In this proof, we adopt the convention that each appearance of the symbol
\(*\) represents a
different positive constant.
For the benefit of the reader that may not have worked out
Exercise 1.6.3, we spell out the argument: by writing
\begin{equation*}
\Real(\log(\zeta(s)) = \sum_p \sum_{n=1}^\infty \frac{1}{n}
\cos(\Imag(s) \log p^n) p^{-n \Real(s)}
\end{equation*}
and observing that
\begin{equation*}
3 + 4 \cos \theta + \cos 2 \theta \geq 0 \qquad (\theta \in \RR)\text{,}
\end{equation*}
we deduce that
\begin{equation*}
3 \Real (\log \zeta(\sigma)) + 4 \Real (\log \zeta(\sigma + it))
+ \Real(\log \zeta(\sigma + 2it)) \geq 0 \qquad (\sigma > 1, t \in \RR)\text{.}
\end{equation*}
By contrast, if \(\zeta(1 + it)\) were zero for some \(t\text{,}\) then the sum would tend to \(-\infty\) as \(\sigma \to 1^+\) (because \(\zeta(\sigma + 2it)\) does not diverge and \(4 > 3\)).
We can apply the same argument with \(\log \zeta\) replaced by its negative derivative
\begin{equation*}
- \Real \zeta'(s)/\zeta(s) = \sum_{n=1}^\infty \Lambda(n) n^{-\Real(s)} \cos(\Imag(s) \log n)
\end{equation*}
to obtain an analogous inequality
\begin{equation}
-3 \Real \frac{\zeta'(\sigma)}{\zeta(\sigma)}
-4 \Real \frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}
- \Real \frac{\zeta'(\sigma+2it)}{\zeta(\sigma+2it)}
\geq 0 \qquad (\sigma \gt 1, t \in \RR)\text{.}\tag{8.3.1}
\end{equation}
Suppose now that
\(t\) is the imaginary part of a zero
\(\rho\) of
\(\zeta\text{.}\) To use
(8.3.1) to get information about
\(t\text{,}\) we bound each term on the left-hand side for
\(\sigma\) slightly bigger than 1. For starters, to control the terms of
(8.3.1) involving
\(\sigma\text{,}\) note that since
\(\zeta\) has a simple pole at
\(s=1\text{,}\)
\begin{equation}
-\frac{\zeta'(\sigma)}{\zeta(\sigma)} \lt \frac{1}{\sigma - 1} + *\text{.}\tag{8.3.2}
\end{equation}
To control the terms of
(8.3.1) involving
\(s = \sigma+2it\text{,}\) we adjust
(8.2.5) to get rid of the Gamma factors:
\begin{equation}
-\frac{\zeta'(s)}{\zeta(s)} = \frac{1}{s-1} - B - \frac{1}{2} \log \pi
+ \frac{1}{2} \frac{\Gamma'((s+1)/2)}{\Gamma((s+1)/2)} -
\sum_\rho \left(\frac{1}{s - \rho} + \frac{1}{\rho} \right)\text{.}\tag{8.3.3}
\end{equation}
For
\(1 \leq \Real(s) \leq 2\) and
\(|\Imag(s)| \geq 1\text{,}\) everything on the right-hand side of
(8.3.3) other than the sum over
\(\rho\) is dominated by
\(* \log |\Imag(s)|\text{.}\) Hence taking real parts, we obtain
\begin{equation}
- \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |\Imag(s)| - \sum_\rho \Real \left(
\frac{1}{s - \rho} + \frac{1}{\rho} \right)\text{.}\tag{8.3.4}
\end{equation}
Since \(\Real(\rho) > 0\) and \(\Real(s-\rho) > 0\text{,}\) we also have \(\Real(1/\rho) > 0\) and \(\Real(1/(s-\rho)) > 0\text{,}\) so the sum over \(\rho\) is positive. Hence
\begin{equation}
- \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |\Imag(s)|\text{.}\tag{8.3.5}
\end{equation}
To control the terms of
(8.3.1) involving
\(s = \sigma+it\text{,}\) we bound
\(-\Real \frac{\zeta'(s)}{\zeta(s)}\) by keeping only the summand of
(8.3.4) corresponding to
\(\rho\) (which is valid by the previous argument). Writing
\(\rho = \beta + it\text{,}\) we have
\begin{equation}
- \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |t| - \frac{1}{\sigma - \beta}\text{.}\tag{8.3.6}
\end{equation}
\begin{equation*}
\frac{4}{\sigma - \beta} \lt \frac{3}{\sigma - 1} + * \log |t|\text{.}
\end{equation*}
For \(\sigma = 1 + */(\log |t|)\text{,}\) this yields
\begin{equation*}
\beta \lt 1 - \frac{*}{\log |t|}
\end{equation*}
as desired.
