%! TEX root = AN.tex
% vim: tw=50
% 28/02/2025 09AM

% No lecture next Wednesday.
% Additional lecture on Friday, 21 March.

% Correction from example class:
% The final lemma: $E(x) \vll x^{\half} \log x$.

\subsection{Zero-free region}

\begin{fcprop}[]
  Assuming:
  - $\sigma > 1$ and $t \in \Rbb$
  Then:
  \[
    3 \frac{\zeta'}{\zeta}(\sigma)
    + 4\Re \left( \frac{\zeta'}{\zeta}(\sigma +
    it) \right)
    + \Re \left( \frac{\zeta'}{\zeta}(\sigma +
    2it) \right)
    \le 0
    \tag{$**$}
    \label{lec16eq1}
  .\]
\end{fcprop}

\begin{proof}
  Recall that
  \[
    \frac{\zeta'}{\zeta}(s)
    = -\sum_{n = 1}^{\infty} \frac{\vman(n)}{n^s}
  ,\]
  where $\vman$ is the \gls{vman} function, and
  $\Re(s) > 1$.

  Taking linear combinations, the LHS of
  \eqref{lec16eq1} becomes
  \[
    -\sum_{n = 1}^{\infty} \vman(n) \frac{3 +
    4\Re(n^{it}) + \Re(n^{-2it})}{n^\sigma}
    = -\sum_{n = 1}^{\infty} \vman(n) \frac{3 +
    4\cos(t\log n) + \cos(2 + \log n)}{n^\sigma}
  \]
  ($\Re(n^{iu}) = \cos(u \log n)$).

  We are done by the inequality:
  \[
    3 + 4\cos \alpha + \cos 2\alpha
    = 2(1 + \cos \alpha)^2
    \ge 0
  \]
  for $\alpha \in \Rbb$.
\end{proof}

\begin{fcthm}[Zero-free region]
  There is a constant $c > 0$ such that
  \cloze{$\zeta(\sigma + it) \neq 0$} whenever
  \cloze{$\sigma > 1 - \frac{c}{\log(|t| + 2)}$}.

  In particular, \cloze{$\zeta(s) \neq 0$ for
  $\Re (s) = 1$.}
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/91b5f1b6d73f4841.png}
\end{center}

\begin{proof}
  Let $\sigma \in [1, 2]$ and $t \in \Rbb$.
  Suppose $\zeta(\beta + it)$. uwe know that
  $\beta \le 1$. We know that $\zeta$ has no
  zeroes in some ball $B(1, r)$ for some $r > 0$
  (otherwise the entire function $(s - 1)
  \zeta(s)$ would have an accumulation point for
  its zeros).

  Choosing $c > 0$ small enough, we can assume
  that $|t| \ge r$. By the key inequality for
  $\frac{\zeta'}{\zeta}$,
  \[
    3 \frac{\zeta'}{\zeta}(\sigma)
    + 4\Re \left( \frac{\zeta'}{\zeta}(\sigma +
    it) \right).
    + \Re \left( \frac{\zeta'}{\zeta}(\sigma +
    2it) \right).
    \le 0
  .\]
  Apply partial fraction decomposition of
  $\frac{\zeta'}{\zeta}$. So
  \[
    \frac{\zeta'}{\zeta}(s)
    = -\frac{1}{s - 1}
    + \sum_{|s - \rho| \le \frac{1}{10}}
    \frac{1}{s - \rho}
    + O(\log(|t| + 2))
    \tag{$**$}
    \label{lec16eq2}
  .\]
  ($t = \Im(s)$).

  Since $\Re(\rho) \le 1$ for any zero $\rho$,
  \[
    \Re \frac{1}{\sigma + iu - \rho}
    = \frac{\sigma - \Re(\rho)}{|\sigma + iu -
    \rho|^2}
    \ge 0
  \]
  ($\sigma > 1$).

  Discarding terms, we get
  \[
    -\frac{3}{\sigma - 1}
    + \frac{4}{\sigma - \beta}
    \le C\log(|t| + 2)
  .\]
  Take $\sigma = 1 + \frac{sc}{\log(|t| + 2)}$,
  and assume $\beta \ge 1 - \frac{c}{\log(|t| +
  2)}$ to get
  \[
    -3 \frac{\log(|t| + 2)}{5c} + 4 \frac{\log(|t|
    + 2)}{6c}
    \le C(\log|t| + 2)
  .\]
  Take $c = \frac{1}{16C}$ to get a contradiction.
\end{proof}

\begin{fcthm}[Bounding $zeta' / zeta$]
  Assuming:
  - $c > 0$ sufficiently small
  - $T \ge 0$
  - $\Re(s) \ge -10$
  - $|\Im(s)| \in [T, T + 1]$
  - $s$ is at least distance $\frac{c}{\log(T +
  2)}$ away from any zero or pole
  Then:
  \[
    \left| \frac{\zeta'(s)}{\zeta(s)} \right|
    \vll_c (\log(T + 2))^2
  .\]
\end{fcthm}

\begin{proof}
  If $s = \sigma + it$ with $\sigma > 10$, then
  \begin{align*}
    \left| \frac{\zeta'(s)}{\zeta(s)} \right|
    &= \left| \sum_{n = 1}^{\infty} \vman(n)
    n^{-s} \right| \\
    &\le \sum_{n = 1}^{\infty} \vman(n)
    n^{-\sigma} \\
    &\vll 1
  \end{align*}
  Assume then that $\Re(s) \in [-10, 10]$. Apply
  \eqref{lec16eq2}. Each term satisfies
  \begin{align*}
    \frac{1}{|s - \rho|}
    &\le \frac{\log(T + 2)}{c} \\
    \frac{1}{|s - 1|}
    &\le \frac{\log(T + 2)}{c}
  \end{align*}
  We know that there are $O(\log(T + 2))$ zeros
  with multiplicity having imaginary part $\in [T
  - 2, T + 2]$. The claim follows from triangle
  inequality.
\end{proof}

TODO