\mathcal{A}|,\] where \[V(z)=\prod_{p \mid P(z)}(1-g(p)) .\]
Moreover, through the recurrence property of \(f_{n}(s)\), we could obtain and solve a
differential-difference equation that allows us to compute initial
values for \(F(s)\) and \(f(s)\), \[F(s)=\frac{2 e^{\gamma}}{s} \quad \text { for }
\quad 1 \leqslant s \leqslant 3 \text {, }\] and \[f(s)=\frac{2 e^{\gamma} \log (s-1)}{s} \quad
\text { for } \quad z \leqslant s \leqslant 4 \text {, }\] where
\(\gamma\) is the Euler’s constant.