Proof of Zero density of the Riemann Zeta function

Question

I am currently trying to understand the proof of the simplest zero-density theorem for $\zeta$ there is, namely $$N(T+1)-N(T-1) \ll \log(T),$$ where $T>2$ and $N(T) = \#\{\rho \in \mathcal N : |\Im(\rho)| < T \}$ with $\mathcal N$ the set of non-trivial zeroes of $\zeta$. I consulted three books on analytic number theory so far (by Brüdern, Vaugan and Tenenbaum) and in all three of those the proof is (more or less) the same: We use Jensen's Formula to obtain $$\int_0^1 \log | \zeta(2+iT+re(\theta))| d\theta = \log|\zeta(2+iT)| + \sum_{\rho \in \mathcal N \text{ and } |\rho - (2+iT)| < r} \log \left(\frac r {|\rho-(2+iT)|} \right)$$ for some $r \in [3,4]$ and $e(\theta) = \text{e}^{2\pi i \theta}$. In this formula, it is relatively easy to see that the sum on the right hand side is $\approx N(T+1)-N(T-1)$. So far so good. It is now stated that a bound of the form $$\zeta(\sigma + it) \ll t^k$$ for $t>1$, $\sigma \in \mathbb R$ and some $k$ of fitting size implies the bound $$\log|\zeta(s)| \ll \log |t|.$$ Tenenbaum even explicitly mentions the bound $$\log |\xi(s)| \ll |s|\log|s| \text{ (as |s| }\rightarrow \infty \text{ for }\Re s \geq \tfrac 12) $$ for the complete Riemann-Zeta function in his book Introduction to analytic and probabilistic number theory as Formula (41) on page 152. It is clear that these bounds imply the desired result.

I could agree with the authors if they stated (and used) these bounds as $\log |\zeta(s)| \leq O(\log |s|)$, but in general these bounds are clearly wrong (as long as I've not completely lost it)! $\zeta$ (and $\xi$) have zeroes with arbitrarily large imaginary part, and $\log |\zeta|$ ($\log |\xi|$ resp.) has to be large near those.

Am I missing something? It might well be that I am misinterpreting some notations here, it's hard to believe that all three (or even one) of these authors would have published this without thinking about my issue.

Answer 1

Copying Conrad's comment so that more people see the method:

For $T$ large and $r=3$

Jensen's formula gives $$\int_0^1 \log | \zeta(2+iT+re^{2i\pi \theta})| d\theta = \log|\zeta(2+iT)| + \sum_{\zeta(\rho)=0, |\rho - (2+iT)| < r} \log \left(\frac r {|\rho-(2+iT)|} \right)$$ The RHS is $\ge a(N(T+1)-N(T))$ and it is $$\le \int_0^1\max(0, \log | \zeta(2+iT+re^{2i\pi \theta})|) d\theta=\int_0^1 O(\log(T^3))d\theta=O(\log T)$$ where $\zeta(2+iT+re^{2i\pi \theta})=O(T^3)$ follows from $\zeta(s)=\frac{s}{s-1}+s\int_1^\infty(\lfloor x\rfloor-x)x^{-s-1}dx$ and say the functional equation.

Answer 2

It follows from

$$ |\zeta^2(2)\zeta^4(2+iT)\zeta(2+2iT)|\ge1 $$

and $|\zeta(2+it)|\le\zeta(2)$ for all $t\in\mathbb R$ that

$$ 1/\zeta(2+iT)=\mathcal O(1) $$

This implies that $-\log\zeta(2+iT)$ is both upper bounded and lower bounded.

Consequently, by Jensen's inequality the number of roots within the region $|s-2-iT|\le r<R$ is bounded above by

$$ n(r)\le{\max_{|z|=R}|\log\zeta(2+iT+z)|-\log\zeta(2+iT)\over R/r} $$

Now, it follows from Titchmarsh's book on Riemann zeta, we have $\log\zeta(2+iT+z)\ll\log T$. Choosing proper $R$ and $r$, we deduce

$$ N(T+1)-N(T-1)\ll\log T $$