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% 14/02/2025 09AM

  We need to translate the condition $\rho_1 = 1$. Note that
  \begin{align*}
    \sum_{\substack{m \le z \\ m \equiv 0
    \pmod{c}}} \mob(m) \zeta_m
    &= \sum_{d \mid P(z)} \rho_d g(d)
    \ub{\sum_{\substack{m \mid d \\ c \mid m}}
    \mob(m)}_{\sum_{m' \mid \frac{d}{c}} \mob(c)
    \mob(m') = \mob(d) \indicator{d = e}} \\
    &= \mob(e) \rho_e g(e)
  \end{align*}
  Hence,
  \[
    \rho_e = \frac{\mu(e)}{g(e)} \sum_{\substack{m
    \le 2 \\ m \equiv 0 \pmod{c}}} \mob(m) \zeta_m
  .\]
  Now,
  \[
    \rho_1 = 1 = \sum_{m \le z} \mob(m) \zeta_m
  .\]
  By Cauchy-Schwarz, we then get
  \[
    \left( \sum_{m \le z} h(m)^{-1} \zeta_m^2
    \right) \left( \sum_{m \le z} \mob(m)^2 h(m)
    \right) \ge \left( \sum_{m \le z} \mob(m)
    \zeta_m \right)^2 = 1
  .\]
  Hence
  \[
    \sum_{m \le z} h(m)^{-1} \zeta_m^2
    \ge \frac{1}{\sum_{m \le z} \mob(m)^2 h(m)}
    = \frac{1}{\sum_{m \le z} h(m)}
  .\]
  Equality holds for $T_m = \frac{h(m)}{G(z)}$,
  where $G(z) = \sum_{m \le z} h(m)$. We now check
  that with these $\zeta_m$, $\rho_d = 0$ for $d >
  z$.

  Note that
  \[
    \rho_c = \frac{\mob(c)}{g(c) G(z)}
    \sum_{\substack{m \le z \\ m \equiv \pmod{c}}}
    \mob(m) h(m)
  .\]
  Hence, $\rho_c = 0$ for $c > 2$.

  This proves Claim 1.

  Now we prove Claim 2 ($|\rho_c| \le 1$).

  Note that any $m$ has at most one representation
  as $m = em'$, where $e \mid d$, $(m', d) = 1$
  (for any $d \in \Nbb$).

  Now,
  \begin{align*}
    G(z)
    &\ge \sum_{c \mid d} \sum_{\substack{m' \le \frac{z}{c}
    \\ (m', d) = 1}} h(cm') \\
    &= \sum_{c \mid d} h(c) \sum_{\substack{m' \le
    \frac{z}{c} \\ (m', d) = 1}} h(m') \\
    &\ge \sum_{c \mid d} h(e) \sum_{\substack{m' \le
    \frac{z}{d} \\ (m', d) = 1}} h(m')
  \end{align*}
  Now,
  \[
    \rho_d = \frac{\mob(d)^2 h(d)}{g(d) G(z)}
    \sum_{\substack{m' \le \frac{z}{d} \\ (m', d)
    = 1}} \mob(m') h(m')
  .\]
  Substituting the lower bound for $G(z)$,
  \[
    |\rho_d| \le \frac{h(d)}{g(d) \sum_{c \mid d}
    h(e)} = 1
  ,\]
  since $1 \conv h = \frac{h}{g}$.
\end{proof}

\begin{fclemma}[]
  Assuming:
  - $z \ge 3$
  - $g : \Nbb \to [0, 1]$ \gls{mult}
  - for some $K, A \in \Rbb$ we have
  \[
    \sum_{p \le z} g(p) \log p
    \le \kappa \log z + A
  .\]
  Then:
  \[
    \frac{1}{\sum_{m \le z} h(m)}
    \le 2 \prod_{p \le z^{1 / (e\kappa + 1)}} (1 - g(p))
  ,\]
  where $h$ is defined in terms of $g$ as in
  Selberg's sieve.
\end{fclemma}

\begin{proof}
  Note that for any $c \in (0, 1)$,
  \[
    \sum_{m \le z} h(m)
    \ge \sum_{\substack{m \le z \\ m \mid P(z^c)}}
    h(m) = G(z, c)
  .\]
  Then
  \begin{align*}
    \prod_{\substack{p \le z^c \\ p \in
    \mathcal{P}}} (1 - g(p))^{-1} - G(z, c)
    &= \prod_{\substack{p \le z^c \\ p \in
    \mathcal{P}}} (1+  h(p)) - \sum_{\substack{m
    \le z \\ m \mid P(z^c)}} h(m) \\
    &= \sum_{\substack{m > z \\ m \mid P(z^c)}}
    h(m)
  \end{align*}
  By Rankin's trick,
  \begin{align*}
    1 - \prod_{p \le z^c} (1 - g(p)) G(z, c)
    &= \prod_{p \le z^c} (1 - g(p))
    \sum_{\substack{m > z \\ m \mid P(z^c)}} h(m)
    \\
    &\le \prod_{p \le z^c} (1 - g(p))
    z^{-\frac{\lambda}{\log z}} \sum_{m \mid P(z^c)}
    h(m) m^{\frac{\lambda}{\log z}}
    &&\text{(for any $\lambda > 0$)} \\
    &= \prod_{p \le z^c} (1 - g(p)) e^{-\lambda}
    \prod_{p \le z^c} (1 + h(p)
    p^{\frac{\lambda}{\log z}}) \\
    &= e^{-\lambda} \prod_{p \le z^c} (1 - g(p) +
    g(p) p^{\frac{\lambda}{\log z}} \\
    &\le e^{-\lambda} \exp
    \left( \sum_{p \le z^c}
    \ub{(p^{\frac{\lambda}{\log z}} - 1)}_{\le
    \frac{\lambda}{\log z} (\log p)
    p^{\frac{\lambda}{\log z}}} \right) \\
    &\le \exp \left( -\lambda +
    \frac{\lambda}{\log z} \sum_{p \le z^c}
    g(p)(\log p) p^{\frac{\lambda}{\log z}} \right) 
    &&(1 + t \le e^t) \\
    &\le \exp \left( -\lambda + ce^{c\lambda}
    \lambda \kappa + \lambda e^{c\lambda}
    \frac{A}{\log z} \right)
  \end{align*}
  Choose $c = \frac{1}{\lambda}$ and $\lambda =
  e\kappa + 1$ to get the claim.
\end{proof}