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.

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    $\begingroup$ The point is that inferior bounds are trivial eg by definition $N(T+1)-N(T) \ge 0$ so you need only superior bounds and the logarithm of the absolute value is negative near zeroes.... ( more generally Jensen implies that the integral of $\log |f|$ increases on increasing circles for analytic functions, hence one needs upper bounds only when estimating number of zeroes) $\endgroup$
    – Conrad
    Commented Mar 16, 2021 at 21:29
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    $\begingroup$ As for the comment about $<<, O$ here the context is real and means upper bounds only for the reasons above (in other words a negative number is $O(1)$ by definition - agree notation a bit abusive but it is clear from the context) $\endgroup$
    – Conrad
    Commented Mar 16, 2021 at 21:39
  • $\begingroup$ @Conrad I just realized this while brushing my teeth! Nevertheless, thanks a lot. I feel relieved now :) $\endgroup$ Commented Mar 16, 2021 at 21:43
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    $\begingroup$ Happy to be of help; on circles holomorphic functions cannot be too small in a definite sense measured by the integral of the logarithm of the absolute value unless they are hugely big too and this last cannot happen if for example the function is holomorphic on the circle ( meaning in a neighborhood); this is a subtle property that is best appreciated in Hardy space context $\endgroup$
    – Conrad
    Commented Mar 16, 2021 at 21:57
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    $\begingroup$ Remarkably I asked a similar question (with the same mischaracterization of $\log \lvert s \rvert$ in particular) on MO recently, and Conrad was the one who helped guide me in the right direction there too. $\endgroup$
    – davidlowryduda
    Commented Mar 16, 2021 at 23:40

2 Answers 2


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.


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

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


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